Character Theory of Finite Groups

Character Theory of Finite Groups

Pure and Applied Mathematics A Series of Monographs and Textbooks Editors Samuel Ellenberg and Hyman Bass Columbia University, New York

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CHARACTER THEORY OF FINITE GROUPS

1. MARTIN ISMCS Department of Mathematics University of Wisconsin Madison, Wisconsin

@

1926

ACADEMIC PRESS

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COPYRIGHT 0 1976, BY ACADEMIC PRESS, INC. ALL RIGHTS RESERVED. NO PART OF THIS PUBLICATION MAY BE REPRODUCED OR TRANSMITTED m ANY FORM OR BY ANY MEANS, ELECTRONIC OR MECHANICAL, INCLUDING PHOTOCOPY, RECORDING, OR ANY INFORMATION STORAGE AND RETRIEVAL SYSTEM, WITHOUT PERMISSION IN WRITING FROM THE PUBLISHER.

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Library of Congress Cataloging in Publication Data Isaacs, Irving Martin, (date) Character theory of finite groups. (Pure and applied mathematics, a series of monographs and textbooks; 69) Bibliography: p. 1. Finite groups. 2. Characters of groups. I. 11. Series. tle. QA3.P8 [QA171] 510'.8s ISBN 0- 12-374550-0

A M S (MOS) 1970 Subject Classifications: 2OC15,2OC05, 2OC25, and 2OC20

PRINTED IN THE UNITED STATES OF AMERICA

Ti[512.24] 75-19652

Contents

PREFACE ACKNOWLEDGMENTS NOTATION

1. Algebras, modules, and representations 2.

vii

ix xi

1

Problems

11

Group representations and characters

13 29

Problems

3. Characters and integrality Problems

4. Products of characters Problems

5. Induced characters Problems

6. Normal subgroups Problems

7. T.I. sets and exceptional characters Problems

33 44 47 59 62 73

78 95

99 121 V

Contents

vi

8. Brauer’s theorem Problems

9. Changing the field Problems

10. The Schur index Problems

11. Projective representations Problems

12. Character degrees Problems

13. Character correspondence Problems

14. Linear groups Problems

15. Changing the characteristic Problems

126 141 144 156 160 171 174 194 198 216

219 237 240 260 262 285

Appendix Some character tables

287

Bibliographic notes

292

References

295

INDEX

297

Preface

Character theory provides a powerful tool for proving theorems about finite groups. In fact, there are some important results, such as Frobenius’ theorem, for which no proof without characters is known. (Until fairly recently, Burnside’sp”q” theorem was another outstanding example of this.) Although a significant part of this book deals with techniques for applying characters to “pure” group theory, an even larger part is devoted to the properties of characters themselves and how these properties reflect and are reflected in the structure of the group. The reader will need to know some basic finite group theory: the Sylow theorems and how to use them and some elementary properties of permutation groups and solvable and nilpotent groups. A knowledge of additional topics such as transfer and the Schur-Zassenhaus theorem would be helpful at a few points but is not essential. The other prerequisites are Galois theory and some familiarity with rings. In summary, the content of a first-year graduate algebra course should provide sufficient preparation. Chapter 1 consists of ring theoretic preliminaries, and Chapters 2-6 and 8 contain the basic material of character theory. Chapter 7 is concerned with one of the more important techniques for the application of characters to group theory. The emphasis in all chapters except 1,9, 10, and 15 is on characters over the complex numbers rather than on modules and representations over other fields. In Chapter 9, irreducible representations over arbitrary fields are considered; and in Chapter 10, this is specialized to subfields of the complex numbers. Chapter 15 is an introduction (and only that) to Brauer’s vii

viii

Preface

theory of blocks and “modular characters” and so is concerned with connections between complex characters and characteristic p-representations. The remaining chapters concern more specialized topics. Chapter 12 deals with the connections between the set of degrees of the irreducible characters and the structure of a group. Chapter 13 is essentially an exposition of a paper of Glauberman, and Chapter 14 contains some “classical” results on complex linear groups and a small sampling of more recent developments. Following each chapter there is a selection of problems. In addition to some routine exercises, these include examples, further results, and extensions and variations of theorems in the text. It is hoped that the reader will find that these problems enhance his understanding (and enjoyment) of character theory.

Acknowledgments

At various times during the preparation of this book I was supported by research grants from the National Science Foundation and a research fellowship from the Sloan Foundation. These organizations deserve (and hereby receive) my sincere thanks. Finally, I thank my teacher, Richard Brauer, both for introducing me to character theory and for his large part in developing the subject which I find so fascinating.

ix

This Page Intentionally Left Blank

Notation

algebra of n x n matrices over F linear transformation of the A-module J( induced by x E A {XV I X E A) regular A-module internal direct sum see Definition 1.12 see Lemma 1.13 the Jacobson radical, Problem 1.4 conjugacy class of g symmetric group of degree n transpose of the matrix X the order of g see Definition 2.26 restriction of to H see Problem 2.3 the algebra homomorphism Z(@[G]) -,C induced by x x-'y- ' x y , the commutator see Lemma 4.4 and the discussion preceding it see Definition 4.20 see Definition 5.1 induced character stabilizer of c( in permutation representation inertia group, Definition 6.10

x

xi

Notation

Xii

determinantal order of X , equals the order of det x in group of linear characters Qk the field Q(ezni/k) OP(G),OP’(G) minimal normal subgroups of p-power index and index prime to p ring of R-linear combinations of Y E Irr(G) RC91 “I” {9~R[Y]19(1)= 0) P(G, Jf‘) ~[{~H~IHE*)I gn,gn,,g p , gg, see the discussion following Lemma 8.18 see the discussion at beginning of Chapter 9 f E the field generated over F by the values of x F(X) the Schur index, Definition 10.1 %(X) Z(G, A), B(G, A), H(G, A ) see the discussion preceding Theorem 11.7 the Schur multiplier, Definition 11.12 M(G) A the group of linear characters of A Ch(G IS), Irr(G19) see the discussion preceding Definition 11.23 Hgfor g E G coreG(H) c.d.(G) M1) I x E Irr(G)l vanishing-off subgroup, (x E G I~ ( x #) 0) V01) d.l.(G) derived length of G F(G) Fitting subgroup of G b(G) max(c.d.(G)) O,(G), OJG) maximal normal subgroup of order a power of p , prime to p the set of S-invariant x E Irr(G) Irrs(G) Frattini subgroup of G IBr(G) the set of irreducible Brauer characters of G 4, decomposition numbers, Definition 15.9 the projective character associated with cp E IBr(G) % the set of p-blocks of G BUG) the idempotent and algebra homomorphism of Z ( F [ G ] ) e ~AB, associated with B E Bl(G) the set of defect groups for the class .X 4.K) 2 the sum of the elements of .K in F[G] coefficient of 2 in eB UB(W the set of defect groups for B E BI(G) W) defect of B E Bl(G) d(B)

4x1

n

1 Algebras, modules, and representations

Character theory provides a means of applying ring theoretic techniques to the study of finite groups. Although much of the theory can be developed in other ways, it seems more natural to approach characters via rings (or more accurately, algebras). The purpose of this chapter is to provide the reader with the ring theoretic prerequisites needed in the rest of the book. Many of the results in this chapter are true in more general contexts than those considered here. Nevertheless, an effort has been made to avoid excess generality and to prove only that which will be needed later. (1.1) DEFINITION Let F be a field and let A be an F-vector space which is also a ring with 1. Suppose for all c E F and x, y E A , that (cx)y = c(xy) = x(cy).

Then A is an F-algebra. We mention some examples of algebras over a field F:

(a) M,(F) is the algebra of n x n matrices over the field F. (b) Let I/ be an F-vector space. Then End(V), the set of F-linear transformations of I! is an F-algebra under the following conventions. If x, y E End(I/), then x y is defined by (u)xy = ((u)x)y and if c E F , then cx is . course, (u)(x + y) = (u)x + (u)y. defined by (u)(cx) = ( c u ) ~Of This is a good place to digress briefly to discuss some notational conventions which will be used throughout this book. In writing scalar multiplication in a vector space, the scalar may be written on either side of the vector 1

Chapter 1

2

to which it is applied. Similarly, functions are written on whichever side is convenient, but the rule for function composition is always “fg means do f first, then g.” Because of this rule, in those situations where function composition is important [as in example (b)], it will usually be convenient to write functions on the right. We now return to another example of an algebra, the one of primary importance for us: the group algebra. (c) Let G be a finite group. Then F [ G ] is the set of “formal” sums {C,,Ga,gla, E F } . The structure of an F-vector space is given to F[G] in the obvious way and the element of F [ G ] for which a, = 1 and ah = 0 if h # g is identified with g. This identification embeds G into F [ C ] and in fact G is a basis for F [ G ] . One result of this identification is to give a new meaning for a,g. We may now view this expression not only as a formal sum, but also as an actual sum, a linear combination of the basis vectors. Finally, to define multiplication on F [ G ] , we multiply the basis vectors according to their group multiplication and extend linearly to all of F [ G ] . It is routine to check that this defines the structure of an F-algebra on F[G].

1

The construction of F [ G ] suggests a general method of constructing algebras which should be mentioned. Let A be a finite dimensional F-algebra with F-basis u l , . . . , u,. We have then uiuj = Cijkuk, where cijk E F are the multiplication constants of A with respect to the basis {ui}. It is clear that these constants determine the algebra, so that any n-dimensional algebra may be specified by prescribing n3 constants cijkE F . Of course, only a small subset of all possible sets of constants define an algebra since most sets of constants define multiplications that turn out to be nonassociative. From now on, the word “algebra” in this book will mean a finite dimensional algebra. We make a few observations and definitions before going on to prove anything. Let A be an F-algebra. Then F . 1 = {cl Ic E F } is a subalgebra of A contained in the center Z ( A ) since (cl)x = c(lx) = c(x1) = x(c1) for x E A . It is sometimes convenient to identify F with F 1 and thus to view F as a subalgebra of A . If I is a left or right ideal of A as a ring and x E I and c E F , we have cx = (c1)x = x(c1) E I and I is a subspace. (If we had not required algebras to contain 1, then ideals would not automatically be subspaces.) If I is an ideal (this means two-sided), then A/I has the structure of an Falgebra in a natural manner. If A and B are F-algebras and cp is a ring homomorphism from A to B with cp(1) = 1, it is not necessarily true that cp is an F-linear transformation.

1

+

(1.2)

DEFINITION

satisfies

Let A and B be F-algebras. Suppose that cp: A

-+

B

Algebras, modules, and representations

3

(a) cp(XY) = cp(X)cp(Y)for all x,Y E A ; (b) cp(1) = 1; (c) cp is an F-linear transformation. Then cp is an algebra homomorphism or an F-homomorphism. (1.3) DEFINITION Let A be an F-algebra and let V be a finite dimensional F-vector space. Suppose for every u E V and x E A that a unique ux E V is defined. Assume for all x, y E A, u, w E r! and c E F that

(a) ( u + w)x = ux (b) U(X + y) = ux

(4

+ wx, + OY,

(VX)Y = V(XY),

(d) ( C U ) X = C(UX) = U(CX), (e) u l = u. Then V is an A-module. Let V be an A-module. Each x E A defines a map xy: V -+ V by u H ux. By (a) and part of (d) of the definition of a module, x y E End( V).By (b), (c), (e),and part of (d),the map x H x y is an algebra homomorphism A -+ End( V ) . Its image is denoted by A". Some important examples of modules are the following. If A E End( V ) , then V is an A-module in a natural way. If A = M,(F), then the row space of dimension n over F is an A-module under matrix multiplication. If A is any algebra, then A itself is an A-module under right multiplication. This module is called the regular A-module and is denoted by A". If V is an A-module and W s V is an A-invariant subspace, then W is a submodule of V Thus the right ideals of A are exactly the submodules of A". If W is a submodule of r! then the space V W becomes an A-module in the usual manner. Note that if I is a proper ideal of A, then the objects A / I , A"/I,and (AII)"are all defined and all different, being respectively an algebra, an A-module, and the regular (A/I)-module. However, Ao/I is an A-module which is annihilated by I (i.e., if u E A"/I and x E I, then ux = 0), and so it may be viewed as an (A/I)-module. As such it becomes (AII)". If V and W are A-modules, a linear transformation cp: V -+ W is an A-homomorphism if cp(ux) = cp(u)x for all u E I/ and x E A. An A-isomorphism is an A-homomorphism which is one-to-one and onto and, as is the usual situation, if V and W are A-isomorphic, they are exactly the same "as seen by A." For instance, in this situation they are annihilated by exactly the same set of elements of A . The set Horn,( r/; W )of A-homomorphisms from V to W has the structure of an F-space by (ccp)(u)= c(cpu) for c E F and (cp + $)(u) = ~ ( u +) 9(u). In addition HornA(K V ) is a ring [remember, cp9 is defined by ( u ) p 9 = (ucp)9], and in fact Hom,(X V ) is an F-algebra. It is exactly the centralizer

4

Chapter 1

of AV in End( V )and is denoted EA(V).If E = EA(V ) ,then V is an E-module and EAV) 2 A". Later in this chapter we shall find a sufficient condition for equality here. (1.4) DEFINITION Let V be a nonzero A-module. Then V is irreducible if its only submodules are 0 and I*:

It is obvious that if cp E Hom,(V, W ) then ker cp and im cp are submodules of V and W respectively. The following important lemma is now immediate. (1.5) LEMMA (Schur) If V and W are irreducible A-modules, then every nonzero element of HornA(V, W )has an inverse in HomA(WV).

An immediate consequence of Schur's lemma is that if V is an irreducible A-module, then EA(V ) is a division algebra, i.e., every nonzero element is invertible. (1.6) COROLLARY Let F be algebraically closed, A an F-algebra, and V an irreducible A-module. Then EA(V)= Fa 1, the set of scalar multiplications on 1.: Proof Clearly, F . 1 c EA(V).Let 9 E EA(V).Then 9 is a linear transformation of a finite dimensional vector space V over F and so has an eigenvalue 1. Then 9 - A1 E EA(V)and is not invertible. Thus 9 - 11 = 0 and 9 = 11 E F . 1 as claimed. I

To justify the study of irreducible modules we remark that for certain algebras, every module is a direct sum of irreducible ones, and thus to know all irreducibles is to know all modules for these algebras. It happens that group algebras over fields of characteristic zero are among these fortunate algebras. This will follow from Maschke's theorem which will be proved shortly. (1.7) DEFINITION Let V be an A-module. Suppose for every submodule W c V, there exists another submodule U G V such that V = W + U (the dot indicating that the sum is direct). Then V is a completely reducible module. Observe that irreducible modules are completely reducible as are all modules over fields (i.e., vector spaces). (1.8) DEFINITION An algebra A is semisimple if its regular module, A" is completely reducible.

(1.9) THEOREM (Muschke) Let G be a finite group and F a field whose characteristic does not divide I GI. Then every F[G]-module is completely reducible.

Algebras, modules, and representations

5

Proof Let V be an F[G]-module with submodule W Let U o be an F-subspace of V which is complementary to W i.e., V = W U o . Let cp be the projection map of V onto W with respect to U o . Now cp is a linear transformation which is not necessarily an F[G]-homomorphism. The object of the remainder of the proof is to modify cp in order to create,anF[G]-homomorphism. Define 9: V W by

+

--f

Clearly, 9 is F-linear. To show that 9 is an FIG]-homomorphism, we compute

since hg runs over G as g does for fixed h. If w E W we have wg E W for all g E G and thus cp(wg) = wg. It follows that $(w) = w. Now let U = ker 9, an F[G]-submodule of K We have 9(u) E W so that S(9)u))= S(u) and S(u - S(u)) = S(u) - S(u) = 0. Thus u = S(u) + (u - S ( ~ ) )W E + U and V = W + U . Finally, if W E W n U , we have w = S(w) = 0, so that V = W -I U and the proof is complete. I

A consequence of Maschke's theorem is that F[G] is semisimple if char(F),+'(GJ. The converse of this statement is also true and the reader is referred to the problems for a proof. (1.10) THEOREM Let V be an Amodule. Then V is completely reducible iffit is a sum of irreducible submodules. Proof Suppose V = C V, where the V, are irreducible. Let W G K By finite dimensionality, choose U E V maximal such that W n U = 0. We claim that W + U = I/: Otherwise, we may choose V, $ W + U and thus (W + U ) n V, = 0 by the irreducibility of V,. It follows that W n ( U + V,) = 0 and this violates the maximality of U . Thus V = W U and V is completely reducible. Conversely, suppose V is completely reducible and let S be the sum of all of the irreducible submodules of K If S < c! we may write V = S T with T # 0. By finite dimensionality, T contains an irreducible submodule which is a contradiction since T n S = 0. I

+

+

(1.1 1) LEMMA Let V be an A-module and suppose V = C V,where the V, are irreducible submodules. Then V is the direct sum of some of the c s . Prosf Choose W E V maximal with the property that W is the direct sum of some V,'s. If W < then V , $ W for some a. Since V, is irreducible,

Chapter 1

6

+

we have W n V , = 0 and W, = W V, > W This violates the maximality of W and we conclude that W = V as desired. I We see from the previous two results that the completely reducible modules are exactly the direct sums of irreducible modules. It follows that in order to know all modules for a group algebra over a field of characteristic 0, it suffices to know all irreducible modules. (1.12) DEFINITION Let V be a completely reducible A-module and let M be an irreducible A-module. The M-homogeneous part of C: denoted M ( V ) , is the sum of all those submodules of V which are isomorphic to M .

Observe that if M z N , then M ( V ) = N(V). As will be shown shortly, if M and N are nonisomorphic irreducible A-modules, then M ( V ) n N ( V ) = 0. If V has no submodules isomorphic to M , then M ( V ) = 0.

2

(1.13) LEMMA Let V = . i4( be a direct sum of A-modules with l4( irreducible for all i. Let M be any irreducible A-module. Then (a) M ( V ) is an EA(V)-submodule of V ; (b) M ( V ) = {Kl E M } ; (c) The number nM( V )of & which are isomorphic to M is an invariant of r/: independent of the given direct sum decomposition.

1

w

Proof (a) Let 9 E EA(V ) . We need to show M ( V)9 c M ( V ) . It suffices to show that if W E V and W E M , then W9 c M ( V ) . This is sufficient because M ( V ) is the sum of such W If W9 = 0, there is nothing to prove; if W9 # 0, then since W9 is a homomorphic image of the irreducible module W, we have W9 E W z M . Thus W9 c M ( V ) . (b) Clearly, { U:I E M } c M ( V ) .Let ni be the projection map of V onto Now let W c r/: W E M . If nj(W) # 0, then nLW) = Wj and W z W,.Thus n,( W) c { K I z M } for all j . However,’ W c nJ(W) andhenceW E: ~ { ~ / ; I ~ ~ M } . I t f o l l o w s t h a t M ( V ) c ~ { ~ : l ~ ~ (c) By (b), we have dim M ( V ) = nM( V ) dim M and it is immediate that nM( V ) is an invariant. I

w..

1

1

zj

By part (b)of Lemma 1.13 it follows that a completely reducible module is the direct sum of its M-homogeneous components for distinct M . In particular, M ( V ) n N ( V ) = 0 if M $ N . We wish to discuss “all irreducible A-modules.” This poses some difficulties. First, we really mean “all isomorphism classes of irreducible A-modules.”It would be convenient to have a representative set of A-modules. By this we mean a set A ( A ) of irreducible A-modules with the property that every irreducible A-module is isomorphic to exactly one element of &(A). Since modules are external to the algebra, it is not clear how one can find a

7

Algebras, modules, and representations

representative set for a given algebra A . To do this we need to produce a set of A-modules large enough to contain copies of all irreducibles. The next lemma shows that the set of homomorphic images of A" suffices. (1.14) LEMMA Let A be an F-algebra. Then every irreducible A-module is isomorphic to a factor module of A". If A is semisimple,then every irreducible A-module is isomorphic to a submodule of A".

Proof Let V be an irreducible A-module, choose 0 # u E r/: and define 9: A + V by 9 ( x ) = ux. Then 9 is clearly F-linear and 9 ( x y ) = uxy = 9(x)y so that 9 E HomA(Ao,V ) . Now, u E im 9 E V and hence im 9 = V since V is irreducible. Let W = ker 9. Then V A"/W If A is semisimple, then A" = W U and A"/W z U . The proof is complete.

+

Now fix a representative set &(A) of irreducible A-modules. By part (b) MeAAI(A)M(V) for every completely reof Lemma 1.13, we have V = ducible A-module K Suppose A is a semisimple algebra. We can then apply the above to A" and write A" = M(A").It turns out that M(A")is actually a two-sided ideal of A . For notational convenience, we write M ( A ) for M(A")in what follows.

c.

c.

(1.15) THEOREM (Wedderburn) Let A be a semisimple algebra and let M be an irreducible A-module. Then

(a) (b) (c) (d)

M ( A ) is a minimal ideal of A ; if W is irreducible, then it is annihilated by M ( A ) unless W z M , the map x H x M is one-to-one from M ( A ) onto A M c End(M); A ( A ) is a finite set.

Proof If x E A , the map 9,: y H x y satisfies 9, E E,(A"). Therefore, by Lemma l.l3(a), x M ( A ) = M(A)9, E M ( A ) and M ( A ) is a left ideal. Since M ( A ) is a submodule of A", it follows that it is an ideal of A. Minimality will follow after (c)is proved. If W is an irreducible A-module with W M , then W ( A )n M ( A ) = 0 by Lemma l.l3(b). Since W ( A )and M ( A ) are ideals, we have W ( A ) M ( A )= 0. By Lemma 1.14, A" has a submodule W, z W and W, E W(A), so that M ( A ) annihilates W,. Since W z W,, they have the same annihilator in A and (b) follows. By (b),it follows that x w = 0 if x E M ( A )and M $ W It now follows from that for y E A, we have the direct sum decomposition A = . MEAN(A)M(A) y , = x M ,where x is the component of y in M ( A ) .We conclude that the map x H x Mmaps M ( A )onto A M .If x E M ( A )and x M = 0, then it follows from (b) that x annihilates every irreducible, and hence every completely reducible A-module. Thus x = lx E A"x = 0 and (c) is proved.

+

1

Chapter 1

8

To show the minimality of M(A), let I < M ( A ) be an ideal of A. Now M ( A ) is a sum of submodules isomorphic to M and thus there exists M , c M(A), M , z M , M , $Z I . As M , n I < M , and M , z M is irreducible, we have M , n I = 0. Thus M , I c M, n I = 0 and I annihilates M , and hence also M . Therefore, if x E I , we have x, = 0. However, x H x M is one-to-one for x E M ( A ) and thus x = 0. We conclude that I = 0 and the minimality of M ( A ) is proved. Finally, M ( A ) # 0 for every M by Lemma 1.14, and yet A = ME./I(A$4(A) is finite dimensional. It follows that I&(A)I is finite and the proof is complete. I

c.

Observe that each M ( A ) is actually an algebra, its unit element being the M(A). Since the component of 1 in M ( A ) under the decomposition A = map x H xy is an algebra homomorphism from A to A M ,it follows from (c) of the theorem that the restriction of this map to M ( A ) is an algebra isomorphism from M ( A ) onto A M . Since M , ( A ) M , ( A ) = 0 for M , g M z , it follows that every ideal of the algebra M ( A ) is in fact an ideal of A. We conclude from the minimality of M ( A ) as an ideal of A , that M ( A ) is a simple algebra, i.e., it has no nontrivial proper ideals. Thus the preceding theorem asserts (among other things) that a semisimple algebra is a direct sum of simple algebras. This is a somewhat more usual statement of Wedderburn’s theorem. (It is also true that every simple algebra is semisimple. This follows from Problem 1.5.) To review the situation now: group algebras over fields of characteristic zero are semisimple; semisimple algebras are direct sums of ideals M ( A ) and M ( A )is naturally isomorphic to A,. What remains is to study the simple algebras A M .This is the purpose of the “double centralizer” theorem which follows. The proof we give here is based on an idea of M. Rieffel. It should be remarked that the hypothesis that A is semisimple is actually superfluous since the theorem is really about A , which is automatically semisimple when M is completely reducible. See Problem 1.6.

1

+

(1.16) THEOREM (Double Centralizer) Let A be a semisimple algebra and let M be an irreducible A-module. Let D = E,(M). Then ED(M) = A,.

Proof It is no loss to replace M by an isomorphic module, and so by Lemma 1.14, we may assume that M c A”. Let I = M ( A ) so that M c I . It is clear that A , E E,(M) so we prove the reverse inclusion. Let 9 E ED(M) so that (ma)9 = (rn9)a for a E D. If m E M , define a,: M + A by (x)a, = mx. Since m E M is a right ideal of A , we have mx E M and a,: M + M . If a E A and x E M , we have (xa)a, = m(xa) = (mx)a = (xcY,)~.It follows that a, E E,(M) = D. Thus for m, n E M , we have (*)

(mn)9 = (na,)9

= (n9)a, = m(n9).

Algebras, modules,and representations

9

Now fix n E M with n # 0 and let e be the unit element of I. We have AnA E I and by the minimality of the ideal I, we have e E Z = AnA and e = ainbi for suitable a,, b, E A. If rn E M, we have

1

rn = me = m

1ainbi = C (rnai)(nbi).

Since ma,E M and nb, E M, Equation (*) yields for all m E M that

1((rnai)(nbi))9= 1(mai)((nbi)9)= m 1ai((nbi)9). AM where u = 1 ai((nbi)9).The proof is complete. I

m9 = Thus 9 = uy

E

(1.17) COROLLARY Let A be a semisimple algebra over an algebraically closed field F and let M be an irreducible A-module. Then

(a) A M = End(M); (b) dim(&) = dim(M(A))= dim(M)'; (c) n,(A") = dim(M). Furthermore, if A ( A ) is a representative set of irreducible A-modules, then

(4 dimM = 1yM,AnrcA)dim(M)2, (e) dim(Z(A))= l A ( A ) l . Proof

By Corollary 1.6, E,(M) = F . 1 for irreducible M and thus A M =

EF,,(M) = End(M).By elementary linear algebra, End(M) 1MAF),the algebra of d x d matrices, where d = dim M . Thus dim A M = dim(End(M)) = d2. Since M ( A ) A M , we have (b). Now M ( A ) is the direct sum of n,(A") isomorphic copies of M, thus d2 = dim M ( A ) = n,(A") dim M = dn,(A"), and (c) follows. Since A = - MEAAI(A+W(A), (d) is immediate from (b). Finally, let ZM = Z(M(A)). Now Z(A,) = A M n E,(M) = A M n Fa 1 = F . 1. Thus dim ZM = dim(Z(A,)) = 1. Clearly, MEAAI(A)ZM c Z(A) and dim(1 2") = I&(A) I. However, if z E Z(A), write z = u', aME M(A). If u E M(A), then uaM = uz = zu = aMu since the distinct M(A)'s annihilate each other. Thus aM E ZM and C ZM = Z(A). The proof is complete. I

1

c- c

Although the word "representations" constitutes one third of the title of this chapter, so far nothing has been said about them. In fact, that isn't really true, because as we shall see, representations are just a different way of looking at modules. (1.18) DEFINITION Let A be an F-algebra. A representation of A is an algebra homomorphism 3E: A -,M,(F). The integer n is the degree of X. Two representations X, 9 of degree n are similar if there exists a nonsingular n x n matrix P, such that X(a) = P - q ( a ) P for all a E A.

10

Chapter 1

Clearly, similarity is an equivalence relation among representations. Also, if 1 ' ) is a representation of degree n and P is any nonsingular n x n matrix, then the formula X(a) = P - "1)(u)Pdefines a new representation X. It is easy to build modules from representations and representations from modules. If X is a representation of degree n of the F-algebra A, let V be the n-dimensional row vector space over F. If u E Vand X is any n x n matrix over F, then uX E K Define ua = uX(a) for a E A . It is routine to check that this gives the structure of an A-module to K Conversely, if M is an A-module, choose an F-basis for M and let X(a) be the matrix of aMwith respect to this basis. It is now easy to check that X is a representation. Note that a different choice of basis might give a different representation (and usually does). Starting with a representation X, constructing the module V as above, and then choosing the appropriate basis for V and constructing the corresponding representation will result in the original representation X. Suppose V and W are A-modules and 9 E Horn,( K W ) .How does this situation look from the representation point of view? Choose bases G, and G, for V and W . Since 9 is a linear transformation from V to K it has a matrix P with respect to the given bases. Note that P is m x n where m = dim V and n = dim W Now, the fact that (ua)9 = (&)a for all u E V and a E A yields X(a)P = P'1)(a), where X and 9 are the representations given by V and W with respect to the bases G, and G,. If 9 is a module isomorphism, then P is nonsingular and the above matrix equation yields that X and 1 ' ) are similar representations. In particular, the different representations arising from a given module with different choices of basis are all similar. The above reasoning may be reversed to show that if the representations arising from V and W with respect to bases G, and 8, are similar, then V and W are in fact isomorphic modules. It follows that there is a natural one-to-one correspondence between isomorphism classes of A-modules and similarity classes of representations of A. If V is an A-module and W < V is a proper nonzero submodule, choose a basis 8, for W and extend this to 8,, a basis for K Number G, so that the last m vectors are G,, where m = dim W Let X be the representation of A corresponding to V with respect to the basis G, and let 1 ' ) be the representation corresponding to W with respect to the basis G, . It is then easy to see for a E A that X(a) has the form

Furthermore, 3 is a representation corresponding to V/W Note that U is a function from A into (n - m) x m matrices (where n = dim V ) ,but U is not a representation.

Problems

11

The representation 3 is said to be in reduced form and one similar to X is reducible.Thus the irreducible representations correspond to the irreducible modules. If there exists a submodule U E V in the above situation, with V = W + U , then the basis for W may be extended to V by adjoining to it a basis for U . When this is done, the result is that U(a) = 0 for all a E A. It follows from this discussion that if X is any representation corresponding to a completely reducible module, then X is similar to a representation in block diagonal form, where each of the blocks is an irreducible representation. Problems

(1.1) Let I/ be an A-module. Show that V is completely reducible iff the intersection of all of the maximal submodules of V is trivial. Hint To prove “if,” embed V into a sum of irreducible modules. Recall that our definition of “module” requires finite dimensionality. Note This problem is “dual” to Theorem 1.10 which implies that V is completely reducible iff it is the sum of its minimal submodules. Theorem 1.10is true much more generally than we have proved it. It holds for arbitrary modules over rings. Problem 1.1, however, is not valid in this greater generality. A counterexample is the regular module of the ring of integers E .

(1.2) Let 3 and 9 be representations of an F-algebra, A. A nonzero matrix P is said to intertwine 3 and 9 if PX(a) = g(a)P for all a E A. Assume 3E and 9 are irreducible. (a) If P intertwines 3 and 9, show that P is square and nonsingular. (b) Assume that F is algebraically closed and that P and Q both intertwine 3 and 9. Show that Q = AP for some 1 E F.

(1.3) Show that an algebra A is semisimple iff every A-module is completely reducible. (1.4) Let A be an algebra. For A-module I! let d ( V ) = { a E A ( V a = O}. Let J ( A ) = nMEA.II(A)LC4(M), where &(A) is a representative set of irreducible A-modules. Show

(a) (b) (c) (d)

d ( V ) is an ideal of A for all V I/J(A) < V for every nonzero A-module I/: J(A)” = 0 for some integer n. If I is a right ideal of A and I“‘ = 0 for some m,then I E J(A). Note The ideal J ( A ) is called the Jacobson radical of A ; d ( V ) is the annihilator of V and a right ideal I with I” = 0 is said to be nilpotent.

12

Chapter 1

(1.5) Prove that the following are equivalent for the algebra A . (a) (b) (c) (d)

J ( A ) = 0. A has no nonzero nilpotent right ideals. A has no nonzero nilpotent ideals. A is semisimple.

Hint If V is irreducible, then ideals of A .

a(V ) is an intersection of maximal right

(1.6) Let A be an algebra and I/ a completely reducible A-module. Show that the algebra AV is semisimple. N o t e A consequence of Problem 1.6 is that the hypothesis that A is semisimple in the Double Centralizer Theorem 1.16may be dropped since this theorem is really about AM for an irreducible module M .

(1.7) Let A be an algebra and V an irreducible A-module. Show that I J K 4 V ) l = 1. (1.8) Let G be a group, H s G a subgroup and F a field with characteristic prime to I G : HI.Let V be an F[G]-module with submodule M! Suppose that there exists U o E K an F[H]-submodule such that V = W U o . Show that there exists an F[G]-submodule U c V with V = W U .

+

+

Note This generalization of Maschke’s Theorem 1.9 is due to D. G. Higman.

(1.9) Let G be a group and F a field of characteristic p . Suppose p I 1 GI and show that F[G] is not semisimple. Hint

(CgeCg)’ = 0.

(1.10) Let M be an A-module. Show that M is completely reducible iff M J ( A ) = 0. Hint

A / J ( A )is semisimple.

2

Group representations and characters

Let G be a finite group and let F be a field. Suppose 3 is a representation of F [ G ] with degree n. Since 3 is an algebra homomorphism, X ( 1 ) = I, the identity matrix. It follows for g E G that 3 ( g ) is nonsingular and 3 ( g ) - ’ = 3 ( g - ’). If we restrict the function 3 to G c FCG], we obtain a group homomorphism from G into the general linear group GL(n, F), that is, the multiplicative group of nonsingular n x n matrices over F .

(2.1) DEFINITION Let F be a field and G a group. Then an F-representation of G is a homomorphism 3? G + GL(n, F ) for some integer n. We have seen that a representation of F[G] determines an F-representation of G by restriction. Conversely,an F-representation 3, of G determines a representation 3 of F [ G ] by linear extension. That is,

XCC a,g) = c a,Xo(g).

We shall usually use the same symbol to denote both an F-representation of G and the corresponding representation of FCG]. Also, the adjectives “similar” and “irreducible” will be applied to F-representations of G as if they were the corresponding representations of FCG]. Some caution is necessary here since if F c E , a larger field, and 3 is an F-representation of G, then X is automatically an E-representation. It is entirely possible, however, that X is irreducible as an F-representation, and yet is reducible as an E-representation. We will explore this situation in some depth in Chapter 9. One further triviality which should be mentioned now is the following. If N -a G and X is an F-representation of G with N G ker X, then there is a 13

Chapter 2

14

unique F-representation of G/N defined by X(Ng) = X(g). This formula can also be used to define the representation X if is given. Note that X is irreducible iff f is. We shall often fail to distinguish between X and E. The trouble with representations is that they contain too much information. If X is an F-representation of G of degree n, then for each element of G we have n2 entries in X(g). Some of this data is clearly redundant because it distinguishes between similar representations. The idea behind character theory is to throw away most of the information and to savejust enough to be useful. This is done by calculating the traces (that is, the sums of the diagonal entries) of the matrices in question. Recall that if A and B are any two n x n matrices over a field, then tr(AB) = tr(BA).

x

(2.2) DEFINITION Let X be an F-representation of G. Then the F-character X of G aforded by X is the function given by X(g) = tr X(g). As is the case with F-representations of G , we may view F-characters as functions on all of F [ G ] . Note that if the characteristic char(F) # 0, then the constant function 0 is an F-character. On the other hand, if char(F) = 0, then 0 is definitely not an F-character because ~ ( 1 = ) deg X, where X is an F-representation of G which affords X. In this case, we say that ~ ( 1 is) the degree of X. Most F-characters of a group G are not homomorphisms of any kind. However, if 1is a homomorphism from G into the multiplicative group of F, then X(g) = (1(g)) is an F-representation of G of degree 1 which affords 1 as its character. Characters of degree 1 are called linear characters. In particular, the function lGwith constant value 1 on G is a linear F-character. It is called the principal F-character. (2.3) LEMMA (a) Similar F-representations of G afford equal characters. (b) Characters are constant on the conjugacy classes of a group. Proof If P is nonsingular then tr(P-'A . P ) = tr(P. P - ' A ) = tr(A). Both (a) and (b) follow from this observation. To see (b), observe that X(h-'gh) = X(h)-'X(g)X(h) if X is a representation of G , and hence tr(X(h-'gh)) = tr(X(g)). I We make one further general observation. If X and tions of G , then

is also an F-representation. Since tr 3@)= tr X(g) the set of F-characters of G is closed under addition.

9 are F-representa-

+ tr g(g),it follows that

Group representations and characters

15

We now restrict our attention to the special case that the field F = C, the complex numbers. We emphasize, however, that the subfield consisting of the algebraic elements in C would work exactly as well, and in fact with only minor modifications, most of what follows works for any algebraically closed field of characteristic not dividing I GI. Let us establish some notation. Fix a finite group G and choose a representative set of irreducible @[GI-modules, &(C[G]) = {MI,.. . , Mk}. Choose a basis in each M i and let Xi be the resulting representation of @[GI. Let xi be the character afforded by X i . It follows that the set Irr(G) = {x,,. . . ,xk} is the set of all irreducible @-charactersof G (that is, characters afforded by irreducible representations). Henceforth, the word “character” will mean C-character unless otherwise stated. Since sums of characters are characters, it follows that x = n i x i is a character whenever the n, are nonnegative integers which are not all zero. Conversely, if x is any character of G afforded by a representation 3 corresponding to amodule we can decompose I/ into a direct sum of irreducible modules. It follows that x is the sum of the corresponding irreducible characters. We have, in fact, x = C nM,(V ) x i . Corollary 1.17(d) asserts that dim(C[G]) = Cf,,(dim M i ) 2 . Since dim C[G] = IGI and dim M i = deg Xi = xi(l), we obtain the fundamental formula

c:=

k

IGI =

C xi(112. i= 1

It seems natural at this point to ask how we can determine the integer k purely group theoretically, without looking at representations. By Corollary 1.17(e),we have k = I&(C[G])I = dim Z(C[G]). (2.4) THEOREM Let XI, X 2 , . .., X , be the conjugacy classes of a group G. Let K , = x E C[G]. Then the K , form a basis for Z(C[G]) and if K , K , = a i j v K v ,then the multiplication constants aijv are nonnegative integers.

c xxejyi

Proof It is clear that the K i lie in Z(C[G]). Moreover, they are linearly independent because they are sums of disjoint sets of elements. If z = 1a,g E Z(C[G]) and h E G, we have z = h-’zh = 1a,gh. Comparing the coefficients of gh on both sides, we obtain ugh = a,. In other words, the coefficients a, have the constant value ai for all g E X i . It follows that z = a i K i and thus the K , span Z(C[G]). To find aijv,pick g E X,. Then aijvis the coefficient of g in K i K j .From the definition of multiplication in a group algebra, this is 1 {(x, y ) ( xE Xi, y E X j , xy = g} I. Since aijv is the cardinality of a set, it is a nonnegative integer. I

c

Chapter 2

16

(2.5) COROLLARY The number k of similarity classes of irreducible representations of G is equal to the number of conjugacy classes of G. (2.6) COROLLARY The group G is abelian iff every irreducible character is linear. Proof Let k be the number of classes of G. Then k = I GI iff G is abelian. Now IGI = ~ ~ ( 1 and ) ' xi(l) 2 1 for all i. It follows that k = IGI iff x i ( l ) = 1 for all i. The proof is complete. I

xf=,

We have not yet proved that the xi are distinct. To see this, we introduce a little more notation. From the results of Chapter 1 we have the direct sum k

1 Mi(C[G]).

C[G] =

*

i= 1

c

Let 1 = e, with e, E Mi(C[G]).Since M,(C[G]) annihilates the module M i if i # j , we have Xi(ej) = 0 in this case. It follows that Xi(ei) = Xi(l) = 1. Therefore, xi(ej) = 0 if i # j and xi(ei)= xi(l) # 0 and we conclude that the xi are distinct as functions on @[GI.Thus the xi are also distinct as functions on G. We may now restate two of the earlier results in a slightly more convenient form. (2.7) COROLLARY Let G be a group. Then IIrr(G)I equals the number of conjugacy classes of G and

c

=

x E Irr(G)

IGI.

For certain very small groups, the information contained in Corollary 2.7 is sufficient to determine the irreducible character degrees. For instance, if G = CJ, the symmetric group on three symbols, then G has exactly three conjugacy classes and (GI = 6. It follows that the x i ( l ) are 1,1, and 2. Actually, the preceding argument with the e;s yields more than the fact that the 2;s are distinct. A classfunction on a group G is a function, cp: G + C which is constant on conjugacy classes. All characters are class functions. (2.8) THEOREM Every class function cp of G can be uniquely expressed in the form cp =

c

a,x,

x E Irr(G)

where a, E C. Furthermore, cp is a character iff all of the a, are nonnegative integers and cp # 0.

17

Group representations and characters

Proof The set of class functions of G forms a vector space over C whose dimension is the number of classes of G . We claim that Irr(G) is a basis for this space. Since IIrr(G)I = k = number of classes, it suffices to show that if aixi = 0, then each ai = 0. This is immediate by evaluation at e,. The second statement has already been proved. 1

1

If x = nixiis a character, then those xi with n, > 0 are called the irreducible constituents of x. In general, if is a character such that x is also a character or is zero, then JI is called a constituent of x. An important consequence of Theorem 2.8 is the following.

+

+

(2.9) COROLLARY Let X and '2) be @-representationsof a group G . Then X and '2) are similar iff they afford equal characters. Proof We already know that similar representations afford equal characters. Let V and W be @[GI-modules corresponding to X and '2) respectively. If these Then 3E affords the character nM,(V)xiand '2) affords nMI(W)xi. characters are equal, it follows that nM,(V)= nM,(W)for all i. Therefore, V and W are isomorphic to identical direct sums of irreducible C[G]modules and thus are isomorphic and X and '2) are similar as required. 1

1

1

For a particular group G, the irreducible characters are usually presented in a character table: a (square) array of complex numbers whose rows correspond to the xi and whose columns correspond to the classes .Xi. An example of a character table is the accompanying one for the symmetric group Z4 on four symbols. g:

1 (1 2) (1 2)(34) ( 1 2 34) (1 2 3)

ICl(9)I:

1

4 6

8 3

XI:

1 1

1 -1

2 3 3

0 1

1 1 2

IC(9)I: 24

xz: x3: x4:

xs:

-1

-1 -1

4 6

3 8

1

1 1

0 -1 1

-1 0 0

-1

In this table, the classes are denoted by writing a representative element g in its cyclic notation. (The reader is reminded that in a symmetric group,

two elements are conjugate iff they have the same cycle structure.) The sizes of the centralizer C(g) and of the corresponding conjugacy class Cl(g) are

Chapter 2

18

given for convenience, although they are not, properly speaking, a part of the table. Further character tables are given in the Appendix (page 287). It is rather difficult to describe the process by which these tables are constructed. Usually, various combinations of ad hoc arguments and general theorems are necessary. As the student learns some of these theorems, he is urged to try to construct some character tables on his own. The important point is that it is very much easier to construct a character table than it is to construct representations. Theorem 2.8 tells us that every class function is a linear combination of irreducible characters. For example, if G = C, and q ( g ) is defined to be the number of points moved by g E G , then cp is a class function and is a linear combination of xl,xz,.. . ,xs. Thus the row (0,2,4,4,3)is a linear combination of the five rows of the given table. It turns out that there is an easy method for computing the coefficients, practically by inspection from the table. This is done by using the so called "orthogonality relations," which we are about to derive. These relations are also extremely useful in the construction of character tables. Before we leave C4 we should comment on the fact that all of the character values, which a priori are only known to be complex numbers, in fact turn out to be integers. (Of course the xi(l), being the degrees of representations, are always positive integers.) It is not always true that all character values are ordinary integers, although frequently a large fraction of them are. It is true, however, for all symmetric groups. All of this should become clear later.

The key to the orthogonality relations is to compute explicitly the coefficients of the group elements in the e,'s in terms of the characters. To do this we use the character p of G afforded by a representation corresponding to the regular module @ [ G ] " .This regular character p will be computed in two ways. (2.10)

LEMMA

If g E G and g # 1, then p ( g ) = 0. Also p(1) = I GI.

Proof We must choose a basis for CCG]" to obtain a corresponding representation. We simply take G , in some ordering, as the basis and let % be the corresponding representation. If %(g) = (aij),then aij = 0 unless g i g = g j , in which case aij = 1. Since p ( g ) is the number of g i satisfying g i g = g i , the lemma follows. I

Since p is a character of G, it may be expressed as an integer linear combination of the xi.We do this explicitly.

(2.11)

LEMMA

p =

cf=lX i ( l ) X i .

Group representations and characters

19

Proof If V is any @[GI-module,it may be decomposed as a direct sum of irreducibles. The character afforded by a corresponding representation is C nM,(i/)xi. Now by Corollary 1.17(c) we know that nM,(C[G]") = dim M i = xi( 1). The result follows. I It is suggested that the reader use Lemma 2.1 1 and the character table of C4 to compute p explicitly for this group, and check the result against

Lemma 2.10. (2.12)

THEOREM

ei = (l/lGl) X

B Exi(1)XiW')g. ~

1

Proof Write ei = a,g. By Lemma 2.10, we have p(eig- ') = a,[ GI. Lemma 2.11 thus yields aglGI =

1 X,U)XJ@;g-'). j

Since

we have xJ(eig-') = xi(g- ' ) h i j ,where the Kronecker aij is 0 or 1 depending on whether i and j are unequal or equal. We now have ag I G I =

and the result follows.

xi(l)xi(g- '1

I

(2.13) THEOREM (Generalized Orthogonality Relation) The following holds for every h E G.

Proof The ei lie in trivially intersecting ideals of C[G] and thus eiej = 0 if i # j. Since 1 = ej, multiplication by e, yields ei2 = ei. We now substitute the formula of Theorem 2.12 into the equation eiej = aijei and compare the coefficients of the group elements on both sides. The coefficient of a fixed h E G on the right hand side is (hij/lG l)xi(l)xi(h-') and that on the left-hand side is

1

The result now follows by equating these expressions and substituting h for h-'. I

Chapter 2

20

By taking h = 1 in Theorem 2.13 we obtain (2.14)

(First Orthogonality Relation)

COROLLARY

Since the expression x ( g - ' ) has come up several times, we digress for a while to discuss its connection with x(g) and some related questions. (2.15) LEMMA Let X be a representation of G affording the character x and let g E G. Let n = o(g), the order of g . Then (a) X ( g ) is similar to a diagonal matrix diag(e,, . . .,e f ) ; (b) = 1; (c) x ( g ) = C E i and I x(g)I 5 ~ ( 1 ) ;

(4 x b - ')

=

m.

Proof The restriction of X to the cyclic group ( 9 ) is a representation of ( 9 ) and hence it is no loss to assume G = ( 9 ) . By Maschke's theorem and

other results of Chapter 1, it follows that X is similar to a representation in block diagonal form, with irreducible representations of G appearing as the diagonal blocks. Since G = ( 9 ) is abelian, Corollary 2.6 asserts that its irreducible representations have degree 1, and thus X is similar to a diagonal representation. Now (a) follows, and we may assume that X is diagonal. We Therefore (b)is proved. It follows have I = X(g") = X(g)" = diag(e,", . . . ,q"). eiI I (eil = f = ~ ( 1 )It. is clear that x(g) = ci so that leil = 1 and that (c) follows. Now X @ - ' ) = X(g)-' = diag(e,-', . ..,e f - ') so that x(g-') = C e i - ' . Since leil = 1, we have ei-' = 4 and x(g-') = x(s).The proof is complete. I

11

1

Combining Corollary 2.14 and Lemma 2.15(d), we obtain 1 -

IGI

1xi(g)Xj(B) =

Jij.

geG

This suggests the following definition. (2.16)

DEFINITION

Let rp and 9 be class functions on a group G. Then

is the inner product of cp and 9. Some of the obvious properties of this "inner product" are (a) Crp, 91 = [%cp1; (b) [rp, rp] > 0 unless rp = 0;

Group representations and characters

21

(4 C C l c p I + c2cp2,91 = ClCcp19 91 + C2Ccp2, 91; (4 C% c19, + C 2 Q 2 1 = a - c p , 911 + CzC% 921. Therefore, [ , ] has all of the properties usually used to define an inner product in linear algebra and analysis. (In fact, this makes the space of class functions into a finite dimensional Hilbert space.) We know that Irr(G) is a basis for the space of class functions and it is the content of the orthogonality relation that it is, in fact, an orthonormal basis, that is,

[xi, xjl

= 6ij.

This yields the promised method for expressing an arbitrary class function in terms of the irreducible characters; for if [cp, xi] = c,, then cp = cixi. Another application of the inner product is to determine instantaneously whether or not a given character is irreducible.

1

+

(2.17) COROLLARY Let x and be (not necessarily irreducible) characters of G. Then [x, 4b] = [+, x] is a nonnegative integer. Also x is irreducible iff

cx, XI = 1.

1

+ 1

Proof We have x = nixi and = mixi, with all ni and m i nonnegative integers. Then [x, $1 = nimi = [$, x] and [x,x] = ni2. The result is now immediate, since x is irreducible exactly when one ni = 1 and all other n, = 0. I

1

The following “second orthogonality relation” is derived from the first and so imposes no new necessary condition for an array of complex numbers to be a character table. Nevertheless, it is often extremely useful in the construction of character tables and in the extraction of information from them.

(2.18)

THEOREM

(Second Orthogonality Relation) Let g, h E G. Then

1

X(S)X(h)= 0

x ~lrr(G)

if g is not conjugate to h in G . Otherwise, the sum is equal to IC(g)I. Proof Let gl, g2, . . . , g k be representatives of the conjugacy classes of G. Let X be the k x k matrix whose ( i , j ) entry is xi(gj).(In other words, X is the character table, viewed as a matrix.) Let D be the diagonal matrix.with entries hijI .Xi I where X iis the conjugacy class Cl(g,).The first orthogonality relation asserts that k

Chapter 2

22

This system of k2 equations may be replaced by the single matrix equation lGlI = XDXT, where I is the identity matrix and the superscript denotes transpose. Since a right inverse for a square matrix is necessarily also a left inverse, this yields lGlI = D X T X . We now write this as a system of equations and obtain

Since 1 G I/ I X i 1 = I C(gi)1, this yields

which is the desired result.

I

In the character table for C4 that was given earlier, the size of each conjugacy class was given. We see now that this information is derivable from the body of the table. It was given only for ease of computation of inner products. As a check on our results so far, observe what happens if we take h = 1 in the second orthogonality relation. The reader should recognize what results as a combination of Lemmas 2.10 and 2.1 1. Let E c C be the field of algebraic numbers. By Lemma 2.15, all of the character values x(g) E E for x E Irr(G) and g E G . How is Irr(G),which is a set of functions G + E related to the set of irreducible E-characters of G, which we denote by IrrE(G)?In fact these sets are equal. Suppose X is an irreducible E-representation of G which affords x E IrrE(G). Now X may be viewed as a C-representation and thus x is a character (that is, a @-character)of G . Since the entire development of character theory up to this point would have been the same over E as over C and since x E IrrE(G),it follows that [x, x] = 1. However x is a C-character of G and it follows from Corollary 2.17 that x E Irr(G). Thus Irr,(G) E Irr(G). We also have that both IrrdG) and Irr(G) have the same cardinality, namely that of the set of conjugacy classes of G . Therefore IrrdG) = Irr(G). The point of this digression is to suggest that there is something “absolute” about a character table. It is not entirely an artifact of our choice of the particular field C. Another consequence of this argument which is sometimes useful is that if x E Irr(G),then x is afforded by an E-representation of G . This type of consideration will be discussed much more fully in Chapter 9. Until then, we resume our convention that “character” means “@-character.”

Group representations and characters

23

A great deal of information about a group can be recovered from its character table. In particular, all of the normal subgroups of G can be found. A normal subgroup is a union of conjugacy classes and the word “found” in the preceding sentence means that those sets of conjugacy classes whose unions form subgroups can be listed. In particular, the orders of all normal subgroups and the inclusion relations among them can be determined.

(2.19) LEMMA Let X be a C-representation of G which affords the character x. Then g E ker 3E iff x(g) = ~ ( 1 ) . Proof If g E ker X, then X(g) = I = X( 1) and X(g) = X( 1). Conversely, by ) E~ + ..- + E ~ where , gi is a root of unity andf = ~ ( 1 ) . Lemma 2.15, ~ ( g= Since I E~ I = 1, the equation x(g) =f forces E~ = 1 for all i. Now 3(g) is similar to diag(e,, . . . ,E/) = I and therefore X(g) = Z and the proof is complete. I (2.20) X(1)l.

DEFINITION

Let x be a character of G . Then ker

x = {g E GI X(g) =

(2.21) LEMMA Let x be a character of G with x = nixi for Then ker x = n{ker xilni > 0). Also n{ker xil 1 Ii Ik} = 1.

xi E Irr(G).

Proof Since IXi(g)l I~ ~ ( 1by) Lemma 2.15, x(g) = ~ ( 1 forces ) xi(g) = xi(l) whenever ni # 0. The reverse inclusion is trivial and the first assertion follows. To prove the second statement, consider the regular character p. By Lemma 2.10, ker p = 1 and the result follows. I The normal subgroups Ni = ker xi can be found by inspection from the character table of a group G. We claim that every normal subgroup is the intersection of some of the N i s and thus can be found from the character table. To see this, let N 4 G and let % be the regular representation of the group G/N so that ker % = N/N. Now view % as a representation of G with kernel N and let x be the corresponding character of G. Then N = ker % = ker x = n{Ni I CX, xi1 0). Given a normal subgroup N of G (where “given” means listing the classes .Xi which it contains), we may calculate IN I from the character table using IN1 = {IXiIIXic N}. (Recall that ICl(g)l = IG: C@)I and IC(g)I = I X(g)1’ so that I .XiI is determined from the character table.) It follows from this discussion that G is simple iff ker x = 1 for all nonprincipal x E Irr(G) and therefore simplicity of a group can be easily determined from its character table.

+

xXEIrr(G)

Chapter 2

24

The group G is solvable iff it has a chain of normal subgroups, 1 = M , c M , G . . . c M , = G such that I M i : M i - I is a prime power for all i, 1 I i In. Since the M ican be located and their orders determined from the character table of G, it follows that the table determines solvability or nonsolvability of G. This may be a good place to remark that the character table of G does not determine G up to isomorphism. In fact if p is any prime, there are two nonisomorphic nonabelian groups of order p 3 . These two groups have identical character tables. Let N 4 G.It seems natural to ask whether the character tables of N and of GIN can be calculated from that of G. The answer is “no” for N but “yes” for G I N . We have already observed that there is a one-to-one correspondence between representations of GIN and representations of G with kernel containing N. Furthermore, under this correspondence, irreducible representations correspond to irreducible representations. This situation may be interpreted in terms of characters as follows.

(2.22)

LEMMA

Let N

4

G.

(a) If x is a character of G and N E ker x, then x is constant on cosets of N in G and the function 2 on GIN defined by f ( N g ) = x(g) is a character of GIN.

(b) If 2 is a character of G I N , then the function x defined by x(g) = f(Ng) is a character of G. (c) In both (a) and (b), x E Irr(G) iff 2 E Irr(G/N). Usually, we shall identify x and f . Under this identification, we have Irr(G/N) = {x E Irr(G)IN c ker x}. To demonstrate what is happening here, let us consider the example G = C, and N the normal subgroup of order 4. The classes of G which are contained in N are the identity and C1((1 2)(3 4)). Referring to the character table on p. 17, we see that the irreducible characters xiof G with N E ker xiare x,,xz,and x3 and so under above identification, we have Irr(GIN) = { X I , 1 2 7 x31. In order to write the character table for GIN we need to know its conjugacy classes. If X is a class of G, then its image in G I N , is a class of GIN. However, distinct classes of G may have equal images in GIN. This poses no problem since ifg, h E G, then Band Fare conjugate in = GIN iff x(g) = x(L) for all x E Irr(G/N). (The second orthogonality relation, applied to GIN proves this.) The part of the character table of G = C, corresponding to the characters of GIN is

z,

Group representations and characters

x1:

I

12:

1

x3:

2

25

I

1 1

-1

1

-I

0

2

0

I

1

-I

Observe that columns 1 and 3 are identical, as are columns 2 and 4. Deleting repeats, we obtain the character table 1

I 2

1

1

I

-1

0

-1

for GIN. (In this case GIN z Z3.) The preceding discussion provides a second method for computing IN I from the character table of G. Namely, by using the fact that

IG : N I = =

{x(1)’ I z E Irr(GIN))

1 ( X ( I ) ~ I ~ Irr(G) (E

and

N

G

ker I}.

By Corollary 2.7, a group is abelian iff all of its irreducible characters are linear. It follows that given N u G. the character table of G determines whether or not GIN is abelian. There is no known way to determine from the table whether or not N is abelian.

(2.23)

COROLLARY

Let G be a group with commutator subgroup G’. Then

(a) G’ = n{ker >.lAEIrr(G),A(1) = I } ; (b) 1 G : G’I = the number of linear characters of G. Proof’ If A is a linear character of G, then iL is a homomorphism into the abelian multiplicative group of @. It follows that G’ c ker 1. Since G/G’ is abelian, all z~1rr(G/G’)are linear and thus Irr(G/G’) = {A E Irr(G)(i,(I ) = 1). (This equality, of course, depends on the identification of characters of G/G’ with characters of G.) Finally, for any N a G, we have N = n{ker xIx E Irr(G) and N G ker x} and hence (a) follows. The number of linear characters of G is equal to the total number of irreducible characters of the abelian group G/G‘ and hence equals IG/C’I. The proof is now complete. I

The information in Corollary 2.23(b) is useful for finding the set of character degrees of a group G. For instance, if G is a nonabelian group of order 27, then I G : G’I = 9 and G has exactly 11 conjugacy classes. By the

Chapter 2

26

preceding, G has exactly nine linear characters and two nonlinear irreducible characters x and $. We have 27 = [GI = 9

+ ~ ( 1 ) ’+ $(1)’.

Since the only way that 27 - 9 = 18 can be written as a sum of two squares is 3’ + 3’, it follows that x( 1) = 3 = $( 1). Before leaving the discussion of character tables and factor groups, we mention an amusing result. The following could be proved without characters but it is somewhat tricky to do so. (2.24)

COROLLARY

Let g E G and N -=I G. Then )CG,,(Ng)I 5 lCG(g)l.

Proof From the second orthogonality relation, we have ICG/,(Ng)I

=

c

x E Irr(G/N)

1

x E Irr(G)

IXWg)l2 =

c {IX@)IZlXEIrr(G)9N

IX(g)l’ = lcG(g)l*

E

ker x>

I

We now discuss the connections between characters and the center of a group. (2.25) LEMMA Let X be an irreducible @-representation of G of degree n. Suppose A is an n x n matrix over C which commutes with X(g) for all g E G . Then A = a1 for some a E C. Proof Let M be the n-dimensional row space over @ so that M is an irreducible @[GI-module via me a = m3(a) for m E M and a E @[GI. Let 9: M + M be defined by m9 = mA. Then

and 9 = a . 1 for some a E C. The result follows. (2.26)

DEFINITION

I

Let x be a character of G . Then Z(x) = (9 E GI I~ ( gI )=

X(1)). If H !G G and X is a representation of G, then its restriction to H,denoted X,, is a representation of H.Similarly, the restriction xH of a character x of G to H is a character of H and we can write XH=

n#$

9 E Irr(H)

for suitable integers ng . Note that if xH E Irr(H), then x E Irr(G).Of course, the converse of this statement is false.

27

Group representations and characters

(2.27) LEMMA Let x be a character of G and let Z X be a representation of G which affords x. Then (a) (b) (c) (d) (e)

=

Z(x) andf =

x( 1). Let

Z = {gEG1X(g)=dforsomecE@}; Z is a subgroup of G ; xz = f A for some linear character I of Z ; Z/ker x is cyclic; Z/ker x E Z(G/ker x).

Furthermore, if x E Irr(G), then

(f)

Z/ker

x = Z(G/ker x).

Pro@ By Lemma 2.15, X(g) is similar to diag(e,, . . .,cS), with leil = 1,

1 I i I f . Since x(g) = C c i r it follows that Ix(g)I =J' iff all ci are equal. Since the only matrix similar to EIis c l itself, conclusion (a) follows. Define the function k Z -,C by X@) = A(g)I for g EZ. It follows for

g, h E Z that X(gh) = A(g)l(h)land hence Z is a subgroup and il is a hornomorphism (linear character) of Z. We have that x(g) = fi.(g) for g E 2 and (b) and (c) have been proved. Clearly, ker x = ker il and thus Z/ker x is isomorphic to the image of 1, a finite multiplicative subgroup of the field C. This subgroup is necessarily cyclic and (d) follows. Also, ker x = ker X and X(Z) c Z(X(G)) and (e) is an immediate consequence. Finally, if g(ker x) E Z(G/ker x), then X(g) E Z(X(G)).If x E Irr(G), then by Lemma 2.25, we conclude that X(g) = d for some E E C. Now (f)followsfrom (a) and the proof is complete. 1 (2.28)

COROLLARY

Let G be a group. Then Z(G) = ncZcx)Ix E I ~ ~ ( G ) ) .

Proof Since (Z(G) ker x)/ker x E Z(G/ker x), it follows from Lemma 2.27(f) that Z(G) E Z(x). Conversely, suppose g E Z(x) for every x E Irr(G). It follows that g(ker(x))E Z(G/ker x) and thus for any x E G, the commutator b , x ] = g-'x-'gxEkerx.

Thus b, x] E n{ker xIx E Irr(G)} = 1 and g commutes with x. Since x E G was arbitrary, we have g E Z(G). 1 It is apparent from Corollary 2.28 that Z(G) can be located from the character table of G. It follows that it can be determined from the table whether or not G is nilpotent. This is done by finding Z(G), then finding the character table of G/Z(G), and iterating this process. The sequence of subgroups of G which results is the upper central series and G is nilpotent iff this sequence reaches G.

Chapter 2

28

Some information about character degrees can be obtained using Lemma 2.27(c). We need a lemma first. (2.29)

LEMMA

Let H E G and let x be a character of G. Then

I IG:HICx,xI with equality iff x(g) = 0 for all g E G - H. CXH3XHI

Proof

We have IHI[xH,xH1

=

xIx(h)125 xIX(g)1’=IGI[X,X1

hsH

IJEC

since Ix(g)1’ 2 0 for g E G - H . Equality thus holds iff x(g) = 0 for all g E G - H. The result follows. I (2.30) COROLLARY Let x E Irr(G). Then x( 1)’ I1 G : Z(x)I. Equality occurs iff x vanishes on G - Z(x). Proof

x( l)’[A,

A]

By Lemma 2.27(c),we have xzcx, = x(l)A and thus [ x ~ ( ~xztx,] ,, x( 1)’. Therefore

=

=

x(U2 I IG:Z(x)ICx,xI = IG:Z(x)I with equality iff x vanishes on G - Z(x).

I

We already knew, of course, that x( 1)’ II GI for x E Irr(G). We now have the slight improvement that x( 1)’ I I G : Z(G) I. Equality can occur here, and when it does, Z(x) = Z(G) and x vanishes on G - Z(G). It has been conjectured that only in a solvable group is it possible to have x( 1)’ = I G : Z(G)I with x E Irr(G). As of this writing, the question is still open. Observe that the nonabelian groups of order 27 which were discussed earlier give examples where equality occurs. (2.31) THEOREM Suppose that I G Z(x)I = x(1I2.

x E Irr(G) and that G/Z(x) is abelian. Then

Proof It suffices to prove that x vanishes on G - Z(x). Let g E G - Z(x). Then by Lemma 2.27(f),we have that thereexists h E G withg-lh- ‘ g h # ker 2. However, since G/Z(x) is abelian, we have g- ‘ h - ‘ g h = z E Z(x). Now, if X is a representation of G which affords x,then X(z) = EZ and E # 1 since z 4 ker X. We have X(gz) = X@)X(z) = EX(g) and thus x(gz) = q ( g ) . However, gz = h-’gh and so x(gz) = x(g). Since Ex(g) = x(g) and E # 1, we have x(g) = 0 as desired. I A character x of G is said to befaithful if ker x = 1. Every group has a faithful character, namely its regular character p ; but not every group has a faithful irreducible character.

Problems

29

(2.32) THEOREM (a) If G has a faithful irreducible character, then Z(G) is cyclic. (b) If G is a p-group and Z(G) is cyclic, then G has a faithful irreducible character. Proof (a) Let x ~ I r r ( G be ) faithful. By Lemma 2.27(f), Z(G) = Z(x) and by part (d) of that lemma, Z(x) is cyclic. (b) Since G is a p-group, it follows that if 1 # N 4 G, then N n Z(G) # 1. Now let Z be the unique subgroup of order p in the cyclic group Z(G), so that Z E N for every nontrivial normal subgroup N of G. Since n{ker xIx E Irr(G)} = 1, it follows that Z $ ker x for some x E Irr(G). We conclude that ker x = 1 and the proof is complete. Problem 2.19 provides an example to show that the full converse of Theorem 2.32(a)is not true. Problems (2.1) (a) Let 3E be an irreducible F-representation of G over an arbitrary X ( g ) = 0 unless 3 is the principal representation. field. Show that (b) Let H c G and g E G be such that all elements of the coset H g are conjugate in G. Let x be a C-character of G such that [ x H , l H ] = 0. Show that x(g) = 0.

xgEG

3 ( h g ) , where 3 affords x. Hint (b) Compute the trace of ChEH

In the following, all characters are over C. (2.2) (a) Let x be a character of G. Show that x is afforded by a representation X such that all entries of X(g) for all g E G lie in some field F E C with ( F : Q ( < a. (b) Let E = elni/",where n = I G 1 and let x be a character of G. (Note that x(g)EQ[E] for all g E G by Lemma 2.15.) Let CJ be an automorphism of the field Q [ E ] and define xu: G 4 C by f ( g ) = x(g)d. Show that xu is a character and that xu E Irr(G) iff x E Irr(G). (2.3) Let x be a character of G. Define det affording x and set (det x ) ( g ) = det

x: G

4

C as follows. Choose 3E

w.

Show that det x is a uniquely defined linear character of G. (2.4) (a) Let G be a nonabelian group of order 8. Show that G has a unique ) 2, ~ ( z=) -2, and nonlinear irreducible character x. Show that ~ ( 1 = ~ ( x= ) 0, where Z E G' - (1) and x E G - G'.

Chapter 2

30

(b) If G g D,, show that det x # 1,. (c) If G g Q , , show that det x = 1,. Hint 2.15.

Show that ker(det x) contains all elements of order 4.Use Lemma

Note Although D, and Q , have identical character tables, thelmt+p det: Irr(G) Irr(G) is not the same for both groups.

(2.5) (a) Find a real representation of D8which affords the charabtm x of Problem 2.4(a). (b) Show that this cannot be done for the group Q,. (2.6) Let x, $ be characters of G. Define &::G

4

Chyi (%$)@)=,x(g)$(g).

(a) If $(1) = 1, show that x$ is a thracter. (b) If $(1) = 1, show that x$ E Irr(6) iff2 ~'In(ff). (c) If $ = (that is, @(g)= m).and,z(l) >'l,&ow'that x@ $ Irr(G). Note In Chapter 4 we will showithatgyhis always a character. (2.7) Let G be abelian and write G =kr(G).

e

(a) Show that is an abelian Lgroup under the multiplication of Problem 2.6. (b) If H c G, let H I = {A€ e(H-ciker A}.Showithat Iis 0 bijection from the set of subgroups of G onto the&t>ofsubgroups(-of:e. (c) Show that G r G. Hints There is a natural isomorphism rbf ;G onto 6.!Use ithis for (b). For (c), use the fundamental theorem of abelian groups.

(2.8) Let x be a faithful character of G. Show that H s G isabelianiff e w y irreducible constituent of xH is linear. (2.9) (a) Let x be a character of an abelian group A . Show

c 1x(4I2 2 IAIx(1).

xsA

(b) Let A E G with A abelian and I G: A I = n. Show that x(1) I n for all x E Irr(G). (2.10) Suppose G = A i n A j = 1ifiZj.

A i , where the Ai are abelian subgroups of G and

) 1, then ~ ( 1 2 ) I GJ/(n - 1). (a) Let x E Irr(G). Show that if ~ ( 1> (b) If G is nonabelian, then 1 Ail 5 n - 1 for each i and n - 1 2 (1 GI)''z. Hints For (a), bound Ix(g)I2 using Problem 2.9(a). For (b), use Problem 2.9(b).

xBEG

Problems

31

(2.11) Let g E G. Show that g is conjugate to g - in G iff x(g) is real for all characters x of G. Note An element of a group which is conjugate to its inverse is called a real element. If G has any real elements other than 1, then G must necessarily have even order.

(2.12) Let I GI = n and let g E G. Show that x(g) is rational for every character x of G iff g is conjugate to g" for every integer m with (m, n) = 1. Hints Let E be a primitive nth root of 1 in C and let E = QC.53. Let Y be the Galois group of E over Q. Given (m, n) = 1, show that there exists 0 E Y with x(g") = x(g)" for all g E G and all characters 2. Conversely, for every 0 E 9, there is an m such that this formula holds.

(2.13) Let I G' I = p , a prime. Assume that G

E

Z(G). Show that

~(1)'= IG: Z(G)l

for every nonlinear x E Irr(G). (2.14) Let H E G' n Z(G) be cyclic of order n and let m be the maximum of the orders of the elements of G / H . Assume that n is a prime power and show that I G I 2 n'm. Hints Choose x ~ I r r ( G )with H n ker x = 1. Let 3, = det x (as in Problem 2.3). We have xH = x(1)p with p E Irr(H). Using A and p, show that ~ ( 12 ) n. Finish the proof using Problem 2.9(b). Note The assumption on n can be removed using Corollary 5.4.

(2.15) Let that

x E Irr(G) be faithful and suppose H E G and xH E Irr(H). Show CG(H)= Z(G).

(2.16) Let H E G and let x be a (possibly reducible) character of G which vanishes on G - H. Assume either that H = 1 or that G is abelian. Show that I G : H I divides ~ ( 1 ) . Hint Let 3, be an irreducible constituent of xH. Under either hypothesis, find p E Irr(G) with p H = A. Compute b,p] and conclude I G : HI I [xH, A]. Note A natural common generalization of the situations H = 1 and G is abelian is H E Z(G). Is the result true under the hypothesis H E Z(G)?

-=

(2.17) Let A G be abelian and assume there exists x E Irr(G) with ~ ( 1= ) 1 G : A 1. Show that G has a nontrivial normal abelian subgroup. Hint

Show that

x vanishes on G - A.

Chapter 2

32

(2.18) Let A -a G and suppose A = C,(a) for every a # 1, a E A . Assume further that G / A is abelian. Show that G has exactly (IAI - l)/lG: A1 nonlinear irreducible characters and that these all have degree equal to 1G:AIandvanishonG - A . Hints Let k = I Irr(G)I. By counting classes, show that

k I1

+ (IAI - l ) / I G : A l + (IGI - l A l ) / l A l .

Using characters, show that k 2 IG: A1

+ (]GI - IG:A I ) / I G :AI'.

Use Problem 2.9(b). Note The group G of Problem 2.18 is a special case of a Frobenius group. The character theory of such groups will be discussed more fully later. Observe that although the hypotheses of 2.18 are very special, this situation does arise frequently. Some examples are A4 and nonabelian groups of order pq where p and q are primes with p I (q - 1).

(2.19) Let E = (xl, x2,x3,x4) be an elementary abelian group of order 16. Let P = (y) be cyclic of order 3. Let P act on E by XIY = x2,

x2y

=

x1x2,

xgy =

x4,

x4y

= X3X4.

Let G be the semidirect product E x P. Show that Z(G) = 1 but that G does not have a faithful irreducible character. Hint The smallest possi'ble degree for a faithful character of E is 4.

(2.20) Let X and 9 be irreducible @-representations of G and define the functions a&) and bijg)by X(g) = (a&)) and 9(g)= (b,jg)). Write Spqrs =

C apq(g)brs(g - ')* BEG

Show that S,,,, = 0 if X and 9 are not similar. If X = 9,show that Spqrs= 0 unless p = s and q = r in which case S,,,, = I GI /deg X. Hint Let P,, = CgEG X(g)E,,g(g-') where E,, is the (deg X) x (deg 9) matrix with all entries zero except the (4, r ) entry which equals one. Note that XP,, = P,,9. Use Schur's lemma. For the last statement use Lemma 2.25 and compute tr(P,,). Note The results of this problem are called the Schur relations. They can be used to give another proof of the orthogonality relations.

3

Characters and integrality

One of the most celebrated applications of character theory to pure group theory is Burnside's theorem which asserts that a group with order divisible by at most two primes is solvable. The proof of this theorem (and much of the rest of character theory) depends on properties of algebraic integers. We begin by establishing some of the mod basic of these properties. (3.1) DEFINITION An algebraic integer is a complex number which is a root of a polynomial of the form x"

+ a,+X"-' + . + ao, * *

where ai E 72 for 0 5 i I n - 1.

(3.2) of Z.

LEMMA

The rational algebraic integers are precisely the elements

Proof If a E Z,then a is a root of the polynomial x - a and thus is an algebraic integer. Conversely, let r/s be an algebraic integer with r, s E Z.We may assume that (r, s) = 1. We have

+

(r/s)" a,- l(r/s)"-

+ . + a. *

=

0.

Now multiply by s" and rearrange terms to obtain rn = - s ( ~ , , - ~ r ~+- 'a,-2srn-2 +

+ aosn-').

We conclude that slr". However, since (r, s) = 1, this yields s = f 1 and r/s E Zas desired. I 33

Chapter 3

34

Frequently, the word “integer” is used to mean an algebraic integer, and the elements of Z are referred to as “rational integers.” One of the most important properties of the set of algebraic integers is that it is a ring. In other words, sums and products of integers are integers. This fact seems surprising from the definition, but it is not hard to prove indirectly. This we proceed to do. (3.3) LEMMA Let X = {al, . . .,ak} be a finite set of algebraic integers. Then there exists a ring S satisfying (a) Z E S E C : (b) X E S ; (c) there exists a finite subset, Y of S such that every element of S is a Z-linear combination of elements of Y . Proof The integer ai satisfies an equation of the form a!‘ = f;.(ai),

wheref;. is a polynomial of degree ni - 1 with coefficients in Z. Let Y = {a;la;2 . . a r l O Iri I ni - l } and let S be the set of all Z-linear combinations of elements of Y . Using the equation a? = f;.(ai),any power of ai may be written as a Zlinear combination of 1, ai, a t , . . . ,a;‘- It follows from this that the product of any two elements of Y lies in S and hence S is a ring. All of the properties claimed for S are now clear. I

’.

Condition (c) of the above lemma may be paraphrased by saying that S is finitely generated as a Z-module. We now prove a strong converse to Lemma 3.3. (3.4) THEOREM Let S be a ring with Z E S E C. Suppose that S is finitely generated as a Z-module. Then every element of S is an algebraic integer. Proof Let s E S and let Y = {yl, . . . , y.} E S have the property that every element of S is a Z-linear combination of elements of Y . We then have syi

=

aijyj

i

for all i, with a i j € Z . Let A be the matrix (aij)and let u be the column, col(y,, . . . ,y,,). Then Av = sv and thus s is a root of the polynomial f(x) = det(xZ - A). It follows that s is an algebraic integer and the proof is complete.

I

35

Characters and integrality

(3.5) COROLLARY Sums and products of algebraic integers are algebraic integers. Proof Let a and /3 be algebraic integers. By Lemma 3.3, there exists a ring S with Z c S c C such that a, /3 E S and S is finitely generated as a Z-module. Since a + /3 and afi E S , it follows from Theorem 3.4 that they are algebraic integers. I

(3.6) COROLLARY Let x be a character of a group G. Then x(g) is an algebraic integer for all g E G. Proof By Lemma 2.15, we know that x(g) = + + where the are roots of a polynomial of the form x" - 1, and therefore are algebraic integers. The result now follows. I E,

We can now see the reason for the assertion made in Chapter 2 that all of the entries in the character table of the symmetric group C , lie in Z. If g E C , and rn is relatively prime to o(g),then g m and g have identical cycle structures and therefore these elements are conjugate in C,. It follows from Problem 2.12 that x(g) is rational for all x E Irr(C,,). Since x(g) is an algebraic integer, Lemma 3.2 yields that x(g) E Zas claimed.

Let G be a group and x E Irr(G).We wish to define a function o depending on x, from the center of the group algebra @[GI into C. Let X be any representation which affords x. If z E Z(C[G]), then we may conclude from Lemma 2.25 that X(z) = EIfor some E E C. Observe that since the only matrix similar to EZ is EZ itself, the complex number E does not depend on the choice of the particular representation affording 2. We now define o by setting o ( z ) = E. In other words X(z) = W(Z)Z

for all z E Z(C[G]). We shall often write w = oxin order to emphasize the dependence of o on x. Since SE is an algebra homomorphism, it is easy to see that o is also a homomorphism. In particular, o is C-linear and hence to determine o on Z(C[G]), it suffices to calculate its values on a basis. Such a basis is given by the class sums fore the conjugacy classes of G. Let X be a class with sum K E @[GI and let g E X . Calculation of traces in the equation X(K) = w(K)Z yields x(l)o(K) = x(K) =

c x(x)

XE

x

=

I.xlx(g)

36

Chapter 3

and thus

Note that it follows from this formula that the functions w, aredetermined by the character table of G. (3.7) THEOREM Let x ~ I r r ( G and ) let K be a class sum in C[G]. Then o , ( K ) is an algebraic integer. Proof Let XI, . . . , Xk be the classes of G, with corresponding class a i j v K v where sums K 1 , .. . , K k . By Theorem 2.4, we have K i K j = aijvE Z. Since o = owis an algebra homomorphism from Z(C[G]) to C, we have

cy

o ( K i M K j ) = C aijvw(Kv)* V

Let S be the set of all Z-linear combinations of the o(Ki).It follows that S is closed under multiplication. Since ~ ( 1 = ) 1, it follows that Z c S c C and Theorem 3.4 applies. All of the elements of S are therefore algebraic integers and the proof is complete. I It should be emphasized that the fact that x(g)1 Cl(g)I /x(1) is an algebraic integer does not follow from the fact that x(g) is integral since division of an integer by an integer does not usually result in an integer. We proceed now toward Burnside's solvability theorem. The essence of the argument is contained in the next result. (3.8) THEOREM (Burnside) Let x E Irr(G) and let X be a conjugacy class of G with g E X , Suppose that (x(l), 1 X I ) = 1. Then either g E Z(x) or else x(g) = 0. Proof We know that x@) I X [/I( 1) is an algebraic integer. Since (x(1), I X I ) = 1, we may choose rational integers u and v so that ux( 1) + v I X I = 1.

Thus

is an algebraic integer. Since ux(g) is also integral, it follows that o! = x(g)/x(1) is an algebraic integer. Suppose that g $Z(x), so that I x(g)1 ~ ( 1 )and lo![ <

-=

1.

Now let n = o@)and let E be the splitting field for the polynomial x" - 1 over Q in C so that o!E E . Let Y be the Galois group of E over Q. Since x(g) is a ) of unity, so is x(g)6 for each u E Y. It follows that I x@)" I IX( 1) sum of ~ ( 1roots

Characters and integrality

31

and (au[ I 1 for c E Y . We have then

I naq

< 1.

U€g

For each c E 3, ct' satisfies the same rational polynomials that ci satisfies and hence is integral. Therefore p = a' is an algebraic integer. However, p is clearly fixed by all E 9 and therefore /?E Q by elementary Galois theory. It follows from Lemma 3.2 that p E Z. Since 1 fl 1 .c 1, we have fl = 0 and hence a' = 0 for some c. Therefore 0 = ct = x(g)/x(l) and x(g) = 0. The proof is complete. 1

fl

(3.9) THEOREM Let G be a nonabelian simple group. Then { 1) is the only conjugacy class of G which has prime power size. Proof Suppose g E G , I Cl(g)I = p", and g # 1. Let x E Irr(G), x # lG. Then ker x = 1 since G is simple and Z(x) = Z(G) = 1 since G is nonabelian. Thus if p$x( l), then x(g) = 0 by Theorem 3.8. Now

0 = P(g) =

c

x(l)x(g) = 1 +

,x~lrr(G)

c

x(l)x(g).

,xeIrr(G): plx(1)

We have - 1 = pa, where

the sum being taken over x E Irr(G) where p I x( 1). It follows that ct = - l/p is an algebraic integer and this violates Lemma 3.2. I

(3.10) THEOREM Let IG( = pa$, where p and q are primes. Then G is solvable. Proof Use induction on ( G I . We may assume ( G I > 1 and choose a maximal proper normal subgroup N . If N > 1, then by the inductive hypothesis, N and GIN are solvable and thus G is solvable and the result follows. Suppose then N = 1, so that G is simple. Let P # 1 be a Sylow subgroup of G . We may choose g E Z(P), g # 1. Then ICl(g)l = IG: C(g)I divides I G : P 1, which is a prime power. It now follows from Theorem 3.9 that the simple group G is abelian and the proof is complete. I We now obtain some strong results about the degrees of the irreducible characters of a group G . The fact is that x( 1)I I G : Z(x)I for x E Irr(G). We shall first prove the weaker statement that the irreducible character degrees divide the group order. This proof is much less complicated and serves to motivate the stronger proof.

Chapter 3

38

(3.11)

THEOREM

Let x E Irr(G). Then x( 1)I IG I

Proqf From the first orthogonality relation we have IGI = Cx(s)x(g-'). G

We wish to rewrite this equation in terms of w x .Let X l ,X,, . . ., Xk be the classes of G, with class sums Ki and representative elements gi.We have then k

IGI =

k

C IXiIx(gi)x(gi-l) = i1 AIMKi)x(gi-l), i= 1 =1

where w = o x This . yields which is an algebraic integer. Since I G I/x( 1) is rational, it lies in Z and the result follows. I (3.12)

THEOREM

Let x ~ I r r ( G )Then . x(1)IIG: Z(x)l.

Proof Since x may be viewed as a character of G/ker x, it is no loss to assume that ker x = 1. Under this assumption, Z(G) = Z(x). For x, ~ E G define , x = y if there exists Z E Z= Z(G) such that x is conjugate to yz. It is easy to check that = is an equivalence relation and thus ) constant as x runs partitions G into equivalenceclasses. We claim that I ~ ( xI is over one of these classes. To see this, observe that xz = x(l)A, where A is a faithful linear character of 2, and that ~ ( y z=) A(z)x(y)for z E Z and y E G. If x = y, then xg = yz for some z and ~ ( x = ) A(z)x(y). Since I A(z)l = 1, the claim follows. Let Vl, V,, . . . ,V, be those (=))-classeson which x does not vanish. We have then r

( G I = CIx(g)12 = EG

C I%iIMgi)I', i= 1

where the gi are representatives for the V i . We claim JVil = ~ C l ( g i ) ~ ~ Z ~ . Clearly, every x E 'Xi is of the form yz where y E Cl(gi)and z E 2. It suffices to show that all of these elements yz are distinct. Suppose that ylzl = y 2 z 2 , y,, y, E Cl(gi), and zl, z2 E Z. Then X(Y1)4Zl) =

x(y,V(z,)

and x(yl) = x(y2) = x(gi) # 0. Thus l(zl) = A(z,) and hence z1 = z 2 since A is faithful on 2. Thus y , = y, and the claim is established. We have now IGI = C IViIIx(gi)12= C IC~(S~)IX(S~)X(S~-')I~I = X(l)w(Ki)X(gi- IZI 9

Characters and integrality

39

where K i = ~ x E C , ( B iand ) ~ w = w,. It follows that r

I G:ZI/X(~)=

1dKi)X(gi- ‘1,

i= 1

an algebraic integer which is a rational number. The result now follows. As a combined application of Theorems 3.8 and 3.12, we prove the following.

(3.13) THEOREM Let G have a faithful irreducible character of degree pa, where p is a prime and suppose that a Sylow p-subgroup of G is abelian. Then p“ is the exact power of p dividing I G : Z(G) I.

Proof Let x be the given faithful character of G. By Theorem 3.12, pa = x( 1) divides I G : Z(G) I. Let P E Syl,(G) and let x E P . Thus P c C(x) and hence (x(l), I Cl(x)I) = 1. By Theorem 3.8, ~ ( x = ) 0 if x # Z ( x ) = Z(G). Let Z = P n Z(G) so that x vanishes on P - Z. Now by Problem 2.16, we conclude that IP: ZI Ips. Since P / Z s PZ(G)/Z(G), which is a Sylow subgroup of G/Z(G), the result follows. 1 The above result is typical of a number of theorems about “complex linear groups,” that is, groups of nonsingular matrices over C. In these theorems, one is given the degree n of a finite linear group G and the object is to control the structure of G in terms of n, often under certain additional assumptions such as the irreducibility of G. In Theorem 3.13 we are given a group having a faithful representation of known degree, and this is obviously equivalent to being given a linear group of that degree. If G is a complex linear group, let S be the subgroup consisting of the elements of G which are scalar matrices (that is, of the form &I).Observe that if G is irreducible, then S = Z(G). The group GIS is called the collineation group associated with G. Frequently, the object of a theorem about linear groups of given degree is to obtain information about the associated collineation group. Theorem 3.13 is of this nature. The reason for this situation may be seen from the following. Let G be a linear group of degree n and let C be a group of n x n scalar matrices. Let G* = G C and let S* be the scalar subgroup of G*. The linear group G* may be much larger than G but G/S s G*/S*.

We have already seen several situations in which a character value is forced to be zero. We shall now prove that every nonlinear irreducible character vanishes somewhere. We begin with a preliminary result.

Chapter 3

40

(3.14) LEMMA Let G be a cyclic group and let x be a (possibly reducible) character of G . Let S = {g E GI G = (9)) and assume that ~ ( s )# 0 for all s E S . Then

CIx(s)lZ 2

ISI.

sss

Proof Let n = I GI and let E be the splitting field for the polynomial 1 over Q in C. Let 9 be the Galois group of E over Q. If D E Q and E is an nth root of 1, then e' = E"' for some m E Z, (m,n) = 1. Now ~ ( s )= el + . . . + Ef where E: = 1 and hence ~(s)" = elm + + Efm = ~(s"'). The group 9 is abelian and the restriction of complex conjugation to E is an element of 9.It follows that 2 = (ti)" for all a E E and (r E Q, and thus Ia' 1' = a u 2 = a'tiu = (I a ."1)' Therefore (I ~ ( s 1)"') = I ~(s"')12, where (m,n) = 1 and m depends only on 0. Observe that if s E S and (m,n) = 1, then s"' E S . Also, the map x H x"' is one-to-one on G and therefore effects a permutation of S . It follows that I ~ ( s1' ) is invariant under 9 and hence is rational. Since it is an algebraic integer, it must lie in Z, and since x does not vanish on S , we have

X" -

nscS

Now we use the fact that for any positive real numbers r l , r 2 , . . . ,i k , we have 1 -

k

C ri 2 (nri)'Ik

and we conclude that

and the proof is complete.

I

(3.15) THEOREM (Burnside) Let z ~ I r r ( Gwith ) ~ ( 1> ) 1. Then X(g) = 0 for some g E G. Proof Partition G into equivalence classes by calling two elements of G equivalent if they generate the same cyclic subgroup of G. Assume X(g) # 0 for all g E G. Then by Lemma 3.14, we have CIX(S)I2 2 IS1 scs

for every equivalence class S . Sum this inequality over all equivalence classes of nonidentity elements to obtain

C IX@)I2 2 IGI - 1 8+ 1

41

Characters and integrality

and thus IGI =

1 lX(S)I2 2 IGI - 1 + X(l)2. BEG

This forces ~ ( 1I ) 1 which contradicts the hypothesis.

I

The next topic we shall discuss does not, strictly speaking, depend on algebraic integers. Nevertheless it seems appropriate to include it here. Let G and H be finite groups and suppose C[G] E C [ H ] , where this is a @-algebraisomorphism. What can we infer about the relationship between G and H ? Clearly, [GI= IHJ and there exists a degree-preserving one-to-one correspondence between Irr(G) and Irr(H). We cannot conclude, however, that G and H are isomorphic or even that they have identical character tables. Indeed, if G is abelian, it follows from the results of Chapter 1 that @[GI is the direct sum of I GI copies of C. Thus if G and H are abelian and 1 G I = I H 1, then @[GI E C [ H ] . The situation becomes more interesting if we make the weaker assumption that Z[G] z Z[H], where h[G] represents the group ring of G over Z and may be identified with the ring of H-linear combinations of elements of G in C[G]. It is conjectured that if Z[G] E Z[H] for finite G and H, then G E H . The best result in this direction that has been proved as of this writing is due to A. Whitcomb. It asserts that if G is metabelian and Z[G] z Z[H], then G z H . If Z[G] E E [ H ] , then it is clear that we may view H as a multiplicative subgroup of Z[G] c @[GI and that H spans C[G] over C. In particular, I HI 2 1 GI and hence by symmetry I H I = I G I and H is a basis for C[G]. (3.16) DEFINITION Let H c C[G] be such that H is a multiplicative group which is a basis for C[G]. Suppose that every element of H is a h-linear combination of elements of G. Then H is an integral group basis in C[G]. If H is an integral group basis in CCG], we identify C[G] with C [ H ] in the natural manner. An important result used in studying the isomorphism problem is due to G. Glauberman. Most of the rest of this chapter is devoted to its proof and some consequences. We use the notation C X to denote the sum of the elements of the conjugacy class X , computed in the group algebra. (3.17) THEOREM (Glaubemuzn) Let H be an integral group basis in @[GI. Then there exists a one-to-one correspondence between the sets of conjugacy classes of H and G such that if Y corresponds to X , then I Y I = [ X l a n d C Y= + E X .

Chapter 3

42

We need a lemma. (3.18) LEMMA Let a, fi E Z(@[G]) and view the characters of G as being defined on all of @[GI. Define

1

(a, B> =

x(a)x(P).

I: E Irr(C)

Then (a) (clal + c2a29 8) = C I < % (b) ( A a> = (c) (Ki, K j ) = 0 if i # j;

(a,);

8) + c 2 < a 2 , 8 > ;

(d) (Ki,Ki) = I G I I X i l ;

where the X i are the classes of G and K i = C X i . Proof Statements (a) and (b) are immediate since characters are linear functions. To prove (c) and (d), let x E T i and y E X j .Then x(Ki) = I X i1 ~ ( x ) and x(Kj) = I X j I x ( y ) . The result is now immediate from the second orthogonality relation. I

Proof of Theorem 3.17 Identify C[H] with @[GI so that k = dim(Z(C[G])) is the common number of classes of H and G. Let gl,Y 2 , . . . , pk be the classes of H and XI, X 2 , .. .,.?fk the classes of G. Write K i = EXi and Li = Z Y i . Since the K i are a basis for Z(@[G]), we may write Li =

1c i j K j . i

Since Li is a Z-linear combination of elements of G, we conclude that all cij E Z. The form ( , ) defined in Lemma 3.18 depends only on the algebra @[GI = @[HI and not on the particular groups G and H. This is because the irreducible characters, defined on @[GI, are simply the traces of the irreducible representations of the algebra. We therefore have (1)

IHIIYiI = ( L i , L i ) = C(Cij)’IGIIXjI* j

Since the Li span Z(@[GJ)but no proper subset of { K j } spans this space, it follows that for each j, there exists i with cil # 0. Thus Ci (cij)22 1. Since IH I = I GI, Equation (2) yields

Characters and integrality

43

xi

and thus we have equality and (cij)* = 1 for all j. Therefore, for each j there is a unique i with cij # 0. This defines a map{Xj} + {Yi}.Since for each i there exists j with cij # 0, this map is onto and hence is one-to-one. Because all nonzero cij = k 1, we have Li = SK, when Y iand X j correspond.Finally,Equation(l)yieldsIHllLYil = IGI(XjJandthusIYil= lXjl and the proof is complete. I (3.19) COROLLARY Let H be an integral group basis in C[G].Then there exists an integral group basis H* in C[G] such that H z H* and the class sums of H* equal the class sums of G. Proof

Let 6 be the linear extension of the principal character of G to

@[GI. (This is usually called the augmentation map.) Note that S(C a , g )

1a, and 6 is an algebra homomorphism @[GI

=

C. If h E H, then 6(h)E Z and since G(h)S(h- ') = 6(1) = 1,we have 6(h) = L- 1. Let h* = d(h)h~C[G]for ~ E and H note that 6(h*) = 6(h)2 = 1. Put H* = {h* Ih E H} and note that H* is a group and the map h Hh* is an isomorphism. Clearly, H* is an integral group basis in C[G]. Let 9 be a class of H* so that Z Y = L-ZX for some class X of G by 9 1 . Since Theorem 3.17. Since 6(h*) = 1 for h* E H*, we have 6(X9)= 1 6 ( Z X ) = I X 1, the ambiguous sign above must be positive and the proof is complete. --*

(3.20) COROLLARY Let H be an integral group basis in @[GI. Then G and H have identical character tables. Proof By Corollary 3.19, we may assume that there exists a one-to-one correspondence between the classes of H and the classes of G such that if Y and X correspond, then X 9 = Z X and IY I = 1 X 1, Let x be the trace of an irreducible representation of @[HI = @[GI. Let Y and X correspond and let h E 9 and g E X . It suffices to show that ~ ( h=) x(g). However x(h) I 9 I =

x m w = x ( Z X ) = x@)I

and since 1 9I = I X I , the result follows.

I

I

We laave seen that if Z[G] E Z[H], then C[G] has an integral group basis isomorphic to H.The converse is true but is less obvious. (3.21) THEOREM Let H be an integral group basis in C[G]. Then Z[G] = Z [ H ] in C[G]. Proof Here, Z[Kj denotes the ring of Z-linear combinations of elements of H.(It is isomorphic to the abstract integral group ring.) Since H E Z[G], we have Z[H]c Z[G]. We show that G E Z[XJ. WriteG = { g i l l 5 i In}andH = { h i l l Ii In},wheren = IGI = IHI.

Chapter 3

44

xi

We have hi = aijgj for aijE Z. We shall show that the matrix A = (aij) has an inverse with entries in Zand this will complete the proof. Write h j - = bijgi-l and B = (bij), an integer matrix. Let p be the character of the regular representation of C[G] = C [ H ] so that p(hihj- ') = n6i,i.Now express h i h j - ' as a linear combination ofelements of G and observe that the coefficient of 1 is a i , b V j It . follows that

xi

cy

nsij

= p(hihj-') = n

c V

and thus A B

=

I, the identity matrix. The result now follows. I

Problems

(3.1) Let f(x)

E O[x]

be an algebraic integer and suppose that f ( c t ) = 0, where is irreducible and monic. Show that f (x) E Z[x].

ct

(3.2) Let G be a group, g E G, and let x be a character of G. Suppose Ix(g)I = 1. Show that x(g) is a root of unity.

Hint Let E s C be the splitting field for xn - 1 over Q. For integral with lctl = 1, letfa E Z[x] be the polynomial of Problem 3.1. Show that only finitely many polynomials can arise this way. Do this by bounding the degree and the coefficients of f a .

ct E E,

(3.3) Show that no simple group can have an irreducible character of degree 2. Hint

Problem 2.3 is relevant.

(3.4) Let G be a simple group and suppose x E Irr(G) with ~ ( 1 = ) p , a prime. Show that a Sylow p-subgroup of G has order p .

Hint

If the Sylow p-subgroup P is nonabelian, then Z ( P ) E Z(x).

(3.5) Suppose A G' < G.

G

G is abelian and I G : A 1 is a prime power. Show that

(3.6) Let G be a p-group and suppose x E Irr(G). Show that x( 1)'1 I G : Z(x)l. (3.7) Let x ~ I r r ( G be ) faithful and suppose x(1) = pa for some prime p . Let P E Syl,(G) and suppose that CG(P)$ P. Show that G < G . Hint Let Q _C CG(P)be a p'-subgroup, Q # 1. Show that Q n Z(x) # 1 and consider det x. (3.8) Let x be a (possibly reducible) character of G which is constant on G - { l}. Show that x = a l G bp,, where a, b E Z and p is the regular character of G . Show that if G # ker x, then x( 1) 2 I G I - 1.

+

Problems

45

Hint First show that x

=

al,

+ bp, for some a, b E C.

(3.9) Let XI, . .., x k be the conjugacy classes of a group G and let K . . . , Kk be the corresponding class sums. Choose representatives g iE X i and let aijvbe the integers defined by

,,

K i K j = C aijvK,. V

Show that

Hint

Use the oxand the second orthogonality relation.

Notes This formula shows that the aijvcan be calculated from the character table. Therefore, the character table of G can be used to answer questions such as: Is an element g E X , the product of an element of .Xi with one of X j ? The fact that the aijvare nonnegative rational integers imposes another necessary condition on an array of complex numbers that the array be a character table for some group. Since the K i are a basis for Z(@[G]), it follows that the character table of G determines Z(C[G]) up to algebra isomorphism. (3.10) We write [x, y] for the commutator x- ' y - 'xy of x and y in a group G.

(a) Let g E G and fix x E G. Show that g is conjugate to [x, y] for some yEGiff 2Ix(x)l x(g) o*

c

x E Irr(G)

x(

)

(b) Show that g = [x, y] for some x, y E G iff x(g) 1 -x(') #O.

XElrr(G)

N o t e We already knew that the character table of a group determines the commutator subgroup. Problem 3.10 says more than this, since the commutator subgroup usually does not consist entirely of commutators. (3.11) Let g E G be a commutator. Suppose m E Z, (m,o(g)) = 1. Show that g" is a commutator.

Hint

See the hint to Problem 2.12.

(3.12) Let x E Irr(G) and g. h E G. Show

Chapter 3

46

Hint Let X afford x and let Ki and K j be the class sums in C[G] which contain g and h. Use the fact that X(Ki)X(Kj) = X ( K i K j ) .

(3.13) (Brauer) Let K,,. . . , K kbe the class sums in C[G]. Suppose there exists c E C such that Ki = c Ki. Show G = G .

1

Hint

Let lG #

n

x E Irr(G). Show that w,(Ki) = 0 for some i.

(3.14) (Brauer) In the notation of the previous problem, show that if G = G , then there exists c E Q such that 1K i = c n K i . Hint To show that two elements a, b E ZjC[G]) are equal, it suffices to prove that o,(a) = o,(b) for all x E Irr(G).

(3.15) (Thompson) Let E be a Galois extension of Q with Galois group 9. Let a E E be an algebraic integer with the property that amis real and positive for all c E 9.A theorem of Siege1 (Ann. OfMath.46 (1945)p. 303,Theorem 111) asserts that if a # 1, then

Use this to show that if x E Irr(G) then ~ ( xis) either zero or a root of unity for more than a third of the elements x E G. Hints Mimic the proof of Theorem 3.15.Use Problem 3.2.

(3.16) (Burnside) Let I GI be odd and suppose x E lrr(G) is not principal. Show that x # j . Hint Using orthogonality, show that if x # 1, and x = 2, then x( 1) = 2a for some algebraic integer a.

(3.17) (Burnside) Let I GI be odd and suppose that G has exactly k conjugacy classes. Show that I G J= k mod 16.

Hint

If n is an odd integer, then n2 = 1 mod 8.

4

Products of characters

Let x and $ be characters of G. The fact that x + $ is a character is a triviality. We may define a new class function x$ on G by setting (x$)@) = x(g)$(g). It is true but somewhat less trivial that x$ is a character. [If either x or is linear, this is Problem 2.6(a).] Let V and W be @[GI-modules.We shall construct a new C[G]-module V 0 W called the tensor product of V and W. Choose bases {ulr . . . , u,} for V and {wl,. . .,w,} for W . Let V 0 W be the @-spacespanned by the mn symbols ui @I w j . [More precisely, V 8 W is the set of formal sums of the form ai.(ui0 wj), with aijE @.I If u E V and w E W, suppose u = aiui and w = b j w j .Wedefine u 0 w = aib,{ui 0 wj)E V 0 W.

1

1

c

Note that not every element of V @I W has the form u 0 w for u E I/ and w E W (except in the special case that n or rn = 1). We define an action of G on V 0 W by setting (ui 0 W j ) g = V i g 0 W j g and extending this by linearity to all of V 0 W. The reader should check that if u E V ,w E W , and g E G, then ( u 0 w)g = ug @I wg. It follows that (xgl)gz = x(g,g,) for x E V 0 W and g i E G. Next we give V 0 W the structure of a C[G]-module by extending the action of G by linearity in @[GI. In other words, for x E V 0 W we define

c

a&) = a , ( w ) . It is routine to check that this really makes V 0 W into a C[G]-module. 41

Chapter 4

48

A few words of caution are in order here. If a E C[G], it is not necessarily true that (ui 0 wj)a = via 0 wja. It is for this reason that we first defined the action of G on V O W and then extended it to CCG]. If A.is an arbitrary algebra with modules V and W , it is not generally possible to define the structure of an A-module on V 0 W . We have not yet shown that the C[G]-module V 0 W is determined (up to isomorphism) by V and W , independently of the choice of bases. This is, in fact, not hard to prove for group algebras over any field. We shall leave the general situation to the problems. (4.1) THEOREM Let C[G]-modules V and W afford characters x and $, respectively. Choose bases in V and W and construct V 0 W . Then V 0 W affords the character p j and is independent of the choice of bases.

Proof Let {ui I 1 Ii In} and { w, I 1 Ir Irn} be bases for V and W , respectively, and let g E G . Write m ...

n

uig = x a i j u j

w,g

and

=

j= 1

with a i j ,brsE C. Then x(g) = by V 0 W . Since

aiiand +(g) (ui

1 brsws, d=

=

1

xF= b,, . Let 9 be the character afforded

6 wr)g = 1aijbrs(vj O ws), j ,s

we have

9(g)=

1aiibrr = 1aii C brr = x(g)lc/(g). i, r

i

r

We now know that the character afforded by V 0 W is independent of the choice of bases and the result follows by Corollary 2.9. I (4.2)

COROLLARY

Products of characters are characters.

Corollary 4.2 provides another necessary condition for an array of complex numbers to be a character table. It asserts that the inner product of any row with the product of any two rows is a nonnegative integer. In this connection, the following formula is relevant :

In general, products of irreducible characters are not irreducible. For instance, if x E Irr(G), then f E Irr(G); nevertheless, lGis a constituent of xf since

[xf, l G 1

=

[X? X 1 G 1

= 1.

Products of characters

49

(4.3) THEOREM (Burnside-Brauer) Let x be a faithful character of G and suppose x(g)takes on exactly m different values for g E G.Then every E Irr(G) is a constituent of one of the characters ( x ) j for 0 Ij < m.

+

Note In the special case x = p, the regular character of G, we have m = 2. The theorem is already known to be true in this case.

Proof Let al, ctz,.

. . , a,

be the distinct values taken on by = (9 E GI x(g) = ail.

Gi

x = { l}. Fix t,h E Irr(G) and

Assume al = x(1) so that Gl = ker Gi Il/o.Now for j 2 0, we have

1,

i

If

+ is not a constituent of

x. Let

xj

let

pi =

m

for any j , 0 Ij < m, we have

1i (ai)jpi = 0,

O Ij < m.

The determinant of this system of m equations in the m "unknowns" piis the so-called "Vandermonde determinant" and is equal to kHi<,(ai - ctj) # 0. It follows that all pi = 0. On the other hand, p1 = +(1) # 0 and this contradiction proves the theorem. I Let g E G and let n > 0 be an integer. We ask how many nth roots g has in G. Let 9,(g)

=

I { h ~ G l h "= g31.

Observe that if (I G 1, n) = 1, we may choose an integer m such that nm = 1 mod I G I. Thus if h" = k", then h = h"" = k"" = k and we have $,(g) I1 for all g E G . Since the map h t--t h" is one-to-one on G, it must be onto and it follows that 9,(g) = 1 for all g E G. The situation becomes considerably more interesting if we drop the assumption that n is prime to 1 G 1. Since 9, is clearly a class function on G, we may write 9,

1

=

vn(x)x,

.y~Irr(G)

where v,(x) is a uniquely determined complex number. (4.4)

LEMMA

vn(x)

= (I/ I G I)

1,

EG

x(g").

Chapter 4

50

Proof By the orthogonality relations we have

Finally, replace h by h- to obtain the desired result.

1

Let cp be any class function of G and let n be a positive integer. We define a new function cpcn) by cp(")@) = cp(g"). Note that cp(") is a class function. Lemma 4.4 asserts that v,(x) s [x("), l,] for x E Irr(G). It is not, in general, true that x(") is a character for x E Irr(G), however, it is always a difference of two characters (see problems) and thus v,(x) E Z. (4.5) THEOREM

(Frobenius-Schur) Let x E Irr(G). Then

(a) x") is a difference of characters; (b) vz(x) = 1, - 1, or 0; (c) vZ(x) # 0 iff x is real valued.

Proof Let V be a C[G]-module which affords x and let u l , u z , . . .,u, be a basis for V. Let W = V @ V and define the linear map *: W -+ W by (ui @I uj)* = uj @I ui. Let and

W, = {WEWlw* = w}

WA= {WEWlw* = -w}.

These subspaces are called the symmetric and antisymmetric parts of W , respectively. If w E W , we have w + w* E Ws and:w - w* E WA.Since w=-

w+w* 2

+-w-w* 2

'

it follows that W = Ws + WA. We claim that if w E W and g E G, then (we)* = w*g. It suffices to check this as w runs over the basis ui @I u j , and thus we need to show that

.

(Vig @I ujg)* = 01s 8 DigIn fact, it is true that (x

@ y)*

= y @I x

for all x, y E V. This is seen by expanding x and y in terms of the ui.

Products of characters

51

It now follows that W, and WAare @[GI-submodulesof W and we have

x2 = x S + X A I where xs and xA are the characters afforded by Wsand WA.We compute X A using the basis w i j = vi @ v j - v j Q v i r

i <j ,

for WA. Suppose v i g = 1 airor.We have w EJ. . g =

1

(airajs

r,s

C (airajs - ajrais)wrs*

- ajrais)vr Q 0s =

r<s

Therefore xA@)

=

xaiiajj

-

ajiaij.

icj

This yields

Let X be the representation corresponding to the basis { u i } of I/, so that X(g) = (aij)= M . We have 2xA(g) = (tr W 2- tr(M2) = x ( d 2 - x(g2) since M 2 = X(g2). This yields x”)(g) = x(g)’ - 2xA(g) and thus = - 2xA, a difference of characters, as claimed. Now v2(x) = [x”), 1G] = [x’, 1G] - 2 [ x A , lG]. If is not real valued, then 0 = [x,I] = [x’, 1G] and thus [ x A , lG] = O since xA is a constituent of x’. It follows that v2(x) = 0 in this case. If x is real valued, then 1 = [x2, 1G] thus ha,1G] = 0 or 1 and v2(x) = 1 or - 1. The proof is now complete. I

x‘’)

(4.6)

COROLLARY

x’ x

Let G have exactly t involutions. Then 1+ t =

where v2(x) = 0 if x #

c

v,(x)x(1)9

z~Irr(G)

and vz(x) =

1 if x = z.

Proof This is immediate since 9,(1) = 1

+ t.

I

If N 4 G , we have identified Irr(G/N) with a subset of Irr(G). The following lemma shows that v,,(x) is well defined under this identification. (4.7) LEMMA Let N 4 G and let x E Irr(G) with N E ker 2. Let vnk)be as above and let P,,(x) be the corresponding number computed in GIN. Then vn(X) = PAX)*

Chapter 4

52

Proof We have

(4.8) LEMMA Let x E Irr(G)and let I be a linear character of G with I" = 1,. Then v,,(x) = v,(Ix). Proof Recall that Ax E Irr(G) by Problem 2.6(b). Now

since A(@) = I(g)" = I"(g) = 1.

I

As an application of the Frobenius-Schur theorem, we prove the following. (4.9) THEOREM (Alperin-Feit- Thompson) Let G be a 2-group containing exactly t involutions. If t _= 1 mod 4, then either G is cyclic or 1 G :G'J = 4. Proof If G is abelian, then it clearly must be cyclic. We suppose G is not abelian and show that I G : G I = 4. If Z(G) is not cyclic, choose K E Z(G)elementary abelian of order 4. The set {x E G Ixz = 1) is a union of cosets of K and hence 4 I ( t + l), a contradiction. Therefore, Z(G) is cyclic and G contains the unique minimal subgroup Z of order 2. Since G' > 1, we have Z E G . Also G/Z is not cyclic since G is not abelian. If G/Z satisfies the hypotheses of the theorem, then 1G:GI = I(G/Z):(G/Z)'I = 4 by induction and we are done. We may therefore assume that the number of involutions in G/Z is not = 1 mod 4. We have 1 vz(x)x(l) = t 1 5 2 mod 4

+

x E Irr(G)

and

c

x ~ l r r ( G ) ; Z Ek e r x

u2(x)x(l)

+ 2 mod

4 7

where the second statement follows via Lemma 4.7. We conclude that (*)

c

v2(x)x(1) f 0 mod 4.

xElrr(G);ZSt k e r z

Now let C be the group of linear characters I of G which satisfy 'A = 1,. For x~Irr(G),we have Ax€Irr(G) and since Z c ker A, we conclude that

Products of characters

53

Z $ ker x iff Z $ ker(Ix). Therefore C permutes {x E Irr(G)1 Z $ ker x} and partitions this set into orbits Oi. By Lemma 4.8, v2(x) is constant on each orbit as is x( 1). Since I Oi I is a power of 2, we conclude from Equation (*) that there exists x E Irr(G)suchthat

(i) z $ ker x, (ii) v2(x) # 0, (iii) x(l)lOl I 2, where 0 is the orbit containing 2. Since Z $ ker x, we have ker x = 1. Since G is not abelian, ~ ( 1>) 1 and hence x( 1) = 2 and 10I = 1 . Therefore, Ax = x for all I. E C. Now let F = WG), the Frattini subgroup. If g E G - F, then there exists I E C for which I ( g ) # 1 and it follows that x(g) = 0. By Lemma 2.29, we now have 4 = x(1)’ 2 [XF, XF] = IG: F I .

Since G is not cyclic, I G :F I 2 4 and we have equality above. This forces

xF = 2p, where p is a faithful linear character of F.

Since v2(x) # 0, we know that x is real valued. Therefore, p is also, and so IF1 1 2 . Thus IGI I 8 and hence G E & or Q 8 . In either case, IG: G‘I = 4 as desired. I A theorem of 0. Taussky [Theorem 111 11.9(a)in Huppert] asserts that the only nonabelian 2-groups G for which IG : G’I = 4 are the dihedral, semidihedral, and generalized quaternion groups. In each of these groups, the number of involutions is in fact = 1 mod 4.

As another application of Corollary 4.6, we prove some results of Brauer and Fowler that were originally obtained in a different way.

(4.10) LEMMA Let a,, a 2 ,. . . ,a, be arbitrary real numbers. Then

Proof

The well-known Schwarz inequality asserts that

Cc

aibi)’ I

CC ai2)(Cbi2)

for real ai and bi . The lemma follows by setting all bi = 1.

I

Chapter 4

54

(4.11) THEOREM Assume [GI = g is even and that G contains exactly t involutions. Let a = (g - l)/t. Then (a) There exists x E G, x # 1, with I G : C(x)I < a2. (b) There exists real X E Irr(G), x # l G ,with x(1) Ia.

Proof

By Corollary 4.6, 0 < (g - l)/a = t I

1 x(1), X E Y

where Y = {xEIrr(G)Ix = j

and

x#

lG}.

In particular, s = I Y I # 0. Now t2 I

>'

1x(1) (9

by Lemma 4.10. It follows that

Is

c x(1)2I s(g - 1)

g - 1 I sa2

and thus sa2 2 x ~ ( 1 ) ~ . Y

Therefore, x( 1) I a for some x E 9, proving (b). Since s < k - 1, where k = IIrr(G)I is the total number of conjugacy classes of G , we have

(k - 1)a2 2 g

- 1

and hence some nonidentity class of G has size I a2. This proves (a).

I

Recall that an element x E G is said to be real if x is conjugate in G to x-'. It is a fact (which will be proved later) that the number of classes of real elements of G is equal to the number of real irreducible characters. Assuming this, it is immediate from the inequality saz 2 g - 1 in the above proof that statement (a)can be strengthened to guarantee that a real xEGexistswithx# landIG:C(x)I < a 2 . (4.12) COROLLARY Let n be a positive integer. There exist at most finitely many simple groups containing an involution with centralizer of order n. Proof Let G be such a group with 1 GI = g. Then G contains at least g/n involutions and hence a < n in the notation of Theorem 4.1 1. By (a) of that theorem, there exists x E G with 1 < 1 G : C(x)I < n2. Therefore, G is isomorphic to a subgroup of the alternating group An2- The result follows. I

Products of characters

55

(4.13) COROLLARY Let G have even order, g > 2. Then G contains a proper subgroup of order > ( g ) ' / 3 . Proof We use induction on 1 GI.First assume Z(G) = Z > 1. If I G : Z I is even, then there exists H / Z < G/Zwith IG:HI

=

I(G/Z):(H/Z)I < IG/Z12/3< g2/3

and the result follows. If I G:ZI is odd, then G has a central Sylow 2-subgroup and thus has a normal 2-complement by a transfer theorem of Burnside. (Or see Theorem 5.6.) It follows that G has a subgroup of index 2 and the result follows. Now assume Z(G) = 1 and let a be as in Theorem 4.11. Then 1 < IG:C(x)I I u2

for some x E G and we are done if a' < g 2 / 3 .Assume, then, that a 2 g1l3and let T be an involution in G . Then 1 < I G : C(t)lI (g - l ) / u < g/a I g2l3 and the result follows.

I

We now wish to discuss the question of which real x E Irr(G) satisfy vz(x) = 1. The answer is interesting but does not seem to be very important in the applications of the Frobenius-Schur theorem. The solution to this problem involves some matrix theory. Let U and V be C[G]-modules with bases { u l , . . . ,u,} and { u l , . . . , v,,,}, respectively. An element w E U 0 V is uniquely of the form

+

w

=

C aij(ui O u j ) Lj

and this defines'the n x m matrix (aij).We write M(w) = (aij).For g E G, we compute M(wg). Let X and 9be representations of G corresponding to U and V respectively, with respect to the given bases. Write

X@) = (brs13

9(s)= ( c p q ) .

Then (ui

O uj)g

=

uig O v j g

=

C s. 4

Thus

o vq).

bi,~jq(~,

Chapter 4

58

and, therefore, W w g ) = WTM(w)%J(g). (4.14) THEOREM Let x E Irr(G) be real valued and let 3 be a representation which affords 2. Then there exists a nonzero matrix M such that X S ( g ) T M w= M

for all g E G. Furthermore, for any such matrix, MT= v,(x)M.

Proof Let V be a @[GI-module corresponding to X and write W = V @I V. Let M : W -+ M,(@) be as previously stated so that W w g ) = X@)Tww)X@)

for W E W andgEG. Now W affords the character x2 and b2,1G] = [x, x] = 1 and hence there is a one-dimensional space of G-fixed points, W, c W. The matrices M satisfying X@)TMX(g)= M are precisely the M ( x ) for X E W,. In the proof of Theorem 4.5, we had a map *: W --* W and we decomposed W = ws w,, where ws= {w E W I w* = w } and w, = {w E Wlw* = -w}. Also, the proof of Theorem 4.5 showed that vz(x) = 1 iff bA, 1G-j = 0, where xA is the character afforded by W,. We must have one of the following two situations:

+

(i) WO (ii) WO

= K,

[XA,

wS, [ X A ,

1G1 = 1, vZ(x) = -1; l G 1 = 0, V Z ( d = 1*

Thus if x E W,, we have x* = v,(x)x. Now, for W E W we clearly have M(w*) = M ( w ) ~Thus . for have

X E

W, we

M W T = V,(X)M(X) and the result follows. (4.15)

I

COROLLARY Suppose x E Irr(G) is afforded by a real representation.

Then v,(x) =

+ 1.

Proof Let X be an R-representation which affords x and let

M

=

c X(h)TX(h).

heG

It is clear that MT = M and X@)TMX(g) = M for all g E G. All that remains is to show that M # 0. For any n x n real matrix X,is it clear that the diagonal entries of XTXare 2 0, and at least one of them is > O unless X = 0. The result now follows. I

Products of characters

57

By Problem 2.5(b), we know that it is possible that a real-valued irreducible character x is not afforded by any real representation. Such characters are exactly those for which v 2 ( x ) = - 1. The proof of this seems to require some nontrivial matrix theory. If M is a square, complex matrix, we write M * = RT.and say M is unitary if M M * = I and M is normal if M M * = M * M . Astandard theorem of linear algebra asserts that if M is normal, then there exists a unitary matrix, U , such that U - ' M U is diagonal. (4.16) LEMMA Let D be a diagonal matrix. Then D = E 2 for some diagonal matrix E such that every matrix which commutes with D also commutes with E .

Proof Suppose a',. . . ,a, are distinct complex numbers and pl,. . . , p, E @ are arbitrary. Then there exists a polynomial f such that f ( a i ) = p i . We may therefore choose f such that f(a)2 = a for every diagonal entry a of D.Then E = f ( D ) has the desired properties. 1 (4.17) THEOREM Let X be a @-representation of a group, G. Then X is similar to a representation 'I) such that 'I)(g)is unitary for all g E G.

Proof Let M = LEG X(g)*X(g).Then M* = M and M is normal. We may choose a unitary matrix U such that U - M U is diagonal. Since U - = U *, we have

'

'

(U-'X(g)U)*

=

U-'X(g)*U

and

U-'MU =

3El(g)*Xl(g), #eG

where 3E1 = U-lXU. It is therefore no loss to assume that M is diagonal. The diagonal entries of X(g)*X(g)are all real and nonnegative and since X(1) = I , it follows that the diagonal entries of M are positive. We may therefore write M = P2where P is a nonsingular real diagonal matrix. Since X(g)*MX(g)= M for all g E G, we have ( P - 'X(g)*P)(PX(g)P-1) = I . Since (PX(g)P-')* = P-'X(g)*P, it follows that 'I) = PXP-' is the desired representat ion. 1 LEMMA Let P*P = I and P T = P for a square matrix P . Then P = QQ- for some matrix Q.

(4.18)

Proof Since P is unitary, it is normal and we may write U - ' P U = is unitary, D and E are diagonal, and every matrix which commutes with D also commutes with E . Now DT= D, P T = P , and UT = U-'.Thisyields D = DT = ( U - ' P I ! J ) ~= u ' - ' p B = U - ' U D u - ' u .

D

= E2, where U

Chapter 4

58

Therefore, U - ' 8 commutes with D and hence also with E . It follows that E = 8 - ' U E U - ' B and U E U - ' = BEB-'. From P*P = I, we obtain I = D*D = DD and it follows that the diagonal entries of D and hence of E have absolute value 1 and E = E - I . Now let Q = U E U - ' . Then

0Q-l = ( V E U - ' ) ( U E U - ' ) = (UEU-I)' and the proof is complete.

=

UDU-' = P

I

(4.19) THEOREM Let x E Irr(G)and suppose vz(x) = 1.Then x is afforded by some real representation of G. Proqf By Theorem 4.17,

x is afforded by a representation X, such that x = 1,X and f are similar and we may

X(g) is unitary for all g E G. Since

choose P such that P - ' K P = X. Now - X(g)TPX(g)= X(g)-'X(g)P = P. Since v2(x) = 1, it follows that P T= P by Theorem 4.14. From the equation 3E(g)P = PX(g), we obtain P*5(9)* = X(g)*P* and since X(g)* = X(g)-', this yields P*X(s)-' = X(g)-'P*. Thus XP* = P*x and XP*P = P*XP = P*PX.

By Schur's lemma, it follows that P*P = aI for some a. Clearly, a is real and positive and we may therefore replace P by P / d i 2 and assume P*P = 1. Now Lemma 4.18 applies to yield P = QQ-' for some Q and we have QQ-'X

=

xQQ-'

or Q-'XQ = Q-'XQ. In other words, = Q-'XQ is a real representation which affords x and the proof is complete. I We may summarize the situation as follows for x E Irr(G):

x x

(a) vz(x) = 0 iff is not real; (b) v2(x) = 1 iff is afforded by a real representation; (c) vz(x) = - 1 iff is real but is not afforded by any real representation.

x

One further remark should be made. If x E Irr(G) and vz(x) = - 1, then ~ ( 1 must ) be even. This may be seen as follows. The nonzero matrix M in Theorem 4.14 is nonsingular by Schur's lemma. If v2(x) = - 1 then MT =

- M and det(M) = det(MT)= det( - M ) = ( - l)x(l) det M . It follows that (- l)x('' = 1 and x( 1) is even.

Problems

59

We close this chapter with a discussion of the characters of direct products. (4.20) DEFINITION Let G = H x K and let cp and 9 be class functions on H and K , respectively. Define x = cp x 9 by ~ ( h k=) cp(h)9(k)for h E H and kEK. Observe that cp x 9 is a class function of G. If cp is a character of H , then under the isomorphism H G/K,there is a corresponding character i$ of G with K E ker 8 and @(hk) = cp(h).Similarly, if 9 is a character of K , there is a corresponding character 8 of G, with 8(hk) = 9(k). It follows that cp x 9 = $8 is a character of G. (4.21) THEOREM Let G = H x K . Then the characters cp x 9 for cp E Irr(H) and 9 E Irr(K) are exactly the irreducible characters of G. Proof

Let cp,

'pi E Irr(H)

and 9,

s1E Irr(K). Let x = cp x

9 and

xi

=

cpl x 9,.Then

It follows that the cp x 9 are all distinct and irreducible. Now

c

fpE Irr(H); 9E Irr(K)

(cp x 9)(1)2= ccp(1)29(1)2= 'p.

9

=

and thus the cp x 9 are all of Irr(G).

IHllKl= IGI,

I

Problems

(4.1) Let x and $ be characters of G. Express det(x$) in terms of det det $. Hint

x and

It suffices to assume that G is cyclic.

(4.2) (Garrison) Let 2- be a class of G which is not contained in any proper normal subgroup. Let K be the corresponding class sum in C [ G ] and let m be the number of distinct values of o,(K) for x E Irr(G). Show that every element of G can be written as a product of fewer than m elements of X .

Chapter 4

60

Hint Mimic the proof of Theorem 4.3 using the second orthogonality relation.

(4.3) Let G = H x K. Let rp E Irr(H) and 9 E Irr(K) be faithful. Show that rp x 9 is faithful iff (IZ(H)I, IZ(K)I) = 1. (4.4) Suppose G

=

HK with H

c C(K).

(a) Let x ~ I r r ( G ) Show . that

xH

9(l)rp and

=

rp E Irr(H) and 9 E Irr(K). (b) Let cp E Irr(H) and 9 E Irr(K) and suppose that

xK = rp(1)9 for

some

q H n K and Q H n K have a common constituent. Show that there exists a unique x ~ I r r ( G )with x H = W b a n d x , = (P(1)9. Hint G is a homomorphic image of H x K .

(4.5) Let G = H x A where A is abelian and let n > 0 with ( IHI,n ) = 1.

Show that f " ) E Irr(G) for every x E Irr(G).

(4.6) Let n > 0 and assume that x(") E Irr(G) for every x E Irr(G). Show that G = H x A , where A is abelian and (JHJ, n ) = 1.

Hints (a) Let d = (IGl, n). Show that it is no loss to assume that (IGIld, 4 = 1. (b) Let A = n X E I r r ( G )ker x("). Show that A = {g E Gig" = 1) and IAl = d. (c) L e t H = r){kerxIxEIrr(G),x(")= l,}.ShowIG:HI=d.

(4.7) Let V be a @[GI-modulewith basis {u,, , . . , u,} and let p be a prime. Let W = V @ @ V(ptimes)anddefineaon W so that (x, @ @ xp)ol = x2 @ x3 @ . . - @ x p @ x1 for all x i E I/. Let E be a primitive pth root of 1 in C. Let W,= { w E WI W' = E W } . Show that W, is a @[GI-submodule of W which affords the character x1 = (xp - ~ ( ~ ) ) /where p, x is the character afforded by V . Conclude that x("') is a difference of characters for all rn 2 1. Hints Let W = {uil 6 . . . 0 ui, Jnotall i j are equal}. For w E W , let

w=

+ & - I w " + E-2Wa2 +

. .. + & - ( P -

1)wap-l

.

Now (a) permutes W into orbits of size p. Let W , be a set of representatives of these orbits. Then {Wlw E W , } is a basis for W,. (4.8) Let G be a p-group with Frattini factor group of order 2 p 2 " - ' . Show that the number of elements of order p in G is 3 - 1 mod p".

Hint Mimic the proof of Theorem 4.9. Use the fact that integer (Problem 4.7).

v,k) is an

Problems

(4.9) If x E Irr(G), then I v&) exists for any n > 2.

61

I I 1. Show that no absolute bound on v,(x)

Hint There exist groups of prime exponent p # 2 and of exponent 4 with irreducible characters of arbitrarily large degree.

(4.10) Let r! W, V,, and W, be F[G]-modules for an arbitrary field F . Fix a particular F-basis in each of these modules and construct V 0 W and V, 0 W, as was done for @[GI. Now suppose a: V + V, and fi: W --* W, are module homomorphisms. Show that there exists a module homomorphism y: V 0 W + V, 0 W,such that y(u Q w) = a(u) 0 fi(w) for u E V and w E W . Conclude that V 0 W is determined up to module isomorphism independently of the choice of bases. (4.11) Let G be simple and let S E Syl,(G) be elementary abelian of order q. Suppose S = C,(x) for all x, 1 # x E S . Show that v&) = 1 for all x E Irr(G) and that ~ ( x=) ~ ( 1 )- q for x # lGand x E S. Hint First show that G contains exactly IGl/q involutions and that the remaining elements of G have odd order. This is done without characters using the fact that if x, y are nonconjugate involutions, then JZ((x,y))l is even.

Note This problem is continued as Problem 5.20 where it is proved that - l)(q + 1).

IGI =

(4.12) Let x, $, 9 E Irr(G). Show that [&, 93 I9(1).

5

Induced characters

Let H c G be a subgroup. Given a character x of G, we obtained the character xH of H by restriction. In this chapter we study an approximately dual process, where characters of G are induced from characters of H . (5.1) DEFINITION Let H E G and let cp be a class function of H . Then cpG, the induced class function on G , is given by

where cpo is defined by cp"(h) = q ( h ) if h E H and cp"(y) = 0 if y 4 H . Observe that cpG is really a class function on G and that cpG( 1) = I G : H I cp( 1). Another useful formula for cpG(g) is obtained by choosing a transversal T for the right cosets of H in G (that is, a set of representatives for these cosets). It is then easy to see that PG(d=

1cpo(tSt- ').

reT

(5.2) LEMMA (Frobenius Reciprocity) Let H c G and suppose that cp is a class function on H and that 9 is a class function on G . Then [P,

Proof We have

62

$HI =

[VG, 91*

Induced characters

Setting y

= xgx-'

63

and observing that 9(g) = $(y), we obtain

(5.3) COROLLARY Let H cpc is a character of G.

E

G and suppose cp is a character of H . Then

Proof Let z ~ I r r ( G )Then . zH is a character of H and thus [qc, [cp, XH] is a nonnegative integer. This proves the result since qG # 0.

x] = I

(5.4) COROLLARY Let H E G and suppose cp E Irr(H). Then there exists x E Irr(G) such that cp is a constituent of xH. Proof Take x to be an irreducible constituent of cpG. Then 0#

and the result follows.

ccpc9

XI = [I%XHI

I

(5.5) COROLLARY Let G be abelian and suppose H I E Irr(H) is the restriction of some p E Irr(G).

E

G. Then every

Induction of characters is closely related to the transfer map from a group into an abelian factor group of a subgroup. Many of the results provable by transfer can also be proved using induced characters. We give one example of this now.

(5.6) THEOREM Let G have an abelian Sylow p-subgroup. Then

I G' " Z(G)I is not divisible by p . Proof Assume the contrary and choose U E G n Z(G) of order p , with U c P E Syl,(G). Let I E Irr(U), A # l,, and choose p E Irr(P) with p, = 1.Now write

and observe that pG(1) = p( 1)I G : P I = I G : PI is prime to p . Therefore, there exists x E Irr(G) such that a, # 0 and pY~(1).We have 0

z

CPG, XI

=

b? XPI

and thus I is a constituent of xu. Since U _c Z(G), we have xu = x(l)A and = Ax('). (See Problem 2.3.)Since U c G', we have (det x)" = 1, and (det x), thus = 1,. Since p$x(l), I # l,, and I U I = p , this is a contradiction and proves the theorem. I

Chapter 5

64

Explicit computation of induced characters is extremely useful for the construction of character tables, despite the fact that cpG is usually reducible even if cp is irreducible. [Note that if cp is reducible, then cpG is necessarily reducible since (cpl + q2)' = cplG + cp2'.] Given H E G, cp a character of H and g E G, an efficient way to compute cpc(g) explicitly is to choose representatives, x l , . . . ,x , for the classes of H contained in Cl(g) in G and to use the formula

where it is understood that cpG(g) = 0 if H n Cl(g) = 0.This formula is immediate from the definition of cpG since as x runs over G, x g x - = x i for exactly Ic&)I values of x . We shall now construct the character table of G = A , . There are five conjugacy classes; one each of elements of order 1,2, and 3 and two of elements of order 5. It is convenient to label these classes 1,2, 3, 5 , , and 52 in the table. Let x1 = 1,. We have: ClassK

1

3

2

51

) C ( x ) l f o r x E Z 60 4 3 5 1x1: 1 15 20 12 x l : l l l l

52

5 12 1

Now let H = A4 E G and compute (lH)'.We have (1H)':

5

1 2 0 0

using the above formula. [For instance, if o@) = 3, there are two classes in H of elements conjugate in G to g. If x 1 and x 2 are representatives of these classes, we have l&i) = 1 and Ic&)I = 3 = IC H ( x iI)and thus (IH)'@) = 2 as listed.] Now [(IH)', 1G] = [IH, 1H] = 1 by Frobenius reciprocity and thus (lH)' - 1, is a character. We call this character x4. Thus 1,: 4 0 1 -1 -1. Direct computation now yields [x4, x4] = 1 and we have found an irreducible character. Next, let 1 be a nonprincipal linear character of H = A , so that 1(x) = 1 if o(x) = 1 or 2 and 1(x) = w = e2ni/3for one class of elements of order 3 in H and L(x) = w2 for the other class. We get xs=1G: 5 1 -1 0 0 x4 = ( 1 H ) ' -

since w

+ w2 = - 1. We compute [I,,x,]

=

1 and so x, E Irr(G).

Induced characters

65

Now let K be a cyclic subgroup of G of order 5 and let ,u E Irr(K), ,u # 1,. The class 5 , of G contains a pair of inverse elements of K , say x and x - ’ . Choose ,u so that ,u(x) = E = eZUi/’. Since x 2 and x - lie in the class 52 we have

’

p:

12 0 0

+ E4

&

&’

+ E3.

Computation yields LuG,x5] = 1 and thus ,uG-

xs:

7

1

-1

&+

E4

E2

+ E3

is a character. Further computation yields that LuG- x s , x4] = 1 and hence -1

3

X3=pG-Xs-X4:

0

& + E 4 + 1

&’+E3+1

is a character and [ x 3 , x3] = 1 so that x 3 is irreducible. Finally, replacing ,u above, by v E Irr(K) with v(x) = E’ yields $3:

3

-1

0

E2+E3+f

&+E4+1

is irreducible and we have found all five irreducible characters of G. We now consider the module theoretic interpretation of induced characters.

+

(5.7) DEFINITION Let I/ be a F[G]-module. Suppose V = * * wk where the N( are subspaces of V which are transitively permuted by the wk is an imprimitivity decomposition of V . action of G. Then V = W, - . If V is irreducible and has no proper imprimitivity decomposition (that is, with k > l), then V is a primitive F[G]-module.

+ +

+ +

W, is an imprimitivity decomWe emphasize that if V = W, position of V with k > 1, then the are definitely not F[G]-submodules of V . In this situation, let H = { g E GI W,h = W,},then W, is a F[H]-module.

+ +

(5.8) THEOREM Let V = Wl wk be an imprimitivity decomposition of the F[G]-module V . Assume F G @. Let H be the stabilizer of W,. Suppose V affords the character x of G and Wl affords the character 9 of H . Then x = SG.

Proof Let T be a right transversal for H in G and write W = W,. Then Wt runs over the N( as t runs over T and V=

1 . Wt. teT

Choose a basis, w,, . . .,w, for W so that {w,t I 1 Ii Im,t E T } is a basis for V . Let g E G. We compute x(g) using this basis.

Chapter 5

66

For t E T , the contribution of the basis vectors wit to x(g) will be zero unless Wtg = Wt, that is, tgt-' E H .Assuming tgt-' = h E H , write w i h = aiJwi so that aii = 9(tgt-'). Now, (wit)g = wiht = aijwjt and hence the contribution of wit to x(g) is aii.Therefore, the total contribution of the w i t , 1 Ii Im, is s ( t g t - ' ) . It follows that x ( g ) = Z t s T 8"(tgt-') = sG(g) as desired. 1

EY=,

1

1

(5.9) THEOREM Let H c G and let W be an F[H]-module. Then there exists an F[G]-module V having an imprimitivity decomposition V = . where H is the stabilizer of W, and W, W as F[H]-modules.

1 w,

Proof Let V be the external direct sum of I G : HI copies of W as an F-vectorspace. Let T be a right transversal for H in G and assume 1 E T . Let W Q t, for t E T , denote subspaces of V , each isomorphic to W , such that V= rcT(W0 t). Fix isomorphisms a,: W + W Q t and let w 0 t denote a,(w)so that W 0 t = { w @ t Iw E W } . Now for w E W and g E G , define w Q g E V as follows. Write g = ht, with h E H and t E T , and set w Q g = wh Q t. Note that for w E W , x E H , and g E G , we have wx Q g = w Q xg. (The usual notation for what we have constructed is V = W Q FrH,F[G].) We now let G act on V by defining (w Q t)g = w Q t g for g E G. Observe that (w Q g l ) g 2 = w Q g1g2 for all g , , g 2 E G. Extending this definition by linearity gives V the structure of an F[G]-module and

c.

V= C*WQt fsT

is an imprimitivity decomposition. Clearly the stabilizer of W Q 1 is H and the map a,: W + W Q I is an isomorphism of F[H]-modules. The proof is complete. 1 If s E H t , then W Q s = W 0 t in the above construction and it is reasonably clear that the module V is independent of the choice of coset representatives. We write V = WG.Note, however, that we have not yet proved that W Gis the unique (up to F[G]-isomorphism) F[G]-module which satisfies the conclusion of Theorem 5.9. In the special case that F = @, let W afford the character 9. Then by Theorem 5.8, it follows that any @[GI-modulewhich satisfies the conclusion of 5.9 affordsx = SG.This proves uniqueness in this case. The general case is left to the problems. Note that Theorems 5.8 and 5.9 provide an alternate proof of the fact that induced characters are characters. Another consequence of these results is that if x E Irr(G), then x is afforded by a primitive module iff x # SG for any character 9 of a proper subgroup of G. Such x are called primitive characters.

67

Induced characters

(5.10) DEFINITION Let x be a character of G . Then x is monomial if x =,'A where A is a linear character of some (not necessarily proper) subgroup of G. The group G is an M-group if every x E Irr(G) is monomial. Note that if x is monomial, then by Theorems 5.8 and 5.9, x is afforded by a module which has an imprimitivity decomposition into subspaces of dimension 1. It follows that x is afforded by a representation X with the property that each row and column of X(g) has exactly one nonzero entry for each g E G . (5.11)

LEMMA

Let 9 be a character of H E G. Then ker(9') =

n(ker 9)".

xsc

Proof Let x = 9.' Then g E ker x iff

1 P(Xgx-1) = 1 9(1). xoG

xsG

Since IS"(xgx-')I I 9(1), we conclude that g Eker x iff 9"(xgx-') = 9(1) for all x E G. This happens iff g E (ker 9)" for all x and the proof is complete. I (5.12) THEOREM Let G be an M-group and let 1 = f l < f2 < . < fk be the distinct degrees of the irreducible characters of G. Let 1~ E Irr(G) with ~ ( 1= ) fi. Then G") E ker x, where G(')denotes the ith term of the derived series of G.

Proof If i = 1, then xis linear and G c ker x.Assume i > 1and work by , $(1) = f j for j < i and induction on i. If t,b E Irr(G) and t,b(l) < ~ ( l )then G"-') 5 G") E ker $.I Since G is an M-group, we may choose H < G and linear 1E Irr(H) with x = AG. Now [(lH)', lG]= [IH, 1H] = 1 and thus (lH)' is not irreducible. If t,b is any irreducible constituent of (lH)', then $(1) < (lH)'(l) = 1 G:HI = AC(1) = ~ ( 1and ) thus G('-') c ker +. It follows that G"-" E ker((lH)') C H by Lemma 5.11. Now G'" G H E ker 1and since G(') -a G , we have (3'')

c

n(ker A)"

= ker

x

X€G

and the proof is complete.

(5.13)

COROLLARY

I

(Taketa) Let G be an M-group. Then G is solvable.

It is not the case that all solvable groups are M-groups. [The smallest counterexample is SL(2, 3) of order 24.1 In Chapter 6 we shall discuss some sufficientconditions. There is no known characterization of M-groups other than in terms of characters.

Chapter 5

68

We shall now discuss some of the connections between character theory and the theory of permutation groups. Suppose the group G acts on the finite set R. In other words, for each a E R and g E G, an element as E R is defined such that 01' = a and (ae)h= ashfor a E R and g, h E G. For a E R, we shall use the notation G, = {g E GI as = a} so that I G : G, 1 = I01, where 0 is the orbit containing a. Let G act on R. For g E G, let x@) = I {a E R I as = a} I. The nonnegative integer valued function x is called the permutation character associated with the action. To see that x really is a character, let V be a C-space with a basis identified with R. Let G act on V by permuting the basis. This makes V into a C[G]-module called a permutation module. It is clear that V affords x.

(5.14) LEMMA Let G act transitively on R, let a E R and let H = G,. Then (lH)' is the permutation character of the action. Proof Let V be the permutation module with basis R. Then V

=

C . s,nC/? is an imprimitivity decomposition for V The result is now immediate from Theorem 5.8.

I

(5.15) COROLLARY Let G act on R with permutation character x. Suppose R decomposes into exactly k orbits under the action of G. Then [x, 1G] = k.

u:=l

Proof Write R = Oi where the Loi are orbits. Let xi be the permutation character of G on Oi so that x = xi. For a E Lo,, we have xi = ( 1 ~ ~ by ) ' Lemma 5.14. Thus [Xi, lc] = [(l~~)',1 ~ = 1 [lcm,1 ~ =~ 1.1Therefore, [x, lG] = 1[&,1G] = k and the proof is complete. I

c

Corollary 5.15 can also be proved as follows. Let Oi be as above with aEO1.ThenIOiI = IGI/IG,I.NowletY= {(a,g)(aER,gEG,ae=a}and 9 'I two ways. We have compute 1

E X @=) IYI = aCe R lGal = eeG

i

IGI C IGal = C 1 - = C IGI = klGl aeOi asor loil i i

and the result follows.

(5.16) COROLLARY Let G act transitively on R with permutation character 2. Suppose that a E R and that G, has exactly r orbits on R.(One of these is {a}.) Then [x, x] = r. Proof We have

by 5.15, 5.14 and Frobenius reciprocity.

I

Induced characters

69

The integer r in Corollary 5.16 is called the rank of the transitive action. Recall that G is doubly transitive (or 2-transitve) on R if In(2 2 and G , is transitive on R - {a}, that is, if r = 2. COROLLARY Let G act on R with permutation character x. Then the action is 2-transitive iff x = 1, + $ where $ E Irr(G) and $ # 1,.

(5.17)

Corollary 5.17 is often useful for finding irreducible characters. For instance, it shows that the symmetric group C, has an irreducible character of degree n - 1 for n 2 2, and that this character restricts irreducibly to A , for n 2 4. We now consider the question of finding (not necessarily normal) subgroups of a group G from information about its characters. Suppose we wish to prove that G has a subgroup H of fairly small index. This could be done if the character (lH),could be recognized as such. Because (lH),is a transitive permutation character (of the action on the set of right cosets of H), there are a number of necessary conditions it must satisfy. (5.18) THEOREM Let H G G and let x = (lH),. Then (a) x(1)IIGI; (b) cx, $1 441 1 for $ E W G ) ; (c) [x, lG1 = (d) x@) is a nonnegative rational integer for all g E G ; (e) x(g) 5 x(gm)for g E G and integers m ; (f) xk?) = 0 ifo(s)YIGl/x(1); (g) x(g)I Cl(g)I /x(1)is integral for all g E G . Proof Most of these assertions are immediate from the fact that transitive permutation character. Statement (b) follows since

cx, $1

x is a

c(lH)G,$1 = C l H , $HI 5 $(1)Statement (f) follows since IHI = IG(/x(l) and if o@)$IHI, then no conjugate of g lies in H and so (lH),(g) = 0. To prove (g),let R be the set of right cosets of H so that x is the character of the action of G on R. Let X = Cl(g) and let Y = {(a, x)Ia E R, x E X , and ax = cr}. Since x is constant on X , we have =

IXIx(g) = IYI =

C IX

Gal-

aen

Since all G, are conjugate in G, I X n G , I = m is independent of a and we have I XI x(g) = m IR 1 = mx(l), proving the result. I The necessary conditions given in Theorem 5.18 are definitely not sufficient to guarantee that x is a permutation character. For example, the

Chapter 5

10

simple group M z 2 ,of order 443,520, has an irreducible character of degree 55. The character obtained by adding the principal character to it satisfies the conditions (a)-(g) and yet M z 2 has no subgroup of index 56. The following result can often be used to find proper subgroups of a group. THEOREM (Erauer) Let x be a character of G with [x, lG] = 0. Let A , B 5 G and suppose

(5.19)

Then A and B generate a proper subgroup of G. Proof Let U be a C[G]-module which affords x and let V , W and Y be the subspaces of fixed points of U under A , B and A n B respectively. Then I/ E Y, W c Y,and

dim V

+ dim W = [xA, lA] + [xB, IS] >

[xAnB,

1AnBl

=

dim Y.

It follows that V n W # 0 and thus ( A , B ) has nontrivial fixed points on U . Since [x, lC]= 0, we conclude that ( A , B ) < G . I An example showing how Theorem 5.19 can be used is the following proofthat A6 is the unique simplegroup ofits order. [Since I A6 I = I PSL(2,9) I, it follows that A6 Z‘ PSL(2,9).] As is often the case when trying to identify a particular group up to isomorphism, there is a great deal of tedious and detailed work involved. Since the result hardly seems to be worth that much effort, the proof which follows is somewhat more sketchy than most in this book. (5.20)

Then G

THEOREM Z‘ A s .

Let G be simple and suppose IGI = 360 = 23 .3’

a 5 .

Proof Suppose Go G G with I G : Go I = k > 1. Then G is isomorphic to a subgroup of the alternating group A k .Therefore k 2 6 and the result follows if we can find Go with k = 6. The only divisors of 360 which are = 1 mod 3 are 1, 4, 10, and 40. Let P ~ s y l , ( G ) and N = NG(P). We conclude that IG: NI = 10 or 40. By Burnside’s transfer theorem, N > P and hence I G : N I = 10. Now suppose P c M < G. We claim M c N . If not, then 1 < J M :M n N ( I10 and hence I M : M n NI = 4 or 10 by Sylow’s theorem. Since the Sylow 3-subgroups of G generate all of G , we have I Syl,(M) I < 10 a n d t h u s I M : M n N I = 4 a n d I G : M I d i v i d e s lO.ThuslG:Ml= 10and M n N = P . Now application of Burnside’s theorem to M yields K Q M with IK I = 4. Thus 8 I(NG(K)Jand P s NG(K) so that I G : NG(K)I I 5. This contradiction shows M 5 N as claimed.

Induced characters

11

Now let P # PI ~ s y l , ( G ) and set D = P n P , . Then PI G N(D) so that N(D) $ N . However, P E N(D) and we conclude that N(D) = G so thatD = 1. Next we claim that G contains no element of order 6. If x is an even permutation on 10 points and o(x) = 6, then x must contain a 2-cycle and x2 fixes at least two points. Since no element of G of order 3 lies in two distinct Sylow 3-subgroups, the claim follows. In particular, NIP acts faithfully on P . Since the automorphism group of a cyclic group of order 9 has order 6, it follows that P is elementary abelian. Furthermore, each involution in N inverts all elements of P . Now P = C(y) for each 1 # y E P and it follows that N has exactly six conjugacy classes and thus exactly two nonlinear irreducible characters each of degree 4. If 1, # x E Irr(G),then xN must have one of these as a constituent since N' $ ker 1.Therefore ~ ( 1 2 ) 4. Let x E Irr(G) with ~ ( 1 odd. ) Since 4$~(1),it follows that xN has a linear constituent I and thus x is a constituent of I,. Therefore, ~ ( 15 ) IC(l)= 10 and if x # 1, we have ~ ( 1 = ) 5 or 9. We will show that G has an irreducible character of degree 5. From ~ X E , r r ( C )~ ( 1= ) 360 ~ = 0 mod 4, it follows that there exist at least three nonprincipal irreducible characters of odd degree. It suffices to show that there is at most one of degree 9. In fact we show that if x E Irr(G) and x( 1) = 9, then x is a constituent of ( lN)'. Indeed, x is a constituent of I' for some linear character I of N . Thus'A = x + p, where p( 1) = 1 and thus p = 1,. Since [pN,I] # 0, it follows that I = IN as claimed. Now fix x E Irr(G) with x(1) = 5. Then xN has a linear constituent and so [xp, lP] # 0. Let v be a nonprincipal linear constituent of xp. Since every element of P is real, x p is real valued and [xp, v ] = [xp, i]. Let U = ker v, so that I U I = 3. It follows that [xu, l,] 2 3 and hence by Theorem 5.19, any two conjugates of U generate a proper subgroup of G . Note that we can find ten conjugates of U such that no two of them centralize each other. Let V be a conjugate of U which does not centralize U , and let H = ( U , V ) < G. If I G : HI = 6, we are done, so assume that I H I < 60. Now U E Syl,(H) or else P E H and thus H c N . This would imply V c P which is not the case. Since U , V E Syl,(H), it follows that 3 x 4 or 3 x 10 divides I HI and thus I HI = 12,24, or 30. If I H I = 30, then there exists S u H with 1 S I = 5. Therefore, I G : N(S)I = 1 mod 5 and divides I G : HI = 12. Thus I G : N(S)I = 6. We may assume that this does not occur and hence [ H I = 12 or 24. The only group of order 24 which is generated by elements of order 3 has a normal subgroup of order 8 and thus contains elements of order 6. We conclude I H ( = 12 and there exists K u H, elementary of order 4. Next, we claim [xH, lH] = 2. To see this, we use the fact that the unique

Chapter 5

12

nonlinear irreducible character 9 of H has degree 3 and is real. It follows that [S,, l,] = 1 and thus xH = 9 + A1 + A, where (Ai), = 1,. It follows that Izi = l Hand the claim is established. Also bo,l,] = 3. Now choose W conjugate to U with W $ H and W not commuting with U. Let L = ( U , W). We may assume ILJ < 60 and thus IL( = 12 and H n L = U . Since [xH, fH] = 2 = [xL, lL], Theorem 5.19 applies again to show R = (H, L ) < G. Since R $tN , it follows that 94'IR I. Reasoning as before, J R1 # 24. Since 12I J RI, this forces JR1 = 60 and the proof is complete. 1 It is often convenient to be able to write a character of a group in terms of characters induced from linear characters of subgroups of a specified type. For example, an important theorem of Brauer (which we will prove in Chapter 8) asserts that every character is a Z-linear combination of characters induced from linear characters of nilpotent subgroups. For now we consider only rational valued characters and cyclic subgroups. We prove the following. (5.21)

THEOREM

(Artin) Let x be a rational valued character of G . Then

where H runs over the cyclic subgroups of G and aH E Z. The proof of Artin's theorem depends on a well-known result of algebraic number theory, namely that the cyclotomicpolynomials are irreducible over Q. In other words, for each positive integer n, all of the primitive nth roots of unity are conjugate over Q. (This fact is also needed for Problem 3.1 1.) (5.22) LEMMA Let x be a rational valued character of G and let x, y E G with (x) = (y). Then ~ ( x= ) ~(y). Proof Let n = o(x) = o(y) and let E be a primitive nth root of unity. We have y = xm,with (m,n) = 1 and thus E"' is a primitive nth root of unity and E~ = E= for some automorphism 0 of Q[E]. Now ~ ( x = ) zfL1] ei, where each ci is an nth root of unity and hence is a power of E. We have x(y) = x(x") =

1

&irn

=

1

E;

= x(x)b.

Since ~ ( xis) rational, we have ~ ( x ) "= ~ ( x and ) the proof is complete.

1

Proof of5.21 Define an equivalence relation = on G by x = y if (x) and (y) are conjugate in G . It follows from Lemma 5.22 that x is constant on the equivalence classes under = and thus x is a Z-linear combination of the characteristic functions of these classes.

Problems

13

Let V1, g2,. . . , %, be the distinct (-)-classes of G and let Qi be the characteristic function of W i so that Qi(x) = 1 if x E g i and Qi(x) = 0, otherwise. Choose representatives x i E V i and let H i = ( x i ) and ni = IH i I. We have

I gi I

(1)

=

N(Hi)I d n i ) .

IG

We prove by induction on ni that

IN(Hi)IQi = C ajnj(1Hj)'

(2)

j

for suitable aj E Z where aj = 0 unless H j is conjugate to a subgroup of H i . The result will then follow since IN(Hi)I divides IN(Hj)Iwhen aj # 0. If ni = 1, then W i= { l } and I G lai = (lH,)' and (2) holds in this case. Suppose ni > 1. Write (1Hi)'

=

1bj@j

and compute the coefficientsbj as follows using Frobenius reciprocity and the fact that [ @ j , @k] = d j k l g j l / l G l : bjIwjI/IGI =

C(lHJG, QjI

=

C I H i , (@j)HJ*

( Q j ) H i = 0 unless H , is conjugate to a subgroup of Hi. In that case, takes on the value 1 on cp(nj)elements of H i and vanishes elsewhere.Thus [ l H i ,(Qj)H,] = cp(nj)/niand we conclude that

Now

(Qj)Hi

bj = l G l d n j ) / n i l ~ j l= IN(Hj)I/ni

using (1). Therefore, ni(1HJ'

=

C IN(Hj)I@j* i

wherej runs over the set of subscripts for which H j is conjugate to a subgroup of Hi.Solving for (N(Hi)IQiand applying the inductive hypothesis yields (2) and proves the theorem.

(5.23) COROLLARY Let x be any rational valued character of G. Then I G I x = aHlHG,where H runs over the cyclic subgroups of G and aH E Z.

CH

Problems (5.1) Let H E K E G and suppose that cp is a class function of H . Show that (cpK)G = 'p'.

(5.2) Let H , K E G with HK = G and suppose cp is a class function of H. Show that ((P')~ = (cpHn#.

Chapter 5

14

(5.3) Let H E G and suppose cp is a class function of H and $ is a class function of G. Show that (cpt,!~~)~= cpG$. Note

Problems (5.1)-(5.3) will be used frequently.

(5.4) Let b(G) = max{x(l)lxEIrr(G)}. If H E G, show that b(H) 5 b(G) I

IG:HIb(H). N o t e Since H is abelian iff b(H) = 1, the inequality b(G) I I G : HI b(H) generalizes Problem 2.9(b).

(5.5) Let F be an arbitrary field and let H E G. Let W be an F[H]-module. Show that there is a unique (up to F[G]-isomorphism) F[G]-module V such that V has an imprimitivity decomposition V = - &, where H is the stabilizer of W, and W z W, as F[H]-modules.

1

Notes By Theorem 5.9, V z W G .In the following problem, if V is an F[G]-module and H E G, we use the notation VHto denote the module V viewed as an F[H]-module.

(5.6) ( M a c k e y ) Let H , K sentatives so that

c G and let T be a set of double coset repreG

=

u

HtK

~ E T

is a disjoint union. Let W be an F[H]-module for an arbitrary field F. Show that (W G ) K =

*

( wH'

t€T

Note

Mackey's theorem generalizes Problem 5.2.

(5.7) Let H E G and suppose t,b is a character of H. Let K E G and assume ($G)K E Irr(K). Show that H K = G. N o t e Problem 5.7 is an immediate consequence of Problem 5.6. It can, however, be done without using modules. The key step is to show that [(I)~)~,($Hn,#] # 0 and then conclude that I G: HI I I K : K n H I .

(5.8) Let x be a monomial character of G and suppose K G G with xK E Irr(K). Show that xK is a monomial character of K . (5.9) Suppose that F is a subfield of C and let H E G. Suppose that 9 is a character of H which is afforded by an F-representation. Show that gG is afforded by an F-representation. N o t e It follows that every monomial character of G is afforded by a Q[E]-representation, where E is a primitive nth root of unity, n being the

Problems

75

exponent of G (that is, the least common multiple of the orders of the elements of G). Clearly, an arbitrary character of G has values in Q[e]. In Chapter 10 we shall prove the theorem of Brauer which asserts that every character of G is afforded by a Q[E]-representation. (5.10) Let H E G and view @[HI c C[G]. Let I be a right ideal of C[H] and let J = IC[G]. Suppose that I affords 9 (viewing I as a C[H]-module). Show that J affords 9.' (5.11) Let N N G kerx. Hint

4

G and

x E Irr(G). Suppose that [xN, lN] # 0. Show that

Use Lemma 5.1 1.

(5.12) Let cp' = x E Irr(G) with cp E Irr(H). Show that Z(x) E H. (5.13) (L. Solomon) Show that each row sum in a character table is a nonnegative rational integer. Hint acter.

Let G act on G by conjugation. Consider the permutation char-

Note The column sums are also integers. They can be negative. (5.14) Let G be nonabelian and letf

=

min{X(l)lx E Irr(G), x(1) > l}. Show

(a) If I G I I f, then G G Z(G), (b) if I [G, G'] I I f, then G is abelian, (c) if H s G and IG: HI I.f, then G' c H . Hint

For (a). If x E G, then ICl(x)I I I G' I.

(5.15) Let H E G and suppose cp is a character of H with det cp x = cpG and show (det x ) =~ lG.

=

l H . Let

(5.16) Let H be a maximal subgroup of G and let x = (lH)'. Let nonprincipal irreducible constituent of x. Show ker i+b = ker x.

i+b be a

(5.17) Let H E G and let x = (lH)'. Fix a positive integer n. For g E G, let m(g) = [Q~), l,,,] and define 9(g) = nm(g).Show that 9 is a character of G. Hint Let R be the set of right cosets of H in G. Fix a set S with IS I = n and let A be the set of all functions R -+ S . Then G acts on A.

Note This result provides another necessary condition to add to the list in Theorem 5.18.

Chapter 5

16

(5.18) Let G be a group. Write a(n) = I { g E G lo(g) = n} I. Then the polynomial f ( x ) = (I/ I G I )

c a(m)xIGl/m m

takes on integer values whenever x E Z.

Hint Use Problem 5.17. (5.19) Let G be doubly transitive on R and let H Show that H is transitive on R.

c G with IG: HI < IRI.

(5.20) Assume the notation and hypotheses of Problem 4.11 and show that IGI = 4(4 - 1)(4 + 1). Hints Let N = N(S) and show IN1 = 4(4 - 1). Let 1 ~ I r r ( N with ) 1# lN and S c ker 1. Let x E S , x # 1. Show 1'(x) = 1(l). Write '1 = E Irr(G) a, x and

1,

1(1)

=

c a,(x(l)

-

4 ) = I G : Nl 4 1 ) - 4

c

a,.

Finish the problem by showing that if a, # 0, then x(1) 2 4 + 1 so that z a , I 1(l).To show x(1) 2 4 + 1, compute [x,, l,] and use the fact that 11 > 0 and x is real, then [x,, l,] 2 2. since bN, Note Using the result of this problem, it is not very hard to prove that G z SL(2, q), a theorem originally due to Brauer, Suzuki, and Wall.

(5.21) Let P(G) denote the set of all Z-linear combinations of characters of of G of the form lHGfor H c G. Show that P(G) is a ring.

(5.22) Let x be a rational valued character of G and let n&) be the least ) . N = ker x so that x may be positive integer such that n G ( x ) x ~ P ( G Let viewed as a character of GIN. Show that rick) = ~ G , N ( Z ) .

Note The existence of nG(x) is guaranteed by Corollary 5.23.

Hint If nx = C aH(I,)', then nx = C aH(lNH)G. Show this by writing (lH)' = (1NH)' + <(w, where < ( H ) is a character of G with the property that N is not contained in the kernel of any of its irreducible constituents. (5.23) Let x be any character of G . Show that IGIx =

1ad',

where 1runs over the set of linear characters of cyclic subgroups and aAE Z.

Hint Use Problem 5.3.

Problems

11

(5.24) Let G be a doubly transitive permutation group on R and let a, fi E 0, with a # p. Let cp E Irr(G,) and assume that cpGaB E 1rr(Gap).Show that ccpc, OG1 5 2. Hint

Use Problem 5.6.

(5.25) Assume that every Sylow subgroup of G has a faithful irreducible character. Show that G has one also. Hint Let S be the product of all of the minimal normal subgroups of G. (This is called the socle of G.) Then S is a direct product of simple groups. Find 9 E Irr(S) such that ker 9 contains no nontrivial normal subgroup of G.

(5.26) (Passman) Let X E G - { l } be a subset and let n = I X I. Assume for all primes p < n, that a Sylow p-subgroup of G is cyclic. Show that there exists x E Irr(G) such that X n ker(X) = 125. Hints Reduce to the case that G is abelian by observing that it suffices to assume that G = (X) and that if [x, y] 4 ker(X), then both x and y 4 ker 2. In the abelian case, let N E G be maximal such that N n X = 0.Show that GIN is cyclic.

Note If n = 2, the hypothesis on Sylow subgroups is vacuously satisfied.

b

Normal subgroups

Let z ~ I r r ( G and ) H E G. In general, very little can be said about the restriction z H The . situation is quite different if H is normal in G. Let H 4 G. If 9 is a class function of H and g E G, we define Sg: H -+ C by P ( h ) = 9(ghg- '). We say that gg is conjugate to 9 in G .

(6.1) Then

LEMMA

Let H 4 G and let cp, 9 be class functions of H and x, y E G.

(a) cp" is a class function; (b) (cp")' = cpxy; (4 Ccp", 9 7 = [cp, 91; , = [ z H cp] , for class functions x of G ; (d) [ z H cp"] cp" is a character if cp is. (e) Proof (a) If h and k are conjugate in H , then so are hg and kg for all g E G. Take g = x-'. (b) Compute. [Note that had we defined gg(h)to be 9(hg), then (b)would fail.] (c) The two sums are identical since xhx- runs over H as h does. (d) We have (zH)X= z H .Apply (c). (e) Let X afford cp and define X" by X"(h) = X(xhx-I). Check that X" is a representation which affords cp". 1

It follows from Lemma 6.1 that G permutes Irr(H) by g: 9 H Sg. Note that H acts trivially and thus G/H permutes Irr(H). 18

Normal subgroups

79

(6.2) THEOREM (Chflord) Let H u G and let x~Irr(G).Let 9 be an irreducible constituent of zH and suppose 9 = gl,9,, . . . , 9, are the distinct conjugates of 9 in G. Then t

XH =

where e = [xH, 91.

e

19i, i= 1

Proof We compute (sG)". For h E H , we have

xxeG

= 9" and hence if since xhx-' E H for all x E G. Thus IH 1(9G)H cp E Irr(H) and cp # {Si}, we have 0 = 9", cp] and therefore [(9G)H,cp] = 0. Since x is a constituent of gG by Frobenius reciprocity, it follows that [zH, 431 = 0. Thus all irreducible constituents of zH are among the Giand xH = [xH, 9i]9i. However, [xH, Si] = [xH, 93 by Lemma 6.l(d) and the proof is complete. I

[c

Theorem 6.2 is so important that we digress to consider another proof in a more general, module theoretic setting. If X is an F-representation of H 4 G for an arbitrary field F , we define the conjugate representation Xg for g E G by P ( h ) = X(ghg- '). Note that Xgis an F-representation which is irreducible iff X is. Also, if X , and X, are F-representations of H , then Xlg and Xzgare similar iff Xl and X, are. (6.3) DEFINITION Let H u G and let Wl and W, be F[H]-modules. Then Wl is conjugate to W, if there exist bases for the K. such that the corresponding F-representations of H are conjugate. (6.4) LEMMA Let V be an F[G]-module and let H u G. Suppose W E V is an F[a-submodule. (a) If g E G, then Wg is an F[H]-submodule of V and is conjugate to W . (b) If M is an F[H]-module conjugate to W , then M z Wg for some g E G. (c) If U E V is an F[H]-submodule isomorphic to W , then Ug 2 Wg as F[H]-modules. Proqf We have (Wg)h= W(ghg-')g = Wg since ghg-' E H . Thus Wg is an F[H]-submodule. Let w l , w,, ..., w, be a basis for W. Then w,g, . . . ,w,g is a basis for Wg. Let X and I)be the F-representations of H corresponding to Wand Wg respectively, with respect to these bases. We claim that r) = f i g .

Chapter 6

80

If X(ghg-')

= (uij),we

have

and thus (wig)h =

1ai,(wjg)* j

Therefore, g(h)= (aij) = Xg(h),establishing the claim. The proof of (a) is now complete. If M is conjugate to W, then M corresponds to the representation X g for some gEG. Since Wg corresponds to the same representation, we have Wg r M yproving (b). Finally, if U z W then with respect to a suitable basis, U corresponds to X . Thus Ug and Wg both correspond to X g and (c)follows. I (6.5) THEOREM (Chflord) Let H 4 G and let V be an irreducible F [ G ] module. Let W be any irreducible F[HJ-submodule of V. Then are irreducible F[H]-submodules of V . (a) V = 1K where the (b) Each is of the form Wgi for some gi E G and thus is conjugate to W. (c) In the notation of Lemma 1.13, nw(V) = nM(V)for every F[H]module M conjugate to W. Note Another way of saying (c)is that each isomorphism class of F [ H ] modules conjugate to W is represented equally often among the &. Proof of Theorem 6.5 Clearly,

xBEG Wg is G-invariant and hence

V= c w g fl=G

by the irreducibility of V. The F[H]-modules Wg are conjugate to W and hence are irreducible. By Lemma 1.11, it follows that V is the direct sum of some of the Wg. This proves (a) and (b). Now let M be an F[H]-module conjugate to W. Since dim(M) = dim(W ) , it follows from Lemma 1.13(b) that it suffices to show that dim(M(V)) = dim(W(V)) in order to prove that nM(V)= nw(V). Since M is conjugate to W , it follows by Lemma 6.4(b) that M s Wg for some g E G. Now 6.4(c) yields W(V)g E M ( V ) and M(V)g-' c W(V). Thus W(V)g = M ( V ) and the result follows. I (6.6) COROLLARY Let H Q G and let V be an irreducible F[G]-module for some field F. Then when viewed as an F[H]-module, V is completely reducible. Proof Use Theorems 1.10 and 6.5.

I

Normal subgroups

81

Note that Corollary 6.6 is independent of Maschke's theorem and is valid even if char(F) divides IHI. Also note that Theorem 6.2 is a consequence of Theorem 6.5.

We now return to the study of @-characters and obtain some consequences of Theorem 6.2. (6.7) COROLLARY Let H a G and suppose that x E Irr(G)and [x,,1H] # 0. Then H E ker x.

1

Proof We have X, = e S i , where 9, = ,1 and the gi are conjugate. Since (1,# = 1, for all g E G, we have xH = ~(l)l,and H c ker x. I Corollary 6.7 can also be obtained as a consequence of Lemma 5.11 (see Problem 5.11) since x is a constituent of (l,)G and ker(l,JG = n H g = H . An interesting class of problems in character theory arises from considerI x E Irr(G)}are related. We ing how the structure of a group G and the set (~(1) give one result of this type as an application of Theorem 6.2. Many more results like this will be found in Chapter 12. (6.8) LEMMA Let H Q G and suppose x ~ I r r ( G )and 9 ~ I r r ( H )with CXH, 91 # 0.Then 9(1)1~(1). e

Proof Since P(1)= 9(1),we conclude that ~(1) = et9(1) where gi as in Theorem 6.2 with 9 = 9,. I

xH =

is a power of the prime p for every x E Irr(G). (6.9) THEOREM Suppose ~(1) Then G has a normal abelian p-complement. Proof If N -a G , then by Lemma 6.8,9(1)is a power of p for all 9 E Irr(N). Working by induction on I G 1, we see that it suffices to find a normal subgroup N of index p , since the normal abelian p-complement for N will be one for G. We have IGI = 1G:G'I 1 ~(1)'.

+

Xelrr(G); ~ ( 1 > )1

If G is abelian, the result is trivial. Otherwise, some x E Irr(G) is nonlinear and thus p I I GI. Since the last sum is divisible by p , we see that p I IG : G'I. It follows that G has a normal subgroup of index p and the proof is complete. I The converse of Theorem 6.9 is true and will be proved later in this chapter.

Chapter 6

82

(6.10)

DEFINITION

Let H -a G and let 9 E Irr(H). Then Z G ( ~ ) = {g E GIs8 = 8 }

is the inertia group of 9 in G . Since ZG(9)is the stabilizer of 9 in the action of G on Irr(H), it follows that it is a subgroup and thatlG(9)ZJH . Also I G : ZG(9)I is the size of the orbit of 9 and so in the formula t

of Theorem 6.2, we have t = I G : 1,(9) I. In particular, t divides I G : H 1. It turns out that e divides I G : HI also, but that is much more difficult to show. We shall prove some special cases in this chapter and the general result will be proved in Chapter 11, using projective representations. The followingresult is of fundamental importance in the character theory of normal subgroups. (6.11)

THEOREM

Let H

4 G, 9 E Irr(H), and

d = {$ E Irr(T)I[ $ H , 93 # 01,

B=

T = ZG(9).Let

{x E W G ) I [ X H , 91 f 01.

Then

xT

(a) If $ E d ,then $" is irreducible; (b) The map $ H 9' is a bijection of d onto 9; (c) If $' = 2, with $ E d ,then $ is the unique irreducible constituent of which lies in d ; , [I)~, 93 = [ x H , 93. (d) If $" = x, with $ ~ dthen

Proof Let $ E d and choose any irreducible constituent x of $". Then $ is a constituent of zT and since 9 is a constituent of t+hH,we conclude x E 9. Let 9 = 9,,9,,. . . ,9, be the distinct G-conjugates of 9. Then t = I G : TI and xH = e C:= si for some integer e. Since 9 is invariant in T, we conclude from Theorem 6.2 that = f9 for some5 Since $ is a constituent of x T , we have f Ie. Therefore et9( 1)

=

x( 1) I $"( 1) = t$( 1) = ftS(1) Ieta( 1)

and hence equality holds throughout. In particular, x(1) conclude that x = $'so that (a) follows. Also [XH,

91 = e = f

= [$H?

91

=

$'(1)

and we

Normal subgroups

83

and (d) is proved. Statement (c) follows from the last equality since if $, 9, # $, and $, is a constituent of xT, then CXH,

91 2

E d,

+ $1)H, 91 = C9H191 + C($l)H, 91 > [$HI 91,

a contradiction. The map in (b)is well defined by (a)and its image lies in g by (d).It is oneto-one by (c).To prove that it maps onto 93, let x E 9. Since 9 is a constituent of xH, there must be some irreducible constituent $ of xT with [ $ H , 91 # 0. Thus $ E d and x is a constituent of $". Therefore x = $" and the proof is complete. I (6.12) COROLLARY Let x E Irr(G) be primitive. Then for every N a G, xN is a multiple of an irreducible character of N . Proof We have

xN = e I:=9i,where

the

are distinct and t =

1 G : IG(S1)I. By Theorem 6.1 1, x = $G for some $ E Irr(IG(91)).Since x is primitive, it follows that IG(gl)= G and t complete. I

=

1. Thus xN

=

e 9 , and the proof is

The converse of Corollary 6.12 is false. For instance the irreducible character of A , of degree 5 is imprimitive and yet the conclusion of the corollary is (trivially)valid for this character. Characters which are multiples of an irreducible are called homogeneous characters. Irreducible characters whose restriction to every normal subgroup is homogeneous are often called quasi-primitive characters. (6.13) COROLLARY Suppose G has a faithful primitive character and let A a G be abelian. Then A E Z(G). Proof Let x E Irr(G) be primitive and faithful. Then linear character II. Thus A E Z(x) = Z(G). I (6.14)

COROLLARY

xA = eII for some

Every nilpotent group is an M-group.

Proof Let G be nilpotent and let x E Irr(G).Choose H 5 G minimal, such that there exists $ E Irr(H),with x = $'. By transitivity of induction (Problem 5.1), $ is a primitive character of H. It follows that $ is a faithful primitive character of H/ker($) = H and thus every normal abelian subgroup of R is central. Since every nilpotent group contains a normal self-centralizing subgroup, we conclude that B is abelian. Then $ is linear and the proof is complete. I We shall soon give a much more general sufficient condition for a group to be an M-group. The preceding result is included here simply to illustrate the usefulness of Theorem 6.1 1.

Chapter 6

04

(6.15)

THEOREM

for all x E Irr(G).

(Zto)

Let A

Q

G be abelian. Then ~ ( 1 dividies ) IG :A 1

Proof Let rl E Irr(A), with [xA, A] # 0, and let T = IG(rl).Then for some $ E Irr(T), we have x = $' and $ A = el. Thus A c Z($) and hence $(1) divides 1 T : A I by Theorem 3.12. Since x(1) = 1 G : TI $(l), we conclude that ~ ( 1divides ) IG : A I and the proof is complete. I Note that the converse of Theorem 6.9 is immediate from Theorem 6.15. Therefore, the purely group theoretic condition that G has an abelian normal p-complement is precisely equivalent to the condition that ~ ( 1is) a power of p for all x E Irr(G).

1

Suppose N Q G and 9 E Irr(N). Write 9' = eixi for xi E Irr(G) and ei > 0. Then (xi)N= ei sj, where 9 = 9,, . . . , 9, are the distinct conjugates of 9 in G. Let T = I'(9). By Theorem 6.11 it follows that 9T= C with $ i E Irr(T) and $iG = xi.It follows that for the purpose of investigating the nature of the integers ei, it suffices to assume T = G, that is, that 9 is invariant in G. For the remainder of this discussion we assume that 9 is invariant. We have 9' = C eixi and (xJN = ei9.

&,

Therefore, xi( 1) = e, 9(1) and so that

C e i 2 = 1G:NJ. It was remarked earlier (without proof) that ei divides IG : N I. (Note that if N is abelian then ei = x i ( l ) and in this case we know that eiI IG: NI by Theorem 6.15.) We see that in some respects, the integers ei behave like character degrees for GIN. If they were really character degrees, then one of them, say e l , would equal 1.That would mean that ( x ~=) 9.~In other words, that 9 is extendible to G. (Note that if 9 is extendible then it is automatically invariant.) A consequence of the next result is that if some ei = 1, then the ei are exactly the degrees of the irreducible characters of GIN. (6.16) THEOREM Let N Q G and let cp, 9 E Irr(N) be invariant in G. Assume (p9 is irreducible and that 9 extends to x E Irr(G). Let 9' = {B E Irr(G)I [cp', PI # 01 and 9-= {$ E Irr(G)I [(cp9)', $1 # O}. Then fi I-+ Bx defines a bijection of Y onto 9.

Normal subgroups

85

Proof We have ( ( P ' ) ~ is a multiple of cp and hence ((P')~ = I G : N I cp by comparing degrees. Thus [cp', cp'] = [cp, (p')N] = I G : N I [cp, cp] = 1 G :N 1. Similarly, [((p9)', ( ( P S ) ~=] 1 G : N I. Also, by Problem 5.3, ( ~ 9 ) ' = cp'x. Now write cp' = es p. We have [q', cp'] = IG : N 1 = [cp'x, 4 ~ ~ and hence

xaEs

c epz 8

=

1 q e , [ P x , rxl. PSYEY

Since Cpx,yx] 2 0 and [fix, f ? ~ ]2 1 and all ea > 0, this forces [px, yx] = 0 iff?# y and [ f ? ~f?x] , = 1. Thus the fix are distinct irreducible characters for distinct p. These are all of the irreducible constituents of c p ' ~ = (~9)'. The proof is complete. I What is probably the most important special case of Theorem 6.16 is when cp = 1,. (Gallugher) Let N Q G and let x E Irr(G) be such that the characters fix for p E Irr(G/N) are irreducible, distinct for distinct fi and are all of the irreducible constituents of 9'. (6.17)

COROLLARY

xN = 9 E Irr(N). Then

Proof This is exactly Theorem 6.16 in the case cp = 1,.

Note that in Corollary 6.17 we have identified Irr(G/N) with a subset of Irr(G). In this situation, we see that the e, are exactly the p(1) for fl E Irr(G/N).

There are other situations where we have control of the ei's. The following "going down" theorem is useful in studying the characters of solvable groups. THEOREM Let K/L be an abelian chief factor of G. (That is, K, L Q G and no M Q G exists with L M c K.) Suppose 9 E Irr(K) is invariant in G. Then one of the following holds:

(6.18)

-=

(a) 9, E Irr(L); (b) Q L = ecpforsomecpEIrr(L)andeZ=1K:LI; 'pi where the ' p i E Irr(L) are distinct and t = I K : LI . (c) 9, = Proof Let cp be an irreducible constituent of 9, and let T = Z,(cp). Since 9 is invariant in G, every G-conjugate of cp is a constituent of QL and hence is K-conjugate to cp. It follows that I G : TI = (K : K n TI and hence K T = G. Since K/L is abelian, K n T 4 K T = G and thus either K n T = K or K n T = L. If K n T = L, then 9,= cpi, where t = )K:LI, 'pl = cp, and the cpi are distinct. Thus 9( 1) = e I K : L I cp( 1). Since 9 is a constituent of cpK, we have 9( 1) IIK : L I cp( 1) and therefore e = 1. This is situation (c).

c:=l

x1

Chapter 6

86

Now assume K n T = K so that cp is invariant in K and 9, = ecp for some e. Let 1 E Irr(K/L). Since 1is linear, 29 E Irr(K). Also (As),= 9, = ecp. Suppose that all of the characters 19 are distinct as 1runs over Irr(K/L). Each of these IK :LI characters is an irreducible constituent of cpK with multiplicity e and we have e l K : L19(1) IcpK(l)= I K : Llcp(1). Therefore, e2cp(1) = e q 1 ) Icp(1) and e = 1. This is situation (a). In the remaining case, 19 = p9 for some 1,p E Irr(K/L) with 1# p. Let U = ker(1p). Then L E U < K and 9 vanishes on K - U . Since 9 is invariant in G, it follows that 9 vanishes on K - Ug for all g E G. Since U g= L, we conclude that 9 vanishes on K - L. By Lemma 2.29 we have IK:LI=IK:LI[9,9]=[9,,9,] =e2

ngcc

and the proof is complete. (6.19)

COROLLARY

Let N

x E Irr(G). Then either (a) (b)

I G with IG: NI = p, a prime. Suppose

( I

zN is irreducible or zN = Cp= 9,, where the siare distinct and irreducible.

Proof Take K = G and L = N and apply Theorem 6.18. Case (b) of that theorem cannot occur since p is not a square. I

(6.20) COROLLARY Let N 4 G and suppose IG : NI = p, a prime. Let 9 E Irr(N) be invariant in G. Then 9 is extendible to G. Proof Let x be an irreducible constituent of 9.' Then xN Comparison with Corollary 6.19 yields e = 1. I

= e9 for some e.

Actually, Corollary 6.20 would still be true if the hypothesis that 1 G : N 1 is prime were weakened to GIN cyclic. This stronger result will be proved in Chapter 11. (6.21) DEFINITION Let N (IG and let x E Irr(G). Then x is a relative Mcharacter with respect to N if there exists H with N E H E G and E Irr(H) such that $' = x and t,hN E Irr(N). If every x E Irr(G) is a relative M-character with respect to N, then G is a relatioe M-group with respect to N.

+

Note that x E Irr(G) is a relative M-character with respect to 1 iff it is a monomial character, and G is a relative M-group with respect to 1 iff it is an

Normal subgroups

81

M-group. Also, it is clear that if G is a relative M-group with respect to N, then GIN is an M-group. The converse of the last statement is false. If G = SL(2,3) and N = Z(G), then GIN r A4, which is an M-group. However, G is not a relative M-group with respect to N since G has an irreducible character of degree 2 and has no subgroup of index 2 as would be required by the definition. (6.22) THEOREM Suppose N 4 G and GIN is solvable. Suppose, furthermore, that every chief factor of every subgroup of GIN has nonsquare order. Then G is a relative M-group with respect to N.

Note The hypotheses on GIN, above, are automatically satisfied if GIN is nilpotent or supersolvable since then all chief factors of subgroups have prime order. Proof of Theorem 6.22 Let x E Irr(G). We must show that x is a relative M-character with respect to N. If xNis irreducible, this is clearly the case and hence we assume xN reduces. Let K Q G be minimal such that K 2 N and zK is irreducible. Then K > N and we may choose L 4 G, L zN such that KIL is a chief factor of G. In particular, KIL is abelian of nonsquare order. We apply Theorem 6.18 to the chief factor KIL and the G-invariant character xx E Irr(K). Since xL is reducible and I K : LI is nonsquare, we conclude that xL = (pi, where the (pieIrr(L) are distinct and t = IK:LI > 1. Let T = IG(cpl)2 L 2 N. By Theorem 6.11, we conclude that x = $" for some $ E Irr(T). Since TIN < GIN, we may apply induction on 1 G : N I to conclude that T is a relative M-group with respect to N and that $ = ST for some 9 E Irr(H) where N c H E T and QN E Irr(N). Now x = ' )I = (9T)G = 9' by Problem 5.1 and the proof is complete.

Note that Corollary 6.14 follows immediately from this result as does the somewhat more general fact that all supersolvable groups are M-groups. We prove a still more general sufficient condition. (6.23) THEOREM Let N Q G and suppose that all Sylow subgroups of N are abelian. Assume that G is solvable and is a relative M-group with respect to N. Then G is an M-group. Proof Let x E Irr(G). Since x is a relative M-character with respect to N, choose H 5 G with N c H, $ E Irr(H), I)NE Irr(N) and $" = x. Now choose U c H, minimal such that there exists 9 E Irr(U), with aH = $. By Problem 5.1, 9" = (9G)H = $" = x and it suffices to prove that 9 is linear. is irreducible, it follows Let M = U n N. Since (SNU)H= $ and that $ N U = gN" and thus = $N E Irr(N). By Problem 5.2, (9# = E Irr(N) and hence 8 M E Irr(M).

Chapter 6

88

By the minimality of U and Problem 5.1,s is a primitive character of U . Let K = ker 9, U = U / K and R = M K / K . Then 9 is a faithful primitive character of U and R has all of its Sylow subgroups abelian. Now let Z = Z(R)u 0. If Z < M, let A/Z be a chief factor of with A c R.Then A/Z is a p-group for some prime p. Let P E Syl,(A). Then P is abelian and A = PZ, so that A is abelian. Since A Q U , we conclude from Corollary 6.13 that A c Z(U). This contradicts A > Z. We conclude that Z = M, and hence R is abelian. Since 9, is irreducible, so is sMK. Thus Q M K E Irr(M) and hence 9( 1) = 1. The proof is complete. I We introduce some notation. If x is a character of G, let det x = A be the uniquely defined linear character of Problem 2.3. Now write o(x) = o(I),the order of A as an element of the group of linear characters of G. (We call o(x) the deterrninantal order of x.) Since I G : ker I I = o(I), it follows that o(x) divides I G 1. Now let N u G and suppose 9 E Irr(N) is invariant in G. We wish to find sufficient conditions that 9 be extendible to G. If ( I G : N I,9(1)) = 1, then 9 is extendible iff det 9 is extendible. We shall prove this now under the additional hypothesis that GIN is solvable and defer the general proof to Chapter 8. As will be seen, there is a gain in replacing the problem of extending 9 by that of extending the linear character, det 9. (6.24) LEMMA Let N u G and suppose 9 E Irr(N) is extendible to G and that (I G : N I, 9( 1)) = 1. Let I = det 9 and let p be an extension of I to G. Then there exists a unique extension x of 9 to G such that det x = p.

Proof Let q be an extension of 9 to G and let v = det q so that vN = A. Let pC = u SO that uN = IN and hence ulC:NI= I,. Writef = 9(1) so that det(ubq)= ubS det q = ubfv for b E Z.Since (.L I G : N I) = I, we can choose b E h such that bf = 1 mod1 G: NI and thus abS = u. Let x = ubq. Then det x = ubfv = uv = p. Since uN = lN,we have xN = q N = 9 as desired. To prove uniqueness, suppose xo is an extension of 9 with det(xo) = p. By Corollary 6.17, we have xo = xfi for some fi E Irr(G/N). Since xo(l) = 9(1) = ~ ( l we ) , have p(1) = 1 and p = det(Xo)= f i f det

x

= fifp

and fiS = 1., Since (f, IG : N I) = 1, this forces fi is now complete. 8

=

1, and xo = x. The proof

(6.25) THEOREM Let N u G and suppose GIN is solvable. Let 9 E Irr(N) be invariant in G and suppose (9(1), I G : N I ) = 1. Then 9 is extendible to G iff det 9 is extendible to G.

Normal subgroups

89

Proof If 9 is extendible, then obviously so is det 9. We prove the converse. Let ,u be an extension of 1 = det $ to G. We work by induction on I G : N I. (We may assume G > N . ) Let Go be a maximal normal subgroup of G with Go 2 N . Since I extends to pG0,the inductive hypothesis guarantees that 9 is extendible to Go. By Lemma 6.24, there is a unique extension, xo E Irr(G,) with det(xo)= pG0. We claim xo is invariant in G. For y E G , we have ( ( x ~ )=~gg ) ~= 9 and det((xo)g)= det(,yo)g= = pG0.The uniqueness of xo now yields ( x o ) g = xo and establishes the claim. Since GIN is solvable, IG: G o [ is prime and by Corollary 6.20, xo is extendible to G . The proof is now complete. I We now discuss some sufficient conditions for extending linear characters. A version of the following theorem is true without the assumption of linearity. The general result will be derived from this special case in Chapter 11.

(6.26) THEOREM Let N 4 G and suppose 1 is a linear character of N which is invariant in G. For each prime p ( o ( I ) ,choose H , c G with H,/N E Syl,(G/N), and assume that I is extendible to H,. Then I is extendible to G. Note If p $ I G : N 1, then H , = N and 1 is automatically extendible to H , . It is only necessary, therefore, to check the hypothesis for primes dividing IG: N ) . Proof Let rn = o(I).For each p I rn, we may choose I,, a power of 1,such that 1 = 1, and o(I,) is a power of p . We shall show that I , is extendible to ,up E Irr(G). Then ,u = ,upis an extension of 1. Since I , is a power of 1,which is extendible to H i , it follows that 1, is extendible to H , . Also 1, is invariant in G. We now see that it is no loss to assume that rn is a power of p. Let v be an extension of I to H,.Since p $ I G : H,I and v(1) = 1, it follows that p$vG(l) and hence there exists an irreducible constituent x of vG with PYX(1). Let x u ) = .f We have [xHp,v] # 0 and hence [zN,I ] # 0. Since A is invariant in G, ~ If.Let 6 = det x. Since p $ f we conclude that xN = fI and thus (det x ) = and rn is a power of p , we may choose b E Z withfb = 1 mod m.Then (b’), = I f b = I and the proof is complete. I

n

n

(6.27) COROLLARY Let N a G and suppose I is a linear character of N which is invariant in G. Assume (I G : N 1, o(1)) = 1. Then 1 has a unique extension ,u to G with the property that ( I G : N 1, o(,u)) = 1. In fact, o(p) = o(I).

Chapter 6

90

Proof The existence of an extension is immediate from Theorem 6.26 since the hypotheses are trivially satisfied. Let v be an extension of A and choose b E Z, with b IG : N I = 1 mod (o(I)). Let p = vblG:NI.Then p N = AblG:NI = I. In particular, o(p) 2 o(A). On the other hand, (v'(")~= l N , so that v0(') E Irr(G/N) and ,yG)= (,,O(W)lG:NI

=1

G.

Therefore, o(p) = o(I). For uniqueness, suppose z is an extension of A with (o(t),1 G : N I ) = 1. Then (o(p?),I G : N I) = 1 and (p?), = l Nso that p? E Irr(G/N). It follows that p? = lGandp = zasdesired. I (6.28) COROLLARY Let N Q G with GIN solvable and suppose 9 E Irr(N) is invariant in G. Assume (I G : N 1, o(9)9(1)) = 1. Then 9 has a unique extension, x E Irr(G) with (I G : N I, &)) = 1. In fact o(x) = o(9). Proof By Corollary 6.27, let p be the unique extension of I = det 9 to G with (I G : N 1, o(p)) = 1. By Theorem 6.25 and Lemma 6.24, let x be the unique extension of 9 to G with det x = p. Then o(x) = o(p) = o(I) = o(9). If xo extends 9 and (I G : N 1, o(xo))= 1, let po = det(xo).Thus

(IG: NI, O ( P 0 ) ) = 1 and hence po = p by the uniqueness of p. Then of x. This completes the proof. I

xo = x by the uniqueness

Note that since o(8)and 9( 1) divide IN 1, the condition that (~G:N~,~(9)9= ( 1 1) ) in Corollary 6.28 will be automatic if (I G : N I, IN I) = 1. The hypothesis of solvability in Corollary 6.28 will be removed in Chapter 8. We shall give one further extendibility criterion now. It can be used when GIN is a p-group for p # 2. Unlike the previous results, this condition is independent of character degree. (6.29) DEFINITION Let x be a character of G and letp be a prime. Then x is p-rational if there exists an integer r with p y r , such that ~ ( g E) Q[eZni1'] for every g E G. We use the notation Q k = Q[e2ni'k] for 1 I k E H . As is well known, n Ql = Q d , where d = (k,1). By Lemma 2.15, it follows that x@) E QIGl for every character x of G and every g E G. Write I G 1 = n = Pam, with p y m. It follows that x is p-rational iff all of its values lie in Q., Let %,,(G) denote the Galois group %(Q,,/Q,). If x is a character of G, let xu be defined by xu@) = x@)" for CT E s(Q,/Q). Then x is p-rational iff xu = x for all cr E %@ , ) m o t e that by Problem 2.2, xu is necessarily a character. In particular, Sp(G)permutes Irr(G).] Qk

Normal subgroups

91

From Galois theory we have 9(O,,/O,) z 9(Opa/O),where as above n = mp' and p x m . This isomorphism is the restriction map. It follows if p # 2 and a > 0, that elni', is not invariant under 9(Q,,/Q,) and hence if p # 2 and 1 is a linear character of G with p 1 o(1), then 1 is not p-rational. Also, when p # 2, we have Sp(G) 9(O,,/O) is cyclic. We have one more general remark. If H G G, then QIHl E O,, and so x" is defined for characters x of H with (r E 9,(G). It follows that x is p-rational iff x" = x for all (r E gP(G). (6.30) THEOREM Let N 4 G with GIN a p-group, p # 2. Suppose 9 E Irr(N) is invariant in G and p-rational. Then sG has a unique p-rational irreducible constituent x. Furthermore, xN = 9.

Proof Use induction on IG : N I. We suppose I G : N I > 1. Let K 4 G with N E K and IK : N I = p . By Corollary 6.20, 9 can be extended to q E Irr(K). Since p # 2, %,(K) is cyclic and we choose a generator 6.Now q" E Irr(K) and ( v " ) ~= 9" = 9 since 9 is p-rational. By Corollary 6.17, we have q" = q1 for some (linear) 1E Irr(K/N). If 1= l,, then q" = q and q is p-rational. Suppose that this is not the case. Then p = o(1) and since p # 2 we have 1"# 1 and so 1"= 1" for some m E Z, with m f 1 mod p. Choose b E Z with (1 - m)b = 1 mod p and let = Pq. Since AN = l N ,we have $N = qN = 9. Also, mb 1 = b mod p and thus

*

*"

+

= (pqy =

p + 1 q

=

pq=

*

and $ is p-rational. Thus in any case, 9 has a p-rational extension, t,b E Irr(K). If cp E Irr(K) is any p-rational extension of 9, then cp = $p for some unique p E Irr(K/N) by Corollary 6.17. We have cp = cp" =,($p)b = @'p" = $p' and thus p" = p. Since p # 2, it follows that p 8 o(p) and thus p = 1,. Therefore cp =

**

Now let g E G. Then (+g)" = (Ic/")g = I,P and S8 = 9 so that I,P is a p rational extension of 9 to K and by the preceding paragraph, we have i,b8 = $ and t,b is invariant in G . Since I G : K I I G : N 1, the inductive hypothesis guarantees that $G has a unique p-rational irreducible constituent x and that xK = $ so that xN = 9. Let xo be any p-rational irreducible constituent of 9'. We show xo = x. Let cp be an irreducible constituent of (xO),. Since ( x ~ is) a~multiple of 9, it follows that cpN = 9 and cp = $v for some v E Irr(K/N). Since K/N c Z(G/N), we have v8 = v for g E G and thus cpB = $ 8 ~ 8 = $v = cp and cp is invariant in G. It follows that (xO), = ecp for some integer e and thus cp is p-rational since cp(k) = ( l/e)xo(k) for k E K . Therefore cp = $ and xo is a constituent of $'. It follows that xo = x and the proof is complete.

-=

Chapter 6

92

We can use some of our results on character extendibility to prove Tate’s theorem. This result, which was originally proved in an entirely different way, serves as a “booster” for transfer theory, as will be explained. We define OP(G)to be the unique minimal normal subgroup of G such that G/OP(G)is a p-group (for the fixed prime p). Similarly, let AP(G)be the unique minimal normal subgroup of G such that G/Ap(G)is an abelian pgroup. [Note that AP(G) = GOP(G).] Let P E Sylp(G)and let N 2 P. Then N is said to control p-transfer if N / A P ( N )z G/AP(G), or equivalently, N n Ap(G) = AP(N).Several of the standard transfer theorems assert that under suitable hypotheses, certain subgroups N control p-transfer. [Usually, N = N(W) for some subgroup, W a P.] Tate’s theorem guarantees that whenever N / A P ( N )z G/AP(G), then also N / O p ( N )z G/OP(G). (6.31) THEOREM (Tute) Let PeSyl,(G) and N N n AP(G) = AP(N).Then N n OP(G)= Op(N).

zP.

Suppose that

G

h

Proof (Thompson) Since N zP, we have NOP(G)= G . Let U = N n OP(G)so that U a N and N / U z G/OP(G)is a p-group. Thus U 1 Op(N).Assume U > OP(N)and choose V a N , V 2 OP(N)such that U / V is a chief factor of N . Let I be a nonprincipal linear character of U / V . Since U / V is a chief factor of the p-group N / O P ( N )we , conclude that U / V G Z ( N / V ) and hence I is invariant in N . Let 9 = If x is an irreducible constituent of 9 and x E N , then since I” = I , we have Since P

c N , it follows that we may write

as A runs over the sums of the orbits of Irr(OP(G))under the action of P. Now 9(1) = I OP(G): U I = I G : N I is prime to p. It follows that for some A, we have p $ aAand p $ A(1). Write A = CXEO x where 0 is an orbit. Since x(1)

Normal subgroups

93

is constant for x E 0, we conclude that p Y 10I. Since 0 is an orbit under the p-group P , we have 10I = 1 and A = x, where x E Irr(OP(G)),x is invariant and p Y CS, X I . under p , P Y Since POP(G)= G and x is invariant under P , it follows that x is invariant in G. We claim that xis extendible to G. By Corollary 6.28, it suffices to show that p $ x(l)o(x). Now OP(G) has a normal subgroup, ker(det x) = K , such that IOP(G): KI = o(x) and OP(G)/K is abelian. Since Op(Op(G))= OP(G),it follows that Op(G)has no nontrivial p-factor group and thus p Y o(x). Since p $ x( l), Corollary 6.28 does apply. Let II/ be an extension of x to G and write

xu)

$N=

We have

c b,C%

4

1

bqcp* g~Irr(N)

= CII/a, 4 =

[xu,4 = Ex, 91,

v

which is prime to p . We may therefore choose rp E Irr(N) with p &' [p,, I ] . In particular, [qU,l ] # 0 and since I E Irr(U) is invariant in N , we conclude that rp, = eI and thus cp(1) = e = [ q u , l ] is prime to p . Since I/ c ker I and 'pu = e l , we conclude that I/ c ker cp and cp E Irr(N/V). Since N / V is a p-group, cp( 1) is a power of p and hence cp( 1) = 1. Now N/ker cp is abelian and thus AP(N)c ker cp. However, U = N n OP(G) c N n AP(G) = AP(N)and therefore U 5 ker cp. Since cp, = e l , it follows that I = lU. This contradicts the choice of I and completes the proof of the theorem. I

We discuss one further topic in this chapter. Let N 4 G so that G permutes Irr(N). It is also true that G permutes the set of conjugacy classes of N and it is natural to consider the relationship between these two actions. It is not the case that they are necessarily permutation isomorphic although they are closely related. (6.32) THEOREM (Bruuer) Let A be a group which acts on Irr(G) and on the set of conjugacy classes of G. Assume that x(g) = x"(g") for all x E Irr(G), a E A and g E G; where go is an element of Cl(g)". Then for each a E A , the number of fixed irreducible characters of G is equal to the number of fixed classes.

Proof Let xi and X j be the irreducible characters and conjugacy classes of G for 1 I i, j I k. Choose gj E X j and write g; = g j if .Xi' = X j . Let X = (xi(gj)),the character table of G, viewed as a matrix.

Chapter 6

94

For a E A, let P(a) = (pij), where p i j = 0 unless :x = xj, in which case p i j = 1. Similarly, define Q(a) = (qij),where qij = 1 if Xi" = X j and is zero otherwise. The (u, u) entry of the matrix P(a)X is

1i

Puixibo)

since only when XQ(a)is

xi =

= xu%")

is pui # 0. Similarly, the (u, u) entry of the matrix

1

xubj)qju

=

xu(S:-')

j

since only when g j = gz-' is qju# 0. The hypothesis of the theorem now implies that P(a)X = XQ(a). Thus Q(a) = X-'P(a)X since the orthogonality relations guarantee that X is nonsingular. We conclude that tr P(a) = tr Q(a). Since tr P(a)is the number of x ~ I r r ( Gwhich ) a fixes and tr Q(a) is the number of a-fixed conjugacy classes, the proof is complete. I (6.33) COROLLARY Under the hypotheses of Theorem 6.32 the numbers of orbits in the actions of A on the irreducible characters and conjugacy classes of G are equal. Proof Apply Corollary 5.15 to the result of Theorem 6.32.

I

We may apply Theorem 6.32 to obtain information about the characters of "Frobenius groups." (6.34) THEOREM Let N 1 # x E N . Then

4

G and assume that C,(X)

_C

N for every

(a) For cp E Irr(N), with cp # l,, we have ZG(cp) = N and cpG E Irr(G). (b) For x E Irr(G) with N $4 ker x, we have x = cpG for some cp E Irr(N). Proof Let cp~Irr(N),cp # 1,. To show that cp'~Irr(G), it suffices by Theorem 6.11 to show that Z,(cp) = N . In order to prove (a), therefore, it suffices to show that no element g E G - N can normalize any nontrivial conjugacy class of N and then apply Theorem 6.32. Suppose then, g E G - N normalizes X , a class of N . Let x E X . Then x BE X and thus xu = x" for some n E N . Therefore gn- E C(x). Since gn- 4 N and x E N , the hypothesis yields x = 1 and thus X = { 1). This proves (a). Now let x E Irr(G) with N $4 ker x. Choose an irreducible constituent rp of xN with cp # 1,. Then x is a constituent of cp' which is irreducible and so cpN = x. The proof is now complete. I

'

Problems

95

It turns out that the groups G satisfying the hypotheses of Theorem 6.34 and for which N c G are exactly the "Frobenius groups" which are discussed at some length in the next chapter. Problems

(6.1) Let N a G and 9 E Irr(N). Show that 9'

E Irr(G) iff

Z'(9) = N .

(6.2) Let N a G and suppose GIN is abelian. Let C be the group of linear characters of GIN so that C acts on Irr(G)by multiplication. (See Problem 2.6.) Let 9 E Irr(N). Show that S

p' =

f C xi, i= 1

wherefis an integer and the xi E Irr(G) constitute an orbit under C . Hint Let of GIN.

x E Irr(G). Then (zN)'

= px,

where p is the regular character

(6.3) Let N a G and let x E Irr(G)and 9 E Irr(N) with [xN,93 # 0. Show that the following are equivalent:

xN = e9,withe' = ( G :NI; x vanishes on G - N and 9 is invariant in G; x is the unique irreducible constituent of 9' and 9 is invariant in G. Note In the situation of this problem, we say that x and 9 are fully

(a) (b) (c)

ramiJed with respect to GIN. (6.4) Define %(G) = {H E GI cp' E Irr(G) for some cp E Irr(H)}. Let H(G) = n S ( G ) . Show that if G is an M-group, then H(G) is abelian. Note By Problem 5.12,Z(G) E H(G) for all groups, G. Of course, H(G) is characteristic in G. (6.5) Let H(G) be as in the preceding problem. Show that H(G) centralizes NIN' for every N a G and conclude that if G is solvable, then H(G) is nilpotent. Note In fact we may conclude that if G is solvable, then H(G)' E Z(H(G)).This is so because a group which is nilpotent of class > 2 necessarily has a characteristic abelian noncentral subgroup; namely, the next to the last nontrivial term in its lower central series. Hint If cp E Irr(N) with N a G, then H(G) c Z,(cp). [N, H(G)] E N'.

Use this to show that

96

Chapter 6

(6.6) Let G be solvable and assume that every x E Irr(G) is quasi-primitive. Show that G is abelian. (6.7) Let N u G and assume that GIN is solvable. Let x E Irr(G) and 9 E Irr(N), with [xN, 91 # 0. Show that x(1)/9(1)divides I G : N 1. N o t e The conclusion of this problem is valid even if GIN is not solvable.

(6.8) Suppose that G has exactly one nonlinear irreducible character. Show that G’ is an elementary abelian p-group. (6.9) (Dornhofl) Let G be an M-group and suppose N q G with (INI, 1G:NI) = 1. Show that N is an M-group. Hint Let s ~ I r r ( N ) Find . H 5 G and L E Irr(H), L(1) = 1 such that [(ARH),, 91 # 0 and ,IG E Irr(G). Use Problem 6.7 to show that L”’(1) = 9(1) and then use Problem 5.8.

(6.10) Let N 4 G with (I G : N 1, IN I) = 1. Suppose that every subgroup of G / N is an M-group. Show that G is a relative M-group with respect to N . (6.11) Let A Q G with A abelian. Suppose x E Irr(G) is a monomial character. Show that x is a relative M-character with respect to A. Hint Write x = LG, with L a linear character of H c G . Then every irreducible constituent of (LA,+# has multiplicity 1. N o t e Since Problem 6.11 is false if A is not abelian, it does not follow (and is not true) that an M-group is necessarily a relative M-group with respect to every normal subgroup.

(6.12) Let K/L be an abelian chief factor of G . Let cp E Irr(L) and T = Z,(cp). Assume that K T = G. Show that one of the following occurs: (a) cpK E Irr(K); (b) cpK = e 9 for some 9 E Irr(K), where e z = I K : L I. si, where the 9 i Irr(K) ~ are distinct and t = I K : LI. (c) cp‘ = Hint Use Problem 6.2 and the ideas in the proof of Theorem 6.18. (6.13) Show that the number of real classes of a group G is equal to the number of real valued x E Irr(G). Note It is easy to see that if I GI is odd, then 1 is the only real element. It follows that Problem 6.13 generalizes Problem 3.16.

(6.14) Let F be a field with Q s F 5 C. Say that g E G is an F-element if x(g) E F for every x E Irr(G). Let G be a p-group with p # 2 and show that the

Problems

91

number of classes of F-elements of G is equal to the number of x E Irr(G) with values in F. Hint Let E be a primitive I G 1 th root of unity. Let 9 be the Galois group of F[E] over F . Define actions of 9 on Irr(G) and on the set of classes of G.

Note Problem 6.14 is false without some assumption on G. Counterexampleswith G a 2-group and G of odd order have been found by Thompson and Dade respectively. (6.15) Let F be an arbitrary field and let N 4 G . View F [ N ] E F [ G ] . Show that J ( F [ N ] ) E J(F[G]). (See Problem 1.4.) (6.16) Let P be a p-group and Q a q-group, where p and q are distinct odd primes and P E Aut(Q). Show that 1 PI < $1 Q I by carrying out the following steps.

(a) It suffices to assume that Q is elementary abelian, that is, is an F[P]module with F = GF(q). (b) It suffices to assume that Q is an irreducible F[P]-module. (c) We may assume that P is not cyclic, so that by theorems on p-groups, it follows that there exists A u P with A elementary abelian of order p 2 . (d) Q is imprimitive as an F[P]-module and we may choose H E P, I P:H I = p , with Q = f= each an F[H]-module with all I 4 I equal. (e) Let N i be the kernel of the action of H on &. Then 1

w, w

IPI I p n l H : N J . (f) Complete the proof by induction. Hint Clifford's theorem (6.5)is used in (d). Show that F [ A ] cannot have a faithful irreducible module.

Notes The result of Problem 6.16 may be stated as a theorem in arithmetic. If Q is elementary abelian of order q", then IAut(Q)I = g: (4" - 4') and hence this number is not divisible by p" if p" > iq". The induction in the previous step (f) would not go through to prove the weaker theorem that I PI < IQ I.

n

(6.17) Let N u G with GIN cyclic. Let 9 E Irr(N) be invariant in G and assume that (9(1),I H : N I ) = 1. Show that 9 is extendible to G. Hint

Show that G/ker(det 9) is abelian.

Note The result of Problem 6.17 is true without the assumption that ( 9 ( l ) , ~ G : N=~ )1.

98

Chapter 6

(6.18) Let N Q G and suppose G = NH with N n H = 1. Let 9 E Irr(N) be invariant in G and assume (9(l), I G : N I ) = 1. If H is solvable, show that 9 is extendible to G: Note The result of Problem 6.18 is true without the assumption that H is solvable. However, even if H is abelian, the condition (9(1),IG: N I ) = 1 cannot be removed. (6.19) Let N Q G with GIN a p-group and let 9 E Irr(N) be invariant and p rational and assume p y o(9) and p y9(1).Let x be the extension of 9 to G with p ,j' o ( ~(which ) exists by Corollary 6.28.) Show that x is p-rational. (6.20) (Thompson) Let EP(G)denote the minimal normal subgroup of G such that G/Ep(G) is an elementary abelian p-group. Suppose N c G with p y IG : N I and EP(N)= Ep(G)n N. Show that O p ( N ) = OP(G) n N . Hints Sharpen the proof of Theorem 6.31 as follows. Choose I so that I p = 1,. Let Y = Y(QlGI/Qp)and let A E Sylp(Y).Now A permutes Irr(H) for all H c G and I is invariant under A. Now proceed with the proof of Theorem 6.31 but arrange matters so that 2, $ and rp are invariant under A. (6.21) Let 1 = H, -a HI Q Q H, = G. Assume that H i / H i - l is non) 2". abelian. Show that there exists x E Irr(G) with ~ ( 1 2

Hint Use Corollary 6.17.

?

T.I.sets and exceptional characters

Suppose we know that G has a subgroup H with certain specified properties and assume that we have some information about how H is embedded in G. How can we draw conclusions about G? We might try inducing the irreducible characters of H to G . Usually this gives little information since if I H 1 is much smaller than I G I, the characters 9‘ tend to have large numbers of irreducible constituents with large multiplicities, even if 9 E Irr(H). In certain situations, however, one can find a difference of two characters 9, - 9, of H where (9, - 9,)G = x1 - x2 and x, and x2 are under control. In these situations, information about Irr(G) can be obtained. The earliest example of the use of this technique is in the proof of Frobenius’ theorem. We shall give this before discussing any of the later refinements of these ideas. In Theorem 6.34 we considered groups G having a normal subgroup N such that C,(X) c N for all 1 # x E N. It follows immediately using Sylow’s theorem and the fact that nontrivial p-groups have nontrivial centers that (IN 1, I G : N I) = 1. By the Schur-Zassenhaus theorem we conclude that there exists H c G with NH = G and N n H = 1. In this situation, let g E G - H . Write g = xn with x E H and 1 # n E N. If y E H n Hg, then y E H” and y = h” for some h E H. Since y E H , we have [h, n] = h-’h” = h - ’ y E H. Since N 4 G, we have [h, n] E H n N = 1 and h E C(n) E N. Thus h = 1 and y = 1. We conclude that H n H B = 1. (7.1) DEFINITION Let H C G , with 1 c H c G . Assume that H n Hg= 1 whenever g E G - H. Then H is a Frobenius complement in G . A group which contains a Frobenius complement is called a Frobenius group. 99

Chapter 7

100

We have proved in the foregoing that groups satisfying the hypotheses of Theorem 6.34 with 1 < N < G are Frobenius groups. Frobenius' theorem is the converse of this. (7.2) THEOREM (Frobenius) Let G be a Frobenius group with complement H . Then there exists N-u G with H N = G and H n N = 1.

The fact that the group N of Theorem 7.2 satisfies the condition that CG(x)c N for all 1 # x E N is not difficult to prove and is left to the reader.

Before beginning the proof of Theorem 7.2, we mention the curious fact that it is trivial to find N . What is hard is to prove that N is a subgroup. (7.3)

LEMMA

Let H be a Frobenius complement in G. Let

(

)

N = G- UHx ~ { l } . xcG

Then IN1 = IG: H I . If M a G with M n H = 1, then M G N . Proof Since H = N,(H), there are I G : HI distinct subgroups of the form H". These contain exactly I G : HI ( 1 HI - 1) nonidentity elements. The remaining elements of G constitute the set N . We have I N ) = IGI - IG:HI(IHI - 1 ) = ]GI- IGI

+ 1G:HJ= 1G:HI.

If M -u G and M n H = 1, then M n H" = 1 for all x E G and thus M c N . The proof is complete. I We mention that except for some special cases, no proof of Theorem 7.2 is known which does not use characters. (7.4) LEMMA Let H be a Frobenius complement in G. Let 9 be a class = 9. function of H which satisfies 9(1) = 0. Then (9G)H Proof Let 1 # h E H . Then

c

P ( h ) = (l/(HI) $"(xhx-'). XEG

If S"(xhx-') # 0, then 1 # xhx-' E H n H x - ' and x E H . Then $"(xhx-') 9(h). We have P ( h ) = (l/IHI) 1 9 ( h ) = 9(h).

=

X P H

Since SG(1) = I G : HI $( 1) = 0, the proof is complete.

I

The proof of Theorem 7.2 may be motivated as follows. Assuming the theorem is true, let x E Irr(G) with N c ker x. Then xH E Irr(H). Now given xH = cp E Irr(H) we try to find x E Irr(G). We do this for each cp E Irr(H) and check that n k e r x is the desired normal subgroup.

T. I. sets and exceptional characters

101

Proof of Theorem 7 . 2 Let 1, # cp E Irr(H) and write 9 = cp - cp(l)lH so that S(1) = 0. Now [9', 8'1 = [S,(sG)H] = [9, $1 by Lemma 7.4. Thus [S', 8'1 = 1 + ( ~ ( 1 )Now ~ . [S', lc] = [ S , 1H] = -cp(l). We may therefore write 9' = cp* - cp(l)lG,where cp* is a class function of G , [q*, 1G] = 0, and 1 cp( 1)2 = [q*,cp*] + ( ~ ( l so ) ~that , [q*,cp*] = 1. Since 9 is a difference of characters, so is'9 and hence cp* is a difference of characters also. Since [cp*, cp*] = 1, it follows that fcp* E Irr(G). Furthermore, if h E H , then

+

cp*(h) = $'(A)

+ cp(1) = 9(h) + cp(1) = cp(h).

In particular, cp*(l) > 0 and thus cp* E Irr(G). For every nonprincipal cp E Irr(H), we have now chosen an extension, cp* E Irr(G).Let M = ker cp*. If x E M n H , then cp(x) = cp*(x) = cp*( 1) = cp(1) for all cp E Irr(H) and thus x = 1. By Lemma 7.3, M c N . Conversely, if g E G lies in no conjugate of H , then

n,

cp*@) - cp( 1) =.P ( g ) = 0

and g Eker cp*. It follows that M = N and hence the normal subgroup M satisfies IMI = 1 G : H I . We have l M H l = I M J J H I= I G : H I ( H I = (GI and the result follows. The normal subgroup whose existence is guaranteed by Theorem 7.2 is called the Frobenius kernel of G . By Lemma 7.3, it is uniquely determined by H . We mention without proof that in fact a Frobenius group has a unique conjugacy class of complements and a unique kernel. An entirely equivalent version of Frobenius' theorem may be stated as follows.

(7.5) COROLLARY Let G be a transitive permutation group on R with character x. Assume x(g) _< 1 for all g E G with g # 1. Then the set {g E GI x(g) = 0} u { l } is a transitive normal subgroup. Proof Let CI E R and assume I R 1 2 2. If there exists any g E G, g # 1with x(g) # 0, then G, is a Frobenius complement. By Theorem 7.2. and Lemma 7.3, {g ~ G l x ( g = ) 0} u (1) = N is the Frobenius kernel. It is transitive since NG, = G . I

We shall now discuss some of the ways that the ideas in Frobenius' proof have been extended and used in other contexts. We need to introduce some terminology.

(7.6) DEFINITION Let X E G be a subset. Then X is a T I . set (trivial intersection set) if for every g E G, either X g= X or X g n X c (1).

Chapter 7

102

(7.7) LEMMA Let X be a T.I. set in G and let cp and 9 be class functions on N = NG(X).Assume that cp and 9 vanish on N - X and that 9(1) = 0. Then sG(x) = $(x) for all X E Xand [SG, ( p G ] = [a, cp].

S"(yxy- I). If 9"(yxy- ') # Proof Let x E X.Then sG(x) = (1/ I N I ) 0, thenyxy- E X n XY-' and yxy- # 1.It follows that y E N and 9"(yxy- ') = 9(x). The first statement now follows. We have [P,cpG] = [(9G)N, cp]. Since cp vanishes on N - X and (SG)N- 9 vanishes on X, we conclude that [(SG)N - 9,cp] = 0 and hence [SG, S"] = [a, cp] as claimed. I Because of the requirement that 9(1) = 0, the preceding lemma cannot be applied if 9 is a character. We usually take 9 to be a difference of two characters. Such a difference is called a generalized character. Note that 9 is a generalized character of G iff [ S , x] EZ for all x~Irr(G).Also, the set of generalized characters of G is a ring. We are now ready to consider groups whose Sylow 2-subgroup P is generalized quaternion. If I PI 2 16, we shall prove the theorem of Brauer and Suzuki which asserts (among other things) that such a group cannot be simple. This theorem is also true if I PI = 8 but it is more difficult to prove in that case. We shall use the following facts about a generalized quaternion group P : (a) P has a cyclic subgroup of index 2; (b) IP: P'I = 4; (4 IZ(P)l = 2; (d) noncyclic subgroups of P are themselves generalized quaternion; (e) P contains a unique involution. We shall need to use the result of Problem 3.9 in the proof. (7.8) THEOREM (Brauer-Suzuki) Let P E Syl,(G) be generalized quaternion with I PI 2 16. Then there exists N Q G with (NI odd and such that GIN has a normal subgroup of order 2.

Note that possibly N = 1 in 7.8. We first prove the following weaker version. The full result will then follow easily. (7.9) THEOREM Let P E Syl,(G) be generalized quaternion with I PI 2 16. Then there exists M -=I G such that I M I is even and G / M is nonabelian.

Proof Let H c P be cyclic with IP: HI = 2. We have P' E H and = IPI14 2 4, so that P' > Z ( P ) and hence C,(P') = H.Let C = C,(P') and N = N,(P'). Now P E Syl,(N) and C Q N so that H = P n C E Syl,(C). Since a Sylow 2-subgroup of C is cyclic, it follows (for instance from Burnside's transfer theorem) that C has a normal 2-complement K and K -a N.

IP'I

T.I. sets and exceptional characters

103

Now N/C is isomorphic to a subgroup of Aut(P'). Since P' is a cyclic 2-group, it follows that N/C is a 2-group. We therefore conclude that N = KP. Since C = K H we have I N : CI = 2. Let U c P' with IP':UI = 2. Let X = C-UK. We claim that X is a T.I. set and N = N(X). Now C / K H is cyclic and U K / K is its unique subgroup of order equal to I U I. It follows for y E C that y E X iff o b K ) in C / K exceeds I U I. We conclude that y E X iff (2 1 U I) I ob). Now IP'I = 2 )U I and P' -a C. Since P' contains all elements of C of then P' c order 2 I U 1, it follows that P' E (x) for all x E X. If x E X n XB, (x) and (P')gE (x). Since IP'I = l(P')gl, we have P' = and g E N. Since clearly N E N(X), it now follows, as claimed, that X is a T.I. set and N = N(X). Since IP' I 2 4,we also have 4 Io(x) for every x E X. Now C/UK is cyclic of order 4. Let I be a linear character of C with ker I = U K . Let 9 = AN - (IC),. Since ker IN = UK G ker(l,-),, we conclude that 9 vanishes on U K and in particular 9(1) = 0. Clearly, 9 vanishes on N - C and hence 9 vanishes on N - X. We may therefore apply Lemma 7.7 with cp = 9 to conclude that [S", 93' = [S, 93. To compute [S, 93, observe that (1,-), = 1, + p where ker p = C . We claim that IN E Irr(N). Otherwise, AN is a sum of linear characters and P' E N' G ker IN = UK, which is not the case. Thus 9 = AN - p - 1, and [a, 93 = 3. NOW [gG, sG] = 3 and [sG,1G] = [S, IN] = - 1. It fOllOWS that SG =

f x l fxZ

-

1G,

where xl, x2 E Irr(G) are not principal. Since S'(1) = 9(1) = 0, we conclude that the signs above are opposite and without loss we may write

SG = x1 - x z Since 9'(g) (1)

=

- 16.

0 unless g is conjugate to an element of X, we conclude

Xlb) - x2b) = 1

if 4Yo@)-

Since P has a unique involution, it follows that G has a unique conjugacy class of involutions. Let .XIbe this class. Define the class function cp on G by = I{(x,y)lx,YE-fl,

XY =

g>l*

If p(g) # 0, then g = xy for involutions x and y and hence gx = yx = g - '. If o(g) is even, let a be the involution in (9). Then (x, a) is an abelian, noncyclic group of order 4. Since P has no such subgroup, neither does G by Sylow's theorem. We conclude that cp(g) = 0 if o@) is even. Thus cp(g)(xl(g) - x2(g) - 1) = 0 for all g E G. Therefore (2)

ccp,

x1

- x z - 1G1 = 0.

Chapter 7

104

In @[GI, let K , be the class sum corresponding to the conjugacy class X,. Then in the notation of Problem 3.9,

1

K l K , = a11vKw andifgEX,,thencp@) = u l l v . By Problem 3.9 we have for g E X , and x E X1that cp(d = all, = (I.xl12/1GI)

c

2-

x(x) x(g)/x(1)*

.y~lrr(G)

Since ~ ( xand ) cp(g) are real, we may rewrite this equation as ( l G l / l ~ l 1 2 ) c p=

c

(x(x)2/x(1))x

x E Irr(G)

and

(IGI/l~l12)Ccp,X1= x(x)2/x(1) for all x E Irr(G). We conclude from Equation (2) that xl(x)2/x1(1)- x2(x)2/x2(1)= 1. From Equation (1)we have x2(x) = xl(x) - 1 since 4 $ o(x) and also ~ ~ (=1 ) xl(l) - 1. Substitution into the preceding equation yields x1(x)2/x1(1)- (Xl(X) - 1)2/(xl(l) - 1) = 1.

Simplifying this, we obtain (Xl(X) - x1(1))2= 0 and we conclude that x E ker xl. Since o(x) = 2, we have lker x1 I is even. Now, ~ ~ (=1 1)+ ~ ~ (212) and thus G/ker x1 is nonabelian. The proof is now complete with M = ker XI. I Proofof Theorem 7.8 Let U be the (normal) subgroup generated by all of the involutions in G. If U has a cyclic Sylow 2-subgroup, then U has a normal 2-complement N and N Q G. In this case, U / N Q GIN and U / N is a cyclic 2-group. The result follows. Assume then that the Sylow 2-subgroups of U are noncyclic. We derive a contradiction. Since 8 I I U 1, we may choose V, with U c V 5 U P such that I V : U I I2 and 16I I V I. Now Theorem 7.9 applies to V and we may choose M Q V with V / M nonabelian and I M I even. Since a Sylow 2-subgroup of V contains a unique involution, it follows that all involutions of V are conjugate. Since M -a V contains an involution, it follows that M contains all of the involutions of V and hence U G M . Thus I V :MI II V : UI I 2 and this contradicts V / M being nonabelian and completes the proof. I

T. I. sets and exceptional characters

105

By the Brauer-Fowler theorem (Corollary 4.12), there are at most finitely many nonisomorphic simple groups which contain an involution whose centralizer is isomorphic to some given group C. Much work has been done in recent years to find all of the simple groups corresponding to various specific C . A key step is to find all possible orders of these simple groups and this often involves character theoretic techniques related to those in this chapter. As an illustration of this we prove the following. (7.10) THEOREM Let G = G‘ and suppose z E G is an involution with C,(T) dihedral of order 8. Then 1 GI = 168 or 360. We need a lemma. (7.11) LEMMA (Thompson) Let S ~ S y l , ( c )and suppose M G S with IS : M 1 = 2. Let z E S be an involution which is not conjugate in G to any element of M . Then z # G’. Proof Let G act by right multiplication on SZ = { M g J g EG } . Then = M g , then gzg-’ E M , which is not the case. Thus z has no fixed points on SZ and since I G : S I is odd, it follows that t induces an odd permutation on 0.Therefore, there exists A 4 G with 1 G : A 1 = 2 and z 4 A. The result follows. I

(SZI = 2(G:S ( . Now if Mgr

Proof of Theorem 7.10 Let D = C,(z) and D E s~Syl,(G). Then Z(S) E Z(D) = (z) and hence S E C(z) = D.Thus D E Syl,(G). Let M G D be cyclic of order 4 so that z is the unique involution in M . Since G = G‘, Lemma 7.11 guarantees that every involution in G is conjugate to z. Now M is a T.I. set in G since if M n M” # 1, then Z E M ”and thus z = zx and x E D. Since M -=I D we have M = M”. This argument also shows that D = NG(M). Since I Dis Let ;1be a faithful linear character of M and let 9 = (1, irreducible, it follows that [9, $1 = 3. Also, $(1) = 0 and 9 vanishes on D - M . Thus by Lemma 7.7 we have [sG, sG] = 3 and also (19‘)~= 9. Since 9‘(1) = $(1) = 0, it follows that we may write 9‘ = 1, x - t+h, where x, t+h E Irr(G). Calculation in D yields 4 = 9(z) = sG(z) and we have

+

(1)

0=1

+ x(1) - $(1),

4=1

+ x(z) - $(z).

Let X denote the (unique) conjugacy class of involutions in G and define the class function cp by

cp(d = I {(x, Y)lX, Y E .x,X Y

= s} I.

Chapter 7

106

If xy = g for involutions x and y, then gx = g- and conversely, if x E X , x # g, and g" = g-l, then y = xg is an involution. It follows that cp(g) = I { x E X I X# g, gx = g-'}l. If 1 # g E M and g" = g-l, then tX= z and x E D. We conclude that cp(g) = 4 for 1 # g E M. Reasoning as in the proof of Theorem 7.9 and using Problem 3.9, we have

Since IX I = I GI /8 this yields

Also, [S", cp] = [(lM M

- { l}, this yields

[s",

- A), qM].Since cp has the constant value 4 on cp] = (1 i ) + (2) (1 - i ) = 4. We conclude that

+

+

Write a = ~ ( 1 )and b = ~ ( t )so that $(1) = a + 1 and Equations (1). By the second orthogonality relation we have 8 = IC(Z)I2 1

+ x(t2) + $(t)'

=1

$(t)

=

b

-

3 by

+ b2 + (b - 3)2.

Since b E Z, we conclude that b = 1 or 2. Assume b = 1. Equation (2)yields 2' = I GI [1

+ (l/a) - (4/(a + l))] and

+ l)/(a - 1)2.

IGI = 2 ' 4 ~

Now 2 I a(a + 1) but 24 $ I GI and we conclude that Z3 I (a - 1). Therefore 22 $ a(a + 1) and since 23I I G 1 we have 24 8 (a - 1). No odd prime divisor of a - 1 can divide 2'42 + 1) and we conclude that a - 1 is a power of 2 and thus a = 9. This yields I GI = 22 x 9 x 10 = 360. Now assume b = 2. Equation (2) yields 2* = I G I [l + (4/a) - (l/(a + l))] and I GI = 2 ' 4 1 + l)/(a + 2)2. Reasoning exactly as above, we conclude that a 2 = 8 and (GI = 22 x 6 x 7 = 168. The proof is complete. I

+

We mention that GL(3, 2) g PSL(2, 7) is the unique group G of order 168 with G = G' and A6 2 PSL(2,9) is the unique one of order 360. We go back to the problem of obtaining information about the irreducible characters of a group from information about a subgroup. Let

T. I. sets and exceptional characters

101

9'E Irr(N). We introduce the notation Z[Y] for the set of Z-linear combin(Thus Z[Irr(N)] is the set of generalized characters ations of elements of 9. of N.) Suppose N E G and that we can find a map *: Z[9] + Z[Irr(G)] such that * is Z-linear and that [cp*, 9*] = [q,93 for all cp, 9 E Z[Y]. (Such a map is called a linear isometry.) In that case we have [x*, x*] = 1 for x E Y and thus fx* E Irr(G). Write E ( X ) = f 1 so that E ( X ) X * E Irr(G). The characters E(X)X* are called the exceptional characters associated with 9and *. They are in one-to-one correspondence with 9. How might such a map * be constructed? An easy example of a linear map ' 9 This map, however, is rarely Z[9] + Z[Irr(G)] is the induction map 9 +. an isometry on Z[sP]. By Lemma 7.7, we see that there are situations when induction is an isometry on Z[9]" = (9 E Z[9JI9(1) = O}. This occurs, for instance, if N = N(X),where X is a T.I. set and 9 = { X E Irr(N)Ix vanishes on N - X}.The problem then is to extend a linear isometry from Z[sP]" to all of ZC9-J. (7.12) DEFINITION Let N c G and Y E Irr(N) with 1 9 12 2. Suppose z) is coherent if z:Z[Y]" --+ Z[Irr(G)]" is a linear isometry. We say that (9, z can be extended to a linear isometry * defined on Z[9]. z) is coherent, we simply say 9 is If s is the induction map and (9, coherent. It should be emphasized that even in this case, the map * usually is not induction. The prototypical example of coherence is where N is a Frobenius complement in G . There, Irr(N) is coherent; and the proof of that fact is the essence of the proof of Frobenius' theorem. z) is coherent, the map * is not always uniquely defined. Nevertheless If (9, the set of exceptional characters E ( X ) X * is uniquely determined by (9, z). 7)be coherent and let * be a linear isometry which (7.13) LEMMA Let (9, extends 7.Then there exists E = _+ 1 such that EX* E Irr(G) for all x E 9'.The functionf: 9 -+ Irr(G) defined byf(x) = EX* is one-to-one. The image off is {I(/ E Irr(G)I[s',I(/] # 0 for some 9E Z [ 9 ] " } .

Proof For x E 9, [x*, x*] = 1, and we may choose E ( X ) = f1 so that E Irr(G). We claim E ( ~ J= ~ ( 5 for ) all 5 E 9.We have 9 = x(l)< <(1)xE Z[9]" and thus

E(X)X*

x( 1)(*

- ((1)x*

=

9'E Z[Irr(G)]".

Evaluation at 1 yields 0 = x(l)<*(l)- 5(1)~*(1)and hence 5*(1) and ~ * ( 1 ) have the same sign. Thus E ( X ) = E(<) as claimed. The foregoing also shows that [S', f(x)] # 0 for some 9EZ[Y]". Conversely, suppose $eIrr(G) and [s', $1 # 0, where 9 ~ Z [ 9 ] " . Then

Chapter 7

108

9 = CxeYa , ~and 9' = C a,x*. Therefore 0 # [x*, $1 = c [ f ( x ) , $1 for some x E Y and hencef(x) = 9. To show thatfis one-to-one, it suffices to show that * has trivial kernel. This follows since if 9* = 0, then 0 = [9*, 9*] = [ S , $1and hence 9 = 0. I

Suppose N E G, Y E Irr(N), (91 > 1 and z: Z[9']" + Z[Irr(G)]" is a z) linear isometry. We seek conditions on 9 sufficient to guarantee that (9, is coherent. One such condition is that all x E Y have equal degrees. Although this is not hard to prove directly, we shall derive it as a corollary of the following more general result of Feit. The main point of Feit's theorem may be summarized as follows. List the degrees ~ ( 1for ) x E Y in increasing order. Assume that the smallest of these divides all of the others. Then Y is coherent if the degrees do not increase too rapidly. (7.14) THEOREM (Feit) Let N c G, Y G Irr(N) and let z: Z[Y]" + Z[Irr(G)]" be a linear isometry. Suppose Y = Y ou {x}, where (Yo,z) is coherent. Assume that there exists $ E Y osuch that $(1)I x( 1) and

t)is coherent. Then (9,

Proof Let *: Z I Y O ]+ Z[Irr(G)] be a linear isometry which extends z on ZIYO]". We shall define x* so that the map * can be extended to all of Z[Y] by linearity. Let ~ ( 1= ) d $ ( l ) so that x - d$ E Z[Y]". Define A by

(x - d $ r

(1)

=

A - d$*

+

b,t*, re Y

O

where [A, t*] = 0 for all E Y o .Of course, A is a (possibly 0) generalized character of G and all b, E Z.We shall show that [A, A] = 1 and all b, = 0. Now 1 + d2 = [(x - d$), (x - d$)I = [(x - d*Y9 (x - d$Yl and thus 1

+ d 2 = [A, A] +

b:

+ (b, - a)'.

5*11r

Therefore [A,A]

+ Cb,'=

1 + 2db,.

5

Furthermore, since z maps into Z[Irr(G)]", we have (x - d$)'(I) = 0 and thus (3)

0=

1b<<*(l)- d$*(l) + A(1). r

109

T. I. sets and exceptional characters

For

5 E Y olet Z = $(1)5 - 5(1)$ E ZIYO]". Then [Er, (X - d$y] = [Z, ( X

- d$)] = dl(1).

Therefore dt(1) = [Z',

1b,q*]

- 5(l)b,

= $(l)b,

since E' = $(1)5*

-

- d[E', $*]

+ [E',A]

+ d5(1) + 0

5(1)$*. We conclude that W b , = 5(l)b,.

Now define p = b,/$(l). We have then b, = P5U)

(4)

for all 5 E Y o . Also, since Z(1) = 0, we have $(1)5*(1) = 5(1)$*(1)

and we may write <*(I) = .5(1)

(5)

for all 5 E Y oand fixed a # 0. By hypothesis we have

Suppose p # 0. Then by (4) we obtain 2 db,2 = p2(2 d$(l)') < p2 1

and therefore 1

+ 2 db,2

=

1be2

c

I b,'.

Now (2) yields [A, A]

+ 2 dbG2I2 db,.

Since b, is an integer # O (since p # 0) and [A, A] 2 0, we obtain A = 0 and b, = 1. Thus A(1) = 0 and (3), (5), and (4) yield =

Since b,

=

1, p

=

C b,5(1) = P C

1/$(1) and thus =

This contradicts (6) and proves p = 0.

c 5(112.

Chapter 7

110

It now follows that all b, = 0 and (1) yields (X

From (2) we have [A,A]

=

- d$)1

=

A - d$*.

1 and we extend

* to Z [ Y ]

linearly by defining

X* = A.

We now show that * is a linear isometry on Z[Y] and that it extends z. If 9 E ZCY]", with [ S , X ] = a, we can write 9 = a(x - d$)

+ CP,

where cp E ZIYO]". Then 9* = U ( X - d$)*

+ ~ p *= ~

( -x d$)'

+ ~ p '= 9'

and thus * extends z. To show that * is an isometry, it sufficesto show [p*, v*] = Lu, v] for p, v E Y. If p, v E Y o we , already know this. Since [A, = 0 for E Y oand [A, A] = 1, the result now follows. I

c*]

<

9 ' 1 2 2. Suppose (7.15) COROLLARY Let N G G and Y c Irr(N) with 1 z:ZEsp]"+ Z[Irr(G)] is a linear isometry. Suppose that all X E Y have equal z) is coherent. degrees. Then (9, Proof Use induction on n = 1 Y I. Let [(Xl

- X2Y9 (Xl

- X2)'l = [ ( X l

xl, x2 E Y be distinct. Then - XZ),

(Xl -

X2)l = 2

and since (xl - x2)l(l) = 0, we may write (xi - x2)' = a - p where a, p E Irr(G) are distinct. If n = 2, define xl* = a and x2* = p and extend by } thus * linearity to Z [ Y ] . In this case Z[Y]" = {a(xl - x 2 ) ( a ~ Zand agrees with z on Z[sP]". If n = 3, let x3 €9- {xl, x2}. Reasoning as above, (xi - x3)1 = p - v where p, v E Irr(G) and p # v. Also

[(Xi - X2)1,

(Xi

- X3YI

=1

and we conclude that either p = a or v = p but not both. If p = a, define xi* = a, x2* = p, and x3* = v. If v = /I, define xi* = - p, x2* = -a, and x3* = - p . In either case, extend * linearly to Z[Y] and check that * agrees with z on xi - x2 and on x1 - x3 and that * is an isometry. Since

Z[9I0

= {a(Xi

- Xz)

+ b(X1 - Xdla, b E Z},

the result follows in this case. Suppose n > 3. Write Y = Y ou {x}, where 3 I lYol = n - 1. By the inductive hypothesis, (Yo,z) is coherent. Choose $ E Y oand observe that

T. I. sets and exceptional characters

111

x(

$( 1) 1 1) since $( 1) = ~ ( 1 )Now .

T) is coherent by Theorem 7.14. and thus (9,

I

The most common applications of coherence are in the situation of a “tamely imbedded” subgroup as defined in the paper of Feit and Thompson on the solvability of groups of odd order. For the purposes of this book, we give a considerably more restrictive definition which, nevertheless, is important in applications. Let K

G. Assume

(7.16)

DEFINITION

(a) (b) (c) Then K

K is a T.I. set in G; K < NG(K); C,(k) E K for all 1 # k E K .

S

is a T.I.F.N. subgroup of G.

The “F.N.” in the foregoing definition stands for “Frobenius normalizer.” Note that if N = N,(K), where K is T.I.F.N. in G, then N is a Frobenius group and K is the Frobenius kernel. By Theorem 6.34, the irreducible characters of N are thus of two types, those with kernel containing K and those induced from nonprincipal q E Irr(K). If Y = { x E Irr(N)I K $2 ker x}, then the object of what follows is to prove that 9is coherent. Results of Feit and Sibley do, in fact, prove this in most cases. Note that if K c G satisfies condition (a) of Definition 7.16, then (c) is equivalent to C,(k) E K for all 1 # k E K , where N = NG(K). An important theorem of Thompson (not involving characters) asserts that Frobenius kernels are necessarily nilpotent. The nilpotence of T.I.F.N. subgroups will be assumed in what follows. (The reader may simply consider this to be an additional hypothesis in Definition 7.16.) A more elementary fact about a T.I.F.N. subgroup K of G is that ( I G : K I , IKl) = 1, that is, K is a Hall subgroup. To see this, let p l l K l , let P E Syl,(K) and suppose P c S E Syl,(G). Then Z(S) E C(P) E K by part (c) of 7.16. Now Z(S) # 1 and so S c C(Z(S))s K , again by 7.16(c).Thus

P Y lG:Kl. We mention some situations where T.I.F.N. subgroups arise. If P E Syl,(G), IP I = p and P = C,(P) NG(P),then Pis T.I.F.N. This happens, for instance in permutation groups of degree p and in the groups PSL(2, p). A doubly transitive permutation group is a Zassenhaus group if some nonidentity element fixes two points but none fixes three. Let G be a Zassenhaus group on a set R and let a, p E R with a # p. Then G , is a Frobenius

-=

Chapter 7

112

group with complement GaD.(We are assuming 1521 > 2 to avoid triviality.) Let K be the Frobenius kernel of G,. The nonidentity elements of K are exactly those elements of G which fix ct and no other point. It follows easily that K is T.I.F.N. in G and G, = NG(K). (7.17) LEMMA Let K be T.I.F.N. in G and let N = NJK) and 9 '= {x E Irr(N)IK $ ker x}. Then induction defines a linear isometry Z[Y]" + Z[Irr(G)]". Also, if 9 E Z[Y]", then

=

9.

Proof If x E 9, then x = tN for some 5 E Irr(K) by Theorem 6.34.Thus x vanishes on N - K . If cp, ~ E Z [ Y ] "it, follows that cp, 9 vanish on N - K and Lemma 7.7 yields [cp', 93 ' = [cp, 91 and the first assertion is proved. Also, ($'), = 9, by Lemma 7.7. Since 9 vanishes N - K , it suffices to = 9. However, show that 9' vanishes on N - K in order to prove that (9G)N no element of N - K can be G-conjugate to an element of K since if X E N- K and o(x)IIKI, then K c ( K , x) and ( K , x) = K ( x ) so that I (K, x) : K I is not relatively prime to I K I. This contradicts the fact that (I G : K I, I K I ) = 1. It now follows from the definition of 9' and the fact that 9 vanishes on N - K that 9' also vanishes on N - K and the proof is complete. I (7.18) COROLLARY (Brauer-Suzuki) Let K be an abelian T.I.F.N. subgroupofG.LetN = N,(K)ande = )N:KI.LetY = {x€Irr(N)IK $ kerx}. Then either (a) 1 Y I = 1, I K I = 1 + e and K is an elementary abelian p-group or (b) Y is coherent and there exists a one-to-one functionf: Y + Irr(G) and E = k 1 such that (x - t)G = E ( ~ ( x )- f(t))for all x, 5 E 9. Proof All of the orbits of nonprincipal irreducible characters of K under the action of N have size e. Thus x( 1 ) = e for all x E Y and 1 Y I = ( I K I - l)/e. If 1 9 12 2, then Corollary 7.15 yields that Y is coherent. In this case, (b) follows from Lemma 7.13. If 1 9 ' 1= 1, then lKl - 1 = e and the nonidentity elements of K are all conjugate in N . It follows that they all have the same prime order and (a) follows. I

The theorems of Feit and Sibley generalize Corollary 7.18 by dropping the assumption that K is abelian. If lKl - 1 # e, they prove that either Y is coherent or K is a 2-group. Before proceeding with the proofs, we discuss some implications of coherence in the T.I.F.N. situation. (7.19) LEMMA Let K E G be T.I.F.N. and let N = NG(K). Suppose 9" E {x E Irr(N)IK $ ker x} and that 9" is coherent. Let * denote an iso-

T. I. sets and exceptional characters

113

metry ZCX] -+ Z[Irr(G)] such that a = C X E d x(1)x. Then

* extends induction on ZCX]".Let

(a) If @ E Irr(G) and @ 4 { _+x* I x E X},then yjN = aa + 9, where a is rational and [S, x] = 0 for all x E X. (b) ( ( x * )~ x)/x(1) is independent of x E %. (c) ( x * ) ~ x = act + 9 for x E X,where a is rational and [9, 51 = 0 for all 5 E 3. Proof (a) Let x. 5 E 3.Then c@N,

x(l)( - 5(l)xl

=

=

[$*(x(l)< - 5(1)x)c1 [ ~x(l)t* , - 5(1)x*I = 0.

Thus x( 1) [@N, 51 = {( 1) [@N, x] and hence [t,bN, x]/x( 1) = a is independent of x E 3.Therefore, t,bN = act + 9, where [S, x] = 0 for all x E 22". Thus (a)is proved. (b) Let x, 5 E X.Then A = x(l)l - <(l)x E Z[X]" and so (Ac)N = A by Lemma 7.17. However, A' = A* = x(l)t*- <(l)x* and thus x(1)5 - 5U)x = X(1)(5*)N - 5(1)(x*h

and W)((X*)N - x) = X(1)((5*)N - 5).

Thus (b) follows. (c) Let 5, q E 22" and write A = <(l)q - q(1)t. Then C(X*)N,

A1 =

cx*, AG1 = cx*, A*l = cx,A1

so that [(x*)N - x, A] = 0. It now follows as in (a) that ( x * ) ~ x = aa where a = [ ( x * ) ~ x, 5]/5(1) and [5, 93 = 0 for 5 E 22". I

+ 9,

THEOREM Let K c G be T.I.F.N. and let N = N,(K). Suppose 9' = {x E Irr(N))K $ ker x} is coherent and let d c Irr(G) be the corresponding set of exceptional characters. Let * be an isometry Z [ 9 ] + Z[S] which extends induction on Z[9'p1° and let E = f1 so that ex* E 8 for x E 9. Then

(7.20)

(a) I = {t,b E Irr(G)J@is not constant on K - (1)). (b) If @ = EX* E 6, then @K - exK is constant on K - (1). This constant has the form rnx(l)/ I N : K 1 for some rn E Zwhich is independent of the choice OfXE9.

(c) If g E G is not conjugate to an element of K - { l}, then @(g)/@(1) is independent of the choice of $ E I . (d) If x E 9, then x*(l)/x(l) is independent of the choice of x.

Chapter 7

114

Proof We apply Lemma 7.19 with 3 = 9. Then a = pN - P N / g where p N and PNIK are regular characters. In particular, a is constant on K - { 1). Also, if 9 is a generalized character of N and [9, x] = 0, for all x E 9,then K is in the kernel of every irreducible constituent of 9 and hence SK is constant. Now, if $ 4 I , then 7.19(a) yields ~ + = b ~act + 9 and $ is constant on K - (1). If $ E 8, then 7.19(c) yields E $ ~= x + aa + 9 where E$ = x*. However, x E Y cannot be constant on K - {I), otherwise, by Problem 3.8, every nonprincipal irreducible character of K is a constituent of x K . This would force IY 1 = 1 which is not the case. Thus t,hN is not constant on K - { 1) and (a) is proved. The first part of (b) is immediate from 7.19(c). To evaluate the constant, choose t E Y with t(1)= 1 N: K I, that is, 5 = I N for some nonprincipal - tKis a generalized character ofK with constant linear I E Irr(K). Now (t*)K 'is arbitrary, 7.19(b) value m on K - (1). It follows that m E Z. If x E 9 yields that m'/x(l) = m/t(l) where m' is the constant value taken on by ( x * ) ~- xK on K - (1). Since t(1) = I N : K I , we have m' = mx(l)/IN:KI and (b) follows. Let 1, t E Y so that t(l)x* - x(l)t* = (t(l)x - ~(1)<)', which vanishes on elements g E G not conjugate to elements of K - (1). Thus t(l)x*(g) - x(l)t*(g) = 0 and (c) follows. Finally, (d) is immediate from 7.19(b). The proof is complete. I We now begin work toward the theorems of Feit and Sibley. (7.21) LEMMA Let N be a Frobenius group with Frobenius kernel K. Suppose IN: K I is even. Then K is abelian.

Proof Let t E N be an involution. Then t # K since (I K I, N : K I) = 1 and thus x' # x for x E K - (1). Now map K + K by x H x- 'x'. This map isone-to-onesinceifx-'x' = y - ' y r , thenyx-' = (yx-')'andthusyx-' = 1. Therefore the map is onto and every ~ E has K the form u-lu'. Thus xt = ( U - l ur )t = (u')-'u = x-'. We now have (xy)' = (xy)-' = y - ' x - ' = y'x' = (yx)'. Thus xy = yx for x, y E K and the result follows. I (7.22) LEMMA Let P E Syl,,(G) with p # 2 and suppose P is T.I.F.N. in G and is nonabelian. Let N = NG(P) and let 1 < Z c Z(P) with Z Q N. Let % = ( x E Irr(N)IZ $ ker x}. Then % is coherent. Note By Corollary 7.15, the lemma is also true if P is abelian except inthedegeneratecasethat131 = LInthatcaseZ = PandIPI = 1N:PI 1.

+

Proof of Lemma 7.22 Let e = 1N:PI. By Lemma 7.21, e is odd and thus IN I is odd. Since Z > 1, 3 # Let t E 3 have minimum possible degree

a.

T. I. sets and exceptional characters

115

andletX0 = { x ~ % I x ( l= ) 5(1)}.SinceINJisodd,f # 5andthusl%oI 2 2. By Corollary 7.15, Sois coherent. Now define sets Zifor i > 0 by 3 = %i-l u {xi},wherethexiarechosenfrom 3- Tosothatxi(l)I~ ~ + ~ ( l ) . E ... E Xk= X We use induction on i to show that Thus Z0 E is coherent. Now the degrees of the x E % are all of the form ep" and so 5( 1)1 x( 1)for all x E %. It follows by Theorem 7.14 that in order to show that X i is coherent, it suffices to show for i 2 1 that

C

X(l)xi(l) <

~(1)'.

XEbl-I

We have

IN1 =

c

x(l)2 ~~lrr(N)

=

J N : Z I+

1 x(1)2 X E b

and so I P: 2 I divides EXES x( 1)2.Thus e2 IP: 2 I

x( 1)2.

divides X E b

If x E %, then x = SN for some 9 E Irr(P) and 9( 1)2 I I P: 2 I by Corollary 2.30. divides e2 IP:ZI and hence ~ ~ (divides 1 ) ~ CXEb ~(1)~. Thus ~ ( 1=) e29(1)2 ~ Also, xi(1)2 I xA1 )2 for j 2 i and we conclude that xi(1)2divides - I x( 1)2 . In particular, since 25(1) < p5( 1) I xi(l),we have

1,

XXl)xi(l) < Xi(112

1- ~ ( 1 ) ~

XEbi

and the result follows.

1

I

(7.23) LEMMA Let N E G and 9 'E Irr(N). Let a: Z[9']" -+ Z[Irr(G)]" be a linear isometry and let 3, Y c_ Y with Tn % = 521 and (%, a) and (%, a) coherent. Let 0 and z be isometries on Z[%l and it[%], respectively, where 0 and z are as implied by coherence. Then (a) [xu, 17 = 0 for all x E % and q E Y. (b) Suppose xo E %and qo E Y with (Xo(1ho - rlo(1)xo)"= Xo(1)?or - ?O(l)XO"* Then % u

is coherent.

Proof For p, v E 9, write A@, v) = p( l)v - v( l)p E Z[9']". Let 6, E = k 1 so that 6x" and ~ q 'E Irr(G) for x E % and q E %. To prove (a), suppose [xu, q7 # 0. Then 6x" = 9 = eqr for some 9 E Irr(G). Pick x1 E % - { x } and q l E Y - { q } and write t,b = 6x1" and 5 = &qlr.Then t,b # 9 # 5 and

Chapter 7

116

we have 0 = [A(x, Xl), A h

11)l= [ 4 x ,

x 1 Y 9 Ah9

rlJl

&X(X(1)$- Xl(l)@,(rl(1)t - rll(l)@l EJ(Xl(l)ill(l)+ x(lh(l)C$, 51). Since [$, t] 2 0, this is impossible, and (a) is proved. Now fix xo E % and qo E 9 and assume A(xo, qoY = xo(1)qo' - ~ o ( ~ ) x o " . Define*on%u%byX* = X"forxEXandq* = q'forqE%andextend*by linearity to Q[% u Y]. It follows from (a) and the fact that 0 and z are isometries, that * is an isometry and *: Z[% u Y] + Z[Irr(G)]. It thus suffices to show that * agrees with the linear extension of 01 to Q[% u Y]". However, * agrees with 01 on Q[%]", Q[Y]" and A(xo, qo). Adimension count shows that these span Q [ 2 u Y]' over Q and the result follows. = =

(7.24) THEOREM (Sibley) Let P E Syl,(G) be T.I.F.N. in G with p # 2. Let N = N,(P) and Y = (xEIrr(N)IP $ kerx}. Let e = 1 N : P I . Then one of the following occurs. 9 1= 1, P is elementary abelian, and [PI= e + 1. (a) 1 (b) Y is coherent. Proof If P is abelian, this is included in Corollary 7.18. Assume that P is nonabelian so that e is odd by Lemma 7.21, and thus IN1 is odd. Let 2 = P'nZ(P)> l a n d l e t Y = { q E Y l Z $ k e rq } . Let %= ( X E Y O J P ' G ~ ~ ~ X } , Then Y is coherent by Lemma 7.22 and every x E % satisfies x(1) = e. Since x # j for x E 2, we have % coherent by Corollary 7.15. Since Z s P', we have % n Y Most of the proof is devoted to showing that % u % is coherent. Let *:Z[X] + Z[Irr(G)] be a linear isometry which extends induction on if"%]".Similarly, let z be an isometry on Z[Y] which extends induction. Since % n Y Lemma 7.23 yields [x*, q7 = 0 for x E 9"and q E Y. Let 01 = (l/e) qEg q(1)q. Since e I q( 1) for q E Y, a is a character of N . For x E %, x* # fq' for any q E Y and Lemma 7.19(a) yields

=a.

=a,

( x * h = 9,

+ a,a,

where 9, is a generalized character of N such that [9, q] = 0 for q and a, E Q. Since all ~ ( 1are ) equal for x E % Lemma 7.19(b) yields that

9, - x + a,a is independent of x E 3. It follows that 9, - x = A and a, dependent of x E % and (x*)N = x A UO~. (1) The next several paragraphs are devoted to proving that a E Z. (X*)N -

x

E

Y

=

+ +

=a

are in-

T. I. sets and exceptional characters

117

Since Z is contained in the kernel of x E 3"and of every irreducible constituent of A, we have for z E Z that ~ ( z=) X ( 1)and A(z) = A( 1) and hence Equation (1) yields

x*( 1 ) - x*(z) = a(a(1) - a(z)).

+

We also have pN = PN/Z eu where pN and PN/Z are regular characters. F o r z E Z - (1)wethushave

lNI

= PN(l)

- PN(Z)

= e(a(l) - a(z))

and hence

(2)

x*(l) - x*(z) = alNl/e = alPI.

WenowcompareX*(l)andX*(z)inadifferentway.Let X o , X , , ... bethe conjugacy classes of G, numbered so that (a) Jf-0 = (11; (b) X i n Z # @ i f f i I r ; (c) X i n P f @ i f f i < s . Note that since P is a T.I. set with N = N(P),we have X i n P is a conjugacy class of N for i I sand X i n P c Z -a N for i I r. Write K i E C[G] for the class sum corresponding to X i and K i K j = aij,,Kp,where aijuE Z. Now fix i, j I r. Let $ E Irr(G) and let o(K,) = $(x) I X,,I/$( 1)for x E X, as in Chapter 3. Let R c C be the ring of algebraic integers so that o(K,) E R . Write $(1) = mp' with p ,i'm, and let 4 = IP l/p*. We have

1

o ( K i b ( K j )=

1aijfio(Kp)* P

For p > s, we claim that w(K,) E qR. From the T.I.F.N. property and it follows that IPI divides IX,,l and thbs the fact that X,, n P m(o(K,)/q)E R. Since also q(w(K,)/q)E R and (m, q) = 1, it follows that o ( K , ) / q E R as claimed. Thus

=a,

o ( K i ) w ( K j= )

S

1 aijro(K,)mod qR. p=O

Now let r < ,u I s and x E X,, n P . Let C = C,(x) and let R = {(u, u)lu E X i , U E X j , uu = x}. Then aiju= IRl and C acts on R by (u, u)' = (u', uc). If c E C - (1) and (u, u)' = (u, u), then u, u E C&) 5 P and so u e X i n P s Z and similarly U E Z .However uu = X E X and , X,, n Z =@. This contradiction shows that all orbits of C on R have size IC( and thus I CI divides aijp.

Chapter 7

118

We have C,(X) 5 P and so C,(X) = C and I P : C I divides I X , 1. It now follows that rn(aijpo(Kp)/q) E R and since q(aijpo(K,)/q) E R, we have aijrw(K,)E qR and r

w(Ki)w(Kj)=

aijpw(Kp) mod qR. p=O

Now suppose $ E Irr(G) is such that 1(1 is constant on 2 - { l}. Then all o ( K p ) are equal (to w, say) for 1 5 ,u 5 r and w(Ko)= 1. Also write aij = u i j p ,so that

I:=,

wz

= aijO+ aijw mod qR.

Since 1 N I is odd, the nonidentity elements of 2 are not conjugate in N (and hence not in G either) to their inverses. Thus a, = 0. Assume X 2is the class of inverses of Xl so that alz0= lX1I = IGl/lPl. We thus have

(3)

allw

E

w2 = lGl/lPl

+ a 1 2 wmod qR.

In the special case that $ = l,, we have w = lGl/lPl and q = IPI so that a1,lGI/IPI and a,,

= lGl/lPl + a12lGl/lPl mod IPI

= 1 + a12mod IPI. Now (3) yields (1

+ aI2)w= lGl/lPl + a12wmod qR

and (4)

w

= IGl/lPI mod qR.

We apply this to the character $ = EX* E Irr(G) with E = f1 and Note that $ is constant on 2 - { l} by (2) and thus (4)applies. We have for z E 2 - (1) that x*(z)IG:PI/x*(l) = w

x E X.

= IG:PI mod qR.

Since I PI divides qx*( l), this yields x*(z)IG:PJ = X*(l)IG:PI mod JPIR. Now (2) yields IG:PI-IPla = IG:PI(x*(l) - x * ( z ) ) E lPlR and hence I G : P ~ u E RSince . also lPla = x*(l) - x*(z)ERand (IG:PI, [ P I ) = 1, we have a E R n Q = Z as desired. Now let x, x1 E X and q E 3.Note that 1) = e divides q(1) and we write c = q(l)/e. Let cp = cx - q E Z [9']".We have

x(

CqG,xi*] = Ccp,

( ~ 1 * ) ~= 1

[v, X I

+ A + am].

T. I. sets and exceptional characters

Since [x,a] = 0 = [q, x1

119

+ A], this yields

ccx, x1 + A1 - 4% a]. However, [q, a] = c by definition of a and since a E Z, we have c [ [qc, xi*]. This calculation also shows that ccpc,

x1*l

=

CvG,x*l for x1 #

=c

+ CvC,Xi*]

x. This yields cpG = c(l

+ b)x* + cb C <* - r, 3;€ + x 5E

where b E h and [r,5*] We also have

= 0 for

all 5 E X.

CVG,vCl =

cv, cp1 = 1 + c2

and thus there are two possibilities: (i) b = 0 and [r,r] = 1 ;or (ii) b = - 1,131= 2 and [I-, 1-1 = 1. In situation (ii),we can replace * by ** where x:* = - xz*andxt* = - X1 * for X = {xi, x2}. The result of this change is to put us into situation (i). We thus assume that (cx - q)' = cx* - r. Now let q1 E Y - { q } so that

cv, ?(l)vll

?1(1) =

-

rlt(1)rI

rl(1)Vri' - 1 1 ( l ) r 7 Vl(1)rl' - v1(1)1717.

=

CVG9

=

D-9

Thus again we have two possibilities: (i) (ii)

= q'; or

r = -qlr

and q(1) = ql(l).

If r # q', then ql' = -r for every q1 E Y - { q } and thus IYyl = 2 and Y = {q, q l } with q(1) = q l ( l ) . In this situation, we can redefine T and thus we may assume that (i) occurs. Thus (cx - q ) G

=

cx* - q'

and hence X u ?Y is coherent by Lemma 7.23(b). ~(1)'= IN1 - IN:ZI To complete the proof, we observe that EVES/ and so is divisible by 1 P : 2 I and hence by ez 1 P :Z I. Let $ E Y - (Xu Y). Then as in Lemma 7.22, we have $ = sN,with 9 E Irr(P) and 9(1)' II P : Z I. Thus

Chapter 7

120

Let x E S so that ~ ( 1 = ) e and $(1) 2 px(1) > 2x(1) since $ 4 3 and so 9(1) 2 p . Thus 2x(1)*(1) < W)’ 5 5(1)’.

c

CEZU9

Since 5? u Y is coherent and x(l)l$(l) for all $ E 9, repeated application of Theorem 7.14 yields that Y is coherent and the proof is complete. I

(7.25) THEOREM (Feit-Sibley) Let K E G be T.I.F.N. with N = N,(K), e = IN:KI and Y = { x E Irr(N)IK $ ker x}. Then one of the following occurs. (a) IYI = 1, I K I = e 1 and K is an elementary abelian p-group. (b) K E Syl,(G) and IK:K’I < 4e’. (c) Y is coherent.

+

Proof If K is abelian, either (a) or (c) occurs by Corollary 7.18. Assume then, that K is nonabelian and so e is odd. If 1K:K’I = e + 1, then K / K ‘ is an elementary abelian p-group and thus K is a p-group. Since e + 1 is even p = 2 and (b) follows. Assume therefore, that I { x E Y I K’ E ker x } I 2 2. This set is thus coherent by Corollary 7.15. Let L Q N with L E K’ minimal L G ker x } is coherent and assume that Y is not such that 0 = { x E 9’1 coherent so that L > 1. Let M Q N be such that L / M is a chief factor of N . Since

{ x E YI M

E ker

x}

is not coherent, repeated application of Theorem 7.14 yields $ E Irr(N) such that

2eJ1(1) 2

c~(1)~ 1N:LI - 1N:KI =

= e(lK:LI

- 1)

xEZ

and M c ker $. Let Z / M = Z ( K / M ) .Then Z n L > M and hence L E 2. Also $ = GN for some 9 E Irr(K) and 9(1)’ II K : 21 so that $(1)’ I e’ I K : 21. Thus 4e41K:2I 2 4e2J1(1)’ 2 e’(1K:LI - 1)’. Now write a = IK :L ( and b = 12 : LI.We have blK :ZI = IK :LI = a and

4e’a = 4e’bIK:ZI 2 b(lK:LI - 1)’= b(a - 1)’. If 4e’ I b(a - 2), this would yield ba(a - 2) 2 b(a - l)’, which is not the case. Thus b(a - 2) < 4e’. If Z = L, then Z ( K / M ) = L / M is a p-group for some prime and thus KIM is a p-group. Since M c K‘, it follows that K is a p-group. By Theorem 7.24, we have p = 2. Also, a - 2 < 4e’ so that 1 K : K ’ )I J K :L J = a 5 4e’

+ 1.

Problems

121

Since 1 K : K' I is a power of 2 and e > 1 is odd, we have 1 K :K' I < 4e2 and (b) holds. If 2 > L, then we must have e J ( J 2 : L J- 1) and b 2 e + 1. Also, KIM is nonabelian and thus 2 < K and e I ( I K : 2 I - 1). Thus

+ l)b 2 (e + 1)' 2 e2 + 2. 4e2 > b(a - 2) 2 (e + l)e2 2 4e2 since e 2 3. This a 2 (e

Thus proves the theorem.

I

contradiction

We remark that situation (b)can actually occur with Y not coherent. The inequality I K :K'I < 4e2 can be sharpened. In fact, the groups in which (b) occurs have been classified. Problems

(7.1) Let N 4 G , H E G, with NH = G and N n H following are equivalent: (a) (b) (c) (d) (e) (f)

=

1. Show that the

C,(n) c N for all 1 # n E N; C,(n) = 1 for all 1 # n E N ; C,(h) c H for all 1 # h E H ; Every x E G - N is conjugate to an element of H ; If 1 # h E H , then h is conjugate to every element of Nh. H is a Frobenius complement in G.

Note Problem 7.1 does not involve characters. It is included to acquaint the reader with some elementary properties of Frobenius groups. Much deeper information is known. (7.2) (Wielandt) Let H c G with M a H and suppose that H n H" G M whenever x H . Show that there exists N -a G with N H = G and N n H = M.

+

Hint Note that H theorem.

-M

is a T.I. set. Mimic the proof of Frobenius'

(7.3) Let H G G and 5 E Irr(H). Suppose (5 - t(l)l,)' = 9 and [ S , 93 = 1 + <(1)2. Show that there exists N a G with N n H = ker 5 and every x E G - N conjugate to some element of H . (7.4) Let H < G and suppose induction to G is an isometry on Z[Irr(H)]". Show that H is a Frobenius complement in G. (7.5) Let N c G and Y s Irr(N). Assume that induction to G is an isometry on Z[Y]" and 1 9 ' 1 2 2. Suppose Y o= {< E Y 15 is extendible to G } # 0.

Chapter 7

122

Let d = g.c.d.({((l)I( E Y o } )and assume d(((1) for all t E Y . Show that Y o= Y ,Y is coherent, and that every ( E Y has a unique extension to G . (7.6) Let P E Syl,(G) with I PI = p , C,(P) = P, and I NG(P): PI = e. Show that G has at most e + (p - l)/e irreducible characters with degree not divisible by p . Hints Note that P is T.I.F.N. in G if e > 1. If e < p corresponding set of exceptional characters of G . Show where 1 # x E P. If ~ ( x= ) 0, then p I x( 1).

-

1, let & be the

xXElI ~ ( x1’ ) 2 p - e,

Note It is a consequence of some of R. Brauer’s deep results in block theory that in the situation of Problem 7.6, the upper bound is always attained. (7.7) In the situation of the preceding problem, assume that G has exactly e + (p - l)/e irreducible characters of p’-degree. Let P = (x). If x E Irr(G) ) - L O , or 1 is not exceptional (in particular if e = p - l), show that ~ ( x= and that ~ ( 1 = ) ~ ( xmod ) p . If e < p - 1 and & is the corresponding set of ~ ( x ) .Show that cr = f 1 and ~ ( 1 = ) exceptional characters, let CJ = -eamodpforxE&.

xXEd

Note The results of Problem 7.7 together with the equation

1

x(1)x(x) = 0

x~Irr(G)

and the facts that x(l)l I GI and ~ ( 1=) IGI ~ provide a powerful tool for computing the degrees of the irreducible characters of a group which satisfies the hypotheses of Problem 7.6.

9 1> 2. Prove that (7.8) In the situation of Lemma 7.13, assume that 1 and fare uniquely determined by (9, z).

*, E,

(7.9) Let K E G be T.I.F.N. Let x be a (possibly reducible) character of G which is constant on K - { l}.Assume that 9’= {$ E Irr(N,(K))( K $ ker $} is coherent. Show that the multiplicity with which each exceptional character appears in x is proportional to its degree. (7.10) Let K be T.I.F.N. in G and assume that Y = {$ E Irr(N(K))IK $ ker $}

is coherent. Let 9 be the corresponding set of exceptional characters ker x. Show that M n K = 1. of G . Let M =

nXEY

(7.11) In the situation of Problem 7.10, show that either M x E 5 or else M K Q G.

=

ker x for all

Problems

123

Hint Suppose x E Y and U = ker x > M . Then U / M is a n-group where TI is the set of prime divisors of I K I and so U 5 M K . Use the Frattini argument. (7.12) Let K be T.I.F.N. in G and let e = IN(K):K I. Suppose that G has a faithful irreducible character of degree <2e. Show that K is abelian. (7.13) In the situation of Problem 7.12, assume that K an elementary abelian p-group.

+ G. Show that K is

Hint Let A be a linear character of K with 1 K : ker ill = p . Show that there exists faithful x E Irr(G) with x(1) < 2e and [xK, 11 # 0. Use Problem 7.11. (7.14) Let G be a simple Zassenhaus group in which the stabilizer of two points has even order e. Let the degree of G (the number of points permuted) be k + 1. Show that k is a prime power.

Hint Count involutions to prove that G has a nonprincipal irreducible character of degree < 2e. Note Problem 7.14 is true without the assumption that e is even. This is a theorem of Feit and the case where e is odd is proved in his book.

(7.15) Let N c G and Y c Irr(N). Suppose that z: Z [ Y ] " + Z[Irr(G)]" is 9 ' 12 3 and that for every x, @ E 9, we have a linear isometry. Assume that 1 (x(l)$ - t,h(l)x)' = a9 - b q for some cp, 9 E Irr(G) with a, b E Z. Show that (9,~) is coherent.

Hint

Suppose x, $, q E Y are distinct. Write

(x(1)$ - w ) x ) t = a9 - b, (vl(l)lc/ - W ) r ) ' = C P - dv with a, b, c, d > 0. Then exactly one of 9 = p or q = v holds. (7.16) Let N E G, Y c Irr(N) and suppose t:Z [ 9 ] ' -, Z[Irr(G)]" is a linear isometry. Let Y1,9, c Y such that Y1n Y 2 Assume either that 19, I 2 3 or that Yl = {$, x} with $(l) = ~(1). Show that (Yl u Y,, z) is coherent.

#a.

Hint Let&: Yi -, Irr(G) and ci = +_ 1 be such that cifi defines an appropriate extension of z on Z[Yi]. If IYil 2 3 for i = 1, 2, use the hint for Problem 7.15 to show that c1 = c2 andfl(x) = f2(x) for x E Y1 n 9,.

(7.17) Let K E G be T.I.F.N. with N = N,(K) and Y = (x E Irr(N))K $ ker x}. Suppose that $ is an exceptional character of G corresponding to that the values of $ and x generate the same field over Q.

x. Show

Chapter 7

124

(7.18) Let K , , K , E G be T.I.F.N. with N i = NG(Ki)and

9, = {x E Irr(Ni)1 K i $ ker x} and assume that the Y i are coherent with bi the corresponding sets of exceptional characters. Suppose 8, n 8, # 0. Show that bl = 8, and K , is conjugate to K,. Hint

To show that K z

=

Kle, it suffices to show that xgE K , for some

x E K,,x # 1. Use the fact that the K i are nilpotent to prove this.

(7.19) Suppose IGI is odd and that C,(x) is abelian for every 1 # X E G. Assume that G is neither abelian nor a Frobenius group and show that every nonprincipal x E Irr(G) is exceptional for some T.I.F.N. subgroup. Note A group with all C(x) abelian for 1 # x E G is called a CA-group. Problem 7.19 is a step in M. Suzuki’s determination of all simple CA-groups. (7.20) Let K E G be T.I.F.N. with lKl even. Let N = N,(K) < G and Y = {$ E Irr(N)IK $ ker $}. Assume that Y is coherent. By carrying out the following steps, show that G has a nontrivial normal subgroup of odd order. (a) Show that G has a unique conjugacy class of involutions and that N contains exactly e = IN : K 1 involutions. (b) Let $ E Y and let $* be the corresponding exceptional character of G. For 1 # x E K we have $*(x) = &I&) + a,, ,where a,, E Zis independent of x. Show that VZ($*)

=

w($) + ($*(I)

- W)- a,)/lKI

where v, is as in Theorem 4.5. (c) Show that ($*)K = I//K + a#lK. (d) Show that there exists $ E Y with ker($*) # 1. (e) Complete the proof. Hints (a) See the hint for Problem 4.11. (b) Let d,= { x E G I # ~ 1 , ~ ’= l}; d , = {x E GlxZ # 1, x is conjugate to an element of K } ; d 3 = G - (diu dz). Compute $*(xz) separately on each of d,, d,,and d 3 .Note that C X d 3 $*V) = C x s d 3 $*(XI. ( 4 Let m,,= W*U) - 4 ( 1 ) - ~ * ) / l K l . Use (b) to study the behavior of mJ,as $ varies over Y and conclude that m,,= 0. Use the fact that m,,is proportional to $(1). (d) Note that K is not elementary abelian.

1

Problems

125

Notes A corollary of Problem 7.20 is that a simple Zassenhaus group of odd degree has degree 1 + 2". This is another special case of Feit's theorem. There do exist simple groups with T.I.F.N. subgroups of even order. If G = SL(2, 29, with n 2 2, then the Sylow 2-subgroup K of G is T.I.F.N. It is elementary abelian of order 2". In this case, coherence fails because IN(K):KI = 2" - 1 and the set Y contains only one character. If G = Sz(2") with odd n > 1 [the Suzuki simple group of order

(2" - 1)(2'")(22" + l)], then again the Sylow 2-subgroup is T.I.F.N. Here, case (b) of Theorem 7.25 holds. (7.21) (Sibley) In the situation of Theorem 7.20, let e = I N : K ( and k = 1 K I. If m is as in 7.20 (b), show that

-el(&

-t 1) < m

< e/(& - 1).

In particular, if K is not an elementary abelian p-group, show that m Hints

= 0.

Let 1 # x E K and use the inequality

where d is the set of exceptional characters. Derive that

2m -t e > m2(k

-

l)/e

(7.22) Let K c G be T.I.F.N. with N = N(K) and Y

=

{XEIrr(N)IK $ ker x }

coherent. Let I,$ E Irr(G) with K $ ker I,$. (a) If $ is nonexceptional, show that IKl I $(1) (b) If I,$ is exceptional, show that I K I < $( 1)'.

+

+ 1 - [+K,

lK].

Hints (b) Reduce to the case that t,bN = x A, where x E Y and either A = 0 or K c ker A. If A # 0, appeal to Problem 7.21. If A = 0, pick an irreducible constituent, q # 1, of t+b$ and apply part (a) to q.

8

Brauer’s theorem

There are many ways in which class functions of a group can be constructed. For instance, we could define 9(g) = ING((g))l or $@) = 1 if g2 is conjugate to some fixed x E G, and 9(g) = 0 otherwise. It is perhaps too much to expect that such arbitrarily defined functions will turn out to be characters. (For instance, in the second example given, 9(1) = 0 if x # 1.) One might hope, however, that a well chosen 9 will be a generalized character, that is, 9 E Z[Irr(G)], or at least that it is an R-linear combination of irreducible characters for some specified ring R with Z E R E C. (We denote the set of these R-generalized characters by R[Irr(G)].) How can one decide if a given class function 9 is an R-generalized character? Of course, if one knows Irr(G), the answer is easy: simply check that [9, x] E R for all x E Irr(G). A more typical situation is that one has enough information about some family M of subgroups of G so that it can be shown that 9, E R[Irr(H)] for every H E M.The main point of Brauer’s theorem is that for suitable families M the last relationship is sufficient to guarantee that 9 E R[Irr(G)]. For instance, this will be true if M is the collection of all nilpotent subgroups of G. (8.1) DEFINITION Let R be a ring with Z E R E C and let M be a family of subgroups of G. (a) BR(G,M )is the set of class functions 9 of G such that 9, E R[Irr(H)] for all H E M. (b) 9 R ( G ,M )is the set of R-linear combinations of characters t,bG for t,b E Irr(H), H E M. 126

Brauer’s theorem

127

If R = Z,we delete the subscripts and write W(G,X ) and 9 ( G , X ) . (8.2) LEMMA Let X be a collection of subgroups of G and let R be a ring with Z G R c C. Then

(a) 9 ( G , X ) E XR(G, X ) E R[Irr(G)] c WR(G, X). (b) WR(G,X ) is a ring in which 9 R ( G ,X ) is an ideal. (Addition and multiplication are pointwise.) Proof The containments in (a) are all obvious from the definitions. To prove (b) observe that Z[Irr(H)] is a ring for every H E X .Since Z G R, it follows that R[Irr(H)] is a ring. If cp, 9 E WR(G,X ) ,then ( ( P ~ ) H= ~ P 9,H E R[Irr(H)l

and hence cp9 E WR(G, X ) ,which is, therefore, a ring. To prove that 9 R ( G , X ) is an ideal, we use the fact that if M is a class function of H E G and p is a class function of G , then (aPH)‘ = a‘p. This is immediate from the definition of induction (and appears as Problem 5.3). Let (PEX,(G, 2) and 9 e W R ( G , X ) . Then cp = (1,9(~,)‘ with E R[Irr(H)]. NOW

xHEz

( ~= 9

CH (I,9(H))‘s

=

CH (I,9(H)QG.

Since I,9(H)9H E R[Irr(H)], it follows that (p9 E X R ( G ,X ) and the proof is complete. I It follows from Lemma 8.2 that in order to prove that YR(G, X ) = R[Irr(G)], it suffices to show that 1, E 9 ( G , X ) .Furthermore, if this can be done for some family X , it follows that

XR(G, X ) = R[Irr(G)] = 9 R ( G , X ) for all R with Z G R G C. (8.3) DEFINITION (Brauer) A group E is p-elementary (where p is a prime) if E is the direct product of a cyclic group and a p-group. We say that E is elementary if it is p-elementary for some prime. (8.4) THEOREM (Brauer) Let Z

cR

G

C,where R is a ring.

(a) ‘ Aclass function 9 of G is an R-generalized character iff 9, E R[Irr(E)] for every elementary E E G. (b) Every x E Irr(G) is a Z-linear combination of characters of the form AG for linear characters A of elementary subgroups of G . Let 8 be the set of elementary subgroups of G. Statement (a) of Brauer’s theorem is exactly the assertion that R[Irr(G)] = WR(G, 8).This part of the result is often called the “characterization of characters.”

Chapter 8

128

Every E E b is nilpotent and hence is an M-group by Corollary 6.14. If cp E Irr(E), then cp = IIE, where II is a linear character of some E , G E. Since

Eo E b and cpG =,'A it follows that Y(G, 8)is exactly the set of Z-linear combinations of I' for linear I E Irr(E) with E E b. Therefore, statement (b) of Brauer's theorem amounts to the assertion that Z[Irr(G)] = 9(G,8).This is often called the "theorem on induced characters." By Lemma 8.2, both parts of Brauer's Theorem 8.4will be proved when we show that 1' E Y(G, 8).We work toward this goal. The proof given here is based on an idea of B. Banaschewski. It is suggested that the reader also consult the Brauer-Tate paper for another approach. (8.5) LEMMA (Banaschewski) Let S be a nonempty finite set and let R be a ring of Z-valued functions defined on S (with pointwise addition and multiplication). If the function 1, with constant value 1 does not lie in R , then there exists x E S and a prime p , such that p dividesf(x) for everyf E R .

Proof For each x E S , let I, = {f(x)l f E R } . Then I, is an additive subgroup of Z. If for some x E S , we have I, < Z,then I , E (p) for some prime and the result follows. Assume then, that I, = Zfor every x E S. For each x, we may therefore choosef, E R withf,(x) = 1. Thusf, - 1, vanishes at x and (f,- 1,) = 0.Expanding this product yields an expression for 1, as a linear combination of products of the functions f,. Thus 1, E R. I

nx,,

To obtain a ring to which we can apply Lemma 8.5, we consider permutation characters of G , that is, characters of the form (1,)' for subgroups H E G. (8.6) LEMMA Let H , K E G . Then (l,)'(lK)' U c H and integers a, 2 0.

=

aU(lU)'for subgroups

Proof Write 9 = (lK)' so that (l,)'(lK)' = 9(1,)' = (9,)' by Problem 5.3. By Lemma 5.14, 9 is the permutation character of G acting on the set of right cosets of K.It follows that 9, is a permutation character of H and hence 9, = aU(l# for subgroups U c H and integers a, 2 0. Since ((l,),)' = (lU)', the result follows. I (8.7) COROLLARY The set of Z-linear combinations of characters of G of the form (1,)' is a ring P(G). Let X be a collection of subgroups of G with the property that if K c H E X , then K E X . Let P(G, X ) denote the set of Z-linear combinations of characters of the form (1,)' with H E X . Then P(G, X ) is an ideal of P(G).

To apply Corollary 8.7, we define a class of groups more general than elementary groups.

Brauer's theorem

129

(8.8) DEFINITION A group H is p-quasi-elementary if H has a cyclic normal p-complement for the prime p . We say that H is quasi-elementary if it is p quasi-elementary for some p . Note that subgroups of p-quasi-elementary groups are themselves p quasi-elementary. Thus in the notation of Corollary 8.7, P(G, X ) is closed under multiplication, where X is either the set of all quasi-elementary subgroups of G or is the set of all p-quasi-elementary subgroups of G for some fixed prime p . (8.9) LEMMA Let x E G and let p be a prime. Then there exists a p-quasielementary subgroup, H E G, such that (lH)'(X) is not divisible by p . Proof Let C be the p-complement in the group (x). (Possibly C = 1 or C = ( x ) . ) Let N = NG(C), so that x E N . Since ( x ) / C is a p-group, we may choose H / C E Syl,(N/C) with ( x ) G H. Then C is a normal p-complement for

H and H is p-quasi-elementary. Now (lH)' is the permutation character of G acting on the right cosets of H, and so ( l J G ( x ) = I{HyIHyx = H y , y € G } ( . If H y x = Hy, we have x y - ' E H and hence C y - G H. However, C is the unique p-complement in H and hence CY-' = C and y E N . We therefore need to count the number of fixed points in the action of ( x ) on the cosets of H in N . Since C 4 N , and C E H, we see that C is in the kernel of the action of N on the cosets of H in N . Since ( x ) / C is a p-group, it follows that the number of nonfixed cosets is divisible by p and hence (lH)'(x) 3 IN :HI mod p . By the choice of H, p$I N : HI and the proof is complete. I (8.10) THEOREM (L. Solomon) Let X be the set of quasi-elementary subgroups of G and X pthe set of p-quasi-elementary subgroups for some prime p . Then (a) 1G E P(G, XI. (b) mlG E P(G, X p )for some m E Zwith p y m . Proof By Corollary 8.7, P(G, X ) is a ring of Z-valued functions on G . If l G $ P(G, X), then by Lemma 8.5, there exists x E G and a prime p with plcp(x) for all cp E P(G, X ) .This contradicts Lemma 8.9 and so (a) is proved. For (b) let R = {cp + nplGl cp E P(G, X,), n E Z}. Then R is a ring. If there exists x E G and a prime q, with q I cp(x) for all cp E R, then since P I G E R, we have q = p and thus we have a contradiction to Lemma 8.9. By Lemma 8.5, we conclude that l G €R and hence ( 1 - np)lGEP(G, X p )for some n EZ. This completes the proof. I

Only part (a) of Theorem 8.10 is needed to prove Brauer's theorem; however, part (b) is useful for certain refinements of the result. Recall that we

Chapter 8

130

need 1, E Y(G, 8).The method of proof is essentially to use 8.10 to reduce the problem to quasi-elementary groups. We need a lemma. (8.11) LEMMA Suppose G = CP where C u G, P is a p-group and ~ $ 1CI . Let 1 be a linear character of C which is invariant in G and suppose C,(P) E ker 1.Then 1 = 1,.

Proof Let K = ker 1.Since 1is linear, it takes on distinct constant values on the distinct cosets of K in C . Since 1is invariant under the action of P,it follows that P normalizes each coset K x for x E C . In the conjugation action of P on K x , the number of moved points is divisible by p . Since ~ $ 1K 1, we have pYlKxl and thus Kx n Cc(P)# Qr. Since C,(P) c_ K by hypothesis, we conclude that K x = K and thus K = C and 1= 1, as claimed. I Proof of Theorem 8.4 As was remarked following the statement of the theorem, it suffices to prove that 1, E Y(G, 8).We use induction on I GI. We may therefore suppose that 1, E Y(H,8,) whenever H < G, where 8, is the set of elementary subgroups of H . Thus Z[Irr(H)] = Y(H, a,) for these H by Lemma 8.2. By transitivity of induction (Problem 5.1), it follows for cp EY(H,a,), that cpG E Y(G, a,) c Y(G, 8).Thus for all H < G and cp E Irr(H), we have cpG E Y(G, 8)and it will suffice to show that 1, is a Zlinear combination of characters induced from proper subgroups. By Solomon's Theorem 8.10(a) we are done if G is not quasi-elementary and so we may assume that G has the cyclic normal p-complement C . Let P E Syl,(G) and 2 = C,(P).Since we may clearly assume that G is not elementary, we have Z < C and E = PZ < G . Write (lE)' = 1, E,where z is a possibly reducible character of G. We shall show that every irreducible constituent of E is induced from a proper subgroup and thus 1, = (1,)' - z E Y(G, 8 )as desired. Let x be an irreducible constituent of E.Now CE = G and C n E = Z so that

+

1,

+ z,

= ((lE)G),

= (lZ)C

by Problem 5.2. Thus 1 = c(lz)c, 1,l = c1,

+ =c, 1c1

and hence [E,, l,] = 0. Thus hc,l,] = 0. Let 1be an irreducible constituent of x,. Then 1 # 1,. However, Z 4 G and so Z E ker((1,)') and hence 2 E ker x. We therefore have 2 G ker 1 and thus by Lemma 8.11, 1is not invariant in G. Let T = I,&) < G . By Theorem 6.1 1, x = $' for some $ E Irr(T). The result now follows. I The following is a useful special case of part (a) of Brauer's theorem.

Brauer 's theorem

(8.12) COROLLARY (a) (b) (c)

131

Let x be a class function of G . Suppose

zEis a generalized character for every elementary E s G ;

cx,XI = 1;

X U ) 2 0.

Then x E Irr(G).

1

Proof Using (a) and Theorem 8.4(a), we have x = a t < for 4 E Irr(G) and uL:E Z.From (b) we conclude that at most one uL:# 0 and x = f for some 5 E Irr(G). Then (c) yields the result. I

<

Before going on to reap some of the harvest of applications of Brauer's theorem, we digress briefly to derive the ''local'' version of the result. In the Brauer-Tate proof, this is obtained as an intermediate result. (8.13) COROLLARY Let 8, be the set of p-elementary subgroups of G for some fixed prime p. Then mlGEy(G, 8,) for some m E Zwith ptm. Proqf By Theorem 8.10(b),there exists m E Zwith p t m such that m l Gis a Z-linear combination of characters of the form (1,)' for p-quasi-elementary H E G . Let 8, denote the set of elementary subgroups of H. If H is p-quasielementary, then E, E E, and by transitivity of induction, we have (Iff)' The result now follows.

E 9 ( G , 8,)

s 9(G,

I

We shall use Brauer's theorem to remove the solvability hypothesis from Theorem 6.25 on extending characters. To do this we need part (c) of the following result. [Part (b) will be used in Chapter 13.1 (8.14) LEMMA Let N a G and let z ~ I r r ( G )with $ E Irr(G). For each coset N g of N in G compute W g ) = U/IN I)

1 WZ). xENg

We have (a) If [$N, 91 # 0, then q E Irr(G/N). (b) If [ $ N , 91 = 0, then q ( N g ) = 0 for all 9. (c) If x = $, then q ( N g ) = 1 for all 9.

xN = SeIrr(N). Let

Chapter 8

132

Proof Conjugates of Ng in GIN all have the form Ngh = h-'(Ng)h for h E G. It follows that q is a class function on GIN. We may thus write q = 1b,<, where 5 runs over Irr(G/N) and b, E @. Now

<

<

where in the last expression, we view E Irr(G) and use the fact that as such, is constant on cosets of N. We thus have b, = [$, x t ] . Now by Corollary 6.17, x l E Irr(G) and the characters x< are distinct for distinct 5 E Irr(G/N). Therefore q = C.b,< satisfies q = 0 unless $ = x< for some 5 E Irr(G/N) in which case q = <. By Corollary 6.17, $ = xt for some iff [I)~, 91 # 0. The result now follows.

<

(8.15) THEOREM (Gallagher) Let N Q G with 9 E Irr(N) invariant in G. Suppose (9(1), 1 G : N I) = 1. Then 9 is extendible to G iff det 9 is extendible.

Proof The "only if" part of the assertion is trivial. Assume 1 = det 9 is extendible to p E Irr(G). If N c H c G with H/N solvable, then 9 is extendible to H by Theorem 6.25. By Lemma 6.24, there is a unique extension x ( ~ )such , that det(x(a,) = pH. Define the function x on G as follows. If g E G, let H = (N, g) so that H/N is cyclic. Set x(g) = x(,&). We shall use Corollary 8.12 to show that x E Irr(G). First we establish that x is a class function. If g, x E G, let H = (N, g), so that H" = (N, 9"). Define $ on H" by $(h") = ~ ( ~ ) for ( h h) E H. Since the map h I+ h" is an isomorphism and x ( ~E )Irr(H), we have $ E Irr(H"). Also, (det $)(h") = (det(x(H,))(h)= p(h) = p(h"),so that det $ = pHr.Furthermore, if n E N, we have $(n) = X(H)(nX-')= 9(nx-')= P ( n ) = 9(n). It now follows from the uniqueness of x ( ~ that ~ ) ,xtHX) = $ and hence

x@") = x(Hx)(gX)= $(9") = X(H)@) = x(9) and x is a class function. Next, let E E G be elementary and set H = NE so that H/N E/(N n E) is nilpotent and xQ) is defined. We shall show that the restriction xH = x ( ~ and thus xE = (x& which is a character of E. If g E H,then (N, g) = K c H. Clearly, (x(*JK is an extension of 9 to K and det((Z(&) = pK so that ( x ( H ) ) K = x(K). Therefore, x(g) = x(&) = x(H)(g), hence XH = x(H) as claimed, and xE is a character. Next, we compute [x,x]. For each coset Ng of N in G, we have

=

c Ix(x)12

X E N ~

=

1

xeNg

IX(H)(X)I2,

)

Brauer's theorem

133

where H = ( N , g) = ( N , x) for all x ~ N g Since . ( x ~ ~=, 9) ~~I r r ( N ) , Lemma 8.14(c)yields (xtH)(x)I2= I N ( and it follows that 1x(x)12 = I G : N I . I N I = IGIand[x,~] = 1. Finally, ~ ( 1 = ) 9(1) > 1 and hence x E Irr(G) by Corollary 8.12. Clearly, xN = 9 and the proof is complete. I

xxeNg

xxaG

(8.16) COROLLARY Let N Q G and 9 E Irr(N) with 9 invariant in G . Suppose (I G : N ],o(9)9(1)) = 1. Then 9 has a unique extension, x E Irr(G) with ( I G : N I, o(x)) = 1. In fact, o(x) = o(9). In particular, this holds if ( I G : N 1, I N 1 ) = 1. Proof This is immediate from Corollary 6.27 and Theorem 8.15. (See the proof of Corollary 6.28.) 1 We have already seen several conditions sufficient to guarantee that a character value is zero. The following is a powerful one.

(8.17) THEOREM Let x E Irr(G ) and suppose p$( I G I /x( 1)) for some prime p. Then x(g) = 0 whenever p I o(g). Proof Define 9 on G by 9(g) = x(g) if p$'o(g) and 9(g) = 0 if p(o@).We shall use Brauer's theorem to show that 9 is a generalized character. Let E c G be elementary. Since E is nilpotent, we may write E = P x Q, where P is a p-group and p$I Q I. If x E E and p$o(x), we have x E Q and so 9 vanishes on E - Q and 9, = xp. Let $ E Irr(E). We have

SinceIEl = IPIIQI,weconcludethat IPI[~,,$]EZ. Now let w = ' w x be the algebra homomorphism Z(C[G]) -+ C associated with x so that o ( K ) = x(g)I X I /x( l), where X is the conjugacy class containing g and K = C X E @[GI. We shall write w@)for the algebraic integer w(K), so that x(g) = x(lk&)/ I I = x(l)O(g) I cG(g) I/ I I

Now

Since P E CG(x)for x E Q, we have

Chapter 8

134

which is an algebraic integer. Since I P 1 [ Q E , $1 E Z,we have [ Q E , $3 E Q and thus ( I G I I Q I /x( 1))[19E,$1 E Z.Since I G 1 I Q 1 /x(1) E Z and is relatively prime to I P 1, we conclude that [QE, $1 E Z and hence QE is a generalized character of E. Brauer's theorem now yields that 9 is a generalized character of G . In particular Cs, xl E Z.Now [S, XI= (1/1 G I) C lx(g)I2 IpA'o(g)), and so 0 < [9,x] I[x,x] = 1. We conclude that [a, X] = [x,x] and

The result now follows. A character

I

x E Irr(G) is said to have p-defect zero if p does not divide

IGl/X(l)*

In Chapter 2 we discussed the question of what information can be obtained about a group from a knowledge of its character table. Since D 8 and Q8 have the same table, one cannot determine the orders of the elements in the various classes. Nevertheless, it is possible to determine the sets of prime divisors of these orders. To do this, we shall apply Theorem 8.4(a) in a situation where R # Z. If g E G , we shall use the notation n(g) to denote the set of prime divisors of o(g). LEMMA Let g E G and let n be a set of primes. Then there exist unique x, y E G with

(8.18)

(a) g = xy = yx; (b) n(x) c n and n(y) n n = 0. Furthermore, x, y E (9).

Proof Write o(g) = mn such that every prime divisor of m is in n and no prime divisor of n is in n. Then (m,n) = 1 and we have km + In = 1 for some k , 1E Z.Let x = g'" and y = gkm so that g = xy = yx and x, y E (9). Since x m = 1 = y", (b) follows. Then Now suppose g = uu = vu with n(u) E n and n(u) n n = 0. u c C(g) c C(x)and hence n(xu- ') c n(x) u n(u) c n. Similarly, n(y-'v) E .(u) u 4Y),

so that n(y-'u) n ~ ( X U - ' ) = @. Since y-lo = xu-', we have xu-l 1 = y-'v and uniqueness follows. I If g, n, x, and y are as in Lemma 8.18, we write x n = { p } , we write x = gp and y = gpf.

= g n and

=

y = gx,.If

Brauer 's theorem

135

(8.19) LEMMA Let 1 GI = mn with (m,n) = 1 and let n be the set of prime divisors of m. Let g E G, with g = g,., and define 9 on G by

9(x) =

{i

if x,. if x,.

is conjugate to g, is not conjugate to g.

Let R be the ring of algebraic integers in Q [ E ] ,where root of 1. Then 9 is an R-generalized character.

E

is a primitive nth

Proof Let E c G be elementary. It suffices to show that Q E E R[Irr(E)]. Since E is nilpotent, we may write E = H x K , where H is a n-group and K is a n'-group. Then for x E E, we have x = uv, u E H,u E K and thus u = x, and v = x,.. If g is not conjugate to any element of K , then QE = 0 and there is nothing to prove. Suppose then, that g l , g 2 , . . . , gr are the distinct elements of K which are conjugate to g. Then the elements x E E with x,. conjugate to g are exactly the elements of the cosets Hg,, 1 Ii I r. Let $ E Irr(E). By Theorem 4.21, we have $ = 5 x r] for some 5 E Irr(H) and r] E Irr(K). We have

Since

xuEH

= 0 unless

5

=

l H , we have either

Now IEl = I K I IHI and IK I In and thus [ Q E , lies in R . The result now follows. I

[QE,

$1 = 0 or

$1 is a multiple of 1r1(9i)which

(8.20) THEOREM Let R be the ring of algebraic integers in Q [ E ] , where E is a primitive I GI th root of 1. Let p E Z be a prime and let I be a maximal ideal of R with p E I . Let x, y E G. Then x,, and y p , are conjugate in G iff ~ ( x=) ~(y) mod I for every x E Irr(G). Proof Write I G I = pan, with p t n , and define 9 on G by 9(g) = n if g,, is conjugate to x,. and 9(g) = 0, otherwise. By Lemma 8.19, 9 is an R-generalized character of G. ) ~(y)mod I for every x ~ I r r ( G ) .It follows that 9(x) Suppose ~ ( x = = 9(y) mod I . Now 9(x) = n. If 90.1) = 0, then n = 0 mod I and n E I . Since p E I and p y n , it follows that 1 = (p, n) E I and this is a contradiction. Thus 9(y) = n and hence y,. is conjugate to x,,, as desired. ) x(g) Conversely, we may suppose x,, = g = y,, and we show that ~ ( x = = ~ ( ymod ) I . Write x<,, = Ai, where the Ai are linear characters. Since g E (x), it suffices to show that A(g) = A(x)mod I for linear A.

1

Chapter 8

136

Now x = gh, where h = x, E (x) and A(x) = A(g)A(h). Let A(h) = 6 so that 6," = 1 for some m. For a E R, let a* denote its image in R/I which is a field of characteristic p . We have O* = (S*Y" - 1* = (6* - 1*)," and thus 6* - 1* = O* and 6 = 1 mod I . It follows that A(x) = Ah)mod I and the result follows. I Note that by Theorem 8.20, it follows that if x,, and y,, are conjugate and

x E Irr(G), then ~ ( x -) ~ ( ylies ) in every maximal ideal of R which contains p . (8.21) THEOREM (G. Higman) Let X I , X 2 ,. . ., X k be the conjugacy classes of G. Then the character table of G determines the sets of primes n,,= n(gp)for g, E X,,.

{xil

Proof Let Irr(G) = 1 s i < k } . We are given the complex numbers aiP= xi(g,) for g, E X,,. If 1 E X,,, then v is the unique integer 1 Iv 5 k, such that ai,E Z and ai,2 I ai,I for all i, p and hence we know n, = 0.For notational simplicity, we now assume 1 E XI. Next we compute (ail)2= I GI and let R be the ring of algebraic in-

xi

tegers in Q[E], where E is a primitive I GI th root of 1. For each prime, p 11 GI, we choose a maximal ideal of R, I , 2 pR. Construct the equivalence reon {pi 1 < p < k} by setting p v if ail, = ai, mod I , for all lation i, 1 < i s k. By Theorem 8.20, p v iff the elements of X,, and X , have v, then n,, u { p } = n, u { p } . conjugate p'-parts. In particular, if p Furthermore, given p, there exists v, with p 4 n,, such that p v. To see this, take g E X,, and choose v so that g, E X,. Let n be a set of prime divisors of G. Using induction on 1 n I, we show how = {pin,, = n}. We have 9@= (1). If n # 0,write to construct the set 9, n = no u { p } , where p # no. By the preceding remarks, it follows that

-,

9,= { p l p $ 9 , , The proof is now complete.

-,

and p

-, -,

-,v

for some

-,

VE~'~,,}.

I

Transfer theory is a tool commonly used for producing normal subgroups of a group, especially normal p-complements or normal n-complements. Many of these transfer theorems can also be proved using characters, and in particular using Brauer's theorem. An example of a typical transfer theorem is the following: Let H be a Hall subgroup of G (that is, (I H I, 1 G : H I ) = 1) and suppose that H is nilpotent and that if x, y E H are any two elements which are conjugate in G, then x and y are already conjugate in H . Then H has a normal complement in G. In the above situation, let n be the set of prime divisors of IH I. Using the assumption that H is nilpotent, it is not too hard to prove that if U c G is a

Brauer's theorem

137

nilpotent n-subgroup, then U is conjugate to a subgroup of H . Because of this, the following result is a generalization of the above transfer theorem. (8.22) THEOREM (Brauer-Suzuki) Let H be a Hall subgroup of G and suppose that whenever two elements of H are conjugate in G, then they are already conjugate in H . Assume for every elementary subgroup E E G, that if I E I I I H 1, then E is conjugate to a subgroup of H . Then there exists K 4 G with H K = G and H n K = 1. Proof Let n be the set of prime divisors of 1 H 1, so that by hypothesis, every elementary n-subgroup of G is conjugate to a subgroup of H . Let 9 be a class function of H . We define a class function 9 on G as follows. For g E G, (gn) is an elementary n-group and hence g n is conjugate to some element x E H. Set 8(g) = 9(x) and observe that this is well defined since if g n is also conjugate to y E H , then x and y are conjugate in H by hypothesis and so 9(x) = qy). For x E H , let p(x)denote the number of elements g E G with gx conjugate to x. Let x l , x2, . . . ,x, be representatives for the conjugacy classes of H . If 9 and cp are class functions of H , we have

Now let R be a ring with h E R c C and suppose that 9 is an R-generalized character of H . We claim that 9 is an R-generalized character of G.To prove this, let E E G be elementary. We show that (o), is an R-generalized character of E . Write E = U x V , where U is a n-group and V is a n'-group. We may assume that U E H. If u E U and u E V , then (uu), = u and hence &u) = 9(u). Since U E H , we may write 9, = a#$ for $ E Irr(U) and a# E R. It follows that (S), = a,($ x lv), which is an R-generalized character of E. Thus 9 is an Rgeneralized character of G by Brauer's theorem as claimed. , E is a primitive Now let R be the ring of algebraic integers in O [ E ] where I H I th root of 1. For 1 5 i I t, define = cp(xi)cp, so that 9i(xj)= 0 if i # j and &(xi)= I CH(xi)I by the second orthogonality relation. Also, Si is an R-generalized character of H and hence [&, 1G] E R and

1

xvEIrr(H)

We conclude that ,u(xi)l CH(xi)l/lG I is a positive rational number which is an algebraic integer. It follows that it is a positive rational integer and thus p(xi) I cH(xi)I / I

I2

Chapter 8

138

and ,u(xi)2 I GI / 1 CH(xj) I for all i. We now have

Therefore, equality holds throughout and &xi) = I GI /I CH(xi) I. Thus Equation (*) yields

Now suppose x E Irr(H). By taking R = Z, we conclude that f is a gener23 ,= [x,X] = 1 and f ( 1 ) = ~ ( 1 > ) 0, it alized character of G. Since [I follows that f E Irr(G). Let K = n{ker(f)IX E Irr(H)}. Then K Q G. If u E H n K , then ~ ( u =) f ( u ) = f ( 1 ) = ~ ( 1 for ) all x E Irr(H) and thus u = 1. On the other hand, if p $ s , P E Syl,(G), and U E P, then u, = 1 and f ( v ) = ~ ( 1 = ) f ( 1 ) for all x E Irr(H), so that u E K . Thus P E K and hence I K I 2 IG :H I . It follows that H K = G. 1 To demonstrate the power of Theorem 8.22, we show how easily Burnside’s transfer theorem follows from it. (8.23) COROLLARY Let P E Syl,(G) and suppose P exists K a G with K P = G and K n P = 1.

E

Z(N(P)).Then there

Proqf If E c G is a p-group, then E is conjugate to some subgroup of P by Sylow’s theorems. It therefore suffices to assume x, y E P are conjugate in G and show that x and y are conjugate in P . We have then, y = xB for some g E G . Since P is abelian, P E C,(y) and P B E CG(xe) = C,(y). Thus PBc= P for some CEC,(X~) and hence gc E N ( P ) E C(x)by hypothesis. Now y = xB = (xe)f= xBc= x and the proof is complete. 1 We also remark that Theorem 8.22 provides an alternate proof of Frobenius’ Theorem 7.2. It is routine to check that a Frobenius complement necessarily satisfies the hypotheses of Theorem 8.22.

Our next results depend on Corollary 8.13 where we restrict attention to p-elementary subgroups for some fixed p. (8.24) THEOREM (Dude) Let N Q G with GIN a p-group. Let 9 E Irr(N) be invariant in G. Then there exists a p-elementary subgroup E E G , and

Brauer ’s theorem

139

cp E Irr(E n N), such that

(a) NE = G ; (b) cp is invariant in E ; (c) [QE N , cp] is prime to p.

Proof By Corollary 8.13, we have

where $ runs over irreducible characters of p-elementary subgroups, u, E E and pym. If E is p-elementary and \cI E Irr(E), we compute [P, (14‘)~] for G-invariant class functions /? of N. Now (/?G)N = 1 G : N 18 and (PEN)N= I EN : N I/?. Since P‘ and PEN both vanish on EN - N, we have =IG:ENI~~~.NOW

[P, ($‘)N]

=

[P‘, $‘I

= [@‘)EN,

where the latter equality follows since $‘ = ( $ E N ) G . This yields

[P, ($‘)NI

=

I G : EN 1 [DEN,

: EN 1 [/?, ($ENIN] : EN I ($EnN)Nl,

I =1 =

where we have used Problem 5.2 to obtain the last equality. We apply this with /? = 99. For each $ in Equation (*), write $ E Irr(E,) for p-elementary E, E G . We obtain m = [@,mlN] =

C aJM,

($‘)J

Q =

C,

I G :NE, I C93, ($e*

n NY”.

Since pXm, there exists $ such that p does not divide I G : NE I [ S o , ( $ E n N)N], where we have written E = E,. Since GIN is a p-group and ~ $ 1G :NE I , we conclude that NE = G , and (a)follows. Now write L = E n N so that W L Y 1 = C9LSL9

$L1

= CSL. $LQLI

is prime to p . Since 9 is invariant in G we have QL is invariant in E and we may write 9, = eAA, where A runs over sums of orbits of the action of E on Irr(L) and e A E Z.Write y = $ L 9 L so that y is invariant in E. We have

xA

f9L, $L9LI

=

x A

eACA,

71

is prime to p . Choose A such that p$eA[A, y ] and write A = cpl where the cpi E Irr(L) and are conjugate under E.

+ - .. -k q t ,

Chapter 8

140

Since y is invariant under E, all [qi,y] are equal and [A, y] = t[cpl, y]. However t is a divisor of I E : L I = I G : N I and so is a power of p . Since p$[A, y], we have t = 1 and cpl = cp is invariant in E. This is (b). Finally, [QL,cp] = e, is prime to p and the proof is complete. I An interesting special case is when N = G. (8.25) COROLLARY Let x E Irr(G) and let p be a fixed prime. Then there exists p-elementary E c G and cp E Irr(E) with [xE, cp] f 0 mod p . Examples exist which show that in Corollary 8.25, E cannot always be taken to be a p-group. Dade's Theorem 8.24 has a nice application to the question of character extendibility. (8.26) THEOREM Let N a G with GIN a p-group. Let P ESyl,(G) and assume that P n N c Z ( P ) and that every linear character of P n N is extendible to P. Then every G-invariant irreducible character of N is extendible to G . Proof Let 9 ~ I r r ( N be ) invariant in G. By Theorem 8.24, choose a p-elementary subgroup E c G, with EN = G , and cp E Irr(E n N) such that p#[% n N cpl. We have E = Q x C , where Q is a p-group and C is cyclic. We may assume that C is a p'-group so that E n N = (Q n N) x C and we write cp = a x I where a E Irr(Q n N) and I E Irr(C). We may assume Q c P so that Q n N c P n N, which is abelian. It follows that a is extendible to P n N and thence to P by hypothesis. Therefore a has an extension 6 E Irr(Q) and @ = 6 x 3, is an extension of cp to E. Now since EN = G, we have ( @ G ) N = (GE ,,JN = cpN and hence 9

W G ) N , 91 = bN, 91 = ccp, 9 E n N 1 , which is prime to p . Therefore, there exists an irreducible constituent x of with p $ [ x N , 91. Since 9 is invariant in G , it follows that xN = e9 with p$e. Now e divides I G : N I by Problem 6.7 and since GIN is a p-group it follows that e = 1 and x extends 9. (Note that Problem 6.7 follows immediately from Corollary 6.19 by induction.) I We remark that in the notation of Theorem 8.26, the hypothesis on P will automatically be satisfied if P is abelian. More generally, it suffices for P'nN=l. It has been mentioned that if N 4 G and 9 E Irr(N) is invariant in G, then it sufficesto show that 9 is extendible to the inverse images in G of the Sylow subgroups of GIN in order to prove that 9 is extendible to G. It is for this reason that it is useful to obtain results like Theorem 8.26 which give sufficient conditions for extendibility when the factor group is a p-group.

Problems

141

Problems

(8.1) Let Z c R G C, where R is a ring and the additive group (Z, + ) is a direct summand of (R, +). Let X be a family of subgroups of G. Show that Y R ( G ,A?)n Z[Irr(G)] = 9 ( G , X ) . Hint Write R = Z + S . If 9 = for all x E Irr(G).

1a*$',

where all aJ,E S , then [S, x] E S

Note The Brauer-Tate proof of Brauer's theorem uses this fact applied to R = Z [ E ] ,where E is a root of unity.

(8.2) Let d be the collection of abelian subgroups of the group G. Show that the following statements are equivalent: (a) 1, E W,4; (b) W(G,d )= Z[Irr(G)]; (c) every Sylow subgroup of G is abelian. Hint If a Sylow p-subgroup of G is nonabelian, consider the function 9 = (l/p)p, where p is the regular character of G .

(8.3) Let x be a character of G . For g E G and prime p, write g = gqgp,in the notation introduced following Lemma 8.18. Define the class function x p by x,@) = x(gp.). Show that ( x , ) ~is a character for every nilpotent N E G . Show that if G is not nilpotent then x p is not a character of G for some x E Irr(G) and some prime p. Note In fact, x p is a character for every x E Irr(G) iff G has a normal Sylow p-subgroup. Of course, by Brauer's theorem, x p is always a generalized character.

(8.4) Let x and t,b be characters of G and suppose IG I = mn, where (myn) = 1. Let

where the sum is taken over those g E G such that o(g)I m. Show that u/m E Z. (8.5) Let X E Irr(G) and suppose IGI = mn with (m,n) = 1. Assume that ~ ( x= ) 0 for all 1 # x E G such that x" = 1. Suppose y E G and y" # 1. Show that ~ ( y=) 0. (8.6) Let G and H be groups with classes .Xiand LYi respectively and irreducible characters xi and t,bi, respectively. Assume that whenever g E .Xi and h E 2Zi we have xJ(g) = +,(h) for all i and j . (In short, G and H have identical character tables.) Let x E Z(P) where P E Syl,(G). Suppose x E .Xi. Let y E Yi. Show o(x) = o(y).

Chapter 8

142

Hint If o(y) > o(x) and Z E H ,with (y) = ( z ) and y p = z p , show that zEYi and conclude that p 11 Y i I. See the hint for Problem 2.12. (8.7) Suppose G and H have identical character tables and that G is solvable and has an abelian Sylow p-subgroup. Show that G and H have isomorphic Sylow p-subgroups.

Hint If G has no nontrivial normal p’-subgroup, then its Sylow p subgroup is normal. Use Problem 8.6. (8.8) (Dade) Let N

4

H c G. Assume the following:

(a) ( I H :N I , 1 G : H I ) = 1 . (b) If x,xgE H for some g E G, then there exists h E H and n E N with XQ = xhn. (c) If E G G is elementary and (IEJ, I G : H I ) = 1, then Eg E H for some g E G. Show that there exists M u G such that H M = G and H n M = N . Hint Mimic the proof of Theorem 8.22. Consider only those class functions 9 of H which are constant on cosets of N . Note Many of the important “transfer” theorems follow from Problem 8.8, for instance, the Focal Subgroup Theorem: Let P E Syl,(G) and let N = (x- lxglx E P , xgE P , g E G ) . Then Ap(G) n P =N. Here, AP(G) denotes the normal subgroup of G minimal such that the factor group is an abelian p-group. Problem 8.8 yields M 4 G with M P = G and M n P = N . It follows that M 2 AP(G). The reverse inclusion follows from N c_ G c_ AP(G).

(8.9) Show that a group G is quasielementary iff 1, cannot be written in the , proper subgroups H , with aHE Z. form a H ( l H ) for

1

1

Hint For “only if”: Assume 1, = aH(IH),,where the H run over representatives for the conjugacy classes of proper subgroups. Let C be a cyclic normal p-complement for G. Choose H , minimal, such that pk(a,,IG : H o I), and let 3, be a faithful linear character of C/(C n Ho). Con~ ) ~ , sider the numbers [ ( L I , ( ~ ~ ) 4. Note Problem 8.9 essentially says that Theorem 8.10(a) is the best possible. (8.10) Show that the integers m in Theorem 8.10(b) and Corollary 8.13 can be taken to be the p‘-part of G I.

Problems

143

(8.11) Let x E Irr(G) be a character of defect zero with respect to the prime p. Let Q G G, with p$lQl and let P E Syl,(CG(Q)). Show that (l/IPI)xQis a character of Q. (8.12) Let P 4 G, where P is a p-group, and let x E Irr(G/P) be of defect zero in G/P with respect to p. Let 9 E Irr(P) be invariant in G . Define the function $ on G by

44s) = 0

*(s)= I % p ) x ( s p , )

if if

g p $ P g pE

p.

(a) Show that $ is a generalized character of G. (b) If G = PCG(P),show that $ E Irr(G).

(8.13) Let N 4 G be a Hall 7c-subgroup.Define the function v on N by v(x) = I {y E C,(x) I y is of d-order} 1. Show that v is a generalized character of N . Hint If 9 is a G-invariant character of N (which is not necessarily irreducible), define 9 on G by kl)= S(s3.

Show that 9 is a generalized character of G and compute [J, 1G].

9 Changing the field

So far we have restricted our attention almost exclusively to characters, representations, and modules over the complex numbers. In this chapter we digress from our study of the properties of Irr(G) to consider irreducible group representations over arbitrary fields. (In prime characteristic, this falls far short of the general case, since not all representations are completely reducible.) In particular, if F G E is a field extension we explore the connections between the irreducible E-representations and F-representations of a group. In prime characteristic, we shall see that this situation is (surprisingly) under somewhat better control than it is in characteristic zero, which will be more fully considered in Chapter 10. Let F c E and let X be an F-representation of a group G. Then X maps G into a group of nonsingular matrices over F that, of course, are also nonsingular over E . We may, therefore, view X as an E-representation of G. As such we denote it by XE. (The superscript thus merely indicates a change in point of view.) If X1 and X, are similar F-representations, then ElEand X2E are similar and it follows that if 3 corresponds to the F[G]-module V , then there is a uniquely defined (up to isomorphism) E[G]-module V Ethat corresponds to XE. (For those readers familiar with tensor products, we remark that V E z V 0 FE.)We shall not, however, need to refer to V Eagain, since it is usually easier to work with the representation XE. The F-representation X may be extended by linearity to obtain a representation of F[G], which we shall continue to call X. Under this convention, the E[G]-representation XE is an extension of the F[G]-representation X. 144

Changing the field

145

If XE is irreducible, then clearly so is 3.However, X E may well reduce, even if X is irreducible. To illustrate what can happen, we consider two examples, both for the field extension R E @. Let G = (9) be cyclic of order 3 and let X be the R-representation defined by X@)=

(

O

-1

1).

-1

Then X" affords the character I + 1,where I is a faithful linear character of G and so X" is reducible. Since I is not real valued, it is not afforded by any R-representation and it follows that X is irreducible over R. Now let Qs = ( a , b ) be the quaternion group of order 8, where o(a) = 4 = o(b).Define

It is not hard to check that this defines a representation of Qs . (In fact this is the representation whose module is the quaternion algebra W = R -t Ri + R j + Rk with respect to the basis (1, i, j, k } , where a = i and b = j act by right multiplication.) Now 3" affords the character 2 ~where , x E Irr(Qs) and ~ ( 1= ) 2. By Problem 2.5(b), x is not afforded by any R-representation of Q s and thus X is irreducible over R. (This can also be deduced from the fact that W is a division algebra.) (9.1) DEFINITION Let 3 be an F-representation of G. Then X is absolutely irreducible if XE is irreducible for every field E 2 F. (9.2) THEOREM Let X be an irreducible F-representation of G with degree n. The following are equivalent. (a) X is absolutely irreducible. (b) XE is irreducible for every finite degree extension E 2 F. (c) The centralizer of X(G)in the matrix ring M,(F) consists of scalar matrices. (4 W C G I )= M,(F). Proof That (a) implies (b) is trivial. Now assume (b) and let M E M,(F), with MX(g)= X(g)M for all g E G. Let E be a finite degree extension of F, chosen so that M has an eigenvalue, I E E. Since XE defines an irreducible representation of E[G] and M - IZ is a singular matrix centralizing its image, it follows from Schur's Lemma 1.5 that M - I 1 = 0. Thus (c) follows. That (c) implies (d) is immediate from the Double Centralizer Theorem 1.16. Finally, assume (d) and let L 2 F. Since every M E M,(L) is an L-linear

Chapter 9

146

combination of matrices in M,(F), it follows that X(L[G]) = M,(L). Thus for every L representation 9 similar to X, we have g ( L [ G ] )= M,(L) and so 9 cannot be in reduced form. Thus 3 is irreducible over L and the proof is complete. I (9.3) DEFINITION The field F is a splitting j e l d for G if every irreducible F-representation of G is absolutely irreducible. (9.4) COROLLARY If F is algebraically closed, then F is a splitting field for every group. Proof Since F has no proper finite degree extension, condition (b) of Theorem 9.2 holds for every irreducible F-representation of G . 1

Suppose F is a splitting field for G and E 2 F. Then every irreducible F-representation X determines an irreducible E-representation XE. If {Xi} is a set of representatives for the similarity classes of irreducible F-representations of G, then { X i E } is a complete set of representatives for the irreducible E-representations of G. In order to prove this, we need to discuss the “irreducible constituents” of a possibly reducible representation. If V is an F[G]-module, then a composition series for V is a chain of submodules V = V, > V, > ... > v, = 0 such that each 6- ,/K is an irreducible module. The modules K- J K are the factors of the series. The Jordan-Holder theorem asserts that the factors of any two composition series for V are the same up ta isomorphism (and counting multiplicities).An irreducible module W isomorphic to a factor of some (and hence all) composition series for V is called an irreducible constituent of V . If W is isomorphic to an irreducible submodule of V , then W is a constituent of V . We call it a bottom constituent in that case. Similarly any irreducible homomorphic image of V is a constituent. These are the top constituents of V . If V is completely reducible [e.g., if char(F)$I G 11, then every irreducible constituent of V is both a top and a bottom constituent. (Caution: the converse is false.) All of the preceding remarks may be translated into the language of representations. If X is an F-representation of G corresponding to the F[G]module V , then X is similar to a representation 3 in triangular block form

* \

where the irreducible representations

3i correspond to the factors E- ,/E.

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147

Representations similar to the Ji are the irreducible constituents of X;those similar to 3, being top constituents and those similar to 3, being bottom constituents. (Of course some of the other irreducible constituents of X may also be top or bottom constituents since j is not necessarily the only representation in triangular block form to which X is similar.) Note that the character afforded by a representation is the sum of the characters of all of the irreducible constituents (counting multiplicities). We emphasize that by the Jordan-Holder theorem, the representation X has only finitely many irreducible constituents (up to similarity),namely those which appear in any single triangular block representation similar to X.

(9.5)

COROLLARY

Let F be a field and G a group.

(a) Every irreducible F-representation of G is a top constituent of the regular F-representation %. (b) There exist only finitely many similarity classes of irreducible F-representations of G. (c) If E 2 F and 'I) is an irreducible E-representation of G, then 'I) is a constituent of X E for some irreducible F-representation X. Proof Statement (a) is immediate from Lemma 1.14 and (b) follows from (a) via the Jordan-Holder theorem. Let 3 be any F-representation of G . The irreducible constituents of 3Emay be found by taking the irreducible conStituents Xiof 3 and then finding the irreducible constituents of the XF.Now (c) follows by applying this remark to 3 = %. I

The following result is a useful tool for establishing the similarity of two F-representations of G and for other purposes. Let X be an irreducible representation of F[G] and let X(b) = X(a) and 'I)@) = 0 for every irreducible F[G]-representation 9 which is not similar to X. (9.6)

THEOREM

a E F[G]. Then there exists b E F[G] such that

Proof Let (Xi} be a set of representatives for the similarity classes of irreducible F[G]-representations. Let Ii = {x E F[G] I Xi(x) = 0) so that Zi is an ideal of F[G] and in the notation of Problem 1.4, J(F[G]) = nli.Let A = F[G]/(nZ,), so that each X i may be viewed as a representation of the algebra A . As such, the Xi are irreducible and pairwise nonsimilar. In particular, J ( A ) = 0 and by Problem 1.5, A is semisimple. By Theorem 1.15, A has minimal ideals M i , such that X,{Mi) = 0 i f j # i and Xi(Mi)= Xi(A) = X,(F[G]). Now suppose X = XI.Choose b in the inverse image in F[G] of M , with X(b) = X(a). The result follows. I (9.7) COROLLARY Let X and 'I) be irreducible F-representations of G. Suppose F E E and that XE and qEhave a common irreducible constituent. Then X is similar to 'I).

Chapter 9

148

Proof Let 3 be an irreducible constituent of X E and of 9".If 3and 9 are not similar, view them as representations of F[G] and choose b E F[G] with X(b) = X(1) and '1)(b)= 0. It follows that if U is any E[G]-representation similar to X", then U(b)is the identity matrix and hence 3(b) is the identity matrix. A similar argument, working with g", shows that 3(b) = 0 and this contradiction proves the result. I Note that Corollary 9.7 asserts that the representation 3 in Corollary 93 c ) is unique up to similarity. (9.8) COROLLARY Let F be a splitting field for G and let {Xi} be a set of representatives for the similarity classes of irreducible F-representations. Suppose E 3 F. Then E is a splitting field and {X:} is a set of representatives for the irreducible E-representations of G. Proof Since the Xi are absolutely irreducible, the XF are absolutely irreducible. They are pairwise nonsimilar by Corollary 9.7. Finally, suppose 1 ' ) is any irreducible E-representation. By Corollary 9.5(c), '1) is a constituent 3 for some i. Since X: is irreducible, it is similar to '1) and the proof is of : complete. I (9.9) THEOREM Let E be a splitting field for G and let F c E. Then F is a splitting field iff every irreducible E-representation of G is similar to '1)" for some F-representation '1). Proof Suppose F is a splitting field. Then by Corollary 9.8, every irreducible E-representation is as desired. Conversely, let 3 be any irreducible F-representation and let X be an irreducible constituent of 3".By hypothesis, there exists an irreducible F-representation 9, such that '1)" is similar to X and hence g Eand 3Ehave an irreducible constituent in common. By Corollary 9.7,'1) is similar to 3,and thus 3"is absolutely irreducible. It follows that 3 is absolutely irreducible since the only F-matrices which centralize all 3 ( g ) are scalar. The proof is complete. I (9.10) COROLLARY Let F be any field and G a group. Then some finite degree extension of F is a splitting field for G.

Proof Let F be the algebraic closure of F, so that F is a splitting field. Let {Xi}be a set of representatives for the similarity classes of irreducible F-representations. By Corollary 9.5(b), I {Xi}I co and hence only finitely many elements of F occur as entries in any of the matrices Xi@) for g E G. Adjoin all of these elements to F so as to obtain the field E. Since F is algebraic over F, it follows that IE : FI < co. Since each X imay be viewed as an E-representation of G, it follows from Theorem 9.9 that E is a splitting field. I

-=

Changing the field

149

(9.11) COROLLARY Let Q G F E C. Then F is a splitting field for G iff every x E Irr(G) is afforded by some F-representation.

Proof If F is a splitting field and x E Irr(G), choose a C-representation X that affords x. By Theorem 9.9,X is similar to '2)' for some F-representation '2). Then 'I) affords x. Conversely, suppose that every x E Irr(G) is afforded by some F-representation. Let X be an irreducible C-representation and let 'I) be an F representation which affords the same character. Then by the linear independence of Irr(G) and using the fact that C has characteristic zero, it follows that X is the unique irreducible constituent of '2)' and occurs with multiplicity 1. Thus 'I)' is similar to 3. The result now follows from Theorem 9.9. I We shall now discuss the character theory of a group over an arbitrary splitting field. The following result is actually true without the assumption that E is a splitting field. We shall prove the more general fact later. (9.12) LEMMA Let E be a splitting field for G. Then the characters of nonsimilar irreducible E-representations of G are nonzero, distinct, and linearly independent over E.

Proof Let {Xi} be a set of representatives for the similarity classes of irreducible E-representations, and let xi be the character afforded by Xi. View xi as being defined on all of ECG]. Since Xi(E[G]) is a full matrix ring over E, we may choose ai E E[G] with zi(ai)= 1. By Theorem 9.6, we may assume that xXai) = 0 if i # j. The result is now immediate. I If E is a splitting field for G, we shall use the notation Irr,(G) to denote the set of characters of the (absolutely) irreducible E-representations of G. The point of the next result is that in some sense Irr,(G) is not as dependent on the particular field E as the notation would indicate. (9.13) LEMMA Let E be a splitting field for G and let x E IrrdG). Suppose K G E is a subfield which contains x(g) for all g E G and let F 2 K be another splitting field for G. Then x E IrrF(G).

Proof We may replace F by a K-isomorphic copy and assume that E, F E L for some field L. By Corollary 9.8, L is a splitting field for G and Irr,(G) = Irr,(G) = Irr,(G). The result follows. I The next result is of great importance in studying representations in prime characteristic. Its proof depends on Wedderburn's theorem which asserts that finite division rings are commutative. As the quaternion group of order 8 shows, Theorem 9.14 would be false in characteristic zero.

130

Chapter 9

(9.14) THEOREM Let X be an absolutely irreducible E-representation of G , where E is of prime characteristic. Suppose that X affords the character x and that x(g) E F for all g E G , where F is some subfield of E. Then X is similar to g Efor some absolutely irreducible F-representation 9. Proof First we consider the crucial special case that 1 E 1 < co.Working by induction on I E 1, it is no loss to assume that F is a maximal subfield of E. Let A be the set of all F-linear combinations of the matrices X(g) for g E G. Then A is an F-subalgebra of the matrix ring M,(E), where n is the degree of X. If a E Z(A),then a is a matrix over E which commutes with all X(g). Since X is absolutely irreducible, a is a scalar matrix and thus F - 1 E Z(A) C E 1. It follows that Z(A) is a finite integral domain, and hence is a field. Since F is a maximal sufield of E, we conclude that either Z(A) = F 1 or Z(A) = E 1. If Z ( A ) = E 1, then A = E - A = X(E[G]). Since X is absolutely irreducible, we conclude that A = M,(E) and thus if a E E - F , we can choose a E A with tr(a) = a. However, a is an F-linear combination of matrices of the form X(g), each of which has trace in F . This contradiction shows that Z(A) = F . 1. Next, we claim that the F-algebra A is algebra isomorphic to the full matrix algebra Mk(F)for some integer k. To show this, we first establish that A is semisimple. By Problem 1.5, it suffices to show that A has no nonzero nilpotent ideals. If I is a nilpotent ideal of A , then E . I is a nilpotent ideal in E . A = X(E[G]) = M,(E). It follows that I = 0. (Any nonzero matrix has a multiple which is not nilpotent.) By Theorem 1.15, the semisimple F-algebra A is a direct sum of minimal ideals. Since dim,(Z(A)) = 1, it follows that there is only one direct summand and hence in the notation of 1.15, A = M ( A ) z A M for some irreducible A-module M . Let D = E,(M), so that D is a division ring by Schur’s Lemma 1.5. Since dim,(M) < co and IF I < co,we have ID 1 < co and thus D is commutative by Wedderburn’s theorem. By the Double Centralizer Theorem 1.16, we have AM = E,(M) and thus D c Z(AM)z F . Since D contains the scalar multiplications, 1 . F , of F on M , we conclude that D = 1 . F . Thus if k = dim,(M), then A z A M = E,(M) = E,(M) z M,(F) and we have an F-algebra isomorphism 9 : A + Mk(F). Thus 9 = 19 : F [ G ] + Mk(F) is a representation of FCG]. It is absolutely irreducible since it maps onto Mk(F). We claim that g Eis similar to X. To see this, let 3 be an irreducible Frepresentation such that X is a constituent of 3E.If 3 is not similar to 9,use Theorem 9.6 to choose b EF[G] such that 3 ( b ) = 0 and g(b) # 0. Since X is a constituent of 3E, we have X(b) = 0 and thus g(b) = (X(b))9 = 0. This contradiction shows that 3 is similar to 3 and thus X is a constituent of gE. Since 9 is absolutely irreducible, the claim follows and the theorem is proved when E is finite.

Changing the field

151

We now consider the general case where E may be infinite. Since we may replace E by a larger field, it is no loss to assume that E is algebraically closed. Let K E E be the prime field and let L 2 K be a splitting field with IL : K I < 00 (by Corollary 9.10). Since E is algebraically closed, we may assume L E E. Since L is a splitting field, X is similar to 3Efor some Lrepresentation 3 and 3 affords x which takes values in L n F. Since I L I < 00, the first part of the proof yields an absolutely irreducible ( L n F)-representation 9,such that g Lis similar to 3 and hence 9JEis similar to X. Now g F is the desired F-representation and the proof is complete. I Theorem 9.14 remains true if f is irreducible but not absolutely irreducible. This will be proved in Corollary 9.23. By the exponent of G we mean the least positive integer n, such that g" = 1 for all g E G. Clearly n I GI.

I

(9.15) COROLLARY Let G have exponent ~tand assume the polynomial x" - 1 splits into linear factors in the field F . If F has prime characteristic, then it is a splitting field for G. Proof Let E 2 F be a splitting field and let x E IrrE(G).Then x(g) is a sum of nth roots of unity and hence x(g) E F. The result follows by Theorems 9.14 and 9.9. I An entirely different proof shows that Corollary 9.15 also holds in characteristic zero. This result of R. Brauer depends on his theorem on induced characters and will be proved in Chapter 10.

Let E be any field, CT an automorphism of E, and X an E-representation of G. We can apply CT to every entry in the matrix X(g) for every g E G. What results is a new representation, denoted Xu, which may or may not be similar to X. Similarly, if 9 : G + E is a function, we write 9"(g) = (9(g))". Suppose X affords the E-character x and that x takes values in F E E. If z E Aut(F), it is not obvious that X* is an E-character of G. (9.16) LEMMA Let E be a splitting field for G and let x E IrrE(G).Suppose x(g)E F E E for all g E G and let z E Aut(F). Then x' E IrrE(G).

Proof Let E be an algebraic closure for E and let F s E be an algebraic closure for F. Then Irr,(G) = IrrE(G) = Irrp(G) by Corollary 9.8. Since z is extendible to an automorphism of F, the result follows. I Let K E E be a field extension and suppose x is an E-character of G . We write Kk)to denote the subfield of E generated by K and the character values x(g) for g E G. Note that K ( x )is contained in a splitting field for a polynomial

152

Chapter 9

of the form x" - 1 over K . Since this polynomial yields a Galois extension with an abelian Galois group, it follows that K(x) is a finite degree Galois extension of K and the Galois group B(K(x)/K)is abelian. Assume that E is a splitting field for G and that F c E. If x, $ E IrrE(G),we say that x and $ are Galois conjugate over F if F(x) = F($) and there exists ~EB(F(x)/F) such that xu = $. It is clear that this defines an equivalence relation on IrrE(G). (9.17) LEMMA Let E be a splitting field for G and let x E IrrE(G).Let Y be the equivalence class of x with respect to Galois conjugacy over F where F G E. We have

(a) If K E E, K ( x )= K , and o E B ( K / K n F), then xu E 9. (b) If F , c F and I) E 9, then $ is Galois conjugate to x over F , . (4 IYI = IF(x):FI. Proof (a) By 9.16, xu E IrrE(G).Now ( K n F ) ( x ) 3 K n F is a normal extension and so ( K n F)(x)E K is invariant under o. It is thus no loss to assume that K = ( K n F ) ( x )and so K E F ( x )and no proper subfield of F(x) contains both F and K . It follows by Galois theory that restiction maps B(F(x)/F)onto B(K/K n F ) and so xu = xr for some ~EB(F(~)/F). Now F ( f ) = F(x) and so xu = xT E 9. (b) This is immediate by applying (a) to F , and taking K = F(x). (c) We have that Y is the orbit of x under B(F(x)/F).By definition of F(x), the stabilizer of x in this group is trivial and so IYI = IB(F(x)/F)I

=IF(x):FI. I Now let F be any field and let X be an irreducible 8'-representation of G. Let E 2 F be a splitting field for G . What does X E look like? We shall prove that it is completely reducible; that all irreducible constituents occur with equal multiplicity; that the characters of these constituents constitute a Galois conjugacy class over F and that the common multiplicity is 1 except possibly when F has characteristic zero. (9.18) LEMMA Let F E E with I E : F ( = n < 00 and let V be an irreducible E[G]-module corresponding to the E-representation X. Then V may be viewed as an F[G]-module and as such let it correspond to the F-representation 3. Then (a) deg 3 = n deg X. (b) 3 has a unique (up to similarity) irreducible constituent. It is the F-representation 9 such that X is a constituent of 'I)E. (c) If X affords the E-character x and F(x) = F , then 3 affords nx. Proof We may certainly view V as an F-space. Let ul, v 2 , . . ., v, be an E-basis for V and let e l , e 2 , . . . ,en be an F-basis for E . It is routine to check

Changing the field

153

that { v i e j } is an F-basis for V and so V is finite dimensional over F , 3 is defined, and (a) follows. Let 'I) be a (unique up to similarity) irreducible F-representation of G such that X is a constituent of gE.By Theorem 9.6, choose b E F[G] c E[G], such that 'I)@) = 'I)(l) and 'I)&) = 0 for every irreducible F-representation 'I)o not similar to 'I). Since X is a constituent of 'I)E and 'I)@) is an identity matrix, then so is X(b) an identity matrix. Thus b acts as an identity on V and hence 3(b) is an identity matrix and 'I)@)# 0 for every irreducible constituent giof 3. Thus every 'I)i is similar to r) and (b) is proved. v j a i j ,where aij E E and x(g) = uii. For (c),let g E G and write uig = Now write

cj

for 1 5 p, v I n and (Vie&

We have

xi

pijpvE F. Let 3 afford the character +. Then =

C

and

UjevPijpv

+(g) =

1

Biipp.

i,lr

j.v

X i aii = x(g) E F , we conclude that

and since we are assuming that

1 i

aii

=

1 p..ww i

for each p, 1 I p 5 n. The result now follows.

I

(9.19) COROLLARY Let F E E with I E : FI = n < 00. Let X be an irreducible E-representation of G and let 'I) be an irreducible F-representation such that X is a constituent of 'I)E. Then deg 'I) divides n(deg X).

Proof Let 3 be an F-representation obtained by viewing an E[G]module corresponding to X as an F[G]-module. By Corollary 9.7 and Lemma 9.18(b),we conclude that 'I) is the unique irreducible constituent of 3 and so deg 'I) divides deg 3. The result now follows from 9.18(a). I The following corollary is what survives of Theorem 9.14 when the hypothesis of prime characteristic is dropped. (9.20) COROLLARY Let X be an absolutely irreducible E-representation of G which affords the character x. Let F E E be such that F(x) = F . Then there exists an irreducible F-representation 'I), such that X is the unique (up to similarity)irreducible constituent of gE.In particular, 'I) affords mx for some integer m.

154

Chapter 9

Proof If F has prime charactersitic, then by Theorem 9.14, there exists 'I) such that 'I)' is similar to X and there is nothing to prove. Assume then that char(F) = 0. It will suffice to show for some positive integer n that the character nx is afforded by some F-representation 3. This is sufficient since by the linear independence of Irr,(G) for a splitting field L 2 E, it follows that every irreducible constituent of 3Laffords x and so is similar to XL.We may thus take 'I) to be any irreducible constituent of 3 and quote Corollary 9.7 to complete the proof. To produce 3, let K 2 F be a splitting field with I K : F I = n 00. By Lemma 9.13, x E IrrK(G).Let 3 be the F-representation which results by taking a K[G]-module which affords x and viewing it as an F[G]-module. Then 3 affords nx by Lemma 9.18(c) and the result follows. I

-=

Of course it follows in the above situation that if gois any irreducible F-representation such that X is a constituent of (go)',then is similar to 'I) and hence X is the unique irreducible constituent of (go)'. (9.21) THEOREM Let F E E , where E is a splitting field for G . Let 'I) be an irreducible F-representation of G. Then (a) The irreducible constituents of 'I)' all occur with equal multiplicity m. (b) If char(E) # 0, then in = 1. (c) The characters xi E Irr,(G) afforded by the irreducible constituents of 'I)' constitute a Galois conjugacy class over F and so the fields F(xi)are all equal. (d) Let L = F(xi). The irreducible constituents of 'I)L occur with multiplicity 1. (e) If 3 is any irreducible constituent of 'I)L then 3' has a unique irreducible constituent. Its multiplicity is m. (f) 'I)'-and 'I)' are completely reducible.

Proof Let X be an irreducible constituent of 'I)' and suppose X affords

x E IrrE(G).Let L = F(x) and let 3 be an irreducible constituent of 'I)'-such

that X is a constituent of 3'. By Corollary 9.20, 3 is the unique irreducible constituent of 3".Let m be its multiplicity, so that 3 affords the character mx. If char(E) # 0, then Theorem 9.14 yields m = 1 and 3 is absolutely irreducible. Let x = xl,x2,. . . ,xn be the distinct Galois conjugates of x over F so that n = I L :F I by Lemma 9.17(c).For r~ E S(L/F), form the L-representation 3" defined by 3"@) = &)". As CT runs over Y(L/F)we obtain n representations 3" affording the characters mXi, 1 I i I n. Since m = 1 when char(E) # 0, the mxi are distinct in all cases and thus the 3" are pairwise nonsimilar. Also,

Changing the field

155

each (3")E has a unique irreducible constituent and it occurs with multiplicity m. This follows from the characters if char@) = 0 and it holds in prime characteristic since then all of the 3"are absolutely irreducible. We claim that the n = I L : FI representations 3" are exactly the irreducible constituents of g' and that each occurs with multiplicity 1. Since the irreducible constituents of (3")E and (3')E are nonsimilar if (T # 7,statements (a)-(e) will follow when the claim is established. Since 3 is a constituent of 'I)L and ('I)')" = 'I)L for Q E Q(L/F),it follows that 3"is a constituent of 'I)L for every (T. Therefore n(deg 3)Ideg 'I) since n = I Q(L/F)I. By Corollary 9.19, deg 'I) divides n(deg 3)and we thus have equality. Therefore the 3" are the only irreducible constituents of 'I)L and each has multiplicity 1 as claimed. All that remains is to show that 'I)L and 'I)E are completely reducible. This follows from Maschke's Theorem 1.9 if char(E) = 0 so we assume char(E) # 0. We may assume without loss that 3 is a bottom constituent of 'I)' (since some constituent certainly is). Since (9')" = 'I)', it follows that 3" is a bottom constituent for every oeQ(L/F).Now let V be an L[G]-module corresponding to 'I)' and let W be the sum of all of the irreducible submodules of V . By Theorem 1.10, W is completely reducible and it suffices to show W = I/. Since every irreducible constituent of g Lis a bottom constituent, every composition factor of V occurs as a composition factor of W. Since the composition factors of V all have multiplicity 1, V / W is necessarily trivial. The proof for 'I)" is similar since 3" is irreducible in this case. I We obtain some consequences now. The first generalizes Theorem 9.14.

(9.22) COROLLARY Let F be any field. Then the characters of nonsimilar irreducible F-representations of G are nonzero, distinct and linearly independent over F. Proof Let E 2 F be a splitting field for G. By Theorem 9.21, the characters of nonsimilar irreducible F-representations of G are nonzero multiples of sums of disjoint subsets of IrrE(G).The result follows from 9.12 I

(9.23) COROLLARY Let F c E be fields of prime characteristic. Let X be an irreducible E-representation of G which affords the character 2. Let 'I) be an irreducible F-representation such that X is a constituent of g E .Then deg 'I) = I F ( x ) : F I deg X. In particular, if F(x) = F , then X is similar to 'I)E. Proof Let L E? E be a splitting field for G and let c ~ I r r ~ ( G be) the character of an irreducible constituent of XL. Let Y and y be the Galois conjugacy classes of I over E and F respectively. Since E has prime characteristic, it suffices by Theorem 9.21 to show that Iy I = IF(x) :F I I Y I.

Chapter 9

156

By Lemma 9.17(b),we have Y E 9-and thus x = C Y takes on values in F(5) and so F ( x ) c F(5). Since 19-1= IF(5) : F I by 9.17(c),we must show that I Y I = I F(5): F(x)I . Let 9 = Y(F(l)/F(x)).If n E 9, then

and it follows from the linear independence of Irr,(G) and Lemma 9.16 that 9 permutes 9. Since only the identity of 9 can fix 5,this yields 1 9 12 1%) = I F(5) :F(x) I. However, F(x) E E n F(5)and thus IYI = IE(5):El = IF(5):E n F(5)l I IF(C):F(x)I

and the proof is complete.

I

Problems

(9.1) Let F c E be fields and let V be a finite dimensional E-space. Let W c V be an F-subspace. We write V = W * , E provided V = WE and dim,(V) = dim,(W). (a) If V = W *, E and U c W is an F-subspace, show that WE = u *F E. (b) If X is an F-representation of G, show that dim,(X(F[G])) = dimE(XE( E [GI)). Note The symbol *, denotes an internal tensor product.

(9.2) Let X be an F-representation of G and let E 2 F. Let V be an E [ G ] module corresponding to XE. (a) Show that V = W *, E for some G-invariant, F-subspace W E V such that the F[G]-module W corresponds to X. (b) Let U c W be an F[G]-submodule. Let U and W/U correspond to the F-representations 'I) and 3, respectively. Show that WE is an E[G]submodule of V which corresponds to g Eand that V/UE corresponds to 3E. (9.3) Let W be an F[G]-module and let F E E with I E :FI = n < 00. Let W Ebe an E[G]-module corresponding to XE where X is an F-representation corresponding to W. Now view W Eas an F-space. Show that the resulting F[G]-module is isomorphic to the direct sum of n copies of W . Note It follows from the Krull-Schmidt theorem that if V and W are F[G]-modules and V 0 V 0 . . . 0 V 2 W 0 W 0 . + .0 W , where each direct sum has n terms, then V z W. We conclude via Problem 9.3 that if X and 'I) are F-representations of G and XE is similar to qE, then X is similar to 'I), provided I E : F I c 03.

Problems

157

(9.4) Let F G E be a field extension of possibly infinite degree. Let X and 9 be F-representations of G and assume XE and ?lEare similar. (a) Show that there exist matrices M , , M 2 , . . .,M , over F and elements

el, e 2 , . . . ,e, E E such that X(g)Mi = M i 9 ( g )for all g E G and 1 Ii Ik and such that elM, . . . ekMk is nonsingular over E. (b) If IF 1 > deg X, show that X is similar to 9.

+

+

Hint [For (b)] Let f ( x l , x 2 , . . . , xk) be a nonzero polynomial over F and assume that the degree off in x i is < 1 F 1 for each i. Then there exists U l , U 2 , . . . , ak E F such thatf(u,, U z , . . . ,a,) # 0. (9.5) Under the hypotheses of Problem 9.4, show that without assuming anything about IF I or I E : F I.

X and 9 are similar

Hint Combine the results of Problem 9.4(b) and the note following Problem 9.3. (9.6) Let F c E and let V be an irreducible E[G]-module which affords the character x. Assume that E = F(x). Show that V is irreducible when viewed as an F[G]-module. (9.7) Let F have prime characteristic and let X be an irreducible F-representation of G. Let D be the centralizer of X(G) in the matrix algebra M,(F) where n = deg X. Show that D is a field.

Hint Let E 2 F be a splitting field and consider the centralizer of X E in M,(E). Use Theorems 9.21 and 9.2. (9.8) In the situation of Problem 9.7, let x be the character of an irreducible constituent of XE where E z F is a splitting field for G. Show that the natural isomorphism between F and F . 1, the scalar matrices in M,(F), extends to an isomorphism F ( x ) 2 D.

Hint Let L = F(x) and let I/ be an L[G]-module affording 2. View I/ as an F[G]-module. Then D Z EFLG1(I/). Use Problem 9.6. (9.9) Let X be an F-representation of G and assume X E is completely reducible for some E 2 F . Show that X is completely reducible.

Hint Consider X(J(F[G])). Use Problem 1.10. (9.10) Let 35 be a completely reducible F-representation of G and let E

2 F.

Show that X E is completely reducible.

Note The analogous statement for algebras other than group algebras does not hold, in general. (9.11) Let F E E and view F[G] E ECG]. Show that J(E[G]) J(F[G]) * F E (in the notation of Problem 9.1).

=

Chapter 9

158

Hint To obtain J(E[G]) E (J(F[G])E, use Problems 9.2(b), 9.10, and 1.10 to argue that E[G]"/(J(F[G]))E is a completely reducible E[G]-module.

(9.12) If X is an F-representation of G let s(X) denote dim,(X(F[G])). Let 'I)l, . . . , 'I)k be the distinct irreducible constituents of X. Show that

Ci s('I)i)I ~ ( 3 ) with equality if X is completely reducible. (9.13) Let X be an irreducible F-representation of G. Let 'I) be an irreducible constituent of X", where E 2 F is a splitting field. Show that rns(3) = deg(X) deg('I)),where s(X) is as in Problem 9.12 and rn is the multiplicity of 'I) as a constituent of 3". (9.14) Let X be an irreducible F-representation of G and let D be the centralizer of X(F[G]) in the matrix ring M,(F), where n = deg(X). Let d = dim@). Show that s(3) = n2/d. Hint Let V be an F[G]-module corresponding to X so that V becomes a vector space over the division ring D. Express s(X) = dim,(X(F[G])) in terms of dm i& )' and d. Use Theorem 1.16.

Note Applying Problem 9.14 in the situation of Problem 9.13 we obtain rnn2/d = ms(X) = n deg('I))and rnn = d deg('I)).Since n = rn deg('I))) F ( x ):F 1, where x E Irr,(G) is afforded by 'I), we obtaind = rn2 I F(x) :F I. Compare this with Problem 9.8. This formula can be used to generalize Problem 9.8 to the characteristic zero case.

(9.15) Let X be an irreducible F-representation of G and let x be the character afforded by an irreducible constituent 'I) of X", where E z F is a splitting field for G. Let D be the centralizer of X(F[G]) as in Problem 9.14. Show that the isomorphism F E Fa 1 E D extends to an isomorphism F ( x ) E Z(D). Hints Let L = F(x) and let 3 be the irreducible constituent of X L such that 9 is a constituent of 3".Let V be an L[G]-module corresponding to 3. Let Do = E L l G j ( V ) and D, = EFrGl(V).Check that Do c D1 E D. Use the note following Problem 9.14 to show that Do = D1. Let Z = Z(Do) Z L - 1. Observe that Do = EZrGl(V). Let '123 be the Z representation corresponding to V viewed as a Z[G]-module, and compute the multiplicity of 'I) as a constituent of '123". Now use the note following Problem 9.14 again to show Z = L . 1.

(9.16) Let X be an irreducible F-representation of G and let E 2 F. Show that g E G lies in ker(X)iff g E ker('I)) for every irreducible constituent 'I) of X".

Problems

159

(9.17) Let F have prime characteristic p. Show that O,(G) (the largest normal p-subgroup of G ) is the intersection of the kernels of the irreducible F-representations of G. (9.18) Let k = char(F). (a) If k # 2, show that Qs has a unique (up to similarity) nonlinear irreducible F-representation. (b) If k # 0, show that F is a splitting field for Qs. (c) If k = 0, show that F is a splitting field for Q s iff - 1 is a sum of two squares in F.

(9.19) Let F E E, where E is a splitting field for G. Does there necessarily exist a splitting field K such that F E K E E and I K : FI < co? Consider zero and nonzero characteristic separately. Hint Consider F = Q and E = Q(a, fl) E @, where a, transcendentals. Take G = Qs .

fl are suitable

(9.20) Let G be cyclic of order n and let F have characteristic not dividing n. Let E = F(E),where E is a primitive nth root of unity. Let rn = I E : F I . Show that every faithful irreducible F-representation of G has degree m and that there are exactly cp(n)/rn similarity classes of such representations. Note If IF1 = q < 00, then rn can be characterized as the least positive integer such that n l(qm- 1). (9.21) Let G = GL(n, p") be the full group of nonsingular n x n matrices over GF(p'). Show that G has an irreducible representation of degree ne over WP). (9.22) Let F G E be fields of prime characteristic and let 3E be an E-representation of G. Assume that 3E affords a character x such that F ( x ) = F. Do not assume X is irreducible. (a) Show that x is afforded by some F-representation. (b) Find an example where 3E is not similar to g Efor any F-representation 9.

10 The Schur index

The main question considered in this chapter is the following. If x E Irr(G), then for which fields F E C is x afforded by an F-representation? If F E C is not one of these fields, we wish to measure the extent to which x fails to be afforded over F. This and the results of Chapter 9 (and, in particular, Theorem 9.21) suggest the following definition. (10.1)

DEFINITION

Let F E E , where E is any splitting field for G . Let

x E IrrE(G).Choose an irreducible E-representation 3 which affords x and an ' ) such that 3 is a constituent of ?IE. Then the irreducible F-representation 1 is the Schur index of x over F. It is multiplicity of X as a constituent of denoted by r n , ( ~ ) . Note that given x as above, representations X and '1) do exist and are unique up to similarity and so r n , ( ~ )is well defined. Actually, rnF(x)does not really depend on E. If L 2 Fk)is another splitting field, then x E IrrJG) by Lemma 9.13. A routine argument shows that rnF(x) is the same when computed in E or in L. By Theorem 9.21(b), Schur indices are always trivial (that is, equal to 1) in prime characteristic. For this reason we now restrict our attention to the characteristic zero case. In fact, it is really no loss to confine .ourselves to subfields of the complex numbers, C. Many important results about Schur indices appear to depend on deep facts about division algebras and number theory. Nevertheless, as this chapter will demonstrate, much can be done by elementary means. In particular, some of the results of Chapter 8 will prove invaluable. 160

The Schur index

161

We collect some facts. (10.2)

COROLLARY

Let x E Irr(G) and F E @. We have

(a) r n F ( X ) = r n F ( l ) ( X ) . (b) Let 9 'be the Galois conjugacy class of x over F. Then m , ( ~ ) ( 9 z ') is the character of an irreducible F-representation of G. (c) If E is the character of any F-representation, then m , ( ~ )divides

E,XI. (d) r n , ( ~ )is the smallest integer rn such that r n X is afforded by an F(x)representation. (e) rn,(X) is the unique integer rn such that r n X is afforded by an irreducible F(X)-representation. (f) If F E E E @, then rnE(X) divides m,(~). (g) If F c E c C and I E : F I = n < co,then rn,(X) divides nrnE(X). (h) Q ( X ) divides ~ ( 1 ) .

Proof Part (a) follows from Theorem 9.21(e) and part (b) is a consequence of 9.21(a, c). Let 9be the F-representation in (b).Then 9 is the unique (up to similarity) irreducible F-representation whose character $ satisfies [$, X] # 0. Since WI,(X) = [$, 23, part (c) follows. To prove (d) and (e), we may assume (by (a))that F = F(x). Now 9'= {x} and the representation 1) of the preceding paragraph affords r n X . Parts (d) and (e) are now immediate. Now let F E E E @. Any character afforded by an F-representation is also afforded by an E-representation. Now (f) follows from (b) and (c). Assume IE :FI = n < CO. Then no = IE(x) : = IE : E n F(x)I which divides n. Since rn,(X)X is afforded by an &)-representation, we conclude from Lemma 9.18(c)that no rnE(X)X is afforded by an F(X)-representation.Now (c) and (a)yield (g). Finally, let p be the character of the regular F-representation of G . Then [ p , X] = ~ ( 1and ) (h) follows from (c). The proof is complete. I

A useful method for obtaining information about Schur indices is to use induced representations. Let F E C and H c G. If 9 is a character of H which is afforded by an F-representation of H,then SG is afforded by an F-representation of G. (See Theorems 5.8 and 5.9.) This idea underlies much of the remainder of the chapter. We use it to prove the following celebrated result. (10.3) THEOREM (Brauer) Let G have exponent n and let F = Q(e2ni/n). Then F is a splitting field for G and every ;c E Irr(G) is afforded by an F representation.

Chapter 10

162

Proof The two conclusions are equivalent by Corollary 9.11. Let x~Irr(G).Since F ( x ) = F , it suffices by 10.2(d) to show that mF(X) = 1. By Brauer's Theorem 8.4 we may write

where 1 runs over linear characters of subgroups of G and al E Z. Now a linear I E Irr(H) is obviously afforded by an F-representation of H and hence '1 is afforded by an F-representation of G. By 10.2(c),we have mF(X)1 [AG, X ] and since

it follows that mF(X) = 1 and the proof is complete.

I

We wish to exploit the ideas in the above proof in order to obtain some more delicate information about the Schur index.

(10.4) LEMMA Let H c G and F E C. Suppose $ E Irr(H) and Then m F ( X ) divides IF(XI$1 :F(X),'I$C XI.

x E Irr(G).

Proof We may replace F by E = F(x). This is so since m,(X) = mF(X) by 10.2(a)and mE($)lrnF($)by 10.2(f) (and of course F(x, +) = E($)). Thus we assume F = F(x). Let 9 'be the Galois conjugacy class of $ over F so that mF($) C Y is afforded by an F-representation of G. Therefore rn&) divides [qG* XI

mF($) v c y

by 10.2(c). Now if q E 9, then $ = rf' for some u E '3(F($)/F). We have x = X" and

Cr", XI = C(VG)", X"1

=

WG,XI

and thus mF(X)divides mF($)1 Y I [$', x ] . Since 1 9 ' 1 = IF($) : F 1, the proof is complete. I The next theorem is a variation on a result of Brauer and Witt. It allows us to analyze mF(X)one prime at a time in terms of certain sections of a group with sharply defined properties. (A section of G is a factor group, H / K , where K 4 H E G.) (10.5)

DEFINITION

Let F G @. Then (H, X,9) is an F-triple provided

(a) H is a group, X 4 H, X = C,(X); (b) 9 E Irr(H) is faithful; (c) the irreducible (linear) constituents of 9, are Galois conjugate over F(9).

The Schur index

163

(10.6) COROLLARY Let ( H , X , 9) be an F-triple and let I be a linear constituent of 9,. Then (a) (b) (c) (d)

I is faithful and X is cyclic; I&) = X , AH = 9 and F(9) E F(1); 9, is afforded by an irreducible F(9)-representation; H / X z Y(F(A)/F(9)).

Proof By part (c) of the definition, all linear constituents of 9, have the same kernel. Thus ker I c ker 9 = 1 and (a)follows. Since I takes on distinct values at distinct elements of X , it follows that if I" = I , then h E C ( X ) = X . Thus X = I&) and'A is irreducible by Theorem 6.11(a). Now (b) follows. Also 9, is the sum of 9(1) = J H: X I distinct linear characters. These must constitute the full Galois conjugacyclass of over F(9)since if 0 E 9(F(I)/F(9)), then [A', = [ I , 9.J # 0. Since mF&) = 1, Corollary 10.2(b)asserts that 9, is afforded by an irreducible F(9)-representation and (c) is proved. Finally, for each h E H , there exists a unique 0"E %(F(I)/F(9))such that = ( j l k - ' ) a h = I"k"h. Therefore, gkh = gkO,, A"-' = I"". Now = and we have a homomorphism, a: H + 3(F(I)/F(9)),defined by ~ ( h=) o h . Now ker tl = I&) = X . Since it was shown above that the H-conjugates of I constitute a full Galois conjugacy class over F(9),it follows that ct maps onto %(F(I)/F(9))and the proof is complete. I

(10.7) THEOREM Let x ~ I r r ( G and ) F E C. Assume pa divides m&) for some prime p . Then there exists an F-triple ( H , X , 9) such that

(a) H is a section of G ; (b) P" I mF(8); (c) H / X is a p-group; (4 PYI F(X, 9): F(X)I * Proof By Lemma 9.17(b), it follows that if ( H , X,9) is an E-triple for some E 2 F , then it is an F-triple. By 10.2(f, a) we may therefore replace F by F(x) and so we assume F = F(x). Use induction on I G I. If there exists K c G and JI E Irr(K) such that p$[+', x] and p$ I F(+) : F 1, then by Lemma 10.4 it follows that pa I mF(+)and we may apply the inductive hypothesis to K with respect to and obtain an F-triple ( H , X , 9) which satisfies (a), (b), and (c) and such that pklF(JI,9) : F(+)I. Since also p$IF(+) : F 1, we conclude that pyI F(9): F ( and the proof is complete in this case. We therefore assume that no such pair ( K , JI) exists. By Solomon's Theorem 8.7 we can write 1, = uQ(lQ)', where uQ E Z and Q runs over if, the set of quasi-elementary subgroups of G . It follows that

+

Chapter 10

164

and thus

CX, XI = 1~ Q C ( X Q ) XI. ~, Choose Q E Z such that p does not divide [(xQ)', x] = [xQ, xQ]. If $ is any irreducible constituent of xQ, then since F(2) = F , every element of the 1=

Galois conjugacy class of $ over F has equal multiplicity as a constituent of

xQ and we may write xQ = 1bAA, where bAE Z and A runs over sums of

Galois conjugacy classes over F in Irr(Q). Thus p does not divide bA2[A, A] and we choose A such that p,f'bA'[A, A]. Write A = 9, where 9 is a Galois conjugacy class over F and let $ E 9. Then bA = [xQ, $1 = [x, is not divisible by p and [A, A] = 1 9 1= IF($): FI is not divisible by p . By the result of the second paragraph of the proof, we conclude that G = Q is quasi-elementary. Thus G has a cyclic normal q-complement C for some prime q. By Theorem 6.15 we have ~ ( 1 is) a power of q. Now if a = 0, the result is trivial, taking H = X = 1, and so we assume a > 0 and p I rnF(x). Since rnF(x)Ix( 1) by 10.2(h), it follows that q = p and G/C is a p-group. Let X a G, with X 3 C be chosen maximal such that X is abelian. By the inductive hypothesis applied to G/ker x, we may assume that x is faithful. We shall show that (G, X,x) is an F-triple to complete the proof. We have X E CG(X)Q G. If X CG(X),then since G/X is a p-group, there exists U Q G with X c U G C,(X) and I U : X I = p . Thus U is abelian and this contradicts the choice of X . Thus part (a) of Definition 10.5 is established. Now let A be a linear constituent of xx and let S = {g E GI Ae is Galois conjugate to A over F } . If sl, s2 E S , there exist cl,cz E $(F(A)/F) with Asi = 1"'. It follows that A"'"' = A"'"' and thus slsz E S and S is a subgroup of G. Let T = ZG(A) E S . By Theorem 6.1l(b), there exists a unique q E Irr(T) such that qG = x and [ q x , A] # 0. Let $ = qs so that $' = x and $ E Irr(S). We claim that F(*) = F. Let c E %(F($, A)/F). Then xu = x and so [A", xx] = [A, xx] # 0. Thus A" = Ae for some g E G and since A" is Galois conjugate to A over F , we have g E S . Since Galois conjugate characters have the same inertia groups, we have T a S and so ( v " ) ~ '- E Irr(T ) .Now

-=

a g - l

((? )

G

) = (rl")G = xu = x

and

[((v')e-

I)X,

11 = C(rt")X,

ngl= [(?")x,

A"] # 0.

By the uniqueness of q we conclude that ( u ~ ) ~=- 'q and so

p = (q")S

= (qB)S = qs = $

The Schur index

165

since g E S . Since F($, A) is a Galois extension of F, it follows that $ takes values in F and thus F($) = F as claimed. By the result of the second paragraph of the proof applied to (S, $), it follows that S = G. Thus condition (c) of Definition 10.5 is satisfied and the proof is complete. I Before proceeding to our major applications of Theorem 10.7 we derive two easier consequences which demonstrate its power. These are included in Theorem 10.9. (10.8) LEMMA Let H c G and suppose H has a complement in G . Let cp E Irr(H) and suppose cpG = x E Irr(G). Then mF(x)divides cp(1) for every F c @. Proof Let U E G with UH = G and U n H = 1. Since(lu)Gisafforded by an F-representation, we have mF(x)I [( 1u)G, x]. Now [( 1u)G, x] = [( 1,'),, cpG] = [((lu)G)H, cp]. However, ((lu)G)H = ( l U n J H= p H , the regular character of H. Thus [(lu)G,x] = [ p H , cp] = cp(1) and the result follows. I

(10.9) THEOREM Let x E Irr(G) and assume p Im&) for some F E @ and prime p . Then the Sylow p-subgroups of G are not elementary abelian and pm,(x) divides I G I. Proof Let pa be the p-part of mF(x)so that a > 0. Since mF(x)Ix(l),the second statement will follow if pa+ divides I G I. Let (H, X, 9) be the F-triple whose existence is guaranteed by Theorem 10.7.Thus 9 = AH for some linear A E Irr(X) by Corollary 10.6(b). Since rnF(9)> 1, we conclude from Lemma 10.8 that X is not complemented in H. Since H/X is a p-group, it follows that a Sylow p-subgroup of H is not elementary abelian. Also, if P E Syl,(H), then P n X # 1 and so 9(1) = 1H:XI < IPI. Since p" I ~ ~ (<9(1), 9 ) we have pa < [PI and so p " + ' l l H I . The result now follows. I

There is an important case when we can decide whether or not mF(9) = 1 for an F-triple (H, X, 9).If H/X is cyclic, the problem "reduces" to a question , define in field theory. If E 2 F is a Galois field extension and ~ E Ewe N,,,(a) = a', The image of this norm map is clearly contained in F.

uoEg(,/,)

(10.10) THEOREM Let (H, X, 9) be an F-triple for some F 5 C and assume H = XC, where C is a cyclic group. Let A be a linear constituent of 9, and write E = F(1) and K = F(9). Let m = I X n C 1 and let E be a primitive rnth root of unity in @. Then E E K . Also, m,(9) = 1 iff E lies in the image of N E I K . Proof Since X n C G Z ( H ) and 9 E Irr(H) is faithful, we can choose a generator y of X n C, such that 9(y) = &(I) and thus E E F(9)= K . Write I X : X n CI = sand1H:XI = IC:X n CI = tandchoosegeneratorsxand c for X and C such that xs = y = c'. Note that A(x)s = A(y) = E. By Corollary

Chapter 10

166

10.6 we have t = 9(1) = I%(E/K)I = 1E:KI.

Write xC= x' for some integer r. By Definition 10.5 there exists a E 9 ( E / K )such that A'-' = 1".It follows that the orbits of I under C and under (a) are identical. Since C is transitive on the t distinct linear constituents of 9, we conclude that I(a)l 2 t = I9(E/K)I and thus %(E/K) = (a). Therefore, NEIK(a) = aa"auz

aUf-'

for a E E . Also, 1(x)r = I(xr) = Ac-'(x) = L(x)lr. Let V be the (unique up to isomorphism) irreducible K[X]-module which affords 9,. (See Corollary 10.6(c).)Now mF(9)= 1 iff 9 is afforded by a KCHI-module (by Corollary 10.2(d))and this happens iff there exists a K-linear action of C on V such that for v E V we have (a) u . c ' = ve; (b) ( ( u * c - ~ ) * x ) *=c u * x ' . (The sufficiency of this condition involves an easy generators and relations argument on H.) View E as an E [ X ] module affording I so that P . u = p1(u) for u E X. With this same action of X on E, we now view E as a KCXI-module of dimension I E : K 1 = t. By Lemma 9.18(b)it follows that V is an irreducible constituent of the K[X]-module E. (This is because I is a constituent of g E where 9 is a K-representation corresponding to V.) Since dim,( V ) = $( 1) = t = dim,(E), it follows that V E E as K[X] modules. We conclude that mF($)= 1 iff there exists a K-linear transformation c^ of E such that for all p E E (a') E' = PE; (b') ((P * E - ' ) I ( x* E) )= P(A(x))l. Now suppose that rnF(9)= 1 and let E be as above. Write 6 = I(x), u = 1 . E and let n 2 0 be an integer. We show that (6"). E = UPby induction on n. This is clear for n = 0. For n > 0, the inductive hypothesis and (b') yield 1"). c^- ')6). c^ = Ma("- 1 )'d' = ad"* (6").c^ = as claimed. Since 6" = 6' we have (6") - 2 = ~(6")".Since E = K(L) and 6 = L(x), the 6" span E over K . The map + ap" for /3 E E is K-linear and agrees with E on the 6". We conclude that p - E = ap" for all P E E. Now (a') yields & = 1 . (2)' = uu"u"z . . . a"t- - NE/K(a) as desired.

The Schur index

167

Conversely, suppose NEIK(a) = E for some tl E E. Define c^ on E by b . 2 = up" and observe that E is K-linear. Check that (a') and (b') are satisfied. Thus rnF(9)= 1 and the proof is complete. (10.11) LEMMA Let p be a prime and let 1 # k E Z.Assume that p I(k - 1) and if p = 2, assume 41(k - 1). Write

f(e) = ( k p " - l)/(k - 1 ) for 0 I e E Z.Then p' is the exact power of p dividingf(e). Proof Sincef(0) = 1, we assume e 2 1 and use induction on e. Since = 1 ( k - l)f(e - l), we have

kPe-'

+

kP' = 1

+ p(k - l)f(e - 1)

for suitable integers m,n. Thus

and it suffices to show that the second and third terms in the brackets are divisible by p2. Since p ( ( k - l), this is clear if p # 2. If p = 2, the hypothesis that 41(k - 1) yields the result. I We next give our principal result about the Schur index. Note that it generalizes Brauer's Theorem 10.3. (10.12) THEOREM (Goldschrnidt-lsaacs) Let x E Irr(G), where G has exponent n and let E be a primitive nth root of unity in C. Let F E C and assume that %(F(z)/F(x))has a cyclic Sylow p-subgroup P. Then p,+"m,(x) except possibly when p = 2, P > 1 and r$ F(x).

J-1

Proof We may replace F by F ( x ) and assume F(x) = F. Suppose p I r n F ( X ) and let (H, X,9) be the F-triple whose existence is given by Theorem 10.7 so that pJrnF(9).Let I be a linear constituent of 9x so that F(x) = F c F(9) c F(1) c F(E).By 10.6(d),H/X z B(F(L)/F(S))which is a section of %(F(&)/F(x)).Since H/Xis a p-group, it must be cyclic. Also, since 9(1) > 1, we have H > X and we conclude that P > 1. There exists a cyclic p-group C with XC = H.Let I C I = p' and I C n XI = pb and write E = F(1) and K = F(9). Since ~ ( 9#) 1, Theorem 10.10 yields that no primitive pbth root of unity lies in the image of N E I Kbut , that K does contain such a root. Since NEIK(l) = 1, we have b > 0 and K contains a primitive pth root of unity.

Chapter 10

168

Let y be a primitive p'th root of unity in C. Then I K(y) : K 1 divides p'-b and K(y) c F(E).Also I E : K I = I H :X 1 = pcPband E G F(E).Since Q(F(e)/K) is abelian and has a cyclic Sylow p-subgroup, its subgroups of p-power index are linearly ordered by inclusion. The fundamental theorem of Galois theory now yields that K(y) E E (since IK(y) : K I II E : K I) and so y E E. We compute NEJK(y). Write t = pCPb. If y E K , then NEIK(y)= yf, which is a primitive pbth root of unity, a contradiction. Thus y 4 K. Write Q ( E / K ) = (o), where o(a) = t and y" = y k for some integer k # 1. Then yUi = yki and where q = (k' - l)/(k - 1). If pc-b is the exact p-power divisor of q, then y4 is a primitive pbth root of unity which is not the case. By Lemma 10.1 1 then, p$(k - 1) if p # 2 and 4 k ( k - 1) if p = 2. Let 6 E K be a primitive path root of unity with a chosen as large as possible. Then 0 < b Ia. Since y 4 K , we have a < c and 6 E (7) so that 6 = 6" = dk and p"J(k - 1). Since a > 0, we have p = 2 and a = 1 so that -4 K . The proof is complete. I (10.13) COROLLARY (Fong) Let G have exponent n = mp", where p is prime and p$m. Suppose F c C and F contains a primitive mth root of unity. F , in which case Let x E Irr(G). Then m&) = 1 unless p = 2 and m F ( X ) 5 2*

PI$

Proof Ifp = 2 and -&F, let E = F[-] so that I E : FI = 2 and mF(X)I (2m,(x)) by Corollary 10.2(g).Thus it suffices to assume that EF if p = 2 and to prove mF(X) = 1. Let E be a primitive nth root of unity in C. The hypotheses on F now is cyclic. If q is a prime divisor of mF(X),then Theorem guarantee that Q(F(&)/F) F . Thus p # 2,4$m, and m&) is a power of 2. 10.12 yields q = 2 and - 4 We have 4$n and thus a Sylow 2-subgroup of G is elementary abelian and 2$mF(X) by Theorem 10.9. The result follows. I

J-1

(10.14) COROLLARY (Roquette) Let G be a p-group and z ~ I r r ( G ) . Let F s C. Then m&) = 1 unless p = 2 and 4 F in which case m F ( d I2. Proof Immediate from 10.13. I Note that taking G to be the quaternion group of order 8 shows that the exceptional cases in the three preceding results can, in fact, occur. (10.15) COROLLARY (L. Solomon) Let k be the product of the distinct prime divisors of [ G I and let F E C. Assume that F contains a primitive (24th root of unity. Then mF(z) = 1 for all x E Irr(G).

0

The Schur index

169

J-1

Proof Note that if 2 I I G 1, then E F. Let E be a primitive nth root of unity, where n is the exponent of G . Since every prime divisor of n divides k , the hypothesis on F guarantees that %@'(&)IF)is cyclic. If x E Irr(G) and p is a prime divisor of rnF(x),then Theorem 10.12 yields p = 2 and $ F. Since rnF(x)(X(l) we conclude that 2 I I G I and we have a contradiction.

J-1

Most of the above results are directed to showing that Schur indices are small. We have not yet seen an example to show that rnF(x)can be greater than 2. In fact every positive integer can occur as a Schur index. Here, we shall settle for an indication of how to prove that every prime can occur. We need two facts from number theory

(a) Let F = a(&), where E is a root of unity. Let R = Z [ E ] and let I be a proper ideal of the ring R. Let a E F . Then LY = u/u for some u, u E R with not both u, u E I . (b) (Dirichlet) Let a, b E Z with (a, b) = 1. Then there exists a prime of the form ak + b for some k E Z.

+

Let p be a prime. By (b) above, choose a prime q = p2k + ( p 1) for some k. Then p l ( q - 1) but p2&q - 1). We now construct the semidirect product G = QP, where Q is cyclic of order q and P is cyclic of order p 2 and acts nontrivially (but not faithfully) on Q. There exists faithful x E Irr(G) with ~ ( 1= ) p . We claim that rn&) = p.

(10.16) THEOREM Let G = X P , where X Q G is cyclic of order p q , P +i G is cyclic of order p 2 and I GI = p 2 q for primes p and q such that p2,i'(q - 1). Then there exists faithful x E Irr(G) such that rn&) = p . Proof Note that X = C,(X). Let A be a faithful linear character of X and let x = .'A Then x E Irr(G), x( 1) = p , and (G, X,x) is a Q-triple. Since m&)Ip by 10.2(h),it suffices to show that m&) # 1. Let K = Q(x), E = Q(A), and w E K be a primitive pth root of unity. By Theorem 10.10, it suffices to show that w is not in the image of N E I K . Let v be a primitive qth root of unity in E and write E = o v so that E is a primitive (Pq)th root and E = Q(E). Let R = Z [ E ] and let I 2 qR be a maximal ideal of R. Since R/I is a field of characteristic q, the only qth root of unity in R/I is 1. Thus if 0 E Y(E/K),then v = 1 E 'v mod I and since w E K , we have E~ = wv' = ov = E mod I . Since R = Z [ E ] ,it follows that r' = r mod I for all r E R . Since I E :K I = p , we conclude that NEIK(r)= r p mod I for all r E R . Suppose, by way of contradiction, that w = NEl,(a) for some a E E . By Fact (a), write a = b/c, where b, c E R but not both b, c E I . We conclude that

Chapter 10

170

bP = NEl,(b) = NE/K(c)NE/K(cI) = cpw mod I . Since not both b, c E I , we conclude that o* = (b*/c*)p,where *: R --+ R / f is the natural homomorphism. Now 1 + x + x2 + . . -+ xP-' = npZ: (x - a')and setting x = 1, we see that 1 - o divides p in R. Since p* # 0 in R/I, we conclude that o* # 1. Thus b*/c* is a primitive p 2 root of unity in RII. Since p I ( q - I), there exists a E Z such that a* is a primitive pth root of unity in R / I . Since ( c * ) ~= 1, we have E* = (a*)k for some k and hence E E Z + I . Thus R = Z [ E ] = Z + I and R/I E Z/(Z n I) = Z/qZ. Since p 2 X ( q - l), Z/qZ does not contain a primitive p 2 root of unity and this contradiction completes the proof. I Suppose F c E c @, with IE: F ( = n < co and let x ~ I r r ( G with ) F(x) = F. By 10.2(g),mF(x) divides nmE(x). In particular, if is afforded by an E-

x

representation, then m&) = l and so m& divides I E : F I. This suggests the question of finding minimal fields E 2 F over which x is afforded.

(10.17) THEOREM Let x ~ I r r ( Gand ) F E C with F(x)= F. Let m = mF(X). Then there exists E z F such that x is afforded by an E-representation and I E : F J = m. Proof Let V be an irreducible F[G]-module which affords mX (by Corollary 10.2(b)).Let D = EF[GI(V),the centralizer ring of V , so that D is a division ring by Schur's lemma. Let E be any maximal subfield of D.Then E acts on V and V may be viewed as an E-space. Since E commutes with the action of G, we may view V as an E[G]-module. Now EEIGi(V)is the centralizer of E in D. The maximality of E thus yields EErG1(V) = E and thus the E[G]-module V corresponds to an absolutely irreducible E-representation 3E by Theorem 9.2(a, c). Let 3 be the F-representation corresponding to the F[G]-module V. By Lemma 9.18, 3E is a constituent of 3".Since 3 affords rnx and X is absolutely irreducible, we conclude that X affords X. Finally, by 9.18(a) we conclude that m = I E : F I and the proof is complete. I

We can, of course, find an F-isomorphism of E into C in the above situation, and so the result remains true if we add the condition that E c @. However, it is not always true that if F G L E C, where L is a splitting field for G and x E Irr(G)with F(x)= F, then there exists a field E with F E E G L, 1 E :F 1 = m&) and such that x is afforded by an E-representation. We close this chapter by stating a few more facts about the Schur index. The proofs of some of these seem to require a fairly deep knowledge of number theory and division algebras. (a) (Fein) If integer n.

x E Irr(G), then m&)

divides n[~",1G] for every positive

Problems

171

(b) (Brauer-Speiser) (Corollary of (a)) If x E Irr(G) is real valued, then ma&) 5 2. (c) (Fein-Yumada) If x E Irr(G), then m&) divides the exponent of G and mQ(x)2divides I G I.

Problems (10.1) Let H

E

G, 9 E Irr(H) and x E Irr(G). Suppose F E C.

(a) If xH = 9, show that m&) 1 rnF(9)and mF(9) 5 1 G :H 1 mF(x). (b) If ' 9 = x, show that mF(9) I rnF(x) and m&) IIG : H 1 mF(9). (10.2) Let N 4 G and cp E Irr(N). Let F irreducible constituent of cp" and let

5 C and

put H = I,(cp). Let 9 be an that mF(x) divides

x = ,!JG. Show

I N G ( H ) : 1 md9)* Hint Consider {g E G I cpff and cp are Galois conjugate over F(x)}. (10.3) Let x E Irr(G) and F E C. Define I&) to be the greatest common divisor of the integers nL = IF(x, 1): F(x)I [A', x] as I runs over linear ) but that examples exist characters of subgroups of G. Show that m F ( ~1 IF(x) where m F ( X ) <

Hint

For the example, take G = Q s .

(10.4) Let x E Irr(G) and F E C. We say that x is F-semiprimitive if there does not exist H c G and $I E Irr(H) such that $IG = x and F($) = F(x).

(a) If x is F-semiprimitive and N -=I G, show that the irreducible constituents of zN are Galois conjugate over F(x). E F if I GI is even. Show that (b) Let G be nilpotent and assume x E Irr(G) is F-semiprimitive iff ~ ( 1 =) 1. (c) In the situation of (b) and the notation of Problem 10.3, show that IF(x) = 1 for all E Irr(G).

J-1

x

Hint A noncyclic p-group in which every normal abelian subgroup is cyclic is necessarily a 2-group and contains a cyclic maximal subgroup. this problem provides an alternate (and more Note Since mF(x)IIF(x), elementary) proof of Corollary 10.14. Since IF(x) can, in general, exceed Q(x), this result strengthens Corollary 10.14 slightly. (10.5) Let F E C be a field in which - 1 is a sum of two squares. Let G be a 2-group and x E Irr(G). Show mF(x) = 1. Hint Reduce to the case where x(1) = 2 and Qs E G.

Chapter 10

172

Note This strengthens Corollary 10.14 in a different direction and suggests a strengthening of Theorem 10.12. In fact, B. Fein has proved that the exceptional case in Theorem 10.12 can only occur if - 1 is not a sum of two squares in F.

(10.6) Let x ~ I r r ( G and ) F c C. Let H = G x G x ... x G be a direct . that product of copies of G and let $ = x x x x ... x x ~ I r r ( H )Show mF($) divides mF(2). Note In fact if H is the product of mF(x) copies of G, then rnF($) = 1. Conversely, if mF($) = 1 and H is. the product of n copies of G, then mF(x)ln provided IF : Q I c co.

(10.7) (Fein) Let H be the product of n copies of G and let x and $ be as in Problem 10.6. Show that ma($) divides [x", 1G]. Note Problem 10.7 and the note preceding it prove that m&) divides n[x", l G ] for every n > 0 and x E Irr(G).

(10.8) Let H = G x G and let x~Irr(G).Let $ = x x i ~ I r r ( H ) Show . that ma($) = 1. (10.9) Let G = Qs x E , and let x E Irr(G) be faithful. Show that m&) = 1. (10.10) Let G F c C.

=

H x K and $ E Irr(H) and 9 E Irr(K). Let x = $ x 9. Let

(a) Show that mF(x)divides mF($)mF(Q). (b) Show that equality occurs in (a) provided (mF($), 9(1)1F(9) : F I) = 1 and(mF(s),$(l)IF($):FI) = 1. (c) Let p , q be primes such that p$(q - 1) and q$(p - 1). Show that p q occurs as a Schur index. (10.11) Let x E Irr(G) and p I m&) for some odd prime p . Show that G contains an element of order pq, where q is some prime such that p I (q - 1). (10.12) Let p be an odd prime and let P be a nonabelian p-group of order p 3 and exponent p . It is possible to find H E Aut(P) such that H z Qs and C,(H) = Z ( P ) = C,(t), where t is the involution in H. Let G = P H , the semidirect product. Let 9 E Irr(P) with 9(1) = p .

(a) Show that 9 can be extended to 9 E Irr(G) such that $(t) = 1 1 and

$ ( t ) = p mod 4.

(b) Show that [g, x] = (p - 9(t))/4, where x E Irr(H), ~ ( 1 = ) 2. (c) Conclude that -1 is a sum of two squares in Q(eZ""P) if p or 5 mod 8.

=3

Problems

173

Hints Obtain 9 via Corollary 6.28. Compute 9 ( t ) by working in N = ( P , t ) . Note that ICN(t)l= 2 p and that the p - 1 Galois conjugates of 8, are all equal at t . If p = 3 or 5 mod 8, show that m&) = 1, where F = Q,(~~W ). P

Note Conversely, if -1 is a sum of two squares in Q(e2"i'P), then p f 7 mod 8.Some primes = 1 mod 8 can occur, however. (10.13) Let P ~ s y l , ( G ) be abelian of exponent pa. Show that p"$m,(x) for x E Irr(G). Hint If X Q G is cyclic and G/X is a p-group, then there exists V such that I/ n X = 1 and IG: VXI c pa.

E

G

(10.14) Show that the following two statements can be added to the conclusions of Theorem 10.7 provided p # 2. (e) If Y is the p-complement in X , then C,(Y) = X . (f) If U ~Sylp(X)and P E S ~ I J H ) ,then U E @(P), the Frattini subgroup. Hints If C,(Y) > U , show that H has a noncyclic normal abelian p subgroup. If M E P and M U = P , then QMMYis irreducible.

(10.15) Suppose m&) = x( 1) for some x E Irr(G) and F c C. (a) If H E G , show that all irreducible constituents of xH are Galois conjugate over F(x) and that rnF(9)= 9(1) for each such constituent 9. (b) If x # l,, show that G is not simple. (c) If 2)(~(1)or E F , show that G/ker x is solvable.

0

Hint

Use Lemma 10.4.

(10.16) Suppose that m&) = x(1) for all x E Irr(G) with F E C. Show that every subgroup of G is normal.

11 Projective representations

Let N 4 G and suppose 9 E Irr(N) is invariant in G . For each irreducible constituent x of 9‘ we have that xN = e(x)9, where e(x), the ramification, is a positive integer. In Chapter 6 we obtained some information about these mysterious integers. If 9 is extendible to G [which is equivalent to saying that some e(x) = 13, then the e(x)’s are exactly the degrees of the irreducible characters of GIN. (See Corollary 6.17.) We shall see that, in general, the e(x)’s are the degrees of irreducible “projective representations” of GIN. (11.1) DEFINITION Let G be a group and F a field. Let f :G -+ GL(n, F ) be such that for every g , h E G , there exists a scalar a(g, h) E F such that f(g)fi(h)= %7h)a(g,h). Then f is a projective F-representation of G . Its degree is n and the function a: G x G -+ F is the associatedfactor set of X. Note that the “factor set” a has nonzero values and is uniquely determined by X. Both of these observations follow from the fact that the matrices X ( g ) are nonsingular. Let Z(n, F ) c GL(n, F ) be the group of nonzero scalar matrices. [Note that Z(n, F ) = Z(GL(n, F)).] By definition, PGL(n, F ) = GL(n, F)/Z(n,F ) is the projective general linear group. If X is a projective F-representation of G of degree n, then the composition of f with the canonical homomorphism GL(n, F ) -+ PGL(n, F ) is a homomorphism G -+ PGL(n, F). Conversely, if n: G -+ PGL(n, F ) is any homomorphism, we can define a projective representation f of G by setting X@) equal to any element of the coset n(g) of Z(n, F ) in GL(n, F). 174

Projective representations

175

Before proceeding with the study of projective representations, we digress to give an instance of how they can arise in the study of ordinary representations. (1 1.2) THEOREM Let N 4 G and suppose 'I) is an irreducible C-representation of N whose character is invariant in G. Then there exists a projective C-representation X of G such that for all n E N and g E G we have

(a) W n ) = 'D(n); (b) X(W) = X(n)X(g); (4 = X(g)X(n), Furthermore, if X, is another projective representation satisfying (a), (b), and (c),then X,@) = X(g)p(g) for some function p: G + C", which is constant on cosets of N . Proof For g E G and n E N , write 'I)(gng- ') = 'I))"(n).Since 'I) affords a are similar representations G-invariant character, we conclude that 'I) and 'I))" of N . Now choose a transversal T for N in G (that is, a set of coset representatives).Take 1 E T . For each t E T , choose a nonsingular matrix P , such that Pt'I)Pt-' = 9'.Take P1= I. Since every element of G is uniquely of the form nt for n E N and t E T we can define 3 on G by X(nt) = 'I)(n)P,.Properties (a) and (b) are immediate and (c) follows since X(nt)X(m) = 'I)(n)~,'I)(rn) = 9(n)g'(rn)~~ = 'I)(ntmt- ')Pt = 3E(ntmtt ) = X ( n t . m).

Properties (a), (b), and (c) yield X(d'I)(n) =

= X(gng-' - 9 ) =

'I)(gns-')w

and

W ' I ) ( n ) X ( g ) - ' = 'I)(sns-'1 for all g E G and n E N . If A is any nonsingular matrix such that A'I)(n)A-' = 'I)(gng-')

for all n E N , then A - 'X(g) commutes with all g(n)for n E N and thus A - 'X(g) is a scalar matrix by Corollary 1.6. If Xo also satisfies (a), (b),and (c), we may take A = X,@) and conclude that Xo(g) = X(g)p(g) for some p(g) E C '. Also X(g)X(h)'I)(n)X(h)- 'X(g)-

'

= X(g)'I)(hnh-')X@)- = 'D(ghnh-lg-1).

Comparing this with

'

X(gh)'I)(n)X(gh)- = g(ghnh- ' 9 - ')

Chapter 11

176

yields X@)X(h) = X(gh)a(g, h) for some a(g, h) E C " and thus X is a projective representation. All that remains now is to check that p is constant on cosets of N . We have W M S ) P ( S ) = X o ( n ) X o ( d = fi,(ng) =

Since X(n)X(g)

= X(ng)

~(ns)dw).

is nonsingular, the result follows.

I

Of course, the above argument does not really require that 9 be a complex representation. Using Theorem 9.2 and Lemma 9.12, any absolutely irreducible representation will do. We begin our study of projective representations by considering the associated factor sets. (11.3) LEMMA Let a: G x G -, F " be the associated factor set of a projective F-representation of G. Then 4 x y , zb(x, y) = 4x9 Y Z M Y , 4

for all x, y, z E G. Proof Let X be the projective representation. We have W X ( Y ) X ( Z ) = %v(z)a(x,

Y ) = fi(xyz)a(xy, z)a(x, Y).

Also X ( x ) X ( y ) X ( z ) = X ( x ) X ( y M Y , 2 ) = X(xyz)a(x, Y M Y , 2).

The result now follows because all of the matrices are nonsingular.

I

(11.4) DEFINITION Let A be a possibly infinite abelian group and let G be any group. Then an A-factor set of G is a function a:G x G -, A such that

Z(XY, Z M X , Y ) = a@, Y M Y , 2 )

for all x, y, z E G. Thus the factor set of a projective F-representation is an F -factor set where F denotes the multiplicative group of F. Conversely, we shall show that every F -factor set is associated with a projective F-representation. To do this we introduce the "twisted group algebra." Let G be a finite group and F a field. Let a be an F -factor set of G. Let F"[G] be the F-vectorspace with basis {g I g E G}. (That is, there is a specified basis of F"[G] which is in one-to-one correspondence with G.) Define multiplication in F"[G] by 0 - h = $a(g, h) and extend via the distributive law. To establish that the multiplication thus defined is associative,it sufficesto check it on the basis elements g for g E G. That it holds there is immediate from the

Projective representations

177

definition of a factor set. The finite dimensional algebra F"[G] is the twisted group algebra with respect to a. Note that if a is the trivial F -factor set, that is, a(g, h) = 1 for all g , h, we can identify F"[G] with F [ G ] . In order to see that F"[G] has a 1, we need the following. (11.5) LEMMA Let a be an A-factor set of G. Then a(1, x) for all x E G.

=

a(1, 1) = a(x, 1)

Proof We have a(1 . 1, x)a(l, 1) = a(1, 1 .x)a(l,x).

Canceling a(1, x) yields a(1, 1) = a(1, x) for x E G. That a(1, 1) = a(x, 1) follows symmetrically. I Let a be an F -factor set of G and let v = a(1, 1)- E F . Now (vi)g = vgu(1, g ) = and similarly g(vT) = 3. Thus v i is the unit element in F"[G]. It is now immediate that the elements 3 E F"[G] for g E G all have inverses. Now let a be an F -factor set of G and let 9 be any representation of the algebra F"[G]. Define X(g) = g(Q).Then XE@)is nonsingular and X(g)fi(h)= 9(S)?Io(ti) = 5) = 9 ( i i w g , h)) = X@h)a(g,h) so that X is a projective representation of G with factor set a. Conversely, if X is a projective F-representation of G with factor set a, we can define a representation 9 of F"[G] by setting g(g)= X ( g ) and extending by linearity. In other words, the projective F-representations of G having factor set a are in a natural one-to-one correspondence with the representations of the twisted group algebra F"[G]. The situation is analogous to the connection between ordinary representations and the ordinary group algebra. Exactly as in the case with ordinary representations, we define two projective representations X and 9J to be similar if 9 = P-'XP for some nonsingular matrix P . Also X is irreducible if it is not similar to a projective representation in the form

(; :). Note that similar projective representations have equal factor sets and correspond to similar representations of the appropriate twisted group algebra. Also, irreducible projective representations correspond to irreducible representations of the algebra. Since every finite dimensional algebra has irreducible representations, we have proved the following. (11.6) COROLLARY Let c1 be an F -factor set of G. Then G has irreducible projective F-representations with factor set a.

Chapter 11

178

The set of A-factor sets of G forms a group under pointwise multiplication. In the language of group cohomology, this group is denoted Z2(G, A), the group of “2-cocycles.” If p: G --+ A is an arbitrary function, we can define 6(p):G x G A by

It is routine to check that B(p) is a factor set. Note that 6 is a homomorphism from the group of A-valued functions on G (with pointwise multiplication) into Z2(G, A). The image of 6 is the subgroup BZ(G,A) c Z2(G, A) which is called the group of “2-coboundaries.” The factor group Z 2 ( G , A)/B2(G,A) can be identified with the second cohomology group H2(G, A). (More generally in cohomology theory, one assumes that G acts on A. Here we are considering only the “trivial action’’ case.) The superscript 2 above refers to the fact that we are discussing functions of two variables on G . Since that will be the only situation considered here, we write Z ( G , A) for the group of A-factor sets B(G, A ) for the image of 6 and we define H ( G , A) = Z(G, A)/B(G, A). We say that two A-factor sets of G are equivalent if they are congruent mod B(G, A). Thus H(G, A) is the set of equivalence classes of A-factor sets on G. If X is a projective F-representation on G with factor set a, and p: G --+ F is any function, define ‘I) = Xp by g ( g ) = X(g)p(g).It is trivial to check that ‘I) is a projective representation of G with factor set fl = a6(p). Thus 3 and ‘I) have equivalent factor sets. If X and ‘I) are projective F-representations of G , we say that X and ‘I) are equivalent if 9 is similar to Xp for some function p: G --+ F ’. This is easily seen to define an equivalence relation which preserves irreduciblility. We can now give a necessary and sufficient condition for an invariant irreducible character of a normal subgroup to be extendible. (11.7) THEOREM Let N Q G and let 9 E Irr(N) be invariant in G. Let ‘I) be a representation affording 9 and let X be a projective representation of G satisfying conditions (a),(b),and (c) of Theorem 11.2. Let a be the factor set of X. Define p E Z(G/N, C ”) by p ( g N , h N ) = a(g, h). Then p is well-defined and H(G/N, @ ” ) depends only on 9. Also, 9 is extendible to G its image iff B = 1.

BE

Proof For rn, n E N and g, h E G, we have

cr(gn, hm)X(gnhrn) = X(gn)X(hrn) = X(g)X(nhm)

Projective representations

119

using 11.2(b)and (c). Furthermore, X(g)X(nhm) = X(g)X(hnhm)= X(g)X(h)X(nhrn) = a(g, h)X(gh)X(n"m) = a(g, h)X(ghnhrn)

by 11.2(c). Since X(gnhrn) = X(ghnhrn) is nonsingular, we have a(gn, hm) = a(g, h) and P is well defined. That P is a factor set is clear. If Xo were chosen in place of X, we have Xo = Xp where p: G C" is constant on cosets of N so that we can define v(gN) = p(g). If Xo has factor set go, then a,(& h) = 4 9 , h)P(g)P(h)P(gh)= P(gN, hN)v(gN)v(hN)v(ghN)and hence p is independent of the choice of X. If P q P - were chosen in place of 9,we can replace X by PXP- which leaves a, P and fl unchanged. Thus fl is uniquely determined once 9 is given. If 9 is extendible to G , we can choose X and 9 so that X is a representation. Thus a = 1 and hence p = 1. Conversely, if fl = 1, there exists v: GIN --f C" such that --f

'

P(gN, h N ) = v(gN)v(hN)v(ghN)-

Define p: G + C " by p(g) = v(gN). Now define X,@) X, is a projective representation of G with factor set go(g, h) =

a(g, h)P(g)-'P(h)-'P(gh)

=

X(g)p(g)- so that

= 1

and thus Xo is a representation of G. Furthermore, since X(1) = g(1) = I , we have 1 = a(1, 1) = p(l)p(l)p(l)-l = p(1). Thus p(n) = v(1) = p(1) = 1 for all n E N and X,(n) = X(n)p(n)-' = Thus Xo is an extension of 9 and the proof is complete. I

9(n).

A word of caution about Theorem 11.7 is appropriate. In the notation of the theorem, if CI E B(G, C ") so that cl = 1 in H(G, C "), it does not follow that 9 is extendible to G . For instance, take G = Qs,N = Z(G), and 9 the nonprincipal linear character of N . Then 9 is not extendible to G and yet H(G, C " ) = 1. (See Problem 11.18.) The theory of projective representations is closely related to that of central extensions. (11.8) DEFINITION A central extension of a group G is a (possibly infinite) group r together with a homomorphism of r onto G such that ker n E Z(T).

Chapter 11

180

(11.9) LEMMA Let (r,n) be a central extension of G with A = ker E. Let X be a set of coset representatives for A in r and write X = {x,lg E G } , where x(x,) = 9 . Define a: G X G A by X , X h = a(g, h)Xgh. Then c( E Z(G, A). Furthermore, the equivalence class of a is independent of the choice of X . Proqf That a is an A-factor set follows by computing X g X h X k two ways, using the associative law in r. If Y = {y,} is another set of coset representatives, then y, = p(g)x, for some p(g) E A . Now y gYh

=

pU(g)p(h)XgXh= &)p(h)a(g, hkgh = p(g)/dh)P(gh)-'a(g, h)ygh

and the result follows.

I

If A and U are abelian groups and I E Hom(A, U ) , then for each A(a) by A(Cr)(g,h) = l ( a ( g , h)). It is routine to check that A(a)E Z ( G , V).

a E Z(G, A), we define

(11.10) COROLLARY Let (r,n) be a finite central extension of G and let A , X = { x , } and a be as in Lemma 11.9. Let 9 be an (ordinary) F-representation of r such that the restriction ?IA is the scalar representation AZ for some 3, E Hom(A, F "). Define X(g) = 9 ( x , ) for g E G. Then X is a projective F representation of G with factor set 3,(a).Furthermore, W Y )= WY))P(Y)

for all y E r,where p: r + F is the function defined by p(y) = 3,(yx$,). Also X is irreducible iff 9 is and the equivalence class of X is independent of the choice of coset representatives X . Proof We have X(g)X(h)= g(xg)(I)(xh)= g ( a ( g , h)Xgh) = n(a(g, h))X(gh) and 3 is a projective representation with factor set y = ax,, where g = n(y) and a E A . Thus

A@). If y E r, we have

9b) = X ( g ) W = ~(.b))4Yx,:,,. In particular, X(G) and 9(r)span the same vectorspace of matrices over F and the assertion about irreducibility follows. Finally, if X1 is the projective representation determined by an alternate choice of coset representatives, we have

XlMY)) = 9 ( Y ) P 1 b r 1 = 3wY)MY)Plb)-1 and X and X, are equivalent.

I

Projective representations

181

Note that if '2) is an irreducible C-representation (or any absolutely irreducible F-representation), then the condition that be a scalar representation in Corollary 11.10 is automatically satisfied. Henceforth we shall consider only C-representations. (11.11) DEFINITION Let (I-, n) be a finite central extension of G. Let X be a projective @-representation of G. We say that X can be lifed to r if there exists an ordinary representation '2) of r and a function p: r + C " such that

2x4 = X(n(x))PL(x) for all x E r. Furthermore, (r,n) has the projective lifting property for G if every projective @-representationof G can be lifted to r. Note that if X is lifted to the representation r) on r,then O Ais necessarily a scalar representation and by Corollary 11.10 we can construct a projective representation X1 of G by choosing coset representatives for ker n in r.Then ~ r ( n ( x ) ) p 1 (= x )'2)(x)= X(n(x))Ax)

and hence X , is equivalent to X.Thus if (r,n) has the projective lifting property for G, then all projective @-representationsof G are equivalent to ones obtained via the construction in Corollary 11.10. We shall prove a theorem of Schur which asserts that every finite group has a finite central extension with the projective lifting property. The point of Schur's theorem is that it allows us to apply what we know about ordinary representations to the study of projective representations. For instance, in the situation of Corollary 11.10, if 9 is irreducible and F = 6,then we have deg X = deg '2) divides 1 r : A 1 = I GI by Theorem 3.12. Thus a consequence of Schur's theorem is that the degree of every irreducible projective C-representation of G divides I G I. (11.12) DEFINITION The Schur multiplier of G is the group H(G, C"). It is denoted M(G). For a finite abelian group A we use the notation A^ to denote the group Irr(A).If (r,n)is a central extension of G with A = ker n finite, we construct a homomorphism q: A^ --t M(G) as follows. Choose a set X of coset representatives for A in r and write X = {x,lg E G}, where n(xJ = g. Let a E Z ( G , A ) be defined by X g X h = a(g, h)X,h as in Lemma 11.9. For A E A define ?(A) = where A(a)E Z(G, C " ) is defined by A(cr)(g, h) = A(a(g,h)) and the bar denotes the canonical map Z(G, C ") -+ H(G, C " ) = M(G). Note that q is a homomorphism. The map q: A+ M ( G ) is independent of the choice of coset representatives X . This follows since another choice would yield a factor set p E Z(G, A), which is equivalent to a by Lemma 11.9. Since up- E B(G, A ) , it

AT),

Chapter 11

182

follows that I(a)I(!)-' = I(aj?-') E B(G, C " ) . Thus I(a)and I(!) are equivalent in Z(G, C " ) and n(a) = n(p). We shall call the unique homomorphism q: A + M(G)constructed above the standard map. (11.13) THEOREM Let (r,A) be a finite central extension of G and let q be the associated standard map. Let X be a projective C-representation of G with factor set y. Then X can be lifted to r iff 7 lies in the image of q. In particular, (r,A) has the projective lifting property iff q maps onto M(G). Proof: Let A = ker A and let X = { x g l gE G } be a set of coset representatives for A in r with n(xg)= g. Write X g X h = a(g, h ) x g h so that a E Z(G, A). Now suppose 7 = u(A)forsome I E A^. Then I ( a )is equivalent to y and we have W g , h)) = Y(9,h)P(g)P(h)p(gh)for some function p: G + @ " . Define '2) on r by '2)(axg)= A(a)X(g)p(g)for a E A and g E G. We have '2)(xg)'2)(xh)

= X(g)X(h)p(g)p(h)= X ( g h h ( g , h)p(g)p(h) = il(a(g, h))X(gh)/&h) =

'2)(a(g?h)xgh) = % ) ( x g x h ) . Since '2)(axg)= I(u)'2)(xg),it follows that '2) is a representation. Thus '2) lifts x to r. Conversely, if X can be lifted, we have '2)(x) = X(n(x))p(x)for some representation '2) of r and function p: r -+ @. Thus X( 1) = p(1)- '%)(1) and for every a E A we have '2)(a) = X(l)p(a) = p(a)p(l)-"2)(1) and so A(a) = p(a)p(1)- is a linear character of A. Now write v(g) = p(xg).We have

'

Thus and I(a)is equivalent to y. Therefore q(A) = IT)= 7

and the proof is complete.

I

Given an arbitrary finite group G we shall show that M(G) is finite and construct a central extension (r,n)of G with A = ker A such that the standard map q: A -+ M(G) is an isomorphism. This will thus be a group with the projective lifting property for G with smallest possible order, namely I G 1 IM(G)I, Such a group r is called a Schur representation group for G. An abelian group Q is divisible if for every x E Q and positive integer n there exists y E Q, with y" = x. For instance,F is divisible if F is algebraically closed.

Projective representations

183

(11.14) LEMMA Let A be an (infinite)abelian group and let Q G A with Q divisible. Assume I A : Q I < 00. Then Q is complemented in A . Proof Use induction on IA : QI. We may assume A > Q and choose a E A - Q. Let n = o(aQ)in A / Q and let u = a" E Q. Let u E Q with u" = u by divisibility and let b = au-' so that b" = 1. Since aQ = bQ, it follows that n = o(bQ) in A/Q and thus ( b ) n Q = 1. Now let A = A / ( b ) so Q = Q ( b ) / ( b ) satisfies 0 z Q and [ A : IA : Q ( b ) I < I A : Q I. By the inductive hypothesis, Q is complemented in A and thus there exists B E A with B n Q ( b ) = ( b ) and QB = A. Now Q n B = Q n Q ( b ) n B = Q n ( b ) = 1 and the proof is complete. I

el -

The above result remains true without the assumption that IA : QI is finite. We will not need that more general fact, however. (11.15) THEOREM Let F be an algebraically closed field and G a finite group. Then H ( G , F ") is finite and each of its elements has order dividing I GI. Furthermore, B(G, F " ) is complemented in Z(G, F "). Proof First, we argue that B(G, F ") is divisible. If /3 E B(G, F " ) and n is a positive integer, write /3 = 6(p)for some function p: G + F " . For each g E G, choose v(g)EF" such that v(g)" = p(g). Then 6(v)" = 601) = /3. We can thus apply Lemma 11.14 when we show that IH(G, F")I < 00. Let a E Z(G, F " ) and define =

nah?, 4.

xeG

For fixed g , h E G, we have a(g,hx)a(h,x) = a(gh, x)a(g,h). Taking the product over all x E G yields P@)P(h) = Ic(gh)a(g,hYC'

and thus alGl E B(G, F "). This shows that H(G, F ")has exponent dividing I GI. Now let U = {crEZ(G,FX)lalGt = l}. For a E Z ( G , F X ) , let A = (B(G, F " ) , a). By the result of the previous paragraph I A : B(G, F")I divides I G 1. Thus by Lemma 11.14, B(G, F ") is complemented in A and the complement is a subgroup of U. Thus a E B(G, F " ) U and hence B(G, F " ) U = Z(G, F " ) . Now every element of U is a function from G x G into { y E F lylGl = l}. Since this is a finite set, it follows that I U I < 03 and thus IH(G,F")I = IB(G,F")U:B(G,F")I I I U ( <

and the proof is complete.

00

I

Next we need a general method for constructing central extensions.

Chapter 11

184

(11.16) LEMMA Let A be an arbitrary abelian group and let aEZ(G, A). Then there exists a central extension (r,n) of G with ker n = A and such that a set of coset representatives {x,lg E G } exists with n(x,) = g and X & , = a@,h)Xgh* Proof Let = G x A as a set and define (9, a)@, b) = (gh, a(g, h)ab). That this multiplication is associative follows from the fact that a is an Afactor set. Let a(1, l)-l = z E A. Then (1, z)(g,a ) = (9, a(1, g)za) = (g, a ) by Lemma 11.5. Also (9- I, a- ' & I - I, 91- 'z)(g,a) = (LZ) and hence r is a group with 1 = (1, z). Clearly n: --t G defined by n(g, a ) = g is a homomorphism with ker n = A" = ((1, a)la E A } . That A" E Z(G) follows since a(1, g ) = a(g, 1) by Lemma 11.5. Since (1, za)(l, zb) = (1, zab), it follows that a t ) (1, za) defines an isomorphism A z A". We identify A and A' via this isomorphism. Let X = {(g,l)lgE G}. Then n maps X one-to-one and onto G and so X is a set of coset representatives for A in r. Now (9, l)(h, 1) = (gh,a(g, h)) = (1, za(g, h))(gh,1) by Lemma 11.5. Since ( 1 , za(g, h)) = a(g, h) under our identification, the proof is complete. I

(11.17) THEOREM (Schur) Given G, there exists a finite central extension (r,A) which has the projective lifting property for G . Furthermore, (r,n)can be chosen such that ker n = A Z M ( G ) and the standard map A^ --t M ( G )is an isomorphism. Proof Let M be a complement for B(G, C " ) in Z(G, @ " ) by Theorem 11.15. Let A = M. Define a(g, h) E A by a(g, h)(y)= y(g, h) for y E M . It is clear that in fact a(g, h) E M = A . Next (a(gh,k)a(g,h))(y)= y ( g k k)y(g,h) using the definition of multiplication in M.Similarly, (a(g,hk)a(h,W ( y ) = y(g, hk)y(h, k ) and since y runs over a set of factor sets, it follows that a E Z(G, A). Now let (r,n) and X = {x,lg E G } be as in Lemma 11.16 so that X 8 X h = a(g, h)Xgh. Let q: A^ + M ( G )be the standard map. We show that q maps onto. For 7 E M(G) = Z(G, C ")/B(G,C "), there exists y E M n 7 since M complements B(G, C") in Z(G, C"). Now define A on A = M to be the evaluation map at y. Note that A E A^. Now

447,h)) = a(g, N y ) = y(g, N

Projective representations

185

so that A(a)= y. Therefore q(A) = &j = 7 as desired. By Theorem 11.13, it follows that G. Also, IAl

=

lAl 2

r has the projective lifting property for

Iq(A)I = IM(G)I = [MI = ( A t

and so q must be one-to-one and A^ z M ( G ) . However A s 2.7(c) and the proof is complete. I

A^ by Problem

(11.18) COROLLARY Let 3 be an irreducible projective @-representation of G. Then deg(3) divides I GI.

Proof See the discussion preceding Definition 11.12.

I

Before going on to exploit Schur's theorem, we give another result which is useful in the computation of M(G).We have defined a Schur representation group of G to be a minimal central extension with the projective lifting property. The next theorem will yield another characterization. (1 1.19) THEOREM Let (r,TI) be a finite central extension of G with A = ker TI and let q: A^ + M ( G ) be the standard map. Let A , = A n r'.Then ker q

=

{ A E A I A , E ker A}.

In particular, q is one-to-one iff A E r'.

Proof Let X

be as usual and write X,Xh = a(g, h)xgh with q. Then 1 = q(A) = I(cr)and so A(M) E B(G, C ").

= {x,lg E G }

M E Z(G, A). Suppose A E ker

Thus A(a(g9 h)) = P(g)P(h)P(g@-

for some function p: G + @ '. Now define 2 on check that

r by I ( a x , ) = A(a)p(g) and

l(xg)x(xh)= l(x,xh).

(This is essentially the same calculation as in the proof of Theorem 11.13.) It follows that 1is a linear character of r which extends A. Now r' E ker 2 and so A , E ker A. Conversely, let A , E ker A. Then A is extendible to A, E Irr(Ar'/I-') by A,(ax) = A(a) for x E r'.Since T/T' is abelian, A. is extendible to A.l E Irr(r/r'). Now

A 1( x g ) A

1( x h )

=

A 1 (xgx h )

=

1 (Xgh).

Chapter 11

186

Define p(g) = Al(xg).Then A(a(g9

and the result follows.

h)) = P(g)P(h)P(gh)-

I

(11.20) COROLLARY Let

(r,n) be

a finite central extension of G with

A = ker n.

(a) If A E r',then A is isomorphic to a subgroup of M(G). (b) Assume IAl = IM(G)I. Then A E r' iff r has the projective lifting property for G . In this case M ( G ) z A . Proof (a) follows since A^ 2 A and (b)is immediate from Theorems 1 1.19 and 11.13. I

Thus we see that r is a Schur representation group for G iff T / A z G for some A c Z(T) such that I A I = I M(G)I and A c .'-I (11.21) COROLLARY Let p be a prime divisor of IM(G)I. Then a Sylow p-subgroup of G is noncyclic.

Proof Let r be a Schur representation group for G (which exists by Theorem 11.17). We may assume that T/A = G with A c Z(T). Let P / A E Syl,(G) and assume P / A is cyclic. Since A is central, it follows that P is abelian and thus r has an abelian Sylow p-subgroup. By Theorem 5.6, p$ I r' n Z(T)I. Since A c r' by Corollary 11.20, we have p$ I A I. Thus PXIM(G)I. I The next corollary is quite important. It can, of course, be proved without all of the complex machinery which we have developed. (11.22) COROLLARY Let N a G with GIN cyclic and let

Q E Irr(N)

be

invariant in G . Then 9 is extendible to G. Proof By Corollary 11.21, M ( G / N ) = H ( G / N , C " ) is trivial. The result now follows from Theorem 11.7. I

The above theory, together with some rather technical computations yields a tool which is very useful when studying the characters of groups with normal subgroups. Let N a G and let SeIrr(N) be invariant in G. Under these hypotheses we say that (G, N , 9) is a character triple. The analysis of this situation is much easier if 9 is linear. We shall use a Schur representation group for GIN to replace (G, N , 9) by another character triple (r,A , A) in which T/A z G / N and A is linear. Of course, this would not be of much value unless we knew that the character theory of was somehow closely tied to

Projective representations

187

the character theory of G. For instance, we would want ,Ir and sGto have the same numbers of irreducible constituents with the same ramifications. We define below the notion of "isomorphism" of character triples. This rather complicated definition makes precise some of the ways in which the character theory of two character triples can be related. We first introduce some notation. If (G, N , 9) is a character triple, let Ch(G 19) denote the set of (possibly reducible) characters x of G such that zN is a multiple of 9. Let Irr(G 19) be the irreducible characters among these, so that Irr(G 19) is the set of irreducible constituents of SG. Note that if N c H c G , then (H, N , 9) is a character triple and xH E Ch(H 19) whenever x E Ch(G 19). If T : U + I/ is an isomorphism of groups and cp E Irr( U), let cp' E Irr( V ) denote the corresponding character, so that cp'(u') = cp(u). (1 1.23) DEFINITION Let (G, N , 9) and (r,M , cp) be character triples and let T: G / N + T / M be an isomorphism. For N c H E G, let H' denote the inverse image in r of r ( H / N ) . For every such H , suppose there exists a map a H :Ch(H19) -+ Ch(H'1cp) such that the following conditions hold for H , K with N c K c H 5 G and x, $ECh(H19). (a) + $1 = OH(X) + O H ( $ ) ; (b) [x,$1 = LaH(X), OH($)] ; (c) o K ( x d = (aH(x))Kt; (d) ~ H ( x P=) aH(x)P' for B E IrdH/N).

denote the union of the maps o H Then . (z, a) is an isomorphism from G, N , 9) to (r,M , cp). Let

a'

Note that if (z, a) is an isomorphism from (G, N , 9) to (r,M , cp), then aH is determined by its restriction to Irr(H 19). (This follows from (a).)By (b) it follows that aH maps Irr(H 19) one-to-one into Irr(H'1cp). Therefore, to construct an isomorphism (T,a) it suffices to define aHon Irr(H 19) to be oneto-one, then extend the definition by (a) and check that (c) and (d) hold for x E Irr(H 19). (11.24) LEMMA Let (z, a):(G, N , 9) 3 (r,M , cp) be an isomorphism of character triples. Then oHis a bijection of Ch(H 19) onto Ch(H'1 cp) for all H with N c H c G. Furthermore, x(1)/9(1)= aH(x)(l)/cp(l) for all x E Ch(H 19). Proof If oH(xl)= for x i € Ch(H19), we have [xi,$1 = [oH(xi),OH($)] is independent of i for all $ E Irr(H 19). It follows that x1 = xz and hence aHis one-to-one. For xECh(H19) write e(x) = x(1)/9(1)and similarly set e(q) = q(l)/cp(l) for E Ch(H'1 cp). Note that aN(9)E Irr(M I cp) and so aN(9)= cp. We have

188

Chapter 11

xN = e(x)9and qM = e(q)cp and thus ~ ( o H (=x ( )~ H~( x ) ) M= o N ( x N ) = ajv(e(~)9) = e(~)cp and thus e(aH(x))= e(x)as claimed. By Frobenius reciprocity, we have

SH =

1

4x)x

~~lrr(H1B)

1

1

and comparing degrees yields e(x)'9( 1) = IH :N I9(1) so that e(# = I H : N I where x runs over Irr(H 19). Similarly e(q)' = IH' : MI = I H : N I for q E Irr(H'1 cp). Since 0,, maps Irr(H I 9) one-to-one into Irr(H'1 cp), we have 1 H : N I = Ce(x)'

=

1

Ce(aH(X))'ICe(q)' = ( H : N I .

It follows that every q E Irr(H'1 cp) is of the form a&) The result now follows. j

for some x E Irr(H 19).

(11.25) COROLLARY Isomorphism is an equivalence relation on character triples.

Proof: The reflexive and transitive properties are obvious. If (T,6): (G,

N , 9) -,

o-,M , cp)

is an isomorphism, then o,,: Ch(H 19) -, Ch(K Icp) is one-to-one and onto ~ (o,,)-' for where K = H'. We can, therefore, define 0 - l by ( o - ' ) = M E K G r where H = K ' - * . It is routine to check that (T-',

a-'):

(r,M , cp) -, (G, N , 9)

is an isomorphism. j

A nearly trivial example of an isomorphism of character triples is given by the following result. (11.26) LEMMA Let (G, N , 9) be a character triple and let p: G + be an onto homomorphism with ker p E ker 9. Let M = p ( N )and let cp E Irr(M) be the character corresponding to 9 E Irr(N/ker p). Then (G, N , 9) and (r,M , cp) are isomorphic character triples.

Proqf We have T:G / N -, T / M defined naturally from p. For N E H E G and x E Ch(H 19) we obtain ker p E ker x and we may view x as a character of H/ker p. Let a&) be the corresponding character of p ( H ) z H/ker p. Check that (T, a) is the desired isomorphism. j

Another example of an isomorphism of character triples is provided by the following.

Projective representations

189

( 1 1.27) LEMMA Let (G, N , cp) be a character triple and let r] E Irr(G) be such that q N cp = 9 E Irr(N). For N E H E G, define oH:Ch(H I cp) + Ch(H 19)by aH($)= $qH. Let i: GIN -+ GIN be the identity map. Then

(i, 4:(G, N, cp) is an isomorphism of character triples.

+

(G, N , 9)

Proof' It is clear that oH does map Ch(H 19)to Ch(H 19) and that properties (a), (c), and (d) of Definition 11.23 hold. It thus suffices to show that oH maps Irr(H1cp) one-to-one into Irr(H 19).This follows from Theorem 6.16. I Given invariant A, 9 E Irr(N) for N a G, with A(1) = 1, it follows that (G, N , A) and (G, N , 9) are isomorphic if 9A-I is extendible to G. (11.28) THEOREM Let (G, N , 9) be a character triple and let (r,n) be a finite central extension of GIN having the projective lifting property. Let A = ker n. Then (G, N , 9) is isomorphic to (r,A , A) for some A E A^.

Proof' By Theorem 11.7, the triple (G, N , 9) determines an element has the projective lifting property for GIN, we can find 1 E A^ such that r](A) = /?' where r] is the standard map (Theorem 1 1.13). Now let G* c G x r be defined by G* = {(g, x)la = n(x)},where = g N , the image of g in GIN. Note that G* is a subgroup of G x r. Let L = N x A and observe that L 4 G*. Define 9* and A* on L by 9*(n, a) = $(n) and A*@, a) = A(a). Note that 9*, A* E Irr(L) are invariant in G*. We have projection homomorphisms pG: G* -, G and py: G* + r.These maps are onto and ker pG = 1 x A E ker 9* and ker p,- = N x 1 E ker A*. It follows by Lemma 11.26 that (G*, L, 9*) is isomorphic to (G, N , 9) and (G*, L, A*) is isomorphic to (r,A, A). By Corollary 11.25, it suffices to show that (G*, L, A*) and (G*, L, 9*) are isomorphic and by Theorem 11.27 we will be done when we show that 9*(A*)- is extendible to G*. Let 9 be a representation of N affording 9 and let 3E be a projective representation of G as in Theorem 11.2. Let c( be the factor set of G belonging to 3E and let p be the corresponding factor set of GIN as in Theorem 11.7. Let {xglijE G I N } be a set of coset representatives for A in r with n(xg)= 8. Take xi = 1. Write xgxfi= y(a, h)xgfiso that y E Z ( G / N , A). Since r](A) = 8- ', we have

P E H(G/N, C " )= M ( G / N ) . Since

a

'

4 y ) B E B(GIN7

1

and hence

wa,m 4 g , h) = 47)' -

for some function v: GIN

-, C '.

-

v(a4

Chapter 11

190

Every element of G* is uniquely of the-form (g, axg)for g E G and a E A . Define 3 on G* by 3(9, ax,) = w 4 4 -

'v(8).

We compute 3(9 ,axg)3(h,bxd

=

X(gh)4ab)- 'a(9, h)v(8)v(h)

and

3(gh, aby(8, &#I;)

m-

h ) w -'W8, 'v(8@. Since these two expressions are equal, we conclude that 3 is a representation of G*. Now y(T, f) = 1 and a(1, 1) = 1 and it follows that u(ii) = v(1) = 1 for n E N. Therefore 3(n, a) = g(n)A(a)- and 3Laffords 9*(A*)-I . The proof is now complete. I =m

'

We now consider some applications. . 9 ~ I r r ( N be ) a con(11.29) COROLLARY Let N a G and z ~ I r r ( G ) Let stituent of z N .Then x( 1)/9(1) divides I G : N I. Proof Let T = ZG(8), the inertia group, and let II/eIrr(T) such that = x and t,hN = e9 (Theorem 6.11). Since X( 1) = I G : T I $(l), it suffices to show that $(1)/9(1) divides I T : N I. Let (r,A , A) be a character triple isomorphic to ( T ,N, 9) with 1linear. Let ( E Irr(rlA) correspond to t+h E Irr(T 19). Then II/( 1)/9(1) = c( 1 )/A( 1) = c( 1) by Lemma 11.24. Since A E Z(c) we have c(1) divides I l- : A I = I T : N 1 by Theorem 3.12. The result follows. I

II/'

Note that by repeated application of this result we can weaken the hypothesis that N is normal in G and assume only that N is subnormal, that is, that there exist subgroups Ni such that N a N l a ... a N k a G . In particular, if N is subnormal and abelian then since 9(1) = 1 we obtain the following generalization of Ito's Theorem 6.15. (11.30) COROLLARY Let N E G be subnormal and abelian. Then ~ ( 1 ) divides I G :N I for every x E Irr(G). (11.31) COROLLARY Let N a G and let 9 E Irr(N) be invariant in G. Suppose for every Sylow subgroup PIN of GIN that 9 is extendible to P. Then 8 is extendible to G. Proof Let (r,A , A) be a character triple isomorphic to (G, N, 9) and with A linear. If N c H G G and A E K E r are such that H and K correspond,

Projective representations

191

then 9 is extendible to H iff 1 is extendible to K . This follows from Lemma 11.24 since extendibility of 9 is equivalent to existence of x E Irr(H 19) with x( 1)/9(1) = 1 and similarly for extendibility of A. Thus 1 is extendible to the inverse image in r of every Sylow subgroup of T/A. By Theorem 6.26, we conclude that 1is extendible to and hence 9 is extendible to G, as desired. I We can use Corollary 11.31 to obtain an alternate proof of Gallagher’s Theorem 8.15, a proof independent of the results of Chapter 8. We are assuming that (G, N, 9) is a character triple in which (9(1),I G :N I) = 1 and that det(9) is extendible to G. By Theorem 6.25, 9 is extendible to H for every H with N c H E G and H/N solvable. The result now follows from Corollary 11.31. The following combines several of our extendibility criteria. (11.32) THEOREM Let N Q G and 9 E Irr(N), with 9 invariant in G. Assume for every prime divisor p of (9(1)0(9),I G : N I) that a Sylow p-subgroup of G is abelian or if p # 2, that 9 is p-rational. Then 9 is extendible to G . Proof By Corollary 11.31, it suffices to assume that GIN is a p-group for some prime p. If a Sylow p-subgroup of G is abelian, then 9 is extendible to G by Theorem 8.26. If p # 2 and 9 is p-rational, then 9 is extendible by Theorem 6.30. In the remaining case, ( 1 G : N 1, o(9)9(1))= 1 and Corollary 6.28 yields the result. I As another application of the theory of projective representations, we prove a theorem of T. Berger about the characters of solvable groups. Recall that x E Irr(G) is said to be quasi-primitive if xN is homogeneous (that is, a multiple of an irreducible) for every N 4 G. Primitive characters are necessarily quasi-primitive and, in general, the converse is false.

(1 1.33) THEOREM (Berger) Let G be solvable and suppose quasi-primitive. Then x is primitive.

x E Irr(G) is

In order to prove Berger’s theorem, we shall exploit the fact that there exists a central extension of G with the projective lifting property. We break the proof into several intermediate results. The first of these is independent of projective representations. (11.34) THEOREM Let L E H < G with L Q G, H maximal and G/L solvable. Let 9 E Irr(H) such that 9‘ = x and 9,. are irreducible. Then there exists M Q G such that xM is not homogeneous and M 2 L. Proof We may assume without loss that L = core,(H), the largest normal subgroup of G contained in H . Let K / L be a chief factor of G so that

192

Chapter 11

K / L is an abelian p-group and K $ H. Thus G = KH by the maximality of H. Since K/L is abelian, we have H n K a K and since H n K Q H we conclude that H n K Q G and thus H n K = L. Let C = CG(K/L). Then C -a G and so C n H Q H. However, K E N(C n H) and hence C n H a G. The maximality of L forces L = C n H . We have C = K(C n H) = K . If K = G then H = L Q G and since 9' E Irr(G) we have ZG(9)= L by Problem 6.1 and thus xr. is not homogeneous. We assume, therefore, that K < G and let M/K be a chief factor of G so that M/K is an abelian q-group for some prime q. If q = p , then (KIL)n Z(M/L) # 1 and the minimality of K / L yields K/L c Z(M/L). This contradicts C,(K/L) = K and we conclude P # 4. We suppose that xr. and 2, are homogeneous and write xu = eq for some q E Irr(M). Now x( 1)/9( 1) = I G : HI = I K : L I is a power of p . Since 9, E Irr(L), it is a constituent of qr.and hence q( 1)/9(1)is an integer which divides x( 1)/9(1). Therefore q( 1)/9(1) is a power of p . Since M/K is a q-group for q # p , we conclude that y~~ E Irr(K) by Theorem 6.18 or Corollary 11.29. Since xL is homogeneous, it is a multiple of 9, and thus qL is also. Let R = M n H and let z be any irreducible constituent of q,. Then [ z L ,QL] # 0 and hence z = QR/? for some /? E Irr(R/L) by Corollary 6.17. Since M = K R and K n R = L, we can find y E Irr(M/K) such that 7, = p. Now

However, since MH = G and M n H = R, we have

and thus zM = eqy. Since q is a constituent of T, by Frobenius reciprocity, we have [q, q y ] # 0. Since q K E Irr(K), Corollary 6.17 yields y = 1, and hence p = 1, and z = 9,. Thus q R is homogeneous and since = eq we have qR = e9, and q(1) = e9(1). This yields 9(1)1M : RI = ((9,),)(1) = e29(1) and [q,, qR] = e2 = IM: RI. It follows that q vanishes on M - R by Lemma 2.29. As q is G-invariant, it vanishes on M - Rg for all g E G and hence vanishes on M - O R g = M - L. In particular, q vanishes on M - K and so I M : KI = [qK, qK] by Lemma 2.29. This contradicts q K E Irr(K) and completes the proof. I (11.35) LEMMA Let (7,a): (G, N , 9) + (r,M, cp) be an isomorphism of character triples and let N c H c G. Suppose $ is a character of H. Then $ E Ch(H 19) iff t,hG E Ch(G 19). Also, if $ E Ch(H I$), we have aG($') = (a€I($))r.

Projective representations

193

Proof The first assertion is immediate from Frobenius reciprocity. To prove that oC($G)= when $ E Ch(H IS), it suffices to check that

XI = [(%($)KXI for all x E Irr(T I cp). Since aG maps Irr(G 19) onto Irr(T I cp) by Lemma 11.24, we have y, = aG(t)for some t E Irr(G 19). Now CoG($Gb

o,(t)l

CII/G, 51 = 5H1 = Ca,($h aH(tH11 = [a,($), = =

and the result follows.

[$?

C(d$)F'

(OG(t))Hl

OG(t)I

I Z c Z(T) be

(11.36) LEMMA Let such that r has the projective lifting property for G = r / Z with respect to the natural homomorphism K: r + G . Suppose K -a T, H E r7HK = r and H n K 2 Z. Let 9 E Irr(H/Z) and assume H n K E Z(9). Then there exists x E Irr(T) such that K c Z(x) and zH = 19 for some linear 1 E Irr(H). Proof Let r) be a representation of H which affords 9. Let L = H n K -a H. Since L E Z(9), g Lis a scalar representation and we can apply Corollary 11.10 to H/(L n ker 9) and construct a projective representation I of H/L such that g(h)= X(hL)p(h) for some function p: H -,C '. Now let z: G --t H/L be the composite of the natural maps

G

=

T/Z + T/K

=

HK/K 2 H/L

so that z(hk2) = hL for h E H and k E K. Let I*be the projective representation of G obtained by composing z with X so that X*(hkZ) = I(hL) = CD(h)p(h)-I. Now lift X* to the representation 3 of r so that 3 ( x ) = X*(xZ)v(x)for x E r, where v: r + C '. This yields

3Wk) = r)(h)p(h)- ' v ( W for h E H and k E K. Let 3 afford X. Restricting the above to H, we have 3 ( h ) = r)(h)p(h)-lv(h) and

hence L(h) = p(h)-'v(h) defines a linear character 1 ~ I r r ( Hand ) X, = 91.Thus x E Irr(T). For k E K we have

3 ( k )= which is a scalar matrix. Thus K

1)PL(1)- v(k)7

c Z(X)and the proof is complete. I

The next result is a generalization of Theorem 11.34 which includes Berger's Theorem 11.33.

Chapter 11

104

H < G with L Q G and G/L solvable. Let 9 E Irr(H), with gG = x E Irr(G). Then there exists M -a G such that M 2 L and xu is not homogeneous. (11.37) THEOREM Let L E

Proof We may assume that there do not exist subgroups Lo E H o < G and 9, E Irr(H,) such that L < Lo Q G and (90)G= x. We may also assume that H is maximal in G. We have L = core@). We assume that xL is homogeneous and write xL = acp with cp E Irr(L) so that (G, L, cp) is a character triple and 9 E Irr(H I cp). It follows from Lemma 11.35 that the hypotheses of the theorem remain invariant if (G, L, cp) is replaced by an isomorphic triple, and similarly for the conclusion. Therefore, by Theorem 11.28, we may assume that cp(1) = 1. Thus L E Z(x) and in particular, L E Z(9). Let (r,n) be a finite central extension of G having the projective lifting property. Let Z = ker n and identify G with T/Z via n.For subgroups U E G, let U* denote the inverse image of U in r so that U*/Z = U and G* = r. Let K/L be a chief factor of G so that K/L is abelian and K $ H.Thus K H = G and K n H -a K H so that K n H = L. We have 9 E Irr(H*/Z) and L* E Z(9). Also K*H* = r and K* n H* = L* 2 Z . Lemma 11.36 thus applies and yields q E Irr(T) such that AqH*= 9 for some linear I E Irr(H*) and K* c Z(q). We have x = gr = (AqH*)r= Arq

and hence Ar E Irr(T). Since AL, E Irr(L*), Theorem 11.34 applies and we can find M z L* with M Q r and (Ar)M not homogeneous. If M = L*, let v1 and v2 be distinct irreducible constituents of Let p be an irreducible constituent of qL*.Since L* E K* E Z(q) we have p( 1) = 1 and thus pv, and pv2 are distinct irreducible constituents of ( ~ 2 ~ ) ~ ~ = xL.. This is a contradiction since xL* is homogeneous. Therefore, M > L*. Since (Ir), is not homogeneous, Theorem 6.11 yields a subgroup T z M with T < r a n d $ E Irr(T) such that t+br = Ir. Thus

x

= qIr = q*r = (qT*)r.

Since Z E ker x we have Z E ker(q,$) and thus in G, x is induced from the proper subgroup T / Z < G. Also, T/Z z M/Z > L and this contradicts the first sentence of the proof. It follows that xL is not homogeneous and the proof is complete. I Problems (1 1.1) Let a, P E Z(G, F " ) , where F is a field. If a and fi are equivalent, show that F"[G] E F B [ G ] .

Problems

195

(11.2) Let a E Z(G, A ) and let g E G. Show that a(x, y ) = a(y, x ) for x, Y E (9).

Note In general, x y = yx does not imply that a(x, y ) = a(y, x).

(11.3) Let a E Z(G, A ) and let x , y E G commute. Assume that A is divisible. Show that a(x, y ) = a(y, x) iff the restriction of a to ( x , y ) is equivalent to the trivial factor set. Hirit Use Lemma 11.16. (11.4) Let a E Z(G, A). Say that g E G is a-special if a(g, c ) = a(c, g) for every c E C,@). Show that if g is a-special, then so is every conjugate of g in G.

Hint Let (r,n) be as in Lemma 11.16 with respect to G, A , and a. Then n(x) is a-special iff n(C,(x)) = C,(n(x)). Note If a and P are equivalent, then g E G is a-special iff it is P-special. Thus one can speak of &special elements for & E H(G, A).

(11.5) Let a E Z(G, A ) and let X be a conjugacy class of G. Show that X consists of a-special elements (see Problem 11.4) iff there exists a function p: %+ A such that P@)a(g, h) = Il(sh)a(h,sh)

for all g E X and h E G. Hint

See the hint for Problem 11.4. Consider the conjugacy class of x g

r, where xg is the coset representative of A in r corresponding to g. (11.6) Let (r,n) be a finite central extension of G with ker n = A. Choose a set of coset representatives for A in r and let a E Z(G, A ) be as in Lemma 11.9. in

Let 1E 2.For each x E Irr(TIA), choose a representation r) affording x and construct the projective @-representation X as in Lemma 11.10. Show that this defines a bijection of Irr(l-11) onto the set of similarity classes of irreducible projective @-representationsof G with factor set 1(a). , Show that C"[GI is semisimple. (11.7) Let ~ E Z ( G@").

Hint Show that (dim M)' = I G 1, where M runs over a representative set of irreducible @"[GI-modules.Use Problem 11.6.

(11.8) Let a E Z(G, F " ) . Show that dim Z(F"[G])is equal to the number of conjugacy classes of a-special elements in G. Hint

Use Problem 11.5.

(11.9) Let (G, N , 9)be a character triple. If E GIN, say that g is 9-special if 9 is extendible to ( N , g, c ) for all c E G with b,c] E N . Check that this is well

Chapter 11

196

defined. Let E H ( G / N , C " ) be as in Theorem 11.7. Show that B E GIN is 9-special iff it is 8-special. (See the note following Problem 11.4.) Note It is clear (without appeal to Problem 11.4) that conjugates of $-special elements in GIN are 9-special. (11.10) (Gallagher) Let (G, N , 9) be a character triple. Show that I Irr(G 19)I

is the number of 9-special conjugacy classes of GIN. Hint

Several of the previous problems are relevant.

(11.11) The character triple (G, N , 9) is fully ramijied if there exists E Irr(G 19) such that (~(1)/9(1))~ = I G : N I. Suppose (G, N , 9) is fully rami-

fied and G > N . Show that no cyclic subgroup of GIN can be its own centralizer in GIN. Hint

Use Problem 11.10.

(11.12) Let (G, N , 9) be a character triple. For x, y E G with [x, y] E N , define the complex number ((x, y>>as follows: Let $ be an extension of 9 to H = ( N , y ) . Write @' = $2 for 2 E Irr(H/N) and put ((x, y>>= 2(y). Show that (( ,>> is uniquely defined and that

(a) ((x, y>> = (xn, ym>>for n, m E N ; (b) ( ( X l X 2 , Y>>= e l , Y > > < X 2 Y > whenever cx1, YI, cx29 Y l E N ; (c) ((x, y>> = 9([x, y]) if 9 is linear. Note By (a) we can view (( ,>> as being defined on commuting pairs of elements of GIN. (11.13) Let (G, N , 9) be a character triple. If N E H E G , @ E Ch(H19) and = g N E GIN we define @# E Ch(H819) by $#(hs) = @(h).Check that this is well defined. We say that the isomorphism.

(r, 4 : (G, N , 9)

+

(r,M , rp)

a

is strong provided (oH(@))1(a) = C T ~ ( @ #for ) all E GIN, all H with N E H c G and all $ E Ch(H 19). Show that the isomorphism constructed in Theorem 11.28 is strong. (11.14) In the notation of Problem 11.12, show that ((x, y>> = ((y, x>-'

and <x, YlY,>> = <x, Y l > > < X ,

Y2>> if

cx, Yll, 1x9 y21 E N .

Hint Viewing (( ,>> as defined on commuting pairs of elements of GIN, it is invariant under strong isomorphisms of character triples. (See Problem 11.13.)

Problems

197

(1 1.1 5 ) Let (G, N , 9) be a character triple. (a) If g E G and ((9, x) # 1 for some x E G, show that x(g) = 0 for all

x E Irr(G 19).

(b) If g E GIN, show that g is $-special iff ((9, x> = 1 for all x E G such that ((9,x) is defined. See Problems 11.12 and 11.9 for definitions. (1 1.16) Let E be elementary abelian of order p". Show that M ( E )is elementary abelian of order pncn-')/,. Hint

Consider a Schur representation group for E to get IM(E)I

- pn(n-1)12.For equality, let { x i [1 I i I n } be a generating set for E . < Define aij: E x E -+ C by u i j ( nx p a p , x p b ~=) where E is a primitive

n

E'I~J,

pth root of unity. (11.17) Let G = A , . Show that IM(G)I

=

2.

Note In fact, IM(A,)I = 2 for all n 2 4 except for n = 6, 7, where the Schur multipliers have order 6.

(1 1.18) Let G = C H , where H and C are cyclic, C 4 G, and H n C = Z(G). Show that M(G) = 1.

r is a Schur representation group for G, show that I G' 1 = I r'I. (1 1.19) Let (I?,, n,) and (r,,n2)be Schur representation groups for G. Define r G rl x r2by r = {(x, y ) I n , ( ~=) n2(y)}and show that Yl E r'z r;. Hint Let Ai = ker x i and define n: r + G by x(x, y) = n,(x), so that ker 'IZ = A , x A , and (r,n) is a central extension of G. Use Theorem 11.19 to show that Ir'I IIr; I. Note that IF:r' n ( A , x A , ) ) = IG'I. Note One cannot conclude that r, z r2as the two nonabelian groups of order p 3 show. If G = G then T i = r; and r, z T2 in this case. Hint

If

12 Character degrees

Let c.d.(G) denote* the set {x(l)lx €Irr(G)}, where G is a group. In this chapter we consider what can be said about G when c.d.(G) is known. We have already seen in Theorem 6.9 that if everyf E c.d.(G) is a power of the prime p then G has an abelian normal p-complement. The first results in this chapter consider a weaker hypothesis, namely that p divides f for every f E c.d.(G) withf > 1. Fix a prime p . Write Y ( G ) = {x~Irr(G)lpYx(l)and pYo(x)I. Let s(G) = C Y ( G ) X ( W Then IOP(G)IE s(G) mod p . (12.1)

THEOREM

Proof Let N = OP(G). Since Op(N)= N , N can have no irreducible character with determinantal order divisible by p . Thus Irr(N) = 9“) u {$E1rr(N)Ipl$(l)} and hence I N I = E@slrr(N)$(l)* s(N) mod P. It will therefore suffice to show that s ( N ) = s(G) mod p . Now G acts on Y ( N ) and the resulting orbits are of p-power size. Let 9,= { $ E Y ( N ) I $is G-invariant}. Since all characters in an orbit have $(1)2 mod p . We show that reequal degrees, it follows that s ( N ) = striction defines a one-to-one map of Y ( G )onto Y oand this will complete the proof. If x E Y ( G ) then ( ~ ( l )I G , : N I) = 1 and so xN E Irr(N) by Problem 6.7. Thus xN E Y o Conversely, . if $ E 9, then by Corollary 6.28, $ is extendible to G and a unique extension of $ lies in Y(G), The result now follows. 1

x9EY0

* The initials c.d. stand for “character degrees.” 198

Character degrees

199

(Thompson) Suppose p 1 x( 1) for every nonlinear x E Irr(G), where p is a prime. Then G has a normal p-complement. Proof In the notation of Theorem 12.1, all x E Y ( G )are linear and have

(1 2.2) COROLLARY

kernels which contain OP’(G). It follows that Y ( G )= Irr(G/G’Op’(G))and s(G) = IG: G’OP‘(G)I.Thus p#s(G) and hence by 12.1, p#IOP(G)I. Thus

OP(G)is a normal p-complement for G .

I

The following lemma is very useful for inductive proofs of theorems giving information about G when c.d.(G) is known. (12.3) LEMMA Let G be solvable and assume that G is the unique minimal normal subgroup of G . Then all nonlinear irreducible characters of G have equal degree f and one of the following situations obtains: (a) G is a p-group, Z(G) is cyclic and G/Z(G) is elementary abelian of order f (b) G is a Frobenius group with an abelian Frobenius complement of orderf: Also, G’ is the Frobenius kernel and is an elementary abelian p-group.

’.

Proof If Z(G) # 1, we must have that Z(G) is a cyclic p-group and G c Z(G)with 1 G 1 = p . Every x E Irr(G) with ~ ( 1>) 1 is faithful and satisfies ~ ( 1 ) ’= I G : Z(G)I by Theorem 2.31. If x, y E G , then since [x, y ] E Z(G), we have [ x p ,y ] = [x, y I p = 1 since I G’I = p . Thus x p E Z(G) for all x E G . This completes the proof in situation (a). Now suppose that Z(G) = 1. Certainly, G’ is an elementary abelian p-group for some prime p . Choose 41 [GI, a prime different from p and let Q E Syl,(G). Then G‘Q Q G and it follows by the Frattini argument that G = G“ where N = NG(Q). Now N n G ’ a N since G Q G and N n G Q G’ since G’ is abelian. Thus N n G ’ a G. Now Q-#I G and so N < G and N 8 G’. By the minimality of G‘, it follows that N n G = 1 and thus N is abelian. If 1 # X E N , then N normalizes C,.(X) since N centralizes x. Since C,.(x) Q G’ we have C,.(X) Q NG‘ = G . If C,,(X) # 1, then x centralizes G’ and it follows that x E Z(G), a contradiction. It follows that C,.(X) = 1 and thus G is a Frobenius group with kernel G‘ and complement N by Problem 7.1. We conclude that all nonlinear x E Irr(G) are of the form I‘ for linear I E Irr(G’). Thus I G : G’I = IN I is the common degree of all nonlinear irreducible characters of G . The proof is complete. I The way Lemma 12.3 is applied in practice is the following. Given nonabelian G , let K a G be maximal such that G / K is nonabelian. Then (GIK)’

Chapter 12

200

is the unique minimal normal subgroup of G/K. Thus either G/K is nonsolvable or else satisfies the hypotheses of Lemma 12.3. If x is a character of G, we introduce the notation V(x) = (g E G I x(g) # 0). Then V(x), the uanishing-o~subgroup,is the smallest subgroup, V E G such that x vanishes on G - V . Of course, V(x) a G. The relevance of V(x) to character degrees is this. Suppose x E Irr(G) and V(x) c N a G. Then by Lemma 2.29, we have [x,,XN] = IG : NI. Now write 1, = e Si where the Si E Irr(N) are distinct and of equal degree (Theorem 6.2). We have I G : N I = [x,,XN] = e2t and x(1) = et$(l), where 9 is one of the Si. It follows that J G: N19(1)* divides ~ ( 1 ) In ~ . particular, I G : N I 5 x( 1)2 and x( 1) is divisible by every prime divisor of I G : N I. The following result is useful when K a G and G/K satisfies the conditions of case (b) of Lemma 12.3.

c:=

(12.4) THEOREM Let K a G be such that G/K is a Frobenius group with kernel N/K, an elementary abelian p-group. Let E Irr(N). Then one of the following holds.

(a) I G : N I $( 1) E c.d.(G). (b) V($) E K and thus IN : K I divides $(1)2. Proof For I€Irr(N/K), let T ( I ) denote ZG($I), so that T(A)2 N. If T ( I )= N for some I , then (I$)GE Irr(G) and hence I G : N I $(1) E c.d.(G). We suppose then, that T ( i )> N for all I . Let W = V($)K so that K E W G N and let S = NG(W)2 N. Since ZG($) normalizes V($), we have ZG($) E S . However, V($) = V(I$) for all I E Irr(N/K) since the I are linear, and it follows that T ( I )c S for all I . Now view N/K as an F[S/N]-module, where F is the field with p elements. Since G/K is a Frobenius group, we have IGIN I divides IN/K I - 1 and thus p$ I GIN I. In particular, p$ I SIN I and hence N/K is completely reducible as an F[S/N]-module by Maschke's Theorem 1.9. Since W/K is a submodule, we can write N/K = (W/K) x (U/K), where S normalizes U . Now suppose W > K so that U < N. Let I E I ~ ~ ( N / with U ) I # 1,. Since G/K is a Frobenius group, we have = , N and thus Z,(I) = N. We claim that there exist distinct Therefore, IIrr(N/U)I 2 1 + IS/". 2, p E Irr(N/U) with T ( I )n T ( p )> N . If not, then since N < T ( I )E S for all I E Irr(N/U) we have

ISIN1 - 1 2

c (I T(I)/NI

- 1) 2

IIrr(N/U)I > IS/N I>

where the sum runs over 2~1rr(N/U).This contradiction shows that I , p E Irr(N/U) exist with T ( I )n T ( p )> N and I # p, as claimed. Now let x E (T(1) n T(p))- N and write v = I p . Then I" = pxvxand

$1

=

($ny = $"I"

= ($"pX)vX = ($p)Xv" = $pv"

Character degrees

201

and thus $v = $v" and we have V($) c ker(Vv"). Also U E ker(ijv")and thus N = V($)U E ker(Vvx) so that ijvx = 1, and vx = v. Since 2 # p, we have v # 1, and I,(v) = N. This contradicts x q! N and thus proves that W = K and V($) E K as desired. That this implies IN : K I I11/(1)2 follows from the remarks preceding the statement of the theorem. I In particular, in the situation of Theorem 12.4, we have for $ E Irr(N) that either I G : N I $(1) E c.d.(G) or p I $(1). This most important consequence of 12.4 could have been obtained with somewhat less work than was needed to prove the full strength of the theorem.

(12.5) THEOREM Let c.d.(G) = {1, m}. Then at least one of the following occurs. (a) G has an abelian normal subgroup of index m. (b) m = pe for a prime p and G is the direct product of a p-group and an abelian group.

Proof Suppose there exists K a G such that G / K satisfies conclusion (b) of Lemma 12.3. Let N/K = (G/K)' so that IG : N 1 = m and N/K is a p-group. Since G/K is a Frobenius group, m I ( I N/K I - 1) and so pym. We claim that N is abelian. Let $ E Irr(N) and let x be an irreducible constituent of $'. Then ~ ( 1 = ) 1 or m and $(1) I x( 1). In particular, p$$( 1). By Theorem 12.4, m$(1)E c.d.(G) and thus $(1) = 1. This establishes the claim and situation (a) of the theorem holds in this case. We now suppose that no K d G as above exists. Let n be the set of prime divisors of m. By Corollary 12.2, G has a normal p-complement for every p E n. If 1x1 > 1, then G has no irreducible character of p-power degree and hence G/OP(G)is abelian for all p . It follows that G' is a d-group. Since the elements of c.d.(G') divide elements of c.d.(G),we conclude that c.d.(G) = { 1 ) and G' is abelian. In particular, G is solvable.Now let K a G be maximal such that G/K is nonabelian. Thus G/K satisfies the hypotheses of Lemma 12.3.By assumption, we are in case (a) of the lemma and this contradicts m not being a prime power. We may now assume that 7c = { p } . Let A be the normal p-complement of G so that A is abelian. Let 1E Irr(A) and let T = IG(l).If $ is any irreducible constituent of ,'A then by Clifford's theorem, IG : TI divides $(1), and thus I G : TI I m. Now ( 1 T)G is not irreducible and has degree I m. It follows that all of its irreducible constituents are linear and thus G' E ker((1.)') E T . (Note that we have just done Problem 5.14(c).) Now if D = ZG(2), then A E Z(g) for all 5 eIrr(D) and thus A E Z(D). Thus D = A x P for P E Syl,(D). Since G' E D, we have D a G and hence P a G. If G E P then [G, A] c A n P = 1 so that A E Z(G)

nAE,rr(A)

Chapter 12

202

the result follows. Otherwise, let K 2 P, K Q G with K maximal such,that G / K is nonabelian. Since G is solvable, the hypotheses of Lemma 12.3 are satisfied. By assumption we are in case (a)of the lemma and G / K is a p-group. Thus K 2 OP(G)= A. We have K I> AP = D 2 G' and this contradicts G / K being nonabelian and completes the proof. I (12.6) COROLLARY If Ic.d.(G)I = 2, then G' is abelian. Proof Let c.d.(G) = 11, m}. If A Q G with J G :A J = m then G / A can have no nonlinear irreducible characters and so is abelian. Thus if case (a)of Theorem 12.5 holds, we are done. In case (b) of 12.5, G is nilpotent and hence is an M-group by Corollary 6.14. The result follows by Theorem 5.12. I

In the case that m is a prime we can sharpen Theorem 12.5 to give a necessary and sufficient condition on G that c.d.(G) = (1, m } . The following results prove more than is needed for this and will be used again. Theorem 12.7 is actually a generalization of Theorem 6.16. (12.7) THEOREM Let N 4 G and suppose 9,, s2 E Irr(N) are invariant in G and s19, E Irr(N). Let xi be an irreducible constituent of (Qi)' for i = 1, 2 and let $ be an irreducible constituent of x1x2.Then $(1)x1(1) 2 x2(1)91(1)2. Proof We have 0 # [x1x2, $3 = [x2, $jl]and thus x2 is a constituent and it follows that of $jl.Also ( x 2 ) N = (~~(1)/$~(1))9~ x2(1)/92(1)

5

C92, ( $ N ) ( X I ) N l

=

[Q2(X1)N,

$N].

However, ( x ~ =) (~1(1)/91(1))Ql ~ and this yields

Now 9,9, is the unique irreducible constituent of and thus [9,9,, $N] $(1)/91(1)92(1).Substitute this in the above and simplify to obtain the result. I =

(12.8) COROLLARY Let N u G and suppose fi E Irr(G) with N c Z(P). Let 9 ~ I r r ( N ) Then . there exists an integer b such that bS(l)~c.d.(G)and b2 2 P(l)t, where t = I G : 1,(9) I. Proof Let T =

and let y be an irreducible constituent of P T . Then

p is a constituent of yG and so p(1) I; y(1)t. Also, N c Z(y) and we write yN = y ( l N Let 5 be an irreducible constituent of ST and let q be an irreducible

Character degrees

203

constituent of (y. Since 91 E Irr(N), Theorem 12.7, applies and so <(l)rl(l)2

Y(1PW2.

Thus TG(l)qG(l)2 y(l)t28(1)22 p(l)t9(1)2.

eG,

Now T = ZG(9)= ZG(9A)and hence qG E Irr(G) by Theorem 6.1 1. Let x be whichever of (‘, qGhas larger degree. Then x( 1) = b9( 1) for some integer b and b29(1)2= ~ ( 1 2 ) ~(‘(l)qc(1) 2 /3(l)t9(1)2 and the result follows.

I

(12.9) COROLLARY Let c.d.(G) = (1, p } , where p is a prime. Then there exists abelian A -a G with I G : A I = p or p 2 .

Proof By Theorem 12.5, we may assume that G is a p-group. Let K Q G be maximal such that G / K is nonabelian and let Z / K = Z ( G / K ) .By Lemma 12.3 it follows that I G : 21 = p 2 . Let /?E Irr(G/K) with /3( 1) = p and let 9 E Irr(Z). Then since Z E Z(p), Corollary 12.8 yields an integer b such that b2 2 p(1) = p and b9(1) E c.d.(G). This forces 9(1) = 1. Thus Z is abelian and the proof is complete. I (12.10)

LEMMA

Let A E G be abelian and let b = max(c.d.(G)). Then

(l/IAI)x

2

ICG(a)I

.€A

Proof Since 1 C,(U)I =

x

(l/IAl)

ICG(a)I

1,I~ ( al 2 )for x =

a

IGI/b*

E Irr(G), we

(l/IAl) 1 IX(a)I2 = x

a

have

1

[XA, XA1.

X

However, xA is the sum of ~ ( 1 linear ) characters and hence Thus

[ z A ,X A ] 2 ~ ( 1 ) .

1X(l)* Furthermore, I G I = 1,~ ( 1 I ) ~b cxX( 1) and thus 1,~ ( 1 2 ) I GI / b and the (l/IAl)

ICG(a)I 2

a

result follows.

X

I

(12.11) THEOREM Let G be nonabelian and let p be a prime. Then c.d.(G) = (1, p} iff one of the following holds. (a) There exists abelian A (b) I G : Z(G)I = p 3 .

4

G with I G : A I = p .

Proof If (a) holds, then ~ ( 1Ip) for every x E Irr(G) by Ito’s Theorem 6.15. Since G is nonabelian, c.d.(G) = { 1, p}. If I G : Z(G)I = p 3 , then ~ ( 1 ) ’ I p 3 by

Chapter 12

204

Corollary 2.30 and x(l)lp3 by 6.15 (or 3.12). Again it follows that c.d.(G) = (1, P I . Conversely, suppose c.d.(G) = { 1, p } and assume that (a) is false. We may assume that G is a p-group. By Corollary 12.9, there exists abelian A 4 G with I G :A I = p z . By Lemma 12.10, we have ( l / I A I ) x ICG(a)l 2 IGl/p. aeA

Now A acts by conjugation on G - A and by Corollary 5.15, the number of orbits of this action is ( l / ~ ~ ~ ) ~ ( ~ -c IGA I )( a2) (IG\/P) ~ aeA

It follows that the average size CI of these orbits satisfies a

(IGl - lAl)/((lGl/P) - I A I ) = P

+ 1 < PZ

and so A has an orbit of size 1 or p on G - A . Since G does not have an abelian subgroup of index p , A = C G ( A ) and thus there exists x E G - A in an orbit of size p . Let K = ( A , x ) . Then Z(K) = C , ( X ) has index p in A and index p 3 in G. We shall show that Z(G) = Z(K) to complete the proof. ) xK is irreducible, then Z c Z(x) and Let Z = Z(K). If z ~ I r r ( G and [G, Z ] c ker 1.On the other hand, if xK reduces, then all irreducible constituents are linear and K' c ker x. Suppose [G, Z ] > 1 . We also have K' > 1 but K' n [G, Z ] G ker x for every x E Irr(G) so that K' n [G, Z ] = 1. Now K'[G, Z ] is the direct product of two nontrivial groups and so has an irreducible character 9 with K' $ ker 9 and [G, Z ] $ ker 9. Let x be an irreducible constituent of aG.Then K' $ ker x and [G, Z ] $Z ker x. This contradiction shows that [G, Z ] = 1 and completes the proof. I We remark that the last several sentences of the proof could be replaced by an appeal to Problem 5.26. Next we refine Theorem 12.5 in a somewhat different direction. Namely, if c.d.(G) = (1, m} and G has no abelian normal subgroup of index m, then the nilpotence class of G is I3. (12.12)

LEMMA

Let A-=I G with A abelian and G/A cyclic. Then IAl =

I G I I A n Z(G)I.

Proof Let G/A = ( A g ) and let a: A + A be defined by a(a) = a-'ag. Then a is a homomorphism and ker n = c A ( g ) = A n Z(G). Let I be the image of a. Then g E N(I) and so I -=I G. Since G = ( A , g ) and g centralizes A mod I , it follows that G/I is abelian and G' c I. Clearly, I E G and hence IAl = lker ollll = IA n Z(G)IIGI and the proof is complete.

I

Character degrees

205

(12.13) LEMMA Let c.d.(G) = { 1, rn} and suppose A , B index m with A # B. Then IG’l Irn and G E Z(G).

c G are abelian of

Proof If K c G with IG:KI Irn, then G E K and hence K a G by Problem 5.14(c). In particular, A a G. Let A < K E G, then K a G and thus K‘ 4 G. If K‘ < G , let 1,. # p E Irr(G’/K’) and let x be an irreducible constituent of pG. Now K‘ c ker x and so xK has a linear constituent 3, and x(1) I (‘A 1) = IG : K I < m. Thus x( 1) = 1 and G‘ G ker x. It follows that G G ker p, contradicting the choice of p. Therefore K‘ = G . Since A # B,let b E B - A and let K = ( A , b ) . Thus KIA is cyclic and IG’I = IK’I = J A: A n Z(K)I by Lemma 12.12. However, A

I?

Z(K) 2 A n B

and so [ A : A n Z(K)I I I A : A n BI II G : BI = rn and the first assertion is proved. For g E G, the conjugacy class of g is contained in the coset gG‘ and so has size II G’ I Irn. Thus IG : C(g)I Im and G‘ G C(g) by the first sentence of the proof. Sinceg is arbitrary, we have G‘ E Z(G)and the proof is complete. I (12.14) THEOREM Let c.d.(G) = (1, m} and suppose that G has no abelian normal subgroup of index m. Then [G, G ] E Z(G), that is, G is nilpotent of class 1 3 .

Proof By Theorem 12.5 we may assume that G is a p-group. If G has a faithful irreducible character x then x = IG for linear 3, E Irr(K) and K E G since G is an M-group. We have I G : K I = x( 1) = m and so K a G by Problem 5.14(c). Thus all irreducible constituents of xK are linear and K’ E ker x = 1, a contradiction. Thus G has no faithful irreducible character and hence Z(G) is not cyclic. Let Z , , Z,, Z, E Z(G) be distinct subgroups of order p . If G/Zi is of nilpotence class 5 3 , then [G, G, G, G] c Zi. If this happens for two distinct Zi, we conclude that [G, G, G , G] = 1 and we are done. Assume then that G/Z, and G/Z, (say) do not have class I3. Working by induction on I G 1, it follows that there exist A , B a G with 1 G : A I = m = I G : B 1 and A’ c 2, and B’ c Z, . Let Z = Z , Z , so that AZ/Z and BZ/Z are abelian. Since G/Z is nonabelian, we have m E c.d.(G/Z) and so I G : AZI 2 m and 1G:BZI 2 m. It follows that Z c A n B. If A/Z = B/Z, then A’ c Z, n Z, = 1 and A is abelian, a contradiction. Thus A/Z # BIZ and Lemma 12.13 applies to yield GZ/Z C Z(G/Z). Thus [G’, G] E Z c Z(G) and the proof is complete. I

Chapter 12

206

We now consider groups G for which Ic.d.(G)I = 3. Although the information obtained is not as detailed as when Ic.d.(G)I = 2, we do prove a result analogous to Corollary 12.6. (12.15) THEOREM Let Ic.d.(G)I = 3. Then derived length I 3.

G“‘ = 1, that is, G is solvable of

Before proceeding with the proof of this result we make some general remarks. As the group A, shows, we cannot conclude from Ic.d.(G)I = 4 that G is solvable. It has been conjectured by G. Seitz that for solvable G, d.l.(G) I Ic.d.(G)I. (Here, d.l.(G) is the derived length of G ; the smallest the kth commutator subgroup of G, is trivial.) By integer k for which GCk), Theorem 5.12, this conjecture holds for M-groups. It has also been proved when Ic.d.(G)l = 4 (S. Garrison) and when I GI is odd (T. Berger). It is known that for any solvable group, d.l.(G) I3 I c.d.(G)I. (12.16) LEMMA Let N a G with ( I N

I, 1 G :N I) = 1. Then

I c.d.(N)I I I c.d.(G) 1. Also if G/N is supersolvable and N is solvable with d.l.(N) I 1 c.d.(N)I, then d.l.(G) I I c.d.(G)I. Proof Let x be the set of prime divisors of IN I. If x E Irr(G), let 9 be an irreducible constituent of xN. By Corollary 11.29, X( 1)/9( 1) divides 1 G : N I. Since 9(1) divides IN I, it follows that 9(1) is exactly the x-part of ~ ( 1 ) Since . every 9 E Irr(N) arises this way, we see that c.d.(N)is exactly the set of x-parts of the elements of c.d.(G). The first assertion follows. Now assume that N is solvable and GIN is supersolvable. It follows that GIN’ is an M-group by Theorems 6.22 and 6.23 and thus d.l.(G/N’) I Ic.d.(G/N’)I by Theorem 5.12. We may assume that N’ > 1 and observe that d.l.(G) Id.l.(G/N’)

+ d.l.(N’) I Ic.d.(G/N’)I + d.l.(N) - 1 I Ic.d.(G/N’)J+ Ic.d.(N)I - 1,

where the last inequality follows from the assumption that d.l.(N) I Ic.d.(N)I. Now everyf E c.d.(G/N’) divides 1G:NI by Ito’s Theorem 6.15 and thus the x-part off is trivial. We conclude from the first part of the proof that I c.d.(N)I I I c.d.(G)I - Ic.d.(G/N’)I 1 and the result follows. I

+

Proofof Theorem 12.Z5 Let c.d.(G) = (1, m, n}. If (m, n) # 1, then G has a proper normal p-complement N for some prime p by Corollary 12.2. By Lemma 12.16, Ic.d.(N)I I 3 and so N is solvable and d.l.(N) I Ic.d.(N)I by induction if Ic.d.(N)I = 3 and by Corollary 12.6 if Ic.d.(N) I < 3. In this case we are done by Lemma 12.16.

Character degrees

201

Assume now that (n,m) = 1. Suppose there exists K -a G with G/K solvable and nonabelian. We may assume that G/K satisfies the hypotheses of Lemma 12.3.If G/K is a p-group then (say) n is a power of p . Let x E Irr(G) with x(1) = m. Since pkm, we have xKeIrr(K)and thus by Gallagher's theorem (Corollary 6.17),xfl E Irr(G) for fl E Irr(G/K). This is a contradiction since mn 4: c.d.(G). Therefore G/K is a Frobenius group with kernel N/K = (G/K)', where N/K is a p-group and IG : N I = n (say). Suppose $ E Irr(N) with $(1) > 1. Since n$( 1) $ c.d.(G), we conclude from Theorem 12.4 that p I $(1). Let x be an irreducible constituent of $". Then plx(1) and since p k n , we have x(1) = m. Thus xN is irreducible since (IG : NI, ~(1))= 1. We conclude that xN = $ and $(1) = m. Thus c.d.(N) = (1, m } and d.l.(N) I 2 by Corollary 12.6. Since GIN is abelian, we have d.l.(G) I 3 as desired. ) 1, then G/ker x Suppose now that no such K exists. If x E Irr(G)with ~ ( 1 # is nonabelian and thus is nonsolvable. However Ic.d.(G/ker x)I I 3 and working by induction on IGI we must have ker x = 1. Therefore, every nonlinear irreducible character of G is faithful. Now let n m and let x E Irr(G) with x(1) = n. By Theorem 4.3, each irreducible character of G is a constituent of x' for suitable integers t. Let t be minimal such that x' has an irreducible constituent $ of degree m. Thus [ x t , $1 # 0 for some irreducible constituent t of x'- '. Now r(l) # 1 or else x t is irreducible forcing 25 = $ which is not the case. Also t(l)# m by the minimality oft. Thus t(l)= n. Since [xA, $1 = 0 for linear A, we conclude that [A, $21 = 0 and $2 has no linear constituents. Thus

-=

b

a

$1 = iC ti + jC qj, =l =1

<

where ti,qj E Irr(G), &(l)= n, q j l ) = m and a 2 1 since is one of the ti. Comparing degrees yields mn = an + bm and since (m,n) = 1, we have m I a. Since b 2 0 and a > 0, this yields a = m and b = 0 and $2 = x y = ti. We claim that each ti is of the form A i l for some linear character, Ai. It sufficesto find a linear constituent of xti. Suppose (for some i ) that xti has no linear constituent. Then

x r

xti

=

S

xj

j=l

+

11C/k k=l

where xj, t,hk E Irr(G), xX1) = n and t,bk(l) = in. Thus nz = m + sm and n Is. However, 0 # [$I, ti] = [$, xti] and 1C/ is one of the $k. Thus s 2 1 and hence s 2 n. Since r 2 0, this yields nz 2 nm and n 2 m, a contradiction. We now have ti = A i j for linear Ai and thus @b, = ti = 2 Ai and hence ($G')(XG')

= miG'*

Chapter 12

208

Since ker $ = 1, we have $(x) # m for x # 1 and thus j ( x ) = 0 for all x E G - {l}.It follows that [IG,,1Gt-j # 0 and thus G' c ker j . Thus contradiction completes the proof. [ Another technique for studying character degrees is based on the following lemma. (12.17) LEMMA (Garrison) Let H E G and let 9 E Irr(H). Suppose for every irreducible constituent x of 9' that zH = 9. Then V(9) -a G. Proof We have (9G)His a multiple of 9 and so (9G)H= I G :H 19. Let h E H with 9(h) # 0 and let S = {x E GI h" E H } . By definition of 9' we have

(1/IHI)C9(hX)= 9'(h) = IG:HI9(h). XES

However, since 9'(h") = $'(h), it follows for x E S that 9(h") = 9(h). This yields IS19(h) = IG19(h) and since 9(h) # 0, we have IS1 = [ G I and S = G. Thus he E H for all g E G and 9(hg) = 9(h) # 0. Thus G leaves the generating set for V(9) invariant under conjugation and the result follows. [ (12.18)

LEMMA

Let x E Irr(G) and let ker x < N

4

G. Then

ker x < N n V(x). Proof We may assume ker x = 1. If V(x) n N = 1 then x vanishes on N - (1) and so [ x N , lN] # 0. This forces N E ker x, a contradiction. [

In the following, F(G) denotes the Fitting subgroup of G , the (unique) largest normal nilpotent subgroup. (12.19) THEOREM (Broline-Garrison) Let x E Irr(G)and K = ker x. Either of the following conditions guarantees the existence of $ E Irr(G) with $(1) > ~ ( 1and ) ker $ < K :

(a) K 4 (b) K = F(G), G/K is solvable and K < G. Proof Suppose H is a maximal subgroup of G and that K H = G. Then 9 ~ I r r ( H )If. $ is an irreducible constituent of 9'such that t+hH # 9, then \C/H reduces and $(1) > 9(1) = x( 1). Also, we cannot have (ker $)H = G or else t,hH would be irreducible. The maximality of H thus yields ker $ E H and since 9 is a constituent of we have ker $ E ker 9 = H n K < K . Thus the result follows in this situation and we may suppose that whenever K H = G for a maximal subgroup H E G, the character 9 = xH satisfies the hypotheses of Lemma 12.17. In particular, V(9) -a G.

xH

=

Character degrees

209

Now we choose L -a G as follows. If K $ F(G),take L = K . If K c F(G), then we are in situation (b) and F(G) = K < G. Here, G / K is solvable and we take L > K with L/K an elementary abelian chief factor of G. Thus L is not nilpotent and we choose a nonnormal Sylow subgroup P of L. Note that in the case K = F(G) and L > K , we must have L = KP. By the Frattini argument, G = LNG(P) = KNG(P) and NG(P) < G. Let H 2 NG(P) be a maximal subgroup of G . Thus G = HK and we let 9 = xH. Then V ( 9 )4 G and thus V ( 9 )n L 4 G. Now K n H = ker 9 c V ( 9 ) and so in the case K = L, we have P c K n H E L n V(9).In the case that L/K is a chief factor of G, we have ( L n H)/(ker 9)is a chief factor of H and thus L n H E V ( 9 )by Lemma 12.18. Thus in any case, P c L n V ( 9 )4 G and P is Sylow in L n V(9). The Frattini argument now yields G = ( L n V(S))N,(P)c H . This is a contradiction and proves the theorem. I (12.20) COROLLARY Let X E Irr(G). If either ~ ( 1 = ) max(c.d.(G)) or ker x is minimal among kernels of irreducible characters of G then ker x is nilpotent. (12.21) COROLLARY (Garrison) Let G be solvable and let Ic.d.(G)I = n. Then there exist N i 4 G with ~ = N , ~ N ~ E . * . E N , , = G such that N i + J N i is nilpotent for 0 I i < n.

Proof Let N , = F(G), the Fitting subgroup. Then c.d.(G/N,) E c.d.(G) and if N1 < G, then c.d.(G/N,) does not contain the largest element of c.d.(G) by Theorem 12.19. In this case, Ic.d.(G/N,)I < n and the result follows by induction on 1 GI. I Suppose A c G is abelian. By Problem 5.4 (or 2.9(b)),I G : A I is an upper bound for c.d.(G).Conversely, suppose we know max(c.d.(G)).Can we conclude that there exists an abelian subgroup with bounded index in G? We can, although it is certainly not true that there necessarily exists abelian A E G with I G : 41 = max(c.d.(G)). We use the notation b(G) = max(c.d.(G)). Note that if H E G and $ E Irr(H), then $ is a constituent of zH for some x E Irr(G), and thus ) b(G) $(1 I ~ ( 1 I and hence b(H) I b(G). (12.22) LEMMA Let b(G) = b. Then there exists x E G - (1) such that I G :C,(x)I I b2.

Chapter 12

210

Proof Let k = I Irr(G)I - 1. Then

I GI - 1

= Cx(1)’ I

kb’,

where the sum runs over nonprincipal x E Irr(G). Let

m = min{IG:C,(x)llxEG,x # l}. Then I GI - 1 2 mk since each of the k nonidentity conjugacy classes has size 2 m . We now have kb’ 2 mk and the result follows. I

(12.23) THEOREM Let b(G) = b. Then G has an abelian subgroup of index I(b!)’. Proof Use induction on b. If b = 1, the result is trivial and if b = 2, then c.d.(G) = {1,2} and we are done by Corollary 12.9. Assume then, that b 2 3. Let K Q G be maximal such that G/K is nonabelian. It follows that b(K) Ib/2 since if $ E Irr(K) with $(1) > b/2, then necessarily $ = xK for some x~Irr(G).By Corollary 6.17, we have p x ~ I r r ( G )for P€Irr(G/K). Since G/K is nonabelian, we may choose p with p(1) 2 2 and this yields p( 1)x(1) > b, a contradiction. By Lemma 12.3,we have three main cases to consider namely: G/K is nonsolvable, G/K is a Frobenius group and G/K is a p-group.

Case I G/K is nonsolvable. Since b(G/K) Ib, Lemma 12.22 yields such that x # 1 and

x E G/K

I

: cG,K(x)l I bZ*

Let C/K = CG&). If b(C) I b - 1, then there exists abelian A c C with IC:AI I((b - l)!)’andsoIG:AI = IG:CIIC:AI I @!)’andwearedone. Assume that b(C) = b and let $ E Irr(C) with $(1) = b. Then every irreducible constituent x of $G satisfies x(1) = b and thus xc = $. It follows by Lemma 12.17 that V($) Q G and thus KV($)aG. If KV($) > K, then G/KV($) is abelian and so C Q G since KV($) E C. If Z/K = Z ( C / K ) then Z Q G. Also, x E Z/K and so Z > K. It follows that G/Z is abelian and G/K is solvable, a contradiction. We conclude that V($) E K and hence IC :K 1 I $(1)’ Ib’ by the remarks preceding Theorem 12.4. Thus IG:KI 5 b4. By the inductive hypothesis, K has an abelian subgroup of index < ( [ b / 2 ] ! ) 2For . b 2 5, we have ([b/2]!)’b4 I(b!)’ and the result follows. Assume then, that b I4. If 2 4 c.d.(G/K), then c.d.(G/K) c { 1, 3 , 4 } which contradicts the nonsolvability of G/K by Theorem 12.15. Thus we may choose 9 E Irr(G/K) with 9(1) = 2. We have G/ker 9 is nonabelian and hence ker 9 = K. Let MIK = (GJK)’. Then M’ $Z K = ker 9 and hence SM is

Character degrees

211

irreducible. Pick y E M/K of order 2. Since y E ker(det(9)), it follows that y E Z(9) = Z(G/K) = 1, a contradiction. This completes case 1. Case2 G/K is a Frobenius group. Let N/K = (G/K)’, the Frobenius kernel. By Lemma 12.3, I G : N I E c.d.(G) and so IG : N I I b. If b(N) Ib/2, then application of the inductive hypothesis to N yields an abelian subgroup A C N with IN : A1 I ([b/2]!)’. Since b([b/2]!)’ I (b!)2we are done. Suppose then, that b(N) > b/2 and let $ E Irr(N) with $(1) > b/2. Then $(l)IG : NI $c.d.(G) and by Theorem 12.4 it follows that $(1)’ 2 I N : KI and hence I G : K I I b3. Since b 2 3, we have ([b/2]!)’b3 I (b!)’ and we are done in case 2 by applying the inductive hypothesis to K. Case 3 G/K is a p-group. Let Z/K = Z(G/K). By Lemma 12.3, I G : Z I = p(l)’, where p E Irr(G)and Z E Z(p).In particular, I G : Z I I b2.If b(Z) > b/2, pick $ E Irr(Z) with $(1) > b/2. Then $ = xz for some x E Irr(G). It follows from Theorem 12.7 or 6.16 that p(1)$(1) E c.d.(G) and this is a contradiction since 8(1)$(1)> b. Thus b(Z) I b/2 and the result follows as in previous cases since ([l1/2]!)~b’ I ( b ! ) 2The . proof is now complete. I

It is apparent in the above proof that in cases 2 and 3, the inequalities obtained are far from being best possible. The limiting factor in this proof is the nonsolvable case 1. If one assumes that G is solvable it is possible to obtain a better bound. It is not known what the best possible bound is either in the general situation or for solvable groups. The following result provides a tool which can be used to find an abelian subgroup in G of index Ib(G)4 when G has an abelian normal subgroup with nilpotent factor group. (Compare Theorem 12.24 with part (a) of Theorem 12.19.)

(12.24) THEOREM (Broline) Let x E Irr(G) and let K = ker x and F/K = F(G/K). Suppose F is not nilpotent. Then there exists $ E Irr(G) with ker $ < K.

+I

Proof Let Q E Syl,(F) with Q F and let H 2 NG(Q)be a maximal subgroup. Now QK 4 G since F/K is nilpotent and normal in G/K and hence HK = G by the Frattini argument. Thus 9 = xH E Irr(H). Since ker(9‘) E H, we have ker(9‘) @ K. Let $ be an irreducible constituent of 9‘ with K $Z ker $ and let L = ker $. If L E H, then L E ker 9 = H n K c K and we are done. Suppose then, that L $tH so that LH = G and i,hH is irreducible. Then $H = 9 and L n H = ker 9 = K n H. Write N = ker 9. Now K n L -a G and K > K n L 2 K n H . It follows that

(K n L)H c K H = G and thus K n L c H. Therefore K n L = K n H

=

N

Chapter 12

212

We have

( F : F n H I = 1G:HI = 1L:NI

=

1KL:KI.

Since Q E F n H and Q E Syl,(F), it follows that q$IKL/KI. Since KQ/K a G / K , we conclude that the commutator [ K L / K , KQ/K] = 1 and thus [L, Q] c K . Since L -a G, we have [L, Q] G K n L = N and hence L c N(NQ). However, NQ = (H n K)Q = H n K Q a H since KQ a G. Therefore, G = LH E N(NQ) and NQ argument yields G = NN(Q) c H, a contradiction. I

G. The Frattini

(12.25) LEMMA Let G = 1 and let m = ) G : F ( G ) I .Then m I b(G) and b(F(G))Ib(G)/m. Proqf Let x E Irr(G) be such that ker x is minimal. Since G is a relative M-group with respect to G which is abelian, there exists H 2 G and linear I E Irr(H) such that I' = x. Also, H 2 ker x. Since H a G, all irreducible constituents of xH are linear and thus H/ker x is abelian. Thus H/ker x c F(G/ker x) and hence H is nilpotent by Theorem 12.24. In particular, H E F(G)and m I I G : HI = x( 1) I b(G). Now let 9~1rr(F(G))and let J/ be an irreducible constituent of 9.' Choose x E Irr(G) with minimal ker x c ker $ and let H be as above with IG : HI I b(G),H/ker x abelian and H c F(G).We have H c ker x c ker $ and thus $, has linear constituents. Thus 9, has linear constituents and hence 9(1) I 1 F(G) :H 1 = IG :H 1 / m I b(G)/mand the proof is complete. 1

(12.26) THEOREM Suppose U a G is abelian and G/U is nilpotent. Then there exists abelian A E G such that I G :A I I b(G)4. Proof Use induction on b = b(G). We may assume b > 1 and so G is not abelian. First, suppose that every nilpotent factor group of G is abelian. Then G' c U , G = 1 and Lemma 12.25 applies. Since G is not nilpotent, F(G) G and b(F(G))I b/m < b, where m = IG : F(G)I 5 b. By the inductive hypothesis, there exists abelian A E F(G) with I F(G): A I I ( b / n ~ ) ~ and hence 1 G :A I I b4/m3 I b4. Now suppose that G does have a nonabelian nilpotent factor group. Choose K a G, maximal such that G / K is nonabelian and nilpotent. By Lemma 12.3, G / K is a p-group and 1 G :Z I = f 2 , where Z / K = Z ( G / K )and 2 = Z(p)for some fl E Irr(G) with p(1) = f . By Corollary 12.8, we conclude that b(Z) Ib/f'12 < b. By the inductive hypothesis, there exists abelian A c Z , with J Z : A lIb(Z)4 Ib 4 / f 2 . Thus ( G : A I = 1 G : Z I I Z : A I Ib4 and the proof is complete. I

-=

Character degrees

213

Next we discuss the "p-structure" of G, where p is a prime which is in some sense large when compared to b(G). The objective here is to show that I G : O,(G) I is not divisible by too large a power of p. We prove that if b(G) < p , then p#IG: O,(G)I and that if b(G) < p3I2, then p2#IG: O,(G)J.

(12.27) LEMMA Let G act transitively on a set R with IRJ > 1. Let H = G, for some a E R. Then the average size of the orbits of H on R - { a } is I b(G). Proof Let 9 = (l,,)', the permutation character of G on R. Write 9 = l G + 1a , ~ , where the sum runs over the nonprincipal irreducible constituents of 9. The number of orbits t of H on R - { a } is given by t = [ 1 H , QH] - 1 = [g, 91 - 1 = ax2. Now IR - { a } I = 9(1) - 1 = 1 a , ~ ( 1 )I b a,, where b = b(G). Thus the average orbit size s is given by

c

(12.28) COROLLARY If G has more than one Sylow p-subgroup, then there exist distinct P, Q E Syl,(G) such that b(G) 2 ING(P):NG(P) n NG(Q)I2 IP:P n Q I . Proof Apply Lemma 12.27 to the conjugation action of G on Syl,(G). Let P E Syl,(G) and H = NG(P).Then some orbit of H on Syl,(G) - {P}must have size Ib(G). Let Q be in such an orbit. Then b(G) 2 ( H: NH(Q)I and we have the first inequality. Also

IH:NH(Q)I2 IP:NAQ)I and since NJQ) = P n Q, the result follows.

1

(12.29) THEOREM Let p be a prime and let b(G) < p. Then G has a normal abelian Sylow p-subgroup.

-+

Proof Let P E Syl,(G). If P G, then by Corollary 12.28, we can find Q E Syl,(G) such that Q # P and I P :P n Q I 5 b(G) < p. This is a contradiction and shows P Q G. That P is abelian follows since everyf E c.d.(P) is a power of p satisfying f < p. The proof is complete. I

(12.30) LEMMA (Burnside) Let P E Syl,(G) and let X, Y E P be normal subsets of P which are conjugate in G. Then X and Y are conjugate in NG(P). Proof Suppose Y = Xg so that P G N(Y) and P* c N(X)g= N(X8) Sylow's theorem in N(Y), we have Pgu= P for some UEN(Y) and gu E N(P). However, X@' = Y" = Y and the proof is complete. I = N(Y). By

(12.31)

LEMMA

Let H E G with IG : HI = p, a prime. Then OP'(H) -a G.

Chapter 12

214

Proof Let K be the kernel of the action of G on the right cosets of H (acting by right multiplication).Then K c H and K u G. Also 1 G :K I divides p ! and so I H :K I divides (p - l)! and is prime to p . Thus Op'(H)E K 4 H and so OP'(H)= Op'(K)u G since K 4 G . I

(12.32)

THEOREM

Suppose b(G) < p312 for

some prime p . Then

p2YIG: Op(G)I.

Proof We may assume that O,(G) = 1. Let P ~ s y l , ( G ) and assume 2 p 2 . Let N = N,(P). By Corollary 12.28,choose Q # P, with Q E Syl,(G) such that p3/'>b(G)2IN:NN(Q)I2IP:PnQI

I PI

a n d s e t D = P n Q . T h e n I P : D I =p.LetM=N,(D)?P.NowP$NN(Q) and so "(Q) < PNN(Q) E N n M . It follows that p312 > I N : N N ( Q ) 1 2 p l N : N n M I

and IN : N n MI < p112.Note that M < G. We consider the action (by right multiplication) of M on R = { M x I x E G } . Suppose r of the orbits of M on R - { M } have size

If M x E R,, then DD" = D"D and x E M N M .

Assuming (*), let us complete the proof. We have 10 - {M}I 2 p2s so that Lemma 12.27 yields p312 > b(G) 2 IR - { M }l/(r

+ s) 2 p2s/(r + s).

If s = 0, then M x E R, for every x E G - M and hence DD" = D"D for all x E G by (*). Thus (D" I x E G ) is a p-group that is normal in G, a contradiction. Thus s > 0 and p112 < (r + s)/s I 1 + r. Now if M x E R , , then x E M N M because of (*) and hence the orbit of M x under M contains an element of the form M n for n E N . The number of distinct M n for n E N is IN :N n MI < p1j2and thus at most p112M-orbits of R contain an element of the form Mn. These include the trivial orbit { M } and so we have 1 r I p112.This contradicts a previous inequality. We now work to establish (*). Let M x E Ro so that x 4 M and D # D". If either of D or D" normalizes the other, then DD" = D"D is a p-group properly containing D. Since I P : DI = p it follows that DD"E Syl,(G) and I DD" : D I = p so that D -a DD" and DD" c M . Since also P c M we have P = (DD")mfor some m E M by Sylow's theorem. Now D", D"" u P and

+

Character degrees

215

hence Lemma 12.30 yields D" = D""" for some n E N. Thus xmnm-' E N ( D ) = M and x E MNM as desired. The remaining case is where D $ M" and D" $ M. Let W = M n M" so that WD and WD" are groups. Since M x E R, , we have 1 M : M n M" 1 < p 2 and hence IDW: WI I IM: WI < p 2 and I D W : WI = p . Similarly I D"W : W I = p . By Lemma 12.31, Op'(W )is normal in both W D and WD". Write K = Op'(W)and L = N,(K). Since IM : WI < p 2 and I DW : W I = p it follows that I M : D WI < p and DW contains a full Sylow p-subgroup of M and hence of G. It follows that DKIK and D X K / K are Sylow subgroups of LIK. Also, D"K # DK since DK E M and D" $ M. Thus ISyl,(L/K)I > 1 and b ( L / K ) 2 p by Theorem 12.29. By Corollary 12.8, b(L)2 ab(K) for some a with a2 2 b ( L / K ) 2 p . Thus p3" > b(L)2 p%(K) and b(K) < p. Thus K has a normal Sylow p subgroup U by Theorem 12.29. Now U D E Syl,(M) and UD u W D since U is characteristic in K 4 WD. Thus I M : N,(UD)I II M : D W I < p and it follows that U D u M. Thus I Syl,(M) I = 1 which is a contradiction since P,Q E Syl,(M). This completes the proof. I We close this chapter by considering the opposite of the stituation with which we began. Suppose nof E c.d.(G)is divisible by the prime p . A sufficient condition for this to happen is that G has a normal abelian Sylow p-subgroup. (This follows by Ito's Theorem 6.15.) It is conjectured that this condition is also necessary. The next result shows that to prove the conjecture, it would suffice to check simple groups. (12.33) THEOREM Suppose G does not have a normal abelian Sylow p subgroup and that no element of c.d.(G) is divisible by p . Then G has a nonabelian simple composition factor S of order divisible by p such that no element of c.d.(S) is divisible by p . Proof If N u G and $ E Irr(N), choose x E Irr(G) with [xN, $3 # 0. Then $(1)I x( 1) and so p$$( 1). In particular, if N E Syl,(G), then N is necessarily abelian and thus G does not have a normal Sylow p-subgroup. Also, if G is simple, the result is trivial and we assume G is not simple. Let N be a maximal normal subgroup of G . Working by induction on 1 G 1, we may assume that N has a normal Sylow p-subgroup P.Then G / P does not have a normal Sylow p-subgroup and if P > 1 we complete the proof by applying the inductive hypothesis to G / P .Suppose then, that P = 1. We must have p 11 G : N I and since GIN is simple we can take S = G/N unless GIN is abelian, that is, I GIN I = p . We suppose this is the case. Let Q E Syl,(G) so that I Q I = p .

Chapter 12

216

If $ E Irr(N), then $" cannot be irreducible since p I $"( 1). Thus I,($) > N and hence I&) = G and $ is invariant in G. Thus Q acts trivially on Irr(G). It follows by Brauer's Theorem 6.32 that Q acts trivially on the set of conjugacy classes of N. If X is a class of N then Q permutes X . Since py IN I, we have pk I X I and thus Q fixes an element of X . It follows that C = CN(Q) meets every conjugacy class of N. Thus N = UxEN C" and this yields IN1 - 1 5 IN:N,(C)I(ICI - 1) I IN:CI(ICI - 1) = IN1 - 1N:CI. Thus IN : C JI 1 and C = N. We conclude that Q

4

G, a contradiction.

I

(12.34) COROLLARY (Zto) Let G be solvable. Then G has a normal abelian Sylow p-subgroup iff every element of c.d.(G) is relatively prime to p .

Problems (12.1) The normal subgroups Ni 4 G are a Sylow tower if 1 = N O S N1 G . . . G Nk= G

and N,, ,IN, is a Sylow subgroup of GINi for each i. 0 I i c k. Suppose for every m,n E c.d.(G),either m 1 n or n I rn. Show that G has a Sylow tower. (12.2) Suppose that every f E c.d.(G) is a power of the integer m. Assume that m is not a prime power. Show that there exists abelian A G G with I G :A I = b(G) and that such an A is necessarily normal in G.

Hint This generalizes part of Theorem 12.5. Mimic the proof of that theorem. (12.3) Suppose that G is solvable and that for every m,n E c.d.(G) with m # n, we have (m,n) = 1. Show that Ic.d.(G)I 5 3. (12.4) Let G be solvable with b(G) = b > 1 and suppose that G has no factor group which is a nonabelian p-group. Show that there exists L E G and an integer r with 2 I r < b such that b(L) I b/r and 1 G : LI I br.

(12.5) Let G be solvable with b(G) = b. Show that G has an abelian subfor a suitable constant k independent of b. group of index I kb'"g2(b) (12.6) Let c.d.(G) = (1, p"}, where p is a prime and e > 1. If a Sylow p subgroup of G is nonabelian, show that G is nilpotent.

Hint Use the fact that abelian Frobenius complements are cyclic to show that if G is not nilpotent, then there exists abelian H Q G with I G :HI = pe and G/H cyclic. Now let G/K be as in Lemma 12.3(a) and consider H K . (12.7) Let G be solvable with b = b(G). Let p be a prime.

Problems

217

(a) Show that there exists proper K u G and integer e 2 0 such that p'I4b(K) Ib and p'+ ' y I G : K I. (b) Show that if pf I I G : O,(G) 1, then pf I b4. (12.8) Let Gff= 1 and assume that all Sylow subgroups of G are abelian. Show that b(G) = I G : A I for some abelian A u G. (12.9) Let G be solvable and suppose that all Sylow subgroups of G are abelian. Show that I G : F(G)I Ib(G)'.

Hints F(G) = CG(F(G)). Let A E Irr(F(G)) be such that I G : ZG(A)l is maximal. Let T = and s = IG: TI. Show that b(T/F(G))Ib/s and let F/F(G) = F(T/F(G)).Show b(F) I s. Use Corollary 11.32. (12.10) Suppose that everyf E c.d.(G) divides pe where p is a prime. Show that there exists abelian A G G with index dividing p4'. (12.11) Suppose P E Irr(G) with P(1) = f and that Z = Z(P) satisfies IG: ZI = f Z . Let Z c H E G.

-=

(a) If I G : HI > f, show that b(H) b(G). (b) If I G : HI = fand x E Irr(H) with ~ ( 1 = ) b(G), show that V(X)G Z. (12.12) (a) Suppose A c G is abelian and IG: CG(A)I> b(G). Show that there exists A . c A such that [ A: A , [ I b(G)' and CG(Ao)> CG(A). (b) If G is a p-group, improve part (a) to read I A :A . I I b(G).

Hint

Use Lemma 12.10.

Show that there exists a functionfdefined on positive integers such that for any group G if b(G) = b, then there exists H c G with I G :HI I b and I H : Z ( H )I If(b).

(12.13)

Hint

Use repeated applications of Problem 12.12.

(12.14) Let G be solvable and let p be a prime. Suppose p 2 Y f for all f E c.d.(G).Show that either Op(G)> 1 or a Sylow p-subgroup of G is abelian.

Hint In a minimal counterexample, let M be a minimal normal subgroup. Show that I Op(G/M)I= p. Now show that O,(G/M) is a direct factor of a Sylow subgroup of G / M . Produce the other factor by considering IG(A) for suitable I E Irr(M). (12.15) Let N -a G with GIN a p-group and p # 2. Let Q E Irr(N) be invariant in G. Suppose that every irreducible constituent of 9' has degree 1p9(1). Show that b(G/N) I p.

Hints

Extend 9 to 9~ Irr(H) with N E H and IG :HI= p . For Conclude that there exists linear

cp E Irr(H/N) with cp(1) = p , consider (I$)'.

Chapter 12

218

p E Irr(H/N) such that cp” = cpy for x E G , with p independent of the choice of cp. Use this and the hypothesis p # 2 to show that cp is invariant in G.

Note

If p = 2, then Problem 12.15 is actually false.

(12.16) Let G be a p-group and suppose b(G) = p 2 . If p # 2, show that n{ker xlx(1) = p 2 } = 1.

Hint

Use Problem 12.15.

Note Passman has conjectured that if G is any p-group with b(G) = pe and e < p , then n{ker xlx(1) = p‘} = 1. He has proved this when G has class 2.

13 Character correspondence

We have already seen several examples of the following situation (for instance, Theorems 6.11 and 6.16). We are given H , a subgroup or factor group of G. Subsets Y G Irr(H) and F c Irr(G)are specified and it is proved that there exists a “natural” one-to-one correspondence between Y and F. Here, the word “natural” is intended to mean that the correspondence is uniquely described by some general rule and thus more is being said than merely that I Y I = 1 Y I. We shall not attempt to give a precise definition of naturalness. Most of this chapter is devoted to the study of a particular character correspondence which was discovered by G. Glauberman. We introduce some notation. Let S and G be groups such that S acts on G. [That is, we are given a homomorphism S + Aut(G).] In this situation, we can construct the semidirect product of G by S so that G a r, S G r, GS = r, G n S = 1 and the given action of S on G is the action by conjugation in r. (In fact these properties characterize the semidirect product.) If x is a character of G and s E S, then as usual we define the character x” of G by xs(gs) = x(g). Then S permutes Irr(G). We write Irr,(G) = {x E Irr(G)Ix” = x for all s E S } . (13.1) THEOREM (Gluuberman) For every pair of groups (G, S ) such that S is solvable and acts on G and (I GI, IS I) = 1, there exists a uniquely defined one-to-one map n(G, S): Irrs(G) + Irr(CG(S)).These maps satisfy the followingproperties: (a) If T 4 S and B = CG(T),then n(C, T ) maps Irr,(G) onto Irrs(B). (b) In the situation of (a), n(G, S ) = n(G, T)n(B,S / T ) . 219

Chapter 13

220

(c) Suppose S is a p-group and C = C,(S). Let x E Irr,(G) and $ = (x)n(G, S). Then $ is the unique irreducible constituent of xc such that PkCXc, $1. In the situation of (a) in the theorem, note that B is S-invariant since T Q S. Thus S acts on B and in fact S/T acts on B. Therefore n(B, S / T ) is defined on 1rrs,&3) = Irr,(B) which is the image of Irr,(G) under n(G, T). Also, n(B, S / T ) maps to the irreducible characters of C,(S/T) = CB(S)= C,(S). Thus the equation in (b) makes sense. Also note that if T = S in (a), then B = C,(S) and Irr,(B) = Irr(J3).Thus this special case of (a) asserts that n(G, S)always maps Irrs(G)onto Irr(CG(S)). There is enough information in the statement of Theorem 13.1 to determine n uniquely. Thus suppose that nO(G,s) is defined whenever (G, s) satisfies the hypotheses of 13.1 and that no satisfies (a), (b), and (c). We claim that nO(G,S) = n(G, S).By 13.1(c),this is certainly the case if S is a p-group so we may work by induction on JSIand assume that S has composite order. Let T 4 S have prime index and let B = C,( T). Then n(B, S/T) = no(& S / T ) and n(G, T) = no(G, T) and 13.l(b) yields n(G, S) = no(G, S). The preceding argument suggests how to construct the map n(G, S); namely, prove that if S is a cyclic p-group and x E Irr,(G), then xc does have a unique irreducible constituent /3 such that [xc, /3] f 0 mod p , where C = C,(S). Define n(G, S) in this case by (x)n(G,S) = /3. For general solvable S, define n(C, S) by working along a composition series for S. There are numerous technical difficultieswith this approach, not the least of which is to show that the map constructed is independent of the composition series. The key to overcoming these difficulties is to find a uniform definition for n(G, S) for all cyclic S. Following Glauberman, this is what we shall do. We establish some notation which will be used repeatedly. (13.2) HYPOTHESIS Let S act on G and suppose (IGl, IS\) = 1. Let C = C,(S) and let r be the semidirect product r = GS. (13.3) LEMMA Assume the situation in 13.2 and let x E Irr,(G). Then there exists a unique extension of x to r such that @(a),ISl) = 1. Also, f is the unique extension such that S E ker(det

a

a).

Proof The first statement is just Corollary 8.16. Since o(&) divides both IS1 and o(f), we have o(jzs) = 1 and S E ker(det If $ is an extension of x with S c ker(det $), then o($) = o(det $) and divides Ir :ker(det $)I which is prime to IS I. Thus b,t = and the proof is complete.

a).

We call the character of Lemma 13.3 the canonical extension of x. For positive integers n, we write Q, = Q(E),where E is a primitive nth root of unity. If (m,n) = 1, it is well known from Galois theory that Q, n Q,

Character correspondence = Q.

22 1

Thus if a E Q, is invariant under the Galois group %(Q,/Q,),

then

UEQ.

If M a H with [MI = m and I H : M I = n such that (m,n ) = 1, we shall Note that 9 ( H / M )permutes use the notation %(HIM)to denote %(Q,/Q,). Irr(H) (by Problem 2.2 or Lemma 9.16). Suppose 9 E Irr(M) is extendible to H. Since %(HIM)fixes 9, it permutes the set of extensions of 9 to H. In the situation of 13.2, if x E IrrS(G),and 2 is the canonical extension of x to r, then for z E 9(r/G),(I)' is an extension of x and o ( r )= o(2). Thus a. = 2 and hence has values in QIGland 2, has values in Qlcl n Qlsl = Q. We have thus proved the following corollary. (13.4)

COROLLARY

In the notation of Lemma 13.3, as is rational valued.

The following strengthens this. (13.5) LEMMA Assume Hypothesis 13.2. Let x€IrrS(G) and let 2 be the canonical extension of x to r. Note that CS = C x S. For each irreducible constituent p of xc there exists a (possibly reducible) character $Bof S such that fcs = (p x $@). This equation uniquely determines the Also, the $B are rational valued.

cs

Proof We have Irr(CS) = { p x cp I p E Irr(C), cp E Irr(S)}. Write fcs = a,,(P x cp). Set t+hB= aa,,,cp and observe that =0 unless [xc, p] # 0. Thus fcs = Is (p x $,J, where the sum runs over those P E Irr(C), which are constituents of xc. Also, this equation uniquely determines the $ i s . 4 If z E 9(T/G) then (2,. = and p' = p. Thus

ce

cs,cp

acs = ((f)')CS =

c (P

$@

x ($BY).

B

Thus $Bis invariant under %(T/G) and hence has values in values also lie in Qlsl, the result follows. I

QIGI.

Since its

(13.6) THEOREM Assume Hypothesis 13.2 and that S is cyclic. Then for each x E Irrs(G), there exist unique p E Irr(C) and E = k 1 such that ~ ( c s=) EP(c)for all c E C and all generators s of S , where is the canonical extension of x to r. Also, P is a constituent of xc and the map x H j? is one-to-one.

c

Proof Write aCs= p x $@asin Lemma 13.5and fix a generators of S. Write $(c) = ~ ( c s for ) c E C. Then 9 = $&)P. In particular, 9 is a class function on C. We claim that [S, 93 = 1. Let T be a set of representatives for the right cosets of C in G. If t,, t 2 E T and x E (CS)~' n (CS)", then there exist cl, c2 E C with C1r1s''

= (CIS)" =

Is

x = (C2S)'2

= C 2 fZ $ 2 .

Chapter 13

222

We thus have two factorizations of x into products of commuting elements of orders dividing IGI and ISI. By Lemma 8.18, s" = st2 and thus t 1 t 2 - l E C(s) = C since S = (s). Thus t l = t,. Hence the sets (Cs)'are disjoint for distinct ~ E and T Iu,(Cs)'I = 1TIICI = 1GI = IGsl. Since s'sC'EG, we have (Cs)' E Gs and hence Gs = (Cs)' is a disjoint union. We have

ut

IClC9, 91 =

1 lf(X)l2 = 1 If(x')12 xecs

xscs

for all t E T . It follows that ITIICIC9,91

=

c

lli(X)l2.

xeGs

By Lemma 8.14(c),the latter sum equals I GI = I T 11 C I and thus [9, 91 = 1 as claimed. We now have

However, $&) is a rational algebraic integer by Lemma 13.5 and so lies in Z. Thus $&s) is nonzero for some unique /3 and for that p, $&) = E = f 1. This yields f(cs) = E/?(c)for c E C. This equation clearly determines E and p uniquely. We now show that p is independent of the choice of the generator s of S. If S = (so), then so = sm for some m with (m,ISl) = 1. Thus there exists an = A(so) for all automorphism B of the field Q,,, such that I'(s) = I E Irr(S). Therefore 0 # $&)" $&so) and hence replacing s by so yields the same p. Finally, suppose xl,x, E Irr,(G) determine the same character p so that fi(cs) = ciP(c)for c E C and i = 1,2. Since Gs = ~ ( C S )it' ,follows that fl(gs) = c l g 2 Z2(gs)for all g E G and thus by Lemma 8.14(b),it follows that j

XI

=

(f1)G

= (f2)G =

x2

and the map x H fi is one-to-one. The proof is complete.

I

(13.7) DEFINITION Assume Hypothesis 13.2 and that S is cyclic. Construct maps y(G, S): Irr,(G) + Irr(C)ande(G,S): Irr,(G) -,{ - 1, 1) by(x)y(G,S ) = p and (x)E(G,S ) = E = f 1 where f(cs) = E ~ ( cfor ) c E C , (s) = S , and 2 is the canonical extension of x to r. It will turn out that the map x(G, S ) of Theorem 13.1 equals y(G, S ) when y(G, S ) is defined, that is, for cyclic S . The map y(G, S): Irr,(G) + Irr(C) of Definition 13.7 is one-to-one by 13.6. It is also onto. One way to prove this is to show that I Irrs(G)I = I Irr(C)I.

Character correspondence

223

Since S is cyclic, it follows from Brauer's Theorem 6.32 that I Irrs(G)I is equal to the number of S-invariant conjugacy classes of G. It is true that each Sinvariant class of G intersects C nontrivially and, in fact, the interesction is a single conjugacy class of C. It follows that the number of S-invariant classes of G is equal to the total number of conjugacy classes of C, and hence equals I Irr(C)I. Thus y(G. S) maps onto. The assertion about S-invariant classes of G in the preceding paragraph follows from the Schur-Zassenhaus theorem. We digress from the discussion of characters in order to give a proof.

(13.8) LEMMA (Glauberman) Let S act on G with (IS(,[ G I ) = 1. Assume that one of S or G is solvable. Let S and G both act on a set R such that

(a) (a . g ) . s = ( a .s) g" for all a E R, g E G , and s E S. (b) G is transitive on R. a

Then S fixes a point of R. Proof Let r = GS, the semidirect product. For g s E r and ~ E R , define a . ( g s ) = ( a . g ) s. Condition (a) above guarantees that this is an action. Pick a ER and let H = r,. Since IRI = IG : G n HI = Ir :HI by (b), it follows that IH : G n HI = ISI. By the existence part of the SchurZassenhaus theorem, let T be a complement for G n H in H . Then I T I = I S I and T is a complement for G in r. Now the conjugacy part of the Schur-Zassenhaus theorem yields S = T" for some x E r. Thus S E H" and S fixes a . x E R. The proof is complete. I 3

(13.9) COROLLARY In the situation of Lemma 13.8, the set of S-fixed points of R is an orbit under the action of C,(S). Proof If a E R is fixed by S and c E C,(S), then ( a . c) s = ( a . s) . c" = and a . c is S-fixed. Now suppose a, PER are S-fixed. Let X = { g E GI a . g = p}. Then X is a left coset of G, and is S-invariant. Let G , act on X by right multiplication. Note that G , is S-invariant and is transitive onX.ForxEX,gEG,andsES,wehave(x.g)-s = (xg)s = xsgs = ( x . s ) . g s and Lemma 13.8 applies to the actions of S on G, and S and G , on X . Thus S fixes a point x E X . Then x E C,(S) and a . x = p. The proof is complete. I M.C

(13.10) COROLLARY Assume Hypothesis 13.2 and that at least one of G or S is solvable. Then X I+ X n C defines a bijection from the set of S-invariant conjugacy classes of G onto the set of conjugacy classes of C. Proof Let X be an S-invariant class of G. The conjugation action of G on X is transitive. For k E X , g E G, and s E S , we have

( k * g ) * s= (g-'kg)S

=

(g-')"k"gS = ( k * ~ ) * g ' .

Chapter 13

224

By Lemma 13.8 we conclude that X n C # (21 and by 13.9, X n C is a class of C. Since the classes of G are disjoint, the map X I+ X n C is one-to-one. If C E C , then the G-class of c is S-invariant. It follows that the map XHX n Cis onto. I (13.11) Irr(C).

COROLLARY

The map y(G, S) of Definition 13.7 maps Irrs(G) onto

Assume Hypothesis 13.2 and that S is solvable. Let Y be a composition series for S with 9’:l = S o d S 1 d * * * d S k = S .

Let Ci = C,(Si) so that G = CO2 C1

Z

* . . Z Ck

=

C.

Also, S i + G N,(Ci) for 0 I i < k and we view S i + l/Si as acting on Ci. Since Si+ JSi is cyclic, yi =

y(Ci, Si+ 1/Si)

is defined for each i, 0 Ii < k. We have y i : Irrsi

yi-

+

,(Ci) -, Irr(Ci+ l).

Each yi is one-to-one and maps onto Irr(Ci+l)and so we can define Irr(Ci+1)-, Irr(Ci).

l:

(13.12) DEFINITION Assume Hypothesis 13.2 with S solvable. Let Y be a = composition series for S and use the above notation. Put X(G, 9) (Irr(C))y;?1y;-12 yl-lyo-l c Irr(G) and let n(G, 9): %(G, 9) -, Irr(C) be given by n(G, 9’) = yo yl . ykThus X(G, 9’) is the largest set on which the composite function yoyl . - . 7,’- is defined. Since each y i is one-to-one, n(G, 9’)is one-to-one and by construction of X(G, 9’), we see that n(G, 9) maps X(G, 9) onto Irr(C). Our = Irrs(G), that n(G, 9’)is independent objective now is to show that %(G, 9) of the choice of Y and that for cyclic S, n(G, 9’) = y(G, 9’) We . obtain these results by considering the case that S is a p-group. +

(13.13) LEMMA Let Z G R c C, where R is a ring containing the values of all 5 E Irr(G). Let 9 be a generalized character of G with values in an ideal I of R. Assume I n Z c pZ for some prime p not dividing I GI. Then p divides [S, 51 for all 5 E Irr(G).

Character correspondence

Proof We have I G I [9, 51 = S(g)t(g) the result follows since pk I G I. I

225 EI

1

n h. Thus p I G 1 [S,

51 and

(13.14) THEOREM Assume Hypothesis 13.2 and that S is a p-group. Let x~Irr,(G).Then there exists a unique p ~ I r r ( C )with , [xc, p] f 0 mod p . Furthermore (a) Cxc, PI = f 1 mod p ; (b) if S is cyclic, then P = (x)y(G, S ) and (x)E(G,S ) = [xc, P] mod p . (c) If 9 'is a composition series for S , then X(G, 9')= IrrdG) and (x)n(G,9')= P. In particular, n(G, 9) is independent of the choice of 9'. Proof First assume that S is cyclic and let fi = (x)y(G,S ) and E = (y)c(G,S ) so that f(cs) = EP(c)for c E S and S = (s), where f is the canonical extension of x to r. Let R be the ring of algebraic integers in Qlrl and let I be a maximal ideal of R with p E I . In the notation of Theorem 8.20, we have ( C S ) ~ , = c and thus ) f(c) = f(cs) mod I . that theorem yields ~ ( c = We therefore have ~ ( c=) E&) mod I for all c E C and thus xc - EP is a generalized character of C with values in I . Since 1 # I and PEI,we have I n Z = pZ and since pJ' IC 1, Lemma 13.13 yields [x, - EP, 51 = 0 mod p for all 5 E Irr(C). It follows that [xc, 51 = 0 mod p for 5 # P and [xc, P] = E mod p . When S is cyclic, this proves everything but (c). If 1st = p , then 9':1 a S and X(G,9')= Irr,(G) and n(G, S ) = y(G, S). In particular, part (c) of the theorem holds when IS I = p . Now assume IS I > p and drop the assumption that S is cyclic. Work by induction on IS I. Let

9':1 = so4 * * * 4 Sk = S and write T = s k -

and F : 1 = & a ... a SkPl= T.

Let B = C,(T). Then n(G, 9) = n(G, F)y(B,S/T). Also, %(G, 9')is the inverse image in X(G, Y) of Irr,,,(B) = Irr,(B) under the map n(G, F):X ( G , F )-,Irr(B).

By the inductive hypothesis applied to T , we have x E Irr,(G) E Irr,(G) xB 9 is a character of B or is zero and [xB, 53 = [(f)*, 5") = [xB, 57 and thus [xB, 5 E Irrs(B). In particular, x E X ( G , 9'). If P E Irr(C), we have

= X(G, F ) and = p9 & 5, where 5 = (x)n(G,9). If s E S, we have 5"] $ 0 mod p . Thus 5 = 5" and

cxc,

PI = c(P9 f 5)c,PI =

*

CtC,

PI mod P

Chapter 13

226

and since 5 E Irrs,T(B)and S/T is cyclic, it follows from the first part of the proof that there is a unique p ~ I r r ( C )with [tc,p] f Omod p , namely p = (&O,S/T). Thus f l = (x)n(G,9')is unique in Irr(C) such that [xc, p] $ 0 mod p and in fact [xc, p] 5 f[&-,p] = f1 mod p . We already have Irr,(G) E X ( G , 9'). Now suppose $ E X(G, 9')E X ( G , Y) = IrrT(G). Let q

($)n(G,Y) so that q E Irrs(B). Let s E S. Then 0 f [~,b~,q] = = [ ( @ ) B , q] mod p . Since T 4 S and $ E IrrT(G),it follows that 1c/" E Irr,(G) = X(G, Y)and thus q = ($')n(G, Y) since [ ( @ ) B , q] f 0 mod p . Since n(G, 9) is one-to-one, we have $ = $' and thus $ E Irrs(G). We have now shown that X ( G , 9')= Irr,(G) and have given a description of the map n(G,9') which is independent of 9'. This completes the proof. I =

[(lc/")B, q']

Assume Hypothesis 13.2 and that S is cyclic so that y(G, S): Irr,(G)

-+

Irr(C)

is defined. Suppose a: r + rl is an isomorphism and that a(G) = G1 and a(S) = S , . Then a(C) = C1 = C,,(Sl) and y(G1, S , ) : Irrsl(Gl) + Irr(Cl) is defined. Because y(G, S ) is uniquely defined, independently of any arbitrary choices, it is clear that if (X)Y(G,S ) = P, then (Xl)Y(Gl, Sl) = P1, where x1 and correspond to x and p via the isomorphism, a. That is, xl(g") = x(g) and pl(cu) = p(c) for g E G and c E C . (Recall that the computation of (x)y(G, S ) requires choosing a generator s of S , but that the result is independent of this choice.) An important special case of this invariance under isomorphism of y(G, S ) is when a E Aut(T) and G" = G and S" = S . In that case we have (x")y(G,S ) = ((X)Y(G, S)P.

(13.15) LEMMA Assume Hypothesis 13.2 and let T -a S with T cyclic and B = C,(T). Then y(G, 7')maps Irr,(G) onto Irr,(B).

Proof Since Irr,(G) E IrrT(G), y(G, T ) is defined on Irrs(G). Let H = G T a r. If S E S , then s defines an automorphism of H with Gs = G and TS = T . Therefore, by the above discussion, we have (x"y(G, T ) = ((X)Y(G, T))S

for all x E IrrT(G).It is immediate that y(G, T) maps Irr,(G) into Irr,(B). Since y(G, 7') maps onto Irr(B) and is one-to-one, the result follows. I

Character correspondence

221

(13.16) THEOREM Assume Hypothesis 13.2 and that S is cyclic. Let p be a prime and let T be the p-complement in S. Let B = C,(T). Then y(G, S ) = Y(G,T)Y(B,S / T )on Irrs(G). Proof If p,i"S(, then T = S, B = C, and y(C, 1) is the identity map on Irr(C) (as is clear from Definition 13.7).The result is thus trivial in this case and we assume p I IS I. In particular, p$ I G I. Let x E Irr,(G) c IrrT(G)and let 5 = (x)y(G,T )E Irr(B).Let /3 = (x)y(G,S). By Lemma 13.15,t E Irrs(B) = 1rrs&3). We must show that P = (&o,S/T). Let Po = (<)y(B,S/T). Since SIT is a p-group, Theorem 13.14 yields tc= pcp &,,Po, where cp is a character (or is zero) and c0 = & 1. Let R be the ring of algebraic integers in Qlrl and let I be a maximal ideal containingp. We have then t(c) = E ~ P ~ (mod c ) I for c E C. By definition of y(G, T ) , we have f(bt) = ~ t ( b )where , T = ( t ) , b E B, E = f1 and 2 is the canonical extension of x to r. Applying this to c E C c B yields

+

f ( c t ) = E&)

= c ~ ~ B ~ ( c ) mI o d

for any generator, t of T . Now let S = (s) and let t = ( s ) ~ .in the notation of Theorem 8.20. Then ( t ) = T and (cs),,, = ct for c E C. Thus by 8.20, we have ~(cs= ) f(ct)mod I

and thus ~ ( c s )= E E ~ ~ ? ~ mod ( c ) I.

By definition of y(G, S), we have ~ ( c s= ) SP(c) for c E C with 6 = k 1. Thus P(c) = S E E ~ P ~ mod ( C ) I for all c E C. By Lemma 13.13, [P - &xOPO,P] is divisible by p and thus P = Po as desired.

I

(13.17) COROLLARY Assume Hypothesis 13.2 and that S is cyclic with IS1 = p q , where p and q are primes. Let Y be a composition series for S. Then X(G, 9') = Irr,(G) and n(G, Y )= y(G, S).

Proof If p ' = q, the result is immediate from Theorem 13.14(b)and (c). Assume then that p # q. Write Y : 1 u T u S . We may assume I TI = q. Now Lemma 13.15yields that X(G, 9) = Irrs(G). Theorem 13.16 asserts that n(G, 9)= y(G, TMB, S/T) = y(G, S )

where B = C,(T). The proof is complete. (13.18) THEOREM Assume Hypothesis 13.2 with S solvable. Let Y and F be composition series for S. Then X(G, Y )= X(G, F) and n(G, 9) = WJ,TI.

Chapter 13

228

Proof Use induction on the composition length k of S. If k = 1 then Y = Y and there is nothing to prove so assume k > 1. Write 9 1

=s,

= S O a . . . d S k

Y: 1 = T o 4 * . -Tka= s. Let Y* and Y* be the composition series for S k - l and Tk-1 respectively, obtained by deleting S from Y and Y. Consider the case that S k - 1 = & - l . Let B = CG(Sk-l). Then Irr,(B) = (Irr(C))y(B,s/sk-1)- and so

'

%(G, 9) = (Irrs(B))n(G,Y * ) -

and %(G, Y) = (IrrS(B))n(G,Y*)-l.

By the inductive hypothesis, n(G, Y * )= n(G, Y*) and hence X(G, 9) =

X(G,Y). Also n(G, Y )= n(G, Y*)y(B,s/sk-1)

=

n(G, Y * ) y ( B ,S/&- 1) = n(G, F).

Assume now that s k - # &- and let M = s k - n &- 1. Let A be a composition series for M and extend A to composition series Y oand Yofor S which run through s k - and &- ',respectively. By the preceding paragraph,

X(G, Y o )= X(G, 9) and

n(G, Y o )= n(G, 9).

We may thus replace Y by Y oand similarly replace 9by P.We may now assume that Si = for i I k - 2 and that A!: I = So 4

.

* *

4

Sk - 2 = Ma

Let

9 14 S k - l / M 4 S/M,

1 4 &-JM

4

S/M

and let D = CG(M).It follows that

X(G, 9') = ( X ( D ,S))n(G,A)-' and

X(G, Y) = ( X ( D ,.T))n(G,A)-'.

Also andI, n(G, Y) = n(G, A)n(D,Y). n(G, 9) = n(G, &) 9) @ Therefore, it suffices to prove that X ( D , 9) = %(D,3)

and

n(D, 9)= n(D, y).

We may therefore assume that M = 1and D = G. Thus S = S,is abelian of order p q for primes p and q. If p = q, then %(G, Y )= Irr,(G) = X(G, Y)

and

x

n(G, 9) = n(G, Y)

&-

Character correspondence

229

by Theorem 13.14. If p # q, then G is cyclic and Corollary 13.17 yields the result. The proof is complete. I Assume Hypothesis 13.2 with solvable S and choose a composition series Y for S. By Theorem 13.18, %(G, 9) is independent of the particular com. we can position series and so we can write %(G, S ) = %(G, 9)Similarly, Then X ( G , S ) and n(G, S ) are unambiguously write n(G, S ) = n(G, 9). defined. We have %(G, S ) 5 Irr(G) and n(G, S ) is a bijection of X(G, S ) onto Irr(C). As in the discussion preceding Lemma 13.15, suppose a: + rl is an isomorphism and let G , = a(G), S , = o(S),and C , = a(C) = CG,(sl). Then a induces bijections Irr(G) + Irr(G,) and Irr(C) + Irr(C,). Since %(G, S ) is uniquely defined, we have %(G,, S , ) is the image of %(G, S ) in Irr(G,) and p = (x)n(G,S ) implies that p1 = OI1)n(G,,S,), where XI(#"' = x(g) and pl(c") = p(c). In particular, if (T E Aut(r) and G' = G and S" = S, then (T leaves %(G, S ) setwise invariant and ( f ) n ( G ,S ) = ((x)n(G,S)J' for x E %(G, S). (13.19) COROLLARY Assume Hypothesis 13.2, with S solvable, and let %(G, S ) and n(G, S ) be as above. Then %(G, S ) = Irr,(G). Also, if T a S and B = CG(T),then

(a) n(G, T )maps Irr,(G) onto Irrs(B); (b) n(G, S ) = n(G, T)n(B,S/T). Proof If T -a S and B = CG(T),use a composition series for S , which runs through T in order to construct n(G, S). Then (b) is immediate and %(G, S ) = (%(B,S/T))n(G, T)-'

Thus (a) will follow once we prove the first statement. We show that %(G, S ) = Irr,(G) by induction on the composition length k of S . If k = 1 then %(G, S ) = (Irr(C))y(G,S)-' = IrrdG). Suppose then, , that k > 1 and let 7 ' 4 S with 1 c T < S . Let H = G T a r. For ~ E Swe have Gs = G and T s = T and hence by the discussion preceding the statement of the corollary, we see that (XS).(G, T ) = ((X)X(G,T))S for XE%(G, T ) = IrrT(G). Since n(G, T ) is a bijection from Irr,(G) onto Irr(B), where B = CG(T),we see that n(G, T ) carries the S-invariant characters in Irr,(G) onto Irrs(B). Since Irr,(G) c IrrT(G),it follows that

Irrs(G) = (Irrs(B))n(G,T)-' = %(B, S/T)n(G,T)-' = %(G, S), where the second equality is by the inductive hypothesis applied to S/T.

I

(13.20) DEFINITION Assume Hypothesis 13.2 with S solvable. Then the Glauberman map is the map n(G, S): Irrs(G) + Irr(C) constructed above.

Chapter 13

230

We have now completed the proof of Theorem 13.1. The Glauberman map satisfies conditions (a) and (b) of 13.1 by Corollary 13.19. It satisfies condition (c) by Theorem 13.14(c).Also, by Theorem 13.14(a),we have the following. (13.21) COROLLARY Let n(G, S) be the Glauberman map with S a p-group. Let C = C,(S) and let x E Irrs(G) and fl = (x)n(G,S). Then [xc, fl] E f 1 mod p . There is one further loose end. (13.22) COROLLARY Assume Hypothesis 13.2 with S cyclic. Then n(G, S ) = y(G, S). Proof Use induction on I S I. Let p I I S I. If S is a p-group, the result follows from Theorem 13.14(b).Assume that S is not a p-group and let T be the p complement in S and B = C,(T). Then n(G, T ) = y(G, 7’)and n(B, S / T ) = y(B, S / T ) by the inductive hypothesis. The result now follows from Corollary 13.19(b)(or Theorem 13.l(b)) and Theorem 13.16. I

Let S act on G. Then S permutes Irr(G) and S permutes the set Cl(G) of conjugacy classes of G. By Brauer’s Theorem 6.32, the permutation characters of S on Irr(G) and Cl(G) are equal and it is natural to ask if these actions are permutation isomorphic. That is, does there exist a bijection a: Irr(G) -+ Cl(G)such that a ( f ) = a(x)”for all x E Irr(G)and s E S? In general, the answer is no. However, if (I G 1, IS I) = 1 and S is solvable, it follows via Glauberman’s Theorem 13.1 that the actions of S on Irr(G) and CI(G) are permutation isomorphic. (13.23) LEMMA Let the group S permute two sets R and A. Suppose that for every T c S , the number of fixed points of T on R equals the number on A. Then R and A are permutation isomorphic. Proof We prove the existence of a bijection a: R -P A such that a(w . s) = I R 1. (Note that taking T = 1 yields IRl = IAI.) Let T E S be maximal such that T has a fixed point on R. (Possibly T = S.) Let T fix o E R and AEA. By the maximality of T , we have S , = T = S,. Let 0, be the orbit of o and 0,the orbit of 3, under S . Write i2 = 0, v R, and A = On v A, where the unions are disjoint. Map ao: 0, -P On by a o ( o .s) = A. s and check that a. is well-defined, one-to-one, onto and that ao(v.s) = ao(v).s for all v E 0, and s E S. Since every H E S has equal numbers of fixed points on 0, and O n , it follows that H has equal numbers of fixed points on R, and A1. By the a(w) . s for all w E R and s E S by induction on

Character correspondence

231

inductive hypothesis, there exists a permutation isomorphism a l : R, Now define a on R by combining a. and a l . 1 (13.24)

THEOREM

+ Al.

Let S act on G with S solvable and ( [ G I , 1st)= 1. Then

(a) S fixes the same numbers of irreducible characters and conjugacy classes of G. (b) The actions of S on Irr(G) and Cl(G) are permutation isomorphic. Proof It suffices to prove (a) since (b) follows via Lemma 13.23 by application of (a) to all subgroups of S. Since n(C, S) maps Irr,(G) one-to-one and onto Irr(C), it follows that IIrr,(G)I

=

IIrr(C)I = ICl(C)I.

By Corollary 13.10, intersection defines a bijection from the set of S-fixed conjugacy classes of G onto Cl(C). The result now follows. I Theorem 13.24 becomes false if the hypothesis that ([GI, ISl) = 1 is dropped. (See Problem 13.16 for an example.)

Suppose we continue to assume that (IGl, IS[)= 1, but drop the assumption that S is solvable. If S is nonsolvable, then 211S( by the FeitThompson theorem and thus 24'IGI. Thus G is solvable, again by FeitThompson. In this situation, where Hypothesis 13.2 is satisfied with G solvable of odd order, it is possible to construct a natural character correspondence from Irr,(G) onto Irr(C) by a method entirely different from Glauberman's. (And thus Theorem 13.24 remains valid without the hypothesis that S is solvable.) We shall describe the map Irr,(G) + Irr(C) in this case but without giving the proof since the only known proofs are too long to include. The key step is the following. (13.25) THEOREM Assume Hypothesis 13.2 and that G is solvable of odd order. Let C c H c G. Suppose that there exist S-invariant normal subgroups, K and L of G such that (a) L G K and K / L is abelian; (b) G = K C ; (c) H = LC. Then for each x E Irr,(G), there exists a unique i,b E Irrs(H) such that is odd. The map x H $ is a bijection from Irr,(G) onto Irrs(H). Proof Omitted.

I

[zH,$3

Chapter 13

232

To construct the correspondence between Irr,(G) and Irr(C) we construct a chain of subgroups G = Co > C1 >

> Ck

=

C

and apply Theorem 13.25 to obtain maps Irrs(Ci)+ Irrs(Ci+ The composition of these maps is the desired correspondence. Assume Hypothesis 13.2 and that G is solvable of odd order. We consider define the set &' of S-invariant subgroups H with C 5 H E G. For H E 2, H* = [ H , S]'C. Since [ H , S ] Q H S it follows that [ H , S]' Q H S and thus H* E &'. If H > C, then [ H , S] > 1 and [ H , S ] > [ H , S]' by solvability. Since [ H * , S ] c [ H , S]', it follows that H* < H. We now define the subgroups Ci E &' by Co = G and C i + = (C,)*and we have the desired chain of subgroups. We must now check that the hypotheses of Theorem 13.25are satisfied by taking H = G*. Let K = [G, S ] and L = [ G , S]'. Then K and L are normal S-invariant subgroups of G and K / L is abelian. That G = K C follows fairly easily from Glauberman's Lemma 13.8. Condition (c) of Theorem 13.25 is automatic from the definition of G*. It has been conjectured for every group G and prime p that if N = N,(P) for P E Syl,(G), then the numbers of irreducible characters of p'-degree of G and of N are equal. (For simple groups G, this conjecture is due to McKay.) Using Glauberman's Theorem 13.1, we prove a result which includes the special case of this conjecture when G has a normal p-complement. (13.26) THEOREM Let G = K H with K = 1. Let N = NG(H) and put

Q

G , H solvable, and ( [ H I ,IKI)

9- = {XEIrr(G)I(IHI,X(1))= 1) and =

{rt EIrr")I(IHI, rt(1)) = 11.

Then there exists a uniquely defined bijection of 9- onto Y. Proof We have N = ( N n K ) H and the commutator

[ N n K,Z-Zl s H n K

=

1

so that N n K = C = CK(H)and N = C x H . It follows that ??l =

{fi

x AlfiEIrr(C), AEIrr(H), A(1) = 1).

Now if x E f,let 9 be an irreducible constituent of xK. Then I G :1,(9) I divides both ~ ( 1 and ) I G :K I = I H I . Thus 1,(9) = G and 9 E IrrH(K). Let

Character correspondence

233

x = 95 for some unique we must have ((1) = 1. Also, x uniquely determines 9 by xK = 9 and thus determines 9 and 5. We can now map % to tiY by x H ((9)n(K,H))x tH.Since 9 be the canonical extension of 9 to G. Then

( ~1rr(G/K) by Gallagher's theorem (Corollary 6.17). Since ( ~ ( l IHI) ) , = 1,

restriction defines a bijection of Irr(G/K) onto Irr(H) and since 95 E S for every 9~Irr,(K) and linear (~1rr(G/K),it follows that we have mapped X onto tiY. The map is one-to-one since the q E tiY are uniquely of the form pxn.

I

Suppose S acts on G and that N 4 G is S-invariant. Also, assume that

(I G : N 1, IS I) = 1. We consider a pair of "dual" questions: (a) Let x E Irr,(G). Does there exist 9 E Irr,(N) with [zN,91 # O? (b) Let 9 E IrrJN). Does there exist x E Irr,(G) with [SC, x ] # O? Both questions can be answered in the affirmative. By the Feit-Thompson theorem, at least one of S and GIN is solvable and we assume this. Also, note that if S is a p-group, both facts can be proved relatively easily by counting arguments. (13.27)

THEOREM

Let S act on G and leave N

4

G invariant. Assume that

( 1 S 1, I G : N I) = 1 and that one of S or GIN is solvable. Let x E Irrs(G). Then

zN has an S-invariant irreducible constituent. Proof Let R = (9 E Irr(N)I [xN, 93 # 0).

Then GIN permutes SZ transivively and since x is S-invariant, S permutes R. Also, the action of S on G induces an action on GIN. Let 9 E 0, s E S , and g E G. For x E N, we have (98)yXS)= 9"x) = S(gxg-1)

and 1) = qgxg - 1). (P)"(xx") = 9"(gSxS(gS)-

The hypotheses of Glauberman's Lemma 13.8 are thus satisfied and the result follows. I The situation of question (b) is more difficult and interesting. First we consider the case where GIN is solvable. (13.28) THEOREM Let S act on G and leave N 4 G invariant. Assume that (IS 1, I G :N I) = 1 and that G / N is solvable. Let 9 E Irr,(N). Then 9' has an S-invariant irreducible constituent. Proof First assume that GIN is abelian, and let A be the group of linear characters of GIN. Let R = { x E Irr(G)I [S", x] # O}. If x E R and ilE A , then

Chapter 13

234

xN. This defines an action of A on R. We claim that this action is transitive. = [ ( x ~ )$1 ~ ,and $ is an irreducible Let x, $ E R. Then 0 # [xN, constituent of (xN)' = ( ~ ~= ~ 1 ( 1~~ ) However, '). ~ (1~)'= A and thus (xN)' = ~ 3 ,Since . xA E Irr(G) for each 3, E A , it follows that I) = x/z for some 3, and A is transitive on R as claimed. Since 9 is S-invariant, it follows that S permutes R and also S acts on the group A since (3,pL)s= Asps,If x E R, 3, E A , and s E S , we clearly have x3, E R since (xA), =

xleA

1

and the hypotheses of Glauberman's Lemma 13.8 are satisfied since 1 A I = I G : N I and thus (I A 1, IS I) = 1. The result follows in this case. We now assume that GIN is nonabelian and work by induction on I G :N I. Let M / N = (G/N)'. Then M 4 G is S-invariant and M < G since GIN is solvable. By the inductive hypothesis, g M has an S-invariant irreducible constituent $ and by the first part of the proof, $G has an S-invariant irreducible constituent. The result now follows since 9' = (sM)'. 1

To handle the case that GIN is not solvable,we appeal to the Glauberman correspondence, Theorem 13.1. We first restrict attention to the situation where ([GI, ISl) = 1. (13.29) THEOREM Assume Hypothesis 13.2 and that S is solvable. Let N -a r with N E G . Let x E Irr,(G) and 9 E Irr,(N). Write = (x)n(G,S) and cp = (9)7c(N,S). Then [9', x] # 0 iff [qC,51 # 0.

<

Proof We first consider the case that S is a p-group. Then [xc, <] f 0 mod p and every irreducible constituent of xc other than 5 occurs with multiplicity divisible by p . Since xc = (xc)c ,, N , it follows that (1)

CxcnN,(PI

1

CXC,~IC~C~N,VI~O~P-

Now write xN = bAA, where A runs over sums of orbits of the action of S on Irr(N). If A = 0 where 0 is such an orbit, then [ACnN, cp] = IOI[qcnN, cp] for q~ 0. If 101 > 1, then p 1 ) 0 1 and [ACnN, cp] = Omod p . Thus k n N , c p1 = [XN,q I [ q C n N , Cp] mod p .

1

m Irrs 0"

However,for q E Irrs(N),we have [qc that is, unless q = 9. Thus (2)

CXCnN,

N,

cp]

= 0 mod p unless cp = (q)n(N,S),

CPI E C X N , 91C9cnN,CPImod P .

Since [xc, 53 f 0 f [QCnN, cp], comparison of Equations (1) and (2) yields that [xN, 91 = 0 mod p iff [ t C n N cp], E 0 mod p . Since N 4 G and

Character correspondence

235

~ $ 1GI it follows that [xN, 91 = 0 mod p iff [ x N , 91 = 0. Similarly, [CcnN, cp] = 0 mod p iff [ScnN, cp] = 0. The result thus follows if S is a p-group. To complete the proof, we use induction on IS I. We may choose T S Q

with SIT a nontrivial p-group. Let B = C,(T). By Theorem 13.1, we have n(G, S ) = n(G, T)n(B,S/T) and similarly, n(N, S ) = z ( N , T)n(B n N , S/T). Now [ 9 ' , x] # 0 iff [((9)n(N,T))B,(x)n(G,T ) ]# 0 by the inductive hypothesis. The result now follows by the first part of the proof. 1 (13.30) COROLLARY Let S act on G with S solvable and ( I S I, 1 GI) = 1. Let N Q G be S-invariant. Let 9 E Irrs(N). Then ' 9 has an S-invariant irreducible constituent.

Proof Let C = CG(S)and cp = (9)n(N,S ) so that cp E Irr(C n N ) . Let 5 be an irreducible constituent of cpc and let x = (cp)n(G,S ) - ' . Then x E Irr,(G) and [ 9 ' , x] # 0 by Theorem 13.29. The proof is complete. 1 To complete our analysis of question (b), we need to consider the case where S is solvable and ( IS 1, I G I ) # 1. Of course, we continue to assume that (IS(,J G :N I ) = 1. First, we observe that it is no loss to assume that 9 is invariant in G since if U = ZG(9),then S leaves U invariant. Now, if we can find an S-invariant irreducible constituent $ of 9",then $ ' ~ 1 r r ~ ( G )as desired. We shall finish the proof by an appeal to the theory of projective representations of Chapter 11. The following proves slightly more than we need.

c G Q with N 4 I- and ( I l- : G I, I G : N I ) = 1. Assume that one of I'/G or GIN is solvable. Let 9 E Irr(N) be invariant in I'. Then ' 9 has some I'-invariant irreducible constituent. ( 13.31) THEOREM Let N

Proof By Theorem 11.28, we can find a character triple (rl, N , , 9') and an isomorphism (z, 0):(r,N , 9)+ (I-', N1,9,)with S1(l)= 1. Let G, = G' so that the isomorphism z: r / N + T J N , carries GIN to G1/N,.Suppose we can find €Irr(G119') with $1 invariant in rl. Let $ ~ I r r ( G 1 9 )with oC($)= $'. We claim that $ is invariant in r. To see this, let x be an irreducible constituent of $r so that x E Irr(l-1 9). Let x1 = ar(x). Then (xl)', = e$' for some integer e and it follows that xc = e$. Thus $ is invariant as desired. The argument of the preceding paragraph shows that it is no loss to assume that 9 is linear. We may thus factor 9 = I p where (o(A),1 G : N I) = 1 and (o(p),I r : G I) = 1and A and p are powers of 9. Thus I and pare invariant in r. By Corollary 6.27, I has a unique extension, E Irr(G) such that o(1) = o(I).Because of the uniqueness, it follows that 1is invariant in r. Suppose we can find a r-invariant irreducible constituent $ of p'. Then $1E Irr(G) is r-invariant and 9 = p i is a constituent of ($i)N.

x

Chapter 13

236

We may therefore assume that 9 = p, that is, that 9 is linear and I r : GI ) = 1. Also, we may replace r by r/ker 9 and thus assume that 9 is faithful. Thus IN I = o(9) is relatively prime to I : GI and hence (IGl, Ir : GI) = 1. By the Schur-Zassenhaus theorem, we can find S c r with SG = and S n G = 1. Thus S acts on G and leaves N and 9 invariant. Also, one of S or GIN is solvable. Now Corollary 13.30 or Theorem 13.28 yields an S-invariant irreducible constituent of SG. The result follows. I

(o(9),

To close the chapter, we obtain some further information in a special case of Theorem 13.6. (13.32) THEOREM Assume Hypothesis 13.2 and that S is cyclic. Also suppose that C = C,(X) for all x E S - { l}. Let x E Irr,(G) and fi = (x)n(G,s). Let 2 be the canonical extension of x to r. Then there exists E = & 1 and p E Irr(S) with p2 = 1, such that

(a) ~ ( c x = ) E/?(c)~(x) for all x E S - { 1) ; (b) xc = IS 19 + ~ f iwhere , 9 is a character of C or is zero; (4 (x(1) - Efi(l))/lSl = k E Z . (d) as = kp, + ~ f i ( l ) pwhere , ps is the regular character. (e) p # 1, iff IS1 is even and k is odd. Proof If 1 < T 5 S then n(G, S ) = n(G, T ) = y(G, T ) and hence ~ ( c x ) = ffi(c) for all x E S - { 1) where the sign is independent of c. It follows that in the notation of Lemma 13.5, we have t,ba(x)= f 1 for all x # 1 and $Jx) = Oforx # 1 andfi # rpEIrr(C). Write $ = $ f l .We work to express $ in terms of Irr(S). Let A1, 1, E Irr(S)

and compute “A1

- 121,

$1

=

(l/lSl)

c

f(Al(4

-

A,(x)).

xes; X f l

Since IAl(x) - A,(x)l I 2 for x E S, this yields IC(A1

- A213 $11 I2(lSl - l)/lSl < 2.

Therefore, the multiplicities with which I , and I , occur in $ differ by at most 1. It follows that $ = ap, f p where a E Z and p is a sum of distinct linear characters of S with p( 1) I IS I /2. Also, p(x) = k $(x) = & 1 for 1 # x E S and hence

P(1) = b , P I

=

(l/lsl)(cl(l)2+ IS1 - 1)

+ IS[ - 1 = 0. It follows that p(1) = 1 and p ~ I r r ( S ) . and ~ ( 1-) ISlp(1) ~ . that if I S ( = 2, then Define E = f 1 by the equation $ = ap, + ~ p (Note

Problems

237

neither E nor p is uniquely defined.) Now ~ ( c x= ) /?(c)@(x)= E / ? ( c ) ~ ( xfor ) 1 # x E S and (a) is proved. Also p ( x ) = & 1 for x E S so that p 2 = 1,. Since all @, for cp # /? vanish on S - { 1 }, each is a multiple of p, and hence (cp x @JC is a multiple of IS I cp. Since (/? x @& = a IS I /? E/?, statement (b) follows and (c) is immediate from (b). Also (d) now follows. We now prove (e). Since p2 = l,, it follows that p = 1, if IS1 is odd. Suppose then, that JSI is even and let T E Irr(S) with r2 = 1, but z # 1,. (This A = z. By (d) it follows that uniquely defines z.) Now det(p,) =

+

nlEIrr(,)

1,

=

det(f,) = zkpeB(').

Since 2 I I SI, we have 24'/?( 1) and therefore p and the proof is complete. I

= T ~ .Statement (e) now

follows

(13.33) COROLLARY Let (r,N , 9) be a character triple and let N c G 4 r. Suppose T/G is cyclic and that TIN is a Frobenius group with kernel GIN. Assume that 3, = ex for x E Irr(G) (and thus e2 = IG : N I ) . Then I r : GI divides e - E for some E = k 1. Proof We may replace (r,N , 9) by an isomorphic character triple and assume 9(1) = 1. Write 9 = pull, where (o(A),IG : N I) = 1 and (o(p), I r : GI) = 1 and A and p are powers of 9 and thus invariant in r and L is extendible to G by Corollary 6.27. Let v be an extension of A. Then pG = (A$), = ( ~ ~= 9vQG ) =~e(vx).

Since v x E Irr(G) we may replace 9 by p and assume (o($),I I : GI) = 1. We may also assume that ker 9 = 1 so that N c Z(T) and (I GI, l- : GI) = 1. Let S be a complement for G in r. If 1 # x E S, then C,, (x) = 1 and it follows that N = C,(X)and we are in the situation of Theorem 13.32. (Note that x E Irr,(G) since x is the unique irreducible constituent of sG.)The result now follows from 13.32(c). I Problems

(13.1) Assume Hypothesis 13.2 with S solvable. Prove the following facts by using the statement of Theorem 13.1, but do not appeal to any of the results used in constructing the Glauberman map.

(a) If x E Irr,(G) then (x)n(G,S ) is a constituent of xc. (b) If/?= (x)n(G,S), then Q(x) = QW). (c) If C = G , then n(G,S): Irr(G) + Irr(G) is the identity map. (13.2) In the situation of Theorem 13.1, let Show that x( 1) divides I G : C I/?(1). Hint

/? = (z)n(G,S ) for x E Irr,(G).

In the case that S is cyclic, consider o = mi as in Chapter 3.

Chapter 13

238

(13.3) In the situation of Theorem 13.1, let N -=I G be S-invariant with NC = G. Let 9 E Irrs(N). Show that Ic((9)n(N7s))= I G ( ~ n ) c. (13.4) Let S act on G and leave N a G invariant. Assume that ( 1 S 1, I G : N I ) = 1 and that one of S or GIN is solvable. Suppose CGIN(S) = 1. If x E Irrs(G), show that there exists a unique $ E Irr,(N) such that [xN,$3 # 0. Write (x)6 = $. Show that 6 maps Irr,(G) onto Irrs(N). (13.5) Assume Hypothesis 13.2 and that S is solvable. Let C N 4 r.

G

N

E

G with

(a) Show that CG,N(S)= 1. (b) Let 6: Irr,(G) -+ Irr,(N) be as in Problem 13.4. Show that 6 is oneto-one. (c) Show that 6n(N, S ) = n(G, S). Hint (For (a)) Use Glauberman’s Lemma 13.8. (13.6) In the situation of Theorem 13.1, assume that G is solvable. Let

x E Irr,(G) and fi = (x)n(G,S). Show that fi( 1) divides x( 1).

Hints Let N < G be normal, S-invariant and maximal such. Then either NC = G or N 2 C. Use Problem 13.3 or 13.5(c)and induction on I GI. (13.7) In Problem 13.4, assume that GIN is solvable. Show that 6 is one-toone. Hint Use induction on I G : N 1. (13.8) Assume Hypothesis 13.2 and that G is nilpotent. Show that IIrr,(G)J = IIrr(C)I. Do not assume Theorem 13.25. Hint

Use Problem 13.7.

(13.9) Assume Hypothesis 13.2 and that G is solvable. Show that C > 1 iff IIrr,(G)J > 1. Do not assume Theorem 13.25. (13.10) Let N a r with N E Gar and (Ir:GI,I G : N ( ) = 1. Assume that one of T/G or GIN is solvable. Let KIN be a complement for GIN in TIN and assume CG,N(K/N)= 1. Let 9 E Irr(N) be invariant in K. Show that there exists a unique r-invariant x E Irr(G) with [9‘, x] # 0. Hints

Use the argument of Theorem 13.31 to reduce to the case

(I G 1, I r : GI) = 1. Use Problem 13.5(b)if T/G is solvable.

(13.11) Let S be solvable and let H be a group with (ISl, IHI)= 1. Let G =H x H x x H where there are JSI factors and let S act on G by

Problems

239

permuting the factors regularly. Let C

=

C,(S) and note that

c = { ( h , h, . . . , h ) l h E H } Z H . x ~ I r r , ( G ) , then x = 9 x 9 x . . . x 9 for

Also, if (x)n(C,S) = p, show that P((h,h, . . . , h)) = 9(hls').

some s ~ I r r ( H ) .If

(13.12) Let N 4 G and suppose 9E Irr(N)is invariant in G. Suppose = ex for some x E Irr(G). (Thus (G, N, 9)is a fully ramified character triple and e2 = I G : N I .) Assume that S is solvable, acts on G, leaves N and 9 invariant and that ( I G : NI, IS()= 1. Let C/N = CGIN(S). Show that xc = f < with E Irr(C) and,f2 = I G : C I.

<

Hint Use the technique of the proof of Theorem 13.31 to reduce to the case that ( [ G I , IS]) = 1 and 9(1) = 1 with 9faithful. In this case, C = C,(S). Use Theorem 13.1. (13.13) Let SactonGandleave N u Ginvariant.Assumethat(ISI,IG:NI) = 1 and that S acts trivially on GIN. Let Q E Irr(N) and z ~ I r r ( G with ) [xN, 91 # 0. Show that 9 is S-invariant iff x is S-invariant. Do not assume the Feit-Thompson theorem.

Hint In showing that 9E Irr,(N) implies x E Irr,(G), it suffices to assume that G / N is cyclic. (13.14) In the situation of Theorem 13.1, let N u G be S-invariant with NC = G. Let 9E Irr,(N) and cp = (9).n(N,S). Let I = I,($) (SO that I A C = Ic(cp) by Problem 13.3). If $ is an irreducible constituent of 9', show that ($G).(G, S) = ( ( $ ) N I ,

w.

(13.15) Let E be elementary abelian of order p2 for p # 2 and let S be dihedral of order 2p. (a) Define an action of S on E such that Irr,(E) = {lE}but CE(S)> 1. (b) Define an action of S on E such that C,(S) = 1 but IIrr,(E)I > 1.

14

Linear groups

Suppose x is a faithful character of G . In this chapter we are concerned with drawing conclusions about G when given information about x. For instance, we already know that if x is irreducible, then Z(G) is cyclic and that if all irreducible constituents of x are linear, then G is abelian. Another, less trivial example which we have seen is Theorem 3.13. A faithful F-representation of G with degree n is an isomorphism of G with a linear group of degree n over F; in other words, a subgroup of GL(n, F). For our purposes, we will restrict attention to finite linear groups. (It should be pointed out, however, that stating that an infinite group is a linear group imposes a type of finiteness condition on it, that is, it guarantees that the group is not “too badly” infinite.) We shall also restrict attention to complex linear groups. Thus from now on, a “linear group” is a finite subgroup of GL(n, C ) for some n. A group is thus isomorphic to a linear group of degree n iff it has a faithful character of degree n. We say that a linear group is irreducible if the identity map is an irreducible representation. (14.1) THEOREM (Blichfeldt) Let G be a linear group of degree n and let n = { p l p is prime, p > n + l}.Then G has an abelian Hall n-subgroup. We need a lemma. Recall that a character x is p-rational (wherep is prime) if its values lie in Q, for some r with pyr. (This is Definition 6.29.) (14.2) LEMMA Let p # q be primes such that G has no element of order pq. Then each x E Irr(G) is either p-rational or q-rational. 240

Linear groups

241

Proof Let x ~ I r r ( G and ) suppose that x is neither p-rational nor qrational. Then in the notation of the discussion following Definition 6.29, there exists (T E Bp(G)and z E B,(G) with xu # x # x'. Also, by Problem 2.2(b), xu, 'x E Irr(G). If g E G and p$o(g), then x(g) E Q,, where 1 G 1 = p"m and p$m. Thus by definition of Qp(G), we have x(gY = x(g). Similarly, if q#o(g), then x(g)r =

x(s)*

Since G has no element of order pq, it follows for every g E G that either p#o(g) or q#o(g) and we conclude that

C(x - XU), (x - x')l

= 0.

Since [x, x'] = 0 = [xu,x3, we obtain

0 = Cx,XI + Cx", x'l = 1 + Cx", x'l 2 1 and this contradiction proves the result. The following easy lemma is a special case of a more general result due to Schur which we will prove later. (14.3) LEMMA Let x be a faithful p-rational character of G and assume that p divides JGI. Then x(1) 2 p - 1.

Proof Since x is p-rational, it has values in Q, for some r with p#r. Let G with JPI = p . Then xphas values in Q, n Q, = Q. Let B = B(Qp/Q). Then B fixes xp and thus permutes the linear constituents of xp. Since x is faithful, xp has a nonprincipal linear constituent I and I ( x ) # 1, where 1 # x E P . Since B is transitive on the p - 1primitive pth roots of 1,it follows that the images of I under B take on p - 1 different values at x and thus there are at least p - 1 different characters in the orbit of 1 under 9.Since each is a constituent of xp, the result follows. 1

P

E

An observation that is often useful when working with linear groups is that if { K i }is a family of normal subgroups of G with O K i = 1, then G can be isomorphically embedded in the direct product n ( G / K i ) via g H (. . . ,g K i , . . .). Proof of Theorem 14.I Use induction on n and for groups with degree n, induct on 1 GI. Let x be a faithful character of G with ~ ( 1 = ) n. Suppose that x is reducible and write x = x1 + xz. Let K i = ker xi so that G / K i is isomorphic to a linear group of degree xi(l) < n. By the inductive hypothesis, G / K i has an abelian Hall xi-subgroup where xi = { p l p > 1 xi(l)}. Since n E xi, it follows that G / K ihas an abelian Hall x-subgroup H i / K i .If H i < G, then the inductive hypothesis yields an abelian Hall n-subgroup H of H i . Since no prime in A divides I ( G / K i ): (Hi/&) I = I G : H i 1, it follows that H is a

+

Chapter 14

242

Hall n-subgroup of G and we are done if H i< G . We may thus assume that H i= G and G / K i is an abelian n-group for i = 1,2. However, K1 n K z = ker x1 n ker xz = ker x = 1 and thus G is isomorphic to a subgroup of the abelian n-group, ( G I K , ) x (GIK,). The result follows in this case. We now assume that x is irreducible. If H E G is a n-subgroup, let 9 be an irreducible constituent of xH. Then 9( 1) divides I H I and so a prime divisor p of 9(1) satisfies n < p I $( 1) I n and this contradiction shows that 9(1) = 1. Thus xH is a sum of linear characters and since x is faithful, it follows that H is abelian. Suppose there exists M 4 G with I G :MI = p E n. Let H be a Hall nsubgroup of M , which exists by the inductive hypothesis. If H 4 G , let P E Syl,(G). Then ,HP is a Hall n-subgroup of G and we are done. Suppose then, H G and choose a Sylow subgroup Q of H with Q +i G. Then Q is Sylow in M and thus G = MN,(Q) by the Frattini argument. Now H is abelian and so H E NG(Q). We have I G : NG(Q)(= I M : N,(Q)( which divides I M : H I . Thus I G : N,(Q)I involves no primes from n. Since N,(Q) < G, it has a Hall n-subgroup which is one for G. We may thus assume that no such M exists. Next, suppose Z E Z(G) with 12I = p E n. Then xz = x(l)A with o(1) = p . Thus (det(& = A X c 1 ) # 1, since ~ ( 1 < ) p . Therefore, p l o ( ~and ) it follows that p I I G : ker(det x)I. This yields a normal subgroup of index p , a contradiction. (Note that we have reproved part of Theorem 5.6 here.) Thus no such 2 exists. Now let H E G be a n-subgroup of maximum possible order. Suppose H is not a Hall n-subgroup. Then there exists q E n with q I I G : H I . In particular, H # 1 or else H < QeSyl,(G) which violates the maximality of H . Let x E H have prime order p E n and let C = C,(x). Then C < G by the previous paragraph and thus C has a Hall n-subgroup K by the inductive hypothesis. Since H is abelian, we have H c C and thus I HI I I K I. The maximality of IHI yields I HI = I K 1 and H is a Hall n-subgroup of C . In particular, q$ I C :HI and it follows that q I I G : C I. Let x E P E Syl,(G). Then P is abelian and so P _C C(x) = C and p$ I G : C 1. Thus p # q. Since p can be any prime divisor of tH 1, we have q$ I H I. Therefore, q$ I C I since H is a Hall n-subgroup of C . We conclude that x centralizes no element of order q in G. We claim that G contains no element of order pq. Otherwise, there exist commuting y, z E G with o(y) = p and o(z) = q. However, H contains a full Sylow p-subgroup of G since C does and H is a Hall n-subgroup of C . Thus H contains a conjugate of y which we may suppose to be x. Since x centralizes no element of order q, this is a contradiction and proves the claim. Lemma 14.2now yields that x is r-rational for r = p or q. Thus x( 1) 2 r - 1 by Lemma 14.3.This is a contradiction since r E n and the proof is complete.

+

a

Linear groups

243

(14.4) COROLLARY Suppose ~ ( 1 is) prime for every x E Irr(G) with ~ ( 1 > ) 1. Then G is solvable. Proof The hypothesis is inherited by factor groups and by normal subgroups and so we may assume that G is simple. Let z ~ I r r ( G )with minimal ~ ( 1> ) 1. By Problem 3.3, ~ ( 1 = ) p > 2. Let n: = {qlq is prime, q > p 1 }. Since G is simple, x is faithful and Theorem 14.1 yields an abelian Hall mubgroup A E G . Let 1 # a E A . Then A E C,(U) and the only prime divisors of 1 G : C,(U) I are I p 1 and hence IP. If $ E Irr(G) and $(1) $ { 1, p } , then

+

+

($(l),IG:CG(a)I) =

and by Burnside’s Theorem 3.8, we have +(a) = 0 or a E Z($). Since G is simple, Z($) = 1 and hence $(a) = 0. Now

0=

c

t(l)t(a) = 1

<slrr(G)

+P

1

t(4

CEIrr(G); 9 ( 1 ) = p

and hence l/p is an algebraic integer. This contradiction completes the proof. 1 In the situation of Corollary 14.4,Problem 12.3applies since G is solvable. It follows that Ic.d.(G)I I 3 and thus G = 1 by Theorem 12.15.

Our next results concern finding normal subgroups of a linear group of degree n whose order is divisible by a prime which is large when compared with n. (14.5) THEOREM (D.L. Winter) Let G be a solvable irreducible linear group of degree n. Suppose that a Sylow p-subgroup of G is not normal. Then n is divisible by a prime power q > 1 such that q = - 1,0, or 1 mod p. Proof Let x E Irr(G) be faithful with ~ ( 1 = ) n. Suppose G is a counterexample to the theorem with minimum possible order. We argue first that every proper normal subgroup of G has index divisible by p and has a normal Sylow p-subgroup. Let M 4 G be proper and let gl, . . . , 9, be the distinct irreducible constituents of zM. Let m be the common degree of the gi.Then m I n and hence no prime power q > 1 with q E - 1,0, or 1 can divide m. Since I MI < I G 1, it follows that Mlker gihas a normal Sylow p-subgroup. Now n k e r gi G ker x = 1 and thus M is isomorphic to a subgroup of (M/ker S1) x ... x (Mlker 9J

244

Chapter 14

It follows that M has a normal Sylow p-subgroup which is necessarily normal in G. Since G does not have a normal Sylow p-subgroup, we conclude that p divides 1 G : M 1 as claimed. Now let K be a maximal normal subgroup of G and let P E Syl,(K). Then P -a G and I G : K I = p . Since G is not a p-group, we have P < K and we choose L 2 P such that K/L is a chief factor of G. Write I K : L I = a and let S E Syl,(G). Now LS < G and pk I G : LS I. Thus LS Q G. Since K(LS) = G, it follows that C,,,(S) < K / L . However, C,,,(S) u G / L and we conclude that C,,,(S) = 1. Since S is a p-group, it follows that a = 1 mod p . We claim that x, reduces. Otherwise, xLs is irreducible and the minimality of G yields S u LS. Since pyx(1) it follows that xs is a sum of linear constituents and hence S is abelian. Thus S E C,(P) u G and hence py IG : CG(P)I. By the first part of the proof, we conclude that CG(P)= G and P E Z(G). Therefore, L = P x N where N = O,.(L)-a G. Since S Q LS and S n N = 1, it follows that S E CG(N)4 G and thus py IG : CG(N)I. Therefore,C,(N) = G and N c Z(G). Thus L = N P E Z(G) and since xr. E Irr(L), we have n = 1 and G is abelian. This is a contradiction and proves that xr. reduces, as claimed. Since pJ’x(1). we have X , E Irr(K). Since xL $ Irr(L) there are two possibilities by Theorem 6.18. Either x, is a sum of I K : LI = a distinct irreducible constituents or else xL = e$, with $ E Irr(L), and ez = a. In the first case, a > 1 is a prime power dividing n and a = 1 mod p , a contradiction. In the second case, e > 1 is a prime power dividing n and ez = a = 1 mod p . Thus e = & 1 mod p and the proof is complete. 1 (14.6) COROLLARY (Ito) Let G be a solvable linear group of degree n and let p 2 n + 1 be a prime. Suppose that a Sylow p-subgroup of G is not normal. Then G is irreducible, p = n + 1, and n is a power of 2. Proof If G is irreducible, then by Winter’s Theorem 14.5, there exists a prime power q > 1 that divides n such that q = -1, 0, or 1 mod p . Thus p - 1 2 n 2 q 2 p - 1 and hence n = p - 1 is a prime power. We conclude that n is a power of 2 and the proof is complete in this case. If G is reducible, write x = x1 + xz, where x is a faithful character of G of degree n. Let K i = ker xi so that K1 n K 2 = 1 and G is isomorphic to a subgroup of (GIK,) x (GIK,). Each G/Ki is isomorphic to a solvable linear group of degree xi(l) < n. Working by induction on n, it follows that each G/Ki has a normal Sylow p-subgroup and hence so does their direct‘product. Since G does not have a normal Sylow subgroup we have a contradiction and the proof is complete. 1 The hypothesis that G is solvable in Corollary 14.6 can easily be relaxed to the assumption that G is “p-solvable.” We say that G is p-solvable if there

Linear groups

245

exist subgroups N i Q G with 1= N o E N , E . * . GNk= G

and such that each factor N i + l/Ni is either a p-group or of order prime to p . (14.7) COROLLARY Let G be a p-solvable linear group of degree n s p - 1 and suppose that a Sylow p-subgroup of G is not normal. Then n = p - 1 is a power of 2 and G is irreducible. Proof Assume either that G is reducible, n < p - 1 or that n is not a power of 2. Use induction on I GI. Let M be the next to last term in a p solvable series for G so that M c G, G/M is either a p-group or a p'-group and M is p-solvable. By the inductive hypothesis, M has a normal Sylow psubgroup P and thus we may assume that G/M is a p-group. Let Q/P E Syl,(M/P) for some prime q # p . The Frattini argument yields MNG(Q) = G and thus I G :NG(Q)I = I M :N,(Q)I. Since P E Q c N,(Q) and P E Syl,(M), we conclude that p,+' I G :NG(Q)I. Let S E Syl,(NG(Q)). Thus S E Syl,(G). Now SQ is solvable and hence S a SQ by Corollary 14.6. Since I G : Q I is not divisible by q, we conclude that q$I G : NG(S)I. Since IG : NG(S)I is independent of the choice of S E Syl,(G) and q # p is arbitrary, the result follows. I

It is conjectured that Winter's Theorem 14.5 holds for all p-solvable groups. The conjecture is known to hold for n I2p + 1. Unlike the ease with which Corollary 14.7 follows from Corollary 14.6, the facts for n > p seem quite deep. Even the special case where G has a normal p-complement is open if n is significantly larger than 2p. The results which have been obtained all seem to depend heavily on the Glauberman correspondence, and in particular on Theorem 13.14. The proofs are too complicated to give here. Now we drop all hypotheses of solvability or p-solvability. Suppose G is a linear group of degree n and a Sylow p-subgroup of G is not normal. How big can p be? The first bound was established by Blichfeldt and the best possible bound, p I2n + 1, was proved by Feit and Thompson. The Feit-Thompson proof depends on a very deep result of Brauer which gives the bound under the additional assumption that p2# I GI. Here we shall prove p < (n + 1)* and also show how the Feit-Thompson reduction to the case p2$ I GI works. We need some preliminary results. (14.8) LEMMA Let H G G be abelian and let Y be the set of nonprincipal irreducible constituents of (lH)'. Compute u = min{x(l)/[x, (lH)G]Ix E .Y}. Then $(1) 2 u - 1 for every nonlinear $ E Irr(G).

Chapter 14

246

Proof Let $ E Irr(G) with $(1) > 1. Write $3 = l G + a , ~ ,where the sum runs over nonprincipal x E Irr(G). Since H is abelian, we have

1

$(l) 5

[$H,

$HI

lH1

= [$H$H,

=

[lc/$,

(lH)G1

=

+

c a,[X,

(lH)G1.

7€Y

However, [x,(lH)‘] I x(l)/a for x E 9’and this yields

1

$(1) I1 + (l/a) a,x(l) I 1 + (l/a)($(l)z - 1). Thus a($(l) - 1) I $(1)’ - 1 and since $(1) > 1 we obtain a 5 $(1) as desired. I

+1

(14.9) LEMMA Let N 4 G and suppose g E G with N n C(g) = 1. Let x be a character of G such that [xN, IN] = 0. Then x(g) = 0.

a

Proqf Let = gN E GIN. Then CG,&) 2 C,(g)N/N and hence I c G , N ( g ) I 2 I CG(g)N/N I = IcG@) I * Thus # E Irr(G/N)

I$(S)I’ 2

1

15(g)lZ.

{ E Irr(G)

Viewing Irr(G/N) E Irr(G), we conclude that { ( g ) = 0 for N $ ker 5. The result follows. I

5 E Irr(G) with

Note that the above proof is essentially a repetition of the proof of Corollary 2.24. Also, if g and N are as in the Lemma 14.9,then all elements of the coset Ng are conjugate in G and thus the result follows from Problem 2.l(b). (14.10) THEOREM (Feit-Thompson) Let H 5 G be abelian with H n Z(G) = 1 . Assume for every g E G - Z(G) that cG(g) has nontrivial intersection with at most one conjugate of H in G . Let x E Irr(G) with H $ ker x. Then

(a) x(lI2 > I H I I X H , l H l z . Also, if x( 1) > 1 and H is contained in no proper normal subgroup of G, then (b) (1

+ x(1))’ > IHI.

Proof Let N

NG(H) and X = { X E G- Z(G)IC(x)n H > l}.

=

Note that H - { l } G X and that N c N ( X ) . We claim that X is a T.I. set and that N = N ( X ) . In particular, X c N . Let g E G be such that X n Xg # 0. Choose x E X with x8 E X. Then C(x)n H > 1 and so C(x8)n He > 1. Also, C(x8) n H > 1 and hence H = H8 by hypothesis. Thus g E N and X = X g and the claim is established. We have IGl=

IGICX,XI=

+

~ I x @ ) 1 ’ 2 ~ ( 1 )I~G : N I ~ I x ( x ) l Z BEG

xeX

Linear groups

241

and thus

Now write xN = a + P, where [aH, lH] = 0 and PH = P(1)lH.(This is possible since H Q N . ) Since H $ ker x, we have c1 # 0 but we allow the possibility that /3 = 0. Write Z = Z(G). Then lx(z)I2 = IZIx(1)2 and since 2 n X = 0, we conclude from (1) that

xzeZ

(2)

1

Izlx(1)2+ IN1 >

1X(X)l2 =

xcxuz

By Lemma 14.9, we have a(y) = 0 for y E N

(3)

la(x) xeXuZ

Since [a, P]

=

c

144 + P(x)12.

XEXUZ

-

(X u 2 )and thus

+ P(x)12= INI[cc,al + 2INICa,Pl +

c

IP(x)12.

XEXUZ

0 and [a, a] 2 1, (2) and (3) yield IzIx(1)2 + IN1 > IN1 +

c

IP(X)l2.

xsxuz

Because HZ

X v Z , we obtain

E

(4) Write xz = x(1)A. Then Pz = P(1)L and since PH = P(1)lH, we have P(hz) = /3( l)A(z)for h E H and z E 2. Thus 1 P(x) 1’ = P( 1)2for x E H Z and (4) yields P(1)21HIIzI < IzIx(1)2. Thus x( 1)2 > p(1)2I HI and since P(1) = [xH, IH], the proof of (a)is complete. Now assume that no proper normal subgroup of G contains H.Let 9’ be the set of nonprincipal irreducible constituents of (lH)G.If II/ E 9, then H $ ker I) and (a) yields 11/(1)2 > [t,hH, 1H]21HIand IH11’2< a = min{W)/[I), (1H)GlIII/E9’4P). Lemma 14.8 then gives for ~ ( 1 > ) 1 that x(1)

and (b) follows.

+12 a>

)H1”2

I

(14.11) THEOREM (Feit-Thompson-Blichfeldt). Let G be a linear group of degree n and let p be a prime. Assume one of the following.

(a) Every subgroup of G of order not divisible by p2 has a normal Sylow p-subgroup and p > n + 1. (b) p 2 (n 1)2. Then G has a normal Sylow subgroup.

+

Chapter 14

248

Proof Use induction on I G I. We may thus suppose that every proper subgroup of G has a normal Sylow p-subgroup. Let P E Syl,(G) and assume P G. Since P has a faithful character of degree n < p and this character can have no nonlinear irreducible constituents, we conclude that P is abelian. If P n Z(G) # 1, then by Theorem 5.6, we cannot have P n Z(G) c G‘ and thus p l l G : G’J. It follows that there exists N 4 G with ( G : N I = p . Since N has a normal Sylow p-subgroup, we conclude that G is p-solvable. Then G violates Corollary 14.7 since n < p - 1 and P+I G. We conclude that P n Z(G) = 1. Now if 1 # y E P n P9, then P, P g e Syl,(C(y)) and this forces P = Pg since C(y) < G. Thus P is a T.I. set. Now suppose that x E G - Z(G). Then C(x) has a unique Sylow p-subgroup S and if S > 1, then S is contained in a unique conjugate of P. It follows that C(x) has nontrivial intersection with at most one conjugate of P. If P E M 4 G with M < G , then P -a A4 and thus P 4 G , a contradiction. We have now shown that P satisfies the hypotheses of Theorem 14.10(b)and thus I P I < ($( 1) + I)’ for every nonlinear $ E Irr(G). Since G is nonabelian, the given faithful character must have some nonlinear irreducible constituent and we conclude that IPI < (n + 1)’. Thus p < (n + 1)2, contradicting hypothesis (b). We are therefore in the situation of hypothesis (a) and in particular, p211GI. Thus p’ I(PI < (n + 1)2and p < n + 1 which is our final contradiction. I

+

As was mentioned before, Brauer proved that if G is a linear group of degree n and p’$lGl where p > 2n + 1, then a Sylow p-subgroup of G is normal. It then follows from Theorem 14.1l(a) that the hypothesis p’YI GI is unnecessary. There is a great deal of information known about linear groups of degree n in which a Sylow p-subgroup is not normal and n p I2n + 1. This all requires deep results from Brauer’s “modular character” theory and we will not discuss it further here.

-=

There are many ways in which the structure of a group is limited in terms of its degree as a linear group. For example, there exist integer valued functionsi such that for linear groups G of degree n we have: (a) If G is solvable, then d.l.(G) Ifi(n). (b) If G is p-solvable and p“ divides I G : O,(G)1, then e I f2(n). (c) If G is a p-group and I G : WG) I = p“, then e I f3(n). A much more general result of this type is Jordan’s theorem.

Linear groups

249

(14.12) THEOREM (C. Jordan) Let G be a linear group of degree n. Then there exists abelian A Q G such that I G : A 1 < (n !)12"'"'"+I)+ 1) where d k ) denotes the number of primes I k . In fact, the existence of the functions f,, ,f2, and ,f3 mentioned above follows easily from Jordan's theorem, although the best known bounds for f i , f 2 , and.1, are very much better than those which can be derived from the inequality in Theorem 14.12 (or from any other known bound for Jordan's theorem). There is no reason to suppose that the index of an abelian normal subgroup of maximum possible order in a linear group of degree n can be ' ) + l ). (For instance, if n = 2, the anywhere nearly as large as (n!)12"("("+ "correct" bound is 60 rather than 2. 126.)The good bounds for the functions .tare proved by methods independent of Jordan's theorem. We begin work now on a proof of Theorem 14.12 which is due to Frobenius. As the reader will see, this proof has an unusual geometric flavor. Recall that a square complex matrix U is said to be "unitary" if UT= U - Note that the unitary n x n matrices form a subgroup of GL(n, C). Also, if U is unitary, then there exists unitary V such that V - ' U V is diagonal. (More generally, this holds for all normal matrices U , that is, those which satisfy UU* = U * U , where U* = OT.) Also, the diagonal matrix V - l U V is unitary since both U and V are. Write V - U V = diag(l,, . . . , A,) so that the A j are the eigenvalues of U . Since V - ' U V is unitary, it follows that A j - = Aj and this proves that the eigenvalues of a unitary matrix lie on the unit circle.

'.

(14.13) LEMMA Let A and B be n x n complex unitary matrices. Assume that the eigenvalues of B lie in the interior of some arc of length TC on the unit circle. Suppose that A commutes with A - ' B - ' A B . Then A commutes with B. Proof We may conjugate A and B by a unitary matrix so as to diagonalize B and hence we may assume that B = diag(b,, . . . ,b,). Since a permutation matrix is unitary, we may rearrange the bj in any desired order by conjugating both A and B by an appropriate permuation matrix. We may thus write bj = ei9jwhere 9' < 9, I . . . I 9, < 9, 7c. Write A = (apv). We shall show that up, = 0 if b, # b,. This immediately yields that A B = BA. Now write C = A-'B-'AB. We have A-' = A T and B-' = B so that C = ATBAB. Also, since AC = C A we have ,qTBAB = c = A C A - ' = B - ' A B A - ' = BABA7

+

Now evaluate the diagonal entry, cmmof C . We have

C apm6yapmbm = cmm = II

6mamvbvGtv V

Chapter 14

250

and thus P

V

We compare imaginary parts and obtain

C 1 a,,, l2 sin(9,

- Q,,)= C Iamv1' sin($, - 9),.

P

V

Thus for all rn, 1 I rn I n, we have

C (IarnvI' + Iavm12)sin($, - 9),

(*I

= 0.

V

We now use (*) to prove that if Sj # 9 k , then u p = 0. Suppose this is false and choosej minimal such that there exists k with aj < 8, and either a j k # 0 or a k j # 0. Take rn = j in (*). If v Ij,then either 9, = gj in which case sin($, - Sj) = 0 or 9, < Sj in which case avj = 0 = ajv by the minimality of j . Thus (*) yields

C (lajv12+ ( u , ~ (sin(9, ~ ) - Sj) = 0. v> j

Now for v > j we have Sj I 9, c Sj + n and thus sin(9, - Sj) 2 0. Also, lajv12 laVjl22 0. We conclude that for each value v > j, either

+

or sin(9, - Sj) = 0. Now 0 c 9, - Sj < n and hence sin(9, - Sj) > 0. Therefore lajk12+ lakj12= 0 and thus ajk = 0 = a k j . This is a contradiction and completes the proof. I (14.14) LEMMA Let A and B be n x n complex unitary matrices and suppose that the eigenvalues of A lie in an arc of length CT < n on the unit circle. Then the eigenvalues of A - ' B - 'AB lie on the unit circle between - CT and 17.

Proof Certainly A-'B-'AB is unitary and so its eigenvalues lie on the unit circle. Let a be an arc of length CT which contains the eigenvalues of A and write C = B-'AB. We need to find the eigenvalues of A-'C. Let A be an eigenvalue of A - ' C and let the column vector x be a corresponding eigenvector so that A - ' C x = Ax and thus Cx = AAx and XTCx = A(XTAx). Thus it suffices to show that arg(ZTAx)E a and arg(XTCx)E a, where arg(a) = a/lal for 0 # a E @. Since both A and C are unitary with eigenvalues in a it suffices to show that arg(jjTUy)E a whenever y # 0 and U is unitary with eigenvalues in a. Let U and y be such and let V be unitary with V - ' U V = D, a diagonal matrix. Let z = V - ' y so that J'Uy = Z T V T U V z= ZTD2

Linear groups

231

since vT= V - ' . Now ifD = diag(d,, . . .,d,,)andz = col(z,, . . . ,z,,),we have

1

Since the di lie in a and I zi I2 2 0, it follows that I zi I2di lies in the infinite wedge which is the union of all rays from the origin through points of a. Since c < n and some IziI2 is nonzero, it follows that Izi12di # 0 and arg(x I zi I2di) E a as desired. I

x

(14.15) THEOREM (Frobenius) Let G be a linear group, let A , B E G and suppose that the eigenvalues of A and B lie in arcs of length a and p respectively, on the unit circle. Then (a) If fl < n and A commutes with the commutator [A, B], then [ A , B ] = 1. (b) If a < 4 3 and fi < n, then [ A , B] = 1. Proof By Theorem 4.17we can conjugate all of the elements of G by a matrix such that the resulting matrices are all unitary. We may thus suppose that A and B are unitary matrices. Now (a) is just a restatement of Lemma 14.13. To prove (b),conjugate by a unitary matrix which diagonalizes A. We may thus assume that A is diagonal. Construct elements Bi E G by setting Bo = B and Bi = [ A , B i V l ]for i 2 1. By Lemma 14.14,the eigenvalues of Bi lie between - 4 3 and 4 3 on the unit circle for i 2 1 and thus for each i 2 0, they lie in the interior of some arc of length n. Thus by part (a) we see that if Bi+' = 1 for any i 2 0, then Bi = 1. It thus suffices to prove that Bi= 1 for some i. If M is any complex matrix, we define 9(M) = tr(RTM)so that if M = (mJ, we have 9(M) = Imrv12. Thus 9(M) 2 0 and 9(M) = 0 iff M is a zero matrix. If U is unitary, we have

xrv

~ ( u M= ) tr(R' DTU M ) = tr(GTM)= $(M). Now we compute (*)

9(Bi+l - 1) = 9(BiA(Bi+l- 1)) = 9(BiA(A-'Bi-'ABi - 1)) = 9(ABi - B i A ) = 9(A(Bi - 1)- (Bi - 1)A).

Now write Bi - 1 = (b,,,,). Since A is diagonal, we can write A = diag(a,, u 2 , . . . , a,,), where all of the aj lie inside some arc of length n/3. The (p, v ) entry of A(Bi - 1) - (Bi - l)A is arbrv - brvav = b,,(a, - uv).Note that Jar - avl < 1. Equation (*) now yields that

c

C

Ibpv12

where in fact the inequality is strict unless all b,,

= 0.

9(Bi+l - 1) =

r.v

Ibpv121ar - av12 I

r.v

= 9(Bi

-

113

Chapter 14

252

If no Bi= 1, we therefore have

9(B,

-

1) > 9(B, - 1) > 9(Bz

-

1) > * * * ,

Since G is a finite group, there are only finitely many Bi and this is a contradiction. We conclude that Bi= 1 for some i and the proof is complete. I Observe that we have used our standing assumption that the groups we consider are finite twice in the above proof. The first time was in the appeal to Theorem 4.17 which is false for infinite groups. This use of finiteness could have been avoided by assuming that G was contained in the group of unitary matrices. The second application of finiteness in the last paragraph of the proof is more fundamental. For instance, part (b) of the theorem is not valid for the full unitary group. (14.16) THEOREM Let G be a linear group of degree n. Then there exists abelian A a G such that ( H : H n A1 I12"

for every abelian subgroup H c G .

Proof Since G is a finite group, the eigenvalues of every element of G lie on the unit circle. If g E G , let a(g) be the length of the shortest closed arc which contains the eigenvalues of g. Note that a is a class function on G . Put A = (a E Gla(a) < 4 3 ) . Then A a G and A is abelian since the generators of A commute by Theorem 14.15(b). Now let H 5 G with H abelian. By Maschke's Theorem 1.9 we may conjugate all of the matrices in G so as to assume that all h E H are diagonal matrices. Partition the unit circle into twelve half-open arcs, a,, a,, . . . , a,,, each of length n/6. Each element h E H determines a function& from { i I 1 I i I n} into { j (1 I j I 12) by writing h = diag(a,, . . . ,a,) and setting fh(i) = j if ai E aj. Note that there are at most 12" different functions & that can be obtained this way. Suppose x, y e H with f, = 1,. Write x = diag(al, . . . ,a,) and y = diag(/?,, . . . ,1,).Then for each i, ai and pi lie in the same aj and hence aiBi lies strictly between - n/6 and 4 6 on the unit circle. Thus xy- E A and it follows that elements in distinct cosets of A n H in H determine different functions. Thus I H :A n HI I 12" and the proof is complete. I (14.17) LEMMA Let P be a linear p-group of degree n. Then there exists abelian A a P such that I P :A 1 divides n ! Proof By Corollary 6.14, P is an M-group and hence if I) E Irr(P), then there exists H E P with IP: HI = $(1) and such that I)H has a linear constituent. Choose such a subgroup H, for each I) E Irr(P).

Linear groups

253

Let x be a faithful character of P of degree n and let R be the set of all right cosets of all H , for irreducible constituents $ of x. Then I Ql I n and G acts on R by right multiplication. Let A be the kernel of this action. Thus I P : A 1 divides n ! To show that A is abelian, it suffices to show that xA is a sum of linear characters. However, for each irreducible constituent 1(1 of x we have A E H , and thus $ A has a linear constituent. Since A a G , it follows that all irreducible constituents of $ A are linear and the result follows. I Proof of Theorem 14.12 Let A u G be as in Theorem 14.16 so that A is abelian and I H : H n A I I 12" for every abelian subgroup H E G. We claim that I G : A I I (n!)12"("("+')+l). Factor I G : A I into prime powers. For each prime p let r p = 1 G : A I p . By Blichfeldt's Theorem 14.1,there exists an abelian Hall subgroup H , of G for the set of primes > n 1. We have (G: A 1 = J G:H,AJJH,A : A1 and it follows that

+

fl r, = ( H , A : A I = I H , : H ,

n A1 I 12".

p>n+i

For p I n + 1, let P E Syl,(G) and let H , G P be abelian with I P : H , ( dividing n ! (using Lemma 14.17).Then I P : H , I I (n ! ) p and r , = I P A : A l = IPA:H,AlIH,A:Al= I P : P n H , A ( I H , : H , n A l I IP:Hp112"I (n!),12".

Thus

The result now follows.

I

We can also use Frobenius' Theorem 14.15 in a different way. (14.18) THEOREM Let p 2 7 and let G be an irreducible linear group of degree n 2p with n # p . If the Sylow p-subgroups of G are nonabelian, then p2$ 1 G : O,(G) I. Thus in any case, G/O,(G) has abelian Sylow p-subgroups.

-=

Proof Let P€Syl,(G) be nonabelian so that P' n Z ( P ) # 1. Let U E P n Z ( P ) with IUl = p . Let x E Irr(G) be the character of the given faithful representation so that ~ ( 1= ) n. Since P is not abelian, xp has some nonlinear irreducible constituent $, We must have $(1) = p . Since x(1) < 2p, we conclude that xp = $ + A, where A is a sum of linear characters. Since LJ c P', we have U c ker A and since U E Z(P), we also have $" = p p for some linear p E Irr(U). Thus xu = p p + (n - p)l,. Since x is faithful, we have p # l U .

Chapter 14

254

Now pick u E U , with p(u) = eZniip, and note that 2n/p .c n/3 since p 2 7. Therefore, u has only two distinct eigenvalues, namely 1 and eZni’,,and these lie in an arc of length
I:=

We now present the promised generalization of Lemma 14.3. (14.19) THEOREM (Schur) Let G have a faithful p-rational character of degree n and let P E Syl,(G) with I P I = p”. Then i=O

The brackets above denote the greatest integer function and thus the “infinite” sum has only finitely many nonzero terms. If n < p - 1 then all terms in the sum are zero and the result that a = 0 is Lemma 14.3. Note that G is really irrelevant in Theorem 14.19;the result is really about P. Also, a p-rational character of a p-group is necessarily rational valued since the values lie in Q., n Q, where p$r. Suppose x E Irr(P) is faithful and we wish to bound P 1. Obviously,knowl) 1, P could be an unboundedly edge of ~ ( 1is)not sufficient since even if ~ ( 1 = ) the degree of large cyclic group. However, 1 PI is bounded in terms of ~ ( 1and the field extension Q(x) 2 0. If z EZ(P) with o(z) = p , then ~ ( z=) x(~)E, where E is a primitive pth root of 1. Thus E E 0 ( x ) and hence p - 1 divides IQ(x): 01. Since IQ,.: 01 = (p - l)p‘-’, it follows that if P # 1, then 1001): Q I is of the form ( p - 1)p‘ for some integer t 2 0. (14.20) THEOREM Let I P I = pa, where p is a prime and a > 0. Let x E Irr(P) be faithful with ~ ( 1=) f and IQ(x): 0 1 = (p - l)pf.Then a

I (f - l)/(p - 1) + (t + 1)f.

Proof We use induction on f: Iff = 1, then P is cyclic and since x is faithful, it follows that 0 ( x ) = 0,. and thus t = a - 1. In this case, the inequality is actually equality.

Linear groups

255

Now suppose f > 1 . Since P is an M-group, there exists a maximal ,!Ii, subgroup H and 9 E Irr(H) with 9' = x. Thus ( P: HI = p and xH = where the sli are the distinct conjugates of 9 under P. Note that the fields Q(9,) are all equal and that since 9 p = x, we have Q(x) c Q(9). Suppose Q(x) = Q(9). Let K i = ker g i . Thus n K i = 1 and H is isomorphically contained in ( H I K , ) x . . x (HIK,). The IH : K i I are all equal, say to pb. Thus a Ipb +- 1. Since 9(1) = f / p , the inductive hypothesis yields - 1) + ( t + l ) f / p b I( ( . f / p )and hence

z=

a Ipb

+ 1 I(f - p)/(p - 1) + ( t + 1)f + 1 = (f- l)/(p - 1) + (r + 1)f

as required. The remaining case is where Q(x) Q(9). Consider the Galois group 9 = S(Q(S)/O(x)) so that 191 = IQ(9): Q(x)l. In particular, '3 # 1 is a pgroup. Since Y fixes x, it permutes the irreducible constituents si of xH. Since the stabilizer of 9 in 9 is trivial, the orbit of 9 under '3 has size 19 ' 1. It follows that I YI = p and I Q(9) : Q I = (p - 1)p" '. Also, the 9' are %conjugate and hence all ker & are equal. Since n k e r 9' = 1, we conclude that ker 9 = 1. We can now apply the inductive hypothesis to 9 E Irr(H) and obtain

-=

a - 1 I( ( f / P ) -

- 1) f ( t

+ 2 ) f h = (f - l ) / ( P - 1) + ( t + l)f/p I(f - 11/07 - 1)

and the proof is complete.

+ ( t + 1)f-

1

I

We note that if p # 2, it is always possible to choose H in the above proof so that Q(x) = Q(9). The inequality in Theorem 14.20 is best possible.

Proof of Theorem 14.19 Define the function a on positive integers by

Since [ k / ( p - l)pi] Ik / ( p - 1)p' and this inequality is strict for large i, we have 7,

a(k) < (k/(p - 1)) z p - ' = kp/@ - 112 i=O

and the second inequality in the statement of the theorem follows. Since [x + y] 2 [x] + b], we have a(k + I ) 2 a(k) + a(Z). We know that xp is rational valued, where x is the given character of G. If xp = x1 x2, where the xi are rational valued, then working by induction on n we have ai S cr(xi(l)), where IP/ker xi( = pai. By the usual argument, a I a l a2

+ +

Chapter 14

266

and so

a s a(xl(l))+ a(x2(1))5 a(xl(l)+ x2(l))= a(n) as desired. We may now suppose that x p is not the sum of two rational characters. It follows that x p = C‘= t i ,where the tiE Irr(P) constitute an orbit under the Galois group Q = Q(Qpa/Q). Let = t,. Then ker t i = ker 5 for all i and thus 1 = n k e r t i = ker t and t is faithful. The stabilizer in 3 of 5 is S(Q,JQ(t)) and it follows that the orbit size r = I a(<) : Q I = (p - 1)p*for some t 2 0. Letf = ((1) = &(l)for all i so that n = (p - 1)p:f. Now a(n) = 1 p p2 * . * py = ( f - 1y(p - 1) f pf . * * p‘f 2 ( f - l)/(p - 1) (t + 1)f 2 a, where the last inequality is by Theorem 14.20. I

<

+ + + + + + + + +

As an application of Schur’s Theorem 14.19, we prove the following. (14.21) THEOREM Let G be a p-solvable linear group of degree n. Let P E SyI,(G) and IP : O,(G) I = p”. Then a

I

2[n/b

-

1)pil.

i=O

The proof of Theorem 14.21 depends on some standard facts about psolvable groups. The first is nearly trivial, namely that subgroups and factor groups of p-solvable groups are p-solvable. The second fact is a special case of what is often called “Lemma 1.2.3” since that was the designation in the paper of P. Hall and G. Higman where it first appeared. (14.22) LEMMA (Hall-Higman) Let G be p-solvable with O,(G) Then CG(O,,(G)) c O,,(G).

=

1.

Proof Let H = OJG) and C = C,(O,(G)). Then C -a G and C is p-solvable. Let D = O,,(C)so that D 4 G and thus D c H. If H $ C , then C/D is a nontrivial p-solvable group and O,,(C/D) = 1. Thus O,(C/D) > 1and we let E / D = O,(C/D) and P E Syl,(E) so that P > 1. Now D E H c C(C) E N(P) and thus P 4 PD = E 4 G. It follows that P 4 G . Since P > 1 and O,(G) = 1, this is a contradiction and the proof is complete. I

Proof of Theorem 14.21 Let U = O,(G) and H / U = O,,(G/U). By the Schur-Zassenhaus Theorem, U is complemented in H and we may choose a complement K . Let N = NG(K). Since G permutes the set ofcomplements for U in H and H is transitive on this set, it follows that G = N H = N U . Let S E Syl,(N) and put Go = K S . We have N n U E S c Go and I G : U I , = I N : N n UI,= I G , : N n UI,.

Linear groups

251

We claim that N n U = O,(Go)and thus it suffices to assume G = G o . To prove the claim, note that N n U 4 N and so N n U 4 Go and thus N n U c V = O,(Go).Since V 4 G o , K 4 Go and V n K = 1, we have V E C ( K ) and thus V U / U E C ( K U / U )= C ( H / U ) . Since O,(G/U) = 1, Lemma 14.22 yields C ( H / U ) G H / U and hence VUIU E H / U and V U E H . Since VU is a p-group and U E Syl,(H), we have V U = U and thus V E U . Therefore, V E N n U and the claim is established. We may now assume that Go = G so that K is a normal p-complement for G. Let x be a faithful character of G with X( 1) = n. For each irreducible constituent 9 of xk., let $ be the canonical extension of 9 to Z,($) and let $* = (9)' E Irr(G). Note that any field automorphism of QIGlwhich fixes 9 also fixes $*, and it follows that $* is p-rational. Choose a set Y of representatives for the G-orbits of irreducible con9*. Then stituents of zK and define Ic/ =

csEY

n =~ ( 12 )

c 9(l)~G:ZG(9)~1 9 * ( 1 ) =

SEY

=

Ic/(l).

9€Y

Also zK and t,hK have the same sets of irreducible constituents and so ker Ic/ n K = ker x n K = 1. Thus ker $ is a p-group and ker Ic/ G O,(G). Now application of Schur's Theorem 14.19 to G/ker $ yields the desired result. I

cy=o

Write a,@) = [k/(p - l)pi]. With further work one can replace the inequality a I a,(n) in Theorem 14.21 by a 5 a,(pn), where p = 3 if p = 2 and p = (p - l)/p if p # 2 is not of the form 2" + 1. With these improvements, equality can be obtained for all n. If G is solvable, this improvement is due to J. D. Dixon. The general p-solvable case was done by D. L. Winter and depends on the results of Chapter 13. Suppose G is a primitive linear group of degree n. In other words, the identity map is a primitive representation in the sense of Chapter 5. By Corollary 6.13, it follows that every abelian normal subgroup of G is central and thus by Jordan's Theorem 14.12there are only finitely many possiblities for the group G/Z(G). These have been explicitly enumerated for certain small values of n. We give a sample of this type of result although this proof is not typical. (14.23) THEOREM Let G be a primitive linear group of degree 2. Then I G :Z(G)I = 12,24, or 60. Proof Write 2 = Z(G) and let z ~ I r r ( G be ) faithful with ~ ( 1 = ) 2. If H c G is nonabelian, then xH E Irr(H) and it follows that Z ( H ) c Z ( X )= 2. Thus if g E G - 2, then C(g)is abelian.

Chapter 14

258

Let Y be the set of maximal abelian subgroups of G . Then clearly G = UY and Z = nY. If A , B E Y with A n B > Z, then C(A n B) is abelian and contains both A and B. Therefore A = B and thus G is a disjoint union: G = Z U U(A-Z). AEY

For A E Y we have bA, X A ] = 2 and k z , X Z ] = 4. Thus

This yields

= 4121

+2

IA - ZI - 21YI(ZI

AEY

= 4121

+ 2(IGI - 121)- 2slZI

wheres= IYI.WethusobtainIG:ZI = 2s- 2. Now Y is a union of conjugacy classes of subgroups. If A E 9, we want to compute N = NG(A).If A < N then N is nonabelian and zNE Irr(N). However, xA = 3, + p with 3, # p. It follows that IN : I N @ )I = 2 and the restriction of x to ZN(3,) is reducible. Thus Zd3,)is abelian and hence A = ZN(3,). We we have I N(A):A I I2. Also A -#J G . conclude that for all A E 9, We now consider the group G/Z = G of order 2s - 2. Let 9 = { A / Z I A€9) so that 9 partitions C and 1 9 1= s. Also, .Y is a union of conjugacy classes of subgroups of G. Write 9 = F , u F 2u . .. u Y r , where the F i are the distinct conjugacy classes. Let ai be the common size of the subgroups in Fiand let t i = IYiI. If BEY^, then IG:N(B)I= IG:Bl/2 or IG:N(B)I = Ic:BI and hence (1)

t i = (s -

l)/ai

or

ti =

2(s - l)/ai.

We also have

(3)

c

ti(Ui

- 1) = 2(s

- 1) - 1,

and all t i > 1 and ai > 1. We may assume that a, I a, I I a, and that if ai = a i + , then t i 2 t i + l . If a, > 2 then ti(ai - 1) 2 2 C ti = 2s which contradicts (3). Thus a , = 2 and t , = s - 1 or (s - 1)/2. However, if t , = s - 1, then (2) yields

Linear groups

259

t , = 1, a contradiction. Thus t l = (s - 1)/2. If a, = 2, then t , = (s - 1)/2 forcing t 3 = 1, which is impossible. Thus a2 2 3. If a, 2 4, then

2(s - 1) - 1 = C ti(ai - 1) 2 (s - 1)/2

r

+ 3 1ti i=2

+ 3(s - (s - 1)/2) = 2(s - 1) + 3, = (s - 1)/2

a contradiction. Thus a, = 3 and t , Suppose t , = 2(s - 1)/3. Then s 2

tl

=

(s - 1)/3 or 2(s - 1)/3.

+ t 2 = ( S - 1)/2 + 2(s - 1)/3 = 7(s - 1)/6 > s - 1

and hence s = 7(s - 1)/6 which yields s = 7 and 1 G : Z I = 12. We now suppose that t , = (s - 1)/3. Since t , + t , = 5(s - 1)/6 < s, we have r 2 3. If a3 2 6, Equation (3) yields

1 ti(ai - 1) = - 1)/2 + 2(s - 1)/3 + 2 7(s - 1)/6 + 5(s - -

2(s - 1) - 1 =

(S

t1

= 7(s - 1)/6

tl

t2)

+ 5(s - 1)/6 + 5,

a contradiction. Thus a3 = 3,4, or 5. If a3 = 3 = a2 then t 3 I t 2 and so t , s2

Cti(ai - 1) 1=3

+ t, + t3 =

(S

- 1)/2

= (s -

1)/3. This yields

+ ( S - 1)/3 +

(S

- 1)/3

and again s = 7(s - 1)/6 and I G : ZI = 12. Suppose a, = 4. If t , = (s - 1)/2, we have s 2

(S

- 1)/2

+ (S - 1)/3 + (S - 1)/2 = 4(s - 1)/3 > s - 1

and thus s = 4. This yields I GI = 6, which is impossible since a, must divide I GI by Lagrange’s theorem. Thus t , = (s - 1)/4 which yields s2

(S

- 1)/2

+ (S - 1)/3 +

(S

- 1)/4 = 13(s - 1)/12 > s - 1.

Thus s = 13(s - 1)/12 and s = 13 and IG : ZI = 24 in this case. The remaining case is a3 = 5. If t 3 = 2(s - 1)/5, this yields s2

(S

- 1)/2

+ (S - 1)/3 + 2(s - 1)/5 = 37(s - 1)/30 > s - 1.

since this has no integer solution, we have t , = (s - 1)/5 and s 2 31(s

- 1)/30 > s

- 1

and s = 31. This gives I G :Z I = 60 and the proof is complete.

I

Chapter 14

260

We mention that all three cases of Theorem 14.23 can occur. If G = SL(2,3), we get G/Z Z A4 of order 12. If G = GL(2, 3), we get G/Z Z C4 of order 24 and if G = SL(2,5), we get G/Z Z A S of order 60. In fact, A4, C4, and A5 are the only possibilities for G / Z . Problems

(14.1) Let G be a linear group of degree n. Show that there exists another linear group G* of degree n such that G/Z(G) z G*/Z(G*) and det(g*) = 1 for g* E G*. Do this in such a way that G* is irreducible iff G is irreducible. Hint Let I be the n x n identity matrix and let S = { a l l a @, ~ a" = det(g)for some g E G}. Consider GS E GL(n, @).

(14.2) Let x E Irr(G) be p-rational and faithful and assume that x(1) < 4

P

-

1)

for some k I p. Show that the Sylow p-subgroups of G are elementary abelian of order < p k . (14.3) Let G be an irreducible p-solvable linear group of degree pa > 1. Let N = OJG) and U / N = O,(G/N). Show that U is nonabelian. (14.4) Let G be a linear group of degree n and suppose that n is not divisible by any prime power q > 1 such that q = 1, 0, or - 1 mod p ,where p is some prime. Assume that G has a solvable irreducible normal subgroup. Show that a Sylow p-subgroup of G is normal. Hint Let Po E Syl,(S) where S is normal, irreducible, and solvable and consider C = C,(P,) Q G.

(14.5) Let G have a normal p-complement N and assume that a Sylow psubgroup of G is not normal. Suppose that G is a linear group of degree p - 1. Show that N / ( N n Z(G))is a 2-group and thus N is nilpotent. Hint

Let P E Syl,(G). Show that CN(P)c Z ( N ) .

(14.6) Let G be a linear group of degree p - 1, where p is a prime. Suppose there exists p-solvable M Q G such that M does not have a normal Sylow p-subgroup. Show that G is solvable. (14.7) Let G be an irreducible linear group of degree p + 1 where p is an odd prime. Let P E Syl,(G) and suppose P 4 G. Assume that G has a normal p-complement, N. Show that p + 1 is a power of 2. Hints Let G be a minimal counterexample and let q be an odd prime divisor of IN : CN(P)I.Let Q E Syl,(N) and show that "(Q) is abelian.

Problems

261

(14.8) Replace the hypothesis that G has a normal p-complement in Problem 14.7 by the weaker condition that G is p-solvable. Draw the same conclusion. Hints Let U = O,(G) and M / U = O,.(G/U). It is no loss to assume that G / M is a p-group and that OP'(G)= G . If U $tZ(G),let T = I&), where 1 is a linear constituent of xu and x is the given faithful character of G . Show that xT has an irreducible constituent of degree p and that M n T u M . Let G be a linear group which is generated by two elements, x and y. Suppose that each of x and y have only two distinct eigenvalues. Show that the irreducible constituents of G (that is, of the given faithful representation) have degree at most 2.

(14.9)

Hint If V is a nonzero vectorspace over C and V,, V,, V,, and V, are subspaces such that V = 5 + 6 whenever i # j , then there exists a subspace U c V such that dim U = 2 and U n V;. # 0 for all i.

(14.10) Let G be a solvable linear group of degree n. Show that G has a nilpotent normal subgroup of index 5 n !. (14.11) Let p be a prime. Show that for every integer n > 0, there exists a p-solvable linear group G of degree n with I P : O,(G)I 2 p", where P E Syl,(G) and

15 Changing the Characteristic

Some of the deepest and most powerful results in group representation theory involve “modular” representations, that is, representations over fields of prime characteristic. These are important for at least two reasons. First, if K , H 4 G , with K E H,and H / K is an elementary abelian p-group, then H/K may be viewed as an F[G]-module, where F is the field of order p. In this situation, the representation theory can give direct structural information about G . Perhaps an even more important way in which the characteristic p representations of G are relevant is that they can give new information about the characteristic zero situation. This is especially the case when I GI is divisible by p since then it is possible to obtain results which relate the p-subgroups of G with the properties of Irr(G). Our emphasis in this chapter will be on the relationship between the absolutely irreducible characteristic p representations of G and Irr(G). Following R. Brauer, who was the originator of this theory, we shall focus our attention on characters rather than on modules or representations. (In fact, we shall not even mention the indecomposable but not irreducible modules which seem to be crucial for many of the deeper results.) The objective of this chapter is to familiarize the reader with the principal definitionsand the most basic results of the theory. We do not attempt to give a comprehensive treatment of the subject and we shall not prove every fact that is mentioned. We establish some notation which will remain fixed throughout this chapter. Let R be the full ring of algebraic integers in C and let p be a prime. We construct a particular field F of characteristic p by choosing a maximal 262

Changing the characteristic

263

ideal M 2 p R of R and setting F = R / M . (Note that there is a certain amount of arbitrariness here since M is not uniquely determined.) Let * denote the natural homomorphism R -+ F. Since p l * = p* = 0, it follows that char(F) = p as claimed. (15.1) LEMMA Let U = and * be as above. Then

{EE

@I&"'

=1

for some m E Z withpl(m}. Let R, F

(a) U G R ; (b) * maps U isomorphically onto F " ; (c) F is algebraically closed and algebraic over its prime field. Proof Clearly U E R and so * is defined on U . If a E U - { l},then a is a primitive nth root of 1 for some n > 1 withpyn and hence 1

+ x + . . . + xn-'

n

n- 1

=

(xn - l)/(x - 1) =

(x - ai).

i= 1

Setting x = 1, we conclude that 1 - a divides n in R . If a* = 1, then (1 - a)* = 0 and thus n* = 0. Since p* = 0 and ( p , n ) = 1, it follows that 1* = 0, a contradiction. Thus * maps U isomorphically into F". If a E F, then a = a* for some a E R and there exists monicf E Z[x], with f ( a ) = 0. Let K c F be the prime subfield. Then 0 # f* E K [ x ] andf*(a) = f(a)* = 0. Thus F is algebraic over K. To complete the proof, let E be an algebraic extension of F. Then U* c F " E E x and it suffices to show that E x E U* in order to conclude that U* = F " and that F is algebraically closed. Let fl E E " . Then fl is algebraic over F and hence over K and thus p" = 1, where m = IK(p))- 1. Nowpym and so U* contains m roots of x"' - 1. Thus fl E U* and the proof is complete. I Continuing with the above notation, let 3 be an F-representation of a group G . Let Y be the set of p-regular elements of G , that is, elements of order not divisible by p . We define a function cp: Y + C as follows. Let x E Y and let el, E ~ . ,. . , E / . E F " be the eigenvalues of X(x), counting multiplicities. (Thusf = deg 3 and E~ = $(x) where $ is the F-character afforded by 3.) For each i,there exists a unique u iE U such that (ui)*= ci.Let q ( x ) = ui . The function cp: Y + R c C is called the Brauer character of G afforded by X. Note that similar F-representations afford equal Brauer characters and that Brauer characters are constant on conjugacy classes. Both of these statements follow from the fact that X(x) andP-'X(x)P have the same eigenvalues. Also, if x E 9, then cp(x-') = cp(x)since the eigenvalues of 3(x- ') are the reciprocals of those of X(x) and for u E U we have (U)* = (u-')* = (u*)-'.

1

264

Chapter 15

(15.2) LEMMA Let X be an F-representation of G which affords the Brauer character cp and the F-character $. For g E G , let x = g, E 9. Then cp(x)* = *(g)*

Proof We have g = x y , where x, y E (g), p$o(x) and o(y) is a power of p . Replace X by a similar representation so as to assume that X@) is in upper triangular form. Since X(g) = X(x)X(y) it follows that the eigenvalues ai of X(g) can be factored, ai = where E ~ .,. . ,Ef are the eigenvalues of X(x) and = 1. Since o(y) is a power of p , we have si= 1 and ai= E ~ Thus . $(g) = $(x). That *(x) = cp(x)* is immediate from the definition of cp. I

Lemma 15.2 provides one reason why we only bother to define Brauer characters on p-regular elements: this is sufficient to reconstruct the full F-character afforded by X. Some words of caution are appropriate here. Given a group G and a function, cp: Y + @, where Y is the set of p-regular elements of G , it is not always meaningful to ask if cp is a Brauer character unless the ideal M G R is specified or some other additional information is given. Examples exist where cp is a Brauer character with respect to some choice of M and is not one when some other maximal ideal is chosen. Also, if is an automorphism of the complex numbers and cp is a Brauer character of G , then cp" need not also be a Brauer character where cp" is defined by cp"(x) = cp(x)" for x E 9. (15.3) LEMMA Let cp be a Brauer character of G. Then q, the complex conjugate function, is also a Brauer character. Proof Let X be an F-representation affording cp, where F = R I M , as and observe that 9 is an F-represenusual. For g E G, define g(g) = X@tation of G. If E ~ .,. .,Ef are the eigenvalues of X(g), then E ~ - - ' , . . ., E / - 1 are the eigenvalues of X(g)-' = X(g-') and hence of g(g). The result now follows. I

Let X1, , . . ,3, be a set of representatives for the similarity classes of irreducible F-representations of G and let cpi be the Brauer character afforded by X i . We say that the cpi are irreducible Brauer characters and we write IBr(G) = {cpi}. When we use this notation, it is understood that a particular prime p and maximal ideal M have been fixed, It is routine to prove that sums of Brauer characters are Brauer characters and that every Brauer character is of the form nicpi,where the ni E Z are nonnegative and not all zero. To prove that the cpiare linearly independent we need the following fact from algebraic number theory. Although it is not terribly deep, we omit the proof.

Changing the characteristic

265

(15.4) LEMMA Let a,, . . . ,a, E C be algebraic over Q and let I be a proper ideal of R, the ring of algebraic integers. Suppose that not all ai = 0. Then E I. there exists /? E: C such that Pai E R for all i but not all /?ai Proof Omitted.

I

(15.5) THEOREM The irreducible Brauer characters cpi are distinct and linearly independent over C. Also, if aicpi(x)= 0 mod M for all x E Y with ai E R, then all ai = 0 mod M .

c

Proof We prove the second statement first. Since the Xi are absolutely irreducible, we have X,(F[G]) = Mf,(F), whereh = deg Xi. It follows that we can choose bi E F[G], with t,b,(bi) = 1, where t,bi is the character afforded by Xi. Also, by Theorem 9.6, we may suppose that t,bi(bj)= 0 if i # j. We have cr,cpi(x)= 0 mod M for all x E Y and thus ai*t,bi(g)= 0 for all g E G by Lemma 15.2. It follows that Ci ai*t,bi(bj)= 0 for all j , and thus ai* = 0 for all i. This proves the assertion. Now let E be the algebraic closure of Q in C and suppose that aicp, = 0 with a, E E . If not all cq = 0, we apply Lemma 15.4 and choose /? with all /?aiE R but not all Bai E M. Since (/?cci)cpi = 0, this contradicts the first part of the proof. Thus all ai = 0 and the 'pi are linearly independent over E . Since the 'pi have values in E , it follows by elementary linear algebra that they remain linearly independent over any extension field of E . 1

c

1

1

The principal reason that Brauer characters are important is that they provide a link which connects Irr(G) with the characteristic p representations of G. (15.6) THEOREM Let x be an ordinary character of G and let 2 denote the restriction of x to 9, the set of p-regular elements of G. Then f is a Brauer character of G (for any choice of M). In order to prove Theorem 15.6, we need to consider a somewhat larger ring than R. As always, we assume that we have fixed a particular maximal [ /? E R, /? # M} G C and observe that 8 is ideal, M 2 pR. Let 8 = { a / / ?a, a r i n g a n d 8 2 R . L e t A?= { ~ / / ? I ~ E M , / ? EM}.Theni@isanidealof Rr? and every element of 8 -fi has an inverse in 8. It follows that is the unique maximal ideal of r?. We call 8 a ring of local integers for the prime p . We extend the homomorphism *: R -+ F to r? by defining (a//?)*= a*//?*. Note that i@ is the kernel of this extension and M = i@ n R. (15.7) LEMMA (Nakayama) Let r? and fibe as in the preceding and let V be a finitely generated r?-module. Suppose V = VA? U for some submodule U E V . Then U = V .

+

Chapter 15

266

Proof Since V is finitely generated, we can choose ul, u z , . . . ,u, E I/ such that V = via+ U . Do this with the minimal possible n. If n = 0, then U = V and we are done. Suppose n 2 1 and write

c

n

u, =

C u i m i + u, i= 1

where mi EBand u E U.Thus n- 1

u,(l - m,) E

C uiA+ U.

i= 1

Since m, E A,we have 1 - m, E i? - fi and thus 1 - m, is invertible in and E uiA? + U . It follows that V = c ; Z : ui + U and this contradicts the minimality of n and completes the proof. I u,

c;:

Suppose X is a @-representationof G with the property that all entries in the matrices X(g) for g E G lie in a. We can construct an F-representation X* of G by setting X*(g) = X(g)*, that is, we apply * to every entry of X(g). The following includes Theorem 15.6.

(15.8) THEOREM Let X be a @-representationof G. Then there exists a @representation '2) similar to 3 such that all entries in '2)(g) lie in a for all g E G. If '2) is any such representation, then the F-representation '2)* affords the Brauer character 2 where x is the ordinary character afforded by X (and by 2)). Proof Let E be the algebraic closure of Q in C. By the results of Chapter 9, every @-representationof G is similar to one of the form 3" for some Erepresentation 3.It therefore suffices to assume that X is an E-representation and to produce a similar E-representation 9 with entries in r?. Let V be an E[G]-module corresponding to X and let ul, . . . ,u, be an E-basis for V . Let W be the a-span of the finite set {uigI 1 5 i I n, g E G } so that W is a finitely generated a-module which is G-invariant. Let z:

w -+ w/wa

be the natural homomorphism and view W/W@ as an F-vector space via (wz)a* = (wa)z for a E a . Let {wLz} be an F-basis for W / W a so that wj& = W / W A and W = W M + wjl?. By Nakayama's Lemma 15.7, we have W = wj a and thus the w j span V over E since W contains an E-basis for V . We claim that the w j are linearly independent over E. Suppose that wjaj = 0 with aj E E and not all aj = 0. By Lemma 15.4, we can multiply by a suitable /?E E and assume that all aj E R c I? but that not all aj* = 0. Now 0 = wjaj)z = (wjz)aj* and this contradicts the linear independence of the wjz and proves the claim.

(c

c

c

(1

c

Changing the characteristic

267

Now let 9 be the E-representation of G corresponding to V with respect to the basis ( w i } so that '2)(g) = (aij),when w i g = w j a i j .However, w i g E W = w j 8 and so all aij€R".This completes the proof of the first assertion. Now for p-regular BEG, we need to compute the eigenvalues of g * ( g ) = g(g)*. Let f be the characteristic polynomial of g(g) so that .f(x) = det(xZ - g(g)). Then . f &x] ~ and .f* E F[x] is the characteristic polynomial of 'z)(g)*.Writef(x) = n ( x - ,Ii) and note that the eigenvalues li lie in R c 8. Then f *(x) = n ( x - ,Ii*) and hence the eigenvalues of g * ( g ) are the li*.Thus f ( g ) = liis the value of the Brauer character afforded by g*at the element g. The proof is complete. I

1

1

In the foregoing proof, it is not the case that the character x uniquely determines the F-representation g* up to similarity. It is possible that x is afforded by another C-representation 3 with entries in R" such that g*and 3* are not similar over F. Of course, x does uniquely determine the Brauer character 2 and we may write 2 = n,rp where rp runs over IBr(G) and 0 5 n E Z. The coefficientsncpare uniquely determined because of the linear independence of IBr(G) and since n, is the multiplicity of a particular irreducible F-representation as a constituent of ?)*, it follows that g* and 3* have the same irreducible constituents with the same multiplicities. Since ?I and *3* need not be completely reducible, it does not follow that they are similar. (15.9) DEFINITION Let x E Irr(G) and let 2 be the restriction of regular elements of G . Write

li

=

c

x to the p -

d,,'P.

rp E IBr(G)

The uniquely defined nonnegative integers d,, are the decomposition numbers of G for the prime p . We view the decomposition numbers as forming a IIrr(G)I x IIBr(G)I matrix, called the decompositionmatrix. Although the decomposition numbers are not even defined until the maximal ideal M is chosen, it is a fact (whose proof we omit) that the decomposition matrix of G for the prime p is uniquely determined up to permutations of the rows and columns. (15.10) THEOREM The decomposition matrix (dx,) has linearly independent columns. Also, IBr(G) is a basis for the space of @-valuedfunctions defined on p-regular elements of G and constant on conjugacy classes. Proof Let V be the space of p-regular class functions and let W G V be the span of IBr(G). Let U be the span of the columns of (d,,) so that dim U I I IBr(G)I = dim W I dim V ,

Chapter 15

268

where the equality follows from Theorem 15.5. The theorem will follow when we show that dim V Idim U . It therefore suffices to find a one-to-one linear map from the dual space 8 of V into U.The elements of U are columns indexed by x E Irr(G). For a E 8, define u = col(a(2)). To show that u E U , observe that a(f) = ace, d,,cp) = d,,a(cp). Thus u is a linear combination of the col(d,,) for cp E IBr(G). We map P + U by a H col(a(f)). If a@) = 0 for all x E Irr(G), we claim that a = 0. To see this, let 9 E V and extend 9 to a class function 9, of G. Write 9, = a,X so that 9 = a,f and a($) = 0 as desired. The result now follows. 1

c,

c

c

(15.11) COROLLARY The number of conjugacy classes of p-regular elements of G is equal to I IBr(G)I. This is also the number of similarity classes of irreducible K-representations of G for every splitting field K of characteristic p.

Proof The first assertion is immediate from Theorem 15.10. To prove the second statement, we may replace K by its algebraic closure by Corollary 9.8. Assuming that K is algebraically closed, it contains an isomorphic copy of F which is an algebraic closure for E,. By Corollary 9.8 again, we may assume that F = K. The result now follows from the fact that nonsimilar irreducible F-representations afford distinct Brauer characters. 1 (15.12)

COROLLARY

If cp E IBr(G), then there exists x E Irr(G) with d,, # 0.

Note that for each x E Irr(G), it is trivial that there exists cp E IBr(G) with d,, # 0 since 0 # ~ ( 1=) 1 d,,cp(l). The interesting case of this theory is when p I G I. The reason for this is given by the following.

I

(15.13) THEOREM Supposep&'(GI. Then IBr(G) = Irr(G).

Proof In this case, the group algebra F[G] is completely reducible and thus by Corollary 1.17 we have I G I = dim F[G] = (deg Xi)2, where the X iare a set of representatives for the similarity classes of irreducible F-representations. If Xi affords cpi E IBr(G), then deg X i= cpi(l)and hence

c

=

c

'P. c E IBr(G)

c,

c

cp(l)P(1) d,,d,,. x

1,

If cp # p, then d,,d,, 2 0 and if cp = p, then d,,d,, 2 1 by Corollary 15.12. It follows that these inequalities are all equalities. ,(dXJ2= 1 and hence for each cp, there is a unique x We now have 1 such that d,, # 0, and in fact d,, = 1. Since 1 ,d,,d,, = 0 if cp # p, it

Changing the characteristic

269

follows that for each x E Irr(G),there is a unique cp with d,, # 0. We conclude that x = 2 = cp and the result follows. I

A necessary condition that a function 9: Y -+ C be a Brauer character is that for each subgroup H G G with p,/'IHI, the restriction 9, is a Brauer character and hence is an ordinary character of H . This is something which can be checked. It is trivial that every cp E IBr(G) is a @-linearcombination of the Brauer characters 2 for x E G. This may be seen by extending cp to a class function of G. (15.14) THEOREM Let cp€IBr(G). Then cp is a Z-linear combination of

(2 I x E Irr(G)). Proof Extend cp to a class function 9 of G by setting 9(g) = cp@,.). It suffices to show that 9 is a generalized character of G. We appeal to Brauer's Theorem 8.4(a). Let E c G be elementary and write E = P x Q, where P is a p-group and p,/'lQI. IfgE E, write g = xy with ~ E and P ~ E QThen . 9(g) = ~ ( yand ) so 9" = 1, x cpQ. By Theorem 15.13, cpQ is a character of Q and hence 9" is a character of E. It follows that 9 is a generalized character and the proof is complete. I

We digress to show another way in which Theorem 15.13 can be used. (15.15) LEMMA Let E be an algebraically closed field of characteristic not dividing IN 1, where N is a group. Let H act on N and suppose that C&) = 1 for all 1 # n E N . Let 1 ' ) be a nonprincipal irreducible E-representation of N and write gh(nh) = g(n)for n E N and h E H . Then the representations g h are pairwise nonsimilar for h E H .

Proof Let K be the algebraic closure in E of the prime subfield of E. Then '1) = 3" for some irreducible K-representation 3 of N . It suffices to show that the 3hare pairwise nonsimilar by Corollary 9.7 and hence we may assume that K = E. If char(E) = 0, then (up to isomorphism) E E @ and the representations (g)")h are pairwise nonsimilar by Theorem 6.34 and Problem 7.1 which proves this case. Suppose char(E) = p. Choose a maximal ideal, M 2 pR of R and let F = RIM as usual. Then F z E and we may assume F = E . Let 'I) afford H all cpeIBr(N) = Irr(N). By Theorem 6.34, the characters cph for ~ E are distinct and the result follows. I

The following result is often quite useful.

Chapter 15

270

(15.16) THEOREM Let G be a Frobenius group with kernel N and let K be a field with characteristic not dividing IN I. Let V be a K[G]-module and suppose that C,(N) = 0. Let H be a Frobenius complement for G. Then V has a basis which is permuted by H with orbits of size IH 1. Also, if H, 5 H, then dim C,(H,) = I H :H,1 dim C,(H).

Proof The second statement follows from the first since if b is a basis permuted by H , then dim C,(H,) is equal to the number of orbits of the action of H o on b. Since each orbit of H o on b has size I H, I, the assertion follows. To prove the first statement we argue that it suffices to assume that K is algebraically closed. Let X be a K-representation of G corresponding to V and let E 2 K . The condition that C,(N) = 0 is equivalent to the restriction 3, having no principal constituent and this property is inherited by (XE), by Theorem 9.6. The conclusion of the theorem is equivalent to XH being similar to a representation ‘1) in block diagonal form with each block being the regular representation of H. If we can prove that ( X H ) Eand ‘l)Eare similar, then XH and 9 are similar by Problem 9.5. It follows that we may replace K by any extension field and thus we assume that K is algebraically closed. Now let A! be a representative set of irreducible K[N]-modules. Since H acts on the set of similarity classes of K-representations of N , we can define a corresponding action of H on A!. Let A!, E A! be a set of representatives for the H-orbits of nonprincipal K[N]-modules in A!. In the notation of Definition 1.12,let

w=

c M(V)GV.

ME”&

We claim that V = ‘ h E H Wh. This will suffice to prove the result since we obtain a basis for V by choosing any basis for W and taking all H-transu { V } ,where U is a lates. To prove the claim, observe that A! = U h f H principal module. By Lemma 15.15, this union is disjoint. Since V, has no principal constituent, Lemma 1.13 yields

The result now follows from Lemma 6.4.

I

We now resume consideration of the general case where p can divide 1 G I. We introduce the concept of “blocks” which is at the heart of Brauer’s theory. For each x E Irr(G),we have the algebra homomorphism ox:Z(C[G]) + C as in Chapter 3. The function wxis determined by its values on the conjugacy

Changing the characteristic

27 1

-

class sums K , which form a basis for Z(C[G]). Furthermore, the values wx(Ki)lie in R. For x, t,b E Irr(G), write x t,b if ox(Ki)*= a&)* for all i. This clearly establishes an equivalence relation on Irr(G). (15.17) DEFINITION A p-block of G is a subset B E Irr(G) u IBr(G)such that

-

defined (a) B n Irr(G) is an equivalence class under the relation above. (b) B n IBr(G) = (cp E IBr(G)ldx, # 0 for some x E B n Irr(G)}.

-

From its definition, it appears that the equivalence relation on Irr(G) depends on the choice of the maximal ideal M. In fact, this is not true. (15.18) THEOREM Let x, i,b E Irr(G). Then x and t,b lie in the same p-block iff o,(K) - w,(K) lies in every maximal ideal of R which contains pR for every class sum K.

-

Proof The "if" statement is clear. Assume x t,b and fix K. Let a = o x ( K )- w,(K). We shall show that a" E pR for some integer n and the result will follow. Let 8 = 9(QIGI/Q),the Galois group, and let CJ E 8.Let E be a primitive I GJthroot of unity so that E* = E" for some m,with (m,IGl) = 1. Let g E G be in the class with sum equal to K and let L be the sum of the class containing 9". We have x(g)" = x(g"). Also IC(g)l = IC(g")l since ( 9 ) = (9"). It follows that wx(K)"= o,(L) and similarly o,(K)" = a&). It follows that a" = wx(L)- w&) E M since x t,b. Let f(x) = (x - a") E (Q n R)[x] = Z[x]. Since a" EM for all CJ E 8,all of the coefficients offexcept for the leading one lie in M n Z = pZ. Thus 0 = f ( a ) = d mod pR, where n = 181.The proof is now complete. I

naog

-

Although it is clear from the definition that every x E Irr(G) lies in a unique p-block, the analogous statement about IBr(G),though true, is not so obvious. To prove it, we relate the p-blocks of G to the F-algebra, Z(F[G]). P[G] We extend the map *: a -,F to a ring homomorphism *: I?[G] by setting g* = g for g E G. (Here, i?[G] is simply the a-span of the group elements in @[GI.) Since the class sums K i form a basis for Z(C[G]) and their images Ki* E F[G] form a basis for Z(F[G]), it follows that Z(&G]) = a[G] n Z(@[G]) maps onto Z(F[G]) via *. Now suppose x E Irr(G). Then oxmaps Z(&G]) to R" and we can define ox*:Z(F[G]) .+ F by setting a x * ( z * ) = ox(z)* for z E Z(I?[G]). This is well defined since if zl, z2 E Z(R[G]), with zl* = z2*, then z1 - z2 has coefficients in &and thus (wx(zl - z2))* = 0. It is trivial to check that ox*:Z(F[G]) + F is an algebra homomorphism and that if x, t,b E Irr(G), then ox*= o$*iff x t,b; that is, iff x and t,b lie in the same p-block. --f

-

Chapter 15

212

Let Bl(G) be the set of p-blocks of G. If B E Bl(G), let I , = o>,* for H A, is a one-to-one map from Bl(G) into the set of algebra homomorphisms Z ( F [ G ] )+ F .

x E B n Irr(G). Thus B

(15.19) THEOREM Let cp E IBr(G). Then cp lies in a unique p-block B. If ‘I) is an irreducible F-representation which affords cp, then ‘l)(u) = R , ( u ) ~ for all u EZ(F[G]).

Proof’ By Corollary 15.12, d,, # 0 for some x E Irr(G). Let B E Bl(G), with x E B. Then cp E B. Let ‘l) afford cp. We will be done when we show that ‘l)(u) = A,(u)l for all u E Z ( F [ G ] )since this equation uniquely determines i , and thus uniquely determines B. Let X be a @-representation which affords x and which has entries in a (Theorem 15.8). Thus X* affords the Brauer character 2 and by the linear independence of IBr(G), it follows that ‘l) has multiplicity d,, > 0 as a constituent of X*. Now let u E Z ( F [ G ] )and write u = z* for some z E Z ( R [ G ] ) . Then X*(u) = X(z)*

=

(oj,(z)l)* = O>,*(u)l = AB(Ll)I.

Since ’I) is a constituent of X*, the result follows. We define a graph with Irr(G) as its vertex set by linking x, $ E Irr(G) iff there exists cp E IBr(G)such that d,, and d,, are both nonzero. This is called the Bruuer graph. If x and $ are linked bycp, it follows that x and $ lie in the same p-block, namely the unique one which contains cp. This proves the following. ( 1 5.20) COROLLARY Let B be a p-block of G. Then B n Irr(G) is a union of connected components of the Brauer graph.

We shall see that in fact B n Irr(G) is a single connected component. One of the principal benefits of considering blocks in various applications of the theory is that in certain circumstances we can replace equations like

(which arise from the second orthogonality relation) by equations like

where B is a p-block. In particular, this holds whenever x, and y, are not conjugate in G. We shall prove a weak form of this “block orthogonality.” For each cp E IBr(G) we define @ ,

= XE

c

Irr(G)

d,,x.

Changing the characteristic

273

The 0’s are called projective characters of G. (There is no connection with projective representations in the sense of Chapter 11, but there is a connection with projective modules in ring theory. The 0’s are also called “principal indecomposable characters” in the literature.) (15.21) LEMMA Let d c Irr(G) be a union of connected components of the Brauer graph and let W = {cp €IBr(G))d,, # 0 for some X E d } .Let x , y E G with p t o ( x ) . Then

For x E at, we have

Proof

x(x) =

d,,cp(x) (Pea

since d,, = 0 for all p E IBr(G) - W. Also, if cp E 59, then @,(Y)

=

c d,,XCV)

Xed

since d,, Now

=

0 for all 5 E Irr(G) - d.

and the proof is complete.

I

(15.22) COROLLARY For each cp€IBr(G) we have @Jy) = 0 if pJo(y). Furthermore, IPI divides @,(l)where P E Syl,(G).

Proof Let x E G be p-regular and let p (o(y).By the second orthogonality relation we have

By Lemma 15.21 with d

=

Irr(G) and W

=

IBr(G), we conclude that

Since this holds for all p-regular x E G, the linear independence of IBr(G) yields @,b)= 0 for all cp. The second statement follows since [PI[(@JP, lP] = @,(l). 1

(15.23) COROLLARY (Weak Block Orthogonality) Let x, y E G with p$o(x) and p Iok). Let B be a p-block of G. Then

Chapter 15

214

Proof Apply Lemma 15.21 with d = Irr(G) n B and 9# = IBr(G) n B. Since (D,(y) = 0 for all cp, the result is immediate. I Before going on to develop more of the theory of blocks we digress to give an application of what we have already done.

(15.24) THEOREM (Brauer) Let G be a simple group of order pa4*r where p , q, r are distinct primes. Let S E Syl,(G). Then S = C,(S). The p-block containing 1, is called the principal block.

(15.25) LEMMA Let G be simple and let B be the principal p-block of G . Suppose x E B n Irr(G) and x(1) is a power of p . Then x = 1,. Proof Let g E G and let K E C[G] be the sum of the elements in Cl(g), the conjugacy class containing g. Then w , ( K ) = ICl(g)l and since x and 1, lie in the same p-block, we have

) I G I, we have x(1) = 1 Now let P E Sylp(G).If P = 1, then since ~ ( 1divides and by simplicity x = 1,. Assume then, that P > 1 and take g E Z ( P ) - { l}. Then p,+'ICl(g)J and so ICl(g)I $0 mod M since M n Z = p Z . Thus (*) yields x(g) # 0. However ( ~ ( l ) ,ICl(g)l) = 1 and Burnside's Theorem 3.8 yields that g E Z(x). Since g # 1 and G is simple, it follows that x = 1,. I Proof of Theorem 15.24 Suppose C&) > S. Then there exists x E G of order pr or qr. Say o(x) = pr and let B be the principal p-block of G. By Corollary 15.23, we have

We claim that the second sum is zero. If X E B n Irr(G) and x # 1, and q,+'x(l), then rlx(1) by Lemma 15.25. Since rlo(x), we have ~ ( x = ) 0 by Theorem 8.17 and the claim follows. We conclude that - l/q is an algebraic integer and this contradiction completes the proof. I For x E Irr(G), let ex E Z(C[G]) be the idempotent corresponding to By Theorem 2.12 we have

x.

Changing the characteristic

275

Note that in general, e, 4 8 [ G ] (where 8 is a ring of local integers for p ) and so we cannot find idempotents in F [ G ] simply by applying * to e x . Since e,e@ = 0 if x # $, sums of the various e,’s are also idempotents and we shall consider when such a sum lies in 8 [ G ] . Note that if d G Irr(G), then the set d can be recovered from s = e, since d = {x E Irr(G)I se, # 0).

xreB

(1 5.26) THEOREM (Osima) Let d be a connected component of the Brauer graph and letf= e,. Writef = C a,g. Then

cxed

(a) a, = j l / I G l ) C X E d x(l)x(9); (b) a, E R for all g E G ; (c) a, = 0 if plo(g). Proof Statement (a) is immediate from the formula for e , in Theorem 2.12. By Lemma 15.21, we have a, = (l/IGJ)C,,,rp(l)cD,(g), where ?!.# = {cp E IBr(G)Id,, # 0 for some x E d } .If pIo@),then OJg) = 0 by Corollary 15.22 and (c) follows. Now assume that p#o(g). Lemma 15.21 then yields a, = (1/IGI) xOJ1)qT). (PEB

However, I P I divides Oq(1) where P E Syl,(G) and thus O& 1)/1 G I E 8. Since cp(g)E R , it follows that a, E 8 and (b) is proved. I (15.27) THEOREM The connected components of the Brauer graph are exactly the sets Irr(G) n B for p-blocks B. Furthermore, every set d E Irr(G) e, E 8 [ G ] is a union of sets of the form Irr(G) n B. such that

xXEd

Proof We prove the second statement first. If x E Irr(G) is afforded by X, then X(ex) is the identity matrix and X(e@)= 0 if $ # x. Thus w,(e,) = 1 and o,(e+) = 0 for $ # x. Writef = e,. It followsthat x E d iff o,(f)= 1 and otherwise o x ( f )= 0. Thus x E d iff o,(f)*# 0. Now iff E 8[G], then all o,(f)*are equal as x runs over Irr(G) n B for a block B. The assertion follows. Now let d be a connected component of the Brauer graph. By Theorem 15.26, e, E &GI and thus d is a union of sets of the form Irr(G) n B. Since each Irr(G) n B is a union of connected components of the Brauer graph, the result follows. I

Id

xd

We now have three different characterizations of the sets B n Irr(G). In addition to their definition as equivalence classes under -, they are also the connected components of the Brauer graph and they are the minimal e, E &GI. nonempty subsets d c Irr(G) such that The following is a strengthening of Theorem 15.14.

Chapter 15

216

(15.28) LEMMA Let B be a block of G and let cp E IBr(G) n B. Then cp is a Z-linear combination of Brauer characters of the form 2 for x E Irr(G) n B. Proof By Theorem 15.14, there exist b, E Zsuch that cp = Thus cp = 9, + So, where

1

9, = XE

b,f

and

1

9, =

Irr(G) n B

XE

cXEIrr(G) b,i.

b,f.

Irr(G) - B

Now we express 9, and 9, in terms of IBr(G) using the decomposition numbers d,, for p E IBr(G). If x E B, then d,, = 0 when p 4 B and hence 9, is a linear combination of p E B. Similarly, if x 4 B, then d,, = 0 when p E B and hence 9, is a linear combination of p .$ B. The equation cp = 9, + and the linear independence of IBr(G) now yield cp = 9, and the proof is complete. I (15.29)

THEOREM

Let B be a p-block of G . Then

I B n Irr(G)I 2 I B n IBr(G)1. Let x E B n Irr(G).Then the following are equivalent. (a) IB n Irr(G)I = IB n IBr(G)I. (b) Pk(I G I lx( 1)). (c) B n Irr(G) = {x}. Also in this case, B n IBr(G) = {i}.

,

Proof Let D = (&,) be the decomposition matrix and let DB be the submatrix corresponding to the rows and columns indexed by elements of B. For each ~ E Bthe , part of the corresponding column of D outside of D, consists of zeros. Since the columns of D are linearly independent by Theorem 15.10, it follows that the columns of D, are linearly independent and thus DB has at least as many rows as columns. Now assume (a). Then D B is a square nonsingular matrix and we let DB- = (arpx). For fixed x E Irr(G) n B, we have

and hence vanishes on elements of and so x is a linear combination of the arp order divisible by p (Corollary 15.22). If P E Syl,(G), then IP 1 [xp, lP] = x( 1) and (b) follows. Assuming (b) we have e,

=

(x(l)/lGI)eX(S)S4Gl gEG

and so B n Irr(G) = {x} by Theorem 15.27.

Changing the characteristic

211

Finally assume (c). Then 0 c IIBr(G) n BI IIIrr(G) n BI = 1 and (a) follows. Also, if IBr(G) n B = {cp}, then cp = bf for some b E iZ by Lemma 15.28.Thus f = d,,cp = d,,bf and d,,b = 1. It follows that d,, = 1 and the proof is complete. I Note that Theorem 15.29 provides an alternate proof of Theorem 8.17. Also observe that if pk I G 1, then each p-block contains a unique irreducible character. We usexheorem 15.26 to obtain further connections between the p-blocks e , . This Osima of G and Z ( F [ G ] ) . For each B E Bl(G),writef, = idempotent lies in Z ( B [ G ] )and we let e , = f,* E Z ( F [ G ] ) .

lXEBnIrr(G

(15.30) (a) (b) (c) (d) (e) (f) (g)

THEOREM

We have the following.

IB(eB)= 1 and IB,(e,) = 0 for blocks B # B’. The e , are idempotents and eBeW= 0 for B # B’. e, is an F-linear combination of class sums of p-regular classes. C e , = 1. If I,(z) = 0 for all B E Bl(G), then z is nilpotent. The I , are all of the algebra homomorphisms Z(F[G]) + F. Every idempotent of Z ( F [ G ] )is a sum of some of thee,.

Proof (a) For z ~ I r r ( G )we , have ~,(f,) = 1 if X E B and otherwise E B‘, then ox*= IB,and (a) follows: (b) SincefBf,. = dBB’fBwe have eBe,, = d,,.e,. Since Ade,) = 1, we have eB # 0. (c) Immediate from Osima’s Theorem 15.26(c). f B = e , = 1 and thus eB = 1* = 1. (d) (e) Let ‘I) be any irreducible F-representation of G and let ‘I) afford cp E IBr(G) n B. Then ‘I)(z)= I,(z)Z = 0 by Theorem 15.19. Thus z is in the Jacobson radical, J ( F [ G ] )and hence is nilpotent (Problem 1.4). (f) Let I : Z ( F [ G ] ) + F be an algebra homomorphism. Then ker , Ihas codimension 1 and so Z ( F [ G ] ) = ker I + Fa 1 and I is determined by its kernel. If I # I,, then ker I , $ ker I and we can choose Z, E ker I , with I(z,) # 0. Assuming 1 # {I,}, let z = z B . Then I,(z) = 0 for all B and hence z is nilpotent by (e). However, I ( z ) = I(z,) # 0. This is a contradiction. (g) Let e E Z ( F [ G ] ) be an idempotent. Then e = e e , = ee, and it suffices to show that either ee, = 0 or ee, = e B . Suppose ee, # 0. Then since ee, is not nilpotent and IB’(ee,) = 0 for B’ # B, we have I,(ee,) # 0 and thus I,(ee,) = 1 = IB(eB).Thus I,,(eB(l - e)) = 0 for all B’ E BI(G) and since e,(l - e ) = (e,(l - e))’, we have eB(l - e ) = 0 and the result follows. I ~,(f,) = 0. If x

1

1

1

n n

1

1

Chapter 15

218

We now discuss some of the connections between block theory and the collection of p-subgroups of G. We work in the situation of Theorem 15.30. If X is a conjugacy class of G we consider the Sylow p-subgroups of C,(x) . are called the p-defect groups for X . They constitute a for X E XThese conjugacy class of p-subgroups of G. The collection of p-defect groups for X is denoted 6(X). We use the notation ,%? for the sum in F[G] of the elements of the class X so that the 9 form a basis for Z(F[G]). If B E Bl(G), write eB = CaB(X)$ so that uB is a uniquely determined function from the set of classes of G into F. In fact, a B ( X )= ((1/ I G I ) C X e l r r ( G ) B x(l)X(s))*for g E X by Theorem 15.26(a). By 15.30(c) we have uB(X)= 0 if X does not consist of p-regular elements. Since 1 = AB(eB)= aB(X)AB($),itfollows that for each B E Bl(G),there exists at least one class X such that u B ( X )# 0 and &($) # 0. We call such a class a defect class for B.

c

(15.31) THEOREM (Min-Max) Let X be a defect class for B E Bl(G) and let D E 6(X).Let Y be any class of G. (a) If a B ( Y )# 0, then D contains a defect group for 9. (b) If A,@) # 0, then D is contained in a defect group for Y . To prove the min-max theorem, we define the Brauer homomorphism P p . Let P _C G be a p-subgroup. Let N = NG(P) and C = CG(P). We map P P : Z(FCG1) Z(F"1) by --f

P P ( 2 )

=

c

x,

x e X nC

and extend by linearity. Since X n C is a union of classes of N, we do have P P @ ) E Z(F"1). (15.32) LEMMA The map Pp: Z(F[G]) + Z(F[N]) is an algebra homomorphism. Proof If suffices to check that Pp(2@) = Pp(g)Pp(@ for classes X , 9. ForcEC,let d = { ( x , y ) l x ~ X , y ~ 9 , x y = c } a n d d o= {(x,y ) I x E X n C , y E Y n C , xy = c}.

Then I d I* is the coefficientof c in /3p(g@) and I d oI* is the coefficient of c in jp(&pP(L?). It thus sufficesto check that Id1 = Idolmod p. Since P E C(c), it follows that P acts on d by (x, y)" = (xu,y") for u E P. Then d ois exactly the set of fixed points of d under the action of P. The result follows. 1

Changing the characteristic

219

(15.33) LEMMA Let P be a p-subgroup of G and let P p be the corresponding Brauer homomorphism. Let z = u x g . Then PP(z)# 0 iff P is contained in some D E 6(X)with us # 0. Proof The sets 3? n C(P)are disjoint for distinct classes X and thus the nonzero elements of the form P p ( 2 ) are linearly independent. It follows that Pp(z) # 0 iff P P ( 2 ) # 0 for some X with ax # 0. Now PP(g) # 0 iff X n C(P)# Qr and this happens iff P E C(x)for some x E X . Since P G C(x)iff P is contained in some Sylow p-subgroup of C(x), the result follows. I

(15.34) LEMMA Let B E Bl(G) and let 9be a class of G with ~ ~ #(0.9 Let) P E 6 ( 9 )and N = N,(P). Then there exists b E Bl(N)such that

Proof By Lemma 15.33, PP(ee)# 0. Since P p is a homomorphism, e = PP(e,) is a nonzero idempotent inZ(F[N]) and so is not nilpotent. Thus there exists b E Bl(N) with Ab(e)# 0. Let p = P p & . Then p(e,) = Ab(e)# 0 and p is an algebra homomorphism Z(F[G]) + F . Since p(ee) # 0, we have p = A, and the proof is complete. I Proof of Theorem 15.31 If ~ ~#(0, 9 let ) P E 6 ( 9 ) . Then A, = P p A b for some b E Bl(N(P)) by Lemma 15.34. Since X is a defect class, we have 0 # A E ( 2 ) = &(Pp($)) and so Pp(3?) # 0. Thus P is contained in some defect group of X by Lemma 15.33. Now (a) follows. Now apply Lemma 15.34 to X . Since a,(.%) # 0 we have A, = PDAb for some bEBl(N(D)). Suppose A,($) # 0. Then PD($) # 0 and hence D is contained in a defect group of 2’.

(15.35) DEFINITION Let B be a p-block of G. Then the p-defect groups of the defect classes of B are called defect groups of B. The set of these is denoted 6(B). (15.36) COROLLARY Let B E Bl(G). Then 6(B) is a single conjugacy class of subgroups. Proof Let X , and X 2 be defect classes for G. It suffices to show that 6(XJ = 6(X2). Let Di E &Xi). By Theorem 15.31, each of the Di contains a conjugate of the other. The result follows. I

Which p-subgroups of G can be defect groups for blocks? It is a fact (which we shall not prove) that a defect group for a block is necessarily of the form P n Q for some P,Q E Syl,(G). We prove the weaker assertion that O,(G) is contained in every defect group of a block.

Chapter 15

280

(15.37) LEMMA Let P = O,(G). Then P E ker 9 for every irreducible F-representation 9 of G. Proof Since P has a unique p-regular class, the principal representation of P is its unique irreducible F-representation by Corollary 15.11.If 9 is an irreducible F-representation of G, then the restriction ( D p is completely reducible by Corollary 6.6.The result follows from these two facts. I (15.38) THEOREM Let X be a class of G and assume that X n C(O,(G)) = 0.Then 9 is nilpotent. Proof Let P = O,(G) act on X by conjugation and let 8 be an orbit of Let ~ E O If. ~ E O then , y = xu = this action. Then 101 > 1 and so ~1101. x[x, u] for some u E P. Since P 4 G, we have [x, u] E P and so 8 c xP. Now let 9 be an irreducible F-representation of G so that P c ker 9 by Lemma 15.37and thus 9 has the constant value 9(x) on the coset x P . Therefore 9 ( y ) = 10Ig(x) = 0 since p I 101. We thus have g($) = 0 for all irreducible 9 and hence 2 E J ( F [ G ] ) ,the Jacobson radical. It follows that 2 is nilpotent, I @ &,

(15.39)

COROLLARY

Every defect group for a p-block of G contains O,(G).

Proof Let B E Bl(G) and let X be a defect class for B. Then I,($) # 0 and hence s f is not nilpotent. Thus X n C(O,(G)) # fa and it follows that O,(G) E D for every D E 6(X). I (15.40) COROLLARY Let G be p-solvable with OJG) = 1. Then G has a unique p-block. Proof Let B, B‘ E Bl(G). We claim that LB(eB.)= aB,({I}). Since this is independent of B, the result will follow. We have IB(eBt)= C aB,(X)AB($), where the sum runs over classes X . Since IB(l)= 1, it suffices to show that if X # { l}, then either a,.(X) = 0 or I,@) = 0. Let P = O,(G). If A,($) # 0, then X n C(P) # @. However, the HallHigman “Lemma 1.2.3” yields C(P) c P and thus X c P. Since X # {l}, the elements ofX are not p-regular and thus a r ( X ) = 0. The proof is complete. I

If B E Bl(G),let D be a defect group for B and write ID I = pd. We call d the defect of B and write d = d(B). (This is well defined by Corollary 15.36.)We show how to compute d(B) from a knowledge of B n Irr(G) or B n IBr(G). We mention that if m, n E Z with pxn, then mln E R.Thus if p I m, we have m/n E pR E W. (15.41) THEOREM Let I GI = p4m withpxrn and let B E Bl(G)with d(B) = d. Then is the largest power of p which divides all ~(1) for x E Irr(G) n B.

Changing the characteristic

28 1

Proof Let X be a defect class for B and let g E X . Then 1 x1= I GI / I C@)I and so pa-d is the p-part of I X I since ID I = pd for D E Syl,(C(g)).

Let x E B n Irr(G). We have #

= (ox)*(2)

= (x(g)I.xI/x(l))*.

Since X(g) E R and x(g)I X I /x( 1)E a - A?, we conclude that I X I/ ~ ( 14) fi and hence the p-part of 1 X 1 cannot exceed that of ~ ( 1 )Thus . p a - d divides ~ ( 1 ) . The coefficient of g in the Osima idempotentfB is given by a, = (1/lGI)

and thus a, = ( l / l G l )

1

x E B n Irr(G)

1

x(1)xo

@#GG)

v~BnlBr(G)

by Lemma 15.21 (since g is p-regular). However, paI@& 1) by Corollary We have 0 # a#-) = a#* and a, 4 15.22 and thus @Jl)/lG I E R for all cp E IBr(G). We conclude that G)4 for some cp E IBr(G) n B. Now cp(g) is a Z-linear combination of x(g) for x E Irr(G) n B by Lemma IX I/x(l) = c1 E R 15.28 and thus x(s)4 A? for some x E Irr(G) n B. Now and x(s)= ax(I)/ I .XI. It follows that X( 1)/I X I 4 and so the p-part of ~ ( 1 ) cannot exceed p a - d . The proof is complete. I

a.

a

a

(15.42) COROLLARY In the situation of Theorem 15.41, pa-d is the largest power of p which divides all cp(1) for cp E IBr(G) n B. Proof In fact {x(l)lx~Irr(G) n B } and {cp(l)lcp~IBr(G) n B } have the same greatest common divisor. This follows since each x(1) is a Z-linear combination of the ~ ( 1 using ) decomposition numbers and each cp(1) is a Z-linear combination of the ~ ( 1by ) Lemma 15.28. I In connection with Corollary 15.42, we mention that cp(1)need not divide

I GI for cp E IBr(G).

In the situation of Theorem 15.41, if x E B n Irr(G),then the p-part of x(1) can be written in the form p a - d + h , where h 2 0. The integer h is called the height of x. Brauer has conjectured that all x E Irr(G) n B have height zero iff a defect group of B is abelian. Note that if d(B) = 1 and x E B has positive ) thus B n Irr(G) = {x} by Theorem 15.29. height, then pa divides ~ ( 1and This forces d(B) = 0, a contradiction. Thus if d(B) = 1, then all characters in B have height zero. Suppose H E G and b E Bl(H). Let A = A,: Z(F[H]) + F be the corresponding central homomorphism. We construct a linear map:'A Z(F[G]) + F

Chapter 15

282

by setting I G ( J t ) = I(

c

x).

x e X nH

It may be that IG is an algebra homomorphism, in which case IG = Is for some unique B E BlCG). When this happens we say that bG is deJned and we write bG = B. The block bG is called the induced block. Let b E Bl(H) for H E G and suppose bG is defined. Then every defect group for b is contained in a defect group for bG.

(15.43) LEMMA

Proof Let X be a defect class for bc and let D ~ 6 ( b ) We . have = B),where 9runs over the classes ofH contained in X . In particular, X n H # 0and there exists 9 E X n H such that &(p) # 0. By the Min-Max Theorem 15.31, there exists P E d ( 9 ) with D E P . Now P E Syl,(C,(x)) for some x E 9 c X and thus there exists S E Syl,(C,(x)), with S 2 P . Thus D E S E 6(X)= 6(bG). I

oP

&(x

The Brauer homomorphism can be used to give a sufficient condition for induced blocks to be defined. (15.44) LEMMA Let P E G be a p-subgroup and let C ( P ) E H E N ( P ) . Then bG is defined for all b E Bl(H). If b E Bl(H) and B E Bl(G), then bG = B iff A, = f i p & , where f i p is the Brauer homomorphism.

Proof The image of f i p : Z ( F [ G ] )+ Z ( F [ N ( P ) ] )actually lies in Z ( F [ H ] ) . Let I : Z ( F [ H ] )+ F be an algebra homomorphism and let p = BPI:Z ( F [ G ] )+ Z ( F [ H ] )+ F

so that ,u is an algebra homomorphism. We claim that ,u = IG. Let C = C(P)E H and let X be a class of G . Write x = u u, where u = ZxsXnc x. Then ,IG($) = I(u + u) and ~ ( = 2I(fip($)) ) = I(u). We must therefore show that I(u) = 0. Now u is a sum of elements of the form 2,where 9is a class of H such that 9 n C =@. Since P Q H we have C(O,(H)) c C and it follows that S? is nilpotent by Theorem 15.38. Thus I @ ) = 0 and hence A(u) = 0 and IG = ,u as claimed. If 1 = &, then IG = fipI is an algebra homomorphism and bc = B is defined, where B is the unique block such that I , = j p I . I

xxEX

+

Brauer’s first and second “main theorems” concern induced blocks. (15.45) THEOREM (First Main) Let D c G be a p-subgroup and let N = N(D).Then b H bG is a bijection of

{ b E Bl(N)IDE 6(b)}

onto

{ B EBl(G)(D E 6(B)}.

Changing the characteristic

283

(15.46) LEMMA Let P be a p-subgroup of G. Let C = C(P) and N = N(P). Then X w X n C is a bijection of the set of classes of G with p-defect group

P onto the set of classes of N with p-defect group P. Proof Suppose P E ~ ( Xand ) x, EX n C. Then P E Syl,(C(x)) and P E Syl,(C(y)). Write y = xg. Then P, PgE Syl,(C(y)) and hence P = Pgc for some c E C(y). Then gc E N and y = xgC.It follows that 9 = X n C is a class of N . Clearly P E 6(9)and the map X H X n C is one-to-one. If 2’ is a class of N with P E 6(9),let X be the unique class of G which contains 9. Now 9 c C since 2’ n C # 0 and C a N. If x E 9, let P E S E syl,(cG(x)). If S > P, then P < N,(P) = S n N c C,(x). Since S n N is a p-group, this contradicts P E 6(9).We conclude that P = S E 6(X) and X I+ 3’.The proof is complete. I Proof of Theorem 15.45 Write W = { b E Bl(N)ID E 6(b)}.If b E 9, then bG is defined by Lemma 15.44. Let bG = B€Bl(G). Since D ~ 6 ( b )we , have D E P for some P E 6(B)by Lemma 15.43. We claim that D = P. Let 2’be a defect class of b and let X z3 be a class of G. Then X n C = 2’ and D E 6(X)by Lemma 15.46. By Lemma 15.44 we have

A,($) = & ( b D ( g ) ) = A,($) # 0. By the Min-Max Theorem 15.31, it follows that D contains some defect group of B. Thus Pg E D c P for some g E G and hence D = P E 6(B)as desired. Thus block induction maps 93 into { B E Bl(G)ID E 6(B)}. Now let BEBI(G) with D E ~ ( B )Let . X be a defect class for B. Then a e ( X ) # 0 and D E 6(X). Thus L, = p D & for some b E Bl(N)by Lemma 15.34 and hence bG = B by Lemma 15.44. We must show that b E 93. Let P E 6(b) so that D E P by Corollary 15.39 since D a N . By Lemma 15.43, P is contained in some defect group for B and we have D E P c Dg for some g E G. Thus D = P E 6(b)and b E W. with blG = B = b2G.Let 9 be a class of N with Finally, let b,, b2 D zP ~ 6 ( 2 ’ ) If . D > P, then by the min-max theorem, &,@) = 0 = A&.?) If D = P, then by Lemma 15.46,9 = C n X for some class X of G. Then &,(p) = L b i ( P D ( 3 ) ) = A,(.%?) for i = I, 2 and hence the &, agree on all 2for classes 9with defect group contained in D. By the min-max theorem, it fOllOWS that L b 2 ( e b , ) = &,(eb,) = 1 and thus bl = b2. I To state Brauer’s “second main theorem” we need to introduce “generalized decomposition numbers.” (15.47) LEMMA Let n: E G with o(n)= pe and let C = CG(n:).For x E Irr(G) and cp E IBrCC), there exist unique d!& E R n Q, such that

x w for all p-regular x E C.

=

c

d:,cp(x)

veEIBr(C)

Chapter 15

284

1

Proof Write xc = u@I)with u@E Z and t,b E Irr(C). We have t,bzco = t,b(l)pJI for some linear pJlE Irr(Z(C)). Then I)(xn)= I)(x)p@(n) and therefore

x(x4 =

c

a@P@(4d@,cp(x)

@eIrr(C);qieIBr(C)

and so we take d:, = CILEIrr(C) u@p@(n)d@,. Uniqueness follows from the linear independence of IBr(C). I The algebraic integers d:, are called generalized decomposition numbers. Note that if n = 1, then C = G and d:, = dx,. Also, if g E G is arbitrary, we can take R = g p and x = g p , and thus express x(g) in terms of generalized decomposition numbers and Brauer characters. Note that if b is a block of C = C(n) for a p-element n then bc is defined by Lemma 15.44. (Second Main) Let x ~ I r r ( G )and cp€IBr(C), where C = C,(n) for some p-element n E G. Let x E B E Bl(G) and cp E b E BI(C). Then d:, = 0 if bG # B.

(15.48) THEOREM

Proof Omitted.

I

Theorem 15.48 is extremely powerful and useful. We give a few consequences. Let x E Irr(G) and g E G. Suppose g p is not contained in any defect group for the p-block containing x. Then x ( g ) = 0.

(15.49) COROLLARY

Proof Let n = g p and write g = nx, where x E C(n) is p-regular. Then d:,cp(x). If d:, # 0 then cp E b E Bl(C(n))and x E bc. Let Q E d(b). Then n E O,(C(n)) E Q by Corollary 15.39. Also, Lemma 15.43 yields P E d(bc) with Q E P and thus R E P , a contradiction. Thus all d;, = 0 and the result follows. I x(g) =

1,

Note that Corollary 15.49 generalizes Theorem 8.17 since if x has p-defect zero, then the subgroup 1 is the unique defect group for the p-block containing x. If 9 is a class function of G and B E Bl(G), write QB = ~ x E l r r ( C ) n B[S, x]x SO that 9 = C B s B I ( G ) Q B . (15.50) THEOREM Let 9 be a class function of G and let n E G be a p-element. Suppose ~ ( x x=) 0 for all p-regular x E C(n). Then S,(~CX)= 0 for all such x

and all B E Bl(G). Proof We have 0 = 9(nx) =

C

x E Irr(G)

CSY X I X ( X 4 =

c

cg, xldf,cp(x)

~ ~ l r r ( Gq) ~; 1 B r ( C )

Problems

285

for all p-regular x E C ( ~=) C. The linear independence of IBr(C) yields [S, xld;, = 0 for each cp E IBr(C). If cp E IBr(C) n b with b E Bl(C), bC [s,xld;,,, = 0. Theorem 15.48 yields d;, = 0 for x $ bc and thus Now let d = U { b n IBr(C)lb E Bl(C),bc = B } . Then for each cp E d we [S, ~ ] d ; = , Oand thus have CXeB

xXEIrr(G)

xxcIrr(C)

9B(nx) =

1

[9, xld;,'?(x)

I

= 0.

XEB;(PEJB

(15.51) COROLLARY (Block Orthogonality) Let g , h E G be such that g, and h, are not conjugate in G. Then

c X(S)X(h) 0 for every p-block B. Proof Write 9 xXEIrr(G) X(h)x and let n If C(n)is p-regular, 0 by the second then n (nx), is not conjugate to h, in G and hence orthgonality relation. Thus 0 cxeB x(g)X(h) by Theorem 15.50. I =

x~lrr(G)nB

= g,.

=

=

xE

Q(ZX)

=

= &(g) =

Problems

(15.1) Let D be the decomposition matrix for G. The matrix DTD = Cis the Cartan matrix. (The rows and columns of C are indexed by IBr(G).) For cp, Q E IBr(G), let yips = (1/IGI) cp(x)9(x), where 9 'is the set of p-regular elements of G. Define the matrix = (y,J. Show that r = C - ' .

xxeY

Hint Let X, be the part of the character table corresponding to the p-regular classes and let Y be the Brauer character table. Then X o = DY.

(15.2) (a) Let cp, 9 E IBr(G) lie in different p-blocks. Show that

c

Cp(x)W= 0.

X E Y

(b) Let = 0.

x, $ E Irr(G) lie in different p-blocks. Show that zxEY x(x)=)

(15.3) Let I G I = p"m with p$m and let $ E Irr(G) u IBr(G). Define the class function, E$ by E@(x)= p"$(x) if pxo(x) and E#(x)= 0 if p I o(x). Show that E$ is a generalized character of G. Conclude that det(C) is a power of p , where C is the Cartan matrix as in Problem 15.1. (15.4) (a) Show that {O,lcp E IBr(G)} is a basis for the space of class functions on G which vanish on G - 9'. (b) If cp, CL E IBr(G),show that (I/ I G I 1 Ex. @,(x)p(X) = d,, . (15.5) (a) Show that the product of two Brauer characters is a Brauer character.

Chapter 15

286

(b) If cp, p E IBr(G), define E on G by E(x) = Oq(x)p(x)for p-regular x; E(x) = 0 when plo(x). Show that E is a nonnegative integer linear comIv E IBr(G)}. bination of {av (c) If cp, p€IBr(G), show that OqOpis a nonnegative integer combination of {QV I v E IBr(G)}. (15.6) Let %? be the collection of classes of G which are defect classes for blocks of defect d. Show that I { B E Bl(G)Id@) = d } I I I W I. Hint Let A E Z(F[G]) be the span of the 2 for X E %?. Then the restrictions of the algebra homomorphisms & to A are linearly independent for those B with defect d.

(15.7) Let N = O , ( G ) .Show that if x, 9 ~ I r r ( Glie ) in the same p-block, then xN and I(/N have the same irreducible constituents. If n denotes the number of classes of G contained in N , conclude that I Bl(G)1 2 n.

(15.8) Let P E Syl,(G). Show that the number of p-blocks of G with defect group P is equal to the number of p-regular classes of N(P)contained in C(P). Hint

Use Problems 15.6 and 15.7.

(15.9) Let N be a normal p-complement for G and let 5 be the set of orbits of the action of G on Irr(N). For each BEBI(G), let O(B) denote (9 E Irr(N))[xN, 93 # 0 for some x E B n Irr(G)}. Show that B HO(B)is a bijection of Bl(G) onto 9. (15.10) In the situation and notation of Problem 15.9, show that 6(B) = U S E O(B) sY1p(zG(9))-

Hints If x E Irr(G), 9 E Irr(N), and [zN,93 # 0, then w,(K) = o,(K), where K is any conjugacy class sum of G for a class contained in N . If X is a defect class for B and P E sylp(zG(9)) for some 9 E O(B),show that P fixes one of the classes of N contained in X and conclude that P is contained in a defect group for B.

Appendix Some character tables

In the following tables, each conjugacy class is denoted by the order of its elements. If there are more than one class of elements of a given order, they will be distinguished by subscripts. 1. G = C 4 I G I = 24

=

23 x 3

XI:

xz:

x3:

x4: xs:

1 1

2 3 3

1 -1 0 1 -1

1 1

2 -1 -1

1 1 1 -1 -1 0 0 -1 0 1

Note Class 2, is the class of transpositions.

287

Appendix

288

2. G = SL(2, 3) lGl=24=23 x 3

Class: 1 IC@)I: 24 (Cl(g)J: 1

2 24 1

4 4 6

31 6 4

32 61 6 2 6 6 6 4 4 4

1

1

1

1

1

x1:

1

0 w2 1 1 1 w2 0) x3: 1 1 1 3 - 1 0 0 x 4 : 3 0 -1 -1 x5: 2 -2 x6: 2 -2 0 -w -0' x,: 2 - 2 0 - w 2 - w x2:

1

w

w2 w2 w

0 1

0 1

w w2 w2 w

Irrational Entries w = eZnii3.

3. G = A s Z PSL(2,5) Z SL(2,4) (GI = 60 = 2' x 3 x 5

21:

1

x2:

4

x3:

5 3 3

x4: x5:

+

1 0 1 -1 -1

Irrational Entries a1 = (1 $)/2 + c2 + c3,where E = eZnii5.

=1

1 1 -1 0 0 =1

1 -1 0

1 -1 0

a1

a2

a2

al

+ + c4 and a2 = (1 - *)/2 E

Some character tables

289

4. G = PSL(2, 7) 2 GL(3,2) IGI = 168 = 23 x 3 x 7

1 Class: 168 IC(g)l: ICl(g)(: 1

x3: x4:

xs: x6:

4 4 42

1 2

1 0

7, 7 24

3 3 56

1 1 0 -1 - 1 -1 1 0 8 0 0 -1 1 a 3 -1 1 0 1 0 8 3 -1

x 1 : l

xz:

2 8 21

6 7

Zrrational Entries a = (- 1

+ i&2

=E

72 7 24 1 -1 0 1 a a

+ E’ + c4, where E = eZnil7.

= A6 2 PSL(2, 9) IGI = 360 = 23 x 32 x 5

5. G

Class: 1 IC(g)I: 360 ICl(g)I: 1

XI: x2: x3:

x4:

xs: x6:

x7:

=

2 8 45

4 4 90

40

3, 9 40

1 1 1 1 2 5 1 -1 -1 5 1 -1 1 0 9 1 0 1 10 - 2 0 -1 8 0 0 -1 8 0

1 -1 2 0 1 -1 -1

+

Irrational Entries a, = (1 $)/2 E’ e3, where&= eZails.

1

+ +

3, 9

=

1

51

5 72 1 0 0

-1 0

5, 5 72

1 0 0 -1 0

a1 a2

+ E + c4 and a,

a2 a1 = (1

- $)/2

Appendix

290

6. G = SL(2, 8) IGI = 504 = 23 x 32 x 7

Class: 1 IC(g)I: 504 ICl@)I: 1

6

Irrational Entries . P1 = -(6

= e2ni/7 = e2ni19,

7. G

=

2 8 63

3 9 56

9, 9 56

9, 9 56

93 7, 9 7 56 72

72

7 72

73 7 72

tll = + e6, a2 = + and a3 = + E ~ where , + S8), P2 = -(S2 + S7), and P3 = -(a4 + S5), where E

E’

E~

PSL(2, 11) = 2’ x 3 x 5 x 11

IGI = 660

Class: 1 IC(g)I: 660 ICl(g)(: 1

2 3 6 51 52 12 6 6 5 5 55 110 110 132 132

1 x l : l 1 1 2 1 -1 x2: 10 1 1 x3: 10 -2 x4: 11 -1 -1 -1 xs: 12 0 0 0 x6: 12 0 0 0 x7: 5 1 - 1 1 1 - 1 x8: 5 1

+

1 0 0 1

1 0 0 1

a1

M2

a2

a,

0

0 0

0

+

11, 11 60 1 -1 -1 0 1 1

p

p

112 11 60 1 -1 -1 0 1 1

p

P

Irrational Entries ctl = (- 1 $)/2 = E c4, a2 = (- 1 - 3 ) / 2 = where E = elni/’. P = (- 1 i f i ) / 2 = 6 63 64 d5 d9, where 6 = e 2 n i / 1 1 . E’

+ c3,

+

+ + + +

Some character tables

291

8. G = M I , (GI = 7920 = 24 x 3’ x 5 x 11 Class:

1

2 3 6 4 81 82 5 11, 112 8 8 8 48 18 5 11 6 11 1 165 990 990 990 440 1320 1584 720 720

(C(g)J: 7920

ICl(g)(: x1:

xz: x3: x4:

xs: ~ 6 : x7:

xs: xs: 210:

+

1 1 1 1 0 10 2 2 -1 11 3 -1 c1 10 -2 0 0 B 10 -2 0 0 16 0 0 16 0 0 0 44 4 0 1 -1 45 -3 1 55 - 1 - 1

Irrational Entries a = where E = e2ni/11.

E’,

id.fl

1

0 -1

-

c1

a

0 0 0 -1 1 =

(- 1

1 1 2 1 1 -2 -2 -1 0 1

1 -1 0 1 1 0 0 1 0 -1

+ ifi)/2

1 1 0 -1 1 0 0 -1 0 -1 I

1 - 1

=

E

1 -1 0 -1 -1

P

P

P

P 0 1 0

0

0

1

0

0

+ .c3 + E~ + E’

Bibliographic notes

There exist several excellent bibliographies in group theory and character theory (for instance in the books by Huppert [26], Dornhoff [lS], and Curtis and Reiner [123) and so, instead of giving a comprehensive list of publications, we shall only mention some of the items which are directly relevant to the various chapters of this book. General Some other books on characters and representations are Dornhoff [lS], Feit [16], and Curtis and Reiner [123. Each of these has a point of view somewhat different from the others and from this book. As a reference on group theory we mention Huppert [26] which also has an extensive chapter on characters. Finally, we come to Burnside [lo]. This classic, although somewhat difficult for the modern reader, contains a wealth of material. Chapter 1 Further relevant information on rings and algebras can be found in Curtis and Reiner [123 and Herstein [25]. Chapter 2 Other methods exist for obtaining the basic results about characters such as the orthogonality relations. For instance, instead of using the central idempotents of C [ G ] , Feit [16] and Dornhoff [l5) use a matrix approach which results in additional information, namely the Schur relations which appear here as Problem 2.20. Chapter 3 There is, of course, a large literature in algebraic number theory. A reference for those parts of the subject most relevant to group theory is the appropriate chapter in Curtis and Reiner [12]. The proof of 292

Bibliographic notes

293

Theorem 3.12 given here was discovered (independently) by G. Glauberman and the author. For another proof, see (4.2) of [16]. Whitcomb’s result on isomorphism of integer group rings occurs in Whitcomb [37]. Chapter 4 The Brauer-Fowler proof [8] of Theorem 4.1 l(a) does not depend on characters. They obtain a slightly better bound which is of the same order of magnitude. Chapter 5 There is a great deal more to be said about the relationship between permutation groups and representation theory. For instance, see Chapter V of Wielandt [38] and Sections V.20 and V.21 of Huppert [26]. Chapter 6 Clifford’s results appear in Ref. [ I l l . This paper includes significantly more than Theorem 6.5; it also has part of Theorem 6.11 and bears heavily on Chapter 11. The “going down” Theorem 6.18 and its dual Problem 6.12 appears in Isaacs [28), however Corollary 6.19 goes back to Burnside’s book [lo]. A result on relative M-groups which is more general than Theorem 6.22 occurs in Price [35]. Theorems 6.22 and 6.23 give sufficient conditions for a group to be an M-group which generalize a result of Huppert (Satz V.18.4 of Huppert [26]). The connections between the extendibility of 9 and det(9) are due to Gallagher [19]. A proof of a version of Tate’s Theorem 6.3 1 without characters appears as Satz IV.4.7 of Huppert [26) and Thompson’s proof (including Problem 6.20) is found in Ref. [36]. Chapter 7 A proof of Theorem 7.8 for the case that I P I = 8 that does not depend on “modular characters” has recently been discovered by Glauberman [22]. For a proof of Theorem 7.10 without characters, see Bender [2]. A large fraction of the known applications of character theory to “pure” group theory are either directly or indirectly related to the content of this chapter. We mention as examples Sections 28 and 32 of Feit [16]. Chapter 8 Brauer’s Theorem 8.4occurs in Brauer and Tate [9] and in some of Brauer’s earlier papers. Banaschewski’s Lemma 8.5 appears (in a somewhat more complicated form) in Ref. [13. Dade’s Theorem 8.24 and its consequence Theorem 8.26 appears in more general form in Ref. [13]. Chapter 9 An alternate source for much of this material is Curtis and Reiner [121. Chapter 10 The more standard version of Theorem 10.7is due to Brauer and Witt (Theorem 70.28 of [123 or Yamada’s notes [41]). Theorem 10.12 appears in Goldschmidt and Isaacs [24]. What amounts to a special case of Theorem 10.16 occurs in Burnside [101as Exercise 8 on page 319.That every integer can occur as a Schur index was proved by Brauer [5] using groups

294

Bibliographic notes

similar to those of Theorem 10.16. A great deal of further information can be found in Yamada [41]. Chapter 1I Much of the theory relating projective representations with the properties of a character triple was originated by Clifford [ll]. Our more character theoretic setting of Clifford’s work occurs in Isaacs [31]. Berger’s Theorem 11.33 appears in more general form in Ref. [3]. Chapter 12 Much of this material is the work of Passman and the author and appears in Isaacs and Passman [33, 341, and Isaacs [29]. Also relevant are Isaacs [32], Berger [4], and Garrison [20]. Chapter 13 This chapter is taken almost entirely from Glauberman’s paper [21]. Parts of Theorems 13.6 and 13.14 also appear in Isaacs [27]. A proof of Theorem 13.25 can be found in Isaacs [3 11. Chapter 14 The literature on linear groups is very extensive and we mention just a sample. Dixon’s book [14] is a good reference. For information on solvable and p-solvable linear groups, see Winter [39, 401 and Isaacs [30]. We also mention the lecture notes by Feit and Sibley [18] for results without solvability hypotheses. Chapter 15 Brauer’s papers [6] and [7] are good sources for further reading on blocks and Brauer characters. There is also a chapter on the subject in Curtis and Reiner [12]. The material is treated from a different point of view in Part B of Dornhoff [l5] and in Feit’s notes [17]. Goldschmidt’s notes [23] provide a development of the subject along lines similar to those used here.

References

I . B. Banaschewski, On the character rings of finite groups, Canad. J . Math. 15 (1963), 605-61 2. 2. H. Bender, Finite groups with large subgroups, Illinois J. Math. 18 (1974), 223-228. 3. T. R. Berger, Primitive solvable groups, J . Algebra 33 (1975). 9-21. 4. T. R. Berger, Characters and derived length in groups of odd order, J . Algebra (to be published). 5. R. Brauer, Gruppen hearer Substitutionen 11, Math. Z . 31 (1930), 733-747. 6. R. Brauer, Zur Darstellungstheorie der Gruppen endlicher Ordnung I, 11, Math. Z . 63 (1956) 406-444; 72 (1959), 25-46. 7. R. Brauer, Some applications of the theory of blocks of characters I-V, J . Algebra 1 (1964), 152-167; 1 (1964), 307-334; 3 (1966), 225-255; 17 (1971), 489-521 ; 28 (1974), 433-460. 8. R. Brauer and K. A. Fowler, Groups of even order, Ann. Math. (2) 62 (1955),565-583. 9. R. Brauer and J. Tate, On the characters of finite groups, Ann. Math. (2) 62 (1955), 1-7. 10. W. Burnside, Theory ofgroups offinite order (2nd ed. 191 I), reprinted by Dover, New York, 1955. 11. A. H. Clifford, Representations induced in an invariant subgroup, Ann. Math. (2) 38 (1937), 533-550. 12. C. W.Curtis and I. Reiner, Representation theory offinite groups and associative algebras, Wiley (Interscience), New York, 1962. 13. E. C. Dade, Isomorphisms of Clifford extensions, Ann. Math. (2) 92 (1970), 375-433. 14. J. D. Dixon, The structure of linear groups, Van Nostrand-Reinhold, Princeton, New Jersey, 197 1. 15. L. Dornhoff, Group representation theory, Dekker, New York, 1971. 16. W. Feit, Characters offinite groups, Benjamin, New York, 1967. 17. W. Feit, Representations offinite groups, Mimeographed notes, Yale Univ., 1969. 18. W. Feit and D. A. Sibley, Finite linear groups of relatively small degree, Mimeographed notes, Yale Univ., 1974. 19. P. X. Gallagher, Group characters and normal Hall subgroups, Nagoya Math. J . 21 (1962), 223-230. 295

296

References

20. S. C. Garrison, On groups with a small number of character degrees, Ph.D. Thesis, Univ. of Wisconsin, Madison, 1973. 21. G. Glauberman, Correspondences of characters for relatively prime operator groups, Canad. J. Math. 20 (1968), 1465-1488. 22. G. Glauberman, On groups with a quaternion Sylow 2-subgroup, Illinois J. Math. 18 (1974), 60-65. 23. D. M. Goldschmidt (to be published). 24. D. M. Goldschmidt and I. M. Isaacs, Schur indices in finite groups, J. Algebra 33 (1975), 191-199. 25. I. N. Herstein, Noncommutative rings, Carus monograph 15, Math. Assoc. Amer., 1968. 26. B. Huppert, Endliche Gruppen Z, Springer-Verlag, Berlin, 1967. 27. I. M. Isaacs, Extensions of certain linear groups, J. Algebra 4 (1966), 3-12. 28. I. M. Isaacs, Fixed points and characters . . . , Canad. J . Math. 20 (1968), 1315-1320. 29. I. M. Isaacs, Groups having at most three irreducible character degrees, Proc. Amer. Math. SOC.21 (1969), 185-188. 30. I. M. Isaacs, Complexp-solvable linear groups, J . Algebra 24 (1973), 513-530. 31. I. M. Isaacs, Characters of solvable and symplectic groups, Amer. J . Marh. 95 (1973), 594-635. 32. I. M. Isaacs, Character degrees and derived length of a solvable group, Canad. J. Math. 27 (1975), 146-151. 33. I. M. Isaacs and D. S. Passman, A characterization of groups in terms of the degrees of their characters I, 11, Pacific J. Math. 15 (1965), 877-903; 24 (1968), 467-510. 34. I. M. Isaacs and D. S. Passman, Finite groups with small character degrees and large prime divisors 11, Pacific J. Marh. 29 (1969), 311-324. 35. D. Price, A generalization of M-groups, Ph.D. Thesis, Univ. of Chicago, 1971. 36. J. G. Thompson, Normal p-complements and irreducible characters, J. Algebra 14 (1970), 129- 134. 37. A. Whitcomb, The group ring problem, Ph.D. Thesis, Univ. of Chicago, 1971. 38. H. Wielandt, Finite permutarion groups, Academic Press, New York, 1964. 39. D. L. Winter,p-solvable linear groups of finite order, Trans. Amer. Math. SOC.157 (1971), 155-160. 40. D. L. Winter, On the structure of certain p-solvable linear groups 11, J. Algebra 33 (1979, 170-1 90. 41. T. Yamada, The Schur subgroup of the Brauer group, Springer-Verlag, Berlin, 1974.

Index

A

Abelian group, 16, 30, 182-183 extension of characters, 63 Abelian subgroup and character degree, 30,84, 190, 209-212, 2 16-2 17 of linear group, 240,249 Absolutely irreducible representation, 145146 character of, 149-156 Action of group on classes and characters, 93-94, 23&231 number of orbits, 68 p-group on q-group, 97 Algebra, 1 Algebra homomorphism, 3 Algebraic integer, 3340, 44,135, 265 Alperin, J., 52 Antisymmetric part of tensor square, 50, 56 Artin, E., 72 Augmentation map, 43 Automorphism of field, see Field automorphism, Galois group, Galois conjugate characters

B Banashewski, B., 128 Berger, T., 191, 206

Blichfeldt, H. F., 240, 247 Block, 271-286 defect of, 280 defect group of, 279 number of with maximal defect, 286 numbers of characters in, 276 Block orthogonality, 273,285 Bottom constituent, 146-147 Brauer character and character over F, 264 definition, 263 linear independence, 265 from ordinary character, 266-267 Brauer-Fowler theorem, 54-55 Brauer graph, 212, 275 Brauer homomorphism, 278-279,282 Brauer, R., 46,49, 122 characterization of characters, 127 first main theorem, 282 second main theorem, 284 theorem on actions on classes and characters, 93 theorem on groups of order paqbr, 274 theorem on induced characters, 127 theorem on splitting fields, 161 theorem yielding proper subgroups, 70 Brauer-Speiser theorem, 171 Brauer-Suzuki theorem, on coherence, 112 giving normal x-complement, 137 on quaternion Sylow subgroups, 102 297

298

Index

Brauer-Suzuki-Wall theorem, 76 Brauer-Witt theorem, 162 Broline, D., 208, 21 1 Burnside-Brauer theorem, 49 Burnside, W., 36,40, 46,213 pa$ theorem, 37 transfer theorem, 138

C

Canonical extension, 220 Cartan matrix, 285 Center of character, 26-28, 75 of group, 27,29,63 of group algebra, 9, 15,4546, 277, see also Homomorphism, Multiplication constants Central extension, 179-182, 384-386, 195 Centralizer ring of module, 3 4 8 of representation, 26, 145, 157-158 Character, 14, see also Irreducible character afforded over subfield of C, 22,29-30,58, 74, see also Schur index and field automorphism, see Galois group Character correspondence, see Correspondence of characters Character degree, see Degree of character, Set of irreducible character degrees Character table, 17, 21,45, 48, 287-291 construction for A,, 6 4 6 5 of factor group, 24 fields generated by rows and columns, 9697 information from, 22-25, 27, 36,45, 134-136, 141-142 and isomorphism of groups, 24 row sum, 75 Character triple, 186-189, 195-197, 237 Character value, 20,3544, 46, 122 constant on G - { I } , 44 equals zero, 28-29,36,40,133,246,284 equals zero off subgroup, 28, 31,95, see also Vanishing-off subgroup exceptional character, 113, 123 rational, 31 real, 31 symmetric groups, 17-18,35

Characterization of characters, 127 Chief factor, 85, 87, 96 Class function, 16, 126 of direct product, 59 of form x("), 50, 60 induced, 62 Clifford, A. H., 79-80 Coboundary, 178 Cocycle, 178 Coherence, 107-125 Collineation group, 39 Commutator, 45 Commutator subgroup, 25,63,75,96, 204-205 Completely reducible module, 4 7 , 12, 80, 146 Completely reducible representation, 1 I , 154, 157-158 Complex linear group, 240-261, see also Faithful character p-solvable, 245, 256,260-261 solvable, 243-244, 260-261 Composition series, 146 Conjugacy class in factor group, 24 number of, 16,46 number of p-regular, 268 prime power size, 37 in simple group, 37 in symmetric group, 17 Conjugacy class sum in C[C], see Center of group algebra, Homomorphism, Multiplication constants in Z[G], 4 1 4 4 Conjugate class function, 78-79 module and representation, 79-80 Constituent of character, 17 of representation, 146-147 Control of transfer, 92 Correspondence of characters and coprime action, 219-239 and inertia group, 82 and products, 84-85 Cyclic center and faithful character, 29, 32 Cyclic subgroups and induced characters, 72,76 Cyclotomic polynomial, 72

Index

299

D Dade, E. C., 97, 138, 142 Decomposition matrix, 267, 285 Decomposition number, 267 Defect class for block, 278 Defect group of a block, 279 of a class, 278 Defect of a block, 280 Defect zero character, 134, 143, 276, 284 Degree of character, 14, 37-38, see also Set of irreducible character degrees and abelian subgroup, 30, 84, 190 and center, 28, 31, 38,44 and derived series, 67 in group of order 27,25 in M-group, 67 in nilpotent group, 28 in p-block, 274,280-28 1 prime power, 39,44,274 in Z3, 16 whenp2$lGI, 122 Degree of projective representation, 174 Degree of representation, 9 Derived series, 67, 202, 206 Determinant of character, 29, 59, 75, 88 and extendibility, 88-90, 132-133 Determinantal order, 88-90, 133, 198 Difference of characters, 50,60,99, see also Generalized character Dihedral group, 30, 53, 105 Direct product, 59-60, 172 Dirichlet, P. G. L., 169 Divisible abelian group, 182 Division algebra, 4, 149 maximal subfield of, 170 Dornhoff, L., 96 Double centralizer theorem, 8 Double transitivity, 69, 76-77, 111

E Elementary group, 127 End( V ) , 1 Equivalent factor sets and projective representations, 178 Even order group, 5 4 5 5 Exceptional character, 107, 113, 122-125

Exponent of group, 151, 161, 167 Extendible character, 63, 8486, 88-91, 97-98, 132-133, 140, 178, 186, 190-191

F F-triple, 162-163 Factor group centralizer in, 26 characters of, 24, 51, 76, 8 4 8 5 representation of, 14 Factor set, 174, 176-180, 183, 194195 Faithful character, 28-30, 32, 77, 83, 123, see also Complex linear group of direct product, 60 of p-power degree, 39,44 Fein, B., 170-172 Feit-Sibley theorem, 120 Feit-Thompson theorem on abelian T.I. subgroup, 246 on odd order groups, 1 11,231 Feit-Thompson-Blichfeldt theorem, 247 Feit, W., 52, 108, 123, 125 Field automorphism, 29, 151, see also Galois group, Galois conjugate characters Field extension and representations, 144159 and Schur index, 161, 170 Field generated by character values, 151 and corresponding representation, 150, 153, 159 and order of p-group, 254 Field restriction and modules, 152-1 54, 156-157 Fitting subgroup, 208-209,211,217 Focal subgroup theorem, 142 Fong, P., 168 Frattini subgroup, 60, 173 Frobenius complement, 99, 107, 121 Frobenius group, 99-101, 121,200,237 representations of, 94,269-270 Frobenius kernel, 100-101, 111, 114 Frobenius reciprocity, 62 Frobenius-Schur theorem, 50-52, 58 Frobenius theorem on existence of kernel, 100-101 on linear groups, 251 Fully ramified character, 95-96, 196,237, 239

300

Index G

Gallagher, P. X.,85, 132, 191, 196 Galois conjugate characters, 152, 154, 161-162, 173 Galois group, 31, 36,46,9&91, 165,221, see also Field automorphism Garrison, S., 59,206,208-209 General linear group, 13 Generalized character, 102, 107, 141, 143, see also Difference of characters over a ring, 126, 135 Generalized decomposition numbers, 283-284 Generalized orthogonality relations, 19 Generalized quaternion group, 53, 102 GL(n, F), 13, 174 Glauberman, G., 41, 219 Glauberman map, 219,229 Glauberman’s lemma, 223 Going down theorem, 85 Goldschmidt-Isaacs theorem, 167 Group algebra, 2 Group ring over Z, 41,43

H Hall-Higman lemma, 256 Hall subgroup, I 11,240 character degrees of, 206 and extendibility of characters, 90, 133 generalized character of, 143 and M-group, 96 normal complement for, 136-137 Height of a character, 281 Higman, D. G., 12 Higman, G., 136 Homogeneous character, 83 Homogeneous part of module, 6 Homomorphism of algebras, 3 of center of group algebra, 35-36, 59, 271-272,277 of modules, 3 of tensor product, 61

Idempotents, 16, 19,274-277 Imprimitivity decomposition, 6 5 6 6 Induced block, 282 Induced character, 63-77 center of, 75 and cyclic subgroups, 72,76 explicit computation, 64 from inertia group, 82 kernel of, 67 module interpretation, 65 and right ideal of group algebra, 75 Induced class function, 62, 73-74 Induced module, 66.74 Induction and restriction, 62 Inertia group, 82, 95 Inner product, 2&21 Integral group basis, 41-44 Intertwining matrix, I 1 Invariant character, 84-85,91, 138-139, 175, 178,233-236, see also Extendible character, Character triple Involution, centralizer in simple group, 54, 105 as fixed point free automorphism, 114 number of, 51-54 WG),15 Irreducible Brauer character, 264 number of, 268 Irreducible character, 15 over arbitrary field, 149-156 number of, 16,232 Irreducible constituent, see Constituent Irreducible linear group, 240 Irreducible module, 4 Irreducible projective representation, 177 Irreducible representation, 1 1 of cyclic group, 159 of group, 13 structure over arbitrary field, 154 Isometry, 107-110, 113, 115, 121 Isomorphism of character triples, 187-189, 192, 196 Isomorphism of integer group rings, 41 Ito, N., 84, 190,216,244 J

I IBr(G), 264 Ideal, 2, 75, 127, 135, 224

Jacobson radical, 11-12, 97, 157 Jordan-Holder theorem, 146 Jordan’s theorem, 249

Index

301

K Kernel of character, 23-24, 75, 77,81,208-209, 21 1 of induced character, 67 Krull-Schmidt theorem, 156

L Lifting a projective representation, 181-182 Linear character, 14 and abelian group, 16, 30 action of group of, 95 and commutator subgroup, 25 extendibility, 89 Linear group, see Complex linear group, General linear group Linear isometry, see Isometry Local integers, 265

M M-group, 67,86-87,95-96 solvability and derived length, 67 sufficient condition for, 83, 87 McKay, J., 232 Mackey, G. W., 74 Maschke’s theorem, 4 Mathieu group M , , , 70 Maximal subgroup, 75, 191 Min-max theorem, 278 Module, 3 and corresponding representation, 10 finitely generated over Z,34 and normal subgroup, 79-80 Monomial character, 67,74, 86,96 Multiplication constants, 2 of center of group algebra, 15,45

N

Nakayama’s lemma, 265 Nilpotent group, 27,83, 111, 171, 212 character degree in, 28 Nilpotent ideal, 11 Norm map, 165

Normal matrix, 57, 249 Normal subgroup, 78-98 from character table, 23 in linear group, 243-245, 249,260 prime index, 86 restriction of character to, 79 Normal r-complement, 137-138

0 Odd order group, 46,231 Order of character, see Determinantal order Orthogonality relations, 18-22, 273, 285 Orthonormality of Irr(G), 21 Osima idempotent, 277 Osima, M., 275

P p-block, see Block p-defect group, see Defect group p-defect zero, 134, 143,276,284 p-elementary group, 127, 131, 138,140 p-group, 29,60,91,96-98, 168, 218 p-quasi-elementary group, 129, 131 p-rational character, 90, 98, 191, 24&241, 254,260 p-regular element, 263 p-solvable group, 244245, 256,260,280 Partitioned group, 30 Passman, D. S., 77,218 Perfect group, 46 Permutation character, 68, 93, 101 necessary conditions for, 69, 75 ring of, 76, 128 Permutation group, 68,7677, 101, 111 Permutation isomorphism, 230 Permutation module, 68 PGL(n, F), 174 Primitive character, 66, 83, 191 Primitive linear group, 257 Primitive module, 65 Principal block, 274 Principal character, 14 Principal indecomposable character, see Projective character Product of characters, 30,4749, 59-61, 8685,202,285-286

302

Index

Projective character, 273, 285 Projective general linear group, 174 Projective lifting property, 181-182, 193 Projective representation, 174-197

Q Quasi-elementary group, 129-130, 142 Quasi-primitive character, 83, 96, 191 Quaternion algebra, 145 Quaternion group, 30, 145, 159, 168, 171, see also Generalized quaternion group Quotient group, see Factor group

R R-generalized character, 126, 135 Radical, see Jacobson radical Ramification, 174 Rank of permutation group, 69 Rational character, 31, 72-73, 76, see also p-rational character Rational integer, 34 Real character, 46, 50, 54-58,96 Real element, 3 1, 54, 96 Real representation, 30, 56-58 Regular character, 18 Regular module, 3 Regular representation, 18, 147 Relative M-character, 86, 96 Relative M-group, 86, 96 Representation, 9, see also Irreducible representation and corresponding module, 10 of group, 13 over R, 30, 56-58 Representation group, see Schur representation group Representative set of modules, 6 Restriction, 26 connection with induction, 62 to normal subgroup, 79 Rieffel, M., 8 Roots of unity, 20, 44,46, 90-91, 220-221, 263 and splitting field, 161, 165-168 Roquette, P., 168

S Schur, I., theorem on faithful p-rational character, 254 theorem on lifting projectives, 181, 184 Schur index, 16C173 and abelian Sylow subgroup, 165, 173 and direct products, 172 and exponent of group, 161, 167-168 and field extension, 161, 170 Schur multiplier, 181-186, 197 Schur relations, 32 Schur representation group, 182, 185-186, 197 Schur's lemma, 4 Schur-Zassenhaus theorem, 99,223 Schwarz inequality, 53 Second orthogonality relation, 21 Section of group, 162 Seitz, G., 206 Semidihedral group, 53 Semidirect product, 219 Semiprimitive character, 171 Semisimple algebra, 4, 7-9, 12 Set of irreducible character degrees, 81, 198-218,243 and abelian subgroup, 209-212,216217 cardinality three, 206 cardinality two, 201-205, 216 and derived length, 202,206 divisible by p, 199 and Fitting length, 209 maximum, 74,98,203, 209-214,216-218 minimum nonlinear, 75 and p-structure, 213-217 power of p. 8 I , 84 prime top, 215-216 Sibley, D., 116, 120, 125 Siegel, C. L., 46 Similar projective representations, 177 Similar representations, 9, 144, 147, 157 characters of, 14, 17 Simple algebra, 8 Simple group and centralizer of involution, 54 characters of, 23, 44 conjugacy classes of, 37 order pa&, 274 order 360. 70

Index

303

SL(2, 4). 76 Sock, 77 Solomon, L., 75, 129, 168 Solvable group, 24,95, 142, 199, 209,212, 2 16-2 17, see also M-group character correspondence in, 231 complex linear, 243-244 and extendible characters, 88, 90, 98 quasi-primitive characters of, 96, 191, 194 sufficient conditions for, 37, 67, 206, 243 Special element, 195-196 Splitting field, 146, 148-149, 154, 159, 161 Standard map, 182 Strong isomorphism of character triples, 196 Subnormal subgroup, 190 Supersolvable group, 87,206 Suzuki group, 125 Suzuki, M., 102, 112, 124 Sylow subgroup, 77, 114, 116, 122, 138, 142 all abelian, 87, 217 and character degrees, 2 13-2 15 and extendible character, 89, 190-191 of Galois group, 98, 167-168 in linear group, 39, 44,243-245,247,253, 260 and Schur multiplier, 186 strongly self-centralizing, 61, 76 Sylow tower, 216 Symmetric group, 17-18, 35, 69 Symmetric part of tensor square, 50, 56

Top constituent, 146-147 Trace of matrix, 14 Transfer, 63, 92, 136, 138, 142 Transitivity of induction, 73 Transversal, 62 Trivial intersection set, see T.I. set Twisted group algebra, 176-177, 194-195

U Unitary matrix, 57, 249-250

V Values of characters, see Character value Vandermonde determinant, 49 Vanishing-off subgroup, see also character value, 200,208

W Weak block orthogonality, 273 Wedderburn, J. H. M. theorem on finite division rings, 149 theorem on semisimple algebras, 7 Whitcomb, A., 41 Wielandt, H., 121 Winter, D. L., 243,257

T

A 6 8 7

C D E F 6 H 1 J

B 9 O 1 2 3 4 5

T.I. set, 101-102, 107, 11 1,246 T.I.F.N. subgroup, 11 1-125 Tdketa, K., 67 Tamely imbedded subgroup, 11 1 Tate, J., 92, 128 Taussky, O., 53 Tensor power, 60 Tensor product, 4748,50, 55,61, 156 and induced module, 66 Thompson, J. G., 46, 52, 92,97-98, 105, 111, 199

Y Yamada, T., 171

2

Zassenhaus group, 111, 123, 125 Zeros of characters, see Character value, Vanishing-off subgroup

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