de Gruyter Expositions in Mathematics 47

Editors V. P. Maslov, Academy of Sciences, Moscow W. D. Neumann, Columbia University, New York R. O. Wells, Jr., International University, Bremen

Groups of Prime Power Order Volume 2 by

Yakov Berkovich and Zvonimir Janko

≥

Walter de Gruyter · Berlin · New York

Authors Yakov Berkovich Jerusalem str. 53, apt. 15 Afula 18251 Israel E-Mail: [email protected]

Zvonimir Janko Mathematisches Institut Ruprecht-Karls-Universität Heidelberg Im Neuenheimer Feld 288 69120 Heidelberg E-Mail: [email protected]

Mathematics Subject Classification 2000: 20-02, 20D15, 20E07 Key words: Finite p-group theory, minimal nonabelian subgroups, metacyclic subgroups, extraspecial subgroups, equally partitioned groups, p-groups with given maximal subgroups, 2-groups with few cyclic subgroups of given order, Ward’s theorem on quaternion-free groups, 2-groups with small centralizers of an involution, Blackburn’s theorem on minimal nonmetacyclic groups.

앝 Printed on acid-free paper which falls within the guidelines 앪 of the ANSI to ensure permanence and durability.

ISSN 0938-6572 ISBN 978-3-11-020419-3 Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available in the Internet at http://dnb.d-nb.de. 쑔 Copyright 2008 by Walter de Gruyter GmbH & Co. KG, 10785 Berlin, Germany. All rights reserved, including those of translation into foreign languages. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage or retrieval system, without permission in writing from the publisher. Typeset using the authors’ TeX files: Kay Dimler, Müncheberg. Printing and binding: Hubert & Co. GmbH & Co. KG, Göttingen. Cover design: Thomas Bonnie, Hamburg.

Contents

List of definitions and notations

. . . . . . . . . . . . . . . . . . . . . . . . . viii

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

xiv

46

Degrees of irreducible characters of Suzuki p-groups

. . . . . . . . . .

1

47

On the number of metacyclic epimorphic images of finite p-groups . . .

14

48

On 2-groups with small centralizer of an involution, I

. . . . . . . . . .

19

49

On 2-groups with small centralizer of an involution, II . . . . . . . . . .

28

50

Janko’s theorem on 2-groups without normal elementary abelian subgroups of order 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

51

2-groups with self centralizing subgroup isomorphic to E8 . . . . . . . .

52

52

2-groups with 2 -subgroup of small order . . . . . . . . . . . . . . . .

75

53

2-groups G with c2 .G/ D 4 . . . . . . . . . . . . . . . . . . . . . . . .

96

54

2-groups G with cn .G/ D 4, n > 2 . . . . . . . . . . . . . . . . . . . . 109

55

2-groups G with small subgroup hx 2 G j o.x/ D 2n i . . . . . . . . . . 122

56

Theorem of Ward on quaternion-free 2-groups . . . . . . . . . . . . . . 134

57

Nonabelian 2-groups all of whose minimal nonabelian subgroups are isomorphic and have exponent 4 . . . . . . . . . . . . . . . . . . . . . . . 140

58

Non-Dedekindian p-groups all of whose nonnormal subgroups of the same order are conjugate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

59

p-groups with few nonnormal subgroups . . . . . . . . . . . . . . . . . 150

60

The structure of the Burnside group of order 212 . . . . . . . . . . . . . 151

61

Groups of exponent 4 generated by three involutions . . . . . . . . . . . 163

62

Groups with large normal closures of nonnormal cyclic subgroups . . . . 169

63

Groups all of whose cyclic subgroups of composite orders are normal . . 172

vi

Groups of prime power order

64

p-groups generated by elements of given order . . . . . . . . . . . . . . 179

65

A2 -groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188

66

A new proof of Blackburn’s theorem on minimal nonmetacyclic 2-groups 197

67

Determination of U2 -groups . . . . . . . . . . . . . . . . . . . . . . . . 202

68

Characterization of groups of prime exponent . . . . . . . . . . . . . . . 206

69

Elementary proofs of some Blackburn’s theorems

70

Non-2-generator p-groups all of whose maximal subgroups are 2-generator 214

71

Determination of A2 -groups . . . . . . . . . . . . . . . . . . . . . . . . 233

72

An -groups, n > 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248

73

Classification of modular p-groups . . . . . . . . . . . . . . . . . . . . 257

74

p-groups with a cyclic subgroup of index p 2 . . . . . . . . . . . . . . . 274

75

Elements of order 4 in p-groups

76

p-groups with few A1 -subgroups . . . . . . . . . . . . . . . . . . . . . 282

77

2-groups with a self-centralizing abelian subgroup of type .4; 2/ . . . . . 316

78

Minimal nonmodular p-groups

79

Nonmodular quaternion-free 2-groups

. . . . . . . . . . . . . . . . . . 334

80

Minimal non-quaternion-free 2-groups

. . . . . . . . . . . . . . . . . . 356

81

Maximal abelian subgroups in 2-groups . . . . . . . . . . . . . . . . . . 361

82

A classification of 2-groups with exactly three involutions . . . . . . . . 368

83

p-groups G with 2 .G/ or 2 .G/ extraspecial

84

2-groups whose nonmetacyclic subgroups are generated by involutions . 399

85

2-groups with a nonabelian Frattini subgroup of order 16

86

p-groups G with metacyclic 2 .G/

87

2-groups with exactly one nonmetacyclic maximal subgroup . . . . . . . 412

88

Hall chains in normal subgroups of p-groups . . . . . . . . . . . . . . . 437

89

2-groups with exactly six cyclic subgroups of order 4

. . . . . . . . . . . . 209

. . . . . . . . . . . . . . . . . . . . 277

. . . . . . . . . . . . . . . . . . . . . . 323

. . . . . . . . . . . . . 396

. . . . . . . . 402

. . . . . . . . . . . . . . . . . . . 406

. . . . . . . . . . 454

Contents

vii

90

Nonabelian 2-groups all of whose minimal nonabelian subgroups are of order 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463

91

Maximal abelian subgroups of p-groups

92

On minimal nonabelian subgroups of p-groups . . . . . . . . . . . . . . 474

. . . . . . . . . . . . . . . . . 467

Appendix A.16 Some central products . . . . . . . . . . . . . . . . . . . . . . . . . . . 485 A.17 Alternate proofs of characterization theorems of Miller and Janko on 2groups, and some related results . . . . . . . . . . . . . . . . . . . . . . 492 A.18 Replacement theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . 501 A.19 New proof of Ward’s theorem on quaternion-free 2-groups . . . . . . . . 506 A.20 Some remarks on automorphisms . . . . . . . . . . . . . . . . . . . . . 509 A.21 Isaacs’ examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512 A.22 Minimal nonnilpotent groups

. . . . . . . . . . . . . . . . . . . . . . . 516

A.23 Groups all of whose noncentral conjugacy classes have the same size . . 519 A.24 On modular 2-groups

. . . . . . . . . . . . . . . . . . . . . . . . . . . 522

A.25 Schreier’s inequality for p-groups . . . . . . . . . . . . . . . . . . . . . 526 A.26 p-groups all of whose nonabelian maximal subgroups are either absolutely regular or of maximal class . . . . . . . . . . . . . . . . . . . . . . . . 529 Research problems and themes II

. . . . . . . . . . . . . . . . . . . . . . . . 531

Author index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593 Subject index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594

List of definitions and notations

Set theory jM j is the cardinality of a set M (if G is a finite group, then jGj is called its order). x 2 M (x 62 M ) means that x is (is not) an element of a set M . N M (N 6 M ) means that N is (is not) a subset of the set M ; moreover, if M ¤ N M we write N M. ¿ is the empty set. N is called a nontrivial subset of M , if N ¤ ¿ and N M . If N M we say that N is a proper subset of M . M \ N is the intersection and M [ N is the union of sets M and N . If M; N are sets, then N M D fx 2 N j x 62 M g is the difference of N and M . Z is the set (ring) of integers: Z D f0; ˙1; ˙2; : : : g. N is the set of all natural numbers. Q is the set (field) of all rational numbers. R is the set (field) of all real numbers. C is the set (field) of all complex numbers.

Number theory and general algebra p is always a prime number. is a set of primes; 0 is the set of all primes not contained in . m, n, k, r, s are, as a rule, natural numbers. .m/ is the set of prime divisors of m; then m is a -number. np is the p-part of n, n is the -part of n. .m; n/ is the greatest common divisor of m and n. m j n should be read as: m divides n. m − n should be read as: m does not divide n.

ix

List of definitions and notations

GF.p m / is the finite field containing p m elements. F is the multiplicative group of a field F. L.G/ is the lattice of all subgroups of a group G. If n D p1˛1 : : : pk˛k is the standard prime decomposition of n, then .n/ D

Pk

iD1 ˛i .

Groups We consider only finite groups which are denoted, with a pair exceptions, by upper case Latin letters. If G is a group, then .G/ D .jGj/. G is a p-group if jGj is a power of p; G is a -group if .G/ . G is, as a rule, a finite p-group. H G means that H is a subgroup of G. H < G means that H G and H ¤ G (in that case H is called a proper subgroup of G). f1g denotes the group containing only one element. H is a nontrivial subgroup of G if f1g < H < G. H is a maximal subgroup of G if H < G and it follows from H M < G that H D M. H E G means that H is a normal subgroup of G; moreover, if, in addition, H ¤ G we write H GG and say that H is a proper normal subgroup of G. Expressions ‘normal subgroup of G’ and ‘G-invariant subgroup’ are synonyms. H G G is called a nontrivial normal subgroup of G provided H > f1g. H is a minimal normal subgroup of G if (a) H E G; (b) H > f1g; (c) N G G and N < H implies N D f1g. Thus, the group f1g has no minimal normal subgroup. G is simple if it is a minimal normal subgroup of G (so jGj > 1). H is a maximal normal subgroup of G if H < G and G=H is simple. The subgroup generated by all minimal normal subgroups of G is called the socle of G and denoted by Sc.G/. We put, by definition, Sc.f1g/ D f1g. NG .M / D fx 2 G j x 1 M x D M g is the normalizer of a subset M in G. CG .x/ is the centralizer of an element x in G W CG .x/ D fz 2 G j zx D xzg. T CG .M / D x2M CG .x/ is the centralizer of a subset M in G. If A B and A; B E G, then CG .B=A/ D H , where H=A D CG=A .B=A/.

x

Groups of prime power order

A wr B is the wreath product of the ‘passive’ group A and the transitive permutation group B (in what follows we assume that B is regular); B is called the active factor of the wreath product). Then the order of that group is jAjjBj jBj. Aut.G/ is the group of automorphisms of G (the automorphism group of G). Inn.G/ is the group of all inner automorphisms of G. Out.G/ D Aut.G/=Inn.G/, the outer automorphism group of G. If a; b 2 G, then ab D b 1 ab. An element x 2 G inverts a subgroup H G if hx D h1 for all h 2 H . If M G, then hM i D hx j x 2 M i is the subgroup of G generated by M . M x D x 1 M x D fy x j y 2 M g for x 2 G and M G. Œx; y D x 1 y 1 xy D x 1 x y is the commutator of elements x; y of G. If M; N G then ŒM; N D hŒx; y j x 2 M; y 2 N i is a subgroup of G. o.x/ is the order of an element x of G. An element x 2 G is a -element if .o.x// . G is a -group, if .G/ . Obviously, G is a -group if and only if all of its elements are -elements. G 0 is the subgroup generated by all commutators Œx; y, x; y 2 G (i.e., G 0 D ŒG; G), G .2/ D ŒG 0 ; G 0 D G 00 D .G 0 /0 , G .3/ D ŒG 00 ; G 00 D .G 00 /0 and so on. G 0 is called the commutator (or derived) subgroup of G. T Z.G/ D x2G CG .x/ is the center of G. Zi .G/ is the i -th member of the upper central series of G; in particular, Z0 .G/ D f1g, Z1 .G/ D Z.G/. Ki .G/ is the i -th member of the lower central series of G; in particular, K2 .G/ D G 0 . We have Ki .G/ D ŒG; : : : ; G (i 1 times). We set K1 .G/ D G. If G is nonabelian, then .G/=K3 .G/ D Z.G=K3 .G//. M.G/ D hx 2 G j CG .x/ D CG .x p / is the Mann subgroup of a p-group G. Sylp .G/ is the set of p-Sylow subgroups of an arbitrary finite group G. Sn is the symmetric group of degree n. An is the alternating group of degree n †p n is a Sylow p-subgroup of Sp n . GL.n; F / is the set of all nonsingular n n matrices with entries in a field F , the n-dimensional general linear group over F , SL.n; F / D fA 2 GL.n; F / j det.A/ D 1 2 F g, the n-dimensional special linear group over F .

List of definitions and notations

xi

T If H T G, then HG D x2G x 1 H x is the core of the subgroup H in G and H G D H N EG N is the normal closure or normal hull of H in G. Obviously, HG E G. If G is a p-group, then p b.x/ D jG W CG .x/j; b.x/ is said to be the breadth of x 2 G, where G is a p-group; b.G/ D max fb.x/ j x 2 Gg is the breadth of G. ˆ.G/ is the Frattini subgroup of G (= the intersection of all maximal subgroups of G), ˆ.f1g/ D f1g, p d.G/ D jG W ˆ.G/j. i D fH < G j ˆ.G/ H; jG W H j D p i g, i D 1; : : : ; d.G/, where G > f1g. If H < G, then 1 .H / is the set of all maximal subgroups of H . exp.G/ is the exponent of G (the least common multiple of the orders of elements of G). If G is a p-group, then exp.G/ D max fo.x/ j x 2 Gg. k.G/ is the number of conjugacy classes of G (D G-classes), the class number of G. Kx is the G-class containing an element x (sometimes we also write ccl G .x/). Cm is the cyclic group of order m. G m is the direct product of m copies of a group G. A B is the direct product of groups A and B. A B is a central product of groups A and B, i.e., A B D AB with ŒA; B D f1g. Ep m D Cpm is the elementary abelian group of order p m . G is an elementary abelian p-group if and only if it is a p-group > f1g and G coincides with its socle. Next, f1g is elementary abelian for each prime p. A group G is said to be homocyclic if it is a direct product of isomorphic cyclic subgroups (obviously, elementary abelian p-groups are homocyclic). ES.m; p/ is an extraspecial group of order p 1C2m (a p-group G is said to be extraspecial if G 0 D ˆ.G/ D Z.G/ is of order p). Note that for each m 2 N, there are exactly two nonisomorphic extraspecial groups of order p 2mC1 . S.p 3 / is a nonabelian group of order p 3 and exponent p > 2. A special p-group is a nonabelian p-group G such that G 0 D ˆ.G/ D Z.G/ is elementary abelian. Direct products of extraspecial p-groups are special. D2m is the dihedral group of order 2m; m > 2. Some authors consider E22 as the dihedral group D4 . Q2m is the generalized quaternion group of order 2m 23 . SD2m is the semidihedral group of order 2m 24 . Mp m is a nonabelian p-group containing exactly p cyclic subgroups of index p.

xii

Groups of prime power order

cl.G/ is the nilpotence class of a p-group G. dl.G/ is the derived length of a p-group G. CL.G/ is the set of all G-classes. A p-group of maximal class is a nonabelian group G of order p m with cl.G/ D m 1. m .G/ D hx 2 G j o.x/ p m i, m .G/ D hx 2 G j o.x/ D p m i and Ãm .G/ D m hx p j x 2 Gi. A p-group is absolutely regular if jG=Ã1 .G/j < p p . A p-group is thin if it is either absolutely regular or of maximal class. G D A B is a semidirect product with kernel B and complement A. A group G is an extension of N E G by a group H if G=N Š H . A group G splits over N if G D H N with H G and H \ N D f1g (in that case, G is a semidirect product of H and N with kernel N ). H # D H feH g, where eH is the identity element of the group H . If M G, then M # D M feG g. An automorphism ˛ of G is regular (D fixed-point-free) if it induces a regular permutation on G # (a permutation is said to be regular if it has no fixed points). An involution is an element of order 2 in a group. A section of a group G is an epimorphic image of some subgroup of G. If F D GF.p n /, then we write GL.m; p n /; SL.m; p n /; : : : instead of GL.m; F /, SL.m; F /; : : : . cn .G/ is the number of cyclic subgroups of order p n in a p-group G. sn .G/ is the number of subgroups of order p n in a p-group G. en .G/ is the number of subgroups of order p n and exponent p in G. An -group is a p-group G all of whose subgroups of index p n are abelian but G contains a nonabelian subgroup of index p n1 . In particular, A1 -group is a minimal nonabelian p-group for some p. ˛n .G/ is the number of An -subgroups in a p-group G.

Characters and representations Irr.G/ is the set of all irreducible characters of G over C. A character of degree 1 is said to be linear. Lin.G/ is the set of all linear characters of G (obviously, Lin.G/ Irr.G/).

List of definitions and notations

xiii

Irr1 .G/ D Irr.G/ Lin.G/ is the set of all nonlinear irreducible characters of G; n.G/ D jIrr1 .G/j. .1/ is the degree of a character of G, H is the restriction of a character of G to H G. G is the character of G induced from the character of some subgroup of G. N is a character of G defined as follows: .x/ N D .x/ (here wN is the complex conjugate of w 2 C). Irr. / is the set of irreducible constituents of a character of G. If is a character of G, then ker. / D fx 2 G j .x/ D .1/g is the kernel of a character . Z. / D fx 2 G j j .x/j D .1/g is the quasikernel of . If N E G, then Irr.G j N / D f 2 Irr.G/ j N — ker. /g. P h ; i D jGj1 x2G .x/ .x 1 / is the inner product of characters and of G. IG ./ D hx 2 G j x D i is the inertia subgroup of 2 Irr.H / in G, where H G G. 1G is the principal character of G (1G .x/ D 1 for all x 2 G). M.G/ is the Schur multiplier of G. cd.G/ D f .1/ j 2 Irr.G/g. mc.G/ D k.G/=jGj is the measure of commutativity of G. P T.G/ D 2Irr.G/ .1/; f.G/ D T.G/=jGj.

Preface

This is the second part of the book. Sections 48–57, 60, 61, 66, 67, 70, 71, 73–75, 77– 87, 89–92 and Appendix 19 are written by the second author, all other sections – by the first author. This volume contains a number of very strong results on 2-groups due to the second author. All exercises and about all problems are due to the first author. All material of this part is appeared in the book form at the first time. Some outstanding problems of p-group theory are solved in this volume: (i) classification of 2-groups with exactly three involutions, (ii) classification of 2-groups containing exactly one nonmetacyclic maximal subgroup, (iii) classification of 2-groups G containing an involution t such that CG .t / D ht i Q, where Q contains only one involution, (iv) classification of 2-groups all of whose minimal nonabelian subgroups have the same order 8, (v) classification of 2-groups of rank 3 all of whose maximal subgroups are of rank 2, (vi) classification of 2-groups containing selfcentralizing noncyclic abelian subgroups of order 8, and so on. There are, in this part, a number of new proofs of known important results: (a) Blackburn’s classification of minimal nonmetacyclic groups (we presented, in Sec. 66 and 69, two different proofs), (b) classification of p-groups all of whose subgroups of index p 2 are abelian, (c) Ward’s theorem on quaternion-free 2-groups (we presented two different proofs), (d) classification of p-groups with cyclic subgroup of index p 2 , (e) Kazarin’s classification of p-groups all of whose cyclic subgroups of order > p are normal, (f) Iwasawa’s classification of modular p-groups, and so on. Some results proved in this part have no analogs in existing literature: (a) classification of 2-groups all of whose minimal nonabelian subgroups are isomorphic and have order 16, (b) study the 2-groups with at most 1 C p C p 2 minimal nonabelian subgroups,

Preface

xv

(c) classification of 2-groups all of whose nonabelian two-generator subgroups are of maximal class, (d) classification of 2-groups G with j2 .G/j 24 and j2 .G/j D 24 , (e) classification of p-groups G with 2 .G/ or 2 .G/ is metacyclic (extraspecial), (f) classification of 2-groups all of whose nonabelian subgroups are generated by involutions, and so on. As the previous part, this one contains a great number of open problems posed, as a rule, by the first author; some of these problems are solved and solutions are presented below. The first author is indebted to Avinoam Mann for numerous useful discussions and help. The correspondence with Martin Isaacs allowed us to acquaint the reader with a number of his old and new important results. Moreover, Mann and Isaacs familiarized us with a number of their papers prior of publication. Noboru Ito read a number of sections and all appendices and made numerous useful remarks and suggestions. The help of Lev Kazarin was very important and allowed us to improve a number of places of the book, especially, in 46 and 63; he also acquainted the first author with a fragment of his PhD thesis (see 65, 71). The first author also indebted to Gregory Freiman, Marcel Herzog (both at Tel-Aviv University), Moshe Roitman and Izu Vaisman (both at University of Haifa) for help and support. The publication of the book gives us great pleasure. We are grateful to the publishing house of Walter de Gruyter and all who promoted the publication, among of them Prof. M. Hazewinkel, Dr. R. Plato and K. Dimler, for their support and competent handling of the project.

46

Degrees of irreducible characters of Suzuki p-groups

The results of this section are due to I. A. Sagirov [Sag1, Sag2]. Throughout this section we use the following notation: F D GF.p m /, m > 1; is an automorphism of F of order k for some divisor k > 1 of m (recall that the group of automorphisms of F is cyclic of order m whose generator is a 7! ap for all a 2 F); n D m .< m/. k n Let, for example, W a 7! ap (a 2 F). The set of fixed points of is a subfield n F of F satisfying ap D 1 so containing p n elements (for example, if k D m, then F D F0 , the prime subfield of F). Let F be the (cyclic) multiplicative group of F. Next, let Irr1 .G/ denote the set of all nonlinear irreducible characters of G. Put cd.G/ D f .1/ j 2 Irr.G/g. Next, Irr.t/ .G/ denotes the number of characters of degree t in Irr.G/. Definition. The Suzuki p-group Ap .m; / is the set F F with multiplication defined as follows: .a; b/.c; d / D .a C c; b C d C a .c//. Let G D Ap .m; /; then jGj D jFj2 D p 2m . It follows from the definition that elements .a; b/ and .c; d / commute if and only if c .a/ D a .c/, i.e., either a D 0 or .c=a/ D c=a so c D au for some u 2 F . The identity element of G is .0; 0/ and .a; b/1 D .a; b C a .a//. Since so defined multiplication is associative, G indeed is a group. If .c; d / 2 Z.G/ and .a; b/ 2 G, then c D au for u 2 F and all a 2 F which implies c D 0 since k > 1. Thus, Z.G/ D f.0; d / j d 2 Fg. We have Z.G/ Š Ep m . It is easy to prove, by induction that .a; b/n D .na; nb C n2 a .a//. Taking n D p, we get .a; b/p 2 Z.G/; moreover, if p > 2, then exp.G/ D p. 1o . Throughout this subsection p D 2 and G D A.m; / D A2 .m; /. Our aim is to find cd.G/ and the number of irreducible characters of every degree s 2 cd.G/. Theorem 46.1 ([Sag1]). Suppose that p D 2, G D A.m; /, where m > 1 and is an automorphism of the field F D GF.2m / of order k > 1 and n D m . Then one of the k following holds: 1

(a) If k is odd, then cd.G/ D f1; 2 2 .mn/ g. (b) If k D 2, then cd.G/ D f1; 2m=2 g D f1; 2n g. (c) If k > 2 is even, then cd.G/ D f1; 2m=2 ; 2.m=2/n g, jIrr.2m=2 / .G/j D and jIrr.2.m=2/n / .G/j D

.2m 1/22n 2n C1 .

.2m 1/2n 2n C1

2

Groups of prime power order s

If x 2 F then, since is an automorphism of F of order k > 1, then .x/ D x 2 for some nonnegative s independent of x and k .x/ D x, x 2 F. On the other hand, sk sk k .x/ D x 2 so x 2 D x, and this is true for each x 2 F. If x is a primitive element of F, it follows that 2m 1 divides 2sk 1 so m.D nk/ divides sk (see Lemma 46.5 below) hence n divides s, and we conclude that s D nt , where .t; k/ D 1 since nt o. / D k. Thus, .x/ D x 2 . Therefore, the number of automorphisms of F of order k equals '.k/, where './ is the Euler’s totient function. Then fa 2 F j .a/ D nt ag D F is the set of elements a 2 F such that a2 D a and the cardinality of that set is 2n . As we have shown, if a ¤ 0, then CG ..a; b// D f.az; u/ j z 2 F ; u 2 Fg so jCG ..a; b//j D 2mCn . In particular, the size of every noncentral G-class equals 22m D 2mn . If a 2 F f0g, then .a; b/2 D .0; a .a// ¤ .0; 0/ so Z.G/ D 1 .G/. 2mCn Therefore, since exp.G/ D 4, we get 1 .G/ D Ã1 .G/ D ˆ.G/. Lemma 46.2. jG 0 j 2mn . Indeed, jG 0 j is at least the size of a noncentral G-class. However, all noncentral G-classes have the same size 2mn . From the last assertion follows Lemma 46.3. k.G/ D jZ.G/j C

jGjjZ.G/j 2mn

D 2mCn C 2m 2n .

An element 2 F is said to be primitive if the (cyclic) multiplicative group F D hi. Lemma 46.4. A mapping ' ..a; b// D .a; ./b/ is an automorphism of G of order o./ for every element 2 .F /# . Proof. Set ./ D . Then ' ..a; b// D .a; b/, ' ..c; d // D .c; d /, ' ..a; b/.c; d // D ' ..a C c; b C d C a .c// D ..a C c/; .b C d C a .c///: On the other hand, ' ..a; b//' ..c; d // D .a; b/.c; d / D ..a C c/; b C d C a .c// D ..a C c/; .b C d C a .c///: It follows that ' is an endomorphism of G. Next, .a; b/ D ' ..a; b// D .0; 0/ if and only if .a; b/ D .0; 0/ so ' is an automorphism of G since G is finite. Set o./ D d . Then 'd ..a; b// D .d a; . .//d b/ D .a; b/ and, if a ¤ 0, then 'r ..a; b// ¤ .a; b/ for r 2 f1; 2; : : : ; d 1g. It follows that o.' / D d D o./. Lemma 46.5. Given a > 1, m; n, we have .am 1; an 1/ D a.m;n/ 1. Proof. Set .m:n/ D ı and d D .am 1; an 1/. Then there exist u; v 2 N such that ı D mu nv. Clearly, aı 1 divides d . On the other hand, d divides amu 1 and anv 1. Hence d divides the number amu 1 .anv 1/ D anv .amunv 1/ D anv .aı 1/ so d divides aı 1 since .d; a/ D 1, and we conclude that d D aı 1.

46

Degrees of irreducible characters of Suzuki p-groups

3

Lemma 46.6. Let m D nk, where n; k 2 N, and let t 2 N be coprime with k; then .m; nt / D n. (a) Set d D .2m 1; 2nt C 1/. Then d D 1 if k is odd and d D 2n C 1 if k is even. (b) Suppose that 2n C 1 divides 2m 1. Then 2n divides m. Proof. (a) The number d0 D .2m 1; 22nt 1/ equals 2.nk;2nt/ 1 D 2n 1 if k is odd and d0 D 22n 1 if k is even (Lemma 46.5). Since 2nt C 1 divides 22nt 1, the number d divides d0 . We claim that .2nt C 1; 2n 1/ D 1. Indeed, if a prime p divides that number, then 2n 1 .mod p/ so 2nt 1t 1 .mod p/, a contradiction since p (if it exists) must be odd and 2nt 1 .mod p/. Suppose that k is odd. By definition, d divides 2nt C 1 and, by the first paragraph, d divides d0 D 2n 1 so d divides .2nt C 1; 2n 1/ D 1 (see the previous paragraph), and we get d D 1. This proves the first assertion in (a). Now suppose that k is even; then t is odd. In that case, by the first paragraph, d divides d0 D 22n 1 D .2n 1/.2n C 1/ and, by definition, d divides 2nt C 1. Therefore, by the second paragraph, d divides .2n C 1; 2nt C 1/ D 2n C 1 since t is odd. Since 2n C 1 divides d , we get d D 2n C 1. The proof of (a) is complete. (b) Since .2n 1; 2n C 1/ D 1 and 2n 1 divides 2m 1 D 2nk 1, it follows that 22n 1 D .2n 1/.2n C 1/ divides 2m 1 and so 2n divides m (Lemma 46.5). Write ˆ D h' i ( 2 F# ); then jˆ j D o./. If is primitive, we write ˆ D ˆ Lemma 46.7. Let 2 F be primitive; then jˆj D o./ D 2m 1 and: (a) The size of every ˆ-orbit on the set G Z.G/ equals 2m 1. (b) For odd k, there is only one ˆ-orbit on the set Z.G/# . (In that case, ˆ G is a Frobenius group with minimal normal subgroup Z.G/, its kernel.) (c) For even k, the size of every ˆ-orbit on the set Z.G/# equals the number of ˆ-orbits on Z.G/# equals 2n C 1/.

2m 1 2n C1

(in that case,

Proof. (a) For .a; b/ 2 G and i 2 N, we have 'i ..a; b// D .i a; . .//i b/. Let .a; b/ 2 G Z.G/. In that case, a ¤ 0 so 'i ..a; b// D .a; b/ if and only if i D 1, i.e., i is divisible by 2m 1 since is primitive. This proves (a) (b) Now let k be odd and b ¤ 0. Then 'i ..0; b// D .0; . ./i b/ D .0; b/ if and nt only if . .//i D 1. Since ./ D 2 with .t; k/ D 1, we get 1 D . .//i D .2nt C1/i m . Therefore, o./ D 2 1 divides .2nt C 1/i . By Lemma 46.6(a), if k is odd, 2m 1 divides i since .2m 1; 2nt C 1/ D 1, so the ˆ-orbit of .0; b/ contains 2m 1 elements. Then all elements of Z.G/# are ˆ-conjugate, and (b) is proven. nt (c) Let k be even. The minimal positive integer i such that .2 C1/i D 1 equals 2m 1 2n C1 , by Lemma 46.6(a), and this number is the size of the ˆ-orbit of the element .0; b/ 2 Z.G/# . It follows that the number of ˆ-orbits on Z.G/# equals 2n C 1, completing the proof of (c).

4

Groups of prime power order

For 2 F f1F g, ' has no fixed points on .G=Z.G//# (check!) so h' ; G=Z.G/i is a Frobenius group. Corollary 46.8. If k is odd, then G 0 D Z.G/. Proof. Indeed, Z.G/ is a minimal normal subgroup of h' ; Z.G/i, where is a primitive element of F, by Lemma 46.7(b), and G 0 Z.G/ so G 0 D Z.G/. Remark. Suppose that 2 Irr1 .G/, .1/ D 2a and Z0 D Z.G/ \ ker. /, where G D A.m; /. Then jZ.G/ W Z0 j D 2 since the center of G= ker. / is cyclic. It follows that G 0 Z0 D Z.G/, cd.G=Z0 / D f1; 2a g [Isa1, Theorem 2.31] and so jIrr1 .G=Z0 /j D 2mC1 2m D 2m2a . A character 2 Irr1 .G/ determines Z0 uniquely. 22a Theorem 46.9 ([Han]). If G D A.m; / is a Suzuki 2-group, where is an automorphism of F of odd order k > 1, then cd.G/ D f1; 2.mn/=2 g (here n D m ). k Proof. Let 2 F be primitive. Consider the action of the automorphism ' on subgroups of index 2 in Z.G/. By Lemma 46.7(b), h' ; Z.G/i is a Frobenius group and Z.G/# is the unique ' -orbit. By Maschke’s theorem, if 's ¤ id, then that automorphism has no invariant subgroups of index 2 in Z.G/, i.e., h' i acts on the set of such subgroups in a fixed-point-free manner. Since the order of ' equals 2m 1, all 2m 1 subgroups of index 2 in Z.G/ are .ˆ D/h' i-conjugate. Let 2 Irr1 .G/, Z0 D Z.G/ \ ker. /. Then the isomorphism type of G=Z0 independent of , since all subgroups of index 2 in Z.G/ are ˆ-conjugate. It follows that cd.G/ D f1; d g for some d > 1, by the Remark. By Lemma 46.3 and Corollary 1 46.8, we get 2m C .2mCn 2n /d 2 D 22m so d D 2 2 .mn/ . Thus, Theorem 46.1(a) is proven. In what follows, k is even. Lemma 46.10. Let 2 F be primitive and k even. Then: (a) If k > 2, i.e., m D k n > 2n, then G 0 D Z.G/ and the number of ˆ-orbits on m 1 . .G 0 /# equals 2n C 1, and all these orbits have the same size 22n C1 (b) If k D 2, i.e., m D 2n, then either G 0 D Z.G/ and .G 0 /# has 2n C 1 distinct m 1 m D 2n 1 or else jG 0 j D 2n .D 2 2 / and all elements of ˆ-orbits of size 22n C1 .G 0 /# are ˆ-conjugate. Proof. The assertion on the sizes of ˆ-orbits follows from Lemma 46.7(c). Suppose m 1 that the number of ˆ-orbits on .G 0 /# equals t and jG 0 j D 2f . Then 2f 1 D t 22n C1 (see Lemma 46.7(c)). It follows that t 1 D t 2m C 2n 2f Cn 2f . By Lemma 46.2, f m n n so 2n divides t 1. It follows that either t D 1 or t 2n C 1. On the other hand, f m since G 0 Z.G/ and, by the previous paragraph, m 1 D 2f 1 2m 1. It follows that t 2n C 1 so t 2 f1; 2n C 1g. t 22n C1 m 1 Suppose that t D 1. Then, by the previous paragraph, 2f 1 D 22n C1 so 2n .2mn C f n 1/ D 2 .2 C 1/. It follows that then f D n and m n D n, i.e., m D 2n, k D 2. In that case, f D n D 12 m, and we have the second possibility in (b). m 1 D 2m 1 so f D m. Suppose that t D 2n C 1. Then 2f 1 D .2n C 1/ 22n C1

46

5

Degrees of irreducible characters of Suzuki p-groups

In what follows we first assume that G 0 D Z.G/. Then the number of ˆ-orbits on m 1 (Lemma 46.10). .G 0 /# equals 2n C 1 and all these orbits have the same size 22n C1 Lemma 46.11. Suppose that G 0 D Z.G/, 2 F is primitive and k is even. (a) Under ˆ, the set of subgroups of index 2 in Z.G/ is partitioned in 2n C 1 orbits m 1 of size 22n C1 . (b) Under ˆ, the set Irr.G/ f1G g is partitioned in 2n orbits of size 2m 1 and m 1 2n C 1 orbits of size 22n C1 . Proof. Z.G/ has 2m 1 subgroups of index 2 and jIrr.G/j D 2mCn C 2m 2n . (a) By Lemma 46.7, the restriction of ' to Z.G/# is a product of 2n C1 independent m 1 cycles of size 22n C1 . This proves (a) (see the first paragraph in the proof of Lemma 46.9). (b) Define the action of ' on the set Irr.G/ as follows. Let R be a representation affording the character 2 Irr.G/. Set T D R ı '1 , i.e., T .g/ D R.g'1 / for g 2 G. If h 2 G, then since ' 2 Aut.G/, we get T .gh/ D R..gh/'1 / D R.g'1 h'1 / D R.g'1 /R.h'1 / D T .g/T .h/; i.e., T is a representation of G. Clearly, the value of the character ' .g/ of T equals .g'1 / for every g 2 G, and this character is also irreducible. Let CL.G/ D fK1 ; : : : ; Kr g be the set of G-classes, xi 2 Ki , all i , and let X D X.G/ D . i .xj // be the character table of G. Then ' acts on columns and rows of X by permutations 1 and 2 , respectively, in such a way that if one considers these '

' 1

permutations as elements of GL.r; C/, then 2 X D . i .xj // D . i .xj // D X1 . Since the matrix X is nonsingular, we get 1 D X 1 2 X in GL.r; C/. Then, by Vishnevetsky’s lemma [BZ, Lemma 10.4(a)], 1 and 2 are conjugate in the symmetric group Sr and so they have the same cycle structure. Now the result follows m 1 on from Lemma 46.7. Indeed, under ˆ, there are exactly 2n C 1 orbits of size 22n C1 2 D 2n orbits of size 2m 1 on noncentral nonidentity central G-classes and 2 2m 1 m 1 n and 2n orbits of size 2m 1 G-classes. It follows that ˆ has 2 C 1 orbits of size 22n C1 on the set Irr.G/ f1G g. mCn

n

Lemma 46.12. If 2 Irr1 .G/, k is even and .1/ D 2a , then 12 m n a 12 m. Proof. By the Remark, we have 2 Irr1 .G=Z0 /, where Z0 < Z.G/ is of index 2, and cd.G=Z0 / D f1; 2a g so jIrr1 .G=Z0 /j D 2m2a . It follows that m 2a 0 so a 12 m. Let Z1 ; : : : ; Z2n C1 be a transversal of the set of ˆ-orbits on the set M of subgroups of index 2 in Z.G/. By the Remark, cd.G=Zi / D f1; 2ai g, jIrr1 .G=Zi /j D 2m2ai , all i . Next, if Zi and Zi;1 belong to the same ˆ-orbit, then, obviously, G=Zi Š G=Zi;1 . m 1 P2n C1 m2ai (here 2m 1 Since jIrr1 .G/j D 2mCn 2n , we get 2mCn 2n D 22n C1 iD1 2 2n C1 P2n C1 m2a i D 2n .2n C 1/. is the common size of ˆ-orbit) which implies that iD1 2

6

Groups of prime power order

Assume that m 2ai > 2n for some i . Then 2m2ai 22nC1 > 2n .2n C 1/, contrary to the result of the previous paragraph. It follows that m 2ai 2n so ai 12 m n for all i . Theorem 46.13. If k is even and G 0 D Z.G/, then cd.G/ D f1; 2.m=2/n ; 2m=2 g. Next, jIrr.2.m=2n / .G/j D .2m 1/22n =.2n C1/, jIrr.2m=2 / .G/j D .2m 1/2n =.2n C1/. 1 . Proof. Set s D 22n C1 By Lemma 46.11(b), the set Irr.G/ f1G g is partitioned under the action of ˆ in 2n orbits of size 2m 1 and 2n C 1 orbits of size s. Since hˆ; Gi=G 0 is a Frobenius group (see the paragraph following Lemma 46.7), all jG W G 0 j 1 D 2m 1 nonprincipal linear characters of G are ˆ-conjugate. Therefore, under ˆ, there are exactly 2n 1 orbits of size 2m 1 on Irr1 .G/. By Lemma 46.11(a), the set M of subgroups of index 2 in Z.G/ is partitioned in exactly 2n C 1 orbits of size s under ˆ. Then, if Z0 2 M, then 's .Z0 / D Z0 since the order of the restriction of ' to M is s. Suppose that the size of the ˆ-orbit of 2 Irr1 .G/ equals 2m 1 and 2 i Irr1 .G=Z0 / for some Z0 2 M. Then ¤ ' 2 Irr1 .G=Z0 / for 0 < i < 2n C 1 so the number of ˆ-conjugates of equals 2n C 1. Suppose that for some Z0 2 M, Irr1 .G=Z0 / has no characters belonging to a ˆorbit of size s. Let cd.G=Z0 / D f1; 2a g. Then all of the 2m2a characters of the set Irr1 .G=Z0 / are partitioned in hˆi-orbits of the same odd size > 1, which is not the case. It follows that, for each Z0 2 M, there exists in Irr1 .G=Z0 / a character such that its ˆ-orbit has size s. Since the number of such ˆ-orbits is 2n C 1 and, under ˆ, the set M has exactly 2n C 1 orbits of size s (Lemma 46.11(a)), it follows that each set Irr1 .G=Z0 / (where Z0 2 M) has exactly one character belonging to a ˆ-orbit of size s since s.2n C 1/ D 2m 1 is the total number of nonprincipal irreducible characters in ˆ-orbits of size s, by Lemma 46.11(b). Now suppose that the size of the ˆ-orbit of a character 2 Irr1 .G/ is 2m 1 (such exists since jIrr1 .G/j D 2n .2m 1/ > s.2n C 1/), .1/ D 2a and 2 Irr1 .G=Z0 /, where Z0 2 M. By the previous paragraph, jIrr1 .G=Z0 /j D 2m2a D 1 C t .2n C 1/, where t is the number of orbits of characters in Irr1 .G=Z0 / belonging to ˆ-orbits of size ¤ s (there are in Irr1 .G=Z0 / exactly 2n C1 characters conjugate with under ˆ). It follows that 2n C 1 divides 2m2a 1, and so n divides m 2a (Lemma 46.5) and m2a is even (Lemma 46.6(b)). By Lemma 46.12, m 2a 2n m so m2a 2, n n m2a i.e., n 2 f0; 2g. Since G=Z0 has at least two nonlinear irreducible characters, by the previous paragraph, m 2a > 0. Hence, m 2a D 2n so a D 12 m n and m

1 Irr1 .G=Z0 / has exactly 2m2a D 22n characters that lie in t D 2 2n C11 D 22n C1 n m 2 1 distinct ˆ-orbits of size 2 1 and one ˆ-orbit of size s. m 1 We see that there are 22n s D 22n 22n C1 nonlinear irreducible characters of m 1 of degree 2.m=2/n . The sum of squares of their degrees is 2m2n 22n 22n C1 m m 1 1 2m 22n C1 . The sum of squares of degrees of remaining t1 D 2n .2m 1/22n 22n C1 m2a

2n

D G D D

46

Degrees of irreducible characters of Suzuki p-groups

1 2n 22n C1 nonlinear irreducible characters of G is † D 2m .2m 1/ 2m m

7 2m 1 2n C1 D t1 2m , it

1/ 2m 2 2.2n C1 . If 2 Irr1 .G/, then .1/2 jG W Z.G/j D 2m . Since † D follows that all latter characters have the same degree 2m=2 . This means that cd.G/ D f1; 2.m=2/n ; 2m=2 g, and G has exactly 22n s irreducible characters of degree 2.m=2/n and t1 D 2n s irreducible characters of degree 2m=2 . n

m

Corollary 46.14. If k D 2, i.e., m D 2n, then jG 0 j D 2m=2 . Proof. In that case, by Lemma 46.10, either jG 0 j D 2m=2 or else jG 0 j D 2m . In the second case, however, by Theorem 46.13, cd.G/ D f1; 2.m=2/n ; 2m=2 g, which is not the case since m 2n D 0. Theorem 46.15. If k D 2, then cd.G/ D f1; 2m=2 g. Proof. By Corollary 46.14, jG 0 j D 2m=2 , jG W G 0 j D 2mC.m=2/ D 2mCn D jLin.G/j. Since k.G/ D 2mCn C 2m 2n , we get jIrr1 .G/j D jIrr.G/j 2mCn D 2m 2n D 2m=2 .2m=2 1/. By the remark following Corollary 46.8, if 2 Irr1 .G/, then 2 Irr1 .G=Z0 /, where Z0 2 M, G 0 6 Z0 . The number of such subgroups Z0 equals 2m 1 .2m=2 1/ D 2m=2 .2m=2 1/ D jIrr1 .G/j. (Here 2m 1 D jMj and 2m=2 1 is the number of subgroups of index 2 in Z.G/=G 0 ; this explains our formula.) Thus, to every subgroup Z0 corresponds the unique nonlinear irreducible character of G=Z0 . Since jG=Z0 j D 2mC1 and j.G=Z0 /0 j D 2 so G=Z0 is extraspecial, we get cd.G=Z0 / D f1; 2m=2 g. Theorem 46.1 is proven. Corollary 46.16. If k D 2 and 2 Irr1 .G/, then G=Z0 is extraspecial for every subgroup Z0 of index 2 in Z.G/ such that G 0 6 Z0 . 1

Proof. Suppose that 2 Irr1 .G=Z0 /; then .1/ D 2 2 m , by Theorem 46.15. It follows that jZ.G=Z0 /j D 2, so G=Z0 is extraspecial since its center and derived subgroup have order 2. 2o . In this subsection, p > 2. Thus, F D GF.p m /, m 2 N with m > 1, q D p m , is an automorphism of F of order k for some divisor k > 1 of m, m D n and G D k Ap .m; / is the generalized Suzuki p-group of order q 2 . Our aim in this subsection is to prove the following Theorem 46.17 ([Sag2]). If G D Ap .m; / is a p-group of order q 2 , where p > 2, m > 1 is an integer, q D p m , is an automorphism of the field F D GF.q/ of order k > 1 and m D n, then one of the following holds: k (a) If k is odd, then cd.G/ D f1; p .mn/=2 g. (b) If k D 2, then cd.G/ D f1; p m=2 g.

8

Groups of prime power order

(c) If k > 2 is even, then cd.G/ D f1; p m=2 ; p .m=2/n g, jIrr.p m=2 / .G/j D and jIrr.p .m=2/n / .G/j D

p m 1 p n C1

p m 1 n p n C1 p

p 2n .

By the text preceding 1o , exp.G/ D p, .a; b/1 D .a; b C a .a//, Z.G/ D f.0; b/ j b 2 Fg is of order q and the centralizer of every element from G Z.G/ has order p mCn . It follows that k.G/ D p mCn C p m p n so the number of noncentral G-classes is p mCn p n . Next, jG 0 j p mn . We see that Lemmas 46.2, 46.3 and 46.4 are also true for odd p. Argument following Lemma 46.4, is also true for odd p. nt Therefore, for every x 2 F, .x/ D x p for some t coprime with k. Lemma 46.18. (a) Suppose that m D nk, .t; k/ D 1 and .p m 1; p nt C 1/ D d . If k is odd, then d D 2. If k is even, then d D p n C 1. (b) Suppose that k; n 2 N, k even and .p m 1; p n C 1/ D d . Then d D 2 if n is even and d D p .m=2;n/ C 1 if .m=2;n/ is odd.

n .m=2;n/

Proof. (a) Since .k; t / D 1, then d divides .p m 1; p 2nt 1/ D .p m 1; p 2n 1/ D p .m;2n/ 1 (Lemma 46.5). If k is odd then .m; 2n/ D .nk; 2n/ D n so d divides p n 1. If k is even, then .m; 2n/ D 2n so d divides p 2n 1. Let us find d0 D .p nt C 1; p n 1/. Let k be odd; then d divides p n 1 so d divides d0 . If r > 2 is a prime divisor of d0 , then p n 1 .mod r/ so p nt C 1 2 .mod r/, a contradiction. The same argument also works in the case r D 4. It follows that d0 D 2 since d0 is even. Since the even number d divides d0 D 2, we get d D 2. Suppose that k is even; then t is odd and so p n C 1 divides .p m 1; p nt C 1/ D d . Next, p nt C1 D .p n C1/A and p m 1 D .p n C1/B, where A D p n.t1/ p n.t2/ C C p 2n p n C 1 and B D p n.k1/ p n.k2/ C C p n 1. The number A is odd since t is odd, and B is even since k is even. In that case, d D .p n C 1/.A; B/ so p ndC1 D .A; B/ D u is odd. Since d D .p n C 1/u with odd u divides p 2n 1 D .p n 1/.p n C 1/ and .p nt C 1; p n 1/ D 2 (see the previous paragraph!), we get d D p n C 1. (b) Set d D .p m 1; p n C 1/, m D 2m1 , d0 D .p .m1 ;n/ C 1; p n C 1/. Then d divides the number .p m 1; p 2n 1/ D p .m;2n/ 1 D p 2.m1 ;n/ 1 D .p .m1 ;n/ 1/.p .m1 ;n/ C 1/: As in the proof of (a), .p .m1 ;n/ 1; p n C 1/ D 2 and d0 D 2 if n .m1 ;n/

d0 D p C 1 if is odd. Suppose that .mn1 ;n/ is even. Then d 2 f2; 4g since d divides .m1 ;n/

n .m1 ;n/

.p n C 1; p .m1 ;n/ 1/.p n C 1; p .m1 ;n/ C 1/ D 2: Indeed, since n is even, we get p n C 1 2 .mod 4/ so d D 2.

is even and

46

Suppose that s D

9

Degrees of irreducible characters of Suzuki p-groups n .m1 ;n/

is odd and set t D .m1 ; n/. Then

p n C 1 D p ts C 1 D .p t C 1/.p t.s1/ p t.s2/ C C p 2t p t C 1/ so

p n C1 p t C1

is odd.

Let be as in Lemma 46.4 and set ˆ D h i provided is a primitive element of the field F. Lemma 46.19. If is a primitive element of the field F, then: (a) The size of ˆ-orbits on G Z.G/ is p m 1. (b) The size of ˆ-orbits on Z.G/# is 12 .p m 1/ for odd k and

p m 1 p n C1

for even k.

Proof. By definition, i ..a; b// D .i a; . .//i b/. If a ¤ 0, the assertion follows since is primitive. Let a D 0. Then i .0; b/ D .0; . .//i b/ D .0; b/ if and n n only if . .//i D 1. Since ./ D p , we obtain .p C1/i D 1. It follows that o./ D p m 1 divides .p n C 1/i , and the assertion follows from Lemma 46.18(a). Lemma 46.20. Suppose that 2 Irr1 .G/, .1/ D p a , Z0 D Z.G/ \ ker. /. Then jZ.G/ W Z0 j D p, G 0 6 Z0 , cd.G=Z0 / D f1; p a g, and jIrr1 .G=Z0 /j D .p 1/p m2a . Next, every nonlinear irreducible character of G is a character of exactly one group G=Z, where Z is a subgroup of index p in Z.G/ not containing G 0 . Proof. Since Z.G= ker. // is cyclic and Z.G/ is elementary abelian, we get jZ.G/ W Z0 j D p. Let us consider the quotient group G=Z0 . Since is nonlinear and G 0 6 Z0 but G 0 Z.G/, we get G 0 Z0 D Z.G/. Next, j.G=Z0 /0 j D p and exp.G=Z0 / D p. It follows that G=Z0 D .G1 =Z0 / .Z1 =Z0 /, where Z1 =Z0 < Z.G=Z0 / and G1 =Z0 is extraspecial. Now all assertions of the lemma are obvious. Theorem 46.21. If G D Ap .m; /, where an automorphism of F of odd order k > 1, mn then cd.G/ D f1; p 2 g, where n D m . k Proof. Let be a primitive element of F. In that case, there are two ˆ-orbits of size 12 .p m 1/ on Z.G/, where ˆ is a cyclic subgroup generated by , by Lemma 46.19(b). If g 2 G 0 , then the ˆ-orbit of g lies in G 0 since G 0 is characteristic in G. It follows that jG 0 j 1 12 .p m 1/ so jG 0 j D p m since jG 0 j divides p m . It follows that G 0 D Z.G/. By Lemma 46.19(b), there are at most two ˆ-orbits on the set of all subgroups of order p in Z.G/. It follows that there are at most two ˆ-orbits on the set of maximal subgroups of Z.G/. Since G=Z0 Š G=Z0 for every subgroup Z0 of index p in Z.G/ and jcd.G=Z0 /j D 2, by Lemma 46.20, then G has at most two distinct degrees of nonlinear irreducible characters. Assume that cd.G/ D f1; p a ; p b g. Then there are two ˆ-orbits on the set of subp m 1 . In that case, there groups of index p in Z.G/ and the sizes of these orbits are 2.p1/ are

p m 1 .p 2.p1/

1/p m2a D

1 m 2 .p

1/p m2a characters of degree p a in Irr1 .G/

10

Groups of prime power order

p 1 and, similarly, 2.p1/ .p 1/p m2b D 12 .p m 1/p m2b characters of degree p b in Irr1 .G/. It follows that m

1 1 p mCn p n D jIrr1 .G/j D .p m 1/p m2a C .p m 1/p m2b 2 2 so that p n D 12 .p m2a C p m2b / or, what is the same, p nC2aC2b D 12 p m .p 2a C p 2b /. Let, for definiteness, a b. Then p nC2aC2b D 12 p mC2b .p 2a2b 1/. The last equality is possible if and only if p 2a2b 1 D 0. Thus a D b so cd.G/ D f1; p a g. It follows that jGj D p 2m D p m C .p mCn p n /p 2a so a D 21 .m n/. Hence, the Theorem 46.17 is proven for odd k. Lemma 46.22. Suppose that 2 F is primitive and k is even. Then: (a) If k > 2, i.e., m > 2n, then G 0 D Z.G/ and the number of ˆ-orbits on .G 0 /# m 1 equals p n C 1; all these orbits have size pp n C1 . (b) If k D 2, i.e., m D 2n, then either G 0 D Z.G/ and the number of ˆ-orbits m 1 on .G 0 /# equals p n C 1 and all these orbits are of size pp n C1 D p n 1, or m

jG 0 j D p 2 D p n and all elements of .G 0 /# are conjugate under ˆ.

Proof. Repeat, word for word, the proof of Lemma 46.10. Lemma 46.23. Let 2 F be primitive, k even and G 0 D Z.G/. Then Irr.G/ f1G g is partitioned, under action of ˆ, in p n orbits of size p m 1 and p n C 1 orbits of m 1 size pp n C1 . Proof. See the proof of Lemma 46.11(b). Lemma 46.24. Let 2 F be primitive, k even and G 0 D Z.G/. Then the set of all subgroups of Z.G/ of index p is partitioned, under action of ˆ, either in p n C 1 orbits m 1/ p m 1 of size .p n C1/.p1/ or in 12 .p n C 1/ orbits of size .p2.p n C1/.p1/ . Proof. Consider a subgroup P D h.0; x/i of Z.G/ of order p. Let s 2 N be minimal m 1 such that s ..0; x// 2 P . Since o.. /Z.G/ / D pp n C1 and P # is partitioned in p1 t

1 orbits of size t under action of h's i, st D pp n C1 . Then the size of ˆ-orbit containing P equals s. Let P1 D h.0; y/i be another subgroup of order p in Z.G/. Since Z.G/ D f.0; b/ j b 2 Fg and 2 F is primitive, we get .0; y/ D .0; f x/ for some f 2 N. Next, since s ..0; x// 2 P , we get s ..0; x// D .0; . .//s x/ D .0; x/r D .0; rx/ for some r 2 N. Then m

s ..0; y// D .0; . .//s y/ D .0; . .//s f x/ D .0; rf x/ D .0; ry/ D .0; y/r : This means that the size of the ˆ-orbit containing P1 , is also equal to s. Thus, the sizes of all ˆ-orbits on the set of all subgroups of order p in Z.G/ are equal. It follows

46

p m 1 p1 , t.p n C1/ p1 .

from c1 .Z.G// D p m 1 s.p1/

11

Degrees of irreducible characters of Suzuki p-groups

that the number of ˆ-orbits on the set of such subgroups

D Since .p n C 1; p 1/ D 2, we get t 2 fp 1; 12 .p 1/g. equals Hence, the set of all subgroups of order p in Z.G/ is partitioned, under action of ˆ, m 1/ p m 1 either in p n C 1 orbits of size .p n C1/.p1/ or in 12 .p n C 1/ orbits of size .p2.p n C1/.p1/ . Now let Q < G be of index p. Since Q D P1 Pm1 with jPi j D p for all i , then s .Q/ D Q. If for some natural s1 < s we have s1 .Q/ D Q, then by Maschke’s theorem, Z.G/ D Q P , where P is a subgroup of order p and s1 .P / D P , contrary to the minimal choice of s. Now the conclusion of the lemma follows from the previous paragraph. Lemma 46.25. Suppose that 2 Irr1 .G/, .1/ D p a . Then a 12 m. Next, a 1 2 m n if the number of ˆ-orbits on the set of subgroups of index p in Z.G/ equals p n C 1 and a > 12 m n, if the number of ˆ-orbits on the same set equals 12 .p n C 1/. Proof. By Lemma 46.20, 2 Irr1 .G=Z0 /, where Z0 is a subgroup of index p in Z.G/ and jIrr1 .G=Z0 /j D .p 1/p m2a . It follows that a 12 m. Let Z1 ; : : : ; Z t be the set of representatives of ˆ-orbits on the set of maximal subgroups of Z.G/. Then cd.G=Zi / D f1; p ai g, jIrr1 .G=Zi /j D .p 1/p m2ai . Next, any two subgroups of index p in Z.G/ belonging to the same ˆ-orbit, lead to isomorphicPquotient groups. Since jIrr1 .G/j p mCn p n , we get p mCn p n D Pt D m2a t p m 1 m2ai n i. so tp D iD1 p iD .p 1/p t.p1/ Suppose that t D p n C 1. If m 2ai > 2n for some i , then p m2ai p 2nC1 > C 1/ D tp n , a contradiction. Thus, m 2ai 2n so ai 12 m n for all i . Suppose that t D 12 .p n C 1/. If m 2ai 2n for some i , then p m2ai p 2n > 1 n n 1 n 2 p .p C 1/ D tp , a contradiction. Thus, ai > 2 m n for all i .

p n .p n

1

1

Theorem 46.26. If G 0 D Z.G/ and k is even, then cd.G/ D f1; p 2 mn ; p 2 m g. In m 1 m 1 p 2n , jIrr 1 m .G/j D pp n C1 pn . that case, jIrr 1 mn .G/j D pp n C1 .p 2

/

.p 2

/

Proof. By Lemma 46.23, the set Irr.G/ f1G g, under action of ˆ, is partitioned in m 1 p n orbits of size p m 1 and p n C 1 orbits of size pp n C1 . Since jLin.G/j D jG=G 0 j 0 and ˆ .G=G / is a Frobenius group, nonprincipal linear characters lie in the same ˆ-orbit of length p m 1. Therefore, under ˆ, there are on Irr1 .G/ exactly p n 1 orbits of size p m 1. By Lemma 46.21, each character from Irr1 .G/ is contained in Irr1 .G=Z0 /, where Z0 is a subgroup of index p in Z.G/. By Lemma 46.24, the set M m 1 of all pp1 subgroups of index p in Z.G/, under action of ˆ, is partitioned either in

p 1 1/ or in 12 .p n C 1/ orbits of size .p2.p p n C 1 orbits of size .p n C1/.p1/ n C1/.p1/ . n First suppose that the set M is partitioned in p C 1 orbits under action of ˆ. Consider 2 Irr1 .G=Z0 /, Z0 2 M. If the size of the ˆ-orbit of equals p m 1, then Irr1 .G=Z0 / contains exactly .p 1/.p n C 1/ characters from that orbit. If the size m 1 of ˆ-orbit of equals pp n C1 , then Irr1 .G=Z0 / contains exactly p 1 characters from m

m

12

Groups of prime power order

1 that orbit. Note that if, for some Z0 2 M, Irr1 .G=Z0 / has no orbits of size pp n C1 , then n Irr1 .G=Z0 / is partitioned in orbits of sizes .p 1/.p C 1/. Since jIrr1 .G=Z0 /j D p m2a , where p a 2 cd.G=Z0 /, then p n C 1 divides p m2a , which is a contradiction. It follows that for every Z0 2 M, there are in Irr1 .G=Z0 / at least p 1 characters m 1 whose ˆ-orbits have size pp n C1 . Since the number of such characters is p m 1 and m

p m 1 p1 , p m 1 p n C1 .

jMj D

every group G=Z0 has exactly p 1 characters belonging to orbits of

size Let cd.G=Z0 / D f1; p a g and let Irr1 .G=Z0 / contain exactly t .p 1/.p n C 1/ characters belonging to ˆ-orbits of size p m 1. Then .p 1/p m2a D jIrr1 .G=Z0 /j D .p 1/ C t .p 1/.p n C 1/. It follows that p m2a 1 D t .p n C 1/ so p n C 1 divides p m2a 1. It follows that n divides m 2a and m2a is even. By Lemma n 46.25, 12 m n a 12 m so that 0 m2a 2. Suppose that t > 0. Then n m2a jIrr1 .G=Z0 /j > p 1 or m 2a > 0. It follows that n D 2 and so a D 12 m n, jIrr1 .G=Z0 /j D .p 1/p 2n , t D p n 1. Since t D p n 1, Irr1 .G=Z0 / contains representatives of all ˆ-orbits of size p m 1 on Irr1 .G/. Therefore, all characters from a ˆ-orbit of size p m 1 belong to members of M contained in the same ˆ-orbit. m 1 p 2n . The number of remaining irreducible nonlinear Then jIrr 1 mn/ .G/j D pp n C1 .p 2

/

1 p 2n D characters equals t1 D p n .p m 1/ pp n C1

p m 1 n p n C1 p and the sum of squares m 1 m 1 † D p m .p m 1/ pp n 1 p m D pp n 1 p mCn . If 2 Irr1 .G/, m m W Z.G/j D p . Since † D t1 p , all other nonlinear irreducible 1 2m m

of their degrees is then .1/2 jG characters are of degree p . Now suppose that the number of ˆ-orbits on M equals 12 .p n C 1/. Recall that all 2.p m 1/ these orbits are of size .p1/.p n C1/ . If 2 Irr1 .G=Z0 / (Z0 2 M) belongs to the ˆm orbit of size p 1, then that set of characters contains also exactly .p1/ 12 .p n C1/ ˆconjugates with . Similarly, as before, we can show that then Irr1 .G=Z0 / (Z0 2 M) m 1 contains a character representing a ˆ-orbit of size pp n C1 . Suppose that, for some Z0 2 M, the set Irr1 .G=Z0 / has exactly 12 .p 1/ characters m 1 . Then jIrr1 .G=Z0 /j D .p every of which is contained in a ˆ-orbit of size pp n C1 1 1 m2a n D 2 .p 1/ C t .p 1/ 2 .p C 1/, t 0. It follows that 2p m2a D 1/p n 1 C t .p C 1/, which is impossible since p n C 1 is even. Therefore, every group G=Z0 (Z0 2 M) has at least p 1 characters every of which belongs to a ˆ-orbit of m 1 m 1 . Since jMj D pp1 and the number of the above characters is p m 1, size pp n C1 every group G=Z0 (Z0 2 M) has exactly p 1 characters belonging to ˆ-orbits of m 1 size pp n C1 (these characters belong to two different ˆ-orbits). Now let G=Z0 (Z0 2 M) have a character belonging to a ˆ-orbit of size p m 1. n Then, by the above, jIrr1 .G=Z0 /j D .p 1/p m2a D .p 1/ C t .p 1/ p 2C1 so 2.p m2a 1/ D t .p n C 1/. Since m is even then m 2a is also even; therefore, using Lemma 46.18, we can show that either n divides 12 .m 2a/ , p n C 1 divides p m2a 1 and t D

2.p m2a 1/ p n C1

or p D 3, n D 1 and t D 12 .32m2a 1/. However,

46

13

Degrees of irreducible characters of Suzuki p-groups

by hypothesis, n > 1. In the first case we get either 2n m 2a, a 12 m n, contrary to Lemma 46.25, or else m 2a D 0, a D 12 m, t D 0, a contradiction again. Thus, the case where M is partitioned in 12 .p n C1/ orbits under ˆ is impossible. 1

Corollary 46.27. If m D 2n, then jG 0 j D p 2 m . Proof. Assume that G 0 D Z.G/ holds. Then, by Theorem 46.26, we have cd.G/ D 1 1 f1; p 2 mn ; p 2 m g, which is impossible since 12 m n D 0. 1

Theorem 46.28. If m D 2n, then cd.G/ D f1; p 2 m g. 1

1

Proof. By Lemma 46.22 and Corollary 46.27, jG 0 j D p 2 m , jG W G 0 j D p mC 2 m D 1 1 p mCn D jLin.G/j so jIrr1 .G/j D p m p n D p 2 m .p 2 m 1/. Every irreducible nonlinear character of G is a character of G=Z0 , where Z0 2 M and G 0 6 Z0 . The 1m 2

1m 2

1 p p11 D p p11 p 2 m , and every group G=Z0 number of such subgroups is pp1 has at least p 1 nonlinear irreducible characters. Comparing the number jIrr1 .G/j with the number of subgroups of index p in Z.G/ not containing G 0 , we see that every group G=Z0 has exactly p 1 nonlinear irreducible characters. Let cd.G/ D f1; p a g. Then p 1 D jIrr1 .G=Z0 /j D .p 1/p m2a so a D 12 m, completing the proof. m

Theorem 46.17 is proven.

1

47

On the number of metacyclic epimorphic images of finite p-groups

In this section we prove the following Theorem 47.1 ([Ber33]). Let G be a nonmetacyclic p-group of order p m and let n < m. (a) If p D 2 and n > 3, then the number of normal subgroups D of G such that G=D is metacyclic of order 2n , is even. (b) If p > 2 and n > 2, then the number of normal subgroups D of G such that G=D is metacyclic of order p n , is a multiple of p. Remarks. 1. Let G be a two-generator nonmetacyclic 2-group. Then all epimorphic images of G of order 8 are metacyclic. Since the number of normal subgroups of given index in a 2-group G is odd, it follows that Theorem 47.1(a) is not true for n D 3. 2. Let r s t > 0, r C s C t > 3 and G D hai hbi hci, where o.a/ D p r , o.b/ D p s , o.c/ D p t , be an abelian p-group of rank three. Let us check Theorem 47.1 for n D r C s C t 1. Let be the number of subgroups D of G of order p such that G=D is metacyclic. If t > 1, then 1 .G/ ˆ.G/; then D 0. Suppose that s > t D 1. If D < hai hbi, then G=D is nonmetacyclic. If D 6 hai hbi, then G=D is abelian of type .p r ; p s / so metacyclic, and in that case, D p 2 . Now suppose that r > s D 1. Then D D Ãr 1 .G/ is the unique normal subgroup of order p in G such that G=D is not metacyclic so D p 2 C p. Thus, p divides in all cases. 3. Suppose that G D ha; b j a4 D b 4 D c 2 D 1; c D Œa; b; Œa; c D Œb; c D 1i is a nonmetacyclic minimal nonabelian group of order 25 . Let us check Theorem 47.1(a) for n D 4. The group G has exactly seven central subgroups of order 2: X1 D ha2 i;

X2 D hb 2 i;

X3 D ha2 b 2 i;

X5 D ha2 ci;

X6 D hb 2 ci;

X7 D ha2 b 2 ci:

X4 D hci;

Then G=Xi is metacyclic for i D 3; 4; 5; 6 and nonmetacyclic for i D 1; 2; 7. m

4. Suppose that G D ha; b j a2 D b 2 D c 2 D 1; m > 2; c D Œa; b; Œa; c D Œb; c D 1i is a nonmetacyclic minimal nonabelian group of order 2mC2 . Set ˛ D

47

On the number of metacyclic epimorphic images of finite p-groups

15

m1

a2 ; then 1 D fhc˛i; h˛i; hcig is the set of central subgroups of order 2 in G. Then G=hci and G=hc˛i are metacyclic and G=h˛i is not metacyclic. Let Sc.G/ D 1 .Z.G// be the socle of G. Let i denote, for all i such that p i jSc.G/j, the set of subgroups of order p i in Sc.G/. Let M be a set of nonidentity normal subgroups of G. Given H 2 1 [ 2 [ [ fSc.G/g, let ˛.H / be the number of members of the set M containing H . Set jSc.G/j D p t so that t D fSc.G/g. We claim that the following identity holds (see 10): (1)

˛.G/ D jMj D

t X

i .1/i1 p .2/

iD1

X

˛.H /:

H 2i

Indeed, let D 2 M and jD \ Sc.G/j D p k ; then k 1 since D > f1g. For natural numbers u v, let 'u;v denote the number of subgroups of order p v in Ep u . Then the P i contribution of D in the right-hand side of (1) is equal to kiD1 .1/i1 p .2/ 'k;i , and that number equals 1, by Hall’s identity (Theorem 5.2). Since the contribution of D in the left-hand side of (1) is also equal 1, identity () is true. In particular, X ˛.G/ D jMj

(2) ˛.H / .mod p/: H 21

Remark 5. Suppose that G is a noncyclic group of order p m , m > 2 and 1 < n < m. We claim that the number c.G/ of normal subgroups N of G such that G=N is cyclic of order p n , is a multiple of p. One may assume that c.G/ > 0. Let n D m 1. There is N Z.G/ of order p such that G=N is cyclic. Then G is abelian of type .p m1 ; p/ so G D Z N , where Z is cyclic of order p m1 D p n hence c.G/ D p. Now let n < m 1. Take H 2 1 . Suppose that G=H is cyclic; then G is abelian of type .p m1 ; p/. The group G contains exactly p C 1 subgroups of index p n . If N is one of such subgroups, then G=N is cyclic if and only if N 6 ˆ.G/ so, since ˆ.G/ is cyclic, we get c.G/ D p. Next we assume that G is not abelian of type .p m1 ; p/; then G=H is not cyclic for all H 2 1 , therefore, by induction on m, we get ˛.H / 0 .mod p/ so c.G/ 0 .mod p/, by (2). Suppose that G is a group of order p m and p n jG W G 0 j. We claim that then the number .G/ of N G G such that G=N is nonabelian of order p n is a multiple of p. One may assume that .G/ > 0; then n > 2. The number of normal subgroups of given index in G is 1 .mod p/ (Sylow). If N G G has index p n , then G=N is nonabelian if and only if G 0 6 N . Therefore, .G/ 0 .mod p/ (Sylow again). If a 2-group G of order 2m > 23 is not of maximal class, then the number of N G G such that G=N is nonabelian of order 23 is even since jG W G 0 j 23 . Proof of Theorem 47.1. Suppose that M is the set of normal subgroups D of G such that G=D is metacyclic of order p n . One may assume that M ¤ ¿. As above, ˛.H / is the number of members of the set M containing H 2 1 .

16

Groups of prime power order

A. First we consider the most difficult case n D m 1; then M 1 . One may assume that G is nonabelian (Remark 2). We have to prove that p divides jMj.D ˛.G//, i.e., j1 Mj 1 .mod p/. By (2), we have to prove that X (3) ˛.H / 0 .mod p/: ˛.G/

H 21

Let U 2 M and suppose that U — ˆ.G/. Then G D U M , where M 2 1 is metacyclic. If V 2 1 and V — M , then G D V M so V 2 M. If W Z.M / is a member of the set M, then M=W is cyclic and G is abelian of rank 3 so p divides jMj (Remark 2). We see that jMj D j1 j c1 .Z.M // 0 .mod p/. Next we assume that all members of the set M are contained in ˆ.G/. (i) Let p > 2. Then jG=Ã1 .G/j p 3 since G is nonmetacyclic (Theorem 9.11). Take D 2 M; then G=D is metacyclic. Assuming that jG=Ã1 .G/j p 4 , we conclude that G=DÃ1 .G/ is of order p 3 and exponent p so nonmetacyclic, a contradiction. Thus, jG=Ã1 .G/j D p 3 . Take U 2 1 M; then G=U is not metacyclic so j.G=U /=Ã1 .G=U /j p 3 (Theorem 9.11 again) and so U Ã1 .G/. Thus, all members of the set 1 M are contained in Ã1 .G/. Conversely, if V 2 1 is contained in Ã1 .G/, then G=V is not metacyclic. Thus, j1 Mj equals the number of those members of the set 1 that are contained in Ã1 .G/. By Sylow, the last number is 1 .mod p/ so jMj j1 j 1 D ' t;1 1 0 .mod p/. (ii) Suppose that p D 2. Here we have to consider the nonmetacyclic quotient group G=Ã2 .G/ (see Lemma 42.2(b)) of order 24 . (ii1) Assume that jG=Ã2 .G/j D 24 . Then Ã2 .G/ > f1g since m > 4. The number of members of the set 1 contained in Ã2 .G/, is odd (Sylow). Therefore, if all members of the set 1 M are contained in Ã2 .G/, then ˛.G/ D jMj is even since j1 j is odd, so we assume that there is U 2 1 M such that U — Ã2 .G/; then GN D G=U N is nonmetacyclic so N 2 .G/ is nonmetacyclic. It follows from Theorem 44.2 that G=Ã 4 N then G=H.Š G= N HN / is not metacyclic of order its order is 2 . Write HN D Ã2 .G/; 24 and exponent 4. It follows that H D Ã2 .G/ since jG=Ã2 .G/j D 24 , contrary to the choice of U : U — Ã2 .G/. (ii2) Let jG=Ã2 .G/j > 24 . Take D 2 M; then, by Theorem 44.2, D — Ã2 .G/. Write GN D G=D. We claim that jG=Ã2 .G/j 25 . Indeed, G=DÃ2 .G/, as an epimorphic image of groups GN D G=D and G=Ã2 .G/, is metacyclic of exponent 4 so its order is 24 , and our claim on order follows since DÃ2 .G/ D D Ã2 .G/ and jDj D 2. Thus, jG=Ã2 .G/j D 25 , by (ii1). Since G=D is metacyclic, we have jSc.G=D/j 22 so jSc.G/j jDjjSc.G=D/j D 23 , and we conclude that j1 j 2 f1; 3; 7g. Since j1 j is odd, to complete this case, it suffices to show that the number j1 Mj is also odd. But G=D is metacyclic so D — G 0 (Theorem 36.1) and hence G 0 Š .G=D/0 is cyclic. Since all members of the set M are contained in ˆ.G/, we get d.G/ D 2. N (GN is defined in (ii1)). Since G=DÃ2 .G/ is metacyclic, its order is Let HN D Ã2 .G/ 4 N 2 .G/j N D 24 since jG=Ã2 .G/j D 25 . 2 so H D DÃ2 .G/ D D Ã2 .G/ and jG=Ã

47

On the number of metacyclic epimorphic images of finite p-groups

17

Set N0 D fV 2 1 j V Ã2 .G/g; then jN0 j is odd (Sylow) and N0 1 M. One may assume that N0 1 M (otherwise, jMj D j1 j jN0 j is even). In that case, there is U1 2 1 .M [ N0 /; then GN D G=U1 is not metacyclic. Write N then G= N HN is not metacyclic (Theorem 44.2) and, since U1 Ã2 .G/ H HN D Ã2 .G/; and U1 6 Ã2 .G/, we get H1 D U1 Ã2 .G/ and jG=H1 j D 12 jG=Ã2 .G/j D 12 25 D 24 . If V < H1 is a member of the set 1 , then G=V is nonmetacyclic. Let N1 D fU0 2 1 N0 j U0 < H1 g. Then jN0 [ N1 j D jN0 j C jN1 j, the number of members of the set 1 contained in H1 , is odd (Sylow). It follows that the number jN1 j is even since the number jN0 j is odd. Since U1 2 N1 , we obtain jN1 j 2 so jN0 [N1 j 1C2 D 3. Assuming that N0 [N1 1 M, we add to the set N0 [N1 the set N2 (of even cardinality > 0) of new members of the set 1 M. Indeed, if U2 2 1 .N0 [ N1 [ M/, then, writing H2 D U2 Ã2 .G/, we conclude, as above with U1 and H1 , that G=H2 is nonmetacyclic of order 24 . We have H1 \H2 D Ã2 .G/ since Ã2 .G/ H1 \ H2 has index 25 in G and H1 ; H2 are distinct of index 24 in G, so N2 is the set of members of the set 1 .N0 [ N1 / contained in H2 but not in H1 and jN2 j > 0 is even (if Vi 2 Ni , then Vi Ã2 .G/ D Hi , i D 0; 1; 2). Thus, we get jN0 [ N1 [ N2 j 1 C 2 C 2 D 5. We claim that then the set N0 [ N1 [ N2 , which contains exactly five members, coincides with the set 1 M. Indeed, otherwise, we can, acting as above, add to that sum-set at least two new members of the set 1 M, which is impossible since j1 Mj 6: the set M ¤ ¿ and j1 j D 7. Thus, in any case, the number j1 Mj is odd so that the number jMj is even. B. Now let n < m 1. Here we consider cases p D 2 and p > 2 together. We proceed by induction on jGj. Take H 2 1 . If G=H is metacyclic, then ˛.H / 1 .mod p/ (Sylow). However, the number of such H , by part A of the proof, is a multiple of p. Therefore, the contribution of such H in the sum on the right-hand side of formula (2), is a multiple of p. Now suppose that G=H is not metacyclic. Then, by induction, ˛.H / 0 .mod p/. Therefore, the contribution of such H in the sum on the right-hand side of formula (2), is a multiple of p again, Thus, congruence (3) is proven. Supplement to Theorem 47.1. Let 1 k p 1, k < n < m and a p-group G of order p m satisfies jG=Ã1 .G/j > p k . Then the number of normal subgroups D of G such that j.G=D/=Ã1 .G=D/j p k and jG=Dj D p n , is a multiple of p. 2o . Here we prove the following Theorem 47.2. If G be a nonabelian metacyclic p-group, then all representation groups of G are also metacyclic so that M.G/, the Schur multiplier of G, is cyclic. Proof. Let be a representation group of G. By definition, there is in Z./ \ 0 a subgroup M Š M.G/ such that =M Š G. Since G is nonabelian, we get M < 0 . Since =M Š G is metacyclic, is also metacyclic, by Theorem 36.1.

18

Groups of prime power order

The cyclicity of Schur multipliers of metacyclic p-groups is known, their orders are also computed in [Kar, Theorem 10.1.25]. Isaacs has reported (in letter at Jan 10, 2006) that he also proved Theorem 47.3. His proof is based on the following Lemma 47.3 (Isaacs). Let G be a p-group and let Z Z.G/ \ G 0 be of order p. If G=Z is metacyclic, then G is metacyclic. Proof. Let U=Z G G=Z be with cyclic U=Z and G=U . Note that jU=Zj > p since G=Z is not abelian. We want to show that U is cyclic. Now let hxU i D G=U . Since U is abelian and normal in U and G=U is cyclic, we see that G 0 D ŒU; x D hŒu; x j u 2 U i is isomorphic to U=CU .x/ [Isa2, Lemma 12.12]. But Z is contained in CU .x/ and U=Z is cyclic, and hence G 0 is cyclic as an epimorphic image of U=Z. Since G 0 > Z is cyclic, we have Z ˆ.G 0 / ˆ.U / so d.U / D d.U=Z/ D 1 and U is cyclic. (In fact, Theorem 36.1 and Lemma 47.4 are equivalent.)

48

On 2-groups with small centralizer of an involution, I

All results of this section are due to the second author. The main part of this section coincides with [Jan1]. In this section we give the classification of 2-groups containing an involution t such that CG .t / D ht i C2m , m 1. We may assume from the start that m > 1 since, if m D 1, G is a 2-group of maximal class, by Proposition 1.8. Note that case m D 2 was considered in [GLS, Proposition 10.27]. Exercise 1. Let G be a nonabelian 2-group and t an involution in G Z.G/. (a) Prove that jNG .CG .t // W CG .t /j < c1 .CG .t //, where c1 .G/ is the number of subgroups of order 2 (or, what is the same, involutions) in G. (b) Let CG .t / D ht iQ, where Q is either cyclic of order 2m , m > 1, or Q Š Q2m , m 3. Then jNG .CG .t // W CG .t /j D 2, by (a), and (b1) 1 .Z.G// D Z.Q/. (b2) If u is an involution in Q, then t and t u are conjugate in NG .CG .t //. (b3) If t 2 ˆ.G/, then jˆ.G/j > 4. Solution. (a) One may assume that CG .t / E G. Then G acts transitively on the conjugacy class K t containing t . Since jG W CG .t //j > 1 is a power of 2 and c1 .CG .t // > 1 is odd (Sylow), the inequality follows. (b1, b2) Let u 2 Z.G/ be an involution. Then u is a square and t is not, so t and u are not conjugate in G. The result follows since CG .t /.> Z.G// has exactly 3 involutions. (b3) Assume that jˆ.G/j D 4. Then ˆ.G/ Z.G/, by [BZ, Lemma 31.8], a contradiction since t 62 Z.G/. Theorem 48.1 ([Jan1]). Let G be a nonabelian 2-group containing an involution t such that the centralizer CG .t / D ht i C , where C is a cyclic group of order 2m , m 1 (clearly, t 62 Z.G/). Then G has no elementary abelian subgroups of order 16 and G is generated by at most three elements. Next, G has no t -invariant elementary abelian subgroups of order 8.

20

Groups of prime power order

(A1) If G has no elementary abelian subgroups of order 8, then one of the following holds: (a) G 2 fD2n ; SD2n g. Here we have m D 1. (b) G Š M2n . Here we have m D n 2. (c) jG W CG .t /j D 2, t 2 ˆ.G/, Z.G/ is a cyclic subgroup of order 4 not contained in ˆ.CG .t //, G=Z.G/ is dihedral with cyclic subgroup L=Z.G/ of index 2, L is abelian of type .2; 2m /, m 2, t 2 L. If x is an element of maximal order in G L, then hx 2 i D Z.G/. (In that case CG .t / D L.) (d) G has a subgroup S of index 2, where S D AL, the subgroup L is normal in n1 G, L D hb; t j b 2 D t 2 D 1; b t D b 1 ; n 3i Š D2n . A D hai is cyclic of m order 2 , m 2, A \ L D Z.L/, Œa; t D 1, 1 .G/ D 1 .S / D 2 .A/ L. If jG W S j D 2, then there is an element x 2 G S so that t x D t b and CG .t / D ht i hai. Next, G D ha; b; ti if G D S and G D ha; t; xi if S < G. (A2) If G has an elementary abelian subgroup of order 8, then Z.G/ is of order 2, CG .t / D ht i hai, where A D hai has order 2m (m 2), G has a normal subgroup L such as in (A1)(d), A \ L D Z.L/ D hzi, S D AL is a normal subgroup of index 2 in G and G is isomorphic to one of the following groups: m

n1

(e) G D ha; b; t j a2 D b 2 D t 2 D Œa; t D 1; m 3; n 4; b t D m1 n2 i D b2 D z; b a D b 1C2 ; i D n m 2i. We have G D AL D b 1 ; a2 S and the cyclic group hai=hzi acts faithfully on L. m n1 D t 2 D s 2 D Œa; t D 1; m 4; n 5; b t D (f) G D ha; b; t; s j a2 D b 2 m1 n2 i b s D b 1 ; a2q D b2 D z; b a D b 1C2 ; i D n m C 1 2; t s D m2 i 1 t b; s a D a2 b 2 si. Here S D AL is a subgroup of index 2 in G and 1 .S / D 2 .A/L. Also G D hsi S , M D hsiL Š D2nC1 , NG .M / D ha2 iM and s inverts 2 .A/. The order of G is 2mCn and G D ha; t; si. (g) Groups G D F .m; n/ (see Appendix 14). Here we have G D ha; b; t; s j m n1 a2 D b 2 D t 2 D s 2 D Œa; t D Œa; b D 1; m 2; n 3; b t D m1 n2 b s D b 1 ; a2 D b2 D z; t s D t b; as D a1 z v ; v D 0; 1; and if v D 1; then m 3i. Here S D A L and G D hsi S , where jG W S j D 2. We have M D hsiL Š D2nC1 and G D ha; t; si. m

n1

(h) G D ha; b; t; s j a2 D b 2 D t 2 D s 2 D Œa; t D 1; m; n 4; b t D b s D m1 n2 m2 2n3 D b2 D z; t s D t b; b a D bz; as D a1C2 b i. Here we b 1 ; a2 have again G D hsi S so that jG W S j D 2. Finally, M D hsi L Š D2nC1 and G D ha; t; si. Proof. Let G be a 2-group containing a noncentral involution t such that CG .t / D ht i C , where C is cyclic of order 2m , m 1. By Proposition 1.8, if m D 1, then G 2 fD2n ; n 3; SD2n ; n 4g. In what follows we assume that m > 1. Then G is not of maximal class so it contains a normal abelian subgroup U of type .2; 2/ (Lemma 1.4). It follows from the structure of CG .t / that Z.G/ is cyclic of order 2m

48 On 2-groups with small centralizer of an involution, I

21

at most (see Exercise 1(b1)) so T D CG .U / has index 2 in G. Let u be the involution in C ; then u 2 Z.G/ (Exercise 1(b1)). If an element x 2 NG .CG .t // CG .t /, then t x D t u, by Exercise 1(b2). Assume that E is a t -invariant elementary abelian subgroup of G of order 8. Set H D ht; Ei. Since H is not of maximal class, jCE .t /j > 2, by Suzuki’s theorem. Then the elementary abelian subgroup ht i CE .t / of order 8 is contained in (the metacyclic subgroup) CG .t /, which is a contradiction. Hence E does not exist. It follows that every t -invariant abelian subgroup of G is metacyclic. () We examine now the case where t 2 T D CG .U /. Then ht; U i CG .t / so t 2 U since CG .t / has no elementary abelian subgroups of order 8. Also we see that T D CG .U / D ht i C has index 2 in G since U 6 Z.G/, and T D CG .t /. (1) Suppose that U is the only normal abelian subgroup of type .2; 2/ in G. Set 1 .C / D hui D Ãm1 .T / so that U D ht; ui D 1 .T / and u 2 Z.G/. Since Z.G/ is cyclic, hui D 1 .Z.G//. Next, Ã1 .C / Ã1 .G/ ˆ.G/. (1.1) Suppose, in addition, that t 62 ˆ.G/. Then there exists a maximal subgroup M of G such that t 62 M ; in that case, G D ht i M , a semidirect product with kernel M . Setting M0 D T \ M , we get, by the modular law, T D ht i M0 . It follows from the structure of T that M0 Š C2m . If M is cyclic, then, clearly, G Š M2mC2 , and this is the case (b) of the theorem. So assume that M is not cyclic. Since U 6 M since t 2 U and t 62 M , the subgroup M has no G-invariant subgroups of type .2; 2/ (by assumption in (1), U is the unique G-invariant four-subgroup) so M is of maximal class, by Lemma 1.4. Then Z.M / D hui, where ht; ui D U . Let hvi be the cyclic subgroup of order 4 contained in M0 (recall that, by assumption. m 2; obviously, v 2 D u) and let y 2 M M0 ; then v y D v 1 D vv 2 D vu and t y D t u (see Exercise 1(b2)). Hence we have .t v/y D t y v y D t u vu D t v so K D ht vi is a cyclic subgroup of order 4 contained in T but not contained in M0 , and .t v/2 D t 2 v 2 D v 2 D u since, by the choice, v 2 CG .t /. We have CG .t v/ hT; yi D G. It follows that G D M K, the central product, M \ K D Z.M /. This group is then a group in part (d) of the theorem (with S D G and A is of order 4 centralizing L). In the case under consideration, ˆ.G/ D Ã1 .C / so C is normal in G. (1.2) It remains to consider the case t 2 ˆ.G/. In this case ht i Ã1 .C / ˆ.G/ has index 4 in G so there we have equality and hence G is generated by two elements. If m D 2, then ˆ.G/ D U , being a four-group, is contained in Z.G/, by Exercise 1(b3), which is a contradiction. Hence, we must have m 3. There is an element x 2 G T such that x 2 2 ˆ.G/ .Ã1 .C / [ U /. This is due to the fact that Z.G/ is cyclic. Indeed, there exists an element x 2 G T such that x 2 62 C (otherwise, ˆ.G/ D Ã1 .G/ D Ã1 .C /, which is a contradiction). Next, if x 2 2 U , then CG .x 2 / hx; T i D G so Z.G/ is not cyclic since it contains a noncyclic subgroup hu; x 2 i, a contradiction. Our claim about the existence of x is justified. Then o.x 2 / 4 since x 2 62 U D 1 .ˆ.G//. Set C D hai so that m1 a2 D u, T D CG .t / D ht i hai. Also, Ã1 .T / D Ã1 .C / D ha2 i is normal

22

Groups of prime power order

in G since T is, and ˆ.G/ D ht i ha2 i, where o.a2 / D 2m1 4. It follows from the choice of x that ˆ.G/ D ha2 ; x 2 i (however, generators x 2 and a2 are not independent). By Exercise 1(b2), we have t x D t u. If x would centralize a2 , then x centralizes ha2 ; x 2 i D ˆ.G/ and ˆ.G/ contains t so x 2 CG .t / D T , contrary to the choice of x. If y 2 G T , then y 2 2 Z.G/ since T is abelian and generates G together with y. Therefore, x 2 2 Z.G/. Since x 62 CG .a2 /, we get a2 62 Z.G/. It follows that all elements in .G=Z.G// .T =Z.G// are involutions. Since Z.G/ 6 ˆ.T /, it follows that T =Z.G/ is cyclic. In that case, G=Z.G/ is dihedral. Indeed, since d.G/ D 2 and Z.G/ 6 ˆ.G/, we conclude that G=Z.G/ is not abelian of type .2; 2/. If z is an element of maximal order in G T , then z 2 is a generator of Z.G/ since zZ.G/ is a noncentral involution in the dihedral group G=Z.G/. As we saw, jZ.G/j 4. We obtain a group in part (c) of the theorem. () In the rest of the proof we assume that G has a normal abelian subgroup U of type .2; 2/ which does not contain our involution t (if G has two normal abelian subgroups of type .2; 2/, then one of them does not contain t : otherwise t 2 Z.G/, which is not the case since G is nonabelian). Therefore, this case includes the one where G has at least two distinct normal four-subgroups. Set again T D CG .U / so that G D ht i T since t 62 T . By the modular law, G0 D CG .t / D ht i A, where A D CG .t / \ T D CT .t / is cyclic of order 2m , m 2, and so Z D 1 .A/ D CU .t / D 1 .Z.G// is of order 2. We have Z.G/ CG .U / \ CG .t / D T \ CG .t / D CT .t / D A. Set G1 D NG .G0 /, where G0 D CG .t /; then G1 > G0 since G > G0 is nonabelian. Since t and the generator z of 1 .A/ are not conjugate in view of m > 1, t has only two conjugates t and t z in G1 , and we have 1 .G0 / D ht; zi. It follows that 2 D jG1 W CG1 .t /j D jG1 W G0 j (see also Exercise 1(a)) and so G1 D G0 U because G0 U G1 and jG0 U W G0 j D 2 since G0 \ U D 1 .A/. Set D0 D ht i U ; then D0 Š D8 since it is nonabelian of order 8 and generated by involutions, and G1 is the central product G1 D A D0 with A \ D0 D hzi (indeed, A centralizes t and U so ht; U i D D0 ; before we used the product formula). We also have U D hu; zi < D0 for some involution u. Let v be an element of order 4 in D0 and y an element of order 4 in A; then y 2 D v 2 D z, and since x 2 D y 2 v 2 D y 4 D 1, x D yv is an involution in G1 D0 . Since x t D y t v t D yv 1 D yv v 2 D xz, we see that D1 D hx; t i Š D8 . Because A D Z.G1 /, we also have G1 D A D1 , t 2 D1 and D1 \ U D Z.D1 / D hzi. In the rest of the proof we consider a subgroup S of G which is maximal subject to the following four conditions: (i) S contains G1 D A D1 . (ii) S D AL, where L is a normal subgroup of S and L Š D2n , n 3, (iii) A \ L D hzi D Z.L/, and (iv) L \ G1 D D1 Š D8 . We recall that A D hai is of order 2m , m 2, and Z.G/ A D Z.G1 / so Z.G/

48 On 2-groups with small centralizer of an involution, I

23

is cyclic of order 2m , and 1 .A/ D hzi D Z.D1 / D Z.L/ D 1 .Z.G//, t 2 D1 , CG .t / D ht i A. Also G1 contains U D hz; ui which is a normal abelian subgroup of G of type .2; 2/ and ut D uz. Now we act with A on the dihedral group L, n1 where we set L D hb; t j b 2 D t 2 D 1; b t D b 1 i. Here hbi is the unique cyclic subgroup of index 2 in L so hbi is A-admissible (recall that L is normal in S D AL). We have jhbi \ D1 j D 4 (since D1 L, by (iv)), A centralizes D1 (by (i)) and CL .a/ D ht i.CL .a/ \ hbi/, by the modular law, and so CL .a/ is a dihedral subgroup of L of order 8 containing D1 (recall that t 2 D1 ). Looking at Aut.L/ (see Proposition 34.8, where Aut.D2n / is described) we see that A induces on L a cyclic group of automorphisms of order at most 2n3 and so jA W CA .L/j 2n3 . Since S=L > f1g is cyclic, U 6 L and L is generated by involutions, we have 1 .S / D UL. It follows that 1 .S / \ A D hyi, where o.y/ D 4 (by the product formula) with y 2 D z, and we have hyi D 2 .A/. Let us consider at first the case where y induces on L a nonidentity automorphism so that CA .L/ D hzi and the group hai=hzi acts faithfully on L. Then y induces an automorphism of order 2 on L and since L > CL .A/ D1 Š D8 we must have n 4, Œy; t D 1 (recall that the element y 2 A centralizes t ) and b y D bz, .b 2 /y D .bz/2 D b 2 so that CL .y/ D ht; b 2 i Š D2n1 and CL .y/ is a maximal subgroup of L. Let v be an element of order 4 in D1 ; then v 2 hbi. Since U D hz; ui AD1 , we may set u D yv. We have y b D y.D y y 2 D y 1 , and so utb D .uz/b D .yvz/b D y 1 vz D y 1 vy 2 D yv D u so that hz; u; t bi is an elementary abelian subgroup of order 8 since t b is an involution not contained in U . Obviously, C1 .S/ .u/ D CS .u/\1 .S / D huiht b; b 2 i, where ht b; b 2 i Š D2n1 , and since CG .t / D ht iA does not contain an elementary abelian subgroup of order 8, t cannot be fused in G to any involution contained in CS .u/\1 .S / since the centralizer of any involution from CS .u/ \ 1 .S / contains a subgroup isomorphic to E8 . On the other hand, it is easy to compute that any involution in S lies in L [ .CG .u/ \ 1 .S // since 1 .S / D UL. Naturally, t cannot be fused in G to any involution in U GG since t 62 U . It follows that the G-class of t contains at most 2n2 elements so jG W CG .t /j D 2n2 D jS I CS .t /j. Since CG .t / D CS .t /, this forces that S D G. If m D 2, we have G D F .2; n/, stated in part (g) since ht; b 2 i Š D2n1 centralizes A and the involution t b inverts A. i So assume that m 3. Then we may set (see Proposition 34.8) b a D b 1C2 , where m1 i 2 because CL .a/ D1 . Since CL .a/ Š D2i C1 and CL .a2 / Š D2i Cm and m1 m1 D z 2 Z.L/, we get CL .a2 / D L. It follows that i D n m. We have a2 obtained exactly the groups stated in part (e). In what follows we shall assume always, that hyi D 2 .A/ centralizes L so that 1 .S / D L 2 .A/, the central product (of order 2nC1 ). In this case each involution in S is contained in L [ U (see Appendix 16, where we have proved that D2n C4 Š Q2n C4 ); it follows that U as the unique normal abelian subgroup of type .2; 2/ in S , is characteristic in S and so the subgroup L being generated by its own involutions is characteristic in 1 .S / and so in S (see also Appendix 16). If S D G, we get some groups stated in part (d) of the theorem.

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Groups of prime power order

From now on we shall assume, in addition, that S < G. Let W D NG .1 .S //; obviously, S < W . Let Kx denotes the G-class containing an element x 2 G. Since W fuses K t \ 1 .S / with K tb \ 1 .S / and CG .t /; CG .t b/ < S . where both classes are of size 2n2 and are contained in L, we get jW W S j D 2, S is normal in W so L is normal in W since L is characteristic in S , by the above. Suppose that W < G and let g 2 NG .W / W be such that g 2 2 W . We have hK t \ W i ht; t bi D L (indeed, involutions t and t b are not conjugate in L so they generate L), U is normal in G and all involutions in S lie in L [ U (see Appendix 16). Therefore (replacing g with gw, w 2 W , if necessary) one may assume that s D t g 2 W S ; indeed, by assumption W < G and the subgroup 1 .S / is not normal in G). Since s normalizes L and CG .t / \ .hsi L/ D ht; zi, we have hsi L Š D2nC1 : indeed, by Proposition 1.8 (Suzuki’s theorem), hsi L is of maximal class and s 62 L. Now L and Lg are normal in W (since L is characteristic in S G W ), Lg \ S D Lg \ L, Lg S D W , jLg S W S j D jLg W .Lg \ S /j D jLg W .L \ Lg /j D 2 and jLLg j D 2nC1 , and so 2 LLg D hsi L. Hence, .LLg /g D Lg Lg D Lg L D LLg since g 2 2 W and L is normal in W , and so hsi L is the dihedral group of order 2nC1 which is normal in W . We have W D A.hsi L/ and this contradicts the maximality of S D AL (see (i–iv)). Hence we must have W D G in any case. If 1 .S / D 1 .G/, we get the groups stated in part (d) of the theorem. In view of the result of the previous paragraph, in what follows we assume, in addition, that 1 .S / ¤ 1 .G/. Then for each involution s 2 G S , s normalizes L since L is characteristic in 1 .G/ G G, and so M D hsi L Š D2nC1 , by Suzuki’s theorem (see Proposition 1.8) since CG .t / \ M D ht; zi is abelian of type .2; 2/. Because of the maximality of S (see (i–iv)), M is not normal in G (otherwise, A would normalize the dihedral subgroup M containing L properly, A \ M D 1 .A/ and M \ G1 D D1 , against the choice of S and L); moreover, the above argument shows also that M is not A-invariant. But L is normal in G so since t 2 L, we get A0 D CG .L/ D CA .L/ (containing 2 .A/, by assumption) is also normal in G and we have A0 D Z.S / which is of order 4. We look at the structure of GN D G=L which is a group with cyclic subgroup SN D AN D .AL/=L Š A=hzi (bar convention!) of order 2m1 and index 2 and with a subgroup MN D M=L of order 2 which intersects N a semidirect product, AN trivially. Since M is not normal in G, we see that GN D MN A, N 4 so m 3 since AN D jA=1 .A/j < jAj). is nonabelian. We conclude that jAj Now jA0 j 4 and so jAN0 j 2 and AN0 is normal in GN (recall that A0 D Z.S / is N Since G= N AN0 is a subgroup of the characteristic in S so normal in G) and AN0 A. N AN0 is abelian outer automorphism group of the dihedral group L, it follows that G= (see Proposition 34.8). Suppose at first that GN is not of maximal class; in that case, GN Š M2m , where N D .UM /=L and UM D 1 .G/ since 2m D jAj (see Theorem 1.2), so that 1 .G/ U and M are generated by involutions. Since M is not normal in G, it follows that s must invert Z.1 .S // D 2 .A/ and so hz; u; si Š E8 . Namely, if s centralizes 2 .A/, then 1 .G/ D 2 .A/ M , the central product, and each involution in G

48 On 2-groups with small centralizer of an involution, I

25

lies in M [ U and so M would be characteristic in 1 .G/ so normal in G, which contradicts the maximal choice of S since S < AM (see (i–iv)). If we set A D hai, then NG .M / D M ha2 i which follows from the structure of GN Š M2m (see Theorem 1.2). Since 2 .A/ centralizes L but 2 .A/=hzi acts faithfully on M , it follows that ha2 i=hzi acts faithfully on M and so ha2 i=2 .A/ acts faithfully on L and CG .L/ D 2 .A/ D A0 . It follows that Z.G/ < A0 D 2 .G/ is of order 2 and A=2 .A/ acts faithfully on L (since the cyclic group Ã1 .A/=2 .A/, as we saw, acts faithfully on L). We have m 4 since GN Š M2m in view of the fact that GN is not of maximal class, N 8. One may assume that by the assumption of this paragraph; we conclude that jAj s the involution s acts on L D hb; t i as follows: b D b 1 and t s D t b and we set n2 b2 D z, where z generates Z.G/. Replacing a with ar (r is odd), one may assume i a that b D b 1C2 , i 2, because CL .a/ D1 Š D8 . Hence we have CL .a/ Š D2i C1 m2 and so CL .a2 / Š D2i Cm1 Š L Š D2n , where we have taken into account m2 i D A=2 .A/ acts faithfully on L and 2 .A/ centralizes L. Thus, that A=ha2 i C m 1 D n and so i D n m C 1. Since i 2, we must have n m C 1. Because m 4, we have here n 5. It remains to determine the commutator Œa; s. We know that A D hai does not normalize M (indeed, G D AM and M is not normal in G) and so hŒa; N sN i D AN0 which yields Œa; s D a0 l, where a0 , a generator of A0 , has order 2 4, a0 D z and l 2 L. We can also write Œa; s D .a0 z/.zl/ D .a0 a02 /.zl/ D a01 .zl/, m2 and s and so replacing l with zl, if necessary, one may assume that a0 D a2 1 a inverts ha0 i. Hence we get Œa; s D a sas D a0 l and so s D a0 ls. Since, by the previous equality, a0 ls is an involution, we must have l D b j for a suitable integer j . 1 j i Therefore, t a sa D t a0 b s which yields b a D b 12j . On the other hand, b a D b 1C2 and so j 2i1 .mod 2n2 /. This yields m2

(1) s a D a2 (2)

sa

D

b 2

i 1

s or

m2 2i 1 a2 b sz. ni 1

ni 2

However, replacing b with b 0 D b 12 and t with t 0 D t b 2 , we see that all so far obtained relations in this case go into the same relations with b 0 instead of b but the relation (1) goes into relation (2). (Note that L D hb; t i D hb 0 ; t 0 i). Therefore we may choose relation (1). We have determined in this case the structure of G uniquely and this the group given in part (f) of the theorem. N GN 0 is of order 4 and Suppose now that GN D G=L is of maximal class. Since G= N AN0 is abelian, we have jAN W AN0 j 2. G= We consider at first the case AN D AN0 which means that A D CG .L/ and so S D A L (a central product) with A \ L D hzi D Z.L/. Hence A D Z.S / and so A is normal in G. Since Ahsi=hzi is of maximal class and the case ŒA; s hzi is not possible because M D hsi L is not normal in G, we have either as D a1 or as D a1 z, which is possible only when m 3. Our group is isomorphic to a group F .m; n/ as stated in part (g). Also we have here hz; u; si Š E8 . It remains to consider the case where jAN W AN0 j D 2 so that CG .L/ D A0 is of index 2 in A. We have A0 D ha2 i D Z.S / and so a induces an automorphism of

26

Groups of prime power order

order 2 on L with CL .a/ D1 . This yields at once that b a D bz and so n 4. Hence we also have ab D az. Since ŒA; L D hzi and L=hzi is a dihedral group of order 2n1 8, it follows that CS .L=hzi/ Š A hb 0 i, where hb 0 i is a cyclic subgroup of order 4 in L. Hence A hb 0 i is normal in G since L=hzi is G-invariant. If s normalizes A, then as D a1 z , D 0; 1, since Ahsi=hzi is of maximal class. But from st s D t b we get acting on a: asts D atb which yields az D a. Indeed, asts D .a1 z /ts D .a1 z /s D az z D a, and this is a contradiction. Hence s does not normalize A. Since As A hb 0 i and .Ahb 0 i=hb 0 i/hsi is of maximal class, one may set (3) as D a1 b 0 or (4) as D a1C2

m2

b 0.

We must have a D a1 D a. But in case (3) this yields .a1 b 0 /s D a and so az D a, which is a contradiction. Hence we must have relation (4). Replacing a with m2 a1 in (4), we get .a1 /s D .a1 /1C2 .b 0 /1 and other relations in this case are m3 0 2 not changed. Thus one may put b D b in (4). This determines the group G as stated in part (h) of the theorem. Here we have also hz; u; si Š E8 . An inspection of obtained groups shows that in any case G has no elementary abelian subgroups of order 16. Indeed, S has no elementary abelian subgroups of order 8 since S=L is cyclic and L is dihedral. It follows from jG W S j 2 that G has no elementary abelian subgroups of order 16, as desired. It follows from defining relations of G that it is generated by 3 elements. Also in case where G has an elementary abelian group of order 8, we see that Z.G/ has order 2. The proof is complete. s2

Since G of Theorem 48.1 has no normal subgroups isomorphic to E8 , each subgroup of G is generated by 4 elements, by MacWilliams’ theorem [MacW]. (See the proof of this theorem of MacWilliams in 50.) Let G D U wr V , where U Š C2m and V D ht i Š C2 . Then CG .t / D ht i C , where C , the diagonal of U U t , is cyclic of order 2m , Z.G/ D C . Exercise 2. Let L be a subgroup of maximal class in a 2-group G. Suppose that, whenever an involution s 2 G L, then hs; Li is of maximal class. Is G of maximal class? Solution. Yes. If G L has no involutions, G is of maximal class, by Theorem 1.17(a). So suppose that G L has an involution. Assume that G is a counterexample of minimal order. Then jG W Lj > 2 and all maximal subgroups of G containing L are of maximal class. The number of maximal subgroups of G containing L which are of maximal class is even (see 13). Since the number of maximal subgroups of G containing L is odd, we get a contradiction. Exercise 3. Let t be an involution of a 2-group G such that CG .t / D ht i Q, where Q Š Q2m . Prove that (i) G has no normal elementary abelian subgroups of order 8, (ii) jNG .CG .t // W CG .t /j 2.

48 On 2-groups with small centralizer of an involution, I

27

Exercise 4. Let G be a nonabelian p-group, p is odd. Suppose that G has an element t of order p such that CG .t / D ht i C , where C is cyclic of order p m , m > 1. Prove that G has no ht i-invariant subgroups of order p pC1 and exponent p. (Hint. Use Theorem 9.6.)

Problems Problem 1. Classify the 2-groups G containing an abelian subgroup A of type .4; 2/ such that CG .A/ D A. Problem 2. Classify the 2-groups containing a cyclic subgroup Z of order 4 such that CG .Z/ D Z C2 . (For a solution, see 77.) Problem 3. Let a 2-group G admit an automorphism ˛ of order 2 and set CG .˛/ D Q, where Q is either cyclic or of maximal class. Describe the structure of G. (See 51.) Problem 4. Classify the 2-groups G containing an involution t such that CG .t / D ht i M2m . Problem 5 (Janko). Classify the 2-groups G such that CG .E/ D E for some elementary abelian subgroup E of order 8. (For solution, see 51.) Problem 6 (Blackburn). Classify the nonabelian p-groups G, p > 2, containing an element t of order p such that CG .t / D ht i C , where C Š Cp m , m > 1. Problem 7. Classify the 2-groups G containing a nonabelian subgroup D of order 8 such that jCG .D/j D 4. Problem 8. Study the 2-groups G containing an extraspecial subgroup E of order > 23 such that CG .E/ D Z.E/. Problem 9. Study the 2-groups G containing a subgroup M Š M2n such that we have CG .M / D Z.M /.

49

On 2-groups with small centralizer of an involution, II

In the previous section finite 2-groups G have been classified with the property that G has an involution t such that CG .t / D ht i C , where C Š C2m is a cyclic group of order 2m , m 1. There is one more case of the centralizer of an involution in 2-groups which is very important. This is seen from the following “conditionless” Theorem 10.26: For a finite p-group G, one of the following holds: (a) G has no maximal elementary abelian subgroups of order p 2 . (b) j1 .G/j p 2 . (c) There exists in G an element x of order p such that CG .x/ D hxi Q, where Q is cyclic or generalized quaternion. In this section we classify the finite 2-groups G which possess an involution t such that CG .t / D ht i Q, where Q Š Q2m is generalized quaternion of order 2m ; m 3. Of course, the situation becomes more complicated; however, in spite of that, the exposition remains elementary. From the start, it is clear, that such a group G cannot be of maximal class. Then, by the known results, G must contain a normal four-subgroup U . We have two essentially different possibilities according to t 2 U or t … U . In the first case, where t 2 U , we have either CG .t / D G or the order of G is equal to 2mC2 and then we get four classes of 2-groups which will be given in terms of generators and relations (see Theorem 49.1). In the second case, where t 62 U (Theorem 49.2), the situation is more complicated since the order of G is not bounded with the parameter m. In fact, in some cases with given m 3, we get infinitely many 2-groups with the same centralizer of our involution t . The idea of the proof in the second case is to construct certain large subgroup S of the known structure with U CG .t / S and then to show that S is normal in G and that jG=S j 4. This last “closure” argument is the main trick in the proof. The corresponding argument in the previous “cyclic” case, which was treated in the previous subsection, was considerably simpler. As a result, we obtain three different types of 2-groups G, since we have exactly three possibilities for the structure of S . In two of these types, the groups G could possess elementary abelian

49

On 2-groups with small centralizer of an involution, II

29

subgroups of order 16 and in that case the corresponding series of 2-groups will be given in terms of generators and relations (Theorem 49.3). The first member of one of these series is a 2-group of order 27 which is isomorphic to a Sylow 2-subgroup of the simple group J2 of order 604800. Finally, it turns out that the case m D 3 of an ordinary quaternion group Q Q8 is more difficult than the general case m 4. As a direct application of these results and the results in the previous subsection, we obtain a classification of 2-groups which have more than 3 involutions but which do not have an elementary abelian subgroup of order 8 (Theorem 49.4). Such groups have been considered by D. J. Rusin [Rus] by using a heavy cohomological machinery. Our methods are completely elementary. Let G D Q wr C , where Q Š Q2m and C D ht i Š C2 . Then CG .t / D ht i D, where D is the diagonal of the base Q Qt so D Š Q2m . Let G be a 2-group possessing an involution t such that CG .t / D ht i Q, where Q Š Q2m . We claim that then G has no normal elementary subgroups of order 8. Assume that this is false, and let E Š E8 be normal in G. Clearly, t 62 E. Set H D ht i E. Then CH .t / contains an elementary abelian subgroup of order 8, by Proposition 1.8 (Suzuki’s theorem), which is not the case. We see that the group G is one of the groups of 50. It follows from Theorem 50.3 that every subgroup of G is generated by four elements. 1o . The case t 2 U . In this subsection we prove the following Theorem 49.1. Let G be a 2-group containing an involution t such that CG .t / D ht i Q, where Q is a generalized quaternion group of order 2m ; m 3. We assume, in addition, that t is contained in each normal four-subgroup U of G and that G ¤ CG .t /. Then we have the following possibilities: (A1) If t 62 ˆ.G/, then G has a maximal subgroup M so that G D M ht i, where M Š Q2mC1 or M Š SD2mC1 and CM .t / Š Q2m . (A2) If t 2 ˆ.G/, then one of the following assertions holds: (a) m D 3 and G D ht; u; v; w; y j y 2 D t v; w 2 D v 2 D u; u2 D t 2 D Œt; v D Œt; w D 1; Œv; w D u; t y D t u; v y D v 1 ; w y D wt i. We have jGj D 25 , Z.G/ D hui has order 2, ˆ.G/ D ht i hvi is abelian of type .4; 2/, G 0 D hu; t i is a four-group and CG .t / D ht i hv; wi, where hv; wi Š Q8 . Also G D hy; wi and G has exactly three involutions. m1

D t 2 D Œt; a D Œt; b D Œa; x D (b) m 4 and G D ha; b; t; u; xi, where a2 m3 2 2m2 b 1 2 1, b D a D u, a D a , x D a2 , b x D bt , t x D t u. The mC2 , Z.G/ D hui is of order 2, ˆ.G/ D G 0 D ht i ha2 i is order of G is 2 abelian of type .2m2 ; 2/ and G is not an -group since there exist involutions in G ha; b; ti but G has no subgroups isomorphic to E8 . Also, G=ˆ.G/ has order 8 and CG .t / D ht i ha; bi, where ha; bi Š Q2m and G D ha; b; xi.

30

Groups of prime power order m1

(c) m 4 and G D ha; b; t; u; xi, where a2 D t 2 D Œt; a D Œt; b D Œa; x D m2 m3 1, b 2 D a2 D u, ab D a1 , x 2 D a1C2 , b x D bat , t x D t u. The mC2 order of G is 2 ; Z.G/ D hui is of order 2, ˆ.G/ D ht i hai is abelian of type .2m1 ; 2/ and G 0 D hat i is cyclic of order 2m1 . Also, G has exactly 3 involutions , G=ˆ.G/ has order 4, CG .t / D ht i ha; bi, where ha; bi Š Q2m and G D hb; xi. m1

(d) m 5 and G D ha; b; t; u; xi, where a2 D t 2 D Œt; a D Œt; b D x 2 D 1, m3 2 2m2 b 1 x b D a D u, a D a , t D t u, ax D a1C2 t , b x D b. The order of G is 2mC2 , Z.G/ D hui is of order 2, ˆ.G/ D G 0 D ht i ha2 i. Also, G is not an -group but G has no subgroups isomorphic to E8 . We have CG .t / D ht i ha; bi, where ha; bi Š Q2m and G D ha; b; xi, Here we have CG .x/ Š C2 Q2m2 . Proof. Suppose that G satisfies the assumptions of the theorem and let U be a normal four-subgroup of G. By assumption, we have t 2 U . Let T D CG .U /, so that jG W T j D 2. We have T D ht i Q, where Q Š Q2m ; m 3. Set 1 .Q/ D hui and then we have U D ht; ui D 1 .T / and obviously Z.G/ D hui. We consider at first the easy case t … ˆ.G/. Let M be a maximal subgroup of G such that t … M . Then G D ht i M and, by the modular law, CG .t / D ht i M0 , where M0 D T \ M Š Q2m . If M is not of maximal class, then by Lemma 1.4, M contains a G-invariant four-subgroup U0 . But then t … U0 , which contradicts the assumption of our theorem. Thus, M is of maximal class and therefore M Š Q2mC1 or M Š SD2mC1 (see Theorem 1.2). There is an element y 2 M M0 such that y 2 2 hui D Z.M /. We have y 1 ty D ut so that tyt D yu and this determines the action of t on M since all such y’s together with M0 generate M . We have obtained the group G stated in part (A1) of the theorem. We examine now the difficult case t 2 ˆ.G/ and suppose at first that Q Š Q8 so that hui D Z.Q/ D Z.G/ D ˆ.T / and U D ht; ui. It follows from u 2 ˆ.Q/ ˆ.G/ and U 6 Z.G/ that U < ˆ.G/, by Exercise 1(b3). Since ˆ.G/ < T , we get, by the modular law, ˆ.G/ D ht i hvi, where hvi is a cyclic subgroup of order 4 in Q and v 2 D u; then ˆ.G/ is abelian of type .4; 2/ and G is generated by two elements. Because in case of 2-groups ˆ.G/ D Ã1 .G/ and here we have ˆ.G/ > U , it follows that there are elements y; z 2 G T so that y 2 D t v and z 2 D v since t and t u are not squares. Next, hyi 6E G (otherwise, G is metacyclic since G=hyi is cyclic in view of ˆ.G/ 6 hyi). Similarly, hzi 6E G. If z normalizes Q, then hQ; zi is of maximal class and order 16, by Theorem 1.2, a contradiction since ˆ.G/ is not isomorphic to subgroup of a 2-group of maximal class. Thus, y does not normalize Q so Q is not normal in G. It follows that the core hziG D hvi (let us consider the representation of G by permutations of left cosets of hzi). Since c2 .ˆ.G// D 2, hvt i is also normal in G. We have t y D t u D t z , by Exercise 1(b2). It follows from t v D .t v/y D t y v y D t uv y that v y D vu D v 3 D v 1 . Take an element w 2 Q hvi so that Q D hv; wi. Since G=ˆ.G/ is abelian and Q 6E G, hvi G G, we

49

On 2-groups with small centralizer of an involution, II

31

have w y D wv r t for some integer r. If r is odd, then Œw; y D v r t D .vt /r . Thus, modulo hvt i, elements w, y and v are pairwise permutable. Since G D hQ; yi D hv; w; yi and v 2 D w 2 D u and y 2 D vt are contained in hvt i, we see that G=hvt i is elementary abelian, contrary to the fact that ˆ.G/ is noncyclic. Thus, r is even, and we get w y D wt or w y D wut D w 1 t . However, if w y D w 1 t , then replacing y with yv, we get w yv D .w 1 t /v D wt and other relations remain unchanged: t yv D t u, v yv D v 1 , .yv/2 D yvyv D y 2 v y v D t vv 1 v D t v. Hence we may assume from the start that w y D wt . The structure of the group G of order 25 is uniquely determined and this is the group stated in part (A2)(a) of our theorem. We turn now to the general case Q Š Q2m , m 4. Let a be an element of order 2m1 in Q so that hai is the unique cyclic subgroup of index 2 in Q. Here CG .t / D ht i Q D T D CG .U /, where U D ht; ui is the G-invariant four-subgroup in T , hui D Z.Q/ D Z.G/ and hui < hai. Now, T =U Š Q=hui Š D2m1 so that ha; U i=U is the unique cyclic subgroup of T =U of order 2m2 4 and so L D ha; U i D ht i hai is normal in G. If b 2 Q hai, then b 2 D u and ab D a1 . Taking any x 2 G T , we have t x D t u and ax 2 L. If x does not normalize hai, then hai\hax i is an x-invariant subgroup of index 2 in hai since jL W haij D 2 (here we use the product formula). Hence, in any case, ha2 i is a cyclic normal subgroup of order 2m2 4 of G. In particular, the cyclic subgroup hvi of order 4 contained in ha2 i is normal in G. Thus, A D ht ihvi is a normal abelian subgroup of type .4; 2/ in G. We have CG .A/ D L so that G=L acts as an automorphism group of order 4 on A. Since hvi is normal in G and v b D v 1 , so there is an element x in G T with v x D v and also t x D t u D t v 2 . Indeed, CG .v/ is a maximal subgroup of G different of T so as x we can take every element in CG .v/T . Since t b D t , we see that b and x induce two distinct involutory automorphisms on A and therefore G=L is an elementary abelian group of order 4 (recall that b 62 L and t; v 2 L). In particular, we get that ˆ.G/ L. On the other hand, ˆ.G/ ht; a2 i. Hence we have either ˆ.G/ D L D ht; ai or ˆ.G/ D ht; a2 i.< L/. In any case, x 2 2 L since exp.G=L/ D 2 and so x induces an automorphism of order 2 on L such that t x D t u; ha2 ix D ha2 i and x centralizes the subgroup hvi of order 4 which is contained in ha2 i. Therefore, we have ax D aj t i ; where i D 0; 1 and an integer j is odd (otherwise, o.aj t i / < o.a/). Since x normalizes ha2 i and centralizes hvi ha2 i, we have either .a2 /x D a2 or .a2 /x D a2 u (indeed, if x does not centralize a2 , then hx; a2 i=Chxi .a2 / Š M2m1 , by Theorem 1.2); in particular, in the case under consideration, m 5. If .a2 /x D a2 , we obtain ax D at i or ax D aut i . Indeed, .a2 /x D a2j so j 1 .mod 2m2 /; then aj D a or aj D au. If .a2 /x D a2 u, we get ax D avt i or ax D auvt i , where we set m3 v D a2 . Indeed, then j 1 C 2m2 so aj D av or aj D auv. As we established above, we must have in the case where x does not centralize a2 , o.a2 / 8 and so m 5. Let ax D at i or ax D aut i , where m 4 and o.a/ D 2m1 . If ax D aut i , then replacing a with at , we get .at /x D aut i t u D .at /t i . Note that hat; bi D Q0 Š Q2m and also CG .t / D ht i Q0 . Hence we may assume from the start that ax D at i ,

32

Groups of prime power order 2

i D 0; 1. However, if ax D at , then we compute a D ax D .at /x D at t u D au, which gives u D 1 and this is a contradiction. It follows that ax D a. If x 2 D t aj with an integer j , then CG .x/ ha; t aj i D L, which is a contradiction since x 62 L D CG .L/. Hence we have x 2 2 hai. If b x 2 Q, then hxi normalizes Q D ha; bi and since x 2 2 hai, so Qhxi would be a maximal subgroup of G which does not contain t 2 ˆ.G/, a contradiction. It follows that b x D b 0 t , where b 0 is an element of order 4 contained in Q hai. There are exactly two conjugacy classes of elements of order 4 in Q hai with the representatives b and ba. Thus, replacing x with xak for some integer k, we may assume that b x D bt or b x D bat . 2 Let b x D bt . We compute b x D .bt /x D bt t u D b 1 and so x 2 2 hvi, where m3 m3 v is an element of order 4 in hai. Hence, we may set x 2 D a2 or x 2 D a2 u m3 2 2 because these elements are the only elements of order 4 in hai. If x D a u, then we replace x with xt . Then xt 2 G T , xt centralizes hai, b xt D .bt /t D bt and m3 m3 finally .xt /2 D xtxt D x 2 t x t D a2 ut ut D a2 . Hence, we may assume from m3 the start that x 2 D a2 and so the structure of G is uniquely determined in this m4 case. We see that here ˆ.G/ D ht i ha2 i D G 0 and G D ha; b; xi. Also, xa2 t is an involution in G ha; b; t i. We have obtained the group stated in part .A2/.b/. 2 Now let b x D bat . We compute b x D .bat /x D bat at u D ba2 u. On the other m3

1C2m3

hand, by setting v D a2 , we see that b a D b av D ba2 u. Hence, we may set 2 2 x D av or x D avu. Note that the group hai=hui produces by conjugation exactly 2m3 distinct conjugates of b. However, if x 2 D avu, then similarly as before, we replace x with xt so that we obtain .xt /2 D xtxt D x 2 t x t D avut ut D av and all m3 other relations remain unchanged. Hence, we may assume that x 2 D av D a1C2 and the group G is again uniquely determined. Here we have ˆ.G/ D ht i hai D L and our group G D hx; bi is 2-generated. Also, we see that G 0 D hat i and there are no involutions in G T . Therefore G has exactly 3 involutions. We have obtained the group stated in part .A2/.c/. m3 Now let ax D avt i or ax D auvt i , where v D a2 , m 5. If ax D auvt i , then replacing a with at (as above) we get .at /x D auvt i t u D .at /vt i . Hence we may assume from the start that ax D avt i , i D 0; 1: However, if ax D av; then a D 2 ax D .av/x D .av/v D av 2 D au, which gives u D 1 and this is a contradiction. It remains only the case ax D avt: We note that x 2 2 L and CL .x/ D ha4 ; x 2 i, where hvi ha4 i since o.a/ D 2m1 16. All elements in T L are of order 4. There are exactly four conjugate classes of such elements in T L and under the action of hai each such element is conjugate to one of the following four elements: b, ba, bt , bat , where ha; bi D Q Š Q2m . So replacing x with xak for some integer k (note that xak also centralizes v), we may assume that we have one of the following four 2 possibilities: b x D b, b x D bt , b x D ba or b x D bat . If b x D ba, then b x D x 2 2 .ba/ D baavt D ba vt , which is a contradiction since x 2 L < T and Q is normal 2 in T: Similarly, if b x D bat; then b x D .bat /x D bat avt t u D ba2 vut , which is a contradiction since Q is normal in T: If b x D bt; then we replace b with b 0 D ba

49

and x with x 0 D xa2

m4

m4 xa2

33

On 2-groups with small centralizer of an involution, II m3

(which also centralizes v D a2

m4 a2

0

). We compute .b 0 /x D

.ba/ D .bt avt / D bb 1 a2 ba2 av D ba2 a2 av D 1 0 x bv av D ba D b . Hence we may assume from the start that b D b: In this case x 2 2 CL .b/ D ht; ui D U: Since CG .U / D T; we must have x 2 2 hui D Z.G/: If x 2 D u; then replacing x with xt; we get .xt /2 D xtxt D x 2 t x t D ut ut D 1 and other relations remain unchanged: axt D .avt /t D avt and b xt D b t D b, Hence we may assume that x 2 D 1 and so the structure of G is uniquely determined as given in part (A2)(d). m4

m4

m4

m4

2o . The case t 62 U . In this subsection we examine the difficult case, where our 2group G has a normal four-subgroup U such that an involution t 2 G is not contained in U and CG .t / D ht i Q with Q Š Q2m , m 3. In order to formulate our main result, we shall define three types of 2-groups: A.m; n/, B.m; n/ and C.m; n/, m; n 2 N . m1

Definition. Let S D QL be a product of two normal subgroups Q D ha; b j a2 D 4 2 2m2 2 b 1 2n1 2 b D z D 1; a D b D z; a D a i Š Q2m and L D hc; t j c Dt D n2 z 2 D 1; c 2 D z; c t D c 1 i Š D2n , where m 3, n 3 and Q \ L D Z.Q/ D Z.L/ D hzi. If ŒQ; L D 1; then S D Q L is the central product of Q and L which we denote with A.m; n/ for m 3, n 3. If CQ .L/ D hai and t b D t , c b D cz, then so determined group S we denote with B.m; n/ for m 3, n 4. If CQ .L/ D ha2 ; bi Š Q2m1 , m 4 and t a D t , c a D cz, then so determined group S we denote with C.m; n/ for m 4, n 4. Obviously, jGj D 2mCn1 . Theorem 49.2. Let G be a 2-group containing an involution t such that CG .t / D ht i Q, where Q is a generalized quaternion group of order 2m , m 3. We assume, in addition, that G has a normal four-subgroup U such that t … U . Then G has a normal subgroup S containing U CG .t / such that jG=S j 4 and S is isomorphic to one of the groups A.m; n/, B.m; n/ or C.m; n/. If S Š B.m; n/, then we must have S D G. If S Š C.m; n/, then jG=S j 2 and if, in addition, jG=S j D 2, then we have m D n. Finally, if jG=S j D 4, then we must have S Š A.m; m/ and G acts transitively on the set of involutions in S U . Proof. Let E4 Š U G G with t 62 U . Set T D CG .U / and then we have obviously t 62 T . This gives jG W T j D 2 and so G D ht iT . By the modular law, G0 D CG .t / D ht i Q, where Q D CT .t / Š Q2m , m 3, and so Z D hzi D 1 .Q/ D Z.Q/ D CU .t / D Z.G/. Set G1 D NG .G0 /. Since j.UG0 / W G0 j D 2, we have UG0 G1 . It follows from 1 .G0 / D ht; zi that t has exactly two conjugates t and t z in G1 . Therefore, jG1 W CG1 .t /j D jG1 W G0 j D 2 so G1 D UG0 . Set D0 D U ht i so that D0 Š D8 , ŒQ; D0 D f1g with Q \ D0 D hzi and U D hz; ui <

34

Groups of prime power order

D0 so G1 D Q D0 , by the product formula. Let v be an element of order 4 in m1 D0 and let y be an element of order 4 in hai, where Q D ha; b j a2 D b4 D m2 2 2 2 b 1 2 D b D z; a D a i Š Q2m , m 3. Since y D v 2 D z, z D 1; a 2 we get .yv/ D y 2 v 2 D y 4 D 1 so x D yv is an involution in G1 D0 . But x t D .yv/t D yv 1 D yvv 2 D yvz D xz and therefore D1 D hx; t i Š D8 and D1 \ U D Z.D1 / D hzi. We see that CG1 .D1 / D ha; bt i D Q1 Š Q2m because x bt D .yv/bt D y 1 v 1 D yv D x, .bt /2 D z and abt D a1 . Thus we have G1 D Q1 D1 with Q1 Š Q, Q1 \ D1 D hzi, CG .t / D ht i Q D ht i Q1 and U D hz; ui 6 D1 . Denoting again Q1 with Q and bt with b, we have obtained the following initial result. (R) Supposing that U D hz; ui is a normal four-subgroup of our 2-group G possessing an involution t with t … U and CG .t / D ht i Q, where Q Š Q2m , m 3, then G1 D U CG .t / has the following structure. We have G1 D Q D1 , where D1 Š D8 , t 2 D1 , Q \ D1 D hzi D Z.G/ and U 6 D1 . The next step in the proof is to “blow up” the subgroup G1 D U CG .t / as much as possible so that the structure of a large subgroup S containing G1 remains “about” the same as the structure of G1 . In the rest of the proof, we denote with S a subgroup of G of the maximal possible order subject to the following conditions: (i) S G1 D U CG .t / D Q D1 , where Q Š Q2m , m 3, D1 Š D8 , t 2 D1 . (ii) S D QL, where L is normal in S , L Š D2n , n 3, Q \ L D hzi D Z.Q/ D Z.L/, L D1 . (iii) U D hz; ui 6 L. m1

In the sequel, we fix the following notation. We set Q D ha; b j a2 D b4 D 2 2m2 2 b 1 2n1 2 z D 1; a D b D z; a D a i Š Q2m and L D hc; t j c D t D z2 D n2 D z; c t D c 1 i Š D2n , where m 3, n 3 and Q \ L D Z.Q/ D 1; c 2 Z.L/ D hzi. Then we have u D yv, where y is an element of order 4 in hai and v is an element of order 4 in hci. If a4 D 1, then we put y D a. Similarly, if c 4 D 1, then we put v D c. At first we shall determine the structure of S . Act with Q on the dihedral group L. Since Aut.L/=Inn.L/ is abelian of type .2n3 ; 2/ (see Theorem 34.8), so Q0 D ha2 i centralizes L. Also, we know that Q centralizes hv; t i D D1 Š D8 . It follows that either Q centralizes L and then S D Q L Š A.m; n/, m 3, n 3 or M D CQ .L/ is a maximal subgroup of Q, in which case n 4. We have essentially two different possibilities for M . First possibility is M D hai, t b D t , c b D cz and then S Š B.m; n/, m 3, n 4. Second possibility is M D ha2 ; bi Š Q2m1 , m 4, t a D t , c a D cz and then S Š C.m; n/, m 4, n 4. We see that we have in any case ŒQ; L hzi and so Q is a normal subgroup in S . Suppose that S Š B.m; n/. Then S has exactly six conjugacy classes of involutions contained in S U with the representatives t , t c, bv, bav, bt c, bat c. The

49

35

On 2-groups with small centralizer of an involution, II

corresponding centralizers are: CS .t / D CG .t / D ht i Q

with

Q Š Q2m ;

CS .t c/ D ht ci ha; bvi

with

ha; bvi Š D2m ;

CS .bv/ D hbvi hcy; t ci

with

hcy; t ci Š SD2n ;

CS .bav/ D hbavi hcy; t ci

with

hcy; t ci Š SD2n ;

CS .bt c/ D hbt ci hby; ui

with

hby; ui Š D8 ;

CS .bat c/ D hbat ci hbay; ui

with

hbay; ui Š D8 :

m 3;

It follows that t cannot be fused in NG .S / to any of the other five conjugacy classes of involutions in S U . But CS .t / D CG .t / then forces that S D G. We make here the following simple observation. Since t 2 G T , where T D CG .U / and jG W T j D 2, t cannot be conjugate (fused) in G to any involution t 0 which centralizes U: Therefore, with respect to that fusion, it is enough to consider only those involutions in S which act faithfully on U . Now suppose that S Š C.m; n/. Then S has exactly four conjugacy classes of involutions contained in S U acting faithfully on U with the representatives t , t c, bv, and bav. The corresponding centralizers in S are: CS .t / D CG .t / D ht i Q

with

Q Š Q2m ;

CS .t c/ D ht ci hav; bavi

with

hav; bavi Š SD2m ;

CS .bv/ D hbvi hc; yt i

with

hc; yt i Š Q2n ;

CS .bav/ D hbavi hcy; ct i

with

hcy; ct i Š SD2n :

m 4;

We may assume that NG .S / ¤ S (otherwise we are finished). Then NG .S / can fuse the conjugacy class of t only with the conjugacy class of bv under the assumption that m D n and then jNG .S / W S j D 2. Then cclS .t /, the conjugacy class of in S containing t , equals ft c 2i ; 1 i 2m2 g and CS .t c 2i / D ht c 2i i Q so that .CS .t c 2i //0 D ha2 i. Similarly, we have cclS .bv/ D fba2i v; 1 i 2m2 g, CS .ba2i v/ D hba2i vi hc; yt i, where hc; yt i Š Q2m so that .CS .ba2i v//0 D hc 2 i. Hence for any x 2 NG .S / S , we have t x D ba2j v for some j and so, by the above, ha2 ix D hc 2 i. Assume now that NG .S / ¤ G (otherwise we are finished). Since CS .t / D CG .t / and NG .S / already fuses the involutions in cclS .t / with involutions in cclS .bv/, so there is an element s 2 NG .NG .S // NG .S / with s 2 2 NG .S / and t 0 D t s 2 NG .S / S . Indeed, t s 2 cclS .t c/ or t s 2 cclS .bav/ is not possible since a semi-dihedral group is not a subgroup of a generalized quaternion group. However, if t s 2 cclS .t / [ cclS .bv/; then there is an n 2 NG .S / such that t s D t n . But then 1 t sn D t , which contradicts the fact that CG .t / NG .S /. By the above, we have 0 .a2 /t D c 2k with k odd. Since y is an element of order 4 contained in ha2 i and v 0 is an element of order 4 contained in hc 2 i, we get that y t D vz l with l D 0; 1. But

36

Groups of prime power order 0

0

then .yvz l /t D yvz l which gives .uz l /t D uz l , recalling that yv D u. We have proved that t 0 centralizes U D hz; ui and so CG .t 0 / contains a subgroup isomorphic to E8 . This is a contradiction, since t 0 is conjugate in G to t . Hence, we must have NG .S / D G and so jG W S j D 2, as required. Suppose, finally, that S Š A.m; n/, m 3, n 3; i.e. S D Q L is the central product of Q and L. Then S has exactly four conjugacy classes of involutions contained in S U with the representatives t , t c, bv and bav and all these involutions act faithfully on U . The corresponding centralizers are: CS .t / D CG .t / D ht i Q CS .bv/ D hbvi hc; yt i

with Q Š Q2m ; CS .t c/ D ht ci Q;

with hc; yt i Š Q2n and CS .bav/ D hbavi hc; yt i:

Note that all 2n2 conjugates of t in S lie in the dihedral subgroup L D hc; t i and also all 2n2 conjugates of t c in S lie in L. Similarly, all 2m2 conjugates of bv in S lie in the dihedral subgroup K D hbv; bavi Š D2m and also all 2m2 conjugates of bav in S lie in K. These are all 2 2n2 C 2 2m2 involutions in S U . Also, we see that m1 K D ha; bv j a2 D .bv/2 D 1; abv D a1 i. We want to determine all dihedral subgroups of S which are generated by a pair of distinct involutions in S U . We recall that any two distinct involutions always generate a dihedral subgroup. Any two distinct involutions in cclS .t / [ cclS .t c/ generate a dihedral subgroup which is contained in L. Similarly, any two distinct involutions in cclS .bv/ [ cclS .bav/ generate a dihedral subgroup which is contained in K. We note that hci fuses (by conjugation) all involutions in cclS .t / to t and all involutions in cclS .t c/ to t c. Similarly, hai fuses all involutions in cclS .bv/ to bv and all involutions in cclS .bav/ to bav. We have Œc; a D 1 and a centralizes t and both t c and c centralize bv and bav. All this means that it is enough to see what is the order of the following four dihedral subgroups: hbv; t i; hbv; t ci, hbav; t i and hbav; t ci. The computation shows: .bv t /2 D .bv t c/2 D .bav t /2 D .bav t c/2 D z, and so all these four dihedral subgroups have order 8. Suppose that t is not conjugate in NG .S / to any involution in S L. That will certainly happen if (for example) m ¤ n. Supposing, in addition, that S ¤ G, we see that NG .S / fuses t with t c and so jNG .S / W S j D 2, since CG .t / D CS .t /. If NG .S / D G, then we are finished. Therefore, we assume that NG .S / ¤ G. Take an element s 2 G NG .S / such that s normalizes NG .S / and s 2 2 NG .S /. If an involution x 2 NG .S / S , then .cclL .t //x D cclL .t c/ implies that L0 D Lhxi Š D2nC1 : If jCS .x/j D 2m ; then CL .x/ D hzi and CNG .S/ .x/ D hxi CS .x/ imply that CS .x/ covers S=L and so Q normalizes L0 Š D2nC1 and QL0 D NG .S /; which contradicts the maximality of S . Hence there is no such involution x in NG .S /S with jCS .x/j D 2m : Set P D cclS .t / [ cclS .t c/ [ cclS .bv/ [ cclS .bav/ so that hP i D S and P G T . If t 0 D t s 2 NG .S / S , then CNG .S/ .t 0 / D ht 0 i CS .t 0 / and so jCS .t 0 /j D 2m ; a contradiction. If t 0 2 cclS .t / [ cclS .t c/, then there is n 2 NG .S / such that t 0 D t s D t n and so CG .t / 6 NG .S /; a contradiction. Hence t 0 2 cclS .bv/ or t 0 2 cclS .bav/. If NG .S / fuses bv and bav, then all elements in P are conjugate in

49

On 2-groups with small centralizer of an involution, II

37

hyiNG .S / to t; and since hP i D S , there is x 0 2 P such that x0 D .x 0 /s 2 NG .S /S and so CNG .S/ .x0 / D hx0 i CS .x0 / and jCS .x0 /j D 2m , a contradiction. Suppose that NG .S / does not fuse bv and bav. Then jCNG .S/ .bv/j D jCNG .S/ .bav/j D 2mC2 and so n C 2 D m C 1 because t 0 2 cclS .bv/ or t 0 2 cclS .bav/. There is x 0 2 P such that x0 D .x 0 /s 2 NG .S / S and so jCNG .S/ .x0 /j D 2nC2 D 2mC1 which gives jCS .x0 /j D 2m , a contradiction. This gives NG .S / D G and so jG W S j D 2, as required. Now suppose that t is not conjugate in NG .S / to t c 2 L but S ¤ G. Then NG .S / must fuse t to an involution in S L U . Without loss of generality, we may assume that t is fused in NG .S / to bv. This gives jNG .S / W S j D 2 and m D n because CS .bv/ Š C2 Q2n . Supposing NG .S / ¤ G, we take an element s 2 G NG .S / such that s normalizes NG .S / and s 2 2 NG .S /. Set again P D cclS .t / [ cclS .t c/ [ cclS .bv/ [ cclS .bav/ so that hP i D S and P G T . There is an x 0 2 P such that t 0 D .x 0 /s 2 NG .S / S and t 0 acts faithfully on U . Suppose at first that m D n 4. Then L D hc; t i and K D ha; bvi are the only dihedral subgroups of order 2m in S and so for each x 2 NG .S / S we have Lx D K and so hcix D hai. In particular, 0 v t D yz ; D 0; 1. It follows that t 0 centralizes yvz D uz and since z 2 Z.G/, so t 0 centralizes u. Hence, t 0 centralizes U D hz; ui , a contradiction. We get in this case NG .S / D G and so jG W S j D 2, as required. Suppose now that m D n D 3 so that S is extraspecial of order 25 and S Š Q8 D8 . Then t (fused with bv) lies 0 in a conjugacy class of length 4 in NG .S /. We have t t 2 fbv; bvzg and ht; bvi D D Š D8 . Hence Dht 0 i D D0 Š D16 and so t 0 acts fixed-point-free on D hzi. For each involution t0 in S U , we have CS .t0 / D ht0 i Q0 with Q0 Š Q8 . Since t 0 is a conjugate of an involution in S U under the action os s (normalizing NG .S /), it follows that CNG .S/ .t 0 / D ht 0 i CS .t 0 / contains a subgroup isomorphic to Q8 and so CS .t 0 / D Q Š Q8 because Q \D D hzi. Since NNG .S/ .D/ hS; t 0 i D NG .S /, it follows that Q normalizes D0 D Dht 0 i Š D16 and so D0 is normal in NG .S / (since Q covers S=D). Note that t 2 D and so we have obtained a contradiction with the maximality of S . Hence, also in this case, we have NG .S / D G and so jG W S j D 2. It remains to consider the case where t is fused in NG .S / to each involution in S U . This gives at once that m D n and jNG .S / W S j D 4 so that NG .S /=S acts regularly on the four S -classes of involutions which are contained in S U . Let m D n 4. Then S has exactly two dihedral subgroups L D hc; t i and K D ha; bvi of order 2m and L and K contain all involutions from S U . Therefore, for each x 2 NG .S / S , we have either K x D K and Lx D L or K x D L. Supposing again that NG .S / ¤ G, we take an element s 2 NG .NG .S // NG .S / 0 0 with s 2 2 NG .S /. Then we have t 0 D t s 2 NG .S / S . If Lt D L and K t D K, then t 0 fuses the two classes of involutions in L hci and also t 0 fuses the two classes 0 0 of involutions in K hai. This gives that c t D c 1 and at D a1 . In particular, 0 0 we get ut D .yv/t D y 1 v 1 D yz vz D yv D u. Hence, the elementary abelian subgroup hz; u; t 0 i Š E8 is contained in CG .t 0 / and this is a contradiction. 0 0 0 Now assume that Lt D K. Then hcit D hai and, in particular, v t D yz , D 0; 1.

38

Groups of prime power order 0

0

But then .yvz /t D yvz and consequently .uz /t D uz . Hence t 0 centralizes U D hz; ui, which is a contradiction, since t 0 is conjugate in G to t . Finally, we assume that m D n D 3 so that S Š Q8 D8 is an extraspecial group of order 25 . We recall that NG .S / acts transitively on the set of 8 involutions contained in S U . Again assuming that NG .S / ¤ G, we take an element s 2 NG .NG .S / NG .S / with s 2 2 NG .S /. Then we have t 0 D t s 2 NG .S / S . Set M D hsiNG .S / so that jM W NG .S /j D 2. We shall determine V D S M and we note that S < V NG .S /. Consider at first the possibility V < NG .S / which gives jV W S j D 2. Then V D S S s , t 0 D t s 2 S s S and V D S ht 0 i. By the modular law, O where CS .t 0 / D QO Š Q8 . Then CS .Q/ O D DO Š D8 and CG .t 0 / D CV .t 0 / D ht 0 i Q, O Since DO is t 0 -invariant and t 0 acts fixed-point-free S is the central product of QO and D. O 0 i/ Q, O V fuses all four involutions in DO hzi O 0 i Š D16 , V D .Dht on DO hzi, so Dht and other V -classes of involutions in S U have length 2. This contradicts the fact that NG .S /=S acts regularly on the four S -classes of involutions in S U . It follows that V D S M D NG .S /. Since S S s is normal in M , so we get V D S S s D NG .S /. Thus F D S \ S s is a normal subgroup of M of order 8. If F is nonabelian, then S D F CS .F /, where F1 D CS .F / is nonabelian and normal in NG .S /. Since S Š Q8 D8 , so fF; F1 g D fD8 ; Q8 g. But this contradicts the fact that NG .S / acts transitively an all 8 involutions from S U . We have proved that F is abelian. Since Q8 D8 does not possess a subgroup isomorphic to E8 , it follows that F must be of type .4; 2/ and consequently we have z 2 F . Again, because NG .S / acts transitively an all involutions from S U , so we have U F . Furthermore, S 0 D .S s /0 D hzi and so if hf i is a cyclic subgroup of order 4 in F , so NG .hf i/ hS; S s i D NG .S /. Also, we have S1 D CS .f / Š C4 Q8 and jS W S1 j D 2. Hence, S1 contains exactly 6 involutions which are different from z. This implies that there exist involutions in S S1 . Let t1 be an involution in S1 F and let t2 be an involution in S S1 . Then t1 centralizes f but t2 inverts f . This is a contradiction because NG .S / normalizes hf i and NG .S / acts transitively on all involutions in S U . We have proved again that NG .S / D G and so jG W S j D 4, as required. 3o . Groups with subgroup isomorphic to E16 . As an application of Theorems 49.1 and 49.2, we shall determine up to isomorphism all 2-groups with an involution t such that CG .t / D ht iQ, where Q Š Q2m ; m 3 and which have an elementary abelian subgroup E16 of order 16. We obtain exactly two classes of 2-groups and we present them in terms of generators and relations. More precisely, we shall prove the following result. Theorem 49.3. Let G be a 2-group containing an involution t such that CG .t / D ht iQ, where Q is a generalized quaternion group of order 2m , m 3. We assume in addition that G has an elementary abelian subgroup of order 16. Then G is isomorphic to one of the following groups. m1 m1 (E1 ) G D ha; b; z; c; t; x j a2 D b4 D z2 D c 2 D t 2 D x 2 D 1, m 4, m2 m2 2 2 2 1 1 1 a Db Dc D z, b ab D a , t ct D c , bc D cb, bt D t b, a1 ca D

49

39

On 2-groups with small centralizer of an involution, II m3

m3

m3

cz, at D t a, xcx D a, xtx D bc 2 i. Here hz, x,a2 c2 E16 and ha; bi Š Q2m , hc; t i Š D2m , ha; b; c; t i Š C.m; m/. m1

m3

, ba1C2

t ci Š

m1

(E2 ) G D ha; b; z; c; t; x; s j a2 D b4 D z2 D c 2 D t 2 D x 2 D s 2 D 1, m2 m2 2 2 2 1 1 Db Dc D z, b ab D a , t ct D c 1 , bc D cb, bt D t b, m 3, a ac D ca, at D t a, sx D xs, xtx D t c, xcx D c 1 , xbx D ba, xax D a1 , m3 scs D az, st s D bc 2 i. Here ha; bi Š Q2m , hc; t i Š D2m , ha; b; c; ti Š A.m; m/, m3 2m3 hz, x, s, a2 c i Š E16 . Finally, for m D 3, our group G is isomorphic to a Sylow 2-subgroup of the simple group J2 . Proof. Suppose that a 2-group G satisfies the assumptions of Theorem 49.3 and let E be an elementary abelian subgroup of order 16 in G. Using theorems 49.1 and 49.2, we see that we have exactly two possibilities: (E1 ) G has a normal subgroup S Š C.m; m/, m 4 of index 2 in G, jE \ S j D 8 and so G D SE. (E2 ) G has a normal subgroup S Š A.m; m/, m 3 of index 4 in G, E \ S D U is a normal four-subgroup of G and so G D SE. Case .E1 /. Assume that S Š C.m; m/, m 4; is given with generators and relations as in the definition. Let U D hz; yvi be the normal four-subgroup of G, where m3 is an element of order 4 in hci: y is an element of order 4 in hai and v D c 2 In the proof of Theorem 49.2 we have found all five conjugacy classes of involutions (and their centralizers) in S U . Their representatives are the involutions: t , t c, bv, bav and bat c: It follows that t can be fused in G only to an involution in the conjugacy class of bv: We see also that S has exactly two normal subgroups isomorphic to D2m which do not contain any conjugate of bat c W L D ht; ci and K D hbv; ai. Take an involution x 2 E S: Since x cannot fuse t with t c; it follows that x does not normalize the dihedral subgroup L: Hence we have Lx D K: The involution x sends the conjugacy class of t in L onto the conjugacy class of bv in K: Replacing m3 t with t c 2j (j is an integer), we may assume that t x D bv D bc 2 . Also we x k have hci D hai and so replacing c with c (k is an odd integer), we may assume that c x D a: We note that these replacements of generators did not effect the defining relations for S Š C.m; m/: The structure of G is uniquely determined. Case .E2 /. Assume that S Š A.m; m/ is given with generators and relations as in of A.m; n/. Then we have S D Q L; where Q Š Q2m and L Š D2m : Let U D hz; yvi be the normal four-subgroup of G, where y is an element of order 4 in m3 hai and v D c 2 is an element of order 4 in hci: In the proof of Theorem 49.2 we have found all four conjugacy classes of involutions (and their centralizers which are all isomorphic to C2 Q2m ) in S U: Their representatives are the involutions: t , t c, bv, and bav: Since jG W S j D 4; so G fuses all involutions in S U into a single conjugacy class in G: This gives that E \ S D U: Assume for a moment that m 4: Then we have seen in the proof of Theorem 49.2 that S has exactly two subgroups isomorphic to D2m : They are L D ht; ci and

40

Groups of prime power order

K D hbv; ai and both are normal in S: If we set E D U hx; si; where hx; si\S D 1; then the four-subgroup hx; si acts on fL; Kg: We may assume that x normalizes L and K and Ls D K: The involution x must induce outer automorphisms on L and K and so c x D c 1 and ax D a1 since x fuses two conjugacy classes of non-central involutions in L and also in K: Replacing s with sx (if necessary), we may assume that s sends t to an involution in the conjugacy class of bv (instead of bav) in K: m3 Replacing t with t c 2j (where j is an integer) we may assume that t s D bv D bc 2 : Also we have hcis D hai: Let t x D t 0 , where t 0 D t c k with k an odd integer. Then replacing c with c k ; we see that we may assume from the start that t x D t c: Here it m3 might happen that .c k /2 D v 1 D vz and then the previous relation t s D bv is s 1 D bvz: In that case we replace b with b 1 D bz so that changed into t D bv s we may assume again t D bv: Replacing a with al ; where l is an odd integer, we may assume c s D az: (Here we have taken az rather than a so that the last relation looks simpler!) Act on the relation xtx D t c with s: We get x s t s x s D t s az and so since Œx; s D 1 we have xbvx D bvaz or b x v 1 D bavz and finally b x D ba: We note that these replacements of generators did not effect the defining relations for S Š A.m; m/: The structure of G is uniquely determined. In the remaining case m D 3 we have S Š Q8 D8 : We shall determine the group G almost in the same way as above for m 4: The only difference is that here S has exactly six subgroups isomorphic to D8 which do not contain the normal foursubgroup U of G (see the proof of Theorem 49.2). They are: L D ht; t ci; D2 D ht; bavi;

K D hbv; bavi; D3 D ht c; bvi;

D1 D ht; bvi; D4 D ht c; bavi:

In S U lie exactly eight involutions which are distributed in four S -classes: ft; t zg, ft c; t czg, fbv; bvzg, fbav; bavzg. We set again E D U hx; si; where hx; si\S D 1: Then the four-group hx; si acts transitively (and so regularly) on the above four S classes. This follows from the fact that G acts transitively on the eight involutions in S U since CG .t / has index 8 in G: If is any involution in hx; si; then we must have either L D L and K D K or L D K: Indeed, if L D Di ; where i 2 f1; 2; 3; 4g; then L \ Di D ht; zi or ht c; zi and L \ Di is -invariant. But then the S -class ft; t zg or the S -class ft c; t czg is -invariant which is a contradiction. Similarly, if K D Dj ; where j 2 f1; 2; 3; 4g; then the S -class fbv; bvzg or the S -class fbav; bavzg is invariant which is again a contradiction. Hence the four-group hx; si acts on fL; Kg: We may assume that x normalizes L and K and Ls D K: Then we determine the group G uniquely in exactly the same way as above. The proof is complete. 4o . Groups without subgroups isomorphic to E8 . As a direct application of the above results, we shall prove the following Theorem 49.4. Let G be a 2-group which is not of maximal class and such that j1 .G/j > 4. If G has no elementary abelian subgroups of order 8, then G is

49

On 2-groups with small centralizer of an involution, II

41

isomorphic to one of the following groups: (a) A group G from (A1)(c) of Theorem 48.1 with involutions in G T . (b) A group G from (A1)(d) of Theorem 48.1. (c) A group G from (A1) of Theorem 49.1. (d) A group G from (A2)(b) of Theorem 49.1. (e) A group G from (A2)(d) of Theorem 49.1. (f) A group G with a normal subgroup S of index at most 4 in G so that S is isomorphic to the central product of Q2m and D2n , where m > 2, n > 2 and 1 .G/ S (see Theorem 49.2). Proof. Let G be a 2-group which has more than 3 involutions and which is not of maximal class. Also assume that G has no subgroups isomorphic to E8 . Let U be a normal four-subgroup of G and set T D CG .U /. By assumption we have 1 .T / D U and so there is an involution t 2 G T . We have G D ht iT and so CG .t / D ht i Q, where Q D CT .t /. Since t does not centralize U , it follows that Q has only one involution. Therefore, Q is either cyclic or a generalized quaternion group. Such groups G are then described in this and previous subsection. A simple inspection yields the above result. Exercise 1 (Alperin [Alp3]). Suppose that a nonabelian 2-group G has no normal elementary abelian subgroups of order 8. Suppose that G is neither cyclic nor of maximal class. Then one of the following holds: (a) G has a normal abelian subgroup A of type .4; 2/ with CG .A/ abelian of type .2; 2n /, n 2. In that case, G=CG .A/ is isomorphic to a subgroup of Aut.A/ Š D8 . (b) G has a normal abelian subgroup A of type .4; 4/ with metacyclic CG .A/. In that case, jG=CG .A/j 25 . Solution. Let A be the greatest normal noncyclic abelian subgroup of G of exponent 4. Since G is not of maximal class, either A is abelian of type .2; 4/ or .4; 4/ (Lemma 1.4). By Corollary 10.2, 2 .CG .A// D A. It follows from Theorem 41.1 that CG .A/ is metacyclic. Note that Aut.C4 C2 / Š D8 and jAut.C4 C4 /j D 3 25 . It remains to show that, if G has not G-invariant abelian subgroup of type .4; 4/, then CG .A/ is abelian of type .2; 2n / for some n > 1. Indeed, 2 .CG .A// D A is of order 8 so our claim follows from Lemma 42.1. Exercise 2. Let Q Š Q2m , where m 5 and let H be the holomorph of the cyclic group C Š C2n , n > 2. Set G D Q H with Q \ H D Z.Q/. Show that G has a normal elementary abelian subgroup of order 8. Solution (Janko). Let hyi be a normal (cyclic) subgroup of order 4 in Q, let hvi < C be of order 4 and set hzi D Z.G/; then .yv/2 D y 2 v 2 D zz D 1. Let a be an

42

Groups of prime power order

involution in A D Aut.C / such that ha; C i Š M2nC1 (1 .A/ C has exactly three maximal subgroups containing C ; these subgroups are isomorphic to D2nC1 , SD2nC1 and M2nC1 , respectively). Then a centralizes v 2 Z.ha; C i// and D D hz; yv; ai is the desired normal elementary abelian subgroup of order 8 in G. Indeed, V D hz; yvi is a normal four-subgroup of G, since hzi, hyi, hvi are normal in G and o.yv/ D 2. Next, jha; V ij D 8 and W D hz; ai D 1 .ha; C i/ is characteristic in a normal subgroup ha; C i of G. It follows that D D V W is also normal in G. Since a centralizes V , D Š E8 , and we are done. Exercise 3. Let G be a p-group containing an elementary abelian subgroup of order p 3 and let E be the subgroup generated by all elementary abelian subgroups of G of order p 3 . Suppose that G E contains an element x of order p. Then Chx;E i .x/ D hxiQ, where Q is cyclic or generalized quaternion. Exercise 4. Classify the 2-groups G such that (i) 2 .G/ D C M , where jC j D 2 and M is of maximal class and order > 8, (ii) i .G/ D C M , where i D 1 or 2, C Š C4 and M is of maximal class.

50

Janko’s theorem on 2-groups without normal elementary abelian subgroups of order 8

The main part of this section coincides with [Jan5]. Theorem 50.1 yields a deep insight in the structure of 2-groups without normal elementary abelian subgroups of order 8. The proof is fairly difficult, however elementary. In the analogous case, for p > 2, we have to classify the p-groups G without normal subgroup of order p pC1 and exponent p; it seems that this problem is far from a solution. The first important result about the groups of the title is the 4-generator theorem from [MacW] which asserts that any subgroup of such groups can be generated by four elements. However, this result does not say anything more about the structure of such groups. In [Kon3], the groups of the title are determined under the additional assumption that the Frattini subgroup contains an elementary abelian subgroup of order 8. This determination is somewhat unfortunate, because the resulting groups are given in terms of generators and relations without any comments and from these it is difficult to disclose the structure of such groups. In his long paper [Ust], Ustjuzaninov has asserted that the groups G of the title must have a normal metacyclic subgroup N such that G=N is isomorphic to a subgroup of the dihedral group D8 of order 8. But this paper is completely unreadable. With any attempt to read it, computational errors were discovered. However, it turns out that this result is correct after all! We shall give here a relatively short proof of a stronger result. For example, in the case where G=N is isomorphic to D8 , we shall determine completely the structure of N by showing at first that N is either abelian or minimal nonabelian. In our proof the computations are reduced to a minimum. The proof is based on a method of “pushing up” normal metacyclic subgroups of G combined with a very detailed knowledge of Aut.C4 C4 / (Proposition 50.5). We also note that our proof of the 4-generator theorem is character-free, i.e., it is completely elementary. We state now our main result. Theorem 50.1 (Janko [Jan5]). Let G be a 2-group which has no normal subgroups isomorphic to E8 . Suppose that G is neither abelian nor of maximal class. Then G has a normal metacyclic subgroup N such that CG .2 .N // N and one of the following holds:

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Groups of prime power order

(a) jG=N j 4 and either 2 .N / is abelian of type .4; 4/ or N is abelian of type .2j ; 2/, j 2. (b) G=N Š D8 and (b1) N is either abelian of type .2k ; 2kC1 /, k 1 or of type .2l ; 2l /, l 2 or (b2) N is minimal nonabelian, 2 .N / is abelian of type .4; 4/ and more m n m1 precisely N D ha; b j a2 D b 2 D 1; ab D a1C2 i, where m D n with n 3 or m D n C 1 with n 2. We prove here the following easy special case of the above theorem. Theorem 50.2. Suppose that a 2-group G has no normal elementary abelian subgroups of order 8 and has two distinct normal four-subgroups U and V . Then D D U V Š D8 and G is the central product G D D C of D and C with D \ C D Z.D/ and C is either cyclic or of maximal class 6Š D8 . Conversely, if G D D C , where D Š D8 , C is cyclic or of maximal class 6Š D8 and D \ C D Z.D/, then G has no normal subgroups isomorphic to E8 . Proof. By hypothesis, ŒU; V ¤ f1g. Hence jU \V j D 2 which gives D D U V Š D8 . We have C D CG .D/ D CG .U /\ CG .V / so jG W C j D 4. It follows that G D D C with D \ C D Z.G/, where C D CG .D/. If W is a normal four-subgroup in C , then U W would be an elementary abelian normal subgroup of order 8 in G. This is a contradiction and so C is either cyclic or a group of maximal class (Lemma 1.4) which is not isomorphic to D8 . Now assume that G D DC , where D, C and D\C are the same as in the previous paragraph. Clearly, Z.G/ D Z.C / is cyclic. Assume that G has a normal elementary abelian subgroup E of order 8. Since C 6Š D8 and Z.G/ is cyclic, we get E \ C D Z.D/. By the product formula, G D EC . Let U , V be two distinct four-subgroups in D; then they are normal in G. Since CG .U / D U C and CG .V / D V C , we get U 6 E and V 6 E (recall that G D EC ). It follows that E \ D D Z.D/. Then G=D is noncyclic since it contains a four-subgroup DE=D Š E=Z.D/; it follows that C is of maximal class. Clearly, U is the only G-invariant four-subgroup in U C since C 6Š D8 and U C D Z C , where Z is a subgroup of order 2 in U . However, E \ .U C / is a G-invariant four-subgroup. It follows that U < E, which is a contradiction. The proof is complete. In view of Theorem 50.2, in what follows we may confine our approach to the case where G has exactly one normal four-subgroup. Remark 1. Let G D D C be a 2-group from Theorem 50.2 with G ¤ D. Let Z1 be the cyclic subgroup of order 4 in D.Š D8 / and let Z2 be a maximal cyclic subgroup of index 2 in C . Then N D Z1 Z2 is an abelian normal subgroup of type .2j ; 2/, j 2. and G=N is elementary abelian of order 4. Hence this is a special case of groups occurring in Theorem 50.1(a).

50 Janko’s theorem on 2-groups

45

The 4-generator theorem follows as a trivial consequence of Theorem 50.1. Theorem 50.3 (4-generator theorem). Let G be a 2-group which has no normal elementary abelian subgroups of order 8. Then every subgroup U of G is generated by four elements. Proof. By Theorem 50.1, G has a normal metacyclic subgroup N such that G=N is isomorphic to a subgroup of D8 . Since U=.U \ N / is isomorphic to a subgroup of G=N , we have d.U=.U \ N // 2. Also, U \ N is metacyclic and therefore d.U \ N / 2: Hence d.U / 4 and we are done. We prove three important preliminary results. Proposition 50.4. Let G be a 2-group with a metacyclic normal subgroup N . Suppose that N has a G-invariant four-subgroup N0 which is not contained in Z.G/ but there is no G-invariant cyclic subgroup of order 4 contained in N . If N is abelian, it is of type .2n ; 2n / or .2nC1 ; 2n /, where n 1. If N is nonabelian, it is minimal nonabelian m n m1 and moreover N D ha; b j a2 D b 2 D 1; ab D a1C2 i, where n 2 and m D n or m D n C 1. Proof. If N is abelian (of rank 2), the result is clear since N cannot contain a characteristic cyclic subgroup of order 4. Suppose that N 0 ¤ 1. Since N 0 is cyclic, we must have jN 0 j D 2. By Example 10.14, N is minimal nonabelian. Then a result of R´edei m n m1 (Exercise 1.8a) implies that N D ha; b j a2 D b 2 D 1; ab D a1C2 ; m 2; n 1i. If n D 1, then N Š D8 or N Š M2mC1 with m 3. In both cases N has a characteristic cyclic subgroup of order 4, which is a contradiction. Hence we m1 have n 2. It follows that N 0 D ha2 i and Z.N / D ha2 ; b 2 i. If m < n, then n1 2 Z0 D hb i D Ãn2 .Z.N // is a G-invariant subgroup of order 2. In that case, N0 D N 0 Z0 lies in Z.G/, contrary to our assumption. Thus m n. However, if m > n C 1, then Ãn1 .Z.N // is a characteristic cyclic subgroup of N of order 4, which is a contradiction. Thus, m D n or m D n C 1. Remark 2. Let W0 be a normal subgroup of a group W and exp.W0 / D n. Let A Aut.W / stabilize the chain W > W0 f1g. Then exp.A/ divides n. Indeed, take n w 2 W and 2 A. Then w D ww0 for some w0 2 W0 so w D ww0n D w. It follows that n D idW so exp.A/ divides n, as claimed. n

n

Exercise 1. Let W D ha; b j a2 D b 2 D Œa; b D 1; n > 1i be abelian of type .2n ; 2n / and let Y be a subgroup of index 2 in W . Then A D f 2 Aut.W / j Y D Y g is of order 24n3 . In particular, if n D 2, jAj D 25 . Exercise 2. Let W be an abelian group of type .4; 4/ and A the stabilizer of the chain W > 1 .W / > f1g. Then jAj D 24 .

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Groups of prime power order

Exercise 3. Let W be abelian of type .4; 4/. Suppose that 2 Aut.W / stabilizes the chain W > 1 .W /. Then stabilizes the chain W > 1 .W / > f1g. (Hint. Use Proposition 4.9.) Proposition 50.5. The automorphism group Aut.W / of the group W D hu; y j u4 D y 4 D Œu; y D 1i Š C4 C4 is of order 25 3. The subgroup A of Aut.W / of all automorphisms fixing the subgroup Y D hu2 ; yi Š C2 C4 is of order 25 and so is a Sylow 2-subgroup of Aut.W /. We have A D h; ; j 4 D 4 D 2 D Œ 2 ; D 1; Œ; D 2 2 ; Œ; D 2 D Œ; i: where the automorphisms ; ; are induced with: u D u1 y, y D u2 y; u D uy, y D y; u D u1 , y D y. We have A0 D Z.A/ D ˆ.A/ D h 2 ; 2 D i Š E4 and so A is a special 2-group. Set W0 D 1 .W / D hu2 ; y 2 i. Then the stabilizer A0 of the chain W > W0 > 1 is elementary abelian of order 24 (Exercise 2 and Remark 50.2) and A0 D h; D 0 ; ; 2 i D CA .W0 /. The subgroup A0 contains the “special” subset S D f 2 ; ; ; ; g of five automorphisms defined by: u D u;

y D yu2 I

u D uu2 y 2 ;

y D yu2 y 2 I

u D uu2 y 2 ;

y D yu2 I

u D uy 2 ;

y D yu2 y 2 :

If X is any maximal subgroup of A0 , then X \ S is nonempty. Each 2 S has the property that does not invert any element of order 4 in W . In addition, the “superspecial” automorphisms and have also the property that they do not fix (centralize) any element of order 4 in W . We have CA .Y / D h; 2 i Š E4 is normal in A and A=h; 2 i Š D8 . In fact A is a splitting extension of h; 2 i by h; i Š D8 . Finally, if U is any subgroup of A covering A=h; 2 i (i.e. U h; 2 i D A), then U does not normalize any of the six cyclic subgroups of order 4 in W . Proof. The number of elements x, x 0 of order 4 in W such that hx; x 0 i D W (by the product formula, fx; x 0 g is a basis of W ) is 12 8 and so jAut.W /j D 25 3. By Exercise 1, jAj D 25 and therefore A is a Sylow 2-subgroup of Aut.W /. The stabilizer A0 of the chain W > W0 > 1 is elementary abelian (Remark 50.2) of order 24 (Exercise 2) and so jA W A0 j D 2. Any automorphism from A A0 acts faithfully on W =W0 (and so also on W0 ; see Exercise 3) and so we see that A0 D CA .W0 / (recall that W0 D 1 .W /). Defining the automorphisms , , , D 2 , 0 D as above, we verify the defining relations for A. Since , 0 , , 2 all lie in A0 and generate a subgroup of order 24 , we get A0 D h; 0 ; ; 2 i. For any x 2 A A0 , x 2 2 Z.A/ which gives ˆ.A/ Z.A/: We have h 2 ; 2 i ˆ.A/ Z.A/ and so the foursubgroup h 2 ; 2 i is normal in G. On the other hand, the relations for A show that A=h 2 ; 2 i is elementary abelian and h 2 ; 2 i A0 . Hence h 2 ; 2 i D ˆ.A/ D A0 . Also, h 2 ; 2 i Z.A/ < A0 and we check that no element in A0 h 2 ; 2 i lies in Z.A/. Hence Z.A/ D A0 and so A is special.

50 Janko’s theorem on 2-groups

47

Since CA .Y / A0 , so for each 2 CA .Y / we have u D uw0 with w0 2 W0 . It follows that jCA .Y /j 4. On the other hand, h; 2 i CA .Y / and so CA .Y / D h; 2 i Š E4 . Hence h; 2 i is normal in A and since A=h; 2 i acts faithfully on Y Š C4 C2 , it follows that A=h; 2 i Š Aut.C4 C2 / Š D8 . Since is an involution and D 1 , it follows that h; i Š D8 . Because h; i\h; 2 i D 1, A is a splitting extension of the four-group h; 2 i by h; i. Let U be any subgroup of A covering A=h; 2 i. Then we have U=.U \h; 2 i/ Š D8 and so there exists an element ˛ of order 4 in U . Since ˛ 2 A A0 (recall that exp.A0 / D 2), ˛ acts faithfully on W =W0 . But ˛ fixes Y D W0 hyi (since Y is Ainvariant) and so ˛ sends two cyclic subgroups of order 4 in W0 hui onto two cyclic subgroups of order 4 in W0 huyi since W0 hui is not A-invariant. It remains to be shown that U does not fix the cyclic subgroup hyi contained in Y . Suppose that U fixes hyi so that jU W CU .y/j 2. Sine c1 .Y / D 3, we get jU W CU .u2 /j 2 and so CU .Y / D CU .y/ \ CU .u2 / has index 4 in U . This is a contradiction, since jU W .U \ h; 2 i/j D 8 (recall that CA .Y / D h; 2 i). We have proved that U does not fix any cyclic subgroup of order 4 in W . Let X be any maximal subgroup of A0 . Suppose that S \ X is empty, where S D fx1 ; : : : ; x5 g is the “special” subset of automorphisms in A0 with x1 D 2 , x2 D , x3 D , x4 D and x5 D . Then we verify that the 10 products xi xj .1 i < j 5/ give 10 pairwise distinct elements in A0 . But all these products lie in X. This is a contradiction since jXj D 23 . Other statements about “special” and “extraspecial” elements in A0 are self-explanatory. Proposition 50.6. Let a 2-group G have no normal elementary abelian subgroups of order 8. Suppose that G is neither abelian nor of maximal class. The one of the following holds: (a) G has a normal abelian subgroup A of type .4; 2/ with CG .A/ abelian of type .2n ; 2/, n 2. In that case, G=CG .A/ is isomorphic to a subgroup of Aut.A/ Š D8 . If G=CG .A/ Š D8 , then CG .A/ D A. (b) G has a normal abelian subgroup W of type .4; 4/ with metacyclic CG .W /. In that case 2 .CG .W // D W and G=CG .W / is isomorphic to a subgroup of Aut.W /, i.e., jG=CG .W /j 25 (see Proposition 50.5). Proof. Let A be a greatest normal noncyclic abelian subgroup of G of exponent 4. Since G is not of maximal class, A is abelian of type .4; 2/ or .4; 4/. By Corollary 10.2, 2 .CG .A// D A. Then Theorem 41.1 implies that CG .A/ is metacyclic. Suppose that A Š C4 C2 . Then j2 .CG .A//j D 8 and Lemma 42.1 gives that CG .A/ is abelian of type .2n ; 2/, n 2. Suppose in addition that G=CG .A/ Š D8 . If CG .A/ > A, then CG .A/ contains a characteristic cyclic subgroup Z of order 4. We have Z < A and Z is normal in G. But G=CG .A/ Š Aut.A/ and Aut.A/ Š D8 contains an automorphism ˛ which permutes two cyclic subgroups of order 4 in A (Remark 50.3), a contradiction and so we must have CG .A/ D A.

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Groups of prime power order

Proof of Theorem 50.1. Let a 2-group G have no normal elementary abelian subgroups of order 8. Assume that G is neither abelian nor of maximal class. The starting point is Proposition 50.6. If we are in case (a) of Proposition 50.6, then G has a normal abelian subgroup A of type .4; 2/ with N D CG .A/ abelian of type .2j ; 2/, j 2. In that case G=N is isomorphic to a subgroup of D8 and if G=N Š D8 , then N D A and we are done. We assume that we are in case (b) of Proposition 50.6. Then G has a normal abelian subgroup W D hu; y j u4 D y 4 D Œu; y D 1i Š C4 C4 with metacyclic C D CG .W / and 2 .C / D W . Set W0 D 1 .W / D hu2 ; y 2 i so that W0 is a normal four-subgroup of G. Let Y be a normal subgroup of G such that W0 < Y < W . Then Y is abelian of type .4; 2/ and we may choose the generators u and y of W so that Y D hu2 ; yi. Also set D D CG .Y / so that G=D is isomorphic to a subgroup of Aut.Y / Š D8 . It follows from hy 2 i D Ã1 .Y / that y 2 2 Z.G/. Now, G=C is isomorphic to a subgroup of A D h; ; i which is the Sylow 2-subgroup of Aut.W / fixing Y . In what follows we use freely Proposition 50.5 about the structure of A and the action on W together with the notation introduced there. It follows that D=C is elementary abelian of order 4 since D=C induces on W a subgroup of h; 2 i Š E4 . We note that A=h; 2 i Š D8 and W0 Z.G/ if and only if G=C induces on W a subgroup of A0 in which case G=C is elementary abelian of order 24 . If C D D, then G=C is isomorphic to a subgroup of D8 . Suppose in addition that G=C Š D8 . Then W0 6 Z.G/ and G=C induces on W a subgroup U Š D8 of automorphisms which covers A=h; 2 i. By Proposition 50.5, there is no cyclic G-invariant subgroup of order 4 contained in C . Using Proposition 50.4, we see that the structure of the normal metacyclic subgroup C is completely determined, as stated in Theorem 50.1. From now on, let D=C ¤ f1g. Suppose at first that D=C contains a subgroup F=C 2 of order 2 such that F=C induces the automorphism 2 on W , where u D uy 2 , 2 2 y D y and .uy/ D uyy 2 . Since 2 does not invert any element of order 4 in W , it follows that 2 does not invert any element in C W0 . Because 2 2 Z.A/, so F is normal in G. We claim that 2 .F / D W and so, by Theorem 41.1, F is metacyclic. If not, then 2 .F / ¤ W and so 2 .F / 6 C . There exists an element s 2 F C so that o.s/ 4 which gives s 2 2 W0 . It is easy to see that there exist involutions in .W hsi/ W . Indeed, we have y s D y, us D uy 2 . If s 2 D 1, then we are finished. If s 2 D y 2 , then .sy/2 D s 2 y 2 D 1. If s 2 D u2 , then .syu/2 D s 2 .yu/s yu D u2 yuy 2 yu D 1. If s 2 D y 2 u2 , then .su/2 D s 2 us u D y 2 u2 uy 2 u D 1. Hence we may assume that s 2 F C is an involution. For any c 2 C , sc is an involution if and only if .sc/2 D 1 or c s D c 1 , i.e., s inverts c. But s inverts in C only the elements in W0 . (Actually, s centralizes W0 .) It follows that there are exactly 4 involutions in F C and they all lie in the elementary abelian subgroup E D hsi W0 . Thus E is characteristic in F and so E is normal in G, which is a contradiction. We have proved that 2 .F / D W and so F G G is metacyclic. Assume now that jD=C j D 2. If D=C induces the automorphism 2 on W , then by the above, D D F is a normal metacyclic subgroup of G with 2 .D/ D W .

50 Janko’s theorem on 2-groups

49

If jG=Dj 4, then we are done. It remains to consider the case where G=D Š D8 . In this case G=C induces on W a subgroup U of automorphisms which covers A=h; 2 i. In fact, we have in addition U \ h; 2 i D h 2 i. Again by Proposition 50.5, there is no cyclic G-invariant subgroup of order 4 contained in D. In this case, W0 6 Z.G/. By Proposition 50.4, the structure of the normal metacyclic subgroup D is uniquely determined, as stated in Theorem 50.1. It remains to consider here (where jD=C j D 2) the case that D=C induces on W the automorphism or 2 . Note that neither nor 2 is central in A and so NA .hi/ D A0 and NA .h 2 i/ D A0 . Hence G=C induces on W a subgroup of A0 which does not contain 2 . If jG=C j 4, we are done. Hence we may assume that G=C induces on W a maximal subgroup X of A0 which does not contain 2 . In that case G=C Š E8 . By Proposition 50.5, X contains a “special” element 2 f; ; ; g, where does not invert any element of order 4 in W . Let H=C be a subgroup of order 2 in G=C Š E8 so that H=C induces the automorphism D on W , where y D yu2 and u D u. If 2 .H / > W , then there is an element s 2 H C with s 2 2 W0 . By replacing s with sw for a suitable w 2 W , we may assume that s is an involution. Indeed, if s 2 D u2 , then us D u and y s D yu2 imply that .su/2 D s 2 u2 D 1. If s 2 D y 2 , then .syu/2 D s 2 .yu/s yu D y 2 yu2 u yu D 1. If s 2 D y 2 u2 , then .sy/2 D s 2 .y/s y D y 2 u2 yu2 y D 1. Then sc (with c 2 C ) is an involution if and only if s inverts c and so there are exactly 4 involutions in H C and therefore they all lie in E D hsi W0 Š E8 . Hence E is characteristic in H and so E is normal in G, which is a contradiction. Thus 2 .H / D W and so H must be metacyclic. Since G=H Š E4 , we are done. Let K=C be a subgroup of order 2 in G=C Š E8 so that K=C induces the automorphism D on W , where y D yu2 y 2 and u D uu2 y 2 . If 2 .K/ > W , then there is an element s 2 K C with s 2 2 W0 . By replacing s with sw for a suitable w 2 W , we may assume that s is an involution. If s 2 D y 2 , then .su/2 D 1. If s 2 D u2 , then .sy/2 D 1. If s 2 D y 2 u2 , then .suy/2 D 1. Then sc (with c 2 C ) is an involution if and only if s inverts c and so there are exactly 4 involutions in K C and therefore they all lie in E D hsi W0 Š E8 . Hence E is characteristic in K and so E is normal in G, which is a contradiction. Thus 2 .K/ D W and so K must be metacyclic. Since G=K Š E4 , we are done. Let L=C be a subgroup of order 2 in G=C Š E8 so that L=C induces a “superspecial” automorphism 2 f; g on W . Then has the additional property that it does not fix (centralize) any element of order 4 in W . Suppose that 2 .L/ D W so that L is metacyclic. In that case an element l 2 L C commutes with an element of order 4 in W , which contradicts the above property of . Hence there is s 2 L C so that s 2 2 W0 . Again we may assume that s is an involution. If D , then us D uy 2 , y s D yu2 y 2 . In this case, if s 2 D u2 , then .sy/2 D 1. If s 2 D y 2 u2 , then .su/2 D 1 and if s 2 D y 2 , then .suy/2 D 1. If D , then us D uu2 y 2 , y s D yu2 . In this case, if s 2 D u2 y 2 , then .sy/2 D 1. If s 2 D u2 , then .suy/2 D 1 and if s 2 D y 2 , then .su/2 D 1. It follows that L C contains exactly 4 involutions

50

Groups of prime power order

and so E D hsi W0 Š E8 (being characteristic in L ) is normal in G, which is a contradiction. This finishes completely the case jD=C j D 2. We make here the following observation. If G=C Š E16 and G=C induces on W the group A0 of automorphisms, then we get a contradiction. Indeed, we consider a subgroup L=C of order 2 in G=C so that L=C induces the automorphism on W . Then exactly the same proof as above shows that L has an elementary abelian subgroup E of order 8 which contains all involutions in L and so E would be normal in G, which is a contradiction. It remains to consider the case where D=C is a four-group inducing on W the fourgroup h; 2 i. If C < F < D and F=C induces 2 on W , then we have already proved that 2 .F / D W and F is a normal metacyclic subgroup of G. If jG=Dj 2, then jG=F j 4 and we are finished. Hence we may assume that jG=Dj 4 and so G induces on W either the whole group A (in which case G=D Š D8 ) or one of the maximal subgroups of A containing h; 2 i: A. h; 2 ; ; 0 D i D A0 , B. h; 2 ; ; i D hi h; i Š C2 D8 with D 1 , C. h; 2 ; ; i, where 2 D and Œ; D 2 . Case A is impossible, by the previous paragraph. Suppose that we are in case B. We note that h i is normal in h; 2 ; ; i. Let K0 be a subgroup of G such that F < K0 < G and K0 =C induces the cyclic group h i Š C4 on W . Hence K0 =C Š C4 and K0 is normal in G. Suppose that 2 .K0 / > W . Since 2 .F / D W , it follows that there is an element x 2 K0 F so that o.x/ 4 and hxiC D K0 . But then o.x 2 / 2 and x 2 2 F and so x 2 2 W0 C . This is a contradiction since K0 =C Š C4 . Hence 2 .K0 / D W and so K0 is a normal metacyclic subgroup of G with jG=K0 j D 4 and so we are finished in this case. Finally, we assume that we are in case C or G=D Š D8 (in which case G induces the full group A on W ). In any case, there is an element k 2 G such that k induces the automorphism on W and so uk D u1 y and y k D yu2 . Hence k does not normalize any cyclic subgroup of order 4 in W . Also, W0 6 Z.G/. We know that F is a Ginvariant nonabelian metacyclic subgroup with 2 .F / D W . By Proposition 50.4, F m n m1 is minimal nonabelian and we may set: F D ha; b j a2 D b 2 D 1; ab D a1C2 i, where n 2 and m D n or m D n C 1. If n 3, then W ha2 ; b 2 i D Z.F /. This is a contradiction since F=C induces the automorphism 2 on W . It follows that n D 2. But we have F > C W and so m D n C 1 D 3. In particular, W D C and so W is self-centralizing in G. Since CW .a/ D Y D hy; u2 i, we have a2 2 Y W0 . Replacing y with some other element of order 4 in Y (if necessary), we may assume that a2 D y. The structure of F is uniquely determined. We have F D ha; u j a8 D u4 D 1; au D aa4 i and W D hu; yi, where y D a2 and CG .W / D C D W . Set y 2 D z so that hzi D Z.G/ because CW .k/ D hy 2 i. The element k 2 induces the automorphism D 2 on W and so k 2 inverts W . Set S D F hki so that S=F Š C4 and k 4 2 C D W . Since h 2 ; i is normal in A, so S is normal in G.

51

50 Janko’s theorem on 2-groups

Since k 4 2 CW .k/ D hzi, we have k 4 D 1 or k 4 D z. Because 2 .F / D W , it follows that all elements in F W have the order 8. We shall determine the action of k on F . We have ak D aui y j , where i , j are some integers and so ui y j is an arbitrary element in W . We get .a2 /k D y k D yu2 D .ak /2 D .aui y j /2 D aui y j aui y j D a2 .ui y j /a ui y j D y.uz/i y j ui y j D yu2i y 2j z i : Thus u2 D u2i y 2j z i and so .u2 /1i D z iCj . Hence 1i 0 .mod 2/ and i Cj 0 .mod 2/. This gives that we must have i j 1 .mod 2/. Let x be an element in G inducing the automorphism on W . Then x normalizes S , x 2 2 W , y x D y and ux D u1 . Since . 2 / D 2 and D 2 , we may set ax D aum y n and k x D kaup y q , where m; n; p; q are some integers. We get .a2 /x D y x D y D .ax /2 D .aum y n /2 D aum y n aum y n D a2 .um y n /a um y n D y.uy 2 /m y n um y n D yu2m y 2mC2n : Thus u2m y 2mC2n D 1 and so m n 0 .mod 2/. Then the relation ak D x aui y j with i j 1 .mod 2/ should hold under the action of x, i.e. .ax /k D p ax .ux /i .y x /j . Since m and n are even and au D ay 2 and so au D ay 2p , this gives: p yq

.aum y n /kau

D aum y n ui y j D aumi y nCj D .aui y j .u1 y/m y n u2n /au

p yq

D .a.uy 2 /i y j um y m y n /u

p yq

D ay 2p ui y 2i y j um y m y n

D auim y 2pC2iCj CmCn: Hence umi D uim and so u2i2m D 1. Since m 0 .mod 2/, we get u2i D 1 and so i 0 .mod 2/, which contradicts the above statement. Exercise 4. Let W be a homocyclic p-group of exponent 2n and rank d and let L be a subgroup of index 2 in W . Find the order of AutL .G/ D f 2 Aut.W / j L D Lg. Exercise 5. Suppose that a 2-group G has a subgroup isomorphic to E8 . If G has no subgroups isomorphic to C2 D8 , it has an odd number of subgroups isomorphic to E8 .

51

2-groups with self centralizing subgroup isomorphic to E8

In 48, 49 the 2-groups G have been described which have the property that they possess an involution t such that CG .t / D ht i C , where C is either a non-trivial cyclic group or a generalized quaternion group. It is natural to ask what happens if C is isomorphic to one of the other two groups of maximal class, i.e., C is isomorphic to D2n (dihedral group) or to SD2n (semi-dihedral group) (Question 49.4). It is easy to see that in that case the 2-group G has a self-centralizing elementary abelian subgroup E of order 8 but the problem of classifying 2-groups G which possess such a subgroup E is far more general. In this section we classify 2-groups G which possess a self-centralizing elementary abelian subgroup E of order 8. If E is normal in G, then G=E is isomorphic to a subgroup of GL.3; 2/ and therefore G=E is isomorphic to a subgroup of the dihedral group D8 of order 8. We determine here the unique 2-group G such that E D ˆ.G/ (Theorem 51.3). There are exactly five 2-groups G, where E is normal in G and E ˆ.G/. We shall present these groups in terms of generators and relations (Theorems 51.3 and 51.4). All these groups exist because we find some faithful transitive permutation representation of degree 8 or 16 for these groups. We assume in the sequel that E is not normal in G. Then we have two essentially different possibilities according to E ˆ.G/ or E 6 ˆ.G/, where ˆ.G/ is the Frattini subgroup of G. Suppose that E ˆ.G/. Then our first non-trivial result is that G has no normal elementary abelian subgroups of order 8 (Theorem 51.5). We are now in a position to use the main theorem in 50 which allows us to describe the structure of G very accurately (Theorem 51.6). Suppose that E 6 ˆ.G/. In that case, we see easily that for each involution t 2 E ˆ.G/, we have CG .t / D ht iM0 , where M0 is isomorphic to one of the following groups: E4 , D2n .n 3/, or SD2m .m 4/. Then we try to pin down the structure of G by using the structure of CG .t /. Assume in addition that G has normal elementary abelian subgroups of order 8. In that special case we see that M0 is isomorphic to E4 or D8 (Theorem 51.7). An exceptional case is treated in Theorem 51.8. The group G could possess also a normal elementary abelian subgroup B of order 16 such that G=B Š D8 (Theorem 51.9). In all other cases, the structure of G is described

51 2-groups with self centralizing subgroup isomorphic to E8

53

in Theorem 51.10. Finally, the example produces a group of order 28 (in terms of generators and relations) which satisfies the assumptions of Theorem 51.10. It remains to consider the case E 6 ˆ.G/ and G has no normal elementary abelian subgroups of order 8. In this case Theorem 50.1 does not give a very precise information about the structure of G. Therefore, we use here the fusion arguments for involutions which have been introduced in 48,49. In this way we show that G has in most cases a large normal subgroup S with jG=S j 4 and for the structure of S we have exactly seven possibilities (Theorems 51.14 and 51.15). The exceptional cases are treated in Theorems 51.11–51.13. We use here the standard notation introduced in 48. 49. In particular, for x 2 G we denote with cclG .x/ the conjugacy class of x in G. For the sake of completeness, we give here some known results which will be used often in this book. Proposition 51.1. Let be an involutory automorphism acting on an elementary abelian 2-group A. Then we have jCA . /j jA W CA . /j. Proof. Consider the Jordan normal form of the linear transformation induced by on A (considered as a vector space over GF.2/).Then the result follows. Proposition 51.2. Let be an involutory automorphism acting on an abelian 2-group B so that CB . / D W0 1 .B/. Then inverts Ã1 .B/ and B=W0 . 2

Proof. If x 2 B; then .xx / D x x D x x D xx . Hence, xx D w0 with w0 2 W0 and so x D x 1 w0 . This gives .x 2 / D .x /2 D .x 1 w0 /2 D x 2 w0 2 D x 2 and our result follows. 1o . The case E ˆ.G/. In this subsection we suppose that a 2-group G contains a self-centralizing elementary abelian subgroup E of order 8 which is contained in ˆ.G/. Obviously, Z.G/ E and also Z.ˆ.G// E. If jZ.ˆ.G//j D 2 then, by Proposition 1.13, we get that ˆ.G/ is cyclic, which is not the case. It follows that 4 jZ.ˆ.G//j 8. We consider at first the possibility jZ.ˆ.G//j D 8 which gives that Z.ˆ.G// D E and since CG .E/ D E, it follows that E D ˆ.G/ is a normal self-centralizing elementary abelian subgroup of order 8 of G. On the other hand, G=E is an elementary abelian subgroup of D8 and so G=E is a four-group. Thus jGj D 25 . For each x 2 G E, we must have jCE .x/j D 4 (since x acts faithfully on E and hx; Ei is not of maximal class according to Theorem 1.7, Propositions 1.8 and 51.1), CG .x/ D hxiCE .x/ with x 2 2 CE .x/. Otherwise, CG .x/ would cover G=E, which is not possible since E D ˆ.G/. It follows that for each x 2 G E, jcclG .x/j D 4 and so jG 0 j 4. If G 0 D E, then by the Taussky’s theorem (Proposition 1.6), G is of maximal class, which is not the case. Thus jG 0 j D 4. Suppose that jZ.G/j D 4. Then for each x 2 G E we have CG .x/ D hxiZ.G/ which gives x 2 2 Z.G/. But then E D ˆ.G/ D Ã1 .G/ Z.G/, a contradiction. Hence Z.G/ D hzi is of

54

Groups of prime power order

order 2 and Z.G/ < G 0 . Since Z.G/ < G 0 < E, we have ŒG; G 0 D Z.G/, so G is a 2-group of class 3. The group G=G 0 is abelian of type .4; 2/ and so there is an element x 2 G E so that x 2 D v 2 E G 0 . Since x centralizes z and v D x 2 , and E D hv; G 0 i, it follows that x does not centralize G 0 (otherwise x would centralize E). Let y 2 G .Ehxi/ which centralizes G 0 so that y 2 2 G 0 and G D hx; yi. Note that jG W CG .G 0 /j D 2. If Œx; y 2 hzi, then G=hzi is abelian, a contradiction. Hence we may put Œx; y D u, where G 0 D hz; ui. From Œx; y D u we get x y D xu and so v y D .x 2 /y D .x y /2 D .xu/2 D xuxu D x 2 .x 1 ux/u D v.uz/u D vz. Hence we get z y D z, uy D u, v y D vz, x y D xu, x 2 D v, z x D z, ux D uz, v x D v. The relation Œx; y D u gives also y x D yu D uy since y centralizes u 2 G 0 . From this we get .y 2 /x D .y x /2 D .uy/2 D y 2 . Hence x centralizes y 2 2 G 0 . But x acts faithfully on G 0 and so we must have y 2 2 hzi. Suppose now that y 2 D z. Then we replace y with y 0 D yv. We get .y 0 /2 D .yv/2 D yvyv D y 2 v y v D z.vz/v D 1 and other 0 0 0 0 relations remain unchanged z y D z, uy D u, v y D vz, x y D x yv D .xu/v D xu, 2 and so we may assume from the start that y D 1. The structure of G is uniquely determined. It is easy to see that this group G is isomorphic to a subgroup of A8 . It is enough to set x D .2; 7; 6; 8/.4; 5/, y D .1; 2/.3; 6/.4; 7/.5; 8/, and we compute further permutations x 2 D v D .2; 6/.7; 8/, Œv; y D z D .1; 3/.2; 6/.4; 5/.7; 8/, Œx; y D u D .1; 4/.2; 7/.3; 5/.6; 8/. Then we simply check that all above relations are satisfied for these permutations. Since z is represented with a non-trivial permutation, so the induced permutation representation of degree 8 is faithful. Note that all above permutations are even. Hence our group G exists and is isomorphic to a subgroup of A8 . We have proved the following result. Theorem 51.3. Suppose that a 2-group G contains a self-centralizing elementary abelian subgroup E of order 8 and E D ˆ.G/. Then G is uniquely determined and we have G D hx; y; z; u; v j x 4 D y 2 D z 2 D u2 D v 2 D Œz; u D Œz; v D Œz; x D Œz; y D Œu; v D Œu; y D Œv; x D 1; x 2 D v; v y D vz; x y D xu; ux D uzi. Here E D hz; u; vi is a self-centralizing normal elementary abelian subgroup of order 8 of G, E D ˆ.G/, G D hx; yi is of order 25 , G 0 D hz; ui, and Z.G/ D ŒG; G 0 D hzi is of order 2. The group G exists and is isomorphic to a subgroup of A8 . It remains to consider the second possibility jZ.ˆ.G//j D 4. We set Z.ˆ.G// D W0 so that W0 is a G-invariant four-group contained in E. Therefore the subgroup T D CG .W0 / is of index 2 in G. Since CG .E/ D E, so for each t 2 E W0 we have CT .t / D E. Assume that T D G, in which case W0 D Z.G/. Let A be a G-invariant subgroup so that W0 < A ˆ.G/ and jA W W0 j D 2. Then A is abelian of order 8 and G stabilizes the chain A > W0 > 1 so that G=CG .A/ is elementary abelian (of order 4/. In particular, ˆ.G/ CG .A/ and so A Z.ˆ.G//, which is a contradiction. It follows that jG W T j D 2 and so Z.G/ D hzi is of order 2 and we set W0 D hz; ui. For each x 2 G T , ux D uz. Since jˆ.G/j 24 , so jGj 26 . Assume now in addition that E is normal in G. Since G=E is isomorphic to a subgroup of D8 and jGj 26 , so jGj D 26 and G=E Š D8 . The subgroup T

51 2-groups with self centralizing subgroup isomorphic to E8

55

stabilizes the chain E > W0 > 1 and jG W T j D 2, so T =E is a 4-group. We have Z.T / D W0 . For each x 2 T E, x 2 2 W0 since x acts non-trivially on E. Thus ˆ.T / D Ã1 .T / W0 and so T 0 W0 . On the other hand, if t 2 E W0 , then CT .t / D E implies that jcclT .t /j D 4 and so cclT .t / D E W0 and jT 0 j 4. It follows that Z.T / D T 0 D ˆ.T / D W0 and therefore T is a special 2-group of order 25 . Let x; y 2 T E be such that Ex and Ey are two distinct elements in T =E Š E4 . Then M D W0 hx; yi is a maximal subgroup of T not containing E. If M is nonabelian, then Œx; y D v is an involution in W0 . Let t 2 E W0 . We know that CT .t / D E and so Œx; t , Œy; t , and Œxy; t D Œx; t Œy; t are three distinct involutions in W0 . Note that T is of class 2. If, for example, Œx; t D Œy; t , then Œxy; t D Œx; t Œy; t D 1, a contradiction. Therefore, we may assume that Œx; t D v. It follows Œx; ty D Œx; t Œx; y D v 2 D 1. Hence, replacing y with ty 2 Ey, we may assume from the start that Œx; y D 1 and so M D W0 hx; yi is abelian. We claim that M is the unique abelian maximal subgroup of T . Indeed, if M0 ¤ M is another abelian maximal subgroup of T , then M \ M0 Z.T / and jM \ M0 j D 23 , a contradiction. Since T D CG .W0 / and Z.ˆ.G// D W0 , so T is a characteristic subgroup of G. Hence M is also a characteristic subgroup of G. Since G=E Š D8 and D8 has five involutions, there is an element s 2 G T such that s 2 2 E. Note that s normalizes M and so if s 2 2 W0 M , then M hsi is a maximal subgroup of G which does not contain E, a contradiction. Thus, we must have s 2 D t 2 E W0 . Since G=E Š D8 , s acts non-trivially on M=W0 Š T =E Š E4 . Hence, there is an element x 2 M W0 such that y D x s has the property M D W0 hx; yi. Also, x 1 y D x 1 x s D Œx; s 2 ˆ.G/ and so x 2 .x 1 y/ D xy 2 ˆ.G/. Hence ˆ.G/ D 2 Ehxyi and G D hs; xi. Now, y s D x s D x t D xu, where u 2 W0 hzi. Indeed, if x t D txt D xz, then we act on this relation with s. We get t s x s t s D x s z s which gives tyt D yz and then .xy/t D xzyz D xy. But this contradicts the fact that CT .t / D E and so we must have y s D xu, with u 2 W0 hzi. Also note that us D uz since s acts non-trivially on W0 . It remains to determine x 2 and y 2 . There are exactly three possibilities for the structure of the abelian group M . (a) 1 .M / D M0 is of order 23 . Since M0 is normal in G, so x and y D x s are elements of order 4 contained in M M0 . On the other hand, jÃ1 .M /j D 2 and so Ã1 .M / D hzi D Z.G/. This gives x 2 D y 2 D z. If we set s D .2; 5; 8; 12/.4; 6; 10; 11/.7; 9/.13; 15; 14; 16/; x D .1; 2; 3; 4/.5; 13; 6; 14/.7; 8; 9; 10/.11; 15; 12; 16/; then we see (in the same way as in the proof of Theorem 51.3) that G exists in this case and is isomorphic to a subgroup of A16 . (b) 1 .M / D M is of order 24 . Here we have x 2 D y 2 D 1. If we set s D .1; 4; 3; 5/.2; 7; 6; 8/, x D .4; 7/, then we see easily that G exists in this case as a subgroup of S8 . (c) 1 .M / D W0 is of order 22 and so M Š C4 C4 . Here we have Ã1 .M / D W0 D hx 2 ; y 2 i and hx; x s i D M so that we have two possibilities for x 2 and y 2 .

56

Groups of prime power order

(c1) If x 2 D u, then y 2 D uz and so y s D xu D x 1 . This gives y t D y s D 2 .x 1 /s D y 1 and x t D x s D y s D x 1 and so t inverts M . If we set s D .1; 5/.2; 8; 4; 7/.3; 6/, x D .5; 7; 6; 8/, then we see easily that G exists in this case as a subgroup of S8 . (c2) If x 2 D uz, then y 2 D u and then y s D xu D .xuz/z D x 1 z. This gives 2 2 y t D y s D .x 1 z/s D y 1 z and x t D x s D y s D x 1 z and so t does not inverts M . If we set 2

s D .2; 5; 12; 10/.3; 6/.4; 7; 14; 15/.9; 13; 16; 11/; x D .1; 2; 3; 4/.5; 9; 10; 11/.6; 12; 8; 14/.7; 13; 15; 16/; then we see as before that G exists in this case as a subgroup of A16 . We have proved the following result. Theorem 51.4. Suppose that a 2-group G contains a self-centralizing elementary abelian subgroup E of order 8, E < ˆ.G/, and E is normal in G. Then we have G D hx; y; u; z; t; si with the following initial relations u2 D z 2 D t 2 D s 4 D Œx; y D Œx; u D Œx; z D Œy; u D Œy; z D Œu; z D Œu; t D Œz; t D Œz; s D 1;

s 2 D t; us D uz; x s D y; y s D xu:

Here E D ht; z; ui is a self-centralizing elementary abelian normal subgroup of order 8 in G, G=E Š D8 , Z.G/ D hzi is of order 2, G is of order 26 , ˆ.G/ D Ehxyi and G D hs; xi. Also, Z.ˆ.G// D W0 D hz; ui is a four-group, M D hx; y; z; ui is a characteristic abelian subgroup of G of order 24 and exponent at most 4 and M is the unique abelian maximal subgroup of T D CG .W0 /, jG W T j D 2 and T =E Š E4 . There are exactly four possibilities for the structure of G (with two additional relations). (a) x 2 D y 2 D z, M is abelian of type .4; 2; 2/, and the group G is in this case isomorphic to a subgroup of A16 . (b) x 2 D y 2 D 1, M is abelian of type .2; 2; 2; 2/, and the group G is in this case isomorphic to a subgroup of S8 . (c1) x 2 D u, y 2 D uz, M Š C4 C4 , t inverts M , and the group G is in this case isomorphic to a subgroup of S8 . (c2) x 2 D uz, y 2 D u, M Š C4 C4 , t does not invert on M , and the group G is in this case isomorphic to a subgroup of A16 . In what follows, we assume that E is not normal in G. We recall that W0 D Z.ˆ.G// is a four-group and jG W T j D 2, where T D CG .W0 /. We set W0 D hz; ui and we know that Z.G/ D hzi is of order 2. Finally, for each involution t 2 E W0 , CT .t / D E and jGj 26 since jˆ.G/j 24 . Our aim is to prove that G does not possess in this case a normal elementary abelian subgroup of order 8.

51 2-groups with self centralizing subgroup isomorphic to E8

57

Suppose that A1 is a normal elementary abelian subgroup of order 16 in T . Let t 2 E W0 . No involution in E W0 could be contained in A1 and, by Proposition 51.1, jCA1 .t /j 22 . Hence, jCA1 .t /j D 4 and CA1 .t / D W0 so that E \ A1 D W0 . Set C D EA1 and so jC j D 25 with jC W A1 j D 2. Take any a 2 A1 . Then t a is an involution if and only if .t a/2 D 1 or equivalently at D a1 D a. It follows that all involutions in C A1 lie in E W0 . Since hE W0 i D E, we see that E is normal in NG .C /. If C D T , then E would be normal in G, a contradiction. Let D=C be a subgroup of order 2 in NT .C /=C . Then E is normal in D and note that all involutions in E W0 lie in a single conjugacy class in C since CG .t / D E and jC W Ej D 4. But then CD .t / 6 C , a contradiction. We have proved that T has no normal elementary abelian subgroups of order 16. Let A2 be a normal elementary abelian subgroup of order 16 in G. By the previous paragraph, A2 6 T and so B2 D A2 \ T is a normal elementary abelian subgroup of order 8 in G. If W0 6 B2 , then W0 B2 is a normal elementary abelian subgroup of order 16 contained in T , a contradiction. Hence we must have W0 B2 . But then CG .W0 / hT; A2 i D G, which is a contradiction. We have proved that G has no normal elementary abelian subgroups of order 16. Suppose that A0 is a normal elementary abelian subgroup of order 8 in G but A0 6 T . Then AQ0 D A0 \ T is a G-invariant 4-subgroup. If W0 D AQ0 , then CG .W0 / hA0 ; T i D G, a contradiction. Hence W0 ¤ AQ0 and so AQ0 W0 is a Ginvariant elementary abelian subgroup of order 8 contained in T . We have proved that if G has a normal subgroup isomorphic to E8 , then G has also such one which is contained in T . Now assume that G has a normal subgroup A Š E8 . By the previous paragraph, we may assume that A T . Since E is not normal in G, so we get E ¤ A. If W0 6 A, then W0 A would be a G-invariant elementary abelian subgroup of order 16, a contradiction. Hence W0 A and so E \ A D W0 . It is well known that there is a self-centralizing normal abelian subgroup B of T such that A B and B is normal in G. We have 1 .B/ D A and E \ B D E \ A D W0 . Since G=T acts faithfully on W0 (and so on B), B is also self-centralizing in G. For any x 2 B, tx is an involution if and only if t inverts x. Suppose that B D A. In this special case, A is a self-centralizing normal elementary abelian subgroup of order 8 of G and so G=A Š D8 since jGj 26 . Set V0 D EA and note that V0 A contains exactly four involutions and they are contained in E W0 . Hence E is normal in NG .V0 / and so V0 =A does not lie in Z.G=A/ since E is not normal in G. Let R=A be the cyclic subgroup of index 2 in G=A so that we have R \ V0 D A and R is a maximal subgroup of G. Hence E 6 R and this is a contradiction since E ˆ.G/. We have proved that B ¤ A and so jBj 24 . We shall determine now the structure of B. If an involution e 2 A W0 would be a square in B, then t would centralize e, since t inverts on Ã1 .B/ (Proposition 51.2). This is a contradiction. Suppose that all three involutions in W0 are squares in B. Then 2 .B/ D hb1 i hb2 i heiq, where o.b1 / D o.b2 / D 4, e is an involution

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Groups of prime power order

in A W0 , and W0 D hb12 ; b22 i. We act with t 2 E W0 on 2 .B/. By Proposition 51.2, b1t D b11 w with w 2 W0 . This gives b1t D b1 .b12 w/, where b12 w D w1 2 W0 . Similarly, b2t D b2 w2 with w2 2 W0 and so .b1 b2 /t D .b1 b2 /.w1 w2 /. Here w1 ; w2 , and w3 D w1 w2 must be three pairwise distinct involutions in W0 since C2 .B/ .t / D hb12 ; b22 i D W0 . But then e t D ewj with j 2 f1; 2; 3g and so t centralizes one of the elements b1 e, b2 e, or b1 b2 e of order 4, which is a contradiction. We have proved that at most one involution in W0 is a square in B. Hence, we have obtained the following result. The subgroup B .> A/ is abelian of type .2m ; 2; 2/, m 2, and so m1 B D hbi hwi hei, where o.b/ D 2m 4, b 2 D z, hzi D Z.G/ D Ãm1 .B/, W0 D hz; wi, e 2 A W0 . Set V D EB D ht iB, where t is an involution in E W0 D E B. Our aim is to show that V D T . Therefore, we assume V ¤ T and set VQ D NT .V / so that jVQ =V j 2. We note that jBj D 2mC2 .m 2/ and so V B is a normal subset of VQ and jV Bj D 2mC2 D jtBj. If jVQ =V j 4, then jVQ W CVQ .t /j D jVQ W Ej 2mC2 and so jVQ =V j D 4 and all elements in V B are involutions since they are all conjugate in VQ to t . This is not the case since t e is not an involution in view of Œt; e ¤ 1. Hence we must have jVQ =V j D 2. Assume for a moment that VQ D NG .V /. By Theorem 1.7 and Proposition 1.8, G=B is of maximal class and V =B is a non-central subgroup of order 2 in G=B. In fact G=B must be dihedral of order 8 or semidihedral. Let R =B be the cyclic subgroup of index 2 in G=B. Then jG W R j D 2 and R \ V D B. But then E 6 R and so E 6 ˆ.G/, a contradiction. Hence we must have NG .V / > VQ . We set V D NG .V / so that jV W VQ j D 2 and V T D G. If CV .t / D E, then we see (since jV W Ej D 2mC2 ) that all 2mC2 elements in the normal subset V B in V are conjugate to t and so they are all involutions. But this is not the case, since t e (as above) is not an involution. Thus CV .t / D EQ is of order 24 , jEQ W Ej D 2 and EQ \ T D E. Hence all elements y 2 EQ E lie in G T , y 2 2 E and w y D wz, where W0 D hz; wi. Suppose at first that there is y 2 EQ E with y 2 2 W0 . In this case W0 hyi Š D8 and since there are involutions in .W0 hyi/ W0 , we may assume that y 2 D 1. We act with the involution y on A. Since CW0 .y/ D hzi and jCA .y/j 4 (Proposition 51.1), we see that there is an element e 2 A W0 so that Œy; e D 1. We have 1 ¤ Œt; e 2 W0 and since m2 Œt; ey D Œt y ; e y D Œt; e, so Œt; e D z which gives e t D ez. Set v D b 2 so that hvi is the cyclic subgroup of order 4 in hbi. Since e t D ez, we must have v t D vw for an element w 2 W0 hzi. Otherwise, t would centralize ve, which is not possible. In particular, t does not invert v, and so v 62 Ã1 .B/. Recall that, by Proposition 51.2, t inverts Ã1 .B/. But this forces that o.b/ D 4; hvi D hbi; jBj D 24 , and E is normal in V . The last statement follows from the fact that there are only four involutions in V B (since t inverts in B only the involutions in W0 ) and they lie in E W0 . Suppose now that for each y 2 EQ E, y 2 2 E W0 and so we may assume in this case that y 2 D t 2 E W0 . The cyclic group hyi of order 4 acts faithfully on A and hyi acts fixed-point-free on the set A W0 . If e y D ez with e 2 A W0 , 2 then e y D .ez/y D ezz D e and so e t D e, a contradiction. Hence, e y D ew

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2

with w 2 W0 hzi. This gives e y D e t D .ew/y D ewwz D ez. It follows that v t D vw1 with some w1 2 W0 hzi, where again hvi is the cyclic subgroup of order 4 in hbi. Otherwise, t would centralize ve. But this means that t does not invert hvi. As before, this forces that o.b/ D 4; hbi D hvi, jBj D 24 , and E is normal in V . Again, the last statement follows from the fact that there are only four involutions in V B (since t inverts in B only the involutions in W0 ) and these lie in E W0 . In both cases for the structure of EQ D hyiE, we have jBj D 24 , jVQ j D 26 , and VQ normalizes the subset V B containing exactly four involutions. On the other hand, jVQ W CVQ .t /j D jVQ W Ej D 23 which means that V B contains eight conjugates of t , which is the final contradiction in this paragraph. We have proved that we must have V=T and so T D ht iB, where t 2 E W0 . m2 so that o.v/ D 4 and v 2 D z. Since t 2 ˆ.G/ and ˆ.G/ D Set again v D b 2 Ã1 .G/, there is an element x 2 G T such that x 2 D t 0 2 T B. Because t 0 D lt 0 with l 2 B, we have for all y 2 B, y t D y lt D y t and .t 0 /2 2 B. Hence x induces an automorphism of order 4 on 2 .B/ D Ahvi since x 2 induces the same involutory automorphism on 2 .B/ as the automorphism induced by t . Since t acts fixed-point-free on A W0 , so x induces a 4-cycle on A W0 and so e x D ew, where 0 2 w 2 W0 hzi. Indeed, if e x D ez, then e t D e t D e x D .ez/x D ezz D e, which is 2 0 a contradiction. From e x D ew follows e x D e t D e t D .ew/x D .ew/.wz/ D ez. But then we must have v t D vw 0 with w 0 2 W0 hzi. Namely, if v t D vz D v 1 , then .ve/t D .vz/.ez/ D ve, a contradiction. It follows from v t D vw 0 that t does not invert v. By Proposition 51.2, t inverts Ã1 .B/. Hence v is not a square in B and so 2 .B/ D B is of order 24 . Since t inverts in B only the elements in W0 , so there are exactly four involutions in tB and they lie in E W0 . It follows that E x D E and so E is normal in G. This is the final contradiction. We have proved the following major result. Theorem 51.5. Suppose that a 2-group G contains a self-centralizing elementary abelian subgroup E of order 8, E ˆ.G/, and E is not normal in G. Then G has no normal elementary abelian subgroups of order 8. In the rest of this subsection, we study the structure of a 2-group G satisfying the assumptions of Theorem 51.5. We recall that W0 D Z.ˆ.G// is a normal four-subgroup contained in E and jG W T j D 2, where T D CG .W0 /. We set W0 D hz; ui and we know that Z.G/ D hzi is of order 2. Finally, for each involution t 2 E W0 , CT .t / D E and jGj 26 . Since G has no normal elementary abelian subgroups of order 8, we may apply Theorems 50.1 and 50.2. Suppose that G has two distinct normal 4-subgroups. Then Theorem 50.2 implies that G D D C , where D Š D8 , D \ C D Z.D/ and C is either cyclic or of maximal class but not isomorphic to D8 . We get ˆ.G/ C and so E C , which is not possible. We have proved that W0 is the unique normal four-subgroup of G.

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Since G is neither abelian nor of maximal class, we may apply Proposition 50.1. It follows that G has a normal metacyclic subgroup N such that CG .2 .N // N , G=N is isomorphic to a subgroup of D8 and W D 2 .N / is abelian of type .4; 2/ or .4; 4/. In any case, W0 D 1 .N / is the unique normal four-subgroup of G and W0 6 Z.G/. It follows that E \ N D W0 . Since E ˆ.G/, G=N is not elementary abelian. Hence G=N is either cyclic of order 4 or G=N Š D8 . Assume that G=N Š C4 . If N is abelian of type .2j ; 2/, j 2, then G=N acts faithfully on 2 .N / Š C4 C2 . If, in addition, N > 2 .N /, then there is a characteristic cyclic subgroup Z of order 4 contained in 2 .N / so that Z is normal in G. But then acting with G=N on Z, we see that t 2 E W0 centralizes Z T D CG .W0 /, which is a contradiction. Thus, N D 2 .N / and so jGj D 25 , which is again a contradiction. It follows that 2 .N / D W is abelian of type .4; 4/. We want to show that N does not possess a G-invariant cyclic subgroup Z of order 4. Suppose that there is such Z so that Z W T . But jG W CG .Z/j 2 and so E centralizes Z since E ˆ.G/, a contradiction. It follows that we may use Proposition 50.4. Hence in case G=N Š C4 , N is either abelian of type .2n ; 2n / or .2nC1 ; 2n / with n 2 or N is minimal nonabelian and m n m1 i, where m D n with more precisely N D ha; b j a2 D b 2 D 1; ab D a1C2 n 3 or m D n C 1 with n 2. Assume that G=N Š D8 . Then the structure of N is determined by Theorem 50.1. We shall show here that the minimal case with N Š C4 C2 cannot occur. Indeed, suppose that N is abelian of type .4; 2/. Let L=N D Z.G=N / so that jL=N j D 2 and E L since E ˆ.G/. Set W0 D 1 .N / so that W0 D Z.ˆ.G//, W0 < E and since E < ˆ.G/, we get ˆ.G/ D L. By the structure of G=N Š D8 Š Aut.N /, we know that t 2 E W0 inverts N . Thus, all elements in L N are involutions. Set N D hy; w j y 4 D w 2 D Œy; w D 1; y 2 D zi so that W0 D hz; wi, E D ht; z; wi with hzi D Z.G/. Let k be any element in G such that hkiN=N Š C4 . Set K D hkiN . It follows that k 2 2 L N and since k 2 is an involution, so k 4 D 1. By the structure of Aut.N / Š D8 , we have y k D yw, w k D wz. Then we see that Ã1 .K/ D ˆ.K/ D hk 2 ; w; zi. Hence the elementary abelian subgroup hk 2 ; w; zi of order 8 is normal in G. But then the other elementary abelian subgroup hk 2 y; w; zi of order 8 in K must be also normal in G. Note that K has exactly two elementary abelian subgroups of order 8 since there are exactly eight involutions in L N . In particular, E is normal in G, a contradiction. This proves that the case G=N Š D8 with N Š C4 C2 cannot occur. We have proved the following final result of this subsection. Theorem 51.6. Let G be a 2-group which has a self-centralizing elementary abelian subgroup E of order 8. Assume that E ˆ.G/ and E is not normal in G. Then the group G has the following properties. (a) G has no normal elementary abelian subgroups of order 8. (b) G has the unique normal four-subgroup W0 and W0 < E.

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(c) G has a normal metacyclic subgroup N such that 2 .N / D W is abelian of type .4; 4/, CG .W / N , 1 .W / D W0 , and G=N is either cyclic of order 4 or G=N Š D8 . (d) N is either abelian of type .2k ; 2kC1 / or .2k ; 2k /, k 2, or N is minimal m n m1 nonabelian and more precisely N D ha; b j a2 D b 2 D 1; ab D a1C2 i, where m D n with n 3 or m D n C 1 with n 2. 2o . The case E 6 ˆ.G/ and G has a normal elementary abelian subgroup of order 8. We suppose throughout this subsection that our 2-group G contains a self-centralizing non-normal elementary abelian subgroup E of order 8 such that E 6 ˆ.G/. (If E is normal in G, then G=E is isomorphic to a subgroup of D8 .) Also, we assume that G has a normal elementary abelian subgroup of order 8. Let t 2 E ˆ.G/ and let M be a maximal subgroup of G such that t 62 M . Then CG .t / D ht i CM .t /. But M0 D CM .t / contains the four-subgroup E0 D E \ M and CM0 .E0 / D E0 . If E0 ¤ M0 , then Theorem 1.7 and Proposition 1.8 imply that M0 Š D2n or M0 Š SD2n . Let A be a normal elementary abelian subgroup of order 8 in G. If t 2 A, then A is a normal subgroup of CG .t /. If t 62 A, then Proposition 51.1 implies that CA .t / D A0 is of order 4 and so ht i A0 is a normal elementary abelian subgroup of order 8 of CG .t /. In any case, M0 has a normal 4-subgroup and so E0 ¤ M0 implies that M0 Š D8 . We have proved the following result. Theorem 51.7. Let G be a 2-group which has a self-centralizing elementary abelian subgroup E of order 8 which is not contained in ˆ.G/. Assume that G has a normal elementary abelian subgroup of order 8. Then for each t 2 E ˆ.G/, we have either CG .t / D E or CG .t / D ht i M0 , where M0 Š D8 . Suppose that there is an involution t 2 E ˆ.G/ such that CG .t / contains an elementary abelian G-invariant subgroup A of order 8. Since E is not normal in G, so E ¤ A and therefore C D CG .t / D hE; Ai D ht i D, where D Š D8 . The subgroup C has exactly two elementary abelian subgroups of order 8 and so they are A and E. Also, t 2 A \ E. Since CC .A/ D A, so A is self-centralizing in G and therefore G=A is isomorphic to a subgroup of D8 . Obviously, E is normal in NG .C /. But E is not normal in G and so NG .C / ¤ G. This forces that jG W C j 4 and therefore G=A Š D8 . We have proved the following result which describes an exceptional case. Theorem 51.8. Let G be a 2-group which has a self-centralizing non-normal elementary abelian subgroup E of order 8 which is not contained in ˆ.G/. Suppose that there is an involution t 2 E ˆ.G/ such that CG .t / contains an elementary abelian G-invariant subgroup A of order 8. Then A is self-centralizing in G and G=A Š D8 . We assume in the rest of this subsection that there is no involution t 2 E ˆ.G/ such that CG .t / contains an elementary abelian G-invariant subgroup of order 8.

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Suppose that G has a normal elementary abelian subgroup B of order 16. Let t be an involution in E ˆ.G/. By the structure of CG .t /, we have jCB .t /j 4. Then Proposition 51.1 forces that jBj D 16 and CB .t / D W0 is of order 4. Let C be a maximal normal abelian subgroup of G containing B. Then B D 1 .C / and, by the structure of CG .t /, we have CC .t / D CB .t / D W0 . Also, G=C acts faithfully on C . Set W0 D hz1 ; z2 i and let W1 D hu1 ; u2 i be a complement of W0 in B. Set z3 D z1 z2 and u3 D u1 u2 . We claim that we may choose the generators of W0 and W1 in such a way that uti D ui zi for i D 1; 2; 3. Indeed, suppose that there exist two elements u ¤ u0 2 W1 so that ut D uz and .u0 /t D u0 z, where 1 ¤ z 2 W0 . Then uu0 ¤ 1 and .uu0 /t D .uz/.u0 z/ D uu0 , which contradicts the fact that CC .t / D W0 . Hence ut D uz and .u0 /t D u0 z 0 , where z; z 0 are distinct involutions in W0 . This gives .uu0 /t D .uu0 /.zz 0 /, where uu0 ¤ 1 and zz 0 ¤ 1. It remains to set u1 D u, u2 D u0 , u3 D uu0 , z1 D z, z2 D z 0 , z3 D zz 0 , and we see that the above claim is proved. Suppose that C ¤ B. Let v be an element of order 4 contained in C B. Then v 2 2 B and, by Proposition 51.2, .v 2 /t D v 2 D v 2 and so v 2 2 W0 . By Proposition 51.2, v t D v 1 w0 with w0 2 W0 and so v t D v.v 2 w0 / D vzj , where j 2 f1; 2; 3g. But then .vuj /t D vzj uj zj D vuj , which is a contradiction. We have proved that C D B is a self-centralizing normal elementary abelian subgroup of order 16 of G. Set E1 D ht i W0 and L D ht iB. Then all elements in L .E1 [ B/ have order 4 and four involutions in E1 W0 form a single conjugacy class in L since jL W CL .t /j D 4. It follows L0 D Z.L/ D ˆ.L/ D W0 . If L D G, then CG .t / D E1 D E is normal in G. This is a contradiction and so L ¤ G. Set K D NG .L/. Then K normalizes the L-class E1 W0 . Hence CK .t / must cover K=L. Since N D CK .t / D CG .t / Š C2 D8 , so jK=Lj D 2 and N \ L D E1 . We have NG .E1 / hN; Li D K and so K ¤ G by our last assumption. Indeed, if K D G, then N D CG .t / would contain the G-invariant subgroup E1 . Now, Theorem 1.7 and Proposition 1.8 imply that G=B is of maximal class (acting faithfully on B). Also, L=B is a non-central subgroup of order 2 in G=B and so G=B is not a generalized quaternion group. On the other hand, GL.4; 2/ does not possess elements of order 8 and so jG=Bj D 8 and consequently G=B Š D8 . We have Z.G/ CG .t / D N and so Z.G/ Z.N / D ht; z 0 i, where hz 0 i D N 0 < W0 . Hence Z.G/ Z.L/ D W0 and so Z.G/ D ht; z 0 i \ W0 D hz 0 i is of order 2. We have proved the following result. Theorem 51.9. Let G be a 2-group which has a self-centralizing non-normal elementary abelian subgroup E of order 8 which is not contained in ˆ.G/. Assume that there is no involution t 2 E ˆ.G/ such that CG .t / contains an elementary abelian G-invariant subgroup of order 8. Suppose that G has a normal elementary abelian subgroup B of order 16. Then jBj D 16, B is self-centralizing in G, G=B Š D8 , and Z.G/ is of order 2. We suppose also in the rest of this subsection that G has no normal elementary abelian subgroups of order 16.

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Let A be a normal elementary abelian subgroup of order 8 in G. Let B be a maximal normal abelian subgroup of G containing A. Then G=B acts faithfully on B and we have 1 .B/ D A. Let t be any involution in E ˆ.G/. By our assumptions, CG .t / does not contain a G-invariant elementary abelian subgroup of order 8. Hence t 62 B. By Theorem 51.7, we have either CG .t / D E or CG .t / D ht i D, where D Š D8 . By Proposition 51.1, jCA .t /j 4 and so CB .t / D CA .t / D W0 is a four-group. Assume that B D A. Then G=A is isomorphic to a subgroup of D8 . Set ht iA D L and ht i W0 D E1 . All elements in L .E1 [ A/ are of order 4 and so E1 is normal in NG .L/ since hE1 W0 i D E1 . If L is normal in G, then E1 is normal in G. But then CG .t / contains a G-invariant elementary abelian subgroup of order 8, which contradicts our assumptions. It follows that L=A is a non-central subgroup in G=A and so G=A Š D8 . In what follows we shall assume that B ¤ A. By Proposition 51.2, no element in A W0 is a square in B. Assume that all three involutions in W0 are squares in B. Then 2 .B/ D hb1 i hb2 i hei, where o.b1 / D o.b2 / D 4, e 2 A W0 , and W0 D hb12 ; b22 i. By Proposition 51.2, b1t D .b1 /1 w0 with w0 2 W0 and so b1t D b1 .b12 w0 /, where 1 ¤ w1 D b12 w0 2 W0 . Similarly, b2t D b2 w2 with 1 ¤ w2 2 W0 and so .b1 b2 /t D .b1 b2 /.w1 w2 /. Since C2 .B/ .t / D W0 , so w1 ; w2 ; w3 D w1 w2 must be pairwise distinct involutions in W0 . But then e t D ewj with j 2 f1; 2; 3g and so t centralizes one of the elements b1 e; b2 e, or b1 b2 e (all of order 4), which is a contradiction. It follows that exactly one involution in W0 is a square in B. We have proved that B D hbi hwi hei, where m1 D z, hzi Z.G/, W0 D hz; wi, and e 2 A W0 . o.b/ D 2m , m 2, b 2 Set V D ht iB and suppose that CG .t / D ht i D, where D Š D8 . Then W0 is normal in CG .t /, Z.CG .t // D ht; zi and .CG .t //0 D hzi. Hence CG .t / contains an involution u 2 G V so that w u D wz and so W0 hui D D Š D8 . Acting with u on A, we see that jCA .u/j 4 (Proposition 51.1). Since CW0 .u/ D hzi, so u centralizes an involution in A W0 . We may assume e u D e and so CA .u/ D hz; ei. If Œt; e D w0 with w0 2 W0 hzi, then Œt; eu D Œt u ; e u D Œt; e D w0 D w0u D w0 z and so z D 1, a contradiction. Hence, we must have Œt; e D z. If m 3, then t inverts m2 the element v D b 2 of order 4 (Proposition 51.2) and so v t D v 1 D vz. But then .ve/t D .vz/.ez/ D ve, which is a contradiction since CB .t / D W0 . Hence we must have m D 2, o.b/ D 4, b 2 D z, b t D bw with w 2 W0 hzi, and e t D ez. It follows that t inverts in B exactly four elements, namely the elements in W0 (which are actually centralized by t ). Hence there are exactly four involutions in V B and they all lie in E1 D ht i W0 . These four involutions in E1 W0 lie in a single conjugacy class in V since jV W CV .t /j D 4. Set VQ D V CG .t / D V hui so that NG .V / D VQ (since CG .t / must cover NG .V /=V ) and E1 is normal in VQ because hE1 W0 i D E1 . If VQ D G, then CG .t / would contain a G-invariant elementary abelian subgroup E1 of order 8, contrary to our assumptions. Hence VQ ¤ G and then Theorem 1.7 and Proposition 1.8 imply that G=B is of maximal class and VQ =B Š ht; ui Š E4 . Replacing u with ut (if necessary), we may assume that huiB=B D Z.G=B/. We claim that CG .A/ D B. Set CG .A/ D BQ so that BQ B and BQ is normal in G.

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If BQ ¤ B, then BQ huiB. This is a contradiction since w u D wz and so u acts faithfully on A. Hence we have CG .A/ D B and so G=B Š D8 since VQ ¤ G. It remains to consider the case where CG .t / D E D ht iW0 for each t 2 Eˆ.G/. We set again V D ht iB. Since jBj D 2mC2 , m 2, so also jV Bj D jBt j D 2mC2 . Set VQ D NG .V /. If jVQ =V j 4, then jVQ W CVQ .t /j D jVQ W Ej 2mC2 and so all 2mC2 elements in V B are involutions. This is a contradiction since t e is not an involution in view of Œt; e ¤ 1. Therefore we must have jVQ =V j 2. If VQ D G, then jG=Bj D 2 or 4. We assume that G ¤ VQ and so in this case jVQ =V j D 2. Since B is abelian, so the set B0 of elements of B which are inverted by t is a subgroup of B. The number of involutions in V B is equal to jB0 j. Since t does not invert e 2 A W0 , so jB0 j 2mC1 . On the other hand, jcclVQ .t /j D jVQ W Ej D 2mC1 and so jB0 j D 2mC1 , V B has exactly 2mC1 involutions and they lie in a single VQ -class. By Proposition 51.2, Ã1 .B/ D hb 2 i B0 and so B1 D Ã1 .B/W0 D hb 2 ; wi B0 , where w 2 W0 hzi. Since B=B1 Š E4 and hB1 ; ei 6 B0 , so B0 D hB1 ; bi or B0 D hB1 ; bei. Replacing b with be (if necessary), we may assume that t inverts b and so B0 D hB1 ; bi D hb; wi. m2 Since t inverts v D b 2 with v 2 D z, so v t D v 1 D vz. Then e t D ez would t imply .ve/ D .vz/.ez/ D ve, which is a contradiction since CB .t / D W0 D hz; wi. Therefore e t D ew with w 2 W0 hzi. Since NG .V /=B is of order 4 and G ¤ NG .V / D VQ , Theorem 1.7 and Proposition 1.8 imply that G=B is of maximal class. The subgroup V =B (of order 2) is non-central in G=B and so G=B is dihedral or semi-dihedral. In particular, VQ =B is a 4-group. Let x 2 G VQ be such that x normalizes VQ and x 2 2 VQ . Set D D hVQ ; xi so that (by the structure of G=B) D =B Š D8 and therefore we may assume that x 2 2 B. Since V =B is noncentral in D =B, u D t x 2 VQ V . We claim that W0 D hz; wi is normal in G. Obviously Z.G/ CB .t / D W0 . If Z.G/ D W0 , then our claim is clear. Suppose that Z.G/ < W0 so that Z.G/ D hzi is of order 2. Let U be a normal 4-subgroup contained in A and assume that U ¤ W0 . Then we have U > hzi D Z.G/ and so U \W0 D hzi. But then for an element u 2 U hzi, we have ut D uz since CA .t / D W0 . This gives .vu/t D .vz/.uz/ D vu, where v is the cyclic subgroup of order 4 in hbi. This is a contradiction and so W0 D hz; wi is normal in G. Since CA .t / D W0 and x normalizes A and W0 , it follows CA .u/ D W0 . Hence CG .u/ D hui W0 and Z.VQ / D W0 since VQ D Bht; ui. Note that ht; bi Š D2mC1 and t e D t w. There are exactly two V -classes of involutions contained in V B with the representatives t and t b containing 2m elements each. The V -class of t is the set ft w i b 2j g and the V -class of t b is the set ft bw i b 2j g, where i; j are any integers. Now u fuses these two V -classes and so t u D t bw r b 2s for suitable integers r; s. This gives .t u/2 D t ut u D t t u D bw r b 2s D b 2sC1 w r and so o.t u/ D 2mC1 and ht; ui Š D2mC2 . Since u inverts ht ui, so u inverts .t u/2 and w. But h.t u/2 ; wi D B0 and so u inverts B0 . It follows that b u D b 1 and w u D w. Thus t u centralizes B0 D hb; wi. Since t u acts faithfully on B and B D B0 A, t u must act faithfully

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on A. Note that ht uiB=B D Z.D =B/ D Z.G=B/. If BQ D CG .A/ > B, then t u 2 BQ since BQ is normal in G and G=B is of maximal class. This is a contradiction and so CG .A/ D B, D D G, and G=B Š D8 . We have proved the following final result of this subsection. Theorem 51.10. Let G be a 2-group which has a self-centralizing non-normal elementary abelian subgroup E of order 8 which is not contained in ˆ.G/. Assume that there is no involution t 2 E ˆ.G/, such that CG .t / contains an elementary abelian G-invariant subgroup of order 8. Suppose that G has a normal elementary abelian subgroup of order 8 but none of order 16. Then G has a maximal normal abelian subgroup B of type .2m ; 2; 2/, m 1 such that G=B is isomorphic to a non-trivial subgroup of D8 . Example. We give here an example of a 2-group G of order 28 satisfying the assumptions of Theorem 51.10. We set G D hb; z; w; e; t; u; x j b 8 D z 2 D w 2 D e 2 D t 2 D u2 D x 2 D Œb; w D Œb; e D Œw; e D Œt; w D Œu; w D Œe; x D 1; b 4 D z; u D t x ; b t D b 1 ; b u D b 1 ; e t D ew; e u D ewz; .t u/2 D b; b x D b 1 ; w x D wzi. Here B D hb; w; ei is abelian of type .8; 2; 2/ and G=B Š D8 . A coset enumeration computer program assures that such a group G exists! 3o . The case E 6 ˆ.G/ and G has no normal elementary abelian subgroups of order 8. We suppose throughout this subsection that a 2-group G contains a selfcentralizing elementary abelian subgroup E of order 8 which is not contained in ˆ.G/. We assume in addition that G has no normal elementary abelian subgroups of order 8. Let t 2 E ˆ.G/ and M be a maximal subgroup of G such that t 62 M . Then (by modular law) CG .t / D ht i M0 , where M0 D CM .t / contains the four-subgroup E0 D E \M and CM0 .E0 / D M0 . Hence we have either CG .t / D E or (by Theorem 1.7 and Proposition 1.8) M0 Š D2n , n 3 or M0 Š SD2m , m 4. Suppose that G has (at least) two distinct normal 4-subgroups. Then Theorem 50.2 implies that G is the central product G D D C of D Š D8 and C with D \ C D Z.D/ and C is either cyclic or of maximal class different from D8 . However, if C is cyclic or generalized quaternion, then G does not possess a self-centralizing elementary abelian subgroup of order 8. Hence C Š D2n or C Š SD2n , n 4. We have proved the following result. Theorem 51.11. Let G be a 2-group which has a self centralizing elementary abelian subgroup E of order 8 with E 6 ˆ.G/. Assume in addition that G has no normal elementary abelian subgroups of order 8. Then for each involution t 2 E ˆ.G/, we have CG .t / D ht i M0 , where M0 is isomorphic to one of the following groups: E4 , D2n .n 3/, or SD2m .m 4/. If G has (at least) two distinct normal 4-subgroups, then G D D C is the central product of D and C , where D Š D8 , D \ C D Z.D/, and C Š D2n or C Š SD2n with n 4. In view of Theorem 51.11, we assume in the rest of this subsection that G has the unique normal four-subgroup U (see Lemma 1.4).

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Groups of prime power order

Suppose that CG .t / D E for an involution t 2 E ˆ.G/. This is an exceptional case. Since G is neither abelian nor of maximal class, we may apply Theorem 50.1. Let N be a metacyclic normal subgroup of G such that G=N is isomorphic to a subgroup of D8 , W D 2 .N / is abelian of type .4; 4/ or .4; 2/, and CG .W / N . Set W0 D 1 .W / so that W0 is the unique normal four-subgroup of G. Suppose that jE \ N j 2. In that case jE \ N j D 2 and E \ N D E \ W0 . Since CG .t / D E, t 2 E N induces on W an involutory automorphism, where W D CW .t / is of order 2. We may apply Proposition 51.2 which shows that t inverts Ã1 .W / and also on W =W . If W Š C4 C4 , then t inverts Ã1 .W / D W0 contrary to the fact that jCW .t /j D 2. Suppose that W D ha; s j a4 D s 2 D Œa; s D 1; a2 D zi Š C4 C2 . In this case W D CW .t / D hzi D Ã1 .W /. We must have s t D sz and at D a1 z with D 0; 1. If D 1, then at D a1 z D a, a contradiction. If D 0, then at D a1 D az and so .as/t D .az/.sz/ D as, a contradiction. We have proved that in this special case jE \ N j D 4 and so E > W0 , which gives the following result. Theorem 51.12. Let G be a 2-group which has a self-centralizing elementary abelian subgroup E of order 8 with E 6 ˆ.G/. Assume that G has no normal elementary abelian subgroups of order 8. Suppose in addition that G has the unique normal 4subgroup W0 and that for an involution t 2 E ˆ.G/, CG .t / D E. Then we have E > W0 and Theorem 50.1 gives further information about the structure of G. In the rest of this subsection we assume also that there is an involution t 2 E ˆ.G/ such that G ¤ CG .t / D ht i D with D Š D2m , m 3 or D Š SD2n , n 4. Remark. If G is a 2-group possessing an involution t such that CG .t / D ht i D with D Š D2m or D Š SD2m , where m 4, then G has no normal elementary abelian subgroup A of order 8. Indeed, if A is such a subgroup, then Proposition 51.1 implies that jCA .t /j 4 and so CG .t / would contain a normal elementary abelian subgroup of order 8, which is not the case. Let U be the unique normal 4-subgroup of G. Set T D CG .U / so that jG W T j 2. Suppose at first that t 2 U . Since G ¤ CG .t /, we have jG W T j D 2 and T D CG .t / D ht i D with D \ U D hui D Z.D/ D Z.G/ and D Š D2m , m 3 or D Š SD2n , n 4. Let M be a maximal subgroup of G such that t 62 M . Set M0 D T \ M so that CG .t / D ht i M0 and M0 Š D. If M were not of maximal class, then Lemma 1.4 implies that M contains a G-invariant 4-subgroup U0 . Since U0 ¤ U , we have obtained a contradiction. Hence M is of maximal class and therefore M Š D2mC1 or M Š SD2mC1 and M0 Š D Š D2m , m 3. There is an element y 2 M M0 such that y 2 2 hui D Z.M / D Z.M0 /. We get t y D t u so that y t D yu and this determines the action of t on M . We have proved the following result. Theorem 51.13. Let G be a 2-group which has a self-centralizing elementary abelian subgroup E of order 8 with E 6 ˆ.G/. Assume that G has no normal elementary

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abelian subgroups of order 8. Suppose in addition that G has the unique normal 4subgroup U and there is an involution t 2 E ˆ.G/ such that t 2 U and G ¤ CG .t / D ht i D with D Š D2m , m 3 or D Š SD2n , n 4. Then G has a maximal subgroup M such that G D ht iM , where M Š D2mC1 or M Š SD2mC1 and CM .t / Š D2m , m 3. In what follows we assume also that t 62 U , where U is the unique normal foursubgroup of G. Suppose for a moment that t 2 T U , where T D CG .U /. Then CG .t / has the normal elementary abelian subgroup E1 D ht i U of order 8. This forces CG .t / D ht i D, where D Š D8 . Since E1 6 Z.CG .t //, it follows CG .t / 6 T and CT .t / D E1 . In particular, jG W T j D 2 and there is an involution t 0 2 CG .t / T . Set hzi D CU .t 0 / so that z 2 Z.G/. It follows that E2 D ht 0 ; t; zi is another elementary abelian subgroup of order 8 contained in CG .t / and so CG .E2 / D E2 . Also, t 0 is an involution in E2 ˆ.G/. If CG .t 0 / D E2 , then applying Theorem 51.12 we get a contradiction since E2 does not contain U . It follows that G ¤ CG .t 0 / D ht 0 i M0 with M0 Š D2m , m 3 or M0 Š SD2n , n 4. Replacing E with E2 and t with t 0 , we see that we may assume from the start that t 2 G T . It remains to consider only the case t 2 G T , where CG .t / D ht i M0 with M0 Š D2m , m 3 or M0 Š SD2n , n 4. Here T D CG .U /, jG W T j D 2, where U is the unique normal four-subgroup of G. It follows that G D ht iT and so, by the modular law, G0 D CG .t / D ht i D, where D D CT .t / Š M0 . We have hzi D Z.D/ D CU .t / D Z.G/ is of order 2. Set G1 D NG .G0 /. Since j.UG0 / W G0 j D 2, we have UG0 G1 . But Z.G0 / D ht; zi, so t has exactly two conjugates t and t z in G1 . It follows G1 D UG0 . Set D0 D U ht i and so D0 Š D8 and G1 D D D0 with D \ D0 D hzi and U D hz; ui < D0 . We fix in the rest of this subsection the notation for the structure of D. We set m1

D D ha; b j a2

m2

D b 2 D z 2 D 1; a2

D z; ab D a1 z ; D 0; 1i;

where m 3, and if D 1, then m 4. Let v be an element of order 4 in D0 and let y be an element of order 4 in hai. Since y 2 D v 2 D z, so x D yv is an involution in G1 .D0 [ D/. We compute x t D .yv/t D yv 1 D yvz D xz, and so D1 D hx; t i Š D8 and D1 \ U D Z.D1 / D Z.G/ D hzi. We see that CG1 .D1 / D ha; bt i D D Š D because x bt D .yv/bt D y 1 v 1 D yv D x, .bt /2 D 1 and abt D a1 z . Thus G1 D D D1 with D Š D, D \ D1 D hzi, CG .t / D ht i D D ht i D and U D hz; ui 6 D1 . Denoting again D with D and bt with b, we have obtained the following initial configuration. (R) The subgroup G1 D U CG .t / is the central product G1 D D D1 , where D1 Š D8 , t 2 D1 , D \ D1 D hzi D Z.G/, CG .t / D ht i D with D Š D2m or D Š SD2m , m 3, and U 6 D1 , U 6 D. In the rest of this subsection we denote with S a subgroup of G of the maximal possible order subject to the following three conditions.

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(i) S G1 D U CG .t / D D D1 , where D Š D2m or D Š SD2m , m 3, D1 Š D8 , t 2 D1 . (ii) S D DL, where L is normal in S , L Š D2n , n 3, D \ L D hzi D Z.D/ D Z.L/ and L D1 . (iii) U D hz; ui 6 L and U 6 D. We set for the rest of this subsection: n1

L D hc; t j c 2

n2

D t 2 D z 2 D 1; c 2

D z; c t D c 1 ; n 3i Š D2n :

We may set u D yv, where y is an element of order 4 in hai and v is an element of order 4 in hci. If a4 D 1, then we put a D y and if c 4 D 1, then we set c D v. It is now easy to determine all possibilities for the structure of S . Act with D on the dihedral subgroup L. Since Aut.L/=Inn.L/ is abelian of type .2n3 ; 2/, it follows that D 0 D ha2 i centralizes L. Also we know that D centralizes hv; t i D D1 Š D8 , where D1 L and so by the structure of Aut.L/ either D centralizes L (and so S D D L) or M D CD .L/ is a maximal subgroup of D in which case n 4. In that case D=M induces such an involutory automorphism on L so that CL .D=M / D hc 2 ; t i Š D2n1 . In any case, ŒD; L hzi and so D is also normal in S . If D 1 (i.e. D is semidihedral), then we have three possibilities for the maximal subgroup M of D. If D 0 (i.e. D is dihedral), then two distinct maximal subgroups of D are isomorphic and so we have in this case only two possibilities for the maximal subgroup M of D. This gives (together with central products) exactly seven possibilities Si , i D 1; 2; : : : ; 7 for the structure of S and they are given explicitly (in terms of generators and relations) in the following definition. Definition 1. Let S D DL be a product of two normal subgroups m1

D D ha; b j a2

m2

D b 2 D z 2 D 1; a2

D z; ab D a1 z ; D 0; 1; m 3;

and if D 1; then m 4i and n1

L D hc; t j c 2

n2

D t 2 D z 2 D 1; c 2

D z; c t D c 1 ; n 3i Š D2n ;

where D \ L D Z.D/ D Z.L/ D hzi and Œa; t D Œb; t D 1 so that CS .t / D ht i D. Here if D 1, then D Š SD2m , m 4, and if D 0, then D Š D2m , m 3. (1) If D 1 and Œa; c D Œb; c D 1, then S D D L D S1 , m 4, n 3. (2) If D 1, CD .L/ D hai, and c b D cz, then S D S2 , m 4, n 4. (3) If D 1, CD .L/ D ha2 ; bi Š D2m1 , and c a D cz, then S D S3 , m 4, n 4. (4) If D 1, CD .L/ D ha2 ; bai Š Q2m1 , then S D S4 , m 4, n 4. (5) If D 0, CD .L/ D hai, and c b D cz, then S D S5 , m 3, n 4.

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(6) If D 0, CD .L/ D ha2 ; bi Š E4 or D2m1 , and c a D cz, then S D S6 , m 3, n 4. (7) If D 0, and CD .L/ D D, then S D D L D S7 , m 3, n 3. We make here the following simple observation. Since t 2 G T , where T D CG .U / and jG W T j D 2, t cannot be conjugate (fused) in G to any involution t 0 which centralizes U . Therefore, with respect to that fusion, it is enough to consider only those involutions in S which act faithfully on U . We want to show in the sequel that S must be a normal subgroup of G and jG=S j 4. In some cases for the structure of S this is difficult. 4o . The case S Š S1 . We have exactly four conjugacy classes of involutions contained in S U which act faithfully on U with the representatives t , t c, b, and bav. The corresponding centralizers in S are CS .t / D CG .t / D ht i D with D Š SD2m , m 4; CS .t c/ D ht ci D with D Š SD2m m 4; CS .b/ D hbi L with L Š D2n , n 3; CS .bav/ D hbavi hc; tyi with hc; tyi Š Q2n , n 3. We may assume NG .S / ¤ S (otherwise we are finished). Then NG .S / can fuse cclS .t / (the conjugacy class of t in S ) only with cclS .t c/ and so jNG .S / W S j D 2. Obviously, Lhci is a normal subset of involutions in NG .S / and L D hLhcii which implies that L is normal in NG .S /. Suppose that N—G .S / ¤ G (otherwise we are finished). Note that each x in NG .S /S sends (by conjugation) cclS .t / onto cclS .t c/. By the above, NG .S / cannot fuse cclS .b/ and cclS .bav/ to any other conjugate class of involutions in S . This gives that CNG .S/ .b/ covers NG .S /=S and also CNG .S/ .bav/ covers NG .S /=S . In particular jCNG .S/ .b/j D jCNG .S/ .bav/j D 2nC2 . The subgroup K D hb; bavi is dihedral of order 2m and S D KL. Since both K and L are generated by its non-central involutions, we have hcclS .t /, cclS .t c/, cclS .b/, cclS .bav/i D S . Let x be any involution in NG .S /S . Since .cclL .t //x D cclL .t c/, so x induces an outer involutory automorphism on L which inverts hci. Indeed, if t x D t c k .k odd/, then x inverts t .t c k / D c k and so x inverts c. Hence CL .x/ D hzi and so, by Theorem 1.7 and Proposition 1.8, Lhxi D L0 is of maximal class. Since L0 is generated by its involutions, it follows L0 Š D2nC1 . We have NG .S / D S hxi and so CNG .S/ .x/ D hxi CS .x/. On the other hand CL .x/ D hzi and so if jCS .x/j D 2m , then CS .x/ covers S=L which implies that L0 Š D2nC1 is normal in NG .S / D DL0 > S . This contradicts the maximality of S . (See the construction of S .) It follows that there is no involution x 2 NG .S / S with the property jCS .x/j D 2m . Let y 0 be an element in G NG .S / such that y 0 normalizes NG .S / and .y 0 /2 2 0 NG .S /. If x D t y 2 NG .S / S , then CNG .S/ .x/ D hxi CS .x/ Š ht i D and 0 so jCS .x/j D 2m , which is a contradiction. If t y 2 cclS .t / [ cclS .t c/, then there 0 0 1 is an n 2 NG .S / such that t y D t n . But then t y n D t and so CG .t / 6 NG .S /, 0 0 y a contradiction. Hence we must have t 2 cclS .b/ or t y 2 cclS .bav/. We know 0 that jCNG .S/ .b/j D jCNG .S/ .bav/j D 2nC2 and on the other hand jCNG .S/ .t y /j D jCNG .S/ .t /j D 2mC1 . Hence m D n C 1. It follows that for each x 2 .cclS .t / [ cclS .t c/ [ cclS .b/ [ cclS .bav// D P , we have jCNG .S/ .x/j D 2nC2 D 2mC1 .

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Groups of prime power order 0

0

Finally, since S y ¤ S , there is an x 2 P such that x 0 D x y 2 NG .S / S . But then CNG .S/ .x 0 / D hx 0 i CS .x 0 / and CS .x 0 /j D 2m , which is the final contradiction. It follows that NG .S / D G and so in case S Š S1 , the subgroup S is normal in G and jG=S j 2. 5o . The case S Š S2 . We have exactly four conjugacy classes of involutions contained in S U which act faithfully on U with the representatives t , t c, b, and bav. The corresponding centralizers in S are CS .t / D CG .t / D ht i D with D Š SD2m , m 4; CS .t c/ D ht ci ha; bavi with ha; bavi Š SD2m , m 4; CS .b/ D hbi hyc; t i with hyc; t i Š SD2n , n 4; CS .bav/ D hbavi hcy; t ci with hcy; t ci Š SD2n , n 4. We assume NG .S / ¤ S (otherwise we are finished). Set P D cclS .t / [ cclS .t c/ [ cclS .b/ [ cclS .bav/. Obviously, hP i D S and P G T , where T D CG .U / and jG W T j D 2. Suppose also that t is conjugate in NG .S / to t c but t is not conjugate in NG .S / to any involution in S L. Then jNG .S / W S j D 2 and since NG .S / normalizes the set L hci and hL hcii D L, so L is normal in NG .S /. Let x be any involution in NG .S / S . Since .cclL .t //x D cclL .t c/, we get (as in case S Š S1 ) L0 D Lhxi Š D2nC1 . We have CNG .S/ .x/ D hxi CS .x/ and CL .x/ D hzi. If jCS .x/j D 2m , then CS .x/ covers S=L and so L0 is normal in DL0 > S . This contradicts the maximality of S . Hence there is no involution x 2 NG .S / S with jCS .x/j D 2m . Suppose in addition that NG .S / ¤ G (otherwise we are finished). Let y 0 be an element in G NG .S / such that y 0 normalizes NG .S / and .y 0 /2 2 NG .S /. 0 If x D t y 2 NG .S / S , then jCNG .S/ .x/j D 2mC1 and so jCS .x/j D 2m , a 0 contradiction. If t y 2 cclS .t / [ cclS .t c/, then CG .t / 6 NG .S / (as in case S Š S1 ), 0 0 which is a contradiction. Hence we must have t y 2 cclS .b/ or t y 2 cclS .bav/. 0 Since y 0 does not normalize S , there is x 0 2 P such that x0 D .x 0 /y 2 NG .S / S . If NG .S / fuses b with bav, then all involutions in P are fused in hy 0 iNG .S / to t and so jCNG .S/ .x0 /j D 2mC1 and jCS .x0 /j D 2m , a contradiction. If NG .S / does not 0 fuse b with bav, then jCNG .S/ .b/j D jCNG .S/ .bav/j D 2nC2 . Since jCNG .S/ .t y /j D 2mC1 , so n C 2 D m C 1 and therefore jCNG .S/ .x0 /j D 2nC2 D 2mC1 and so again jCS .x0 /j D 2m , a contradiction. Hence we must have NG .S / D G and so we are done in this case. Suppose now that t is not conjugate to t c in NG .S /. Then t must be conjugate in NG .S / to b or to bav. This implies jNG .S / W S j D 2 and m D n 4. Each element x 2 NG .S / S sends .CS .t //0 D ha2 i onto .CS .t x //0 D hc 2 i. Suppose that NG .S / ¤ G and let y 0 be an element in G NG .S / such that y 0 normalizes NG .S / and .y 0 /2 2 NG .S /. Since y 0 does not normalize S , there is an x 0 2 P such 0 that x0 D .x 0 /y 2 NG .S / S , x0 2 G T , and so x0 does not centralize U . On the other hand, x0 sends hyi (contained in ha2 i) onto hvi (contained in hc 2 i). We have y x0 D vz with D 0; 1. But then .yvz /x0 D yvz and so x0 centralizes U D hz; u D yvi. This is a contradiction and so we have NG .S / D G also in this case.

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It remains to consider the case that t is fused in NG .S / to all three involutions t c, b, and bav. In that case m D n 4 and jNG .S / W S j D 4. Note that L and K D hb; bavi are two dihedral normal subgroups both of order 2m of S containing cclS .t /, cclS .t c/, cclS .b/, and cclS .bav/ with NG .S /=S acting regularly on those four S -classes. Hence for each x 2 NG .S / S we have either Lx D L and K x D K or Lx D K. Suppose now that NG .S / ¤ G and let y 0 be an element in G NG .S / 0 such that y 0 normalizes NG .S / and .y 0 /2 2 NG .S /. Then t 0 D t y 2 NG .S / S . 0 Indeed, if t 0 2 S , then t 0 2 P and so there is an n 2 NG .S / such that t 0 D t y D t n . 0 1 0 D t and so CG .t / 6 NG .S /, a contradiction. If Lt D L and But then t y n 0 0 0 K t D K, then .cclL .t //t D cclL .t c/ and .cclK .b//t D cclK .bav/. This gives (as in case S Š S1 ) Lht 0 i Š D2mC1 and Kht 0 i Š D2mC1 . In particular, t 0 inverts hci m3 m3 m3 and on havi. Hence t 0 inverts v 2 hc 2 i and on y 2 h.av/2 i D ha2 i. From 0 0 0 0 y t D y 1 and v t D v 1 , we get ut D .yv/t D y 1 v 1 D yv D u. Hence t 0 centralizes U D hz; ui and so t 0 2 T . But t 2 G T and jG W T j D 2, which 0 0 0 is a contradiction. If Lt D K, then hcit D havi and so v t D yz , D 0; 1. 0 0 This gives .yvz /t D yvz and .uz /t D uz and so t 0 centralizes U , which is a contradiction. It follows that we must have NG .S / D G also in this case. We have proved that in case S Š S2 , the subgroup S is normal in G, jG W S j 4, and if jG=S j D 4, then m D n 4. 6o . Cases S Š S3 and S Š S4 . We have exactly four conjugacy classes of involutions contained in S U which act faithfully on U with the representatives t , t c, b, and bav. The corresponding centralizers in S in case S Š S3 are CS .t / D CG .t / D ht i D with D Š SD2m , m 4; CS .t c/ D ht ci hav; bi with hav; bi Š D2m , m 4; CS .b/ D hbi L with L Š D2n , n 4; CS .bav/ D hbavi hcy; t ci with hcy; t ci Š SD2n , n 4. The centralizers in S in case S Š S4 are CS .t / D CG .t / D ht i D with D Š SD2m , m 4; CS .t c/ D ht ci hav; bai with hav; bai Š Q2m , m 4; CS .b/ D hbi hcy; t i with hcy; t i Š SD2n , n 4; CS .bav/ D hbavi hc; tyi with hc; tyi Š Q2n , n 4. Suppose that S ¤ NG .S /. Then NG .S / can fuse t only with an involution in cclS .bav/ when S Š S3 and only with an involution in cclS .b/ when S Š S4 . This gives m D n and jNG .S /=S j D 2. Set P D cclS .t /[cclS .t c/[cclS .b/[cclS .bav/. Obviously, hP i D S and P G T , where T D CG .U / and jG W T j D 2. Assume that NG .S / ¤ G and let y 0 be an element in G NG .S / such that y 0 normalizes NG .S / and .y 0 /2 2 NG .S /. Since y 0 does not normalize S , there is x 0 2 P such 0 that x0 D .x 0 /y 2 NG .S / S , x0 2 G T and so x0 does not centralize U . Since .cclS .t //x0 D cclS .t x0 /, where t x0 2 cclS .bav/ in case S Š S3 and t x0 2 cclS .b/ in case S Š S4 , so x0 sends .CS .t //0 D ha2 i onto .CS .t x0 //0 D hc 2 i. Indeed, .CS .bav//0 D hc 2 i in case S Š S3 and .CS .b//0 D hc 2 i in case S Š S4 and in both cases hc 2 i is normal in S . We get ha2 ix0 D hc 2 i and so, in particular, y x0 D vz , D 0; 1 and so .yvz /x0 D yvz or .uz /x0 D uz and so x0 centralizes U D

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Groups of prime power order

hz; u D yvi. This is a contradiction and so we must have NG .S / D G. We have obtained in both cases S Š S3 and S Š S4 that jG W S j 2 and if jG W S j D 2, then m D n 4. We have proved the following result. Theorem 51.14. Let G be a 2-group which has a self-centralizing elementary abelian subgroup E of order 8 with E 6 ˆ.G/. We assume that G has the unique normal foursubgroup U and set T D CG .U /. Suppose that there is an involution t 2 E ˆ.G/ such that CG .t / D ht i D with D Š SD2m , m 4. Finally, assume t 62 U and in that case we have proved that we may assume t 2 G T . Then G has a (large) normal subgroup S containing U CG .t / which is isomorphic to one of the groups Si for i D 1; 2; 3; 4 (see Definition 1) so that jG=S j 4. More precisely, (a) if S Š S1 , then jG=S j 2; (b) if S Š S2 , then jG=S j 4 and if jG=S j D 4, then m D n 4; (c) if S Š S3 , then jG=S j 2 and if jG=S j D 2, then m D n 4; (d) if S Š S4 , then jG=S j 2 and if jG=S j D 2, then m D n 4. We have to analyze three remaining cases S5 , S6 , and S7 for the structure of S . 7o . The case S Š S5 . We have exactly four conjugacy classes of involutions contained in S U which act faithfully on U with the representatives t , t c, b, and ba. The corresponding centralizers in S are CS .t / D CG .t / D ht i D with D Š D2m , m 3; CS .t c/ D ht ci ha; bvi with hv; bvi Š Q2m , m 3; CS .b/ D hbi hyc; t i with hyc; t i Š SD2n , n 4; CS .ba/ D hbai hyc; t i with hyc; t i Š SD2n , n 4. We see that t cannot be fused in NG .S / to any of the involutions t c, b, or ba and this implies S D G. By our assumption, G has the unique normal 4-subgroup U and so m 4. 8o . The case S Š S6 . We have exactly four conjugacy classes of involutions contained in S U which act faithfully on U with the representatives t , t c, b, and ba. The corresponding centralizers in S are CS .t / D CG .t / D ht i D with D Š D2m , m 3; CS .t c/ D ht ci hav; bi with hav; bi Š SD2m , m 4 and hav; bi Š D8 if m D 3 (in which case t c centralizes U ), CS .b/ D hbi L with L Š D2n , n 4; CS .ba/ D hbai hcy; t i with hcy; t i Š SD2n , n 4. Suppose that NG .S / ¤ S . Since NG .S / can fuse t only with an involution in cclS .b/, we have m D n 4 and jNG .S /=S j D 2. Set P D cclS .t / [ cclS .t c/ [ cclS .b/ [ cclS .ba/. Obviously, hP i D S and P G T , where T D CG .U / and jG W T j D 2. Assume that NG .S / ¤ G and let y 0 be an element in G NG .S / such that y 0 normalizes NG .S / and .y 0 /2 2 NG .S /. Since y 0 does not normalize S , there 0 is an x 0 2 P such that x0 D .x 0 /y 2 NG .S / S , x0 2 G T and so x0 does not centralize U . Then x0 sends (by conjugation) t onto an S -conjugate of b and so x0 sends .CS .t //0 D ha2 i onto .CS .b//0 D hc 2 i. Indeed, each S -conjugate b 0 of b has the property CS .b 0 / D hb 0 i L because L is normal in S . In particular, y x0 D vz , D 0; 1 and so yvz D uz is centralized by x0 . Hence x0 centralizes U D hz; ui,

51 2-groups with self centralizing subgroup isomorphic to E8

73

which is a contradiction. Hence we must have NG .S / D G which implies that in case S Š S6 we get jG=S j 2 and if jG=S j D 2, then m D n 4. It remains to consider the case S Š S7 which is rather difficult since there are several possibilities for the fusion of the involution t with other involutions in S U which act faithfully on U . 9o . The case S Š S7 . We have exactly four conjugacy classes of involutions contained in S U which act faithfully on U with the representatives t , t c, b, and ba. The corresponding centralizers in S are CS .t / D CG .t / D ht i D with D Š D2m , m 3; CS .t c/ D ht ci D with D Š D2m , m 3; CS .b/ D hbi L with L Š D2n , n 3; CS .ba/ D hbai L with L Š D2n , n 3. Suppose that NG .S / ¤ S (otherwise we are finished). Set P D cclS .t /[cclS .t c/[ cclS .b/ [ cclS .ba/. Obviously, hP i D S and P G T , where T D CG .U / and jG W T j D 2. Suppose also that t is not fused in NG .S / to any involution in S L. Then NG .S / fuses t with t c and so jNG .S / W S j D 2 and L is normal in NG .S / since L hci is a normal subset in NG .S /. Let x be any involution in NG .S / S . Since .cclL .t //x D cclL .t c/, we get L0 D Lhxi Š D2nC1 (as in case S Š S1 ). We have CNG .S/ .x/ D hxi CS .x/ and CL .x/ D hzi D Z.L/ D Z.L0 /. If jCS .x/j D 2m , then CS .x/ covers S=L and so L0 is normal in DL0 > L. This contradicts the maximality of S . Hence there is no involution x 2 NG .S /S such that jCS .x/j D 2m . Suppose in addition that NG .S / ¤ G. Let y 0 be an element in G NG .S / such 0 that y 0 normalizes NG .S / and .y 0 /2 2 NG .S /. If x D t y 2 NG .S / S , then 0 jCNG .S/ .x/j D 2mC1 and so jCS .x/j D 2m , a contradiction. If t y 2 cclS .t / [ 0 cclS .t c/, then (as before) CG .t / 6 NG .S /, a contradiction. Hence t y 2 cclS .b/ or 0 0 t y 2 cclS .ba/. Since y 0 62 NG .S /, there is an x 0 2 P with x0 D .x 0 /y 2 NG .S /S . If NG .S / fuses b and ba, then all involutions in P are fused in hy 0 iNG .S / to t and so jCNG .S/ .x0 /j D 2mC1 and jCS .x0 /j D 2m , a contradiction. If NG .S / does not fuse b 0 and ba, then jCNG .S/ .b/j D jCNG .S/ .ba/j D 2nC2 and since jCNG .S/ .t y /j D 2mC1 , so we get n C 2 D m C 1 and therefore jCNG .S/ .x0 /j D 2nC2 D 2mC1 and so again jCS .x0 /j D 2m , a contradiction. Hence we must have NG .S / D G in this case. Suppose that t is not conjugate to t c in NG .S /. Then t must be conjugate in NG .S / to b or ba. This implies jNG .S / W S j D 2 and m D n 3. We make here the following simple observation. All involutions in S .D [ L/ centralize U and so lie in T . All noncentral involutions in D and L lie in G T . Since S is the central product S D D L of D and L, it follows that D and L are the only dihedral normal subgroups of order 2m of S all of whose noncentral involutions belong to G T . Therefore for each element x 2 NG .S / S , we have Lx D D since t x 2 cclS .b/ or t x 2 cclS .ba/. In particular, hcix D hai. Suppose again that NG .S / ¤ G. Let y 0 be an element in G NG .S / such that y 0 normalizes NG .S / and .y 0 /2 2 NG .S /. Since y 0 62 NG .S /, there is an x 0 2 P such 0 that x0 D .x 0 /y 2 NG .S / S and x0 2 G T and so x0 does not centralize U . Note that hP i D S and P G T . On the other hand (by the above), hcix0 D hai

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and so v x0 D yz , D 0; 1. But then .yvz /x0 D yvz , where yv D u. Hence x0 centralizes U D hz; ui, a contradiction. Thus in this case we have also NG .S / D G. It remains to analyze the case that t is fused in NG .S / to all three involutions t c, b, and ba. In that case m D n 3 and jNG .S / W S j D 4. By the above observation, for each x 2 NG .S / S , we have either Lx D L and D x D D or Lx D D. Suppose now that NG .S / ¤ G. Let y 0 be an element in G NG .S / such that y 0 normalizes 0 0 NG .S / and .y 0 /2 2 NG .S /. Then t 0 D t y 2 NG .S / S . Otherwise, t y 2 P and 0 0 1 so there is an n 2 NG .S / with t y D t n and so t y n D t , which is a contradiction 0 0 0 since y 0 n1 62 NG .S /. If Lt D L and D t D D, then .cclL .t //t D cclL .t c/ 0 and .cclD .b//t D cclD .ba/ imply (as before) that Lht 0 i Š Dht 0 i Š D2mC1 . In 0 0 particular, t 0 inverts v and y which gives ut D .yv/t D y 1 v 1 D yv D u and so t 0 0 centralizes U . This is not possible since t 0 (together with t ) lies in G T . If Lt D D, 0 0 then hcit D hai and so in particular v t D yz , D 0; 1. But then t 0 centralizes 0 yv D u and so t centralizes U , which is again a contradiction. Hence we must have also in this case NG .S / D G. We have proved the following final result. Theorem 51.15. Let G be a 2-group which has a self-centralizing elementary abelian subgroup E of order 8 with E 6 ˆ.G/. We assume that G has the unique normal 4-subgroup U and set T D CG .U /. Suppose that there is an involution t 2 E ˆ.G/ such that CG .t / D ht i D with D Š D2m , m 3. Also, in case D Š D8 we assume in addition that G does not possess a normal elementary abelian subgroup of order 8. Finally, assume t 62 U and in that case we have proved that we may assume t 2 G T . Then G has a (large) normal subgroup S containing U CG .t / which is isomorphic to one of the groups Si for i D 5; 6; 7 (see Definition 1) so that jG=S j 4. More precisely, (a) If S Š S5 , then S D G and m 4, n 4; (b) if S Š S6 , then jG=S j 2 and if jG=S j D 2, then m D n 4; (c) if S Š S7 , then jG=S j 4 and if jG=S j D 4, then m D n 3.

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2-groups with 2 -subgroup of small order

The main results of this section are due to Janko [Jan6]. In Lemma 42.1 finite 2-groups G have been determined with j2 .G/j 8. In this section we do the next step and determine the finite 2-groups G with jGj > 16 and j2 .G/j D 16 (D 2 -groups). All nonmetacyclic 2 -groups will be given in terms of generators and relations. For more general result, see 55. If jGj D 16, then 2 .G/ < G if and only if G 62 fC16 ; C8 C2 ; M16 g. Therefore, we consider in this section only groups of order > 16. If j2 .G/j D 16, then the numbers c1 .G/ and c2 .G/ are small. This observation allows us to generalize the main result of this section (see 55). In subsection 1o we study 2 -groups G which have no normal subgroups isomorphic to E8 . Using the main theorem from 50, we obtain three classes of such 2-groups and one exceptional group of order 25 (Theorem 52.1). In subsection 2o we study nonabelian 2 -groups G with normal subgroup isomorphic to E8 . It is easy to see that then G=E is either cyclic or generalized quaternion. If G=E is cyclic, then we get three classes of groups (Theorem 52.2). If G=E is generalized quaternion and E 6 ˆ.G/, then we get one class of groups (Theorem 52.4). If G=E is generalized quaternion and E ˆ.G/, we get two classes of groups (Theorem 52.5). In subsection 3o we investigate 2-groups G with jGj > 16 which have exactly one subgroup of order 16 and exponent 4. It turns out that a 2-group G has this property if and only if jGj > 16 and j2 .G/j D 16 (Theorem 52.6). It is interesting to note that, for a 2 -group G, there are exactly five possibilities for the structure of 2 .G/: Q8 C4 , Q8 C2 , D8 C2 , C4 C2 C2 , C4 C4 (note that Q8 C4 Š D8 C4 ). In fact, if 2 .G/ Š Q8 C2 or D8 C2 and jGj > 16, then we get exactly one group of order 25 in each case (see Theorem 52.1(d) and Theorem 52.2(a) for n D 2). In other three cases for the structure of 2 .G/ we get infinitely many groups. In subsection 4o we consider a similar problem. In [Ber25, 48] the 2-groups G have been considered which have exactly one abelian subgroup of type .4; 2/. It was shown that either j2 .G/j D 8 (and then G is isomorphic to one of the groups in Lemma 42.1) or G has a self-centralizing elementary abelian subgroup of order 8 (see 51). We improve this result by determining completely the groups of the second possibility (Theorem 52.7). Finally, Theorem 52.8 shows that our result also slightly

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Groups of prime power order

improves the classification of 2-groups with exactly two cyclic subgroups of order 4 given in Theorem 43.4. Lemma A ([Ber25, 48]). Let G be a metacyclic 2-group of order > 24 . If j2 .G/j D 24 , then 2 .G/ Š C4 C4 . Proof. Let jGj D 2m > 24 . Assume that 2 .G/ D H is nonabelian. We want to show that this assumption leads to a contradiction. If H is not minimal nonabelian, then Proposition 10.19 implies that G is of maximal class. But then 2 .G/ D G, a contradiction. Suppose that H has a cyclic subgroup of index 2. Since in that case H is minimal nonabelian, H Š M24 (Theorem 1.2). But then 2 .G/ D 2 .H / is abelian of type .4; 2/ which contradicts the fact that H D 2 .G/. Exercise 1.8A implies H D ha; b j a4 D b 4 D 1; ab D a1 i. It follows that Z.H / D ha2 ; b 2 i D 1 .H / D 1 .G/, H 0 D ha2 i, and Z D hb 2 i is a characteristic subgroup of order 2 in H so Z G G. Indeed, b 2 is a square in H , a2 b 2 is not a square in H (check!), and Z.H / H 0 D fb 2 ; a2 b 2 g. Thus Z is central in G. Since H=Z Š D8 , Proposition 10.19 implies that G=Z is of maximal class. All involutions in G=Z lie in H=Z and so G=Z Š SD16 and jGj D 25 . But there are elements of order 4 in .G=Z/ .H=Z/ whose square lies in Z.G=Z/ D ha2 ; b 2 i=Z. Hence there is an element x 2 G H such that x 2 2 ha2 ; b 2 i and so o.x/ 4, which is the final contradiction. Thus, H Š C4 C4 . 1o . G has no normal elementary abelian subgroups of order 8. In this subsection we prove the following result. Theorem 52.1. Let G be a 2-group of order > 24 satisfying j2 .G/j D 24 . Suppose, in addition, that G has no normal elementary abelian subgroups of order 8. Then we have one of the following four possibilities: (a) G D D C (the central product) with D Š D8 , D \ C D Z.D/, and C is cyclic of order 8. Here we have 2 .G/ Š Q8 C4 Š D8 C4 . (For such groups, see Appendix 16; it is proved there, that also G Š Q8 C .) (b) G is metacyclic of order 25 (in that case, by Lemma A, 2 .G/ Š C4 C4 ). (c) G D QS , where Q Š Q8 , Q is normal in G, S is cyclic of order 16, Q \ S D Z.Q/, and if C is the subgroup of index 2 in S , then CG .Q/ D C . Setting Q D ha; bi and S D hsi, we have as D a1 and b s D ba. Here we have 2 .G/ Š Q8 C4 . (d) The exceptional group G of order 25 has a maximal subgroup M D hui Q, where u is an involution, Q Š Q8 , CG .u/ D M and G is isomorphic to the group A2(a) from Theorem 49.1. We have 2 .G/ D M Š Q8 C2 . Proof. We may assume that G is nonabelian. If G were of maximal class, then 2 .G/ D G, a contradiction since jGj > 24 . Since G is neither abelian nor of maximal class, we may apply main theorem from 50 about 2-groups without nor-

52 2-groups with 2 -subgroup of small order

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mal elementary abelian subgroup of order 8. Then G has a normal metacyclic subgroup N such that CG .2 .N // N , G=N is isomorphic to a subgroup of D8 , and W D 2 .N / is either abelian of type .4; 4/ or N is abelian of type .2j ; 2/, j 2. However, if W Š C4 C4 , then W D 2 .G/ since j2 .W /j D 24 D j2 .G/j, and Theorem 41.1 and Remark 41.2 imply that G is metacyclic. This gives the case (b) of our theorem, by Lemma A. In what follows we assume that G is not metacyclic. We assume in the sequel that N is abelian of type .2j ; 2/, j 2 and so W D 2 .N / is abelian of type .4; 2/. By hypothesis, j2 .G/ W W j D 2. Suppose that G has at least two distinct normal four-subgroups. Then it follows from Theorem 50.2 that G D D C , where D is dihedral of order 8 and C is either cyclic or of maximal class and D \ C D Z.D/. In view of j2 .G/j D 16, C must be cyclic of order 8 since jGj > 24 , and this gives the case (a) of our theorem. Next we assume that G has the unique normal four-subgroup W0 D 1 .W / D 1 .N /. Also we have CG .W / D N (N is abelian of type .2j ; 2/, j 2). Set WQ D 2 .G/ so that WQ \ N D W , jWQ W W j D 2, and WQ is nonabelian (the last assertion follows from CG .W / D N WQ ). If WQ is metacyclic, then Theorem 41.1 and Remark 41.2 imply that G is metacyclic. But then Lemma A gives a contradiction. Thus WQ is nonabelian non-metacyclic. In particular, exp.WQ / D 4 (see Theorem 1.2). Proposition 1.6 gives jWQ =WQ 0 j D 8 and so jWQ 0 j D 2. By Lemma 1.1, jWQ j D 2jWQ 0 jjZ.WQ /j and so jZ.WQ /j D 4 and Z.WQ / < W . First assume that Z.WQ / D hvi Š C4 is cyclic and set z D v 2 , W0 D hz; ui D 1 .W /. Since f1g < WQ 0 < hvi (by the previous paragraph, jWQ j D 2), we have WQ 0 D hzi. For any x 2 WQ W , we have x 2 2 Z.WQ / D hvi (indeed, WQ =Z.WQ / is elementary abelian). But WQ has no elements of order 8 and so x 2 2 hzi and ux D uz since CWQ .u/ D W . It follows that the nonabelian subgroup W0 hxi Š D8 (it is nonabelian since W0 6 Z.WQ /) and so, since D8 has five involutions, we may assume that x is an involution. Then WQ is the central product of hu; xi Š D8 and hvi D Z.WQ / Š C4 with hvi \ hu; xi D hzi and .ux/2 D z. We set Q D hux; vxi Š Q8 and see that WQ D Q hvi is also the central product of Q8 and C4 (see also Appendix 16, subsection 1o ). All six elements in WQ .Q [ hvi/ are involutions and therefore Q is the unique quaternion subgroup of WQ D 2 .G/ and so Q is characteristic in WQ hence normal in G (see also Appendix 16). There are exactly three four-subgroups in WQ and one of them is W0 D hz; ui and this one is normal in G. Since G has exactly one normal four-subgroup, it follows that the other two four-subgroups in WQ are conjugate to each other in G. Set C D CG .Q/. Obviously C is normal in G and C \ WQ D hvi. But hvi is the unique subgroup of order 4 in C (indeed, all subgroups of G of order 4 are contained in WQ ) so C is cyclic, by Proposition 1.3, and WQ C D Q C with Q \ C D hzi. Set Q D ha; bi, where we choose the generator a so that u D av. Since W0 D 1 .W / D hz; ui and hvi D Z.WQ / are both normal in G, it follows that hai . W / is normal in G (indeed, W has exactly three subgroups of order 4). Every four-subgroup L is normal in WQ . Indeed, L is characteristic in LZ.WQ /, which is normal in WQ ; it follows that all four-subgroups are normal in Q C . The four-

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Groups of prime power order

subgroups hbv; zi and hbav; zi must be conjugate in G since they are not normal in G and are contained in WQ . Q C / (WQ has exactly three four-subgroups and exactly one of them is normal in G). Therefore Q C < G so jG W .Q C /j D 2, by the product formula, and G=.Q C / induces an outer automorphism ˛ on Q normalizing hai and sending hbi onto hbai. A Sylow 2-subgroup of Aut.Q/ is isomorphic to D8 . It follows that G=C Š D8 and so there is an element s 2 G .Q C / such that s 2 2 C and b s D ba. If as D a, then .ba/s D ba2 D b 1 and so ba D .b s /1 D .ba/1 , which is a contradiction. It follows that as D a1 . It remains to determine s 2 . Set hsiC D S and we see that C is a cyclic subgroup of index 2 in S . Since o.s/ 8 (indeed, all elements of G of order 4 are contained in WQ Q C ), we have o.s 2 / 4 and therefore jZ.S /j 4. The same argument shows that all elements in S C have order 8. It follows from Theorem 1.2 that S is cyclic. We may set s 2 D c, where hci D C . We compute .sb/2 D sbsb D s 2 b s b D cbab D cabzb D cab 2 z D cazz D ca: If jC j D 4, then C D hvi and so ca is an involution. But then o.sb/ D 4, by the displayed formula, and this is a contradiction since sb 62 WQ . Thus jC j 8 and so o.s/ 16. We have obtained the groups stated in part (c) of our theorem. Now suppose that Z.WQ / Š E4 , the four-group, so that Z.WQ / D W0 D 1 .W / D 1 .N / D hz; ui, where z D v 2 and W D hui hvi Š C2 C4 . If WQ is minimal nonabelian, then the fact that WQ is nonmetacyclic implies with the help of Exercise 1.8A that WQ D ha; b j Œa; b D c; a4 D b 2 D c 2 D Œa; c D Œb; c D 1i. We see that Z.WQ / D ˆ.WQ / D ha2 ; ci Š E4 and WQ has exactly three maximal subgroups ha2 ; c; bi Š E8 , ha; ci Š C4 C2 , hba; ci Š C4 C2 since .ba/2 D aŒa; ba D aca D a2 c. Hence ha2 ; c; bi is normal in G, contrary to the assumption of this subsection. We have proved that WQ is not minimal nonabelian. Let D be a minimal nonabelian subgroup of WQ (of order 8). If u 2 Z.WQ / D, then WQ D hui D and ˆ.WQ / D WQ 0 D ˆ.D/ D Ã1 .D/ D hzi since z D v 2 . Suppose first that D Š D8 . Let t be an involution in WQ W (t exists since c1 .WQ / D 7 > 5 D c1 .D/ > 3 D c1 .W /). Since Œv; t D z D v 2 (hvi is normal in WQ and it is not contained in Z.WQ / Š E4 ), t inverts hvi and so we may set D D hv; t i, where WQ D hui D. Since t inverts 2 .N / D W (indeed, t centralizes 1 .N / and both cyclic subgroups of W of order 4 are t -invariant but not centralized by t ), it follows that CN .t / D W0 D hz; ui. Assume that N W contains an element x of order 8 (i.e., j > 2; by the above, N is abelian of type .2j ; 2/). Then Theorem 51.2 implies x t D x 1 w0 with w0 2 W0 . But then .tx/2 D .txt /x D x t x D x 1 w0 x D w0 and therefore o.tx/ 4 with tx 62 WQ , a contradiction. Thus, x does not exist so N D W is abelian of type .4; 2/. Since t inverts N D W , all eight elements in WQ N D WQ W are involutions: indeed, WQ D ht i W . In particular, no involution in WQ N is a square in G (otherwise, G has an element of order 4 not contained in WQ ) and so G=N Š E4 (indeed, G=N is isomorphic to a subgroup of

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Aut.W / Š D8 and jG W N j > 2) and jGj D 25 . All elements in G WQ must be of order 8 and so for any y 2 G WQ , y 2 2 N W0 D W W0 since G=N Š E4 . Replacing huvi with hvi and v with v 1 , if necessary, we may assume that y 2 D v since c2 .W / D 2. Since CG .N / D N.D hy 2 ; ui/, we get uy D uz, and so y u D yz. Thus hy; ui Š M24 and hy; ui > N . The cyclic group hvi D Z.hy; ui/ of order 4 is normal in G and CG .v/ D hyiN Š M24 . Since ty is also an element of order 8, we have .ty/2 2 N W0 . If .ty/2 2 hvi, then CG .v/ hN; y; tyi D G, a contradiction. Thus .ty/2 2 huvi. If .ty/2 D .uv/1 D .uz/v, then replacing u with uz, we may assume from the start that .ty/2 D uv. We have uv D tyty D ty 2 t y D t vt y . It follows that t y D v 1 t uv D t v uv D t uz, and so y normalizes the elementary abelian subgroup E D hz; u; t i of order 8. Since E is normal in WQ , E is also normal in G D hWQ ; yi, contrary to the assumption of this subsection. Finally, let D Š Q8 so that WQ D hui D, Z.D/ D hzi, z D v 2 , and D \ N D hvi. Take r 2 D hvi. We have r 2 D z with z 2 Z.G/ and so r induces an involutory automorphism on the abelian group N of type .2j ; 2/, j 2. Since r inverts W D hu; vi D 2 .N /, it follows that CN .r/ D W0 D hu; zi. Then Theorem 51.2 implies that r inverts N=W0 . Suppose that W < N (i.e., j > 2) and let x 2 N W . We get x r D x 1 w0 with w0 2 W0 and so .rx/2 D rxrx D r 2 x r x D zx 1 w0 x D zw0 2 W0 . Since rx 62 WQ D 2 .G/ and o.rx/ 4, we get a contradiction. We have proved that N D W D CG .N / is abelian of type .4; 2/. We have jGj 26 . Set S D CG .W0 / so that S WQ and S=N stabilizes the chain N > W0 > f1g since jN=W0 j D 2. Hence S=N is elementary abelian of order 4. Suppose that S > WQ ; then S=N is a four-group. Let SQ =N be a subgroup of order 2 in S=N distinct from WQ =N . Thus SQ \ WQ D N . Take an element y 2 SQ N so that o.y/ D 8 and therefore y 2 2 N W0 . But then y centralizes hy 2 ; W0 i D N , a contradiction. Hence S D WQ D CG .W0 /, jG W WQ j D 2 and jGj D 25 . In particular, CG .u/ D huiD with D Š Q8 . Note that the involution u is contained in the unique normal four-subgroup W0 of G and CG .u/ ¤ G. Then our group G is isomorphic to the group A2(a) of the Theorem 49.1. The proof is complete. 2o . G has a normal E8 . Throughout this subsection we assume that G is a nonabelian 2-group of order > 24 satisfying j2 .G/j D 24 . We assume in addition that G has a normal elementary abelian subgroup E of order 8. In that case, G=E has exactly one subgroup 2 .G/=E of order 2 so G=E is either cyclic of order 4 or generalized quaternion (see Proposition 1.3). In the case where G=E is cyclic, G is determined in Theorem 52.2. In the case where G=E is generalized quaternion and E 6 ˆ.G/, G is determined in Theorem 52.4. Finally, if G=E is generalized quaternion and E ˆ.G/, then G is determined in Theorem 52.5. Theorem 52.2. Let G be a nonabelian 2-group of order > 24 satisfying j2 .G/j D 24 . Suppose, in addition, that G has a normal subgroup E Š E8 such that G=E is cyclic of order 2n , n 2. Then one of the following holds:

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Groups of prime power order nC1

n

(a) G D ha; e1 j a2 D e12 D 1; n 2; a2 D z; Œa; e1 D e2 ; e22 D Œe1 ; e2 D 1; Œa; e2 D zi is a group of order 2nC3 , Z.G/ D ha4 i, G 0 D hz; e2 i, ˆ.G/ D ha2 ; e2 i, E D hz; e1 ; e2 i is a normal elementary abelian subgroup of order 8, n1 and 2 .G/ D Eha2 i is abelian of type .4; 2; 2/ for n 3, and 2 .G/ Š C2 D8 for n D 2. (b) G Š M2nC2 C2 with n 2, and 2 .G/ is abelian of type .4; 2; 2/. nC1

(c) G D ha; e1 j a2 D e12 D 1; n 2; Œa; e1 D e2 ; e22 D Œe1 ; e2 D Œa; e2 D 1i being minimal nonabelian, jGj D 2nC3 , ˆ.G/ D ha2 ; e2 i, G 0 D he2 i, and n1 n 2 .G/ D Eha2 i is abelian of type .4; 2; 2/, where E D ha2 ; e1 ; e2 i is a normal elementary abelian subgroup of order 8. Proof. By assumption, G D Ehai for some a 2 G E; let jG=Ej D 2n > 2. Obviously G does not split over E (otherwise, if G D T E, then 2 .G/ D 2 .T / E is of order 25 > 24 ) and so o.a/ D 2nC1 and hai \ E D hzi is of order 2, where n z D a2 . Note that a Sylow 2-subgroup of Aut.E/ is isomorphic to D8 , so a induces an automorphism of order 4 on E hence a4 2 Z.G/. It is easy to check, that if CE .a/ is a four-subgroup, then a induces an involutory automorphism on E. Indeed, then a stabilizes the chain f1g < CE .a/ < E with factors of exponent 2. Suppose that a induces an automorphism of order 4 on E. In that case, by what has been said in the previous paragraph, CE .a/ D hzi so that Z.G/ D ha4 i hzi and CG .E/ D Z.G/E. There are involutions e1 ; e2 2 E such that hz; e1 ; e2 i D E and e1a D e1 e2 , e2a D e2 z. Then CE .a2 / D hz; e2 i D G 0 , G D ha; e1 i, ˆ.G/ D ha2 ; e2 i, n1 2 .G/ D Eha2 i. We have determined the groups stated in part (a). Suppose that a induces an involutory automorphism on E. We have CE .a/ D E1 D hz; e2 i Š E4 (see Exercise 19, below), Z.G/ D ha2 ; e2 i and G 0 D Œa; E is a subgroup of order 2 in E1 (by Lemma 1.1, since G has an abelian subgroup ha2 ; Ei of index 2 and Z.G/ is of index 4 in G). If G 0 D hzi, then hai is normal in G. Let e1 2 E E1 . Then ha; e1 i Š M2nC2 and G D he2 i ha; e1 i which is the group stated in part (b) of our theorem. If G 0 D he2 i, where e2 2 E1 hzi and e1 2 E E1 , then Œa; e1 D e2 and this group is stated in part (c) of our theorem. In the rest of this subsection we assume that G=E Š Q2n ; n 3. Then CG .E/ > E, since a Sylow 2-subgroup of Aut.E/ is isomorphic to D8 . Therefore WQ D 2 .G/ centralizes E, where WQ =E D Z.G=E/ (it is important that j2 .G/j D 24 ), WQ is abelian of type .4; 2; 2/ and hzi D Ã1 .WQ / is of order 2. All elements in WQ E are of order 4 and so no involution in E hzi is a square in G. Indeed, if x 2 2 E hzi, then o.x/ D 4 and x 62 WQ since z is the unique nontrivial square on WQ , and this is a contradiction. We first prove the following useful result. Lemma 52.3. Let, as above, G=E Š Q2n , where G is an 2 -group. We have CG .E/ WQ D 2 .G/, CG .E/=E is cyclic of order 2, and G=CG .E/ is isomorphic to C2 ; E4 or D8 .

52 2-groups with 2 -subgroup of small order

81

Proof. Suppose that CG .E/ contains a subgroup H > E such that H=E Š Q8 . Then Z.H=E/ D WQ =E. Let A=E and B=E be two distinct cyclic subgroups of order 4 in H=E. Then A and B are abelian maximal subgroups of H . It follows that A \ B D WQ Z.H / and therefore WQ D Z.H / is of order 24 . Using Lemma 1.1, we get 26 D jH j D 2jZ.H /jjH 0 j, which gives jH 0 j D 2. Since 1 .H / D E, we get H 0 E and H=E is abelian, a contradiction. It follows that CG .E/=E must be a normal cyclic subgroup of G=E and so G=CG .E/ is isomorphic to a nonidentity epimorphic image of Q2n of order 8 such that G=CG .E/ 6Š Q8 (the last condition is superfluous if n > 3). We conclude that G=CG .E/ 2 fC2 ; E4 ; D8 g. Theorem 52.4. Let G be a 2-group of order > 24 satisfying j2 .G/j D 24 . Suppose, in addition, that G has a normal elementary abelian subgroup E of order 8 such that G=E Š Q2n , n 3, and E 6 ˆ.G/. Then we have n

n1

G D ha; b; e j a2 D b 8 D e 2 D 1; n 3; a2

D b 4 D z; ab D a1 ;

Œe; b D 1; Œe; a D z ; D 0; 1i: Here G is a group of order 2nC3 , G 0 D ha2 i, ˆ.G/ D ha2 ; b 2 i is abelian of type n2 .2n1 ; 2/. The subgroup E D he; a2 b 2 ; zi is a normal elementary abelian subgroup of order 8 in G and G=E Š Q2n . If D 0, then Z.G/ D he; b 2 i is abelian of type .4; 2/. If D 1, then Z.G/ D hb 2 i Š C4 . Finally, 2 .G/ D Ehb 2 i is abelian of type .4; 2; 2/. Proof. Let WQ =E D Z.G=E/; then WQ D 2 .G/ since j2 .G/j D 16 and the subgroup WQ has order 16 and exponent 4. It follows from Aut.E/ D GL.3; 2/ that a Sylow 2-subgroup of Aut.E/ is dihedral of order 8 so CG .E/ > E. We conclude that WQ is abelian of type .4; 2; 2/. Since E 6 ˆ.G/, there is a maximal subgroup M of G such that E 6 M ; then G D EM . We have E0 D E \M Š E4 (obviously, E0 is normal in G since E and M are), WQ0 D WQ \ M is abelian of type .4; 2/, WQ0 D 2 .M /, and M=E0 Š Q2n , n 3. If ˆ.WQ 0 / D hzi, then z 2 E0 . By Lemma 42.1, M is isomorphic to a group (c) in that proposition. Hence M is metacyclic with a cyclic normal subgroup hai of order 2n and the cyclic quotient group M=hai of order 4 so that n1 there is an element b of order 8 in M .E0 hai/ with b 2 2 WQ0 E0 , b 4 D z D a2 , n2 hbi \ hai D hzi, WQ0 D hb 2 iha2 i ˆ.M /. We have used the following facts: 1 .WQ / D E, Ã1 .WQ / D hzi, hai \ hbi is of order 2 (by the product formula), ˆ.G/ D Ã1 .G/. We have (see Lemma 42.1(c)) ˆ.M / D ha2 ; b 2 i WQ0 , M 0 D ha2 i, ab D a1 . Also, Z.M / D hb 2 i is cyclic of order 4 and so E0 hai D hb 2 ; ai is abelian of index 2 in M , and CM .E0 / D E0 hai (again by Lemma 42.1(c)). The element b induces an involutory automorphism on E (because b 2 2 Z.G/). The element b does not centralize E0 since b 62 ha; E0 i D CM .E0 /. For every u 2 E0 f1; zg, we have hb; ui Š M16 (we have jhb; E0 ij D 16 so E0 normalizes hbi). We conclude that Z.hb; E0 i/ D hzi. The subgroup WQ D hb 2 ; Ei is abelian so b induces an involutory automorphism on E. Therefore, by Exercise 19, Z D CE .b/ is of order 4. By the

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Groups of prime power order

above, Z ¤ E0 so there is an element e 2 E E0 such that b centralizes e. Set n2 u D a2 b 2 2 E0 hzi and we have ub D uz. We act with the cyclic group hai on E. By the above, ha; E0 i is abelian. It follows from jE W E0 j D 2 that hai stabilizes the chain E > E0 > f1g and so a2 centralizes E and e a D ey with y 2 E0 . This gives ae D ay. We may set ab D a1 . Then we act on b 1 ab D a1 with e. It follows b 1 ayb D .ay/1 and so a1 y b D a1 y which gives y b D y and y 2 hzi. If y D 1, then G D hei M . We set ae D az , where D 0; 1. The group G is completely determined. Theorem 52.5. Let G be a 2-group of order > 24 satisfying j2 .G/j D 24 . Suppose, in addition, that G has a normal elementary abelian subgroup E of order 8 such that G=E Š Q2n , n 3, and E ˆ.G/. Then one of the following holds: (a)

n

n1

G D ha; b j a2 D b 8 D 1; a2 b

1

2

2

a Da

n2

D b 4 D z; b 2 D a2

e;

ue ; D 0; 1;

u D e D Œe; u D Œe; z D Œu; z D 1; e a D ez; ua D u; e b D ez; ub D uz; n 3; and if D 1; then n 4i: Here G is a group of order 2nC3 , G 0 D hue i ha2 i is abelian of type .2n1 ; 2/, ˆ.G/ D Eha2 i, where E D hz; e; ui is a normal elementary abelian subgroup of order 8 in G with G=E Š Q2n . Finally, Z.G/ D hzi is of order 2, and 2 .G/ D n2 Eha2 i is abelian of type .4; 2; 2/. n

n1

(b) G D ha; b j a2 D b 8 D 1; n 3; a2 n2

b 2 D a2

D b 4 D z;

u; ab D a1 e;

u2 D e 2 D Œe; u D Œe; z D Œu; z D Œa; e D Œa; u D Œb; e D 1; ub D uzi: Here G is a group of order 2nC3 , G 0 D ha2 ei is cyclic of order 2n1 , ˆ.G/ D Eha2 i, where E D hz; e; ui is a normal elementary abelian subgroup of order 8 in G with G=E Š Q2n and CG .E/ D Ehai. Finally, Z.G/ D he; b 2 i, and 2 .G/ D Ehb 2 i is abelian of type .4; 2; 2/. Proof. We have in this case ˆ.G/ WQ D 2 .G/, WQ =E D Z.G=E/ and WQ is abelian of type .4; 2; 2/ since CG .E/ > E (a Sylow 2-subgroup of Aut.E/ is isomorphic to D8 ) and jWQ j D 24 . Hence hzi D Ã1 .WQ / is a central subgroup of order 2 contained in E. By Lemma 52.3, since every normal cyclic subgroup of G=E is contained in a cyclic subgroup of index 2 in G=E, there is a subgroup M of index 2 in G such that M=E is cyclic and CG .E/ M . (If n > 3, then M is unique since G=E Š Q2n has only one cyclic subgroup of index 2. However, if n D 3

52 2-groups with 2 -subgroup of small order

83

and CG .E/ D WQ , then we have three possibilities for M since G=E Š Q8 has exactly three cyclic subgroups of index 2.) Let a be an element in M E such that ha; Ei D M . Then o.a/ D 2n since jM=Ej D 2n1 and M does not split over E n2 (indeed, all involutions of G lie in E). Hence a0 D a2 2 WQ E, a02 D z 2 E. By the structure of G=E Š Q2n , all elements in .G=E/ .M=E/ have order 4, so we have for each x 2 G M , x 2 2 WQ E, x 4 D z, x induces an involutory automorphism on E since CG .E/ M and x 62 M , and so CE .x/ Š E4 (see Exercise 19). CE .x/ > hzi, and CWQ .x/ D hx 2 iCE .x/ is abelian of type .4; 2/. It follows that all elements in G M have the same order 8. We have M 0 < E (recall that M=E is cyclic) so jM 0 j 4, and M 0 D Œhai; E. For each ai e 2 M , where e 2 E and i is an integer, we get .ai e/2 D ai eai e D a2i .ai eai /e D a2i Œai ; e and so ˆ.M / D Ã1 .M / D ha2 iM 0 . We have exactly the following four cases for the action of the cyclic subgroup hai of order 2n on E: (1) a induces an automorphism of order 4 on E. (2) a induces an automorphism of order 2 on E and ŒE; a D hui, where u 2 E hzi. (3) a induces an automorphism of order 2 on E and ŒE; a D hzi. (4) a centralizes E (in that case, obviously, M is abelian). Case (1). Since a induces an automorphism of order 4 on E, we get n 4 (recall n2 that a2 centralizes E) and CE .a/ D hzi (see, in the first paragraph, the proof of the equality CG .x/ Š E4 ). In this case M 0 is a four-subgroup contained in E, M 0 > hzi (if hzi 6 M 0 , then hai \ M 0 D f1g and a centralizes some element in .M 0 /# , which is not the case) and for each eQ 2 E M 0 , eQ a D eQ uQ with some uQ 2 M 0 hzi and uQ a D uz. Q We see that ha; M 0 i Š M2nC1 so CE .a2 / D M 0 (this is true again since a2 induces an involutory automorphism on E). Since jG W CG .M 0 /j D 2, it follows that CG .M 0 / D Eha2 ; b 0 i, where b 0 2 G M and Eha2 ; b 0 i is a maximal subgroup of G. Set b D b 0 a so that uQ b D uz Q for each uQ 2 M 0 hzi and therefore CE .b/ D he; zi 0 for some e 2 E M (see the proof of Theorem 52.4), e a D eu with u 2 M 0 hzi, ua D uz, and ub D uz. Since G=E Š Q2n , b 1 ab D a1 y with y 2 E. We have two subcases according to y 2 E M 0 or y 2 M 0 . Subcase (1a). Here y D e 0 2 E M 0 and so b 1 ab D a1 e 0 . We compute ab D .ab /b D .a1 e 0 /b D .a1 e 0 /1 .e 0 /b D e 0 a.e 0 /b D a.e 0 /a .e 0 /b D au: 2

Indeed, if e 0 D ez . D 0; 1/, then .e 0 /a D e 0 u, .e 0 /b D e 0 , and if e 0 D euz . D 0; 1/, then .e 0 /a D e a ua z D euuzz D e 0 uz, .e 0 /b D e 0 z. On the other hand, 2 b 2 D a0 x with x 2 E. But M 0 hai D hu; ai is isomorphic to M2nC1 and so ab D au D aa0 x D ax implies that x 2 E M 0 . We may set x D eu0 with u0 2 M 0 and so b 2 D a0 eu0 . We compute b 1 a2 b D .a1 e 0 /2 D a1 e 0 a1 e 0 D .a1 e 0 a/a2 e 0 D .e 0 /a .e 0 /a a2 D u0 za2 ; 2

84

Groups of prime power order

since .e 0 /a D e 0 u0 .u0 2 M 0 hzi/ and .e 0 /a D e 0 z. Therefore .a4 /b D ..a2 /b /2 D n2 .u0 za2 /2 D a4 because CE .a2 / D M 0 . Since a0 D a2 2 ha4 i (recall that b 1 2 n 4), we get a0 D a0 . Obviously, b centralizes b D a0 eu0 .u0 2 M 0 / and so a0 eu0 D .a0 eu0 /b D a01 e.u0 /b D a0 ze.u0 /b which gives .u0 /b D u0 z and u0 2 M 0 hzi. Therefore u0 D uz . D 0; 1/ and b 2 D a0 euz . On the other hand, 2 au D ab D aa0 euz D aeu D a.a1 eua/eu D a.eu/.uz/eu D auz, which is a contradiction. Subcase (1b). Here b 1 ab D a1 y with y 2 M 0 . This gives .a2 /b D .a1 y/2 D 2 a1 ya1 y D .a1 ya/a2 y D y a y a a2 D yz ya2 D z a2 . D 0; 1/, and so .a4 /b D .z a2 /2 D a4 . Since n 4, we get a0 2 ha4 i so a0b D a01 . If b 2 2 M 0 ha0 i, then S D M 0 ha; bi is a maximal subgroup of G not containing E since S \ E D M 0 , a contradiction (recall, that, by hypothesis, E ˆ.G/). Thus b 2 D a0 e0 with e0 2 E M 0 . Since b centralizes b 2 , we get a0 e0 D .a0 e0 /b D a01 e0b D a0 ze0b and so e0b D e0 z. Hence e0 D euz . D 0; 1/ and b 2 D a0 euz . From b 1 ab D a1 y with y 2 M 0 , we get b 2 ab 2 D .a1 y/b D .ab /1 y b D yay b D a.a1 ya/y b D a.yz /.yz / D a, where D 0 if y 2 hzi and D 1 if y 2 M 0 hzi. On the other hand, we compute 2

a D ab D aa0 euz D aeu D euaeu D a.a1 eua/eu D a.eu/.uz/eu D auz; 2

which is a contradiction. Case (2). Here, by assumption, a induces an automorphism of order 2 on E and M 0 D ŒE; a D hui, where u 2 E hzi. It follows hu; zi Z.G/. For each eQ 2 E hu; zi, eQ a D eu. Q Since ˆ.M / D hui ha2 i and ˆ.G/ E, there exists n2 b 2 G M such that b 2 D a0 e, where e 2 E hu; zi and a0 D a2 . We have ab D a1 y with y 2 E (see the last two sentences of the first paragraph of Case 2 (1)). This gives ab D .a1 y/b D .ab /1 y b D yay b D ay a y b D aa0 e D ae D eae D a.a1 ea/e D aeue D au. Hence y a y b D u. If y 2 hu; zi Z.G/, then u D y a y b D y 2 D 1, a contradiction. Therefore y D es with s 2 hu; zi. But then u D y a y b D .es/a .es/b D .eus/.e b s/ D uee b and so e b D e. Hence b centralizes E D he; u; zi, a contradiction. Case (3). Here a induces an automorphism of order 2 on E and M 0 D ŒE; a D hzi. It follows ˆ.M / D ha2 i ha0 i and so ha0 i is normal in G since the cyclic subgroup ˆ.M / is G-invariant. Since Ã1 .G/ D ˆ.G/ WQ D Eha0 i, it follows ha0 ; x 2 j x 2 G M i D WQ . Since hx 2 i is normal in G for each x 2 G M and x 4 D z (indeed, M=M 0 D M=hzi is an abelian maximal subgroup of G=M 0 ), we get WQ =hzi Z.G=hzi/. Set U D CE .a/ Š E4 and obviously Z.M / D U ha2 i. Also, U D Z.M / \ E is normal in G. Suppose that U Z.G/. Then the fourgroup G=Eha2 i acts as the full stability group of the chain E > U > f1g since G=Eha2 i acts faithfully on E. (Note that CG .E/ M .) This is a contradiction since E=hzi < WQ =hzi Z.G=hzi/. Hence for each x 2 G M and each uQ 2 U hzi, we have uQ x D uz. Q There exists b 2 G M such that b 2 D a0 e with e 2 E U .

52 2-groups with 2 -subgroup of small order

85

Also, ab D a1 y with y 2 E. Then we compute ab D .a1 y/b D yay b D a.a1 ya/y b D ay a y b D aa0 e D eae D a.a1 ea/e D aeze D az. This gives 2

(1) y a y b D z. On the other hand, b centralizes b 2 and so a0 e D .a0 e/b D a0b e b which gives (2) a0 e D a0b e b . We compute further b 1 a2 b D .a1 y/.a1 y/ D .a1 ya/.a2 ya2 /a2 D y a ya2 , and so we get (3) .a2 /b D y a ya2 . There are two subcases according to y 2 U or y 2 E U . Subcase (3a). Suppose that y 2 U D CE .a/. Then (1) implies y b D yz and so y D u 2 U hzi and b 1 ab D a1 u. From (3) we get .a2 /b D y a ya2 D yya2 D a2 and so a0b D a01 since a0 is a power of a2 . From (2) follows e b D a0b a0 e D a02 e D ez and we have determined the group stated in part (a) of our theorem for D 0. Subcase (3b). Suppose that y 2 E U , where ab D a1 y. From (1) we get yzy b D z and so y b D y. The relation (3) implies .a2 /b D a2 z. We compute, for each x 2 E and each integer i , .b.a2 /i x/2 D b 2 z and .ba.a2 /i x/2 D b 2 yz , where ; D 0; 1. Hence the square of each element in G M lies in hb 2 i D ha0 ei and in ha0 eyi D hb 2 yi. Since ˆ.G/ E, we must have y D eu with some u 2 U hzi. From y b D y follows eu D e b uz (since ub D uz) and so e b D ez. Then (2) implies a0 e D a0b ez and so a0b D a0 z D a01 . From .a2 /b D a2 z follows that ha2 i > ha0 i which gives n 4. We get the group stated in part (a) with D 1 and n 4. Case (4). Here a centralizes E and so M is abelian of type .2n ; 2; 2/, n 3; and n2 ha0 i is normal in G, where a0 D a2 . Since ˆ.G/ WQ D Eha0 i, there exists b 2 G M such that b 2 D a0 u, where u is an involution in E hzi (recall that all elements in G M have the same order 8). Indeed, if such b does not exist, we get b 2 D a0 or a03 for all b 2 G M . Then ˆ.G/ D Ã1 .G/ E, a contradiction. Since CG .b 2 / hM; bi D G, we get b 2 2 Z.G/, ha0 ; b 2 i D ha0 i hui is abelian of type .4; 2/ and ˆ.G/ ha2 ; ui. Again ˆ.G/ WQ and ha2 ; b 2 i WQ implies that there exists c 2 G M such that c 2 2 WQ ha0 ; b 2 i and (as before) c 2 2 Z.G/. If a0 2 Z.G/, then ha0 ; b 2 ; c 2 i D WQ lies in Z.G/ contrary to Lemma 52.3. Hence a0 62 Z.G/ and therefore a0b D a01 and ub D uz (recall that u D a01 b 2 62 Z.G/). The subgroup Z.G/ D hb 2 ; c 2 i is abelian of type .4; 2/. Lemma 1.1 implies jGj D 2nC3 D 2jZ.G/jjG 0 j and so jG 0 j D 2n1 . Since j.G=E/0 j D 2n2 and ˆ.G/ > G 0 > hzi, G 0 is cyclic of order 2n1 , by the modular law. Set ab D a1 e with e 2 E. If e 2 ha0 ; ui D ha0 ; b 2 i, then ha; bi, a metacyclic maximal subgroup of G, does not contain E, a contradiction. Thus e 2 E hz; ui and we compute .ba/2 D baba D b 2 .b 1 ab/a D b 2 .a1 e/a D b 2 e, Since both b 2 and b 2 e lie in Z.G/, it follows e 2 Z.G/ and therefore Z.G/ D he; b 2 i. Finally, G 0 D ha2 ei is cyclic of order 2n1 . We get the group from part (b).

86

Groups of prime power order

3o . 2-groups with exactly one subgroup of order 24 and exponent 4. Theorem 52.6. The following two statements for a 2-group G of order > 24 are equivalent: (a) j2 .G/j D 24 and (b) G has exactly one subgroup of order 24 and exponent 4. Proof. Suppose that (a) holds. The results of Subsections 1 and 2 imply that 2 .G/ is isomorphic to one of the following groups Q8 C4 , Q8 C2 , D8 C2 , C4 C2 C2 , C4 C4 . In particular, exp.2 .G// D 4 and so .b/ holds. Suppose now that (b) holds. Let H be the unique subgroup of order 24 and exponent 4 in G, where jGj > 24 . We want to show that H D 2 .G/. Suppose that this is false. Then there exist elements of order 4 in G H , where H is a characteristic subgroup of G. In particular, there is an element a 2 G H such that o.a/ 4 and a2 2 H . Set HQ D H hai so that jHQ j D 25 . Since H is neither cyclic nor a 2-group of maximal class, there exists a G-invariant four-subgroup W0 contained in H , by Lemma 1.4. Let x be any element in HQ H with o.x/ 4. Then hx; W0 i is a subgroup of order 24 in HQ and so there exists a maximal subgroup M of HQ containing hx; W0 i. Since exp.M \ H / 4 and M D hM \ H; xi, we have 2 .M / D M . But jM j D 24 and M ¤ H , so M is either elementary abelian or exp.M / D 8. Suppose that exp.M / D 8. Then M is a nonabelian group of order 24 with a cyclic subgroup of index 2. Since M has the normal four-subgroup W0 , it follows that M is not of maximal class. Thus M Š M24 . But 2 .M24 / is abelian of type .4; 2/ which contradicts the above fact that 2 .M / D M . We have proved that M Š E24 . In particular, x must be an involution and exp.HQ / D 4 because for each y 2 HQ , y 2 2 M . Hence all elements in HQ H are of order 4 and so (by the above) all these elements must be involutions. In that case, every involution x 2 HQ H inverts H . Let k be any element of order 4 in H . Since k x D k 1 , hk; xi D D Š D8 . Let MQ be a maximal subgroup of HQ containing D. Then jMQ j D 24 , exp.MQ / D 4, MQ ¤ H , and this is a final contradiction. 4o . Generalization of Theorem 43.4. Here we improve Theorem 43.4 and some results of [Ber25, 48]. In Theorem 52.7 we classify the 2-groups with exactly on abelian subgroups of type .4; 2/; the proof is independent of the proof of Theorem 43.4. The following remark shows that there are few 2-groups without abelian subgroups of type .4; 2/. Remarks. 1. Let us classify the 2-groups without abelian subgroup of type .4; 2/. Of course, if G is cyclic, elementary abelian or of maximal class, it has no abelian subgroups of type .4; 2/. So in what follows suppose that G is of exponent > 2 and is neither cyclic nor of maximal class. Let A be a cyclic subgroup of order 4 in G. Then CG .A/ A has no involutions. It follows that CG .A/ is cyclic (Proposition 1.3). By Suzuki’s theorem (Proposition 1.8), jCG .A/j > 4. By Lemma 1.4, G has a normal abelian subgroup R of type .2; 2/. Since Z.G/ < CG .A/, we get R \ CG .A/ is of

52 2-groups with 2 -subgroup of small order

87

order 2. It follows that H D RCG .A/ Š M2n for some n > 3. Let B be a subgroup of order 4 in Z.H /. In that case, RB is abelian of type .4; 2/, which is not the case. Thus, our G is elementary abelian, cyclic or of maximal class. 2. Let G be a p-group without abelian subgroup of type .p 2 ; p/, p > 2, exp.G/ > p. Then, as in Remark 1, we obtain exp.G/ D p 2 and CG .x/ D hxi so G is of maximal class (Proposition 1.8). In particular, jGj p 2p1 (Theorem 9.6). We claim that jGj p pC1 . Assume that this is false. Let G1 be a fundamental subgroup of G. Then exp.G1 / > p and Z.G1 / is noncyclic. In that case, obviously, G1 contains an abelian subgroup of type .p 2 ; p/, a contradiction. Thus, jGj p pC1 . It is easy to check that a Sylow p-subgroup of the symmetric group of degree p 2 has no abelian subgroups of type .p 2 ; p/. Theorem 52.7. Let G be a 2-group containing exactly one abelian subgroup of type .4; 2/. Then one of the following holds: (a) j2 .G/j D 8 and G is isomorphic to one of the metacyclic groups (a), (b) or (c) in Lemma 42.1. (b) G Š C2 D2nC1 , n 2. D t 2 D 1; b t D b 1C2 u; u2 D Œu; t D 1; b u D (c) G D hb; t j b 2 n n 1C2 b ; n 2i. Here jGj D 2nC3 , Z.G/ D hb 2 i is of order 2, ˆ.G/ D n hb 2 ; ui < hb; ui Š M2nC2 , E D hb 2 ; u; t i Š E8 is self-centralizing in G, n 2 2 .G/ D hui hb ; t i Š C2 D2nC1 , G 0 D hb 2 ; ui Š E4 in case n D 2, and G 0 D hb 2 ui Š C2n for n 3. Finally, the group G for n D 2 (of order 25 ) is isomorphic to the group (a) in Theorem 52.2 for n D 2 (since 2 .G/ Š C2 D8 ). nC1

n1

Proof. Let G be a 2-group with exactly one abelian subgroup A of type .4; 2/; then A is characteristic in G. Set C D CG .A/. Then C is normal in G and G=C is isomorphic to a subgroup of Aut.A/ Š D8 . We claim that 2 .C / D A. Indeed, let y 2 C A such that o.y/ 4 and y 2 2 A. Then Ahyi is an abelian subgroup of order 24 and exponent 4. But then Ahyi contains an abelian subgroup of type .4; 2/ distinct from A, a contradiction. Thus 2 .C / D A, as claimed. Lemma 42.1 implies that C must be abelian of type .2n ; 2/, n 2. If 2 .G/ D 2 .C / D A, then G is metacyclic and G is isomorphic to one of the groups in Lemma 42.1. We assume from now on that 2 .G/ > A. Set U D 1 .A/ and hzi D ˆ.A/ so that z 2 Z.G/ and U is a normal four-subgroup of G. Let a 2 G C be such that o.a/ 4 and a2 2 C . Obviously, a2 2 U since U D 1 .C /. Let D D haiC ; then jD W C j D 2. Since D is a nonabelian group (of order 24 ) containing a normal four-subgroup U , it follows that D is not of maximal class. If o.a/ D 4, then a does not centralize U (otherwise U hai would be abelian of type .4; 2/ distinct from A) and so U hai Š D8 . But then there exists an involution in .U hai/U . In any case the coset C a . Ua/ contains involutions and let t be one of them. If t centralizes an element s of order 4 in C , then hs; t i is an abelian subgroup of type .4; 2/ distinct from A, a

88

Groups of prime power order

contradiction. Hence CC .t / is of exponent 2 and (since D is not of maximal class) CC .t / D U (Proposition 1.8) and E D CD .t / D U ht i Š E8 . It is easy to check that, for each x 2 D C D C t , CC .x/ D U . Hence x 2 2 U and hU; xi is elementary abelian (of order 8). Thus, all elements in D C are involutions and therefore t inverts C . Let B be a cyclic subgroup of index 2 in C and an involution u 2 C B. Then hB; t i Š D2nC1 and D Š C2 D2nC1 , n 2, where C2 D hui. Obviously, we may choose a 2 G C so that D D ha; C i G G. Suppose that G=C has a cyclic subgroup K=C of order 4. Then K > D since D=C , in the case under consideration, is contained in ˆ.G=C /, and let k 2 K be such that K D hk; C i. We have k 2 2 D C and so k 2 is an involution. It follows o.k/ D 4 and so hk; zi is an abelian subgroup of type .4; 2/ distinct from A since hki \ C D f1g, and this is a contradiction. We have proved that G=C is elementary abelian. If jG=C j D 2, then G D D and we are done. It remains to study the possibility G=C Š E4 . By the above, if y 2 G D and o.y/ 4, then y 2 2 C and y is an involution inverting C (indeed, as in the second paragraph of the proof, all elements in hy; C i C are involutions). But then yt 62 C and yt centralizes C , a contradiction (C D CG .A/ is self centralizing in G). Hence, for each element y 2 G D, o.y/ 8 and y 2 2 C . This gives 2 .G/ D D, CG .t / D E D ht i1 .A/ and so E is a self-centralizing elementary abelian subgroup of order 8 in G. For each y 2 G D, we have y 2 2 C and o.y 2 / 4 and so y centralizes a cyclic subgroup of order 4 in A. Since y does not centralize A, it follows that y does not centralize U D 1 .A/ since y 62 C D CG .A/. Hence D D CG .U / and so Z.G/ D hzi D ˆ.A/ is of order 2. Since 2 .hyiC / D A and hyiC is nonabelian, Lemma 42.1 implies that hyiC is isomorphic to a group (a) or (c) of that lemma. Assume that there is b 2 G D such that hbiC is isomorphic to a group of Lemma 42.1(c). In particular, jhbiC j 25 , C is the unique abelian maximal subgroup of hbiC (since Z.hbiC / Š C4 ), o.b/ D 8, Z.hbiC / D hb 2 i < A, hb 4 i D hzi D ˆ.A/, C contains a cyclic subgroup hai of index 2 such that hai is normal in hbiC , o.a/ 23 , A \ hai Š C4 , hbi \ hai D hzi, and ab D a1 . Also, we see that for each y 2 .hbiC / C , CC .y/ D hb 2 i Š C4 , and y is of order 8. Since t inverts C , we get abt D a and so jCC .bt /j 23 . This implies that hbt iC Š M2nC2 because CC .bt / Z.hbt iC /. Replacing b with bt (if necessary), we may assume from the start that there is an element b 2 G D such that hbiC Š M2nC2 .n 2/, which is a group in part (a) of Lemma 42.1. We have o.b/ D 2nC1 , hbi is a cyclic subgroup of index 2 in hbiC , n n z D b 2 , U D hz; ui D 1 .hbiC /, and ub D uz. This gives also b u D bz D b 1C2 , n1 D v; ui, C D hb 2 i hui. We see that Z.hbiC / D hb 2 i Š C2n , A D hb 2 CC .bt / D hvui Š C4 and so Z.hbt iC / D hvui. It follows that hbt iC Š M24 for n D 2 and hbt iC is isomorphic to a group (c) of Lemma 42.1 for n > 2. In any case, .bt /2 2 hvui hzi and so .bt /2 D vu or .bt /2 D v 1 u D vuz. Replacing u with uz n1 (if necessary), we may assume that .bt /2 D vu, where v D b 2 . From this crucial

52 2-groups with 2 -subgroup of small order

relation follows b t D b 1C2 theorem is proved.

n1

89

u. The structure of G D hb; t i is determined and our

Theorem 52.8. The following two statements for a 2-group G are equivalent: (a) c2 .G/ D 4; (b) G has exactly one abelian subgroup of type .4; 2/. Proof. Suppose that G is a 2-group having exactly two cyclic subgroups U; V of order 4. Then jG W NG .U /j 2 and so V NG .U /. Similarly, U NG .V /. We get ŒU; V U \ V . Since A D hU; V i has exactly two cyclic subgroups of order 4, A must be abelian of type .4; 2/. But A is generated by its two cyclic subgroups of order 4 and so (b) holds. Let (b) hold. Then the group G is completely determined by Theorem 52.7. Looking at 2 .G/, we see that G has exactly two cyclic subgroups of order 4. Exercise 1. Let G be a 2-group, exp.G/ 2n with n > 1. Let H < G contains all cyclic subgroups of G of order 2n . If x 2 H with o.x/ D 2n , then n1 .CG .x// H. Exercise 2. Let G be a 2-group with 2 .G/j 25 . Prove that dl.G/, the derived length of G, is at most 5. Solution. If G has no normal elementary abelian subgroups of order 8, then the dl.G/ 4, by 50. Now let E be a normal elementary abelian subgroup of G of order 8. Then j1 .G=E/j 4 so dl.G=E/ 4, and we are done. Exercise 3. Let G be a 2-group and n > 3. Denote by Mn the set of noncyclic subgroups of order 2n in G which are abelian of type .2n1 ; 2/ or M2n . Study the structure of G in the case jMn j D 1. Exercise 4. Study the 2-groups without abelian subgroups (i) of type .4; 4/, (ii) of type .4; 2; 2/. Exercise 5. Suppose that a noncyclic p-group G has no abelian subgroups of type .p 2 ; p/. If exp.G/ > p, then either G is a 2-group of maximal class or p > 2 and G is of maximal class and order p pC1 . Exercise 6. Let G be a noncyclic p-group of order > p 4 and exponent > p, p > 2. Prove that if G has no subgroups of order p 4 and exponent p 2 , then G has a cyclic subgroup of index p. 5o . 2-groups G with j3 .G/j D 25 . The ultimate goal is to classify finite 2groups G of order > 26 with j3 .G/j D 26 (see #525 in Research problems and themes I). However, if 3 .G/ 6 ˆ.G/, there is a maximal subgroup M of G such that j3 .M /j 25 and without the exact knowledge of the structure of M there is no chance to determine G in that case. All results of this subsection are due to Z. Bozikov [Boz].

90

Groups of prime power order

In this subsection we determine up to isomorphism all 2-groups G with j3 .G/j 25 and G > 3 .G/. The cases j3 .G/j D 23 and j3 .G/j D 24 are easy (Proposition 52.9). If j3 .G/j D 25 , we show first that exp.3 .G// D 23 and j2 .G/j D 24 (Proposition 52.10). This allows us to use the classification of 2-groups G with j2 .G/j D 24 and jGj > 24 given in first two subsections. Indeed, in Theorem 52.11 we consider nonmetacyclic 2-groups G and get our result almost immediately. In Theorem 52.12 we consider metacyclic 2-groups G with the above property. This case is difficult since in 50 there is no single information for metacyclic groups but we have to determine the groups G up to isomorphism also in this case. In conclusion, we classify the 2-groups G containing exactly one subgroup of order 5 2 and exponent 8 (see #437 in Research problems and themes I). We show that such groups satisfy j3 .G/j D 25 . In fact, we get a more general result (Theorem 52.14). 5.1o . For the convenience we prove two preliminary results (Propositions 52.9 and 52.10). Proposition 52.9. Let G be a 2-group with G > 3 .G/. If j3 .G/j 23 , then j3 .G/j D 23 and G is cyclic. If j3 .G/j D 24 , then 3 .G/ is abelian of type .8; 2/ and G is either abelian of type .2m ; 2/ or G Š M2mC1 , m 4. Proof. Set 3 .G/ D H . Assume that G is noncyclic. If G is of maximal class, then H 2 .G/ D G, which is a contradiction. Therefore, G has a normal abelian subgroup R of type .2; 2/. By hypothesis, H=R is cyclic of order 4 so G=R cyclic. It follows that R D 1 .G/ so G contains a cyclic subgroup of index 2. Now the result follows from Theorem 1.2. Proposition 52.10. Let G be a 2-group with G > 3 .G/. If j3 .G/j D 25 , then exp.3 .G// D 23 and j2 .G/j D 24 . Proof. Set 3 .G/ D H so that jH j D 25 and G > H . If H is cyclic, then 3 .H / < H , a contradiction. If H is of maximal class so G is since c1 .G/ D c1 .H / (see Theorem 1.17); then H 2 .G/ D G, a contradiction. If exp.H / D 24 , then (Theorem 1.2) H is either abelian of type .16; 2/ or H Š M32 . In any case, 3 .H / < H which is again a contradiction. We have proved that exp.H / D 23 (since exp.H / 22 is obviously impossible). Since H is neither cyclic nor of maximal class, there is a G-invariant four-subgroup R contained in H . We assume (by way of contradiction) that H D 2 .G/. Then 2 .G=R/ D H=R is of order 23 and jG=Rj > 23 . Lemma 42.1 implies that H=R is abelian of type .4; 2/. In particular, H 0 R and H 0 is elementary abelian. If the class cl.H / of H is 2, then for each x; y 2 H with o.x/ 4, o.y/ 4, we get .xy/4 D x 4 y 4 Œy; x6 D 1, which implies that exp.H / 4, a contradiction. Thus (since H is not of maximal class) cl.H / D 3 and so H 0 D R and R 6 Z.H /. We set C D CH .R/ and D D CG .R/. Since jH W C j D 2, D covers G=H and G D HD.

52 2-groups with 2 -subgroup of small order

91

If exp.C / 4, then taking x 2 D C such that x 2 2 C , we get o.x/ 8, a contradiction. It follows exp.C / D 8. Let a 2 C with o.a/ D 8. Then C D Rhai, jR \ haij D 2, and so C is abelian of type .8; 2/. Since 3 .D/ D C , our Proposition 52.9 implies that D is either abelian of type .2m ; 2/ or D Š M2mC1 , m 4. In any case, ˆ.D/ is cyclic of order 2m1 and so ˆ.D/ contains a characteristic cyclic subgroup Z of order 8. It follows that Z C and Z is normal in G. But jH=Zj D 4 and so H 0 Z. This contradicts the fact that H 0 D R and jR \ Zj D 2. We have proved that 2 .G/ < H . On the other hand, R < 2 .G/ and so j2 .G/j D 8 or 16. Set K D 2 .G/ and assume jKj D 8. Then Lemma 42.1 shows that either j3 .G/j D 24 (in cases (a) or (b)) or 3 .G/ D G (in case (c)). This is a contradiction and so jKj D 16 and we are done. 5.2o . Now let G be a nonmetacyclic 2-group of order > 25 with j3 .G/j D 25 . We set H D 3 .G/ and apply Proposition 52.10. We see that exp.H / D 23 and 2 .G/ D K is of order 24 so that jH W Kj D 2. By Theorem 41.1, K is nonmetacyclic. We are in a position to use the first part of this section. Since jGj > 25 , we get either K Š Q8 C4 or K Š C4 C2 C2 . All elements in H K are of order 8. Assume first that K D Q Z, where Q Š Q8 , Z Š C4 , and Q \ Z D Z.Q/. All elements in K .Q [ Z/ are involutions and so Q is the unique quaternion subgroup in K. Thus Q is normal in G and so also C D CG .Q/ is normal in G. Since jG=.QC /j 2 and jGj > 25 , we have jC j 23 and C \ K D Z. Also, 2 .C / D Z and so Lemma 42.1 implies that C is cyclic and H D Q 3 .C /. Suppose that jG=.QC /j D 2. Since G=C Š D8 , there is an element c 2 G .QC / such that c 2 2 C . Set P D C hci. Again, 2 .P / D Z and so P D hci is cyclic. But hci induces on Q an “outer” involutory automorphism and so we may set Q D ha; bi so that ac D a1 and b c D ba. Assume now K Š C4 C2 C2 so that E D 1 .K/ is a normal elementary abelian subgroup of order 8 in G. We have Ã1 .K/ D hzi is a central subgroup of order 2 in G. Obviously, K=E is the unique subgroup of order 2 in G=E and so G=E is either cyclic (of order 8) or generalized quaternion. Suppose that G=E is generalized quaternion and let L=E be a cyclic subgroup of index 2 in G=E. For each x 2 G L, x 2 2 K E and so o.x/ D 8 which forces 3 .G/ D G, a contradiction. We have proved that G=E is cyclic. Let a 2 G K be such that hai covers G=E. Then hai is cyclic of order 2m , m 4, G D Ehai, K \ hai Š C4 , and E \ hai D hzi. We may set E D he; u; zi so that we have the following possibilities for the action of hai on E: (i) e a D eu, ua D uz (here a induces an automorphism of order 4 on E); (ii) e a D e, ua D uz (here G D hei ha; ui Š C2 M2mC1 ; (iii) e a D eu, ua D u (here G is minimal nonabelian); (iv) ŒE; a D 1 (here G is abelian of type .2m ; 2; 2/).

92

Groups of prime power order

We have proved the following result. Theorem 52.11 ([Boz]). Let G be a non-metacyclic 2-group of order > 25 with j3 .G/j D 25 . Then one of the following holds. (a) G D QP , where Q D ha; bi is a normal quaternion subgroup of G, P D hci is cyclic of order 2m , m 4, Q \ P D Z.Q/ and c either centralizes Q or ac D a1 and b c D ba. (b) G D EP , where E D he; u; zi is a normal elementary abelian subgroup of order 8, P D hai is cyclic of order 2m , m 4, and E \ P D hzi with z D m1 . For the action of hai on E we have one of the following possibilities: a2 (i) e a D eu, ua D uz, and here a induces an automorphism of order 4 on E; (ii) e a D e, ua D uz, and here G Š C2 M2mC1 ; (iii) e a D eu, ua D u, and here G is minimal nonabelian; (iv) e a D e, ua D u, and here G is abelian of type .2m ; 2; 2/. 5.3o . Now we assume that G is a metacyclic 2-group of order > 25 with j3 .G/j D We set H D 3 .G/. By Proposition 52.10, exp.H / D 23 and 2 .G/ D K is of order 24 . The first part of this section implies that K Š C4 C4 and all elements in H K are of order 8. It follows that R D 1 .K/ is a normal 4-subgroup. Suppose that CH .R/ D K. Since CG .R/ covers G=H , there is an element x 2 CG .R/ H with x 2 2 K. But then o.x/ 8, a contradiction. We have proved that R Z.H /. Let Y =K be a subgroup of order 2 in G=K. Since Y 3 .G/, we get Y D H . Hence H=K is the unique subgroup of order 2 in G=K. This implies that G=K is either cyclic (of order 4) or generalized quaternion. We study first the case, where G=K is cyclic. Let a 2 G K be such that hai covers G=K. Then hai \ H Š C23 , hai \ K Š C22 , and hai \ R D hzi Z.G/. Thus a is of m3 m2 order 2m , m 4, and we set s D a2 and v D a2 so that s 2 D v, v 2 D z, and o.s/ D 8. Set C D CG .K/. Since G=C is cyclic and acts faithfully on K, we have jG=C j 4. For the structure of Aut.C4 C4 / see Proposition 50.5. Also, G 0 K and G 0 is cyclic which implies that jG 0 j 4. If G 0 D f1g, then G is abelian of type .2m ; 4/, m 4. Suppose jG 0 j D 2. Then Exercise 10.13 implies that G is minimal nonabelian. Since jGj > 8, Exercise 1.8a together with the fact that j3 .G/j D 25 gives G D m m1 ha; b j a2 D b 4 D 1; m 4; ab D a1C2 or b a D b 1 i. 0 Suppose jG j D 4 and jG W C j D 2. Then a induces an involutory automorphism on K, H D hK; si C , and so H is abelian of type .8; 4/. If a centralizes K=R, then G 0 R. But G 0 is cyclic and so jG 0 j 2, a contradiction. Hence a acts nontrivially on K=R and so a acts also non-trivially on R D Ã1 .K/. There is an element w 2 K R such that w 2 D u 2 R hzi, w a D w 0 , .w 0 /2 D uz, K D hwi hw 0 i, 2 and ua D uz. Since a induces an involutory automorphism on K, .w 0 /a D w a D w 25 .

93

52 2-groups with 2 -subgroup of small order

and therefore CK .a/ D hww 0 i D hvi with .ww 0 /2 D z. From w a w 1 D Œa; w 1 D w 0 w 1 D w 0 wu D ww 0 u follows that G 0 D hww 0 ui Š C4 , .ww 0 u/2 D z, and G 0 6 Z.G/. From the above, ww 0 D vz . D 0; 1/ and set l D swuC1 . Then l 2 D s 2 w 2 D vu D ww 0 uz 2 G 0 , .vu/2 D z, and so hli Š C8 is normal in G with hli \ hai D hzi. We compute l a D .swuC1 /a D sw 0 uC1 z C1 D s 1 s 2 w 0 uC1 z C1 D s 1 vw 0 uC1z C1 D s 1 .ww 0 z /w 0 uC1 z C1 D s 1 w 1 w 2 .w 0 /2 uC1 z D s 1 w 1 uuzuC1z D s 1 w 1 uC1 D l 1 : m

m1

D We have obtained the following metacyclic group G D ha; l j a2 D l 8 D 1, a2 m3 l 4 , l a D l 1 , m 4i, where H D hl; si D hl; a2 i is abelian of type .8; 4/. It remains to study here the case jG 0 j D 4 and jG W C j D 4. Hence a induces on K an automorphism of order 4 and so, in particular, a acts nontrivially on R. This implies that CK .a/ D hvi, where hv 2 i D hzi D CR .a/. We may set K D hwi hw 0 i, where w a D w 0 , w 2 D u 2 R hzi, .w 0 /2 D uz, ua D uz, and .w 0 /a D ws0 with s0 2 R. We have s0 ¤ 1 (otherwise a induces on K an involutory automorphism). Since .ww 0 /2 D w 2 .w 0 /2 D uuz D z, it follows that ww 0 2 Rhvi. We compute .ww 0 /a D w 0 ws0 D .ww 0 /s0 , which implies hww 0 i ¤ hvi and so hww 0 ui D hvi. Here we have used the fact that Rhvi contains exactly two cyclic subgroups of order 4 and they are hvi and hvui. This gives (since a centralizes v) ww 0 u D .ww 0 u/a D w 0 ws0 uz, and consequently s0 D z. Hence .w 0 /a D wz and so the action of hai on K is uniquely determined. We see that Œa; w 1 D w a w 1 D w 0 w 1 D w 0 wu D ww 0 u and so G 0 D hww 0 ui. But hww 0 ui D hvi hai and so hai is normal in G. Also, hwi induces on hai an automorphism of order 4 with hwi \ hai D 1 and jhai W Chai .w/j D 4. This determines uniquely the structure of our “splitting” metacyclic group G D m m2 m3 i, where H D hw; a2 i. For ha; w j a2 D w 4 D 1; m 4; aw D a1C2 m D 4, H is nonabelian and for m > 4, H is abelian of type .8; 4/. Here G 0 D m2 ha2 i Z.G/ D ha4 i. We have to study the difficult case, where G=K is generalized quaternion. Since a generalized quaternion group is not a subgroup of Aut.K/ (see Proposition 50.5), we have CG .K/ H and consequently H is abelian of type .8; 4/. Note that H=K D Z.G=K/. We set L=K D .G=K/0 D ˆ.G=K/ so that L=K is cyclic containing H=K. We have G=L Š E4 and since G is metacyclic (and so d.G/ D 2), we get L D ˆ.G/ and G 0 is cyclic. On the other hand, G 0 covers L=K and all elements in H K are of order 8. Hence G 0 \ H Š C8 , G 0 \ K Š C4 , and jL W G 0 j D 4. Again, since G is metacyclic, there is a cyclic normal subgroup S of G such that G 0 < S and jS W G 0 j D 2. If S L, then jS \ Kj D 8. But a maximal subgroup S \ K of K is not cyclic since exp.K/ D 4. This contradiction shows that S \ L D G 0 . Set T D KS D LS so that T =K is a cyclic subgroup of index 2 in G=K. For each

94

Groups of prime power order

x 2 G T , we have x 2 2 H K and so all elements in G T are of order 16. Now, ˆ.T / L and ˆ.T / hG 0 ; Ã1 .K/ D Ri since G 0 D Ã1 .S /. But jL W .G 0 R/j D 2 and T is noncyclic and so ˆ.T / D G 0 R. Then F D ˆ.T / \ H is abelian of type .8; 2/ and F1 D ˆ.T / \ K is abelian of type .4; 2/. m3 m2 , v D a2 , and Set S D hai, where o.a/ D 2m , m 4. Also, set s D a2 m1 2 z D a so that z 2 Z.G/, hsi D S \ H , hvi D S \ K, and hzi D S \ R. Obviously, hsi and hvi are normal subgroups in G. Since ˆ.G/ D L, there is an element b 2 G T (of order 16) such that b 2 2 H K and b 2 62 F . Then b 2 D ws, where w 2 K F1 and K D hwi hvi. We set w 2 D u and so R D 1 .K/ D hu; zi. We compute b 4 D w 2 s 2 D uv and b 8 D z. Hence G D haihbi, where hai is a cyclic normal subgroup of order 2m .m 4/ of G, hbi is cyclic of order 16, and hai \ hbi D hzi (of order 2). It remains to determine the action of hbi on hai. Now, hbi induces a cyclic automorphism group on hai so that b inverts hai=hvi since G=K is generalized quaternion. Thus ab D a1 v0 with v0 2 hvi and so .a4 /b D .ab /4 D .a1 v0 /4 D a4 . In particular, b inverts hvi. On the other hand, b centralizes b 4 D uv and 2 so uv D .uv/b D ub v b D ub v 1 , which gives ub D uz. We compute ab D .a1 v0 /b D .a1 v0 /1 v01 D av02 D az , D 0; 1. Since b 2 D sw, this gives 2 asw D aw D az and au D aw D .az /w D az z D a and so Œa; u D 1. Replacing a with au we get o.au/ D o.a/ D 2m and .au/b D a1 v0 uz D .au/1 v0 z. m2 Hence replacing a with au (if necessary), we may assume that v0 D v D a2 or 2m v0 D 1. We have obtained exactly two possibilities for our group G D ha; b j a D m1 m2 D b 8 ; ab D a1C2 ; D 0; 1i. The following result b 16 D 1; m 4; a2 was proved. Theorem 52.12 ([Boz]). Let G be a metacyclic 2-group of order > 25 with j3 .G/j D 25 . Then one of the following holds. (a) G is abelian of type .2m ; 4/, m 4. m

(b) G D ha; b j a2 nonabelian.

m1

D b 4 D 1; m 4; ab D a1C2

or b a D b 1 i is minimal

(c) G D ha; l j a2 D l 8 D 1; m 4; a2 D l 4 ; l a D l 1 i, where 3 .G/ D m3 hl; a2 i is abelian of type .8; 4/, G 0 D hl 2 i Š C4 , Z.G/ D ha2 i Š C2m1 , 0 and G 6 Z.G/. m

m1

m

m2

i; where 3 .G/ D (d) G D ha; w j a2 D w 4 D 1; m 4; aw D a1C2 m3 m2 i, G 0 D ha2 i Š C4 , Z.G/ D ha4 i, and so G 0 Z.G/. Also, for hw; a2 m D 4, 3 .G/ is nonabelian and for m > 4, 3 .G/ is abelian of type .8; 4/. (e) G D ha; b j a2 D b 16 D 1; m 4; a2 D b 8 ; ab D a1C2 ; D m3 2 2 0; 1i, where 3 .G/ D hb ; a i is abelian of type .8; 4/ and G=2 .G/ Š Q2m1 . If D 0, then jG W CG .2 .G//j D 2 and if D 1, then jG W CG .2 .G//j D 4. m

m1

m2

52 2-groups with 2 -subgroup of small order

95

5.4o . We prove here the following result. Theorem 52.13 ([Boz]). Let G be a 2-group containing exactly one subgroup of order 25 and exponent 8. Then we have j3 .G/j D 25 . In fact, with the similar proof, we get the following more general result. Theorem 52.14 ([Boz]). Let G be a 2-group containing exactly one subgroup of order 2nC2 and exponent 2n , where n 2 is a fixed integer. Then we have jn .G/j D 2nC2 . Proof. Let H be the unique subgroup of order 2nC2 and exponent 2n .n 2/. Then H is a characteristic subgroup of G and H n .G/. Suppose that the theorem is false. Then there exists an element a 2 G H of order 2n with a2 2 H . Set HQ D H hai so that jHQ j D 2nC3 . Since H is neither cyclic nor of maximal class, H has a G-invariant four-subgroup R. We have jRhaij 2nC2 and so there is a maximal subgroup M of HQ such that M Rhai. We have exp.M \ H / 2n and since M D .M \ H /hai, we get n .M / D M , jM j D 2nC2 , M ¤ H , and exp.M / 2nC1 . The uniqueness of H forces exp.M / D 2nC1 so M has a cyclic subgroup of index 2. But jM j 24 (since n 2) and M has the normal 4-subgroup R. This implies that M is not of maximal class and so M is either abelian of type .2nC1 ; 2/ or M Š M2nC2 . In both cases n .M / < M and this contradicts our result in the previous paragraph. The theorem is proved.

53

2-groups G with c2 .G / D 4

For a finite 2-group G and a fixed integer n 1 we denote with cn .G/ the number of cyclic subgroups of order 2n . The starting point is the following result: if a 2-group G is neither cyclic nor of maximal class and n > 1, then cn .G/ is even. Thus, for any 2-group G we have c2 .G/ D 1 if and only if G is either cyclic or dihedral and c2 .G/ D 3 if and only if G 2 fQ8 ; SD16 g. In this section we shall determine (up to isomorphism) all 2-groups G of order > 24 with c2 .G/ D 4. If, in addition, j2 .G/j D 24 , then such groups G have been determined in 52. In the sequel we assume that j2 .G/j > 24 . If moreover G has a quaternion subgroup, then we get an infinite class of 2-groups and they have the properties 2 .G/ D G and jZ.G/j D 2 (Theorem 53.6). Therefore we assume also in what follows that G has no quaternion subgroups. Then we show first that A D hU1 ; U2 ; U3 ; U4 i Š C4 C2 C2 , where fU1 ; U2 ; U3 ; U4 g is the set of four cyclic subgroups of order 4 in G (Theorem 53.7). It is interesting to note here that each subgroup Ui turns out to be normal in G. If in addition 2 .G/ D G, then G D Bht i, where B is abelian of type .2m ; 2; 2/, m 2, and t is an involution inverting B and here Z.G/ Š E8 (Theorem 53.8). If jG W 2 .G/j 4, then jG W 2 .G/j D 4 and G is a uniquely determined group of order 27 with jZ.G/j D 2 (Theorem 53.9). Finally, if jG W 2 .G/j D 2, then we get an infinite class of 2-groups G with the properties jGj 27 , G 0 6 1 .A/, and Z.G/ Š E4 (Theorem 53.10(a)) and we get an exceptional group G of order 26 with Z.G/ Š E4 and G 0 1 .A/ (Theorem 53.10(b)). All groups will be given in terms of generators and relations. The above exceptional groups of orders 26 and 27 have peculiar structure but they exist as subgroups of the alternating group A16 . Exercise 1 (see Theorem 6.10). Let A be an abelian 2-group of type .2n ; 2; : : : ; 2/ and of order 2nCd 1 , n > 1, d > 1. Prove that jAut.A/j D 2nCd 2 .2d 2/.2d 22 / : : : .2d 2d 1 /. We need the following information about the structure of Aut.C4 C2 C2 /. Proposition 53.1. Let S 2 Syl2 .Aut.A//, where A Š C4 C2 C2 . Let F be the stability group of the chain A > 1 .A/ > f1g. Then F Š E8 is normal in Aut.A/ and S D D F , where F \ D D f1g and D Š D8 . If X is any subgroup of S such that X \ F D ht i is of order 2, then t is either a square in X or t 62 ˆ.X/.

53

2-groups G with c2 .G/ D 4

97

Proof. By Exercise 1 with n D 2 and d D 3, we get jAut.A/j D 8 6 4 D 26 3 and so jS j D 26 . Let A D hv; u; ei, where o.v/ D 4, o.u/ D 2 D o.e/. The stability group F of the chain A > 1 .A/ > f1g is normal in Aut.A/, F Š E8 , and so F S . Consider the automorphisms ˛ and ˇ of A given by v ˛ D v, u˛ D uz, e ˛ D eu; v ˇ D v, uˇ D uz, e ˇ D e, where z D v 2 . Then o.˛/ D 4, o.ˇ/ D 2, ˛ ˇ D ˛ 1 , and so D D h˛; ˇi Š D8 with D \ F D f1g. Since jFDj D 26 D jS j, we may set S D D F. Let X S be such that X \ F D ht i is of order 2 and assume that t is not a square in X. Since F is normal in S , ht i Z.X/ and X=ht i is isomorphic to a subgroup of S=F Š D8 . If X=ht i is elementary abelian, then ˆ.X/ ht i and therefore ˆ.X/ D f1g since t is not a square in X; in that case t 62 ˆ.X/. If X=ht i is cyclic of order 4, then, since X is not cyclic, we get t 62 ˆ.X/.D Ã1 .X//. Finally, suppose that X=ht i Š D8 . Let U=ht i be the cyclic subgroup of index 2 in X=ht i. Then U is noncyclic so t 62 ˆ.U / D Ã1 .U /. For each x 2 X U , x 2 2 ht i (indeed, hx; t i=ht i is of order 2), and so (by our assumption) x 2 D 1. It follows that ˆ.X/ D Ã1 .U /. Hence in any case t 62 ˆ.X/. Theorems 52.7 and 52.8 imply the following Proposition 53.2. Let K be a 2-group possessing exactly two cyclic subgroups of order 4 and assume that neither of them is a characteristic subgroup of K. Then one of the following holds: (a) K Š C4 C2 ; (b) K Š D8 C2 ; (c) K D hb; t j b 8 D t 2 D 1; b t D bu; u2 D Œu; t D 1; b u D bz; z D b 4 i and this is a group of order 25 with 2 .K/ D hb 2 ; t i hui Š D8 C2 . The following two propositions describe the results from 52 (see Theorems 52.4 and 52.5 for the Proposition 53.3 and Theorem 52.2 for Proposition 53.4) adopted for an application in the proofs of Theorems 53.9 and 53.10. Since the notation introduced in these propositions will be used in the proofs (which are quite involved), the explicit statement of these results is unavoidable. Proposition 53.3. Let H be a 2-group of order > 24 and suppose that A D 2 .H / Š C4 C2 C2 . We set E D 1 .A/ Š E8 , Ã1 .A/ D hzi, B D CH .A/. Then CH .E/=E is cyclic and so B is abelian and H=E is either cyclic or generalized quaternion. Suppose that H=E Š Q2n , n 3, and set E D hz; u; ei, H D hE; a; bi, n1 D z, where hE; ai=E is a cyclic subgroup of index 2 in H=E, o.a/ D 2n , a2 n2 2 2 a D v is of order 4, b 2 A E, o.b/ D 8, and A D hE; vi. Then we have one of the following possibilities: (a) If E 6 ˆ.H /, then b 2 D uv, Œa; u D 1, Œa; e D z , D 0; 1, Œb; e D 1, ub D uz, and ab D a1 . If D 0, then B D hE; ai, and if D 1, then B D hE; a2 i. (b) If E ˆ.H / and ŒE; a ¤ f1g, then b 2 D ev, Œa; u D 1, Œa; e D z, Œb; e D z, Œb; u D z, ab D a1 ue , D 0; 1, and if D 1, then n 4. Here B D hE; a2 i.

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Groups of prime power order

(c) If E ˆ.H / and ŒE; a D f1g, then b 2 D uv, Œb; e D 1, ub D uz, and ab D a1 e. Here B D hE; ai. Proposition 53.4. Let H be a nonabelian 2-group of order > 24 and suppose that A D 2 .H / Š C4 C2 C2 . We set E D 1 .A/ Š E8 , Ã1 .A/ D hzi, and B D CH .A/. Suppose that H=E Š C2n1 , n 3, and set E D hz; u; ei, H D hE; ai, n1 n2 where o.a/ D 2n , a2 D z, a2 D v is of order 4, and A D hE; vi. Then we have one of the following possibilities: (a) If jŒE; aj D 4, then Œa; e D u, Œa; u D z, n 4, and B D hE; a4 i. (b) If jŒE; aj D 2 and ŒE; a D hzi, then Œa; e D 1, Œa; u D z, and B D hE; a2 i. (c) If jŒE; aj D 2 and ŒE; a ¤ hzi, then Œa; e D u, Œa; u D 1, and B D hE; a2 i. Let G be a 2-group of order > 24 with c2 .G/ D 4. Then j2 .G/j 24 . If j2 .G/j D 24 , then from results of 52 follows that 2 .G/ is isomorphic to Q8 C4 or C4 C2 C2 and the following result follows at once. Theorem 53.5. Let G be a 2-group of order > 24 with c2 .G/ D 4 and j2 .G/j D 24 . Then one of the following holds: (a) 2 .G/ Š Q8 C4 and G is isomorphic to a group (a) or (c) of Theorem 52.1. (b) 2 .G/ Š C4 C2 C2 and G is isomorphic to a group described in Propositions 53.3 and 53.4 or G Š C2m C2 C2 , m 2. In the rest of this section we assume that j2 .G/j > 24 . First we prove the following easy special result. Theorem 53.6. Let G be a 2-group of order > 24 with c2 .G/ D 4 and j2 .G/j > 24 . If G has a subgroup Q Š Q8 , then Q G G, C D CG .Q/ is cyclic of order 2n , n 2, G D .Q C /ht i, where t is an involution such that Qht i Š SD24 and ht iC Š D2nC1 . We have jZ.G/j D 2, jGj D 2nC3 and 2 .G/ D G. Proof. If Q were not normal in G, there is g 2 G such that Q ¤ Qg . But then jQ \ Qg j 4 and so Q [ Qg contains at least five cyclic subgroups of order 4, a contradiction. Hence Q is normal in G. If CG .Q/ Q, then jGj 24 , a contradiction. Therefore C D CG .Q/ > Z.Q/ and C is normal in G. If t0 is an involution in C Z.Q/, then c2 .Q ht0 i/ D 6, a contradiction. Since C cannot be generalized quaternion (otherwise c2 .G/ 6), C is cyclic of order 2n , n 2. We see that c2 .Q C / D 4 and 2 .Q C / D Q 2 .C / is of order 24 . But j2 .G/j > 24 and so there is an involution t 2 G .QC /, G D .QC /ht i and t induces an “outer” automorphism on Q. It follows Qht i Š SD24 . Since there are no elements of order 4 in G .QC /, CQC .t / is elementary abelian. Hence CC .t / D Z.Q/ and so ht i C is of maximal class. The only possibility is that ht i C is dihedral of order 2nC1 . The structure of G is completely determined.

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2-groups G with c2 .G/ D 4

99

Remark. Here we will give another proof of Theorem 53.6, retaining the same notation. Since G is not of maximal class, C D CG .Q/ 6 Q, by Proposition 10.17. If z is an involution in C Z.Q/, then c2 .Q hzi/ D 6 > 4 D c2 .G/, a contradiction. If c2 .C / > 1, then, obviously, c2 .QC / 5 > 4 D c2 .G/. Thus, c1 .C / D c2 .C / D 1 so, by Theorem 1.17(b), C is cyclic, C \ Q D Z.Q/. We have c2 .Q 2 .C // D 4 D c2 .G/ so Q 2 .C / is normal in G. Since Q is characteristic in Q 2 .C / (Appendix 16), it is normal in G. Since jAut.Q/j2 D 8, we have jG W .Q C /j 2. Since j2 .G/j > 16 D jQ 2 .C /j, there exists an involution t in G .Q C /, and we get G D ht i .QC /. Using the condition c2 .G/ D 4, we get ht i Q Š SD16 and ht i C Š D2n for some n 2 N. The group G is completely determined. Next we assume also that G has no subgroups isomorphic to Q8 . The subgroup generated by four cyclic subgroups of order 4 will be determined in our next basic result. Theorem 53.7. Let G be a 2-group of order > 24 with c2 .G/ D 4 and j2 .G/j > 24 . Suppose that G has no subgroups isomorphic to Q8 . Then A D hU1 ; U2 ; U3 ; U4 i is abelian of type .4; 2; 2/, where U1 ; U2 ; U3 ; U4 < G are cyclic of order 4. Proof. We first show the following easy fact. If a cyclic subgroup V of order 4 normalizes another cyclic subgroup U of order 4, then jU \ V j D 2 and hU; V i is abelian of type .4; 2/. Indeed, if U \ V D f1g, then j.U V /0 j 2 and so U V is metacyclic of order 24 and class 2. This implies that exp.U V / D 4 and 1 .U V / is abelian of type .2; 2/. If N is a maximal subgroup of U V containing U , then all eight elements in .U V / N are of order 4 and so c2 .U V / > 4, a contradiction. Hence jU \ V j D 2 and jU V j D 8. But U V cannot be nonabelian, i.e., quaternion or dihedral, and so U V is abelian of type .4; 2/. If each Ui is normal in G, then (by the above) A D hU1 ; U2 ; U3 ; U4 i is abelian of type .4; 2; 2/, and we are done. Therefore we may assume that U1 is not normal in G. Set K D NG .U1 / and we have 2 jG W Kj 4. By Theorem 1.17(b), c2 .K/ is even. Let M be a subgroup of G such that K < M , jM W Kj D 2 and jG W M j 2. For each m 2 M K, U1m D U2 ¤ U1 and NG .U2 / D K. By the above, A0 D hU1 ; U2 i Š C4 C2 and A0 is normal in M . If there is a further cyclic subgroup U3 of order 4 contained in K, then U3 6 A0 , U3 normalizes U1 and U2 and so, by the above, U3 centralizes U1 and U2 and jA0 \ U3 j D 2. Hence A0 U3 is abelian of type .4; 2; 2/ and we are done. We assume in the sequel that K has exactly two cyclic subgroups U1 and U2 of order 4. Since U1 and U2 are conjugate in M , it follows that neither U1 nor U2 is a characteristic subgroup of K. By Proposition 53.2, we have exactly three possibilities for the structure of K. Assume (by way of contradiction) that K > A0 and set L D 2 .K/. By Lemma 42.1, jLj > 23 (if jLj D 23 , then Z.L/ is a characteristic cyclic subgroup of order 4 in

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Groups of prime power order

K, which is not the case). Therefore, by Proposition 53.2, we have L Š D8 C2 and L is normal in M . If K > L, then jK W Lj D 2 (Proposition 53.2(c)) and all elements in K L are of order 8. In any case, by Proposition 53.2, CK .A0 / D A0 and K=A0 is elementary abelian of order 2 or 4. Suppose that K > L. Then M=A0 Š Aut.A0 / Š D8 since jM=A0 j D 8. Since L=A0 D ˆ.M=A0 /, there is an element k 2 M K such that k 2 2 L A0 (if k 2 2 A0 for all k 2 M K, then A0 ˆ.M /, a contradiction since M=A0 Š D8 ). All elements in L A0 are involutions and so k is an element of order 4. In fact, all 16 elements in .hkiA0 / L are of order 4. But then c2 .hkiA0 / 8, a contradiction. We have K D L Š D8 C2 ; in that case, jM j D 25 . Assume that M=A0 Š C4 . Let x 2 M K. If o.x/ D 8, then hxi \ K is cyclic of order 4, say U1 . Then NG .U1 / hx; Ki > K, which is a contradiction. Thus, M K has no elements of order 8. Suppose that o.x/ D 2. Then hx; A0 i D K since M=A0 is cyclic; in that case x 2 K, again a contradiction. Thus, in the case under consideration, all 16 elements in M K are of order 4, a contradiction. Thus M=A0 Š E4 . For each m 2 M K, U1m D U2 and so m2 2 1 .A0 / and m is an element of order 2 or 4. Since c2 .A0 / D 2, it follows that M K contains at most four elements of order 4. Hence there exists an involution s 2 M K since jM Kj D 16 (in fact, there are at least 12 involutions in M K). Let t be an involution in K A0 . We set A0 D ha; u j a4 D u2 D Œa; u D 1; a2 D zi and so 1 .A0 / D hu; zi and at D a1 , ut D u, as D au, us D u, because U1s D U2 and s induces an involutory automorphism on A0 . Since s inverts (actually centralizes) exactly the elements in 1 .A0 /, it follows that A0 s has exactly four involutions. Hence the other four elements in A0 s are of order 4. This means that we have found all four elements of order 4 in M K. Hence the coset A0 .st / consists only of involutions. But then the involution st must invert A0 . On the other hand, we have ast D .au/t D a1 u, which is a contradiction. We have proved that K D A0 . Since jGj > 24 , it follows jG W M j D 2 and jGj D 25 . In this case jG W Kj D 4 and so fU1 ; U2 ; U3 ; U4 g forms a single conjugacy class in G. This implies that NG .A0 / D M (otherwise, A0 is normal in G so U1 and U3 are not conjugate in G) and M A0 contains exactly four elements of order 4, and the other four elements in M A0 are involutions. Since j2 .G/j > 24 , there are involutions in G M . Let m be an involution in M A0 . We set again A0 D ha; u j a4 D u2 D Œa; u D 1; a2 D zi. Since M is not of maximal class (see Proposition 1.8), U1m D U2 , and m induces an involutory automorphism on A0 , we may set am D a u ( D ˙1) and um D u. Thus CA0 .m/ D 1 .A0 / D hu; zi and so E D hu; z; mi D 1 .M / (indeed, jhu; z; mi jhu; zij D 4 is the number of involutions in M K D M A0 ) is a normal elementary abelian subgroup of order 8 in G. Let be an involution in G M . If does not centralize E, then E would contain some elements of order 4, a contradiction since M \ E M \ M D ¿. Thus hE; i D E hi is an elementary abelian maximal subgroup of G. But then exp.G/ D 4 and so all elements in G M must be involutions. It follows that

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2-groups G with c2 .G/ D 4

101

inverts M and so M is abelian. In this case, as it is easy to check since jM j D 24 , M is abelian of type .4; 2; 2/, and the theorem is proved. In the rest of this section we set A D hU1 ; U2 ; U3 ; U4 i Š C4 C2 C2 , where fU1 ; U2 ; U3 ; U4 g is the set of four cyclic subgroups of order 4 in G, E D 1 .A/ Š E8 , hzi D Ã1 .A/ Z.G/, B D CG .A/. Let v 2 A E so that v 2 D z. If there is an involution s 2 B A, then sv is an element of order 4 in B A, a contradiction. Hence A D 2 .B/ and Proposition 53.3 implies that B=E is cyclic and so B is abelian of type .2m ; 2; 2/, m 2. It follows that CG .B/ D B. If c 2 B with o.c/ D 2m , then m1 m2 c2 D hzi, B D Ehci, E \ hci D hzi, A D Ehc 2 i. Since j2 .G/j > 24 , there exists an involution t 2 G B. If there is an element x 2 Bt with o.x/ > 2, then o.x/ 8 since B \ Bx D ¿, and so o.x 2 / 4 and x 2 2 B. Thus CB .t / D CB .x/ is not elementary abelian and so Bt would contain elements of order 4, a contradiction (namely, if y is an element of order 4 in CB .t /, then yt 2 Bt is of order 4). Hence all elements in Bt are involutions and consequently t inverts B. If there is an involution t 0 2 G Bht i, then t 0 also inverts B and so t t 0 2 G B would centralize B, a contradiction since CG .B/ D B. Thus D D Bht i D 2 .G/. If 2 .G/ D G, the structure of G is determined and we have proved the following Theorem 53.8. Let G be a 2-group of order > 24 with c2 .G/ D 4 and j2 .G/j > 24 . Suppose that G has no subgroups isomorphic to Q8 . Then D D 2 .G/ D Bht i, where B is abelian of type .2m ; 2; 2/, m 2, and t is an involution inverting B. We have Z.D/ D E D 1 .B/, CG .t / D ht i E D F , and so F is a selfcentralizing elementary abelian subgroup of order 16. If 2 .G/ D G, the structure of G is completely determined. In what follows we assume, in addition, that G > 2 .G/. We show next that CG .E/ D D D 2 .G/. Indeed, suppose that CG .E/ > D. Since CG .E/ stabilizes the chain A > E > f1g in view of jA W Ej D 2, it follows that CG .E/=B .D CG .E/=CG .A// is elementary abelian and 4 jCG .E/=Bj 8. Let F0 =B be a subgroup of order 2 in CG .E/=B such that D \ F0 D B. Then 2 .F0 / D A (otherwise, c2 .F0 / > 4 D c2 .G/) and F0 is nonabelian of order > 24 since F0 6 B D CG .A/). Proposition 53.3 implies, however, that F0 =E is cyclic and so F0 is abelian, a contradiction. We have proved that CG .E/ D D .D 2 .G/). Hence X D G=B is a subgroup of Aut.A/ (see Proposition 53.1) and if F1 .Š E8 / is the stability group of A > E > f1g (note that F1 < Aut.A/), then X \ F1 D D=B is of order 2. Suppose that the involution in D=B is a square in X. Then there exists x 2 G such that x 2 D t 0 2 D B. But t 0 is an involution (indeed, all elements in D B, by the above, are involutions) and so x is an element of order 4 in G B, a contradiction since A B. Proposition 53.1 implies that there exists a maximal subgroup H of G containing B such that H \ D D B (indeed, by the above proposition, D=B 6 ˆ.G=B/), and so G=B D .D=B/ .H=B/ since D is normal in G. Note that 2 .H / D A (indeed, H \2 .G/ D H \D D B and 2 .B/ D A) and H > B is nonabelian of order > 24 ,

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Groups of prime power order

where B D CG .A/ D CH .A/. In this situation we shall apply Propositions 53.3 and 53.4 which describe all possibilities for the structure of H . Also we shall use freely the notation introduced in these propositions. In particular, 2 jH=Bj 4 and so G=B.D .D=B/ .H=B// is abelian of order 4. We also set D D Bht i, where t is an involution inverting B. It is possible at this stage to eliminate the possibility (c) of Proposition 53.3 and the possibilities (a) and (c) of Proposition 53.4 (thus, we intend to consider these three possibilities using the notation introduced in corresponding parts). Suppose that H is isomorphic to a group (c) of Proposition 53.3. Consider the subgroup K D Bht bi, where .t b/2 2 B and 2 .K/ D A. We compute atb D .a1 /b D .ab /1 D .a1 e/1 D ae and so t b centralizes B=E and jB=Ej 4. Hence K=E cannot be generalized quaternion and so K=E is cyclic. It follows that ht bi covers K=E which gives .t b/2 D ai s, where s 2 E and i is an odd integer. On the other hand, t b centralizes .t b/2 and so ai s D .ai s/tb D .ai s/b D .a1 e/i s b D ai es b . Thus s b D se and Œb; s D e. This is a contradiction since Œb; E D hzi according to Proposition 53.3(c). Suppose that H is isomorphic to a group (a) of Proposition 53.4; in that case, H=B Š C4 since H D hE; ai and B D hE; a4 i. Consider the subgroup K D Bht ai. Since G=B Š C2 C4 , it follows K=B Š C4 and 2 .K/ D A. We compute .a4 /ta D .a4 /1 and so if jB=Ej 4, K=E is nonabelian and so K=E is generalized quaternion. But K=E has the cyclic factor group .K=E/=.B=E/ Š K=B of order 4, a contradiction. Hence we must have B D A, n D 4, jGj D 27 , and K=E is cyclic of order 8. Since ht ai covers K=E, we get .t a/4 D sv with s 2 E. But t a centralizes .t a/4 and so we get (recall that v D a4 ) sv D .sv/ta D .sv 1 /a D s a v 1 D s a vz which implies s a D sz. This gives s D uz with D 0; 1, and .t a/4 D uz v. On the other hand, .t a/2 2 .Bha2 i/ B, and so .t a/2 D a2 s 0 , where s 0 2 A. Indeed, G=B is abelian and so .t a/2 D t 2 a2 s 0 with s 0 2 B D A. We compute 2 uz v D .t a/4 D .a2 s 0 /2 D a2 s 0 a2 s 0 D a4 .a2 s 0 a2 /s 0 D v.s 0 /a s 0 , which gives 2 .s 0 /a D .s 0 /1 uz and Œa2 ; .s 0 /1 D u.z .s 0 /2 / with z .s 0 /2 2 hzi. This is a contradiction since, according to Proposition 53.4(a), Œa2 ; A D hzi. Suppose that H is isomorphic to a group (c) of Proposition 53.4. Here B D hE; a2 i and we consider the subgroup K D Bht ai. We see .a2 /ta D .a2 /1 and so t a inverts B=E. Note that 2 .K/ D A and so in case jB=Ej 4, K=E is generalized quaternion and consequently .t a/2 2 A E (since A=E D Z.K=E/). If jB=Ej < 4, then B D A and again .t a/2 2 A E (since o.t a/ 8). It follows .t a/2 D sv with n2 s 2 E and v D a2 . Since t a centralizes .t a/2 , we get sv D .sv/ta D .sv 1 /a D s a v 1 D s a vz. This gives s a D sz and Œa; s D z. But Proposition 53.4(c) implies Œa; E D hui ¤ hzi, a contradiction. We are now ready to prove the following deep result. Theorem 53.9. Let G be a 2-group of order > 24 with c2 .G/ D 4 and j2 .G/j > 24 . Suppose that G has no subgroups isomorphic to Q8 and jG W 2 .G/j 4. Then we

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2-groups G with c2 .G/ D 4

103

have jG W 2 .G/j D 4 and G is a uniquely determined group of order 27 : G D ha; b; t j a8 D b 8 D t 2 D 1; a2 D v; a4 D z; b 2 D ev; ab D a1 u; e 2 D u2 D Œe; v D Œu; v D Œe; u D Œa; u D Œt; e D Œt; u D 1; e a D ez; e b D ez; ub D uz; v t D v 1 ; at D eva1 ; b t D euvb 1 i: Here E D hz; u; ei is a normal elementary abelian subgroup of order 8. Four cyclic subgroups of order 4 in G generate the abelian subgroup A D hE; vi of type .4; 2; 2/ which is self-centralizing in G and each cyclic subgroup of order 4 is normal in G. We have D D 2 .G/ D Aht i, where the involution t inverts A and CG .t / D E ht i Š E16 is self-centralizing in G. We have ˆ.G/ D A, all elements in G D are of order 8, and Z.G/ D hzi is of order 2. For one of the four maximal subgroups M of G which do not contain D we have M=E Š Q8 and E < ˆ.M / (so this M is generated by two elements) but for the other three such maximal subgroups M we have M=E Š Q8 and E 6 ˆ.M / (so these M are not generated by two elements). For each of the three maximal subgroups N of G which contain D we have N=E Š D8 . The group G exists as a subgroup of the alternating group A16 and 16 is the smallest degree for any faithful permutation representation of G. Proof. Since G=B D .D=B/.H=B/ we get G=2 .G/ D G=D Š H=B, so we have to examine here only the cases, where H is isomorphic to a group (b) of Proposition 53.3 or to a group (a) with D 1 of Proposition 53.3. In these cases we have jH=Bj D 4 and so we have already proved that jG W 2 .G/j D 4. Suppose first that H is isomorphic to a group (b) of Proposition 53.3. Note that in this case H=E Š Q2n , n 3, E ˆ.H /, and if n D 3, then D 0, i.e., ab D a1 ue D a1 u. We have here G=B Š E8 . Consider the subgroup L D Bht ai, where .t a/2 2 B and 2 .L/ D \ L A so 2 .L/ D A (recall that Aht i D D D 2 .G/ and involution t inverts A). We compute .a2 /ta D .a2 /a D .a2 /1 , and so t a inverts B=E (recall that B D hE; a2 i). If n 4, then jB=Ej 4, L=E is generalized quaternion, and consequently .t a/2 2 A E (since o.t a/ 8). If n D 3, then B D A and o.t a/ 8 implies again .t a/2 2 A E. Thus .t a/2 D sv, where s 2 E and n2 v D a2 . From this relation we get sv D t at a D at a so at D sva1 . Since t a centralizes .t a/2 , we have sv D .sv/ta D .sv 1 /a D s a v 1 D s a vv 2 D s a vz and s a D sz, which gives s D eu0 with u0 2 hu; zi since E D he; u; zi. On the other hand, set B1 D Bhai D Ehai and consider the subgroup L1 D B1 ht bi; then 2 .L1 / D A since L ¤ D D 2 .G/ and A 2 .L/. We compute, taking into account the above obtained equalities, atb D .sva1 /b D s b v b .ab /1 D .s b v b e u/a D a0 a, where D 0; 1 and a0 D s b v b e u 2 AE. If n 4, then B1 =E is cyclic of order 8 and so L1 =E Š M2n . This is a contradiction since L1 =E must be either cyclic or generalized quaternion. It follows (see Proposition 53.3(b)) n D 3, B D A, D 0, a2 D v, b 2 D ve, ab D a1 u, and jGj D 27 . We get v b D .a2 /b D

104

Groups of prime power order

.ab /2 D .a1 u/2 D a2 D v 1 , and so v b D v 1 . Since .t b/2 2 B D A and o.t b/ 8 (indeed, t b 2 G D D G 2 .G/), we have .t b/2 D s0 v, where s0 2 E. Since t b centralizes .t b/2 , we get s0 v D .s0 v/tb D .s0 v 1 /b D s0b v, and therefore s0b D s0 , which implies s0 2 heu; zi. We get s0 v D t bt b D b t b and b t D s0 vb 1 . The above elements s; s0 2 E are connected with a relation which we obtain in the following way. We act with t (by conjugation) on the relation b 1 ab D a1 u and get (taking into account the above obtained equalities) .s0 vb 1 /1 .sva1 /.s0 vb 1 / D .sva1 /1 u; bs0 sa1 s0 vb 1 D av 1 su; s0 sa1 s0 v D b 1 .av 1 su/b D a1 uvs b uz D a1 vs b z; .as0 sa1 /s0 v D vs b z; and since s0a relation (1)

1

D s0a , s a

1

D sz, we get s0a szs0 v D vs b z, and this implies the desired s0a s b D s0 s:

Note that s D eu0 with u0 2 hu; zi and s0 2 heu; zi. If s0 D euz ˛ .˛ D 0; 1/, then (1) gives s b D sz and so s D ez ˇ .ˇ D 0; 1/. If s0 D z ˛ .˛ D 0; 1/, then (1) gives s b D s and so s D euz ˇ .ˇ D 0; 1/. We consider again the relations at D sva1 , b t D s0 vb 1 , where s; s0 are the above elements in E. Replacing t with t ue, we get atue D .sva1 /ue D sva1 z D .sz/va1 , b tue D .s0 vb 1 /ue D s0 vb 1 . Replacing t with t u, we get atu D .sva1 /u D sva1 , b tu D .s0 vb 1 /u D .s0 z/vb 1 . The above facts show that in our expressions for s and s0 we may choose ˛ D ˇ D 0. Hence we get either s D e and s0 D eu or s D eu and s0 D 1. We have obtained exactly two groups G1 and G2 (of order 27 ) which differ only in the last two relations: (a) G1 with relations at D eva1 , ; b t D euvb 1 ; (b) G2 with relations at D euva1 , b t D vb 1 . For both groups G1 and G2 we have obtained the following common relations: (c) a8 D b 8 D t 2 D 1, a2 D v, a4 D z, b 2 D ev, ab D a1 u, e 2 D u2 D Œe; v D Œu; v D Œe; u D Œa; u D Œt; e D Œt; u D 1, e a D ez, e b D ez, ub D uz, v t D v 1 . It is possible to show that the groups G1 and G2 are isomorphic. In the group G1 we first obtain by computation (using the relations for G1 ) (2)

.ab/t D zv.ab/1 :

Then in G1 we replace the elements a; v; b; e; t with a0 D a1 , v 0 D v 1 , b 0 D ab, e 0 D eu, t 0 D t e, and see that these new elements satisfy the same common relations

53

2-groups G with c2 .G/ D 4 0

105 0

(c). But from the relations (a) and (2), we get .a0 /t D e 0 uv 0 .a0 /1 , .b 0 /t D v 0 .b 0 /1 , which are the last two relations (b) for the group G2 . Hence the groups G1 and G2 are isomorphic and we have obtained the unique group G D G1 of order 27 as stated in our theorem. We have ˆ.G/ D A. In order to investigate the structure of seven maximal subgroups M of G, we first see a2 D v, b 2 D ev, .ab/2 D euzv, .t a/2 D ev, .t b/2 D euv, and .t ab/2 D zv. If M D hA; a; bi D H , then ˆ.M / D A Š C4 C2 C2 and so E ˆ.M /. If M D hA; a; t bi, then ˆ.M / D hv; eui Š C4 C2 . If M D hA; b; t ai, then ˆ.M / D hv; ei Š C4 C2 . If M D hA; t a; t bi, then ˆ.M / D hev; ui Š C4 C2 . Hence in the last three cases we have E 6 ˆ.M /. For these four maximal subgroups M (which do not contain D D Aht i) we have M=E Š Q8 . For the other three maximal subgroups N (which contain D) we see that N=E Š D8 . We get a faithful permutation representation of degree 16 of G in the following way. We set: a D .1; 2; 3; 4; 5; 6; 7; 8/.9; 16; 12; 13; 11; 14; 10; 15/; b D .1; 9; 3; 10; 5; 11; 7; 12/.2; 13; 8; 14; 6; 15; 4; 16/; t D .3; 7/.4; 8/.10; 12/.14; 16/: We see that the above permutations a; b; t are even and check that they satisfy the defining relations for G. The obtained representation is faithful since z D a4 D .1; 5/.2; 6/.3; 7/.4; 8/.9; 11/.14; 16/.10; 12/.13; 15/ ¤ 1: It remains to show that ı.G/, the minimal degree of a faithful permutation representation of G (minimal representation), equals 16 (in the previous paragraph we have showed that ı.G/ 16). Assume that this is false. Since Z.G/ is cyclic, every minimal representation of G, is transitive so ı.G/ is a power of 2. Then G is isomorphic to a 2-subgroup of Sı.G/ so ı.G/ D 8. Moreover, G Š †3 2 Syl2 .S23 /. However, c1 .G/ D c1 .2 .G// D jD Aj C c1 .A/ D 23 (note that c1 .A/ D 7). On the other hand, c1 .†3 / D jD8 j C Œc1 .D8 / C 12 1 D 8 C 62 1 D 43 > 23, and this is a contradiction. Thus, ı.G/ D 16. In the second half of the proof we investigate the remaining possibility, where H is isomorphic to a group (a) with D 1 of Proposition 53.3. Note that in this case H=E Š Q2n , n 3, and E 6 ˆ.H /. In exactly the same way as in the first part of the proof we get n D 3, B D A, jGj D 27 , a2 D v, b 2 D vu, and ab D a1 . Also we get at D sva1 where s 2 E, s a D sz, s D eu0 with u0 2 hu; zi, b t D s0 vb 1 , where s0 2 E, s0b D s0 , and s0 2 he; zi. We conjugate the relation b 1 ab D a1 with the element t and obtain .b t /1 at b t D t 1 .a / which gives (as in the first part of the proof) the following connection between s and s0 : (3)

s0a s b D s0 sz:

106

Groups of prime power order

If s0 D ez ˛ .˛ D 0; 1/, then (3) gives s b D s and s D ez ˇ .ˇ D 0; 1/. If s0 D z ˛ .˛ D 0; 1/, then (3) gives s b D sz and so s D euz ˇ .ˇ D 0; 1/. Replacing t with t e, we get ate D .sva1 /e D .sz/va1 , b te D .s0 vb 1 /e D s0 vb 1 . Replacing t with t u, we get atu D .sva1 /u D sva1 , b tu D .s0 vb 1 /u D .s0 z/vb 1 . This shows that in our expressions for s and s0 we may choose ˛ D ˇ D 0 and so we have either s D e and s0 D e or s D eu and s0 D 1. We have obtained again exactly two new groups G1 and G2 (of order 27 ) which differ only in the last two relations: (a’) G1 with relations at D eva1 , b t D evb 1 ; (b’) G2 with relations at D euva1 , b t D vb 1 . It is possible to show that the new groups G1 and G2 are isomorphic. In G1 we replace the elements e, b, t with e 0 D eu, b 0 D ab, t 0 D t u, respectively, and see that the common relations for G1 and G2 remain valid. But from the relations (a’) we 0 0 get .a/t D e 0 uv.a/1 , .b 0 /t D v.b 0 /1 , which are the last two relations (b’) for the group G2 . Hence the groups G1 and G2 are isomorphic and we have to study further only the new group G D G1 . We consider the maximal subgroup H D hA; b; t ai and see that ˆ.H / D huv; ev; vi D A. We have 2 .H / D A, H =E Š Q8 , and E ˆ.H /. But this is exactly the starting point for the first part of the proof with the subgroup H instead of H . It follows that our new group G D G1 must be isomorphic to the unique group of order 27 stated in our theorem. Remark. Let us show that every subgroup T G of order 24 contains z (here G is the group of Theorem 53.9). Since Z.G/ D hzi, it suffices to show that TG D f1g. Assume that this is false. Then jG W T j D 8 and G is isomorphic to a Sylow 2-subgroup of the symmetric group S8 . As the proof of Theorem 53.9 shows, G and a Sylow 2subgroup of S8 have distinct numbers of involutions, and this is a contradiction. Thus, z 2 T. In our next result the remaining 2-groups G with c2 .G/ D 4 will be determined. Theorem 53.10. Let G be a 2-group of order > 24 with c2 .G/ D 4 and j2 .G/j > 24 . Suppose that G has no subgroups isomorphic to Q8 and jG W 2 .G/j D 2. Then we have the following two possibilities: (a) G D hb; e; t i with b 8 D e 2 D t 2 D 1; n2

a2

D v;

n1

a2

.t b/2 D a;

D z;

n

a2 D 1; n 3;

b 2 D uv; u2 D 1

Œb; e D Œa; e D Œa; u D Œu; e D Œt; e D Œt; u D 1; ub D uz;

ab D a1 ;

at D a1 :

53

2-groups G with c2 .G/ D 4

107

Here jGj D 2nC4 , n 3, G D hei ha; b; t i, and E D hz; u; ei is a normal elementary abelian subgroup of order 8 in G. Four cyclic subgroups of order 4 generate the abelian subgroup A D hE; vi of type .4; 2; 2/ and each cyclic subgroup of order 4 is normal in G. We have B D CG .A/ D hE; ai is abelian of type .2n ; 2; 2/, 2 .G/ D Bht i is of order 2nC3 , where the involution t inverts B, and CG .t / D E ht i Š E16 is self-centralizing in G. We have ˆ.G/ D hui hai Š C2 C2n , G 0 6 E, Z.G/ D he; zi Š E4 . Finally, .Eha; bi/=E Š Q2n and .Eht bi/=E Š C2n . (b) G D ha; t; ei with a8 D e 2 D t 2 D 1;

a2 D v;

a4 D z;

u2 D Œa; e D Œu; e D Œt; e D Œt; u D 1;

.t a/2 D uv; ua D uz:

Here jGj D 26 and E D hz; u; ei is a normal elementary abelian subgroup of order 8 in G. Four cyclic subgroups of order 4 generate the abelian subgroup A D hE; vi of type .4; 2; 2/ and each cyclic subgroup of order 4 is normal in G. We have A D CG .A/, 2 .G/ D Aht i is a maximal subgroup of G, where the involution t inverts A. We have ˆ.G/ D hu; vi Š C4 C2 , G 0 D hz; ui E, Z.G/ D he; zi Š E4 . This exceptional group exists as a subgroup of A16 . The subgroup hei is a direct factor of G. The minimal degree of a faithful permutation representation of G equals 10 (obviously, this representation is intransitive: every permutation representation of a 2-group G such that its degree does not equal to a power of 2, is intransitive). Proof. Here H is isomorphic to a group (a) with D 0 of Proposition 53.3 or to a group (b) of Proposition 53.4 since these are the only remaining possibilities with jH=Bj D 2, where B D CG .A/ and G=2 .G/ Š H=B. Suppose that H is isomorphic to a group (a) with D 0 of Proposition 53.3. Here CG .A/ D B D hE; ai is abelian of type .2n ; 2; 2/, n 3, and H=E Š Q2n . We have atb D .a1 /b D a and so looking at K D Bht bi with 2 .K/ D A, we get that K=E must be cyclic. Thus ht bi covers K=E and so .t b/2 D ai s, where s 2 E and i is an odd integer. Replacing a with ai we may assume from the start .t b/2 D as which gives b t D asb 1 . Since t b centralizes .t b/2 we obtain as D .as/tb D .a1 s/b D as b , which gives s b D s and so s 2 he; zi. We compute b tu D .asb 1 /u D asb 1 z D a.sz/b 1 , and so replacing t with the involution t u (if necessary), we may assume that s D 1 or s D e. Suppose that s D e and so .t b/2 D ae. But then replacing a with a0 D ae, all other relations remain unchanged and the last relation is transformed into .t b/2 D a0 . Hence we may assume from the start that .t b/2 D a and our group G is uniquely determined as stated in part (a) of our theorem. Suppose, finally, that H is isomorphic to a group (b) of Proposition 53.4. Here CG .A/ D B D hE; a2 i is abelian of type .2n1 ; 2; 2/, n 3, and H=E Š C2n1 . We set K D Bht ai and see that .a2 /ta D .a2 /1 and therefore t a inverts B=E. If jB=Ej 4, then 2 .K/ D A implies that K=E is generalized quaternion. But then replacing H with K, we know that such groups have been determined before.

108

Groups of prime power order

It follows that we may assume that jB=Ej D 2 and so B D A, K=E Š C4 , and jGj D 26 . Hence o.t a/ D 8 and .t a/2 D sv, where s 2 E and at D sva1 . Since t a centralizes .t a/2 , we get sv D .sv/ta D .sv 1 /a D s a v 1 D s a vz, which gives s a D sz and so s D ue 0 with e 0 2 he; zi. We note that atu D .sva1 /u D .sz/va1 and so replacing t with t u (if necessary), we may set s D u or s D ue. However, if s D ue, then .t a/2 D uev. In this case we replace u with u0 D eu and see that all other relations remain valid (with u0 instead of u) and only the last relation is transformed in .t a/2 D u0 . Hence we may assume from the start that .t a/2 D u. Our group G is uniquely determined as stated in part (b) of our theorem. We get a faithful representation of degree 16 of G in the following way. We set a D .1; 2; 3; 4; 5; 6; 7; 8/.9; 10; 11; 12; 13; 14; 15; 16/; e D .1; 9/.2; 10/.3; 11/.4; 12/.5; 13/.6; 14/.7; 15/.8; 16/; t D .2; 6/.3; 7/.10; 14/.11; 15/: We see that the permutations a; e; t are even and they satisfy the defining relations for G. The obtained representation is faithful since the central permutations z D a4 , e, and ez are nontrivial. Since e 62 ˆ.G/, we get G D hei M for some maximal subgroup M of G. The subgroup Z.M / D hzi is cyclic so a faithful permutation representation of minimal degree ı.M / is transitive, i.e., ı.M / is a power of 2. We claim that ı.M / D 8. Clearly, ı.M / 8. Set U D hai and V D hu; e; t i. Since U \ V D f1g, we have ı.M / jM W U j C jM W V j D 4 C 4 D 8. It follows that ı.M / D 8. We get ı.G/ D ı.M hei/ ı.M / C ı.hai/ D 8 C 2 D 10. Since, by [Ber26, page 740], ı.G/ ı.hai hei/ D ı.hai/ C ı.hei/ D o.a/ C o.e/ D 8 C 2 D 10; we get ı.G/ D 10. The proof is complete. Problem. Classify the 2-groups G such that c2 .G/ 6 0 .mod 4/. In particular, classify the 2-groups G with c2 .G/ D 6.

54

2-groups G with cn.G / D 4, n > 2

In this section we study a 2-group G with cn .G/ D 4, n > 2 [Jan8]. If fU1 ; U2 ; U3 ; U4 g is the set of four cyclic subgroups of order 2n , then we describe first the structure of the subgroup X D hU1 ; U2 ; U3 ; U4 i (Theorems 54.1 and 54.2). It is interesting to note that always jXj D 2nC2 . If G has a normal elementary abelian subgroup of order 8, the structure of G is described in Theorem 54.6. If G has no normal elementary abelian subgroups of order 8, the structure of G is described in Theorem 54.7. The proofs of these theorems are quite involved and therefore we prepare the stage with Propositions 54.3, 54.4, and Theorem 54.5 which are also of independent interest. If cn .G/ D 2, then the 2-groups are known (Corollary 43.6). The next interesting and important case is cn .G/ D 4 (see Research Problems #188 and #425). However, this problem is essentially more difficult. Throughout this section we assume that G is a 2-group with cn .G/ D 4, n > 2. We consider first the case that G has a normal subgroup isomorphic to E8 . Theorem 54.1. Let G be a 2-group with cn .G/ D 4, n > 2, and suppose that G has a normal subgroup E Š E8 . Let fU1 ; U2 ; U3 ; U4 g be the set of four cyclic subgroups of order 2n in G. Then X D hU1 ; U2 ; U3 ; U4 i D EU1 , where E \ U1 > f1g and so jXj D 2nC2 . Moreover, 2 .X/ Š C4 C2 C2 or 2 .X/ Š D8 C2 in which case n D 3. Proof. Assume that for each cyclic subgroup U of order 2n , U \ E D 1. Let U D hai be one of them. Then .EU /=E Š C2n and all 2nC2 elements in .EU / .Eha2 i/ are of order 2n . Indeed, if x 2 .EU / .Eha2 i/ and o.x/ D 2nC1 , then o.x 2 / D 2n and 2nC2 hx 2 i \ E ¤ f1g, a contradiction. But then cn .G/ '.2 n / D 8 which contradicts our assumption (here './ is Euler’s totient function; '.2n / is the number of generators of a cyclic group of order 2n ). We have proved that there is a cyclic subgroup hbi of order 2n with jhbi \ Ej D 2 n1 n2 so that .Ehbi/=E Š C2n1 . Let H D Ehbi, z D b 2 , and v D b 2 ; then hbi \ E D hzi. Assume that CH .E/ D E. Then n D 3, H=E Š C4 acts faithfully on E, and we may set E D he; u; zi, e b D eu, ub D uz. The structure of H is uniquely determined, H D hb; ei. We have uv D .ub /b D .uz/b D uzz D u, e v D .eu/b D eu uz D ez; v e D eve D ve v e D veze D vz D vv 2 D v 1 . It follows that Ehvi D he; vi hui Š D8 C2 . Let us check that 2 .H / D Ehvi, ˆ.H / D

110

Groups of prime power order

hui hvi Š C4 C2 . We have .be/2 D b 2 e b e D veue D vu 2 ˆ.G/. Since v D b 2 2 ˆ.G/ we get u D v 1 vu 2 ˆ.G/. It follows from d.H / D 2 that ˆ.G/ has order 23 so is abelian, and the second isomorphism in the displayed formula is proved. It remains to prove the first equality in the displayed formula. Assume that it is false. Then there exists an element y of order 4 in H .Ehvi/. We may assume that y 2 fbe; bu; beu; bez; buz; beuzg. However, it is easy to check that all these six elements have order 8. Thus, 2 .H / D Ehvi Š D8 C2 . and we see that all 16 elements in H .Ehvi/ are of order 8. Thus c3 .H / D c3 .G/ D 4 and therefore, by the product formula, X D EU1 , where U1 D hbi. Suppose now that CH .E/ > E. Then A D Ehvi is abelian of type .4; 2; 2/, all elements in A E are of order 4, and Ã1 .A/ D hzi. For each element y 2 n2 H .Ehb 2 i/, we have y 2 2 A E and so o.y/ D 2n . Hence all 2nC1 elements 2nC1 in H .Ehb 2 i/ are of order 2n which gives cn .H / D '.2 n / D 4 D cn .G/. We get again X D EU1 and 2 .X/ D Ehvi Š C4 C2 C2 . We assume now that G is a nonabelian 2-group which has no normal subgroups isomorphic to E8 . Since G is not of maximal class, we may apply the main theorem in 50 to conclude that G has a normal subgroup W which is either abelian of type .4; 4/ with CG .W / being metacyclic or W is abelian of type .4; 2/ with CG .W / being abelian of type .2j ; 2/, j 2. Theorem 54.2. Let G be a nonabelian 2-group with cn .G/ D 4, n > 2, and suppose that G has no normal elementary abelian subgroups of order 8. Let fU1 ; U2 ; U3 ; U4 g be the set of four cyclic subgroups of order 2n . Then the subgroup X D hU1 ; U2 ; U3 ; U4 i is of order 2nC2 and for the structure of X we have the following possibilities: (a) n > 2, X D W U1 with W \ U1 Š C4 , where W is an abelian normal subgroup of type .4; 4/. We have 2 .X/ D W and X is metacyclic. (b) n > 2, X D QU1 , where Q Š Q8 is a normal quaternion subgroup of X, Q \ U1 D Z.Q/, U1 D hbi Š C2n , and b either centralizes Q or b induces on Q an involutory outer automorphism in which case n > 3. (c) n D 3, jXj D 25 , and X has a self-centralizing (in X) abelian normal subgroup A of type .4; 2/. Furthermore, 2 .X/ Š Q8 C2 or 2 .X/ Š D8 C2 and so (according to 52) the group X is uniquely determined in each of the two possibilities for 2 .X/. Proof. By Proposition 50.6, G has a normal abelian subgroup W of exponent 4 such that either W is of type .4; 4/ or of type .4; 2/ and such that 2 .CG .W // D W ; in the first case CG .W / is metacyclic and in the second case CG .W / is abelian and has a cyclic subgroup of index 2. (i) We examine first the possibility that G has a normal subgroup W Š C4 C4 , where CG .W / is metacyclic. Set W0 D 1 .W /, H D W U1 , where U1 D hai is a cyclic subgroup of order 2n , n > 2. The structure of Aut.W / is described in

54

2-groups G with cn .G/ D 4, n > 2

111

Proposition 50.5. In particular, exp.Aut.W /2 / D 4, where Aut.W /2 2 Syl2 .Aut.W // (moreover, Aut.W /2 is special) and if ˛ ¤ ˇ are two involutions in Aut.W /2 which do not stabilize the chain W > W0 > 1, then h˛; ˇi is a four-group. We want to show that W \U1 Š C4 . So suppose that jW \U1 j 2. If W \U1 D 1, n1 then CH .W / contains the involution z D a2 since jU1 j > 4 D exp.Aut.W //2 and W0 hzi Š E8 , contrary to the fact that CH .W / is metacyclic. Hence we have n2 W \ U1 D hzi is of order 2. Set v D a2 so that v 2 D z. Since 2 .CG .W // D W , v does not centralize W . Hence in this case U1 =hzi acts faithfully on W , where U1 D hvi. This gives o.a/ D 8, n D 3, a2 D v, H=W Š C4 and jH j D 26 . We shall determine the structure of H D W U1 , where U1 D hai. Since a induces an automorphism of order 4 on W , a cannot stabilize the chain W > W0 > f1g (Remark 50.2). Set W D hxi hyi, x 2 D z, y 2 D u; then W0 D hzi hui. Assume that a stabilizes the chain W0 > f1g. Then x a is an element containing z so x a 2 xW0 . Similarly, y a 2 yW0 . We conclude that a stabilizes the chain W > W0 > f1g, which is not the case. Thus, we have ua D uz. If w 2 W is such that w 2 D u, then setting w a D y, we get y 2 D .w a /2 D .w 2 /a D .ua / D uz and W D hwi hyi with .wy/2 D z. Since W0 hai Š M24 , we get c3 .W0 hai/ D 2. It remains to determine the element y a D ws, where s 2 W0 . We compute 2

w v D w a D y a D ws;

2

y v D y a D .ws/a D w a s a D ys a ;

and so s ¤ 1 since a2 D v induces an involutory automorphism on W . Computing again .a.wy//2 D awyawy D a2 .wy/a wy D vywswy D vzs, we see that o.awy/ D 8 since o.vzs/ D 4 noting that W0 hvi is abelian of type .4; 2/. Since .wy/a D .wy/s, we conclude that W0 hwyihai contains exactly four cyclic subgroups of order 8. Therefore all other elements in aW must be of orders 2 or 4. We compute .aw/2 D awaw D a2 w a w D vyw; .vyw/2 D v 2 .yw/v .yw/ D zys a ws.yw/ D s a s: It follows that s a s D 1, and so s a D s (otherwise aw would be of order 8). Since s 2 Z.ha; W0 i/# D fzg (recall, that ha; W0 i Š M24 ), we get s D z, and so y a D wz and .wy/a D wyz D .wy/1 since z D .wy/2 . The structure of H is uniquely determined and X D hU1 ; U2 ; U3 ; U4 i D W0 hwyihai is a normal subgroup of G. The group X is an extension of the normal abelian subgroup W0 hwyi of type .4; 2/ by C4 and .W0 hwyi/ \ hai D hzi. We have ua D uz and .wy/a D .wy/1 and so v D a2 centralizes W0 hwyi. Therefore A D hu; wy; vi is abelian of type .4; 2; 2/ and 1 .X/ D 1 .A/ D hu; z; wyvi Š E8 is normal in G. This is a contradiction. We have proved that for each cyclic subgroup U of order 2n we have W \ U Š C4 . n3 Set again H D W U1 so that we have H=W Š C2n2 . Set U1 D hai, y D a2 , y 2 D v, and y 4 D z. Assume that CH .W / > W . Then W hyi is abelian of type .8; 4/ and so all elements in .W hyi/ W are of order 8; in particular, all elements in the coset W y are of

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order 8. For each x 2 H .W ha2 i/, we have x 2 2 W y since all elements in n3 2 n2 2 .H=W / .W ha i=W / have order 2 , and so o.x / D 8 and o.x/ D 2n . We jH jjW haij 2nC1 D 2n1 D 4 D cn .G/ and so X D H D W U1 as stated in get cn .H / '.2n / part (a) of our theorem. We have in this case 2 .X/ D W . Until the end of part (i), we may assume that CH .W / D W . Since exp.Aut.W /2 / D 4, where Aut.W /2 is a Sylow 2-subgroup of Aut.W /, jH=W j 4, and so n D 3 or 4. Set W0 D 1 .W / and assume that the involutory automorphism of W induced by y stabilizes the chain W > W0 > f1g. (This will certainly happen if n D 4 because in that case y D a2 .) Let w 2 W be such that W D hwi hvi. We have w y D ws, where s 2 W0 and we compute .y.w i v j //2 D yw i v j yw i v j D y 2 .w i v j /y w i v j D v.ws/i v j w i v j D vs i w 2i v 2j ; where i , j are any integers. Since o.vs i w 2i v 2j / D 4, all elements in yW are of order 8. If n D 4, then for each x 2 H .hyiW /, x 2 2 yW and so c4 .H / D 32 W 8 D 4. If n D 3, then H D hyiW and so c3 .H / D 4. Hence we get again X D H D W U1 and 2 .X/ D W , as stated in part (a). It remains to consider the case CH .W / D W , where the involutory automorphism of W induced by y does not stabilize the chain W > W0 > f1g. By the above, n D 3, H D hyiW , and if we set W0 D hu; zi, then uy D uz and z 2 Z.G/. If w 2 W is such that w 2 D u, then setting w 0 D w y , we get .w 0 /2 D uz and W D hwihw 0 i. Note that CW .y/ D hvi and so .ww 0 /y D ww 0 implies ww 0 D v ˙1 . But replacing w with w 1 (if necessary), we may assume from the start that ww 0 D v and so w 0 D w 1 v. Since W0 hyi Š M24 , we get c3 .W0 hyi/ D 2 and we claim that c3 .H / D 2. We compute (for any s 2 W0 ) .yws/2 D y 2 .ws/y ws D vw 0 s y ws D v 2 s y s D zs y s; .yw 0 s/2 D y 2 .w 0 s/y w 0 s D vws y w 0 s D v 2 s y s D zs y s: Since o.zs y s/ 2, the above claim is proved. Hence there exists an element y 0 of order 8 in G H such that v 0 D .y 0 /2 2 W and y 0 induces on W an involutory automorphism which does not stabilize the chain W > W0 > f1g. Otherwise, we get (as above) c3 .hy 0 iW / D 4, which is a contradiction (since c3 .H / D 2). Since 0 z 2 Z.G/, we get uy D uz and therefore .y 0 /4 D z and v 0 D .y 0 /2 2 W0 hvi. Hence both y and y 0 normalize the abelian subgroup A D W0 hvi of type .4; 2/. Since W0 hy 0 i Š M24 , we get c3 .W0 hy 0 i/ D 2. It is easy to see that CG .W / D W . Indeed, if W < S CG .W / and jS W W j D 2, then S is abelian metacyclic and so S is of type .8; 4/ and c3 .S / D 4, a contradiction. Hence y and y 0 induce on W two distinct involutory automorphisms which do not stabilize the chain W > W0 > f1g. By our remark about Aut.W / at the beginning of the proof, yy 0 induces an involutory automorphism on W which

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113

obviously stabilizes the chain W > W0 > f1g. Since the coset W .yy 0 / cannot contain elements of order 8, we get .yy 0 /2 2 W0 A. The subgroup Ahy; y 0 i is of order 25 and since c3 .Ahy; y 0 i/ D 4, we get X D Ahy; y 0 i. Suppose that B D Ahyy 0 i is abelian. Then exp.B/ D 4 and all 16 elements in X B are of order 8. Since X D hU1 ; U2 ; U3 ; U4 i is normal in G, B is normal in G, and so (by assumption) 1 .B/ D 1 .A/ D W0 Š E4 and B is of type .4; 4/. (If 1 .B/ Š E8 , then G would possess a normal elementary abelian subgroup of order 8.) But uy D uz and so y does not stabilize the chain B > W0 > f1g. By the above, c2 .Bhyi/ D 2 which is a contradiction since Bhyi D X D hU1 ; U2 ; U3 ; U4 i. Hence CX .A/ D A and we have obtained a group stated in part (c) with X=A Š E4 . (ii) It remains to examine the second possibility, where G has a normal abelian subgroup W1 of type .4; 2/ such that N D CG .W1 / is abelian of type .2j ; 2/, j 2, and G=N is isomorphic to a subgroup of D8 . If j > 2, then G=N is elementary abelian of order 4. Indeed, if j > 2, then N contains a characteristic subgroup Z Š C4 . Since Z < W1 and Z is normal in G, it is easy to see that the nontrivial elements in G=N induce on W1 only involutory automorphisms. We set W0 D 1 .W1 / Š E4 and hzi D Ã1 .W1 / Z.G/. Suppose that G=N is not elementary abelian. Then j D 2, W1 D N D CG .W1 /, and G=N Š C4 or G=N Š D8 . In that case n 4. Suppose that U G and U Š C24 . Then U=.U \ W1 / Š C4 acts faithfully on W1 . But U centralizes the cyclic subgroup U \ W1 Š C4 , a contradiction. It follows n D 3. If L > W1 is such that L=W1 Š C4 , then (by the structure of Aut.C4 C2 /) 1 .L=W1 / D L0 =W1 inverts W1 . In particular, L0 does not contain a cyclic subgroup of order 8. Suppose first that L contains a cyclic subgroup U1 D hai of order 8. Then U1 covers L=W1 and U1 \ W1 D hzi is of order 2, where z D a4 and L0 D W1 hvi with v D a2 inverting W1 . For each w 2 W1 we compute .vw/2 D v 2 w v w D v 2 w 1 w D v 2 D z, and so all elements in L0 W1 D vW1 are of order 4. For each x 2 L L0 , x 2 2 L0 W1 and so all 16 elements in L L0 are of order 8. Hence c3 .L/ D 4 and so X D L is a group of order 25 as stated in part (c) of our theorem with X=W1 Š C4 . We may assume that exp.L/ < 8. In that case G=W1 Š D8 . Let U1 G and U1 D ha1 i Š C8 . Then U1 \ W1 D hvi Š C4 , where v D a12 . We may set W1 D hv; u j v 4 D u2 D Œv; u D 1; v 2 D zi, and then v a1 D v, ua1 D uz (since a1 does not centralize W1 ). Obviously, M D W1 U1 D ha1 ; ui Š M24 and so c3 .M / D 2. Hence there is another cyclic subgroup U2 D ha2 i in G which is not contained in M and U2 \ W1 Š C4 . Then a2 must induce on W1 an involutory automorphism which is distinct from that one induced by a1 . The only possibility is .vu/a2 D vu, ua2 D uz, and so v a2 D vz D v 1 , with U2 \ W1 D hvui. Again, P D W1 U2 D ha2 ; ui Š M24 and so c3 .P / D 2. Since a1 a2 inverts W1 , ha1 ; a2 i induces a four-group of automorphisms on W1 and therefore .MP W1 /=W1 Š E4 , where MP W1 D ha1 ; a2 iW1 . Note that c3 .ha1 a2 iW1 / D 0 (by a remark above). We get X D MP W1 and so we have obtained a group of order 25 stated in part (c) of our theorem with X=W1 Š E4 .

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It remains to consider the case where G=N is elementary abelian (of order 4) with N abelian of type .2j ; 2/, j 2. Assume first j D 2. In that case we have again N D CG .W1 / D W1 Š C4 C2 . Let U1 G with U1 Š C8 . Then U1 \ W1 Š C4 and (as before) W1 U1 Š M24 which gives c2 .W1 U1 / D 2. Hence there is U2 G with U2 Š C8 and U2 6 W1 U1 . Then again W1 U2 Š M24 and c2 .W1 U2 / D 2. We have G=W1 Š E4 and we have obtained again a group X D G of order 25 as stated in part (c). j We assume now j 3 and set N D hu; a j u2 D a2 D Œu; a D 1i. Then put j 2 v D a2 and z D v 2 . We see that W1 D hu; vi, z 2 Z.G/, hvi is a characteristic subgroup of N , and so hvi is normal in G. Obviously, n j C 1. Assume first that n D j C 1. Let x 2 G N be such that o.x/ D 2n D 2j C1 . Then x centralizes x 2 2 N and o.x 2 / D 2j . In particular, x centralizes hvi and so ux D uz (since x does not centralize W1 ). We get hxiN D hx; ui Š M2nC1 and so c2 .hxiN / D 2. It follows that G=N Š E4 and there is another element y 2 G .hxiN / with o.y/ D 2n . We get again that y centralizes hvi and uy D uz. But then xy centralizes W1 and xy 2 G N , a contradiction. We have proved that n j . Then cn .N / D 2. Take a0 2 hai N with o.a0 / D 2n . Then A D hui ha0 i is an abelian normal subgroup of type .2n ; 2/ and cn .A/ D cn .N / D 2. There is b0 2 G N with o.b0 / D 2n . Since o.b02 / D 2n1 , it follows that b02 2 A, b0 centralizes a cyclic subgroup of order 4 in W1 and so ub0 D uz. Also, hu; b0 i Š M2nC1 and so cn .hu; b0 i/ D 2. We get X D Ahb0 i. We see that hb02 i (of order 2n1 ) is contained in Z.X/. Set Z D Z.X/ and assume first that Z > hb02 i. Then Z is a cyclic subgroup of order 2n contained in A and hb0 iZ is abelian of order 2nC1 . Since hb0 iZ is not cyclic (because it contains two distinct cyclic subgroups Z A and hb0 i 6 A of order 2n ), there is an involution t 0 2 .hb0 iZ/ A. Since t 0 acts faithfully on W0 , it follows that W0 ht 0 i D D Š D8 and X D D Z. There is a subgroup Q Š Q8 contained in X (Appendix 16) so that X D Q Z with Q \ Z D Z.Q / D hzi and Z Š C2n . We have obtained a group X of order 2nC2 as stated in part (b) of our theorem. Assume now that Z.X/ D hb02 i is cyclic of order 2n1 . Suppose at the moment that 2 .X/ D W1 . In that case we can apply the classification result of (Lemma 42.1) in which all 2-groups X with jXj > 23 and j2 .X/j 23 are determined. Since jX W Z.X/j D 23 , X is not isomorphic to M2nC2 . Hence X is isomorphic to the metacyclic group G from part (c) of the above lemma. Since Z.G / Š C4 , it follows that n D 3 and jXj D jG j D 25 . But in that case c3 .G / D 6, which is a contradiction. We have proved that 2 .X/ > W1 . Hence there is an element x0 2 X A with o.x0 / 4. Since o.x02 / 2, it follows that x02 2 W0 and D D hW0 ; x0 i Š D8 because ux0 D uz. Thus X0 D D hb02 i is the central product of D Š D8 and Z.X/ D hb02 i, where D \ hb02 i D Z.D/ D hzi. Since jX0 j D 2nC1 , X0 is a maximal subgroup of X and obviously exp.X0 / D 2n1 . All 2nC1 elements in X X0 must be of order 2n (since cn .X/ D 4). There exists (as above) a quaternion subgroup Q X0 so that X0 D Q hb02 i. Since Q is a characteristic subgroup in X0

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(because it is a unique quaternion subgroup of X0 ), Q is normal in X and X D Qhb0 i. From Z.X/ D hb02 i follows that CX .Q/ D hb02 i and so b0 induces on Q D hx; yi an outer involutory automorphism with x b0 D y and .xy/b0 D .xy/1 . We compute .b0 x/2 D b02 x b0 x D b02 .yx/. If n D 3, then b02 .yx/ is an involution. In that case o.b0 x/ D 4, which is a contradiction (since all elements in X X0 must be of order 23 ). Hence we must have n > 3. In that case o.b0 x/ D 2n , as required. We have obtained a group stated in part (b) of our theorem. It remains to determine 2 .X/, where X is a group (of order 25 ) given in part (c) of our theorem with A being a self-centralizing (in X) abelian normal subgroup of type .4; 2/. We know that Aut.A/ Š D8 and Z.Aut.A// inverts A. Since jX=Aj D 4 and X=A acts faithfully on A, there is a subgroup B .> A/ of order 24 such that B=A inverts A. Then B A cannot contain elements of order 8 and so exp.B/ D 4. Since c3 .X/ D 4, all 16 elements in X B must be of order 8 and so B D 2 .X/. But B is nonabelian and so using the results in 52, we see that B is isomorphic to one of the following groups: Q8 C2 , D8 C2 , and Q8 C4 . It is easy to see that the last case cannot occur. Indeed, if B Š Q8 C4 , then, by 52, X Š Q8 C8 . But in that case A is not self-centralizing. Using again the results in 52, we see that in each of the first two cases the group X is uniquely determined. Proposition 54.3 (see 18). Let G be a 2-group possessing exactly one Ls -subgroup H of order 2sCm with a fixed integer m 2. Then G is either an Ls -group or a Us -group with respect to the kernel R D 1 .H /. Proof. Set R D 1 .H / so that R Š E2s , s 2, H=R is cyclic of order 2m , m 2, and R is normal in G. Set S=R D 1 .H=R/. Then we have S=R ˆ.H=R/ ˆ.G=R/ and 1 .S / D R. We use induction on jGj. If H D G, we are done. In the sequel let H < G. Let H M < G, where M is maximal in G. By induction, M is either an Ls -group or a Us -group with respect to the kernel R D 1 .H /. Note that the Frattini subgroup of an arbitrary p-group is not of maximal class. Since ˆ.G=R/ M=R, ˆ.G=R/ is cyclic (of order 2 ) or abelian of type .2; 2/, where the last case is possible only if M=R Š D8 . However, if ˆ.G=R/ is abelian of type .2; 2/, then a result of Nekrasov (see Proposition 4.9) implies ˆ.G=R/ Z.G=R/, and then M=R is abelian, a contradiction. Thus ˆ.G=R/ is cyclic and so S=R D 1 .ˆ.G=R//. Let F=R < G=R be cyclic of order 2m . Now, 1 .F=R/ ˆ.F=R/ ˆ.G=R/ and so 1 .F=R/ D S=R which gives 1 .F / D 1 .S / D R. It follows that F is an Ls -group of order 2sCm . By assumption, F D H and so G=R has exactly one cyclic subgroup of order 2m . By Exercise 1.104, G=R is either cyclic or of maximal class. If G=R is cyclic, 1 .G/ D 1 .S / D R and G is an Ls -group. Suppose that G=R is of maximal class and let T =R be a cyclic subgroup of index 2 in G=R. Since S=R ˆ.G=R/ < T =R, it follows S=R D 1 .T =R/ and 1 .T / D 1 .S / D R. This means that G is a Us -group with the kernel R and we are done.

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Proposition 54.4. Let G be a 2-group with cn .G/ D 2, n > 2. Then G is an L2 -group or a U2 -group. Proof. Since G is neither cyclic nor of maximal class, it has a normal abelian subgroup R of type .2; 2/. Let U1 be a cyclic subgroup of order 2n . Set H D U1 R and assume that R \ U1 D f1g. Let K=R be the subgroup of index 2 in H=R and let Z=R be the n1 subgroup of order 2 in K=R. Then Z Š E8 and for each x 2 H K, x 2 2 ZR n nC1 n nC1 n1 which gives o.x/ D 2 . But then cn .H / 2 W '.2 / D 2 W 2 D 4, a contradiction. Hence jU1 \Rj D 2 and so H is an L2 -group of order 2nC1 , i.e., H is either abelian of type .2n ; 2/ or H Š M2nC1 . In any case H is generated by its two cyclic subgroups of order 2n and so H is the unique L2 -subgroup of G of order 2nC1 . By Proposition 54.3, G is either an L2 -group or a U2 -group with the kernel R. Theorem 54.5. Let X be a 2-group of order > 24 with 2 .X/ Š D8 C2 or 2 .X/ Š Q8 C2 . Then jXj D 25 and for each of the two possibilities for 2 .X/, X is uniquely determined and is given explicitly in Theorems 52.2(a) and 52.1(d). In both cases ˆ.X/ is abelian of type .4; 2/, X 0 D 1 .ˆ.X// Š E4 , Z.X/ D Ã1 .ˆ.X// is of order 2, and c3 .X/ D 4. Also, each maximal subgroup of X is nonabelian and therefore A D ˆ.X/ is self-centralizing in X. Suppose that G is a 2-group with G > X and c3 .G/ D 4. Then we have the following possibilities: (a) CG .A/ D A, G=A Š D8 , and G has a normal elementary abelian group of order 8. (b) CG .A/ D C Š C4 C4 , C is self-centralizing in G, and G=C Š E4 or D8 . Proof. The first part of this proposition is a direct consequence of the results in 52. Suppose that X has an abelian maximal subgroup. Then the formula jXj D 2jZ.X/jjX 0 j (see Lemma 1.1) gives a contradiction (since jXj D 25 ). We have to determine the structure of G, where G > X and c3 .G/ D 4 D c3 .X/. This implies that X is normal in G. If CG .A/ D A, then G > X and Aut.A/ Š D8 imply G=A Š D8 . We have 2 .X/ > A, j2 .X/ W Aj D 2, and 2 .X/=A D Z.G=A/. Since X=A Š E4 , there is an element x 2 G X such that x 2 2 2 .X/ A. But o.x 2 / 4 and x is not of order 8. Hence x 2 D t is an involution which gives 2 .X/ Š D8 C2 and Z.2 .X// D 1 .A/ D X 0 . Thus E D ht i 1 .A/ is an elementary abelian normal subgroup (of order 8) of X. Since x centralizes t and normalizes 1 .A/ D X 0 , it follows that E G G, as required. We have to study the case C D CG .A/ > A so that we get C \ X D A, C is normal in G, .XC /=C Š X=A Š E4 , and G=C Š E4 or D8 . It remains to determine the structure of C . Since C has no elements of order 8, we get exp.C / D 4. Let s be an element of order 8 in X. Then s 2 2 A D ˆ.X/, D D Ahsi is of order 24 , and 1 .A/ D 1 .D/ is a normal 4-subgroup of D. Since every maximal subgroup of X is nonabelian, it follows that D Š M24 and c3 .D/ D 2. Set K D C hsi D CD.

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Because K \ X D D, we have c3 .K/ D 2 and D is the unique L2 -subgroup of K of order 24 . By Proposition 54.3, K (of order 25 ) is an L2 -group or a U2 -group with the kernel R D 1 .A/ D 1 .D/. If K is an L2 -group, then K Š M2m , m > 4, a contradiction. Indeed, M2m is minimal nonabelian contrary to the fact that K contains the nonabelian proper subgroup D. We have proved that K is a U2 -group with the kernel R D 1 .A/, where K=R is of maximal class. Let T =R be a cyclic subgroup of index 2 in K=R. Then jT =Rj 4 and 1 .T / D R. Let T1 =R be the cyclic subgroup of order 4 in T =R. Then T1 is an L2 -subgroup of K of order 24 and so T1 D D. If T1 ¤ T , then CT .R/ T1 and so T1 D D would be abelian, a contradiction. Hence T1 D T D D and so jKj D 25 . Since jK W C j D 2, C is of order 24 . But A Z.C / and so C is abelian. Assume that E0 D 1 .C / > 1 .A/ so that E0 is a normal elementary abelian subgroup of order 8 in G. Consider the subgroup X0 D E0 hsi, where E0 \ hsi D hs 4 i is of order 2. We see (as in the proof of Theorem 54.1) that c3 .X0 / D 4. This implies X0 X, a contradiction. We have proved that C is of rank 2 and so C Š C4 C4 (since exp.C / D 4). Our proposition is proved. Next we shall classify 2-groups G appearing in Theorems 54.1 and 54.2. Theorem 54.6. Let G be a 2-group with cn .G/ D 4, n > 2, and suppose that G has a normal elementary abelian subgroup E of order 8. Let fU1 ; U2 ; U3 ; U4 g be the set of four cyclic subgroups of order 2n . Then X D hU1 ; U2 ; U3 ; U4 i D EU1 , where E \ U1 ¤ 1 and we have the following possibilities. (a) 2 .X/ Š C4 C2 C2 and G=E is either cyclic (of order 2n1 ) or G=E is of maximal class (and order 2n ) and for n D 3 the last group must be dihedral. (b) n D 3, 2 .X/ Š D8 C2 , and one of the following holds: (b1) G D X, which is a uniquely determined group of order 25 ; (b2) G has a self-centralizing abelian normal subgroup A of type .4; 2/ such that G=A Š D8 ; (b3) G has a self-centralizing abelian normal subgroup W of type .4; 4/ such that G=W Š E4 or D8 . Proof. Theorem 54.1 implies that X D EU1 with E \ U1 ¤ 1 and 2 .X/ Š C4 C2 C2 or 2 .X/ Š D8 C2 , where in the second case n D 3. Suppose that 2 .X/ Š C4 C2 C2 . Then 1 .X/ D E Š E23 and X=E Š C2n1 with jX=Ej 4. Hence X is an L3 -group of order 2nC2 . Since X is generated by its cyclic subgroups of order 2n , it follows that X is the unique L3 -subgroup of order 2nC2 in G. By Proposition 54.3, G is either an L3 -group or a U3 -group with respect to the kernel R D 1 .X/ D E. In particular, G=E is either cyclic (of order 2n1 ) or G=E is of maximal class (and order 2n ). If in the second case n D 3, then G=E must be dihedral. Indeed, if n D 3 and G=E is of maximal class but not dihedral, then there

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is an element x 2 G X such that x 2 2 2 .X/ E, where 2 .X/=E D Z.G=E/. But then o.x/ D 8, a contradiction. Assume now that 2 .X/ Š D8 C2 in which case n D 3. Our result follows at once from Theorem 54.5. Theorem 54.7. Let G be a 2-group with cn .G/ D 4, n > 2, and suppose that G does not have a normal elementary abelian subgroup of order 8. Let fU1 ; U2 ; U3 ; U4 g be the set of four cyclic subgroups of order 2n . Then the subgroup X D hU1 ; U2 ; U3 ; U4 i is of order 2nC2 and we have the following possibilities. (a) X D W U1 with W \ U1 Š C4 , where W is an abelian normal subgroup of type .4; 4/, 2 .X/ D W , X is metacyclic, CG .W / is metacyclic and one of the following holds: (a1) n > 3, and G=W is either cyclic (of order 2n2 ) or G=W is of maximal class (and order 2n1 ) and for n D 4 the last group must be dihedral; (a2) n D 3, G has a normal metacyclic subgroup N such that CG .W / N , X N , and G=N is isomorphic to a subgroup of D8 . Moreover, CG .W /=W is cyclic, N=W is either cyclic or generalized quaternion, and in the second case G=N is elementary abelian of order 4. (b) n > 2, X D QU1 , where Q Š Q8 is a normal quaternion subgroup of G, G D QP with Q \ P D Z.Q/, U1 P , CG .Q/ P , jP W CG .Q/j 2, and P is either cyclic or of maximal class. (c) n D 3, jXj D 25 , 2 .X/ Š Q8 C2 or 2 .X/ Š D8 C2 and if G > X, then G has a self-centralizing normal abelian subgroup W of type .4; 4/ with G=W Š E4 or D8 . Proof. We may assume that G is nonabelian. (i) We consider first the groups G from Theorem 54.2(a). In this case X D W U1 with U1 Š C2n , W \ U1 Š C4 , where W is an abelian normal subgroup of type .4; 4/ and 2 .X/ D W . Also, X and CG .W / are metacyclic. Assume that n > 3. Then X=W is cyclic of order 2n2 4. Set S=W D 1 .X=W / so that S=W ˆ.X=W / ˆ.G=W /. Since S=W stabilizes the chain W > 1 .W / > f1g, it follows (as in the proof of Theorem 54.2) that all elements in S W are of order 8. Indeed, set U1 \ S D hd i, where d is an element of order 8. For each w 2 W , w d D wl with l 2 1 .W /. Then .dw/2 D dwdw D d 2 w d w D d 2 wlw D d 2 w 2 l, where w 2 l 2 1 .W / and so o.dw/ D 8. Also, X is generated by its elements of order 2n . We want to show that G=W is either cyclic or of maximal class and we use an induction on jGj. If X D G, we are done. In the sequel let X < G. Let X M < G, where M is maximal in G. By induction, M=W is either cyclic or of maximal class. We have ˆ.G=W / M=W and since ˆ.G=W / cannot be of maximal class

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2-groups G with cn .G/ D 4, n > 2

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(Burnside), ˆ.G=W / is either cyclic (of order 2) or abelian of type .2; 2/ (the last case being possible only if M=W Š D8 ). However, if ˆ.G=W / Š E4 , then ˆ.G=W / Z.G=W / (see Proposition 4.9). But then M=W is abelian, a contradiction. Thus ˆ.G=W / is cyclic and so S=W D 1 .ˆ.G=W //. Let F=W be a cyclic subgroup of order 2n2 in G=W . Now, 1 .F=W / ˆ.F=W / ˆ.G=W / and so 1 .F=W / D S=W . But all elements in S W are of order 8 and so cn .F / D 4 and F is generated by its elements of order 2n . Indeed, let n3 F0 =W D ˆ.F=W / and x 2 F F0 . Then x 2 2 S W and so o.x/ D 2n . This gives F D X and so X=W is the unique cyclic subgroup of order 2n2 in G=W . As in the proof of Proposition 54.3, G=W is either cyclic (of order 2n2 ) or G=W is of maximal class (and order 2n1 ) and for n D 4 the last group (obviously) must be dihedral. It remains to consider the case n D 3. In this case jXj D 25 and all 16 elements in X W are of order 8. Suppose first that X is nonabelian. Then CG .W / \ X D W and CG .W / is metacyclic. If CG .W / > W , then CG .W / (being metacyclic) has elements of order 8 (which are not contained in X), a contradiction. We have proved that CG .W / D W . By Theorem 50.1, the group G has a normal metacyclic subgroup N such that W N , 2 .N / D W , and G=N is isomorphic to a subgroup of D8 and we assume that N is maximal subject to these conditions. In that case N > W . Indeed, if N D W , then G=W is isomorphic to a subgroup of D8 . But X is a normal metacyclic subgroup of G, W X, 2 .X/ D W , and G=X (being a factor-group of D8 ) is also isomorphic to a subgroup of D8 . This contradicts the maximality of N and so N > W . In that case there are elements of order 8 in N and so X N . Suppose that Y =W is a subgroup of order 2 in N=W . Since 2 .Y / D W , all elements in Y W are of order 8. Hence Y D X and so N=W has exactly one subgroup of order 2. By the structure of Aut.W / (Proposition 50.5), we see that Aut.W / has no quaternion subgroups. Thus N=W is cyclic (of order 4). We have obtained all properties stated in part (a2) of our theorem. We consider now the remaining case, where X D W U1 is abelian of order 25 (and of type .8; 4/). According to Theorem 50.1, G has a normal metacyclic subgroup N such that 2 .N / D W , C D CG .W / N , and G=N is isomorphic to a subgroup of D8 . Obviously, X C . Suppose that Y =W is a subgroup of order 2 in N=W . Since 2 .Y / D W , all elements in Y W are of order 8. Hence Y D X and so N=W has exactly one subgroup of order 2. It follows that N=W is either cyclic or generalized quaternion. We claim that C =W is cyclic. Suppose false. Then C =W is generalized quaternion. Let C0 =W Š Q8 be a quaternion subgroup in C =W , where X < C0 (because X=W is the unique subgroup of order 2 in N=W ). Let Z1 =W and Z2 =W be two distinct cyclic subgroups of order 4 in C0 =W . Then Z1 and Z2 are both abelian, hZ1 ; Z2 i D C0 , and Z1 \ Z2 D X. This implies that X Z.C0 / and so X D Z.C0 /. Since C0 is metacyclic, C00 is cyclic and C00 covers X=W . All elements in X W are of order 8, so jC00 j D 8. But then the formula (see Lemma 1.1) jC0 j D 27 D 2jZ.C0 /jjC00 j gives

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a contradiction. We have proved that CG .W /=W is cyclic. Thus CG .W / D CG .X/ is abelian. We have to consider the case, where N=W is generalized quaternion. We set K=W D .N=W /0 D ˆ.N=W / so that K=W is cyclic and Z.N=W / D X=W . We have N=K Š E4 and since N is metacyclic (and therefore d.N / D 2), we get K D ˆ.N / and N 0 is cyclic. On the other hand, N 0 covers K=W and all elements in X W are of order 8. Hence N 0 \ X Š C8 , N 0 \ W Š C4 , and jK W N 0 j D 4. Again, since N is metacyclic, there is a cyclic normal subgroup S of N such that N 0 < S and jS W N 0 j D 2. If S K, then jS \ W j D 8. But a maximal subgroup S \ W of W is not cyclic since S \ W ˆ.W / D 1 .W / Š E4 . This contradiction shows that S \ K D N 0 . Set T D W S D KS so that T =W is a cyclic subgroup of index 2 in N=W . For each x 2 N T , we have x 2 2 X W and so all elements in N T are of order 16. Also, T =S Š W =.S \ W / Š C4 . Now, ˆ.T / K and ˆ.T / hN 0 ; 1 .W /i since N 0 D Ã1 .S / and 1 .W / D ˆ.W /. Since jK W .N 0 1 .W //j D 2 and T is not cyclic, we get ˆ.T / D N 0 1 .W /, F D ˆ.T / \ X is abelian of type .8; 2/, and F1 D ˆ.T / \ W is abelian of type .4; 2/. m3 m2 Set S D hai, where o.a/ D 2m , m 4. Also set s D a2 , v D a2 , and m1 2 zDa , so that z 2 Z.G/, hsi D S \ X, hvi D S \ W , and hzi D S \ 1 .W /. Obviously, hsi and hvi are normal subgroups of G. Since ˆ.N / D K, there is an element b 2 N T (of order 16) such that b 2 2 X W and b 2 62 F . Then b 2 D ws, where w 2 W F1 and W D hwi hvi. We set w 2 D u and so 1 .W / D hu; zi. We compute b 4 D w 2 s 2 D uv and b 8 D z. Hence N D haihbi, where hai is a cyclic normal subgroup of order 2m .m 4/ of N and hbi is cyclic of order 16 with hai \ hbi D hzi (of order 2). It remains to determine the action of hbi on hai. Now, hbi induces a cyclic automorphism group on hai so that b inverts hai=hvi (since N=W is generalized quaternion). Thus ab D a1 v0 with v0 2 hvi and so .a4 /b D .ab /4 D .a1 v0 /4 D a4 . In particular, b inverts hvi. On the other hand, b centralizes b 4 D uv. We compute uv D .uv/b D ub v b D ub v 1 , and so ub D uz. The last relation is crucial and shows that b does not stabilize the chain W > 1 .W / > f1g. If A (of order 25 ) is a Sylow 2-subgroup of Aut.W / and B is the full stabilizer group of the chain W > 1 .W / > f1g, then B is elementary abelian of order 24 and jA=Bj D 2 (see Proposition 50.5). We note that CG .W / N and CG .W /=W is a cyclic normal subgroup of N=W and so CG .W / T . Hence, if L is the set of all y 2 G which stabilize the chain W > 1 .W / > f1g, then the subgroup L covers G=N (since b 62 L) and so G=N (being isomorphic to a subgroup of D8 ) is elementary abelian of order 4. We have obtained all properties stated in part (a2) of our theorem. (ii) We consider now the groups G from Theorem 54.2(b). In this case n > 2, X D QU1 , where Q Š Q8 is a normal quaternion subgroup of X, Q \ U1 D Z.Q/, U1 D hbi Š C2n , and b either centralizes Q or b induces on Q an involutory outer automorphism in which case n > 3.

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Assume first that b centralizes Q. In that case X D Q hbi with Q \ hbi D n2 Z.Q/ D hzi. Set v D b 2 so that 2 .X/ D Q hvi. Since Q is the unique quaternion subgroup in 2 .X/, it follows that Q is characteristic in X and so Q is a normal subgroup in G. Set C D CG .Q/ so that X \ C D U1 D hbi, C is normal in G, and jG W .Q C /j 2. Since U1 is the unique cyclic subgroup of order 2n in C , it follows that C is either cyclic or of maximal class. If G D Q C , we are done. Suppose that jG W .Q C /j D 2. Since G=C Š D8 (a Sylow 2-subgroup of Aut.Q8 //, there is an element d 2 G .QC / with d 2 2 C . Set P D C hd i so that G D QP and P \ X D U1 . Since U1 is the unique cyclic subgroup of order 2n in P , we get again that P is either cyclic or of maximal class. It remains to consider the case, where b induces on Q an outer involutory automorphism (and in that case n > 3). Obviously, Q is the unique quaternion subgroup of Qhb 2 i D Q hb 2 i, where Q \ hb 2 i D hzi. We have exp.Qhb 2 i/ D 2n1 and all elements in X .Qhb 2 i/ are of order 2n . Hence Q is also the unique quaternion subgroup of X and so Q is normal in G. Again set C D CG .Q/ so that C is normal in G and X \ C D hb 2 i. Also, jG W .QC /j D 2 and so G D XC . Set P D C hbi. We have jP W C j D 2, P \ X D hbi D U1 , and G D QP with Q \ P D hzi. Since U1 is the unique cyclic subgroup of order 2n in P , it follows that P is either cyclic or of maximal class and we are done. (iii) Finally, we consider the groups G from Theorem 54.2(c). In this case n D 3, jXj D 25 , and 2 .X/ Š Q8 C2 or 2 .X/ Š D8 C2 . Since G has no normal subgroups isomorphic to E8 , Proposition 54.5 gives at once the result stated in part (c) of our theorem.

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2-groups G with small subgroup hx 2 G j o.x/ D 2ni

In 52, 2-groups G with j2 .G/j D 16 are classified. In this section we consider essentially more difficult problem to classify the 2-groups in which all elements of order 4 generate the subgroup of order 16. In what follows we suppose that G is a 2-group of order > 24 all of whose elements of order 4 generate a subgroup H of order 24 . In that case, obviously, H is not of maximal class so it has no cyclic subgroups of index 2 (see Theorem 1.2). It is obvious that, in the case under consideration, c2 .G/ 4. Indeed, if the above assertion is false, then c2 .G/ D 2, by Theorem 1.17(b). In that case, by 52, the elements of order 4 generate the abelian subgroup of type .4; 2/, contrary to our hypothesis (this also follows from 43). If c2 .G/ D 4, then H Š Q8 C4 or H Š C4 C2 C2 and the group G was completely determined in 52. If c2 .G/ D 5, then G is of maximal class (Theorem 1.17(b)) and so H Š Q16 and (since jGj > 24 ) G Š SD32 . It remains to examine the case c2 .G/ D 6 since there is no 2-group G with c2 .G/ D 7 (Theorem 1.17(b)). The 2-groups G with j2 .G/j D 24 have been determined in 52, and so we may also assume j2 .G/j > 24 . Exercise 1. Let G be a 2-group, H D hx 2 G j o.x/ D 4i. If exp.Z.H // D 2, then CG .H / D Z.H /. Solution. Assume that CG .H / > Z.H /. Let x 2 CG .H / Z.H / be such that x 2 2 Z.H /. Then o.x/ 4 so x is an involution since x 62 H . If a 2 H is of order 4, then o.ax/ D 4 and ax 62 H , a contradiction. Let G be a 2-group and let 2 .G/ D hx 2 G j o.x/ D 22 i. Suppose that j2 .G/j D 24 and c2 .G/ > 4; then c2 .G/ 6 (Theorem 1.17(b)) so G contains at least twelve elements of order 4. Then c2 .G/ D 6 since c1 .G/ 3. Theorem 55.1 ([Jan8]). Let G be a 2-group of order > 24 all of whose elements of order 4 generate a subgroup H of order 24 , i.e., H D 2 .G/. Assume, in addition, that c2 .G/ D 6 and j2 .G/j > 24 . Then we have the following possibilities: (a) H Š Q8 C2 and G Š SD16 C2 . (b) H Š ha; b j a4 D b 4 D 1; ab D a1 i is metacyclic and G D hb; t j b 4 D t 2 D 1; b t D ab; a4 D 1; ab D a1 ; at D a1 i. Here jGj D 25 , H D ha; bi, ˆ.G/ D ha; b 2 i Š C4 C2 , 2 .G/ D G, Z.G/ D ha2 ; b 2 i Š E4 .

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(c) H Š C4 C4 and G has a metacyclic maximal subgroup M such that 2 .M / D H , G D M ht i, where t is an involution with CM .t / D 1 .M / Š E4 . Proof. Since H is neither cyclic nor of maximal class, H has a normal four-subgroup H0 (Lemma 1.4) and so all twelve elements in H H0 are of order 4. If H is abelian, then H0 D 1 .H / implies that H D C4 C4 . Suppose that H is nonabelian. Then H=H0 is elementary abelian and so ˆ.H / H0 . Suppose, in addition, that H is not minimal nonabelian and let Q be a nonabelian subgroup of order 8. Since H has only three involutions, we have Q Š Q8 . If CH .Q/ Q, then H is of maximal class (Proposition 10.17), a contradiction. Hence H D Q Z, where jZj D 4 and Q \ Z D Z.Q/. If Z is cyclic, then c2 .H / D 4, a contradiction. Hence Z Š E4 and H Š Q8 C2 . Now suppose that H is minimal nonabelian. Then d.H / D 2 so ˆ.H / D H0 and jH 0 j D 2. In that case, H=H 0 is abelian of type .4; 2/. Let Z=H 0 be a direct factor of H=H 0 of order 2. Since H=Z is cyclic of order 4, we get Z ¤ ˆ.H / D 1 .H / so Z is cyclic. Let Z D hai and H=Z D hbZi; then hbi \ hai D f1g. Since Aut.Z/ Š C2 , we get H D ha; b j a4 D b 4 D 1; ab D a1 i. We have to determine the structure of G for each of the above three possibilities for the structure of H . (i) Assume that H D Q hui, where Q Š Q8 and u is an involution. Set hzi D Z.Q/ so that Z.H / D hu; zi. There are exactly three maximal subgroups of H containing Z.H / and they are abelian of type .4; 2/. One of them, say hu; ai (a 2 Q Z.Q/), is normal in G (consider the action of G by conjugation on the set of above three maximal subgroups of H ). If b 2 Q hai, then the other two such abelian subgroups of type .4; 2/ are hu; bi D hui hbi and hu; abi D hui habi. Since 2 .G/ > H , there is an involution t 2 G H since H contains all elements of order 4 in G. Let hu; xi, where x 2 Q Z.Q/, be any t -invariant abelian subgroup of type .4; 2/ in H . If t acts nontrivially on Z.H /, then ht; Z.H /i Š D8 has a cyclic subgroup of order 4 that is not contained in H , a contradiction. Thus t centralizes Z.H /. Since CH .t / must be elementary abelian (otherwise, .ht i CH .t // CH .t / has an element of order 4), we get CH .t / D Z.H / and so t acts fixed-point-free on hu; xi Z.H /. If x t D xs with s 2 Z.H / hzi, then .tx/2 D .txt /x D x t x D xsx D sx 2 D sz 2 Z.H /# , and so o.tx/ D 4, a contradiction since tx 62 H . Hence x t D x 1 and so t inverts the subgroup hu; xi. In particular, t inverts hu; ai. If t normalizes hu; bi, then t also normalizes hu; abi. But in that case t inverts hu; bi and hu; abi and so t inverts H D hu; ai [ hu; bi [ hu; abi. This implies that H is abelian (Burnside) which is not the case. Hence we must have hu; bit D hu; abi. Let M D NG .hu; bi/ so that H M , jG W M j D 2 and G D M ht i (see the last sentence of the previous paragraph). By the previous paragraph, the set M H has no involutions so 2 .M / D H since exp.H / D 4 and H contains all elements of order 4 from G. Assume that M > H so that we can apply the results of 52 (see Theorem 52.1(d)). It follows that M is a uniquely determined group of order 25

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which is given explicitly in Theorem 49.1, (A2)(a)). There is an element y 2 M H of order 8 (indeed, exp.M / D 8 and H D 2 .M /) such that y 2 D ua, uy D uz, ay D a1 , b y D bu. Hence y acts nontrivially on Z.H / and y normalizes all three abelian subgroups of type .4; 2/ in H . We have G=H Š E4 since M=H and H ht i=H are two distinct subgroups of order 2 in G=H . Hence .yt /2 2 H and we consider the subgroup N D H hyt i of order 25 . Since yt acts nontrivially on Z.H / (t centralizes Z.H /), it follows that there are no involutions in N H (indeed, if w is an involution in N H , then hw; Z.H /i Š D8 contains an element of order 4 which is not contained in H , a contradiction). Thus H D 2 .N / and so, by the above, N is uniquely determined and therefore N Š M . In particular, N normalizes all three abelian subgroups of type .4; 2/ in H . Since M ¤ N , we get that G D hM; N i normalizes all three abelian subgroups of type .4; 2/ in H (see the last sentence of the previous paragraph), a contradiction. We have proved that H D M and so G D H ht i. Since hu; bit D hu; abi, we have b t D abs0 with s0 2 Z.H /. Hence t normalizes Q D hb; b t i Š Q8 and H D hui Q . Replacing Q with Q , we may assume from the start that t normalizes Q and so t induces on Q an outer automorphism (indeed, hbi is not t -invariant). We get Qht i D D Š SD24 (if D 6Š SD24 , then cl.D/ D 2; in that case, hbi G D so t -invariant, which is not the case) and G D hui D. We have obtained the possibility (a). (ii) Now suppose that H D ha; b j a4 D b 4 D 1; ab D a1 i is metacyclic of order 16. Here ˆ.H / D Z.H / D ha2 ; b 2 i D 1 .H / Š E4 , H 0 D ha2 i, all elements in H Z.H / are of order 4 and H=H 0 is abelian of type .4; 2/. It follows that H is minimal nonabelian. Therefore all three maximal subgroups Wi , i D 1; 2; 3, of H are abelian of type .4; 2/: W1 D hZ.H /; ai with Ã1 .W1 / D ha2 i; W2 D hZ.H /; bi with Ã1 .W2 / D hb 2 i; W3 D hZ.H /; abi with Ã1 .W3 / D hb 2 i since .ab/2 D ab 2 ab D ab 2 a1 D b 2 . It follows that a2 b 2 is not a square in H.D W1 [ W2 [ W3 / and W1 is normal in G. (Note, however, that all involutions in a homocyclic 2-group of composite exponent are squares; see, for example, Theorem 6.1.) Hence ha2 i D H 0 , hb 2 i, and ha2 b 2 i are characteristic subgroups in H and therefore they are normal in G. This gives Z.H / Z.G/. In fact we have Z.H / D Z.G/, by Exercise 1. Since 2 .G/ > H , there is an involution t 2 G H . By Exercise 1, CH .t / D Z.H / D Z.G/. If t normalizes a maximal subgroup Wi of H , then t inverts Wi , by Exercise 1. We know that t normalizes W1 and so t inverts W1 and therefore at D a1 : If t normalizes W2 , then t also normalizes W3 and consequently t inverts H (see the previous paragraph). But then H is abelian, which is not the case. We have proved that W2t D W3 . Let K D NG .W2 / .D NG .W3 // so that H K, jG W Kj D 2, and G D Kht i (see the last sentence of the previous paragraph). But then the set K H has no involutions (otherwise, that involution inverts H so H is abelian, which is not the case) and so 2 .K/ D H . Since H is metacyclic, it follows from Theorem 41.1 and Remark 41.2 that K is also metacyclic. If K > H , then [Ber25, Remark R48.2]

55 2-groups G with small subgroup hx 2 G j o.x/ D 2n i

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implies that H Š C4 C4 , which is not the case. Hence we have K D H and so G D H ht i is of order 25 . Since W2t D W3 , we have b t 2 fab; .ab/1 ; .ab/a2 ; .ab/1 a2 g. Note that a b D a2 b and so .ab/a D a.a2 b/ D .ab/a2 ;

..ab/1 /a D ..ab/a /1 D ..ab/a2 /1 D .ab/1 a2 :

Hence, replacing t with the involution t a (if necessary), we see that we may assume that b t 2 fab; .ab/1 D .ab/b 2 D ab 1 g. If b t D ab 1 , then we replace a with a0 D ab 2 . We get b t D ab 1 D ab 3 D .ab 2 /b D a0 b, .a0 /b D .a0 /1 , .a0 /t D .a0 /1 , and see that one may assume from the start (writing again a instead of a0 ) that b t D ab. The structure of G is uniquely determined as stated in part (b). (iii) Finally, assume that H D hai hbi Š C4 C4 . Set C D CG .H /. By Exercise 1, 2 .C / D H and therefore C is metacyclic. Since, by assumption, 2 .G/ > H , there is an involution t 2 G C . Suppose that the coset C t contains an element y which is not an involution. Then o.y/ 8 and o.y 2 / 4. In particular, y o.y/=4 D d is an element of order 4 in H . We have y D ct with c 2 C . But then d D d y D d ct D d t , contrary to Exercise 1. Hence, all element in C t are involutions. This implies that t inverts C and C is abelian (Burnside). Since 2 .C / D H , we conclude that C is of rank 2. Suppose that t 0 is an involution in G .C ht i/. Then, by the above, t 0 inverts C . But then t t 0 2 G C and t t 0 centralizes C , a contradiction. Thus C ht i D 2 .G/. Set D D 2 .G/. We have CG .t / D ht i H0 D CD .t /, where H0 D 1 .H / D ha2 ; b 2 i. Indeed, if CG .t / 6 2 .G/, then there is an element y 0 2 CG .t / CD .t / such that .y 0 /2 2 CD .t / D 1 .H / so o.y 0 / 4 and y 0 2 H , which is a contradiction. The last result gives an upper bound for the order of G=C . The conjugacy class of t in G is locked in D C and so there are at most jD C j D jC j conjugates of t in G. This gives 18 jGj D jG W CG .t /j jC j and so jG=C j 8. Also note that D=C is a normal subgroup of order 2 in G=C . Suppose that the involution in D=C is a square in G=C . Then there is an element k 2 G D such that k 2 2 D C . But all elements in D C are involutions and so k is an element of order 4, a contradiction. It follows that G=C is abelian and D=C 6 ˆ.G=C /. There is a maximal subgroup M of G such that D \ M D C . Since 2 .M / D H , it follows that M is metacyclic. We have G D M ht i, where CM .t / D 1 .M / D 1 .H / Š E4 . We have obtained a group stated in part (c) of our theorem which is now completely proved. Let n > 1 and let G be a p-group. Denote n .G/ D hx 2 G j o.x/ D p n i. We have n .G/ D f1g if exp.G/ < p n . Proposition 55.2 (Berkovich). Let G be an irregular p-group. Suppose that we have j2 .G/j D p pC1 < p m D jGj. Then the following assertions hold: (a) If p D 2, then G Š SD24 and H Š Q8 (see Theorems 43.4 and 52.8). In the sequel we assume that p > 2.

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(b) If 2 .G/ D 2 .G/, then G is not of maximal class and 2 .G/ is regular (see Lemma 42.1). Moreover, G is a group of Lemma 42.1. (c) If G is of maximal class and p > 3, then 2 .G/ D G1 , where G1 is the fundamental subgroup of G. In this case every irregular maximal subgroup of G contains exactly p subgroups of order p p and exponent p, and G contains exactly p 2 such subgroups. Next, m D p C 2. (d) If p D 3 and G is of maximal class, then 2 .G/ D 2 .G1 / is metacyclic, all nonmetacyclic subgroups of order 33 are nonabelian and their number equals 3m3 . In the sequel we assume that G is not of maximal class. (e) If 2 .G/ is regular, then 2 .G/ D 2 .G/ (see (b)). In the sequel we assume that 2 .G/ < 2 .G/. (f) Suppose that p > 3. Let t be an element of order p in G 2 .G/ and set D D ht; 2 .G/i. Then d.D/ D 3, exactly p 2 maximal subgroups of D are of maximal class; other maximal subgroups of D are of exponent p. (g) All pC1 maximal subgroups of 2 .G/ are normal in G. ˆ.2 .G// is contained in a G-invariant subgroup L of order p p and exponent p such that L 6 2 .G/. Proof. (a) Let p D 2. Then c2 .G/ D c2 .2 .G// 2 f2; 3g. If c2 .G/ D 3, then G is of maximal class by Theorem 1.17(b). In that case, G Š SD24 . If c2 .G/ D 2, then such G are described in Theorems 43.4 and 52.8. In what follows we assume that p > 2. (b) The p-groups with j2 .G/j D p pC1 are classified in Lemma 42.1. In all cases, 2 .G/ is regular. (c) Let G be of maximal class and p > 3. In this case, j2 .G1 /j p pC1 since j1 .G1 /j p p1 and G=1 .G1 / has no cyclic subgroups of index p (see Theorems 9.5 and 9.6). It follows that 2 .G/ D 2 .G1 / and j2 .G1 /j D p pC1 < p 2p2 since p > 3. In that case, G1 D 2 .G1 / D 2 .G/. If M is an irregular maximal subgroup of G, then 2 .M / D M \ 2 .G/ < M and 2 .M / is the unique absolutely regular maximal subgroup of M . We conclude that M has exactly p distinct maximal subgroups of exponent p. Since ˆ.G/, the intersection of any two distinct maximal subgroups of G, is absolutely regular, all remaining assertions of (c) are true. (d) Now let G be a 3-group of maximal class and order 35 . Since, 2 .G1 / D 2 .G/ is of order 34 , we get 2 .G/ D 2 .G1 /. The remaining assertions in (d) are trivial since the number of irregular subgroups of order 34 in G equals 3m4 (Exercise 9.27) and every such subgroup contains exactly three subgroups of order 33 and exponent 3 (see the proof of (c)). All subgroups of order 33 and exponent 3 are nonabelian since m > 4 (this is an easy consequence of Theorems 9.5 and 9.6). (e) follows from Theorem 7.2. In what follows we assume that 2 .G/ < 2 .G/. (f, g) By hypothesis, G 2 .G/ has an element t of order p. Set D D ht; 2 .G/i. Assume that D is of maximal class. Let D1 be the fundamental subgroup of D (see 9, Theorems 9.5 and 9.6). Since p > 3 and D1 is absolutely regular of width p 1 and

55 2-groups G with small subgroup hx 2 G j o.x/ D 2n i

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order p pC1 , we get 2 .D1 / D 2 .D1 / D D. It follows that c2 .G/ D c2 .2 .G// is not divisible by p p1 so G is of maximal class (Theorem 13.2(b)), contrary to the assumption. Thus, D is not of maximal class. By Theorem 12.12, D contains exactly p 2 subgroups M1 D 2 .G/; : : : ; Mp 2 of maximal class and index p. We have jMi j D p pC1 for all i . All other p C 1 maximal subgroups of D are not generated by two elements so they are regular (Theorem 12.12). Let M be one of regular maximal subgroups of D. Supposing that exp.M / D p 2 , we get 2 .M / D M 6 2 .G/, a contradiction. Thus, all regular maximal subgroups of D have exponent p. We see that N D M \ 2 .G/ is a maximal subgroup of exponent p in 2 .G/. If M1 is another maximal subgroup of exponent p in D, then N1 D M1 \ 2 .G/ is a maximal subgroup of exponent p in 2 .G/. By Theorem 12.12(c), M \ M1 D .D/ and .D/ 6 2 .G/. It follows that N ¤ N1 . We see that the numbers of subgroups of order p p and exponent p is not divisible by p. By hypothesis, 2 .G/ has at least two absolutely regular subgroups of index p. It follows that the number of absolutely regular subgroups of index p in 2 .G/ is not divisible by p. Therefore, all maximal subgroups of 2 .G/ are normal in G. Since the number of normal subgroups of order p p and exponent p in G is 1 .mod p/ (Theorem 13.5), the last assertion in (h) follows. Exercise 2. Study the irregular p-groups G D n .G/ of order p pCn , generated by elements of order p n , n > 3, p > 2. Is it true that exp.G/ D p n ? Exercise 3. Classify the metacyclic 2-groups G such that 2 .G/ D G. Exercise 4. Classify the nonmetacyclic 2-groups G with metacyclic 2 .G/. Theorem 55.3 ([Jan8, Theorem 4.2]). Let G be a 2-group with jn .G/j D 2nC2 , n > 2. Then cn .G/ D 4 or 6. If cn .G/ D 4, then such groups G are classified in 54. If cn .G/ D 6, then n D 3 and the structure of H D 3 .G/ is uniquely determined. We have H D ha; b j a8 D b 8 D 1; ab D a1 ; a4 D b 4 D zi, where jH j D 25 and E D 1 .H / D hz; a2 b 2 D ui Š E4 . If G > H , then jGj D 26 and we have for the structure of G one of the following possibilities: (a) G is the “splitting” metacyclic group G D hw; c j w 16 D c 4 D 1; w c D w 11 i, where H D hwc 1 ; w 2 i and Z.G/ D hw 8 i is of order 2. (b) G D hH; ci with c 2 D z . D 0; 1/, .bc/2 D a, and ac D a1 . Here CG .E/ D hui ha; ci, where ha; ci Š D16 or Q16 for D 0 or 1, respectively. Also, Z.G/ D hzi is of order 2. (c) G is a non-metacyclic U2 -group with the kernel E Š E4 and G=E Š SD16 . More precisely, G D hd; s j d 16 D s 2 D 1; d 4 D v; d 8 D z; d s D d 1 vu; u2 D 1; ud D du; us D uzi, where E D hu; zi and H D hsd; d 2 i with Z.H / D Z.G/ D hvui Š C4 . (This is the group (c) with n D 4 of Theorem 67.3.) All above groups are determined up to isomorphism and they exist.

128

Groups of prime power order

Proof. Let G be a 2-group with jn .G/j D 2nC2 , n > 2. This implies that G is neither cyclic nor of maximal class. By Theorem 1.17(b), cn .G/ is even. If cn .G/ D 2, then Proposition 54.4 gives that jn .G/j D 2nC1 , a contradiction. If cn .G/ D 4, then such groups G have been classified in 54. Hence, we may assume that cn .G/ 6. However, if cn .G/ 8, then the number of elements of order 2n is 8'.2n / D 8 2n1 D 2nC2 , a contradiction. We assume in the sequel that cn .G/ D 6 and set n .G/ D H . We shall determine the structure of H . The number of elements of order 2n in H equals l D 6'.2n / D 6 2n1 D 3 2n D 2nC1 C 2n . Let E be a normal four-subgroup of H and let Z be a cyclic subgroup of order 2n1 in H . It is easy to see that H0 D ZE is of exponent 2n1 . Indeed, H00 < E and so H0 is of class 2 with jH00 j 2. Also, H0 is generated with elements of orders 2n1 and 2n1 4. If x; y 2 H0 and o.x/ 2n1 , n1 n1 n1 o.y/ 2n1 , then .xy/2 D x2 y 2 D 1. This shows that exp.H0 / D 2n1 . nC2 nC1 n n 2 2 D 2 , it follows that jH0 j 2n and so jH0 j D 2n Since jH j l D 2 and Z \ E ¤ f1g. All elements in H H0 are of order 2n . This implies that 2 .H / D 2 .H0 / is of order 8 and so we may use Lemma 42.1. It follows that H must be isomorphic to a group (c) of that proposition. In particular, H is a U2 -group with the kernel E, H=E is generalized quaternion, 2 .H / is abelian of type .4; 2/, and 2 .H /=E D Z.H=E/. If n > 3, then cn .G/ D cn .H / D 2, a contradiction. Hence n D 3, H=E Š Q8 , 2 .H / D H0 , and all 24 elements in H H0 are of order 8. More precisely, H D ha; b j a8 D b 8 D 1; ab D a1 ; a4 D b 4 D zi, and so jH j D 25 , Z.H / D hb 2 i Š C4 , H 0 D ha2 i Š C4 , H0 D 2 .H / D ha2 ; b 2 i, E D 1 .H / D hz; a2 b 2 D ui Š E4 , and CH .E/ D CH .H0 / D H1 D hb 2 ; ai is abelian of type .8; 2/. The following subgroups E, hzi D Ã1 .H0 /, H0 , ha2 i, hb 2 i, and H1 are characteristic in H and so they are normal in G. We assume G > H and we intend to determine the structure of G up to isomorphism. Set C D CG .E/ so that C is a maximal subgroup of G, C > H1 , G D CH D C hbi, C \ H D H1 , and c3 .C / D 2. By Proposition 54.4, C is an L2 -group or a U2 -group with kernel E. Note that U2 -groups contain a unique normal four-subgroup. If C is a U2 -group, then H0 =E is a normal subgroup of order 2 in C =E and therefore H0 =E D Z.C =E/. If C =E in this case is not dihedral, then there is an element y 2 C H1 such that y 2 2 H0 E. But then o.y/ D 8, which is a contradiction (since 3 .G/ D H ). Here we have also used the fact that H1 =E is a normal cyclic subgroup of order 4 in C =E and so H1 =E is contained in a cyclic subgroup of index 2 in C =E. We have proved that C is either an L2 -group (in which case C is abelian of type .2i ; 2/, i 4) or C is a U2 -group with C =E dihedral. In any case, C =E has the unique maximal cyclic subgroup T =E of index 2. Then T is normal in G, T is abelian of type .2j ; 2/, j 3, and T H1 (since H1 =E is the unique cyclic normal subgroup of order 4 in C =E). Since b 2 2 H1 T , it follows that in case jG=T j D 4, C =T and hbiT =T are two distinct subgroups of order 2 in G=T and so G=T Š E4 . In any case, G=T is elementary abelian of order 4. We have CH1 .b/ D hb 2 i Š C4 and so CT .b/ D hb 2 i. Indeed, if b centralizes an element

55 2-groups G with small subgroup hx 2 G j o.x/ D 2n i

129

y 2 T H1 , then o.y/ 16 and so b centralizes an element of order 8 and this one must lie in H1 , a contradiction. Set K D hbiT and assume K > H or equivalently T > H1 . We shall determine the structure of K. Suppose that there is an element k of order 16 in K T . Then k 2 2 T and o.k 2 / D 8. But CT .b/ D CT .k/ and so b centralizes an element of order 8 in T , which contradicts our last result in the previous paragraph. Hence c4 .K/ D c4 .T / D 2. Proposition 54.4 implies that K is either an L2 -group or a U2 -group with the kernel E. But H is a U2 -group with H=E Š Q8 and so K is a U2 -group. We have H0 =E D Z.K=E/ and K=E is of maximal class containing the proper subgroup H=E isomorphic to Q8 . We know that T =E is the cyclic subgroup of index 2 in K=E. If x 2 K T is such that x 2 2 H0 E, then o.x/ D 8 and so x 2 H . Hence, if s 2 K .H [ T /, then s 2 2 E and CT .s/ D hb 2 i implies s 2 2 E \ hb 2 i D hzi. This forces K=E Š SD16 and jT =Ej D 8. Note that SD16 is the unique 2-group of maximal class and order > 8 containing Q8 as its maximal generalized quaternion subgroup. Since s 62 C , it follows us D uz, where E D hz, u D a2 b 2 i. Thus Ehsi Š D8 and so (since D8 has five involutions) we may assume that s is an involution. We may set T D huihd i, where o.d / D 16 and d 8 D z. Also set d 4 D v so that o.v/ D 4 and v 2 D z. Note that E ˆ.H / ˆ.K/ and so the involution s does not normalize hd i (otherwise hd; si would be a maximal subgroup of K not containing E). On the other hand, K=E Š SD16 and so d s D d 1 ve, where e 2 E hzi. Replacing b with b 1 leads to the replacement of u D a2 b 2 with a2 b 2 D a2 b 2 z D uz and so we may assume from the start that e D u and d s D d 1 vu. It is easy to see that K is not metacyclic. We have Ã1 .T / D hd 2 i and so s normalizes hd 2 i. Indeed, .d 2 /s D .d s /2 D .d 1 vu/2 D d 2 v 2 u2 D d 2 z, and so hs; d 2 i Š SD16 and Ehs; d 2 i is a non-metacyclic maximal subgroup of K. We locate H as the subgroup of K generated by elements of order 8 in K. We set b0 D sd and compute b02 D .sds/d D d 1 vud D vu. Hence b0 and d 2 are elements of order 8, hd 2 i is normal in K, hb0 i \ hd 2 i D hzi, and so H D hsd , d 2 i with Z.H / D Z.K/ D hvui Š C4 . Suppose in addition that C > T so that C is a U2 -group with C =E Š D16 , G D KC , and G=T Š E4 . Let c 2 C T . Then c 2 2 E, c inverts T =E, c centralizes E (since C D CG .E/), and d c D d 1 f with f 2 E. We compute .d 2 /c D .d 1 f /2 D d 2 , and so v c D v 1 because v D d 4 2 hd 2 i. Hence d sc D .d 1 vu/c D df v 1 u D dvuzf and so .T hsci/=E Š M24 . On the other hand, .T hsci/ \ H D H1 and therefore c3 .T hsci/ D c3 .H1 / D 2. By Proposition 54.4, T hsci is either an L2 -group or a U2 -group with the kernel E. But .T hsci/=E Š M24 , a contradiction. We have proved that in this case G D K and so it remains to prove the existence of K. We set: d D .1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16/ .17; 21; 24; 25; 28; 20; 23; 26; 27; 29; 31; 32; 30; 18; 19; 22/;

130

Groups of prime power order

and s D .2; 17/.3; 7/.4; 23/.5; 13/.6; 30/.8; 24/.10; 27/ .11; 15/.12; 19/.14; 28/.16; 31/.18; 20/.22; 32/.25; 26/: Then we see that the even permutations d and s (of degree 32) satisfy the defining relations for G D K D hd; si: d 16 D s 2 D 1; u2 D 1;

d s D d 1 vu;

d 4 D v;

d 8 D z;

ud D du;

us D uz:

Since the permutation z D d 8 is non-trivial, z 2 Z.K/, and Z.K/ Š C4 , it follows that the induced permutation representation of G is faithful and so G is realized as a subgroup of the alternating group A32 . Thus, G is the group from (c). It remains to analyze the case K D H or equivalently T D H1 . Since G > H , we have C > T and so C =E Š D8 . We know that H=E Š Q8 , G=T D G=H1 Š E4 , and jGj D 26 . Let c 2 C T . Since C =E Š D8 and T =E is the unique cyclic subgroup of order 4 in C =E, it follows c 2 2 E. Both elements b and c invert T =E and therefore bc centralizes T =E. Set N D T hbci, where .bc/2 2 T . Then N \ H D T and so c3 .N / D c3 .T / D 2. Proposition 54.4 implies that N is either an L2 -group or a U2 -group. But bc centralizes T =E and therefore N=E is abelian. This shows that N is an L2 -group and so N=E Š C23 with 1 .N / D E. Since bc 62 C , ubc D uz and so N Š M25 and .bc/2 is an element of order 8 in T D H1 . Hence .bc/2 D ai or .bc/2 D ai u, where i is an odd integer. Set a0 D ai and we see that (replacing a with a0 ) the relations for H remain unchanged: b 8 D .a0 /8 D 1;

b 4 D .a0 /4 D z;

and .a0 /b D .a0 /1 ;

except possibly .a0 /2 b 2 D uz, where u D a2 b 2 . But in that case we replace b with b 0 D b 1 and obtain the old relations for H : .b 0 /8 D .a0 /8 D 1, .b 0 /4 D .a0 /4 D z, 0 .a0 /b D .a0 /1 and .a0 /2 .b 0 /2 D .a0 /2 b 2 D .a0 /2 b 2 z D u. Hence we may write again a and b instead of a0 and b 0 and so we may assume from the start that: .bc/2 D a or .bc/2 D au, where u D a2 b 2 , and the relations for H D ha; bi remain unchanged. Assume first .bc/2 D au. Then bc centralizes au and so we get au D .au/bc D .a1 uz/c D .a1 /c uz, which gives ac D a1 z and .a2 /c D a2 . Note that c 2 C D CG .E/, where E D hu; zi and so c centralizes u D a2 b 2 . We get u D uc D .a2 b 2 /c D a2 .b 2 /c D a2 b 2 , and so .b 2 /c D zb 2 D b 2 . From .bc/2 D au follows (since c 2 2 E and c 4 D 1) bcbc D au;

cbc D b 1 au;

c 2 .c 1 bc/ D b 1 au;

b c D c 2 b 1 au;

and so noting that b 2 2 Z.H / we get .b 2 /c D c 2 b 1 auc 2 b 1 au D b 2 ; c 2 a1 uz.c 2 /b au D 1;

c 2 z.c 2 /b D 1;

c 2 .b 1 auc 2 b/au D 1;

.c 2 /b D c 2 z;

55 2-groups G with small subgroup hx 2 G j o.x/ D 2n i

131

and so c 2 2 E hzi and therefore we may set c 2 D uz . D 0; 1/ and b c D uz b 1 au. Conjugating the last relation with c 1 we get 1

b D uz .b c /1 a1 zu; bc

1

D uz C1 b 1 au;

bc

1

D uz C1 a1 b 1 u;

bc

1

D b c z:

We compute uzb 1 au D uz.b 1 aub/b 1 D uza1 uzb 1 D ua1 ub 1 D ua1 .a2 b 2 /b 1 D .ua/b D .au/b D .bc/2 b; and noting that .bc/8 D .au/4 D a4 D z we get .bc/c D b c c D uz b 1 auc D z C1 .uzb 1 au/c D z C1 .bc/2 bc D .bc/8.C1/ .bc/3 D .bc/11C8 : Since hbci\hci D f1g (which follows from .bc/8 D .au/4 D a4 D z and c 2 D uz ), we see that G D hbc; ci is a “splitting” metacyclic group with the cyclic normal 1 subgroup hbci of order 16 and a complement hci Š C4 . Since b c D b c z, we get .bc/c

1

1

D b c c D b c zc D .bc/c z D .bc/11C8 .bc/8 D .bc/3C8 : 1

If D 0, then .bc/c D .bc/11 and .bc/c D .bc/3 . If D 1, then .bc/c D .bc/3 1 and .bc/c D .bc/11 . Hence, replacing c with c 1 (if necessary), we may assume from the start that .bc/c D .bc/11 . We set bc D w and so we have obtained the unique metacyclic group G D hw; c j w 16 D c 4 D 1; w c D w 11 i, which was stated in part (a) of our theorem. This group obviously exists because w ! w 11 induces an automorphism of order 4 of the cyclic group hwi of order 16. Assume now .bc/2 D a. Then bc centralizes a and so we get a D abc D .a1 /c which implies ac D a1 and .a2 /c D a2 . Note that c 2 C D CG .E/, where E D hu; zi and so c centralizes u D a2 b 2 . We get u D uc D .a2 b 2 /c D a2 .b 2 /c D a2 b 2

and so .b 2 /c D zb 2 D b 2 :

From .bc/2 D a follows (since c 2 2 E and c 4 D 1) bcbc D a;

cbc D b 1 a;

c 2 .c 1 bc/ D b 1 a;

b c D c 2 b 1 a;

and so noting that b 2 2 Z.H / we get .b 2 /c D c 2 b 1 ac 2 b 1 a D b 2 ; c 2 a1 .c 2 /b a D 1;

c 2 .b 1 ac 2 b/a D 1; .c 2 /b D c 2 ;

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Groups of prime power order

and so c 2 2 hzi. We set c 2 D z . D 0; 1/ and have C D CG .E/ D hui ha; ci. If D 0, then ha; ci Š D16 . If D 1, then ha; ci Š Q16 . Also, Z.G/ D hzi since Z.H / D hb 2 i and .b 2 /c D b 2 . If D 0, then we set (identify): b D .1; 9; 7; 10; 5; 11; 3; 12/.2; 13; 8; 14; 6; 15; 4; 16/; c D .2; 8/.3; 7/.4; 6/.9; 13/.10; 16/.11; 15/.12; 14/: Since b 4 D z is a nontrivial permutation, we have realized the group G D hb; ci as a subgroup of S16 . If D 1, then we set b D .1; 9; 7; 10; 5; 11; 3; 12/.2; 13; 8; 17; 6; 18; 4; 19/ .14; 15; 23; 24; 28; 22; 31; 27/.16; 30; 25; 32; 29; 20; 21; 26/; c D .1; 16; 5; 29/.2; 22; 6; 15/.3; 25; 7; 21/.4; 27; 8; 24/ .9; 14; 11; 28/.10; 31; 12; 23/.13; 20; 18; 30/.17; 32; 19; 26/: Since b 4 D z is a nontrivial permutation, we have obtained our group G D hb; ci as a subgroup of A32 . We have obtained two groups stated in part (b) of our theorem which is now completely proved. In conclusion we prove the following Theorem 55.4. Let G be a p-group such that 2 .G/ is extraspecial. Then 2 .G/ D G. Proof. Suppose the theorem is false, i.e., 2 .G/ < G. We consider cases p D 2 and p > 2 separately. Case 1. Let p D 2. Set E D 2 .G/ and hzi D Z.E/ so that jEj D 22nC1 , n 1, where n is the width of E. Let F be a subgroup of G containing E such that jF W Ej D 2. To get a contradiction, one may assume, without loss of generality, that G D F . We use induction on n. Suppose that n D 1. Then E Š D8 or Q8 . If CG .E/ 6 E, then CG .E/hzi contains elements of order 4, a contradiction. Hence CG .E/ E and then G is of maximal class. But then 2 .G/ D G, a contradiction. We assume now n > 1. Since jEj 25 and exp.E/ D 4, G has no cyclic subgroups of index 2 and so G is not of maximal class. It follows that G has a normal foursubgroup R. We have R < E since 2 .G/ D E. In particular, z 2 R and we may set R D hz; ui for some involution u 2 .E hzi/ so that CG .R/ D CG .u/. Since jE W CE .R/j D 2, it follows that CG .R/ covers G=E. By the structure of E, CE .u/ D hui E0 , where E0 is extraspecial of order 22.n1/C1 . Set F0 D CG .u/ so that jF0 W .hui E0 /j D 2 and consider the factor-group F0 =hui D FN0 (bar convention), where jFN0 W EN0 j D 2 and EN0 is extraspecial of width n 1.

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By induction, there is an element x 2 F0 .hui E0 / such that o.x/ N 4. If N D 4 and so o.x/ N D 2, then x 2 2 hui and so o.x/ 4, a contradiction. Hence o.x/ x 4 2 hui but x 2 62 hui. We have x 2 2 CE .u/. If x 2 is an involution, then o.x/ D 4, a contradiction. Hence x 2 is an element of order 4 in CE .u/ and so x 4 D z, contrary to x 4 2 hui. This completes Case 1. Case 2. Now let p > 2 and let E D 2 .G/ be extraspecial. Since G > E, we have exp.E/ D p 2 . We have E D E1 Em1 Em , where, in case m > 1, E1 ; : : : ; Em1 are nonabelian of order p 3 and exponent p and Em Š Mp 3 (see 4). Then we have S D 1 .E/ D 1 .G/ D E1 : : : Em1 1 .Em /, where exp.S / D p and jE W S j D p (here we use the fact that E is regular; see Theorems 7.1 and 7.2). Set 1 .Em / D R. We have CS .R/ D E1 Em1 R D S so R D Z.S /. It follows that R is normal in G. Set C D CG .R/; then jG W C j D p, S < C and G D EC . Since S D 1 .G/ and 1 .G=S / D E=S is of order p so G=S is cyclic since p > 2 and E=C is the unique subgroup of order p in G=S , we must have E=S C =S . This is a contradiction since R 6 Z.E/. The proof is complete.

56

Theorem of Ward on quaternion-free 2-groups

In this section we present the proof of the following important result of Ward [War]: Theorem 56.1 (H. N. Ward [War]). Let G be a nonabelian quaternion-free 2-group. Then G has a characteristic subgroup of index 2. Recall that a 2-group is quaternion-free if it has no sections isomorphic to Q8 ; that property is inherited by sections. Note that the original proof of that theorem is based essentially on properties of the Burnside group B.4; 2/ (this is a finite two-generator group of exponent 4 of maximal order; by Burnside, jB.4; 2/j D 212 ). The offered proof is shorter and elementary. However, we use ideas of [War]. The offered proof is due to the second author. Let P 2 Syl2 .G/ be the kernel of a Frobenius group G. If P is nonabelian, it has a section isomorphic to Q8 . Indeed, P has no characteristic subgroups of index 2 so the result follows from Theorem 56.1. Let P 2 Syl2 .G/, where G D Sz.22mC1 / is the Suzuki simple group. Since NG .P / is a Frobenius group, P has a section isomorphic to Q8 , by what has been said. Let G be a nonabelian quaternion-free 2-group and let A Aut.G/ be of odd order. Then A has a fixed point on G=ˆ.G/ (use Maschke’s theorem). First we prove the following two auxiliary results. Lemma 56.2. In a Q8 -free 2-group X there are no elements x, y with o.x/ D 2k > 2 and o.y/ D 4 so that x y D x 1 . If D X and D Š D8 , then CX .D/ is elementary abelian. k1

Proof. If y 2 D x 2 , then we have hx; yi Š Q2kC1 . If hxi \ hyi D f1g, then we k1 have hx; yi=hx 2 y 2 i Š Q2kC1 . In both cases hx; yi is not Q8 -free, a contradiction. Suppose that D X, where D D ha; t j a4 D t 2 D 1; at D a1 i. If v is an element of order 4 in CX .D/, then o.t v/ D 4 and t v inverts a, contrary to what has just been proved. Thus, CX .D/ must be elementary abelian. Lemma 56.3. Let X be a Q8 -free 2-group with ˆ.X/ Z.X/. If a; b 2 X, o.a/ D o.b/ D 4, and Œa; b ¤ 1, then .ab/2 D 1. Proof. Set c D Œa; b so that c 2 D Œa; b2 D Œa2 ; b D 1 and therefore c is a central involution in X. Also, a2 and b 2 are central involutions in X. Set W D ha2 c; b 2 ci

56

Theorem of Ward on quaternion-free 2-groups

135

and so W Z.ha; bi/ is of order 4 and exponent 2. In particular, a; b 62 W . We compute ab D aŒa; b D ac D a1 .a2 c/, b a D bŒb; a D bc D b 1 .b 2 c/. It follows from the above equalities that .aW /b D .aW /1 and .bW /a D .bW /1 . Since ha; bi=W is Q8 -free, Lemma 56.2 implies that at least one of aW or bW is an involution. Hence a2 D b 2 c or b 2 D a2 c (indeed, W contains exactly three involutions a2 c, b 2 c, a2 b 2 and a2 ¤ a2 b 2 ¤ b 2 ). In any case a2 b 2 D c. But then .ab/2 D a2 b 2 Œb; a D c 2 D 1. and we are done. Proof of Theorem 56.1. We proceed by induction on jGj. Let K > f1g be a minimal characteristic subgroup of G. If G=K is nonabelian, it has a characteristic subgroup H=K of index 2, by induction; then H is characteristic of index 2 in G. Therefore, we may assume that K D G 0 , i.e., G 0 is the unique minimal characteristic subgroup of G; then G 0 is elementary abelian. Since G 0 \ Z.G/ is a nonidentity characteristic subgroup of G, we get G 0 Z.G/. Thus, the group G is of class 2 and so for all x; y 2 G, we have Œx 2 ; y D Œx; y2 D 1, which gives ˆ.G/ D Ã1 .G/ Z.G/; in m particular, G=Z.G/ is elementary abelian. The map x 7! x 2 is an endomorphism of G if m 2 since (1)

m

m

m

m .2m 1//=2

.xy/2 D x 2 y 2 Œy; x.2

m

m

D x2 y 2 ;

where x; y 2 G. Let exp.G/ D 2nC1 .n 1/. If n 2, G is generated by the n elements of order 2nC1 . For, if x 2 D 1 and o.y/ D 2nC1 , then, taking m D n in (1), we get o.xy/ D 2nC1 and x D .xy/y 1 is a product of two elements of order 2nC1 . If n D 1, we may also assume that G is generated by the elements of order 4. Indeed, let H D H2 .G/ be the subgroup generated by all elements of G of order 4. Since H > f1g, then H < G implies jG W H j D 2 (Straus–Szekeres). As H is characteristic, this is a contradiction. Thus, we may assume that H D G, i.e., G is generated by elements of order 4 D 2nC1 again. We prove our theorem in six steps. Step 1. Suppose that n D 1, i.e., exp.G/ D 4. Denote Z0 D 1 .Z.G//. We claim that if a1 ; a2 ; : : : ; am are elements of G of order 4 such that a1 a2 : : : am 2 Z0 , then 2 D 1. a12 a22 : : : am Indeed, let m be the smallest integer for which a1 ; : : : ; am exist satisfying the hypothesis but not the conclusion of the above statement. Then m > 1. In fact m 3, for if a1 a2 D z 2 Z0 , then a1 D a21 z so a12 D a22 z 2 D a22 and a12 a22 D 1 2 Z0 . Since Œai ; aj 2 G 0 ˆ.G/ 1 .Z.G// D Z0 , elements ai and aj commute modulo Z0 so the hypothesis a1 a2 : : : am 2 Z0 is independent of the ordering of ai ’s. We have for any i ¤ j , .ai aj /2 D ai2 aj2 Œaj ; ai and so Œai ; aj D ai2 aj2 .ai aj /2 . Suppose that Œai ; aj D 1. If, in addition, .ai aj /2 ¤ 1, then o.ai aj / D 4 and .ai aj /2 D ai2 aj2 . Then the two terms ai and aj could be combined to .ai aj / and m lowered by one, contrary to the assumption. Thus if Œai ; aj D 1, then ai2 aj2 D .ai aj /2 D 1. If Œai ; aj ¤ 1, then Lemma 56.3 implies .ai aj /2 D 1. So in any event, Œai ; aj D ai2 aj2 .

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Now let i; j; k be distinct. Then, by the last result, using the collecting process, we obtain, since Ã1 .G/ Z.G/, .ai aj ak /2 D ai2 aj2 ak2 Œai ; aj Œai ; ak Œaj ; ak D ai2 aj2 ak2 ai2 aj2 ai2 ak2 aj2 ak2 D ai2 aj2 ak2 : If then .ai aj ak /2 ¤ 1, ai ; aj ; ak can be combined to .ai aj ak / and m lowered by 2. Therefore ai2 aj2 ak2 D 1 for all triples i; j; k. In particular, m > 3, so all the squares ai2 coincide (for example, it follows from a12 a22 a32 D 1 D a22 a32 a42 that a12 D a42 ). But then Œai ; aj D ai2 aj2 D ai4 D 1 so the elements ai ’s commute, and, since a1 a2 : : : am 2 2 Z0 , we get 1 D .a1 a2 : : : am /2 D a12 a22 : : : am after all. Step 2. Suppose, as in Step 1, that n D 1, i.e., exp.G/ D 4. Denote again Z0 D 1 .Z.G//. We claim that there is the unique homomorphism s W g 7! g s of G into Z0 such that if o.a/ D 4, then as D a2 . In addition, x s D 1 for every x 2 Z0 . (g s is called the “artificial square”of g.) It remains to define s on involutions from G Z0 . For g 2 G, write g D a1 a2 : : : am , where o.ai / D 4 (recall that G is generated 2 by elements of order 4, by the paragraph preceding Step 1). By Step 1, a12 a22 : : : am depends only on g, not of choice of a1 ; : : : ; am . Indeed, if, in addition, g D b1 : : : b t , 1 2 b 2 : : : b 2 D 1 and then a1 : : : am b 1 D 1 2 Z0 so, by Step 1, a12 : : : am t : : : b1 t 1 2 2 2 2 s 2 a1 : : : am D b1 : : : b t , justifying our claim. If one defines g D a12 a22 : : : am , the results stated in the previous paragraph are direct consequences of Step 1. This definition also works in the case o.g/ D 4: since g s depends only on g we must have g s D g 2 . It follows that im.N/ D GN is characteristic in G. n Step 3. We define the endomorphism N of G by xN D x 2 if n > 1 and xN D x s (the artificial square, defined in Step 2) if n D 1; recall that exp.G/ D 2nC1 . It is easy to check, using (1) with m D n if n > 1 and Step 1 if n D 1 that N is a homomorphism. We observe that, if n > 1, then the kernel of N equals n .G/, which is < G, by (1) N for all N D 2. We claim that in any case Œa; b 2 ha; with m D n; in that case, exp.G/ N bi a; b 2 G. e have im.N/ Ã1 .G/ \ 1 .Z.G//. Let n D 1. We have .ab/2 D a2 b 2 Œb; a. Then, if o.a/ D o.b/ D o.ab/ D 4, N If o.ab/ < 4, then, N D aN 2 bN 2 D 1 2 ha; N bi. we get Œa; b D a2 b 2 .ab/2 D aN bab 2 2 N If o.a/ < 4, then independent of the orders of a and b, Œa; b D a b 2 ha; N bi. 2 2 N Œa; b D b .ab/ 2 ha; N bi. Similarly, the same inclusion is true, if o.b/ < 4. Thus, our claim is true for n D 1. Now let n > 1. Suppose first that aN ¤ 1, bN ¤ 1, and aN bN ¤ 1 (or, what is the same, elements a; b; ab have the same order 2nC1 D exp.G/). Then ha4 ; b 4 i Ã2 .G/ < ˆ.G/ Z.G/, exponent of ha; bi=ha4 ; b 4 i is 4, and the orders of a and b equal 4 modulo ha4 ; b 4 i. If Œa; b 62 ha4 ; b 4 i, then Lemma 56.3 implies .ab/2 2 ha4 ; b 4 i. n But taking 2n1 -th powers of the last inclusion and using (1), one obtains .ab/2 2 nC1 nC1 ha2 ; b 2 i D f1g since exp.G/ D 2nC1 , or, what is the same, aN bN D 1, contrary to N (recall the assumption. Therefore, Œa; b 2 ha4 ; b 4 i and since 1 .ha4 ; b 4 i/ D ha; N bi 0 N N D 2) and G is elementary abelian, Œa; b 2 ha; that exp.G/ N bi.

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Theorem of Ward on quaternion-free 2-groups

137

Next let a be an involution and bN ¤ 1 (this means that o.b/ D 2nC1 ). We may N 4 (otherwise the kernel n .G/ of N is a characteristic subgroup assume that jGj N (otherwise, of G of index 2). Hence, for some element c 2 G, we have cN 62 hbi nC1 2 2 N N G D hbi were of order 2); it follows that o.c/ D 2 . In G=hb i (b 2 Z.G/), then, some power of c leads to a central element of order 4. By the second part of Lemma 56.2, G=hb 2 i has no subgroups isomorphic to D8 . Hence, a and b commute modulo hb 2 i, and we have Œa; b 2 hb 2 i and since o.Œa; b/ 2 (recall that exp.G 0 / D 2), N Œa; b 2 hbi. Now let aN D 1, bN ¤ 1 (this means that o.a/ 2n , o.b/ D 2nC1 ), and let a have the minimal order 2m .2 m n/ among such elements for which aN D 1 N D hbi N (case m D 1 was considered in the previous paragraph). but Œa; b 62 ha; N bi m1 N 1 .ha4 ; b 4 i/, where ha4 ; b 4 i Z.G/. If Œa; b 62 ha4 ; b 4 i, then Then ha2 ; bi by Lemma 56.3, .ab/2 2 ha4 ; b 4 i and so (taking again 2n1 -th powers) we get bN D 1bN D aN bN D ab D 1, contrary to the assumption of this paragraph. Hence, since m1 N Again let o.Œa; b/ 2, we get Œa; b 2 1 .ha4 ; b 4 i/ and so Œa; b 2 ha2 ; bi. N N c 2 G be such that cN 62 hbi. Since ac D cN ¤ 1 and cb D cN b ¤ 1, we have, by N and Œc; b 2 hc; N Then Œac; b D Œa; bŒc; b implies the above, Œac; b 2 hc; N bi N bi. N Combining this with Œa; b 2 ha2m1 ; bi, N we get Œa; b 2 hbi, N which Œa; b 2 hc; N bi. m1 m1 m1 2 2 2 N is a contradiction, unless a 2 hc; N bi. But in this case a Dz for some m1 z 2 hb 2 ; c 2 i Z.G/. Then .az/2 D 1 and so o.az/ < 2m and Œa; b D Œaz; b 2 N by minimality of m. This contradiction proves that for any a; b 2 G with aN D 1 hbi, N and bN ¤ 1, we have Œa; b 2 hbi. n nC1 If aN D bN ¤ 1, then Œa; b D Œab; b with ab D aN 2 D .a2 /2 D a2 D 1, so that N by the previous paragraph. Œa; b D Œab; b 2 hbi, Finally, suppose that aN D bN D 1. Let c 2 G be any element with cN ¤ 1. Then Œa; bc 2 ha; N bci D hci N and Œa; c 2 ha; N ci N D hci, N by the above. Hence Œa; bc D N 4, there is d 2 G with dN 62 hci. Œa; bŒa; c gives Œa; b 2 hci. N Since jGj N Then, N D 2), as again, Œa; b 2 hdN i and so Œa; b 2 hdN i \ hci N D f1g (recall that exp.G/ N for all a; b 2 G. needed. All possibilities have been considered and so Œa; b 2 ha; N bi N Obviously, G 0 GN since G 0 is the unique Step 4. We claim now that G 0 D G. minimal characteristic subgroup of G and GN is a nonidentity characteristic subgroup N see Step 2). Note that if G 0 ¤ G, N then of G (this follows from the definition of G; 0 0 N N N jG W G j 4. Assume that this is false; then jG W G j D 2. We have G Š G=K with characteristic subgroup K < G, the kernel of N. Since G 0 K < G is characteristic in G, we get jG W G 0 Kj 4. It follows jGN W GN 0 j D jG=K W .G 0 K=K/j D jG W G 0 Kj 4, as claimed. N Then if aN 2 GN G 0 , we have Œa; b D 1 for all b 2 G. Suppose that G 0 < G. N are not contained in N ab, all involutions in ha; N bi, To see this, note first that if a; N b; 0 0 N G , then Œa; b 2 G \ ha; N bi (using Step 3), and that intersection equals f1g since N D f1; a; N abg. Now suppose that aN 62 G 0 but bN 2 G 0 . Find c 2 G with ha; N bi N b; N D 2). Then ac 62 G 0 , so that cN 62 hG 0 ; ai N (c exists since jGN W G 0 j 4 and exp.G/

138

Groups of prime power order

Œa; c D 1 by what has just been proved. And, Œa; b D Œa; bŒa; c D Œa; bc 2 ha; N bci, N bc 62 G 0 , by the choice. It remains by Step 3. But ha; N bci \ G 0 D f1g. Indeed, a; to show, that the third involution abc N D aN bN cN of ha; N bci is not contained in G 0 . If 0 N this is false, then, since b 2 G , we get ac D aN cN D aN bN cN bN 2 G 0 , contrary to what has been proved. Hence Œa; b D 1. Finally, if aN 62 G 0 , bN 62 G 0 , and aN bN 2 G 0 , then Œa; b D Œa; ab D 1 (the second equality follows, by the previous case). Thus, Œa; b D 1 for all b 2 G. But G is generated by the elements a 2 G with aN 62 G 0 , and G would be abelian, a contradiction. Set T D hx 2 G j xN 62 G 0 i. To justify our claim, we must to prove that G D hT i. Indeed, G is generated by all elements of order 2nC1 D exp.G/. Let gN 2 G 0 with gN ¤ 1 and let aN 2 GN G 0 . Then also ga D gN aN 2 GN G 0 and gN D gaaN so that g D gaat with tN D 1. But then g D .ga/.at / and ga 62 G 0 and at D aN 62 G 0 , i.e., ga; at 2 T . Thus, G D hT i, as claimed. Thus GN D G 0 . Let K be the kernel of N . Then GN D G=K is isomorphic to G 0 and so jG 0 j D jG W Kj > 2; the last inequality holds in view of G is nonabelian and K is characteristic in G. Step 5. Suppose that jG 0 j 8. Then we obtain a characteristic subgroup of index 2 in G as follows. N D 4 and Œa; b D 1 (since exp.G/ N D 2, if such (i) There exist a; b 2 G with jha; N bij N a; b exist, then ha; N bi is a four-group). N D 4 but Œa; b ¤ 1 (we may assume that GN is For, let a and b be such that jha; N bij N and so without loss of generality we may noncyclic). By Step 3, 1 ¤ Œa; b 2 ha; N bi N then replace Œa; b with Œb; a D Œa; b1 D bN 1 D b. N take Œa; b D a. N (If Œa; b D b, N If Œa; b D aN b, then replace Œa; b with Œab; b D Œa; b D ab.) Now take c with N (such c exists in view of jGj N D jG 0 j 8). Then Œa; bc D Œa; bŒa; c D cN 62 ha; N bi aŒa; N c 2 ha; N bci \ ah N a; N ci N D f1; ag. N If Œa; bc D 1, then a and bc are the desired elements. If Œa; bc D a.D N Œa; b/, then Œa; c D 1 so a and c are the desired elements. N for all x 2 G. (ii) Suppose that aN ¤ 1, bN ¤ a, N and Œa; b D 1. Then Œb; x 2 hbi N N N bi. Assume that this intersection For, (by Step 3) Œax; b D Œx; b 2 hax; bi \ hx; N Then hax; bi N D hx; N is a four-group. Since aN ¤ 1, we get ax D bN xN so is not hbi. N bi N contrary to the assumption. Thus, our intersection is hbi, N as claimed. aN D b, (iii) Now let H be the set of elements a 2 G for which Œa; x 2 hai N for all x 2 G. We show that H is a subgroup of G of index 2; it will clearly be characteristic. If a1 ; a2 2 H , then we have to show that a1 a2 2 H . Suppose that bN 62 haN1 ; aN2 i. N \ haN1 ihaN2 i D ha1 a2 i, as needed. Since G Then Œa1 a2 ; b D Œa1 ; bŒa2 ; b 2 ha1 a2 ; bi is generated by the set of all such b’s (noting that GN D G=K is elementary abelian of order jG 0 j 8), we have Œa1 a2 ; b 2 ha1 a2 i for all b 2 G. Thus a1 a2 2 H so H is a characteristic subgroup of G. Obviously, G 0 Z.G/ H . By (i) and (ii), there is an element a 2 H with aN ¤ 1. Indeed, by (i), there exist a; b 2 G such that aN ¤ 1, bN ¤ 1, bN ¤ aN and Œa; b D 1. Then, by (ii), b 2 H # , proving our claim. Suppose for such an a that Œa; b D 1 for some b 2 G. If bN ¤ a, N

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Theorem of Ward on quaternion-free 2-groups

139

then b 2 H by (ii); but if bN D a, N then ba D aN aN D 1 ¤ aN and Œa; ba D Œa; b D 1 and so by (ii) again, ba 2 H . Since a 2 H , so b 2 H in any event. We have proved that CG .a/ H . So if b 2 G H , then Œa; b ¤ 1. But then Œa; b D aN (as a 2 H and o.a/ N D 2). Hence if H ¤ G, then jG W H j D 2. (Suppose there are b1 ; b2 2 G H such that b1 b2 2 G H . Then aN D Œa; b1 b2 D Œa; b1 Œa; b2 D aN aN D 1, a contradiction.) Assume that H D G. Then G is Dedekindian so abelian since it is Q8 -free (see Theorem 1.19). Step 6. Finally suppose that jG 0 j D 4 (by the last paragraph of Step 4. jG 0 j > 2). It cannot be that K, the kernel of N , is central, for jG W Kj D 4 would then imply jG 0 j D 2, by Lemma 1.1. Thus there is c 2 K (or, what is the same, cN D 1) and a 2 G with Œa; c ¤ 1. By Step 3, Œa; c 2 ha; N ci N D ha; N 1i D hai N and so Œa; c D a. N N N Suppose that b 62 hai. N Then Œab; c D Œa; cŒb; c D aŒb; N c 2 habi \ ah N bi, by Step 3, N Since G is generated and so Œab; c D ab. Thus Œb; c D abŒa; c1 D ab aN 1 D b. by all b 2 G with bN 62 hai, N this implies Œg; c D gN for all g 2 G. It follows that jK W .K \ Z.G//j D 2. Indeed, assume that there are c1 ; c2 2 K Z.G/ such that c1 c2 2 K Z.G/ and aN ¤ 1. Then aN D Œa; c1 c2 D Œa; c1 Œa; c2 D aN aN D 1, which is a contradiction. N D G 0 . If Œa; b ¤ 1, we may assume that Œa; b D Let a; b 2 G be such that ha; N bi N then take the elements ab; b so that Œab; b D Œa; b D ab. If a. N (If Œa; b D aN b, N Then Œa; b D b, then take the elements b; a so that Œb; a D Œa; b1 D bN 1 D b.) Œa; bc D Œa; bŒa; c D aN aN D 1 and jha; N bcij D 4 (with c 2 K Z.G/ as in the previous paragraph). Thus we may assume from the start that Œa; b D 1. Now, G D ha; biK D ha; b; ciZ.G/. By the previous paragraph, jK W .K \ Z.G//j D 2. Therefore, Z.G/ < K because otherwise jG W Z.G/j D 4 and then jGj D 2jZ.G/jjG 0 j (Lemma 1.1) would imply jG 0 j D 2, a contradiction (see the last paragraph of Step 4). Hence we have Z.G/ < K and since jK W Z.G/j D 2, it follows that jG W Z.G/j D 8. The fact Œa; b D 1 implies that Z.G/ha; bi is an abelian maximal subgroup of G. Since jG W Z.G/j D 8, we see that Z.G/ha; bi is the unique abelian maximal subgroup of G and therefore it is a characteristic subgroup of index 2. The proof is complete.

57

Nonabelian 2-groups all of whose minimal nonabelian subgroups are isomorphic and have exponent 4

By Proposition 10.28, a nonabelian p-group is generated by its minimal nonabelian subgroups. Therefore it is natural to try to determine the structure of a p-group if the structure of its minimal nonabelian subgroups is known. Here we classify the title groups and note that there are exactly five minimal nonabelian 2-groups of exponent 4 (see Lemma 65.1). All results of this section are due to the second author. Recall that there are exactly five minimal nonabelian 2-groups of exponent 4: D8 , Q8 , H2 D ha; b j a4 D b 4 D 1; ab D a1 i, H16 D ha; t j a4 D t 2 D 1; Œa; t D z; z 2 D Œa; z D Œt; z D 1i and H32 D ha; b j a4 D b 4 D 1; Œa; b D z; z 2 D Œa; z D Œb; z D 1i. It is proved in Appendix 17 (Corollary A.17.3) that, if G is a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to Q8 , then G D Q V , where Q Š Q2n , n 3, and exp.V / 2. In Theorem 10.33 was proved that if G is a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to D8 , then G is generalized dihedral. Here we consider the other three minimal nonabelian 2-groups of exponent 4 and first prove the following two key lemmas. Lemma 57.1. Let G be a nonabelian p-group and let A be a maximal abelian normal subgroup of G. Then for any x 2 G A, there is a 2 A such that Œa; x ¤ 1, Œa; xp D 1, and Œa; x; x D 1 which implies that ha; xi is minimal nonabelian. Therefore, G is generated by its minimal nonabelian subgroups. Proof. Since CG .A/ D A, we have CA .x/ ¤ A and therefore hxiCA .x/ is a proper abelian subgroup of hxiA. Let B be a subgroup of hxiA containing hxiCA .x/ as a subgroup of index p. Then j.A \ B/ W CA .x/j D p, CA .x/ Z.B/ and B 0 CA .x/. Let a 2 .A \ B/ CA .x/ so that ap 2 CA .x/. We get 1 D Œap ; x D Œa; xp . On the other hand, Œa; x 2 CA .x/ and so Œa; x; x D 1. We get ha; xi0 D hŒa; xi and so, by Lemma 65.2(a), ha; xi is minimal nonabelian. Lemma 57.2. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are of exponent 4 and let A be a maximal normal abelian subgroup of G. Then all elements in G A are of order 4 and so either exp.A/ D 2 or exp.A/ D exp.G/.

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Minimal nonabelian subgroups are of exponent 4

141

If x 2 G A with x 2 2 A, then x inverts each element in Ã1 .A/ and in A=1 .A/. If exp.G/ > 4, then either G=A is cyclic of order 4 or G=A Š Q8 . Proof. By Lemma 57.1, all elements in G A are of order 4. Thus, exp.A/ D 2 or exp.A/ D exp.G/. Let x 2 G A with x 2 2 A. Then for each a 2 A we 2 have .aax /x D ax ax D ax a D aax and so aax D w with w 2 CA .x/ and ax D a1 w. We compute .xa/2 D xaxa D x 2 ax a D x 2 w, where o.x 2 / 2. Since o.xa/ 4, we have o.w/ 2 and so x inverts each element of A=1 .A/. Finally, .a2 /x D .ax /2 D .a1 w/2 D a2 and so x inverts each element of Ã1 .A/. If exp .A/ > 4, then G=A does not possess a four-subgroup and so G=A is either cyclic of order at most 4 or G=A Š Q8 . Theorem 57.3. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to H2 D ha; b j a4 D b 4 D 1; ab D a1 i. Then the following holds: (a) If G is of exponent 8, then G has a unique abelian maximal subgroup A. We have exp.A/ 8 and E D 1 .A/ D 1 .G/ D Z.G/ is of order 4. All elements in G A are of order 4 and if v is one of them, then CA .v/ D E and v inverts ˆ.A/ and A=E. (b) If G is of exponent 4, then G D K V , where exp.V / 2 and for the group K we have one of the following possibilities: (b1) K Š H2 is of order 24 ; (b2) K is the minimal nonmetacyclic group of order 25 (see Theorem 66.1(d)); (b3) K is a unique special group of order 26 with Z.K/ Š E4 in which every maximal subgroup is minimal nonmetacyclic of order 25 (from (b2)): K D ha; b; c; d j a4 D b 4 D 1; c 2 D a2 b 2 ; Œa; b D 1; ac D a1 ; b c D a2 b 1 ; d 2 D a2 ; ad D a1 b 2 ; b d D b 1 ; Œc; d D 1iI (b4) K is a splitting extension of B D B1 Bm , m 2, with a cyclic group hbi of order 4, where Bi Š C4 , i D 1; 2; : : : ; m, and b inverts each element of B (and b 2 centralizes B). Proof. Since D8 is not a subgroup of G, 1 .G/ is elementary abelian of order > 2 (because G is not generalized quaternion). Let x 2 G 1 .G/ with o.x/ D 4 and assume that there is a 2 1 .G/ such that Œa; x ¤ 1. Then Œa; x is an involution and we have 1 D Œa; x 2 D Œa; xŒa; xx so that Œa; xx D Œa; x and ha; xi is minimal nonabelian, a contradiction. Here we have used the fact that 1 .B/ Z.B/ for each minimal nonabelian subgroup B G, where B Š H2 . Since 2 .G/ D G, we have proved that 1 .G/ Z.G/. Hence, if A is a maximal normal abelian subgroup of G, then 1 .G/ D 1 .A/ Z.G/, G=A is elementary abelian, 1 .A/ < A, and G A consists of elements of order 4.

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Groups of prime power order

Suppose that there is an element v of order 4 such that CG .v/ is nonabelian and let H D ha; b j a4 D b 4 D 1; ab D a1 i be a minimal nonabelian subgroup of CG .v/, where we set a2 D z, b 2 D u. First suppose that hvi \ H D f1g. Then .av/2 D a2 v 2 D zv 2 62 H , .av/b D ab v D a1 v D .av/z, so that hav; bi is the nonmetacyclic minimal nonabelian group of order 25 and exponent 4, a contradiction. Hence, v 62 H but v 2 2 Z.H / and so v 2 2 fz; u; uzg. If v 2 D z, then i D va is an involution, i 62 H and i b D iz and so i 62 Z.G/, a contradiction. If v 2 D u, then j D vb is an involution, j 62 H and aj D avb D ab D a1 so that j 62 Z.G/, a contradiction. If v 2 D uz, then .vb/2 D v 2 b 2 D .uz/u D z, vb 62 H , and avb D ab D a1 so that hvb; ai Š Q8 , a contradiction. We have proved that the centralizer of each element of order 4 is abelian. In particular, for each x 2 G A, CA .x/ D 1 .A/ and 1 .A/ D Z.G/. If exp.A/ > 4, then Lemma 57.2 implies that jG W Aj D 2 and we have obtained groups in part (a) of our theorem. Indeed, since x 2 G A inverts A=1 .A/ and exp.A/ 8, we have jG 0 j > 2 and so, using Lemma 1.1, we see that A is a unique abelian maximal subgroup of G (otherwise, jG W Z.G/j D 4). From now on we assume that exp.G/ D 4. In that case ˆ.G/ 1 .A/ D Z.G/ and jˆ.G/j 4 (since jˆ.H2 /j D 4). If x; y are elements of order 4 in G with Œx; y ¤ 1, then Œx; y is an involution in Z.G/ and so hx; yi is minimal nonabelian and hx; yi Š H2 implies that in case x 2 ¤ y 2 we have y x 2 fy 1 ; yx 2 g. Considering G=ˆ.G/, we get G D K V , where exp.V / 2 and 1 .K/ D Z.K/ D ˆ.K/ D ˆ.G/. It is easy to determine the structure of G in case jˆ.G/j D 4. Our group K has exactly three involutions and Z.K/ Š E4 is noncyclic. By the results stated in the introduction to 82, K has a metacyclic normal subgroup M such that K=M is elementary abelian of order 4. Since in our case K is of exponent 4, we have jM j 24 and so jKj 26 . If jKj 24 , then K Š H2 . Suppose that jKj D 25 . In this case K is nonmetacyclic (since exp.K/ D 4). If K is not minimal nonmetacyclic, then Theorem 66.1 implies that K must possess a subgroup which is isomorphic to E8 or Q8 . This is not the case and so K is minimal nonmetacyclic of order 25 . Finally, assume that jKj D 26 . Each maximal subgroup of K is minimal nonmetacyclic of order 25 and such a group K is unique according to [CIS, Theorem 2 and Remarks] so K is special with Z.K/ Š E4 . We have obtained the groups stated in parts (b1), (b2) and (b3). It remains to treat the case jˆ.G/j > 4. We note that jA W 1 .A/j D jÃ1 .A/j and let x 2 G A. Consider any element y 2 A with y 2 62 hx 2 i and suppose that y x ¤ y 1 so that y x D yx 2 . Assume that there is v 2 A with v 2 62 hx 2 ; y 2 i. Since y x D y vx (and .vx/2 2 fv 2 ; x 2 g and so .vx/2 ¤ y 2 ), we get x 2 D .vx/2 which gives v x D v 1 . Since .vy/2 62 hx 2 ; y 2 i, we also get .vy/x D .vy/1 . Hence, y x D y 1 for all y 2 A with y 2 62 hx 2 i and so x inverts A. (If z 2 A with z 2 D x 2 , then x inverts zy and so x also inverts z. But in that case hz; xi Š Q8 which cannot happen.) We have proved that in case jA W 1 .A/j 8, each element x 2 G A inverts A and so jG W Aj D 2 and G is as in (b4).

57

Minimal nonabelian subgroups are of exponent 4

143

Assume jA W 1 .A/j D jÃ1 .A/j D 4 so that A D hy; zi1 .A/ and Ã1 .A/ D hy 2 ; z 2 i. Since jˆ.G/j > 4, there is u 2 G A such that u2 62 hy 2 ; z 2 i. By the arguments in the previous paragraph, u inverts A, and so Ahui is as in (b4). Suppose that Ahui ¤ G. Then there is x 2 G .Ahui/ such that Ã1 .Ahxi/ D Ã1 .A/ and we may assume that x 2 D z 2 . If y x D y 1 , then xu 2 G A and xu centralizes y, a contradiction. It follows that y x D yx 2 . From y x D y zx and x 2 ¤ .zx/2 (noting that hz; xi Š H2 ) follows that y 2 D .zx/2 D Œz; x. We compute .yz/x D yx 2 zŒz; x D yx 2 zy 2 D y 1 z 1 D .yz/1 . But then xu centralizes yz, a contradiction. Hence, we must have G D Ahui. Finally, suppose jA W 1 .A/j D 2 so that A D hyi1 .A/. Let x 2 G A, where we may assume that x 2 ¤ y 2 . (Indeed, we have hx; yi Š H2 and so if x 2 D y 2 , then we replace x with x 0 D xy, where .x 0 /2 ¤ y 2 . We may assume that y x D yx 2 (otherwise, if each element in G A inverts A, then jG W Aj D 2 and G would be as in (b1)). Let u 2 G hA; xi with u2 62 hx 2 ; y 2 i. Then we have y u 2 fy 1 ; yu2 g. First suppose that y u D y 1 . Then y xu D .yx 2 /u D y.y 2 x 2 /. But .xu/2 is equal either x 2 u2 (if Œu; x D 1) or u2 or x 2 (if Œu; x ¤ 1 and so hu; xi Š H2 ) and therefore in any case .xu/2 ¤ y 2 . Since hy; xui Š H2 , we get y xu D yy 2 or y xu D y.xu/2 . Since y xu D y.y 2 x 2 /, the only possibility is .xu/2 D y 2 x 2 which is a contradiction. Hence, we must have the second possibility y u D yu2 and we get y xu D .yx 2 /u D y.u2 x 2 / and therefore (because u2 x 2 D y 2 is excluded) x 2 u2 D .xu/2 which implies Œu; x D 1. In that case A1 D hx; ui1 .A/ is an abelian normal subgroup of G with jA1 W 1 .A/j D 4 which leads to one of the former cases, when we consider a maximal normal abelian subgroup A (instead of A) containing A1 . Our theorem is proved. Theorem 57.4. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to H16 D ha; t j a4 D t 2 D 1; Œa; t D z; z 2 D Œa; z D Œt; z D 1i. Then 1 .G/ is a self-centralizing elementary abelian subgroup and G is of exponent 4. Moreover, the centralizer of any element of order 4 is abelian of type .4; 2; : : : ; 2/. Proof. Since D8 is not a subgroup of G, 1 .G/ is elementary abelian (of order > 2). We choose a maximal normal abelian subgroup A of G so that 1 .G/ A. Using Lemma 57.2, we see that all elements in GA are of order 4 which implies ˆ.G/ A. By Lemma 57.1, we have also CG .1 .A// D A. Let v be an element of order 4 such that CG .v/ is nonabelian and let H D ha; t j 4 a D t 2 D 1; Œa; t D z; z 2 D Œa; z D Œt; z D 1i be a minimal nonabelian subgroup of CG .v/. Assume that hvi \ H D f1g so that hH; vi D H hvi is of order 26 . Since .vt /2 D v 2 , it follows that hvt i \ H D f1g and hvt i normalizes H . Consider the subgroup ha; vt i. Since Œa; vt D Œa; t D z and z commutes with a and vt , we have ha; vt i0 D hzi and so S D ha; vt i is minimal nonabelian of order 25 (since S \ H D ha; zi Š C4 C2 and S covers hH; vi=H ), a contradiction. Hence we must have hvi \ H D hv 2 i, where v 2 2 Z.H / and so v 2 2 fz; a2 ; a2 zg. Suppose

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that v 2 D z so that .vt /2 D z and Œa; vt D z which implies that ha; vt i is metacyclic minimal nonabelian of order 24 , a contradiction. Suppose that v 2 D a2 so that va is an involution and Œva; t D z ¤ 1, a contradiction. Similarly, if v 2 D a2 z, then at v is an involution and Œat v; t D z ¤ 1, a contradiction. We have proved that for each element v of order 4, CG .v/ is abelian. Let x; y 2 G be such that x 2 G A, Œx; y 2 D 1 D Œx 2 ; y and Œx; y ¤ 1. Then hx; yi Š H16 and so either y or xy is an involution. Indeed, c D Œx; y 2 A since G=A is abelian. We get 1 D Œx; y 2 D Œx; yŒx; yy , 1 D Œx 2 ; y D Œx; yx Œx; y, and so c x D c 1 , c y D c 1 . Suppose o.c/ > 2 and let c0 be an element of order 4 in hci so that c0x D .c0 /1 D c0 c02 , hc0 ; xi0 D hc02 i and therefore hc0 ; xi is a metacyclic minimal nonabelian subgroup, a contradiction. Hence, c D Œx; y is an involution commuting with x and y so that hx; yi is a minimal nonabelian subgroup isomorphic to H16 . Suppose that A ¤ 1 .A/. Take an element x 2 G A and an element y of order 4 in A. Since CG .y/ is abelian, Œx; y ¤ 1. By Lemma 57.2, x inverts each element in A=1 .A/ and so Œx; y 2 1 .A/. We have Œx 2 ; y D 1 and Œx; y 2 D Œx; yŒx; yy D Œx; y2 D 1. By the above, hx; yi Š H16 and so xy must be an involution, a contradiction. We have proved that A D 1 .A/ and so exp.G/ D 4. For each a 2 G A, we have CG .a/ D ha; CA .a/i. Suppose that this is false. Let b 2 CG .a/ .Ahai/ so that ha; bi Š C4 C4 and ha; bi \ A D ha2 ; b 2 i Š E4 . Indeed, if a2 D b 2 , then ab 2 G A and ab is an involution, a contradiction. Set A0 D CA .a/ so that A0 ¤ A and A0 D CA .b/ (since CG .a/ and CG .b/ are abelian). Let t 2 A A0 and consider the subgroup ha; bt i. We have Œa; bt ¤ 1, o.bt / D 4 and since CA .bt / D CA .b/ D CA .a/ D A0 , we have a2 ; .bt /2 2 A0 and therefore Œa; .bt /2 D Œa2 ; bt D 1. By the above, ha; bt i Š H16 which implies that abt (2 G A) is an involution, a contradiction. Theorem 57.5. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to H32 D ha; b j a4 D b 4 D 1; Œa; b D z; z 2 D Œa; z D Œb; z D 1i. Then 1 .G/ Z.G/ and G is of exponent 4 and class 2. Proof. In exactly the same way as in the first paragraph of the proof of Theorem 57.3, we show that if A is a maximal normal abelian subgroup of G, then 1 .G/ D 1 .A/ Z.G/, 1 .A/ < A and G A consists of elements of order 4. Suppose that exp.A/ > 4. By Lemma 57.2, jG W Aj D 2 and if x 2 G A, then o.x/ D 4 and x inverts each element in Ã1 .A/. Let y 2 Ã1 .A/ with o.y/ D 4. Then y x D y 1 so that hx; yi is metacyclic minimal nonabelian (of order 8 or 16), a contradiction. Hence exp.A/ D 4 and so ˆ.G/ 1 .G/ Z.G/. In conclusion we classify the nonabelian 2-groups all of whose A1 -subgroups are isomorphic to M16 . Theorem 57.6. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to M16 D ha; t j a8 D t 2 D 1; at D a5 i. Then E D 1 .G/

57

Minimal nonabelian subgroups are of exponent 4

145

is elementary abelian and if A is a maximal normal abelian subgroup of G containing E, then A is of type .2s ; 2; : : : ; 2/, s 2, jG W Aj 4, each element x in G A is of order 8, jE W CE .x/j D 2, x inverts each element in A=E and in Ã1 .A/ and we have the following possibilities: (a) s > 2 in which case jG W Aj D 2 and G=E Š Q2s is generalized quaternion of order 2s ; (b) s D 2 in which case either G Š M16 V with exp.V / 2 or G=E Š Q8 and jG W Aj D 4. Proof. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to M16 . Since G does not possess dihedral subgroups, E D 1 .G/ is elementary abelian. Let A be a maximal normal abelian subgroup of G containing E so that E D 1 .A/. By Lemma 57.1, each element in G A is of order 4 or 8. Suppose that x 2 G A and o.x/ D 4. Then x 2 2 E and exp.Ehxi/ D 4. It follows that Ehxi is abelian. But then Ahxi is nonabelian and E Z.Ahxi/. This is a contradiction since in this case M16 cannot be a subgroup of Ahxi. We have proved that all elements in G A are of order 8 and so G=A is elementary abelian (G is metabelian) and exp.A/ 4. We want to determine the structure of A. Let x be a fixed element in G A so that o.x/ D 8, x 2 2 2 .A/ E and we set z D x 4 . By Lemma 57.1, there is a subgroup M Š M16 in Ahxi which contains hxi as a subgroup of index 2. Since all eight elements in M A are of order 8, we have M \ A D M \ 2 .A/ Š C4 C2 and M \E Š E4 . Let t be an involution .M \E/hzi so that t x D t z and M 0 D hzi. Set b D x 2 t so that b 2 D z and b x D .x 2 t /x D .x 2 t /z D bz D b 1 . It follows that each element xs with s 2 A is of order 8 and inverts the cyclic subgroup hbi of order 4. This forces hxsi \ hbi D hzi. Indeed, if hxsi \ hbi D f1g (hxsi cannot contain hbi since xs inverts b), then hb; xsi is metacyclic minimal nonabelian of order 25 , a contradiction. We compute .xs/2 D xsxs D x 2 s x s and z D .xs/4 D x 4 .s x s/2 D z.s x s/2 and so .s x s/2 D 1 and therefore s x s D w 2 E. Thus, s x D s 1 w with w 2 E which shows that x inverts each element of A=E. Further, .s 2 /x D .s x /2 D .s 1 w/2 D s 2 which shows that x also inverts each element in Ã1 .A/. Assume that there is an element a 2 2 .A/E such that a2 ¤ z. First suppose that x centralizes a. In that case, .ab/x D ab x D ab 1 D .ab/z and .ab/2 D a2 b 2 D a2 z ¤ z. Hence hab; xi with habi \ hxi D f1g is minimal nonabelian of order 25 , a contradiction. Now, suppose that ax D a with 1 ¤ 2 E so that Œa; x D . 2 But a D ax D .a/x D ax x D a x and so x D which gives hi D ha; xi0 . It follows that ha; xi is minimal nonabelian of order 25 because hai \ hxi D f1g. We have proved that 2 .A/ is of type .4; 2; : : : ; 2/ which forces that A is of type .2s ; 2; : : : ; 2/, s 2. It is easy to see that jE W CE .x/j D 2. Indeed, set E0 D CE .x/ and suppose jE W E0 j 4. Since x induces on E an involutory automorphism, we know that x centralizes E=E0 and so ŒE; x E0 and jŒE; xj D jE=E0 j, where E0 hzi. In

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Groups of prime power order

that case, there is e 2 E E0 such that Œe; x 2 E0 hzi. Then hx; ei0 D hŒe; xi and therefore hx; ei is nonmetacyclic minimal nonabelian of order 25 , a contradiction. We have proved that CE .x/ is a hyperplane of E for each x 2 G A. First suppose that s > 2 or equivalently, exp.A/ > 4. We know that each element x 2 G A inverts Ã1 .A/ and here exp.Ã1 .A// 4. This forces jG W Aj D 2 noting that G=A is elementary abelian. Since A=E Š C2s1 and jG=Ej D 2s , s 3, G=E cannot be cyclic (otherwise, an element y 2 G A would be of order 2sC1 , contrary to the fact that all elements in G A are of order 8). On the other hand, 2 .A/=E is a unique subgroup of order 2 in G=E because 2 .A/ D 2 .G/. We have proved that G=E Š Q2s . It remains to consider the case s D 2 so that A D 2 .A/ D 2 .G/ and jA W Ej D 2 since A is of type .4; 2; : : : ; 2/. For any x 2 G A, hxi covers A=E and hx 4 i D Ã1 .A/. Let M Š M16 be a minimal nonabelian subgroup in Ahxi containing hxi, where we have again used Lemma 57.1. Since jE W CE .x/j D 2, we get Ahxi D M V with V E. If Ahxi D G, we are done. Suppose that jG W Aj 4 and we note that G=A is elementary abelian. Hence G=E is not cyclic and the fact that 2 .G/ D A shows that A=E is a unique subgroup of order 2 in G=E. It follows that G=E is generalized quaternion. But then d.G=E/ D 2 and so G=A must be elementary abelian of order 4 which forces that G=E Š Q8 and jG W Aj D 4.

58

Non-Dedekindian p-groups all of whose nonnormal subgroups of the same order are conjugate

A non-Dedekindian p-group in which all nonnormal subgroups of the same order are conjugate is termed a CO-group. The second author classified CO-groups (see Theorem 58.3); this solves Problem 1261. Below we offer another proof of that nice result which is due to the first author. A p-group G is termed a CCO-group if it is not Dedekindian and all its nonnormal cyclic subgroups of the same order are conjugate. Obviously, a CO-group is a CCO-group. Lemma 58.1. Let G D ha; bi be a minimal nonabelian p-group as in Lemma 65.1 (see also Exercise 1.8a). If all nonnormal cyclic subgroups of G of the same order are conjugate, then G Š Mp t . Proof. Write A D hai and B D hbi. Let H 6E G; then H \ G 0 D f1g and H G D H G0. m

n

Let G D ha; b j ap D b p D c p D 1; Œa; b D c; Œa; c D Œb; c D 1i and assume that m n. Let n .A/ D hxi. Then B and hbxi of the same order are neither G-invariant nor conjugate since 1 .BG / ¤ 1 .hbxiG / so G is not a CCO-group. m n m1 Let G D ha; b j ap D b p D 1; ab D a1Cp i. If n D 1, then G Š Mp mC1 satisfies the condition. Now let n > 1. If n m, set n .A/ D hxi; then B and hbxi of the same order p n are neither G-invariant nor conjugate. Now assume that n > m. Then B and habi of the same order p n are neither G-invariant nor conjugate. Indeed, m m .ab/p D b p so o.ab/ D p n , habi \ A D f1g, and we conclude that habi is not normal in G (otherwise, G D ha; abi is abelian). Since 1 .habiG / ¤ 1 .BG /, habi and B are not conjugate in G so G is not a CCO-group. Lemma 58.2. If a p-group G is a CCO-group and jG 0 j D p, then G Š Mp t . Proof. In view of Lemma 58.1, one may assume that G has a proper minimal nonabelian subgroup, say B; then B G G since B 0 D G 0 . In that case, G D B C where C D CG .B/ (Lemma 4.2) and we conclude that B is either Dedekindian or CCOgroup; in the first case B Š Q8 , in the second case B Š Mp t (Lemma 58.1). The size of every conjugate class of non-G-invariant cyclic subgroups equals p.

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Groups of prime power order

(i) Let B D hx; y j x 4 D 1; y 2 D x 2 ; x y D x 3 i Š Q8 . We get exp.C / > 2 since G is not Dedekindian. Let L D hli C be cyclic of order 4. Write H D B L; then H G G. If B \ L D Z.B/, then H has exactly 6 > 2 nonnormal subgroups of order 2 so they are not conjugate in G, a contradiction. Thus, B \ L D f1g; then H D B L has 3 > 2 distinct nonnormal cyclic subgroups hxli, hyli, hxyli of the same order 4 so they are not conjugate, a contradiction. Thus, B has no subgroups isomorphic to Q8 . t1 t2 D v p D 1; uv D u1Cp i Š Mp t . Assume that (ii) Now let B D hu; v j up X D hxi < G of order p is such that X 6 B; then X is not conjugate with hvi so X GG and F D B X has exactly p 2 > p noncentral so non-G-invariant subgroups of order p; then G is not a CCO-group, a contradiction. Thus, 1 .G/ D 1 .B/ Š Ep 2 so C has exactly one subgroup of order p; in that case C is cyclic, by (i). Then there is a cyclic W D hwi C such that W 6 B and w p 2 Z.B/. Let A < B be cyclic of index p. Then AW is noncyclic abelian with cyclic subgroup A of index p so, by basic theorem on abelian groups, AW D A Y , where jY j D p and Y 6 B has order p, contrary to what has just been proved. Theorem 58.3 (Janko). If a p-group G is a CO-group, then G Š Mp t . Proof (Berkovich). We use induction on jGj. In view of Lemma 58.2, one may assume that jG 0 j > p. By Theorem 1.23, there is K D G 0 \ Z.G/ of order p such that G=K is non-Dedekindian. Since G=K is a CO-group, we get G=K Š Mp t , by induction. Since K < G 0 ˆ.G/, we get d.G/ D d.G=K/ D 2. Let A=K and B=K be two distinct cyclic subgroups of index p in G=K; then A and B are abelian maximal subgroups of G. It follows that A \ B D Z.G/ so jG 0 j D p (Lemma 1.1), contrary to the assumption. Corollary 58.4. Suppose that all nonnormal subgroups of the same order of a nilpotent non-Dedekindian group G are conjugate. Then G D P C , where P 2 Sylp .G/ is non-Dedekindian and P Š Mp n , C is cyclic. Theorem 58.5 (Berkovich). If G is a non-Dedekindian p-group all of whose nonnormal cyclic subgroups of the same order belong to the same conjugate class of size p, then G Š Mp t . Proof. We use induction on jGj. In view of Lemma 58.1, one may assume that jG 0 j > p. Assume that G is a 2-group of maximal class. Then jGj > 8. If G Š Q2n n > 3, then G has > 2 nonnormal cyclic subgroups of order 4. If G 6Š Q2n , then G has > 2 nonnormal subgroups of order 2. Thus, G is not a 2-group of maximal class. By Theorem 1.23, there is in G 0 \ Z.G/ a subgroup R of order p such that G=R is non-Dedekindian. Let X=R < G=R be nonnormal cyclic. Assume that X is noncyclic. Then X D Xi R, where Xi is cyclic of index p in X, i D 1; : : : ; p. The subgroup Xi is nonnormal in G (otherwise, X D Xi R G G). By hypothesis, the

58

Nonnormal subgroups of the same order are conjugate

149

subgroups X1 ; : : : ; Xp form the conjugate class in G so X D X1 : : : Xp G G, a contradiction. Thus, X must be cyclic. By hypothesis, there are in G exactly p nonnormal cyclic subgroups of order jXj. Therefore, there are exactly p nonnormal cyclic subgroups of order jX=Rj in G=R so G=R satisfies the hypothesis. By induction, the non-Dedekindian group G=R Š Mp n . Then G=R has two distinct cyclic subgroups A=R and B=R of index p. In that case, A and B are two distinct abelian maximal subgroups of G so A \ B D Z.G/. Then jG 0 j D p1 jG W Z.G/j D p (Lemma 1.1), contrary to the assumption.

59

p-groups with few nonnormal subgroups

By Lemma 58.1, if all nonnormal subgroups of a minimal nonabelian p-group G are conjugate, then G Š Mp t . Of course, Theorem 59.1 follows from Theorem 58.3. Theorem 59.1 ([Sch3]). If all nonnormal subgroups of a p-group G are conjugate, then G Š Mp nC1 . Proof (compare with [Sch3]). Let H < G be nonnormal; then H is cyclic since it is not generated by their (G-invariant) maximal subgroups. Set jH j D p s . If s D 1, then G Š Mp nC1 (Theorem 1.23). Now we assume that s > 1. We use induction on jGj. We have R D 1 .H / G G. By induction, G=R Š Mp n ; then jH=Rj D p. Since R ˆ.H / < ˆ.G/, we get d.G/ D d.G=R/ D 2. Let U=R and V =R be distinct cyclic subgroups of index p in G=R; then U and V are abelian subgroups of index p in G so ˆ.G/ D U \ V D Z.G/ and G is minimal nonabelian. Then, by the remark preceding the theorem, G Š Mp nC1 . 2

m

Given m > 1, let H2;m D ha; b j ap D b p D 1; ab D a1Cp i be a metacyclic minimal nonabelian group of order p 2Cm . We suggest to the reader, using the above approach, to prove the following fairly deep Theorem 59.2 ([Sch4]). If a group G is nilpotent and has exactly two conjugate classes of non-invariant subgroups, then one of the following holds: (a) G Š D8 . (b) G Š Q16 . (c) p D 2 and G Š H2;m . (d) G Š Mp n Cq , where a prime q ¤ p. Note that the proof presented in [Sch4], is not full. For the proof of this theorem, see [Ber26].

60

The structure of the Burnside group of order 212

I. N. Sanov [Sano] has proved that each finitely generated group of exponent 4 is finite. W. Burnside [Bur4] has shown that each two-generated group of exponent 4 is a homomorphic image of a certain group G D B.4; 2/ of order at most 212 which we call the Burnside group. In this section, written by the second author, we determine completely the structure of the Burnside group G by presenting this group in a very convenient form from which each structural question about that group can be easily answered which is important for applications (Theorem D). For example, we show that ˆ.G/ D 1 .G/ is a special 2-group of order 210 with Z.ˆ.G// Š E25 . The group G has exactly 12 maximal elementary abelian subgroups. Each of them is of order 26 and the intersection of any two of them is equal Z.ˆ.G//. If M; N; K are the maximal subgroups of G, then d.M / D d.N / D d.K/ D 3, cl.M / D cl.N / D cl.K/ D 4 and M 0 D N 0 D K 0 is abelian of order 27 and type .4; 4; 2; 2; 2/. The subgroup G 0 is of order 28 and class 2, d.G 0 / D 5, 1 .G 0 / D Z.ˆ.G// D Z.G 0 /, G 00 Š E4 , G 00 Z.G/ and Z.G/ Š E8 . Next, jAut.G/j D 221 3 (Theorem C). Theorem A (see [Hup, p. 300]). Let G be a group of exponent 4 generated with an element v of order 4 and an involution t . Then G 0 D hŒv 1 ; t ; Œv; t i is abelian of order 8 and so jGj 26 . Proof. We have .t v/3 D .t v/1 D v 1 t and .t v 1 /3 D .t v 1 /1 D vt hence Œv; t Œv 1 ; t D .v 1 t vt /.vt v 1 t / D v 1 .t v/3 v 2 t D v 1 .v 1 t /v 2 t D .v 2 t /2 D Œv 2 ; t ; Œv 1 ; t Œv; t D .vt v 1 t /.v 1 t vt / D v.t v 1 /3 v 2 t D v.vt /v 2 t D .v 2 t /2 D Œv 2 ; t ; and so: (1)

Œv; t Œv 1 ; t D Œv 1 ; t Œv; t D Œv 2 ; t D .v 2 t /2 :

Hence the subgroup P D hŒv 1 ; t ; Œv; t i is abelian of exponent 4 and order 8 since Œv 1 ; t Œv; t is of order 2 and so P cannot be isomorphic to C4 C4 . Also, P G 0 and it is easy to see that P is normal in G. Indeed, from 1 D Œv i ; t 2 D

152

Groups of prime power order

Œv i ; t Œv i ; t t follows Œv i ; t t D Œv i ; t 1 for each integer i (mod 4). From 1 D Œv 1 v; t D Œv 1 ; t v Œv; t follows Œv 1 ; t v D Œv; t 1 and from Œv 2 ; t D Œv; t v Œv; t follows (using (1)) Œv; t v D Œv 2 ; t Œv; t 1 D Œv 1 ; t . Since G=P is abelian, we get G 0 D hŒv 1 ; t ; Œv; t i and so jGj 26 . Theorem B (see [Hup, p. 300]). Let G D ha; bi be a group of exponent 4 such that 1 1 1 Œa2 ; b 2 D 1. Then jGj 29 and ha2 iG D ha2 ; .a2 /b ; .a2 /b a i is elemen1 1 1 1 1 1 tary abelian of order 8 (where .a2 /b a b D a2 .a2 /b .a2 /b a ). Simi1 1 1 larly, hb 2 iG D hb 2 ; .b 2 /a ; .b 2 /a b i is elementary abelian of order 8 (where 1 1 1 1 1 1 .b 2 /a b a D b 2 .b 2 /a .b 2 /a b ). Proof. Since Œa2 ; b 2 D .a2 b 2 /2 D 1, we get a2 b 2 a2 b 2 D 1 and a2 D b 2 a2 b 2 and so ba2 b 3 D b.b 2 a2 b 2 /b 3 D b 3 a2 b:

() Now, .a2 .a2 /b

1

/2 D .a2 ba2 b 3 /2 D a2 .ba2 b 3 / .a2 ba2 b 3 / D a2 b 3 a2 b a2 ba2 b b 2 D a2 b 3 .a3 b/3 b 2 D a2 b 3 .a2 b/1 b 2 D a2 b 3 b 1 a2 b 2 D .a2 b 2 /2 D 1; 1

1 1

which implies Œa2 ; .a2 /b D 1 and so also Œa2 ; .a2 /b a D 1. Further, using (), baba D a3 b 3 a3 b 3 (which is a consequence of .ba/4 D 1) and .a3 b/3 D .a3 b/1 D b 3 a, we get: .a2 /b

1 a 1 b 1

D baba ab 3 a3 b 3 D a3 b 3 a3 b 3 ab 3 a3 b 3 D a2 .ab 3 a3 b 3 /2 D a2 ..baba3 /1 /2 D a2 .baba3 /2 D a2 baba3 baba3 D a2 b.a2 a3 /ba3 b.a3 a2 /ba3 D a2 ba2 .a3 b/3 b 3 a2 ba3 D a2 ba2 .b 3 a/b 3 a2 ba3 D a2 ba2 b 3 a.b 3 a2 b/a3 D a2 .a2 /b

1

a.ba2 b 3 /a3 D a2 .a2 /b

1 1 1

1

1

.a2 /b

1 a 1

:

1 1

D a2 .a2 /b .a2 /b a and so each element We have obtained .a2 /b a b 1 1 1 2 2 conjugate to a is contained in ha ; .a2 /b ; .a2 /b a i which gives ha2 iG D 1 1 1 1 1 1 ha2 ; .a2 /b ; .a2 /b a i. Further, a2 commutes with .a2 /b and .a2 /b a and 1 1 1 so a2 2 Z.ha2 iG /. But then .a2 /b and .a2 /b a are also contained in Z.ha2 iG / and so ha2 iG is elementary abelian of order 8. By Theorem A, G=ha2 iG is of order 26 and so jGj 29 . The second half of the theorem is obtained by interchanging a and b. Theorem C. Let G be the group given with: G D ha; b j a4 D b 4 D .ab/4 D .a2 b/4 D .ab 2 /4 D .a1 b/4 D .a2 b 2 /4 D .b a b/4 D .ab a/4 D 1i. Then G is of

60

The structure of the Burnside group of order 212

153

order 212 and exponent 4 and so G is the Burnside group. The group G has exactly k.G/ D 88 conjugate classes. The automorphism group of G is of order 221 3. Proof. This theorem (apart from the last statement) is proved by using a computer (W. Lempken at the University of Essen). We may define our group G also in the form: G D ha; b j x 4 D 1 for all x 2 Gi. Then each map ˛ of fa; bg into G such that ha˛ ; b ˛ i D G induces an automorphism of G. Hence jAut.G/j D jAut.G=ˆ.G/jjˆ.G/j2 D 6 220 . The Burnside group G is given in Theorem C in terms of generators a; b and 9 relations. However, from this presentation it is extremely difficult to pin down the structure of all subgroups and factor-groups of G. Therefore we give in the next theorem the group G in terms of 12 generators and 78 relations which will be very convenient to answer any structural question about that group. We shall also deduce many important properties of the Burnside group G by using this new presentation. In the sequel, we use from Theorem C only the fact that the Burnside group G is of order 212 . Theorem D (Complete determination of the structure of the Burnside group). Let G D ha; bi be the Burnside group of order 212 as defined in Theorem C. Then G has the following properties: (a) ˆ.G/ D 1 .G/ is a special group of order 210 with Z.ˆ.G// D .ˆ.G//0 D ˆ.ˆ.G// Š E25 . (b) If M; N; K are maximal subgroups of G, then they are of class 4 and they are generated by three elements but not by two elements. In addition, M 0 D N 0 D K 0 is abelian of order 27 and type .4; 4; 2; 2; 2/, Z.G/ Š E8 , Z.G/ D Z.M / < Z.ˆ.G//, K3 .M / Š E24 , K3 .M / < Z.ˆ.G//, K4 .M / Š E4 , K4 .M / < Z.G/, and an automorphism of order 3 of G acts transitively on fM; N; Kg. (c) The commutator group G 0 is of order 28 and class 2, G=G 0 Š C4 C4 , 1 .G 0 / D Z.ˆ.G// D Z.G 0 /, ˆ.G 0 / D Z.G/, G 00 Z.G/, G 00 Š E4 , and M 0 D K3 .G/ is an abelian maximal subgroup of G 0 . In addition, K4 .G/ D Z.ˆ.G//, E4 Š K5 .G/ D K4 .M / Z.G/, where M is maximal in G, and so G is of class 5. Finally, Z2 .G/ D Z.ˆ.G// Š E25 , Z3 .G/ D K3 .G/ D M 0 (abelian of type .4; 4; 2; 2; 2/), and Z4 .G/ D ˆ.G/. (d) The group G has exactly 12 maximal elementary abelian subgroups. Each of them is of order 26 and the intersection of any two of them is equal Z.ˆ.G// Š E25 . Hence the number of involutions in G is 31 C 12 32 D 415 and each abelian subgroup of G is of rank 6. (e) The special group ˆ.G/ D 1 .G/ is generated with five involutions d; e; m; n; y and we may set Z.ˆ.G// D hc; g; h; i; j i Š E25 so that: Œd; e D ch; Œe; n D cgh;

Œd; m D c; Œe; y D ig;

Œd; n D g;

Œd; y D i cgh;

Œe; m D h;

Œm; n D cg;

Œm; y D jc;

Œn; y D jg:

154

Groups of prime power order

We have G D ha; bi with a2 D d , b 2 D m, .a1 b 1 /2 D y, where the action of a1 and b 1 on ˆ.G/ is given with: ca

1

D g; g a

da

1

D d;

cb

1

d

b 1

1

D c;

ha

1

D cgh;

ia

1

D i ch; j a

1

D jgh;

ea

1

D ei;

ma

1

D n;

na

1

D mc;

ya

1

D de mnychij;

D h; g b

1

D cgh;

hb

1

D c;

ib

1

D igh; j b

1

D jcg;

D e;

e

b 1

D dc;

b 1

m

D m;

n

b 1

D nj;

y

b 1

D de mnygi:

We have determined completely the structure of our group G. We have here M D ˆ.G/hai, N D ˆ.G/hbi, K D ˆ.G/ha1 b 1 i, where M 0 D N 0 D K 0 D K3 .G/ D Z.ˆ.G//hde; mni, K3 .M / D hc; g; h; i i, K5 .G/ D K4 .M / D hcg; chi, G 0 D M 0 he myi, G 00 D hcg; chi, and Z.G/ D Z.M / D hcg; ch; cij i Š E8 . Also, hciG D hc; g; hi Š E8 , ha2 iG D hd; e; i; g; hi and hb 2 iG D hm; n; j; g; hi are subgroups of orders 26 which have its commutator subgroup of order 2. (f) The factor-group H D G=Z.ˆ.G// (of order 27 ) is the group of maximal possible order having the exponent 4, being generated with two elements, and having the property that the squares of any two elements commute. We have ˆ.H / D 1 .H / Š E25 , Z.H / Š E4 , Z.H / < H 0 Š E23 , and ŒH; H 0 D Z.H / so that H is of class 3. The class number k.H / of H is 26. Proof. Let G D ha; bi be the Burnside group (as defined in Theorem C) so that G is the free group of exponent 4 generated with two elements. We have jGj D 212 and we want to determine the exact structure of G. If Œa2 ; b 2 D 1, then by Theorem B we get jGj 29 , a contradiction. Hence 2 2 Œa ; b D .a2 b 2 /2 D c is an involution and ha2 ; b 2 i Š D8 with Z.ha2 ; b 2 i/ D hci so that a and b are elements of order 4 and c commutes with a2 and b 2 . 1 We first determine the structure of hciG . We set h D c b and claim that Œc; h D 1. Indeed, since b 2 D b 3 b 3 and .b 3 a2 /3 D .b 3 a2 /1 D a2 b, we get: s1 D ch D cc b

1

D .a2 b 2 /2 b.a2 b 2 /2 b 3 D a2 b 2 a2 b 2 ba2 b 2 a2 b 2 b 3

D a2 .b 3 b 3 /a2 b 3 a2 .b 3 b 3 /a2 b D a2 b 3 .b 3 a2 /.b 3 a2 /.b 3 a2 /a2 b 3 a2 b D a2 b 3 a2 b a2 b 3 a2 b D .a2 b 3 a2 b/2 and so (2)

s1 D ch D .a2 b 3 a2 b/2 D .a2 .a2 /b /2 :

Hence .ch/2 D 1 and so Œc; h D 1. Also, s1b hc D ch D s1 , and so s1 commutes with b.

1

1

D .cc b /b

1

1

2

1

D cb cb D cb c D

60

The structure of the Burnside group of order 212

155

It is possible to show that s1 also commutes with a. Using (2), a2 b 3 ab 3 a2 D a.ab 3 ab 3 ab 3 /ba D a.ab 3 /3 ba D a.ab 3 /1 ba D aba3 ba; .a3 bab/2 D .b 3 a3 b 3 a/2 D .b 3 a3 b 3 a/2 ; b 3 a3 b 3 a3 b 3 D .b 3 a3 /3 a D .b 3 a3 /1 a D aba: we get (noting that a2 D a3 a3 ): s1 s1a

1 b

D a2 b 3 a2 b a2 b 3 .a2 b b 3 a a2 /b 3 a2 b a2 b 3 a2 b a3 b D a2 b 3 a2 b.a2 b 3 ab 3 a2 /ba2 b 3 a2 ba3 b D a2 b 3 .a2 b aba3 ba b/a2 b 3 a2 ba3 b D a2 b 3 a3 .a3 bab/2 a2 b 3 a2 ba3 b D a2 b 3 a3 .b 3 a3 b 3 a b 3 a3 b 3 a/a2 b 3 a2 ba3 b D a2 .b 3 a3 b 3 a3 b 3 a b 3 a3 b 3 a3 b 3 /a2 ba3 b D a2 aba a aba a2 ba3 b D .a3 b/4 D 1: 1

Hence s1 also commutes with a1 b which implies that s1 D ch D cc b is contained in Z.G/. Applying the automorphism ˛ which interchanges a and b (noting that c ˛ D 1 1 Œb 2 ; a2 D Œa2 ; b 2 D c), we get Œc; c a D 1 and s2 D cc a 2 Z.G/. We set 1 c a D g so that hch; cgi Z.G/, where hcg; chi is elementary abelian of order 4. Since c commutes with h and g, we get ch cg D hg and so .hg/2 D 1 and therefore h also commutes with g and hc; g; hi is elementary abelian of order 8. We have hb

1

D c b D c;

ha

1

D .c ch/a

1

D c a ch D gch D cgh;

gb

1

D .c cg/b

1

D c b cg D hcg D cgh;

2

ga

1

2

D c a D c;

1

1

which implies that hc; g; hi is a normal subgroup of G and so hc; g; hi D hciG . By Theorem B, G=hciG is of order 29 . Since jGj D 212 , we have hciG Š E8 and G=hciG is of order 29 . Also, hcg; chi Š E4 and hcg; chi Z.G/. We have also determined the action of a and b on hc; g; hi Š E8 . 1 1 We determine now the structure of ha2 iG . Set a2 D d , e D .a2 /b D d b , 1 1 1 f D .a2 /b a D e a and S D hc; g; hihd; e; f i. By Theorem B, S=hc; g; hi is a normal elementary abelian subgroup of order 8 of G=hc; g; hi with ha2 iG S . By Theorem A, jG=S j 26 . Since G=hc; g; hi is of order 29 , we have S=hc; g; hi Š E8 and jG=S j D 26 . We note that hs1 ; s2 i is a maximal subgroup of hc; g; hi and hs1 ; s2 i Š Z.G/.

156

Groups of prime power order

We know that a2 D d centralizes c and so d centralizes hc; s1 ; s2 i D hciG . 1 1 From Œc; a2 D 1 follows Œc b ; .a2 /b D 1 and so Œh; e D 1 which gives that 1 1 e centralizes hh; s1 ; s2 i D hciG . From Œc; e D 1 follows Œc a ; e a D 1 and so Œg; f D 1 which gives that f centralizes hg; s1 ; s2 i D hciG . We have proved that hc; g; hi Z.S / and so S is of class 2. From (2) follows s1 D Œa2 ; .a2 /b and conjugating this relation with b 1 gives 1 s1 D Œ.a2 /b ; a2 so that (since s1 D ch is an involution) Œe; d D Œd; e D s1 D ch. 1 Conjugating the last relation with a1 we get Œe a ; d D s1 and so Œf; d D Œd; f D 1 ch. By Theorem B, f b D def l for some l 2 hc; g; hi. From Œd; f D s1 follows conjugating with b 1 : Œd b

1

;f b

1

D s1

and so

Œe; def l D Œe; d Œe; f D s1 ;

which implies s1 Œe; f D s1 and Œe; f D 1. We get Œd; ef D Œd; eŒd; f D s1 s1 D 1 and so ef 2 Z.hd; e; f i/ which gives: hd; e; f i D hef i hd; ei Š C2 D8 S D hc; g; hi hd; e; f i;

with where

hd; e; f i0 D hs1 i; hc; g; hi \ hd; e; f i D hs1 i;

so that S 0 D hs1 i and Z.S / D hc; g; h; ef i Š E24 . It remains to show that ha2 iG D S . For that purpose we have to determine exactly 1 the element f b . First we note that c commutes with a2 and b 2 . From Œa2 ; b 2 D c follows a2 b 2 a2 b 2 D c and a2 D cb 2 a2 b 2 and so ba2 b 3 D b cb 2 a2 b 2 b 3 D .bcb 3 /.b b 2 a2 b 2 b 3 / D c b

1

.b 3 a2 b/;

which gives: ba2 b 3 D h.b 3 a2 b/:

() We compute: fb

1

D .a2 /b

1 a 1 b 1

D baba2 b 3 a3 b 3

(using baba D a3 b 3 a3 b 3 which follows from .baba/.baba/ D 1) D a3 b 3 a3 b 3 a b 3 a3 b 3 D a2 .ab 3 a3 b 3 /.ab 3 a3 b 3 / D a2 .ab 3 a3 b 3 /2 D a2 .baba3 /2 D a2 .baba3 /2 D a2 baba3 baba3 D a2 b.a2 a3 /ba3 b.a3 a2 /ba3 D a2 ba2 .a3 b/.a3 b/.a3 b/b 3 a2 ba3

The structure of the Burnside group of order 212

60

157

(since .a3 b/3 D .a3 b/1 D b 3 a and using ()) D a2 ba2 b 3 a .b 3 a2 b/a3 D a2 .ba2 b 3 /a hba2 b 3 a3 D a2 .ba2 b 3 /.aha3 /.ab a2 b 3 a3 / D a2 .a2 /b

1

1

ha .a2 /b

1 a 1

1

D deha f D def cgh;

1

where we have used ha D cgh (which was proved before) and cgh 2 Z.S /. 1 1 We have obtained the desired relation f b D def cgh. We compute f a D 2 e a D e d D eŒe; d D ech and so fb

1 a 1

D .def cgh/a

1

D df .ech/gc.cgh/ D df ec D def c: 1

Now, hd D a2 ; e; f i ha2 iG and so Œd; e D ch 2 ha2 iG . From f b D def cgh 1 1 follows that cgh 2 ha2 iG and from f b a D def c follows that c 2 ha2 iG . Hence hc; g; hi ha2 iG and so S D hd; e; f ihc; g; hi D ha2 iG . Let ˛ be the automorphism of order 2 of G such that a˛ D b and b ˛ D a. Note that c ˛ D Œb 2 ; a2 D Œa2 ; b 2 D c and g ˛ D h, h˛ D g so that .hciG /˛ D hciG D hc; g; hi. Set hb 2 iG D T so that T D S ˛ , jT j D 26 , T > hc; g; hi, T =hc; g; hi Š E8 . We have U D S T D ha2 ; b 2 iG and G=U is generated with two involutions and so jG=U j 23 . Since jU j 29 , we must have jG=U j D 23 and then jU j D 29 , 1 1 1 S \ T D hc; g; hi, and G=U Š D8 . Set b 2 D m, .b 2 /a D n, .b 2 /a b D p so that m D d ˛ , n D e ˛ , p D f ˛ and ˛ transports the relations for S into the relations for T . We get hm; n; pi D hnpi hm; ni Š C2 D8 , where Œn; p D 1, Œm; n D Œm; p D s2 D cg and T D hm; n; pi hc; g; hi with hm; n; pi \ hc; g; hi D 1 hs2 i, Z.T / D hc; g; h; npi Š Z24 , T 0 D hs2 i. Also, from f b D def cgh follows 1 p a D mnpcgh. Since hc; g; hi Z.S / and hc; g; hi Z.T /, we have hc; g; hi Z.U /. But ˆ.U / S \ T D hc; g; hi (noting that U=S Š U=T Š E8 ) and so U is of class 2. We shall determine completely the structure of U D S T D ha2 ; b 2 iG . This is done by the following computation (using all the previous relations): fa

1

D .a2 /b

1 a 2

2

D e a D e d D eŒe; d D ech;

2

d m D .a2 /b D a2 c D dc; 2

f m D f b D .f b en D e

ab 2 a1

1

D .e

D .f h c h/a dn D dm

a1

/b

1

a1

1

D .def cgh/b

a2 b 2 a1

/

D .f c/a

D d ama

1

1

1

D e dc def cgh h cgh c D f h;

D .f d /ma

1

D .f ch/ma

1

D echg;

D .a2 /ma

1

D d ma

1

D .dc/a

1

D dc a

1

D dg;

158

Groups of prime power order

and b 1

dp D dn

D d bnb

D .echg h/b 2

e m D e b D .e b ep D e

bnb 1

1

f p D f bnb

/b

1

1

b 1 nb 1

D d nb

D .f

a1

D .def h/b

1

1

1 nb 1

D .eh/nb

1

D eh;

D .e m /b

D .dg/b

/

D .dc/b

D dc h cgh D dg;

a2 ma1

1 nb 1

1

2

1 nb 1

1

D .defgh/nb

D .f m /b 1

D .eh/b

1 nb 1

D ecgh;

D .ech/d ma

D f cgh;

D .f b /b

D .def cgh c/nb

1

1 nb 1

1

D .eh/a

1

1

D .dc/b

D .e /

ama1

D e ma

1

2

D .d b /b

D .ecg/b

b2

D .dc c/nb fn Df

1

1

1

D .ech c h/ma

1 nb 1

D .f h/b

1

1 nb 1

D .dg echg f cgh g h/b

1

D e dc def cgh c D f cgh:

From the above relations we get at once the desired commutators which determine completely the structure of U D S T : Œd; m D c;

Œd; n D g;

Œd; p D g;

Œe; m D h;

Œe; p D cgh;

Œf; m D h;

Œf; n D cgh;

Œf; p D cgh:

Œe; n D cgh;

In particular, we see that U 0 D ˆ.U / D hc; g; hi. We know that hc; g; hi Z.U /, ef 2 Z.S / and np 2 Z.T /. In addition, we compute (noting that U is of class 2): Œef; m D Œe; mŒf; m D h h D 1;

Œef; n D 1;

Œef; p D 1;

Œd; np D 1;

Œe; np D 1;

Œf; np D 1;

and so ef; np 2 Z.U /. Therefore Z.U / hc; g; h; ef; npi. We show that we have in fact Z.U / D hc; g; h; ef; npi Š E25 . Indeed, we set U1 D hc; g; hihd; e; m; ni, U2 D hef; npi Š E4 so that U D U1 U2 , where U10 D ˆ.U1 / D hc; g; hi Z.U1 / and we claim that hc; g; hi D Z.U1 /. Suppose that w D d ˛ e ˇ m nı 2 Z.U1 / .˛; ˇ; ; ı D 0; 1/ which implies 1 D Œw; d D Œe; d ˇ Œm; d Œn; d ı D .ch/ˇ c g ı D c ˇ C g ı hˇ and so ˇ D D ı D 0 and w D d ˛ . From 1 D Œw; e D Œd; e˛ D c ˛ h˛ follows ˛ D 0 and so w D 1. We have proved that Z.U1 / D hc; g; hi and so Z.U / D hc; g; h; ef; npi. The factor-group G=U Š D8 is generated with involutions a1 U and b 1 U so that x D a1 b 1 must be an element of order 4 with hxi \ U D f1g. Set y D x 2 so

60

The structure of the Burnside group of order 212

159

that U hyi=U D Z.G=U / D ˆ.G=U /. It follows that ˆ.G/ D U hyi. By Theorem A, the set G ˆ.G/ cannot contain involutions and so (since 1 .U / D U ) we have 1 .G/ D ˆ.G/. We have hciG D hc; g; hi D hc; s1 ; s2 i, where hs1 ; s2 i Z.G/, and so CG .c/ D CG .hc; g; hi/ which shows that CG .c/ is a normal subgroup of G. On the other hand, CG .c/ U and jG W CG .c/j D 4 (since the four conjugates of c are contained in hc; g; hi hs1 ; s2 i), which implies that CG .c/ D U hyi D ˆ.G/ and so c y D c and hc; g; hi Z.ˆ.G//. We determine now the action of G=U on U by giving the actions of the elements a1 and b 1 (we use a1 and b 1 rather than a and b because of some technical reasons) on the elements c; g; h; d; e; f; m; n; p of U . We get (by using all previous relations): ca

1

D g;

ga

1

D c a D c;

ha

1

D cgh;

da

1

D d;

ea

1

D f;

fa

1

D ..a2 /b

ma

1

D n;

na

1

D .b 2 /a D b 2 Œb 2 ; a2 D mc;

pa

1

D mnpcgh;

cb

1

D h;

gb

1

D cgh;

hb

1

D c b D c;

db

1

D e;

eb

1

D d b D d m D d Œd; m D dc;

fb

1

D def cgh;

mb

1

D m;

nb

1

D p;

pb

1

D ..b 2 /a /b D .ma /m D nm D nŒn; m D ncg:

2

1

2

/a D e d D eŒe; d D ech;

2

2

2

1

1

2

Since y D x 2 D .a1 b 1 /2 , we get from the above relations the action of y on U . We know that y centralizes hc; g; hi and we get: d y D def cgh;

e y D fg;

f y D eg;

my D mnpc;

ny D pg;

p y D ng:

Since .ef /y D ef and .np/y D np, we get that Z.U / D hc; g; h; ef; npi Z.ˆ.G//. From the above relations we also get: Œd; y D .ef /cgh;

Œe; y D .ef /g;

Œf; y D .ef /g;

Œm; y D .np/c;

Œn; y D .np/g;

Œp; y D .np/g:

Our relations also show that ˆ.G/=Z.U / is abelian and so elementary abelian. On the other hand, we see that ef; np 2 .ˆ.G//0 and so .ˆ.G//0 D Z.U /. If Z.ˆ.G// > Z.U /, then ˆ.G/ D U Z.ˆ.G// which implies that .ˆ.G//0 D U 0 D hc; g; hi, a

160

Groups of prime power order

contradiction. Thus Z.ˆ.G// D Z.U / Š E25 and so ˆ.G/ is a special group of order 210 , as claimed in part (a) of our theorem. 1 1 It remains to determine the elements y a and y b . This will complete the determination of the action of a1 and b 1 on the special group ˆ.G/. Since x D a1 b 1 and y D x 2 , we get ya

1

D .x 2 /a

1

1

D .x a /2 D .b 1 a1 /2 D b 1 a1 b 1 a1

D b 1 .a1 b 1 a1 b 1 /b D b 1 yb D y b D .y b /b 2

D .Œm; yy/b D yb

1

1

1

D ..np/cy/b

1

D .y.np/c/b

1

1

D yb

D .y m /b

1

1

p ncg h

.np/cgh;

1

and so y a D y b .np/cgh. On the other hand, x a D a1 .a1 b 1 /a D a2 b 1 a D a2 b 2 .ba/ D a2 b 2 .a1 b 1 /1 D d mx 1 , and so y a D .x 2 /a D .x a /2 D d mx 1 d mx 1 D d m.d m/x x 2 D d m.d m/x y: 1

1

Conjugating the last relation with a1 , we get d a ma .d m/a and so using our previous relations we get: d n .f mnpcgh/ y a

1

ya

D y;

1

1 b 1 a 1

ya

1

D y,

D .pnmf nd /ycgh:

Using our commutator relations in ˆ.G/, we get pnmf nd D demng.ef /.np/, and 1 1 so y a D de mnych.ef /.np/. Putting the last relation in (3), we get also y b D de mnyg.ef /. We have determined completely the structure of the Burnside group G D ha; bi. We set ef D i and np D j so that Z.ˆ.G// D hc; g; h; i; j i Š E25 and the special group ˆ.G/ is generated with five involutions d; e; m; n; y with a2 D d , b 2 D m, .a1 b 1 /2 D y and the commutator relations and the action of a1 and b 1 on ˆ.G/ as given in part (e) of our theorem. In the rest of the proof we shall use only this presentation of the group G as given in part (e) of our theorem. We show now that Z.G/ D hcg; ch; cij i Š E8 . Indeed, G=U Š D8 and Z.G=U / D U hyi=U and so Z.G/ U hyi D ˆ.G/. It follows that Z.G/ Z.ˆ.G//. On the other hand, hcg; chi Z.G/. We compute: .cij /a

1

D g i ch jgh D cij;

.cij /b

1

D h igh jcg D cij;

and so cij 2 Z.G/. But hi; j i is a complement of hcg; ch; cij i in Z.ˆ.G// and ia

1

D i ch;

ja

1

D jgh;

.ij /a

1

D ijcg;

which shows that hi; j i \ Z.G/ D f1g and therefore Z.G/ D hcg; ch; cij i Š E8 .

60

The structure of the Burnside group of order 212

161

We examine now the non-trivial cosets of Z.ˆ.G// in ˆ.G/. Each such coset consists either of involutions or only of elements of order 4. Using the defining relations for ˆ.G/, we find out that there are exactly 12 such cosets which consist of involutions and they are the cosets with representatives from the set: V D fd; e; m; n; y; de m; de n; dey; d mn; e mn; mny; de mnyg: Also we see that if v; w 2 V , v ¤ w, then o.vw/ D 4. This shows that G has exactly 12 maximal elementary abelian subgroups. Each of them is of order 26 and the intersection of any two of them is equal Z.ˆ.G//. In order to determine the structure of various subgroups of G which contain the subgroup Z.ˆ.G//, it is convenient first to determine the structure of H D G=Z.ˆ.G// of order 27 . Since Z.ˆ.G// D .ˆ.G//0 , H is obviously the group of maximal possible order having the exponent 4, being generated with two elements, and having the property that the squares of any two elements commute. The images of elements a; b; d; e; m; n; y of G in H D G=Z.ˆ.G// we denote again with the same symbols. Then we have E D 1 .H / D ˆ.H / D hd; e; m; n; yi Š E25 , H D ha; bi with a2 D d , b 2 D m, .a3 b 3 /2 D y, and the action of a and b on E is given with: d a D d;

e a D e;

ma D n;

na D m;

y a D de mny;

d b D e;

e b D d;

mb D m;

nb D n;

y b D de mny:

We have CE .a/ D hd; e; mni, CE .b/ D hde; m; ni, CE .ab/ D hde; mn; yi, Z.H / D hde; mni Š E4 , .Ehai/0 D .Ehbi/0 D .Ehabi/0 D hde; mni D Z.H /, Ehai=Z.H /, Ehbi=Z.H /, Ehabi=Z.H / are abelian groups of type .4; 2; 2/. It follows that the maximal subgroups of H are of rank 3. Also, it is easy to see from these information that the number of conjugate classes in H is k.H / D 26. Indeed, 4 elements in Z.H / give 4 classes of size 1, 12 elements in d Z.H / [ mZ.H / [ yZ.H / give 6 classes of size 2, the remaining 16 elements in E give 4 classes of size 4, and 96 D 27 25 elements in H E give 12 classes of size 8 each. We compute: y D .a1 b 1 /2 D a1 b 1 a1 b 1 D .a1 b 1 ab/ b 1 a1 a1 b 1 D Œa; bb 1 a2 b 1 D Œa; b d b b 2 D Œa; be m; so that Œa; b D e my. This gives H 0 D hde; mn; e myi Š E8 and H=H 0 Š C4 C4 . Since .emy/a D e n de mny D d my D .emy/de; .emy/b D d m de mny D e ny D .emy/mn; it follows that ŒH; H 0 D Z.H / D hde; mni, H is of class 3, and H=ŒH; H 0 is the non-metacyclic minimal nonabelian group of order 25 and exponent 4. We have obtained the results stated in part (f) of our theorem.

162

Groups of prime power order

We return now to our group G and consider the structure of maximal subgroups M D ˆ.G/hai, N D ˆ.G/hbi, K D ˆ.G/ha1 b 1 i. By the preceding paragraph, M 0 D N 0 D K 0 D Z.ˆ.G//hde; mni and M=M 0 is an abelian group of type .4; 2; 2/ so that d.M / D 3. Let be the automorphism of G of order 3 induced with a D b, b D a1 b 1 . Then acts transitively on fM; N; Kg and therefore it is enough to determine the structure of M . By the preceding paragraph, G 0 D Z.ˆ.G//hde; mn; e myi D M 0 he myi, G=G 0 Š C4 C4 , and M 0 =Z.ˆ.G// D Z.H / D K3 .H /, H D G=Z.ˆ.G// and therefore M 0 D Z.ˆ.G//hde; mni is a characteristic subgroup of G. It follows from d.G/ D 2 that G 0 =K3 .G/ is cyclic and G 0 D K3 .G/hŒa; bi. But 1 D .ab/4 a4 b 4 Œb; a6 .mod K3 .G// so that Œb; a2 2 K3 .G/ and jG 0 =K3 .G/j D 2. This gives that K3 .G/ D Z.ˆ.G//hde; mni D M 0 . We compute: Œde; mn D Œd; mŒd; nŒe; mŒe; n D c g h cgh D 1; .de/2 D d 2 e 2 Œe; d D ch;

.mn/2 D cg;

.de mn/2 D gh;

which shows that M 0 D K3 .G/ is abelian of type .4; 4; 2; 2; 2/ with ˆ.M 0 / D hcg; chi Š E4 . Since CG .c/ D ˆ.G/, we have Z.M / Z.ˆ.G//. But Chi;j i .a1 / D f1g and so Z.M / D Z.G/ D hcg; ch; cij i, where we have used the fact that hi; j i is a complement of Z.G/ in Z.ˆ.G//. Since .de/a

1

D .de/i;

.de/y D .de/ch;

.mn/y D .mn/cg; .mn/n D .mn/cg;

.de/n D .de/ch; .mn/a i

a1

D i ch;

j

1

a1

D .mn/g; D jgh;

we get that hc; g; h; i i Š E24 is normal in M , M 0 =hc; g; h; i i Z.M=hc; g; h; i i/ and 1 1 K3 .M / D hc; g; h; i i. Finally, hc; g; h; i i Z.ˆ.G//, c a D c.cg/, g a D g.cg/, 1 1 ha D h.cg/, i a D i.ch/ so that K4 .M / D hcg; chi Z.G/ and M is of class 4. We determine now the structure of G 0 D Z.ˆ.G//hde; mn; e myi, where G=G 0 Š C4 C4 . We know that K3 .G/ D Z.ˆ.G//hde; mni is an abelian maximal subgroup of G 0 , where 1 .K3 .G// D Z.ˆ.G//. But we see that the elements: e my; de e my D d my; mn e my e ny; de mn e my d ny

.mod Z.ˆ.G///

are all of order 4 and so 1 .G 0 / D Z.ˆ.G//. We compute Œde; e my D ch and Œmn; e my D cg and so G 00 D hcg; chi Z.G/. From jG 0 j D 2jZ.G 0 /jjG 00 j follows jZ.G 0 /j D 25 and so Z.G 0 / D Z.ˆ.G///. Also, .de/2 D ch, .mn/2 D cg and .emy/2 D cghij , which gives ˆ.G 0 / D hcg; ch; cghij i D hcg; ch; cij i D Z.G/. We have M 0 D K3 .G/ and K3 .M / D ŒM; M 0 D ŒM; K3 .G/ D hc; g; h; i i, 1 .mn/b D .mn/j , and so K4 .G/ hc; g; h; i; j i D Z.ˆ.G//. On the other hand, we know that K3 .G/=Z.ˆ.G// D Z.G=Z.ˆ.G/// D Z.H / and so K4 .G/ D 1 1 1 Z.ˆ.G//. We have i b D i.gh/, j a D j.gh/, j b D j.cg/, and so we see that K5 .G/ D hcg; chi Z.G/, K6 .G/ D f1g, and G is of class 5. Finally, we see that Z2 .G/ D Z.ˆ.G// Š E25 , Z3 .G/ D K3 .G/ D M 0 (abelian of type .4; 4; 2; 2; 2/), and Z4 .G/ D ˆ.G/. Our theorem is proved.

61

Groups of exponent 4 generated by three involutions

It is known that a group generated with two non-commuting involutions is dihedral. But the problem to determine groups generated with three involutions is already very difficult. In this section we classify groups of exponent 4 which are generated with three involutions. Theorem 61.1. Let G be a group of exponent 4 with d.G/ D 3 so that G is generated with involutions a; b; c and at least one of ab; ac; bc is an involution. Then jGj 26 and one of the following holds: (a) G Š C2 D8 , where jGj D 24 . (b) G is a splitting extension of the abelian group H D hv; w j v 4 D w 4 D Œv; w D 1; v 2 D w 2 i of type .4; 2/ with the four-group ha; bi, where v a D w, v b D v 1 , w b D w 1 . We have jGj D 25 and G D ha; b; ci with c D bv. (c) G is a splitting extension of the nonmetacyclic minimal nonabelian group H D hv; w j v 4 D w 4 D .wv/2 D Œw 2 ; v D 1i of order 24 with the four-group ha; bi, where v a D w, v b D v 1 , w b D w 1 . We have jGj D 26 and G D ha; b; ci with c D bv. The group G is isomorphic to a Sylow 2-subgroup of the alternating group A8 . Proof. Let G be a group of exponent 4 with d.G/ D 3 so that G is generated with three involutions a; b; c. If a; b; c pairwise commute, then G Š E8 , a contradiction. If a commutes with b and c but Œb; c ¤ 1, then G D hai hb; ci Š C2 D8 . Assume now that Œa; b D .ab/2 D 1 but G is not isomorphic to C2 D8 . This implies Œb; c ¤ 1 and Œa; c ¤ 1. Set v D bc so that o.v/ D 4 and D D hb; ci D hb; vi Š D8 with hv 2 i D Z.D/. Also, ha; ci Š D8 and so o.ac/ D 4. Set w D v a . Since b inverts v and a centralizes b, it follows that b inverts hwi. Also, w a D v and so the four-group ha; bi normalizes the subgroup H D hv; wi. Hence G D H ha; bi. We compute .ac/2 D .aca/c D c a c D .bv/a bv D .bwb/v D w 1 v and so o.w 1 v/ D 2. From .av/4 D 1 follows .av/2 D .ava/v D v a v D wv and

164

Groups of prime power order

.wv/2 D 1. The last relation gives 1 D wvwv D w 2 .w 1 vw 1 /w 2 v D w 2 .w 1 vw 1 v/v 1 w 2 v D w 2 .w 1 v/2 v 1 w 2 v D Œw 2 ; v: If hwi D hvi, then H D hvi and G D Dhai. There is an involution i 2 ha; bi hbi which centralizes hvi and so G D hi i D, contrary to our assumption. We have hvi ¤ hwi and since Œw 2 ; v D 1, we get w 2 2 Z.H /. Consider HN D H=hw 2 i (bar convention) so that HN D hw; N vi. N Suppose v 2 D w 2 which implies that HN is generated with two distinct involutions vN and w. N We know that .wv/2 D 1 and N so wv D wN vN is also an involution. Hence H Š E4 . Suppose that H is nonabelian. Since hvi and hwi are two distinct cyclic subgroups of order 4 in H , we get H Š Q8 . But we know that .wv/2 D 1, a contradiction. Hence H is abelian of type .4; 2/. We have G D H ha; bi and so the group G is a splitting extension of the abelian group H D hv; w j v 4 D w 4 D Œv; w D 1; v 2 D w 2 i of type .4; 2/ with the four-group ha; bi, where v a D w, v b D v 1 , w b D w 1 . We get jGj D 25 and G D ha; b; ci with c D bv. Assume now that v 2 ¤ w 2 so that hvi \ hwi D f1g. We have o.w/ N D 2, o.v/ N D4 2 4 N and o.wN v/ N D 2 (since .wv/ D 1 ). Hence H Š D8 and so jH j D 2 . The group H is not of maximal class since H possesses the abelian subgroup hw 2 ihvi of type .4; 2/. Hence H is of class 2 and therefore 1 D Œw 2 ; v D Œw; v2 D Œw; v 2 , so that Œw; v is an involution. Since Œw; v 2 Z.H /, we get that H=hŒw; vi is abelian. Consequently, H 0 D hŒw; vi and so H is minimal nonabelian with Z.H / D hw 2 ; v 2 i Š E4 . Also, 1 D .wv/2 D w 2 v 2 Œv; w, and so Œv; w D Œw; v D w 2 v 2 . Since wv 62 Z.H / D hw 2 ; v 2 i, we get hw 2 ; v 2 ; wvi Š E8 and so H is non-metacyclic. The group G is a splitting extension of the non-metacyclic minimal nonabelian group H D hv; w j v 4 D w 4 D .wv/2 D Œw 2 ; v D 1i of order 24 with the four-group ha; bi, where v a D w, v b D v 1 , w b D w 1 . We have ha; bi \ H D f1g and so jGj D 26 and G D ha; b; ci with c D bv and Z.G/ D hv 2 w 2 i D h.bc/2 ..bc/2 /a i. Indeed, it is easy to see that the involutions in ha; bi Š E4 induces on H only outer automorphisms. No element x 2 H could invert an element y of order 4 in H since Œw; v D w 2 v 2 is not a square in H . Also, hy x i \ hyi hy 2 i since y 2 2 ˆ.H / D Z.H /. The group G is isomorphic to a Sylow 2-subgroup of the alternating group A8 . Set D D hv; bi Š D8 so that .v 2 /a D w 2 and therefore v 2 62 DG . No subgroup of order 2 in D hvi is normal in D. Hence DG D f1g and so G has a faithful transitive permutation representation of degree 8. If we set: a D .1; 2/.3; 4/.5; 6/.7; 8/;

b D .4; 5/.3; 6/;

c D .2; 3/.6; 7/;

then we see that the permutations a; b; c satisfy all the above relations for G and .bc/2 ..bc/2 /a D .1; 8/.2; 7/.3; 6/.4; 5/ ¤ 1 and so the obtained permutation representation is faithful. Since these permutations are even, we are done.

61

Groups of exponent 4 generated by three involutions

165

Theorem 61.2. Let G be the group given with G D ha; b; c j a2 D b 2 D c 2 D .ab/4 D .bc/4 D .ca/4 D 1; .abc/4 D .c.ab/2 /4 D .b.ac/2 /4 D .acab/4 D 1i: Then G is of order 210 and exponent 4. Hence each group of exponent 4 which is generated with three involutions is a epimorphic image of the above group of order 210 . Proof. This theorem is proved by using a computer (W. Lempken of the University of Essen). In what follows we shall determine the structure of the group G D ha; b; ci of order 210 as defined in Theorem 61.2 so that G is the free group of exponent 4 generated with three involutions a; b; c. We set .ab/2 D d , d c D e, e a D f , and f b D g. We shall determine the structure of the normal subgroup S D hd iG hd; e; f; gi, where d; e; f; g are some involutions in ˆ.G/. Since ha; bi Š D8 , the involution d D Œa; b commutes with a and b. From the above follows e c D d , f a D e and g b D f . We compute .de/2 D .abab cababc/2 D .ababc ababc/2 D .ababc/4 D 1; and so Œd; e D 1. From the last relation we obtain (conjugating with the element a and then with the element b): Œd; e a D Œd; f D 1, Œd; f b D Œd; g D 1. Further computation gives (noting that .ab/2 D .ba/2 and .bac/3 D .bac/1 D cab): .ef /2 D .e aea/2 D .ea/4 D 1; .fg/2 D .f bf b/2 D .f b/4 D 1; .eg/2 D .cbabac bacbabacab/2 D .cba.bac bac bac/cbacab/2 D .cba cab cba cab/2 D .cbacab/4 D 1; and so Œe; f D 1, Œf; g D 1 and Œe; g D 1. We have proved that the subgroup hd; e; f; gi is elementary abelian of order 24 . It is possible to show that the subgroup hd; e; f; gi is normal in G and so we get S D hd iG D hd; e; f; gi. Indeed, we compute (noting that .ca/3 D .ca/1 D ac, .ab/3 D ba, .cbab/3 D .cbab/1 D babc): ff c D acababca cacababcac D acabab.ca/3 babcac D acabab ac babcac D ac.ababab/bcbabcac D ac ba bcbabcac D a.cbab/.cbab/.cbab/babac D a babc babac D .ab/2 c.ba/2 c D d d c D de;

166

Groups of prime power order

and so ff c D de which gives f c D def . Also we get: 2

g a D .d c /aba D e aba D e ababb D .e .ab/ /b D .e d /b D e b ; and so g a D e b . We compute (noting that .ab/2 D .ba/2 and bcbcb D .bc/3 c D .bc/1 c D cbc): ee b D caba.bc bcb/abacb D caba.cbc/abacb D .cabacb/2 ; and similarly (noting that baca baca ba D .baca/3 ac D .baca/1 ac D acabac), fg D acba.baca bacaba/bcab D acba acabac bcab D .acbcab/2 D .acbcab/2 D .bacbca/2 : Using the above relations we get (noting that .ac/2 D .ca/2 , .bca/2 D .bca/2 D .acb/2 , .cab/2 D .bac/2 , and .b.ac/2 /3 D .b.ac/2 /1 D .ac/2 b): ee b fg D .cabacb cabacb/.bacbca bacbca/ D cabac.bca bca cab cab/acbca D cabac.acb/2 .bac/2 acbca D cabacacbacbbacbacacbca D ca .b.ac/2 b.ac/2 b.ac/2 /bca D ca .ac/2 b bca D ca.ca/2 ca D .ca/4 D 1; and so e b D efg D g a . Finally, we get (using .ab/3 D ba): g c e D cbacbabac.abc cabab/c D cba cba bac.ab/3 c D .cba/2 .bac/2 D .cba/2 .bac/2 D .abc/2 .cab/2 D abcabccabcab D abc.ab/2 cab D .ab/2 bac.ab/2 cab D dg; and so g c D deg. We have proved by now: d a D d;

e a D f;

f a D e;

g a D efg;

d b D d;

e b D efg;

f b D g;

g b D f;

d c D e;

e c D d;

f c D def;

g c D deg:

This shows that hd; e; f; gi is normal in G and so S D hd; e; f; gi D hd iG is elementary abelian of order 24 . By Theorem 61.1, G=S is of order 26 . Since jGj D 210 (Theorem 61.2), we have S Š E24 and jG=S j D 26 . From the above results we get at once that hef; egi Z.G/. The subgroup hd; ei is a complement of hef; egi in S and Z.G/ \ hd; ei D f1g. This gives Z.G/ \ S D hef; egi.

61

Groups of exponent 4 generated by three involutions

167

Define the automorphism of G induced with a D b, b D c, c D a. Set d D h D .bc/2 ;

e D i D ..bc/2 /a ;

f D j D ..bc/2 /ab ;

g D k D ..bc/2 /abc

so that T D hhiG D hh; i; j; ki Š E24 and transports the above relations for the elements in S into the relations for the elements in T . In particular, ha D i , i a D h, hb D h, i b D j , hc D h, i c D ij k. We want to determine S \T . For that purpose we compute (noting that .bc/3 D cb, .cba/2 D .abc/2 , .ca/3 D ac, .bc/3 D cb): ef i k D cababc acaba.bca abcbc/a cbabcbcabc D cababcacaba.cb acba/bcbcabc D cababcacaba abcabc bcbcabc D cabab.ca/3 .bc/3 abc D cabab ac cb abc D c.ab/4 c D 1; and similarly (noting that we have .bc/2 D .cb/2 , .cbab/2 D .babc/2 , .ab/3 D ba, .bac/3 D cab): egj k D cababc bacababcab bacbcbab cbabcbcabc D cababcbacaba.cbabcbab/cbcabc D cababcbac.aba bab/cbabc cbcabc D cababc.bac ba cbac/abc D cababc cab abc D c.ab/4 c D 1; which gives ef D i k, eg D j k and so S \T hef; egi Š E4 and i b D j D i.ij / D i.fg/, i c D i.j k/ D i.eg/. We have jS T j 26 . By Theorem 61.1, jG=S T j 24 and so the fact that jGj D 210 gives U D S T is of order 26 and G=U Š C2 D8 . This implies that y D .ca/2 62 U and S \ T D hef; egi D hi k; j ki Z.G/. Note that hh; i i is a complement of S \ T in T and we compute: d h D d bcbc D d.fg/;

d i D d abcbca D d.fg/;

e h D e bcbc D e.fg/;

e i D e abcbca D e.fg/;

and so Œd; h D Œd; i D Œe; h D Œe; i D fg. We have U 0 S \ T Z.G/ and so U is of class 2 and, by the above, U 0 D hfgi and U D hd; e; f; gihh; i i ˆ.G/. Since y D .ca/2 62 U and G=G 0 is generated with involutions, we have ˆ.G/ D G 0 D 1 .ˆ.G// D U hyi. We note that fg; eg; ef 2 Z.G/, hg D ghŒh; g D ghŒh; e.eg/ D ghŒh; e D ghfg

168

Groups of prime power order

and Œd; hi D Œd; hŒd; i D .fg/2 D 1 so y b D .caca/b D ch ad ch ad D caha dchad D caidchad D caci c d c had D caci.eg/ehad D cacighad D .caca/i a g a ha d D y h.ef /gi d D y.ef /.hg/id D y.ef /gh.fg/id D y.ef /.fg/g.hi /d D yedhi; and hence we have y b D ydehi , y a D y and y c D y. Set fg D z1 , eg D z2 so that E4 Š hz1 ; z2 i Z.G/, d y D d caca D dz1 z2 ;

e y D ez1 z2 ;

hy D hz2 ;

i y D iz2 ;

Z.U / D hz1 ; z2 ; de; hi i Š E24 , U 0 D hz1 i. Obviously we have Z.ˆ.G// Z.U / and since .de/y D de, .hi /y D hi , we get hde; hi i Z.ˆ.G// and so Z.ˆ.G// D Z.U /. By the above, z1 ; z2 2 .ˆ.G//0 and since the factor-group ˆ.G/=hz1 ; z2 i is abelian, we get G 00 D .ˆ.G//0 D hz1 ; z2 i. Also, hde; hi i is a complement of hz1 ; z2 i in Z.ˆ.G//, Z.G/ ˆ.G/ and .de/a D df , .hi /b D hj , .dehi /a D df hi which all imply that hz1 ; z2 i D Z.G/. Our relations give at once that K3 .G/ D Z.ˆ.G// D hz1 ; z2 ; de; hi i Š E24 and K4 .G/ D hz1 ; z2 i D Z.G/ so that G is of class 4. We have proved the following Theorem 61.3. Let G be the free group of exponent 4 which is generated with three involutions a; b; c. Then G is of order 210 and has the following properties: (a) ˆ.G/ D G 0 D 1 .G 0 / is of order 27 and class 2 with Z.ˆ.G// Š E24 and .ˆ.G//0 D G 00 D ˆ.ˆ.G// Š E4 . (b) We have K3 .G/ D ŒG; G 0 D Z.ˆ.G// and K4 .G/ D .ˆ.G//0 D Z.G/ so that G is of class 4. (c) The group ˆ.G/ is generated with five involutions d; e; h; i; y so that .ˆ.G//0 D hz1 ; z2 i Š E4 and Œd; e D Œh; i D 1, Œd; h D Œd; i D Œe; h D Œe; i D z1 , Œd; y D Œe; y D z1 z2 , Œh; y D Œi; y D z2 , where hz1 ; z2 i D Z.G/ and Z.ˆ.G// D hz1 ; z2 ; de; hi i Š E24 . We have G D ha; b; ci with .ab/2 D d , .bc/2 D h, .ca/2 D y, where the action of a; b; c on ˆ.G/ is given with: d a D d;

e a D ez1 z2 ;

ha D i;

d b D d;

e b D ez1 ;

hb D h;

d c D e;

e c D d;

hc D h;

i a D h;

y a D y;

i c D iz2 ;

y c D y:

62

Groups with large normal closures of nonnormal cyclic subgroups

Let H be a nonnormal subgroup of a p-group; then jG W H G j p, where H G is a normal closure of H in G. Therefore, it is natural to classify the p-groups G such that jG W H G j D p for all nonnormal subgroups H < G. Such G are classified by the second author but all proofs in this section are due to the first author. Theorem 62.1 (Z. Janko). Let G be a non-Dedekindian p-group. Suppose that jG W H G j D p for all nonnormal cyclic subgroups H of G. Then one and only one of the following holds: (a) jGj D p 3 . 2

2

(b) G D ha; b j ap D b p D 1; ab D a1Cp i is the unique nonabelian metacyclic group of order p 4 and exponent p 2 . (c) G is a 2-group of maximal class. (d) p D 2, G D H Z is a semidirect product with cyclic kernel Z of order > 4 and a cyclic complement H of order 4. Write R D 1 .H /. We have Z.G/ D R 1 .Z/ Š E4 , G=R is of maximal class not isomorphic with generalized quaternion group, and G=G 0 is abelian of type .4; 2/. We have G D n n1 hh, z j h4 D z 2 D 1, n > 2, z h D z 1C2 , D 0; 1i. It is trivial that groups (a), (b) and (c) of Theorem 62.1 satisfy the hypothesis. For example, if N G G is of index > 2, where G is 2-group of maximal class, then N ˆ.G/, and this implies that groups of (c) satisfy the hypothesis. Now let G be as in Theorem 62.1(d) and let X < G be nonnormal cyclic. Let us prove that jG W X G j D 2. It follows from 1 .G/ D Z.G/ that jXj > 2. The group G has no proper subgroups which are generalized quaternion (Proposition 10.19(a)) hence 1 .G/ < X G (otherwise, X G is cyclic so X G G). If 1 .H / D R < X, then X=R is a nonnormal cyclic subgroup of the 2-group of maximal class G=R (since j.G=R/ W .G 0 =R/j D 4) so jG W X G j D 2, by (c). Now we assume that R 6 X. Suppose that XR D X R 6E G; then XR=R 6E G=R so jG W .XR/G j D 2. We have .XR/G D X G R D X G since R < 1 .G/ < X G , so jG W X G j D 2. Now assume that XR G G. Then XR=R T =R, where T =R is a cyclic subgroup of index 2 in G=R. We have T D V R and ˆ.V / D ˆ.T / G G so all cyclic subgroups of T of

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Groups of prime power order

order 2k for all k < n, where jV j D 2n (n > 2), are G-invariant, so jXj D 2n ; then jG W Xj D 4. In that case, X G D T so jG W X G j D 2, as required. Lemma 62.2. Let G be a non-Dedekindian minimal nonabelian p-group of order > p 3 . If jG W H G j D p for all nonnormal H < G, then G is the unique nonabelian metacyclic group of order p 4 and exponent p 2 which we denote Hp;2 . Proof. Let H < G be nonnormal cyclic; then H G D H G 0 so jG W H j D p 2 . If the nonmetacyclic G is as in Lemma 65.1(a), then o.a/ D o.b/ D p so jGj D p 3 , contrary to the hypothesis. Thus, G is metacyclic as in Lemma 65.1(b) so G D B A, where A D hai G G, B D hbi and jAj D p 2 , by the above. Assume that n > 2. Write 2 .B/ D hxi, F D hxai; then jF j D p 2 . We have .xa/b D xa1Cp 62 F so F is not G-invariant; then F G D F G 0 has index p n1 > p in G, a final contradiction. Proof of Theorem 62.1. Let H < G be nonnormal cyclic. Since H G Hˆ.G/ < G, we get jG W Hˆ.G/j D p so H 6 ˆ.G/ and d.G/ D 2. Hence all cyclic subgroups of ˆ.G/ are G-invariant so ˆ.G/ is Dedekindian. Since ˆ.G/ has no G-invariant subgroups isomorphic to Q8 (Lemma 1.4), it is abelian. If jG 0 j D p, then G is minimal nonabelian (Lemma 65.2(a)) so G is as in Theorem 62.1(b), by Lemma 62.2. Now suppose that jG 0 j > p so G is not minimal nonabelian. (i) Let ˆ.G/ be cyclic. If p > 2, then ˆ.G/ D Z.G/ and jG 0 j D p (Theorem 4.4), a contradiction. Thus, p D 2; then ˆ.G/ D Ã1 .G/ so G has a cyclic subgroup of index 2 hence G is of maximal class. Now suppose that ˆ.G/ is noncyclic. (ii) Let p > 2. Suppose that ˆ.G/ D U V and U > f1g is cyclic (by (i), V > f1g and, by the above, U; V are G-invariant). Take x 2 G. By induction, Œx; ˆ.G/ U \ V D f1g so ˆ.G/ Z.G/. Then G is minimal nonabelian, a contradiction. (iii) Thus, p D 2 and ˆ.G/ is noncyclic abelian. Assume that G is not metacyclic. Then, by Schreier inequality (see Appendix 25) and Corollary 36.6, there is M 2 1 such that d.M / D 3. Write GN D G=ˆ.M /; then GN of order 24 is neither abelian (in N < MN of order 2 such view of d.G/ D 2 < d.M /) nor of maximal class so there is L N N N N that L 6 Z.G/. Since G is not of maximal class, we get jG 0 j D 2, by Taussky, so N GN D L N GN 0 has index 4 in G, N a contradiction (our hypothesis is inherited by nonL Dedekindian epimorphic images of G). Thus, G is metacyclic so G=Z is cyclic for a cyclic Z GG. Since G is not of maximal class, we get 1 .G/ Š E4 so 1 .G/ Z.G/ in view of G=1 .G/ > 2. Since G is not of maximal class, we get jG=G 0 j 8 (Taussky). Let R < G 0 be of index 2; then G=R is minimal nonabelian and nonDedekindian since d.G=R/ D 2 and jG=R/ > 8 (Lemma 62.2). Therefore, by Lemma 62.2, G=R Š H2;2 so G=G 0 is abelian of type .4; 2/. It follows that jG W Zj D 4 since Z < G0. Let M < G be minimal nonabelian; then jM j > 8 (Proposition 10.19) so M is as in Lemma 65.1(b); then M D H1 Z1 is a semidirect product with cyclic factors Z1

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and H1 and H1 is not normal in M so in G. However, M 0 D 1 .Z1 / < G 0 and G 0 is cyclic so H1 \ G 0 D f1g. Since 2 D jG W H1G j, we get H1G D H1 G 0 ; then jH1 j D 4 since 1 .G/ Z.G/, and G=G 0 is abelian of type .4; 2/, H1 \Z D f1g since G 0 < Z and Z is cyclic. It follows that G D H1 Z, a semidirect product. Since Ã1 .H1 / G G as a subgroup of ˆ.G/, then A D Ã1 .H1 / Z is abelian of index 2 in G so 4 D 1 0 0 2 jG W G j D jZ.G/j (Lemma 1.1). Since jG=Ã1 .H1 / W .G=Ã1 .H1 // j D 4, the group G=Ã1 .H / is of maximal class (Taussky); moreover, it is not generalized quaternion since G=1 .G/ D .H1 1 .G/=1 .G// .Z1 .G/=1 .G// is a semidirect product. Next, jZj > 4 (otherwise, G is as in (b)). If G=1 .G/ is generalized quaternion, then j2 .G/j D 8 so G is a group of Lemma 42.1(c); then Z.G/ is cyclic, a contradiction. Thus, G is as stated in (d).

63

Groups all of whose cyclic subgroups of composite orders are normal

Definition 1. If all cyclic subgroups of a p-group G of composite orders are normal, then G is said to be a Kp -group. The property Kp is inherited by sections. Below we classify Kp -groups and prove that the property (Kp ) is equivalent to the following condition: Whenever f1g < B < A G, where A is cyclic, then NG .B/ D NG .A/ [Kaz1]. All proofs, apart of the proof of Supplement to Theorems 63.1 and 63.3, are due to the first author. Proposition 63.1. If G is a nonabelian Kp -group, p > 2, and exp.G/ > p, then jG 0 j D p. Proof. Let Z < G be cyclic of order p 2 and C0 D 1 .Z/; then C0 G G. Assume that K=C0 < G=C0 is nonnormal of order p; then K is abelian of type .p; p/. Set H D KZ; then jH j D p 3 . Since p > 2, H has exactly p cyclic subgroups of order p 2 and they generate H so H E G. Then 1 .H / D K G G, a contradiction. Lemma 63.2. Let a 2-group G be a K2 -group. Then: (a) If L G G is of order 2 and G=L Š Q8 , then L is a direct factor of G. (b) If T < G is cyclic of order 2e , e > 1, then G=T 6Š Q8 . Proof. (a) By Theorem 1.2, G has no cyclic subgroups of index 2 so L 6 G 0 , by Taussky’s theorem. Then, if Z=L < G=L is cyclic of order 4, then Z is abelian of type .4; 2/ and 1 .Z/ D Z.G/ so all subgroups of order 2 are normal in G, and we conclude that G is Dedekindian. Then, by Theorem 1.20, L is a direct factor of G. (b) Assume that G=T Š Q8 . We use induction on jGj. Let L < T be of order 2; then jT =Lj D 2, by induction, so, by (a), G=L D .T =L/.Q=L/, where Q=L Š Q8 . By (a), Q D L Q1 , where Q1 Š Q8 . Since Q1 Š Q8 is generated by cyclic subgroups of order 4, it is normal in G so G D T Q1 . Since all subgroups of order 2 are normal in G, G is Dedekindian, contrary to Theorem 1.20. Recall that the Hp -subgroup Hp .G/ of a group G is defined as hx 2 G j o.x/ ¤ pi. By Burnside, a nontrivial H2 -subgroup has index 2 in G. If G is a (Kp )-group, then Hp .G/ centralizes G 0 . Note that if jG W H2 .G/j D 2, then exp.Z.G// D 2.

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Theorem 63.3 ([Kaz1]). If G is a nonabelian K2 -group, then either (a) jG 0 j D 2 or (b) H2 .G/ has index 2 in G. Proof. Suppose that G is a counterexample of minimal order. As we noticed, H2 .G/ centralizes G 0 . Since jG W H2 .G/j 2, G 0 < H2 .G/. If jG W H2 .G/j D 2, we have case (b). Now we assume that G D H2 .G/. Then, by what has just been said, cl.G/ D 2. (i) Let jG 0 j > 4. If C < G 0 is of order 2, then, by induction, G=C has an abelian subgroup A=C of index 2 such that all elements in .G=C / .A=C / are involutions since j.G=C /0 j D jG 0 =C j > 2. Let x 2 G A; then x 2 2 A. Assume that o.x/ > 2; then X D hxiGG. In that case, jXC =C j D 2 so x 2 2 C and G=C D .A=C /.X=C / is abelian so C D G 0 is of order 2, which is not the case. Thus, if jG 0 j > 4, we obtain case (b). It remains to consider the case where G 0 is either cyclic of order 4 or abelian of type .2; 2/. (ii) Let G 0 D hci Š C4 . We have c D Œx; y for some x; y 2 G. Set H D hx; yi; then H 0 D G 0 D hci is a central subgroup of order 4. We see that (a nonabelian K2 -group) H of composite exponent is not a group from the conclusion (indeed, since exp.Z.H // > 2, H coincides with its H2 -subgroup; in addition, jH 0 j D 4 > 2); therefore, H D G, by induction. If x; y are involutions, then G is dihedral of class 2 so jG 0 j D 2, a contradiction. Therefore, one may assume that o.x/ > 2; in that case, G D hxihyi is metacyclic. Next, G has no cyclic subgroups of index 2 (otherwise, since G is of class 2, jG 0 j D 2). Then 1 .G/ is of type .2; 2/ and G has no nonabelian subgroups of order 8 (this follows from Proposition 10.19). Therefore, if K=1 .G/ is a subgroup of order 2 in G=1 .G/ then K is abelian of type .4; 2/ so it is normal in G since K D hy j o.y/ D 4i. It follows that G=1 .G/ is nonabelian Dedekindian (nonabelian since G 0 is cyclic of order 4 > 2). Then G=1 .G/ Š Q8 since it is metacyclic (Theorem 1.20). By Taussky’s theorem, G=G 0 is abelian of type .4; 2/. Since G 0 Z.G/, G is minimal nonabelian. In that case, jG 0 j D 2, a contradiction. (iii) It remains to consider the case where G 0 Š E4 . Let Z < G be cyclic of order 2 D exp.G/; then G=Z is nonabelian since G 0 — Z. By Lemma 63.2(b), G=Z has no subgroups isomorphic to Q8 . It follows from Theorem 1.20 that G=Z contains a nonnormal subgroup L=Z of order 2. Since L 6E G, the subgroup L is not generated by cyclic subgroups of composite orders so L is dihedral, by Theorem 1.2 (indeed, jL W Zj D 2 and Z is cyclic). Since L0 is cyclic and exp.G 0 / D 2, we get jL0 j D 2 hence jLj D 8, jZj D 4 and e D 2. Since G D H2 .G/ is nonabelian, there are two cyclic subgroups, say U and V , of order 4 that generate a nonabelian subgroup Q D U V . In that case, Q Š Q8 and Q G G. Since G 0 6 Q, the quotient group G=Q is nonabelian so it contains a cyclic subgroup M=Q of order 4 .D exp.G//. Set M D hx; Qi; then X D hxi is of order 4 and M D Q X. Since M is not Dedekindian (Theorem 1.20) and all subgroups of order 2 are normal in M , it contains a nonnormal cyclic subgroup of order 4, a final contradiction. e

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Groups of prime power order

Supplement to Theorems 63.1 and 63.3 ([Kaz1]). Let G be a (Kp )-group. If there holds exp.G/ > p and jG 0 j D p, then ˆ.G/ is cyclic. Proof (Kazarin, personal communication). Assume that ˆ.G/ is noncyclic; then we have jGj > p 3 and exp.G/ exp.G=G 0 / > p. (i) If x; y 2 G, then 1 D Œx; yp D Œx; y p so y p 2 Z.G/ and ˆ.G/ D G 0 Ã1 .G/ Z.G/. (ii) Let C < G be cyclic of order > p. Assume that G 0 6 C ; then G=C is nonabelian, C \ G 0 D f1g so ŒC; G C \ G 0 D f1g and hence C Z.G/. By Lemma 63.2(b), G=C has no subgroups isomorphic to Q8 so it is not Dedekindian. Therefore, there is a nonnormal B=C < G=C of order p; B is abelian of type .jC j; p/ since C Z.G/. Then B G G since B D hx 2 B j o.x/ > pi, contrary to the choice of B. Thus, every cyclic subgroup of G of composite order contains G 0 . (iii) Let G be minimal nonabelian. Then G=G 0 D haG 0 i hbG 0 i is abelian of type .p m ; p n / with m n; then m > 1. Let o.aG 0 / D p m , o.bG 0 / D p n . By (ii), G 0 hai so G 0 is not a maximal cyclic subgroup of G so G is metacyclic and G D hbi hai (Lemma 65.1). By (ii), n D 1 so ˆ.G/ is cyclic. (iv) If x; y 2 G are elements of composite orders, then H D hx; yi is either abelian or minimal nonabelian and in both cases it contains a cyclic subgroup of index p. Indeed, suppose that H is abelian. Let o.x/ o.y/. Then H D hxi hzi so, by (ii), o.z/ D p, i.e., H has a cyclic subgroup hxi of index p. Now suppose that H is nonabelian. Since hxi G G, H is metacyclic. Since ˆ.H / Z.H /, by (i), and H=ˆ.H / Š Ep 2 , H is minimal nonabelian. By (iii), H has a cyclic subgroup of index p so ˆ.H / is cyclic. Thus, in any case, hx p ; y p i is cyclic for all x; y 2 G. Let x 2 G be of maximal order. Then, by (iv), Ã1 .G/ hxi so, by (ii), ˆ.G/ D Ã1 .G/G 0 hxi whence ˆ.G/ is cyclic. Thus, Theorems 63.1 and 63.3 can be formulated as follows. Theorem 63.4 ([Kaz1]). If G is a Kp -group, then one of the following holds: (a) G is abelian, (b) exp.G/ D p, (c) jG 0 j D p and ˆ.G/ is cyclic, (d) p D 2 and jG W H2 .G/j D 2. The above four groups satisfy the hypothesis. Definition 2. A p-group G is said to be an Np -group if every cyclic subgroup A of composite order satisfies the following condition: Whenever f1g < B < A, then NG .B/ D NG .A/. (Clearly, subgroups of Np -groups are Np -groups.) Theorem 63.5 ([Kaz]). Let G be a nonabelian p-group, exp.G/ > p > 2. Then the following assertions are equivalent:

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(a) G is an Np -group, (b) G is a Kp -group. Proof. Obviously, (b) implies (a). Now suppose that an Np -group G is not a Kp group. Then G has a nonnormal cyclic subgroup A of composite order so, in view of Theorem 1.2, G has no cyclic subgroups of index p. By hypothesis, (i) Every nonidentity subgroup of A is not normal in G, i.e., AG D f1g. Suppose that G is a counterexample of minimal order. In that case, every proper subgroup of G of composite exponent is a Kp -group so the normalizer of every nonnormal cyclic subgroup of G of composite order is maximal in G since this normalizer is an Kp -group; thus N D NG .A/ 2 1 . Then, by Proposition 63.1, jN 0 j p. Let M 2 1 fN g; then A 6 M (otherwise, A is normal in M so in G). If jA \ M j > p, then, by induction, NG .M \ A/ MN D G so AG A \ M > f1g, contrary to (i). Thus jM \ Aj D p so jAj D p 2 . Thus (ii) Let A be a nonnormal cyclic subgroup of G of composite order. Then jAj D p 2 and A is contained in exactly one maximal subgroup of G. Suppose that jG 0 j D p. Then G 0 6 A so G 0 A D G 0 A G G. In that case, ˆ.G 0 A/ > f1g is contained in A and normal in G, contrary to (i). Thus, if jG 0 j D p. the theorem is true. In what follows we assume that (iii) jG 0 j > p. (iv) G has at most one abelian subgroup of index p. Indeed, otherwise, we have jG W Z.G/j D p 2 so jG 0 j D p, by Lemma 1.1. Let R N be of composite exponent. Since jN 0 j p and p > 2, R is regular. It follows that R is generated by cyclic subgroups of composite orders so R is normal in N (by induction, N is a Kp -group). It follows that N=A is abelian since it is Dedekindian and p > 2 so N 0 < A. By (i), N 0 D f1g so N is abelian. Thus, by (iv), (v) N is the unique abelian maximal subgroup of G. Moreover, the normalizer of any nonnormal cyclic subgroup of composite order is an abelian maximal subgroup of G so coincides with N . It follows that (vi) All nonnormal cyclic subgroups of orders > p in G are contained in N . Let B G G be cyclic of order > p. Assume that B 6 N , or, what is the same, B does not normalize A; then B G G, by (vi). Then, AB is not a Kp -group so it is not an Np -group and we get, by induction, AB D G. Thus, G is metacyclic. By (i), A \ B D f1g, and, obviously, CN .B/ D Z.G/ (recall that N is the unique abelian maximal subgroup of G). Note that jB W .B \ N /j D p and B \ N Z.G/. Since AG D f1g, we get Z.G/ < B so B \ N D Z.G/. Then G=Z.G/ is abelian of type .p 2 ; p/. We have Z.G/ D ˆ.B/ ˆ.G/. Let U=Z.G/ and V =Z.G/ be distinct cyclic subgroups of order p 2 in G=Z.G/. Then U and V are distinct abelian maximal subgroups of G, contrary to (v). Thus, B < N so all cyclic subgroups of

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Groups of prime power order

composite orders are contained in N , i.e., N D Hp .G/. This is a contradiction since hG N i D hG Hp .G/i D G. Thus, (vii) All cyclic subgroups of G of composite orders are contained in N D Hp .G/. Let B G G be normal cyclic of composite order. Take x 2 G N and set H D hx; Bi; then H is metacyclic. It follows from p > 2 that H is generated by its cyclic subgroups of maximal order so H N , by (vii), contrary to the choice of x. Thus, (viii) All cyclic subgroups of composite orders are not normal in G. We see that the normalizer of every cyclic subgroup of composite order is abelian so coincides with N . It follows that N is the unique maximal subgroup of G of composite exponent, by (ii). Moreover, 2 .N / D N , by (ii), so exp.G/ D p 2 . Let M 2 1 fN g; then exp.M / D p, by (viii). In that case N contains the elementary abelian subgroup N \ M of index p. Then N D A E, where E is elementary abelian. In that case, ˆ.N / D ˆ.A/ > f1g is characteristic in N so normal in G, contrary to (i). Thus, all cyclic subgroups of composite orders are normal in G so G is a Kp -group. Theorem 63.6 ([Kaz1]). Let G be a nonabelian 2-group, and let exp.G/ > 2. Then the following assertions are equivalent: (a) G is a N2 -group, (b) G is a K2 -group. Proof. Clearly, (b) implies (a). It remains to prove the reverse implication. Suppose that G is a counterexample of minimal order. Then G has a nonnormal cyclic subgroup B D hbi of composite order. As above, () jXj D 4 and XG D f1g, where X < G is nonnormal cyclic of order > 2. Let B < N , where jG W N j D 2. Then N is a K2 -group, by induction, so N D NG .B/ and N is the unique maximal subgroup of G containing B. Let a 2 G N be such that o.a/ is as large as possible; then B a ¤ B. If 1 .B a / D 1 .B/, then 1 .B/a D 1 .B/ and NG .1 .B// hN; ai D G, contrary to (). Thus, 1 .B a / ¤ 1 .B/ so B \ B a D f1g. Since B; B a N , these two subgroups are normal in N so H D BB a D B B a is abelian of type .4; 4/. Since B 6E ha; Bi, we get G D ha; Bi D ha; H i, by induction. Set A D hai; then H G G since a2 2 N , G D AH , G=H is cyclic. Since H is abelian, A \ H G G. If o.a/ D 2, then G is a K2 -group (indeed, then H is an H2 -subgroup of G). In what follows we assume that o.a/ > 2. If A \ H > f1g, then A G G since A \ H G G (see ()), and so, by induction, G D BA is a semidirect product with kernel A since B \ A D f1g, by (). In that case, G D B A is metacyclic. Suppose that A \ H D f1g. Then A is not normal in G since G is nonabelian so CG .H / D H , by (). In that case, the abelian subgroup NG .A/ D A Z.G/ is

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maximal in G, by induction, so Z.G/ is a subgroup of index 2 in H ; then B \ Z.G/ > f1g, contrary to (). Thus, it remains to consider the following two cases. (i) Let G D B A, a semidirect product with kernel A D hai Š C2n (n > 1), B D hbi Š C4 . As above, N D NG .B/, a maximal subgroup of G, is abelian of type .22 ; 2n1 /, CG .A/ D A and N \ A D Z.G/ has index 2 in A. Then G=Z.G/ is abelian of type .4; 2/ so G is minimal nonabelian. In that case, Z.G/ D ˆ.G/ and f1g < B \ Z.G/ BG , contrary to (). (ii) Let G D A H , a semidirect product with abelian kernel H of type .4; 4/, A D hai Š C4 , H D B B a , where B D hbi Š C4 . Set NG .A/ D M ; then jG W M j D 2, M \ H D Z.G/ is abelian of type .4; 2/ so BG B \ Z.G/ > f1g, contrary to (). It follows from the above that epimorphic images of Np -groups are Np -groups or have exponent p. Let G be a p-group with exp.G/ > p. If every subgroup of composite exponent is normal in a p-group G, then either jG 0 j p or G Š D24 [Man18]. Obviously, G is a Kp -group. Suppose that jG 0 j > p. Then p D 2 and A D H2 .G/ has index 2 in G, by Theorem 63.4. If A is cyclic, we are done, by Theorem 1.2. Assume that A is not cyclic. Since G 0 > f1g, we get exp.A/ > 2. Let Z be a cyclic subgroup of order 4 in A. By hypothesis, all subgroups are normal in G=Z, i.e., G=Z is Dedekindian. By Lemma 63.2(b), G=Z has no subgroups isomorphic to Q8 so G=Z is abelian. Let Z1 ¤ Z be another cyclic subgroup of order 4 in A (see Theorem 1.17(b)). Then G=Z1 is abelian, by what has just been proved. It follows that G 0 Z \ Z1 , so jG 0 j 2. Proposition 63.7 ([Li3] (compare with Theorem 58.5)). Suppose that all nonnormal cyclic subgroups of a nonabelian p-group G, p > 2, are conjugate. Then G Š Mp n . Proof. Let T < G be a nonnormal cyclic subgroup of G. Set jT j D p t . Let K be the set of all nonnormal cyclic subgroups of G; then jKj is a power of p. (i) Let t D 1. (i1) Assume that exp.G/ D p. Set jGj D pm , jZ.G/j D p z . Then jKj D D p z .1 C p C C p mz1 /. Since m z 1 > 0, jKj is not a power of p so not all members of the set K are conjugate in G, a contradiction. p m p z p1

(i2) Thus, exp.G/ > p. Then G is a Kp -group so, by Theorem 63.1, jG 0 j D p, and G is regular since p > 2. Let j1 .G/j D p k and j1 .Z.G//j D p . Then the number k p D p .1CpCp k 1 /, which of nonnormal subgroups of order p in G equals p p1 0 is a power of p, so k D C 1. Since jG j D p, it follows that there are exactly p subgroups conjugate with T in G. It follows that D 1 so k D 2. Then jG=Ã1 .G/j D j1 .G/j D p 2 so G is metacyclic. It follows that G is minimal nonabelian (Lemma 65.2(a)). Then, by Lemma 58.1, G Š Mp n .

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Groups of prime power order

(ii) Now let t > 1. Set T0 D 1 .T /; then T0 TG G G so T0 is contained in all nonnormal cyclic subgroups of G. Set GN D G=T0 . Let UN < GN be nonnormal cyclic. If U is noncyclic, then U D T0 U0 and U0 is not normal in G. Then T0 — U0 , contrary to what has just been said. Thus, U is cyclic so conjugate with T . It follows that all nonnormal cyclic subgroups of GN are conjugate, i.e., GN satisfies the hypothesis. Therefore, by induction, GN Š Mp n . We also have T0 ˆ.T / ˆ.G/ N D 2. If A=T0 ; B=T0 < G=T0 be two distinct cyclic subgroups of so d.G/ D d.G/ index p, then A; B 2 1 are distinct abelian so A \ B D Z.G/, and we conclude that G is minimal nonabelian. By Lemma 58.1, G Š Mp n . However, all nonnormal cyclic subgroups of Mp n have order p < p t , so case t > 1 is impossible. Exercise 1. Suppose that all nonnormal cyclic subgroups of a nonabelian 2-group G, are conjugate. Then G Š M2n , i.e., Theorem 63.1 is also true for p D 2. Exercise 2. Let a group G be nonabelian of exponent p. Then the number of nonnormal subgroups of order p in G is not a power of p. There are in the book a number of results on cyclic subgroups of p-groups. Problem. Classify the non-Dedekindian p-groups in which any two nonnormal cyclic subgroups of the same order are conjugate.

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p-groups generated by elements of given order

1o . In what follows, G is a group of order p m , k .G/ D hx 2 G j o.x/ D p k i, where p k exp.G/. Then k .G/ k .G/ and k .G/ is the least subgroup H P i of G subjecting ck .H / D ck .G/. We have 1 iD0 '.p /ci .G/ D jGj, where './ is Euler’s totient function. Given k > 0, set solk .G/ D jfx 2 G j o.x/ p k gj. Then P solk .G/ D kiD0 '.p i /ci .G/. For definition of Ls - and U2 -groups, see 17, 18, 67. A 2-group G is said to be a Un -group, if there is E2n Š R G G such that G=R is of maximal class with cyclic subgroup T =R of index 2 and 1 .T / D R. It is easy to show that R contains all normal elementary abelian subgroups of G and all elements in the set G T have orders 8. A subgroup R is called the kernel of G. For the sake of completeness, we collected in Lemma 64.1 some known facts. We use this lemma until end of the book. Lemma 64.1. Let G be a p-group. (a) (Theorems 7.1 and 7.2) If G is regular of exponent p e and k e, then we have exp.k .G// D p k . All p-groups of class < p are regular. Groups of exponent p are regular. Regular 2-groups are abelian. (b) (Theorem 12.1) If G has no normal subgroups of order p p and exponent p, it is either absolutely regular or of maximal class. If an irregular p-group G has an absolutely regular maximal subgroup H , then G is either of maximal class or G D H 1 .G/, where j1 .G/j D p p . (c) (Theorems 9.5 and 9.6) A p-group of maximal class and order > p p is irregular. A p-group of maximal class and order p m contains only one normal subgroup of order p i for 1 i m 2. (d) (Theorem 9.6, Lemma 12.3) A p-group of maximal class and order > p pC1 has no normal subgroups of order p p and exponent p. For such G, we have c1 .G/ 1 C p C C p p2 .mod p p / and c2 .G/ p p2 .mod p p1 /. If p 2 < p k exp.G/, then ck .G/ D 1 if p D 2 (see Lemma 64.10, below) and p p1 divides ck .G/ if p > 2. (e) (Theorem 9.6) An irregular p-group G of maximal class has an absolutely regular subgroup G1 of index p such that jn .G1 /j D p .p1/n for p n < exp.G1 / D exp.G/. If, in addition, jGj > p pC1 , then Z.G1 / is noncyclic (G1 is called the fundamental subgroup of G); all other maximal subgroups of G are of maximal class. If 2 < k exp.G/, then k .G/ G1 .

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(f) (Theorems 13.2 and 13.5) If G is neither absolutely regular nor of maximal class, then c1 .G/ 1CpC Cp p1 .mod p p / and, for k > 1, ck .G/ 0 .mod p p1 /. The number of subgroups of order p p and exponent p in G is 1 .mod p/. (g) (Theorem 13.19) If G is an irregular p-group of maximal class and H G is of order > p p , then H is either absolutely regular or of maximal class. (h) If A G G is of order p p1 and exponent p, G is a p-group of maximal class and R < G is of order p p and exponent p, then A < R. Next, A is contained in every irregular subgroup of G. (i) (Proposition 10.17) If B G is nonabelian of order p 3 such that CG .B/ < B, then G is of maximal class. (j) (Corollary 18.7) Suppose that an irregular p-group G is not of maximal class, k > 2. Then ck .G/ 0 .mod p p /, unless G is an Lp - or U2 -group. In particular, in that case, solk .G/ 0 .mod p pC2 /. (k) Let C < H < G, where G is a 2-group which is not of maximal class, jH W C j D 2, C is cyclic. Then the number of subgroups of G of order jH j, that are of maximal class, is even. (l) (Theorems 41.1 and 66.1) If G is minimal nonmetacyclic, then one of the following holds: (i) G is of order p 3 and exponent p, (ii) G is of order 34 and class 3, j1 .G/j D 9, (iii) G D C2 Q8 , (iv) G D Q8 C4 Š D8 C8 of order 24 , (v) G is special group of order 25 with jZ.G/j D 22 . Thus, if 2 .G/ is metacyclic then G is metacyclic. (m) (Lemma 44.1, Corollary 44.6, Theorem 9.11) A p-group G is metacyclic if one of the following quotient groups is metacyclic: G=ˆ.G 0 /K3 .G/, G=ˆ.G 0 /, G=K3 .G/. If p > 2 and jG=Ã1 .G/ p 2 , then G is metacyclic (Huppert). (n) (Theorem 44.5) A 2-group G is metacyclic if G and all its maximal subgroups are two-generator. (o) (Corollary 44.9) If G=Ã2 .G/ is metacyclic then G is metacyclic. (p) ([Bla5, Theorem 4.2]) If a p-group G, p > 2, and all its maximal subgroups are two-generator, then G is either metacyclic or Ã1 .G/ D K3 .G/ is of index p 3 in G (in the last case, jG W G 0 j D p 2 ). (q) (Tuan; see Lemma 1.1) If G is nonabelian and A < G is abelian of index p, then jGj D pjG 0 jjZ.G/j. (r) (Lemma 4.2). If jG 0 j D p, then G D .A1 As /Z.G/, where A1 ; : : : ; As are minimal nonabelian; then G=Z.G/ is elementary abelian. (s) (Taussky; Proposition 1.6) If G is a nonabelian 2-group with jG W G 0 j D 4, then G is of maximal class, i.e, dihedral, semidihedral or generalized quaternion. (t) If a nonabelian p-group G has a cyclic subgroup of index p, then G is either Mp n or p D 2 and G is of maximal class (see (s)).

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181

(u) (see Exercise 1.69) If U; V 2 1 are distinct, then jG 0 W U 0 V 0 j p. (v) (Lemma 1.4) Let M be a G-invariant subgroup of ˆ.G/. If Z.M / is cyclic, M is also cyclic. In particular, M cannot be nonabelian of order p 3 . (w) (Exercise 9.13) If a p-group G of maximal class, p > 2, has a subgroup H with d.H / > p 1, then G is isomorphic to a Sylow p-subgroup of Sp 2 . (x) (5) c1 .G/ 1 C p .mod p 2 / and, for k > 1, ck .G/ 0 .mod p/, unless G is either cyclic or a 2-group of maximal class. Under the last exceptions, G has a normal abelian subgroup of type .p; p/. (y) If hU 0 j U 2 1 i < G 0 , then d.G/ D 2. Let us prove parts (g), (h), (y) of Lemma 64.1. Proof of Lemma 64.1(g). Let jGj > p pC1 . If jGj D p pC2 , the result follows from Theorem 9.6. Now let jGj > p pC2 and H — G1 , where G1 is the fundamental subgroup of G. Let H < M 2 1 ; then M.¤ G1 / is of maximal class (Lemma J(e)) and M1 D M \ G1 D ˆ.G/ is the fundamental subgroup of M since jM j > p pC1 . Then, by induction on jGj, H is of maximal class since H 6 M1 . Proof of Lemma 64.1(h). We have A D 1 .ˆ.G//. One may assume that jGj > p pC1 (if jGj D p pC1 , then A D ˆ.G/). Let R < G be either of order p p and exponent p or irregular. Since j1 .R \ G1 /j D p p1 , the result follows. Proof of Lemma 64.1(y). Set D D hU 0 j U 2 1 i. Since, by hypothesis, D < G 0 , there exist x; y 2 G such that Œx; y 62 D. Assume that H D hx; yi M 2 1 ; then M 0 6 D, a contradiction. Thus, H D G. 2o . We begin with the following Definition 1 (Ito). Let G be a group of order p m . Then G is called one-stepped if there exist m elements x1 ; : : : ; xm of order p in G such that jhx1 ; : : : ; xi ij D p i for i D 1; : : : ; m. We consider the identity group f1g as one-stepped. Lemma 64.2. A p-group G is one-stepped if and only if 1 .G/ D G. Proof. We have to prove that if 1 .G/ D G, then G is one-stepped. Let H G be one-stepped of maximal order. Assume that H < G. Set N D NG .H /. If N H has an element x of order p, then H1 D hx; H i is one-stepped of order > jH j, a contradiction. Thus, H D 1 .N /, i.e., H is characteristic in N . Then N D G > H D 1 .G/ D G, a contradiction. Lemma 64.3. Suppose that A < G are one-stepped p-groups and jG W Aj D p n . Then there are x1 ; : : : ; xn 2 G A of order p such that jAi W Ai1 j D p, where A0 D A, Ai D hA; x1 ; : : : ; xi i for i D 1; : : : ; n.

182

Groups of prime power order

Proof. Suppose that Ai has constructed. Assuming that Ai < G, we have to construct AiC1 . Set Ni D NG .Ai /. As in the proof of Lemma 64.2, there is xiC1 2 Ni Ai of order p. Set AiC1 D hxiC1 ; Ai i. Then An D G. A series A D A0 < A1 < < An D G, constructed in Lemma 64.3, we call the Ito series or (Ito 1-series; see below) connecting A D A0 with G. 3o . In this subsection we give some estimates of the order of a p-group G D k .G/ in terms of ck .G/. For p-groups of maximal class and order > p pC1 these estimates are best possible. 1 Set .G/ D Œ p1p c1 .G/ and, for k > 1, set k .G/ D Œ p p1 ck .G/. Theorem 64.4. Suppose that a one-stepped p-group G is not a group of maximal class. Then jGj p pC1C .G/ . Proof. We use induction on jGj. One may assume that jGj D p m > p pC1 (otherwise, there is nothing to prove). If G is regular, then exp.G/ D p (Lemma 64.1(a)), c1 .G/ D 1 C p C C p m1 , and we have to check only that 1 m1 .1 C p C C p / D 1 C p C .1 C p C C p mp1 /: m 1CpC pp This is true since 1 C p C C p mp1 1 C m p 1 D m p. Now let G be irregular. Then there is R G G of order p p and exponent p (Lemma 64.1(b)) and x 2 G R of order p. Set A D A0 D hx; Ri. Let A D A0 < A1 < < An D G be an Ito series connecting A D A0 with G. Then jGj D jA0 j p n D p pC1Cn so it suffices to prove that n .G/. For i D 0; 1; : : : ; n 1 we have c1 .AiC1 / > c1 .Ai / and, since AiC1 is not of maximal class in view of R G Ai and jAiC1 j > p pC1 (Lemma 64.1(d)), we get c1 .AiC1 / c1 .Ai / .mod p p / (Lemma 64.1(f)), so c1 .G/ c1 .A0 / C n p p .1 C p C C p p1 / C n p p : However, by Lemma 64.1(f), c1 .G/ D 1CpC Cp p1 C .G/ p p so n .G/. Corollary 64.5. If a p-group G is not of maximal class, then j1 .G/j p pC1C .G/ so, if p D 2 and c1 .G/ D 4k C 3, then j1 .G/j 23Ck . Let G be one-stepped of maximal class and order p m > p pC1 and let G1 < G be the fundamental subgroup. Then R D 1 .G1 / G G is of order p p1 and exponent p. Let x 2 G R be of order p; then H D hx; Ri is of order p p and exponent p (Lemma 64.1(a)). By Lemma 64.1(g), NG .H / is of maximal class and order p pC1 so there are exactly jG W NG .H /j D p mp1 conjugates with H in G. All these conjugates contain exactly c1 .R/ C .c1 .H / c1 .R//p mp1 D 1 C p C C p p2 C p m2

64

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183

distinct subgroups of order p and they generate M 2 1 (take into account the above formula!). Since G is one-stepped, there is y 2 G M of order p. Setting H1 D hy; Ri, we, as above, obtain at least p m2 new subgroups of order p that together with R generate M1 2 1 fM g. Let G be the number of one-stepped members in the set 1 . Then (see Lemma 12.3) we get 1 C p C C p p2 C .G/ p p D c1 .G/ G p m2 : It follows that p m

p pC2 G

.G/. Since G is one-stepped, p G > 1 so we obtain

Theorem 64.6. Let G be a one-stepped p-group of maximal class and order > p pC1 . Then jGj p pC1 .G/ or, what is the same. jGj p c01 .G/, where c01 .G/ is the number of subgroups of order p not contained in G1 . Definition 2. A p-group G is said to be k-stepped if it has elements x1 ; : : : ; x t , all of order p k , such that G D hx1 ; : : : ; x t i and jAi W Ai1 j > 1, where Ai D hx1 ; : : : ; xi i and xi 2 NG .Ai1 / for i D 1; : : : ; t . If k .A/ D A D A0 < G and there are elements x1 ; : : : ; xn , all of order p k , such that, denoting Ai D hA0 ; x1 ; : : : ; xi i, we have xi 2 NG .Ai1 / and An D G, then the series A0 < A1 < < An D G is said to be an Ito k-series connecting A0 with G. If A0 < A1 < < An D G is an Ito k-series connecting A0 with G, then jG W A0 j p k n . As in Lemma 64.2, G is k-stepped if and only if k .G/ D G. If A D A0 is a k-stepped subgroup of a k-stepped p-group G, then there exists an Ito k-series A0 < A1 < < An < G connecting A0 with G. Lemma 64.7. Let R GG be of order p p and exponent p and let G=R be cyclic of order > p. Then exp.j .G// D p j for all j with p j p e D exp.G/. Next, e .G/ D G, i.e., G is e-stepped. If G=R is of order p e , then ce .G/ D p p . If G=R is of order p e1 (in that case 1 .G/ D R), then ce .G/ D p p1 . Proof. It suffices to show that exp.1 .G// D p. However, this follows from Lemma 17.4(c). Let k > 2 and suppose that a p-group G D k .G/ is not absolutely regular (by Theorem 13.19, G is not of maximal class); then there is R G G of order p p and exponent p (Lemma 64.1(b)). Take in G an element x0 of order p k and set A0 D hx0 ; Ri; then p pCk1 jA0 j jRjp k D p pCk . By Lemma 64.7, A0 is k-stepped. Let A0 < A1 < < An D G be an Ito k-series connecting A0 with G. Since ck .A0 / p p1 and ck .A1 / > ck .A0 / p p1 , A1 is not an Lp -group. Next, if p D 2 and k > 3, then A1 is not a U2 -group. Then, if p > 2, then ck .A1 / p p (Lemma 64.1(j)), and this is also true for p D 2. It follows that ck .AiC1 / > ck .Ai /, and, by Lemma 64.1(j), ck .AiC1 / ck .Ai / .mod p p /, i D 1; : : : ; n 1, so we have p

k .G/p p1 D ck .G/ ck .A1 / C .n h 1/p p i np , hence k .G/ np, and we conclude that jGj jA0 jp k n p

pCkCk

1 p k .G/

. Thus, we have

184

Groups of prime power order

Theorem 64.8. Suppose that k > 2 and a k-stepped p-group G of order p m is neither absolutely regular nor of maximal class. Then m p C k C pk k .G/. Corollary 64.9. Let k > 2 and let a p-group G be h i neither absolutely regular nor of maximal class. Then jk .G/j p

pCkCk

1 p k .G/

.

The case k D 2 we consider for p-groups of maximal class only. Let G D k .G/ be a p-group of maximal class, k > 1, jGj D p m p 2p and p > 2; then k D 2 (Lemma 64.1(e)) and exp.G/ > p 2 . Let c02 .G/ be the number of cyclic subgroups of order p 2 in G not contained in G1 . Set R D 1 .G1 /, where G1 < G is fundamental; then jRj D p p1 . Let x 2 G G1 be of order p 2 . Set A D hx; Ri; then A is absolutely regular of order p p since x p 2 Z.G1 / (Theorem 13.19). Set N D NG .A/. Since N 6 G1 and jN j > p p , N is of maximal class (Lemma 64.1(g)). Since A 6E G, we get N < G so A is not characteristic in N and hence 2 .N / > A. Then jN W Aj D p (every normal subgroup of index > p in N is characteristic) so there are exactly jG W N j D p mp1 subgroups conjugate with A in G; their intersection equals R so their contribution in c02 .G/ is equal to p mp1 c2 .A/ D p mp1 p p2 D p m3 . These G-conjugates of A generate M 2 1 (every normal subgroup of index p 2 in G is contained in ˆ.G/ < G1 ). We have G1 \ M D ˆ.G/. Since m 2p, we get 2 .G1 / ˆ.G/. Since G is 2-stepped, there is y 2 G M of order p 2 ; by the above, also y 2 G G1 . Set A1 D hy; Ri; then A \ A1 D R. As above, the G-conjugates of A1 generate M1 2 1 fM g; their intersection equals R again. Let G be the number of 2-stepped members in the set 3 1 . Then c02 .G/ D G p m3 so jGj D p m D pG c02 .G/. If p D 2, then G Š Q2m since 2 .G/ D G, and equality is attained. Thus we have Supplement to Theorem 64.8. If G is a 2-stepped p-group of maximal class and 3 order p2p , then jGj D pG c02 .G/, where G is the number of 2-stepped members in the set 1 . Theorem 64.10. Let a p-group G, c1 .G/ D 1 C p C p 2 and exp.1 .G// > p. Then p D 2, 1 .G/ D D8 C2 is of order 24 and one of the following holds: (a) G D D C , where D8 Š D, C is cyclic of order 4, D \ C D Z.D/. (b) G D DC , where D8 Š D G G, C is nonnormal cyclic of index 4 in G, D \ C D Z.D/ and CG .D/ D Z.G/ has index 2 in C , G=CG .D/ Š D8 . (c) G D DQ, where D8 Š D G G, Q is a nonnormal generalized quaternion group of index 4 in G, D \ Q D Z.D/, CG .D/ is a cyclic subgroup of index 2 in Q, G=CG .D/ Š D8 . If L is a cyclic subgroup of order 4 in Q such that L 6 CG .D/, then DL Š SD24 . Proof. Let p > 2. Let Ep 2 Š R G G and let x 2 G R be of order p in G R; then H D hx; Ri is of order p 3 and exponent p. Since c1 .G/ D c1 .H /, 1 .G/ D H is of order p 3 and exponent p, a contradiction. For p D 2, the assertion coincides with Theorem 43.9.

64

185

p-groups generated by elements of given order

4o . In this subsection we study p-groups G with small ck .G/. Proposition 64.11. Let G be a p-group, p > 2, with 1 .G/ D G, c1 .G/ D 1 C p C C p p and exp.G/ > p. Then: (a) G is irregular of order p pC2 ; all members of the set 1 have exponent p 2 . (b) d.G/ D 3, ˆ.G/ D G 0 is of order p p1 and exponent p, cl.G/ D p. (c) G=Ã1 .G/ is of order p pC1 and exponent p, Ã1 .G/ D Kp .G/. (d) c2 .G/ D p p . (e) 1 D fM1 ; : : : ; Mp 2 ; T1 ; : : : ; TpC1 g, where M1 ; : : : ; Mp 2 are of maximal class, T1 ; : : : ; TpC1 are not generated by two elements so regular and have expoTpC1 nent p 2 . Next, .G/ D iD1 Ti has exponent p 2 and index p 2 in G, where .G/=K3 .G/ D Z.G=K3 .G//. (f) G has exactly p C 1 subgroups of order p p and exponent p, all those subgroups contain ˆ.G/ so normal in G. (g) Exactly p 2 subgroups of G of order p p , containing ˆ.G/, have exponent p 2 . (h) If L is a subgroup of order p p and exponent p in G, then exactly p maximal subgroups of G, containing L, are of maximal class. (i) Let S 2 2 be of exponent p 2 . If S ¤ .G/ (see (e)), then S is contained in exactly p irregular members of the set 1 . Proof. By Theorems 12.3 and 7.2(b), G is neither of maximal class nor absolutely regular so there is R G G of order p p and exponent p (Theorem 12.1(a)). (a) By Theorem 64.1(a), G is irregular. By Theorem 64.4, jGj p pC1C .G/ D p pC2 , since .G/ D Œ p1p c1 .G/ D 1; here Œx is the integer part of x 2 R. Since 1 .G/ D G, we get G=R Š Ep 2 so exp.G/ D p 2 . We also have jGj D p pC2 . Assume that exp.H / D p for some H 2 1 . Then c1 .H / D c1 .G/ so .G D/ 1 .G/ D H has exponent p, a contradiction. Thus, all members of the set 1 have exponent p2 . (b) If x 2 G R is of order p, then exp.hx; Ri/ > p, by (a), so M D hx; Ri 2 1 is of maximal class (Theorem 7.1); then d.G/ D 3 and ˆ.G/ D G 0 D ˆ.M / D M 0 has exponent p, by (a) and Theorem 12.12(a), and we conclude that cl.G/ D p. (c) follows from Theorem 12.12(b). (d) follows from (a) and the hypothesis. Indeed, c2 .G/ D

jGj.p1/c1 .G/1 '.p 2/

D pp .

(e) The first assertion follows from Theorem 12.12(c). Now assume exp..G// D 1 p. Then c2 .Ti / D p.p1/ .jTi j j.G/j/ D p p1 for i D 1; : : : ; p C 1 so c2 .G/ D PpC1 p1 > p p , contrary to (d). Thus, exp..G// D p 2 . iD1 c2 .Ti / D .p C 1/p (f) Assume that S < G of order p p and exponent p is not normal in G; then ˆ.G/ D G 0 6 S so H D Sˆ.G/ 2 1 . It follows from Theorem 7.2(b) and (a) that

186

Groups of prime power order

H is irregular so it is of maximal class; in that case, ˆ.G/ D ˆ.H / (compare orders!) so H D S 62 1 , a contradiction. If t is the number of subgroups of order p p and exponent p in G, then 1CpC Cp p1 Cp p D c1 .G/ D c1 .ˆ.G//Ctp p1 D 1CpC Cp p2 Ctp p1 so t D p C 1. (g) follows from (a), (b) and (f). (h,i) By (e), the intersection of two distinct regular maximal subgroups of G coincides with .G/. Let L ¤ .G/ be a normal subgroup of index p 2 in G; then L is contained in at most one regular maximal subgroup of G and G=L Š Ep 2 since 1 .G/ D G. Let D be a G-invariant subgroup of index p 2 in L. Set C D CG .L=D/. Let L < H C , where H 2 1 ; then H is regular. It follows that L is contained in exactly one regular member of the set 1 , and the proof is complete. Exercise 1. Suppose that G D 1 .G/ is an irregular p-group containing exactly p C1 subgroups of order p p and exponent p. Then c1 .G/ D 1 C p C C p p . If p D 2, then 1 .G/ D Q8 C4 . Exercise 2. Let G be a p-group of maximal class, p > 2. Then 1 .G/ D G or 2 .G/ D G. (Hint. 1 .G/2 .G/ D 2 .G/ D G. Since any two normal subgroups of G of distinct orders are incident, we are done.) Proposition 64.12. Suppose that a p-group G, p > 2, is such that ck .G/ D p p2 for k > 1. Then one of the following holds: (i) G is regular with cyclic G=k1 .G/ and jk1 .G/j D p pCk3 . (ii) k D 2 and G is of maximal class and order p pC1 having exactly p subgroups of order p p and exponent p. Proof. (i) Let G be regular and jk .G/j D p a , jk1 .G/j D p b . Then p p2 D ck .G/ D

p ab 1 bkC1 jk .G/j jk1 .G/j D : p p1 p k1 .p 1/

It follows that a b D 1 so jk .G/=k1 .G/j D p, and we conclude that the quotient group G=k1 .G/ is cyclic. Then p 2 D b k C 1 so jk1 .G/j D p b D p pCk3 . (ii) Now let G be irregular. By Lemma 64.1(f), G is of maximal class. Assume that jGj > p pC1 . Let G1 be the fundamental subgroup of G; then jk1 .G1 /j D p .k1/.p1/ . If k > 2, then ck .G1 / D

jk .G1 /j jk1 .G1 /j p .k1/.p1/C1 p .k1/.p1/ ; '.p k / .p 1/p k1

p .k1/.p2/ > p p2 D ck .G/;

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187

a contradiction. Thus, k D 2. Then j2 .G1 /j p .p1/C2 D p pC1 so c2 .G1 /

p pC1 p p1 D p p2 .p C 1/ > p p2 D c2 .G/; p.p 1/

again a contradiction. Thus, jGj D p pC1 . The second assertion in (ii) is checked easily.

65

A2-groups

A p-group G is said to be an An -group if it contains a nonabelian subgroup of index p n1 but all its subgroups of index p n are abelian. Lemma 65.1 (= Exercise 1.8a (Redei)). Let G be a minimal nonabelian p-group (= A1 -group). Then G D ha; bi and one of the following holds: m

n

(a) ap D b p D c p D 1, Œa; b D c, Œa; c D Œb; c D 1, jGj D p mCnC1 , G D hbi .hai hci/ D hai .hbi hci/ is nonmetacyclic. m

n

(b) ap D b p D 1, ab D a1Cp

m1

, jGj D p mCn and G D hbi hai is metacyclic.

(c) a4 D 1, a2 D b 2 , ab D a1 , G Š Q8 . We have jG 0 j D p and H=Ã1 .H /j p 3 for H G. If jGj > p 3 and j1 .G/j p 2 , then G is metacyclic. The group G is nonmetacyclic if and only if G 0 is a maximal cyclic subgroup of G. Next, jG=Ã1 .G/j p 3 with equality if and only if G is from (a) and p > 2. If, in (a), u 2 G ˆ.G/, then hui is not normal in G. Suppose that G 0 < L < G, where L is cyclic of order p 2 . By Theorem 6.1, L=G 0 C =G 0 , where C =G 0 is a cyclic direct factor of G=G 0 ; in particular, G=C is cyclic. It remains to show that C is cyclic. We get G 0 D ˆ.L/ ˆ.C /. It follows that C =ˆ.C / as an epimorphic image of a cyclic group C =ˆ.L/, is cyclic so C is cyclic and G is metacyclic. The last assertion of Lemma 65.1 is a partial case of the following general fact. If a nonmetacyclic G D hu; vi, then hui 6E G. The aim of this section is to clear up the subgroup and normal structure of A2 groups. Using comparatively easy arguments, we obtain a lot of information on A2 groups which is difficult to read out from their defining relations. Some results which we shall prove in the sequel, are contained in the thesis of Lev Kazarin (unpublished), however, the proofs presented below, as a rule, are new. The length of [She], also devoted to classification of A2 -groups, indicates the degree of the difficulty of this problem. Every nonabelian group of order p 4 is either an A1 - or A2 -group. Therefore, in what follows, we assume that jGj D p m > p 4 . A2 -groups belong to a wider class of p-groups all of whose nonabelian subgroups are normal. Lemma 65.2. Let G be a nonabelian p-group. Then: (a) If G 0 1 .Z.G// and d.G/ D 2, then G is an A1 -group.

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(b) If G is metacyclic with jG 0 j D p 2 , then G is an A2 -group. (c) If G has two distinct abelian maximal subgroups, then jG 0 j D p. (d) Suppose that jH 0 j p for all H 2 1 . Then jG 0 j p 3 and G 0 is abelian. If F; H 2 1 are distinct and F 0 H 0 , then jG 0 j p 2 . If jG 0 j D p 3 , then G has no abelian maximal subgroups, and if, in addition, d.G/ > 2, then d.G/ D 3 and Ep 3 Š G 0 Z.G/. (e) If G 0 \ Z.G/ is cyclic and jH 0 j p for all H 2 1 , then jG 0 j p 2 . If, in addition, G 0 6 Z.G/, then d.G/ D 2 and G=.G 0 \ Z.G// is an A1 -group. Proof. (a) If a; b 2 G, then 1 D Œa; bp D Œa; b p so Ã1 .G/ Z.G/. Then ˆ.G/ D G 0 Ã1 .G/ Z.G/ so Z.G/ D ˆ.G/ has index p 2 in G. Hence, all members of the set 1 are abelian so G is an A1 -group. (b) By Lemma 65.1, G is not an A1 -group. Let L D Ã1 .G 0 / and H 2 1 ; then L < G 0 < H . By (a), G=L is an A1 -group so H=L is abelian; then, again by (a), H is either abelian or an A1 -group. Thus, G is an A2 -group. (c) follows from Lemma 64.1(u). (d) Let F; H 2 1 be distinct. Then jG 0 j pjF 0 H 0 j p 3 (Lemma 64.1(u)). If, in addition, F 0 H 0 , then jG 0 j p 2 . Now let jG 0 j D p 3 . Since F 0 6 H 0 and H 0 6 F 0 , 1 has no abelian members. Let, in addition, d.G/ > 2. Then A0 ¤ B 0 for distinct A; B 2 1 so c1 .G 0 / j1 j 1 C p C p 2 , and we get d.G 0 / D 3, Ep 3 Š G 0 Z.G/. Therefore, since 1 C p C p 2 D c1 .G 0 / D j1 j, we get d.G/ D 3. (e) We have D D hH 0 j H 2 1 i 1 .G 0 \ Z.G// so jDj D p since 1 .G 0 \ Z.G// is cyclic. Then G=D is either abelian or A1 -group so j.G=D/0 j p (Lemma 65.1), and we get jG 0 j p 2 . Let, in addition, G 0 6 Z.G/. Then G=D is nonabelian so it is an A1 -group. Since D < ˆ.G/, we get d.G/ D d.G=D/ D 2. Corollary 65.3. A metacyclic p-group G is an A2 -group if an only if jG 0 j D p 2 . Remarks. 1. Let us prove that if a p-group G has the cyclic derived subgroup G 0 , then all elements of G 0 are commutators. We have G 0 D hŒx; yi for some x; y 2 G. One may assume that G D hx; yi. By Lemma 64.1(u), there is H 2 1 such that H 0 D Ã1 .G 0 /. By induction, all elements of H 0 are commutators. By [BZ, Theorem 3.27], all generators of G 0 , i.e., members of the set G 0 H 0 , are commutators, completing the proof. 2. Let us prove that if G of order p 4 and exponent p has no nontrivial direct factors, it is of class 3. Let H be an A1 -subgroup of G; then jH j D p 3 . If Z.G/ 6 H , then G D H C , where C < Z.G/ is of order p such that C 6 H . Thus, Z.G/ < H so Z.G/ is of order p. Since CG .H / D Z.G/, we get CG .H / < H . Then G is of maximal class (Proposition 10.17). Lemma 65.4. Let G be an A2 -group of order p m > p 4 . (a) If K G, then jK=Ã1 .K/j p 4 , and this estimate is best possible.

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(b) If K < G with jK=Ã1 .K/j D p 4 , then Ep 4 Š 1 .K/ G G and, if 1 .K/ H 2 1 , then H is abelian. If, in addition, jG 0 j > p, then jG 0 j D p 2 , jG W Z.G/j D p 3 and G=1 .K/ is cyclic. (c) If K G G and G=K is generated by subgroups of index p 2 , then K Z.G/. In particular, if d.G/ D 3, then ˆ.G/ Z.G/. (d) Let jG 0 j D p. Then either G D H C , jC j D p or G D H Z, where H 2 1 is an A1 -subgroup and Z is cyclic of order p 2 . (e) Ã2 .G/ Z.G/. If p > 2 and G is not metacyclic, then Ã1 .G/ Z.G/ and jG=Z.G/j p 3 . (f) Let jG=Ã1 .G/j D p 4 , m > 5 and G 0 6 Z.G/. Then p > 2, G 0 Š Ep 2 , cl.G=Ã1 .G// D 3, Ã1 .G/ Z.G/ is cyclic, G=G 0 Š Cp m3 Cp so d.G/ D 2. If L=G 0 is a direct factor of G=G 0 of order p, then L Š Ep 3 and G D Z L is a semidirect product of L with a cyclic subgroup Z of order p m3 hence Ã1 .G/ D Ã1 .Z/, 1 .G/ D L 1 .Ã1 .G// Š Ep 4 , G 0 \ Ã1 .G/ D f1g and G is an L4 -group (see 17, 18). Next, G=.G 0 \ Z.G// is an A1 -group. (g) If jG 0 j D p 3 and d.G/ D 2, then p > 2 and jGj D p 5 . (h) If G=Z.G/ is nonabelian, then G=.G 0 \ Z.G// is an A1 -group. (i) Let G be metacyclic. Then G=Z.G/ is of order p 4 and exponent p 2 . If the number jG=Z.G/j D p 3 , then p D 2 and G=Z.G/ Š D8 . Now suppose that jG=Z.G/j D p 4 . If p D 2, then G=Z.G/ is abelian of type .4; 4/. If p > 2, then G=Z.G/ can be abelian or nonabelian of exponent p 2 (see examples following the lemma) (j) If G has no normal elementary abelian subgroups of order p 3 , then it is either metacyclic or minimal nonmetacyclic. Proof. (a) In view of Lemma 65.1, one may assume that K < G. Suppose that jK=Ã1 .K/j > p 3 . Then K is not a subgroup of an A1 -group (Lemma 65.1) so it is abelian. Considering K \ A, where A 2 1 is nonabelian, we get j1 .K/j p 4 (Lemma 65.1), and we are done since jK=Ã1 .K/j D j1 .K/j. If A2 -group G D U C , where U is a nonmetacyclic A1 -group of order > p 3 and jC j D p, then 1 .G/ is of order p 4 and exponent p. (b) Let jK=Ã1 .K/j D p 4 for K < G. Then every member of the set 1 containing K is abelian (Lemma 65.1) so Ep 4 Š 1 .K/ G G. Let, in addition, jG 0 j > p. Then G=1 .K/ is cyclic since the set 1 has exactly one abelian member (Lemma 65.2(c)). By Lemma 65.2(d), jG 0 j D p 2 so jG W Z.G/j D p 3 (Lemma 64.1(q)). (c) Let R=K < G=K be of index p 2 in G=K. Then R is abelian so R CG .K/ and K Z.G/ since G D hR < G j K < R; jG W Rj D p 2 i. Now the second assertion follows. (d) This follows from Lemma 64.1(r).

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(e) If there is K G G such that G=K is of order p 3 and exponent p, then K Z.G/, by (c), so we are done since Ã2 .G/ K. Now assume that such K does not exist. Then G=Ã1 .G/ Š Ep 2 . If Ã1 .G/ is cyclic, then p D 2, m D 4 and so Ã2 .G/ D Z.G/ (Lemma 64.1(t)). Now let Ã1 .G/ be noncyclic. Then Ã1 .G/ has a G-invariant subgroup T such that Ã1 .G/=T Š Ep 2 . In that case, Ã2 .G/ T and the result follows from (c). (f) Since d.G/ 3, we get p > 2. By (c), Ã1 .G/ < Z.G/. Next, G=Ã1 .G/ has an abelian subgroup A=Ã1 .G/ of index p; d.A/ d.A=Ã1 .G// D 3 so A is abelian (Lemma 65.1). Then jG 0 j D p 2 (Lemma 64.1(u)) so jG=Z.G/j D p 3 (Lemma 64.1(q)). By Corollary 65.3, G 0 Š Ep 2 since G is nonmetacyclic. By the above, A=Ã1 .G/ is the unique abelian maximal subgroup of G=Ã1 .G/, and we conclude that d.G=Ã1 .G// D 2; then d.G/ D 2 since Ã1 .G/ < ˆ.G/. It follows that cl.G=Ã1 .G// D 3 (Remark 2). Thus, G=Z.G/ is nonabelian so G 0 6 Z.G/. Then, by (c), G=G 0 is not generated by subgroups of index p 2 so it is abelian of type .p m3 ; p/. Let L=G 0 be a direct factor of G=G 0 of order p; then L is abelian of order p 3 since jG W Lj D p m3 > p 2 . We have G=L Š Cp m3 so there exists a cyclic Z < G with G D Z L (semidirect product) since Z \ L D f1g in view of jG=Ã1 .G/j D p 4 . We also see that Ã1 .Z/ D Ã1 .G/ (compare indices!) so exp.L/ D p, L Š Ep 3 . Then G 0 \ Ã1 .G/ L \ Ã1 .Z/ D f1g. Since a Sylow p-subgroup of Aut.L/ is nonabelian of order p 3 and exponent p, we get 1 .G/ D L 1 .Ã1 .G// Š Ep 4 (recall that m > 5). By remark, preceding the lemma, cl.G=Ã1 .G// D 3. Next, the last assertion follows from Lemma 65.2(a). (g) All members of the set 1 are A1 -groups (Lemma 65.2(d)) so they are generated by two elements (Lemma 65.1). By Lemma 64.1(n), p > 2. In that case, jG W G 0 j D p 2 (Lemma 64.1(p)) so jGj D jG W G 0 jjG 0 j D p 2 p 3 D p 5 . (h) Since G 0 6 Z.G/, we get jG 0 j p 2 , and, by (c), d.G/ D 2. Then T D 0 G \ Z.G/ has index p in G 0 . Indeed, this is true, if jG 0 j D p 2 . If jG 0 j D p 3 , there are nonabelian F; H 2 1 such that F 0 ¤ H 0 (Lemma 64.1(u)) so F 0 H 0 D T has index p in G 0 . Then G=T is an A1 -group since jG=T /0 j D p and d.G/ D 2 (Lemma 65.2(a,e)). (i) Since G is not an A1 -group, we have jG W Z.G/j > p 2 . Assume that jG W Z.G/j D p 3 . Then G=Z.G/ has exactly one cyclic subgroup of index p (otherwise, jG W Z.G/j D p 2 ) so G=Z.G/ Š D8 . Since Ã2 .G/ Z.G/, by (e), we get jG=Z.G/j p 4 . Suppose that p D 2 and GN D G=Z.G/ is nonabelian of order 24 . Then GN D ha; b j a4 D b 4 D 1; ab D a1 i. Set UN D hb 2 i. Then G=U Š D8 , and so U Z.G/, by (c), a contradiction. Thus, in that case, G=Z.G/ is abelian. (j) Let H 2 1 . Since 1 .H / G G, we get j1 .H /j p 2 so H is metacyclic (Lemma 65.1). Not all groups of Lemma 65.4(d) are A2 -groups. Indeed, let G D H C , where H D ha; b j a4 D b 4 D 1; ab D a1 i, C D hc j c 4 D 1i; H \ C D 1 .C /. The group G is an A2 -group if and only if c 2 D a2 .

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The group G D ha; b j a8 D b 4 D 1; Œa; b D a2 i is a metacyclic A2 -group with G=Z.G/ Š D8 . n 2 n2 The group G D ha; b j ap D b p D 1; Œa; b D ap i. n > 3, is metacyclic with G=Z.G/ Š Cp 2 Cp 2 . 3 2 Let p > 2 and G D ha; b j ap D b p D 1; Œa; b D ap i. Then G 0 D hap i, 2 Z.G/ D hap i, G=Z.G/ is nonabelian of order p 4 . The above three groups realize all possibilities of Lemma 65.4(i). Let G be a metacyclic 2-group with G=Z.G/ Š D8 . Then jG 0 j D 4 (Lemma 64.1(u)) so G is an A2 -group (Corollary 65.3). If G is a metacyclic p-group such that G=Z.G/ is of order p 4 and exponent p 2 , then G is an A2 -group. Indeed, if H 2 1 , then H=Z.G/ is abelian of type .p 2 ; p/ so H is either abelian or minimal nonabelian since d.H / D 2. Lemma 65.5. Let an A2 -group G of order > p 4 have no abelian maximal subgroups, and d.G/ D 2. Then: (a) If p D 2, then G is metacyclic. (b) If p > 2 and G is nonmetacyclic, then jGj D p 5 . K3 .G/ D Ã1 .G/ D Z.G/ is of order p 2 and all subgroups of index p 2 in G contain Z.G/. Proof. (a) follows from Lemmas 65.1 and 64.1(n). (b) The quotient group G=Ã1 .G/ is nonabelian of order p 3 and exponent p, Ã1 .G/ D K3 .G/, jG W G 0 j D p 2 (Lemma 64.1(p)) and Ã1 .G/ D Z.G/ (Lemma 65.4(c)). If T < G is of index p 2 , then T Z.G/ is abelian so Z.G/ < T . Next, we have jGj D jG W G 0 jjG 0 j p 5 so jGj D p 5 . Lemma 65.6. Let G be a group of order p m > p 4 and all members of the set 1 are generated by two elements. (a) If Z.G/ D ˆ.G/ has index p 3 in G, then G is an A2 -group. (b) If d.G/ D 2 and Z.G/ D Ã1 .G/ has index p 3 in G, then G is an A2 -group. Proof. (a) If H 2 1 , then d.H / D 2 hence all maximal subgroups of H are members of the set 2 so abelian, and H is either abelian or minimal nonabelian. (b) Given H 2 1 , it follows from d.H / D 2 that all maximal subgroups of H contain Z.G/ (since Z.G/ ˆ.G/) so abelian. Thus, again H is either abelian or minimal nonabelian. Theorem 65.7. Suppose that G is an A2 -group of order > p 4 . (a) If G 0 is cyclic of order > p, then G is metacyclic and jG 0 j D p 2 . (b) Suppose that jG 0 j D p 3 . Then every subgroup of index p 2 in G contains Z.G/. If d.G/ D 2, then p > 2 and jG W Z.G/j D p 3 . If d.G/ D 3, then ˆ.G/ D Z.G/ and G 0 Š Ep 3 . If p D 2, then d.G/ D 3 so G 0 Z.G/ is elementary abelian.

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(c) If p D 2, G is nonmetacyclic and D 2 1 is abelian, then G 0 Z.G/. (d) If G is nonmetacyclic, then G 0 is elementary abelian. (e) If G is nonmetacyclic, then jG=Z.G/j p 3 . (f) If G is nonmetacyclic and A; B 2 1 are distinct nonabelian, then Z.A/ D Z.B/ Z.G/. (g) If p D 2 and G=Z.G/ Š D8 , then G is metacyclic. (h) Suppose that G is nonmetacyclic and the set 1 has exactly one abelian member A; then jG 0 j D p 2 . If G 0 6 Z.G/, then p > 2 and G=Z.G/ is nonabelian of order p 3 and exponent p. If G 0 Z.G/, then G=Z.G/ Š Ep 3 . Proof. (a) Since G 0 is cyclic, all nonabelian maximal subgroups of G have the same derived subgroup D 1 .G 0 /. Then, by Lemma 64.1(u), jG 0 j D p 2 . Let a nonabelian F 2 1 . Since F 0 < G 0 < F , then F is metacyclic (Lemma 65.1). If all members of the set 1 are metacyclic, then G is also metacyclic (Theorem 69.1). Suppose that there is a nonmetacyclic A 2 1 ; then A is abelian. We have 1 .A/ Š Ep 3 since d.A/ D 3: F \ A is metacyclic of index p in A. Since all members of the set 1 which contain 1 .A/, are nonmetacyclic so abelian, G=1 .A/ is cyclic. Then G 0 < 1 .A/ so G 0 Š Ep 2 , a contradiction. Thus, A does not exist so, by what has been said before, G is metacyclic. (b) If T < G is of index p 2 , then T Z.G/ D T and jG W Z.G/j > p 2 since T is abelian and the set 1 has no abelian members (Lemma 65.2(d)); therefore, if H 2 1 , then Z.G/ ˆ.H / ˆ.G/. Suppose that d.G/ D 2. Since G is nonmetacyclic (Corollary 65.3 and (a)), we get p > 2 (Lemma 64.1(n) since all members of the set 1 are two-generator) and jG=Ã1 .G/j p 3 (Lemma 64.1(m)), Ã1 .G/ Z.G/ (Lemma 65.4(c)) and jG W Z.G/j D p 3 . Now let d.G/ D 3. Then G 0 ˆ.G/ Z.G/ (Lemma 65.4(c)) so ˆ.G/ D Z.G/ (Lemma 64.1(q)), exp.G 0 / exp.G=Z.G// D p, and we have G 0 Š Ep 3 . Let, in addition, p D 2. Then d.G/ D 3 since all members of the set 1 are two generator (Lemma 64.1(n)), and Œx; y2 D Œx; y 2 D 1 for x; y 2 G, so exp.G 0 / D 2 since G 0 is abelian (Burnside). (c) Assuming G 0 6 Z.G/, we get jG 0 j D 4 (Lemma 64.1(u)), G 0 Š E4 , by (a), and d.G/ D 2 (Lemma 65.4(c)). By Lemma 65.2(c), D is the unique abelian member of the set 1 . Since G=Z.G/ is nonabelian of order 8 (Lemma 64.1(q)), it has exactly one cyclic subgroup of index 2 so G=Z.G/ Š D8 . Let H 2 1 fDg; then H \ D D ˆ.G/. If a 2 H ˆ.G/ and x 2 D ˆ.G/, then G D ha; xi. We have o.Œa; x/ D 2.D exp.G 0 // and hŒa; xi 6E G (otherwise, G=hŒa; xi is abelian so jG 0 j D 2). Next, a2 2 D so 1 D Œa2 ; x D a2 .x 1 ax/2 D a2 .aŒa; x/2 D a2 a2 Œa; x2 ŒŒa; x; a D ŒŒa; x; a since ha; Œa; xi. H / has class at most 2 (Lemma 65.1) and exp.G 0 / D 2. Then a centralizes G 0 D hG 0 \ Z.G/; Œa; xi. It follows from G 0 < D that CG .G 0 / ha; Di D G, contrary to the assumption.

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(d) Assume that exp.G 0 / > p; then G 0 is abelian (Lemma 64.1(v)) of type .p 2 ; p/ since G 0 is noncyclic, by (a). By Lemma 65.1, H 0 1 .G 0 / < G 0 for all H 2 1 so d.G/ D 2 (Lemma 64.1(y)). By Lemma 65.2(d), all members of the set 1 as A1 -groups are two-generator so p > 2 and K3 .G/ D Ã1 .G/ has index p 3 in G hence jG W G 0 j D p 2 (Corollary 36.6). Since Ã1 .G/ Z.G/ (Lemma 65.4(c)), we get Ã1 .G/ D Z.G/. There is H 2 1 such that H 0 ¤ Ã1 .G 0 /. Then .G=H 0 /0 D G 0 =H 0 is cyclic of order p 2 so G=H 0 is metacyclic, by (a). This is a contradiction since G=Z.G/ is nonmetacyclic and H 0 Z.G/. (e) Assume that jG=Z.G/j > p 3 . Then p D 2, exp.G=Z.G// D 4 (Lemma 65.4(e)), G has no epimorphic images isomorphic to D8 or E8 (Lemma 65.4(c)) so d.G/ D 2. By Lemma 64.1(n), there exists A 2 1 with d.A/ > 2; then A is abelian (Lemma 65.1). Let B 2 1 be nonabelian; then Z.B/ D ˆ.B/ < ˆ.G/ < A so CG .Z.B// AB D G. It follows that jG W Z.G/j jG W Z.B/j D jG W BjjB W Z.B/j D 8, a contradiction. (f) By (e), G=Z.G/j p3 . Assume that Z.A/ ¤ Z.B/. Then Z.A/Z.B/. ˆ.G// has index p 2 in G so d.G/ D 2 and Z.G/ < ˆ.G/ (< since A is nonabelian); moreover, jG=Z.G/ D p 3 . It follows that Z.A/ D Z.G/. Similarly, Z.B/ D Z.G/. Then Z.A/ D Z.B/, and G is not a counterexample. (Clearly, (e) follows from (f).) (g) Assume that G=Z.G/ Š D8 and G is nonmetacyclic. Then G 0 6 Z.G/. If D=Z.G/ is the cyclic subgroup of order 4 in G=Z.G/, then D 2 1 is abelian. Then, by (c), G 0 Z.G/, a contradiction. (h) By Lemma 65.4(d), jG 0 j > p so jG 0 j D p 2 (Lemma 64.1(u)). First assume that 6 Z.G/. By (c), p > 2. By (e) and Lemma 65.4(e), G=Z.G/ is nonabelian of order p 3 and exponent p. Now let G 0 Z.G/. By Lemma 64.1(q), jG W Z.G/j D pjG 0 j D p 3 so G=Z.G/ Š Ep 3 . G0

Probably, (a) and (d) are most important parts of Theorem 65.7. We use those parts in the study of A3 - and A4 -groups in 72. Theorem 65.8. Let a two-generator A2 -group G be a 2-group. Then G is metacyclic. Proof. Assume that an A2 -group G is a nonmetacyclic two-generator 2-group. Then there is A 2 1 with d.A/ > 2 (Lemma 64.1(n)) so d.A/ D 3, by Schreier’s theorem (Appendix 25). Since A is not an A1 -group, it is abelian. Let 1 D fA; M; N g; then M and N are nonabelian (if M is abelian, then G is an A1 -group) so A1 -subgroups. We have Z.M / D ˆ.M / < ˆ.G/ < A so CG .Z.M // AM D G whence Z.M / D Z.G/ since Z.G/ < ˆ.G/. Similarly, Z.N / D Z.G/. Since a nonelementary abelian group G=Z.G/ of order 8 has two four-subgroups, we get G=Z.G/ Š D8 . By Lemma 65.4(c), if U G G is such that G=U Š D8 , then U D Z.G/. Since jG 0 j > 2 (Lemma 65.2(a)), A is the unique abelian member of the set 1 (Lemma 65.2(c)). Write GN D G=Ã1 .A/. Then GN is nonabelian of order 16 since d.A/ D 3 > 2 D N D 4, GN is not of maximal class. Then, by Proposition 10.17, GN d.G/. Since exp.G/

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has no nonabelian subgroups of order 8 (otherwise, d.G/ > 2) so it is an A1 -group. N D AN (Lemma 65.3) so Z.G/ N Š E4 . Let RN < Z.G/ N be of It follows that 1 .G/ 0 N N N N N N order 2 and R ¤ G ; then G=R Š D8 since a nonabelian group G=R has a subgroup N RN Š E4 . By the previous paragraph, R D Z.G/. Let B=R < G=R be cyclic of A= order 4; then B 2 1 fAg is abelian, a contradiction. Lemma 65.9. Let G be a nonmetacyclic p-group and R < G 0 be G-invariant of order p. If G=R is minimal nonabelian and all nonabelian maximal subgroups of G are two-generator, then G is an A2 -group and p > 2. Proof. By hypothesis, jG 0 j > p so G is not an A1 -group. We also have d.G/ D d.G=R/ D 2. Therefore, in view of Theorem 65.8, we have to prove that G is an A2 -group. Let M 2 1 be nonabelian. Then d.M / D 2 and M=R (as a maximal subgroup of G=R) is abelian which implies M 0 D R. Hence M is an A1 -group (Lemma 65.2(a)) and so G is an A2 -group. Theorem 65.10 ([CP] (for p D 2), [ZAX]). Suppose that a nonabelian p-group G is neither minimal nonabelian nor metacyclic nor minimal nonmetacyclic. If all nonabelian subgroups of G are metacyclic, then one and only one of the following holds: (a) G D M C , where M 6Š Q8 is a metacyclic A1 -group and jC j D p. (b) p > 2, d.G/ D 2, G D 1 .G/C , where 1 .G/ Š Ep 3 , C is a cyclic subgroup of index p 2 in G, CG D Ã1 .C / D Z.G/ is of index p 3 in G. Proof. Let us check that groups of (a) and (b) satisfy the hypothesis. This is clear for groups from (a). Now let G be as in (b) and V 2 1 . If 1 .G/ V , then V D 1 .G/Ã1 .C / so V is abelian since 1 .G/ is abelian and Ã1 .C / D Z.G/. Now assume that 1 .G/ 6 V . Then 1 .V / D V \ 1 .G/ Š Ep 2 and V =1 .V / is cyclic hence V has a cyclic subgroup of index p and so metacyclic. Now, assuming that G satisfies the hypothesis, we have to prove that G is either as in (a) or in (b). By hypothesis, there are in G two maximal subgroups M and A such that M is nonabelian so metacyclic and A is nonmetacyclic so abelian; then d.A/ > 2 and d.G/ d.M / C 1 D 2 C 1 D 3. Since M \ A is a metacyclic maximal subgroup of A, we get d.A/ D 3. Set E D 1 .A/; then Ep 3 Š E G G. By the product formula, G D ME so M \ E Š Ep 2 . All maximal subgroups of G containing E, are nonmetacyclic so abelian, hence d.G=E/ D d.M=.M \ E// 2 (Exercise 1.6(a) and Lemma 65.1). In what follows, A; M and E are such as defined in this paragraph. Let d.G=E/ D 2. Then there is a maximal subgroup B=E < G=E with B ¤ A so E A \ B D Z.G/ since B is abelian. If x 2 E M , then G D M X, where X D hxi. Let a noncyclic N < M be maximal. Then N X is abelian (otherwise, d.N / 2 so d.N X/ 3). Thus, M is a metacyclic A1 -group so G is as in (a). Now suppose that G=E is cyclic; then G 0 < E. (i) Let jG=Ej D p; then jM j D p 3 . If CG .M / < M , then G is of maximal class (Proposition 1.17) and p > 2 since G is not metacyclic, and E is the unique

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abelian maximal subgroup of G (Lemma 65.2(c)) so exp.G/ D p 2 since M 2 1 is metacyclic. If E D Z.G/ L, then LG D f1g so G is isomorphic to a subgroup of exponent p 2 of a Sylow p-subgroup of the symmetric group Sp 2 ; then G has a subgroup ¤ A of order p 3 and exponent p which is nonabelian and nonmetacyclic, a contradiction. Thus, G D M Z.G/. If Z.G/ is noncyclic, then G D M L is as in (a) (in that case, M 6Š Q8 since G is not minimal nonmetacyclic). If Z.G/ is cyclic, then, since jZ.G/j D p 2 , we get G D EZ.G/, by the product formula, so G is abelian, a contradiction. (ii) Now suppose that G=E is cyclic of order > p; then G 0 < E so jG 0 j p 2 . We have 1 .G=1 .G// < A=1 .G/ so 1 .G/ D 1 .A/ D E. Since M=.M \ E/ D M=1 .M / Š G=E is cyclic, we get M Š Mp n , n > 3, since M is nonabelian and has a cyclic subgroup of index p (Theorem 1.2). Thus, all nonabelian maximal subgroups of G are isomorphic to Mp n . (It follows that G is an A2 -group so one can use the classification of A2 -groups, however we prefer to present independent, more elementary, proof.) Let d.G/ D 2; then Z.G/ < ˆ.G/ and, since G is not an A1 -group, we get 0 G 6 Z.G/ (Lemma 65.2(a)) so G 0 Š Ep 2 ; then cl.G/ D 3, A is the unique abelian maximal subgroup of G and jG W Z.G/j D pjG 0 j D p 3 (Lemma 64.1(q)). Since G 0 < M , we get G 0 D 1 .M / so M=G 0 is a cyclic subgroup of index p of the abelian group G=G 0 . Since Z.G/ < M , then Z.G/ D Z.M / (compare indices!) so Z.G/ is cyclic. Assume that there is a cyclic U=Z.G/ of index p in G=Z.G/. Then U is abelian and metacyclic so U ¤ A, a contradiction. Thus, exp.G=Z.G// D p so G=Z.G/ is nonabelian of exponent p; then p > 2. We have Z.G/ < C < M , where C is cyclic of index p in M . Since jG W C j D p 2 , we get G D EC , and C is not normal in G since G 0 is noncyclic. Thus, G is as in (b). Now we assume that d.G/ D 3. Then G=G 0 has no cyclic subgroups of index p so jG 0 j D p, and we get jG W Z.G/j D pjG 0 j D p 2 (Lemma 64.1(q)). Since jA W Z.G/j D p and d.A/ D 3, the subgroup Z.G/ is noncyclic. By what has been proved already, M Š Mp n . In that case, 1 .Z.G// 6 M since Z.M / is cyclic. If L < Z.G/ of order p is not contained in M , then G D M L so G is as in (a). Exercise 1. Let G be a nonabelian p-group and d.G/ > 2. Prove that all members of the set 2 are abelian if and only if d.G/ D 3 and ˆ.G/ Z.G/. Exercise 2 (Janko). Let A < G be a maximal normal abelian subgroup of a nonabelian p-group G. Prove that, for x 2 G A, there exists a 2 A such that hx; ai is an A1 group. Exercise 3. Let a p-group G be neither abelian nor minimal nonabelian and let A1 .G/ be the set of all minimal nonabelian subgroups of G. Prove that if G is not generated by any proper subset of the set A1 .G/, then p D 2, G is an A2 -group and jA1 .G/j D 2. (Hint. Use Lemma 76.5. Try to give an independent proof.) Problem. Classify the p-groups all of whose nonnormal subgroups are abelian.

66

A new proof of Blackburn’s theorem on minimal nonmetacyclic 2-groups

Here we present a new proof, due to Janko, of Blackburn’s theorem on minimal nonmetacyclic 2-groups avoiding his tedious calculations. It is difficult to estimate the crucial role of this theorem in our book. Only case p D 2 is considered below since, in case p > 2, the original proof is easy but nonelementary. Theorem 66.1 (Blackburn). Suppose that G is a minimal nonmetacyclic 2-group. Then G is one of the following groups: (a) E8 . (b) The direct product C2 Q8 . (c) The central product Q8 C4 Š D8 C4 of order 24 . (d) The group G D ha; b; ci with a4 D b 4 D Œa; b D 1, c 2 D a2 , ac D ab 2 , b c D ba2 , where G is special of order 25 with exp.G/ D 4, 1 .G/ D G 0 D Z.G/ D ˆ.G/ D ha2 ; b 2 i Š E4 , M D hai hbi Š C4 C4 is the unique abelian maximal subgroup of G, and all other six maximal subgroups of G are isomorphic to X D hx; y j x 4 D y 4 D 1; x y D x 1 i (which is the metacyclic minimal nonabelian group of order 24 and exponent 4). Proof [Jan8]. Suppose that G is a minimal nonmetacyclic 2-group. Then Theorem 44.5 implies d.G/ D 3. If jGj 23 , then ˆ.G/ D f1g and so G Š E8 . Clearly, if G is abelian then a consideration of 1 .G/ shows that G D 1 .G/ Š E8 . Suppose that jGj D 24 . Since d.G/ D 3, G is neither abelian nor minimal nonabelian. Let Q be a nonabelian subgroup of order 8 in G. Since G is not of maximal class, Proposition 10.17 implies G D Q Z with jZj D 4 and Q \ Z D Z.Q/. If Z Š E4 , then Q Š Q8 (because in case Q Š D8 the group G would contain a (proper) subgroup isomorphic to E8 which is nonmetacyclic). Thus, G Š C2 Q8 . If Z Š C4 , then G Š Q8 C4 Š D8 C4 with Q8 \ C4 D Z.Q/. In what follows we assume that jGj 25 . Assume that G contains a nonabelian subgroup D of order 23 . If CG .D/ D, then, by Proposition 10.17, G is of maximal class and so G is metacyclic, a contradiction. Hence CG .D/ 6 D and take in CG .D/ a subgroup U of order 4 containing Z.D/. Then jD U j D 24 and D U is not metacyclic since d.D U / D 3, a contradiction. We have proved that every subgroup of order 8 in G is abelian.

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Groups of prime power order

Now suppose that jGj D 25 . Since d.G/ D 3, it follows that jˆ.G/j D 4. Assume ˆ.G/ 6 Z.G/. Then there exists an element x 2 G CG .ˆ.G//; then hˆ.G/; xi is nonabelian of order 8 since x 2 2 ˆ.G/, contrary to the result of the previous paragraph. Hence ˆ.G/ Z.G/. On the other hand, G 0 ˆ.G/ and suppose that jG 0 j D 2. Then G=G 0 is abelian and minimal nonmetacyclic of order 24 , contrary to the result of the first paragraph of the proof. Hence G 0 D ˆ.G/. For any x; y 2 G, we have Œx; y2 D Œx 2 ; y D 1 since G is of class 2, and so ˆ.G/ D G 0 Š E4 . In particular, exp.G/ D 4 and 1 .G/ D ˆ.G/. Assume that Z.G/ > ˆ.G/ so that jZ.G/j D 23 (recall that jGj D 25 ) and G has an abelian subgroup of index 2. But then Lemma 64.1(q) gives jGj D 2jZ.G/jjG 0 j D 26 , a contradiction. We have proved that 1 .G/ D G 0 D Z.G/ D ˆ.G/ Š E4 , and so G is a special group of order 25 . In fact, we shall show that the structure of G is uniquely determined. There are elements x1 ; x2 2 G G 0 such that Œx1 ; x2 D t1 , where t1 is an involution in G 0 . Then X D hx1 ; x2 i is a maximal subgroup of G since it is nonabelian so of order 24 , and X 0 D ht1 i since X=ht1 i is abelian. Since G=ht1 i is nonabelian in view of jG 0 j D 4, there is x3 2 G X such that (interchanging x1 and x2 if necessary) Œx1 ; x3 D t2 2 G 0 ht1 i. We have ht1 ; t2 i D G 0 and G D hX; x3 i D hx1 ; x2 ; G 0 ; x3 i D hx1 ; x2 ; x3 i. For Œx2 ; x3 we have one of the following possibilities: (1) Œx2 ; x3 D 1. In that case, hx2 ; x3 ; G 0 i is an abelian maximal subgroup of G. (2) Œx2 ; x3 D t1 and then Œx1 x3 ; x2 D Œx1 ; x2 Œx3 ; x2 D t1 t1 D 1. In that case, hx1 x3 ; x2 ; G 0 i is an abelian maximal subgroup of G. (3) Œx2 ; x3 D t2 and then Œx1 x2 ; x3 D Œx1 ; x3 Œx2 ; x3 D t2 t2 D 1. In that case, hx1 x2 ; x3 ; G 0 i is an abelian maximal subgroup of G. (4) Œx2 ; x3 D t1 t2 and then Œx1 x2 ; x2 x3 D Œx1 ; x2 x3 Œx2 ; x2 x3 D t1 t2 t1 t2 D 1. In that case hx1 x2 ; x2 x3 ; G 0 i is an abelian maximal subgroup of G. We see that at least one maximal subgroup M of G is abelian and so (being metacyclic of exponent 4) M Š C4 C4 . Since jZ.G/j D 4, M is also the unique abelian maximal subgroup of G, by Lemma 64.1(q). Take an element c 2 G M so that c 2 is an involution (in G 0 ). Such element exists (otherwise, hc; G 0 i Š E8 since exp.G/ D 4; this argument shows that 1 .G/ D G 0 ). Since Z.G/ D G 0 , we get CM .c/ D G 0 and c stabilizes the chain M > G 0 > f1g. Let a 2 M G 0 be an element (of order 4) with a2 D c 2 (recall that every involution in M is a square of an element from M ). If c normalizes hai, then ha; ci Š Q8 is nonabelian, a contradiction. Hence ac D at , where t 2 G 0 hai D G 0 ha2 i. Let b 2 M G 0 be such that b 2 D t . We get M D hai hbi and ac D ab 2 . Since Œb; c 2 G 0 hb 2 i (as above), it follows that either b c D ba2 or b c D b.a2 b 2 /. However, if b c D b.a2 b 2 /, then .cb/2 D cbcb D c 2 b c b D a2 b.a2 b 2 /b D 1. This is a contradiction since cb 62 G 0 and 1 .G/ D G 0 . Hence b c D ba2 and the structure of G is uniquely determined as stated in the theorem.

66 A new proof of Blackburn’s theorem on minimal nonmetacyclic 2-groups

199

It remains to show that jGj < 26 . Assume that G is a minimal counterexample. Then, considering an appropriate quotient group G=N , where N is a subgroup of order 2 in G 0 \ Z.G/, we get jGj D 26 . Let N be a subgroup of order 2 in G 0 \ Z.G/. Then G=N is isomorphic to the group (d) of our theorem. Set W =N D .G=N /0 D ˆ.G=N / Š E4 so that W D G 0 D ˆ.G/ is of order 8. Then W is abelian, by Burnside; moreover, W is abelian of type .4; 2/ since W is metacyclic and W =N Š E4 . We get N D ˆ.W / D Ã1 .W / D hzi and set W0 D 1 .W /. It is easily seen that W0 6 Z.G/. Indeed, if W0 Z.G/, then we consider a subgroup N0 of order 2 in W0 Š E4 with N0 \ N D f1g so that W =N0 Š C4 . On the other hand, G=N0 is also isomorphic to the group (d) of our theorem and therefore .G=N0 /0 Š E4 coincides with W =N0 Š C4 , a contradiction. We have exp.G/ 23 because G=W Š E8 and exp.W / D 4. Also, Z.G/ N and Z.G/ W since Z.G=N / D W =N . It follows that either Z.G/ D N or Z.G/ Š C4 with N < Z.G/ < W . Set C D CG .W0 / so that jG W C j D 2. If x 2 G C , then x 2 2 ˆ.G/ D W . If x 2 2 W0 , then hx; W0 i Š D8 is nonabelian of order 8, a contradiction. Thus x 2 2 W W0 is an element of order 4 and so all elements in G C are of order 8. In particular, exp.G/ D 23 . Since C D CG .W0 / is metacyclic, there are no involutions in C W0 and so W0 D 1 .G/. By the structure of G=N , G has exactly six maximal subgroups M1 ; M2 ; : : : ; M6 such that Mi =N Š X .i D 1; 2; : : : ; 6/, where X D hx; y j x 4 D y 4 D 1; x y D x 1 i is the metacyclic minimal nonabelian group of order 24 and exponent 4. For the seventh maximal subgroup M7 of G we have M7 =N Š C4 C4 . Hence M7 is either abelian of type .23 ; 22 / or M70 D N in which case M7 is minimal nonabelian (Lemma 65.2(a)) with exp.M7 / D 23 . In both cases W0 Z.M7 / and so M7 D C D CG .W0 / (indeed, if M7 is minimal nonabelian, W0 D 1 .M7 / and M7 =W0 is noncyclic so W0 Z.M7 /). We shall determine the structure of Mi , i 2 f1; 2; : : : ; 6g. We have ˆ.Mi / D W since Mi =W Š E4 and Mi is metacyclic. Also, Mi0 W and Mi0 > f1g is cyclic since Mi is metacyclic. Suppose jMi0 j D 2. The case Mi0 D N is not possible since in that case Mi =N would be abelian. Hence Mi0 ¤ N and so W0 D Mi0 N . But then CG .W0 / Mi M7 D G hence W0 Z.Mi /, contrary to what has been proved already. Hence Mi0 is cyclic of order 4 and W > Mi0 > N . Since CG .W0 / D M7 , we have W0 6 Z.Mi / for i < 7. Because Mi , i < 7, is metacyclic and exp.Mi / D 23 , Mi possesses a cyclic normal subgroup Zi > Mi0 with jZi j D 23 and Mi =Zi Š C4 . If Mi splits over Zi , then Mi D Zi Ui with Zi \ Ui D f1g and Ui Š C4 . But then 1 .Ui / W0 and 1 .Ui / centralizes Zi since Aut.Zi / Š E4 . In that case W0 Z.Mi /, a contradiction. Hence Mi does not split over Zi . Let ai 2 Mi Zi be such that hai i covers Mi =Zi . Since exp.Mi / D 23 , we have hai i \ Zi D N D hzi. Then ai2 is of order 4, ai2 2 W W0 , and ai2 62 Mi0 since ai2 62 Zi , by the choice, and Mi0 < Zi . Set hzi i D Zi so that ziai D zi1 z , D 0; 1 (recall that z 2 N # /. We have 2 .Mi / D W and so Mi is uniquely determined according to Lemma 42.1. In

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Groups of prime power order

particular, we may assume D 0 and we have Z.Mi / D hai2 i < W . We see that Mi0 and Z.Mi / are two distinct cyclic subgroups of order 4 in W . On the other hand, c2 .W / D 2 and let W1 and W2 be two cyclic subgroups of order 4 in W . For a given i < 7, one of these two subgroups is equal to Z.Mi / and the other one is equal to Mi0 . Therefore, we have (interchanging W1 and W2 if necessary) W1 D Z.Mi / D Z.Mj / for certain i ¤ j , i; j 2 f1; 2; : : : ; 6g. We get CG .W1 / hMi ; Mj i D G and so W1 D Z.G/. In that case W1 D Z.Mi / and W2 D Mi0 for all i < 7. It follows that all seven maximal subgroups of G=W2 are abelian so G=W2 is abelian. Then E4 < G 0 W2 Š C4 , a final contradiction. For another proof of Theorem 66.1, see Theorem 69.1. Lemma 66.2 (= Theorem 1.25). Suppose that all nonnormal subgroups of a non-Dedekindian 2-group G have order 2. Then one of the following holds: (a) G Š M2n ; (b) G D Z G0 , where Z is cyclic and G0 D D8 ; (c) G D D8 Q8 . Exercise 1. Classify the p-groups all of whose subgroups of index p 2 are absolutely regular. Exercise 2 ([Pas]). Let G be a non-Dedekindian 2-group all of whose nonnormal subgroups have the same order 2n > 2. Then either G is metacyclic or n D 2. Solution. As in Supplement to Theorem 1.25, E4 Š 1 .G/ Z.G/, unless Q Š Q16 , and all nonnormal subgroups of G are cyclic. Suppose that G is nonmetacyclic. It follows from Theorem 66.1 that if M is a minimal nonmetacyclic subgroup of G then M is either the group of Theorem 66.1(d) or M Š Q8 C2 or is the group of Theorem 66.1(d). In the first case, M has a subgroup X Š ha; b j a4 D b 4 D 1; ab D a1 i. Since X has a nonnormal subgroup hbi of order 4, we get n D 2. In the second case, G has a subgroup Q Š Q8 , and Q G G since Q is noncyclic. Set C D CG .Q/. Assume that C has a cyclic subgroup Z of order 4. If Q \ Z > f1g, then QC D Q Z Š D8 Z has a nonnormal subgroup of order 2 (see Appendix 16, Subsection 1o ), a contradiction, since n > 1. If Q \ Z D f1g, then Q Z has a nonnormal cyclic subgroup of order 4 (Theorem 1.20) since 1 .Q Z/ Z.G/, and we get n D 2. Thus, one may assume that C is elementary abelian. Next, G=C is isomorphic to a 2-subgroup of Aut.Q/ Š S4 that contains a subgroup which is isomorphic to Q=Z.Q/ Š E4 . If G=C Š E4 , then G D Q C D Q E, where E is a subgroup of index 2 in C ; in that case G is Dedekindian, a contradiction. Thus, G=C Š D8 . If x 2 G C is such that hxC i is not normal in G=C , then the subgroup hxi is not normal in G. Since o.x 2 / exp.C / D 2 and n > 1, we get o.x/ D 4 so n D 2. Exercise 3. Classify the groups of Exercise 2 containing a subgroup Q Š Q8 . Exercise 4. If a 2-group of Exercise 2 is metacyclic, then either G Š Q16 or G D m n m1 ha; b j a2 D b 2 D 1; ab D a1C2 i, n m (in the second case, G is minimal nonabelian).

66 A new proof of Blackburn’s theorem on minimal nonmetacyclic 2-groups

201

Exercise 5. Suppose that a nonmetacyclic 2-group G of Exercise 2 has a subgroup such as the group of Theorem 66.1(d). If Z is a nonnormal subgroup of G of order 4 then, all nonnormal subgroups of G=1 .Z/ have the same order 2. Solution. Clearly, Z is cyclic. Suppose that H=1 .Z/ 6E G=1 .Z/; then H 6E G. Since G is nonmetacyclic, we get jH j D 4 (Exercise 2) so jH=Zj D 2. Exercise 6. Suppose that a 2-group G of Exercise 2 has a nonnormal (cyclic) subgroup Z. Then G=ZG is such as in Lemma 66.2. If G=ZG Š M2t , then G is metacyclic. Hint. It remains prove that if G=ZG Š M2n , then G is metacyclic. Assume that this is false. Then jZj D 4, by Exercise 2. The group G=ZG has two distinct cyclic subgroups A=ZG and B=ZG of index 2. The subgroups A and B are abelian so A \ B D Z.G/. Since ZG D ˆ.Z/ ˆ.G/, we get dG/ D d.G=ZG / D 2, and we conclude that G is minimal nonabelian. As above, 1 .G/ Š E4 so G is metacyclic and G is not a counterexample. Exercise 7. Study the non-Dedekindian 2-groups G all of whose nonnormal abelian subgroups are cyclic of the same order 2n > 2. (The groups Q2n , n > 3, satisfy this condition.) Hint. We have 1 .Z.G// D 1 .G/ Š E4 , unless G is a generalized quaternion group (indeed, given U < G nonnormal cyclic, let us consider the abelian U 1 .Z.G//). All cyclic subgroups of G=1 .G/ are normal so this group is Dedekindian. Assume that G=1 .G/ is nonabelian. Then G=1 .G/ contains a subgroup L=1 .G/ Š Q8 (Theorem 1.20). Since all subgroups of L containing 1 .G/, are abelian, we get jL W Z.L/j D 4 so jL0 j D 2. It follows that L0 < 1 .G/, a contradiction since L=1 .G/ is nonabelian. Thus, G=1 .G/ is abelian. Now assume that G has a subgroup Q Š Q8 and set C D CG .Q/. If G is metacyclic, it is generalized quaternion (Proposition 1.19). The subgroup C is elementary abelian of order 4 since G is not of maximal class (Proposition 1.17). It follows that QC D Q L, where jLj D 2. Since all noncyclic abelian subgroups of QC are normal in G and QC is generated by these subgroups, we get QC GG. It follows that G=C Š E4 since G=C is isomorphic to a subgroup of Aut.Q/ Š S4 containing a subgroup Š Q=Z.Q/ Š E4 . In that case, G D Q E, where E is a subgroup of index 2 in C , so G is Dedekindian, contrary to the hypothesis. Thus, G has no subgroups isomorphic to Q8 . It follows that if M is a minimal nonmetacyclic subgroup of G, then M is isomorphic to a group of Lemma 66.1(d). Problem. Classify the nonmetacyclic 2-groups all of whose minimal nonmetacyclic subgroups have order 16.

67

Determination of U2-groups

Recall that a 2-group G is said to be a U2 -group if it contains a normal four-subgroup R such that G=R is of maximal class and, provided T =R is cyclic of index 2 in G=R, then 1 .T / D R (see 17, 18). The subgroup R is said to be the kernel of G. The kernel R is the unique normal four-subgroup of G. Theorem 67.1 (Janko [Jan8]). Let G be a U2 -group with d.G/ D 3. Then there is M 2 1 of maximal class, and one of the following holds: (a) G D M C2 ; (b) G D M C4 with M \ C4 D Z.M /; (c) G D M hui, where u is an involution inducing on M an involutory “outer” automorphism such that CM .u/ D M1 is of maximal class, jM W M1 j D 2, and for each x 2 M M1 , x u D xz with hzi D Z.M / (in particular, z 2 D 1). Proof. Let G be a U2 -group with the kernel R and d.G/ D 3. Then G possesses a maximal subgroup M such that R 6 M since, by hypothesis, R 6 ˆ.G/. In that case, M \ R D hzi is of order 2 and M=hzi Š G=R so M is nonabelian. Next, M has no G-invariant abelian subgroups of type .2; 2/ since R 6 M so M is of maximal class (Lemma 1.4). Since jM j > 23 , M=hzi Š G=R is dihedral. Let u 2 R hzi. If R Z.G/, then G D hui M and we are done. Suppose R 6 Z.G/ and set M1 D CM .R/ D CM .u/ so that jM W M1 j D 2. If M1 is cyclic, then take an element v of order 4 in M1 and x 2 M M1 . Since R# D fu; z; uzg and u and z are not G-conjugate, we have .uv/x D ux v x D .uz/v x D .uz/v 1 D .uz/.vz/ D uv and so C D huvi Š C4 and C Z.G/ since hM M1 i D M and CM D G. In this case G D M C with C \M D hzi D Z.M / and we have obtained the group from (b). Now suppose that M1 is of maximal class. Then u induces on M an outer involutory automorphism such that M1 D CM .u/ is a maximal subgroup of M and for each x 2 M M1 , x u D xz. In the rest of this section we assume that G is a U2 -group with the kernel R and d.G/ D 2. Then G=R is of maximal class and order 2n , n 3, and ˆ.G/ R. Let T =R (of order 2n1 ) be a cyclic subgroup of index 2 in G=R and let hai be a cyclic subgroup of T which covers T =R. Since 1 .T / D R, hai is of order 2n and hai \ R D hzi is of order 2. Hence T is either abelian of type .2n ; 2/ or T Š M2nC1 .

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67 Determination of U2 -groups

In any case, ˆ.T / D ha2 i, z 2 ha2 i, ha2 i is normal in G, and z 2 Z.G/. Let n2 v D a2 so that o.v/ D 4, v 2 D z, A D Rhvi D 2 .T / is abelian of type .4; 2/, and hvi D A \ ˆ.T / is normal in G. Hence A=R D Z.G=R/ and so for each x 2 G T , x 2 2 A since G=R is of maximal class. Since Ã1 .G/ D ˆ.G/ D Rha2 i is of exponent 2n1 , we get hai 6 ˆ.G/ so, if hai is normal in G, G is metacyclic. If G=R Š Q2n , then for each x 2 G T , x 2 2 A R. All elements in A R are of order 4 and so o.x/ D 8. In this case 2 .G/ D 2 .T / D A and so G is metacyclic and is isomorphic to a group (c) of Lemma 42.1. Therefore, we may assume in the sequel that G=R 6Š Q2n . In particular, T =R is the unique cyclic subgroup of index 2 in G=R. Conversely, assume that G is metacyclic. Then G 0 is cyclic, G 0 ˆ.G/ D Rha2 i, jG 0 \ Rj D 2, and G 0 covers .Rha2 i/=R. Thus jG 0 j D 2n1 . Let S be a cyclic normal subgroup of G such that G 0 < S and jS W G 0 j D 2. But then jS j D 2n and .RS /=R is a cyclic subgroup of index 2 in G=R. The uniqueness of T =R implies RS D T . But cn .T / D 2, and so hai is normal in G. We have proved that (under our assumptions) G is metacyclic if and only if hai is normal G. We shall use the above notation in the rest of this section. Theorem 67.2. Let G be a metacyclic U2 -group with the kernel R. Then one of the following holds. (a) G D ha; b j a2 D b 4 D 1; ab D a1 z ; D 0; 1; z D a2 where R D hb 2 ; zi D Z.G/ and G=R Š D2n . n

n1

; n 3i,

(b) G is isomorphic to a group (c) of Lemma 42.1 and here G=R Š Q2n and Z.G/ Š C4 . (c) G D ha; b j a2 D b 4 D 1; ab D a1C2 ; z D a2 R D hb 2 ; zi, G=R Š SD2n , and Z.G/ D hzi Š C2 . n

n2

n1

; n 4i, where

Proof. Suppose that G is a metacyclic U2 -group with the kernel R, where G=R 6Š Q2n (for the case G=R Š Q2n , see the second paragraph following the proof of Theorem 67.1). Then hai is normal in G, by the paragraph preceding the theorem. Since G is not of maximal class, R D 1 .G/. Assume that G=R is dihedral. Take x 2 G T ; then x 2 2 R so H D hx; Ri is of order 8 and abelian (otherwise, G would be of maximal class). It follows that CG .R/ hG T i D G. We have obtained the group (a). Suppose now that G=R Š SD2n , n 4. There is b 2 G T with b 2 2 R. If 2 b 2 hzi, then ha; bi is a maximal subgroup of G, by the product formula so R < ˆ.G/ < ha; bi. However, this is a contradiction since we must have ha; bi=R D G=R so that ha; bi D G. Thus b 2 D u 2 R hzi. We have, in view of the structure of n2 G=R that ab D a1 vz . D 0; 1/, where v D a2 (recall that hai is normal in G). We have .a4 /b D .ab /4 D .a1 vz /4 D a4 , and so v b D v 1 since v 2 ha4 i 2 (recall that hai is normal in G). We compute further ab D au D .a1 vz /b D .a1 vz /1 v 1 z D av 2 D az, so au D az and therefore T Š M2nC1 . If D 1, 0 then we replace b with b 0 D bu; then .b 0 /2 D b 2 D u and ab D abu D .a1 vz/u D

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Groups of prime power order

a1 zvz D a1 v. Hence, we may assume from the start that ab D a1 v, and we have obtained the group (c). Theorem 67.3. Let G be a nonmetacyclic U2 -group with d.G/ D 2 and the kernel R. Then one of the following holds: n n1 (a) G D ha; b j a2 D 1; a2 D z; b 2 D z ; D 0; 1; ab D a1 u; u2 D Œu; a D Œu; b D 1; n 3i; where R D hu; zi D Z.G/ and G=R Š D2n . D z; at D a1 u; u2 D 1; Œu; a D (b) G D ha; t j a2 D t 2 D 1; a2 Œu; t D z; n 3i, where R D hu; zi and G=R Š D2n . If n D 3, then Z.G/ D ha2 i Š C4 and if n > 3, then Z.G/ D hzi Š C2 . n

n1

D z; a2 D v; at D a1 vu; u2 D (c) G D ha; t j a2 D t 2 D 1; a2 t Œu; a D 1; u D uz; n 4i, where R D hu; zi, G=R Š SD2n , and Z.G/ D hvui Š C4 . n

n

n1

n1

n2

n2

D z; a2 D v; b 2 D z ; D 0; 1; ab D (d) G D ha; b j a2 D 1; a2 1 2 a a vu; u D Œu; b D 1; u D uz; n 4i, where R D hu; zi, G=R Š SD2n and Z.G/ D hzi Š C2 . Proof. Suppose that G is a nonmetacyclic U2 -group with the kernel R and d.G/ D 2. In that case G=R 6Š Q2n (see the proof of Theorem 67.2) so G=R has the unique cyclic subgroup T =R of index 2. In that case, T is either abelian of type .2n ; 2/ or T Š M2nC1 . Let hai be a cyclic subgroup of index 2 in T ; then hai is not normal in G (see the paragraph preceding Theorem 67.2). Let z be an involution in hai; then z 2 Z.G/. (i) Suppose that G=R Š D2n and T is abelian. If x 2 G T , then x 2 2 R. Since Ã1 .G/ D ˆ.G/ R and Ã1 .T / does not contain R, there is b0 2 G T with b02 D u0 2 R hzi and so R Z.G/. But hai is not normal in G and so ab0 D a1 u0 z . D 0; 1/. We have .b0 a/2 D b0 ab0 a D b02 ab0 a D u0 a1 u0 z a D z . Hence, setting b0 a D b and u0 z D u, we have b 2 D z and ab D a1 u. We have obtained group (a). (ii) Now suppose that G=R Š D2n and T Š M2nC1 . In that case, R 6 Z.G/. Since 1 .G=R/ D G=R, it follows that there is a subgroup H=R of order 2 such that H 6 T and H is nonabelian. Then H Š D8 . Let t 2 H R be an involution. Then at D a1 u with u 2 R hzi, ut D uz, and ua D uz. We have obtained the group (b). (iii) It remains to consider the case where G=R Š SD2n , n 4. If Q=R is the generalized quaternion subgroup of index 2 in G=R, then .T \ Q/=R is a cyclic subgroup of index 2 in Q=R and 1 .T \ Q/ D R. Thus Q is a U2 -group with the kernel R. Since Q=R is generalized quaternion, Q is metacyclic in view of 2 .Q/ D 2 .T / is of order 8 (see Lemma 42.1), and the same lemma implies that Z.Q/ Š C4 . In particular, R 6 Z.G/. Suppose, in addition, that T is abelian. Then CG .R/ D T . If D=R is the dihedral subgroup of index 2 in G=R, then 1 .D \ T / D R and so D is a U2 -group with

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67 Determination of U2 -groups

R 6 Z.D/. In particular, there is an element x 2 D T such that x 2 2 R and Rhxi Š D8 . It follows that there is an involution t 2 D T acting faithfully on R (see (ii)) and, since hai is not normal in G, we get at D a1 vu with u 2 R hzi, ut D uz, Œu; a D 1. We have obtained the group (c). Finally, suppose that T is nonabelian, i.e., T Š M2nC1 so that hai acts non-trivially on R. Then, in view of R 6 Z.G/, M D CG .R/ is a maximal subgroup of G, where M=R is noncyclic so dihedral, T \ M D ha2 iR, ˆ.T / D ha2 i is normal in G, 1 .T \ M / D R, and so M is a U2 -group. Since R Z.M /, M=R is dihedral. Indeed, if Q=R is the generalized quaternion subgroup of index 2 in G=R, then we know that Q is metacyclic. Since d.G/ D d.T / D d.Q/ D 2 and G is, by assumption, nonmetacyclic, Theorem 44.5 implies that d.M / D 3. Hence M is isomorphic to a group (a) of Theorem 67.1. In particular, there is b 2 M T such that b 2 2 hzi. We set b 2 D z , D 0; 1. Finally, b centralizes R, hai acts non-trivially on R, and ab D a1 vu with u 2 R hzi. We have obtained the group (d). Corollary 67.4 (Berkovich). Suppose that a 2-group G of rank three has order 2nC2 and class n > 2. Then one of the following holds: (a) G D H1 C , where H1 is of maximal class and jC j D 2 (three groups); all four subgroups of maximal class and index 2 are isomorphic. (b) G D H1 C , where H1 is of maximal class and C D Z.G/ is cyclic of order 4, C \ H1 D Z.H1 /. In what follows we assume that Z.G/ D Z.H1 / so Z.G/ is of order 2. (c) G D ha; b; x j a2 D b 2 D x 2 D 1; ab D a1 ; ax D a1C2 n1 a2 bi. Here H1 D ha; bi Š D2nC1 . n

n1

; bx D

(d) G D ha; b; x j a2 D x 2 D 1; a2 D b 2 ; ab D a1 ; ax D a1C2 n1 a2 bi. Here H1 D ha; bi Š Q2nC1 .

n1

; bx D

(e) G D ha; b; x j a2 D b 2 D x 2 D 1; ab D a1C2 n1 a2 bi. Here H1 D ha; bi Š SD2nC1 .

n1

; bx D

n

n

n1

n1

; ax D a1C2

Proof. Let K D K3 .G/; then jG 0 W Kj D 2. We have jZ.G=K/j D 4 and G=K is not minimal nonabelian. Let H=K < G=K be minimal nonabelian; then G D K.G/, where .G/=K D Z.G=K/. It follows from Theorem 1.40 that cl.H / D cl.G/ so H is of maximal class. By Lemma 1.4, G has a normal subgroup R of type .2; 2/. Since jH j > 8, we get R 6 H so G D HR. Since G=R Š H=.H \ R/ Š D2n , we conclude that G is a U2 -group. Now the result follows from Theorem 67.1. Exercise. Let G be a group of order p m , cl.G/ D m 2 and G=G 0 Š Ep 3 . Let L be a G-invariant subgroup of index p in G 0 . Let H=L < G=L be nonabelian of index p. Prove that H is of maximal class.

68

Characterization of groups of prime exponent

n Recall Sn that the set † D fAi g1 of subgroups of a group G is a partition of G if G D 1 Ai and Ai \ Aj D f1g for i ¤ j . In what follows we assume that † is a nontrivial partition of a group G (this means that n > 1 and Ai > f1g for all i ). Subgroups Ai are called components of †. If all components of † have equal order, then G is said to be equally partitioned [Isa8]. It follows from Cauchy’s lemma that, for equally partitioned G, we have .A1 / D .G/. In this section we characterize groups of prime exponent as equally partitioned groups proving the following Isaacs’ result [Isa8]: a group G is equally partitioned by a nontrivial partition † if and only if G is of prime exponent. Note that this theorem is not a consequence of known results on partitioned groups.

Lemma 68.1. Let † be a nontrivial partition of a group G and x; y 2 G # with xy D yx. If x and y lie in different components of †, then o.x/ D o.y/ is a prime. Proof. Suppose that o.x/ < o.y/. Then .xy/o.x/ D y o.x/ ¤ 1, and so xy and y lie in the same H 2 †. Then x D .xy/y 1 2 H , which is not the case. Thus, o.x/ D o.y/. If n > 1 is a proper divisor of o.x/, the nonidentity commuting elements x n and y of different orders lie in different components of †, contrary to what has just been proved. Thus o.x/ D o.y/ is a prime. If a component H 2 † does not contain Z.G/, then exp.H / D p for some prime p. Indeed, if x 2 H # and z 2 Z.G/ H , then x and xz commute and lie in different components of † so o.x/ D o.xz/ is a prime, say p (Lemma 68.1); then o.z/ D p. Assume that y 2 H # is of prime order q ¤ p. In that case, o.y/ D o.yz/ D q so o.z/ D q ¤ p, a contradiction. It follows that, if a p-group G of exponent > p admits a nontrivial partition † and Z.G/ H 2 †, then H is the unique component of † of composite exponent so H contains the Hudges subgroup Hp .G/ of G. Lemma 68.2 ([Isa8]). Let † be a nontrivial partition of an equally partitioned group G and ¿ ¤ X G # . Then there exists H 2 † such that X z 6 H for all z 2 G. Proof. Suppose that the lemma is false and for each H 2 †, choose XH , a Gconjugate of X, with XH H . Let NH D NH .XH / so that H contains at least jH W NH j conjugates of X. Let N D NG .X/, g D jGj, h D jH j. We have jG W N j D jG W NG .XH /j jG W NH j D jG W H j jH W NH j

68

Characterization of groups of prime exponent

207

(the first equality follows since X and XH are conjugate in G), and hence jH W NH j g1 g gh g g1 h g jG W N j. Since h1 h D h.h1/ > 0, we get h < h1 D j†j. Now jG W N j is the number of conjugates of X in G and thus, since † is a partition, we get jG W N j

X

jH W NH j D j†jjH W NH j j†j

H 2†

so

g h

j†j D

g1 , h1

h jG W N j; g

a contradiction.

Lemma 68.3 ([Isa8]). Let † be a nontrivial partition of an equally partitioned group G. Then every element of G # has a prime order. Proof. Suppose that x 2 G has a composite order and let K D Kx be the conjugacy G-class of x. By Lemma 68.2, there exists H 2 † such that x z 62 H for all z 2 G so that K \ H D ¿. By Lemma 68.1, no element of H # centralizes any element of K. Thus, H acts semi-regularly by conjugation on K and hence jH j divides jKj. Since 1 2 H and 1 62 K it follows that jH j < jKj. Now pick F 2 † with x 2 F . Since jF j D jH j < jKj, the set K F is nonempty, and, obviously, this set is F -invariant. By Lemma 68.1, F acts on K F semi-regularly via conjugation so that jH j D jF j divides jK F j > 0. Since K D .K F / [ .K \ F / is a partition, jF j divides jK \ F j < jF j, which is a contradiction. Thus elements of G # have prime orders. Exercise 1. Suppose that U is a nontrivial normal p-subgroup of G, where p is the largest prime divisor of jGj. Assume that every element of G # has prime order and let P 2 Sylp .G/. Then either P D G or jG W P j is prime (and so P G G). Solution. Take q 2 .G/fpg and Q 2 Sylq .G/. Then QU is a Frobenius group with kernel U so jQj D q. It follows that P G G. Indeed, if r is a minimal prime divisor of jGj, then jGjr D r so G is r-nilpotent (Burnside). Applying this to the r-normal complement of G, we at last prove our claim. If P < G, then G is a Frobenius group with kernel P and complement of square free order without elements of composite order. It follows that p 0 -Hall subgroup of G is of prime order [BZ, Chapter 10]. Exercise 2. Suppose that every element of G # is of prime order. Let P 2 Sylp .G/, where p is the largest prime divisor of G. Then P is a TI-set, i.e., P \ P x D f1g for x 2 G NG .P /, unless P E G. Solution. Assume that f1g < D D P \ P x , where P ¤ P x and jDj is as large as possible. Applying Exercise 1 to NG .D/ whose Sylow p-subgroup is not normal (Burnside), we obtain a contradiction. Now we are ready to prove the main result of this section. Theorem 68.4 ([Isa8]). Let † be a nontrivial partition of an equally partitioned group G. Then exp.G/ D p for some prime p.

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Groups of prime power order

Proof. One may assume that j.G/j > 1 (Lemma 68.3). Let p be the largest prime divisor of jGj and P 2 Sylp .G/. By Lemma 68.3 and Exercise 2, P is a TI-set since P 6E G. By Lemma 68.3 and Exercise 1, NG .P / D C P , where either C D f1g or jC j D q, a prime. In the first case, G D P H is a Frobenius group with kernel H and complement P . Then jP j D p, a prime. In that case, G is not equally partitioned (by Lemma 68.2: take there X D P # ). Thus, NG .P / D C P is a Frobenius group with kernel P and complement C of prime order q ¤ p. Let jGj D g, jP j D p b , jH j D h for H 2 † and p ˛ D hp , the p-part of h. By Lemma 68.2, given x 2 P # , there exists F 2 † such that x z 62 F for all z 2 G, so ˛ < b since jF j D jH j D h (Sylow). Since P is a TI-set, jP \ H j D p ˛ for all H 2 † such that P \ H ¤ f1g (note that a Sylow p-subgroup of any component H of † is also a TI-subset in H ). By Lemma 68.2, one may choose H 2 † with H \ C x D f1g for all x 2 G. Take P0 2 Sylp .H /. We may assume that P0 < P . Since P is a TI-set, we must have NH .P0 / NG .P / D C P . It follows that NH .P0 / D P0 C0 , where p − jC0 j. Assume that y 2 C0# . Then f1g < P0 P \ P y so P y D P and so C0 is conjugate with C in NG .P /, by Sylow’s theorem, contrary to the choice of H . It follows that C0 D f1g so NH .P0 / D P0 . By assumption, .g/ D .h/ so, to complete the proof, it suffices to prove that H is a p-group. Assume that this is false. Then H D P0 Q0 is a Frobenius group with kernel Q0 and complement P0 since P0 is a TI-subgroup in H (see Frobenius’ theorem in [BZ, Chapter 10]). Since P0 has only one subgroup of order p and its exponent is p, we get jP0 j D p, i.e., ˛ D 1. Since Q0 is nilpotent (Thompson), it is a q-group for some prime q, by Lemma 68.3. Then G is a fp; qggroup, and so G is solvable, by Burnside’s two-prime theorem. We have f1g < Q1 D Oq .G/ since P is a nonnormal TI-subgroup of G (in view of the existence of H ). Since P Q1 is a Frobenius group, we get jP j D p D jP0 j, a contradiction since 1 D ˛ < b. Thus, H is a p-subgroup. For more elementary proof of Theorem 68.4, see [Isa8]. Exercise 3. Suppose that a 2-group G of exponent > 2 admits a nontrivial partition. Then G D Ahbi, where o.b/ D 2 and b inverts A. (Hint. Use Lemma 68.1 and the paragraph following it.) Problem 1 (Isaacs). Does there exist a group partitioned by proper subgroups of equal order not all of which are abelian? Problem 2. Let † be a nontrivial partition of G. Study the structure of G if orders of components of † form a chain under divisibility. Problem 3. Classify all nontrivial partitions of elementary abelian p-groups.

69

Elementary proofs of some Blackburn’s theorems

In this section we present elementary proofs of some Blackburn’s theorems; these proofs are due to the first author. 1o . Blackburn has classified minimal nonmetacyclic p-groups. For a simpler proof in the case p D 2, due to the second author, see Theorem 66.1. Here we offer another elementary proof for all p. The p-groups, p > 2, without normal elementary abelian subgroups of order p 3 were classified by Blackburn (see Theorem 13.7). The proof of Theorem 13.7 was based on the deep theorem 12.1(a). Here we offer another proof based on Theorem 12.12 which is elementary for p D 3 (see Remark 1.) Remark 1. The proof of Theorem 12.12(a) is elementary. Now let p D 3. In the proof of part (c) of that theorem we use the following fact: If G is a 3-group of maximal class and order 3m > 33 , then jG=Ã1 .G/j D 33 . It suffices to prove this for m D 4. By Lemma 64.1(m), jG=Ã1 .G/j > 32 , since G is nonmetacyclic. It is known that groups of exponent 3 are of class 2, so jG=Ã1 .G/j D 33 . Similarly, using Theorem 9.5. it is easy to show that, if a nonabelian group G of exponent 3 is generated by two elements, then its order is 33 . Theorem 69.1. If G is a minimal nonmetacyclic p-group, then, one of the following holds: (a) G is of order p 3 and exponent p; (b) G is a group of maximal class and order 34 ; (c) p D 2 and jGj 25 ; if jGj D 25 , then G is an A2 -group. Proof. We use induction on jGj. One may assume that jGj > p 3 . Clearly, we have jG=Ã1 .G/j p 3 (otherwise, all maximal subgroups of G are nonmetacyclic). (i) Let p > 2. Obviously, G has no subgroups of order p 3 and exponent p. Then G is irregular (otherwise, jG=Ã1 .G/j D j1 .G/j D p 2 and G is metacyclic, by Lemma 64.1(m)). Therefore, if jGj D p 4 , then cl.G/ D 3 so p D 3 (Lemma 64.1(a)). Now suppose that jGj > p 4 . Let R Ã1 .G/ \ Z.G/ be of order p. By Lemma 64.1(m), G=R is nonmetacyclic so minimal nonmetacyclic; then p D 3 and G=R is of maximal class and order 34 , by induction. We get jGj D 35 and jG W Ã1 .G/j D 33 (Remark 1). Let H=R be maximal in G=R; then H=R is either abelian of type .9; 3/ or isomorphic to M33 . Since H is metacyclic of order 34 , jH 0 j 3 so H is either abelian or A1 group (Lemma 65.2(a)). Since R ˆ.H / ˆ.G/, we conclude that all members of

210

Groups of prime power order

the set 1 are either abelian or A1 -groups so G is an A2 -group. By Lemma 64.1(p), jG W G 0 j D 32 hence G 0 Š E33 (Theorem 65.7(d)) so nonmetacyclic, a contradiction. (ii) Now assume that p D 2; then d.G/ D 3, by Lemma 64.1(n). Since all maximal subgroups of an (abelian) group G=G 0 are 2-generator and d.G=G 0 / D 3, we conclude that G=G 0 Š E8 so G 0 D ˆ.G/. Take H 2 1 . N 24 . Since d.G/ N D 3, GN is neither metacyclic nor Set GN D G=K3 .G/; then jGj 0 N since cl.G/ N D 2. In view of jHN W .Z.G/ N \ HN /j an A1 -group. Next, GN Z.G/ jHN W GN 0 j D 4 and d.HN / D 2, HN is either abelian or A1 -group (Lemma 65.2(a)). It follows that GN is a (nonmetacyclic) A2 -group. By Theorem 65.7(d), GN 0 is elementary N D 4. Therefore, we abelian; in particular, jGN 0 j 4 (GN 0 is metacyclic!) and exp.G/ 1 2 0 5 N so jGj N 2 . have 2 jGN j D 8 jGj 5 N D GN 0 .Š E4 /, by N Suppose that jGj D 2 ; then jGN 0 j D 4. In that case, Z.G/ 0 0 N > GN , then jGN j D 2). Then all (noncyclic!) Lemmas 4.9 and 64.1(q) (indeed, if Z.G/ N D GN 0 (otherwise, d.AZ. N G// N > 2, which is subgroups AN of GN of order 8 contain Z.G/ 0 N N N N N N N not the case since AZ.G/ < G/ so A G G. Since G D ˆ.G/, we conclude that GN is special. Assume that K D K3 .G/ > f1g. By the above, GN D G=K is an A2 -group of order 25 . Suppose that jGj D 25 . Assume that G is not an A2 -group. Then G contains a nonabelian subgroup H of order 8. By Lemma 64.1(i), CG .H / contains a subgroup F of order 4 since G is not of maximal class. Taking F so that H \ F D Z.H /, we get H F 2 1 and d.H F / D 3, a contradiction. Thus, if jGj D 25 , then G is a special A2 -group. N D 24 . We claim that this is impossible. Let R be a G-invariant Suppose that jGj subgroup of index 2 in K; then cl.G=R/ D 3. One may assume that R D f1g: then jGj D 25 , contrary to the previous paragraph. It remains to consider the case where K > f1g and jG=Kj D 25 . We claim that this is impossible. Let R be a G-invariant subgroup of index 2 in K. To obtain a contradiction, one may assume that R D f1g; then jGj D 26 . Since cl.G/ D 3 and d.G/ D 3, G is not an A2 -group (Lemma 65.4(c)). It follows that K D K3 .G/ is the unique minimal normal subgroup of G. Therefore, there exists a nonabelian subgroup H of index 4 in G. We have K < H (otherwise, d.K H / > 2). Since E4 Š Z.G=K/ < H=K, H is not of maximal class. There are the following two possibilities for (the metacyclic subgroup) H : either H Š M24 or H D ha; b j a4 D b 4 D 1; ab D a1 i. Since GN 0 < HN , we conclude that H G G. Assume that H Š M24 . We have c3 .H / D 2 and H G G so, if L is a cyclic subgroup of H of index 2, then jG W NG .L/j 2. Let H < M NG .L/ with jM W H j D 2. There are two possibilities for M=L: M=L 2 fC4 ; E4 g. If M=L Š C4 , then CM .L/ D L since H is the unique subgroup of order 16 in M containing L and H is nonabelian. Then M=L Š C4 is a subgroup of Aut.L/ Š E4 , a contradiction. Now suppose that M=L Š E4 . Let F=L be a subgroup of order 2 in M=L with

69 Elementary proofs of some Blackburn’s theorems

211

F=L ¤ H=L. Since K < L < F , F=K G G=K (see the end of the second paragraph of (ii)). We have G 0 D H \F D L is cyclic. This is a contradiction since G 0 =K Š E4 . Now assume that H D ha; b j a4 D b 4 D 1; ab D a1 i. Since H has exactly two nonidentity squares (a2 2 K and b 2 ) and K is characteristic in H , we see that all subgroups of order 2 are characteristic in H so normal in G. This is a contradiction since K is the unique normal subgroup of G of order 2. The proof is complete. The groups of Theorem 69.1(c) are described in Theorem 66.1. 2o . Here we offer another proof of Theorem 13.7. As follows from the proof of Theorem 13.7, it suffices to prove Theorem 69.3, below. First we prove the following Lemma 69.2. Let G be a p-group, p > 2 and c1 .G/ D 1 C p. Then G is either metacyclic or a 3-group of maximal class. Proof. Let G be a counterexample of minimal order. Then every proper subgroup of G is either metacyclic or a 3-group of maximal class, by induction. Let H G be minimal nonmetacyclic; then H is a 3-group of maximal class and order 34 (Theorem 69.1) so H < G. Let B be a nonabelian subgroup of order 33 in H (B exists since H is not minimal nonabelian); then CG .B/ 6 B (Lemma 64.1(i)). Let U be a subgroup of order 9 in CG .B/ with B \ U D Z.B/; then BU < G is neither metacyclic nor of maximal class, a contradiction. It follows from Lemma 69.2 the following new Proof of the last assertion of Lemma 64.1(m) = Theorem 9.11. We have to prove that, if p > 2 and jG=Ã1 .G/j p 2 , then G is metacyclic. Indeed, in that case, G is regular, by Lemma 64.1(a). If jG=Ã1 .G/j D p, then G is cyclic. Now let jG=Ã1 .G/j D p 2 ; then j1 .G/j D p 2 (Lemma 64.1(a)) so c1 .G/ D 1 C p. By Lemma 69.2, G is either metacyclic or an irregular 3-group of maximal class. In the second case, however, jG=Ã1 .G/j D 33 , by Remark 1. Remark 2. If a p-group G has an absolutely regular maximal subgroup A and irregular subgroup M of maximal class, then G is of maximal class. This coincides with Proposition 12.13. Theorem 69.3. If a p-group G, p > 2. has no normal subgroups of order p 3 and exponent p. then G is either metacyclic or a 3-group of maximal class. Proof. Suppose that G is a counterexample of minimal order. Then G is irregular (Lemma 64.1(a,m)). By Theorem 10.4, G has no elementary abelian subgroups of order p 3 so it has no subgroups of order p 4 and exponent p. Let Ep 2 Š R G G (Lemma 64.1(x)). By Lemma 69.2, T D CG .R/ is metacyclic since 1 .T / D R, and so T 2 1 . By Lemma 69.2 again, there exists x 2 G R of order p; then B D hx; Ri is of order p 3 and exponent p (Lemma 64.1(a)). Since B 6E G, there

212

Groups of prime power order

exists y 2 NG .B/ B of order p (see the proof of Lemma 64.2). Set H D hy; Bi; then exp.H / > p so p D 3, H is of maximal class and order 34 (Lemma 64.1(a)). Existence of T and H shows: G is of maximal class (Remark 2). We are ready to prove Blackburn’s Theorem 13.7. Theorem 69.4 (Blackburn). Suppose that a group G of order p m , m > 3, p > 2, has no subgroups Š Ep 3 . If G is neither metacyclic nor a 3-group of maximal class, then G D UE, where E D 1 .G/ is nonabelian of order p 3 and exponent p, U is cyclic, Z.G/ C . Proof. By Theorem 69.3, G has a normal (nonabelian) subgroup E of order p 3 and exponent p. By Theorem 10.4, G has no elementary abelian subgroups of order p 3 so it has no subgroups of order p 4 and exponent p. Therefore, setting C D CG .E/ we see that subgroups of order p in C lie in C \ E D Z.E/ so C is cyclic. Next, C G G and G=C is isomorphic to a p-subgroup of Aut.E/ (which is nonabelian of order p 3 and exponent p) containing a subgroup EC =C Š E=.E \C / Š Ep 2 . If G=C Š Ep 2 , then G D E C , and we are done since G is of class 2 so 1 .G/ D E. Now let jG=C j D p 3 ; then G=C is nonabelian of order p 3 and exponent p. Let Ep 2 Š R < E be normal in G and set F D CG .R/; then F 2 1 since E 6 CG .R/. Since 1 .F / D R, F is metacyclic (Lemma 69.2). Assume that there is x 2 G E of order p. Then H D hx; Ei is of exponent > p so p D 3 and H is of maximal class and order 34 . Existence of F and H shows that G is of maximal class (Remark 2), a contradiction. Thus, 1 .G/ D E. Let U=C < G=C be a subgroup of order p such that U=C 6 CE=C . Since 1 .U / D U \ E D Z.E/, U has only one subgroup of order p, namely Z.E/, so U is cyclic; clearly, G D EU . Since CG .C / EU D G, the proof is complete. Corollary 69.5. Let A be a p-group of operators of a nonmetacyclic p-group G, p > 2, jGj > p 3 . If all proper A-invariant subgroups of G of order p 3 are metacyclic, then G is a 3-group of maximal class. Indeed, by Theorem 10.4, G has no elementary abelian subgroups of order p 3 . Now the result follows from Theorem 69.4. Exercise 1. Let G be a noncyclic p-group. Suppose that for each H G G, there is h 2 H such that hhx j x 2 Gi D H . Prove that G is of maximal class. Solution. We use induction on jGj. One may assume that G is nonabelian and jGj > p 3 . Clearly, jG W G 0 j D p 2 . Indeed, if this is not true, let H=G 0 < G=G 0 be of type .p; p/. Then there is no h 2 H such that hhx j x 2 Gi D H . Next, the factors of the lower central series of G apart of the first one, are cyclic and the factors of the upper central series of G apart of the last one, are cyclic. One may assume that jGj > p 3 . Let R be a minimal normal subgroup of G. Then, by induction, G=R is of maximal class. Assuming that G is not of maximal class, we conclude that Z.G/ Š Cp 2 and,

69 Elementary proofs of some Blackburn’s theorems

213

by Lemma 1.4, G has a normal abelian subgroup L of type .p; p/. We have R < L so Z.G=R/ Š Ep 2 , and G=R is not of maximal class, a contradiction. Exercise 2. If a 3-group G of order > 34 is nonmetacyclic but all its minimal nonmetacyclic subgroups have the same order 34 , then G is of maximal class with 1 .G/ Š E9 . (Hint. Use Theorem 69.4.) Exercise 3. If G is a minimal nonmetacyclic group of order 25 , then (a) G is special with jG 0 j D 4, (b) the set 1 has exactly one abelian member. Solution. (a) By Theorem 69.1, G=G 0 Š E8 so G 0 D ˆ.G/. Assume that G has a nonabelian subgroup H of order 8. Since G is not of maximal class, we get CG .H / 6 H (Lemma 64.1(i)). If F is a subgroup of order 4 in CG .H / with Z.H / < F , then d.HF / D 3, a contradiction. Thus, G is an A2 -group so G 0 Š E4 (Lemma 65.7(d)). We have CG .G 0 / hH j H 2 2 i so G 0 Z.G/. If G 0 < Z.G/, then by Lemma 64.1(q), jG 0 j D 2, a contradiction. It follows that G is special. (b) Assume that the set 1 has no abelian members. Let Zi , i D 1; 2; 3, be all subgroups of order 2 in G and let Ai D fT 2 1 j T 0 D Zi g. Since G=Zi has exactly three abelian subgroups of index 2, we have jAi j D 3, i D 1; 2; 3, and assume that all Ai are non-empty. Obviously, the sets A1 ; A2 ; A3 are pairwise disjoint, a contradiction since jA1 [ A2 [ A3 j D 3 C 3 C 3 D 9 > j1 j. Thus, 1 has exactly one abelian member since jG 0 j D 4.

70

Non-2-generator p-groups all of whose maximal subgroups are 2-generator

We begin with a study of p-groups of the title by proving the following five theorems. We use results on A2 -groups from 65. The following theorem is essential in determination of A2 -groups (see 71). For p > 2, Theorem 70.1 was proved by Blackburn (however, another proof is presented below). Theorem 70.1. Let G be a nonabelian p-group with d.G/ D 3. Suppose that all members of the set 1 are generated by two elements. If G is of class 2, then it is an A2 -group of order p 6 and one of the following holds: (a) jGj D p 4 and either G D E C , where E is nonabelian of order p 3 , C Š Cp 2 , E \ C D Z.E/ or G D Q Z with Q Š Q8 and jZj D 2. (b) G D ha; b; c j a4 D b 4 D Œa; b D 1; c 2 D a2 ; ac D ab 2 ; b c D ba2 i is the minimal nonmetacyclic group of order 25 , G is special, 1 .G/ D G 0 D Z.G/ D ˆ.G/ D ha2 ; b 2 i Š E4 , ha; bi Š C4 C4 is the unique abelian maximal subgroup of G. All subgroups of G of order 8 contain Z.G/ D G 0 . (c) jGj D p 5 , Ep 2 Š ˆ.G/ D G 0 D Z.G/ < 1 .G/ Š Ep 3 so G is special, Exactly p 2 members of the set 1 , not containing 1 .G/, are metacyclic and exactly one of them is abelian and p C 1 other members of the set 1 , containing 1 .G/, are nonmetacyclic minimal nonabelian. More precisely, G D ha; b; ci with 2

2

ap D b p D c p D Œb; c D 1; c p D x y ı ;

Œa; b D x;

Œa; c D y;

bp D x˛ y ˇ ;

x p D y p D Œa; x D Œa; y D Œb; x D Œb; y D Œc; x D Œc; y D 1;

where in case p D 2 we have ˛ D 0, ˇ D D ı D 1, and in case p > 2, the number 4ˇ C .ı ˛/2 is a quadratic non-residue mod p. Here G 0 D hx; yi, 1 .G/ D G 0 hai, and hb; ci Š Cp 2 Cp 2 is the unique abelian member of the set 1 . All subgroups of G of order p 3 contain Z.G/ D G 0 . (d) p D 2, G D ha; b; ci is of order 26 and a4 D b 4 D c 4 D 1;

Œa; b D c 2 ;

Œa; c D b 2 c 2 ;

Œb; c D a2 b 2 ;

Œa2 ; b D Œa2 ; c D Œb 2 ; a D Œb 2 ; c D Œc 2 ; a D Œc 2 ; b D 1;

70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 215

where G 0 D ha2 ; b 2 ; c 2 i D Z.G/ D ˆ.G/ D 1 .G/ Š E23 , so G is special and all members of the set 1 are nonmetacyclic minimal nonabelian. If H < G is of order 24 , then H G 0 . Proof. Let G satisfy the assumptions of the theorem; then G is not an A1 -group (Lemma 65.1) so jGj > p 3 . Since G=G 0 is of rank 3 and all maximal subgroups of G=G 0 are two-generator, we get G=G 0 Š Ep 3 and so G 0 D ˆ.G/. For any a; b 2 G, Œa; bp D Œap ; b D 1 since G is of class 2, and so G 0 is elementary abelian. If jGj D p 4 and E < G is an A1 -subgroup, then G D EZ.G/ (Lemma 64.1(i)), so G is from (a). From now on we assume that jGj p 5 and so jG 0 j D p13 jGj p 2 . If the set 1 has an abelian member, then jGj D pjZ.G/jjG 0 j (Lemma 64.1(q)) implies jG W Z.G/j p 3 . If the set 1 has no abelian members, then jG W Z.G/j p 3 again. Since jG W G 0 j D p 3 and G 0 Z.G/, we get Z.G/ D G 0 D ˆ.G/ so G is special and exp.G/ p 2 . Let H 2 1 . Then jH W Z.G/j D p 2 D p d.H / so Z.G/ D ˆ.H / and H is either abelian or an A1 -subgroup (Lemma 65.2(a)). Thus G is an A2 -group so jG 0 j p 3 (Lemma 65.2(d)) and jGj D jG W G 0 jjG 0 j p 6 . Since G has a minimal nonabelian subgroup H 2 1 , then exp.H / p 2 so exp.G/ D p 2 . (i) Assume first that jGj D p 5 ; then G 0 Š Ep 2 . If each H 2 1 is metacyclic, then G is minimal nonmetacyclic. By Lemma 64.1(l), p D 2 and G is the uniquely determined group of order 25 given in (b) (see Theorem 66.1). In what follows we assume that G is not minimal nonmetacyclic. Let H 2 1 be nonmetacyclic. Since d.H / D 2, H must be an A1 -group and so H D ha; b j 2 ap D b p D 1; Œa; b D c; c p D Œc; a D Œc; b D 1i; we have E D 1 .H / D hap ; b; ci Š Ep 3 (Lemma 65.1). Assume that there is x 2 G E of order p. Then D D hx; Ei 2 1 is neither abelian (since d.D/ D 2) nor an A1 -group (Lemma 65.1), a contradiction. Thus, E D 1 .G/ Š Ep 3 . Let T1 =E; : : : ; TpC1 =E be all maximal subgroups of G=E. Then T1 ; : : : ; TpC1 are nonmetacyclic A1 -groups (since d.Ti / D 2) isomorphic to H (Lemma 65.1). Suppose that M 2 1 does not contain E. Then 1 .M / D E \ M Š Ep 2 and so M is metacyclic since it is either abelian or an A1 -group (Lemma 65.1). Let X < G and jXj D p 3 .> p 2 D exp.G//; then X is noncyclic. If Z.G/ 6 Xthen d.XZ.G// > 2 and XZ.G/ 2 1 , a contradiction. Hence .G 0 D/Z.G/ < X and so X is abelian and normal in G. N Œa; N ci. N ci; N b; N c; N Œb; N Hence G 0 .Š Ep 2 / is generated Let G D ha; N b; N then G 0 D hŒa; N N c by two of these elements, say xN D Œa; N b and yN D Œa; N c. N If Œb; N D xN yN , let

b D bN aN , c D cN aN . We compute (recall that G is of class 2) N cŒ N a N b; N Œa; N c N D xN yN xN yN D 1: Œb; c D ŒbN aN ; cN aN D Œb; Then A D hb; c; G 0 i 2 1 is abelian and so A E D 1 .G/ (otherwise, d.A/ D 3). Hence A D hbi hci Š Cp 2 Cp 2 with 1 .A/ D G 0 D Ã1 .A/. Take an element

216

Groups of prime power order

a 2 E A (of order p) so that G D ha; b; ci. Setting x D Œa; b; y D Œa; c, we get G 0 D hx; yi D hb p ; c p i since x ¤ 1 ¤ y ¤ x. Set b p D x ˛ y ˇ , c p D x y ı . The subgroup L D ha; b c ; G 0 i 2 1 for any integers ; , unless

0 .mod p/ since a; b; c independent modulo G 0 . Since, for such and , we have Œa; b c D x y ¤ 1, L is nonabelian, In that case, d.L/ D 2 so ˆ.L/ D G 0 D Z.L/, and G 0 is generated by elements .b c /p D x ˛C y ˇ Cı and Œa; b ˇc D ˇ ˇ ˇ Cı ˇ x y , which are linear independent since G 0 Š Ep 2 . Hence ˇ ˛C ˇ 0 .mod p/ only if 0 .mod p/. This gives (1)

ˇ 2 C .ı ˛/ 2 0 only if 0

.mod p/:

From (1) we get ˇ 6 0 .mod p/ (otherwise, D 1, D 0 is a nontrivial solution of (1)) and 6 0 .mod p/ (otherwise, D 0, D 1 is a nontrivial solution of (1)). (i1) Assume that p > 2. We compute (2)

.2ˇ C .ı ˛//2 .4ˇ C .ı ˛/2 /2 D 4ˇ.ˇ 2 C .ı ˛/ 2 /:

Using (1), we see (because ˇ 6 0 .mod p/) that the right hand side of (2) vanishes .mod p/ only if 0 .mod p/, and so we have the same for the left hand side of (2): (3)

.2ˇ C .ı ˛//2 .4ˇ C .ı ˛/2 /2 only if 0

.mod p/

.mod p/:

Hence 4ˇ C .ı ˛/2 is a quadratic non-residue .mod p/. Indeed, if 4ˇ C .ı

2 .mod p/, then we set D 1 in (3) and solve the congruence .2ˇ C .ı 2 ˛// 2 .mod p/ for , which gives a contradiction (since for D 1 we have obtained a nontrivial solution of (3) in and ). (i2) Suppose that p D 2. Then ˇ 1 .mod 2/ and so relation (1) becomes ˛/2

(4)

2 C .ı C ˛/ C 2 0

.mod 2/ only if 0

.mod 2/:

If ı C ˛ 0 .mod 2/, then 1 .mod 2/ satisfies (4), a contradiction. Hence ı C ˛ 1 .mod 2/ and we get b 2 D x ˛ y, c 2 D xy 1C˛ . If ˛ 0 .mod 2/, then (recall that o.x/ D o.y/ D 2) (5)

b 2 D y;

c 2 D xy:

b 2 D xy;

c 2 D x:

If ˛ 1 .mod 2/, then (6)

However, interchanging b and c and also x and y, we obtain from (6) the relations (5). Hence we may assume from the start that ˛ D 0, ˇ D D ı D 1 and our group G is uniquely determined. We have obtained the groups from (c).

70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 217

(ii) Now suppose that jGj D p 6 so that ˆ.G/ D G 0 Š Ep 3 . Each H 2 1 is nonmetacyclic and minimal nonabelian since it is 2-generator, and so one may set 2 2 H D hb; c j b p D c p D 1; Œb; c D z; z p D Œb; z D Œc; z D 1i, where ˆ.H / D hb p ; c p ; zi D ˆ.G/ D G 0 Š Ep 3 , and 1 .H / D ˆ.H / implies 1 .G/ D G 0 D ˆ.G/ D Z.G/ and we again conclude that G is special. (ii1) First assume that p > 2. Then Ã1 .H / D hb p ; c p i and H 0 D hzi, where z 62 Ã1 .H /. All p 2 C p C 1 members of the set 1 are isomorphic to H (Lemma 65.1) and their derived subgroups (of order p) are pairwise distinct (Lemma 65.2(d)). Since G is regular and jG W Ã1 .G/j D p 3 (if jG=Ã1 .G/j > p 3 , then G=Ã1 .G/ has a maximal subgroup that is not generated by two elements), we get Ã1 .G/ D G 0 D fx p j x 2 Gg (Lemma 64.1(a)). In particular, there exists a 2 G H with ap D z. Set b p D x, c p D y. We have G D ha; H i D ha; b; ci and G 0 D hx; y; zi (by Lemma 65.2(d), all elements of G 0 are commutators). Set Œa; b D x ˛ y ˇ z , Œa; c D x y ı z , and recall that Œb; c D z. We shall construct a maximal subgroup K of G with d.K/ > 2. To do this we shall do some purely number-theoretic computations using heavily the oddness of p. First, we solve the equation t 2 2 . 2 4ˇ/ C 2. 2.ı ˛// C 2 .2 C 4 / .mod p/

()

for t , , with , not both 0 .mod p/. If r D 2 C 4 0 .mod p/, we set D 1, D 0. If r D 2 C 4 6 0 .mod p/, we put D 1 and set 2.ı ˛/ D s, 2 4ˇ D u, so that () can be rewritten in the simplified form ()

t 2 r. C r 1 s/2 r 1 s 2 C u .mod p/;

where r, s, u are constant numbers mod p and , t run through all integers mod p. The left hand side of () runs through 1C.p1/=2 D .pC1/=2 distinct values (quadratic residues) mod p. Now, C r 1 s runs through all numbers mod p, . C r 1 s/2 runs through .p C 1/=2 distinct values (quadratic residues) mod p so that the right hand side of ./ runs through .p C 1/=2 distinct values (mod p) (take into account that r 1 s 2 Cu is a constant number modulo p). Since .pC1/=2C.pC1/=2 D pC1 > p, it follows that there are values for t and which satisfy (), as required. Since at least one of the integers , is not a multiple of p, we can solve t

N C C .mod p/ for , N . 2. N / N Substituting this into ./, we obtain (after reforming) N N / N C .ˇ C ı/ .˛ C / 0 . C C N /. or in the determinant form: ˇ ˇ ˇ C C N N ˛ C ˇ C ı ˇˇ ˇ ˇ ˇ 0 1 N N ˇ ˇ ˇ ˇ 0

.mod p/;

.mod p/:

218

Groups of prime power order N

We now define the elements a D ab N c , b D b c of G, where p does not divide at least one of the integers , . Let K D ha ; b ; G 0 i; then K 2 1 . It follows from ˆ.K/ D Ã1 .K/K 0 that ˆ.K/ D hŒa ; b ; .a /p ; .b /p i. We have N

N ˛C ˇ Cı Œa ; b D z CC x y ;

N

.a /p D zx N y ;

.b /p D x y ;

where x, y, z are determined in the first paragraph of this part (ii1). From the vanishing of the above determinant, we see that “linear independent” elements x, y, z cannot be expressed as “linear combinations” of Œa ; b , .a /p , .b /p and so ˆ.K/ < G 0 D hx; y; zi so jK W ˆ.K/j > p 2 or, what is the same, d.K/ > 2, contrary to the hypothesis of the theorem. (ii2) Now assume that p D 2. Let G 0 x, G 0 y be two distinct involutions in G=G 0 . Then X D hx; G 0 i and Y D hy; G 0 i are distinct abelian subgroups of type .4; 2; 2/. We have ˆ.X/ D hx 2 i and ˆ.Y / D hy 2 i. Assume that x 2 D y 2 . Then X=ˆ.X/ Š E8 Š Y =ˆ.Y / D Y =ˆ.X/ and X=ˆ.X/ ¤ Y =ˆ.X/. This is a contradiction since 1 .G=ˆ.X// Š E8 . Thus, x 2 ¤ y 2 . Hence, the seven nontrivial elements in G=G 0 produce seven pairwise distinct squares in G 0 which are the seven involutions in G 0 . In particular, each involution in G 0 is a square in G. Let H 2 1 be fixed. We may set H D ha; b j a4 D b 4 D 1; Œa; b D z; z 2 D Œa; z D Œb; z D 1i (Lemma 65.1(a)). Set a2 D x, b 2 D y so that G 0 D hxihyihzi and .ab/2 D a2 b a b D xbzb D xb 2 z D xyz. There exists c 2 G H such that c 2 D z (see the previous paragraph; note that z is not a square in H ). Then G D hH; ci D ha; b; ci. We claim that ac ¤ ca. Assume that this is false. Then CG .c/ ha; Z.G/; ci, and the subgroup at the right hand side is abelian of index 2 in G, a contradiction since the set 1 has no abelian members (Lemma 65.2(d)). It follows that ha; ci is nonabelian. Since G is an A2 -group, we get ha; ci 2 1 . Since ˆ.ha; ci/ D ha2 D x; c 2 D z; Œa; ci D G 0 , we must have Œa; c 2 G 0 hx; zi and so Œa; c D x ˛ yz ˇ . Similarly, considering the maximal subgroup hb; ci, we get Œb; c 2 G 0 hy; zi and so Œb; c D xy z ı . Consider the subgroup K D ha; bci. Assume that a bc D bc a. It follows bca D ab c D baz c so c a D cz D cc 2 D c 1 and ha; ci has order 16 so is not a member of the set 1 , a contradiction. Since G is an A2 -group and d.G/ D 3, we get K 2 1 . We have ˆ.K/ D ha2 D x; .bc/2 D xy 1C z 1Cı ; Œa; bc D x ˛ yz 1Cˇ i Š E23 ; and since ˆ.K/ D G 0 D hx; y; zi, we have ˇ ˇ ˇ1 0 0 ˇˇ ˇ ˇ1 1 C 1 C ı ˇ D 1 ˇ ˇ ˇ˛ 1 1 C ˇˇ and so (7)

ˇ C C ˇ C ı D 1:

70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 219

Considering the maximal subgroup L D hab; ci, we get ˆ.L/ D hxyz; z; x 1C˛ y 1C z ˇ Cı i Š E23 ; ˇ ˇ 1 1 ˇ ˇ 0 0 ˇ ˇ1 C ˛ 1 C

and so

ˇ 1 ˇˇ 1 ˇˇ D 1; ˇ C ıˇ

which gives ˛ C D 1:

(8)

Considering M D hb; aci 2 1 , we get ˆ.M / D hy; x 1C˛ yz 1Cˇ ; xy z 1Cı i so ˛ C ı C ˛ı C ˇ D 1:

(9)

Finally, consider the maximal subgroup N D hab; aci. We get ˆ.N / D h.ab/2 D xyz; .ac/2 D x 1C˛ yz 1Cˇ ; Œab; ac D x 1C˛ y 1C z 1Cˇ Cı i; and so ˛ C ˛ı C ˇ D 1:

(10)

Relations (7–10) have exactly two solutions: (A) ˛ D 0, ˇ D 1, D 1, ı D 0, which implies Œa; c D yz, Œb; c D xy, and (B) ˛ D 1, ˇ D 0, D 0, ı D 1, which implies Œa; c D xy, Œb; c D xz. However, interchanging a and b and also a2 D x and b 2 D y, we obtain from the solution (A) the solution (B). Hence we may assume from the start that we have solution (A) and so the group G of order 26 is uniquely determined as stated in part (d). Assume that F is a subgroup of G of order 24 and F 6 G 0 . We have F G 0 < G so M D F G 0 2 1 . But then M=.F \ G 0 / Š E23 so d.M / 3, a contradiction. Thus, G 0 is contained in all subgroups of G of order 24 . The proof is complete. Theorem 70.2. Suppose that all maximal subgroups of a p-group G are generated by two elements but d.G/ D 3. If G is of class > 2, then p D 2, jGj 27 , and G=K3 .G/ is isomorphic to the group of order 26 of Theorem 70.1(d). If, in addition, jGj D 27 , then the class of G is 3 and G is uniquely determined, namely, G D ha; b; ci, where a4 D b 4 D c 4 D k 2 D Œa; k D Œb; k D Œc; k D 1; Œa; b D c 2 ; Œb 2 ; a D 1;

Œa; c D b 2 c 2 ; Œb 2 ; c D k;

Œb; c D a2 b 2 ;

Œc 2 ; a D k;

Œa2 ; b D k;

Œa2 ; c D k;

Œc 2 ; b D 1:

Here hki D K3 .G/ D ŒG; G 0 D Z.G/ is of order 2, ha2 ; b 2 ; c 2 ; ki D G 0 D ˆ.G/ D 1 .G/ Š E24 , and G exists as a subgroup of the alternating group A16 .

220

Groups of prime power order

Proof. Suppose that G is a p-group of class > 2, d.G/ > 2, and all members of the set 1 are generated by two elements. Then K3 .G/ ¤ f1g and G=K3 .G/ is isomorphic to a group of Theorem 70.1. In particular, jGj p 5 . Let jGj D p 5 and assume that G has a nonabelian subgroup S of order p 3 . Since d.G/ D 3, G is neither of maximal class nor an A1 -group. Hence CG .S / 6 S (Lemma 64.1(i)) and let F > Z.S / be a subgroup of order p 2 in CG .S / containing Z.S /. Then SF D S F 2 1 and d.SF / D 3, a contradiction. Thus S does not exist so G is an A2 -group. But then jG=ˆ.G/j D p 3 and so ˆ.G/ Z.G/ (Lemma 65.4(c)), so G is of class 2, a contradiction. Thus, jGj p 6 . (i) Let jGj D p 6 . If jG=K3 .G/j D p 4 , then take in K3 .G/ a G-invariant subgroup L of index p. But then jG=Lj D p 5 and G=L satisfies the hypothesis, contrary to the previous paragraph. Hence jG=K3 .G/j D p 5 and so G=K3 .G/ is isomorphic to a group (b) or (c) of Theorem 70.1. Also, K D K3 .G/ is of order p. We have G 0 =K D Z.G=K/ Š Ep 2 and G=G 0 Š Ep 3 . The subgroup K is the unique minimal normal subgroup of G. Indeed, let L be a minimal normal subgroup of G distinct from K. Then L < Z.G/ and so L < G 0 . We have d.G=L/ D 3 and each maximal subgroup of G=L is generated by two elements. By the above, G=L is of class 2 and so ŒG; G 0 L. This is a contradiction since ŒG; G 0 D K and K \ L D f1g. The group G is not an A2 -group (otherwise, d.G/ D 3 would imply G 0 Z.G/, by Lemma 65.4(c)). Therefore G has a nonabelian subgroup H of order p 4 . We have H K (otherwise, KH D K H 2 1 and d.KH / > 2). By Theorem 70.1, applied to G=K, H > G 0 and so H is normal in G. Also, G=K is an A2 -group and so H=K Š Ep 2 H 0 D K. Assume that H is not minimal nonabelian. Then H (of class 2) contains a nonabelian subgroup B of order p 3 , and we have d.H / > 2 since H D BCG .B/ (Lemma 64.1(i)) so H=H 0 D H=K Š Ep 3 , G=K is isomorphic to a group (c) of Theorem 70.1 and H=K D 1 .G=K/, i.e., H is the uniquely determined subgroup of index p 2 in G. Let H < M 2 1 . Since d.M / D 2, we get ˆ.M / D ˆ.G/ D G 0 so all maximal subgroups of M are normal in G. Any nonabelian p-group has exactly 0, 1 or p C 1 abelian subgroups of index p (Exercise 1.6(a)) so M has a nonabelian maximal subgroup F ¤ H . As above, K < F . Since H is the unique A2 -subgroup of index p 2 in G, by the above, F is an A1 -subgroup and F is normal in G. Since F=K ¤ H=K D 1 .G=K/, the subgroup F=K is abelian of type .p 2 ; p/. Assume that F is not metacyclic. If p > 2, then Ã1 .F / is a normal subgroup of G of order p (Theorem 9.11), distinct from K, contrary to what has proved already. Now let p D 2. Then Z.F / Š E4 (Z.F / is noncyclic since Z.F / D ˆ.G/ D Ã1 .F /) and F 0 D K. Let U D CG .Z.F //; then U 2 1 since Z.G/ is cyclic. Let K1 ¤ K be a subgroup of order 2 in Z.F /. Then U=K1 contains a nonabelian subgroup F=K1 of order 8. Since d.U=K1 / D 2, it follows that U=K1 is of maximal class (Lemma 64.1(i)). By Taussky’s theorem, K1 6 U 0 so U 0 Š C4 . We have F < U so F 0 < U 0 < F and U 0 is cyclic of order 4. It follows that F is metacyclic (Lemma 65.1). Thus, all subgroups

70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 221

of G of index p 2 which are different from H , are either abelian or metacyclic. We conclude that if G has an A2 -subgroup of index p 2 , then G has a normal metacyclic minimal nonabelian subgroup F of the same index. Let p > 2. It follows from G 0 < F that G 0 is abelian of type .p 2 ; p/. As we know, Ã1 .G/ D ˆ.G/ D G 0 . By Theorem 13.7 (or Theorem 69.4), G has a normal subgroup E Š Ep 3 . It follows from the structure of G=K that E1 .H /=K/ 1 .G=K/ D H=K so E 1 .H /, and we conclude that M=E is cyclic of order p 2 (indeed, M has at most one abelian maximal subgroup so, in view of p > 2, it has at least p 1 2 nonabelian metacyclic maximal subgroups). Then CM .E/ > E since p 2 does not divide exp.Aut.E//. Since H is the unique maximal subgroup of M containing E, we get H CM .E/ so H is abelian, which is a contradiction. Thus, if p > 2, all nonabelian subgroups of G of index p 2 are A1 -groups. Now let p D 2. Assume that exp.F / D 4, where F ¤ H is metacyclic of order 16. Then all subgroups of order 2 in F are characteristic so normal in G, a contradiction since Z.G/ is cyclic. Thus, F Š M24 . This case we shall consider in (i1). (i1) Assume that p D 2 and F Š M24 . We have c3 .F / D 2 and F G G. Therefore, if L is a cyclic subgroup of F of index 2, then jG W NG .L/j 2. Let F < M NG .L/ with jM W F j D 2. If M=L Š C4 , then CM .L/ D L since F=L is the unique subgroup of order 2 in M=L. But in this case M=L is isomorphic to a subgroup of Aut.L/ Š E4 , a contradiction. Hence, we have M=L Š E4 . Let R=L be a subgroup of order 2 in M=L with R=L ¤ F=L. Since K < L < R and jR=Kj D 23 , Theorem 70.1 forces R > G 0 . But then R \ F D L G 0 and so G 0 is cyclic. This is a contradiction since G 0 =K Š E4 . Thus, if p D 2, then G has no nonabelian metacyclic subgroups of order 24 . It follows that then G has no A2 -subgroups of index 4 (otherwise, by the above, it has a nonabelian metacyclic subgroup of order 24 ). (i2) Assume that p D 2 and H is a nonmetacyclic A1 -group of exponent 4. Then ˆ.H / D Z.H / D L K, where jLj D 2, Z.H / is normal in G and H=Z.H / Š E4 . Set U D CG .L/ D CG .Z.H //. Then jG W U j D 2 since L 6 Z.G/ (recalling that K is the unique minimal normal subgroup of G). In that case H=L is a (proper) nonabelian subgroup of order 8 in U=L which implies that U=L is not an A1 -group. By hypothesis, d.U / D 2 and so L < ˆ.U / and CU .H=L/ < H=L. It follows that U=L is of maximal class (Lemma 64.1(i)), and so .U=L/0 Š C4 . By Taussky’s theorem, L 6 U 0 so U 0 Š C4 . However, H < U and H 0 < U 0 < H so H is metacyclic (Lemma 65.1), contrary to the assumption. Thus, if p D 2, then jGj > 26 (otherwise, G is an A2 -group so of class 2). (i3) Now we assume that p > 2. Then, as we have been proved already, all nonabelian subgroups of G of order p 4 are either abelian or A1 -groups. Let H be a nonabelian subgroup of index p 2 in G. Assume that H is a nonmetacyclic A1 -group. Then H=K is abelian of type .p 2 ; p/ and Ã1 .H / > f1g (see Lemma 65.1) is a normal subgroup of G not containing K D H 0 , a contradiction since K D K3 .G/ is the unique minimal normal subgroup of G.

222

Groups of prime power order

Hence, H is metacyclic, H=K is abelian of type .p 2 ; p/ and GN D G=K is isomorphic N so that EN Š Ep 3 and E (the inverse to a group of Theorem 70.1(c). Set EN D 1 .G/ 4 N image of E in G) of order p , is abelian since it is not an A1 -group in view of d.E/ N and Z.G/ N < 1 .G/, N we get 3 (see the paragraph, preceding (i1)). Since HN Z.G/ N and jH \ Ej D p 3 . But H is metacyclic and so H \ E is abelian of EN \ HN D Z.G/ type .p 2 ; p/ and therefore E is abelian of type .p 2 ; p; p/. Set E1 D 1 .E/ so that E1 Š Ep 3 , E1 \ H Š Ep 2 , and E1 > K. The quotient group G=E1 is noncyclic since N EN Š Ep 2 is its epimorphic image. Let A=E1 and B=E1 be two distinct subgroups G= of order p in G=E1 so that .AB/=E1 Š Ep 2 . Note that Theorem 70.1(c) implies that N BN being order p 3 are normal in G. N Since A and B are nonmetacyclic of order p 4 , A, they must be abelian. Therefore CG .E1 / D AB 2 1 (note that E1 6 Z.G/ because of Z.G/ is cyclic). Since d.AB/ D 2, AB is nonabelian. But AB has two distinct abelian maximal subgroups and so j.AB/0 j D 2 (Lemma 65.2(c)), and we get .AB/0 D K. It follows that AB is a nonmetacyclic A1 -group with Z.AB/ D ˆ.AB/ D E1 (Lemma 65.2(a)). Then Ã1 .AB/ Š Ep 2 is normal in G and Ã1 .AB/ \ K D f1g, a contradiction since K is the unique minimal normal subgroup of G and Ã1 .AB/ is normal in G. Thus, jGj > p 6 for all p. (ii) Let jGj p 7 . As above, set K D K3 .G/; it follows from Theorem 70.1 that jG=Kj p 6 . Assume that jG=Kj p 5 . Take a G-invariant subgroup L < K with jK W Lj D p. We have jG=Lj p 6 and G=L satisfies the assumptions of our theorem. By (i), we have a contradiction since L > f1g. Hence jG=Kj D p 6 ; then p D 2 (Theorem 70.1), and G=K is isomorphic to the group of Theorem 70.1(d). Let, in addition, jGj D 27 so that jKj D 2. Assume that L is another minimal normal subgroup of G. Then jG=Lj D 26 so, by (i), G=L is of class 2. Then G, as a subgroup of .G=K/ .G=L/, is also of class 2, a contradiction. Thus, K is the unique minimal normal subgroup of G; in particular, Z.G/ is cyclic. Let H be an arbitrary subgroup of order 25 in G. Since exp.G/ 23 (by the above and Theorem 70.1), H is not cyclic. If K 6 H , then KH D K H 2 1 and d.KH / 3, a contradiction. Hence H > K and so, by Theorem 70.1(d), H > G 0 D ˆ.G/ and H is normal in G. Since G=K is an A2 -group, H=K is abelian in view of j.G=K/ W .H=K/j D 4. Set G 0 D W . By Theorem 70.1(d), 1 .G=K/ D W =K D Z.G=K/ Š E23 . If A=K < W =K is of order 2, then A is normal in G so W CG .A/ since jAut.A/j2 D 2. Thus, W D hA j K < A < W; jAj D 4i Z.W / so W is abelian. It follows that W 2 fE24 ; C4 C2 C2 g. By Theorem 70.1(d), G D ha; b; ci and a; b; c are not involutions modulo K since ˆ.G=K/ D 1 .G=K/ so that hx; W i is of order 25 and class 2 for x 2 G W since hx; W i=K is abelian (this fact will be used in the sequel) and ha2 ; b 2 ; c 2 ; Ki D G 0 .D W /, ()

Œa; b c 2 ;

Œa; c b 2 c 2 ;

Œb; c a2 b 2

.mod K/:

70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 223

Also, for any x; y 2 G, Œx; y2 D Œy; x2 D Œy; x2 ;

Œx 2 ; y1 D Œx 2 ; y;

Œx 2 ; y 1 D Œx 2 ; y;

Œx 2 ; y D x 2 y 1 x 2 y D x 2 .y 1 xy/2 D x 2 .xŒx; y/2 D x 2 x 2 Œx; y2 ŒŒx; y; x D Œx; y2 Œx; y; x: Using the above identity and the relations for G=K (see ()), we get Œa2 ; c D Œa; c2 ŒŒa; c; a D Œa; c2 Œb 2 c 2 k; a D Œa; c2 Œb 2 ; aŒc 2 ; a; where k 2 K Z.G/ and so (11)

Œa2 ; c D Œa; c2 Œb 2 ; aŒc 2 ; a:

In the same way we obtain the following relations: (12)

Œb 2 ; c D Œb; c2 Œb; c; b D Œb; c2 Œa2 b 2 k; b D Œb; c2 Œa2 ; b;

(13)

Œb 2 ; a D Œb; a2 Œb; a; b D Œa; b2 Œc 2 k; b D Œa; b2 Œc 2 ; b;

(14)

Œc 2 ; a D Œc; a2 Œc; a; c D Œc; a2 Œb 2 c 2 k; c D Œa; c2 Œb 2 ; c;

(15)

Œa2 ; b D Œa; b2 Œa; b; a D Œa; b2 Œc 2 k; a D Œa; b2 Œc 2 ; a;

(16)

Œc 2 ; b D Œc; b2 Œc; b; c D Œc; b2 Œa2 b 2 k; c D Œb; c2 Œa2 ; cŒb 2 ; c:

For any x; y; z 2 G we get (noting that W is abelian and Œx; y 2 2 Œx; Ã1 .G/ D Œx; G 0 K Z.G/) Œx; y 1 D Œx; yy 2 D Œx; y 2 Œx; yy

2

D Œx; y 2 Œx; y

since G 0 is abelian and y 2 2 Ã1 .G/ D G 0 , and so Œx; y 1 ; z D ŒŒx; y 2 Œx; y; z D Œx; y; z. Hence from the Hall–Witt identity Œa; b 1 ; cb Œb; c 1 ; ac Œc; a1 ; ba D 1 we get Œa; b; cb Œb; c; ac Œc; a; ba D 1 so Œc 2 ; cb Œa2 b 2 ; ac Œb 2 c 2 ; ba D 1, by ./, or, taking into account that Œb 2 ; a; Œc 2 ; b 2 K Z.G/, we get (17)

Œb 2 ; aŒc 2 ; b D 1; or, what is the same, Œb 2 ; a D Œc 2 ; b;

It follows from (17) and (13) that Œa; b2 D 1 so Œa2 ; b D Œc 2 ; a, by (15). It follows from (15) and (12) that Œc 2 ; a D Œa; c2 Œb 2 ; c D Œa; c2 Œb; c2 Œa2 ; b D Œa; c2 Œb; c2 Œc 2 ; a so, canceling, we get Œa; c2 D Œb; c2 . Then, by (16), (12), (14), (11) we get Œc 2 ; b D Œb; c2 Œa2 ; cŒb 2 ; c D Œa2 ; bŒa2 ; c D Œc 2 ; aŒa2 ; c D Œa; c2 Œb 2 ; aŒc 2 ; aŒc 2 ; a D Œa; c2 Œc 2 ; b

224

Groups of prime power order

so Œa; c2 D 1. Thus, Œa; b2 D Œa; c2 D Œb; c2 D 1 and one can rewrite (11–17) as follows: (18)

Œb 2 ; a D Œa2 ; bŒa2 ; c;

(19)

Œb 2 ; c D Œa2 ; b;

(20)

Œc 2 ; a D Œa2 ; b;

(21)

Œc 2 ; b D Œa2 ; bŒa2 ; c:

In particular, since the abelian group W D G 0 D hŒa; b; Œa; c; Œb; c; Ki is generated by elements of order 2, we get W Š E24 . Since K D hki is the unique minimal normal subgroup of G and K Z.G/ W , we get Z.G/ D K. If Œa2 ; b D Œa2 ; c D 1, then the relations (18–21) and () imply that W D ha2 ; b 2 ; c 2 ; Ki Z.G/, a contradiction. It follows that we have exactly three possibilities for the values of Œa2 ; b and Œa2 ; c (below K D hk j k 2 D 1i): (G1 )

Œa2 ; b D k;

Œa2 ; c D kI

(G2 )

Œa2 ; b D k;

Œa2 ; c D 1I

(G3 )

Œa2 ; b D 1;

Œa2 ; c D k:

In all three cases we have (see ()): Œa; b c 2 , Œa; c b 2 c 2 , Œb; c a2 b 2 .mod K/. In case (G1 ) we substitute Œa2 ; b D k, Œa2 ; c D k in relations (18–21) and obtain Œb 2 ; a D 1;

Œb 2 ; c D k;

Œc 2 ; a D k;

Œc 2 ; b D 1:

If Œa; b D c 2 k, then we replace a with a0 D a1 D aa2 and get Œa0 ; b D Œaa2 ; b D Œa; ba Œa2 ; b D Œa; bk D c 2 k 2 D c 2 ; 2

and so writing again a instead of a0 , we may assume from the start that (˛)

Œa; b D c 2 :

If Œa; c D b 2 c 2 k, then we replace c with c 0 D c 1 so that the relation (˛) remains valid in view of o.c/ D 4, and we get Œa; c 0 D Œa; cc 2 D Œa; c 2 Œa; cc D kŒa; c D k 2 b 2 c 2 D b 2 .c 0 /2 ; 2

and so we may assume from the start that (ˇ)

Œa; c D b 2 c 2 :

Finally, in case Œb; c D a2 b 2 k, we replace b with b 0 D b 1 . Then the relation (˛) 2 remains unchanged: Œa; b 0 D Œa; bb 2 D Œa; b 2 Œa; bb D Œa; b, and also the relation

70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 225

(ˇ) remains unchanged and we get Œb 0 ; c D Œbb 2 ; c D Œb; cb Œb 2 ; c D Œb; ck D a2 b 2 k 2 D a2 .b 0 /2 ; 2

and so we may assume from the start that Œb; c D a2 b 2 :

( )

The structure of G D G1 is uniquely determined. If we set (identify) a D .1; 2/.3; 13; 7; 9/.4; 10; 8; 11/.5; 6/.12; 16/.14; 15/; b D .1; 3/.2; 9/.4; 14; 8; 12/.5; 7/.6; 13/.10; 16; 11; 15/; c D .1; 4/.2; 10; 6; 11/.3; 12/.5; 8/.7; 14/.9; 15; 13; 16/; we see that the permutations a; b; c satisfy all relations for G1 . Since k D Œa2 ; b D .1; 5/.2; 6/.3; 7/.4; 8/.9; 13/.10; 11/.12; 14/.15; 16/ and hki D Z.G/, we see that our permutations a; b; c induce a faithful permutation representation from G1 into the alternating group A16 . In case (G2 ) we substitute Œa2 ; b D k, Œa2 ; c D 1 in relations (18–21) and obtain Œb 2 ; a D k;

Œb 2 ; c D k;

Œc 2 ; a D k;

Œc 2 ; b D k:

If Œa; b D c 2 k, then we replace a with a1 . If Œa; c D b 2 c 2 k, then we replace c with c 1 . Finally, in case Œb; c D a2 b 2 k, we replace a with a1 and b with b 1 . Doing so, we can assume (as before) from the start that Œa; b D c 2 , Œa; c D b 2 c 2 , Œb; c D a2 b 2 . The structure of G D G2 is uniquely determined. Similarly, the permutations: a D .1; 2/.3; 9; 7; 10/.13; 16; 14; 15/.4; 12/.5; 6/.8; 11/; b D .1; 3/.5; 7/.2; 9; 6; 10/.11; 15/.12; 16/.4; 13; 8; 14/; c D .1; 4/.2; 11; 6; 12/.5; 8/.9; 16/.10; 15/.3; 13; 7; 14/; induce a faithful permutation representation from G2 into A16 . In case (G3 ) we substitute Œa2 ; b D 1, Œa2 ; c D k in the relations (18) to (21) and obtain Œb 2 ; a D k, Œb 2 ; c D 1, Œc 2 ; a D 1, Œc 2 ; b D k. If Œa; b D c 2 k, then we replace b with b 1 . If Œa; c D b 2 c 2 k, then we replace a with a1 . Finally, in case Œb; c D a2 b 2 k, we replace c with c 1 . Doing so, we can assume from the start that Œa; b D c 2 , Œa; c D b 2 c 2 , Œb; c D a2 b 2 . The structure of G D G3 is uniquely determined. The permutations: a D .1; 2/.4; 14; 8; 11/.12; 15; 13; 16/.3; 10/.5; 6/.7; 9/; b D .1; 3/.4; 13/.2; 9; 6; 10/.5; 7/.8; 12/.11; 16; 14; 15/; c D .1; 4/.3; 12; 7; 13/.2; 11/.5; 8/.6; 14/.9; 16; 10; 15/; induce a faithful permutation representation from G3 into A16 .

226

Groups of prime power order

Considering G1 ; G2 ; G3 as above subgroups of A16 , we see that the permutation .2; 4; 11; 6; 8; 12; 9; 7; 14; 10/.3; 13/ conjugates G1 onto G2 , and the permutation .2; 11; 8; 6; 14; 7; 10; 4/.3; 9; 13; 12/ conjugates G1 onto G3 . In particular, G2 and G3 are isomorphic to G D G1 . Since K < ˆ.G/, G satisfies the hypothesis since G=K does. The proof is complete. Thus, groups of the title are classified for p > 2 (see also [Bla1, Theorem 3.1]). Theorem 70.3. Let G be a p-group with d.G/ > 2, whose maximal subgroups are generated by two elements. If G is of class 3, then p D 2, 27 jGj 28 and G=K3 .G/ is isomorphic to the group of order 26 of Theorem 70.1(d). If jGj D 27 , then jK3 .G/j D 2 and G is isomorphic to the group of Theorem 70.2. If jGj D 28 , then K3 .G/ D Z.G/ Š E4 , G 0 D ˆ.G/ D 1 .G/ Š E25 and such groups exist. Proof. Suppose that all maximal subgroups of a p-group G are generated by two elements but d.G/ D 3. Assume, in addition, that G is of class 3. By Theorem 70.2, p D 2, jGj 27 , f1g < K D K3 .G/ Z.G/, and G=K is isomorphic to the group of order 26 of Theorem 70.1(d). It follows that G 0 D ˆ.G/;

G=G 0 Š E8 ;

G 0 =K Š E8 ;

ŒG 0 ; G D K;

1 .G=K/ D Z.G=K/ D G 0 =K: Let x 2 G 0 and g 2 G. Then Œx; g 2 K Z.G/ and so Œx; g2 D Œx 2 ; g D 1 since x 2 2 K. Hence the abelian group K is generated by involutions and so K is elementary abelian. Let S be a maximal subgroup in K. Then G=S (of order 27 ) is isomorphic to the group of Theorem 70.2. In particular, Z.G=S / D K=S and so K D Z.G/. Also, G 0 =S is elementary abelian. Hence ˆ.G 0 / S for any maximal subgroup S of K. Since K is elementary abelian and noncyclic, we get ˆ.G 0 / D f1g and so 1 .G/ D G 0 is elementary abelian. We have G D ha; b; ci, ha2 ; b 2 ; c 2 iK D G 0 and for each maximal subgroup S of K, G=S is isomorphic to the group of Theorem 70.2. Suppose that jKj 23 . Then there exists a maximal subgroup S of K which contains hŒa2 ; b; Œa2 ; ci. But then a2 S is contained in Z.G=S /. This is a contradiction since a2 62 K and Z.G=S / D K=S . We have proved that jKj 4. A group G of order 28 of Theorem 70.3 exists. For example, such a group is G D ha; b; c j a4 D b 4 D c 4 D k 2 D l 2 D m2 D klm D 1; Œa; k D Œb; k D Œc; k D 1; Œa; l D Œb; l D Œc; l D 1; Œa; b D c 2 ; Œa; c D b 2 c 2 ; Œb; c D a2 b 2 ; Œa2 ; b D m; Œa2 ; c D k; Œb 2 ; a D l; Œb 2 ; c D m; Œc 2 ; a D m; Œc 2 ; b D li:

70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 227

Here hk; li D Z.G/ D K3 .G/ Š E4 and G 0 D ˆ.G/ D ha2 ; b 2 ; c 2 ; k; li D 1 .G/ Š E25 . We get a faithful permutation representation of G by setting: a D .1; 2/.3; 21; 22; 15/.4; 23; 10; 18/.5; 8/.6; 11/.7; 17/.9; 14; 12; 26/ .13; 16; 25; 27/.19; 30; 20; 29/.24; 32; 31; 28/; b D .1; 3/.2; 14; 11; 15/.4; 24; 25; 20/.5; 9/.6; 12/.7; 22/.8; 21; 17; 26/ .10; 19; 13; 31/.16; 29; 27; 28/.18; 32; 23; 30/; c D .1; 4/.2; 16; 17; 18/.3; 19; 12; 20/.5; 10/.6; 13/.7; 25/.8; 27; 11; 23/ .9; 24; 22; 31/.14; 32; 21; 30/.15; 28; 26; 29/: In this way we see that the group G exists as a subgroup of A32 . In view of K < ˆ.G/, G satisfies the hypothesis since G=K does. Theorem 70.4. Let G be a p-group with d.G/ > 2 all of whose maximal subgroups are generated by two elements. If G is of class > 2, then p D 2 and one of the following holds: (i) G=K4 .G/ is isomorphic to the uniquely determined group of order 27 given in Theorem 70.2. (ii) H D G=K4 .G/ is isomorphic to one of the groups of order 28 described in Theorem 70.3 and more precisely, H D ha; b; ci with a4 D b 4 D c 4 D k 2 D l 2 D m2 D klm D 1;

Œa; k D Œb; k D Œc; k D 1;

Œa; l D Œb; l D Œc; l D 1; Œa; b D c 2 k ; Œa; c D b 2 c 2 k ; Œb; c D a2 b 2 ; Œa2 ; b D m; Œa2 ; c D k; Œb 2 ; a D l; Œb 2 ; c D m; Œc 2 ; a D m; Œc 2 ; b D l; where D 0; 1. These two groups (for D 0 and D 1) exist as subgroups of the alternating group A32 and they are not isomorphic. We have in both cases K3 .H / D Z.H / D hk; li Š E4 and H 0 D ˆ.H / D 1 .H / D ha2 ; b 2 ; c 2 ; k; li Š E25 . In case D 1, H possesses an automorphism of order 7 which acts fixed-point-freely on H=Z.H / and so H=Z.H / is isomorphic to the Suzuki group of order 26 (given in Theorem 70.1(d)). Proof. We may assume that K4 .G/ D f1g so that G is of class 3. Using Theorems 70.2 and 70.3, we see that G is either isomorphic to the group of order 27 given in Theorem 70.2 or a group of order 28 described in Theorem 70.3. It remains to determine completely the structure of G in the second case. In that case G=K3 .G/ is the special group of order 26 isomorphic to the group of Theorem 70.1(d), K3 .G/ D Z.G/ Š E4 , and G 0 D ˆ.G/ D 1 .G/ Š E25 . We have G D ha; b; ci, where a; b; c are elements of order 4 in G G 0 so that ha2 ; b 2 ; c 2 i K3 .G/ D G 0 and Œa; b c 2 , Œa; c b 2 c 2 , Œb; c a2 b 2 .mod K3 .G//. Also, for each maximal subgroup S of

228

Groups of prime power order

K3 .G/, jS j D 2 and G=S is isomorphic to the group of order 27 of Theorem 70.2. In particular, Z.G=S / D Z.G/=S is of order 2. We claim that each of the commutators Œa2 ; b, Œa2 ; c, Œb 2 ; a, Œb 2 ; c, Œc 2 ; a, Œc 2 ; b are distinct from 1. Indeed, if (for example) Œa2 ; b D 1, then we consider a maximal subgroup S of Z.G/ which contains Œa2 ; c. Then a2 S 2 Z.G=S /, contrary to the fact that Z.G=S / D Z.G/=S (and noting that a2 62 Z.G/). Set Œa2 ; b D m and Œa2 ; c D k. If m D k, then considering G=hmi, we have a hmi 2 Z.G=hmi/, a contradiction. It follows that k ¤ m and so K3 .G/ D hk; mi since K3 .G/ Š E4 . Set l D km so that k, l, m are the three distinct involutions in Z.G/ D K3 .G/. We compute (with some z 2 Z.G/): 2

Œb 2 ; c D Œbb; c D Œb; cb Œb; c D .a2 b 2 z/b a2 b 2 z D .a2 /b a2 D Œb; a2 D Œa2 ; b D m; Œc 2 ; a D Œc; ac Œc; a D .b 2 c 2 z/c b 2 c 2 z D .b 2 /c b 2 D Œc; b 2 D m; Œc 2 ; b D Œc; bc Œc; b D .a2 b 2 z/c a2 b 2 z D .a2 /c a2 .b 2 /c b 2 D Œc; a2 Œc; b 2 D km D l; Œb 2 ; a D Œb; ab Œb; a D .c 2 z/b c 2 z D .c 2 /b c 2 D Œb; c 2 D l: Hence we have obtained the following relations:

(˛ )

Œa2 ; b D m; Œa2 ; c D k;

Œb 2 ; a D l;

Œb 2 ; c D m; Œc 2 ; a D m; Œc 2 ; b D l;

where Z.G/ D K3 .G/ D hk; li Š E4 and m D kl. Also, we may set (˛ )

Œa; bc 2 D k p l q ;

Œa; cb 2 c 2 D k r l s ;

Œb; ca2 b 2 D k t l u ;

where p, q, r, s, t , u are some integers mod 2. For any x 2 G G 0 , Œx; G 0 Z.G/ and so hx; G 0 i is of class 2. This fact will be used many times. For any x; y 2 G G 0 and any g1 ; g2 2 G 0 , we have Œ.xg1 /2 ; yg2 D Œ.xg1 /2 ; y D Œx 2 g12 Œg1 ; x; y D Œx 2 ; y: Therefore, if we replace a; b; c with a0 D aa2.qCsCu/ , b 0 D bb 2.sCu/ , c 0 D cc 2s , respectively, then the relations (˛ ) remain preserved. However, the relations (˛ )

70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 229

are simplified, as the following computation shows: Œa0 ; b 0 .c 0 /2 D Œaa2.qCsCu/ ; bb 2.sCu/ c 2 D Œa; bb 2.sCu/ a

2.qCsCu/

Œa2.qCsCu/ ; bb 2.sCu/ c 2

D Œa; bb 2.sCu/ Œa2.qCsCu/ ; bb 2.sCu/ c 2 ; Œa; b 2.sCu/ Œa; bŒa2.qCsCu/ ; b 2.sCu/ Œa2.qCsCu/ ; bc 2 D Œa; b 2 sCu Œa; bŒa2 ; bqCsCu c 2 D .Œa; bc 2 /l sCu mqCsCu D k p l q l sCu k qCsCu l qCsCu D k pCqCsCu; and we get in a similar way Œa0 ; c 0 .b 0 /2 .c 0 /2 D k qCrCu , Œb 0 ; c 0 .a0 /2 .b 0 /2 D k sCtCu . Writing again a, b, c (instead of a0 , b 0 , c 0 ), we see that the relations (˛ ) are simplified into: (˛ )

Œa; bc 2 D k ;

Œa; cb 2 c 2 D k ;

Œb; ca2 b 2 D k ;

where , , are some integers mod 2. In this way we have obtained eight possibilities G1 ; G2 ; : : : ; G8 for the structure of G depending on the values of these integers: (G1 )

D 0;

D 0;

D 0I

(G2 )

D 0;

D 0;

D 1I

(G3 )

D 0;

D 1;

D 0I

(G4 )

D 0;

D 1;

D 1I

(G5 )

D 1;

D 0;

D 0I

(G6 )

D 1;

D 0;

D 1I

(G7 )

D 1;

D 1;

D 1I

(G8 )

D 1;

D 1;

D 0:

We replace now a; b; c with (ˇ )

a D bb 2 ;

b D c;

c D aca2 b 2 l;

respectively, and verify first that the relations (˛ ) remain unchanged. For example: Œ.a /2 ; c D Œb 2 ; aca2 b 2 l D Œb 2 ; ac D Œb 2 ; cŒb 2 ; ac D ml c D ml D k: We want to see what happens with the relations (˛ ) for arbitrary integers , , .mod 2/. First we compute: .ac/2 D a.ca/c D a.acŒc; a/c D a.acb 2 c 2 k /c D a2 .cb 2 /c 3 k D a2 .b 2 cm/c 3 k D a2 b 2 k C1 l;

230

Groups of prime power order

which gives .c /2 D .aca2 b 2 l/2 D .ac/2 .a2 b 2 l/2 Œa2 b 2 l; ac D a2 b 2 k C1 lŒa2 b 2 ; cŒa2 b 2 ; ac D a2 b 2 k C1 lŒa2 ; cŒb 2 ; cŒb 2; ac D a2 b 2 k C1 l: Then it is easy to transform the relations (˛ ): Œa ; b .c /2 D Œbb 2; ca2 b 2 k C1 l D Œb; cb Œb 2 ; ca2 b 2 k C1 l 2

D .Œb; ca2 b 2 /mk C1 l D k mk C1 l D k C ; and we get in a similar way: Œa ; c .b /2 .c /2 D k CCC1 and

Œb ; c .a /2 .b /2 D k C1 :

These results show that our group Gi defined with the constants , , in the relations (˛ ) is isomorphic with the group Gj defined with the constants C , C C C 1, C 1. In particular, the group G1 defined with D D D 0 is isomorphic to the group G4 defined with the constants 0; 1; 1. Further, the group G4 with D 0, D 1, D 1 is isomorphic with G3 defined with the constants 0 ,1, 0 and this one is isomorphic with G5 . Then G5 is isomorphic with G2 , G2 is isomorphic with G6 and G6 is isomorphic with G7 . However, our replacement (ˇ ) sends G8 onto G8 and we verify that in fact (ˇ ) induces an automorphism of order 7 on G8 which acts fixed-point-freely on G8 =Z.G8 /. Hence G=Z.G/ is isomorphic to the Suzuki group of order 26 given in Theorem 70.1(d). The group G1 is exactly the group occurring as an example in the proof of Theorem 70.3, where we have found a faithful permutation representation of G1 of degree 32 (with even permutations). We get a faithful permutation representation of G8 by setting: a D .1; 2/.3; 14; 21; 25/.4; 18; 10; 27/.5; 8/.6; 11/.7; 17/ .9; 24; 12; 15/.13; 26; 23; 16/.19; 30; 20; 29/.22; 32; 31; 28/I b D .1; 3/.2; 14; 11; 15/.4; 22; 23; 20/.5; 9/.6; 12/.7; 21/ .8; 24; 17; 25/.10; 19; 13; 31/.16; 29; 26; 28/.18; 32; 27; 30/I c D .1; 4/.2; 16; 17; 18/.3; 19; 12; 20/.5; 10/.6; 13/.7; 23/ .8; 26; 11; 27/.9; 22; 21; 31/.14; 32; 24; 30/.15; 28; 25; 29/: In this way, we see that also the group G8 exists as a subgroup of A32 . Finally, considering G1 and G8 as subgroups of A32 , W. Lempken (Institut f¨ur experimentelle Mathematik, Essen) has shown that the groups G1 and G8 are not isomorphic.

70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 231

Let G be a 2-group of class > 3 with d.G/ > 2 but d.H / D 2 for all H 2 1 . Then, it is possible to prove that we must have jGj 29 and we exhibit here an example of such a group G of order 29 and class 4, where G D ha; b; ci with the following defining relations: a8 D b 4 D c 8 D 1;

a4 D c 4 D u;

Œa2 ; b D m;

Œa2 ; c D k;

Œb 2 ; a D lu;

Œb 2 ; c D um;

Œa; c D b 2 c 2 ;

m2 D k 2 D 1;

Œb; c D a2 b 2 ;

Œm; a D 1;

Œm; b D 1;

Œk; c D 1;

Œl; a D 1;

Œa; u D Œb; u D Œc; u D 1; km D l;

Œc 2 ; a D um;

Œc 2 ; b D l;

Œa2 ; b 2 D 1;

Œm; c D u; Œl; b D u;

l 2 D 1; Œa; b D c 2 ;

Œa2 ; c 2 D 1;

Œk; a D 1;

Œb 2 ; c 2 D u;

Œk; b D u;

Œl; c D u:

Here G 0 D ha2 ; b 2 ; c 2 ; k; li is of order 26 , K4 .G/ D Z.G/ D hui is of order 2, K3 .G/ D hk; l; ui Š E8 , and Z.G 0 / D ha2 ; k; li is abelian of type .4; 2; 2/. Finally, we have shown that this group G exists as a subgroup of the symmetric group of degree 64. We can improve Theorem 70.4 in case (i). Theorem 70.5. Let G be a p-group with d.G/ > 2 all of whose maximal subgroups are generated by two elements. If G is of class > 2, then p D 2, and if G=K4 .G/ is isomorphic to the group of order 27 given in Theorem 70.2 (case (i) of Theorem 70.4), then K4 .G/ D f1g. Proof. Assume that K4 .G/ ¤ f1g. To get a contradiction we may assume jK4 .G/j D 2 (by considering a suitable quotient group of G). Then G=K4 .G/ is isomorphic to the group of order 27 of Theorem 70.2. We may set G D ha; b; ci, where a4 b 4 c 4 k 2 1; Œa; c b 2 c 2 ;

Œa; k Œb; k Œc; k 1; Œb; c a2 b 2 ;

Œa2 ; c k;

Œb 2 ; a 1;

Œc 2 ; a k;

Œc 2 ; b 1

Œa; b c 2 ; Œa2 ; b k; Œb 2 ; c k;

.mod K4 .G//:

Also we may set K4 .G/ D hui Z.G/. Then K3 .G/ D hk; ui, W D G 0 D ˆ.G/ D ha2 ; b 2 ; c 2 ; k; ui is of order 25 and class 2 since W =K4 .G/ Š E24 . Since CG .K3 .G// is a subgroup of index 2 in G, we get K3 .G/ Z.W /. From Œa2 ; b k .mod K4 .G// with k 62 K4 .G/ follows Œa2 ; b D k0 with k0 2 K3 .G/ K4 .G/. From a4 1 .mod K4 .G// follows a4 2 K4 .G/ Z.G/ and so 2

2

1 D Œa4 ; b D Œa2 a2 ; b D Œa2 ; ba Œa2 ; b D k0a k0 D k02 ; since a2 2 W and K3 .G/ D hk0 ; ui Z.W /. Hence k0 is an involution and so we have proved that K3 .G/ Š E4 .

232

Groups of prime power order

Using Œc; b 2 D kui and Œc 2 ; b D uj , where i; j are some integers (mod 2), we compute Œc 2 ; b 2 D Œcc; b 2 D Œc; b 2 c Œc; b 2 D .kui /c kui D k c k D c 1 kck D Œc; k; since k is an involution. On the other hand, we get Œc 2 ; b 2 D Œc 2 ; bb D Œc 2 ; bŒc 2 ; bb D uj .uj /b D .uj /2 D 1; and so Œc; k D 1. Using Œa; b 2 2 K4 .G/ Z.G/ and Œa2 ; b D kui with an integer i .mod 2/, we compute Œa2 ; b 2 D Œaa; b 2 D Œa; b 2 a Œa; b 2 D Œa; b 2 2 D 1; Œa2 ; b 2 D Œa2 ; bb D Œa2 ; bŒa2 ; bb D kui .kui /b D kk b D Œk; b; and so Œk; b D 1. Finally, using Œa; c 2 D kuj and Œa2 ; c D kul with some integers j , l .mod 2/, we get Œa2 ; c 2 D Œaa; c 2 D Œa; c 2 a Œa; c 2 D .kuj /a kuj D k a k D Œa; k; Œa2 ; c 2 D Œa2 ; cc D Œa2 ; cŒa2 ; cc D kul .kul /c D kk c D Œk; c: Hence (by the above) Œa; k D Œk; c D 1. We have obtained Œk; a D Œk; b D Œk; c D 1, and so K3 .G/ D hk; ui Z.G/. This is a contradiction since G was of class 4. The group G of Theorem 70.2 of order 27 is an A4 -group since ha2 ; bi is a nonabelian subgroup of order 24 (this subgroup is minimal nonabelian nonmetacyclic) and G has no nonabelian subgroups of order 8. Obviously, D8 is not a subgroup of G since 1 .G/ D E24 D G 0 . Also, Q D Q8 is not a subgroup of G (otherwise, Q \ G 0 D C2 and so S D G 0 Q is a maximal subgroup of G and so d.S / D 2 so ˆ.S / D G 0 and, by Taussky’s theorem, S is of maximal class, a contradiction). Both groups G of Theorem 70.4 of order 28 are A5 -groups since ha2 ; bi is a nonabelian subgroup of order 24 (this subgroup is minimal nonabelian nonmetacyclic) and G has no nonabelian subgroups of order 8. Obviously, D8 is not a subgroup of G since 1 .G/ D E25 D G 0 . Also, Q D Q8 is not a subgroup of G (otherwise, Q \ G 0 D C2 and so T D G 0 Q is a maximal subgroup of G hence d.T / D 2; then ˆ.T / D G 0 so, by Taussky’s theorem, T is of maximal class, a contradiction). The above two paragraphs show that if a 2-group G is such that d.G/ D 3 but d.H / D 2 for all H 2 1 and the class of G is > 2, then G is an An -group with n < 5 if and only if G is the unique group of order 27 from Theorem 70.2, and this G is an A4 -group.

71

Determination of A2-groups

We recall that a p-group G is called an An -group if every subgroup of index p n is abelian but at least one subgroup of index p n1 is nonabelian. Here we determine A2 -groups of order > p 4 up to isomorphism in terms of generators and relations. This determination will follow five propositions covering all A2 -groups. Proposition 71.1. Suppose that G is an A2 -group of order > p 4 . Then jG 0 j D p if and only if G has at least two distinct abelian maximal subgroups and in that case one of the following holds: (i) G D H Cp , where H is minimal nonabelian. (ii) G D ha; b; ci, where m

n

2

ap D b p D c p D 1;

m n 1; m 2;

Œa; b D d;

c p D d;

Œa; d D Œb; d D Œa; c D Œb; c D 1: Here H D ha; bi is nonmetacyclic minimal nonabelian, H 0 D G 0 D hd i, G D H C , where C D hci is cyclic of order p 2 and H \ C D hd i D hc p i D H 0 . Proof. Let G be an A2 -group of order > p 4 with jG 0 j D p. By Lemmas 65.2(c) and 65.4(d), jG 0 j D p if and only if the set 1 has at least two abelian members. In that case the set 1 has a nonabelian member H so that G D H Z.G/, where Z.G/ \ H D Z.H /. We have jH j > p 3 , H 0 D G 0 , Z.H / D ˆ.H / D ˆ.G/, and jH W Z.H /j D p 2 since H is an A1 -group. Clearly, jZ.G/ W Z.H /j D p. Suppose that there is an element c 2 Z.G/ H of order p. Then G D H hci and it is clear that all such groups are A2 -groups. We assume in the sequel that 1 .Z.G// H ; then d.Z.H // D d.Z.G// so jZ.H / W ˆ.Z.H //j D jZ.G/ W ˆ.Z.G//j D pjZ.H / W ˆ.Z.G//j (note that ˆ.Z.H // < ˆ.Z.G// Z.H / since Z.H / is maximal in Z.G/) and so jˆ.Z.G// W ˆ.Z.H //j D p. Hence, if Z.G/=ˆ.Z.G// Š Ep d , then Z.G/=ˆ.Z.H // Š Cp 2 Ep d 1 . This implies that for each c 2 Z.G/ Z.H /, we have c p 2 Z.H / ˆ.Z.H //. (i) Assume, in addition, that H is metacyclic; then, by Lemma 65.1(b), we have m n m1 m1 H D ha; b j ap D b p D 1, m 2, n 1, ab D a1Cp i, where H 0 D hap i mCn mCnC1 p p and jH j D p , m C n > 3, jGj D p . We have Z.H / D ha i hb i D 2 2 2 2 ˆ.H / and for each c 2 Z.G/ H , c p 2 hap ; b p i hap ; b p i since hap ; b p i D ˆ.Z.H //.

234

Groups of prime power order

If n D 1, then H Š Mp mC1 , m > 2, and Z.H / D hap i is cyclic. In this case, replacing an element c 2 Z.G/ Z.H / with c i , i 6 0 .mod p/, if necessary, we may m1 assume that c p D ap and so .ca/p D c p ap D 1. Since .ca/b D cab D .ca/ap , 3 it follows that hca; bi is nonabelian of order p . This is a contradiction, since jGj p 5 and G is an A2 -group. We have proved that n 2. Suppose that there exists an element x 2 G H of order p. Since d.Z.G// D d.Z.H //, we get x 62 Z.G/ and so Œa; x ¤ 1 or Œb; x ¤ 1. If Œa; x ¤ 1, then (noting that hai is normal in G) ha; xi is nonabelian of order p mC1 . This is a contradiction since jGj D p mCnC1 and n 2. Hence we must have Œa; x D 1 and Œb; x ¤ 1. In that case hŒb; xi D G 0 D H 0 and so H D hb; xi is a nonmetacyclic A1 -group of order p nC2 . Hence H must be a maximal subgroup of G containing Z.H / D ˆ.G/ but not containing Z.G/ (every member of the set 1 , containing Z.G/, is abelian!), and so jGj D p nC3 . We get m D 2 and G D H Z.G/, where H is a nonmetacyclic A1 -group. This case will be considered in part (ii) of the proof. Hence assume that 1 .G/ H . By the paragraph, preceding (i), for each c 2 Z.G/ Z.H /, we get c p 2 Z.H / ˆ.Z.H //, where Z.H / D ˆ.H / D Ã1 .H / since H is metacyclic. If there is h 2 H such that c p D hp , then the abelian subgroup hh; ci is not cyclic since hhi and hci are two distinct cyclic subgroups of hh; ci of the same order. Since hh; ci \ H D hhi, there is an element of order p in hh; ci H , contrary to 1 .G/ H . We conclude that hc p i is a maximal cyclic subgroup of H . Since c p 2 Z.H /, the quotient group H=hc p i is noncyclic so c p ˆ.H / D Ã1 .H / since H is metacyclic. It follows that not every element in Ã1 .H / is a p-th power of an element in H so H is irregular so p D 2 since H is of class 2 (Lemma 64.1(a)), and c 2 is not a square in H . Set d D Œa; b ¤ 1, where d 2 D 1, and compute for any integers i; j : .ai b j /2 D a2i b 2j Œb j ; ai D a2i b 2j d ij . If i or j is even, then a2i b 2j D .ai b j /2 is a square in H . Therefore c 2 D a2k b 2l , where both k and l are odd. Consider the nonabelian subgroup S D ha; b l ci (recall that c 2 Z.G/). We have .b l c/2 D b 2l c 2 D b 2l a2k b 2l D a2k , and so jS j D 2mC1 . This is a contradiction, since jG W S j D p 2 > p in view of jGj D 2mCnC1 and n 2. (ii) It remains to consider the case where H is a nonmetacyclic A1 -group of order p 4 . We set m

n

H D ha; b j ap D b p D 1; Œa; b D d; d p D Œa; d D Œb; d D 1i; where we may assume that m n 1. Here jH j D p mCnC1 and so jGj D p mCnC2 . Since jH j p 4 , must be m 2. We have Z.H / D ˆ.H / D hap i hb p i hd i and 2 2 ˆ.Z.H // D hap i hb p i since o.d / D p. Also note that hd i is a maximal cyclic subgroup in H and for each c 2 Z.G/ Z.H /, c p 2 Z.H / ˆ.Z.H //. Note that H \ Z.G/ D Z.H /. We want to show that there exists c 2 Z.G/ Z.H / so that 1 ¤ c p 2 hd i. Assume first that n D 1 so that Z.H / D hap i hd i and for an element c 2 Z.G/ Z.H /, c p D aip d j ; recall that o.c/ > p). If i 0 .mod p/, then j 6 0

71

Determination of A2 -groups

235 0

2

.mod p/ (otherwise, o.c/ D p) and we may set c p D ai p d j . Then we compute 0 0 2 0 .ai p c/p D ai p c p D d j ¤ 1, and we are done since ai p c 2 Z.G/ Z.H / D Z.G/ H . If i 6 0 .mod p/, then S D hai c; bi is nonabelian and .ai c/p D aip c p D aip aip d j D d j . Hence p o.ai c/ p 2 . If o.ai c/ D p 2 , then j 6 0 .mod p/ and hai ci hd j i D G 0 . In any case, jS j D p 3 . This is a contradiction since jG W S j p 2 and G is an A2 -group. Thus, we may also assume that n 2. We have for c 2 Z.G/ Z.H /, c p D ip jp k a b d , where at least one of the integers i; j; k is 6 0 .mod p/ since c p 62 ˆ.Z.H //. If i 0 .mod p/ and j 0 .mod p/, then k 6 0 .mod p/ and ai ; b j 2 Z.H / < Z.G/ so .ai b j c/p D aip b jp c p D aip b jp aip b jp d k D d k ¤ 1; and we are done since ai b j c 2 Z.G/ Z.H /. Now assume that one of the integers i; j is 0 .mod p/ and the other one is 6 0 .mod p/. Note that a and b occur symmetrically since we have m 2 and n 2. Interchanging a and b (if necessary), we may assume that i 6 0 .mod p/ but j 0 .mod p/. In that case T D hai b j c; bi is nonabelian and since b j ; c 2 Z.G/ we get .ai b j c/p D aip b jp c p D d k . Thus p o.ai b j c/ p 2 and if o.ai b j c/ D p 2 , then k 6 0 .mod p/ and hai b j ci hd k i D G 0 . In any case, jT j D p nC2 . This is a contradiction since jG W T j D p m > p. It remains to study the possibility i 6 0 .mod p/ and j 6 0 .mod p/. We consider again the nonabelian subgroup T D hai b j c; bi. If p > 2, then .ai b j c/p D aip b jp c p D d k . If p D 2, then .ai b j c/2 D .ai b j /2 c 2 D a2i b 2j Œb; aij c 2 D a2i b 2j d ij a2i b 2j d k D d ij Ck : In any case, we have jT j D p nC2 , a contradiction since jG W T j D p m > p. We have proved that there exists c 2 Z.G/ H such that 1 ¤ c p 2 hd i. Replacing c with another generator of hci, we may assume that c p D d . The structure of G is uniquely determined. Proposition 71.2. Let G be a metacyclic A2 -group of order > p 4 . Then G 0 Š Cp 2 and we have one of the following possibilities: m n m1 , n 1, D 0; 1, m C n 5, (a) G D ha; b j ap D 1, m 3, b p D ap m2 ab D a1Cp i, where in case p D 2, m 4. (b) p D 2, G D ha; b j a8 D 1, b 2 D a4 , D 0; 1, n 2, ab D a1C4 , D 0; 1i. n

Proof. By Corollary 65.3, a metacyclic p-group G is an A2 -group if and only if G 0 Š Cp 2 . Let G0 be the subgroup of order p in G 0 ; then G=G0 is an A1 -group (Lemma

236

Groups of prime power order

65.2(a)). But jG=G0 j > p 3 and so G=G0 is a “splitting” metacyclic group (Lemma 65.1(b,c)). We have G D AB, where A G 0 , A \ B D G0 , A=G0 (of order p 2 ) and B=G0 are cyclic. Since G0 < G 0 < A and G 0 is cyclic, A does not split over G0 . Hence A is a cyclic normal subgroup of order p m , m 3, and we set A D hai. n m1 Let b 2 B be such that hbi covers A=G0 so that b p 2 G0 D hap i, where n p D jB=G0 j, n 1. Changing a generator of hai if necessary, one may assume n m1 b p D ap with D 0; 1. Now, hbi induces an automorphism group of order p 2 on A so that ŒA; B Š Cp 2 . In that case our result follows at once. Proposition 71.3. Let G be a nonmetacyclic A2 -group of order > p 4 possessing exactly one abelian maximal subgroup. Then G 0 Š Ep 2 and assume, in addition, that G 0 6 Z.G/. In that case p > 2, d.G/ D 2, K3 .G/ is of order p, G=K3 .G/ is nonmetacyclic minimal nonabelian, and one of the following holds: (i) G D ha; bi, where n

ap D b p D 1; n 3;

c p D Œa; c D 1;

Œa; b D c;

Œb; c D b p

n1

:

Here jGj D p nC2 , G 0 D hc; b p i Š Ep 2 , K3 .G/ D hb p i, M D ha; c; b p i 2 1 is abelian of type .p; p; p n1 /, and all members of the set 1 fM g are metacyclic. n1

n1

(ii) G D ha; bi, where n

ap D b p D 1; n 2;

Œa; b D c;

Œb; c D d;

c p D d p D Œa; c D Œd; a D Œd; b D 1: Here jGj D p nC3 , G 0 D hc; d i Š Ep 2 , K3 .G/ D hd i, M D ha; c; d; b p i 2 1 is abelian of type .p; p; p; p n1 /. (iii) G D ha; bi, where 2

n

ap D b p D 1; n 2;

Œa; b D c;

Œc; b D asp ;

c p D Œa; c D 1;

and s D 1 or s is a fixed quadratic non-residue mod p. Here jGj D p nC3 , G 0 D hc; ap i Š Ep 2 , K3 .G/ D hap i, M D ha; c; b p i 2 1 is abelian of type .p; p 2 ; p n1 /, and all members of the set 1 are nonmetacyclic. Proof. Lemma 65.2(d,e), Lemma 65.4(c) and Theorem 65.7(c,d) imply that G 0 Š Ep 2 , d.G/ D 2 and p > 2. Since G 0 6 Z.G/, we have jK3 .G/j D p and G=K3 .G/ is a nonmetacyclic A1 -group (Lemmas 65.2(e) and 64.1(m)). By the structure of G=K3 .G/, there are cyclic subgroups A=K3 .G/ > f1g and B=K3 .G/ > f1g of G=K3 .G/ such that hA; Bi D G;

A \ G 0 D B \ G 0 D A \ B D .G 0 A/ \ B D .G 0 B/ \ A D K3 .G/:

71

Determination of A2 -groups

237

Let A D ha; K3 .G/i and B D hb; K3 .G/i. The commutator Œa; b is of order p since ab ¤ ba and exp.G 0 / D p and hŒa; bi \ K3 .G/ D f1g since G=K3 .G/ is nonabelian. Since G 0 6 Z.G/, both A and B cannot centralize G 0 . To fix ideas, assume that B does not centralize G 0 , which implies ŒB; G 0 D K3 .G/. Since G 0 B < G is nonabelian, we must have jG W .G 0 B/j D p. But .G 0 B/ \ A D K3 .G/ implies jAj D p 2 , by the product formula. If also A does not centralize G 0 , then G 0 A is nonabelian and so G 0 A 2 1 . In that case jGj D pjG 0 Aj D p p 3 D p 4 , a contradiction. We have proved that A centralizes G 0 . Let B1 B be such that K3 .G/ B1 and jB W B1 j D p. Let M 2 1 be abelian. Since G 0 < M and G 0 6 Z.G/, we have M D CG .G 0 /. On the other hand, A centralizes G 0 and so A < M . Finally, B1 ˆ.G/ < M and so M D G 0 AB1 (recall that jG W B1 G 0 j D p 2 ) and A 6 B1 G 0 since G D hA; Bi). Since jGj > p 4 and jAG 0 j D p 3 , we have jB1 j p 2 and so jBj p 3 . (i) Suppose first that B does not split over K3 .G/, i.e., B is cyclic. Set B D hbi n1 so that o.b/ D p n with n 3 and hb p i D K3 .G/. The subgroup BG 0 Š Mp nC1 is maximal in G since .BG 0 /A D G, .BG 0 / \ A D K3 .G/ and jAj D p 2 . By Lemma 64.1(p), d.M / > 2 (otherwise, all members of the set 1 are two-generator which is not the case since a nonmetacyclic A1 -group has a maximal subgroup that is not generated by two elements; see Lemma 65.1(a)); moreover, d.M / D 3 (consider M \ .BG 0 /) and so 1 .M / Š Ep 3 . By the product formula, G D B1 .M /. It follows from jABj < jGj and hA; Bi D G that B is not normal in G. Therefore, there is a 2 1 .M / that does not normalize B. One can take A D ha; K3 .G/i. Set Œa; b D c, where c is an element of order p in G 0 K3 .G/. Since BG 0 Š Mp nC1 , n1

one may assume that b c D b 1Cp (if necessary, changing b by another element of order p n in BG 0 ). Thus we have obtained the group (i) of our proposition. (ii) Suppose that B splits over K3 .G/ so that B D K3 .G/ hbi, where o.b/ D p n , n 2. Consider first the subcase (ii1) where A also splits over K3 .G/ so that A D K3 .G/ hai with o.a/ D p. Then Œa; b D c, where c 2 G 0 K3 .G/ is of order p and Œb; c D d , where hd i D K3 .G/. We have obtained the group (ii) of our proposition. (ii2) It remains to consider the subcase where A does not split over K3 .G/ so that A D hai is cyclic of order p 2 and hap i D K3 .G/. We have again Œa; b D c, where c 2 G 0 K3 .G/ is of order p and Œc; b D asp , where s 6 0 .mod p/. For any i 6 0 .mod p/, we replace b with b 0 D b i . Then Œa; b 0 D Œa; b i D Œa; bi D c i D c 0 2 with a suitable element 2 K3 .G/ Z.G/. Hence Œc 0 ; b 0 D Œc i ; b i D Œc; bi D 2 0 ai sp D as p , where s 0 D i 2 s 6 0 .mod p/. It follows that we may assume (running through all i 6 0 .mod p/) that either s 0 D 1 or s 0 is a (fixed) quadratic non-residue mod p. Writing again b; c; s instead of b 0 ; c 0 ; s 0 , we have obtained the groups stated in part (iii) of our proposition. Proposition 71.4. Let G be a nonmetacyclic A2 -group of order > p 4 possessing exactly one abelian maximal subgroup. Then G 0 Š Ep 2 and assume, in addition, that

238

Groups of prime power order

G 0 Z.G/. In that case d.G/ D 3, Z.G/ D ˆ.G/, and one of the following holds: (a) If G has no normal elementary abelian subgroups of order p 3 , then p D 2 and G D ha; b; c j a4 D b 4 D Œa; b D 1; c 2 D a2 ; ac D ab 2 ; b c D ba2 i is the minimal nonmetacyclic group of order 25 . Here G is a special group of exponent 4, 1 .G/ D G 0 D Z.G/ D ˆ.G/ D ha2 ; b 2 i Š E4 , C4 C4 Š M D hai hbi 2 1 is abelian, and all other six members of the set 1 are metacyclic of exponent 4. (b) If G has a normal elementary abelian subgroup E of order p 3 , then E D 1 .G/ and one of the following holds: (b1) G D ha; b; d i with ˛

2

ap D b p D d p D 1; Œa; b D d p a

p ˛1

;

˛ 2; Œa; d D ap

Œd; b D 1;

˛1

;

where D 0 unless p D 2 and ˛ D 2 in which case D 1. Here jGj D p ˛C3 ;

G 0 D hd p ; ap

E D 1 .G/ D hb; d p ; a

˛1

p ˛1

i Š Ep 2 ;

i 6 Z.G/;

and hb; d; ap i 2 1 is abelian of type .p; p 2 ; p ˛1 /. (b2) G D ha; b; ci with ˛

2

2

ap D b p D c p D Œb; c D 1; Œa; b D x;

Œa; c D y;

˛ 1; ˛0

bp D x yˇ ;

c p D x y ı ;

x p D y p D Œa; x D Œa; y D Œb; x D Œb; y D Œc; x D Œc; y D 1; where in case p D 2, ˛ 0 D 0, ˇ D D ı D 1, and in case p > 2, 4ˇ C .ı ˛ 0 /2 is a quadratic non-residue mod p. We have jGj D p ˛C4 , ˛1 G 0 D hx; yi Š Ep 2 , E D 1 .G/ D G 0 hap i. Let A 2 1 be abelian. Then in case ˛ D 1, A D hb; ci Š Cp 2 Cp 2 and E 6 Z.G/ and in case ˛ 2, A D hap ; b; ci is abelian of type .p ˛1 ; p 2 ; p 2 / and E Z.G/. Proof. Let G be a nonmetacyclic A2 -group of order > p 4 having exactly one abelian maximal subgroup. Then we have G 0 Š Ep 2 (Lemmas 64.1(q), 65.4(d) and Theorem 65.7(a)), and we assume here, in addition, that G 0 Z.G/. It follows from Lemma 65.2(a) that d.G/ D 3. Next, Z.G/ D ˆ.G/ (Lemma 65.4(c)), G=Z.G/ Š Ep 3 , and if H ¤ K 2 1 are both nonabelian, then Z.H / D Z.K/ D Z.G/ (Theorem 65.7(f)). It follows that if x; y 2 G, then jhx; yij p o.x/o.y/ (since d.G/ D 3, hx; yi is either abelian or an A1 -subgroup). Also, if G has no normal elementary abelian subgroups of order p 3 , then G is minimal nonmetacyclic (Lemma 65.4(j)). In the last case, p D 2,

71

Determination of A2 -groups

239

jGj D 25 , and G is the group stated in part (a) of our proposition (Lemma 64.1(l) and Theorem 66.1). From now on we assume that the group G has a normal subgroup E Š Ep 3 . By Lemma 65.4(k), either 1 .G/ D E or E < 1 .G/ D K Š Ep 4 . Suppose that the second case occurs. Every maximal subgroup of G containing K must be abelian (Lemma 65.1) and G=K > f1g is cyclic (Lemma 65.4(b)). Let G D Khai; then K \ hai D D is of order p. Indeed, jDj p. Assume that D D f1g. Then, if hai < T 2 1 , then 1 .T / h1 .hai/; T \ Ki is of order p 4 , so T is abelian (Lemma 65.1). Since K 6 T , the set 1 has two distinct abelian members, contrary to the assumption. Thus, p s D jhaij p 2 , D Z.G/, and jGj D p sC3 . Since Khap i is abelian, hap i Z.G/. We have G 0 Š Ep 2 , G 0 < K, and G 0 Z.G/ (by assumption). Since G=Z.G/ Š Ep 3 and G 0 D Z.G/, we get G 0 > D (otherwise, d.G=G 0 / D 2 < 3 D d.G/). Therefore, G 0 hai is an abelian normal subgroup of G of type .p s ; p/, s 2, and Z.G/ D G 0 hap i. We have CK .a/ < Z.G/ and so CK .a/ D G 0 and CG .a/ D G 0 hai. The subgroup G 0 hai has exactly p cyclic subgroups of order p s . Since jG W .G 0 hai/j D p 2 , we get NK .hai/ > G 0 . Let x 2 NK .hai/ G 0 so that x (of order p) induces an automorphism of order p on hai. But then ha; xi is a nonabelian subgroup of order p sC1 so its index in G equals p 2 , a contradiction. We have proved that 1 .G/ D E Š Ep 3 . Since G 0 < E, the quotient group G=E is abelian. Since d.G=G 0 / D 3, G=E is also noncyclic. In what follows A denotes the unique abelian maximal subgroup of G. (i) First assume that E 6 ˆ.G/ .D Z.G//; then d.G=E/ D 2. If A does not contain E, then G D AE, 1 .A/ D A \ E D G 0 Š Ep 2 and so A is metacyclic. Hence each maximal subgroup of G is generated by two elements (Lemma 65.1) and d.G/ D 3. Since G is of class 2 (by hypothesis), Theorem 70.1 shows that G is isomorphic to a group (b2) for ˛ D 1 in our proposition. We may assume that E < A and so each maximal subgroup of G which does not contain E is nonabelian metacyclic (Lemma 65.1, since all members of the set 1 fAg are A1 -groups). Let M be a nonabelian maximal subgroup of G containing E (there are exactly p such subgroups since d.G=E/ D 2). Then M is a nonmetacyclic A1 group with Z.M / D ˆ.M / D ˆ.G/ D Z.G/ (Theorem 65.7(f)). By Lemma 65.1, we have, since E 6 Z.M / (otherwise, E Z.G/, which is not the case since d.G=E/ D 2 < 3 D d.G=Z.G//), ˛

M D ha; b j ap D b p D 1; ˛ 2; Œa; b D c; c p D Œa; c D Œb; c D 1i; where hci D M 0 ; Z.M / D Z.G/ D hci hap i;

b 2 E G0; jGj D p ˛C3 ;

M D hbi .hci hai/; G 0 D hap

˛1

i hci:

In particular, G=E has a cyclic subgroup M=E of index p. Now, ha; ci . G 0 / is an abelian normal subgroup of type .p ˛ ; p/ and it has exactly p cyclic subgroups of

240

Groups of prime power order

index p which are not normal in M (see the last assertion of Lemma 65.1). Therefore, NM .hai/ D ha; ci and so N D NG .hai/ must cover G=M since N 6 M in view of jG W N j D c˛ .ha; ci/ D p D jG W M j (here c˛ .X/ is the number of cyclic subgroups of order p ˛ in a p-group X). Thus N 2 1 with M \ N D ha; ci and so E \ N D G 0 . Therefore N is nonabelian metacyclic, CG .a/ D ha; ci, N is an A1 group, and so it is “splitting” metacyclic (Lemma 65.1). We have f1g < N 0 < hai\G 0 ˛1 and so N 0 D hap i Ã1 .N / (Lemma 65.1). It follows that Z.G/ D hap ; ci D Z.N / D ˆ.N / D Ã1 .N /, and so there exists d 2 N M D N ha; ci such that ˛1 d p 2 G 0 hap i and o.d / D p 2 (note that 1 .N / ha; ci). We have N D ha; d i and N=hai is cyclic. Next, CG .E/ D A. ˛2 If ˛ > 2, then working in the abelian subgroup hd; ap i Š Cp 2 Cp 2 , we may choose an element d 0 2 hd; ap i M such that .d 0 /p D c and so we may assume from the start that d p D c D Œa; b. If u 2 G is of order p 2 , then jhu; Eij D p 4 so hu; Ei is abelian, and we conclude that u 2 CG .E/ D A. It follows that 2 .G/ D Ehu 2 G j o.u/ D p 2 i A so Œd; b D 1. Replacing d with d 0 D d i (i 6 0 0 i ˛1 .mod p/), we may assume that ad D ad D a1Cp . But then we also replace b with b 0 D b i and c with c 0 D c i and obtain .d 0 /p D .d p /i D c i D c 0 D Œa; b 0 and the relation Œb 0 ; d 0 D 1 remains valid. Writing again d; b; c instead of d 0 ; b 0 ; c 0 , respectively, we obtain exactly the relations stated in part (b1) for ˛ > 2 of our proposition. It remains to investigate the case ˛ D 2, where jGj D p 5 , G 0 D Z.G/ D ˆ.G/ Š Ep 2 , E D 1 .G/ Š Ep 3 , exp.G/ D p 2 . For each x 2 N ha; ci, x p 2 G 0 hap i (otherwise, ha; xi would be nonabelian of order p 3 ; here N D NG .hai). Since G D EN , we have A D E.A \ N /, by the modular law. Note that A \ N is abelian of type .p 2 ; p/ so it is generated by elements of order p 2 . Since N D hai.A \ N / (indeed, M D haiE is nonabelian so a 62 CG .E/ D A), there exists d 2 .A \ N / 1 .A \ N / such that ha; d i is nonabelian. It follows that d p 62 hap i (otherwise, ha; d i is nonabelian of order p 3 , contrary to the fact that G is an A2 -group). Replacing d with another generator of hd i if necessary, we may assume that ad D a1Cp . Set d p D c i ajp , where i 6 0 .mod p/. We replace b with b 0 D b i and c with c 0 D c i , so that we get Œa; b 0 D Œa; b i D Œa; bi D c i D c 0 and d p D c 0 ajp . Thus, we may assume from the start that Œa; b D c and c D ajp d p with j 2 Z. We shall determine the integer j .mod p/. Consider the nonabelian maximal subgroup S D hb; ad i, where is any integer (recall that d.G/ D 3). The subgroup S is a nonmetacyclic (otherwise, b 2 Ã1 .S / D ˆ.S /) A1 -group containing E with Z.S / D Z.G/ D G 0 D hap ; d p i D h.ad /p ; Œad ; bi. If p > 2, then ˛2

.ad /p D ap .d p / ;

Œad ; b D Œa; b D c D ajp d p ;

ˇ ˇ ˇ ˇ and so we must have (since the displayed two elements are independent) ˇ j1 1 ˇ 6 0 .mod p/, which gives 1 j 6 0 .mod p/ for every integer . If j 6 0 .mod p/, then we can solve the congruence j 1 .mod p/ for and get a contradiction.

71

Determination of A2 -groups

241

Hence j 0 .mod p/ and we may set in this case j D 0; then c D ajp d p D d p . If p D 2, then taking D 1, we have .ad /2 D a2 d 2 Œd; a D a2 d 2 a2 D d 2 ;

Œad; b D Œa; b D c D a2j d 2 ˇ ˇ (by the above, ad D a1Cp for any p), and we must have ˇ j0 11 ˇ 6 0 .mod 2/, which gives j 6 0 .mod 2/ and so we may set in this case j D 1; then c D a2 d 2 . We have obtained the relations stated in part (b1) for ˛ D 2 of our proposition. (ii) Suppose that Ep 3 Š E D 1 .G/ Z.G/ D ˆ.G/. Let H 2 1 be nonabelian. Then E < H and so H is a nonmetacyclic A1 -group: ˛

ˇ

H D ha; b j ap D b p D 1; ˛1

ˇ1

Œa; b D c;

c p D Œa; c D Œb; c D 1i;

where E D hap ; b p ; ci. Since E Z.G/ D Z.H / D ˆ.H / (Theorem 65.7(f)), we must have ˛ 2 and ˇ 2 so jGj D pjH j D p ˛Cˇ C2 p 6 . Let G0 < G 0 be of order p such that G0 ¤ H 0 . Since .G=G0 /0 D G 0 =G0 is of order p and d.G=G0 / D 3, a nonabelian group G=G0 is an A2 -group of order p 5 with the commutator group of order p, where H=G0 is an A1 -subgroup. By Proposition 71.1, there exists d 2 G H such that d p 2 G 0 . Hence o.d / D p 2 and d 62 Z.G/ since d 62 H and 1 .G/ D E < H . It follows that d does not centralize a or b. Without loss of generality, we assume that Œd; a ¤ 1 and so nonabelian hd; ai 2 1 in view of d.G/ D 3. Since o.d / D p 2 and o.a/ D p ˛ , we have jhd; aij p ˛C2C1 in view of o.Œd; a/ D p, and therefore jGj p ˛C4 . Hence jGj D p ˛Cˇ C2 p ˛C4 and so ˇ 2. This gives ˇ D 2 and jGj D p ˛C4 with ˛ 2. We have Z.G/ D Ehap i. Assume first that ˛ > 2; then jGj > p 6 . In that case, any two elements x; y 2 G of orders p 2 commute (otherwise, hx; yi 2 1 , since d.G/ D 3, is nonabelian with jhx; yij p o.x/o.y/ p 5 , a contradiction). Since Œb; d D 1 in view of o.b/ D o.d / D p 2 , we get hb; d i D hbi hd i. Assume that this is false. Then hb; d i D hbi hd1 i with o.d1 / D p. Since b 2 H , d 62 H , we get d1 62 H , a contradiction: 1 .G/ D E < H . Consider K D ha; d i, where K 2 1 is nonabelian containing E (by assumption, E < ˆ.G/). Let G0 ¤ K 0 be a subgroup of order p in G 0 . Then applying Proposition 71.1 on G=G0 , we see that there is b0 2 G K with b0p 2 G 0 and o.b0 / D p 2 . Since Œd; b0 D 1 and there are no elements of order p in G K, we see (as before) that hd; b0 i D hd i hb0 i. But G D ha; d; b0 i, b0 62 Z.G/, and so Œa; b0 ¤ 1. Replacing H with H0 D ha; b0 i (¤ K) and b with b0 , we may assume from the start that b p 2 G 0 . Hence, we may assume that Ã1 .hb; d i/ D 1 .hb; d i/ D G 0 . Since jGj D p ˛C4 and GN D G=G 0 is generated N dN ; aN (bar convention) of orders at most p; p; p ˛ , respectively, and jGj N D p ˛C2 , by b; ˛ 0 we get o.a/ N D p which gives hai \ G D f1g and so hai \ hb; d i D f1g since 1 .hb; d i/ D G 0 . Suppose that ˛ D 2; then jGj D p 6 , exp.G/ D p 2 , E D Z.G/ D ˆ.G/ D 1 .G/. Let X < G 0 be of order p. Since d.G=X/ D 3 and .G=X/0 is of order p, G=X is an A2 -group of order p 5 , and so we may apply again Proposition 71.1. There is M 2 1

242

Groups of prime power order

such that M=X is an A1 -group and there is c 2 G M such that c p 2 G 0 and ŒG; c X. If c p 2 X, then hci (of order p 2 ; recall that 1 .G/ < M ) is normal in G. Since c 62 Z.G/ D E, there is an element l of order p 2 with Œl; c ¤ 1 and so hŒl; ci D hc p i. But then jhl; cij o.l/o.c/ p 4 , a contradiction. Thus c p 2 G 0 X. We consider now the subgroup Y D hc p i of order p in G 0 and apply Proposition 71.1 to G=Y . As before, there is an element b 2 G such that b p 2 G 0 Y . It follows hb p ; c p i D G 0 . The abelian group hb; ci=G 0 is generated by two elements of order p and so jhb; cij p 4 . Thus hb; ci is abelian and since hbi \ hci D f1g, we have S D hb; ci Š Cp 2 Cp 2 and S is normal in G. But, E D ˆ.G/ D G 0 Ã1 .G/ and so there is a 2 G such that ap 2 E G 0 . Hence hai \ S D f1g and so S hai D G since o.a/ D p 2 . In any case (for all ˛ 2) we have obtained the following configuration (where in ˛ case ˛ > 2 we write c instead of d ). We have G D hai .hbi hci/, with ap D 1, 2 2 ˛ 2, b p D c p D 1, Œb; c D 1, where G is a semidirect product with kernel hbi hci Š Cp 2 Cp 2 , G 0 D hb p ; c p i, and A D hap ; b; ci is the unique abelian maximal subgroup of G (of type .p ˛1 ; p 2 ; p 2 /). p Set Œa; b D x, Œa; c D y. Then G 0 D hx; yi D hb p ; c p i ( Z.G/) and ˇ 0so ˇb D 0 ˇ ˇ x ˛ y ˇ and c p D x y ı , where, since b p and c p are independent, we get ˇ ˛ ˇı ˇ 6 0 .mod p/. The subgroup ha; b c i is nonabelian if and only if Œa; b c D x y ¤ 1 which occurs if and only if both and are not multiples of p. In that case L D ha; b c i is a nonmetacyclic A1 -subgroup of index p in G. Since hx y i D L0 and L0 is a 0 p p maximal cyclic subgroup in L, we have G ˇ 0 D h.b c / ˇ; x y i, where .b c / D 0 ˇ ˇ ˇ Cı .b p / .c p / D x ˛ C y ˇ Cı . Hence, ˇ ˛ C ˇ 0 .mod p/ only if

0 .mod p/. This gives (1)

ˇ 2 C .ı ˛ 0 / 2 0 .mod p/

only if 0 .mod p/. From (1) follows (setting D 1, D 0 or D 0, D 1) ˇ 6 0 .mod p/ and 6 0 .mod p/. Suppose first p > 2. Then we compute (using (1)) .2ˇ C.ı˛ 0 //2 .4ˇ C.ı˛ 0 /2 /2 D 4ˇ.ˇ 2 C.ı˛ 0 / 2 / 0 .mod p/ only if 0 .mod p/. Hence (2)

.2ˇ C .ı ˛ 0 //2 .4ˇ C .ı ˛ 0 /2 /2

.mod p/

only if 0 .mod p/. From (2) follows at once that 4ˇ C .ı ˛ 0 /2 is a quadratic non-residue mod p. Indeed, if 4ˇ C .ı ˛ 0 /2 is a quadratic residue modulo p, then we set D 1 in (2) and then we solve the congruence .2ˇ C .ı ˛ 0 //2

4ˇ C .ı ˛ 0 /2 .mod p/ for and get a contradiction.

71

Determination of A2 -groups

243

Suppose now that p D 2. We know already that ˇ D D 1. Then (1) yields (3)

2 C .ı C ˛ 0 / C 2 0

.mod 2/

only if 0 .mod 2/. Setting D D 1 in (3), we get ı C ˛ 0 D 1. This gives 0 0 b 2 D x ˛ y and c 2 D xy ˛ C1 , where ˛ 0 D 0; 1. Hence we have the following two possibilities: (4)

b 2 D y;

c 2 D xy;

(5)

b 2 D xy;

c 2 D x:

However, interchanging b and c and also x and y, we see that the relations Œa; b D x and Œa; c D y remain unchanged but (4) goes onto (5). Hence, we may assume that we have relations (4) and so we get ˛ 0 D 0 and ˇ D D ı D 1. Our group G is uniquely determined. Proposition 71.5. Let G be a nonmetacyclic A2 -group of order > p 4 all of whose maximal subgroups are nonabelian. Then we have one of the following possibilities: (a) d.G/ D 3, p D 2, and G D ha; b; ci with a4 D b 4 D c 4 D 1;

Œa; b D c 2 ;

Œa; c D b 2 c 2 ;

Œb; c D a2 b 2 ;

Œa2 ; b D Œa2 ; c D Œb 2 ; a D Œb 2 ; c D Œc 2 ; a D Œc 2 ; b D 1; where G 0 D ha2 ; b 2 ; c 2 i D Z.G/ D ˆ.G/ D 1 .G/ Š E23 , exp.G/ D 4 and all members of the set 1 are nonmetacyclic A1 -groups. (From Theorem 70.4 follows that G is isomorphic to the Suzuki 2-group of order 26 .) (b) d.G/ D 2, p > 2, G is of order p 5 : 2

2

G D ha; x j ap D x p D 1; Œa; x D b; Œa; b D y1 ; Œx; b D y2 ; b p D y1p D y2p D Œa; y1 D Œx; y1 D Œa; y2 D Œx; y2 D 1; ap D y1˛ y2ˇ ; x p D y1 y2ı i; where in case p > 3, 4ˇ C .ı ˛/2 is a quadratic non-residue mod p. Here G 0 D hb; y1 ; y2 i D 1 .G/ Š Ep 3 , Z.G/ D K3 .G/ D Ã1 .G/ D hy1 ; y2 i Š Ep 2 . Proof. Let G be a nonmetacyclic A2 -group of order > p 4 all of whose maximal subgroups are nonabelian and therefore they are generated by two elements. Suppose first that d.G/ D 3. Then G 0 Z.G/ (Lemma 65.4(a)). By Theorem 70.1, G is a uniquely determined group of order 26 as stated in part (a) of our proposition. Now suppose that d.G/ D 2. If p D 2, then G is metacyclic (Lemma 64.1(n)), a contradiction. Hence p > 2; then Ã1 .G/ D K3 .G/, jG W G 0 j D p 2 (Lemma 64.1(m))

244

Groups of prime power order

so jG 0 j D p 3 and jGj D p 5 (Lemma 65.2(d)). Since G=K3 .G/ is generated by its subgroups of index p 2 , K3 .G/ Z.G/ (Lemma 65.4(c)) and so G is of class 3. We have K3 .G/ D Z.G/ Š Ep 2 and G 0 Š Ep 3 (Lemma 65.7(d)). Each M 2 1 is a nonmetacyclic A1 -group with G 0 D 1 .M / Š Ep 3 and so ˆ.G/ D G 0 D 1 .G/ and exp.G/ D p 2 (Lemma 65.1). Let G D ha; xi; then a; x 2 G ˆ.G/ D G 1 .G/ and therefore o.a/ D o.b/ D p 2 . We have G 0 D hŒa; x; K3 .G/i and so Œa; x D b 2 G 0 K3 .G/ which gives o.b/ D p. Set Œa; b D y1 and Œx; b D y2 , where y1 ; y2 2 K3 .G/ D Z.G/. But ha; biZ.G/ and hx; biZ.G/ are members of the set 1 and so they are nonabelian. Hence o.y1 / D o.y2 / D p (Lemma 65.1) and ha; biZ.G/ D ha; bi, hx; biZ.G/ D hx; bi. Since ha; bi and hx; bi are distinct members of the set 1 , we have hy1 i ¤ hy2 i (Lemma 65.2(d)). It follows hy1 ; y2 i D Z.G/. Since ap ; x p 2 Z.G/, we get ap D y1˛ y2ˇ and x p D y1 y2ı . Suppose that p > 3. Then G is regular (Lemma 64.1(a)) and since G 0 is elementary abelian, we get that G is p-abelian, i.e., .lk/p D l p k p for all l; k 2 G (see 7). Since ha; xi D G, we have in this case hap ; x p i D Ã1 .G/ D Z.G/ D hy1 ; y2 i. Consider the subgroup S D hb; a x i (S < G since b 2 G 0 ; it follows that S is of class 2). Obviously, a x 2 G G 0 if and only if not both and are divisible by p. In that case Œa x ; b D Œa; b Œx; b D y1 y2 ¤ 1, and so S is nonabelian; then S 2 1 hence E < S . It follows that S is nonmetacyclic. Here we have used the fact that Œa; b; Œx; b 2 Z.G/. Since Z.G/ D Z.S / and S 0 is a maximal cyclic subgroup of S , we have Z.S / D Z.G/ D hy1 ; y2 i D h.a x /p ; Œa x ; bi. We compute ˛C ˇ Cı y2

.a x /p D .ap / .x p / D y1 and so ()

and Œa x ; b D y1 y2 ;

ˇ ˇ ˇ˛ C ˇ C ıˇ ˇ ˇ 0 .mod p/ ˇ ˇ

only if 0 .mod p/. From () we get (as in the proof of Proposition 71.4) that 4ˇ C .ı ˛/2 is a quadratic non-residue modulo p. Our proposition is proved and all A2 -groups are determined. The classification of A2 -groups is contained in Propositions 71.1–71.5. The subgroup structure of A2 -groups is given in 65. It is impossible to compare our proof with Kazarin’s proof since the last one is not published and only sketched in his PhD Thesis. The paper [She], also devoted to A2 -groups, is not accessible. In any case, our proof is based on other ideas. Indeed, the above authors were forced to use tables [HS]. Besides, Kazarin used NazarovaRoiter’s classification of p-groups with abelian subgroup of index p so his proof is not elementary. Instead of this, we use Theorem 70.1 which is new for p D 2. Exercise 1. Let a p-group G D A C , where C Š Cp 2 , is an A2 -group. Describe the intersection A \ C . Consider in detail the case when A is metacyclic.

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245

Exercise 2. Classify the metacyclic A1 -groups G of order 2n such that there is an involution x 2 G which is not square. Exercise 3. Find the A2 -groups G such that jH G W H j D p for all nonnormal H < G. Exercise 4. Find the A2 -groups G such that exp.H G / D exp.H / for all H < G. The following three theorems are due to the first author. Theorem 71.6 ([BJ2]). Let G be a nonmetacyclic two-generator 2-group. Then the number of two-generator members of the set 1 is even. Proof. By hypothesis, G is nonabelian. If A 2 1 , then Ã2 .G/ Ã1 .A/ D ˆ.A/ so d.A=Ã2 .G// D d.A/. Therefore, without loss of generality, one may assume that Ã2 .G/ D f1g; then exp.G/ D 4. Let 1 .G/ D fA; B; C g. Since G is not metacyclic, one may assume that d.C / D 3 (Appendix 25 and Theorem 44.5). Assume that the theorem is false. Then we may assume that d.A/ D 2 and d.B/ D 3 (see Appendix 25). Let R be a G-invariant subgroup of index 2 in G 0 . Then G=R is minimal nonabelian (Lemma 65.2(a)) and nonmetacyclic (Theorem 36.1). Since exp.G=R/ D 4, we get jG=Rj 25 (Lemma 65.1). If jG=Rj D 25 , the set 1 has no two-generator members (Lemma 65.1), contrary to the assumption. Thus, jG=Rj D 24 so jG=G 0 j D 8. One may assume that C =R D 1 .G=R/. Since B=R is abelian of type .4; 2/ so two-generator and d.B/ D 3, we get R ¤ ˆ.B/. Clearly, ˆ.B/ is a G-invariant subgroup of index 8 in B. Write GN D G=ˆ.B/; then GN is not of maximal N D 2. Since jG=G 0 j D 8, we get class since it contains BN Š E8 . It follows that cl.G/ jG 0 W ˆ.B/j D 2. Since G 0 has only one G-invariant subgroup of index 2 (Remark 36.1), we conclude that ˆ.B/ D R. This is a contradiction since B=R is abelian of type .4; 2/ and exp.B=ˆ.B// D 2. Thus, the number of two-generator members of the set 1 is even, as was to be shown. Theorem 71.7. Suppose that all nonabelian maximal subgroups of a 2-group G are metacyclic and jGj > 24 . Then one of the following holds: (a) G is abelian. (b) G is metacyclic. (c) G is an A1 -group. (d) G is minimal nonmetacyclic. (e) G D M C , where M is a metacyclic A1 -group and jC j D 2. Proof. Assume that G is not of types (a)–(d). Then, by Theorem 71.6, d.G/ > 2 and there is in G a maximal subgroup A that is not two-generator; in that case, by hypothesis, A is abelian. By assumption, there is in G a nonabelian maximal subgroup M ; then M is metacyclic so d.G/ D 3. Since M \A is a metacyclic maximal subgroup of A, we get d.A/ D 3. Write E D 1 .A/; then E G G and E Š E8 . By hypothesis, all maximal subgroups of G that contain E, are abelian. Since G has at most three abelian maximal subgroups, we get d.G=E/ 2. (i) Suppose that d.G=E/ D 2. Let B=E be a maximal subgroup of G=E and B ¤ A. Then B is abelian so A \ B D Z.G/ has index 4 in G; in that case, jG 0 j D 2 (Lemma 64.1(q)) and E Z.G/. It follows from jM 0 j D 2 and d.M / D 2 that M is

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an A1 -group. Let C < E be a subgroup of order 2 not contained in 1 .M / .Š E4 /; then G D M C is as stated in (e). (ii) Now let G=E be cyclic; then E 6 Z.G/ since G is nonabelian. Since d.G/ D 3, we get jG 0 j D 2. We have E D 1 .G/ since 1 .G/ < A in view of G=E is cyclic of order > 2. Let, as above, M is a nonabelian maximal subgroup of G. By Lemma 65.2(a), M is an A1 -subgroup; moreover, M Š M2n , n > 3, since 1 .M / D M \ E Š E4 and M=1 .M / Š G=E is cyclic of order > 2, and we conclude that Z.M / is cyclic. If Z.G/ is noncyclic, then G D M C , where jC j D 2, and G is as stated in (e). Next assume that Z.G/ is cyclic. We get jG W Z.G/j D 2jG 0 j D 4 (Lemma 64.1(q)). Then, by the product formula, G D EZ.G/ so G is abelian, a final contradiction. Theorem 71.8. Suppose that a two-generator nonabelian p-group G contains an abelian maximal subgroup A. Set R D hx p j x 2 G Ai. Then R Z.G/ and G=R is of maximal class, unless G is an A1 -group. Proof. Recall that if A is an abelian maximal subgroup of a nonabelian p-group G and jG W G 0 j D p 2 , then G is of maximal class (this is proved by induction with help of Lemma 64.1(q)). We have R Z.G/ \ ˆ.G/ since CG .x p / hx; Ai for all x 2 G A. Write GN D G=R; then GN is noncyclic. Since all elements of the set GN AN N hGN Ai N D G. N It follows that GN 0 D ˆ.G/ N so have the same order p, we get 1 .G/ N GN 0 Š Ep 2 since d.G/ N D d.G/ D 2. Suppose that GN 0 D f1g. N Then R D ˆ.G/ that G= so G=Z.G/ Š Ep 2 . In that case, all maximal subgroups of G are abelian so G is an N then GN is nonabelian. In that case, as we have A1 -group. Now suppose that GN 0 > f1g; noticed, GN is of maximal class.

Problems Problem 1. Classify the p-groups G such that, whenever H < G is minimal nonabelian and H Š H1 < G, then H1 D H . Problem 2. Classify the p-groups all of whose A2 -subgroups are isomorphic. Moreover, study the p-groups all of whose A1 -subgroups of the same order are isomorphic. Problem 3. Classify the non-Dedekindian p-groups all of whose nonnormal subgroups are abelian. Problem 4. Classify the p-groups all of whose A2 -subgroups are metacyclic. Problem 5. Let G be a p-group all of whose maximal subgroups are A2 -groups. Is it true that jGj is bounded? Problem 6. Classify the p-groups with at most two A2 -subgroups. (Note that pgroups with exactly one A2 -subgroup are not classified.)

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247

Problem 7. Describe the automorphism groups of all A1 - and A2 -groups. Problem 8. Study the p-groups G such that, whenever A; B 2 1 are distinct, then A \ B is either abelian or minimal nonabelian. Problem 9. Study the p-groups all of whose A2 -subgroups are not two-generator. Problem 10. Study the subgroup structure of A A, where A is an A1 -group. Problem 11. Classify the p-groups G which are lattice isomorphic with B, where B is (i) an A1 -group, (ii) an A2 -group. Problem 12. Study the p-groups G such that, whenever A < G is an A1 -subgroup, then A < B G, where B is an A2 -subgroup. Problem 13. Classify the p-groups of exponent > p all of whose A2 -groups have exponent p (then these subgroups have order p 4 ). Problem 14. Study the p-groups G such that ˆ.G/ is an A1 -group (A2 -group). Problem 15. Study the p-groups G all of whose subgroups of fixed order p n are A2 groups. Is it possible to estimate jGj in terms of n? Problem 16. Classify the p-groups with minimal nonabelian subgroup of index p.

72

An-groups, n > 2

In this section we study An -groups with n > 2. A p-group G is said to be an An group if all its subgroups of index p n are abelian but G has a nonabelian subgroup of index p n1 . Main results of this section are due to the first author. 1o . First we classify An -groups G with G 0 Š Cp n , n > 1 (for n D 2, see Corollary 65.3). Proposition 72.1. (a) Let G be a metacyclic p-group with jG 0 j D p n . Then G is an An -group. (b) Let G be an An -group, n > 1, with cyclic derived subgroup G 0 of order p n . Then G is metacyclic and jG 0 j D p n . Proof. We use induction on jGj. (a) If n D 1, then G is an A1 -group (Lemma 65.2(a)). Now suppose that n > 1. Let L D Ã1 .G 0 /. ˆ.G//. By Lemma 65.2(a), G=L is minimal nonabelian so, if H 2 1 , then H=L is abelian, and we have H 0 L. By induction, H is an As -group with s < n. By Lemma 64.1(u), there is F 2 1 with jG 0 W F 0 j D p so F is an An1 -group since jF 0 j D p n1 . It follows that G is an An -group. (b) Set jG 0 j D p s , s n. In view of Lemma 65.2(a,b) and Theorem 65.7(a), one may assume that n > 2. Take H 2 1 . If H 0 D G 0 , then, by induction, H is an As group, a contradiction since s n. Therefore, if G 0 D hŒx; yi, then G D hx; yi, i.e., d.G/ D 2. By Lemma 64.1(u), there is F 2 1 with jG 0 W F 0 j D p so, by induction, F is an As1 -group and so s 1 n1 since G is an An -group. It follows that s D n and F is metacyclic, by induction since it is an An1 -group with cyclic F 0 of order p n1 > p. Let L D Ã1 .G 0 / and T D Ã2 .G 0 /. If H 2 1 , then H=L is abelian since G=L is an A1 -group (Lemma 65.2(a)). By Lemma 65.1, G=T is not an A1 -group. Then G=T has a maximal subgroup A=T that is nonabelian so .A=T /0 > f1g. Since A0 and G 0 are cyclic, we get T < A0 so, by the above, jA0 j D p n1 . By induction, A is metacyclic. Then A=T is an A1 -group (Lemma 65.2(a)). Thus, every maximal subgroup of G=T is either abelian or an A1 -group so G=T is an A2 -group. Since j.G=T /0 j D p 2 , G=T is metacyclic, by Lemma 65.7(a). Now, by Lemma 64.1(o), G is metacyclic since T Ã2 .G/.

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Let G be a nonabelian metacyclic group of order p m and p k D min fjAj j A < G; A0 > f1gg. Then G is an Am.k1/ -group (indeed, all subgroups of G of order p k1 are abelian and G has a nonabelian subgroup A of order p k ) so jG 0 j D p m.k1/ (Proposition 72.1(a)) whence jG=G 0 j D p k1 . In that case, G 0 < Z < G, where Z is cyclic of index p k2 in G. In particular, if k D 3, then G has a cyclic subgroup of index p 32 D p so it is either a 2-group of maximal class or p > 2 and m D 3 (Lemma 64.1(t)). The last result coincides with Proposition 10.19. Let p > 2 and suppose that a metacyclic p-group G of order p m > p 4 contains a normal nonabelian subgroup H of order p 4 and exponent p 2 . Then H D 2 .G/ so all subgroups of G of order p 3 are abelian. It follows that G is an Am3 -group so jG 0 j D p m3 (Proposition 72.1) and G has a normal cyclic subgroup Z of index p 2 , G 0 < Z, such that G=Z is cyclic. We have Z.H / Š Ep 2 . Let L be a subgroup of order p in Z.H /, L ¤ H 0 . Then jG W CG .L/j p. Since CG .L/=L contains a nonabelian subgroup H=L of order p 3 , we get CG .L/=L D H=L (Lemma 64.1(t)). In that case, jG W H j D p so jGj D pjG W H j D p 5 , and Z.H / 6 Z.G/ so Z.G/ is cyclic. Let G be a metacyclic p-group and Mp 4 Š H < G. In that case, G has no cyclic subgroup of index p (Lemma 64.1(t)). Then G is an Am3 -group so jG 0 j D p m3 and G has a normal cyclic subgroup Z ¤ ˆ.G/ of index p 2 . Exercise. Describe the nonabelian metacyclic 2-groups of order 2m > 25 without normal subgroups of order 24 and exponent 4. Solution. Suppose that G has no cyclic subgroups of index 2 (otherwise, G satisfies the condition). Then 1 .G/ Š E4 . Write GN D G=1 .G/. Then GN has no normal abelian subgroups of type .2; 2/ so it is of maximal class (if GN is cyclic, then G has a cyclic subgroup of index 2 so Š M2m ). Suppose that GN is dihedral. By Proposition N D G, N it follows that 10.17, all subgroups of G of order 8 are abelian. Since 1 .G/ 1 .G/ Z.G/. If G=1 .G/ is of maximal class, then G has no normal subgroups of order 24 and exponent 4 since any such subgroup must contain 1 .G/. In the case under consideration, G is a U2 -group (see 67). 2o . In this subsection and subsection 3o we estimate jG 0 j, where G is an An -group, n D 3; 4. If a p-group G is an A3 -group and U; V 2 1 are distinct, then jG 0 j pjU 0 V 0 j p 7 (Lemmas 65.2(d) and 64.1(u)). Proposition 72.2 yields essentially better estimate, and that estimate is best possible. Proposition 72.2. If a p-group G is an A3 -group, then jG 0 j p 4 . For p D 5, there exists an A3 -group G with jG 0 j D 54 . Proof. Assume that jG 0 j > p 4 . Take U 2 1 . Then jU 0 j p 3 and, if jU 0 j D p 3 , then U 0 Š Ep 3 (Theorem 65.7(a,d)). (i) We claim that G 0 is regular. This is the case if jG W G 0 j > p 2 since then G 0 is abelian. Let jG W G 0 j D p 2 and let G 0 be nonabelian; then p > 2 (Taussky’s theorem)

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and G 0 is an A1 -group. In that case, G 0 is regular since cl.G/ 2 (Lemmas 65.1 and 64.1(a)); moreover, G 0 is metacyclic (Theorem 44.12). (ii) We claim that if H G, then jH=Ã1 .H /j p 5 , and this inequality is strong if H < G. Indeed, there is an A2 -subgroup A 2 1 . Then jA=Ã1 .A/j p 4 (Lemma 65.4(a)) so jG=Ã1 .G/j p 5 . Take H < G such that jH=Ã1 .H /j p 5 . By Lemmas 65.1 and 65.4(b), H is abelian so 1 .H / is elementary abelian of order p 5 . Then all members of the set 1 containing 1 .H /, are abelian (Lemmas 65.1 and 65.4(b)). By assumption, jG 0 j > p 4 > p so H is the unique member of the set 1 containing 1 .H / (Lemma 65.2(c)), and we conclude that G=1 .H / is cyclic whence G 0 < 1 .H /. Considering H \ F , where F 2 1 is nonabelian, we get j1 .H /j p 5 so jG 0 j < p 5 , a contradiction. (If G D S.p 3 / Ep 2 , where S.p 3 / is nonabelian of order p 3 and exponent p > 2, then G is an A3 -group, jGj D p 5 and exp.G/ D p.) (iii) Assume that jG 0 j > p 5 . Let U; V 2 1 be distinct; then jU 0 V 0 j p1 jG 0 j p 5 (Lemma 64.1(u)) so one may assume that jU 0 j D p 3 ; then jV 0 j p 2 . If jV 0 j D p 3 , then exp.U 0 V 0 / D p since exp.U 0 / D exp.V 0 / D p (Theorem 65.7(d), Lemma 64.1(a) and (i)), and jU 0 V 0 j p 5 , contrary to (ii). Thus, jV 0 j D p 2 ; then U 0 \ V 0 D f1g, by the product formula. By (ii), since jU 0 V 0 j p 5 , we get exp.U 0 V 0 / > p so V 0 is cyclic; then V is metacyclic (Theorem 65.7(a)), a contradiction since Ep 3 Š U 0 < G 0 ˆ.G/ < V . Thus, jG 0 j D p 5 . Then the set 1 has no abelian members, by Lemmas 65.2(d) and 64.1(u) (otherwise, if A 2 1 is abelian and H 2 1 fAg, then jH 0 j p 3 so jG 0 j pjA0 H 0 j p 4 , contrary to the assumption). (iv) Assume that there are distinct U; V 2 1 such that exp.U 0 / D exp.V 0 / D p. Then, using (i) and (ii), we get: U 0 V 0 is of order p 4 and exponent p so U 0 V 0 < G 0 , and exp.G 0 / D p 2 since jG 0 j D p 5 . In that case, U=U 0 V 0 and V =U 0 V 0 are cyclic (Lemma 65.4(b)). Thus, the nonabelian group G=U 0 V 0 has two distinct cyclic subgroups U=U 0 V 0 and V =U 0 V 0 of index p so it is either ordinary quaternion or Š Mp n (Lemma 64.1(t)). In the first case jG W G 0 j D 4 so G is a 2-group of maximal class (Lemma 64.1(s)), a contradiction. Assume that G=U 0 V 0 Š Mp n . Let H=U 0 V 0 be the noncyclic subgroup of index p in G=U 0 V 0 . In that case, H is neither A1 - nor A2 -group (Lemmas 65.1 and 65.4(b)) so H is abelian, a contradiction (see the last sentence of part (iii)). Now we are ready to complete the proof. By (iv), there is V 2 1 such that exp.V 0 / > p; then V is metacyclic and V 0 is cyclic of order p 2 (Theorem 65.7(d,a)). Take U 2 1 fV g. Since U 0 < V , we get j1 .U 0 /j p 2 so jU 0 j p 2 (Lemma 65.1 and Theorem 65.7(a,d)). Then U 0 \ V 0 D f1g (Lemmas 65.2(d) and 64.1(u)). Since U 0 V 0 D U 0 V 0 < V and V is metacyclic, U 0 Š Cp 2 (Corollary 65.3 and Lemma 64.1(u)). Thus, all members of the set 1 are metacyclic. Since G is nonmetacyclic (Proposition 72.1(a)), we get, jG 0 j p 2 (Lemma 64.1(l)), a contradiction. Below we present an A3 -group G with jG 0 j D p 4 for p D 5 (this example was constructed by the second author).

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251

Let G be the group of order 56 with the following generators and relations: hb; h j b 25 D h25 D 1; Œh; b D a; a5 D 1; Œa; b D c; c 5 D 1; Œb; c D d; d 5 D 1; Œc; a D Œh; d D Œb; d D Œh; c D 1; Œh; a D s; s 5 D 1; Œh; s D Œb; s D 1; d D h5 ; b 5 D si: Here G 0 D ha; c; d; si Š E54 , K3 .G/ D hc; d; si Š E53 , K4 .G/ D hd i is of order 5, and Z.G/ D hd; si Š E52 . All maximal subgroups of G are A2 -groups and so G is an A3 -group. Finally, we have shown that this group G exists as a subgroup of the symmetric group of degree 625. In the following proposition we investigate A3 -groups with derived subgroup of order p 4 in some detail. We show that then G 0 is nonmetacyclic. Note that jGj p 6 with strong inequality if p D 2 (Taussky’s theorem). Proposition 72.3. Suppose that a p-group G is an A3 -group with jG 0 j D p 4 . (a) G 0 is abelian. (b) If exp.G 0 / D p, then p > 2, G=G 0 is abelian of type .p n ; p/ (n 1) and G 0 Š Ep 4 . If, in addition, n > 1, then 1 .G/ D G 0 . (c) G 0 is not metacyclic. Proof. By Proposition 72.1, G 0 is noncyclic so G is not metacyclic. (i) Assume that G 0 is metacyclic. Let us prove that then p D 2, the set 1 has only one nonmetacyclic member and G has no normal elementary abelian subgroups of order 8. Since the set 1 has a nonmetacyclic member (Lemma 64.1(l)), there is U 2 1 such that exp.U 0 / p (Theorem 65.7(a,d)); then jU 0 j j1 .G 0 /j D p 2 so U 0 Z.G 0 / (indeed, every G-invariant subgroup L of order p 2 in G 0 is contained in Z.G 0 /: consider CG .L/). Take V 2 1 fU g. By Lemma 64.1(u), jG 0 W U 0 V 0 j p so jU 0 V 0 j p 3 hence exp.U 0 V 0 / > p since G 0 is metacyclic. It follows that exp.V 0 / > p. Since V 0 is abelian in view of jV 0 j p 3 , U 0 V 0 is also abelian (recall that U 0 Z.G 0 /), V 0 is cyclic of order > p (Theorem 65.7(d)) so V is a metacyclic A2 -group and jV 0 j D p 2 (Corollary 65.3). It follows that all members of the set 1 are nonabelian. If E G G, where E is of order p 3 and exponent p, then there is only one maximal subgroup of G containing E, by the previous paragraph, so G=E is cyclic. Then G 0 < E, a contradiction since jG 0 j D p 4 > jEj. Thus, E does not exist. Assume that p > 2. Then, by Theorem 13.7 (see also Theorem 69.4), G is a 3-group of maximal class and jGj D 32 jG 0 j D 36 . In that case, G has a nonabelian subgroup of index 33 (Theorem 9.6), a contradiction. Thus, p D 2. (ii) Assume that G 0 is nonabelian; then jG W G 0 j D p 2 and G 0 D ˆ.G/ is an A1 -group. By Theorem 44.12, G 0 must be metacyclic so p D 2, by (i); then G is of

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maximal class (Taussky’s theorem) so Z.G 0 / is cyclic, a contradiction (Lemma 1.4). Thus, G 0 is abelian. (iii) Let exp.G 0 / D p. Then, by (ii), G 0 Š Ep 4 . If jG W G 0 j D p 2 , then p > 2 (Lemma 64.1(s)). Now assume that jG W G 0 j > p 2 and assume, in addition, that G=G 0 has two distinct noncyclic maximal subgroups U=G 0 and V =G 0 . Then U is either abelian or an A2 group with the derived subgroup of order p, the same is true for V (Lemma 65.4(b)). Then jG 0 j p 3 (Lemma 64.1(u)), a contradiction. Thus, G=G 0 has exactly p cyclic subgroups of index p, so it is abelian of type .p n ; p/ with n > 1. Let H=G 0 be the noncyclic maximal subgroup of G=G 0 . Then H is either abelian or an A2 -group with jH 0 j D p. If F 2 1 fH g, then jF 0 j p 2 (Lemmas 64.1(u) and 65.2(d)) and so F is an A2 -group. If jF 0 j D p 3 , then F has no maximal abelian subgroups (Lemma 65.2(d)). But, if F1 is a maximal subgroup of F containing G 0 , then F1 must be abelian (Lemma 65.1). This contradiction shows that jF 0 j D p 2 and then jH 0 j D p. Looking at the determination of A2 -groups (see 71), we see that F must be isomorphic to a group (ii) of Proposition 71.3. In particular, p > 2. (H must be isomorphic with a group of Proposition 71.1(i); then H must be a nonmetacyclic A1 -group). Assume that there is x 2 G G 0 of order p. Then D D hx; G 0 i Š Ep 5 since jG W Dj p 2 and 1 .D/ D D. If D M 2 1 , then M is abelian (Lemmas 65.1 and 65.4(a)). Take N 2 1 fM g. By Lemma 64.1(u) applied to M; N.2 1 /, jN 0 j p 3 . Considering N \ M and using Lemma 65.4(b), we conclude that N=G 0 is cyclic and jN 0 j p 2 , contrary to the result of the previous sentence. Thus, x does not exist so 1 .G/ D G 0 . (iv) By (i), p D 2 and the set 1 has only one nonmetacyclic member H and G has no normal elementary abelian subgroups of order 8. Take V 2 1 fH g; then V is metacyclic so jV 0 j 4 (Corollary 65.3 and Lemma 65.1). Then H is nonabelian (apply Lemma 64.1(u) to H; V 2 1 ). If H is an A1 -group, then 1 .H / Š E8 G G (Lemma 65.1), contrary to (i). If H is an A2 -group with jH 0 j D 2, then 1 .H / Š E8 or E16 (Proposition 72.1), contrary to (i). Hence, H must be an A2 -group of Proposition 71.4. If H is minimal nonmetacyclic of order 25 , then jGj D 2jH j D 26 so jG W G 0 j D 4 and G is of maximal class (Lemma 64.1(s)), a contradiction. In other cases of Proposition 71.4, 1 .H / Š E8 , a final contradiction. 3o . It follows from Proposition 72.2 and Lemma 64.1(u) that if a p-group G is an A4 -group, then jG 0 j p 9 . In this subsection we improve essentially that estimate. Since every p-group of order < p 7 is an Ai -group with i 4, in what follows we consider only A4 -groups of order p 7 ; then our A4 -group is not of maximal class (Theorems 9.5 and 9.6). Lemma 72.4. Suppose that a p-group G of order > p 6 is an A4 -group with nonabelian G 0 . Let jG W G 0 j D p i ; then i 2 f2; 3g.

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(a) If i D 2, then p > 2 and G 0 is either metacyclic or an A2 -subgroup with d.G/ D 3. (b) If i D 3, then G 0 is a metacyclic A1 -subgroup. Proof. We have jG W G 0 j D p i , i 2 f2; 3g. If i D 3, then G 0 is an A1 -group so d.G 0 / D 2 (Lemma 65.1); then G 0 is metacyclic (Theorem 44.12). Now suppose that i D 2. If p D 2, then G is of maximal class (Lemma 64.1(s)), a contradiction. Thus, p > 2. Assume that G 0 is nonmetacyclic; then d.G 0 / > 2 (Theorem 44.12) and G 0 is an A2 -subgroup with d.G/ D 3. Proposition 72.5. If a p-group G is an A4 -group, then jG 0 j p 7 . Proof. Assume that jG 0 j D p 9 . For U 2 1 , we have jU 0 j D p 4 so U 0 is nonmetacyclic, abelian (Lemma 64.1(u), Propositions 72.2, 72.3 and 72.1); in particular, all members of the set 1 are A3 -groups (Lemmas 65.1 and 65.2(c)). If V 2 1 fU g, then U 0 \ V 0 D f1g (Lemma 64.1(u)) so U 0 V 0 D U 0 V 0 is abelian of order p 8 . Set L D 1 .U 0 V 0 /; then j1 .L/j p 6 (Proposition 72.3) and L < U . If A < U is nonabelian subgroup, then jA \ Lj p 5 , contrary to Lemmas 65.1 and 65.4(b). Thus, jG 0 j p 8 . Next we assume that jG 0 j D p 8 . If U; V; W 2 1 are distinct with jU 0 j jV 0 j jW 0 j, then using Lemma 64.1(u) twice, we get jU 0 j D jV 0 j D p 4 and jU 0 \ V 0 j p since jU 0 V 0 j p1 jG 0 j D p 7 . Then U and V are both A3 -groups and U 0 and V 0 are nonmetacyclic abelian (Proposition 72.3(a,c)) so cl.U 0 V 0 / 2 (Fitting’s lemma), hence, if p > 2, then U 0 V 0 is regular (Lemma 64.1(a)). Now let p D 2. If U 0 V 0 D G 0 , then G 0 is abelian since then G 0 D U 0 V 0 (here we use Proposition 72.3(a) again). Suppose that U 0 V 0 < G 0 . Then jG W U 0 V 0 j > jG W G 0 j 8, by Lemma 64.1(s), so U 0 V 0 is abelian again. Set L D 1 .U 0 V 0 / so that exp.L/ D p. By the previous paragraph, jLj p 5 . If U1 is a maximal subgroup of U containing L, then U1 must be abelian (Lemma 65.4(b)). Since jU 0 j > p, U=L is cyclic (indeed, U1 is the unique abelian maximal subgroup of U ), and so U 0 < L. Hence exp.U 0 / D p and Proposition 72.3(c) implies that U 0 D 1 .U /. This is a contradiction since U 0 < L < U and exp.L/ D p. Now we improve the estimate of Proposition 72.5. Theorem 72.6. If a p-group G is an A4 -group, then jG 0 j p 6 . Proof. Assume, by the way of contradiction, that a p-group G is an A4 -group with jG 0 j D p 7 . By Propositions 72.2, 72.3(a) and Lemma 64.1(u), p 2 jU 0 j p 4 and U 0 is abelian for all U 2 1 . (i) We claim that the subgroup G 0 is nonmetacyclic and then d.G 0 / > 2 (Theorem 44.12). Suppose that this is false. By Proposition 72.1, jU 0 j D p 3 for all U 2 1 and so U 0 V 0 D U 0 V 0 for distinct U; V 2 1 (Lemma 64.1(u)). It follows that each

254

Groups of prime power order

U 0 is cyclic (otherwise, U 0 V 0 is nonmetacyclic) and so U is metacyclic (Proposition 72.1(b)). But then G is minimal nonmetacyclic (Proposition 72.1(a)), contrary to Lemma 64.1(l). (ii) If jG=G 0 j > p 2 , then G 0 is abelian and p 3 j1 .G 0 /j p 4 . Indeed, since d.G 0 / > 2, by (i) and Theorem 44.12, G 0 is not an A1 -group and so G 0 must be abelian. Since G=G 0 is noncyclic, there is U 2 1 such that U=G 0 is noncyclic. Let V1 and V2 be two distinct maximal subgroups of U containing G 0 . If j1 .G 0 /j > p 4 , then V1 and V2 cannot be A1 -groups or A2 -groups (Lemmas 65.1 and 65.4(a,b)) and so they are abelian. It follows that jU 0 j p (Lemma 64.1(q)), a contradiction. Thus j1 .G 0 /j p 4 , and j1 .G 0 /j p 3 follows from d.G 0 / > 2 and commutativity of G 0 . (iii) If G 0 is nonabelian, then G=G 0 Š Ep 2 , by (ii), so p > 2 since G is not of maximal class (Taussky’s theorem), and G 0 is an A2 -group with d.G 0 / D 3, by (i). Hence G 0 is isomorphic to a group of Proposition 71.1 or Proposition 71.4(b). In any case, 1 .G 0 / Š Ep 3 or Ep 4 , which follows from 71. (iv) If U 2 1 and jU 0 j D p 4 , then U 0 is abelian of type .p 2 ; p; p/. If V 2 1 and jV 0 j D p 3 , then V 0 is abelian noncyclic since V is nonmetacyclic, in view of G 0 < V . Indeed, U 0 is abelian nonmetacyclic (Proposition 72.3). Suppose that U 0 Š Ep 4 . Then Proposition 72.3(c) (noting that jU j D p1 jGj > p 6 ) implies that 1 .U / D U 0 . Take W 2 1 fU g. Then U 0 ˆ.U / ˆ.G/ < W so W is nonmetacyclic. Next, jU 0 W 0 j p1 jG 0 j D p 6 (Lemma 64.1(u)) and we must have W 0 U , 1 .W 0 / U 0 (see (iii)) so jW 0 j p 3 . By Proposition 72.1(b), W 0 is noncyclic so jU 0 \ W 0 j p 2 . But then jU 0 \ W 0 j D p 2 and jW 0 j D p 4 (Lemma 64.1(u)). Since W 0 is abelian and nonmetacyclic (Proposition 72.3(c)), we have j1 .W 0 /j p 3 and so 1 .W 0 / — U 0 , contrary to what has just been proved. (v) We claim that j1 .G 0 /j p 4 . Indeed, assume that j1 .G 0 /j p 5 . Then is not an Ai -group, i D 1; 2 (Lemmas 65.1 and 65.4(a,b)) so G 0 is abelian and E D 1 .G 0 / is elementary abelian of order p 5 . By (ii), G=G 0 Š Ep 2 so p > 2 (Taussky’s theorem). For each U 2 1 , p 2 jU 0 j p 4 (Lemma 64.1(u)) so U is an A3 -group (indeed, the nonabelian U cannot be an Ai -group, i 2, since E < U ; see Lemma 65.4(a,b)). If U=E is not cyclic, U has a maximal subgroup U0 ¤ G 0 containing E (note that G 0 is maximal in U ). Then U0 must be abelian and so jU 0 j p (Lemma 65.2(c)), a contradiction. Hence U=E is cyclic; then U 0 < E and so U 0 is elementary abelian for each U 2 1 . Then (iv) forces that U 0 Š Ep 3 for each U 2 1 and U 0 V 0 D U 0 V 0 for any two distinct U; V 2 1 . In particular, E D 1 .G 0 / is elementary abelian of order p 6 . Let an A2 -subgroup A be maximal in U . But in that case jA \ Ej p 5 since G 0 < U , contrary to Lemma 65.4(a,b). G0

(vi) We claim that, for each U 2 1 , U 0 is abelian of exponent p 2 . Also, 1 .G 0 / Š Ep 4 and, if G 0 is nonabelian, then G 0 D H Cp , where H is a nonmetacyclic A1 group of order p 6 . Indeed, assume that there is U 2 1 with U 0 Š Ep 3 , and take V 2 1 fU g. Since jU 0 V 0 j p 6 (Lemma 64.1(u)), we have jV 0 W .U 0 \ V 0 /j p 3 . If jV 0 j D p 3 , then U 0 \ V 0 D f1g and V 0 is noncyclic, by Proposition 72.1(b) and (i),

72 An -groups, n > 2

255

since G 0 < V . If jV 0 j D p 4 , then jU 0 \V 0 j p (Lemma 64.1(u)) and 1 .V 0 / Š Ep 3 , by (iv). In any case, j1 .U 0 V 0 /j p 5 , contrary to (v). Suppose that W 2 1 with W 0 Š Ep 2 and take V0 2 1 fW g. Then jV00 j D p 4 , W 0 \ V00 D f1g (Lemma 64.1(u)), so again j1 .W 0 V00 /j p 5 , by (iv), which is a contradiction. We have proved that if U 2 1 and jU 0 j D p 2 , then U 0 Š Cp 2 and if jU 0 j D p 3 , then U 0 Š Cp 2 Cp . This together with (iv) proves our first assertion on exp.U 0 /. Suppose that U 2 1 is such that jU 0 j is as large as possible. Let jU 0 j D p 4 ; then U 0 is abelian of type .p 2 ; p; p/, by (iv). Let V 2 1 fU g be such that jV 0 W .U 0 \ V 0 /j p 2 . If jV 0 j D p 2 , then U 0 \ V 0 D f1g and we have j1 .U 0 V 0 /j p 4 . If jV 0 j D p 3 , then jU 0 \ V 0 j p and so there are elements of order p in V 0 U 0 since V 0 is abelian of type .p 2 ; p/, by the previous paragraph. If jV 0 j D p 4 , then jU 0 \ V 0 j p 2 (Lemma 64.1(u)) and again there are elements of order p in V 0 U 0 since V 0 is abelian of type .p 2 ; p; p/, by (iv). In any case, we get j1 .U 0 V 0 /j p 4 . Suppose now that jU 0 j D p 3 . If W 2 1 fU g, then jW 0 j D p 3 , U 0 W 0 D U 0 W 0 (Lemma 64.1(u)) and both U 0 and W 0 are noncyclic ((i) and Proposition 72.1). Hence we get j1 .U 0 W 0 /j p 4 again. These results together with (iii) and (v) give that 1 .G 0 / Š Ep 4 . If G 0 is nonabelian, then using (iii), we see that G 0 must be isomorphic to a group of Proposition 71.1(i) and we are done. (vii) The group G is an irregular 3-group of order 39 . Each maximal subgroup of G is two-generator and Ã1 .G/ D K3 .G/ is of index 33 in G. In particular, G=G 0 Š E9 . Indeed, let U 2 1 and let A < U be maximal. Since 1 .G 0 / Š Ep 4 , we have jA \ 1 .G 0 /j p 3 since G 0 < U , and so A is nonmetacyclic. This implies that A0 is elementary abelian (Lemma 65.1 and Theorem 65.7(d)) and so A0 1 .U 0 / < U 0 since exp.U 0 / D p 2 , by (vi), and we get d.U / D 2 (Lemma 64.1(y)). Thus, all members of the set 1 are two-generator. Assume that d.G/ > 2. Using Theorem 70.1 and 70.2, we get in case p > 2 that jGj p 5 , a contradiction. If p D 2, then it follows from jGj 210 that cl.G/ > 2. The quotient group G=K4 .G/ is of order 27 or 28 . If jG=K4 .G/j D 28 , then we know that G=K4 .G/ is an A5 -group, a contradiction (see the text following Theorem 70.5). If jG=K4 .G/j D 27 , then Theorem 70.5 implies that K4 .G/ D f1g, a contradiction. We have proved that d.G/ D 2. Since G is not metacyclic, we get p > 2 (Lemma 64.1(n)). By Lemma 64.1(p), Ã1 .G/ D K3 .G/ is of index p 3 in G. On the other hand, j1 .G 0 /j D p 4 > jG=Ã1 .G/j so G is irregular (Lemma 64.1(a)). This implies that p D 3 (Theorem 9.8(a)), G=G 0 Š E9 and jGj D 39 . (viii) Let U; V 2 1 be distinct. Then jU 0 V 0 j 13 jG 0 j D 36 (Lemma 64.1(u)) and we claim that jU 0 V 0 j D 36 . Since H D K3 .G/ D ŒG; G 0 U 0 V 0 , we have H D U 0 V 0 D Ã1 .G/ in view of jG W H j D 33 . Also, 1 .G 0 / D 1 .H / and H is abelian of exponent 32 . Finally, G=H is nonabelian of order 33 and exponent 3. Suppose that there are distinct U; V 2 1 such that U 0 V 0 D G 0 . We may assume that jU 0 j D 34 and V 0 covers G 0 =U 0 . If jV 0 j D 33 , then G 0 D U 0 V 0 . But then, by (iv), j1 .G 0 /j 35 , contrary to (v). If jV 0 j D 34 , then jU 0 \ V 0 j D 3. But j1 .V 0 /j D 33 and again we get j1 .G 0 /j 35 , a contradiction. Hence, for each distinct U; V 2 1 ,

256

Groups of prime power order

we have jU 0 V 0 j D 36 . Since U 0 V 0 G G and G 0 =.U 0 V 0 / Z.G=.U 0 V 0 //, we have H D ŒG; G 0 U 0 V 0 . By (vii), H D U 0 V 0 D Ã1 .G/ (recall that jG W G 0 j D 32 ). Since j1 .U 0 V 0 /j 34 , we have j1 .U 0 V 0 /j D 34 , by (v), and so E D 1 .H / D 1 .G 0 / Š E34 . If H would be nonabelian, then H is an A1 -group, contrary to Lemma 65.1. Thus, H is abelian and so exp.H / D 32 since exp.U 0 / D exp.V 0 / D 32 and H D U 0 V 0. We are now able to get the final contradiction. Take U 2 1 . Since G=H is nonabelian of order 33 and exponent 3, we get U=H Š E9 , where H D K3 .G/ D Ã1 .G/. On the other hand, d.U / D 2 (by (vii)) and so ˆ.U / D H . Let X be any of the four maximal subgroups of U . Then jX W H j D 3 and E D 1 .H / Š E34 is a normal subgroup of X. Suppose that jX 0 j 32 so that X is an A2 -subgroup. By the results of 71, X must be isomorphic to a group of Proposition 71.3(ii). In particular, X (of order 37 ) has the unique abelian maximal subgroup of type .33 ; 3; 3; 3/. This is a contradiction since H is a unique maximal abelian subgroup of X and exp.H / D 32 (by (viii)). We have proved that jX 0 j 3 for any maximal subgroup X of U . This implies that jU 0 j 33 for any U 2 1 (Lemma 64.1(u)). We have proved that for each U 2 1 , U 0 is abelian of type .32 ; 3/ and so H D 0 U V 0 , where V 2 1 fU g. This implies that H is abelian of type .32 ; 32 ; 3; 3/ and each maximal subgroup X of U must be an A2 -group with jX 0 j D 3 (Lemmas 65.1 and 64.1(u)). By the results of 71, X must be the following group of Proposition 71.1(i): 3

2

X D ha; b j a3 D b 3 D 1; Œa; b D c; c 3 D Œa; c D Œb; c D 1i C3 ; where 1 .X/ D ha3 ; b 3 ; ci C3 D E D 1 .H / and X 0 D hci is a maximal cyclic subgroup in X (see Lemma 65.1). We have U 0 < H and set U 0 \ E D E0 so that E0 Š E9 . Also set he0 i D Ã1 .U 0 / so that he0 i is a subgroup of order 3 in E0 and he0 i is not a maximal cyclic subgroup in H . Let X1 , X2 , X3 , X4 be all maximal subgroups of U and note that each one is isomorphic to the above group X. Then X10 , X20 , X30 , X40 must be four distinct subgroups of order 3 in E0 (Lemma 64.1(u)). It follows that for some i 2 f1; 2; 3; 4g, Xi0 D he0 i. This is a contradiction since he0 i is not a maximal cyclic subgroup in H (and H is contained in Xi ); see the last sentence of the previous paragraph. 2

73

Classification of modular p-groups

We recall that a p-group G is modular if and only if any two subgroups of G are permutable (for general groups there is another definition). Sections of modular pgroups are modular. A nonabelian Dedekindian p-group G is a modular 2-group of the form G D Q E, where Q Š Q8 and exp.E/ 2 (Theorem 1.20). Such groups are called Hamiltonian. Nonmodular p-groups of order p 3 are D8 and the nonabelian group S.p 3 / of order p 3 and exponent p > 2. If G is a minimal nonmodular p-group, then there is N G G such that G=N is either S.p 3 / (p > 2) or D8 (see Theorem 44.13). The aim of this section is to correct the original Iwasawa’s proof [Iwa] of the theorem on the structure of modular p-groups. ((Iwasawa’s proof contains essential gaps. Two gaps were filled by Napolitani [Nap], remaining ones – by the second author, who wrote this corrected proof. There exists another proof of Iwasawa’s theorem, due to R. Schmidt [Sch, Chapter 2]; that proof is based on entirely other ideas and is essentially shorter.) Our proof is self-contained. The following proposition is obvious. Proposition 73.1. Let G be a modular p-group. Then 1 .G/ is elementary abelian. In particular, modular p-groups, p > 2, are powerful (apply Lemma 73.1 to G=Ã1 .G/; see [LM] and 26). It follows from Theorem A.24.4 that non-Hamiltonian modular 2-groups G are also powerful (see [LM] and 26.2). Indeed, by that theorem, G is Q8 -free. One may assume that exp.G/ D 4. Assume that G has a minimal nonabelian subgroup H . Then 8 < jH j < 32 since exp.H / D 4. Using Lemma 65.1, we see that H has a nonabelian epimorphic image of order 8, contrary to what has just been said. Thus, H does not exist so G is abelian. Proposition 73.2. Let G be a modular p-group. Then exp.n .G// p n and the group n .G/=n1 .G/ is elementary abelian for all n. Use induction on n and Proposition 73.1. By Theorem 44.13, a p-group G is modular if and only if it is D8 -free if p D 2 and S.p 3 /-free if p > 2. Remark. A two-generator modular p-group G is metacyclic. Assume that G is not metacyclic. In view of Lemma 64.1(m), one may assume that K3 .G/ˆ.G 0 / D f1g;

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Groups of prime power order

then G is minimal nonabelian such as in Lemma 65.1(a) with m C n > 2, jGj D p mCnC1 . Then G D ha; bi D haihbi has order p mCn < p mCnC1 D jGj, a contradiction. Since metacyclic p-groups, p > 2, are modular, it follows from the remark that a p-group G, p > 2, is modular if and only if every its two-generator subgroup is metacyclic. It is interesting to classify the 2-groups satisfying this property. Note that dihedral 2-groups are metacyclic but nonmodular. The group M2n is modular. A central product of two modular p-group is not necessary modular (for example, the group Q23 C4 is not modular. Proposition 73.3. Nonabelian modular groups G of order p 4 are: (a) Mp 4 , (b) Mp 3 2

2

Cp , p > 2, (c) G D ha; b j ap D b p D 1; ab D a1Cp i, p > 2, (d) Q8 C2 . Proposition 73.4. Suppose that G is a modular 2-group containing a normal elementary abelian subgroup E such that G=E Š Q8 . Then G D Q E with Q Š Q8 and so G is Hamiltonian. Proposition 73.5. Let G be a modular 2-group which is not Hamiltonian. Then each quotient group i .G/=i2 .G/ is abelian. In particular, 2 .G/ is abelian. Proof. Let G be a modular 2-group of exponent 4. If G does not contain a quaternion subgroup, then Proposition 73.3 implies that any two elements of G commute and so G is abelian. Suppose that Q8 Š Q G. In that case, G is Hamiltonian, by Theorem A.24.4, and we are done. The result follows since all sections of G are modular. Proposition 73.6. Let G be a modular p-group of exponent p . 2/ which is not Hamiltonian. Then .uv/p

()

1

D up

1

vp

1

for any u; v 2 G, i.e., G is p 1 -abelian. Proof. (i) Suppose that D 2. If p D 2, then G D 2 .G/ is abelian (Proposition 73.5). Suppose that p > 2 and let a; b 2 G. Then H D ha; bi D haihbi of order p p 4 is metacyclic (Lemma 64.1(m)), jH 0 j D p so .ab/p D ap b p Œb; a. 2 / D ap b p . The general case follows by induction on . Let > 2. Since G=1 .G/ is of 2 2 p 2 exponent p 1 , we have, for x; y 2 G, .xy/p D xp y z, where z 2 2 2 1 .G/. Since x p ; yp 2 2 .G/ and exp.2 .G// D p 2 , we get .xy/p

1

D ..xy/p D xp

1

2

yp

/p D .x p

1

2

zp D xp

yp

1

2

yp

z/p

1

:

73 Classification of modular p-groups

259

Proposition 73.7. Let G be a non-Hamiltonian modular group of exponent p . 2/. Set j˛ .G/ W ˛1 .G/j D p !˛ for ˛ D 1; : : : ; . Take elements a1 ; : : : ; a! (of order p ) of G so that they form .mod 1 .G// a basis for the elementary abelian p p group .G/= 1 .G/. Now, a1 ; : : : ; a! can be extended .mod 2 .G// to a p p basis a1 ; : : : ; a! ; a! C1 ; : : : ; a!1 of 1 .G/= 2 .G/ and so on. Every element of G can be written in a unique way as a product of powers of these elements a1 ; a2 ; : : : . In other words, the elements a1 ; a2 ; : : : form a basis of G. Looking at the series 1 .G/; : : : ; .G/, we see that the constants o.a1 /; o.a2 /; : : : are uniquely determined (up to the ordering). Next, if a1 ; : : : ; ad is so constructed basis, then Q d iD1 o.ai / D jGj. Proof. Working in G= 2 .G/, we may assume for a moment that 2 .G/ D f1g p are so that exp.G/ D p 2 . We have to show that (commuting) elements a1p ; : : : ; a! n! p n1 p ! linearly independent in 1 .G/ .D 1 .G/ Š Ep 1 /. Indeed, if a1 : : : a! D n! 1 for some integers n1 ; : : : ; n! , then, by Proposition 73.6, we get .a1n1 : : : a! /p D n! 1, and so a1n1 : : : a! 1 .mod 1 .G//, and we conclude that ni 0 .mod p/. p p Hence a1 ; : : : ; a! are linearly independent in 1 .G/ and so they are extendible p ; a! C1 ; : : : ; a!1 of 1 .G/ .D 1 .G//. to a basis a1p ; : : : ; a! Now, in general, (without assuming D 2 ) we consider the elements p2

2

p

p p ; a! C1 ; : : : ; a! 2 2 .G/ a1 ; : : : ; a! 1

and working in 1 .G/= 3 .G/ (of exponent p 2 ) we again show that these elements are linearly independent modulo 3 .G/ and therefore they are extendible with elements a!1 C1 ; : : : ; a!2 2 2 .G/ in such a way that 2

2

p p p ; a! ; : : : ; a! ; a!1 C1 ; : : : ; a!2 a1p ; : : : ; a! 1 C1

considered modulo 3 .G/, form a basis of 2 .G/= 3 .G/. Then in the obvious way we continue with the construction of a “basis” a1 ; a2 ; : : : of G. Since for any two “basis” elements ai and aj we have hai ihaj i D haj ihai i, every element of G can be written as a product of powers of elements a1 ; a2 ; : : : . Moreover, each element of G can be written in a unique way as a product of powers of elements a1 ; a2 ; : : : (in that order). Indeed, by the construction of these elements, we have o.a1 / o.a2 / D jGj and so the elements a1 ; a2 ; : : : form a basis of G. Proposition 73.8. Let G be a modular p-group which is not Hamiltonian. Let a 2 1 .G/# (this assumption is essential: it cannot be avoided even in the case where G is the abelian group of type .p 3 ; p/)). Then there is a basis a1 ; : : : ; ar of G such that a 2 ha1 i. Proof. We use induction on jGj. Let a1 ; : : : ; ar be a basis of G such that o.a1 / o.a2 / o.ar /. Let a 2 1 .G/# . Set ha1 ; : : : ; ar1 i D G1 so that G1 is a

260

Groups of prime power order

complement of har i in G. If a 2 1 .har i/, we are done. If a 2 G1 , then, by induction, there is a basis b1 ; : : : ; br 1 of G1 such that a 2 hb1 i. But then b1 ; : : : ; br1 ; ar is a basis of G and we are done. It remains to consider the case a D cd D dc, where c 2 1 .G1 /# and d 2 1 .har i/# . By induction, there is a basis c1 ; : : : ; cr1 of G1 such that c 2 hcr1 i. Set o.ar / D p s . If s D 1, then we replace ar with a. Since hai \ G1 D f1g and o.a/ D o.ar / D p, c1 ; : : : ; cr 1 ; a is a basis of G and we are done. Suppose s 2 s1 and replace ar with ar0 so that d D .ar0 /p . Then c1 ; : : : ; cr1 ; ar0 is also a basis of G. Since o.a1 / o.a2 / o.ar / (and these numbers are invariants of G), we have also o.cr 1 / o.ar0 / D p s . Thus, there is an element l 2 hcr1 i of order p s s1 such that l p D c. Since hl; ar0 i is a modular subgroup of exponent p s (noting that exp.s .G// p s ), s1 s1 s1 it follows from Proposition 73.6 that .lar0 /p D l p .ar0 /p D cd D a, and so o.lar0 / D o.ar0 / D p s . On the other hand, hc1 ; : : : ; cr1 ; lar0 i D G and G1 \ hlar0 i D f1g. Hence c1 ; : : : ; cr 1 ; lar0 is a basis of G and a 2 hlar0 i. Proposition 73.9. Let G be a modular p-group which is not Hamiltonian. If jG 0 j D p, m n m1 then G D G1 G2 , where G1 D ha1 ; a2 j a1p D a2p D 1; a1a2 D a11Cp ; m> 1; n 1i and G2 is abelian of exponent p m1 with m > 2 if p D 2. Hence m1 A D ha1 ; G2 i is an abelian normal subgroup of G D hA; a2 i and aa2 D a1Cp for all a 2 A. Proof. By Proposition 73.8, we may choose a basis a1 ; a2 ; : : : ; as of G so that G 0 ha1 i. Since hai ; aj i \ G 0 hai ; aj i \ ha1 i D 1 .i; j > 1/, we see that ha2 ; : : : ; as i is abelian. Let a2 be an element of the smallest order among those basis elements a2 ; : : : ; as which do not commute with a1 . By Lemma 65.2(a), ha1 ; a2 i is minimal p m1 nonabelian. So, in view of Lemma 65.1, we may put Œa1 ; a2 D z, where z D a1 and o.a1 / D p m , m 2. (We get ha1 i ¤ G 0 since G is nonabelian.) We have Œa1 ; ai D z ki with ki 2 f0; 1; : : : ; p 1g .i > 2/, where (by assumption) o.ai / o.a2 / whenever ki ¤ 0. Then we replace the basis elements a3 ; : : : ; as with the elements a2ki ai D bi .3 i s/. We have o.bi / D o.ai / and Œa1 ; bi D Œa1 ; a2ki ai D Œa1 ; a2 ki Œa1 ; ai D z ki z ki D 1 and so the abelian subgroup hb3 ; : : : ; bs i centralizes ha1 ; a2 i. It is clear that fa1 ; a2 ; b3 ; : : : ; bs g is also a basis of G. Indeed, ha1 ; a2 ; b3 ; : : : ; bs i D G and jGj D o.a1 /o.a2 /o.a3 / : : : o.as / D o.a1 /o.a2 /o.b3 / : : : o.bs /: Denoting b3 ; : : : ; bs again with a3 ; : : : ; as , we may assume from the start that G D G1 G2 , where G1 D ha1 ; a2 i and G2 D ha3 ; : : : ; as i. For i 3 we have .a1 ai /a2 D a1a2 ai D .a21 a1 a2 /ai D a1 .a11 a21 a1 a2 /ai D a1 zai D z.a1 ai /:

261

73 Classification of modular p-groups

Now, .G 0 / A D ha1 ; a3 ; : : : ; as i G G, G D Aha2 i, and A \ ha2 i D f1g. Consider the subgroup H D ha1 ai ; a2 i .i > 2/ which is nonabelian. Since H D ha1 ai iha2 i and ha1 ai i A, we have H \ A D ha1 ai i and so a2 normalizes ha1 ai i. Thus p m1 2 ha1 ai i. This implies .a1 ai /a2 D z.a1 ai / D .a1 ai /x .x integer/ and so z D a1 m1 that the order of ai .i > 2/ is at most p . For each a 2 A, we have a D a1r b with an integer r and b 2 G2 so that 1Cp m1 r

aa2 D .a1r b/a2 D .a1 D .a1r b/1Cp

m1

/ b D .a1r /1Cp

D a1Cp

m1

m1

b 1Cp

m1

:

If p D 2 and m D 2, then K D ha1 ; a2 j a14 D a22 D 1; a1a2 D a13 D a11 i is not n1 modular, since K=ha22 i Š D8 . Hence, if p D 2, then m > 2. Our proposition is proved. n

Proposition 73.10. Let G be a modular p-group which is not Hamiltonian. Then each subgroup and each factor group is modular and is not Hamiltonian. Subgroups and epimorphic images of modular p-groups are modular so the assertion follows since, by Theorem A.24.4, G is Q8 -free. s

Proposition 73.11. Let t; x 2 G, where exp.G/ D p . If x t D x 1Cp z, where z 2 1 .Z.G//# and 1 s < 1 is a natural number which is 2 if p D 2, then xt

p 1s

D x 1Cp

1

. In particular, if o.x/ p 1 , then Œx; t p i

Proof. By induction on i , we first prove that x t D x .1Cp xt

i C1

D .x .1Cp

p 1s

s /i

s

z i /t D .x 1Cp z/.1Cp

s /i

s /i

1s

z i . Indeed, we have

z i D x .1Cp

s p 1s

s /i C1

D x .1Cp / , since s < 1 and so z p This gives x t It remains to prove the following congruence: (1)

.1 C p s /p

1s

1 C p 1

D 1.

z iC1 :

1s

D 1.

.mod p /:

According to P. Roquette, everything depends on the following formula: (2)

.a C bp s /p ap C ap1 bp sC1

.mod p sC2 /;

where a; b are integers, p prime, s 1 and, if p D 2, then s 2. Indeed, we get ! X p pi i si s p p p1 s a (3) bp C b p C b p p sp : .a C bp / D a C pa i 2ip1

Therefore each member of the above sum is 0 .mod p 1Csi / and so 0 .mod p sC2 / since 1 C si s C 2 in view of i 2. For the last term on the right-hand side of (3)

262

Groups of prime power order

we have sp s C 2 for p 3 and this holds also for p D 2 because in that case we have assumed s 2. Thus, the last term is also 0 .mod p sC2 /, and (2) is proven. It follows from (2) that z 1 C ps

(4)

.mod p sC1 / implies z p 1 C p sC1

.mod p sC2 /:

To prove (4) we set z D 1 C p s C bp sC1 D a C bp sC1 , where a D 1 C p s and b is an integer. From (2) follows (with s C 1 instead of s and noting that always s C 1 2) z p D .a C bp sC1 /p ap C ap1 bp sC2 .mod p sC3 / and so z p ap .1 C p s /p .mod p sC2 /. Again using (2) (for a D b D 1), we get .1 C p s /p 1 C p sC1 .mod p sC2 /, proving (4). With iteration we get from (4) (with k D 1; 2; 3; : : : ) z 1 C ps

(5)

k

.mod p sC1 / implies z p 1 C p sCk

.mod p sCkC1 /:

If we set in (5) z D 1 C p s , k D 1 s, where s < 1, we get (1). Proposition 73.12. Let G be a non-Hamiltonian modular p-group. Then each element i of Ãi .G/ .i 1/ can be written as ap for some a 2 G. Proof. We prove first this result for i D 1, working by induction on jGj. Let x 2 p p p Ã1 .G/ so that x D a1 a2 : : : as with some ai 2 G. Now, G=Ã2 .G/ is of exponent 2 p and so, using Proposition 73.6, we get x D .a1 a2 : : : as /p l where l 2 Ã2 .G/. p2 p2 p2 We set a1 a2 : : : as D b, H D hb; Ã1 .G/i, and l D c1 c2 : : : cr with some c1 ; c2 ; : : : ; cr 2 G. Thus x D b p .c1p /p : : : .crp /p 2 Ã1 .H /. We have H ¤ G, unless G is cyclic, so x D ap for some a 2 H , by induction. pi

pi

Assume that i > 1 and use induction on i . Let y 2 Ãi .G/ so that y D b1 : : : b t D i 1

i 1

i 1

/p for some b1 ; : : : ; b t 2 G. Here bjp 2 Ãi1 .G/, j D .b1p /p : : : .b p t 1; : : : ; t , so y 2 Ã1 .Ãi1 .G//. By the above, y D c p with c 2 Ãi1 .G/. By i 1 i and so y D d p . induction, c D d p Proposition 73.13. Let G be a non-Hamiltonian modular p-group. If we have G D i i ha1 ; : : : ; ar i, then Ãi .G/ D ha1p ; : : : ; arp i. Proof. One may assume that fai gr1 is a minimal basis. First let i D 1. We have ha1 i : : : har i D G so by the product formula applied r 1 times, jG W ha1p i : : : harp ij D p r , Since p r D jG W ˆ.G/j D jG W Ã1 .G/j, we are done. p i 1

p i 1

Suppose that i > 2. By induction on i , Ãi1 .G/ D ha1 ; : : : ; ar i. If y 2 i 1 i Ãi1 .G/, then y D ap ; a 2 G (Proposition 73.12) and so y p D ap 2 Ãi .G/. This gives Ã1 .Ãi1 .G// Ãi .G/. On the other hand, if g 2 Ãi .G/ , then by i i 1 Proposition 73.12, g D b p D .b p /p ; b 2 G, and so g 2 Ã1 .Ãi1 .G// which pi pi gives Ãi .G/ D Ã1 .Ãi1 .G//. By the above, this gives Ãi .G/ D ha1 ; : : : ; ar i.

263

73 Classification of modular p-groups

Proposition 73.14 (compare with [Sch, Lemma 2.3.4]). Let G be a p-group and let A G G be abelian and such that G=A D htAi is cyclic, all subgroups of A are normal in G and, if p D 2, when 2 .A/ Z.G/. Then G is modular. Proof. Suppose that G is a counterexample of minimal order. Then, by induction, H < G is modular since H=.H \ A/ Š AH=A. G=A/ is cyclic and H satisfies the hypothesis with respect to H \ A. Hence, G is minimal nonmodular. Then, by Theorem 44.18, there exists N GG with G=N 2 fD8 ; S.p 3 /g. Since G=AN is a cyclic epimorphic image of G=N , we have jG W AN j p. But every subgroup of AN=N is normal in G=N , and this implies that p D 2 and AN=N is the cyclic subgroup of order 4 in G=N Š D8 . Hence A=.A \ N / Š C4 so A D .A \ N /hui, and A is modular, by the above. Let G D Ahvi. Then G D hu; vi since hu; vi is nonmodular since it has an epimorphic image Š D8 . Since v 2 2 A, W u ! uv is an automorphism of order 2. Let o.u/ D 2m , o.v/ D 2n . By Theorem 1.2, we have to consider one of the following three possibilities: (i) uv D u1 . Then u2 tion.

m2

m2

D .u2

/v D u2

m2

, m > 2. Then u2 (ii) uv D u1C2 m1 m2 / D 1, a contradiction. u2 .1C2 m1

m2

m2

D .u2

m1

so u2

D 1, a contradicm2 .1C2m1 /

/v D u2

so

m1

, m > 2, i.e., G Š M2mC1 . In that case, G is modular, contrary (iii) uv D u1C2 to the assumption. Theorem 73.15 ([Iwa]). A non-Hamiltonian p-group G is modular if and only if it contains an abelian normal subgroup N with cyclic quotient group G=N and there exists an element t in G with G D hN; t i and a positive integer s which is at least 2 in s the case p D 2 such that at D a1Cp for all a 2 N . Proof (Iwasawa–Napolitani–Janko). If G is as in the theorem, it is modular, by Proposition 73.14. Suppose, conversely, that G is a modular p-group which is not Hamiltonian. This property is inherited by sections (Proposition 73.10), so we may assume that the theo1 rem is true for groups of smaller order. By Proposition 73.6, the mapping g 7! g p , where p D exp.G/, is a homomorphism of G. The image of this homomorphism is Ã 1 .G/ ¤ 1. There is a G-invariant subgroup Z D hzi of order p in Ã 1 .G/. By induction, G=Z contains an abelian normal subgroup N=Z with properties stated in s the theorem. There is an element t 2 G such that G D hN; t i and at D a1Cp z k for all a 2 N , where k 2 N [ f0g depends on a. There are two major possibilities for the structure of N . The subgroup N is either (I) abelian or (II) nonabelian. Case (I). N is abelian. Here z 2 Ã 1 .G/ is the p 1 -th power of an element c in 1 G, z D c p . There are two possibilities for the position of c in G.

264

Groups of prime power order

(˛) It is possible to choose c 2 N . Since o.c/ D p , c is an element of a basis of G (Proposition 73.7). In particular, hci has a complement T in G which is generated by remaining elements of those basis of G that contains c. We have G D T hci with T \hci D f1g and so N D .T \N /hci, by the modular law, and T covers G=N since T N T hci D G. The quotient group T =.T \ N / Š G=N is cyclic, and we choose t1 2 T \ .N t / so that T D hT \ N; t1 i; then G D hN; t1 i. Since t 1 t1 2 N and N is abelian, t and t1 induce the same s automorphism on N . In view of Z \ T D f1g, we have at D at1 D a1Cp for all a 2 T \ N. s If also c t D c 1Cp , then the fact that N D .T \ N / hci is abelian implies s x t D x 1Cp for all x 2 N and the theorem is proved. Therefore assume that s s 1 c t D c 1Cp z k D c 1Cp Ckp , where k 6 0 .mod p/. We may assume that k 2 f1; 2; : : : ; p 1g and so p k 2 f1; 2; : : : ; p 1g. We claim that Ã 1 .T \N / D f1g. Otherwise, there is a 2 T \N of order p . We have hai \ hci D f1g and hcai \ hci D f1g. Both elements ca and c are of order p

and they, clearly, are “linearly independent” modulo 1 .G/. Hence, by the process of constructing a basis of G in Proposition 73.7, there is a basis of G containing both c and ca. Let T1 be a complement of hci in G containing ca. By the preceding s s s s argument, .ca/t D .ca/1Cp and so .ca/t D c t at D c t a1Cp D c 1Cp a1Cp , which s gives c t D c 1Cp , contrary to what has been assumed in the previous paragraph. We have proved that Ã 1 .T \N / D f1g or, what is the same, exp.T \N / p 1 . Hence, if s 1, then G=hzi is abelian. Assuming that G is nonabelian (otherwise, there is nothing to prove), we conclude that G 0 D hzi is of order p and so Proposition 73.9 implies the validity of the theorem. It remains to examine the case s < 1 (and s 2 for p D 2). We use Proposition p 1s

73.11 and get x t D x 1Cp , whenever x t D x 1Cp z 0 with z 0 2 Z. From this relation we get, iterating (p k) times, xt

1

.pk/p 1s

D x .t

z

p 1s /pk

D x 1C.pk/p Thus, setting t D t 1C.pk/p

ct D ct

.pk/p 1s t

D .c 1Cp

s Ckp 1

1s

1 C pk 2

.

1/ /pk

/p 2.1/ C D x 1kp 1 :

, we get

D .c 1kp /.1kp

D x .1Cp

1

1 /

/t D .c t /.1kp

D c 1Cp

1 /

s Ckp 1 kp 1 kp 1Cs k 2 p 22

s

D c 1Cp : Finally, for each a 2 T \ N (noting that ap t .pk/p

1s

1kp 1 t

1

.a / D .a / D a D a y 2 N . Since G D hN; t i, we are done. t

t

D 1 holds) we obtain also at

1Cp s

. Thus y

t

D y

1Cp s

D

for each

265

73 Classification of modular p-groups

(ˇ) It is not possible to choose c in N . We claim that hc; N i D G. Indeed, if hc; N i < G, then we can choose a genph ph erator t1 of ht i such that t1 c .mod N / with h > 0. Hence c D bt1 , where b 2 N . It follows b D ct1p and then applying Proposition 73.6 we get b p h

p h 1 .ct1 /p

1 p hC1 c p t1

1

p 1

D

D D c D z, where b 2 N and o.b/ D p , a contradiction. It follows that a suitable generator c0 of hci operates on N in the same way as t and so we may assume that c itself operates on N as t . In what follows we shall identify t 1 and c. This means that we shall assume that o.t / D p , t p D z, and so for each s 1 t 1Cp kp t with k 2 f0; 1; : : : ; p 1g depending on a. a 2 N we have a D a Let fa1 ; a2 ; : : : ; ar g be a basis of N such that Z D hzi ha1 i (such a basis exists 1Cp s or by Proposition 73.8). For each ai , i D 1; 2; : : : ; r, we have either ait D ai 1Cp s ki p 1 t t , ki 6 0 .mod p/. ai D ai

s

We shall prove that it is possible to assume that ait D ai1Cp for i 3. For this purpose, we order the basis elements a2 ; : : : ; ar so that: 1Cp s

(1ˇ) there exists such that ait D ai

if and only if i D C 1; : : : ; r;

(2ˇ) if > 1, then o.a2 / o.aj / for each j with 2 j . If 2, the above statement is obviously true. Supposing > 2, we observe that for each i , 2 i , fa2 ; a2q ai g (q integer) is a basis of ha2 ; ai i. Since s

.a2q ai /t D .a2t /q ait D .a21Cp t k2 p q

s

D .a2 /1Cp t k2 qp

1

1

1Cp s ki p 1

ai

t

s

/q .ai1Cp /ki p q

1

s

D .a2 ai /1Cp t .k2 qCki /p

1

and k2 6 0 .mod p/, we can solve the congruence equation k2 q C ki 0 .mod p/, s in q D qi (depending on i ) so that .a2qi ai /t D .a2qi ai /1Cp . Hence the set fa1 ; a2 ; a2q3 a3 ; : : : ; a2q a ; a C1 ; : : : ; ar g is a basis with the desired property. We consider first the following special possibility: 1Cp s

p 1

1

p 1

t k2 p , k2 6 0 .mod p/, a2 ¤ 1, and ai D 1 for all (ˇ1 ) a2t D a2 i 3. We shall prove that in this case either the theorem holds or we can choose our abelian normal subgroup N D ha1 ; a2 ; : : : ; ar i in such a way that 1Cp s also a2t D a2 . Here we distinguish two subcases (i) s 1 and (ii) s < 1. 1Cp s

1

(i) s 1. If a1t D a1 t k1 p with k1 6 0 .mod p/, then we multiply a2 qN with a1 , where qN is a solution of the congruence equation k1 qN C k2 0 .mod p/ and qN we replace the basis fa1 ; a2 ; : : : ; ar g of N with the basis fa1 ; aO 2 D a1 a2 ; a3 ; : : : ; ar g

1

1 noting that o.a1 / p (since in our case .ˇ/, z ha1 i is not a p -th power of

266

Groups of prime power order

any element in N ). Then we compute qN

1Cp s k1 p 1 qN 1Cp s k2 p 1 t / a2 t

.aO 2 /t D .a1 a2 /t D .a1

N / 1Cp .k1 qCk D a1q.1Cp a2 t N 2 /p s

s

1Cp s

1), then G 0 D ht

D .a1qN a2 /1Cp D .aO 2 /1Cp ; s

D a1 (noting that s 1 and a1

p 1 k2 p 1 ha2 t i

i or G 0 D

s

p 1

as required. If, however, a1t D a1 p 1

1

D

depending whether s > 1 or 1

s D 1. Here we have also used our assumption that aip D 1 and the fact that 1Cp s ait D ai for all i 3. But then our theorem follows from Proposition 73.9. (ii) s < 1. We replace here the subgroup N with the subgroup N D ha1 ; a2 ; 1s with ˛ being a solution of the congruence ˛k2

a3 ; : : : ; ar i, where a2 D a2˛ t p 1 1 D tp D z. We note that 1 .mod p/. Note that ˛ 6 0 .mod p/ and t ˛k2 p s s 1 D z k2 2 Z.G/ and a2p (s 1) is of order p 1 and therefore a2p and t k2 p 1s commute (Proposition 73.11). Then we compute tp .a2 /t D .a2˛ t p

1s

˛.1Cp s /

D a2

s

/t D .a2t /˛ t p

t ˛k2 p

D a2˛ a2˛p t p

1

1s

.t p

The subgroup H D ha2 ; t p p 1s

a2t

1Cp 1

D a2

and so

.a2 /p D .a2˛ t p s

s

1s

1s

1s

˛p s

t

D a2˛ a2 t p s

/p D .a2˛ t p 1

/ t

1

1s

tp

1s s

/.a2˛p .t p

1s

s

/p /:

i is of order p. Therefore,

˛p s

s

/p D a2 .t p

1s

1Cp s k2 p 1 ˛ p 1s

D .a2

i is of class 2. Indeed, by Proposition 73.11,

D ha2p

H0

1s

D a2˛p .t p

tp

1s

1s

s

/p Œt p

1s

ps

; a2˛ . 2 /

s

/p ;

s since (in any case) p divides p2 . Substituting this last result in the above relation, s s we get .a2 /t D a2 .a2 /p D .a2 /1Cp . 1s commutes with each ai , i ¤ 2, The subgroup N is abelian. Indeed, t p because each such ai is of order p 1 (Proposition 73.11). Also, G D hN ; t i, .N /t D N , and so N is normal in G with G=N cyclic. We observe that a2 is of order p and so ha2 i\ha1 ; a3 ; : : : ; ar i D f1g, which implies that fa1 ; a2 ; a3 ; : : : ; ar g 1 1s p 1 D .a2˛ t p / D is a basis of N . Indeed, by Proposition 73.6, .a2 /p p 1 ˛ p 22s

.a2

/ t

˛p 1 ha2 i

˛p 1

D a2

is an element of order p (˛ 6 0 .mod p/) and

\ ha1 ; a3 ; : : : ; ar i D f1g, where we have used the fact that 2 2 s . All the statements in case .ˇ1 / are proved. Indeed, a2 is of order p , all other basis p 1

elements ai of N are of order p 1 and so Ã 1 .N / D ha2 Hence z 2 ha1 i cannot be a p 1 -th power of any element of N .

i is of order p.

267

73 Classification of modular p-groups

In view of what was proved so far in the case .ˇ/, it remains to consider the following two possibilities: .ˇ2 / There exists an ai , i 2, such that aip

1

s

¤ 1 and ait D ai1Cp . 1Cp s

for j 3. (We are .ˇ3 / All ai are of order p 1 and, in addition, ajt D aj

1 in case .ˇ/ and so also a1 is of order p since z 2 ha1 i and z is not a p 1 -th power of any element in N .) Assume that the condition .ˇ2 / is satisfied. Hence there exists an ai , i 2, such 1Cp s

p 1

1

and ai ¤ 1. Here t p D z 2 ha1 i and therefore ht i \ hai i D that ait D ai 1 1 1 p 1 p 1 p ¤ 1 and ai ¤ 1, we get .t ai /p D t p ai ¤ 1 so f1g. Since t ht ai i \ hai i D f1g. It follows that elements ai ; t ai are contained in a basis of G since ai and t ai are both of order p and they are “linearly independent” mod 1 .G/ (see the proof of Proposition 73.7). Therefore, there exists a complement T of hai i in G containing t ai . We have N D .N \ T / hai i and t ai operates in the same way as t on N \ T . If t ai does not transform each element of N \ T into its .1 C p s /-th s N 1 with kN 6 0 power, then there exists aN 2 N \ T such that aN tai D aN t D aN 1Cp t kp 1 1 1 1 p 2 T . Since .t ai /p D t p ai 2 T , it follows that .mod p/, and so t p aip

1

s

2 T , a contradiction. Hence at D a1Cp for each a 2 T \ N and ait D ai1Cp . s

s

But N D .T \ N / hai i is abelian and so x t D x 1Cp for all x 2 N and we are done. Assume, finally, that the condition .ˇ3 / is satisfied. We distinguish here two subcases: (i) s < 1 and (ii) s 1. 1Cp s

1

t k1 p , k1 6 0 .mod p/ and ait D (i) Suppose that s < 1. If a1t D a1 s 1s , where ˛ is such that ai1Cp for i 2, then we replace a1 with a1 D a1˛ t p ˛k1 1 .mod p/. Applying Proposition 73.11, we prove that N D ha1 ; a2 ; : : : ; ar i s is an abelian normal subgroup of G with G D hN ; t i, .a1 /t D .a1 /1Cp , and ait D s 1Cp for i 2 (where fa1 ; a2 ; : : : ; ar g is not necessarily a basis of N ). Indeed, ai 1s commutes with each ai , i D 1; : : : ; r, and so N is abelian. It remains to tp compute .a1 /t D .a1˛ t p

1s

/t D .a1t /˛ t p

s

1

s

1s

D .a1˛ /1Cp t ˛k1 p D .a1˛ /1Cp .t p 1Cp s

tp

1s

1s

s

s

D .a11Cp t k1 p s

D .a1˛ /1Cp t p

/1Cp D .a1˛ t p

1s

1

1

tp

/˛ t p

1s

1s

/1Cp D .a1 /1Cp : s

s

t k2 p with k2 6 0 .mod p/, then we replace a2 with a2 D If a2t D a2 1s , where ˛ is such that ˛k2 1 .mod p/. Applying Proposition 73.11 a2˛ t p again, we see that N D ha1 ; a2 ; a3 ; : : : ; ar i is an abelian normal subgroup of G with G D hN ; t i (where fa1 ; a2 ; a3 ; : : : ; ar g is not necessarily a basis of N ). It remains 1

268

Groups of prime power order

only to compute .a2 /t D .a2˛ t p

1s s

D a2˛.1Cp / t p

˛.1Cp s / ˛k2 p 1 p 1s

/t D a2 1

tp

t

1s

t

D .a2˛ t p

1s

/1Cp D .a2 /1Cp : s

s

Then our theorem either holds or we have the case considered already in the previous paragraph (noting that the order of a2 is p 1 ). (ii) Suppose that s 1. Since o.ai / p 1 for all i , then G is either abelian or G 0 D hzi D Z is of order p. Then the theorem follows from Proposition 73.9. Case (II). N is nonabelian. The idea here is to reduce this case to Case (I). By Proposition 73.9, N has the following structure. There is a basis fa1 ; a2 ; : : : ; ar g of N such that pm

a1

pn

D a2 D 1;

ai aj D aj ai

1Cp m1

a1a2 D a1

m > 1; n 1;

p m1

for i; j 3;

D 1 for

ai

N D ha1 ; a2 i ha3 ; : : : ; ar i;

; i 3;

and if p D 2; then m > 2:

p m1

kp m1

Here Z D ha1 i D N 0 Z.G/ and, for each a 2 N , we have at D a1Cp a1 , where k .mod p/ depends on a. ˇ Replacing t by a suitable t a1˛ a2 (˛ and ˇ are integers .mod p/), we may assume s

a1t D a11Cp

(6)

s

s

and a2t D a21Cp :

1Cp s k1 p m1 1Cp s k2 p m1 a1 and a2t D a2 a1 , where k1 ; k2 2 m1 p m1 a2 a1 f0; 1; : : : ; p 1g. It follows from a1 D a1 a1 that a2 D a2 a1p , and these ˇ a2 a1˛ 1Cˇp m1 ˛p m1 and a2 D a2 a1 . We shall choose ˛ and two relations imply a1 D a1 ˇ so that ˛ k2 .mod p/ and ˇ k1 .mod p/. Then we compute (noting that

Indeed, we have a1t D a1

s 1) ˇ

ta1˛ a2

s

D .a11Cp a1k1 p

a1

m1

ˇ

/a2 D .a11Cˇp

1Cp s C.ˇ Ck1 /p m1

D a1 ˇ

ta1˛ a2

and similarly a2

s

D .a21Cp a1k2 p

m1

m1

s

/1Cp a1k1 p

1Cp s

D a1

s

˛ ˇ

.˛Ck2 /p m1

/a1 a2 D a21Cp a1

Since Z Ã 1 .G/, where p D exp.G/, a generator z D a1p p m1

th power of some element c. Suppose that a1 Then (7)

a1p

m1

D cp

1

m1

D a1e1 p

1

Dc

a2e2 p

p 1

1

m1

s

D a21Cp .

of Z is a p 1 -

and c D a1e1 a2e2 : : : arer t f0 .

t f0 p

1

:

269

73 Classification of modular p-groups p 1

p 1

¤ 1. Then m D . If s D 1 and a2

Suppose that a1

G0

1 D ha1p i p 1 (i) If a1

is of order p and our theorem holds by Proposition 73.9. p 1

¤ 1, s D 1, and a2 t 1

From (6) we get a1

D

1p 1 a1 ,

a

a1 2 D a1a2 t p 1

1

¤ 1, then we replace a2 by a2 D a2 t 1 .

and so p 1 p 1 a1

p 1 t 1

D .a1 a1

D a1 a1

/

D a1 ; 1Cp 1 ki p 1 a1

2 Z.G/. Assume that, for an i 3, we have ait D ai

since a1

k p 1 ai a1 i

D 1, then

with ki 6 0 .mod p/. This gives Œai ; t D p 1 ha1 i.

k p 1 a1 i

D

and so H D hai ; t i

contains Z D But Z is normal in G and so H 0 D Z is of order p. By Lemma 65.2(a), H is minimal nonabelian and by Remark, following Proposition 73.2, H is p 1 metacyclic. Therefore 1 .H / Š Ep 2 which implies 1 .H / D h1 .hai i/; a1 i: In particular, 1 .ht i/ 1 .H /. Also Œai ; t 1 D Œai ; t 1 D a1ki p

1 ait

D

k p 1 ai a1 i .

a

Now consider the subgroup K D hai ; a2 i. We have ai 2 D aia2 t

k ai a1 i

k a1 i

p 1

1

p 1

1

, and so

D ait

1

D

and so Œai ; a2 D , which is a generator of Z. Thus K contains Z and since Z is normal in G, we have K 0 D Z. It follows that K is also minimal 1 nonabelian and metacyclic and we have 1 .K/ D h1 .hai i/; a1p i D 1 .H /: In 1 D particular, 1 .ha2 i/ 2 1 .H /. We compute, using Proposition 73.6, .a2 /p .a2 t 1 /p

1

p 1 p 1

D a2

t

and since t p

1

p 1 h1 .hai i/; a1 i.

p 1

2 1 .H /, we get 1 ¤ a2

2

1 .H / D This is a contradiction since fa1 ; : : : ; ar g is a basis of N . We have proved that t centralizes ha3 ; : : : ; ar i. This implies that the subgroup N D ha1 ; a2 ; a3 ; : : : ; ar i is abelian. Since t normalizes ha2 i, it follows that t centralizes a2p

1

. We compute 1Cp 1 1

.a2 /t D .a2 t 1 /t D a2t t 1 D a2 p 1

D a2 a2

D .a2 /1Cp p 1

1

t

tp

1

p 1

D .a2 t 1 /a2

;

D .a2 t 1 /p D a2 t p . Also, G D hN ; t i. If t p D 1, since .a2 /p 1 t 1Cp for each x 2 N and then t normalizes the abelian subgroup N and x D x 1 we are done. Suppose that t p ¤ 1. Note that t induces an automorphism of order 1 1 p on N and so t p centralizes N . In particular, t p 2 Z.G/. Hence A D N ht p i 1 is a normal abelian subgroup of G, G D Aht i, and for each a 2 A, at a1Cp 1 1 .mod ht p i/. Finally, t p 2 Ã 1 .G/ and this is exactly the situation studied in Case (I). 1

1

1

1

270

Groups of prime power order p 1

(ii) Suppose that a1 ¤ 1 .m D / and s < 1. In this case we replace 1s . Since o.ai / p 1 for i > 2, Proposition 73.11 implies a2 by a2 D a2 t p p 1s

p 1

that ha2 ; a3 ; : : : ; ar i is abelian. Again, Proposition 73.11 gives a1t and conjugating this relation with t p a1t

1s

p 1 a1

and so

a

a1 2 D a1a2 t

p a1t

1s

p 1s

1p 22

D a1

p 1s

D

(noting that

1p 1 a1 .

D .a11Cp

1

/t

p 1 a1

D a1 a1

,

2 Z.G/) we get a1 D

We compute

p 1s

D .a11p

1

/1Cp

1

D a1

since 2 2 and so N D ha1 ; a2 ; a3 ; : : : ; ar i is an abelian subgroup of G. It remains to compute .a2 /t . 1s Let us consider the subgroup H D ha2 ; t p i. Since, by Proposition 73.11, p 1s

1

1

a2t D a21Cp , it follows that H 0 D ha2p i is of order p. By Proposition 1s ps 73.11, a2 (s 1 and if p D 2 then s 2) commutes with t p and so working in the subgroup H of class 2, we get .a2 /1Cp D .a2 t p s

D a2 t p

1s

1s

/1Cp D a2 t p s

s

a2p t p

On the other hand, .a2 /t D .a2 t p s 1 . we get finally .a2 /t D .a2 /1Cp t p

1s

We have G D hN ; t i. If t p

1

1

s

.a2 t p

D a21Cp t p t

1s

1s

1Cp s p 1s

/t D a2

p 1

2 ha1

1s

t p

s

/p D

1

:

, and so by the above

i, then the abelian subgroup N is nor1

mal in G and we have reduced this case to Case (I). So suppose t p 62 ha1p i. By 1s 1s centralizes a3 ; : : : ; ar and t p induces an automorProposition 73.11, t p 1 1s p s phism of order p on ha1 i and ha2 i and so t p D .t p / 2 Z.G/. Hence 1 p N D hN ; t i is a normal abelian subgroup of G, G D hN ; t i, and, for each 1 s 1 x 2 N , we get x t D x 1Cp t kx p a1lx p , where kx , lx are integers depending on x. Also, t and a1 are both of order p and they are “linearly independent” modulo 1 .G/. Indeed, if t n1 a1n2 2 1 .G/ (n1 , n2 integers), then (Proposi1

1

p 1

tion 73.6) .t p /n1 .a1 /n2 D 1 and so n1 n2 0 .mod p/. Hence there is a basis a1 ; t; : : : of G (Proposition 73.7). In particular, there is a complement T of ha1 i in G containing t . Since ha1 i N , we get N D ha1 i .N \ T /, by the 1 s 1 with modular law. Take a 2 N \ T and assume that at D a1Cp t ka p a1la p p 1

2 T , a contradiction. Hence, for each a 2 N \ T , la 6 0 .mod p/. Then a1 s 1 t 1Cp s ka p 1 t . But N is abelian and so x t D x 1Cp t kx p for all we get a D a 1 1 p p x 2 N . Since ht i 2 Z.G/ and t 2 Ã 1 .G/, we have again reduced this case to Case (I).

271

73 Classification of modular p-groups p 1

(iii) Suppose, finally, that a1 p m1

(8)

D 1. Then from relation (7) follows

e p 1 f0 p 1

D a22

a1

t

tp

;

1

¤ 1;

f0 6 0

.mod p/ 1

since fa1 ; a2 g is a basis of the minimal abelian subgroup ha1 ; a2 i. Also, t p 2 p m1 p 1 ; 1 .ha2 i/i but t 62 1 .ha2 i/. 1 .ha1 ; a2 i/ D ha1 Denote p l D jha1 ; a2 ; t i W ha1 ; a2 ij and note that (6) implies that t normalizes ha2 i l p (and ha1 i) and so t p 2 Nha1 ;a2 i .ha2 i/ D ha1 i ha2 i. Replacing a1 with a1i .i 6 0 .mod p//, we may assume (writing again a1 instead of a1i ) that t p D a1p a2hp ; where k 1; h 6 0 .mod p/: k

l

(9) pk

f

hp f

pf

Since Œa1 ; a2 D 1 and 1 .ht i/ 6 1 .ha2 i/, we get o.a2 l o.t p /

k o.a1p /.

l o.t p /

p k

/ o.a1

k o.a1p /

/ and so

D On the other hand, D and D and so m

l D m k. By assumption, > m, where p D o.a1 /, and so l > k 1. If f D 0, then we consider the normal abelian subgroup M D ha1 ; a3 ; : : : ; ar i of pk l G. Since a2h D a1 t p , h 6 0 .mod p/, we have G D hM; t i. For each x 2 M , m1

p l

m1

p mk

x t D x 1Cp a1kx p , where (8) gives a1p D .a2e2 t f0 /p 2 Z.G/. This case is reduced to Case (I). Therefore we assume in the sequel that f 1. k f l The element t centralizes t p D a1p a2hp , where h 6 0 .mod p/, k 1, f 1. On the other hand, t normalizes ha1 i and ha2 i and so (since fa1 ; a2 g is a pk pf basis of the subgroup ha1 ; a2 i) t centralizes ha1 i and ha2 i. This implies that t p s

p k1

centralizes ha1

l1

p f 1

i and ha2

i. We have t p

p k1

centralizes ha1 l 2. Thus t p We consider an element (10)

1

p f 1

; a2

g D tp

l1

l1

2 ht p i since l > k 1 and so

i.

xp k1 yp f 1 a2

a1

and show that there exist integers x 6 0 .mod p/ and y 6 0 .mod p/ such that l1 ha1 ; a2 i contains elements of order p.) Note that g p D 1. (Hence the coset t p k1 l1 xp yp f 1 p centralizes ha1 ; a2 i and from (9) follows t (11)

l

p k hp f

t p a1 a2

k1

D 1: f 1

k1

f 1

We get from (10) in any case g p D t p .a1xp a2yp /p , and so if Œa1p ; a2p D 1, then we take x D 1 and y D h so that (with the help of (11)) follows g p D k f k1 f 1 l t p a1p a2hp D 1. If Œa1p ; a2p ¤ 1, then k D f D 1 (since ha1 ; a2 i is minimal nonabelian) and Œa1 ; a2 p D 1. If p > 2, then ha1 ; a2 i is p-abelian and so setting k f l again x D 1 and y D h we get g p D t p a1p a2hp D 1. However, if p D 2, then m1 Œa1 ; a2 D a12 is an involution in Z.G/ and here we set x D 1 2m2 (noting that l

272

Groups of prime power order

in this case m > 2 according to Proposition 73.9) and y D h. We obtain in that case also l

m2

g 2 D t 2 .a112

m1 C2m1

l

m2 2.12m2 / 2h a2 Œa2 ; a1 .12 /h

l

a2h /2 D t 2 a1

D t 2 a122

a22h D 1;

m2

m1

. since Œa2 ; a1 .12 /h D Œa2 ; a1 D a12 p lk ; a1 i noting that l > k 1. Using ProposiWe consider the subgroup K D ht l1 p k1 tion 73.13 and (9), we get Ãk1 .K/ D ht p ; a1 i and l

p k hp f a2

pk

Ãk .K/ D ht p ; a1 i D ha1

pk

pk

pf

; a1 i D ha1 ; a2 i Z.ha1 ; a2 i/:

Since K=Ãk .K/ is of exponent p k , Proposition 73.6 implies that K=Ãk .K/ is p k1 abelian. Take the element tp

(12)

l1

xp k1

a1

2 Ãk1 .K/

where x 6 0 .mod p/ is the integer from (10). Using Proposition 73.12, we see that lk there exists an element g D t ˛p a1ˇ (˛, ˇ integers) in K such that .g /p

(13)

k1

D tp

l1

xp k1

a1

:

On the other hand, using the fact that K=Ãk .K/ is p k1 -abelian, we get (14)

.g /p

k1

D .t ˛p

lk

a1ˇ /p

k1

D t ˛p

l1

a1ˇp

k1

s;

where (15)

p k ıp f a2

s D a1

2 Ãk .K/ .; ı integers/:

Using (12), (14), and (15), we get (16)

xp k1

a1

D t .˛1/p

l1

p k1 .ˇ Cp/ ıp f a2

a1

:

l1

It follows that t .˛1/p 2 ha1 ; a2 i and so ˛1 0 .mod p/. We may set ˛1 D d (d integer), where ˛ 6 0 .mod p/ and together with (14) we get (17)

t .˛1/p

l1

D t dp D a1dp a2hdp : l

k

f

From (16) and (17) (noting that x 6 0 .mod p/) follows also ˇ 6 0 .mod p/ since fa1 ; a2 g is a basis of ha1 ; a2 i. k1 yp f 1 Substituting (13) in (15), we get g D .g /p a2 and so the element g of order p (not being contained in ha1 ; a2 i) is contained in the subgroup H D hg ; a2 i.

273

73 Classification of modular p-groups

On the other hand, H is modular with d.H / D 2 and so j1 .H /j D p 2 and therefore n1 n1 1 .H / D hg; a2p i, where 1 .ha2 i/ D ha2p i. From a1a2 D a11Cp s a21Cp

D compute a2t

we get

t ˛p

a2g D a2 p m1

since a1

lk

m1

˛p a2t

ˇ

a1

follows a2a1 D a2 a1p

lk

D

a2u

m1

a

m1

. From

with a certain integer u 6 0 .mod p/. Then we

ˇ

ˇ

D .a2u /a1 D .a2 1 /u D .a2 a1ˇp ˇ up m1

2 Z.G/. Hence a1

ˇ

and a2 1 D a2 a1ˇp

a

m1

/u D a2u a1ˇ up

m1

;

2 H and ˇu 6 0 .mod p/ implies a1

p m1

2

is a contradiction since

n1 hg; a2p i

\

m1 n1 H . Thus a1p 2 1 .H / D hg; a2p i. This p n1 i. The theorem is proved. ha1 ; a2 i D ha2

Minimal nonmodular p-groups are described in [Jan1]. Corollary 73.16 (v. d. Waall). Every non-Dedekindian modular 2-group has a characteristic subgroup of index 2. Proof. By Theorem A.24.4, G is Q8 -free. Now the result follows from Ward’s Theorem 56.1. As follows from v. d. Waall’s result, if p > 2 and G is nonabelian modular, it contains a characteristic subgroup of index p. Exercise. Let H be a powerful 2-group. If a 2-group G is lattice isomorphic with H via , then G is also powerful. (Hint. Use Proposition 73.5.)

74

p-groups with a cyclic subgroup of index p 2

The title groups have been determined in [Nin] (see also the old paper [HT1] by Hua and Tuan), where these groups are given in terms of generators and relations without any comments about its subgroup structure and therefore this result was not very useful for applications. We shall classify here the title groups in a structural form. If a group G of order 24 has no cyclic subgroups of index 4, it is elementary abelian. Therefore, it suffices to classify the 2-groups with cyclic subgroup of index 4, which have the order > 24 . Theorem 74.1. Let G be a nonabelian group of order p m , p > 2, m > 3, and exponent p m2 . Then one and only one of the following holds: (a) G is metacyclic, jG=Ã2 .G/j D p 4 . (b) m > 4, G is an L3 -group. (c) m D 4, G is regular, 1 .G/ is of order p 3 and exponent p. (d) m D 4, G is irregular, p D 3, G is of maximal class. Proof (Berkovich). Let Z < G be a cyclic subgroup of index p 2 in G. Suppose that G has a normal subgroup E of order p 3 and exponent p; then G D EZ with jE \ Zj D p. We know that exp.Aut.E//p D p and so jG W CG .E/j p. Therefore, if m > 4, then 1 .G/ D E and G is an L3 -group. Now let m D 4. If G is regular, then j1 .G/j 2 fp 2 ; p 3 g. In the first case, G is metacyclic (Lemma 64.1(a,m). In the second case, 1 .G/ is of exponent p (Lemma 64.1(a)). Next suppose that G is irregular. Then p D 3 (Lemma 64.1(a)). (Note that a 3-group of maximal class and order 34 has a cyclic subgroup of index 32 , since it is irregular.) In what follows we assume that m > 4 and G has no normal subgroups of order p 3 and exponent p. According to Theorem 13.7 (see also Theorem 69.4), G is either metacyclic or a 3-group of maximal class. Let us consider these possibilities. Let G be metacyclic. In that case, G is regular. Since G has no cyclic subgroups of index p, the subgroup Ã1 .G/ is noncyclic. Therefore, jG=Ã2 .G/j D p 4 . Let us show that Ã2 .G/ is cyclic. If not, G=Ã3 .G/ is of order p 6 and exponent p 3 , which is not the case since exp.G/ D p m2 . Conversely, if G is metacyclic with jG=Ã2 .G/j D p 4 and 2 Ã2 .G/ is cyclic, it has a cyclic subgroup of index p 2 since Ã2 .G/ D fx p j x 2 Gg (Lemma 64.1(a)).

74

p-groups with a cyclic subgroup of index p 2

275

Now let G be a 3-group of maximal class, m > 4. Then G1 , the fundamental subgroup of G, is metacyclic and exp.G1 / D exp.G/ (see 9). It follows that G1 has a cyclic subgroup of index 3. In that case, Ã1 .G1 / is cyclic of index p 2 in G1 so jÃ1 .G/j 32 . This is a contradiction since a p-group of maximal class and order > p 3 , p > 2, has no normal cyclic subgroups of order > p. Theorem 74.2. Let G be a nonabelian group of order 2m , m > 4, and exponent 2m2 . Then one of the following holds: (a) G is an L3 -group. (b) G is the uniquely determined group of order 25 with 2 .G/ Š D8 C2 . The group G has a normal elementary abelian self centralizing subgroup E D hz; u; ei of order 8 and an element a of order 8 such that G D Ehai, where a4 D z, e a D eu, and ua D z. Here ˆ.G/ D ha2 ; ui is abelian of type .4; 2/, G 0 D 1 .ˆ.G// Š E4 , Z.G/ D Ã1 .ˆ.G// is of order 2, and c3 .G/ D 4. (c) G is a U2 -group (all such groups are completely determined in 67). If m > 5, then cm2 .G/ D 2. (d) G D W Z with W \ Z Š C4 , where W is an abelian normal subgroup of type .4; 4/ and Z Š C2m2 . We have W D 2 .G/ and G is metacyclic. Also, 2 jZ W CZ .W /j 4, (since exp.Aut.W //2 4; see 33) and cm2 .G/ D 4. (e) G D QZ, where Q Š Q8 is a normal subgroup of G, Q\Z D Z.Q/, Z D hbi Š C2m2 and b either centralizes Q or b induces on Q an involutory outer automorphism in which case m > 5. Also we have cm2 .G/ D 4. (f) G is the uniquely determined group of order 25 with 2 .G/ D ha; bi hui, where ha; bi Š Q8 and u is an involution with CG .u/ D 2 .G/. Let hzi D Z.Q/; then a2 D b 2 D z. There is an element y of order 8 in G such that y 2 D ua, uy D uz, ay D a1 , b y D bu. Here ˆ.G/ D hy 2 ; ui is abelian of type .4; 2/, G 0 D 1 .ˆ.G// Š E4 , Z.G/ D Ã1 .G/ is of order 2, and c3 .G/ D 4. Proof (Janko). Suppose that G has a normal subgroup E Š E8 ; then G D EZ with Z Š C2m2 and E \ Z Š C2 . If K=E is the subgroup of order 2 in G=E, then 1 .G/ K. Therefore, if K is abelian, we get 1 .G/ D E, and so G is an L3 group. Now let CG .E/ D E. Since exp.Aut.E//2 D 4, we get jGj D 25 and Z D hai is of order 8. In that case, jZ.G/j D 2. Therefore, setting z D a4 , one can choose e; u 2 E so that E D he; u; zi; e a D eu and ua D uz. The structure of G is determined and we get the group (b) stated in the theorem. From now on we assume that G has no normal elementary abelian subgroups of order 8. Since G is neither cyclic nor of maximal class, it has a normal four-subgroup W0 . Suppose that G=W0 is cyclic (of order 2m2 ), then G D W0 S with S Š C2m2 and W0 \ S D f1g since G has no cyclic subgroups of index 2. But then W0 1 .S / Š E8 is normal in G, contrary to our assumption. Hence, G=W0 is noncyclic. Let Z < G be cyclic of order 2m2 . Then W0 \ Z Š C2 and so W0 Z is maximal

276

Groups of prime power order

in G. It follows that G=W0 has a cyclic subgroup .W0 Z/=W0 of order 2m3 4 and index 2. Set F=W0 D ˆ.G=W0 / so that F < W0 Z and F=W0 is cyclic of order 2. Since CW0 Z .W0 / F (indeed, jW0 Z W CW0 Z .W0 /j 2), it follows that 1 .W0 Z/ D W0 . Let M=W0 be any cyclic subgroup of index 2 in G=W0 . Since M > F and 1 .M / 1 .F / D W0 , we get 1 .M / D W0 . If G=W0 is of maximal class, then G is an U2 -group and all such groups have been determined in 67. Note, that if m > 5, then all cyclic subgroups of order 2m2 are contained in the L2 -subgroup W0 Z so c2m2 .G/ D c2m2 .W0 Z/ D 2. We may assume that G=W0 is not of maximal class. It follows that G=W0 is either abelian of type .2m3 ; 2/ or G=W0 Š M2m2 in which case m > 5 (Lemma 64.1(t)). In any case, G=W0 has exactly two cyclic subgroups M1 =W0 and M2 =W0 of index 2 and the third maximal subgroup M3 =W0 is abelian of type .2m3 ; 2/. Hence M3 cannot contain cyclic subgroups of index 2. But 1 .Mi / D W0 and so Mi is either abelian of type .2m2 ; 2/ or Mi Š M2m1 , i D 1; 2. In any case, cm2 .Mi / D 2 and two cyclic subgroups of order 2m2 in Mi generate Mi , i D 1; 2. Since exp..M1 \ M2 /=W0 / D 2m4 , it follows that M1 \ M2 does not contain any cyclic subgroup of order 2m2 . We get (see the proof of Theorem 1.10) cm2 .G/ D cm2 .M1 / C cm2 .M2 / D 2C2 D 4 and G is generated by its four cyclic subgroups of order 2m2 . We are now in a position to use Theorem 54.2. The groups (d) and (a) of our theorem correspond to possibilities (a) and (b) of Theorem 54.2, respectively. In case (c) of Theorem 54.2, we have here only the possibility jGj D 25 with 2 .G/ Š Q8 C2 (since in case 2 .G/ D D8 C2 the group G would have a normal elementary abelian subgroup of order 8). This leads to the group (f) of our theorem, and we are done. Exercise. Let p > 2 and let G be a nonmetacyclic group of order p m > p 4 with nonnormal cyclic subgroup L of order p m2 . Suppose that G is not minimal nonabelian. Prove that G has a maximal subgroup H Š Mp m1 . Solution. We have p > 2. The quotient group G=LG is isomorphic to a Sylow psubgroup of the symmetric group Sp 2 so LG is cyclic of order > p since m > 4. It follows from Theorem 9.6 that G is not a p-group of maximal class. Then, by Theorem 69.3, G has a normal subgroup R of order p 3 and exponent p. Since G has a cyclic subgroup of index p 2 , then 1 .G/ D R, G=R is cyclic so exp.k .G// p k for all k 2 N. By Proposition 10.28, G is generated by minimal nonabelian subgroups. It follows that G has a minimal normal subgroup H of exponent p m2 . By hypothesis, H < G. Then H has a cyclic subgroup of index p, and the result follows from Theorem 1.2.

75

Elements of order 4 in p-groups

All results of this section are due to the second author. If G is a finite p-group and n 2 N a fixed natural number, then we define n .G/ D hx 2 G j o.x/ D p n i. It is a known fact that 2 .G/ has a strong influence on the structure of a finite 2-group G. For example, if 2 .G/ is metacyclic, then the 2-group G is also metacyclic (this follows from the classification of minimal nonmetacyclic p-groups; see Theorem 66.1). In 52, all 2-groups G with the property j2 .G/j D 24 are classified. If j2 .G/j 23 , then we get for a 2-group G four infinite classes of groups (see Lemma 42.1). In this section we consider for the first time the case, where for 2 .G/ we could have an infinite class of 2-groups. More precisely, we assume that G is a finite 2-group with 2 .G/ Š C2 D, where D is any 2-group of maximal class. Then either G D 2 .G/ or jG W 2 .G/j D 2 and the structure of G is uniquely determined (Theorem 75.1). In fact, the most difficult part of the proof is to show that jG W 2 .G/j D 2. The exact determination of the structure of G is obtained by applying a classification of so called U2 -groups given in 67. A 2-group H is said to be a Us -group (s 2) with respect to the kernel R if H has a normal elementary abelian subgroup R of order 2s , H=R is of maximal class and whenever T =R is a cyclic subgroup of index 2 in H=R, then 1 .T / D R. We shall determine also the structure of a finite 2-group G with 2 .G/ Š C2 Q2n , where Q2n is a generalized quaternion group of order 2n , n 4 (see ‘Research problems and themes I’, #563). If 2 .G/ > 2 .G/, then we show that G D 2 .G/ Š C2 SD2nC1 , where SD2nC1 is the semidihedral group of order 2nC1 (Theorem 75.2). The case n D 3 was treated in 55. Finally, we show that a finite p-group G, all of whose noncyclic subgroups H have the property H D 1 .H /, is either cyclic or of exponent p or p D 2 and G is a dihedral group D2n , n 3 (Theorem 75.3). Here the case p > 2 is almost trivial but some proving must be done in case p D 2; see also Exercise 1.113. Theorem 75.1. Let G be a 2-group with 2 .G/ D hui D, where u is an involution, D is any 2-group of maximal class, and jDj D 2n , n 4. If G ¤ 2 .G/, then jG W 2 .G/j D 2, D is dihedral or generalized quaternion, CG .u/ D 2 .G/, G is a nonmetacyclic U2 -group with respect to the kernel R D hui Z.D/, G=R Š SD2n , n1 Z.G/ D Z.D/, and d.G/ D 2. More precisely, we set D D ha0 ; b j .a0 /2 D n2 n3 1; .a0 /2 D z; .a0 /2 D v; b 2 D z ; D 0; 1; .a0 /b D .a0 /1 i, and then

278

Groups of prime power order

there is an element a 2 G 2 .G/ such that a2 D a0 , ua D uz, ab D a1 vu, and this determines the structure of G uniquely. (For n < 4, such groups G have been determined in Lemma 42.1.) Proof. Set H D 2 .G/ so that H D hui D with jDj D 2n , n 4. Let Z be the unique cyclic subgroup of index 2 in D, where jZj D 2n1 8. Since D has no normal four-subgroups, it follows that G has no normal elementary abelian subgroups of order 8. We know that D has no normal abelian subgroups of type .4; 2/ and so H (and also G) has no normal abelian subgroups of type .4; 4/. Set hzi D 1 .Z/ so that hzi D Z.D/. Let hvi be the cyclic subgroup of order 4 in Z, where v 2 D z. Since hvi Ã1 .H / D Ã1 .Z/, it follows that hvi is normal in G and so z 2 Z.G/. Obviously, W0 D hui hzi D Z.H / and so W0 is normal in G. Therefore, W D W0 hvi is normal in G, W0 D 1 .W /, and W is a normal abelian subgroup of G of type .4; 2/. We see that W0 is the unique normal four-subgroup in H and then W is the unique normal abelian subgroup of H of type .4; 2/. It follows that W0 is the unique normal four-subgroup in G and W is the unique normal abelian subgroup of G of type .4; 2/. We are now in a position to use Proposition 50.6(a). It follows that N D CG .W / is abelian of type .2m ; 2/, where m n 1, N \ H D W0 Z D hui Z, G=N is isomorphic to a subgroup of D8 , and W D 2 .N /. Let r be an element in D Z so that r 2 2 hzi and v r D v 1 . Hence r induces an involutory automorphism on the abelian group N and r inverts W D 2 .N /. This implies that CN .r/ D W0 D 1 .W / and so, using Theorems 67.1, 67.2 and 67.3, we see that r inverts N=W0 . Suppose that N ¤ W0 Z. Take an element x 2 N .W0 Z/. We have x r D x 1 w0 with some w0 2 W0 . But then .rx/2 D rxrx D r 2 x r x D r 2 x 1 w0 x D r 2 w0 2 W0 , and so o.rx/ 4. This is a contradiction since rx 62 H D 2 .G/. We have proved that N D W0 Z. Since hvi is normal in G, v r D v 1 , and ur D u for any r 2 D Z, it follows that CG .v/ covers G=H . If G ¤ H , there is s 2 CG .v/H with us D uz, v s D v, so that G=.W0 Z/ is a four-group of automorphisms of W induced with hs; ri. From now on we assume G ¤ H . In that case jG W H j D 2, CG .W / D W0 Z, G=.W0 Z/ Š E4 , and for each x 2 G H , ux D uz so that CG .u/ D huiD D H . It remains to determine the structure of G. Since S D CG .v/ covers G=H and CH .v/ D W0 Z, it follows that S is a nonabelian maximal subgroup of G with S \ H D W0 Z. Hence 2 .S / D W and so we may apply Lemma 42.1 for p D 2. It follows that n1 n2 either S Š M2nC1 or S D ha; b j a2 D b 8 D 1; ab D a1 ; a2 D b 4 i, n 4. In the second case, hb 2 i D hvi D Z.S /; ha2 i D S 0 ; ˆ.S / D Z.S /ha2 i; 1 .S / D 1 .W / D hu; zi; n2

and CS .1 .S // D hb 2 ; ai is the unique abelian maximal subgroup of S z D a2 and so hb 2 ; ai D W0 Z D hui Z. Since W0 Z contains exactly two cyclic subgroups Z1 , Z2 of index 2, one of them, say Z1 D hai, is inverted by the element b 2 S .W0 Z/. But Z1 \ Z2 hvi and so b inverts hvi, contrary to S D CG .v/.

75 Elements of order 4 in p-groups

279

We have proved that S Š M2nC1 . In particular, S has a cyclic subgroup S1 of index 2 containing hvi and so S=W0 is a cyclic subgroup of index 2 in G=W0 . But G=W0 contains DW0 =W0 D H=W0 Š D=hzi, where D=hzi is a group of maximal class. Thus G=W0 is of maximal class and order 2n (n 4) and so S=W0 is the unique cyclic subgroup of index 2 in G=W0 . Since 1 .S / D W0 , it follows that G is a U2 -group with the kernel W0 . We are now in the position to use the classification of U2 -groups given in 67 (see Theorems 67.1 and 67.3). It follows that only the group G given in Theorem 67.3(d) satisfies our assumptions: n

n1

n2

G D ha; b j a2 D 1; a2

D z; a2

D v; b 2 D z ; D 0; 1;

ab D a1 vu; u2 D Œu; b D 1; ua D uz; n 4i; where W0 D hu; zi, G=W0 Š SD2n , Z.G/ D hzi Š C2 , G is non-metacyclic, and d.G/ D 2. Also, CG .u/ D hui D D 2 .G/, where D D ha2 ; bi. For D 0, D is dihedral and for D 1, D is generalized quaternion. Our theorem is proved. Of course, groups of Theorem 75.1 have no normal elementary abelian subgroups of order 8. Therefore, to prove that theorem, we must find all groups of 50 satisfying the hypothesis of Theorem 75.1. However, such proof will be not easier than the presented proof. Theorem 75.2. Let G be a 2-group with 2 .G/ D Q2n C2 , where n 4. If 2 .G/ ¤ 2 .G/, then 2 .G/ D G Š SD2nC1 C2 . (If n D 3, then such groups G have been determined in Theorem 55.1.) Proof. Set H D 2 .G/ D hui Q, where u is an involution and Q Š Q2n , n 4. Assume 2 .G/ ¤ H . Let hai be the unique cyclic subgroup of index 2 in Q and we set n1

Q D ha; b j a2

n2

D b 4 D 1; a2

n3

D z; a2

D v; b 2 D z; ab D a1 i;

where Q1 D ha2 ; bi and Q2 D ha2 ; abi are the other two maximal subgroups of Q and Q1 Š Q2 Š Q2n1 . Also, Z.H / D hu; zi is normal in G. Since hvi ha2 i D Ã1 .H /, hvi and ha2 i are normal subgroups in G and hzi Z.G/. Since 2 .G/ ¤ H , there exists an involution t 2 G H and we set K D H ht i. If ut D uz, then Z.H /ht i Š D8 and so o.ut / D 4, a contradiction since the element ut of order 4 is not contained in H D 2 .G/. Hence t centralizes Z.H /. Since t cannot commute with any element of order 4 in H , we get CH .t / D Z.H / D hu; zi. We act with t on the abelian group hui hai and apply 55. It follows that t inverts the quotient group hu; ai=Z.H / and so at D a1 s with s 2 Z.H /. We compute .t a/2 D .t at /a D a1 sa D s, and so s D 1 (since t a cannot be of order 4 in view t a 62 H ) and at D a1 . Hence t inverts the maximal subgroup hu; ai of H containing Z.H /.

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Groups of prime power order

Suppose that .Q1 Z.H //t D Q1 Z.H /. In that case b t D ba2i uj , where i , j are suitable integers. Working in Q, there is ak 2 hai (k is an integer) such that k k k b a D ba2i and so b ta D .ba2i uj /a D ba2i a2i uj D buj , where t 0 D t ak is an involution in K H . We compute .t 0 b/2 D .t 0 bt 0 /b D buj b D uj z, and so o.t 0 b/ D 4, a contradiction. Hence .Q1 Z.H //t D Q2 Z.H /. It follows b t D aba2r us (r; s are some integers) and so hb; b t i D Q Š Q Š Q2n with K D hui .Q ht i/, where Q ht i Š SD2nC1 . Let N D NG .Q1 Z.H // so that jG W N j D 2, H N , and G D N ht i. Suppose N ¤ H . By the above argument, N H cannot contain involutions and so 2 .N / D H . By Theorem 75.1, we get jN W H j D 2 and N is a uniquely determined group. In particular, by the structure of N , if y 2 N H , .Q1 Z.H //y D Q2 Z.H /, contrary to N D NG .Q1 Z.H //. Hence, N D H and so G D H ht i D K and we are done. Let G be a 2-group. Of course, 2 .2 .G// D 2 .G/. Therefore, if 2 .G/ D C2 D, where D is of maximal class and order > 8, then D is a generalized quaternion group. Remark. Let, as in Theorem 75.1, 2 .G/ D C2 D, where D is a 2-group of maximal class. If D is dihedral, then c2 .G/ D 2, and such groups are classified (see 43). It follows from the results of 54 that the 2-groups G with 2 .G/ D E4 D, where D is dihedral, also known. Similarly, if n > 3 and n1 .G/ D C2 M2n or n1 .G/ D C2 M2n , then cn1 .G/ D 4, and the classification of such groups follows from results of 55. Theorem 75.3. Let G be a p-group all of whose noncyclic subgroups are generated by its elements of order p. Then one of the following holds: G is cyclic, G is of exponent p, G is a dihedral 2-group D2n of order 2n . Proof. Suppose that p > 2 and assume that G is neither cyclic nor of exponent p. Let Z be a maximal cyclic subgroup of order p s , s 2. Since Z ¤ G, there is a subgroup Y > Z with jY W Zj D p. It follows that Y is either abelian of type .p s ; p/ or Y Š Mp sC1 . In any case, 1 .Y / Š Ep 2 so 1 .Y / < Y , a contradiction. From now on assume p D 2. We may also assume that G is not cyclic, exp.G/ 4, and G is not of maximal class. Let R be a normal four-subgroup of G. If CG .R/ has an element v of order 4, then Rhvi contains an abelian subgroup of type .4; 2/, a contradiction. Hence H D CG .R/ is elementary abelian and jG W H j D 2. We have exp.G/ D 4 and let s 2 G H be an element of order 4 so that s 2 2 H . Since G has no abelian subgroups of type .4; 2/, it follows that hsi is a maximal abelian subgroup of G. Then, by Suzuki (Proposition 1.8), G is of maximal class, contrary to the assumption. Problem. Classify the 2-groups G satisfying one of the following conditions: (i) 2 .G/ D E D, where E is elementary abelian and D is of maximal class. (ii) 2 .G/ D E D, where E is elementary abelian and D is of maximal class.

75 Elements of order 4 in p-groups

(iii) n1 .G/ D E M2n , where n > 3 and E is elementary abelian. (iv) n .G/ D E C2n , where n > 1 and E is elementary abelian. (v) n .G/ D E C2n , where n > 1 and E is elementary abelian.

281

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p-groups with few A1-subgroups

1o . Recall that G is an An -group if it contains a nonabelian subgroup of index p n1 but all its subgroups of index p n are abelian (see 72). For example, a p-group G of maximal class and order p m is an Am2 -group since it contains a nonabelian subgroup of order p 3 (see Theorems 9.5 and 9.6). Let ˛n .G/ be the number of An -subgroups in a p-group G. If G is an An -group, then ˛i .G/ > 0 for i D 1; : : : ; n 1, ˛n .G/ D 1 and ˛j .G/ D 0 for j > n. For H < G, we set ˇ1 .G; H / D ˛1 .G/ ˛1 .H /. For example, if H < G is abelian, then ˇ1 .G; H / D ˛1 .G/. If H < G, then iH denotes the set of all members of the set i which contain H . Given H < G, let i .H / D fU < H j ˆ.H / U; jH W U j D p i g. If G is a p-group of maximal class containing an abelian subgroup of index p, then ˛1 .G/ D p m3 . It is fairly difficult to compute ˛1 .G/ even for groups with not very complicated structure. Below we compute ˛1 .G/ for three families of groups. Let X be a group and '2 .X/ the number of noncommuting ordered pairs .x; y/ 2 X X such that hx; yi D X (it follows that '2 .X/ > 0 if and only if X is nonabelian and two-generator). Let k.X/ be the class number of X. We claim that (see 2; this identity is due to Mann) X (1) '2 .H / D jXj.jXj k.X//: H X

Indeed, let fK1 ; : : : ; Kr g (here r D k.X/) be the set of conjugacy classes of X. Then the number of commuting ordered pairs of G equals (2)

r r X X X jXj D rjXj D jXjk.X/ . jCX .x/j/ D jKi j jKi j iD1 x2Ki

iD1

so the number of noncommuting ordered pairs of elementsP of X is identical with 2 jXj jXjk.X/. On the other hand, that number also equals H X '2 .H /. Examples. 1. Let us find ˛1 .G/, where G is extraspecial of order p 2mC1 . We have k.G/ D jZ.G/j C jGZ.G/j D p 2m C p 1. Let E G be nonabelian and twop 0 0 generator. Then E D G and E=E 0 Š Ep 2 since E=E 0 G=E 0 D G=G 0 Š Ep 2m . It follows that jEj D p 3 so E is an A1 -subgroup. We have '2 .E/ D .jE ˆ.E/j/.jEj pjˆ.E/j/ D .p 3 p/.p 3 p 2 / D p 3 .p 2 1/.p 1/:

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283

It follows from (1) that ˛1 .G/'2 .E/ D ˛1 .G/p 3 .p 2 1/.p 1/ D p 2mC1 .p 2mC1 p 2m p C 1/ D p 2mC1 .p 1/.p 2m 1/; and we get ˛1 .G/ D p 2m2 ˛1 .G/ D p 2 .p 2 C 1/.

p 2m 1 . p 2 1

In particular, if jGj D p 5 , i.e., m D 2, then

2. Let G be a nonabelian group of order p m all of whose nonabelian two-generator m3 .p m k.G// subgroups have the same order p 3 . Then, using (1), we get ˛1 .G/ D p .p1/.p 2 1/ . 3. Let G D S Ep n , where S is nonabelian of order p 3 . Then jGj D p nC3 , k.G/ D p n .p 2 C p 1/. If A < G is an A1 -subgroup, then jAj D p 3 . Therefore, by the n nC3 n .p 2 Cp1/ D p 2n . formula in Example 2, we get ˛1 .G/ D p Œp .p 2p 1/.p1/ Epimorphic images of an An -group are Ak -groups with k n. Indeed, if N G G, where G is an An -group, then all subgroups of index p n in G=N are abelian. The main result of this section is the following Theorem A ([Ber30]). Let G be a nonabelian p-group. If ˛1 .G/ p 2 C p C 1, then G is an An -group, n 2 f1; 2; 3g. Exercise 1. Prove that if G is metacyclic, distinct F; H 2 1 are Ar -, As -groups respectively, and r s, then G is an AsC1 -group. Solution. Since G 0 is cyclic, we get F 0 H 0 . It follows from the above, Lemma 64.1(u) that jG 0 j D pjH 0 j D p sC1 so G is an AsC1 -group, by Theorem 72.1. Exercise 2. Let G be a nonabelian p-group. Suppose that its maximal subgroups M1 ; : : : ; Mk contain all A1 -subgroups of G. Is it true that k p? Exercise 3. Suppose that maximal subgroups M1 ; : : : ; MT p together contain all A1 p subgroups of a nonabelian p-group G. Is it true that jG W . iD1 Mi /j D p 2 ? Exercise 4. Let G D M C , where M is nonabelian of order p 3 and C is cyclic of order p 2 . Prove that G has an A1 -subgroup of order p 4 and find the number of such subgroups in G. 2o . Proof of Theorem A. We begin with the following Remark 1. Let G be neither abelian nor an A1 -group and let F1 2 1 be nonabelian. We claim that ˇ.G; F1 / p 1. By Exercise 1.6(a), the set 1 has at least p nonabelian members. Let F2 2 1 fF1 g be nonabelian and set D D F1 \ F2 . Since D — Z.G/, at least p members of the set 1D , say F1 ; : : : ; Fp , are nonabelian. Let Ui Fi be an A1 -subgroup such that Ui — D, i D 1; : : : ; p (such Ui exists, by Proposition 10.28). Then U1 ; : : : ; Up are distinct A1 -subgroups of G and U2 ; : : : ; Up — F1 so ˇ1 .G; F1 / p 1. (For more detailed description of such groups, see Lemma 76.5.)

284

Groups of prime power order

Lemma 76.1 (= Exercise 1.8a (L. Redei)). Let G be an A1 -group. Then G D ha; bi, Z.G/ D ˆ.G/, jG 0 j D p and one of the following holds: m

n

m

n

(a) ap D b p D c p D 1, Œa; b D c, Œa; c D Œb; c D 1, jGj D p mCnC1 , G D hbi .hai hci/ D hai .hbi hci/ (semidirect products with kernels in brackets) is nonmetacyclic. Here G 0 D hci, Z.G/ D ˆ.G/ D hap i hb p i hci. If m C n > 2, then 1 .G/ D h1 .hai/; 1 .hbi/; ci Š Ep 3 . m1

, jGj D p mCn is metacyclic. Here (b) ap D b p D 1, m > 1, ab D a1Cp m1 m1 n1 0 p p p i, Z.G/ D ˆ.G/ D ha i hb i, 1 .G/ D hap ; b p i Š Ep 2 . If G D ha 1 .G/ — Z.G/, then n D 1 so G Š Mp mC1 . (c) a4 D 1, a2 D b 2 , ab D a1 , G Š Q8 . If j1 .G/j p 2 , then G is metacyclic. The group G is nonmetacyclic if and only if G 0 is a maximal cyclic subgroup of G. Next, jG=Ã1 .G/j p 3 with equality if and only if G is from (a) and p > 2. If, in (a), u 2 G ˆ.G/, then hui 6E G. All members of the set 1 have ranks at most 3. If N is normal in G and G=N is not cyclic, then N Z.G/. In what follows G is a nonabelian group of order p m . Let us prove that if, in Lemma 76.1, G 0 is not a maximal cyclic subgroup of G, then G is metacyclic. Let G 0 < L < G, where L is cyclic of order p 2 . By Theorem 6.1, L=G 0 C =G 0 , where C =G 0 is a cyclic direct factor of G=G 0 ; in particular, G=C is cyclic. It remains to show that C is cyclic. We get G 0 D ˆ.L/ ˆ.C /. It follows that C =ˆ.C / as an epimorphic image of a cyclic group C =G 0 , is cyclic. In that case. C is also cyclic, as was to be shown. In what follows G is a nonabelian group of order p m . Lemma 76.2. Suppose that a p-group G is neither abelian nor an A1 -group and 1 D fH1 ; : : : ; Hp ; Ag, where A is abelian. Then H10 D D Hp0 and G=H10 is an A1 group so jG 0 j D pjH10 j. In particular, d.Hi / 3 for all i . If, in addition, d.Hi / D 2 for i D 1; : : : ; p, and L is a G-invariant subgroup of index p in H10 , then G=L is an A2 -group. Proof. By Remark 1, H1 ; : : : ; Hp are nonabelian. Let jH10 j jHp0 j. Since .H1 =H10 / \ .A=H10 / Z.G=H10 /, we get G=H10 =Z.G=H10 / Š Ep 2 so all maximal subgroups of G=H10 must be abelian so H10 D D Hp0 since d.G/ D 2. Assume that G=H10 is abelian; then G 0 D H10 . Let L be a G-invariant subgroup of index p in G 0 . Then, by Lemma 65.2(a), G=L is an A1 -group since d.G=L/ D 2, so H10 L < G 0 D H 0 , a contradiction. Thus, G=H10 is an A1 -group so jG 0 j D pjH10 j. By Lemma 76.1, d.Hi / D d.Hi =H10 / 3. Next suppose that d.Hi / D 2 for i D 1; : : : ; p and L is taken as above; then Hi =L is an A1 -group (Lemma 65.2(a)) so G=L is an A2 -group. Lemma 76.3. If all members of the set 2 are abelian, then d.G/ 3. If, in addition, d.G/ D 3, then ˆ.G/ Z.G/.

76 p-groups with few A1 -subgroups

285

Proof. Suppose that d.G/ > 3. Take F 2 1 ; then F contains at least p 2 C p C 1 (abelian) members of the set 2 so it is abelian (Exercise 1.6(a)). In that case, since G is not two-generator, it is abelian (Lemma 76.1), contrary to the hypothesis. Now suppose that d.G/ D 3. Then CG .ˆ.G// hH j H 2 2 i D G, completing the proof. In what follows we make use of the following fact: If N is a normal subgroup of a p-group G, jG=N j > p 2 and G=N is generated by subgroups of index p 2 , whose inverse images in G are abelian, then N Z.G/. Indeed, CG .N / hH < G j N < H; jG W H j D p 2 i D G. Lemma 76.4. Let G be a nonmetacyclic A2 -group of order p m > p 4 . Then (a) d.G/ 3, jG 0 j p 3 , exp.G 0 / D p, G=Z.G/ is of order p 3 and exponent p. (b) ˛1 .G/ 2 fp; p C 1; p 2 ; p 2 C p; p 2 C p C 1g. (c) If ˛1 .G/ < p 2 , then p > 2, d.G/ D 2, and cl.G/ D 3. (c1) If ˛1 .G/ D p, then G=.G 0 \ Z.G// is an A1 -group, G 0 Š Ep 2 . (c2) If ˛1 .G/ D p C 1, then jGj D p 5 , G 0 D 1 .G/ Š Ep 3 , Z.G/ D K3 .G/ D Ã1 .G/ Š Ep 2 . (d) If ˛1 .G/ D p 2 , then d.G/ D 3, G=Z.G/ Š Ep 2 , jG 0 j D p. (e) If ˛1 .G/ D p 2 C p, then d.G/ D 3, G 0 Š Ep 2 , G 0 Z.G/ D ˆ.G/. (f) If ˛1 .G/ D p 2 C p C 1, then p D 2, jGj D 26 , G is special with G 0 D Z.G/ D ˆ.G/ D 1 .G/ Š E8 . Proof. (a) Two inequalities follow from Lemmas 76.1 and 64.1(u). Next, exp.G 0 / D p (Theorem 65.7(d)). If d.G/ D 3, then ˆ.G/ Z.G/, by the paragraph preceding the lemma. If p > 2, then jG=Ã1 .G/j > p 2 (Theorem 9.11) so G=Ã1 .G/ is generated by subgroups of index p 2 , and we get Ã1 .G/ Z.G/. Hence, the last assertion of (a) holds for p > 2. Now suppose that p D 2 and d.G/ D 2. However, this case does not occur according to Propositions 71.1 to 71.5. Indeed, the results of 71 imply that an A2 -group of order 2m > 24 with d.G/ D 2 must be metacyclic (this is also true for m 4, by Lemma 64.1(i)). Assertions (b) to (f) are direct consequences of 71. Indeed, if G is a group of Proposition 71.1, then ˛1 .G/ D p 2 , d.G/ D 3, G=Z.G/ Š Ep 2 , and jG 0 j D p. If G is a group of Proposition 71.3, then ˛1 .G/ D p, p > 2, d.G/ D 2, cl.G/ D 3, N D 2 and jGN 0 j D G 0 Š Ep 2 , and GN D G=.G 0 \ Z.G// is an A1 -group since d.G/ p (Lemma 65.2(a)). If G is a group of Proposition 71.4, then ˛1 .G/ D p 2 C p, d.G/ D 3, G 0 Š Ep 2 , and G 0 Z.G/ D ˆ.G/. If G is a group of Proposition 71.5 (a), then ˛1 .G/ D p 2 C p C 1, p D 2, d.G/ D 3, jGj D 26 , G is special with G 0 D Z.G/ D ˆ.G/ D 1 .G/ Š E8 ; moreover, G is isomorphic to a Sylow 2-subgroup of the Suzuki simple group Sz.8/. If G is a group of Proposition 71.5(b),

286

Groups of prime power order

then p > 2, d.G/ D 2, cl.G/ D 3, ˛1 .G/ D p C 1, jGj D p 5 , G 0 D 1 .G/ Š Ep 3 , Z.G/ D K3 .G/ D Ã1 .G/ Š Ep 2 . Our proof is complete since the A2 -groups of Proposition 71.2 are metacyclic. Let us give a proof, independent of 71, that if G is a nonmetacyclic A2 -group with ˛1 .G/ < p 2 , then cl.G/ D 3. It follows from Exercise 1.6(a) that d.G/ D 2. Assume that cl.G/ D 2. Then G 0 Z.G/. Since G is nonmetacyclic, we get exp.G 0 / D p, by Lemma 76.4(a). In that case, G is an A1 -group (Lemma 65.2(a)), a contradiction. It remains to prove that cl.G/ D 3. This is the case if jG 0 j D p 2 . Now let jG 0 j D p 3 (see Lemma 76.4(a)); then ˛1 .G/ D p C 1 and G is nonmetacyclic with all two-generator members of the set 1 since the latter has no abelian members (Lemma 64.1(u)). It follows that G=Ã1 .G/ is nonabelian of order p 3 and exponent p and Ã1 .G/ D K3 .G/ so jG W G 0 j D p 2 (Lemma 64.1(p)), and we get jGj D jG W G 0 jjG 0 j D p 5 . Since all subgroups of order p 3 that contain K3 .G/, are abelian (G is an A2 -group!) and generate G, we get K3 .G/ D Z.G/ since jG W Z.G/j > p 2 , and so cl.G/ D 3. Remarks. Suppose that A; B 2 1 are distinct. 2. If A is abelian and B an A1 -group, then jG=Z.G/j p 3 . If, in addition, A is the unique abelian member of the set 1 , then G=Z.G/ is either of order p 3 and exponent p or G=Z.G/ Š D8 . Indeed, jG 0 j p 2 (Lemmas 64.1(u) and 76.1) so jG W Z.G/j p 3 (Lemma 64.1(q)). If A is the unique abelian member of the set 1 , then jG W Z.G/j D p 3 so G=Z.G/ has at most one cyclic subgroup of index p, and the last assertion follows. 3. If A; B are A1 -subgroups, then jG=Z.G/j p 4 . Indeed, put D D A \ B. Then Z.A/ D ˆ.A/ ˆ.G/ A \ B D D and similarly Z.B/ < D so, comparing orders, we conclude that Z.A/ and Z.B/ are maximal in D (Lemma 76.1). It follows that U D Z.A/ \ Z.B/ has index at most p 2 in D and U Z.G/ since CG .U / AB D G, and we get jG=Z.G/j jG=U j jG=DjjD=U j p 4 . The following lemma is the key result. Lemma 76.5. Suppose that H is a nonabelian maximal subgroup of a p-group G. Then ˇ1 .G; H / p 1. If ˇ1 .G; H / D p 1, then the following holds: (a) d.G/ D 2, 1 D fH1 D H; H2 ; : : : ; Hp ; A D HpC1 g, where A is abelian and all members of the set 1 fH; Ag are A1 -groups. (b) H10 D D Hp0 is of order p. (c) G=H10 is an A1 -group so jG 0 j D p 2 . We also have d.H1 / 3. If G is nonmetacyclic, then G=H10 is also nonmetacyclic. (d) G=Z.G/ is either nonabelian of order p 3 and exponent p or G=Z.G/ Š D8 , Z.Hi / D Z.G/ and Hi =G 0 is cyclic, i D 2; : : : ; p. In particular, cl.G/ D 3. (e) If G 0 Š Cp 2 , then H2 ; : : : ; Hp ; A are metacyclic and G has no normal subgroups of order p 3 and exponent p. If p > 2, then G is metacyclic (in that case, G is an A2 -group, by Corollary 65.3).

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(f) If G 0 Š Ep 2 , then cl.G/ D 3. In what follows we assume that G is not an A2 -group: then jGj D p m > p 4 . Since jG 0 j D p 2 , G is nonmetacyclic. (g) d.H / D 3. In what follows we assume, in addition, that p D 2. Then, since G is not of maximal class, we get jG W G 0 j > 4, by Taussky’s theorem. (h) d.A/ D 2 so A is metacyclic. (i) G has no normal elementary abelian subgroups of order 8. (j) H2 is metacyclic. (k) U D 1 .H2 / D 1 .A/ < ˆ.G/ is the unique normal abelian four-subgroup of G. (l) If Z.G/ is cyclic, then H2 Š M2m1 and G 0 6 Z.G/. In that case, if GN D G=H 0 N Š E8 and T D Q L, where L Š C4 , Q Š Q8 , and TN D 1 .HN /, then TN D 1 .G/ 0 Q GG and G=Q is cyclic so G < Q is cyclic of order 4. Since 1 .G/ D T , it follows that G has exactly 7 involutions. Next, G=G 0 is abelian with cyclic subgroup of index 2 and G has no elementary abelian subgroups of order 8. (m) Suppose that Z.G/ is noncyclic. Then G 0 Š C4 is a maximal cyclic subgroup of G. Let T be as in (l). Then jT j D 16 and T D Q L. where Q is nonabelian of order 8. Next, H2 =G 0 is cyclic since G 0 6 Z.G/ D Z.H / D ˆ.H / and H2 has no cyclic subgroups of index 2. The quotient group A=G 0 is also cyclic. (n) G 0 is a maximal cyclic subgroup of G. Proof. The set 1 .H / \ 2 contains a member N ; then N 6 Z.G/ since H is nonabelian. Set 1N D fH1 D H; : : : ; Hp ; HpC1 g. Since at most one member of the set 1N is abelian, one may assume that H1 ; : : : ; Hp are nonabelian. By Remark 1, ˇ1 .G; H / p 1. Next we suppose that ˇ1 .G; H / D p 1. (a–c) If Ui Hi is an A1 -subgroup not contained in N (see Theorem 10.28), then U2 ; : : : ; Up are pairwise distinct. It follows that UpC1 does not exist so HpC1 D A is Then N D H \ A is also abelian, and we get p 1 D ˇ1 .G; H / Pabelian. p ˛ .H 1 i /. It follows that ˛1 .Hi / D 1 (i D 2; : : : ; p) so H2 ; : : : ; Hp are A1 iD2 groups and the subgroups H1 ; : : : ; Hp together contain all A1 -subgroups of G. Assume that F 2 1 1N . The intersection F \ Hi is abelian since Hi is an A1 subgroup, i D 2; : : : ; p. Therefore, if F is nonabelian, then all A1 -subgroups of F are contained in H1 . In that case, F H1 (Theorem 10.28), a contradiction. Thus, all members of the set 1 1N are abelian. Assuming that d.G/ > 2, we see that the set 1 has at least .p 2 C p C 1/ p D p 2 C 1 > p C 1 abelian members, contrary to Exercise 1.6(a). Thus, 1 D 1N D fH1 ; : : : ; Hp ; Ag so d.G/ D 2 and ˆ.G/ D N , completing the proof of (a). Now (b) follows from (a) and Lemma 76.2. By Lemma 76.2 again, G=H10 is an A1 -group and hence jG 0 j D pjH10 j D p 2 . In that

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case, d.H1 / D d.H1 =H10 / 3 (Lemma 76.1). The last assertion in (c) follows from Theorem 36.1. (d) It follows from d.G/ D 2 that jG W Z.G/j > p 2 . By Remark 2, G=Z.G/ is either nonabelian of order p 3 and exponent p or G=Z.G/ Š D8 since A is the unique abelian member of the set 1 . Now let i 2 f2; : : : ; pg. In view of d.G/ D 2, we get Z.G/ ˆ.G/ < Hi so Z.G/ Z.Hi /, and the equality jZ.G/j D jZ.Hi /j implies Z.G/ D Z.Hi / .D ˆ.Hi //. Therefore, since G 0 6 Z.G/, then Hi =G 0 is cyclic. (e) Suppose that G 0 Š Cp 2 . Then Hi is metacyclic since Hi0 is not a maximal cyclic subgroup in Hi in view of Hi0 < G 0 , i D 2; : : : ; p (Lemma 76.1). Assume that G has a normal subgroup R of order p 3 and exponent p. For i D 2; : : : ; p, we get Hi R D G so G=.R \Hi / D .Hi =.R \Hi //.R=.R \Hi //, and we conclude that Hi =.R \Hi / is cyclic; then G=.R \ Hi / is abelian. On the other hand, G=.R \ Hi / is nonabelian since Cp 2 Š G 0 6 R \ Hi Š Ep 2 , and this is a contradiction. Thus, R does not exist so A is metacyclic. Let p > 2. Since G 0 is cyclic, G is regular (Theorem 7.1(c)) so p 2 D j1 .G/j D jG=Ã1 .G/j whence G is metacyclic (Theorem 9.11). (f) If G 0 Š Ep 2 , then G 0 6 Z.G/ ((a) and Lemma 65.2(a)) so cl.G/ D 3. (g) By hypothesis, H is not an A1 -group (otherwise, G is an A2 -group). By Lemma 65.2(a), d.H / > 2. Since G=H 0 is an A1 -group, we get d.H / 3, and our claim is proved. In what follows we assume that p D 2 so jG W G 0 j > 4 since G is not of maximal class (Taussky). (h) By Theorem 71.6, the number of two-generator members in the set 1 is even. Since j1 j D 3 and d.H2 / D 2 < 3 D d.H /, by (g), we get d.A/ D 2, so A is metacyclic. Clearly, A is noncyclic. (i) Assume that E8 Š E GG. Then E \A D 1 .A/, by (h). We have CG .E \A/ D EA D G so E \ A Z.G/. Since E 6 A 2 1 and j1 j D 3, the number of maximal subgroups of G=E is < 3 so G=E is cyclic. In that case, A D L Z, where jLj D 2 and Z is cyclic so A=L is a cyclic subgroup of index 2 in G=L (we have L < 1 .A/ Z.G/). Since L < Z.G/ and G=Z.G/ Š D8 , G=L is of maximal class (Theorem 1.2). However, G=L has a normal subgroup E=L Š E4 and jG=Lj 24 , and this is a contradiction. Thus, E does not exist. (j) Assume that H2 is nonmetacyclic. Then 1 .H2 / Š E8 (Lemma 65.1) and 1 .H2 / G G, contrary to (i). (k) Since the two-generator 2-group G has no cyclic subgroups of index 2, the subgroup ˆ.G/ is noncyclic. We have H2 \ A D ˆ.G/ is noncyclic so U D 1 .H2 / D 1 .A/ Š E4 . Assume that V ¤ U is a G-invariant four-subgroup. Then V — H2 ; A so G=V is cyclic. Since G has no cyclic subgroups of index 2 and jGj > 24 , we get 1 .G/ D U V Š E8 , contrary to (i). Thus, V does not exist. Let GN D G=H 0 and TN D 1 .HN /; then TN Š E8 is normal in GN since HN is abelian N D TN so of rank 3, by (g). Since GN is minimal nonabelian, by (c), we get 1 .G/ 5 N TN Š G=T is cyclic since T 6 A; H2 , 1 .G/ T . Assume that jGj > 2 . Then G=

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by (h) and (j). Since d.T / D 3, it follows that T is nonabelian, by (i). By (k), U < T , where U is the unique normal abelian four-subgroup of G. Let Q < T be minimal nonabelian; then jQj D 8, T D QZ.T / (Lemma 64.1(i)) and Z.T / G G since T G G. (l) Suppose that Z.G/ is cyclic. Then the noncyclic abelian subgroup ˆ.G/ has a cyclic subgroup Z.G/ of index 2 (recall that 8jZ.G/j D jGj D 4jˆ.G/j and Z.G/ < ˆ.G/ since d.G/ D 2) so 1 .ˆ.G// D U 6 Z.H2 / (otherwise, CG .U / H2 A D G and E4 Š U Z.G/, contrary to the assumption) so H2 Š M2m1 (Lemma 76.1). Let T be defined as in the previous paragraph. Then Z.T / is cyclic (otherwise, since Z.T / G G, we get Z.T / D U , by (k), and CG .U / TA D G; then Z.G/ is noncyclic, a contradiction), so, by the product formula, H D QZ.G/, where Q Š Q8 (Appendix 16) since Q — A; H2 . Since Q is the unique subgroup isomorphic to Q8 in T (Appendix 16), we get Q G G, and then G=Q is cyclic since d.G/ D 2 and Q 6 A. In that case, G 0 < Q is of order 4 so G 0 Š C4 and G 0 6 Z.G/ D Z.H2 /. Then G=G 0 , as abelian group of type .jG=Qj; 2/, has exactly two cyclic subgroups H2 =G 0 and A=G 0 of index 2 (H=G 0 is not cyclic since d.H / D 3). Assume that E8 Š E < G. Then E \ A D 1 .A/ D U so CG .U / EA D G, contrary to the cyclicity of Z.G/. Thus, G has no elementary abelian subgroups of order 8. By the above and Appendix 16, T has exactly 7 involutions so G has also 7 involutions since 1 .G/ D T (see Theorem 43.10). (m) Suppose that Z.G/ is noncyclic; then U D 1 .Z.G//. We have G 0 ¤ U (otherwise, G is an A1 -group, by (a) and Lemma 65.2(a), a contradiction). Since U is the unique normal four-subgroup of G, we get G 0 Š C4 . Let T =H 0 D 1 .H=H 0 / be as in (l); then T =H 0 Š E8 (see the proof of (l)), T is nonabelian and T D QZ.T /, where Q is nonabelian of order 8 (Lemma 64.1(i)). As above, G=T is cyclic so G 0 < T . Since U < T \ Z.G/, we get T D Q L, where jLj D 2. We know that 1 .G/ D 1 .T /. Since Z.T / D U ¤ G 0 , we get G 0 6 Z.T / so G 0 6 Z.G/ hence cl.G/ D 3. Since U < H2 , H2 has no cyclic subgroups of index 2 (otherwise, 1 .H2 / D U Z.G/ so H2 is abelian). Next, H2 =G 0 is cyclic since G 0 6 Z.G/ D Z.H / D ˆ.H /. Since G=G 0 is abelian with cyclic subgroup of index 2, it follows that A=G 0 is cyclic (otherwise, H=G 0 is cyclic so H is metacyclic, which is a contradiction). (n) By (l) and (m), G 0 Š C4 . Assume that G 0 < Z, where Z Š C8 . Then < Z=H 0 so G 0 =H 0 is not a maximal cyclic subgroup of the A1 -group G=H 0 . It follows that G=H 0 is metacyclic (Lemma 76.1). In that case, G is metacyclic, by Theorem 36.1, which is a contradiction. Thus, G 0 < G is maximal cyclic. G 0 =H 0

Remarks. 4. Let G be a p-group and let members H1 ; : : : ; Hk , k > 1, of the set 1 contain together all A1 -subgroups of G. Suppose that there exists A 2 1 fH1 ; : : : ; Hk g such that A \ Hi is abelian for all i D 2; : : : ; k. Then A is abelian. Assuming that this is false, we can take an A1 -subgroup U A such that U 6 H1 (Proposition 10.28). Since, for i D 2; : : : ; k, also U 6 Hi (recall that A \ Hi is abelian, by hypothesis), we get a contradiction. 5. For a nonabelian p-group G, the following two conditions are equivalent: (a) G

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is of maximal class with abelian subgroup A of index p. (b) Whenever H G is nonabelian of order p k , then ˛1 .H / D p k3 . In the proof we use induction on jGj. Let us prove that (a) ) (b). By Fitting’s lemma, all members of the set 1 fAg are of maximal class so, by Hall’s enumeration principle and induction, ˛1 .G/ D

X

˛1 .H / D p p m13 D p m3 :

H 21 fAg

Next, let L < G be nonabelian. Then L H 2 1 and the subgroup H of maximal class has an abelian subgroup H \A of index p so, by induction, L is of maximal class and ˛1 .L/ D p l3 , where jLj D p l . Now we prove that (b) ) (a). In that case, all proper nonabelian subgroups of G are of maximal class, by induction. Take T G, where T is an A1 -subgroup of G. Setting jT j D p k , we get 1 D ˛1 .T / D p k3 so k D 3, i.e., all A1 -subgroups of G have the same order p 3 . Assume that G is not of maximal class. Then we get CH .T / 6 T (Lemma 64.1(i)). Let F be a subgroup of order p 2 in CH .T / such that Z.T / < F . Then jF T j D p 4 and ˛1 .F T / D p 2 ¤ p D p 43 , a contradiction. Thus, G is of maximal class. Next, let R G G be of order p 2 and set U D CG .R/; then jG W U j D p and, by induction, U is abelian since it is not of maximal class. 6. Let d.G/ D 3 and ˆ.G/ 6 Z.G/; then the set 2 has a nonabelian member N (Lemma 76.3). We claim that the set 2 contains at least p 2 nonabelian members. Let A1 ; A2 2 2 be distinct abelian; then A D A1 A2 2 1 contains exactly p C 1 members of the set 2 and all of them are abelian since ˆ.G/ D A1 \ A2 Z.A/. Assume that B 2 2 1 .A/ is abelian. Then CG .ˆ.G// AB D G, contrary to the hypothesis. Thus, 1 .A/ \ 2 is the set of all abelian members of the set 2 so the last set contains exactly j2 j .p C 1/ D p 2 nonabelian members. Thus, in our case, the set 2 has p 2 , p 2 C p or p 2 C p C 1 nonabelian members. 7. Let G be a nonabelian p-group with d D d.G/ > 3. Let T D fT1 ; : : : ; Tr g 2 and M D fM1 ; : : : ; Ms g 1 be the sets of nonabelian members of the sets 2 , 1 , respectively (T ¤ ¿, by Lemma 76.3, M ¤ ¿, by Exercise 1.6(a)). We claim that r > .p 1/s. Let i be the number of members of the set M, containing Ti , i D 1; : : : ; r, and let j be the number of members of the set T contained in Mj , j D 1; : : : ; s. Then, by double counting,

1 C C r D 1 C C s ; and, in view of i D p C 1 for all i , that formula can be rewritten as follows: (3)

.p C 1/r D 1 C C s :

Note that Mj contains exactly p d 2 C p d 3 C C p C 1 members of the set 2 . Since, by Exercise 1.6(a), the set 1 .Mj / contains at most p C 1 abelian members, we

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get j p d 2 C C p 2 for j D 1; : : : ; s, and we deduce from (3) the following inequality: .p C 1/r .p d 2 C C p 2 /s. We get r

p2 .p 2 1/ C 1 p d 2 C C p 2 s sD s > .p 1/s; pC1 pC1 pC1

as was to be shown. By Exercise 1.6(a), we have s j1 j.pC1/ D p d 1 C Cp 2 . 8. Given a set M of subgroups of a p-group G, let ˛1 .M/ denote the number of A1 -subgroups that contain together members of the set M. Suppose that G is neither abelian nor an A1 -group. We claim that if, for each K 2 2 , we have ˛1 .1K / pC1, then G is an A2 -group. By hypothesis, there is K 2 2 which is not contained in Z.G/. Indeed, this is the case if d.G/ D 2. If d.G/ > 2, our claim is obvious. Set 1K D fH1 ; : : : ; HpC1 g and let ˛1 .H1 / ˛1 .HpC1 /. Since at most one member of the set 1K is abelian, H1 ; : : : ; Hp are nonabelian. Assume that ˛1 .H1 / > 1; then ˛1 .H1 / D p, by Remark 1 and hypothesis. We get p C 1 ˛1 .1K / D ˛1 .H1 / C

pC1 X

ˇ1 .Hi ; K/ D p C

iD2

pC1 X

ˇ1 .Hi ; K/

iD2

so ˇ1 .H2 ; K/ D 1 and we get p D 2 and ˇ1 .H3 ; K/ D 0 (Remark 1) whence H3 is abelian, and we conclude that K is abelian; in that case, H2 is an A1 -group. Thus, all members of the set 2 are abelian so d.G/ 3 (Lemma 76.3). Assume that G is not an A2 -group. If d.G/ D 2, then 1 D 1K D fH1 ; H2 ; A/, where ˛1 .H1 / D 2, ˛1 .H2 / D 1 and A is abelian. In this case, ˇ1 .G; H1 / D 1 so jH10 j D 2 (Lemma 76.5), and the equality ˛1 .H1 / D 2 is impossible (Lemma 76.4(c)), a contradiction. Now assume that d.G/ D 3; then j1 .H1 / \ 2 j D p C 1 > 1 so jH10 j D 2 (Lemma 64.1(q)), and again we get a contradiction. Lemma 76.6. Suppose that ˛1 .G/ > 1. Then ˛1 .G/ p. (a) If ˛1 .G/ D p, then G is an A2 -group with d.G/ D 2 and jG 0 j D p 2 . If G is nonmetacyclic, it is of class 3. (b) If ˛1 .G/ D p C1, then G is an A2 -group with d.G/ D 2. If G is not metacyclic, then p > 2, G is of order p 5 with G 0 Š Ep 3 and of class 3. Proof. The inequality ˛1 .G/ p follows from Remark 1. By Remark 8, G is an A2 -group. All remaining assertions follows from Lemma 76.4(c). Lemma 76.7. Let N 2 2 and 1N D fH1 ; : : : ; HpC1 g. Then: PpC1 ˇ1 .Hi ; N /. (a) ˛1 .G/ ˛1 .N / C iD1 (b) If N is nonabelian, then ˇ1 .G; N / p 2 1. (c) If ˛1 .G/ < p 2 , then NG .K/=K has only one subgroup of order p for each nonabelian K < G.

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Proof. (a) is obvious, (b) follows from (a) and Remark 1. It remains to prove (c). Assume that K < G is nonabelian and NG .K/=K contains an abelian subgroup L=K of type .p; p/; then 1 C ˇ1 .L; K/ ˛1 .K/ C ˇ1 .L; K/ D ˛1 .L/ ˛1 .G/ p 2 1 so ˇ1 .L; K/ < p 2 1, contrary to (b). Thus, L=K does not exist so NG .K/=K has only one subgroup of order p so it is either cyclic or generalized quaternion. Lemma 76.8. Suppose that G is an An -group, n > 2. (a) If d.G/ > 2 then ˛1 .G/ p 2 . (b) If d.G/ D 3 and ˆ.G/ D Z.G/, then ˛1 .G/ 2p 2 C p 1. (c) If d.G/ D 3 and jG W Z.G/j D p 2 , then ˛1 .G/ 2p 2 1. Proof. (a) There is N 6 Z.G/ for P some N 2 2 since hM j M 2 2 i D G. If all such N are abelian, then ˛1 .G/ D H 21 ˛1 .H / p.j1 j .p C 1// p 3 > p 2 (Exercise 1.6(a) and Lemma 76.6). If N is nonabelian, then ˛1 .G/ p 2 (Lemma 76.7(b)). (b) If L 2 1 is nonabelian, then jL W Z.L/j D jL W Z.G/j D p 2 so jL0 j D p (Lemma 64.1(q)). Suppose that L is neither abelian nor an A1 -group (such an L exists since n > 2). Then, by Lemma 76.4, ˛1 .L/ ˛1 .H / p 2 , where H L is an A2 -subgroup. In view of jG W Z.G/j D p 3 , the set 1 has at most one abelian member (Lemma 64.(u)) so this set has at least p 2PC p nonabelian members, and, by Hall’s enumeration principle, ˛1 .G/ D ˛1 .L/ C M 21 fLg ˛1 .M / p 2 C p 2 C p 1 D 2p 2 C p 1 (we have here equality if and only if ˛1 .L/ D p 2 and all nonabelian members of the set 1 fLg are A1 -groups). (c) In our case, jG 0 j D p and all members of the set 2 are abelian, the set 1 has exactly p 2 nonabelian members (Remark 6). If L 2 1 is neither abelian nor an A1 -group, then ˛1 .L/ p 2 (see the proof of (b)) so we get ˛1 .G/ D ˛1 .L/ C P 2 2 2 M 21 fLg ˛1 .M / p C .p 1/ D 2p 1. Theorem 76.9. A p-group G with 1 < ˛1 .G/ < p 2 is a two-generator A2 -group. In particular, ˛1 .G/ 2 fp; p C 1g. Proof. Suppose that G is a counterexample of minimal order; then p 2 > ˛1 .G/ > p C 1 (Lemma 76.6(a,b)) so p > 2. By Lemma 76.8(a), d.G/ D 2 so ˆ.G/ > Z.G/. By Lemma 76.7(c), ˆ.G/ is abelian. By Proposition 10.28 and induction, G is an A3 group. Then there is H 2 1 , which is an A2 -group, and so, by Proposition 10.28, Lemmas 76.4 and 76.6, d.H / D 2, jH 0 j > p and ˛1 .H / 2 fp; p C 1g. (i) Assume that A; U 2 1 , where A is abelian and U is an A1 -group. Then G=Z.G/ is of order p 3 and exponent p (Remark 2), and Z.G/ < ˆ.G/ < H . In that case, jH W Z.G/j D p 2 so jH 0 j D p (Lemma 64.1(q)), contrary to what has been said in the previous paragraph. Thus, a pair .A; U / does not exist. Let F1 ; : : : ; Fs ; V1 ; : : : ; Vu be all nonabelian members of the set 1 , where ˛1 .Fi / D p and ˛1 .Vk / D 1 (since ˆ.G/ is abelian, the set 1 has no members H with ˛1 .H / D p C 1, by Lemma 76.4). Since the set 1 has at most one abelian member (Lemma 64.1(u)), we get s C u j1 j 1 D p.

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(ii) Assume that A 2 1P is abelian. Then u D 0, by (i), so s D p and, since ˆ.G/ 2 is abelian, p 2 > ˛1 .G/ D p iD1 ˛1 .Fi / D p p D p , a contradiction. Thus, all members of the set 1 are nonabelian, i.e., s C u D p C 1. As in the previous paragraph, s < p so u 2. Then G and all members of the set 1 are two-generator so G is either metacyclic or G=K3 .G/ is nonabelian of order p 3 and exponent p and then jG W G 0 j D p 2 (Lemma 64.1(p)). (ii1) Assume that G is metacyclic. Then V1 ; V2 2 1 are distinct A1 -subgroups so G is an A2 -group (Exercise 1), a contradiction. Thus, G is nonmetacyclic. (ii2) Since u > 1, then jG 0 j pjV10 V20 j D p 3 so jGj D jG W G 0 jjG 0 j p 5 . Since G is not an A2 -group, we get jGj D p 5 . In that case, G has a nonabelian subgroup of order p 3 . Since all nonabelian groups of order p 3 are A1 -groups, we get by Proposition 2.3, ˛1 .G/ p 2 , a final contradiction. Let G be a p-group of maximal class and order p 5 with abelian subgroup of index p. Then ˛1 .G/ D p 2 and G is an A3 -group. Remark 9. Let a p-group G be neither abelian nor an A1 -group. If, for each K 2 2 , we have ˛1 .1K / p 2 , then G is either an A2 - or A3 -group. Indeed, let H1 2 1 be nonabelian and take K 2 1 .H1 / \ 2 . Set 1K D fH1 ; : : : ; HpC1 g. Since at most one member of the set 1K is abelian, we get ˛1 .H1 / < p 2 , i.e., H1 is either A1 - or A2 -group (Theorem 76.9) so G is an An -group, n 2 f2; 3g. Theorem 76.10. If ˛1 .G/ D p 2 , then G is an An -group, n 2 f2; 3g. Next suppose that G is an A3 -group. Then jG 0 j D p 3 and ˛1 .H / 2 f1; pg for every nonabelian H < G. (a) If the set 2 has a nonabelian member N , then N is an A1 -group and one of the following holds: (a1) G is metacyclic with ˛1 .H / D p for all H 2 1 , i.e., all members of the set 1 are A2 -groups. (a2) d.G/ D 3, 1 D fF1 ; : : : ; Fp 2 Cp ; Ag, where ˛1 .Fi / D p for all i and A is abelian, Z.G/ D Z.N / is of index p 4 and the set 2 has exactly p 2 nonabelian members which are A1 -groups. (b) If all members of the set 2 are abelian, then d.G/ D 2 and one of the following holds: (b1) G is metacyclic, 1 D fH1 ; : : : ; Hp ; Ag, where ˛1 .Hi / D p for all i , H10 D D Hp0 , A is abelian. (b2) 1 D fH1 ; : : : ; Hp ; Ag, where ˛1 .Hi / D p, H10 D D Hp0 is of order p 2 , A is abelian, G 0 is noncyclic, G=H10 is an A1 -group. (b3) p D 2, 1 D fH1 ; H2 ; H3 g, ˛.H1 / D 2, ˛.H2 / D ˛.H3 / D 1. We have jG=Z.G/j D 16.

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Groups of prime power order

Proof. Suppose that G is not an A2 -group; then G is an A3 -group (Proposition 10.28 and Theorem 76.9). (a) Suppose that the set 2 has a nonabelian member N . Let 1N D fH1 ; H2 ; : : : ; HpC1 g; then all members of the set 1N are neither abelian nor A1 -groups. Using Remark 1, we get p 2 D ˛1 .G/ ˛1 .N / C

pC1 X

ˇ1 .Hi ; N / 1 C .p 1/.p C 1/ D p 2 ;

iD1

and hence ˛1 .N / D 1, ˇ1 .Hi ; N / D p 1, ˛1 .Hi / D p, d.Hi / D 2 and jHi0 j D p 2 for all i , H1 ; : : : ; HpC1 are A2 -groups (Remark 1, Lemmas 74.4 and 76.6(a)) and these p C 1 subgroups contain together all A1 -subgroups of G. If G is metacyclic, it is as in (a1). Next we assume that G is nonmetacyclic. We also have d.G/ d.H1 / C 1 D 3. Suppose, in addition, that d.G/ D 2 so N D ˆ.G/. Then G and all members of the set 1 are two-generator (Lemma 76.6(a)). Since G is nonmetacyclic, we get p > 2 and jG W G 0 j D p 2 (Lemma 64.1(p)). By the above and Lemma 76.6(a), Hi has the unique abelian maximal subgroup Ai , all i . Then the subgroups A1 ¤ A2 are normal in G; therefore, A D A1 A2 is of class at most 2 (Fitting’s lemma). By Lemma 64.1(l), G is not minimal nonmetacyclic so we may assume that H1 is not metacyclic. Since A > A1 and cl.G/ cl.H1 / > 2 (Lemma 76.4(c)), we get A 2 1 . Then A1 D H1 \ A D ˆ.G/ D N , a contradiction since N is nonabelian. Thus, d.G/ D 3. Then, by Remark 6, the set 2 has at least p 2 nonabelian members, say L1 ; : : : ; Lp 2 , and all Li are A1 -groups since G is an A3 -group and jG W Li j D p 2 . Since ˛1 .G/ D p 2 , all A1 -subgroups of G are members of the set 2 so each nonabelian H 2 1 is an A2 -group. If U1 ; : : : ; UpC1 are all abelian members of the set 2 , then A D U1 U2 2 1 , by the product formula, and jA0 j p (Lemma 64.1(q)) so A is not an A2 -group since ˛1 .A/ < ˛1 .G/ D p 2 (Lemma 76.4). It follows that A is abelian; moreover, A is the unique abelian member of the set 1 (Lemma 76.8(b)). Thus, 1 D fF1 ; : : : ; Fp 2 Cp ; Ag, where ˛1 .Fi / D p and jFi0 j D p 2 (Lemma 76.6). Next, in view of d.Fi / D 2, we get Z.Fi / < ˆ.Fi / D ˆ.G/ < A; then CG .Z.Fi // Fi A D G so Z.Fi / Z.G/ and, by Lemma 64.1(q), jFi W Z.Fi /j D pjFi0 j D p 3 . Thus, jG W Z.G/j p 4 . By Lemma 64.1(q), jG W Z.G/j D pjG 0 j pjF10 j p 3 . Since G=Z.G/ is noncyclic and A is the unique abelian member of the set 1 , it follows that Z.G/ Fi for some i . By the above, Z.Fi / Z.G/ so Z.G/ D Z.Fi /, and we get jG W Z.G/j D jG W Fi jjFi W Z.G/j D p 4 so jG 0 j D p1 jG W Z.G/j D p 3 and G is as stated in (a2) since, if N < Fi , then Z.N / D ˆ.N / < ˆ.G/ < A so CG .Z.N // AN D G hence Z.N / Z.G/ and jG W Z.N /j D p 4 D jG W Z.G/j. 2 P(b) Suppose that all members of the set 2 are abelian. Then p D ˛1 .G/ D M 21 ˛1 .M / since the intersection of any two distinct members of the set 1 is abelian (here we use Hall’s enumeration principle). It follows from Remark 6 and

76 p-groups with few A1 -subgroups

295

Lemma 76.8(b,c) that d.G/ D 2 (indeed, if d.G/ D 3, then ˆ.G/ Z.G/, by Lemma 76.3, and, by Lemma 76.8(b,c), ˛1 .G/ 2p 2 1 > p 2 ). Let fH1 ; : : : ; Hs ; V1 ; : : : ; Vu g be the set of nonabelian members of the set 1 , where ˛1 .Hi / D p, ˛1 .Vj / D 1 (since ˆ.G/ 2 2 is abelian, the set 1 has no member H with ˛1 .H / D p C 1; see Lemma 76.4). By assumption, s > 0. Next, jG 0 j jHi0 j D p 2 so the set 1 has at most one abelian member (Lemma 64.1(q)). We have p 2 D ˛1 .G/ D sp C u sp C .p C 1 s/ D s.p 1/ C .p C 1/. If s < p, then p 2 .p 1/2 C p C 1 D p 2 p C 2 so p D 2, s D 1, u D 2. Then ˛1 .H1 / D 2 so cl.H1 / D 3 (Lemma 76.4). By Remark 3, jG=Z.G/j p 4 . Since Z.G/ < H1 and cl.H1 / D 3, we get jH1 W Z.G/j > p 2 so jG=Z.G/j D p 4 . Assume that jG 0 j D p 2 . Then G=V10 is an A1 -group (Lemma 65.2(a)) hence H1 =V10 is abelian and H10 D V10 is of order p, a contradiction. Since jG 0 j 2jV10 V20 j 8, we get jG 0 j D 8 so G is as in (b3). Thus, s p. Since s < p C 1, we get s D p. It follows from p 2 D ˛1 .G/ D sp C u D p 2 C u that u D 0. Thus, 1 D fH1 ; : : : ; Hp ; Ag, where A is abelian and ˛1 .Hi / D p for i D 1; : : : ; p. Suppose that G is metacyclic. Then jG W Z.G/j D pjG 0 j D p 4 (Lemma 64.1(q) and Theorem 72.1) and G is as stated in (b1). Suppose that G is not metacyclic. Then G=Hi0 is an A1 -group for all i (Lemma 76.2) whence jG 0 j D pjH10 j D p 3 . By Proposition 72.1(b), G 0 is noncyclic so G is as stated in (b2). Theorem 76.11. Suppose that G is a p-group with p 2 < ˛1 .G/ p 2 C p 1. Then G is a two-generator A3 -group with jG 0 j > p and one of the following holds: (a) 1 D fH1 ; H2 ; : : : ; Hp ; Ag, where A is abelian, H1 is an A2 -group with d.H1 / D 3, ˛1 .H1 / D p 2 , ˛1 .Hi / D 1, i D 2; : : : ; p, H10 D D Hp0 is of order p, G=H10 is an A1 -group so jG 0 j D p 2 , ˛1 .G/ D p 2 C p 1. (b) 1 D fH1 ; : : : ; Hp ; Bg, where ˛1 .Hi / D p, i D 1; : : : ; p, ˛1 .B/ D 1, ˛1 .G/ D p 2 C 1. If G is nonmetacyclic, then p > 2, jG W G 0 j D p 2 and jGj 2 fp 5 ; p 6 g. (c) G is metacyclic, 1 D fF1 ; : : : ; Fs ; H1 ; : : : ; Hu g, where s C u D p C 1, ˛1 .Fi / D p, s 2, ˛1 .Hj / D p C 1, ˛1 .ˆ.G// D 1 and jG 0 j D p 3 . Proof. By Lemma 76.4(b), G is not an A2 -group. Assume that G is not an A3 -group. Then there exists a nonabelian H 2 1 which is an An -group, n 3. By Theorem 76.9, ˛1 .H / p 2 . Then, by Lemma 76.5, p 2 C p 1 ˛1 .G/ D ˛1 .H / C ˇ1 .G; H / p 2 C p 1 so ˛1 .G/ D p 2 Cp1 and ˛1 .H / D p 2 , ˇ1 .G; H / D p1 so jH 0 j D p. Let L < H be an A2 -subgroup. Since jL0 j D p, we get, by Lemma 76.4, ˛1 .L/ p 2 D ˛1 .H /, contrary to Proposition 10.28. Thus, H is an A2 -group so G is an A3 -group. This argument also shows that the set 1 has no member U with ˛1 .U / > p 2 .

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Groups of prime power order

If U 2 1 , ˛1 .U / D p 2 , then ˇ1 .G; U / D p 1 so, by Lemma 76.5, G is as in (a). Next we assume that ˛1 .U / < p 2 for all U 2 1 ; then d.G/ d.U / C 1 D 3 (Lemma 76.1 and Theorem 76.9). Let 1 D fF1 ; : : : ; Fs ; H1 ; : : : ; H t ; B1 ; : : : ; Bu ; A1 ; : : : ; Av g;

s C t > 0;

where ˛1 .Fi / D p, ˛1 .Hj / D p C 1 (see Lemma 76.6), ˛1 .Bk / D 1 and Al are abelian. Since s C t > 0 and jFi0 j; jHj0 j > p, we get v 1 (Lemma 64.1(u)). We have () If t > 0, then u D v D 0 since the set 1 .H1 / has no abelian members. () If d.G/ D 2, then uv D 0. Indeed, if uv > 0, then all maximal subgroups of G=B10 are abelian, by Exercise 1.6(a), a contradiction since jT 0 j p 2 for T 2 fF1 ; : : : ; Fs ; H1 ; : : : ; H t g (Lemma 76.4). (i) Suppose that d.G/ D 2. Then uv D 0, by (). Put N D ˆ.G/. (i1) Suppose, in addition, that N is nonabelian so it is an A1 -group since jG W N j D p 2 . Then s C t D j1 j D p C 1, X p 2 C p 2 ˇ1 .G; N / D ˇ1 .M; N / D s.p 1/ C tp M 21

D s.p 1/ C .p C 1 s/p D p 2 C p s; and we get s 2 and t p 1. Let Ui be (the unique so normal in G) abelian maximal subgroup of Fi , i D 1; 2, and set A D U1 U2 . Since F1 \ F2 D ˆ.G/ is nonabelian, we get U1 ¤ U2 . Then A .> U1 / is of class 2 (Fitting). If A 2 1 , then ˆ.G/ D A\F1 D A1 is abelian, a contradiction. Thus, A D G, so cl.G/ D 2. In that case, all members of the set 1 are metacyclic (Lemma 76.4) so G is also metacyclic (Lemma 64.1(e,f,h)), and G is as in (c). (i2) In what follows we assume that N D ˆ.G/ is abelian; then t D 0. If s D p C1, then ˛1 .G/ D sp D .p C 1/p > p 2 C p 1, a contradiction. Thus, u C v > 0. Since jG 0 j jF10 j D p 2 , we get v 1 (Lemma 64.1(u)). If v D 1, then u D 0, by (); then s D p and we get p 2 C 1 ˛1 .G/ D sp D p 2 , a contradiction. Thus, v D 0 so s C u D p C 1 and u D p C 1 s > 0. We have X ˛1 .M / D ps C u D .p 1/s C p C 1 p 2 C p 1: p 2 C 1 ˛1 .G/ D M 21 1 , and so s D p and u D 1. In that case, It follows that p s p C 1 p1 2 ˛1 .G/ D ps C u D p C 1, 1 D fF1 ; : : : ; Fp ; Bg, where ˛1 .Fi / D p, ˛1 .B/ D 1. Thus, G and all members of the set 1 are two-generator so G is either metacyclic or p > 2 and G=K3 .G/ is nonabelian of order p 3 and exponent p (Lemma 64.1(n,p)). Suppose that G is not metacyclic; then jG W G 0 j D p 2 . In that case, jG 0 j pjF10 B 0 j p 4 so jGj D jG W G 0 jjG 0 j p 6 . Since G is an A3 -group, we get jGj 2 fp 5 ; p 6 g so G is as in (b).

76 p-groups with few A1 -subgroups

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(ii) Now let d.G/ D 3. By Lemma 76.8(b,c), ˆ.G/ — Z.G/. Then, by Lemma 76.3 and the last sentence of Remark 6, the set 2 has exactly p 2 nonabelian members which are A1 -groups so this set has two distinct abelian members U1 and U2 . Then M D U1 U2 2 1 , by the product formula and j1 .M / \ 2 j D p C 1. It follows from U1 \ U2 Z.M / that all members of the set 1 .M / \ 2 are abelian so the set 2 has exactly p C 1 abelian members, say U1 ; : : : ; UpC1 . We have jM 0 j p (Lemma 64.1(u)) and 2 D fL1 ; : : : ; Lp 2 ; U1 ; : : : ; UpC1 g, where all Li are A1 -groups, all Ui < M . Since Li \ M D ˆ.G/ is abelian for all i , we get ˛1 .M / C p 2 ˛1 .G/ p 2 C p 1 so ˛1 .M / p 1 and M is either abelian or A1 -group (Remark 1). (ii1) Suppose that M is abelian. Then v D 1 (Lemma 64.1(u)) and so t D 0. Also, u D 0 since all p C 1 abelian members of the set 2 lie in M . Hence, s D j1 j v D p 2 C p. By the hypothesis and Hall’s enumeration principle (Theorem 5.2), X X ˛1 .H /p ˛1 .H / D ps p p 2 D p.p 2 Cp/p 3 D p 2 ; p 2 < ˛1 .G/ D H 21

H 22

a contradiction. Thus, v D 0. (ii2) Now let M be an A1 -group; then u D 1 since all Ui < M . We get t D 0, by (), so s D j1 j u D p 2 C p. Therefore, by Hall’s enumeration principle, X X ˛1 .H / p ˛1 .H / p 2 C 1 ˛1 .G/ D H 21 2

H 22 2

D 1 C sp p p D 1 C .p C p/p p 3 D p 2 C 1: Note that d.G/ D 3 but all members of the set 1 are two-generator. We have cl.G/ cl.F1 / D 3. Then p D 2 (Theorem 70.1) and G=K4 .G/ is of order 27 or 28 (Theorem 70.4). However, by remarks following Theorem 70.5, the above groups of order 27 and 28 are A4 -groups and A5 -groups, respectively, a contradiction. Lemma 76.12. Let G be an A4 -group and let R 2 2 be an A2 -group. Then ˛1 .G/ > p 2 C p C 1, unless p D 2 and G satisfies the following conditions: (a) d.G/ D 3, ˛1 .G/ D 22 C 2 C 1 D 7. (b) 2 D fR; R1 ; R2 ; R3 ; A1 ; A2 ; A3 g, where ˛1 .R/ D 4, jR0 j D 2, d.R/ D 3, R1 ; R2 ; R3 are A1 -groups and A1 ; A2 ; A3 are abelian. (c) 1 D fF1 ; F2 ; F3 ; H1 ; H2 ; H3 ; Ag, where A is abelian and (c1) 1 .Fi / D fR; Ri ; Ai g, ˛1 .Fi / D 22 C1 D 5, jFi0 j D 4, R0 D Ri0 , Fi =R0 is an A1 -group, i.e., Fi is a group of Theorem 76.11(a), i D 1; 2; 3. (c2) 1 .H1 / D fA1 ; R2 ; R3 g, 1 .H2 / D fA2 ; R3 ; R1 g, 1 .H3 / D fA3 ; R1 ; R2 g so ˛1 .Hi / D 2 and jHi0 j D 4 hence Hi is an A2 -group, i D 1; 2; 3. (d) Z.G/ < ˆ.G/, jG 0 j D 8, jG=Z.G/j D 16, Z.G/ D Z.Ri /, i D 1; 2; 3. All nonabelian members of the set 1 are two-generator.

298

Groups of prime power order

Proof. Let 1R D fF1 ; : : : ; FpC1 g; then F1 ; : : : ; FpC1 are A3 -groups. By Theorem 76.9, ˛1 .Fi / p 2 for all i . By Lemma 76.6, ˛1 .R/ p. If ˛1 .R/ D p, then ˇ1 .Fi ; R/ p 2 p for all i , and we are done since ˛1 .G/ ˛1 .R/ C

pC1 X

ˇ1 .Fi ; R/ p C .p 2 p/.p C 1/ D p 3 > p 2 C p C 1:

iD1

Now suppose that ˛1 .R/ D p C 1. Then ˛1 .Fi / p 2 C p, by Theorems 76.10 and 76.11, so ˇ1 .Fi ; R/ p 2 1 for all i , and we are done since ˛1 .G/ ˛1 .R/C

pC1 X

ˇ1 .Fi ; R/ p C1C.p 2 1/.p C1/ D p 3 Cp 2 > p 2 Cp C1:

iD1

If ˛1 .R/ > p C 1, then ˛1 .R/ p 2 (Theorem 72.1 and 76.4). Since Fi is an A3 -group with R 2 1 .Fi /, we get ˇ1 .Fi ; R/ p 1 (Lemma 76.5), and so

(4)

˛1 .G/ ˛1 .R/ C

pC1 X

ˇ1 .Fi ; R/ p 2 C .p 1/.p C 1/

iD1 2

D 2p 1 p 2 C p C 1: If p > 2, then ˛1 .G/ 2p 2 1 > p 2 C p C 1, and we are done in this case. Now let 2p 2 1 D p 2 C p C 1; then p D 2 and ˛1 .G/ D 7. In that case, as it follows from (4), ˛1 .R/ D 4 and ˇ1 .Fi ; R/ D 1 so ˛1 .Fi / D 5 for i D 1; 2; 3. By Lemma 76.5 applied to the pair R < Fi , we get jR0 j D 2, d.Fi / D 2 and 1 .Fi / D fR; Ri ; Ai /, where Ri is an A1 -group with R0 D Ri0 and Ai is abelian, jFi0 j D 4, i.e., Fi is a group of Theorem 76.11(a), i D 1; 2; 3. By Lemma 76.5, d.G/ 1 C d.F1 / D 3. Assume that d.G/ D 2. Since all members of the set 1 are also two-generator, G is metacyclic since p D 2 (Lemma 64.1(n)). In that case, jG 0 j D 24 so jFi0 j D 23 (Theorem 72.1), contrary to what has been said in the previous paragraph. S3Thus, d.G/ D 3. Then ˆ.Fi / D ˆ.G/ and so fR; R1 ; R2 ; R3 ; A1 ; A2 ; A3 g D iD1 1 .Fi / D 2 , where A1 ; A2 ; A3 are abelian so ˆ.G/ is abelian, ˛1 .Ri / D 1 (i D 1; 2; 3). Set A D A1 A2 ; then A 2 1 and A3 < A (if A3 — A then CG .ˆ.G// AA3 D G so ˆ.G/ Z.G/, a contradiction since jR W ˆ.G/j D 2 and R is nonabelian). Since F1 ; F2 ; F3 together contain all A1 -subgroups of G and A \ Fi is abelian for i D 2; 3, then A is abelian (Remark 4). Set 1 1R D fH1 ; H2 ; H3 ; Ag. One may assume that Ai < Hi , i D 1; 2; 3. Since Hi \ A > ˆ.G/ is an abelian maximal subgroup of Hi and Hi \ A 2 2 , then Ri — Hi since Ri Ai D Fi ¤ Hi , i D 1; 2; 3. Since R — Hi and j1 .Hi / \ 2 j D 3, i D 1; 2; 3, we get H1 D fA1 ; R2 ; R3 g;

H2 D fA2 ; R1 ; R3 g;

H3 D fA3 ; R1 ; R2 g:

76 p-groups with few A1 -subgroups

299

Now, R, R1 , R2 , R3 together contain all A1 -subgroups of G. Each of the four A1 subgroups S1 , S2 , S3 , S4 of R satisfies Sj ˆ.G/ D R, j D 1; 2; 3; 4, since ˆ.G/ is an abelian maximal subgroup of R, and so, in view of R — Hi , no one of S1 ; S2 ; S3 ; S4 is contained in Hi , i D 1; 2; 3. Thus, ˛1 .Hi / D 2, i D 1; 2; 3. By Lemma 76.6(a), Hi is an A2 -group with d.Hi / D 2 and jHi0 j D 4, i D 1; 2; 3. Hence, all nonabelian members of the set 1 are two-generator. By Lemma 64.1(u), jG 0 j 2jH10 A0 j D 8 and so Lemma 64.1(q) implies that jG W Z.G/j D 2jG 0 j 24 . On the other hand, G D Ri A with A \ Ri D ˆ.G/. We have Z.Ri / D ˆ.Ri / < ˆ.G/ < A, jˆ.G/ W Z.Ri /j D 2, by the product formula, and CG .Z.Ri // Ri A D G. Thus, Z.Ri / Z.G/, i D 1; 2; 3. Assume that jG W Z.G/j < 24 ; then jG W Z.G/j D 23 . In that case, all members of the set 1 , containing Z.G/, must be abelian or A1 -subgroups since nonabelian members of the set 1 are two-generator. Since A is the unique abelian member of the set 1 , this set also contains an A1 -group, contrary to what has been said about 1 . Thus, Z.G/ D Z.Ri / (i D 1; 2; 3) so jG W Z.G/j D 24 and jG 0 j D 8 (Lemma 64.1(q)). Definition 1. A 2-group G of Lemma 76.12 is called a 2-group of type G7;1 . We shall see (see Appendix, below) that groups of type G7;1 do not exist. Lemma 76.13. Let a p-group G be an An -group, n > 2, let H1 2 1 and ˇ1 .G; H1 / D p; then H1 is nonabelian (Lemma 76.6). In that case, d.G/ D 2 so 1 D fH1 ; : : : ; HpC1 g, and one of the following holds: (a) If H2 is neither abelian nor an A1 -group, then p D 2, ˛1 .H2 / D 2, H3 is abelian, H10 D H20 is of order 4, G=H20 is an A1 -group so jG 0 j D 8 and jG=Z.G/j D 16. 0 (b) All p members of the set 1 fH1 g are A1 -groups, H10 D H20 D D HpC1 and G=H10 is an A1 -group so jG 0 j D p 2 . 0 (c) All p members of the set 1 fH1 g are A1 -groups and H20 ; : : : ; HpC1 are 0 pairwise distinct. Set Q D H2 : : : HpC1 ; then Q Š Ep 2 , G=Q is an A1 -group so H10 Q and jG 0 j D p 3 .

Proof. Let R 2 1 .H1 / \ 2 and set 1R D fH1 ; H2 ; : : : ; HpC1 g. Since H1 is nonabelian (Lemma 76.6(a)), we get R — Z.G/ so the set 1R has at most one abelian member. (i) Suppose that R is nonabelian; then all subgroups Hi are nonabelian and we have PpC1ˇ1 .Hi ; R/ p 1 for i > 1 (Remark 1), and we get p D ˇ1 .G; H1 / iD2 ˇ1 .Hi ; R/ p.p 1/ so that p D 2 and ˇ1 .Hi ; R/ D 1 for i D 2; 3. Therefore, by Lemma 76.5, applied to the pair R < Hi , i D 2; 3, we have jR0 j D 2;

d.Hi / D 2;

1 .H3 / D fR; L3 ; A3 g;

jHi0 j D 4;

1 .H2 / D fR; L2 ; A2 g;

300

Groups of prime power order

where ˛1 .Li / D 1 and A is abelian. Since, in view of ˇ.G; H1 / D 2, H1 , H2 , H3 contain together all A1 -subgroups of G, the members of the set 1 are not A1 -groups. Next, A2 ; A3 G G. Set A D A2 A3 ; then cl.A/ 2 (Fitting). Since cl.G/ cl.H2 / D 3 > 2, we get A 2 1 . Then A is abelian since A \ Hi D Ai is abelian for i D 2; 3 (Remark 4). Therefore, in view of jG 0 j jH20 j D 4, A is the unique abelian member of the set 1 (Lemma 64.1(u)). Since R 6 A, we get d.G/ > 2 so d.G/ D 3 since d.H2 / D 2; then j1 .Hi / \ 2 j D 3, i D 1; 2; 3, hence 1 .H2 / [ 1 .H3 / 2 . We have Hi \ A D Ai 2 2 and Ai is the unique abelian member of the set 1 .Hi /, i D 2; 3. Also put A1 D H1 \ A.2 2 / and 1Ai D fHi ; Fi ; Ag, i D 1; 2; 3. It follows that 1 D fH1 ; H2 ; H3 ; F1 ; F2 ; F3 ; Ag. Since F1 \ H1 D A1 and ˛1 .F1 / D 2 D ˇ1 .G; H1 /, F1 is an A2 -group (Lemma 76.6). For i D 2; 3, set Si D F1 \ Hi ; then Si 2 2 and, since Si ¤ A1 , the unique abelian member of the set 1 .F1 /, Si is an A1 -subgroup (Lemma 76.5). We also have Si ¤ R, i D 2; 3, and S2 ¤ S3 so 1 .F1 / \ 2 D fS2 ; S3 ; A1 g. One may assume, without loss of generality, that 1S2 D fF1 ; H2 ; F2 g; then 1S3 D fF1 ; H3 ; F3 g. By Lemma 76.5, d.Hi / D d.Fj / D 2, i D 2; 3 and j D 1; 2; 3. Let 1 .H1 / \ 2 D fR; U; A1 g. Then H1 \ Fi D U for i D 2; 3 so 1U D fH1 ; F2 ; F3 g. By the above, 2 D fR; S2 ; S3 ; U; A1 ; A2 ; A3 g and A1 D H1 \ F1 \ A; R D H1 \ H2 \ H3 ;

A2 D H2 \ F2 \ A; U D H1 \ F2 \ F3 ;

A3 D H3 \ F3 \ A; S2 D H2 \ F1 \ F2 ;

S3 D H3 \ F1 \ F3 : By Lemma 64.1(u), jG 0 j 2jH20 A0 j D 8. Assume that jG 0 j D 4. Then all nonabelian maximal subgroups of G have the same derived subgroup which is equal to G 0 . By Lemma 64.1(q), jG W Z.G/j D 8. If K=Z.G/ is a maximal subgroup of G=Z.G/ such that K ¤ A, then jK 0 j D 2 (Lemma 64.1(q)), a contradiction. Thus, jG 0 j D 8 so jG=Z.G/j D 16 (Lemma 64.1(q)). If Z.G/ — H 2 1 , then G D H Z.G/ so jG 0 j D jH 0 j D 4, which is a contradiction. Thus, Z.G/ < ˆ.G/. Since R \ U D ˆ.G/, we get jU 0 j D 2 (Lemma 64.1(q)). Let us consider the quotient group GN D G=Z.G/ which is of rank 3 and order N SN2 , SN3 and UN are four-groups so GN has at least 9 involutions. 16. The subgroups R, N D 4, it follows that GN is nonabelian. Let DN be a minimal nonabelian Since exp.G/ N Then, by Lemma 64.1(i), GN D DZ. N G/, N and now it is easy to see that subgroup of G. N N G Š D8 C2 (see Appendix 16). Then G contains exactly two distinct elementary N Then K 2 1 is abelian subgroups of order 8. Let KN Š E8 be such that KN ¤ A. nonabelian of rank 3, contrary to what has been proved already. Thus, all members of the set 1 .H1 / \ 2 are abelian.

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(ii) Since R is abelian, we have p D ˇ1 .G; H1 /

pC1 X iD2

ˇ1 .Hi ; R/ D

pC1 X

˛1 .Hi /:

iD2

Thus, for nonabelian H 2 1R fH1 g we have ˛1 .H / 2 f1; pg so d.H / D 2 (Lemmas 76.1 and 76.6), and so d.G/ 1 C d.H / D 3. (ii1) Assume that H2 is neither abelian nor an A1 -group. Then ˛1 .H2 / D p D ˇ1 .G; H1 / (Lemma 76.6) so Hi is abelian for i > 2. Next, H2 is an A2 -subgroup and jG 0 j jH20 j D p 2 (Lemma 76.6) so the set 1 has at most one abelian member (Lemma 64.1(u)). In that case, p D 2 and 1R D fH1 ; H2 ; H3 D Ag; here A is abelian, ˛1 .H2 / D 2 and jH20 j D 4 (Lemmas 76.6 and 76.4). Assume that d.G/ D 2. Then H10 D H20 has order 4 and G=H20 is an A1 -group (Lemmas 76.6 and 76.2) so jG 0 j D 8, G=Z.G/j D 2jG=G 0 j D 16, and G is as in (a). Now suppose that d.G/ D 3. Then j2 \ 1 .H1 /j D 2 C 1 D 3. Take S 2 .2 \ 1 .H1 // fRg; then S is abelian, by (i). Set 1S D fH1 ; F2 ; F3 g. Since H1 and H2 together contain all A1 -subgroups of G, F2 and F3 are not A1 -subgroups (they are nonabelian since A is the unique abelian maximal subgroup of G). Since Fi \ H1 D S .2 1 .H1 // is abelian, we get 1 < ˛1 .Fi / ˇ1 .G; H1 / D 2 so ˛1 .Fi / D 2 and Fi is an A2 -group, i D 2; 3 (Lemma 76.6(a)). Then F2 and F3 together contain 4 distinct A1 -subgroups and all of them are not contained in H1 (indeed, F2 \ H1 D S D F3 \ H1 is abelian), i.e., ˇ1 .G; H1 / 4 > 2, contrary to the hypothesis. Thus, all nonabelian members of the set 1R fH1 g are A1 -groups. It follows that all nonabelian members of the set 1 fH1 g are A1 -groups. P (ii2) Assume that d.G/ D 3. Then p D ˇ1 .G; H1 / D F 21 fH1 g ˛1 .F / 2 2 p 1 since the set 1 fH1 g has at least p 1 nonabelian members (Lemma 64.1(q)) which are A1 -subgroups. However, p 2 1 > p, and we get a contradiction. Thus, d.G/ D 2. Then H2 ; : : : ; HpC1 are A1 -groups, by (ii1) and hypothesis. (ii3) Let H20 D H30 .Š Cp ). In that case, CG .H20 / H2 H3 D G so H20 Z.G/. Then all maximal subgroups of the quotient group G=H20 are abelian (Exercise 1.6(a)) so G=H20 is nonabelian hence an A1 -group (Lemma 65.2(a)). We get 0 ; H10 D D HpC1

jG 0 j D j.G=H20 /0 jjH20 j D p p D p 2 ;

and G is as in (b). 0 (ii4) Now let H20 ; : : : ; HpC1 be all distinct. Set Q D H20 H30 . Then, as above, all 0 DQŠ maximal subgroups of the quotient group G=Q are abelian so H20 : : : HpC1 0 0 0 0 Ep 2 and H1 < Q. To fix ideas, assume that H2 ¤ H1 . Then H1 =H2 is nonabelian and ˇ1 .G=H20 ; H1 =H20 / D p1 so, by Lemma 76.5, jG 0 j D j.G=H20 /0 jjH20 j D p 2 p D p 3 and G=Q is an A1 -group. Thus, G is as in (c).

302

Groups of prime power order

Now we are ready to complete the proof of Theorem A for odd p. Theorem 76.14. Let G be a p-group, p > 2. Suppose that ˛1 .G/ 2 fp 2 C p; p 2 C p C 1g. Then G is an An -group, n 2 f2; 3g. Proof. Assume that the theorem is false. Then there is an Ak -group M 2 1 with k > 2, and so ˛1 .M / p 2 (Theorem 76.9), hence ˇ1 .G; M / p C 1. Assume that jM 0 j D p; then d.M / > 2 (Lemma 65.2(a)). Let S < M be an A2 -subgroup. Then, by Lemma 76.4, ˛1 .S / p 2 . If S 62 1 .M /, then, by Lemma 76.5, ˛1 .M / ˛1 .S / C 2.p 1/ p 2 C p C .p 2/ p 2 C p C 1 so, by Theorem 10.28, ˛1 .G/ > ˛1 .M / p 2 C p C 1, a contradiction. Thus, S 2 1 .M /. By Lemma 76.5, ˇ1 .M; S / p and ˇ1 .G; M / p 1, so ˛1 .G/ ˛1 .S / C ˇ1 .G; M / C ˇ1 .M; S / p 2 C p C p 1 > p 2 C p C 1 since p > 2, a contradiction. It follows that (A ) jM 0 j p 2 so ˇ1 .G; M / p (Lemma 76.5). Therefore, p 2 C p C 1 ˛1 .G/ D ˛1 .M / C ˇ1 .G; M / ˛1 .M / C p; hence p 2 ˛1 .M / p 2 C 1 and so M is an A3 -group of Theorem 76.10 or Theorem 76.11; then G is an A4 -group. By Lemma 76.12, since p > 2, we have (B ) Any member of the set 2 is either abelian or an A1 -group. (i) Let ˛1 .M / D p 2 ; then M is a group of Theorem 76.10 so jM 0 j D p 3 , and ˇ1 .G; M / 2 fp; p C 1g. By (B ), M is not a group of Theorem 76.10(a1) since the set 1 .M / \ 2 has no members which are A2 -groups. (i1) Let M be a group of Theorem 76.10(a), i.e., d.M / D 3, jM 0 j D p 3 , 1 .M / D fF1 ; : : : ; Fp 2 2Cp ; Ag, where ˛1 .Fi / D p for all i and A is abelian. By (B ), Fi 62 2 , i D 1; : : : ; p 2 , so j1 .M / \ 2 j D 1 hence A D ˆ.G/ and d.G/ D 2. Let 1 D fM1 D M; M2 ; : : : ; MpC1 g; then at most one member of the set 1 is abelian and PpC1 ˇ1 .G; M / D iD2 ˛1 .Mi / since ˆ.G/ D A is abelian. Let ˛1 .G/ D p 2 C p; then ˇ1 .G; M / D p so jM 0 j p 2 (Lemma 76.13), contrary to the first paragraph of (i). Thus, we have ˛1 .G/ D p 2 C p C 1. In view of ˇ1 .G; M / D p C 1 > p D jfM2 ; : : : ; MpC1 gj, one may assume that M2 is neither abelian nor an A1 -group (Dirichlet’s principle); then ˛1 .M2 / D p. It follows from p C 1 D ˇ1 .G; M / D ˛1 .M2 / C

pC1 X iD3

˛1 .Mi / D p C

pC1 X iD3

˛1 .Mi /

76 p-groups with few A1 -subgroups

303

that j1 j 3 members of the set 1 are abelian and exactly one its member, say M3 , is an A1 -group. Since jG 0 j jM 0 j D p 3 > p, the set 1 has at most one abelian member (Lemma 64.1(u)) so we get p D 3. Then 1 D fM1 D M; M2 ; M3 ; M4 D Ag, where A is abelian. However, by Lemma 76.2, we get M30 D M 0 , a contradiction since jM 0 j D p 3 > p D jM30 j. (i2) Let M be a group of Theorem 76.10(b1,b2), i.e., d.M / D 2, 1 .M / D fF1 ; : : : ; Fp ; Ag, where ˛1 .Fi / D p, F10 D D Fp0 has order p 2 , A is abelian. It follows from (B ) that Fi 62 1 .M / \ 2 for all i so A D ˆ.G/ and d.G/ D 2. Let 1 D fM1 D M; : : : ; MpC1 g. As in (i1), one has to consider two possibilities: ˛1 .G/ 2 fp 2 C p; p 2 C p C 1g. Let ˛1 .G/ D p 2 C p; then ˇ1 .G; M / D p so jM 0 j p 2 (Lemma 76.13), contrary to the first paragraph of (i). Now let ˛1 .G/ D p 2 CpC1. In that case, ˇ1 .G; M / D pC1. Therefore, as in (i1), one may assume that M2 is neither abelian nor an A1 -group. In that case, as in (i1), we get p D 3 and 1 D fM1 D M; M2 ; M3 ; M4 D Ag, where A is abelian, ˛1 .M3 / D 1. By Lemma 76.2, we get M30 D M 0 , a contradiction since jM 0 j p 2 > p D jM30 j. All the above, together with (A ) and (B ), yields (C ) ˛1 .M / D p 2 C 1, i.e., M is a group of Theorem 76.11(b). (ii) In view of (C ), Theorem 76.11(b) and Lemma 76.5, we must have ˇ1 .G; M / p so that ˛1 .G/ D p 2 C p C 1; then ˇ1 .G; M / D p. By Theorem 76.11(b), d.M / D 2, 1 .M / D fF1 ; : : : ; Fp ; Bg, where ˛1 .Fi / D p, i D 1; : : : ; p, ˛1 .B/ D 1. By (B ), ˆ.G/ D B so d.G/ D 2. Set 1 D fM D M1 ; : : : ; MpC1 g; then ˇ1 .Mi ; B/ p 1, i > 1 (Lemma 76.5). We have ˛1 .G/ D ˛1 .M / C

pC1 X

ˇ1 .Mi ; B/ p 2 C 1 C .p 1/p > p 2 C p C 1;

iD2

a final contradiction. Theorem 76.15. A 2-group G satisfying ˛1 .G/ D 6 is an An -group, n 2 f2; 3g. Proof. Assume that this is false. Then, by Theorems 76.9–76.11, G is an A4 -group so there is an A3 -group H 2 1 . Since ˛1 .H / 4 (Theorem 76.9) and ˇ1 .G; H / > 0 (Proposition 10.28), we get ˛1 .H / 2 f4; 5g so ˇ1 .G; H / 2 f2; 1g. (i) Let ˛1 .H / D 4; then ˇ1 .G; H / D 2, jH 0 j 4 (Lemma 76.13). On the other hand, H is a group of Theorem 76.10, and so jH 0 j D 8, a contradiction. (ii) Let ˛1 .H / D 5. Then H is a group of Theorem 76.11 so jH 0 j 4. However, ˇ1 .G; H / D 1 so jH 0 j D 2 (Lemma 76.5), a final contradiction.

Appendix by Z. Janko: Nonexistence of groups of types G7;1 and G7;2 Theorem A.1. 2-groups of type G7;1 do not exist.

304

Groups of prime power order

Proof. Let G be a 2-group of type G7;1 , i.e., G is a 2-group of Lemma 76.12 with ˛1 .G/ D 7. We shall use freely the notation and the results of Lemma 76.12. By Lemma 76.4(c), Hi is metacyclic for i D 1; 2; 3 since p D 2 and ˛1 .Hi / D 2. Here we note that jGj 26 and so jHi j 25 since R 2 2 and R is an A2 -group and so jRj 24 . In particular, R1 , R2 , R3 , A1 , A2 , A3 are also metacyclic. Next, R is nonmetacyclic since d.R/ D 3. Since A1 is a maximal subgroup of A and d.A1 / D 2, we have d.A/ 3. Assume that d.A/ D 2. In that case, all members of the set 1 are two-generator so we can use the results of 70. By Theorem 70.1, cl.G/ > 2. Then Theorem 70.2 implies that .G=K3 .G//0 Š E8 . But K3 .G/ ¤ f1g and so jG 0 j > 8, contrary to Lemma 76.12(d). Hence we have d.A/ D 3 and so 1 .A/ Š E8 . Since Hi is metacyclic and jHi j 25 (i D 1; 2; 3), we can use Proposition 71.2. It follows that Hi0 Š C4 (Exercise 1) and Hi is isomorphic to one of the following groups: m

n

m1

(5) Hi D ha; b j a2 D 1; m 4; b 2 D a2 ab D a

1C2m2

(6) Hi D ha; b j a8 D 1; b

2n

; n 2; D 0; 1;

i;

D a4 ; n 2; D 0; 1; ab D a1C4 ; D 0; 1i:

However, in case (5) we get Ã1 .Hi / D ha2 ; b 2 i (this subgroup is abelian) and we see that ha; b 2 i, ha2 ; bi, ha2 ; b 2 ihabi are three maximal subgroups of Hi and they are all A1 -groups, contrary to ˛1 .Hi / D 2 (Lemma 76.12). (Indeed, to check that a nonabelian metacyclic p-group is an A1 -group, is suffices, in view of Lemma 65.2(a), to show that its derived subgroup has order p. For example, if, in (5), K D ha; b 2 i, 2 m2 /2 m1 m1 D a1C2 so K 0 D ha2 i has order 2 and K is an A1 then ab D a.1C2 group.) We have proved that Hi must be isomorphic to a group given in (6). Here we distinguish two essentially different cases D 0 (splitting case) and D 1 (nonsplitting case). (i) We consider first the splitting case D 0 so that H1 D ha; b j a8 D b 2 D 1; n 2; ab D a1 z ; z D a4 ; D 0; 1i; n

n1

and v D a2 so that v 2 D z and u are involutions. Since and we also set u D b 2 nC3 jH1 j D 2 , we have jGj D 2nC4 , n 2. But A1 D hb 2 i hai is the unique abelian maximal subgroup of H1 and so A \ H1 D A1 . The fact that d.A/ D 3 implies that there is an involution t 2 A A1 so that 1 .A/ D hu; z; t i because 1 .H1 / D hu; zi < 1 .A/. We have Z.H1 / D hb 2 ; zi D Z.G/ (indeed, Z.G/ H1 and jZ.G/j D jZ.H1 /j) and ˆ.H1 / D hb 2 i hvi D ˆ.G/ since ˆ.H1 / ˆ.G/ and n1 d.G/ D 3; in particular, u D b 2 2 Z.G/. Also, H10 D hvi Š C4 is normal in G and 0 jG j D 8 (Lemma 76.12). Since ˆ.G/ D hb 2 ihvi and H1 > G 0 > hvi, it follows, by the modular law, that G 0 D huihvi is abelian of type .2; 4/. Note that R1 , R2 , R3 are

76 p-groups with few A1 -subgroups

305

metacyclic of order 2nC2 because 1 .H1 / D fA1 ; R2 ; R3 g, 1 .H2 / D fA2 ; R1 ; R3 g and H1 and H2 are metacyclic of order 2nC3 . We have G D hH1 ; t i D ha; b; t i, hŒa; bi D hvi, Œa; t D 1 (recall that a; t 2 A) and, since G=hvi is nonabelian, we get Œb; t 2 G 0 hvi. On the other hand, 1 .A/ D hu; z; t i is normal in G and so Œb; t D t b t 2 1 .A/. It follows that Œb; t D uz ı with ı 2 f0; 1g since Œb; t D b 1 b t 2 H1 \ 1 .A/ D 1 .H1 / and Œb; t 2 G 0 hvi (recall that z D v 2 ). Also note that R 2 2 is an A2 -group of order 2nC2 and therefore four A1 -subgroups S1 , S2 , S3 , S4 which are contained in R are of order 2nC1 . Since ˛1 .G/ D 7, fS1 ; S2 ; S3 ; S4 ; R1 ; R2 ; R3 g is the set of all A1 -subgroups in G. In particular, all A1 -subgroups of G of order < 2nC2 are contained in R. Let ı D 1 so that Œb; t D uz. Since uz 2 hb 2 ; zi D Z.G/, huzi G G and so hb; t i=huzi is abelian which implies that hb; t i0 D huzi Š C2 . It follows that hb; t i n1 is an A1 -subgroup (Lemma 65.2(a)) containing hb 2 D u; z; t i Š E8 and so hb; t i nC2 is nonmetacyclic. Also, hb; t i is of order 2 (since hu; z; t i \ hbi D hui) and so hb; t i is of order 2nC2 and therefore hb; t i is one of R1 , R2 , R3 (see the last sentence of the previous paragraph). This is a contradiction since R1 , R2 , R3 are metacyclic. n1 Suppose that ı D 0; then Œb; t D u D b 2 is an involution, and therefore we have either hb; t i Š M2nC1 .n > 2/ or hb; t i Š D8 .n D 2/. It follows that hb; t i is an A1 -subgroup of order 2nC1 and so hb; t i < R. On the other hand, R > ˆ.G/ and so R ˆ.G/hb; t i. But ˆ.G/hb; t i 2 1 . This is a contradiction since R 2 2 . (ii) We consider now the non-splitting case D 1 so that H1 D ha; b j a8 D 1; b 2 D a4 D z; n 2; ab D a1 z ; D 0; 1i n

n1

and we set v D a2 , w D b 2 and u D vw. We see again that A1 D hb 2 ; ai D A\H1 is the unique abelian maximal subgroup in H1 . Since d.A/ D 3 and 1 .A1 / D 1 .H1 / D hu; zi, there is an involution t 2 AA1 such that 1 .A/ D hu; z; t i Š E8 . Also, jH1 j D 2nC3 , jGj D 2nC4 and the metacyclic A1 -subgroups R1 , R2 , R3 are of order 2nC2 . Since R 2 2 and R is an A2 -group with ˛1 .R/ D 4, four A1 -subgroups S1 , S2 , S3 , S4 , which are contained in R, are of order 2nC1 . We have Z.H1 / D hb 2 i Š C2n , Z.H1 / D Z.G/, ˆ.G/ D ˆ.H1 / D hb 2 ihvi since ˆ.H1 / ˆ.G/ and d.G/ D 3. Also, H10 D hvi Š C4 is normal in G, jG 0 j D 8 (Lemma 76.12) and so G 0 D hv; wi D hui hvi is abelian of type .4; 2/. We have G D hH1 ; t i D ha; b; t i, hŒa; bi D hvi, where t 2 1 .A/ 1 .H1 /, Œa; t D 1 and, since G=hvi is nonabelian (recall that o.v/ D 4 < 8 D jG 0 j), we get Œb; t 2 G 0 hvi. On the other hand, 1 .A/ D hu; z; t i is normal in G and so b 1 b t D Œb; t D t b t 2 1 .A/ \ H1 D 1 .H1 / D 1 .G 0 / D hui hzi: Since z D v 2 2 hvi, it follows that Œb; t D u or uz. Note that o.b/ D 2nC1 8 and so, taking into account that ub D .vw/b D v 1 w D vwz D uz, we have ˆ.G/ D n1 D w, v D uw 1 and so v 2 hb; ui. hb 2 ; vi < hb; ui Š M2nC2 noting that b 2 nC2 Therefore hb; ui is an A1 -subgroup of order 2 contained in H1 and consequently

306

Groups of prime power order

hb; ui D R2 or R3 . Since t normalizes hb; ui, we have jhb; t i W hb; uij D 2 and so hb; t i is of order 2nC3 which implies that hb; t i is a maximal subgroup of G with hb; t i0 hz; ui and so hb; t i0 D hz; ui Š E4 . It follows that hb; t i is a nonmetacyclic member of the set 1 . Hence hb; t i D F2 or F3 and so hb; t i is an A3 -group. On the other hand, AQ D hb 2 i hui ht i is an abelian nonmetacyclic subgroup of order 2nC2 in hb; t i. Hence AQ is the unique abelian maximal subgroup of hb; t i and so AQ D A2 or A3 (see, in Lemma 76.12, lists of the members of the sets 1 .F2 / and 1 .F3 /). This is a contradiction since A1 , A2 , A3 are metacyclic. Definition 2. A 2-group G is said to be a group of type G7;2 , if it satisfies the following conditions: (1G7;2 ) d.G/ D 2, 1 D fH1 ; H2 ; H3 g, ˛1 .H1 / D 4, ˛1 .H2 / D 2, ˛1 .H3 / D 1. (2G7;2 ) 1 .H1 / D fF1 ; : : : ; F6 ; Ag, where ˛1 .Fi / D 2, i 6, A D ˆ.G/ is abelian, jH10 j D 8. We have ˇ1 .G; H1 / D 3 so ˛1 .G/ D 7. (3G7;2 ) jG W Z.G/j D 25 , Z.G/ D Z.H1 / < ˆ.H1 /. (4G7;2 ) H10 D H20 H30 , G=H10 is an A1 -group so jG 0 j D 24 , ˇ1 .G=H20 ; H1 =H20 / D 1, ˇ1 .G=H30 ; H1 =H30 / D 2. It follows from Definition 2 that a 2-group G of type G7;2 , if it exists, is an A4 group. It will be proved in the Appendix that groups of type G7;2 do not exist. Theorem A.2. Groups of type G7;2 do not exist. Proof. We prove first that a two-generator 2-group X which is an A2 -group is metacyclic. Indeed, if jXj > 24 , then the result follows from Lemma 76.4(c). So let jXj D 24 and let Y be a nonabelian subgroup of order 23 . Since d.X/ D 2, we have CX .Y / Y . It follows from Lemma 64.1(i) that X is of maximal class and so X is metacyclic and we are done. Let G be a 2-group of type G7;2 , where we use the notation from Definition 2. By Lemma 76.6(a), H2 ; F1 ; : : : ; F6 are all two-generator A2 -groups. Therefore, the groups H2 ; F1 ; : : : ; F6 are all metacyclic. But A D ˆ.G/ < H2 and so A is also metacyclic. Hence H1 is minimal nonmetacyclic and so, by Theorem 66.1, H1 is an A2 -group of order 25 . This is a contradiction since H1 contains a proper subgroup F1 which is an A2 -group. Now we are ready to prove Theorem 76.16. If a 2-group G satisfies ˛1 .G/ D 7, then G is an An -group, n 2 f2; 3g. Proof. Let G be a counterexample of minimal order. If H 2 1 , then ˛1 .H / < ˛1 .G/ D 7, so, by Lemma 76.6 and Theorems 76.9–76.11, 76.15, H is either abelian or an An -group, n 3. Therefore, G is an A4 -group so one can choose H 2 1 so that H is an A3 -group. In that case, 4 ˛1 .H / 6 (Theorem 76.9) so H is one of

76 p-groups with few A1 -subgroups

307

the A3 -groups of Theorems 76.10, 76.11 or 76.15. Since G7;1 -groups do not exist, by Appendix, it follows from Lemma 76.12 that (i) The set 2 has no members which are A2 -groups. It follows that H is not a group of Theorems 76.10(a1) and 76.11(c). (ii) Let H 2 1 be a group of Theorem 76.10(a2,b1,b2). In that case, ˛1 .H / D 4, jH 0 j D 8, the set 1 .H / has exactly one abelian member A and all other its members are A2 -groups. It follows from (i) that A D ˆ.G/ so, in view of jG W Aj D 22 , we get d.G/ D 2, and then Z.G/ < ˆ.G/ D A. Set 1 D fH1 D H; H2 ; H3 g; then A 2 1 .Hi /, i D 1; 2; 3, so ˛1 .H2 / C ˛1 .H3 / D ˇ1 .G; H / D ˛1 .G/ ˛1 .H / D 7 4 D 3: One may assume that ˛1 .H2 / D 2 and ˛1 .H3 / D 1 (˛1 .H2 / ¤ 3 since the set 1 .H2 / has one abelian member). Thus, H2 is an A2 -group (Lemma 76.6(a)) and H3 is an A1 -group. We have d.H2 / D 2 D d.H3 / (Lemmas 76.1 and 76.6(a)). Assume that H is a group from Theorem 76.10(b1,b2); then d.H / D 2. In that case, by the above, G and all members of the set 1 are two-generator so G is metacyclic (Lemma 64.1(n)). Since H2 is an A2 -group and H3 is an A1 -group, Exercise 1 shows that G is an A3 -group, a contradiction. Thus, H is a group from Theorem 76.10(a2). Then d.H / D 3, jH 0 j D 8, 1 .H / D fF1 ; : : : ; F6 ; Ag, where ˛1 .Fi / D 2 and d.Fi / D 2 for i 6 and A is abelian, jH W Z.H /j D 2jH 0 j D 16 (Lemma 64.1(q)). By 65, H2 =Z.H2 / 2 fD8 ; C4 C4 g since d.H2 / D 2. Suppose that H2 =Z.H2 / Š D8 . Then Z.H2 / ˆ.H2 / ˆ.G/ D A since d.H2 / D 2, and so jA W Z.H2 /j D 4. Since H3 is an A1 -group, we get Z.H3 / D ˆ.H3 / < ˆ.G/ D A and jA W ˆ.H3 /j D 2. Then Z.H2 / \ Z.H3 / Z.G/ and jA W Z.G/j 8 so jG W Z.G/j D jG W AjjA W Z.G/j 25 . By the above, jH W Z.G/j D 24 so Z.G/ D Z.H / has index 25 in G. Now suppose that H2 =Z.H2 / is abelian of type .4; 4/. We have Z.H2 / < ˆ.H2 / < ˆ.G/ < H3 . Assume that Z.H2 / 6 Z.H3 /. Then H3 =Z.H2 / is cyclic since Z.H3 / D ˆ.H3 / (Lemma 76.1). Thus, G=Z.H2 / has the cyclic subgroup H3 =Z.H2 / of index 2 and the abelian subgroup H2 =Z.H2 / of type .4; 4/ and index 2, which is impossible. It follows that Z.H2 / Z.H3 / so Z.H2 / Z.G/, and we get jG W Z.G/j D jG W H2 jjH2 W Z.H2 /j 2 24 D 25 . Since Z.G/ < ˆ.G/ < H2 , we get Z.H2 / D Z.G/ and so jG W Z.G/j D 25 . Comparing the orders, we get Z.G/ D Z.H /. Since jH 0 j D 8 and jFi0 j D 4, the equality Fi Z.G/ D H is impossible for i 6. It follows that Z.G/ < Fi for all i . In particular, Z.G/ < ˆ.H /. We see that G is a group of type G7;2 . By the Appendix, groups of type G7;2 do not exist, and this is a contradiction. (iii) Now let ˛1 .H / D 22 C 1, i.e., H is a group of Theorem 76.11. As we have noticed in (i), H is not a group of Theorem 76.11(c). We have jH 0 j > 2 (Theorem 76.11) and ˇ1 .G; H / D 2.

308

Groups of prime power order

(iii1) Let H 2 1 be a group of Theorem 76.11(a), i.e., d.H / D 2, jH 0 j D 4, 1 .H / D fF1 ; F2 ; Ag, where d.F1 / D 3, ˛1 .F1 / D 4, jF10 j D 2, ˛1 .F2 / D 1 and A is abelian. If d.G/ D 3, then F1 2 1 .H / 2 , contrary to (i). Thus, d.G/ D 2 so 1 D fH1 D H; H2 ; H3 g. We have ˆ.G/ 2 fF2 ; Ag, by (i). Assume that F2 D ˆ.G/. Then 7 D ˛1 .G/ D ˛1 .H1 / C

3 X iD2

ˇ1 .Hi ; F2 / D 5 C

3 X

ˇ1 .Hi ; F2 /

iD2

so ˇ1 .H2 ; F2 / C ˇ1 .H3 ; F2 / D 2. By Remark 1, we get ˇ1 .Hi ; F2 / D 1, i D 2; 3. In that case, ˛1 .Hi / D 2 so jHi0 j D 4, d.Hi / D 2, i D 2; 3 (Lemma 76.6(a)). Thus, G and all members of the set 1 are two-generator so G is metacyclic (Lemma 64.1(n)). Then, by Exercise 1, applied to H2 and H3 , G is an A3 -group so G is not a counterexample. Therefore, we must have ˆ.G/ D A. Then we have ˛1 .H2 / C ˛1 .H3 / D 2. If ˛1 .Hi / D 1, i D 2; 3, then G and all members of the set 1 are two-generator so G is metacyclic (Lemma 64.1(n)). In that case, by Exercise 1, G must be A2 -group so it is not a counterexample. Thus, ˛1 .H2 / D 2 so jH20 j D 4 and H3 is abelian. By Lemma 76.2, H 0 D H20 and G=H 0 is an A1 -group. It follows that jG 0 j D 2jH 0 j D 8 so jG W Z.G/j D 16 (Lemma 64.1(q)). Since d.G/ D 2, we get Z.G/ < ˆ.G/ < Hi so jHi W Z.G/j D 8, i D 1; 2; 3. Since H1 =Z.G/ is noncyclic and two-generator. we get Z.G/ < ˆ.H1 / < Fi , i D 1; 2. Then jFi W Z.G/j D 4, i D 1; 2. Since d.F1 / D 3 and Z.F1 / > ˆ.F1 /, we get, by Lemma 76.8(c), ˛1 .F1 / 2 22 1 D ˛1 .G/, contrary to Proposition 10.28. (iii2) Let H 2 1 be a group from Theorem 76.11(b). Retaining the notation of Theorem 76.11, we get, in view of (i), ˆ.G/ D B so d.G/ D 2. Set 1 D fH1 D H; H2 ; H3 g. Then, as above, ˇ1 .Hi ; R/ D 1, and we conclude that ˛1 .Hi / D 2 and d.Hi / D 2 for i D 2; 3 (Lemma 76.6(a)). In that case, G and all members of the set 1 are two-generator so G is metacyclic (Lemma 64.1(n)). Then, by Exercise 1, applied to H2 and H3 , G is an A3 -group so G is not a counterexample. (iv) It remains to consider the case where H 2 1 is a group of Theorem 76.15, i.e., ˛1 .H / D 6. In that case, ˇ1 .G; H / D 1 so jH 0 j D 2 and jG 0 j D 4, d.G/ D 2, 1 D fH1 D H; H2 ; Ag, where H2 is an A1 -group and A is abelian (Lemma 76.5). By Lemma 64.1(q), jG W Z.G/j D 8 and, since Z.G/ < ˆ.G/ < H , we get jH W Z.G/j D 4 so jH W Z.H /j D 4. Since H is not an A1 -group, we get d.H / D 3. Then, by Lemma 76.8(c), ˛1 .H / 2 22 1 > 6, a final contradiction. Theorem A follows immediately from Lemma 76.6, Theorems 76.9–76.11, 76.14– 76.16 and the Appendix. If G is a 2-group of maximal class and order 26 , then ˛1 .G/ D 8 and G is an A4 -group. I do not know if, for odd p, there exists an A4 -group G with ˛1 .G/ D .p 2 C p C 1/ C 1.

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If G is a nonabelian p-group such that for each K 2 2 , we have ˛1 .1K / < p 2 C 2p C1, then G is an An -group, n 4. Assume that this is false. Then there is H 2 1 such that H is an Ak -group with k 4. By Theorem A, ˛1 .H / p 2 C p C 2. If K K 2 1 .H / is abelian, we have Pthen, since the set 1 has 2at most one abelian member, K ˛1 .1 / D ˛1 .H / C M 2 K fH g ˛1 .M / .p C p C 2/ C .p 1/ D p 2 C 2p C 1 1 (Remark 1), a contradiction. If K is nonabelian, then, by Remark 1, X ˇ1 .M; K/ ˛1 .1K / D ˛1 .H / C M 21K fH g

.p 2 C p C 2/ C p.p 1/ D 2p 2 C 2 > p 2 C 2p C 1; a final contradiction. 3o . A lower estimate of ˛1 .G/ in terms of d.G/. To facilitate the proof of Theorem B, we begin with the following remarks. Remarks. 10. Let G be a nonabelian p-group, d.G/ D 3 and H 2 1 . Let us prove that ˇ1 .G; H / p.p 1/. This is the case if H is abelian since then ˇ1 .G; H / D ˛1 .G/ p 2 > p.p 1/ (Theorem 76.9). Next we assume that H is nonabelian. If ˛1 .G/ D p 2 , then ˛1 .H / p (Theorem 76.10) so ˇ1 .G; H / p 2 p D p.p 1/. Now let ˛1 .G/ p 2 C 1. The result is true if ˛1 .H / p C 1. So, in view of Lemma 76.6 and Theorem 76.9, one may assume that ˛1 .H / p 2 . .H / \ 2 . Assuming that R is nonabelian, we get, using Remark 1, Take R 2 1P ˇ1 .G; H / F 2 R fH g ˇ1 .F; R/ p.p 1/ since all p C 1 members of the set 1

1R are nonabelian. If R is abelian,Pthen the set 1R has at least p nonabelian members so, by Lemma 76.6, ˇ1 .G; H / F 2 R fH g ˛1 .F / p 2 .p 1/p. 1

11. Suppose that G is a p-group such d D d.G/ 3. We claim that if H 2 1 is nonabelian, then ˇ1 .G; H / p d 2 .p 1/. We use induction on d . Remark 10 is the basis of induction so assume that d > 3. If X 2 1 , then d.X/ d 1 3. Since j1 .H / \ 2 j D p d 2 C C p C 1 > p C 1, the set 1 .H / \ 2 has a nonabelian member R (Exercise 1.6(a)). Then all members of the set 1R D fH1 D H; : : : ; HpC1 g are nonabelian. Let d.Hi / D di , where di d 1. By induction, ˇ1 .Hi ; R/ p di 2 .p 1/ p d 3 .p 1/, i > 1. Therefore, ˇ1 .G; H / PpC1 d 3 .p 1/ p D p d 2 .p 1/, as required. iD2 ˇ1 .Hi ; R/ p Theorem B. Suppose that a p-group G is neither abelian nor an A1 -group and d D d.G/. Then ˛1 .G/ p d 1 . Proof. By Lemma 76.6 and Theorem 76.9, the theorem is true for d 3 so we may assume in what follows that d > 3. If X 2 1 , then d.X/ d 1. We proceed by induction on d . If H 2 1 is nonabelian and such that d.H / d , then, by induction, ˛1 .H / p d 1 , and we are done. Therefore, one may assume that, for all nonabelian H 2 1 , we have d.H / < d ; then, for such H we

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have d.H / D d 1. 3/. It follows from Lemma 76.3 that there is a nonabelian R 2 2 . Set 1R D fH1 ; : : : ; HpC1 g. By induction, ˛1 .Hi / p d 2 and, by Remark 11, ˇ1 .Hi ; R/ p d 3 .p 1/ for all i . Therefore, ˛1 .G/ ˛1 .H1 / C P pC1 d 2 C p d 3 .p 1/ p D p d 1 . iD2 ˇ1 .Hi ; R/ p 4o . Groups of exponent p. We assume that all groups, considered in this subsection, have exponent p. If G is an A1 -group of exponent p, then jGj D p 3 . Now let G be an A2 -group; then jGj D p 4 . If the set 1 has exactly one abelian member, then cl.G/ D 3 and ˛1 .G/ D p. If the set 1 has two distinct abelian members, then G D S C and ˛1 .G/ D p 2 , where jC j D p. Remarks. 12. Suppose that G is a nonabelian group of exponent p and order p 5 and H 2 1 . By Proposition 2.3, p 2 j ˛1 .H /, p 2 j ˛1 .G/ so p 2 j ˇ1 .G; H /. 13. We claim that if G is nonabelian of order p 6 and exponent p, then ˛1 .G/ p 3 . Let R 2 2 be such that R ¤ Z.G/. Set 1R D fH1 ; : : : ; HpC1 g. If R is nonabelian, P then, by Remark 12 and Proposition 10.28, ˛1 .G/ ˛1 .R/ C pC1 iD1 ˇ1 .Hi ; R/ > p 2 .p C 1/ > p 3 . If R is abelian, one may assume that H1 ; : : : ; Hp are nonabelian so ˛1 .Hi / p 2 .i D 1; : : : ; p/, by Proposition 2.3. In that case we have ˛1 .G/ P p 2 3 iD1 ˛1 .Hi / p p D p . Now let R D Z.G/. Then G D A E, where A is an A1 -group and E Š Ep 3 so, by Remark 3, ˛1 .G/ D p 6 . Theorem C. If G is nonabelian of order p m and exponent p, m > 3, then: (a) If H 2 1 , then ˇ1 .G; H / p m4 .p 1/. (b) ˛1 .G/ p m3 . (c) If ˛1 .G/ D p m3 , then G is of maximal class with abelian subgroup of index p and m p. Proof. In all three cases we proceed by induction on m. (a) One may assume that m > 4, in view of Remark 1. Let H 2 1 be nonabelian and R 2 1 .H / \ 2 . Then the set 1R D fH1 D H; : : : ; HpC1 g has at most one abelian member. Suppose that PpHpC1 is abelian; then R is also abelian. In that case, by induction, ˇ1 .G; H / iD2 ˛1 .Hi / p .m1/3 .p 1/ D p m4 .p 1/. Now suppose that the set 1R has no abelian members. Then, by induction, we get PpC1 ˇ1 .G; H / iD2 ˇ1 .Hi ; R/ p .m1/4 .p 1/ p D p m4 .p 1/, and (a) is proven. (b) In view of Remarks 12 and 13, one may assume that m > 6. There is R 2 2 such that R ¤ Z.G/ since G is neither abelian nor A1 -group. Set 1R D fH1 ; : : : ; HpC1 g:

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Suppose that R is abelian. One may assume in that case that H1 ; : : : ; Hp are nonabelian. By induction, we have ˛1 .G/ D

p X

˛1 .Hi / p .m1/3 p D p m3 :

iD1

Now let R be nonabelian. Then, by induction and (a), we have ˛1 .G/ D ˛1 .R/ C

pC1 X

ˇ1 .Hi ; R/ p .m2/3 C p .m1/4 .p 1/.p C 1/ D p m3 ;

iD1

completing the proof of (b). (c) Let ˛1 .G/ D p m3 . As in (b), we prove that there is an abelian A 2 1 . (i) Let jZ.G/j D p. Then G is of maximal class (Remark 5) and, by Theorem 9.5, since exp.G/ D p, we must have m p. It is easy to check that then indeed ˛1 .G/ D p m3 (see Remark 5). Next we assume that jZ.G/j > p. (ii) As in Remark 13, we have jG W Z.G/j > p 2 . Therefore, there is a nonabelian H1 2 1 containing Z.G/ as a subgroup of index > p. Take Z.G/ < R 2 1 .H1 / \ 2 and set 1R D fH1 ; : : : ; HpC1 g. One may assume that H1 ; : : : ; Hp are nonabelian. It follows from Z.G/ < Hi and jZ.G/j > p that Hi is not of maximal class so, by induction,P ˛1 .Hi / > p m4 . If HpC1 is abelian, then R is also abelian, and we have m4 p D p m3 , contrary to the hypothesis. If H ˛1 .G/ p pC1 is iD1 ˛1 .Hi / > p nonabelian, then, by (a), we get ˛1 .G/ ˛1 .H1 / C

pC1 X

ˇ1 .Hi ; R/ > p m4 C p m5 .p 1/ p D p m3 ;

iD2

a contradiction. 5o . M3 -groups. A p-group G is said to be an M3 -group, if all its A1 -subgroups have the same order p 3 . The groups of maximal class with abelian subgroup of index p are M3 -groups (Remark 5). If an A2 -group G is also an M3 -group, then jGj D p 4 and ˛1 .G/ 2 fp; p 2 g. Theorem C1 . Let G be a nonabelian M3 -group of order p m and exponent p, m > 3. Then: (a) If H 2 1 , then ˇ1 .G; H / p m4 .p 1/. (b) ˛1 .G/ p m3 . (c) If, in addition, ˛1 .G/ D p m3 , then G is of maximal class with abelian subgroup of index p.

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Groups of prime power order

Theorem C2 . Let G be a nonabelian group of order p m . Suppose that G has an A1 subgroup of order p a but has no A1 -subgroups of order > p a . Then ˛1 .G/ p ma . If, in addition, ˛1 .G/ D p ma , then all A1 -subgroups of G have the same order p a . The proofs are omitted since they are repetitions of the proof of Theorem C. For p D 2, see 90. See also Problem 155. 6o . p-groups with few conjugate classes of A1 -subgroups. Given a p-group G, let 1 .G/ denote the number of conjugate classes of A1 -subgroups in G. Given H 2 1 , we define (i) ˇN1 .G; H / is the number of conjugate G-classes of A1 subgroups not contained in H , (ii) N 1 .H / is the number of G-classes of A1 -subgroups contained in H . We have N 1 .H / 1 .H / and, as a rule, the strong inequality holds. Obviously, N 1 .G/ D 1 .G/. Theorem 76.17 (compare with Lemma 76.5). Let G be a p-group and let H 2 1 be nonabelian. Then ˇN1 .G; H / p 1. If ˇN1 .G; H / D p 1, then: (a) d.G/ D 2 and 1 D fH1 D H; : : : ; Hp ; Ag, where A is the unique abelian member of the set 1 . (b) H10 D D Hp0 , G=H10 is an A1 -group so jG 0 j D pjH10 j and d.Hi / 3, i p. (c) If jHi0 j D p for some i > 1, then H2 ; : : : ; Hp are A1 -groups and ˇ1 .G; H / D p 1. Lemma 76.18. Let G be a p-group and let a nonabelian H 2 1 be not an A1 subgroup. Suppose that N 1 .H / D 1. Then (a) H has no proper nonabelian Ginvariant subgroups and the set 1 .H / has exactly one abelian member, (b) d.G/ D 2, (c) jH 0 j > p. Proof. Assume that a nonabelian A 2 1 .H / is G-invariant and let U A be an A1 -subgroup. By Proposition 10.28, H has an A1 -subgroup V that is not contained in A. Since U G A, U and V are not conjugate in G, a contradiction. This proves the first assertion of (a). Assume that jH 0 j D p; then d.H / > 2 since H is not an A1 -group (Lemma 65.2(a)). Let U < H be an A1 -subgroup. Then U 0 D H 0 so U G H . It follows that H D NG .U / so jG W NG .U /j D p. In that case, H contains exactly p subgroups conjugate with U in G so, by hypothesis, ˛1 .H / D p. However, by Theorem B, ˛1 .H / p 2 , contrary to what has just been said. Thus, jH 0 j > p, proving (c). The set 1 .H / has exactly one abelian member (otherwise, jH 0 j D p), and this completes the proof of (a). Then, since all members of the set 1 .H / \ 2 are abelian, we get j1 .H / \ 2 j D 1 (Lemma 64.1(q)), hence d.G/ D 2, completing the proof of (b). Proof of Theorem 76.17. Let N 2 2 \ 1 .H /; then N 6 Z.G/ since H is nonabelian. Set 1N D fH1 D H; : : : ; HpC1 g. Since the set 1N has at most one abelian member, one may assume that H1 ; : : : ; Hp are nonabelian. By Proposition 10.28,

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there exists an A1 -subgroup Bi Hi such that Bi 6 N , i D 2; : : : ; p. For i ¤ j , we have BiG N D Hi ¤ Hj D BjG N so BiG ¤ BjG , and we conclude that Bi and Bj are not conjugate in G. Thus, ˇN1 .G; H / jfB2 ; : : : ; Bp gj D p 1. Now let ˇN1 .G; H / D p 1. For i D 2; : : : ; p, all A1 -subgroups of Hi that are not contained in N , are G-conjugate with Bi so HpC1 must be abelian (otherwise, ˇN1 .G; H / > p 1); then N is also abelian. In that case, N 1 .Hi / D 1 for i D 2; : : : ; p so, by Lemma 76.18(b), d.G/ D 2; then 1 D 1N and N D ˆ.G/. By Lemma 76.2, H10 D D Hp0 and G=H10 is an A1 -subgroup so jG 0 j D pjH10 j and, since Hi0 ˆ.Hi /, we get d.Hi / D d.Hi =Hi0 / 3 for i D 2; : : : ; p (Lemma 76.1). Now, (c) follows easily from Lemmas 65.2(a) and 76.18. If G is a group of maximal class and order p n > p 3 with abelian subgroup of index N p, then 1 .G/ D p and, if H 2 1 is nonabelian, then ˇ.G; H / D p 1. Corollary 76.19 (compare with Lemma 76.6(b)). Let G be a p-group. If 1 .G/ D p C 1, then d.G/ D 2 and ˆ.G/ is abelian. Proof. Since G is neither abelian nor an A1 -group, there exists R 2 2 such that R 6 Z.G/ (Lemma 76.3). Set 1R D fH1 ; : : : ; HpC1 g; then at most one member of the set 1R is abelian. Suppose that H1 is nonabelian; then ˇN1 .G; H1 / p, by hypothesis. If ˇN1 .G; H1 / D p 1, then d.G/ D 2 and R D ˆ.G/ is abelian (Theorem 76.17). Now we assume that ˇN1 .G; H1 / D p; then N 1 .H1 / D 1 (otherwise, 1 .G/ > p C 1). In that case, d.G/ D 2 and H1 has no proper nonabelian G-invariant subgroups (Lemma 76.18(a,b)) so ˆ.G/, as a proper G-invariant subgroup of H1 , is abelian. Proposition 76.20. Suppose that a p-group G is neither abelian nor an A1 -group. Let all proper nonabelian normal subgroups of G be members of the set 1 . Then d.G/ 3 and one of the following holds: (a) d.G/ D 2 and either cl.G/ D 2 or G=G 0 has a cyclic subgroup of index p. (b) d.G/ D 3. If R is a G-invariant subgroup of index p in G 0 , then G=R is an A2 -group. Proof. Suppose that d.G/ > 2. By hypothesis, all members of the set 2 are abelian so ˆ.G/ Z.G/ and d.G/ D 3 (Lemma 76.3). Let R be a G-invariant subgroup of index p in G 0 . We claim that G=R is an A2 -group. Without loss of generality, one may assume that R D f1g; then jG 0 j D p. It follows from d.G/ D 3 that G is not an A1 -group (Lemma 76.1). Let B < G be an A1 -subgroup. In view of B 0 D G 0 we get B G G so B 2 1 , by hypothesis, and G is an A2 -group. Now suppose that d.G/ D 2. Let M=G 0 < G=G 0 be of index p 2 . Then, by hypothesis, M is abelian. It follows that M CG .G 0 /. If G is not of class 2, then G=G 0 is not generated by subgroups of index p 2 so it has a cyclic subgroup of index p. Proposition 76.21. Let G be a nonabelian p-group.

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Groups of prime power order

(a) (Compare with Theorem 76.9) If d.G/ > 2. then 1 .G/ p 2 . (b) If d.G/ > 3, then 1 .G/ > p 2 C p C 1. Proof. (a) Suppose that all members of the set 2 are abelian. Then d.G/ D 3 and ˆ.G/ Z.G/ (Lemma 76.3). Let fM1 ; : : : ; Ms g be the set of all nonabelian members of the set 1 ; then s p 2 (Exercise 1.6(a)). It remains to show that 1 .G/ s. Let Ai Mi be an A1 -subgroup; then Mi D Ai ˆ.G/ is the unique member of the set 1 , containing Ai , i D 1; : : : ; s. It follows that A1 ; : : : ; As are pairwise not conjugate in G, and we are done in this case. Now suppose that the set 2 has a nonabelian member N . Take T 2 1 .N / \ 3 ; then T — Z.G/. Since G=T is generated by any p C 2 subgroups of order p, it follows that T is contained in at most p C 1 abelian members of the set 2 . In that case, there are p 2 distinct subgroups L1 =T; : : : ; Lp 2 =T of order p in G=T such that L1 ; : : : ; Lp 2 are nonabelian; obviously, Li 2 2 for all i . Let Ai Li be an A1 subgroup not contained in T (Proposition 10.28), i D 1; : : : ; p 2 ; then Li D Ai T . Since A1 ; : : : ; Ap 2 are pairwise not conjugate in G, we get 1 .G/ p 2 . (b) Let d.G/ > 3. Suppose that all members of the set 3 are abelian. Then all members of the set 4 are contained in Z.G/ so d.G/ D 4 (otherwise, G D hH j H 2 4 i is abelian). If jG W Z.G/j D p 2 , then G D BZ.G/ for each A1 -subgroup B < G so all A1 -subgroups are normal in G. Therefore, 1 .G/ D ˛1 .G/ p 3 (Theorem B). Since, p 3 > p 2 C p C 1, we are done in this case. Thus, jG W Z.G/j 2 fp 3 ; p 4 g. It follows that at least j3 j1 D p 3 Cp 2 Cp members of the set 3 are not contained in Z.G/. Let T1 ; T2 and T3 be those distinct members of the set 3 that are not contained in Z.G/. Since any p C 2 distinct subgroups of order p generate G=Ti , it follows that at least p 2 members of the set 2Ti are nonabelian, i D 1; 2; 3. Let i ¤ j , T i; j 3 and K 2 2Ti \ 2 j . Then K D Ti Tj is determined uniquely. It follows S3 that the set M D iD1 2Ti contains at least 3p 2 3 distinct nonabelian members. Let M0 D fH1 ; : : : ; Hk g be the set of all nonabelian members in the set M, where k 3p 2 3, and let Ls Hs be an A1 -subgroup, s k. Since the intersection of any two distinct members of the set M0 is abelian, Hs in the unique member of the set M0 containing Ls , s k. It follows that L1 ; : : : ; Lk are not pairwise G-conjugate so 1 .G/ k 3p 2 3 > p 2 C p C 1. Now suppose that the set 3 has a nonabelian member T . Set 2T D fH1 ; : : : ; Hk g, k D p 2 C p C 1. Let Li Hi be an A1 -subgroup not contained in T , i D 1; : : : ; k (Li exists, by Proposition 10.28), and let L T be an A1 -subgroup. We have Li T D Hi for i k. Since L; L1 ; : : : ; Lk are pairwise not conjugate in G, we get 1 .G/ k C 1 > k D p 2 C p C 1. Let G be a nonabelian p-group. Given M 1 , let 1 .M/ be the number of conjugate G-classes of A1 -subgroups contained in the members of the set M (obviously, 1 .1 / D 1 .G/, unless G is an A1 -group).

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Remarks. 14. Suppose that every A1 -subgroup is contained in the unique maximal subgroup of a nonabelian p-group G. Then ˆ.G/ is abelian and one and only one of the following holds: (i) d.G/ D 2, (ii) d.G/ D 3 and ˆ.G/ Z.G/. Indeed, the groups from (i) and (ii) satisfy the hypothesis (if H < G is an A1 -subgroup, then, in both cases, Hˆ.G/ is the unique maximal subgroup of G containing H ). Now let G satisfy the hypothesis. Then all members of the set 2 are abelian. By Lemma 76.3, d.G/ 3 and, if d.G/ D 3, then ˆ.G/ Z.G/. 15. Suppose that G is a nonabelian p-group with d.G/ > 2. We claim that the following assertions are equivalent: (a) 1 .1K / p for all K 2 2 , (b) G is an A2 -group with ˛1 .G/ D p 2 . Let (b) holds. Then all nonabelian members of the set 1 are A1 -subgroups and jG W Z.G/j D p 2 (Lemma 76.4(d)). If K 2 2 fZ.G/g, then KZ.G/ 2 1 so 1 .1K / D p and (b) ) (a). It remains to prove the reverse implication. Assume that H1 2 1 is neither abelian nor minimal nonabelian. Then there is a nonabelian K1 2 1 .H1 /. In that case, ˛1 .1K // p C 2 > p, a contradiction. Thus H1 does not exist so G is A2 -group. Exercise 5. Let G be a p-group and let the set 1 have exactly p nonabelian members. Then: (a) 1 D fH1 ; : : : ; Hp ; Ag, where A is abelian. (b) H10 D D Hp0 has index p in G 0 . Exercise 6. Let G be a nonabelian two-generator p-group and jG W G 0 j D p n > p 2 . Prove that if G has only one normal subgroup of index p n , then G=G 0 is abelian of type .p n1 ; p/ and exp.G/ p n . (Hint. Consider the quotient group G=R, where R is a G-invariant subgroup of index p in G 0 . Use Lemma 65.2(a).)

77

2-groups with a self-centralizing abelian subgroup of type .4; 2/

The results of this section are taken from [Jan9]. Here we classify finite 2-groups G which possess an abelian subgroup A Š C4 C2 such that CG .A/ D A. If A is normal in G, then G=A is isomorphic to a subgroup of Aut.A/ Š D8 . Therefore we may assume in the sequel that such a subgroup A is not normal in G. We investigate first the case where A ˆ.G/ and prove the following basic result. Theorem 77.1. Let G be a 2-group which possesses a self-centralizing abelian subgroup A of type .4; 2/ but G does not possess any self-centralizing abelian normal subgroup of type .4; 2/. If A ˆ.G/, then G has no normal elementary abelian subgroups of order 8. Proof. We assume that A ˆ.G/. Let W0 D 1 .A/ so that W0 Š E4 and Z.ˆ.G// < A (the inclusion is strong since A is not normal in G). It follows from jG W CG .W0 /j 2 that W0 Z.ˆ.G//. Since Z.ˆ.G// < A, it follows that W0 D Z.ˆ.G// so W0 is a normal four-subgroup in G. We claim that W0 is the unique normal four-subgroup of G. Indeed, if W1 were another normal four-subgroup of G, then A does not centralize W1 and so CG .W1 / is a maximal subgroup of G not containing A, contrary to our assumption that A ˆ.G/. Since A < ˆ.G/ and jG=ˆ.G/j 4, we get jGj 26 . Assume that W0 Z.G/. Let B be a normal subgroup of G such that W0 < B ˆ.G/ and jB W W0 j D 2. Then G stabilizes the chain B > W0 > f1g and so G=CG .B/ is elementary abelian. In particular, ˆ.G/ CG .B/ which contradicts the fact that Z.ˆ.G// D W0 . Hence jG W CG .W0 /j D 2 and so Z.G/ D hzi < W0 . Set W0 D hz; ui and T D CG .W0 /. For each x 2 G T , ux D uz and for each a 2 A W0 , CT .a/ D CT .A/ D A since A D ha; W0 i and W0 Z.T /. Suppose that A1 is a normal elementary abelian subgroup of order 16 of T D CG .W0 /. Take an element a 2 A W0 . Since a2 2 W0 Z.T /, the element a induces on A1 an automorphism of order 2 and so jCA1 .a/j 4 since ha; A1 i is not of maximal class (Proposition 1.8). Since CT .a/ D A, we get CA1 .a/ D W0 and A \ A1 D W0 . Set C D AA1 so that jC j D 25 , Z.C / D W0 and jC W A1 j D 2. Set hsi D Ã1 .A/ < W0 and we see that all four elements in A W0 are square roots of s in C . Let aa1 (a 2 A W0 , a1 2 A1 ) be any square root of s in C . Then we have

77 2-groups with a self-centralizing abelian subgroup of type .4; 2/

317

s D .aa1 /2 D aa1 aa1 D a2 .a1 a1 a/a1 D sa1a a1 , and so a1a D a1 which implies that a1 2 W0 hence aa1 2 A W0 . It follows that four elements in A W0 are the only square roots of s in C and they form a single conjugate class in C (it follows from CT .a/ D A that CC .a/ has index 4 in C ). Since T centralizes s, it follows that AW0 is a conjugacy class in NT .C / so hA W0 i D A is normal in NT .C / which implies C D T (in view of W0 Z.T / we have jT W Aj D jT W CT .A/j 4 D jC W Aj) and jGj D 2jT j D 2jC j D 26 . Since W0 Z.T / < A, we conclude that Z.T / D W0 . It follows that A1 is the unique abelian subgroup of index 2 in T so A is characteristic in T ; then A1 is normal in G. It follows from the previous sentence that T =W0 Š E23 (otherwise, T contains an abelian subgroup of index 2 and exponent 4, which is not the case). By Lemma 1.1, jT 0 j D 4 so T 0 D W0 D ˆ.G/ D Z.T /, and T is special. Let b be any element in C A1 D T A1 so that b D a1 a for an a 2 A W0 and a1 2 A1 (recall that A \ A1 D W0 ). Hence CA1 .b/ D CA1 .a/ D W0 . In particular, b 2 2 W0 and so four elements in hW0 ; bi W0 are either involutions (in case b 2 D 1) or square roots of the involution b 2 . Conversely, let bx (x 2 A1 ) be any square root of b 2 or an involution if b 2 D 1 in C A1 . Then b 2 D .bx/2 D bxbx D b 2 .b 1 xb/x D b 2 x b x, which implies x b D x and so x 2 CT .b/ D W0 . Hence four elements in hW0 ; bi W0 are all possible square roots of b 2 if hW0 ; bi Š C4 C2 (or involutions if hW0 ; bi Š E8 ) in T A1 . By Exercises 17–19 in 10 and Theorem 1.17(a), each group of order 25 contains at most 19 involutions, unless it is elementary abelian or isomorphic to D8 E4 . It follows that T A1 contains at least 12 elements of order 4 and so, by what has been proved already, the set of squares of these elements is W0# . Thus, each involution in W0 has exactly four square roots in T A1 and, in addition, there are exactly four involutions in T A1 . Let z 2 W0 be the central involution in G. By the previous paragraph, there are exactly four square roots of z in C and if c is one of them, then hW0 ; ci Š C4 C2 and all square roots of z in C D T form the set hW0 ; ci W0 . Hence hW0 ; ci W0 is a normal subset in G and so hW0 ; ci is normal in G in view of z 2 Z.G/. Since Z.T / D W0 , A1 is the unique abelian subgroup of index 2 in T so CT .c/ D hW0 ; ci. Thus hW0 ; ci is a self-centralizing (in G) normal abelian subgroup of G of type .4; 2/, contrary to the hypothesis. We have proved that T D CG .W0 / has no normal elementary abelian subgroups of order 16. Let U be a noncyclic abelian normal subgroup of G. Then W0 U (because of the uniqueness of W0 ) so U CG .W0 / D T . By the previous paragraph, U 6Š E24 so G has no normal elementary abelian subgroups of order 16. Suppose that E8 Š E G G. By the previous paragraph, E T , and E > W0 , and so A \ E D W0 . Let B be a maximal G-invariant abelian subgroup containing E. Since B is noncyclic, we have B T . We know that CG .B/ D B. Suppose that B D E. Then G=E, as a subgroup of order jG W Ej D 8 of Aut.E/ Š GL.3; 2/ Š PSL.2; 7/, is isomorphic to D8 . Set V0 D AE so that V0 =E D ˆ.G=E/

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Groups of prime power order

and therefore V0 is normal in G. Indeed, if M=E is a maximal subgroup of G=E, then A M since A ˆ.G/, and so V0 =E D AE=E < M=E, proving our claim. Since T stabilizes the chain E > W0 > f1g and CT .E/ D E, it follows that T =E Š E4 . Take an element a 2 A W0 so that four elements in A W0 are square roots of a2 ¤ 1. Let ae 0 .e 0 2 E/ be any square root of a2 in V0 . Then we have a2 D .ae 0 /2 D ae 0 ae 0 D a2 .e 0 /a e 0 , which gives .e 0 /a D e 0 and so e 0 2 W0 since CT .a/ D A. It follows that A W0 is the set of all square roots of a2 in V0 and so A D hA W0 i is normal in T . Since A is not normal in G, we have for every x 2 G T , Ax ¤ A. Since Ax V0 and A \ Ax D W0 , we have V0 E D Ea D .A W0 / [ .Ax W0 / and so x sends four elements in A W0 onto four elements in Ax W0 . But G=E Š D8 and so (since T =E Š E4 and V0 =E D ˆ.G=E/) there is y 2 G T with y 2 2 V0 E. Since V0 is nonabelian, A, Ax and E are all abelian subgroups of index 2 in V0 . Since hy 2 ; W0 i equals either A or Ax , one of these subgroups is normal in G since it is normalized by hy; T i D G; then both of them are normal in G, and this is a contradiction. We have proved that B > E and so jBj 24 . Since 1 .B/ D E (recall that G has no normal elementary abelian subgroups of order 24 ), B is an abelian group of rank 3. Take an element a 2 A W0 . Since a2 2 W0 Z.T /, a induces an involutory automorphism on B with CB .a/ D W0 1 .B/ (since a 62 B in view of CT .a/ D A). Applying Proposition 51.2, we see that a inverts Ã1 .B/ and B=W0 . Suppose that some e 2 E W0 .D E A/ is a square in B. Then a inverts (centralizes) e .2 Ã1 .B// so e centralizes ha; W0 i D A, a contradiction. Suppose that each involution in W0 is a square in B. Then 2 .B/ D hb1 i hb2 i hb3 i, where o.b1 / D o.b2 / D 4, b3 is an involution in E W0 , and hb12 ; b22 i D W0 . Since a inverts B=W0 , we get b1a D b11 w with w 2 W0 and so b1a D b11 w D b1 .b12 w/ D b1 w1 , where w1 2 W0 . Similarly, b2a D b2 w2 and b3a D b3 w3 with w2 ; w3 2 W0 . Since C2 .B/ .a/ D W0 , w1 , w2 , w3 must be three distinct involutions in W0 (otherwise, if, for example, w1 D w2 , then .b1 b2 /a D b1 b2 , which is not the case). But then .b1 b2 b3 /a D .b1 b2 b3 /.w1 w2 w3 / D b1 b2 b3 2 2 .B/ W0 , a contradiction. Hence we have B D hbi hwi hei, where o.b/ D 2m ; m > 1;

m1

b2

D z;

hzi D Ãm1 .B/ D Z.G/; m2

W0 D hz; wi;

and e 2 E W0 . Also we set v D b 2 so that o.v/ D 4, v 2 D z, and 2 .B/ D hE; vi is abelian of type .4; 2; 2/. Set V D AB and assume that V < T (V T since A; B < T ). Set VQ D NT .V / so that jVQ W V j 2. Since B is normal in G, V B D aB (a 2 A W0 ) is a normal subset in V D NG .V / VQ . Set a2 D w 0 2 W0 so that four elements in A W0 are square roots of w 0 in V B. An element ab 0 2 V B (b 0 2 B) is a square root of w 0 if and only if w 0 D .ab 0 /2 D ab 0 ab 0 D a2 .b 0 /a b 0 D w 0 .b 0 /a b 0 or .b 0 /a D .b 0 /1 . If .b 0 /a D .b 0 /1 and .b 00 /a D .b 00 /1 (b 0 ; b 00 2 B), then .b 0 b 00 /a D .b 0 /a .b 00 /a D .b 0 /1 .b 00 /1 D .b 0 b 00 /1 , and so the set B0 of elements of B which are inverted by a is a subgroup of B. Therefore, the number of square roots of w 0 in V B equals jB0 j. It follows from CT .a/ D A that B0 \ E D W0 . We have jBj D 2mC2 .m > 1/ (recall

77 2-groups with a self-centralizing abelian subgroup of type .4; 2/

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that B is abelian of type .2m ; 2; 2/) so jV Bj D 2mC2 and therefore jB0 j 2mC1 . All conjugates of a in VQ lie in V B and all these conjugates are square roots of w 0 since VQ T and T centralizes w 0 2 W0 . Since CVQ .a/ D A, we get jVQ W Aj 2mC1 . But jV j D 2mC3 and so jVQ j D 2mC4 , jVQ W V j D 2, and jB0 j D 2mC1 . It follows that B0 covers B=E and so we may choose b 2 B0 such that B0 D hbi hwi, where m1 w 2 W0 hzi (since b 2 D z). Suppose for a moment that VQ D NT .V / D NG .V /. Then looking at G=B, we get, by Proposition 1.8, that G=B is of maximal class and V =B is a noncentral subgroup of order 2 in G=B. Let R=B be a cyclic subgroup of index 2 in G=B. Then R is a maximal subgroup of G and R \ V D B. In particular, A 6 R, contrary to our assumption A ˆ.G/. We have proved that V D NG .V / > VQ D NT .V / so that jV W VQ j D 2, G D T V , and T \ V D T \ NG .V / D NT .V / D VQ . Assume that CV .a/ D A D CVQ .a/ and take an element y 2 V VQ so that y 2 G T . Then all 2mC2 elements in V B form a single conjugate class in V . Since w 0 has exactly 2mC1 square roots in V B, it follows that w 0 D a2 ¤ z (since hzi D Z.G/) and so .w 0 /y D w 0 z. Hence y sends 2mC1 square roots of w 0 in V B onto 2mC1 square roots of w 0 z in V B. Since e 62 B0 (e 2 E W0 ), we have .ae/2 D w 0 z D aeae D a2 e a e D w 0 e a e, and so e a D ez. On the other hand, a m2 inverts B0 D hb; wi and so setting v D b 2 , we get v 2 D z and v a D v 1 D vz. But then .ev/a D ezvz D ev, contrary to CB .a/ D W0 . We have proved that AQ D CV .a/ > A, where jAQ W Aj D 2 and AQ \ T D A. Take an element y 2 AQ A so that y acts non-trivially on W0 . But y centralizes m1 a and so y centralizes a2 D w 0 which gives w 0 D z D b 2 , where hzi D Z.G/. Suppose y 2 2 W0 so that hW0 ; yi Š D8 . In that case there are involutions in hW0 ; yi W0 and so we may assume that y is an involution. Act with the involution y on E. Since y acts non-trivially on W0 and jCE .y/j D 4, there is an involution e 0 2 E W0 with .e 0 /y D e 0 . We have 1 ¤ Œa; e 0 2 W0 and so Œa; e 0 y D Œay ; .e 0 /y D Œa; e 0 , which gives Œa; e 0 D z or .e 0 /a D e 0 z. But a inverts B0 and so v a D v 1 D vz m2 (where v D b 2 ) which gives .ve 0 /a D vze 0 z D ve 0 62 W0 . This is a contradiction since CB .a/ D W0 . We have proved that for each y 2 AQ A, y 2 2 A W0 and so we may assume that 2 y D a. Since a2 D z centralizes E, it follows that y induces on E an automorphism of order 4 and so CE .y/ D hzi. Hence for an e 2 E W0 , we get e y D ew0 with 2 y w0 2 W0 hzi and w0 D w0 z. This gives e a D e y D .ew0 /y D .ew0 /.w0 z/ D ez. On the other hand, a inverts B0 D hb; w0 i and so v a D v 1 D vz. This gives .ve/a D vzez D ve 62 W0 , contrary to CB .a/ D W0 . The contradiction in the previous paragraph shows that we must have V D AB D m2 T . We set again a2 D w 0 2 W0 .a 2 A W0 / and v D b 2 .m > 1/ so that 2 v D z, where hzi D Z.G/ and B D hE; bi. We have jG=Bj D 4. If G=B Š E4 , then ˆ.G/ B and this contradicts our assumption A ˆ.G/. Hence G=B Š C4 and G=B acts faithfully on E (since T D Bhai and a acts non-trivially on E). If

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Groups of prime power order

y 2 G T , then for an element e 2 E W0 , we have e y D ew with w 2 W0 hzi 2 and w y D wz. Then we compute e y D .ew/y D .ew/.wz/ D ez. But y 2 2 T B and T D Bhai and so a acts in the same way on E as the element y 2 which implies e a D ez. By Proposition 51.2, a inverts Ã1 .B/. If m > 2, then v 2 Ã1 .B/ and so v a D v 1 D vz. We get in that case .ev/a D ezvz D ev which contradicts to CB .a/ D W0 . We have proved that we must have m D 2 and so b D v, jBj D 24 , and jGj D 26 . The argument of the previous paragraph shows that a does not invert any element (of order 4) in B E and so the subgroup B0 of all elements of B inverted by a is equal B0 D W0 . Indeed, if a inverts an element s 2 B E, then s a D s 1 D sz and so .es/a D ezsz D es, contrary to CB .a/ D W0 . Hence a2 D w 0 has exactly jB0 j D 4 square roots in T B D V B D .AB/ B and they all lie in A W0 . Since hA W0 i D A, A is normal in T D AB. For each x 2 T B, x 2 2 W0 since CB .x/ D CB .a/ D W0 , and x (acting in the same way on B as the element a) inverts on B exactly the elements of W0 . It follows that x 2 has exactly four square roots in T B. Hence each involution in W0 has exactly four square roots in T B and (since jT Bj D 16) T B contains exactly four involutions. Let a0 2 T B so that .a0 /2 D z. Set A D W0 ha0 i Š C4 C2 and A is normal in G. Indeed, hzi D Z.G/ and so G normalizes the subset f.W0 ha0 i/ W0 g of all square roots of z in T B. We have CG .W0 / D T and CT .a0 / D ha0 iCB .a0 /. But CB .a0 / D CB .a/ D W0 and so A is a self-centralizing abelian normal subgroup of type .4; 2/ in G. This is a final contradiction and our theorem is proved. In our next result we shall determine the structure of the groups appearing in Theorem 77.1. Theorem 77.2. Let G be a 2-group which possesses a self-centralizing abelian subgroup A of type .4; 2/ but G does not possess any self-centralizing abelian normal subgroup of type .4; 2/. If A ˆ.G/, then G has the following properties: (i) G has no normal elementary abelian subgroups of order 8. (ii) G has the unique normal four-subgroup W0 D 1 .A/. (iii) G has a normal metacyclic subgroup N such that 2 .N / D W is abelian of type .4; 4/, CG .W / N , 1 .W / D W0 , CG .W0 / N , jG W CG .W0 /j D 2, and G=N Š C4 or D8 . (iv) N is either abelian of type .2k ; 2kC1 / or .2k ; 2k /, k 2, or N is minimal nonm n m1 abelian and more precisely N D ha; b j a2 D b 2 D 1; ab D a1C2 i, where either m D n with n 3 or m D n C 1 with n 2. Proof. Suppose that our 2-group G satisfies all the assumptions of our theorem together with A ˆ.G/. By Theorem 77.1, G has no normal elementary abelian subgroups of order 8. By the first three paragraphs of the proof of Theorem 77.1, we know that W0 D Z.ˆ.G// D 1 .A/ is the unique normal four-subgroup of G,

77 2-groups with a self-centralizing abelian subgroup of type .4; 2/

321

jG W CG .W0 /j D 2, Z.G/ is of order 2, and jGj 26 . Set W0 D hz; ui, T D CG .W0 /, where hzi D Z.G/. For each t 2 A W0 , CT .t / D A. We apply now the results of 50, since G is neither abelian nor of maximal class. It follows that G possesses a normal metacyclic subgroup N such that CG .2 .N // N , G=N is isomorphic to a subgroup of D8 and W D 2 .N / is abelian of type .4; 2/ or .4; 4/. In any case, W0 D 1 .W / D 1 .N / is the unique normal four-subgroup of G and W0 6 Z.G/. If A N , then A W D 2 .N /. Since CG .A/ D A, we have A D W and then A is normal in G, a contradiction. Hence A 6 N and so A \ N D W0 . Since A ˆ.G/, G=N is not elementary abelian. Hence G=N is isomorphic to C4 or D8 . If T D CG .W0 / does not contain N , then T covers G=N and G=N acts faithfully on W (since CG .W / N ). In that case CG .W0 /=CN .W0 / cannot contain elements of order 4 (since that group centralizes W0 D 1 .W /) and so G=N is elementary abelian, a contradiction. Hence T D CG .W0 / N . Assume that G=N Š C4 . If N is abelian of type .2j ; 2/, j 2, then G=N acts faithfully on 2 .N / Š C4 C2 . If N > 2 .N /, then there is a characteristic cyclic subgroup Z Š C4 of N so that Z is normal in G. But then jG W CG .Z/j 2 and therefore A (being contained in ˆ.G/) centralizes Z, a contradiction. Thus N D 2 .N / is a normal abelian self-centralizing subgroup of type .4; 2/ in G, a contradiction. We have proved that 2 .N / D W is abelian of type .4; 4/. Suppose that Z is a G-invariant cyclic subgroup of order 4 contained in N . But then again A centralizes Z , a contradiction. Hence there is no such Z and so we may apply Proposition 50.4. It follows that N is either abelian of type .2n ; 2n / or .2nC1 ; 2n / with m n n 2 or N is minimal nonabelian and more precisely N D ha; b j a2 D b 2 D m1 1; ab D a1C2 i; where either m D n with n 3 or m D n C 1 with n 2. Assume that G=N Š D8 . The structure of N in that case is already determined by Theorem 50.1. The minimal case N Š C4 C2 cannot occur because in that case N would be a self-centralizing normal abelian subgroup of type .4; 2/ of G, a contradiction. Finally, we consider the case where A 6 ˆ.G/. Theorem 77.3. Suppose that G is a 2-group that possesses a self-centralizing abelian subgroup A of type .4; 2/. If 1 .A/ 6 ˆ.G/, then G possesses an involution t such that CG .t / D ht i D, where D is isomorphic to one of the following groups: C4 , D8 , Q2n , n 3, or SD2m , m 4. Such groups G have been classified in 48, 49, and 51. Proof. Suppose that 1 .A/ 6 ˆ.G/. There is a maximal subgroup M of G such that A M contains an involution t . It follows that A0 D A \ M Š C4 and CG .t / D ht i D, where D D CM .t / A0 . We have CD .A0 / D A0 and so (by a well-known result of M. Suzuki) either D D A0 Š C4 or D is a 2-group of maximal class. In the second case D is isomorphic to D8 , Q2n , n 3 or SD2m , m 4.

322

Groups of prime power order

Theorem 77.4. Suppose that G is a 2-group that possesses a self-centralizing abelian subgroup A of type .4; 2/. If W0 D 1 .A/ ˆ.G/ but A 6 ˆ.G/, then for any a 2 AW0 , CG .a/ D haiM0 , jhai\M0 j D 2, where M0 D W0 or M0 is isomorphic to one of the following groups: D2n , n 3 or SD2m , m 4 and G > CG .a/. Proof. Suppose that M is a maximal subgroup of G which does not contain A. Then A \ M D W0 D 1 .A/. Let a 2 A W0 so that G D M hai and CG .a/ D haiCM .a/ with CM0 .W0 / D W0 , where M0 D CM .a/. By a result of M. Suzuki, either M0 D W0 or M0 is of maximal class. In the second case, M0 is isomorphic to one of the following groups: D2n , n 3, or SD2m , m 4. If G D CG .a/, then ˆ.G/ D ˆ.CG .a// D ˆ.M0 / is cyclic, contrary to our assumption. Exercise. Describe the subgroup structure of the holomorph of the abelian group of type .4; 2/.

78

Minimal nonmodular p-groups

The main results of this section are taken from [Jan10]. 1o . We know that a 2-group is modular if and only if it is D8 -free (Theorem 44.13). Here we classify minimal nonmodular 2-groups G, i.e., nonmodular 2-groups all of whose proper subgroups are modular. Hence there is N G G such that G=N Š D8 free but each proper subgroup of G is D8 -free. We shall use freely all results on modular p-groups from 73 and Appendix 24. In this section a 2-group is said to be Hamiltonian if it is nonabelian and all its subgroups are normal. A Dedekindian 2group is Hamiltonian if it is nonabelian. A metacyclic 2-group H is called ordinary metacyclic with respect to A if H possesses a cyclic normal subgroup A such that H=A is cyclic and H centralizes A=Ã2 .A/, or what is the same, H=Ã2 .A/ is abelian. In other words, a metacyclic 2-group is ordinary if and only if it is powerful (see 26). Below we use freely the following facts: () If minimal nonmodular 2-group G is of maximal class and order 24 , then G Š Q24 . Moreover, a nonabelian 2-group G with cyclic subgroup of index 2 is modular if and only if G 2 fQ8 ; M2n g. It follows that any two involutions of G are permutable so 1 .G/ is elementary abelian. () [Wil] Let G be a Q8 -free modular 2-group. Then d.1 .G// D d.G/. If, in addition, d.G/ D 2, then G is ordinary metacyclic. () Let N be a normal subgroup of a 2-group G such that G=N Š D8 . If L=N is the unique cyclic subgroup of index 2 in G=N and x 2 G L, then x 2 2 N . Indeed, all elements if .G=N / .L=N / are involutions. Until the end of this subsection, G is a minimal nonmodular 2-group. In that case, G contains two non-permutable cyclic subgroups A and B. Since the subgroup hA; Bi is nonmodular, we get G D hA; Bi so d.G/ D 2. By hypothesis, G has a normal subgroup N such that G=N Š D8 . It follows from d.G/ D 2 D d.G=N / that N ˆ.G/. The following proposition clears up the structure of N . Remark 1. Suppose that G is a modular 2-group of order 16 containing a subgroup H Š E8 . Then, G is Q8 -free in view of the existence of H . Since G has no subgroups Š D8 , it is either abelian or minimal nonabelian. Assume that G is nonabelian. Then G D ha; b j a4 D b 2 D c 2 D 1; c D Œa; b; Œa; c D Œb; c D 1i and G=ha2 i Š D8 , a contradiction. Thus, G is abelian.

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Groups of prime power order

Proposition 78.1. We have d.N / 2, so N is metacyclic (Theorem 44.13). Proof. Suppose that d.N / 3. Then N possesses a G-invariant subgroup R such that N=R Š E8 . The quotient group G=R is also minimal nonmodular. We want to obtain a contradiction. To this end, we may assume that R D f1g; then N Š E8 . Let S=N be any subgroup of order 2 in G=N . Then S (being modular) is abelian, by Remark 1, so S centralizes N . On the other hand, G=N Š D8 is generated by its subgroups of order 2, and so N Z.G/. Assume that Z.G/ > N ; then Z.G/=N D Z.G=N / D ˆ.G=N / so Z.G/ D ˆ.G/. In that case, G is minimal nonabelian. Then 1 .G/ D N and jG 0 j D 2 (Lemma 65.1) so G 0 < N , a contradiction since G=N Š D8 is nonabelian. Thus, N D Z.G/. Let L=N be the unique cyclic subgroup of index 2 in G=N . Then L is abelian since N D Z.G/. If L D N L1 with L1 Š C4 , then Ã1 .L/ D Ã1 .L1 / is of order 2 and so Ã1 .L/ Z.G/, a contradiction since Ã1 .L/ 6 N D Z.G/. Hence L does not split over N and so L D NC , where C Š C8 and C \ N D C0 is of order 2. We have ˆ.L/ D ˆ.C / D C1 Š C4 , where C0 < C1 , and ˆ.G/ D C1 N is abelian of type .4; 2; 2/. For each x 2 G L, x 2 2 N , by (), and so there is b 2 G L with b 2 2 N C0 (otherwise, ˆ.G/ D Ã1 .G/ D C1 ). Since CG .C1 / D L, Aut.C1 / Š C2 and C1 is normal in G, it follows that b inverts C1 . In that case, D D hC1 ; bi is (nonabelian) metacyclic of order 24 and exponent 22 so D=hb 2 i Š D8 , a contradiction. Proposition 78.1 is a variant of Theorem 44.13. It follows from Theorem 44.13 that the subgroup N of Proposition 78.1 is metacyclic. Remark 2. Let, as in the following four propositions, the subgroup N be cyclic. We claim that then Ã2 .G/ D ˆ.N /. One may assume that N > f1g. In view of Ã2 .G/ Ã2 .N /, it suffices to show that exp.G=Ã2 .N // D 8. Without loss of generality, one may assume that Ã2 .N / D f1g, i.e., N is cyclic of order 4. Then ˆ.G/ is abelian of order 8. One may assume that ˆ.G/ is abelian of type .4; 2/ (otherwise, G has a cyclic subgroup of index 2, contrary to ()). Since ˆ.G/ D Ã1 .G/ is not generated by involutions, there exists x 2 G such that o.x 2 / D 4; then o.x/ D 8, and we are done. It follows from the obtained result and Lemma 64.1(m) that G is metacyclic if and only if G=ˆ.N / is metacyclic. Proposition 78.2. Suppose that N is cyclic and some proper subgroup of G is not Q8 free. Then G is isomorphic to Q24 or to the uniquely determined group X of order 25 with 2 .X/ Š Q8 C2 given in 52. Proof. Suppose that N is cyclic and G has a maximal subgroup M which is not Q8 free. Since M is modular, it follows that M is Hamiltonian (73 or Appendix 24), i.e., M D Q E with Q Š Q8 and exp.E/ 2. In particular, exp.M / D 4 and Ã1 .M / D ˆ.M / D Ã1 .Q/. We have N < M since N ˆ.G/ so jN j 4; in that case, jGj D jG=N jjN j 8 4 D 32. If jG 0 j D 2, it follows from d.G/ D 2 that G

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325

is minimal nonabelian (Lemma 65.2(a)), a contradiction since M < G is nonabelian. Hence jG 0 j 4 so G has at most one abelian maximal subgroup (Lemma 65.2(c)). (i) Suppose that jN j D 4; then jGj D 32 and jˆ.G/j D 8. Let L=N Š C4 be the unique cyclic subgroup of index 2 in G=N .Š D8 /. By (), G is not of maximal class. In that case, the metacyclic subgroup L is noncyclic (Lemma 64.1(t)) so jÃ1 .L/j D 1 4 jLj D 4 and therefore L is not Hamiltonian. Suppose that L is abelian. We have Ã1 .L/ > Ã1 .N / D Ã1 .M /. Let K be the maximal subgroup of G distinct from M and L. We know that K must be nonabelian and, since G=N Š D8 , we get K=N Š E4 . By (), K is not of maximal class. Since N Š C4 does not lie in Z.M /, we have CG .N / D L. Then jK W CK .N /j D 2 so K 6Š M16 (otherwise, N D ˆ.K/ D Z.K/), and we conclude (Theorem 1.2) that exp.K/ D 4. Take k 2 K CK .N /; then k 2 2 N , in view of K=N Š E4 , and therefore hN; ki Š Q8 . It follows that K is Hamiltonian (73 or Appendix 24) and so Ã1 .K/ D Ã1 .N / < Ã1 .L/. Hence ˆ.G/ D Ã1 .G/ D Ã1 .L/ is of order 4, a contradiction since jˆ.G/j D 8. We have proved that L is nonabelian. In particular, N D hni 6 Z.L/ since L=N is cyclic, and if we set L D hN; li, then nl D n1 and l 4 2 hn2 i. If o.l/ D 4, then L=hl 2 i Š D8 , a contradiction. Hence o.l/ D 8 and so L Š M16 with hl 4 i D hn2 i D L0 and ˆ.L/ D Z.L/ D hl 2 i Š C4 . Note that 2 .L/ D N ˆ.L/ is abelian of type .4; 2/. Set K D CG .N / so that K is the maximal subgroup of G distinct from M and L and (noting that hl 2 i > hn2 i) we get (see also Exercise 1.133) ˆ.G/ D ˆ.M /ˆ.L/ˆ.K/ D hn2 ihl 2 iˆ.K/ D hl 2 iˆ.K/. Since K=N Š E4 , we have ˆ.K/ N . But ˆ.G/ > N and so we must have ˆ.K/ D N (otherwise, ˆ.G/ D hl 2 i is of order 4). It follows that the modular subgroup K has a cyclic subgroup of index 2 and K is Q8 -free since K cannot be Hamiltonian (because jKj D 16 and jÃ1 .K/j D 4; see 73 and Appendix 24). Hence K is either abelian of type .8; 2/ or K Š M16 . In any case, 2 .K/ D ˆ.L/N D ˆ.G/ is abelian of type .4; 2/. It follows that 2 .G/ D M Š Q8 C2 and consequently G is the uniquely determined group of order 25 described in 52. (ii) Now let jN j D 2. In that case, jGj D 16 and jG 0 j D 4 so jG W G 0 j D 4. By Taussky’s theorem, G is of maximal class so, by (), G Š Q16 . It is easy to check that the group G of Proposition 78.2 of order 25 is the group F from Appendix 24. Indeed, take u 2 G M and put U D hui; then jU j D 8 since M D 2 .G/. Assume that UQ D QU . Then UQ is of maximal class (Theorem 1.2), contrary to (). It follows that U \ Q is normal in G and G=.U \ Q/ Š D8 . By Appendix 24, G Š F . Proposition 78.3. Suppose that N > f1g is cyclic and all proper subgroups of G are Q8 -free. Then Ã2 .G/ D ˆ.N / and G=ˆ.N / is minimal nonabelian of order 24 and exponent 4. Thus G=ˆ.N / is isomorphic to one of the following groups: (a) hx; y j x 4 D y 2 D 1; Œx; y D z; z 2 D Œx; z D Œy; z D 1i is nonmetacyclic, (b) hx; y j x 4 D y 4 D 1; x y D x 1 i is metacyclic.

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Proof. By Remark 2, Ã2 .G/ D ˆ.N / so G=Ã2 .G/ is 2-generator nonabelian of exponent 4 and order 24 . Therefore, by hypothesis, G=Ã2 .G/ is minimal nonabelian. By Lemma 65.1 that G=Ã2 .G/ D G=ˆ.N / is isomorphic to the group (a) or (b). In the next two propositions we determine the groups G of Proposition 78.3. Proposition 78.4. Suppose that N > f1g is cyclic and all proper subgroups of G are Q8 -free. If G=ˆ.N / is as in Proposition 78.3(a), then G has a normal elementary abelian subgroup E D hn; z; t i of order 8 such that G=E is cyclic. We set G D hE; xi, where o.x/ D 2sC1 , s 1, E \ hxi D hni, Œt; x D z, Œz; x D n , D 0; 1, and G D hx; t i. We have G=hx 2 i Š D8 , ˆ.G/ D hx 2 i hzi, 1 .G/ D E, and G is Q8 -free. If D 0, then G is minimal nonabelian nonmetacyclic. If D 1, then s 2, G 0 D hz; ni Š E4 and Z.G/ D hx 4 i. Proof. Let M=ˆ.N / D 1 .G=ˆ.N //.Š E8 /; then N is a maximal cyclic subgroup of M . Set jN j D 2s , s 1. Let S=N be any subgroup of order 2 in M=N . Since M is D8 -free and Q8 -free, S cannot be of maximal class. It follows that S is either abelian of type .2s ; 2/ or S Š M2sC1 (s > 2). In any case, there exists an involution in S N . Hence 1 .M / covers M=N Š E4 and, since M is modular, E D 1 .M / is elementary abelian of order 8, by the product formula. It follows that E is normal in G and E \ N D 1 .N / D hni Z.G/. In particular, ˆ.G/ D N 1 .ˆ.G// is abelian of type .2s ; 2/, s 1 since 1 .ˆ.G// Š E4 centralizes ˆ.G/. Take an involution t 2 E ˆ.G/. Let K ¤ M be a maximal subgroup of G such that K=N Š E4 ; then K \ M D ˆ.G/ since d.G/ D 2. Suppose that N is a maximal cyclic subgroup of K. Then, by the argument of the previous paragraph, 1 .K/ covers K=N and so there is an involution r 2 K M since E8 Š 1 .K/ ¤ 1 .M / .D E/; then r 62 E. Since G has no elementary abelian subgroups of order 24 , there is in E an involution u such that ru ¤ ur. Then hr; ui is dihedral, a contradiction. We have proved that there is an element x 2 K M such that hx 2 i D N and so o.x/ D 2sC1 , s 1. Since hx; t i D G (indeed, involutions xN and tN lie in different maximal subgroups of G=N Š D8 ) and t 2 E ˆ.G/, we get G D Ehxi with E \ hxi D 1 .N / D hni Š C2 and so G=E is cyclic of order 2s and jGj D 2sC3 . In particular, G 0 < E and so jG 0 j 2 f2; 4g. Since d.G/ D 2 and G is not minimal nonabelian, we get jG 0 j D 4 (Lemma 65.2(a)). We have Œx; t ¤ 1 since hx; t i D G is nonabelian, so z D Œx; t is an involution in .E \ ˆ.G// hni since hxi is not normal in G. Indeed, if hxi were normal in G, then G=ˆ.N / D ht ˆ.N /i hxi=ˆ.N / is metacyclic, which is not the case. It follows that E D hn; z; t i, where z 2 ˆ.G/, and hxi is not normal in G. Thus NG .hxi/ is a maximal subgroup of G and so z 2 ˆ.G/ NG .hxi/ and therefore z normalizes hxi. Since hx; zi cannot be of maximal class, by (), we have either Œx; z D 1 or Œx; z D n (in which case hx; zi Š M2sC1 , s > 1). Let us consider the first possibility. We have CG .z/ hE; xi D G so z 2 Z.G/ \ G 0 . Then G=hzi has a cyclic subgroup hx; zi=hzi of index 2. It follows that then G=hzi is neither abelian nor isomorphic to

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M2sC1 (otherwise, G=hzi has two distinct cyclic subgroups of index 2 so G has two distinct abelian maximal subgroups, and we get jG 0 j < 4, a contradiction). It follows (Lemma 64.1(t)) that G=hzi is of maximal class so it is isomorphic to Q16 , by (), contrary to the existence of E. Now let hx; zi Š M2sC1 , s > 1. Then hxi induces an automorphism of order 4 on E. Let u be an involution in G E. Then, by Remark 1, F D Ehui Š E16 . But s1 s1 G=E is cyclic and so F=E D .Ehx 2 i/=E and so x 2 is an element of order 4 contained in F E, a contradiction. We have proved that 1 .G/ D E. Proposition 78.5. Suppose that N > f1g is cyclic and all proper subgroups of G are Q8 -free. If G=ˆ.N / is metacyclic, then G is also metacyclic and we have one of the following possibilities: sC1

(i) G D hx; y j x 4 D y 2 abelian and N D hy 2 i.

D 1; s 1; x y D x 1 i, where G is minimal non-

(ii) G D hx; a j x 2 D a8 D 1; s 2; x 2 D a4 ; ax D a1 i, where G is an A2 -group with N D hx 2 i, Z.G/ D N (see 65), G 0 D ha2 i Š C4 and G is of class 3. sC1

s

In both cases (i) and (ii), G is a minimal non-Q8 -free 2-group. Proof. By Remark 2, ˆ.N / D Ã2 .G/ and G is also metacyclic. Set jN j D 2s , s 1. If s D 1, then G is isomorphic to the group (b) of Proposition 78.3 and we are done. Now we assume that s 2. Let S=N be a subgroup of order 2 in G=N . By (), S is not of maximal class so S is either cyclic of order 2sC1 or S is abelian of type .2s ; 2/ or s > 2 and S Š M2sC1 . If ˆ.G/ is cyclic, then G has a cyclic subgroup of index 2, contrary to (). Also, M2sC1 , s > 2, having cyclic center, cannot be the Frattini subgroup (Burnside). Taking S D ˆ.G/, we see that ˆ.G/ is abelian of type .2s ; 2/, s 2. Let 1 .N / D hni so that hni ˆ.N / \ Z.G/. For any subgroup S=N of order 2 in G=N , by the previous paragraph, S=hni is abelian so centralizes N=hni. Since G=N is generated by its subgroups of order 2, we get N=hni Z.G=hni/. Suppose for a moment that G is minimal nonabelian. Since ˆ.G/ is abelian of type sC1 s .2 ; 2/, we get at once (Lemma 65.1): G D hx; y j x 4 D y 2 D 1; s 1; x y D 1 2 x i, where N D hy i; this is a group of part (i). In what follows we assume that G is not minimal nonabelian. In particular, jG 0 j 4 (Lemma 65.2(a)). We will determine the structure of all three maximal subgroups of G. Let M be a maximal subgroup of G such that M=N Š E4 . If N is a maximal cyclic subgroup of M , then for each subgroup S=N of order 2 of M=N , there is an involution in S N . Hence 1 .M / covers M=N and, since M is D8 -free and Q8 -free, 1 .M / is elementary abelian and 1 .M / \ N D hni so that 1 .M / Š E8 , by the product formula. This is a contradiction since G is metacyclic. It follows that N is not a maximal cyclic subgroup of M . Let M0 be a maximal cyclic subgroup of M containing N so that M0 Š C2sC1 is of index 2 in M . By (), M is not of maximal class and

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so M is either abelian of type .2sC1 ; 2/ or M Š M2sC2 , s 2 (Lemma 64.1(t)). In any case, N Z.M / and M=hni is abelian since M 0 hni. Let K (¤ M ) be another maximal subgroup of G with K=N Š E4 . Then K is either abelian of type .2sC1 ; 2/ or K Š M2sC2 , s 2, and again N Z.K/ and K=hni is abelian. We get CG .N / MK D G so N Z.G/. Since G is not minimal nonabelian, we get ˆ.G/ ¤ Z.G/. We have proved that N D Z.G/. Let L=N be the unique cyclic subgroup of index 2 in G. Then L is abelian and using Lemma 64.1(q), we get jG 0 j D 4. By Lemma 64.1(u), L is the unique abelian maximal subgroup of G and so M Š K Š M2sC2 with M 0 D K 0 D hni. In particular, G is an A2 -group (since jG 0 j D 4, this also follows from Corollary 65.3). We have G 0 > hni and, since G 0 6 N D Z.G/ and jˆ.G/=N j D 2, we get ˆ.G/ D NG 0 , N \ G 0 D hni D 1 .N / and cl.G/ D 3. Since G is metacyclic, there exists a cyclic normal subgroup Z of order 8 such that Z > G 0 . But N \Z D N \G 0 D hni and so N Z D L which determines the structure of the maximal subgroup L and shows that L does not split over N . It follows that L is abelian of type .2s ; 2/. Set Z D hai. By (), we get hx 2 i D N for each x 2 G L. Hence G D Zhxi with Z \ hxi D hni for a fixed x 2 G L (since 1 .G/ D 1 .L/, x is not an involution). In view of jG 0 j D 4 and G 0 < Z, we get either ax D a1 or ax D a1 n, where n D a4 . However, if ax D a1 n, then we replace Z D hai with Z D haui, where u 2 N is such that u2 D n. Then we have .au/x D a1 nu D a1 u1 D .au/1 . Since h.au/2 i D ha2 i D G 0 , we may assume from the start that ax D a1 and so the structure of G is completely determined. Remark 3. Let G be a 2-group and let N be a G-invariant metacyclic subgroup of ˆ.G/. We claim that N is ordinary metacyclic. One may assume that N is nonabelian; then N has no cyclic subgroups of index 2 since Z.N / must be noncyclic. There exists a maximal cyclic subgroup A of N such that N 0 < A and N=A is cyclic. We have to prove that N=Ã2 .A/ is abelian. This is the case if jA W N 0 j > 2. Now we assume that jA W N 0 j D 2 and obtain a contradiction. Under our assumption, N=N 0 is abelian with cyclic subgroup of index 2 (indeed, by the choice of A and Frobenius–Stickelberger theorem on abelian groups, A=N is a direct factor of N=N 0 ). In particular, N has no epimorphic images which is abelian of type .4; 4/. Now consider the quotient group NN D N=Ã2 .N /. Since N has no cyclic subgroups of index 2, we get jNN j D 16. By assumption, NN is nonabelian so NN D hx; y j x 4 D y 4 D 1; x y D x 1 i. In that case, NN =hy 2 i Š D8 has a cyclic center, a contradiction since N=Ã2 .N / ˆ.G=Ã2 .N //. In the rest of this subsection we consider the case d.N / D 2. Proposition 78.6. Suppose that d.N / D 2. Then G=ˆ.N / is the minimal nonabelian nonmetacyclic group of order 25 and exponent 4. In particular, G=ˆ.N / has the unique epimorphic image isomorphic to Q8 . Each maximal subgroup of G is Q8 -free and N is ordinary metacyclic.

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Proof. We want to determine the structure of G=ˆ.N /. Since G=ˆ.N / is also minimal nonmodular, we may assume for a moment that ˆ.N / D f1g so that N Š E4 and jGj D 25 . Let S=N be any subgroup of order 2 in G=N . Since S 6Š D8 , S is abelian and so N Z.G/ since such subgroups S generate G. Suppose that Z.G/ D N . Let L=N be the unique cyclic subgroup of index 2 in G=N ; then L is abelian. If L D N R with R Š C4 , then Ã1 .L/ D Ã1 .R/ 6 N and Ã1 .L/ Z.G/, contrary to our assumption. Hence L D NL1 with L1 Š C8 and L0 D L1 \ N Š C2 ; thus, L is abelian of type .8; 2/. We have ˆ.L/ D ˆ.L1 / Š C4 , where ˆ.L/ > L0 . For each x 2 G L, x 2 2 N , by (), and ˆ.G/ D Ã1 .G/ D ˆ.L/N . This implies that there exists b 2 G L such that b 2 2 N L0 . Assume that this is false. Then all elements in .G=L0 / .L=L0 / are involutions so G=L0 is generated by involutions and, since G=L0 is not dihedral, it is not generated by two involutions. Since G=L0 is minimal nonmodular, all its involutions commute so G=L0 is elementary abelian; then d.G/ 4, a contradiction. Since ˆ.L/ 6 Z.G/, we get CG .ˆ.L// D L so b inverts ˆ.L/. But then D D hˆ.L/; bi is nonabelian metacyclic of order 24 and exponent 4; in that case, D=hb 2 i Š D8 , a contradiction. We have proved that Z.G/ > N and so Z.G/ D ˆ.G/. It follows that each maximal subgroup of G is abelian and so G is minimal nonabelian. In particular, jG 0 j D 2 and since G 0 covers Z.G/=N D .G=N /0 , we have Z.G/ D N G 0 is elementary abelian of order 8. It follows that G is the uniquely determined minimal nonabelian nonmetacyclic group of order 25 and exponent 4: G D ha; b j a4 D b 4 D 1; Œa; b D c; c 2 D Œa; c D Œb; c D 1i, where Z.G/ D ha2 ; b 2 ; ci, G 0 D hci, and G=ha2 c; b 2 ci is the unique quotient group of G which is isomorphic to Q8 . In particular, G is not Q8 -free. We return now to the remaining case ˆ.N / > f1g. Assume that N is not Q8 -free. Then N (being modular) is Hamiltonian (see 73 and Appendix 24). But d.N / D 2 and so N Š Q8 . This is a contradiction since Z.N / is cyclic and N < ˆ.G/ (Burnside). We have proved that N is Q8 -free and so N=Ã2 .N / must be abelian. By Remark 3, N is ordinary metacyclic. Suppose that a maximal subgroup M of G is not Q8 -free. Then M (being modular) is Hamiltonian and so M D Q E, Q Š Q8 , exp.E/ 2. In particular, ˆ.M / is of order 2 and exp.M / D 4. In view of N < ˆ.G/ < M , we get exp.N / D 4 and N is abelian of type .4; 2/ since N is metacyclic. In that case, jGj D 26 . We get ˆ.M / D Ã1 .M / D Ã1 .N / D ˆ.N / and so M=ˆ.N / is an elementary abelian subgroup of order 16 in the minimal nonabelian group G=ˆ.N /, contrary to Lemma 65.1. Thus, all maximal subgroups of G are Q8 -free. Proposition 78.7. Suppose that d.N / D 2. Then for each maximal subgroup M of G we have d.M / D 3. Also, ˆ.N / D Ã2 .G/, E D 1 .G/ D 1 .ˆ.G// Š E8 , E Z.ˆ.G//, and either G=E Š Q8 with 2 .G/ D ˆ.G/ being abelian of type .4; 2; 2/ or G=E is noncyclic and ordinary metacyclic. Proof. By (), 1 .G/ is elementary abelian.

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Groups of prime power order

Set F D ˆ.G/ so we have F=ˆ.N / D ˆ.G=ˆ.N // D 1 .G=ˆ.N // Š E8 (Proposition 78.6). Since ˆ.N / ˆ.F /, we get ˆ.N / D ˆ.F /. Thus d.F / D 3 and F is D8 -free and, by Proposition 78.6, Q8 -free, E D 1 .F / Š E8 (by ()) is normal in G. Let M be any maximal subgroup of G so that M=ˆ.N / is abelian of type .4; 2; 2/, ˆ.N / ˆ.M / and so d.M / D 3. But M is also D8 -free and Q8 -free and therefore, by (), 1 .M / Š E8 which implies 1 .M / D 1 .F / D 1 .G/ since 1 .G/ 1 .M / in view of 1 .G=N / D 1 .M=N / (Lemma 65.3). We have ˆ.N / Ã2 .G/ (Proposition 78.6). On the other hand, exp.G=Ã2 .G// D 4 and so each maximal subgroup of G=Ã2 .G/, being modular and Q8 -free, is abelian. Thus G=Ã2 .G/ is minimal nonabelian of exponent 4 and so jG=Ã2 .G/j 25 . It follows Ã2 .G/ D ˆ.N /. If G=E is not D8 -free, then there is a normal subgroup N of G such that E N and G=N Š D8 . By Proposition 78.1, N must be metacyclic, a contradiction. Hence G=E is D8 -free so it is modular. Suppose that G=E is not Q8 -free. Then G=E is Hamiltonian (73 or Appendix 24; by the previous paragraph, G=E is modular). Since d.G=E/ D 2, we get G=E Š Q8 so 2 .G/ D ˆ.G/ since E D 1 .G/. On the other hand, G=E cannot act faithfully on E since Aut.E/ has no subgroups Š Q8 , and so CG .E/ ˆ.G/. In particular, ˆ.G/ is abelian of type .4; 2; 2/. We assume that G=E is Q8 -free. In that case, by (), G=E is ordinary metacyclic but noncyclic since ˆ.G/ E. There is a cyclic normal subgroup S=E of G=E with the cyclic factor-group G=S . Let s 2 S be such that S D hE; si and let r 2 G S be such that G D hS; ri. Since E ˆ.G/, we have G D hr; si. Since S D hE; si is a proper subgroup of G, it follows that S is D8 -free and, by Proposition 78.6, Q8 -free. Since S=hsiS is a subgroup of the symmetric group SjSjWhsij D S4 , whose Sylow 2-subgroup is isomorphic to D8 , that quotient group is abelian and contains a subgroup isomorphic S=.S hsi/ Š E4 so hsi is normal in S and s induces an automorphism of order 2 on E which implies that s 2 centralizes E. Since hE; ri < G, we get (as in the previous paragraph) that r 2 centralizes E. On the other hand, ˆ.G/ D hE; r 2 ; s 2 i and so we get again E Z.ˆ.G//. We summarize our results in a somewhat different form. Theorem 78.8 (Janko). Let G be a minimal nonmodular 2-group of order > 25 . Then each proper subgroup of G is Q8 -free and G=Ã2 .G/ is minimal nonabelian of order 24 or 25 . (a) Suppose that jG=Ã2 .G/j D 24 . If N is any normal subgroup of G such that G=N Š D8 , then N is cyclic. If G=Ã2 .G/ is nonmetacyclic, then G is Q8 -free and 1 .G/ Š E8 with G=1 .G/ cyclic. If G=Ã2 .G/ is metacyclic, then G is also metacyclic and G is either minimal nonabelian or an A2 -group. (b) Suppose that jG=Ã2 .G/j D 25 . Then G=Ã2 .G/ is nonmetacyclic, G is not Q8 -free and 1 .G/ Š E8 with G=1 .G/ Š Q8 or G=1 .G/ is ordinary meta-

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cyclic (but not cyclic). Moreover, if N is any normal subgroup of G such that G=N Š D8 , then N is ordinary metacyclic but noncyclic. Remark 4. Let us change the last paragraph of the proof of Proposition 78.1 by the following argument. Let A be a minimal nonabelian subgroup of G. Let jAj > 8. If Z.A/ is cyclic, setting N D .N \ A/ N1 , we get G D A N1 , d.G/ D 4, a contradiction. Let Z.A/ Š E4 . Then A is metacyclic without cyclic subgroup of index 2. Since A is D8 -free, jAj D 25 and G D A N1 , where N < N1 is of order 2. In that case, G is modular, a contradiction. If Z.A/ Š E8 , then A is D8 -free so A D G. In that case, G 0 < N , which is a contradiction. Now let jAj D 8; then A Š Q8 . By Lemma 1.1, jG 0 j D 4. If L is as in the last paragraph of the proof of Proposition 78.1, then G 0 < L and G 0 6 N D 1 .L/ so G 0 Š C4 . Then A0 D 1 .G 0 / and AN=A0 Š E16 , a contradiction since G=A0 is minimal nonabelian in view of j.G=A/0 D G 0 =A0 Š C2 (Lemma 65.2(a)) so must be j1 .G=A0 /j 8 (Lemma 65.1). 2o . We recall that a p-group G is modular if and only if any subgroups X and Y of G are permutable, i.e., XY D YX. We turn now to the case p > 2. Proposition 78.9. Let G be a modular p-group with p > 2 and d.G/ D 2. Then G is metacyclic. Proposition 78.10. Let G be a minimal nonmodular p-group, p > 2, which is generated by two subgroups A and B of order p. Then G Š S.p 3 / (the nonabelian group of order p 3 and exponent p). Proof. Since G is a p-group, G1 D hAG i and G2 D hB G i are proper normal subgroups of G and so G1 and G2 are modular. It follows that G1 and G2 are elementary abelian. But hG1 ; Bi D hA; Bi D hG2 ; Ai D G, and so G1 and G2 are two distinct maximal subgroups of G. By Lemma 64.1(u), we have jG 0 j D p and G 0 G1 \ G2 . Thus G=G 0 is abelian and G=G 0 is generated by elementary abelian subgroups G1 =G 0 and G2 =G 0 . Hence G=G 0 is elementary abelian and d.G/ D 2 implies that G=G 0 Š Ep 2 . Thus, G Š S.p 3 / since the metacyclic nonabelian group of order p 3 is modular. Proposition 78.11 (see Theorem 44.13). Let G be a minimal nonmodular p-group. Then G possesses a normal subgroup N such that d.N / 2, N Ã1 .G/, and G=N is a nonmodular group of order p 3 . If p D 2, then G=N Š D8 and if p > 2, then G=N Š S.p 3 /, N D Ã1 .G/, and N is metacyclic. The subgroup N is metacyclic. In particular, if p > 2, then G is nonmetacyclic. Proof. There are non-permutable cyclic subgroups hai and hbi. It follows G D ha; bi and so d.G/ D 2. Since hap ; b p i ˆ.G/, the subgroups E D hap ; bi and F D ha; b p i are proper subgroups of G. Hence E and F are modular and so E D hap ihbi, F D haihb p i, and G D hE; F i. Set N D hap ihb p i so that jE W N j D jF W N j D p. It follows that N is normal in G, N Ã1 .G/, and d.N / 2. It remains to determine

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Groups of prime power order

N where GN is a minimal nonmodular p-group the structure of GN D G=N D ha; N bi, generated by elements aN and bN of order p. If p D 2, then GN is dihedral, and, by (), GN Š D8 . If p > 2, then Proposition 78.10 implies that GN Š S.p 3 /. In that case we have N D Ã1 .G/ and Proposition 78.9 implies that N is metacyclic. Proposition 78.12. Let G be a minimal nonmodular p-group, p > 2, with jGj > p 4 . Then 1 .G/ is elementary abelian of order p 3 . Proof. Let A and B be subgroups of order p in G such that AB ¤ BA. Then G D hA; Bi and so Proposition 78.10 implies that G Š S.p 3 /, a contradiction. We have proved that AB D BA and so hA; Bi is abelian. Hence 1 .G/ is elementary abelian. Assume that each proper subgroup of G is metacyclic. By Proposition 78.11, G is nonmetacyclic and so jGj p 4 (Theorem 69.1 for p > 2), a contradiction. Let M be a nonmetacyclic maximal subgroup of G. Since M is modular, Proposition 78.9 implies that d.M / 3. By Proposition 78.11, G is not a 3-group of maximal class (since it has a maximal subgroup H which satisfies H=Ã1 .H / Š S.33 /, so nonmodular. Therefore, if j1 .G/j < p 3 , then, by Theorem 13.7, G must be metacyclic so modular (Proposition 78.11), a contradiction. Thus, 1 .G/j p 3 . Remark 5. Minimal nonmodular p-group of exponent p is isomorphic to S.p 3 /. Indeed, p > 2. All proper subgroups of G are elementary abelian so G Š S.p 3 /, by Lemma 65.3. Theorem 78.13. Let G be a minimal nonmodular p-group, p > 2, with jGj > p 4 . If Ã1 .G/ is cyclic, then 1 .G/ Š Ep 3 and G=1 .G/ is cyclic of order p 2 (i.e. G is an L3 -group; see 17, 18). Proof. By Proposition 78.11, G=Ã1 .G/ Š S.p 3 /. By assumption, N D Ã1 .G/ is cyclic. By Proposition 78.12, E D 1 .G/ is elementary abelian of order p 3 . But jE \ N j D p and E does not cover G=N , and so E Š Ep 3 , by the product formula. On the other hand, there is a 2 G N with hap i D N . Since jG W haij D p 2 and jhai \ Ej D p, we get G D hE; ai, by the product formula, and we are done. Proposition 78.14. Let G be a minimal nonmodular p-group, p > 2, with jGj > p 4 . Suppose that d.Ã1 .G// D 2 and let M be any maximal subgroup of G. Then d.M / 3. Proof. Suppose that this is false. Let M be a maximal subgroup of G with d.M / 4. Set N D Ã1 .G/ so that M=N Š Ep 2 and N=ˆ.N / Š Ep 2 . Since ˆ.N / ˆ.M /, we must have ˆ.M / D ˆ.N / so that M=ˆ.N / Š Ep 4 . We shall study the structure of G=ˆ.N / (which is also minimal nonmodular of order > p 4 and exponent p 2 ) and so we may assume ˆ.N / D f1g which implies M Š Ep 4 . Since 1 .G/ is elementary abelian, we have M D 1 .G/. If x 2 G M , then x p 2 Z.G/# . There is y 2 G M such that y p 2 N hx p i since N D Ã1 .G/; then again y p 2 Z.G/# . Since N Š Ep 2 , we get N Z.G/. If Z.G/ > N , then Z.G/=N D Z.G=N / D ˆ.G=N / and

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333

so Z.G/ D ˆ.G/. But then G is minimal nonabelian. By Lemma 65.1, we get j1 .G/j p 3 , a contradiction. We have proved that Z.G/ D N . By Lemma 1.1, jGj D p 5 D pjZ.G/jjG 0 j and so jG 0 j D p 2 . Since G=N Š S.p 3 /, G 0 ¤ N.D Z.G//, G 0 \ Z.G/ Š Cp and G 0 < M so that CG .G 0 / D M . Since N D Ã1 .G/, where is v 2 G M such that v p 2 N G 0 . It follows that the subgroup H D hG 0 ; vi is nonabelian. We claim that H=hvi Š S.p 3 /. Assume that this is false. Then that quotient group is N elementary abelian. Set GN D G=hvi; then CGN .GN 0 / hMN ; vi N D GN so GN D Z.G/. 0 N N N N Since G=Z.G/ Š Ep 2 , the group G is minimal nonabelian so jG j D p (Lemma 65.1), a contradiction. We have proved that each maximal subgroup of G is generated by three elements. Theorem 78.15 (Janko). Let G be a minimal nonmodular p-group, p > 2, with jGj > p 4 . Then Ã1 .G/ is metacyclic and G=Ã1 .G/ Š S.p 3 / (nonabelian group of order p 3 and exponent p). If Ã1 .G/ is noncyclic, then ˆ.G/ D Ã1 .G/ Cp , 1 .ˆ.G// D 1 .G/ Š Ep 3 , G=1 .G/ is metacyclic and for each maximal subgroup M of G we have d.M / D 3. Proof. Set N D Ã1 .G/ .< ˆ.G// and suppose that d.N / D 2. By Proposition 78.12, 1 .G/ is elementary abelian of order p 3 and 1 .G/ \ N Š Ep 2 . Since 1 .G/ does not cover G=N Š S.p 3 /, N 1 .G/ is contained in a maximal subgroup M of G. By Proposition 78.14, d.M / 3 and the modularity of M implies d.M / D d.1 .M // [Suz1]. This implies 1 .G/ Š Ep 3 and .N 1 .G//=N D ˆ.G=N /. Thus ˆ.G/ D N 1 .G/ and so for each maximal subgroup X of G, we have d.X/ D 3 since X ˆ.G/ and d.X/ D d.1 .X// D d.1 .G//. We know that Aut.1 .G// does not possess an automorphism of order p 2 (see, for example, 33). There are elements a; b 2 G such that N D hap ihb p i and ap and b p centralize 1 .G/. Hence ˆ.G/ D N Z with jZj D p. If G=1 .G/ is nonmodular, then (Proposition 78.11) there is a normal subgroup K of G with K 1 .G/, G=K Š S.p 3 /, and d.K/ 2. This is a contradiction since d.K/ D d.1 .K// D 3. Hence G=1 .G/ is modular and since d.G=1 .G// 2, G=1 .G/ is metacyclic and our theorem is proved.

79

Nonmodular quaternion-free 2-groups

Modular Q8 -free 2-groups are classified in [Iwa] (see 73). Here we classify nonmodular Q8 -free 2-groups. The original proof of the corresponding classification theorem, given in [Wil2], depends on the structure theory of powerful 2-groups. In addition, in the proof of Lemma 10 and 13 in [Wil2] there are some gaps. Our new proof of the classification theorem is completely elementary and does not involve powerful 2groups. Nevertheless, the proof is very involved and reaches probably a deepest result ever proved in the finite 2-group theory. We first prove some easy preliminary results. Then we state the Main Theorem 79.7 and afterwards we describe in great detail the groups appearing in the Main Theorem. Propositions 79.8 to 79.11, describing these groups, are also of independent interest since they are needed by applying the Main Theorem in future investigations. After that the proof of the Main Theorem follows. Lemma 79.1. In a Q8 -free 2-group X there are no elements x; y with o.x/ D 2k > 2 and o.y/ D 4 so that x y D x 1 . If D X and D Š D8 , then CX .D/ is elementary abelian. k1

Proof. If y 2 D x 2 , then hx; yi Š Q2kC1 . If hxi \ hyi D f1g, then we have k1 hx; yi=hx 2 y 2 i Š Q2kC1 . Suppose that D X, where D D ha; t j a4 D t 2 D 1; at D a1 i Š D8 . if v is an element of order 4 in CX .D/, then o.t v/ D 4 and t v inverts a, a contradiction. Hence CX .D/ must be elementary abelian. Lemma 79.2. Let X be a Q8 -free 2-group with elements a and b of order 4 such that Œa; b 2 D Œa2 ; b D 1. If Œa; b ¤ 1, then ha; bi is minimal nonabelian nonmetacyclic of order 24 and therefore Œa; b D a2 b 2 and ab is an involution. Proof. Without loss of generality, one may assume that X D ha; bi holds. We have ha2 ; b 2 i Z.X/. Set Œa; b D c; then c ¤ 1. We compute 1 D Œa2 ; b D Œa; ba Œa; b D c a c

and

1 D Œa; b 2 D Œa; bŒa; bb D cc b :

By Lemma 79.1, c must be an involution and, by the displayed equalities, Œc; a D Œc; b D 1. Hence hci is normal in X and X=hci is abelian. It follows that X 0 D hci and so X, by Lemma 65(a), is minimal nonabelian (and so of class 2) and therefore exp.X/ D 4. By assumption, X 6Š Q8 , and X 6Š D8 since D8 has only one cyclic

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335

subgroup of order 4. We have proved that jXj 24 . Considering X=hci, we conclude that jha; bij 25 . On the other hand, c D Œa; b, a2 , and b 2 are central involutions in X. Set V D 2 ha c; b 2 ci so that V Z.ha; bi/ and jV j D 4. We consider X=V and compute ab D aŒa; b D ac D a1 .a2 c/;

b a D bŒb; a D bc D b 1 .b 2 c/:

Since X=V is Q8 -free, Lemma 79.1 implies that at least one of a2 or b 2 is contained in V . Hence a2 D b 2 c or b 2 D a2 c and so in any case c D a2 b 2 2 V since a2 c b 2 c D a2 b 2 . It follows from the displayed equalities that X=V Š E4 so jXj D jV jjE4 j D 24 . Lemma 79.3 (see 78 and especially Proposition 78.4). Let G be a Q8 -free minimal nonmodular 2-group of order > 8. Then G has a normal elementary abelian subgroup E D 1 .G/ D hn; z; t i of order 8 with G=E cyclic. There is an element x 2 G E of order 2sC1 , s 1, such that G D hE; xi, E \ hxi= hni, and t x D t z, z x D zn , D 0; 1, where in case D 1 we must have s > 1 and we have in that case G 0 D hn; zi Š E4 and Z.G/ D hx 4 i. If D 0, then G is a minimal nonabelian nonmetacyclic group so E Š E8 . In any case, hx 2 i is normal in G, G=hx 2 i Š D8 , and ˆ.G/ D hx 2 ; zi is abelian of type .2s ; 2/. Lemma 79.4. Let V be a minimal non-quaternion-free 2-group. Then there is a normal subgroup U of V such that V =U Š Q8 and U < ˆ.V / so that d.V / D 2. We have ˆ.V /=U D Z.V =U / so that for each x 2 V ˆ.V /, x 2 2 ˆ.V / U . In particular, there are no involutions in V ˆ.V /. We use very often the following Lemma 79.5 (= Lemma 64.1(u)). If A and B are two distinct maximal subgroups of a p-group G, then jG 0 W .A0 B 0 /j p. For the sake completeness we also state the Iwasawa’s result in a suitable form. Proposition 79.6 ([Iwa] and 73). A 2-group G is modular if and only if G is D8 free. A 2-group G is modular and Q8 -free if and only if G possesses a normal abelian subgroup A with cyclic G=A and there is an element g 2 G and an integer s 2 such s that G D hA; gi and ag D a1C2 for all a 2 A (and so, if exp.G/ 4, then G is abelian). Main Theorem 79.7 (B. Wilkens). A finite 2-group G is nonmodular and quaternionfree if and only if G is one of the following groups: (a) (Wilkens group of type (a) with respect to N ) G is a semidirect product hxi N , where N is a maximal abelian normal subgroup of G with exp.N / > 2 and, if t is the involution in hxi, then every element in N is inverted by t . (b) (Wilkens group of type (b) with respect to N ) G D N hxi, where N is a maximal elementary abelian normal subgroup of G and hxi is not normal in G.

336

Groups of prime power order

(c) (Wilkens group of type (c) with respect to N , x, t ) G D hN; x; t i, where N is an elementary abelian normal subgroup of G and t is an involution with ŒN; t D 1. k If o.xN / D 2k , then G=N Š M2kC1 , k 3, and x 2 ¤ 1; furthermore k1 k1 Œx 2 ; N D 1 and ht; x 2 i Š D8 . We analyze now in great detail the above Wilkens groups of types (a), (b), and (c). In what follows we call these groups Wx -groups, where x 2 fa; b; cg so, for example, Wb -group is a Wilkens group of type (b). Remark 1. A nonabelian 2-group G D ht i N is said to be generalized dihedral with base N , if t is an involution and N is an (abelian) subgroup of index 2 in G such that t inverts N . We claim that if also G D ht1 i N1 is generalized dihedral with base N1 , then N1 D N . Assume that this is false. We have N \ N1 D Z.G/ and jG W .N \ N1 /j D 4, G D N [ N t is a partition and all elements of the coset N t are involutions and invert N . Therefore, if x 2 N1 \ N t , then x centralizes and inverts N \ N ˛ so exp.N \ N1 / D 2. It follows that N1 D hx; N \ N1 i is elementary abelian, which is a contradiction since exp.N1 / > 2. In particular, the base of G is characteristic in G. Proposition 79.8. Let G be a Wa -group with respect to N . Then 1 .G/ D N ht i, where t is an involution in G N inverting N and N is characteristic in 1 .G/ and so in G. If G is a Wa -group with respect to N1 , then N D N1 . Also, G is not D8 free but G is Q8 -free. If z 2 1 .Z.G//, then G=hzi is either abelian or a Wa - or Wb -group. Proof. By hypothesis, G is a semidirect product hxi N , where N is a maximal abelian normal subgroup of G with exp.N / > 2 and, if t is the involution in hxi, then every element in N is inverted by t . Since G=N is cyclic, we have 1 .G/ N ht i. The coset N t consist of involutions and so 1 .G/ D N ht i. It follows that 1 .G/ is a generalized dihedral group with respect N . By Remark 1, N is the unique base of 1 .G/ so it is characteristic in 1 .G/ and in G. Thus, if G is also a Wa -group with respect to N1 , then N1 D N . Since t inverts N and exp.N / > 2, G is not D8 -free. Suppose that G is not Q8 -free. Let V be a minimal non-Q8 -free subgroup of G so that V has a normal subgroup U with V =U Š Q8 and ˆ.V / > U ; it follows that d.V / D 2. Since V 6 N and G=N is cyclic, we see that V =.V \ N / > f1g is cyclic. Let t 0 be an element in V N such that .t 0 /2 2 N . Then N t 0 is the involution in G=N and so all elements in N t 0 are involutions. In particular, all elements in the set S D .V \ N /t 0 are involutions. By Lemma 79.4, S ˆ.V / and so also hS i D .V \ N /ht 0 i ˆ.V /. But then V =ˆ.V /, as an epimorphic image of V =.V \ N /, is cyclic, a contradiction. Let z 2 Z.G/ be an involution. We want to determine the structure of G=hzi. We know that G D hxi N is a semidirect product. Let t be the involution in hxi. We have z 2 N and t inverts N and so t inverts N=hzi. If exp.N=hzi/ > 2, then G=hzi is a Wa -group.

79

Nonmodular quaternion-free 2-groups

337

Let exp.N=hzi/ D 2. Set E D ht iN . Then E=hzi D 1 .G=hzi/ is a maximal elementary abelian normal subgroup of G=hzi. If hx; zi D hxi hzi is not normal in G, then G=hzi is a Wb -group. Suppose that hx; zi is normal in G. Then G 0 hx; zi \ N D hzi so G 0 D hzi. In that case, G=hzi is abelian, and we are done. Remark 2. Let a 2-group G of exponent > 2 have two distinct elementary abelian subgroups E and E1 of index 2. Then G D D L, where D Š D8 and exp.L/ divides 2. Since exp.G/ > 2, G is nonabelian. It follows that E \ E1 D Z.G/ has index 4 in G. Let A be a minimal abelian subgroup of G; then jA W .A \ Z.G//j > 2 so G D AZ.G/, by the product formula. In that case, A \ Z.G/ D Z.A/. If Z.G/ D Z.A/ L, then G D A L and exp.L/ divides 2. Since A has two distinct elementary abelian subgroup of index 2, then A Š D8 (Lemma 65.1). Proposition 79.9. (i) A 2-group G is a Wb -group with respect to E if and only if E is a maximal normal elementary abelian subgroup E such that G=E is cyclic and G is not D8 -free. (ii) Let G be a Wb -group with respect to E. Then G is Q8 -free. We have j1 .G/ W Ej 2. If j1 .G/ W Ej D 2, then G has exactly two maximal normal elementary abelian subgroups E and E1 and we have 1 .G/ D EE1 . In that case, if G is also a Wb -group with respect to E1 , then j1 .G/ W E1 j D 2 and 1 .G/ Š D8 E2s . Let z 2 1 .Z.G//. Then G=hzi is either abelian or a Wb -group or G=hzi Š D F , where exp.F / 2 and either D Š D8 or D Š M2n , n 4 (in which case G=hzi is modular and nonabelian). Finally, if G is any 2-group with an elementary abelian normal subgroup E0 such that G D hE0 ; yi (and so G=E0 is cyclic) and hyi is not normal in G, then G is a Wb –group with respect to any maximal elementary abelian normal subgroup E of G containing E0 . Proof. (i) Let G be a nonmodular 2-group possessing a maximal elementary abelian normal subgroup E such that G=E is cyclic. We have G D hE; xi for some x 2 G and if hxi 6E G, then G is a Wb -group. Assume that hxi is normal in G. In that case, G 0 hxi \ E and, since G is nonmodular (and so nonabelian), G 0 D hxi\E D hzi Š C2 . We have jG W CG .x/j D 2 since jG W CG .x/j jG 0 j for each x 2 G. We set E1 D CE .x/ so that jE W E1 j D 2 and E1 Z.G/. Let t be an involution in E E1 and let V be a complement of hzi in E1 so that G D V hx; t i, by the product formula. If jG=Ej D 2s > 2, then hx; t i Š M2sC2 , s 2 since a 2-group of maximal class has no cyclic epimorphic s images of order 2s > 2, and so for each a 2 A D hxi V , at D a1C2 since exp.V / divides 2. But Proposition 79.6 implies that G is modular, a contradiction. Hence jG=Ej D 2 and hx; t i Š D8 . In that case xQ D xt is an involution in G E, G D Ehxi, Q hxi Q is not normal in G, and so G is a Wb -group. (ii) Conversely, let G be a Wb -group with respect to E so that G D hE; gi, where E is a maximal normal elementary abelian subgroup of G and hgi is not normal in G.

338

Groups of prime power order

Set Z D hgi \ E so that jZj 2 and Z Z.G/. Set S D NG .hgi/ so that S ¤ G and S \ E < E. Since G D hgiS , we get NG .S / D hgiNE .S /, by the modular law, so, in view of NE .S / > E \ S , there is an involution n 2 E S normalizing S . Since Œn; g 2 Œn; S and Œn; g 2 ŒE; g E, we get Œn; g 2 S \ E and 1 ¤ u D Œn; g 62 hgi since n 62 S D NG .hgi/. We have Œn; g D ng 1 ng D nng D u. 2 On the other hand, ˆ.S / D hg 2 i and so hg 2 i is normal in hS; ni so that ng D nz with z 2 Z. Hence hgi normalizes hn; ng ; Zi and acts nontrivially on the four-group hn; ng ; Zi=Z, where hg 2 i Z. It follows that hn; gi=hg 2 i Š D8 and so G is not D8 -free. We claim that G is Q8 -free. Indeed, if V is a minimal non-Q8 -free subgroup of G, then, by Lemma 79.4, there are no involutions in V ˆ.V / so that ˆ.V / V \ E. But then V =ˆ.V / is cyclic since V =.V \ E/ is, a contradiction. Set W =E D 1 .G=E/ so that 1 .G/ W and j1 .G/ W Ej jW W Ej D 2. Suppose that j1 .G/ W Ej D 2 so that 1 .G/ D W and there is an involution t 2 W E. Since E is a maximal normal elementary abelian subgroup of G, W is not elementary abelian so ht i is not normal in W , and we conclude that W is a Wb -group with respect to E. Let t1 2 W E be an involution, t1 ¤ t . Then tE D t1 E since W W Ej D 2. In that case t t1 2 E is involution so all involutions in W E commute with t . Set E1 D CW .t / so that E1 E is the set of all involutions in W E, by what has just been proved. Since hE1 Ei D E1 , E1 is normal in G and E and E1 are the only maximal normal elementary abelian subgroups of G. If G is also a Wb -group with respect to E1 , then G=E1 must be cyclic and so jW W E1 j D 2 and in that case W D 1 .G/ Š D8 E2s , by Remark 2. Let z be a central involution in G, where G D hE; xi and hxi is not normal in G; then z 2 E since E is a maximal elementary abelian subgroup of G. If z 2 hxi, then hxi=hzi is not normal in G=hzi so G=hzi is a Wb -group with respect to E=hzi. Suppose that z 62 hxi. If hx; zi is not normal in G, then again G=hzi is a Wb -group. Assume that hx; zi D hxihzi is normal in G. If hxi\E D f1g, then hx; zi\E D hzi and G 0 hx; zi \ E D hzi and so G=hzi is abelian. Assume that hxi \ E ¤ f1g so that hx; zi \ E D 1 .hxi/ hzi Š E4 and suppose that G=hzi D GN is nonabelian. Then GN D hxi N EN with hxi N \ EN Š C2 and both hxi N and EN are normal in GN so that 0 N N N G D hxi N \ E. Let tN be an involution in E which does not centralize hxi. N If o.x/ N D 4, N > 4, then hx; N tNi Š M2n , n 4 since hx; N tNi \ EN Š E4 then hx; N tNi Š D8 and if o.x/ N htNi VN , where .hxi N VN D C N .x/ (see Theorem 1.2). We have EN D .hxi N \ E/ N \ E/ E N N N N so that G D V hx; N t i and we are done. Proposition 79.10. Let G be a Wc -group with respect to N , x, t . Then G is Q8 k1 free but is not D8 -free. We have 1 .G/ D hx 2 ; t iN Š D8 E2s (k 3) and G D 1 .G/hxi so that G=1 .G/ is cyclic of order 4. Also, N is the unique maximal normal elementary abelian subgroup of G. No subgroup of order 8 in hxi k is normal in G. The involution z D x 2 lies in G 0 \ Z.G/ and G=hzi is a Wb -group. If z 0 2 1 .Z.G// and z 0 ¤ z, then z 0 2 N and G=hz 0 i is a Wc -group.

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Nonmodular quaternion-free 2-groups

339

Proof. By hypothesis, G D hN; x; t i, where N is an elementary abelian normal subgroup of G and t is an involution with ŒN; t D 1. If o.xN / D 2k , then G=N Š k k1 k1 M2kC1 , k 3, and x 2 ¤ 1; furthermore Œx 2 ; N D 1 and ht; x 2 i Š D8 so that k1 and z D a2 . We have G is not D8 -free. We set a D x 2 .G=N /0 D 1 .Z.G=N // D .haiN /=N

and 1 .G=N / D .ht; aiN /=N D W =N;

where W D ht; aiN and Z.W / D N . Each involution in G is contained in W and W D 1 .W / and so 1 .G/ D W Š D8 E2s (Remark 2) and G D 1 .G/hxi so that G=1 .G/ is cyclic of order 4. By the structure of G=N , if X is a normal subgroup of G with N X W , then X 2 fN; W; haiN g, where haiN is abelian of type .4; 2; : : : ; 2/ so that N is a maximal normal elementary abelian subgroup of G. Let N1 be any maximal normal elementary abelian subgroup of G; then N1 < 1 .G/ D W . Assume that N1 ¤ N . Since N1 does not cover W =N (since all elements in .haiN / N are of order 4), we get j.N N1 / W N j D 2, N N1 G G and so (by the above) N N1 D haiN , a contradiction. Thus, N is the unique maximal normal elementary abelian subgroup of G so it is characteristic in G. Let Y hxi with jY j 8 and assume that t normalizes Y . Since t inverts hai and hai < Y , it follows that Y ht i is of maximal class and order 2m , m 4, and so .Y ht i/=hzi Š D2m1 is isomorphic to a proper subgroup of G=N Š M2n , n 4, which is not the case. Thus, Y is not normal in G. Let G be not Q8 -free and let V G be minimal non-Q8 -free. By Lemma 79.4, there are no involutions in V ˆ.V / and so ˆ.V / V \ N . On the other hand, ˆ.V / is contained in the maximal subgroup hxiN of G and note that 1 .hxiN / D N (since N \ hxi D hzi and a centralizes N ). It follows that 1 .ˆ.V // D V \ N . Note that G=N has exactly three involutions: Na, N t , and N.at /, where all elements in the coset Na are of order 4 and all elements in cosets N t and N.at / are involutions. Let .V \ N /s (s 2 V ) be an involution in V =.V \ N /. Then N s is an involution in G=N . If all elements in the coset N s are involutions, then s 2 ˆ.V /, contrary to the above fact that 1 .ˆ.V // D V \N . It follows that N s D Na and so .V \N /s D .Na/\V is the unique involution in V =.V \ N /. Since G=N is Q8 -free, we get that V =.V \ N / is cyclic. But then V =ˆ.V / is also cyclic, a contradiction. N Since 1 .G/ D 1 .G/=hzi is an We shall determine the structure of G=hzi D G. N GN D 1 .G/hxi, N and hxi N is not normal in elementary abelian normal subgroup of G, N N G (noting that z 2 hxi and hxi is not normal in G), G is a Wb -group. Let z 0 be an involution in Z.G/ and z 0 ¤ z. Then z 0 2 N and hz 0 i \ ha; t i D f1g so that G=hz 0 i is not D8 -free. Obviously, G=hz 0 i is a Wc -group. Proposition 79.11. Let G be one of the Wilkens groups. Suppose that there is an involution z 2 Z.G/ such that G=hzi is modular (i.e., D8 -free). (i) If G=hzi is abelian, then G is a Wb -group and more precisely: G D D E n where exp.E/ 2 and D D hx; t j x 2 D t 2 D 1, n 1, Œx; t D z, z 2 D

340

Groups of prime power order

Œx; z D Œt; z D 1i. (If n D 1, then D Š D8 , and if n > 1, then D is minimal nonabelian nonmetacyclic with 1 .D/ 6 Z.D/.) (ii) If G=hzi is nonabelian, then there is another involution z 0 2 Z.G/ such that hz 0 i is a characteristic subgroup of G and G=hz 0 i is a Wb -group. Proof. (i) Suppose that G=hzi is abelian. Then G 0 D hzi and by Propositions 79.8, 79.9, and 79.10, G is either a Wa - or Wb -group. Suppose that G is a Wa -group. Then G D hxi N (a semidirect product), where N is a maximal normal abelian subgroup of G with exp.N / > 2 and if t is the involution in hxi, then t inverts each element of N . Suppose n 2 N with o.n/ > 2. Then hn; t i is dihedral and so n2 2 hn; t i0 . It follows that n2 2 hzi and therefore N=hzi is elementary abelian with Ã1 .N / D hzi and jN W 1 .N /j D 2. Suppose hxi > ht i and let v 2 hxi with v 2 D t . Let n 2 N with o.n/ D 4. We have Œn; v ¤ 1 and so 2 Œn; v D z which gives nv D nz. But then nt D nv D .nz/v D nv z v D nzz D n since z 2 Z.G/. This is a contradiction and so hxi D ht i. If n 2 N with o.n/ D 4, then D D hn; t i Š D8 . Let E be a complement of hzi in 1 .N /, where t centralizes 1 .N /. We get G D ht i N D D E, where D Š D8 and exp.E/ 2. Suppose that G is a Wb -group. Then G D hN; xi, where N is a maximal normal elementary abelian subgroup of G and hxi is not normal in G. We have z 2 N and G 0 D hzi. If z 2 hxi \ N , then hxi is normal in G, a contradiction. Hence z 62 hxi and hx; zi D hxi hzi is normal in G. Since hx; zi contains exactly two cyclic subgroups hxi and hxzi of index 2 not containing hzi, we have jG W NG .hxi/j D 2. There is t 2 N NG .hxi/ such that x t D xz. Also note that NG .hxi/ centralizes hxi (since G 0 D hzi). Let E be a complement of hz; hxi \ N i in NN .hxi/. Then G D D E, n where D D hx; t j x 2 D t 2 D 1; n 1; Œx; t D z; z 2 D Œx; z D Œt; z D 1i. (ii) Suppose that G=hzi is nonabelian. By Propositions 79.8, 79.9, and 79.10, G is a Wb -group. Then G D hN; xi, where N is a maximal normal elementary abelian subgroup of G and hxi is not normal in G. If z 2 hxi \ N , then the fact that hxi is not normal in G gives that hxi=hzi is not normal in G=hzi and so G=hzi is a Wb -group, a contradiction. Hence z 62 hxi. If hx; zi is not normal in G, then again G=hzi is a Wb -group, a contradiction. Hence hx; zi D hxi hzi is normal in G. Assume first that hxi \ N D f1g. Then hx; zi \ N D hzi and G 0 hx; zi \ N D hzi and so G=hzi is abelian, a contradiction. Hence hxi \ N > f1g and so hx; zi \ N D 1 .hxi/ hzi. Since G 0 hx; zi \ N and G=hzi is nonabelian (by assumption), we get 1 ¤ jG 0 j 4 and G 0 6 hzi. We have 1 .hxi/ Z.G/. Set 1 .hxi/ D hx0 i and S D NG .hxi/ so that hx0 ; zi Z.G/, jG W S j D 2 and jN W .S \ N /j D 2 because hxi is not normal in G and the abelian normal subgroup hx; zi has exactly two cyclic subgroups hxi and hxzi of index 2. Therefore, we have Œx; s D z or Œx; s D x0 z for an s 2 N S and so ˆ.G/ D hx 2 ; Œhxi; N i D hx 2 i hzi. Suppose o.x/ 8 so that ˆ.ˆ.G// D hx 4 i hx0 i and hx0 i is a characteristic subgroup of G. But hxi=hx0 i is not normal in G=hx0 i (since hxi is not normal in G) and so G=hx0 i is a Wb -group, and we are done in this case.

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Suppose o.x/ D 4 so that x 2 D x0 . If hxi 6 Z.S /, then there is an involution t in S \ N which inverts hxi and so hx; t i Š D8 . But then G=hzi is not D8 -free, a contradiction: G=hzi is modular. Hence hxi is central in S and so G 0 D hŒx; si D hx0 zi (since in case G 0 D hŒx; si D hzi, G=hzi would be abelian). But then hx; si is the minimal nonabelian nonmetacyclic group of order 24 with hx; si0 D hx0 zi and hx; si=hzi Š D8 , contrary to our assumption that G=hzi is modular. Lemma 79.12. Let G be a Wb -group with respect to N . Suppose in addition that 1 .G/ D N . Then for each element g 2 G such that G D hN; gi, N \ hgi D hg0 i is of order 2 and G=hg0 i is also a Wb -group (and so G=hg0 i is nonmodular). Proof. It is enough to show that hgi is not normal in G (because then hgi=hg0 i is also not normal in G=hg0 i). Suppose false. Then G 0 N \ hgi D hg0 i and so G 0 D hg0 i. We have jG W CG .g/j D 2 and let t be an involution in N CN .g/. If o.g/ D 4, then hg; t i Š D8 and so gt is an involution in G N , a contradiction. Hence o.g/ > 4 and hg; t i Š M2n , n 4. If V is a complement of hg0 i in CN .g/, then G D V hg; t i. But then G is modular (see Proposition 79.6), a contradiction.

Proof of the Main Theorem Let G be a nonmodular quaternion-free 2-group of a smallest possible order which is not isomorphic to any Wilkens group. Hence any proper nonmodular subgroup and any proper nonmodular factor group is isomorphic to a Wilkens group. We shall study such a minimal counter-example G and our purpose is to show that such a group G does not exist. (i) There is a central involution z of G such that G=hzi is nonmodular (and so G=hzi is isomorphic to a Wilkens group). Suppose that this is false. Then for each z 2 1 .Z.G//, G=hzi is modular. Suppose that z0 2 1 .Z.G// is such that G=hz0 i is modular. Since G is nonmodular, there is a minimal nonmodular subgroup K of G which is isomorphic to a group of Lemma 79.3. Obviously, K is a Wb -group and so K ¤ G. Since G=hz0 i is modular, we have z0 2 K. If K Š D8 , then hz0 i D 1 .Z.G// D K 0 . Suppose that K has a normal elementary abelian subgroup E D hn; z; t i of order 8 such that K D hE; xi, o.x/ D 2sC1 , s 1, E \ hxi D hni, t x D t z, z x D zn , D 0; 1, 1 .K/ D E, and in case D 1 we have s > 1 and Z.K/ D hx 4 i. If D 1, then we must have z0 D n. But then K=hz0 i is nonmodular since K=hx 2 i Š D8 . Hence we have D 0 in which case K is minimal nonabelian nonmetacyclic with Z.K/ D hx 2 ; zi D ˆ.K/ and K 0 D hzi. Since K=hz0 i is modular (and K=hni is nonmodular), we have either z0 D z (and then K=hz0 i is abelian) or z0 D zn (in which case s > 1 and K=hzni Š M2sC2 ). Let H be a maximal subgroup of G containing K. Since H is nonmodular and H ¤ G, H is a Wilkens group. Since H=hz0 i is modular, we may use Proposition 79.11. If H=hz0 i is nonabelian, then there is another involution z00 in Z.H / such that hz00 i is a characteristic subgroup in H and H=hz00 i is nonmodular. But then z00 2 Z.G/

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and G=hz00 i is nonmodular, contrary to our assumption. Hence H=hz0 i must be abelian and so K=hz0 i is also abelian. In particular, z0 D z, where hzi D K 0 . In any case 1 .Z.G// D hzi is of order 2. By Proposition 79.11(a), we have H D D E0 , where exp.E0 / 2 and m

D D hy; t j y 2 D t 2 D 1; m 1; Œy; t D z; z 2 D Œz; y D Œz; t D 1i: If m D 1, then D Š D8 and if m > 1, then D is minimal nonabelian nonmetacyclic with E8 Š 1 .D/ 6 Z.D/ and z D z0 , where hzi D D 0 D H 0 D 1 .Z.G//. Suppose m D 1. Then Z.H / D hzi E0 is elementary abelian. If jE0 j 4, then acting with an element x 2 G H on Z.H /, we see that jCZ.H / .x/j 4 and CZ.H / .x/ Z.G/, contrary to the fact that 1 .Z.G// is of order 2. Hence jE0 j 2. If D D H Š D8 , then CG .D/ D would imply that G is of maximal class and then G=hzi Š D8 , a contradiction. If D D H Š D8 and CG .D/ 6 D, then Lemma 79.1 implies that G Š D8 C2 , contrary to j1 .Z.G//j D 2. Hence we must have H D D ht i, where t is an involution with CG .t / D H . Since H=hzi is elementary abelian, exp.G=hzi/ 4 and therefore G=hzi is abelian since G=hzi is modular (and Q8 -free) of exponent 4. In particular, D is normal in G. We have CH .D/ D hz; t i. If CG .D/ > hz; t i, then CG .t / D G, a contradiction. Hence CG .D/ D hz; t i and Aut.D8 / Š D8 implies that G=hz; t i Š D8 , a contradiction: G=hzi is modular. Suppose m > 1. Here Z.D/ D ˆ.D/ D hy 2 ; zi is abelian of type .2m1 ; 2/ and Z.H / D hy 2 ; zi E0 so that Ã1 .Z.H // D hy 4 i. If m > 2, then 1 .hy 4 i/ is of order 2 and 1 .hy 4 i/ Z.G/, contrary to the fact that 1 .Z.G// D hzi. Thus we have m D 2, jDj D 24 , and Z.H / D hy 2 ; zi E0 is elementary abelian. Suppose that E0 ¤ f1g. Then acting with an element x 2 G H on Z.H /, we get jCZ.H / .x/j 4 and CZ.H / .x/ Z.G/, a contradiction. It follows D D H and so jGj D 25 . If x 2 G H is of order 8, then x 4 2 Z.H / hzi (with hzi D H 0 ) since ˆ.H / D Z.H / and z is not a square in H . But then Z.H / Z.G/, a contradiction. Hence exp.G/ D 4 and the fact that G=hzi is modular gives that G=hzi is abelian. It follows G 0 D H 0 D hzi. Since H D D D hy; t i and jG W CG .y/j D jG W CG .t /j D 2, we get jG W CG .H /j 4. But jH W CH .H /j D jH W Z.H /j D 4 and so CG .H / must cover G=H . But then E4 Š Z.H / Z.G/, a final contradiction. (ii) The factor group G=hzi (z 2 1 .Z.G//) is not isomorphic to a Wa -group. Suppose that this is false. Then G=hzi has a maximal normal abelian subgroup N=hzi of exponent > 2 such that G=N is cyclic of order 2 and if L=N D 1 .G=N /, then for each element x 2 L N , x 2 2 hzi, and x inverts each element of N=hzi. If all elements in L N are involutions, then each y 2 L N inverts each element in N which implies that N is abelian of exponent > 2, N is a maximal normal abelian subgroup of G and if G D hN; gi, then G is a semidirect product of N and hgi and the involution in hgi inverts N . Thus, G is a Wa -group, a contradiction. Hence, there is v 2 L N with v 2 D z. Let n 2 N be of order 8. Then nv D n1 z ( D 0; 1) and therefore .n2 /v D 1 2 .n z / D n2 , contrary to Lemma 79.1. Thus, exp.N / D exp.N=hzi/ D 4.

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Suppose that N is abelian. Let s and l be elements of order 4 in N such that hsi \ hli D f1g. In that case o.sl/ D 4 and using Lemma 79.1, we get s v D s 1 z, l v D l 1 z and consequently .sl/v D s 1 zl 1 z D .sl/1 , a contradiction. Hence N is abelian of type .4; 2; : : : ; 2/ and since exp.N=hzi/ D 4, z 62 Ã1 .N /. We set Ã1 .N / D ht i with t ¤ z and so jN W 1 .N /j D 2, where each element in N 1 .N / is of order 4 and has square equal to t . Also, ht i D Ã1 .N / is central in G. Suppose that a 2 N 1 .N / is such that a2 D t and (by Lemma 79.1) av D a1 z D a.zt /. By Lemma 79.2, hv; ai is minimal nonabelian nonmetacyclic of order 24 with Œv; a D zt and va is an involution. Suppose that v does not commute with an involution u 2 1 .N /. Then uv D uz, o.au/ D 4, and .au/v D a1 zuz D .au/1 , contrary to Lemma 79.1. It follows that CN .v/ D 1 .N /, Œhvi; N D ht zi, L0 D ht zi, 1 .N / D Z.L/, ˆ.L/ D hz; t i, where ht i D Ã1 .N /. We have L N D v1 .N / [ .va/1 .N /, where all elements in v1 .N / are of order 4 and their squares are equal z and all elements in .va/1 .N / are involutions. Thus E D hvai1 .N / D 1 .L/ is elementary abelian of index 2 in L and also E D 1 .G/ (since G=N is cyclic) is the unique maximal normal elementary abelian subgroup of G. Now, L=E is a normal subgroup of order 2 in G=E with cyclic factor group G=L. Thus G=E is abelian. If G=E were cyclic, then the fact that G is not D8 -free gives that G is a Wb -group, a contradiction. Thus G=E is abelian of type .2s ; 2/, s 1. If s > 1, then we set K=E D 1 .G=E/ Š E4 . Since K L, K is not D8 -free and so K < G implies that K must be a Wilkens group. But E D 1 .K/ and K=E is noncyclic, a contradiction (see Propositions 79.8 to 79.10). Hence s D 1, G=E Š E4 , and so exp.G/ D 4. We have G=N Š C4 and so for each x 2 G L, x 2 2 L N and so x 2 2 E N . Since the square of each element in G is contained in hzi or ht i or in E N , it follows that t z is not a square in G. Hence 1 .G=ht zi/ D E=ht zi. If G=ht zi were nonmodular, then G=ht zi must be a Wilkens group and then .G=ht zi/=.E=ht zi/ Š G=E must be cyclic, a contradiction. It follows that G=ht zi is modular and since exp.G/ D 4, G=ht zi is abelian and so G 0 D ht zi. Suppose that x 2 G L is such that x 2 D E N . Then Œv; x 2 ht zi and so Œv; x 2 D Œv; x2 D 1. But then v centralizes E and since L D hE; vi, we get that L is abelian, a contradiction. Thus, N is nonabelian and so N 0 D hzi. The subgroup N is nonmodular because a Q8 -free modular 2-group of exponent 4 is abelian. Let S be a minimal nonabelian subgroup of N . Then S 0 D hzi and S is normal in N . Since v inverts N=hzi, v normalizes S and so S is normal in L D hN; vi. Since exp.S / D 4 and S is Q8 -free, it follows that either S Š D8 or S is minimal nonabelian nonmetacyclic of order 24 . If S Š D8 , then Ã1 .S / D hzi and N 0 D hzi implies that CN .S / covers N=S . In that case, Lemma 79.1 implies that CN .S / is elementary abelian and so Ã1 .N / D hzi, contrary to exp.N=hzi/ > 2. It follows that we have the second possibility: S D ha; t j a4 D t 2 D 1; Œa; t D z; z 2 D Œa; z D Œt; z D 1i: We put b D at and compute b 2 D a2 t 2 Œt; a D a2 z, so that S 0 D hzi, o.b/ D 4,

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hai\hbi D f1g, and S D ha; bi with Œa; b D a2 b 2 D z. Again, since N 0 D S 0 D hzi, we get that CN .S / covers N=S and CN .S / \ S D Z.S / D ˆ.S / D hz; a2 i. Suppose that y 2 CN .S / is of order 4. Since hai \ hbi D f1g, there is an s 2 fa; bg so that hsi \ hyi D f1g and o.sy/ D 4. We get .sy/v D s 1 zy 1 z D s 1 y 1 D .sy/1 , contrary to Lemma 79.1. Thus, CN .S / is elementary abelian and N D S CN .S /, S \ CN .S / D Z.S / and so the structure of N is completely determined. We act with hvi on S D ha; bi and get (using Lemma 79.1) av D a1 z D a.a2 z/, v b D b 1 z D ba2 , which together with v 2 D z determines uniquely the structure of T D S hvi. Suppose that v does not centralize CN .S / D Z.N /. Then there is an involution s 2 CN .S / S such that s v D sz. Then o.as/ D 4 and .as/v D a1 zsz D .as/1 , contrary to Lemma 79.1. Hence CN .S / D CN .T / and L D T CL .T / with T \ CL .T / D Z.T / D hz; a2 i. We have 1 .S / D hz; a2 ; abi Š E8 , hz; a2 i D Z.T /, .av/2 D avav D av 2 v 1 av D aza1 z D 1; .ab/av D .abz/v D a1 zb 1 zz D a1 b 1 z D aa2 bb 2 z D aba2 .a2 z/z D ab: Hence F D hz; a2 ; ab; avi Š E24 is an elementary abelian maximal subgroup of T and so from T 0 hz; a2 i and jT j D 25 D 2jT 0 jjZ.T /j follows that T 0 D Z.T / D hz; a2 i. Finally, T =hz; a2 i is elementary abelian and therefore Z.T / D T 0 D ˆ.T / Š E4 and so T is a special group of order 25 . For each x 2 T F , CF .x/ D Z.T / and so the set T0 D T F has exactly four square roots of z, four square roots of a2 , four square roots of za2 and so T0 must contain exactly four involutions in T S . If t0 is one of them, then CF .t0 / D Z.T /, and F and hz; a2 ; t0 i Š E8 are the only maximal normal elementary abelian subgroups of T (containing all involutions of T ). We have CN .S / D CN .T / D Z.L/ and so if we set U D F Z.L/ and V D hz; a2 ; t0 iZ.L/, then L D U V , U \ V D Z.L/, U and V are the only maximal normal elementary abelian subgroup of L and they are of distinct orders and so U and V are normal in G and jU W Z.L/j D 4, jV W Z.L/j D 2, L0 D ˆ.L/ D hz; a2 i. For each t0 2 V Z.L/, CU .t0 / D Z.L/ and so both U an V are self-centralizing in L. Also, U is the unique abelian maximal subgroup of L (otherwise, by a result of A. Mann, jL0 j 2). Now, G=N is cyclic, L=N D 1 .G=N /, and so 1 .G/ D L, which is a Wb -group with respect to U . In particular, L < G. If G=U is cyclic, then (since G is not D8 -free) G is a Wb -group, a contradiction. Hence G=U is noncyclic. But L=U Z.G=U / and G=L is cyclic (since L N and G=N is cyclic) and so G=U is abelian of type .2n ; 2/. Set K=U D 1 .G=U / Š E4 . If G ¤ K, then K is a Wilkens group with 1 .K/ D L. By the structure of L, L has no abelian maximal subgroups of exponent > 2 and so K is not a Wa -group. Also, jK W 1 .K/j D 2 and so K is not a Wc -group. Hence K must be a Wb -group with respect to U . But K=U Š E4 , a contradiction. Hence G D K and so G=U Š E4

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implies exp.G/ D 4. Since 1 .G/ D L, all elements in G L are of order 4 and if x 2 G L, then x 2 2 U N , where N D hZ.L/; ab; ai and U \ N D Z.L/ habi. If CG .V / > V , then CL .V / D V implies that there is y 2 G L with y 2 2 V , contrary to y 2 2 U N . We have proved that CG .V / D V and so G=V acts faithfully on V . We have G=V Š .U hxi/=Z.L/, where U=Z.L/ Š E4 and hxiZ.L/=.Z.L/ Š C4 / with x 2 G L. Thus G=V Š D8 or G=V Š C4 C2 . But L=V Š U=Z.L/ Š E4 is a four-subgroup in G=V and for each x 2 G L, x 2 2 U N so that hxi \ V D f1g. Hence all elements in .G=V / .L=V / are of order 4 and so G=V Š C4 C2 . If L0 D hz; a2 i D Z.L/, then V Š E8 . But G=V Š C4 C2 cannot act faithfully on V Š E8 . We have proved that Z.L/ > L0 and so jZ.L/j 8. We act with hxi on Z.L/, where x 2 G L. Since x 2 2 L, hxi induces an automorphism of order 2 on Z.L/. Since jZ.L/j 8, it follows that jCZ.L/ .x/j 4. Suppose that x centralizes an involution u 2 Z.L/ L0 so that u 2 Z.G/. Since G=hui is nonabelian and of exponent 4, G=hui must be nonmodular. Thus G=hui is a Wilkens group with 1 .G=hui/ D L=hui because 1 .G/ D L and u is not a square in G. Then U=hui and V =hui are the only maximal normal elementary abelian subgroups of G=hui and both G=U and G=V are noncyclic and so G=hui cannot be a Wb -group. Also, G=hui cannot be a Wa -group since 1 .G=hui/ D L=hui and jG=Lj D 2. If G=hui is a Wa -group, then (by the first part of the proof of (ii)) L must possess a nonabelian maximal subgroup N0 containing hui such that N00 D hui. This is a contradiction since u 2 Z.L/ L0 . We have proved that for each x 2 G L, CZ.L/ .x/ L0 . Since jCZ.L/ .x/j 4, we must have CZ.L/ .x/ D L0 D Z.G/ D hz; a2 i. Suppose that x 2 GL is such that x 2 2 U N . We have U D hz; a2 ; ab; aviZ.L/ 2 and so .ab/x D ab. On the other hand, x 2 2 L N and x 2 inverts N=hzi. Thus 2 2 ax D a1 z , b x D b 1 z .; D 0; 1/, and so, noting that b 2 D a2 z we get ab D .ab/x D a1 z b 1 z D a1 b 1 z C D aa2 bb 2 z C D abz 1CC ; 2

which gives C 1 .mod 2/ and so D 1 or D 1. 2 Suppose D 1. Then ax D a1 z D a.a2 z/ and we apply Lemma 79.2 in 2 N We have (using bar convention) o.x/ the group G=ha zi D G. N D o.a/ N D 4 and 2 2 2 N D 1 D Œx; N aN . If Œx; N a N D 1, then Œx; a 2 ha zi (and hx; ai is of class 2 ŒxN ; a since hx; ai0 ha2 zi) and so Œx 2 ; a D Œx; a2 D 1, a contradiction. Thus Œx; N a N ¤1 and so (by Lemma 79.2) o.xa/ D 2. Hence .xa/2 2 ha2 zi, contrary to the fact that xa 2 G L and .xa/2 2 U N . 2 N Let D 1. Then b x D b 1 z D b.a2 / and we apply Lemma 79.2 to G=ha2 i D G. 2 N 2 2 N N N We have o.x/ N D o.b/ D 4 and ŒxN ; b D 1 D Œx; N b . If Œx; N b D 1, then Œx; b 2 ha i (and hx; bi is of class 2 since hx; bi0 ha2 i) and so Œx 2 ; b D Œx; b2 D 1, a N ¤ 1 and so (by Lemma 79.2) o.xb/ D 2. Hence .xb/2 2 contradiction. Thus Œx; N b 2 ha i, contrary to the fact that xb 2 G L and .xb/2 2 U N . Our claim (ii) is proved.

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(iii) G=hzi (z 2 1 .Z.G//) is not isomorphic to a Wb -group. Suppose this is false. Then G=hzi has a maximal normal elementary abelian subgroup N=hzi so that G=N is cyclic and G=hzi is not D8 -free . If N is elementary abelian, then G is a Wb -group with respect to N , a contradiction. Hence N is not elementary abelian and so Ã1 .N / D hzi and G ¤ N . Set L=N D 1 .G=N /. (˛) Let N be abelian. Then jN W 1 .N /j D 2 and for each a 2 N 1 .N /, a2 D z. Also, hN 1 .N /i D N , 1 .G/ L, and 1 .G=hzi/ L=hzi. Let v 2 L N with v 2 D z and let x 2 N 1 .N /. By Lemma 79.2, Œv; x D 1 and so v centralizes N . But then L is abelian with Ã1 .L/ D hzi and L G G, a contradiction: N=hzi is a maximal normal elementary abelian subgroup of G=hzi. Thus, for each v 2 L N , v 2 ¤ z. Suppose that for each x 2 L N , x 2 2 N 1 .N /. If G D hN; gi, then hgi covers G=1 .N /. But then G=1 .N / is cyclic and (since G is not D8 -free) G is a Wb -group, a contradiction. We have proved that there is x 2 LN with x 2 2 1 .N /. Suppose that there are no involutions in L N . There is x 2 L N such that x 2 2 1 .N / hzi. Suppose that x does not commute with an a 2 N 1 .N /. By Lemma 79.2, xa is an involution, a contradiction. Hence x commutes with all elements in N 1 .N / and therefore L is abelian of type .4; 4; 2; : : : ; 2/. We have 1 .L/ D 1 .N / D 1 .G/ and L=1 .L/ Š E4 . Since G is nonabelian, we have L < G. Because N=1 .N / is a normal subgroup of order 2 in G=1 .N / and G=N is cyclic, G=1 .N / is abelian of type .2s ; 2/, s 2. For each y 2 L N , y 2 D x 2 or y 2 D x 2 z. Hence, if g 2 G is such that G D hN; gi, then 1 .hgi/ D hx 2 i or 1 .hgi/ D hx 2 zi and so Ã1 .L/ D hz; x 2 i Z.G/. We may assume (by a suitable notation) that hgi hxi. Set M D hgi1 .N / so that M is a maximal subgroup of G. Suppose that G=hx 2 i is nonmodular so that it is a Wilkens group. But 1 .G=hx 2 i/ D S0 =hx 2 i, where S0 D 1 .N /hxi. On the other hand, G=S0 is noncyclic (since L=S0 and M=S0 are two nontrivial cyclic subgroups of G=S0 with .L=S0 /\.M=S0 / D f1g). This is a contradiction and so G=hx 2 i is modular. Since 1 .G=hzi/ D N=hzi, we may use Lemma 79.12 and we see that G=hx 2 ; zi is nonmodular. This is not possible since G=hx 2 i is modular. Thus, there are involutions in L N . Let t be an involution in L N . If t centralizes an element a 2 N 1 .N /, then t a 2 L N and .t a/2 D z, a contradiction. Thus, CN .t / 1 .N /. For any x 2 L N , CN .x/ D CN .t / and so x 2 2 1 .N /. It follows that exp.L/ D 4. Suppose that t does not centralize 1 .N /. Let w be a fixed involution in 1 .N / CN .t / and let a be a fixed element in N 1 .N /. We have w t D wu with 1 ¤ u 2 CN .t / and .t w/2 D .t wt /w D wuw D u and so u 2 CN .t /hzi (since t w 2 LN ). Since CN .t w/ D CN .t / < 1 .N /, we have Œt w; a ¤ 1. By Lemma 79.2, .t w/a is an involution. We get 1 D .t wa/2 D t wat wa D w t at wa D wuat wa D uaat , and so at D a1 u D aa2 u D a.uz/. We consider the factor group L=huzi D LN and we see that o.a/ N D 4, o.tN/ D 2, Œa; N tN D 1, aN 2 D zN . Also, w t D wu D wz.uz/ tN gives wN D wN zN so that hw; N tNi Š D8 with Z.hw; N tNi/ D hzi. N Hence hw; N tNihai N is the N We have proved central product D8 C4 , contrary to Lemma 79.1 (applied in L).

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that t centralizes 1 .N / so that S D ht i 1 .N / is an elementary abelian maximal subgroup of L. Since L is nonabelian and exp.L/ D 4, L is nonmodular. Let a 2 N 1 .N /. Then by Proposition 51.2, at D a1 s, s 2 1 .N /. If s D 1, then t inverts N , G=N is cyclic and so G would be a Wa -group, a contradiction. Thus, s ¤ 1 and .t a/2 D .t at /a D a1 sa D s and so s 2 1 .N / hzi. Set t a D y so that y 2 D s and since CN .y/ D CN .t /, we have for each n 2 CN .t / D 1 .N /, .ny/2 D n2 y 2 D s and Œt; N D hzsi D L0 . The group L has exactly three abelian maximal subgroups: S D 1 .G/, N , and 1 .N /hyi, where only S is elementary abelian and all three are normal in G with Ã1 .1 .N /hyi/ D hsi, s 2 1 .N / hzi. We have 1 .N / D Z.L/ and z; s and zs are central involutions in G. If G D L, then G would be a Wb -group, a contradiction. Thus L < G. Since s1 G=N is cyclic of order 2s , s 2, we have G D hN; gi with g 2 2 L N . If s1 g2 2 L N S , then hgi covers G=S , G=S is cyclic and so G would be a Wb s1 group with respect to S , a contradiction. It follows that g 2 2 S 1 .N /. If L is not maximal in G, then there is a subgroup K > L with jK W Lj D 2 and so K is a Wilkens group. Since S D 1 .K/ and K=S Š E4 , we have a contradiction. s2 Indeed, .S hg 2 i/=S and L=S are two distinct subgroups of order 2 in K=S . Hence jG W Lj D 2 and for each g 2 G L, g 2 2 S 1 .N /. N where hzsi D L0 . We have We apply now Lemma 79.2 in the group G=hzsi D G, N D o.a/ N D 4 and since for some elements g 2 G L and a 2 N 1 .N /, o.g/ 2 ag D a.zs/, we get Œa; N gN 2 D 1 D ŒaN 2 ; g. N If Œg; N a N D 1, then Œg; a 2 hzsi and so Œg 2 ; a D Œg; a2 D 1, a contradiction. Hence Œg; N a N ¤ 1 and so (by Lemma 79.2) 2 o.gN a/ N D 2 which gives .ga/ 2 hzsi, contrary to ga 2 G L and (by the above) .ga/2 2 S 1 .N /. We have proved that N must be nonabelian. (ˇ) Let N be nonabelian. This case is very difficult. We have Ã1 .N / D N 0 D hzi. Let D be a minimal nonabelian subgroup of N so that D 0 D hzi. Since d.D/ D 2 and D=hzi is elementary abelian, we have D=hzi Š E4 and therefore D Š D8 . The subgroup D is normal in N and since D 0 D N 0 , CN .D/ covers N=D. By Lemma 79.1, CN .D/ is elementary abelian. We have CN .D/ D Z.N /, N D DZ.N / with D \ Z.N / D Z.D/ D hzi so that Z.N / is normal in G. Set D D ha; u j a4 D u2 D 1; a2 D z; au D a1 i so that A D ha; Z.N /i, E1 D hu; Z.N /i, and E2 D hau; Z.N /i are all abelian maximal subgroups of N , where A is of exponent 4 (all elements in A Z.N / are of order 4, Ã1 .A/ D hzi), and E1 and E2 are both elementary abelian. Thus A is normal in G. All elements in N A are involutions, Z.N / D 1 .A/, N is a Wa -group with respect to A and also a Wb -group with respect to E1 and E2 . Since N=A is a normal subgroup of order 2 in G=A and G=N is cyclic, it follows that G=A is abelian. If G=A were cyclic, then we have G D hA; gi with some g 2 G and since 1 .hgi/ is of order 2 and inverts A (of exponent > 2), G is a Wa -group, a contradiction. It follows that G=A is abelian of type .2m ; 2/, m 1, and L=A Š E4 , where L=N D 1 .G=N /. Also note that N=Z.N / Š E4 .

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Suppose that for each l 2 L N , l 2 2 A Z.N /. This is equivalent to assuming that L=Z.N / Š C4 C2 which also implies that both E1 and E2 are normal in L. If G D hN; gi, then either E1g D E2 (in which case G > L and G=Z.N / Š M2n , g n 4, and G is a Wc -group) or E1 D E1 , E1 is normal in G, hgi covers G=E1 and since G is not D8 -free, G is a Wb -group. In both cases we have a contradiction. Thus, L=Z.N / is not isomorphic to C4 C2 and so L=Z.N / is either elementary abelian or L=Z.N / Š D8 . In any case, there is l 2 L N with l 2 2 Z.N /. (ˇ1) Let L=Z.N / Š D8 . Then E1x D E2 for x 2 L N , and so G D L. Indeed, G=N acts on the set fE1 ; E2 g and so, if G > L, then L would normalize E1 . Suppose in addition that there are no involutions in L N . Let l 2 L N with l 2 2 Z.N / so that l 2 ¤ 1 and o.l/ D 4. Let a0 be any element in A Z.N / so that a02 D z. If Œl; a0 ¤ 1, then la0 is an involution in L N , a contradiction. Thus l centralizes each element in A Z.N / and since hA Z.N /i D A, Ahli is an abelian maximal subgroup of G D L. Also, Z.N / D Z.G/ (since Z.L=Z.N // D A=Z.N /) and so we may use the relation jGj D 2jG 0 jjZ.G/j which gives jG 0 j D 4. But G 0 covers A=Z.N / D .G=Z.N //0 and so G 0 Š C4 is a cyclic subgroup of order 4 in A inverted by u (since u inverts A) and so we have G 0 hui Š D8 . We may assume G 0 D hai < D Š D8 so that D is normal in G. By Lemma 79.1, CG .D/ is elementary abelian and so CG .D/ cannot cover G=N (since there are no involutions in G N ). Hence CG .D/ D Z.N / and so l 2 G N induces an outer automorphism on D. Hence al D a1 , contrary to Lemma 79.1. We have proved that there are involutions in G N and let t be one of them so that 1 .G/ D G. Since G=Z.N / Š D8 , there is k 2 G N such that k 2 2 A Z.N /. Our ultimate goal is to show that Z.N / D Z.G/. Suppose Z.N / ¤ Z.G/. Assume there is s 2 G N with s 2 D z and let a0 2 A Z.N /. If Œs; a0 ¤ 1, then (Lemma 79.2) Œs; a0 D s 2 a02 D zz D 1, a contradiction. Thus s centralizes A Z.N / and hA Z.N /i D A and so Z.N / Z.G/. Since Z.G=Z.N // D A=Z.N /, we have Z.N / D Z.G/. This is a contradiction and so there is no s 2 G N with s 2 D z. In particular, the involution t does not centralize any element in A Z.N /. Now, Aht i < G maximal. If x 2 At is of order 8, then o.x 2 / D 4 and x 2 2 A Z.N /. But then t centralizes x 2 (noting that CA .t / D CA .x/), a contradiction. Hence exp.Aht i/ D 4 and since Aht i is nonabelian, Aht i is nonmodular and therefore Aht i must be a Wilkens group. If X is an abelian maximal subgroup of Aht i distinct from A, then X \ A Z.N / (recalling that t does not commute with any element in A Z.N /) and since jA W .X \ A/j D 2, we get X Z.N / and so Z.N / D Z.G/, a contradiction. Hence A is the unique abelian maximal subgroup of Aht i. If all elements in .Aht i/ A are involutions, then t inverts each element of A and so t centralizes Z.N / and then Z.N / D Z.G/, a contradiction. It follows that there is an element c of order 4 in At such that c 2 2 Z.N / hzi. But Œc; a ¤ 1 (since t does not centralize a) and so (by Lemma 79.2) ac is an involution for each a 2 AZ.N /. Since t does not centralize Z.N / (otherwise Z.N / D Z.G/), Z.N /ht i contains less than 2jZ.N /j 1 involutions. All jZ.N /j elements ca (a 2 A Z.N /) are involutions and

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so Aht i contains at least 2jZ.N /j 1 involutions. This shows that 1 .Aht i/ D Aht i. Since A (of exponent > 2) is the unique abelian maximal subgroup of Aht i, Aht i must be a Wa -group. In that case t inverts A and so t centralizes Z.N / and Z.N / D Z.G/, a contradiction. We have proved that Z.N / D Z.G/ and so Z.N /hki D Ahki is an abelian maximal subgroup of G (noting that k 2 2 A Z.N / and so Z.N /hki=Z.N / is cyclic). Using the relation jGj D 2jG 0 jjZ.G/j we get jG 0 j D 4 and so G 0 Š C4 (since G 0 covers A=Z.N /). We may assume (as before) that G 0 D and so D is normal in G with CG .D/ D Z.N / (since CG .D/ is elementary abelian and jG W Z.G/j D 8). The involution t induces an outer automorphism on D and so Dht i Š D24 . It follows that G D .Dht i/ E2m for some m 1, which is a Wa -group, a contradiction. (ˇ2) We have proved that we must have L=Z.N / Š E8 and so exp.L/ D 4. In that case we prove first that there are involutions in L N . Suppose false. If v is any element in L N and a0 2 A Z.N /, then 1 ¤ v 2 2 Z.N / and a02 D z so that we may apply Lemma 79.2. If Œv; a0 ¤ 1, then va0 is an involution, a contradiction. Hence L D hL N i centralizes A D hA Z.N /i. In particular, A Z.N / which contradicts the fact that Z.N / < A. We have proved that there are involutions in LN and so L D 1 .L/ D 1 .G/ (noting that G=N is cyclic and L=N D 1 .G=N /). Assume CG .D/ > Z.N / so that CG .D/ covers L=N and (Lemma 79.1) CG .D/ is elementary abelian. In that case, L Š D8 E2m and Ã1 .L/ D hzi which contradicts our assumption that N=hzi is a maximal normal elementary abelian subgroup of G=hzi. Hence CG .D/ D Z.N /. If D were normal in L, then L=Z.N / Š D8 since Aut.D/ Š D8 . This is a contradiction and so NG .D/ D N . In particular, Z.L/ Z.N / and Z.N / > hzi. (ˇ2a) We assume that G > L. Then L D 1 .L/ (being nonmodular) is a Wa - or Wb -group. In particular, L has an abelian maximal subgroup B. Since B \ N is an abelian maximal subgroup of N , we get B \ N 2 fA; E1 ; E2 g. Hence B \ N Z.N / and so Z.N / Z.L/. By the result in the previous paragraph, we get Z.N / D Z.L/. Since jL W Z.L/j D 8, B is the unique abelian maximal subgroup of L and so B is normal in G. Using the relation jLj D 2jL0 jjZ.L/j, we get jL0 j D 4. Since L0 Z.L/, L0 Š E4 and L0 > hzi. It is easy to see that jG W Lj D 2. Suppose that this is false. Let K < G be such that jK W Lj D 2. Since L D 1 .K/, K must be a Wa - or Wb -group. By the uniqueness of B in L, follows that K=B is a cyclic group (of order 4). Since N < L and G=N is cyclic, G=L is cyclic. But L=B is a normal subgroup of order 2 in G=B and so G=B is abelian. We have L=B 1 .G=B/. On the other hand, G=L is cyclic and K=L D 1 .G=L/ so that 1 .G=B/ K=B. Since K=B Š C4 , we get 1 .G=B/ L=B and so 1 .G=B/ D L=B. Hence G=B is cyclic. If L is of type (a) (in the case exp.B/ > 2), then all elements in L B are involutions and so G is also a Wa -group, a contradiction. If L is of type (b), then B must be elementary abelian and (since G is not D8 -free) G is also a Wb -group, a contradiction. We have proved that jG W Lj D 2 and G=B Š E4 (because in case G=B Š C4 ,

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G would be a Wa - or Wb -group). But G=N Š C4 and so for each x 2 G L, x2 2 B N . We assume first that exp.B/ > 2. Then L is a Wa -group. In that case B=Z.N / Š E4 , all elements in L B are involutions and 1 .B/ D Z.N /. Indeed, if jB W 1 .B/j D 2, then acting by conjugation with an involution in L B on B, we see that jL0 j D 2, a contradiction. Hence we must have 1 .B/ D Z.N / so that B is abelian of type .4; 4; 2; : : : ; 2/, B \ N D A, u inverts B, each element in B N is of order 4, and if k 2 B N , then k 2 D z0 2 L0 hzi (noting that k u D k 1 and so k 2 2 L0 ) and L0 ha; ki D ha; ki Š C4 C4 (if k 2 D z, then ak would be an involution in B N , contrary to 1 .B/ D Z.N /). Let x 2 G L so that x 2 2 B N and we may assume that x 2 D k, where k 2 D z0 2 L0 hzi. In particular, CG .z0 / hL; xi D G and so L0 Z.G/. Let x 0 be an arbitrary element in G L so that .x 0 /2 D k 0 2 B N and .k 0 /2 D z 0 2 L0 hzi. Consider the factor N D o.x 0 / D 4 and ŒaN 2 ; x 0 D 1 D Œa; N .x 0 /2 and so group GN D G=hz 0 i. We have o.a/ 0 2 0 0 we may use Lemma 79.2. If Œa; N x ¤ 1, then o.ax N / D 2 and so .ax / 2 hz 0 i N , a contradiction. Hence we must have Œa; N x 0 D 1 and so Œa; x 0 2 h.x 0 /4 i. It follows that Œa; x D z0 . D 0; 1/. Consider the element y D xu 2 G L; then Œa; y D Œa; xu D Œa; uŒa; xu D zz0 ¤ 1. On the other hand, Œa; y 2 hy 4 i, y 4 2 L0 hzi. Thus D 1 and y 4 D zz0 . Finally, consider the factor group GQ D G=hzz0 i so that Q D 4 and Œy; Q and apply again Lemma 79.2. Since o.y/ Q D o.k/ Q kQ 2 D 1 D ŒyQ 2 ; k Q y Q ¤ 1. Œk; x D 1, we get Œk; y D Œk; xu D Œk; uŒk; xu D k 2 D z0 , and so Œk; 2 Q Thus o.k y/ Q D 2 and so .ky/ 2 hzz0 i N . This is a contradiction since ky 2 G L. We study now the case exp.B/ D 2 and L is a Wb -group. We may assume that B \ N D E1 . Since G=B Š E4 , exp.G/ D 4. Let t be an involution in B N . Since B E1 D Z.N /hui, t centralizes u 2 D. But NL .D/ D N and so Œa; t 2 L0 hzi. Suppose that L0 Z.G/. Let x 2 G L so that x 2 2 B N and (by the above) Œa; x 2 D z0 2 L0 hzi. Consider the factor group G=hz0 i D GN so that o.a/ N D o.x/ N D 4. If Œx; N a N D 1, then Œx; a 2 hz0 i. But hx; ai0 D hz0 i and so hx; ai is of class 2. Thus Œx 2 ; a D Œx; a2 D 1, a contradiction. Hence Œx; N a N ¤ 1 and so (by Lemma 79.2) o.xN a/ N D 2, which gives .xa/2 2 hz0 i, contrary to xa 2 G L and .xa/2 2 B N . We have proved that L0 6 Z.G/. Again, let x 2 G L so that hxi induces an involutory automorphism on Z.L/ D Z.N /. Suppose Z.N / > L0 . Then there is an involution z 0 2 Z.L/ L0 centralized by x and so z 0 2 Z.G/. Since G=hz 0 i is nonmodular (noting that D \ hz 0 i D f1g), G=hz 0 i is a Wilkens group. All squares of elements of G lie either in B-N or in L0 . Indeed, let as .s 2 B/ be any element in L B. Then .as/2 D a2 .a1 sas/ D zŒa; s 2 L0 . Therefore z 0 is not a square of any element in G which implies L=hz 0 i D 1 .G=hz 0 i/. Let y 2 L be such that Œy; L hz 0 i. But hz 0 i \ L0 D f1g and so Œy; L D f1g and therefore y 2 Z.L/. Hence Z.L=hz 0 i/ D Z.L/=hz 0 i. Since jL W Z.L/j D 8, the elementary abelian group B=hz 0 i is the unique abelian maximal subgroup of L=hz 0 i. It follows that G=hz 0 i must be a Wb -group with respect to B=hz 0 i. But then G=B must be cyclic, a contradiction. We have proved that L0 D Z.L/ and so jGj D 26 . Now,

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A D hai hz0 i D haiL0 Š C4 C2 is normal in G and is self-centralizing in L (since B is the unique abelian maximal subgroup of L). If CG .A/ 6 L, then there is g 2 CG .A/ L such that g 2 2 A N , contrary to g 2 2 B N . Thus A is self-centralizing in G and so G=A Š D8 since Aut.A/ Š D8 . On the other hand, N=A is a normal subgroup of order 2 in G=A and G=N Š C4 so that G=A is abelian. This is a contradiction. We have proved that the case G > L is not possible. (ˇ2b) It remains to study the case G D L D 1 .G/. Since G=Z.N / Š E8 , exp.G/ D 4. If for each x 2 G N , x 2 2 hzi, then G=hzi is elementary abelian, contrary to our assumption that N=hzi is a maximal normal elementary abelian subgroup of G=hzi. Hence there is k 2 G N such that k 2 2 Z.N / hzi. It follows that k 2 2 Z.G/ and so hzi < Z.G/ Z.N /. Let z 0 2 Z.G/ hzi. Since G=hz 0 i is nonmodular (noting that D \ hz 0 i D f1g with D Š D8 ), it follows that G=hz 0 i is a Wilkens group with 1 .G=hz 0 i/ D G=hz 0 i and so G=hz 0 i cannot be a Wc -group. Hence G=hz 0 i must be a Wb -group by our previous result (ii). Thus, G has a maximal subgroup N1 containing z 0 such that N1 =hz 0 i is a maximal normal elementary abelian subgroup of G=hz 0 i. By our previous result (iii)(˛), hz 0 i D Ã1 .N1 / D N10 (since N1 must be nonabelian). In particular, z 0 is a square of an element in G N and z 0 is a commutator in G. Conversely, if k 2 G N , then k 2 2 Z.G/ so that ˆ.G/ Z.G/. Hence Z.G/ D G 0 D ˆ.G/ and so G is a special group. Now, N \ N1 is a maximal subgroup of N and since Ã1 .N \ N1 / hzi \ hz 0 i D f1g, N \ N1 is elementary abelian and so N \ N1 D E1 (or E2 ) containing Z.N / and (by the structure of N1 Š N ) Z.N1 / is a subgroup of index 2 in N \ N1 . But Z.N / \ Z.N1 / Z.G/ and so jZ.N / W Z.G/j 2. Let jZ.N / W Z.G/j D 2 and let s be an involution in Z.N / Z.G/. Let t be an involution in G N . We have Œt; s D s0 with s0 2 Z.G/ and s0 ¤ 1 since Z.N / ¤ Z.G/. If n 2 N , then Œt n; s D Œt; sŒn; s D Œt; s D s0 2 Z.G/. Hence, for each x 2 G N , Œx; s D s0 , where s0 is a fixed involution in Z.G/. Take an involution z 0 2 Z.G/ hzi. Let N1 be a maximal subgroup of G such that hz 0 i D Ã1 .N1 / D N10 . Then N \ N1 is an elementary abelian maximal subgroup of N containing Z.N /. If f1 2 N1 N , then Œf1 ; s D z 0 D s0 . Let N2 (¤ N1 ) be a maximal subgroup of G such that hzz 0 i D Ã1 .N2 / D N20 . Then again, N \ N2 Z.N / and if f2 2 N2 N , then Œf2 ; s D zz 0 D s0 . Hence zz 0 D z 0 and so z D 1, a contradiction. We have proved that Z.N / D Z.G/. Suppose that Z.G/ possesses a four-subgroup hz1 ; z2 i such that hz1 ; z2 i \ hzi D f1g (which is equivalent with the assumption jZ.G/j 8). Let k1 ; k2 2 G N be such that k12 D z1 and k22 D z2 . Suppose that k1 centralizes all elements (of order 4) in A Z.N /. Then k1 centralizes A D Z.N /hai D hA Z.N /i and so Ahk1 i is an abelian maximal subgroup of G. Using a result of A. Mann (Lemma 79.5 with respect to maximal subgroups Ahk1 i and N ), we get jG 0 j 4. But G 0 D Z.G/ is of order 8, a contradiction. We may assume that Œa; k1 ¤ 1 and so (Lemma 79.2) Œa; k1 D a2 k12 D zz1 . We have either Œa; k2 D 1 or Œa; k2 ¤ 1 in which case (Lemma 79.2) Œa; k2 D a2 k22 D zz2 . We have Œa; k1 k2 D Œa; k1 Œa; k2 D zz1 Œa; k2 , and so either Œa; k1 k2 D zz1

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or Œa; k1 k2 D zz1 zz2 D z1 z2 . But the element k1 k2 is contained in N and so Œa; k1 k2 2 hzi, a contradiction. We have proved that we must have Z.N / D Z.G/ Š E4 and so jGj D 25 . Let 0 z 2 Z.G/ hzi and let k1 ; k2 2 G N be such that k12 D z 0 and k22 D zz 0 . Suppose Œk1 ; k2 D 1 so that hk1 ; k2 i Š C4 C4 , k1 k2 2 N and .k1 k2 /2 D k12 k22 D z and therefore we may assume that k1 k2 D a. Hence hk1 ; k2 i is an abelian maximal subgroup of G normalized by u, where u inverts each element in hk1 ; k2 i \ N D hai hz 0 i Š C4 C2

and

G D hk1 ; k2 ihui:

We know that there are involutions in GN and so there is an element k 2 hk1 ; k2 iN such that uk is an involution. This gives .uk/2 D ukuk D 1, k u D k 1 and so u inverts each element of the abelian group hk1 ; k2 i. In this case G is a Wa -group, a contradiction. Hence Œk1 ; k2 ¤ 1 and using Lemma 79.2 we get Œk1 ; k2 D k12 k22 D z 0 .zz 0 / D z, o.k1 k2 / D 2, k1 k2 2 N . Because k1 k2 62 Z.hk1 ; k2 i/, we may assume k1 k2 D u (an involution in N A). Since Œk1 ; u D Œk1 ; k1 k2 D Œk1 ; k2 D z (which gives uk1 D uz) and D D ha; ui Š D8 is not normal in G, we have Œa; k1 ¤ 1. By Lemma 79.2, Œa; k1 D a2 k12 D zz 0 and k1 a is an involution in G N . We have uk1 a D .uz/a D .uz/z D u. Hence hu; k1 ai Š E4 with hu; k1 ai \ Z.G/ D f1g and so hu; k1 a; Z.G/i Š E16 . Since G is not D8 -free, it follows that G is a Wb -group. This is our final contradiction and so our statement (iii) is completely proved. (iv) The factor group G=hzi (z 2 1 .Z.G//) is not isomorphic to a Wc -group. Suppose that the assertion (iv) is is false. Then G=hzi is a Wc -group. Hence we may set 1 .G=hzi/ D H=hzi (implying that 1 .G/ H ) so that H D H1 H2 , where H1 and H2 are normal subgroups in H with H1 \ H2 D hzi, H1 =hzi Š D8 , H2 =hzi is elementary abelian and G=H is cyclic of order 4. Let Z=hzi be the unique cyclic subgroup of index 2 in H1 =hzi and set Z.H1 =hzi/ D Z0 =hzi so that jZ W Z0 j D 2 and jZ0 W hzij D 2. Let Z0 H2 D N , ZH2 D A, so that Z.H=hzi/ D N=hzi and N=hzi is the unique maximal normal elementary abelian subgroup of G=hzi, G=N Š M2n , n 4, A=N D .G=N /0 , H=N D 1 .G=N / Š E4 . If E1 =N and E2 =N are other two subgroups of order 2 in H=N (distinct from A=N ), then E1G D E2 , E1 =hzi and E2 =hzi are elementary abelian and A=hzi is abelian of type .4; 2; : : : ; 2/. Also, NG .E1 / D NG .E2 / > H and jG W NG .E1 /j D 2. Finally, G possesses a subgroup S such that G D HS , H \ S D Z, and S=hzi is cyclic so that S is abelian. In fact, S is either cyclic or abelian of type .2n ; 2/, n 4. Also, S is cyclic if and only if Z0 is cyclic (since Z0 =hzi D 1 .S=hzi/). We have H 0 Z0 and H 0 covers Z0 =hzi. (˛) Suppose Z0 (and so also S ) is cyclic. We have H1 =hzi Š D8 and .H1 =hzi/0 D Z0 =hzi. Since H10 covers Z0 =hzi and Z0 is cyclic, we get that H10 D Z0 is of index 4 in H1 . By a very well-known result of O. Taussky, H1 is of maximal class. Since H1 is Q8 -free, we get H1 Š D24 . An involution t 2 H1 Z inverts Z and so H3 D hZ0 ; t i Š D8 . The subgroup H3 is normal in H1 and ŒH3 ; H2 H1 \ H2 D hzi implies that H3 is normal in H .

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Nonmodular quaternion-free 2-groups

353

Since CH1 .H3 / D hzi and H1 =hzi Š D8 Š Aut.H3 /, we get that CH .H3 / covers H=H1 . By Lemma 79.1, CH .H3 / is elementary abelian. In particular, 1 .H / D H D 1 .G/. Since H is nonmodular, H is a Wilkens group with 1 .H / D H and so H is either a Wa - or Wb -group. It follows that H has an abelian maximal subgroup AQ which is unique (by a result of A. Mann) since H 0 H10 and jH10 j D 4 (see Lemma 79.5). In particular, AQ is normal in G. Also, AQ \ H1 D Z Š C8 since Q > 2. If Z is the unique abelian maximal subgroup of H1 Š D24 and so exp.A/ Q Let t 2 H1 Z, then H must be a Wa -group and the involution t inverts on A. Q Q Q Q Q N D 2 .A/ D Z0 1 .A/ (since A=Z is elementary abelian) so that N is normal in G, and NQ =hzi is a normal elementary abelian subgroup of G=hzi. By the uniqueness of N=hzi, NQ N and so AQ D Z NQ ZN D A and therefore AQ D A is abelian of type .8; 2; : : : ; 2/. Let G1 > H be such that jG1 W H j D 2. Hence G1 ¤ G and G D HS1 , where S1 D S \ G1 . It follows that G1 is also a Wilkens group with 1 .G1 / D H . Since jG1 W H j < 4, G1 is of type (a) or (b). But A D AQ is the unique abelian maximal subgroup of H and exp.A/ > 2. Thus G1 must be a Wa -group with respect to A and so G1 =A Š C4 . On the other hand, we know that S1 \ H D Z. Hence, if x 2 S1 H , then x 2 2 Z A. This is a contradiction since G1 =A Š C4 and therefore x 2 2 H A. (ˇ) We have Z0 Š E4 and so S splits over hzi. We set Z D haihzi so that Z0 D ha2 ihzi and replacing a with az (if necessary), we may put S D hsi hzi with 2 .hsi/ D hai D S \ H . If jH10 j D 4, then jH1 W H10 j D 4, H10 D Z0 , and (by a result of O. Taussky) H1 is of maximal class. In that case Z0 would be cyclic, a contradiction. We have proved that jH10 j D 2 and so Z0 D H10 hzi. We set H10 D hz0 i so that z0 is a central element in H . But S is abelian, G D HS , and S \ H D Z > Z0 D hz; z0 i. It follows CG .hz; z0 i/ hH; S i D G and so hz; z0 i Z.G/. We show that there are exactly two possibilities for the structure of H1 . Since H1 =Z0 Š E4 , we have exp.H1 / D 4. Suppose that H1 is minimal nonabelian. If H1 is metacyclic (of exponent 4), then we know that H1 is not Q8 -free. Thus H1 must be nonmetacyclic and we know that there is only one such minimal nonabelian group of order 24 and exponent 4. In particular, there is an element b 2 H1 Z such that b 2 D z, Œa; b D a2 b 2 D a2 z D z0 , where H10 D hz0 i, z0 is not a square in H1 , and ab is an involution so that 1 .H1 / D hz; z0 ; abi. Suppose now that H1 is not minimal nonabelian. Then there is a subgroup D Š D8 in H1 which covers H1 =hzi. Thus H1 D D hzi and H10 D D 0 D hz0 i. We have D \ Z D hai or D \ Z D hazi and all elements in H1 Z are involutions inverting ha; zi. Replacing a with az (if necessary), we may assume that D \ Z D hai. If t is an involution in D Z, then D D ha; t i, where z0 D a2 is a square in H1 . Suppose that H2 is nonabelian. Then H20 D hzi since H2 =hzi is elementary abelian. Let H4 be a minimal nonabelian subgroup of H2 so that H40 D hzi, H4 =hzi is elementary abelian and d.H4 / D 2. Thus H4 =hzi Š E4 and so H4 Š D8 . The subgroup H4 is normal in H2 and H2 centralizes H4 =hzi. We have ŒH1 ; H4 H1 \ H2 D hzi

354

Groups of prime power order

and so H1 also centralizes H4 =hzi and H4 is normal in H . Thus, H centralizes H4 =hzi and therefore there is no h 2 H inducing an outer automorphism on H4 (because otherwise such an element h would act nontrivially on H4 =hzi). It follows that CH .H4 / covers H=H4 . But H=H4 is nonabelian since H4 H2 and H=H2 Š H1 =hzi Š D8 . This contradicts Lemma 79.1. We have proved that H2 must be abelian and so H2 is either abelian of type .4; 2; : : : ; 2/ or elementary abelian. In any case, N D Z0 H2 D hz0 i H2 since z0 2 Z.G/. (ˇ1) Suppose that H2 is abelian of type .4; 2; : : : ; 2/, where Ã1 .N / D Ã1 .H2 / D hzi. Set E D 1 .H2 / so that 1 .N / D hz0 i E and all elements in H2 E are of order 4. Let h be an arbitrary element in H2 E so that h2 D z. We consider first the possibility that H1 is minimal nonabelian nonmetacyclic. Let x be any element of order 4 in H1 so that x 2 D z or x 2 D zz0 , where hz0 i D H10 . Suppose that Œx; h ¤ 1. By Lemma 79.2, Œx; h D x 2 h2 D x 2 z. On the other hand, Œx; h H1 \ H2 D hzi and so Œx; h D z which gives x 2 D 1, a contradiction. Hence Œx; h D 1 and since H1 is generated by its elements of order 4 (noting that 1 .H1 / Š E8 ) and H2 D hH2 Ei, we get ŒH1 ; H2 D f1g. We apply Lemma 79.1 N is the in the factor group HN D H=hzz0 i. We have HN1 Š D8 and hH1 ; hi D HN1 hhi N N N N N central product of H1 with hhi Š C4 , where H1 \ hhi D Z.H1 /, a contradiction. We consider now the possibility, where H1 D D hzi with D D ha; t j a4 D t 2 D 1; at D a1 i;

a2 D z0 ;

and

hz0 i D D 0 :

If Œa; h ¤ 1, then Œa; h 2 H1 \ H2 D hzi and so Œa; h D z. On the other hand, Lemma 79.2 implies Œa; h D a2 h2 D z0 z, a contradiction. Hence Œa; h D 1 and so a centralizes H2 D hH2 Ei. It follows that A D ZH2 D hai H2 is an abelian maximal subgroup of H with exp.A/ D 4 and A is normal in G. We have 1 .H1 / D H1 and so 1 .H / contains the maximal subgroup H1 E of H , where E D 1 .H2 / with jH2 W Ej D 2. If Œt; h D 1, then D D ha; t i Š D8 centralizes hhi Š C4 , contrary to Lemma 79.1. Hence Œt; h ¤ 1 and since Œt; h 2 H1 \ H2 D hzi, we get Œt; h D z with z D h2 . Thus, ht; hi Š D8 and so t h is an involution in H .H1 E/. We have proved that 1 .H / D H and since 1 .G/ H , we get also 1 .G/ D H . Also, H 0 D hz; z0 i Š E4 and therefore, by a result of A. Mann (Lemma 79.5), A is the unique abelian maximal subgroup of H . Take a subgroup G1 of G with H < G1 < G and jG1 W H j D 2. It follows that G1 must be a Wa -group with 1 .G1 / D H since jG1 W H j D 2 and A is the unique abelian maximal subgroup of H (with exp.A/ > 2). In that case G1 =A is cyclic. On the other hand, setting S1 D S \ G1 , we have G1 D HS1 and S1 \ H D Z. Thus, if g 2 S1 H , then g 2 2 Z A. This contradicts the fact that G1 =A Š C4 . (ˇ2) We have proved that H2 (and so also N D hz0 i H2 ) is elementary abelian. Suppose first that H1 is minimal nonabelian nonmetacyclic. In this case z0 is not a square in H1 , where hz0 i D H10 . There is b 2 H1 Z with b 2 D z, t D ab is an involution, and a2 D zz0 . Now, A D ZN D haiN is normal in G and .ht iN /s D

79

355

Nonmodular quaternion-free 2-groups

hbiN (recalling that S D hsi hzi) since G=N Š M2n , n 4, and so G acts nontrivially on the four-group H=N . The subgroup 1 .H / contains the maximal subgroup ht iN of H , where 1 .H1 / D hz; z0 ; t i Š E8 and N \ H1 D hz; z0 i. If t centralizes H2 , then ht iN is elementary abelian. But .ht iN /s D hbiN and hbiN is not elementary abelian, a contradiction. Hence t does not centralize H2 and so ŒH2 ; t D hzi since ŒH2 ; t H1 \ H2 D hzi. It follows that ht iN is nonabelian and so hbiN is also nonabelian. In particular, H 0 D hz; z0 i Š E4 , ŒH2 ; b D hzi, and so hbi is normal in H2 hbi. Let u be an involution in H2 with Œb; u D z so that hb; ui Š D8 and therefore bu is an involution. But bu 2 H .ht iN / and so 1 .H / D H . It follows that H must be a Wa - or Wb -group. In that case H must possess an abelian maximal subgroup M which is also unique (by a result of A. Mann since jH 0 j D 4). We have M H 0 D hz; z0 i. If M does not contain H2 , then M covers H=H2 which is nonabelian (since H=H2 Š H1 =hzi Š D8 ), a contradiction. Hence M H2 and so M hH 0 ; H2 i D N . Since ht iN and hbiN are nonabelian, we get that M D A D haiN is abelian (of exponent 4). Thus H is a Wa -group with respect to A and so the involution t must invert each element in A. In particular, at D a1 D aa2 D a.zz0 /. This is a contradiction since H10 D hz0 i. It remains to investigate the case, where H1 D D hzi with D D ha; t j a4 D t 2 D 1; at D a1 i;

a2 D z0 ;

hz0 i D D 0

and S D hsi hzi;

hsi \ H D hai;

.ht iN /s D hat iN

since G=N Š M2n , n 4. Obviously, in this case 1 .H / D H D 1 .G/. Suppose that t does not centralize H2 . Then Œt; H2 H1 \ H2 D hzi and so Œt; H2 D hzi and H 0 D hz; z0 i Š E4 . Then ht iN and hat iN D .ht iN /s are nonabelian. But H is a Wilkens group with 1 .H / D H and so H must have an abelian maximal subgroup U which is unique (by a result of A. Mann). If U does not contain H2 , U covers H=H2 and this is a contradiction since H=H2 Š D8 . Hence U hH 0 ; H2 i D N and so U D A D haiN is abelian of exponent 4. Thus, H is a Wa -group with respect to A. In particular, the involution t inverts each element in A and so t centralizes H2 , a contradiction. Hence t centralizes H2 and so ht iN is elementary abelian. In that case hat iN D .ht iN /s is also elementary abelian. In particular, D D ht; at i Š D8 centralizes H2 and so H D D H2 . It follows that G is a Wc -group, a contradiction. Our statement (iv) is proved. We have proved that the nonmodular factor group G=hzi (z 2 1 .Z.G//) (according to our statement (i)) is not isomorphic to any Wilkens group (according to (ii), (iii), and (iv)). This is a final contradiction and so the Main Theorem is proved.

80

Minimal non-quaternion-free 2-groups

Here we classify minimal non-quaternion-free 2-groups. A 2-group G is minimal non-quaternion-free if G is not quaternion-free but each proper subgroup of G is quaternion-free. Recall that a 2-group G is modular if and only if G is D8 -free. The main theorem is a consequence of results of the previous section. Theorem 80.1. Let G be a minimal non-quaternion-free 2-group. Then G possesses a unique normal subgroup N such that G=N Š Q8 . We have N < ˆ.G/ and so G=ˆ.G/ Š E4 and 1 .G/ ˆ.G/. If R is any G-invariant subgroup of index 2 in N , then H2 D G=R is the minimal nonabelian metacyclic group of order 24 and exponent 4: H2 D ha; b j a4 D b 4 D 1; ab D a1 i, where H2 =ha2 b 2 i Š Q8 , X=hb 2 i Š D8 , and X=ha2 i Š C4 C2 . In particular, if jGj > 8, then G has a normal subgroup S such that G=S Š D8 and so G is nonmodular. Proof. The group G possesses a normal subgroup N such that G=N Š Q8 . Suppose that A is a maximal subgroup of G not containing N . Then AN D G so A=.A\N / Š G=N Š Q8 , a contradiction. Thus, N < ˆ.G/ so d.G/ D 2. It follows from 1 .G=N / D ˆ.G=N / that 1 .G/ ˆ.G/. We may assume jGj > 8. Let R be any G-invariant subgroup of index 2 in N . We shall determine the structure of G=R. For that purpose we may assume R D f1g so that jN j D 2, N < ˆ.G/, N Z.G/, jˆ.G/j D 4, jGj D 24 and ˆ.G/=N D Z.G=N / D .G=N /0 , where G=N Š Q8 . By Theorem 1.2, G has no cyclic subgroups of index 2 so ˆ.G/ is a four-subgroup. In that case, N < Z.G/ since G is not of maximal class so ˆ.G/ D Z.G/ and G is minimal nonabelian. Since 1 .G/ D ˆ.G/ is a four-subgroup, G is metacyclic (of exponent 4). Since H2 D ha; b j a4 D b 4 D 1; ab D a1 i is the unique nonabelian metacyclic group of order 16 and exponent 4, we get G Š H2 . Next we do not assume that R D f1g. Let M ¤ N be a G-invariant subgroup of index 2 in ˆ.G/; then L D M \ N is of index 2 in N , by the product formula. By the previous paragraph, G=L Š H2 so G=M 2 fC4 C2 ; D8 g. In particular, G is nonmodular. This argument shows that N is the unique G-invariant subgroup of index 8 in G such that G=N Š Q8 . Let G be a minimal non-Q8 -free 2-group of order > 8 such that each proper subgroup of G is modular. Then G is a minimal nonmodular 2-group and such groups

80

Minimal non-quaternion-free 2-groups

357

have been classified in 78. Therefore, we may assume in the sequel that G has a maximal subgroup H which is nonmodular. Since H is Q8 -free, we are in a position to apply Theorem 79.7 classifying nonmodular Q8 -free 2-groups. Also, we assume that the reader is familiar with Propositions 79.8, 79.9 and 79.10 describing the Wilkens groups of types (a), (b), and (c), which appear in Theorem 79.7. Theorem 80.2. Let G be a minimal non-quaternion-free 2-group which has a nonmodular proper subgroup. Then G has one of the following properties: (i) 1 .G/ D ˆ.G/ D Aht i, where A is a maximal normal abelian subgroup of G with exp.A/ > 2, t is an involution inverting each element in A, and G=A Š Q8 , D8 , or C4 C2 . If G=A Š D8 or C4 C2 , then A is abelian of type .4; 2; : : : ; 2/. (ii) 1 .G/ D EE1 , where E ¤ E1 are the only maximal normal elementary abelian subgroups of 1 .G/ and j1 .G/ W Ej D 2, j1 .G/ W E1 j 2. We have either 1 .G/ D ˆ.G/ (and then G=1 .G/ Š E4 ) or G=1 .G/ Š Q8 . If G=1 .G/ Š Q8 , then also j1 .G/ W E1 j D 2 which implies 1 .G/ Š D8 E2s . (iii) 1 .G/ D E is elementary abelian and G=E is isomorphic to Q8 , M2n , n 4, or C2m C2 , m 1. If G=E Š M2n or C2m C2 with m 2, then 2 .G/ is abelian of type .4; 4; 2; : : : ; 2/. Proof. Let G be a minimal non-quaternion-free 2-group possessing a maximal subgroup H which is nonmodular. It follows that H is a Wilkens group of type (a), (b), or (c). In particular, H=1 .H / is cyclic, where 1 .H / D 1 .G/ since 1 .G/ ˆ.G/. If 1 .G/ is not elementary abelian, then we know (by the structure of the Wilkens group H ) that 1 .G/ is nonmodular and so in that case each maximal subgroup M of G is a Wilkens group. It follows that M=1 .G/ is cyclic, where 1 .M / D 1 .G/. In that case, X D G=1 .G/ Š E4 or Q8 . Here we have used a trivial fact that noncyclic 2-group X, all of whose maximal subgroups are cyclic, must be isomorphic to E4 or Q8 . (i) Suppose that H is a Wilkens group of type (a). Then H is a semidirect product H D hxi A, where A is a maximal normal abelian subgroup of H with exp.A/ > 2 and if t is the involution in hxi, then t inverts each element of A. We have 1 .H / D 1 .G/ D Aht i and A is a characteristic subgroup in H (Proposition 79.8) and so A is normal in G. By the previous paragraph, G=1 .G/ Š E4 or Q8 . However, if G=1 .G/ Š Q8 , then G=A (having a cyclic subgroup H=A of index 2) is of maximal class. But such a group does not have a proper factor group isomorphic to Q8 . Hence G=1 .G/ Š E4 and so 1 .G/ D ˆ.G/. Since d.G=A/ D 2, we get G=A Š Q8 , D8 or C4 C2 . Suppose that A is not a maximal normal abelian subgroup of G. Let B be a maximal normal abelian subgroup of G containing A so that B \ H D A, jB W Aj D 2, and G D hxi B with hxi \ B D f1g. Let b 2 B A and we compute .bb t /t D b t b D

358

Groups of prime power order

bb t 2 A, since b t 2 B A so that bb t D s 2 1 .A/. If s D 1, b t D b 1 and so t inverts each element of the abelian group B. It follows that G D hxi B is a Wilkens group of type (a). In that case G would be Q8 -free (Proposition 79.8), a contradiction. Hence s ¤ 1 and .bt /2 D bt bt D bb t D s shows that o.bt / D 4. Let a be an element of order 4 in A. We have abt D a1 and ha; bt i ¤ G (since a 2 ˆ.G/) and so ha; bt i is Q8 -free, contrary to Lemma 79.1. We have proved that A is a maximal normal abelian subgroup of G. Suppose that G=A Š D8 or C4 C2 . In that case there is a maximal subgroup K of G such that K=A Š E4 and 1 .K/ D 1 .G/. Then K is a Wilkens group of type (a) or (b) since jK W 1 .K/j D 2 (and so K cannot be a Wilkens group of type (c)). Suppose that K is a Wilkens group of type (a) with respect to a maximal normal abelian subgroup A1 of K with exp.A1 / > 2. We know that A1 1 .K/ D 1 .G/, j1 .G/ W A1 j D 2 and K=A1 is cyclic. Since K=A is noncyclic, we have A1 ¤ A. All elements in A1 A are involutions and if t0 2 A1 A, then t0 inverts and centralizes each element in A \ A1 and so A \ A1 is elementary abelian. It follows that exp.A1 / D 2, a contradiction. We have proved that K is a Wilkens group of type (b) with respect to a maximal normal elementary abelian subgroup E of K. We know that j1 .K/ W Ej D 2 (Proposition 79.9). Since 1 .K/ D 1 .G/, we have jA W A \ Ej D 2 and so A is abelian of type .4; 2; : : : ; 2/. (ii) Assume that H is a Wilkens group of type (b) with respect to E and j1 .H / W Ej D 2, where H=E is cyclic. We have 1 .H / D 1 .G/ and 1 .G/ ˆ.G/ so that 1 .G/ is nonmodular. Indeed, 1 .G/ has exactly two maximal normal elementary abelian subgroups E and E1 , where 1 .G/ D EE1 and so 1 .G/ is a Wilkens group of type (b) (and so nonmodular). By the first paragraph of the proof, G=1 .G/ Š E4 or Q8 . Suppose that G=1 .G/ Š Q8 . It is easy to see that E is not normal in G. Suppose false. Since G=E has a cyclic subgroup H=E of index 2 and G=1 .G/ Š Q8 , we get that G=E must be of maximal class. But there is no 2-group of maximal class having Q8 as a proper homomorphic image. Hence E is not normal in G and so for each x 2 G H , E x D E1 . In particular, j1 .G/ W E1 j D 2 and F D E \ E1 D Z.1 .G//. Take e 2 E F , e1 2 E1 F so that D D he; e1 i Š D8 and if V is a complement of hŒe; e1 i in F , then 1 .G/ D D V Š D8 E2s . (iii) Suppose that H is a Wilkens group of type (b) with respect to E and E D 1 .H / so that E D 1 .G/ (since E ˆ.G/), E is normal in G and G=E is noncyclic with the cyclic subgroup H=E of index 2. Let F=E be a proper subgroup of G=E such that F=E Š E4 . Then F is abelian. Indeed, since F ¤ G, F is Q8 -free. If F is not D8 -free, then F must be a Wilkens group. But then F=1 .F / must be cyclic. This is a contradiction since 1 .F / D E. It follows that F is D8 -free. Since exp.F / 4, F must be abelian (Proposition 79.6). Since each proper subgroup of G=E is Q8 -free, G=E cannot be semidihedral or Q2m with m 4. Suppose that G=E Š D2n , n 3. In that case G=E is generated by its four-subgroups and so E Z.G/. This is a contradiction because in that case H

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Minimal non-quaternion-free 2-groups

359

would be abelian (noting that H=E is cyclic). We have proved that G=E is isomorphic to Q8 , M2n , n 4, or C2m C2 , m 1. Suppose that G=E is not isomorphic to Q8 or C2 C2 . Set F=E D 1 .G=E/ so that F=E Š E4 . By the above, F is abelian. Obviously, F D 2 .G/ is abelian of type .4; 4; 2; : : : ; 2/. (iv) Finally, assume that H is a Wilkens group of type (c). We have 1 .H / D 1 .G/ Š D8 E2s and H=1 .G/ is cyclic of order 4 (Proposition 79.10). The subgroup 1 .G/ has exactly three abelian maximal subgroups E1 , E2 , A, where E1 and E2 are elementary abelian and A is abelian of type .4; 2; : : : ; 2/. By the structure of H , E1 and E2 are not normal in H and H=E1 \E2 Š M2n , n 4. By the first paragraph of the proof, G=1 .G/ Š Q8 since jH=1 .G/j 4. Thus H=E1 \ E2 Š M24 . On the other hand, 1 .G/ is normal in G and so NG .E1 / D NG .E2 / D K is a maximal subgroup of G distinct from H . Since K 1 .G/, K is nonmodular and therefore K is a Wilkens group. But K cannot be a Wilkens group of type (c) since E1 and E2 are normal in K. Hence K is either a Wilkens group of type (a) with respect to A or K is a Wilkens group of type (b) with respect to E1 or E2 . The group G with such a maximal subgroup K has been considered in (i) and (ii) and so we do not get here new possibilities for the structure of G. Our theorem is proved. Let H2 D ha; b j a4 D b 4 D 1; ab D a1 i; H1 D ha; b j a4 D b 4 D 1; c D Œa; b; c 2 D 1; Œa; c D Œb; c D 1i: Exercise 1. Let G D H1 . Then 1, a2 , b 2 are only squares in G. In particular, a2 b 2 is not a square. Solution. Let g D ai b j c k 2 G. Then g 2 D .ai b j /2 . If i is even, then g 2 D b 2j 2 f1; b 2 g. If j is even, then g 2 D a2i 2 f1; a2 g. If i and j are odd, then g 2 D .ab/2 D ab 2 ab D ab 2 ac D a2 b 2 c. Exercise 2. If G D H1 , then G=ha2 b 2 i Š H2 . Solution. Since G=ha2 b 2 i is nonabelian of exponent 4, it suffices to show that it is metacyclic. Note that a2 b 2 2 ˆ.G/. Let Ai , i D 1; 2; 3, be all maximal subgroups of G; then these subgroups are abelian of type .4; 2; 2/. It follows that ha2 b 2 i is a direct factor of Ai , i D 1; 2; 3, so all maximal subgroups of G=ha2 b 2 i are abelian of type .4; 2/. Then, by Lemma 65.1, G=ha2 b 2 i is metacyclic, as was to be shown. Exercise 3. Classify Q8 -free minimal nonabelian 2-groups. (Hint. G is Q8 -free if and only if G=Ã2 .G/ is metacyclic.) Exercise 4. (a) A direct product of two Q8 -free 2-groups is not necessarily Q8 -free. (Hint. (a) The group D8 C4 has an epimorphic image isomorphic to Q8 C4 .)

360

Groups of prime power order

(b) A direct product of two D8 -free 2-groups is not necessarily D8 -free. (Hint. The group Q8 C4 has an epimorphic image isomorphic to D8 C4 .) (c) Suppose that G is Q8 -free. If D8 Š D G, then CG .D/ is elementary abelian. (d) Suppose that G is D8 -free. If Q8 Š Q G, then CG .Q/ is elementary abelian. Problem 1. Study the structure of nonabelian H2 -free 2-groups. Problem 2. Study the structure of minimal non-H2 -free 2-groups. Let S.p 3 / be a nonabelian group of order p 3 and exponent p; then p > 2. Obviously, S.p 3 /-free p-groups are metacyclic (see Theorem 9.11). It is easy to check that if G is a minimal non-S.p 3 /-free p-group, then p D 3 and G is a minimal nonmetacyclic group of order 34 (see Theorem 69.1).

81

Maximal abelian subgroups in 2-groups

Abelian subgroups in 2-groups G play an important role. Therefore, it is not very surprising that our assumption that every two distinct maximal abelian subgroups have cyclic intersection determines completely the structure of G. We obtain five classes of 2-groups with this property. More precisely, we prove here the following result. Theorem 81.1. Let G be a nonabelian 2-group in which any two distinct maximal abelian subgroups have cyclic intersection. Then Z.G/ is cyclic, each abelian subgroup of G is of rank at most 2, the intersection of any two distinct maximal abelian subgroups is equal Z.G/, and G has (at least) one abelian subgroup of index 2. Moreover, G is isomorphic to one of the following groups. (a) Group of maximal class (dihedral, semidihedral or generalized quaternion). n1

(b) M2n D ha; t j a2

n2

D t 2 D 1; n 4; at D a1C2

i.

(c) G D D C (central product), where D Š D2n , n 3, is dihedral of order 2n , C Š C2m , m 2, is cyclic of order 2m and D \ C D Z.D/. m

n1

D 1; n (d) G D hx; t j .xt /2 D a; a2 D t 2 D 1; m 2; x 2 D ab; b 2 t 1 x 2m1 2n2 3; b D b ; Œa; x D Œa; t D 1; t D t b; a Db i, where jGj D 2mCn , 0 m 2, n 3, Z.G/ D hai Š C2m , G D hbi Š C2n1 , and M D hx; ai is a unique abelian maximal subgroup of G. We have CG .t / D ht i hai Š C2 C2m and hb; t i Š D2n . D h2 ; hg D h1 i, (e) G D hg; h j g 2 D h2 D 1; m 3; n 3; g 2 mCn1 2n1 where G is metacyclic, jGj D 2 since hgi \ hhi D hg i Š C2 . Also, 2 0 2 2 Z.G/ D hg i Š C2n1 , G D hh i Š C2m1 and M D hh; g i is a unique abelian maximal subgroup of G. n

m

n1

m1

The more general problem to determine the structure of a nonabelian p-group G such that A \ B D Z.G/ for any two distinct maximal abelian subgroups A and B is very difficult. First we show that a p-group G has this property if and only if CG .x/ is abelian for each x 2 G Z.G/ (Theorem 81.2). Then we show that such a 2-group G has either an abelian subgroup of index 2 or G is of class 2 and G 0 is elementary abelian (Theorem 81.3). In Corollary 81.5 we get a new result for an arbitrary finite 2-group. We also classify 2-groups G such that A=Z.G/ is cyclic for each maximal abelian subgroup A of G (a problem of Heineken–Mann). It is surprising that such groups

362

Groups of prime power order

have the property that CG .x/ is abelian for each element x 2 G Z.G/. Then we may use our Theorems 81.2 and 81.3 to classify such groups (Theorem 81.4). In this classification we distinguish the cases, where G has an abelian subgroup of index 2 and the case where jG W Aj > 2 for each maximal abelian subgroup A of G. Proof of Theorem 81.1. Let G be a nonabelian 2-group in which any two distinct maximal abelian subgroups have cyclic intersection. Since Z.G/ is contained in each maximal abelian subgroup of G, it follows that Z.G/ is cyclic. Suppose that G possesses an elementary abelian subgroup E of order 8. Let A be a maximal abelian subgroup containing E and set F D 1 .A/ so that E F . Let B G be such that A < B and jB W Aj D 2 and let x 2 B A. Then x 2 2 A and therefore x induces on F an automorphism of order 2. It follows that jCF .x/j 4 and the abelian subgroup CF .x/hxi is contained in a maximal abelian subgroup C which is distinct from A since x 62 A. But A \ C CF .x/ and so A \ C is noncyclic, a contradiction. We have proved that each abelian subgroup of G is of rank at most 2. We may assume that G is not of maximal class (case (a) of Theorem 81.1) and so G possesses a normal four-subgroup U . Set M D CG .U / so that jG W M j D 2 since Z.G/ is cyclic. Let A be a maximal abelian subgroup of G which contains U so that A M . Suppose that A ¤ M and let y 2 M A be such that y 2 2 A. Let B be a maximal abelian subgroup of G containing the abelian subgroup U hyi. Then B ¤ A (since y 62 A) and A \ B U , a contradiction. We have proved that whenever U is a normal four-subgroup of G, then M D CG .U / is an abelian maximal subgroup of G. If x is any element in G M , then CM .x/ D Z.G/ and Z.G/hxi is a maximal abelian subgroup of G. Thus, the intersection of any two distinct maximal abelian subgroups of G is equal to Z.G/ and this statement also holds for 2-groups of maximal class. Suppose that G has more than one normal four-subgroup. By Theorem 50.2, G D D C with D Š D8 , D \ C D Z.D/ and C is either cyclic of order 4 or C is of maximal class (distinct from D8 ). Let U be a four-subgroup in D so that U is normal in G. By the above, CG .U / is abelian and so C must be cyclic. We have obtained a group stated in part (c) of Theorem 81.1. In the sequel we assume that G has a unique normal four-subgroup U and set M D CG .U / so that M is an abelian maximal subgroup of rank 2 with 1 .M / D U . (i) First assume 2 .G/ 6 M . Then there is an element y 2 G M of order 4 so that y 2 2 U . We have U hyi Š D8 and so there is an involution t 2 G M . Since t does not centralize U and M is abelian of rank 2, we get CG .t / D ht i CM .t /, where CM .t / is cyclic of order 2m , m 2. Indeed, if m D 1, then G is of maximal class. Also, we have t 62 ˆ.G/, G has no elementary abelian subgroups of order 8 and G is not isomorphic to M2s , s 4, since M2s has only three involutions. We are now in a position to use Theorem 48.1. It follows that G has a subgroup S of index 2, n1 D t 2 D 1; b t D b 1 i Š D2n , where S D AL, L is normal in G, L D hb; t j b 2 n 3, A D hai Š C2m , m 2, A \ L D Z.L/ D hzi, Œa; t D 1, CG .t / D ht i hai,

81 Maximal abelian subgroups in 2-groups

363

1 .G/ D 1 .S / D 2 .A/ L, 2 .A/ \ L D Z.L/ and if jG W S j D 2, then there is an element x 2 G S such that t x D t b. Since hbi is a unique cyclic subgroup of index 2 in L, hbi is normal in G. Set B D 2 .A/ L D 1 .G/,

Editors V. P. Maslov, Academy of Sciences, Moscow W. D. Neumann, Columbia University, New York R. O. Wells, Jr., International University, Bremen

Groups of Prime Power Order Volume 2 by

Yakov Berkovich and Zvonimir Janko

≥

Walter de Gruyter · Berlin · New York

Authors Yakov Berkovich Jerusalem str. 53, apt. 15 Afula 18251 Israel E-Mail: [email protected]

Zvonimir Janko Mathematisches Institut Ruprecht-Karls-Universität Heidelberg Im Neuenheimer Feld 288 69120 Heidelberg E-Mail: [email protected]

Mathematics Subject Classification 2000: 20-02, 20D15, 20E07 Key words: Finite p-group theory, minimal nonabelian subgroups, metacyclic subgroups, extraspecial subgroups, equally partitioned groups, p-groups with given maximal subgroups, 2-groups with few cyclic subgroups of given order, Ward’s theorem on quaternion-free groups, 2-groups with small centralizers of an involution, Blackburn’s theorem on minimal nonmetacyclic groups.

앝 Printed on acid-free paper which falls within the guidelines 앪 of the ANSI to ensure permanence and durability.

ISSN 0938-6572 ISBN 978-3-11-020419-3 Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available in the Internet at http://dnb.d-nb.de. 쑔 Copyright 2008 by Walter de Gruyter GmbH & Co. KG, 10785 Berlin, Germany. All rights reserved, including those of translation into foreign languages. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage or retrieval system, without permission in writing from the publisher. Typeset using the authors’ TeX files: Kay Dimler, Müncheberg. Printing and binding: Hubert & Co. GmbH & Co. KG, Göttingen. Cover design: Thomas Bonnie, Hamburg.

Contents

List of definitions and notations

. . . . . . . . . . . . . . . . . . . . . . . . . viii

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

xiv

46

Degrees of irreducible characters of Suzuki p-groups

. . . . . . . . . .

1

47

On the number of metacyclic epimorphic images of finite p-groups . . .

14

48

On 2-groups with small centralizer of an involution, I

. . . . . . . . . .

19

49

On 2-groups with small centralizer of an involution, II . . . . . . . . . .

28

50

Janko’s theorem on 2-groups without normal elementary abelian subgroups of order 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

51

2-groups with self centralizing subgroup isomorphic to E8 . . . . . . . .

52

52

2-groups with 2 -subgroup of small order . . . . . . . . . . . . . . . .

75

53

2-groups G with c2 .G/ D 4 . . . . . . . . . . . . . . . . . . . . . . . .

96

54

2-groups G with cn .G/ D 4, n > 2 . . . . . . . . . . . . . . . . . . . . 109

55

2-groups G with small subgroup hx 2 G j o.x/ D 2n i . . . . . . . . . . 122

56

Theorem of Ward on quaternion-free 2-groups . . . . . . . . . . . . . . 134

57

Nonabelian 2-groups all of whose minimal nonabelian subgroups are isomorphic and have exponent 4 . . . . . . . . . . . . . . . . . . . . . . . 140

58

Non-Dedekindian p-groups all of whose nonnormal subgroups of the same order are conjugate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

59

p-groups with few nonnormal subgroups . . . . . . . . . . . . . . . . . 150

60

The structure of the Burnside group of order 212 . . . . . . . . . . . . . 151

61

Groups of exponent 4 generated by three involutions . . . . . . . . . . . 163

62

Groups with large normal closures of nonnormal cyclic subgroups . . . . 169

63

Groups all of whose cyclic subgroups of composite orders are normal . . 172

vi

Groups of prime power order

64

p-groups generated by elements of given order . . . . . . . . . . . . . . 179

65

A2 -groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188

66

A new proof of Blackburn’s theorem on minimal nonmetacyclic 2-groups 197

67

Determination of U2 -groups . . . . . . . . . . . . . . . . . . . . . . . . 202

68

Characterization of groups of prime exponent . . . . . . . . . . . . . . . 206

69

Elementary proofs of some Blackburn’s theorems

70

Non-2-generator p-groups all of whose maximal subgroups are 2-generator 214

71

Determination of A2 -groups . . . . . . . . . . . . . . . . . . . . . . . . 233

72

An -groups, n > 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248

73

Classification of modular p-groups . . . . . . . . . . . . . . . . . . . . 257

74

p-groups with a cyclic subgroup of index p 2 . . . . . . . . . . . . . . . 274

75

Elements of order 4 in p-groups

76

p-groups with few A1 -subgroups . . . . . . . . . . . . . . . . . . . . . 282

77

2-groups with a self-centralizing abelian subgroup of type .4; 2/ . . . . . 316

78

Minimal nonmodular p-groups

79

Nonmodular quaternion-free 2-groups

. . . . . . . . . . . . . . . . . . 334

80

Minimal non-quaternion-free 2-groups

. . . . . . . . . . . . . . . . . . 356

81

Maximal abelian subgroups in 2-groups . . . . . . . . . . . . . . . . . . 361

82

A classification of 2-groups with exactly three involutions . . . . . . . . 368

83

p-groups G with 2 .G/ or 2 .G/ extraspecial

84

2-groups whose nonmetacyclic subgroups are generated by involutions . 399

85

2-groups with a nonabelian Frattini subgroup of order 16

86

p-groups G with metacyclic 2 .G/

87

2-groups with exactly one nonmetacyclic maximal subgroup . . . . . . . 412

88

Hall chains in normal subgroups of p-groups . . . . . . . . . . . . . . . 437

89

2-groups with exactly six cyclic subgroups of order 4

. . . . . . . . . . . . 209

. . . . . . . . . . . . . . . . . . . . 277

. . . . . . . . . . . . . . . . . . . . . . 323

. . . . . . . . . . . . . 396

. . . . . . . . 402

. . . . . . . . . . . . . . . . . . . 406

. . . . . . . . . . 454

Contents

vii

90

Nonabelian 2-groups all of whose minimal nonabelian subgroups are of order 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463

91

Maximal abelian subgroups of p-groups

92

On minimal nonabelian subgroups of p-groups . . . . . . . . . . . . . . 474

. . . . . . . . . . . . . . . . . 467

Appendix A.16 Some central products . . . . . . . . . . . . . . . . . . . . . . . . . . . 485 A.17 Alternate proofs of characterization theorems of Miller and Janko on 2groups, and some related results . . . . . . . . . . . . . . . . . . . . . . 492 A.18 Replacement theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . 501 A.19 New proof of Ward’s theorem on quaternion-free 2-groups . . . . . . . . 506 A.20 Some remarks on automorphisms . . . . . . . . . . . . . . . . . . . . . 509 A.21 Isaacs’ examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512 A.22 Minimal nonnilpotent groups

. . . . . . . . . . . . . . . . . . . . . . . 516

A.23 Groups all of whose noncentral conjugacy classes have the same size . . 519 A.24 On modular 2-groups

. . . . . . . . . . . . . . . . . . . . . . . . . . . 522

A.25 Schreier’s inequality for p-groups . . . . . . . . . . . . . . . . . . . . . 526 A.26 p-groups all of whose nonabelian maximal subgroups are either absolutely regular or of maximal class . . . . . . . . . . . . . . . . . . . . . . . . 529 Research problems and themes II

. . . . . . . . . . . . . . . . . . . . . . . . 531

Author index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593 Subject index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594

List of definitions and notations

Set theory jM j is the cardinality of a set M (if G is a finite group, then jGj is called its order). x 2 M (x 62 M ) means that x is (is not) an element of a set M . N M (N 6 M ) means that N is (is not) a subset of the set M ; moreover, if M ¤ N M we write N M. ¿ is the empty set. N is called a nontrivial subset of M , if N ¤ ¿ and N M . If N M we say that N is a proper subset of M . M \ N is the intersection and M [ N is the union of sets M and N . If M; N are sets, then N M D fx 2 N j x 62 M g is the difference of N and M . Z is the set (ring) of integers: Z D f0; ˙1; ˙2; : : : g. N is the set of all natural numbers. Q is the set (field) of all rational numbers. R is the set (field) of all real numbers. C is the set (field) of all complex numbers.

Number theory and general algebra p is always a prime number. is a set of primes; 0 is the set of all primes not contained in . m, n, k, r, s are, as a rule, natural numbers. .m/ is the set of prime divisors of m; then m is a -number. np is the p-part of n, n is the -part of n. .m; n/ is the greatest common divisor of m and n. m j n should be read as: m divides n. m − n should be read as: m does not divide n.

ix

List of definitions and notations

GF.p m / is the finite field containing p m elements. F is the multiplicative group of a field F. L.G/ is the lattice of all subgroups of a group G. If n D p1˛1 : : : pk˛k is the standard prime decomposition of n, then .n/ D

Pk

iD1 ˛i .

Groups We consider only finite groups which are denoted, with a pair exceptions, by upper case Latin letters. If G is a group, then .G/ D .jGj/. G is a p-group if jGj is a power of p; G is a -group if .G/ . G is, as a rule, a finite p-group. H G means that H is a subgroup of G. H < G means that H G and H ¤ G (in that case H is called a proper subgroup of G). f1g denotes the group containing only one element. H is a nontrivial subgroup of G if f1g < H < G. H is a maximal subgroup of G if H < G and it follows from H M < G that H D M. H E G means that H is a normal subgroup of G; moreover, if, in addition, H ¤ G we write H GG and say that H is a proper normal subgroup of G. Expressions ‘normal subgroup of G’ and ‘G-invariant subgroup’ are synonyms. H G G is called a nontrivial normal subgroup of G provided H > f1g. H is a minimal normal subgroup of G if (a) H E G; (b) H > f1g; (c) N G G and N < H implies N D f1g. Thus, the group f1g has no minimal normal subgroup. G is simple if it is a minimal normal subgroup of G (so jGj > 1). H is a maximal normal subgroup of G if H < G and G=H is simple. The subgroup generated by all minimal normal subgroups of G is called the socle of G and denoted by Sc.G/. We put, by definition, Sc.f1g/ D f1g. NG .M / D fx 2 G j x 1 M x D M g is the normalizer of a subset M in G. CG .x/ is the centralizer of an element x in G W CG .x/ D fz 2 G j zx D xzg. T CG .M / D x2M CG .x/ is the centralizer of a subset M in G. If A B and A; B E G, then CG .B=A/ D H , where H=A D CG=A .B=A/.

x

Groups of prime power order

A wr B is the wreath product of the ‘passive’ group A and the transitive permutation group B (in what follows we assume that B is regular); B is called the active factor of the wreath product). Then the order of that group is jAjjBj jBj. Aut.G/ is the group of automorphisms of G (the automorphism group of G). Inn.G/ is the group of all inner automorphisms of G. Out.G/ D Aut.G/=Inn.G/, the outer automorphism group of G. If a; b 2 G, then ab D b 1 ab. An element x 2 G inverts a subgroup H G if hx D h1 for all h 2 H . If M G, then hM i D hx j x 2 M i is the subgroup of G generated by M . M x D x 1 M x D fy x j y 2 M g for x 2 G and M G. Œx; y D x 1 y 1 xy D x 1 x y is the commutator of elements x; y of G. If M; N G then ŒM; N D hŒx; y j x 2 M; y 2 N i is a subgroup of G. o.x/ is the order of an element x of G. An element x 2 G is a -element if .o.x// . G is a -group, if .G/ . Obviously, G is a -group if and only if all of its elements are -elements. G 0 is the subgroup generated by all commutators Œx; y, x; y 2 G (i.e., G 0 D ŒG; G), G .2/ D ŒG 0 ; G 0 D G 00 D .G 0 /0 , G .3/ D ŒG 00 ; G 00 D .G 00 /0 and so on. G 0 is called the commutator (or derived) subgroup of G. T Z.G/ D x2G CG .x/ is the center of G. Zi .G/ is the i -th member of the upper central series of G; in particular, Z0 .G/ D f1g, Z1 .G/ D Z.G/. Ki .G/ is the i -th member of the lower central series of G; in particular, K2 .G/ D G 0 . We have Ki .G/ D ŒG; : : : ; G (i 1 times). We set K1 .G/ D G. If G is nonabelian, then .G/=K3 .G/ D Z.G=K3 .G//. M.G/ D hx 2 G j CG .x/ D CG .x p / is the Mann subgroup of a p-group G. Sylp .G/ is the set of p-Sylow subgroups of an arbitrary finite group G. Sn is the symmetric group of degree n. An is the alternating group of degree n †p n is a Sylow p-subgroup of Sp n . GL.n; F / is the set of all nonsingular n n matrices with entries in a field F , the n-dimensional general linear group over F , SL.n; F / D fA 2 GL.n; F / j det.A/ D 1 2 F g, the n-dimensional special linear group over F .

List of definitions and notations

xi

T If H T G, then HG D x2G x 1 H x is the core of the subgroup H in G and H G D H N EG N is the normal closure or normal hull of H in G. Obviously, HG E G. If G is a p-group, then p b.x/ D jG W CG .x/j; b.x/ is said to be the breadth of x 2 G, where G is a p-group; b.G/ D max fb.x/ j x 2 Gg is the breadth of G. ˆ.G/ is the Frattini subgroup of G (= the intersection of all maximal subgroups of G), ˆ.f1g/ D f1g, p d.G/ D jG W ˆ.G/j. i D fH < G j ˆ.G/ H; jG W H j D p i g, i D 1; : : : ; d.G/, where G > f1g. If H < G, then 1 .H / is the set of all maximal subgroups of H . exp.G/ is the exponent of G (the least common multiple of the orders of elements of G). If G is a p-group, then exp.G/ D max fo.x/ j x 2 Gg. k.G/ is the number of conjugacy classes of G (D G-classes), the class number of G. Kx is the G-class containing an element x (sometimes we also write ccl G .x/). Cm is the cyclic group of order m. G m is the direct product of m copies of a group G. A B is the direct product of groups A and B. A B is a central product of groups A and B, i.e., A B D AB with ŒA; B D f1g. Ep m D Cpm is the elementary abelian group of order p m . G is an elementary abelian p-group if and only if it is a p-group > f1g and G coincides with its socle. Next, f1g is elementary abelian for each prime p. A group G is said to be homocyclic if it is a direct product of isomorphic cyclic subgroups (obviously, elementary abelian p-groups are homocyclic). ES.m; p/ is an extraspecial group of order p 1C2m (a p-group G is said to be extraspecial if G 0 D ˆ.G/ D Z.G/ is of order p). Note that for each m 2 N, there are exactly two nonisomorphic extraspecial groups of order p 2mC1 . S.p 3 / is a nonabelian group of order p 3 and exponent p > 2. A special p-group is a nonabelian p-group G such that G 0 D ˆ.G/ D Z.G/ is elementary abelian. Direct products of extraspecial p-groups are special. D2m is the dihedral group of order 2m; m > 2. Some authors consider E22 as the dihedral group D4 . Q2m is the generalized quaternion group of order 2m 23 . SD2m is the semidihedral group of order 2m 24 . Mp m is a nonabelian p-group containing exactly p cyclic subgroups of index p.

xii

Groups of prime power order

cl.G/ is the nilpotence class of a p-group G. dl.G/ is the derived length of a p-group G. CL.G/ is the set of all G-classes. A p-group of maximal class is a nonabelian group G of order p m with cl.G/ D m 1. m .G/ D hx 2 G j o.x/ p m i, m .G/ D hx 2 G j o.x/ D p m i and Ãm .G/ D m hx p j x 2 Gi. A p-group is absolutely regular if jG=Ã1 .G/j < p p . A p-group is thin if it is either absolutely regular or of maximal class. G D A B is a semidirect product with kernel B and complement A. A group G is an extension of N E G by a group H if G=N Š H . A group G splits over N if G D H N with H G and H \ N D f1g (in that case, G is a semidirect product of H and N with kernel N ). H # D H feH g, where eH is the identity element of the group H . If M G, then M # D M feG g. An automorphism ˛ of G is regular (D fixed-point-free) if it induces a regular permutation on G # (a permutation is said to be regular if it has no fixed points). An involution is an element of order 2 in a group. A section of a group G is an epimorphic image of some subgroup of G. If F D GF.p n /, then we write GL.m; p n /; SL.m; p n /; : : : instead of GL.m; F /, SL.m; F /; : : : . cn .G/ is the number of cyclic subgroups of order p n in a p-group G. sn .G/ is the number of subgroups of order p n in a p-group G. en .G/ is the number of subgroups of order p n and exponent p in G. An -group is a p-group G all of whose subgroups of index p n are abelian but G contains a nonabelian subgroup of index p n1 . In particular, A1 -group is a minimal nonabelian p-group for some p. ˛n .G/ is the number of An -subgroups in a p-group G.

Characters and representations Irr.G/ is the set of all irreducible characters of G over C. A character of degree 1 is said to be linear. Lin.G/ is the set of all linear characters of G (obviously, Lin.G/ Irr.G/).

List of definitions and notations

xiii

Irr1 .G/ D Irr.G/ Lin.G/ is the set of all nonlinear irreducible characters of G; n.G/ D jIrr1 .G/j. .1/ is the degree of a character of G, H is the restriction of a character of G to H G. G is the character of G induced from the character of some subgroup of G. N is a character of G defined as follows: .x/ N D .x/ (here wN is the complex conjugate of w 2 C). Irr. / is the set of irreducible constituents of a character of G. If is a character of G, then ker. / D fx 2 G j .x/ D .1/g is the kernel of a character . Z. / D fx 2 G j j .x/j D .1/g is the quasikernel of . If N E G, then Irr.G j N / D f 2 Irr.G/ j N — ker. /g. P h ; i D jGj1 x2G .x/ .x 1 / is the inner product of characters and of G. IG ./ D hx 2 G j x D i is the inertia subgroup of 2 Irr.H / in G, where H G G. 1G is the principal character of G (1G .x/ D 1 for all x 2 G). M.G/ is the Schur multiplier of G. cd.G/ D f .1/ j 2 Irr.G/g. mc.G/ D k.G/=jGj is the measure of commutativity of G. P T.G/ D 2Irr.G/ .1/; f.G/ D T.G/=jGj.

Preface

This is the second part of the book. Sections 48–57, 60, 61, 66, 67, 70, 71, 73–75, 77– 87, 89–92 and Appendix 19 are written by the second author, all other sections – by the first author. This volume contains a number of very strong results on 2-groups due to the second author. All exercises and about all problems are due to the first author. All material of this part is appeared in the book form at the first time. Some outstanding problems of p-group theory are solved in this volume: (i) classification of 2-groups with exactly three involutions, (ii) classification of 2-groups containing exactly one nonmetacyclic maximal subgroup, (iii) classification of 2-groups G containing an involution t such that CG .t / D ht i Q, where Q contains only one involution, (iv) classification of 2-groups all of whose minimal nonabelian subgroups have the same order 8, (v) classification of 2-groups of rank 3 all of whose maximal subgroups are of rank 2, (vi) classification of 2-groups containing selfcentralizing noncyclic abelian subgroups of order 8, and so on. There are, in this part, a number of new proofs of known important results: (a) Blackburn’s classification of minimal nonmetacyclic groups (we presented, in Sec. 66 and 69, two different proofs), (b) classification of p-groups all of whose subgroups of index p 2 are abelian, (c) Ward’s theorem on quaternion-free 2-groups (we presented two different proofs), (d) classification of p-groups with cyclic subgroup of index p 2 , (e) Kazarin’s classification of p-groups all of whose cyclic subgroups of order > p are normal, (f) Iwasawa’s classification of modular p-groups, and so on. Some results proved in this part have no analogs in existing literature: (a) classification of 2-groups all of whose minimal nonabelian subgroups are isomorphic and have order 16, (b) study the 2-groups with at most 1 C p C p 2 minimal nonabelian subgroups,

Preface

xv

(c) classification of 2-groups all of whose nonabelian two-generator subgroups are of maximal class, (d) classification of 2-groups G with j2 .G/j 24 and j2 .G/j D 24 , (e) classification of p-groups G with 2 .G/ or 2 .G/ is metacyclic (extraspecial), (f) classification of 2-groups all of whose nonabelian subgroups are generated by involutions, and so on. As the previous part, this one contains a great number of open problems posed, as a rule, by the first author; some of these problems are solved and solutions are presented below. The first author is indebted to Avinoam Mann for numerous useful discussions and help. The correspondence with Martin Isaacs allowed us to acquaint the reader with a number of his old and new important results. Moreover, Mann and Isaacs familiarized us with a number of their papers prior of publication. Noboru Ito read a number of sections and all appendices and made numerous useful remarks and suggestions. The help of Lev Kazarin was very important and allowed us to improve a number of places of the book, especially, in 46 and 63; he also acquainted the first author with a fragment of his PhD thesis (see 65, 71). The first author also indebted to Gregory Freiman, Marcel Herzog (both at Tel-Aviv University), Moshe Roitman and Izu Vaisman (both at University of Haifa) for help and support. The publication of the book gives us great pleasure. We are grateful to the publishing house of Walter de Gruyter and all who promoted the publication, among of them Prof. M. Hazewinkel, Dr. R. Plato and K. Dimler, for their support and competent handling of the project.

46

Degrees of irreducible characters of Suzuki p-groups

The results of this section are due to I. A. Sagirov [Sag1, Sag2]. Throughout this section we use the following notation: F D GF.p m /, m > 1; is an automorphism of F of order k for some divisor k > 1 of m (recall that the group of automorphisms of F is cyclic of order m whose generator is a 7! ap for all a 2 F); n D m .< m/. k n Let, for example, W a 7! ap (a 2 F). The set of fixed points of is a subfield n F of F satisfying ap D 1 so containing p n elements (for example, if k D m, then F D F0 , the prime subfield of F). Let F be the (cyclic) multiplicative group of F. Next, let Irr1 .G/ denote the set of all nonlinear irreducible characters of G. Put cd.G/ D f .1/ j 2 Irr.G/g. Next, Irr.t/ .G/ denotes the number of characters of degree t in Irr.G/. Definition. The Suzuki p-group Ap .m; / is the set F F with multiplication defined as follows: .a; b/.c; d / D .a C c; b C d C a .c//. Let G D Ap .m; /; then jGj D jFj2 D p 2m . It follows from the definition that elements .a; b/ and .c; d / commute if and only if c .a/ D a .c/, i.e., either a D 0 or .c=a/ D c=a so c D au for some u 2 F . The identity element of G is .0; 0/ and .a; b/1 D .a; b C a .a//. Since so defined multiplication is associative, G indeed is a group. If .c; d / 2 Z.G/ and .a; b/ 2 G, then c D au for u 2 F and all a 2 F which implies c D 0 since k > 1. Thus, Z.G/ D f.0; d / j d 2 Fg. We have Z.G/ Š Ep m . It is easy to prove, by induction that .a; b/n D .na; nb C n2 a .a//. Taking n D p, we get .a; b/p 2 Z.G/; moreover, if p > 2, then exp.G/ D p. 1o . Throughout this subsection p D 2 and G D A.m; / D A2 .m; /. Our aim is to find cd.G/ and the number of irreducible characters of every degree s 2 cd.G/. Theorem 46.1 ([Sag1]). Suppose that p D 2, G D A.m; /, where m > 1 and is an automorphism of the field F D GF.2m / of order k > 1 and n D m . Then one of the k following holds: 1

(a) If k is odd, then cd.G/ D f1; 2 2 .mn/ g. (b) If k D 2, then cd.G/ D f1; 2m=2 g D f1; 2n g. (c) If k > 2 is even, then cd.G/ D f1; 2m=2 ; 2.m=2/n g, jIrr.2m=2 / .G/j D and jIrr.2.m=2/n / .G/j D

.2m 1/22n 2n C1 .

.2m 1/2n 2n C1

2

Groups of prime power order s

If x 2 F then, since is an automorphism of F of order k > 1, then .x/ D x 2 for some nonnegative s independent of x and k .x/ D x, x 2 F. On the other hand, sk sk k .x/ D x 2 so x 2 D x, and this is true for each x 2 F. If x is a primitive element of F, it follows that 2m 1 divides 2sk 1 so m.D nk/ divides sk (see Lemma 46.5 below) hence n divides s, and we conclude that s D nt , where .t; k/ D 1 since nt o. / D k. Thus, .x/ D x 2 . Therefore, the number of automorphisms of F of order k equals '.k/, where './ is the Euler’s totient function. Then fa 2 F j .a/ D nt ag D F is the set of elements a 2 F such that a2 D a and the cardinality of that set is 2n . As we have shown, if a ¤ 0, then CG ..a; b// D f.az; u/ j z 2 F ; u 2 Fg so jCG ..a; b//j D 2mCn . In particular, the size of every noncentral G-class equals 22m D 2mn . If a 2 F f0g, then .a; b/2 D .0; a .a// ¤ .0; 0/ so Z.G/ D 1 .G/. 2mCn Therefore, since exp.G/ D 4, we get 1 .G/ D Ã1 .G/ D ˆ.G/. Lemma 46.2. jG 0 j 2mn . Indeed, jG 0 j is at least the size of a noncentral G-class. However, all noncentral G-classes have the same size 2mn . From the last assertion follows Lemma 46.3. k.G/ D jZ.G/j C

jGjjZ.G/j 2mn

D 2mCn C 2m 2n .

An element 2 F is said to be primitive if the (cyclic) multiplicative group F D hi. Lemma 46.4. A mapping ' ..a; b// D .a; ./b/ is an automorphism of G of order o./ for every element 2 .F /# . Proof. Set ./ D . Then ' ..a; b// D .a; b/, ' ..c; d // D .c; d /, ' ..a; b/.c; d // D ' ..a C c; b C d C a .c// D ..a C c/; .b C d C a .c///: On the other hand, ' ..a; b//' ..c; d // D .a; b/.c; d / D ..a C c/; b C d C a .c// D ..a C c/; .b C d C a .c///: It follows that ' is an endomorphism of G. Next, .a; b/ D ' ..a; b// D .0; 0/ if and only if .a; b/ D .0; 0/ so ' is an automorphism of G since G is finite. Set o./ D d . Then 'd ..a; b// D .d a; . .//d b/ D .a; b/ and, if a ¤ 0, then 'r ..a; b// ¤ .a; b/ for r 2 f1; 2; : : : ; d 1g. It follows that o.' / D d D o./. Lemma 46.5. Given a > 1, m; n, we have .am 1; an 1/ D a.m;n/ 1. Proof. Set .m:n/ D ı and d D .am 1; an 1/. Then there exist u; v 2 N such that ı D mu nv. Clearly, aı 1 divides d . On the other hand, d divides amu 1 and anv 1. Hence d divides the number amu 1 .anv 1/ D anv .amunv 1/ D anv .aı 1/ so d divides aı 1 since .d; a/ D 1, and we conclude that d D aı 1.

46

Degrees of irreducible characters of Suzuki p-groups

3

Lemma 46.6. Let m D nk, where n; k 2 N, and let t 2 N be coprime with k; then .m; nt / D n. (a) Set d D .2m 1; 2nt C 1/. Then d D 1 if k is odd and d D 2n C 1 if k is even. (b) Suppose that 2n C 1 divides 2m 1. Then 2n divides m. Proof. (a) The number d0 D .2m 1; 22nt 1/ equals 2.nk;2nt/ 1 D 2n 1 if k is odd and d0 D 22n 1 if k is even (Lemma 46.5). Since 2nt C 1 divides 22nt 1, the number d divides d0 . We claim that .2nt C 1; 2n 1/ D 1. Indeed, if a prime p divides that number, then 2n 1 .mod p/ so 2nt 1t 1 .mod p/, a contradiction since p (if it exists) must be odd and 2nt 1 .mod p/. Suppose that k is odd. By definition, d divides 2nt C 1 and, by the first paragraph, d divides d0 D 2n 1 so d divides .2nt C 1; 2n 1/ D 1 (see the previous paragraph), and we get d D 1. This proves the first assertion in (a). Now suppose that k is even; then t is odd. In that case, by the first paragraph, d divides d0 D 22n 1 D .2n 1/.2n C 1/ and, by definition, d divides 2nt C 1. Therefore, by the second paragraph, d divides .2n C 1; 2nt C 1/ D 2n C 1 since t is odd. Since 2n C 1 divides d , we get d D 2n C 1. The proof of (a) is complete. (b) Since .2n 1; 2n C 1/ D 1 and 2n 1 divides 2m 1 D 2nk 1, it follows that 22n 1 D .2n 1/.2n C 1/ divides 2m 1 and so 2n divides m (Lemma 46.5). Write ˆ D h' i ( 2 F# ); then jˆ j D o./. If is primitive, we write ˆ D ˆ Lemma 46.7. Let 2 F be primitive; then jˆj D o./ D 2m 1 and: (a) The size of every ˆ-orbit on the set G Z.G/ equals 2m 1. (b) For odd k, there is only one ˆ-orbit on the set Z.G/# . (In that case, ˆ G is a Frobenius group with minimal normal subgroup Z.G/, its kernel.) (c) For even k, the size of every ˆ-orbit on the set Z.G/# equals the number of ˆ-orbits on Z.G/# equals 2n C 1/.

2m 1 2n C1

(in that case,

Proof. (a) For .a; b/ 2 G and i 2 N, we have 'i ..a; b// D .i a; . .//i b/. Let .a; b/ 2 G Z.G/. In that case, a ¤ 0 so 'i ..a; b// D .a; b/ if and only if i D 1, i.e., i is divisible by 2m 1 since is primitive. This proves (a) (b) Now let k be odd and b ¤ 0. Then 'i ..0; b// D .0; . ./i b/ D .0; b/ if and nt only if . .//i D 1. Since ./ D 2 with .t; k/ D 1, we get 1 D . .//i D .2nt C1/i m . Therefore, o./ D 2 1 divides .2nt C 1/i . By Lemma 46.6(a), if k is odd, 2m 1 divides i since .2m 1; 2nt C 1/ D 1, so the ˆ-orbit of .0; b/ contains 2m 1 elements. Then all elements of Z.G/# are ˆ-conjugate, and (b) is proven. nt (c) Let k be even. The minimal positive integer i such that .2 C1/i D 1 equals 2m 1 2n C1 , by Lemma 46.6(a), and this number is the size of the ˆ-orbit of the element .0; b/ 2 Z.G/# . It follows that the number of ˆ-orbits on Z.G/# equals 2n C 1, completing the proof of (c).

4

Groups of prime power order

For 2 F f1F g, ' has no fixed points on .G=Z.G//# (check!) so h' ; G=Z.G/i is a Frobenius group. Corollary 46.8. If k is odd, then G 0 D Z.G/. Proof. Indeed, Z.G/ is a minimal normal subgroup of h' ; Z.G/i, where is a primitive element of F, by Lemma 46.7(b), and G 0 Z.G/ so G 0 D Z.G/. Remark. Suppose that 2 Irr1 .G/, .1/ D 2a and Z0 D Z.G/ \ ker. /, where G D A.m; /. Then jZ.G/ W Z0 j D 2 since the center of G= ker. / is cyclic. It follows that G 0 Z0 D Z.G/, cd.G=Z0 / D f1; 2a g [Isa1, Theorem 2.31] and so jIrr1 .G=Z0 /j D 2mC1 2m D 2m2a . A character 2 Irr1 .G/ determines Z0 uniquely. 22a Theorem 46.9 ([Han]). If G D A.m; / is a Suzuki 2-group, where is an automorphism of F of odd order k > 1, then cd.G/ D f1; 2.mn/=2 g (here n D m ). k Proof. Let 2 F be primitive. Consider the action of the automorphism ' on subgroups of index 2 in Z.G/. By Lemma 46.7(b), h' ; Z.G/i is a Frobenius group and Z.G/# is the unique ' -orbit. By Maschke’s theorem, if 's ¤ id, then that automorphism has no invariant subgroups of index 2 in Z.G/, i.e., h' i acts on the set of such subgroups in a fixed-point-free manner. Since the order of ' equals 2m 1, all 2m 1 subgroups of index 2 in Z.G/ are .ˆ D/h' i-conjugate. Let 2 Irr1 .G/, Z0 D Z.G/ \ ker. /. Then the isomorphism type of G=Z0 independent of , since all subgroups of index 2 in Z.G/ are ˆ-conjugate. It follows that cd.G/ D f1; d g for some d > 1, by the Remark. By Lemma 46.3 and Corollary 1 46.8, we get 2m C .2mCn 2n /d 2 D 22m so d D 2 2 .mn/ . Thus, Theorem 46.1(a) is proven. In what follows, k is even. Lemma 46.10. Let 2 F be primitive and k even. Then: (a) If k > 2, i.e., m D k n > 2n, then G 0 D Z.G/ and the number of ˆ-orbits on m 1 . .G 0 /# equals 2n C 1, and all these orbits have the same size 22n C1 (b) If k D 2, i.e., m D 2n, then either G 0 D Z.G/ and .G 0 /# has 2n C 1 distinct m 1 m D 2n 1 or else jG 0 j D 2n .D 2 2 / and all elements of ˆ-orbits of size 22n C1 .G 0 /# are ˆ-conjugate. Proof. The assertion on the sizes of ˆ-orbits follows from Lemma 46.7(c). Suppose m 1 that the number of ˆ-orbits on .G 0 /# equals t and jG 0 j D 2f . Then 2f 1 D t 22n C1 (see Lemma 46.7(c)). It follows that t 1 D t 2m C 2n 2f Cn 2f . By Lemma 46.2, f m n n so 2n divides t 1. It follows that either t D 1 or t 2n C 1. On the other hand, f m since G 0 Z.G/ and, by the previous paragraph, m 1 D 2f 1 2m 1. It follows that t 2n C 1 so t 2 f1; 2n C 1g. t 22n C1 m 1 Suppose that t D 1. Then, by the previous paragraph, 2f 1 D 22n C1 so 2n .2mn C f n 1/ D 2 .2 C 1/. It follows that then f D n and m n D n, i.e., m D 2n, k D 2. In that case, f D n D 12 m, and we have the second possibility in (b). m 1 D 2m 1 so f D m. Suppose that t D 2n C 1. Then 2f 1 D .2n C 1/ 22n C1

46

5

Degrees of irreducible characters of Suzuki p-groups

In what follows we first assume that G 0 D Z.G/. Then the number of ˆ-orbits on m 1 (Lemma 46.10). .G 0 /# equals 2n C 1 and all these orbits have the same size 22n C1 Lemma 46.11. Suppose that G 0 D Z.G/, 2 F is primitive and k is even. (a) Under ˆ, the set of subgroups of index 2 in Z.G/ is partitioned in 2n C 1 orbits m 1 of size 22n C1 . (b) Under ˆ, the set Irr.G/ f1G g is partitioned in 2n orbits of size 2m 1 and m 1 2n C 1 orbits of size 22n C1 . Proof. Z.G/ has 2m 1 subgroups of index 2 and jIrr.G/j D 2mCn C 2m 2n . (a) By Lemma 46.7, the restriction of ' to Z.G/# is a product of 2n C1 independent m 1 cycles of size 22n C1 . This proves (a) (see the first paragraph in the proof of Lemma 46.9). (b) Define the action of ' on the set Irr.G/ as follows. Let R be a representation affording the character 2 Irr.G/. Set T D R ı '1 , i.e., T .g/ D R.g'1 / for g 2 G. If h 2 G, then since ' 2 Aut.G/, we get T .gh/ D R..gh/'1 / D R.g'1 h'1 / D R.g'1 /R.h'1 / D T .g/T .h/; i.e., T is a representation of G. Clearly, the value of the character ' .g/ of T equals .g'1 / for every g 2 G, and this character is also irreducible. Let CL.G/ D fK1 ; : : : ; Kr g be the set of G-classes, xi 2 Ki , all i , and let X D X.G/ D . i .xj // be the character table of G. Then ' acts on columns and rows of X by permutations 1 and 2 , respectively, in such a way that if one considers these '

' 1

permutations as elements of GL.r; C/, then 2 X D . i .xj // D . i .xj // D X1 . Since the matrix X is nonsingular, we get 1 D X 1 2 X in GL.r; C/. Then, by Vishnevetsky’s lemma [BZ, Lemma 10.4(a)], 1 and 2 are conjugate in the symmetric group Sr and so they have the same cycle structure. Now the result follows m 1 on from Lemma 46.7. Indeed, under ˆ, there are exactly 2n C 1 orbits of size 22n C1 2 D 2n orbits of size 2m 1 on noncentral nonidentity central G-classes and 2 2m 1 m 1 n and 2n orbits of size 2m 1 G-classes. It follows that ˆ has 2 C 1 orbits of size 22n C1 on the set Irr.G/ f1G g. mCn

n

Lemma 46.12. If 2 Irr1 .G/, k is even and .1/ D 2a , then 12 m n a 12 m. Proof. By the Remark, we have 2 Irr1 .G=Z0 /, where Z0 < Z.G/ is of index 2, and cd.G=Z0 / D f1; 2a g so jIrr1 .G=Z0 /j D 2m2a . It follows that m 2a 0 so a 12 m. Let Z1 ; : : : ; Z2n C1 be a transversal of the set of ˆ-orbits on the set M of subgroups of index 2 in Z.G/. By the Remark, cd.G=Zi / D f1; 2ai g, jIrr1 .G=Zi /j D 2m2ai , all i . Next, if Zi and Zi;1 belong to the same ˆ-orbit, then, obviously, G=Zi Š G=Zi;1 . m 1 P2n C1 m2ai (here 2m 1 Since jIrr1 .G/j D 2mCn 2n , we get 2mCn 2n D 22n C1 iD1 2 2n C1 P2n C1 m2a i D 2n .2n C 1/. is the common size of ˆ-orbit) which implies that iD1 2

6

Groups of prime power order

Assume that m 2ai > 2n for some i . Then 2m2ai 22nC1 > 2n .2n C 1/, contrary to the result of the previous paragraph. It follows that m 2ai 2n so ai 12 m n for all i . Theorem 46.13. If k is even and G 0 D Z.G/, then cd.G/ D f1; 2.m=2/n ; 2m=2 g. Next, jIrr.2.m=2n / .G/j D .2m 1/22n =.2n C1/, jIrr.2m=2 / .G/j D .2m 1/2n =.2n C1/. 1 . Proof. Set s D 22n C1 By Lemma 46.11(b), the set Irr.G/ f1G g is partitioned under the action of ˆ in 2n orbits of size 2m 1 and 2n C 1 orbits of size s. Since hˆ; Gi=G 0 is a Frobenius group (see the paragraph following Lemma 46.7), all jG W G 0 j 1 D 2m 1 nonprincipal linear characters of G are ˆ-conjugate. Therefore, under ˆ, there are exactly 2n 1 orbits of size 2m 1 on Irr1 .G/. By Lemma 46.11(a), the set M of subgroups of index 2 in Z.G/ is partitioned in exactly 2n C 1 orbits of size s under ˆ. Then, if Z0 2 M, then 's .Z0 / D Z0 since the order of the restriction of ' to M is s. Suppose that the size of the ˆ-orbit of 2 Irr1 .G/ equals 2m 1 and 2 i Irr1 .G=Z0 / for some Z0 2 M. Then ¤ ' 2 Irr1 .G=Z0 / for 0 < i < 2n C 1 so the number of ˆ-conjugates of equals 2n C 1. Suppose that for some Z0 2 M, Irr1 .G=Z0 / has no characters belonging to a ˆorbit of size s. Let cd.G=Z0 / D f1; 2a g. Then all of the 2m2a characters of the set Irr1 .G=Z0 / are partitioned in hˆi-orbits of the same odd size > 1, which is not the case. It follows that, for each Z0 2 M, there exists in Irr1 .G=Z0 / a character such that its ˆ-orbit has size s. Since the number of such ˆ-orbits is 2n C 1 and, under ˆ, the set M has exactly 2n C 1 orbits of size s (Lemma 46.11(a)), it follows that each set Irr1 .G=Z0 / (where Z0 2 M) has exactly one character belonging to a ˆ-orbit of size s since s.2n C 1/ D 2m 1 is the total number of nonprincipal irreducible characters in ˆ-orbits of size s, by Lemma 46.11(b). Now suppose that the size of the ˆ-orbit of a character 2 Irr1 .G/ is 2m 1 (such exists since jIrr1 .G/j D 2n .2m 1/ > s.2n C 1/), .1/ D 2a and 2 Irr1 .G=Z0 /, where Z0 2 M. By the previous paragraph, jIrr1 .G=Z0 /j D 2m2a D 1 C t .2n C 1/, where t is the number of orbits of characters in Irr1 .G=Z0 / belonging to ˆ-orbits of size ¤ s (there are in Irr1 .G=Z0 / exactly 2n C1 characters conjugate with under ˆ). It follows that 2n C 1 divides 2m2a 1, and so n divides m 2a (Lemma 46.5) and m2a is even (Lemma 46.6(b)). By Lemma 46.12, m 2a 2n m so m2a 2, n n m2a i.e., n 2 f0; 2g. Since G=Z0 has at least two nonlinear irreducible characters, by the previous paragraph, m 2a > 0. Hence, m 2a D 2n so a D 12 m n and m

1 Irr1 .G=Z0 / has exactly 2m2a D 22n characters that lie in t D 2 2n C11 D 22n C1 n m 2 1 distinct ˆ-orbits of size 2 1 and one ˆ-orbit of size s. m 1 We see that there are 22n s D 22n 22n C1 nonlinear irreducible characters of m 1 of degree 2.m=2/n . The sum of squares of their degrees is 2m2n 22n 22n C1 m m 1 1 2m 22n C1 . The sum of squares of degrees of remaining t1 D 2n .2m 1/22n 22n C1 m2a

2n

D G D D

46

Degrees of irreducible characters of Suzuki p-groups

1 2n 22n C1 nonlinear irreducible characters of G is † D 2m .2m 1/ 2m m

7 2m 1 2n C1 D t1 2m , it

1/ 2m 2 2.2n C1 . If 2 Irr1 .G/, then .1/2 jG W Z.G/j D 2m . Since † D follows that all latter characters have the same degree 2m=2 . This means that cd.G/ D f1; 2.m=2/n ; 2m=2 g, and G has exactly 22n s irreducible characters of degree 2.m=2/n and t1 D 2n s irreducible characters of degree 2m=2 . n

m

Corollary 46.14. If k D 2, i.e., m D 2n, then jG 0 j D 2m=2 . Proof. In that case, by Lemma 46.10, either jG 0 j D 2m=2 or else jG 0 j D 2m . In the second case, however, by Theorem 46.13, cd.G/ D f1; 2.m=2/n ; 2m=2 g, which is not the case since m 2n D 0. Theorem 46.15. If k D 2, then cd.G/ D f1; 2m=2 g. Proof. By Corollary 46.14, jG 0 j D 2m=2 , jG W G 0 j D 2mC.m=2/ D 2mCn D jLin.G/j. Since k.G/ D 2mCn C 2m 2n , we get jIrr1 .G/j D jIrr.G/j 2mCn D 2m 2n D 2m=2 .2m=2 1/. By the remark following Corollary 46.8, if 2 Irr1 .G/, then 2 Irr1 .G=Z0 /, where Z0 2 M, G 0 6 Z0 . The number of such subgroups Z0 equals 2m 1 .2m=2 1/ D 2m=2 .2m=2 1/ D jIrr1 .G/j. (Here 2m 1 D jMj and 2m=2 1 is the number of subgroups of index 2 in Z.G/=G 0 ; this explains our formula.) Thus, to every subgroup Z0 corresponds the unique nonlinear irreducible character of G=Z0 . Since jG=Z0 j D 2mC1 and j.G=Z0 /0 j D 2 so G=Z0 is extraspecial, we get cd.G=Z0 / D f1; 2m=2 g. Theorem 46.1 is proven. Corollary 46.16. If k D 2 and 2 Irr1 .G/, then G=Z0 is extraspecial for every subgroup Z0 of index 2 in Z.G/ such that G 0 6 Z0 . 1

Proof. Suppose that 2 Irr1 .G=Z0 /; then .1/ D 2 2 m , by Theorem 46.15. It follows that jZ.G=Z0 /j D 2, so G=Z0 is extraspecial since its center and derived subgroup have order 2. 2o . In this subsection, p > 2. Thus, F D GF.p m /, m 2 N with m > 1, q D p m , is an automorphism of F of order k for some divisor k > 1 of m, m D n and G D k Ap .m; / is the generalized Suzuki p-group of order q 2 . Our aim in this subsection is to prove the following Theorem 46.17 ([Sag2]). If G D Ap .m; / is a p-group of order q 2 , where p > 2, m > 1 is an integer, q D p m , is an automorphism of the field F D GF.q/ of order k > 1 and m D n, then one of the following holds: k (a) If k is odd, then cd.G/ D f1; p .mn/=2 g. (b) If k D 2, then cd.G/ D f1; p m=2 g.

8

Groups of prime power order

(c) If k > 2 is even, then cd.G/ D f1; p m=2 ; p .m=2/n g, jIrr.p m=2 / .G/j D and jIrr.p .m=2/n / .G/j D

p m 1 p n C1

p m 1 n p n C1 p

p 2n .

By the text preceding 1o , exp.G/ D p, .a; b/1 D .a; b C a .a//, Z.G/ D f.0; b/ j b 2 Fg is of order q and the centralizer of every element from G Z.G/ has order p mCn . It follows that k.G/ D p mCn C p m p n so the number of noncentral G-classes is p mCn p n . Next, jG 0 j p mn . We see that Lemmas 46.2, 46.3 and 46.4 are also true for odd p. Argument following Lemma 46.4, is also true for odd p. nt Therefore, for every x 2 F, .x/ D x p for some t coprime with k. Lemma 46.18. (a) Suppose that m D nk, .t; k/ D 1 and .p m 1; p nt C 1/ D d . If k is odd, then d D 2. If k is even, then d D p n C 1. (b) Suppose that k; n 2 N, k even and .p m 1; p n C 1/ D d . Then d D 2 if n is even and d D p .m=2;n/ C 1 if .m=2;n/ is odd.

n .m=2;n/

Proof. (a) Since .k; t / D 1, then d divides .p m 1; p 2nt 1/ D .p m 1; p 2n 1/ D p .m;2n/ 1 (Lemma 46.5). If k is odd then .m; 2n/ D .nk; 2n/ D n so d divides p n 1. If k is even, then .m; 2n/ D 2n so d divides p 2n 1. Let us find d0 D .p nt C 1; p n 1/. Let k be odd; then d divides p n 1 so d divides d0 . If r > 2 is a prime divisor of d0 , then p n 1 .mod r/ so p nt C 1 2 .mod r/, a contradiction. The same argument also works in the case r D 4. It follows that d0 D 2 since d0 is even. Since the even number d divides d0 D 2, we get d D 2. Suppose that k is even; then t is odd and so p n C 1 divides .p m 1; p nt C 1/ D d . Next, p nt C1 D .p n C1/A and p m 1 D .p n C1/B, where A D p n.t1/ p n.t2/ C C p 2n p n C 1 and B D p n.k1/ p n.k2/ C C p n 1. The number A is odd since t is odd, and B is even since k is even. In that case, d D .p n C 1/.A; B/ so p ndC1 D .A; B/ D u is odd. Since d D .p n C 1/u with odd u divides p 2n 1 D .p n 1/.p n C 1/ and .p nt C 1; p n 1/ D 2 (see the previous paragraph!), we get d D p n C 1. (b) Set d D .p m 1; p n C 1/, m D 2m1 , d0 D .p .m1 ;n/ C 1; p n C 1/. Then d divides the number .p m 1; p 2n 1/ D p .m;2n/ 1 D p 2.m1 ;n/ 1 D .p .m1 ;n/ 1/.p .m1 ;n/ C 1/: As in the proof of (a), .p .m1 ;n/ 1; p n C 1/ D 2 and d0 D 2 if n .m1 ;n/

d0 D p C 1 if is odd. Suppose that .mn1 ;n/ is even. Then d 2 f2; 4g since d divides .m1 ;n/

n .m1 ;n/

.p n C 1; p .m1 ;n/ 1/.p n C 1; p .m1 ;n/ C 1/ D 2: Indeed, since n is even, we get p n C 1 2 .mod 4/ so d D 2.

is even and

46

Suppose that s D

9

Degrees of irreducible characters of Suzuki p-groups n .m1 ;n/

is odd and set t D .m1 ; n/. Then

p n C 1 D p ts C 1 D .p t C 1/.p t.s1/ p t.s2/ C C p 2t p t C 1/ so

p n C1 p t C1

is odd.

Let be as in Lemma 46.4 and set ˆ D h i provided is a primitive element of the field F. Lemma 46.19. If is a primitive element of the field F, then: (a) The size of ˆ-orbits on G Z.G/ is p m 1. (b) The size of ˆ-orbits on Z.G/# is 12 .p m 1/ for odd k and

p m 1 p n C1

for even k.

Proof. By definition, i ..a; b// D .i a; . .//i b/. If a ¤ 0, the assertion follows since is primitive. Let a D 0. Then i .0; b/ D .0; . .//i b/ D .0; b/ if and n n only if . .//i D 1. Since ./ D p , we obtain .p C1/i D 1. It follows that o./ D p m 1 divides .p n C 1/i , and the assertion follows from Lemma 46.18(a). Lemma 46.20. Suppose that 2 Irr1 .G/, .1/ D p a , Z0 D Z.G/ \ ker. /. Then jZ.G/ W Z0 j D p, G 0 6 Z0 , cd.G=Z0 / D f1; p a g, and jIrr1 .G=Z0 /j D .p 1/p m2a . Next, every nonlinear irreducible character of G is a character of exactly one group G=Z, where Z is a subgroup of index p in Z.G/ not containing G 0 . Proof. Since Z.G= ker. // is cyclic and Z.G/ is elementary abelian, we get jZ.G/ W Z0 j D p. Let us consider the quotient group G=Z0 . Since is nonlinear and G 0 6 Z0 but G 0 Z.G/, we get G 0 Z0 D Z.G/. Next, j.G=Z0 /0 j D p and exp.G=Z0 / D p. It follows that G=Z0 D .G1 =Z0 / .Z1 =Z0 /, where Z1 =Z0 < Z.G=Z0 / and G1 =Z0 is extraspecial. Now all assertions of the lemma are obvious. Theorem 46.21. If G D Ap .m; /, where an automorphism of F of odd order k > 1, mn then cd.G/ D f1; p 2 g, where n D m . k Proof. Let be a primitive element of F. In that case, there are two ˆ-orbits of size 12 .p m 1/ on Z.G/, where ˆ is a cyclic subgroup generated by , by Lemma 46.19(b). If g 2 G 0 , then the ˆ-orbit of g lies in G 0 since G 0 is characteristic in G. It follows that jG 0 j 1 12 .p m 1/ so jG 0 j D p m since jG 0 j divides p m . It follows that G 0 D Z.G/. By Lemma 46.19(b), there are at most two ˆ-orbits on the set of all subgroups of order p in Z.G/. It follows that there are at most two ˆ-orbits on the set of maximal subgroups of Z.G/. Since G=Z0 Š G=Z0 for every subgroup Z0 of index p in Z.G/ and jcd.G=Z0 /j D 2, by Lemma 46.20, then G has at most two distinct degrees of nonlinear irreducible characters. Assume that cd.G/ D f1; p a ; p b g. Then there are two ˆ-orbits on the set of subp m 1 . In that case, there groups of index p in Z.G/ and the sizes of these orbits are 2.p1/ are

p m 1 .p 2.p1/

1/p m2a D

1 m 2 .p

1/p m2a characters of degree p a in Irr1 .G/

10

Groups of prime power order

p 1 and, similarly, 2.p1/ .p 1/p m2b D 12 .p m 1/p m2b characters of degree p b in Irr1 .G/. It follows that m

1 1 p mCn p n D jIrr1 .G/j D .p m 1/p m2a C .p m 1/p m2b 2 2 so that p n D 12 .p m2a C p m2b / or, what is the same, p nC2aC2b D 12 p m .p 2a C p 2b /. Let, for definiteness, a b. Then p nC2aC2b D 12 p mC2b .p 2a2b 1/. The last equality is possible if and only if p 2a2b 1 D 0. Thus a D b so cd.G/ D f1; p a g. It follows that jGj D p 2m D p m C .p mCn p n /p 2a so a D 21 .m n/. Hence, the Theorem 46.17 is proven for odd k. Lemma 46.22. Suppose that 2 F is primitive and k is even. Then: (a) If k > 2, i.e., m > 2n, then G 0 D Z.G/ and the number of ˆ-orbits on .G 0 /# m 1 equals p n C 1; all these orbits have size pp n C1 . (b) If k D 2, i.e., m D 2n, then either G 0 D Z.G/ and the number of ˆ-orbits m 1 on .G 0 /# equals p n C 1 and all these orbits are of size pp n C1 D p n 1, or m

jG 0 j D p 2 D p n and all elements of .G 0 /# are conjugate under ˆ.

Proof. Repeat, word for word, the proof of Lemma 46.10. Lemma 46.23. Let 2 F be primitive, k even and G 0 D Z.G/. Then Irr.G/ f1G g is partitioned, under action of ˆ, in p n orbits of size p m 1 and p n C 1 orbits of m 1 size pp n C1 . Proof. See the proof of Lemma 46.11(b). Lemma 46.24. Let 2 F be primitive, k even and G 0 D Z.G/. Then the set of all subgroups of Z.G/ of index p is partitioned, under action of ˆ, either in p n C 1 orbits m 1/ p m 1 of size .p n C1/.p1/ or in 12 .p n C 1/ orbits of size .p2.p n C1/.p1/ . Proof. Consider a subgroup P D h.0; x/i of Z.G/ of order p. Let s 2 N be minimal m 1 such that s ..0; x// 2 P . Since o.. /Z.G/ / D pp n C1 and P # is partitioned in p1 t

1 orbits of size t under action of h's i, st D pp n C1 . Then the size of ˆ-orbit containing P equals s. Let P1 D h.0; y/i be another subgroup of order p in Z.G/. Since Z.G/ D f.0; b/ j b 2 Fg and 2 F is primitive, we get .0; y/ D .0; f x/ for some f 2 N. Next, since s ..0; x// 2 P , we get s ..0; x// D .0; . .//s x/ D .0; x/r D .0; rx/ for some r 2 N. Then m

s ..0; y// D .0; . .//s y/ D .0; . .//s f x/ D .0; rf x/ D .0; ry/ D .0; y/r : This means that the size of the ˆ-orbit containing P1 , is also equal to s. Thus, the sizes of all ˆ-orbits on the set of all subgroups of order p in Z.G/ are equal. It follows

46

p m 1 p1 , t.p n C1/ p1 .

from c1 .Z.G// D p m 1 s.p1/

11

Degrees of irreducible characters of Suzuki p-groups

that the number of ˆ-orbits on the set of such subgroups

D Since .p n C 1; p 1/ D 2, we get t 2 fp 1; 12 .p 1/g. equals Hence, the set of all subgroups of order p in Z.G/ is partitioned, under action of ˆ, m 1/ p m 1 either in p n C 1 orbits of size .p n C1/.p1/ or in 12 .p n C 1/ orbits of size .p2.p n C1/.p1/ . Now let Q < G be of index p. Since Q D P1 Pm1 with jPi j D p for all i , then s .Q/ D Q. If for some natural s1 < s we have s1 .Q/ D Q, then by Maschke’s theorem, Z.G/ D Q P , where P is a subgroup of order p and s1 .P / D P , contrary to the minimal choice of s. Now the conclusion of the lemma follows from the previous paragraph. Lemma 46.25. Suppose that 2 Irr1 .G/, .1/ D p a . Then a 12 m. Next, a 1 2 m n if the number of ˆ-orbits on the set of subgroups of index p in Z.G/ equals p n C 1 and a > 12 m n, if the number of ˆ-orbits on the same set equals 12 .p n C 1/. Proof. By Lemma 46.20, 2 Irr1 .G=Z0 /, where Z0 is a subgroup of index p in Z.G/ and jIrr1 .G=Z0 /j D .p 1/p m2a . It follows that a 12 m. Let Z1 ; : : : ; Z t be the set of representatives of ˆ-orbits on the set of maximal subgroups of Z.G/. Then cd.G=Zi / D f1; p ai g, jIrr1 .G=Zi /j D .p 1/p m2ai . Next, any two subgroups of index p in Z.G/ belonging to the same ˆ-orbit, lead to isomorphicPquotient groups. Since jIrr1 .G/j p mCn p n , we get p mCn p n D Pt D m2a t p m 1 m2ai n i. so tp D iD1 p iD .p 1/p t.p1/ Suppose that t D p n C 1. If m 2ai > 2n for some i , then p m2ai p 2nC1 > C 1/ D tp n , a contradiction. Thus, m 2ai 2n so ai 12 m n for all i . Suppose that t D 12 .p n C 1/. If m 2ai 2n for some i , then p m2ai p 2n > 1 n n 1 n 2 p .p C 1/ D tp , a contradiction. Thus, ai > 2 m n for all i .

p n .p n

1

1

Theorem 46.26. If G 0 D Z.G/ and k is even, then cd.G/ D f1; p 2 mn ; p 2 m g. In m 1 m 1 p 2n , jIrr 1 m .G/j D pp n C1 pn . that case, jIrr 1 mn .G/j D pp n C1 .p 2

/

.p 2

/

Proof. By Lemma 46.23, the set Irr.G/ f1G g, under action of ˆ, is partitioned in m 1 p n orbits of size p m 1 and p n C 1 orbits of size pp n C1 . Since jLin.G/j D jG=G 0 j 0 and ˆ .G=G / is a Frobenius group, nonprincipal linear characters lie in the same ˆ-orbit of length p m 1. Therefore, under ˆ, there are on Irr1 .G/ exactly p n 1 orbits of size p m 1. By Lemma 46.21, each character from Irr1 .G/ is contained in Irr1 .G=Z0 /, where Z0 is a subgroup of index p in Z.G/. By Lemma 46.24, the set M m 1 of all pp1 subgroups of index p in Z.G/, under action of ˆ, is partitioned either in

p 1 1/ or in 12 .p n C 1/ orbits of size .p2.p p n C 1 orbits of size .p n C1/.p1/ n C1/.p1/ . n First suppose that the set M is partitioned in p C 1 orbits under action of ˆ. Consider 2 Irr1 .G=Z0 /, Z0 2 M. If the size of the ˆ-orbit of equals p m 1, then Irr1 .G=Z0 / contains exactly .p 1/.p n C 1/ characters from that orbit. If the size m 1 of ˆ-orbit of equals pp n C1 , then Irr1 .G=Z0 / contains exactly p 1 characters from m

m

12

Groups of prime power order

1 that orbit. Note that if, for some Z0 2 M, Irr1 .G=Z0 / has no orbits of size pp n C1 , then n Irr1 .G=Z0 / is partitioned in orbits of sizes .p 1/.p C 1/. Since jIrr1 .G=Z0 /j D p m2a , where p a 2 cd.G=Z0 /, then p n C 1 divides p m2a , which is a contradiction. It follows that for every Z0 2 M, there are in Irr1 .G=Z0 / at least p 1 characters m 1 whose ˆ-orbits have size pp n C1 . Since the number of such characters is p m 1 and m

p m 1 p1 , p m 1 p n C1 .

jMj D

every group G=Z0 has exactly p 1 characters belonging to orbits of

size Let cd.G=Z0 / D f1; p a g and let Irr1 .G=Z0 / contain exactly t .p 1/.p n C 1/ characters belonging to ˆ-orbits of size p m 1. Then .p 1/p m2a D jIrr1 .G=Z0 /j D .p 1/ C t .p 1/.p n C 1/. It follows that p m2a 1 D t .p n C 1/ so p n C 1 divides p m2a 1. It follows that n divides m 2a and m2a is even. By Lemma n 46.25, 12 m n a 12 m so that 0 m2a 2. Suppose that t > 0. Then n m2a jIrr1 .G=Z0 /j > p 1 or m 2a > 0. It follows that n D 2 and so a D 12 m n, jIrr1 .G=Z0 /j D .p 1/p 2n , t D p n 1. Since t D p n 1, Irr1 .G=Z0 / contains representatives of all ˆ-orbits of size p m 1 on Irr1 .G/. Therefore, all characters from a ˆ-orbit of size p m 1 belong to members of M contained in the same ˆ-orbit. m 1 p 2n . The number of remaining irreducible nonlinear Then jIrr 1 mn/ .G/j D pp n C1 .p 2

/

1 p 2n D characters equals t1 D p n .p m 1/ pp n C1

p m 1 n p n C1 p and the sum of squares m 1 m 1 † D p m .p m 1/ pp n 1 p m D pp n 1 p mCn . If 2 Irr1 .G/, m m W Z.G/j D p . Since † D t1 p , all other nonlinear irreducible 1 2m m

of their degrees is then .1/2 jG characters are of degree p . Now suppose that the number of ˆ-orbits on M equals 12 .p n C 1/. Recall that all 2.p m 1/ these orbits are of size .p1/.p n C1/ . If 2 Irr1 .G=Z0 / (Z0 2 M) belongs to the ˆm orbit of size p 1, then that set of characters contains also exactly .p1/ 12 .p n C1/ ˆconjugates with . Similarly, as before, we can show that then Irr1 .G=Z0 / (Z0 2 M) m 1 contains a character representing a ˆ-orbit of size pp n C1 . Suppose that, for some Z0 2 M, the set Irr1 .G=Z0 / has exactly 12 .p 1/ characters m 1 . Then jIrr1 .G=Z0 /j D .p every of which is contained in a ˆ-orbit of size pp n C1 1 1 m2a n D 2 .p 1/ C t .p 1/ 2 .p C 1/, t 0. It follows that 2p m2a D 1/p n 1 C t .p C 1/, which is impossible since p n C 1 is even. Therefore, every group G=Z0 (Z0 2 M) has at least p 1 characters every of which belongs to a ˆ-orbit of m 1 m 1 . Since jMj D pp1 and the number of the above characters is p m 1, size pp n C1 every group G=Z0 (Z0 2 M) has exactly p 1 characters belonging to ˆ-orbits of m 1 size pp n C1 (these characters belong to two different ˆ-orbits). Now let G=Z0 (Z0 2 M) have a character belonging to a ˆ-orbit of size p m 1. n Then, by the above, jIrr1 .G=Z0 /j D .p 1/p m2a D .p 1/ C t .p 1/ p 2C1 so 2.p m2a 1/ D t .p n C 1/. Since m is even then m 2a is also even; therefore, using Lemma 46.18, we can show that either n divides 12 .m 2a/ , p n C 1 divides p m2a 1 and t D

2.p m2a 1/ p n C1

or p D 3, n D 1 and t D 12 .32m2a 1/. However,

46

13

Degrees of irreducible characters of Suzuki p-groups

by hypothesis, n > 1. In the first case we get either 2n m 2a, a 12 m n, contrary to Lemma 46.25, or else m 2a D 0, a D 12 m, t D 0, a contradiction again. Thus, the case where M is partitioned in 12 .p n C1/ orbits under ˆ is impossible. 1

Corollary 46.27. If m D 2n, then jG 0 j D p 2 m . Proof. Assume that G 0 D Z.G/ holds. Then, by Theorem 46.26, we have cd.G/ D 1 1 f1; p 2 mn ; p 2 m g, which is impossible since 12 m n D 0. 1

Theorem 46.28. If m D 2n, then cd.G/ D f1; p 2 m g. 1

1

Proof. By Lemma 46.22 and Corollary 46.27, jG 0 j D p 2 m , jG W G 0 j D p mC 2 m D 1 1 p mCn D jLin.G/j so jIrr1 .G/j D p m p n D p 2 m .p 2 m 1/. Every irreducible nonlinear character of G is a character of G=Z0 , where Z0 2 M and G 0 6 Z0 . The 1m 2

1m 2

1 p p11 D p p11 p 2 m , and every group G=Z0 number of such subgroups is pp1 has at least p 1 nonlinear irreducible characters. Comparing the number jIrr1 .G/j with the number of subgroups of index p in Z.G/ not containing G 0 , we see that every group G=Z0 has exactly p 1 nonlinear irreducible characters. Let cd.G/ D f1; p a g. Then p 1 D jIrr1 .G=Z0 /j D .p 1/p m2a so a D 12 m, completing the proof. m

Theorem 46.17 is proven.

1

47

On the number of metacyclic epimorphic images of finite p-groups

In this section we prove the following Theorem 47.1 ([Ber33]). Let G be a nonmetacyclic p-group of order p m and let n < m. (a) If p D 2 and n > 3, then the number of normal subgroups D of G such that G=D is metacyclic of order 2n , is even. (b) If p > 2 and n > 2, then the number of normal subgroups D of G such that G=D is metacyclic of order p n , is a multiple of p. Remarks. 1. Let G be a two-generator nonmetacyclic 2-group. Then all epimorphic images of G of order 8 are metacyclic. Since the number of normal subgroups of given index in a 2-group G is odd, it follows that Theorem 47.1(a) is not true for n D 3. 2. Let r s t > 0, r C s C t > 3 and G D hai hbi hci, where o.a/ D p r , o.b/ D p s , o.c/ D p t , be an abelian p-group of rank three. Let us check Theorem 47.1 for n D r C s C t 1. Let be the number of subgroups D of G of order p such that G=D is metacyclic. If t > 1, then 1 .G/ ˆ.G/; then D 0. Suppose that s > t D 1. If D < hai hbi, then G=D is nonmetacyclic. If D 6 hai hbi, then G=D is abelian of type .p r ; p s / so metacyclic, and in that case, D p 2 . Now suppose that r > s D 1. Then D D Ãr 1 .G/ is the unique normal subgroup of order p in G such that G=D is not metacyclic so D p 2 C p. Thus, p divides in all cases. 3. Suppose that G D ha; b j a4 D b 4 D c 2 D 1; c D Œa; b; Œa; c D Œb; c D 1i is a nonmetacyclic minimal nonabelian group of order 25 . Let us check Theorem 47.1(a) for n D 4. The group G has exactly seven central subgroups of order 2: X1 D ha2 i;

X2 D hb 2 i;

X3 D ha2 b 2 i;

X5 D ha2 ci;

X6 D hb 2 ci;

X7 D ha2 b 2 ci:

X4 D hci;

Then G=Xi is metacyclic for i D 3; 4; 5; 6 and nonmetacyclic for i D 1; 2; 7. m

4. Suppose that G D ha; b j a2 D b 2 D c 2 D 1; m > 2; c D Œa; b; Œa; c D Œb; c D 1i is a nonmetacyclic minimal nonabelian group of order 2mC2 . Set ˛ D

47

On the number of metacyclic epimorphic images of finite p-groups

15

m1

a2 ; then 1 D fhc˛i; h˛i; hcig is the set of central subgroups of order 2 in G. Then G=hci and G=hc˛i are metacyclic and G=h˛i is not metacyclic. Let Sc.G/ D 1 .Z.G// be the socle of G. Let i denote, for all i such that p i jSc.G/j, the set of subgroups of order p i in Sc.G/. Let M be a set of nonidentity normal subgroups of G. Given H 2 1 [ 2 [ [ fSc.G/g, let ˛.H / be the number of members of the set M containing H . Set jSc.G/j D p t so that t D fSc.G/g. We claim that the following identity holds (see 10): (1)

˛.G/ D jMj D

t X

i .1/i1 p .2/

iD1

X

˛.H /:

H 2i

Indeed, let D 2 M and jD \ Sc.G/j D p k ; then k 1 since D > f1g. For natural numbers u v, let 'u;v denote the number of subgroups of order p v in Ep u . Then the P i contribution of D in the right-hand side of (1) is equal to kiD1 .1/i1 p .2/ 'k;i , and that number equals 1, by Hall’s identity (Theorem 5.2). Since the contribution of D in the left-hand side of (1) is also equal 1, identity () is true. In particular, X ˛.G/ D jMj

(2) ˛.H / .mod p/: H 21

Remark 5. Suppose that G is a noncyclic group of order p m , m > 2 and 1 < n < m. We claim that the number c.G/ of normal subgroups N of G such that G=N is cyclic of order p n , is a multiple of p. One may assume that c.G/ > 0. Let n D m 1. There is N Z.G/ of order p such that G=N is cyclic. Then G is abelian of type .p m1 ; p/ so G D Z N , where Z is cyclic of order p m1 D p n hence c.G/ D p. Now let n < m 1. Take H 2 1 . Suppose that G=H is cyclic; then G is abelian of type .p m1 ; p/. The group G contains exactly p C 1 subgroups of index p n . If N is one of such subgroups, then G=N is cyclic if and only if N 6 ˆ.G/ so, since ˆ.G/ is cyclic, we get c.G/ D p. Next we assume that G is not abelian of type .p m1 ; p/; then G=H is not cyclic for all H 2 1 , therefore, by induction on m, we get ˛.H / 0 .mod p/ so c.G/ 0 .mod p/, by (2). Suppose that G is a group of order p m and p n jG W G 0 j. We claim that then the number .G/ of N G G such that G=N is nonabelian of order p n is a multiple of p. One may assume that .G/ > 0; then n > 2. The number of normal subgroups of given index in G is 1 .mod p/ (Sylow). If N G G has index p n , then G=N is nonabelian if and only if G 0 6 N . Therefore, .G/ 0 .mod p/ (Sylow again). If a 2-group G of order 2m > 23 is not of maximal class, then the number of N G G such that G=N is nonabelian of order 23 is even since jG W G 0 j 23 . Proof of Theorem 47.1. Suppose that M is the set of normal subgroups D of G such that G=D is metacyclic of order p n . One may assume that M ¤ ¿. As above, ˛.H / is the number of members of the set M containing H 2 1 .

16

Groups of prime power order

A. First we consider the most difficult case n D m 1; then M 1 . One may assume that G is nonabelian (Remark 2). We have to prove that p divides jMj.D ˛.G//, i.e., j1 Mj 1 .mod p/. By (2), we have to prove that X (3) ˛.H / 0 .mod p/: ˛.G/

H 21

Let U 2 M and suppose that U — ˆ.G/. Then G D U M , where M 2 1 is metacyclic. If V 2 1 and V — M , then G D V M so V 2 M. If W Z.M / is a member of the set M, then M=W is cyclic and G is abelian of rank 3 so p divides jMj (Remark 2). We see that jMj D j1 j c1 .Z.M // 0 .mod p/. Next we assume that all members of the set M are contained in ˆ.G/. (i) Let p > 2. Then jG=Ã1 .G/j p 3 since G is nonmetacyclic (Theorem 9.11). Take D 2 M; then G=D is metacyclic. Assuming that jG=Ã1 .G/j p 4 , we conclude that G=DÃ1 .G/ is of order p 3 and exponent p so nonmetacyclic, a contradiction. Thus, jG=Ã1 .G/j D p 3 . Take U 2 1 M; then G=U is not metacyclic so j.G=U /=Ã1 .G=U /j p 3 (Theorem 9.11 again) and so U Ã1 .G/. Thus, all members of the set 1 M are contained in Ã1 .G/. Conversely, if V 2 1 is contained in Ã1 .G/, then G=V is not metacyclic. Thus, j1 Mj equals the number of those members of the set 1 that are contained in Ã1 .G/. By Sylow, the last number is 1 .mod p/ so jMj j1 j 1 D ' t;1 1 0 .mod p/. (ii) Suppose that p D 2. Here we have to consider the nonmetacyclic quotient group G=Ã2 .G/ (see Lemma 42.2(b)) of order 24 . (ii1) Assume that jG=Ã2 .G/j D 24 . Then Ã2 .G/ > f1g since m > 4. The number of members of the set 1 contained in Ã2 .G/, is odd (Sylow). Therefore, if all members of the set 1 M are contained in Ã2 .G/, then ˛.G/ D jMj is even since j1 j is odd, so we assume that there is U 2 1 M such that U — Ã2 .G/; then GN D G=U N is nonmetacyclic so N 2 .G/ is nonmetacyclic. It follows from Theorem 44.2 that G=Ã 4 N then G=H.Š G= N HN / is not metacyclic of order its order is 2 . Write HN D Ã2 .G/; 24 and exponent 4. It follows that H D Ã2 .G/ since jG=Ã2 .G/j D 24 , contrary to the choice of U : U — Ã2 .G/. (ii2) Let jG=Ã2 .G/j > 24 . Take D 2 M; then, by Theorem 44.2, D — Ã2 .G/. Write GN D G=D. We claim that jG=Ã2 .G/j 25 . Indeed, G=DÃ2 .G/, as an epimorphic image of groups GN D G=D and G=Ã2 .G/, is metacyclic of exponent 4 so its order is 24 , and our claim on order follows since DÃ2 .G/ D D Ã2 .G/ and jDj D 2. Thus, jG=Ã2 .G/j D 25 , by (ii1). Since G=D is metacyclic, we have jSc.G=D/j 22 so jSc.G/j jDjjSc.G=D/j D 23 , and we conclude that j1 j 2 f1; 3; 7g. Since j1 j is odd, to complete this case, it suffices to show that the number j1 Mj is also odd. But G=D is metacyclic so D — G 0 (Theorem 36.1) and hence G 0 Š .G=D/0 is cyclic. Since all members of the set M are contained in ˆ.G/, we get d.G/ D 2. N (GN is defined in (ii1)). Since G=DÃ2 .G/ is metacyclic, its order is Let HN D Ã2 .G/ 4 N 2 .G/j N D 24 since jG=Ã2 .G/j D 25 . 2 so H D DÃ2 .G/ D D Ã2 .G/ and jG=Ã

47

On the number of metacyclic epimorphic images of finite p-groups

17

Set N0 D fV 2 1 j V Ã2 .G/g; then jN0 j is odd (Sylow) and N0 1 M. One may assume that N0 1 M (otherwise, jMj D j1 j jN0 j is even). In that case, there is U1 2 1 .M [ N0 /; then GN D G=U1 is not metacyclic. Write N then G= N HN is not metacyclic (Theorem 44.2) and, since U1 Ã2 .G/ H HN D Ã2 .G/; and U1 6 Ã2 .G/, we get H1 D U1 Ã2 .G/ and jG=H1 j D 12 jG=Ã2 .G/j D 12 25 D 24 . If V < H1 is a member of the set 1 , then G=V is nonmetacyclic. Let N1 D fU0 2 1 N0 j U0 < H1 g. Then jN0 [ N1 j D jN0 j C jN1 j, the number of members of the set 1 contained in H1 , is odd (Sylow). It follows that the number jN1 j is even since the number jN0 j is odd. Since U1 2 N1 , we obtain jN1 j 2 so jN0 [N1 j 1C2 D 3. Assuming that N0 [N1 1 M, we add to the set N0 [N1 the set N2 (of even cardinality > 0) of new members of the set 1 M. Indeed, if U2 2 1 .N0 [ N1 [ M/, then, writing H2 D U2 Ã2 .G/, we conclude, as above with U1 and H1 , that G=H2 is nonmetacyclic of order 24 . We have H1 \H2 D Ã2 .G/ since Ã2 .G/ H1 \ H2 has index 25 in G and H1 ; H2 are distinct of index 24 in G, so N2 is the set of members of the set 1 .N0 [ N1 / contained in H2 but not in H1 and jN2 j > 0 is even (if Vi 2 Ni , then Vi Ã2 .G/ D Hi , i D 0; 1; 2). Thus, we get jN0 [ N1 [ N2 j 1 C 2 C 2 D 5. We claim that then the set N0 [ N1 [ N2 , which contains exactly five members, coincides with the set 1 M. Indeed, otherwise, we can, acting as above, add to that sum-set at least two new members of the set 1 M, which is impossible since j1 Mj 6: the set M ¤ ¿ and j1 j D 7. Thus, in any case, the number j1 Mj is odd so that the number jMj is even. B. Now let n < m 1. Here we consider cases p D 2 and p > 2 together. We proceed by induction on jGj. Take H 2 1 . If G=H is metacyclic, then ˛.H / 1 .mod p/ (Sylow). However, the number of such H , by part A of the proof, is a multiple of p. Therefore, the contribution of such H in the sum on the right-hand side of formula (2), is a multiple of p. Now suppose that G=H is not metacyclic. Then, by induction, ˛.H / 0 .mod p/. Therefore, the contribution of such H in the sum on the right-hand side of formula (2), is a multiple of p again, Thus, congruence (3) is proven. Supplement to Theorem 47.1. Let 1 k p 1, k < n < m and a p-group G of order p m satisfies jG=Ã1 .G/j > p k . Then the number of normal subgroups D of G such that j.G=D/=Ã1 .G=D/j p k and jG=Dj D p n , is a multiple of p. 2o . Here we prove the following Theorem 47.2. If G be a nonabelian metacyclic p-group, then all representation groups of G are also metacyclic so that M.G/, the Schur multiplier of G, is cyclic. Proof. Let be a representation group of G. By definition, there is in Z./ \ 0 a subgroup M Š M.G/ such that =M Š G. Since G is nonabelian, we get M < 0 . Since =M Š G is metacyclic, is also metacyclic, by Theorem 36.1.

18

Groups of prime power order

The cyclicity of Schur multipliers of metacyclic p-groups is known, their orders are also computed in [Kar, Theorem 10.1.25]. Isaacs has reported (in letter at Jan 10, 2006) that he also proved Theorem 47.3. His proof is based on the following Lemma 47.3 (Isaacs). Let G be a p-group and let Z Z.G/ \ G 0 be of order p. If G=Z is metacyclic, then G is metacyclic. Proof. Let U=Z G G=Z be with cyclic U=Z and G=U . Note that jU=Zj > p since G=Z is not abelian. We want to show that U is cyclic. Now let hxU i D G=U . Since U is abelian and normal in U and G=U is cyclic, we see that G 0 D ŒU; x D hŒu; x j u 2 U i is isomorphic to U=CU .x/ [Isa2, Lemma 12.12]. But Z is contained in CU .x/ and U=Z is cyclic, and hence G 0 is cyclic as an epimorphic image of U=Z. Since G 0 > Z is cyclic, we have Z ˆ.G 0 / ˆ.U / so d.U / D d.U=Z/ D 1 and U is cyclic. (In fact, Theorem 36.1 and Lemma 47.4 are equivalent.)

48

On 2-groups with small centralizer of an involution, I

All results of this section are due to the second author. The main part of this section coincides with [Jan1]. In this section we give the classification of 2-groups containing an involution t such that CG .t / D ht i C2m , m 1. We may assume from the start that m > 1 since, if m D 1, G is a 2-group of maximal class, by Proposition 1.8. Note that case m D 2 was considered in [GLS, Proposition 10.27]. Exercise 1. Let G be a nonabelian 2-group and t an involution in G Z.G/. (a) Prove that jNG .CG .t // W CG .t /j < c1 .CG .t //, where c1 .G/ is the number of subgroups of order 2 (or, what is the same, involutions) in G. (b) Let CG .t / D ht iQ, where Q is either cyclic of order 2m , m > 1, or Q Š Q2m , m 3. Then jNG .CG .t // W CG .t /j D 2, by (a), and (b1) 1 .Z.G// D Z.Q/. (b2) If u is an involution in Q, then t and t u are conjugate in NG .CG .t //. (b3) If t 2 ˆ.G/, then jˆ.G/j > 4. Solution. (a) One may assume that CG .t / E G. Then G acts transitively on the conjugacy class K t containing t . Since jG W CG .t //j > 1 is a power of 2 and c1 .CG .t // > 1 is odd (Sylow), the inequality follows. (b1, b2) Let u 2 Z.G/ be an involution. Then u is a square and t is not, so t and u are not conjugate in G. The result follows since CG .t /.> Z.G// has exactly 3 involutions. (b3) Assume that jˆ.G/j D 4. Then ˆ.G/ Z.G/, by [BZ, Lemma 31.8], a contradiction since t 62 Z.G/. Theorem 48.1 ([Jan1]). Let G be a nonabelian 2-group containing an involution t such that the centralizer CG .t / D ht i C , where C is a cyclic group of order 2m , m 1 (clearly, t 62 Z.G/). Then G has no elementary abelian subgroups of order 16 and G is generated by at most three elements. Next, G has no t -invariant elementary abelian subgroups of order 8.

20

Groups of prime power order

(A1) If G has no elementary abelian subgroups of order 8, then one of the following holds: (a) G 2 fD2n ; SD2n g. Here we have m D 1. (b) G Š M2n . Here we have m D n 2. (c) jG W CG .t /j D 2, t 2 ˆ.G/, Z.G/ is a cyclic subgroup of order 4 not contained in ˆ.CG .t //, G=Z.G/ is dihedral with cyclic subgroup L=Z.G/ of index 2, L is abelian of type .2; 2m /, m 2, t 2 L. If x is an element of maximal order in G L, then hx 2 i D Z.G/. (In that case CG .t / D L.) (d) G has a subgroup S of index 2, where S D AL, the subgroup L is normal in n1 G, L D hb; t j b 2 D t 2 D 1; b t D b 1 ; n 3i Š D2n . A D hai is cyclic of m order 2 , m 2, A \ L D Z.L/, Œa; t D 1, 1 .G/ D 1 .S / D 2 .A/ L. If jG W S j D 2, then there is an element x 2 G S so that t x D t b and CG .t / D ht i hai. Next, G D ha; b; ti if G D S and G D ha; t; xi if S < G. (A2) If G has an elementary abelian subgroup of order 8, then Z.G/ is of order 2, CG .t / D ht i hai, where A D hai has order 2m (m 2), G has a normal subgroup L such as in (A1)(d), A \ L D Z.L/ D hzi, S D AL is a normal subgroup of index 2 in G and G is isomorphic to one of the following groups: m

n1

(e) G D ha; b; t j a2 D b 2 D t 2 D Œa; t D 1; m 3; n 4; b t D m1 n2 i D b2 D z; b a D b 1C2 ; i D n m 2i. We have G D AL D b 1 ; a2 S and the cyclic group hai=hzi acts faithfully on L. m n1 D t 2 D s 2 D Œa; t D 1; m 4; n 5; b t D (f) G D ha; b; t; s j a2 D b 2 m1 n2 i b s D b 1 ; a2q D b2 D z; b a D b 1C2 ; i D n m C 1 2; t s D m2 i 1 t b; s a D a2 b 2 si. Here S D AL is a subgroup of index 2 in G and 1 .S / D 2 .A/L. Also G D hsi S , M D hsiL Š D2nC1 , NG .M / D ha2 iM and s inverts 2 .A/. The order of G is 2mCn and G D ha; t; si. (g) Groups G D F .m; n/ (see Appendix 14). Here we have G D ha; b; t; s j m n1 a2 D b 2 D t 2 D s 2 D Œa; t D Œa; b D 1; m 2; n 3; b t D m1 n2 b s D b 1 ; a2 D b2 D z; t s D t b; as D a1 z v ; v D 0; 1; and if v D 1; then m 3i. Here S D A L and G D hsi S , where jG W S j D 2. We have M D hsiL Š D2nC1 and G D ha; t; si. m

n1

(h) G D ha; b; t; s j a2 D b 2 D t 2 D s 2 D Œa; t D 1; m; n 4; b t D b s D m1 n2 m2 2n3 D b2 D z; t s D t b; b a D bz; as D a1C2 b i. Here we b 1 ; a2 have again G D hsi S so that jG W S j D 2. Finally, M D hsi L Š D2nC1 and G D ha; t; si. Proof. Let G be a 2-group containing a noncentral involution t such that CG .t / D ht i C , where C is cyclic of order 2m , m 1. By Proposition 1.8, if m D 1, then G 2 fD2n ; n 3; SD2n ; n 4g. In what follows we assume that m > 1. Then G is not of maximal class so it contains a normal abelian subgroup U of type .2; 2/ (Lemma 1.4). It follows from the structure of CG .t / that Z.G/ is cyclic of order 2m

48 On 2-groups with small centralizer of an involution, I

21

at most (see Exercise 1(b1)) so T D CG .U / has index 2 in G. Let u be the involution in C ; then u 2 Z.G/ (Exercise 1(b1)). If an element x 2 NG .CG .t // CG .t /, then t x D t u, by Exercise 1(b2). Assume that E is a t -invariant elementary abelian subgroup of G of order 8. Set H D ht; Ei. Since H is not of maximal class, jCE .t /j > 2, by Suzuki’s theorem. Then the elementary abelian subgroup ht i CE .t / of order 8 is contained in (the metacyclic subgroup) CG .t /, which is a contradiction. Hence E does not exist. It follows that every t -invariant abelian subgroup of G is metacyclic. () We examine now the case where t 2 T D CG .U /. Then ht; U i CG .t / so t 2 U since CG .t / has no elementary abelian subgroups of order 8. Also we see that T D CG .U / D ht i C has index 2 in G since U 6 Z.G/, and T D CG .t /. (1) Suppose that U is the only normal abelian subgroup of type .2; 2/ in G. Set 1 .C / D hui D Ãm1 .T / so that U D ht; ui D 1 .T / and u 2 Z.G/. Since Z.G/ is cyclic, hui D 1 .Z.G//. Next, Ã1 .C / Ã1 .G/ ˆ.G/. (1.1) Suppose, in addition, that t 62 ˆ.G/. Then there exists a maximal subgroup M of G such that t 62 M ; in that case, G D ht i M , a semidirect product with kernel M . Setting M0 D T \ M , we get, by the modular law, T D ht i M0 . It follows from the structure of T that M0 Š C2m . If M is cyclic, then, clearly, G Š M2mC2 , and this is the case (b) of the theorem. So assume that M is not cyclic. Since U 6 M since t 2 U and t 62 M , the subgroup M has no G-invariant subgroups of type .2; 2/ (by assumption in (1), U is the unique G-invariant four-subgroup) so M is of maximal class, by Lemma 1.4. Then Z.M / D hui, where ht; ui D U . Let hvi be the cyclic subgroup of order 4 contained in M0 (recall that, by assumption. m 2; obviously, v 2 D u) and let y 2 M M0 ; then v y D v 1 D vv 2 D vu and t y D t u (see Exercise 1(b2)). Hence we have .t v/y D t y v y D t u vu D t v so K D ht vi is a cyclic subgroup of order 4 contained in T but not contained in M0 , and .t v/2 D t 2 v 2 D v 2 D u since, by the choice, v 2 CG .t /. We have CG .t v/ hT; yi D G. It follows that G D M K, the central product, M \ K D Z.M /. This group is then a group in part (d) of the theorem (with S D G and A is of order 4 centralizing L). In the case under consideration, ˆ.G/ D Ã1 .C / so C is normal in G. (1.2) It remains to consider the case t 2 ˆ.G/. In this case ht i Ã1 .C / ˆ.G/ has index 4 in G so there we have equality and hence G is generated by two elements. If m D 2, then ˆ.G/ D U , being a four-group, is contained in Z.G/, by Exercise 1(b3), which is a contradiction. Hence, we must have m 3. There is an element x 2 G T such that x 2 2 ˆ.G/ .Ã1 .C / [ U /. This is due to the fact that Z.G/ is cyclic. Indeed, there exists an element x 2 G T such that x 2 62 C (otherwise, ˆ.G/ D Ã1 .G/ D Ã1 .C /, which is a contradiction). Next, if x 2 2 U , then CG .x 2 / hx; T i D G so Z.G/ is not cyclic since it contains a noncyclic subgroup hu; x 2 i, a contradiction. Our claim about the existence of x is justified. Then o.x 2 / 4 since x 2 62 U D 1 .ˆ.G//. Set C D hai so that m1 a2 D u, T D CG .t / D ht i hai. Also, Ã1 .T / D Ã1 .C / D ha2 i is normal

22

Groups of prime power order

in G since T is, and ˆ.G/ D ht i ha2 i, where o.a2 / D 2m1 4. It follows from the choice of x that ˆ.G/ D ha2 ; x 2 i (however, generators x 2 and a2 are not independent). By Exercise 1(b2), we have t x D t u. If x would centralize a2 , then x centralizes ha2 ; x 2 i D ˆ.G/ and ˆ.G/ contains t so x 2 CG .t / D T , contrary to the choice of x. If y 2 G T , then y 2 2 Z.G/ since T is abelian and generates G together with y. Therefore, x 2 2 Z.G/. Since x 62 CG .a2 /, we get a2 62 Z.G/. It follows that all elements in .G=Z.G// .T =Z.G// are involutions. Since Z.G/ 6 ˆ.T /, it follows that T =Z.G/ is cyclic. In that case, G=Z.G/ is dihedral. Indeed, since d.G/ D 2 and Z.G/ 6 ˆ.G/, we conclude that G=Z.G/ is not abelian of type .2; 2/. If z is an element of maximal order in G T , then z 2 is a generator of Z.G/ since zZ.G/ is a noncentral involution in the dihedral group G=Z.G/. As we saw, jZ.G/j 4. We obtain a group in part (c) of the theorem. () In the rest of the proof we assume that G has a normal abelian subgroup U of type .2; 2/ which does not contain our involution t (if G has two normal abelian subgroups of type .2; 2/, then one of them does not contain t : otherwise t 2 Z.G/, which is not the case since G is nonabelian). Therefore, this case includes the one where G has at least two distinct normal four-subgroups. Set again T D CG .U / so that G D ht i T since t 62 T . By the modular law, G0 D CG .t / D ht i A, where A D CG .t / \ T D CT .t / is cyclic of order 2m , m 2, and so Z D 1 .A/ D CU .t / D 1 .Z.G// is of order 2. We have Z.G/ CG .U / \ CG .t / D T \ CG .t / D CT .t / D A. Set G1 D NG .G0 /, where G0 D CG .t /; then G1 > G0 since G > G0 is nonabelian. Since t and the generator z of 1 .A/ are not conjugate in view of m > 1, t has only two conjugates t and t z in G1 , and we have 1 .G0 / D ht; zi. It follows that 2 D jG1 W CG1 .t /j D jG1 W G0 j (see also Exercise 1(a)) and so G1 D G0 U because G0 U G1 and jG0 U W G0 j D 2 since G0 \ U D 1 .A/. Set D0 D ht i U ; then D0 Š D8 since it is nonabelian of order 8 and generated by involutions, and G1 is the central product G1 D A D0 with A \ D0 D hzi (indeed, A centralizes t and U so ht; U i D D0 ; before we used the product formula). We also have U D hu; zi < D0 for some involution u. Let v be an element of order 4 in D0 and y an element of order 4 in A; then y 2 D v 2 D z, and since x 2 D y 2 v 2 D y 4 D 1, x D yv is an involution in G1 D0 . Since x t D y t v t D yv 1 D yv v 2 D xz, we see that D1 D hx; t i Š D8 . Because A D Z.G1 /, we also have G1 D A D1 , t 2 D1 and D1 \ U D Z.D1 / D hzi. In the rest of the proof we consider a subgroup S of G which is maximal subject to the following four conditions: (i) S contains G1 D A D1 . (ii) S D AL, where L is a normal subgroup of S and L Š D2n , n 3, (iii) A \ L D hzi D Z.L/, and (iv) L \ G1 D D1 Š D8 . We recall that A D hai is of order 2m , m 2, and Z.G/ A D Z.G1 / so Z.G/

48 On 2-groups with small centralizer of an involution, I

23

is cyclic of order 2m , and 1 .A/ D hzi D Z.D1 / D Z.L/ D 1 .Z.G//, t 2 D1 , CG .t / D ht i A. Also G1 contains U D hz; ui which is a normal abelian subgroup of G of type .2; 2/ and ut D uz. Now we act with A on the dihedral group L, n1 where we set L D hb; t j b 2 D t 2 D 1; b t D b 1 i. Here hbi is the unique cyclic subgroup of index 2 in L so hbi is A-admissible (recall that L is normal in S D AL). We have jhbi \ D1 j D 4 (since D1 L, by (iv)), A centralizes D1 (by (i)) and CL .a/ D ht i.CL .a/ \ hbi/, by the modular law, and so CL .a/ is a dihedral subgroup of L of order 8 containing D1 (recall that t 2 D1 ). Looking at Aut.L/ (see Proposition 34.8, where Aut.D2n / is described) we see that A induces on L a cyclic group of automorphisms of order at most 2n3 and so jA W CA .L/j 2n3 . Since S=L > f1g is cyclic, U 6 L and L is generated by involutions, we have 1 .S / D UL. It follows that 1 .S / \ A D hyi, where o.y/ D 4 (by the product formula) with y 2 D z, and we have hyi D 2 .A/. Let us consider at first the case where y induces on L a nonidentity automorphism so that CA .L/ D hzi and the group hai=hzi acts faithfully on L. Then y induces an automorphism of order 2 on L and since L > CL .A/ D1 Š D8 we must have n 4, Œy; t D 1 (recall that the element y 2 A centralizes t ) and b y D bz, .b 2 /y D .bz/2 D b 2 so that CL .y/ D ht; b 2 i Š D2n1 and CL .y/ is a maximal subgroup of L. Let v be an element of order 4 in D1 ; then v 2 hbi. Since U D hz; ui AD1 , we may set u D yv. We have y b D y.D y y 2 D y 1 , and so utb D .uz/b D .yvz/b D y 1 vz D y 1 vy 2 D yv D u so that hz; u; t bi is an elementary abelian subgroup of order 8 since t b is an involution not contained in U . Obviously, C1 .S/ .u/ D CS .u/\1 .S / D huiht b; b 2 i, where ht b; b 2 i Š D2n1 , and since CG .t / D ht iA does not contain an elementary abelian subgroup of order 8, t cannot be fused in G to any involution contained in CS .u/\1 .S / since the centralizer of any involution from CS .u/ \ 1 .S / contains a subgroup isomorphic to E8 . On the other hand, it is easy to compute that any involution in S lies in L [ .CG .u/ \ 1 .S // since 1 .S / D UL. Naturally, t cannot be fused in G to any involution in U GG since t 62 U . It follows that the G-class of t contains at most 2n2 elements so jG W CG .t /j D 2n2 D jS I CS .t /j. Since CG .t / D CS .t /, this forces that S D G. If m D 2, we have G D F .2; n/, stated in part (g) since ht; b 2 i Š D2n1 centralizes A and the involution t b inverts A. i So assume that m 3. Then we may set (see Proposition 34.8) b a D b 1C2 , where m1 i 2 because CL .a/ D1 . Since CL .a/ Š D2i C1 and CL .a2 / Š D2i Cm and m1 m1 D z 2 Z.L/, we get CL .a2 / D L. It follows that i D n m. We have a2 obtained exactly the groups stated in part (e). In what follows we shall assume always, that hyi D 2 .A/ centralizes L so that 1 .S / D L 2 .A/, the central product (of order 2nC1 ). In this case each involution in S is contained in L [ U (see Appendix 16, where we have proved that D2n C4 Š Q2n C4 ); it follows that U as the unique normal abelian subgroup of type .2; 2/ in S , is characteristic in S and so the subgroup L being generated by its own involutions is characteristic in 1 .S / and so in S (see also Appendix 16). If S D G, we get some groups stated in part (d) of the theorem.

24

Groups of prime power order

From now on we shall assume, in addition, that S < G. Let W D NG .1 .S //; obviously, S < W . Let Kx denotes the G-class containing an element x 2 G. Since W fuses K t \ 1 .S / with K tb \ 1 .S / and CG .t /; CG .t b/ < S . where both classes are of size 2n2 and are contained in L, we get jW W S j D 2, S is normal in W so L is normal in W since L is characteristic in S , by the above. Suppose that W < G and let g 2 NG .W / W be such that g 2 2 W . We have hK t \ W i ht; t bi D L (indeed, involutions t and t b are not conjugate in L so they generate L), U is normal in G and all involutions in S lie in L [ U (see Appendix 16). Therefore (replacing g with gw, w 2 W , if necessary) one may assume that s D t g 2 W S ; indeed, by assumption W < G and the subgroup 1 .S / is not normal in G). Since s normalizes L and CG .t / \ .hsi L/ D ht; zi, we have hsi L Š D2nC1 : indeed, by Proposition 1.8 (Suzuki’s theorem), hsi L is of maximal class and s 62 L. Now L and Lg are normal in W (since L is characteristic in S G W ), Lg \ S D Lg \ L, Lg S D W , jLg S W S j D jLg W .Lg \ S /j D jLg W .L \ Lg /j D 2 and jLLg j D 2nC1 , and so 2 LLg D hsi L. Hence, .LLg /g D Lg Lg D Lg L D LLg since g 2 2 W and L is normal in W , and so hsi L is the dihedral group of order 2nC1 which is normal in W . We have W D A.hsi L/ and this contradicts the maximality of S D AL (see (i–iv)). Hence we must have W D G in any case. If 1 .S / D 1 .G/, we get the groups stated in part (d) of the theorem. In view of the result of the previous paragraph, in what follows we assume, in addition, that 1 .S / ¤ 1 .G/. Then for each involution s 2 G S , s normalizes L since L is characteristic in 1 .G/ G G, and so M D hsi L Š D2nC1 , by Suzuki’s theorem (see Proposition 1.8) since CG .t / \ M D ht; zi is abelian of type .2; 2/. Because of the maximality of S (see (i–iv)), M is not normal in G (otherwise, A would normalize the dihedral subgroup M containing L properly, A \ M D 1 .A/ and M \ G1 D D1 , against the choice of S and L); moreover, the above argument shows also that M is not A-invariant. But L is normal in G so since t 2 L, we get A0 D CG .L/ D CA .L/ (containing 2 .A/, by assumption) is also normal in G and we have A0 D Z.S / which is of order 4. We look at the structure of GN D G=L which is a group with cyclic subgroup SN D AN D .AL/=L Š A=hzi (bar convention!) of order 2m1 and index 2 and with a subgroup MN D M=L of order 2 which intersects N a semidirect product, AN trivially. Since M is not normal in G, we see that GN D MN A, N 4 so m 3 since AN D jA=1 .A/j < jAj). is nonabelian. We conclude that jAj Now jA0 j 4 and so jAN0 j 2 and AN0 is normal in GN (recall that A0 D Z.S / is N Since G= N AN0 is a subgroup of the characteristic in S so normal in G) and AN0 A. N AN0 is abelian outer automorphism group of the dihedral group L, it follows that G= (see Proposition 34.8). Suppose at first that GN is not of maximal class; in that case, GN Š M2m , where N D .UM /=L and UM D 1 .G/ since 2m D jAj (see Theorem 1.2), so that 1 .G/ U and M are generated by involutions. Since M is not normal in G, it follows that s must invert Z.1 .S // D 2 .A/ and so hz; u; si Š E8 . Namely, if s centralizes 2 .A/, then 1 .G/ D 2 .A/ M , the central product, and each involution in G

48 On 2-groups with small centralizer of an involution, I

25

lies in M [ U and so M would be characteristic in 1 .G/ so normal in G, which contradicts the maximal choice of S since S < AM (see (i–iv)). If we set A D hai, then NG .M / D M ha2 i which follows from the structure of GN Š M2m (see Theorem 1.2). Since 2 .A/ centralizes L but 2 .A/=hzi acts faithfully on M , it follows that ha2 i=hzi acts faithfully on M and so ha2 i=2 .A/ acts faithfully on L and CG .L/ D 2 .A/ D A0 . It follows that Z.G/ < A0 D 2 .G/ is of order 2 and A=2 .A/ acts faithfully on L (since the cyclic group Ã1 .A/=2 .A/, as we saw, acts faithfully on L). We have m 4 since GN Š M2m in view of the fact that GN is not of maximal class, N 8. One may assume that by the assumption of this paragraph; we conclude that jAj s the involution s acts on L D hb; t i as follows: b D b 1 and t s D t b and we set n2 b2 D z, where z generates Z.G/. Replacing a with ar (r is odd), one may assume i a that b D b 1C2 , i 2, because CL .a/ D1 Š D8 . Hence we have CL .a/ Š D2i C1 m2 and so CL .a2 / Š D2i Cm1 Š L Š D2n , where we have taken into account m2 i D A=2 .A/ acts faithfully on L and 2 .A/ centralizes L. Thus, that A=ha2 i C m 1 D n and so i D n m C 1. Since i 2, we must have n m C 1. Because m 4, we have here n 5. It remains to determine the commutator Œa; s. We know that A D hai does not normalize M (indeed, G D AM and M is not normal in G) and so hŒa; N sN i D AN0 which yields Œa; s D a0 l, where a0 , a generator of A0 , has order 2 4, a0 D z and l 2 L. We can also write Œa; s D .a0 z/.zl/ D .a0 a02 /.zl/ D a01 .zl/, m2 and s and so replacing l with zl, if necessary, one may assume that a0 D a2 1 a inverts ha0 i. Hence we get Œa; s D a sas D a0 l and so s D a0 ls. Since, by the previous equality, a0 ls is an involution, we must have l D b j for a suitable integer j . 1 j i Therefore, t a sa D t a0 b s which yields b a D b 12j . On the other hand, b a D b 1C2 and so j 2i1 .mod 2n2 /. This yields m2

(1) s a D a2 (2)

sa

D

b 2

i 1

s or

m2 2i 1 a2 b sz. ni 1

ni 2

However, replacing b with b 0 D b 12 and t with t 0 D t b 2 , we see that all so far obtained relations in this case go into the same relations with b 0 instead of b but the relation (1) goes into relation (2). (Note that L D hb; t i D hb 0 ; t 0 i). Therefore we may choose relation (1). We have determined in this case the structure of G uniquely and this the group given in part (f) of the theorem. N GN 0 is of order 4 and Suppose now that GN D G=L is of maximal class. Since G= N AN0 is abelian, we have jAN W AN0 j 2. G= We consider at first the case AN D AN0 which means that A D CG .L/ and so S D A L (a central product) with A \ L D hzi D Z.L/. Hence A D Z.S / and so A is normal in G. Since Ahsi=hzi is of maximal class and the case ŒA; s hzi is not possible because M D hsi L is not normal in G, we have either as D a1 or as D a1 z, which is possible only when m 3. Our group is isomorphic to a group F .m; n/ as stated in part (g). Also we have here hz; u; si Š E8 . It remains to consider the case where jAN W AN0 j D 2 so that CG .L/ D A0 is of index 2 in A. We have A0 D ha2 i D Z.S / and so a induces an automorphism of

26

Groups of prime power order

order 2 on L with CL .a/ D1 . This yields at once that b a D bz and so n 4. Hence we also have ab D az. Since ŒA; L D hzi and L=hzi is a dihedral group of order 2n1 8, it follows that CS .L=hzi/ Š A hb 0 i, where hb 0 i is a cyclic subgroup of order 4 in L. Hence A hb 0 i is normal in G since L=hzi is G-invariant. If s normalizes A, then as D a1 z , D 0; 1, since Ahsi=hzi is of maximal class. But from st s D t b we get acting on a: asts D atb which yields az D a. Indeed, asts D .a1 z /ts D .a1 z /s D az z D a, and this is a contradiction. Hence s does not normalize A. Since As A hb 0 i and .Ahb 0 i=hb 0 i/hsi is of maximal class, one may set (3) as D a1 b 0 or (4) as D a1C2

m2

b 0.

We must have a D a1 D a. But in case (3) this yields .a1 b 0 /s D a and so az D a, which is a contradiction. Hence we must have relation (4). Replacing a with m2 a1 in (4), we get .a1 /s D .a1 /1C2 .b 0 /1 and other relations in this case are m3 0 2 not changed. Thus one may put b D b in (4). This determines the group G as stated in part (h) of the theorem. Here we have also hz; u; si Š E8 . An inspection of obtained groups shows that in any case G has no elementary abelian subgroups of order 16. Indeed, S has no elementary abelian subgroups of order 8 since S=L is cyclic and L is dihedral. It follows from jG W S j 2 that G has no elementary abelian subgroups of order 16, as desired. It follows from defining relations of G that it is generated by 3 elements. Also in case where G has an elementary abelian group of order 8, we see that Z.G/ has order 2. The proof is complete. s2

Since G of Theorem 48.1 has no normal subgroups isomorphic to E8 , each subgroup of G is generated by 4 elements, by MacWilliams’ theorem [MacW]. (See the proof of this theorem of MacWilliams in 50.) Let G D U wr V , where U Š C2m and V D ht i Š C2 . Then CG .t / D ht i C , where C , the diagonal of U U t , is cyclic of order 2m , Z.G/ D C . Exercise 2. Let L be a subgroup of maximal class in a 2-group G. Suppose that, whenever an involution s 2 G L, then hs; Li is of maximal class. Is G of maximal class? Solution. Yes. If G L has no involutions, G is of maximal class, by Theorem 1.17(a). So suppose that G L has an involution. Assume that G is a counterexample of minimal order. Then jG W Lj > 2 and all maximal subgroups of G containing L are of maximal class. The number of maximal subgroups of G containing L which are of maximal class is even (see 13). Since the number of maximal subgroups of G containing L is odd, we get a contradiction. Exercise 3. Let t be an involution of a 2-group G such that CG .t / D ht i Q, where Q Š Q2m . Prove that (i) G has no normal elementary abelian subgroups of order 8, (ii) jNG .CG .t // W CG .t /j 2.

48 On 2-groups with small centralizer of an involution, I

27

Exercise 4. Let G be a nonabelian p-group, p is odd. Suppose that G has an element t of order p such that CG .t / D ht i C , where C is cyclic of order p m , m > 1. Prove that G has no ht i-invariant subgroups of order p pC1 and exponent p. (Hint. Use Theorem 9.6.)

Problems Problem 1. Classify the 2-groups G containing an abelian subgroup A of type .4; 2/ such that CG .A/ D A. Problem 2. Classify the 2-groups containing a cyclic subgroup Z of order 4 such that CG .Z/ D Z C2 . (For a solution, see 77.) Problem 3. Let a 2-group G admit an automorphism ˛ of order 2 and set CG .˛/ D Q, where Q is either cyclic or of maximal class. Describe the structure of G. (See 51.) Problem 4. Classify the 2-groups G containing an involution t such that CG .t / D ht i M2m . Problem 5 (Janko). Classify the 2-groups G such that CG .E/ D E for some elementary abelian subgroup E of order 8. (For solution, see 51.) Problem 6 (Blackburn). Classify the nonabelian p-groups G, p > 2, containing an element t of order p such that CG .t / D ht i C , where C Š Cp m , m > 1. Problem 7. Classify the 2-groups G containing a nonabelian subgroup D of order 8 such that jCG .D/j D 4. Problem 8. Study the 2-groups G containing an extraspecial subgroup E of order > 23 such that CG .E/ D Z.E/. Problem 9. Study the 2-groups G containing a subgroup M Š M2n such that we have CG .M / D Z.M /.

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On 2-groups with small centralizer of an involution, II

In the previous section finite 2-groups G have been classified with the property that G has an involution t such that CG .t / D ht i C , where C Š C2m is a cyclic group of order 2m , m 1. There is one more case of the centralizer of an involution in 2-groups which is very important. This is seen from the following “conditionless” Theorem 10.26: For a finite p-group G, one of the following holds: (a) G has no maximal elementary abelian subgroups of order p 2 . (b) j1 .G/j p 2 . (c) There exists in G an element x of order p such that CG .x/ D hxi Q, where Q is cyclic or generalized quaternion. In this section we classify the finite 2-groups G which possess an involution t such that CG .t / D ht i Q, where Q Š Q2m is generalized quaternion of order 2m ; m 3. Of course, the situation becomes more complicated; however, in spite of that, the exposition remains elementary. From the start, it is clear, that such a group G cannot be of maximal class. Then, by the known results, G must contain a normal four-subgroup U . We have two essentially different possibilities according to t 2 U or t … U . In the first case, where t 2 U , we have either CG .t / D G or the order of G is equal to 2mC2 and then we get four classes of 2-groups which will be given in terms of generators and relations (see Theorem 49.1). In the second case, where t 62 U (Theorem 49.2), the situation is more complicated since the order of G is not bounded with the parameter m. In fact, in some cases with given m 3, we get infinitely many 2-groups with the same centralizer of our involution t . The idea of the proof in the second case is to construct certain large subgroup S of the known structure with U CG .t / S and then to show that S is normal in G and that jG=S j 4. This last “closure” argument is the main trick in the proof. The corresponding argument in the previous “cyclic” case, which was treated in the previous subsection, was considerably simpler. As a result, we obtain three different types of 2-groups G, since we have exactly three possibilities for the structure of S . In two of these types, the groups G could possess elementary abelian

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On 2-groups with small centralizer of an involution, II

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subgroups of order 16 and in that case the corresponding series of 2-groups will be given in terms of generators and relations (Theorem 49.3). The first member of one of these series is a 2-group of order 27 which is isomorphic to a Sylow 2-subgroup of the simple group J2 of order 604800. Finally, it turns out that the case m D 3 of an ordinary quaternion group Q Q8 is more difficult than the general case m 4. As a direct application of these results and the results in the previous subsection, we obtain a classification of 2-groups which have more than 3 involutions but which do not have an elementary abelian subgroup of order 8 (Theorem 49.4). Such groups have been considered by D. J. Rusin [Rus] by using a heavy cohomological machinery. Our methods are completely elementary. Let G D Q wr C , where Q Š Q2m and C D ht i Š C2 . Then CG .t / D ht i D, where D is the diagonal of the base Q Qt so D Š Q2m . Let G be a 2-group possessing an involution t such that CG .t / D ht i Q, where Q Š Q2m . We claim that then G has no normal elementary subgroups of order 8. Assume that this is false, and let E Š E8 be normal in G. Clearly, t 62 E. Set H D ht i E. Then CH .t / contains an elementary abelian subgroup of order 8, by Proposition 1.8 (Suzuki’s theorem), which is not the case. We see that the group G is one of the groups of 50. It follows from Theorem 50.3 that every subgroup of G is generated by four elements. 1o . The case t 2 U . In this subsection we prove the following Theorem 49.1. Let G be a 2-group containing an involution t such that CG .t / D ht i Q, where Q is a generalized quaternion group of order 2m ; m 3. We assume, in addition, that t is contained in each normal four-subgroup U of G and that G ¤ CG .t /. Then we have the following possibilities: (A1) If t 62 ˆ.G/, then G has a maximal subgroup M so that G D M ht i, where M Š Q2mC1 or M Š SD2mC1 and CM .t / Š Q2m . (A2) If t 2 ˆ.G/, then one of the following assertions holds: (a) m D 3 and G D ht; u; v; w; y j y 2 D t v; w 2 D v 2 D u; u2 D t 2 D Œt; v D Œt; w D 1; Œv; w D u; t y D t u; v y D v 1 ; w y D wt i. We have jGj D 25 , Z.G/ D hui has order 2, ˆ.G/ D ht i hvi is abelian of type .4; 2/, G 0 D hu; t i is a four-group and CG .t / D ht i hv; wi, where hv; wi Š Q8 . Also G D hy; wi and G has exactly three involutions. m1

D t 2 D Œt; a D Œt; b D Œa; x D (b) m 4 and G D ha; b; t; u; xi, where a2 m3 2 2m2 b 1 2 1, b D a D u, a D a , x D a2 , b x D bt , t x D t u. The mC2 , Z.G/ D hui is of order 2, ˆ.G/ D G 0 D ht i ha2 i is order of G is 2 abelian of type .2m2 ; 2/ and G is not an -group since there exist involutions in G ha; b; ti but G has no subgroups isomorphic to E8 . Also, G=ˆ.G/ has order 8 and CG .t / D ht i ha; bi, where ha; bi Š Q2m and G D ha; b; xi.

30

Groups of prime power order m1

(c) m 4 and G D ha; b; t; u; xi, where a2 D t 2 D Œt; a D Œt; b D Œa; x D m2 m3 1, b 2 D a2 D u, ab D a1 , x 2 D a1C2 , b x D bat , t x D t u. The mC2 order of G is 2 ; Z.G/ D hui is of order 2, ˆ.G/ D ht i hai is abelian of type .2m1 ; 2/ and G 0 D hat i is cyclic of order 2m1 . Also, G has exactly 3 involutions , G=ˆ.G/ has order 4, CG .t / D ht i ha; bi, where ha; bi Š Q2m and G D hb; xi. m1

(d) m 5 and G D ha; b; t; u; xi, where a2 D t 2 D Œt; a D Œt; b D x 2 D 1, m3 2 2m2 b 1 x b D a D u, a D a , t D t u, ax D a1C2 t , b x D b. The order of G is 2mC2 , Z.G/ D hui is of order 2, ˆ.G/ D G 0 D ht i ha2 i. Also, G is not an -group but G has no subgroups isomorphic to E8 . We have CG .t / D ht i ha; bi, where ha; bi Š Q2m and G D ha; b; xi, Here we have CG .x/ Š C2 Q2m2 . Proof. Suppose that G satisfies the assumptions of the theorem and let U be a normal four-subgroup of G. By assumption, we have t 2 U . Let T D CG .U /, so that jG W T j D 2. We have T D ht i Q, where Q Š Q2m ; m 3. Set 1 .Q/ D hui and then we have U D ht; ui D 1 .T / and obviously Z.G/ D hui. We consider at first the easy case t … ˆ.G/. Let M be a maximal subgroup of G such that t … M . Then G D ht i M and, by the modular law, CG .t / D ht i M0 , where M0 D T \ M Š Q2m . If M is not of maximal class, then by Lemma 1.4, M contains a G-invariant four-subgroup U0 . But then t … U0 , which contradicts the assumption of our theorem. Thus, M is of maximal class and therefore M Š Q2mC1 or M Š SD2mC1 (see Theorem 1.2). There is an element y 2 M M0 such that y 2 2 hui D Z.M /. We have y 1 ty D ut so that tyt D yu and this determines the action of t on M since all such y’s together with M0 generate M . We have obtained the group G stated in part (A1) of the theorem. We examine now the difficult case t 2 ˆ.G/ and suppose at first that Q Š Q8 so that hui D Z.Q/ D Z.G/ D ˆ.T / and U D ht; ui. It follows from u 2 ˆ.Q/ ˆ.G/ and U 6 Z.G/ that U < ˆ.G/, by Exercise 1(b3). Since ˆ.G/ < T , we get, by the modular law, ˆ.G/ D ht i hvi, where hvi is a cyclic subgroup of order 4 in Q and v 2 D u; then ˆ.G/ is abelian of type .4; 2/ and G is generated by two elements. Because in case of 2-groups ˆ.G/ D Ã1 .G/ and here we have ˆ.G/ > U , it follows that there are elements y; z 2 G T so that y 2 D t v and z 2 D v since t and t u are not squares. Next, hyi 6E G (otherwise, G is metacyclic since G=hyi is cyclic in view of ˆ.G/ 6 hyi). Similarly, hzi 6E G. If z normalizes Q, then hQ; zi is of maximal class and order 16, by Theorem 1.2, a contradiction since ˆ.G/ is not isomorphic to subgroup of a 2-group of maximal class. Thus, y does not normalize Q so Q is not normal in G. It follows that the core hziG D hvi (let us consider the representation of G by permutations of left cosets of hzi). Since c2 .ˆ.G// D 2, hvt i is also normal in G. We have t y D t u D t z , by Exercise 1(b2). It follows from t v D .t v/y D t y v y D t uv y that v y D vu D v 3 D v 1 . Take an element w 2 Q hvi so that Q D hv; wi. Since G=ˆ.G/ is abelian and Q 6E G, hvi G G, we

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On 2-groups with small centralizer of an involution, II

31

have w y D wv r t for some integer r. If r is odd, then Œw; y D v r t D .vt /r . Thus, modulo hvt i, elements w, y and v are pairwise permutable. Since G D hQ; yi D hv; w; yi and v 2 D w 2 D u and y 2 D vt are contained in hvt i, we see that G=hvt i is elementary abelian, contrary to the fact that ˆ.G/ is noncyclic. Thus, r is even, and we get w y D wt or w y D wut D w 1 t . However, if w y D w 1 t , then replacing y with yv, we get w yv D .w 1 t /v D wt and other relations remain unchanged: t yv D t u, v yv D v 1 , .yv/2 D yvyv D y 2 v y v D t vv 1 v D t v. Hence we may assume from the start that w y D wt . The structure of the group G of order 25 is uniquely determined and this is the group stated in part (A2)(a) of our theorem. We turn now to the general case Q Š Q2m , m 4. Let a be an element of order 2m1 in Q so that hai is the unique cyclic subgroup of index 2 in Q. Here CG .t / D ht i Q D T D CG .U /, where U D ht; ui is the G-invariant four-subgroup in T , hui D Z.Q/ D Z.G/ and hui < hai. Now, T =U Š Q=hui Š D2m1 so that ha; U i=U is the unique cyclic subgroup of T =U of order 2m2 4 and so L D ha; U i D ht i hai is normal in G. If b 2 Q hai, then b 2 D u and ab D a1 . Taking any x 2 G T , we have t x D t u and ax 2 L. If x does not normalize hai, then hai\hax i is an x-invariant subgroup of index 2 in hai since jL W haij D 2 (here we use the product formula). Hence, in any case, ha2 i is a cyclic normal subgroup of order 2m2 4 of G. In particular, the cyclic subgroup hvi of order 4 contained in ha2 i is normal in G. Thus, A D ht ihvi is a normal abelian subgroup of type .4; 2/ in G. We have CG .A/ D L so that G=L acts as an automorphism group of order 4 on A. Since hvi is normal in G and v b D v 1 , so there is an element x in G T with v x D v and also t x D t u D t v 2 . Indeed, CG .v/ is a maximal subgroup of G different of T so as x we can take every element in CG .v/T . Since t b D t , we see that b and x induce two distinct involutory automorphisms on A and therefore G=L is an elementary abelian group of order 4 (recall that b 62 L and t; v 2 L). In particular, we get that ˆ.G/ L. On the other hand, ˆ.G/ ht; a2 i. Hence we have either ˆ.G/ D L D ht; ai or ˆ.G/ D ht; a2 i.< L/. In any case, x 2 2 L since exp.G=L/ D 2 and so x induces an automorphism of order 2 on L such that t x D t u; ha2 ix D ha2 i and x centralizes the subgroup hvi of order 4 which is contained in ha2 i. Therefore, we have ax D aj t i ; where i D 0; 1 and an integer j is odd (otherwise, o.aj t i / < o.a/). Since x normalizes ha2 i and centralizes hvi ha2 i, we have either .a2 /x D a2 or .a2 /x D a2 u (indeed, if x does not centralize a2 , then hx; a2 i=Chxi .a2 / Š M2m1 , by Theorem 1.2); in particular, in the case under consideration, m 5. If .a2 /x D a2 , we obtain ax D at i or ax D aut i . Indeed, .a2 /x D a2j so j 1 .mod 2m2 /; then aj D a or aj D au. If .a2 /x D a2 u, we get ax D avt i or ax D auvt i , where we set m3 v D a2 . Indeed, then j 1 C 2m2 so aj D av or aj D auv. As we established above, we must have in the case where x does not centralize a2 , o.a2 / 8 and so m 5. Let ax D at i or ax D aut i , where m 4 and o.a/ D 2m1 . If ax D aut i , then replacing a with at , we get .at /x D aut i t u D .at /t i . Note that hat; bi D Q0 Š Q2m and also CG .t / D ht i Q0 . Hence we may assume from the start that ax D at i ,

32

Groups of prime power order 2

i D 0; 1. However, if ax D at , then we compute a D ax D .at /x D at t u D au, which gives u D 1 and this is a contradiction. It follows that ax D a. If x 2 D t aj with an integer j , then CG .x/ ha; t aj i D L, which is a contradiction since x 62 L D CG .L/. Hence we have x 2 2 hai. If b x 2 Q, then hxi normalizes Q D ha; bi and since x 2 2 hai, so Qhxi would be a maximal subgroup of G which does not contain t 2 ˆ.G/, a contradiction. It follows that b x D b 0 t , where b 0 is an element of order 4 contained in Q hai. There are exactly two conjugacy classes of elements of order 4 in Q hai with the representatives b and ba. Thus, replacing x with xak for some integer k, we may assume that b x D bt or b x D bat . 2 Let b x D bt . We compute b x D .bt /x D bt t u D b 1 and so x 2 2 hvi, where m3 m3 v is an element of order 4 in hai. Hence, we may set x 2 D a2 or x 2 D a2 u m3 2 2 because these elements are the only elements of order 4 in hai. If x D a u, then we replace x with xt . Then xt 2 G T , xt centralizes hai, b xt D .bt /t D bt and m3 m3 finally .xt /2 D xtxt D x 2 t x t D a2 ut ut D a2 . Hence, we may assume from m3 the start that x 2 D a2 and so the structure of G is uniquely determined in this m4 case. We see that here ˆ.G/ D ht i ha2 i D G 0 and G D ha; b; xi. Also, xa2 t is an involution in G ha; b; t i. We have obtained the group stated in part .A2/.b/. 2 Now let b x D bat . We compute b x D .bat /x D bat at u D ba2 u. On the other m3

1C2m3

hand, by setting v D a2 , we see that b a D b av D ba2 u. Hence, we may set 2 2 x D av or x D avu. Note that the group hai=hui produces by conjugation exactly 2m3 distinct conjugates of b. However, if x 2 D avu, then similarly as before, we replace x with xt so that we obtain .xt /2 D xtxt D x 2 t x t D avut ut D av and all m3 other relations remain unchanged. Hence, we may assume that x 2 D av D a1C2 and the group G is again uniquely determined. Here we have ˆ.G/ D ht i hai D L and our group G D hx; bi is 2-generated. Also, we see that G 0 D hat i and there are no involutions in G T . Therefore G has exactly 3 involutions. We have obtained the group stated in part .A2/.c/. m3 Now let ax D avt i or ax D auvt i , where v D a2 , m 5. If ax D auvt i , then replacing a with at (as above) we get .at /x D auvt i t u D .at /vt i . Hence we may assume from the start that ax D avt i , i D 0; 1: However, if ax D av; then a D 2 ax D .av/x D .av/v D av 2 D au, which gives u D 1 and this is a contradiction. It remains only the case ax D avt: We note that x 2 2 L and CL .x/ D ha4 ; x 2 i, where hvi ha4 i since o.a/ D 2m1 16. All elements in T L are of order 4. There are exactly four conjugate classes of such elements in T L and under the action of hai each such element is conjugate to one of the following four elements: b, ba, bt , bat , where ha; bi D Q Š Q2m . So replacing x with xak for some integer k (note that xak also centralizes v), we may assume that we have one of the following four 2 possibilities: b x D b, b x D bt , b x D ba or b x D bat . If b x D ba, then b x D x 2 2 .ba/ D baavt D ba vt , which is a contradiction since x 2 L < T and Q is normal 2 in T: Similarly, if b x D bat; then b x D .bat /x D bat avt t u D ba2 vut , which is a contradiction since Q is normal in T: If b x D bt; then we replace b with b 0 D ba

49

and x with x 0 D xa2

m4

m4 xa2

33

On 2-groups with small centralizer of an involution, II m3

(which also centralizes v D a2

m4 a2

0

). We compute .b 0 /x D

.ba/ D .bt avt / D bb 1 a2 ba2 av D ba2 a2 av D 1 0 x bv av D ba D b . Hence we may assume from the start that b D b: In this case x 2 2 CL .b/ D ht; ui D U: Since CG .U / D T; we must have x 2 2 hui D Z.G/: If x 2 D u; then replacing x with xt; we get .xt /2 D xtxt D x 2 t x t D ut ut D 1 and other relations remain unchanged: axt D .avt /t D avt and b xt D b t D b, Hence we may assume that x 2 D 1 and so the structure of G is uniquely determined as given in part (A2)(d). m4

m4

m4

m4

2o . The case t 62 U . In this subsection we examine the difficult case, where our 2group G has a normal four-subgroup U such that an involution t 2 G is not contained in U and CG .t / D ht i Q with Q Š Q2m , m 3. In order to formulate our main result, we shall define three types of 2-groups: A.m; n/, B.m; n/ and C.m; n/, m; n 2 N . m1

Definition. Let S D QL be a product of two normal subgroups Q D ha; b j a2 D 4 2 2m2 2 b 1 2n1 2 b D z D 1; a D b D z; a D a i Š Q2m and L D hc; t j c Dt D n2 z 2 D 1; c 2 D z; c t D c 1 i Š D2n , where m 3, n 3 and Q \ L D Z.Q/ D Z.L/ D hzi. If ŒQ; L D 1; then S D Q L is the central product of Q and L which we denote with A.m; n/ for m 3, n 3. If CQ .L/ D hai and t b D t , c b D cz, then so determined group S we denote with B.m; n/ for m 3, n 4. If CQ .L/ D ha2 ; bi Š Q2m1 , m 4 and t a D t , c a D cz, then so determined group S we denote with C.m; n/ for m 4, n 4. Obviously, jGj D 2mCn1 . Theorem 49.2. Let G be a 2-group containing an involution t such that CG .t / D ht i Q, where Q is a generalized quaternion group of order 2m , m 3. We assume, in addition, that G has a normal four-subgroup U such that t … U . Then G has a normal subgroup S containing U CG .t / such that jG=S j 4 and S is isomorphic to one of the groups A.m; n/, B.m; n/ or C.m; n/. If S Š B.m; n/, then we must have S D G. If S Š C.m; n/, then jG=S j 2 and if, in addition, jG=S j D 2, then we have m D n. Finally, if jG=S j D 4, then we must have S Š A.m; m/ and G acts transitively on the set of involutions in S U . Proof. Let E4 Š U G G with t 62 U . Set T D CG .U / and then we have obviously t 62 T . This gives jG W T j D 2 and so G D ht iT . By the modular law, G0 D CG .t / D ht i Q, where Q D CT .t / Š Q2m , m 3, and so Z D hzi D 1 .Q/ D Z.Q/ D CU .t / D Z.G/. Set G1 D NG .G0 /. Since j.UG0 / W G0 j D 2, we have UG0 G1 . It follows from 1 .G0 / D ht; zi that t has exactly two conjugates t and t z in G1 . Therefore, jG1 W CG1 .t /j D jG1 W G0 j D 2 so G1 D UG0 . Set D0 D U ht i so that D0 Š D8 , ŒQ; D0 D f1g with Q \ D0 D hzi and U D hz; ui <

34

Groups of prime power order

D0 so G1 D Q D0 , by the product formula. Let v be an element of order 4 in m1 D0 and let y be an element of order 4 in hai, where Q D ha; b j a2 D b4 D m2 2 2 2 b 1 2 D b D z; a D a i Š Q2m , m 3. Since y D v 2 D z, z D 1; a 2 we get .yv/ D y 2 v 2 D y 4 D 1 so x D yv is an involution in G1 D0 . But x t D .yv/t D yv 1 D yvv 2 D yvz D xz and therefore D1 D hx; t i Š D8 and D1 \ U D Z.D1 / D hzi. We see that CG1 .D1 / D ha; bt i D Q1 Š Q2m because x bt D .yv/bt D y 1 v 1 D yv D x, .bt /2 D z and abt D a1 . Thus we have G1 D Q1 D1 with Q1 Š Q, Q1 \ D1 D hzi, CG .t / D ht i Q D ht i Q1 and U D hz; ui 6 D1 . Denoting again Q1 with Q and bt with b, we have obtained the following initial result. (R) Supposing that U D hz; ui is a normal four-subgroup of our 2-group G possessing an involution t with t … U and CG .t / D ht i Q, where Q Š Q2m , m 3, then G1 D U CG .t / has the following structure. We have G1 D Q D1 , where D1 Š D8 , t 2 D1 , Q \ D1 D hzi D Z.G/ and U 6 D1 . The next step in the proof is to “blow up” the subgroup G1 D U CG .t / as much as possible so that the structure of a large subgroup S containing G1 remains “about” the same as the structure of G1 . In the rest of the proof, we denote with S a subgroup of G of the maximal possible order subject to the following conditions: (i) S G1 D U CG .t / D Q D1 , where Q Š Q2m , m 3, D1 Š D8 , t 2 D1 . (ii) S D QL, where L is normal in S , L Š D2n , n 3, Q \ L D hzi D Z.Q/ D Z.L/, L D1 . (iii) U D hz; ui 6 L. m1

In the sequel, we fix the following notation. We set Q D ha; b j a2 D b4 D 2 2m2 2 b 1 2n1 2 z D 1; a D b D z; a D a i Š Q2m and L D hc; t j c D t D z2 D n2 D z; c t D c 1 i Š D2n , where m 3, n 3 and Q \ L D Z.Q/ D 1; c 2 Z.L/ D hzi. Then we have u D yv, where y is an element of order 4 in hai and v is an element of order 4 in hci. If a4 D 1, then we put y D a. Similarly, if c 4 D 1, then we put v D c. At first we shall determine the structure of S . Act with Q on the dihedral group L. Since Aut.L/=Inn.L/ is abelian of type .2n3 ; 2/ (see Theorem 34.8), so Q0 D ha2 i centralizes L. Also, we know that Q centralizes hv; t i D D1 Š D8 . It follows that either Q centralizes L and then S D Q L Š A.m; n/, m 3, n 3 or M D CQ .L/ is a maximal subgroup of Q, in which case n 4. We have essentially two different possibilities for M . First possibility is M D hai, t b D t , c b D cz and then S Š B.m; n/, m 3, n 4. Second possibility is M D ha2 ; bi Š Q2m1 , m 4, t a D t , c a D cz and then S Š C.m; n/, m 4, n 4. We see that we have in any case ŒQ; L hzi and so Q is a normal subgroup in S . Suppose that S Š B.m; n/. Then S has exactly six conjugacy classes of involutions contained in S U with the representatives t , t c, bv, bav, bt c, bat c. The

49

35

On 2-groups with small centralizer of an involution, II

corresponding centralizers are: CS .t / D CG .t / D ht i Q

with

Q Š Q2m ;

CS .t c/ D ht ci ha; bvi

with

ha; bvi Š D2m ;

CS .bv/ D hbvi hcy; t ci

with

hcy; t ci Š SD2n ;

CS .bav/ D hbavi hcy; t ci

with

hcy; t ci Š SD2n ;

CS .bt c/ D hbt ci hby; ui

with

hby; ui Š D8 ;

CS .bat c/ D hbat ci hbay; ui

with

hbay; ui Š D8 :

m 3;

It follows that t cannot be fused in NG .S / to any of the other five conjugacy classes of involutions in S U . But CS .t / D CG .t / then forces that S D G. We make here the following simple observation. Since t 2 G T , where T D CG .U / and jG W T j D 2, t cannot be conjugate (fused) in G to any involution t 0 which centralizes U: Therefore, with respect to that fusion, it is enough to consider only those involutions in S which act faithfully on U . Now suppose that S Š C.m; n/. Then S has exactly four conjugacy classes of involutions contained in S U acting faithfully on U with the representatives t , t c, bv, and bav. The corresponding centralizers in S are: CS .t / D CG .t / D ht i Q

with

Q Š Q2m ;

CS .t c/ D ht ci hav; bavi

with

hav; bavi Š SD2m ;

CS .bv/ D hbvi hc; yt i

with

hc; yt i Š Q2n ;

CS .bav/ D hbavi hcy; ct i

with

hcy; ct i Š SD2n :

m 4;

We may assume that NG .S / ¤ S (otherwise we are finished). Then NG .S / can fuse the conjugacy class of t only with the conjugacy class of bv under the assumption that m D n and then jNG .S / W S j D 2. Then cclS .t /, the conjugacy class of in S containing t , equals ft c 2i ; 1 i 2m2 g and CS .t c 2i / D ht c 2i i Q so that .CS .t c 2i //0 D ha2 i. Similarly, we have cclS .bv/ D fba2i v; 1 i 2m2 g, CS .ba2i v/ D hba2i vi hc; yt i, where hc; yt i Š Q2m so that .CS .ba2i v//0 D hc 2 i. Hence for any x 2 NG .S / S , we have t x D ba2j v for some j and so, by the above, ha2 ix D hc 2 i. Assume now that NG .S / ¤ G (otherwise we are finished). Since CS .t / D CG .t / and NG .S / already fuses the involutions in cclS .t / with involutions in cclS .bv/, so there is an element s 2 NG .NG .S // NG .S / with s 2 2 NG .S / and t 0 D t s 2 NG .S / S . Indeed, t s 2 cclS .t c/ or t s 2 cclS .bav/ is not possible since a semi-dihedral group is not a subgroup of a generalized quaternion group. However, if t s 2 cclS .t / [ cclS .bv/; then there is an n 2 NG .S / such that t s D t n . But then 1 t sn D t , which contradicts the fact that CG .t / NG .S /. By the above, we have 0 .a2 /t D c 2k with k odd. Since y is an element of order 4 contained in ha2 i and v 0 is an element of order 4 contained in hc 2 i, we get that y t D vz l with l D 0; 1. But

36

Groups of prime power order 0

0

then .yvz l /t D yvz l which gives .uz l /t D uz l , recalling that yv D u. We have proved that t 0 centralizes U D hz; ui and so CG .t 0 / contains a subgroup isomorphic to E8 . This is a contradiction, since t 0 is conjugate in G to t . Hence, we must have NG .S / D G and so jG W S j D 2, as required. Suppose, finally, that S Š A.m; n/, m 3, n 3; i.e. S D Q L is the central product of Q and L. Then S has exactly four conjugacy classes of involutions contained in S U with the representatives t , t c, bv and bav and all these involutions act faithfully on U . The corresponding centralizers are: CS .t / D CG .t / D ht i Q CS .bv/ D hbvi hc; yt i

with Q Š Q2m ; CS .t c/ D ht ci Q;

with hc; yt i Š Q2n and CS .bav/ D hbavi hc; yt i:

Note that all 2n2 conjugates of t in S lie in the dihedral subgroup L D hc; t i and also all 2n2 conjugates of t c in S lie in L. Similarly, all 2m2 conjugates of bv in S lie in the dihedral subgroup K D hbv; bavi Š D2m and also all 2m2 conjugates of bav in S lie in K. These are all 2 2n2 C 2 2m2 involutions in S U . Also, we see that m1 K D ha; bv j a2 D .bv/2 D 1; abv D a1 i. We want to determine all dihedral subgroups of S which are generated by a pair of distinct involutions in S U . We recall that any two distinct involutions always generate a dihedral subgroup. Any two distinct involutions in cclS .t / [ cclS .t c/ generate a dihedral subgroup which is contained in L. Similarly, any two distinct involutions in cclS .bv/ [ cclS .bav/ generate a dihedral subgroup which is contained in K. We note that hci fuses (by conjugation) all involutions in cclS .t / to t and all involutions in cclS .t c/ to t c. Similarly, hai fuses all involutions in cclS .bv/ to bv and all involutions in cclS .bav/ to bav. We have Œc; a D 1 and a centralizes t and both t c and c centralize bv and bav. All this means that it is enough to see what is the order of the following four dihedral subgroups: hbv; t i; hbv; t ci, hbav; t i and hbav; t ci. The computation shows: .bv t /2 D .bv t c/2 D .bav t /2 D .bav t c/2 D z, and so all these four dihedral subgroups have order 8. Suppose that t is not conjugate in NG .S / to any involution in S L. That will certainly happen if (for example) m ¤ n. Supposing, in addition, that S ¤ G, we see that NG .S / fuses t with t c and so jNG .S / W S j D 2, since CG .t / D CS .t /. If NG .S / D G, then we are finished. Therefore, we assume that NG .S / ¤ G. Take an element s 2 G NG .S / such that s normalizes NG .S / and s 2 2 NG .S /. If an involution x 2 NG .S / S , then .cclL .t //x D cclL .t c/ implies that L0 D Lhxi Š D2nC1 : If jCS .x/j D 2m ; then CL .x/ D hzi and CNG .S/ .x/ D hxi CS .x/ imply that CS .x/ covers S=L and so Q normalizes L0 Š D2nC1 and QL0 D NG .S /; which contradicts the maximality of S . Hence there is no such involution x in NG .S /S with jCS .x/j D 2m : Set P D cclS .t / [ cclS .t c/ [ cclS .bv/ [ cclS .bav/ so that hP i D S and P G T . If t 0 D t s 2 NG .S / S , then CNG .S/ .t 0 / D ht 0 i CS .t 0 / and so jCS .t 0 /j D 2m ; a contradiction. If t 0 2 cclS .t / [ cclS .t c/, then there is n 2 NG .S / such that t 0 D t s D t n and so CG .t / 6 NG .S /; a contradiction. Hence t 0 2 cclS .bv/ or t 0 2 cclS .bav/. If NG .S / fuses bv and bav, then all elements in P are conjugate in

49

On 2-groups with small centralizer of an involution, II

37

hyiNG .S / to t; and since hP i D S , there is x 0 2 P such that x0 D .x 0 /s 2 NG .S /S and so CNG .S/ .x0 / D hx0 i CS .x0 / and jCS .x0 /j D 2m , a contradiction. Suppose that NG .S / does not fuse bv and bav. Then jCNG .S/ .bv/j D jCNG .S/ .bav/j D 2mC2 and so n C 2 D m C 1 because t 0 2 cclS .bv/ or t 0 2 cclS .bav/. There is x 0 2 P such that x0 D .x 0 /s 2 NG .S / S and so jCNG .S/ .x0 /j D 2nC2 D 2mC1 which gives jCS .x0 /j D 2m , a contradiction. This gives NG .S / D G and so jG W S j D 2, as required. Now suppose that t is not conjugate in NG .S / to t c 2 L but S ¤ G. Then NG .S / must fuse t to an involution in S L U . Without loss of generality, we may assume that t is fused in NG .S / to bv. This gives jNG .S / W S j D 2 and m D n because CS .bv/ Š C2 Q2n . Supposing NG .S / ¤ G, we take an element s 2 G NG .S / such that s normalizes NG .S / and s 2 2 NG .S /. Set again P D cclS .t / [ cclS .t c/ [ cclS .bv/ [ cclS .bav/ so that hP i D S and P G T . There is an x 0 2 P such that t 0 D .x 0 /s 2 NG .S / S and t 0 acts faithfully on U . Suppose at first that m D n 4. Then L D hc; t i and K D ha; bvi are the only dihedral subgroups of order 2m in S and so for each x 2 NG .S / S we have Lx D K and so hcix D hai. In particular, 0 v t D yz ; D 0; 1. It follows that t 0 centralizes yvz D uz and since z 2 Z.G/, so t 0 centralizes u. Hence, t 0 centralizes U D hz; ui , a contradiction. We get in this case NG .S / D G and so jG W S j D 2, as required. Suppose now that m D n D 3 so that S is extraspecial of order 25 and S Š Q8 D8 . Then t (fused with bv) lies 0 in a conjugacy class of length 4 in NG .S /. We have t t 2 fbv; bvzg and ht; bvi D D Š D8 . Hence Dht 0 i D D0 Š D16 and so t 0 acts fixed-point-free on D hzi. For each involution t0 in S U , we have CS .t0 / D ht0 i Q0 with Q0 Š Q8 . Since t 0 is a conjugate of an involution in S U under the action os s (normalizing NG .S /), it follows that CNG .S/ .t 0 / D ht 0 i CS .t 0 / contains a subgroup isomorphic to Q8 and so CS .t 0 / D Q Š Q8 because Q \D D hzi. Since NNG .S/ .D/ hS; t 0 i D NG .S /, it follows that Q normalizes D0 D Dht 0 i Š D16 and so D0 is normal in NG .S / (since Q covers S=D). Note that t 2 D and so we have obtained a contradiction with the maximality of S . Hence, also in this case, we have NG .S / D G and so jG W S j D 2. It remains to consider the case where t is fused in NG .S / to each involution in S U . This gives at once that m D n and jNG .S / W S j D 4 so that NG .S /=S acts regularly on the four S -classes of involutions which are contained in S U . Let m D n 4. Then S has exactly two dihedral subgroups L D hc; t i and K D ha; bvi of order 2m and L and K contain all involutions from S U . Therefore, for each x 2 NG .S / S , we have either K x D K and Lx D L or K x D L. Supposing again that NG .S / ¤ G, we take an element s 2 NG .NG .S // NG .S / 0 0 with s 2 2 NG .S /. Then we have t 0 D t s 2 NG .S / S . If Lt D L and K t D K, then t 0 fuses the two classes of involutions in L hci and also t 0 fuses the two classes 0 0 of involutions in K hai. This gives that c t D c 1 and at D a1 . In particular, 0 0 we get ut D .yv/t D y 1 v 1 D yz vz D yv D u. Hence, the elementary abelian subgroup hz; u; t 0 i Š E8 is contained in CG .t 0 / and this is a contradiction. 0 0 0 Now assume that Lt D K. Then hcit D hai and, in particular, v t D yz , D 0; 1.

38

Groups of prime power order 0

0

But then .yvz /t D yvz and consequently .uz /t D uz . Hence t 0 centralizes U D hz; ui, which is a contradiction, since t 0 is conjugate in G to t . Finally, we assume that m D n D 3 so that S Š Q8 D8 is an extraspecial group of order 25 . We recall that NG .S / acts transitively on the set of 8 involutions contained in S U . Again assuming that NG .S / ¤ G, we take an element s 2 NG .NG .S / NG .S / with s 2 2 NG .S /. Then we have t 0 D t s 2 NG .S / S . Set M D hsiNG .S / so that jM W NG .S /j D 2. We shall determine V D S M and we note that S < V NG .S /. Consider at first the possibility V < NG .S / which gives jV W S j D 2. Then V D S S s , t 0 D t s 2 S s S and V D S ht 0 i. By the modular law, O where CS .t 0 / D QO Š Q8 . Then CS .Q/ O D DO Š D8 and CG .t 0 / D CV .t 0 / D ht 0 i Q, O Since DO is t 0 -invariant and t 0 acts fixed-point-free S is the central product of QO and D. O 0 i/ Q, O V fuses all four involutions in DO hzi O 0 i Š D16 , V D .Dht on DO hzi, so Dht and other V -classes of involutions in S U have length 2. This contradicts the fact that NG .S /=S acts regularly on the four S -classes of involutions in S U . It follows that V D S M D NG .S /. Since S S s is normal in M , so we get V D S S s D NG .S /. Thus F D S \ S s is a normal subgroup of M of order 8. If F is nonabelian, then S D F CS .F /, where F1 D CS .F / is nonabelian and normal in NG .S /. Since S Š Q8 D8 , so fF; F1 g D fD8 ; Q8 g. But this contradicts the fact that NG .S / acts transitively an all 8 involutions from S U . We have proved that F is abelian. Since Q8 D8 does not possess a subgroup isomorphic to E8 , it follows that F must be of type .4; 2/ and consequently we have z 2 F . Again, because NG .S / acts transitively an all involutions from S U , so we have U F . Furthermore, S 0 D .S s /0 D hzi and so if hf i is a cyclic subgroup of order 4 in F , so NG .hf i/ hS; S s i D NG .S /. Also, we have S1 D CS .f / Š C4 Q8 and jS W S1 j D 2. Hence, S1 contains exactly 6 involutions which are different from z. This implies that there exist involutions in S S1 . Let t1 be an involution in S1 F and let t2 be an involution in S S1 . Then t1 centralizes f but t2 inverts f . This is a contradiction because NG .S / normalizes hf i and NG .S / acts transitively on all involutions in S U . We have proved again that NG .S / D G and so jG W S j D 4, as required. 3o . Groups with subgroup isomorphic to E16 . As an application of Theorems 49.1 and 49.2, we shall determine up to isomorphism all 2-groups with an involution t such that CG .t / D ht iQ, where Q Š Q2m ; m 3 and which have an elementary abelian subgroup E16 of order 16. We obtain exactly two classes of 2-groups and we present them in terms of generators and relations. More precisely, we shall prove the following result. Theorem 49.3. Let G be a 2-group containing an involution t such that CG .t / D ht iQ, where Q is a generalized quaternion group of order 2m , m 3. We assume in addition that G has an elementary abelian subgroup of order 16. Then G is isomorphic to one of the following groups. m1 m1 (E1 ) G D ha; b; z; c; t; x j a2 D b4 D z2 D c 2 D t 2 D x 2 D 1, m 4, m2 m2 2 2 2 1 1 1 a Db Dc D z, b ab D a , t ct D c , bc D cb, bt D t b, a1 ca D

49

39

On 2-groups with small centralizer of an involution, II m3

m3

m3

cz, at D t a, xcx D a, xtx D bc 2 i. Here hz, x,a2 c2 E16 and ha; bi Š Q2m , hc; t i Š D2m , ha; b; c; t i Š C.m; m/. m1

m3

, ba1C2

t ci Š

m1

(E2 ) G D ha; b; z; c; t; x; s j a2 D b4 D z2 D c 2 D t 2 D x 2 D s 2 D 1, m2 m2 2 2 2 1 1 Db Dc D z, b ab D a , t ct D c 1 , bc D cb, bt D t b, m 3, a ac D ca, at D t a, sx D xs, xtx D t c, xcx D c 1 , xbx D ba, xax D a1 , m3 scs D az, st s D bc 2 i. Here ha; bi Š Q2m , hc; t i Š D2m , ha; b; c; ti Š A.m; m/, m3 2m3 hz, x, s, a2 c i Š E16 . Finally, for m D 3, our group G is isomorphic to a Sylow 2-subgroup of the simple group J2 . Proof. Suppose that a 2-group G satisfies the assumptions of Theorem 49.3 and let E be an elementary abelian subgroup of order 16 in G. Using theorems 49.1 and 49.2, we see that we have exactly two possibilities: (E1 ) G has a normal subgroup S Š C.m; m/, m 4 of index 2 in G, jE \ S j D 8 and so G D SE. (E2 ) G has a normal subgroup S Š A.m; m/, m 3 of index 4 in G, E \ S D U is a normal four-subgroup of G and so G D SE. Case .E1 /. Assume that S Š C.m; m/, m 4; is given with generators and relations as in the definition. Let U D hz; yvi be the normal four-subgroup of G, where m3 is an element of order 4 in hci: y is an element of order 4 in hai and v D c 2 In the proof of Theorem 49.2 we have found all five conjugacy classes of involutions (and their centralizers) in S U . Their representatives are the involutions: t , t c, bv, bav and bat c: It follows that t can be fused in G only to an involution in the conjugacy class of bv: We see also that S has exactly two normal subgroups isomorphic to D2m which do not contain any conjugate of bat c W L D ht; ci and K D hbv; ai. Take an involution x 2 E S: Since x cannot fuse t with t c; it follows that x does not normalize the dihedral subgroup L: Hence we have Lx D K: The involution x sends the conjugacy class of t in L onto the conjugacy class of bv in K: Replacing m3 t with t c 2j (j is an integer), we may assume that t x D bv D bc 2 . Also we x k have hci D hai and so replacing c with c (k is an odd integer), we may assume that c x D a: We note that these replacements of generators did not effect the defining relations for S Š C.m; m/: The structure of G is uniquely determined. Case .E2 /. Assume that S Š A.m; m/ is given with generators and relations as in of A.m; n/. Then we have S D Q L; where Q Š Q2m and L Š D2m : Let U D hz; yvi be the normal four-subgroup of G, where y is an element of order 4 in m3 hai and v D c 2 is an element of order 4 in hci: In the proof of Theorem 49.2 we have found all four conjugacy classes of involutions (and their centralizers which are all isomorphic to C2 Q2m ) in S U: Their representatives are the involutions: t , t c, bv, and bav: Since jG W S j D 4; so G fuses all involutions in S U into a single conjugacy class in G: This gives that E \ S D U: Assume for a moment that m 4: Then we have seen in the proof of Theorem 49.2 that S has exactly two subgroups isomorphic to D2m : They are L D ht; ci and

40

Groups of prime power order

K D hbv; ai and both are normal in S: If we set E D U hx; si; where hx; si\S D 1; then the four-subgroup hx; si acts on fL; Kg: We may assume that x normalizes L and K and Ls D K: The involution x must induce outer automorphisms on L and K and so c x D c 1 and ax D a1 since x fuses two conjugacy classes of non-central involutions in L and also in K: Replacing s with sx (if necessary), we may assume that s sends t to an involution in the conjugacy class of bv (instead of bav) in K: m3 Replacing t with t c 2j (where j is an integer) we may assume that t s D bv D bc 2 : Also we have hcis D hai: Let t x D t 0 , where t 0 D t c k with k an odd integer. Then replacing c with c k ; we see that we may assume from the start that t x D t c: Here it m3 might happen that .c k /2 D v 1 D vz and then the previous relation t s D bv is s 1 D bvz: In that case we replace b with b 1 D bz so that changed into t D bv s we may assume again t D bv: Replacing a with al ; where l is an odd integer, we may assume c s D az: (Here we have taken az rather than a so that the last relation looks simpler!) Act on the relation xtx D t c with s: We get x s t s x s D t s az and so since Œx; s D 1 we have xbvx D bvaz or b x v 1 D bavz and finally b x D ba: We note that these replacements of generators did not effect the defining relations for S Š A.m; m/: The structure of G is uniquely determined. In the remaining case m D 3 we have S Š Q8 D8 : We shall determine the group G almost in the same way as above for m 4: The only difference is that here S has exactly six subgroups isomorphic to D8 which do not contain the normal foursubgroup U of G (see the proof of Theorem 49.2). They are: L D ht; t ci; D2 D ht; bavi;

K D hbv; bavi; D3 D ht c; bvi;

D1 D ht; bvi; D4 D ht c; bavi:

In S U lie exactly eight involutions which are distributed in four S -classes: ft; t zg, ft c; t czg, fbv; bvzg, fbav; bavzg. We set again E D U hx; si; where hx; si\S D 1: Then the four-group hx; si acts transitively (and so regularly) on the above four S classes. This follows from the fact that G acts transitively on the eight involutions in S U since CG .t / has index 8 in G: If is any involution in hx; si; then we must have either L D L and K D K or L D K: Indeed, if L D Di ; where i 2 f1; 2; 3; 4g; then L \ Di D ht; zi or ht c; zi and L \ Di is -invariant. But then the S -class ft; t zg or the S -class ft c; t czg is -invariant which is a contradiction. Similarly, if K D Dj ; where j 2 f1; 2; 3; 4g; then the S -class fbv; bvzg or the S -class fbav; bavzg is invariant which is again a contradiction. Hence the four-group hx; si acts on fL; Kg: We may assume that x normalizes L and K and Ls D K: Then we determine the group G uniquely in exactly the same way as above. The proof is complete. 4o . Groups without subgroups isomorphic to E8 . As a direct application of the above results, we shall prove the following Theorem 49.4. Let G be a 2-group which is not of maximal class and such that j1 .G/j > 4. If G has no elementary abelian subgroups of order 8, then G is

49

On 2-groups with small centralizer of an involution, II

41

isomorphic to one of the following groups: (a) A group G from (A1)(c) of Theorem 48.1 with involutions in G T . (b) A group G from (A1)(d) of Theorem 48.1. (c) A group G from (A1) of Theorem 49.1. (d) A group G from (A2)(b) of Theorem 49.1. (e) A group G from (A2)(d) of Theorem 49.1. (f) A group G with a normal subgroup S of index at most 4 in G so that S is isomorphic to the central product of Q2m and D2n , where m > 2, n > 2 and 1 .G/ S (see Theorem 49.2). Proof. Let G be a 2-group which has more than 3 involutions and which is not of maximal class. Also assume that G has no subgroups isomorphic to E8 . Let U be a normal four-subgroup of G and set T D CG .U /. By assumption we have 1 .T / D U and so there is an involution t 2 G T . We have G D ht iT and so CG .t / D ht i Q, where Q D CT .t /. Since t does not centralize U , it follows that Q has only one involution. Therefore, Q is either cyclic or a generalized quaternion group. Such groups G are then described in this and previous subsection. A simple inspection yields the above result. Exercise 1 (Alperin [Alp3]). Suppose that a nonabelian 2-group G has no normal elementary abelian subgroups of order 8. Suppose that G is neither cyclic nor of maximal class. Then one of the following holds: (a) G has a normal abelian subgroup A of type .4; 2/ with CG .A/ abelian of type .2; 2n /, n 2. In that case, G=CG .A/ is isomorphic to a subgroup of Aut.A/ Š D8 . (b) G has a normal abelian subgroup A of type .4; 4/ with metacyclic CG .A/. In that case, jG=CG .A/j 25 . Solution. Let A be the greatest normal noncyclic abelian subgroup of G of exponent 4. Since G is not of maximal class, either A is abelian of type .2; 4/ or .4; 4/ (Lemma 1.4). By Corollary 10.2, 2 .CG .A// D A. It follows from Theorem 41.1 that CG .A/ is metacyclic. Note that Aut.C4 C2 / Š D8 and jAut.C4 C4 /j D 3 25 . It remains to show that, if G has not G-invariant abelian subgroup of type .4; 4/, then CG .A/ is abelian of type .2; 2n / for some n > 1. Indeed, 2 .CG .A// D A is of order 8 so our claim follows from Lemma 42.1. Exercise 2. Let Q Š Q2m , where m 5 and let H be the holomorph of the cyclic group C Š C2n , n > 2. Set G D Q H with Q \ H D Z.Q/. Show that G has a normal elementary abelian subgroup of order 8. Solution (Janko). Let hyi be a normal (cyclic) subgroup of order 4 in Q, let hvi < C be of order 4 and set hzi D Z.G/; then .yv/2 D y 2 v 2 D zz D 1. Let a be an

42

Groups of prime power order

involution in A D Aut.C / such that ha; C i Š M2nC1 (1 .A/ C has exactly three maximal subgroups containing C ; these subgroups are isomorphic to D2nC1 , SD2nC1 and M2nC1 , respectively). Then a centralizes v 2 Z.ha; C i// and D D hz; yv; ai is the desired normal elementary abelian subgroup of order 8 in G. Indeed, V D hz; yvi is a normal four-subgroup of G, since hzi, hyi, hvi are normal in G and o.yv/ D 2. Next, jha; V ij D 8 and W D hz; ai D 1 .ha; C i/ is characteristic in a normal subgroup ha; C i of G. It follows that D D V W is also normal in G. Since a centralizes V , D Š E8 , and we are done. Exercise 3. Let G be a p-group containing an elementary abelian subgroup of order p 3 and let E be the subgroup generated by all elementary abelian subgroups of G of order p 3 . Suppose that G E contains an element x of order p. Then Chx;E i .x/ D hxiQ, where Q is cyclic or generalized quaternion. Exercise 4. Classify the 2-groups G such that (i) 2 .G/ D C M , where jC j D 2 and M is of maximal class and order > 8, (ii) i .G/ D C M , where i D 1 or 2, C Š C4 and M is of maximal class.

50

Janko’s theorem on 2-groups without normal elementary abelian subgroups of order 8

The main part of this section coincides with [Jan5]. Theorem 50.1 yields a deep insight in the structure of 2-groups without normal elementary abelian subgroups of order 8. The proof is fairly difficult, however elementary. In the analogous case, for p > 2, we have to classify the p-groups G without normal subgroup of order p pC1 and exponent p; it seems that this problem is far from a solution. The first important result about the groups of the title is the 4-generator theorem from [MacW] which asserts that any subgroup of such groups can be generated by four elements. However, this result does not say anything more about the structure of such groups. In [Kon3], the groups of the title are determined under the additional assumption that the Frattini subgroup contains an elementary abelian subgroup of order 8. This determination is somewhat unfortunate, because the resulting groups are given in terms of generators and relations without any comments and from these it is difficult to disclose the structure of such groups. In his long paper [Ust], Ustjuzaninov has asserted that the groups G of the title must have a normal metacyclic subgroup N such that G=N is isomorphic to a subgroup of the dihedral group D8 of order 8. But this paper is completely unreadable. With any attempt to read it, computational errors were discovered. However, it turns out that this result is correct after all! We shall give here a relatively short proof of a stronger result. For example, in the case where G=N is isomorphic to D8 , we shall determine completely the structure of N by showing at first that N is either abelian or minimal nonabelian. In our proof the computations are reduced to a minimum. The proof is based on a method of “pushing up” normal metacyclic subgroups of G combined with a very detailed knowledge of Aut.C4 C4 / (Proposition 50.5). We also note that our proof of the 4-generator theorem is character-free, i.e., it is completely elementary. We state now our main result. Theorem 50.1 (Janko [Jan5]). Let G be a 2-group which has no normal subgroups isomorphic to E8 . Suppose that G is neither abelian nor of maximal class. Then G has a normal metacyclic subgroup N such that CG .2 .N // N and one of the following holds:

44

Groups of prime power order

(a) jG=N j 4 and either 2 .N / is abelian of type .4; 4/ or N is abelian of type .2j ; 2/, j 2. (b) G=N Š D8 and (b1) N is either abelian of type .2k ; 2kC1 /, k 1 or of type .2l ; 2l /, l 2 or (b2) N is minimal nonabelian, 2 .N / is abelian of type .4; 4/ and more m n m1 precisely N D ha; b j a2 D b 2 D 1; ab D a1C2 i, where m D n with n 3 or m D n C 1 with n 2. We prove here the following easy special case of the above theorem. Theorem 50.2. Suppose that a 2-group G has no normal elementary abelian subgroups of order 8 and has two distinct normal four-subgroups U and V . Then D D U V Š D8 and G is the central product G D D C of D and C with D \ C D Z.D/ and C is either cyclic or of maximal class 6Š D8 . Conversely, if G D D C , where D Š D8 , C is cyclic or of maximal class 6Š D8 and D \ C D Z.D/, then G has no normal subgroups isomorphic to E8 . Proof. By hypothesis, ŒU; V ¤ f1g. Hence jU \V j D 2 which gives D D U V Š D8 . We have C D CG .D/ D CG .U /\ CG .V / so jG W C j D 4. It follows that G D D C with D \ C D Z.G/, where C D CG .D/. If W is a normal four-subgroup in C , then U W would be an elementary abelian normal subgroup of order 8 in G. This is a contradiction and so C is either cyclic or a group of maximal class (Lemma 1.4) which is not isomorphic to D8 . Now assume that G D DC , where D, C and D\C are the same as in the previous paragraph. Clearly, Z.G/ D Z.C / is cyclic. Assume that G has a normal elementary abelian subgroup E of order 8. Since C 6Š D8 and Z.G/ is cyclic, we get E \ C D Z.D/. By the product formula, G D EC . Let U , V be two distinct four-subgroups in D; then they are normal in G. Since CG .U / D U C and CG .V / D V C , we get U 6 E and V 6 E (recall that G D EC ). It follows that E \ D D Z.D/. Then G=D is noncyclic since it contains a four-subgroup DE=D Š E=Z.D/; it follows that C is of maximal class. Clearly, U is the only G-invariant four-subgroup in U C since C 6Š D8 and U C D Z C , where Z is a subgroup of order 2 in U . However, E \ .U C / is a G-invariant four-subgroup. It follows that U < E, which is a contradiction. The proof is complete. In view of Theorem 50.2, in what follows we may confine our approach to the case where G has exactly one normal four-subgroup. Remark 1. Let G D D C be a 2-group from Theorem 50.2 with G ¤ D. Let Z1 be the cyclic subgroup of order 4 in D.Š D8 / and let Z2 be a maximal cyclic subgroup of index 2 in C . Then N D Z1 Z2 is an abelian normal subgroup of type .2j ; 2/, j 2. and G=N is elementary abelian of order 4. Hence this is a special case of groups occurring in Theorem 50.1(a).

50 Janko’s theorem on 2-groups

45

The 4-generator theorem follows as a trivial consequence of Theorem 50.1. Theorem 50.3 (4-generator theorem). Let G be a 2-group which has no normal elementary abelian subgroups of order 8. Then every subgroup U of G is generated by four elements. Proof. By Theorem 50.1, G has a normal metacyclic subgroup N such that G=N is isomorphic to a subgroup of D8 . Since U=.U \ N / is isomorphic to a subgroup of G=N , we have d.U=.U \ N // 2. Also, U \ N is metacyclic and therefore d.U \ N / 2: Hence d.U / 4 and we are done. We prove three important preliminary results. Proposition 50.4. Let G be a 2-group with a metacyclic normal subgroup N . Suppose that N has a G-invariant four-subgroup N0 which is not contained in Z.G/ but there is no G-invariant cyclic subgroup of order 4 contained in N . If N is abelian, it is of type .2n ; 2n / or .2nC1 ; 2n /, where n 1. If N is nonabelian, it is minimal nonabelian m n m1 and moreover N D ha; b j a2 D b 2 D 1; ab D a1C2 i, where n 2 and m D n or m D n C 1. Proof. If N is abelian (of rank 2), the result is clear since N cannot contain a characteristic cyclic subgroup of order 4. Suppose that N 0 ¤ 1. Since N 0 is cyclic, we must have jN 0 j D 2. By Example 10.14, N is minimal nonabelian. Then a result of R´edei m n m1 (Exercise 1.8a) implies that N D ha; b j a2 D b 2 D 1; ab D a1C2 ; m 2; n 1i. If n D 1, then N Š D8 or N Š M2mC1 with m 3. In both cases N has a characteristic cyclic subgroup of order 4, which is a contradiction. Hence we m1 have n 2. It follows that N 0 D ha2 i and Z.N / D ha2 ; b 2 i. If m < n, then n1 2 Z0 D hb i D Ãn2 .Z.N // is a G-invariant subgroup of order 2. In that case, N0 D N 0 Z0 lies in Z.G/, contrary to our assumption. Thus m n. However, if m > n C 1, then Ãn1 .Z.N // is a characteristic cyclic subgroup of N of order 4, which is a contradiction. Thus, m D n or m D n C 1. Remark 2. Let W0 be a normal subgroup of a group W and exp.W0 / D n. Let A Aut.W / stabilize the chain W > W0 f1g. Then exp.A/ divides n. Indeed, take n w 2 W and 2 A. Then w D ww0 for some w0 2 W0 so w D ww0n D w. It follows that n D idW so exp.A/ divides n, as claimed. n

n

Exercise 1. Let W D ha; b j a2 D b 2 D Œa; b D 1; n > 1i be abelian of type .2n ; 2n / and let Y be a subgroup of index 2 in W . Then A D f 2 Aut.W / j Y D Y g is of order 24n3 . In particular, if n D 2, jAj D 25 . Exercise 2. Let W be an abelian group of type .4; 4/ and A the stabilizer of the chain W > 1 .W / > f1g. Then jAj D 24 .

46

Groups of prime power order

Exercise 3. Let W be abelian of type .4; 4/. Suppose that 2 Aut.W / stabilizes the chain W > 1 .W /. Then stabilizes the chain W > 1 .W / > f1g. (Hint. Use Proposition 4.9.) Proposition 50.5. The automorphism group Aut.W / of the group W D hu; y j u4 D y 4 D Œu; y D 1i Š C4 C4 is of order 25 3. The subgroup A of Aut.W / of all automorphisms fixing the subgroup Y D hu2 ; yi Š C2 C4 is of order 25 and so is a Sylow 2-subgroup of Aut.W /. We have A D h; ; j 4 D 4 D 2 D Œ 2 ; D 1; Œ; D 2 2 ; Œ; D 2 D Œ; i: where the automorphisms ; ; are induced with: u D u1 y, y D u2 y; u D uy, y D y; u D u1 , y D y. We have A0 D Z.A/ D ˆ.A/ D h 2 ; 2 D i Š E4 and so A is a special 2-group. Set W0 D 1 .W / D hu2 ; y 2 i. Then the stabilizer A0 of the chain W > W0 > 1 is elementary abelian of order 24 (Exercise 2 and Remark 50.2) and A0 D h; D 0 ; ; 2 i D CA .W0 /. The subgroup A0 contains the “special” subset S D f 2 ; ; ; ; g of five automorphisms defined by: u D u;

y D yu2 I

u D uu2 y 2 ;

y D yu2 y 2 I

u D uu2 y 2 ;

y D yu2 I

u D uy 2 ;

y D yu2 y 2 :

If X is any maximal subgroup of A0 , then X \ S is nonempty. Each 2 S has the property that does not invert any element of order 4 in W . In addition, the “superspecial” automorphisms and have also the property that they do not fix (centralize) any element of order 4 in W . We have CA .Y / D h; 2 i Š E4 is normal in A and A=h; 2 i Š D8 . In fact A is a splitting extension of h; 2 i by h; i Š D8 . Finally, if U is any subgroup of A covering A=h; 2 i (i.e. U h; 2 i D A), then U does not normalize any of the six cyclic subgroups of order 4 in W . Proof. The number of elements x, x 0 of order 4 in W such that hx; x 0 i D W (by the product formula, fx; x 0 g is a basis of W ) is 12 8 and so jAut.W /j D 25 3. By Exercise 1, jAj D 25 and therefore A is a Sylow 2-subgroup of Aut.W /. The stabilizer A0 of the chain W > W0 > 1 is elementary abelian (Remark 50.2) of order 24 (Exercise 2) and so jA W A0 j D 2. Any automorphism from A A0 acts faithfully on W =W0 (and so also on W0 ; see Exercise 3) and so we see that A0 D CA .W0 / (recall that W0 D 1 .W /). Defining the automorphisms , , , D 2 , 0 D as above, we verify the defining relations for A. Since , 0 , , 2 all lie in A0 and generate a subgroup of order 24 , we get A0 D h; 0 ; ; 2 i. For any x 2 A A0 , x 2 2 Z.A/ which gives ˆ.A/ Z.A/: We have h 2 ; 2 i ˆ.A/ Z.A/ and so the foursubgroup h 2 ; 2 i is normal in G. On the other hand, the relations for A show that A=h 2 ; 2 i is elementary abelian and h 2 ; 2 i A0 . Hence h 2 ; 2 i D ˆ.A/ D A0 . Also, h 2 ; 2 i Z.A/ < A0 and we check that no element in A0 h 2 ; 2 i lies in Z.A/. Hence Z.A/ D A0 and so A is special.

50 Janko’s theorem on 2-groups

47

Since CA .Y / A0 , so for each 2 CA .Y / we have u D uw0 with w0 2 W0 . It follows that jCA .Y /j 4. On the other hand, h; 2 i CA .Y / and so CA .Y / D h; 2 i Š E4 . Hence h; 2 i is normal in A and since A=h; 2 i acts faithfully on Y Š C4 C2 , it follows that A=h; 2 i Š Aut.C4 C2 / Š D8 . Since is an involution and D 1 , it follows that h; i Š D8 . Because h; i\h; 2 i D 1, A is a splitting extension of the four-group h; 2 i by h; i. Let U be any subgroup of A covering A=h; 2 i. Then we have U=.U \h; 2 i/ Š D8 and so there exists an element ˛ of order 4 in U . Since ˛ 2 A A0 (recall that exp.A0 / D 2), ˛ acts faithfully on W =W0 . But ˛ fixes Y D W0 hyi (since Y is Ainvariant) and so ˛ sends two cyclic subgroups of order 4 in W0 hui onto two cyclic subgroups of order 4 in W0 huyi since W0 hui is not A-invariant. It remains to be shown that U does not fix the cyclic subgroup hyi contained in Y . Suppose that U fixes hyi so that jU W CU .y/j 2. Sine c1 .Y / D 3, we get jU W CU .u2 /j 2 and so CU .Y / D CU .y/ \ CU .u2 / has index 4 in U . This is a contradiction, since jU W .U \ h; 2 i/j D 8 (recall that CA .Y / D h; 2 i). We have proved that U does not fix any cyclic subgroup of order 4 in W . Let X be any maximal subgroup of A0 . Suppose that S \ X is empty, where S D fx1 ; : : : ; x5 g is the “special” subset of automorphisms in A0 with x1 D 2 , x2 D , x3 D , x4 D and x5 D . Then we verify that the 10 products xi xj .1 i < j 5/ give 10 pairwise distinct elements in A0 . But all these products lie in X. This is a contradiction since jXj D 23 . Other statements about “special” and “extraspecial” elements in A0 are self-explanatory. Proposition 50.6. Let a 2-group G have no normal elementary abelian subgroups of order 8. Suppose that G is neither abelian nor of maximal class. The one of the following holds: (a) G has a normal abelian subgroup A of type .4; 2/ with CG .A/ abelian of type .2n ; 2/, n 2. In that case, G=CG .A/ is isomorphic to a subgroup of Aut.A/ Š D8 . If G=CG .A/ Š D8 , then CG .A/ D A. (b) G has a normal abelian subgroup W of type .4; 4/ with metacyclic CG .W /. In that case 2 .CG .W // D W and G=CG .W / is isomorphic to a subgroup of Aut.W /, i.e., jG=CG .W /j 25 (see Proposition 50.5). Proof. Let A be a greatest normal noncyclic abelian subgroup of G of exponent 4. Since G is not of maximal class, A is abelian of type .4; 2/ or .4; 4/. By Corollary 10.2, 2 .CG .A// D A. Then Theorem 41.1 implies that CG .A/ is metacyclic. Suppose that A Š C4 C2 . Then j2 .CG .A//j D 8 and Lemma 42.1 gives that CG .A/ is abelian of type .2n ; 2/, n 2. Suppose in addition that G=CG .A/ Š D8 . If CG .A/ > A, then CG .A/ contains a characteristic cyclic subgroup Z of order 4. We have Z < A and Z is normal in G. But G=CG .A/ Š Aut.A/ and Aut.A/ Š D8 contains an automorphism ˛ which permutes two cyclic subgroups of order 4 in A (Remark 50.3), a contradiction and so we must have CG .A/ D A.

48

Groups of prime power order

Proof of Theorem 50.1. Let a 2-group G have no normal elementary abelian subgroups of order 8. Assume that G is neither abelian nor of maximal class. The starting point is Proposition 50.6. If we are in case (a) of Proposition 50.6, then G has a normal abelian subgroup A of type .4; 2/ with N D CG .A/ abelian of type .2j ; 2/, j 2. In that case G=N is isomorphic to a subgroup of D8 and if G=N Š D8 , then N D A and we are done. We assume that we are in case (b) of Proposition 50.6. Then G has a normal abelian subgroup W D hu; y j u4 D y 4 D Œu; y D 1i Š C4 C4 with metacyclic C D CG .W / and 2 .C / D W . Set W0 D 1 .W / D hu2 ; y 2 i so that W0 is a normal four-subgroup of G. Let Y be a normal subgroup of G such that W0 < Y < W . Then Y is abelian of type .4; 2/ and we may choose the generators u and y of W so that Y D hu2 ; yi. Also set D D CG .Y / so that G=D is isomorphic to a subgroup of Aut.Y / Š D8 . It follows from hy 2 i D Ã1 .Y / that y 2 2 Z.G/. Now, G=C is isomorphic to a subgroup of A D h; ; i which is the Sylow 2-subgroup of Aut.W / fixing Y . In what follows we use freely Proposition 50.5 about the structure of A and the action on W together with the notation introduced there. It follows that D=C is elementary abelian of order 4 since D=C induces on W a subgroup of h; 2 i Š E4 . We note that A=h; 2 i Š D8 and W0 Z.G/ if and only if G=C induces on W a subgroup of A0 in which case G=C is elementary abelian of order 24 . If C D D, then G=C is isomorphic to a subgroup of D8 . Suppose in addition that G=C Š D8 . Then W0 6 Z.G/ and G=C induces on W a subgroup U Š D8 of automorphisms which covers A=h; 2 i. By Proposition 50.5, there is no cyclic G-invariant subgroup of order 4 contained in C . Using Proposition 50.4, we see that the structure of the normal metacyclic subgroup C is completely determined, as stated in Theorem 50.1. From now on, let D=C ¤ f1g. Suppose at first that D=C contains a subgroup F=C 2 of order 2 such that F=C induces the automorphism 2 on W , where u D uy 2 , 2 2 y D y and .uy/ D uyy 2 . Since 2 does not invert any element of order 4 in W , it follows that 2 does not invert any element in C W0 . Because 2 2 Z.A/, so F is normal in G. We claim that 2 .F / D W and so, by Theorem 41.1, F is metacyclic. If not, then 2 .F / ¤ W and so 2 .F / 6 C . There exists an element s 2 F C so that o.s/ 4 which gives s 2 2 W0 . It is easy to see that there exist involutions in .W hsi/ W . Indeed, we have y s D y, us D uy 2 . If s 2 D 1, then we are finished. If s 2 D y 2 , then .sy/2 D s 2 y 2 D 1. If s 2 D u2 , then .syu/2 D s 2 .yu/s yu D u2 yuy 2 yu D 1. If s 2 D y 2 u2 , then .su/2 D s 2 us u D y 2 u2 uy 2 u D 1. Hence we may assume that s 2 F C is an involution. For any c 2 C , sc is an involution if and only if .sc/2 D 1 or c s D c 1 , i.e., s inverts c. But s inverts in C only the elements in W0 . (Actually, s centralizes W0 .) It follows that there are exactly 4 involutions in F C and they all lie in the elementary abelian subgroup E D hsi W0 . Thus E is characteristic in F and so E is normal in G, which is a contradiction. We have proved that 2 .F / D W and so F G G is metacyclic. Assume now that jD=C j D 2. If D=C induces the automorphism 2 on W , then by the above, D D F is a normal metacyclic subgroup of G with 2 .D/ D W .

50 Janko’s theorem on 2-groups

49

If jG=Dj 4, then we are done. It remains to consider the case where G=D Š D8 . In this case G=C induces on W a subgroup U of automorphisms which covers A=h; 2 i. In fact, we have in addition U \ h; 2 i D h 2 i. Again by Proposition 50.5, there is no cyclic G-invariant subgroup of order 4 contained in D. In this case, W0 6 Z.G/. By Proposition 50.4, the structure of the normal metacyclic subgroup D is uniquely determined, as stated in Theorem 50.1. It remains to consider here (where jD=C j D 2) the case that D=C induces on W the automorphism or 2 . Note that neither nor 2 is central in A and so NA .hi/ D A0 and NA .h 2 i/ D A0 . Hence G=C induces on W a subgroup of A0 which does not contain 2 . If jG=C j 4, we are done. Hence we may assume that G=C induces on W a maximal subgroup X of A0 which does not contain 2 . In that case G=C Š E8 . By Proposition 50.5, X contains a “special” element 2 f; ; ; g, where does not invert any element of order 4 in W . Let H=C be a subgroup of order 2 in G=C Š E8 so that H=C induces the automorphism D on W , where y D yu2 and u D u. If 2 .H / > W , then there is an element s 2 H C with s 2 2 W0 . By replacing s with sw for a suitable w 2 W , we may assume that s is an involution. Indeed, if s 2 D u2 , then us D u and y s D yu2 imply that .su/2 D s 2 u2 D 1. If s 2 D y 2 , then .syu/2 D s 2 .yu/s yu D y 2 yu2 u yu D 1. If s 2 D y 2 u2 , then .sy/2 D s 2 .y/s y D y 2 u2 yu2 y D 1. Then sc (with c 2 C ) is an involution if and only if s inverts c and so there are exactly 4 involutions in H C and therefore they all lie in E D hsi W0 Š E8 . Hence E is characteristic in H and so E is normal in G, which is a contradiction. Thus 2 .H / D W and so H must be metacyclic. Since G=H Š E4 , we are done. Let K=C be a subgroup of order 2 in G=C Š E8 so that K=C induces the automorphism D on W , where y D yu2 y 2 and u D uu2 y 2 . If 2 .K/ > W , then there is an element s 2 K C with s 2 2 W0 . By replacing s with sw for a suitable w 2 W , we may assume that s is an involution. If s 2 D y 2 , then .su/2 D 1. If s 2 D u2 , then .sy/2 D 1. If s 2 D y 2 u2 , then .suy/2 D 1. Then sc (with c 2 C ) is an involution if and only if s inverts c and so there are exactly 4 involutions in K C and therefore they all lie in E D hsi W0 Š E8 . Hence E is characteristic in K and so E is normal in G, which is a contradiction. Thus 2 .K/ D W and so K must be metacyclic. Since G=K Š E4 , we are done. Let L=C be a subgroup of order 2 in G=C Š E8 so that L=C induces a “superspecial” automorphism 2 f; g on W . Then has the additional property that it does not fix (centralize) any element of order 4 in W . Suppose that 2 .L/ D W so that L is metacyclic. In that case an element l 2 L C commutes with an element of order 4 in W , which contradicts the above property of . Hence there is s 2 L C so that s 2 2 W0 . Again we may assume that s is an involution. If D , then us D uy 2 , y s D yu2 y 2 . In this case, if s 2 D u2 , then .sy/2 D 1. If s 2 D y 2 u2 , then .su/2 D 1 and if s 2 D y 2 , then .suy/2 D 1. If D , then us D uu2 y 2 , y s D yu2 . In this case, if s 2 D u2 y 2 , then .sy/2 D 1. If s 2 D u2 , then .suy/2 D 1 and if s 2 D y 2 , then .su/2 D 1. It follows that L C contains exactly 4 involutions

50

Groups of prime power order

and so E D hsi W0 Š E8 (being characteristic in L ) is normal in G, which is a contradiction. This finishes completely the case jD=C j D 2. We make here the following observation. If G=C Š E16 and G=C induces on W the group A0 of automorphisms, then we get a contradiction. Indeed, we consider a subgroup L=C of order 2 in G=C so that L=C induces the automorphism on W . Then exactly the same proof as above shows that L has an elementary abelian subgroup E of order 8 which contains all involutions in L and so E would be normal in G, which is a contradiction. It remains to consider the case where D=C is a four-group inducing on W the fourgroup h; 2 i. If C < F < D and F=C induces 2 on W , then we have already proved that 2 .F / D W and F is a normal metacyclic subgroup of G. If jG=Dj 2, then jG=F j 4 and we are finished. Hence we may assume that jG=Dj 4 and so G induces on W either the whole group A (in which case G=D Š D8 ) or one of the maximal subgroups of A containing h; 2 i: A. h; 2 ; ; 0 D i D A0 , B. h; 2 ; ; i D hi h; i Š C2 D8 with D 1 , C. h; 2 ; ; i, where 2 D and Œ; D 2 . Case A is impossible, by the previous paragraph. Suppose that we are in case B. We note that h i is normal in h; 2 ; ; i. Let K0 be a subgroup of G such that F < K0 < G and K0 =C induces the cyclic group h i Š C4 on W . Hence K0 =C Š C4 and K0 is normal in G. Suppose that 2 .K0 / > W . Since 2 .F / D W , it follows that there is an element x 2 K0 F so that o.x/ 4 and hxiC D K0 . But then o.x 2 / 2 and x 2 2 F and so x 2 2 W0 C . This is a contradiction since K0 =C Š C4 . Hence 2 .K0 / D W and so K0 is a normal metacyclic subgroup of G with jG=K0 j D 4 and so we are finished in this case. Finally, we assume that we are in case C or G=D Š D8 (in which case G induces the full group A on W ). In any case, there is an element k 2 G such that k induces the automorphism on W and so uk D u1 y and y k D yu2 . Hence k does not normalize any cyclic subgroup of order 4 in W . Also, W0 6 Z.G/. We know that F is a Ginvariant nonabelian metacyclic subgroup with 2 .F / D W . By Proposition 50.4, F m n m1 is minimal nonabelian and we may set: F D ha; b j a2 D b 2 D 1; ab D a1C2 i, where n 2 and m D n or m D n C 1. If n 3, then W ha2 ; b 2 i D Z.F /. This is a contradiction since F=C induces the automorphism 2 on W . It follows that n D 2. But we have F > C W and so m D n C 1 D 3. In particular, W D C and so W is self-centralizing in G. Since CW .a/ D Y D hy; u2 i, we have a2 2 Y W0 . Replacing y with some other element of order 4 in Y (if necessary), we may assume that a2 D y. The structure of F is uniquely determined. We have F D ha; u j a8 D u4 D 1; au D aa4 i and W D hu; yi, where y D a2 and CG .W / D C D W . Set y 2 D z so that hzi D Z.G/ because CW .k/ D hy 2 i. The element k 2 induces the automorphism D 2 on W and so k 2 inverts W . Set S D F hki so that S=F Š C4 and k 4 2 C D W . Since h 2 ; i is normal in A, so S is normal in G.

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50 Janko’s theorem on 2-groups

Since k 4 2 CW .k/ D hzi, we have k 4 D 1 or k 4 D z. Because 2 .F / D W , it follows that all elements in F W have the order 8. We shall determine the action of k on F . We have ak D aui y j , where i , j are some integers and so ui y j is an arbitrary element in W . We get .a2 /k D y k D yu2 D .ak /2 D .aui y j /2 D aui y j aui y j D a2 .ui y j /a ui y j D y.uz/i y j ui y j D yu2i y 2j z i : Thus u2 D u2i y 2j z i and so .u2 /1i D z iCj . Hence 1i 0 .mod 2/ and i Cj 0 .mod 2/. This gives that we must have i j 1 .mod 2/. Let x be an element in G inducing the automorphism on W . Then x normalizes S , x 2 2 W , y x D y and ux D u1 . Since . 2 / D 2 and D 2 , we may set ax D aum y n and k x D kaup y q , where m; n; p; q are some integers. We get .a2 /x D y x D y D .ax /2 D .aum y n /2 D aum y n aum y n D a2 .um y n /a um y n D y.uy 2 /m y n um y n D yu2m y 2mC2n : Thus u2m y 2mC2n D 1 and so m n 0 .mod 2/. Then the relation ak D x aui y j with i j 1 .mod 2/ should hold under the action of x, i.e. .ax /k D p ax .ux /i .y x /j . Since m and n are even and au D ay 2 and so au D ay 2p , this gives: p yq

.aum y n /kau

D aum y n ui y j D aumi y nCj D .aui y j .u1 y/m y n u2n /au

p yq

D .a.uy 2 /i y j um y m y n /u

p yq

D ay 2p ui y 2i y j um y m y n

D auim y 2pC2iCj CmCn: Hence umi D uim and so u2i2m D 1. Since m 0 .mod 2/, we get u2i D 1 and so i 0 .mod 2/, which contradicts the above statement. Exercise 4. Let W be a homocyclic p-group of exponent 2n and rank d and let L be a subgroup of index 2 in W . Find the order of AutL .G/ D f 2 Aut.W / j L D Lg. Exercise 5. Suppose that a 2-group G has a subgroup isomorphic to E8 . If G has no subgroups isomorphic to C2 D8 , it has an odd number of subgroups isomorphic to E8 .

51

2-groups with self centralizing subgroup isomorphic to E8

In 48, 49 the 2-groups G have been described which have the property that they possess an involution t such that CG .t / D ht i C , where C is either a non-trivial cyclic group or a generalized quaternion group. It is natural to ask what happens if C is isomorphic to one of the other two groups of maximal class, i.e., C is isomorphic to D2n (dihedral group) or to SD2n (semi-dihedral group) (Question 49.4). It is easy to see that in that case the 2-group G has a self-centralizing elementary abelian subgroup E of order 8 but the problem of classifying 2-groups G which possess such a subgroup E is far more general. In this section we classify 2-groups G which possess a self-centralizing elementary abelian subgroup E of order 8. If E is normal in G, then G=E is isomorphic to a subgroup of GL.3; 2/ and therefore G=E is isomorphic to a subgroup of the dihedral group D8 of order 8. We determine here the unique 2-group G such that E D ˆ.G/ (Theorem 51.3). There are exactly five 2-groups G, where E is normal in G and E ˆ.G/. We shall present these groups in terms of generators and relations (Theorems 51.3 and 51.4). All these groups exist because we find some faithful transitive permutation representation of degree 8 or 16 for these groups. We assume in the sequel that E is not normal in G. Then we have two essentially different possibilities according to E ˆ.G/ or E 6 ˆ.G/, where ˆ.G/ is the Frattini subgroup of G. Suppose that E ˆ.G/. Then our first non-trivial result is that G has no normal elementary abelian subgroups of order 8 (Theorem 51.5). We are now in a position to use the main theorem in 50 which allows us to describe the structure of G very accurately (Theorem 51.6). Suppose that E 6 ˆ.G/. In that case, we see easily that for each involution t 2 E ˆ.G/, we have CG .t / D ht iM0 , where M0 is isomorphic to one of the following groups: E4 , D2n .n 3/, or SD2m .m 4/. Then we try to pin down the structure of G by using the structure of CG .t /. Assume in addition that G has normal elementary abelian subgroups of order 8. In that special case we see that M0 is isomorphic to E4 or D8 (Theorem 51.7). An exceptional case is treated in Theorem 51.8. The group G could possess also a normal elementary abelian subgroup B of order 16 such that G=B Š D8 (Theorem 51.9). In all other cases, the structure of G is described

51 2-groups with self centralizing subgroup isomorphic to E8

53

in Theorem 51.10. Finally, the example produces a group of order 28 (in terms of generators and relations) which satisfies the assumptions of Theorem 51.10. It remains to consider the case E 6 ˆ.G/ and G has no normal elementary abelian subgroups of order 8. In this case Theorem 50.1 does not give a very precise information about the structure of G. Therefore, we use here the fusion arguments for involutions which have been introduced in 48,49. In this way we show that G has in most cases a large normal subgroup S with jG=S j 4 and for the structure of S we have exactly seven possibilities (Theorems 51.14 and 51.15). The exceptional cases are treated in Theorems 51.11–51.13. We use here the standard notation introduced in 48. 49. In particular, for x 2 G we denote with cclG .x/ the conjugacy class of x in G. For the sake of completeness, we give here some known results which will be used often in this book. Proposition 51.1. Let be an involutory automorphism acting on an elementary abelian 2-group A. Then we have jCA . /j jA W CA . /j. Proof. Consider the Jordan normal form of the linear transformation induced by on A (considered as a vector space over GF.2/).Then the result follows. Proposition 51.2. Let be an involutory automorphism acting on an abelian 2-group B so that CB . / D W0 1 .B/. Then inverts Ã1 .B/ and B=W0 . 2

Proof. If x 2 B; then .xx / D x x D x x D xx . Hence, xx D w0 with w0 2 W0 and so x D x 1 w0 . This gives .x 2 / D .x /2 D .x 1 w0 /2 D x 2 w0 2 D x 2 and our result follows. 1o . The case E ˆ.G/. In this subsection we suppose that a 2-group G contains a self-centralizing elementary abelian subgroup E of order 8 which is contained in ˆ.G/. Obviously, Z.G/ E and also Z.ˆ.G// E. If jZ.ˆ.G//j D 2 then, by Proposition 1.13, we get that ˆ.G/ is cyclic, which is not the case. It follows that 4 jZ.ˆ.G//j 8. We consider at first the possibility jZ.ˆ.G//j D 8 which gives that Z.ˆ.G// D E and since CG .E/ D E, it follows that E D ˆ.G/ is a normal self-centralizing elementary abelian subgroup of order 8 of G. On the other hand, G=E is an elementary abelian subgroup of D8 and so G=E is a four-group. Thus jGj D 25 . For each x 2 G E, we must have jCE .x/j D 4 (since x acts faithfully on E and hx; Ei is not of maximal class according to Theorem 1.7, Propositions 1.8 and 51.1), CG .x/ D hxiCE .x/ with x 2 2 CE .x/. Otherwise, CG .x/ would cover G=E, which is not possible since E D ˆ.G/. It follows that for each x 2 G E, jcclG .x/j D 4 and so jG 0 j 4. If G 0 D E, then by the Taussky’s theorem (Proposition 1.6), G is of maximal class, which is not the case. Thus jG 0 j D 4. Suppose that jZ.G/j D 4. Then for each x 2 G E we have CG .x/ D hxiZ.G/ which gives x 2 2 Z.G/. But then E D ˆ.G/ D Ã1 .G/ Z.G/, a contradiction. Hence Z.G/ D hzi is of

54

Groups of prime power order

order 2 and Z.G/ < G 0 . Since Z.G/ < G 0 < E, we have ŒG; G 0 D Z.G/, so G is a 2-group of class 3. The group G=G 0 is abelian of type .4; 2/ and so there is an element x 2 G E so that x 2 D v 2 E G 0 . Since x centralizes z and v D x 2 , and E D hv; G 0 i, it follows that x does not centralize G 0 (otherwise x would centralize E). Let y 2 G .Ehxi/ which centralizes G 0 so that y 2 2 G 0 and G D hx; yi. Note that jG W CG .G 0 /j D 2. If Œx; y 2 hzi, then G=hzi is abelian, a contradiction. Hence we may put Œx; y D u, where G 0 D hz; ui. From Œx; y D u we get x y D xu and so v y D .x 2 /y D .x y /2 D .xu/2 D xuxu D x 2 .x 1 ux/u D v.uz/u D vz. Hence we get z y D z, uy D u, v y D vz, x y D xu, x 2 D v, z x D z, ux D uz, v x D v. The relation Œx; y D u gives also y x D yu D uy since y centralizes u 2 G 0 . From this we get .y 2 /x D .y x /2 D .uy/2 D y 2 . Hence x centralizes y 2 2 G 0 . But x acts faithfully on G 0 and so we must have y 2 2 hzi. Suppose now that y 2 D z. Then we replace y with y 0 D yv. We get .y 0 /2 D .yv/2 D yvyv D y 2 v y v D z.vz/v D 1 and other 0 0 0 0 relations remain unchanged z y D z, uy D u, v y D vz, x y D x yv D .xu/v D xu, 2 and so we may assume from the start that y D 1. The structure of G is uniquely determined. It is easy to see that this group G is isomorphic to a subgroup of A8 . It is enough to set x D .2; 7; 6; 8/.4; 5/, y D .1; 2/.3; 6/.4; 7/.5; 8/, and we compute further permutations x 2 D v D .2; 6/.7; 8/, Œv; y D z D .1; 3/.2; 6/.4; 5/.7; 8/, Œx; y D u D .1; 4/.2; 7/.3; 5/.6; 8/. Then we simply check that all above relations are satisfied for these permutations. Since z is represented with a non-trivial permutation, so the induced permutation representation of degree 8 is faithful. Note that all above permutations are even. Hence our group G exists and is isomorphic to a subgroup of A8 . We have proved the following result. Theorem 51.3. Suppose that a 2-group G contains a self-centralizing elementary abelian subgroup E of order 8 and E D ˆ.G/. Then G is uniquely determined and we have G D hx; y; z; u; v j x 4 D y 2 D z 2 D u2 D v 2 D Œz; u D Œz; v D Œz; x D Œz; y D Œu; v D Œu; y D Œv; x D 1; x 2 D v; v y D vz; x y D xu; ux D uzi. Here E D hz; u; vi is a self-centralizing normal elementary abelian subgroup of order 8 of G, E D ˆ.G/, G D hx; yi is of order 25 , G 0 D hz; ui, and Z.G/ D ŒG; G 0 D hzi is of order 2. The group G exists and is isomorphic to a subgroup of A8 . It remains to consider the second possibility jZ.ˆ.G//j D 4. We set Z.ˆ.G// D W0 so that W0 is a G-invariant four-group contained in E. Therefore the subgroup T D CG .W0 / is of index 2 in G. Since CG .E/ D E, so for each t 2 E W0 we have CT .t / D E. Assume that T D G, in which case W0 D Z.G/. Let A be a G-invariant subgroup so that W0 < A ˆ.G/ and jA W W0 j D 2. Then A is abelian of order 8 and G stabilizes the chain A > W0 > 1 so that G=CG .A/ is elementary abelian (of order 4/. In particular, ˆ.G/ CG .A/ and so A Z.ˆ.G//, which is a contradiction. It follows that jG W T j D 2 and so Z.G/ D hzi is of order 2 and we set W0 D hz; ui. For each x 2 G T , ux D uz. Since jˆ.G/j 24 , so jGj 26 . Assume now in addition that E is normal in G. Since G=E is isomorphic to a subgroup of D8 and jGj 26 , so jGj D 26 and G=E Š D8 . The subgroup T

51 2-groups with self centralizing subgroup isomorphic to E8

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stabilizes the chain E > W0 > 1 and jG W T j D 2, so T =E is a 4-group. We have Z.T / D W0 . For each x 2 T E, x 2 2 W0 since x acts non-trivially on E. Thus ˆ.T / D Ã1 .T / W0 and so T 0 W0 . On the other hand, if t 2 E W0 , then CT .t / D E implies that jcclT .t /j D 4 and so cclT .t / D E W0 and jT 0 j 4. It follows that Z.T / D T 0 D ˆ.T / D W0 and therefore T is a special 2-group of order 25 . Let x; y 2 T E be such that Ex and Ey are two distinct elements in T =E Š E4 . Then M D W0 hx; yi is a maximal subgroup of T not containing E. If M is nonabelian, then Œx; y D v is an involution in W0 . Let t 2 E W0 . We know that CT .t / D E and so Œx; t , Œy; t , and Œxy; t D Œx; t Œy; t are three distinct involutions in W0 . Note that T is of class 2. If, for example, Œx; t D Œy; t , then Œxy; t D Œx; t Œy; t D 1, a contradiction. Therefore, we may assume that Œx; t D v. It follows Œx; ty D Œx; t Œx; y D v 2 D 1. Hence, replacing y with ty 2 Ey, we may assume from the start that Œx; y D 1 and so M D W0 hx; yi is abelian. We claim that M is the unique abelian maximal subgroup of T . Indeed, if M0 ¤ M is another abelian maximal subgroup of T , then M \ M0 Z.T / and jM \ M0 j D 23 , a contradiction. Since T D CG .W0 / and Z.ˆ.G// D W0 , so T is a characteristic subgroup of G. Hence M is also a characteristic subgroup of G. Since G=E Š D8 and D8 has five involutions, there is an element s 2 G T such that s 2 2 E. Note that s normalizes M and so if s 2 2 W0 M , then M hsi is a maximal subgroup of G which does not contain E, a contradiction. Thus, we must have s 2 D t 2 E W0 . Since G=E Š D8 , s acts non-trivially on M=W0 Š T =E Š E4 . Hence, there is an element x 2 M W0 such that y D x s has the property M D W0 hx; yi. Also, x 1 y D x 1 x s D Œx; s 2 ˆ.G/ and so x 2 .x 1 y/ D xy 2 ˆ.G/. Hence ˆ.G/ D 2 Ehxyi and G D hs; xi. Now, y s D x s D x t D xu, where u 2 W0 hzi. Indeed, if x t D txt D xz, then we act on this relation with s. We get t s x s t s D x s z s which gives tyt D yz and then .xy/t D xzyz D xy. But this contradicts the fact that CT .t / D E and so we must have y s D xu, with u 2 W0 hzi. Also note that us D uz since s acts non-trivially on W0 . It remains to determine x 2 and y 2 . There are exactly three possibilities for the structure of the abelian group M . (a) 1 .M / D M0 is of order 23 . Since M0 is normal in G, so x and y D x s are elements of order 4 contained in M M0 . On the other hand, jÃ1 .M /j D 2 and so Ã1 .M / D hzi D Z.G/. This gives x 2 D y 2 D z. If we set s D .2; 5; 8; 12/.4; 6; 10; 11/.7; 9/.13; 15; 14; 16/; x D .1; 2; 3; 4/.5; 13; 6; 14/.7; 8; 9; 10/.11; 15; 12; 16/; then we see (in the same way as in the proof of Theorem 51.3) that G exists in this case and is isomorphic to a subgroup of A16 . (b) 1 .M / D M is of order 24 . Here we have x 2 D y 2 D 1. If we set s D .1; 4; 3; 5/.2; 7; 6; 8/, x D .4; 7/, then we see easily that G exists in this case as a subgroup of S8 . (c) 1 .M / D W0 is of order 22 and so M Š C4 C4 . Here we have Ã1 .M / D W0 D hx 2 ; y 2 i and hx; x s i D M so that we have two possibilities for x 2 and y 2 .

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(c1) If x 2 D u, then y 2 D uz and so y s D xu D x 1 . This gives y t D y s D 2 .x 1 /s D y 1 and x t D x s D y s D x 1 and so t inverts M . If we set s D .1; 5/.2; 8; 4; 7/.3; 6/, x D .5; 7; 6; 8/, then we see easily that G exists in this case as a subgroup of S8 . (c2) If x 2 D uz, then y 2 D u and then y s D xu D .xuz/z D x 1 z. This gives 2 2 y t D y s D .x 1 z/s D y 1 z and x t D x s D y s D x 1 z and so t does not inverts M . If we set 2

s D .2; 5; 12; 10/.3; 6/.4; 7; 14; 15/.9; 13; 16; 11/; x D .1; 2; 3; 4/.5; 9; 10; 11/.6; 12; 8; 14/.7; 13; 15; 16/; then we see as before that G exists in this case as a subgroup of A16 . We have proved the following result. Theorem 51.4. Suppose that a 2-group G contains a self-centralizing elementary abelian subgroup E of order 8, E < ˆ.G/, and E is normal in G. Then we have G D hx; y; u; z; t; si with the following initial relations u2 D z 2 D t 2 D s 4 D Œx; y D Œx; u D Œx; z D Œy; u D Œy; z D Œu; z D Œu; t D Œz; t D Œz; s D 1;

s 2 D t; us D uz; x s D y; y s D xu:

Here E D ht; z; ui is a self-centralizing elementary abelian normal subgroup of order 8 in G, G=E Š D8 , Z.G/ D hzi is of order 2, G is of order 26 , ˆ.G/ D Ehxyi and G D hs; xi. Also, Z.ˆ.G// D W0 D hz; ui is a four-group, M D hx; y; z; ui is a characteristic abelian subgroup of G of order 24 and exponent at most 4 and M is the unique abelian maximal subgroup of T D CG .W0 /, jG W T j D 2 and T =E Š E4 . There are exactly four possibilities for the structure of G (with two additional relations). (a) x 2 D y 2 D z, M is abelian of type .4; 2; 2/, and the group G is in this case isomorphic to a subgroup of A16 . (b) x 2 D y 2 D 1, M is abelian of type .2; 2; 2; 2/, and the group G is in this case isomorphic to a subgroup of S8 . (c1) x 2 D u, y 2 D uz, M Š C4 C4 , t inverts M , and the group G is in this case isomorphic to a subgroup of S8 . (c2) x 2 D uz, y 2 D u, M Š C4 C4 , t does not invert on M , and the group G is in this case isomorphic to a subgroup of A16 . In what follows, we assume that E is not normal in G. We recall that W0 D Z.ˆ.G// is a four-group and jG W T j D 2, where T D CG .W0 /. We set W0 D hz; ui and we know that Z.G/ D hzi is of order 2. Finally, for each involution t 2 E W0 , CT .t / D E and jGj 26 since jˆ.G/j 24 . Our aim is to prove that G does not possess in this case a normal elementary abelian subgroup of order 8.

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Suppose that A1 is a normal elementary abelian subgroup of order 16 in T . Let t 2 E W0 . No involution in E W0 could be contained in A1 and, by Proposition 51.1, jCA1 .t /j 22 . Hence, jCA1 .t /j D 4 and CA1 .t / D W0 so that E \ A1 D W0 . Set C D EA1 and so jC j D 25 with jC W A1 j D 2. Take any a 2 A1 . Then t a is an involution if and only if .t a/2 D 1 or equivalently at D a1 D a. It follows that all involutions in C A1 lie in E W0 . Since hE W0 i D E, we see that E is normal in NG .C /. If C D T , then E would be normal in G, a contradiction. Let D=C be a subgroup of order 2 in NT .C /=C . Then E is normal in D and note that all involutions in E W0 lie in a single conjugacy class in C since CG .t / D E and jC W Ej D 4. But then CD .t / 6 C , a contradiction. We have proved that T has no normal elementary abelian subgroups of order 16. Let A2 be a normal elementary abelian subgroup of order 16 in G. By the previous paragraph, A2 6 T and so B2 D A2 \ T is a normal elementary abelian subgroup of order 8 in G. If W0 6 B2 , then W0 B2 is a normal elementary abelian subgroup of order 16 contained in T , a contradiction. Hence we must have W0 B2 . But then CG .W0 / hT; A2 i D G, which is a contradiction. We have proved that G has no normal elementary abelian subgroups of order 16. Suppose that A0 is a normal elementary abelian subgroup of order 8 in G but A0 6 T . Then AQ0 D A0 \ T is a G-invariant 4-subgroup. If W0 D AQ0 , then CG .W0 / hA0 ; T i D G, a contradiction. Hence W0 ¤ AQ0 and so AQ0 W0 is a Ginvariant elementary abelian subgroup of order 8 contained in T . We have proved that if G has a normal subgroup isomorphic to E8 , then G has also such one which is contained in T . Now assume that G has a normal subgroup A Š E8 . By the previous paragraph, we may assume that A T . Since E is not normal in G, so we get E ¤ A. If W0 6 A, then W0 A would be a G-invariant elementary abelian subgroup of order 16, a contradiction. Hence W0 A and so E \ A D W0 . It is well known that there is a self-centralizing normal abelian subgroup B of T such that A B and B is normal in G. We have 1 .B/ D A and E \ B D E \ A D W0 . Since G=T acts faithfully on W0 (and so on B), B is also self-centralizing in G. For any x 2 B, tx is an involution if and only if t inverts x. Suppose that B D A. In this special case, A is a self-centralizing normal elementary abelian subgroup of order 8 of G and so G=A Š D8 since jGj 26 . Set V0 D EA and note that V0 A contains exactly four involutions and they are contained in E W0 . Hence E is normal in NG .V0 / and so V0 =A does not lie in Z.G=A/ since E is not normal in G. Let R=A be the cyclic subgroup of index 2 in G=A so that we have R \ V0 D A and R is a maximal subgroup of G. Hence E 6 R and this is a contradiction since E ˆ.G/. We have proved that B ¤ A and so jBj 24 . We shall determine now the structure of B. If an involution e 2 A W0 would be a square in B, then t would centralize e, since t inverts on Ã1 .B/ (Proposition 51.2). This is a contradiction. Suppose that all three involutions in W0 are squares in B. Then 2 .B/ D hb1 i hb2 i heiq, where o.b1 / D o.b2 / D 4, e is an involution

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in A W0 , and W0 D hb12 ; b22 i. We act with t 2 E W0 on 2 .B/. By Proposition 51.2, b1t D b11 w with w 2 W0 . This gives b1t D b1 .b12 w/, where b12 w D w1 2 W0 . Similarly, b2t D b2 w2 with w2 2 W0 and so .b1 b2 /t D .b1 b2 /.w1 w2 /. Here w1 ; w2 , and w3 D w1 w2 must be three pairwise distinct involutions in W0 since C2 .B/ .t / D hb12 ; b22 i D W0 . But then e t D ewj with j 2 f1; 2; 3g and so t centralizes one of the elements b1 e, b2 e, or b1 b2 e of order 4, which is a contradiction. We have proved that at most one involution in W0 is a square in B. Hence, we have obtained the following result. The subgroup B .> A/ is abelian of type .2m ; 2; 2/, m 2, and so m1 B D hbi hwi hei, where o.b/ D 2m 4, b 2 D z, hzi D Z.G/ D Ãm1 .B/, W0 D hz; wi, e 2 A W0 . Set V D EB D ht iB, where t is an involution in E W0 D E B. Our aim is to show that V D T . Therefore, we assume V ¤ T and set VQ D NT .V / so that jVQ =V j 2. We note that jBj D 2mC2 .m 2/ and so V B is a normal subset of VQ and jV Bj D 2mC2 D jtBj. If jVQ =V j 4, then jVQ W CVQ .t /j D jVQ W Ej 2mC2 and so jVQ =V j D 4 and all elements in V B are involutions since they are all conjugate in VQ to t . This is not the case since t e is not an involution in view of Œt; e ¤ 1. Hence we must have jVQ =V j D 2. Assume for a moment that VQ D NG .V /. By Theorem 1.7 and Proposition 1.8, G=B is of maximal class and V =B is a non-central subgroup of order 2 in G=B. In fact G=B must be dihedral of order 8 or semidihedral. Let R =B be the cyclic subgroup of index 2 in G=B. Then jG W R j D 2 and R \ V D B. But then E 6 R and so E 6 ˆ.G/, a contradiction. Hence we must have NG .V / > VQ . We set V D NG .V / so that jV W VQ j D 2 and V T D G. If CV .t / D E, then we see (since jV W Ej D 2mC2 ) that all 2mC2 elements in the normal subset V B in V are conjugate to t and so they are all involutions. But this is not the case, since t e (as above) is not an involution. Thus CV .t / D EQ is of order 24 , jEQ W Ej D 2 and EQ \ T D E. Hence all elements y 2 EQ E lie in G T , y 2 2 E and w y D wz, where W0 D hz; wi. Suppose at first that there is y 2 EQ E with y 2 2 W0 . In this case W0 hyi Š D8 and since there are involutions in .W0 hyi/ W0 , we may assume that y 2 D 1. We act with the involution y on A. Since CW0 .y/ D hzi and jCA .y/j 4 (Proposition 51.1), we see that there is an element e 2 A W0 so that Œy; e D 1. We have 1 ¤ Œt; e 2 W0 and since m2 Œt; ey D Œt y ; e y D Œt; e, so Œt; e D z which gives e t D ez. Set v D b 2 so that hvi is the cyclic subgroup of order 4 in hbi. Since e t D ez, we must have v t D vw for an element w 2 W0 hzi. Otherwise, t would centralize ve, which is not possible. In particular, t does not invert v, and so v 62 Ã1 .B/. Recall that, by Proposition 51.2, t inverts Ã1 .B/. But this forces that o.b/ D 4; hvi D hbi; jBj D 24 , and E is normal in V . The last statement follows from the fact that there are only four involutions in V B (since t inverts in B only the involutions in W0 ) and they lie in E W0 . Suppose now that for each y 2 EQ E, y 2 2 E W0 and so we may assume in this case that y 2 D t 2 E W0 . The cyclic group hyi of order 4 acts faithfully on A and hyi acts fixed-point-free on the set A W0 . If e y D ez with e 2 A W0 , 2 then e y D .ez/y D ezz D e and so e t D e, a contradiction. Hence, e y D ew

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2

with w 2 W0 hzi. This gives e y D e t D .ew/y D ewwz D ez. It follows that v t D vw1 with some w1 2 W0 hzi, where again hvi is the cyclic subgroup of order 4 in hbi. Otherwise, t would centralize ve. But this means that t does not invert hvi. As before, this forces that o.b/ D 4; hbi D hvi, jBj D 24 , and E is normal in V . Again, the last statement follows from the fact that there are only four involutions in V B (since t inverts in B only the involutions in W0 ) and these lie in E W0 . In both cases for the structure of EQ D hyiE, we have jBj D 24 , jVQ j D 26 , and VQ normalizes the subset V B containing exactly four involutions. On the other hand, jVQ W CVQ .t /j D jVQ W Ej D 23 which means that V B contains eight conjugates of t , which is the final contradiction in this paragraph. We have proved that we must have V=T and so T D ht iB, where t 2 E W0 . m2 so that o.v/ D 4 and v 2 D z. Since t 2 ˆ.G/ and ˆ.G/ D Set again v D b 2 Ã1 .G/, there is an element x 2 G T such that x 2 D t 0 2 T B. Because t 0 D lt 0 with l 2 B, we have for all y 2 B, y t D y lt D y t and .t 0 /2 2 B. Hence x induces an automorphism of order 4 on 2 .B/ D Ahvi since x 2 induces the same involutory automorphism on 2 .B/ as the automorphism induced by t . Since t acts fixed-point-free on A W0 , so x induces a 4-cycle on A W0 and so e x D ew, where 0 2 w 2 W0 hzi. Indeed, if e x D ez, then e t D e t D e x D .ez/x D ezz D e, which is 2 0 a contradiction. From e x D ew follows e x D e t D e t D .ew/x D .ew/.wz/ D ez. But then we must have v t D vw 0 with w 0 2 W0 hzi. Namely, if v t D vz D v 1 , then .ve/t D .vz/.ez/ D ve, a contradiction. It follows from v t D vw 0 that t does not invert v. By Proposition 51.2, t inverts Ã1 .B/. Hence v is not a square in B and so 2 .B/ D B is of order 24 . Since t inverts in B only the elements in W0 , so there are exactly four involutions in tB and they lie in E W0 . It follows that E x D E and so E is normal in G. This is the final contradiction. We have proved the following major result. Theorem 51.5. Suppose that a 2-group G contains a self-centralizing elementary abelian subgroup E of order 8, E ˆ.G/, and E is not normal in G. Then G has no normal elementary abelian subgroups of order 8. In the rest of this subsection, we study the structure of a 2-group G satisfying the assumptions of Theorem 51.5. We recall that W0 D Z.ˆ.G// is a normal four-subgroup contained in E and jG W T j D 2, where T D CG .W0 /. We set W0 D hz; ui and we know that Z.G/ D hzi is of order 2. Finally, for each involution t 2 E W0 , CT .t / D E and jGj 26 . Since G has no normal elementary abelian subgroups of order 8, we may apply Theorems 50.1 and 50.2. Suppose that G has two distinct normal 4-subgroups. Then Theorem 50.2 implies that G D D C , where D Š D8 , D \ C D Z.D/ and C is either cyclic or of maximal class but not isomorphic to D8 . We get ˆ.G/ C and so E C , which is not possible. We have proved that W0 is the unique normal four-subgroup of G.

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Since G is neither abelian nor of maximal class, we may apply Proposition 50.1. It follows that G has a normal metacyclic subgroup N such that CG .2 .N // N , G=N is isomorphic to a subgroup of D8 and W D 2 .N / is abelian of type .4; 2/ or .4; 4/. In any case, W0 D 1 .N / is the unique normal four-subgroup of G and W0 6 Z.G/. It follows that E \ N D W0 . Since E ˆ.G/, G=N is not elementary abelian. Hence G=N is either cyclic of order 4 or G=N Š D8 . Assume that G=N Š C4 . If N is abelian of type .2j ; 2/, j 2, then G=N acts faithfully on 2 .N / Š C4 C2 . If, in addition, N > 2 .N /, then there is a characteristic cyclic subgroup Z of order 4 contained in 2 .N / so that Z is normal in G. But then acting with G=N on Z, we see that t 2 E W0 centralizes Z T D CG .W0 /, which is a contradiction. Thus, N D 2 .N / and so jGj D 25 , which is again a contradiction. It follows that 2 .N / D W is abelian of type .4; 4/. We want to show that N does not possess a G-invariant cyclic subgroup Z of order 4. Suppose that there is such Z so that Z W T . But jG W CG .Z/j 2 and so E centralizes Z since E ˆ.G/, a contradiction. It follows that we may use Proposition 50.4. Hence in case G=N Š C4 , N is either abelian of type .2n ; 2n / or .2nC1 ; 2n / with n 2 or N is minimal nonabelian and m n m1 i, where m D n with more precisely N D ha; b j a2 D b 2 D 1; ab D a1C2 n 3 or m D n C 1 with n 2. Assume that G=N Š D8 . Then the structure of N is determined by Theorem 50.1. We shall show here that the minimal case with N Š C4 C2 cannot occur. Indeed, suppose that N is abelian of type .4; 2/. Let L=N D Z.G=N / so that jL=N j D 2 and E L since E ˆ.G/. Set W0 D 1 .N / so that W0 D Z.ˆ.G//, W0 < E and since E < ˆ.G/, we get ˆ.G/ D L. By the structure of G=N Š D8 Š Aut.N /, we know that t 2 E W0 inverts N . Thus, all elements in L N are involutions. Set N D hy; w j y 4 D w 2 D Œy; w D 1; y 2 D zi so that W0 D hz; wi, E D ht; z; wi with hzi D Z.G/. Let k be any element in G such that hkiN=N Š C4 . Set K D hkiN . It follows that k 2 2 L N and since k 2 is an involution, so k 4 D 1. By the structure of Aut.N / Š D8 , we have y k D yw, w k D wz. Then we see that Ã1 .K/ D ˆ.K/ D hk 2 ; w; zi. Hence the elementary abelian subgroup hk 2 ; w; zi of order 8 is normal in G. But then the other elementary abelian subgroup hk 2 y; w; zi of order 8 in K must be also normal in G. Note that K has exactly two elementary abelian subgroups of order 8 since there are exactly eight involutions in L N . In particular, E is normal in G, a contradiction. This proves that the case G=N Š D8 with N Š C4 C2 cannot occur. We have proved the following final result of this subsection. Theorem 51.6. Let G be a 2-group which has a self-centralizing elementary abelian subgroup E of order 8. Assume that E ˆ.G/ and E is not normal in G. Then the group G has the following properties. (a) G has no normal elementary abelian subgroups of order 8. (b) G has the unique normal four-subgroup W0 and W0 < E.

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(c) G has a normal metacyclic subgroup N such that 2 .N / D W is abelian of type .4; 4/, CG .W / N , 1 .W / D W0 , and G=N is either cyclic of order 4 or G=N Š D8 . (d) N is either abelian of type .2k ; 2kC1 / or .2k ; 2k /, k 2, or N is minimal m n m1 nonabelian and more precisely N D ha; b j a2 D b 2 D 1; ab D a1C2 i, where m D n with n 3 or m D n C 1 with n 2. 2o . The case E 6 ˆ.G/ and G has a normal elementary abelian subgroup of order 8. We suppose throughout this subsection that our 2-group G contains a self-centralizing non-normal elementary abelian subgroup E of order 8 such that E 6 ˆ.G/. (If E is normal in G, then G=E is isomorphic to a subgroup of D8 .) Also, we assume that G has a normal elementary abelian subgroup of order 8. Let t 2 E ˆ.G/ and let M be a maximal subgroup of G such that t 62 M . Then CG .t / D ht i CM .t /. But M0 D CM .t / contains the four-subgroup E0 D E \ M and CM0 .E0 / D E0 . If E0 ¤ M0 , then Theorem 1.7 and Proposition 1.8 imply that M0 Š D2n or M0 Š SD2n . Let A be a normal elementary abelian subgroup of order 8 in G. If t 2 A, then A is a normal subgroup of CG .t /. If t 62 A, then Proposition 51.1 implies that CA .t / D A0 is of order 4 and so ht i A0 is a normal elementary abelian subgroup of order 8 of CG .t /. In any case, M0 has a normal 4-subgroup and so E0 ¤ M0 implies that M0 Š D8 . We have proved the following result. Theorem 51.7. Let G be a 2-group which has a self-centralizing elementary abelian subgroup E of order 8 which is not contained in ˆ.G/. Assume that G has a normal elementary abelian subgroup of order 8. Then for each t 2 E ˆ.G/, we have either CG .t / D E or CG .t / D ht i M0 , where M0 Š D8 . Suppose that there is an involution t 2 E ˆ.G/ such that CG .t / contains an elementary abelian G-invariant subgroup A of order 8. Since E is not normal in G, so E ¤ A and therefore C D CG .t / D hE; Ai D ht i D, where D Š D8 . The subgroup C has exactly two elementary abelian subgroups of order 8 and so they are A and E. Also, t 2 A \ E. Since CC .A/ D A, so A is self-centralizing in G and therefore G=A is isomorphic to a subgroup of D8 . Obviously, E is normal in NG .C /. But E is not normal in G and so NG .C / ¤ G. This forces that jG W C j 4 and therefore G=A Š D8 . We have proved the following result which describes an exceptional case. Theorem 51.8. Let G be a 2-group which has a self-centralizing non-normal elementary abelian subgroup E of order 8 which is not contained in ˆ.G/. Suppose that there is an involution t 2 E ˆ.G/ such that CG .t / contains an elementary abelian G-invariant subgroup A of order 8. Then A is self-centralizing in G and G=A Š D8 . We assume in the rest of this subsection that there is no involution t 2 E ˆ.G/ such that CG .t / contains an elementary abelian G-invariant subgroup of order 8.

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Groups of prime power order

Suppose that G has a normal elementary abelian subgroup B of order 16. Let t be an involution in E ˆ.G/. By the structure of CG .t /, we have jCB .t /j 4. Then Proposition 51.1 forces that jBj D 16 and CB .t / D W0 is of order 4. Let C be a maximal normal abelian subgroup of G containing B. Then B D 1 .C / and, by the structure of CG .t /, we have CC .t / D CB .t / D W0 . Also, G=C acts faithfully on C . Set W0 D hz1 ; z2 i and let W1 D hu1 ; u2 i be a complement of W0 in B. Set z3 D z1 z2 and u3 D u1 u2 . We claim that we may choose the generators of W0 and W1 in such a way that uti D ui zi for i D 1; 2; 3. Indeed, suppose that there exist two elements u ¤ u0 2 W1 so that ut D uz and .u0 /t D u0 z, where 1 ¤ z 2 W0 . Then uu0 ¤ 1 and .uu0 /t D .uz/.u0 z/ D uu0 , which contradicts the fact that CC .t / D W0 . Hence ut D uz and .u0 /t D u0 z 0 , where z; z 0 are distinct involutions in W0 . This gives .uu0 /t D .uu0 /.zz 0 /, where uu0 ¤ 1 and zz 0 ¤ 1. It remains to set u1 D u, u2 D u0 , u3 D uu0 , z1 D z, z2 D z 0 , z3 D zz 0 , and we see that the above claim is proved. Suppose that C ¤ B. Let v be an element of order 4 contained in C B. Then v 2 2 B and, by Proposition 51.2, .v 2 /t D v 2 D v 2 and so v 2 2 W0 . By Proposition 51.2, v t D v 1 w0 with w0 2 W0 and so v t D v.v 2 w0 / D vzj , where j 2 f1; 2; 3g. But then .vuj /t D vzj uj zj D vuj , which is a contradiction. We have proved that C D B is a self-centralizing normal elementary abelian subgroup of order 16 of G. Set E1 D ht i W0 and L D ht iB. Then all elements in L .E1 [ B/ have order 4 and four involutions in E1 W0 form a single conjugacy class in L since jL W CL .t /j D 4. It follows L0 D Z.L/ D ˆ.L/ D W0 . If L D G, then CG .t / D E1 D E is normal in G. This is a contradiction and so L ¤ G. Set K D NG .L/. Then K normalizes the L-class E1 W0 . Hence CK .t / must cover K=L. Since N D CK .t / D CG .t / Š C2 D8 , so jK=Lj D 2 and N \ L D E1 . We have NG .E1 / hN; Li D K and so K ¤ G by our last assumption. Indeed, if K D G, then N D CG .t / would contain the G-invariant subgroup E1 . Now, Theorem 1.7 and Proposition 1.8 imply that G=B is of maximal class (acting faithfully on B). Also, L=B is a non-central subgroup of order 2 in G=B and so G=B is not a generalized quaternion group. On the other hand, GL.4; 2/ does not possess elements of order 8 and so jG=Bj D 8 and consequently G=B Š D8 . We have Z.G/ CG .t / D N and so Z.G/ Z.N / D ht; z 0 i, where hz 0 i D N 0 < W0 . Hence Z.G/ Z.L/ D W0 and so Z.G/ D ht; z 0 i \ W0 D hz 0 i is of order 2. We have proved the following result. Theorem 51.9. Let G be a 2-group which has a self-centralizing non-normal elementary abelian subgroup E of order 8 which is not contained in ˆ.G/. Assume that there is no involution t 2 E ˆ.G/ such that CG .t / contains an elementary abelian G-invariant subgroup of order 8. Suppose that G has a normal elementary abelian subgroup B of order 16. Then jBj D 16, B is self-centralizing in G, G=B Š D8 , and Z.G/ is of order 2. We suppose also in the rest of this subsection that G has no normal elementary abelian subgroups of order 16.

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Let A be a normal elementary abelian subgroup of order 8 in G. Let B be a maximal normal abelian subgroup of G containing A. Then G=B acts faithfully on B and we have 1 .B/ D A. Let t be any involution in E ˆ.G/. By our assumptions, CG .t / does not contain a G-invariant elementary abelian subgroup of order 8. Hence t 62 B. By Theorem 51.7, we have either CG .t / D E or CG .t / D ht i D, where D Š D8 . By Proposition 51.1, jCA .t /j 4 and so CB .t / D CA .t / D W0 is a four-group. Assume that B D A. Then G=A is isomorphic to a subgroup of D8 . Set ht iA D L and ht i W0 D E1 . All elements in L .E1 [ A/ are of order 4 and so E1 is normal in NG .L/ since hE1 W0 i D E1 . If L is normal in G, then E1 is normal in G. But then CG .t / contains a G-invariant elementary abelian subgroup of order 8, which contradicts our assumptions. It follows that L=A is a non-central subgroup in G=A and so G=A Š D8 . In what follows we shall assume that B ¤ A. By Proposition 51.2, no element in A W0 is a square in B. Assume that all three involutions in W0 are squares in B. Then 2 .B/ D hb1 i hb2 i hei, where o.b1 / D o.b2 / D 4, e 2 A W0 , and W0 D hb12 ; b22 i. By Proposition 51.2, b1t D .b1 /1 w0 with w0 2 W0 and so b1t D b1 .b12 w0 /, where 1 ¤ w1 D b12 w0 2 W0 . Similarly, b2t D b2 w2 with 1 ¤ w2 2 W0 and so .b1 b2 /t D .b1 b2 /.w1 w2 /. Since C2 .B/ .t / D W0 , so w1 ; w2 ; w3 D w1 w2 must be pairwise distinct involutions in W0 . But then e t D ewj with j 2 f1; 2; 3g and so t centralizes one of the elements b1 e; b2 e, or b1 b2 e (all of order 4), which is a contradiction. It follows that exactly one involution in W0 is a square in B. We have proved that B D hbi hwi hei, where m1 D z, hzi Z.G/, W0 D hz; wi, and e 2 A W0 . o.b/ D 2m , m 2, b 2 Set V D ht iB and suppose that CG .t / D ht i D, where D Š D8 . Then W0 is normal in CG .t /, Z.CG .t // D ht; zi and .CG .t //0 D hzi. Hence CG .t / contains an involution u 2 G V so that w u D wz and so W0 hui D D Š D8 . Acting with u on A, we see that jCA .u/j 4 (Proposition 51.1). Since CW0 .u/ D hzi, so u centralizes an involution in A W0 . We may assume e u D e and so CA .u/ D hz; ei. If Œt; e D w0 with w0 2 W0 hzi, then Œt; eu D Œt u ; e u D Œt; e D w0 D w0u D w0 z and so z D 1, a contradiction. Hence, we must have Œt; e D z. If m 3, then t inverts m2 the element v D b 2 of order 4 (Proposition 51.2) and so v t D v 1 D vz. But then .ve/t D .vz/.ez/ D ve, which is a contradiction since CB .t / D W0 . Hence we must have m D 2, o.b/ D 4, b 2 D z, b t D bw with w 2 W0 hzi, and e t D ez. It follows that t inverts in B exactly four elements, namely the elements in W0 (which are actually centralized by t ). Hence there are exactly four involutions in V B and they all lie in E1 D ht i W0 . These four involutions in E1 W0 lie in a single conjugacy class in V since jV W CV .t /j D 4. Set VQ D V CG .t / D V hui so that NG .V / D VQ (since CG .t / must cover NG .V /=V ) and E1 is normal in VQ because hE1 W0 i D E1 . If VQ D G, then CG .t / would contain a G-invariant elementary abelian subgroup E1 of order 8, contrary to our assumptions. Hence VQ ¤ G and then Theorem 1.7 and Proposition 1.8 imply that G=B is of maximal class and VQ =B Š ht; ui Š E4 . Replacing u with ut (if necessary), we may assume that huiB=B D Z.G=B/. We claim that CG .A/ D B. Set CG .A/ D BQ so that BQ B and BQ is normal in G.

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Groups of prime power order

If BQ ¤ B, then BQ huiB. This is a contradiction since w u D wz and so u acts faithfully on A. Hence we have CG .A/ D B and so G=B Š D8 since VQ ¤ G. It remains to consider the case where CG .t / D E D ht iW0 for each t 2 Eˆ.G/. We set again V D ht iB. Since jBj D 2mC2 , m 2, so also jV Bj D jBt j D 2mC2 . Set VQ D NG .V /. If jVQ =V j 4, then jVQ W CVQ .t /j D jVQ W Ej 2mC2 and so all 2mC2 elements in V B are involutions. This is a contradiction since t e is not an involution in view of Œt; e ¤ 1. Therefore we must have jVQ =V j 2. If VQ D G, then jG=Bj D 2 or 4. We assume that G ¤ VQ and so in this case jVQ =V j D 2. Since B is abelian, so the set B0 of elements of B which are inverted by t is a subgroup of B. The number of involutions in V B is equal to jB0 j. Since t does not invert e 2 A W0 , so jB0 j 2mC1 . On the other hand, jcclVQ .t /j D jVQ W Ej D 2mC1 and so jB0 j D 2mC1 , V B has exactly 2mC1 involutions and they lie in a single VQ -class. By Proposition 51.2, Ã1 .B/ D hb 2 i B0 and so B1 D Ã1 .B/W0 D hb 2 ; wi B0 , where w 2 W0 hzi. Since B=B1 Š E4 and hB1 ; ei 6 B0 , so B0 D hB1 ; bi or B0 D hB1 ; bei. Replacing b with be (if necessary), we may assume that t inverts b and so B0 D hB1 ; bi D hb; wi. m2 Since t inverts v D b 2 with v 2 D z, so v t D v 1 D vz. Then e t D ez would t imply .ve/ D .vz/.ez/ D ve, which is a contradiction since CB .t / D W0 D hz; wi. Therefore e t D ew with w 2 W0 hzi. Since NG .V /=B is of order 4 and G ¤ NG .V / D VQ , Theorem 1.7 and Proposition 1.8 imply that G=B is of maximal class. The subgroup V =B (of order 2) is non-central in G=B and so G=B is dihedral or semi-dihedral. In particular, VQ =B is a 4-group. Let x 2 G VQ be such that x normalizes VQ and x 2 2 VQ . Set D D hVQ ; xi so that (by the structure of G=B) D =B Š D8 and therefore we may assume that x 2 2 B. Since V =B is noncentral in D =B, u D t x 2 VQ V . We claim that W0 D hz; wi is normal in G. Obviously Z.G/ CB .t / D W0 . If Z.G/ D W0 , then our claim is clear. Suppose that Z.G/ < W0 so that Z.G/ D hzi is of order 2. Let U be a normal 4-subgroup contained in A and assume that U ¤ W0 . Then we have U > hzi D Z.G/ and so U \W0 D hzi. But then for an element u 2 U hzi, we have ut D uz since CA .t / D W0 . This gives .vu/t D .vz/.uz/ D vu, where v is the cyclic subgroup of order 4 in hbi. This is a contradiction and so W0 D hz; wi is normal in G. Since CA .t / D W0 and x normalizes A and W0 , it follows CA .u/ D W0 . Hence CG .u/ D hui W0 and Z.VQ / D W0 since VQ D Bht; ui. Note that ht; bi Š D2mC1 and t e D t w. There are exactly two V -classes of involutions contained in V B with the representatives t and t b containing 2m elements each. The V -class of t is the set ft w i b 2j g and the V -class of t b is the set ft bw i b 2j g, where i; j are any integers. Now u fuses these two V -classes and so t u D t bw r b 2s for suitable integers r; s. This gives .t u/2 D t ut u D t t u D bw r b 2s D b 2sC1 w r and so o.t u/ D 2mC1 and ht; ui Š D2mC2 . Since u inverts ht ui, so u inverts .t u/2 and w. But h.t u/2 ; wi D B0 and so u inverts B0 . It follows that b u D b 1 and w u D w. Thus t u centralizes B0 D hb; wi. Since t u acts faithfully on B and B D B0 A, t u must act faithfully

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on A. Note that ht uiB=B D Z.D =B/ D Z.G=B/. If BQ D CG .A/ > B, then t u 2 BQ since BQ is normal in G and G=B is of maximal class. This is a contradiction and so CG .A/ D B, D D G, and G=B Š D8 . We have proved the following final result of this subsection. Theorem 51.10. Let G be a 2-group which has a self-centralizing non-normal elementary abelian subgroup E of order 8 which is not contained in ˆ.G/. Assume that there is no involution t 2 E ˆ.G/, such that CG .t / contains an elementary abelian G-invariant subgroup of order 8. Suppose that G has a normal elementary abelian subgroup of order 8 but none of order 16. Then G has a maximal normal abelian subgroup B of type .2m ; 2; 2/, m 1 such that G=B is isomorphic to a non-trivial subgroup of D8 . Example. We give here an example of a 2-group G of order 28 satisfying the assumptions of Theorem 51.10. We set G D hb; z; w; e; t; u; x j b 8 D z 2 D w 2 D e 2 D t 2 D u2 D x 2 D Œb; w D Œb; e D Œw; e D Œt; w D Œu; w D Œe; x D 1; b 4 D z; u D t x ; b t D b 1 ; b u D b 1 ; e t D ew; e u D ewz; .t u/2 D b; b x D b 1 ; w x D wzi. Here B D hb; w; ei is abelian of type .8; 2; 2/ and G=B Š D8 . A coset enumeration computer program assures that such a group G exists! 3o . The case E 6 ˆ.G/ and G has no normal elementary abelian subgroups of order 8. We suppose throughout this subsection that a 2-group G contains a selfcentralizing elementary abelian subgroup E of order 8 which is not contained in ˆ.G/. We assume in addition that G has no normal elementary abelian subgroups of order 8. Let t 2 E ˆ.G/ and M be a maximal subgroup of G such that t 62 M . Then (by modular law) CG .t / D ht i M0 , where M0 D CM .t / contains the four-subgroup E0 D E \M and CM0 .E0 / D M0 . Hence we have either CG .t / D E or (by Theorem 1.7 and Proposition 1.8) M0 Š D2n , n 3 or M0 Š SD2m , m 4. Suppose that G has (at least) two distinct normal 4-subgroups. Then Theorem 50.2 implies that G is the central product G D D C of D Š D8 and C with D \ C D Z.D/ and C is either cyclic or of maximal class different from D8 . However, if C is cyclic or generalized quaternion, then G does not possess a self-centralizing elementary abelian subgroup of order 8. Hence C Š D2n or C Š SD2n , n 4. We have proved the following result. Theorem 51.11. Let G be a 2-group which has a self centralizing elementary abelian subgroup E of order 8 with E 6 ˆ.G/. Assume in addition that G has no normal elementary abelian subgroups of order 8. Then for each involution t 2 E ˆ.G/, we have CG .t / D ht i M0 , where M0 is isomorphic to one of the following groups: E4 , D2n .n 3/, or SD2m .m 4/. If G has (at least) two distinct normal 4-subgroups, then G D D C is the central product of D and C , where D Š D8 , D \ C D Z.D/, and C Š D2n or C Š SD2n with n 4. In view of Theorem 51.11, we assume in the rest of this subsection that G has the unique normal four-subgroup U (see Lemma 1.4).

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Groups of prime power order

Suppose that CG .t / D E for an involution t 2 E ˆ.G/. This is an exceptional case. Since G is neither abelian nor of maximal class, we may apply Theorem 50.1. Let N be a metacyclic normal subgroup of G such that G=N is isomorphic to a subgroup of D8 , W D 2 .N / is abelian of type .4; 4/ or .4; 2/, and CG .W / N . Set W0 D 1 .W / so that W0 is the unique normal four-subgroup of G. Suppose that jE \ N j 2. In that case jE \ N j D 2 and E \ N D E \ W0 . Since CG .t / D E, t 2 E N induces on W an involutory automorphism, where W D CW .t / is of order 2. We may apply Proposition 51.2 which shows that t inverts Ã1 .W / and also on W =W . If W Š C4 C4 , then t inverts Ã1 .W / D W0 contrary to the fact that jCW .t /j D 2. Suppose that W D ha; s j a4 D s 2 D Œa; s D 1; a2 D zi Š C4 C2 . In this case W D CW .t / D hzi D Ã1 .W /. We must have s t D sz and at D a1 z with D 0; 1. If D 1, then at D a1 z D a, a contradiction. If D 0, then at D a1 D az and so .as/t D .az/.sz/ D as, a contradiction. We have proved that in this special case jE \ N j D 4 and so E > W0 , which gives the following result. Theorem 51.12. Let G be a 2-group which has a self-centralizing elementary abelian subgroup E of order 8 with E 6 ˆ.G/. Assume that G has no normal elementary abelian subgroups of order 8. Suppose in addition that G has the unique normal 4subgroup W0 and that for an involution t 2 E ˆ.G/, CG .t / D E. Then we have E > W0 and Theorem 50.1 gives further information about the structure of G. In the rest of this subsection we assume also that there is an involution t 2 E ˆ.G/ such that G ¤ CG .t / D ht i D with D Š D2m , m 3 or D Š SD2n , n 4. Remark. If G is a 2-group possessing an involution t such that CG .t / D ht i D with D Š D2m or D Š SD2m , where m 4, then G has no normal elementary abelian subgroup A of order 8. Indeed, if A is such a subgroup, then Proposition 51.1 implies that jCA .t /j 4 and so CG .t / would contain a normal elementary abelian subgroup of order 8, which is not the case. Let U be the unique normal 4-subgroup of G. Set T D CG .U / so that jG W T j 2. Suppose at first that t 2 U . Since G ¤ CG .t /, we have jG W T j D 2 and T D CG .t / D ht i D with D \ U D hui D Z.D/ D Z.G/ and D Š D2m , m 3 or D Š SD2n , n 4. Let M be a maximal subgroup of G such that t 62 M . Set M0 D T \ M so that CG .t / D ht i M0 and M0 Š D. If M were not of maximal class, then Lemma 1.4 implies that M contains a G-invariant 4-subgroup U0 . Since U0 ¤ U , we have obtained a contradiction. Hence M is of maximal class and therefore M Š D2mC1 or M Š SD2mC1 and M0 Š D Š D2m , m 3. There is an element y 2 M M0 such that y 2 2 hui D Z.M / D Z.M0 /. We get t y D t u so that y t D yu and this determines the action of t on M . We have proved the following result. Theorem 51.13. Let G be a 2-group which has a self-centralizing elementary abelian subgroup E of order 8 with E 6 ˆ.G/. Assume that G has no normal elementary

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abelian subgroups of order 8. Suppose in addition that G has the unique normal 4subgroup U and there is an involution t 2 E ˆ.G/ such that t 2 U and G ¤ CG .t / D ht i D with D Š D2m , m 3 or D Š SD2n , n 4. Then G has a maximal subgroup M such that G D ht iM , where M Š D2mC1 or M Š SD2mC1 and CM .t / Š D2m , m 3. In what follows we assume also that t 62 U , where U is the unique normal foursubgroup of G. Suppose for a moment that t 2 T U , where T D CG .U /. Then CG .t / has the normal elementary abelian subgroup E1 D ht i U of order 8. This forces CG .t / D ht i D, where D Š D8 . Since E1 6 Z.CG .t //, it follows CG .t / 6 T and CT .t / D E1 . In particular, jG W T j D 2 and there is an involution t 0 2 CG .t / T . Set hzi D CU .t 0 / so that z 2 Z.G/. It follows that E2 D ht 0 ; t; zi is another elementary abelian subgroup of order 8 contained in CG .t / and so CG .E2 / D E2 . Also, t 0 is an involution in E2 ˆ.G/. If CG .t 0 / D E2 , then applying Theorem 51.12 we get a contradiction since E2 does not contain U . It follows that G ¤ CG .t 0 / D ht 0 i M0 with M0 Š D2m , m 3 or M0 Š SD2n , n 4. Replacing E with E2 and t with t 0 , we see that we may assume from the start that t 2 G T . It remains to consider only the case t 2 G T , where CG .t / D ht i M0 with M0 Š D2m , m 3 or M0 Š SD2n , n 4. Here T D CG .U /, jG W T j D 2, where U is the unique normal four-subgroup of G. It follows that G D ht iT and so, by the modular law, G0 D CG .t / D ht i D, where D D CT .t / Š M0 . We have hzi D Z.D/ D CU .t / D Z.G/ is of order 2. Set G1 D NG .G0 /. Since j.UG0 / W G0 j D 2, we have UG0 G1 . But Z.G0 / D ht; zi, so t has exactly two conjugates t and t z in G1 . It follows G1 D UG0 . Set D0 D U ht i and so D0 Š D8 and G1 D D D0 with D \ D0 D hzi and U D hz; ui < D0 . We fix in the rest of this subsection the notation for the structure of D. We set m1

D D ha; b j a2

m2

D b 2 D z 2 D 1; a2

D z; ab D a1 z ; D 0; 1i;

where m 3, and if D 1, then m 4. Let v be an element of order 4 in D0 and let y be an element of order 4 in hai. Since y 2 D v 2 D z, so x D yv is an involution in G1 .D0 [ D/. We compute x t D .yv/t D yv 1 D yvz D xz, and so D1 D hx; t i Š D8 and D1 \ U D Z.D1 / D Z.G/ D hzi. We see that CG1 .D1 / D ha; bt i D D Š D because x bt D .yv/bt D y 1 v 1 D yv D x, .bt /2 D 1 and abt D a1 z . Thus G1 D D D1 with D Š D, D \ D1 D hzi, CG .t / D ht i D D ht i D and U D hz; ui 6 D1 . Denoting again D with D and bt with b, we have obtained the following initial configuration. (R) The subgroup G1 D U CG .t / is the central product G1 D D D1 , where D1 Š D8 , t 2 D1 , D \ D1 D hzi D Z.G/, CG .t / D ht i D with D Š D2m or D Š SD2m , m 3, and U 6 D1 , U 6 D. In the rest of this subsection we denote with S a subgroup of G of the maximal possible order subject to the following three conditions.

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(i) S G1 D U CG .t / D D D1 , where D Š D2m or D Š SD2m , m 3, D1 Š D8 , t 2 D1 . (ii) S D DL, where L is normal in S , L Š D2n , n 3, D \ L D hzi D Z.D/ D Z.L/ and L D1 . (iii) U D hz; ui 6 L and U 6 D. We set for the rest of this subsection: n1

L D hc; t j c 2

n2

D t 2 D z 2 D 1; c 2

D z; c t D c 1 ; n 3i Š D2n :

We may set u D yv, where y is an element of order 4 in hai and v is an element of order 4 in hci. If a4 D 1, then we put a D y and if c 4 D 1, then we set c D v. It is now easy to determine all possibilities for the structure of S . Act with D on the dihedral subgroup L. Since Aut.L/=Inn.L/ is abelian of type .2n3 ; 2/, it follows that D 0 D ha2 i centralizes L. Also we know that D centralizes hv; t i D D1 Š D8 , where D1 L and so by the structure of Aut.L/ either D centralizes L (and so S D D L) or M D CD .L/ is a maximal subgroup of D in which case n 4. In that case D=M induces such an involutory automorphism on L so that CL .D=M / D hc 2 ; t i Š D2n1 . In any case, ŒD; L hzi and so D is also normal in S . If D 1 (i.e. D is semidihedral), then we have three possibilities for the maximal subgroup M of D. If D 0 (i.e. D is dihedral), then two distinct maximal subgroups of D are isomorphic and so we have in this case only two possibilities for the maximal subgroup M of D. This gives (together with central products) exactly seven possibilities Si , i D 1; 2; : : : ; 7 for the structure of S and they are given explicitly (in terms of generators and relations) in the following definition. Definition 1. Let S D DL be a product of two normal subgroups m1

D D ha; b j a2

m2

D b 2 D z 2 D 1; a2

D z; ab D a1 z ; D 0; 1; m 3;

and if D 1; then m 4i and n1

L D hc; t j c 2

n2

D t 2 D z 2 D 1; c 2

D z; c t D c 1 ; n 3i Š D2n ;

where D \ L D Z.D/ D Z.L/ D hzi and Œa; t D Œb; t D 1 so that CS .t / D ht i D. Here if D 1, then D Š SD2m , m 4, and if D 0, then D Š D2m , m 3. (1) If D 1 and Œa; c D Œb; c D 1, then S D D L D S1 , m 4, n 3. (2) If D 1, CD .L/ D hai, and c b D cz, then S D S2 , m 4, n 4. (3) If D 1, CD .L/ D ha2 ; bi Š D2m1 , and c a D cz, then S D S3 , m 4, n 4. (4) If D 1, CD .L/ D ha2 ; bai Š Q2m1 , then S D S4 , m 4, n 4. (5) If D 0, CD .L/ D hai, and c b D cz, then S D S5 , m 3, n 4.

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(6) If D 0, CD .L/ D ha2 ; bi Š E4 or D2m1 , and c a D cz, then S D S6 , m 3, n 4. (7) If D 0, and CD .L/ D D, then S D D L D S7 , m 3, n 3. We make here the following simple observation. Since t 2 G T , where T D CG .U / and jG W T j D 2, t cannot be conjugate (fused) in G to any involution t 0 which centralizes U . Therefore, with respect to that fusion, it is enough to consider only those involutions in S which act faithfully on U . We want to show in the sequel that S must be a normal subgroup of G and jG=S j 4. In some cases for the structure of S this is difficult. 4o . The case S Š S1 . We have exactly four conjugacy classes of involutions contained in S U which act faithfully on U with the representatives t , t c, b, and bav. The corresponding centralizers in S are CS .t / D CG .t / D ht i D with D Š SD2m , m 4; CS .t c/ D ht ci D with D Š SD2m m 4; CS .b/ D hbi L with L Š D2n , n 3; CS .bav/ D hbavi hc; tyi with hc; tyi Š Q2n , n 3. We may assume NG .S / ¤ S (otherwise we are finished). Then NG .S / can fuse cclS .t / (the conjugacy class of t in S ) only with cclS .t c/ and so jNG .S / W S j D 2. Obviously, Lhci is a normal subset of involutions in NG .S / and L D hLhcii which implies that L is normal in NG .S /. Suppose that N—G .S / ¤ G (otherwise we are finished). Note that each x in NG .S /S sends (by conjugation) cclS .t / onto cclS .t c/. By the above, NG .S / cannot fuse cclS .b/ and cclS .bav/ to any other conjugate class of involutions in S . This gives that CNG .S/ .b/ covers NG .S /=S and also CNG .S/ .bav/ covers NG .S /=S . In particular jCNG .S/ .b/j D jCNG .S/ .bav/j D 2nC2 . The subgroup K D hb; bavi is dihedral of order 2m and S D KL. Since both K and L are generated by its non-central involutions, we have hcclS .t /, cclS .t c/, cclS .b/, cclS .bav/i D S . Let x be any involution in NG .S /S . Since .cclL .t //x D cclL .t c/, so x induces an outer involutory automorphism on L which inverts hci. Indeed, if t x D t c k .k odd/, then x inverts t .t c k / D c k and so x inverts c. Hence CL .x/ D hzi and so, by Theorem 1.7 and Proposition 1.8, Lhxi D L0 is of maximal class. Since L0 is generated by its involutions, it follows L0 Š D2nC1 . We have NG .S / D S hxi and so CNG .S/ .x/ D hxi CS .x/. On the other hand CL .x/ D hzi and so if jCS .x/j D 2m , then CS .x/ covers S=L which implies that L0 Š D2nC1 is normal in NG .S / D DL0 > S . This contradicts the maximality of S . (See the construction of S .) It follows that there is no involution x 2 NG .S / S with the property jCS .x/j D 2m . Let y 0 be an element in G NG .S / such that y 0 normalizes NG .S / and .y 0 /2 2 0 NG .S /. If x D t y 2 NG .S / S , then CNG .S/ .x/ D hxi CS .x/ Š ht i D and 0 so jCS .x/j D 2m , which is a contradiction. If t y 2 cclS .t / [ cclS .t c/, then there 0 0 1 is an n 2 NG .S / such that t y D t n . But then t y n D t and so CG .t / 6 NG .S /, 0 0 y a contradiction. Hence we must have t 2 cclS .b/ or t y 2 cclS .bav/. We know 0 that jCNG .S/ .b/j D jCNG .S/ .bav/j D 2nC2 and on the other hand jCNG .S/ .t y /j D jCNG .S/ .t /j D 2mC1 . Hence m D n C 1. It follows that for each x 2 .cclS .t / [ cclS .t c/ [ cclS .b/ [ cclS .bav// D P , we have jCNG .S/ .x/j D 2nC2 D 2mC1 .

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Groups of prime power order 0

0

Finally, since S y ¤ S , there is an x 2 P such that x 0 D x y 2 NG .S / S . But then CNG .S/ .x 0 / D hx 0 i CS .x 0 / and CS .x 0 /j D 2m , which is the final contradiction. It follows that NG .S / D G and so in case S Š S1 , the subgroup S is normal in G and jG=S j 2. 5o . The case S Š S2 . We have exactly four conjugacy classes of involutions contained in S U which act faithfully on U with the representatives t , t c, b, and bav. The corresponding centralizers in S are CS .t / D CG .t / D ht i D with D Š SD2m , m 4; CS .t c/ D ht ci ha; bavi with ha; bavi Š SD2m , m 4; CS .b/ D hbi hyc; t i with hyc; t i Š SD2n , n 4; CS .bav/ D hbavi hcy; t ci with hcy; t ci Š SD2n , n 4. We assume NG .S / ¤ S (otherwise we are finished). Set P D cclS .t / [ cclS .t c/ [ cclS .b/ [ cclS .bav/. Obviously, hP i D S and P G T , where T D CG .U / and jG W T j D 2. Suppose also that t is conjugate in NG .S / to t c but t is not conjugate in NG .S / to any involution in S L. Then jNG .S / W S j D 2 and since NG .S / normalizes the set L hci and hL hcii D L, so L is normal in NG .S /. Let x be any involution in NG .S / S . Since .cclL .t //x D cclL .t c/, we get (as in case S Š S1 ) L0 D Lhxi Š D2nC1 . We have CNG .S/ .x/ D hxi CS .x/ and CL .x/ D hzi. If jCS .x/j D 2m , then CS .x/ covers S=L and so L0 is normal in DL0 > S . This contradicts the maximality of S . Hence there is no involution x 2 NG .S / S with jCS .x/j D 2m . Suppose in addition that NG .S / ¤ G (otherwise we are finished). Let y 0 be an element in G NG .S / such that y 0 normalizes NG .S / and .y 0 /2 2 NG .S /. 0 If x D t y 2 NG .S / S , then jCNG .S/ .x/j D 2mC1 and so jCS .x/j D 2m , a 0 contradiction. If t y 2 cclS .t / [ cclS .t c/, then CG .t / 6 NG .S / (as in case S Š S1 ), 0 0 which is a contradiction. Hence we must have t y 2 cclS .b/ or t y 2 cclS .bav/. 0 Since y 0 does not normalize S , there is x 0 2 P such that x0 D .x 0 /y 2 NG .S / S . If NG .S / fuses b with bav, then all involutions in P are fused in hy 0 iNG .S / to t and so jCNG .S/ .x0 /j D 2mC1 and jCS .x0 /j D 2m , a contradiction. If NG .S / does not 0 fuse b with bav, then jCNG .S/ .b/j D jCNG .S/ .bav/j D 2nC2 . Since jCNG .S/ .t y /j D 2mC1 , so n C 2 D m C 1 and therefore jCNG .S/ .x0 /j D 2nC2 D 2mC1 and so again jCS .x0 /j D 2m , a contradiction. Hence we must have NG .S / D G and so we are done in this case. Suppose now that t is not conjugate to t c in NG .S /. Then t must be conjugate in NG .S / to b or to bav. This implies jNG .S / W S j D 2 and m D n 4. Each element x 2 NG .S / S sends .CS .t //0 D ha2 i onto .CS .t x //0 D hc 2 i. Suppose that NG .S / ¤ G and let y 0 be an element in G NG .S / such that y 0 normalizes NG .S / and .y 0 /2 2 NG .S /. Since y 0 does not normalize S , there is an x 0 2 P such 0 that x0 D .x 0 /y 2 NG .S / S , x0 2 G T , and so x0 does not centralize U . On the other hand, x0 sends hyi (contained in ha2 i) onto hvi (contained in hc 2 i). We have y x0 D vz with D 0; 1. But then .yvz /x0 D yvz and so x0 centralizes U D hz; u D yvi. This is a contradiction and so we have NG .S / D G also in this case.

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It remains to consider the case that t is fused in NG .S / to all three involutions t c, b, and bav. In that case m D n 4 and jNG .S / W S j D 4. Note that L and K D hb; bavi are two dihedral normal subgroups both of order 2m of S containing cclS .t /, cclS .t c/, cclS .b/, and cclS .bav/ with NG .S /=S acting regularly on those four S -classes. Hence for each x 2 NG .S / S we have either Lx D L and K x D K or Lx D K. Suppose now that NG .S / ¤ G and let y 0 be an element in G NG .S / 0 such that y 0 normalizes NG .S / and .y 0 /2 2 NG .S /. Then t 0 D t y 2 NG .S / S . 0 Indeed, if t 0 2 S , then t 0 2 P and so there is an n 2 NG .S / such that t 0 D t y D t n . 0 1 0 D t and so CG .t / 6 NG .S /, a contradiction. If Lt D L and But then t y n 0 0 0 K t D K, then .cclL .t //t D cclL .t c/ and .cclK .b//t D cclK .bav/. This gives (as in case S Š S1 ) Lht 0 i Š D2mC1 and Kht 0 i Š D2mC1 . In particular, t 0 inverts hci m3 m3 m3 and on havi. Hence t 0 inverts v 2 hc 2 i and on y 2 h.av/2 i D ha2 i. From 0 0 0 0 y t D y 1 and v t D v 1 , we get ut D .yv/t D y 1 v 1 D yv D u. Hence t 0 centralizes U D hz; ui and so t 0 2 T . But t 2 G T and jG W T j D 2, which 0 0 0 is a contradiction. If Lt D K, then hcit D havi and so v t D yz , D 0; 1. 0 0 This gives .yvz /t D yvz and .uz /t D uz and so t 0 centralizes U , which is a contradiction. It follows that we must have NG .S / D G also in this case. We have proved that in case S Š S2 , the subgroup S is normal in G, jG W S j 4, and if jG=S j D 4, then m D n 4. 6o . Cases S Š S3 and S Š S4 . We have exactly four conjugacy classes of involutions contained in S U which act faithfully on U with the representatives t , t c, b, and bav. The corresponding centralizers in S in case S Š S3 are CS .t / D CG .t / D ht i D with D Š SD2m , m 4; CS .t c/ D ht ci hav; bi with hav; bi Š D2m , m 4; CS .b/ D hbi L with L Š D2n , n 4; CS .bav/ D hbavi hcy; t ci with hcy; t ci Š SD2n , n 4. The centralizers in S in case S Š S4 are CS .t / D CG .t / D ht i D with D Š SD2m , m 4; CS .t c/ D ht ci hav; bai with hav; bai Š Q2m , m 4; CS .b/ D hbi hcy; t i with hcy; t i Š SD2n , n 4; CS .bav/ D hbavi hc; tyi with hc; tyi Š Q2n , n 4. Suppose that S ¤ NG .S /. Then NG .S / can fuse t only with an involution in cclS .bav/ when S Š S3 and only with an involution in cclS .b/ when S Š S4 . This gives m D n and jNG .S /=S j D 2. Set P D cclS .t /[cclS .t c/[cclS .b/[cclS .bav/. Obviously, hP i D S and P G T , where T D CG .U / and jG W T j D 2. Assume that NG .S / ¤ G and let y 0 be an element in G NG .S / such that y 0 normalizes NG .S / and .y 0 /2 2 NG .S /. Since y 0 does not normalize S , there is x 0 2 P such 0 that x0 D .x 0 /y 2 NG .S / S , x0 2 G T and so x0 does not centralize U . Since .cclS .t //x0 D cclS .t x0 /, where t x0 2 cclS .bav/ in case S Š S3 and t x0 2 cclS .b/ in case S Š S4 , so x0 sends .CS .t //0 D ha2 i onto .CS .t x0 //0 D hc 2 i. Indeed, .CS .bav//0 D hc 2 i in case S Š S3 and .CS .b//0 D hc 2 i in case S Š S4 and in both cases hc 2 i is normal in S . We get ha2 ix0 D hc 2 i and so, in particular, y x0 D vz , D 0; 1 and so .yvz /x0 D yvz or .uz /x0 D uz and so x0 centralizes U D

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Groups of prime power order

hz; u D yvi. This is a contradiction and so we must have NG .S / D G. We have obtained in both cases S Š S3 and S Š S4 that jG W S j 2 and if jG W S j D 2, then m D n 4. We have proved the following result. Theorem 51.14. Let G be a 2-group which has a self-centralizing elementary abelian subgroup E of order 8 with E 6 ˆ.G/. We assume that G has the unique normal foursubgroup U and set T D CG .U /. Suppose that there is an involution t 2 E ˆ.G/ such that CG .t / D ht i D with D Š SD2m , m 4. Finally, assume t 62 U and in that case we have proved that we may assume t 2 G T . Then G has a (large) normal subgroup S containing U CG .t / which is isomorphic to one of the groups Si for i D 1; 2; 3; 4 (see Definition 1) so that jG=S j 4. More precisely, (a) if S Š S1 , then jG=S j 2; (b) if S Š S2 , then jG=S j 4 and if jG=S j D 4, then m D n 4; (c) if S Š S3 , then jG=S j 2 and if jG=S j D 2, then m D n 4; (d) if S Š S4 , then jG=S j 2 and if jG=S j D 2, then m D n 4. We have to analyze three remaining cases S5 , S6 , and S7 for the structure of S . 7o . The case S Š S5 . We have exactly four conjugacy classes of involutions contained in S U which act faithfully on U with the representatives t , t c, b, and ba. The corresponding centralizers in S are CS .t / D CG .t / D ht i D with D Š D2m , m 3; CS .t c/ D ht ci ha; bvi with hv; bvi Š Q2m , m 3; CS .b/ D hbi hyc; t i with hyc; t i Š SD2n , n 4; CS .ba/ D hbai hyc; t i with hyc; t i Š SD2n , n 4. We see that t cannot be fused in NG .S / to any of the involutions t c, b, or ba and this implies S D G. By our assumption, G has the unique normal 4-subgroup U and so m 4. 8o . The case S Š S6 . We have exactly four conjugacy classes of involutions contained in S U which act faithfully on U with the representatives t , t c, b, and ba. The corresponding centralizers in S are CS .t / D CG .t / D ht i D with D Š D2m , m 3; CS .t c/ D ht ci hav; bi with hav; bi Š SD2m , m 4 and hav; bi Š D8 if m D 3 (in which case t c centralizes U ), CS .b/ D hbi L with L Š D2n , n 4; CS .ba/ D hbai hcy; t i with hcy; t i Š SD2n , n 4. Suppose that NG .S / ¤ S . Since NG .S / can fuse t only with an involution in cclS .b/, we have m D n 4 and jNG .S /=S j D 2. Set P D cclS .t / [ cclS .t c/ [ cclS .b/ [ cclS .ba/. Obviously, hP i D S and P G T , where T D CG .U / and jG W T j D 2. Assume that NG .S / ¤ G and let y 0 be an element in G NG .S / such that y 0 normalizes NG .S / and .y 0 /2 2 NG .S /. Since y 0 does not normalize S , there 0 is an x 0 2 P such that x0 D .x 0 /y 2 NG .S / S , x0 2 G T and so x0 does not centralize U . Then x0 sends (by conjugation) t onto an S -conjugate of b and so x0 sends .CS .t //0 D ha2 i onto .CS .b//0 D hc 2 i. Indeed, each S -conjugate b 0 of b has the property CS .b 0 / D hb 0 i L because L is normal in S . In particular, y x0 D vz , D 0; 1 and so yvz D uz is centralized by x0 . Hence x0 centralizes U D hz; ui,

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which is a contradiction. Hence we must have NG .S / D G which implies that in case S Š S6 we get jG=S j 2 and if jG=S j D 2, then m D n 4. It remains to consider the case S Š S7 which is rather difficult since there are several possibilities for the fusion of the involution t with other involutions in S U which act faithfully on U . 9o . The case S Š S7 . We have exactly four conjugacy classes of involutions contained in S U which act faithfully on U with the representatives t , t c, b, and ba. The corresponding centralizers in S are CS .t / D CG .t / D ht i D with D Š D2m , m 3; CS .t c/ D ht ci D with D Š D2m , m 3; CS .b/ D hbi L with L Š D2n , n 3; CS .ba/ D hbai L with L Š D2n , n 3. Suppose that NG .S / ¤ S (otherwise we are finished). Set P D cclS .t /[cclS .t c/[ cclS .b/ [ cclS .ba/. Obviously, hP i D S and P G T , where T D CG .U / and jG W T j D 2. Suppose also that t is not fused in NG .S / to any involution in S L. Then NG .S / fuses t with t c and so jNG .S / W S j D 2 and L is normal in NG .S / since L hci is a normal subset in NG .S /. Let x be any involution in NG .S / S . Since .cclL .t //x D cclL .t c/, we get L0 D Lhxi Š D2nC1 (as in case S Š S1 ). We have CNG .S/ .x/ D hxi CS .x/ and CL .x/ D hzi D Z.L/ D Z.L0 /. If jCS .x/j D 2m , then CS .x/ covers S=L and so L0 is normal in DL0 > L. This contradicts the maximality of S . Hence there is no involution x 2 NG .S /S such that jCS .x/j D 2m . Suppose in addition that NG .S / ¤ G. Let y 0 be an element in G NG .S / such 0 that y 0 normalizes NG .S / and .y 0 /2 2 NG .S /. If x D t y 2 NG .S / S , then 0 jCNG .S/ .x/j D 2mC1 and so jCS .x/j D 2m , a contradiction. If t y 2 cclS .t / [ 0 cclS .t c/, then (as before) CG .t / 6 NG .S /, a contradiction. Hence t y 2 cclS .b/ or 0 0 t y 2 cclS .ba/. Since y 0 62 NG .S /, there is an x 0 2 P with x0 D .x 0 /y 2 NG .S /S . If NG .S / fuses b and ba, then all involutions in P are fused in hy 0 iNG .S / to t and so jCNG .S/ .x0 /j D 2mC1 and jCS .x0 /j D 2m , a contradiction. If NG .S / does not fuse b 0 and ba, then jCNG .S/ .b/j D jCNG .S/ .ba/j D 2nC2 and since jCNG .S/ .t y /j D 2mC1 , so we get n C 2 D m C 1 and therefore jCNG .S/ .x0 /j D 2nC2 D 2mC1 and so again jCS .x0 /j D 2m , a contradiction. Hence we must have NG .S / D G in this case. Suppose that t is not conjugate to t c in NG .S /. Then t must be conjugate in NG .S / to b or ba. This implies jNG .S / W S j D 2 and m D n 3. We make here the following simple observation. All involutions in S .D [ L/ centralize U and so lie in T . All noncentral involutions in D and L lie in G T . Since S is the central product S D D L of D and L, it follows that D and L are the only dihedral normal subgroups of order 2m of S all of whose noncentral involutions belong to G T . Therefore for each element x 2 NG .S / S , we have Lx D D since t x 2 cclS .b/ or t x 2 cclS .ba/. In particular, hcix D hai. Suppose again that NG .S / ¤ G. Let y 0 be an element in G NG .S / such that y 0 normalizes NG .S / and .y 0 /2 2 NG .S /. Since y 0 62 NG .S /, there is an x 0 2 P such 0 that x0 D .x 0 /y 2 NG .S / S and x0 2 G T and so x0 does not centralize U . Note that hP i D S and P G T . On the other hand (by the above), hcix0 D hai

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Groups of prime power order

and so v x0 D yz , D 0; 1. But then .yvz /x0 D yvz , where yv D u. Hence x0 centralizes U D hz; ui, a contradiction. Thus in this case we have also NG .S / D G. It remains to analyze the case that t is fused in NG .S / to all three involutions t c, b, and ba. In that case m D n 3 and jNG .S / W S j D 4. By the above observation, for each x 2 NG .S / S , we have either Lx D L and D x D D or Lx D D. Suppose now that NG .S / ¤ G. Let y 0 be an element in G NG .S / such that y 0 normalizes 0 0 NG .S / and .y 0 /2 2 NG .S /. Then t 0 D t y 2 NG .S / S . Otherwise, t y 2 P and 0 0 1 so there is an n 2 NG .S / with t y D t n and so t y n D t , which is a contradiction 0 0 0 since y 0 n1 62 NG .S /. If Lt D L and D t D D, then .cclL .t //t D cclL .t c/ 0 and .cclD .b//t D cclD .ba/ imply (as before) that Lht 0 i Š Dht 0 i Š D2mC1 . In 0 0 particular, t 0 inverts v and y which gives ut D .yv/t D y 1 v 1 D yv D u and so t 0 0 centralizes U . This is not possible since t 0 (together with t ) lies in G T . If Lt D D, 0 0 then hcit D hai and so in particular v t D yz , D 0; 1. But then t 0 centralizes 0 yv D u and so t centralizes U , which is again a contradiction. Hence we must have also in this case NG .S / D G. We have proved the following final result. Theorem 51.15. Let G be a 2-group which has a self-centralizing elementary abelian subgroup E of order 8 with E 6 ˆ.G/. We assume that G has the unique normal 4-subgroup U and set T D CG .U /. Suppose that there is an involution t 2 E ˆ.G/ such that CG .t / D ht i D with D Š D2m , m 3. Also, in case D Š D8 we assume in addition that G does not possess a normal elementary abelian subgroup of order 8. Finally, assume t 62 U and in that case we have proved that we may assume t 2 G T . Then G has a (large) normal subgroup S containing U CG .t / which is isomorphic to one of the groups Si for i D 5; 6; 7 (see Definition 1) so that jG=S j 4. More precisely, (a) If S Š S5 , then S D G and m 4, n 4; (b) if S Š S6 , then jG=S j 2 and if jG=S j D 2, then m D n 4; (c) if S Š S7 , then jG=S j 4 and if jG=S j D 4, then m D n 3.

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The main results of this section are due to Janko [Jan6]. In Lemma 42.1 finite 2-groups G have been determined with j2 .G/j 8. In this section we do the next step and determine the finite 2-groups G with jGj > 16 and j2 .G/j D 16 (D 2 -groups). All nonmetacyclic 2 -groups will be given in terms of generators and relations. For more general result, see 55. If jGj D 16, then 2 .G/ < G if and only if G 62 fC16 ; C8 C2 ; M16 g. Therefore, we consider in this section only groups of order > 16. If j2 .G/j D 16, then the numbers c1 .G/ and c2 .G/ are small. This observation allows us to generalize the main result of this section (see 55). In subsection 1o we study 2 -groups G which have no normal subgroups isomorphic to E8 . Using the main theorem from 50, we obtain three classes of such 2-groups and one exceptional group of order 25 (Theorem 52.1). In subsection 2o we study nonabelian 2 -groups G with normal subgroup isomorphic to E8 . It is easy to see that then G=E is either cyclic or generalized quaternion. If G=E is cyclic, then we get three classes of groups (Theorem 52.2). If G=E is generalized quaternion and E 6 ˆ.G/, then we get one class of groups (Theorem 52.4). If G=E is generalized quaternion and E ˆ.G/, we get two classes of groups (Theorem 52.5). In subsection 3o we investigate 2-groups G with jGj > 16 which have exactly one subgroup of order 16 and exponent 4. It turns out that a 2-group G has this property if and only if jGj > 16 and j2 .G/j D 16 (Theorem 52.6). It is interesting to note that, for a 2 -group G, there are exactly five possibilities for the structure of 2 .G/: Q8 C4 , Q8 C2 , D8 C2 , C4 C2 C2 , C4 C4 (note that Q8 C4 Š D8 C4 ). In fact, if 2 .G/ Š Q8 C2 or D8 C2 and jGj > 16, then we get exactly one group of order 25 in each case (see Theorem 52.1(d) and Theorem 52.2(a) for n D 2). In other three cases for the structure of 2 .G/ we get infinitely many groups. In subsection 4o we consider a similar problem. In [Ber25, 48] the 2-groups G have been considered which have exactly one abelian subgroup of type .4; 2/. It was shown that either j2 .G/j D 8 (and then G is isomorphic to one of the groups in Lemma 42.1) or G has a self-centralizing elementary abelian subgroup of order 8 (see 51). We improve this result by determining completely the groups of the second possibility (Theorem 52.7). Finally, Theorem 52.8 shows that our result also slightly

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Groups of prime power order

improves the classification of 2-groups with exactly two cyclic subgroups of order 4 given in Theorem 43.4. Lemma A ([Ber25, 48]). Let G be a metacyclic 2-group of order > 24 . If j2 .G/j D 24 , then 2 .G/ Š C4 C4 . Proof. Let jGj D 2m > 24 . Assume that 2 .G/ D H is nonabelian. We want to show that this assumption leads to a contradiction. If H is not minimal nonabelian, then Proposition 10.19 implies that G is of maximal class. But then 2 .G/ D G, a contradiction. Suppose that H has a cyclic subgroup of index 2. Since in that case H is minimal nonabelian, H Š M24 (Theorem 1.2). But then 2 .G/ D 2 .H / is abelian of type .4; 2/ which contradicts the fact that H D 2 .G/. Exercise 1.8A implies H D ha; b j a4 D b 4 D 1; ab D a1 i. It follows that Z.H / D ha2 ; b 2 i D 1 .H / D 1 .G/, H 0 D ha2 i, and Z D hb 2 i is a characteristic subgroup of order 2 in H so Z G G. Indeed, b 2 is a square in H , a2 b 2 is not a square in H (check!), and Z.H / H 0 D fb 2 ; a2 b 2 g. Thus Z is central in G. Since H=Z Š D8 , Proposition 10.19 implies that G=Z is of maximal class. All involutions in G=Z lie in H=Z and so G=Z Š SD16 and jGj D 25 . But there are elements of order 4 in .G=Z/ .H=Z/ whose square lies in Z.G=Z/ D ha2 ; b 2 i=Z. Hence there is an element x 2 G H such that x 2 2 ha2 ; b 2 i and so o.x/ 4, which is the final contradiction. Thus, H Š C4 C4 . 1o . G has no normal elementary abelian subgroups of order 8. In this subsection we prove the following result. Theorem 52.1. Let G be a 2-group of order > 24 satisfying j2 .G/j D 24 . Suppose, in addition, that G has no normal elementary abelian subgroups of order 8. Then we have one of the following four possibilities: (a) G D D C (the central product) with D Š D8 , D \ C D Z.D/, and C is cyclic of order 8. Here we have 2 .G/ Š Q8 C4 Š D8 C4 . (For such groups, see Appendix 16; it is proved there, that also G Š Q8 C .) (b) G is metacyclic of order 25 (in that case, by Lemma A, 2 .G/ Š C4 C4 ). (c) G D QS , where Q Š Q8 , Q is normal in G, S is cyclic of order 16, Q \ S D Z.Q/, and if C is the subgroup of index 2 in S , then CG .Q/ D C . Setting Q D ha; bi and S D hsi, we have as D a1 and b s D ba. Here we have 2 .G/ Š Q8 C4 . (d) The exceptional group G of order 25 has a maximal subgroup M D hui Q, where u is an involution, Q Š Q8 , CG .u/ D M and G is isomorphic to the group A2(a) from Theorem 49.1. We have 2 .G/ D M Š Q8 C2 . Proof. We may assume that G is nonabelian. If G were of maximal class, then 2 .G/ D G, a contradiction since jGj > 24 . Since G is neither abelian nor of maximal class, we may apply main theorem from 50 about 2-groups without nor-

52 2-groups with 2 -subgroup of small order

77

mal elementary abelian subgroup of order 8. Then G has a normal metacyclic subgroup N such that CG .2 .N // N , G=N is isomorphic to a subgroup of D8 , and W D 2 .N / is either abelian of type .4; 4/ or N is abelian of type .2j ; 2/, j 2. However, if W Š C4 C4 , then W D 2 .G/ since j2 .W /j D 24 D j2 .G/j, and Theorem 41.1 and Remark 41.2 imply that G is metacyclic. This gives the case (b) of our theorem, by Lemma A. In what follows we assume that G is not metacyclic. We assume in the sequel that N is abelian of type .2j ; 2/, j 2 and so W D 2 .N / is abelian of type .4; 2/. By hypothesis, j2 .G/ W W j D 2. Suppose that G has at least two distinct normal four-subgroups. Then it follows from Theorem 50.2 that G D D C , where D is dihedral of order 8 and C is either cyclic or of maximal class and D \ C D Z.D/. In view of j2 .G/j D 16, C must be cyclic of order 8 since jGj > 24 , and this gives the case (a) of our theorem. Next we assume that G has the unique normal four-subgroup W0 D 1 .W / D 1 .N /. Also we have CG .W / D N (N is abelian of type .2j ; 2/, j 2). Set WQ D 2 .G/ so that WQ \ N D W , jWQ W W j D 2, and WQ is nonabelian (the last assertion follows from CG .W / D N WQ ). If WQ is metacyclic, then Theorem 41.1 and Remark 41.2 imply that G is metacyclic. But then Lemma A gives a contradiction. Thus WQ is nonabelian non-metacyclic. In particular, exp.WQ / D 4 (see Theorem 1.2). Proposition 1.6 gives jWQ =WQ 0 j D 8 and so jWQ 0 j D 2. By Lemma 1.1, jWQ j D 2jWQ 0 jjZ.WQ /j and so jZ.WQ /j D 4 and Z.WQ / < W . First assume that Z.WQ / D hvi Š C4 is cyclic and set z D v 2 , W0 D hz; ui D 1 .W /. Since f1g < WQ 0 < hvi (by the previous paragraph, jWQ j D 2), we have WQ 0 D hzi. For any x 2 WQ W , we have x 2 2 Z.WQ / D hvi (indeed, WQ =Z.WQ / is elementary abelian). But WQ has no elements of order 8 and so x 2 2 hzi and ux D uz since CWQ .u/ D W . It follows that the nonabelian subgroup W0 hxi Š D8 (it is nonabelian since W0 6 Z.WQ /) and so, since D8 has five involutions, we may assume that x is an involution. Then WQ is the central product of hu; xi Š D8 and hvi D Z.WQ / Š C4 with hvi \ hu; xi D hzi and .ux/2 D z. We set Q D hux; vxi Š Q8 and see that WQ D Q hvi is also the central product of Q8 and C4 (see also Appendix 16, subsection 1o ). All six elements in WQ .Q [ hvi/ are involutions and therefore Q is the unique quaternion subgroup of WQ D 2 .G/ and so Q is characteristic in WQ hence normal in G (see also Appendix 16). There are exactly three four-subgroups in WQ and one of them is W0 D hz; ui and this one is normal in G. Since G has exactly one normal four-subgroup, it follows that the other two four-subgroups in WQ are conjugate to each other in G. Set C D CG .Q/. Obviously C is normal in G and C \ WQ D hvi. But hvi is the unique subgroup of order 4 in C (indeed, all subgroups of G of order 4 are contained in WQ ) so C is cyclic, by Proposition 1.3, and WQ C D Q C with Q \ C D hzi. Set Q D ha; bi, where we choose the generator a so that u D av. Since W0 D 1 .W / D hz; ui and hvi D Z.WQ / are both normal in G, it follows that hai . W / is normal in G (indeed, W has exactly three subgroups of order 4). Every four-subgroup L is normal in WQ . Indeed, L is characteristic in LZ.WQ /, which is normal in WQ ; it follows that all four-subgroups are normal in Q C . The four-

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subgroups hbv; zi and hbav; zi must be conjugate in G since they are not normal in G and are contained in WQ . Q C / (WQ has exactly three four-subgroups and exactly one of them is normal in G). Therefore Q C < G so jG W .Q C /j D 2, by the product formula, and G=.Q C / induces an outer automorphism ˛ on Q normalizing hai and sending hbi onto hbai. A Sylow 2-subgroup of Aut.Q/ is isomorphic to D8 . It follows that G=C Š D8 and so there is an element s 2 G .Q C / such that s 2 2 C and b s D ba. If as D a, then .ba/s D ba2 D b 1 and so ba D .b s /1 D .ba/1 , which is a contradiction. It follows that as D a1 . It remains to determine s 2 . Set hsiC D S and we see that C is a cyclic subgroup of index 2 in S . Since o.s/ 8 (indeed, all elements of G of order 4 are contained in WQ Q C ), we have o.s 2 / 4 and therefore jZ.S /j 4. The same argument shows that all elements in S C have order 8. It follows from Theorem 1.2 that S is cyclic. We may set s 2 D c, where hci D C . We compute .sb/2 D sbsb D s 2 b s b D cbab D cabzb D cab 2 z D cazz D ca: If jC j D 4, then C D hvi and so ca is an involution. But then o.sb/ D 4, by the displayed formula, and this is a contradiction since sb 62 WQ . Thus jC j 8 and so o.s/ 16. We have obtained the groups stated in part (c) of our theorem. Now suppose that Z.WQ / Š E4 , the four-group, so that Z.WQ / D W0 D 1 .W / D 1 .N / D hz; ui, where z D v 2 and W D hui hvi Š C2 C4 . If WQ is minimal nonabelian, then the fact that WQ is nonmetacyclic implies with the help of Exercise 1.8A that WQ D ha; b j Œa; b D c; a4 D b 2 D c 2 D Œa; c D Œb; c D 1i. We see that Z.WQ / D ˆ.WQ / D ha2 ; ci Š E4 and WQ has exactly three maximal subgroups ha2 ; c; bi Š E8 , ha; ci Š C4 C2 , hba; ci Š C4 C2 since .ba/2 D aŒa; ba D aca D a2 c. Hence ha2 ; c; bi is normal in G, contrary to the assumption of this subsection. We have proved that WQ is not minimal nonabelian. Let D be a minimal nonabelian subgroup of WQ (of order 8). If u 2 Z.WQ / D, then WQ D hui D and ˆ.WQ / D WQ 0 D ˆ.D/ D Ã1 .D/ D hzi since z D v 2 . Suppose first that D Š D8 . Let t be an involution in WQ W (t exists since c1 .WQ / D 7 > 5 D c1 .D/ > 3 D c1 .W /). Since Œv; t D z D v 2 (hvi is normal in WQ and it is not contained in Z.WQ / Š E4 ), t inverts hvi and so we may set D D hv; t i, where WQ D hui D. Since t inverts 2 .N / D W (indeed, t centralizes 1 .N / and both cyclic subgroups of W of order 4 are t -invariant but not centralized by t ), it follows that CN .t / D W0 D hz; ui. Assume that N W contains an element x of order 8 (i.e., j > 2; by the above, N is abelian of type .2j ; 2/). Then Theorem 51.2 implies x t D x 1 w0 with w0 2 W0 . But then .tx/2 D .txt /x D x t x D x 1 w0 x D w0 and therefore o.tx/ 4 with tx 62 WQ , a contradiction. Thus, x does not exist so N D W is abelian of type .4; 2/. Since t inverts N D W , all eight elements in WQ N D WQ W are involutions: indeed, WQ D ht i W . In particular, no involution in WQ N is a square in G (otherwise, G has an element of order 4 not contained in WQ ) and so G=N Š E4 (indeed, G=N is isomorphic to a subgroup of

52 2-groups with 2 -subgroup of small order

79

Aut.W / Š D8 and jG W N j > 2) and jGj D 25 . All elements in G WQ must be of order 8 and so for any y 2 G WQ , y 2 2 N W0 D W W0 since G=N Š E4 . Replacing huvi with hvi and v with v 1 , if necessary, we may assume that y 2 D v since c2 .W / D 2. Since CG .N / D N.D hy 2 ; ui/, we get uy D uz, and so y u D yz. Thus hy; ui Š M24 and hy; ui > N . The cyclic group hvi D Z.hy; ui/ of order 4 is normal in G and CG .v/ D hyiN Š M24 . Since ty is also an element of order 8, we have .ty/2 2 N W0 . If .ty/2 2 hvi, then CG .v/ hN; y; tyi D G, a contradiction. Thus .ty/2 2 huvi. If .ty/2 D .uv/1 D .uz/v, then replacing u with uz, we may assume from the start that .ty/2 D uv. We have uv D tyty D ty 2 t y D t vt y . It follows that t y D v 1 t uv D t v uv D t uz, and so y normalizes the elementary abelian subgroup E D hz; u; t i of order 8. Since E is normal in WQ , E is also normal in G D hWQ ; yi, contrary to the assumption of this subsection. Finally, let D Š Q8 so that WQ D hui D, Z.D/ D hzi, z D v 2 , and D \ N D hvi. Take r 2 D hvi. We have r 2 D z with z 2 Z.G/ and so r induces an involutory automorphism on the abelian group N of type .2j ; 2/, j 2. Since r inverts W D hu; vi D 2 .N /, it follows that CN .r/ D W0 D hu; zi. Then Theorem 51.2 implies that r inverts N=W0 . Suppose that W < N (i.e., j > 2) and let x 2 N W . We get x r D x 1 w0 with w0 2 W0 and so .rx/2 D rxrx D r 2 x r x D zx 1 w0 x D zw0 2 W0 . Since rx 62 WQ D 2 .G/ and o.rx/ 4, we get a contradiction. We have proved that N D W D CG .N / is abelian of type .4; 2/. We have jGj 26 . Set S D CG .W0 / so that S WQ and S=N stabilizes the chain N > W0 > f1g since jN=W0 j D 2. Hence S=N is elementary abelian of order 4. Suppose that S > WQ ; then S=N is a four-group. Let SQ =N be a subgroup of order 2 in S=N distinct from WQ =N . Thus SQ \ WQ D N . Take an element y 2 SQ N so that o.y/ D 8 and therefore y 2 2 N W0 . But then y centralizes hy 2 ; W0 i D N , a contradiction. Hence S D WQ D CG .W0 /, jG W WQ j D 2 and jGj D 25 . In particular, CG .u/ D huiD with D Š Q8 . Note that the involution u is contained in the unique normal four-subgroup W0 of G and CG .u/ ¤ G. Then our group G is isomorphic to the group A2(a) of the Theorem 49.1. The proof is complete. 2o . G has a normal E8 . Throughout this subsection we assume that G is a nonabelian 2-group of order > 24 satisfying j2 .G/j D 24 . We assume in addition that G has a normal elementary abelian subgroup E of order 8. In that case, G=E has exactly one subgroup 2 .G/=E of order 2 so G=E is either cyclic of order 4 or generalized quaternion (see Proposition 1.3). In the case where G=E is cyclic, G is determined in Theorem 52.2. In the case where G=E is generalized quaternion and E 6 ˆ.G/, G is determined in Theorem 52.4. Finally, if G=E is generalized quaternion and E ˆ.G/, then G is determined in Theorem 52.5. Theorem 52.2. Let G be a nonabelian 2-group of order > 24 satisfying j2 .G/j D 24 . Suppose, in addition, that G has a normal subgroup E Š E8 such that G=E is cyclic of order 2n , n 2. Then one of the following holds:

80

Groups of prime power order nC1

n

(a) G D ha; e1 j a2 D e12 D 1; n 2; a2 D z; Œa; e1 D e2 ; e22 D Œe1 ; e2 D 1; Œa; e2 D zi is a group of order 2nC3 , Z.G/ D ha4 i, G 0 D hz; e2 i, ˆ.G/ D ha2 ; e2 i, E D hz; e1 ; e2 i is a normal elementary abelian subgroup of order 8, n1 and 2 .G/ D Eha2 i is abelian of type .4; 2; 2/ for n 3, and 2 .G/ Š C2 D8 for n D 2. (b) G Š M2nC2 C2 with n 2, and 2 .G/ is abelian of type .4; 2; 2/. nC1

(c) G D ha; e1 j a2 D e12 D 1; n 2; Œa; e1 D e2 ; e22 D Œe1 ; e2 D Œa; e2 D 1i being minimal nonabelian, jGj D 2nC3 , ˆ.G/ D ha2 ; e2 i, G 0 D he2 i, and n1 n 2 .G/ D Eha2 i is abelian of type .4; 2; 2/, where E D ha2 ; e1 ; e2 i is a normal elementary abelian subgroup of order 8. Proof. By assumption, G D Ehai for some a 2 G E; let jG=Ej D 2n > 2. Obviously G does not split over E (otherwise, if G D T E, then 2 .G/ D 2 .T / E is of order 25 > 24 ) and so o.a/ D 2nC1 and hai \ E D hzi is of order 2, where n z D a2 . Note that a Sylow 2-subgroup of Aut.E/ is isomorphic to D8 , so a induces an automorphism of order 4 on E hence a4 2 Z.G/. It is easy to check, that if CE .a/ is a four-subgroup, then a induces an involutory automorphism on E. Indeed, then a stabilizes the chain f1g < CE .a/ < E with factors of exponent 2. Suppose that a induces an automorphism of order 4 on E. In that case, by what has been said in the previous paragraph, CE .a/ D hzi so that Z.G/ D ha4 i hzi and CG .E/ D Z.G/E. There are involutions e1 ; e2 2 E such that hz; e1 ; e2 i D E and e1a D e1 e2 , e2a D e2 z. Then CE .a2 / D hz; e2 i D G 0 , G D ha; e1 i, ˆ.G/ D ha2 ; e2 i, n1 2 .G/ D Eha2 i. We have determined the groups stated in part (a). Suppose that a induces an involutory automorphism on E. We have CE .a/ D E1 D hz; e2 i Š E4 (see Exercise 19, below), Z.G/ D ha2 ; e2 i and G 0 D Œa; E is a subgroup of order 2 in E1 (by Lemma 1.1, since G has an abelian subgroup ha2 ; Ei of index 2 and Z.G/ is of index 4 in G). If G 0 D hzi, then hai is normal in G. Let e1 2 E E1 . Then ha; e1 i Š M2nC2 and G D he2 i ha; e1 i which is the group stated in part (b) of our theorem. If G 0 D he2 i, where e2 2 E1 hzi and e1 2 E E1 , then Œa; e1 D e2 and this group is stated in part (c) of our theorem. In the rest of this subsection we assume that G=E Š Q2n ; n 3. Then CG .E/ > E, since a Sylow 2-subgroup of Aut.E/ is isomorphic to D8 . Therefore WQ D 2 .G/ centralizes E, where WQ =E D Z.G=E/ (it is important that j2 .G/j D 24 ), WQ is abelian of type .4; 2; 2/ and hzi D Ã1 .WQ / is of order 2. All elements in WQ E are of order 4 and so no involution in E hzi is a square in G. Indeed, if x 2 2 E hzi, then o.x/ D 4 and x 62 WQ since z is the unique nontrivial square on WQ , and this is a contradiction. We first prove the following useful result. Lemma 52.3. Let, as above, G=E Š Q2n , where G is an 2 -group. We have CG .E/ WQ D 2 .G/, CG .E/=E is cyclic of order 2, and G=CG .E/ is isomorphic to C2 ; E4 or D8 .

52 2-groups with 2 -subgroup of small order

81

Proof. Suppose that CG .E/ contains a subgroup H > E such that H=E Š Q8 . Then Z.H=E/ D WQ =E. Let A=E and B=E be two distinct cyclic subgroups of order 4 in H=E. Then A and B are abelian maximal subgroups of H . It follows that A \ B D WQ Z.H / and therefore WQ D Z.H / is of order 24 . Using Lemma 1.1, we get 26 D jH j D 2jZ.H /jjH 0 j, which gives jH 0 j D 2. Since 1 .H / D E, we get H 0 E and H=E is abelian, a contradiction. It follows that CG .E/=E must be a normal cyclic subgroup of G=E and so G=CG .E/ is isomorphic to a nonidentity epimorphic image of Q2n of order 8 such that G=CG .E/ 6Š Q8 (the last condition is superfluous if n > 3). We conclude that G=CG .E/ 2 fC2 ; E4 ; D8 g. Theorem 52.4. Let G be a 2-group of order > 24 satisfying j2 .G/j D 24 . Suppose, in addition, that G has a normal elementary abelian subgroup E of order 8 such that G=E Š Q2n , n 3, and E 6 ˆ.G/. Then we have n

n1

G D ha; b; e j a2 D b 8 D e 2 D 1; n 3; a2

D b 4 D z; ab D a1 ;

Œe; b D 1; Œe; a D z ; D 0; 1i: Here G is a group of order 2nC3 , G 0 D ha2 i, ˆ.G/ D ha2 ; b 2 i is abelian of type n2 .2n1 ; 2/. The subgroup E D he; a2 b 2 ; zi is a normal elementary abelian subgroup of order 8 in G and G=E Š Q2n . If D 0, then Z.G/ D he; b 2 i is abelian of type .4; 2/. If D 1, then Z.G/ D hb 2 i Š C4 . Finally, 2 .G/ D Ehb 2 i is abelian of type .4; 2; 2/. Proof. Let WQ =E D Z.G=E/; then WQ D 2 .G/ since j2 .G/j D 16 and the subgroup WQ has order 16 and exponent 4. It follows from Aut.E/ D GL.3; 2/ that a Sylow 2-subgroup of Aut.E/ is dihedral of order 8 so CG .E/ > E. We conclude that WQ is abelian of type .4; 2; 2/. Since E 6 ˆ.G/, there is a maximal subgroup M of G such that E 6 M ; then G D EM . We have E0 D E \M Š E4 (obviously, E0 is normal in G since E and M are), WQ0 D WQ \ M is abelian of type .4; 2/, WQ0 D 2 .M /, and M=E0 Š Q2n , n 3. If ˆ.WQ 0 / D hzi, then z 2 E0 . By Lemma 42.1, M is isomorphic to a group (c) in that proposition. Hence M is metacyclic with a cyclic normal subgroup hai of order 2n and the cyclic quotient group M=hai of order 4 so that n1 there is an element b of order 8 in M .E0 hai/ with b 2 2 WQ0 E0 , b 4 D z D a2 , n2 hbi \ hai D hzi, WQ0 D hb 2 iha2 i ˆ.M /. We have used the following facts: 1 .WQ / D E, Ã1 .WQ / D hzi, hai \ hbi is of order 2 (by the product formula), ˆ.G/ D Ã1 .G/. We have (see Lemma 42.1(c)) ˆ.M / D ha2 ; b 2 i WQ0 , M 0 D ha2 i, ab D a1 . Also, Z.M / D hb 2 i is cyclic of order 4 and so E0 hai D hb 2 ; ai is abelian of index 2 in M , and CM .E0 / D E0 hai (again by Lemma 42.1(c)). The element b induces an involutory automorphism on E (because b 2 2 Z.G/). The element b does not centralize E0 since b 62 ha; E0 i D CM .E0 /. For every u 2 E0 f1; zg, we have hb; ui Š M16 (we have jhb; E0 ij D 16 so E0 normalizes hbi). We conclude that Z.hb; E0 i/ D hzi. The subgroup WQ D hb 2 ; Ei is abelian so b induces an involutory automorphism on E. Therefore, by Exercise 19, Z D CE .b/ is of order 4. By the

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Groups of prime power order

above, Z ¤ E0 so there is an element e 2 E E0 such that b centralizes e. Set n2 u D a2 b 2 2 E0 hzi and we have ub D uz. We act with the cyclic group hai on E. By the above, ha; E0 i is abelian. It follows from jE W E0 j D 2 that hai stabilizes the chain E > E0 > f1g and so a2 centralizes E and e a D ey with y 2 E0 . This gives ae D ay. We may set ab D a1 . Then we act on b 1 ab D a1 with e. It follows b 1 ayb D .ay/1 and so a1 y b D a1 y which gives y b D y and y 2 hzi. If y D 1, then G D hei M . We set ae D az , where D 0; 1. The group G is completely determined. Theorem 52.5. Let G be a 2-group of order > 24 satisfying j2 .G/j D 24 . Suppose, in addition, that G has a normal elementary abelian subgroup E of order 8 such that G=E Š Q2n , n 3, and E ˆ.G/. Then one of the following holds: (a)

n

n1

G D ha; b j a2 D b 8 D 1; a2 b

1

2

2

a Da

n2

D b 4 D z; b 2 D a2

e;

ue ; D 0; 1;

u D e D Œe; u D Œe; z D Œu; z D 1; e a D ez; ua D u; e b D ez; ub D uz; n 3; and if D 1; then n 4i: Here G is a group of order 2nC3 , G 0 D hue i ha2 i is abelian of type .2n1 ; 2/, ˆ.G/ D Eha2 i, where E D hz; e; ui is a normal elementary abelian subgroup of order 8 in G with G=E Š Q2n . Finally, Z.G/ D hzi is of order 2, and 2 .G/ D n2 Eha2 i is abelian of type .4; 2; 2/. n

n1

(b) G D ha; b j a2 D b 8 D 1; n 3; a2 n2

b 2 D a2

D b 4 D z;

u; ab D a1 e;

u2 D e 2 D Œe; u D Œe; z D Œu; z D Œa; e D Œa; u D Œb; e D 1; ub D uzi: Here G is a group of order 2nC3 , G 0 D ha2 ei is cyclic of order 2n1 , ˆ.G/ D Eha2 i, where E D hz; e; ui is a normal elementary abelian subgroup of order 8 in G with G=E Š Q2n and CG .E/ D Ehai. Finally, Z.G/ D he; b 2 i, and 2 .G/ D Ehb 2 i is abelian of type .4; 2; 2/. Proof. We have in this case ˆ.G/ WQ D 2 .G/, WQ =E D Z.G=E/ and WQ is abelian of type .4; 2; 2/ since CG .E/ > E (a Sylow 2-subgroup of Aut.E/ is isomorphic to D8 ) and jWQ j D 24 . Hence hzi D Ã1 .WQ / is a central subgroup of order 2 contained in E. By Lemma 52.3, since every normal cyclic subgroup of G=E is contained in a cyclic subgroup of index 2 in G=E, there is a subgroup M of index 2 in G such that M=E is cyclic and CG .E/ M . (If n > 3, then M is unique since G=E Š Q2n has only one cyclic subgroup of index 2. However, if n D 3

52 2-groups with 2 -subgroup of small order

83

and CG .E/ D WQ , then we have three possibilities for M since G=E Š Q8 has exactly three cyclic subgroups of index 2.) Let a be an element in M E such that ha; Ei D M . Then o.a/ D 2n since jM=Ej D 2n1 and M does not split over E n2 (indeed, all involutions of G lie in E). Hence a0 D a2 2 WQ E, a02 D z 2 E. By the structure of G=E Š Q2n , all elements in .G=E/ .M=E/ have order 4, so we have for each x 2 G M , x 2 2 WQ E, x 4 D z, x induces an involutory automorphism on E since CG .E/ M and x 62 M , and so CE .x/ Š E4 (see Exercise 19). CE .x/ > hzi, and CWQ .x/ D hx 2 iCE .x/ is abelian of type .4; 2/. It follows that all elements in G M have the same order 8. We have M 0 < E (recall that M=E is cyclic) so jM 0 j 4, and M 0 D Œhai; E. For each ai e 2 M , where e 2 E and i is an integer, we get .ai e/2 D ai eai e D a2i .ai eai /e D a2i Œai ; e and so ˆ.M / D Ã1 .M / D ha2 iM 0 . We have exactly the following four cases for the action of the cyclic subgroup hai of order 2n on E: (1) a induces an automorphism of order 4 on E. (2) a induces an automorphism of order 2 on E and ŒE; a D hui, where u 2 E hzi. (3) a induces an automorphism of order 2 on E and ŒE; a D hzi. (4) a centralizes E (in that case, obviously, M is abelian). Case (1). Since a induces an automorphism of order 4 on E, we get n 4 (recall n2 that a2 centralizes E) and CE .a/ D hzi (see, in the first paragraph, the proof of the equality CG .x/ Š E4 ). In this case M 0 is a four-subgroup contained in E, M 0 > hzi (if hzi 6 M 0 , then hai \ M 0 D f1g and a centralizes some element in .M 0 /# , which is not the case) and for each eQ 2 E M 0 , eQ a D eQ uQ with some uQ 2 M 0 hzi and uQ a D uz. Q We see that ha; M 0 i Š M2nC1 so CE .a2 / D M 0 (this is true again since a2 induces an involutory automorphism on E). Since jG W CG .M 0 /j D 2, it follows that CG .M 0 / D Eha2 ; b 0 i, where b 0 2 G M and Eha2 ; b 0 i is a maximal subgroup of G. Set b D b 0 a so that uQ b D uz Q for each uQ 2 M 0 hzi and therefore CE .b/ D he; zi 0 for some e 2 E M (see the proof of Theorem 52.4), e a D eu with u 2 M 0 hzi, ua D uz, and ub D uz. Since G=E Š Q2n , b 1 ab D a1 y with y 2 E. We have two subcases according to y 2 E M 0 or y 2 M 0 . Subcase (1a). Here y D e 0 2 E M 0 and so b 1 ab D a1 e 0 . We compute ab D .ab /b D .a1 e 0 /b D .a1 e 0 /1 .e 0 /b D e 0 a.e 0 /b D a.e 0 /a .e 0 /b D au: 2

Indeed, if e 0 D ez . D 0; 1/, then .e 0 /a D e 0 u, .e 0 /b D e 0 , and if e 0 D euz . D 0; 1/, then .e 0 /a D e a ua z D euuzz D e 0 uz, .e 0 /b D e 0 z. On the other hand, 2 b 2 D a0 x with x 2 E. But M 0 hai D hu; ai is isomorphic to M2nC1 and so ab D au D aa0 x D ax implies that x 2 E M 0 . We may set x D eu0 with u0 2 M 0 and so b 2 D a0 eu0 . We compute b 1 a2 b D .a1 e 0 /2 D a1 e 0 a1 e 0 D .a1 e 0 a/a2 e 0 D .e 0 /a .e 0 /a a2 D u0 za2 ; 2

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Groups of prime power order

since .e 0 /a D e 0 u0 .u0 2 M 0 hzi/ and .e 0 /a D e 0 z. Therefore .a4 /b D ..a2 /b /2 D n2 .u0 za2 /2 D a4 because CE .a2 / D M 0 . Since a0 D a2 2 ha4 i (recall that b 1 2 n 4), we get a0 D a0 . Obviously, b centralizes b D a0 eu0 .u0 2 M 0 / and so a0 eu0 D .a0 eu0 /b D a01 e.u0 /b D a0 ze.u0 /b which gives .u0 /b D u0 z and u0 2 M 0 hzi. Therefore u0 D uz . D 0; 1/ and b 2 D a0 euz . On the other hand, 2 au D ab D aa0 euz D aeu D a.a1 eua/eu D a.eu/.uz/eu D auz, which is a contradiction. Subcase (1b). Here b 1 ab D a1 y with y 2 M 0 . This gives .a2 /b D .a1 y/2 D 2 a1 ya1 y D .a1 ya/a2 y D y a y a a2 D yz ya2 D z a2 . D 0; 1/, and so .a4 /b D .z a2 /2 D a4 . Since n 4, we get a0 2 ha4 i so a0b D a01 . If b 2 2 M 0 ha0 i, then S D M 0 ha; bi is a maximal subgroup of G not containing E since S \ E D M 0 , a contradiction (recall, that, by hypothesis, E ˆ.G/). Thus b 2 D a0 e0 with e0 2 E M 0 . Since b centralizes b 2 , we get a0 e0 D .a0 e0 /b D a01 e0b D a0 ze0b and so e0b D e0 z. Hence e0 D euz . D 0; 1/ and b 2 D a0 euz . From b 1 ab D a1 y with y 2 M 0 , we get b 2 ab 2 D .a1 y/b D .ab /1 y b D yay b D a.a1 ya/y b D a.yz /.yz / D a, where D 0 if y 2 hzi and D 1 if y 2 M 0 hzi. On the other hand, we compute 2

a D ab D aa0 euz D aeu D euaeu D a.a1 eua/eu D a.eu/.uz/eu D auz; 2

which is a contradiction. Case (2). Here, by assumption, a induces an automorphism of order 2 on E and M 0 D ŒE; a D hui, where u 2 E hzi. It follows hu; zi Z.G/. For each eQ 2 E hu; zi, eQ a D eu. Q Since ˆ.M / D hui ha2 i and ˆ.G/ E, there exists n2 b 2 G M such that b 2 D a0 e, where e 2 E hu; zi and a0 D a2 . We have ab D a1 y with y 2 E (see the last two sentences of the first paragraph of Case 2 (1)). This gives ab D .a1 y/b D .ab /1 y b D yay b D ay a y b D aa0 e D ae D eae D a.a1 ea/e D aeue D au. Hence y a y b D u. If y 2 hu; zi Z.G/, then u D y a y b D y 2 D 1, a contradiction. Therefore y D es with s 2 hu; zi. But then u D y a y b D .es/a .es/b D .eus/.e b s/ D uee b and so e b D e. Hence b centralizes E D he; u; zi, a contradiction. Case (3). Here a induces an automorphism of order 2 on E and M 0 D ŒE; a D hzi. It follows ˆ.M / D ha2 i ha0 i and so ha0 i is normal in G since the cyclic subgroup ˆ.M / is G-invariant. Since Ã1 .G/ D ˆ.G/ WQ D Eha0 i, it follows ha0 ; x 2 j x 2 G M i D WQ . Since hx 2 i is normal in G for each x 2 G M and x 4 D z (indeed, M=M 0 D M=hzi is an abelian maximal subgroup of G=M 0 ), we get WQ =hzi Z.G=hzi/. Set U D CE .a/ Š E4 and obviously Z.M / D U ha2 i. Also, U D Z.M / \ E is normal in G. Suppose that U Z.G/. Then the fourgroup G=Eha2 i acts as the full stability group of the chain E > U > f1g since G=Eha2 i acts faithfully on E. (Note that CG .E/ M .) This is a contradiction since E=hzi < WQ =hzi Z.G=hzi/. Hence for each x 2 G M and each uQ 2 U hzi, we have uQ x D uz. Q There exists b 2 G M such that b 2 D a0 e with e 2 E U .

52 2-groups with 2 -subgroup of small order

85

Also, ab D a1 y with y 2 E. Then we compute ab D .a1 y/b D yay b D a.a1 ya/y b D ay a y b D aa0 e D eae D a.a1 ea/e D aeze D az. This gives 2

(1) y a y b D z. On the other hand, b centralizes b 2 and so a0 e D .a0 e/b D a0b e b which gives (2) a0 e D a0b e b . We compute further b 1 a2 b D .a1 y/.a1 y/ D .a1 ya/.a2 ya2 /a2 D y a ya2 , and so we get (3) .a2 /b D y a ya2 . There are two subcases according to y 2 U or y 2 E U . Subcase (3a). Suppose that y 2 U D CE .a/. Then (1) implies y b D yz and so y D u 2 U hzi and b 1 ab D a1 u. From (3) we get .a2 /b D y a ya2 D yya2 D a2 and so a0b D a01 since a0 is a power of a2 . From (2) follows e b D a0b a0 e D a02 e D ez and we have determined the group stated in part (a) of our theorem for D 0. Subcase (3b). Suppose that y 2 E U , where ab D a1 y. From (1) we get yzy b D z and so y b D y. The relation (3) implies .a2 /b D a2 z. We compute, for each x 2 E and each integer i , .b.a2 /i x/2 D b 2 z and .ba.a2 /i x/2 D b 2 yz , where ; D 0; 1. Hence the square of each element in G M lies in hb 2 i D ha0 ei and in ha0 eyi D hb 2 yi. Since ˆ.G/ E, we must have y D eu with some u 2 U hzi. From y b D y follows eu D e b uz (since ub D uz) and so e b D ez. Then (2) implies a0 e D a0b ez and so a0b D a0 z D a01 . From .a2 /b D a2 z follows that ha2 i > ha0 i which gives n 4. We get the group stated in part (a) with D 1 and n 4. Case (4). Here a centralizes E and so M is abelian of type .2n ; 2; 2/, n 3; and n2 ha0 i is normal in G, where a0 D a2 . Since ˆ.G/ WQ D Eha0 i, there exists b 2 G M such that b 2 D a0 u, where u is an involution in E hzi (recall that all elements in G M have the same order 8). Indeed, if such b does not exist, we get b 2 D a0 or a03 for all b 2 G M . Then ˆ.G/ D Ã1 .G/ E, a contradiction. Since CG .b 2 / hM; bi D G, we get b 2 2 Z.G/, ha0 ; b 2 i D ha0 i hui is abelian of type .4; 2/ and ˆ.G/ ha2 ; ui. Again ˆ.G/ WQ and ha2 ; b 2 i WQ implies that there exists c 2 G M such that c 2 2 WQ ha0 ; b 2 i and (as before) c 2 2 Z.G/. If a0 2 Z.G/, then ha0 ; b 2 ; c 2 i D WQ lies in Z.G/ contrary to Lemma 52.3. Hence a0 62 Z.G/ and therefore a0b D a01 and ub D uz (recall that u D a01 b 2 62 Z.G/). The subgroup Z.G/ D hb 2 ; c 2 i is abelian of type .4; 2/. Lemma 1.1 implies jGj D 2nC3 D 2jZ.G/jjG 0 j and so jG 0 j D 2n1 . Since j.G=E/0 j D 2n2 and ˆ.G/ > G 0 > hzi, G 0 is cyclic of order 2n1 , by the modular law. Set ab D a1 e with e 2 E. If e 2 ha0 ; ui D ha0 ; b 2 i, then ha; bi, a metacyclic maximal subgroup of G, does not contain E, a contradiction. Thus e 2 E hz; ui and we compute .ba/2 D baba D b 2 .b 1 ab/a D b 2 .a1 e/a D b 2 e, Since both b 2 and b 2 e lie in Z.G/, it follows e 2 Z.G/ and therefore Z.G/ D he; b 2 i. Finally, G 0 D ha2 ei is cyclic of order 2n1 . We get the group from part (b).

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Groups of prime power order

3o . 2-groups with exactly one subgroup of order 24 and exponent 4. Theorem 52.6. The following two statements for a 2-group G of order > 24 are equivalent: (a) j2 .G/j D 24 and (b) G has exactly one subgroup of order 24 and exponent 4. Proof. Suppose that (a) holds. The results of Subsections 1 and 2 imply that 2 .G/ is isomorphic to one of the following groups Q8 C4 , Q8 C2 , D8 C2 , C4 C2 C2 , C4 C4 . In particular, exp.2 .G// D 4 and so .b/ holds. Suppose now that (b) holds. Let H be the unique subgroup of order 24 and exponent 4 in G, where jGj > 24 . We want to show that H D 2 .G/. Suppose that this is false. Then there exist elements of order 4 in G H , where H is a characteristic subgroup of G. In particular, there is an element a 2 G H such that o.a/ 4 and a2 2 H . Set HQ D H hai so that jHQ j D 25 . Since H is neither cyclic nor a 2-group of maximal class, there exists a G-invariant four-subgroup W0 contained in H , by Lemma 1.4. Let x be any element in HQ H with o.x/ 4. Then hx; W0 i is a subgroup of order 24 in HQ and so there exists a maximal subgroup M of HQ containing hx; W0 i. Since exp.M \ H / 4 and M D hM \ H; xi, we have 2 .M / D M . But jM j D 24 and M ¤ H , so M is either elementary abelian or exp.M / D 8. Suppose that exp.M / D 8. Then M is a nonabelian group of order 24 with a cyclic subgroup of index 2. Since M has the normal four-subgroup W0 , it follows that M is not of maximal class. Thus M Š M24 . But 2 .M24 / is abelian of type .4; 2/ which contradicts the above fact that 2 .M / D M . We have proved that M Š E24 . In particular, x must be an involution and exp.HQ / D 4 because for each y 2 HQ , y 2 2 M . Hence all elements in HQ H are of order 4 and so (by the above) all these elements must be involutions. In that case, every involution x 2 HQ H inverts H . Let k be any element of order 4 in H . Since k x D k 1 , hk; xi D D Š D8 . Let MQ be a maximal subgroup of HQ containing D. Then jMQ j D 24 , exp.MQ / D 4, MQ ¤ H , and this is a final contradiction. 4o . Generalization of Theorem 43.4. Here we improve Theorem 43.4 and some results of [Ber25, 48]. In Theorem 52.7 we classify the 2-groups with exactly on abelian subgroups of type .4; 2/; the proof is independent of the proof of Theorem 43.4. The following remark shows that there are few 2-groups without abelian subgroups of type .4; 2/. Remarks. 1. Let us classify the 2-groups without abelian subgroup of type .4; 2/. Of course, if G is cyclic, elementary abelian or of maximal class, it has no abelian subgroups of type .4; 2/. So in what follows suppose that G is of exponent > 2 and is neither cyclic nor of maximal class. Let A be a cyclic subgroup of order 4 in G. Then CG .A/ A has no involutions. It follows that CG .A/ is cyclic (Proposition 1.3). By Suzuki’s theorem (Proposition 1.8), jCG .A/j > 4. By Lemma 1.4, G has a normal abelian subgroup R of type .2; 2/. Since Z.G/ < CG .A/, we get R \ CG .A/ is of

52 2-groups with 2 -subgroup of small order

87

order 2. It follows that H D RCG .A/ Š M2n for some n > 3. Let B be a subgroup of order 4 in Z.H /. In that case, RB is abelian of type .4; 2/, which is not the case. Thus, our G is elementary abelian, cyclic or of maximal class. 2. Let G be a p-group without abelian subgroup of type .p 2 ; p/, p > 2, exp.G/ > p. Then, as in Remark 1, we obtain exp.G/ D p 2 and CG .x/ D hxi so G is of maximal class (Proposition 1.8). In particular, jGj p 2p1 (Theorem 9.6). We claim that jGj p pC1 . Assume that this is false. Let G1 be a fundamental subgroup of G. Then exp.G1 / > p and Z.G1 / is noncyclic. In that case, obviously, G1 contains an abelian subgroup of type .p 2 ; p/, a contradiction. Thus, jGj p pC1 . It is easy to check that a Sylow p-subgroup of the symmetric group of degree p 2 has no abelian subgroups of type .p 2 ; p/. Theorem 52.7. Let G be a 2-group containing exactly one abelian subgroup of type .4; 2/. Then one of the following holds: (a) j2 .G/j D 8 and G is isomorphic to one of the metacyclic groups (a), (b) or (c) in Lemma 42.1. (b) G Š C2 D2nC1 , n 2. D t 2 D 1; b t D b 1C2 u; u2 D Œu; t D 1; b u D (c) G D hb; t j b 2 n n 1C2 b ; n 2i. Here jGj D 2nC3 , Z.G/ D hb 2 i is of order 2, ˆ.G/ D n hb 2 ; ui < hb; ui Š M2nC2 , E D hb 2 ; u; t i Š E8 is self-centralizing in G, n 2 2 .G/ D hui hb ; t i Š C2 D2nC1 , G 0 D hb 2 ; ui Š E4 in case n D 2, and G 0 D hb 2 ui Š C2n for n 3. Finally, the group G for n D 2 (of order 25 ) is isomorphic to the group (a) in Theorem 52.2 for n D 2 (since 2 .G/ Š C2 D8 ). nC1

n1

Proof. Let G be a 2-group with exactly one abelian subgroup A of type .4; 2/; then A is characteristic in G. Set C D CG .A/. Then C is normal in G and G=C is isomorphic to a subgroup of Aut.A/ Š D8 . We claim that 2 .C / D A. Indeed, let y 2 C A such that o.y/ 4 and y 2 2 A. Then Ahyi is an abelian subgroup of order 24 and exponent 4. But then Ahyi contains an abelian subgroup of type .4; 2/ distinct from A, a contradiction. Thus 2 .C / D A, as claimed. Lemma 42.1 implies that C must be abelian of type .2n ; 2/, n 2. If 2 .G/ D 2 .C / D A, then G is metacyclic and G is isomorphic to one of the groups in Lemma 42.1. We assume from now on that 2 .G/ > A. Set U D 1 .A/ and hzi D ˆ.A/ so that z 2 Z.G/ and U is a normal four-subgroup of G. Let a 2 G C be such that o.a/ 4 and a2 2 C . Obviously, a2 2 U since U D 1 .C /. Let D D haiC ; then jD W C j D 2. Since D is a nonabelian group (of order 24 ) containing a normal four-subgroup U , it follows that D is not of maximal class. If o.a/ D 4, then a does not centralize U (otherwise U hai would be abelian of type .4; 2/ distinct from A) and so U hai Š D8 . But then there exists an involution in .U hai/U . In any case the coset C a . Ua/ contains involutions and let t be one of them. If t centralizes an element s of order 4 in C , then hs; t i is an abelian subgroup of type .4; 2/ distinct from A, a

88

Groups of prime power order

contradiction. Hence CC .t / is of exponent 2 and (since D is not of maximal class) CC .t / D U (Proposition 1.8) and E D CD .t / D U ht i Š E8 . It is easy to check that, for each x 2 D C D C t , CC .x/ D U . Hence x 2 2 U and hU; xi is elementary abelian (of order 8). Thus, all elements in D C are involutions and therefore t inverts C . Let B be a cyclic subgroup of index 2 in C and an involution u 2 C B. Then hB; t i Š D2nC1 and D Š C2 D2nC1 , n 2, where C2 D hui. Obviously, we may choose a 2 G C so that D D ha; C i G G. Suppose that G=C has a cyclic subgroup K=C of order 4. Then K > D since D=C , in the case under consideration, is contained in ˆ.G=C /, and let k 2 K be such that K D hk; C i. We have k 2 2 D C and so k 2 is an involution. It follows o.k/ D 4 and so hk; zi is an abelian subgroup of type .4; 2/ distinct from A since hki \ C D f1g, and this is a contradiction. We have proved that G=C is elementary abelian. If jG=C j D 2, then G D D and we are done. It remains to study the possibility G=C Š E4 . By the above, if y 2 G D and o.y/ 4, then y 2 2 C and y is an involution inverting C (indeed, as in the second paragraph of the proof, all elements in hy; C i C are involutions). But then yt 62 C and yt centralizes C , a contradiction (C D CG .A/ is self centralizing in G). Hence, for each element y 2 G D, o.y/ 8 and y 2 2 C . This gives 2 .G/ D D, CG .t / D E D ht i1 .A/ and so E is a self-centralizing elementary abelian subgroup of order 8 in G. For each y 2 G D, we have y 2 2 C and o.y 2 / 4 and so y centralizes a cyclic subgroup of order 4 in A. Since y does not centralize A, it follows that y does not centralize U D 1 .A/ since y 62 C D CG .A/. Hence D D CG .U / and so Z.G/ D hzi D ˆ.A/ is of order 2. Since 2 .hyiC / D A and hyiC is nonabelian, Lemma 42.1 implies that hyiC is isomorphic to a group (a) or (c) of that lemma. Assume that there is b 2 G D such that hbiC is isomorphic to a group of Lemma 42.1(c). In particular, jhbiC j 25 , C is the unique abelian maximal subgroup of hbiC (since Z.hbiC / Š C4 ), o.b/ D 8, Z.hbiC / D hb 2 i < A, hb 4 i D hzi D ˆ.A/, C contains a cyclic subgroup hai of index 2 such that hai is normal in hbiC , o.a/ 23 , A \ hai Š C4 , hbi \ hai D hzi, and ab D a1 . Also, we see that for each y 2 .hbiC / C , CC .y/ D hb 2 i Š C4 , and y is of order 8. Since t inverts C , we get abt D a and so jCC .bt /j 23 . This implies that hbt iC Š M2nC2 because CC .bt / Z.hbt iC /. Replacing b with bt (if necessary), we may assume from the start that there is an element b 2 G D such that hbiC Š M2nC2 .n 2/, which is a group in part (a) of Lemma 42.1. We have o.b/ D 2nC1 , hbi is a cyclic subgroup of index 2 in hbiC , n n z D b 2 , U D hz; ui D 1 .hbiC /, and ub D uz. This gives also b u D bz D b 1C2 , n1 D v; ui, C D hb 2 i hui. We see that Z.hbiC / D hb 2 i Š C2n , A D hb 2 CC .bt / D hvui Š C4 and so Z.hbt iC / D hvui. It follows that hbt iC Š M24 for n D 2 and hbt iC is isomorphic to a group (c) of Lemma 42.1 for n > 2. In any case, .bt /2 2 hvui hzi and so .bt /2 D vu or .bt /2 D v 1 u D vuz. Replacing u with uz n1 (if necessary), we may assume that .bt /2 D vu, where v D b 2 . From this crucial

52 2-groups with 2 -subgroup of small order

relation follows b t D b 1C2 theorem is proved.

n1

89

u. The structure of G D hb; t i is determined and our

Theorem 52.8. The following two statements for a 2-group G are equivalent: (a) c2 .G/ D 4; (b) G has exactly one abelian subgroup of type .4; 2/. Proof. Suppose that G is a 2-group having exactly two cyclic subgroups U; V of order 4. Then jG W NG .U /j 2 and so V NG .U /. Similarly, U NG .V /. We get ŒU; V U \ V . Since A D hU; V i has exactly two cyclic subgroups of order 4, A must be abelian of type .4; 2/. But A is generated by its two cyclic subgroups of order 4 and so (b) holds. Let (b) hold. Then the group G is completely determined by Theorem 52.7. Looking at 2 .G/, we see that G has exactly two cyclic subgroups of order 4. Exercise 1. Let G be a 2-group, exp.G/ 2n with n > 1. Let H < G contains all cyclic subgroups of G of order 2n . If x 2 H with o.x/ D 2n , then n1 .CG .x// H. Exercise 2. Let G be a 2-group with 2 .G/j 25 . Prove that dl.G/, the derived length of G, is at most 5. Solution. If G has no normal elementary abelian subgroups of order 8, then the dl.G/ 4, by 50. Now let E be a normal elementary abelian subgroup of G of order 8. Then j1 .G=E/j 4 so dl.G=E/ 4, and we are done. Exercise 3. Let G be a 2-group and n > 3. Denote by Mn the set of noncyclic subgroups of order 2n in G which are abelian of type .2n1 ; 2/ or M2n . Study the structure of G in the case jMn j D 1. Exercise 4. Study the 2-groups without abelian subgroups (i) of type .4; 4/, (ii) of type .4; 2; 2/. Exercise 5. Suppose that a noncyclic p-group G has no abelian subgroups of type .p 2 ; p/. If exp.G/ > p, then either G is a 2-group of maximal class or p > 2 and G is of maximal class and order p pC1 . Exercise 6. Let G be a noncyclic p-group of order > p 4 and exponent > p, p > 2. Prove that if G has no subgroups of order p 4 and exponent p 2 , then G has a cyclic subgroup of index p. 5o . 2-groups G with j3 .G/j D 25 . The ultimate goal is to classify finite 2groups G of order > 26 with j3 .G/j D 26 (see #525 in Research problems and themes I). However, if 3 .G/ 6 ˆ.G/, there is a maximal subgroup M of G such that j3 .M /j 25 and without the exact knowledge of the structure of M there is no chance to determine G in that case. All results of this subsection are due to Z. Bozikov [Boz].

90

Groups of prime power order

In this subsection we determine up to isomorphism all 2-groups G with j3 .G/j 25 and G > 3 .G/. The cases j3 .G/j D 23 and j3 .G/j D 24 are easy (Proposition 52.9). If j3 .G/j D 25 , we show first that exp.3 .G// D 23 and j2 .G/j D 24 (Proposition 52.10). This allows us to use the classification of 2-groups G with j2 .G/j D 24 and jGj > 24 given in first two subsections. Indeed, in Theorem 52.11 we consider nonmetacyclic 2-groups G and get our result almost immediately. In Theorem 52.12 we consider metacyclic 2-groups G with the above property. This case is difficult since in 50 there is no single information for metacyclic groups but we have to determine the groups G up to isomorphism also in this case. In conclusion, we classify the 2-groups G containing exactly one subgroup of order 5 2 and exponent 8 (see #437 in Research problems and themes I). We show that such groups satisfy j3 .G/j D 25 . In fact, we get a more general result (Theorem 52.14). 5.1o . For the convenience we prove two preliminary results (Propositions 52.9 and 52.10). Proposition 52.9. Let G be a 2-group with G > 3 .G/. If j3 .G/j 23 , then j3 .G/j D 23 and G is cyclic. If j3 .G/j D 24 , then 3 .G/ is abelian of type .8; 2/ and G is either abelian of type .2m ; 2/ or G Š M2mC1 , m 4. Proof. Set 3 .G/ D H . Assume that G is noncyclic. If G is of maximal class, then H 2 .G/ D G, which is a contradiction. Therefore, G has a normal abelian subgroup R of type .2; 2/. By hypothesis, H=R is cyclic of order 4 so G=R cyclic. It follows that R D 1 .G/ so G contains a cyclic subgroup of index 2. Now the result follows from Theorem 1.2. Proposition 52.10. Let G be a 2-group with G > 3 .G/. If j3 .G/j D 25 , then exp.3 .G// D 23 and j2 .G/j D 24 . Proof. Set 3 .G/ D H so that jH j D 25 and G > H . If H is cyclic, then 3 .H / < H , a contradiction. If H is of maximal class so G is since c1 .G/ D c1 .H / (see Theorem 1.17); then H 2 .G/ D G, a contradiction. If exp.H / D 24 , then (Theorem 1.2) H is either abelian of type .16; 2/ or H Š M32 . In any case, 3 .H / < H which is again a contradiction. We have proved that exp.H / D 23 (since exp.H / 22 is obviously impossible). Since H is neither cyclic nor of maximal class, there is a G-invariant four-subgroup R contained in H . We assume (by way of contradiction) that H D 2 .G/. Then 2 .G=R/ D H=R is of order 23 and jG=Rj > 23 . Lemma 42.1 implies that H=R is abelian of type .4; 2/. In particular, H 0 R and H 0 is elementary abelian. If the class cl.H / of H is 2, then for each x; y 2 H with o.x/ 4, o.y/ 4, we get .xy/4 D x 4 y 4 Œy; x6 D 1, which implies that exp.H / 4, a contradiction. Thus (since H is not of maximal class) cl.H / D 3 and so H 0 D R and R 6 Z.H /. We set C D CH .R/ and D D CG .R/. Since jH W C j D 2, D covers G=H and G D HD.

52 2-groups with 2 -subgroup of small order

91

If exp.C / 4, then taking x 2 D C such that x 2 2 C , we get o.x/ 8, a contradiction. It follows exp.C / D 8. Let a 2 C with o.a/ D 8. Then C D Rhai, jR \ haij D 2, and so C is abelian of type .8; 2/. Since 3 .D/ D C , our Proposition 52.9 implies that D is either abelian of type .2m ; 2/ or D Š M2mC1 , m 4. In any case, ˆ.D/ is cyclic of order 2m1 and so ˆ.D/ contains a characteristic cyclic subgroup Z of order 8. It follows that Z C and Z is normal in G. But jH=Zj D 4 and so H 0 Z. This contradicts the fact that H 0 D R and jR \ Zj D 2. We have proved that 2 .G/ < H . On the other hand, R < 2 .G/ and so j2 .G/j D 8 or 16. Set K D 2 .G/ and assume jKj D 8. Then Lemma 42.1 shows that either j3 .G/j D 24 (in cases (a) or (b)) or 3 .G/ D G (in case (c)). This is a contradiction and so jKj D 16 and we are done. 5.2o . Now let G be a nonmetacyclic 2-group of order > 25 with j3 .G/j D 25 . We set H D 3 .G/ and apply Proposition 52.10. We see that exp.H / D 23 and 2 .G/ D K is of order 24 so that jH W Kj D 2. By Theorem 41.1, K is nonmetacyclic. We are in a position to use the first part of this section. Since jGj > 25 , we get either K Š Q8 C4 or K Š C4 C2 C2 . All elements in H K are of order 8. Assume first that K D Q Z, where Q Š Q8 , Z Š C4 , and Q \ Z D Z.Q/. All elements in K .Q [ Z/ are involutions and so Q is the unique quaternion subgroup in K. Thus Q is normal in G and so also C D CG .Q/ is normal in G. Since jG=.QC /j 2 and jGj > 25 , we have jC j 23 and C \ K D Z. Also, 2 .C / D Z and so Lemma 42.1 implies that C is cyclic and H D Q 3 .C /. Suppose that jG=.QC /j D 2. Since G=C Š D8 , there is an element c 2 G .QC / such that c 2 2 C . Set P D C hci. Again, 2 .P / D Z and so P D hci is cyclic. But hci induces on Q an “outer” involutory automorphism and so we may set Q D ha; bi so that ac D a1 and b c D ba. Assume now K Š C4 C2 C2 so that E D 1 .K/ is a normal elementary abelian subgroup of order 8 in G. We have Ã1 .K/ D hzi is a central subgroup of order 2 in G. Obviously, K=E is the unique subgroup of order 2 in G=E and so G=E is either cyclic (of order 8) or generalized quaternion. Suppose that G=E is generalized quaternion and let L=E be a cyclic subgroup of index 2 in G=E. For each x 2 G L, x 2 2 K E and so o.x/ D 8 which forces 3 .G/ D G, a contradiction. We have proved that G=E is cyclic. Let a 2 G K be such that hai covers G=E. Then hai is cyclic of order 2m , m 4, G D Ehai, K \ hai Š C4 , and E \ hai D hzi. We may set E D he; u; zi so that we have the following possibilities for the action of hai on E: (i) e a D eu, ua D uz (here a induces an automorphism of order 4 on E); (ii) e a D e, ua D uz (here G D hei ha; ui Š C2 M2mC1 ; (iii) e a D eu, ua D u (here G is minimal nonabelian); (iv) ŒE; a D 1 (here G is abelian of type .2m ; 2; 2/).

92

Groups of prime power order

We have proved the following result. Theorem 52.11 ([Boz]). Let G be a non-metacyclic 2-group of order > 25 with j3 .G/j D 25 . Then one of the following holds. (a) G D QP , where Q D ha; bi is a normal quaternion subgroup of G, P D hci is cyclic of order 2m , m 4, Q \ P D Z.Q/ and c either centralizes Q or ac D a1 and b c D ba. (b) G D EP , where E D he; u; zi is a normal elementary abelian subgroup of order 8, P D hai is cyclic of order 2m , m 4, and E \ P D hzi with z D m1 . For the action of hai on E we have one of the following possibilities: a2 (i) e a D eu, ua D uz, and here a induces an automorphism of order 4 on E; (ii) e a D e, ua D uz, and here G Š C2 M2mC1 ; (iii) e a D eu, ua D u, and here G is minimal nonabelian; (iv) e a D e, ua D u, and here G is abelian of type .2m ; 2; 2/. 5.3o . Now we assume that G is a metacyclic 2-group of order > 25 with j3 .G/j D We set H D 3 .G/. By Proposition 52.10, exp.H / D 23 and 2 .G/ D K is of order 24 . The first part of this section implies that K Š C4 C4 and all elements in H K are of order 8. It follows that R D 1 .K/ is a normal 4-subgroup. Suppose that CH .R/ D K. Since CG .R/ covers G=H , there is an element x 2 CG .R/ H with x 2 2 K. But then o.x/ 8, a contradiction. We have proved that R Z.H /. Let Y =K be a subgroup of order 2 in G=K. Since Y 3 .G/, we get Y D H . Hence H=K is the unique subgroup of order 2 in G=K. This implies that G=K is either cyclic (of order 4) or generalized quaternion. We study first the case, where G=K is cyclic. Let a 2 G K be such that hai covers G=K. Then hai \ H Š C23 , hai \ K Š C22 , and hai \ R D hzi Z.G/. Thus a is of m3 m2 order 2m , m 4, and we set s D a2 and v D a2 so that s 2 D v, v 2 D z, and o.s/ D 8. Set C D CG .K/. Since G=C is cyclic and acts faithfully on K, we have jG=C j 4. For the structure of Aut.C4 C4 / see Proposition 50.5. Also, G 0 K and G 0 is cyclic which implies that jG 0 j 4. If G 0 D f1g, then G is abelian of type .2m ; 4/, m 4. Suppose jG 0 j D 2. Then Exercise 10.13 implies that G is minimal nonabelian. Since jGj > 8, Exercise 1.8a together with the fact that j3 .G/j D 25 gives G D m m1 ha; b j a2 D b 4 D 1; m 4; ab D a1C2 or b a D b 1 i. 0 Suppose jG j D 4 and jG W C j D 2. Then a induces an involutory automorphism on K, H D hK; si C , and so H is abelian of type .8; 4/. If a centralizes K=R, then G 0 R. But G 0 is cyclic and so jG 0 j 2, a contradiction. Hence a acts nontrivially on K=R and so a acts also non-trivially on R D Ã1 .K/. There is an element w 2 K R such that w 2 D u 2 R hzi, w a D w 0 , .w 0 /2 D uz, K D hwi hw 0 i, 2 and ua D uz. Since a induces an involutory automorphism on K, .w 0 /a D w a D w 25 .

93

52 2-groups with 2 -subgroup of small order

and therefore CK .a/ D hww 0 i D hvi with .ww 0 /2 D z. From w a w 1 D Œa; w 1 D w 0 w 1 D w 0 wu D ww 0 u follows that G 0 D hww 0 ui Š C4 , .ww 0 u/2 D z, and G 0 6 Z.G/. From the above, ww 0 D vz . D 0; 1/ and set l D swuC1 . Then l 2 D s 2 w 2 D vu D ww 0 uz 2 G 0 , .vu/2 D z, and so hli Š C8 is normal in G with hli \ hai D hzi. We compute l a D .swuC1 /a D sw 0 uC1 z C1 D s 1 s 2 w 0 uC1 z C1 D s 1 vw 0 uC1z C1 D s 1 .ww 0 z /w 0 uC1 z C1 D s 1 w 1 w 2 .w 0 /2 uC1 z D s 1 w 1 uuzuC1z D s 1 w 1 uC1 D l 1 : m

m1

D We have obtained the following metacyclic group G D ha; l j a2 D l 8 D 1, a2 m3 l 4 , l a D l 1 , m 4i, where H D hl; si D hl; a2 i is abelian of type .8; 4/. It remains to study here the case jG 0 j D 4 and jG W C j D 4. Hence a induces on K an automorphism of order 4 and so, in particular, a acts nontrivially on R. This implies that CK .a/ D hvi, where hv 2 i D hzi D CR .a/. We may set K D hwi hw 0 i, where w a D w 0 , w 2 D u 2 R hzi, .w 0 /2 D uz, ua D uz, and .w 0 /a D ws0 with s0 2 R. We have s0 ¤ 1 (otherwise a induces on K an involutory automorphism). Since .ww 0 /2 D w 2 .w 0 /2 D uuz D z, it follows that ww 0 2 Rhvi. We compute .ww 0 /a D w 0 ws0 D .ww 0 /s0 , which implies hww 0 i ¤ hvi and so hww 0 ui D hvi. Here we have used the fact that Rhvi contains exactly two cyclic subgroups of order 4 and they are hvi and hvui. This gives (since a centralizes v) ww 0 u D .ww 0 u/a D w 0 ws0 uz, and consequently s0 D z. Hence .w 0 /a D wz and so the action of hai on K is uniquely determined. We see that Œa; w 1 D w a w 1 D w 0 w 1 D w 0 wu D ww 0 u and so G 0 D hww 0 ui. But hww 0 ui D hvi hai and so hai is normal in G. Also, hwi induces on hai an automorphism of order 4 with hwi \ hai D 1 and jhai W Chai .w/j D 4. This determines uniquely the structure of our “splitting” metacyclic group G D m m2 m3 i, where H D hw; a2 i. For ha; w j a2 D w 4 D 1; m 4; aw D a1C2 m D 4, H is nonabelian and for m > 4, H is abelian of type .8; 4/. Here G 0 D m2 ha2 i Z.G/ D ha4 i. We have to study the difficult case, where G=K is generalized quaternion. Since a generalized quaternion group is not a subgroup of Aut.K/ (see Proposition 50.5), we have CG .K/ H and consequently H is abelian of type .8; 4/. Note that H=K D Z.G=K/. We set L=K D .G=K/0 D ˆ.G=K/ so that L=K is cyclic containing H=K. We have G=L Š E4 and since G is metacyclic (and so d.G/ D 2), we get L D ˆ.G/ and G 0 is cyclic. On the other hand, G 0 covers L=K and all elements in H K are of order 8. Hence G 0 \ H Š C8 , G 0 \ K Š C4 , and jL W G 0 j D 4. Again, since G is metacyclic, there is a cyclic normal subgroup S of G such that G 0 < S and jS W G 0 j D 2. If S L, then jS \ Kj D 8. But a maximal subgroup S \ K of K is not cyclic since exp.K/ D 4. This contradiction shows that S \ L D G 0 . Set T D KS D LS so that T =K is a cyclic subgroup of index 2 in G=K. For each

94

Groups of prime power order

x 2 G T , we have x 2 2 H K and so all elements in G T are of order 16. Now, ˆ.T / L and ˆ.T / hG 0 ; Ã1 .K/ D Ri since G 0 D Ã1 .S /. But jL W .G 0 R/j D 2 and T is noncyclic and so ˆ.T / D G 0 R. Then F D ˆ.T / \ H is abelian of type .8; 2/ and F1 D ˆ.T / \ K is abelian of type .4; 2/. m3 m2 , v D a2 , and Set S D hai, where o.a/ D 2m , m 4. Also, set s D a2 m1 2 z D a so that z 2 Z.G/, hsi D S \ H , hvi D S \ K, and hzi D S \ R. Obviously, hsi and hvi are normal subgroups in G. Since ˆ.G/ D L, there is an element b 2 G T (of order 16) such that b 2 2 H K and b 2 62 F . Then b 2 D ws, where w 2 K F1 and K D hwi hvi. We set w 2 D u and so R D 1 .K/ D hu; zi. We compute b 4 D w 2 s 2 D uv and b 8 D z. Hence G D haihbi, where hai is a cyclic normal subgroup of order 2m .m 4/ of G, hbi is cyclic of order 16, and hai \ hbi D hzi (of order 2). It remains to determine the action of hbi on hai. Now, hbi induces a cyclic automorphism group on hai so that b inverts hai=hvi since G=K is generalized quaternion. Thus ab D a1 v0 with v0 2 hvi and so .a4 /b D .ab /4 D .a1 v0 /4 D a4 . In particular, b inverts hvi. On the other hand, b centralizes b 4 D uv and 2 so uv D .uv/b D ub v b D ub v 1 , which gives ub D uz. We compute ab D .a1 v0 /b D .a1 v0 /1 v01 D av02 D az , D 0; 1. Since b 2 D sw, this gives 2 asw D aw D az and au D aw D .az /w D az z D a and so Œa; u D 1. Replacing a with au we get o.au/ D o.a/ D 2m and .au/b D a1 v0 uz D .au/1 v0 z. m2 Hence replacing a with au (if necessary), we may assume that v0 D v D a2 or 2m v0 D 1. We have obtained exactly two possibilities for our group G D ha; b j a D m1 m2 D b 8 ; ab D a1C2 ; D 0; 1i. The following result b 16 D 1; m 4; a2 was proved. Theorem 52.12 ([Boz]). Let G be a metacyclic 2-group of order > 25 with j3 .G/j D 25 . Then one of the following holds. (a) G is abelian of type .2m ; 4/, m 4. m

(b) G D ha; b j a2 nonabelian.

m1

D b 4 D 1; m 4; ab D a1C2

or b a D b 1 i is minimal

(c) G D ha; l j a2 D l 8 D 1; m 4; a2 D l 4 ; l a D l 1 i, where 3 .G/ D m3 hl; a2 i is abelian of type .8; 4/, G 0 D hl 2 i Š C4 , Z.G/ D ha2 i Š C2m1 , 0 and G 6 Z.G/. m

m1

m

m2

i; where 3 .G/ D (d) G D ha; w j a2 D w 4 D 1; m 4; aw D a1C2 m3 m2 i, G 0 D ha2 i Š C4 , Z.G/ D ha4 i, and so G 0 Z.G/. Also, for hw; a2 m D 4, 3 .G/ is nonabelian and for m > 4, 3 .G/ is abelian of type .8; 4/. (e) G D ha; b j a2 D b 16 D 1; m 4; a2 D b 8 ; ab D a1C2 ; D m3 2 2 0; 1i, where 3 .G/ D hb ; a i is abelian of type .8; 4/ and G=2 .G/ Š Q2m1 . If D 0, then jG W CG .2 .G//j D 2 and if D 1, then jG W CG .2 .G//j D 4. m

m1

m2

52 2-groups with 2 -subgroup of small order

95

5.4o . We prove here the following result. Theorem 52.13 ([Boz]). Let G be a 2-group containing exactly one subgroup of order 25 and exponent 8. Then we have j3 .G/j D 25 . In fact, with the similar proof, we get the following more general result. Theorem 52.14 ([Boz]). Let G be a 2-group containing exactly one subgroup of order 2nC2 and exponent 2n , where n 2 is a fixed integer. Then we have jn .G/j D 2nC2 . Proof. Let H be the unique subgroup of order 2nC2 and exponent 2n .n 2/. Then H is a characteristic subgroup of G and H n .G/. Suppose that the theorem is false. Then there exists an element a 2 G H of order 2n with a2 2 H . Set HQ D H hai so that jHQ j D 2nC3 . Since H is neither cyclic nor of maximal class, H has a G-invariant four-subgroup R. We have jRhaij 2nC2 and so there is a maximal subgroup M of HQ such that M Rhai. We have exp.M \ H / 2n and since M D .M \ H /hai, we get n .M / D M , jM j D 2nC2 , M ¤ H , and exp.M / 2nC1 . The uniqueness of H forces exp.M / D 2nC1 so M has a cyclic subgroup of index 2. But jM j 24 (since n 2) and M has the normal 4-subgroup R. This implies that M is not of maximal class and so M is either abelian of type .2nC1 ; 2/ or M Š M2nC2 . In both cases n .M / < M and this contradicts our result in the previous paragraph. The theorem is proved.

53

2-groups G with c2 .G / D 4

For a finite 2-group G and a fixed integer n 1 we denote with cn .G/ the number of cyclic subgroups of order 2n . The starting point is the following result: if a 2-group G is neither cyclic nor of maximal class and n > 1, then cn .G/ is even. Thus, for any 2-group G we have c2 .G/ D 1 if and only if G is either cyclic or dihedral and c2 .G/ D 3 if and only if G 2 fQ8 ; SD16 g. In this section we shall determine (up to isomorphism) all 2-groups G of order > 24 with c2 .G/ D 4. If, in addition, j2 .G/j D 24 , then such groups G have been determined in 52. In the sequel we assume that j2 .G/j > 24 . If moreover G has a quaternion subgroup, then we get an infinite class of 2-groups and they have the properties 2 .G/ D G and jZ.G/j D 2 (Theorem 53.6). Therefore we assume also in what follows that G has no quaternion subgroups. Then we show first that A D hU1 ; U2 ; U3 ; U4 i Š C4 C2 C2 , where fU1 ; U2 ; U3 ; U4 g is the set of four cyclic subgroups of order 4 in G (Theorem 53.7). It is interesting to note here that each subgroup Ui turns out to be normal in G. If in addition 2 .G/ D G, then G D Bht i, where B is abelian of type .2m ; 2; 2/, m 2, and t is an involution inverting B and here Z.G/ Š E8 (Theorem 53.8). If jG W 2 .G/j 4, then jG W 2 .G/j D 4 and G is a uniquely determined group of order 27 with jZ.G/j D 2 (Theorem 53.9). Finally, if jG W 2 .G/j D 2, then we get an infinite class of 2-groups G with the properties jGj 27 , G 0 6 1 .A/, and Z.G/ Š E4 (Theorem 53.10(a)) and we get an exceptional group G of order 26 with Z.G/ Š E4 and G 0 1 .A/ (Theorem 53.10(b)). All groups will be given in terms of generators and relations. The above exceptional groups of orders 26 and 27 have peculiar structure but they exist as subgroups of the alternating group A16 . Exercise 1 (see Theorem 6.10). Let A be an abelian 2-group of type .2n ; 2; : : : ; 2/ and of order 2nCd 1 , n > 1, d > 1. Prove that jAut.A/j D 2nCd 2 .2d 2/.2d 22 / : : : .2d 2d 1 /. We need the following information about the structure of Aut.C4 C2 C2 /. Proposition 53.1. Let S 2 Syl2 .Aut.A//, where A Š C4 C2 C2 . Let F be the stability group of the chain A > 1 .A/ > f1g. Then F Š E8 is normal in Aut.A/ and S D D F , where F \ D D f1g and D Š D8 . If X is any subgroup of S such that X \ F D ht i is of order 2, then t is either a square in X or t 62 ˆ.X/.

53

2-groups G with c2 .G/ D 4

97

Proof. By Exercise 1 with n D 2 and d D 3, we get jAut.A/j D 8 6 4 D 26 3 and so jS j D 26 . Let A D hv; u; ei, where o.v/ D 4, o.u/ D 2 D o.e/. The stability group F of the chain A > 1 .A/ > f1g is normal in Aut.A/, F Š E8 , and so F S . Consider the automorphisms ˛ and ˇ of A given by v ˛ D v, u˛ D uz, e ˛ D eu; v ˇ D v, uˇ D uz, e ˇ D e, where z D v 2 . Then o.˛/ D 4, o.ˇ/ D 2, ˛ ˇ D ˛ 1 , and so D D h˛; ˇi Š D8 with D \ F D f1g. Since jFDj D 26 D jS j, we may set S D D F. Let X S be such that X \ F D ht i is of order 2 and assume that t is not a square in X. Since F is normal in S , ht i Z.X/ and X=ht i is isomorphic to a subgroup of S=F Š D8 . If X=ht i is elementary abelian, then ˆ.X/ ht i and therefore ˆ.X/ D f1g since t is not a square in X; in that case t 62 ˆ.X/. If X=ht i is cyclic of order 4, then, since X is not cyclic, we get t 62 ˆ.X/.D Ã1 .X//. Finally, suppose that X=ht i Š D8 . Let U=ht i be the cyclic subgroup of index 2 in X=ht i. Then U is noncyclic so t 62 ˆ.U / D Ã1 .U /. For each x 2 X U , x 2 2 ht i (indeed, hx; t i=ht i is of order 2), and so (by our assumption) x 2 D 1. It follows that ˆ.X/ D Ã1 .U /. Hence in any case t 62 ˆ.X/. Theorems 52.7 and 52.8 imply the following Proposition 53.2. Let K be a 2-group possessing exactly two cyclic subgroups of order 4 and assume that neither of them is a characteristic subgroup of K. Then one of the following holds: (a) K Š C4 C2 ; (b) K Š D8 C2 ; (c) K D hb; t j b 8 D t 2 D 1; b t D bu; u2 D Œu; t D 1; b u D bz; z D b 4 i and this is a group of order 25 with 2 .K/ D hb 2 ; t i hui Š D8 C2 . The following two propositions describe the results from 52 (see Theorems 52.4 and 52.5 for the Proposition 53.3 and Theorem 52.2 for Proposition 53.4) adopted for an application in the proofs of Theorems 53.9 and 53.10. Since the notation introduced in these propositions will be used in the proofs (which are quite involved), the explicit statement of these results is unavoidable. Proposition 53.3. Let H be a 2-group of order > 24 and suppose that A D 2 .H / Š C4 C2 C2 . We set E D 1 .A/ Š E8 , Ã1 .A/ D hzi, B D CH .A/. Then CH .E/=E is cyclic and so B is abelian and H=E is either cyclic or generalized quaternion. Suppose that H=E Š Q2n , n 3, and set E D hz; u; ei, H D hE; a; bi, n1 D z, where hE; ai=E is a cyclic subgroup of index 2 in H=E, o.a/ D 2n , a2 n2 2 2 a D v is of order 4, b 2 A E, o.b/ D 8, and A D hE; vi. Then we have one of the following possibilities: (a) If E 6 ˆ.H /, then b 2 D uv, Œa; u D 1, Œa; e D z , D 0; 1, Œb; e D 1, ub D uz, and ab D a1 . If D 0, then B D hE; ai, and if D 1, then B D hE; a2 i. (b) If E ˆ.H / and ŒE; a ¤ f1g, then b 2 D ev, Œa; u D 1, Œa; e D z, Œb; e D z, Œb; u D z, ab D a1 ue , D 0; 1, and if D 1, then n 4. Here B D hE; a2 i.

98

Groups of prime power order

(c) If E ˆ.H / and ŒE; a D f1g, then b 2 D uv, Œb; e D 1, ub D uz, and ab D a1 e. Here B D hE; ai. Proposition 53.4. Let H be a nonabelian 2-group of order > 24 and suppose that A D 2 .H / Š C4 C2 C2 . We set E D 1 .A/ Š E8 , Ã1 .A/ D hzi, and B D CH .A/. Suppose that H=E Š C2n1 , n 3, and set E D hz; u; ei, H D hE; ai, n1 n2 where o.a/ D 2n , a2 D z, a2 D v is of order 4, and A D hE; vi. Then we have one of the following possibilities: (a) If jŒE; aj D 4, then Œa; e D u, Œa; u D z, n 4, and B D hE; a4 i. (b) If jŒE; aj D 2 and ŒE; a D hzi, then Œa; e D 1, Œa; u D z, and B D hE; a2 i. (c) If jŒE; aj D 2 and ŒE; a ¤ hzi, then Œa; e D u, Œa; u D 1, and B D hE; a2 i. Let G be a 2-group of order > 24 with c2 .G/ D 4. Then j2 .G/j 24 . If j2 .G/j D 24 , then from results of 52 follows that 2 .G/ is isomorphic to Q8 C4 or C4 C2 C2 and the following result follows at once. Theorem 53.5. Let G be a 2-group of order > 24 with c2 .G/ D 4 and j2 .G/j D 24 . Then one of the following holds: (a) 2 .G/ Š Q8 C4 and G is isomorphic to a group (a) or (c) of Theorem 52.1. (b) 2 .G/ Š C4 C2 C2 and G is isomorphic to a group described in Propositions 53.3 and 53.4 or G Š C2m C2 C2 , m 2. In the rest of this section we assume that j2 .G/j > 24 . First we prove the following easy special result. Theorem 53.6. Let G be a 2-group of order > 24 with c2 .G/ D 4 and j2 .G/j > 24 . If G has a subgroup Q Š Q8 , then Q G G, C D CG .Q/ is cyclic of order 2n , n 2, G D .Q C /ht i, where t is an involution such that Qht i Š SD24 and ht iC Š D2nC1 . We have jZ.G/j D 2, jGj D 2nC3 and 2 .G/ D G. Proof. If Q were not normal in G, there is g 2 G such that Q ¤ Qg . But then jQ \ Qg j 4 and so Q [ Qg contains at least five cyclic subgroups of order 4, a contradiction. Hence Q is normal in G. If CG .Q/ Q, then jGj 24 , a contradiction. Therefore C D CG .Q/ > Z.Q/ and C is normal in G. If t0 is an involution in C Z.Q/, then c2 .Q ht0 i/ D 6, a contradiction. Since C cannot be generalized quaternion (otherwise c2 .G/ 6), C is cyclic of order 2n , n 2. We see that c2 .Q C / D 4 and 2 .Q C / D Q 2 .C / is of order 24 . But j2 .G/j > 24 and so there is an involution t 2 G .QC /, G D .QC /ht i and t induces an “outer” automorphism on Q. It follows Qht i Š SD24 . Since there are no elements of order 4 in G .QC /, CQC .t / is elementary abelian. Hence CC .t / D Z.Q/ and so ht i C is of maximal class. The only possibility is that ht i C is dihedral of order 2nC1 . The structure of G is completely determined.

53

2-groups G with c2 .G/ D 4

99

Remark. Here we will give another proof of Theorem 53.6, retaining the same notation. Since G is not of maximal class, C D CG .Q/ 6 Q, by Proposition 10.17. If z is an involution in C Z.Q/, then c2 .Q hzi/ D 6 > 4 D c2 .G/, a contradiction. If c2 .C / > 1, then, obviously, c2 .QC / 5 > 4 D c2 .G/. Thus, c1 .C / D c2 .C / D 1 so, by Theorem 1.17(b), C is cyclic, C \ Q D Z.Q/. We have c2 .Q 2 .C // D 4 D c2 .G/ so Q 2 .C / is normal in G. Since Q is characteristic in Q 2 .C / (Appendix 16), it is normal in G. Since jAut.Q/j2 D 8, we have jG W .Q C /j 2. Since j2 .G/j > 16 D jQ 2 .C /j, there exists an involution t in G .Q C /, and we get G D ht i .QC /. Using the condition c2 .G/ D 4, we get ht i Q Š SD16 and ht i C Š D2n for some n 2 N. The group G is completely determined. Next we assume also that G has no subgroups isomorphic to Q8 . The subgroup generated by four cyclic subgroups of order 4 will be determined in our next basic result. Theorem 53.7. Let G be a 2-group of order > 24 with c2 .G/ D 4 and j2 .G/j > 24 . Suppose that G has no subgroups isomorphic to Q8 . Then A D hU1 ; U2 ; U3 ; U4 i is abelian of type .4; 2; 2/, where U1 ; U2 ; U3 ; U4 < G are cyclic of order 4. Proof. We first show the following easy fact. If a cyclic subgroup V of order 4 normalizes another cyclic subgroup U of order 4, then jU \ V j D 2 and hU; V i is abelian of type .4; 2/. Indeed, if U \ V D f1g, then j.U V /0 j 2 and so U V is metacyclic of order 24 and class 2. This implies that exp.U V / D 4 and 1 .U V / is abelian of type .2; 2/. If N is a maximal subgroup of U V containing U , then all eight elements in .U V / N are of order 4 and so c2 .U V / > 4, a contradiction. Hence jU \ V j D 2 and jU V j D 8. But U V cannot be nonabelian, i.e., quaternion or dihedral, and so U V is abelian of type .4; 2/. If each Ui is normal in G, then (by the above) A D hU1 ; U2 ; U3 ; U4 i is abelian of type .4; 2; 2/, and we are done. Therefore we may assume that U1 is not normal in G. Set K D NG .U1 / and we have 2 jG W Kj 4. By Theorem 1.17(b), c2 .K/ is even. Let M be a subgroup of G such that K < M , jM W Kj D 2 and jG W M j 2. For each m 2 M K, U1m D U2 ¤ U1 and NG .U2 / D K. By the above, A0 D hU1 ; U2 i Š C4 C2 and A0 is normal in M . If there is a further cyclic subgroup U3 of order 4 contained in K, then U3 6 A0 , U3 normalizes U1 and U2 and so, by the above, U3 centralizes U1 and U2 and jA0 \ U3 j D 2. Hence A0 U3 is abelian of type .4; 2; 2/ and we are done. We assume in the sequel that K has exactly two cyclic subgroups U1 and U2 of order 4. Since U1 and U2 are conjugate in M , it follows that neither U1 nor U2 is a characteristic subgroup of K. By Proposition 53.2, we have exactly three possibilities for the structure of K. Assume (by way of contradiction) that K > A0 and set L D 2 .K/. By Lemma 42.1, jLj > 23 (if jLj D 23 , then Z.L/ is a characteristic cyclic subgroup of order 4 in

100

Groups of prime power order

K, which is not the case). Therefore, by Proposition 53.2, we have L Š D8 C2 and L is normal in M . If K > L, then jK W Lj D 2 (Proposition 53.2(c)) and all elements in K L are of order 8. In any case, by Proposition 53.2, CK .A0 / D A0 and K=A0 is elementary abelian of order 2 or 4. Suppose that K > L. Then M=A0 Š Aut.A0 / Š D8 since jM=A0 j D 8. Since L=A0 D ˆ.M=A0 /, there is an element k 2 M K such that k 2 2 L A0 (if k 2 2 A0 for all k 2 M K, then A0 ˆ.M /, a contradiction since M=A0 Š D8 ). All elements in L A0 are involutions and so k is an element of order 4. In fact, all 16 elements in .hkiA0 / L are of order 4. But then c2 .hkiA0 / 8, a contradiction. We have K D L Š D8 C2 ; in that case, jM j D 25 . Assume that M=A0 Š C4 . Let x 2 M K. If o.x/ D 8, then hxi \ K is cyclic of order 4, say U1 . Then NG .U1 / hx; Ki > K, which is a contradiction. Thus, M K has no elements of order 8. Suppose that o.x/ D 2. Then hx; A0 i D K since M=A0 is cyclic; in that case x 2 K, again a contradiction. Thus, in the case under consideration, all 16 elements in M K are of order 4, a contradiction. Thus M=A0 Š E4 . For each m 2 M K, U1m D U2 and so m2 2 1 .A0 / and m is an element of order 2 or 4. Since c2 .A0 / D 2, it follows that M K contains at most four elements of order 4. Hence there exists an involution s 2 M K since jM Kj D 16 (in fact, there are at least 12 involutions in M K). Let t be an involution in K A0 . We set A0 D ha; u j a4 D u2 D Œa; u D 1; a2 D zi and so 1 .A0 / D hu; zi and at D a1 , ut D u, as D au, us D u, because U1s D U2 and s induces an involutory automorphism on A0 . Since s inverts (actually centralizes) exactly the elements in 1 .A0 /, it follows that A0 s has exactly four involutions. Hence the other four elements in A0 s are of order 4. This means that we have found all four elements of order 4 in M K. Hence the coset A0 .st / consists only of involutions. But then the involution st must invert A0 . On the other hand, we have ast D .au/t D a1 u, which is a contradiction. We have proved that K D A0 . Since jGj > 24 , it follows jG W M j D 2 and jGj D 25 . In this case jG W Kj D 4 and so fU1 ; U2 ; U3 ; U4 g forms a single conjugacy class in G. This implies that NG .A0 / D M (otherwise, A0 is normal in G so U1 and U3 are not conjugate in G) and M A0 contains exactly four elements of order 4, and the other four elements in M A0 are involutions. Since j2 .G/j > 24 , there are involutions in G M . Let m be an involution in M A0 . We set again A0 D ha; u j a4 D u2 D Œa; u D 1; a2 D zi. Since M is not of maximal class (see Proposition 1.8), U1m D U2 , and m induces an involutory automorphism on A0 , we may set am D a u ( D ˙1) and um D u. Thus CA0 .m/ D 1 .A0 / D hu; zi and so E D hu; z; mi D 1 .M / (indeed, jhu; z; mi jhu; zij D 4 is the number of involutions in M K D M A0 ) is a normal elementary abelian subgroup of order 8 in G. Let be an involution in G M . If does not centralize E, then E would contain some elements of order 4, a contradiction since M \ E M \ M D ¿. Thus hE; i D E hi is an elementary abelian maximal subgroup of G. But then exp.G/ D 4 and so all elements in G M must be involutions. It follows that

53

2-groups G with c2 .G/ D 4

101

inverts M and so M is abelian. In this case, as it is easy to check since jM j D 24 , M is abelian of type .4; 2; 2/, and the theorem is proved. In the rest of this section we set A D hU1 ; U2 ; U3 ; U4 i Š C4 C2 C2 , where fU1 ; U2 ; U3 ; U4 g is the set of four cyclic subgroups of order 4 in G, E D 1 .A/ Š E8 , hzi D Ã1 .A/ Z.G/, B D CG .A/. Let v 2 A E so that v 2 D z. If there is an involution s 2 B A, then sv is an element of order 4 in B A, a contradiction. Hence A D 2 .B/ and Proposition 53.3 implies that B=E is cyclic and so B is abelian of type .2m ; 2; 2/, m 2. It follows that CG .B/ D B. If c 2 B with o.c/ D 2m , then m1 m2 c2 D hzi, B D Ehci, E \ hci D hzi, A D Ehc 2 i. Since j2 .G/j > 24 , there exists an involution t 2 G B. If there is an element x 2 Bt with o.x/ > 2, then o.x/ 8 since B \ Bx D ¿, and so o.x 2 / 4 and x 2 2 B. Thus CB .t / D CB .x/ is not elementary abelian and so Bt would contain elements of order 4, a contradiction (namely, if y is an element of order 4 in CB .t /, then yt 2 Bt is of order 4). Hence all elements in Bt are involutions and consequently t inverts B. If there is an involution t 0 2 G Bht i, then t 0 also inverts B and so t t 0 2 G B would centralize B, a contradiction since CG .B/ D B. Thus D D Bht i D 2 .G/. If 2 .G/ D G, the structure of G is determined and we have proved the following Theorem 53.8. Let G be a 2-group of order > 24 with c2 .G/ D 4 and j2 .G/j > 24 . Suppose that G has no subgroups isomorphic to Q8 . Then D D 2 .G/ D Bht i, where B is abelian of type .2m ; 2; 2/, m 2, and t is an involution inverting B. We have Z.D/ D E D 1 .B/, CG .t / D ht i E D F , and so F is a selfcentralizing elementary abelian subgroup of order 16. If 2 .G/ D G, the structure of G is completely determined. In what follows we assume, in addition, that G > 2 .G/. We show next that CG .E/ D D D 2 .G/. Indeed, suppose that CG .E/ > D. Since CG .E/ stabilizes the chain A > E > f1g in view of jA W Ej D 2, it follows that CG .E/=B .D CG .E/=CG .A// is elementary abelian and 4 jCG .E/=Bj 8. Let F0 =B be a subgroup of order 2 in CG .E/=B such that D \ F0 D B. Then 2 .F0 / D A (otherwise, c2 .F0 / > 4 D c2 .G/) and F0 is nonabelian of order > 24 since F0 6 B D CG .A/). Proposition 53.3 implies, however, that F0 =E is cyclic and so F0 is abelian, a contradiction. We have proved that CG .E/ D D .D 2 .G/). Hence X D G=B is a subgroup of Aut.A/ (see Proposition 53.1) and if F1 .Š E8 / is the stability group of A > E > f1g (note that F1 < Aut.A/), then X \ F1 D D=B is of order 2. Suppose that the involution in D=B is a square in X. Then there exists x 2 G such that x 2 D t 0 2 D B. But t 0 is an involution (indeed, all elements in D B, by the above, are involutions) and so x is an element of order 4 in G B, a contradiction since A B. Proposition 53.1 implies that there exists a maximal subgroup H of G containing B such that H \ D D B (indeed, by the above proposition, D=B 6 ˆ.G=B/), and so G=B D .D=B/ .H=B/ since D is normal in G. Note that 2 .H / D A (indeed, H \2 .G/ D H \D D B and 2 .B/ D A) and H > B is nonabelian of order > 24 ,

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Groups of prime power order

where B D CG .A/ D CH .A/. In this situation we shall apply Propositions 53.3 and 53.4 which describe all possibilities for the structure of H . Also we shall use freely the notation introduced in these propositions. In particular, 2 jH=Bj 4 and so G=B.D .D=B/ .H=B// is abelian of order 4. We also set D D Bht i, where t is an involution inverting B. It is possible at this stage to eliminate the possibility (c) of Proposition 53.3 and the possibilities (a) and (c) of Proposition 53.4 (thus, we intend to consider these three possibilities using the notation introduced in corresponding parts). Suppose that H is isomorphic to a group (c) of Proposition 53.3. Consider the subgroup K D Bht bi, where .t b/2 2 B and 2 .K/ D A. We compute atb D .a1 /b D .ab /1 D .a1 e/1 D ae and so t b centralizes B=E and jB=Ej 4. Hence K=E cannot be generalized quaternion and so K=E is cyclic. It follows that ht bi covers K=E which gives .t b/2 D ai s, where s 2 E and i is an odd integer. On the other hand, t b centralizes .t b/2 and so ai s D .ai s/tb D .ai s/b D .a1 e/i s b D ai es b . Thus s b D se and Œb; s D e. This is a contradiction since Œb; E D hzi according to Proposition 53.3(c). Suppose that H is isomorphic to a group (a) of Proposition 53.4; in that case, H=B Š C4 since H D hE; ai and B D hE; a4 i. Consider the subgroup K D Bht ai. Since G=B Š C2 C4 , it follows K=B Š C4 and 2 .K/ D A. We compute .a4 /ta D .a4 /1 and so if jB=Ej 4, K=E is nonabelian and so K=E is generalized quaternion. But K=E has the cyclic factor group .K=E/=.B=E/ Š K=B of order 4, a contradiction. Hence we must have B D A, n D 4, jGj D 27 , and K=E is cyclic of order 8. Since ht ai covers K=E, we get .t a/4 D sv with s 2 E. But t a centralizes .t a/4 and so we get (recall that v D a4 ) sv D .sv/ta D .sv 1 /a D s a v 1 D s a vz which implies s a D sz. This gives s D uz with D 0; 1, and .t a/4 D uz v. On the other hand, .t a/2 2 .Bha2 i/ B, and so .t a/2 D a2 s 0 , where s 0 2 A. Indeed, G=B is abelian and so .t a/2 D t 2 a2 s 0 with s 0 2 B D A. We compute 2 uz v D .t a/4 D .a2 s 0 /2 D a2 s 0 a2 s 0 D a4 .a2 s 0 a2 /s 0 D v.s 0 /a s 0 , which gives 2 .s 0 /a D .s 0 /1 uz and Œa2 ; .s 0 /1 D u.z .s 0 /2 / with z .s 0 /2 2 hzi. This is a contradiction since, according to Proposition 53.4(a), Œa2 ; A D hzi. Suppose that H is isomorphic to a group (c) of Proposition 53.4. Here B D hE; a2 i and we consider the subgroup K D Bht ai. We see .a2 /ta D .a2 /1 and so t a inverts B=E. Note that 2 .K/ D A and so in case jB=Ej 4, K=E is generalized quaternion and consequently .t a/2 2 A E (since A=E D Z.K=E/). If jB=Ej < 4, then B D A and again .t a/2 2 A E (since o.t a/ 8). It follows .t a/2 D sv with n2 s 2 E and v D a2 . Since t a centralizes .t a/2 , we get sv D .sv/ta D .sv 1 /a D s a v 1 D s a vz. This gives s a D sz and Œa; s D z. But Proposition 53.4(c) implies Œa; E D hui ¤ hzi, a contradiction. We are now ready to prove the following deep result. Theorem 53.9. Let G be a 2-group of order > 24 with c2 .G/ D 4 and j2 .G/j > 24 . Suppose that G has no subgroups isomorphic to Q8 and jG W 2 .G/j 4. Then we

53

2-groups G with c2 .G/ D 4

103

have jG W 2 .G/j D 4 and G is a uniquely determined group of order 27 : G D ha; b; t j a8 D b 8 D t 2 D 1; a2 D v; a4 D z; b 2 D ev; ab D a1 u; e 2 D u2 D Œe; v D Œu; v D Œe; u D Œa; u D Œt; e D Œt; u D 1; e a D ez; e b D ez; ub D uz; v t D v 1 ; at D eva1 ; b t D euvb 1 i: Here E D hz; u; ei is a normal elementary abelian subgroup of order 8. Four cyclic subgroups of order 4 in G generate the abelian subgroup A D hE; vi of type .4; 2; 2/ which is self-centralizing in G and each cyclic subgroup of order 4 is normal in G. We have D D 2 .G/ D Aht i, where the involution t inverts A and CG .t / D E ht i Š E16 is self-centralizing in G. We have ˆ.G/ D A, all elements in G D are of order 8, and Z.G/ D hzi is of order 2. For one of the four maximal subgroups M of G which do not contain D we have M=E Š Q8 and E < ˆ.M / (so this M is generated by two elements) but for the other three such maximal subgroups M we have M=E Š Q8 and E 6 ˆ.M / (so these M are not generated by two elements). For each of the three maximal subgroups N of G which contain D we have N=E Š D8 . The group G exists as a subgroup of the alternating group A16 and 16 is the smallest degree for any faithful permutation representation of G. Proof. Since G=B D .D=B/.H=B/ we get G=2 .G/ D G=D Š H=B, so we have to examine here only the cases, where H is isomorphic to a group (b) of Proposition 53.3 or to a group (a) with D 1 of Proposition 53.3. In these cases we have jH=Bj D 4 and so we have already proved that jG W 2 .G/j D 4. Suppose first that H is isomorphic to a group (b) of Proposition 53.3. Note that in this case H=E Š Q2n , n 3, E ˆ.H /, and if n D 3, then D 0, i.e., ab D a1 ue D a1 u. We have here G=B Š E8 . Consider the subgroup L D Bht ai, where .t a/2 2 B and 2 .L/ D \ L A so 2 .L/ D A (recall that Aht i D D D 2 .G/ and involution t inverts A). We compute .a2 /ta D .a2 /a D .a2 /1 , and so t a inverts B=E (recall that B D hE; a2 i). If n 4, then jB=Ej 4, L=E is generalized quaternion, and consequently .t a/2 2 A E (since o.t a/ 8). If n D 3, then B D A and o.t a/ 8 implies again .t a/2 2 A E. Thus .t a/2 D sv, where s 2 E and n2 v D a2 . From this relation we get sv D t at a D at a so at D sva1 . Since t a centralizes .t a/2 , we have sv D .sv/ta D .sv 1 /a D s a v 1 D s a vv 2 D s a vz and s a D sz, which gives s D eu0 with u0 2 hu; zi since E D he; u; zi. On the other hand, set B1 D Bhai D Ehai and consider the subgroup L1 D B1 ht bi; then 2 .L1 / D A since L ¤ D D 2 .G/ and A 2 .L/. We compute, taking into account the above obtained equalities, atb D .sva1 /b D s b v b .ab /1 D .s b v b e u/a D a0 a, where D 0; 1 and a0 D s b v b e u 2 AE. If n 4, then B1 =E is cyclic of order 8 and so L1 =E Š M2n . This is a contradiction since L1 =E must be either cyclic or generalized quaternion. It follows (see Proposition 53.3(b)) n D 3, B D A, D 0, a2 D v, b 2 D ve, ab D a1 u, and jGj D 27 . We get v b D .a2 /b D

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Groups of prime power order

.ab /2 D .a1 u/2 D a2 D v 1 , and so v b D v 1 . Since .t b/2 2 B D A and o.t b/ 8 (indeed, t b 2 G D D G 2 .G/), we have .t b/2 D s0 v, where s0 2 E. Since t b centralizes .t b/2 , we get s0 v D .s0 v/tb D .s0 v 1 /b D s0b v, and therefore s0b D s0 , which implies s0 2 heu; zi. We get s0 v D t bt b D b t b and b t D s0 vb 1 . The above elements s; s0 2 E are connected with a relation which we obtain in the following way. We act with t (by conjugation) on the relation b 1 ab D a1 u and get (taking into account the above obtained equalities) .s0 vb 1 /1 .sva1 /.s0 vb 1 / D .sva1 /1 u; bs0 sa1 s0 vb 1 D av 1 su; s0 sa1 s0 v D b 1 .av 1 su/b D a1 uvs b uz D a1 vs b z; .as0 sa1 /s0 v D vs b z; and since s0a relation (1)

1

D s0a , s a

1

D sz, we get s0a szs0 v D vs b z, and this implies the desired s0a s b D s0 s:

Note that s D eu0 with u0 2 hu; zi and s0 2 heu; zi. If s0 D euz ˛ .˛ D 0; 1/, then (1) gives s b D sz and so s D ez ˇ .ˇ D 0; 1/. If s0 D z ˛ .˛ D 0; 1/, then (1) gives s b D s and so s D euz ˇ .ˇ D 0; 1/. We consider again the relations at D sva1 , b t D s0 vb 1 , where s; s0 are the above elements in E. Replacing t with t ue, we get atue D .sva1 /ue D sva1 z D .sz/va1 , b tue D .s0 vb 1 /ue D s0 vb 1 . Replacing t with t u, we get atu D .sva1 /u D sva1 , b tu D .s0 vb 1 /u D .s0 z/vb 1 . The above facts show that in our expressions for s and s0 we may choose ˛ D ˇ D 0. Hence we get either s D e and s0 D eu or s D eu and s0 D 1. We have obtained exactly two groups G1 and G2 (of order 27 ) which differ only in the last two relations: (a) G1 with relations at D eva1 , ; b t D euvb 1 ; (b) G2 with relations at D euva1 , b t D vb 1 . For both groups G1 and G2 we have obtained the following common relations: (c) a8 D b 8 D t 2 D 1, a2 D v, a4 D z, b 2 D ev, ab D a1 u, e 2 D u2 D Œe; v D Œu; v D Œe; u D Œa; u D Œt; e D Œt; u D 1, e a D ez, e b D ez, ub D uz, v t D v 1 . It is possible to show that the groups G1 and G2 are isomorphic. In the group G1 we first obtain by computation (using the relations for G1 ) (2)

.ab/t D zv.ab/1 :

Then in G1 we replace the elements a; v; b; e; t with a0 D a1 , v 0 D v 1 , b 0 D ab, e 0 D eu, t 0 D t e, and see that these new elements satisfy the same common relations

53

2-groups G with c2 .G/ D 4 0

105 0

(c). But from the relations (a) and (2), we get .a0 /t D e 0 uv 0 .a0 /1 , .b 0 /t D v 0 .b 0 /1 , which are the last two relations (b) for the group G2 . Hence the groups G1 and G2 are isomorphic and we have obtained the unique group G D G1 of order 27 as stated in our theorem. We have ˆ.G/ D A. In order to investigate the structure of seven maximal subgroups M of G, we first see a2 D v, b 2 D ev, .ab/2 D euzv, .t a/2 D ev, .t b/2 D euv, and .t ab/2 D zv. If M D hA; a; bi D H , then ˆ.M / D A Š C4 C2 C2 and so E ˆ.M /. If M D hA; a; t bi, then ˆ.M / D hv; eui Š C4 C2 . If M D hA; b; t ai, then ˆ.M / D hv; ei Š C4 C2 . If M D hA; t a; t bi, then ˆ.M / D hev; ui Š C4 C2 . Hence in the last three cases we have E 6 ˆ.M /. For these four maximal subgroups M (which do not contain D D Aht i) we have M=E Š Q8 . For the other three maximal subgroups N (which contain D) we see that N=E Š D8 . We get a faithful permutation representation of degree 16 of G in the following way. We set: a D .1; 2; 3; 4; 5; 6; 7; 8/.9; 16; 12; 13; 11; 14; 10; 15/; b D .1; 9; 3; 10; 5; 11; 7; 12/.2; 13; 8; 14; 6; 15; 4; 16/; t D .3; 7/.4; 8/.10; 12/.14; 16/: We see that the above permutations a; b; t are even and check that they satisfy the defining relations for G. The obtained representation is faithful since z D a4 D .1; 5/.2; 6/.3; 7/.4; 8/.9; 11/.14; 16/.10; 12/.13; 15/ ¤ 1: It remains to show that ı.G/, the minimal degree of a faithful permutation representation of G (minimal representation), equals 16 (in the previous paragraph we have showed that ı.G/ 16). Assume that this is false. Since Z.G/ is cyclic, every minimal representation of G, is transitive so ı.G/ is a power of 2. Then G is isomorphic to a 2-subgroup of Sı.G/ so ı.G/ D 8. Moreover, G Š †3 2 Syl2 .S23 /. However, c1 .G/ D c1 .2 .G// D jD Aj C c1 .A/ D 23 (note that c1 .A/ D 7). On the other hand, c1 .†3 / D jD8 j C Œc1 .D8 / C 12 1 D 8 C 62 1 D 43 > 23, and this is a contradiction. Thus, ı.G/ D 16. In the second half of the proof we investigate the remaining possibility, where H is isomorphic to a group (a) with D 1 of Proposition 53.3. Note that in this case H=E Š Q2n , n 3, and E 6 ˆ.H /. In exactly the same way as in the first part of the proof we get n D 3, B D A, jGj D 27 , a2 D v, b 2 D vu, and ab D a1 . Also we get at D sva1 where s 2 E, s a D sz, s D eu0 with u0 2 hu; zi, b t D s0 vb 1 , where s0 2 E, s0b D s0 , and s0 2 he; zi. We conjugate the relation b 1 ab D a1 with the element t and obtain .b t /1 at b t D t 1 .a / which gives (as in the first part of the proof) the following connection between s and s0 : (3)

s0a s b D s0 sz:

106

Groups of prime power order

If s0 D ez ˛ .˛ D 0; 1/, then (3) gives s b D s and s D ez ˇ .ˇ D 0; 1/. If s0 D z ˛ .˛ D 0; 1/, then (3) gives s b D sz and so s D euz ˇ .ˇ D 0; 1/. Replacing t with t e, we get ate D .sva1 /e D .sz/va1 , b te D .s0 vb 1 /e D s0 vb 1 . Replacing t with t u, we get atu D .sva1 /u D sva1 , b tu D .s0 vb 1 /u D .s0 z/vb 1 . This shows that in our expressions for s and s0 we may choose ˛ D ˇ D 0 and so we have either s D e and s0 D e or s D eu and s0 D 1. We have obtained again exactly two new groups G1 and G2 (of order 27 ) which differ only in the last two relations: (a’) G1 with relations at D eva1 , b t D evb 1 ; (b’) G2 with relations at D euva1 , b t D vb 1 . It is possible to show that the new groups G1 and G2 are isomorphic. In G1 we replace the elements e, b, t with e 0 D eu, b 0 D ab, t 0 D t u, respectively, and see that the common relations for G1 and G2 remain valid. But from the relations (a’) we 0 0 get .a/t D e 0 uv.a/1 , .b 0 /t D v.b 0 /1 , which are the last two relations (b’) for the group G2 . Hence the groups G1 and G2 are isomorphic and we have to study further only the new group G D G1 . We consider the maximal subgroup H D hA; b; t ai and see that ˆ.H / D huv; ev; vi D A. We have 2 .H / D A, H =E Š Q8 , and E ˆ.H /. But this is exactly the starting point for the first part of the proof with the subgroup H instead of H . It follows that our new group G D G1 must be isomorphic to the unique group of order 27 stated in our theorem. Remark. Let us show that every subgroup T G of order 24 contains z (here G is the group of Theorem 53.9). Since Z.G/ D hzi, it suffices to show that TG D f1g. Assume that this is false. Then jG W T j D 8 and G is isomorphic to a Sylow 2-subgroup of the symmetric group S8 . As the proof of Theorem 53.9 shows, G and a Sylow 2subgroup of S8 have distinct numbers of involutions, and this is a contradiction. Thus, z 2 T. In our next result the remaining 2-groups G with c2 .G/ D 4 will be determined. Theorem 53.10. Let G be a 2-group of order > 24 with c2 .G/ D 4 and j2 .G/j > 24 . Suppose that G has no subgroups isomorphic to Q8 and jG W 2 .G/j D 2. Then we have the following two possibilities: (a) G D hb; e; t i with b 8 D e 2 D t 2 D 1; n2

a2

D v;

n1

a2

.t b/2 D a;

D z;

n

a2 D 1; n 3;

b 2 D uv; u2 D 1

Œb; e D Œa; e D Œa; u D Œu; e D Œt; e D Œt; u D 1; ub D uz;

ab D a1 ;

at D a1 :

53

2-groups G with c2 .G/ D 4

107

Here jGj D 2nC4 , n 3, G D hei ha; b; t i, and E D hz; u; ei is a normal elementary abelian subgroup of order 8 in G. Four cyclic subgroups of order 4 generate the abelian subgroup A D hE; vi of type .4; 2; 2/ and each cyclic subgroup of order 4 is normal in G. We have B D CG .A/ D hE; ai is abelian of type .2n ; 2; 2/, 2 .G/ D Bht i is of order 2nC3 , where the involution t inverts B, and CG .t / D E ht i Š E16 is self-centralizing in G. We have ˆ.G/ D hui hai Š C2 C2n , G 0 6 E, Z.G/ D he; zi Š E4 . Finally, .Eha; bi/=E Š Q2n and .Eht bi/=E Š C2n . (b) G D ha; t; ei with a8 D e 2 D t 2 D 1;

a2 D v;

a4 D z;

u2 D Œa; e D Œu; e D Œt; e D Œt; u D 1;

.t a/2 D uv; ua D uz:

Here jGj D 26 and E D hz; u; ei is a normal elementary abelian subgroup of order 8 in G. Four cyclic subgroups of order 4 generate the abelian subgroup A D hE; vi of type .4; 2; 2/ and each cyclic subgroup of order 4 is normal in G. We have A D CG .A/, 2 .G/ D Aht i is a maximal subgroup of G, where the involution t inverts A. We have ˆ.G/ D hu; vi Š C4 C2 , G 0 D hz; ui E, Z.G/ D he; zi Š E4 . This exceptional group exists as a subgroup of A16 . The subgroup hei is a direct factor of G. The minimal degree of a faithful permutation representation of G equals 10 (obviously, this representation is intransitive: every permutation representation of a 2-group G such that its degree does not equal to a power of 2, is intransitive). Proof. Here H is isomorphic to a group (a) with D 0 of Proposition 53.3 or to a group (b) of Proposition 53.4 since these are the only remaining possibilities with jH=Bj D 2, where B D CG .A/ and G=2 .G/ Š H=B. Suppose that H is isomorphic to a group (a) with D 0 of Proposition 53.3. Here CG .A/ D B D hE; ai is abelian of type .2n ; 2; 2/, n 3, and H=E Š Q2n . We have atb D .a1 /b D a and so looking at K D Bht bi with 2 .K/ D A, we get that K=E must be cyclic. Thus ht bi covers K=E and so .t b/2 D ai s, where s 2 E and i is an odd integer. Replacing a with ai we may assume from the start .t b/2 D as which gives b t D asb 1 . Since t b centralizes .t b/2 we obtain as D .as/tb D .a1 s/b D as b , which gives s b D s and so s 2 he; zi. We compute b tu D .asb 1 /u D asb 1 z D a.sz/b 1 , and so replacing t with the involution t u (if necessary), we may assume that s D 1 or s D e. Suppose that s D e and so .t b/2 D ae. But then replacing a with a0 D ae, all other relations remain unchanged and the last relation is transformed into .t b/2 D a0 . Hence we may assume from the start that .t b/2 D a and our group G is uniquely determined as stated in part (a) of our theorem. Suppose, finally, that H is isomorphic to a group (b) of Proposition 53.4. Here CG .A/ D B D hE; a2 i is abelian of type .2n1 ; 2; 2/, n 3, and H=E Š C2n1 . We set K D Bht ai and see that .a2 /ta D .a2 /1 and therefore t a inverts B=E. If jB=Ej 4, then 2 .K/ D A implies that K=E is generalized quaternion. But then replacing H with K, we know that such groups have been determined before.

108

Groups of prime power order

It follows that we may assume that jB=Ej D 2 and so B D A, K=E Š C4 , and jGj D 26 . Hence o.t a/ D 8 and .t a/2 D sv, where s 2 E and at D sva1 . Since t a centralizes .t a/2 , we get sv D .sv/ta D .sv 1 /a D s a v 1 D s a vz, which gives s a D sz and so s D ue 0 with e 0 2 he; zi. We note that atu D .sva1 /u D .sz/va1 and so replacing t with t u (if necessary), we may set s D u or s D ue. However, if s D ue, then .t a/2 D uev. In this case we replace u with u0 D eu and see that all other relations remain valid (with u0 instead of u) and only the last relation is transformed in .t a/2 D u0 . Hence we may assume from the start that .t a/2 D u. Our group G is uniquely determined as stated in part (b) of our theorem. We get a faithful representation of degree 16 of G in the following way. We set a D .1; 2; 3; 4; 5; 6; 7; 8/.9; 10; 11; 12; 13; 14; 15; 16/; e D .1; 9/.2; 10/.3; 11/.4; 12/.5; 13/.6; 14/.7; 15/.8; 16/; t D .2; 6/.3; 7/.10; 14/.11; 15/: We see that the permutations a; e; t are even and they satisfy the defining relations for G. The obtained representation is faithful since the central permutations z D a4 , e, and ez are nontrivial. Since e 62 ˆ.G/, we get G D hei M for some maximal subgroup M of G. The subgroup Z.M / D hzi is cyclic so a faithful permutation representation of minimal degree ı.M / is transitive, i.e., ı.M / is a power of 2. We claim that ı.M / D 8. Clearly, ı.M / 8. Set U D hai and V D hu; e; t i. Since U \ V D f1g, we have ı.M / jM W U j C jM W V j D 4 C 4 D 8. It follows that ı.M / D 8. We get ı.G/ D ı.M hei/ ı.M / C ı.hai/ D 8 C 2 D 10. Since, by [Ber26, page 740], ı.G/ ı.hai hei/ D ı.hai/ C ı.hei/ D o.a/ C o.e/ D 8 C 2 D 10; we get ı.G/ D 10. The proof is complete. Problem. Classify the 2-groups G such that c2 .G/ 6 0 .mod 4/. In particular, classify the 2-groups G with c2 .G/ D 6.

54

2-groups G with cn.G / D 4, n > 2

In this section we study a 2-group G with cn .G/ D 4, n > 2 [Jan8]. If fU1 ; U2 ; U3 ; U4 g is the set of four cyclic subgroups of order 2n , then we describe first the structure of the subgroup X D hU1 ; U2 ; U3 ; U4 i (Theorems 54.1 and 54.2). It is interesting to note that always jXj D 2nC2 . If G has a normal elementary abelian subgroup of order 8, the structure of G is described in Theorem 54.6. If G has no normal elementary abelian subgroups of order 8, the structure of G is described in Theorem 54.7. The proofs of these theorems are quite involved and therefore we prepare the stage with Propositions 54.3, 54.4, and Theorem 54.5 which are also of independent interest. If cn .G/ D 2, then the 2-groups are known (Corollary 43.6). The next interesting and important case is cn .G/ D 4 (see Research Problems #188 and #425). However, this problem is essentially more difficult. Throughout this section we assume that G is a 2-group with cn .G/ D 4, n > 2. We consider first the case that G has a normal subgroup isomorphic to E8 . Theorem 54.1. Let G be a 2-group with cn .G/ D 4, n > 2, and suppose that G has a normal subgroup E Š E8 . Let fU1 ; U2 ; U3 ; U4 g be the set of four cyclic subgroups of order 2n in G. Then X D hU1 ; U2 ; U3 ; U4 i D EU1 , where E \ U1 > f1g and so jXj D 2nC2 . Moreover, 2 .X/ Š C4 C2 C2 or 2 .X/ Š D8 C2 in which case n D 3. Proof. Assume that for each cyclic subgroup U of order 2n , U \ E D 1. Let U D hai be one of them. Then .EU /=E Š C2n and all 2nC2 elements in .EU / .Eha2 i/ are of order 2n . Indeed, if x 2 .EU / .Eha2 i/ and o.x/ D 2nC1 , then o.x 2 / D 2n and 2nC2 hx 2 i \ E ¤ f1g, a contradiction. But then cn .G/ '.2 n / D 8 which contradicts our assumption (here './ is Euler’s totient function; '.2n / is the number of generators of a cyclic group of order 2n ). We have proved that there is a cyclic subgroup hbi of order 2n with jhbi \ Ej D 2 n1 n2 so that .Ehbi/=E Š C2n1 . Let H D Ehbi, z D b 2 , and v D b 2 ; then hbi \ E D hzi. Assume that CH .E/ D E. Then n D 3, H=E Š C4 acts faithfully on E, and we may set E D he; u; zi, e b D eu, ub D uz. The structure of H is uniquely determined, H D hb; ei. We have uv D .ub /b D .uz/b D uzz D u, e v D .eu/b D eu uz D ez; v e D eve D ve v e D veze D vz D vv 2 D v 1 . It follows that Ehvi D he; vi hui Š D8 C2 . Let us check that 2 .H / D Ehvi, ˆ.H / D

110

Groups of prime power order

hui hvi Š C4 C2 . We have .be/2 D b 2 e b e D veue D vu 2 ˆ.G/. Since v D b 2 2 ˆ.G/ we get u D v 1 vu 2 ˆ.G/. It follows from d.H / D 2 that ˆ.G/ has order 23 so is abelian, and the second isomorphism in the displayed formula is proved. It remains to prove the first equality in the displayed formula. Assume that it is false. Then there exists an element y of order 4 in H .Ehvi/. We may assume that y 2 fbe; bu; beu; bez; buz; beuzg. However, it is easy to check that all these six elements have order 8. Thus, 2 .H / D Ehvi Š D8 C2 . and we see that all 16 elements in H .Ehvi/ are of order 8. Thus c3 .H / D c3 .G/ D 4 and therefore, by the product formula, X D EU1 , where U1 D hbi. Suppose now that CH .E/ > E. Then A D Ehvi is abelian of type .4; 2; 2/, all elements in A E are of order 4, and Ã1 .A/ D hzi. For each element y 2 n2 H .Ehb 2 i/, we have y 2 2 A E and so o.y/ D 2n . Hence all 2nC1 elements 2nC1 in H .Ehb 2 i/ are of order 2n which gives cn .H / D '.2 n / D 4 D cn .G/. We get again X D EU1 and 2 .X/ D Ehvi Š C4 C2 C2 . We assume now that G is a nonabelian 2-group which has no normal subgroups isomorphic to E8 . Since G is not of maximal class, we may apply the main theorem in 50 to conclude that G has a normal subgroup W which is either abelian of type .4; 4/ with CG .W / being metacyclic or W is abelian of type .4; 2/ with CG .W / being abelian of type .2j ; 2/, j 2. Theorem 54.2. Let G be a nonabelian 2-group with cn .G/ D 4, n > 2, and suppose that G has no normal elementary abelian subgroups of order 8. Let fU1 ; U2 ; U3 ; U4 g be the set of four cyclic subgroups of order 2n . Then the subgroup X D hU1 ; U2 ; U3 ; U4 i is of order 2nC2 and for the structure of X we have the following possibilities: (a) n > 2, X D W U1 with W \ U1 Š C4 , where W is an abelian normal subgroup of type .4; 4/. We have 2 .X/ D W and X is metacyclic. (b) n > 2, X D QU1 , where Q Š Q8 is a normal quaternion subgroup of X, Q \ U1 D Z.Q/, U1 D hbi Š C2n , and b either centralizes Q or b induces on Q an involutory outer automorphism in which case n > 3. (c) n D 3, jXj D 25 , and X has a self-centralizing (in X) abelian normal subgroup A of type .4; 2/. Furthermore, 2 .X/ Š Q8 C2 or 2 .X/ Š D8 C2 and so (according to 52) the group X is uniquely determined in each of the two possibilities for 2 .X/. Proof. By Proposition 50.6, G has a normal abelian subgroup W of exponent 4 such that either W is of type .4; 4/ or of type .4; 2/ and such that 2 .CG .W // D W ; in the first case CG .W / is metacyclic and in the second case CG .W / is abelian and has a cyclic subgroup of index 2. (i) We examine first the possibility that G has a normal subgroup W Š C4 C4 , where CG .W / is metacyclic. Set W0 D 1 .W /, H D W U1 , where U1 D hai is a cyclic subgroup of order 2n , n > 2. The structure of Aut.W / is described in

54

2-groups G with cn .G/ D 4, n > 2

111

Proposition 50.5. In particular, exp.Aut.W /2 / D 4, where Aut.W /2 2 Syl2 .Aut.W // (moreover, Aut.W /2 is special) and if ˛ ¤ ˇ are two involutions in Aut.W /2 which do not stabilize the chain W > W0 > 1, then h˛; ˇi is a four-group. We want to show that W \U1 Š C4 . So suppose that jW \U1 j 2. If W \U1 D 1, n1 then CH .W / contains the involution z D a2 since jU1 j > 4 D exp.Aut.W //2 and W0 hzi Š E8 , contrary to the fact that CH .W / is metacyclic. Hence we have n2 W \ U1 D hzi is of order 2. Set v D a2 so that v 2 D z. Since 2 .CG .W // D W , v does not centralize W . Hence in this case U1 =hzi acts faithfully on W , where U1 D hvi. This gives o.a/ D 8, n D 3, a2 D v, H=W Š C4 and jH j D 26 . We shall determine the structure of H D W U1 , where U1 D hai. Since a induces an automorphism of order 4 on W , a cannot stabilize the chain W > W0 > f1g (Remark 50.2). Set W D hxi hyi, x 2 D z, y 2 D u; then W0 D hzi hui. Assume that a stabilizes the chain W0 > f1g. Then x a is an element containing z so x a 2 xW0 . Similarly, y a 2 yW0 . We conclude that a stabilizes the chain W > W0 > f1g, which is not the case. Thus, we have ua D uz. If w 2 W is such that w 2 D u, then setting w a D y, we get y 2 D .w a /2 D .w 2 /a D .ua / D uz and W D hwi hyi with .wy/2 D z. Since W0 hai Š M24 , we get c3 .W0 hai/ D 2. It remains to determine the element y a D ws, where s 2 W0 . We compute 2

w v D w a D y a D ws;

2

y v D y a D .ws/a D w a s a D ys a ;

and so s ¤ 1 since a2 D v induces an involutory automorphism on W . Computing again .a.wy//2 D awyawy D a2 .wy/a wy D vywswy D vzs, we see that o.awy/ D 8 since o.vzs/ D 4 noting that W0 hvi is abelian of type .4; 2/. Since .wy/a D .wy/s, we conclude that W0 hwyihai contains exactly four cyclic subgroups of order 8. Therefore all other elements in aW must be of orders 2 or 4. We compute .aw/2 D awaw D a2 w a w D vyw; .vyw/2 D v 2 .yw/v .yw/ D zys a ws.yw/ D s a s: It follows that s a s D 1, and so s a D s (otherwise aw would be of order 8). Since s 2 Z.ha; W0 i/# D fzg (recall, that ha; W0 i Š M24 ), we get s D z, and so y a D wz and .wy/a D wyz D .wy/1 since z D .wy/2 . The structure of H is uniquely determined and X D hU1 ; U2 ; U3 ; U4 i D W0 hwyihai is a normal subgroup of G. The group X is an extension of the normal abelian subgroup W0 hwyi of type .4; 2/ by C4 and .W0 hwyi/ \ hai D hzi. We have ua D uz and .wy/a D .wy/1 and so v D a2 centralizes W0 hwyi. Therefore A D hu; wy; vi is abelian of type .4; 2; 2/ and 1 .X/ D 1 .A/ D hu; z; wyvi Š E8 is normal in G. This is a contradiction. We have proved that for each cyclic subgroup U of order 2n we have W \ U Š C4 . n3 Set again H D W U1 so that we have H=W Š C2n2 . Set U1 D hai, y D a2 , y 2 D v, and y 4 D z. Assume that CH .W / > W . Then W hyi is abelian of type .8; 4/ and so all elements in .W hyi/ W are of order 8; in particular, all elements in the coset W y are of

112

Groups of prime power order n3

order 8. For each x 2 H .W ha2 i/, we have x 2 2 W y since all elements in n3 2 n2 2 .H=W / .W ha i=W / have order 2 , and so o.x / D 8 and o.x/ D 2n . We jH jjW haij 2nC1 D 2n1 D 4 D cn .G/ and so X D H D W U1 as stated in get cn .H / '.2n / part (a) of our theorem. We have in this case 2 .X/ D W . Until the end of part (i), we may assume that CH .W / D W . Since exp.Aut.W /2 / D 4, where Aut.W /2 is a Sylow 2-subgroup of Aut.W /, jH=W j 4, and so n D 3 or 4. Set W0 D 1 .W / and assume that the involutory automorphism of W induced by y stabilizes the chain W > W0 > f1g. (This will certainly happen if n D 4 because in that case y D a2 .) Let w 2 W be such that W D hwi hvi. We have w y D ws, where s 2 W0 and we compute .y.w i v j //2 D yw i v j yw i v j D y 2 .w i v j /y w i v j D v.ws/i v j w i v j D vs i w 2i v 2j ; where i , j are any integers. Since o.vs i w 2i v 2j / D 4, all elements in yW are of order 8. If n D 4, then for each x 2 H .hyiW /, x 2 2 yW and so c4 .H / D 32 W 8 D 4. If n D 3, then H D hyiW and so c3 .H / D 4. Hence we get again X D H D W U1 and 2 .X/ D W , as stated in part (a). It remains to consider the case CH .W / D W , where the involutory automorphism of W induced by y does not stabilize the chain W > W0 > f1g. By the above, n D 3, H D hyiW , and if we set W0 D hu; zi, then uy D uz and z 2 Z.G/. If w 2 W is such that w 2 D u, then setting w 0 D w y , we get .w 0 /2 D uz and W D hwihw 0 i. Note that CW .y/ D hvi and so .ww 0 /y D ww 0 implies ww 0 D v ˙1 . But replacing w with w 1 (if necessary), we may assume from the start that ww 0 D v and so w 0 D w 1 v. Since W0 hyi Š M24 , we get c3 .W0 hyi/ D 2 and we claim that c3 .H / D 2. We compute (for any s 2 W0 ) .yws/2 D y 2 .ws/y ws D vw 0 s y ws D v 2 s y s D zs y s; .yw 0 s/2 D y 2 .w 0 s/y w 0 s D vws y w 0 s D v 2 s y s D zs y s: Since o.zs y s/ 2, the above claim is proved. Hence there exists an element y 0 of order 8 in G H such that v 0 D .y 0 /2 2 W and y 0 induces on W an involutory automorphism which does not stabilize the chain W > W0 > f1g. Otherwise, we get (as above) c3 .hy 0 iW / D 4, which is a contradiction (since c3 .H / D 2). Since 0 z 2 Z.G/, we get uy D uz and therefore .y 0 /4 D z and v 0 D .y 0 /2 2 W0 hvi. Hence both y and y 0 normalize the abelian subgroup A D W0 hvi of type .4; 2/. Since W0 hy 0 i Š M24 , we get c3 .W0 hy 0 i/ D 2. It is easy to see that CG .W / D W . Indeed, if W < S CG .W / and jS W W j D 2, then S is abelian metacyclic and so S is of type .8; 4/ and c3 .S / D 4, a contradiction. Hence y and y 0 induce on W two distinct involutory automorphisms which do not stabilize the chain W > W0 > f1g. By our remark about Aut.W / at the beginning of the proof, yy 0 induces an involutory automorphism on W which

54

2-groups G with cn .G/ D 4, n > 2

113

obviously stabilizes the chain W > W0 > f1g. Since the coset W .yy 0 / cannot contain elements of order 8, we get .yy 0 /2 2 W0 A. The subgroup Ahy; y 0 i is of order 25 and since c3 .Ahy; y 0 i/ D 4, we get X D Ahy; y 0 i. Suppose that B D Ahyy 0 i is abelian. Then exp.B/ D 4 and all 16 elements in X B are of order 8. Since X D hU1 ; U2 ; U3 ; U4 i is normal in G, B is normal in G, and so (by assumption) 1 .B/ D 1 .A/ D W0 Š E4 and B is of type .4; 4/. (If 1 .B/ Š E8 , then G would possess a normal elementary abelian subgroup of order 8.) But uy D uz and so y does not stabilize the chain B > W0 > f1g. By the above, c2 .Bhyi/ D 2 which is a contradiction since Bhyi D X D hU1 ; U2 ; U3 ; U4 i. Hence CX .A/ D A and we have obtained a group stated in part (c) with X=A Š E4 . (ii) It remains to examine the second possibility, where G has a normal abelian subgroup W1 of type .4; 2/ such that N D CG .W1 / is abelian of type .2j ; 2/, j 2, and G=N is isomorphic to a subgroup of D8 . If j > 2, then G=N is elementary abelian of order 4. Indeed, if j > 2, then N contains a characteristic subgroup Z Š C4 . Since Z < W1 and Z is normal in G, it is easy to see that the nontrivial elements in G=N induce on W1 only involutory automorphisms. We set W0 D 1 .W1 / Š E4 and hzi D Ã1 .W1 / Z.G/. Suppose that G=N is not elementary abelian. Then j D 2, W1 D N D CG .W1 /, and G=N Š C4 or G=N Š D8 . In that case n 4. Suppose that U G and U Š C24 . Then U=.U \ W1 / Š C4 acts faithfully on W1 . But U centralizes the cyclic subgroup U \ W1 Š C4 , a contradiction. It follows n D 3. If L > W1 is such that L=W1 Š C4 , then (by the structure of Aut.C4 C2 /) 1 .L=W1 / D L0 =W1 inverts W1 . In particular, L0 does not contain a cyclic subgroup of order 8. Suppose first that L contains a cyclic subgroup U1 D hai of order 8. Then U1 covers L=W1 and U1 \ W1 D hzi is of order 2, where z D a4 and L0 D W1 hvi with v D a2 inverting W1 . For each w 2 W1 we compute .vw/2 D v 2 w v w D v 2 w 1 w D v 2 D z, and so all elements in L0 W1 D vW1 are of order 4. For each x 2 L L0 , x 2 2 L0 W1 and so all 16 elements in L L0 are of order 8. Hence c3 .L/ D 4 and so X D L is a group of order 25 as stated in part (c) of our theorem with X=W1 Š C4 . We may assume that exp.L/ < 8. In that case G=W1 Š D8 . Let U1 G and U1 D ha1 i Š C8 . Then U1 \ W1 D hvi Š C4 , where v D a12 . We may set W1 D hv; u j v 4 D u2 D Œv; u D 1; v 2 D zi, and then v a1 D v, ua1 D uz (since a1 does not centralize W1 ). Obviously, M D W1 U1 D ha1 ; ui Š M24 and so c3 .M / D 2. Hence there is another cyclic subgroup U2 D ha2 i in G which is not contained in M and U2 \ W1 Š C4 . Then a2 must induce on W1 an involutory automorphism which is distinct from that one induced by a1 . The only possibility is .vu/a2 D vu, ua2 D uz, and so v a2 D vz D v 1 , with U2 \ W1 D hvui. Again, P D W1 U2 D ha2 ; ui Š M24 and so c3 .P / D 2. Since a1 a2 inverts W1 , ha1 ; a2 i induces a four-group of automorphisms on W1 and therefore .MP W1 /=W1 Š E4 , where MP W1 D ha1 ; a2 iW1 . Note that c3 .ha1 a2 iW1 / D 0 (by a remark above). We get X D MP W1 and so we have obtained a group of order 25 stated in part (c) of our theorem with X=W1 Š E4 .

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It remains to consider the case where G=N is elementary abelian (of order 4) with N abelian of type .2j ; 2/, j 2. Assume first j D 2. In that case we have again N D CG .W1 / D W1 Š C4 C2 . Let U1 G with U1 Š C8 . Then U1 \ W1 Š C4 and (as before) W1 U1 Š M24 which gives c2 .W1 U1 / D 2. Hence there is U2 G with U2 Š C8 and U2 6 W1 U1 . Then again W1 U2 Š M24 and c2 .W1 U2 / D 2. We have G=W1 Š E4 and we have obtained again a group X D G of order 25 as stated in part (c). j We assume now j 3 and set N D hu; a j u2 D a2 D Œu; a D 1i. Then put j 2 v D a2 and z D v 2 . We see that W1 D hu; vi, z 2 Z.G/, hvi is a characteristic subgroup of N , and so hvi is normal in G. Obviously, n j C 1. Assume first that n D j C 1. Let x 2 G N be such that o.x/ D 2n D 2j C1 . Then x centralizes x 2 2 N and o.x 2 / D 2j . In particular, x centralizes hvi and so ux D uz (since x does not centralize W1 ). We get hxiN D hx; ui Š M2nC1 and so c2 .hxiN / D 2. It follows that G=N Š E4 and there is another element y 2 G .hxiN / with o.y/ D 2n . We get again that y centralizes hvi and uy D uz. But then xy centralizes W1 and xy 2 G N , a contradiction. We have proved that n j . Then cn .N / D 2. Take a0 2 hai N with o.a0 / D 2n . Then A D hui ha0 i is an abelian normal subgroup of type .2n ; 2/ and cn .A/ D cn .N / D 2. There is b0 2 G N with o.b0 / D 2n . Since o.b02 / D 2n1 , it follows that b02 2 A, b0 centralizes a cyclic subgroup of order 4 in W1 and so ub0 D uz. Also, hu; b0 i Š M2nC1 and so cn .hu; b0 i/ D 2. We get X D Ahb0 i. We see that hb02 i (of order 2n1 ) is contained in Z.X/. Set Z D Z.X/ and assume first that Z > hb02 i. Then Z is a cyclic subgroup of order 2n contained in A and hb0 iZ is abelian of order 2nC1 . Since hb0 iZ is not cyclic (because it contains two distinct cyclic subgroups Z A and hb0 i 6 A of order 2n ), there is an involution t 0 2 .hb0 iZ/ A. Since t 0 acts faithfully on W0 , it follows that W0 ht 0 i D D Š D8 and X D D Z. There is a subgroup Q Š Q8 contained in X (Appendix 16) so that X D Q Z with Q \ Z D Z.Q / D hzi and Z Š C2n . We have obtained a group X of order 2nC2 as stated in part (b) of our theorem. Assume now that Z.X/ D hb02 i is cyclic of order 2n1 . Suppose at the moment that 2 .X/ D W1 . In that case we can apply the classification result of (Lemma 42.1) in which all 2-groups X with jXj > 23 and j2 .X/j 23 are determined. Since jX W Z.X/j D 23 , X is not isomorphic to M2nC2 . Hence X is isomorphic to the metacyclic group G from part (c) of the above lemma. Since Z.G / Š C4 , it follows that n D 3 and jXj D jG j D 25 . But in that case c3 .G / D 6, which is a contradiction. We have proved that 2 .X/ > W1 . Hence there is an element x0 2 X A with o.x0 / 4. Since o.x02 / 2, it follows that x02 2 W0 and D D hW0 ; x0 i Š D8 because ux0 D uz. Thus X0 D D hb02 i is the central product of D Š D8 and Z.X/ D hb02 i, where D \ hb02 i D Z.D/ D hzi. Since jX0 j D 2nC1 , X0 is a maximal subgroup of X and obviously exp.X0 / D 2n1 . All 2nC1 elements in X X0 must be of order 2n (since cn .X/ D 4). There exists (as above) a quaternion subgroup Q X0 so that X0 D Q hb02 i. Since Q is a characteristic subgroup in X0

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(because it is a unique quaternion subgroup of X0 ), Q is normal in X and X D Qhb0 i. From Z.X/ D hb02 i follows that CX .Q/ D hb02 i and so b0 induces on Q D hx; yi an outer involutory automorphism with x b0 D y and .xy/b0 D .xy/1 . We compute .b0 x/2 D b02 x b0 x D b02 .yx/. If n D 3, then b02 .yx/ is an involution. In that case o.b0 x/ D 4, which is a contradiction (since all elements in X X0 must be of order 23 ). Hence we must have n > 3. In that case o.b0 x/ D 2n , as required. We have obtained a group stated in part (b) of our theorem. It remains to determine 2 .X/, where X is a group (of order 25 ) given in part (c) of our theorem with A being a self-centralizing (in X) abelian normal subgroup of type .4; 2/. We know that Aut.A/ Š D8 and Z.Aut.A// inverts A. Since jX=Aj D 4 and X=A acts faithfully on A, there is a subgroup B .> A/ of order 24 such that B=A inverts A. Then B A cannot contain elements of order 8 and so exp.B/ D 4. Since c3 .X/ D 4, all 16 elements in X B must be of order 8 and so B D 2 .X/. But B is nonabelian and so using the results in 52, we see that B is isomorphic to one of the following groups: Q8 C2 , D8 C2 , and Q8 C4 . It is easy to see that the last case cannot occur. Indeed, if B Š Q8 C4 , then, by 52, X Š Q8 C8 . But in that case A is not self-centralizing. Using again the results in 52, we see that in each of the first two cases the group X is uniquely determined. Proposition 54.3 (see 18). Let G be a 2-group possessing exactly one Ls -subgroup H of order 2sCm with a fixed integer m 2. Then G is either an Ls -group or a Us -group with respect to the kernel R D 1 .H /. Proof. Set R D 1 .H / so that R Š E2s , s 2, H=R is cyclic of order 2m , m 2, and R is normal in G. Set S=R D 1 .H=R/. Then we have S=R ˆ.H=R/ ˆ.G=R/ and 1 .S / D R. We use induction on jGj. If H D G, we are done. In the sequel let H < G. Let H M < G, where M is maximal in G. By induction, M is either an Ls -group or a Us -group with respect to the kernel R D 1 .H /. Note that the Frattini subgroup of an arbitrary p-group is not of maximal class. Since ˆ.G=R/ M=R, ˆ.G=R/ is cyclic (of order 2 ) or abelian of type .2; 2/, where the last case is possible only if M=R Š D8 . However, if ˆ.G=R/ is abelian of type .2; 2/, then a result of Nekrasov (see Proposition 4.9) implies ˆ.G=R/ Z.G=R/, and then M=R is abelian, a contradiction. Thus ˆ.G=R/ is cyclic and so S=R D 1 .ˆ.G=R//. Let F=R < G=R be cyclic of order 2m . Now, 1 .F=R/ ˆ.F=R/ ˆ.G=R/ and so 1 .F=R/ D S=R which gives 1 .F / D 1 .S / D R. It follows that F is an Ls -group of order 2sCm . By assumption, F D H and so G=R has exactly one cyclic subgroup of order 2m . By Exercise 1.104, G=R is either cyclic or of maximal class. If G=R is cyclic, 1 .G/ D 1 .S / D R and G is an Ls -group. Suppose that G=R is of maximal class and let T =R be a cyclic subgroup of index 2 in G=R. Since S=R ˆ.G=R/ < T =R, it follows S=R D 1 .T =R/ and 1 .T / D 1 .S / D R. This means that G is a Us -group with the kernel R and we are done.

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Proposition 54.4. Let G be a 2-group with cn .G/ D 2, n > 2. Then G is an L2 -group or a U2 -group. Proof. Since G is neither cyclic nor of maximal class, it has a normal abelian subgroup R of type .2; 2/. Let U1 be a cyclic subgroup of order 2n . Set H D U1 R and assume that R \ U1 D f1g. Let K=R be the subgroup of index 2 in H=R and let Z=R be the n1 subgroup of order 2 in K=R. Then Z Š E8 and for each x 2 H K, x 2 2 ZR n nC1 n nC1 n1 which gives o.x/ D 2 . But then cn .H / 2 W '.2 / D 2 W 2 D 4, a contradiction. Hence jU1 \Rj D 2 and so H is an L2 -group of order 2nC1 , i.e., H is either abelian of type .2n ; 2/ or H Š M2nC1 . In any case H is generated by its two cyclic subgroups of order 2n and so H is the unique L2 -subgroup of G of order 2nC1 . By Proposition 54.3, G is either an L2 -group or a U2 -group with the kernel R. Theorem 54.5. Let X be a 2-group of order > 24 with 2 .X/ Š D8 C2 or 2 .X/ Š Q8 C2 . Then jXj D 25 and for each of the two possibilities for 2 .X/, X is uniquely determined and is given explicitly in Theorems 52.2(a) and 52.1(d). In both cases ˆ.X/ is abelian of type .4; 2/, X 0 D 1 .ˆ.X// Š E4 , Z.X/ D Ã1 .ˆ.X// is of order 2, and c3 .X/ D 4. Also, each maximal subgroup of X is nonabelian and therefore A D ˆ.X/ is self-centralizing in X. Suppose that G is a 2-group with G > X and c3 .G/ D 4. Then we have the following possibilities: (a) CG .A/ D A, G=A Š D8 , and G has a normal elementary abelian group of order 8. (b) CG .A/ D C Š C4 C4 , C is self-centralizing in G, and G=C Š E4 or D8 . Proof. The first part of this proposition is a direct consequence of the results in 52. Suppose that X has an abelian maximal subgroup. Then the formula jXj D 2jZ.X/jjX 0 j (see Lemma 1.1) gives a contradiction (since jXj D 25 ). We have to determine the structure of G, where G > X and c3 .G/ D 4 D c3 .X/. This implies that X is normal in G. If CG .A/ D A, then G > X and Aut.A/ Š D8 imply G=A Š D8 . We have 2 .X/ > A, j2 .X/ W Aj D 2, and 2 .X/=A D Z.G=A/. Since X=A Š E4 , there is an element x 2 G X such that x 2 2 2 .X/ A. But o.x 2 / 4 and x is not of order 8. Hence x 2 D t is an involution which gives 2 .X/ Š D8 C2 and Z.2 .X// D 1 .A/ D X 0 . Thus E D ht i 1 .A/ is an elementary abelian normal subgroup (of order 8) of X. Since x centralizes t and normalizes 1 .A/ D X 0 , it follows that E G G, as required. We have to study the case C D CG .A/ > A so that we get C \ X D A, C is normal in G, .XC /=C Š X=A Š E4 , and G=C Š E4 or D8 . It remains to determine the structure of C . Since C has no elements of order 8, we get exp.C / D 4. Let s be an element of order 8 in X. Then s 2 2 A D ˆ.X/, D D Ahsi is of order 24 , and 1 .A/ D 1 .D/ is a normal 4-subgroup of D. Since every maximal subgroup of X is nonabelian, it follows that D Š M24 and c3 .D/ D 2. Set K D C hsi D CD.

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Because K \ X D D, we have c3 .K/ D 2 and D is the unique L2 -subgroup of K of order 24 . By Proposition 54.3, K (of order 25 ) is an L2 -group or a U2 -group with the kernel R D 1 .A/ D 1 .D/. If K is an L2 -group, then K Š M2m , m > 4, a contradiction. Indeed, M2m is minimal nonabelian contrary to the fact that K contains the nonabelian proper subgroup D. We have proved that K is a U2 -group with the kernel R D 1 .A/, where K=R is of maximal class. Let T =R be a cyclic subgroup of index 2 in K=R. Then jT =Rj 4 and 1 .T / D R. Let T1 =R be the cyclic subgroup of order 4 in T =R. Then T1 is an L2 -subgroup of K of order 24 and so T1 D D. If T1 ¤ T , then CT .R/ T1 and so T1 D D would be abelian, a contradiction. Hence T1 D T D D and so jKj D 25 . Since jK W C j D 2, C is of order 24 . But A Z.C / and so C is abelian. Assume that E0 D 1 .C / > 1 .A/ so that E0 is a normal elementary abelian subgroup of order 8 in G. Consider the subgroup X0 D E0 hsi, where E0 \ hsi D hs 4 i is of order 2. We see (as in the proof of Theorem 54.1) that c3 .X0 / D 4. This implies X0 X, a contradiction. We have proved that C is of rank 2 and so C Š C4 C4 (since exp.C / D 4). Our proposition is proved. Next we shall classify 2-groups G appearing in Theorems 54.1 and 54.2. Theorem 54.6. Let G be a 2-group with cn .G/ D 4, n > 2, and suppose that G has a normal elementary abelian subgroup E of order 8. Let fU1 ; U2 ; U3 ; U4 g be the set of four cyclic subgroups of order 2n . Then X D hU1 ; U2 ; U3 ; U4 i D EU1 , where E \ U1 ¤ 1 and we have the following possibilities. (a) 2 .X/ Š C4 C2 C2 and G=E is either cyclic (of order 2n1 ) or G=E is of maximal class (and order 2n ) and for n D 3 the last group must be dihedral. (b) n D 3, 2 .X/ Š D8 C2 , and one of the following holds: (b1) G D X, which is a uniquely determined group of order 25 ; (b2) G has a self-centralizing abelian normal subgroup A of type .4; 2/ such that G=A Š D8 ; (b3) G has a self-centralizing abelian normal subgroup W of type .4; 4/ such that G=W Š E4 or D8 . Proof. Theorem 54.1 implies that X D EU1 with E \ U1 ¤ 1 and 2 .X/ Š C4 C2 C2 or 2 .X/ Š D8 C2 , where in the second case n D 3. Suppose that 2 .X/ Š C4 C2 C2 . Then 1 .X/ D E Š E23 and X=E Š C2n1 with jX=Ej 4. Hence X is an L3 -group of order 2nC2 . Since X is generated by its cyclic subgroups of order 2n , it follows that X is the unique L3 -subgroup of order 2nC2 in G. By Proposition 54.3, G is either an L3 -group or a U3 -group with respect to the kernel R D 1 .X/ D E. In particular, G=E is either cyclic (of order 2n1 ) or G=E is of maximal class (and order 2n ). If in the second case n D 3, then G=E must be dihedral. Indeed, if n D 3 and G=E is of maximal class but not dihedral, then there

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is an element x 2 G X such that x 2 2 2 .X/ E, where 2 .X/=E D Z.G=E/. But then o.x/ D 8, a contradiction. Assume now that 2 .X/ Š D8 C2 in which case n D 3. Our result follows at once from Theorem 54.5. Theorem 54.7. Let G be a 2-group with cn .G/ D 4, n > 2, and suppose that G does not have a normal elementary abelian subgroup of order 8. Let fU1 ; U2 ; U3 ; U4 g be the set of four cyclic subgroups of order 2n . Then the subgroup X D hU1 ; U2 ; U3 ; U4 i is of order 2nC2 and we have the following possibilities. (a) X D W U1 with W \ U1 Š C4 , where W is an abelian normal subgroup of type .4; 4/, 2 .X/ D W , X is metacyclic, CG .W / is metacyclic and one of the following holds: (a1) n > 3, and G=W is either cyclic (of order 2n2 ) or G=W is of maximal class (and order 2n1 ) and for n D 4 the last group must be dihedral; (a2) n D 3, G has a normal metacyclic subgroup N such that CG .W / N , X N , and G=N is isomorphic to a subgroup of D8 . Moreover, CG .W /=W is cyclic, N=W is either cyclic or generalized quaternion, and in the second case G=N is elementary abelian of order 4. (b) n > 2, X D QU1 , where Q Š Q8 is a normal quaternion subgroup of G, G D QP with Q \ P D Z.Q/, U1 P , CG .Q/ P , jP W CG .Q/j 2, and P is either cyclic or of maximal class. (c) n D 3, jXj D 25 , 2 .X/ Š Q8 C2 or 2 .X/ Š D8 C2 and if G > X, then G has a self-centralizing normal abelian subgroup W of type .4; 4/ with G=W Š E4 or D8 . Proof. We may assume that G is nonabelian. (i) We consider first the groups G from Theorem 54.2(a). In this case X D W U1 with U1 Š C2n , W \ U1 Š C4 , where W is an abelian normal subgroup of type .4; 4/ and 2 .X/ D W . Also, X and CG .W / are metacyclic. Assume that n > 3. Then X=W is cyclic of order 2n2 4. Set S=W D 1 .X=W / so that S=W ˆ.X=W / ˆ.G=W /. Since S=W stabilizes the chain W > 1 .W / > f1g, it follows (as in the proof of Theorem 54.2) that all elements in S W are of order 8. Indeed, set U1 \ S D hd i, where d is an element of order 8. For each w 2 W , w d D wl with l 2 1 .W /. Then .dw/2 D dwdw D d 2 w d w D d 2 wlw D d 2 w 2 l, where w 2 l 2 1 .W / and so o.dw/ D 8. Also, X is generated by its elements of order 2n . We want to show that G=W is either cyclic or of maximal class and we use an induction on jGj. If X D G, we are done. In the sequel let X < G. Let X M < G, where M is maximal in G. By induction, M=W is either cyclic or of maximal class. We have ˆ.G=W / M=W and since ˆ.G=W / cannot be of maximal class

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(Burnside), ˆ.G=W / is either cyclic (of order 2) or abelian of type .2; 2/ (the last case being possible only if M=W Š D8 ). However, if ˆ.G=W / Š E4 , then ˆ.G=W / Z.G=W / (see Proposition 4.9). But then M=W is abelian, a contradiction. Thus ˆ.G=W / is cyclic and so S=W D 1 .ˆ.G=W //. Let F=W be a cyclic subgroup of order 2n2 in G=W . Now, 1 .F=W / ˆ.F=W / ˆ.G=W / and so 1 .F=W / D S=W . But all elements in S W are of order 8 and so cn .F / D 4 and F is generated by its elements of order 2n . Indeed, let n3 F0 =W D ˆ.F=W / and x 2 F F0 . Then x 2 2 S W and so o.x/ D 2n . This gives F D X and so X=W is the unique cyclic subgroup of order 2n2 in G=W . As in the proof of Proposition 54.3, G=W is either cyclic (of order 2n2 ) or G=W is of maximal class (and order 2n1 ) and for n D 4 the last group (obviously) must be dihedral. It remains to consider the case n D 3. In this case jXj D 25 and all 16 elements in X W are of order 8. Suppose first that X is nonabelian. Then CG .W / \ X D W and CG .W / is metacyclic. If CG .W / > W , then CG .W / (being metacyclic) has elements of order 8 (which are not contained in X), a contradiction. We have proved that CG .W / D W . By Theorem 50.1, the group G has a normal metacyclic subgroup N such that W N , 2 .N / D W , and G=N is isomorphic to a subgroup of D8 and we assume that N is maximal subject to these conditions. In that case N > W . Indeed, if N D W , then G=W is isomorphic to a subgroup of D8 . But X is a normal metacyclic subgroup of G, W X, 2 .X/ D W , and G=X (being a factor-group of D8 ) is also isomorphic to a subgroup of D8 . This contradicts the maximality of N and so N > W . In that case there are elements of order 8 in N and so X N . Suppose that Y =W is a subgroup of order 2 in N=W . Since 2 .Y / D W , all elements in Y W are of order 8. Hence Y D X and so N=W has exactly one subgroup of order 2. By the structure of Aut.W / (Proposition 50.5), we see that Aut.W / has no quaternion subgroups. Thus N=W is cyclic (of order 4). We have obtained all properties stated in part (a2) of our theorem. We consider now the remaining case, where X D W U1 is abelian of order 25 (and of type .8; 4/). According to Theorem 50.1, G has a normal metacyclic subgroup N such that 2 .N / D W , C D CG .W / N , and G=N is isomorphic to a subgroup of D8 . Obviously, X C . Suppose that Y =W is a subgroup of order 2 in N=W . Since 2 .Y / D W , all elements in Y W are of order 8. Hence Y D X and so N=W has exactly one subgroup of order 2. It follows that N=W is either cyclic or generalized quaternion. We claim that C =W is cyclic. Suppose false. Then C =W is generalized quaternion. Let C0 =W Š Q8 be a quaternion subgroup in C =W , where X < C0 (because X=W is the unique subgroup of order 2 in N=W ). Let Z1 =W and Z2 =W be two distinct cyclic subgroups of order 4 in C0 =W . Then Z1 and Z2 are both abelian, hZ1 ; Z2 i D C0 , and Z1 \ Z2 D X. This implies that X Z.C0 / and so X D Z.C0 /. Since C0 is metacyclic, C00 is cyclic and C00 covers X=W . All elements in X W are of order 8, so jC00 j D 8. But then the formula (see Lemma 1.1) jC0 j D 27 D 2jZ.C0 /jjC00 j gives

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a contradiction. We have proved that CG .W /=W is cyclic. Thus CG .W / D CG .X/ is abelian. We have to consider the case, where N=W is generalized quaternion. We set K=W D .N=W /0 D ˆ.N=W / so that K=W is cyclic and Z.N=W / D X=W . We have N=K Š E4 and since N is metacyclic (and therefore d.N / D 2), we get K D ˆ.N / and N 0 is cyclic. On the other hand, N 0 covers K=W and all elements in X W are of order 8. Hence N 0 \ X Š C8 , N 0 \ W Š C4 , and jK W N 0 j D 4. Again, since N is metacyclic, there is a cyclic normal subgroup S of N such that N 0 < S and jS W N 0 j D 2. If S K, then jS \ W j D 8. But a maximal subgroup S \ W of W is not cyclic since S \ W ˆ.W / D 1 .W / Š E4 . This contradiction shows that S \ K D N 0 . Set T D W S D KS so that T =W is a cyclic subgroup of index 2 in N=W . For each x 2 N T , we have x 2 2 X W and so all elements in N T are of order 16. Also, T =S Š W =.S \ W / Š C4 . Now, ˆ.T / K and ˆ.T / hN 0 ; 1 .W /i since N 0 D Ã1 .S / and 1 .W / D ˆ.W /. Since jK W .N 0 1 .W //j D 2 and T is not cyclic, we get ˆ.T / D N 0 1 .W /, F D ˆ.T / \ X is abelian of type .8; 2/, and F1 D ˆ.T / \ W is abelian of type .4; 2/. m3 m2 Set S D hai, where o.a/ D 2m , m 4. Also set s D a2 , v D a2 , and m1 2 zDa , so that z 2 Z.G/, hsi D S \ X, hvi D S \ W , and hzi D S \ 1 .W /. Obviously, hsi and hvi are normal subgroups of G. Since ˆ.N / D K, there is an element b 2 N T (of order 16) such that b 2 2 X W and b 2 62 F . Then b 2 D ws, where w 2 W F1 and W D hwi hvi. We set w 2 D u and so 1 .W / D hu; zi. We compute b 4 D w 2 s 2 D uv and b 8 D z. Hence N D haihbi, where hai is a cyclic normal subgroup of order 2m .m 4/ of N and hbi is cyclic of order 16 with hai \ hbi D hzi (of order 2). It remains to determine the action of hbi on hai. Now, hbi induces a cyclic automorphism group on hai so that b inverts hai=hvi (since N=W is generalized quaternion). Thus ab D a1 v0 with v0 2 hvi and so .a4 /b D .ab /4 D .a1 v0 /4 D a4 . In particular, b inverts hvi. On the other hand, b centralizes b 4 D uv. We compute uv D .uv/b D ub v b D ub v 1 , and so ub D uz. The last relation is crucial and shows that b does not stabilize the chain W > 1 .W / > f1g. If A (of order 25 ) is a Sylow 2-subgroup of Aut.W / and B is the full stabilizer group of the chain W > 1 .W / > f1g, then B is elementary abelian of order 24 and jA=Bj D 2 (see Proposition 50.5). We note that CG .W / N and CG .W /=W is a cyclic normal subgroup of N=W and so CG .W / T . Hence, if L is the set of all y 2 G which stabilize the chain W > 1 .W / > f1g, then the subgroup L covers G=N (since b 62 L) and so G=N (being isomorphic to a subgroup of D8 ) is elementary abelian of order 4. We have obtained all properties stated in part (a2) of our theorem. (ii) We consider now the groups G from Theorem 54.2(b). In this case n > 2, X D QU1 , where Q Š Q8 is a normal quaternion subgroup of X, Q \ U1 D Z.Q/, U1 D hbi Š C2n , and b either centralizes Q or b induces on Q an involutory outer automorphism in which case n > 3.

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Assume first that b centralizes Q. In that case X D Q hbi with Q \ hbi D n2 Z.Q/ D hzi. Set v D b 2 so that 2 .X/ D Q hvi. Since Q is the unique quaternion subgroup in 2 .X/, it follows that Q is characteristic in X and so Q is a normal subgroup in G. Set C D CG .Q/ so that X \ C D U1 D hbi, C is normal in G, and jG W .Q C /j 2. Since U1 is the unique cyclic subgroup of order 2n in C , it follows that C is either cyclic or of maximal class. If G D Q C , we are done. Suppose that jG W .Q C /j D 2. Since G=C Š D8 (a Sylow 2-subgroup of Aut.Q8 //, there is an element d 2 G .QC / with d 2 2 C . Set P D C hd i so that G D QP and P \ X D U1 . Since U1 is the unique cyclic subgroup of order 2n in P , we get again that P is either cyclic or of maximal class. It remains to consider the case, where b induces on Q an outer involutory automorphism (and in that case n > 3). Obviously, Q is the unique quaternion subgroup of Qhb 2 i D Q hb 2 i, where Q \ hb 2 i D hzi. We have exp.Qhb 2 i/ D 2n1 and all elements in X .Qhb 2 i/ are of order 2n . Hence Q is also the unique quaternion subgroup of X and so Q is normal in G. Again set C D CG .Q/ so that C is normal in G and X \ C D hb 2 i. Also, jG W .QC /j D 2 and so G D XC . Set P D C hbi. We have jP W C j D 2, P \ X D hbi D U1 , and G D QP with Q \ P D hzi. Since U1 is the unique cyclic subgroup of order 2n in P , it follows that P is either cyclic or of maximal class and we are done. (iii) Finally, we consider the groups G from Theorem 54.2(c). In this case n D 3, jXj D 25 , and 2 .X/ Š Q8 C2 or 2 .X/ Š D8 C2 . Since G has no normal subgroups isomorphic to E8 , Proposition 54.5 gives at once the result stated in part (c) of our theorem.

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2-groups G with small subgroup hx 2 G j o.x/ D 2ni

In 52, 2-groups G with j2 .G/j D 16 are classified. In this section we consider essentially more difficult problem to classify the 2-groups in which all elements of order 4 generate the subgroup of order 16. In what follows we suppose that G is a 2-group of order > 24 all of whose elements of order 4 generate a subgroup H of order 24 . In that case, obviously, H is not of maximal class so it has no cyclic subgroups of index 2 (see Theorem 1.2). It is obvious that, in the case under consideration, c2 .G/ 4. Indeed, if the above assertion is false, then c2 .G/ D 2, by Theorem 1.17(b). In that case, by 52, the elements of order 4 generate the abelian subgroup of type .4; 2/, contrary to our hypothesis (this also follows from 43). If c2 .G/ D 4, then H Š Q8 C4 or H Š C4 C2 C2 and the group G was completely determined in 52. If c2 .G/ D 5, then G is of maximal class (Theorem 1.17(b)) and so H Š Q16 and (since jGj > 24 ) G Š SD32 . It remains to examine the case c2 .G/ D 6 since there is no 2-group G with c2 .G/ D 7 (Theorem 1.17(b)). The 2-groups G with j2 .G/j D 24 have been determined in 52, and so we may also assume j2 .G/j > 24 . Exercise 1. Let G be a 2-group, H D hx 2 G j o.x/ D 4i. If exp.Z.H // D 2, then CG .H / D Z.H /. Solution. Assume that CG .H / > Z.H /. Let x 2 CG .H / Z.H / be such that x 2 2 Z.H /. Then o.x/ 4 so x is an involution since x 62 H . If a 2 H is of order 4, then o.ax/ D 4 and ax 62 H , a contradiction. Let G be a 2-group and let 2 .G/ D hx 2 G j o.x/ D 22 i. Suppose that j2 .G/j D 24 and c2 .G/ > 4; then c2 .G/ 6 (Theorem 1.17(b)) so G contains at least twelve elements of order 4. Then c2 .G/ D 6 since c1 .G/ 3. Theorem 55.1 ([Jan8]). Let G be a 2-group of order > 24 all of whose elements of order 4 generate a subgroup H of order 24 , i.e., H D 2 .G/. Assume, in addition, that c2 .G/ D 6 and j2 .G/j > 24 . Then we have the following possibilities: (a) H Š Q8 C2 and G Š SD16 C2 . (b) H Š ha; b j a4 D b 4 D 1; ab D a1 i is metacyclic and G D hb; t j b 4 D t 2 D 1; b t D ab; a4 D 1; ab D a1 ; at D a1 i. Here jGj D 25 , H D ha; bi, ˆ.G/ D ha; b 2 i Š C4 C2 , 2 .G/ D G, Z.G/ D ha2 ; b 2 i Š E4 .

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(c) H Š C4 C4 and G has a metacyclic maximal subgroup M such that 2 .M / D H , G D M ht i, where t is an involution with CM .t / D 1 .M / Š E4 . Proof. Since H is neither cyclic nor of maximal class, H has a normal four-subgroup H0 (Lemma 1.4) and so all twelve elements in H H0 are of order 4. If H is abelian, then H0 D 1 .H / implies that H D C4 C4 . Suppose that H is nonabelian. Then H=H0 is elementary abelian and so ˆ.H / H0 . Suppose, in addition, that H is not minimal nonabelian and let Q be a nonabelian subgroup of order 8. Since H has only three involutions, we have Q Š Q8 . If CH .Q/ Q, then H is of maximal class (Proposition 10.17), a contradiction. Hence H D Q Z, where jZj D 4 and Q \ Z D Z.Q/. If Z is cyclic, then c2 .H / D 4, a contradiction. Hence Z Š E4 and H Š Q8 C2 . Now suppose that H is minimal nonabelian. Then d.H / D 2 so ˆ.H / D H0 and jH 0 j D 2. In that case, H=H 0 is abelian of type .4; 2/. Let Z=H 0 be a direct factor of H=H 0 of order 2. Since H=Z is cyclic of order 4, we get Z ¤ ˆ.H / D 1 .H / so Z is cyclic. Let Z D hai and H=Z D hbZi; then hbi \ hai D f1g. Since Aut.Z/ Š C2 , we get H D ha; b j a4 D b 4 D 1; ab D a1 i. We have to determine the structure of G for each of the above three possibilities for the structure of H . (i) Assume that H D Q hui, where Q Š Q8 and u is an involution. Set hzi D Z.Q/ so that Z.H / D hu; zi. There are exactly three maximal subgroups of H containing Z.H / and they are abelian of type .4; 2/. One of them, say hu; ai (a 2 Q Z.Q/), is normal in G (consider the action of G by conjugation on the set of above three maximal subgroups of H ). If b 2 Q hai, then the other two such abelian subgroups of type .4; 2/ are hu; bi D hui hbi and hu; abi D hui habi. Since 2 .G/ > H , there is an involution t 2 G H since H contains all elements of order 4 in G. Let hu; xi, where x 2 Q Z.Q/, be any t -invariant abelian subgroup of type .4; 2/ in H . If t acts nontrivially on Z.H /, then ht; Z.H /i Š D8 has a cyclic subgroup of order 4 that is not contained in H , a contradiction. Thus t centralizes Z.H /. Since CH .t / must be elementary abelian (otherwise, .ht i CH .t // CH .t / has an element of order 4), we get CH .t / D Z.H / and so t acts fixed-point-free on hu; xi Z.H /. If x t D xs with s 2 Z.H / hzi, then .tx/2 D .txt /x D x t x D xsx D sx 2 D sz 2 Z.H /# , and so o.tx/ D 4, a contradiction since tx 62 H . Hence x t D x 1 and so t inverts the subgroup hu; xi. In particular, t inverts hu; ai. If t normalizes hu; bi, then t also normalizes hu; abi. But in that case t inverts hu; bi and hu; abi and so t inverts H D hu; ai [ hu; bi [ hu; abi. This implies that H is abelian (Burnside) which is not the case. Hence we must have hu; bit D hu; abi. Let M D NG .hu; bi/ so that H M , jG W M j D 2 and G D M ht i (see the last sentence of the previous paragraph). By the previous paragraph, the set M H has no involutions so 2 .M / D H since exp.H / D 4 and H contains all elements of order 4 from G. Assume that M > H so that we can apply the results of 52 (see Theorem 52.1(d)). It follows that M is a uniquely determined group of order 25

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which is given explicitly in Theorem 49.1, (A2)(a)). There is an element y 2 M H of order 8 (indeed, exp.M / D 8 and H D 2 .M /) such that y 2 D ua, uy D uz, ay D a1 , b y D bu. Hence y acts nontrivially on Z.H / and y normalizes all three abelian subgroups of type .4; 2/ in H . We have G=H Š E4 since M=H and H ht i=H are two distinct subgroups of order 2 in G=H . Hence .yt /2 2 H and we consider the subgroup N D H hyt i of order 25 . Since yt acts nontrivially on Z.H / (t centralizes Z.H /), it follows that there are no involutions in N H (indeed, if w is an involution in N H , then hw; Z.H /i Š D8 contains an element of order 4 which is not contained in H , a contradiction). Thus H D 2 .N / and so, by the above, N is uniquely determined and therefore N Š M . In particular, N normalizes all three abelian subgroups of type .4; 2/ in H . Since M ¤ N , we get that G D hM; N i normalizes all three abelian subgroups of type .4; 2/ in H (see the last sentence of the previous paragraph), a contradiction. We have proved that H D M and so G D H ht i. Since hu; bit D hu; abi, we have b t D abs0 with s0 2 Z.H /. Hence t normalizes Q D hb; b t i Š Q8 and H D hui Q . Replacing Q with Q , we may assume from the start that t normalizes Q and so t induces on Q an outer automorphism (indeed, hbi is not t -invariant). We get Qht i D D Š SD24 (if D 6Š SD24 , then cl.D/ D 2; in that case, hbi G D so t -invariant, which is not the case) and G D hui D. We have obtained the possibility (a). (ii) Now suppose that H D ha; b j a4 D b 4 D 1; ab D a1 i is metacyclic of order 16. Here ˆ.H / D Z.H / D ha2 ; b 2 i D 1 .H / Š E4 , H 0 D ha2 i, all elements in H Z.H / are of order 4 and H=H 0 is abelian of type .4; 2/. It follows that H is minimal nonabelian. Therefore all three maximal subgroups Wi , i D 1; 2; 3, of H are abelian of type .4; 2/: W1 D hZ.H /; ai with Ã1 .W1 / D ha2 i; W2 D hZ.H /; bi with Ã1 .W2 / D hb 2 i; W3 D hZ.H /; abi with Ã1 .W3 / D hb 2 i since .ab/2 D ab 2 ab D ab 2 a1 D b 2 . It follows that a2 b 2 is not a square in H.D W1 [ W2 [ W3 / and W1 is normal in G. (Note, however, that all involutions in a homocyclic 2-group of composite exponent are squares; see, for example, Theorem 6.1.) Hence ha2 i D H 0 , hb 2 i, and ha2 b 2 i are characteristic subgroups in H and therefore they are normal in G. This gives Z.H / Z.G/. In fact we have Z.H / D Z.G/, by Exercise 1. Since 2 .G/ > H , there is an involution t 2 G H . By Exercise 1, CH .t / D Z.H / D Z.G/. If t normalizes a maximal subgroup Wi of H , then t inverts Wi , by Exercise 1. We know that t normalizes W1 and so t inverts W1 and therefore at D a1 : If t normalizes W2 , then t also normalizes W3 and consequently t inverts H (see the previous paragraph). But then H is abelian, which is not the case. We have proved that W2t D W3 . Let K D NG .W2 / .D NG .W3 // so that H K, jG W Kj D 2, and G D Kht i (see the last sentence of the previous paragraph). But then the set K H has no involutions (otherwise, that involution inverts H so H is abelian, which is not the case) and so 2 .K/ D H . Since H is metacyclic, it follows from Theorem 41.1 and Remark 41.2 that K is also metacyclic. If K > H , then [Ber25, Remark R48.2]

55 2-groups G with small subgroup hx 2 G j o.x/ D 2n i

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implies that H Š C4 C4 , which is not the case. Hence we have K D H and so G D H ht i is of order 25 . Since W2t D W3 , we have b t 2 fab; .ab/1 ; .ab/a2 ; .ab/1 a2 g. Note that a b D a2 b and so .ab/a D a.a2 b/ D .ab/a2 ;

..ab/1 /a D ..ab/a /1 D ..ab/a2 /1 D .ab/1 a2 :

Hence, replacing t with the involution t a (if necessary), we see that we may assume that b t 2 fab; .ab/1 D .ab/b 2 D ab 1 g. If b t D ab 1 , then we replace a with a0 D ab 2 . We get b t D ab 1 D ab 3 D .ab 2 /b D a0 b, .a0 /b D .a0 /1 , .a0 /t D .a0 /1 , and see that one may assume from the start (writing again a instead of a0 ) that b t D ab. The structure of G is uniquely determined as stated in part (b). (iii) Finally, assume that H D hai hbi Š C4 C4 . Set C D CG .H /. By Exercise 1, 2 .C / D H and therefore C is metacyclic. Since, by assumption, 2 .G/ > H , there is an involution t 2 G C . Suppose that the coset C t contains an element y which is not an involution. Then o.y/ 8 and o.y 2 / 4. In particular, y o.y/=4 D d is an element of order 4 in H . We have y D ct with c 2 C . But then d D d y D d ct D d t , contrary to Exercise 1. Hence, all element in C t are involutions. This implies that t inverts C and C is abelian (Burnside). Since 2 .C / D H , we conclude that C is of rank 2. Suppose that t 0 is an involution in G .C ht i/. Then, by the above, t 0 inverts C . But then t t 0 2 G C and t t 0 centralizes C , a contradiction. Thus C ht i D 2 .G/. Set D D 2 .G/. We have CG .t / D ht i H0 D CD .t /, where H0 D 1 .H / D ha2 ; b 2 i. Indeed, if CG .t / 6 2 .G/, then there is an element y 0 2 CG .t / CD .t / such that .y 0 /2 2 CD .t / D 1 .H / so o.y 0 / 4 and y 0 2 H , which is a contradiction. The last result gives an upper bound for the order of G=C . The conjugacy class of t in G is locked in D C and so there are at most jD C j D jC j conjugates of t in G. This gives 18 jGj D jG W CG .t /j jC j and so jG=C j 8. Also note that D=C is a normal subgroup of order 2 in G=C . Suppose that the involution in D=C is a square in G=C . Then there is an element k 2 G D such that k 2 2 D C . But all elements in D C are involutions and so k is an element of order 4, a contradiction. It follows that G=C is abelian and D=C 6 ˆ.G=C /. There is a maximal subgroup M of G such that D \ M D C . Since 2 .M / D H , it follows that M is metacyclic. We have G D M ht i, where CM .t / D 1 .M / D 1 .H / Š E4 . We have obtained a group stated in part (c) of our theorem which is now completely proved. Let n > 1 and let G be a p-group. Denote n .G/ D hx 2 G j o.x/ D p n i. We have n .G/ D f1g if exp.G/ < p n . Proposition 55.2 (Berkovich). Let G be an irregular p-group. Suppose that we have j2 .G/j D p pC1 < p m D jGj. Then the following assertions hold: (a) If p D 2, then G Š SD24 and H Š Q8 (see Theorems 43.4 and 52.8). In the sequel we assume that p > 2.

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(b) If 2 .G/ D 2 .G/, then G is not of maximal class and 2 .G/ is regular (see Lemma 42.1). Moreover, G is a group of Lemma 42.1. (c) If G is of maximal class and p > 3, then 2 .G/ D G1 , where G1 is the fundamental subgroup of G. In this case every irregular maximal subgroup of G contains exactly p subgroups of order p p and exponent p, and G contains exactly p 2 such subgroups. Next, m D p C 2. (d) If p D 3 and G is of maximal class, then 2 .G/ D 2 .G1 / is metacyclic, all nonmetacyclic subgroups of order 33 are nonabelian and their number equals 3m3 . In the sequel we assume that G is not of maximal class. (e) If 2 .G/ is regular, then 2 .G/ D 2 .G/ (see (b)). In the sequel we assume that 2 .G/ < 2 .G/. (f) Suppose that p > 3. Let t be an element of order p in G 2 .G/ and set D D ht; 2 .G/i. Then d.D/ D 3, exactly p 2 maximal subgroups of D are of maximal class; other maximal subgroups of D are of exponent p. (g) All pC1 maximal subgroups of 2 .G/ are normal in G. ˆ.2 .G// is contained in a G-invariant subgroup L of order p p and exponent p such that L 6 2 .G/. Proof. (a) Let p D 2. Then c2 .G/ D c2 .2 .G// 2 f2; 3g. If c2 .G/ D 3, then G is of maximal class by Theorem 1.17(b). In that case, G Š SD24 . If c2 .G/ D 2, then such G are described in Theorems 43.4 and 52.8. In what follows we assume that p > 2. (b) The p-groups with j2 .G/j D p pC1 are classified in Lemma 42.1. In all cases, 2 .G/ is regular. (c) Let G be of maximal class and p > 3. In this case, j2 .G1 /j p pC1 since j1 .G1 /j p p1 and G=1 .G1 / has no cyclic subgroups of index p (see Theorems 9.5 and 9.6). It follows that 2 .G/ D 2 .G1 / and j2 .G1 /j D p pC1 < p 2p2 since p > 3. In that case, G1 D 2 .G1 / D 2 .G/. If M is an irregular maximal subgroup of G, then 2 .M / D M \ 2 .G/ < M and 2 .M / is the unique absolutely regular maximal subgroup of M . We conclude that M has exactly p distinct maximal subgroups of exponent p. Since ˆ.G/, the intersection of any two distinct maximal subgroups of G, is absolutely regular, all remaining assertions of (c) are true. (d) Now let G be a 3-group of maximal class and order 35 . Since, 2 .G1 / D 2 .G/ is of order 34 , we get 2 .G/ D 2 .G1 /. The remaining assertions in (d) are trivial since the number of irregular subgroups of order 34 in G equals 3m4 (Exercise 9.27) and every such subgroup contains exactly three subgroups of order 33 and exponent 3 (see the proof of (c)). All subgroups of order 33 and exponent 3 are nonabelian since m > 4 (this is an easy consequence of Theorems 9.5 and 9.6). (e) follows from Theorem 7.2. In what follows we assume that 2 .G/ < 2 .G/. (f, g) By hypothesis, G 2 .G/ has an element t of order p. Set D D ht; 2 .G/i. Assume that D is of maximal class. Let D1 be the fundamental subgroup of D (see 9, Theorems 9.5 and 9.6). Since p > 3 and D1 is absolutely regular of width p 1 and

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order p pC1 , we get 2 .D1 / D 2 .D1 / D D. It follows that c2 .G/ D c2 .2 .G// is not divisible by p p1 so G is of maximal class (Theorem 13.2(b)), contrary to the assumption. Thus, D is not of maximal class. By Theorem 12.12, D contains exactly p 2 subgroups M1 D 2 .G/; : : : ; Mp 2 of maximal class and index p. We have jMi j D p pC1 for all i . All other p C 1 maximal subgroups of D are not generated by two elements so they are regular (Theorem 12.12). Let M be one of regular maximal subgroups of D. Supposing that exp.M / D p 2 , we get 2 .M / D M 6 2 .G/, a contradiction. Thus, all regular maximal subgroups of D have exponent p. We see that N D M \ 2 .G/ is a maximal subgroup of exponent p in 2 .G/. If M1 is another maximal subgroup of exponent p in D, then N1 D M1 \ 2 .G/ is a maximal subgroup of exponent p in 2 .G/. By Theorem 12.12(c), M \ M1 D .D/ and .D/ 6 2 .G/. It follows that N ¤ N1 . We see that the numbers of subgroups of order p p and exponent p is not divisible by p. By hypothesis, 2 .G/ has at least two absolutely regular subgroups of index p. It follows that the number of absolutely regular subgroups of index p in 2 .G/ is not divisible by p. Therefore, all maximal subgroups of 2 .G/ are normal in G. Since the number of normal subgroups of order p p and exponent p in G is 1 .mod p/ (Theorem 13.5), the last assertion in (h) follows. Exercise 2. Study the irregular p-groups G D n .G/ of order p pCn , generated by elements of order p n , n > 3, p > 2. Is it true that exp.G/ D p n ? Exercise 3. Classify the metacyclic 2-groups G such that 2 .G/ D G. Exercise 4. Classify the nonmetacyclic 2-groups G with metacyclic 2 .G/. Theorem 55.3 ([Jan8, Theorem 4.2]). Let G be a 2-group with jn .G/j D 2nC2 , n > 2. Then cn .G/ D 4 or 6. If cn .G/ D 4, then such groups G are classified in 54. If cn .G/ D 6, then n D 3 and the structure of H D 3 .G/ is uniquely determined. We have H D ha; b j a8 D b 8 D 1; ab D a1 ; a4 D b 4 D zi, where jH j D 25 and E D 1 .H / D hz; a2 b 2 D ui Š E4 . If G > H , then jGj D 26 and we have for the structure of G one of the following possibilities: (a) G is the “splitting” metacyclic group G D hw; c j w 16 D c 4 D 1; w c D w 11 i, where H D hwc 1 ; w 2 i and Z.G/ D hw 8 i is of order 2. (b) G D hH; ci with c 2 D z . D 0; 1/, .bc/2 D a, and ac D a1 . Here CG .E/ D hui ha; ci, where ha; ci Š D16 or Q16 for D 0 or 1, respectively. Also, Z.G/ D hzi is of order 2. (c) G is a non-metacyclic U2 -group with the kernel E Š E4 and G=E Š SD16 . More precisely, G D hd; s j d 16 D s 2 D 1; d 4 D v; d 8 D z; d s D d 1 vu; u2 D 1; ud D du; us D uzi, where E D hu; zi and H D hsd; d 2 i with Z.H / D Z.G/ D hvui Š C4 . (This is the group (c) with n D 4 of Theorem 67.3.) All above groups are determined up to isomorphism and they exist.

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Groups of prime power order

Proof. Let G be a 2-group with jn .G/j D 2nC2 , n > 2. This implies that G is neither cyclic nor of maximal class. By Theorem 1.17(b), cn .G/ is even. If cn .G/ D 2, then Proposition 54.4 gives that jn .G/j D 2nC1 , a contradiction. If cn .G/ D 4, then such groups G have been classified in 54. Hence, we may assume that cn .G/ 6. However, if cn .G/ 8, then the number of elements of order 2n is 8'.2n / D 8 2n1 D 2nC2 , a contradiction. We assume in the sequel that cn .G/ D 6 and set n .G/ D H . We shall determine the structure of H . The number of elements of order 2n in H equals l D 6'.2n / D 6 2n1 D 3 2n D 2nC1 C 2n . Let E be a normal four-subgroup of H and let Z be a cyclic subgroup of order 2n1 in H . It is easy to see that H0 D ZE is of exponent 2n1 . Indeed, H00 < E and so H0 is of class 2 with jH00 j 2. Also, H0 is generated with elements of orders 2n1 and 2n1 4. If x; y 2 H0 and o.x/ 2n1 , n1 n1 n1 o.y/ 2n1 , then .xy/2 D x2 y 2 D 1. This shows that exp.H0 / D 2n1 . nC2 nC1 n n 2 2 D 2 , it follows that jH0 j 2n and so jH0 j D 2n Since jH j l D 2 and Z \ E ¤ f1g. All elements in H H0 are of order 2n . This implies that 2 .H / D 2 .H0 / is of order 8 and so we may use Lemma 42.1. It follows that H must be isomorphic to a group (c) of that proposition. In particular, H is a U2 -group with the kernel E, H=E is generalized quaternion, 2 .H / is abelian of type .4; 2/, and 2 .H /=E D Z.H=E/. If n > 3, then cn .G/ D cn .H / D 2, a contradiction. Hence n D 3, H=E Š Q8 , 2 .H / D H0 , and all 24 elements in H H0 are of order 8. More precisely, H D ha; b j a8 D b 8 D 1; ab D a1 ; a4 D b 4 D zi, and so jH j D 25 , Z.H / D hb 2 i Š C4 , H 0 D ha2 i Š C4 , H0 D 2 .H / D ha2 ; b 2 i, E D 1 .H / D hz; a2 b 2 D ui Š E4 , and CH .E/ D CH .H0 / D H1 D hb 2 ; ai is abelian of type .8; 2/. The following subgroups E, hzi D Ã1 .H0 /, H0 , ha2 i, hb 2 i, and H1 are characteristic in H and so they are normal in G. We assume G > H and we intend to determine the structure of G up to isomorphism. Set C D CG .E/ so that C is a maximal subgroup of G, C > H1 , G D CH D C hbi, C \ H D H1 , and c3 .C / D 2. By Proposition 54.4, C is an L2 -group or a U2 -group with kernel E. Note that U2 -groups contain a unique normal four-subgroup. If C is a U2 -group, then H0 =E is a normal subgroup of order 2 in C =E and therefore H0 =E D Z.C =E/. If C =E in this case is not dihedral, then there is an element y 2 C H1 such that y 2 2 H0 E. But then o.y/ D 8, which is a contradiction (since 3 .G/ D H ). Here we have also used the fact that H1 =E is a normal cyclic subgroup of order 4 in C =E and so H1 =E is contained in a cyclic subgroup of index 2 in C =E. We have proved that C is either an L2 -group (in which case C is abelian of type .2i ; 2/, i 4) or C is a U2 -group with C =E dihedral. In any case, C =E has the unique maximal cyclic subgroup T =E of index 2. Then T is normal in G, T is abelian of type .2j ; 2/, j 3, and T H1 (since H1 =E is the unique cyclic normal subgroup of order 4 in C =E). Since b 2 2 H1 T , it follows that in case jG=T j D 4, C =T and hbiT =T are two distinct subgroups of order 2 in G=T and so G=T Š E4 . In any case, G=T is elementary abelian of order 4. We have CH1 .b/ D hb 2 i Š C4 and so CT .b/ D hb 2 i. Indeed, if b centralizes an element

55 2-groups G with small subgroup hx 2 G j o.x/ D 2n i

129

y 2 T H1 , then o.y/ 16 and so b centralizes an element of order 8 and this one must lie in H1 , a contradiction. Set K D hbiT and assume K > H or equivalently T > H1 . We shall determine the structure of K. Suppose that there is an element k of order 16 in K T . Then k 2 2 T and o.k 2 / D 8. But CT .b/ D CT .k/ and so b centralizes an element of order 8 in T , which contradicts our last result in the previous paragraph. Hence c4 .K/ D c4 .T / D 2. Proposition 54.4 implies that K is either an L2 -group or a U2 -group with the kernel E. But H is a U2 -group with H=E Š Q8 and so K is a U2 -group. We have H0 =E D Z.K=E/ and K=E is of maximal class containing the proper subgroup H=E isomorphic to Q8 . We know that T =E is the cyclic subgroup of index 2 in K=E. If x 2 K T is such that x 2 2 H0 E, then o.x/ D 8 and so x 2 H . Hence, if s 2 K .H [ T /, then s 2 2 E and CT .s/ D hb 2 i implies s 2 2 E \ hb 2 i D hzi. This forces K=E Š SD16 and jT =Ej D 8. Note that SD16 is the unique 2-group of maximal class and order > 8 containing Q8 as its maximal generalized quaternion subgroup. Since s 62 C , it follows us D uz, where E D hz, u D a2 b 2 i. Thus Ehsi Š D8 and so (since D8 has five involutions) we may assume that s is an involution. We may set T D huihd i, where o.d / D 16 and d 8 D z. Also set d 4 D v so that o.v/ D 4 and v 2 D z. Note that E ˆ.H / ˆ.K/ and so the involution s does not normalize hd i (otherwise hd; si would be a maximal subgroup of K not containing E). On the other hand, K=E Š SD16 and so d s D d 1 ve, where e 2 E hzi. Replacing b with b 1 leads to the replacement of u D a2 b 2 with a2 b 2 D a2 b 2 z D uz and so we may assume from the start that e D u and d s D d 1 vu. It is easy to see that K is not metacyclic. We have Ã1 .T / D hd 2 i and so s normalizes hd 2 i. Indeed, .d 2 /s D .d s /2 D .d 1 vu/2 D d 2 v 2 u2 D d 2 z, and so hs; d 2 i Š SD16 and Ehs; d 2 i is a non-metacyclic maximal subgroup of K. We locate H as the subgroup of K generated by elements of order 8 in K. We set b0 D sd and compute b02 D .sds/d D d 1 vud D vu. Hence b0 and d 2 are elements of order 8, hd 2 i is normal in K, hb0 i \ hd 2 i D hzi, and so H D hsd , d 2 i with Z.H / D Z.K/ D hvui Š C4 . Suppose in addition that C > T so that C is a U2 -group with C =E Š D16 , G D KC , and G=T Š E4 . Let c 2 C T . Then c 2 2 E, c inverts T =E, c centralizes E (since C D CG .E/), and d c D d 1 f with f 2 E. We compute .d 2 /c D .d 1 f /2 D d 2 , and so v c D v 1 because v D d 4 2 hd 2 i. Hence d sc D .d 1 vu/c D df v 1 u D dvuzf and so .T hsci/=E Š M24 . On the other hand, .T hsci/ \ H D H1 and therefore c3 .T hsci/ D c3 .H1 / D 2. By Proposition 54.4, T hsci is either an L2 -group or a U2 -group with the kernel E. But .T hsci/=E Š M24 , a contradiction. We have proved that in this case G D K and so it remains to prove the existence of K. We set: d D .1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16/ .17; 21; 24; 25; 28; 20; 23; 26; 27; 29; 31; 32; 30; 18; 19; 22/;

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Groups of prime power order

and s D .2; 17/.3; 7/.4; 23/.5; 13/.6; 30/.8; 24/.10; 27/ .11; 15/.12; 19/.14; 28/.16; 31/.18; 20/.22; 32/.25; 26/: Then we see that the even permutations d and s (of degree 32) satisfy the defining relations for G D K D hd; si: d 16 D s 2 D 1; u2 D 1;

d s D d 1 vu;

d 4 D v;

d 8 D z;

ud D du;

us D uz:

Since the permutation z D d 8 is non-trivial, z 2 Z.K/, and Z.K/ Š C4 , it follows that the induced permutation representation of G is faithful and so G is realized as a subgroup of the alternating group A32 . Thus, G is the group from (c). It remains to analyze the case K D H or equivalently T D H1 . Since G > H , we have C > T and so C =E Š D8 . We know that H=E Š Q8 , G=T D G=H1 Š E4 , and jGj D 26 . Let c 2 C T . Since C =E Š D8 and T =E is the unique cyclic subgroup of order 4 in C =E, it follows c 2 2 E. Both elements b and c invert T =E and therefore bc centralizes T =E. Set N D T hbci, where .bc/2 2 T . Then N \ H D T and so c3 .N / D c3 .T / D 2. Proposition 54.4 implies that N is either an L2 -group or a U2 -group. But bc centralizes T =E and therefore N=E is abelian. This shows that N is an L2 -group and so N=E Š C23 with 1 .N / D E. Since bc 62 C , ubc D uz and so N Š M25 and .bc/2 is an element of order 8 in T D H1 . Hence .bc/2 D ai or .bc/2 D ai u, where i is an odd integer. Set a0 D ai and we see that (replacing a with a0 ) the relations for H remain unchanged: b 8 D .a0 /8 D 1;

b 4 D .a0 /4 D z;

and .a0 /b D .a0 /1 ;

except possibly .a0 /2 b 2 D uz, where u D a2 b 2 . But in that case we replace b with b 0 D b 1 and obtain the old relations for H : .b 0 /8 D .a0 /8 D 1, .b 0 /4 D .a0 /4 D z, 0 .a0 /b D .a0 /1 and .a0 /2 .b 0 /2 D .a0 /2 b 2 D .a0 /2 b 2 z D u. Hence we may write again a and b instead of a0 and b 0 and so we may assume from the start that: .bc/2 D a or .bc/2 D au, where u D a2 b 2 , and the relations for H D ha; bi remain unchanged. Assume first .bc/2 D au. Then bc centralizes au and so we get au D .au/bc D .a1 uz/c D .a1 /c uz, which gives ac D a1 z and .a2 /c D a2 . Note that c 2 C D CG .E/, where E D hu; zi and so c centralizes u D a2 b 2 . We get u D uc D .a2 b 2 /c D a2 .b 2 /c D a2 b 2 , and so .b 2 /c D zb 2 D b 2 . From .bc/2 D au follows (since c 2 2 E and c 4 D 1) bcbc D au;

cbc D b 1 au;

c 2 .c 1 bc/ D b 1 au;

b c D c 2 b 1 au;

and so noting that b 2 2 Z.H / we get .b 2 /c D c 2 b 1 auc 2 b 1 au D b 2 ; c 2 a1 uz.c 2 /b au D 1;

c 2 z.c 2 /b D 1;

c 2 .b 1 auc 2 b/au D 1;

.c 2 /b D c 2 z;

55 2-groups G with small subgroup hx 2 G j o.x/ D 2n i

131

and so c 2 2 E hzi and therefore we may set c 2 D uz . D 0; 1/ and b c D uz b 1 au. Conjugating the last relation with c 1 we get 1

b D uz .b c /1 a1 zu; bc

1

D uz C1 b 1 au;

bc

1

D uz C1 a1 b 1 u;

bc

1

D b c z:

We compute uzb 1 au D uz.b 1 aub/b 1 D uza1 uzb 1 D ua1 ub 1 D ua1 .a2 b 2 /b 1 D .ua/b D .au/b D .bc/2 b; and noting that .bc/8 D .au/4 D a4 D z we get .bc/c D b c c D uz b 1 auc D z C1 .uzb 1 au/c D z C1 .bc/2 bc D .bc/8.C1/ .bc/3 D .bc/11C8 : Since hbci\hci D f1g (which follows from .bc/8 D .au/4 D a4 D z and c 2 D uz ), we see that G D hbc; ci is a “splitting” metacyclic group with the cyclic normal 1 subgroup hbci of order 16 and a complement hci Š C4 . Since b c D b c z, we get .bc/c

1

1

D b c c D b c zc D .bc/c z D .bc/11C8 .bc/8 D .bc/3C8 : 1

If D 0, then .bc/c D .bc/11 and .bc/c D .bc/3 . If D 1, then .bc/c D .bc/3 1 and .bc/c D .bc/11 . Hence, replacing c with c 1 (if necessary), we may assume from the start that .bc/c D .bc/11 . We set bc D w and so we have obtained the unique metacyclic group G D hw; c j w 16 D c 4 D 1; w c D w 11 i, which was stated in part (a) of our theorem. This group obviously exists because w ! w 11 induces an automorphism of order 4 of the cyclic group hwi of order 16. Assume now .bc/2 D a. Then bc centralizes a and so we get a D abc D .a1 /c which implies ac D a1 and .a2 /c D a2 . Note that c 2 C D CG .E/, where E D hu; zi and so c centralizes u D a2 b 2 . We get u D uc D .a2 b 2 /c D a2 .b 2 /c D a2 b 2

and so .b 2 /c D zb 2 D b 2 :

From .bc/2 D a follows (since c 2 2 E and c 4 D 1) bcbc D a;

cbc D b 1 a;

c 2 .c 1 bc/ D b 1 a;

b c D c 2 b 1 a;

and so noting that b 2 2 Z.H / we get .b 2 /c D c 2 b 1 ac 2 b 1 a D b 2 ; c 2 a1 .c 2 /b a D 1;

c 2 .b 1 ac 2 b/a D 1; .c 2 /b D c 2 ;

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Groups of prime power order

and so c 2 2 hzi. We set c 2 D z . D 0; 1/ and have C D CG .E/ D hui ha; ci. If D 0, then ha; ci Š D16 . If D 1, then ha; ci Š Q16 . Also, Z.G/ D hzi since Z.H / D hb 2 i and .b 2 /c D b 2 . If D 0, then we set (identify): b D .1; 9; 7; 10; 5; 11; 3; 12/.2; 13; 8; 14; 6; 15; 4; 16/; c D .2; 8/.3; 7/.4; 6/.9; 13/.10; 16/.11; 15/.12; 14/: Since b 4 D z is a nontrivial permutation, we have realized the group G D hb; ci as a subgroup of S16 . If D 1, then we set b D .1; 9; 7; 10; 5; 11; 3; 12/.2; 13; 8; 17; 6; 18; 4; 19/ .14; 15; 23; 24; 28; 22; 31; 27/.16; 30; 25; 32; 29; 20; 21; 26/; c D .1; 16; 5; 29/.2; 22; 6; 15/.3; 25; 7; 21/.4; 27; 8; 24/ .9; 14; 11; 28/.10; 31; 12; 23/.13; 20; 18; 30/.17; 32; 19; 26/: Since b 4 D z is a nontrivial permutation, we have obtained our group G D hb; ci as a subgroup of A32 . We have obtained two groups stated in part (b) of our theorem which is now completely proved. In conclusion we prove the following Theorem 55.4. Let G be a p-group such that 2 .G/ is extraspecial. Then 2 .G/ D G. Proof. Suppose the theorem is false, i.e., 2 .G/ < G. We consider cases p D 2 and p > 2 separately. Case 1. Let p D 2. Set E D 2 .G/ and hzi D Z.E/ so that jEj D 22nC1 , n 1, where n is the width of E. Let F be a subgroup of G containing E such that jF W Ej D 2. To get a contradiction, one may assume, without loss of generality, that G D F . We use induction on n. Suppose that n D 1. Then E Š D8 or Q8 . If CG .E/ 6 E, then CG .E/hzi contains elements of order 4, a contradiction. Hence CG .E/ E and then G is of maximal class. But then 2 .G/ D G, a contradiction. We assume now n > 1. Since jEj 25 and exp.E/ D 4, G has no cyclic subgroups of index 2 and so G is not of maximal class. It follows that G has a normal foursubgroup R. We have R < E since 2 .G/ D E. In particular, z 2 R and we may set R D hz; ui for some involution u 2 .E hzi/ so that CG .R/ D CG .u/. Since jE W CE .R/j D 2, it follows that CG .R/ covers G=E. By the structure of E, CE .u/ D hui E0 , where E0 is extraspecial of order 22.n1/C1 . Set F0 D CG .u/ so that jF0 W .hui E0 /j D 2 and consider the factor-group F0 =hui D FN0 (bar convention), where jFN0 W EN0 j D 2 and EN0 is extraspecial of width n 1.

55 2-groups G with small subgroup hx 2 G j o.x/ D 2n i

133

By induction, there is an element x 2 F0 .hui E0 / such that o.x/ N 4. If N D 4 and so o.x/ N D 2, then x 2 2 hui and so o.x/ 4, a contradiction. Hence o.x/ x 4 2 hui but x 2 62 hui. We have x 2 2 CE .u/. If x 2 is an involution, then o.x/ D 4, a contradiction. Hence x 2 is an element of order 4 in CE .u/ and so x 4 D z, contrary to x 4 2 hui. This completes Case 1. Case 2. Now let p > 2 and let E D 2 .G/ be extraspecial. Since G > E, we have exp.E/ D p 2 . We have E D E1 Em1 Em , where, in case m > 1, E1 ; : : : ; Em1 are nonabelian of order p 3 and exponent p and Em Š Mp 3 (see 4). Then we have S D 1 .E/ D 1 .G/ D E1 : : : Em1 1 .Em /, where exp.S / D p and jE W S j D p (here we use the fact that E is regular; see Theorems 7.1 and 7.2). Set 1 .Em / D R. We have CS .R/ D E1 Em1 R D S so R D Z.S /. It follows that R is normal in G. Set C D CG .R/; then jG W C j D p, S < C and G D EC . Since S D 1 .G/ and 1 .G=S / D E=S is of order p so G=S is cyclic since p > 2 and E=C is the unique subgroup of order p in G=S , we must have E=S C =S . This is a contradiction since R 6 Z.E/. The proof is complete.

56

Theorem of Ward on quaternion-free 2-groups

In this section we present the proof of the following important result of Ward [War]: Theorem 56.1 (H. N. Ward [War]). Let G be a nonabelian quaternion-free 2-group. Then G has a characteristic subgroup of index 2. Recall that a 2-group is quaternion-free if it has no sections isomorphic to Q8 ; that property is inherited by sections. Note that the original proof of that theorem is based essentially on properties of the Burnside group B.4; 2/ (this is a finite two-generator group of exponent 4 of maximal order; by Burnside, jB.4; 2/j D 212 ). The offered proof is shorter and elementary. However, we use ideas of [War]. The offered proof is due to the second author. Let P 2 Syl2 .G/ be the kernel of a Frobenius group G. If P is nonabelian, it has a section isomorphic to Q8 . Indeed, P has no characteristic subgroups of index 2 so the result follows from Theorem 56.1. Let P 2 Syl2 .G/, where G D Sz.22mC1 / is the Suzuki simple group. Since NG .P / is a Frobenius group, P has a section isomorphic to Q8 , by what has been said. Let G be a nonabelian quaternion-free 2-group and let A Aut.G/ be of odd order. Then A has a fixed point on G=ˆ.G/ (use Maschke’s theorem). First we prove the following two auxiliary results. Lemma 56.2. In a Q8 -free 2-group X there are no elements x, y with o.x/ D 2k > 2 and o.y/ D 4 so that x y D x 1 . If D X and D Š D8 , then CX .D/ is elementary abelian. k1

Proof. If y 2 D x 2 , then we have hx; yi Š Q2kC1 . If hxi \ hyi D f1g, then we k1 have hx; yi=hx 2 y 2 i Š Q2kC1 . In both cases hx; yi is not Q8 -free, a contradiction. Suppose that D X, where D D ha; t j a4 D t 2 D 1; at D a1 i. If v is an element of order 4 in CX .D/, then o.t v/ D 4 and t v inverts a, contrary to what has just been proved. Thus, CX .D/ must be elementary abelian. Lemma 56.3. Let X be a Q8 -free 2-group with ˆ.X/ Z.X/. If a; b 2 X, o.a/ D o.b/ D 4, and Œa; b ¤ 1, then .ab/2 D 1. Proof. Set c D Œa; b so that c 2 D Œa; b2 D Œa2 ; b D 1 and therefore c is a central involution in X. Also, a2 and b 2 are central involutions in X. Set W D ha2 c; b 2 ci

56

Theorem of Ward on quaternion-free 2-groups

135

and so W Z.ha; bi/ is of order 4 and exponent 2. In particular, a; b 62 W . We compute ab D aŒa; b D ac D a1 .a2 c/, b a D bŒb; a D bc D b 1 .b 2 c/. It follows from the above equalities that .aW /b D .aW /1 and .bW /a D .bW /1 . Since ha; bi=W is Q8 -free, Lemma 56.2 implies that at least one of aW or bW is an involution. Hence a2 D b 2 c or b 2 D a2 c (indeed, W contains exactly three involutions a2 c, b 2 c, a2 b 2 and a2 ¤ a2 b 2 ¤ b 2 ). In any case a2 b 2 D c. But then .ab/2 D a2 b 2 Œb; a D c 2 D 1. and we are done. Proof of Theorem 56.1. We proceed by induction on jGj. Let K > f1g be a minimal characteristic subgroup of G. If G=K is nonabelian, it has a characteristic subgroup H=K of index 2, by induction; then H is characteristic of index 2 in G. Therefore, we may assume that K D G 0 , i.e., G 0 is the unique minimal characteristic subgroup of G; then G 0 is elementary abelian. Since G 0 \ Z.G/ is a nonidentity characteristic subgroup of G, we get G 0 Z.G/. Thus, the group G is of class 2 and so for all x; y 2 G, we have Œx 2 ; y D Œx; y2 D 1, which gives ˆ.G/ D Ã1 .G/ Z.G/; in m particular, G=Z.G/ is elementary abelian. The map x 7! x 2 is an endomorphism of G if m 2 since (1)

m

m

m

m .2m 1//=2

.xy/2 D x 2 y 2 Œy; x.2

m

m

D x2 y 2 ;

where x; y 2 G. Let exp.G/ D 2nC1 .n 1/. If n 2, G is generated by the n elements of order 2nC1 . For, if x 2 D 1 and o.y/ D 2nC1 , then, taking m D n in (1), we get o.xy/ D 2nC1 and x D .xy/y 1 is a product of two elements of order 2nC1 . If n D 1, we may also assume that G is generated by the elements of order 4. Indeed, let H D H2 .G/ be the subgroup generated by all elements of G of order 4. Since H > f1g, then H < G implies jG W H j D 2 (Straus–Szekeres). As H is characteristic, this is a contradiction. Thus, we may assume that H D G, i.e., G is generated by elements of order 4 D 2nC1 again. We prove our theorem in six steps. Step 1. Suppose that n D 1, i.e., exp.G/ D 4. Denote Z0 D 1 .Z.G//. We claim that if a1 ; a2 ; : : : ; am are elements of G of order 4 such that a1 a2 : : : am 2 Z0 , then 2 D 1. a12 a22 : : : am Indeed, let m be the smallest integer for which a1 ; : : : ; am exist satisfying the hypothesis but not the conclusion of the above statement. Then m > 1. In fact m 3, for if a1 a2 D z 2 Z0 , then a1 D a21 z so a12 D a22 z 2 D a22 and a12 a22 D 1 2 Z0 . Since Œai ; aj 2 G 0 ˆ.G/ 1 .Z.G// D Z0 , elements ai and aj commute modulo Z0 so the hypothesis a1 a2 : : : am 2 Z0 is independent of the ordering of ai ’s. We have for any i ¤ j , .ai aj /2 D ai2 aj2 Œaj ; ai and so Œai ; aj D ai2 aj2 .ai aj /2 . Suppose that Œai ; aj D 1. If, in addition, .ai aj /2 ¤ 1, then o.ai aj / D 4 and .ai aj /2 D ai2 aj2 . Then the two terms ai and aj could be combined to .ai aj / and m lowered by one, contrary to the assumption. Thus if Œai ; aj D 1, then ai2 aj2 D .ai aj /2 D 1. If Œai ; aj ¤ 1, then Lemma 56.3 implies .ai aj /2 D 1. So in any event, Œai ; aj D ai2 aj2 .

136

Groups of prime power order

Now let i; j; k be distinct. Then, by the last result, using the collecting process, we obtain, since Ã1 .G/ Z.G/, .ai aj ak /2 D ai2 aj2 ak2 Œai ; aj Œai ; ak Œaj ; ak D ai2 aj2 ak2 ai2 aj2 ai2 ak2 aj2 ak2 D ai2 aj2 ak2 : If then .ai aj ak /2 ¤ 1, ai ; aj ; ak can be combined to .ai aj ak / and m lowered by 2. Therefore ai2 aj2 ak2 D 1 for all triples i; j; k. In particular, m > 3, so all the squares ai2 coincide (for example, it follows from a12 a22 a32 D 1 D a22 a32 a42 that a12 D a42 ). But then Œai ; aj D ai2 aj2 D ai4 D 1 so the elements ai ’s commute, and, since a1 a2 : : : am 2 2 Z0 , we get 1 D .a1 a2 : : : am /2 D a12 a22 : : : am after all. Step 2. Suppose, as in Step 1, that n D 1, i.e., exp.G/ D 4. Denote again Z0 D 1 .Z.G//. We claim that there is the unique homomorphism s W g 7! g s of G into Z0 such that if o.a/ D 4, then as D a2 . In addition, x s D 1 for every x 2 Z0 . (g s is called the “artificial square”of g.) It remains to define s on involutions from G Z0 . For g 2 G, write g D a1 a2 : : : am , where o.ai / D 4 (recall that G is generated 2 by elements of order 4, by the paragraph preceding Step 1). By Step 1, a12 a22 : : : am depends only on g, not of choice of a1 ; : : : ; am . Indeed, if, in addition, g D b1 : : : b t , 1 2 b 2 : : : b 2 D 1 and then a1 : : : am b 1 D 1 2 Z0 so, by Step 1, a12 : : : am t : : : b1 t 1 2 2 2 2 s 2 a1 : : : am D b1 : : : b t , justifying our claim. If one defines g D a12 a22 : : : am , the results stated in the previous paragraph are direct consequences of Step 1. This definition also works in the case o.g/ D 4: since g s depends only on g we must have g s D g 2 . It follows that im.N/ D GN is characteristic in G. n Step 3. We define the endomorphism N of G by xN D x 2 if n > 1 and xN D x s (the artificial square, defined in Step 2) if n D 1; recall that exp.G/ D 2nC1 . It is easy to check, using (1) with m D n if n > 1 and Step 1 if n D 1 that N is a homomorphism. We observe that, if n > 1, then the kernel of N equals n .G/, which is < G, by (1) N for all N D 2. We claim that in any case Œa; b 2 ha; with m D n; in that case, exp.G/ N bi a; b 2 G. e have im.N/ Ã1 .G/ \ 1 .Z.G//. Let n D 1. We have .ab/2 D a2 b 2 Œb; a. Then, if o.a/ D o.b/ D o.ab/ D 4, N If o.ab/ < 4, then, N D aN 2 bN 2 D 1 2 ha; N bi. we get Œa; b D a2 b 2 .ab/2 D aN bab 2 2 N If o.a/ < 4, then independent of the orders of a and b, Œa; b D a b 2 ha; N bi. 2 2 N Œa; b D b .ab/ 2 ha; N bi. Similarly, the same inclusion is true, if o.b/ < 4. Thus, our claim is true for n D 1. Now let n > 1. Suppose first that aN ¤ 1, bN ¤ 1, and aN bN ¤ 1 (or, what is the same, elements a; b; ab have the same order 2nC1 D exp.G/). Then ha4 ; b 4 i Ã2 .G/ < ˆ.G/ Z.G/, exponent of ha; bi=ha4 ; b 4 i is 4, and the orders of a and b equal 4 modulo ha4 ; b 4 i. If Œa; b 62 ha4 ; b 4 i, then Lemma 56.3 implies .ab/2 2 ha4 ; b 4 i. n But taking 2n1 -th powers of the last inclusion and using (1), one obtains .ab/2 2 nC1 nC1 ha2 ; b 2 i D f1g since exp.G/ D 2nC1 , or, what is the same, aN bN D 1, contrary to N (recall the assumption. Therefore, Œa; b 2 ha4 ; b 4 i and since 1 .ha4 ; b 4 i/ D ha; N bi 0 N N D 2) and G is elementary abelian, Œa; b 2 ha; that exp.G/ N bi.

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Next let a be an involution and bN ¤ 1 (this means that o.b/ D 2nC1 ). We may N 4 (otherwise the kernel n .G/ of N is a characteristic subgroup assume that jGj N (otherwise, of G of index 2). Hence, for some element c 2 G, we have cN 62 hbi nC1 2 2 N N G D hbi were of order 2); it follows that o.c/ D 2 . In G=hb i (b 2 Z.G/), then, some power of c leads to a central element of order 4. By the second part of Lemma 56.2, G=hb 2 i has no subgroups isomorphic to D8 . Hence, a and b commute modulo hb 2 i, and we have Œa; b 2 hb 2 i and since o.Œa; b/ 2 (recall that exp.G 0 / D 2), N Œa; b 2 hbi. Now let aN D 1, bN ¤ 1 (this means that o.a/ 2n , o.b/ D 2nC1 ), and let a have the minimal order 2m .2 m n/ among such elements for which aN D 1 N D hbi N (case m D 1 was considered in the previous paragraph). but Œa; b 62 ha; N bi m1 N 1 .ha4 ; b 4 i/, where ha4 ; b 4 i Z.G/. If Œa; b 62 ha4 ; b 4 i, then Then ha2 ; bi by Lemma 56.3, .ab/2 2 ha4 ; b 4 i and so (taking again 2n1 -th powers) we get bN D 1bN D aN bN D ab D 1, contrary to the assumption of this paragraph. Hence, since m1 N Again let o.Œa; b/ 2, we get Œa; b 2 1 .ha4 ; b 4 i/ and so Œa; b 2 ha2 ; bi. N N c 2 G be such that cN 62 hbi. Since ac D cN ¤ 1 and cb D cN b ¤ 1, we have, by N and Œc; b 2 hc; N Then Œac; b D Œa; bŒc; b implies the above, Œac; b 2 hc; N bi N bi. N Combining this with Œa; b 2 ha2m1 ; bi, N we get Œa; b 2 hbi, N which Œa; b 2 hc; N bi. m1 m1 m1 2 2 2 N is a contradiction, unless a 2 hc; N bi. But in this case a Dz for some m1 z 2 hb 2 ; c 2 i Z.G/. Then .az/2 D 1 and so o.az/ < 2m and Œa; b D Œaz; b 2 N by minimality of m. This contradiction proves that for any a; b 2 G with aN D 1 hbi, N and bN ¤ 1, we have Œa; b 2 hbi. n nC1 If aN D bN ¤ 1, then Œa; b D Œab; b with ab D aN 2 D .a2 /2 D a2 D 1, so that N by the previous paragraph. Œa; b D Œab; b 2 hbi, Finally, suppose that aN D bN D 1. Let c 2 G be any element with cN ¤ 1. Then Œa; bc 2 ha; N bci D hci N and Œa; c 2 ha; N ci N D hci, N by the above. Hence Œa; bc D N 4, there is d 2 G with dN 62 hci. Œa; bŒa; c gives Œa; b 2 hci. N Since jGj N Then, N D 2), as again, Œa; b 2 hdN i and so Œa; b 2 hdN i \ hci N D f1g (recall that exp.G/ N for all a; b 2 G. needed. All possibilities have been considered and so Œa; b 2 ha; N bi N Obviously, G 0 GN since G 0 is the unique Step 4. We claim now that G 0 D G. minimal characteristic subgroup of G and GN is a nonidentity characteristic subgroup N see Step 2). Note that if G 0 ¤ G, N then of G (this follows from the definition of G; 0 0 N N N jG W G j 4. Assume that this is false; then jG W G j D 2. We have G Š G=K with characteristic subgroup K < G, the kernel of N. Since G 0 K < G is characteristic in G, we get jG W G 0 Kj 4. It follows jGN W GN 0 j D jG=K W .G 0 K=K/j D jG W G 0 Kj 4, as claimed. N Then if aN 2 GN G 0 , we have Œa; b D 1 for all b 2 G. Suppose that G 0 < G. N are not contained in N ab, all involutions in ha; N bi, To see this, note first that if a; N b; 0 0 N G , then Œa; b 2 G \ ha; N bi (using Step 3), and that intersection equals f1g since N D f1; a; N abg. Now suppose that aN 62 G 0 but bN 2 G 0 . Find c 2 G with ha; N bi N b; N D 2). Then ac 62 G 0 , so that cN 62 hG 0 ; ai N (c exists since jGN W G 0 j 4 and exp.G/

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Groups of prime power order

Œa; c D 1 by what has just been proved. And, Œa; b D Œa; bŒa; c D Œa; bc 2 ha; N bci, N bc 62 G 0 , by the choice. It remains by Step 3. But ha; N bci \ G 0 D f1g. Indeed, a; to show, that the third involution abc N D aN bN cN of ha; N bci is not contained in G 0 . If 0 N this is false, then, since b 2 G , we get ac D aN cN D aN bN cN bN 2 G 0 , contrary to what has been proved. Hence Œa; b D 1. Finally, if aN 62 G 0 , bN 62 G 0 , and aN bN 2 G 0 , then Œa; b D Œa; ab D 1 (the second equality follows, by the previous case). Thus, Œa; b D 1 for all b 2 G. But G is generated by the elements a 2 G with aN 62 G 0 , and G would be abelian, a contradiction. Set T D hx 2 G j xN 62 G 0 i. To justify our claim, we must to prove that G D hT i. Indeed, G is generated by all elements of order 2nC1 D exp.G/. Let gN 2 G 0 with gN ¤ 1 and let aN 2 GN G 0 . Then also ga D gN aN 2 GN G 0 and gN D gaaN so that g D gaat with tN D 1. But then g D .ga/.at / and ga 62 G 0 and at D aN 62 G 0 , i.e., ga; at 2 T . Thus, G D hT i, as claimed. Thus GN D G 0 . Let K be the kernel of N . Then GN D G=K is isomorphic to G 0 and so jG 0 j D jG W Kj > 2; the last inequality holds in view of G is nonabelian and K is characteristic in G. Step 5. Suppose that jG 0 j 8. Then we obtain a characteristic subgroup of index 2 in G as follows. N D 4 and Œa; b D 1 (since exp.G/ N D 2, if such (i) There exist a; b 2 G with jha; N bij N a; b exist, then ha; N bi is a four-group). N D 4 but Œa; b ¤ 1 (we may assume that GN is For, let a and b be such that jha; N bij N and so without loss of generality we may noncyclic). By Step 3, 1 ¤ Œa; b 2 ha; N bi N then replace Œa; b with Œb; a D Œa; b1 D bN 1 D b. N take Œa; b D a. N (If Œa; b D b, N If Œa; b D aN b, then replace Œa; b with Œab; b D Œa; b D ab.) Now take c with N (such c exists in view of jGj N D jG 0 j 8). Then Œa; bc D Œa; bŒa; c D cN 62 ha; N bi aŒa; N c 2 ha; N bci \ ah N a; N ci N D f1; ag. N If Œa; bc D 1, then a and bc are the desired elements. If Œa; bc D a.D N Œa; b/, then Œa; c D 1 so a and c are the desired elements. N for all x 2 G. (ii) Suppose that aN ¤ 1, bN ¤ a, N and Œa; b D 1. Then Œb; x 2 hbi N N N bi. Assume that this intersection For, (by Step 3) Œax; b D Œx; b 2 hax; bi \ hx; N Then hax; bi N D hx; N is a four-group. Since aN ¤ 1, we get ax D bN xN so is not hbi. N bi N contrary to the assumption. Thus, our intersection is hbi, N as claimed. aN D b, (iii) Now let H be the set of elements a 2 G for which Œa; x 2 hai N for all x 2 G. We show that H is a subgroup of G of index 2; it will clearly be characteristic. If a1 ; a2 2 H , then we have to show that a1 a2 2 H . Suppose that bN 62 haN1 ; aN2 i. N \ haN1 ihaN2 i D ha1 a2 i, as needed. Since G Then Œa1 a2 ; b D Œa1 ; bŒa2 ; b 2 ha1 a2 ; bi is generated by the set of all such b’s (noting that GN D G=K is elementary abelian of order jG 0 j 8), we have Œa1 a2 ; b 2 ha1 a2 i for all b 2 G. Thus a1 a2 2 H so H is a characteristic subgroup of G. Obviously, G 0 Z.G/ H . By (i) and (ii), there is an element a 2 H with aN ¤ 1. Indeed, by (i), there exist a; b 2 G such that aN ¤ 1, bN ¤ 1, bN ¤ aN and Œa; b D 1. Then, by (ii), b 2 H # , proving our claim. Suppose for such an a that Œa; b D 1 for some b 2 G. If bN ¤ a, N

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139

then b 2 H by (ii); but if bN D a, N then ba D aN aN D 1 ¤ aN and Œa; ba D Œa; b D 1 and so by (ii) again, ba 2 H . Since a 2 H , so b 2 H in any event. We have proved that CG .a/ H . So if b 2 G H , then Œa; b ¤ 1. But then Œa; b D aN (as a 2 H and o.a/ N D 2). Hence if H ¤ G, then jG W H j D 2. (Suppose there are b1 ; b2 2 G H such that b1 b2 2 G H . Then aN D Œa; b1 b2 D Œa; b1 Œa; b2 D aN aN D 1, a contradiction.) Assume that H D G. Then G is Dedekindian so abelian since it is Q8 -free (see Theorem 1.19). Step 6. Finally suppose that jG 0 j D 4 (by the last paragraph of Step 4. jG 0 j > 2). It cannot be that K, the kernel of N , is central, for jG W Kj D 4 would then imply jG 0 j D 2, by Lemma 1.1. Thus there is c 2 K (or, what is the same, cN D 1) and a 2 G with Œa; c ¤ 1. By Step 3, Œa; c 2 ha; N ci N D ha; N 1i D hai N and so Œa; c D a. N N N Suppose that b 62 hai. N Then Œab; c D Œa; cŒb; c D aŒb; N c 2 habi \ ah N bi, by Step 3, N Since G is generated and so Œab; c D ab. Thus Œb; c D abŒa; c1 D ab aN 1 D b. by all b 2 G with bN 62 hai, N this implies Œg; c D gN for all g 2 G. It follows that jK W .K \ Z.G//j D 2. Indeed, assume that there are c1 ; c2 2 K Z.G/ such that c1 c2 2 K Z.G/ and aN ¤ 1. Then aN D Œa; c1 c2 D Œa; c1 Œa; c2 D aN aN D 1, which is a contradiction. N D G 0 . If Œa; b ¤ 1, we may assume that Œa; b D Let a; b 2 G be such that ha; N bi N then take the elements ab; b so that Œab; b D Œa; b D ab. If a. N (If Œa; b D aN b, N Then Œa; b D b, then take the elements b; a so that Œb; a D Œa; b1 D bN 1 D b.) Œa; bc D Œa; bŒa; c D aN aN D 1 and jha; N bcij D 4 (with c 2 K Z.G/ as in the previous paragraph). Thus we may assume from the start that Œa; b D 1. Now, G D ha; biK D ha; b; ciZ.G/. By the previous paragraph, jK W .K \ Z.G//j D 2. Therefore, Z.G/ < K because otherwise jG W Z.G/j D 4 and then jGj D 2jZ.G/jjG 0 j (Lemma 1.1) would imply jG 0 j D 2, a contradiction (see the last paragraph of Step 4). Hence we have Z.G/ < K and since jK W Z.G/j D 2, it follows that jG W Z.G/j D 8. The fact Œa; b D 1 implies that Z.G/ha; bi is an abelian maximal subgroup of G. Since jG W Z.G/j D 8, we see that Z.G/ha; bi is the unique abelian maximal subgroup of G and therefore it is a characteristic subgroup of index 2. The proof is complete.

57

Nonabelian 2-groups all of whose minimal nonabelian subgroups are isomorphic and have exponent 4

By Proposition 10.28, a nonabelian p-group is generated by its minimal nonabelian subgroups. Therefore it is natural to try to determine the structure of a p-group if the structure of its minimal nonabelian subgroups is known. Here we classify the title groups and note that there are exactly five minimal nonabelian 2-groups of exponent 4 (see Lemma 65.1). All results of this section are due to the second author. Recall that there are exactly five minimal nonabelian 2-groups of exponent 4: D8 , Q8 , H2 D ha; b j a4 D b 4 D 1; ab D a1 i, H16 D ha; t j a4 D t 2 D 1; Œa; t D z; z 2 D Œa; z D Œt; z D 1i and H32 D ha; b j a4 D b 4 D 1; Œa; b D z; z 2 D Œa; z D Œb; z D 1i. It is proved in Appendix 17 (Corollary A.17.3) that, if G is a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to Q8 , then G D Q V , where Q Š Q2n , n 3, and exp.V / 2. In Theorem 10.33 was proved that if G is a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to D8 , then G is generalized dihedral. Here we consider the other three minimal nonabelian 2-groups of exponent 4 and first prove the following two key lemmas. Lemma 57.1. Let G be a nonabelian p-group and let A be a maximal abelian normal subgroup of G. Then for any x 2 G A, there is a 2 A such that Œa; x ¤ 1, Œa; xp D 1, and Œa; x; x D 1 which implies that ha; xi is minimal nonabelian. Therefore, G is generated by its minimal nonabelian subgroups. Proof. Since CG .A/ D A, we have CA .x/ ¤ A and therefore hxiCA .x/ is a proper abelian subgroup of hxiA. Let B be a subgroup of hxiA containing hxiCA .x/ as a subgroup of index p. Then j.A \ B/ W CA .x/j D p, CA .x/ Z.B/ and B 0 CA .x/. Let a 2 .A \ B/ CA .x/ so that ap 2 CA .x/. We get 1 D Œap ; x D Œa; xp . On the other hand, Œa; x 2 CA .x/ and so Œa; x; x D 1. We get ha; xi0 D hŒa; xi and so, by Lemma 65.2(a), ha; xi is minimal nonabelian. Lemma 57.2. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are of exponent 4 and let A be a maximal normal abelian subgroup of G. Then all elements in G A are of order 4 and so either exp.A/ D 2 or exp.A/ D exp.G/.

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If x 2 G A with x 2 2 A, then x inverts each element in Ã1 .A/ and in A=1 .A/. If exp.G/ > 4, then either G=A is cyclic of order 4 or G=A Š Q8 . Proof. By Lemma 57.1, all elements in G A are of order 4. Thus, exp.A/ D 2 or exp.A/ D exp.G/. Let x 2 G A with x 2 2 A. Then for each a 2 A we 2 have .aax /x D ax ax D ax a D aax and so aax D w with w 2 CA .x/ and ax D a1 w. We compute .xa/2 D xaxa D x 2 ax a D x 2 w, where o.x 2 / 2. Since o.xa/ 4, we have o.w/ 2 and so x inverts each element of A=1 .A/. Finally, .a2 /x D .ax /2 D .a1 w/2 D a2 and so x inverts each element of Ã1 .A/. If exp .A/ > 4, then G=A does not possess a four-subgroup and so G=A is either cyclic of order at most 4 or G=A Š Q8 . Theorem 57.3. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to H2 D ha; b j a4 D b 4 D 1; ab D a1 i. Then the following holds: (a) If G is of exponent 8, then G has a unique abelian maximal subgroup A. We have exp.A/ 8 and E D 1 .A/ D 1 .G/ D Z.G/ is of order 4. All elements in G A are of order 4 and if v is one of them, then CA .v/ D E and v inverts ˆ.A/ and A=E. (b) If G is of exponent 4, then G D K V , where exp.V / 2 and for the group K we have one of the following possibilities: (b1) K Š H2 is of order 24 ; (b2) K is the minimal nonmetacyclic group of order 25 (see Theorem 66.1(d)); (b3) K is a unique special group of order 26 with Z.K/ Š E4 in which every maximal subgroup is minimal nonmetacyclic of order 25 (from (b2)): K D ha; b; c; d j a4 D b 4 D 1; c 2 D a2 b 2 ; Œa; b D 1; ac D a1 ; b c D a2 b 1 ; d 2 D a2 ; ad D a1 b 2 ; b d D b 1 ; Œc; d D 1iI (b4) K is a splitting extension of B D B1 Bm , m 2, with a cyclic group hbi of order 4, where Bi Š C4 , i D 1; 2; : : : ; m, and b inverts each element of B (and b 2 centralizes B). Proof. Since D8 is not a subgroup of G, 1 .G/ is elementary abelian of order > 2 (because G is not generalized quaternion). Let x 2 G 1 .G/ with o.x/ D 4 and assume that there is a 2 1 .G/ such that Œa; x ¤ 1. Then Œa; x is an involution and we have 1 D Œa; x 2 D Œa; xŒa; xx so that Œa; xx D Œa; x and ha; xi is minimal nonabelian, a contradiction. Here we have used the fact that 1 .B/ Z.B/ for each minimal nonabelian subgroup B G, where B Š H2 . Since 2 .G/ D G, we have proved that 1 .G/ Z.G/. Hence, if A is a maximal normal abelian subgroup of G, then 1 .G/ D 1 .A/ Z.G/, G=A is elementary abelian, 1 .A/ < A, and G A consists of elements of order 4.

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Groups of prime power order

Suppose that there is an element v of order 4 such that CG .v/ is nonabelian and let H D ha; b j a4 D b 4 D 1; ab D a1 i be a minimal nonabelian subgroup of CG .v/, where we set a2 D z, b 2 D u. First suppose that hvi \ H D f1g. Then .av/2 D a2 v 2 D zv 2 62 H , .av/b D ab v D a1 v D .av/z, so that hav; bi is the nonmetacyclic minimal nonabelian group of order 25 and exponent 4, a contradiction. Hence, v 62 H but v 2 2 Z.H / and so v 2 2 fz; u; uzg. If v 2 D z, then i D va is an involution, i 62 H and i b D iz and so i 62 Z.G/, a contradiction. If v 2 D u, then j D vb is an involution, j 62 H and aj D avb D ab D a1 so that j 62 Z.G/, a contradiction. If v 2 D uz, then .vb/2 D v 2 b 2 D .uz/u D z, vb 62 H , and avb D ab D a1 so that hvb; ai Š Q8 , a contradiction. We have proved that the centralizer of each element of order 4 is abelian. In particular, for each x 2 G A, CA .x/ D 1 .A/ and 1 .A/ D Z.G/. If exp.A/ > 4, then Lemma 57.2 implies that jG W Aj D 2 and we have obtained groups in part (a) of our theorem. Indeed, since x 2 G A inverts A=1 .A/ and exp.A/ 8, we have jG 0 j > 2 and so, using Lemma 1.1, we see that A is a unique abelian maximal subgroup of G (otherwise, jG W Z.G/j D 4). From now on we assume that exp.G/ D 4. In that case ˆ.G/ 1 .A/ D Z.G/ and jˆ.G/j 4 (since jˆ.H2 /j D 4). If x; y are elements of order 4 in G with Œx; y ¤ 1, then Œx; y is an involution in Z.G/ and so hx; yi is minimal nonabelian and hx; yi Š H2 implies that in case x 2 ¤ y 2 we have y x 2 fy 1 ; yx 2 g. Considering G=ˆ.G/, we get G D K V , where exp.V / 2 and 1 .K/ D Z.K/ D ˆ.K/ D ˆ.G/. It is easy to determine the structure of G in case jˆ.G/j D 4. Our group K has exactly three involutions and Z.K/ Š E4 is noncyclic. By the results stated in the introduction to 82, K has a metacyclic normal subgroup M such that K=M is elementary abelian of order 4. Since in our case K is of exponent 4, we have jM j 24 and so jKj 26 . If jKj 24 , then K Š H2 . Suppose that jKj D 25 . In this case K is nonmetacyclic (since exp.K/ D 4). If K is not minimal nonmetacyclic, then Theorem 66.1 implies that K must possess a subgroup which is isomorphic to E8 or Q8 . This is not the case and so K is minimal nonmetacyclic of order 25 . Finally, assume that jKj D 26 . Each maximal subgroup of K is minimal nonmetacyclic of order 25 and such a group K is unique according to [CIS, Theorem 2 and Remarks] so K is special with Z.K/ Š E4 . We have obtained the groups stated in parts (b1), (b2) and (b3). It remains to treat the case jˆ.G/j > 4. We note that jA W 1 .A/j D jÃ1 .A/j and let x 2 G A. Consider any element y 2 A with y 2 62 hx 2 i and suppose that y x ¤ y 1 so that y x D yx 2 . Assume that there is v 2 A with v 2 62 hx 2 ; y 2 i. Since y x D y vx (and .vx/2 2 fv 2 ; x 2 g and so .vx/2 ¤ y 2 ), we get x 2 D .vx/2 which gives v x D v 1 . Since .vy/2 62 hx 2 ; y 2 i, we also get .vy/x D .vy/1 . Hence, y x D y 1 for all y 2 A with y 2 62 hx 2 i and so x inverts A. (If z 2 A with z 2 D x 2 , then x inverts zy and so x also inverts z. But in that case hz; xi Š Q8 which cannot happen.) We have proved that in case jA W 1 .A/j 8, each element x 2 G A inverts A and so jG W Aj D 2 and G is as in (b4).

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Minimal nonabelian subgroups are of exponent 4

143

Assume jA W 1 .A/j D jÃ1 .A/j D 4 so that A D hy; zi1 .A/ and Ã1 .A/ D hy 2 ; z 2 i. Since jˆ.G/j > 4, there is u 2 G A such that u2 62 hy 2 ; z 2 i. By the arguments in the previous paragraph, u inverts A, and so Ahui is as in (b4). Suppose that Ahui ¤ G. Then there is x 2 G .Ahui/ such that Ã1 .Ahxi/ D Ã1 .A/ and we may assume that x 2 D z 2 . If y x D y 1 , then xu 2 G A and xu centralizes y, a contradiction. It follows that y x D yx 2 . From y x D y zx and x 2 ¤ .zx/2 (noting that hz; xi Š H2 ) follows that y 2 D .zx/2 D Œz; x. We compute .yz/x D yx 2 zŒz; x D yx 2 zy 2 D y 1 z 1 D .yz/1 . But then xu centralizes yz, a contradiction. Hence, we must have G D Ahui. Finally, suppose jA W 1 .A/j D 2 so that A D hyi1 .A/. Let x 2 G A, where we may assume that x 2 ¤ y 2 . (Indeed, we have hx; yi Š H2 and so if x 2 D y 2 , then we replace x with x 0 D xy, where .x 0 /2 ¤ y 2 . We may assume that y x D yx 2 (otherwise, if each element in G A inverts A, then jG W Aj D 2 and G would be as in (b1)). Let u 2 G hA; xi with u2 62 hx 2 ; y 2 i. Then we have y u 2 fy 1 ; yu2 g. First suppose that y u D y 1 . Then y xu D .yx 2 /u D y.y 2 x 2 /. But .xu/2 is equal either x 2 u2 (if Œu; x D 1) or u2 or x 2 (if Œu; x ¤ 1 and so hu; xi Š H2 ) and therefore in any case .xu/2 ¤ y 2 . Since hy; xui Š H2 , we get y xu D yy 2 or y xu D y.xu/2 . Since y xu D y.y 2 x 2 /, the only possibility is .xu/2 D y 2 x 2 which is a contradiction. Hence, we must have the second possibility y u D yu2 and we get y xu D .yx 2 /u D y.u2 x 2 / and therefore (because u2 x 2 D y 2 is excluded) x 2 u2 D .xu/2 which implies Œu; x D 1. In that case A1 D hx; ui1 .A/ is an abelian normal subgroup of G with jA1 W 1 .A/j D 4 which leads to one of the former cases, when we consider a maximal normal abelian subgroup A (instead of A) containing A1 . Our theorem is proved. Theorem 57.4. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to H16 D ha; t j a4 D t 2 D 1; Œa; t D z; z 2 D Œa; z D Œt; z D 1i. Then 1 .G/ is a self-centralizing elementary abelian subgroup and G is of exponent 4. Moreover, the centralizer of any element of order 4 is abelian of type .4; 2; : : : ; 2/. Proof. Since D8 is not a subgroup of G, 1 .G/ is elementary abelian (of order > 2). We choose a maximal normal abelian subgroup A of G so that 1 .G/ A. Using Lemma 57.2, we see that all elements in GA are of order 4 which implies ˆ.G/ A. By Lemma 57.1, we have also CG .1 .A// D A. Let v be an element of order 4 such that CG .v/ is nonabelian and let H D ha; t j 4 a D t 2 D 1; Œa; t D z; z 2 D Œa; z D Œt; z D 1i be a minimal nonabelian subgroup of CG .v/. Assume that hvi \ H D f1g so that hH; vi D H hvi is of order 26 . Since .vt /2 D v 2 , it follows that hvt i \ H D f1g and hvt i normalizes H . Consider the subgroup ha; vt i. Since Œa; vt D Œa; t D z and z commutes with a and vt , we have ha; vt i0 D hzi and so S D ha; vt i is minimal nonabelian of order 25 (since S \ H D ha; zi Š C4 C2 and S covers hH; vi=H ), a contradiction. Hence we must have hvi \ H D hv 2 i, where v 2 2 Z.H / and so v 2 2 fz; a2 ; a2 zg. Suppose

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Groups of prime power order

that v 2 D z so that .vt /2 D z and Œa; vt D z which implies that ha; vt i is metacyclic minimal nonabelian of order 24 , a contradiction. Suppose that v 2 D a2 so that va is an involution and Œva; t D z ¤ 1, a contradiction. Similarly, if v 2 D a2 z, then at v is an involution and Œat v; t D z ¤ 1, a contradiction. We have proved that for each element v of order 4, CG .v/ is abelian. Let x; y 2 G be such that x 2 G A, Œx; y 2 D 1 D Œx 2 ; y and Œx; y ¤ 1. Then hx; yi Š H16 and so either y or xy is an involution. Indeed, c D Œx; y 2 A since G=A is abelian. We get 1 D Œx; y 2 D Œx; yŒx; yy , 1 D Œx 2 ; y D Œx; yx Œx; y, and so c x D c 1 , c y D c 1 . Suppose o.c/ > 2 and let c0 be an element of order 4 in hci so that c0x D .c0 /1 D c0 c02 , hc0 ; xi0 D hc02 i and therefore hc0 ; xi is a metacyclic minimal nonabelian subgroup, a contradiction. Hence, c D Œx; y is an involution commuting with x and y so that hx; yi is a minimal nonabelian subgroup isomorphic to H16 . Suppose that A ¤ 1 .A/. Take an element x 2 G A and an element y of order 4 in A. Since CG .y/ is abelian, Œx; y ¤ 1. By Lemma 57.2, x inverts each element in A=1 .A/ and so Œx; y 2 1 .A/. We have Œx 2 ; y D 1 and Œx; y 2 D Œx; yŒx; yy D Œx; y2 D 1. By the above, hx; yi Š H16 and so xy must be an involution, a contradiction. We have proved that A D 1 .A/ and so exp.G/ D 4. For each a 2 G A, we have CG .a/ D ha; CA .a/i. Suppose that this is false. Let b 2 CG .a/ .Ahai/ so that ha; bi Š C4 C4 and ha; bi \ A D ha2 ; b 2 i Š E4 . Indeed, if a2 D b 2 , then ab 2 G A and ab is an involution, a contradiction. Set A0 D CA .a/ so that A0 ¤ A and A0 D CA .b/ (since CG .a/ and CG .b/ are abelian). Let t 2 A A0 and consider the subgroup ha; bt i. We have Œa; bt ¤ 1, o.bt / D 4 and since CA .bt / D CA .b/ D CA .a/ D A0 , we have a2 ; .bt /2 2 A0 and therefore Œa; .bt /2 D Œa2 ; bt D 1. By the above, ha; bt i Š H16 which implies that abt (2 G A) is an involution, a contradiction. Theorem 57.5. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to H32 D ha; b j a4 D b 4 D 1; Œa; b D z; z 2 D Œa; z D Œb; z D 1i. Then 1 .G/ Z.G/ and G is of exponent 4 and class 2. Proof. In exactly the same way as in the first paragraph of the proof of Theorem 57.3, we show that if A is a maximal normal abelian subgroup of G, then 1 .G/ D 1 .A/ Z.G/, 1 .A/ < A and G A consists of elements of order 4. Suppose that exp.A/ > 4. By Lemma 57.2, jG W Aj D 2 and if x 2 G A, then o.x/ D 4 and x inverts each element in Ã1 .A/. Let y 2 Ã1 .A/ with o.y/ D 4. Then y x D y 1 so that hx; yi is metacyclic minimal nonabelian (of order 8 or 16), a contradiction. Hence exp.A/ D 4 and so ˆ.G/ 1 .G/ Z.G/. In conclusion we classify the nonabelian 2-groups all of whose A1 -subgroups are isomorphic to M16 . Theorem 57.6. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to M16 D ha; t j a8 D t 2 D 1; at D a5 i. Then E D 1 .G/

57

Minimal nonabelian subgroups are of exponent 4

145

is elementary abelian and if A is a maximal normal abelian subgroup of G containing E, then A is of type .2s ; 2; : : : ; 2/, s 2, jG W Aj 4, each element x in G A is of order 8, jE W CE .x/j D 2, x inverts each element in A=E and in Ã1 .A/ and we have the following possibilities: (a) s > 2 in which case jG W Aj D 2 and G=E Š Q2s is generalized quaternion of order 2s ; (b) s D 2 in which case either G Š M16 V with exp.V / 2 or G=E Š Q8 and jG W Aj D 4. Proof. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to M16 . Since G does not possess dihedral subgroups, E D 1 .G/ is elementary abelian. Let A be a maximal normal abelian subgroup of G containing E so that E D 1 .A/. By Lemma 57.1, each element in G A is of order 4 or 8. Suppose that x 2 G A and o.x/ D 4. Then x 2 2 E and exp.Ehxi/ D 4. It follows that Ehxi is abelian. But then Ahxi is nonabelian and E Z.Ahxi/. This is a contradiction since in this case M16 cannot be a subgroup of Ahxi. We have proved that all elements in G A are of order 8 and so G=A is elementary abelian (G is metabelian) and exp.A/ 4. We want to determine the structure of A. Let x be a fixed element in G A so that o.x/ D 8, x 2 2 2 .A/ E and we set z D x 4 . By Lemma 57.1, there is a subgroup M Š M16 in Ahxi which contains hxi as a subgroup of index 2. Since all eight elements in M A are of order 8, we have M \ A D M \ 2 .A/ Š C4 C2 and M \E Š E4 . Let t be an involution .M \E/hzi so that t x D t z and M 0 D hzi. Set b D x 2 t so that b 2 D z and b x D .x 2 t /x D .x 2 t /z D bz D b 1 . It follows that each element xs with s 2 A is of order 8 and inverts the cyclic subgroup hbi of order 4. This forces hxsi \ hbi D hzi. Indeed, if hxsi \ hbi D f1g (hxsi cannot contain hbi since xs inverts b), then hb; xsi is metacyclic minimal nonabelian of order 25 , a contradiction. We compute .xs/2 D xsxs D x 2 s x s and z D .xs/4 D x 4 .s x s/2 D z.s x s/2 and so .s x s/2 D 1 and therefore s x s D w 2 E. Thus, s x D s 1 w with w 2 E which shows that x inverts each element of A=E. Further, .s 2 /x D .s x /2 D .s 1 w/2 D s 2 which shows that x also inverts each element in Ã1 .A/. Assume that there is an element a 2 2 .A/E such that a2 ¤ z. First suppose that x centralizes a. In that case, .ab/x D ab x D ab 1 D .ab/z and .ab/2 D a2 b 2 D a2 z ¤ z. Hence hab; xi with habi \ hxi D f1g is minimal nonabelian of order 25 , a contradiction. Now, suppose that ax D a with 1 ¤ 2 E so that Œa; x D . 2 But a D ax D .a/x D ax x D a x and so x D which gives hi D ha; xi0 . It follows that ha; xi is minimal nonabelian of order 25 because hai \ hxi D f1g. We have proved that 2 .A/ is of type .4; 2; : : : ; 2/ which forces that A is of type .2s ; 2; : : : ; 2/, s 2. It is easy to see that jE W CE .x/j D 2. Indeed, set E0 D CE .x/ and suppose jE W E0 j 4. Since x induces on E an involutory automorphism, we know that x centralizes E=E0 and so ŒE; x E0 and jŒE; xj D jE=E0 j, where E0 hzi. In

146

Groups of prime power order

that case, there is e 2 E E0 such that Œe; x 2 E0 hzi. Then hx; ei0 D hŒe; xi and therefore hx; ei is nonmetacyclic minimal nonabelian of order 25 , a contradiction. We have proved that CE .x/ is a hyperplane of E for each x 2 G A. First suppose that s > 2 or equivalently, exp.A/ > 4. We know that each element x 2 G A inverts Ã1 .A/ and here exp.Ã1 .A// 4. This forces jG W Aj D 2 noting that G=A is elementary abelian. Since A=E Š C2s1 and jG=Ej D 2s , s 3, G=E cannot be cyclic (otherwise, an element y 2 G A would be of order 2sC1 , contrary to the fact that all elements in G A are of order 8). On the other hand, 2 .A/=E is a unique subgroup of order 2 in G=E because 2 .A/ D 2 .G/. We have proved that G=E Š Q2s . It remains to consider the case s D 2 so that A D 2 .A/ D 2 .G/ and jA W Ej D 2 since A is of type .4; 2; : : : ; 2/. For any x 2 G A, hxi covers A=E and hx 4 i D Ã1 .A/. Let M Š M16 be a minimal nonabelian subgroup in Ahxi containing hxi, where we have again used Lemma 57.1. Since jE W CE .x/j D 2, we get Ahxi D M V with V E. If Ahxi D G, we are done. Suppose that jG W Aj 4 and we note that G=A is elementary abelian. Hence G=E is not cyclic and the fact that 2 .G/ D A shows that A=E is a unique subgroup of order 2 in G=E. It follows that G=E is generalized quaternion. But then d.G=E/ D 2 and so G=A must be elementary abelian of order 4 which forces that G=E Š Q8 and jG W Aj D 4.

58

Non-Dedekindian p-groups all of whose nonnormal subgroups of the same order are conjugate

A non-Dedekindian p-group in which all nonnormal subgroups of the same order are conjugate is termed a CO-group. The second author classified CO-groups (see Theorem 58.3); this solves Problem 1261. Below we offer another proof of that nice result which is due to the first author. A p-group G is termed a CCO-group if it is not Dedekindian and all its nonnormal cyclic subgroups of the same order are conjugate. Obviously, a CO-group is a CCO-group. Lemma 58.1. Let G D ha; bi be a minimal nonabelian p-group as in Lemma 65.1 (see also Exercise 1.8a). If all nonnormal cyclic subgroups of G of the same order are conjugate, then G Š Mp t . Proof. Write A D hai and B D hbi. Let H 6E G; then H \ G 0 D f1g and H G D H G0. m

n

Let G D ha; b j ap D b p D c p D 1; Œa; b D c; Œa; c D Œb; c D 1i and assume that m n. Let n .A/ D hxi. Then B and hbxi of the same order are neither G-invariant nor conjugate since 1 .BG / ¤ 1 .hbxiG / so G is not a CCO-group. m n m1 Let G D ha; b j ap D b p D 1; ab D a1Cp i. If n D 1, then G Š Mp mC1 satisfies the condition. Now let n > 1. If n m, set n .A/ D hxi; then B and hbxi of the same order p n are neither G-invariant nor conjugate. Now assume that n > m. Then B and habi of the same order p n are neither G-invariant nor conjugate. Indeed, m m .ab/p D b p so o.ab/ D p n , habi \ A D f1g, and we conclude that habi is not normal in G (otherwise, G D ha; abi is abelian). Since 1 .habiG / ¤ 1 .BG /, habi and B are not conjugate in G so G is not a CCO-group. Lemma 58.2. If a p-group G is a CCO-group and jG 0 j D p, then G Š Mp t . Proof. In view of Lemma 58.1, one may assume that G has a proper minimal nonabelian subgroup, say B; then B G G since B 0 D G 0 . In that case, G D B C where C D CG .B/ (Lemma 4.2) and we conclude that B is either Dedekindian or CCOgroup; in the first case B Š Q8 , in the second case B Š Mp t (Lemma 58.1). The size of every conjugate class of non-G-invariant cyclic subgroups equals p.

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Groups of prime power order

(i) Let B D hx; y j x 4 D 1; y 2 D x 2 ; x y D x 3 i Š Q8 . We get exp.C / > 2 since G is not Dedekindian. Let L D hli C be cyclic of order 4. Write H D B L; then H G G. If B \ L D Z.B/, then H has exactly 6 > 2 nonnormal subgroups of order 2 so they are not conjugate in G, a contradiction. Thus, B \ L D f1g; then H D B L has 3 > 2 distinct nonnormal cyclic subgroups hxli, hyli, hxyli of the same order 4 so they are not conjugate, a contradiction. Thus, B has no subgroups isomorphic to Q8 . t1 t2 D v p D 1; uv D u1Cp i Š Mp t . Assume that (ii) Now let B D hu; v j up X D hxi < G of order p is such that X 6 B; then X is not conjugate with hvi so X GG and F D B X has exactly p 2 > p noncentral so non-G-invariant subgroups of order p; then G is not a CCO-group, a contradiction. Thus, 1 .G/ D 1 .B/ Š Ep 2 so C has exactly one subgroup of order p; in that case C is cyclic, by (i). Then there is a cyclic W D hwi C such that W 6 B and w p 2 Z.B/. Let A < B be cyclic of index p. Then AW is noncyclic abelian with cyclic subgroup A of index p so, by basic theorem on abelian groups, AW D A Y , where jY j D p and Y 6 B has order p, contrary to what has just been proved. Theorem 58.3 (Janko). If a p-group G is a CO-group, then G Š Mp t . Proof (Berkovich). We use induction on jGj. In view of Lemma 58.2, one may assume that jG 0 j > p. By Theorem 1.23, there is K D G 0 \ Z.G/ of order p such that G=K is non-Dedekindian. Since G=K is a CO-group, we get G=K Š Mp t , by induction. Since K < G 0 ˆ.G/, we get d.G/ D d.G=K/ D 2. Let A=K and B=K be two distinct cyclic subgroups of index p in G=K; then A and B are abelian maximal subgroups of G. It follows that A \ B D Z.G/ so jG 0 j D p (Lemma 1.1), contrary to the assumption. Corollary 58.4. Suppose that all nonnormal subgroups of the same order of a nilpotent non-Dedekindian group G are conjugate. Then G D P C , where P 2 Sylp .G/ is non-Dedekindian and P Š Mp n , C is cyclic. Theorem 58.5 (Berkovich). If G is a non-Dedekindian p-group all of whose nonnormal cyclic subgroups of the same order belong to the same conjugate class of size p, then G Š Mp t . Proof. We use induction on jGj. In view of Lemma 58.1, one may assume that jG 0 j > p. Assume that G is a 2-group of maximal class. Then jGj > 8. If G Š Q2n n > 3, then G has > 2 nonnormal cyclic subgroups of order 4. If G 6Š Q2n , then G has > 2 nonnormal subgroups of order 2. Thus, G is not a 2-group of maximal class. By Theorem 1.23, there is in G 0 \ Z.G/ a subgroup R of order p such that G=R is non-Dedekindian. Let X=R < G=R be nonnormal cyclic. Assume that X is noncyclic. Then X D Xi R, where Xi is cyclic of index p in X, i D 1; : : : ; p. The subgroup Xi is nonnormal in G (otherwise, X D Xi R G G). By hypothesis, the

58

Nonnormal subgroups of the same order are conjugate

149

subgroups X1 ; : : : ; Xp form the conjugate class in G so X D X1 : : : Xp G G, a contradiction. Thus, X must be cyclic. By hypothesis, there are in G exactly p nonnormal cyclic subgroups of order jXj. Therefore, there are exactly p nonnormal cyclic subgroups of order jX=Rj in G=R so G=R satisfies the hypothesis. By induction, the non-Dedekindian group G=R Š Mp n . Then G=R has two distinct cyclic subgroups A=R and B=R of index p. In that case, A and B are two distinct abelian maximal subgroups of G so A \ B D Z.G/. Then jG 0 j D p1 jG W Z.G/j D p (Lemma 1.1), contrary to the assumption.

59

p-groups with few nonnormal subgroups

By Lemma 58.1, if all nonnormal subgroups of a minimal nonabelian p-group G are conjugate, then G Š Mp t . Of course, Theorem 59.1 follows from Theorem 58.3. Theorem 59.1 ([Sch3]). If all nonnormal subgroups of a p-group G are conjugate, then G Š Mp nC1 . Proof (compare with [Sch3]). Let H < G be nonnormal; then H is cyclic since it is not generated by their (G-invariant) maximal subgroups. Set jH j D p s . If s D 1, then G Š Mp nC1 (Theorem 1.23). Now we assume that s > 1. We use induction on jGj. We have R D 1 .H / G G. By induction, G=R Š Mp n ; then jH=Rj D p. Since R ˆ.H / < ˆ.G/, we get d.G/ D d.G=R/ D 2. Let U=R and V =R be distinct cyclic subgroups of index p in G=R; then U and V are abelian subgroups of index p in G so ˆ.G/ D U \ V D Z.G/ and G is minimal nonabelian. Then, by the remark preceding the theorem, G Š Mp nC1 . 2

m

Given m > 1, let H2;m D ha; b j ap D b p D 1; ab D a1Cp i be a metacyclic minimal nonabelian group of order p 2Cm . We suggest to the reader, using the above approach, to prove the following fairly deep Theorem 59.2 ([Sch4]). If a group G is nilpotent and has exactly two conjugate classes of non-invariant subgroups, then one of the following holds: (a) G Š D8 . (b) G Š Q16 . (c) p D 2 and G Š H2;m . (d) G Š Mp n Cq , where a prime q ¤ p. Note that the proof presented in [Sch4], is not full. For the proof of this theorem, see [Ber26].

60

The structure of the Burnside group of order 212

I. N. Sanov [Sano] has proved that each finitely generated group of exponent 4 is finite. W. Burnside [Bur4] has shown that each two-generated group of exponent 4 is a homomorphic image of a certain group G D B.4; 2/ of order at most 212 which we call the Burnside group. In this section, written by the second author, we determine completely the structure of the Burnside group G by presenting this group in a very convenient form from which each structural question about that group can be easily answered which is important for applications (Theorem D). For example, we show that ˆ.G/ D 1 .G/ is a special 2-group of order 210 with Z.ˆ.G// Š E25 . The group G has exactly 12 maximal elementary abelian subgroups. Each of them is of order 26 and the intersection of any two of them is equal Z.ˆ.G//. If M; N; K are the maximal subgroups of G, then d.M / D d.N / D d.K/ D 3, cl.M / D cl.N / D cl.K/ D 4 and M 0 D N 0 D K 0 is abelian of order 27 and type .4; 4; 2; 2; 2/. The subgroup G 0 is of order 28 and class 2, d.G 0 / D 5, 1 .G 0 / D Z.ˆ.G// D Z.G 0 /, G 00 Š E4 , G 00 Z.G/ and Z.G/ Š E8 . Next, jAut.G/j D 221 3 (Theorem C). Theorem A (see [Hup, p. 300]). Let G be a group of exponent 4 generated with an element v of order 4 and an involution t . Then G 0 D hŒv 1 ; t ; Œv; t i is abelian of order 8 and so jGj 26 . Proof. We have .t v/3 D .t v/1 D v 1 t and .t v 1 /3 D .t v 1 /1 D vt hence Œv; t Œv 1 ; t D .v 1 t vt /.vt v 1 t / D v 1 .t v/3 v 2 t D v 1 .v 1 t /v 2 t D .v 2 t /2 D Œv 2 ; t ; Œv 1 ; t Œv; t D .vt v 1 t /.v 1 t vt / D v.t v 1 /3 v 2 t D v.vt /v 2 t D .v 2 t /2 D Œv 2 ; t ; and so: (1)

Œv; t Œv 1 ; t D Œv 1 ; t Œv; t D Œv 2 ; t D .v 2 t /2 :

Hence the subgroup P D hŒv 1 ; t ; Œv; t i is abelian of exponent 4 and order 8 since Œv 1 ; t Œv; t is of order 2 and so P cannot be isomorphic to C4 C4 . Also, P G 0 and it is easy to see that P is normal in G. Indeed, from 1 D Œv i ; t 2 D

152

Groups of prime power order

Œv i ; t Œv i ; t t follows Œv i ; t t D Œv i ; t 1 for each integer i (mod 4). From 1 D Œv 1 v; t D Œv 1 ; t v Œv; t follows Œv 1 ; t v D Œv; t 1 and from Œv 2 ; t D Œv; t v Œv; t follows (using (1)) Œv; t v D Œv 2 ; t Œv; t 1 D Œv 1 ; t . Since G=P is abelian, we get G 0 D hŒv 1 ; t ; Œv; t i and so jGj 26 . Theorem B (see [Hup, p. 300]). Let G D ha; bi be a group of exponent 4 such that 1 1 1 Œa2 ; b 2 D 1. Then jGj 29 and ha2 iG D ha2 ; .a2 /b ; .a2 /b a i is elemen1 1 1 1 1 1 tary abelian of order 8 (where .a2 /b a b D a2 .a2 /b .a2 /b a ). Simi1 1 1 larly, hb 2 iG D hb 2 ; .b 2 /a ; .b 2 /a b i is elementary abelian of order 8 (where 1 1 1 1 1 1 .b 2 /a b a D b 2 .b 2 /a .b 2 /a b ). Proof. Since Œa2 ; b 2 D .a2 b 2 /2 D 1, we get a2 b 2 a2 b 2 D 1 and a2 D b 2 a2 b 2 and so ba2 b 3 D b.b 2 a2 b 2 /b 3 D b 3 a2 b:

() Now, .a2 .a2 /b

1

/2 D .a2 ba2 b 3 /2 D a2 .ba2 b 3 / .a2 ba2 b 3 / D a2 b 3 a2 b a2 ba2 b b 2 D a2 b 3 .a3 b/3 b 2 D a2 b 3 .a2 b/1 b 2 D a2 b 3 b 1 a2 b 2 D .a2 b 2 /2 D 1; 1

1 1

which implies Œa2 ; .a2 /b D 1 and so also Œa2 ; .a2 /b a D 1. Further, using (), baba D a3 b 3 a3 b 3 (which is a consequence of .ba/4 D 1) and .a3 b/3 D .a3 b/1 D b 3 a, we get: .a2 /b

1 a 1 b 1

D baba ab 3 a3 b 3 D a3 b 3 a3 b 3 ab 3 a3 b 3 D a2 .ab 3 a3 b 3 /2 D a2 ..baba3 /1 /2 D a2 .baba3 /2 D a2 baba3 baba3 D a2 b.a2 a3 /ba3 b.a3 a2 /ba3 D a2 ba2 .a3 b/3 b 3 a2 ba3 D a2 ba2 .b 3 a/b 3 a2 ba3 D a2 ba2 b 3 a.b 3 a2 b/a3 D a2 .a2 /b

1

a.ba2 b 3 /a3 D a2 .a2 /b

1 1 1

1

1

.a2 /b

1 a 1

:

1 1

D a2 .a2 /b .a2 /b a and so each element We have obtained .a2 /b a b 1 1 1 2 2 conjugate to a is contained in ha ; .a2 /b ; .a2 /b a i which gives ha2 iG D 1 1 1 1 1 1 ha2 ; .a2 /b ; .a2 /b a i. Further, a2 commutes with .a2 /b and .a2 /b a and 1 1 1 so a2 2 Z.ha2 iG /. But then .a2 /b and .a2 /b a are also contained in Z.ha2 iG / and so ha2 iG is elementary abelian of order 8. By Theorem A, G=ha2 iG is of order 26 and so jGj 29 . The second half of the theorem is obtained by interchanging a and b. Theorem C. Let G be the group given with: G D ha; b j a4 D b 4 D .ab/4 D .a2 b/4 D .ab 2 /4 D .a1 b/4 D .a2 b 2 /4 D .b a b/4 D .ab a/4 D 1i. Then G is of

60

The structure of the Burnside group of order 212

153

order 212 and exponent 4 and so G is the Burnside group. The group G has exactly k.G/ D 88 conjugate classes. The automorphism group of G is of order 221 3. Proof. This theorem (apart from the last statement) is proved by using a computer (W. Lempken at the University of Essen). We may define our group G also in the form: G D ha; b j x 4 D 1 for all x 2 Gi. Then each map ˛ of fa; bg into G such that ha˛ ; b ˛ i D G induces an automorphism of G. Hence jAut.G/j D jAut.G=ˆ.G/jjˆ.G/j2 D 6 220 . The Burnside group G is given in Theorem C in terms of generators a; b and 9 relations. However, from this presentation it is extremely difficult to pin down the structure of all subgroups and factor-groups of G. Therefore we give in the next theorem the group G in terms of 12 generators and 78 relations which will be very convenient to answer any structural question about that group. We shall also deduce many important properties of the Burnside group G by using this new presentation. In the sequel, we use from Theorem C only the fact that the Burnside group G is of order 212 . Theorem D (Complete determination of the structure of the Burnside group). Let G D ha; bi be the Burnside group of order 212 as defined in Theorem C. Then G has the following properties: (a) ˆ.G/ D 1 .G/ is a special group of order 210 with Z.ˆ.G// D .ˆ.G//0 D ˆ.ˆ.G// Š E25 . (b) If M; N; K are maximal subgroups of G, then they are of class 4 and they are generated by three elements but not by two elements. In addition, M 0 D N 0 D K 0 is abelian of order 27 and type .4; 4; 2; 2; 2/, Z.G/ Š E8 , Z.G/ D Z.M / < Z.ˆ.G//, K3 .M / Š E24 , K3 .M / < Z.ˆ.G//, K4 .M / Š E4 , K4 .M / < Z.G/, and an automorphism of order 3 of G acts transitively on fM; N; Kg. (c) The commutator group G 0 is of order 28 and class 2, G=G 0 Š C4 C4 , 1 .G 0 / D Z.ˆ.G// D Z.G 0 /, ˆ.G 0 / D Z.G/, G 00 Z.G/, G 00 Š E4 , and M 0 D K3 .G/ is an abelian maximal subgroup of G 0 . In addition, K4 .G/ D Z.ˆ.G//, E4 Š K5 .G/ D K4 .M / Z.G/, where M is maximal in G, and so G is of class 5. Finally, Z2 .G/ D Z.ˆ.G// Š E25 , Z3 .G/ D K3 .G/ D M 0 (abelian of type .4; 4; 2; 2; 2/), and Z4 .G/ D ˆ.G/. (d) The group G has exactly 12 maximal elementary abelian subgroups. Each of them is of order 26 and the intersection of any two of them is equal Z.ˆ.G// Š E25 . Hence the number of involutions in G is 31 C 12 32 D 415 and each abelian subgroup of G is of rank 6. (e) The special group ˆ.G/ D 1 .G/ is generated with five involutions d; e; m; n; y and we may set Z.ˆ.G// D hc; g; h; i; j i Š E25 so that: Œd; e D ch; Œe; n D cgh;

Œd; m D c; Œe; y D ig;

Œd; n D g;

Œd; y D i cgh;

Œe; m D h;

Œm; n D cg;

Œm; y D jc;

Œn; y D jg:

154

Groups of prime power order

We have G D ha; bi with a2 D d , b 2 D m, .a1 b 1 /2 D y, where the action of a1 and b 1 on ˆ.G/ is given with: ca

1

D g; g a

da

1

D d;

cb

1

d

b 1

1

D c;

ha

1

D cgh;

ia

1

D i ch; j a

1

D jgh;

ea

1

D ei;

ma

1

D n;

na

1

D mc;

ya

1

D de mnychij;

D h; g b

1

D cgh;

hb

1

D c;

ib

1

D igh; j b

1

D jcg;

D e;

e

b 1

D dc;

b 1

m

D m;

n

b 1

D nj;

y

b 1

D de mnygi:

We have determined completely the structure of our group G. We have here M D ˆ.G/hai, N D ˆ.G/hbi, K D ˆ.G/ha1 b 1 i, where M 0 D N 0 D K 0 D K3 .G/ D Z.ˆ.G//hde; mni, K3 .M / D hc; g; h; i i, K5 .G/ D K4 .M / D hcg; chi, G 0 D M 0 he myi, G 00 D hcg; chi, and Z.G/ D Z.M / D hcg; ch; cij i Š E8 . Also, hciG D hc; g; hi Š E8 , ha2 iG D hd; e; i; g; hi and hb 2 iG D hm; n; j; g; hi are subgroups of orders 26 which have its commutator subgroup of order 2. (f) The factor-group H D G=Z.ˆ.G// (of order 27 ) is the group of maximal possible order having the exponent 4, being generated with two elements, and having the property that the squares of any two elements commute. We have ˆ.H / D 1 .H / Š E25 , Z.H / Š E4 , Z.H / < H 0 Š E23 , and ŒH; H 0 D Z.H / so that H is of class 3. The class number k.H / of H is 26. Proof. Let G D ha; bi be the Burnside group (as defined in Theorem C) so that G is the free group of exponent 4 generated with two elements. We have jGj D 212 and we want to determine the exact structure of G. If Œa2 ; b 2 D 1, then by Theorem B we get jGj 29 , a contradiction. Hence 2 2 Œa ; b D .a2 b 2 /2 D c is an involution and ha2 ; b 2 i Š D8 with Z.ha2 ; b 2 i/ D hci so that a and b are elements of order 4 and c commutes with a2 and b 2 . 1 We first determine the structure of hciG . We set h D c b and claim that Œc; h D 1. Indeed, since b 2 D b 3 b 3 and .b 3 a2 /3 D .b 3 a2 /1 D a2 b, we get: s1 D ch D cc b

1

D .a2 b 2 /2 b.a2 b 2 /2 b 3 D a2 b 2 a2 b 2 ba2 b 2 a2 b 2 b 3

D a2 .b 3 b 3 /a2 b 3 a2 .b 3 b 3 /a2 b D a2 b 3 .b 3 a2 /.b 3 a2 /.b 3 a2 /a2 b 3 a2 b D a2 b 3 a2 b a2 b 3 a2 b D .a2 b 3 a2 b/2 and so (2)

s1 D ch D .a2 b 3 a2 b/2 D .a2 .a2 /b /2 :

Hence .ch/2 D 1 and so Œc; h D 1. Also, s1b hc D ch D s1 , and so s1 commutes with b.

1

1

D .cc b /b

1

1

2

1

D cb cb D cb c D

60

The structure of the Burnside group of order 212

155

It is possible to show that s1 also commutes with a. Using (2), a2 b 3 ab 3 a2 D a.ab 3 ab 3 ab 3 /ba D a.ab 3 /3 ba D a.ab 3 /1 ba D aba3 ba; .a3 bab/2 D .b 3 a3 b 3 a/2 D .b 3 a3 b 3 a/2 ; b 3 a3 b 3 a3 b 3 D .b 3 a3 /3 a D .b 3 a3 /1 a D aba: we get (noting that a2 D a3 a3 ): s1 s1a

1 b

D a2 b 3 a2 b a2 b 3 .a2 b b 3 a a2 /b 3 a2 b a2 b 3 a2 b a3 b D a2 b 3 a2 b.a2 b 3 ab 3 a2 /ba2 b 3 a2 ba3 b D a2 b 3 .a2 b aba3 ba b/a2 b 3 a2 ba3 b D a2 b 3 a3 .a3 bab/2 a2 b 3 a2 ba3 b D a2 b 3 a3 .b 3 a3 b 3 a b 3 a3 b 3 a/a2 b 3 a2 ba3 b D a2 .b 3 a3 b 3 a3 b 3 a b 3 a3 b 3 a3 b 3 /a2 ba3 b D a2 aba a aba a2 ba3 b D .a3 b/4 D 1: 1

Hence s1 also commutes with a1 b which implies that s1 D ch D cc b is contained in Z.G/. Applying the automorphism ˛ which interchanges a and b (noting that c ˛ D 1 1 Œb 2 ; a2 D Œa2 ; b 2 D c), we get Œc; c a D 1 and s2 D cc a 2 Z.G/. We set 1 c a D g so that hch; cgi Z.G/, where hcg; chi is elementary abelian of order 4. Since c commutes with h and g, we get ch cg D hg and so .hg/2 D 1 and therefore h also commutes with g and hc; g; hi is elementary abelian of order 8. We have hb

1

D c b D c;

ha

1

D .c ch/a

1

D c a ch D gch D cgh;

gb

1

D .c cg/b

1

D c b cg D hcg D cgh;

2

ga

1

2

D c a D c;

1

1

which implies that hc; g; hi is a normal subgroup of G and so hc; g; hi D hciG . By Theorem B, G=hciG is of order 29 . Since jGj D 212 , we have hciG Š E8 and G=hciG is of order 29 . Also, hcg; chi Š E4 and hcg; chi Z.G/. We have also determined the action of a and b on hc; g; hi Š E8 . 1 1 We determine now the structure of ha2 iG . Set a2 D d , e D .a2 /b D d b , 1 1 1 f D .a2 /b a D e a and S D hc; g; hihd; e; f i. By Theorem B, S=hc; g; hi is a normal elementary abelian subgroup of order 8 of G=hc; g; hi with ha2 iG S . By Theorem A, jG=S j 26 . Since G=hc; g; hi is of order 29 , we have S=hc; g; hi Š E8 and jG=S j D 26 . We note that hs1 ; s2 i is a maximal subgroup of hc; g; hi and hs1 ; s2 i Š Z.G/.

156

Groups of prime power order

We know that a2 D d centralizes c and so d centralizes hc; s1 ; s2 i D hciG . 1 1 From Œc; a2 D 1 follows Œc b ; .a2 /b D 1 and so Œh; e D 1 which gives that 1 1 e centralizes hh; s1 ; s2 i D hciG . From Œc; e D 1 follows Œc a ; e a D 1 and so Œg; f D 1 which gives that f centralizes hg; s1 ; s2 i D hciG . We have proved that hc; g; hi Z.S / and so S is of class 2. From (2) follows s1 D Œa2 ; .a2 /b and conjugating this relation with b 1 gives 1 s1 D Œ.a2 /b ; a2 so that (since s1 D ch is an involution) Œe; d D Œd; e D s1 D ch. 1 Conjugating the last relation with a1 we get Œe a ; d D s1 and so Œf; d D Œd; f D 1 ch. By Theorem B, f b D def l for some l 2 hc; g; hi. From Œd; f D s1 follows conjugating with b 1 : Œd b

1

;f b

1

D s1

and so

Œe; def l D Œe; d Œe; f D s1 ;

which implies s1 Œe; f D s1 and Œe; f D 1. We get Œd; ef D Œd; eŒd; f D s1 s1 D 1 and so ef 2 Z.hd; e; f i/ which gives: hd; e; f i D hef i hd; ei Š C2 D8 S D hc; g; hi hd; e; f i;

with where

hd; e; f i0 D hs1 i; hc; g; hi \ hd; e; f i D hs1 i;

so that S 0 D hs1 i and Z.S / D hc; g; h; ef i Š E24 . It remains to show that ha2 iG D S . For that purpose we have to determine exactly 1 the element f b . First we note that c commutes with a2 and b 2 . From Œa2 ; b 2 D c follows a2 b 2 a2 b 2 D c and a2 D cb 2 a2 b 2 and so ba2 b 3 D b cb 2 a2 b 2 b 3 D .bcb 3 /.b b 2 a2 b 2 b 3 / D c b

1

.b 3 a2 b/;

which gives: ba2 b 3 D h.b 3 a2 b/:

() We compute: fb

1

D .a2 /b

1 a 1 b 1

D baba2 b 3 a3 b 3

(using baba D a3 b 3 a3 b 3 which follows from .baba/.baba/ D 1) D a3 b 3 a3 b 3 a b 3 a3 b 3 D a2 .ab 3 a3 b 3 /.ab 3 a3 b 3 / D a2 .ab 3 a3 b 3 /2 D a2 .baba3 /2 D a2 .baba3 /2 D a2 baba3 baba3 D a2 b.a2 a3 /ba3 b.a3 a2 /ba3 D a2 ba2 .a3 b/.a3 b/.a3 b/b 3 a2 ba3

The structure of the Burnside group of order 212

60

157

(since .a3 b/3 D .a3 b/1 D b 3 a and using ()) D a2 ba2 b 3 a .b 3 a2 b/a3 D a2 .ba2 b 3 /a hba2 b 3 a3 D a2 .ba2 b 3 /.aha3 /.ab a2 b 3 a3 / D a2 .a2 /b

1

1

ha .a2 /b

1 a 1

1

D deha f D def cgh;

1

where we have used ha D cgh (which was proved before) and cgh 2 Z.S /. 1 1 We have obtained the desired relation f b D def cgh. We compute f a D 2 e a D e d D eŒe; d D ech and so fb

1 a 1

D .def cgh/a

1

D df .ech/gc.cgh/ D df ec D def c: 1

Now, hd D a2 ; e; f i ha2 iG and so Œd; e D ch 2 ha2 iG . From f b D def cgh 1 1 follows that cgh 2 ha2 iG and from f b a D def c follows that c 2 ha2 iG . Hence hc; g; hi ha2 iG and so S D hd; e; f ihc; g; hi D ha2 iG . Let ˛ be the automorphism of order 2 of G such that a˛ D b and b ˛ D a. Note that c ˛ D Œb 2 ; a2 D Œa2 ; b 2 D c and g ˛ D h, h˛ D g so that .hciG /˛ D hciG D hc; g; hi. Set hb 2 iG D T so that T D S ˛ , jT j D 26 , T > hc; g; hi, T =hc; g; hi Š E8 . We have U D S T D ha2 ; b 2 iG and G=U is generated with two involutions and so jG=U j 23 . Since jU j 29 , we must have jG=U j D 23 and then jU j D 29 , 1 1 1 S \ T D hc; g; hi, and G=U Š D8 . Set b 2 D m, .b 2 /a D n, .b 2 /a b D p so that m D d ˛ , n D e ˛ , p D f ˛ and ˛ transports the relations for S into the relations for T . We get hm; n; pi D hnpi hm; ni Š C2 D8 , where Œn; p D 1, Œm; n D Œm; p D s2 D cg and T D hm; n; pi hc; g; hi with hm; n; pi \ hc; g; hi D 1 hs2 i, Z.T / D hc; g; h; npi Š Z24 , T 0 D hs2 i. Also, from f b D def cgh follows 1 p a D mnpcgh. Since hc; g; hi Z.S / and hc; g; hi Z.T /, we have hc; g; hi Z.U /. But ˆ.U / S \ T D hc; g; hi (noting that U=S Š U=T Š E8 ) and so U is of class 2. We shall determine completely the structure of U D S T D ha2 ; b 2 iG . This is done by the following computation (using all the previous relations): fa

1

D .a2 /b

1 a 2

2

D e a D e d D eŒe; d D ech;

2

d m D .a2 /b D a2 c D dc; 2

f m D f b D .f b en D e

ab 2 a1

1

D .e

D .f h c h/a dn D dm

a1

/b

1

a1

1

D .def cgh/b

a2 b 2 a1

/

D .f c/a

D d ama

1

1

1

D e dc def cgh h cgh c D f h;

D .f d /ma

1

D .f ch/ma

1

D echg;

D .a2 /ma

1

D d ma

1

D .dc/a

1

D dc a

1

D dg;

158

Groups of prime power order

and b 1

dp D dn

D d bnb

D .echg h/b 2

e m D e b D .e b ep D e

bnb 1

1

f p D f bnb

/b

1

1

b 1 nb 1

D d nb

D .f

a1

D .def h/b

1

1

1 nb 1

D .eh/nb

1

D eh;

D .e m /b

D .dg/b

/

D .dc/b

D dc h cgh D dg;

a2 ma1

1 nb 1

1

2

1 nb 1

1

D .defgh/nb

D .f m /b 1

D .eh/b

1 nb 1

D ecgh;

D .ech/d ma

D f cgh;

D .f b /b

D .def cgh c/nb

1

1 nb 1

1

D .eh/a

1

1

D .dc/b

D .e /

ama1

D e ma

1

2

D .d b /b

D .ecg/b

b2

D .dc c/nb fn Df

1

1

1

D .ech c h/ma

1 nb 1

D .f h/b

1

1 nb 1

D .dg echg f cgh g h/b

1

D e dc def cgh c D f cgh:

From the above relations we get at once the desired commutators which determine completely the structure of U D S T : Œd; m D c;

Œd; n D g;

Œd; p D g;

Œe; m D h;

Œe; p D cgh;

Œf; m D h;

Œf; n D cgh;

Œf; p D cgh:

Œe; n D cgh;

In particular, we see that U 0 D ˆ.U / D hc; g; hi. We know that hc; g; hi Z.U /, ef 2 Z.S / and np 2 Z.T /. In addition, we compute (noting that U is of class 2): Œef; m D Œe; mŒf; m D h h D 1;

Œef; n D 1;

Œef; p D 1;

Œd; np D 1;

Œe; np D 1;

Œf; np D 1;

and so ef; np 2 Z.U /. Therefore Z.U / hc; g; h; ef; npi. We show that we have in fact Z.U / D hc; g; h; ef; npi Š E25 . Indeed, we set U1 D hc; g; hihd; e; m; ni, U2 D hef; npi Š E4 so that U D U1 U2 , where U10 D ˆ.U1 / D hc; g; hi Z.U1 / and we claim that hc; g; hi D Z.U1 /. Suppose that w D d ˛ e ˇ m nı 2 Z.U1 / .˛; ˇ; ; ı D 0; 1/ which implies 1 D Œw; d D Œe; d ˇ Œm; d Œn; d ı D .ch/ˇ c g ı D c ˇ C g ı hˇ and so ˇ D D ı D 0 and w D d ˛ . From 1 D Œw; e D Œd; e˛ D c ˛ h˛ follows ˛ D 0 and so w D 1. We have proved that Z.U1 / D hc; g; hi and so Z.U / D hc; g; h; ef; npi. The factor-group G=U Š D8 is generated with involutions a1 U and b 1 U so that x D a1 b 1 must be an element of order 4 with hxi \ U D f1g. Set y D x 2 so

60

The structure of the Burnside group of order 212

159

that U hyi=U D Z.G=U / D ˆ.G=U /. It follows that ˆ.G/ D U hyi. By Theorem A, the set G ˆ.G/ cannot contain involutions and so (since 1 .U / D U ) we have 1 .G/ D ˆ.G/. We have hciG D hc; g; hi D hc; s1 ; s2 i, where hs1 ; s2 i Z.G/, and so CG .c/ D CG .hc; g; hi/ which shows that CG .c/ is a normal subgroup of G. On the other hand, CG .c/ U and jG W CG .c/j D 4 (since the four conjugates of c are contained in hc; g; hi hs1 ; s2 i), which implies that CG .c/ D U hyi D ˆ.G/ and so c y D c and hc; g; hi Z.ˆ.G//. We determine now the action of G=U on U by giving the actions of the elements a1 and b 1 (we use a1 and b 1 rather than a and b because of some technical reasons) on the elements c; g; h; d; e; f; m; n; p of U . We get (by using all previous relations): ca

1

D g;

ga

1

D c a D c;

ha

1

D cgh;

da

1

D d;

ea

1

D f;

fa

1

D ..a2 /b

ma

1

D n;

na

1

D .b 2 /a D b 2 Œb 2 ; a2 D mc;

pa

1

D mnpcgh;

cb

1

D h;

gb

1

D cgh;

hb

1

D c b D c;

db

1

D e;

eb

1

D d b D d m D d Œd; m D dc;

fb

1

D def cgh;

mb

1

D m;

nb

1

D p;

pb

1

D ..b 2 /a /b D .ma /m D nm D nŒn; m D ncg:

2

1

2

/a D e d D eŒe; d D ech;

2

2

2

1

1

2

Since y D x 2 D .a1 b 1 /2 , we get from the above relations the action of y on U . We know that y centralizes hc; g; hi and we get: d y D def cgh;

e y D fg;

f y D eg;

my D mnpc;

ny D pg;

p y D ng:

Since .ef /y D ef and .np/y D np, we get that Z.U / D hc; g; h; ef; npi Z.ˆ.G//. From the above relations we also get: Œd; y D .ef /cgh;

Œe; y D .ef /g;

Œf; y D .ef /g;

Œm; y D .np/c;

Œn; y D .np/g;

Œp; y D .np/g:

Our relations also show that ˆ.G/=Z.U / is abelian and so elementary abelian. On the other hand, we see that ef; np 2 .ˆ.G//0 and so .ˆ.G//0 D Z.U /. If Z.ˆ.G// > Z.U /, then ˆ.G/ D U Z.ˆ.G// which implies that .ˆ.G//0 D U 0 D hc; g; hi, a

160

Groups of prime power order

contradiction. Thus Z.ˆ.G// D Z.U / Š E25 and so ˆ.G/ is a special group of order 210 , as claimed in part (a) of our theorem. 1 1 It remains to determine the elements y a and y b . This will complete the determination of the action of a1 and b 1 on the special group ˆ.G/. Since x D a1 b 1 and y D x 2 , we get ya

1

D .x 2 /a

1

1

D .x a /2 D .b 1 a1 /2 D b 1 a1 b 1 a1

D b 1 .a1 b 1 a1 b 1 /b D b 1 yb D y b D .y b /b 2

D .Œm; yy/b D yb

1

1

1

D ..np/cy/b

1

D .y.np/c/b

1

1

D yb

D .y m /b

1

1

p ncg h

.np/cgh;

1

and so y a D y b .np/cgh. On the other hand, x a D a1 .a1 b 1 /a D a2 b 1 a D a2 b 2 .ba/ D a2 b 2 .a1 b 1 /1 D d mx 1 , and so y a D .x 2 /a D .x a /2 D d mx 1 d mx 1 D d m.d m/x x 2 D d m.d m/x y: 1

1

Conjugating the last relation with a1 , we get d a ma .d m/a and so using our previous relations we get: d n .f mnpcgh/ y a

1

ya

D y;

1

1 b 1 a 1

ya

1

D y,

D .pnmf nd /ycgh:

Using our commutator relations in ˆ.G/, we get pnmf nd D demng.ef /.np/, and 1 1 so y a D de mnych.ef /.np/. Putting the last relation in (3), we get also y b D de mnyg.ef /. We have determined completely the structure of the Burnside group G D ha; bi. We set ef D i and np D j so that Z.ˆ.G// D hc; g; h; i; j i Š E25 and the special group ˆ.G/ is generated with five involutions d; e; m; n; y with a2 D d , b 2 D m, .a1 b 1 /2 D y and the commutator relations and the action of a1 and b 1 on ˆ.G/ as given in part (e) of our theorem. In the rest of the proof we shall use only this presentation of the group G as given in part (e) of our theorem. We show now that Z.G/ D hcg; ch; cij i Š E8 . Indeed, G=U Š D8 and Z.G=U / D U hyi=U and so Z.G/ U hyi D ˆ.G/. It follows that Z.G/ Z.ˆ.G//. On the other hand, hcg; chi Z.G/. We compute: .cij /a

1

D g i ch jgh D cij;

.cij /b

1

D h igh jcg D cij;

and so cij 2 Z.G/. But hi; j i is a complement of hcg; ch; cij i in Z.ˆ.G// and ia

1

D i ch;

ja

1

D jgh;

.ij /a

1

D ijcg;

which shows that hi; j i \ Z.G/ D f1g and therefore Z.G/ D hcg; ch; cij i Š E8 .

60

The structure of the Burnside group of order 212

161

We examine now the non-trivial cosets of Z.ˆ.G// in ˆ.G/. Each such coset consists either of involutions or only of elements of order 4. Using the defining relations for ˆ.G/, we find out that there are exactly 12 such cosets which consist of involutions and they are the cosets with representatives from the set: V D fd; e; m; n; y; de m; de n; dey; d mn; e mn; mny; de mnyg: Also we see that if v; w 2 V , v ¤ w, then o.vw/ D 4. This shows that G has exactly 12 maximal elementary abelian subgroups. Each of them is of order 26 and the intersection of any two of them is equal Z.ˆ.G//. In order to determine the structure of various subgroups of G which contain the subgroup Z.ˆ.G//, it is convenient first to determine the structure of H D G=Z.ˆ.G// of order 27 . Since Z.ˆ.G// D .ˆ.G//0 , H is obviously the group of maximal possible order having the exponent 4, being generated with two elements, and having the property that the squares of any two elements commute. The images of elements a; b; d; e; m; n; y of G in H D G=Z.ˆ.G// we denote again with the same symbols. Then we have E D 1 .H / D ˆ.H / D hd; e; m; n; yi Š E25 , H D ha; bi with a2 D d , b 2 D m, .a3 b 3 /2 D y, and the action of a and b on E is given with: d a D d;

e a D e;

ma D n;

na D m;

y a D de mny;

d b D e;

e b D d;

mb D m;

nb D n;

y b D de mny:

We have CE .a/ D hd; e; mni, CE .b/ D hde; m; ni, CE .ab/ D hde; mn; yi, Z.H / D hde; mni Š E4 , .Ehai/0 D .Ehbi/0 D .Ehabi/0 D hde; mni D Z.H /, Ehai=Z.H /, Ehbi=Z.H /, Ehabi=Z.H / are abelian groups of type .4; 2; 2/. It follows that the maximal subgroups of H are of rank 3. Also, it is easy to see from these information that the number of conjugate classes in H is k.H / D 26. Indeed, 4 elements in Z.H / give 4 classes of size 1, 12 elements in d Z.H / [ mZ.H / [ yZ.H / give 6 classes of size 2, the remaining 16 elements in E give 4 classes of size 4, and 96 D 27 25 elements in H E give 12 classes of size 8 each. We compute: y D .a1 b 1 /2 D a1 b 1 a1 b 1 D .a1 b 1 ab/ b 1 a1 a1 b 1 D Œa; bb 1 a2 b 1 D Œa; b d b b 2 D Œa; be m; so that Œa; b D e my. This gives H 0 D hde; mn; e myi Š E8 and H=H 0 Š C4 C4 . Since .emy/a D e n de mny D d my D .emy/de; .emy/b D d m de mny D e ny D .emy/mn; it follows that ŒH; H 0 D Z.H / D hde; mni, H is of class 3, and H=ŒH; H 0 is the non-metacyclic minimal nonabelian group of order 25 and exponent 4. We have obtained the results stated in part (f) of our theorem.

162

Groups of prime power order

We return now to our group G and consider the structure of maximal subgroups M D ˆ.G/hai, N D ˆ.G/hbi, K D ˆ.G/ha1 b 1 i. By the preceding paragraph, M 0 D N 0 D K 0 D Z.ˆ.G//hde; mni and M=M 0 is an abelian group of type .4; 2; 2/ so that d.M / D 3. Let be the automorphism of G of order 3 induced with a D b, b D a1 b 1 . Then acts transitively on fM; N; Kg and therefore it is enough to determine the structure of M . By the preceding paragraph, G 0 D Z.ˆ.G//hde; mn; e myi D M 0 he myi, G=G 0 Š C4 C4 , and M 0 =Z.ˆ.G// D Z.H / D K3 .H /, H D G=Z.ˆ.G// and therefore M 0 D Z.ˆ.G//hde; mni is a characteristic subgroup of G. It follows from d.G/ D 2 that G 0 =K3 .G/ is cyclic and G 0 D K3 .G/hŒa; bi. But 1 D .ab/4 a4 b 4 Œb; a6 .mod K3 .G// so that Œb; a2 2 K3 .G/ and jG 0 =K3 .G/j D 2. This gives that K3 .G/ D Z.ˆ.G//hde; mni D M 0 . We compute: Œde; mn D Œd; mŒd; nŒe; mŒe; n D c g h cgh D 1; .de/2 D d 2 e 2 Œe; d D ch;

.mn/2 D cg;

.de mn/2 D gh;

which shows that M 0 D K3 .G/ is abelian of type .4; 4; 2; 2; 2/ with ˆ.M 0 / D hcg; chi Š E4 . Since CG .c/ D ˆ.G/, we have Z.M / Z.ˆ.G//. But Chi;j i .a1 / D f1g and so Z.M / D Z.G/ D hcg; ch; cij i, where we have used the fact that hi; j i is a complement of Z.G/ in Z.ˆ.G//. Since .de/a

1

D .de/i;

.de/y D .de/ch;

.mn/y D .mn/cg; .mn/n D .mn/cg;

.de/n D .de/ch; .mn/a i

a1

D i ch;

j

1

a1

D .mn/g; D jgh;

we get that hc; g; h; i i Š E24 is normal in M , M 0 =hc; g; h; i i Z.M=hc; g; h; i i/ and 1 1 K3 .M / D hc; g; h; i i. Finally, hc; g; h; i i Z.ˆ.G//, c a D c.cg/, g a D g.cg/, 1 1 ha D h.cg/, i a D i.ch/ so that K4 .M / D hcg; chi Z.G/ and M is of class 4. We determine now the structure of G 0 D Z.ˆ.G//hde; mn; e myi, where G=G 0 Š C4 C4 . We know that K3 .G/ D Z.ˆ.G//hde; mni is an abelian maximal subgroup of G 0 , where 1 .K3 .G// D Z.ˆ.G//. But we see that the elements: e my; de e my D d my; mn e my e ny; de mn e my d ny

.mod Z.ˆ.G///

are all of order 4 and so 1 .G 0 / D Z.ˆ.G//. We compute Œde; e my D ch and Œmn; e my D cg and so G 00 D hcg; chi Z.G/. From jG 0 j D 2jZ.G 0 /jjG 00 j follows jZ.G 0 /j D 25 and so Z.G 0 / D Z.ˆ.G///. Also, .de/2 D ch, .mn/2 D cg and .emy/2 D cghij , which gives ˆ.G 0 / D hcg; ch; cghij i D hcg; ch; cij i D Z.G/. We have M 0 D K3 .G/ and K3 .M / D ŒM; M 0 D ŒM; K3 .G/ D hc; g; h; i i, 1 .mn/b D .mn/j , and so K4 .G/ hc; g; h; i; j i D Z.ˆ.G//. On the other hand, we know that K3 .G/=Z.ˆ.G// D Z.G=Z.ˆ.G/// D Z.H / and so K4 .G/ D 1 1 1 Z.ˆ.G//. We have i b D i.gh/, j a D j.gh/, j b D j.cg/, and so we see that K5 .G/ D hcg; chi Z.G/, K6 .G/ D f1g, and G is of class 5. Finally, we see that Z2 .G/ D Z.ˆ.G// Š E25 , Z3 .G/ D K3 .G/ D M 0 (abelian of type .4; 4; 2; 2; 2/), and Z4 .G/ D ˆ.G/. Our theorem is proved.

61

Groups of exponent 4 generated by three involutions

It is known that a group generated with two non-commuting involutions is dihedral. But the problem to determine groups generated with three involutions is already very difficult. In this section we classify groups of exponent 4 which are generated with three involutions. Theorem 61.1. Let G be a group of exponent 4 with d.G/ D 3 so that G is generated with involutions a; b; c and at least one of ab; ac; bc is an involution. Then jGj 26 and one of the following holds: (a) G Š C2 D8 , where jGj D 24 . (b) G is a splitting extension of the abelian group H D hv; w j v 4 D w 4 D Œv; w D 1; v 2 D w 2 i of type .4; 2/ with the four-group ha; bi, where v a D w, v b D v 1 , w b D w 1 . We have jGj D 25 and G D ha; b; ci with c D bv. (c) G is a splitting extension of the nonmetacyclic minimal nonabelian group H D hv; w j v 4 D w 4 D .wv/2 D Œw 2 ; v D 1i of order 24 with the four-group ha; bi, where v a D w, v b D v 1 , w b D w 1 . We have jGj D 26 and G D ha; b; ci with c D bv. The group G is isomorphic to a Sylow 2-subgroup of the alternating group A8 . Proof. Let G be a group of exponent 4 with d.G/ D 3 so that G is generated with three involutions a; b; c. If a; b; c pairwise commute, then G Š E8 , a contradiction. If a commutes with b and c but Œb; c ¤ 1, then G D hai hb; ci Š C2 D8 . Assume now that Œa; b D .ab/2 D 1 but G is not isomorphic to C2 D8 . This implies Œb; c ¤ 1 and Œa; c ¤ 1. Set v D bc so that o.v/ D 4 and D D hb; ci D hb; vi Š D8 with hv 2 i D Z.D/. Also, ha; ci Š D8 and so o.ac/ D 4. Set w D v a . Since b inverts v and a centralizes b, it follows that b inverts hwi. Also, w a D v and so the four-group ha; bi normalizes the subgroup H D hv; wi. Hence G D H ha; bi. We compute .ac/2 D .aca/c D c a c D .bv/a bv D .bwb/v D w 1 v and so o.w 1 v/ D 2. From .av/4 D 1 follows .av/2 D .ava/v D v a v D wv and

164

Groups of prime power order

.wv/2 D 1. The last relation gives 1 D wvwv D w 2 .w 1 vw 1 /w 2 v D w 2 .w 1 vw 1 v/v 1 w 2 v D w 2 .w 1 v/2 v 1 w 2 v D Œw 2 ; v: If hwi D hvi, then H D hvi and G D Dhai. There is an involution i 2 ha; bi hbi which centralizes hvi and so G D hi i D, contrary to our assumption. We have hvi ¤ hwi and since Œw 2 ; v D 1, we get w 2 2 Z.H /. Consider HN D H=hw 2 i (bar convention) so that HN D hw; N vi. N Suppose v 2 D w 2 which implies that HN is generated with two distinct involutions vN and w. N We know that .wv/2 D 1 and N so wv D wN vN is also an involution. Hence H Š E4 . Suppose that H is nonabelian. Since hvi and hwi are two distinct cyclic subgroups of order 4 in H , we get H Š Q8 . But we know that .wv/2 D 1, a contradiction. Hence H is abelian of type .4; 2/. We have G D H ha; bi and so the group G is a splitting extension of the abelian group H D hv; w j v 4 D w 4 D Œv; w D 1; v 2 D w 2 i of type .4; 2/ with the four-group ha; bi, where v a D w, v b D v 1 , w b D w 1 . We get jGj D 25 and G D ha; b; ci with c D bv. Assume now that v 2 ¤ w 2 so that hvi \ hwi D f1g. We have o.w/ N D 2, o.v/ N D4 2 4 N and o.wN v/ N D 2 (since .wv/ D 1 ). Hence H Š D8 and so jH j D 2 . The group H is not of maximal class since H possesses the abelian subgroup hw 2 ihvi of type .4; 2/. Hence H is of class 2 and therefore 1 D Œw 2 ; v D Œw; v2 D Œw; v 2 , so that Œw; v is an involution. Since Œw; v 2 Z.H /, we get that H=hŒw; vi is abelian. Consequently, H 0 D hŒw; vi and so H is minimal nonabelian with Z.H / D hw 2 ; v 2 i Š E4 . Also, 1 D .wv/2 D w 2 v 2 Œv; w, and so Œv; w D Œw; v D w 2 v 2 . Since wv 62 Z.H / D hw 2 ; v 2 i, we get hw 2 ; v 2 ; wvi Š E8 and so H is non-metacyclic. The group G is a splitting extension of the non-metacyclic minimal nonabelian group H D hv; w j v 4 D w 4 D .wv/2 D Œw 2 ; v D 1i of order 24 with the four-group ha; bi, where v a D w, v b D v 1 , w b D w 1 . We have ha; bi \ H D f1g and so jGj D 26 and G D ha; b; ci with c D bv and Z.G/ D hv 2 w 2 i D h.bc/2 ..bc/2 /a i. Indeed, it is easy to see that the involutions in ha; bi Š E4 induces on H only outer automorphisms. No element x 2 H could invert an element y of order 4 in H since Œw; v D w 2 v 2 is not a square in H . Also, hy x i \ hyi hy 2 i since y 2 2 ˆ.H / D Z.H /. The group G is isomorphic to a Sylow 2-subgroup of the alternating group A8 . Set D D hv; bi Š D8 so that .v 2 /a D w 2 and therefore v 2 62 DG . No subgroup of order 2 in D hvi is normal in D. Hence DG D f1g and so G has a faithful transitive permutation representation of degree 8. If we set: a D .1; 2/.3; 4/.5; 6/.7; 8/;

b D .4; 5/.3; 6/;

c D .2; 3/.6; 7/;

then we see that the permutations a; b; c satisfy all the above relations for G and .bc/2 ..bc/2 /a D .1; 8/.2; 7/.3; 6/.4; 5/ ¤ 1 and so the obtained permutation representation is faithful. Since these permutations are even, we are done.

61

Groups of exponent 4 generated by three involutions

165

Theorem 61.2. Let G be the group given with G D ha; b; c j a2 D b 2 D c 2 D .ab/4 D .bc/4 D .ca/4 D 1; .abc/4 D .c.ab/2 /4 D .b.ac/2 /4 D .acab/4 D 1i: Then G is of order 210 and exponent 4. Hence each group of exponent 4 which is generated with three involutions is a epimorphic image of the above group of order 210 . Proof. This theorem is proved by using a computer (W. Lempken of the University of Essen). In what follows we shall determine the structure of the group G D ha; b; ci of order 210 as defined in Theorem 61.2 so that G is the free group of exponent 4 generated with three involutions a; b; c. We set .ab/2 D d , d c D e, e a D f , and f b D g. We shall determine the structure of the normal subgroup S D hd iG hd; e; f; gi, where d; e; f; g are some involutions in ˆ.G/. Since ha; bi Š D8 , the involution d D Œa; b commutes with a and b. From the above follows e c D d , f a D e and g b D f . We compute .de/2 D .abab cababc/2 D .ababc ababc/2 D .ababc/4 D 1; and so Œd; e D 1. From the last relation we obtain (conjugating with the element a and then with the element b): Œd; e a D Œd; f D 1, Œd; f b D Œd; g D 1. Further computation gives (noting that .ab/2 D .ba/2 and .bac/3 D .bac/1 D cab): .ef /2 D .e aea/2 D .ea/4 D 1; .fg/2 D .f bf b/2 D .f b/4 D 1; .eg/2 D .cbabac bacbabacab/2 D .cba.bac bac bac/cbacab/2 D .cba cab cba cab/2 D .cbacab/4 D 1; and so Œe; f D 1, Œf; g D 1 and Œe; g D 1. We have proved that the subgroup hd; e; f; gi is elementary abelian of order 24 . It is possible to show that the subgroup hd; e; f; gi is normal in G and so we get S D hd iG D hd; e; f; gi. Indeed, we compute (noting that .ca/3 D .ca/1 D ac, .ab/3 D ba, .cbab/3 D .cbab/1 D babc): ff c D acababca cacababcac D acabab.ca/3 babcac D acabab ac babcac D ac.ababab/bcbabcac D ac ba bcbabcac D a.cbab/.cbab/.cbab/babac D a babc babac D .ab/2 c.ba/2 c D d d c D de;

166

Groups of prime power order

and so ff c D de which gives f c D def . Also we get: 2

g a D .d c /aba D e aba D e ababb D .e .ab/ /b D .e d /b D e b ; and so g a D e b . We compute (noting that .ab/2 D .ba/2 and bcbcb D .bc/3 c D .bc/1 c D cbc): ee b D caba.bc bcb/abacb D caba.cbc/abacb D .cabacb/2 ; and similarly (noting that baca baca ba D .baca/3 ac D .baca/1 ac D acabac), fg D acba.baca bacaba/bcab D acba acabac bcab D .acbcab/2 D .acbcab/2 D .bacbca/2 : Using the above relations we get (noting that .ac/2 D .ca/2 , .bca/2 D .bca/2 D .acb/2 , .cab/2 D .bac/2 , and .b.ac/2 /3 D .b.ac/2 /1 D .ac/2 b): ee b fg D .cabacb cabacb/.bacbca bacbca/ D cabac.bca bca cab cab/acbca D cabac.acb/2 .bac/2 acbca D cabacacbacbbacbacacbca D ca .b.ac/2 b.ac/2 b.ac/2 /bca D ca .ac/2 b bca D ca.ca/2 ca D .ca/4 D 1; and so e b D efg D g a . Finally, we get (using .ab/3 D ba): g c e D cbacbabac.abc cabab/c D cba cba bac.ab/3 c D .cba/2 .bac/2 D .cba/2 .bac/2 D .abc/2 .cab/2 D abcabccabcab D abc.ab/2 cab D .ab/2 bac.ab/2 cab D dg; and so g c D deg. We have proved by now: d a D d;

e a D f;

f a D e;

g a D efg;

d b D d;

e b D efg;

f b D g;

g b D f;

d c D e;

e c D d;

f c D def;

g c D deg:

This shows that hd; e; f; gi is normal in G and so S D hd; e; f; gi D hd iG is elementary abelian of order 24 . By Theorem 61.1, G=S is of order 26 . Since jGj D 210 (Theorem 61.2), we have S Š E24 and jG=S j D 26 . From the above results we get at once that hef; egi Z.G/. The subgroup hd; ei is a complement of hef; egi in S and Z.G/ \ hd; ei D f1g. This gives Z.G/ \ S D hef; egi.

61

Groups of exponent 4 generated by three involutions

167

Define the automorphism of G induced with a D b, b D c, c D a. Set d D h D .bc/2 ;

e D i D ..bc/2 /a ;

f D j D ..bc/2 /ab ;

g D k D ..bc/2 /abc

so that T D hhiG D hh; i; j; ki Š E24 and transports the above relations for the elements in S into the relations for the elements in T . In particular, ha D i , i a D h, hb D h, i b D j , hc D h, i c D ij k. We want to determine S \T . For that purpose we compute (noting that .bc/3 D cb, .cba/2 D .abc/2 , .ca/3 D ac, .bc/3 D cb): ef i k D cababc acaba.bca abcbc/a cbabcbcabc D cababcacaba.cb acba/bcbcabc D cababcacaba abcabc bcbcabc D cabab.ca/3 .bc/3 abc D cabab ac cb abc D c.ab/4 c D 1; and similarly (noting that we have .bc/2 D .cb/2 , .cbab/2 D .babc/2 , .ab/3 D ba, .bac/3 D cab): egj k D cababc bacababcab bacbcbab cbabcbcabc D cababcbacaba.cbabcbab/cbcabc D cababcbac.aba bab/cbabc cbcabc D cababc.bac ba cbac/abc D cababc cab abc D c.ab/4 c D 1; which gives ef D i k, eg D j k and so S \T hef; egi Š E4 and i b D j D i.ij / D i.fg/, i c D i.j k/ D i.eg/. We have jS T j 26 . By Theorem 61.1, jG=S T j 24 and so the fact that jGj D 210 gives U D S T is of order 26 and G=U Š C2 D8 . This implies that y D .ca/2 62 U and S \ T D hef; egi D hi k; j ki Z.G/. Note that hh; i i is a complement of S \ T in T and we compute: d h D d bcbc D d.fg/;

d i D d abcbca D d.fg/;

e h D e bcbc D e.fg/;

e i D e abcbca D e.fg/;

and so Œd; h D Œd; i D Œe; h D Œe; i D fg. We have U 0 S \ T Z.G/ and so U is of class 2 and, by the above, U 0 D hfgi and U D hd; e; f; gihh; i i ˆ.G/. Since y D .ca/2 62 U and G=G 0 is generated with involutions, we have ˆ.G/ D G 0 D 1 .ˆ.G// D U hyi. We note that fg; eg; ef 2 Z.G/, hg D ghŒh; g D ghŒh; e.eg/ D ghŒh; e D ghfg

168

Groups of prime power order

and Œd; hi D Œd; hŒd; i D .fg/2 D 1 so y b D .caca/b D ch ad ch ad D caha dchad D caidchad D caci c d c had D caci.eg/ehad D cacighad D .caca/i a g a ha d D y h.ef /gi d D y.ef /.hg/id D y.ef /gh.fg/id D y.ef /.fg/g.hi /d D yedhi; and hence we have y b D ydehi , y a D y and y c D y. Set fg D z1 , eg D z2 so that E4 Š hz1 ; z2 i Z.G/, d y D d caca D dz1 z2 ;

e y D ez1 z2 ;

hy D hz2 ;

i y D iz2 ;

Z.U / D hz1 ; z2 ; de; hi i Š E24 , U 0 D hz1 i. Obviously we have Z.ˆ.G// Z.U / and since .de/y D de, .hi /y D hi , we get hde; hi i Z.ˆ.G// and so Z.ˆ.G// D Z.U /. By the above, z1 ; z2 2 .ˆ.G//0 and since the factor-group ˆ.G/=hz1 ; z2 i is abelian, we get G 00 D .ˆ.G//0 D hz1 ; z2 i. Also, hde; hi i is a complement of hz1 ; z2 i in Z.ˆ.G//, Z.G/ ˆ.G/ and .de/a D df , .hi /b D hj , .dehi /a D df hi which all imply that hz1 ; z2 i D Z.G/. Our relations give at once that K3 .G/ D Z.ˆ.G// D hz1 ; z2 ; de; hi i Š E24 and K4 .G/ D hz1 ; z2 i D Z.G/ so that G is of class 4. We have proved the following Theorem 61.3. Let G be the free group of exponent 4 which is generated with three involutions a; b; c. Then G is of order 210 and has the following properties: (a) ˆ.G/ D G 0 D 1 .G 0 / is of order 27 and class 2 with Z.ˆ.G// Š E24 and .ˆ.G//0 D G 00 D ˆ.ˆ.G// Š E4 . (b) We have K3 .G/ D ŒG; G 0 D Z.ˆ.G// and K4 .G/ D .ˆ.G//0 D Z.G/ so that G is of class 4. (c) The group ˆ.G/ is generated with five involutions d; e; h; i; y so that .ˆ.G//0 D hz1 ; z2 i Š E4 and Œd; e D Œh; i D 1, Œd; h D Œd; i D Œe; h D Œe; i D z1 , Œd; y D Œe; y D z1 z2 , Œh; y D Œi; y D z2 , where hz1 ; z2 i D Z.G/ and Z.ˆ.G// D hz1 ; z2 ; de; hi i Š E24 . We have G D ha; b; ci with .ab/2 D d , .bc/2 D h, .ca/2 D y, where the action of a; b; c on ˆ.G/ is given with: d a D d;

e a D ez1 z2 ;

ha D i;

d b D d;

e b D ez1 ;

hb D h;

d c D e;

e c D d;

hc D h;

i a D h;

y a D y;

i c D iz2 ;

y c D y:

62

Groups with large normal closures of nonnormal cyclic subgroups

Let H be a nonnormal subgroup of a p-group; then jG W H G j p, where H G is a normal closure of H in G. Therefore, it is natural to classify the p-groups G such that jG W H G j D p for all nonnormal subgroups H < G. Such G are classified by the second author but all proofs in this section are due to the first author. Theorem 62.1 (Z. Janko). Let G be a non-Dedekindian p-group. Suppose that jG W H G j D p for all nonnormal cyclic subgroups H of G. Then one and only one of the following holds: (a) jGj D p 3 . 2

2

(b) G D ha; b j ap D b p D 1; ab D a1Cp i is the unique nonabelian metacyclic group of order p 4 and exponent p 2 . (c) G is a 2-group of maximal class. (d) p D 2, G D H Z is a semidirect product with cyclic kernel Z of order > 4 and a cyclic complement H of order 4. Write R D 1 .H /. We have Z.G/ D R 1 .Z/ Š E4 , G=R is of maximal class not isomorphic with generalized quaternion group, and G=G 0 is abelian of type .4; 2/. We have G D n n1 hh, z j h4 D z 2 D 1, n > 2, z h D z 1C2 , D 0; 1i. It is trivial that groups (a), (b) and (c) of Theorem 62.1 satisfy the hypothesis. For example, if N G G is of index > 2, where G is 2-group of maximal class, then N ˆ.G/, and this implies that groups of (c) satisfy the hypothesis. Now let G be as in Theorem 62.1(d) and let X < G be nonnormal cyclic. Let us prove that jG W X G j D 2. It follows from 1 .G/ D Z.G/ that jXj > 2. The group G has no proper subgroups which are generalized quaternion (Proposition 10.19(a)) hence 1 .G/ < X G (otherwise, X G is cyclic so X G G). If 1 .H / D R < X, then X=R is a nonnormal cyclic subgroup of the 2-group of maximal class G=R (since j.G=R/ W .G 0 =R/j D 4) so jG W X G j D 2, by (c). Now we assume that R 6 X. Suppose that XR D X R 6E G; then XR=R 6E G=R so jG W .XR/G j D 2. We have .XR/G D X G R D X G since R < 1 .G/ < X G , so jG W X G j D 2. Now assume that XR G G. Then XR=R T =R, where T =R is a cyclic subgroup of index 2 in G=R. We have T D V R and ˆ.V / D ˆ.T / G G so all cyclic subgroups of T of

170

Groups of prime power order

order 2k for all k < n, where jV j D 2n (n > 2), are G-invariant, so jXj D 2n ; then jG W Xj D 4. In that case, X G D T so jG W X G j D 2, as required. Lemma 62.2. Let G be a non-Dedekindian minimal nonabelian p-group of order > p 3 . If jG W H G j D p for all nonnormal H < G, then G is the unique nonabelian metacyclic group of order p 4 and exponent p 2 which we denote Hp;2 . Proof. Let H < G be nonnormal cyclic; then H G D H G 0 so jG W H j D p 2 . If the nonmetacyclic G is as in Lemma 65.1(a), then o.a/ D o.b/ D p so jGj D p 3 , contrary to the hypothesis. Thus, G is metacyclic as in Lemma 65.1(b) so G D B A, where A D hai G G, B D hbi and jAj D p 2 , by the above. Assume that n > 2. Write 2 .B/ D hxi, F D hxai; then jF j D p 2 . We have .xa/b D xa1Cp 62 F so F is not G-invariant; then F G D F G 0 has index p n1 > p in G, a final contradiction. Proof of Theorem 62.1. Let H < G be nonnormal cyclic. Since H G Hˆ.G/ < G, we get jG W Hˆ.G/j D p so H 6 ˆ.G/ and d.G/ D 2. Hence all cyclic subgroups of ˆ.G/ are G-invariant so ˆ.G/ is Dedekindian. Since ˆ.G/ has no G-invariant subgroups isomorphic to Q8 (Lemma 1.4), it is abelian. If jG 0 j D p, then G is minimal nonabelian (Lemma 65.2(a)) so G is as in Theorem 62.1(b), by Lemma 62.2. Now suppose that jG 0 j > p so G is not minimal nonabelian. (i) Let ˆ.G/ be cyclic. If p > 2, then ˆ.G/ D Z.G/ and jG 0 j D p (Theorem 4.4), a contradiction. Thus, p D 2; then ˆ.G/ D Ã1 .G/ so G has a cyclic subgroup of index 2 hence G is of maximal class. Now suppose that ˆ.G/ is noncyclic. (ii) Let p > 2. Suppose that ˆ.G/ D U V and U > f1g is cyclic (by (i), V > f1g and, by the above, U; V are G-invariant). Take x 2 G. By induction, Œx; ˆ.G/ U \ V D f1g so ˆ.G/ Z.G/. Then G is minimal nonabelian, a contradiction. (iii) Thus, p D 2 and ˆ.G/ is noncyclic abelian. Assume that G is not metacyclic. Then, by Schreier inequality (see Appendix 25) and Corollary 36.6, there is M 2 1 such that d.M / D 3. Write GN D G=ˆ.M /; then GN of order 24 is neither abelian (in N < MN of order 2 such view of d.G/ D 2 < d.M /) nor of maximal class so there is L N N N N that L 6 Z.G/. Since G is not of maximal class, we get jG 0 j D 2, by Taussky, so N GN D L N GN 0 has index 4 in G, N a contradiction (our hypothesis is inherited by nonL Dedekindian epimorphic images of G). Thus, G is metacyclic so G=Z is cyclic for a cyclic Z GG. Since G is not of maximal class, we get 1 .G/ Š E4 so 1 .G/ Z.G/ in view of G=1 .G/ > 2. Since G is not of maximal class, we get jG=G 0 j 8 (Taussky). Let R < G 0 be of index 2; then G=R is minimal nonabelian and nonDedekindian since d.G=R/ D 2 and jG=R/ > 8 (Lemma 62.2). Therefore, by Lemma 62.2, G=R Š H2;2 so G=G 0 is abelian of type .4; 2/. It follows that jG W Zj D 4 since Z < G0. Let M < G be minimal nonabelian; then jM j > 8 (Proposition 10.19) so M is as in Lemma 65.1(b); then M D H1 Z1 is a semidirect product with cyclic factors Z1

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and H1 and H1 is not normal in M so in G. However, M 0 D 1 .Z1 / < G 0 and G 0 is cyclic so H1 \ G 0 D f1g. Since 2 D jG W H1G j, we get H1G D H1 G 0 ; then jH1 j D 4 since 1 .G/ Z.G/, and G=G 0 is abelian of type .4; 2/, H1 \Z D f1g since G 0 < Z and Z is cyclic. It follows that G D H1 Z, a semidirect product. Since Ã1 .H1 / G G as a subgroup of ˆ.G/, then A D Ã1 .H1 / Z is abelian of index 2 in G so 4 D 1 0 0 2 jG W G j D jZ.G/j (Lemma 1.1). Since jG=Ã1 .H1 / W .G=Ã1 .H1 // j D 4, the group G=Ã1 .H / is of maximal class (Taussky); moreover, it is not generalized quaternion since G=1 .G/ D .H1 1 .G/=1 .G// .Z1 .G/=1 .G// is a semidirect product. Next, jZj > 4 (otherwise, G is as in (b)). If G=1 .G/ is generalized quaternion, then j2 .G/j D 8 so G is a group of Lemma 42.1(c); then Z.G/ is cyclic, a contradiction. Thus, G is as stated in (d).

63

Groups all of whose cyclic subgroups of composite orders are normal

Definition 1. If all cyclic subgroups of a p-group G of composite orders are normal, then G is said to be a Kp -group. The property Kp is inherited by sections. Below we classify Kp -groups and prove that the property (Kp ) is equivalent to the following condition: Whenever f1g < B < A G, where A is cyclic, then NG .B/ D NG .A/ [Kaz1]. All proofs, apart of the proof of Supplement to Theorems 63.1 and 63.3, are due to the first author. Proposition 63.1. If G is a nonabelian Kp -group, p > 2, and exp.G/ > p, then jG 0 j D p. Proof. Let Z < G be cyclic of order p 2 and C0 D 1 .Z/; then C0 G G. Assume that K=C0 < G=C0 is nonnormal of order p; then K is abelian of type .p; p/. Set H D KZ; then jH j D p 3 . Since p > 2, H has exactly p cyclic subgroups of order p 2 and they generate H so H E G. Then 1 .H / D K G G, a contradiction. Lemma 63.2. Let a 2-group G be a K2 -group. Then: (a) If L G G is of order 2 and G=L Š Q8 , then L is a direct factor of G. (b) If T < G is cyclic of order 2e , e > 1, then G=T 6Š Q8 . Proof. (a) By Theorem 1.2, G has no cyclic subgroups of index 2 so L 6 G 0 , by Taussky’s theorem. Then, if Z=L < G=L is cyclic of order 4, then Z is abelian of type .4; 2/ and 1 .Z/ D Z.G/ so all subgroups of order 2 are normal in G, and we conclude that G is Dedekindian. Then, by Theorem 1.20, L is a direct factor of G. (b) Assume that G=T Š Q8 . We use induction on jGj. Let L < T be of order 2; then jT =Lj D 2, by induction, so, by (a), G=L D .T =L/.Q=L/, where Q=L Š Q8 . By (a), Q D L Q1 , where Q1 Š Q8 . Since Q1 Š Q8 is generated by cyclic subgroups of order 4, it is normal in G so G D T Q1 . Since all subgroups of order 2 are normal in G, G is Dedekindian, contrary to Theorem 1.20. Recall that the Hp -subgroup Hp .G/ of a group G is defined as hx 2 G j o.x/ ¤ pi. By Burnside, a nontrivial H2 -subgroup has index 2 in G. If G is a (Kp )-group, then Hp .G/ centralizes G 0 . Note that if jG W H2 .G/j D 2, then exp.Z.G// D 2.

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Theorem 63.3 ([Kaz1]). If G is a nonabelian K2 -group, then either (a) jG 0 j D 2 or (b) H2 .G/ has index 2 in G. Proof. Suppose that G is a counterexample of minimal order. As we noticed, H2 .G/ centralizes G 0 . Since jG W H2 .G/j 2, G 0 < H2 .G/. If jG W H2 .G/j D 2, we have case (b). Now we assume that G D H2 .G/. Then, by what has just been said, cl.G/ D 2. (i) Let jG 0 j > 4. If C < G 0 is of order 2, then, by induction, G=C has an abelian subgroup A=C of index 2 such that all elements in .G=C / .A=C / are involutions since j.G=C /0 j D jG 0 =C j > 2. Let x 2 G A; then x 2 2 A. Assume that o.x/ > 2; then X D hxiGG. In that case, jXC =C j D 2 so x 2 2 C and G=C D .A=C /.X=C / is abelian so C D G 0 is of order 2, which is not the case. Thus, if jG 0 j > 4, we obtain case (b). It remains to consider the case where G 0 is either cyclic of order 4 or abelian of type .2; 2/. (ii) Let G 0 D hci Š C4 . We have c D Œx; y for some x; y 2 G. Set H D hx; yi; then H 0 D G 0 D hci is a central subgroup of order 4. We see that (a nonabelian K2 -group) H of composite exponent is not a group from the conclusion (indeed, since exp.Z.H // > 2, H coincides with its H2 -subgroup; in addition, jH 0 j D 4 > 2); therefore, H D G, by induction. If x; y are involutions, then G is dihedral of class 2 so jG 0 j D 2, a contradiction. Therefore, one may assume that o.x/ > 2; in that case, G D hxihyi is metacyclic. Next, G has no cyclic subgroups of index 2 (otherwise, since G is of class 2, jG 0 j D 2). Then 1 .G/ is of type .2; 2/ and G has no nonabelian subgroups of order 8 (this follows from Proposition 10.19). Therefore, if K=1 .G/ is a subgroup of order 2 in G=1 .G/ then K is abelian of type .4; 2/ so it is normal in G since K D hy j o.y/ D 4i. It follows that G=1 .G/ is nonabelian Dedekindian (nonabelian since G 0 is cyclic of order 4 > 2). Then G=1 .G/ Š Q8 since it is metacyclic (Theorem 1.20). By Taussky’s theorem, G=G 0 is abelian of type .4; 2/. Since G 0 Z.G/, G is minimal nonabelian. In that case, jG 0 j D 2, a contradiction. (iii) It remains to consider the case where G 0 Š E4 . Let Z < G be cyclic of order 2 D exp.G/; then G=Z is nonabelian since G 0 — Z. By Lemma 63.2(b), G=Z has no subgroups isomorphic to Q8 . It follows from Theorem 1.20 that G=Z contains a nonnormal subgroup L=Z of order 2. Since L 6E G, the subgroup L is not generated by cyclic subgroups of composite orders so L is dihedral, by Theorem 1.2 (indeed, jL W Zj D 2 and Z is cyclic). Since L0 is cyclic and exp.G 0 / D 2, we get jL0 j D 2 hence jLj D 8, jZj D 4 and e D 2. Since G D H2 .G/ is nonabelian, there are two cyclic subgroups, say U and V , of order 4 that generate a nonabelian subgroup Q D U V . In that case, Q Š Q8 and Q G G. Since G 0 6 Q, the quotient group G=Q is nonabelian so it contains a cyclic subgroup M=Q of order 4 .D exp.G//. Set M D hx; Qi; then X D hxi is of order 4 and M D Q X. Since M is not Dedekindian (Theorem 1.20) and all subgroups of order 2 are normal in M , it contains a nonnormal cyclic subgroup of order 4, a final contradiction. e

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Groups of prime power order

Supplement to Theorems 63.1 and 63.3 ([Kaz1]). Let G be a (Kp )-group. If there holds exp.G/ > p and jG 0 j D p, then ˆ.G/ is cyclic. Proof (Kazarin, personal communication). Assume that ˆ.G/ is noncyclic; then we have jGj > p 3 and exp.G/ exp.G=G 0 / > p. (i) If x; y 2 G, then 1 D Œx; yp D Œx; y p so y p 2 Z.G/ and ˆ.G/ D G 0 Ã1 .G/ Z.G/. (ii) Let C < G be cyclic of order > p. Assume that G 0 6 C ; then G=C is nonabelian, C \ G 0 D f1g so ŒC; G C \ G 0 D f1g and hence C Z.G/. By Lemma 63.2(b), G=C has no subgroups isomorphic to Q8 so it is not Dedekindian. Therefore, there is a nonnormal B=C < G=C of order p; B is abelian of type .jC j; p/ since C Z.G/. Then B G G since B D hx 2 B j o.x/ > pi, contrary to the choice of B. Thus, every cyclic subgroup of G of composite order contains G 0 . (iii) Let G be minimal nonabelian. Then G=G 0 D haG 0 i hbG 0 i is abelian of type .p m ; p n / with m n; then m > 1. Let o.aG 0 / D p m , o.bG 0 / D p n . By (ii), G 0 hai so G 0 is not a maximal cyclic subgroup of G so G is metacyclic and G D hbi hai (Lemma 65.1). By (ii), n D 1 so ˆ.G/ is cyclic. (iv) If x; y 2 G are elements of composite orders, then H D hx; yi is either abelian or minimal nonabelian and in both cases it contains a cyclic subgroup of index p. Indeed, suppose that H is abelian. Let o.x/ o.y/. Then H D hxi hzi so, by (ii), o.z/ D p, i.e., H has a cyclic subgroup hxi of index p. Now suppose that H is nonabelian. Since hxi G G, H is metacyclic. Since ˆ.H / Z.H /, by (i), and H=ˆ.H / Š Ep 2 , H is minimal nonabelian. By (iii), H has a cyclic subgroup of index p so ˆ.H / is cyclic. Thus, in any case, hx p ; y p i is cyclic for all x; y 2 G. Let x 2 G be of maximal order. Then, by (iv), Ã1 .G/ hxi so, by (ii), ˆ.G/ D Ã1 .G/G 0 hxi whence ˆ.G/ is cyclic. Thus, Theorems 63.1 and 63.3 can be formulated as follows. Theorem 63.4 ([Kaz1]). If G is a Kp -group, then one of the following holds: (a) G is abelian, (b) exp.G/ D p, (c) jG 0 j D p and ˆ.G/ is cyclic, (d) p D 2 and jG W H2 .G/j D 2. The above four groups satisfy the hypothesis. Definition 2. A p-group G is said to be an Np -group if every cyclic subgroup A of composite order satisfies the following condition: Whenever f1g < B < A, then NG .B/ D NG .A/. (Clearly, subgroups of Np -groups are Np -groups.) Theorem 63.5 ([Kaz]). Let G be a nonabelian p-group, exp.G/ > p > 2. Then the following assertions are equivalent:

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(a) G is an Np -group, (b) G is a Kp -group. Proof. Obviously, (b) implies (a). Now suppose that an Np -group G is not a Kp group. Then G has a nonnormal cyclic subgroup A of composite order so, in view of Theorem 1.2, G has no cyclic subgroups of index p. By hypothesis, (i) Every nonidentity subgroup of A is not normal in G, i.e., AG D f1g. Suppose that G is a counterexample of minimal order. In that case, every proper subgroup of G of composite exponent is a Kp -group so the normalizer of every nonnormal cyclic subgroup of G of composite order is maximal in G since this normalizer is an Kp -group; thus N D NG .A/ 2 1 . Then, by Proposition 63.1, jN 0 j p. Let M 2 1 fN g; then A 6 M (otherwise, A is normal in M so in G). If jA \ M j > p, then, by induction, NG .M \ A/ MN D G so AG A \ M > f1g, contrary to (i). Thus jM \ Aj D p so jAj D p 2 . Thus (ii) Let A be a nonnormal cyclic subgroup of G of composite order. Then jAj D p 2 and A is contained in exactly one maximal subgroup of G. Suppose that jG 0 j D p. Then G 0 6 A so G 0 A D G 0 A G G. In that case, ˆ.G 0 A/ > f1g is contained in A and normal in G, contrary to (i). Thus, if jG 0 j D p. the theorem is true. In what follows we assume that (iii) jG 0 j > p. (iv) G has at most one abelian subgroup of index p. Indeed, otherwise, we have jG W Z.G/j D p 2 so jG 0 j D p, by Lemma 1.1. Let R N be of composite exponent. Since jN 0 j p and p > 2, R is regular. It follows that R is generated by cyclic subgroups of composite orders so R is normal in N (by induction, N is a Kp -group). It follows that N=A is abelian since it is Dedekindian and p > 2 so N 0 < A. By (i), N 0 D f1g so N is abelian. Thus, by (iv), (v) N is the unique abelian maximal subgroup of G. Moreover, the normalizer of any nonnormal cyclic subgroup of composite order is an abelian maximal subgroup of G so coincides with N . It follows that (vi) All nonnormal cyclic subgroups of orders > p in G are contained in N . Let B G G be cyclic of order > p. Assume that B 6 N , or, what is the same, B does not normalize A; then B G G, by (vi). Then, AB is not a Kp -group so it is not an Np -group and we get, by induction, AB D G. Thus, G is metacyclic. By (i), A \ B D f1g, and, obviously, CN .B/ D Z.G/ (recall that N is the unique abelian maximal subgroup of G). Note that jB W .B \ N /j D p and B \ N Z.G/. Since AG D f1g, we get Z.G/ < B so B \ N D Z.G/. Then G=Z.G/ is abelian of type .p 2 ; p/. We have Z.G/ D ˆ.B/ ˆ.G/. Let U=Z.G/ and V =Z.G/ be distinct cyclic subgroups of order p 2 in G=Z.G/. Then U and V are distinct abelian maximal subgroups of G, contrary to (v). Thus, B < N so all cyclic subgroups of

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Groups of prime power order

composite orders are contained in N , i.e., N D Hp .G/. This is a contradiction since hG N i D hG Hp .G/i D G. Thus, (vii) All cyclic subgroups of G of composite orders are contained in N D Hp .G/. Let B G G be normal cyclic of composite order. Take x 2 G N and set H D hx; Bi; then H is metacyclic. It follows from p > 2 that H is generated by its cyclic subgroups of maximal order so H N , by (vii), contrary to the choice of x. Thus, (viii) All cyclic subgroups of composite orders are not normal in G. We see that the normalizer of every cyclic subgroup of composite order is abelian so coincides with N . It follows that N is the unique maximal subgroup of G of composite exponent, by (ii). Moreover, 2 .N / D N , by (ii), so exp.G/ D p 2 . Let M 2 1 fN g; then exp.M / D p, by (viii). In that case N contains the elementary abelian subgroup N \ M of index p. Then N D A E, where E is elementary abelian. In that case, ˆ.N / D ˆ.A/ > f1g is characteristic in N so normal in G, contrary to (i). Thus, all cyclic subgroups of composite orders are normal in G so G is a Kp -group. Theorem 63.6 ([Kaz1]). Let G be a nonabelian 2-group, and let exp.G/ > 2. Then the following assertions are equivalent: (a) G is a N2 -group, (b) G is a K2 -group. Proof. Clearly, (b) implies (a). It remains to prove the reverse implication. Suppose that G is a counterexample of minimal order. Then G has a nonnormal cyclic subgroup B D hbi of composite order. As above, () jXj D 4 and XG D f1g, where X < G is nonnormal cyclic of order > 2. Let B < N , where jG W N j D 2. Then N is a K2 -group, by induction, so N D NG .B/ and N is the unique maximal subgroup of G containing B. Let a 2 G N be such that o.a/ is as large as possible; then B a ¤ B. If 1 .B a / D 1 .B/, then 1 .B/a D 1 .B/ and NG .1 .B// hN; ai D G, contrary to (). Thus, 1 .B a / ¤ 1 .B/ so B \ B a D f1g. Since B; B a N , these two subgroups are normal in N so H D BB a D B B a is abelian of type .4; 4/. Since B 6E ha; Bi, we get G D ha; Bi D ha; H i, by induction. Set A D hai; then H G G since a2 2 N , G D AH , G=H is cyclic. Since H is abelian, A \ H G G. If o.a/ D 2, then G is a K2 -group (indeed, then H is an H2 -subgroup of G). In what follows we assume that o.a/ > 2. If A \ H > f1g, then A G G since A \ H G G (see ()), and so, by induction, G D BA is a semidirect product with kernel A since B \ A D f1g, by (). In that case, G D B A is metacyclic. Suppose that A \ H D f1g. Then A is not normal in G since G is nonabelian so CG .H / D H , by (). In that case, the abelian subgroup NG .A/ D A Z.G/ is

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maximal in G, by induction, so Z.G/ is a subgroup of index 2 in H ; then B \ Z.G/ > f1g, contrary to (). Thus, it remains to consider the following two cases. (i) Let G D B A, a semidirect product with kernel A D hai Š C2n (n > 1), B D hbi Š C4 . As above, N D NG .B/, a maximal subgroup of G, is abelian of type .22 ; 2n1 /, CG .A/ D A and N \ A D Z.G/ has index 2 in A. Then G=Z.G/ is abelian of type .4; 2/ so G is minimal nonabelian. In that case, Z.G/ D ˆ.G/ and f1g < B \ Z.G/ BG , contrary to (). (ii) Let G D A H , a semidirect product with abelian kernel H of type .4; 4/, A D hai Š C4 , H D B B a , where B D hbi Š C4 . Set NG .A/ D M ; then jG W M j D 2, M \ H D Z.G/ is abelian of type .4; 2/ so BG B \ Z.G/ > f1g, contrary to (). It follows from the above that epimorphic images of Np -groups are Np -groups or have exponent p. Let G be a p-group with exp.G/ > p. If every subgroup of composite exponent is normal in a p-group G, then either jG 0 j p or G Š D24 [Man18]. Obviously, G is a Kp -group. Suppose that jG 0 j > p. Then p D 2 and A D H2 .G/ has index 2 in G, by Theorem 63.4. If A is cyclic, we are done, by Theorem 1.2. Assume that A is not cyclic. Since G 0 > f1g, we get exp.A/ > 2. Let Z be a cyclic subgroup of order 4 in A. By hypothesis, all subgroups are normal in G=Z, i.e., G=Z is Dedekindian. By Lemma 63.2(b), G=Z has no subgroups isomorphic to Q8 so G=Z is abelian. Let Z1 ¤ Z be another cyclic subgroup of order 4 in A (see Theorem 1.17(b)). Then G=Z1 is abelian, by what has just been proved. It follows that G 0 Z \ Z1 , so jG 0 j 2. Proposition 63.7 ([Li3] (compare with Theorem 58.5)). Suppose that all nonnormal cyclic subgroups of a nonabelian p-group G, p > 2, are conjugate. Then G Š Mp n . Proof. Let T < G be a nonnormal cyclic subgroup of G. Set jT j D p t . Let K be the set of all nonnormal cyclic subgroups of G; then jKj is a power of p. (i) Let t D 1. (i1) Assume that exp.G/ D p. Set jGj D pm , jZ.G/j D p z . Then jKj D D p z .1 C p C C p mz1 /. Since m z 1 > 0, jKj is not a power of p so not all members of the set K are conjugate in G, a contradiction. p m p z p1

(i2) Thus, exp.G/ > p. Then G is a Kp -group so, by Theorem 63.1, jG 0 j D p, and G is regular since p > 2. Let j1 .G/j D p k and j1 .Z.G//j D p . Then the number k p D p .1CpCp k 1 /, which of nonnormal subgroups of order p in G equals p p1 0 is a power of p, so k D C 1. Since jG j D p, it follows that there are exactly p subgroups conjugate with T in G. It follows that D 1 so k D 2. Then jG=Ã1 .G/j D j1 .G/j D p 2 so G is metacyclic. It follows that G is minimal nonabelian (Lemma 65.2(a)). Then, by Lemma 58.1, G Š Mp n .

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Groups of prime power order

(ii) Now let t > 1. Set T0 D 1 .T /; then T0 TG G G so T0 is contained in all nonnormal cyclic subgroups of G. Set GN D G=T0 . Let UN < GN be nonnormal cyclic. If U is noncyclic, then U D T0 U0 and U0 is not normal in G. Then T0 — U0 , contrary to what has just been said. Thus, U is cyclic so conjugate with T . It follows that all nonnormal cyclic subgroups of GN are conjugate, i.e., GN satisfies the hypothesis. Therefore, by induction, GN Š Mp n . We also have T0 ˆ.T / ˆ.G/ N D 2. If A=T0 ; B=T0 < G=T0 be two distinct cyclic subgroups of so d.G/ D d.G/ index p, then A; B 2 1 are distinct abelian so A \ B D Z.G/, and we conclude that G is minimal nonabelian. By Lemma 58.1, G Š Mp n . However, all nonnormal cyclic subgroups of Mp n have order p < p t , so case t > 1 is impossible. Exercise 1. Suppose that all nonnormal cyclic subgroups of a nonabelian 2-group G, are conjugate. Then G Š M2n , i.e., Theorem 63.1 is also true for p D 2. Exercise 2. Let a group G be nonabelian of exponent p. Then the number of nonnormal subgroups of order p in G is not a power of p. There are in the book a number of results on cyclic subgroups of p-groups. Problem. Classify the non-Dedekindian p-groups in which any two nonnormal cyclic subgroups of the same order are conjugate.

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p-groups generated by elements of given order

1o . In what follows, G is a group of order p m , k .G/ D hx 2 G j o.x/ D p k i, where p k exp.G/. Then k .G/ k .G/ and k .G/ is the least subgroup H P i of G subjecting ck .H / D ck .G/. We have 1 iD0 '.p /ci .G/ D jGj, where './ is Euler’s totient function. Given k > 0, set solk .G/ D jfx 2 G j o.x/ p k gj. Then P solk .G/ D kiD0 '.p i /ci .G/. For definition of Ls - and U2 -groups, see 17, 18, 67. A 2-group G is said to be a Un -group, if there is E2n Š R G G such that G=R is of maximal class with cyclic subgroup T =R of index 2 and 1 .T / D R. It is easy to show that R contains all normal elementary abelian subgroups of G and all elements in the set G T have orders 8. A subgroup R is called the kernel of G. For the sake of completeness, we collected in Lemma 64.1 some known facts. We use this lemma until end of the book. Lemma 64.1. Let G be a p-group. (a) (Theorems 7.1 and 7.2) If G is regular of exponent p e and k e, then we have exp.k .G// D p k . All p-groups of class < p are regular. Groups of exponent p are regular. Regular 2-groups are abelian. (b) (Theorem 12.1) If G has no normal subgroups of order p p and exponent p, it is either absolutely regular or of maximal class. If an irregular p-group G has an absolutely regular maximal subgroup H , then G is either of maximal class or G D H 1 .G/, where j1 .G/j D p p . (c) (Theorems 9.5 and 9.6) A p-group of maximal class and order > p p is irregular. A p-group of maximal class and order p m contains only one normal subgroup of order p i for 1 i m 2. (d) (Theorem 9.6, Lemma 12.3) A p-group of maximal class and order > p pC1 has no normal subgroups of order p p and exponent p. For such G, we have c1 .G/ 1 C p C C p p2 .mod p p / and c2 .G/ p p2 .mod p p1 /. If p 2 < p k exp.G/, then ck .G/ D 1 if p D 2 (see Lemma 64.10, below) and p p1 divides ck .G/ if p > 2. (e) (Theorem 9.6) An irregular p-group G of maximal class has an absolutely regular subgroup G1 of index p such that jn .G1 /j D p .p1/n for p n < exp.G1 / D exp.G/. If, in addition, jGj > p pC1 , then Z.G1 / is noncyclic (G1 is called the fundamental subgroup of G); all other maximal subgroups of G are of maximal class. If 2 < k exp.G/, then k .G/ G1 .

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Groups of prime power order

(f) (Theorems 13.2 and 13.5) If G is neither absolutely regular nor of maximal class, then c1 .G/ 1CpC Cp p1 .mod p p / and, for k > 1, ck .G/ 0 .mod p p1 /. The number of subgroups of order p p and exponent p in G is 1 .mod p/. (g) (Theorem 13.19) If G is an irregular p-group of maximal class and H G is of order > p p , then H is either absolutely regular or of maximal class. (h) If A G G is of order p p1 and exponent p, G is a p-group of maximal class and R < G is of order p p and exponent p, then A < R. Next, A is contained in every irregular subgroup of G. (i) (Proposition 10.17) If B G is nonabelian of order p 3 such that CG .B/ < B, then G is of maximal class. (j) (Corollary 18.7) Suppose that an irregular p-group G is not of maximal class, k > 2. Then ck .G/ 0 .mod p p /, unless G is an Lp - or U2 -group. In particular, in that case, solk .G/ 0 .mod p pC2 /. (k) Let C < H < G, where G is a 2-group which is not of maximal class, jH W C j D 2, C is cyclic. Then the number of subgroups of G of order jH j, that are of maximal class, is even. (l) (Theorems 41.1 and 66.1) If G is minimal nonmetacyclic, then one of the following holds: (i) G is of order p 3 and exponent p, (ii) G is of order 34 and class 3, j1 .G/j D 9, (iii) G D C2 Q8 , (iv) G D Q8 C4 Š D8 C8 of order 24 , (v) G is special group of order 25 with jZ.G/j D 22 . Thus, if 2 .G/ is metacyclic then G is metacyclic. (m) (Lemma 44.1, Corollary 44.6, Theorem 9.11) A p-group G is metacyclic if one of the following quotient groups is metacyclic: G=ˆ.G 0 /K3 .G/, G=ˆ.G 0 /, G=K3 .G/. If p > 2 and jG=Ã1 .G/ p 2 , then G is metacyclic (Huppert). (n) (Theorem 44.5) A 2-group G is metacyclic if G and all its maximal subgroups are two-generator. (o) (Corollary 44.9) If G=Ã2 .G/ is metacyclic then G is metacyclic. (p) ([Bla5, Theorem 4.2]) If a p-group G, p > 2, and all its maximal subgroups are two-generator, then G is either metacyclic or Ã1 .G/ D K3 .G/ is of index p 3 in G (in the last case, jG W G 0 j D p 2 ). (q) (Tuan; see Lemma 1.1) If G is nonabelian and A < G is abelian of index p, then jGj D pjG 0 jjZ.G/j. (r) (Lemma 4.2). If jG 0 j D p, then G D .A1 As /Z.G/, where A1 ; : : : ; As are minimal nonabelian; then G=Z.G/ is elementary abelian. (s) (Taussky; Proposition 1.6) If G is a nonabelian 2-group with jG W G 0 j D 4, then G is of maximal class, i.e, dihedral, semidihedral or generalized quaternion. (t) If a nonabelian p-group G has a cyclic subgroup of index p, then G is either Mp n or p D 2 and G is of maximal class (see (s)).

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(u) (see Exercise 1.69) If U; V 2 1 are distinct, then jG 0 W U 0 V 0 j p. (v) (Lemma 1.4) Let M be a G-invariant subgroup of ˆ.G/. If Z.M / is cyclic, M is also cyclic. In particular, M cannot be nonabelian of order p 3 . (w) (Exercise 9.13) If a p-group G of maximal class, p > 2, has a subgroup H with d.H / > p 1, then G is isomorphic to a Sylow p-subgroup of Sp 2 . (x) (5) c1 .G/ 1 C p .mod p 2 / and, for k > 1, ck .G/ 0 .mod p/, unless G is either cyclic or a 2-group of maximal class. Under the last exceptions, G has a normal abelian subgroup of type .p; p/. (y) If hU 0 j U 2 1 i < G 0 , then d.G/ D 2. Let us prove parts (g), (h), (y) of Lemma 64.1. Proof of Lemma 64.1(g). Let jGj > p pC1 . If jGj D p pC2 , the result follows from Theorem 9.6. Now let jGj > p pC2 and H — G1 , where G1 is the fundamental subgroup of G. Let H < M 2 1 ; then M.¤ G1 / is of maximal class (Lemma J(e)) and M1 D M \ G1 D ˆ.G/ is the fundamental subgroup of M since jM j > p pC1 . Then, by induction on jGj, H is of maximal class since H 6 M1 . Proof of Lemma 64.1(h). We have A D 1 .ˆ.G//. One may assume that jGj > p pC1 (if jGj D p pC1 , then A D ˆ.G/). Let R < G be either of order p p and exponent p or irregular. Since j1 .R \ G1 /j D p p1 , the result follows. Proof of Lemma 64.1(y). Set D D hU 0 j U 2 1 i. Since, by hypothesis, D < G 0 , there exist x; y 2 G such that Œx; y 62 D. Assume that H D hx; yi M 2 1 ; then M 0 6 D, a contradiction. Thus, H D G. 2o . We begin with the following Definition 1 (Ito). Let G be a group of order p m . Then G is called one-stepped if there exist m elements x1 ; : : : ; xm of order p in G such that jhx1 ; : : : ; xi ij D p i for i D 1; : : : ; m. We consider the identity group f1g as one-stepped. Lemma 64.2. A p-group G is one-stepped if and only if 1 .G/ D G. Proof. We have to prove that if 1 .G/ D G, then G is one-stepped. Let H G be one-stepped of maximal order. Assume that H < G. Set N D NG .H /. If N H has an element x of order p, then H1 D hx; H i is one-stepped of order > jH j, a contradiction. Thus, H D 1 .N /, i.e., H is characteristic in N . Then N D G > H D 1 .G/ D G, a contradiction. Lemma 64.3. Suppose that A < G are one-stepped p-groups and jG W Aj D p n . Then there are x1 ; : : : ; xn 2 G A of order p such that jAi W Ai1 j D p, where A0 D A, Ai D hA; x1 ; : : : ; xi i for i D 1; : : : ; n.

182

Groups of prime power order

Proof. Suppose that Ai has constructed. Assuming that Ai < G, we have to construct AiC1 . Set Ni D NG .Ai /. As in the proof of Lemma 64.2, there is xiC1 2 Ni Ai of order p. Set AiC1 D hxiC1 ; Ai i. Then An D G. A series A D A0 < A1 < < An D G, constructed in Lemma 64.3, we call the Ito series or (Ito 1-series; see below) connecting A D A0 with G. 3o . In this subsection we give some estimates of the order of a p-group G D k .G/ in terms of ck .G/. For p-groups of maximal class and order > p pC1 these estimates are best possible. 1 Set .G/ D Œ p1p c1 .G/ and, for k > 1, set k .G/ D Œ p p1 ck .G/. Theorem 64.4. Suppose that a one-stepped p-group G is not a group of maximal class. Then jGj p pC1C .G/ . Proof. We use induction on jGj. One may assume that jGj D p m > p pC1 (otherwise, there is nothing to prove). If G is regular, then exp.G/ D p (Lemma 64.1(a)), c1 .G/ D 1 C p C C p m1 , and we have to check only that 1 m1 .1 C p C C p / D 1 C p C .1 C p C C p mp1 /: m 1CpC pp This is true since 1 C p C C p mp1 1 C m p 1 D m p. Now let G be irregular. Then there is R G G of order p p and exponent p (Lemma 64.1(b)) and x 2 G R of order p. Set A D A0 D hx; Ri. Let A D A0 < A1 < < An D G be an Ito series connecting A D A0 with G. Then jGj D jA0 j p n D p pC1Cn so it suffices to prove that n .G/. For i D 0; 1; : : : ; n 1 we have c1 .AiC1 / > c1 .Ai / and, since AiC1 is not of maximal class in view of R G Ai and jAiC1 j > p pC1 (Lemma 64.1(d)), we get c1 .AiC1 / c1 .Ai / .mod p p / (Lemma 64.1(f)), so c1 .G/ c1 .A0 / C n p p .1 C p C C p p1 / C n p p : However, by Lemma 64.1(f), c1 .G/ D 1CpC Cp p1 C .G/ p p so n .G/. Corollary 64.5. If a p-group G is not of maximal class, then j1 .G/j p pC1C .G/ so, if p D 2 and c1 .G/ D 4k C 3, then j1 .G/j 23Ck . Let G be one-stepped of maximal class and order p m > p pC1 and let G1 < G be the fundamental subgroup. Then R D 1 .G1 / G G is of order p p1 and exponent p. Let x 2 G R be of order p; then H D hx; Ri is of order p p and exponent p (Lemma 64.1(a)). By Lemma 64.1(g), NG .H / is of maximal class and order p pC1 so there are exactly jG W NG .H /j D p mp1 conjugates with H in G. All these conjugates contain exactly c1 .R/ C .c1 .H / c1 .R//p mp1 D 1 C p C C p p2 C p m2

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183

distinct subgroups of order p and they generate M 2 1 (take into account the above formula!). Since G is one-stepped, there is y 2 G M of order p. Setting H1 D hy; Ri, we, as above, obtain at least p m2 new subgroups of order p that together with R generate M1 2 1 fM g. Let G be the number of one-stepped members in the set 1 . Then (see Lemma 12.3) we get 1 C p C C p p2 C .G/ p p D c1 .G/ G p m2 : It follows that p m

p pC2 G

.G/. Since G is one-stepped, p G > 1 so we obtain

Theorem 64.6. Let G be a one-stepped p-group of maximal class and order > p pC1 . Then jGj p pC1 .G/ or, what is the same. jGj p c01 .G/, where c01 .G/ is the number of subgroups of order p not contained in G1 . Definition 2. A p-group G is said to be k-stepped if it has elements x1 ; : : : ; x t , all of order p k , such that G D hx1 ; : : : ; x t i and jAi W Ai1 j > 1, where Ai D hx1 ; : : : ; xi i and xi 2 NG .Ai1 / for i D 1; : : : ; t . If k .A/ D A D A0 < G and there are elements x1 ; : : : ; xn , all of order p k , such that, denoting Ai D hA0 ; x1 ; : : : ; xi i, we have xi 2 NG .Ai1 / and An D G, then the series A0 < A1 < < An D G is said to be an Ito k-series connecting A0 with G. If A0 < A1 < < An D G is an Ito k-series connecting A0 with G, then jG W A0 j p k n . As in Lemma 64.2, G is k-stepped if and only if k .G/ D G. If A D A0 is a k-stepped subgroup of a k-stepped p-group G, then there exists an Ito k-series A0 < A1 < < An < G connecting A0 with G. Lemma 64.7. Let R GG be of order p p and exponent p and let G=R be cyclic of order > p. Then exp.j .G// D p j for all j with p j p e D exp.G/. Next, e .G/ D G, i.e., G is e-stepped. If G=R is of order p e , then ce .G/ D p p . If G=R is of order p e1 (in that case 1 .G/ D R), then ce .G/ D p p1 . Proof. It suffices to show that exp.1 .G// D p. However, this follows from Lemma 17.4(c). Let k > 2 and suppose that a p-group G D k .G/ is not absolutely regular (by Theorem 13.19, G is not of maximal class); then there is R G G of order p p and exponent p (Lemma 64.1(b)). Take in G an element x0 of order p k and set A0 D hx0 ; Ri; then p pCk1 jA0 j jRjp k D p pCk . By Lemma 64.7, A0 is k-stepped. Let A0 < A1 < < An D G be an Ito k-series connecting A0 with G. Since ck .A0 / p p1 and ck .A1 / > ck .A0 / p p1 , A1 is not an Lp -group. Next, if p D 2 and k > 3, then A1 is not a U2 -group. Then, if p > 2, then ck .A1 / p p (Lemma 64.1(j)), and this is also true for p D 2. It follows that ck .AiC1 / > ck .Ai /, and, by Lemma 64.1(j), ck .AiC1 / ck .Ai / .mod p p /, i D 1; : : : ; n 1, so we have p

k .G/p p1 D ck .G/ ck .A1 / C .n h 1/p p i np , hence k .G/ np, and we conclude that jGj jA0 jp k n p

pCkCk

1 p k .G/

. Thus, we have

184

Groups of prime power order

Theorem 64.8. Suppose that k > 2 and a k-stepped p-group G of order p m is neither absolutely regular nor of maximal class. Then m p C k C pk k .G/. Corollary 64.9. Let k > 2 and let a p-group G be h i neither absolutely regular nor of maximal class. Then jk .G/j p

pCkCk

1 p k .G/

.

The case k D 2 we consider for p-groups of maximal class only. Let G D k .G/ be a p-group of maximal class, k > 1, jGj D p m p 2p and p > 2; then k D 2 (Lemma 64.1(e)) and exp.G/ > p 2 . Let c02 .G/ be the number of cyclic subgroups of order p 2 in G not contained in G1 . Set R D 1 .G1 /, where G1 < G is fundamental; then jRj D p p1 . Let x 2 G G1 be of order p 2 . Set A D hx; Ri; then A is absolutely regular of order p p since x p 2 Z.G1 / (Theorem 13.19). Set N D NG .A/. Since N 6 G1 and jN j > p p , N is of maximal class (Lemma 64.1(g)). Since A 6E G, we get N < G so A is not characteristic in N and hence 2 .N / > A. Then jN W Aj D p (every normal subgroup of index > p in N is characteristic) so there are exactly jG W N j D p mp1 subgroups conjugate with A in G; their intersection equals R so their contribution in c02 .G/ is equal to p mp1 c2 .A/ D p mp1 p p2 D p m3 . These G-conjugates of A generate M 2 1 (every normal subgroup of index p 2 in G is contained in ˆ.G/ < G1 ). We have G1 \ M D ˆ.G/. Since m 2p, we get 2 .G1 / ˆ.G/. Since G is 2-stepped, there is y 2 G M of order p 2 ; by the above, also y 2 G G1 . Set A1 D hy; Ri; then A \ A1 D R. As above, the G-conjugates of A1 generate M1 2 1 fM g; their intersection equals R again. Let G be the number of 2-stepped members in the set 3 1 . Then c02 .G/ D G p m3 so jGj D p m D pG c02 .G/. If p D 2, then G Š Q2m since 2 .G/ D G, and equality is attained. Thus we have Supplement to Theorem 64.8. If G is a 2-stepped p-group of maximal class and 3 order p2p , then jGj D pG c02 .G/, where G is the number of 2-stepped members in the set 1 . Theorem 64.10. Let a p-group G, c1 .G/ D 1 C p C p 2 and exp.1 .G// > p. Then p D 2, 1 .G/ D D8 C2 is of order 24 and one of the following holds: (a) G D D C , where D8 Š D, C is cyclic of order 4, D \ C D Z.D/. (b) G D DC , where D8 Š D G G, C is nonnormal cyclic of index 4 in G, D \ C D Z.D/ and CG .D/ D Z.G/ has index 2 in C , G=CG .D/ Š D8 . (c) G D DQ, where D8 Š D G G, Q is a nonnormal generalized quaternion group of index 4 in G, D \ Q D Z.D/, CG .D/ is a cyclic subgroup of index 2 in Q, G=CG .D/ Š D8 . If L is a cyclic subgroup of order 4 in Q such that L 6 CG .D/, then DL Š SD24 . Proof. Let p > 2. Let Ep 2 Š R G G and let x 2 G R be of order p in G R; then H D hx; Ri is of order p 3 and exponent p. Since c1 .G/ D c1 .H /, 1 .G/ D H is of order p 3 and exponent p, a contradiction. For p D 2, the assertion coincides with Theorem 43.9.

64

185

p-groups generated by elements of given order

4o . In this subsection we study p-groups G with small ck .G/. Proposition 64.11. Let G be a p-group, p > 2, with 1 .G/ D G, c1 .G/ D 1 C p C C p p and exp.G/ > p. Then: (a) G is irregular of order p pC2 ; all members of the set 1 have exponent p 2 . (b) d.G/ D 3, ˆ.G/ D G 0 is of order p p1 and exponent p, cl.G/ D p. (c) G=Ã1 .G/ is of order p pC1 and exponent p, Ã1 .G/ D Kp .G/. (d) c2 .G/ D p p . (e) 1 D fM1 ; : : : ; Mp 2 ; T1 ; : : : ; TpC1 g, where M1 ; : : : ; Mp 2 are of maximal class, T1 ; : : : ; TpC1 are not generated by two elements so regular and have expoTpC1 nent p 2 . Next, .G/ D iD1 Ti has exponent p 2 and index p 2 in G, where .G/=K3 .G/ D Z.G=K3 .G//. (f) G has exactly p C 1 subgroups of order p p and exponent p, all those subgroups contain ˆ.G/ so normal in G. (g) Exactly p 2 subgroups of G of order p p , containing ˆ.G/, have exponent p 2 . (h) If L is a subgroup of order p p and exponent p in G, then exactly p maximal subgroups of G, containing L, are of maximal class. (i) Let S 2 2 be of exponent p 2 . If S ¤ .G/ (see (e)), then S is contained in exactly p irregular members of the set 1 . Proof. By Theorems 12.3 and 7.2(b), G is neither of maximal class nor absolutely regular so there is R G G of order p p and exponent p (Theorem 12.1(a)). (a) By Theorem 64.1(a), G is irregular. By Theorem 64.4, jGj p pC1C .G/ D p pC2 , since .G/ D Œ p1p c1 .G/ D 1; here Œx is the integer part of x 2 R. Since 1 .G/ D G, we get G=R Š Ep 2 so exp.G/ D p 2 . We also have jGj D p pC2 . Assume that exp.H / D p for some H 2 1 . Then c1 .H / D c1 .G/ so .G D/ 1 .G/ D H has exponent p, a contradiction. Thus, all members of the set 1 have exponent p2 . (b) If x 2 G R is of order p, then exp.hx; Ri/ > p, by (a), so M D hx; Ri 2 1 is of maximal class (Theorem 7.1); then d.G/ D 3 and ˆ.G/ D G 0 D ˆ.M / D M 0 has exponent p, by (a) and Theorem 12.12(a), and we conclude that cl.G/ D p. (c) follows from Theorem 12.12(b). (d) follows from (a) and the hypothesis. Indeed, c2 .G/ D

jGj.p1/c1 .G/1 '.p 2/

D pp .

(e) The first assertion follows from Theorem 12.12(c). Now assume exp..G// D 1 p. Then c2 .Ti / D p.p1/ .jTi j j.G/j/ D p p1 for i D 1; : : : ; p C 1 so c2 .G/ D PpC1 p1 > p p , contrary to (d). Thus, exp..G// D p 2 . iD1 c2 .Ti / D .p C 1/p (f) Assume that S < G of order p p and exponent p is not normal in G; then ˆ.G/ D G 0 6 S so H D Sˆ.G/ 2 1 . It follows from Theorem 7.2(b) and (a) that

186

Groups of prime power order

H is irregular so it is of maximal class; in that case, ˆ.G/ D ˆ.H / (compare orders!) so H D S 62 1 , a contradiction. If t is the number of subgroups of order p p and exponent p in G, then 1CpC Cp p1 Cp p D c1 .G/ D c1 .ˆ.G//Ctp p1 D 1CpC Cp p2 Ctp p1 so t D p C 1. (g) follows from (a), (b) and (f). (h,i) By (e), the intersection of two distinct regular maximal subgroups of G coincides with .G/. Let L ¤ .G/ be a normal subgroup of index p 2 in G; then L is contained in at most one regular maximal subgroup of G and G=L Š Ep 2 since 1 .G/ D G. Let D be a G-invariant subgroup of index p 2 in L. Set C D CG .L=D/. Let L < H C , where H 2 1 ; then H is regular. It follows that L is contained in exactly one regular member of the set 1 , and the proof is complete. Exercise 1. Suppose that G D 1 .G/ is an irregular p-group containing exactly p C1 subgroups of order p p and exponent p. Then c1 .G/ D 1 C p C C p p . If p D 2, then 1 .G/ D Q8 C4 . Exercise 2. Let G be a p-group of maximal class, p > 2. Then 1 .G/ D G or 2 .G/ D G. (Hint. 1 .G/2 .G/ D 2 .G/ D G. Since any two normal subgroups of G of distinct orders are incident, we are done.) Proposition 64.12. Suppose that a p-group G, p > 2, is such that ck .G/ D p p2 for k > 1. Then one of the following holds: (i) G is regular with cyclic G=k1 .G/ and jk1 .G/j D p pCk3 . (ii) k D 2 and G is of maximal class and order p pC1 having exactly p subgroups of order p p and exponent p. Proof. (i) Let G be regular and jk .G/j D p a , jk1 .G/j D p b . Then p p2 D ck .G/ D

p ab 1 bkC1 jk .G/j jk1 .G/j D : p p1 p k1 .p 1/

It follows that a b D 1 so jk .G/=k1 .G/j D p, and we conclude that the quotient group G=k1 .G/ is cyclic. Then p 2 D b k C 1 so jk1 .G/j D p b D p pCk3 . (ii) Now let G be irregular. By Lemma 64.1(f), G is of maximal class. Assume that jGj > p pC1 . Let G1 be the fundamental subgroup of G; then jk1 .G1 /j D p .k1/.p1/ . If k > 2, then ck .G1 / D

jk .G1 /j jk1 .G1 /j p .k1/.p1/C1 p .k1/.p1/ ; '.p k / .p 1/p k1

p .k1/.p2/ > p p2 D ck .G/;

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a contradiction. Thus, k D 2. Then j2 .G1 /j p .p1/C2 D p pC1 so c2 .G1 /

p pC1 p p1 D p p2 .p C 1/ > p p2 D c2 .G/; p.p 1/

again a contradiction. Thus, jGj D p pC1 . The second assertion in (ii) is checked easily.

65

A2-groups

A p-group G is said to be an An -group if it contains a nonabelian subgroup of index p n1 but all its subgroups of index p n are abelian. Lemma 65.1 (= Exercise 1.8a (Redei)). Let G be a minimal nonabelian p-group (= A1 -group). Then G D ha; bi and one of the following holds: m

n

(a) ap D b p D c p D 1, Œa; b D c, Œa; c D Œb; c D 1, jGj D p mCnC1 , G D hbi .hai hci/ D hai .hbi hci/ is nonmetacyclic. m

n

(b) ap D b p D 1, ab D a1Cp

m1

, jGj D p mCn and G D hbi hai is metacyclic.

(c) a4 D 1, a2 D b 2 , ab D a1 , G Š Q8 . We have jG 0 j D p and H=Ã1 .H /j p 3 for H G. If jGj > p 3 and j1 .G/j p 2 , then G is metacyclic. The group G is nonmetacyclic if and only if G 0 is a maximal cyclic subgroup of G. Next, jG=Ã1 .G/j p 3 with equality if and only if G is from (a) and p > 2. If, in (a), u 2 G ˆ.G/, then hui is not normal in G. Suppose that G 0 < L < G, where L is cyclic of order p 2 . By Theorem 6.1, L=G 0 C =G 0 , where C =G 0 is a cyclic direct factor of G=G 0 ; in particular, G=C is cyclic. It remains to show that C is cyclic. We get G 0 D ˆ.L/ ˆ.C /. It follows that C =ˆ.C / as an epimorphic image of a cyclic group C =ˆ.L/, is cyclic so C is cyclic and G is metacyclic. The last assertion of Lemma 65.1 is a partial case of the following general fact. If a nonmetacyclic G D hu; vi, then hui 6E G. The aim of this section is to clear up the subgroup and normal structure of A2 groups. Using comparatively easy arguments, we obtain a lot of information on A2 groups which is difficult to read out from their defining relations. Some results which we shall prove in the sequel, are contained in the thesis of Lev Kazarin (unpublished), however, the proofs presented below, as a rule, are new. The length of [She], also devoted to classification of A2 -groups, indicates the degree of the difficulty of this problem. Every nonabelian group of order p 4 is either an A1 - or A2 -group. Therefore, in what follows, we assume that jGj D p m > p 4 . A2 -groups belong to a wider class of p-groups all of whose nonabelian subgroups are normal. Lemma 65.2. Let G be a nonabelian p-group. Then: (a) If G 0 1 .Z.G// and d.G/ D 2, then G is an A1 -group.

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(b) If G is metacyclic with jG 0 j D p 2 , then G is an A2 -group. (c) If G has two distinct abelian maximal subgroups, then jG 0 j D p. (d) Suppose that jH 0 j p for all H 2 1 . Then jG 0 j p 3 and G 0 is abelian. If F; H 2 1 are distinct and F 0 H 0 , then jG 0 j p 2 . If jG 0 j D p 3 , then G has no abelian maximal subgroups, and if, in addition, d.G/ > 2, then d.G/ D 3 and Ep 3 Š G 0 Z.G/. (e) If G 0 \ Z.G/ is cyclic and jH 0 j p for all H 2 1 , then jG 0 j p 2 . If, in addition, G 0 6 Z.G/, then d.G/ D 2 and G=.G 0 \ Z.G// is an A1 -group. Proof. (a) If a; b 2 G, then 1 D Œa; bp D Œa; b p so Ã1 .G/ Z.G/. Then ˆ.G/ D G 0 Ã1 .G/ Z.G/ so Z.G/ D ˆ.G/ has index p 2 in G. Hence, all members of the set 1 are abelian so G is an A1 -group. (b) By Lemma 65.1, G is not an A1 -group. Let L D Ã1 .G 0 / and H 2 1 ; then L < G 0 < H . By (a), G=L is an A1 -group so H=L is abelian; then, again by (a), H is either abelian or an A1 -group. Thus, G is an A2 -group. (c) follows from Lemma 64.1(u). (d) Let F; H 2 1 be distinct. Then jG 0 j pjF 0 H 0 j p 3 (Lemma 64.1(u)). If, in addition, F 0 H 0 , then jG 0 j p 2 . Now let jG 0 j D p 3 . Since F 0 6 H 0 and H 0 6 F 0 , 1 has no abelian members. Let, in addition, d.G/ > 2. Then A0 ¤ B 0 for distinct A; B 2 1 so c1 .G 0 / j1 j 1 C p C p 2 , and we get d.G 0 / D 3, Ep 3 Š G 0 Z.G/. Therefore, since 1 C p C p 2 D c1 .G 0 / D j1 j, we get d.G/ D 3. (e) We have D D hH 0 j H 2 1 i 1 .G 0 \ Z.G// so jDj D p since 1 .G 0 \ Z.G// is cyclic. Then G=D is either abelian or A1 -group so j.G=D/0 j p (Lemma 65.1), and we get jG 0 j p 2 . Let, in addition, G 0 6 Z.G/. Then G=D is nonabelian so it is an A1 -group. Since D < ˆ.G/, we get d.G/ D d.G=D/ D 2. Corollary 65.3. A metacyclic p-group G is an A2 -group if an only if jG 0 j D p 2 . Remarks. 1. Let us prove that if a p-group G has the cyclic derived subgroup G 0 , then all elements of G 0 are commutators. We have G 0 D hŒx; yi for some x; y 2 G. One may assume that G D hx; yi. By Lemma 64.1(u), there is H 2 1 such that H 0 D Ã1 .G 0 /. By induction, all elements of H 0 are commutators. By [BZ, Theorem 3.27], all generators of G 0 , i.e., members of the set G 0 H 0 , are commutators, completing the proof. 2. Let us prove that if G of order p 4 and exponent p has no nontrivial direct factors, it is of class 3. Let H be an A1 -subgroup of G; then jH j D p 3 . If Z.G/ 6 H , then G D H C , where C < Z.G/ is of order p such that C 6 H . Thus, Z.G/ < H so Z.G/ is of order p. Since CG .H / D Z.G/, we get CG .H / < H . Then G is of maximal class (Proposition 10.17). Lemma 65.4. Let G be an A2 -group of order p m > p 4 . (a) If K G, then jK=Ã1 .K/j p 4 , and this estimate is best possible.

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(b) If K < G with jK=Ã1 .K/j D p 4 , then Ep 4 Š 1 .K/ G G and, if 1 .K/ H 2 1 , then H is abelian. If, in addition, jG 0 j > p, then jG 0 j D p 2 , jG W Z.G/j D p 3 and G=1 .K/ is cyclic. (c) If K G G and G=K is generated by subgroups of index p 2 , then K Z.G/. In particular, if d.G/ D 3, then ˆ.G/ Z.G/. (d) Let jG 0 j D p. Then either G D H C , jC j D p or G D H Z, where H 2 1 is an A1 -subgroup and Z is cyclic of order p 2 . (e) Ã2 .G/ Z.G/. If p > 2 and G is not metacyclic, then Ã1 .G/ Z.G/ and jG=Z.G/j p 3 . (f) Let jG=Ã1 .G/j D p 4 , m > 5 and G 0 6 Z.G/. Then p > 2, G 0 Š Ep 2 , cl.G=Ã1 .G// D 3, Ã1 .G/ Z.G/ is cyclic, G=G 0 Š Cp m3 Cp so d.G/ D 2. If L=G 0 is a direct factor of G=G 0 of order p, then L Š Ep 3 and G D Z L is a semidirect product of L with a cyclic subgroup Z of order p m3 hence Ã1 .G/ D Ã1 .Z/, 1 .G/ D L 1 .Ã1 .G// Š Ep 4 , G 0 \ Ã1 .G/ D f1g and G is an L4 -group (see 17, 18). Next, G=.G 0 \ Z.G// is an A1 -group. (g) If jG 0 j D p 3 and d.G/ D 2, then p > 2 and jGj D p 5 . (h) If G=Z.G/ is nonabelian, then G=.G 0 \ Z.G// is an A1 -group. (i) Let G be metacyclic. Then G=Z.G/ is of order p 4 and exponent p 2 . If the number jG=Z.G/j D p 3 , then p D 2 and G=Z.G/ Š D8 . Now suppose that jG=Z.G/j D p 4 . If p D 2, then G=Z.G/ is abelian of type .4; 4/. If p > 2, then G=Z.G/ can be abelian or nonabelian of exponent p 2 (see examples following the lemma) (j) If G has no normal elementary abelian subgroups of order p 3 , then it is either metacyclic or minimal nonmetacyclic. Proof. (a) In view of Lemma 65.1, one may assume that K < G. Suppose that jK=Ã1 .K/j > p 3 . Then K is not a subgroup of an A1 -group (Lemma 65.1) so it is abelian. Considering K \ A, where A 2 1 is nonabelian, we get j1 .K/j p 4 (Lemma 65.1), and we are done since jK=Ã1 .K/j D j1 .K/j. If A2 -group G D U C , where U is a nonmetacyclic A1 -group of order > p 3 and jC j D p, then 1 .G/ is of order p 4 and exponent p. (b) Let jK=Ã1 .K/j D p 4 for K < G. Then every member of the set 1 containing K is abelian (Lemma 65.1) so Ep 4 Š 1 .K/ G G. Let, in addition, jG 0 j > p. Then G=1 .K/ is cyclic since the set 1 has exactly one abelian member (Lemma 65.2(c)). By Lemma 65.2(d), jG 0 j D p 2 so jG W Z.G/j D p 3 (Lemma 64.1(q)). (c) Let R=K < G=K be of index p 2 in G=K. Then R is abelian so R CG .K/ and K Z.G/ since G D hR < G j K < R; jG W Rj D p 2 i. Now the second assertion follows. (d) This follows from Lemma 64.1(r).

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(e) If there is K G G such that G=K is of order p 3 and exponent p, then K Z.G/, by (c), so we are done since Ã2 .G/ K. Now assume that such K does not exist. Then G=Ã1 .G/ Š Ep 2 . If Ã1 .G/ is cyclic, then p D 2, m D 4 and so Ã2 .G/ D Z.G/ (Lemma 64.1(t)). Now let Ã1 .G/ be noncyclic. Then Ã1 .G/ has a G-invariant subgroup T such that Ã1 .G/=T Š Ep 2 . In that case, Ã2 .G/ T and the result follows from (c). (f) Since d.G/ 3, we get p > 2. By (c), Ã1 .G/ < Z.G/. Next, G=Ã1 .G/ has an abelian subgroup A=Ã1 .G/ of index p; d.A/ d.A=Ã1 .G// D 3 so A is abelian (Lemma 65.1). Then jG 0 j D p 2 (Lemma 64.1(u)) so jG=Z.G/j D p 3 (Lemma 64.1(q)). By Corollary 65.3, G 0 Š Ep 2 since G is nonmetacyclic. By the above, A=Ã1 .G/ is the unique abelian maximal subgroup of G=Ã1 .G/, and we conclude that d.G=Ã1 .G// D 2; then d.G/ D 2 since Ã1 .G/ < ˆ.G/. It follows that cl.G=Ã1 .G// D 3 (Remark 2). Thus, G=Z.G/ is nonabelian so G 0 6 Z.G/. Then, by (c), G=G 0 is not generated by subgroups of index p 2 so it is abelian of type .p m3 ; p/. Let L=G 0 be a direct factor of G=G 0 of order p; then L is abelian of order p 3 since jG W Lj D p m3 > p 2 . We have G=L Š Cp m3 so there exists a cyclic Z < G with G D Z L (semidirect product) since Z \ L D f1g in view of jG=Ã1 .G/j D p 4 . We also see that Ã1 .Z/ D Ã1 .G/ (compare indices!) so exp.L/ D p, L Š Ep 3 . Then G 0 \ Ã1 .G/ L \ Ã1 .Z/ D f1g. Since a Sylow p-subgroup of Aut.L/ is nonabelian of order p 3 and exponent p, we get 1 .G/ D L 1 .Ã1 .G// Š Ep 4 (recall that m > 5). By remark, preceding the lemma, cl.G=Ã1 .G// D 3. Next, the last assertion follows from Lemma 65.2(a). (g) All members of the set 1 are A1 -groups (Lemma 65.2(d)) so they are generated by two elements (Lemma 65.1). By Lemma 64.1(n), p > 2. In that case, jG W G 0 j D p 2 (Lemma 64.1(p)) so jGj D jG W G 0 jjG 0 j D p 2 p 3 D p 5 . (h) Since G 0 6 Z.G/, we get jG 0 j p 2 , and, by (c), d.G/ D 2. Then T D 0 G \ Z.G/ has index p in G 0 . Indeed, this is true, if jG 0 j D p 2 . If jG 0 j D p 3 , there are nonabelian F; H 2 1 such that F 0 ¤ H 0 (Lemma 64.1(u)) so F 0 H 0 D T has index p in G 0 . Then G=T is an A1 -group since jG=T /0 j D p and d.G/ D 2 (Lemma 65.2(a,e)). (i) Since G is not an A1 -group, we have jG W Z.G/j > p 2 . Assume that jG W Z.G/j D p 3 . Then G=Z.G/ has exactly one cyclic subgroup of index p (otherwise, jG W Z.G/j D p 2 ) so G=Z.G/ Š D8 . Since Ã2 .G/ Z.G/, by (e), we get jG=Z.G/j p 4 . Suppose that p D 2 and GN D G=Z.G/ is nonabelian of order 24 . Then GN D ha; b j a4 D b 4 D 1; ab D a1 i. Set UN D hb 2 i. Then G=U Š D8 , and so U Z.G/, by (c), a contradiction. Thus, in that case, G=Z.G/ is abelian. (j) Let H 2 1 . Since 1 .H / G G, we get j1 .H /j p 2 so H is metacyclic (Lemma 65.1). Not all groups of Lemma 65.4(d) are A2 -groups. Indeed, let G D H C , where H D ha; b j a4 D b 4 D 1; ab D a1 i, C D hc j c 4 D 1i; H \ C D 1 .C /. The group G is an A2 -group if and only if c 2 D a2 .

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The group G D ha; b j a8 D b 4 D 1; Œa; b D a2 i is a metacyclic A2 -group with G=Z.G/ Š D8 . n 2 n2 The group G D ha; b j ap D b p D 1; Œa; b D ap i. n > 3, is metacyclic with G=Z.G/ Š Cp 2 Cp 2 . 3 2 Let p > 2 and G D ha; b j ap D b p D 1; Œa; b D ap i. Then G 0 D hap i, 2 Z.G/ D hap i, G=Z.G/ is nonabelian of order p 4 . The above three groups realize all possibilities of Lemma 65.4(i). Let G be a metacyclic 2-group with G=Z.G/ Š D8 . Then jG 0 j D 4 (Lemma 64.1(u)) so G is an A2 -group (Corollary 65.3). If G is a metacyclic p-group such that G=Z.G/ is of order p 4 and exponent p 2 , then G is an A2 -group. Indeed, if H 2 1 , then H=Z.G/ is abelian of type .p 2 ; p/ so H is either abelian or minimal nonabelian since d.H / D 2. Lemma 65.5. Let an A2 -group G of order > p 4 have no abelian maximal subgroups, and d.G/ D 2. Then: (a) If p D 2, then G is metacyclic. (b) If p > 2 and G is nonmetacyclic, then jGj D p 5 . K3 .G/ D Ã1 .G/ D Z.G/ is of order p 2 and all subgroups of index p 2 in G contain Z.G/. Proof. (a) follows from Lemmas 65.1 and 64.1(n). (b) The quotient group G=Ã1 .G/ is nonabelian of order p 3 and exponent p, Ã1 .G/ D K3 .G/, jG W G 0 j D p 2 (Lemma 64.1(p)) and Ã1 .G/ D Z.G/ (Lemma 65.4(c)). If T < G is of index p 2 , then T Z.G/ is abelian so Z.G/ < T . Next, we have jGj D jG W G 0 jjG 0 j p 5 so jGj D p 5 . Lemma 65.6. Let G be a group of order p m > p 4 and all members of the set 1 are generated by two elements. (a) If Z.G/ D ˆ.G/ has index p 3 in G, then G is an A2 -group. (b) If d.G/ D 2 and Z.G/ D Ã1 .G/ has index p 3 in G, then G is an A2 -group. Proof. (a) If H 2 1 , then d.H / D 2 hence all maximal subgroups of H are members of the set 2 so abelian, and H is either abelian or minimal nonabelian. (b) Given H 2 1 , it follows from d.H / D 2 that all maximal subgroups of H contain Z.G/ (since Z.G/ ˆ.G/) so abelian. Thus, again H is either abelian or minimal nonabelian. Theorem 65.7. Suppose that G is an A2 -group of order > p 4 . (a) If G 0 is cyclic of order > p, then G is metacyclic and jG 0 j D p 2 . (b) Suppose that jG 0 j D p 3 . Then every subgroup of index p 2 in G contains Z.G/. If d.G/ D 2, then p > 2 and jG W Z.G/j D p 3 . If d.G/ D 3, then ˆ.G/ D Z.G/ and G 0 Š Ep 3 . If p D 2, then d.G/ D 3 so G 0 Z.G/ is elementary abelian.

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(c) If p D 2, G is nonmetacyclic and D 2 1 is abelian, then G 0 Z.G/. (d) If G is nonmetacyclic, then G 0 is elementary abelian. (e) If G is nonmetacyclic, then jG=Z.G/j p 3 . (f) If G is nonmetacyclic and A; B 2 1 are distinct nonabelian, then Z.A/ D Z.B/ Z.G/. (g) If p D 2 and G=Z.G/ Š D8 , then G is metacyclic. (h) Suppose that G is nonmetacyclic and the set 1 has exactly one abelian member A; then jG 0 j D p 2 . If G 0 6 Z.G/, then p > 2 and G=Z.G/ is nonabelian of order p 3 and exponent p. If G 0 Z.G/, then G=Z.G/ Š Ep 3 . Proof. (a) Since G 0 is cyclic, all nonabelian maximal subgroups of G have the same derived subgroup D 1 .G 0 /. Then, by Lemma 64.1(u), jG 0 j D p 2 . Let a nonabelian F 2 1 . Since F 0 < G 0 < F , then F is metacyclic (Lemma 65.1). If all members of the set 1 are metacyclic, then G is also metacyclic (Theorem 69.1). Suppose that there is a nonmetacyclic A 2 1 ; then A is abelian. We have 1 .A/ Š Ep 3 since d.A/ D 3: F \ A is metacyclic of index p in A. Since all members of the set 1 which contain 1 .A/, are nonmetacyclic so abelian, G=1 .A/ is cyclic. Then G 0 < 1 .A/ so G 0 Š Ep 2 , a contradiction. Thus, A does not exist so, by what has been said before, G is metacyclic. (b) If T < G is of index p 2 , then T Z.G/ D T and jG W Z.G/j > p 2 since T is abelian and the set 1 has no abelian members (Lemma 65.2(d)); therefore, if H 2 1 , then Z.G/ ˆ.H / ˆ.G/. Suppose that d.G/ D 2. Since G is nonmetacyclic (Corollary 65.3 and (a)), we get p > 2 (Lemma 64.1(n) since all members of the set 1 are two-generator) and jG=Ã1 .G/j p 3 (Lemma 64.1(m)), Ã1 .G/ Z.G/ (Lemma 65.4(c)) and jG W Z.G/j D p 3 . Now let d.G/ D 3. Then G 0 ˆ.G/ Z.G/ (Lemma 65.4(c)) so ˆ.G/ D Z.G/ (Lemma 64.1(q)), exp.G 0 / exp.G=Z.G// D p, and we have G 0 Š Ep 3 . Let, in addition, p D 2. Then d.G/ D 3 since all members of the set 1 are two generator (Lemma 64.1(n)), and Œx; y2 D Œx; y 2 D 1 for x; y 2 G, so exp.G 0 / D 2 since G 0 is abelian (Burnside). (c) Assuming G 0 6 Z.G/, we get jG 0 j D 4 (Lemma 64.1(u)), G 0 Š E4 , by (a), and d.G/ D 2 (Lemma 65.4(c)). By Lemma 65.2(c), D is the unique abelian member of the set 1 . Since G=Z.G/ is nonabelian of order 8 (Lemma 64.1(q)), it has exactly one cyclic subgroup of index 2 so G=Z.G/ Š D8 . Let H 2 1 fDg; then H \ D D ˆ.G/. If a 2 H ˆ.G/ and x 2 D ˆ.G/, then G D ha; xi. We have o.Œa; x/ D 2.D exp.G 0 // and hŒa; xi 6E G (otherwise, G=hŒa; xi is abelian so jG 0 j D 2). Next, a2 2 D so 1 D Œa2 ; x D a2 .x 1 ax/2 D a2 .aŒa; x/2 D a2 a2 Œa; x2 ŒŒa; x; a D ŒŒa; x; a since ha; Œa; xi. H / has class at most 2 (Lemma 65.1) and exp.G 0 / D 2. Then a centralizes G 0 D hG 0 \ Z.G/; Œa; xi. It follows from G 0 < D that CG .G 0 / ha; Di D G, contrary to the assumption.

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(d) Assume that exp.G 0 / > p; then G 0 is abelian (Lemma 64.1(v)) of type .p 2 ; p/ since G 0 is noncyclic, by (a). By Lemma 65.1, H 0 1 .G 0 / < G 0 for all H 2 1 so d.G/ D 2 (Lemma 64.1(y)). By Lemma 65.2(d), all members of the set 1 as A1 -groups are two-generator so p > 2 and K3 .G/ D Ã1 .G/ has index p 3 in G hence jG W G 0 j D p 2 (Corollary 36.6). Since Ã1 .G/ Z.G/ (Lemma 65.4(c)), we get Ã1 .G/ D Z.G/. There is H 2 1 such that H 0 ¤ Ã1 .G 0 /. Then .G=H 0 /0 D G 0 =H 0 is cyclic of order p 2 so G=H 0 is metacyclic, by (a). This is a contradiction since G=Z.G/ is nonmetacyclic and H 0 Z.G/. (e) Assume that jG=Z.G/j > p 3 . Then p D 2, exp.G=Z.G// D 4 (Lemma 65.4(e)), G has no epimorphic images isomorphic to D8 or E8 (Lemma 65.4(c)) so d.G/ D 2. By Lemma 64.1(n), there exists A 2 1 with d.A/ > 2; then A is abelian (Lemma 65.1). Let B 2 1 be nonabelian; then Z.B/ D ˆ.B/ < ˆ.G/ < A so CG .Z.B// AB D G. It follows that jG W Z.G/j jG W Z.B/j D jG W BjjB W Z.B/j D 8, a contradiction. (f) By (e), G=Z.G/j p3 . Assume that Z.A/ ¤ Z.B/. Then Z.A/Z.B/. ˆ.G// has index p 2 in G so d.G/ D 2 and Z.G/ < ˆ.G/ (< since A is nonabelian); moreover, jG=Z.G/ D p 3 . It follows that Z.A/ D Z.G/. Similarly, Z.B/ D Z.G/. Then Z.A/ D Z.B/, and G is not a counterexample. (Clearly, (e) follows from (f).) (g) Assume that G=Z.G/ Š D8 and G is nonmetacyclic. Then G 0 6 Z.G/. If D=Z.G/ is the cyclic subgroup of order 4 in G=Z.G/, then D 2 1 is abelian. Then, by (c), G 0 Z.G/, a contradiction. (h) By Lemma 65.4(d), jG 0 j > p so jG 0 j D p 2 (Lemma 64.1(u)). First assume that 6 Z.G/. By (c), p > 2. By (e) and Lemma 65.4(e), G=Z.G/ is nonabelian of order p 3 and exponent p. Now let G 0 Z.G/. By Lemma 64.1(q), jG W Z.G/j D pjG 0 j D p 3 so G=Z.G/ Š Ep 3 . G0

Probably, (a) and (d) are most important parts of Theorem 65.7. We use those parts in the study of A3 - and A4 -groups in 72. Theorem 65.8. Let a two-generator A2 -group G be a 2-group. Then G is metacyclic. Proof. Assume that an A2 -group G is a nonmetacyclic two-generator 2-group. Then there is A 2 1 with d.A/ > 2 (Lemma 64.1(n)) so d.A/ D 3, by Schreier’s theorem (Appendix 25). Since A is not an A1 -group, it is abelian. Let 1 D fA; M; N g; then M and N are nonabelian (if M is abelian, then G is an A1 -group) so A1 -subgroups. We have Z.M / D ˆ.M / < ˆ.G/ < A so CG .Z.M // AM D G whence Z.M / D Z.G/ since Z.G/ < ˆ.G/. Similarly, Z.N / D Z.G/. Since a nonelementary abelian group G=Z.G/ of order 8 has two four-subgroups, we get G=Z.G/ Š D8 . By Lemma 65.4(c), if U G G is such that G=U Š D8 , then U D Z.G/. Since jG 0 j > 2 (Lemma 65.2(a)), A is the unique abelian member of the set 1 (Lemma 65.2(c)). Write GN D G=Ã1 .A/. Then GN is nonabelian of order 16 since d.A/ D 3 > 2 D N D 4, GN is not of maximal class. Then, by Proposition 10.17, GN d.G/. Since exp.G/

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has no nonabelian subgroups of order 8 (otherwise, d.G/ > 2) so it is an A1 -group. N D AN (Lemma 65.3) so Z.G/ N Š E4 . Let RN < Z.G/ N be of It follows that 1 .G/ 0 N N N N N N order 2 and R ¤ G ; then G=R Š D8 since a nonabelian group G=R has a subgroup N RN Š E4 . By the previous paragraph, R D Z.G/. Let B=R < G=R be cyclic of A= order 4; then B 2 1 fAg is abelian, a contradiction. Lemma 65.9. Let G be a nonmetacyclic p-group and R < G 0 be G-invariant of order p. If G=R is minimal nonabelian and all nonabelian maximal subgroups of G are two-generator, then G is an A2 -group and p > 2. Proof. By hypothesis, jG 0 j > p so G is not an A1 -group. We also have d.G/ D d.G=R/ D 2. Therefore, in view of Theorem 65.8, we have to prove that G is an A2 -group. Let M 2 1 be nonabelian. Then d.M / D 2 and M=R (as a maximal subgroup of G=R) is abelian which implies M 0 D R. Hence M is an A1 -group (Lemma 65.2(a)) and so G is an A2 -group. Theorem 65.10 ([CP] (for p D 2), [ZAX]). Suppose that a nonabelian p-group G is neither minimal nonabelian nor metacyclic nor minimal nonmetacyclic. If all nonabelian subgroups of G are metacyclic, then one and only one of the following holds: (a) G D M C , where M 6Š Q8 is a metacyclic A1 -group and jC j D p. (b) p > 2, d.G/ D 2, G D 1 .G/C , where 1 .G/ Š Ep 3 , C is a cyclic subgroup of index p 2 in G, CG D Ã1 .C / D Z.G/ is of index p 3 in G. Proof. Let us check that groups of (a) and (b) satisfy the hypothesis. This is clear for groups from (a). Now let G be as in (b) and V 2 1 . If 1 .G/ V , then V D 1 .G/Ã1 .C / so V is abelian since 1 .G/ is abelian and Ã1 .C / D Z.G/. Now assume that 1 .G/ 6 V . Then 1 .V / D V \ 1 .G/ Š Ep 2 and V =1 .V / is cyclic hence V has a cyclic subgroup of index p and so metacyclic. Now, assuming that G satisfies the hypothesis, we have to prove that G is either as in (a) or in (b). By hypothesis, there are in G two maximal subgroups M and A such that M is nonabelian so metacyclic and A is nonmetacyclic so abelian; then d.A/ > 2 and d.G/ d.M / C 1 D 2 C 1 D 3. Since M \ A is a metacyclic maximal subgroup of A, we get d.A/ D 3. Set E D 1 .A/; then Ep 3 Š E G G. By the product formula, G D ME so M \ E Š Ep 2 . All maximal subgroups of G containing E, are nonmetacyclic so abelian, hence d.G=E/ D d.M=.M \ E// 2 (Exercise 1.6(a) and Lemma 65.1). In what follows, A; M and E are such as defined in this paragraph. Let d.G=E/ D 2. Then there is a maximal subgroup B=E < G=E with B ¤ A so E A \ B D Z.G/ since B is abelian. If x 2 E M , then G D M X, where X D hxi. Let a noncyclic N < M be maximal. Then N X is abelian (otherwise, d.N / 2 so d.N X/ 3). Thus, M is a metacyclic A1 -group so G is as in (a). Now suppose that G=E is cyclic; then G 0 < E. (i) Let jG=Ej D p; then jM j D p 3 . If CG .M / < M , then G is of maximal class (Proposition 1.17) and p > 2 since G is not metacyclic, and E is the unique

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abelian maximal subgroup of G (Lemma 65.2(c)) so exp.G/ D p 2 since M 2 1 is metacyclic. If E D Z.G/ L, then LG D f1g so G is isomorphic to a subgroup of exponent p 2 of a Sylow p-subgroup of the symmetric group Sp 2 ; then G has a subgroup ¤ A of order p 3 and exponent p which is nonabelian and nonmetacyclic, a contradiction. Thus, G D M Z.G/. If Z.G/ is noncyclic, then G D M L is as in (a) (in that case, M 6Š Q8 since G is not minimal nonmetacyclic). If Z.G/ is cyclic, then, since jZ.G/j D p 2 , we get G D EZ.G/, by the product formula, so G is abelian, a contradiction. (ii) Now suppose that G=E is cyclic of order > p; then G 0 < E so jG 0 j p 2 . We have 1 .G=1 .G// < A=1 .G/ so 1 .G/ D 1 .A/ D E. Since M=.M \ E/ D M=1 .M / Š G=E is cyclic, we get M Š Mp n , n > 3, since M is nonabelian and has a cyclic subgroup of index p (Theorem 1.2). Thus, all nonabelian maximal subgroups of G are isomorphic to Mp n . (It follows that G is an A2 -group so one can use the classification of A2 -groups, however we prefer to present independent, more elementary, proof.) Let d.G/ D 2; then Z.G/ < ˆ.G/ and, since G is not an A1 -group, we get 0 G 6 Z.G/ (Lemma 65.2(a)) so G 0 Š Ep 2 ; then cl.G/ D 3, A is the unique abelian maximal subgroup of G and jG W Z.G/j D pjG 0 j D p 3 (Lemma 64.1(q)). Since G 0 < M , we get G 0 D 1 .M / so M=G 0 is a cyclic subgroup of index p of the abelian group G=G 0 . Since Z.G/ < M , then Z.G/ D Z.M / (compare indices!) so Z.G/ is cyclic. Assume that there is a cyclic U=Z.G/ of index p in G=Z.G/. Then U is abelian and metacyclic so U ¤ A, a contradiction. Thus, exp.G=Z.G// D p so G=Z.G/ is nonabelian of exponent p; then p > 2. We have Z.G/ < C < M , where C is cyclic of index p in M . Since jG W C j D p 2 , we get G D EC , and C is not normal in G since G 0 is noncyclic. Thus, G is as in (b). Now we assume that d.G/ D 3. Then G=G 0 has no cyclic subgroups of index p so jG 0 j D p, and we get jG W Z.G/j D pjG 0 j D p 2 (Lemma 64.1(q)). Since jA W Z.G/j D p and d.A/ D 3, the subgroup Z.G/ is noncyclic. By what has been proved already, M Š Mp n . In that case, 1 .Z.G// 6 M since Z.M / is cyclic. If L < Z.G/ of order p is not contained in M , then G D M L so G is as in (a). Exercise 1. Let G be a nonabelian p-group and d.G/ > 2. Prove that all members of the set 2 are abelian if and only if d.G/ D 3 and ˆ.G/ Z.G/. Exercise 2 (Janko). Let A < G be a maximal normal abelian subgroup of a nonabelian p-group G. Prove that, for x 2 G A, there exists a 2 A such that hx; ai is an A1 group. Exercise 3. Let a p-group G be neither abelian nor minimal nonabelian and let A1 .G/ be the set of all minimal nonabelian subgroups of G. Prove that if G is not generated by any proper subset of the set A1 .G/, then p D 2, G is an A2 -group and jA1 .G/j D 2. (Hint. Use Lemma 76.5. Try to give an independent proof.) Problem. Classify the p-groups all of whose nonnormal subgroups are abelian.

66

A new proof of Blackburn’s theorem on minimal nonmetacyclic 2-groups

Here we present a new proof, due to Janko, of Blackburn’s theorem on minimal nonmetacyclic 2-groups avoiding his tedious calculations. It is difficult to estimate the crucial role of this theorem in our book. Only case p D 2 is considered below since, in case p > 2, the original proof is easy but nonelementary. Theorem 66.1 (Blackburn). Suppose that G is a minimal nonmetacyclic 2-group. Then G is one of the following groups: (a) E8 . (b) The direct product C2 Q8 . (c) The central product Q8 C4 Š D8 C4 of order 24 . (d) The group G D ha; b; ci with a4 D b 4 D Œa; b D 1, c 2 D a2 , ac D ab 2 , b c D ba2 , where G is special of order 25 with exp.G/ D 4, 1 .G/ D G 0 D Z.G/ D ˆ.G/ D ha2 ; b 2 i Š E4 , M D hai hbi Š C4 C4 is the unique abelian maximal subgroup of G, and all other six maximal subgroups of G are isomorphic to X D hx; y j x 4 D y 4 D 1; x y D x 1 i (which is the metacyclic minimal nonabelian group of order 24 and exponent 4). Proof [Jan8]. Suppose that G is a minimal nonmetacyclic 2-group. Then Theorem 44.5 implies d.G/ D 3. If jGj 23 , then ˆ.G/ D f1g and so G Š E8 . Clearly, if G is abelian then a consideration of 1 .G/ shows that G D 1 .G/ Š E8 . Suppose that jGj D 24 . Since d.G/ D 3, G is neither abelian nor minimal nonabelian. Let Q be a nonabelian subgroup of order 8 in G. Since G is not of maximal class, Proposition 10.17 implies G D Q Z with jZj D 4 and Q \ Z D Z.Q/. If Z Š E4 , then Q Š Q8 (because in case Q Š D8 the group G would contain a (proper) subgroup isomorphic to E8 which is nonmetacyclic). Thus, G Š C2 Q8 . If Z Š C4 , then G Š Q8 C4 Š D8 C4 with Q8 \ C4 D Z.Q/. In what follows we assume that jGj 25 . Assume that G contains a nonabelian subgroup D of order 23 . If CG .D/ D, then, by Proposition 10.17, G is of maximal class and so G is metacyclic, a contradiction. Hence CG .D/ 6 D and take in CG .D/ a subgroup U of order 4 containing Z.D/. Then jD U j D 24 and D U is not metacyclic since d.D U / D 3, a contradiction. We have proved that every subgroup of order 8 in G is abelian.

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Groups of prime power order

Now suppose that jGj D 25 . Since d.G/ D 3, it follows that jˆ.G/j D 4. Assume ˆ.G/ 6 Z.G/. Then there exists an element x 2 G CG .ˆ.G//; then hˆ.G/; xi is nonabelian of order 8 since x 2 2 ˆ.G/, contrary to the result of the previous paragraph. Hence ˆ.G/ Z.G/. On the other hand, G 0 ˆ.G/ and suppose that jG 0 j D 2. Then G=G 0 is abelian and minimal nonmetacyclic of order 24 , contrary to the result of the first paragraph of the proof. Hence G 0 D ˆ.G/. For any x; y 2 G, we have Œx; y2 D Œx 2 ; y D 1 since G is of class 2, and so ˆ.G/ D G 0 Š E4 . In particular, exp.G/ D 4 and 1 .G/ D ˆ.G/. Assume that Z.G/ > ˆ.G/ so that jZ.G/j D 23 (recall that jGj D 25 ) and G has an abelian subgroup of index 2. But then Lemma 64.1(q) gives jGj D 2jZ.G/jjG 0 j D 26 , a contradiction. We have proved that 1 .G/ D G 0 D Z.G/ D ˆ.G/ Š E4 , and so G is a special group of order 25 . In fact, we shall show that the structure of G is uniquely determined. There are elements x1 ; x2 2 G G 0 such that Œx1 ; x2 D t1 , where t1 is an involution in G 0 . Then X D hx1 ; x2 i is a maximal subgroup of G since it is nonabelian so of order 24 , and X 0 D ht1 i since X=ht1 i is abelian. Since G=ht1 i is nonabelian in view of jG 0 j D 4, there is x3 2 G X such that (interchanging x1 and x2 if necessary) Œx1 ; x3 D t2 2 G 0 ht1 i. We have ht1 ; t2 i D G 0 and G D hX; x3 i D hx1 ; x2 ; G 0 ; x3 i D hx1 ; x2 ; x3 i. For Œx2 ; x3 we have one of the following possibilities: (1) Œx2 ; x3 D 1. In that case, hx2 ; x3 ; G 0 i is an abelian maximal subgroup of G. (2) Œx2 ; x3 D t1 and then Œx1 x3 ; x2 D Œx1 ; x2 Œx3 ; x2 D t1 t1 D 1. In that case, hx1 x3 ; x2 ; G 0 i is an abelian maximal subgroup of G. (3) Œx2 ; x3 D t2 and then Œx1 x2 ; x3 D Œx1 ; x3 Œx2 ; x3 D t2 t2 D 1. In that case, hx1 x2 ; x3 ; G 0 i is an abelian maximal subgroup of G. (4) Œx2 ; x3 D t1 t2 and then Œx1 x2 ; x2 x3 D Œx1 ; x2 x3 Œx2 ; x2 x3 D t1 t2 t1 t2 D 1. In that case hx1 x2 ; x2 x3 ; G 0 i is an abelian maximal subgroup of G. We see that at least one maximal subgroup M of G is abelian and so (being metacyclic of exponent 4) M Š C4 C4 . Since jZ.G/j D 4, M is also the unique abelian maximal subgroup of G, by Lemma 64.1(q). Take an element c 2 G M so that c 2 is an involution (in G 0 ). Such element exists (otherwise, hc; G 0 i Š E8 since exp.G/ D 4; this argument shows that 1 .G/ D G 0 ). Since Z.G/ D G 0 , we get CM .c/ D G 0 and c stabilizes the chain M > G 0 > f1g. Let a 2 M G 0 be an element (of order 4) with a2 D c 2 (recall that every involution in M is a square of an element from M ). If c normalizes hai, then ha; ci Š Q8 is nonabelian, a contradiction. Hence ac D at , where t 2 G 0 hai D G 0 ha2 i. Let b 2 M G 0 be such that b 2 D t . We get M D hai hbi and ac D ab 2 . Since Œb; c 2 G 0 hb 2 i (as above), it follows that either b c D ba2 or b c D b.a2 b 2 /. However, if b c D b.a2 b 2 /, then .cb/2 D cbcb D c 2 b c b D a2 b.a2 b 2 /b D 1. This is a contradiction since cb 62 G 0 and 1 .G/ D G 0 . Hence b c D ba2 and the structure of G is uniquely determined as stated in the theorem.

66 A new proof of Blackburn’s theorem on minimal nonmetacyclic 2-groups

199

It remains to show that jGj < 26 . Assume that G is a minimal counterexample. Then, considering an appropriate quotient group G=N , where N is a subgroup of order 2 in G 0 \ Z.G/, we get jGj D 26 . Let N be a subgroup of order 2 in G 0 \ Z.G/. Then G=N is isomorphic to the group (d) of our theorem. Set W =N D .G=N /0 D ˆ.G=N / Š E4 so that W D G 0 D ˆ.G/ is of order 8. Then W is abelian, by Burnside; moreover, W is abelian of type .4; 2/ since W is metacyclic and W =N Š E4 . We get N D ˆ.W / D Ã1 .W / D hzi and set W0 D 1 .W /. It is easily seen that W0 6 Z.G/. Indeed, if W0 Z.G/, then we consider a subgroup N0 of order 2 in W0 Š E4 with N0 \ N D f1g so that W =N0 Š C4 . On the other hand, G=N0 is also isomorphic to the group (d) of our theorem and therefore .G=N0 /0 Š E4 coincides with W =N0 Š C4 , a contradiction. We have exp.G/ 23 because G=W Š E8 and exp.W / D 4. Also, Z.G/ N and Z.G/ W since Z.G=N / D W =N . It follows that either Z.G/ D N or Z.G/ Š C4 with N < Z.G/ < W . Set C D CG .W0 / so that jG W C j D 2. If x 2 G C , then x 2 2 ˆ.G/ D W . If x 2 2 W0 , then hx; W0 i Š D8 is nonabelian of order 8, a contradiction. Thus x 2 2 W W0 is an element of order 4 and so all elements in G C are of order 8. In particular, exp.G/ D 23 . Since C D CG .W0 / is metacyclic, there are no involutions in C W0 and so W0 D 1 .G/. By the structure of G=N , G has exactly six maximal subgroups M1 ; M2 ; : : : ; M6 such that Mi =N Š X .i D 1; 2; : : : ; 6/, where X D hx; y j x 4 D y 4 D 1; x y D x 1 i is the metacyclic minimal nonabelian group of order 24 and exponent 4. For the seventh maximal subgroup M7 of G we have M7 =N Š C4 C4 . Hence M7 is either abelian of type .23 ; 22 / or M70 D N in which case M7 is minimal nonabelian (Lemma 65.2(a)) with exp.M7 / D 23 . In both cases W0 Z.M7 / and so M7 D C D CG .W0 / (indeed, if M7 is minimal nonabelian, W0 D 1 .M7 / and M7 =W0 is noncyclic so W0 Z.M7 /). We shall determine the structure of Mi , i 2 f1; 2; : : : ; 6g. We have ˆ.Mi / D W since Mi =W Š E4 and Mi is metacyclic. Also, Mi0 W and Mi0 > f1g is cyclic since Mi is metacyclic. Suppose jMi0 j D 2. The case Mi0 D N is not possible since in that case Mi =N would be abelian. Hence Mi0 ¤ N and so W0 D Mi0 N . But then CG .W0 / Mi M7 D G hence W0 Z.Mi /, contrary to what has been proved already. Hence Mi0 is cyclic of order 4 and W > Mi0 > N . Since CG .W0 / D M7 , we have W0 6 Z.Mi / for i < 7. Because Mi , i < 7, is metacyclic and exp.Mi / D 23 , Mi possesses a cyclic normal subgroup Zi > Mi0 with jZi j D 23 and Mi =Zi Š C4 . If Mi splits over Zi , then Mi D Zi Ui with Zi \ Ui D f1g and Ui Š C4 . But then 1 .Ui / W0 and 1 .Ui / centralizes Zi since Aut.Zi / Š E4 . In that case W0 Z.Mi /, a contradiction. Hence Mi does not split over Zi . Let ai 2 Mi Zi be such that hai i covers Mi =Zi . Since exp.Mi / D 23 , we have hai i \ Zi D N D hzi. Then ai2 is of order 4, ai2 2 W W0 , and ai2 62 Mi0 since ai2 62 Zi , by the choice, and Mi0 < Zi . Set hzi i D Zi so that ziai D zi1 z , D 0; 1 (recall that z 2 N # /. We have 2 .Mi / D W and so Mi is uniquely determined according to Lemma 42.1. In

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Groups of prime power order

particular, we may assume D 0 and we have Z.Mi / D hai2 i < W . We see that Mi0 and Z.Mi / are two distinct cyclic subgroups of order 4 in W . On the other hand, c2 .W / D 2 and let W1 and W2 be two cyclic subgroups of order 4 in W . For a given i < 7, one of these two subgroups is equal to Z.Mi / and the other one is equal to Mi0 . Therefore, we have (interchanging W1 and W2 if necessary) W1 D Z.Mi / D Z.Mj / for certain i ¤ j , i; j 2 f1; 2; : : : ; 6g. We get CG .W1 / hMi ; Mj i D G and so W1 D Z.G/. In that case W1 D Z.Mi / and W2 D Mi0 for all i < 7. It follows that all seven maximal subgroups of G=W2 are abelian so G=W2 is abelian. Then E4 < G 0 W2 Š C4 , a final contradiction. For another proof of Theorem 66.1, see Theorem 69.1. Lemma 66.2 (= Theorem 1.25). Suppose that all nonnormal subgroups of a non-Dedekindian 2-group G have order 2. Then one of the following holds: (a) G Š M2n ; (b) G D Z G0 , where Z is cyclic and G0 D D8 ; (c) G D D8 Q8 . Exercise 1. Classify the p-groups all of whose subgroups of index p 2 are absolutely regular. Exercise 2 ([Pas]). Let G be a non-Dedekindian 2-group all of whose nonnormal subgroups have the same order 2n > 2. Then either G is metacyclic or n D 2. Solution. As in Supplement to Theorem 1.25, E4 Š 1 .G/ Z.G/, unless Q Š Q16 , and all nonnormal subgroups of G are cyclic. Suppose that G is nonmetacyclic. It follows from Theorem 66.1 that if M is a minimal nonmetacyclic subgroup of G then M is either the group of Theorem 66.1(d) or M Š Q8 C2 or is the group of Theorem 66.1(d). In the first case, M has a subgroup X Š ha; b j a4 D b 4 D 1; ab D a1 i. Since X has a nonnormal subgroup hbi of order 4, we get n D 2. In the second case, G has a subgroup Q Š Q8 , and Q G G since Q is noncyclic. Set C D CG .Q/. Assume that C has a cyclic subgroup Z of order 4. If Q \ Z > f1g, then QC D Q Z Š D8 Z has a nonnormal subgroup of order 2 (see Appendix 16, Subsection 1o ), a contradiction, since n > 1. If Q \ Z D f1g, then Q Z has a nonnormal cyclic subgroup of order 4 (Theorem 1.20) since 1 .Q Z/ Z.G/, and we get n D 2. Thus, one may assume that C is elementary abelian. Next, G=C is isomorphic to a 2-subgroup of Aut.Q/ Š S4 that contains a subgroup which is isomorphic to Q=Z.Q/ Š E4 . If G=C Š E4 , then G D Q C D Q E, where E is a subgroup of index 2 in C ; in that case G is Dedekindian, a contradiction. Thus, G=C Š D8 . If x 2 G C is such that hxC i is not normal in G=C , then the subgroup hxi is not normal in G. Since o.x 2 / exp.C / D 2 and n > 1, we get o.x/ D 4 so n D 2. Exercise 3. Classify the groups of Exercise 2 containing a subgroup Q Š Q8 . Exercise 4. If a 2-group of Exercise 2 is metacyclic, then either G Š Q16 or G D m n m1 ha; b j a2 D b 2 D 1; ab D a1C2 i, n m (in the second case, G is minimal nonabelian).

66 A new proof of Blackburn’s theorem on minimal nonmetacyclic 2-groups

201

Exercise 5. Suppose that a nonmetacyclic 2-group G of Exercise 2 has a subgroup such as the group of Theorem 66.1(d). If Z is a nonnormal subgroup of G of order 4 then, all nonnormal subgroups of G=1 .Z/ have the same order 2. Solution. Clearly, Z is cyclic. Suppose that H=1 .Z/ 6E G=1 .Z/; then H 6E G. Since G is nonmetacyclic, we get jH j D 4 (Exercise 2) so jH=Zj D 2. Exercise 6. Suppose that a 2-group G of Exercise 2 has a nonnormal (cyclic) subgroup Z. Then G=ZG is such as in Lemma 66.2. If G=ZG Š M2t , then G is metacyclic. Hint. It remains prove that if G=ZG Š M2n , then G is metacyclic. Assume that this is false. Then jZj D 4, by Exercise 2. The group G=ZG has two distinct cyclic subgroups A=ZG and B=ZG of index 2. The subgroups A and B are abelian so A \ B D Z.G/. Since ZG D ˆ.Z/ ˆ.G/, we get dG/ D d.G=ZG / D 2, and we conclude that G is minimal nonabelian. As above, 1 .G/ Š E4 so G is metacyclic and G is not a counterexample. Exercise 7. Study the non-Dedekindian 2-groups G all of whose nonnormal abelian subgroups are cyclic of the same order 2n > 2. (The groups Q2n , n > 3, satisfy this condition.) Hint. We have 1 .Z.G// D 1 .G/ Š E4 , unless G is a generalized quaternion group (indeed, given U < G nonnormal cyclic, let us consider the abelian U 1 .Z.G//). All cyclic subgroups of G=1 .G/ are normal so this group is Dedekindian. Assume that G=1 .G/ is nonabelian. Then G=1 .G/ contains a subgroup L=1 .G/ Š Q8 (Theorem 1.20). Since all subgroups of L containing 1 .G/, are abelian, we get jL W Z.L/j D 4 so jL0 j D 2. It follows that L0 < 1 .G/, a contradiction since L=1 .G/ is nonabelian. Thus, G=1 .G/ is abelian. Now assume that G has a subgroup Q Š Q8 and set C D CG .Q/. If G is metacyclic, it is generalized quaternion (Proposition 1.19). The subgroup C is elementary abelian of order 4 since G is not of maximal class (Proposition 1.17). It follows that QC D Q L, where jLj D 2. Since all noncyclic abelian subgroups of QC are normal in G and QC is generated by these subgroups, we get QC GG. It follows that G=C Š E4 since G=C is isomorphic to a subgroup of Aut.Q/ Š S4 containing a subgroup Š Q=Z.Q/ Š E4 . In that case, G D Q E, where E is a subgroup of index 2 in C , so G is Dedekindian, contrary to the hypothesis. Thus, G has no subgroups isomorphic to Q8 . It follows that if M is a minimal nonmetacyclic subgroup of G, then M is isomorphic to a group of Lemma 66.1(d). Problem. Classify the nonmetacyclic 2-groups all of whose minimal nonmetacyclic subgroups have order 16.

67

Determination of U2-groups

Recall that a 2-group G is said to be a U2 -group if it contains a normal four-subgroup R such that G=R is of maximal class and, provided T =R is cyclic of index 2 in G=R, then 1 .T / D R (see 17, 18). The subgroup R is said to be the kernel of G. The kernel R is the unique normal four-subgroup of G. Theorem 67.1 (Janko [Jan8]). Let G be a U2 -group with d.G/ D 3. Then there is M 2 1 of maximal class, and one of the following holds: (a) G D M C2 ; (b) G D M C4 with M \ C4 D Z.M /; (c) G D M hui, where u is an involution inducing on M an involutory “outer” automorphism such that CM .u/ D M1 is of maximal class, jM W M1 j D 2, and for each x 2 M M1 , x u D xz with hzi D Z.M / (in particular, z 2 D 1). Proof. Let G be a U2 -group with the kernel R and d.G/ D 3. Then G possesses a maximal subgroup M such that R 6 M since, by hypothesis, R 6 ˆ.G/. In that case, M \ R D hzi is of order 2 and M=hzi Š G=R so M is nonabelian. Next, M has no G-invariant abelian subgroups of type .2; 2/ since R 6 M so M is of maximal class (Lemma 1.4). Since jM j > 23 , M=hzi Š G=R is dihedral. Let u 2 R hzi. If R Z.G/, then G D hui M and we are done. Suppose R 6 Z.G/ and set M1 D CM .R/ D CM .u/ so that jM W M1 j D 2. If M1 is cyclic, then take an element v of order 4 in M1 and x 2 M M1 . Since R# D fu; z; uzg and u and z are not G-conjugate, we have .uv/x D ux v x D .uz/v x D .uz/v 1 D .uz/.vz/ D uv and so C D huvi Š C4 and C Z.G/ since hM M1 i D M and CM D G. In this case G D M C with C \M D hzi D Z.M / and we have obtained the group from (b). Now suppose that M1 is of maximal class. Then u induces on M an outer involutory automorphism such that M1 D CM .u/ is a maximal subgroup of M and for each x 2 M M1 , x u D xz. In the rest of this section we assume that G is a U2 -group with the kernel R and d.G/ D 2. Then G=R is of maximal class and order 2n , n 3, and ˆ.G/ R. Let T =R (of order 2n1 ) be a cyclic subgroup of index 2 in G=R and let hai be a cyclic subgroup of T which covers T =R. Since 1 .T / D R, hai is of order 2n and hai \ R D hzi is of order 2. Hence T is either abelian of type .2n ; 2/ or T Š M2nC1 .

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67 Determination of U2 -groups

In any case, ˆ.T / D ha2 i, z 2 ha2 i, ha2 i is normal in G, and z 2 Z.G/. Let n2 v D a2 so that o.v/ D 4, v 2 D z, A D Rhvi D 2 .T / is abelian of type .4; 2/, and hvi D A \ ˆ.T / is normal in G. Hence A=R D Z.G=R/ and so for each x 2 G T , x 2 2 A since G=R is of maximal class. Since Ã1 .G/ D ˆ.G/ D Rha2 i is of exponent 2n1 , we get hai 6 ˆ.G/ so, if hai is normal in G, G is metacyclic. If G=R Š Q2n , then for each x 2 G T , x 2 2 A R. All elements in A R are of order 4 and so o.x/ D 8. In this case 2 .G/ D 2 .T / D A and so G is metacyclic and is isomorphic to a group (c) of Lemma 42.1. Therefore, we may assume in the sequel that G=R 6Š Q2n . In particular, T =R is the unique cyclic subgroup of index 2 in G=R. Conversely, assume that G is metacyclic. Then G 0 is cyclic, G 0 ˆ.G/ D Rha2 i, jG 0 \ Rj D 2, and G 0 covers .Rha2 i/=R. Thus jG 0 j D 2n1 . Let S be a cyclic normal subgroup of G such that G 0 < S and jS W G 0 j D 2. But then jS j D 2n and .RS /=R is a cyclic subgroup of index 2 in G=R. The uniqueness of T =R implies RS D T . But cn .T / D 2, and so hai is normal in G. We have proved that (under our assumptions) G is metacyclic if and only if hai is normal G. We shall use the above notation in the rest of this section. Theorem 67.2. Let G be a metacyclic U2 -group with the kernel R. Then one of the following holds. (a) G D ha; b j a2 D b 4 D 1; ab D a1 z ; D 0; 1; z D a2 where R D hb 2 ; zi D Z.G/ and G=R Š D2n . n

n1

; n 3i,

(b) G is isomorphic to a group (c) of Lemma 42.1 and here G=R Š Q2n and Z.G/ Š C4 . (c) G D ha; b j a2 D b 4 D 1; ab D a1C2 ; z D a2 R D hb 2 ; zi, G=R Š SD2n , and Z.G/ D hzi Š C2 . n

n2

n1

; n 4i, where

Proof. Suppose that G is a metacyclic U2 -group with the kernel R, where G=R 6Š Q2n (for the case G=R Š Q2n , see the second paragraph following the proof of Theorem 67.1). Then hai is normal in G, by the paragraph preceding the theorem. Since G is not of maximal class, R D 1 .G/. Assume that G=R is dihedral. Take x 2 G T ; then x 2 2 R so H D hx; Ri is of order 8 and abelian (otherwise, G would be of maximal class). It follows that CG .R/ hG T i D G. We have obtained the group (a). Suppose now that G=R Š SD2n , n 4. There is b 2 G T with b 2 2 R. If 2 b 2 hzi, then ha; bi is a maximal subgroup of G, by the product formula so R < ˆ.G/ < ha; bi. However, this is a contradiction since we must have ha; bi=R D G=R so that ha; bi D G. Thus b 2 D u 2 R hzi. We have, in view of the structure of n2 G=R that ab D a1 vz . D 0; 1/, where v D a2 (recall that hai is normal in G). We have .a4 /b D .ab /4 D .a1 vz /4 D a4 , and so v b D v 1 since v 2 ha4 i 2 (recall that hai is normal in G). We compute further ab D au D .a1 vz /b D .a1 vz /1 v 1 z D av 2 D az, so au D az and therefore T Š M2nC1 . If D 1, 0 then we replace b with b 0 D bu; then .b 0 /2 D b 2 D u and ab D abu D .a1 vz/u D

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Groups of prime power order

a1 zvz D a1 v. Hence, we may assume from the start that ab D a1 v, and we have obtained the group (c). Theorem 67.3. Let G be a nonmetacyclic U2 -group with d.G/ D 2 and the kernel R. Then one of the following holds: n n1 (a) G D ha; b j a2 D 1; a2 D z; b 2 D z ; D 0; 1; ab D a1 u; u2 D Œu; a D Œu; b D 1; n 3i; where R D hu; zi D Z.G/ and G=R Š D2n . D z; at D a1 u; u2 D 1; Œu; a D (b) G D ha; t j a2 D t 2 D 1; a2 Œu; t D z; n 3i, where R D hu; zi and G=R Š D2n . If n D 3, then Z.G/ D ha2 i Š C4 and if n > 3, then Z.G/ D hzi Š C2 . n

n1

D z; a2 D v; at D a1 vu; u2 D (c) G D ha; t j a2 D t 2 D 1; a2 t Œu; a D 1; u D uz; n 4i, where R D hu; zi, G=R Š SD2n , and Z.G/ D hvui Š C4 . n

n

n1

n1

n2

n2

D z; a2 D v; b 2 D z ; D 0; 1; ab D (d) G D ha; b j a2 D 1; a2 1 2 a a vu; u D Œu; b D 1; u D uz; n 4i, where R D hu; zi, G=R Š SD2n and Z.G/ D hzi Š C2 . Proof. Suppose that G is a nonmetacyclic U2 -group with the kernel R and d.G/ D 2. In that case G=R 6Š Q2n (see the proof of Theorem 67.2) so G=R has the unique cyclic subgroup T =R of index 2. In that case, T is either abelian of type .2n ; 2/ or T Š M2nC1 . Let hai be a cyclic subgroup of index 2 in T ; then hai is not normal in G (see the paragraph preceding Theorem 67.2). Let z be an involution in hai; then z 2 Z.G/. (i) Suppose that G=R Š D2n and T is abelian. If x 2 G T , then x 2 2 R. Since Ã1 .G/ D ˆ.G/ R and Ã1 .T / does not contain R, there is b0 2 G T with b02 D u0 2 R hzi and so R Z.G/. But hai is not normal in G and so ab0 D a1 u0 z . D 0; 1/. We have .b0 a/2 D b0 ab0 a D b02 ab0 a D u0 a1 u0 z a D z . Hence, setting b0 a D b and u0 z D u, we have b 2 D z and ab D a1 u. We have obtained group (a). (ii) Now suppose that G=R Š D2n and T Š M2nC1 . In that case, R 6 Z.G/. Since 1 .G=R/ D G=R, it follows that there is a subgroup H=R of order 2 such that H 6 T and H is nonabelian. Then H Š D8 . Let t 2 H R be an involution. Then at D a1 u with u 2 R hzi, ut D uz, and ua D uz. We have obtained the group (b). (iii) It remains to consider the case where G=R Š SD2n , n 4. If Q=R is the generalized quaternion subgroup of index 2 in G=R, then .T \ Q/=R is a cyclic subgroup of index 2 in Q=R and 1 .T \ Q/ D R. Thus Q is a U2 -group with the kernel R. Since Q=R is generalized quaternion, Q is metacyclic in view of 2 .Q/ D 2 .T / is of order 8 (see Lemma 42.1), and the same lemma implies that Z.Q/ Š C4 . In particular, R 6 Z.G/. Suppose, in addition, that T is abelian. Then CG .R/ D T . If D=R is the dihedral subgroup of index 2 in G=R, then 1 .D \ T / D R and so D is a U2 -group with

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67 Determination of U2 -groups

R 6 Z.D/. In particular, there is an element x 2 D T such that x 2 2 R and Rhxi Š D8 . It follows that there is an involution t 2 D T acting faithfully on R (see (ii)) and, since hai is not normal in G, we get at D a1 vu with u 2 R hzi, ut D uz, Œu; a D 1. We have obtained the group (c). Finally, suppose that T is nonabelian, i.e., T Š M2nC1 so that hai acts non-trivially on R. Then, in view of R 6 Z.G/, M D CG .R/ is a maximal subgroup of G, where M=R is noncyclic so dihedral, T \ M D ha2 iR, ˆ.T / D ha2 i is normal in G, 1 .T \ M / D R, and so M is a U2 -group. Since R Z.M /, M=R is dihedral. Indeed, if Q=R is the generalized quaternion subgroup of index 2 in G=R, then we know that Q is metacyclic. Since d.G/ D d.T / D d.Q/ D 2 and G is, by assumption, nonmetacyclic, Theorem 44.5 implies that d.M / D 3. Hence M is isomorphic to a group (a) of Theorem 67.1. In particular, there is b 2 M T such that b 2 2 hzi. We set b 2 D z , D 0; 1. Finally, b centralizes R, hai acts non-trivially on R, and ab D a1 vu with u 2 R hzi. We have obtained the group (d). Corollary 67.4 (Berkovich). Suppose that a 2-group G of rank three has order 2nC2 and class n > 2. Then one of the following holds: (a) G D H1 C , where H1 is of maximal class and jC j D 2 (three groups); all four subgroups of maximal class and index 2 are isomorphic. (b) G D H1 C , where H1 is of maximal class and C D Z.G/ is cyclic of order 4, C \ H1 D Z.H1 /. In what follows we assume that Z.G/ D Z.H1 / so Z.G/ is of order 2. (c) G D ha; b; x j a2 D b 2 D x 2 D 1; ab D a1 ; ax D a1C2 n1 a2 bi. Here H1 D ha; bi Š D2nC1 . n

n1

; bx D

(d) G D ha; b; x j a2 D x 2 D 1; a2 D b 2 ; ab D a1 ; ax D a1C2 n1 a2 bi. Here H1 D ha; bi Š Q2nC1 .

n1

; bx D

(e) G D ha; b; x j a2 D b 2 D x 2 D 1; ab D a1C2 n1 a2 bi. Here H1 D ha; bi Š SD2nC1 .

n1

; bx D

n

n

n1

n1

; ax D a1C2

Proof. Let K D K3 .G/; then jG 0 W Kj D 2. We have jZ.G=K/j D 4 and G=K is not minimal nonabelian. Let H=K < G=K be minimal nonabelian; then G D K.G/, where .G/=K D Z.G=K/. It follows from Theorem 1.40 that cl.H / D cl.G/ so H is of maximal class. By Lemma 1.4, G has a normal subgroup R of type .2; 2/. Since jH j > 8, we get R 6 H so G D HR. Since G=R Š H=.H \ R/ Š D2n , we conclude that G is a U2 -group. Now the result follows from Theorem 67.1. Exercise. Let G be a group of order p m , cl.G/ D m 2 and G=G 0 Š Ep 3 . Let L be a G-invariant subgroup of index p in G 0 . Let H=L < G=L be nonabelian of index p. Prove that H is of maximal class.

68

Characterization of groups of prime exponent

n Recall Sn that the set † D fAi g1 of subgroups of a group G is a partition of G if G D 1 Ai and Ai \ Aj D f1g for i ¤ j . In what follows we assume that † is a nontrivial partition of a group G (this means that n > 1 and Ai > f1g for all i ). Subgroups Ai are called components of †. If all components of † have equal order, then G is said to be equally partitioned [Isa8]. It follows from Cauchy’s lemma that, for equally partitioned G, we have .A1 / D .G/. In this section we characterize groups of prime exponent as equally partitioned groups proving the following Isaacs’ result [Isa8]: a group G is equally partitioned by a nontrivial partition † if and only if G is of prime exponent. Note that this theorem is not a consequence of known results on partitioned groups.

Lemma 68.1. Let † be a nontrivial partition of a group G and x; y 2 G # with xy D yx. If x and y lie in different components of †, then o.x/ D o.y/ is a prime. Proof. Suppose that o.x/ < o.y/. Then .xy/o.x/ D y o.x/ ¤ 1, and so xy and y lie in the same H 2 †. Then x D .xy/y 1 2 H , which is not the case. Thus, o.x/ D o.y/. If n > 1 is a proper divisor of o.x/, the nonidentity commuting elements x n and y of different orders lie in different components of †, contrary to what has just been proved. Thus o.x/ D o.y/ is a prime. If a component H 2 † does not contain Z.G/, then exp.H / D p for some prime p. Indeed, if x 2 H # and z 2 Z.G/ H , then x and xz commute and lie in different components of † so o.x/ D o.xz/ is a prime, say p (Lemma 68.1); then o.z/ D p. Assume that y 2 H # is of prime order q ¤ p. In that case, o.y/ D o.yz/ D q so o.z/ D q ¤ p, a contradiction. It follows that, if a p-group G of exponent > p admits a nontrivial partition † and Z.G/ H 2 †, then H is the unique component of † of composite exponent so H contains the Hudges subgroup Hp .G/ of G. Lemma 68.2 ([Isa8]). Let † be a nontrivial partition of an equally partitioned group G and ¿ ¤ X G # . Then there exists H 2 † such that X z 6 H for all z 2 G. Proof. Suppose that the lemma is false and for each H 2 †, choose XH , a Gconjugate of X, with XH H . Let NH D NH .XH / so that H contains at least jH W NH j conjugates of X. Let N D NG .X/, g D jGj, h D jH j. We have jG W N j D jG W NG .XH /j jG W NH j D jG W H j jH W NH j

68

Characterization of groups of prime exponent

207

(the first equality follows since X and XH are conjugate in G), and hence jH W NH j g1 g gh g g1 h g jG W N j. Since h1 h D h.h1/ > 0, we get h < h1 D j†j. Now jG W N j is the number of conjugates of X in G and thus, since † is a partition, we get jG W N j

X

jH W NH j D j†jjH W NH j j†j

H 2†

so

g h

j†j D

g1 , h1

h jG W N j; g

a contradiction.

Lemma 68.3 ([Isa8]). Let † be a nontrivial partition of an equally partitioned group G. Then every element of G # has a prime order. Proof. Suppose that x 2 G has a composite order and let K D Kx be the conjugacy G-class of x. By Lemma 68.2, there exists H 2 † such that x z 62 H for all z 2 G so that K \ H D ¿. By Lemma 68.1, no element of H # centralizes any element of K. Thus, H acts semi-regularly by conjugation on K and hence jH j divides jKj. Since 1 2 H and 1 62 K it follows that jH j < jKj. Now pick F 2 † with x 2 F . Since jF j D jH j < jKj, the set K F is nonempty, and, obviously, this set is F -invariant. By Lemma 68.1, F acts on K F semi-regularly via conjugation so that jH j D jF j divides jK F j > 0. Since K D .K F / [ .K \ F / is a partition, jF j divides jK \ F j < jF j, which is a contradiction. Thus elements of G # have prime orders. Exercise 1. Suppose that U is a nontrivial normal p-subgroup of G, where p is the largest prime divisor of jGj. Assume that every element of G # has prime order and let P 2 Sylp .G/. Then either P D G or jG W P j is prime (and so P G G). Solution. Take q 2 .G/fpg and Q 2 Sylq .G/. Then QU is a Frobenius group with kernel U so jQj D q. It follows that P G G. Indeed, if r is a minimal prime divisor of jGj, then jGjr D r so G is r-nilpotent (Burnside). Applying this to the r-normal complement of G, we at last prove our claim. If P < G, then G is a Frobenius group with kernel P and complement of square free order without elements of composite order. It follows that p 0 -Hall subgroup of G is of prime order [BZ, Chapter 10]. Exercise 2. Suppose that every element of G # is of prime order. Let P 2 Sylp .G/, where p is the largest prime divisor of G. Then P is a TI-set, i.e., P \ P x D f1g for x 2 G NG .P /, unless P E G. Solution. Assume that f1g < D D P \ P x , where P ¤ P x and jDj is as large as possible. Applying Exercise 1 to NG .D/ whose Sylow p-subgroup is not normal (Burnside), we obtain a contradiction. Now we are ready to prove the main result of this section. Theorem 68.4 ([Isa8]). Let † be a nontrivial partition of an equally partitioned group G. Then exp.G/ D p for some prime p.

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Groups of prime power order

Proof. One may assume that j.G/j > 1 (Lemma 68.3). Let p be the largest prime divisor of jGj and P 2 Sylp .G/. By Lemma 68.3 and Exercise 2, P is a TI-set since P 6E G. By Lemma 68.3 and Exercise 1, NG .P / D C P , where either C D f1g or jC j D q, a prime. In the first case, G D P H is a Frobenius group with kernel H and complement P . Then jP j D p, a prime. In that case, G is not equally partitioned (by Lemma 68.2: take there X D P # ). Thus, NG .P / D C P is a Frobenius group with kernel P and complement C of prime order q ¤ p. Let jGj D g, jP j D p b , jH j D h for H 2 † and p ˛ D hp , the p-part of h. By Lemma 68.2, given x 2 P # , there exists F 2 † such that x z 62 F for all z 2 G, so ˛ < b since jF j D jH j D h (Sylow). Since P is a TI-set, jP \ H j D p ˛ for all H 2 † such that P \ H ¤ f1g (note that a Sylow p-subgroup of any component H of † is also a TI-subset in H ). By Lemma 68.2, one may choose H 2 † with H \ C x D f1g for all x 2 G. Take P0 2 Sylp .H /. We may assume that P0 < P . Since P is a TI-set, we must have NH .P0 / NG .P / D C P . It follows that NH .P0 / D P0 C0 , where p − jC0 j. Assume that y 2 C0# . Then f1g < P0 P \ P y so P y D P and so C0 is conjugate with C in NG .P /, by Sylow’s theorem, contrary to the choice of H . It follows that C0 D f1g so NH .P0 / D P0 . By assumption, .g/ D .h/ so, to complete the proof, it suffices to prove that H is a p-group. Assume that this is false. Then H D P0 Q0 is a Frobenius group with kernel Q0 and complement P0 since P0 is a TI-subgroup in H (see Frobenius’ theorem in [BZ, Chapter 10]). Since P0 has only one subgroup of order p and its exponent is p, we get jP0 j D p, i.e., ˛ D 1. Since Q0 is nilpotent (Thompson), it is a q-group for some prime q, by Lemma 68.3. Then G is a fp; qggroup, and so G is solvable, by Burnside’s two-prime theorem. We have f1g < Q1 D Oq .G/ since P is a nonnormal TI-subgroup of G (in view of the existence of H ). Since P Q1 is a Frobenius group, we get jP j D p D jP0 j, a contradiction since 1 D ˛ < b. Thus, H is a p-subgroup. For more elementary proof of Theorem 68.4, see [Isa8]. Exercise 3. Suppose that a 2-group G of exponent > 2 admits a nontrivial partition. Then G D Ahbi, where o.b/ D 2 and b inverts A. (Hint. Use Lemma 68.1 and the paragraph following it.) Problem 1 (Isaacs). Does there exist a group partitioned by proper subgroups of equal order not all of which are abelian? Problem 2. Let † be a nontrivial partition of G. Study the structure of G if orders of components of † form a chain under divisibility. Problem 3. Classify all nontrivial partitions of elementary abelian p-groups.

69

Elementary proofs of some Blackburn’s theorems

In this section we present elementary proofs of some Blackburn’s theorems; these proofs are due to the first author. 1o . Blackburn has classified minimal nonmetacyclic p-groups. For a simpler proof in the case p D 2, due to the second author, see Theorem 66.1. Here we offer another elementary proof for all p. The p-groups, p > 2, without normal elementary abelian subgroups of order p 3 were classified by Blackburn (see Theorem 13.7). The proof of Theorem 13.7 was based on the deep theorem 12.1(a). Here we offer another proof based on Theorem 12.12 which is elementary for p D 3 (see Remark 1.) Remark 1. The proof of Theorem 12.12(a) is elementary. Now let p D 3. In the proof of part (c) of that theorem we use the following fact: If G is a 3-group of maximal class and order 3m > 33 , then jG=Ã1 .G/j D 33 . It suffices to prove this for m D 4. By Lemma 64.1(m), jG=Ã1 .G/j > 32 , since G is nonmetacyclic. It is known that groups of exponent 3 are of class 2, so jG=Ã1 .G/j D 33 . Similarly, using Theorem 9.5. it is easy to show that, if a nonabelian group G of exponent 3 is generated by two elements, then its order is 33 . Theorem 69.1. If G is a minimal nonmetacyclic p-group, then, one of the following holds: (a) G is of order p 3 and exponent p; (b) G is a group of maximal class and order 34 ; (c) p D 2 and jGj 25 ; if jGj D 25 , then G is an A2 -group. Proof. We use induction on jGj. One may assume that jGj > p 3 . Clearly, we have jG=Ã1 .G/j p 3 (otherwise, all maximal subgroups of G are nonmetacyclic). (i) Let p > 2. Obviously, G has no subgroups of order p 3 and exponent p. Then G is irregular (otherwise, jG=Ã1 .G/j D j1 .G/j D p 2 and G is metacyclic, by Lemma 64.1(m)). Therefore, if jGj D p 4 , then cl.G/ D 3 so p D 3 (Lemma 64.1(a)). Now suppose that jGj > p 4 . Let R Ã1 .G/ \ Z.G/ be of order p. By Lemma 64.1(m), G=R is nonmetacyclic so minimal nonmetacyclic; then p D 3 and G=R is of maximal class and order 34 , by induction. We get jGj D 35 and jG W Ã1 .G/j D 33 (Remark 1). Let H=R be maximal in G=R; then H=R is either abelian of type .9; 3/ or isomorphic to M33 . Since H is metacyclic of order 34 , jH 0 j 3 so H is either abelian or A1 group (Lemma 65.2(a)). Since R ˆ.H / ˆ.G/, we conclude that all members of

210

Groups of prime power order

the set 1 are either abelian or A1 -groups so G is an A2 -group. By Lemma 64.1(p), jG W G 0 j D 32 hence G 0 Š E33 (Theorem 65.7(d)) so nonmetacyclic, a contradiction. (ii) Now assume that p D 2; then d.G/ D 3, by Lemma 64.1(n). Since all maximal subgroups of an (abelian) group G=G 0 are 2-generator and d.G=G 0 / D 3, we conclude that G=G 0 Š E8 so G 0 D ˆ.G/. Take H 2 1 . N 24 . Since d.G/ N D 3, GN is neither metacyclic nor Set GN D G=K3 .G/; then jGj 0 N since cl.G/ N D 2. In view of jHN W .Z.G/ N \ HN /j an A1 -group. Next, GN Z.G/ jHN W GN 0 j D 4 and d.HN / D 2, HN is either abelian or A1 -group (Lemma 65.2(a)). It follows that GN is a (nonmetacyclic) A2 -group. By Theorem 65.7(d), GN 0 is elementary N D 4. Therefore, we abelian; in particular, jGN 0 j 4 (GN 0 is metacyclic!) and exp.G/ 1 2 0 5 N so jGj N 2 . have 2 jGN j D 8 jGj 5 N D GN 0 .Š E4 /, by N Suppose that jGj D 2 ; then jGN 0 j D 4. In that case, Z.G/ 0 0 N > GN , then jGN j D 2). Then all (noncyclic!) Lemmas 4.9 and 64.1(q) (indeed, if Z.G/ N D GN 0 (otherwise, d.AZ. N G// N > 2, which is subgroups AN of GN of order 8 contain Z.G/ 0 N N N N N N N not the case since AZ.G/ < G/ so A G G. Since G D ˆ.G/, we conclude that GN is special. Assume that K D K3 .G/ > f1g. By the above, GN D G=K is an A2 -group of order 25 . Suppose that jGj D 25 . Assume that G is not an A2 -group. Then G contains a nonabelian subgroup H of order 8. By Lemma 64.1(i), CG .H / contains a subgroup F of order 4 since G is not of maximal class. Taking F so that H \ F D Z.H /, we get H F 2 1 and d.H F / D 3, a contradiction. Thus, if jGj D 25 , then G is a special A2 -group. N D 24 . We claim that this is impossible. Let R be a G-invariant Suppose that jGj subgroup of index 2 in K; then cl.G=R/ D 3. One may assume that R D f1g: then jGj D 25 , contrary to the previous paragraph. It remains to consider the case where K > f1g and jG=Kj D 25 . We claim that this is impossible. Let R be a G-invariant subgroup of index 2 in K. To obtain a contradiction, one may assume that R D f1g; then jGj D 26 . Since cl.G/ D 3 and d.G/ D 3, G is not an A2 -group (Lemma 65.4(c)). It follows that K D K3 .G/ is the unique minimal normal subgroup of G. Therefore, there exists a nonabelian subgroup H of index 4 in G. We have K < H (otherwise, d.K H / > 2). Since E4 Š Z.G=K/ < H=K, H is not of maximal class. There are the following two possibilities for (the metacyclic subgroup) H : either H Š M24 or H D ha; b j a4 D b 4 D 1; ab D a1 i. Since GN 0 < HN , we conclude that H G G. Assume that H Š M24 . We have c3 .H / D 2 and H G G so, if L is a cyclic subgroup of H of index 2, then jG W NG .L/j 2. Let H < M NG .L/ with jM W H j D 2. There are two possibilities for M=L: M=L 2 fC4 ; E4 g. If M=L Š C4 , then CM .L/ D L since H is the unique subgroup of order 16 in M containing L and H is nonabelian. Then M=L Š C4 is a subgroup of Aut.L/ Š E4 , a contradiction. Now suppose that M=L Š E4 . Let F=L be a subgroup of order 2 in M=L with

69 Elementary proofs of some Blackburn’s theorems

211

F=L ¤ H=L. Since K < L < F , F=K G G=K (see the end of the second paragraph of (ii)). We have G 0 D H \F D L is cyclic. This is a contradiction since G 0 =K Š E4 . Now assume that H D ha; b j a4 D b 4 D 1; ab D a1 i. Since H has exactly two nonidentity squares (a2 2 K and b 2 ) and K is characteristic in H , we see that all subgroups of order 2 are characteristic in H so normal in G. This is a contradiction since K is the unique normal subgroup of G of order 2. The proof is complete. The groups of Theorem 69.1(c) are described in Theorem 66.1. 2o . Here we offer another proof of Theorem 13.7. As follows from the proof of Theorem 13.7, it suffices to prove Theorem 69.3, below. First we prove the following Lemma 69.2. Let G be a p-group, p > 2 and c1 .G/ D 1 C p. Then G is either metacyclic or a 3-group of maximal class. Proof. Let G be a counterexample of minimal order. Then every proper subgroup of G is either metacyclic or a 3-group of maximal class, by induction. Let H G be minimal nonmetacyclic; then H is a 3-group of maximal class and order 34 (Theorem 69.1) so H < G. Let B be a nonabelian subgroup of order 33 in H (B exists since H is not minimal nonabelian); then CG .B/ 6 B (Lemma 64.1(i)). Let U be a subgroup of order 9 in CG .B/ with B \ U D Z.B/; then BU < G is neither metacyclic nor of maximal class, a contradiction. It follows from Lemma 69.2 the following new Proof of the last assertion of Lemma 64.1(m) = Theorem 9.11. We have to prove that, if p > 2 and jG=Ã1 .G/j p 2 , then G is metacyclic. Indeed, in that case, G is regular, by Lemma 64.1(a). If jG=Ã1 .G/j D p, then G is cyclic. Now let jG=Ã1 .G/j D p 2 ; then j1 .G/j D p 2 (Lemma 64.1(a)) so c1 .G/ D 1 C p. By Lemma 69.2, G is either metacyclic or an irregular 3-group of maximal class. In the second case, however, jG=Ã1 .G/j D 33 , by Remark 1. Remark 2. If a p-group G has an absolutely regular maximal subgroup A and irregular subgroup M of maximal class, then G is of maximal class. This coincides with Proposition 12.13. Theorem 69.3. If a p-group G, p > 2. has no normal subgroups of order p 3 and exponent p. then G is either metacyclic or a 3-group of maximal class. Proof. Suppose that G is a counterexample of minimal order. Then G is irregular (Lemma 64.1(a,m)). By Theorem 10.4, G has no elementary abelian subgroups of order p 3 so it has no subgroups of order p 4 and exponent p. Let Ep 2 Š R G G (Lemma 64.1(x)). By Lemma 69.2, T D CG .R/ is metacyclic since 1 .T / D R, and so T 2 1 . By Lemma 69.2 again, there exists x 2 G R of order p; then B D hx; Ri is of order p 3 and exponent p (Lemma 64.1(a)). Since B 6E G, there

212

Groups of prime power order

exists y 2 NG .B/ B of order p (see the proof of Lemma 64.2). Set H D hy; Bi; then exp.H / > p so p D 3, H is of maximal class and order 34 (Lemma 64.1(a)). Existence of T and H shows: G is of maximal class (Remark 2). We are ready to prove Blackburn’s Theorem 13.7. Theorem 69.4 (Blackburn). Suppose that a group G of order p m , m > 3, p > 2, has no subgroups Š Ep 3 . If G is neither metacyclic nor a 3-group of maximal class, then G D UE, where E D 1 .G/ is nonabelian of order p 3 and exponent p, U is cyclic, Z.G/ C . Proof. By Theorem 69.3, G has a normal (nonabelian) subgroup E of order p 3 and exponent p. By Theorem 10.4, G has no elementary abelian subgroups of order p 3 so it has no subgroups of order p 4 and exponent p. Therefore, setting C D CG .E/ we see that subgroups of order p in C lie in C \ E D Z.E/ so C is cyclic. Next, C G G and G=C is isomorphic to a p-subgroup of Aut.E/ (which is nonabelian of order p 3 and exponent p) containing a subgroup EC =C Š E=.E \C / Š Ep 2 . If G=C Š Ep 2 , then G D E C , and we are done since G is of class 2 so 1 .G/ D E. Now let jG=C j D p 3 ; then G=C is nonabelian of order p 3 and exponent p. Let Ep 2 Š R < E be normal in G and set F D CG .R/; then F 2 1 since E 6 CG .R/. Since 1 .F / D R, F is metacyclic (Lemma 69.2). Assume that there is x 2 G E of order p. Then H D hx; Ei is of exponent > p so p D 3 and H is of maximal class and order 34 . Existence of F and H shows that G is of maximal class (Remark 2), a contradiction. Thus, 1 .G/ D E. Let U=C < G=C be a subgroup of order p such that U=C 6 CE=C . Since 1 .U / D U \ E D Z.E/, U has only one subgroup of order p, namely Z.E/, so U is cyclic; clearly, G D EU . Since CG .C / EU D G, the proof is complete. Corollary 69.5. Let A be a p-group of operators of a nonmetacyclic p-group G, p > 2, jGj > p 3 . If all proper A-invariant subgroups of G of order p 3 are metacyclic, then G is a 3-group of maximal class. Indeed, by Theorem 10.4, G has no elementary abelian subgroups of order p 3 . Now the result follows from Theorem 69.4. Exercise 1. Let G be a noncyclic p-group. Suppose that for each H G G, there is h 2 H such that hhx j x 2 Gi D H . Prove that G is of maximal class. Solution. We use induction on jGj. One may assume that G is nonabelian and jGj > p 3 . Clearly, jG W G 0 j D p 2 . Indeed, if this is not true, let H=G 0 < G=G 0 be of type .p; p/. Then there is no h 2 H such that hhx j x 2 Gi D H . Next, the factors of the lower central series of G apart of the first one, are cyclic and the factors of the upper central series of G apart of the last one, are cyclic. One may assume that jGj > p 3 . Let R be a minimal normal subgroup of G. Then, by induction, G=R is of maximal class. Assuming that G is not of maximal class, we conclude that Z.G/ Š Cp 2 and,

69 Elementary proofs of some Blackburn’s theorems

213

by Lemma 1.4, G has a normal abelian subgroup L of type .p; p/. We have R < L so Z.G=R/ Š Ep 2 , and G=R is not of maximal class, a contradiction. Exercise 2. If a 3-group G of order > 34 is nonmetacyclic but all its minimal nonmetacyclic subgroups have the same order 34 , then G is of maximal class with 1 .G/ Š E9 . (Hint. Use Theorem 69.4.) Exercise 3. If G is a minimal nonmetacyclic group of order 25 , then (a) G is special with jG 0 j D 4, (b) the set 1 has exactly one abelian member. Solution. (a) By Theorem 69.1, G=G 0 Š E8 so G 0 D ˆ.G/. Assume that G has a nonabelian subgroup H of order 8. Since G is not of maximal class, we get CG .H / 6 H (Lemma 64.1(i)). If F is a subgroup of order 4 in CG .H / with Z.H / < F , then d.HF / D 3, a contradiction. Thus, G is an A2 -group so G 0 Š E4 (Lemma 65.7(d)). We have CG .G 0 / hH j H 2 2 i so G 0 Z.G/. If G 0 < Z.G/, then by Lemma 64.1(q), jG 0 j D 2, a contradiction. It follows that G is special. (b) Assume that the set 1 has no abelian members. Let Zi , i D 1; 2; 3, be all subgroups of order 2 in G and let Ai D fT 2 1 j T 0 D Zi g. Since G=Zi has exactly three abelian subgroups of index 2, we have jAi j D 3, i D 1; 2; 3, and assume that all Ai are non-empty. Obviously, the sets A1 ; A2 ; A3 are pairwise disjoint, a contradiction since jA1 [ A2 [ A3 j D 3 C 3 C 3 D 9 > j1 j. Thus, 1 has exactly one abelian member since jG 0 j D 4.

70

Non-2-generator p-groups all of whose maximal subgroups are 2-generator

We begin with a study of p-groups of the title by proving the following five theorems. We use results on A2 -groups from 65. The following theorem is essential in determination of A2 -groups (see 71). For p > 2, Theorem 70.1 was proved by Blackburn (however, another proof is presented below). Theorem 70.1. Let G be a nonabelian p-group with d.G/ D 3. Suppose that all members of the set 1 are generated by two elements. If G is of class 2, then it is an A2 -group of order p 6 and one of the following holds: (a) jGj D p 4 and either G D E C , where E is nonabelian of order p 3 , C Š Cp 2 , E \ C D Z.E/ or G D Q Z with Q Š Q8 and jZj D 2. (b) G D ha; b; c j a4 D b 4 D Œa; b D 1; c 2 D a2 ; ac D ab 2 ; b c D ba2 i is the minimal nonmetacyclic group of order 25 , G is special, 1 .G/ D G 0 D Z.G/ D ˆ.G/ D ha2 ; b 2 i Š E4 , ha; bi Š C4 C4 is the unique abelian maximal subgroup of G. All subgroups of G of order 8 contain Z.G/ D G 0 . (c) jGj D p 5 , Ep 2 Š ˆ.G/ D G 0 D Z.G/ < 1 .G/ Š Ep 3 so G is special, Exactly p 2 members of the set 1 , not containing 1 .G/, are metacyclic and exactly one of them is abelian and p C 1 other members of the set 1 , containing 1 .G/, are nonmetacyclic minimal nonabelian. More precisely, G D ha; b; ci with 2

2

ap D b p D c p D Œb; c D 1; c p D x y ı ;

Œa; b D x;

Œa; c D y;

bp D x˛ y ˇ ;

x p D y p D Œa; x D Œa; y D Œb; x D Œb; y D Œc; x D Œc; y D 1;

where in case p D 2 we have ˛ D 0, ˇ D D ı D 1, and in case p > 2, the number 4ˇ C .ı ˛/2 is a quadratic non-residue mod p. Here G 0 D hx; yi, 1 .G/ D G 0 hai, and hb; ci Š Cp 2 Cp 2 is the unique abelian member of the set 1 . All subgroups of G of order p 3 contain Z.G/ D G 0 . (d) p D 2, G D ha; b; ci is of order 26 and a4 D b 4 D c 4 D 1;

Œa; b D c 2 ;

Œa; c D b 2 c 2 ;

Œb; c D a2 b 2 ;

Œa2 ; b D Œa2 ; c D Œb 2 ; a D Œb 2 ; c D Œc 2 ; a D Œc 2 ; b D 1;

70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 215

where G 0 D ha2 ; b 2 ; c 2 i D Z.G/ D ˆ.G/ D 1 .G/ Š E23 , so G is special and all members of the set 1 are nonmetacyclic minimal nonabelian. If H < G is of order 24 , then H G 0 . Proof. Let G satisfy the assumptions of the theorem; then G is not an A1 -group (Lemma 65.1) so jGj > p 3 . Since G=G 0 is of rank 3 and all maximal subgroups of G=G 0 are two-generator, we get G=G 0 Š Ep 3 and so G 0 D ˆ.G/. For any a; b 2 G, Œa; bp D Œap ; b D 1 since G is of class 2, and so G 0 is elementary abelian. If jGj D p 4 and E < G is an A1 -subgroup, then G D EZ.G/ (Lemma 64.1(i)), so G is from (a). From now on we assume that jGj p 5 and so jG 0 j D p13 jGj p 2 . If the set 1 has an abelian member, then jGj D pjZ.G/jjG 0 j (Lemma 64.1(q)) implies jG W Z.G/j p 3 . If the set 1 has no abelian members, then jG W Z.G/j p 3 again. Since jG W G 0 j D p 3 and G 0 Z.G/, we get Z.G/ D G 0 D ˆ.G/ so G is special and exp.G/ p 2 . Let H 2 1 . Then jH W Z.G/j D p 2 D p d.H / so Z.G/ D ˆ.H / and H is either abelian or an A1 -subgroup (Lemma 65.2(a)). Thus G is an A2 -group so jG 0 j p 3 (Lemma 65.2(d)) and jGj D jG W G 0 jjG 0 j p 6 . Since G has a minimal nonabelian subgroup H 2 1 , then exp.H / p 2 so exp.G/ D p 2 . (i) Assume first that jGj D p 5 ; then G 0 Š Ep 2 . If each H 2 1 is metacyclic, then G is minimal nonmetacyclic. By Lemma 64.1(l), p D 2 and G is the uniquely determined group of order 25 given in (b) (see Theorem 66.1). In what follows we assume that G is not minimal nonmetacyclic. Let H 2 1 be nonmetacyclic. Since d.H / D 2, H must be an A1 -group and so H D ha; b j 2 ap D b p D 1; Œa; b D c; c p D Œc; a D Œc; b D 1i; we have E D 1 .H / D hap ; b; ci Š Ep 3 (Lemma 65.1). Assume that there is x 2 G E of order p. Then D D hx; Ei 2 1 is neither abelian (since d.D/ D 2) nor an A1 -group (Lemma 65.1), a contradiction. Thus, E D 1 .G/ Š Ep 3 . Let T1 =E; : : : ; TpC1 =E be all maximal subgroups of G=E. Then T1 ; : : : ; TpC1 are nonmetacyclic A1 -groups (since d.Ti / D 2) isomorphic to H (Lemma 65.1). Suppose that M 2 1 does not contain E. Then 1 .M / D E \ M Š Ep 2 and so M is metacyclic since it is either abelian or an A1 -group (Lemma 65.1). Let X < G and jXj D p 3 .> p 2 D exp.G//; then X is noncyclic. If Z.G/ 6 Xthen d.XZ.G// > 2 and XZ.G/ 2 1 , a contradiction. Hence .G 0 D/Z.G/ < X and so X is abelian and normal in G. N Œa; N ci. N ci; N b; N c; N Œb; N Hence G 0 .Š Ep 2 / is generated Let G D ha; N b; N then G 0 D hŒa; N N c by two of these elements, say xN D Œa; N b and yN D Œa; N c. N If Œb; N D xN yN , let

b D bN aN , c D cN aN . We compute (recall that G is of class 2) N cŒ N a N b; N Œa; N c N D xN yN xN yN D 1: Œb; c D ŒbN aN ; cN aN D Œb; Then A D hb; c; G 0 i 2 1 is abelian and so A E D 1 .G/ (otherwise, d.A/ D 3). Hence A D hbi hci Š Cp 2 Cp 2 with 1 .A/ D G 0 D Ã1 .A/. Take an element

216

Groups of prime power order

a 2 E A (of order p) so that G D ha; b; ci. Setting x D Œa; b; y D Œa; c, we get G 0 D hx; yi D hb p ; c p i since x ¤ 1 ¤ y ¤ x. Set b p D x ˛ y ˇ , c p D x y ı . The subgroup L D ha; b c ; G 0 i 2 1 for any integers ; , unless

0 .mod p/ since a; b; c independent modulo G 0 . Since, for such and , we have Œa; b c D x y ¤ 1, L is nonabelian, In that case, d.L/ D 2 so ˆ.L/ D G 0 D Z.L/, and G 0 is generated by elements .b c /p D x ˛C y ˇ Cı and Œa; b ˇc D ˇ ˇ ˇ Cı ˇ x y , which are linear independent since G 0 Š Ep 2 . Hence ˇ ˛C ˇ 0 .mod p/ only if 0 .mod p/. This gives (1)

ˇ 2 C .ı ˛/ 2 0 only if 0

.mod p/:

From (1) we get ˇ 6 0 .mod p/ (otherwise, D 1, D 0 is a nontrivial solution of (1)) and 6 0 .mod p/ (otherwise, D 0, D 1 is a nontrivial solution of (1)). (i1) Assume that p > 2. We compute (2)

.2ˇ C .ı ˛//2 .4ˇ C .ı ˛/2 /2 D 4ˇ.ˇ 2 C .ı ˛/ 2 /:

Using (1), we see (because ˇ 6 0 .mod p/) that the right hand side of (2) vanishes .mod p/ only if 0 .mod p/, and so we have the same for the left hand side of (2): (3)

.2ˇ C .ı ˛//2 .4ˇ C .ı ˛/2 /2 only if 0

.mod p/

.mod p/:

Hence 4ˇ C .ı ˛/2 is a quadratic non-residue .mod p/. Indeed, if 4ˇ C .ı

2 .mod p/, then we set D 1 in (3) and solve the congruence .2ˇ C .ı 2 ˛// 2 .mod p/ for , which gives a contradiction (since for D 1 we have obtained a nontrivial solution of (3) in and ). (i2) Suppose that p D 2. Then ˇ 1 .mod 2/ and so relation (1) becomes ˛/2

(4)

2 C .ı C ˛/ C 2 0

.mod 2/ only if 0

.mod 2/:

If ı C ˛ 0 .mod 2/, then 1 .mod 2/ satisfies (4), a contradiction. Hence ı C ˛ 1 .mod 2/ and we get b 2 D x ˛ y, c 2 D xy 1C˛ . If ˛ 0 .mod 2/, then (recall that o.x/ D o.y/ D 2) (5)

b 2 D y;

c 2 D xy:

b 2 D xy;

c 2 D x:

If ˛ 1 .mod 2/, then (6)

However, interchanging b and c and also x and y, we obtain from (6) the relations (5). Hence we may assume from the start that ˛ D 0, ˇ D D ı D 1 and our group G is uniquely determined. We have obtained the groups from (c).

70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 217

(ii) Now suppose that jGj D p 6 so that ˆ.G/ D G 0 Š Ep 3 . Each H 2 1 is nonmetacyclic and minimal nonabelian since it is 2-generator, and so one may set 2 2 H D hb; c j b p D c p D 1; Œb; c D z; z p D Œb; z D Œc; z D 1i, where ˆ.H / D hb p ; c p ; zi D ˆ.G/ D G 0 Š Ep 3 , and 1 .H / D ˆ.H / implies 1 .G/ D G 0 D ˆ.G/ D Z.G/ and we again conclude that G is special. (ii1) First assume that p > 2. Then Ã1 .H / D hb p ; c p i and H 0 D hzi, where z 62 Ã1 .H /. All p 2 C p C 1 members of the set 1 are isomorphic to H (Lemma 65.1) and their derived subgroups (of order p) are pairwise distinct (Lemma 65.2(d)). Since G is regular and jG W Ã1 .G/j D p 3 (if jG=Ã1 .G/j > p 3 , then G=Ã1 .G/ has a maximal subgroup that is not generated by two elements), we get Ã1 .G/ D G 0 D fx p j x 2 Gg (Lemma 64.1(a)). In particular, there exists a 2 G H with ap D z. Set b p D x, c p D y. We have G D ha; H i D ha; b; ci and G 0 D hx; y; zi (by Lemma 65.2(d), all elements of G 0 are commutators). Set Œa; b D x ˛ y ˇ z , Œa; c D x y ı z , and recall that Œb; c D z. We shall construct a maximal subgroup K of G with d.K/ > 2. To do this we shall do some purely number-theoretic computations using heavily the oddness of p. First, we solve the equation t 2 2 . 2 4ˇ/ C 2. 2.ı ˛// C 2 .2 C 4 / .mod p/

()

for t , , with , not both 0 .mod p/. If r D 2 C 4 0 .mod p/, we set D 1, D 0. If r D 2 C 4 6 0 .mod p/, we put D 1 and set 2.ı ˛/ D s, 2 4ˇ D u, so that () can be rewritten in the simplified form ()

t 2 r. C r 1 s/2 r 1 s 2 C u .mod p/;

where r, s, u are constant numbers mod p and , t run through all integers mod p. The left hand side of () runs through 1C.p1/=2 D .pC1/=2 distinct values (quadratic residues) mod p. Now, C r 1 s runs through all numbers mod p, . C r 1 s/2 runs through .p C 1/=2 distinct values (quadratic residues) mod p so that the right hand side of ./ runs through .p C 1/=2 distinct values (mod p) (take into account that r 1 s 2 Cu is a constant number modulo p). Since .pC1/=2C.pC1/=2 D pC1 > p, it follows that there are values for t and which satisfy (), as required. Since at least one of the integers , is not a multiple of p, we can solve t

N C C .mod p/ for , N . 2. N / N Substituting this into ./, we obtain (after reforming) N N / N C .ˇ C ı/ .˛ C / 0 . C C N /. or in the determinant form: ˇ ˇ ˇ C C N N ˛ C ˇ C ı ˇˇ ˇ ˇ ˇ 0 1 N N ˇ ˇ ˇ ˇ 0

.mod p/;

.mod p/:

218

Groups of prime power order N

We now define the elements a D ab N c , b D b c of G, where p does not divide at least one of the integers , . Let K D ha ; b ; G 0 i; then K 2 1 . It follows from ˆ.K/ D Ã1 .K/K 0 that ˆ.K/ D hŒa ; b ; .a /p ; .b /p i. We have N

N ˛C ˇ Cı Œa ; b D z CC x y ;

N

.a /p D zx N y ;

.b /p D x y ;

where x, y, z are determined in the first paragraph of this part (ii1). From the vanishing of the above determinant, we see that “linear independent” elements x, y, z cannot be expressed as “linear combinations” of Œa ; b , .a /p , .b /p and so ˆ.K/ < G 0 D hx; y; zi so jK W ˆ.K/j > p 2 or, what is the same, d.K/ > 2, contrary to the hypothesis of the theorem. (ii2) Now assume that p D 2. Let G 0 x, G 0 y be two distinct involutions in G=G 0 . Then X D hx; G 0 i and Y D hy; G 0 i are distinct abelian subgroups of type .4; 2; 2/. We have ˆ.X/ D hx 2 i and ˆ.Y / D hy 2 i. Assume that x 2 D y 2 . Then X=ˆ.X/ Š E8 Š Y =ˆ.Y / D Y =ˆ.X/ and X=ˆ.X/ ¤ Y =ˆ.X/. This is a contradiction since 1 .G=ˆ.X// Š E8 . Thus, x 2 ¤ y 2 . Hence, the seven nontrivial elements in G=G 0 produce seven pairwise distinct squares in G 0 which are the seven involutions in G 0 . In particular, each involution in G 0 is a square in G. Let H 2 1 be fixed. We may set H D ha; b j a4 D b 4 D 1; Œa; b D z; z 2 D Œa; z D Œb; z D 1i (Lemma 65.1(a)). Set a2 D x, b 2 D y so that G 0 D hxihyihzi and .ab/2 D a2 b a b D xbzb D xb 2 z D xyz. There exists c 2 G H such that c 2 D z (see the previous paragraph; note that z is not a square in H ). Then G D hH; ci D ha; b; ci. We claim that ac ¤ ca. Assume that this is false. Then CG .c/ ha; Z.G/; ci, and the subgroup at the right hand side is abelian of index 2 in G, a contradiction since the set 1 has no abelian members (Lemma 65.2(d)). It follows that ha; ci is nonabelian. Since G is an A2 -group, we get ha; ci 2 1 . Since ˆ.ha; ci/ D ha2 D x; c 2 D z; Œa; ci D G 0 , we must have Œa; c 2 G 0 hx; zi and so Œa; c D x ˛ yz ˇ . Similarly, considering the maximal subgroup hb; ci, we get Œb; c 2 G 0 hy; zi and so Œb; c D xy z ı . Consider the subgroup K D ha; bci. Assume that a bc D bc a. It follows bca D ab c D baz c so c a D cz D cc 2 D c 1 and ha; ci has order 16 so is not a member of the set 1 , a contradiction. Since G is an A2 -group and d.G/ D 3, we get K 2 1 . We have ˆ.K/ D ha2 D x; .bc/2 D xy 1C z 1Cı ; Œa; bc D x ˛ yz 1Cˇ i Š E23 ; and since ˆ.K/ D G 0 D hx; y; zi, we have ˇ ˇ ˇ1 0 0 ˇˇ ˇ ˇ1 1 C 1 C ı ˇ D 1 ˇ ˇ ˇ˛ 1 1 C ˇˇ and so (7)

ˇ C C ˇ C ı D 1:

70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 219

Considering the maximal subgroup L D hab; ci, we get ˆ.L/ D hxyz; z; x 1C˛ y 1C z ˇ Cı i Š E23 ; ˇ ˇ 1 1 ˇ ˇ 0 0 ˇ ˇ1 C ˛ 1 C

and so

ˇ 1 ˇˇ 1 ˇˇ D 1; ˇ C ıˇ

which gives ˛ C D 1:

(8)

Considering M D hb; aci 2 1 , we get ˆ.M / D hy; x 1C˛ yz 1Cˇ ; xy z 1Cı i so ˛ C ı C ˛ı C ˇ D 1:

(9)

Finally, consider the maximal subgroup N D hab; aci. We get ˆ.N / D h.ab/2 D xyz; .ac/2 D x 1C˛ yz 1Cˇ ; Œab; ac D x 1C˛ y 1C z 1Cˇ Cı i; and so ˛ C ˛ı C ˇ D 1:

(10)

Relations (7–10) have exactly two solutions: (A) ˛ D 0, ˇ D 1, D 1, ı D 0, which implies Œa; c D yz, Œb; c D xy, and (B) ˛ D 1, ˇ D 0, D 0, ı D 1, which implies Œa; c D xy, Œb; c D xz. However, interchanging a and b and also a2 D x and b 2 D y, we obtain from the solution (A) the solution (B). Hence we may assume from the start that we have solution (A) and so the group G of order 26 is uniquely determined as stated in part (d). Assume that F is a subgroup of G of order 24 and F 6 G 0 . We have F G 0 < G so M D F G 0 2 1 . But then M=.F \ G 0 / Š E23 so d.M / 3, a contradiction. Thus, G 0 is contained in all subgroups of G of order 24 . The proof is complete. Theorem 70.2. Suppose that all maximal subgroups of a p-group G are generated by two elements but d.G/ D 3. If G is of class > 2, then p D 2, jGj 27 , and G=K3 .G/ is isomorphic to the group of order 26 of Theorem 70.1(d). If, in addition, jGj D 27 , then the class of G is 3 and G is uniquely determined, namely, G D ha; b; ci, where a4 D b 4 D c 4 D k 2 D Œa; k D Œb; k D Œc; k D 1; Œa; b D c 2 ; Œb 2 ; a D 1;

Œa; c D b 2 c 2 ; Œb 2 ; c D k;

Œb; c D a2 b 2 ;

Œc 2 ; a D k;

Œa2 ; b D k;

Œa2 ; c D k;

Œc 2 ; b D 1:

Here hki D K3 .G/ D ŒG; G 0 D Z.G/ is of order 2, ha2 ; b 2 ; c 2 ; ki D G 0 D ˆ.G/ D 1 .G/ Š E24 , and G exists as a subgroup of the alternating group A16 .

220

Groups of prime power order

Proof. Suppose that G is a p-group of class > 2, d.G/ > 2, and all members of the set 1 are generated by two elements. Then K3 .G/ ¤ f1g and G=K3 .G/ is isomorphic to a group of Theorem 70.1. In particular, jGj p 5 . Let jGj D p 5 and assume that G has a nonabelian subgroup S of order p 3 . Since d.G/ D 3, G is neither of maximal class nor an A1 -group. Hence CG .S / 6 S (Lemma 64.1(i)) and let F > Z.S / be a subgroup of order p 2 in CG .S / containing Z.S /. Then SF D S F 2 1 and d.SF / D 3, a contradiction. Thus S does not exist so G is an A2 -group. But then jG=ˆ.G/j D p 3 and so ˆ.G/ Z.G/ (Lemma 65.4(c)), so G is of class 2, a contradiction. Thus, jGj p 6 . (i) Let jGj D p 6 . If jG=K3 .G/j D p 4 , then take in K3 .G/ a G-invariant subgroup L of index p. But then jG=Lj D p 5 and G=L satisfies the hypothesis, contrary to the previous paragraph. Hence jG=K3 .G/j D p 5 and so G=K3 .G/ is isomorphic to a group (b) or (c) of Theorem 70.1. Also, K D K3 .G/ is of order p. We have G 0 =K D Z.G=K/ Š Ep 2 and G=G 0 Š Ep 3 . The subgroup K is the unique minimal normal subgroup of G. Indeed, let L be a minimal normal subgroup of G distinct from K. Then L < Z.G/ and so L < G 0 . We have d.G=L/ D 3 and each maximal subgroup of G=L is generated by two elements. By the above, G=L is of class 2 and so ŒG; G 0 L. This is a contradiction since ŒG; G 0 D K and K \ L D f1g. The group G is not an A2 -group (otherwise, d.G/ D 3 would imply G 0 Z.G/, by Lemma 65.4(c)). Therefore G has a nonabelian subgroup H of order p 4 . We have H K (otherwise, KH D K H 2 1 and d.KH / > 2). By Theorem 70.1, applied to G=K, H > G 0 and so H is normal in G. Also, G=K is an A2 -group and so H=K Š Ep 2 H 0 D K. Assume that H is not minimal nonabelian. Then H (of class 2) contains a nonabelian subgroup B of order p 3 , and we have d.H / > 2 since H D BCG .B/ (Lemma 64.1(i)) so H=H 0 D H=K Š Ep 3 , G=K is isomorphic to a group (c) of Theorem 70.1 and H=K D 1 .G=K/, i.e., H is the uniquely determined subgroup of index p 2 in G. Let H < M 2 1 . Since d.M / D 2, we get ˆ.M / D ˆ.G/ D G 0 so all maximal subgroups of M are normal in G. Any nonabelian p-group has exactly 0, 1 or p C 1 abelian subgroups of index p (Exercise 1.6(a)) so M has a nonabelian maximal subgroup F ¤ H . As above, K < F . Since H is the unique A2 -subgroup of index p 2 in G, by the above, F is an A1 -subgroup and F is normal in G. Since F=K ¤ H=K D 1 .G=K/, the subgroup F=K is abelian of type .p 2 ; p/. Assume that F is not metacyclic. If p > 2, then Ã1 .F / is a normal subgroup of G of order p (Theorem 9.11), distinct from K, contrary to what has proved already. Now let p D 2. Then Z.F / Š E4 (Z.F / is noncyclic since Z.F / D ˆ.G/ D Ã1 .F /) and F 0 D K. Let U D CG .Z.F //; then U 2 1 since Z.G/ is cyclic. Let K1 ¤ K be a subgroup of order 2 in Z.F /. Then U=K1 contains a nonabelian subgroup F=K1 of order 8. Since d.U=K1 / D 2, it follows that U=K1 is of maximal class (Lemma 64.1(i)). By Taussky’s theorem, K1 6 U 0 so U 0 Š C4 . We have F < U so F 0 < U 0 < F and U 0 is cyclic of order 4. It follows that F is metacyclic (Lemma 65.1). Thus, all subgroups

70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 221

of G of index p 2 which are different from H , are either abelian or metacyclic. We conclude that if G has an A2 -subgroup of index p 2 , then G has a normal metacyclic minimal nonabelian subgroup F of the same index. Let p > 2. It follows from G 0 < F that G 0 is abelian of type .p 2 ; p/. As we know, Ã1 .G/ D ˆ.G/ D G 0 . By Theorem 13.7 (or Theorem 69.4), G has a normal subgroup E Š Ep 3 . It follows from the structure of G=K that E1 .H /=K/ 1 .G=K/ D H=K so E 1 .H /, and we conclude that M=E is cyclic of order p 2 (indeed, M has at most one abelian maximal subgroup so, in view of p > 2, it has at least p 1 2 nonabelian metacyclic maximal subgroups). Then CM .E/ > E since p 2 does not divide exp.Aut.E//. Since H is the unique maximal subgroup of M containing E, we get H CM .E/ so H is abelian, which is a contradiction. Thus, if p > 2, all nonabelian subgroups of G of index p 2 are A1 -groups. Now let p D 2. Assume that exp.F / D 4, where F ¤ H is metacyclic of order 16. Then all subgroups of order 2 in F are characteristic so normal in G, a contradiction since Z.G/ is cyclic. Thus, F Š M24 . This case we shall consider in (i1). (i1) Assume that p D 2 and F Š M24 . We have c3 .F / D 2 and F G G. Therefore, if L is a cyclic subgroup of F of index 2, then jG W NG .L/j 2. Let F < M NG .L/ with jM W F j D 2. If M=L Š C4 , then CM .L/ D L since F=L is the unique subgroup of order 2 in M=L. But in this case M=L is isomorphic to a subgroup of Aut.L/ Š E4 , a contradiction. Hence, we have M=L Š E4 . Let R=L be a subgroup of order 2 in M=L with R=L ¤ F=L. Since K < L < R and jR=Kj D 23 , Theorem 70.1 forces R > G 0 . But then R \ F D L G 0 and so G 0 is cyclic. This is a contradiction since G 0 =K Š E4 . Thus, if p D 2, then G has no nonabelian metacyclic subgroups of order 24 . It follows that then G has no A2 -subgroups of index 4 (otherwise, by the above, it has a nonabelian metacyclic subgroup of order 24 ). (i2) Assume that p D 2 and H is a nonmetacyclic A1 -group of exponent 4. Then ˆ.H / D Z.H / D L K, where jLj D 2, Z.H / is normal in G and H=Z.H / Š E4 . Set U D CG .L/ D CG .Z.H //. Then jG W U j D 2 since L 6 Z.G/ (recalling that K is the unique minimal normal subgroup of G). In that case H=L is a (proper) nonabelian subgroup of order 8 in U=L which implies that U=L is not an A1 -group. By hypothesis, d.U / D 2 and so L < ˆ.U / and CU .H=L/ < H=L. It follows that U=L is of maximal class (Lemma 64.1(i)), and so .U=L/0 Š C4 . By Taussky’s theorem, L 6 U 0 so U 0 Š C4 . However, H < U and H 0 < U 0 < H so H is metacyclic (Lemma 65.1), contrary to the assumption. Thus, if p D 2, then jGj > 26 (otherwise, G is an A2 -group so of class 2). (i3) Now we assume that p > 2. Then, as we have been proved already, all nonabelian subgroups of G of order p 4 are either abelian or A1 -groups. Let H be a nonabelian subgroup of index p 2 in G. Assume that H is a nonmetacyclic A1 -group. Then H=K is abelian of type .p 2 ; p/ and Ã1 .H / > f1g (see Lemma 65.1) is a normal subgroup of G not containing K D H 0 , a contradiction since K D K3 .G/ is the unique minimal normal subgroup of G.

222

Groups of prime power order

Hence, H is metacyclic, H=K is abelian of type .p 2 ; p/ and GN D G=K is isomorphic N so that EN Š Ep 3 and E (the inverse to a group of Theorem 70.1(c). Set EN D 1 .G/ 4 N image of E in G) of order p , is abelian since it is not an A1 -group in view of d.E/ N and Z.G/ N < 1 .G/, N we get 3 (see the paragraph, preceding (i1)). Since HN Z.G/ N and jH \ Ej D p 3 . But H is metacyclic and so H \ E is abelian of EN \ HN D Z.G/ type .p 2 ; p/ and therefore E is abelian of type .p 2 ; p; p/. Set E1 D 1 .E/ so that E1 Š Ep 3 , E1 \ H Š Ep 2 , and E1 > K. The quotient group G=E1 is noncyclic since N EN Š Ep 2 is its epimorphic image. Let A=E1 and B=E1 be two distinct subgroups G= of order p in G=E1 so that .AB/=E1 Š Ep 2 . Note that Theorem 70.1(c) implies that N BN being order p 3 are normal in G. N Since A and B are nonmetacyclic of order p 4 , A, they must be abelian. Therefore CG .E1 / D AB 2 1 (note that E1 6 Z.G/ because of Z.G/ is cyclic). Since d.AB/ D 2, AB is nonabelian. But AB has two distinct abelian maximal subgroups and so j.AB/0 j D 2 (Lemma 65.2(c)), and we get .AB/0 D K. It follows that AB is a nonmetacyclic A1 -group with Z.AB/ D ˆ.AB/ D E1 (Lemma 65.2(a)). Then Ã1 .AB/ Š Ep 2 is normal in G and Ã1 .AB/ \ K D f1g, a contradiction since K is the unique minimal normal subgroup of G and Ã1 .AB/ is normal in G. Thus, jGj > p 6 for all p. (ii) Let jGj p 7 . As above, set K D K3 .G/; it follows from Theorem 70.1 that jG=Kj p 6 . Assume that jG=Kj p 5 . Take a G-invariant subgroup L < K with jK W Lj D p. We have jG=Lj p 6 and G=L satisfies the assumptions of our theorem. By (i), we have a contradiction since L > f1g. Hence jG=Kj D p 6 ; then p D 2 (Theorem 70.1), and G=K is isomorphic to the group of Theorem 70.1(d). Let, in addition, jGj D 27 so that jKj D 2. Assume that L is another minimal normal subgroup of G. Then jG=Lj D 26 so, by (i), G=L is of class 2. Then G, as a subgroup of .G=K/ .G=L/, is also of class 2, a contradiction. Thus, K is the unique minimal normal subgroup of G; in particular, Z.G/ is cyclic. Let H be an arbitrary subgroup of order 25 in G. Since exp.G/ 23 (by the above and Theorem 70.1), H is not cyclic. If K 6 H , then KH D K H 2 1 and d.KH / 3, a contradiction. Hence H > K and so, by Theorem 70.1(d), H > G 0 D ˆ.G/ and H is normal in G. Since G=K is an A2 -group, H=K is abelian in view of j.G=K/ W .H=K/j D 4. Set G 0 D W . By Theorem 70.1(d), 1 .G=K/ D W =K D Z.G=K/ Š E23 . If A=K < W =K is of order 2, then A is normal in G so W CG .A/ since jAut.A/j2 D 2. Thus, W D hA j K < A < W; jAj D 4i Z.W / so W is abelian. It follows that W 2 fE24 ; C4 C2 C2 g. By Theorem 70.1(d), G D ha; b; ci and a; b; c are not involutions modulo K since ˆ.G=K/ D 1 .G=K/ so that hx; W i is of order 25 and class 2 for x 2 G W since hx; W i=K is abelian (this fact will be used in the sequel) and ha2 ; b 2 ; c 2 ; Ki D G 0 .D W /, ()

Œa; b c 2 ;

Œa; c b 2 c 2 ;

Œb; c a2 b 2

.mod K/:

70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 223

Also, for any x; y 2 G, Œx; y2 D Œy; x2 D Œy; x2 ;

Œx 2 ; y1 D Œx 2 ; y;

Œx 2 ; y 1 D Œx 2 ; y;

Œx 2 ; y D x 2 y 1 x 2 y D x 2 .y 1 xy/2 D x 2 .xŒx; y/2 D x 2 x 2 Œx; y2 ŒŒx; y; x D Œx; y2 Œx; y; x: Using the above identity and the relations for G=K (see ()), we get Œa2 ; c D Œa; c2 ŒŒa; c; a D Œa; c2 Œb 2 c 2 k; a D Œa; c2 Œb 2 ; aŒc 2 ; a; where k 2 K Z.G/ and so (11)

Œa2 ; c D Œa; c2 Œb 2 ; aŒc 2 ; a:

In the same way we obtain the following relations: (12)

Œb 2 ; c D Œb; c2 Œb; c; b D Œb; c2 Œa2 b 2 k; b D Œb; c2 Œa2 ; b;

(13)

Œb 2 ; a D Œb; a2 Œb; a; b D Œa; b2 Œc 2 k; b D Œa; b2 Œc 2 ; b;

(14)

Œc 2 ; a D Œc; a2 Œc; a; c D Œc; a2 Œb 2 c 2 k; c D Œa; c2 Œb 2 ; c;

(15)

Œa2 ; b D Œa; b2 Œa; b; a D Œa; b2 Œc 2 k; a D Œa; b2 Œc 2 ; a;

(16)

Œc 2 ; b D Œc; b2 Œc; b; c D Œc; b2 Œa2 b 2 k; c D Œb; c2 Œa2 ; cŒb 2 ; c:

For any x; y; z 2 G we get (noting that W is abelian and Œx; y 2 2 Œx; Ã1 .G/ D Œx; G 0 K Z.G/) Œx; y 1 D Œx; yy 2 D Œx; y 2 Œx; yy

2

D Œx; y 2 Œx; y

since G 0 is abelian and y 2 2 Ã1 .G/ D G 0 , and so Œx; y 1 ; z D ŒŒx; y 2 Œx; y; z D Œx; y; z. Hence from the Hall–Witt identity Œa; b 1 ; cb Œb; c 1 ; ac Œc; a1 ; ba D 1 we get Œa; b; cb Œb; c; ac Œc; a; ba D 1 so Œc 2 ; cb Œa2 b 2 ; ac Œb 2 c 2 ; ba D 1, by ./, or, taking into account that Œb 2 ; a; Œc 2 ; b 2 K Z.G/, we get (17)

Œb 2 ; aŒc 2 ; b D 1; or, what is the same, Œb 2 ; a D Œc 2 ; b;

It follows from (17) and (13) that Œa; b2 D 1 so Œa2 ; b D Œc 2 ; a, by (15). It follows from (15) and (12) that Œc 2 ; a D Œa; c2 Œb 2 ; c D Œa; c2 Œb; c2 Œa2 ; b D Œa; c2 Œb; c2 Œc 2 ; a so, canceling, we get Œa; c2 D Œb; c2 . Then, by (16), (12), (14), (11) we get Œc 2 ; b D Œb; c2 Œa2 ; cŒb 2 ; c D Œa2 ; bŒa2 ; c D Œc 2 ; aŒa2 ; c D Œa; c2 Œb 2 ; aŒc 2 ; aŒc 2 ; a D Œa; c2 Œc 2 ; b

224

Groups of prime power order

so Œa; c2 D 1. Thus, Œa; b2 D Œa; c2 D Œb; c2 D 1 and one can rewrite (11–17) as follows: (18)

Œb 2 ; a D Œa2 ; bŒa2 ; c;

(19)

Œb 2 ; c D Œa2 ; b;

(20)

Œc 2 ; a D Œa2 ; b;

(21)

Œc 2 ; b D Œa2 ; bŒa2 ; c:

In particular, since the abelian group W D G 0 D hŒa; b; Œa; c; Œb; c; Ki is generated by elements of order 2, we get W Š E24 . Since K D hki is the unique minimal normal subgroup of G and K Z.G/ W , we get Z.G/ D K. If Œa2 ; b D Œa2 ; c D 1, then the relations (18–21) and () imply that W D ha2 ; b 2 ; c 2 ; Ki Z.G/, a contradiction. It follows that we have exactly three possibilities for the values of Œa2 ; b and Œa2 ; c (below K D hk j k 2 D 1i): (G1 )

Œa2 ; b D k;

Œa2 ; c D kI

(G2 )

Œa2 ; b D k;

Œa2 ; c D 1I

(G3 )

Œa2 ; b D 1;

Œa2 ; c D k:

In all three cases we have (see ()): Œa; b c 2 , Œa; c b 2 c 2 , Œb; c a2 b 2 .mod K/. In case (G1 ) we substitute Œa2 ; b D k, Œa2 ; c D k in relations (18–21) and obtain Œb 2 ; a D 1;

Œb 2 ; c D k;

Œc 2 ; a D k;

Œc 2 ; b D 1:

If Œa; b D c 2 k, then we replace a with a0 D a1 D aa2 and get Œa0 ; b D Œaa2 ; b D Œa; ba Œa2 ; b D Œa; bk D c 2 k 2 D c 2 ; 2

and so writing again a instead of a0 , we may assume from the start that (˛)

Œa; b D c 2 :

If Œa; c D b 2 c 2 k, then we replace c with c 0 D c 1 so that the relation (˛) remains valid in view of o.c/ D 4, and we get Œa; c 0 D Œa; cc 2 D Œa; c 2 Œa; cc D kŒa; c D k 2 b 2 c 2 D b 2 .c 0 /2 ; 2

and so we may assume from the start that (ˇ)

Œa; c D b 2 c 2 :

Finally, in case Œb; c D a2 b 2 k, we replace b with b 0 D b 1 . Then the relation (˛) 2 remains unchanged: Œa; b 0 D Œa; bb 2 D Œa; b 2 Œa; bb D Œa; b, and also the relation

70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 225

(ˇ) remains unchanged and we get Œb 0 ; c D Œbb 2 ; c D Œb; cb Œb 2 ; c D Œb; ck D a2 b 2 k 2 D a2 .b 0 /2 ; 2

and so we may assume from the start that Œb; c D a2 b 2 :

( )

The structure of G D G1 is uniquely determined. If we set (identify) a D .1; 2/.3; 13; 7; 9/.4; 10; 8; 11/.5; 6/.12; 16/.14; 15/; b D .1; 3/.2; 9/.4; 14; 8; 12/.5; 7/.6; 13/.10; 16; 11; 15/; c D .1; 4/.2; 10; 6; 11/.3; 12/.5; 8/.7; 14/.9; 15; 13; 16/; we see that the permutations a; b; c satisfy all relations for G1 . Since k D Œa2 ; b D .1; 5/.2; 6/.3; 7/.4; 8/.9; 13/.10; 11/.12; 14/.15; 16/ and hki D Z.G/, we see that our permutations a; b; c induce a faithful permutation representation from G1 into the alternating group A16 . In case (G2 ) we substitute Œa2 ; b D k, Œa2 ; c D 1 in relations (18–21) and obtain Œb 2 ; a D k;

Œb 2 ; c D k;

Œc 2 ; a D k;

Œc 2 ; b D k:

If Œa; b D c 2 k, then we replace a with a1 . If Œa; c D b 2 c 2 k, then we replace c with c 1 . Finally, in case Œb; c D a2 b 2 k, we replace a with a1 and b with b 1 . Doing so, we can assume (as before) from the start that Œa; b D c 2 , Œa; c D b 2 c 2 , Œb; c D a2 b 2 . The structure of G D G2 is uniquely determined. Similarly, the permutations: a D .1; 2/.3; 9; 7; 10/.13; 16; 14; 15/.4; 12/.5; 6/.8; 11/; b D .1; 3/.5; 7/.2; 9; 6; 10/.11; 15/.12; 16/.4; 13; 8; 14/; c D .1; 4/.2; 11; 6; 12/.5; 8/.9; 16/.10; 15/.3; 13; 7; 14/; induce a faithful permutation representation from G2 into A16 . In case (G3 ) we substitute Œa2 ; b D 1, Œa2 ; c D k in the relations (18) to (21) and obtain Œb 2 ; a D k, Œb 2 ; c D 1, Œc 2 ; a D 1, Œc 2 ; b D k. If Œa; b D c 2 k, then we replace b with b 1 . If Œa; c D b 2 c 2 k, then we replace a with a1 . Finally, in case Œb; c D a2 b 2 k, we replace c with c 1 . Doing so, we can assume from the start that Œa; b D c 2 , Œa; c D b 2 c 2 , Œb; c D a2 b 2 . The structure of G D G3 is uniquely determined. The permutations: a D .1; 2/.4; 14; 8; 11/.12; 15; 13; 16/.3; 10/.5; 6/.7; 9/; b D .1; 3/.4; 13/.2; 9; 6; 10/.5; 7/.8; 12/.11; 16; 14; 15/; c D .1; 4/.3; 12; 7; 13/.2; 11/.5; 8/.6; 14/.9; 16; 10; 15/; induce a faithful permutation representation from G3 into A16 .

226

Groups of prime power order

Considering G1 ; G2 ; G3 as above subgroups of A16 , we see that the permutation .2; 4; 11; 6; 8; 12; 9; 7; 14; 10/.3; 13/ conjugates G1 onto G2 , and the permutation .2; 11; 8; 6; 14; 7; 10; 4/.3; 9; 13; 12/ conjugates G1 onto G3 . In particular, G2 and G3 are isomorphic to G D G1 . Since K < ˆ.G/, G satisfies the hypothesis since G=K does. The proof is complete. Thus, groups of the title are classified for p > 2 (see also [Bla1, Theorem 3.1]). Theorem 70.3. Let G be a p-group with d.G/ > 2, whose maximal subgroups are generated by two elements. If G is of class 3, then p D 2, 27 jGj 28 and G=K3 .G/ is isomorphic to the group of order 26 of Theorem 70.1(d). If jGj D 27 , then jK3 .G/j D 2 and G is isomorphic to the group of Theorem 70.2. If jGj D 28 , then K3 .G/ D Z.G/ Š E4 , G 0 D ˆ.G/ D 1 .G/ Š E25 and such groups exist. Proof. Suppose that all maximal subgroups of a p-group G are generated by two elements but d.G/ D 3. Assume, in addition, that G is of class 3. By Theorem 70.2, p D 2, jGj 27 , f1g < K D K3 .G/ Z.G/, and G=K is isomorphic to the group of order 26 of Theorem 70.1(d). It follows that G 0 D ˆ.G/;

G=G 0 Š E8 ;

G 0 =K Š E8 ;

ŒG 0 ; G D K;

1 .G=K/ D Z.G=K/ D G 0 =K: Let x 2 G 0 and g 2 G. Then Œx; g 2 K Z.G/ and so Œx; g2 D Œx 2 ; g D 1 since x 2 2 K. Hence the abelian group K is generated by involutions and so K is elementary abelian. Let S be a maximal subgroup in K. Then G=S (of order 27 ) is isomorphic to the group of Theorem 70.2. In particular, Z.G=S / D K=S and so K D Z.G/. Also, G 0 =S is elementary abelian. Hence ˆ.G 0 / S for any maximal subgroup S of K. Since K is elementary abelian and noncyclic, we get ˆ.G 0 / D f1g and so 1 .G/ D G 0 is elementary abelian. We have G D ha; b; ci, ha2 ; b 2 ; c 2 iK D G 0 and for each maximal subgroup S of K, G=S is isomorphic to the group of Theorem 70.2. Suppose that jKj 23 . Then there exists a maximal subgroup S of K which contains hŒa2 ; b; Œa2 ; ci. But then a2 S is contained in Z.G=S /. This is a contradiction since a2 62 K and Z.G=S / D K=S . We have proved that jKj 4. A group G of order 28 of Theorem 70.3 exists. For example, such a group is G D ha; b; c j a4 D b 4 D c 4 D k 2 D l 2 D m2 D klm D 1; Œa; k D Œb; k D Œc; k D 1; Œa; l D Œb; l D Œc; l D 1; Œa; b D c 2 ; Œa; c D b 2 c 2 ; Œb; c D a2 b 2 ; Œa2 ; b D m; Œa2 ; c D k; Œb 2 ; a D l; Œb 2 ; c D m; Œc 2 ; a D m; Œc 2 ; b D li:

70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 227

Here hk; li D Z.G/ D K3 .G/ Š E4 and G 0 D ˆ.G/ D ha2 ; b 2 ; c 2 ; k; li D 1 .G/ Š E25 . We get a faithful permutation representation of G by setting: a D .1; 2/.3; 21; 22; 15/.4; 23; 10; 18/.5; 8/.6; 11/.7; 17/.9; 14; 12; 26/ .13; 16; 25; 27/.19; 30; 20; 29/.24; 32; 31; 28/; b D .1; 3/.2; 14; 11; 15/.4; 24; 25; 20/.5; 9/.6; 12/.7; 22/.8; 21; 17; 26/ .10; 19; 13; 31/.16; 29; 27; 28/.18; 32; 23; 30/; c D .1; 4/.2; 16; 17; 18/.3; 19; 12; 20/.5; 10/.6; 13/.7; 25/.8; 27; 11; 23/ .9; 24; 22; 31/.14; 32; 21; 30/.15; 28; 26; 29/: In this way we see that the group G exists as a subgroup of A32 . In view of K < ˆ.G/, G satisfies the hypothesis since G=K does. Theorem 70.4. Let G be a p-group with d.G/ > 2 all of whose maximal subgroups are generated by two elements. If G is of class > 2, then p D 2 and one of the following holds: (i) G=K4 .G/ is isomorphic to the uniquely determined group of order 27 given in Theorem 70.2. (ii) H D G=K4 .G/ is isomorphic to one of the groups of order 28 described in Theorem 70.3 and more precisely, H D ha; b; ci with a4 D b 4 D c 4 D k 2 D l 2 D m2 D klm D 1;

Œa; k D Œb; k D Œc; k D 1;

Œa; l D Œb; l D Œc; l D 1; Œa; b D c 2 k ; Œa; c D b 2 c 2 k ; Œb; c D a2 b 2 ; Œa2 ; b D m; Œa2 ; c D k; Œb 2 ; a D l; Œb 2 ; c D m; Œc 2 ; a D m; Œc 2 ; b D l; where D 0; 1. These two groups (for D 0 and D 1) exist as subgroups of the alternating group A32 and they are not isomorphic. We have in both cases K3 .H / D Z.H / D hk; li Š E4 and H 0 D ˆ.H / D 1 .H / D ha2 ; b 2 ; c 2 ; k; li Š E25 . In case D 1, H possesses an automorphism of order 7 which acts fixed-point-freely on H=Z.H / and so H=Z.H / is isomorphic to the Suzuki group of order 26 (given in Theorem 70.1(d)). Proof. We may assume that K4 .G/ D f1g so that G is of class 3. Using Theorems 70.2 and 70.3, we see that G is either isomorphic to the group of order 27 given in Theorem 70.2 or a group of order 28 described in Theorem 70.3. It remains to determine completely the structure of G in the second case. In that case G=K3 .G/ is the special group of order 26 isomorphic to the group of Theorem 70.1(d), K3 .G/ D Z.G/ Š E4 , and G 0 D ˆ.G/ D 1 .G/ Š E25 . We have G D ha; b; ci, where a; b; c are elements of order 4 in G G 0 so that ha2 ; b 2 ; c 2 i K3 .G/ D G 0 and Œa; b c 2 , Œa; c b 2 c 2 , Œb; c a2 b 2 .mod K3 .G//. Also, for each maximal subgroup S of

228

Groups of prime power order

K3 .G/, jS j D 2 and G=S is isomorphic to the group of order 27 of Theorem 70.2. In particular, Z.G=S / D Z.G/=S is of order 2. We claim that each of the commutators Œa2 ; b, Œa2 ; c, Œb 2 ; a, Œb 2 ; c, Œc 2 ; a, Œc 2 ; b are distinct from 1. Indeed, if (for example) Œa2 ; b D 1, then we consider a maximal subgroup S of Z.G/ which contains Œa2 ; c. Then a2 S 2 Z.G=S /, contrary to the fact that Z.G=S / D Z.G/=S (and noting that a2 62 Z.G/). Set Œa2 ; b D m and Œa2 ; c D k. If m D k, then considering G=hmi, we have a hmi 2 Z.G=hmi/, a contradiction. It follows that k ¤ m and so K3 .G/ D hk; mi since K3 .G/ Š E4 . Set l D km so that k, l, m are the three distinct involutions in Z.G/ D K3 .G/. We compute (with some z 2 Z.G/): 2

Œb 2 ; c D Œbb; c D Œb; cb Œb; c D .a2 b 2 z/b a2 b 2 z D .a2 /b a2 D Œb; a2 D Œa2 ; b D m; Œc 2 ; a D Œc; ac Œc; a D .b 2 c 2 z/c b 2 c 2 z D .b 2 /c b 2 D Œc; b 2 D m; Œc 2 ; b D Œc; bc Œc; b D .a2 b 2 z/c a2 b 2 z D .a2 /c a2 .b 2 /c b 2 D Œc; a2 Œc; b 2 D km D l; Œb 2 ; a D Œb; ab Œb; a D .c 2 z/b c 2 z D .c 2 /b c 2 D Œb; c 2 D l: Hence we have obtained the following relations:

(˛ )

Œa2 ; b D m; Œa2 ; c D k;

Œb 2 ; a D l;

Œb 2 ; c D m; Œc 2 ; a D m; Œc 2 ; b D l;

where Z.G/ D K3 .G/ D hk; li Š E4 and m D kl. Also, we may set (˛ )

Œa; bc 2 D k p l q ;

Œa; cb 2 c 2 D k r l s ;

Œb; ca2 b 2 D k t l u ;

where p, q, r, s, t , u are some integers mod 2. For any x 2 G G 0 , Œx; G 0 Z.G/ and so hx; G 0 i is of class 2. This fact will be used many times. For any x; y 2 G G 0 and any g1 ; g2 2 G 0 , we have Œ.xg1 /2 ; yg2 D Œ.xg1 /2 ; y D Œx 2 g12 Œg1 ; x; y D Œx 2 ; y: Therefore, if we replace a; b; c with a0 D aa2.qCsCu/ , b 0 D bb 2.sCu/ , c 0 D cc 2s , respectively, then the relations (˛ ) remain preserved. However, the relations (˛ )

70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 229

are simplified, as the following computation shows: Œa0 ; b 0 .c 0 /2 D Œaa2.qCsCu/ ; bb 2.sCu/ c 2 D Œa; bb 2.sCu/ a

2.qCsCu/

Œa2.qCsCu/ ; bb 2.sCu/ c 2

D Œa; bb 2.sCu/ Œa2.qCsCu/ ; bb 2.sCu/ c 2 ; Œa; b 2.sCu/ Œa; bŒa2.qCsCu/ ; b 2.sCu/ Œa2.qCsCu/ ; bc 2 D Œa; b 2 sCu Œa; bŒa2 ; bqCsCu c 2 D .Œa; bc 2 /l sCu mqCsCu D k p l q l sCu k qCsCu l qCsCu D k pCqCsCu; and we get in a similar way Œa0 ; c 0 .b 0 /2 .c 0 /2 D k qCrCu , Œb 0 ; c 0 .a0 /2 .b 0 /2 D k sCtCu . Writing again a, b, c (instead of a0 , b 0 , c 0 ), we see that the relations (˛ ) are simplified into: (˛ )

Œa; bc 2 D k ;

Œa; cb 2 c 2 D k ;

Œb; ca2 b 2 D k ;

where , , are some integers mod 2. In this way we have obtained eight possibilities G1 ; G2 ; : : : ; G8 for the structure of G depending on the values of these integers: (G1 )

D 0;

D 0;

D 0I

(G2 )

D 0;

D 0;

D 1I

(G3 )

D 0;

D 1;

D 0I

(G4 )

D 0;

D 1;

D 1I

(G5 )

D 1;

D 0;

D 0I

(G6 )

D 1;

D 0;

D 1I

(G7 )

D 1;

D 1;

D 1I

(G8 )

D 1;

D 1;

D 0:

We replace now a; b; c with (ˇ )

a D bb 2 ;

b D c;

c D aca2 b 2 l;

respectively, and verify first that the relations (˛ ) remain unchanged. For example: Œ.a /2 ; c D Œb 2 ; aca2 b 2 l D Œb 2 ; ac D Œb 2 ; cŒb 2 ; ac D ml c D ml D k: We want to see what happens with the relations (˛ ) for arbitrary integers , , .mod 2/. First we compute: .ac/2 D a.ca/c D a.acŒc; a/c D a.acb 2 c 2 k /c D a2 .cb 2 /c 3 k D a2 .b 2 cm/c 3 k D a2 b 2 k C1 l;

230

Groups of prime power order

which gives .c /2 D .aca2 b 2 l/2 D .ac/2 .a2 b 2 l/2 Œa2 b 2 l; ac D a2 b 2 k C1 lŒa2 b 2 ; cŒa2 b 2 ; ac D a2 b 2 k C1 lŒa2 ; cŒb 2 ; cŒb 2; ac D a2 b 2 k C1 l: Then it is easy to transform the relations (˛ ): Œa ; b .c /2 D Œbb 2; ca2 b 2 k C1 l D Œb; cb Œb 2 ; ca2 b 2 k C1 l 2

D .Œb; ca2 b 2 /mk C1 l D k mk C1 l D k C ; and we get in a similar way: Œa ; c .b /2 .c /2 D k CCC1 and

Œb ; c .a /2 .b /2 D k C1 :

These results show that our group Gi defined with the constants , , in the relations (˛ ) is isomorphic with the group Gj defined with the constants C , C C C 1, C 1. In particular, the group G1 defined with D D D 0 is isomorphic to the group G4 defined with the constants 0; 1; 1. Further, the group G4 with D 0, D 1, D 1 is isomorphic with G3 defined with the constants 0 ,1, 0 and this one is isomorphic with G5 . Then G5 is isomorphic with G2 , G2 is isomorphic with G6 and G6 is isomorphic with G7 . However, our replacement (ˇ ) sends G8 onto G8 and we verify that in fact (ˇ ) induces an automorphism of order 7 on G8 which acts fixed-point-freely on G8 =Z.G8 /. Hence G=Z.G/ is isomorphic to the Suzuki group of order 26 given in Theorem 70.1(d). The group G1 is exactly the group occurring as an example in the proof of Theorem 70.3, where we have found a faithful permutation representation of G1 of degree 32 (with even permutations). We get a faithful permutation representation of G8 by setting: a D .1; 2/.3; 14; 21; 25/.4; 18; 10; 27/.5; 8/.6; 11/.7; 17/ .9; 24; 12; 15/.13; 26; 23; 16/.19; 30; 20; 29/.22; 32; 31; 28/I b D .1; 3/.2; 14; 11; 15/.4; 22; 23; 20/.5; 9/.6; 12/.7; 21/ .8; 24; 17; 25/.10; 19; 13; 31/.16; 29; 26; 28/.18; 32; 27; 30/I c D .1; 4/.2; 16; 17; 18/.3; 19; 12; 20/.5; 10/.6; 13/.7; 23/ .8; 26; 11; 27/.9; 22; 21; 31/.14; 32; 24; 30/.15; 28; 25; 29/: In this way, we see that also the group G8 exists as a subgroup of A32 . Finally, considering G1 and G8 as subgroups of A32 , W. Lempken (Institut f¨ur experimentelle Mathematik, Essen) has shown that the groups G1 and G8 are not isomorphic.

70 Non-2-generator p-groups all of whose maximal subgroups are 2-generator 231

Let G be a 2-group of class > 3 with d.G/ > 2 but d.H / D 2 for all H 2 1 . Then, it is possible to prove that we must have jGj 29 and we exhibit here an example of such a group G of order 29 and class 4, where G D ha; b; ci with the following defining relations: a8 D b 4 D c 8 D 1;

a4 D c 4 D u;

Œa2 ; b D m;

Œa2 ; c D k;

Œb 2 ; a D lu;

Œb 2 ; c D um;

Œa; c D b 2 c 2 ;

m2 D k 2 D 1;

Œb; c D a2 b 2 ;

Œm; a D 1;

Œm; b D 1;

Œk; c D 1;

Œl; a D 1;

Œa; u D Œb; u D Œc; u D 1; km D l;

Œc 2 ; a D um;

Œc 2 ; b D l;

Œa2 ; b 2 D 1;

Œm; c D u; Œl; b D u;

l 2 D 1; Œa; b D c 2 ;

Œa2 ; c 2 D 1;

Œk; a D 1;

Œb 2 ; c 2 D u;

Œk; b D u;

Œl; c D u:

Here G 0 D ha2 ; b 2 ; c 2 ; k; li is of order 26 , K4 .G/ D Z.G/ D hui is of order 2, K3 .G/ D hk; l; ui Š E8 , and Z.G 0 / D ha2 ; k; li is abelian of type .4; 2; 2/. Finally, we have shown that this group G exists as a subgroup of the symmetric group of degree 64. We can improve Theorem 70.4 in case (i). Theorem 70.5. Let G be a p-group with d.G/ > 2 all of whose maximal subgroups are generated by two elements. If G is of class > 2, then p D 2, and if G=K4 .G/ is isomorphic to the group of order 27 given in Theorem 70.2 (case (i) of Theorem 70.4), then K4 .G/ D f1g. Proof. Assume that K4 .G/ ¤ f1g. To get a contradiction we may assume jK4 .G/j D 2 (by considering a suitable quotient group of G). Then G=K4 .G/ is isomorphic to the group of order 27 of Theorem 70.2. We may set G D ha; b; ci, where a4 b 4 c 4 k 2 1; Œa; c b 2 c 2 ;

Œa; k Œb; k Œc; k 1; Œb; c a2 b 2 ;

Œa2 ; c k;

Œb 2 ; a 1;

Œc 2 ; a k;

Œc 2 ; b 1

Œa; b c 2 ; Œa2 ; b k; Œb 2 ; c k;

.mod K4 .G//:

Also we may set K4 .G/ D hui Z.G/. Then K3 .G/ D hk; ui, W D G 0 D ˆ.G/ D ha2 ; b 2 ; c 2 ; k; ui is of order 25 and class 2 since W =K4 .G/ Š E24 . Since CG .K3 .G// is a subgroup of index 2 in G, we get K3 .G/ Z.W /. From Œa2 ; b k .mod K4 .G// with k 62 K4 .G/ follows Œa2 ; b D k0 with k0 2 K3 .G/ K4 .G/. From a4 1 .mod K4 .G// follows a4 2 K4 .G/ Z.G/ and so 2

2

1 D Œa4 ; b D Œa2 a2 ; b D Œa2 ; ba Œa2 ; b D k0a k0 D k02 ; since a2 2 W and K3 .G/ D hk0 ; ui Z.W /. Hence k0 is an involution and so we have proved that K3 .G/ Š E4 .

232

Groups of prime power order

Using Œc; b 2 D kui and Œc 2 ; b D uj , where i; j are some integers (mod 2), we compute Œc 2 ; b 2 D Œcc; b 2 D Œc; b 2 c Œc; b 2 D .kui /c kui D k c k D c 1 kck D Œc; k; since k is an involution. On the other hand, we get Œc 2 ; b 2 D Œc 2 ; bb D Œc 2 ; bŒc 2 ; bb D uj .uj /b D .uj /2 D 1; and so Œc; k D 1. Using Œa; b 2 2 K4 .G/ Z.G/ and Œa2 ; b D kui with an integer i .mod 2/, we compute Œa2 ; b 2 D Œaa; b 2 D Œa; b 2 a Œa; b 2 D Œa; b 2 2 D 1; Œa2 ; b 2 D Œa2 ; bb D Œa2 ; bŒa2 ; bb D kui .kui /b D kk b D Œk; b; and so Œk; b D 1. Finally, using Œa; c 2 D kuj and Œa2 ; c D kul with some integers j , l .mod 2/, we get Œa2 ; c 2 D Œaa; c 2 D Œa; c 2 a Œa; c 2 D .kuj /a kuj D k a k D Œa; k; Œa2 ; c 2 D Œa2 ; cc D Œa2 ; cŒa2 ; cc D kul .kul /c D kk c D Œk; c: Hence (by the above) Œa; k D Œk; c D 1. We have obtained Œk; a D Œk; b D Œk; c D 1, and so K3 .G/ D hk; ui Z.G/. This is a contradiction since G was of class 4. The group G of Theorem 70.2 of order 27 is an A4 -group since ha2 ; bi is a nonabelian subgroup of order 24 (this subgroup is minimal nonabelian nonmetacyclic) and G has no nonabelian subgroups of order 8. Obviously, D8 is not a subgroup of G since 1 .G/ D E24 D G 0 . Also, Q D Q8 is not a subgroup of G (otherwise, Q \ G 0 D C2 and so S D G 0 Q is a maximal subgroup of G and so d.S / D 2 so ˆ.S / D G 0 and, by Taussky’s theorem, S is of maximal class, a contradiction). Both groups G of Theorem 70.4 of order 28 are A5 -groups since ha2 ; bi is a nonabelian subgroup of order 24 (this subgroup is minimal nonabelian nonmetacyclic) and G has no nonabelian subgroups of order 8. Obviously, D8 is not a subgroup of G since 1 .G/ D E25 D G 0 . Also, Q D Q8 is not a subgroup of G (otherwise, Q \ G 0 D C2 and so T D G 0 Q is a maximal subgroup of G hence d.T / D 2; then ˆ.T / D G 0 so, by Taussky’s theorem, T is of maximal class, a contradiction). The above two paragraphs show that if a 2-group G is such that d.G/ D 3 but d.H / D 2 for all H 2 1 and the class of G is > 2, then G is an An -group with n < 5 if and only if G is the unique group of order 27 from Theorem 70.2, and this G is an A4 -group.

71

Determination of A2-groups

We recall that a p-group G is called an An -group if every subgroup of index p n is abelian but at least one subgroup of index p n1 is nonabelian. Here we determine A2 -groups of order > p 4 up to isomorphism in terms of generators and relations. This determination will follow five propositions covering all A2 -groups. Proposition 71.1. Suppose that G is an A2 -group of order > p 4 . Then jG 0 j D p if and only if G has at least two distinct abelian maximal subgroups and in that case one of the following holds: (i) G D H Cp , where H is minimal nonabelian. (ii) G D ha; b; ci, where m

n

2

ap D b p D c p D 1;

m n 1; m 2;

Œa; b D d;

c p D d;

Œa; d D Œb; d D Œa; c D Œb; c D 1: Here H D ha; bi is nonmetacyclic minimal nonabelian, H 0 D G 0 D hd i, G D H C , where C D hci is cyclic of order p 2 and H \ C D hd i D hc p i D H 0 . Proof. Let G be an A2 -group of order > p 4 with jG 0 j D p. By Lemmas 65.2(c) and 65.4(d), jG 0 j D p if and only if the set 1 has at least two abelian members. In that case the set 1 has a nonabelian member H so that G D H Z.G/, where Z.G/ \ H D Z.H /. We have jH j > p 3 , H 0 D G 0 , Z.H / D ˆ.H / D ˆ.G/, and jH W Z.H /j D p 2 since H is an A1 -group. Clearly, jZ.G/ W Z.H /j D p. Suppose that there is an element c 2 Z.G/ H of order p. Then G D H hci and it is clear that all such groups are A2 -groups. We assume in the sequel that 1 .Z.G// H ; then d.Z.H // D d.Z.G// so jZ.H / W ˆ.Z.H //j D jZ.G/ W ˆ.Z.G//j D pjZ.H / W ˆ.Z.G//j (note that ˆ.Z.H // < ˆ.Z.G// Z.H / since Z.H / is maximal in Z.G/) and so jˆ.Z.G// W ˆ.Z.H //j D p. Hence, if Z.G/=ˆ.Z.G// Š Ep d , then Z.G/=ˆ.Z.H // Š Cp 2 Ep d 1 . This implies that for each c 2 Z.G/ Z.H /, we have c p 2 Z.H / ˆ.Z.H //. (i) Assume, in addition, that H is metacyclic; then, by Lemma 65.1(b), we have m n m1 m1 H D ha; b j ap D b p D 1, m 2, n 1, ab D a1Cp i, where H 0 D hap i mCn mCnC1 p p and jH j D p , m C n > 3, jGj D p . We have Z.H / D ha i hb i D 2 2 2 2 ˆ.H / and for each c 2 Z.G/ H , c p 2 hap ; b p i hap ; b p i since hap ; b p i D ˆ.Z.H //.

234

Groups of prime power order

If n D 1, then H Š Mp mC1 , m > 2, and Z.H / D hap i is cyclic. In this case, replacing an element c 2 Z.G/ Z.H / with c i , i 6 0 .mod p/, if necessary, we may m1 assume that c p D ap and so .ca/p D c p ap D 1. Since .ca/b D cab D .ca/ap , 3 it follows that hca; bi is nonabelian of order p . This is a contradiction, since jGj p 5 and G is an A2 -group. We have proved that n 2. Suppose that there exists an element x 2 G H of order p. Since d.Z.G// D d.Z.H //, we get x 62 Z.G/ and so Œa; x ¤ 1 or Œb; x ¤ 1. If Œa; x ¤ 1, then (noting that hai is normal in G) ha; xi is nonabelian of order p mC1 . This is a contradiction since jGj D p mCnC1 and n 2. Hence we must have Œa; x D 1 and Œb; x ¤ 1. In that case hŒb; xi D G 0 D H 0 and so H D hb; xi is a nonmetacyclic A1 -group of order p nC2 . Hence H must be a maximal subgroup of G containing Z.H / D ˆ.G/ but not containing Z.G/ (every member of the set 1 , containing Z.G/, is abelian!), and so jGj D p nC3 . We get m D 2 and G D H Z.G/, where H is a nonmetacyclic A1 -group. This case will be considered in part (ii) of the proof. Hence assume that 1 .G/ H . By the paragraph, preceding (i), for each c 2 Z.G/ Z.H /, we get c p 2 Z.H / ˆ.Z.H //, where Z.H / D ˆ.H / D Ã1 .H / since H is metacyclic. If there is h 2 H such that c p D hp , then the abelian subgroup hh; ci is not cyclic since hhi and hci are two distinct cyclic subgroups of hh; ci of the same order. Since hh; ci \ H D hhi, there is an element of order p in hh; ci H , contrary to 1 .G/ H . We conclude that hc p i is a maximal cyclic subgroup of H . Since c p 2 Z.H /, the quotient group H=hc p i is noncyclic so c p ˆ.H / D Ã1 .H / since H is metacyclic. It follows that not every element in Ã1 .H / is a p-th power of an element in H so H is irregular so p D 2 since H is of class 2 (Lemma 64.1(a)), and c 2 is not a square in H . Set d D Œa; b ¤ 1, where d 2 D 1, and compute for any integers i; j : .ai b j /2 D a2i b 2j Œb j ; ai D a2i b 2j d ij . If i or j is even, then a2i b 2j D .ai b j /2 is a square in H . Therefore c 2 D a2k b 2l , where both k and l are odd. Consider the nonabelian subgroup S D ha; b l ci (recall that c 2 Z.G/). We have .b l c/2 D b 2l c 2 D b 2l a2k b 2l D a2k , and so jS j D 2mC1 . This is a contradiction, since jG W S j D p 2 > p in view of jGj D 2mCnC1 and n 2. (ii) It remains to consider the case where H is a nonmetacyclic A1 -group of order p 4 . We set m

n

H D ha; b j ap D b p D 1; Œa; b D d; d p D Œa; d D Œb; d D 1i; where we may assume that m n 1. Here jH j D p mCnC1 and so jGj D p mCnC2 . Since jH j p 4 , must be m 2. We have Z.H / D ˆ.H / D hap i hb p i hd i and 2 2 ˆ.Z.H // D hap i hb p i since o.d / D p. Also note that hd i is a maximal cyclic subgroup in H and for each c 2 Z.G/ Z.H /, c p 2 Z.H / ˆ.Z.H //. Note that H \ Z.G/ D Z.H /. We want to show that there exists c 2 Z.G/ Z.H / so that 1 ¤ c p 2 hd i. Assume first that n D 1 so that Z.H / D hap i hd i and for an element c 2 Z.G/ Z.H /, c p D aip d j ; recall that o.c/ > p). If i 0 .mod p/, then j 6 0

71

Determination of A2 -groups

235 0

2

.mod p/ (otherwise, o.c/ D p) and we may set c p D ai p d j . Then we compute 0 0 2 0 .ai p c/p D ai p c p D d j ¤ 1, and we are done since ai p c 2 Z.G/ Z.H / D Z.G/ H . If i 6 0 .mod p/, then S D hai c; bi is nonabelian and .ai c/p D aip c p D aip aip d j D d j . Hence p o.ai c/ p 2 . If o.ai c/ D p 2 , then j 6 0 .mod p/ and hai ci hd j i D G 0 . In any case, jS j D p 3 . This is a contradiction since jG W S j p 2 and G is an A2 -group. Thus, we may also assume that n 2. We have for c 2 Z.G/ Z.H /, c p D ip jp k a b d , where at least one of the integers i; j; k is 6 0 .mod p/ since c p 62 ˆ.Z.H //. If i 0 .mod p/ and j 0 .mod p/, then k 6 0 .mod p/ and ai ; b j 2 Z.H / < Z.G/ so .ai b j c/p D aip b jp c p D aip b jp aip b jp d k D d k ¤ 1; and we are done since ai b j c 2 Z.G/ Z.H /. Now assume that one of the integers i; j is 0 .mod p/ and the other one is 6 0 .mod p/. Note that a and b occur symmetrically since we have m 2 and n 2. Interchanging a and b (if necessary), we may assume that i 6 0 .mod p/ but j 0 .mod p/. In that case T D hai b j c; bi is nonabelian and since b j ; c 2 Z.G/ we get .ai b j c/p D aip b jp c p D d k . Thus p o.ai b j c/ p 2 and if o.ai b j c/ D p 2 , then k 6 0 .mod p/ and hai b j ci hd k i D G 0 . In any case, jT j D p nC2 . This is a contradiction since jG W T j D p m > p. It remains to study the possibility i 6 0 .mod p/ and j 6 0 .mod p/. We consider again the nonabelian subgroup T D hai b j c; bi. If p > 2, then .ai b j c/p D aip b jp c p D d k . If p D 2, then .ai b j c/2 D .ai b j /2 c 2 D a2i b 2j Œb; aij c 2 D a2i b 2j d ij a2i b 2j d k D d ij Ck : In any case, we have jT j D p nC2 , a contradiction since jG W T j D p m > p. We have proved that there exists c 2 Z.G/ H such that 1 ¤ c p 2 hd i. Replacing c with another generator of hci, we may assume that c p D d . The structure of G is uniquely determined. Proposition 71.2. Let G be a metacyclic A2 -group of order > p 4 . Then G 0 Š Cp 2 and we have one of the following possibilities: m n m1 , n 1, D 0; 1, m C n 5, (a) G D ha; b j ap D 1, m 3, b p D ap m2 ab D a1Cp i, where in case p D 2, m 4. (b) p D 2, G D ha; b j a8 D 1, b 2 D a4 , D 0; 1, n 2, ab D a1C4 , D 0; 1i. n

Proof. By Corollary 65.3, a metacyclic p-group G is an A2 -group if and only if G 0 Š Cp 2 . Let G0 be the subgroup of order p in G 0 ; then G=G0 is an A1 -group (Lemma

236

Groups of prime power order

65.2(a)). But jG=G0 j > p 3 and so G=G0 is a “splitting” metacyclic group (Lemma 65.1(b,c)). We have G D AB, where A G 0 , A \ B D G0 , A=G0 (of order p 2 ) and B=G0 are cyclic. Since G0 < G 0 < A and G 0 is cyclic, A does not split over G0 . Hence A is a cyclic normal subgroup of order p m , m 3, and we set A D hai. n m1 Let b 2 B be such that hbi covers A=G0 so that b p 2 G0 D hap i, where n p D jB=G0 j, n 1. Changing a generator of hai if necessary, one may assume n m1 b p D ap with D 0; 1. Now, hbi induces an automorphism group of order p 2 on A so that ŒA; B Š Cp 2 . In that case our result follows at once. Proposition 71.3. Let G be a nonmetacyclic A2 -group of order > p 4 possessing exactly one abelian maximal subgroup. Then G 0 Š Ep 2 and assume, in addition, that G 0 6 Z.G/. In that case p > 2, d.G/ D 2, K3 .G/ is of order p, G=K3 .G/ is nonmetacyclic minimal nonabelian, and one of the following holds: (i) G D ha; bi, where n

ap D b p D 1; n 3;

c p D Œa; c D 1;

Œa; b D c;

Œb; c D b p

n1

:

Here jGj D p nC2 , G 0 D hc; b p i Š Ep 2 , K3 .G/ D hb p i, M D ha; c; b p i 2 1 is abelian of type .p; p; p n1 /, and all members of the set 1 fM g are metacyclic. n1

n1

(ii) G D ha; bi, where n

ap D b p D 1; n 2;

Œa; b D c;

Œb; c D d;

c p D d p D Œa; c D Œd; a D Œd; b D 1: Here jGj D p nC3 , G 0 D hc; d i Š Ep 2 , K3 .G/ D hd i, M D ha; c; d; b p i 2 1 is abelian of type .p; p; p; p n1 /. (iii) G D ha; bi, where 2

n

ap D b p D 1; n 2;

Œa; b D c;

Œc; b D asp ;

c p D Œa; c D 1;

and s D 1 or s is a fixed quadratic non-residue mod p. Here jGj D p nC3 , G 0 D hc; ap i Š Ep 2 , K3 .G/ D hap i, M D ha; c; b p i 2 1 is abelian of type .p; p 2 ; p n1 /, and all members of the set 1 are nonmetacyclic. Proof. Lemma 65.2(d,e), Lemma 65.4(c) and Theorem 65.7(c,d) imply that G 0 Š Ep 2 , d.G/ D 2 and p > 2. Since G 0 6 Z.G/, we have jK3 .G/j D p and G=K3 .G/ is a nonmetacyclic A1 -group (Lemmas 65.2(e) and 64.1(m)). By the structure of G=K3 .G/, there are cyclic subgroups A=K3 .G/ > f1g and B=K3 .G/ > f1g of G=K3 .G/ such that hA; Bi D G;

A \ G 0 D B \ G 0 D A \ B D .G 0 A/ \ B D .G 0 B/ \ A D K3 .G/:

71

Determination of A2 -groups

237

Let A D ha; K3 .G/i and B D hb; K3 .G/i. The commutator Œa; b is of order p since ab ¤ ba and exp.G 0 / D p and hŒa; bi \ K3 .G/ D f1g since G=K3 .G/ is nonabelian. Since G 0 6 Z.G/, both A and B cannot centralize G 0 . To fix ideas, assume that B does not centralize G 0 , which implies ŒB; G 0 D K3 .G/. Since G 0 B < G is nonabelian, we must have jG W .G 0 B/j D p. But .G 0 B/ \ A D K3 .G/ implies jAj D p 2 , by the product formula. If also A does not centralize G 0 , then G 0 A is nonabelian and so G 0 A 2 1 . In that case jGj D pjG 0 Aj D p p 3 D p 4 , a contradiction. We have proved that A centralizes G 0 . Let B1 B be such that K3 .G/ B1 and jB W B1 j D p. Let M 2 1 be abelian. Since G 0 < M and G 0 6 Z.G/, we have M D CG .G 0 /. On the other hand, A centralizes G 0 and so A < M . Finally, B1 ˆ.G/ < M and so M D G 0 AB1 (recall that jG W B1 G 0 j D p 2 ) and A 6 B1 G 0 since G D hA; Bi). Since jGj > p 4 and jAG 0 j D p 3 , we have jB1 j p 2 and so jBj p 3 . (i) Suppose first that B does not split over K3 .G/, i.e., B is cyclic. Set B D hbi n1 so that o.b/ D p n with n 3 and hb p i D K3 .G/. The subgroup BG 0 Š Mp nC1 is maximal in G since .BG 0 /A D G, .BG 0 / \ A D K3 .G/ and jAj D p 2 . By Lemma 64.1(p), d.M / > 2 (otherwise, all members of the set 1 are two-generator which is not the case since a nonmetacyclic A1 -group has a maximal subgroup that is not generated by two elements; see Lemma 65.1(a)); moreover, d.M / D 3 (consider M \ .BG 0 /) and so 1 .M / Š Ep 3 . By the product formula, G D B1 .M /. It follows from jABj < jGj and hA; Bi D G that B is not normal in G. Therefore, there is a 2 1 .M / that does not normalize B. One can take A D ha; K3 .G/i. Set Œa; b D c, where c is an element of order p in G 0 K3 .G/. Since BG 0 Š Mp nC1 , n1

one may assume that b c D b 1Cp (if necessary, changing b by another element of order p n in BG 0 ). Thus we have obtained the group (i) of our proposition. (ii) Suppose that B splits over K3 .G/ so that B D K3 .G/ hbi, where o.b/ D p n , n 2. Consider first the subcase (ii1) where A also splits over K3 .G/ so that A D K3 .G/ hai with o.a/ D p. Then Œa; b D c, where c 2 G 0 K3 .G/ is of order p and Œb; c D d , where hd i D K3 .G/. We have obtained the group (ii) of our proposition. (ii2) It remains to consider the subcase where A does not split over K3 .G/ so that A D hai is cyclic of order p 2 and hap i D K3 .G/. We have again Œa; b D c, where c 2 G 0 K3 .G/ is of order p and Œc; b D asp , where s 6 0 .mod p/. For any i 6 0 .mod p/, we replace b with b 0 D b i . Then Œa; b 0 D Œa; b i D Œa; bi D c i D c 0 2 with a suitable element 2 K3 .G/ Z.G/. Hence Œc 0 ; b 0 D Œc i ; b i D Œc; bi D 2 0 ai sp D as p , where s 0 D i 2 s 6 0 .mod p/. It follows that we may assume (running through all i 6 0 .mod p/) that either s 0 D 1 or s 0 is a (fixed) quadratic non-residue mod p. Writing again b; c; s instead of b 0 ; c 0 ; s 0 , we have obtained the groups stated in part (iii) of our proposition. Proposition 71.4. Let G be a nonmetacyclic A2 -group of order > p 4 possessing exactly one abelian maximal subgroup. Then G 0 Š Ep 2 and assume, in addition, that

238

Groups of prime power order

G 0 Z.G/. In that case d.G/ D 3, Z.G/ D ˆ.G/, and one of the following holds: (a) If G has no normal elementary abelian subgroups of order p 3 , then p D 2 and G D ha; b; c j a4 D b 4 D Œa; b D 1; c 2 D a2 ; ac D ab 2 ; b c D ba2 i is the minimal nonmetacyclic group of order 25 . Here G is a special group of exponent 4, 1 .G/ D G 0 D Z.G/ D ˆ.G/ D ha2 ; b 2 i Š E4 , C4 C4 Š M D hai hbi 2 1 is abelian, and all other six members of the set 1 are metacyclic of exponent 4. (b) If G has a normal elementary abelian subgroup E of order p 3 , then E D 1 .G/ and one of the following holds: (b1) G D ha; b; d i with ˛

2

ap D b p D d p D 1; Œa; b D d p a

p ˛1

;

˛ 2; Œa; d D ap

Œd; b D 1;

˛1

;

where D 0 unless p D 2 and ˛ D 2 in which case D 1. Here jGj D p ˛C3 ;

G 0 D hd p ; ap

E D 1 .G/ D hb; d p ; a

˛1

p ˛1

i Š Ep 2 ;

i 6 Z.G/;

and hb; d; ap i 2 1 is abelian of type .p; p 2 ; p ˛1 /. (b2) G D ha; b; ci with ˛

2

2

ap D b p D c p D Œb; c D 1; Œa; b D x;

Œa; c D y;

˛ 1; ˛0

bp D x yˇ ;

c p D x y ı ;

x p D y p D Œa; x D Œa; y D Œb; x D Œb; y D Œc; x D Œc; y D 1; where in case p D 2, ˛ 0 D 0, ˇ D D ı D 1, and in case p > 2, 4ˇ C .ı ˛ 0 /2 is a quadratic non-residue mod p. We have jGj D p ˛C4 , ˛1 G 0 D hx; yi Š Ep 2 , E D 1 .G/ D G 0 hap i. Let A 2 1 be abelian. Then in case ˛ D 1, A D hb; ci Š Cp 2 Cp 2 and E 6 Z.G/ and in case ˛ 2, A D hap ; b; ci is abelian of type .p ˛1 ; p 2 ; p 2 / and E Z.G/. Proof. Let G be a nonmetacyclic A2 -group of order > p 4 having exactly one abelian maximal subgroup. Then we have G 0 Š Ep 2 (Lemmas 64.1(q), 65.4(d) and Theorem 65.7(a)), and we assume here, in addition, that G 0 Z.G/. It follows from Lemma 65.2(a) that d.G/ D 3. Next, Z.G/ D ˆ.G/ (Lemma 65.4(c)), G=Z.G/ Š Ep 3 , and if H ¤ K 2 1 are both nonabelian, then Z.H / D Z.K/ D Z.G/ (Theorem 65.7(f)). It follows that if x; y 2 G, then jhx; yij p o.x/o.y/ (since d.G/ D 3, hx; yi is either abelian or an A1 -subgroup). Also, if G has no normal elementary abelian subgroups of order p 3 , then G is minimal nonmetacyclic (Lemma 65.4(j)). In the last case, p D 2,

71

Determination of A2 -groups

239

jGj D 25 , and G is the group stated in part (a) of our proposition (Lemma 64.1(l) and Theorem 66.1). From now on we assume that the group G has a normal subgroup E Š Ep 3 . By Lemma 65.4(k), either 1 .G/ D E or E < 1 .G/ D K Š Ep 4 . Suppose that the second case occurs. Every maximal subgroup of G containing K must be abelian (Lemma 65.1) and G=K > f1g is cyclic (Lemma 65.4(b)). Let G D Khai; then K \ hai D D is of order p. Indeed, jDj p. Assume that D D f1g. Then, if hai < T 2 1 , then 1 .T / h1 .hai/; T \ Ki is of order p 4 , so T is abelian (Lemma 65.1). Since K 6 T , the set 1 has two distinct abelian members, contrary to the assumption. Thus, p s D jhaij p 2 , D Z.G/, and jGj D p sC3 . Since Khap i is abelian, hap i Z.G/. We have G 0 Š Ep 2 , G 0 < K, and G 0 Z.G/ (by assumption). Since G=Z.G/ Š Ep 3 and G 0 D Z.G/, we get G 0 > D (otherwise, d.G=G 0 / D 2 < 3 D d.G/). Therefore, G 0 hai is an abelian normal subgroup of G of type .p s ; p/, s 2, and Z.G/ D G 0 hap i. We have CK .a/ < Z.G/ and so CK .a/ D G 0 and CG .a/ D G 0 hai. The subgroup G 0 hai has exactly p cyclic subgroups of order p s . Since jG W .G 0 hai/j D p 2 , we get NK .hai/ > G 0 . Let x 2 NK .hai/ G 0 so that x (of order p) induces an automorphism of order p on hai. But then ha; xi is a nonabelian subgroup of order p sC1 so its index in G equals p 2 , a contradiction. We have proved that 1 .G/ D E Š Ep 3 . Since G 0 < E, the quotient group G=E is abelian. Since d.G=G 0 / D 3, G=E is also noncyclic. In what follows A denotes the unique abelian maximal subgroup of G. (i) First assume that E 6 ˆ.G/ .D Z.G//; then d.G=E/ D 2. If A does not contain E, then G D AE, 1 .A/ D A \ E D G 0 Š Ep 2 and so A is metacyclic. Hence each maximal subgroup of G is generated by two elements (Lemma 65.1) and d.G/ D 3. Since G is of class 2 (by hypothesis), Theorem 70.1 shows that G is isomorphic to a group (b2) for ˛ D 1 in our proposition. We may assume that E < A and so each maximal subgroup of G which does not contain E is nonabelian metacyclic (Lemma 65.1, since all members of the set 1 fAg are A1 -groups). Let M be a nonabelian maximal subgroup of G containing E (there are exactly p such subgroups since d.G=E/ D 2). Then M is a nonmetacyclic A1 group with Z.M / D ˆ.M / D ˆ.G/ D Z.G/ (Theorem 65.7(f)). By Lemma 65.1, we have, since E 6 Z.M / (otherwise, E Z.G/, which is not the case since d.G=E/ D 2 < 3 D d.G=Z.G//), ˛

M D ha; b j ap D b p D 1; ˛ 2; Œa; b D c; c p D Œa; c D Œb; c D 1i; where hci D M 0 ; Z.M / D Z.G/ D hci hap i;

b 2 E G0; jGj D p ˛C3 ;

M D hbi .hci hai/; G 0 D hap

˛1

i hci:

In particular, G=E has a cyclic subgroup M=E of index p. Now, ha; ci . G 0 / is an abelian normal subgroup of type .p ˛ ; p/ and it has exactly p cyclic subgroups of

240

Groups of prime power order

index p which are not normal in M (see the last assertion of Lemma 65.1). Therefore, NM .hai/ D ha; ci and so N D NG .hai/ must cover G=M since N 6 M in view of jG W N j D c˛ .ha; ci/ D p D jG W M j (here c˛ .X/ is the number of cyclic subgroups of order p ˛ in a p-group X). Thus N 2 1 with M \ N D ha; ci and so E \ N D G 0 . Therefore N is nonabelian metacyclic, CG .a/ D ha; ci, N is an A1 group, and so it is “splitting” metacyclic (Lemma 65.1). We have f1g < N 0 < hai\G 0 ˛1 and so N 0 D hap i Ã1 .N / (Lemma 65.1). It follows that Z.G/ D hap ; ci D Z.N / D ˆ.N / D Ã1 .N /, and so there exists d 2 N M D N ha; ci such that ˛1 d p 2 G 0 hap i and o.d / D p 2 (note that 1 .N / ha; ci). We have N D ha; d i and N=hai is cyclic. Next, CG .E/ D A. ˛2 If ˛ > 2, then working in the abelian subgroup hd; ap i Š Cp 2 Cp 2 , we may choose an element d 0 2 hd; ap i M such that .d 0 /p D c and so we may assume from the start that d p D c D Œa; b. If u 2 G is of order p 2 , then jhu; Eij D p 4 so hu; Ei is abelian, and we conclude that u 2 CG .E/ D A. It follows that 2 .G/ D Ehu 2 G j o.u/ D p 2 i A so Œd; b D 1. Replacing d with d 0 D d i (i 6 0 0 i ˛1 .mod p/), we may assume that ad D ad D a1Cp . But then we also replace b with b 0 D b i and c with c 0 D c i and obtain .d 0 /p D .d p /i D c i D c 0 D Œa; b 0 and the relation Œb 0 ; d 0 D 1 remains valid. Writing again d; b; c instead of d 0 ; b 0 ; c 0 , respectively, we obtain exactly the relations stated in part (b1) for ˛ > 2 of our proposition. It remains to investigate the case ˛ D 2, where jGj D p 5 , G 0 D Z.G/ D ˆ.G/ Š Ep 2 , E D 1 .G/ Š Ep 3 , exp.G/ D p 2 . For each x 2 N ha; ci, x p 2 G 0 hap i (otherwise, ha; xi would be nonabelian of order p 3 ; here N D NG .hai). Since G D EN , we have A D E.A \ N /, by the modular law. Note that A \ N is abelian of type .p 2 ; p/ so it is generated by elements of order p 2 . Since N D hai.A \ N / (indeed, M D haiE is nonabelian so a 62 CG .E/ D A), there exists d 2 .A \ N / 1 .A \ N / such that ha; d i is nonabelian. It follows that d p 62 hap i (otherwise, ha; d i is nonabelian of order p 3 , contrary to the fact that G is an A2 -group). Replacing d with another generator of hd i if necessary, we may assume that ad D a1Cp . Set d p D c i ajp , where i 6 0 .mod p/. We replace b with b 0 D b i and c with c 0 D c i , so that we get Œa; b 0 D Œa; b i D Œa; bi D c i D c 0 and d p D c 0 ajp . Thus, we may assume from the start that Œa; b D c and c D ajp d p with j 2 Z. We shall determine the integer j .mod p/. Consider the nonabelian maximal subgroup S D hb; ad i, where is any integer (recall that d.G/ D 3). The subgroup S is a nonmetacyclic (otherwise, b 2 Ã1 .S / D ˆ.S /) A1 -group containing E with Z.S / D Z.G/ D G 0 D hap ; d p i D h.ad /p ; Œad ; bi. If p > 2, then ˛2

.ad /p D ap .d p / ;

Œad ; b D Œa; b D c D ajp d p ;

ˇ ˇ ˇ ˇ and so we must have (since the displayed two elements are independent) ˇ j1 1 ˇ 6 0 .mod p/, which gives 1 j 6 0 .mod p/ for every integer . If j 6 0 .mod p/, then we can solve the congruence j 1 .mod p/ for and get a contradiction.

71

Determination of A2 -groups

241

Hence j 0 .mod p/ and we may set in this case j D 0; then c D ajp d p D d p . If p D 2, then taking D 1, we have .ad /2 D a2 d 2 Œd; a D a2 d 2 a2 D d 2 ;

Œad; b D Œa; b D c D a2j d 2 ˇ ˇ (by the above, ad D a1Cp for any p), and we must have ˇ j0 11 ˇ 6 0 .mod 2/, which gives j 6 0 .mod 2/ and so we may set in this case j D 1; then c D a2 d 2 . We have obtained the relations stated in part (b1) for ˛ D 2 of our proposition. (ii) Suppose that Ep 3 Š E D 1 .G/ Z.G/ D ˆ.G/. Let H 2 1 be nonabelian. Then E < H and so H is a nonmetacyclic A1 -group: ˛

ˇ

H D ha; b j ap D b p D 1; ˛1

ˇ1

Œa; b D c;

c p D Œa; c D Œb; c D 1i;

where E D hap ; b p ; ci. Since E Z.G/ D Z.H / D ˆ.H / (Theorem 65.7(f)), we must have ˛ 2 and ˇ 2 so jGj D pjH j D p ˛Cˇ C2 p 6 . Let G0 < G 0 be of order p such that G0 ¤ H 0 . Since .G=G0 /0 D G 0 =G0 is of order p and d.G=G0 / D 3, a nonabelian group G=G0 is an A2 -group of order p 5 with the commutator group of order p, where H=G0 is an A1 -subgroup. By Proposition 71.1, there exists d 2 G H such that d p 2 G 0 . Hence o.d / D p 2 and d 62 Z.G/ since d 62 H and 1 .G/ D E < H . It follows that d does not centralize a or b. Without loss of generality, we assume that Œd; a ¤ 1 and so nonabelian hd; ai 2 1 in view of d.G/ D 3. Since o.d / D p 2 and o.a/ D p ˛ , we have jhd; aij p ˛C2C1 in view of o.Œd; a/ D p, and therefore jGj p ˛C4 . Hence jGj D p ˛Cˇ C2 p ˛C4 and so ˇ 2. This gives ˇ D 2 and jGj D p ˛C4 with ˛ 2. We have Z.G/ D Ehap i. Assume first that ˛ > 2; then jGj > p 6 . In that case, any two elements x; y 2 G of orders p 2 commute (otherwise, hx; yi 2 1 , since d.G/ D 3, is nonabelian with jhx; yij p o.x/o.y/ p 5 , a contradiction). Since Œb; d D 1 in view of o.b/ D o.d / D p 2 , we get hb; d i D hbi hd i. Assume that this is false. Then hb; d i D hbi hd1 i with o.d1 / D p. Since b 2 H , d 62 H , we get d1 62 H , a contradiction: 1 .G/ D E < H . Consider K D ha; d i, where K 2 1 is nonabelian containing E (by assumption, E < ˆ.G/). Let G0 ¤ K 0 be a subgroup of order p in G 0 . Then applying Proposition 71.1 on G=G0 , we see that there is b0 2 G K with b0p 2 G 0 and o.b0 / D p 2 . Since Œd; b0 D 1 and there are no elements of order p in G K, we see (as before) that hd; b0 i D hd i hb0 i. But G D ha; d; b0 i, b0 62 Z.G/, and so Œa; b0 ¤ 1. Replacing H with H0 D ha; b0 i (¤ K) and b with b0 , we may assume from the start that b p 2 G 0 . Hence, we may assume that Ã1 .hb; d i/ D 1 .hb; d i/ D G 0 . Since jGj D p ˛C4 and GN D G=G 0 is generated N dN ; aN (bar convention) of orders at most p; p; p ˛ , respectively, and jGj N D p ˛C2 , by b; ˛ 0 we get o.a/ N D p which gives hai \ G D f1g and so hai \ hb; d i D f1g since 1 .hb; d i/ D G 0 . Suppose that ˛ D 2; then jGj D p 6 , exp.G/ D p 2 , E D Z.G/ D ˆ.G/ D 1 .G/. Let X < G 0 be of order p. Since d.G=X/ D 3 and .G=X/0 is of order p, G=X is an A2 -group of order p 5 , and so we may apply again Proposition 71.1. There is M 2 1

242

Groups of prime power order

such that M=X is an A1 -group and there is c 2 G M such that c p 2 G 0 and ŒG; c X. If c p 2 X, then hci (of order p 2 ; recall that 1 .G/ < M ) is normal in G. Since c 62 Z.G/ D E, there is an element l of order p 2 with Œl; c ¤ 1 and so hŒl; ci D hc p i. But then jhl; cij o.l/o.c/ p 4 , a contradiction. Thus c p 2 G 0 X. We consider now the subgroup Y D hc p i of order p in G 0 and apply Proposition 71.1 to G=Y . As before, there is an element b 2 G such that b p 2 G 0 Y . It follows hb p ; c p i D G 0 . The abelian group hb; ci=G 0 is generated by two elements of order p and so jhb; cij p 4 . Thus hb; ci is abelian and since hbi \ hci D f1g, we have S D hb; ci Š Cp 2 Cp 2 and S is normal in G. But, E D ˆ.G/ D G 0 Ã1 .G/ and so there is a 2 G such that ap 2 E G 0 . Hence hai \ S D f1g and so S hai D G since o.a/ D p 2 . In any case (for all ˛ 2) we have obtained the following configuration (where in ˛ case ˛ > 2 we write c instead of d ). We have G D hai .hbi hci/, with ap D 1, 2 2 ˛ 2, b p D c p D 1, Œb; c D 1, where G is a semidirect product with kernel hbi hci Š Cp 2 Cp 2 , G 0 D hb p ; c p i, and A D hap ; b; ci is the unique abelian maximal subgroup of G (of type .p ˛1 ; p 2 ; p 2 /). p Set Œa; b D x, Œa; c D y. Then G 0 D hx; yi D hb p ; c p i ( Z.G/) and ˇ 0so ˇb D 0 ˇ ˇ x ˛ y ˇ and c p D x y ı , where, since b p and c p are independent, we get ˇ ˛ ˇı ˇ 6 0 .mod p/. The subgroup ha; b c i is nonabelian if and only if Œa; b c D x y ¤ 1 which occurs if and only if both and are not multiples of p. In that case L D ha; b c i is a nonmetacyclic A1 -subgroup of index p in G. Since hx y i D L0 and L0 is a 0 p p maximal cyclic subgroup in L, we have G ˇ 0 D h.b c / ˇ; x y i, where .b c / D 0 ˇ ˇ ˇ Cı .b p / .c p / D x ˛ C y ˇ Cı . Hence, ˇ ˛ C ˇ 0 .mod p/ only if

0 .mod p/. This gives (1)

ˇ 2 C .ı ˛ 0 / 2 0 .mod p/

only if 0 .mod p/. From (1) follows (setting D 1, D 0 or D 0, D 1) ˇ 6 0 .mod p/ and 6 0 .mod p/. Suppose first p > 2. Then we compute (using (1)) .2ˇ C.ı˛ 0 //2 .4ˇ C.ı˛ 0 /2 /2 D 4ˇ.ˇ 2 C.ı˛ 0 / 2 / 0 .mod p/ only if 0 .mod p/. Hence (2)

.2ˇ C .ı ˛ 0 //2 .4ˇ C .ı ˛ 0 /2 /2

.mod p/

only if 0 .mod p/. From (2) follows at once that 4ˇ C .ı ˛ 0 /2 is a quadratic non-residue mod p. Indeed, if 4ˇ C .ı ˛ 0 /2 is a quadratic residue modulo p, then we set D 1 in (2) and then we solve the congruence .2ˇ C .ı ˛ 0 //2

4ˇ C .ı ˛ 0 /2 .mod p/ for and get a contradiction.

71

Determination of A2 -groups

243

Suppose now that p D 2. We know already that ˇ D D 1. Then (1) yields (3)

2 C .ı C ˛ 0 / C 2 0

.mod 2/

only if 0 .mod 2/. Setting D D 1 in (3), we get ı C ˛ 0 D 1. This gives 0 0 b 2 D x ˛ y and c 2 D xy ˛ C1 , where ˛ 0 D 0; 1. Hence we have the following two possibilities: (4)

b 2 D y;

c 2 D xy;

(5)

b 2 D xy;

c 2 D x:

However, interchanging b and c and also x and y, we see that the relations Œa; b D x and Œa; c D y remain unchanged but (4) goes onto (5). Hence, we may assume that we have relations (4) and so we get ˛ 0 D 0 and ˇ D D ı D 1. Our group G is uniquely determined. Proposition 71.5. Let G be a nonmetacyclic A2 -group of order > p 4 all of whose maximal subgroups are nonabelian. Then we have one of the following possibilities: (a) d.G/ D 3, p D 2, and G D ha; b; ci with a4 D b 4 D c 4 D 1;

Œa; b D c 2 ;

Œa; c D b 2 c 2 ;

Œb; c D a2 b 2 ;

Œa2 ; b D Œa2 ; c D Œb 2 ; a D Œb 2 ; c D Œc 2 ; a D Œc 2 ; b D 1; where G 0 D ha2 ; b 2 ; c 2 i D Z.G/ D ˆ.G/ D 1 .G/ Š E23 , exp.G/ D 4 and all members of the set 1 are nonmetacyclic A1 -groups. (From Theorem 70.4 follows that G is isomorphic to the Suzuki 2-group of order 26 .) (b) d.G/ D 2, p > 2, G is of order p 5 : 2

2

G D ha; x j ap D x p D 1; Œa; x D b; Œa; b D y1 ; Œx; b D y2 ; b p D y1p D y2p D Œa; y1 D Œx; y1 D Œa; y2 D Œx; y2 D 1; ap D y1˛ y2ˇ ; x p D y1 y2ı i; where in case p > 3, 4ˇ C .ı ˛/2 is a quadratic non-residue mod p. Here G 0 D hb; y1 ; y2 i D 1 .G/ Š Ep 3 , Z.G/ D K3 .G/ D Ã1 .G/ D hy1 ; y2 i Š Ep 2 . Proof. Let G be a nonmetacyclic A2 -group of order > p 4 all of whose maximal subgroups are nonabelian and therefore they are generated by two elements. Suppose first that d.G/ D 3. Then G 0 Z.G/ (Lemma 65.4(a)). By Theorem 70.1, G is a uniquely determined group of order 26 as stated in part (a) of our proposition. Now suppose that d.G/ D 2. If p D 2, then G is metacyclic (Lemma 64.1(n)), a contradiction. Hence p > 2; then Ã1 .G/ D K3 .G/, jG W G 0 j D p 2 (Lemma 64.1(m))

244

Groups of prime power order

so jG 0 j D p 3 and jGj D p 5 (Lemma 65.2(d)). Since G=K3 .G/ is generated by its subgroups of index p 2 , K3 .G/ Z.G/ (Lemma 65.4(c)) and so G is of class 3. We have K3 .G/ D Z.G/ Š Ep 2 and G 0 Š Ep 3 (Lemma 65.7(d)). Each M 2 1 is a nonmetacyclic A1 -group with G 0 D 1 .M / Š Ep 3 and so ˆ.G/ D G 0 D 1 .G/ and exp.G/ D p 2 (Lemma 65.1). Let G D ha; xi; then a; x 2 G ˆ.G/ D G 1 .G/ and therefore o.a/ D o.b/ D p 2 . We have G 0 D hŒa; x; K3 .G/i and so Œa; x D b 2 G 0 K3 .G/ which gives o.b/ D p. Set Œa; b D y1 and Œx; b D y2 , where y1 ; y2 2 K3 .G/ D Z.G/. But ha; biZ.G/ and hx; biZ.G/ are members of the set 1 and so they are nonabelian. Hence o.y1 / D o.y2 / D p (Lemma 65.1) and ha; biZ.G/ D ha; bi, hx; biZ.G/ D hx; bi. Since ha; bi and hx; bi are distinct members of the set 1 , we have hy1 i ¤ hy2 i (Lemma 65.2(d)). It follows hy1 ; y2 i D Z.G/. Since ap ; x p 2 Z.G/, we get ap D y1˛ y2ˇ and x p D y1 y2ı . Suppose that p > 3. Then G is regular (Lemma 64.1(a)) and since G 0 is elementary abelian, we get that G is p-abelian, i.e., .lk/p D l p k p for all l; k 2 G (see 7). Since ha; xi D G, we have in this case hap ; x p i D Ã1 .G/ D Z.G/ D hy1 ; y2 i. Consider the subgroup S D hb; a x i (S < G since b 2 G 0 ; it follows that S is of class 2). Obviously, a x 2 G G 0 if and only if not both and are divisible by p. In that case Œa x ; b D Œa; b Œx; b D y1 y2 ¤ 1, and so S is nonabelian; then S 2 1 hence E < S . It follows that S is nonmetacyclic. Here we have used the fact that Œa; b; Œx; b 2 Z.G/. Since Z.G/ D Z.S / and S 0 is a maximal cyclic subgroup of S , we have Z.S / D Z.G/ D hy1 ; y2 i D h.a x /p ; Œa x ; bi. We compute ˛C ˇ Cı y2

.a x /p D .ap / .x p / D y1 and so ()

and Œa x ; b D y1 y2 ;

ˇ ˇ ˇ˛ C ˇ C ıˇ ˇ ˇ 0 .mod p/ ˇ ˇ

only if 0 .mod p/. From () we get (as in the proof of Proposition 71.4) that 4ˇ C .ı ˛/2 is a quadratic non-residue modulo p. Our proposition is proved and all A2 -groups are determined. The classification of A2 -groups is contained in Propositions 71.1–71.5. The subgroup structure of A2 -groups is given in 65. It is impossible to compare our proof with Kazarin’s proof since the last one is not published and only sketched in his PhD Thesis. The paper [She], also devoted to A2 -groups, is not accessible. In any case, our proof is based on other ideas. Indeed, the above authors were forced to use tables [HS]. Besides, Kazarin used NazarovaRoiter’s classification of p-groups with abelian subgroup of index p so his proof is not elementary. Instead of this, we use Theorem 70.1 which is new for p D 2. Exercise 1. Let a p-group G D A C , where C Š Cp 2 , is an A2 -group. Describe the intersection A \ C . Consider in detail the case when A is metacyclic.

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245

Exercise 2. Classify the metacyclic A1 -groups G of order 2n such that there is an involution x 2 G which is not square. Exercise 3. Find the A2 -groups G such that jH G W H j D p for all nonnormal H < G. Exercise 4. Find the A2 -groups G such that exp.H G / D exp.H / for all H < G. The following three theorems are due to the first author. Theorem 71.6 ([BJ2]). Let G be a nonmetacyclic two-generator 2-group. Then the number of two-generator members of the set 1 is even. Proof. By hypothesis, G is nonabelian. If A 2 1 , then Ã2 .G/ Ã1 .A/ D ˆ.A/ so d.A=Ã2 .G// D d.A/. Therefore, without loss of generality, one may assume that Ã2 .G/ D f1g; then exp.G/ D 4. Let 1 .G/ D fA; B; C g. Since G is not metacyclic, one may assume that d.C / D 3 (Appendix 25 and Theorem 44.5). Assume that the theorem is false. Then we may assume that d.A/ D 2 and d.B/ D 3 (see Appendix 25). Let R be a G-invariant subgroup of index 2 in G 0 . Then G=R is minimal nonabelian (Lemma 65.2(a)) and nonmetacyclic (Theorem 36.1). Since exp.G=R/ D 4, we get jG=Rj 25 (Lemma 65.1). If jG=Rj D 25 , the set 1 has no two-generator members (Lemma 65.1), contrary to the assumption. Thus, jG=Rj D 24 so jG=G 0 j D 8. One may assume that C =R D 1 .G=R/. Since B=R is abelian of type .4; 2/ so two-generator and d.B/ D 3, we get R ¤ ˆ.B/. Clearly, ˆ.B/ is a G-invariant subgroup of index 8 in B. Write GN D G=ˆ.B/; then GN is not of maximal N D 2. Since jG=G 0 j D 8, we get class since it contains BN Š E8 . It follows that cl.G/ jG 0 W ˆ.B/j D 2. Since G 0 has only one G-invariant subgroup of index 2 (Remark 36.1), we conclude that ˆ.B/ D R. This is a contradiction since B=R is abelian of type .4; 2/ and exp.B=ˆ.B// D 2. Thus, the number of two-generator members of the set 1 is even, as was to be shown. Theorem 71.7. Suppose that all nonabelian maximal subgroups of a 2-group G are metacyclic and jGj > 24 . Then one of the following holds: (a) G is abelian. (b) G is metacyclic. (c) G is an A1 -group. (d) G is minimal nonmetacyclic. (e) G D M C , where M is a metacyclic A1 -group and jC j D 2. Proof. Assume that G is not of types (a)–(d). Then, by Theorem 71.6, d.G/ > 2 and there is in G a maximal subgroup A that is not two-generator; in that case, by hypothesis, A is abelian. By assumption, there is in G a nonabelian maximal subgroup M ; then M is metacyclic so d.G/ D 3. Since M \A is a metacyclic maximal subgroup of A, we get d.A/ D 3. Write E D 1 .A/; then E G G and E Š E8 . By hypothesis, all maximal subgroups of G that contain E, are abelian. Since G has at most three abelian maximal subgroups, we get d.G=E/ 2. (i) Suppose that d.G=E/ D 2. Let B=E be a maximal subgroup of G=E and B ¤ A. Then B is abelian so A \ B D Z.G/ has index 4 in G; in that case, jG 0 j D 2 (Lemma 64.1(q)) and E Z.G/. It follows from jM 0 j D 2 and d.M / D 2 that M is

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an A1 -group. Let C < E be a subgroup of order 2 not contained in 1 .M / .Š E4 /; then G D M C is as stated in (e). (ii) Now let G=E be cyclic; then E 6 Z.G/ since G is nonabelian. Since d.G/ D 3, we get jG 0 j D 2. We have E D 1 .G/ since 1 .G/ < A in view of G=E is cyclic of order > 2. Let, as above, M is a nonabelian maximal subgroup of G. By Lemma 65.2(a), M is an A1 -subgroup; moreover, M Š M2n , n > 3, since 1 .M / D M \ E Š E4 and M=1 .M / Š G=E is cyclic of order > 2, and we conclude that Z.M / is cyclic. If Z.G/ is noncyclic, then G D M C , where jC j D 2, and G is as stated in (e). Next assume that Z.G/ is cyclic. We get jG W Z.G/j D 2jG 0 j D 4 (Lemma 64.1(q)). Then, by the product formula, G D EZ.G/ so G is abelian, a final contradiction. Theorem 71.8. Suppose that a two-generator nonabelian p-group G contains an abelian maximal subgroup A. Set R D hx p j x 2 G Ai. Then R Z.G/ and G=R is of maximal class, unless G is an A1 -group. Proof. Recall that if A is an abelian maximal subgroup of a nonabelian p-group G and jG W G 0 j D p 2 , then G is of maximal class (this is proved by induction with help of Lemma 64.1(q)). We have R Z.G/ \ ˆ.G/ since CG .x p / hx; Ai for all x 2 G A. Write GN D G=R; then GN is noncyclic. Since all elements of the set GN AN N hGN Ai N D G. N It follows that GN 0 D ˆ.G/ N so have the same order p, we get 1 .G/ N GN 0 Š Ep 2 since d.G/ N D d.G/ D 2. Suppose that GN 0 D f1g. N Then R D ˆ.G/ that G= so G=Z.G/ Š Ep 2 . In that case, all maximal subgroups of G are abelian so G is an N then GN is nonabelian. In that case, as we have A1 -group. Now suppose that GN 0 > f1g; noticed, GN is of maximal class.

Problems Problem 1. Classify the p-groups G such that, whenever H < G is minimal nonabelian and H Š H1 < G, then H1 D H . Problem 2. Classify the p-groups all of whose A2 -subgroups are isomorphic. Moreover, study the p-groups all of whose A1 -subgroups of the same order are isomorphic. Problem 3. Classify the non-Dedekindian p-groups all of whose nonnormal subgroups are abelian. Problem 4. Classify the p-groups all of whose A2 -subgroups are metacyclic. Problem 5. Let G be a p-group all of whose maximal subgroups are A2 -groups. Is it true that jGj is bounded? Problem 6. Classify the p-groups with at most two A2 -subgroups. (Note that pgroups with exactly one A2 -subgroup are not classified.)

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247

Problem 7. Describe the automorphism groups of all A1 - and A2 -groups. Problem 8. Study the p-groups G such that, whenever A; B 2 1 are distinct, then A \ B is either abelian or minimal nonabelian. Problem 9. Study the p-groups all of whose A2 -subgroups are not two-generator. Problem 10. Study the subgroup structure of A A, where A is an A1 -group. Problem 11. Classify the p-groups G which are lattice isomorphic with B, where B is (i) an A1 -group, (ii) an A2 -group. Problem 12. Study the p-groups G such that, whenever A < G is an A1 -subgroup, then A < B G, where B is an A2 -subgroup. Problem 13. Classify the p-groups of exponent > p all of whose A2 -groups have exponent p (then these subgroups have order p 4 ). Problem 14. Study the p-groups G such that ˆ.G/ is an A1 -group (A2 -group). Problem 15. Study the p-groups G all of whose subgroups of fixed order p n are A2 groups. Is it possible to estimate jGj in terms of n? Problem 16. Classify the p-groups with minimal nonabelian subgroup of index p.

72

An-groups, n > 2

In this section we study An -groups with n > 2. A p-group G is said to be an An group if all its subgroups of index p n are abelian but G has a nonabelian subgroup of index p n1 . Main results of this section are due to the first author. 1o . First we classify An -groups G with G 0 Š Cp n , n > 1 (for n D 2, see Corollary 65.3). Proposition 72.1. (a) Let G be a metacyclic p-group with jG 0 j D p n . Then G is an An -group. (b) Let G be an An -group, n > 1, with cyclic derived subgroup G 0 of order p n . Then G is metacyclic and jG 0 j D p n . Proof. We use induction on jGj. (a) If n D 1, then G is an A1 -group (Lemma 65.2(a)). Now suppose that n > 1. Let L D Ã1 .G 0 /. ˆ.G//. By Lemma 65.2(a), G=L is minimal nonabelian so, if H 2 1 , then H=L is abelian, and we have H 0 L. By induction, H is an As -group with s < n. By Lemma 64.1(u), there is F 2 1 with jG 0 W F 0 j D p so F is an An1 -group since jF 0 j D p n1 . It follows that G is an An -group. (b) Set jG 0 j D p s , s n. In view of Lemma 65.2(a,b) and Theorem 65.7(a), one may assume that n > 2. Take H 2 1 . If H 0 D G 0 , then, by induction, H is an As group, a contradiction since s n. Therefore, if G 0 D hŒx; yi, then G D hx; yi, i.e., d.G/ D 2. By Lemma 64.1(u), there is F 2 1 with jG 0 W F 0 j D p so, by induction, F is an As1 -group and so s 1 n1 since G is an An -group. It follows that s D n and F is metacyclic, by induction since it is an An1 -group with cyclic F 0 of order p n1 > p. Let L D Ã1 .G 0 / and T D Ã2 .G 0 /. If H 2 1 , then H=L is abelian since G=L is an A1 -group (Lemma 65.2(a)). By Lemma 65.1, G=T is not an A1 -group. Then G=T has a maximal subgroup A=T that is nonabelian so .A=T /0 > f1g. Since A0 and G 0 are cyclic, we get T < A0 so, by the above, jA0 j D p n1 . By induction, A is metacyclic. Then A=T is an A1 -group (Lemma 65.2(a)). Thus, every maximal subgroup of G=T is either abelian or an A1 -group so G=T is an A2 -group. Since j.G=T /0 j D p 2 , G=T is metacyclic, by Lemma 65.7(a). Now, by Lemma 64.1(o), G is metacyclic since T Ã2 .G/.

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Let G be a nonabelian metacyclic group of order p m and p k D min fjAj j A < G; A0 > f1gg. Then G is an Am.k1/ -group (indeed, all subgroups of G of order p k1 are abelian and G has a nonabelian subgroup A of order p k ) so jG 0 j D p m.k1/ (Proposition 72.1(a)) whence jG=G 0 j D p k1 . In that case, G 0 < Z < G, where Z is cyclic of index p k2 in G. In particular, if k D 3, then G has a cyclic subgroup of index p 32 D p so it is either a 2-group of maximal class or p > 2 and m D 3 (Lemma 64.1(t)). The last result coincides with Proposition 10.19. Let p > 2 and suppose that a metacyclic p-group G of order p m > p 4 contains a normal nonabelian subgroup H of order p 4 and exponent p 2 . Then H D 2 .G/ so all subgroups of G of order p 3 are abelian. It follows that G is an Am3 -group so jG 0 j D p m3 (Proposition 72.1) and G has a normal cyclic subgroup Z of index p 2 , G 0 < Z, such that G=Z is cyclic. We have Z.H / Š Ep 2 . Let L be a subgroup of order p in Z.H /, L ¤ H 0 . Then jG W CG .L/j p. Since CG .L/=L contains a nonabelian subgroup H=L of order p 3 , we get CG .L/=L D H=L (Lemma 64.1(t)). In that case, jG W H j D p so jGj D pjG W H j D p 5 , and Z.H / 6 Z.G/ so Z.G/ is cyclic. Let G be a metacyclic p-group and Mp 4 Š H < G. In that case, G has no cyclic subgroup of index p (Lemma 64.1(t)). Then G is an Am3 -group so jG 0 j D p m3 and G has a normal cyclic subgroup Z ¤ ˆ.G/ of index p 2 . Exercise. Describe the nonabelian metacyclic 2-groups of order 2m > 25 without normal subgroups of order 24 and exponent 4. Solution. Suppose that G has no cyclic subgroups of index 2 (otherwise, G satisfies the condition). Then 1 .G/ Š E4 . Write GN D G=1 .G/. Then GN has no normal abelian subgroups of type .2; 2/ so it is of maximal class (if GN is cyclic, then G has a cyclic subgroup of index 2 so Š M2m ). Suppose that GN is dihedral. By Proposition N D G, N it follows that 10.17, all subgroups of G of order 8 are abelian. Since 1 .G/ 1 .G/ Z.G/. If G=1 .G/ is of maximal class, then G has no normal subgroups of order 24 and exponent 4 since any such subgroup must contain 1 .G/. In the case under consideration, G is a U2 -group (see 67). 2o . In this subsection and subsection 3o we estimate jG 0 j, where G is an An -group, n D 3; 4. If a p-group G is an A3 -group and U; V 2 1 are distinct, then jG 0 j pjU 0 V 0 j p 7 (Lemmas 65.2(d) and 64.1(u)). Proposition 72.2 yields essentially better estimate, and that estimate is best possible. Proposition 72.2. If a p-group G is an A3 -group, then jG 0 j p 4 . For p D 5, there exists an A3 -group G with jG 0 j D 54 . Proof. Assume that jG 0 j > p 4 . Take U 2 1 . Then jU 0 j p 3 and, if jU 0 j D p 3 , then U 0 Š Ep 3 (Theorem 65.7(a,d)). (i) We claim that G 0 is regular. This is the case if jG W G 0 j > p 2 since then G 0 is abelian. Let jG W G 0 j D p 2 and let G 0 be nonabelian; then p > 2 (Taussky’s theorem)

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and G 0 is an A1 -group. In that case, G 0 is regular since cl.G/ 2 (Lemmas 65.1 and 64.1(a)); moreover, G 0 is metacyclic (Theorem 44.12). (ii) We claim that if H G, then jH=Ã1 .H /j p 5 , and this inequality is strong if H < G. Indeed, there is an A2 -subgroup A 2 1 . Then jA=Ã1 .A/j p 4 (Lemma 65.4(a)) so jG=Ã1 .G/j p 5 . Take H < G such that jH=Ã1 .H /j p 5 . By Lemmas 65.1 and 65.4(b), H is abelian so 1 .H / is elementary abelian of order p 5 . Then all members of the set 1 containing 1 .H /, are abelian (Lemmas 65.1 and 65.4(b)). By assumption, jG 0 j > p 4 > p so H is the unique member of the set 1 containing 1 .H / (Lemma 65.2(c)), and we conclude that G=1 .H / is cyclic whence G 0 < 1 .H /. Considering H \ F , where F 2 1 is nonabelian, we get j1 .H /j p 5 so jG 0 j < p 5 , a contradiction. (If G D S.p 3 / Ep 2 , where S.p 3 / is nonabelian of order p 3 and exponent p > 2, then G is an A3 -group, jGj D p 5 and exp.G/ D p.) (iii) Assume that jG 0 j > p 5 . Let U; V 2 1 be distinct; then jU 0 V 0 j p1 jG 0 j p 5 (Lemma 64.1(u)) so one may assume that jU 0 j D p 3 ; then jV 0 j p 2 . If jV 0 j D p 3 , then exp.U 0 V 0 / D p since exp.U 0 / D exp.V 0 / D p (Theorem 65.7(d), Lemma 64.1(a) and (i)), and jU 0 V 0 j p 5 , contrary to (ii). Thus, jV 0 j D p 2 ; then U 0 \ V 0 D f1g, by the product formula. By (ii), since jU 0 V 0 j p 5 , we get exp.U 0 V 0 / > p so V 0 is cyclic; then V is metacyclic (Theorem 65.7(a)), a contradiction since Ep 3 Š U 0 < G 0 ˆ.G/ < V . Thus, jG 0 j D p 5 . Then the set 1 has no abelian members, by Lemmas 65.2(d) and 64.1(u) (otherwise, if A 2 1 is abelian and H 2 1 fAg, then jH 0 j p 3 so jG 0 j pjA0 H 0 j p 4 , contrary to the assumption). (iv) Assume that there are distinct U; V 2 1 such that exp.U 0 / D exp.V 0 / D p. Then, using (i) and (ii), we get: U 0 V 0 is of order p 4 and exponent p so U 0 V 0 < G 0 , and exp.G 0 / D p 2 since jG 0 j D p 5 . In that case, U=U 0 V 0 and V =U 0 V 0 are cyclic (Lemma 65.4(b)). Thus, the nonabelian group G=U 0 V 0 has two distinct cyclic subgroups U=U 0 V 0 and V =U 0 V 0 of index p so it is either ordinary quaternion or Š Mp n (Lemma 64.1(t)). In the first case jG W G 0 j D 4 so G is a 2-group of maximal class (Lemma 64.1(s)), a contradiction. Assume that G=U 0 V 0 Š Mp n . Let H=U 0 V 0 be the noncyclic subgroup of index p in G=U 0 V 0 . In that case, H is neither A1 - nor A2 -group (Lemmas 65.1 and 65.4(b)) so H is abelian, a contradiction (see the last sentence of part (iii)). Now we are ready to complete the proof. By (iv), there is V 2 1 such that exp.V 0 / > p; then V is metacyclic and V 0 is cyclic of order p 2 (Theorem 65.7(d,a)). Take U 2 1 fV g. Since U 0 < V , we get j1 .U 0 /j p 2 so jU 0 j p 2 (Lemma 65.1 and Theorem 65.7(a,d)). Then U 0 \ V 0 D f1g (Lemmas 65.2(d) and 64.1(u)). Since U 0 V 0 D U 0 V 0 < V and V is metacyclic, U 0 Š Cp 2 (Corollary 65.3 and Lemma 64.1(u)). Thus, all members of the set 1 are metacyclic. Since G is nonmetacyclic (Proposition 72.1(a)), we get, jG 0 j p 2 (Lemma 64.1(l)), a contradiction. Below we present an A3 -group G with jG 0 j D p 4 for p D 5 (this example was constructed by the second author).

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251

Let G be the group of order 56 with the following generators and relations: hb; h j b 25 D h25 D 1; Œh; b D a; a5 D 1; Œa; b D c; c 5 D 1; Œb; c D d; d 5 D 1; Œc; a D Œh; d D Œb; d D Œh; c D 1; Œh; a D s; s 5 D 1; Œh; s D Œb; s D 1; d D h5 ; b 5 D si: Here G 0 D ha; c; d; si Š E54 , K3 .G/ D hc; d; si Š E53 , K4 .G/ D hd i is of order 5, and Z.G/ D hd; si Š E52 . All maximal subgroups of G are A2 -groups and so G is an A3 -group. Finally, we have shown that this group G exists as a subgroup of the symmetric group of degree 625. In the following proposition we investigate A3 -groups with derived subgroup of order p 4 in some detail. We show that then G 0 is nonmetacyclic. Note that jGj p 6 with strong inequality if p D 2 (Taussky’s theorem). Proposition 72.3. Suppose that a p-group G is an A3 -group with jG 0 j D p 4 . (a) G 0 is abelian. (b) If exp.G 0 / D p, then p > 2, G=G 0 is abelian of type .p n ; p/ (n 1) and G 0 Š Ep 4 . If, in addition, n > 1, then 1 .G/ D G 0 . (c) G 0 is not metacyclic. Proof. By Proposition 72.1, G 0 is noncyclic so G is not metacyclic. (i) Assume that G 0 is metacyclic. Let us prove that then p D 2, the set 1 has only one nonmetacyclic member and G has no normal elementary abelian subgroups of order 8. Since the set 1 has a nonmetacyclic member (Lemma 64.1(l)), there is U 2 1 such that exp.U 0 / p (Theorem 65.7(a,d)); then jU 0 j j1 .G 0 /j D p 2 so U 0 Z.G 0 / (indeed, every G-invariant subgroup L of order p 2 in G 0 is contained in Z.G 0 /: consider CG .L/). Take V 2 1 fU g. By Lemma 64.1(u), jG 0 W U 0 V 0 j p so jU 0 V 0 j p 3 hence exp.U 0 V 0 / > p since G 0 is metacyclic. It follows that exp.V 0 / > p. Since V 0 is abelian in view of jV 0 j p 3 , U 0 V 0 is also abelian (recall that U 0 Z.G 0 /), V 0 is cyclic of order > p (Theorem 65.7(d)) so V is a metacyclic A2 -group and jV 0 j D p 2 (Corollary 65.3). It follows that all members of the set 1 are nonabelian. If E G G, where E is of order p 3 and exponent p, then there is only one maximal subgroup of G containing E, by the previous paragraph, so G=E is cyclic. Then G 0 < E, a contradiction since jG 0 j D p 4 > jEj. Thus, E does not exist. Assume that p > 2. Then, by Theorem 13.7 (see also Theorem 69.4), G is a 3-group of maximal class and jGj D 32 jG 0 j D 36 . In that case, G has a nonabelian subgroup of index 33 (Theorem 9.6), a contradiction. Thus, p D 2. (ii) Assume that G 0 is nonabelian; then jG W G 0 j D p 2 and G 0 D ˆ.G/ is an A1 -group. By Theorem 44.12, G 0 must be metacyclic so p D 2, by (i); then G is of

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maximal class (Taussky’s theorem) so Z.G 0 / is cyclic, a contradiction (Lemma 1.4). Thus, G 0 is abelian. (iii) Let exp.G 0 / D p. Then, by (ii), G 0 Š Ep 4 . If jG W G 0 j D p 2 , then p > 2 (Lemma 64.1(s)). Now assume that jG W G 0 j > p 2 and assume, in addition, that G=G 0 has two distinct noncyclic maximal subgroups U=G 0 and V =G 0 . Then U is either abelian or an A2 group with the derived subgroup of order p, the same is true for V (Lemma 65.4(b)). Then jG 0 j p 3 (Lemma 64.1(u)), a contradiction. Thus, G=G 0 has exactly p cyclic subgroups of index p, so it is abelian of type .p n ; p/ with n > 1. Let H=G 0 be the noncyclic maximal subgroup of G=G 0 . Then H is either abelian or an A2 -group with jH 0 j D p. If F 2 1 fH g, then jF 0 j p 2 (Lemmas 64.1(u) and 65.2(d)) and so F is an A2 -group. If jF 0 j D p 3 , then F has no maximal abelian subgroups (Lemma 65.2(d)). But, if F1 is a maximal subgroup of F containing G 0 , then F1 must be abelian (Lemma 65.1). This contradiction shows that jF 0 j D p 2 and then jH 0 j D p. Looking at the determination of A2 -groups (see 71), we see that F must be isomorphic to a group (ii) of Proposition 71.3. In particular, p > 2. (H must be isomorphic with a group of Proposition 71.1(i); then H must be a nonmetacyclic A1 -group). Assume that there is x 2 G G 0 of order p. Then D D hx; G 0 i Š Ep 5 since jG W Dj p 2 and 1 .D/ D D. If D M 2 1 , then M is abelian (Lemmas 65.1 and 65.4(a)). Take N 2 1 fM g. By Lemma 64.1(u) applied to M; N.2 1 /, jN 0 j p 3 . Considering N \ M and using Lemma 65.4(b), we conclude that N=G 0 is cyclic and jN 0 j p 2 , contrary to the result of the previous sentence. Thus, x does not exist so 1 .G/ D G 0 . (iv) By (i), p D 2 and the set 1 has only one nonmetacyclic member H and G has no normal elementary abelian subgroups of order 8. Take V 2 1 fH g; then V is metacyclic so jV 0 j 4 (Corollary 65.3 and Lemma 65.1). Then H is nonabelian (apply Lemma 64.1(u) to H; V 2 1 ). If H is an A1 -group, then 1 .H / Š E8 G G (Lemma 65.1), contrary to (i). If H is an A2 -group with jH 0 j D 2, then 1 .H / Š E8 or E16 (Proposition 72.1), contrary to (i). Hence, H must be an A2 -group of Proposition 71.4. If H is minimal nonmetacyclic of order 25 , then jGj D 2jH j D 26 so jG W G 0 j D 4 and G is of maximal class (Lemma 64.1(s)), a contradiction. In other cases of Proposition 71.4, 1 .H / Š E8 , a final contradiction. 3o . It follows from Proposition 72.2 and Lemma 64.1(u) that if a p-group G is an A4 -group, then jG 0 j p 9 . In this subsection we improve essentially that estimate. Since every p-group of order < p 7 is an Ai -group with i 4, in what follows we consider only A4 -groups of order p 7 ; then our A4 -group is not of maximal class (Theorems 9.5 and 9.6). Lemma 72.4. Suppose that a p-group G of order > p 6 is an A4 -group with nonabelian G 0 . Let jG W G 0 j D p i ; then i 2 f2; 3g.

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(a) If i D 2, then p > 2 and G 0 is either metacyclic or an A2 -subgroup with d.G/ D 3. (b) If i D 3, then G 0 is a metacyclic A1 -subgroup. Proof. We have jG W G 0 j D p i , i 2 f2; 3g. If i D 3, then G 0 is an A1 -group so d.G 0 / D 2 (Lemma 65.1); then G 0 is metacyclic (Theorem 44.12). Now suppose that i D 2. If p D 2, then G is of maximal class (Lemma 64.1(s)), a contradiction. Thus, p > 2. Assume that G 0 is nonmetacyclic; then d.G 0 / > 2 (Theorem 44.12) and G 0 is an A2 -subgroup with d.G/ D 3. Proposition 72.5. If a p-group G is an A4 -group, then jG 0 j p 7 . Proof. Assume that jG 0 j D p 9 . For U 2 1 , we have jU 0 j D p 4 so U 0 is nonmetacyclic, abelian (Lemma 64.1(u), Propositions 72.2, 72.3 and 72.1); in particular, all members of the set 1 are A3 -groups (Lemmas 65.1 and 65.2(c)). If V 2 1 fU g, then U 0 \ V 0 D f1g (Lemma 64.1(u)) so U 0 V 0 D U 0 V 0 is abelian of order p 8 . Set L D 1 .U 0 V 0 /; then j1 .L/j p 6 (Proposition 72.3) and L < U . If A < U is nonabelian subgroup, then jA \ Lj p 5 , contrary to Lemmas 65.1 and 65.4(b). Thus, jG 0 j p 8 . Next we assume that jG 0 j D p 8 . If U; V; W 2 1 are distinct with jU 0 j jV 0 j jW 0 j, then using Lemma 64.1(u) twice, we get jU 0 j D jV 0 j D p 4 and jU 0 \ V 0 j p since jU 0 V 0 j p1 jG 0 j D p 7 . Then U and V are both A3 -groups and U 0 and V 0 are nonmetacyclic abelian (Proposition 72.3(a,c)) so cl.U 0 V 0 / 2 (Fitting’s lemma), hence, if p > 2, then U 0 V 0 is regular (Lemma 64.1(a)). Now let p D 2. If U 0 V 0 D G 0 , then G 0 is abelian since then G 0 D U 0 V 0 (here we use Proposition 72.3(a) again). Suppose that U 0 V 0 < G 0 . Then jG W U 0 V 0 j > jG W G 0 j 8, by Lemma 64.1(s), so U 0 V 0 is abelian again. Set L D 1 .U 0 V 0 / so that exp.L/ D p. By the previous paragraph, jLj p 5 . If U1 is a maximal subgroup of U containing L, then U1 must be abelian (Lemma 65.4(b)). Since jU 0 j > p, U=L is cyclic (indeed, U1 is the unique abelian maximal subgroup of U ), and so U 0 < L. Hence exp.U 0 / D p and Proposition 72.3(c) implies that U 0 D 1 .U /. This is a contradiction since U 0 < L < U and exp.L/ D p. Now we improve the estimate of Proposition 72.5. Theorem 72.6. If a p-group G is an A4 -group, then jG 0 j p 6 . Proof. Assume, by the way of contradiction, that a p-group G is an A4 -group with jG 0 j D p 7 . By Propositions 72.2, 72.3(a) and Lemma 64.1(u), p 2 jU 0 j p 4 and U 0 is abelian for all U 2 1 . (i) We claim that the subgroup G 0 is nonmetacyclic and then d.G 0 / > 2 (Theorem 44.12). Suppose that this is false. By Proposition 72.1, jU 0 j D p 3 for all U 2 1 and so U 0 V 0 D U 0 V 0 for distinct U; V 2 1 (Lemma 64.1(u)). It follows that each

254

Groups of prime power order

U 0 is cyclic (otherwise, U 0 V 0 is nonmetacyclic) and so U is metacyclic (Proposition 72.1(b)). But then G is minimal nonmetacyclic (Proposition 72.1(a)), contrary to Lemma 64.1(l). (ii) If jG=G 0 j > p 2 , then G 0 is abelian and p 3 j1 .G 0 /j p 4 . Indeed, since d.G 0 / > 2, by (i) and Theorem 44.12, G 0 is not an A1 -group and so G 0 must be abelian. Since G=G 0 is noncyclic, there is U 2 1 such that U=G 0 is noncyclic. Let V1 and V2 be two distinct maximal subgroups of U containing G 0 . If j1 .G 0 /j > p 4 , then V1 and V2 cannot be A1 -groups or A2 -groups (Lemmas 65.1 and 65.4(a,b)) and so they are abelian. It follows that jU 0 j p (Lemma 64.1(q)), a contradiction. Thus j1 .G 0 /j p 4 , and j1 .G 0 /j p 3 follows from d.G 0 / > 2 and commutativity of G 0 . (iii) If G 0 is nonabelian, then G=G 0 Š Ep 2 , by (ii), so p > 2 since G is not of maximal class (Taussky’s theorem), and G 0 is an A2 -group with d.G 0 / D 3, by (i). Hence G 0 is isomorphic to a group of Proposition 71.1 or Proposition 71.4(b). In any case, 1 .G 0 / Š Ep 3 or Ep 4 , which follows from 71. (iv) If U 2 1 and jU 0 j D p 4 , then U 0 is abelian of type .p 2 ; p; p/. If V 2 1 and jV 0 j D p 3 , then V 0 is abelian noncyclic since V is nonmetacyclic, in view of G 0 < V . Indeed, U 0 is abelian nonmetacyclic (Proposition 72.3). Suppose that U 0 Š Ep 4 . Then Proposition 72.3(c) (noting that jU j D p1 jGj > p 6 ) implies that 1 .U / D U 0 . Take W 2 1 fU g. Then U 0 ˆ.U / ˆ.G/ < W so W is nonmetacyclic. Next, jU 0 W 0 j p1 jG 0 j D p 6 (Lemma 64.1(u)) and we must have W 0 U , 1 .W 0 / U 0 (see (iii)) so jW 0 j p 3 . By Proposition 72.1(b), W 0 is noncyclic so jU 0 \ W 0 j p 2 . But then jU 0 \ W 0 j D p 2 and jW 0 j D p 4 (Lemma 64.1(u)). Since W 0 is abelian and nonmetacyclic (Proposition 72.3(c)), we have j1 .W 0 /j p 3 and so 1 .W 0 / — U 0 , contrary to what has just been proved. (v) We claim that j1 .G 0 /j p 4 . Indeed, assume that j1 .G 0 /j p 5 . Then is not an Ai -group, i D 1; 2 (Lemmas 65.1 and 65.4(a,b)) so G 0 is abelian and E D 1 .G 0 / is elementary abelian of order p 5 . By (ii), G=G 0 Š Ep 2 so p > 2 (Taussky’s theorem). For each U 2 1 , p 2 jU 0 j p 4 (Lemma 64.1(u)) so U is an A3 -group (indeed, the nonabelian U cannot be an Ai -group, i 2, since E < U ; see Lemma 65.4(a,b)). If U=E is not cyclic, U has a maximal subgroup U0 ¤ G 0 containing E (note that G 0 is maximal in U ). Then U0 must be abelian and so jU 0 j p (Lemma 65.2(c)), a contradiction. Hence U=E is cyclic; then U 0 < E and so U 0 is elementary abelian for each U 2 1 . Then (iv) forces that U 0 Š Ep 3 for each U 2 1 and U 0 V 0 D U 0 V 0 for any two distinct U; V 2 1 . In particular, E D 1 .G 0 / is elementary abelian of order p 6 . Let an A2 -subgroup A be maximal in U . But in that case jA \ Ej p 5 since G 0 < U , contrary to Lemma 65.4(a,b). G0

(vi) We claim that, for each U 2 1 , U 0 is abelian of exponent p 2 . Also, 1 .G 0 / Š Ep 4 and, if G 0 is nonabelian, then G 0 D H Cp , where H is a nonmetacyclic A1 group of order p 6 . Indeed, assume that there is U 2 1 with U 0 Š Ep 3 , and take V 2 1 fU g. Since jU 0 V 0 j p 6 (Lemma 64.1(u)), we have jV 0 W .U 0 \ V 0 /j p 3 . If jV 0 j D p 3 , then U 0 \ V 0 D f1g and V 0 is noncyclic, by Proposition 72.1(b) and (i),

72 An -groups, n > 2

255

since G 0 < V . If jV 0 j D p 4 , then jU 0 \V 0 j p (Lemma 64.1(u)) and 1 .V 0 / Š Ep 3 , by (iv). In any case, j1 .U 0 V 0 /j p 5 , contrary to (v). Suppose that W 2 1 with W 0 Š Ep 2 and take V0 2 1 fW g. Then jV00 j D p 4 , W 0 \ V00 D f1g (Lemma 64.1(u)), so again j1 .W 0 V00 /j p 5 , by (iv), which is a contradiction. We have proved that if U 2 1 and jU 0 j D p 2 , then U 0 Š Cp 2 and if jU 0 j D p 3 , then U 0 Š Cp 2 Cp . This together with (iv) proves our first assertion on exp.U 0 /. Suppose that U 2 1 is such that jU 0 j is as large as possible. Let jU 0 j D p 4 ; then U 0 is abelian of type .p 2 ; p; p/, by (iv). Let V 2 1 fU g be such that jV 0 W .U 0 \ V 0 /j p 2 . If jV 0 j D p 2 , then U 0 \ V 0 D f1g and we have j1 .U 0 V 0 /j p 4 . If jV 0 j D p 3 , then jU 0 \ V 0 j p and so there are elements of order p in V 0 U 0 since V 0 is abelian of type .p 2 ; p/, by the previous paragraph. If jV 0 j D p 4 , then jU 0 \ V 0 j p 2 (Lemma 64.1(u)) and again there are elements of order p in V 0 U 0 since V 0 is abelian of type .p 2 ; p; p/, by (iv). In any case, we get j1 .U 0 V 0 /j p 4 . Suppose now that jU 0 j D p 3 . If W 2 1 fU g, then jW 0 j D p 3 , U 0 W 0 D U 0 W 0 (Lemma 64.1(u)) and both U 0 and W 0 are noncyclic ((i) and Proposition 72.1). Hence we get j1 .U 0 W 0 /j p 4 again. These results together with (iii) and (v) give that 1 .G 0 / Š Ep 4 . If G 0 is nonabelian, then using (iii), we see that G 0 must be isomorphic to a group of Proposition 71.1(i) and we are done. (vii) The group G is an irregular 3-group of order 39 . Each maximal subgroup of G is two-generator and Ã1 .G/ D K3 .G/ is of index 33 in G. In particular, G=G 0 Š E9 . Indeed, let U 2 1 and let A < U be maximal. Since 1 .G 0 / Š Ep 4 , we have jA \ 1 .G 0 /j p 3 since G 0 < U , and so A is nonmetacyclic. This implies that A0 is elementary abelian (Lemma 65.1 and Theorem 65.7(d)) and so A0 1 .U 0 / < U 0 since exp.U 0 / D p 2 , by (vi), and we get d.U / D 2 (Lemma 64.1(y)). Thus, all members of the set 1 are two-generator. Assume that d.G/ > 2. Using Theorem 70.1 and 70.2, we get in case p > 2 that jGj p 5 , a contradiction. If p D 2, then it follows from jGj 210 that cl.G/ > 2. The quotient group G=K4 .G/ is of order 27 or 28 . If jG=K4 .G/j D 28 , then we know that G=K4 .G/ is an A5 -group, a contradiction (see the text following Theorem 70.5). If jG=K4 .G/j D 27 , then Theorem 70.5 implies that K4 .G/ D f1g, a contradiction. We have proved that d.G/ D 2. Since G is not metacyclic, we get p > 2 (Lemma 64.1(n)). By Lemma 64.1(p), Ã1 .G/ D K3 .G/ is of index p 3 in G. On the other hand, j1 .G 0 /j D p 4 > jG=Ã1 .G/j so G is irregular (Lemma 64.1(a)). This implies that p D 3 (Theorem 9.8(a)), G=G 0 Š E9 and jGj D 39 . (viii) Let U; V 2 1 be distinct. Then jU 0 V 0 j 13 jG 0 j D 36 (Lemma 64.1(u)) and we claim that jU 0 V 0 j D 36 . Since H D K3 .G/ D ŒG; G 0 U 0 V 0 , we have H D U 0 V 0 D Ã1 .G/ in view of jG W H j D 33 . Also, 1 .G 0 / D 1 .H / and H is abelian of exponent 32 . Finally, G=H is nonabelian of order 33 and exponent 3. Suppose that there are distinct U; V 2 1 such that U 0 V 0 D G 0 . We may assume that jU 0 j D 34 and V 0 covers G 0 =U 0 . If jV 0 j D 33 , then G 0 D U 0 V 0 . But then, by (iv), j1 .G 0 /j 35 , contrary to (v). If jV 0 j D 34 , then jU 0 \ V 0 j D 3. But j1 .V 0 /j D 33 and again we get j1 .G 0 /j 35 , a contradiction. Hence, for each distinct U; V 2 1 ,

256

Groups of prime power order

we have jU 0 V 0 j D 36 . Since U 0 V 0 G G and G 0 =.U 0 V 0 / Z.G=.U 0 V 0 //, we have H D ŒG; G 0 U 0 V 0 . By (vii), H D U 0 V 0 D Ã1 .G/ (recall that jG W G 0 j D 32 ). Since j1 .U 0 V 0 /j 34 , we have j1 .U 0 V 0 /j D 34 , by (v), and so E D 1 .H / D 1 .G 0 / Š E34 . If H would be nonabelian, then H is an A1 -group, contrary to Lemma 65.1. Thus, H is abelian and so exp.H / D 32 since exp.U 0 / D exp.V 0 / D 32 and H D U 0 V 0. We are now able to get the final contradiction. Take U 2 1 . Since G=H is nonabelian of order 33 and exponent 3, we get U=H Š E9 , where H D K3 .G/ D Ã1 .G/. On the other hand, d.U / D 2 (by (vii)) and so ˆ.U / D H . Let X be any of the four maximal subgroups of U . Then jX W H j D 3 and E D 1 .H / Š E34 is a normal subgroup of X. Suppose that jX 0 j 32 so that X is an A2 -subgroup. By the results of 71, X must be isomorphic to a group of Proposition 71.3(ii). In particular, X (of order 37 ) has the unique abelian maximal subgroup of type .33 ; 3; 3; 3/. This is a contradiction since H is a unique maximal abelian subgroup of X and exp.H / D 32 (by (viii)). We have proved that jX 0 j 3 for any maximal subgroup X of U . This implies that jU 0 j 33 for any U 2 1 (Lemma 64.1(u)). We have proved that for each U 2 1 , U 0 is abelian of type .32 ; 3/ and so H D 0 U V 0 , where V 2 1 fU g. This implies that H is abelian of type .32 ; 32 ; 3; 3/ and each maximal subgroup X of U must be an A2 -group with jX 0 j D 3 (Lemmas 65.1 and 64.1(u)). By the results of 71, X must be the following group of Proposition 71.1(i): 3

2

X D ha; b j a3 D b 3 D 1; Œa; b D c; c 3 D Œa; c D Œb; c D 1i C3 ; where 1 .X/ D ha3 ; b 3 ; ci C3 D E D 1 .H / and X 0 D hci is a maximal cyclic subgroup in X (see Lemma 65.1). We have U 0 < H and set U 0 \ E D E0 so that E0 Š E9 . Also set he0 i D Ã1 .U 0 / so that he0 i is a subgroup of order 3 in E0 and he0 i is not a maximal cyclic subgroup in H . Let X1 , X2 , X3 , X4 be all maximal subgroups of U and note that each one is isomorphic to the above group X. Then X10 , X20 , X30 , X40 must be four distinct subgroups of order 3 in E0 (Lemma 64.1(u)). It follows that for some i 2 f1; 2; 3; 4g, Xi0 D he0 i. This is a contradiction since he0 i is not a maximal cyclic subgroup in H (and H is contained in Xi ); see the last sentence of the previous paragraph. 2

73

Classification of modular p-groups

We recall that a p-group G is modular if and only if any two subgroups of G are permutable (for general groups there is another definition). Sections of modular pgroups are modular. A nonabelian Dedekindian p-group G is a modular 2-group of the form G D Q E, where Q Š Q8 and exp.E/ 2 (Theorem 1.20). Such groups are called Hamiltonian. Nonmodular p-groups of order p 3 are D8 and the nonabelian group S.p 3 / of order p 3 and exponent p > 2. If G is a minimal nonmodular p-group, then there is N G G such that G=N is either S.p 3 / (p > 2) or D8 (see Theorem 44.13). The aim of this section is to correct the original Iwasawa’s proof [Iwa] of the theorem on the structure of modular p-groups. ((Iwasawa’s proof contains essential gaps. Two gaps were filled by Napolitani [Nap], remaining ones – by the second author, who wrote this corrected proof. There exists another proof of Iwasawa’s theorem, due to R. Schmidt [Sch, Chapter 2]; that proof is based on entirely other ideas and is essentially shorter.) Our proof is self-contained. The following proposition is obvious. Proposition 73.1. Let G be a modular p-group. Then 1 .G/ is elementary abelian. In particular, modular p-groups, p > 2, are powerful (apply Lemma 73.1 to G=Ã1 .G/; see [LM] and 26). It follows from Theorem A.24.4 that non-Hamiltonian modular 2-groups G are also powerful (see [LM] and 26.2). Indeed, by that theorem, G is Q8 -free. One may assume that exp.G/ D 4. Assume that G has a minimal nonabelian subgroup H . Then 8 < jH j < 32 since exp.H / D 4. Using Lemma 65.1, we see that H has a nonabelian epimorphic image of order 8, contrary to what has just been said. Thus, H does not exist so G is abelian. Proposition 73.2. Let G be a modular p-group. Then exp.n .G// p n and the group n .G/=n1 .G/ is elementary abelian for all n. Use induction on n and Proposition 73.1. By Theorem 44.13, a p-group G is modular if and only if it is D8 -free if p D 2 and S.p 3 /-free if p > 2. Remark. A two-generator modular p-group G is metacyclic. Assume that G is not metacyclic. In view of Lemma 64.1(m), one may assume that K3 .G/ˆ.G 0 / D f1g;

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Groups of prime power order

then G is minimal nonabelian such as in Lemma 65.1(a) with m C n > 2, jGj D p mCnC1 . Then G D ha; bi D haihbi has order p mCn < p mCnC1 D jGj, a contradiction. Since metacyclic p-groups, p > 2, are modular, it follows from the remark that a p-group G, p > 2, is modular if and only if every its two-generator subgroup is metacyclic. It is interesting to classify the 2-groups satisfying this property. Note that dihedral 2-groups are metacyclic but nonmodular. The group M2n is modular. A central product of two modular p-group is not necessary modular (for example, the group Q23 C4 is not modular. Proposition 73.3. Nonabelian modular groups G of order p 4 are: (a) Mp 4 , (b) Mp 3 2

2

Cp , p > 2, (c) G D ha; b j ap D b p D 1; ab D a1Cp i, p > 2, (d) Q8 C2 . Proposition 73.4. Suppose that G is a modular 2-group containing a normal elementary abelian subgroup E such that G=E Š Q8 . Then G D Q E with Q Š Q8 and so G is Hamiltonian. Proposition 73.5. Let G be a modular 2-group which is not Hamiltonian. Then each quotient group i .G/=i2 .G/ is abelian. In particular, 2 .G/ is abelian. Proof. Let G be a modular 2-group of exponent 4. If G does not contain a quaternion subgroup, then Proposition 73.3 implies that any two elements of G commute and so G is abelian. Suppose that Q8 Š Q G. In that case, G is Hamiltonian, by Theorem A.24.4, and we are done. The result follows since all sections of G are modular. Proposition 73.6. Let G be a modular p-group of exponent p . 2/ which is not Hamiltonian. Then .uv/p

()

1

D up

1

vp

1

for any u; v 2 G, i.e., G is p 1 -abelian. Proof. (i) Suppose that D 2. If p D 2, then G D 2 .G/ is abelian (Proposition 73.5). Suppose that p > 2 and let a; b 2 G. Then H D ha; bi D haihbi of order p p 4 is metacyclic (Lemma 64.1(m)), jH 0 j D p so .ab/p D ap b p Œb; a. 2 / D ap b p . The general case follows by induction on . Let > 2. Since G=1 .G/ is of 2 2 p 2 exponent p 1 , we have, for x; y 2 G, .xy/p D xp y z, where z 2 2 2 1 .G/. Since x p ; yp 2 2 .G/ and exp.2 .G// D p 2 , we get .xy/p

1

D ..xy/p D xp

1

2

yp

/p D .x p

1

2

zp D xp

yp

1

2

yp

z/p

1

:

73 Classification of modular p-groups

259

Proposition 73.7. Let G be a non-Hamiltonian modular group of exponent p . 2/. Set j˛ .G/ W ˛1 .G/j D p !˛ for ˛ D 1; : : : ; . Take elements a1 ; : : : ; a! (of order p ) of G so that they form .mod 1 .G// a basis for the elementary abelian p p group .G/= 1 .G/. Now, a1 ; : : : ; a! can be extended .mod 2 .G// to a p p basis a1 ; : : : ; a! ; a! C1 ; : : : ; a!1 of 1 .G/= 2 .G/ and so on. Every element of G can be written in a unique way as a product of powers of these elements a1 ; a2 ; : : : . In other words, the elements a1 ; a2 ; : : : form a basis of G. Looking at the series 1 .G/; : : : ; .G/, we see that the constants o.a1 /; o.a2 /; : : : are uniquely determined (up to the ordering). Next, if a1 ; : : : ; ad is so constructed basis, then Q d iD1 o.ai / D jGj. Proof. Working in G= 2 .G/, we may assume for a moment that 2 .G/ D f1g p are so that exp.G/ D p 2 . We have to show that (commuting) elements a1p ; : : : ; a! n! p n1 p ! linearly independent in 1 .G/ .D 1 .G/ Š Ep 1 /. Indeed, if a1 : : : a! D n! 1 for some integers n1 ; : : : ; n! , then, by Proposition 73.6, we get .a1n1 : : : a! /p D n! 1, and so a1n1 : : : a! 1 .mod 1 .G//, and we conclude that ni 0 .mod p/. p p Hence a1 ; : : : ; a! are linearly independent in 1 .G/ and so they are extendible p ; a! C1 ; : : : ; a!1 of 1 .G/ .D 1 .G//. to a basis a1p ; : : : ; a! Now, in general, (without assuming D 2 ) we consider the elements p2

2

p

p p ; a! C1 ; : : : ; a! 2 2 .G/ a1 ; : : : ; a! 1

and working in 1 .G/= 3 .G/ (of exponent p 2 ) we again show that these elements are linearly independent modulo 3 .G/ and therefore they are extendible with elements a!1 C1 ; : : : ; a!2 2 2 .G/ in such a way that 2

2

p p p ; a! ; : : : ; a! ; a!1 C1 ; : : : ; a!2 a1p ; : : : ; a! 1 C1

considered modulo 3 .G/, form a basis of 2 .G/= 3 .G/. Then in the obvious way we continue with the construction of a “basis” a1 ; a2 ; : : : of G. Since for any two “basis” elements ai and aj we have hai ihaj i D haj ihai i, every element of G can be written as a product of powers of elements a1 ; a2 ; : : : . Moreover, each element of G can be written in a unique way as a product of powers of elements a1 ; a2 ; : : : (in that order). Indeed, by the construction of these elements, we have o.a1 / o.a2 / D jGj and so the elements a1 ; a2 ; : : : form a basis of G. Proposition 73.8. Let G be a modular p-group which is not Hamiltonian. Let a 2 1 .G/# (this assumption is essential: it cannot be avoided even in the case where G is the abelian group of type .p 3 ; p/)). Then there is a basis a1 ; : : : ; ar of G such that a 2 ha1 i. Proof. We use induction on jGj. Let a1 ; : : : ; ar be a basis of G such that o.a1 / o.a2 / o.ar /. Let a 2 1 .G/# . Set ha1 ; : : : ; ar1 i D G1 so that G1 is a

260

Groups of prime power order

complement of har i in G. If a 2 1 .har i/, we are done. If a 2 G1 , then, by induction, there is a basis b1 ; : : : ; br 1 of G1 such that a 2 hb1 i. But then b1 ; : : : ; br1 ; ar is a basis of G and we are done. It remains to consider the case a D cd D dc, where c 2 1 .G1 /# and d 2 1 .har i/# . By induction, there is a basis c1 ; : : : ; cr1 of G1 such that c 2 hcr1 i. Set o.ar / D p s . If s D 1, then we replace ar with a. Since hai \ G1 D f1g and o.a/ D o.ar / D p, c1 ; : : : ; cr 1 ; a is a basis of G and we are done. Suppose s 2 s1 and replace ar with ar0 so that d D .ar0 /p . Then c1 ; : : : ; cr1 ; ar0 is also a basis of G. Since o.a1 / o.a2 / o.ar / (and these numbers are invariants of G), we have also o.cr 1 / o.ar0 / D p s . Thus, there is an element l 2 hcr1 i of order p s s1 such that l p D c. Since hl; ar0 i is a modular subgroup of exponent p s (noting that exp.s .G// p s ), s1 s1 s1 it follows from Proposition 73.6 that .lar0 /p D l p .ar0 /p D cd D a, and so o.lar0 / D o.ar0 / D p s . On the other hand, hc1 ; : : : ; cr1 ; lar0 i D G and G1 \ hlar0 i D f1g. Hence c1 ; : : : ; cr 1 ; lar0 is a basis of G and a 2 hlar0 i. Proposition 73.9. Let G be a modular p-group which is not Hamiltonian. If jG 0 j D p, m n m1 then G D G1 G2 , where G1 D ha1 ; a2 j a1p D a2p D 1; a1a2 D a11Cp ; m> 1; n 1i and G2 is abelian of exponent p m1 with m > 2 if p D 2. Hence m1 A D ha1 ; G2 i is an abelian normal subgroup of G D hA; a2 i and aa2 D a1Cp for all a 2 A. Proof. By Proposition 73.8, we may choose a basis a1 ; a2 ; : : : ; as of G so that G 0 ha1 i. Since hai ; aj i \ G 0 hai ; aj i \ ha1 i D 1 .i; j > 1/, we see that ha2 ; : : : ; as i is abelian. Let a2 be an element of the smallest order among those basis elements a2 ; : : : ; as which do not commute with a1 . By Lemma 65.2(a), ha1 ; a2 i is minimal p m1 nonabelian. So, in view of Lemma 65.1, we may put Œa1 ; a2 D z, where z D a1 and o.a1 / D p m , m 2. (We get ha1 i ¤ G 0 since G is nonabelian.) We have Œa1 ; ai D z ki with ki 2 f0; 1; : : : ; p 1g .i > 2/, where (by assumption) o.ai / o.a2 / whenever ki ¤ 0. Then we replace the basis elements a3 ; : : : ; as with the elements a2ki ai D bi .3 i s/. We have o.bi / D o.ai / and Œa1 ; bi D Œa1 ; a2ki ai D Œa1 ; a2 ki Œa1 ; ai D z ki z ki D 1 and so the abelian subgroup hb3 ; : : : ; bs i centralizes ha1 ; a2 i. It is clear that fa1 ; a2 ; b3 ; : : : ; bs g is also a basis of G. Indeed, ha1 ; a2 ; b3 ; : : : ; bs i D G and jGj D o.a1 /o.a2 /o.a3 / : : : o.as / D o.a1 /o.a2 /o.b3 / : : : o.bs /: Denoting b3 ; : : : ; bs again with a3 ; : : : ; as , we may assume from the start that G D G1 G2 , where G1 D ha1 ; a2 i and G2 D ha3 ; : : : ; as i. For i 3 we have .a1 ai /a2 D a1a2 ai D .a21 a1 a2 /ai D a1 .a11 a21 a1 a2 /ai D a1 zai D z.a1 ai /:

261

73 Classification of modular p-groups

Now, .G 0 / A D ha1 ; a3 ; : : : ; as i G G, G D Aha2 i, and A \ ha2 i D f1g. Consider the subgroup H D ha1 ai ; a2 i .i > 2/ which is nonabelian. Since H D ha1 ai iha2 i and ha1 ai i A, we have H \ A D ha1 ai i and so a2 normalizes ha1 ai i. Thus p m1 2 ha1 ai i. This implies .a1 ai /a2 D z.a1 ai / D .a1 ai /x .x integer/ and so z D a1 m1 that the order of ai .i > 2/ is at most p . For each a 2 A, we have a D a1r b with an integer r and b 2 G2 so that 1Cp m1 r

aa2 D .a1r b/a2 D .a1 D .a1r b/1Cp

m1

/ b D .a1r /1Cp

D a1Cp

m1

m1

b 1Cp

m1

:

If p D 2 and m D 2, then K D ha1 ; a2 j a14 D a22 D 1; a1a2 D a13 D a11 i is not n1 modular, since K=ha22 i Š D8 . Hence, if p D 2, then m > 2. Our proposition is proved. n

Proposition 73.10. Let G be a modular p-group which is not Hamiltonian. Then each subgroup and each factor group is modular and is not Hamiltonian. Subgroups and epimorphic images of modular p-groups are modular so the assertion follows since, by Theorem A.24.4, G is Q8 -free. s

Proposition 73.11. Let t; x 2 G, where exp.G/ D p . If x t D x 1Cp z, where z 2 1 .Z.G//# and 1 s < 1 is a natural number which is 2 if p D 2, then xt

p 1s

D x 1Cp

1

. In particular, if o.x/ p 1 , then Œx; t p i

Proof. By induction on i , we first prove that x t D x .1Cp xt

i C1

D .x .1Cp

p 1s

s /i

s

z i /t D .x 1Cp z/.1Cp

s /i

s /i

1s

z i . Indeed, we have

z i D x .1Cp

s p 1s

s /i C1

D x .1Cp / , since s < 1 and so z p This gives x t It remains to prove the following congruence: (1)

.1 C p s /p

1s

1 C p 1

D 1.

z iC1 :

1s

D 1.

.mod p /:

According to P. Roquette, everything depends on the following formula: (2)

.a C bp s /p ap C ap1 bp sC1

.mod p sC2 /;

where a; b are integers, p prime, s 1 and, if p D 2, then s 2. Indeed, we get ! X p pi i si s p p p1 s a (3) bp C b p C b p p sp : .a C bp / D a C pa i 2ip1

Therefore each member of the above sum is 0 .mod p 1Csi / and so 0 .mod p sC2 / since 1 C si s C 2 in view of i 2. For the last term on the right-hand side of (3)

262

Groups of prime power order

we have sp s C 2 for p 3 and this holds also for p D 2 because in that case we have assumed s 2. Thus, the last term is also 0 .mod p sC2 /, and (2) is proven. It follows from (2) that z 1 C ps

(4)

.mod p sC1 / implies z p 1 C p sC1

.mod p sC2 /:

To prove (4) we set z D 1 C p s C bp sC1 D a C bp sC1 , where a D 1 C p s and b is an integer. From (2) follows (with s C 1 instead of s and noting that always s C 1 2) z p D .a C bp sC1 /p ap C ap1 bp sC2 .mod p sC3 / and so z p ap .1 C p s /p .mod p sC2 /. Again using (2) (for a D b D 1), we get .1 C p s /p 1 C p sC1 .mod p sC2 /, proving (4). With iteration we get from (4) (with k D 1; 2; 3; : : : ) z 1 C ps

(5)

k

.mod p sC1 / implies z p 1 C p sCk

.mod p sCkC1 /:

If we set in (5) z D 1 C p s , k D 1 s, where s < 1, we get (1). Proposition 73.12. Let G be a non-Hamiltonian modular p-group. Then each element i of Ãi .G/ .i 1/ can be written as ap for some a 2 G. Proof. We prove first this result for i D 1, working by induction on jGj. Let x 2 p p p Ã1 .G/ so that x D a1 a2 : : : as with some ai 2 G. Now, G=Ã2 .G/ is of exponent 2 p and so, using Proposition 73.6, we get x D .a1 a2 : : : as /p l where l 2 Ã2 .G/. p2 p2 p2 We set a1 a2 : : : as D b, H D hb; Ã1 .G/i, and l D c1 c2 : : : cr with some c1 ; c2 ; : : : ; cr 2 G. Thus x D b p .c1p /p : : : .crp /p 2 Ã1 .H /. We have H ¤ G, unless G is cyclic, so x D ap for some a 2 H , by induction. pi

pi

Assume that i > 1 and use induction on i . Let y 2 Ãi .G/ so that y D b1 : : : b t D i 1

i 1

i 1

/p for some b1 ; : : : ; b t 2 G. Here bjp 2 Ãi1 .G/, j D .b1p /p : : : .b p t 1; : : : ; t , so y 2 Ã1 .Ãi1 .G//. By the above, y D c p with c 2 Ãi1 .G/. By i 1 i and so y D d p . induction, c D d p Proposition 73.13. Let G be a non-Hamiltonian modular p-group. If we have G D i i ha1 ; : : : ; ar i, then Ãi .G/ D ha1p ; : : : ; arp i. Proof. One may assume that fai gr1 is a minimal basis. First let i D 1. We have ha1 i : : : har i D G so by the product formula applied r 1 times, jG W ha1p i : : : harp ij D p r , Since p r D jG W ˆ.G/j D jG W Ã1 .G/j, we are done. p i 1

p i 1

Suppose that i > 2. By induction on i , Ãi1 .G/ D ha1 ; : : : ; ar i. If y 2 i 1 i Ãi1 .G/, then y D ap ; a 2 G (Proposition 73.12) and so y p D ap 2 Ãi .G/. This gives Ã1 .Ãi1 .G// Ãi .G/. On the other hand, if g 2 Ãi .G/ , then by i i 1 Proposition 73.12, g D b p D .b p /p ; b 2 G, and so g 2 Ã1 .Ãi1 .G// which pi pi gives Ãi .G/ D Ã1 .Ãi1 .G//. By the above, this gives Ãi .G/ D ha1 ; : : : ; ar i.

263

73 Classification of modular p-groups

Proposition 73.14 (compare with [Sch, Lemma 2.3.4]). Let G be a p-group and let A G G be abelian and such that G=A D htAi is cyclic, all subgroups of A are normal in G and, if p D 2, when 2 .A/ Z.G/. Then G is modular. Proof. Suppose that G is a counterexample of minimal order. Then, by induction, H < G is modular since H=.H \ A/ Š AH=A. G=A/ is cyclic and H satisfies the hypothesis with respect to H \ A. Hence, G is minimal nonmodular. Then, by Theorem 44.18, there exists N GG with G=N 2 fD8 ; S.p 3 /g. Since G=AN is a cyclic epimorphic image of G=N , we have jG W AN j p. But every subgroup of AN=N is normal in G=N , and this implies that p D 2 and AN=N is the cyclic subgroup of order 4 in G=N Š D8 . Hence A=.A \ N / Š C4 so A D .A \ N /hui, and A is modular, by the above. Let G D Ahvi. Then G D hu; vi since hu; vi is nonmodular since it has an epimorphic image Š D8 . Since v 2 2 A, W u ! uv is an automorphism of order 2. Let o.u/ D 2m , o.v/ D 2n . By Theorem 1.2, we have to consider one of the following three possibilities: (i) uv D u1 . Then u2 tion.

m2

m2

D .u2

/v D u2

m2

, m > 2. Then u2 (ii) uv D u1C2 m1 m2 / D 1, a contradiction. u2 .1C2 m1

m2

m2

D .u2

m1

so u2

D 1, a contradicm2 .1C2m1 /

/v D u2

so

m1

, m > 2, i.e., G Š M2mC1 . In that case, G is modular, contrary (iii) uv D u1C2 to the assumption. Theorem 73.15 ([Iwa]). A non-Hamiltonian p-group G is modular if and only if it contains an abelian normal subgroup N with cyclic quotient group G=N and there exists an element t in G with G D hN; t i and a positive integer s which is at least 2 in s the case p D 2 such that at D a1Cp for all a 2 N . Proof (Iwasawa–Napolitani–Janko). If G is as in the theorem, it is modular, by Proposition 73.14. Suppose, conversely, that G is a modular p-group which is not Hamiltonian. This property is inherited by sections (Proposition 73.10), so we may assume that the theo1 rem is true for groups of smaller order. By Proposition 73.6, the mapping g 7! g p , where p D exp.G/, is a homomorphism of G. The image of this homomorphism is Ã 1 .G/ ¤ 1. There is a G-invariant subgroup Z D hzi of order p in Ã 1 .G/. By induction, G=Z contains an abelian normal subgroup N=Z with properties stated in s the theorem. There is an element t 2 G such that G D hN; t i and at D a1Cp z k for all a 2 N , where k 2 N [ f0g depends on a. There are two major possibilities for the structure of N . The subgroup N is either (I) abelian or (II) nonabelian. Case (I). N is abelian. Here z 2 Ã 1 .G/ is the p 1 -th power of an element c in 1 G, z D c p . There are two possibilities for the position of c in G.

264

Groups of prime power order

(˛) It is possible to choose c 2 N . Since o.c/ D p , c is an element of a basis of G (Proposition 73.7). In particular, hci has a complement T in G which is generated by remaining elements of those basis of G that contains c. We have G D T hci with T \hci D f1g and so N D .T \N /hci, by the modular law, and T covers G=N since T N T hci D G. The quotient group T =.T \ N / Š G=N is cyclic, and we choose t1 2 T \ .N t / so that T D hT \ N; t1 i; then G D hN; t1 i. Since t 1 t1 2 N and N is abelian, t and t1 induce the same s automorphism on N . In view of Z \ T D f1g, we have at D at1 D a1Cp for all a 2 T \ N. s If also c t D c 1Cp , then the fact that N D .T \ N / hci is abelian implies s x t D x 1Cp for all x 2 N and the theorem is proved. Therefore assume that s s 1 c t D c 1Cp z k D c 1Cp Ckp , where k 6 0 .mod p/. We may assume that k 2 f1; 2; : : : ; p 1g and so p k 2 f1; 2; : : : ; p 1g. We claim that Ã 1 .T \N / D f1g. Otherwise, there is a 2 T \N of order p . We have hai \ hci D f1g and hcai \ hci D f1g. Both elements ca and c are of order p

and they, clearly, are “linearly independent” modulo 1 .G/. Hence, by the process of constructing a basis of G in Proposition 73.7, there is a basis of G containing both c and ca. Let T1 be a complement of hci in G containing ca. By the preceding s s s s argument, .ca/t D .ca/1Cp and so .ca/t D c t at D c t a1Cp D c 1Cp a1Cp , which s gives c t D c 1Cp , contrary to what has been assumed in the previous paragraph. We have proved that Ã 1 .T \N / D f1g or, what is the same, exp.T \N / p 1 . Hence, if s 1, then G=hzi is abelian. Assuming that G is nonabelian (otherwise, there is nothing to prove), we conclude that G 0 D hzi is of order p and so Proposition 73.9 implies the validity of the theorem. It remains to examine the case s < 1 (and s 2 for p D 2). We use Proposition p 1s

73.11 and get x t D x 1Cp , whenever x t D x 1Cp z 0 with z 0 2 Z. From this relation we get, iterating (p k) times, xt

1

.pk/p 1s

D x .t

z

p 1s /pk

D x 1C.pk/p Thus, setting t D t 1C.pk/p

ct D ct

.pk/p 1s t

D .c 1Cp

s Ckp 1

1s

1 C pk 2

.

1/ /pk

/p 2.1/ C D x 1kp 1 :

, we get

D .c 1kp /.1kp

D x .1Cp

1

1 /

/t D .c t /.1kp

D c 1Cp

1 /

s Ckp 1 kp 1 kp 1Cs k 2 p 22

s

D c 1Cp : Finally, for each a 2 T \ N (noting that ap t .pk/p

1s

1kp 1 t

1

.a / D .a / D a D a y 2 N . Since G D hN; t i, we are done. t

t

D 1 holds) we obtain also at

1Cp s

. Thus y

t

D y

1Cp s

D

for each

265

73 Classification of modular p-groups

(ˇ) It is not possible to choose c in N . We claim that hc; N i D G. Indeed, if hc; N i < G, then we can choose a genph ph erator t1 of ht i such that t1 c .mod N / with h > 0. Hence c D bt1 , where b 2 N . It follows b D ct1p and then applying Proposition 73.6 we get b p h

p h 1 .ct1 /p

1 p hC1 c p t1

1

p 1

D

D D c D z, where b 2 N and o.b/ D p , a contradiction. It follows that a suitable generator c0 of hci operates on N in the same way as t and so we may assume that c itself operates on N as t . In what follows we shall identify t 1 and c. This means that we shall assume that o.t / D p , t p D z, and so for each s 1 t 1Cp kp t with k 2 f0; 1; : : : ; p 1g depending on a. a 2 N we have a D a Let fa1 ; a2 ; : : : ; ar g be a basis of N such that Z D hzi ha1 i (such a basis exists 1Cp s or by Proposition 73.8). For each ai , i D 1; 2; : : : ; r, we have either ait D ai 1Cp s ki p 1 t t , ki 6 0 .mod p/. ai D ai

s

We shall prove that it is possible to assume that ait D ai1Cp for i 3. For this purpose, we order the basis elements a2 ; : : : ; ar so that: 1Cp s

(1ˇ) there exists such that ait D ai

if and only if i D C 1; : : : ; r;

(2ˇ) if > 1, then o.a2 / o.aj / for each j with 2 j . If 2, the above statement is obviously true. Supposing > 2, we observe that for each i , 2 i , fa2 ; a2q ai g (q integer) is a basis of ha2 ; ai i. Since s

.a2q ai /t D .a2t /q ait D .a21Cp t k2 p q

s

D .a2 /1Cp t k2 qp

1

1

1Cp s ki p 1

ai

t

s

/q .ai1Cp /ki p q

1

s

D .a2 ai /1Cp t .k2 qCki /p

1

and k2 6 0 .mod p/, we can solve the congruence equation k2 q C ki 0 .mod p/, s in q D qi (depending on i ) so that .a2qi ai /t D .a2qi ai /1Cp . Hence the set fa1 ; a2 ; a2q3 a3 ; : : : ; a2q a ; a C1 ; : : : ; ar g is a basis with the desired property. We consider first the following special possibility: 1Cp s

p 1

1

p 1

t k2 p , k2 6 0 .mod p/, a2 ¤ 1, and ai D 1 for all (ˇ1 ) a2t D a2 i 3. We shall prove that in this case either the theorem holds or we can choose our abelian normal subgroup N D ha1 ; a2 ; : : : ; ar i in such a way that 1Cp s also a2t D a2 . Here we distinguish two subcases (i) s 1 and (ii) s < 1. 1Cp s

1

(i) s 1. If a1t D a1 t k1 p with k1 6 0 .mod p/, then we multiply a2 qN with a1 , where qN is a solution of the congruence equation k1 qN C k2 0 .mod p/ and qN we replace the basis fa1 ; a2 ; : : : ; ar g of N with the basis fa1 ; aO 2 D a1 a2 ; a3 ; : : : ; ar g

1

1 noting that o.a1 / p (since in our case .ˇ/, z ha1 i is not a p -th power of

266

Groups of prime power order

any element in N ). Then we compute qN

1Cp s k1 p 1 qN 1Cp s k2 p 1 t / a2 t

.aO 2 /t D .a1 a2 /t D .a1

N / 1Cp .k1 qCk D a1q.1Cp a2 t N 2 /p s

s

1Cp s

1), then G 0 D ht

D .a1qN a2 /1Cp D .aO 2 /1Cp ; s

D a1 (noting that s 1 and a1

p 1 k2 p 1 ha2 t i

i or G 0 D

s

p 1

as required. If, however, a1t D a1 p 1

1

D

depending whether s > 1 or 1

s D 1. Here we have also used our assumption that aip D 1 and the fact that 1Cp s ait D ai for all i 3. But then our theorem follows from Proposition 73.9. (ii) s < 1. We replace here the subgroup N with the subgroup N D ha1 ; a2 ; 1s with ˛ being a solution of the congruence ˛k2

a3 ; : : : ; ar i, where a2 D a2˛ t p 1 1 D tp D z. We note that 1 .mod p/. Note that ˛ 6 0 .mod p/ and t ˛k2 p s s 1 D z k2 2 Z.G/ and a2p (s 1) is of order p 1 and therefore a2p and t k2 p 1s commute (Proposition 73.11). Then we compute tp .a2 /t D .a2˛ t p

1s

˛.1Cp s /

D a2

s

/t D .a2t /˛ t p

t ˛k2 p

D a2˛ a2˛p t p

1

1s

.t p

The subgroup H D ha2 ; t p p 1s

a2t

1Cp 1

D a2

and so

.a2 /p D .a2˛ t p s

s

1s

1s

1s

˛p s

t

D a2˛ a2 t p s

/p D .a2˛ t p 1

/ t

1

1s

tp

1s s

/.a2˛p .t p

1s

s

/p /:

i is of order p. Therefore,

˛p s

s

/p D a2 .t p

1s

1Cp s k2 p 1 ˛ p 1s

D .a2

i is of class 2. Indeed, by Proposition 73.11,

D ha2p

H0

1s

D a2˛p .t p

tp

1s

1s

s

/p Œt p

1s

ps

; a2˛ . 2 /

s

/p ;

s since (in any case) p divides p2 . Substituting this last result in the above relation, s s we get .a2 /t D a2 .a2 /p D .a2 /1Cp . 1s commutes with each ai , i ¤ 2, The subgroup N is abelian. Indeed, t p because each such ai is of order p 1 (Proposition 73.11). Also, G D hN ; t i, .N /t D N , and so N is normal in G with G=N cyclic. We observe that a2 is of order p and so ha2 i\ha1 ; a3 ; : : : ; ar i D f1g, which implies that fa1 ; a2 ; a3 ; : : : ; ar g 1 1s p 1 D .a2˛ t p / D is a basis of N . Indeed, by Proposition 73.6, .a2 /p p 1 ˛ p 22s

.a2

/ t

˛p 1 ha2 i

˛p 1

D a2

is an element of order p (˛ 6 0 .mod p/) and

\ ha1 ; a3 ; : : : ; ar i D f1g, where we have used the fact that 2 2 s . All the statements in case .ˇ1 / are proved. Indeed, a2 is of order p , all other basis p 1

elements ai of N are of order p 1 and so Ã 1 .N / D ha2 Hence z 2 ha1 i cannot be a p 1 -th power of any element of N .

i is of order p.

267

73 Classification of modular p-groups

In view of what was proved so far in the case .ˇ/, it remains to consider the following two possibilities: .ˇ2 / There exists an ai , i 2, such that aip

1

s

¤ 1 and ait D ai1Cp . 1Cp s

for j 3. (We are .ˇ3 / All ai are of order p 1 and, in addition, ajt D aj

1 in case .ˇ/ and so also a1 is of order p since z 2 ha1 i and z is not a p 1 -th power of any element in N .) Assume that the condition .ˇ2 / is satisfied. Hence there exists an ai , i 2, such 1Cp s

p 1

1

and ai ¤ 1. Here t p D z 2 ha1 i and therefore ht i \ hai i D that ait D ai 1 1 1 p 1 p 1 p ¤ 1 and ai ¤ 1, we get .t ai /p D t p ai ¤ 1 so f1g. Since t ht ai i \ hai i D f1g. It follows that elements ai ; t ai are contained in a basis of G since ai and t ai are both of order p and they are “linearly independent” mod 1 .G/ (see the proof of Proposition 73.7). Therefore, there exists a complement T of hai i in G containing t ai . We have N D .N \ T / hai i and t ai operates in the same way as t on N \ T . If t ai does not transform each element of N \ T into its .1 C p s /-th s N 1 with kN 6 0 power, then there exists aN 2 N \ T such that aN tai D aN t D aN 1Cp t kp 1 1 1 1 p 2 T . Since .t ai /p D t p ai 2 T , it follows that .mod p/, and so t p aip

1

s

2 T , a contradiction. Hence at D a1Cp for each a 2 T \ N and ait D ai1Cp . s

s

But N D .T \ N / hai i is abelian and so x t D x 1Cp for all x 2 N and we are done. Assume, finally, that the condition .ˇ3 / is satisfied. We distinguish here two subcases: (i) s < 1 and (ii) s 1. 1Cp s

1

t k1 p , k1 6 0 .mod p/ and ait D (i) Suppose that s < 1. If a1t D a1 s 1s , where ˛ is such that ai1Cp for i 2, then we replace a1 with a1 D a1˛ t p ˛k1 1 .mod p/. Applying Proposition 73.11, we prove that N D ha1 ; a2 ; : : : ; ar i s is an abelian normal subgroup of G with G D hN ; t i, .a1 /t D .a1 /1Cp , and ait D s 1Cp for i 2 (where fa1 ; a2 ; : : : ; ar g is not necessarily a basis of N ). Indeed, ai 1s commutes with each ai , i D 1; : : : ; r, and so N is abelian. It remains to tp compute .a1 /t D .a1˛ t p

1s

/t D .a1t /˛ t p

s

1

s

1s

D .a1˛ /1Cp t ˛k1 p D .a1˛ /1Cp .t p 1Cp s

tp

1s

1s

s

s

D .a11Cp t k1 p s

D .a1˛ /1Cp t p

/1Cp D .a1˛ t p

1s

1

1

tp

/˛ t p

1s

1s

/1Cp D .a1 /1Cp : s

s

t k2 p with k2 6 0 .mod p/, then we replace a2 with a2 D If a2t D a2 1s , where ˛ is such that ˛k2 1 .mod p/. Applying Proposition 73.11 a2˛ t p again, we see that N D ha1 ; a2 ; a3 ; : : : ; ar i is an abelian normal subgroup of G with G D hN ; t i (where fa1 ; a2 ; a3 ; : : : ; ar g is not necessarily a basis of N ). It remains 1

268

Groups of prime power order

only to compute .a2 /t D .a2˛ t p

1s s

D a2˛.1Cp / t p

˛.1Cp s / ˛k2 p 1 p 1s

/t D a2 1

tp

t

1s

t

D .a2˛ t p

1s

/1Cp D .a2 /1Cp : s

s

Then our theorem either holds or we have the case considered already in the previous paragraph (noting that the order of a2 is p 1 ). (ii) Suppose that s 1. Since o.ai / p 1 for all i , then G is either abelian or G 0 D hzi D Z is of order p. Then the theorem follows from Proposition 73.9. Case (II). N is nonabelian. The idea here is to reduce this case to Case (I). By Proposition 73.9, N has the following structure. There is a basis fa1 ; a2 ; : : : ; ar g of N such that pm

a1

pn

D a2 D 1;

ai aj D aj ai

1Cp m1

a1a2 D a1

m > 1; n 1;

p m1

for i; j 3;

D 1 for

ai

N D ha1 ; a2 i ha3 ; : : : ; ar i;

; i 3;

and if p D 2; then m > 2:

p m1

kp m1

Here Z D ha1 i D N 0 Z.G/ and, for each a 2 N , we have at D a1Cp a1 , where k .mod p/ depends on a. ˇ Replacing t by a suitable t a1˛ a2 (˛ and ˇ are integers .mod p/), we may assume s

a1t D a11Cp

(6)

s

s

and a2t D a21Cp :

1Cp s k1 p m1 1Cp s k2 p m1 a1 and a2t D a2 a1 , where k1 ; k2 2 m1 p m1 a2 a1 f0; 1; : : : ; p 1g. It follows from a1 D a1 a1 that a2 D a2 a1p , and these ˇ a2 a1˛ 1Cˇp m1 ˛p m1 and a2 D a2 a1 . We shall choose ˛ and two relations imply a1 D a1 ˇ so that ˛ k2 .mod p/ and ˇ k1 .mod p/. Then we compute (noting that

Indeed, we have a1t D a1

s 1) ˇ

ta1˛ a2

s

D .a11Cp a1k1 p

a1

m1

ˇ

/a2 D .a11Cˇp

1Cp s C.ˇ Ck1 /p m1

D a1 ˇ

ta1˛ a2

and similarly a2

s

D .a21Cp a1k2 p

m1

m1

s

/1Cp a1k1 p

1Cp s

D a1

s

˛ ˇ

.˛Ck2 /p m1

/a1 a2 D a21Cp a1

Since Z Ã 1 .G/, where p D exp.G/, a generator z D a1p p m1

th power of some element c. Suppose that a1 Then (7)

a1p

m1

D cp

1

m1

D a1e1 p

1

Dc

a2e2 p

p 1

1

m1

s

D a21Cp .

of Z is a p 1 -

and c D a1e1 a2e2 : : : arer t f0 .

t f0 p

1

:

269

73 Classification of modular p-groups p 1

p 1

¤ 1. Then m D . If s D 1 and a2

Suppose that a1

G0

1 D ha1p i p 1 (i) If a1

is of order p and our theorem holds by Proposition 73.9. p 1

¤ 1, s D 1, and a2 t 1

From (6) we get a1

D

1p 1 a1 ,

a

a1 2 D a1a2 t p 1

1

¤ 1, then we replace a2 by a2 D a2 t 1 .

and so p 1 p 1 a1

p 1 t 1

D .a1 a1

D a1 a1

/

D a1 ; 1Cp 1 ki p 1 a1

2 Z.G/. Assume that, for an i 3, we have ait D ai

since a1

k p 1 ai a1 i

D 1, then

with ki 6 0 .mod p/. This gives Œai ; t D p 1 ha1 i.

k p 1 a1 i

D

and so H D hai ; t i

contains Z D But Z is normal in G and so H 0 D Z is of order p. By Lemma 65.2(a), H is minimal nonabelian and by Remark, following Proposition 73.2, H is p 1 metacyclic. Therefore 1 .H / Š Ep 2 which implies 1 .H / D h1 .hai i/; a1 i: In particular, 1 .ht i/ 1 .H /. Also Œai ; t 1 D Œai ; t 1 D a1ki p

1 ait

D

k p 1 ai a1 i .

a

Now consider the subgroup K D hai ; a2 i. We have ai 2 D aia2 t

k ai a1 i

k a1 i

p 1

1

p 1

1

, and so

D ait

1

D

and so Œai ; a2 D , which is a generator of Z. Thus K contains Z and since Z is normal in G, we have K 0 D Z. It follows that K is also minimal 1 nonabelian and metacyclic and we have 1 .K/ D h1 .hai i/; a1p i D 1 .H /: In 1 D particular, 1 .ha2 i/ 2 1 .H /. We compute, using Proposition 73.6, .a2 /p .a2 t 1 /p

1

p 1 p 1

D a2

t

and since t p

1

p 1 h1 .hai i/; a1 i.

p 1

2 1 .H /, we get 1 ¤ a2

2

1 .H / D This is a contradiction since fa1 ; : : : ; ar g is a basis of N . We have proved that t centralizes ha3 ; : : : ; ar i. This implies that the subgroup N D ha1 ; a2 ; a3 ; : : : ; ar i is abelian. Since t normalizes ha2 i, it follows that t centralizes a2p

1

. We compute 1Cp 1 1

.a2 /t D .a2 t 1 /t D a2t t 1 D a2 p 1

D a2 a2

D .a2 /1Cp p 1

1

t

tp

1

p 1

D .a2 t 1 /a2

;

D .a2 t 1 /p D a2 t p . Also, G D hN ; t i. If t p D 1, since .a2 /p 1 t 1Cp for each x 2 N and then t normalizes the abelian subgroup N and x D x 1 we are done. Suppose that t p ¤ 1. Note that t induces an automorphism of order 1 1 p on N and so t p centralizes N . In particular, t p 2 Z.G/. Hence A D N ht p i 1 is a normal abelian subgroup of G, G D Aht i, and for each a 2 A, at a1Cp 1 1 .mod ht p i/. Finally, t p 2 Ã 1 .G/ and this is exactly the situation studied in Case (I). 1

1

1

1

270

Groups of prime power order p 1

(ii) Suppose that a1 ¤ 1 .m D / and s < 1. In this case we replace 1s . Since o.ai / p 1 for i > 2, Proposition 73.11 implies a2 by a2 D a2 t p p 1s

p 1

that ha2 ; a3 ; : : : ; ar i is abelian. Again, Proposition 73.11 gives a1t and conjugating this relation with t p a1t

1s

p 1 a1

and so

a

a1 2 D a1a2 t

p a1t

1s

p 1s

1p 22

D a1

p 1s

D

(noting that

1p 1 a1 .

D .a11Cp

1

/t

p 1 a1

D a1 a1

,

2 Z.G/) we get a1 D

We compute

p 1s

D .a11p

1

/1Cp

1

D a1

since 2 2 and so N D ha1 ; a2 ; a3 ; : : : ; ar i is an abelian subgroup of G. It remains to compute .a2 /t . 1s Let us consider the subgroup H D ha2 ; t p i. Since, by Proposition 73.11, p 1s

1

1

a2t D a21Cp , it follows that H 0 D ha2p i is of order p. By Proposition 1s ps 73.11, a2 (s 1 and if p D 2 then s 2) commutes with t p and so working in the subgroup H of class 2, we get .a2 /1Cp D .a2 t p s

D a2 t p

1s

1s

/1Cp D a2 t p s

s

a2p t p

On the other hand, .a2 /t D .a2 t p s 1 . we get finally .a2 /t D .a2 /1Cp t p

1s

We have G D hN ; t i. If t p

1

1

s

.a2 t p

D a21Cp t p t

1s

1s

1Cp s p 1s

/t D a2

p 1

2 ha1

1s

t p

s

/p D

1

:

, and so by the above

i, then the abelian subgroup N is nor1

mal in G and we have reduced this case to Case (I). So suppose t p 62 ha1p i. By 1s 1s centralizes a3 ; : : : ; ar and t p induces an automorProposition 73.11, t p 1 1s p s phism of order p on ha1 i and ha2 i and so t p D .t p / 2 Z.G/. Hence 1 p N D hN ; t i is a normal abelian subgroup of G, G D hN ; t i, and, for each 1 s 1 x 2 N , we get x t D x 1Cp t kx p a1lx p , where kx , lx are integers depending on x. Also, t and a1 are both of order p and they are “linearly independent” modulo 1 .G/. Indeed, if t n1 a1n2 2 1 .G/ (n1 , n2 integers), then (Proposi1

1

p 1

tion 73.6) .t p /n1 .a1 /n2 D 1 and so n1 n2 0 .mod p/. Hence there is a basis a1 ; t; : : : of G (Proposition 73.7). In particular, there is a complement T of ha1 i in G containing t . Since ha1 i N , we get N D ha1 i .N \ T /, by the 1 s 1 with modular law. Take a 2 N \ T and assume that at D a1Cp t ka p a1la p p 1

2 T , a contradiction. Hence, for each a 2 N \ T , la 6 0 .mod p/. Then a1 s 1 t 1Cp s ka p 1 t . But N is abelian and so x t D x 1Cp t kx p for all we get a D a 1 1 p p x 2 N . Since ht i 2 Z.G/ and t 2 Ã 1 .G/, we have again reduced this case to Case (I).

271

73 Classification of modular p-groups p 1

(iii) Suppose, finally, that a1 p m1

(8)

D 1. Then from relation (7) follows

e p 1 f0 p 1

D a22

a1

t

tp

;

1

¤ 1;

f0 6 0

.mod p/ 1

since fa1 ; a2 g is a basis of the minimal abelian subgroup ha1 ; a2 i. Also, t p 2 p m1 p 1 ; 1 .ha2 i/i but t 62 1 .ha2 i/. 1 .ha1 ; a2 i/ D ha1 Denote p l D jha1 ; a2 ; t i W ha1 ; a2 ij and note that (6) implies that t normalizes ha2 i l p (and ha1 i) and so t p 2 Nha1 ;a2 i .ha2 i/ D ha1 i ha2 i. Replacing a1 with a1i .i 6 0 .mod p//, we may assume (writing again a1 instead of a1i ) that t p D a1p a2hp ; where k 1; h 6 0 .mod p/: k

l

(9) pk

f

hp f

pf

Since Œa1 ; a2 D 1 and 1 .ht i/ 6 1 .ha2 i/, we get o.a2 l o.t p /

k o.a1p /.

l o.t p /

p k

/ o.a1

k o.a1p /

/ and so

D On the other hand, D and D and so m

l D m k. By assumption, > m, where p D o.a1 /, and so l > k 1. If f D 0, then we consider the normal abelian subgroup M D ha1 ; a3 ; : : : ; ar i of pk l G. Since a2h D a1 t p , h 6 0 .mod p/, we have G D hM; t i. For each x 2 M , m1

p l

m1

p mk

x t D x 1Cp a1kx p , where (8) gives a1p D .a2e2 t f0 /p 2 Z.G/. This case is reduced to Case (I). Therefore we assume in the sequel that f 1. k f l The element t centralizes t p D a1p a2hp , where h 6 0 .mod p/, k 1, f 1. On the other hand, t normalizes ha1 i and ha2 i and so (since fa1 ; a2 g is a pk pf basis of the subgroup ha1 ; a2 i) t centralizes ha1 i and ha2 i. This implies that t p s

p k1

centralizes ha1

l1

p f 1

i and ha2

i. We have t p

p k1

centralizes ha1 l 2. Thus t p We consider an element (10)

1

p f 1

; a2

g D tp

l1

l1

2 ht p i since l > k 1 and so

i.

xp k1 yp f 1 a2

a1

and show that there exist integers x 6 0 .mod p/ and y 6 0 .mod p/ such that l1 ha1 ; a2 i contains elements of order p.) Note that g p D 1. (Hence the coset t p k1 l1 xp yp f 1 p centralizes ha1 ; a2 i and from (9) follows t (11)

l

p k hp f

t p a1 a2

k1

D 1: f 1

k1

f 1

We get from (10) in any case g p D t p .a1xp a2yp /p , and so if Œa1p ; a2p D 1, then we take x D 1 and y D h so that (with the help of (11)) follows g p D k f k1 f 1 l t p a1p a2hp D 1. If Œa1p ; a2p ¤ 1, then k D f D 1 (since ha1 ; a2 i is minimal nonabelian) and Œa1 ; a2 p D 1. If p > 2, then ha1 ; a2 i is p-abelian and so setting k f l again x D 1 and y D h we get g p D t p a1p a2hp D 1. However, if p D 2, then m1 Œa1 ; a2 D a12 is an involution in Z.G/ and here we set x D 1 2m2 (noting that l

272

Groups of prime power order

in this case m > 2 according to Proposition 73.9) and y D h. We obtain in that case also l

m2

g 2 D t 2 .a112

m1 C2m1

l

m2 2.12m2 / 2h a2 Œa2 ; a1 .12 /h

l

a2h /2 D t 2 a1

D t 2 a122

a22h D 1;

m2

m1

. since Œa2 ; a1 .12 /h D Œa2 ; a1 D a12 p lk ; a1 i noting that l > k 1. Using ProposiWe consider the subgroup K D ht l1 p k1 tion 73.13 and (9), we get Ãk1 .K/ D ht p ; a1 i and l

p k hp f a2

pk

Ãk .K/ D ht p ; a1 i D ha1

pk

pk

pf

; a1 i D ha1 ; a2 i Z.ha1 ; a2 i/:

Since K=Ãk .K/ is of exponent p k , Proposition 73.6 implies that K=Ãk .K/ is p k1 abelian. Take the element tp

(12)

l1

xp k1

a1

2 Ãk1 .K/

where x 6 0 .mod p/ is the integer from (10). Using Proposition 73.12, we see that lk there exists an element g D t ˛p a1ˇ (˛, ˇ integers) in K such that .g /p

(13)

k1

D tp

l1

xp k1

a1

:

On the other hand, using the fact that K=Ãk .K/ is p k1 -abelian, we get (14)

.g /p

k1

D .t ˛p

lk

a1ˇ /p

k1

D t ˛p

l1

a1ˇp

k1

s;

where (15)

p k ıp f a2

s D a1

2 Ãk .K/ .; ı integers/:

Using (12), (14), and (15), we get (16)

xp k1

a1

D t .˛1/p

l1

p k1 .ˇ Cp/ ıp f a2

a1

:

l1

It follows that t .˛1/p 2 ha1 ; a2 i and so ˛1 0 .mod p/. We may set ˛1 D d (d integer), where ˛ 6 0 .mod p/ and together with (14) we get (17)

t .˛1/p

l1

D t dp D a1dp a2hdp : l

k

f

From (16) and (17) (noting that x 6 0 .mod p/) follows also ˇ 6 0 .mod p/ since fa1 ; a2 g is a basis of ha1 ; a2 i. k1 yp f 1 Substituting (13) in (15), we get g D .g /p a2 and so the element g of order p (not being contained in ha1 ; a2 i) is contained in the subgroup H D hg ; a2 i.

273

73 Classification of modular p-groups

On the other hand, H is modular with d.H / D 2 and so j1 .H /j D p 2 and therefore n1 n1 1 .H / D hg; a2p i, where 1 .ha2 i/ D ha2p i. From a1a2 D a11Cp s a21Cp

D compute a2t

we get

t ˛p

a2g D a2 p m1

since a1

lk

m1

˛p a2t

ˇ

a1

follows a2a1 D a2 a1p

lk

D

a2u

m1

a

m1

. From

with a certain integer u 6 0 .mod p/. Then we

ˇ

ˇ

D .a2u /a1 D .a2 1 /u D .a2 a1ˇp ˇ up m1

2 Z.G/. Hence a1

ˇ

and a2 1 D a2 a1ˇp

a

m1

/u D a2u a1ˇ up

m1

;

2 H and ˇu 6 0 .mod p/ implies a1

p m1

2

is a contradiction since

n1 hg; a2p i

\

m1 n1 H . Thus a1p 2 1 .H / D hg; a2p i. This p n1 i. The theorem is proved. ha1 ; a2 i D ha2

Minimal nonmodular p-groups are described in [Jan1]. Corollary 73.16 (v. d. Waall). Every non-Dedekindian modular 2-group has a characteristic subgroup of index 2. Proof. By Theorem A.24.4, G is Q8 -free. Now the result follows from Ward’s Theorem 56.1. As follows from v. d. Waall’s result, if p > 2 and G is nonabelian modular, it contains a characteristic subgroup of index p. Exercise. Let H be a powerful 2-group. If a 2-group G is lattice isomorphic with H via , then G is also powerful. (Hint. Use Proposition 73.5.)

74

p-groups with a cyclic subgroup of index p 2

The title groups have been determined in [Nin] (see also the old paper [HT1] by Hua and Tuan), where these groups are given in terms of generators and relations without any comments about its subgroup structure and therefore this result was not very useful for applications. We shall classify here the title groups in a structural form. If a group G of order 24 has no cyclic subgroups of index 4, it is elementary abelian. Therefore, it suffices to classify the 2-groups with cyclic subgroup of index 4, which have the order > 24 . Theorem 74.1. Let G be a nonabelian group of order p m , p > 2, m > 3, and exponent p m2 . Then one and only one of the following holds: (a) G is metacyclic, jG=Ã2 .G/j D p 4 . (b) m > 4, G is an L3 -group. (c) m D 4, G is regular, 1 .G/ is of order p 3 and exponent p. (d) m D 4, G is irregular, p D 3, G is of maximal class. Proof (Berkovich). Let Z < G be a cyclic subgroup of index p 2 in G. Suppose that G has a normal subgroup E of order p 3 and exponent p; then G D EZ with jE \ Zj D p. We know that exp.Aut.E//p D p and so jG W CG .E/j p. Therefore, if m > 4, then 1 .G/ D E and G is an L3 -group. Now let m D 4. If G is regular, then j1 .G/j 2 fp 2 ; p 3 g. In the first case, G is metacyclic (Lemma 64.1(a,m). In the second case, 1 .G/ is of exponent p (Lemma 64.1(a)). Next suppose that G is irregular. Then p D 3 (Lemma 64.1(a)). (Note that a 3-group of maximal class and order 34 has a cyclic subgroup of index 32 , since it is irregular.) In what follows we assume that m > 4 and G has no normal subgroups of order p 3 and exponent p. According to Theorem 13.7 (see also Theorem 69.4), G is either metacyclic or a 3-group of maximal class. Let us consider these possibilities. Let G be metacyclic. In that case, G is regular. Since G has no cyclic subgroups of index p, the subgroup Ã1 .G/ is noncyclic. Therefore, jG=Ã2 .G/j D p 4 . Let us show that Ã2 .G/ is cyclic. If not, G=Ã3 .G/ is of order p 6 and exponent p 3 , which is not the case since exp.G/ D p m2 . Conversely, if G is metacyclic with jG=Ã2 .G/j D p 4 and 2 Ã2 .G/ is cyclic, it has a cyclic subgroup of index p 2 since Ã2 .G/ D fx p j x 2 Gg (Lemma 64.1(a)).

74

p-groups with a cyclic subgroup of index p 2

275

Now let G be a 3-group of maximal class, m > 4. Then G1 , the fundamental subgroup of G, is metacyclic and exp.G1 / D exp.G/ (see 9). It follows that G1 has a cyclic subgroup of index 3. In that case, Ã1 .G1 / is cyclic of index p 2 in G1 so jÃ1 .G/j 32 . This is a contradiction since a p-group of maximal class and order > p 3 , p > 2, has no normal cyclic subgroups of order > p. Theorem 74.2. Let G be a nonabelian group of order 2m , m > 4, and exponent 2m2 . Then one of the following holds: (a) G is an L3 -group. (b) G is the uniquely determined group of order 25 with 2 .G/ Š D8 C2 . The group G has a normal elementary abelian self centralizing subgroup E D hz; u; ei of order 8 and an element a of order 8 such that G D Ehai, where a4 D z, e a D eu, and ua D z. Here ˆ.G/ D ha2 ; ui is abelian of type .4; 2/, G 0 D 1 .ˆ.G// Š E4 , Z.G/ D Ã1 .ˆ.G// is of order 2, and c3 .G/ D 4. (c) G is a U2 -group (all such groups are completely determined in 67). If m > 5, then cm2 .G/ D 2. (d) G D W Z with W \ Z Š C4 , where W is an abelian normal subgroup of type .4; 4/ and Z Š C2m2 . We have W D 2 .G/ and G is metacyclic. Also, 2 jZ W CZ .W /j 4, (since exp.Aut.W //2 4; see 33) and cm2 .G/ D 4. (e) G D QZ, where Q Š Q8 is a normal subgroup of G, Q\Z D Z.Q/, Z D hbi Š C2m2 and b either centralizes Q or b induces on Q an involutory outer automorphism in which case m > 5. Also we have cm2 .G/ D 4. (f) G is the uniquely determined group of order 25 with 2 .G/ D ha; bi hui, where ha; bi Š Q8 and u is an involution with CG .u/ D 2 .G/. Let hzi D Z.Q/; then a2 D b 2 D z. There is an element y of order 8 in G such that y 2 D ua, uy D uz, ay D a1 , b y D bu. Here ˆ.G/ D hy 2 ; ui is abelian of type .4; 2/, G 0 D 1 .ˆ.G// Š E4 , Z.G/ D Ã1 .G/ is of order 2, and c3 .G/ D 4. Proof (Janko). Suppose that G has a normal subgroup E Š E8 ; then G D EZ with Z Š C2m2 and E \ Z Š C2 . If K=E is the subgroup of order 2 in G=E, then 1 .G/ K. Therefore, if K is abelian, we get 1 .G/ D E, and so G is an L3 group. Now let CG .E/ D E. Since exp.Aut.E//2 D 4, we get jGj D 25 and Z D hai is of order 8. In that case, jZ.G/j D 2. Therefore, setting z D a4 , one can choose e; u 2 E so that E D he; u; zi; e a D eu and ua D uz. The structure of G is determined and we get the group (b) stated in the theorem. From now on we assume that G has no normal elementary abelian subgroups of order 8. Since G is neither cyclic nor of maximal class, it has a normal four-subgroup W0 . Suppose that G=W0 is cyclic (of order 2m2 ), then G D W0 S with S Š C2m2 and W0 \ S D f1g since G has no cyclic subgroups of index 2. But then W0 1 .S / Š E8 is normal in G, contrary to our assumption. Hence, G=W0 is noncyclic. Let Z < G be cyclic of order 2m2 . Then W0 \ Z Š C2 and so W0 Z is maximal

276

Groups of prime power order

in G. It follows that G=W0 has a cyclic subgroup .W0 Z/=W0 of order 2m3 4 and index 2. Set F=W0 D ˆ.G=W0 / so that F < W0 Z and F=W0 is cyclic of order 2. Since CW0 Z .W0 / F (indeed, jW0 Z W CW0 Z .W0 /j 2), it follows that 1 .W0 Z/ D W0 . Let M=W0 be any cyclic subgroup of index 2 in G=W0 . Since M > F and 1 .M / 1 .F / D W0 , we get 1 .M / D W0 . If G=W0 is of maximal class, then G is an U2 -group and all such groups have been determined in 67. Note, that if m > 5, then all cyclic subgroups of order 2m2 are contained in the L2 -subgroup W0 Z so c2m2 .G/ D c2m2 .W0 Z/ D 2. We may assume that G=W0 is not of maximal class. It follows that G=W0 is either abelian of type .2m3 ; 2/ or G=W0 Š M2m2 in which case m > 5 (Lemma 64.1(t)). In any case, G=W0 has exactly two cyclic subgroups M1 =W0 and M2 =W0 of index 2 and the third maximal subgroup M3 =W0 is abelian of type .2m3 ; 2/. Hence M3 cannot contain cyclic subgroups of index 2. But 1 .Mi / D W0 and so Mi is either abelian of type .2m2 ; 2/ or Mi Š M2m1 , i D 1; 2. In any case, cm2 .Mi / D 2 and two cyclic subgroups of order 2m2 in Mi generate Mi , i D 1; 2. Since exp..M1 \ M2 /=W0 / D 2m4 , it follows that M1 \ M2 does not contain any cyclic subgroup of order 2m2 . We get (see the proof of Theorem 1.10) cm2 .G/ D cm2 .M1 / C cm2 .M2 / D 2C2 D 4 and G is generated by its four cyclic subgroups of order 2m2 . We are now in a position to use Theorem 54.2. The groups (d) and (a) of our theorem correspond to possibilities (a) and (b) of Theorem 54.2, respectively. In case (c) of Theorem 54.2, we have here only the possibility jGj D 25 with 2 .G/ Š Q8 C2 (since in case 2 .G/ D D8 C2 the group G would have a normal elementary abelian subgroup of order 8). This leads to the group (f) of our theorem, and we are done. Exercise. Let p > 2 and let G be a nonmetacyclic group of order p m > p 4 with nonnormal cyclic subgroup L of order p m2 . Suppose that G is not minimal nonabelian. Prove that G has a maximal subgroup H Š Mp m1 . Solution. We have p > 2. The quotient group G=LG is isomorphic to a Sylow psubgroup of the symmetric group Sp 2 so LG is cyclic of order > p since m > 4. It follows from Theorem 9.6 that G is not a p-group of maximal class. Then, by Theorem 69.3, G has a normal subgroup R of order p 3 and exponent p. Since G has a cyclic subgroup of index p 2 , then 1 .G/ D R, G=R is cyclic so exp.k .G// p k for all k 2 N. By Proposition 10.28, G is generated by minimal nonabelian subgroups. It follows that G has a minimal normal subgroup H of exponent p m2 . By hypothesis, H < G. Then H has a cyclic subgroup of index p, and the result follows from Theorem 1.2.

75

Elements of order 4 in p-groups

All results of this section are due to the second author. If G is a finite p-group and n 2 N a fixed natural number, then we define n .G/ D hx 2 G j o.x/ D p n i. It is a known fact that 2 .G/ has a strong influence on the structure of a finite 2-group G. For example, if 2 .G/ is metacyclic, then the 2-group G is also metacyclic (this follows from the classification of minimal nonmetacyclic p-groups; see Theorem 66.1). In 52, all 2-groups G with the property j2 .G/j D 24 are classified. If j2 .G/j 23 , then we get for a 2-group G four infinite classes of groups (see Lemma 42.1). In this section we consider for the first time the case, where for 2 .G/ we could have an infinite class of 2-groups. More precisely, we assume that G is a finite 2-group with 2 .G/ Š C2 D, where D is any 2-group of maximal class. Then either G D 2 .G/ or jG W 2 .G/j D 2 and the structure of G is uniquely determined (Theorem 75.1). In fact, the most difficult part of the proof is to show that jG W 2 .G/j D 2. The exact determination of the structure of G is obtained by applying a classification of so called U2 -groups given in 67. A 2-group H is said to be a Us -group (s 2) with respect to the kernel R if H has a normal elementary abelian subgroup R of order 2s , H=R is of maximal class and whenever T =R is a cyclic subgroup of index 2 in H=R, then 1 .T / D R. We shall determine also the structure of a finite 2-group G with 2 .G/ Š C2 Q2n , where Q2n is a generalized quaternion group of order 2n , n 4 (see ‘Research problems and themes I’, #563). If 2 .G/ > 2 .G/, then we show that G D 2 .G/ Š C2 SD2nC1 , where SD2nC1 is the semidihedral group of order 2nC1 (Theorem 75.2). The case n D 3 was treated in 55. Finally, we show that a finite p-group G, all of whose noncyclic subgroups H have the property H D 1 .H /, is either cyclic or of exponent p or p D 2 and G is a dihedral group D2n , n 3 (Theorem 75.3). Here the case p > 2 is almost trivial but some proving must be done in case p D 2; see also Exercise 1.113. Theorem 75.1. Let G be a 2-group with 2 .G/ D hui D, where u is an involution, D is any 2-group of maximal class, and jDj D 2n , n 4. If G ¤ 2 .G/, then jG W 2 .G/j D 2, D is dihedral or generalized quaternion, CG .u/ D 2 .G/, G is a nonmetacyclic U2 -group with respect to the kernel R D hui Z.D/, G=R Š SD2n , n1 Z.G/ D Z.D/, and d.G/ D 2. More precisely, we set D D ha0 ; b j .a0 /2 D n2 n3 1; .a0 /2 D z; .a0 /2 D v; b 2 D z ; D 0; 1; .a0 /b D .a0 /1 i, and then

278

Groups of prime power order

there is an element a 2 G 2 .G/ such that a2 D a0 , ua D uz, ab D a1 vu, and this determines the structure of G uniquely. (For n < 4, such groups G have been determined in Lemma 42.1.) Proof. Set H D 2 .G/ so that H D hui D with jDj D 2n , n 4. Let Z be the unique cyclic subgroup of index 2 in D, where jZj D 2n1 8. Since D has no normal four-subgroups, it follows that G has no normal elementary abelian subgroups of order 8. We know that D has no normal abelian subgroups of type .4; 2/ and so H (and also G) has no normal abelian subgroups of type .4; 4/. Set hzi D 1 .Z/ so that hzi D Z.D/. Let hvi be the cyclic subgroup of order 4 in Z, where v 2 D z. Since hvi Ã1 .H / D Ã1 .Z/, it follows that hvi is normal in G and so z 2 Z.G/. Obviously, W0 D hui hzi D Z.H / and so W0 is normal in G. Therefore, W D W0 hvi is normal in G, W0 D 1 .W /, and W is a normal abelian subgroup of G of type .4; 2/. We see that W0 is the unique normal four-subgroup in H and then W is the unique normal abelian subgroup of H of type .4; 2/. It follows that W0 is the unique normal four-subgroup in G and W is the unique normal abelian subgroup of G of type .4; 2/. We are now in a position to use Proposition 50.6(a). It follows that N D CG .W / is abelian of type .2m ; 2/, where m n 1, N \ H D W0 Z D hui Z, G=N is isomorphic to a subgroup of D8 , and W D 2 .N /. Let r be an element in D Z so that r 2 2 hzi and v r D v 1 . Hence r induces an involutory automorphism on the abelian group N and r inverts W D 2 .N /. This implies that CN .r/ D W0 D 1 .W / and so, using Theorems 67.1, 67.2 and 67.3, we see that r inverts N=W0 . Suppose that N ¤ W0 Z. Take an element x 2 N .W0 Z/. We have x r D x 1 w0 with some w0 2 W0 . But then .rx/2 D rxrx D r 2 x r x D r 2 x 1 w0 x D r 2 w0 2 W0 , and so o.rx/ 4. This is a contradiction since rx 62 H D 2 .G/. We have proved that N D W0 Z. Since hvi is normal in G, v r D v 1 , and ur D u for any r 2 D Z, it follows that CG .v/ covers G=H . If G ¤ H , there is s 2 CG .v/H with us D uz, v s D v, so that G=.W0 Z/ is a four-group of automorphisms of W induced with hs; ri. From now on we assume G ¤ H . In that case jG W H j D 2, CG .W / D W0 Z, G=.W0 Z/ Š E4 , and for each x 2 G H , ux D uz so that CG .u/ D huiD D H . It remains to determine the structure of G. Since S D CG .v/ covers G=H and CH .v/ D W0 Z, it follows that S is a nonabelian maximal subgroup of G with S \ H D W0 Z. Hence 2 .S / D W and so we may apply Lemma 42.1 for p D 2. It follows that n1 n2 either S Š M2nC1 or S D ha; b j a2 D b 8 D 1; ab D a1 ; a2 D b 4 i, n 4. In the second case, hb 2 i D hvi D Z.S /; ha2 i D S 0 ; ˆ.S / D Z.S /ha2 i; 1 .S / D 1 .W / D hu; zi; n2

and CS .1 .S // D hb 2 ; ai is the unique abelian maximal subgroup of S z D a2 and so hb 2 ; ai D W0 Z D hui Z. Since W0 Z contains exactly two cyclic subgroups Z1 , Z2 of index 2, one of them, say Z1 D hai, is inverted by the element b 2 S .W0 Z/. But Z1 \ Z2 hvi and so b inverts hvi, contrary to S D CG .v/.

75 Elements of order 4 in p-groups

279

We have proved that S Š M2nC1 . In particular, S has a cyclic subgroup S1 of index 2 containing hvi and so S=W0 is a cyclic subgroup of index 2 in G=W0 . But G=W0 contains DW0 =W0 D H=W0 Š D=hzi, where D=hzi is a group of maximal class. Thus G=W0 is of maximal class and order 2n (n 4) and so S=W0 is the unique cyclic subgroup of index 2 in G=W0 . Since 1 .S / D W0 , it follows that G is a U2 -group with the kernel W0 . We are now in the position to use the classification of U2 -groups given in 67 (see Theorems 67.1 and 67.3). It follows that only the group G given in Theorem 67.3(d) satisfies our assumptions: n

n1

n2

G D ha; b j a2 D 1; a2

D z; a2

D v; b 2 D z ; D 0; 1;

ab D a1 vu; u2 D Œu; b D 1; ua D uz; n 4i; where W0 D hu; zi, G=W0 Š SD2n , Z.G/ D hzi Š C2 , G is non-metacyclic, and d.G/ D 2. Also, CG .u/ D hui D D 2 .G/, where D D ha2 ; bi. For D 0, D is dihedral and for D 1, D is generalized quaternion. Our theorem is proved. Of course, groups of Theorem 75.1 have no normal elementary abelian subgroups of order 8. Therefore, to prove that theorem, we must find all groups of 50 satisfying the hypothesis of Theorem 75.1. However, such proof will be not easier than the presented proof. Theorem 75.2. Let G be a 2-group with 2 .G/ D Q2n C2 , where n 4. If 2 .G/ ¤ 2 .G/, then 2 .G/ D G Š SD2nC1 C2 . (If n D 3, then such groups G have been determined in Theorem 55.1.) Proof. Set H D 2 .G/ D hui Q, where u is an involution and Q Š Q2n , n 4. Assume 2 .G/ ¤ H . Let hai be the unique cyclic subgroup of index 2 in Q and we set n1

Q D ha; b j a2

n2

D b 4 D 1; a2

n3

D z; a2

D v; b 2 D z; ab D a1 i;

where Q1 D ha2 ; bi and Q2 D ha2 ; abi are the other two maximal subgroups of Q and Q1 Š Q2 Š Q2n1 . Also, Z.H / D hu; zi is normal in G. Since hvi ha2 i D Ã1 .H /, hvi and ha2 i are normal subgroups in G and hzi Z.G/. Since 2 .G/ ¤ H , there exists an involution t 2 G H and we set K D H ht i. If ut D uz, then Z.H /ht i Š D8 and so o.ut / D 4, a contradiction since the element ut of order 4 is not contained in H D 2 .G/. Hence t centralizes Z.H /. Since t cannot commute with any element of order 4 in H , we get CH .t / D Z.H / D hu; zi. We act with t on the abelian group hui hai and apply 55. It follows that t inverts the quotient group hu; ai=Z.H / and so at D a1 s with s 2 Z.H /. We compute .t a/2 D .t at /a D a1 sa D s, and so s D 1 (since t a cannot be of order 4 in view t a 62 H ) and at D a1 . Hence t inverts the maximal subgroup hu; ai of H containing Z.H /.

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Groups of prime power order

Suppose that .Q1 Z.H //t D Q1 Z.H /. In that case b t D ba2i uj , where i , j are suitable integers. Working in Q, there is ak 2 hai (k is an integer) such that k k k b a D ba2i and so b ta D .ba2i uj /a D ba2i a2i uj D buj , where t 0 D t ak is an involution in K H . We compute .t 0 b/2 D .t 0 bt 0 /b D buj b D uj z, and so o.t 0 b/ D 4, a contradiction. Hence .Q1 Z.H //t D Q2 Z.H /. It follows b t D aba2r us (r; s are some integers) and so hb; b t i D Q Š Q Š Q2n with K D hui .Q ht i/, where Q ht i Š SD2nC1 . Let N D NG .Q1 Z.H // so that jG W N j D 2, H N , and G D N ht i. Suppose N ¤ H . By the above argument, N H cannot contain involutions and so 2 .N / D H . By Theorem 75.1, we get jN W H j D 2 and N is a uniquely determined group. In particular, by the structure of N , if y 2 N H , .Q1 Z.H //y D Q2 Z.H /, contrary to N D NG .Q1 Z.H //. Hence, N D H and so G D H ht i D K and we are done. Let G be a 2-group. Of course, 2 .2 .G// D 2 .G/. Therefore, if 2 .G/ D C2 D, where D is of maximal class and order > 8, then D is a generalized quaternion group. Remark. Let, as in Theorem 75.1, 2 .G/ D C2 D, where D is a 2-group of maximal class. If D is dihedral, then c2 .G/ D 2, and such groups are classified (see 43). It follows from the results of 54 that the 2-groups G with 2 .G/ D E4 D, where D is dihedral, also known. Similarly, if n > 3 and n1 .G/ D C2 M2n or n1 .G/ D C2 M2n , then cn1 .G/ D 4, and the classification of such groups follows from results of 55. Theorem 75.3. Let G be a p-group all of whose noncyclic subgroups are generated by its elements of order p. Then one of the following holds: G is cyclic, G is of exponent p, G is a dihedral 2-group D2n of order 2n . Proof. Suppose that p > 2 and assume that G is neither cyclic nor of exponent p. Let Z be a maximal cyclic subgroup of order p s , s 2. Since Z ¤ G, there is a subgroup Y > Z with jY W Zj D p. It follows that Y is either abelian of type .p s ; p/ or Y Š Mp sC1 . In any case, 1 .Y / Š Ep 2 so 1 .Y / < Y , a contradiction. From now on assume p D 2. We may also assume that G is not cyclic, exp.G/ 4, and G is not of maximal class. Let R be a normal four-subgroup of G. If CG .R/ has an element v of order 4, then Rhvi contains an abelian subgroup of type .4; 2/, a contradiction. Hence H D CG .R/ is elementary abelian and jG W H j D 2. We have exp.G/ D 4 and let s 2 G H be an element of order 4 so that s 2 2 H . Since G has no abelian subgroups of type .4; 2/, it follows that hsi is a maximal abelian subgroup of G. Then, by Suzuki (Proposition 1.8), G is of maximal class, contrary to the assumption. Problem. Classify the 2-groups G satisfying one of the following conditions: (i) 2 .G/ D E D, where E is elementary abelian and D is of maximal class. (ii) 2 .G/ D E D, where E is elementary abelian and D is of maximal class.

75 Elements of order 4 in p-groups

(iii) n1 .G/ D E M2n , where n > 3 and E is elementary abelian. (iv) n .G/ D E C2n , where n > 1 and E is elementary abelian. (v) n .G/ D E C2n , where n > 1 and E is elementary abelian.

281

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p-groups with few A1-subgroups

1o . Recall that G is an An -group if it contains a nonabelian subgroup of index p n1 but all its subgroups of index p n are abelian (see 72). For example, a p-group G of maximal class and order p m is an Am2 -group since it contains a nonabelian subgroup of order p 3 (see Theorems 9.5 and 9.6). Let ˛n .G/ be the number of An -subgroups in a p-group G. If G is an An -group, then ˛i .G/ > 0 for i D 1; : : : ; n 1, ˛n .G/ D 1 and ˛j .G/ D 0 for j > n. For H < G, we set ˇ1 .G; H / D ˛1 .G/ ˛1 .H /. For example, if H < G is abelian, then ˇ1 .G; H / D ˛1 .G/. If H < G, then iH denotes the set of all members of the set i which contain H . Given H < G, let i .H / D fU < H j ˆ.H / U; jH W U j D p i g. If G is a p-group of maximal class containing an abelian subgroup of index p, then ˛1 .G/ D p m3 . It is fairly difficult to compute ˛1 .G/ even for groups with not very complicated structure. Below we compute ˛1 .G/ for three families of groups. Let X be a group and '2 .X/ the number of noncommuting ordered pairs .x; y/ 2 X X such that hx; yi D X (it follows that '2 .X/ > 0 if and only if X is nonabelian and two-generator). Let k.X/ be the class number of X. We claim that (see 2; this identity is due to Mann) X (1) '2 .H / D jXj.jXj k.X//: H X

Indeed, let fK1 ; : : : ; Kr g (here r D k.X/) be the set of conjugacy classes of X. Then the number of commuting ordered pairs of G equals (2)

r r X X X jXj D rjXj D jXjk.X/ . jCX .x/j/ D jKi j jKi j iD1 x2Ki

iD1

so the number of noncommuting ordered pairs of elementsP of X is identical with 2 jXj jXjk.X/. On the other hand, that number also equals H X '2 .H /. Examples. 1. Let us find ˛1 .G/, where G is extraspecial of order p 2mC1 . We have k.G/ D jZ.G/j C jGZ.G/j D p 2m C p 1. Let E G be nonabelian and twop 0 0 generator. Then E D G and E=E 0 Š Ep 2 since E=E 0 G=E 0 D G=G 0 Š Ep 2m . It follows that jEj D p 3 so E is an A1 -subgroup. We have '2 .E/ D .jE ˆ.E/j/.jEj pjˆ.E/j/ D .p 3 p/.p 3 p 2 / D p 3 .p 2 1/.p 1/:

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283

It follows from (1) that ˛1 .G/'2 .E/ D ˛1 .G/p 3 .p 2 1/.p 1/ D p 2mC1 .p 2mC1 p 2m p C 1/ D p 2mC1 .p 1/.p 2m 1/; and we get ˛1 .G/ D p 2m2 ˛1 .G/ D p 2 .p 2 C 1/.

p 2m 1 . p 2 1

In particular, if jGj D p 5 , i.e., m D 2, then

2. Let G be a nonabelian group of order p m all of whose nonabelian two-generator m3 .p m k.G// subgroups have the same order p 3 . Then, using (1), we get ˛1 .G/ D p .p1/.p 2 1/ . 3. Let G D S Ep n , where S is nonabelian of order p 3 . Then jGj D p nC3 , k.G/ D p n .p 2 C p 1/. If A < G is an A1 -subgroup, then jAj D p 3 . Therefore, by the n nC3 n .p 2 Cp1/ D p 2n . formula in Example 2, we get ˛1 .G/ D p Œp .p 2p 1/.p1/ Epimorphic images of an An -group are Ak -groups with k n. Indeed, if N G G, where G is an An -group, then all subgroups of index p n in G=N are abelian. The main result of this section is the following Theorem A ([Ber30]). Let G be a nonabelian p-group. If ˛1 .G/ p 2 C p C 1, then G is an An -group, n 2 f1; 2; 3g. Exercise 1. Prove that if G is metacyclic, distinct F; H 2 1 are Ar -, As -groups respectively, and r s, then G is an AsC1 -group. Solution. Since G 0 is cyclic, we get F 0 H 0 . It follows from the above, Lemma 64.1(u) that jG 0 j D pjH 0 j D p sC1 so G is an AsC1 -group, by Theorem 72.1. Exercise 2. Let G be a nonabelian p-group. Suppose that its maximal subgroups M1 ; : : : ; Mk contain all A1 -subgroups of G. Is it true that k p? Exercise 3. Suppose that maximal subgroups M1 ; : : : ; MT p together contain all A1 p subgroups of a nonabelian p-group G. Is it true that jG W . iD1 Mi /j D p 2 ? Exercise 4. Let G D M C , where M is nonabelian of order p 3 and C is cyclic of order p 2 . Prove that G has an A1 -subgroup of order p 4 and find the number of such subgroups in G. 2o . Proof of Theorem A. We begin with the following Remark 1. Let G be neither abelian nor an A1 -group and let F1 2 1 be nonabelian. We claim that ˇ.G; F1 / p 1. By Exercise 1.6(a), the set 1 has at least p nonabelian members. Let F2 2 1 fF1 g be nonabelian and set D D F1 \ F2 . Since D — Z.G/, at least p members of the set 1D , say F1 ; : : : ; Fp , are nonabelian. Let Ui Fi be an A1 -subgroup such that Ui — D, i D 1; : : : ; p (such Ui exists, by Proposition 10.28). Then U1 ; : : : ; Up are distinct A1 -subgroups of G and U2 ; : : : ; Up — F1 so ˇ1 .G; F1 / p 1. (For more detailed description of such groups, see Lemma 76.5.)

284

Groups of prime power order

Lemma 76.1 (= Exercise 1.8a (L. Redei)). Let G be an A1 -group. Then G D ha; bi, Z.G/ D ˆ.G/, jG 0 j D p and one of the following holds: m

n

m

n

(a) ap D b p D c p D 1, Œa; b D c, Œa; c D Œb; c D 1, jGj D p mCnC1 , G D hbi .hai hci/ D hai .hbi hci/ (semidirect products with kernels in brackets) is nonmetacyclic. Here G 0 D hci, Z.G/ D ˆ.G/ D hap i hb p i hci. If m C n > 2, then 1 .G/ D h1 .hai/; 1 .hbi/; ci Š Ep 3 . m1

, jGj D p mCn is metacyclic. Here (b) ap D b p D 1, m > 1, ab D a1Cp m1 m1 n1 0 p p p i, Z.G/ D ˆ.G/ D ha i hb i, 1 .G/ D hap ; b p i Š Ep 2 . If G D ha 1 .G/ — Z.G/, then n D 1 so G Š Mp mC1 . (c) a4 D 1, a2 D b 2 , ab D a1 , G Š Q8 . If j1 .G/j p 2 , then G is metacyclic. The group G is nonmetacyclic if and only if G 0 is a maximal cyclic subgroup of G. Next, jG=Ã1 .G/j p 3 with equality if and only if G is from (a) and p > 2. If, in (a), u 2 G ˆ.G/, then hui 6E G. All members of the set 1 have ranks at most 3. If N is normal in G and G=N is not cyclic, then N Z.G/. In what follows G is a nonabelian group of order p m . Let us prove that if, in Lemma 76.1, G 0 is not a maximal cyclic subgroup of G, then G is metacyclic. Let G 0 < L < G, where L is cyclic of order p 2 . By Theorem 6.1, L=G 0 C =G 0 , where C =G 0 is a cyclic direct factor of G=G 0 ; in particular, G=C is cyclic. It remains to show that C is cyclic. We get G 0 D ˆ.L/ ˆ.C /. It follows that C =ˆ.C / as an epimorphic image of a cyclic group C =G 0 , is cyclic. In that case. C is also cyclic, as was to be shown. In what follows G is a nonabelian group of order p m . Lemma 76.2. Suppose that a p-group G is neither abelian nor an A1 -group and 1 D fH1 ; : : : ; Hp ; Ag, where A is abelian. Then H10 D D Hp0 and G=H10 is an A1 group so jG 0 j D pjH10 j. In particular, d.Hi / 3 for all i . If, in addition, d.Hi / D 2 for i D 1; : : : ; p, and L is a G-invariant subgroup of index p in H10 , then G=L is an A2 -group. Proof. By Remark 1, H1 ; : : : ; Hp are nonabelian. Let jH10 j jHp0 j. Since .H1 =H10 / \ .A=H10 / Z.G=H10 /, we get G=H10 =Z.G=H10 / Š Ep 2 so all maximal subgroups of G=H10 must be abelian so H10 D D Hp0 since d.G/ D 2. Assume that G=H10 is abelian; then G 0 D H10 . Let L be a G-invariant subgroup of index p in G 0 . Then, by Lemma 65.2(a), G=L is an A1 -group since d.G=L/ D 2, so H10 L < G 0 D H 0 , a contradiction. Thus, G=H10 is an A1 -group so jG 0 j D pjH10 j. By Lemma 76.1, d.Hi / D d.Hi =H10 / 3. Next suppose that d.Hi / D 2 for i D 1; : : : ; p and L is taken as above; then Hi =L is an A1 -group (Lemma 65.2(a)) so G=L is an A2 -group. Lemma 76.3. If all members of the set 2 are abelian, then d.G/ 3. If, in addition, d.G/ D 3, then ˆ.G/ Z.G/.

76 p-groups with few A1 -subgroups

285

Proof. Suppose that d.G/ > 3. Take F 2 1 ; then F contains at least p 2 C p C 1 (abelian) members of the set 2 so it is abelian (Exercise 1.6(a)). In that case, since G is not two-generator, it is abelian (Lemma 76.1), contrary to the hypothesis. Now suppose that d.G/ D 3. Then CG .ˆ.G// hH j H 2 2 i D G, completing the proof. In what follows we make use of the following fact: If N is a normal subgroup of a p-group G, jG=N j > p 2 and G=N is generated by subgroups of index p 2 , whose inverse images in G are abelian, then N Z.G/. Indeed, CG .N / hH < G j N < H; jG W H j D p 2 i D G. Lemma 76.4. Let G be a nonmetacyclic A2 -group of order p m > p 4 . Then (a) d.G/ 3, jG 0 j p 3 , exp.G 0 / D p, G=Z.G/ is of order p 3 and exponent p. (b) ˛1 .G/ 2 fp; p C 1; p 2 ; p 2 C p; p 2 C p C 1g. (c) If ˛1 .G/ < p 2 , then p > 2, d.G/ D 2, and cl.G/ D 3. (c1) If ˛1 .G/ D p, then G=.G 0 \ Z.G// is an A1 -group, G 0 Š Ep 2 . (c2) If ˛1 .G/ D p C 1, then jGj D p 5 , G 0 D 1 .G/ Š Ep 3 , Z.G/ D K3 .G/ D Ã1 .G/ Š Ep 2 . (d) If ˛1 .G/ D p 2 , then d.G/ D 3, G=Z.G/ Š Ep 2 , jG 0 j D p. (e) If ˛1 .G/ D p 2 C p, then d.G/ D 3, G 0 Š Ep 2 , G 0 Z.G/ D ˆ.G/. (f) If ˛1 .G/ D p 2 C p C 1, then p D 2, jGj D 26 , G is special with G 0 D Z.G/ D ˆ.G/ D 1 .G/ Š E8 . Proof. (a) Two inequalities follow from Lemmas 76.1 and 64.1(u). Next, exp.G 0 / D p (Theorem 65.7(d)). If d.G/ D 3, then ˆ.G/ Z.G/, by the paragraph preceding the lemma. If p > 2, then jG=Ã1 .G/j > p 2 (Theorem 9.11) so G=Ã1 .G/ is generated by subgroups of index p 2 , and we get Ã1 .G/ Z.G/. Hence, the last assertion of (a) holds for p > 2. Now suppose that p D 2 and d.G/ D 2. However, this case does not occur according to Propositions 71.1 to 71.5. Indeed, the results of 71 imply that an A2 -group of order 2m > 24 with d.G/ D 2 must be metacyclic (this is also true for m 4, by Lemma 64.1(i)). Assertions (b) to (f) are direct consequences of 71. Indeed, if G is a group of Proposition 71.1, then ˛1 .G/ D p 2 , d.G/ D 3, G=Z.G/ Š Ep 2 , and jG 0 j D p. If G is a group of Proposition 71.3, then ˛1 .G/ D p, p > 2, d.G/ D 2, cl.G/ D 3, N D 2 and jGN 0 j D G 0 Š Ep 2 , and GN D G=.G 0 \ Z.G// is an A1 -group since d.G/ p (Lemma 65.2(a)). If G is a group of Proposition 71.4, then ˛1 .G/ D p 2 C p, d.G/ D 3, G 0 Š Ep 2 , and G 0 Z.G/ D ˆ.G/. If G is a group of Proposition 71.5 (a), then ˛1 .G/ D p 2 C p C 1, p D 2, d.G/ D 3, jGj D 26 , G is special with G 0 D Z.G/ D ˆ.G/ D 1 .G/ Š E8 ; moreover, G is isomorphic to a Sylow 2-subgroup of the Suzuki simple group Sz.8/. If G is a group of Proposition 71.5(b),

286

Groups of prime power order

then p > 2, d.G/ D 2, cl.G/ D 3, ˛1 .G/ D p C 1, jGj D p 5 , G 0 D 1 .G/ Š Ep 3 , Z.G/ D K3 .G/ D Ã1 .G/ Š Ep 2 . Our proof is complete since the A2 -groups of Proposition 71.2 are metacyclic. Let us give a proof, independent of 71, that if G is a nonmetacyclic A2 -group with ˛1 .G/ < p 2 , then cl.G/ D 3. It follows from Exercise 1.6(a) that d.G/ D 2. Assume that cl.G/ D 2. Then G 0 Z.G/. Since G is nonmetacyclic, we get exp.G 0 / D p, by Lemma 76.4(a). In that case, G is an A1 -group (Lemma 65.2(a)), a contradiction. It remains to prove that cl.G/ D 3. This is the case if jG 0 j D p 2 . Now let jG 0 j D p 3 (see Lemma 76.4(a)); then ˛1 .G/ D p C 1 and G is nonmetacyclic with all two-generator members of the set 1 since the latter has no abelian members (Lemma 64.1(u)). It follows that G=Ã1 .G/ is nonabelian of order p 3 and exponent p and Ã1 .G/ D K3 .G/ so jG W G 0 j D p 2 (Lemma 64.1(p)), and we get jGj D jG W G 0 jjG 0 j D p 5 . Since all subgroups of order p 3 that contain K3 .G/, are abelian (G is an A2 -group!) and generate G, we get K3 .G/ D Z.G/ since jG W Z.G/j > p 2 , and so cl.G/ D 3. Remarks. Suppose that A; B 2 1 are distinct. 2. If A is abelian and B an A1 -group, then jG=Z.G/j p 3 . If, in addition, A is the unique abelian member of the set 1 , then G=Z.G/ is either of order p 3 and exponent p or G=Z.G/ Š D8 . Indeed, jG 0 j p 2 (Lemmas 64.1(u) and 76.1) so jG W Z.G/j p 3 (Lemma 64.1(q)). If A is the unique abelian member of the set 1 , then jG W Z.G/j D p 3 so G=Z.G/ has at most one cyclic subgroup of index p, and the last assertion follows. 3. If A; B are A1 -subgroups, then jG=Z.G/j p 4 . Indeed, put D D A \ B. Then Z.A/ D ˆ.A/ ˆ.G/ A \ B D D and similarly Z.B/ < D so, comparing orders, we conclude that Z.A/ and Z.B/ are maximal in D (Lemma 76.1). It follows that U D Z.A/ \ Z.B/ has index at most p 2 in D and U Z.G/ since CG .U / AB D G, and we get jG=Z.G/j jG=U j jG=DjjD=U j p 4 . The following lemma is the key result. Lemma 76.5. Suppose that H is a nonabelian maximal subgroup of a p-group G. Then ˇ1 .G; H / p 1. If ˇ1 .G; H / D p 1, then the following holds: (a) d.G/ D 2, 1 D fH1 D H; H2 ; : : : ; Hp ; A D HpC1 g, where A is abelian and all members of the set 1 fH; Ag are A1 -groups. (b) H10 D D Hp0 is of order p. (c) G=H10 is an A1 -group so jG 0 j D p 2 . We also have d.H1 / 3. If G is nonmetacyclic, then G=H10 is also nonmetacyclic. (d) G=Z.G/ is either nonabelian of order p 3 and exponent p or G=Z.G/ Š D8 , Z.Hi / D Z.G/ and Hi =G 0 is cyclic, i D 2; : : : ; p. In particular, cl.G/ D 3. (e) If G 0 Š Cp 2 , then H2 ; : : : ; Hp ; A are metacyclic and G has no normal subgroups of order p 3 and exponent p. If p > 2, then G is metacyclic (in that case, G is an A2 -group, by Corollary 65.3).

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(f) If G 0 Š Ep 2 , then cl.G/ D 3. In what follows we assume that G is not an A2 -group: then jGj D p m > p 4 . Since jG 0 j D p 2 , G is nonmetacyclic. (g) d.H / D 3. In what follows we assume, in addition, that p D 2. Then, since G is not of maximal class, we get jG W G 0 j > 4, by Taussky’s theorem. (h) d.A/ D 2 so A is metacyclic. (i) G has no normal elementary abelian subgroups of order 8. (j) H2 is metacyclic. (k) U D 1 .H2 / D 1 .A/ < ˆ.G/ is the unique normal abelian four-subgroup of G. (l) If Z.G/ is cyclic, then H2 Š M2m1 and G 0 6 Z.G/. In that case, if GN D G=H 0 N Š E8 and T D Q L, where L Š C4 , Q Š Q8 , and TN D 1 .HN /, then TN D 1 .G/ 0 Q GG and G=Q is cyclic so G < Q is cyclic of order 4. Since 1 .G/ D T , it follows that G has exactly 7 involutions. Next, G=G 0 is abelian with cyclic subgroup of index 2 and G has no elementary abelian subgroups of order 8. (m) Suppose that Z.G/ is noncyclic. Then G 0 Š C4 is a maximal cyclic subgroup of G. Let T be as in (l). Then jT j D 16 and T D Q L. where Q is nonabelian of order 8. Next, H2 =G 0 is cyclic since G 0 6 Z.G/ D Z.H / D ˆ.H / and H2 has no cyclic subgroups of index 2. The quotient group A=G 0 is also cyclic. (n) G 0 is a maximal cyclic subgroup of G. Proof. The set 1 .H / \ 2 contains a member N ; then N 6 Z.G/ since H is nonabelian. Set 1N D fH1 D H; : : : ; Hp ; HpC1 g. Since at most one member of the set 1N is abelian, one may assume that H1 ; : : : ; Hp are nonabelian. By Remark 1, ˇ1 .G; H / p 1. Next we suppose that ˇ1 .G; H / D p 1. (a–c) If Ui Hi is an A1 -subgroup not contained in N (see Theorem 10.28), then U2 ; : : : ; Up are pairwise distinct. It follows that UpC1 does not exist so HpC1 D A is Then N D H \ A is also abelian, and we get p 1 D ˇ1 .G; H / Pabelian. p ˛ .H 1 i /. It follows that ˛1 .Hi / D 1 (i D 2; : : : ; p) so H2 ; : : : ; Hp are A1 iD2 groups and the subgroups H1 ; : : : ; Hp together contain all A1 -subgroups of G. Assume that F 2 1 1N . The intersection F \ Hi is abelian since Hi is an A1 subgroup, i D 2; : : : ; p. Therefore, if F is nonabelian, then all A1 -subgroups of F are contained in H1 . In that case, F H1 (Theorem 10.28), a contradiction. Thus, all members of the set 1 1N are abelian. Assuming that d.G/ > 2, we see that the set 1 has at least .p 2 C p C 1/ p D p 2 C 1 > p C 1 abelian members, contrary to Exercise 1.6(a). Thus, 1 D 1N D fH1 ; : : : ; Hp ; Ag so d.G/ D 2 and ˆ.G/ D N , completing the proof of (a). Now (b) follows from (a) and Lemma 76.2. By Lemma 76.2 again, G=H10 is an A1 -group and hence jG 0 j D pjH10 j D p 2 . In that

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case, d.H1 / D d.H1 =H10 / 3 (Lemma 76.1). The last assertion in (c) follows from Theorem 36.1. (d) It follows from d.G/ D 2 that jG W Z.G/j > p 2 . By Remark 2, G=Z.G/ is either nonabelian of order p 3 and exponent p or G=Z.G/ Š D8 since A is the unique abelian member of the set 1 . Now let i 2 f2; : : : ; pg. In view of d.G/ D 2, we get Z.G/ ˆ.G/ < Hi so Z.G/ Z.Hi /, and the equality jZ.G/j D jZ.Hi /j implies Z.G/ D Z.Hi / .D ˆ.Hi //. Therefore, since G 0 6 Z.G/, then Hi =G 0 is cyclic. (e) Suppose that G 0 Š Cp 2 . Then Hi is metacyclic since Hi0 is not a maximal cyclic subgroup in Hi in view of Hi0 < G 0 , i D 2; : : : ; p (Lemma 76.1). Assume that G has a normal subgroup R of order p 3 and exponent p. For i D 2; : : : ; p, we get Hi R D G so G=.R \Hi / D .Hi =.R \Hi //.R=.R \Hi //, and we conclude that Hi =.R \Hi / is cyclic; then G=.R \ Hi / is abelian. On the other hand, G=.R \ Hi / is nonabelian since Cp 2 Š G 0 6 R \ Hi Š Ep 2 , and this is a contradiction. Thus, R does not exist so A is metacyclic. Let p > 2. Since G 0 is cyclic, G is regular (Theorem 7.1(c)) so p 2 D j1 .G/j D jG=Ã1 .G/j whence G is metacyclic (Theorem 9.11). (f) If G 0 Š Ep 2 , then G 0 6 Z.G/ ((a) and Lemma 65.2(a)) so cl.G/ D 3. (g) By hypothesis, H is not an A1 -group (otherwise, G is an A2 -group). By Lemma 65.2(a), d.H / > 2. Since G=H 0 is an A1 -group, we get d.H / 3, and our claim is proved. In what follows we assume that p D 2 so jG W G 0 j > 4 since G is not of maximal class (Taussky). (h) By Theorem 71.6, the number of two-generator members in the set 1 is even. Since j1 j D 3 and d.H2 / D 2 < 3 D d.H /, by (g), we get d.A/ D 2, so A is metacyclic. Clearly, A is noncyclic. (i) Assume that E8 Š E GG. Then E \A D 1 .A/, by (h). We have CG .E \A/ D EA D G so E \ A Z.G/. Since E 6 A 2 1 and j1 j D 3, the number of maximal subgroups of G=E is < 3 so G=E is cyclic. In that case, A D L Z, where jLj D 2 and Z is cyclic so A=L is a cyclic subgroup of index 2 in G=L (we have L < 1 .A/ Z.G/). Since L < Z.G/ and G=Z.G/ Š D8 , G=L is of maximal class (Theorem 1.2). However, G=L has a normal subgroup E=L Š E4 and jG=Lj 24 , and this is a contradiction. Thus, E does not exist. (j) Assume that H2 is nonmetacyclic. Then 1 .H2 / Š E8 (Lemma 65.1) and 1 .H2 / G G, contrary to (i). (k) Since the two-generator 2-group G has no cyclic subgroups of index 2, the subgroup ˆ.G/ is noncyclic. We have H2 \ A D ˆ.G/ is noncyclic so U D 1 .H2 / D 1 .A/ Š E4 . Assume that V ¤ U is a G-invariant four-subgroup. Then V — H2 ; A so G=V is cyclic. Since G has no cyclic subgroups of index 2 and jGj > 24 , we get 1 .G/ D U V Š E8 , contrary to (i). Thus, V does not exist. Let GN D G=H 0 and TN D 1 .HN /; then TN Š E8 is normal in GN since HN is abelian N D TN so of rank 3, by (g). Since GN is minimal nonabelian, by (c), we get 1 .G/ 5 N TN Š G=T is cyclic since T 6 A; H2 , 1 .G/ T . Assume that jGj > 2 . Then G=

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by (h) and (j). Since d.T / D 3, it follows that T is nonabelian, by (i). By (k), U < T , where U is the unique normal abelian four-subgroup of G. Let Q < T be minimal nonabelian; then jQj D 8, T D QZ.T / (Lemma 64.1(i)) and Z.T / G G since T G G. (l) Suppose that Z.G/ is cyclic. Then the noncyclic abelian subgroup ˆ.G/ has a cyclic subgroup Z.G/ of index 2 (recall that 8jZ.G/j D jGj D 4jˆ.G/j and Z.G/ < ˆ.G/ since d.G/ D 2) so 1 .ˆ.G// D U 6 Z.H2 / (otherwise, CG .U / H2 A D G and E4 Š U Z.G/, contrary to the assumption) so H2 Š M2m1 (Lemma 76.1). Let T be defined as in the previous paragraph. Then Z.T / is cyclic (otherwise, since Z.T / G G, we get Z.T / D U , by (k), and CG .U / TA D G; then Z.G/ is noncyclic, a contradiction), so, by the product formula, H D QZ.G/, where Q Š Q8 (Appendix 16) since Q — A; H2 . Since Q is the unique subgroup isomorphic to Q8 in T (Appendix 16), we get Q G G, and then G=Q is cyclic since d.G/ D 2 and Q 6 A. In that case, G 0 < Q is of order 4 so G 0 Š C4 and G 0 6 Z.G/ D Z.H2 /. Then G=G 0 , as abelian group of type .jG=Qj; 2/, has exactly two cyclic subgroups H2 =G 0 and A=G 0 of index 2 (H=G 0 is not cyclic since d.H / D 3). Assume that E8 Š E < G. Then E \ A D 1 .A/ D U so CG .U / EA D G, contrary to the cyclicity of Z.G/. Thus, G has no elementary abelian subgroups of order 8. By the above and Appendix 16, T has exactly 7 involutions so G has also 7 involutions since 1 .G/ D T (see Theorem 43.10). (m) Suppose that Z.G/ is noncyclic; then U D 1 .Z.G//. We have G 0 ¤ U (otherwise, G is an A1 -group, by (a) and Lemma 65.2(a), a contradiction). Since U is the unique normal four-subgroup of G, we get G 0 Š C4 . Let T =H 0 D 1 .H=H 0 / be as in (l); then T =H 0 Š E8 (see the proof of (l)), T is nonabelian and T D QZ.T /, where Q is nonabelian of order 8 (Lemma 64.1(i)). As above, G=T is cyclic so G 0 < T . Since U < T \ Z.G/, we get T D Q L, where jLj D 2. We know that 1 .G/ D 1 .T /. Since Z.T / D U ¤ G 0 , we get G 0 6 Z.T / so G 0 6 Z.G/ hence cl.G/ D 3. Since U < H2 , H2 has no cyclic subgroups of index 2 (otherwise, 1 .H2 / D U Z.G/ so H2 is abelian). Next, H2 =G 0 is cyclic since G 0 6 Z.G/ D Z.H / D ˆ.H /. Since G=G 0 is abelian with cyclic subgroup of index 2, it follows that A=G 0 is cyclic (otherwise, H=G 0 is cyclic so H is metacyclic, which is a contradiction). (n) By (l) and (m), G 0 Š C4 . Assume that G 0 < Z, where Z Š C8 . Then < Z=H 0 so G 0 =H 0 is not a maximal cyclic subgroup of the A1 -group G=H 0 . It follows that G=H 0 is metacyclic (Lemma 76.1). In that case, G is metacyclic, by Theorem 36.1, which is a contradiction. Thus, G 0 < G is maximal cyclic. G 0 =H 0

Remarks. 4. Let G be a p-group and let members H1 ; : : : ; Hk , k > 1, of the set 1 contain together all A1 -subgroups of G. Suppose that there exists A 2 1 fH1 ; : : : ; Hk g such that A \ Hi is abelian for all i D 2; : : : ; k. Then A is abelian. Assuming that this is false, we can take an A1 -subgroup U A such that U 6 H1 (Proposition 10.28). Since, for i D 2; : : : ; k, also U 6 Hi (recall that A \ Hi is abelian, by hypothesis), we get a contradiction. 5. For a nonabelian p-group G, the following two conditions are equivalent: (a) G

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is of maximal class with abelian subgroup A of index p. (b) Whenever H G is nonabelian of order p k , then ˛1 .H / D p k3 . In the proof we use induction on jGj. Let us prove that (a) ) (b). By Fitting’s lemma, all members of the set 1 fAg are of maximal class so, by Hall’s enumeration principle and induction, ˛1 .G/ D

X

˛1 .H / D p p m13 D p m3 :

H 21 fAg

Next, let L < G be nonabelian. Then L H 2 1 and the subgroup H of maximal class has an abelian subgroup H \A of index p so, by induction, L is of maximal class and ˛1 .L/ D p l3 , where jLj D p l . Now we prove that (b) ) (a). In that case, all proper nonabelian subgroups of G are of maximal class, by induction. Take T G, where T is an A1 -subgroup of G. Setting jT j D p k , we get 1 D ˛1 .T / D p k3 so k D 3, i.e., all A1 -subgroups of G have the same order p 3 . Assume that G is not of maximal class. Then we get CH .T / 6 T (Lemma 64.1(i)). Let F be a subgroup of order p 2 in CH .T / such that Z.T / < F . Then jF T j D p 4 and ˛1 .F T / D p 2 ¤ p D p 43 , a contradiction. Thus, G is of maximal class. Next, let R G G be of order p 2 and set U D CG .R/; then jG W U j D p and, by induction, U is abelian since it is not of maximal class. 6. Let d.G/ D 3 and ˆ.G/ 6 Z.G/; then the set 2 has a nonabelian member N (Lemma 76.3). We claim that the set 2 contains at least p 2 nonabelian members. Let A1 ; A2 2 2 be distinct abelian; then A D A1 A2 2 1 contains exactly p C 1 members of the set 2 and all of them are abelian since ˆ.G/ D A1 \ A2 Z.A/. Assume that B 2 2 1 .A/ is abelian. Then CG .ˆ.G// AB D G, contrary to the hypothesis. Thus, 1 .A/ \ 2 is the set of all abelian members of the set 2 so the last set contains exactly j2 j .p C 1/ D p 2 nonabelian members. Thus, in our case, the set 2 has p 2 , p 2 C p or p 2 C p C 1 nonabelian members. 7. Let G be a nonabelian p-group with d D d.G/ > 3. Let T D fT1 ; : : : ; Tr g 2 and M D fM1 ; : : : ; Ms g 1 be the sets of nonabelian members of the sets 2 , 1 , respectively (T ¤ ¿, by Lemma 76.3, M ¤ ¿, by Exercise 1.6(a)). We claim that r > .p 1/s. Let i be the number of members of the set M, containing Ti , i D 1; : : : ; r, and let j be the number of members of the set T contained in Mj , j D 1; : : : ; s. Then, by double counting,

1 C C r D 1 C C s ; and, in view of i D p C 1 for all i , that formula can be rewritten as follows: (3)

.p C 1/r D 1 C C s :

Note that Mj contains exactly p d 2 C p d 3 C C p C 1 members of the set 2 . Since, by Exercise 1.6(a), the set 1 .Mj / contains at most p C 1 abelian members, we

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get j p d 2 C C p 2 for j D 1; : : : ; s, and we deduce from (3) the following inequality: .p C 1/r .p d 2 C C p 2 /s. We get r

p2 .p 2 1/ C 1 p d 2 C C p 2 s sD s > .p 1/s; pC1 pC1 pC1

as was to be shown. By Exercise 1.6(a), we have s j1 j.pC1/ D p d 1 C Cp 2 . 8. Given a set M of subgroups of a p-group G, let ˛1 .M/ denote the number of A1 -subgroups that contain together members of the set M. Suppose that G is neither abelian nor an A1 -group. We claim that if, for each K 2 2 , we have ˛1 .1K / pC1, then G is an A2 -group. By hypothesis, there is K 2 2 which is not contained in Z.G/. Indeed, this is the case if d.G/ D 2. If d.G/ > 2, our claim is obvious. Set 1K D fH1 ; : : : ; HpC1 g and let ˛1 .H1 / ˛1 .HpC1 /. Since at most one member of the set 1K is abelian, H1 ; : : : ; Hp are nonabelian. Assume that ˛1 .H1 / > 1; then ˛1 .H1 / D p, by Remark 1 and hypothesis. We get p C 1 ˛1 .1K / D ˛1 .H1 / C

pC1 X

ˇ1 .Hi ; K/ D p C

iD2

pC1 X

ˇ1 .Hi ; K/

iD2

so ˇ1 .H2 ; K/ D 1 and we get p D 2 and ˇ1 .H3 ; K/ D 0 (Remark 1) whence H3 is abelian, and we conclude that K is abelian; in that case, H2 is an A1 -group. Thus, all members of the set 2 are abelian so d.G/ 3 (Lemma 76.3). Assume that G is not an A2 -group. If d.G/ D 2, then 1 D 1K D fH1 ; H2 ; A/, where ˛1 .H1 / D 2, ˛1 .H2 / D 1 and A is abelian. In this case, ˇ1 .G; H1 / D 1 so jH10 j D 2 (Lemma 76.5), and the equality ˛1 .H1 / D 2 is impossible (Lemma 76.4(c)), a contradiction. Now assume that d.G/ D 3; then j1 .H1 / \ 2 j D p C 1 > 1 so jH10 j D 2 (Lemma 64.1(q)), and again we get a contradiction. Lemma 76.6. Suppose that ˛1 .G/ > 1. Then ˛1 .G/ p. (a) If ˛1 .G/ D p, then G is an A2 -group with d.G/ D 2 and jG 0 j D p 2 . If G is nonmetacyclic, it is of class 3. (b) If ˛1 .G/ D p C1, then G is an A2 -group with d.G/ D 2. If G is not metacyclic, then p > 2, G is of order p 5 with G 0 Š Ep 3 and of class 3. Proof. The inequality ˛1 .G/ p follows from Remark 1. By Remark 8, G is an A2 -group. All remaining assertions follows from Lemma 76.4(c). Lemma 76.7. Let N 2 2 and 1N D fH1 ; : : : ; HpC1 g. Then: PpC1 ˇ1 .Hi ; N /. (a) ˛1 .G/ ˛1 .N / C iD1 (b) If N is nonabelian, then ˇ1 .G; N / p 2 1. (c) If ˛1 .G/ < p 2 , then NG .K/=K has only one subgroup of order p for each nonabelian K < G.

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Proof. (a) is obvious, (b) follows from (a) and Remark 1. It remains to prove (c). Assume that K < G is nonabelian and NG .K/=K contains an abelian subgroup L=K of type .p; p/; then 1 C ˇ1 .L; K/ ˛1 .K/ C ˇ1 .L; K/ D ˛1 .L/ ˛1 .G/ p 2 1 so ˇ1 .L; K/ < p 2 1, contrary to (b). Thus, L=K does not exist so NG .K/=K has only one subgroup of order p so it is either cyclic or generalized quaternion. Lemma 76.8. Suppose that G is an An -group, n > 2. (a) If d.G/ > 2 then ˛1 .G/ p 2 . (b) If d.G/ D 3 and ˆ.G/ D Z.G/, then ˛1 .G/ 2p 2 C p 1. (c) If d.G/ D 3 and jG W Z.G/j D p 2 , then ˛1 .G/ 2p 2 1. Proof. (a) There is N 6 Z.G/ for P some N 2 2 since hM j M 2 2 i D G. If all such N are abelian, then ˛1 .G/ D H 21 ˛1 .H / p.j1 j .p C 1// p 3 > p 2 (Exercise 1.6(a) and Lemma 76.6). If N is nonabelian, then ˛1 .G/ p 2 (Lemma 76.7(b)). (b) If L 2 1 is nonabelian, then jL W Z.L/j D jL W Z.G/j D p 2 so jL0 j D p (Lemma 64.1(q)). Suppose that L is neither abelian nor an A1 -group (such an L exists since n > 2). Then, by Lemma 76.4, ˛1 .L/ ˛1 .H / p 2 , where H L is an A2 -subgroup. In view of jG W Z.G/j D p 3 , the set 1 has at most one abelian member (Lemma 64.(u)) so this set has at least p 2PC p nonabelian members, and, by Hall’s enumeration principle, ˛1 .G/ D ˛1 .L/ C M 21 fLg ˛1 .M / p 2 C p 2 C p 1 D 2p 2 C p 1 (we have here equality if and only if ˛1 .L/ D p 2 and all nonabelian members of the set 1 fLg are A1 -groups). (c) In our case, jG 0 j D p and all members of the set 2 are abelian, the set 1 has exactly p 2 nonabelian members (Remark 6). If L 2 1 is neither abelian nor an A1 -group, then ˛1 .L/ p 2 (see the proof of (b)) so we get ˛1 .G/ D ˛1 .L/ C P 2 2 2 M 21 fLg ˛1 .M / p C .p 1/ D 2p 1. Theorem 76.9. A p-group G with 1 < ˛1 .G/ < p 2 is a two-generator A2 -group. In particular, ˛1 .G/ 2 fp; p C 1g. Proof. Suppose that G is a counterexample of minimal order; then p 2 > ˛1 .G/ > p C 1 (Lemma 76.6(a,b)) so p > 2. By Lemma 76.8(a), d.G/ D 2 so ˆ.G/ > Z.G/. By Lemma 76.7(c), ˆ.G/ is abelian. By Proposition 10.28 and induction, G is an A3 group. Then there is H 2 1 , which is an A2 -group, and so, by Proposition 10.28, Lemmas 76.4 and 76.6, d.H / D 2, jH 0 j > p and ˛1 .H / 2 fp; p C 1g. (i) Assume that A; U 2 1 , where A is abelian and U is an A1 -group. Then G=Z.G/ is of order p 3 and exponent p (Remark 2), and Z.G/ < ˆ.G/ < H . In that case, jH W Z.G/j D p 2 so jH 0 j D p (Lemma 64.1(q)), contrary to what has been said in the previous paragraph. Thus, a pair .A; U / does not exist. Let F1 ; : : : ; Fs ; V1 ; : : : ; Vu be all nonabelian members of the set 1 , where ˛1 .Fi / D p and ˛1 .Vk / D 1 (since ˆ.G/ is abelian, the set 1 has no members H with ˛1 .H / D p C 1, by Lemma 76.4). Since the set 1 has at most one abelian member (Lemma 64.1(u)), we get s C u j1 j 1 D p.

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(ii) Assume that A 2 1P is abelian. Then u D 0, by (i), so s D p and, since ˆ.G/ 2 is abelian, p 2 > ˛1 .G/ D p iD1 ˛1 .Fi / D p p D p , a contradiction. Thus, all members of the set 1 are nonabelian, i.e., s C u D p C 1. As in the previous paragraph, s < p so u 2. Then G and all members of the set 1 are two-generator so G is either metacyclic or G=K3 .G/ is nonabelian of order p 3 and exponent p and then jG W G 0 j D p 2 (Lemma 64.1(p)). (ii1) Assume that G is metacyclic. Then V1 ; V2 2 1 are distinct A1 -subgroups so G is an A2 -group (Exercise 1), a contradiction. Thus, G is nonmetacyclic. (ii2) Since u > 1, then jG 0 j pjV10 V20 j D p 3 so jGj D jG W G 0 jjG 0 j p 5 . Since G is not an A2 -group, we get jGj D p 5 . In that case, G has a nonabelian subgroup of order p 3 . Since all nonabelian groups of order p 3 are A1 -groups, we get by Proposition 2.3, ˛1 .G/ p 2 , a final contradiction. Let G be a p-group of maximal class and order p 5 with abelian subgroup of index p. Then ˛1 .G/ D p 2 and G is an A3 -group. Remark 9. Let a p-group G be neither abelian nor an A1 -group. If, for each K 2 2 , we have ˛1 .1K / p 2 , then G is either an A2 - or A3 -group. Indeed, let H1 2 1 be nonabelian and take K 2 1 .H1 / \ 2 . Set 1K D fH1 ; : : : ; HpC1 g. Since at most one member of the set 1K is abelian, we get ˛1 .H1 / < p 2 , i.e., H1 is either A1 - or A2 -group (Theorem 76.9) so G is an An -group, n 2 f2; 3g. Theorem 76.10. If ˛1 .G/ D p 2 , then G is an An -group, n 2 f2; 3g. Next suppose that G is an A3 -group. Then jG 0 j D p 3 and ˛1 .H / 2 f1; pg for every nonabelian H < G. (a) If the set 2 has a nonabelian member N , then N is an A1 -group and one of the following holds: (a1) G is metacyclic with ˛1 .H / D p for all H 2 1 , i.e., all members of the set 1 are A2 -groups. (a2) d.G/ D 3, 1 D fF1 ; : : : ; Fp 2 Cp ; Ag, where ˛1 .Fi / D p for all i and A is abelian, Z.G/ D Z.N / is of index p 4 and the set 2 has exactly p 2 nonabelian members which are A1 -groups. (b) If all members of the set 2 are abelian, then d.G/ D 2 and one of the following holds: (b1) G is metacyclic, 1 D fH1 ; : : : ; Hp ; Ag, where ˛1 .Hi / D p for all i , H10 D D Hp0 , A is abelian. (b2) 1 D fH1 ; : : : ; Hp ; Ag, where ˛1 .Hi / D p, H10 D D Hp0 is of order p 2 , A is abelian, G 0 is noncyclic, G=H10 is an A1 -group. (b3) p D 2, 1 D fH1 ; H2 ; H3 g, ˛.H1 / D 2, ˛.H2 / D ˛.H3 / D 1. We have jG=Z.G/j D 16.

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Groups of prime power order

Proof. Suppose that G is not an A2 -group; then G is an A3 -group (Proposition 10.28 and Theorem 76.9). (a) Suppose that the set 2 has a nonabelian member N . Let 1N D fH1 ; H2 ; : : : ; HpC1 g; then all members of the set 1N are neither abelian nor A1 -groups. Using Remark 1, we get p 2 D ˛1 .G/ ˛1 .N / C

pC1 X

ˇ1 .Hi ; N / 1 C .p 1/.p C 1/ D p 2 ;

iD1

and hence ˛1 .N / D 1, ˇ1 .Hi ; N / D p 1, ˛1 .Hi / D p, d.Hi / D 2 and jHi0 j D p 2 for all i , H1 ; : : : ; HpC1 are A2 -groups (Remark 1, Lemmas 74.4 and 76.6(a)) and these p C 1 subgroups contain together all A1 -subgroups of G. If G is metacyclic, it is as in (a1). Next we assume that G is nonmetacyclic. We also have d.G/ d.H1 / C 1 D 3. Suppose, in addition, that d.G/ D 2 so N D ˆ.G/. Then G and all members of the set 1 are two-generator (Lemma 76.6(a)). Since G is nonmetacyclic, we get p > 2 and jG W G 0 j D p 2 (Lemma 64.1(p)). By the above and Lemma 76.6(a), Hi has the unique abelian maximal subgroup Ai , all i . Then the subgroups A1 ¤ A2 are normal in G; therefore, A D A1 A2 is of class at most 2 (Fitting’s lemma). By Lemma 64.1(l), G is not minimal nonmetacyclic so we may assume that H1 is not metacyclic. Since A > A1 and cl.G/ cl.H1 / > 2 (Lemma 76.4(c)), we get A 2 1 . Then A1 D H1 \ A D ˆ.G/ D N , a contradiction since N is nonabelian. Thus, d.G/ D 3. Then, by Remark 6, the set 2 has at least p 2 nonabelian members, say L1 ; : : : ; Lp 2 , and all Li are A1 -groups since G is an A3 -group and jG W Li j D p 2 . Since ˛1 .G/ D p 2 , all A1 -subgroups of G are members of the set 2 so each nonabelian H 2 1 is an A2 -group. If U1 ; : : : ; UpC1 are all abelian members of the set 2 , then A D U1 U2 2 1 , by the product formula, and jA0 j p (Lemma 64.1(q)) so A is not an A2 -group since ˛1 .A/ < ˛1 .G/ D p 2 (Lemma 76.4). It follows that A is abelian; moreover, A is the unique abelian member of the set 1 (Lemma 76.8(b)). Thus, 1 D fF1 ; : : : ; Fp 2 Cp ; Ag, where ˛1 .Fi / D p and jFi0 j D p 2 (Lemma 76.6). Next, in view of d.Fi / D 2, we get Z.Fi / < ˆ.Fi / D ˆ.G/ < A; then CG .Z.Fi // Fi A D G so Z.Fi / Z.G/ and, by Lemma 64.1(q), jFi W Z.Fi /j D pjFi0 j D p 3 . Thus, jG W Z.G/j p 4 . By Lemma 64.1(q), jG W Z.G/j D pjG 0 j pjF10 j p 3 . Since G=Z.G/ is noncyclic and A is the unique abelian member of the set 1 , it follows that Z.G/ Fi for some i . By the above, Z.Fi / Z.G/ so Z.G/ D Z.Fi /, and we get jG W Z.G/j D jG W Fi jjFi W Z.G/j D p 4 so jG 0 j D p1 jG W Z.G/j D p 3 and G is as stated in (a2) since, if N < Fi , then Z.N / D ˆ.N / < ˆ.G/ < A so CG .Z.N // AN D G hence Z.N / Z.G/ and jG W Z.N /j D p 4 D jG W Z.G/j. 2 P(b) Suppose that all members of the set 2 are abelian. Then p D ˛1 .G/ D M 21 ˛1 .M / since the intersection of any two distinct members of the set 1 is abelian (here we use Hall’s enumeration principle). It follows from Remark 6 and

76 p-groups with few A1 -subgroups

295

Lemma 76.8(b,c) that d.G/ D 2 (indeed, if d.G/ D 3, then ˆ.G/ Z.G/, by Lemma 76.3, and, by Lemma 76.8(b,c), ˛1 .G/ 2p 2 1 > p 2 ). Let fH1 ; : : : ; Hs ; V1 ; : : : ; Vu g be the set of nonabelian members of the set 1 , where ˛1 .Hi / D p, ˛1 .Vj / D 1 (since ˆ.G/ 2 2 is abelian, the set 1 has no member H with ˛1 .H / D p C 1; see Lemma 76.4). By assumption, s > 0. Next, jG 0 j jHi0 j D p 2 so the set 1 has at most one abelian member (Lemma 64.1(q)). We have p 2 D ˛1 .G/ D sp C u sp C .p C 1 s/ D s.p 1/ C .p C 1/. If s < p, then p 2 .p 1/2 C p C 1 D p 2 p C 2 so p D 2, s D 1, u D 2. Then ˛1 .H1 / D 2 so cl.H1 / D 3 (Lemma 76.4). By Remark 3, jG=Z.G/j p 4 . Since Z.G/ < H1 and cl.H1 / D 3, we get jH1 W Z.G/j > p 2 so jG=Z.G/j D p 4 . Assume that jG 0 j D p 2 . Then G=V10 is an A1 -group (Lemma 65.2(a)) hence H1 =V10 is abelian and H10 D V10 is of order p, a contradiction. Since jG 0 j 2jV10 V20 j 8, we get jG 0 j D 8 so G is as in (b3). Thus, s p. Since s < p C 1, we get s D p. It follows from p 2 D ˛1 .G/ D sp C u D p 2 C u that u D 0. Thus, 1 D fH1 ; : : : ; Hp ; Ag, where A is abelian and ˛1 .Hi / D p for i D 1; : : : ; p. Suppose that G is metacyclic. Then jG W Z.G/j D pjG 0 j D p 4 (Lemma 64.1(q) and Theorem 72.1) and G is as stated in (b1). Suppose that G is not metacyclic. Then G=Hi0 is an A1 -group for all i (Lemma 76.2) whence jG 0 j D pjH10 j D p 3 . By Proposition 72.1(b), G 0 is noncyclic so G is as stated in (b2). Theorem 76.11. Suppose that G is a p-group with p 2 < ˛1 .G/ p 2 C p 1. Then G is a two-generator A3 -group with jG 0 j > p and one of the following holds: (a) 1 D fH1 ; H2 ; : : : ; Hp ; Ag, where A is abelian, H1 is an A2 -group with d.H1 / D 3, ˛1 .H1 / D p 2 , ˛1 .Hi / D 1, i D 2; : : : ; p, H10 D D Hp0 is of order p, G=H10 is an A1 -group so jG 0 j D p 2 , ˛1 .G/ D p 2 C p 1. (b) 1 D fH1 ; : : : ; Hp ; Bg, where ˛1 .Hi / D p, i D 1; : : : ; p, ˛1 .B/ D 1, ˛1 .G/ D p 2 C 1. If G is nonmetacyclic, then p > 2, jG W G 0 j D p 2 and jGj 2 fp 5 ; p 6 g. (c) G is metacyclic, 1 D fF1 ; : : : ; Fs ; H1 ; : : : ; Hu g, where s C u D p C 1, ˛1 .Fi / D p, s 2, ˛1 .Hj / D p C 1, ˛1 .ˆ.G// D 1 and jG 0 j D p 3 . Proof. By Lemma 76.4(b), G is not an A2 -group. Assume that G is not an A3 -group. Then there exists a nonabelian H 2 1 which is an An -group, n 3. By Theorem 76.9, ˛1 .H / p 2 . Then, by Lemma 76.5, p 2 C p 1 ˛1 .G/ D ˛1 .H / C ˇ1 .G; H / p 2 C p 1 so ˛1 .G/ D p 2 Cp1 and ˛1 .H / D p 2 , ˇ1 .G; H / D p1 so jH 0 j D p. Let L < H be an A2 -subgroup. Since jL0 j D p, we get, by Lemma 76.4, ˛1 .L/ p 2 D ˛1 .H /, contrary to Proposition 10.28. Thus, H is an A2 -group so G is an A3 -group. This argument also shows that the set 1 has no member U with ˛1 .U / > p 2 .

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Groups of prime power order

If U 2 1 , ˛1 .U / D p 2 , then ˇ1 .G; U / D p 1 so, by Lemma 76.5, G is as in (a). Next we assume that ˛1 .U / < p 2 for all U 2 1 ; then d.G/ d.U / C 1 D 3 (Lemma 76.1 and Theorem 76.9). Let 1 D fF1 ; : : : ; Fs ; H1 ; : : : ; H t ; B1 ; : : : ; Bu ; A1 ; : : : ; Av g;

s C t > 0;

where ˛1 .Fi / D p, ˛1 .Hj / D p C 1 (see Lemma 76.6), ˛1 .Bk / D 1 and Al are abelian. Since s C t > 0 and jFi0 j; jHj0 j > p, we get v 1 (Lemma 64.1(u)). We have () If t > 0, then u D v D 0 since the set 1 .H1 / has no abelian members. () If d.G/ D 2, then uv D 0. Indeed, if uv > 0, then all maximal subgroups of G=B10 are abelian, by Exercise 1.6(a), a contradiction since jT 0 j p 2 for T 2 fF1 ; : : : ; Fs ; H1 ; : : : ; H t g (Lemma 76.4). (i) Suppose that d.G/ D 2. Then uv D 0, by (). Put N D ˆ.G/. (i1) Suppose, in addition, that N is nonabelian so it is an A1 -group since jG W N j D p 2 . Then s C t D j1 j D p C 1, X p 2 C p 2 ˇ1 .G; N / D ˇ1 .M; N / D s.p 1/ C tp M 21

D s.p 1/ C .p C 1 s/p D p 2 C p s; and we get s 2 and t p 1. Let Ui be (the unique so normal in G) abelian maximal subgroup of Fi , i D 1; 2, and set A D U1 U2 . Since F1 \ F2 D ˆ.G/ is nonabelian, we get U1 ¤ U2 . Then A .> U1 / is of class 2 (Fitting). If A 2 1 , then ˆ.G/ D A\F1 D A1 is abelian, a contradiction. Thus, A D G, so cl.G/ D 2. In that case, all members of the set 1 are metacyclic (Lemma 76.4) so G is also metacyclic (Lemma 64.1(e,f,h)), and G is as in (c). (i2) In what follows we assume that N D ˆ.G/ is abelian; then t D 0. If s D p C1, then ˛1 .G/ D sp D .p C 1/p > p 2 C p 1, a contradiction. Thus, u C v > 0. Since jG 0 j jF10 j D p 2 , we get v 1 (Lemma 64.1(u)). If v D 1, then u D 0, by (); then s D p and we get p 2 C 1 ˛1 .G/ D sp D p 2 , a contradiction. Thus, v D 0 so s C u D p C 1 and u D p C 1 s > 0. We have X ˛1 .M / D ps C u D .p 1/s C p C 1 p 2 C p 1: p 2 C 1 ˛1 .G/ D M 21 1 , and so s D p and u D 1. In that case, It follows that p s p C 1 p1 2 ˛1 .G/ D ps C u D p C 1, 1 D fF1 ; : : : ; Fp ; Bg, where ˛1 .Fi / D p, ˛1 .B/ D 1. Thus, G and all members of the set 1 are two-generator so G is either metacyclic or p > 2 and G=K3 .G/ is nonabelian of order p 3 and exponent p (Lemma 64.1(n,p)). Suppose that G is not metacyclic; then jG W G 0 j D p 2 . In that case, jG 0 j pjF10 B 0 j p 4 so jGj D jG W G 0 jjG 0 j p 6 . Since G is an A3 -group, we get jGj 2 fp 5 ; p 6 g so G is as in (b).

76 p-groups with few A1 -subgroups

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(ii) Now let d.G/ D 3. By Lemma 76.8(b,c), ˆ.G/ — Z.G/. Then, by Lemma 76.3 and the last sentence of Remark 6, the set 2 has exactly p 2 nonabelian members which are A1 -groups so this set has two distinct abelian members U1 and U2 . Then M D U1 U2 2 1 , by the product formula and j1 .M / \ 2 j D p C 1. It follows from U1 \ U2 Z.M / that all members of the set 1 .M / \ 2 are abelian so the set 2 has exactly p C 1 abelian members, say U1 ; : : : ; UpC1 . We have jM 0 j p (Lemma 64.1(u)) and 2 D fL1 ; : : : ; Lp 2 ; U1 ; : : : ; UpC1 g, where all Li are A1 -groups, all Ui < M . Since Li \ M D ˆ.G/ is abelian for all i , we get ˛1 .M / C p 2 ˛1 .G/ p 2 C p 1 so ˛1 .M / p 1 and M is either abelian or A1 -group (Remark 1). (ii1) Suppose that M is abelian. Then v D 1 (Lemma 64.1(u)) and so t D 0. Also, u D 0 since all p C 1 abelian members of the set 2 lie in M . Hence, s D j1 j v D p 2 C p. By the hypothesis and Hall’s enumeration principle (Theorem 5.2), X X ˛1 .H /p ˛1 .H / D ps p p 2 D p.p 2 Cp/p 3 D p 2 ; p 2 < ˛1 .G/ D H 21

H 22

a contradiction. Thus, v D 0. (ii2) Now let M be an A1 -group; then u D 1 since all Ui < M . We get t D 0, by (), so s D j1 j u D p 2 C p. Therefore, by Hall’s enumeration principle, X X ˛1 .H / p ˛1 .H / p 2 C 1 ˛1 .G/ D H 21 2

H 22 2

D 1 C sp p p D 1 C .p C p/p p 3 D p 2 C 1: Note that d.G/ D 3 but all members of the set 1 are two-generator. We have cl.G/ cl.F1 / D 3. Then p D 2 (Theorem 70.1) and G=K4 .G/ is of order 27 or 28 (Theorem 70.4). However, by remarks following Theorem 70.5, the above groups of order 27 and 28 are A4 -groups and A5 -groups, respectively, a contradiction. Lemma 76.12. Let G be an A4 -group and let R 2 2 be an A2 -group. Then ˛1 .G/ > p 2 C p C 1, unless p D 2 and G satisfies the following conditions: (a) d.G/ D 3, ˛1 .G/ D 22 C 2 C 1 D 7. (b) 2 D fR; R1 ; R2 ; R3 ; A1 ; A2 ; A3 g, where ˛1 .R/ D 4, jR0 j D 2, d.R/ D 3, R1 ; R2 ; R3 are A1 -groups and A1 ; A2 ; A3 are abelian. (c) 1 D fF1 ; F2 ; F3 ; H1 ; H2 ; H3 ; Ag, where A is abelian and (c1) 1 .Fi / D fR; Ri ; Ai g, ˛1 .Fi / D 22 C1 D 5, jFi0 j D 4, R0 D Ri0 , Fi =R0 is an A1 -group, i.e., Fi is a group of Theorem 76.11(a), i D 1; 2; 3. (c2) 1 .H1 / D fA1 ; R2 ; R3 g, 1 .H2 / D fA2 ; R3 ; R1 g, 1 .H3 / D fA3 ; R1 ; R2 g so ˛1 .Hi / D 2 and jHi0 j D 4 hence Hi is an A2 -group, i D 1; 2; 3. (d) Z.G/ < ˆ.G/, jG 0 j D 8, jG=Z.G/j D 16, Z.G/ D Z.Ri /, i D 1; 2; 3. All nonabelian members of the set 1 are two-generator.

298

Groups of prime power order

Proof. Let 1R D fF1 ; : : : ; FpC1 g; then F1 ; : : : ; FpC1 are A3 -groups. By Theorem 76.9, ˛1 .Fi / p 2 for all i . By Lemma 76.6, ˛1 .R/ p. If ˛1 .R/ D p, then ˇ1 .Fi ; R/ p 2 p for all i , and we are done since ˛1 .G/ ˛1 .R/ C

pC1 X

ˇ1 .Fi ; R/ p C .p 2 p/.p C 1/ D p 3 > p 2 C p C 1:

iD1

Now suppose that ˛1 .R/ D p C 1. Then ˛1 .Fi / p 2 C p, by Theorems 76.10 and 76.11, so ˇ1 .Fi ; R/ p 2 1 for all i , and we are done since ˛1 .G/ ˛1 .R/C

pC1 X

ˇ1 .Fi ; R/ p C1C.p 2 1/.p C1/ D p 3 Cp 2 > p 2 Cp C1:

iD1

If ˛1 .R/ > p C 1, then ˛1 .R/ p 2 (Theorem 72.1 and 76.4). Since Fi is an A3 -group with R 2 1 .Fi /, we get ˇ1 .Fi ; R/ p 1 (Lemma 76.5), and so

(4)

˛1 .G/ ˛1 .R/ C

pC1 X

ˇ1 .Fi ; R/ p 2 C .p 1/.p C 1/

iD1 2

D 2p 1 p 2 C p C 1: If p > 2, then ˛1 .G/ 2p 2 1 > p 2 C p C 1, and we are done in this case. Now let 2p 2 1 D p 2 C p C 1; then p D 2 and ˛1 .G/ D 7. In that case, as it follows from (4), ˛1 .R/ D 4 and ˇ1 .Fi ; R/ D 1 so ˛1 .Fi / D 5 for i D 1; 2; 3. By Lemma 76.5 applied to the pair R < Fi , we get jR0 j D 2, d.Fi / D 2 and 1 .Fi / D fR; Ri ; Ai /, where Ri is an A1 -group with R0 D Ri0 and Ai is abelian, jFi0 j D 4, i.e., Fi is a group of Theorem 76.11(a), i D 1; 2; 3. By Lemma 76.5, d.G/ 1 C d.F1 / D 3. Assume that d.G/ D 2. Since all members of the set 1 are also two-generator, G is metacyclic since p D 2 (Lemma 64.1(n)). In that case, jG 0 j D 24 so jFi0 j D 23 (Theorem 72.1), contrary to what has been said in the previous paragraph. S3Thus, d.G/ D 3. Then ˆ.Fi / D ˆ.G/ and so fR; R1 ; R2 ; R3 ; A1 ; A2 ; A3 g D iD1 1 .Fi / D 2 , where A1 ; A2 ; A3 are abelian so ˆ.G/ is abelian, ˛1 .Ri / D 1 (i D 1; 2; 3). Set A D A1 A2 ; then A 2 1 and A3 < A (if A3 — A then CG .ˆ.G// AA3 D G so ˆ.G/ Z.G/, a contradiction since jR W ˆ.G/j D 2 and R is nonabelian). Since F1 ; F2 ; F3 together contain all A1 -subgroups of G and A \ Fi is abelian for i D 2; 3, then A is abelian (Remark 4). Set 1 1R D fH1 ; H2 ; H3 ; Ag. One may assume that Ai < Hi , i D 1; 2; 3. Since Hi \ A > ˆ.G/ is an abelian maximal subgroup of Hi and Hi \ A 2 2 , then Ri — Hi since Ri Ai D Fi ¤ Hi , i D 1; 2; 3. Since R — Hi and j1 .Hi / \ 2 j D 3, i D 1; 2; 3, we get H1 D fA1 ; R2 ; R3 g;

H2 D fA2 ; R1 ; R3 g;

H3 D fA3 ; R1 ; R2 g:

76 p-groups with few A1 -subgroups

299

Now, R, R1 , R2 , R3 together contain all A1 -subgroups of G. Each of the four A1 subgroups S1 , S2 , S3 , S4 of R satisfies Sj ˆ.G/ D R, j D 1; 2; 3; 4, since ˆ.G/ is an abelian maximal subgroup of R, and so, in view of R — Hi , no one of S1 ; S2 ; S3 ; S4 is contained in Hi , i D 1; 2; 3. Thus, ˛1 .Hi / D 2, i D 1; 2; 3. By Lemma 76.6(a), Hi is an A2 -group with d.Hi / D 2 and jHi0 j D 4, i D 1; 2; 3. Hence, all nonabelian members of the set 1 are two-generator. By Lemma 64.1(u), jG 0 j 2jH10 A0 j D 8 and so Lemma 64.1(q) implies that jG W Z.G/j D 2jG 0 j 24 . On the other hand, G D Ri A with A \ Ri D ˆ.G/. We have Z.Ri / D ˆ.Ri / < ˆ.G/ < A, jˆ.G/ W Z.Ri /j D 2, by the product formula, and CG .Z.Ri // Ri A D G. Thus, Z.Ri / Z.G/, i D 1; 2; 3. Assume that jG W Z.G/j < 24 ; then jG W Z.G/j D 23 . In that case, all members of the set 1 , containing Z.G/, must be abelian or A1 -subgroups since nonabelian members of the set 1 are two-generator. Since A is the unique abelian member of the set 1 , this set also contains an A1 -group, contrary to what has been said about 1 . Thus, Z.G/ D Z.Ri / (i D 1; 2; 3) so jG W Z.G/j D 24 and jG 0 j D 8 (Lemma 64.1(q)). Definition 1. A 2-group G of Lemma 76.12 is called a 2-group of type G7;1 . We shall see (see Appendix, below) that groups of type G7;1 do not exist. Lemma 76.13. Let a p-group G be an An -group, n > 2, let H1 2 1 and ˇ1 .G; H1 / D p; then H1 is nonabelian (Lemma 76.6). In that case, d.G/ D 2 so 1 D fH1 ; : : : ; HpC1 g, and one of the following holds: (a) If H2 is neither abelian nor an A1 -group, then p D 2, ˛1 .H2 / D 2, H3 is abelian, H10 D H20 is of order 4, G=H20 is an A1 -group so jG 0 j D 8 and jG=Z.G/j D 16. 0 (b) All p members of the set 1 fH1 g are A1 -groups, H10 D H20 D D HpC1 and G=H10 is an A1 -group so jG 0 j D p 2 . 0 (c) All p members of the set 1 fH1 g are A1 -groups and H20 ; : : : ; HpC1 are 0 pairwise distinct. Set Q D H2 : : : HpC1 ; then Q Š Ep 2 , G=Q is an A1 -group so H10 Q and jG 0 j D p 3 .

Proof. Let R 2 1 .H1 / \ 2 and set 1R D fH1 ; H2 ; : : : ; HpC1 g. Since H1 is nonabelian (Lemma 76.6(a)), we get R — Z.G/ so the set 1R has at most one abelian member. (i) Suppose that R is nonabelian; then all subgroups Hi are nonabelian and we have PpC1ˇ1 .Hi ; R/ p 1 for i > 1 (Remark 1), and we get p D ˇ1 .G; H1 / iD2 ˇ1 .Hi ; R/ p.p 1/ so that p D 2 and ˇ1 .Hi ; R/ D 1 for i D 2; 3. Therefore, by Lemma 76.5, applied to the pair R < Hi , i D 2; 3, we have jR0 j D 2;

d.Hi / D 2;

1 .H3 / D fR; L3 ; A3 g;

jHi0 j D 4;

1 .H2 / D fR; L2 ; A2 g;

300

Groups of prime power order

where ˛1 .Li / D 1 and A is abelian. Since, in view of ˇ.G; H1 / D 2, H1 , H2 , H3 contain together all A1 -subgroups of G, the members of the set 1 are not A1 -groups. Next, A2 ; A3 G G. Set A D A2 A3 ; then cl.A/ 2 (Fitting). Since cl.G/ cl.H2 / D 3 > 2, we get A 2 1 . Then A is abelian since A \ Hi D Ai is abelian for i D 2; 3 (Remark 4). Therefore, in view of jG 0 j jH20 j D 4, A is the unique abelian member of the set 1 (Lemma 64.1(u)). Since R 6 A, we get d.G/ > 2 so d.G/ D 3 since d.H2 / D 2; then j1 .Hi / \ 2 j D 3, i D 1; 2; 3, hence 1 .H2 / [ 1 .H3 / 2 . We have Hi \ A D Ai 2 2 and Ai is the unique abelian member of the set 1 .Hi /, i D 2; 3. Also put A1 D H1 \ A.2 2 / and 1Ai D fHi ; Fi ; Ag, i D 1; 2; 3. It follows that 1 D fH1 ; H2 ; H3 ; F1 ; F2 ; F3 ; Ag. Since F1 \ H1 D A1 and ˛1 .F1 / D 2 D ˇ1 .G; H1 /, F1 is an A2 -group (Lemma 76.6). For i D 2; 3, set Si D F1 \ Hi ; then Si 2 2 and, since Si ¤ A1 , the unique abelian member of the set 1 .F1 /, Si is an A1 -subgroup (Lemma 76.5). We also have Si ¤ R, i D 2; 3, and S2 ¤ S3 so 1 .F1 / \ 2 D fS2 ; S3 ; A1 g. One may assume, without loss of generality, that 1S2 D fF1 ; H2 ; F2 g; then 1S3 D fF1 ; H3 ; F3 g. By Lemma 76.5, d.Hi / D d.Fj / D 2, i D 2; 3 and j D 1; 2; 3. Let 1 .H1 / \ 2 D fR; U; A1 g. Then H1 \ Fi D U for i D 2; 3 so 1U D fH1 ; F2 ; F3 g. By the above, 2 D fR; S2 ; S3 ; U; A1 ; A2 ; A3 g and A1 D H1 \ F1 \ A; R D H1 \ H2 \ H3 ;

A2 D H2 \ F2 \ A; U D H1 \ F2 \ F3 ;

A3 D H3 \ F3 \ A; S2 D H2 \ F1 \ F2 ;

S3 D H3 \ F1 \ F3 : By Lemma 64.1(u), jG 0 j 2jH20 A0 j D 8. Assume that jG 0 j D 4. Then all nonabelian maximal subgroups of G have the same derived subgroup which is equal to G 0 . By Lemma 64.1(q), jG W Z.G/j D 8. If K=Z.G/ is a maximal subgroup of G=Z.G/ such that K ¤ A, then jK 0 j D 2 (Lemma 64.1(q)), a contradiction. Thus, jG 0 j D 8 so jG=Z.G/j D 16 (Lemma 64.1(q)). If Z.G/ — H 2 1 , then G D H Z.G/ so jG 0 j D jH 0 j D 4, which is a contradiction. Thus, Z.G/ < ˆ.G/. Since R \ U D ˆ.G/, we get jU 0 j D 2 (Lemma 64.1(q)). Let us consider the quotient group GN D G=Z.G/ which is of rank 3 and order N SN2 , SN3 and UN are four-groups so GN has at least 9 involutions. 16. The subgroups R, N D 4, it follows that GN is nonabelian. Let DN be a minimal nonabelian Since exp.G/ N Then, by Lemma 64.1(i), GN D DZ. N G/, N and now it is easy to see that subgroup of G. N N G Š D8 C2 (see Appendix 16). Then G contains exactly two distinct elementary N Then K 2 1 is abelian subgroups of order 8. Let KN Š E8 be such that KN ¤ A. nonabelian of rank 3, contrary to what has been proved already. Thus, all members of the set 1 .H1 / \ 2 are abelian.

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(ii) Since R is abelian, we have p D ˇ1 .G; H1 /

pC1 X iD2

ˇ1 .Hi ; R/ D

pC1 X

˛1 .Hi /:

iD2

Thus, for nonabelian H 2 1R fH1 g we have ˛1 .H / 2 f1; pg so d.H / D 2 (Lemmas 76.1 and 76.6), and so d.G/ 1 C d.H / D 3. (ii1) Assume that H2 is neither abelian nor an A1 -group. Then ˛1 .H2 / D p D ˇ1 .G; H1 / (Lemma 76.6) so Hi is abelian for i > 2. Next, H2 is an A2 -subgroup and jG 0 j jH20 j D p 2 (Lemma 76.6) so the set 1 has at most one abelian member (Lemma 64.1(u)). In that case, p D 2 and 1R D fH1 ; H2 ; H3 D Ag; here A is abelian, ˛1 .H2 / D 2 and jH20 j D 4 (Lemmas 76.6 and 76.4). Assume that d.G/ D 2. Then H10 D H20 has order 4 and G=H20 is an A1 -group (Lemmas 76.6 and 76.2) so jG 0 j D 8, G=Z.G/j D 2jG=G 0 j D 16, and G is as in (a). Now suppose that d.G/ D 3. Then j2 \ 1 .H1 /j D 2 C 1 D 3. Take S 2 .2 \ 1 .H1 // fRg; then S is abelian, by (i). Set 1S D fH1 ; F2 ; F3 g. Since H1 and H2 together contain all A1 -subgroups of G, F2 and F3 are not A1 -subgroups (they are nonabelian since A is the unique abelian maximal subgroup of G). Since Fi \ H1 D S .2 1 .H1 // is abelian, we get 1 < ˛1 .Fi / ˇ1 .G; H1 / D 2 so ˛1 .Fi / D 2 and Fi is an A2 -group, i D 2; 3 (Lemma 76.6(a)). Then F2 and F3 together contain 4 distinct A1 -subgroups and all of them are not contained in H1 (indeed, F2 \ H1 D S D F3 \ H1 is abelian), i.e., ˇ1 .G; H1 / 4 > 2, contrary to the hypothesis. Thus, all nonabelian members of the set 1R fH1 g are A1 -groups. It follows that all nonabelian members of the set 1 fH1 g are A1 -groups. P (ii2) Assume that d.G/ D 3. Then p D ˇ1 .G; H1 / D F 21 fH1 g ˛1 .F / 2 2 p 1 since the set 1 fH1 g has at least p 1 nonabelian members (Lemma 64.1(q)) which are A1 -subgroups. However, p 2 1 > p, and we get a contradiction. Thus, d.G/ D 2. Then H2 ; : : : ; HpC1 are A1 -groups, by (ii1) and hypothesis. (ii3) Let H20 D H30 .Š Cp ). In that case, CG .H20 / H2 H3 D G so H20 Z.G/. Then all maximal subgroups of the quotient group G=H20 are abelian (Exercise 1.6(a)) so G=H20 is nonabelian hence an A1 -group (Lemma 65.2(a)). We get 0 ; H10 D D HpC1

jG 0 j D j.G=H20 /0 jjH20 j D p p D p 2 ;

and G is as in (b). 0 (ii4) Now let H20 ; : : : ; HpC1 be all distinct. Set Q D H20 H30 . Then, as above, all 0 DQŠ maximal subgroups of the quotient group G=Q are abelian so H20 : : : HpC1 0 0 0 0 Ep 2 and H1 < Q. To fix ideas, assume that H2 ¤ H1 . Then H1 =H2 is nonabelian and ˇ1 .G=H20 ; H1 =H20 / D p1 so, by Lemma 76.5, jG 0 j D j.G=H20 /0 jjH20 j D p 2 p D p 3 and G=Q is an A1 -group. Thus, G is as in (c).

302

Groups of prime power order

Now we are ready to complete the proof of Theorem A for odd p. Theorem 76.14. Let G be a p-group, p > 2. Suppose that ˛1 .G/ 2 fp 2 C p; p 2 C p C 1g. Then G is an An -group, n 2 f2; 3g. Proof. Assume that the theorem is false. Then there is an Ak -group M 2 1 with k > 2, and so ˛1 .M / p 2 (Theorem 76.9), hence ˇ1 .G; M / p C 1. Assume that jM 0 j D p; then d.M / > 2 (Lemma 65.2(a)). Let S < M be an A2 -subgroup. Then, by Lemma 76.4, ˛1 .S / p 2 . If S 62 1 .M /, then, by Lemma 76.5, ˛1 .M / ˛1 .S / C 2.p 1/ p 2 C p C .p 2/ p 2 C p C 1 so, by Theorem 10.28, ˛1 .G/ > ˛1 .M / p 2 C p C 1, a contradiction. Thus, S 2 1 .M /. By Lemma 76.5, ˇ1 .M; S / p and ˇ1 .G; M / p 1, so ˛1 .G/ ˛1 .S / C ˇ1 .G; M / C ˇ1 .M; S / p 2 C p C p 1 > p 2 C p C 1 since p > 2, a contradiction. It follows that (A ) jM 0 j p 2 so ˇ1 .G; M / p (Lemma 76.5). Therefore, p 2 C p C 1 ˛1 .G/ D ˛1 .M / C ˇ1 .G; M / ˛1 .M / C p; hence p 2 ˛1 .M / p 2 C 1 and so M is an A3 -group of Theorem 76.10 or Theorem 76.11; then G is an A4 -group. By Lemma 76.12, since p > 2, we have (B ) Any member of the set 2 is either abelian or an A1 -group. (i) Let ˛1 .M / D p 2 ; then M is a group of Theorem 76.10 so jM 0 j D p 3 , and ˇ1 .G; M / 2 fp; p C 1g. By (B ), M is not a group of Theorem 76.10(a1) since the set 1 .M / \ 2 has no members which are A2 -groups. (i1) Let M be a group of Theorem 76.10(a), i.e., d.M / D 3, jM 0 j D p 3 , 1 .M / D fF1 ; : : : ; Fp 2 2Cp ; Ag, where ˛1 .Fi / D p for all i and A is abelian. By (B ), Fi 62 2 , i D 1; : : : ; p 2 , so j1 .M / \ 2 j D 1 hence A D ˆ.G/ and d.G/ D 2. Let 1 D fM1 D M; M2 ; : : : ; MpC1 g; then at most one member of the set 1 is abelian and PpC1 ˇ1 .G; M / D iD2 ˛1 .Mi / since ˆ.G/ D A is abelian. Let ˛1 .G/ D p 2 C p; then ˇ1 .G; M / D p so jM 0 j p 2 (Lemma 76.13), contrary to the first paragraph of (i). Thus, we have ˛1 .G/ D p 2 C p C 1. In view of ˇ1 .G; M / D p C 1 > p D jfM2 ; : : : ; MpC1 gj, one may assume that M2 is neither abelian nor an A1 -group (Dirichlet’s principle); then ˛1 .M2 / D p. It follows from p C 1 D ˇ1 .G; M / D ˛1 .M2 / C

pC1 X iD3

˛1 .Mi / D p C

pC1 X iD3

˛1 .Mi /

76 p-groups with few A1 -subgroups

303

that j1 j 3 members of the set 1 are abelian and exactly one its member, say M3 , is an A1 -group. Since jG 0 j jM 0 j D p 3 > p, the set 1 has at most one abelian member (Lemma 64.1(u)) so we get p D 3. Then 1 D fM1 D M; M2 ; M3 ; M4 D Ag, where A is abelian. However, by Lemma 76.2, we get M30 D M 0 , a contradiction since jM 0 j D p 3 > p D jM30 j. (i2) Let M be a group of Theorem 76.10(b1,b2), i.e., d.M / D 2, 1 .M / D fF1 ; : : : ; Fp ; Ag, where ˛1 .Fi / D p, F10 D D Fp0 has order p 2 , A is abelian. It follows from (B ) that Fi 62 1 .M / \ 2 for all i so A D ˆ.G/ and d.G/ D 2. Let 1 D fM1 D M; : : : ; MpC1 g. As in (i1), one has to consider two possibilities: ˛1 .G/ 2 fp 2 C p; p 2 C p C 1g. Let ˛1 .G/ D p 2 C p; then ˇ1 .G; M / D p so jM 0 j p 2 (Lemma 76.13), contrary to the first paragraph of (i). Now let ˛1 .G/ D p 2 CpC1. In that case, ˇ1 .G; M / D pC1. Therefore, as in (i1), one may assume that M2 is neither abelian nor an A1 -group. In that case, as in (i1), we get p D 3 and 1 D fM1 D M; M2 ; M3 ; M4 D Ag, where A is abelian, ˛1 .M3 / D 1. By Lemma 76.2, we get M30 D M 0 , a contradiction since jM 0 j p 2 > p D jM30 j. All the above, together with (A ) and (B ), yields (C ) ˛1 .M / D p 2 C 1, i.e., M is a group of Theorem 76.11(b). (ii) In view of (C ), Theorem 76.11(b) and Lemma 76.5, we must have ˇ1 .G; M / p so that ˛1 .G/ D p 2 C p C 1; then ˇ1 .G; M / D p. By Theorem 76.11(b), d.M / D 2, 1 .M / D fF1 ; : : : ; Fp ; Bg, where ˛1 .Fi / D p, i D 1; : : : ; p, ˛1 .B/ D 1. By (B ), ˆ.G/ D B so d.G/ D 2. Set 1 D fM D M1 ; : : : ; MpC1 g; then ˇ1 .Mi ; B/ p 1, i > 1 (Lemma 76.5). We have ˛1 .G/ D ˛1 .M / C

pC1 X

ˇ1 .Mi ; B/ p 2 C 1 C .p 1/p > p 2 C p C 1;

iD2

a final contradiction. Theorem 76.15. A 2-group G satisfying ˛1 .G/ D 6 is an An -group, n 2 f2; 3g. Proof. Assume that this is false. Then, by Theorems 76.9–76.11, G is an A4 -group so there is an A3 -group H 2 1 . Since ˛1 .H / 4 (Theorem 76.9) and ˇ1 .G; H / > 0 (Proposition 10.28), we get ˛1 .H / 2 f4; 5g so ˇ1 .G; H / 2 f2; 1g. (i) Let ˛1 .H / D 4; then ˇ1 .G; H / D 2, jH 0 j 4 (Lemma 76.13). On the other hand, H is a group of Theorem 76.10, and so jH 0 j D 8, a contradiction. (ii) Let ˛1 .H / D 5. Then H is a group of Theorem 76.11 so jH 0 j 4. However, ˇ1 .G; H / D 1 so jH 0 j D 2 (Lemma 76.5), a final contradiction.

Appendix by Z. Janko: Nonexistence of groups of types G7;1 and G7;2 Theorem A.1. 2-groups of type G7;1 do not exist.

304

Groups of prime power order

Proof. Let G be a 2-group of type G7;1 , i.e., G is a 2-group of Lemma 76.12 with ˛1 .G/ D 7. We shall use freely the notation and the results of Lemma 76.12. By Lemma 76.4(c), Hi is metacyclic for i D 1; 2; 3 since p D 2 and ˛1 .Hi / D 2. Here we note that jGj 26 and so jHi j 25 since R 2 2 and R is an A2 -group and so jRj 24 . In particular, R1 , R2 , R3 , A1 , A2 , A3 are also metacyclic. Next, R is nonmetacyclic since d.R/ D 3. Since A1 is a maximal subgroup of A and d.A1 / D 2, we have d.A/ 3. Assume that d.A/ D 2. In that case, all members of the set 1 are two-generator so we can use the results of 70. By Theorem 70.1, cl.G/ > 2. Then Theorem 70.2 implies that .G=K3 .G//0 Š E8 . But K3 .G/ ¤ f1g and so jG 0 j > 8, contrary to Lemma 76.12(d). Hence we have d.A/ D 3 and so 1 .A/ Š E8 . Since Hi is metacyclic and jHi j 25 (i D 1; 2; 3), we can use Proposition 71.2. It follows that Hi0 Š C4 (Exercise 1) and Hi is isomorphic to one of the following groups: m

n

m1

(5) Hi D ha; b j a2 D 1; m 4; b 2 D a2 ab D a

1C2m2

(6) Hi D ha; b j a8 D 1; b

2n

; n 2; D 0; 1;

i;

D a4 ; n 2; D 0; 1; ab D a1C4 ; D 0; 1i:

However, in case (5) we get Ã1 .Hi / D ha2 ; b 2 i (this subgroup is abelian) and we see that ha; b 2 i, ha2 ; bi, ha2 ; b 2 ihabi are three maximal subgroups of Hi and they are all A1 -groups, contrary to ˛1 .Hi / D 2 (Lemma 76.12). (Indeed, to check that a nonabelian metacyclic p-group is an A1 -group, is suffices, in view of Lemma 65.2(a), to show that its derived subgroup has order p. For example, if, in (5), K D ha; b 2 i, 2 m2 /2 m1 m1 D a1C2 so K 0 D ha2 i has order 2 and K is an A1 then ab D a.1C2 group.) We have proved that Hi must be isomorphic to a group given in (6). Here we distinguish two essentially different cases D 0 (splitting case) and D 1 (nonsplitting case). (i) We consider first the splitting case D 0 so that H1 D ha; b j a8 D b 2 D 1; n 2; ab D a1 z ; z D a4 ; D 0; 1i; n

n1

and v D a2 so that v 2 D z and u are involutions. Since and we also set u D b 2 nC3 jH1 j D 2 , we have jGj D 2nC4 , n 2. But A1 D hb 2 i hai is the unique abelian maximal subgroup of H1 and so A \ H1 D A1 . The fact that d.A/ D 3 implies that there is an involution t 2 A A1 so that 1 .A/ D hu; z; t i because 1 .H1 / D hu; zi < 1 .A/. We have Z.H1 / D hb 2 ; zi D Z.G/ (indeed, Z.G/ H1 and jZ.G/j D jZ.H1 /j) and ˆ.H1 / D hb 2 i hvi D ˆ.G/ since ˆ.H1 / ˆ.G/ and n1 d.G/ D 3; in particular, u D b 2 2 Z.G/. Also, H10 D hvi Š C4 is normal in G and 0 jG j D 8 (Lemma 76.12). Since ˆ.G/ D hb 2 ihvi and H1 > G 0 > hvi, it follows, by the modular law, that G 0 D huihvi is abelian of type .2; 4/. Note that R1 , R2 , R3 are

76 p-groups with few A1 -subgroups

305

metacyclic of order 2nC2 because 1 .H1 / D fA1 ; R2 ; R3 g, 1 .H2 / D fA2 ; R1 ; R3 g and H1 and H2 are metacyclic of order 2nC3 . We have G D hH1 ; t i D ha; b; t i, hŒa; bi D hvi, Œa; t D 1 (recall that a; t 2 A) and, since G=hvi is nonabelian, we get Œb; t 2 G 0 hvi. On the other hand, 1 .A/ D hu; z; t i is normal in G and so Œb; t D t b t 2 1 .A/. It follows that Œb; t D uz ı with ı 2 f0; 1g since Œb; t D b 1 b t 2 H1 \ 1 .A/ D 1 .H1 / and Œb; t 2 G 0 hvi (recall that z D v 2 ). Also note that R 2 2 is an A2 -group of order 2nC2 and therefore four A1 -subgroups S1 , S2 , S3 , S4 which are contained in R are of order 2nC1 . Since ˛1 .G/ D 7, fS1 ; S2 ; S3 ; S4 ; R1 ; R2 ; R3 g is the set of all A1 -subgroups in G. In particular, all A1 -subgroups of G of order < 2nC2 are contained in R. Let ı D 1 so that Œb; t D uz. Since uz 2 hb 2 ; zi D Z.G/, huzi G G and so hb; t i=huzi is abelian which implies that hb; t i0 D huzi Š C2 . It follows that hb; t i n1 is an A1 -subgroup (Lemma 65.2(a)) containing hb 2 D u; z; t i Š E8 and so hb; t i nC2 is nonmetacyclic. Also, hb; t i is of order 2 (since hu; z; t i \ hbi D hui) and so hb; t i is of order 2nC2 and therefore hb; t i is one of R1 , R2 , R3 (see the last sentence of the previous paragraph). This is a contradiction since R1 , R2 , R3 are metacyclic. n1 Suppose that ı D 0; then Œb; t D u D b 2 is an involution, and therefore we have either hb; t i Š M2nC1 .n > 2/ or hb; t i Š D8 .n D 2/. It follows that hb; t i is an A1 -subgroup of order 2nC1 and so hb; t i < R. On the other hand, R > ˆ.G/ and so R ˆ.G/hb; t i. But ˆ.G/hb; t i 2 1 . This is a contradiction since R 2 2 . (ii) We consider now the non-splitting case D 1 so that H1 D ha; b j a8 D 1; b 2 D a4 D z; n 2; ab D a1 z ; D 0; 1i n

n1

and we set v D a2 , w D b 2 and u D vw. We see again that A1 D hb 2 ; ai D A\H1 is the unique abelian maximal subgroup in H1 . Since d.A/ D 3 and 1 .A1 / D 1 .H1 / D hu; zi, there is an involution t 2 AA1 such that 1 .A/ D hu; z; t i Š E8 . Also, jH1 j D 2nC3 , jGj D 2nC4 and the metacyclic A1 -subgroups R1 , R2 , R3 are of order 2nC2 . Since R 2 2 and R is an A2 -group with ˛1 .R/ D 4, four A1 -subgroups S1 , S2 , S3 , S4 , which are contained in R, are of order 2nC1 . We have Z.H1 / D hb 2 i Š C2n , Z.H1 / D Z.G/, ˆ.G/ D ˆ.H1 / D hb 2 ihvi since ˆ.H1 / ˆ.G/ and d.G/ D 3. Also, H10 D hvi Š C4 is normal in G, jG 0 j D 8 (Lemma 76.12) and so G 0 D hv; wi D hui hvi is abelian of type .4; 2/. We have G D hH1 ; t i D ha; b; t i, hŒa; bi D hvi, where t 2 1 .A/ 1 .H1 /, Œa; t D 1 and, since G=hvi is nonabelian (recall that o.v/ D 4 < 8 D jG 0 j), we get Œb; t 2 G 0 hvi. On the other hand, 1 .A/ D hu; z; t i is normal in G and so b 1 b t D Œb; t D t b t 2 1 .A/ \ H1 D 1 .H1 / D 1 .G 0 / D hui hzi: Since z D v 2 2 hvi, it follows that Œb; t D u or uz. Note that o.b/ D 2nC1 8 and so, taking into account that ub D .vw/b D v 1 w D vwz D uz, we have ˆ.G/ D n1 D w, v D uw 1 and so v 2 hb; ui. hb 2 ; vi < hb; ui Š M2nC2 noting that b 2 nC2 Therefore hb; ui is an A1 -subgroup of order 2 contained in H1 and consequently

306

Groups of prime power order

hb; ui D R2 or R3 . Since t normalizes hb; ui, we have jhb; t i W hb; uij D 2 and so hb; t i is of order 2nC3 which implies that hb; t i is a maximal subgroup of G with hb; t i0 hz; ui and so hb; t i0 D hz; ui Š E4 . It follows that hb; t i is a nonmetacyclic member of the set 1 . Hence hb; t i D F2 or F3 and so hb; t i is an A3 -group. On the other hand, AQ D hb 2 i hui ht i is an abelian nonmetacyclic subgroup of order 2nC2 in hb; t i. Hence AQ is the unique abelian maximal subgroup of hb; t i and so AQ D A2 or A3 (see, in Lemma 76.12, lists of the members of the sets 1 .F2 / and 1 .F3 /). This is a contradiction since A1 , A2 , A3 are metacyclic. Definition 2. A 2-group G is said to be a group of type G7;2 , if it satisfies the following conditions: (1G7;2 ) d.G/ D 2, 1 D fH1 ; H2 ; H3 g, ˛1 .H1 / D 4, ˛1 .H2 / D 2, ˛1 .H3 / D 1. (2G7;2 ) 1 .H1 / D fF1 ; : : : ; F6 ; Ag, where ˛1 .Fi / D 2, i 6, A D ˆ.G/ is abelian, jH10 j D 8. We have ˇ1 .G; H1 / D 3 so ˛1 .G/ D 7. (3G7;2 ) jG W Z.G/j D 25 , Z.G/ D Z.H1 / < ˆ.H1 /. (4G7;2 ) H10 D H20 H30 , G=H10 is an A1 -group so jG 0 j D 24 , ˇ1 .G=H20 ; H1 =H20 / D 1, ˇ1 .G=H30 ; H1 =H30 / D 2. It follows from Definition 2 that a 2-group G of type G7;2 , if it exists, is an A4 group. It will be proved in the Appendix that groups of type G7;2 do not exist. Theorem A.2. Groups of type G7;2 do not exist. Proof. We prove first that a two-generator 2-group X which is an A2 -group is metacyclic. Indeed, if jXj > 24 , then the result follows from Lemma 76.4(c). So let jXj D 24 and let Y be a nonabelian subgroup of order 23 . Since d.X/ D 2, we have CX .Y / Y . It follows from Lemma 64.1(i) that X is of maximal class and so X is metacyclic and we are done. Let G be a 2-group of type G7;2 , where we use the notation from Definition 2. By Lemma 76.6(a), H2 ; F1 ; : : : ; F6 are all two-generator A2 -groups. Therefore, the groups H2 ; F1 ; : : : ; F6 are all metacyclic. But A D ˆ.G/ < H2 and so A is also metacyclic. Hence H1 is minimal nonmetacyclic and so, by Theorem 66.1, H1 is an A2 -group of order 25 . This is a contradiction since H1 contains a proper subgroup F1 which is an A2 -group. Now we are ready to prove Theorem 76.16. If a 2-group G satisfies ˛1 .G/ D 7, then G is an An -group, n 2 f2; 3g. Proof. Let G be a counterexample of minimal order. If H 2 1 , then ˛1 .H / < ˛1 .G/ D 7, so, by Lemma 76.6 and Theorems 76.9–76.11, 76.15, H is either abelian or an An -group, n 3. Therefore, G is an A4 -group so one can choose H 2 1 so that H is an A3 -group. In that case, 4 ˛1 .H / 6 (Theorem 76.9) so H is one of

76 p-groups with few A1 -subgroups

307

the A3 -groups of Theorems 76.10, 76.11 or 76.15. Since G7;1 -groups do not exist, by Appendix, it follows from Lemma 76.12 that (i) The set 2 has no members which are A2 -groups. It follows that H is not a group of Theorems 76.10(a1) and 76.11(c). (ii) Let H 2 1 be a group of Theorem 76.10(a2,b1,b2). In that case, ˛1 .H / D 4, jH 0 j D 8, the set 1 .H / has exactly one abelian member A and all other its members are A2 -groups. It follows from (i) that A D ˆ.G/ so, in view of jG W Aj D 22 , we get d.G/ D 2, and then Z.G/ < ˆ.G/ D A. Set 1 D fH1 D H; H2 ; H3 g; then A 2 1 .Hi /, i D 1; 2; 3, so ˛1 .H2 / C ˛1 .H3 / D ˇ1 .G; H / D ˛1 .G/ ˛1 .H / D 7 4 D 3: One may assume that ˛1 .H2 / D 2 and ˛1 .H3 / D 1 (˛1 .H2 / ¤ 3 since the set 1 .H2 / has one abelian member). Thus, H2 is an A2 -group (Lemma 76.6(a)) and H3 is an A1 -group. We have d.H2 / D 2 D d.H3 / (Lemmas 76.1 and 76.6(a)). Assume that H is a group from Theorem 76.10(b1,b2); then d.H / D 2. In that case, by the above, G and all members of the set 1 are two-generator so G is metacyclic (Lemma 64.1(n)). Since H2 is an A2 -group and H3 is an A1 -group, Exercise 1 shows that G is an A3 -group, a contradiction. Thus, H is a group from Theorem 76.10(a2). Then d.H / D 3, jH 0 j D 8, 1 .H / D fF1 ; : : : ; F6 ; Ag, where ˛1 .Fi / D 2 and d.Fi / D 2 for i 6 and A is abelian, jH W Z.H /j D 2jH 0 j D 16 (Lemma 64.1(q)). By 65, H2 =Z.H2 / 2 fD8 ; C4 C4 g since d.H2 / D 2. Suppose that H2 =Z.H2 / Š D8 . Then Z.H2 / ˆ.H2 / ˆ.G/ D A since d.H2 / D 2, and so jA W Z.H2 /j D 4. Since H3 is an A1 -group, we get Z.H3 / D ˆ.H3 / < ˆ.G/ D A and jA W ˆ.H3 /j D 2. Then Z.H2 / \ Z.H3 / Z.G/ and jA W Z.G/j 8 so jG W Z.G/j D jG W AjjA W Z.G/j 25 . By the above, jH W Z.G/j D 24 so Z.G/ D Z.H / has index 25 in G. Now suppose that H2 =Z.H2 / is abelian of type .4; 4/. We have Z.H2 / < ˆ.H2 / < ˆ.G/ < H3 . Assume that Z.H2 / 6 Z.H3 /. Then H3 =Z.H2 / is cyclic since Z.H3 / D ˆ.H3 / (Lemma 76.1). Thus, G=Z.H2 / has the cyclic subgroup H3 =Z.H2 / of index 2 and the abelian subgroup H2 =Z.H2 / of type .4; 4/ and index 2, which is impossible. It follows that Z.H2 / Z.H3 / so Z.H2 / Z.G/, and we get jG W Z.G/j D jG W H2 jjH2 W Z.H2 /j 2 24 D 25 . Since Z.G/ < ˆ.G/ < H2 , we get Z.H2 / D Z.G/ and so jG W Z.G/j D 25 . Comparing the orders, we get Z.G/ D Z.H /. Since jH 0 j D 8 and jFi0 j D 4, the equality Fi Z.G/ D H is impossible for i 6. It follows that Z.G/ < Fi for all i . In particular, Z.G/ < ˆ.H /. We see that G is a group of type G7;2 . By the Appendix, groups of type G7;2 do not exist, and this is a contradiction. (iii) Now let ˛1 .H / D 22 C 1, i.e., H is a group of Theorem 76.11. As we have noticed in (i), H is not a group of Theorem 76.11(c). We have jH 0 j > 2 (Theorem 76.11) and ˇ1 .G; H / D 2.

308

Groups of prime power order

(iii1) Let H 2 1 be a group of Theorem 76.11(a), i.e., d.H / D 2, jH 0 j D 4, 1 .H / D fF1 ; F2 ; Ag, where d.F1 / D 3, ˛1 .F1 / D 4, jF10 j D 2, ˛1 .F2 / D 1 and A is abelian. If d.G/ D 3, then F1 2 1 .H / 2 , contrary to (i). Thus, d.G/ D 2 so 1 D fH1 D H; H2 ; H3 g. We have ˆ.G/ 2 fF2 ; Ag, by (i). Assume that F2 D ˆ.G/. Then 7 D ˛1 .G/ D ˛1 .H1 / C

3 X iD2

ˇ1 .Hi ; F2 / D 5 C

3 X

ˇ1 .Hi ; F2 /

iD2

so ˇ1 .H2 ; F2 / C ˇ1 .H3 ; F2 / D 2. By Remark 1, we get ˇ1 .Hi ; F2 / D 1, i D 2; 3. In that case, ˛1 .Hi / D 2 so jHi0 j D 4, d.Hi / D 2, i D 2; 3 (Lemma 76.6(a)). Thus, G and all members of the set 1 are two-generator so G is metacyclic (Lemma 64.1(n)). Then, by Exercise 1, applied to H2 and H3 , G is an A3 -group so G is not a counterexample. Therefore, we must have ˆ.G/ D A. Then we have ˛1 .H2 / C ˛1 .H3 / D 2. If ˛1 .Hi / D 1, i D 2; 3, then G and all members of the set 1 are two-generator so G is metacyclic (Lemma 64.1(n)). In that case, by Exercise 1, G must be A2 -group so it is not a counterexample. Thus, ˛1 .H2 / D 2 so jH20 j D 4 and H3 is abelian. By Lemma 76.2, H 0 D H20 and G=H 0 is an A1 -group. It follows that jG 0 j D 2jH 0 j D 8 so jG W Z.G/j D 16 (Lemma 64.1(q)). Since d.G/ D 2, we get Z.G/ < ˆ.G/ < Hi so jHi W Z.G/j D 8, i D 1; 2; 3. Since H1 =Z.G/ is noncyclic and two-generator. we get Z.G/ < ˆ.H1 / < Fi , i D 1; 2. Then jFi W Z.G/j D 4, i D 1; 2. Since d.F1 / D 3 and Z.F1 / > ˆ.F1 /, we get, by Lemma 76.8(c), ˛1 .F1 / 2 22 1 D ˛1 .G/, contrary to Proposition 10.28. (iii2) Let H 2 1 be a group from Theorem 76.11(b). Retaining the notation of Theorem 76.11, we get, in view of (i), ˆ.G/ D B so d.G/ D 2. Set 1 D fH1 D H; H2 ; H3 g. Then, as above, ˇ1 .Hi ; R/ D 1, and we conclude that ˛1 .Hi / D 2 and d.Hi / D 2 for i D 2; 3 (Lemma 76.6(a)). In that case, G and all members of the set 1 are two-generator so G is metacyclic (Lemma 64.1(n)). Then, by Exercise 1, applied to H2 and H3 , G is an A3 -group so G is not a counterexample. (iv) It remains to consider the case where H 2 1 is a group of Theorem 76.15, i.e., ˛1 .H / D 6. In that case, ˇ1 .G; H / D 1 so jH 0 j D 2 and jG 0 j D 4, d.G/ D 2, 1 D fH1 D H; H2 ; Ag, where H2 is an A1 -group and A is abelian (Lemma 76.5). By Lemma 64.1(q), jG W Z.G/j D 8 and, since Z.G/ < ˆ.G/ < H , we get jH W Z.G/j D 4 so jH W Z.H /j D 4. Since H is not an A1 -group, we get d.H / D 3. Then, by Lemma 76.8(c), ˛1 .H / 2 22 1 > 6, a final contradiction. Theorem A follows immediately from Lemma 76.6, Theorems 76.9–76.11, 76.14– 76.16 and the Appendix. If G is a 2-group of maximal class and order 26 , then ˛1 .G/ D 8 and G is an A4 -group. I do not know if, for odd p, there exists an A4 -group G with ˛1 .G/ D .p 2 C p C 1/ C 1.

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If G is a nonabelian p-group such that for each K 2 2 , we have ˛1 .1K / < p 2 C 2p C1, then G is an An -group, n 4. Assume that this is false. Then there is H 2 1 such that H is an Ak -group with k 4. By Theorem A, ˛1 .H / p 2 C p C 2. If K K 2 1 .H / is abelian, we have Pthen, since the set 1 has 2at most one abelian member, K ˛1 .1 / D ˛1 .H / C M 2 K fH g ˛1 .M / .p C p C 2/ C .p 1/ D p 2 C 2p C 1 1 (Remark 1), a contradiction. If K is nonabelian, then, by Remark 1, X ˇ1 .M; K/ ˛1 .1K / D ˛1 .H / C M 21K fH g

.p 2 C p C 2/ C p.p 1/ D 2p 2 C 2 > p 2 C 2p C 1; a final contradiction. 3o . A lower estimate of ˛1 .G/ in terms of d.G/. To facilitate the proof of Theorem B, we begin with the following remarks. Remarks. 10. Let G be a nonabelian p-group, d.G/ D 3 and H 2 1 . Let us prove that ˇ1 .G; H / p.p 1/. This is the case if H is abelian since then ˇ1 .G; H / D ˛1 .G/ p 2 > p.p 1/ (Theorem 76.9). Next we assume that H is nonabelian. If ˛1 .G/ D p 2 , then ˛1 .H / p (Theorem 76.10) so ˇ1 .G; H / p 2 p D p.p 1/. Now let ˛1 .G/ p 2 C 1. The result is true if ˛1 .H / p C 1. So, in view of Lemma 76.6 and Theorem 76.9, one may assume that ˛1 .H / p 2 . .H / \ 2 . Assuming that R is nonabelian, we get, using Remark 1, Take R 2 1P ˇ1 .G; H / F 2 R fH g ˇ1 .F; R/ p.p 1/ since all p C 1 members of the set 1

1R are nonabelian. If R is abelian,Pthen the set 1R has at least p nonabelian members so, by Lemma 76.6, ˇ1 .G; H / F 2 R fH g ˛1 .F / p 2 .p 1/p. 1

11. Suppose that G is a p-group such d D d.G/ 3. We claim that if H 2 1 is nonabelian, then ˇ1 .G; H / p d 2 .p 1/. We use induction on d . Remark 10 is the basis of induction so assume that d > 3. If X 2 1 , then d.X/ d 1 3. Since j1 .H / \ 2 j D p d 2 C C p C 1 > p C 1, the set 1 .H / \ 2 has a nonabelian member R (Exercise 1.6(a)). Then all members of the set 1R D fH1 D H; : : : ; HpC1 g are nonabelian. Let d.Hi / D di , where di d 1. By induction, ˇ1 .Hi ; R/ p di 2 .p 1/ p d 3 .p 1/, i > 1. Therefore, ˇ1 .G; H / PpC1 d 3 .p 1/ p D p d 2 .p 1/, as required. iD2 ˇ1 .Hi ; R/ p Theorem B. Suppose that a p-group G is neither abelian nor an A1 -group and d D d.G/. Then ˛1 .G/ p d 1 . Proof. By Lemma 76.6 and Theorem 76.9, the theorem is true for d 3 so we may assume in what follows that d > 3. If X 2 1 , then d.X/ d 1. We proceed by induction on d . If H 2 1 is nonabelian and such that d.H / d , then, by induction, ˛1 .H / p d 1 , and we are done. Therefore, one may assume that, for all nonabelian H 2 1 , we have d.H / < d ; then, for such H we

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have d.H / D d 1. 3/. It follows from Lemma 76.3 that there is a nonabelian R 2 2 . Set 1R D fH1 ; : : : ; HpC1 g. By induction, ˛1 .Hi / p d 2 and, by Remark 11, ˇ1 .Hi ; R/ p d 3 .p 1/ for all i . Therefore, ˛1 .G/ ˛1 .H1 / C P pC1 d 2 C p d 3 .p 1/ p D p d 1 . iD2 ˇ1 .Hi ; R/ p 4o . Groups of exponent p. We assume that all groups, considered in this subsection, have exponent p. If G is an A1 -group of exponent p, then jGj D p 3 . Now let G be an A2 -group; then jGj D p 4 . If the set 1 has exactly one abelian member, then cl.G/ D 3 and ˛1 .G/ D p. If the set 1 has two distinct abelian members, then G D S C and ˛1 .G/ D p 2 , where jC j D p. Remarks. 12. Suppose that G is a nonabelian group of exponent p and order p 5 and H 2 1 . By Proposition 2.3, p 2 j ˛1 .H /, p 2 j ˛1 .G/ so p 2 j ˇ1 .G; H /. 13. We claim that if G is nonabelian of order p 6 and exponent p, then ˛1 .G/ p 3 . Let R 2 2 be such that R ¤ Z.G/. Set 1R D fH1 ; : : : ; HpC1 g. If R is nonabelian, P then, by Remark 12 and Proposition 10.28, ˛1 .G/ ˛1 .R/ C pC1 iD1 ˇ1 .Hi ; R/ > p 2 .p C 1/ > p 3 . If R is abelian, one may assume that H1 ; : : : ; Hp are nonabelian so ˛1 .Hi / p 2 .i D 1; : : : ; p/, by Proposition 2.3. In that case we have ˛1 .G/ P p 2 3 iD1 ˛1 .Hi / p p D p . Now let R D Z.G/. Then G D A E, where A is an A1 -group and E Š Ep 3 so, by Remark 3, ˛1 .G/ D p 6 . Theorem C. If G is nonabelian of order p m and exponent p, m > 3, then: (a) If H 2 1 , then ˇ1 .G; H / p m4 .p 1/. (b) ˛1 .G/ p m3 . (c) If ˛1 .G/ D p m3 , then G is of maximal class with abelian subgroup of index p and m p. Proof. In all three cases we proceed by induction on m. (a) One may assume that m > 4, in view of Remark 1. Let H 2 1 be nonabelian and R 2 1 .H / \ 2 . Then the set 1R D fH1 D H; : : : ; HpC1 g has at most one abelian member. Suppose that PpHpC1 is abelian; then R is also abelian. In that case, by induction, ˇ1 .G; H / iD2 ˛1 .Hi / p .m1/3 .p 1/ D p m4 .p 1/. Now suppose that the set 1R has no abelian members. Then, by induction, we get PpC1 ˇ1 .G; H / iD2 ˇ1 .Hi ; R/ p .m1/4 .p 1/ p D p m4 .p 1/, and (a) is proven. (b) In view of Remarks 12 and 13, one may assume that m > 6. There is R 2 2 such that R ¤ Z.G/ since G is neither abelian nor A1 -group. Set 1R D fH1 ; : : : ; HpC1 g:

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Suppose that R is abelian. One may assume in that case that H1 ; : : : ; Hp are nonabelian. By induction, we have ˛1 .G/ D

p X

˛1 .Hi / p .m1/3 p D p m3 :

iD1

Now let R be nonabelian. Then, by induction and (a), we have ˛1 .G/ D ˛1 .R/ C

pC1 X

ˇ1 .Hi ; R/ p .m2/3 C p .m1/4 .p 1/.p C 1/ D p m3 ;

iD1

completing the proof of (b). (c) Let ˛1 .G/ D p m3 . As in (b), we prove that there is an abelian A 2 1 . (i) Let jZ.G/j D p. Then G is of maximal class (Remark 5) and, by Theorem 9.5, since exp.G/ D p, we must have m p. It is easy to check that then indeed ˛1 .G/ D p m3 (see Remark 5). Next we assume that jZ.G/j > p. (ii) As in Remark 13, we have jG W Z.G/j > p 2 . Therefore, there is a nonabelian H1 2 1 containing Z.G/ as a subgroup of index > p. Take Z.G/ < R 2 1 .H1 / \ 2 and set 1R D fH1 ; : : : ; HpC1 g. One may assume that H1 ; : : : ; Hp are nonabelian. It follows from Z.G/ < Hi and jZ.G/j > p that Hi is not of maximal class so, by induction,P ˛1 .Hi / > p m4 . If HpC1 is abelian, then R is also abelian, and we have m4 p D p m3 , contrary to the hypothesis. If H ˛1 .G/ p pC1 is iD1 ˛1 .Hi / > p nonabelian, then, by (a), we get ˛1 .G/ ˛1 .H1 / C

pC1 X

ˇ1 .Hi ; R/ > p m4 C p m5 .p 1/ p D p m3 ;

iD2

a contradiction. 5o . M3 -groups. A p-group G is said to be an M3 -group, if all its A1 -subgroups have the same order p 3 . The groups of maximal class with abelian subgroup of index p are M3 -groups (Remark 5). If an A2 -group G is also an M3 -group, then jGj D p 4 and ˛1 .G/ 2 fp; p 2 g. Theorem C1 . Let G be a nonabelian M3 -group of order p m and exponent p, m > 3. Then: (a) If H 2 1 , then ˇ1 .G; H / p m4 .p 1/. (b) ˛1 .G/ p m3 . (c) If, in addition, ˛1 .G/ D p m3 , then G is of maximal class with abelian subgroup of index p.

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Groups of prime power order

Theorem C2 . Let G be a nonabelian group of order p m . Suppose that G has an A1 subgroup of order p a but has no A1 -subgroups of order > p a . Then ˛1 .G/ p ma . If, in addition, ˛1 .G/ D p ma , then all A1 -subgroups of G have the same order p a . The proofs are omitted since they are repetitions of the proof of Theorem C. For p D 2, see 90. See also Problem 155. 6o . p-groups with few conjugate classes of A1 -subgroups. Given a p-group G, let 1 .G/ denote the number of conjugate classes of A1 -subgroups in G. Given H 2 1 , we define (i) ˇN1 .G; H / is the number of conjugate G-classes of A1 subgroups not contained in H , (ii) N 1 .H / is the number of G-classes of A1 -subgroups contained in H . We have N 1 .H / 1 .H / and, as a rule, the strong inequality holds. Obviously, N 1 .G/ D 1 .G/. Theorem 76.17 (compare with Lemma 76.5). Let G be a p-group and let H 2 1 be nonabelian. Then ˇN1 .G; H / p 1. If ˇN1 .G; H / D p 1, then: (a) d.G/ D 2 and 1 D fH1 D H; : : : ; Hp ; Ag, where A is the unique abelian member of the set 1 . (b) H10 D D Hp0 , G=H10 is an A1 -group so jG 0 j D pjH10 j and d.Hi / 3, i p. (c) If jHi0 j D p for some i > 1, then H2 ; : : : ; Hp are A1 -groups and ˇ1 .G; H / D p 1. Lemma 76.18. Let G be a p-group and let a nonabelian H 2 1 be not an A1 subgroup. Suppose that N 1 .H / D 1. Then (a) H has no proper nonabelian Ginvariant subgroups and the set 1 .H / has exactly one abelian member, (b) d.G/ D 2, (c) jH 0 j > p. Proof. Assume that a nonabelian A 2 1 .H / is G-invariant and let U A be an A1 -subgroup. By Proposition 10.28, H has an A1 -subgroup V that is not contained in A. Since U G A, U and V are not conjugate in G, a contradiction. This proves the first assertion of (a). Assume that jH 0 j D p; then d.H / > 2 since H is not an A1 -group (Lemma 65.2(a)). Let U < H be an A1 -subgroup. Then U 0 D H 0 so U G H . It follows that H D NG .U / so jG W NG .U /j D p. In that case, H contains exactly p subgroups conjugate with U in G so, by hypothesis, ˛1 .H / D p. However, by Theorem B, ˛1 .H / p 2 , contrary to what has just been said. Thus, jH 0 j > p, proving (c). The set 1 .H / has exactly one abelian member (otherwise, jH 0 j D p), and this completes the proof of (a). Then, since all members of the set 1 .H / \ 2 are abelian, we get j1 .H / \ 2 j D 1 (Lemma 64.1(q)), hence d.G/ D 2, completing the proof of (b). Proof of Theorem 76.17. Let N 2 2 \ 1 .H /; then N 6 Z.G/ since H is nonabelian. Set 1N D fH1 D H; : : : ; HpC1 g. Since the set 1N has at most one abelian member, one may assume that H1 ; : : : ; Hp are nonabelian. By Proposition 10.28,

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there exists an A1 -subgroup Bi Hi such that Bi 6 N , i D 2; : : : ; p. For i ¤ j , we have BiG N D Hi ¤ Hj D BjG N so BiG ¤ BjG , and we conclude that Bi and Bj are not conjugate in G. Thus, ˇN1 .G; H / jfB2 ; : : : ; Bp gj D p 1. Now let ˇN1 .G; H / D p 1. For i D 2; : : : ; p, all A1 -subgroups of Hi that are not contained in N , are G-conjugate with Bi so HpC1 must be abelian (otherwise, ˇN1 .G; H / > p 1); then N is also abelian. In that case, N 1 .Hi / D 1 for i D 2; : : : ; p so, by Lemma 76.18(b), d.G/ D 2; then 1 D 1N and N D ˆ.G/. By Lemma 76.2, H10 D D Hp0 and G=H10 is an A1 -subgroup so jG 0 j D pjH10 j and, since Hi0 ˆ.Hi /, we get d.Hi / D d.Hi =Hi0 / 3 for i D 2; : : : ; p (Lemma 76.1). Now, (c) follows easily from Lemmas 65.2(a) and 76.18. If G is a group of maximal class and order p n > p 3 with abelian subgroup of index N p, then 1 .G/ D p and, if H 2 1 is nonabelian, then ˇ.G; H / D p 1. Corollary 76.19 (compare with Lemma 76.6(b)). Let G be a p-group. If 1 .G/ D p C 1, then d.G/ D 2 and ˆ.G/ is abelian. Proof. Since G is neither abelian nor an A1 -group, there exists R 2 2 such that R 6 Z.G/ (Lemma 76.3). Set 1R D fH1 ; : : : ; HpC1 g; then at most one member of the set 1R is abelian. Suppose that H1 is nonabelian; then ˇN1 .G; H1 / p, by hypothesis. If ˇN1 .G; H1 / D p 1, then d.G/ D 2 and R D ˆ.G/ is abelian (Theorem 76.17). Now we assume that ˇN1 .G; H1 / D p; then N 1 .H1 / D 1 (otherwise, 1 .G/ > p C 1). In that case, d.G/ D 2 and H1 has no proper nonabelian G-invariant subgroups (Lemma 76.18(a,b)) so ˆ.G/, as a proper G-invariant subgroup of H1 , is abelian. Proposition 76.20. Suppose that a p-group G is neither abelian nor an A1 -group. Let all proper nonabelian normal subgroups of G be members of the set 1 . Then d.G/ 3 and one of the following holds: (a) d.G/ D 2 and either cl.G/ D 2 or G=G 0 has a cyclic subgroup of index p. (b) d.G/ D 3. If R is a G-invariant subgroup of index p in G 0 , then G=R is an A2 -group. Proof. Suppose that d.G/ > 2. By hypothesis, all members of the set 2 are abelian so ˆ.G/ Z.G/ and d.G/ D 3 (Lemma 76.3). Let R be a G-invariant subgroup of index p in G 0 . We claim that G=R is an A2 -group. Without loss of generality, one may assume that R D f1g; then jG 0 j D p. It follows from d.G/ D 3 that G is not an A1 -group (Lemma 76.1). Let B < G be an A1 -subgroup. In view of B 0 D G 0 we get B G G so B 2 1 , by hypothesis, and G is an A2 -group. Now suppose that d.G/ D 2. Let M=G 0 < G=G 0 be of index p 2 . Then, by hypothesis, M is abelian. It follows that M CG .G 0 /. If G is not of class 2, then G=G 0 is not generated by subgroups of index p 2 so it has a cyclic subgroup of index p. Proposition 76.21. Let G be a nonabelian p-group.

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Groups of prime power order

(a) (Compare with Theorem 76.9) If d.G/ > 2. then 1 .G/ p 2 . (b) If d.G/ > 3, then 1 .G/ > p 2 C p C 1. Proof. (a) Suppose that all members of the set 2 are abelian. Then d.G/ D 3 and ˆ.G/ Z.G/ (Lemma 76.3). Let fM1 ; : : : ; Ms g be the set of all nonabelian members of the set 1 ; then s p 2 (Exercise 1.6(a)). It remains to show that 1 .G/ s. Let Ai Mi be an A1 -subgroup; then Mi D Ai ˆ.G/ is the unique member of the set 1 , containing Ai , i D 1; : : : ; s. It follows that A1 ; : : : ; As are pairwise not conjugate in G, and we are done in this case. Now suppose that the set 2 has a nonabelian member N . Take T 2 1 .N / \ 3 ; then T — Z.G/. Since G=T is generated by any p C 2 subgroups of order p, it follows that T is contained in at most p C 1 abelian members of the set 2 . In that case, there are p 2 distinct subgroups L1 =T; : : : ; Lp 2 =T of order p in G=T such that L1 ; : : : ; Lp 2 are nonabelian; obviously, Li 2 2 for all i . Let Ai Li be an A1 subgroup not contained in T (Proposition 10.28), i D 1; : : : ; p 2 ; then Li D Ai T . Since A1 ; : : : ; Ap 2 are pairwise not conjugate in G, we get 1 .G/ p 2 . (b) Let d.G/ > 3. Suppose that all members of the set 3 are abelian. Then all members of the set 4 are contained in Z.G/ so d.G/ D 4 (otherwise, G D hH j H 2 4 i is abelian). If jG W Z.G/j D p 2 , then G D BZ.G/ for each A1 -subgroup B < G so all A1 -subgroups are normal in G. Therefore, 1 .G/ D ˛1 .G/ p 3 (Theorem B). Since, p 3 > p 2 C p C 1, we are done in this case. Thus, jG W Z.G/j 2 fp 3 ; p 4 g. It follows that at least j3 j1 D p 3 Cp 2 Cp members of the set 3 are not contained in Z.G/. Let T1 ; T2 and T3 be those distinct members of the set 3 that are not contained in Z.G/. Since any p C 2 distinct subgroups of order p generate G=Ti , it follows that at least p 2 members of the set 2Ti are nonabelian, i D 1; 2; 3. Let i ¤ j , T i; j 3 and K 2 2Ti \ 2 j . Then K D Ti Tj is determined uniquely. It follows S3 that the set M D iD1 2Ti contains at least 3p 2 3 distinct nonabelian members. Let M0 D fH1 ; : : : ; Hk g be the set of all nonabelian members in the set M, where k 3p 2 3, and let Ls Hs be an A1 -subgroup, s k. Since the intersection of any two distinct members of the set M0 is abelian, Hs in the unique member of the set M0 containing Ls , s k. It follows that L1 ; : : : ; Lk are not pairwise G-conjugate so 1 .G/ k 3p 2 3 > p 2 C p C 1. Now suppose that the set 3 has a nonabelian member T . Set 2T D fH1 ; : : : ; Hk g, k D p 2 C p C 1. Let Li Hi be an A1 -subgroup not contained in T , i D 1; : : : ; k (Li exists, by Proposition 10.28), and let L T be an A1 -subgroup. We have Li T D Hi for i k. Since L; L1 ; : : : ; Lk are pairwise not conjugate in G, we get 1 .G/ k C 1 > k D p 2 C p C 1. Let G be a nonabelian p-group. Given M 1 , let 1 .M/ be the number of conjugate G-classes of A1 -subgroups contained in the members of the set M (obviously, 1 .1 / D 1 .G/, unless G is an A1 -group).

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Remarks. 14. Suppose that every A1 -subgroup is contained in the unique maximal subgroup of a nonabelian p-group G. Then ˆ.G/ is abelian and one and only one of the following holds: (i) d.G/ D 2, (ii) d.G/ D 3 and ˆ.G/ Z.G/. Indeed, the groups from (i) and (ii) satisfy the hypothesis (if H < G is an A1 -subgroup, then, in both cases, Hˆ.G/ is the unique maximal subgroup of G containing H ). Now let G satisfy the hypothesis. Then all members of the set 2 are abelian. By Lemma 76.3, d.G/ 3 and, if d.G/ D 3, then ˆ.G/ Z.G/. 15. Suppose that G is a nonabelian p-group with d.G/ > 2. We claim that the following assertions are equivalent: (a) 1 .1K / p for all K 2 2 , (b) G is an A2 -group with ˛1 .G/ D p 2 . Let (b) holds. Then all nonabelian members of the set 1 are A1 -subgroups and jG W Z.G/j D p 2 (Lemma 76.4(d)). If K 2 2 fZ.G/g, then KZ.G/ 2 1 so 1 .1K / D p and (b) ) (a). It remains to prove the reverse implication. Assume that H1 2 1 is neither abelian nor minimal nonabelian. Then there is a nonabelian K1 2 1 .H1 /. In that case, ˛1 .1K // p C 2 > p, a contradiction. Thus H1 does not exist so G is A2 -group. Exercise 5. Let G be a p-group and let the set 1 have exactly p nonabelian members. Then: (a) 1 D fH1 ; : : : ; Hp ; Ag, where A is abelian. (b) H10 D D Hp0 has index p in G 0 . Exercise 6. Let G be a nonabelian two-generator p-group and jG W G 0 j D p n > p 2 . Prove that if G has only one normal subgroup of index p n , then G=G 0 is abelian of type .p n1 ; p/ and exp.G/ p n . (Hint. Consider the quotient group G=R, where R is a G-invariant subgroup of index p in G 0 . Use Lemma 65.2(a).)

77

2-groups with a self-centralizing abelian subgroup of type .4; 2/

The results of this section are taken from [Jan9]. Here we classify finite 2-groups G which possess an abelian subgroup A Š C4 C2 such that CG .A/ D A. If A is normal in G, then G=A is isomorphic to a subgroup of Aut.A/ Š D8 . Therefore we may assume in the sequel that such a subgroup A is not normal in G. We investigate first the case where A ˆ.G/ and prove the following basic result. Theorem 77.1. Let G be a 2-group which possesses a self-centralizing abelian subgroup A of type .4; 2/ but G does not possess any self-centralizing abelian normal subgroup of type .4; 2/. If A ˆ.G/, then G has no normal elementary abelian subgroups of order 8. Proof. We assume that A ˆ.G/. Let W0 D 1 .A/ so that W0 Š E4 and Z.ˆ.G// < A (the inclusion is strong since A is not normal in G). It follows from jG W CG .W0 /j 2 that W0 Z.ˆ.G//. Since Z.ˆ.G// < A, it follows that W0 D Z.ˆ.G// so W0 is a normal four-subgroup in G. We claim that W0 is the unique normal four-subgroup of G. Indeed, if W1 were another normal four-subgroup of G, then A does not centralize W1 and so CG .W1 / is a maximal subgroup of G not containing A, contrary to our assumption that A ˆ.G/. Since A < ˆ.G/ and jG=ˆ.G/j 4, we get jGj 26 . Assume that W0 Z.G/. Let B be a normal subgroup of G such that W0 < B ˆ.G/ and jB W W0 j D 2. Then G stabilizes the chain B > W0 > f1g and so G=CG .B/ is elementary abelian. In particular, ˆ.G/ CG .B/ which contradicts the fact that Z.ˆ.G// D W0 . Hence jG W CG .W0 /j D 2 and so Z.G/ D hzi < W0 . Set W0 D hz; ui and T D CG .W0 /. For each x 2 G T , ux D uz and for each a 2 A W0 , CT .a/ D CT .A/ D A since A D ha; W0 i and W0 Z.T /. Suppose that A1 is a normal elementary abelian subgroup of order 16 of T D CG .W0 /. Take an element a 2 A W0 . Since a2 2 W0 Z.T /, the element a induces on A1 an automorphism of order 2 and so jCA1 .a/j 4 since ha; A1 i is not of maximal class (Proposition 1.8). Since CT .a/ D A, we get CA1 .a/ D W0 and A \ A1 D W0 . Set C D AA1 so that jC j D 25 , Z.C / D W0 and jC W A1 j D 2. Set hsi D Ã1 .A/ < W0 and we see that all four elements in A W0 are square roots of s in C . Let aa1 (a 2 A W0 , a1 2 A1 ) be any square root of s in C . Then we have

77 2-groups with a self-centralizing abelian subgroup of type .4; 2/

317

s D .aa1 /2 D aa1 aa1 D a2 .a1 a1 a/a1 D sa1a a1 , and so a1a D a1 which implies that a1 2 W0 hence aa1 2 A W0 . It follows that four elements in A W0 are the only square roots of s in C and they form a single conjugate class in C (it follows from CT .a/ D A that CC .a/ has index 4 in C ). Since T centralizes s, it follows that AW0 is a conjugacy class in NT .C / so hA W0 i D A is normal in NT .C / which implies C D T (in view of W0 Z.T / we have jT W Aj D jT W CT .A/j 4 D jC W Aj) and jGj D 2jT j D 2jC j D 26 . Since W0 Z.T / < A, we conclude that Z.T / D W0 . It follows that A1 is the unique abelian subgroup of index 2 in T so A is characteristic in T ; then A1 is normal in G. It follows from the previous sentence that T =W0 Š E23 (otherwise, T contains an abelian subgroup of index 2 and exponent 4, which is not the case). By Lemma 1.1, jT 0 j D 4 so T 0 D W0 D ˆ.G/ D Z.T /, and T is special. Let b be any element in C A1 D T A1 so that b D a1 a for an a 2 A W0 and a1 2 A1 (recall that A \ A1 D W0 ). Hence CA1 .b/ D CA1 .a/ D W0 . In particular, b 2 2 W0 and so four elements in hW0 ; bi W0 are either involutions (in case b 2 D 1) or square roots of the involution b 2 . Conversely, let bx (x 2 A1 ) be any square root of b 2 or an involution if b 2 D 1 in C A1 . Then b 2 D .bx/2 D bxbx D b 2 .b 1 xb/x D b 2 x b x, which implies x b D x and so x 2 CT .b/ D W0 . Hence four elements in hW0 ; bi W0 are all possible square roots of b 2 if hW0 ; bi Š C4 C2 (or involutions if hW0 ; bi Š E8 ) in T A1 . By Exercises 17–19 in 10 and Theorem 1.17(a), each group of order 25 contains at most 19 involutions, unless it is elementary abelian or isomorphic to D8 E4 . It follows that T A1 contains at least 12 elements of order 4 and so, by what has been proved already, the set of squares of these elements is W0# . Thus, each involution in W0 has exactly four square roots in T A1 and, in addition, there are exactly four involutions in T A1 . Let z 2 W0 be the central involution in G. By the previous paragraph, there are exactly four square roots of z in C and if c is one of them, then hW0 ; ci Š C4 C2 and all square roots of z in C D T form the set hW0 ; ci W0 . Hence hW0 ; ci W0 is a normal subset in G and so hW0 ; ci is normal in G in view of z 2 Z.G/. Since Z.T / D W0 , A1 is the unique abelian subgroup of index 2 in T so CT .c/ D hW0 ; ci. Thus hW0 ; ci is a self-centralizing (in G) normal abelian subgroup of G of type .4; 2/, contrary to the hypothesis. We have proved that T D CG .W0 / has no normal elementary abelian subgroups of order 16. Let U be a noncyclic abelian normal subgroup of G. Then W0 U (because of the uniqueness of W0 ) so U CG .W0 / D T . By the previous paragraph, U 6Š E24 so G has no normal elementary abelian subgroups of order 16. Suppose that E8 Š E G G. By the previous paragraph, E T , and E > W0 , and so A \ E D W0 . Let B be a maximal G-invariant abelian subgroup containing E. Since B is noncyclic, we have B T . We know that CG .B/ D B. Suppose that B D E. Then G=E, as a subgroup of order jG W Ej D 8 of Aut.E/ Š GL.3; 2/ Š PSL.2; 7/, is isomorphic to D8 . Set V0 D AE so that V0 =E D ˆ.G=E/

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Groups of prime power order

and therefore V0 is normal in G. Indeed, if M=E is a maximal subgroup of G=E, then A M since A ˆ.G/, and so V0 =E D AE=E < M=E, proving our claim. Since T stabilizes the chain E > W0 > f1g and CT .E/ D E, it follows that T =E Š E4 . Take an element a 2 A W0 so that four elements in A W0 are square roots of a2 ¤ 1. Let ae 0 .e 0 2 E/ be any square root of a2 in V0 . Then we have a2 D .ae 0 /2 D ae 0 ae 0 D a2 .e 0 /a e 0 , which gives .e 0 /a D e 0 and so e 0 2 W0 since CT .a/ D A. It follows that A W0 is the set of all square roots of a2 in V0 and so A D hA W0 i is normal in T . Since A is not normal in G, we have for every x 2 G T , Ax ¤ A. Since Ax V0 and A \ Ax D W0 , we have V0 E D Ea D .A W0 / [ .Ax W0 / and so x sends four elements in A W0 onto four elements in Ax W0 . But G=E Š D8 and so (since T =E Š E4 and V0 =E D ˆ.G=E/) there is y 2 G T with y 2 2 V0 E. Since V0 is nonabelian, A, Ax and E are all abelian subgroups of index 2 in V0 . Since hy 2 ; W0 i equals either A or Ax , one of these subgroups is normal in G since it is normalized by hy; T i D G; then both of them are normal in G, and this is a contradiction. We have proved that B > E and so jBj 24 . Since 1 .B/ D E (recall that G has no normal elementary abelian subgroups of order 24 ), B is an abelian group of rank 3. Take an element a 2 A W0 . Since a2 2 W0 Z.T /, a induces an involutory automorphism on B with CB .a/ D W0 1 .B/ (since a 62 B in view of CT .a/ D A). Applying Proposition 51.2, we see that a inverts Ã1 .B/ and B=W0 . Suppose that some e 2 E W0 .D E A/ is a square in B. Then a inverts (centralizes) e .2 Ã1 .B// so e centralizes ha; W0 i D A, a contradiction. Suppose that each involution in W0 is a square in B. Then 2 .B/ D hb1 i hb2 i hb3 i, where o.b1 / D o.b2 / D 4, b3 is an involution in E W0 , and hb12 ; b22 i D W0 . Since a inverts B=W0 , we get b1a D b11 w with w 2 W0 and so b1a D b11 w D b1 .b12 w/ D b1 w1 , where w1 2 W0 . Similarly, b2a D b2 w2 and b3a D b3 w3 with w2 ; w3 2 W0 . Since C2 .B/ .a/ D W0 , w1 , w2 , w3 must be three distinct involutions in W0 (otherwise, if, for example, w1 D w2 , then .b1 b2 /a D b1 b2 , which is not the case). But then .b1 b2 b3 /a D .b1 b2 b3 /.w1 w2 w3 / D b1 b2 b3 2 2 .B/ W0 , a contradiction. Hence we have B D hbi hwi hei, where o.b/ D 2m ; m > 1;

m1

b2

D z;

hzi D Ãm1 .B/ D Z.G/; m2

W0 D hz; wi;

and e 2 E W0 . Also we set v D b 2 so that o.v/ D 4, v 2 D z, and 2 .B/ D hE; vi is abelian of type .4; 2; 2/. Set V D AB and assume that V < T (V T since A; B < T ). Set VQ D NT .V / so that jVQ W V j 2. Since B is normal in G, V B D aB (a 2 A W0 ) is a normal subset in V D NG .V / VQ . Set a2 D w 0 2 W0 so that four elements in A W0 are square roots of w 0 in V B. An element ab 0 2 V B (b 0 2 B) is a square root of w 0 if and only if w 0 D .ab 0 /2 D ab 0 ab 0 D a2 .b 0 /a b 0 D w 0 .b 0 /a b 0 or .b 0 /a D .b 0 /1 . If .b 0 /a D .b 0 /1 and .b 00 /a D .b 00 /1 (b 0 ; b 00 2 B), then .b 0 b 00 /a D .b 0 /a .b 00 /a D .b 0 /1 .b 00 /1 D .b 0 b 00 /1 , and so the set B0 of elements of B which are inverted by a is a subgroup of B. Therefore, the number of square roots of w 0 in V B equals jB0 j. It follows from CT .a/ D A that B0 \ E D W0 . We have jBj D 2mC2 .m > 1/ (recall

77 2-groups with a self-centralizing abelian subgroup of type .4; 2/

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that B is abelian of type .2m ; 2; 2/) so jV Bj D 2mC2 and therefore jB0 j 2mC1 . All conjugates of a in VQ lie in V B and all these conjugates are square roots of w 0 since VQ T and T centralizes w 0 2 W0 . Since CVQ .a/ D A, we get jVQ W Aj 2mC1 . But jV j D 2mC3 and so jVQ j D 2mC4 , jVQ W V j D 2, and jB0 j D 2mC1 . It follows that B0 covers B=E and so we may choose b 2 B0 such that B0 D hbi hwi, where m1 w 2 W0 hzi (since b 2 D z). Suppose for a moment that VQ D NT .V / D NG .V /. Then looking at G=B, we get, by Proposition 1.8, that G=B is of maximal class and V =B is a noncentral subgroup of order 2 in G=B. Let R=B be a cyclic subgroup of index 2 in G=B. Then R is a maximal subgroup of G and R \ V D B. In particular, A 6 R, contrary to our assumption A ˆ.G/. We have proved that V D NG .V / > VQ D NT .V / so that jV W VQ j D 2, G D T V , and T \ V D T \ NG .V / D NT .V / D VQ . Assume that CV .a/ D A D CVQ .a/ and take an element y 2 V VQ so that y 2 G T . Then all 2mC2 elements in V B form a single conjugate class in V . Since w 0 has exactly 2mC1 square roots in V B, it follows that w 0 D a2 ¤ z (since hzi D Z.G/) and so .w 0 /y D w 0 z. Hence y sends 2mC1 square roots of w 0 in V B onto 2mC1 square roots of w 0 z in V B. Since e 62 B0 (e 2 E W0 ), we have .ae/2 D w 0 z D aeae D a2 e a e D w 0 e a e, and so e a D ez. On the other hand, a m2 inverts B0 D hb; wi and so setting v D b 2 , we get v 2 D z and v a D v 1 D vz. But then .ev/a D ezvz D ev, contrary to CB .a/ D W0 . We have proved that AQ D CV .a/ > A, where jAQ W Aj D 2 and AQ \ T D A. Take an element y 2 AQ A so that y acts non-trivially on W0 . But y centralizes m1 a and so y centralizes a2 D w 0 which gives w 0 D z D b 2 , where hzi D Z.G/. Suppose y 2 2 W0 so that hW0 ; yi Š D8 . In that case there are involutions in hW0 ; yi W0 and so we may assume that y is an involution. Act with the involution y on E. Since y acts non-trivially on W0 and jCE .y/j D 4, there is an involution e 0 2 E W0 with .e 0 /y D e 0 . We have 1 ¤ Œa; e 0 2 W0 and so Œa; e 0 y D Œay ; .e 0 /y D Œa; e 0 , which gives Œa; e 0 D z or .e 0 /a D e 0 z. But a inverts B0 and so v a D v 1 D vz m2 (where v D b 2 ) which gives .ve 0 /a D vze 0 z D ve 0 62 W0 . This is a contradiction since CB .a/ D W0 . We have proved that for each y 2 AQ A, y 2 2 A W0 and so we may assume that 2 y D a. Since a2 D z centralizes E, it follows that y induces on E an automorphism of order 4 and so CE .y/ D hzi. Hence for an e 2 E W0 , we get e y D ew0 with 2 y w0 2 W0 hzi and w0 D w0 z. This gives e a D e y D .ew0 /y D .ew0 /.w0 z/ D ez. On the other hand, a inverts B0 D hb; w0 i and so v a D v 1 D vz. This gives .ve/a D vzez D ve 62 W0 , contrary to CB .a/ D W0 . The contradiction in the previous paragraph shows that we must have V D AB D m2 T . We set again a2 D w 0 2 W0 .a 2 A W0 / and v D b 2 .m > 1/ so that 2 v D z, where hzi D Z.G/ and B D hE; bi. We have jG=Bj D 4. If G=B Š E4 , then ˆ.G/ B and this contradicts our assumption A ˆ.G/. Hence G=B Š C4 and G=B acts faithfully on E (since T D Bhai and a acts non-trivially on E). If

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Groups of prime power order

y 2 G T , then for an element e 2 E W0 , we have e y D ew with w 2 W0 hzi 2 and w y D wz. Then we compute e y D .ew/y D .ew/.wz/ D ez. But y 2 2 T B and T D Bhai and so a acts in the same way on E as the element y 2 which implies e a D ez. By Proposition 51.2, a inverts Ã1 .B/. If m > 2, then v 2 Ã1 .B/ and so v a D v 1 D vz. We get in that case .ev/a D ezvz D ev which contradicts to CB .a/ D W0 . We have proved that we must have m D 2 and so b D v, jBj D 24 , and jGj D 26 . The argument of the previous paragraph shows that a does not invert any element (of order 4) in B E and so the subgroup B0 of all elements of B inverted by a is equal B0 D W0 . Indeed, if a inverts an element s 2 B E, then s a D s 1 D sz and so .es/a D ezsz D es, contrary to CB .a/ D W0 . Hence a2 D w 0 has exactly jB0 j D 4 square roots in T B D V B D .AB/ B and they all lie in A W0 . Since hA W0 i D A, A is normal in T D AB. For each x 2 T B, x 2 2 W0 since CB .x/ D CB .a/ D W0 , and x (acting in the same way on B as the element a) inverts on B exactly the elements of W0 . It follows that x 2 has exactly four square roots in T B. Hence each involution in W0 has exactly four square roots in T B and (since jT Bj D 16) T B contains exactly four involutions. Let a0 2 T B so that .a0 /2 D z. Set A D W0 ha0 i Š C4 C2 and A is normal in G. Indeed, hzi D Z.G/ and so G normalizes the subset f.W0 ha0 i/ W0 g of all square roots of z in T B. We have CG .W0 / D T and CT .a0 / D ha0 iCB .a0 /. But CB .a0 / D CB .a/ D W0 and so A is a self-centralizing abelian normal subgroup of type .4; 2/ in G. This is a final contradiction and our theorem is proved. In our next result we shall determine the structure of the groups appearing in Theorem 77.1. Theorem 77.2. Let G be a 2-group which possesses a self-centralizing abelian subgroup A of type .4; 2/ but G does not possess any self-centralizing abelian normal subgroup of type .4; 2/. If A ˆ.G/, then G has the following properties: (i) G has no normal elementary abelian subgroups of order 8. (ii) G has the unique normal four-subgroup W0 D 1 .A/. (iii) G has a normal metacyclic subgroup N such that 2 .N / D W is abelian of type .4; 4/, CG .W / N , 1 .W / D W0 , CG .W0 / N , jG W CG .W0 /j D 2, and G=N Š C4 or D8 . (iv) N is either abelian of type .2k ; 2kC1 / or .2k ; 2k /, k 2, or N is minimal nonm n m1 abelian and more precisely N D ha; b j a2 D b 2 D 1; ab D a1C2 i, where either m D n with n 3 or m D n C 1 with n 2. Proof. Suppose that our 2-group G satisfies all the assumptions of our theorem together with A ˆ.G/. By Theorem 77.1, G has no normal elementary abelian subgroups of order 8. By the first three paragraphs of the proof of Theorem 77.1, we know that W0 D Z.ˆ.G// D 1 .A/ is the unique normal four-subgroup of G,

77 2-groups with a self-centralizing abelian subgroup of type .4; 2/

321

jG W CG .W0 /j D 2, Z.G/ is of order 2, and jGj 26 . Set W0 D hz; ui, T D CG .W0 /, where hzi D Z.G/. For each t 2 A W0 , CT .t / D A. We apply now the results of 50, since G is neither abelian nor of maximal class. It follows that G possesses a normal metacyclic subgroup N such that CG .2 .N // N , G=N is isomorphic to a subgroup of D8 and W D 2 .N / is abelian of type .4; 2/ or .4; 4/. In any case, W0 D 1 .W / D 1 .N / is the unique normal four-subgroup of G and W0 6 Z.G/. If A N , then A W D 2 .N /. Since CG .A/ D A, we have A D W and then A is normal in G, a contradiction. Hence A 6 N and so A \ N D W0 . Since A ˆ.G/, G=N is not elementary abelian. Hence G=N is isomorphic to C4 or D8 . If T D CG .W0 / does not contain N , then T covers G=N and G=N acts faithfully on W (since CG .W / N ). In that case CG .W0 /=CN .W0 / cannot contain elements of order 4 (since that group centralizes W0 D 1 .W /) and so G=N is elementary abelian, a contradiction. Hence T D CG .W0 / N . Assume that G=N Š C4 . If N is abelian of type .2j ; 2/, j 2, then G=N acts faithfully on 2 .N / Š C4 C2 . If N > 2 .N /, then there is a characteristic cyclic subgroup Z Š C4 of N so that Z is normal in G. But then jG W CG .Z/j 2 and therefore A (being contained in ˆ.G/) centralizes Z, a contradiction. Thus N D 2 .N / is a normal abelian self-centralizing subgroup of type .4; 2/ in G, a contradiction. We have proved that 2 .N / D W is abelian of type .4; 4/. Suppose that Z is a G-invariant cyclic subgroup of order 4 contained in N . But then again A centralizes Z , a contradiction. Hence there is no such Z and so we may apply Proposition 50.4. It follows that N is either abelian of type .2n ; 2n / or .2nC1 ; 2n / with m n n 2 or N is minimal nonabelian and more precisely N D ha; b j a2 D b 2 D m1 1; ab D a1C2 i; where either m D n with n 3 or m D n C 1 with n 2. Assume that G=N Š D8 . The structure of N in that case is already determined by Theorem 50.1. The minimal case N Š C4 C2 cannot occur because in that case N would be a self-centralizing normal abelian subgroup of type .4; 2/ of G, a contradiction. Finally, we consider the case where A 6 ˆ.G/. Theorem 77.3. Suppose that G is a 2-group that possesses a self-centralizing abelian subgroup A of type .4; 2/. If 1 .A/ 6 ˆ.G/, then G possesses an involution t such that CG .t / D ht i D, where D is isomorphic to one of the following groups: C4 , D8 , Q2n , n 3, or SD2m , m 4. Such groups G have been classified in 48, 49, and 51. Proof. Suppose that 1 .A/ 6 ˆ.G/. There is a maximal subgroup M of G such that A M contains an involution t . It follows that A0 D A \ M Š C4 and CG .t / D ht i D, where D D CM .t / A0 . We have CD .A0 / D A0 and so (by a well-known result of M. Suzuki) either D D A0 Š C4 or D is a 2-group of maximal class. In the second case D is isomorphic to D8 , Q2n , n 3 or SD2m , m 4.

322

Groups of prime power order

Theorem 77.4. Suppose that G is a 2-group that possesses a self-centralizing abelian subgroup A of type .4; 2/. If W0 D 1 .A/ ˆ.G/ but A 6 ˆ.G/, then for any a 2 AW0 , CG .a/ D haiM0 , jhai\M0 j D 2, where M0 D W0 or M0 is isomorphic to one of the following groups: D2n , n 3 or SD2m , m 4 and G > CG .a/. Proof. Suppose that M is a maximal subgroup of G which does not contain A. Then A \ M D W0 D 1 .A/. Let a 2 A W0 so that G D M hai and CG .a/ D haiCM .a/ with CM0 .W0 / D W0 , where M0 D CM .a/. By a result of M. Suzuki, either M0 D W0 or M0 is of maximal class. In the second case, M0 is isomorphic to one of the following groups: D2n , n 3, or SD2m , m 4. If G D CG .a/, then ˆ.G/ D ˆ.CG .a// D ˆ.M0 / is cyclic, contrary to our assumption. Exercise. Describe the subgroup structure of the holomorph of the abelian group of type .4; 2/.

78

Minimal nonmodular p-groups

The main results of this section are taken from [Jan10]. 1o . We know that a 2-group is modular if and only if it is D8 -free (Theorem 44.13). Here we classify minimal nonmodular 2-groups G, i.e., nonmodular 2-groups all of whose proper subgroups are modular. Hence there is N G G such that G=N Š D8 free but each proper subgroup of G is D8 -free. We shall use freely all results on modular p-groups from 73 and Appendix 24. In this section a 2-group is said to be Hamiltonian if it is nonabelian and all its subgroups are normal. A Dedekindian 2group is Hamiltonian if it is nonabelian. A metacyclic 2-group H is called ordinary metacyclic with respect to A if H possesses a cyclic normal subgroup A such that H=A is cyclic and H centralizes A=Ã2 .A/, or what is the same, H=Ã2 .A/ is abelian. In other words, a metacyclic 2-group is ordinary if and only if it is powerful (see 26). Below we use freely the following facts: () If minimal nonmodular 2-group G is of maximal class and order 24 , then G Š Q24 . Moreover, a nonabelian 2-group G with cyclic subgroup of index 2 is modular if and only if G 2 fQ8 ; M2n g. It follows that any two involutions of G are permutable so 1 .G/ is elementary abelian. () [Wil] Let G be a Q8 -free modular 2-group. Then d.1 .G// D d.G/. If, in addition, d.G/ D 2, then G is ordinary metacyclic. () Let N be a normal subgroup of a 2-group G such that G=N Š D8 . If L=N is the unique cyclic subgroup of index 2 in G=N and x 2 G L, then x 2 2 N . Indeed, all elements if .G=N / .L=N / are involutions. Until the end of this subsection, G is a minimal nonmodular 2-group. In that case, G contains two non-permutable cyclic subgroups A and B. Since the subgroup hA; Bi is nonmodular, we get G D hA; Bi so d.G/ D 2. By hypothesis, G has a normal subgroup N such that G=N Š D8 . It follows from d.G/ D 2 D d.G=N / that N ˆ.G/. The following proposition clears up the structure of N . Remark 1. Suppose that G is a modular 2-group of order 16 containing a subgroup H Š E8 . Then, G is Q8 -free in view of the existence of H . Since G has no subgroups Š D8 , it is either abelian or minimal nonabelian. Assume that G is nonabelian. Then G D ha; b j a4 D b 2 D c 2 D 1; c D Œa; b; Œa; c D Œb; c D 1i and G=ha2 i Š D8 , a contradiction. Thus, G is abelian.

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Groups of prime power order

Proposition 78.1. We have d.N / 2, so N is metacyclic (Theorem 44.13). Proof. Suppose that d.N / 3. Then N possesses a G-invariant subgroup R such that N=R Š E8 . The quotient group G=R is also minimal nonmodular. We want to obtain a contradiction. To this end, we may assume that R D f1g; then N Š E8 . Let S=N be any subgroup of order 2 in G=N . Then S (being modular) is abelian, by Remark 1, so S centralizes N . On the other hand, G=N Š D8 is generated by its subgroups of order 2, and so N Z.G/. Assume that Z.G/ > N ; then Z.G/=N D Z.G=N / D ˆ.G=N / so Z.G/ D ˆ.G/. In that case, G is minimal nonabelian. Then 1 .G/ D N and jG 0 j D 2 (Lemma 65.1) so G 0 < N , a contradiction since G=N Š D8 is nonabelian. Thus, N D Z.G/. Let L=N be the unique cyclic subgroup of index 2 in G=N . Then L is abelian since N D Z.G/. If L D N L1 with L1 Š C4 , then Ã1 .L/ D Ã1 .L1 / is of order 2 and so Ã1 .L/ Z.G/, a contradiction since Ã1 .L/ 6 N D Z.G/. Hence L does not split over N and so L D NC , where C Š C8 and C \ N D C0 is of order 2. We have ˆ.L/ D ˆ.C / D C1 Š C4 , where C0 < C1 , and ˆ.G/ D C1 N is abelian of type .4; 2; 2/. For each x 2 G L, x 2 2 N , by (), and so there is b 2 G L with b 2 2 N C0 (otherwise, ˆ.G/ D Ã1 .G/ D C1 ). Since CG .C1 / D L, Aut.C1 / Š C2 and C1 is normal in G, it follows that b inverts C1 . In that case, D D hC1 ; bi is (nonabelian) metacyclic of order 24 and exponent 22 so D=hb 2 i Š D8 , a contradiction. Proposition 78.1 is a variant of Theorem 44.13. It follows from Theorem 44.13 that the subgroup N of Proposition 78.1 is metacyclic. Remark 2. Let, as in the following four propositions, the subgroup N be cyclic. We claim that then Ã2 .G/ D ˆ.N /. One may assume that N > f1g. In view of Ã2 .G/ Ã2 .N /, it suffices to show that exp.G=Ã2 .N // D 8. Without loss of generality, one may assume that Ã2 .N / D f1g, i.e., N is cyclic of order 4. Then ˆ.G/ is abelian of order 8. One may assume that ˆ.G/ is abelian of type .4; 2/ (otherwise, G has a cyclic subgroup of index 2, contrary to ()). Since ˆ.G/ D Ã1 .G/ is not generated by involutions, there exists x 2 G such that o.x 2 / D 4; then o.x/ D 8, and we are done. It follows from the obtained result and Lemma 64.1(m) that G is metacyclic if and only if G=ˆ.N / is metacyclic. Proposition 78.2. Suppose that N is cyclic and some proper subgroup of G is not Q8 free. Then G is isomorphic to Q24 or to the uniquely determined group X of order 25 with 2 .X/ Š Q8 C2 given in 52. Proof. Suppose that N is cyclic and G has a maximal subgroup M which is not Q8 free. Since M is modular, it follows that M is Hamiltonian (73 or Appendix 24), i.e., M D Q E with Q Š Q8 and exp.E/ 2. In particular, exp.M / D 4 and Ã1 .M / D ˆ.M / D Ã1 .Q/. We have N < M since N ˆ.G/ so jN j 4; in that case, jGj D jG=N jjN j 8 4 D 32. If jG 0 j D 2, it follows from d.G/ D 2 that G

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325

is minimal nonabelian (Lemma 65.2(a)), a contradiction since M < G is nonabelian. Hence jG 0 j 4 so G has at most one abelian maximal subgroup (Lemma 65.2(c)). (i) Suppose that jN j D 4; then jGj D 32 and jˆ.G/j D 8. Let L=N Š C4 be the unique cyclic subgroup of index 2 in G=N .Š D8 /. By (), G is not of maximal class. In that case, the metacyclic subgroup L is noncyclic (Lemma 64.1(t)) so jÃ1 .L/j D 1 4 jLj D 4 and therefore L is not Hamiltonian. Suppose that L is abelian. We have Ã1 .L/ > Ã1 .N / D Ã1 .M /. Let K be the maximal subgroup of G distinct from M and L. We know that K must be nonabelian and, since G=N Š D8 , we get K=N Š E4 . By (), K is not of maximal class. Since N Š C4 does not lie in Z.M /, we have CG .N / D L. Then jK W CK .N /j D 2 so K 6Š M16 (otherwise, N D ˆ.K/ D Z.K/), and we conclude (Theorem 1.2) that exp.K/ D 4. Take k 2 K CK .N /; then k 2 2 N , in view of K=N Š E4 , and therefore hN; ki Š Q8 . It follows that K is Hamiltonian (73 or Appendix 24) and so Ã1 .K/ D Ã1 .N / < Ã1 .L/. Hence ˆ.G/ D Ã1 .G/ D Ã1 .L/ is of order 4, a contradiction since jˆ.G/j D 8. We have proved that L is nonabelian. In particular, N D hni 6 Z.L/ since L=N is cyclic, and if we set L D hN; li, then nl D n1 and l 4 2 hn2 i. If o.l/ D 4, then L=hl 2 i Š D8 , a contradiction. Hence o.l/ D 8 and so L Š M16 with hl 4 i D hn2 i D L0 and ˆ.L/ D Z.L/ D hl 2 i Š C4 . Note that 2 .L/ D N ˆ.L/ is abelian of type .4; 2/. Set K D CG .N / so that K is the maximal subgroup of G distinct from M and L and (noting that hl 2 i > hn2 i) we get (see also Exercise 1.133) ˆ.G/ D ˆ.M /ˆ.L/ˆ.K/ D hn2 ihl 2 iˆ.K/ D hl 2 iˆ.K/. Since K=N Š E4 , we have ˆ.K/ N . But ˆ.G/ > N and so we must have ˆ.K/ D N (otherwise, ˆ.G/ D hl 2 i is of order 4). It follows that the modular subgroup K has a cyclic subgroup of index 2 and K is Q8 -free since K cannot be Hamiltonian (because jKj D 16 and jÃ1 .K/j D 4; see 73 and Appendix 24). Hence K is either abelian of type .8; 2/ or K Š M16 . In any case, 2 .K/ D ˆ.L/N D ˆ.G/ is abelian of type .4; 2/. It follows that 2 .G/ D M Š Q8 C2 and consequently G is the uniquely determined group of order 25 described in 52. (ii) Now let jN j D 2. In that case, jGj D 16 and jG 0 j D 4 so jG W G 0 j D 4. By Taussky’s theorem, G is of maximal class so, by (), G Š Q16 . It is easy to check that the group G of Proposition 78.2 of order 25 is the group F from Appendix 24. Indeed, take u 2 G M and put U D hui; then jU j D 8 since M D 2 .G/. Assume that UQ D QU . Then UQ is of maximal class (Theorem 1.2), contrary to (). It follows that U \ Q is normal in G and G=.U \ Q/ Š D8 . By Appendix 24, G Š F . Proposition 78.3. Suppose that N > f1g is cyclic and all proper subgroups of G are Q8 -free. Then Ã2 .G/ D ˆ.N / and G=ˆ.N / is minimal nonabelian of order 24 and exponent 4. Thus G=ˆ.N / is isomorphic to one of the following groups: (a) hx; y j x 4 D y 2 D 1; Œx; y D z; z 2 D Œx; z D Œy; z D 1i is nonmetacyclic, (b) hx; y j x 4 D y 4 D 1; x y D x 1 i is metacyclic.

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Proof. By Remark 2, Ã2 .G/ D ˆ.N / so G=Ã2 .G/ is 2-generator nonabelian of exponent 4 and order 24 . Therefore, by hypothesis, G=Ã2 .G/ is minimal nonabelian. By Lemma 65.1 that G=Ã2 .G/ D G=ˆ.N / is isomorphic to the group (a) or (b). In the next two propositions we determine the groups G of Proposition 78.3. Proposition 78.4. Suppose that N > f1g is cyclic and all proper subgroups of G are Q8 -free. If G=ˆ.N / is as in Proposition 78.3(a), then G has a normal elementary abelian subgroup E D hn; z; t i of order 8 such that G=E is cyclic. We set G D hE; xi, where o.x/ D 2sC1 , s 1, E \ hxi D hni, Œt; x D z, Œz; x D n , D 0; 1, and G D hx; t i. We have G=hx 2 i Š D8 , ˆ.G/ D hx 2 i hzi, 1 .G/ D E, and G is Q8 -free. If D 0, then G is minimal nonabelian nonmetacyclic. If D 1, then s 2, G 0 D hz; ni Š E4 and Z.G/ D hx 4 i. Proof. Let M=ˆ.N / D 1 .G=ˆ.N //.Š E8 /; then N is a maximal cyclic subgroup of M . Set jN j D 2s , s 1. Let S=N be any subgroup of order 2 in M=N . Since M is D8 -free and Q8 -free, S cannot be of maximal class. It follows that S is either abelian of type .2s ; 2/ or S Š M2sC1 (s > 2). In any case, there exists an involution in S N . Hence 1 .M / covers M=N Š E4 and, since M is modular, E D 1 .M / is elementary abelian of order 8, by the product formula. It follows that E is normal in G and E \ N D 1 .N / D hni Z.G/. In particular, ˆ.G/ D N 1 .ˆ.G// is abelian of type .2s ; 2/, s 1 since 1 .ˆ.G// Š E4 centralizes ˆ.G/. Take an involution t 2 E ˆ.G/. Let K ¤ M be a maximal subgroup of G such that K=N Š E4 ; then K \ M D ˆ.G/ since d.G/ D 2. Suppose that N is a maximal cyclic subgroup of K. Then, by the argument of the previous paragraph, 1 .K/ covers K=N and so there is an involution r 2 K M since E8 Š 1 .K/ ¤ 1 .M / .D E/; then r 62 E. Since G has no elementary abelian subgroups of order 24 , there is in E an involution u such that ru ¤ ur. Then hr; ui is dihedral, a contradiction. We have proved that there is an element x 2 K M such that hx 2 i D N and so o.x/ D 2sC1 , s 1. Since hx; t i D G (indeed, involutions xN and tN lie in different maximal subgroups of G=N Š D8 ) and t 2 E ˆ.G/, we get G D Ehxi with E \ hxi D 1 .N / D hni Š C2 and so G=E is cyclic of order 2s and jGj D 2sC3 . In particular, G 0 < E and so jG 0 j 2 f2; 4g. Since d.G/ D 2 and G is not minimal nonabelian, we get jG 0 j D 4 (Lemma 65.2(a)). We have Œx; t ¤ 1 since hx; t i D G is nonabelian, so z D Œx; t is an involution in .E \ ˆ.G// hni since hxi is not normal in G. Indeed, if hxi were normal in G, then G=ˆ.N / D ht ˆ.N /i hxi=ˆ.N / is metacyclic, which is not the case. It follows that E D hn; z; t i, where z 2 ˆ.G/, and hxi is not normal in G. Thus NG .hxi/ is a maximal subgroup of G and so z 2 ˆ.G/ NG .hxi/ and therefore z normalizes hxi. Since hx; zi cannot be of maximal class, by (), we have either Œx; z D 1 or Œx; z D n (in which case hx; zi Š M2sC1 , s > 1). Let us consider the first possibility. We have CG .z/ hE; xi D G so z 2 Z.G/ \ G 0 . Then G=hzi has a cyclic subgroup hx; zi=hzi of index 2. It follows that then G=hzi is neither abelian nor isomorphic to

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M2sC1 (otherwise, G=hzi has two distinct cyclic subgroups of index 2 so G has two distinct abelian maximal subgroups, and we get jG 0 j < 4, a contradiction). It follows (Lemma 64.1(t)) that G=hzi is of maximal class so it is isomorphic to Q16 , by (), contrary to the existence of E. Now let hx; zi Š M2sC1 , s > 1. Then hxi induces an automorphism of order 4 on E. Let u be an involution in G E. Then, by Remark 1, F D Ehui Š E16 . But s1 s1 G=E is cyclic and so F=E D .Ehx 2 i/=E and so x 2 is an element of order 4 contained in F E, a contradiction. We have proved that 1 .G/ D E. Proposition 78.5. Suppose that N > f1g is cyclic and all proper subgroups of G are Q8 -free. If G=ˆ.N / is metacyclic, then G is also metacyclic and we have one of the following possibilities: sC1

(i) G D hx; y j x 4 D y 2 abelian and N D hy 2 i.

D 1; s 1; x y D x 1 i, where G is minimal non-

(ii) G D hx; a j x 2 D a8 D 1; s 2; x 2 D a4 ; ax D a1 i, where G is an A2 -group with N D hx 2 i, Z.G/ D N (see 65), G 0 D ha2 i Š C4 and G is of class 3. sC1

s

In both cases (i) and (ii), G is a minimal non-Q8 -free 2-group. Proof. By Remark 2, ˆ.N / D Ã2 .G/ and G is also metacyclic. Set jN j D 2s , s 1. If s D 1, then G is isomorphic to the group (b) of Proposition 78.3 and we are done. Now we assume that s 2. Let S=N be a subgroup of order 2 in G=N . By (), S is not of maximal class so S is either cyclic of order 2sC1 or S is abelian of type .2s ; 2/ or s > 2 and S Š M2sC1 . If ˆ.G/ is cyclic, then G has a cyclic subgroup of index 2, contrary to (). Also, M2sC1 , s > 2, having cyclic center, cannot be the Frattini subgroup (Burnside). Taking S D ˆ.G/, we see that ˆ.G/ is abelian of type .2s ; 2/, s 2. Let 1 .N / D hni so that hni ˆ.N / \ Z.G/. For any subgroup S=N of order 2 in G=N , by the previous paragraph, S=hni is abelian so centralizes N=hni. Since G=N is generated by its subgroups of order 2, we get N=hni Z.G=hni/. Suppose for a moment that G is minimal nonabelian. Since ˆ.G/ is abelian of type sC1 s .2 ; 2/, we get at once (Lemma 65.1): G D hx; y j x 4 D y 2 D 1; s 1; x y D 1 2 x i, where N D hy i; this is a group of part (i). In what follows we assume that G is not minimal nonabelian. In particular, jG 0 j 4 (Lemma 65.2(a)). We will determine the structure of all three maximal subgroups of G. Let M be a maximal subgroup of G such that M=N Š E4 . If N is a maximal cyclic subgroup of M , then for each subgroup S=N of order 2 of M=N , there is an involution in S N . Hence 1 .M / covers M=N and, since M is D8 -free and Q8 -free, 1 .M / is elementary abelian and 1 .M / \ N D hni so that 1 .M / Š E8 , by the product formula. This is a contradiction since G is metacyclic. It follows that N is not a maximal cyclic subgroup of M . Let M0 be a maximal cyclic subgroup of M containing N so that M0 Š C2sC1 is of index 2 in M . By (), M is not of maximal class and

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so M is either abelian of type .2sC1 ; 2/ or M Š M2sC2 , s 2 (Lemma 64.1(t)). In any case, N Z.M / and M=hni is abelian since M 0 hni. Let K (¤ M ) be another maximal subgroup of G with K=N Š E4 . Then K is either abelian of type .2sC1 ; 2/ or K Š M2sC2 , s 2, and again N Z.K/ and K=hni is abelian. We get CG .N / MK D G so N Z.G/. Since G is not minimal nonabelian, we get ˆ.G/ ¤ Z.G/. We have proved that N D Z.G/. Let L=N be the unique cyclic subgroup of index 2 in G. Then L is abelian and using Lemma 64.1(q), we get jG 0 j D 4. By Lemma 64.1(u), L is the unique abelian maximal subgroup of G and so M Š K Š M2sC2 with M 0 D K 0 D hni. In particular, G is an A2 -group (since jG 0 j D 4, this also follows from Corollary 65.3). We have G 0 > hni and, since G 0 6 N D Z.G/ and jˆ.G/=N j D 2, we get ˆ.G/ D NG 0 , N \ G 0 D hni D 1 .N / and cl.G/ D 3. Since G is metacyclic, there exists a cyclic normal subgroup Z of order 8 such that Z > G 0 . But N \Z D N \G 0 D hni and so N Z D L which determines the structure of the maximal subgroup L and shows that L does not split over N . It follows that L is abelian of type .2s ; 2/. Set Z D hai. By (), we get hx 2 i D N for each x 2 G L. Hence G D Zhxi with Z \ hxi D hni for a fixed x 2 G L (since 1 .G/ D 1 .L/, x is not an involution). In view of jG 0 j D 4 and G 0 < Z, we get either ax D a1 or ax D a1 n, where n D a4 . However, if ax D a1 n, then we replace Z D hai with Z D haui, where u 2 N is such that u2 D n. Then we have .au/x D a1 nu D a1 u1 D .au/1 . Since h.au/2 i D ha2 i D G 0 , we may assume from the start that ax D a1 and so the structure of G is completely determined. Remark 3. Let G be a 2-group and let N be a G-invariant metacyclic subgroup of ˆ.G/. We claim that N is ordinary metacyclic. One may assume that N is nonabelian; then N has no cyclic subgroups of index 2 since Z.N / must be noncyclic. There exists a maximal cyclic subgroup A of N such that N 0 < A and N=A is cyclic. We have to prove that N=Ã2 .A/ is abelian. This is the case if jA W N 0 j > 2. Now we assume that jA W N 0 j D 2 and obtain a contradiction. Under our assumption, N=N 0 is abelian with cyclic subgroup of index 2 (indeed, by the choice of A and Frobenius–Stickelberger theorem on abelian groups, A=N is a direct factor of N=N 0 ). In particular, N has no epimorphic images which is abelian of type .4; 4/. Now consider the quotient group NN D N=Ã2 .N /. Since N has no cyclic subgroups of index 2, we get jNN j D 16. By assumption, NN is nonabelian so NN D hx; y j x 4 D y 4 D 1; x y D x 1 i. In that case, NN =hy 2 i Š D8 has a cyclic center, a contradiction since N=Ã2 .N / ˆ.G=Ã2 .N //. In the rest of this subsection we consider the case d.N / D 2. Proposition 78.6. Suppose that d.N / D 2. Then G=ˆ.N / is the minimal nonabelian nonmetacyclic group of order 25 and exponent 4. In particular, G=ˆ.N / has the unique epimorphic image isomorphic to Q8 . Each maximal subgroup of G is Q8 -free and N is ordinary metacyclic.

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Proof. We want to determine the structure of G=ˆ.N /. Since G=ˆ.N / is also minimal nonmodular, we may assume for a moment that ˆ.N / D f1g so that N Š E4 and jGj D 25 . Let S=N be any subgroup of order 2 in G=N . Since S 6Š D8 , S is abelian and so N Z.G/ since such subgroups S generate G. Suppose that Z.G/ D N . Let L=N be the unique cyclic subgroup of index 2 in G=N ; then L is abelian. If L D N R with R Š C4 , then Ã1 .L/ D Ã1 .R/ 6 N and Ã1 .L/ Z.G/, contrary to our assumption. Hence L D NL1 with L1 Š C8 and L0 D L1 \ N Š C2 ; thus, L is abelian of type .8; 2/. We have ˆ.L/ D ˆ.L1 / Š C4 , where ˆ.L/ > L0 . For each x 2 G L, x 2 2 N , by (), and ˆ.G/ D Ã1 .G/ D ˆ.L/N . This implies that there exists b 2 G L such that b 2 2 N L0 . Assume that this is false. Then all elements in .G=L0 / .L=L0 / are involutions so G=L0 is generated by involutions and, since G=L0 is not dihedral, it is not generated by two involutions. Since G=L0 is minimal nonmodular, all its involutions commute so G=L0 is elementary abelian; then d.G/ 4, a contradiction. Since ˆ.L/ 6 Z.G/, we get CG .ˆ.L// D L so b inverts ˆ.L/. But then D D hˆ.L/; bi is nonabelian metacyclic of order 24 and exponent 4; in that case, D=hb 2 i Š D8 , a contradiction. We have proved that Z.G/ > N and so Z.G/ D ˆ.G/. It follows that each maximal subgroup of G is abelian and so G is minimal nonabelian. In particular, jG 0 j D 2 and since G 0 covers Z.G/=N D .G=N /0 , we have Z.G/ D N G 0 is elementary abelian of order 8. It follows that G is the uniquely determined minimal nonabelian nonmetacyclic group of order 25 and exponent 4: G D ha; b j a4 D b 4 D 1; Œa; b D c; c 2 D Œa; c D Œb; c D 1i, where Z.G/ D ha2 ; b 2 ; ci, G 0 D hci, and G=ha2 c; b 2 ci is the unique quotient group of G which is isomorphic to Q8 . In particular, G is not Q8 -free. We return now to the remaining case ˆ.N / > f1g. Assume that N is not Q8 -free. Then N (being modular) is Hamiltonian (see 73 and Appendix 24). But d.N / D 2 and so N Š Q8 . This is a contradiction since Z.N / is cyclic and N < ˆ.G/ (Burnside). We have proved that N is Q8 -free and so N=Ã2 .N / must be abelian. By Remark 3, N is ordinary metacyclic. Suppose that a maximal subgroup M of G is not Q8 -free. Then M (being modular) is Hamiltonian and so M D Q E, Q Š Q8 , exp.E/ 2. In particular, ˆ.M / is of order 2 and exp.M / D 4. In view of N < ˆ.G/ < M , we get exp.N / D 4 and N is abelian of type .4; 2/ since N is metacyclic. In that case, jGj D 26 . We get ˆ.M / D Ã1 .M / D Ã1 .N / D ˆ.N / and so M=ˆ.N / is an elementary abelian subgroup of order 16 in the minimal nonabelian group G=ˆ.N /, contrary to Lemma 65.1. Thus, all maximal subgroups of G are Q8 -free. Proposition 78.7. Suppose that d.N / D 2. Then for each maximal subgroup M of G we have d.M / D 3. Also, ˆ.N / D Ã2 .G/, E D 1 .G/ D 1 .ˆ.G// Š E8 , E Z.ˆ.G//, and either G=E Š Q8 with 2 .G/ D ˆ.G/ being abelian of type .4; 2; 2/ or G=E is noncyclic and ordinary metacyclic. Proof. By (), 1 .G/ is elementary abelian.

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Groups of prime power order

Set F D ˆ.G/ so we have F=ˆ.N / D ˆ.G=ˆ.N // D 1 .G=ˆ.N // Š E8 (Proposition 78.6). Since ˆ.N / ˆ.F /, we get ˆ.N / D ˆ.F /. Thus d.F / D 3 and F is D8 -free and, by Proposition 78.6, Q8 -free, E D 1 .F / Š E8 (by ()) is normal in G. Let M be any maximal subgroup of G so that M=ˆ.N / is abelian of type .4; 2; 2/, ˆ.N / ˆ.M / and so d.M / D 3. But M is also D8 -free and Q8 -free and therefore, by (), 1 .M / Š E8 which implies 1 .M / D 1 .F / D 1 .G/ since 1 .G/ 1 .M / in view of 1 .G=N / D 1 .M=N / (Lemma 65.3). We have ˆ.N / Ã2 .G/ (Proposition 78.6). On the other hand, exp.G=Ã2 .G// D 4 and so each maximal subgroup of G=Ã2 .G/, being modular and Q8 -free, is abelian. Thus G=Ã2 .G/ is minimal nonabelian of exponent 4 and so jG=Ã2 .G/j 25 . It follows Ã2 .G/ D ˆ.N /. If G=E is not D8 -free, then there is a normal subgroup N of G such that E N and G=N Š D8 . By Proposition 78.1, N must be metacyclic, a contradiction. Hence G=E is D8 -free so it is modular. Suppose that G=E is not Q8 -free. Then G=E is Hamiltonian (73 or Appendix 24; by the previous paragraph, G=E is modular). Since d.G=E/ D 2, we get G=E Š Q8 so 2 .G/ D ˆ.G/ since E D 1 .G/. On the other hand, G=E cannot act faithfully on E since Aut.E/ has no subgroups Š Q8 , and so CG .E/ ˆ.G/. In particular, ˆ.G/ is abelian of type .4; 2; 2/. We assume that G=E is Q8 -free. In that case, by (), G=E is ordinary metacyclic but noncyclic since ˆ.G/ E. There is a cyclic normal subgroup S=E of G=E with the cyclic factor-group G=S . Let s 2 S be such that S D hE; si and let r 2 G S be such that G D hS; ri. Since E ˆ.G/, we have G D hr; si. Since S D hE; si is a proper subgroup of G, it follows that S is D8 -free and, by Proposition 78.6, Q8 -free. Since S=hsiS is a subgroup of the symmetric group SjSjWhsij D S4 , whose Sylow 2-subgroup is isomorphic to D8 , that quotient group is abelian and contains a subgroup isomorphic S=.S hsi/ Š E4 so hsi is normal in S and s induces an automorphism of order 2 on E which implies that s 2 centralizes E. Since hE; ri < G, we get (as in the previous paragraph) that r 2 centralizes E. On the other hand, ˆ.G/ D hE; r 2 ; s 2 i and so we get again E Z.ˆ.G//. We summarize our results in a somewhat different form. Theorem 78.8 (Janko). Let G be a minimal nonmodular 2-group of order > 25 . Then each proper subgroup of G is Q8 -free and G=Ã2 .G/ is minimal nonabelian of order 24 or 25 . (a) Suppose that jG=Ã2 .G/j D 24 . If N is any normal subgroup of G such that G=N Š D8 , then N is cyclic. If G=Ã2 .G/ is nonmetacyclic, then G is Q8 -free and 1 .G/ Š E8 with G=1 .G/ cyclic. If G=Ã2 .G/ is metacyclic, then G is also metacyclic and G is either minimal nonabelian or an A2 -group. (b) Suppose that jG=Ã2 .G/j D 25 . Then G=Ã2 .G/ is nonmetacyclic, G is not Q8 -free and 1 .G/ Š E8 with G=1 .G/ Š Q8 or G=1 .G/ is ordinary meta-

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cyclic (but not cyclic). Moreover, if N is any normal subgroup of G such that G=N Š D8 , then N is ordinary metacyclic but noncyclic. Remark 4. Let us change the last paragraph of the proof of Proposition 78.1 by the following argument. Let A be a minimal nonabelian subgroup of G. Let jAj > 8. If Z.A/ is cyclic, setting N D .N \ A/ N1 , we get G D A N1 , d.G/ D 4, a contradiction. Let Z.A/ Š E4 . Then A is metacyclic without cyclic subgroup of index 2. Since A is D8 -free, jAj D 25 and G D A N1 , where N < N1 is of order 2. In that case, G is modular, a contradiction. If Z.A/ Š E8 , then A is D8 -free so A D G. In that case, G 0 < N , which is a contradiction. Now let jAj D 8; then A Š Q8 . By Lemma 1.1, jG 0 j D 4. If L is as in the last paragraph of the proof of Proposition 78.1, then G 0 < L and G 0 6 N D 1 .L/ so G 0 Š C4 . Then A0 D 1 .G 0 / and AN=A0 Š E16 , a contradiction since G=A0 is minimal nonabelian in view of j.G=A/0 D G 0 =A0 Š C2 (Lemma 65.2(a)) so must be j1 .G=A0 /j 8 (Lemma 65.1). 2o . We recall that a p-group G is modular if and only if any subgroups X and Y of G are permutable, i.e., XY D YX. We turn now to the case p > 2. Proposition 78.9. Let G be a modular p-group with p > 2 and d.G/ D 2. Then G is metacyclic. Proposition 78.10. Let G be a minimal nonmodular p-group, p > 2, which is generated by two subgroups A and B of order p. Then G Š S.p 3 / (the nonabelian group of order p 3 and exponent p). Proof. Since G is a p-group, G1 D hAG i and G2 D hB G i are proper normal subgroups of G and so G1 and G2 are modular. It follows that G1 and G2 are elementary abelian. But hG1 ; Bi D hA; Bi D hG2 ; Ai D G, and so G1 and G2 are two distinct maximal subgroups of G. By Lemma 64.1(u), we have jG 0 j D p and G 0 G1 \ G2 . Thus G=G 0 is abelian and G=G 0 is generated by elementary abelian subgroups G1 =G 0 and G2 =G 0 . Hence G=G 0 is elementary abelian and d.G/ D 2 implies that G=G 0 Š Ep 2 . Thus, G Š S.p 3 / since the metacyclic nonabelian group of order p 3 is modular. Proposition 78.11 (see Theorem 44.13). Let G be a minimal nonmodular p-group. Then G possesses a normal subgroup N such that d.N / 2, N Ã1 .G/, and G=N is a nonmodular group of order p 3 . If p D 2, then G=N Š D8 and if p > 2, then G=N Š S.p 3 /, N D Ã1 .G/, and N is metacyclic. The subgroup N is metacyclic. In particular, if p > 2, then G is nonmetacyclic. Proof. There are non-permutable cyclic subgroups hai and hbi. It follows G D ha; bi and so d.G/ D 2. Since hap ; b p i ˆ.G/, the subgroups E D hap ; bi and F D ha; b p i are proper subgroups of G. Hence E and F are modular and so E D hap ihbi, F D haihb p i, and G D hE; F i. Set N D hap ihb p i so that jE W N j D jF W N j D p. It follows that N is normal in G, N Ã1 .G/, and d.N / 2. It remains to determine

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Groups of prime power order

N where GN is a minimal nonmodular p-group the structure of GN D G=N D ha; N bi, generated by elements aN and bN of order p. If p D 2, then GN is dihedral, and, by (), GN Š D8 . If p > 2, then Proposition 78.10 implies that GN Š S.p 3 /. In that case we have N D Ã1 .G/ and Proposition 78.9 implies that N is metacyclic. Proposition 78.12. Let G be a minimal nonmodular p-group, p > 2, with jGj > p 4 . Then 1 .G/ is elementary abelian of order p 3 . Proof. Let A and B be subgroups of order p in G such that AB ¤ BA. Then G D hA; Bi and so Proposition 78.10 implies that G Š S.p 3 /, a contradiction. We have proved that AB D BA and so hA; Bi is abelian. Hence 1 .G/ is elementary abelian. Assume that each proper subgroup of G is metacyclic. By Proposition 78.11, G is nonmetacyclic and so jGj p 4 (Theorem 69.1 for p > 2), a contradiction. Let M be a nonmetacyclic maximal subgroup of G. Since M is modular, Proposition 78.9 implies that d.M / 3. By Proposition 78.11, G is not a 3-group of maximal class (since it has a maximal subgroup H which satisfies H=Ã1 .H / Š S.33 /, so nonmodular. Therefore, if j1 .G/j < p 3 , then, by Theorem 13.7, G must be metacyclic so modular (Proposition 78.11), a contradiction. Thus, 1 .G/j p 3 . Remark 5. Minimal nonmodular p-group of exponent p is isomorphic to S.p 3 /. Indeed, p > 2. All proper subgroups of G are elementary abelian so G Š S.p 3 /, by Lemma 65.3. Theorem 78.13. Let G be a minimal nonmodular p-group, p > 2, with jGj > p 4 . If Ã1 .G/ is cyclic, then 1 .G/ Š Ep 3 and G=1 .G/ is cyclic of order p 2 (i.e. G is an L3 -group; see 17, 18). Proof. By Proposition 78.11, G=Ã1 .G/ Š S.p 3 /. By assumption, N D Ã1 .G/ is cyclic. By Proposition 78.12, E D 1 .G/ is elementary abelian of order p 3 . But jE \ N j D p and E does not cover G=N , and so E Š Ep 3 , by the product formula. On the other hand, there is a 2 G N with hap i D N . Since jG W haij D p 2 and jhai \ Ej D p, we get G D hE; ai, by the product formula, and we are done. Proposition 78.14. Let G be a minimal nonmodular p-group, p > 2, with jGj > p 4 . Suppose that d.Ã1 .G// D 2 and let M be any maximal subgroup of G. Then d.M / 3. Proof. Suppose that this is false. Let M be a maximal subgroup of G with d.M / 4. Set N D Ã1 .G/ so that M=N Š Ep 2 and N=ˆ.N / Š Ep 2 . Since ˆ.N / ˆ.M /, we must have ˆ.M / D ˆ.N / so that M=ˆ.N / Š Ep 4 . We shall study the structure of G=ˆ.N / (which is also minimal nonmodular of order > p 4 and exponent p 2 ) and so we may assume ˆ.N / D f1g which implies M Š Ep 4 . Since 1 .G/ is elementary abelian, we have M D 1 .G/. If x 2 G M , then x p 2 Z.G/# . There is y 2 G M such that y p 2 N hx p i since N D Ã1 .G/; then again y p 2 Z.G/# . Since N Š Ep 2 , we get N Z.G/. If Z.G/ > N , then Z.G/=N D Z.G=N / D ˆ.G=N / and

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333

so Z.G/ D ˆ.G/. But then G is minimal nonabelian. By Lemma 65.1, we get j1 .G/j p 3 , a contradiction. We have proved that Z.G/ D N . By Lemma 1.1, jGj D p 5 D pjZ.G/jjG 0 j and so jG 0 j D p 2 . Since G=N Š S.p 3 /, G 0 ¤ N.D Z.G//, G 0 \ Z.G/ Š Cp and G 0 < M so that CG .G 0 / D M . Since N D Ã1 .G/, where is v 2 G M such that v p 2 N G 0 . It follows that the subgroup H D hG 0 ; vi is nonabelian. We claim that H=hvi Š S.p 3 /. Assume that this is false. Then that quotient group is N elementary abelian. Set GN D G=hvi; then CGN .GN 0 / hMN ; vi N D GN so GN D Z.G/. 0 N N N N Since G=Z.G/ Š Ep 2 , the group G is minimal nonabelian so jG j D p (Lemma 65.1), a contradiction. We have proved that each maximal subgroup of G is generated by three elements. Theorem 78.15 (Janko). Let G be a minimal nonmodular p-group, p > 2, with jGj > p 4 . Then Ã1 .G/ is metacyclic and G=Ã1 .G/ Š S.p 3 / (nonabelian group of order p 3 and exponent p). If Ã1 .G/ is noncyclic, then ˆ.G/ D Ã1 .G/ Cp , 1 .ˆ.G// D 1 .G/ Š Ep 3 , G=1 .G/ is metacyclic and for each maximal subgroup M of G we have d.M / D 3. Proof. Set N D Ã1 .G/ .< ˆ.G// and suppose that d.N / D 2. By Proposition 78.12, 1 .G/ is elementary abelian of order p 3 and 1 .G/ \ N Š Ep 2 . Since 1 .G/ does not cover G=N Š S.p 3 /, N 1 .G/ is contained in a maximal subgroup M of G. By Proposition 78.14, d.M / 3 and the modularity of M implies d.M / D d.1 .M // [Suz1]. This implies 1 .G/ Š Ep 3 and .N 1 .G//=N D ˆ.G=N /. Thus ˆ.G/ D N 1 .G/ and so for each maximal subgroup X of G, we have d.X/ D 3 since X ˆ.G/ and d.X/ D d.1 .X// D d.1 .G//. We know that Aut.1 .G// does not possess an automorphism of order p 2 (see, for example, 33). There are elements a; b 2 G such that N D hap ihb p i and ap and b p centralize 1 .G/. Hence ˆ.G/ D N Z with jZj D p. If G=1 .G/ is nonmodular, then (Proposition 78.11) there is a normal subgroup K of G with K 1 .G/, G=K Š S.p 3 /, and d.K/ 2. This is a contradiction since d.K/ D d.1 .K// D 3. Hence G=1 .G/ is modular and since d.G=1 .G// 2, G=1 .G/ is metacyclic and our theorem is proved.

79

Nonmodular quaternion-free 2-groups

Modular Q8 -free 2-groups are classified in [Iwa] (see 73). Here we classify nonmodular Q8 -free 2-groups. The original proof of the corresponding classification theorem, given in [Wil2], depends on the structure theory of powerful 2-groups. In addition, in the proof of Lemma 10 and 13 in [Wil2] there are some gaps. Our new proof of the classification theorem is completely elementary and does not involve powerful 2groups. Nevertheless, the proof is very involved and reaches probably a deepest result ever proved in the finite 2-group theory. We first prove some easy preliminary results. Then we state the Main Theorem 79.7 and afterwards we describe in great detail the groups appearing in the Main Theorem. Propositions 79.8 to 79.11, describing these groups, are also of independent interest since they are needed by applying the Main Theorem in future investigations. After that the proof of the Main Theorem follows. Lemma 79.1. In a Q8 -free 2-group X there are no elements x; y with o.x/ D 2k > 2 and o.y/ D 4 so that x y D x 1 . If D X and D Š D8 , then CX .D/ is elementary abelian. k1

Proof. If y 2 D x 2 , then hx; yi Š Q2kC1 . If hxi \ hyi D f1g, then we have k1 hx; yi=hx 2 y 2 i Š Q2kC1 . Suppose that D X, where D D ha; t j a4 D t 2 D 1; at D a1 i Š D8 . if v is an element of order 4 in CX .D/, then o.t v/ D 4 and t v inverts a, a contradiction. Hence CX .D/ must be elementary abelian. Lemma 79.2. Let X be a Q8 -free 2-group with elements a and b of order 4 such that Œa; b 2 D Œa2 ; b D 1. If Œa; b ¤ 1, then ha; bi is minimal nonabelian nonmetacyclic of order 24 and therefore Œa; b D a2 b 2 and ab is an involution. Proof. Without loss of generality, one may assume that X D ha; bi holds. We have ha2 ; b 2 i Z.X/. Set Œa; b D c; then c ¤ 1. We compute 1 D Œa2 ; b D Œa; ba Œa; b D c a c

and

1 D Œa; b 2 D Œa; bŒa; bb D cc b :

By Lemma 79.1, c must be an involution and, by the displayed equalities, Œc; a D Œc; b D 1. Hence hci is normal in X and X=hci is abelian. It follows that X 0 D hci and so X, by Lemma 65(a), is minimal nonabelian (and so of class 2) and therefore exp.X/ D 4. By assumption, X 6Š Q8 , and X 6Š D8 since D8 has only one cyclic

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335

subgroup of order 4. We have proved that jXj 24 . Considering X=hci, we conclude that jha; bij 25 . On the other hand, c D Œa; b, a2 , and b 2 are central involutions in X. Set V D 2 ha c; b 2 ci so that V Z.ha; bi/ and jV j D 4. We consider X=V and compute ab D aŒa; b D ac D a1 .a2 c/;

b a D bŒb; a D bc D b 1 .b 2 c/:

Since X=V is Q8 -free, Lemma 79.1 implies that at least one of a2 or b 2 is contained in V . Hence a2 D b 2 c or b 2 D a2 c and so in any case c D a2 b 2 2 V since a2 c b 2 c D a2 b 2 . It follows from the displayed equalities that X=V Š E4 so jXj D jV jjE4 j D 24 . Lemma 79.3 (see 78 and especially Proposition 78.4). Let G be a Q8 -free minimal nonmodular 2-group of order > 8. Then G has a normal elementary abelian subgroup E D 1 .G/ D hn; z; t i of order 8 with G=E cyclic. There is an element x 2 G E of order 2sC1 , s 1, such that G D hE; xi, E \ hxi= hni, and t x D t z, z x D zn , D 0; 1, where in case D 1 we must have s > 1 and we have in that case G 0 D hn; zi Š E4 and Z.G/ D hx 4 i. If D 0, then G is a minimal nonabelian nonmetacyclic group so E Š E8 . In any case, hx 2 i is normal in G, G=hx 2 i Š D8 , and ˆ.G/ D hx 2 ; zi is abelian of type .2s ; 2/. Lemma 79.4. Let V be a minimal non-quaternion-free 2-group. Then there is a normal subgroup U of V such that V =U Š Q8 and U < ˆ.V / so that d.V / D 2. We have ˆ.V /=U D Z.V =U / so that for each x 2 V ˆ.V /, x 2 2 ˆ.V / U . In particular, there are no involutions in V ˆ.V /. We use very often the following Lemma 79.5 (= Lemma 64.1(u)). If A and B are two distinct maximal subgroups of a p-group G, then jG 0 W .A0 B 0 /j p. For the sake completeness we also state the Iwasawa’s result in a suitable form. Proposition 79.6 ([Iwa] and 73). A 2-group G is modular if and only if G is D8 free. A 2-group G is modular and Q8 -free if and only if G possesses a normal abelian subgroup A with cyclic G=A and there is an element g 2 G and an integer s 2 such s that G D hA; gi and ag D a1C2 for all a 2 A (and so, if exp.G/ 4, then G is abelian). Main Theorem 79.7 (B. Wilkens). A finite 2-group G is nonmodular and quaternionfree if and only if G is one of the following groups: (a) (Wilkens group of type (a) with respect to N ) G is a semidirect product hxi N , where N is a maximal abelian normal subgroup of G with exp.N / > 2 and, if t is the involution in hxi, then every element in N is inverted by t . (b) (Wilkens group of type (b) with respect to N ) G D N hxi, where N is a maximal elementary abelian normal subgroup of G and hxi is not normal in G.

336

Groups of prime power order

(c) (Wilkens group of type (c) with respect to N , x, t ) G D hN; x; t i, where N is an elementary abelian normal subgroup of G and t is an involution with ŒN; t D 1. k If o.xN / D 2k , then G=N Š M2kC1 , k 3, and x 2 ¤ 1; furthermore k1 k1 Œx 2 ; N D 1 and ht; x 2 i Š D8 . We analyze now in great detail the above Wilkens groups of types (a), (b), and (c). In what follows we call these groups Wx -groups, where x 2 fa; b; cg so, for example, Wb -group is a Wilkens group of type (b). Remark 1. A nonabelian 2-group G D ht i N is said to be generalized dihedral with base N , if t is an involution and N is an (abelian) subgroup of index 2 in G such that t inverts N . We claim that if also G D ht1 i N1 is generalized dihedral with base N1 , then N1 D N . Assume that this is false. We have N \ N1 D Z.G/ and jG W .N \ N1 /j D 4, G D N [ N t is a partition and all elements of the coset N t are involutions and invert N . Therefore, if x 2 N1 \ N t , then x centralizes and inverts N \ N ˛ so exp.N \ N1 / D 2. It follows that N1 D hx; N \ N1 i is elementary abelian, which is a contradiction since exp.N1 / > 2. In particular, the base of G is characteristic in G. Proposition 79.8. Let G be a Wa -group with respect to N . Then 1 .G/ D N ht i, where t is an involution in G N inverting N and N is characteristic in 1 .G/ and so in G. If G is a Wa -group with respect to N1 , then N D N1 . Also, G is not D8 free but G is Q8 -free. If z 2 1 .Z.G//, then G=hzi is either abelian or a Wa - or Wb -group. Proof. By hypothesis, G is a semidirect product hxi N , where N is a maximal abelian normal subgroup of G with exp.N / > 2 and, if t is the involution in hxi, then every element in N is inverted by t . Since G=N is cyclic, we have 1 .G/ N ht i. The coset N t consist of involutions and so 1 .G/ D N ht i. It follows that 1 .G/ is a generalized dihedral group with respect N . By Remark 1, N is the unique base of 1 .G/ so it is characteristic in 1 .G/ and in G. Thus, if G is also a Wa -group with respect to N1 , then N1 D N . Since t inverts N and exp.N / > 2, G is not D8 -free. Suppose that G is not Q8 -free. Let V be a minimal non-Q8 -free subgroup of G so that V has a normal subgroup U with V =U Š Q8 and ˆ.V / > U ; it follows that d.V / D 2. Since V 6 N and G=N is cyclic, we see that V =.V \ N / > f1g is cyclic. Let t 0 be an element in V N such that .t 0 /2 2 N . Then N t 0 is the involution in G=N and so all elements in N t 0 are involutions. In particular, all elements in the set S D .V \ N /t 0 are involutions. By Lemma 79.4, S ˆ.V / and so also hS i D .V \ N /ht 0 i ˆ.V /. But then V =ˆ.V /, as an epimorphic image of V =.V \ N /, is cyclic, a contradiction. Let z 2 Z.G/ be an involution. We want to determine the structure of G=hzi. We know that G D hxi N is a semidirect product. Let t be the involution in hxi. We have z 2 N and t inverts N and so t inverts N=hzi. If exp.N=hzi/ > 2, then G=hzi is a Wa -group.

79

Nonmodular quaternion-free 2-groups

337

Let exp.N=hzi/ D 2. Set E D ht iN . Then E=hzi D 1 .G=hzi/ is a maximal elementary abelian normal subgroup of G=hzi. If hx; zi D hxi hzi is not normal in G, then G=hzi is a Wb -group. Suppose that hx; zi is normal in G. Then G 0 hx; zi \ N D hzi so G 0 D hzi. In that case, G=hzi is abelian, and we are done. Remark 2. Let a 2-group G of exponent > 2 have two distinct elementary abelian subgroups E and E1 of index 2. Then G D D L, where D Š D8 and exp.L/ divides 2. Since exp.G/ > 2, G is nonabelian. It follows that E \ E1 D Z.G/ has index 4 in G. Let A be a minimal abelian subgroup of G; then jA W .A \ Z.G//j > 2 so G D AZ.G/, by the product formula. In that case, A \ Z.G/ D Z.A/. If Z.G/ D Z.A/ L, then G D A L and exp.L/ divides 2. Since A has two distinct elementary abelian subgroup of index 2, then A Š D8 (Lemma 65.1). Proposition 79.9. (i) A 2-group G is a Wb -group with respect to E if and only if E is a maximal normal elementary abelian subgroup E such that G=E is cyclic and G is not D8 -free. (ii) Let G be a Wb -group with respect to E. Then G is Q8 -free. We have j1 .G/ W Ej 2. If j1 .G/ W Ej D 2, then G has exactly two maximal normal elementary abelian subgroups E and E1 and we have 1 .G/ D EE1 . In that case, if G is also a Wb -group with respect to E1 , then j1 .G/ W E1 j D 2 and 1 .G/ Š D8 E2s . Let z 2 1 .Z.G//. Then G=hzi is either abelian or a Wb -group or G=hzi Š D F , where exp.F / 2 and either D Š D8 or D Š M2n , n 4 (in which case G=hzi is modular and nonabelian). Finally, if G is any 2-group with an elementary abelian normal subgroup E0 such that G D hE0 ; yi (and so G=E0 is cyclic) and hyi is not normal in G, then G is a Wb –group with respect to any maximal elementary abelian normal subgroup E of G containing E0 . Proof. (i) Let G be a nonmodular 2-group possessing a maximal elementary abelian normal subgroup E such that G=E is cyclic. We have G D hE; xi for some x 2 G and if hxi 6E G, then G is a Wb -group. Assume that hxi is normal in G. In that case, G 0 hxi \ E and, since G is nonmodular (and so nonabelian), G 0 D hxi\E D hzi Š C2 . We have jG W CG .x/j D 2 since jG W CG .x/j jG 0 j for each x 2 G. We set E1 D CE .x/ so that jE W E1 j D 2 and E1 Z.G/. Let t be an involution in E E1 and let V be a complement of hzi in E1 so that G D V hx; t i, by the product formula. If jG=Ej D 2s > 2, then hx; t i Š M2sC2 , s 2 since a 2-group of maximal class has no cyclic epimorphic s images of order 2s > 2, and so for each a 2 A D hxi V , at D a1C2 since exp.V / divides 2. But Proposition 79.6 implies that G is modular, a contradiction. Hence jG=Ej D 2 and hx; t i Š D8 . In that case xQ D xt is an involution in G E, G D Ehxi, Q hxi Q is not normal in G, and so G is a Wb -group. (ii) Conversely, let G be a Wb -group with respect to E so that G D hE; gi, where E is a maximal normal elementary abelian subgroup of G and hgi is not normal in G.

338

Groups of prime power order

Set Z D hgi \ E so that jZj 2 and Z Z.G/. Set S D NG .hgi/ so that S ¤ G and S \ E < E. Since G D hgiS , we get NG .S / D hgiNE .S /, by the modular law, so, in view of NE .S / > E \ S , there is an involution n 2 E S normalizing S . Since Œn; g 2 Œn; S and Œn; g 2 ŒE; g E, we get Œn; g 2 S \ E and 1 ¤ u D Œn; g 62 hgi since n 62 S D NG .hgi/. We have Œn; g D ng 1 ng D nng D u. 2 On the other hand, ˆ.S / D hg 2 i and so hg 2 i is normal in hS; ni so that ng D nz with z 2 Z. Hence hgi normalizes hn; ng ; Zi and acts nontrivially on the four-group hn; ng ; Zi=Z, where hg 2 i Z. It follows that hn; gi=hg 2 i Š D8 and so G is not D8 -free. We claim that G is Q8 -free. Indeed, if V is a minimal non-Q8 -free subgroup of G, then, by Lemma 79.4, there are no involutions in V ˆ.V / so that ˆ.V / V \ E. But then V =ˆ.V / is cyclic since V =.V \ E/ is, a contradiction. Set W =E D 1 .G=E/ so that 1 .G/ W and j1 .G/ W Ej jW W Ej D 2. Suppose that j1 .G/ W Ej D 2 so that 1 .G/ D W and there is an involution t 2 W E. Since E is a maximal normal elementary abelian subgroup of G, W is not elementary abelian so ht i is not normal in W , and we conclude that W is a Wb -group with respect to E. Let t1 2 W E be an involution, t1 ¤ t . Then tE D t1 E since W W Ej D 2. In that case t t1 2 E is involution so all involutions in W E commute with t . Set E1 D CW .t / so that E1 E is the set of all involutions in W E, by what has just been proved. Since hE1 Ei D E1 , E1 is normal in G and E and E1 are the only maximal normal elementary abelian subgroups of G. If G is also a Wb -group with respect to E1 , then G=E1 must be cyclic and so jW W E1 j D 2 and in that case W D 1 .G/ Š D8 E2s , by Remark 2. Let z be a central involution in G, where G D hE; xi and hxi is not normal in G; then z 2 E since E is a maximal elementary abelian subgroup of G. If z 2 hxi, then hxi=hzi is not normal in G=hzi so G=hzi is a Wb -group with respect to E=hzi. Suppose that z 62 hxi. If hx; zi is not normal in G, then again G=hzi is a Wb -group. Assume that hx; zi D hxihzi is normal in G. If hxi\E D f1g, then hx; zi\E D hzi and G 0 hx; zi \ E D hzi and so G=hzi is abelian. Assume that hxi \ E ¤ f1g so that hx; zi \ E D 1 .hxi/ hzi Š E4 and suppose that G=hzi D GN is nonabelian. Then GN D hxi N EN with hxi N \ EN Š C2 and both hxi N and EN are normal in GN so that 0 N N N G D hxi N \ E. Let tN be an involution in E which does not centralize hxi. N If o.x/ N D 4, N > 4, then hx; N tNi Š M2n , n 4 since hx; N tNi \ EN Š E4 then hx; N tNi Š D8 and if o.x/ N htNi VN , where .hxi N VN D C N .x/ (see Theorem 1.2). We have EN D .hxi N \ E/ N \ E/ E N N N N so that G D V hx; N t i and we are done. Proposition 79.10. Let G be a Wc -group with respect to N , x, t . Then G is Q8 k1 free but is not D8 -free. We have 1 .G/ D hx 2 ; t iN Š D8 E2s (k 3) and G D 1 .G/hxi so that G=1 .G/ is cyclic of order 4. Also, N is the unique maximal normal elementary abelian subgroup of G. No subgroup of order 8 in hxi k is normal in G. The involution z D x 2 lies in G 0 \ Z.G/ and G=hzi is a Wb -group. If z 0 2 1 .Z.G// and z 0 ¤ z, then z 0 2 N and G=hz 0 i is a Wc -group.

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Nonmodular quaternion-free 2-groups

339

Proof. By hypothesis, G D hN; x; t i, where N is an elementary abelian normal subgroup of G and t is an involution with ŒN; t D 1. If o.xN / D 2k , then G=N Š k k1 k1 M2kC1 , k 3, and x 2 ¤ 1; furthermore Œx 2 ; N D 1 and ht; x 2 i Š D8 so that k1 and z D a2 . We have G is not D8 -free. We set a D x 2 .G=N /0 D 1 .Z.G=N // D .haiN /=N

and 1 .G=N / D .ht; aiN /=N D W =N;

where W D ht; aiN and Z.W / D N . Each involution in G is contained in W and W D 1 .W / and so 1 .G/ D W Š D8 E2s (Remark 2) and G D 1 .G/hxi so that G=1 .G/ is cyclic of order 4. By the structure of G=N , if X is a normal subgroup of G with N X W , then X 2 fN; W; haiN g, where haiN is abelian of type .4; 2; : : : ; 2/ so that N is a maximal normal elementary abelian subgroup of G. Let N1 be any maximal normal elementary abelian subgroup of G; then N1 < 1 .G/ D W . Assume that N1 ¤ N . Since N1 does not cover W =N (since all elements in .haiN / N are of order 4), we get j.N N1 / W N j D 2, N N1 G G and so (by the above) N N1 D haiN , a contradiction. Thus, N is the unique maximal normal elementary abelian subgroup of G so it is characteristic in G. Let Y hxi with jY j 8 and assume that t normalizes Y . Since t inverts hai and hai < Y , it follows that Y ht i is of maximal class and order 2m , m 4, and so .Y ht i/=hzi Š D2m1 is isomorphic to a proper subgroup of G=N Š M2n , n 4, which is not the case. Thus, Y is not normal in G. Let G be not Q8 -free and let V G be minimal non-Q8 -free. By Lemma 79.4, there are no involutions in V ˆ.V / and so ˆ.V / V \ N . On the other hand, ˆ.V / is contained in the maximal subgroup hxiN of G and note that 1 .hxiN / D N (since N \ hxi D hzi and a centralizes N ). It follows that 1 .ˆ.V // D V \ N . Note that G=N has exactly three involutions: Na, N t , and N.at /, where all elements in the coset Na are of order 4 and all elements in cosets N t and N.at / are involutions. Let .V \ N /s (s 2 V ) be an involution in V =.V \ N /. Then N s is an involution in G=N . If all elements in the coset N s are involutions, then s 2 ˆ.V /, contrary to the above fact that 1 .ˆ.V // D V \N . It follows that N s D Na and so .V \N /s D .Na/\V is the unique involution in V =.V \ N /. Since G=N is Q8 -free, we get that V =.V \ N / is cyclic. But then V =ˆ.V / is also cyclic, a contradiction. N Since 1 .G/ D 1 .G/=hzi is an We shall determine the structure of G=hzi D G. N GN D 1 .G/hxi, N and hxi N is not normal in elementary abelian normal subgroup of G, N N G (noting that z 2 hxi and hxi is not normal in G), G is a Wb -group. Let z 0 be an involution in Z.G/ and z 0 ¤ z. Then z 0 2 N and hz 0 i \ ha; t i D f1g so that G=hz 0 i is not D8 -free. Obviously, G=hz 0 i is a Wc -group. Proposition 79.11. Let G be one of the Wilkens groups. Suppose that there is an involution z 2 Z.G/ such that G=hzi is modular (i.e., D8 -free). (i) If G=hzi is abelian, then G is a Wb -group and more precisely: G D D E n where exp.E/ 2 and D D hx; t j x 2 D t 2 D 1, n 1, Œx; t D z, z 2 D

340

Groups of prime power order

Œx; z D Œt; z D 1i. (If n D 1, then D Š D8 , and if n > 1, then D is minimal nonabelian nonmetacyclic with 1 .D/ 6 Z.D/.) (ii) If G=hzi is nonabelian, then there is another involution z 0 2 Z.G/ such that hz 0 i is a characteristic subgroup of G and G=hz 0 i is a Wb -group. Proof. (i) Suppose that G=hzi is abelian. Then G 0 D hzi and by Propositions 79.8, 79.9, and 79.10, G is either a Wa - or Wb -group. Suppose that G is a Wa -group. Then G D hxi N (a semidirect product), where N is a maximal normal abelian subgroup of G with exp.N / > 2 and if t is the involution in hxi, then t inverts each element of N . Suppose n 2 N with o.n/ > 2. Then hn; t i is dihedral and so n2 2 hn; t i0 . It follows that n2 2 hzi and therefore N=hzi is elementary abelian with Ã1 .N / D hzi and jN W 1 .N /j D 2. Suppose hxi > ht i and let v 2 hxi with v 2 D t . Let n 2 N with o.n/ D 4. We have Œn; v ¤ 1 and so 2 Œn; v D z which gives nv D nz. But then nt D nv D .nz/v D nv z v D nzz D n since z 2 Z.G/. This is a contradiction and so hxi D ht i. If n 2 N with o.n/ D 4, then D D hn; t i Š D8 . Let E be a complement of hzi in 1 .N /, where t centralizes 1 .N /. We get G D ht i N D D E, where D Š D8 and exp.E/ 2. Suppose that G is a Wb -group. Then G D hN; xi, where N is a maximal normal elementary abelian subgroup of G and hxi is not normal in G. We have z 2 N and G 0 D hzi. If z 2 hxi \ N , then hxi is normal in G, a contradiction. Hence z 62 hxi and hx; zi D hxi hzi is normal in G. Since hx; zi contains exactly two cyclic subgroups hxi and hxzi of index 2 not containing hzi, we have jG W NG .hxi/j D 2. There is t 2 N NG .hxi/ such that x t D xz. Also note that NG .hxi/ centralizes hxi (since G 0 D hzi). Let E be a complement of hz; hxi \ N i in NN .hxi/. Then G D D E, n where D D hx; t j x 2 D t 2 D 1; n 1; Œx; t D z; z 2 D Œx; z D Œt; z D 1i. (ii) Suppose that G=hzi is nonabelian. By Propositions 79.8, 79.9, and 79.10, G is a Wb -group. Then G D hN; xi, where N is a maximal normal elementary abelian subgroup of G and hxi is not normal in G. If z 2 hxi \ N , then the fact that hxi is not normal in G gives that hxi=hzi is not normal in G=hzi and so G=hzi is a Wb -group, a contradiction. Hence z 62 hxi. If hx; zi is not normal in G, then again G=hzi is a Wb -group, a contradiction. Hence hx; zi D hxi hzi is normal in G. Assume first that hxi \ N D f1g. Then hx; zi \ N D hzi and G 0 hx; zi \ N D hzi and so G=hzi is abelian, a contradiction. Hence hxi \ N > f1g and so hx; zi \ N D 1 .hxi/ hzi. Since G 0 hx; zi \ N and G=hzi is nonabelian (by assumption), we get 1 ¤ jG 0 j 4 and G 0 6 hzi. We have 1 .hxi/ Z.G/. Set 1 .hxi/ D hx0 i and S D NG .hxi/ so that hx0 ; zi Z.G/, jG W S j D 2 and jN W .S \ N /j D 2 because hxi is not normal in G and the abelian normal subgroup hx; zi has exactly two cyclic subgroups hxi and hxzi of index 2. Therefore, we have Œx; s D z or Œx; s D x0 z for an s 2 N S and so ˆ.G/ D hx 2 ; Œhxi; N i D hx 2 i hzi. Suppose o.x/ 8 so that ˆ.ˆ.G// D hx 4 i hx0 i and hx0 i is a characteristic subgroup of G. But hxi=hx0 i is not normal in G=hx0 i (since hxi is not normal in G) and so G=hx0 i is a Wb -group, and we are done in this case.

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Suppose o.x/ D 4 so that x 2 D x0 . If hxi 6 Z.S /, then there is an involution t in S \ N which inverts hxi and so hx; t i Š D8 . But then G=hzi is not D8 -free, a contradiction: G=hzi is modular. Hence hxi is central in S and so G 0 D hŒx; si D hx0 zi (since in case G 0 D hŒx; si D hzi, G=hzi would be abelian). But then hx; si is the minimal nonabelian nonmetacyclic group of order 24 with hx; si0 D hx0 zi and hx; si=hzi Š D8 , contrary to our assumption that G=hzi is modular. Lemma 79.12. Let G be a Wb -group with respect to N . Suppose in addition that 1 .G/ D N . Then for each element g 2 G such that G D hN; gi, N \ hgi D hg0 i is of order 2 and G=hg0 i is also a Wb -group (and so G=hg0 i is nonmodular). Proof. It is enough to show that hgi is not normal in G (because then hgi=hg0 i is also not normal in G=hg0 i). Suppose false. Then G 0 N \ hgi D hg0 i and so G 0 D hg0 i. We have jG W CG .g/j D 2 and let t be an involution in N CN .g/. If o.g/ D 4, then hg; t i Š D8 and so gt is an involution in G N , a contradiction. Hence o.g/ > 4 and hg; t i Š M2n , n 4. If V is a complement of hg0 i in CN .g/, then G D V hg; t i. But then G is modular (see Proposition 79.6), a contradiction.

Proof of the Main Theorem Let G be a nonmodular quaternion-free 2-group of a smallest possible order which is not isomorphic to any Wilkens group. Hence any proper nonmodular subgroup and any proper nonmodular factor group is isomorphic to a Wilkens group. We shall study such a minimal counter-example G and our purpose is to show that such a group G does not exist. (i) There is a central involution z of G such that G=hzi is nonmodular (and so G=hzi is isomorphic to a Wilkens group). Suppose that this is false. Then for each z 2 1 .Z.G//, G=hzi is modular. Suppose that z0 2 1 .Z.G// is such that G=hz0 i is modular. Since G is nonmodular, there is a minimal nonmodular subgroup K of G which is isomorphic to a group of Lemma 79.3. Obviously, K is a Wb -group and so K ¤ G. Since G=hz0 i is modular, we have z0 2 K. If K Š D8 , then hz0 i D 1 .Z.G// D K 0 . Suppose that K has a normal elementary abelian subgroup E D hn; z; t i of order 8 such that K D hE; xi, o.x/ D 2sC1 , s 1, E \ hxi D hni, t x D t z, z x D zn , D 0; 1, 1 .K/ D E, and in case D 1 we have s > 1 and Z.K/ D hx 4 i. If D 1, then we must have z0 D n. But then K=hz0 i is nonmodular since K=hx 2 i Š D8 . Hence we have D 0 in which case K is minimal nonabelian nonmetacyclic with Z.K/ D hx 2 ; zi D ˆ.K/ and K 0 D hzi. Since K=hz0 i is modular (and K=hni is nonmodular), we have either z0 D z (and then K=hz0 i is abelian) or z0 D zn (in which case s > 1 and K=hzni Š M2sC2 ). Let H be a maximal subgroup of G containing K. Since H is nonmodular and H ¤ G, H is a Wilkens group. Since H=hz0 i is modular, we may use Proposition 79.11. If H=hz0 i is nonabelian, then there is another involution z00 in Z.H / such that hz00 i is a characteristic subgroup in H and H=hz00 i is nonmodular. But then z00 2 Z.G/

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and G=hz00 i is nonmodular, contrary to our assumption. Hence H=hz0 i must be abelian and so K=hz0 i is also abelian. In particular, z0 D z, where hzi D K 0 . In any case 1 .Z.G// D hzi is of order 2. By Proposition 79.11(a), we have H D D E0 , where exp.E0 / 2 and m

D D hy; t j y 2 D t 2 D 1; m 1; Œy; t D z; z 2 D Œz; y D Œz; t D 1i: If m D 1, then D Š D8 and if m > 1, then D is minimal nonabelian nonmetacyclic with E8 Š 1 .D/ 6 Z.D/ and z D z0 , where hzi D D 0 D H 0 D 1 .Z.G//. Suppose m D 1. Then Z.H / D hzi E0 is elementary abelian. If jE0 j 4, then acting with an element x 2 G H on Z.H /, we see that jCZ.H / .x/j 4 and CZ.H / .x/ Z.G/, contrary to the fact that 1 .Z.G// is of order 2. Hence jE0 j 2. If D D H Š D8 , then CG .D/ D would imply that G is of maximal class and then G=hzi Š D8 , a contradiction. If D D H Š D8 and CG .D/ 6 D, then Lemma 79.1 implies that G Š D8 C2 , contrary to j1 .Z.G//j D 2. Hence we must have H D D ht i, where t is an involution with CG .t / D H . Since H=hzi is elementary abelian, exp.G=hzi/ 4 and therefore G=hzi is abelian since G=hzi is modular (and Q8 -free) of exponent 4. In particular, D is normal in G. We have CH .D/ D hz; t i. If CG .D/ > hz; t i, then CG .t / D G, a contradiction. Hence CG .D/ D hz; t i and Aut.D8 / Š D8 implies that G=hz; t i Š D8 , a contradiction: G=hzi is modular. Suppose m > 1. Here Z.D/ D ˆ.D/ D hy 2 ; zi is abelian of type .2m1 ; 2/ and Z.H / D hy 2 ; zi E0 so that Ã1 .Z.H // D hy 4 i. If m > 2, then 1 .hy 4 i/ is of order 2 and 1 .hy 4 i/ Z.G/, contrary to the fact that 1 .Z.G// D hzi. Thus we have m D 2, jDj D 24 , and Z.H / D hy 2 ; zi E0 is elementary abelian. Suppose that E0 ¤ f1g. Then acting with an element x 2 G H on Z.H /, we get jCZ.H / .x/j 4 and CZ.H / .x/ Z.G/, a contradiction. It follows D D H and so jGj D 25 . If x 2 G H is of order 8, then x 4 2 Z.H / hzi (with hzi D H 0 ) since ˆ.H / D Z.H / and z is not a square in H . But then Z.H / Z.G/, a contradiction. Hence exp.G/ D 4 and the fact that G=hzi is modular gives that G=hzi is abelian. It follows G 0 D H 0 D hzi. Since H D D D hy; t i and jG W CG .y/j D jG W CG .t /j D 2, we get jG W CG .H /j 4. But jH W CH .H /j D jH W Z.H /j D 4 and so CG .H / must cover G=H . But then E4 Š Z.H / Z.G/, a final contradiction. (ii) The factor group G=hzi (z 2 1 .Z.G//) is not isomorphic to a Wa -group. Suppose that this is false. Then G=hzi has a maximal normal abelian subgroup N=hzi of exponent > 2 such that G=N is cyclic of order 2 and if L=N D 1 .G=N /, then for each element x 2 L N , x 2 2 hzi, and x inverts each element of N=hzi. If all elements in L N are involutions, then each y 2 L N inverts each element in N which implies that N is abelian of exponent > 2, N is a maximal normal abelian subgroup of G and if G D hN; gi, then G is a semidirect product of N and hgi and the involution in hgi inverts N . Thus, G is a Wa -group, a contradiction. Hence, there is v 2 L N with v 2 D z. Let n 2 N be of order 8. Then nv D n1 z ( D 0; 1) and therefore .n2 /v D 1 2 .n z / D n2 , contrary to Lemma 79.1. Thus, exp.N / D exp.N=hzi/ D 4.

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Suppose that N is abelian. Let s and l be elements of order 4 in N such that hsi \ hli D f1g. In that case o.sl/ D 4 and using Lemma 79.1, we get s v D s 1 z, l v D l 1 z and consequently .sl/v D s 1 zl 1 z D .sl/1 , a contradiction. Hence N is abelian of type .4; 2; : : : ; 2/ and since exp.N=hzi/ D 4, z 62 Ã1 .N /. We set Ã1 .N / D ht i with t ¤ z and so jN W 1 .N /j D 2, where each element in N 1 .N / is of order 4 and has square equal to t . Also, ht i D Ã1 .N / is central in G. Suppose that a 2 N 1 .N / is such that a2 D t and (by Lemma 79.1) av D a1 z D a.zt /. By Lemma 79.2, hv; ai is minimal nonabelian nonmetacyclic of order 24 with Œv; a D zt and va is an involution. Suppose that v does not commute with an involution u 2 1 .N /. Then uv D uz, o.au/ D 4, and .au/v D a1 zuz D .au/1 , contrary to Lemma 79.1. It follows that CN .v/ D 1 .N /, Œhvi; N D ht zi, L0 D ht zi, 1 .N / D Z.L/, ˆ.L/ D hz; t i, where ht i D Ã1 .N /. We have L N D v1 .N / [ .va/1 .N /, where all elements in v1 .N / are of order 4 and their squares are equal z and all elements in .va/1 .N / are involutions. Thus E D hvai1 .N / D 1 .L/ is elementary abelian of index 2 in L and also E D 1 .G/ (since G=N is cyclic) is the unique maximal normal elementary abelian subgroup of G. Now, L=E is a normal subgroup of order 2 in G=E with cyclic factor group G=L. Thus G=E is abelian. If G=E were cyclic, then the fact that G is not D8 -free gives that G is a Wb -group, a contradiction. Thus G=E is abelian of type .2s ; 2/, s 1. If s > 1, then we set K=E D 1 .G=E/ Š E4 . Since K L, K is not D8 -free and so K < G implies that K must be a Wilkens group. But E D 1 .K/ and K=E is noncyclic, a contradiction (see Propositions 79.8 to 79.10). Hence s D 1, G=E Š E4 , and so exp.G/ D 4. We have G=N Š C4 and so for each x 2 G L, x 2 2 L N and so x 2 2 E N . Since the square of each element in G is contained in hzi or ht i or in E N , it follows that t z is not a square in G. Hence 1 .G=ht zi/ D E=ht zi. If G=ht zi were nonmodular, then G=ht zi must be a Wilkens group and then .G=ht zi/=.E=ht zi/ Š G=E must be cyclic, a contradiction. It follows that G=ht zi is modular and since exp.G/ D 4, G=ht zi is abelian and so G 0 D ht zi. Suppose that x 2 G L is such that x 2 D E N . Then Œv; x 2 ht zi and so Œv; x 2 D Œv; x2 D 1. But then v centralizes E and since L D hE; vi, we get that L is abelian, a contradiction. Thus, N is nonabelian and so N 0 D hzi. The subgroup N is nonmodular because a Q8 -free modular 2-group of exponent 4 is abelian. Let S be a minimal nonabelian subgroup of N . Then S 0 D hzi and S is normal in N . Since v inverts N=hzi, v normalizes S and so S is normal in L D hN; vi. Since exp.S / D 4 and S is Q8 -free, it follows that either S Š D8 or S is minimal nonabelian nonmetacyclic of order 24 . If S Š D8 , then Ã1 .S / D hzi and N 0 D hzi implies that CN .S / covers N=S . In that case, Lemma 79.1 implies that CN .S / is elementary abelian and so Ã1 .N / D hzi, contrary to exp.N=hzi/ > 2. It follows that we have the second possibility: S D ha; t j a4 D t 2 D 1; Œa; t D z; z 2 D Œa; z D Œt; z D 1i: We put b D at and compute b 2 D a2 t 2 Œt; a D a2 z, so that S 0 D hzi, o.b/ D 4,

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hai\hbi D f1g, and S D ha; bi with Œa; b D a2 b 2 D z. Again, since N 0 D S 0 D hzi, we get that CN .S / covers N=S and CN .S / \ S D Z.S / D ˆ.S / D hz; a2 i. Suppose that y 2 CN .S / is of order 4. Since hai \ hbi D f1g, there is an s 2 fa; bg so that hsi \ hyi D f1g and o.sy/ D 4. We get .sy/v D s 1 zy 1 z D s 1 y 1 D .sy/1 , contrary to Lemma 79.1. Thus, CN .S / is elementary abelian and N D S CN .S /, S \ CN .S / D Z.S / and so the structure of N is completely determined. We act with hvi on S D ha; bi and get (using Lemma 79.1) av D a1 z D a.a2 z/, v b D b 1 z D ba2 , which together with v 2 D z determines uniquely the structure of T D S hvi. Suppose that v does not centralize CN .S / D Z.N /. Then there is an involution s 2 CN .S / S such that s v D sz. Then o.as/ D 4 and .as/v D a1 zsz D .as/1 , contrary to Lemma 79.1. Hence CN .S / D CN .T / and L D T CL .T / with T \ CL .T / D Z.T / D hz; a2 i. We have 1 .S / D hz; a2 ; abi Š E8 , hz; a2 i D Z.T /, .av/2 D avav D av 2 v 1 av D aza1 z D 1; .ab/av D .abz/v D a1 zb 1 zz D a1 b 1 z D aa2 bb 2 z D aba2 .a2 z/z D ab: Hence F D hz; a2 ; ab; avi Š E24 is an elementary abelian maximal subgroup of T and so from T 0 hz; a2 i and jT j D 25 D 2jT 0 jjZ.T /j follows that T 0 D Z.T / D hz; a2 i. Finally, T =hz; a2 i is elementary abelian and therefore Z.T / D T 0 D ˆ.T / Š E4 and so T is a special group of order 25 . For each x 2 T F , CF .x/ D Z.T / and so the set T0 D T F has exactly four square roots of z, four square roots of a2 , four square roots of za2 and so T0 must contain exactly four involutions in T S . If t0 is one of them, then CF .t0 / D Z.T /, and F and hz; a2 ; t0 i Š E8 are the only maximal normal elementary abelian subgroups of T (containing all involutions of T ). We have CN .S / D CN .T / D Z.L/ and so if we set U D F Z.L/ and V D hz; a2 ; t0 iZ.L/, then L D U V , U \ V D Z.L/, U and V are the only maximal normal elementary abelian subgroup of L and they are of distinct orders and so U and V are normal in G and jU W Z.L/j D 4, jV W Z.L/j D 2, L0 D ˆ.L/ D hz; a2 i. For each t0 2 V Z.L/, CU .t0 / D Z.L/ and so both U an V are self-centralizing in L. Also, U is the unique abelian maximal subgroup of L (otherwise, by a result of A. Mann, jL0 j 2). Now, G=N is cyclic, L=N D 1 .G=N /, and so 1 .G/ D L, which is a Wb -group with respect to U . In particular, L < G. If G=U is cyclic, then (since G is not D8 -free) G is a Wb -group, a contradiction. Hence G=U is noncyclic. But L=U Z.G=U / and G=L is cyclic (since L N and G=N is cyclic) and so G=U is abelian of type .2n ; 2/. Set K=U D 1 .G=U / Š E4 . If G ¤ K, then K is a Wilkens group with 1 .K/ D L. By the structure of L, L has no abelian maximal subgroups of exponent > 2 and so K is not a Wa -group. Also, jK W 1 .K/j D 2 and so K is not a Wc -group. Hence K must be a Wb -group with respect to U . But K=U Š E4 , a contradiction. Hence G D K and so G=U Š E4

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implies exp.G/ D 4. Since 1 .G/ D L, all elements in G L are of order 4 and if x 2 G L, then x 2 2 U N , where N D hZ.L/; ab; ai and U \ N D Z.L/ habi. If CG .V / > V , then CL .V / D V implies that there is y 2 G L with y 2 2 V , contrary to y 2 2 U N . We have proved that CG .V / D V and so G=V acts faithfully on V . We have G=V Š .U hxi/=Z.L/, where U=Z.L/ Š E4 and hxiZ.L/=.Z.L/ Š C4 / with x 2 G L. Thus G=V Š D8 or G=V Š C4 C2 . But L=V Š U=Z.L/ Š E4 is a four-subgroup in G=V and for each x 2 G L, x 2 2 U N so that hxi \ V D f1g. Hence all elements in .G=V / .L=V / are of order 4 and so G=V Š C4 C2 . If L0 D hz; a2 i D Z.L/, then V Š E8 . But G=V Š C4 C2 cannot act faithfully on V Š E8 . We have proved that Z.L/ > L0 and so jZ.L/j 8. We act with hxi on Z.L/, where x 2 G L. Since x 2 2 L, hxi induces an automorphism of order 2 on Z.L/. Since jZ.L/j 8, it follows that jCZ.L/ .x/j 4. Suppose that x centralizes an involution u 2 Z.L/ L0 so that u 2 Z.G/. Since G=hui is nonabelian and of exponent 4, G=hui must be nonmodular. Thus G=hui is a Wilkens group with 1 .G=hui/ D L=hui because 1 .G/ D L and u is not a square in G. Then U=hui and V =hui are the only maximal normal elementary abelian subgroups of G=hui and both G=U and G=V are noncyclic and so G=hui cannot be a Wb -group. Also, G=hui cannot be a Wa -group since 1 .G=hui/ D L=hui and jG=Lj D 2. If G=hui is a Wa -group, then (by the first part of the proof of (ii)) L must possess a nonabelian maximal subgroup N0 containing hui such that N00 D hui. This is a contradiction since u 2 Z.L/ L0 . We have proved that for each x 2 G L, CZ.L/ .x/ L0 . Since jCZ.L/ .x/j 4, we must have CZ.L/ .x/ D L0 D Z.G/ D hz; a2 i. Suppose that x 2 GL is such that x 2 2 U N . We have U D hz; a2 ; ab; aviZ.L/ 2 and so .ab/x D ab. On the other hand, x 2 2 L N and x 2 inverts N=hzi. Thus 2 2 ax D a1 z , b x D b 1 z .; D 0; 1/, and so, noting that b 2 D a2 z we get ab D .ab/x D a1 z b 1 z D a1 b 1 z C D aa2 bb 2 z C D abz 1CC ; 2

which gives C 1 .mod 2/ and so D 1 or D 1. 2 Suppose D 1. Then ax D a1 z D a.a2 z/ and we apply Lemma 79.2 in 2 N We have (using bar convention) o.x/ the group G=ha zi D G. N D o.a/ N D 4 and 2 2 2 N D 1 D Œx; N aN . If Œx; N a N D 1, then Œx; a 2 ha zi (and hx; ai is of class 2 ŒxN ; a since hx; ai0 ha2 zi) and so Œx 2 ; a D Œx; a2 D 1, a contradiction. Thus Œx; N a N ¤1 and so (by Lemma 79.2) o.xa/ D 2. Hence .xa/2 2 ha2 zi, contrary to the fact that xa 2 G L and .xa/2 2 U N . 2 N Let D 1. Then b x D b 1 z D b.a2 / and we apply Lemma 79.2 to G=ha2 i D G. 2 N 2 2 N N N We have o.x/ N D o.b/ D 4 and ŒxN ; b D 1 D Œx; N b . If Œx; N b D 1, then Œx; b 2 ha i (and hx; bi is of class 2 since hx; bi0 ha2 i) and so Œx 2 ; b D Œx; b2 D 1, a N ¤ 1 and so (by Lemma 79.2) o.xb/ D 2. Hence .xb/2 2 contradiction. Thus Œx; N b 2 ha i, contrary to the fact that xb 2 G L and .xb/2 2 U N . Our claim (ii) is proved.

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(iii) G=hzi (z 2 1 .Z.G//) is not isomorphic to a Wb -group. Suppose this is false. Then G=hzi has a maximal normal elementary abelian subgroup N=hzi so that G=N is cyclic and G=hzi is not D8 -free . If N is elementary abelian, then G is a Wb -group with respect to N , a contradiction. Hence N is not elementary abelian and so Ã1 .N / D hzi and G ¤ N . Set L=N D 1 .G=N /. (˛) Let N be abelian. Then jN W 1 .N /j D 2 and for each a 2 N 1 .N /, a2 D z. Also, hN 1 .N /i D N , 1 .G/ L, and 1 .G=hzi/ L=hzi. Let v 2 L N with v 2 D z and let x 2 N 1 .N /. By Lemma 79.2, Œv; x D 1 and so v centralizes N . But then L is abelian with Ã1 .L/ D hzi and L G G, a contradiction: N=hzi is a maximal normal elementary abelian subgroup of G=hzi. Thus, for each v 2 L N , v 2 ¤ z. Suppose that for each x 2 L N , x 2 2 N 1 .N /. If G D hN; gi, then hgi covers G=1 .N /. But then G=1 .N / is cyclic and (since G is not D8 -free) G is a Wb -group, a contradiction. We have proved that there is x 2 LN with x 2 2 1 .N /. Suppose that there are no involutions in L N . There is x 2 L N such that x 2 2 1 .N / hzi. Suppose that x does not commute with an a 2 N 1 .N /. By Lemma 79.2, xa is an involution, a contradiction. Hence x commutes with all elements in N 1 .N / and therefore L is abelian of type .4; 4; 2; : : : ; 2/. We have 1 .L/ D 1 .N / D 1 .G/ and L=1 .L/ Š E4 . Since G is nonabelian, we have L < G. Because N=1 .N / is a normal subgroup of order 2 in G=1 .N / and G=N is cyclic, G=1 .N / is abelian of type .2s ; 2/, s 2. For each y 2 L N , y 2 D x 2 or y 2 D x 2 z. Hence, if g 2 G is such that G D hN; gi, then 1 .hgi/ D hx 2 i or 1 .hgi/ D hx 2 zi and so Ã1 .L/ D hz; x 2 i Z.G/. We may assume (by a suitable notation) that hgi hxi. Set M D hgi1 .N / so that M is a maximal subgroup of G. Suppose that G=hx 2 i is nonmodular so that it is a Wilkens group. But 1 .G=hx 2 i/ D S0 =hx 2 i, where S0 D 1 .N /hxi. On the other hand, G=S0 is noncyclic (since L=S0 and M=S0 are two nontrivial cyclic subgroups of G=S0 with .L=S0 /\.M=S0 / D f1g). This is a contradiction and so G=hx 2 i is modular. Since 1 .G=hzi/ D N=hzi, we may use Lemma 79.12 and we see that G=hx 2 ; zi is nonmodular. This is not possible since G=hx 2 i is modular. Thus, there are involutions in L N . Let t be an involution in L N . If t centralizes an element a 2 N 1 .N /, then t a 2 L N and .t a/2 D z, a contradiction. Thus, CN .t / 1 .N /. For any x 2 L N , CN .x/ D CN .t / and so x 2 2 1 .N /. It follows that exp.L/ D 4. Suppose that t does not centralize 1 .N /. Let w be a fixed involution in 1 .N / CN .t / and let a be a fixed element in N 1 .N /. We have w t D wu with 1 ¤ u 2 CN .t / and .t w/2 D .t wt /w D wuw D u and so u 2 CN .t /hzi (since t w 2 LN ). Since CN .t w/ D CN .t / < 1 .N /, we have Œt w; a ¤ 1. By Lemma 79.2, .t w/a is an involution. We get 1 D .t wa/2 D t wat wa D w t at wa D wuat wa D uaat , and so at D a1 u D aa2 u D a.uz/. We consider the factor group L=huzi D LN and we see that o.a/ N D 4, o.tN/ D 2, Œa; N tN D 1, aN 2 D zN . Also, w t D wu D wz.uz/ tN gives wN D wN zN so that hw; N tNi Š D8 with Z.hw; N tNi/ D hzi. N Hence hw; N tNihai N is the N We have proved central product D8 C4 , contrary to Lemma 79.1 (applied in L).

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that t centralizes 1 .N / so that S D ht i 1 .N / is an elementary abelian maximal subgroup of L. Since L is nonabelian and exp.L/ D 4, L is nonmodular. Let a 2 N 1 .N /. Then by Proposition 51.2, at D a1 s, s 2 1 .N /. If s D 1, then t inverts N , G=N is cyclic and so G would be a Wa -group, a contradiction. Thus, s ¤ 1 and .t a/2 D .t at /a D a1 sa D s and so s 2 1 .N / hzi. Set t a D y so that y 2 D s and since CN .y/ D CN .t /, we have for each n 2 CN .t / D 1 .N /, .ny/2 D n2 y 2 D s and Œt; N D hzsi D L0 . The group L has exactly three abelian maximal subgroups: S D 1 .G/, N , and 1 .N /hyi, where only S is elementary abelian and all three are normal in G with Ã1 .1 .N /hyi/ D hsi, s 2 1 .N / hzi. We have 1 .N / D Z.L/ and z; s and zs are central involutions in G. If G D L, then G would be a Wb -group, a contradiction. Thus L < G. Since s1 G=N is cyclic of order 2s , s 2, we have G D hN; gi with g 2 2 L N . If s1 g2 2 L N S , then hgi covers G=S , G=S is cyclic and so G would be a Wb s1 group with respect to S , a contradiction. It follows that g 2 2 S 1 .N /. If L is not maximal in G, then there is a subgroup K > L with jK W Lj D 2 and so K is a Wilkens group. Since S D 1 .K/ and K=S Š E4 , we have a contradiction. s2 Indeed, .S hg 2 i/=S and L=S are two distinct subgroups of order 2 in K=S . Hence jG W Lj D 2 and for each g 2 G L, g 2 2 S 1 .N /. N where hzsi D L0 . We have We apply now Lemma 79.2 in the group G=hzsi D G, N D o.a/ N D 4 and since for some elements g 2 G L and a 2 N 1 .N /, o.g/ 2 ag D a.zs/, we get Œa; N gN 2 D 1 D ŒaN 2 ; g. N If Œg; N a N D 1, then Œg; a 2 hzsi and so Œg 2 ; a D Œg; a2 D 1, a contradiction. Hence Œg; N a N ¤ 1 and so (by Lemma 79.2) 2 o.gN a/ N D 2 which gives .ga/ 2 hzsi, contrary to ga 2 G L and (by the above) .ga/2 2 S 1 .N /. We have proved that N must be nonabelian. (ˇ) Let N be nonabelian. This case is very difficult. We have Ã1 .N / D N 0 D hzi. Let D be a minimal nonabelian subgroup of N so that D 0 D hzi. Since d.D/ D 2 and D=hzi is elementary abelian, we have D=hzi Š E4 and therefore D Š D8 . The subgroup D is normal in N and since D 0 D N 0 , CN .D/ covers N=D. By Lemma 79.1, CN .D/ is elementary abelian. We have CN .D/ D Z.N /, N D DZ.N / with D \ Z.N / D Z.D/ D hzi so that Z.N / is normal in G. Set D D ha; u j a4 D u2 D 1; a2 D z; au D a1 i so that A D ha; Z.N /i, E1 D hu; Z.N /i, and E2 D hau; Z.N /i are all abelian maximal subgroups of N , where A is of exponent 4 (all elements in A Z.N / are of order 4, Ã1 .A/ D hzi), and E1 and E2 are both elementary abelian. Thus A is normal in G. All elements in N A are involutions, Z.N / D 1 .A/, N is a Wa -group with respect to A and also a Wb -group with respect to E1 and E2 . Since N=A is a normal subgroup of order 2 in G=A and G=N is cyclic, it follows that G=A is abelian. If G=A were cyclic, then we have G D hA; gi with some g 2 G and since 1 .hgi/ is of order 2 and inverts A (of exponent > 2), G is a Wa -group, a contradiction. It follows that G=A is abelian of type .2m ; 2/, m 1, and L=A Š E4 , where L=N D 1 .G=N /. Also note that N=Z.N / Š E4 .

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Suppose that for each l 2 L N , l 2 2 A Z.N /. This is equivalent to assuming that L=Z.N / Š C4 C2 which also implies that both E1 and E2 are normal in L. If G D hN; gi, then either E1g D E2 (in which case G > L and G=Z.N / Š M2n , g n 4, and G is a Wc -group) or E1 D E1 , E1 is normal in G, hgi covers G=E1 and since G is not D8 -free, G is a Wb -group. In both cases we have a contradiction. Thus, L=Z.N / is not isomorphic to C4 C2 and so L=Z.N / is either elementary abelian or L=Z.N / Š D8 . In any case, there is l 2 L N with l 2 2 Z.N /. (ˇ1) Let L=Z.N / Š D8 . Then E1x D E2 for x 2 L N , and so G D L. Indeed, G=N acts on the set fE1 ; E2 g and so, if G > L, then L would normalize E1 . Suppose in addition that there are no involutions in L N . Let l 2 L N with l 2 2 Z.N / so that l 2 ¤ 1 and o.l/ D 4. Let a0 be any element in A Z.N / so that a02 D z. If Œl; a0 ¤ 1, then la0 is an involution in L N , a contradiction. Thus l centralizes each element in A Z.N / and since hA Z.N /i D A, Ahli is an abelian maximal subgroup of G D L. Also, Z.N / D Z.G/ (since Z.L=Z.N // D A=Z.N /) and so we may use the relation jGj D 2jG 0 jjZ.G/j which gives jG 0 j D 4. But G 0 covers A=Z.N / D .G=Z.N //0 and so G 0 Š C4 is a cyclic subgroup of order 4 in A inverted by u (since u inverts A) and so we have G 0 hui Š D8 . We may assume G 0 D hai < D Š D8 so that D is normal in G. By Lemma 79.1, CG .D/ is elementary abelian and so CG .D/ cannot cover G=N (since there are no involutions in G N ). Hence CG .D/ D Z.N / and so l 2 G N induces an outer automorphism on D. Hence al D a1 , contrary to Lemma 79.1. We have proved that there are involutions in G N and let t be one of them so that 1 .G/ D G. Since G=Z.N / Š D8 , there is k 2 G N such that k 2 2 A Z.N /. Our ultimate goal is to show that Z.N / D Z.G/. Suppose Z.N / ¤ Z.G/. Assume there is s 2 G N with s 2 D z and let a0 2 A Z.N /. If Œs; a0 ¤ 1, then (Lemma 79.2) Œs; a0 D s 2 a02 D zz D 1, a contradiction. Thus s centralizes A Z.N / and hA Z.N /i D A and so Z.N / Z.G/. Since Z.G=Z.N // D A=Z.N /, we have Z.N / D Z.G/. This is a contradiction and so there is no s 2 G N with s 2 D z. In particular, the involution t does not centralize any element in A Z.N /. Now, Aht i < G maximal. If x 2 At is of order 8, then o.x 2 / D 4 and x 2 2 A Z.N /. But then t centralizes x 2 (noting that CA .t / D CA .x/), a contradiction. Hence exp.Aht i/ D 4 and since Aht i is nonabelian, Aht i is nonmodular and therefore Aht i must be a Wilkens group. If X is an abelian maximal subgroup of Aht i distinct from A, then X \ A Z.N / (recalling that t does not commute with any element in A Z.N /) and since jA W .X \ A/j D 2, we get X Z.N / and so Z.N / D Z.G/, a contradiction. Hence A is the unique abelian maximal subgroup of Aht i. If all elements in .Aht i/ A are involutions, then t inverts each element of A and so t centralizes Z.N / and then Z.N / D Z.G/, a contradiction. It follows that there is an element c of order 4 in At such that c 2 2 Z.N / hzi. But Œc; a ¤ 1 (since t does not centralize a) and so (by Lemma 79.2) ac is an involution for each a 2 AZ.N /. Since t does not centralize Z.N / (otherwise Z.N / D Z.G/), Z.N /ht i contains less than 2jZ.N /j 1 involutions. All jZ.N /j elements ca (a 2 A Z.N /) are involutions and

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so Aht i contains at least 2jZ.N /j 1 involutions. This shows that 1 .Aht i/ D Aht i. Since A (of exponent > 2) is the unique abelian maximal subgroup of Aht i, Aht i must be a Wa -group. In that case t inverts A and so t centralizes Z.N / and Z.N / D Z.G/, a contradiction. We have proved that Z.N / D Z.G/ and so Z.N /hki D Ahki is an abelian maximal subgroup of G (noting that k 2 2 A Z.N / and so Z.N /hki=Z.N / is cyclic). Using the relation jGj D 2jG 0 jjZ.G/j we get jG 0 j D 4 and so G 0 Š C4 (since G 0 covers A=Z.N /). We may assume (as before) that G 0 D and so D is normal in G with CG .D/ D Z.N / (since CG .D/ is elementary abelian and jG W Z.G/j D 8). The involution t induces an outer automorphism on D and so Dht i Š D24 . It follows that G D .Dht i/ E2m for some m 1, which is a Wa -group, a contradiction. (ˇ2) We have proved that we must have L=Z.N / Š E8 and so exp.L/ D 4. In that case we prove first that there are involutions in L N . Suppose false. If v is any element in L N and a0 2 A Z.N /, then 1 ¤ v 2 2 Z.N / and a02 D z so that we may apply Lemma 79.2. If Œv; a0 ¤ 1, then va0 is an involution, a contradiction. Hence L D hL N i centralizes A D hA Z.N /i. In particular, A Z.N / which contradicts the fact that Z.N / < A. We have proved that there are involutions in LN and so L D 1 .L/ D 1 .G/ (noting that G=N is cyclic and L=N D 1 .G=N /). Assume CG .D/ > Z.N / so that CG .D/ covers L=N and (Lemma 79.1) CG .D/ is elementary abelian. In that case, L Š D8 E2m and Ã1 .L/ D hzi which contradicts our assumption that N=hzi is a maximal normal elementary abelian subgroup of G=hzi. Hence CG .D/ D Z.N /. If D were normal in L, then L=Z.N / Š D8 since Aut.D/ Š D8 . This is a contradiction and so NG .D/ D N . In particular, Z.L/ Z.N / and Z.N / > hzi. (ˇ2a) We assume that G > L. Then L D 1 .L/ (being nonmodular) is a Wa - or Wb -group. In particular, L has an abelian maximal subgroup B. Since B \ N is an abelian maximal subgroup of N , we get B \ N 2 fA; E1 ; E2 g. Hence B \ N Z.N / and so Z.N / Z.L/. By the result in the previous paragraph, we get Z.N / D Z.L/. Since jL W Z.L/j D 8, B is the unique abelian maximal subgroup of L and so B is normal in G. Using the relation jLj D 2jL0 jjZ.L/j, we get jL0 j D 4. Since L0 Z.L/, L0 Š E4 and L0 > hzi. It is easy to see that jG W Lj D 2. Suppose that this is false. Let K < G be such that jK W Lj D 2. Since L D 1 .K/, K must be a Wa - or Wb -group. By the uniqueness of B in L, follows that K=B is a cyclic group (of order 4). Since N < L and G=N is cyclic, G=L is cyclic. But L=B is a normal subgroup of order 2 in G=B and so G=B is abelian. We have L=B 1 .G=B/. On the other hand, G=L is cyclic and K=L D 1 .G=L/ so that 1 .G=B/ K=B. Since K=B Š C4 , we get 1 .G=B/ L=B and so 1 .G=B/ D L=B. Hence G=B is cyclic. If L is of type (a) (in the case exp.B/ > 2), then all elements in L B are involutions and so G is also a Wa -group, a contradiction. If L is of type (b), then B must be elementary abelian and (since G is not D8 -free) G is also a Wb -group, a contradiction. We have proved that jG W Lj D 2 and G=B Š E4 (because in case G=B Š C4 ,

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G would be a Wa - or Wb -group). But G=N Š C4 and so for each x 2 G L, x2 2 B N . We assume first that exp.B/ > 2. Then L is a Wa -group. In that case B=Z.N / Š E4 , all elements in L B are involutions and 1 .B/ D Z.N /. Indeed, if jB W 1 .B/j D 2, then acting by conjugation with an involution in L B on B, we see that jL0 j D 2, a contradiction. Hence we must have 1 .B/ D Z.N / so that B is abelian of type .4; 4; 2; : : : ; 2/, B \ N D A, u inverts B, each element in B N is of order 4, and if k 2 B N , then k 2 D z0 2 L0 hzi (noting that k u D k 1 and so k 2 2 L0 ) and L0 ha; ki D ha; ki Š C4 C4 (if k 2 D z, then ak would be an involution in B N , contrary to 1 .B/ D Z.N /). Let x 2 G L so that x 2 2 B N and we may assume that x 2 D k, where k 2 D z0 2 L0 hzi. In particular, CG .z0 / hL; xi D G and so L0 Z.G/. Let x 0 be an arbitrary element in G L so that .x 0 /2 D k 0 2 B N and .k 0 /2 D z 0 2 L0 hzi. Consider the factor N D o.x 0 / D 4 and ŒaN 2 ; x 0 D 1 D Œa; N .x 0 /2 and so group GN D G=hz 0 i. We have o.a/ 0 2 0 0 we may use Lemma 79.2. If Œa; N x ¤ 1, then o.ax N / D 2 and so .ax / 2 hz 0 i N , a contradiction. Hence we must have Œa; N x 0 D 1 and so Œa; x 0 2 h.x 0 /4 i. It follows that Œa; x D z0 . D 0; 1/. Consider the element y D xu 2 G L; then Œa; y D Œa; xu D Œa; uŒa; xu D zz0 ¤ 1. On the other hand, Œa; y 2 hy 4 i, y 4 2 L0 hzi. Thus D 1 and y 4 D zz0 . Finally, consider the factor group GQ D G=hzz0 i so that Q D 4 and Œy; Q and apply again Lemma 79.2. Since o.y/ Q D o.k/ Q kQ 2 D 1 D ŒyQ 2 ; k Q y Q ¤ 1. Œk; x D 1, we get Œk; y D Œk; xu D Œk; uŒk; xu D k 2 D z0 , and so Œk; 2 Q Thus o.k y/ Q D 2 and so .ky/ 2 hzz0 i N . This is a contradiction since ky 2 G L. We study now the case exp.B/ D 2 and L is a Wb -group. We may assume that B \ N D E1 . Since G=B Š E4 , exp.G/ D 4. Let t be an involution in B N . Since B E1 D Z.N /hui, t centralizes u 2 D. But NL .D/ D N and so Œa; t 2 L0 hzi. Suppose that L0 Z.G/. Let x 2 G L so that x 2 2 B N and (by the above) Œa; x 2 D z0 2 L0 hzi. Consider the factor group G=hz0 i D GN so that o.a/ N D o.x/ N D 4. If Œx; N a N D 1, then Œx; a 2 hz0 i. But hx; ai0 D hz0 i and so hx; ai is of class 2. Thus Œx 2 ; a D Œx; a2 D 1, a contradiction. Hence Œx; N a N ¤ 1 and so (by Lemma 79.2) o.xN a/ N D 2, which gives .xa/2 2 hz0 i, contrary to xa 2 G L and .xa/2 2 B N . We have proved that L0 6 Z.G/. Again, let x 2 G L so that hxi induces an involutory automorphism on Z.L/ D Z.N /. Suppose Z.N / > L0 . Then there is an involution z 0 2 Z.L/ L0 centralized by x and so z 0 2 Z.G/. Since G=hz 0 i is nonmodular (noting that D \ hz 0 i D f1g), G=hz 0 i is a Wilkens group. All squares of elements of G lie either in B-N or in L0 . Indeed, let as .s 2 B/ be any element in L B. Then .as/2 D a2 .a1 sas/ D zŒa; s 2 L0 . Therefore z 0 is not a square of any element in G which implies L=hz 0 i D 1 .G=hz 0 i/. Let y 2 L be such that Œy; L hz 0 i. But hz 0 i \ L0 D f1g and so Œy; L D f1g and therefore y 2 Z.L/. Hence Z.L=hz 0 i/ D Z.L/=hz 0 i. Since jL W Z.L/j D 8, the elementary abelian group B=hz 0 i is the unique abelian maximal subgroup of L=hz 0 i. It follows that G=hz 0 i must be a Wb -group with respect to B=hz 0 i. But then G=B must be cyclic, a contradiction. We have proved that L0 D Z.L/ and so jGj D 26 . Now,

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A D hai hz0 i D haiL0 Š C4 C2 is normal in G and is self-centralizing in L (since B is the unique abelian maximal subgroup of L). If CG .A/ 6 L, then there is g 2 CG .A/ L such that g 2 2 A N , contrary to g 2 2 B N . Thus A is self-centralizing in G and so G=A Š D8 since Aut.A/ Š D8 . On the other hand, N=A is a normal subgroup of order 2 in G=A and G=N Š C4 so that G=A is abelian. This is a contradiction. We have proved that the case G > L is not possible. (ˇ2b) It remains to study the case G D L D 1 .G/. Since G=Z.N / Š E8 , exp.G/ D 4. If for each x 2 G N , x 2 2 hzi, then G=hzi is elementary abelian, contrary to our assumption that N=hzi is a maximal normal elementary abelian subgroup of G=hzi. Hence there is k 2 G N such that k 2 2 Z.N / hzi. It follows that k 2 2 Z.G/ and so hzi < Z.G/ Z.N /. Let z 0 2 Z.G/ hzi. Since G=hz 0 i is nonmodular (noting that D \ hz 0 i D f1g with D Š D8 ), it follows that G=hz 0 i is a Wilkens group with 1 .G=hz 0 i/ D G=hz 0 i and so G=hz 0 i cannot be a Wc -group. Hence G=hz 0 i must be a Wb -group by our previous result (ii). Thus, G has a maximal subgroup N1 containing z 0 such that N1 =hz 0 i is a maximal normal elementary abelian subgroup of G=hz 0 i. By our previous result (iii)(˛), hz 0 i D Ã1 .N1 / D N10 (since N1 must be nonabelian). In particular, z 0 is a square of an element in G N and z 0 is a commutator in G. Conversely, if k 2 G N , then k 2 2 Z.G/ so that ˆ.G/ Z.G/. Hence Z.G/ D G 0 D ˆ.G/ and so G is a special group. Now, N \ N1 is a maximal subgroup of N and since Ã1 .N \ N1 / hzi \ hz 0 i D f1g, N \ N1 is elementary abelian and so N \ N1 D E1 (or E2 ) containing Z.N / and (by the structure of N1 Š N ) Z.N1 / is a subgroup of index 2 in N \ N1 . But Z.N / \ Z.N1 / Z.G/ and so jZ.N / W Z.G/j 2. Let jZ.N / W Z.G/j D 2 and let s be an involution in Z.N / Z.G/. Let t be an involution in G N . We have Œt; s D s0 with s0 2 Z.G/ and s0 ¤ 1 since Z.N / ¤ Z.G/. If n 2 N , then Œt n; s D Œt; sŒn; s D Œt; s D s0 2 Z.G/. Hence, for each x 2 G N , Œx; s D s0 , where s0 is a fixed involution in Z.G/. Take an involution z 0 2 Z.G/ hzi. Let N1 be a maximal subgroup of G such that hz 0 i D Ã1 .N1 / D N10 . Then N \ N1 is an elementary abelian maximal subgroup of N containing Z.N /. If f1 2 N1 N , then Œf1 ; s D z 0 D s0 . Let N2 (¤ N1 ) be a maximal subgroup of G such that hzz 0 i D Ã1 .N2 / D N20 . Then again, N \ N2 Z.N / and if f2 2 N2 N , then Œf2 ; s D zz 0 D s0 . Hence zz 0 D z 0 and so z D 1, a contradiction. We have proved that Z.N / D Z.G/. Suppose that Z.G/ possesses a four-subgroup hz1 ; z2 i such that hz1 ; z2 i \ hzi D f1g (which is equivalent with the assumption jZ.G/j 8). Let k1 ; k2 2 G N be such that k12 D z1 and k22 D z2 . Suppose that k1 centralizes all elements (of order 4) in A Z.N /. Then k1 centralizes A D Z.N /hai D hA Z.N /i and so Ahk1 i is an abelian maximal subgroup of G. Using a result of A. Mann (Lemma 79.5 with respect to maximal subgroups Ahk1 i and N ), we get jG 0 j 4. But G 0 D Z.G/ is of order 8, a contradiction. We may assume that Œa; k1 ¤ 1 and so (Lemma 79.2) Œa; k1 D a2 k12 D zz1 . We have either Œa; k2 D 1 or Œa; k2 ¤ 1 in which case (Lemma 79.2) Œa; k2 D a2 k22 D zz2 . We have Œa; k1 k2 D Œa; k1 Œa; k2 D zz1 Œa; k2 , and so either Œa; k1 k2 D zz1

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or Œa; k1 k2 D zz1 zz2 D z1 z2 . But the element k1 k2 is contained in N and so Œa; k1 k2 2 hzi, a contradiction. We have proved that we must have Z.N / D Z.G/ Š E4 and so jGj D 25 . Let 0 z 2 Z.G/ hzi and let k1 ; k2 2 G N be such that k12 D z 0 and k22 D zz 0 . Suppose Œk1 ; k2 D 1 so that hk1 ; k2 i Š C4 C4 , k1 k2 2 N and .k1 k2 /2 D k12 k22 D z and therefore we may assume that k1 k2 D a. Hence hk1 ; k2 i is an abelian maximal subgroup of G normalized by u, where u inverts each element in hk1 ; k2 i \ N D hai hz 0 i Š C4 C2

and

G D hk1 ; k2 ihui:

We know that there are involutions in GN and so there is an element k 2 hk1 ; k2 iN such that uk is an involution. This gives .uk/2 D ukuk D 1, k u D k 1 and so u inverts each element of the abelian group hk1 ; k2 i. In this case G is a Wa -group, a contradiction. Hence Œk1 ; k2 ¤ 1 and using Lemma 79.2 we get Œk1 ; k2 D k12 k22 D z 0 .zz 0 / D z, o.k1 k2 / D 2, k1 k2 2 N . Because k1 k2 62 Z.hk1 ; k2 i/, we may assume k1 k2 D u (an involution in N A). Since Œk1 ; u D Œk1 ; k1 k2 D Œk1 ; k2 D z (which gives uk1 D uz) and D D ha; ui Š D8 is not normal in G, we have Œa; k1 ¤ 1. By Lemma 79.2, Œa; k1 D a2 k12 D zz 0 and k1 a is an involution in G N . We have uk1 a D .uz/a D .uz/z D u. Hence hu; k1 ai Š E4 with hu; k1 ai \ Z.G/ D f1g and so hu; k1 a; Z.G/i Š E16 . Since G is not D8 -free, it follows that G is a Wb -group. This is our final contradiction and so our statement (iii) is completely proved. (iv) The factor group G=hzi (z 2 1 .Z.G//) is not isomorphic to a Wc -group. Suppose that the assertion (iv) is is false. Then G=hzi is a Wc -group. Hence we may set 1 .G=hzi/ D H=hzi (implying that 1 .G/ H ) so that H D H1 H2 , where H1 and H2 are normal subgroups in H with H1 \ H2 D hzi, H1 =hzi Š D8 , H2 =hzi is elementary abelian and G=H is cyclic of order 4. Let Z=hzi be the unique cyclic subgroup of index 2 in H1 =hzi and set Z.H1 =hzi/ D Z0 =hzi so that jZ W Z0 j D 2 and jZ0 W hzij D 2. Let Z0 H2 D N , ZH2 D A, so that Z.H=hzi/ D N=hzi and N=hzi is the unique maximal normal elementary abelian subgroup of G=hzi, G=N Š M2n , n 4, A=N D .G=N /0 , H=N D 1 .G=N / Š E4 . If E1 =N and E2 =N are other two subgroups of order 2 in H=N (distinct from A=N ), then E1G D E2 , E1 =hzi and E2 =hzi are elementary abelian and A=hzi is abelian of type .4; 2; : : : ; 2/. Also, NG .E1 / D NG .E2 / > H and jG W NG .E1 /j D 2. Finally, G possesses a subgroup S such that G D HS , H \ S D Z, and S=hzi is cyclic so that S is abelian. In fact, S is either cyclic or abelian of type .2n ; 2/, n 4. Also, S is cyclic if and only if Z0 is cyclic (since Z0 =hzi D 1 .S=hzi/). We have H 0 Z0 and H 0 covers Z0 =hzi. (˛) Suppose Z0 (and so also S ) is cyclic. We have H1 =hzi Š D8 and .H1 =hzi/0 D Z0 =hzi. Since H10 covers Z0 =hzi and Z0 is cyclic, we get that H10 D Z0 is of index 4 in H1 . By a very well-known result of O. Taussky, H1 is of maximal class. Since H1 is Q8 -free, we get H1 Š D24 . An involution t 2 H1 Z inverts Z and so H3 D hZ0 ; t i Š D8 . The subgroup H3 is normal in H1 and ŒH3 ; H2 H1 \ H2 D hzi implies that H3 is normal in H .

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Nonmodular quaternion-free 2-groups

353

Since CH1 .H3 / D hzi and H1 =hzi Š D8 Š Aut.H3 /, we get that CH .H3 / covers H=H1 . By Lemma 79.1, CH .H3 / is elementary abelian. In particular, 1 .H / D H D 1 .G/. Since H is nonmodular, H is a Wilkens group with 1 .H / D H and so H is either a Wa - or Wb -group. It follows that H has an abelian maximal subgroup AQ which is unique (by a result of A. Mann) since H 0 H10 and jH10 j D 4 (see Lemma 79.5). In particular, AQ is normal in G. Also, AQ \ H1 D Z Š C8 since Q > 2. If Z is the unique abelian maximal subgroup of H1 Š D24 and so exp.A/ Q Let t 2 H1 Z, then H must be a Wa -group and the involution t inverts on A. Q Q Q Q Q N D 2 .A/ D Z0 1 .A/ (since A=Z is elementary abelian) so that N is normal in G, and NQ =hzi is a normal elementary abelian subgroup of G=hzi. By the uniqueness of N=hzi, NQ N and so AQ D Z NQ ZN D A and therefore AQ D A is abelian of type .8; 2; : : : ; 2/. Let G1 > H be such that jG1 W H j D 2. Hence G1 ¤ G and G D HS1 , where S1 D S \ G1 . It follows that G1 is also a Wilkens group with 1 .G1 / D H . Since jG1 W H j < 4, G1 is of type (a) or (b). But A D AQ is the unique abelian maximal subgroup of H and exp.A/ > 2. Thus G1 must be a Wa -group with respect to A and so G1 =A Š C4 . On the other hand, we know that S1 \ H D Z. Hence, if x 2 S1 H , then x 2 2 Z A. This is a contradiction since G1 =A Š C4 and therefore x 2 2 H A. (ˇ) We have Z0 Š E4 and so S splits over hzi. We set Z D haihzi so that Z0 D ha2 ihzi and replacing a with az (if necessary), we may put S D hsi hzi with 2 .hsi/ D hai D S \ H . If jH10 j D 4, then jH1 W H10 j D 4, H10 D Z0 , and (by a result of O. Taussky) H1 is of maximal class. In that case Z0 would be cyclic, a contradiction. We have proved that jH10 j D 2 and so Z0 D H10 hzi. We set H10 D hz0 i so that z0 is a central element in H . But S is abelian, G D HS , and S \ H D Z > Z0 D hz; z0 i. It follows CG .hz; z0 i/ hH; S i D G and so hz; z0 i Z.G/. We show that there are exactly two possibilities for the structure of H1 . Since H1 =Z0 Š E4 , we have exp.H1 / D 4. Suppose that H1 is minimal nonabelian. If H1 is metacyclic (of exponent 4), then we know that H1 is not Q8 -free. Thus H1 must be nonmetacyclic and we know that there is only one such minimal nonabelian group of order 24 and exponent 4. In particular, there is an element b 2 H1 Z such that b 2 D z, Œa; b D a2 b 2 D a2 z D z0 , where H10 D hz0 i, z0 is not a square in H1 , and ab is an involution so that 1 .H1 / D hz; z0 ; abi. Suppose now that H1 is not minimal nonabelian. Then there is a subgroup D Š D8 in H1 which covers H1 =hzi. Thus H1 D D hzi and H10 D D 0 D hz0 i. We have D \ Z D hai or D \ Z D hazi and all elements in H1 Z are involutions inverting ha; zi. Replacing a with az (if necessary), we may assume that D \ Z D hai. If t is an involution in D Z, then D D ha; t i, where z0 D a2 is a square in H1 . Suppose that H2 is nonabelian. Then H20 D hzi since H2 =hzi is elementary abelian. Let H4 be a minimal nonabelian subgroup of H2 so that H40 D hzi, H4 =hzi is elementary abelian and d.H4 / D 2. Thus H4 =hzi Š E4 and so H4 Š D8 . The subgroup H4 is normal in H2 and H2 centralizes H4 =hzi. We have ŒH1 ; H4 H1 \ H2 D hzi

354

Groups of prime power order

and so H1 also centralizes H4 =hzi and H4 is normal in H . Thus, H centralizes H4 =hzi and therefore there is no h 2 H inducing an outer automorphism on H4 (because otherwise such an element h would act nontrivially on H4 =hzi). It follows that CH .H4 / covers H=H4 . But H=H4 is nonabelian since H4 H2 and H=H2 Š H1 =hzi Š D8 . This contradicts Lemma 79.1. We have proved that H2 must be abelian and so H2 is either abelian of type .4; 2; : : : ; 2/ or elementary abelian. In any case, N D Z0 H2 D hz0 i H2 since z0 2 Z.G/. (ˇ1) Suppose that H2 is abelian of type .4; 2; : : : ; 2/, where Ã1 .N / D Ã1 .H2 / D hzi. Set E D 1 .H2 / so that 1 .N / D hz0 i E and all elements in H2 E are of order 4. Let h be an arbitrary element in H2 E so that h2 D z. We consider first the possibility that H1 is minimal nonabelian nonmetacyclic. Let x be any element of order 4 in H1 so that x 2 D z or x 2 D zz0 , where hz0 i D H10 . Suppose that Œx; h ¤ 1. By Lemma 79.2, Œx; h D x 2 h2 D x 2 z. On the other hand, Œx; h H1 \ H2 D hzi and so Œx; h D z which gives x 2 D 1, a contradiction. Hence Œx; h D 1 and since H1 is generated by its elements of order 4 (noting that 1 .H1 / Š E8 ) and H2 D hH2 Ei, we get ŒH1 ; H2 D f1g. We apply Lemma 79.1 N is the in the factor group HN D H=hzz0 i. We have HN1 Š D8 and hH1 ; hi D HN1 hhi N N N N N central product of H1 with hhi Š C4 , where H1 \ hhi D Z.H1 /, a contradiction. We consider now the possibility, where H1 D D hzi with D D ha; t j a4 D t 2 D 1; at D a1 i;

a2 D z0 ;

and

hz0 i D D 0 :

If Œa; h ¤ 1, then Œa; h 2 H1 \ H2 D hzi and so Œa; h D z. On the other hand, Lemma 79.2 implies Œa; h D a2 h2 D z0 z, a contradiction. Hence Œa; h D 1 and so a centralizes H2 D hH2 Ei. It follows that A D ZH2 D hai H2 is an abelian maximal subgroup of H with exp.A/ D 4 and A is normal in G. We have 1 .H1 / D H1 and so 1 .H / contains the maximal subgroup H1 E of H , where E D 1 .H2 / with jH2 W Ej D 2. If Œt; h D 1, then D D ha; t i Š D8 centralizes hhi Š C4 , contrary to Lemma 79.1. Hence Œt; h ¤ 1 and since Œt; h 2 H1 \ H2 D hzi, we get Œt; h D z with z D h2 . Thus, ht; hi Š D8 and so t h is an involution in H .H1 E/. We have proved that 1 .H / D H and since 1 .G/ H , we get also 1 .G/ D H . Also, H 0 D hz; z0 i Š E4 and therefore, by a result of A. Mann (Lemma 79.5), A is the unique abelian maximal subgroup of H . Take a subgroup G1 of G with H < G1 < G and jG1 W H j D 2. It follows that G1 must be a Wa -group with 1 .G1 / D H since jG1 W H j D 2 and A is the unique abelian maximal subgroup of H (with exp.A/ > 2). In that case G1 =A is cyclic. On the other hand, setting S1 D S \ G1 , we have G1 D HS1 and S1 \ H D Z. Thus, if g 2 S1 H , then g 2 2 Z A. This contradicts the fact that G1 =A Š C4 . (ˇ2) We have proved that H2 (and so also N D hz0 i H2 ) is elementary abelian. Suppose first that H1 is minimal nonabelian nonmetacyclic. In this case z0 is not a square in H1 , where hz0 i D H10 . There is b 2 H1 Z with b 2 D z, t D ab is an involution, and a2 D zz0 . Now, A D ZN D haiN is normal in G and .ht iN /s D

79

355

Nonmodular quaternion-free 2-groups

hbiN (recalling that S D hsi hzi) since G=N Š M2n , n 4, and so G acts nontrivially on the four-group H=N . The subgroup 1 .H / contains the maximal subgroup ht iN of H , where 1 .H1 / D hz; z0 ; t i Š E8 and N \ H1 D hz; z0 i. If t centralizes H2 , then ht iN is elementary abelian. But .ht iN /s D hbiN and hbiN is not elementary abelian, a contradiction. Hence t does not centralize H2 and so ŒH2 ; t D hzi since ŒH2 ; t H1 \ H2 D hzi. It follows that ht iN is nonabelian and so hbiN is also nonabelian. In particular, H 0 D hz; z0 i Š E4 , ŒH2 ; b D hzi, and so hbi is normal in H2 hbi. Let u be an involution in H2 with Œb; u D z so that hb; ui Š D8 and therefore bu is an involution. But bu 2 H .ht iN / and so 1 .H / D H . It follows that H must be a Wa - or Wb -group. In that case H must possess an abelian maximal subgroup M which is also unique (by a result of A. Mann since jH 0 j D 4). We have M H 0 D hz; z0 i. If M does not contain H2 , then M covers H=H2 which is nonabelian (since H=H2 Š H1 =hzi Š D8 ), a contradiction. Hence M H2 and so M hH 0 ; H2 i D N . Since ht iN and hbiN are nonabelian, we get that M D A D haiN is abelian (of exponent 4). Thus H is a Wa -group with respect to A and so the involution t must invert each element in A. In particular, at D a1 D aa2 D a.zz0 /. This is a contradiction since H10 D hz0 i. It remains to investigate the case, where H1 D D hzi with D D ha; t j a4 D t 2 D 1; at D a1 i;

a2 D z0 ;

hz0 i D D 0

and S D hsi hzi;

hsi \ H D hai;

.ht iN /s D hat iN

since G=N Š M2n , n 4. Obviously, in this case 1 .H / D H D 1 .G/. Suppose that t does not centralize H2 . Then Œt; H2 H1 \ H2 D hzi and so Œt; H2 D hzi and H 0 D hz; z0 i Š E4 . Then ht iN and hat iN D .ht iN /s are nonabelian. But H is a Wilkens group with 1 .H / D H and so H must have an abelian maximal subgroup U which is unique (by a result of A. Mann). If U does not contain H2 , U covers H=H2 and this is a contradiction since H=H2 Š D8 . Hence U hH 0 ; H2 i D N and so U D A D haiN is abelian of exponent 4. Thus, H is a Wa -group with respect to A. In particular, the involution t inverts each element in A and so t centralizes H2 , a contradiction. Hence t centralizes H2 and so ht iN is elementary abelian. In that case hat iN D .ht iN /s is also elementary abelian. In particular, D D ht; at i Š D8 centralizes H2 and so H D D H2 . It follows that G is a Wc -group, a contradiction. Our statement (iv) is proved. We have proved that the nonmodular factor group G=hzi (z 2 1 .Z.G//) (according to our statement (i)) is not isomorphic to any Wilkens group (according to (ii), (iii), and (iv)). This is a final contradiction and so the Main Theorem is proved.

80

Minimal non-quaternion-free 2-groups

Here we classify minimal non-quaternion-free 2-groups. A 2-group G is minimal non-quaternion-free if G is not quaternion-free but each proper subgroup of G is quaternion-free. Recall that a 2-group G is modular if and only if G is D8 -free. The main theorem is a consequence of results of the previous section. Theorem 80.1. Let G be a minimal non-quaternion-free 2-group. Then G possesses a unique normal subgroup N such that G=N Š Q8 . We have N < ˆ.G/ and so G=ˆ.G/ Š E4 and 1 .G/ ˆ.G/. If R is any G-invariant subgroup of index 2 in N , then H2 D G=R is the minimal nonabelian metacyclic group of order 24 and exponent 4: H2 D ha; b j a4 D b 4 D 1; ab D a1 i, where H2 =ha2 b 2 i Š Q8 , X=hb 2 i Š D8 , and X=ha2 i Š C4 C2 . In particular, if jGj > 8, then G has a normal subgroup S such that G=S Š D8 and so G is nonmodular. Proof. The group G possesses a normal subgroup N such that G=N Š Q8 . Suppose that A is a maximal subgroup of G not containing N . Then AN D G so A=.A\N / Š G=N Š Q8 , a contradiction. Thus, N < ˆ.G/ so d.G/ D 2. It follows from 1 .G=N / D ˆ.G=N / that 1 .G/ ˆ.G/. We may assume jGj > 8. Let R be any G-invariant subgroup of index 2 in N . We shall determine the structure of G=R. For that purpose we may assume R D f1g so that jN j D 2, N < ˆ.G/, N Z.G/, jˆ.G/j D 4, jGj D 24 and ˆ.G/=N D Z.G=N / D .G=N /0 , where G=N Š Q8 . By Theorem 1.2, G has no cyclic subgroups of index 2 so ˆ.G/ is a four-subgroup. In that case, N < Z.G/ since G is not of maximal class so ˆ.G/ D Z.G/ and G is minimal nonabelian. Since 1 .G/ D ˆ.G/ is a four-subgroup, G is metacyclic (of exponent 4). Since H2 D ha; b j a4 D b 4 D 1; ab D a1 i is the unique nonabelian metacyclic group of order 16 and exponent 4, we get G Š H2 . Next we do not assume that R D f1g. Let M ¤ N be a G-invariant subgroup of index 2 in ˆ.G/; then L D M \ N is of index 2 in N , by the product formula. By the previous paragraph, G=L Š H2 so G=M 2 fC4 C2 ; D8 g. In particular, G is nonmodular. This argument shows that N is the unique G-invariant subgroup of index 8 in G such that G=N Š Q8 . Let G be a minimal non-Q8 -free 2-group of order > 8 such that each proper subgroup of G is modular. Then G is a minimal nonmodular 2-group and such groups

80

Minimal non-quaternion-free 2-groups

357

have been classified in 78. Therefore, we may assume in the sequel that G has a maximal subgroup H which is nonmodular. Since H is Q8 -free, we are in a position to apply Theorem 79.7 classifying nonmodular Q8 -free 2-groups. Also, we assume that the reader is familiar with Propositions 79.8, 79.9 and 79.10 describing the Wilkens groups of types (a), (b), and (c), which appear in Theorem 79.7. Theorem 80.2. Let G be a minimal non-quaternion-free 2-group which has a nonmodular proper subgroup. Then G has one of the following properties: (i) 1 .G/ D ˆ.G/ D Aht i, where A is a maximal normal abelian subgroup of G with exp.A/ > 2, t is an involution inverting each element in A, and G=A Š Q8 , D8 , or C4 C2 . If G=A Š D8 or C4 C2 , then A is abelian of type .4; 2; : : : ; 2/. (ii) 1 .G/ D EE1 , where E ¤ E1 are the only maximal normal elementary abelian subgroups of 1 .G/ and j1 .G/ W Ej D 2, j1 .G/ W E1 j 2. We have either 1 .G/ D ˆ.G/ (and then G=1 .G/ Š E4 ) or G=1 .G/ Š Q8 . If G=1 .G/ Š Q8 , then also j1 .G/ W E1 j D 2 which implies 1 .G/ Š D8 E2s . (iii) 1 .G/ D E is elementary abelian and G=E is isomorphic to Q8 , M2n , n 4, or C2m C2 , m 1. If G=E Š M2n or C2m C2 with m 2, then 2 .G/ is abelian of type .4; 4; 2; : : : ; 2/. Proof. Let G be a minimal non-quaternion-free 2-group possessing a maximal subgroup H which is nonmodular. It follows that H is a Wilkens group of type (a), (b), or (c). In particular, H=1 .H / is cyclic, where 1 .H / D 1 .G/ since 1 .G/ ˆ.G/. If 1 .G/ is not elementary abelian, then we know (by the structure of the Wilkens group H ) that 1 .G/ is nonmodular and so in that case each maximal subgroup M of G is a Wilkens group. It follows that M=1 .G/ is cyclic, where 1 .M / D 1 .G/. In that case, X D G=1 .G/ Š E4 or Q8 . Here we have used a trivial fact that noncyclic 2-group X, all of whose maximal subgroups are cyclic, must be isomorphic to E4 or Q8 . (i) Suppose that H is a Wilkens group of type (a). Then H is a semidirect product H D hxi A, where A is a maximal normal abelian subgroup of H with exp.A/ > 2 and if t is the involution in hxi, then t inverts each element of A. We have 1 .H / D 1 .G/ D Aht i and A is a characteristic subgroup in H (Proposition 79.8) and so A is normal in G. By the previous paragraph, G=1 .G/ Š E4 or Q8 . However, if G=1 .G/ Š Q8 , then G=A (having a cyclic subgroup H=A of index 2) is of maximal class. But such a group does not have a proper factor group isomorphic to Q8 . Hence G=1 .G/ Š E4 and so 1 .G/ D ˆ.G/. Since d.G=A/ D 2, we get G=A Š Q8 , D8 or C4 C2 . Suppose that A is not a maximal normal abelian subgroup of G. Let B be a maximal normal abelian subgroup of G containing A so that B \ H D A, jB W Aj D 2, and G D hxi B with hxi \ B D f1g. Let b 2 B A and we compute .bb t /t D b t b D

358

Groups of prime power order

bb t 2 A, since b t 2 B A so that bb t D s 2 1 .A/. If s D 1, b t D b 1 and so t inverts each element of the abelian group B. It follows that G D hxi B is a Wilkens group of type (a). In that case G would be Q8 -free (Proposition 79.8), a contradiction. Hence s ¤ 1 and .bt /2 D bt bt D bb t D s shows that o.bt / D 4. Let a be an element of order 4 in A. We have abt D a1 and ha; bt i ¤ G (since a 2 ˆ.G/) and so ha; bt i is Q8 -free, contrary to Lemma 79.1. We have proved that A is a maximal normal abelian subgroup of G. Suppose that G=A Š D8 or C4 C2 . In that case there is a maximal subgroup K of G such that K=A Š E4 and 1 .K/ D 1 .G/. Then K is a Wilkens group of type (a) or (b) since jK W 1 .K/j D 2 (and so K cannot be a Wilkens group of type (c)). Suppose that K is a Wilkens group of type (a) with respect to a maximal normal abelian subgroup A1 of K with exp.A1 / > 2. We know that A1 1 .K/ D 1 .G/, j1 .G/ W A1 j D 2 and K=A1 is cyclic. Since K=A is noncyclic, we have A1 ¤ A. All elements in A1 A are involutions and if t0 2 A1 A, then t0 inverts and centralizes each element in A \ A1 and so A \ A1 is elementary abelian. It follows that exp.A1 / D 2, a contradiction. We have proved that K is a Wilkens group of type (b) with respect to a maximal normal elementary abelian subgroup E of K. We know that j1 .K/ W Ej D 2 (Proposition 79.9). Since 1 .K/ D 1 .G/, we have jA W A \ Ej D 2 and so A is abelian of type .4; 2; : : : ; 2/. (ii) Assume that H is a Wilkens group of type (b) with respect to E and j1 .H / W Ej D 2, where H=E is cyclic. We have 1 .H / D 1 .G/ and 1 .G/ ˆ.G/ so that 1 .G/ is nonmodular. Indeed, 1 .G/ has exactly two maximal normal elementary abelian subgroups E and E1 , where 1 .G/ D EE1 and so 1 .G/ is a Wilkens group of type (b) (and so nonmodular). By the first paragraph of the proof, G=1 .G/ Š E4 or Q8 . Suppose that G=1 .G/ Š Q8 . It is easy to see that E is not normal in G. Suppose false. Since G=E has a cyclic subgroup H=E of index 2 and G=1 .G/ Š Q8 , we get that G=E must be of maximal class. But there is no 2-group of maximal class having Q8 as a proper homomorphic image. Hence E is not normal in G and so for each x 2 G H , E x D E1 . In particular, j1 .G/ W E1 j D 2 and F D E \ E1 D Z.1 .G//. Take e 2 E F , e1 2 E1 F so that D D he; e1 i Š D8 and if V is a complement of hŒe; e1 i in F , then 1 .G/ D D V Š D8 E2s . (iii) Suppose that H is a Wilkens group of type (b) with respect to E and E D 1 .H / so that E D 1 .G/ (since E ˆ.G/), E is normal in G and G=E is noncyclic with the cyclic subgroup H=E of index 2. Let F=E be a proper subgroup of G=E such that F=E Š E4 . Then F is abelian. Indeed, since F ¤ G, F is Q8 -free. If F is not D8 -free, then F must be a Wilkens group. But then F=1 .F / must be cyclic. This is a contradiction since 1 .F / D E. It follows that F is D8 -free. Since exp.F / 4, F must be abelian (Proposition 79.6). Since each proper subgroup of G=E is Q8 -free, G=E cannot be semidihedral or Q2m with m 4. Suppose that G=E Š D2n , n 3. In that case G=E is generated by its four-subgroups and so E Z.G/. This is a contradiction because in that case H

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Minimal non-quaternion-free 2-groups

359

would be abelian (noting that H=E is cyclic). We have proved that G=E is isomorphic to Q8 , M2n , n 4, or C2m C2 , m 1. Suppose that G=E is not isomorphic to Q8 or C2 C2 . Set F=E D 1 .G=E/ so that F=E Š E4 . By the above, F is abelian. Obviously, F D 2 .G/ is abelian of type .4; 4; 2; : : : ; 2/. (iv) Finally, assume that H is a Wilkens group of type (c). We have 1 .H / D 1 .G/ Š D8 E2s and H=1 .G/ is cyclic of order 4 (Proposition 79.10). The subgroup 1 .G/ has exactly three abelian maximal subgroups E1 , E2 , A, where E1 and E2 are elementary abelian and A is abelian of type .4; 2; : : : ; 2/. By the structure of H , E1 and E2 are not normal in H and H=E1 \E2 Š M2n , n 4. By the first paragraph of the proof, G=1 .G/ Š Q8 since jH=1 .G/j 4. Thus H=E1 \ E2 Š M24 . On the other hand, 1 .G/ is normal in G and so NG .E1 / D NG .E2 / D K is a maximal subgroup of G distinct from H . Since K 1 .G/, K is nonmodular and therefore K is a Wilkens group. But K cannot be a Wilkens group of type (c) since E1 and E2 are normal in K. Hence K is either a Wilkens group of type (a) with respect to A or K is a Wilkens group of type (b) with respect to E1 or E2 . The group G with such a maximal subgroup K has been considered in (i) and (ii) and so we do not get here new possibilities for the structure of G. Our theorem is proved. Let H2 D ha; b j a4 D b 4 D 1; ab D a1 i; H1 D ha; b j a4 D b 4 D 1; c D Œa; b; c 2 D 1; Œa; c D Œb; c D 1i: Exercise 1. Let G D H1 . Then 1, a2 , b 2 are only squares in G. In particular, a2 b 2 is not a square. Solution. Let g D ai b j c k 2 G. Then g 2 D .ai b j /2 . If i is even, then g 2 D b 2j 2 f1; b 2 g. If j is even, then g 2 D a2i 2 f1; a2 g. If i and j are odd, then g 2 D .ab/2 D ab 2 ab D ab 2 ac D a2 b 2 c. Exercise 2. If G D H1 , then G=ha2 b 2 i Š H2 . Solution. Since G=ha2 b 2 i is nonabelian of exponent 4, it suffices to show that it is metacyclic. Note that a2 b 2 2 ˆ.G/. Let Ai , i D 1; 2; 3, be all maximal subgroups of G; then these subgroups are abelian of type .4; 2; 2/. It follows that ha2 b 2 i is a direct factor of Ai , i D 1; 2; 3, so all maximal subgroups of G=ha2 b 2 i are abelian of type .4; 2/. Then, by Lemma 65.1, G=ha2 b 2 i is metacyclic, as was to be shown. Exercise 3. Classify Q8 -free minimal nonabelian 2-groups. (Hint. G is Q8 -free if and only if G=Ã2 .G/ is metacyclic.) Exercise 4. (a) A direct product of two Q8 -free 2-groups is not necessarily Q8 -free. (Hint. (a) The group D8 C4 has an epimorphic image isomorphic to Q8 C4 .)

360

Groups of prime power order

(b) A direct product of two D8 -free 2-groups is not necessarily D8 -free. (Hint. The group Q8 C4 has an epimorphic image isomorphic to D8 C4 .) (c) Suppose that G is Q8 -free. If D8 Š D G, then CG .D/ is elementary abelian. (d) Suppose that G is D8 -free. If Q8 Š Q G, then CG .Q/ is elementary abelian. Problem 1. Study the structure of nonabelian H2 -free 2-groups. Problem 2. Study the structure of minimal non-H2 -free 2-groups. Let S.p 3 / be a nonabelian group of order p 3 and exponent p; then p > 2. Obviously, S.p 3 /-free p-groups are metacyclic (see Theorem 9.11). It is easy to check that if G is a minimal non-S.p 3 /-free p-group, then p D 3 and G is a minimal nonmetacyclic group of order 34 (see Theorem 69.1).

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Maximal abelian subgroups in 2-groups

Abelian subgroups in 2-groups G play an important role. Therefore, it is not very surprising that our assumption that every two distinct maximal abelian subgroups have cyclic intersection determines completely the structure of G. We obtain five classes of 2-groups with this property. More precisely, we prove here the following result. Theorem 81.1. Let G be a nonabelian 2-group in which any two distinct maximal abelian subgroups have cyclic intersection. Then Z.G/ is cyclic, each abelian subgroup of G is of rank at most 2, the intersection of any two distinct maximal abelian subgroups is equal Z.G/, and G has (at least) one abelian subgroup of index 2. Moreover, G is isomorphic to one of the following groups. (a) Group of maximal class (dihedral, semidihedral or generalized quaternion). n1

(b) M2n D ha; t j a2

n2

D t 2 D 1; n 4; at D a1C2

i.

(c) G D D C (central product), where D Š D2n , n 3, is dihedral of order 2n , C Š C2m , m 2, is cyclic of order 2m and D \ C D Z.D/. m

n1

D 1; n (d) G D hx; t j .xt /2 D a; a2 D t 2 D 1; m 2; x 2 D ab; b 2 t 1 x 2m1 2n2 3; b D b ; Œa; x D Œa; t D 1; t D t b; a Db i, where jGj D 2mCn , 0 m 2, n 3, Z.G/ D hai Š C2m , G D hbi Š C2n1 , and M D hx; ai is a unique abelian maximal subgroup of G. We have CG .t / D ht i hai Š C2 C2m and hb; t i Š D2n . D h2 ; hg D h1 i, (e) G D hg; h j g 2 D h2 D 1; m 3; n 3; g 2 mCn1 2n1 where G is metacyclic, jGj D 2 since hgi \ hhi D hg i Š C2 . Also, 2 0 2 2 Z.G/ D hg i Š C2n1 , G D hh i Š C2m1 and M D hh; g i is a unique abelian maximal subgroup of G. n

m

n1

m1

The more general problem to determine the structure of a nonabelian p-group G such that A \ B D Z.G/ for any two distinct maximal abelian subgroups A and B is very difficult. First we show that a p-group G has this property if and only if CG .x/ is abelian for each x 2 G Z.G/ (Theorem 81.2). Then we show that such a 2-group G has either an abelian subgroup of index 2 or G is of class 2 and G 0 is elementary abelian (Theorem 81.3). In Corollary 81.5 we get a new result for an arbitrary finite 2-group. We also classify 2-groups G such that A=Z.G/ is cyclic for each maximal abelian subgroup A of G (a problem of Heineken–Mann). It is surprising that such groups

362

Groups of prime power order

have the property that CG .x/ is abelian for each element x 2 G Z.G/. Then we may use our Theorems 81.2 and 81.3 to classify such groups (Theorem 81.4). In this classification we distinguish the cases, where G has an abelian subgroup of index 2 and the case where jG W Aj > 2 for each maximal abelian subgroup A of G. Proof of Theorem 81.1. Let G be a nonabelian 2-group in which any two distinct maximal abelian subgroups have cyclic intersection. Since Z.G/ is contained in each maximal abelian subgroup of G, it follows that Z.G/ is cyclic. Suppose that G possesses an elementary abelian subgroup E of order 8. Let A be a maximal abelian subgroup containing E and set F D 1 .A/ so that E F . Let B G be such that A < B and jB W Aj D 2 and let x 2 B A. Then x 2 2 A and therefore x induces on F an automorphism of order 2. It follows that jCF .x/j 4 and the abelian subgroup CF .x/hxi is contained in a maximal abelian subgroup C which is distinct from A since x 62 A. But A \ C CF .x/ and so A \ C is noncyclic, a contradiction. We have proved that each abelian subgroup of G is of rank at most 2. We may assume that G is not of maximal class (case (a) of Theorem 81.1) and so G possesses a normal four-subgroup U . Set M D CG .U / so that jG W M j D 2 since Z.G/ is cyclic. Let A be a maximal abelian subgroup of G which contains U so that A M . Suppose that A ¤ M and let y 2 M A be such that y 2 2 A. Let B be a maximal abelian subgroup of G containing the abelian subgroup U hyi. Then B ¤ A (since y 62 A) and A \ B U , a contradiction. We have proved that whenever U is a normal four-subgroup of G, then M D CG .U / is an abelian maximal subgroup of G. If x is any element in G M , then CM .x/ D Z.G/ and Z.G/hxi is a maximal abelian subgroup of G. Thus, the intersection of any two distinct maximal abelian subgroups of G is equal to Z.G/ and this statement also holds for 2-groups of maximal class. Suppose that G has more than one normal four-subgroup. By Theorem 50.2, G D D C with D Š D8 , D \ C D Z.D/ and C is either cyclic of order 4 or C is of maximal class (distinct from D8 ). Let U be a four-subgroup in D so that U is normal in G. By the above, CG .U / is abelian and so C must be cyclic. We have obtained a group stated in part (c) of Theorem 81.1. In the sequel we assume that G has a unique normal four-subgroup U and set M D CG .U / so that M is an abelian maximal subgroup of rank 2 with 1 .M / D U . (i) First assume 2 .G/ 6 M . Then there is an element y 2 G M of order 4 so that y 2 2 U . We have U hyi Š D8 and so there is an involution t 2 G M . Since t does not centralize U and M is abelian of rank 2, we get CG .t / D ht i CM .t /, where CM .t / is cyclic of order 2m , m 2. Indeed, if m D 1, then G is of maximal class. Also, we have t 62 ˆ.G/, G has no elementary abelian subgroups of order 8 and G is not isomorphic to M2s , s 4, since M2s has only three involutions. We are now in a position to use Theorem 48.1. It follows that G has a subgroup S of index 2, n1 D t 2 D 1; b t D b 1 i Š D2n , where S D AL, L is normal in G, L D hb; t j b 2 n 3, A D hai Š C2m , m 2, A \ L D Z.L/ D hzi, Œa; t D 1, CG .t / D ht i hai,

81 Maximal abelian subgroups in 2-groups

363

1 .G/ D 1 .S / D 2 .A/ L, 2 .A/ \ L D Z.L/ and if jG W S j D 2, then there is an element x 2 G S such that t x D t b. Since hbi is a unique cyclic subgroup of index 2 in L, hbi is normal in G. Set B D 2 .A/ L D 1 .G/,