16
Topicsin FiniteGroups TERENCEM. GAGEN
London Mathematical Society Lecture Note Series.
16
Topics in Finite Groups Terence M. Gagen
CAMBRIDGE UNIVERSITY PRESS CAMBRIDGE LONDON
NEW YORK
MELBOURNE
Published by the Syndics of the Cambridge Cniversity Press The Pitt Buildine:, Tn.:mpington Street. Cambridge CB2 lRP Bentley HOl.:se,
~ ~ ~
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32 East' -th Street.
~ew
York, N. Y. 10022, USA
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Cambridge University Press 1976
Uhrary of Congress Catalogue Card Number: ISB~·
: 521 21002 X
Pr:::ted in Great Britain J.t
the University Printing House, Cambridge
E'..:.an Phillips, University Printer)
75-17116
Contents
page Introduction
vii
Notations
viii
1
Elementary results 1.
Baer's Theorem
3
2.
A theorem of Blackburn
5
3.
A theorem of Bender
7
4.
The Transitivity Theorem
10
5.
The Uniqueness Theorem
12
6.
The case
In(F(H)) I =
I
18
7.
The proof of the Uniqueness Theorem 5. 1
20
8.
30
9.
The Burnside paqb-Theorem, p, q odd ab Matsuyama's proof of the p q -Theorem, p=2
31
10.
A generalization of the Fitting subgroup
34
11.
Groups with abelian Sylow 2-subgroups
38
12.
Preliminary lemmas
40
13.
Properties of A *- groups
47
14.
Proof of the Theorem A, Part I
53
15.
Proof of the Theorem A, Part IT
67
Appendix: p-constraint and p-stability
80
References
85
Introduction
The following material is selected from a course of lectures given at the University of Florida in Gainesville, Florida during 1971/72.
The
reader is expected to have read both Gorensteins' Finite Groups and much of Huppert's Endliche Gruppen 1.
In particular he must be familiar with
the concepts of p-constraint and p- stability in order to begin, although there is a short discussion of these concepts in an appendix here. The topics covered are such that I feel rather diffident about publishing these notes at all.
The title should perhaps be changed to
something like 'Lectures on some results of Bender on finite groups'. No less than three of his major results are studied here and of COurse the classification of A*-groups depends on his 'strongly embedded subgroup' theorem - which is not studied here at alL
I feel that the theorems
and techniques of the papers 'On the uniqueness theorem' and 'On groups with abelian Sylow 2-subgroups' are too important for finite groups and much too original to remain, as at present, accessible only to a very few specialists. I think that I understand the motivation for the abbreviation of the published versions of these two results.
However, though it is
clear that a proof becomes considerably more readable when a two or three page induction can be replaced by the words 'By induction we have', these details must sometime be filled in.
And unfortunately, I think
Dr. Bender has sometimes disguised the deepest and most elegant arguments by this very brevity.
I hope that these notes will serve to make
more of the group theoretical public aware of these incredibly rich results. I must thank here the audience at the University of Florida -
Mark Hale, Karl Keppler, Ray Shepherd and Ernie Shult. tion of Ernie Shult in particular cannot be minimized.
The contribu-
Without him, we
would all have floWldered very soon. December, 1973
Terry Gagen Sydney, Australia vii
Notations
The notation used here is more or less standard.
The reader
should refer to [12] or [15] when in doubt. S(,;:Jl(P)
The set of all self centralizing normal subgroups of P.
S(';:Jl(p)
The set of all self centralizing normal subgroups of a
Sylow p-subgroup. The set of all A-invariant 1T-subgroups of G where
1r
is a
set of primes. The maximal elements of IiI (A, rr). G The number of generators of an elementary abelian subgroup of P of maximal order (amongst all elementary abelian
subgroups of P). b
AB
(A
Gp
A Sylow p-subgroup of G.
°n(G)
The maximal normal 1T-subgroup of C,
oa,n (G)
on(G mod
On (G)
n The smallest normal subgroup of G such that G/O (G) is
a
:b
E
Bl.
17
a set of primes.
0 (G»).
a
17-grOUp.
F(G)
The Fitting subgroup of G.
4'(G)
The Frattini subgroup of G. The following two results are absolutely basic.
1.
The Three Subgroups Lemma If A, B,
[e, A, B] 2.
e ~
~G,
N
e]
~
N, [B,
e, A]
N, then
N.
If P is a p-group of class at most 2, then for all n
all x, y EP,
viii
~
E
Z and for
Elementary results
Definition. a chain P Here [a,
A group of automorphisms of a group P stabilizes
= P 0= ::J P ::J ... ::J P = 1 if 1= = n x] = x-ax for x € P, a € A.
Theorem O. 1. =
::J ... ::J P
1=
Suppose a
€
=
n
1
= 0,
= 1,
... , n-1.
then A is a 1T-grOUp. --- ---
A is a 1i'-automorphism of P.
induction we may assume that [a, PI] where y
1 =
If a group of automorphisms A of a 1T-group P
stabilizes a chain P::J P
Proof.
[A, p,] c P, + l' i
= 1.
Then if x
€
Clearly, by
P, x
a
= xy
P , since [a, P] cP. 1 1 a 2 =2 alal lal It follows that x = xy , ... , x = xy = x. €
Since y is a 1T-element while a is a 1T'-element, we have that y
=1
and [a, P]
= 1.
Thus a
Corollary 0.2. P such that rp, A, A]
Proof.
C p /Q(A)
and A is a 1T-group. I!
!.! A is a 1T'-group of automorphisms = 1, then rp, A] = 1 and so A = 1.
of a 1T-group
A stabilizes the chain P ~ [P, A] ~ [P, A, A]
Lemma O. 3. group P.
=1
= 1.
Let A be a 1T'-grOUP of automorphisms of a
Let Q be an A-invariant normal subgroup of P.
/I
1T-
Then
= Cp(A)Q/Q.
Proof.
Clearly Cp(A)Q/Q
~
C p IQ(A).
Suppose now that xQ is a coset of Q in P which is fixed by A. Let QA act as a group of permutations on xQ where A acts in the obvious way and Q acts by multiplication on the right. transitively on xQ since Q does. Then IAII
=
IQAI/lxQI
=
Then QA acts
Let Al be the stabilizer of a point.
IAI.
By the Schur Zassenhaus-Feit-Thompson Theorem A
I
is con-
1
Thus there exists y E: xQ such that y E: Cp(A). 1/
jugate to A.
Remark.
Note that in every application of O. 3, 0.4 in these
notes, it will be known a priori that at least one of A or P is solvable. Hence the Feit-Thompson Theorem will not be required in the applications of
:.3, 0.4 here. Corollary 0.4.
phis_ills of P. Proof.
Let P be a'll-group, A
Then P
= rp,
~
'II'-group of automor-
A]Cp(A).
[P , A] ~ P and is A- invariant.
Also A centralizes
P ·P. A]. 1/
!!
Corollary O. 5.
A is a
'11' -
group of automorphiSillS of an
;j.belian 17-group P, then P = C p (A) EB rp, A]. Proof. Let
b~
(J
= 0, defined by e =
We show that Cp(A) n [P, A]
be the endomorphism of P
writing P additively.
rh IAI
L: a. aE:A
Clearly
e2 = e. Since pe n ker e = 0, P = pe EB ker e. Now if x E: Cp(A), then xe = a~xa = x and so Cp(A) ~ pe. Finally if [x, a] E: rp, A], then (-x + xa)e = -xe + xe = 0 and so x. a] E: ker e. Thus Cp(A) n rp, A] = O. The result follows from 0.4. 1/
= eb = e
for all b E: A.
Thus
r±r
Lemma 0.6.
(Thompson)
Any p-group P contains a character-
Istic subgroup C such that (a) cl C :::: 2 and C/Z(C) is elementary; (b) rp,
c]
(c) Cp(C)
~
Z(C);
= Z(C);
(d) Any automorphism a
*1
of order prime to p acts non-
tr ivially on C. Proof.
We first show that (c) and C char P together ensure (d).
For suppose that [a, C] Then
[a, C, P] = 1.
2
= 1, where
a is our given p'-automorphism.
Also rC, P, a] <.::. rC, a]
= 1.
Thus rp, a, C] rp, a]
=1
<,; Cp(C) = Z(C)
by (c).
Hence
= 1.
rp, a, a]
By 0.2, rp, a]
= 1.
To show the existence of C we proceed as follows. subgroup A
E S~::J1(P)
(a), (b), (c) hold.
is characteristic in P, then take A
= C.
Clearly
Hence we may suppose that no maximal abelian sub-
group of P is characteristic. abelian subgroup of P. C;D
First if any
Let D be a maximal characteristic
Clearly Cp(D):J D and Cp(D) char P.
= -1\ (Z(P;D))
Let
r Cp(D);D.
Clearly C:J D, and C is a characteristic subgroup of P.
Since
D ;; Z (C) and Z (C) is an abelian characteristic subgroup of P, maximality of D ensures that D
= Z(C).
Clearly C is a characteristic
subgroup of P. (a) Since C;D is elementary abelian, first C /Z(C) is elementary and then rC, c] ;; D ;; Z(C).
Hence cl C ~ 2.
(b) Since C;D;; Z(P;D), [P, C} (c) Suppose that Q = Cp(C) ~ C. Q;D Q
#.
C
P;D and Q;D n C;D
= 1.
~
D = Z(C).
Since· Q n C = Z(C) = D we have
Of course Q
~
Cp(C) ;; Cp(D).
If
D, then Q;D intersects 51 (Z(P ;D)) n Cp(D);D non-trivially. 1
This contradiction completes the proof. 1/
1.
BAER'S THEOREM The Theorem 1. 1 is required in the study of p-stable groups and
a
proof
due to Suzuki is given in [12]. Of course, it follows immedi-
ately from the result of Baer [15] p. 298, this proof being given below as the first proof.
Two other proofs of this result are given, both of which
are interesting and brief.
3
Theorem 1. 1
(R. Baer).
elements in a finite group G.
Let K be a conjugacy class of p-
If (x, y) is a p-group for all x, y
E
K,
then K;; 0p (G). First Proof.
Since (x, y) is a p- group for all x, y
= x-1xg
all g E G, [x, g] g (x, x ). Hence
[g, x, x, .,. , x]
E
K, for
and x are elements of the finite p-group
=1
after a while.
Thus x is a right Engel element and by Theorem Ill, 6. 15 [15], x
E
F(G).
Hence K;; 0p(G). !I
Second Proof
(J. H. WaIters).
example to the Theorem.. Let M
,
M
Let G be a minimal counter
, ••• ,
M
t subgroups of G containing a fixed element x E K. Clearly 0 If t
= 1,
(G)
p
=1
1
2
be all the maximal
since G is a minimal counter example.
then for all y
E
K, (x, y) ~ M
1
,
since (x, y) is a
p-subgroup of G and so is certainly a proper subgroup of G containing x.
Thus K cM. =
by induction.
Let L
1
= M.1 n M.J
Then L
= M1
C
C
G and K
Since L:! G, we have a contradiction.
Thus we have t D
= (K).
>
1.
such that I D I
P
C
=
° (L) P
Among all i, j with i '" j choose is maximal.
Let P be a Sylow p-subgroup
of D containing x. We show that there is no loss of generality in assuming that P is a Sylow p-subgroup of both M. and M .. 1
Sylow p-subgroup of M
J
For suppose that pep., a 1
Then NG(P) n Pi=:> P.
Let M be a maximal k r subgroup of G containing NG{P) C G. Then M n M ~ NG(P) n Pi=:> P k i and so k = i by the choice of i, j. Also NG(P) ~ M and so P is a i Sylow p-subgroup of M Now choose n E NG(P) n Pi - P. Then clearly
r
n f M. and so M~ '" M .. Otherwise M.
P J=
J =
contains x and so Mj quired intersection.
= MZ
J
for some Z.
=P
Take M n M as our rej Z Note that P is a Sylow p-subgroup of both M. and
We derive a contradiction easily now.
4
=
J
By induction
K n M. c 0
(M.) c P CM ..
1= P 1 = = J Hence K n M. C K n M .. 1
=
J
Similarly K n M. C K n M.. J= 1 Thus M. = NG(K n M'» = NG(K n M'» = M., a final contraJ J 1 1 diction. /I Third Proof
(J. Alperin and R. Lyons).
a minimal counter example.
[1] Again let G be
Let P be a Sylow p-subgroup of G. If
(K) is a p-subgroup, then K;; 0 p (G) since K ~ G.
Thus (K) is not a
p-subgroup and so Kt. P.
Let y E K - P and let Q be a Sylow p-sub-
group of G containing y.
Then of course K n P t- K n Q.
Among all Sylow p-subgroups P, Q of G such that K n P t-K nQ X choose P, Q so that IK n P n Q I is maximal. Since p = Q for some XEG, (Knp)x=KnQ and so Knpt.Q, KnQt.p. Let D=(KnpnQ). Suppose D=P cp c ... cp =P where [P.+ :P.]=P. o 1 n 1 1 1 Clearly K n P D.
1
Suppose i is the smallest positive integer such that K n P.t K n D. Let x E (K n P.)-D. Since P. 1
1-
similarly such that y normalizes D. Then (x, y) is a p-group by hypothesis and so (x, y, D) is a p-group also.
Let R be a Sylow p-subgroup of G containing (x, y)D.
Then (x, D) ~ R n P implies that R = P while (y, D) ~ R n Q implies that R = Q. This is a contradiction. /I 2. A THEOREM OF BLACKBURN This theorem duplicates some of the results of [12] - but its proof is so beautiful that it should be included here.
The following lemma is of
crucial importance for many of the results to come. Lemma 2. 1 (J. Thompson). p-group G.
Let a be a p'-automorphism of a
Suppose that X is a p-group of automorphisms of G and
[a, X] = [a, CG(X)] = 1. Proof.
Then a = 1.
Let N;; G be X-invariant such that [a, N] t- 1, but
[a, K] = 1 for all X-invariant proper subgroups K of N.
Then apply the 5
Three Subgroups Lemma, We have
x, a] = 1 because [N, X]
rN,
N
C
and is X invariant.
[X, a, N] = 1. Thus [a, N, XJ ~,
a]
= 1.
= 1,
[N, a]:; CG(X), [N, a, aJ
Lemma 2.2.
By 0.2,
Let a be a lI'-automorphism of a lI-grOUp G and
suppose X <1
= [a,
CG(X)J
= 1.
Then a
= 1.
= G and choose i such that n Let N = NG(X i ). Since Xi +1 ~ N,
Let X
*
= 1.
a, Xi + 1] 1, [a, Xi] a, N] 1. But
*
= 1.
This completes the proof. 1/
=1 N] = 1.
[X., N, a] 1
[X., a, 1
= 1,
Hence [N, a, Xi]
[N, a]
CG(X ) :; CG(X). i Lemma 0.2 implies that [N, a] = 1. 1/ C
Thus [N, a, a]
= 1.
Lemma 2. 3 (N. Blackburn) [6]. ~p-group
P.
~
a p'-automorphism of -Let a be -E be an abelian subgroup of P, maximal of exponent
pn, where n;:; 2 if P is a non-abelian 2-group and no restriction is placed on n otherwise. If [a, E] Proof. If C
Thus E
~
= 1,
then a
Let P be a minimal counter example.
= Cp(E)
C
P, then [a, C]
=1
by induction.
By 2.2, a
Also, since E *- P trivially, (P)E
= [a,
If C((P))
Cp((P))]
=1
C
Choose x
E
C
P.
Thus a centralizes
P, again we have
since E
Thus P has class at most 2 and (P)
C
C((P)).
~
Z(P),
By 2. 2, a
P and consider [x, ~ lpn,
n n n n n n x, x - l]P (p -1) /2 . [x, a ]p -_ (x - 1x alP -_ x - p (x alP [a 6
= 1.
Z(P).
(p) by induction. ra, (P)]
= 1.
=1
again.
a
If P is abelian then of course [x ,x
-1
= 1.
]
On the other hand,
if P is non-abelian, then
n n n-1 n a -l]p (p -1)/2 _x [a, x -p]p (p -1)/2 [x,x since if P
= 2,
But x- P
n 2: 2.
E
(P) ~ Z(P) for all x E P.
Thus in
every case we have
and ~
=1
Since [a, (P)]
By the maximality of E, [x, a] and [P, a, a]
= 1.
2
(P), we see that [x, alP
=1
Q
(P) then a
1--
E
P.
and so a
Let P be a p-group.
of P which centralizes Q
n E
E, for all x
By O. 2, [P , a]
Theorem 2.4. 2-group. If [a,
E
n
=1
Thus
rp,
= 1.
a] ;; E
= 1. /I
If a is a p'-automorphism
unless P is a non-abelian ---
(P)] = 1, then a = 1 without restriction. /I
3. A THEOREM OF BENDER Theorem 3. 1 [2].
Let G be a p-constrained group. If p
assume that the Sylow 2-subgroups of G have class
~
2.
=2
Let E be an
abelian p-subgroup of G which contains every p-element of its centralizer. Then every E-invariant p' -subgroup H of G lies in 0 ,(G). -
--
P
Remark LIfE is a self centralizing normal subgroup of a Sylow p-subgroup of G, then E contains every p-element of its centralizer in G.
For let E
E
S83I(P) where P is a Sylow p-subgroup of G.
Suppose that D;; E is a Sylow p-subgroup of CG(E).
Consider NG(E).
Suppose that Q is a Sylow p-subgroup of NG(E) containing D.
Since
P ;; NG(E), there exists n E NG(E) such that Qn = P. Then Dn ;; P n CG(E) = E. Hence D = E is a Sylow p-subgroup of CG(E). Burnside's Theorem, CG(E) Remark 2.
x 0p,(CG(E)).
Theorem 3. 1 cannot hold without restriction if p
even if G is solvable. fours-group.
=E
Consider for example G
Then O ,(G) 2
= 1,
By
CG(E)
= E,
= GL(2,
3), E
~
= 2,
G a
but there is a subgroup of
7
order 3 which is normalized by E.
Note that a Sylow 2-subgroup of
GL(2, 3) has class 3.
Proof.
Let G be a minimal counter example.
The proof pro-
ceeds by a series of steps. 0p,(G)
1.
= 1.
Otherwise, let a have He
=
°
p
= 1.
,(G)
Let R
=
°
p
= G/Op,(G).
Thus He
=
°
p
~
0p,(RQE) n R
2. 'Q.
51
%
G
= CG(E),
by 0.3, we
,(G).
(G) and let Q *- 1 be a minimal E-invariant p'-sub-
group of G. If RQE C G, then QC
rR, Ql
Since Ca(E)
= 1.
=
°
p
,(RQE) by induction.
Hence
Since CG(R);; R by p-constraint we have
= RQE.
Let S be a QE-invariant subgroup of G minimal with respect to 1. Then S is a special p-group. The argument which verifies
this is standard.
See for example [12].
If S is abelian, then S
= CS(Q)
@
[Q, S] by 0.5.
an E-invariantp-subgroup and so Cp(E) n [Q, S] *- 1. by our hypothesis on E.
But [Q, S] is
Thus En[Q, S]*!
On the other hand [E n [Q, S], Q]
This contradicts CS(Q) n [Q, SJ
~
Q nS
= 1.
= 1.
If S is non-abelian and p is odd, we use a remarkable idea of
Bender, or perhaps of Baer. minimal, S follows: T
= f2 1 (S) = S qua
has exponent p.
Let T be a new group defined as
set.
Every element x is odd.
First by 2.4, since [Q, S] *- 1, and S is
1
E
S has a unique square root x"2
Define a binary operation
0
E
S, since p 1
1
on T as follows x 0 Y = x"2 Yx 2
It is routine to check that T is an elementary abelian group.
QE acts as a group of automorphisms of T.
Also
Since S = T as sets, the
fixed points of both Q and E on T are unchanged. reached a contradiction when S is abelian.
•
But we have already
This same argument can be
applied to TQE. If P
class
=2
and S is non-abelian, first [E, S]
< 2. Thus
=
[S, E, QJ ~ [Z(S), QJ [Q,
8
s,
=1
E] ~ [S, E] ~ Z(S).
~
Z(S) since SE has
Hence [E, Q, S];; Z(S). If [E, Q]
"*
1, then [E, Q] ;; Q stabilizes the chain S ~ Z(S) ~ 1.
Hence [E, Q] ~ CG(S) by 0.1.
Since [E, Q] is an E-invariant subgroup
of Q, minimality of Q ensures that Q
= [E,
Q] ~ CG(S).
This is a
contradiction. Thus we may assume that [E, Q]
= 1.
Since [S, E] C S is
then Q-invariant, minimality of S ensures that [S, E, Q]
= 1.
Also
[E, Q, S] = 1. It follows that [Q, S, E] = [S, E] = 1. Our assumptions on E now give S
~
E and [S, Q]
~
Q nS
= 1.
This contradiction
completes the proof. /1 Lemma 3.2. group G.
Suppose P is a p-subgroup of a p-constrained
Then 0 ,(NG(P)) cO ,(G). --
Proof.
=
P
P
Since G is p-constrained, it follows from O. 3 that
G/O ,(G) is p-constrained. Let G = G/O ,(G) etc. By induction we p p have 0 ,(N-G(P)) cO ,(G) = 1. But clearly 0 '(~G(P)) = 0 ,(C-G(P)) = p = p p p
= 0p,(CG(P)).
0p,(CG(P)) by 0.3.
Thus 0p,(Nc;:(P))
Since
op ,(NG(P)) C= CG(P),
it follows that 0 ,(NG(P)) cO ,(G) in this case.
p = p Hence we may assume that 0 ,(G) = 1. Let M = 0 (G), Q = 0 ,(NG(P)). p p p Since [Q, p] = 1 and [Q, CM(P)] CM n Q = 1 we have [Q, M] = 1 by 2.1.
Hence Q
=1
Remark.
because CG(M)
~
M by p-constraint.
The reader should refer to Lemma 12.5,12. 6 due to
Bender for a far reaching generalization of this result.
!!.
Lemma 3.3.
G is a p-solvable group of odd order and P is
a Sylow p-subgroup of G such that r(P);;; 2, then G has p-length 1. Proof.
o p ,(G) =
1.
Let G be a minimal counter example.
Let R
= 0 p (G).
Clearly
Since G is p-constrained, CG(R) eR.
Let C be a Thompson critical subgroup of R and let D
IG I
is odd and C has a class Since r(P)
S
2, r(D)
~
S
2, D has exponent p.
i
If
IZ(D)
I:>
=
= Q 1 (C).
p2, then D
= Z(D)
Since and if
9
IZ(D) I = p
then any subgroup of type (p, p) containing Z(D) has
centralizer of index
S.-
p. It follows that IDI
only if D is non-abelian of exponent p.
Let
S.- p3. Also D = D/ep(D).
is still a normal p-subgroup of G and so CG(D) ~ R. G/CG(D) ~ GL(2, p).
= p3
Then CG(D)
But
But any odd order subgroup of GL(2, p) has a
normal Sylow p-subgroup. Thus G =
IDI
°p,p ,(G mod CG(D)) !!
Lemma 3.4.
°p,p ,(G).
and so G =
1/
G is a solvable group of odd order and P is a
Sylow p-subgroup of G such that P' is cyclic, then G has p-length 1.
Proof. clearly.
Let R
If ep(R)
1.
Let G be a minimal counter example. First
= 0p(G).
Then CG(R)
~
R since G is solvable.
"* 1, let G = G/ep(R). Since P' is cyclic, G has p-length
Let Qep(R) = 0p,(G mod ep(R)) where Q is a p'-group.
[Q, R]
~
°P ,(G) = 1
ep(R) and so Q centralizes R modulo ep(R).
Hence 0p,(G mod ep(R))
=1
Then
Thus [Q, R]
and G has p-length 1.
Thus we have that R is an elementary abelian p-group. x
E
P.
Then [x, R]
~
P' n R, a cyclic subgroup of order p.
[x, R] ~ Z(P) and [R, x, x]
= 1.
discussion of this see the Appendix p. 80.
4.
Let
Thus
Hence x acts on R with quadratic minimum polynomial. Hall and Higman [x, R] = 1.
= 1.
For a
By the famous Theorem B of
Thus P = R and the Lemma is proved. 1/
THE TRANSITIVITY THEOREM
Included here is a proof of a rather unsatisfactory form of the Thompson Transitivity Theorem.
This is proved completely in [12].
proof given here is shorter but the Theorem is less generaL
The
More
precisely, for the case of odd primes, the Theorem is more general; but for p = 2 it deals only with groups, whose Sylow 2-subgroups are of c lass at most 2. I do not know of any slick way to prove the general result.
The final difficulty arises from the fact that a subgroup can be
self centralizing and normal in one Sylow p-subgroup of a group but conrained in another Sylaw p-subgroup non-normally.
10
The Theorem 3.1 is
se:rr. by
used to overcome this difficulty by replacing the elements of a rather larger class of groups. class
2.
3.
By this means we lose the case p
= 2,
No matter: the result as stated suffices for the results in
these notes. Before stating the Main Theorem, we prove a couple of auxiliary results. Lemma 4. 1.
If P is a p-subgroup of G such that NG(P) is
p-constrained, then CG(P) is also p-constrained. Proof.
First 0p,(NG(P))
= 1.
we may assume that 0pl(NG(P)) Q
= 0 p (C),
R nC
~
R
= °p (N).
C, R n C
~
= 0p,(CG(P)), Let N
Since Q char C
Q and so Q
=R
~
Q
~
= NG(P),
C
Using 0.3,
= CG(P),
QC R n C.
Since
=
n C.
Let x E CC(Q) be a p'-element. x stabilizes the chain R
= N,
<]
clearly.
Then [x, R]
~
R nC
1 and so by O. 1, x E CN(R)
~
= Q.
Thus
R since N
is p-constrained. It follows that CC(Q) is a p-subgroup, normal in C. Thus Cc (R) ~ Q and C is p- constrained. /I
Lemma 4. 2. such that r(A)
Suppose that A is an elementary abelian p-group
= 3. .!!..
a EA such that Cp(a) Proof. P
=
P, Q are A-invariant p'_groups, there exists
"*
1, CQ(a)
"*
1.
Let Y be a subgroup of A of type (P, p).
: v E y#>, there exists v E y# such that Cp(v)
W C A such that W is of type (P, p) and W n
=
w EW such that Cp(w) n Cp(v)
"*
"*
1.
"*
1.
Let
There exists
1, since Cp(v) is W-invariant.
is of type (p, p) and acts on Q. a dv, w># such that CQ(a)
= 1.
Since
Then
Thus there exists an element
~
Since Cp(a)
Cp(w) n Cp(v)
"*
1, we
are done. /I Theorem 4.3.
(Transitivity Theorem) [8].
Let G be a group
in which the normalizer of every non-trivial p-subgroup is p-constrainted.
!!.
p
= 2,
assume that a Sylow p-subgroup of G has class
be an abelian p-subgroup of G such that r(E) tains every p- element of C
= CG(E).
Then
--
°
~
p
I
~
2.
Let E
3 and such that E con(C) acts transitively on
11
the elements of V1*(E, q) where q is a prime, q
* p.
Proof.
Let S , S , ... , St be the 0 ,(C) orbits of elements of 1 2 P VlC(E, q) and suppose that t> 1. Then clearly II1 (E, q) {11.
*
C
Consider now R = S. n S. for subgroups S. E S., S. E S., where 1 J 1 1 J J i '" j and suppose that R is chosen of maximal order. For convenience, write i = 1, j = 2.
Of course, R cS, RC S . 1
Now N=NG(R)::JE.
=
2
Consider N=N(R);R.
Let T.=NnS.::JR. 1
1
Then T. = T.;R, i = 1, 2 are E-invariant, non-trivial, and so by 1
#
1
Lemma 4. 2, there exists e E E
such that CT. (e)
* 1,
i = 1, 2.
By
1
=.3, CT.(e) = CT.(e) and so (CG(e) n Ti)R::J R. By hypothesis and 1
1
Lemma 4. 1, CG(e) = H is p-constrained, and CG(e);; E. Let P. = T. n H. Then P.R::J R. Remember P. C NG(R). 1 1 1 1= P. is E-invariant and by 3.1, P. cO ,(H). 1
1
Let L=R(NnH).
=
Now
P
Since P.cO ,(H) nNco ,(NnH), P.cO ,(L) 1= P 1= P = P because R <J Land R is a p'-group. Thus RP. cO ,(L), i = 1, 2. = 1 = P Let Q. ::J P.R be an E-invariant Sylow q-subgroup of 0 ,(L), 1 = 1 P i = 1, 2. There exists x E CG(E) no ,(L) such that QX = Q by [12], ~. 2. 2.
P
1
2
Since <x> is an E-invariant p'-subgroup of NG(E), which is p-
constrained, 3.1 ensures that x E 0p,(NG(E)), x E 0p,(CG(E)). Let UEV'lC(E, q), U;;Q 1 . Then unS1;;Q 1 rS1;;P1R::JR. By choice of R, U E S . 1 X But U n S ::J QX n S = Q n S ::J P R ::J R. Now since x EC(E), 2= 1 2 2 2= 2 uX is a maximal element of V'l(E, q) clearly. By the choice of R, uX E S2 and so U E S. Thus S = S . 2 1 2 5.
THE UNIQUENESS THEOREM This and the next two sections are devoted to a proof, due to
Bender, of the Uniqueness Theorem 5.1 [3].
Theorem 5. L
Let G be a minimal simple group of odd order.
Let U be an elementary abelian p-subgroup of G of order p3. there is one and only one maximal subgroup of G containing U.
12
Then
Remark.
This major theorem shortens much of the group
theoretic Chapter IV of the Odd Order paper [8] of Feit and Thompson. It should be pointed out that if a group G of odd order contains no ele-
mentary abelian p-subgroup of order p3 for any prime p, then G is In fact, such a group has an ordered Sylow tower, see [8].
not simple.
If there is one and only one maximal subgroup M containing a
subgroup V
~
G, then we call M a uniqueness subgroup.
Theorem 5. 2.
Let G be a finite group, H a maximal local sub-
group of G, F == F(H). 1nl;;; 2.
Let X == CH(F), n == n(F).
Assume that
Choose M a subgroup of F satisfying CF(M) ~ M and let R
be a solvable subgroup of G normalized by MX.
Then for any prime
q E n', Vl (M, q) has only one maximal element and this is a Sylow qR subgroup of 71 (R).
°,
Since 071,(R) ~ 071,(RM) ~ 071,(R), we may replace R by
Proof.
RM and assume that M (i)
~
R.
The proof is divided into a series of steps.
If Y is a n'-subgroup of R normalized by M, then
ynH==1. (The reader should consult Theorem 12. 4(a) for an appr.opriate generalization of this step. ) First Y n H centralizes M since [Y n H, M] But CF(M) ~ M.
Now consider (Y n H)F.
~
Let p E 71.
is a p-element centralizing M , then of course x E F p
p
F(H) n Y == 1.
If x E(Y n H)F
and
x E CF(M ) n CF(M ,) == CF(M) C M. Thus M has the property that it p P == P contains every p-element of its centralizer in (Y n H)F. Since this group is clearly p-constrained, we may apply 3.1 to get Y n H C for all p E 71.
Thus [Y n H, F] == 1 and so Y n H
~
==
° ,(YnH)F) p
CH(F).
Now K == (Y n H)X is a subgroup of Rand;o is a solvable X-invariant group.
Thus Y n H
~
S(X) n H
~
S(H) n CH(F)
~
well known property of Fitting subgroups of solvable groups.
F by the Since Y
is a 71'-group and F is an-group, Y n H == 1. (ii)
If Q EVlR(M, q), q En', then Q~071,(R).
°
We show QC ,(R) for all p E 71. For such a prime p, we have == p M == Mp x L and since 1711) 1 and M ~ CF(M) we must have L ~ 1.
13
Since Mp;; Z(F(H)p)' CG(Mp ) ~ CG(Z(F(H)p)) ~ H. Now . Leo ,(H) n CG(M ) cO ,(CG(M )) cO ,(NG(M )). By 3.2, Leo ,(R) =p p=p p=p p =p since R is solvable and so [L, Q] ;; 0p,(R). We show that [L, Q] = Q. For Q = [L, Q]CQ(L) by 0.4, and CG(L);; H since L contains a nontrivial normal subgroup of H, namely Z(F(H)p')' by step (0.
Thus CQ(L);; QnH=1
Hence Q = [Q, L] cO, (R).
=
11
We have thus shown in (i) and (ii) that every element of
V1 (M, q), q
E 11', lies in 0 , (R). On the other hand, some Sylow q11 R subgroup of 0ll,(R) is certainly M-invariant and the M-invariant Sylow
q-subgToups of 0 ,(R) are conjugate under CG(M) no ,(R) = 1. 11
11
Thus
l!1 (M, q) has a unique maximal element. ;/ R
Theorem 5. 3.
Let G be a minimal simple group of odd order,
H a maximal subgroup of G, V an elementary abelian p-subgroup of F = F(H) such that (i) (ii)
or
Iv! ) p3; Iv I :; p2 -and
V c A
=
Set M = CF(V), lI(F(H)). q
E 11',
l!1 (M, q) G
E
S8JC (p). 3
Assume 1111
> 2.
Then for any prime
= 1.
We show that V1 (M, q) has a unique maximal element. G For then if Q is this unique maximal element of VlG(M, q), since Proof.
NF(M) permutes the maximal elements of V1 (M, q) under conjugation, G it follows that Q is the unique maximal element of VlG(NF(M), q). After a while, since F is nilpotent, Vl (F, q) has a unique maximal element G and then VlG(H, q) has a unique maximal element. But by the maximality of H, Q c H and so Q = 1 because 0 (H) = l.
=
q
The proof that V1 (M, q) has a unique maximal element follows G closely the proof of the Transitivity Theorem 4.3. Suppose that Q, R are maximal elements of V'1 (M, q) and suppose G Q, R have been chosen distinct such that IQ n R I is maximal. If Q n R
"* 1, let K = NG(Q n R). Apply
5.2 with K in place of R.
IIl (M, q) has a unique maximal element S, say. Because K K r Q, K n R ~ V1 (M, q), ;; S. Since S nQ;;K nQ:>Q nR, K if S* E V1 (M, q), S* :; s, choice of Q, R ensures that S* = Q. But
We get
a
14
= R,
again S*
a contradiction.
=1
Thus Q n R
and any two maximal
elements of VlG(M, q) are disjoint. Let x
E
U
#
and apply 5.2 to CG(x). It follows that VlCG(X) (M, q)
has a unique maximal element T.
We show that there exists x
that CQ(X)"* 1, CR(x)"*l. But (CQ(X), CR(X» ;; T. T* ;; T.
Then T* n Q ;; CQ(x) "* 1 whence T*
T* n R :::> C (x)"* 1 and so T*
= R.
E
U such
V1 0(M, q),
But again
This is the required contradiction.
rt I :> p3, we are clearly done by 4.
~
Let T*
= Q.
E
2.
We maytherefore assume that Iu I = p2 and U c A
=
E
se:n
3
(P).
Clearly n (Z(F » c U since otherwise we could replace U by P
1
=
un (Z(F )), an elementary abelian group of order ) p3.
=
P
1
CG(U) ;; CG(n 1 (Z(Fpm;; H. # Now Q = (CQ(X) : x E U >.
As already noticed, CG(x) has a
unique maximal M-invariant q-subgroup, X say. If x CQ(X) "* 1, then letting X* and so x*
= Q.
Thus X
E
Thus
E
U# is such that
VlO(M, q), X* ~ X, we have X* nQ;; C (X)"*1
= CQ(X).
Q
Now A normalizes M
= CF(U).
Hence A permutes the elements of iI1 (M, q) under conjugation. Thus G A;; CG(x) normalizes CQ(x) whenever CQ(x)"* 1. It follows that A Similarly A normalizes R.
normalizes Q.
But by 4. 3, Q andR are conjugate by an element of Gp' (CG(A). Since CG(A);; CG(U) ;; H, there exists h is a q-group.
But h
E
and so Q Since h
E
= R.
~ R.
-
Hence M normalizes It follows that Qh
~R
-h
Now if u E U is such that CQ(u)"* 1, then CR(u ) "* 1. h CG(U), u = u. Hence there is an element u E U such that
CQ(u) "* 1, C (u) "* 1. R Theorem 5.4.
This completes the proof. 11 Let G be a minimal simple group of odd order,
M a subgroup of F(H) containing Z(F(H)),
111 I ~
CH(A) such that (Qh, R)
CG(U) normalizes M.
Qh, R and so M normalizes (Qh, R> h
E
2 and VlG(M, n')
=
{I
J.
11
= lI(F(H»).
Assume that
Then H is the only maximal subgroup
of G which contains M. Proof. tains M. Z (F(H»)
~
Suppose that L is a maximal subgroup of G which con-
Clearly lI(F(L»;; 71 since VlG(M, 71')
= 1.
Because
M, the centralizer of any Hall subgroup of M is contained in
15
M, as we have seen before.
Then if a
~ 11,
M , C CG(M ) fl 0 ,(H) cO ,(CG(M )), where pEa. a == p a == a p Therefore M , cO ,(CG(M )) n Leo ,(CL(M )) for all pEa. a==p p ==p p By 3.2, M , cO ,(L) for all pEa. Hence M t cO ,(L). Thus a==p a==a Ma' ~ CG(F(L)a) and F(L)a ~ CG(M ,) ~ H, for every subset a ~ 7f. If a a == 7f(F(L)) *- 7f, we see that Ma' ~ CG(F(L)) ~ F(L), a a-group. Thus == Ti(F(l)).
Ti
Taking a == 7f - p, we now have M , == M
a
F(O ,(L)) ~ Co (L)(F(O ,(L))). p pt p
o
P
t(L)
C
p ==
0 (L) centralizes
p
Thus by 2.2, [M , 0 ,(L)] == 1 and so p p
H. == But 0p(L) ~ F(L)p ~ Hand CG(Op(L)) ~ L.
C
Hence
o ,(L) cO ,(CG(O (L))) n H C 0 ,(CH(O (L))). By 3.2, 0 ,(L) cO ,(H). P ==p P ==p P P ==p By symmetry, 0 ,(H) cO ,(L) and since 0 t(H) *- 1 we have == p P p H == NG(Op.(H)) == NG(Op,(L)) == L. ;/ Theorem 5. 5.
Let G be a minimal simple group of odd order,
p a prime, H a maximal subgroup of G satisfying 0 ,(H) *- 1, V an p elementary a~elian subgroup of order p2 of G such that CG(x) ~ H for all x E V.
Then H is the only maximal subgroup of G containing
V. Proof.
Let P be a Sylow p-subgroup of H containing V.
Every p'-subgroup of G normalized by V is contained in H. the subgroup (VlG(P, p') P
1
== P no,
Then [P , Q] 1
p,p
C
==
(H),
Q no,
p,p
~H.
Q
Hence
Let E
V1
G
(P, p'),
(H) cO, (H). == p
Q C H. ==
Thus
Since H is a p-constrained, CH(P ) cO, (H). Thus 1 == P ,p Q ~ 0p',p(H) and so Q ~ 0p,(H). It follows that (VlG(P, pt) ~ 0p.(H) ElII (P, p'). Hence (VlG(P, p'» == 0p,(H). G Now NG(P) permutes the elements of V1 (P, pt) under conH jugation and so normalizes (V1 (P, pt» == 0p,(H) *- 1. G
16
~
Thus NG(P) Choose L
IL
(i)
~
(Hi)
Hence P is a Sylow p-subgroup of G.
V, L "* H such that
n HI
is maximal and then
P
I
(ii)
H.
IL IP ILI
is maximal and then is maximal.
Let R be a Sylow p-subgroup of L n H containing V. no loss of generality in assuming R if necessary.
If R
= P,
~
then NG(R)
There is
P replacing V by an H-conjugate
~
H.
If Rep, then Np(R)
=:l
R
and by the choice of L, H is the only maximal subgroup of G containing Np(R).
Thus NG(R)
group of L. We have
~
H in every case.
° ,(L) = H
because
° ,(L) p
is V-invariant.
°
<J R. 8ince L = ,(L) N (8) and L p ,p = p ,(L) ~ H, L "* H, N (8) H. Thus Np (8) = Rand 8"* R. Because G p ILip;;' ING (8) Ip' we may assume that NG (8) ~ L. Now a solvable group of odd order with a Sylow p-subgroup of
8et 8
°
=R
C
p
Clearly R is a Sylow p-sub-
no, (L)
t
rank
:5.
2 has p-length 1 by 3. 3.
r(R) ;; 3.
Thus P
[8] Lemma 8. 4,
~
8ince L does not have p-length 1,
R contains a 3-generated abelian subgroup and by
se::n 3(P)
oF~.
Let A E se;n3 (P).
= 1. The p-stability of L, a A solvable group of odd order, shows that x E 0p(N (8) mod CL (8)) = L 0p',p(L). Thus NA (8) ~ 8 and A ~ S. If x E N (8) ~ L, then [S, x, x]
Now let q be a prime different from p.
The Transitivity
Theorem 4. 3 ensures that if Q E V'la(A, q), n E NG(A), then Qn E lI1 (A, q) G n and even Q El!1a(A, q) because all elements of lI1 (A, q) have the a same order. Moreover, there exists c E 0p,(CG(A)) such that Qnc = Q. Hence NG(A)
= (NG(Q)
n NG(A)) 0p,(CG(A)).
p-subgroup of NG(A) and so there exists m p
m
-1
But p ~ NG(A) is a Sylow - 1
E N (A) such that G
~ NG(A) n NG(Q).
Then Qm is P-invariant.
a
Thus if Q ElI1 (A, q), there exists a
conjugate Qm ElI1 (P, q). We have already seen that (lI1 (P, q) ~Op,(H). a a Thus at least one element of lI1 (A, q) is contained in 0p,(H). But CG(A)
=A x
a
0p,(CG(A)) and 0p,(CG(A)) char CG(A).
V normalizes A, CG(A), 0p,(CG(A)) and so 0p,(CG(A)) ~ H.
Now
Since
17
0p,(CG(A)) acts transitively on the elements of VlG(A, q) and one of them
°P,(H),
lies in
(Vl * (A, q) G
and NG(S) ~ N ([;1G(S, G 6.
THE CASE
1711
2.
cO ,(H). Since S <J A, Vl * (S, G = p = p')) = H, a contradiction. /I
Iw(F(H»)
p')
= 0'p (H)
I=1
The Theorems 5. 2, 5. 3, 5. 4, 5. 5 are directed towards the case When F(H) is a p-group, it was very soon recognized that
2.
similar
r~sults
were implied by the Glauberman ZJ-Theorem.
Let G be a minimal simple group of odd order, H a maximal subgroup of G containing a Sylow p-subgroup P of H for which
se::n 3 (P)
"*~.
Let A
E
se::n 3 (P) and assume that F(H) is a p-group.
The first observation is that P is in fact a Sylow p-subgroup of G and moreover NG(P)
~
H.
Lemma 6. 1. Proof. tained in 0p(H)
VlG(A, p')
= A.
;; CG(F(H)).
E
se::n 3 (P) is con-
Thus Z(F(H));; A and
But by 3.2 0p,(CH(A)) ~ 0p,(H)
=1
and so
By the ZJ-Theorem of Glauberman [9],
Z(J(P)) ~ H.
Hence NG(P);; N(Z(J(P)))
Now take any prime q since CG(A)
{I} and NG(P);; H.
Since H is p-stable, any element A
= F(H)
CG(A) ~ CG(Z(F(H))) ~ H. CG(A)
=
= A,
"* p.
= H.
From the Transitivity Theorem 4. 3,
VlG(A, q) has only one maximal element.
Since
A ~ F(H), V1b(F(H), q) has only one element and then Vlb(H, q) has only element Q.
Since H is maximal in G, Q
Thus VlG(A, q)
=1
and so VlG(A, p')
Lemma 6.2.
Proof.
= 1.
C
=
H and then Q
/I
C
=
°P(H) =
1.
H is a uniqueness subgroup for P.
For if L is another maximal subgroup of G containing
P, 6. 1 implies that
° ,(L) = p
1 and then the ZJ- Theorem ensures that
L = N(Z(J(P))) = H. /I More surprising we see that H is in fact a uniqueness subgroup for A.
18
Lemma 6. 3. Proof. taining A.
H is a uniqueness subgroup for A.
Suppose L is another maximal subgroup of G con-
Choose L in such a way that
IH n L Ip
is maximal.
Let
Q be a Sylow p-subgroup of H n L containing A. If Q is a Sylow p-subgroup of H, then Q = ph, h EH, and so NG(Q) then NG(Q)
~
H by the choice of L.
~
H.
IQI
If
<
IHl p '
Thus in any case Q is a Sylow
p-subgroup of L. Since 0p,(L) = 1 by 6.1, L=N (Z(J(Q))). But then Q is a Sylow G p-subgroup of G and also of H. This contradicts 6. 2. ;l
P
z
Lemma 6.4.
Let V be an elementary abelian subgroup of order
-
of H and suppose that CG(x)
~
#
H for all x E V.
Then H is a
uniqueness subgroup for V. Proof.
Suppose that L is another maximal subgroup of G
containing V and choose L so that
IL
n HI
p
a Sylow p-subgroup of H n L containing V.
is maximal.
Let Q be
Again Q is a Sylow p-
By the ZJ-Theorem L = 0p,(L)N (Z(J(Q))). 1 ,(L) ~ H, N(Z(J(Q))) ~ H. Since INH(Z(J(Q)))I
Since by
subgroup of L. hypothesis
°
p
-
-
p
>
IQI
p
we have a contradiction. ;l Lemma 6. 5. group of order p3, Proof. Let V
z
Let X
Ix I = pZ.
~
Y ~ P where Y is an elementary abelian Then H is a uniqueness subgroup for X.
Let V 1 be a Y- invariant subgroup of
= Cy(V). 1
order at most p.
n 1 (A)
of order p z.
Since Y/Cy(V ) is a subgroup of GL (2, p), it has #
1
Thus
IVzl ~ pZ. If x E VI' CG(X) ~ A, and since H
is a uniqueness subgroup for A, CG(x) #
~
-
H.
By 6.4, H is a uniqueness
subgroupfor VI' Now if XEV Z' CG(x)~VI and so CG(X)~H. By 6.4 again, H is a uniqueness subgroup for V . Finally, if x EX, z CG(X) ~ y ~ Vz and so CG(x) ~ H. By 6.4 yet again, H is a uniqueness subgroup for X. ;l
19
7.
THE PROOF OF THE UNIQUENESS THEOREM 5.1 We commence the final attack on 5. 1 having established in §5 and
§6
uniqueness theorems sufficient for the task.
Thus we fix G a minimal
simple group of odd order, P a Sylow p-subgroup of G for which
"* P,
se~3 (P)
H a maximal subgroup of G containing NG(P), if P is
abelian, and containing NG(Z(P) r, p,) ;; NG(P) otherwise. A
E se~
3
(P) and A
Let
= n 1 (A). 0
We show that H is a uniqueness subgroup for A and establish 5. 1; but note that we will have in fact established that H is a uniqueness
subgroup for every elementary abelian p-subgroup of order p2 which lies in an elementary abelian p-subgroup of P of order p3.
In fact this is
precisely the statement of 6.5 if F(H) is a p-group, while if not, suppose that H is a uniqueness subgroup for A is of type (p, p, p), p2 and let V
2
Ixl = p2.
= Cy(V). 1
Then if x
E
~
P.
Let X
~
#
1
~
Y
~ormalize V -c
Let Y
=
P where Y
n- (A)
of order
1
V , CG(x) ~ A and so CG(x) c H. 1
By 5. 5, H is a uniqueness subgroup for V. 1
=
Now if x
#
E
V , CG(x) 2
= = V1 ~
and so CG(x) C- H. We use here 5.5 since V2 is non-cyclic to get that H is a unique~ss subgroup for V. Finally if x E xi, CG(x) ~ V and 1
so CG(x);; H.
-
1
By 5.5 again, H is a uniqueness subgroup for X.
We thus show that H is a uniqueness subgroup for A.
We say
that a maximal subgroup X of G is of uniqueness type and q is a uniqueness prime if there exists a prime q such that F(X) has an elementary abelian q-subgroup U such that (i)
Iu! )
(u)
IU I = q 2
q3 or and U
=A
C
E
se~ (q). 3
Assume that H is not a uniqueness subgroup for A.
Note that if
F(H) is an r-group where a Sylow r-subgroup R of H satisfies se~
3
(R)
"* P,
then H is a uniqueness subgroup for every elementary
abelian subgroup U of R of order r 2 which lies in an abelian subgroup V or R of type (r, r, r) by 6.5. Suppose on the other hand, /n(F(H)) I ) 2.
Then, if U is an 3 2 elementary abelian r-subgroup of F(H) of o;cter ~ r or of order r contained in an element of V1 (M, n') G
20
=
se~
{1} where M
3
(r), it follows from 5.3 that
= CF(U),
F
= F(H).
Then by 5.4, H is a
Now if V ~ U is a subgroup of order r
uniqueness sUbgroup for M. and x
E
V#, CG(x) ;; M from which it
f~lows that
CG(x)
~
H.
2
Now
apply 5. 5 to get that H is a uniqueness subgroup for V. Thus a subgroup H of uniqueness type is a uniqueness subgroup for very many subgroups of H. The proof of 5. 1 hinges on the study of subgroups of uniqueness type.
First we show that subgroups of uniqueness type exist in G.
Lemma 7. 1.
H is a subgroup of uniqueness type.
Suppose that H is not of uniqueness type.
Proof.
Then F(H)
does not contain an elementary abelian subgroup of order q3 for any prime q and a Sylow p-subgroup of F(H) is cyclic.
For if F(H)p is
non-cyclic, since F(H)p n P ;J P, there exists V;J P of type (p, p), V e F(H).
=
S~~
But every such normal subgroup lies in an element of
p
3
(P).
For let A
type (P, p, p).
E S~Jl
3
(P), A e A a normal subgroup of P of 1
Since CH(V)n A
and is elementary abelian.
=
has order ~ p2, V(CH(V) nA1 ) ~ P
1
If CH(V) nA ~ V, then V lies in a ;; 31
generated abelian normal subgroup of P while if CH(V) n A 1
~
V,
yeA.
=
1
= F(H) q , where q is a prime q"* p. If P does not Q, let R = Q (C) where C is a Thompson subgroup of 1 By 2.4, rp, RJ "* 1. Since r(Q) S 2, r(R) S 2 and so Let R = R/(R). Then P acts ~on-triviallY on R and
Let Q
centralize (See 0.4.) IRI ;;; q3.
H/CH(R) is an odd order subgroup of GL(2, q). (H/CH(R)), is a q-group and so P n H' ~ CH(R). This holds for all q "* p. is cyclic, P n H'
~
Q.
It follows that
Thus P n H' :; CH(Q).
Since H/CH(F(H) ) is abelian because F(H) p
CH(F(H)p) and P n H'
~
CH(F(H))
~
F(H).
If P is abe lian, then fusion of elements of P is controlled by
NG(P) by the well-known Burnside Lemma. 1
(x- x S
g
:x
E
P, g
= ( x -1 x n : x
E
E
G, x
P, n
E
g
E
p)
NG(P)
=S (~H)
Thus the focal subgroup
satisfies )
~
P n H'
~
F(H).
Thus by the application of the transfer we have P n G'
~
F(H).
21
P
Since G has no non-trivial p-factor group, P
~
F(H) is cyclic.
This
is not the case. Hence P' "* 1.
But since P n H' ;; F(H) we must have P n F(H)"*1. g Let Z be the subgroup of order p contained in F(H). If Z ~ H n H g where g E G - H, choose g so that I H n H I is maximal. Let S be a g Sylow p-subgroup of H n H and assume without loss of generality that S:= P. Since Z:SJ H, Z lies in every Sylow p-subgroup of H. If g -1 -l S = P ~ H n H then P, pg ~ H and so P = pg h ,h EH. But then g-l h E-NG(P) ;; H, g EH, a contradiction.
Therefore
se
P.
Let T be a Sylow p-subgroup of NG(S). Then T =:> S and so g NG(T) ;; H by the choice of H . Hence T is a Sylow p-subgroup of NG(S).
Since P n Ht ;; F(H), T'
~
F(H) and so T' is cyclic.
By 3.4,
N (S) has p-length 1. G Thus NG(S) 0p,(NG(S))(NG(S) r NG(T)) by the Frattini argument.
=
Since NG(S) ~ H, NG(T) ;; H, 0p,(NG(S)) ~ H.
But 0p,(NG(S));; CG(S) ;;
CG(Z) ;; H.
C
This contradiction shows that if Z
Now let x E P, x
H n Hg, then g EH. -1
g
E P.
Then Z;; Z(P) and so Z, zg
;; CG(x).
-1
Choose y E CG(z) such that (Z, zg
y> is a p-group and then find
-1
Z E G such that (Z, zg Z E H and then Z, Z Hence x
g
= xy
g
-1
y>z yz
C
=
P.
Then we have first Z, ZZ
E P whence g
-1
yz E H.
Thus g
-1
C
=
P whence
y E H.
-1
g where y-1 gE H since y
E
CG(x).
Thus H controls
p-fusion. Transfer now yields P n G' ;; P n H' ;; F(H).
This contradicts
the simplicity of G. ;/ We devote the next several lemmas to a study of subgroups of uniqueness type, now being certain that they exist.
Fix our notation so
that X is a subgroup of uniqueness type, q is a uniqueness prime. Let B
= n1 (Z 2 (F(X)q )).
Remember that H is not a uniqueness subgroup
for A, otherwise we are done. Lemma 7.2.
Either X is a uniqueness subgroup for every sub-
group of order q2 of B or B is non-abelian of order q3 and X is a 2 uniqueness subgroup for every subgroup of order q of B which is
22
normal in some Sylow q-subgroup of X. Proof.
IB I ?
If
q 4 then every subgroup V of order q
lies in an elementary a;lian subgroup U of order q3 of B.
2
of B
For if
Z(B) ~ V, the result is clear and if Z(B);; V either V case the result is trivial using exp B = q, or Z(B) then V
~
B since B'
= Z(B)
C
= Z(B) in which V, IZ(B) I = q. But ~
and so B/CB(V) has order
q.
Thus
ICB(V) I ;; q3 and we are done. If 1T(F(X))
for M
V.
=
{q}, then 6.5 shows that H is a uniqueness subgroup
= 1, where G Then 5.5 implies that X is a uniqueness subgroup for M.
If 1T(F(X))::l {q}, then 5.4 implies that II1 (M, 1T ' )
= CF(X)(U).
Now if x
E
V# ;; U, CG(x)
~
M and so CG(x);;
x.
Thus X is a unique-
ness subgroup for V by 5. 6. We may therefore assume that
IB I :s q 3.
If B is abelian of
order q3 the above argument applies. We have already remarked that X is a uniqueness subgroup for some q-subgroups of X and so X must contain a Sylow q-subgroup Q of G! Q of type (q, q) lies in an element of
Now any normal subgroup V of S~:rr
3
(Q) by a familiar argument.
Thus by the remark at the beginning of §7 we see that X is a uniqueness This completes the proof. /I
subgroup for V.
Lemma 7. 3.
IV I = q 2,
.!£ V;; Y ;; X, where Y is of type (q, q, q) and
then X is an uniqueness subgroup for V.
Proof.
By 7.2, Y normalizes a q-group U of order q2 and
type (q, q) for which X is an uniqueness subgroup. #
since Y ICy(U) ;; GL(2, q). CG(x) ;; X.
If x
E
Thus
Cy(U) , then CG(x)
~
ICy(U) I 2: q 2 -
U and so
By 6.6 and 5.5, X is a uniqueness subgroup for Cy(U).
Thus X is a uniqueness subgroup for V by 6. 6 and 5. 5 again, since if # x E V , CG(x) ~ y ~ Cy(U) and so CG(x);; x. /I Lemma 7.4.
The uniqueness prime q belonging to X is
different from p. Proof.
This is clear by 7. 3 since X is not a uniqueness sub-
group for A. /I
23
Lemma 7.5.
No non-cyclic subgroup of A centralizes a non-
cyclic subgroup of B.
Proof.
Let DcA
=
be of order p
2
and suppose that
0
V ;; CB (D) is of type (P, p). 3
If B is non-abelian of order q , V;Z(B) is a non-trivial sub-
group of B;Z(B) which is centralized by D.
But the elements of D
induce automorphisms of B which preserve the symplectic form [ , ] : B;Z(B) x B;Z(B) - Z(B) and so these automorphisms have deterThus if B is non-abelian of order q3, then ICB(D) I ;; p2
minant 1.
implies that CB(D) = B and so X is a uniqueness subgroup for CB(D) by 7. 2. If B is not non-abelian of order q 3, then X is a uniqueness subgroup for CB(D) by 6.2 also. We may now apply Lemma 5. 6. if x
E
D#, CG(x) ;; CB(D).
CB(D), CG(x) ;; X.
Note that by 7.4, 0p'(X) = 1, and
Since X is a uniqueness subgroup for
By 5.6, X is a uniqueness subgroup for D, and also
A, and this is a contradiction. // Lemma 7.6.
Let A = n (A) and B = n (Z (F(X) )) as usual. 12 q
-01-
Then (i)
(ii)
CB(A)= 1 and lA I =p3 o 0 A contains a subgroup D of order p2 such that if
o
E = CB(D), then lE I = q and ~G(E);; X (iii)
There exists d
Proof.
E
D
such that CG(d);; X.
By 7.5, if D is any subgroup of A
iCB(D)I~q.
o
or order p2,
If B is non-abelian of order q3, then A
and so A/CA (B);; GL(2, q). hand, if
IcA
o
acts on B = B/Z(B) o Thus lA/CA (B) I ~ p2. On the other
I;;
0
(B) p2, we would have a non-cyclic subgroup of A which o centralizes B and therefore B, a contradiction to 7.5. Therefore
jA I = p3. If A centralizes Z(B), then A will induce a symplectic o o o group of automorphisms of B/Z(B) and so a subgroup V of A of type (P, p) will centralize B/Z(B) and Z(B).
B, a contradiction to 7. 5.
24
It follows that V centralizes
= CA
(Z(B)). Since A iD e aut Z(B), A iD is cyclic and 0=0 o so IDI = p2. Since CB(A ) ~ CB(D) = Z(B) and [Aa' Z(B)] "* 1, we a have C (A ) = 1. Therefore, in this case, E = Z(B)
d
E
D such that B
for B.
~
CG(d).
Then CG(d)
~
X, a uniqueness subgroup
Thus the result is completely proved if B is non-abelian of 3
order q . For the remainder of the proof, we assume that B is not nonabelian of order q3.
Thus X is a uniqueness subgroup for every sub2
group of B of order q . Now the image of any irreducible representation of A on a vector space of characteristic q"* p is cyclic.
Thus each minimal A- invariant
subgroup of Z(B) or B/Z(B) is cyclic. Otherwise A
has a subgroup o of order > p2 which centralizes a subgroup of order q2 of B, contra-
dicting 6.5.
Remember that the fixed points of A on B/Z(B) are just
images of fixed points of A by O. 3. of B of order q2.
Let N be an A-invariant subgroup
[Z(B)I >q2 we may choose Ne Z(B) by
If
=
Maschke's Theorem, while if IZ(B) I
=q
=
we may choose N/Z(B) a
minimal A-invariant subgroup of B/Z(B). As before A/CA (N) ;; GL(2, q) and so lA/CA (N) I ;" p2. o 0 By 7.5 Ic (N) I ~ p and so IA 1= p3. Let NI be a minimal Aa za invariant subgroup of N and define D = CA (NI)' We claim that D o satisfies the conclusion of the Lemma. First if A
centralizes N ,A induces a cyclic group of autoo I 0 2 morphisms on N/N. Thus a subgroup of order p of A centralizes I
N/N
I
and N. I
0
This contradicts 7.5.
CB(A ) e CB(D) n C(A ) 0=
IDI
= p2.
0
In this case
I
some element d
0
= Nl nCB(A ) = 1. Since A ID e o a = E = NI and NG(E) ~ N, while X
ness subgroup for N by 7.2. centralizes N
Therefore D"* A
Thus NG(E)
~
X.
and aut N , I
is a unique-
Finally since D
and induces a cyclic group of automorphisms of N/N E
D centralizes N/'NI and NI'
X is a uniqueness subgroup for N by 7.2.
I
Then N ~ CG(d).
But
Thus CG(d);; X.
This completes the proof. ff
25
Lemma 7. 7.
If R is any p-subgroup of X containing A, then
NG(R) :; X. Let Q* be a maximal element of V1 (A, q) containing B. G Then Q*:= X, because X is a uniqueness subgroup for B. By the Proof.
Transitivity Theorem 4. 3, every element of V1 (A, q) lies in X. G Note that CG(A) ~ CG(D) ~ X by 7.6(iii). Now if R is a psubgroup of X containing A, then (V1 (R, q) ~ X and B ~ (Ii1 (R, q). G G As NG(R) normalizes this subgroup (V1 (R, q), we have NG(R) ~ X G since X is a uniqueness subgroup for B. This completes the proof. /I We are now in a position to contradict the existence of G. We know by 7. 1 that H is a group of uniqueness type and so the results of 6. 4, 7. 5, 7.6, 7. 7 apply to it. Put X
Lemma 7. 8.
=H
and let E, D, B be as defined in 7. 6. 3
If B is not non-abelian of order q , and S
=P
no, (H), then p,p
NH(S) ~ NG(E). Proof.
By 5.6, if CG(d):= H for all d
uniqueness subgroup for D and such that CG(d) ~ H. CG(d).
Then K
=H
A~
is p-stable, A
~
D#, H would be a
Thus there exists an element d ED#
Let K be a maximal subgroup of G containing while E
= CB(D)
~
CG(d) ~ K.
If K is of uniqueness type, since A
get that P:= K.
E
~
K, we may apply 7. 7 and
But A is an abelian normal subgroup of P and since H S.
normalizes B n K
By 7. 7 again applied to K, NG(S) ~
CB(D). If B n K
* CB(D),
~
K.
Thus NG(s)nH
since H is a uniqueness
subgroup for all subgroups of B of type (p, p) by 7.2, it follows that H
= K,
a contradiction. Remember we have assumed B is not non-
abelian of order q3.
Thus B n K
= CB(D)
is normalized by NG(S).
Hence we may assume that K is not of uniqueness type.
We
derive a contradiction. We show first that E centralizes F(K) " For if r is a prime q 1, we note that F(K) has no
different from q such that [E, F(K) ] r
*
r
elementary abelian subgroup of order r 3.
Since A normalizes E, we
may consider the action of EA on F(K).
By a now hopefully familiar
r
argument, using a short trip via the Thompson subgroup, we see that a 26
homomorphic image EA of EA is a subgroup of GL(2, r).
But then
EA, a subgroup of odd order of GL(2, r) is abelian, since the derived subgroup of any such subgroup if an r-group and E
Hence
Thus either E;; CG(F(K)r) or [E, A]
This last possibility does not occur by 7.6 since CB(A ) = 1.
o
centralizes F(K)
q
= 1.
Thus E
I'
Now if F(K) , q
= F(K),
we have a contradiction because then
E ;; CG(F(K)) ~ F(K). Since NG(E);; H by 7.6, E ~ K. E(CG(E) n F(K) ) :::) E, since
q
IEl
Thus EF(K)q n H ~
= q, and EF(K)
q n H, being non-cyclic
and of odd order contains an elementary abelian q-subgroup Q of order q2.
Now we have just shown that [E, F(K)q']
= 1,
while
[F(K) , F(K) I] = 1 obviously. Thus [Q, F(K) ,] = 1. Now q q q F(K\, ;; CG(E) ;; H. If some element of order q in CH(Q) did not lie in Q we would have Q contained in a subgroup of type (q, q, q) of H and by 7.4, H would be a uniqueness subgroup for Q. case since Q
C K. =
We thus have F(K) , q
=H
C
This is not the
normalizes QB and
centralizes the subgroup Q which contains all elements of order q in its centralizer. N(F(K) I):::) B. K
q = = N(F(K)q')
Apply 2.2 and see that [F(K) " B] = 1. Thus q But H is a uniqueness subgroup for B. Hence
;; H unless F(K)q'
=1
and F(K) is a q-group.
F(K) has no subgroup of type (q, q, q). group of F(K\.
Let C be a Thompson sub-
Then CK{C) is a q-group and K/CK{C) ~ GL(2, q)
as usual, where C F(K).
But then
= C/(C).
Thus K has a normal Sylow q-subgroup
But A;; K acts faithfully on F(K).
This is impossible since A
is a 3-generated group and F(K) has no subgroup of type (q, q, q). This completes the proof. 1/ Lemma 7.9.
Without loss of generality there exists z
E
Z(P)
such that CG(z) ~ H. Proof.
Let B
= n 1 (Z 2 (F(H) q ))
of order q 3, then CA (B)
*1
as usual. If B is non-abelian
since A has a subgroup of type (p, p, p).
Also Cp(B) ~ P. If P is abelian, choose z and so CG(z)
~
E
CA(B).
Then CG(Z)~B
H, a uniqueness subgroup for B. If P is non-abelian
27
choose z E P' n Z(P).
Since P /Cp(B) is a p-subgroup of GL(2, q)
where p is odd, P /Cp(B) is abelian. CG(z) ~ B, whence CG(z) ~ H.
Thus z E Cp(B) and so
Hence if B is non-abelian of order q3
we are done. We may therefore apply 7. 8. E
= CB(D)
A
o
2
Ic A
by 7.6 and so
We then have that P normalizes
lE I = q. Thus P' centralizes E. Now [A , E] n
where
I = p3.
(E)
I = p.
Here Aa
= n 1 (A)
By 7.6,
a
If P is abelian then we may choose z holds.
a
as usual.
=d
in 7. 6 and the result
If P' n Z(P) is non-cyclic, then, since by 7.5 the centralizer of
any 2-generated subgroup of A in B has order at most q, we see that CB(P' n Z(P))
= E.
~
Then we may choose D
and then for some d E D we have CG(d) case P n Z(P) is cyclic.
~
H.
P' n Z(P) without any loss We are just left with the
But H was chosen in the beginning of §7 as a
maximal subgroup containing N(Z(P) n PI) Thus if z E Z (P) n P', CG(z) ~ H.
~
NG(P) if P is non-abelian.
This completes the proof. /I
The proof of Theorem 5. 1 now proceeds very quickly. Lemma 7. 10.
If A
g
~ H n H , then g EH.
g Let R be a Sylow p-subgroup of H n H containing A. g 1 Then R "* P, because if P ~ H n H , we have P, pg- ~ H. Thus g-lh -1 P =P for some h EH and so g h E NG(P) ~ H. Therefore g EH, Proof.
Consequently N (R) ~ H. G shows that NG(R) ~ H. /I a contradiction.
Lemma 7.
n.
But an application of 7. 7
If z E P is such that CG(z) ~ H, then z lies in
a unique conjugate of H in G. g Among all g E G such that z EH n H , g EH choose g
Proof. such that
IH n Hg Ip
H n Hg containing z. S:; P.
is maximal.
We may clearly assume without any loss that
Also S"* P since H contains NG(P).
Sylow p-subgroup of NH(S). group of NG(S). such that T
28
Let S be a Sylow p-subgroup of
<j::: 1
Let seT:; P be a
Since T::J S, T must be a Sylow p-sub-
For otherwise we could find a p-subgroup T 1 H and then if x ET
1
- H we would have T
[>
=H
C
T
nH
X
Let U = T no, (NG(S)). p,p Now NG(S) ~ H since S"* P. By 7.10 it follows that A ~ S.
while T::J S. Thus
S NA (S) ::J S. If n ENA(S), we have [S, n, n] = 1 and since
NG(S) is p-stable, n E 0p(N(U) n NG(S) mod CG(U) n NG(S)).
But
CG(U) n NG(S) ;; 0p',p(NG(S)) because NG(S) is p-constrained. NA(S);; 0p"p(NG(S)) n P = U.
Thus
It follows that U::J S and so NG(U);;H,
by the maximality of S. But NG(S) = 0p,(NG(S))(N(U) n NG(S)).
Since 0p,(NG(S));;;
CG(S) ;; CG(z) ;; H we have the desired contradiction, viz. NG(S);; H. This completes the proof. 1/
Lemma 7.12. Proof.
H!H' is a p'-group.
Suppose that a E P and that a
g
E P for some g E G.
Then if z E Z(P) is chosen so that CG(z);; H by Lemma 7.9, we have -1
Z,
zg
- 1
ECG (a). There exists x E CG(a) such that (z, zg
x) is a p-
-1
group.
Thus we can find y E G such that (z, zg
x)Y;;; P.
But then,
-1
z, zy E P implies that y EH by Lemma 7.11. so g-l x EH.
g X Then a = a
-1
g and x- 1 g EH.
Thus zg
x EH and
Thus the largest p-
factor group of G is isomorphic to that of H by an application of the Since G is simple we must have H!H ' a p'-group. completes the proof of Lemma 7. 12. 1/
transfer.
Lemma 7. 13.
pi iH!H' I.
This
Hence H is a uniqueness subgroup
for A. Proof.
For let S = pr:O
normalizes E or
IBI =
In
I (H). By Lemma 7. 8, NH(S) either p,p (Z2(F(H)q))1 =q3. If NH(S);;;NG(E) then 1
NG(S) has a non-trivial p-factor group because NG(E) does. that A does not centralize E by 7. 6(i). H = 0p,(H) NH(S),
H/CH(B) is not a p'-grouP. we have pIIH!H'
This is impossible because
On the other hand, if IBi = q3, letting
we have H/CH(B);;; GL(2, q).
I.
Remember
13 = B/Z(B) 13 and so
Again A does not centralize
But as a subgroup of GL(2, q) with P "* q
1/
29
8.
THE BURNSIDE pqqb-THEOREM, p, q ODD This theorem appears in [10] and the proof is due to Goldschmidt.
However it owes a great deal to the work of Bender, since it relies on ideas developed by him in the proof of the Uniqueness Theorem. Of course, the theorems in §5, §6, §7 are obviously directly applicable since they are concerned with the structure of a minimal simple group of odd order. Theorem 8. 1.
.!!
a b
G is a group of order p q
where p, q are
odd primes, then G is solvable. Proof.
Let G be a minimal counter example.
obViously a minimal simple group of odd order.
Then G is
Let P be a Sylow p-
subgroup, Q a Sylow q-subgroup of G. (i)
V1 (R, r')::::: {1 ) for any Sylow r-subgroup of G. G For if X E V1 (R, r'), there exists a Sylow r'-subgroup R' G
::J
X.
Then XG ::::: ~R' ::::: ~'
and
X
G
C
R'
is a proper normal subgroup of G.
Let H be a maximal subgroup of G. (ii)
If
11T(F(H)) I : : : 2, then F(H) is non-cyclic.
For if r::::: max (P, q), then H::::: CH(F(H)r') clearly and CG(F(H)r) ::::: CG(F(H)) ~ F(H), assuming F(H) is cyclic. Thus a Sylow r-subgroup of H is either cyclic and contained in F(H) or non-abelian and metacyclic. Z :::::
In either case H::::: NG(Z) where
n 1 (F(H)). But Z char H and so a Sylow r-subgroup of H is a r r
Sylow subgroup of G. (Hi)
This contradicts (i).
F(H) is a r-group for all maximal subgroups H.
By 5. 4, H is the only maximal subgroup of G which contains Z(F(H)).
Let V
~ F(H) be a group of type (r, r). Then
CG(x) ;; Z(F(H)) and so CG(x) ~ H. group of G which contains V.
30
if x
E
Vi,
By 5.5, H is only maximal sub-
Then clearly H contains a Sylow r-
subgroup of G, a contradiction to (i). Every Sylow r-subgroup R of G lies in a unique maximal
(iv)
subgroup M of G and every maximal subgroup of G contains a Sylow r-subgroup of G for some r.
= NG(Z(J(R)))
For M (v)
since F(M) is a r-group for some prime r.
Let R be a Sylow r-subgroup of G, M a uniqueness sub-
group containing R.
Then M is a uniqueness subgroup for Z(R).
For suppose Z (R) such that IM
nMI
"* M, M
is maximal.
r
I
C M = I
maximal in G.
I
Let R
Choose M
I be a Sylow r-subgroup of
I
M n M, R ~ Z(R). Conjugating R by a suitable element of M we I I = I may suppose R eR. By (iv) R "* R. Since ING(R ) IR the I = I I I choice of M ensures that NG(R ) C M. Hence R is a Sylow r-subI I = I group of M. But then M contains a Sylow r'-subgroup R', by (iv). I I g X Then if g E G, M = M , x ER since G = R'R, R' cM. I I = I g Thus Z(R) C M <1 G. = gEG I (iv) There exist R ,R Sylow r-subgroups of G such that
I)
I,
n
I
R
I
nR
2
2
=1.
For let M be the uniqueness subgroup for RI' Z(RI)' R
f
2 -
M such that
IR 2
n M I is maximal.
Let R
2
nM
= S.
Choose Without
Hence NG(S);; Z(R I ) and so NG(S);; M if S"* 1. S eR, S c R clearly shows that NG(S) f M. loss, S;; RI' I
2
-
Theorem 8. 1 now follows immediately. r
C
= max (pa,
that R
I
nR
But
qb).
= 1. 2
For we can choose
Then there exists Sylow r-subgroups R ,R such r 2c > paqb IIGI, 2a
Then IGI) IR IIR
=
contradiction! /I
I
2
1=
=
9. MATSUYAMA'S PROOF OF THE paqb-THEOREM, p = 2 Lemma 9. 1.
If G is a p-group and H;; G, then either H <1 G
X
or NG(H);; H "* H, x E G. Proof.
The group H acts on the set S of all conjugates ~"* H
for some x E G by conjugation. If S
I {H) X
u si
= [G
= 0,
then H;1 G.
: NG(H)] "* 1 if H ~ G, pf/sl. X
H "* H such that H
;;
Since
Therefore there exists
NG(H). /I
31
a
Theorem 9.2 [18].
If G is a group of order 2 pb, then G is
solvable.
Proof.
Let G be a minimal counter example.
a minimal simple group.
Clearly G is
By the proof of step (i) in 8. 1, no Sylow r-
subgroup of G normalizes any non-trivial rt-subgroup. If H is a maximal subgroup of G such that 1T(F(H)) = {2, P
(i)
1,
then F(H) contains a subgroup of type (r, r) for some prime r. For otherwise F(H)
is P cyclic. If F(H) is cyclic, then H C CH(F(H) ) and so Z(H ) C 2 p= 2 p= CH(F(H)) C F(H). Thus Z(H) <J Hand Z(H ) char H. It follows that = p p P H is a Sylow p-subgroup of G, a contradiction. 2
is cyclic or quaternion while F(H)
p
If F(H)
2
is quaternion, then since H/CG(F(H) ) is abelian,
p
H~ ~ CG(F(H)p)'
Thus Z(H ) n H~ ~ CG(F(H)) ~ F(H). Let Z be the 2 unique subgroup of order 2 in F(H). Then Z char Z(H ) n Ht char H • 222 Since H = NG(Z), we again have H is a 2-subgroup of G, a contra2 diction. (ii)
If H is a maximal subgroup of G such that 1T(F(H)) = {2, P },
then H is the only maximal subgroup of G containing Z(F(H)).
°
For let K
~
Z be maximal in G, Z = Z(F(H)).
°
Z C (NK(Z )) and so Z C (K) by 3. 2. p= P 2 p= P Hence F(K) C NG(Z ) = H. Similarly F(K) 2
=
Now F(K)2 ~ F(H)
2
C F(K) = 2
(iii)
°2(NH(F(K)p))
P
P
~ F(H)2
First
Thus [Z , F(K) ] = 1. P 2 H.
C
=
by 3.2 again.
By symmetry,
and H = K.
If H is a maximal subgroup of G, then F(H) is an r-sub-
group for some prime r. Let ye F(H) be of type (r, r), using step (i).
=
Then if x
#
E
Y , CG(x)
~
Z(F(H)) and so CG(x)
R :::) Y be a Sylow r-subgroup of H. of G. S
1
Then S =
E
y#).
~
H by (ii).
Let
Let S be a R-invariant rt-subgroup Thus se H. -
is a Sylow rI-subgroup of H containing
S,
Since H = RS
SH = ~Sl = SSl
1
where C S = 1
is
Thus se F(H) t' Hence F(H) , is the = r r unique maximal R-invariant r'-subgroup of G. If R C R where R is a normal r'-subgroup of H.
a Sylow r-subgroup of G, then N
R1
32
=
1
1
(R) normalizes F(H) t and so lies r
in H.
Hence R = R (iv)
M n Z(P)
if-
is a Sylow subgroup of G, a contradiction.
1
There exists a maximal subgroup M such that I, M n Z(Q)
1 where P is a Sylow p-subgroup of G and
if-
Q is a Sylow 2-subgroup of G. For let Z be a conjugacy class of involutions of G containing an element x of Z (Q). 2-group by 1. 1.
There exists y E Z such that (x, y> is not a
Thus xy has order divisible by p.
the unique subgroup of order p in (xy >. M of G containing NG(H). obviously Z(P)
C
=
H.
N (H) G
C
=
Let H~(xy> be
Choose a maximal subgroup
Clearly M n Z(Q)
if-
1, while just as
M, where P is a Sylow p-subgroup containing
We may now complete the proof of 9. 2.
Let M be a maximal
subgroup of G satisfying the conditions of (iv).
By (Hi), F(M) is a
Let R => M
be a Sylow r-subgroup of G and let S => M , be r = r a Sylow r'-subgroup of G. First i Z(R) ~ CG(F(M)r) ;; F(M). Choose
r-group.
=
x EM n Z(S) and let N = (Z(R)X : i E Z>.
Clearly N
C
0 0 =
F(M) is an
Let Q= iZ(R)Y: y EG} = Q u Q u ••• u Q, where Q
r-group.
1
2
S
1
is an (x) orbit in Q. Let N. = (Q'>. Since G = RS, there exists yES such that 1 1 i . Z(R)y E Q. and so N. = (Z(R)Yx : i E Z) = (Z(R)X!y : i E Z) = NY. 1
1
0
Thus N. is an r-group normalized by (x). 1
Choose Z maximal
such that N = (Q. , ... , Q. ) is an r-group normalized by (x). 1 l Z 1 Assume for simplicity that N = (Q1' ... , QZ)' Of course x E NG(N). Let T be a Sylow r-subgroup of G containing' N.
if
N;;J T or there exists
if-
By 9. I, either
N such that NZ ~ NG(N), z ET.
If N <J T then since G = ST, any G- conjugate of x E Z (S) is a G T-conjugate :nd so (x ) ~ NG(N) C G, a contradiction. If NG(N) ~
if
if-
N, z ET, then since Q~ u •.• u Q~ ~ N, there
exists s E S such that Z(R)s:=!'z and Z(R)s Z(R)s E Qi' i> l.
Then Ni
~-NC~N) ~
t
N.
Suppose
x) and;o NG(N)
~
Z(R)s.
Now
NN is an r-group normalized b: x and generated by Q1'···' QZ' Qi' i a contradiction. This completes' e proof. jI
33
10.
A GENERALIZATION OF THE FITTING SUBGROUP
Definition. X.
Let X be any group, F(X), the Fitting subgroup of
We define F*(X)
socle (F(X). CX(F(X)) mod F(X)).
=:
Define E(X) to
be the terminal member of the derived series of F*(X). It is easy to see that F(X)Cx(F(X)) /F(X) has no solvable normal
subgroup.
For then we could choose a p-group P;, CX(F(X)) such that
PF(X)/F(X) is minimal normal in X/F(X) and then PF(X) is a nilpotent normal subgroup of X.
Thus F(X)CX(F(X))/F(X) has no solvable normal
subgroup and its socle is a direct product of non-abelian simple groups. It is easy to see that F*(X)
=:
F(X)E(X) and that CX(F*(X));, F*(X).
Since this is actually the most important property of the group F*(X) being easily true when X is solvable - we verify this in the following Lemma 10. 1.
(a) F*(X)
(b) [F(X), E (X)]
=:
F(X)E(X).
=:
1.
(c) CX(F*(X)) ~ F*(X). Proof.
(a) is clear.
(b) F(X)CX(F(X))/CX(F(X)), being a homomorphic image of a nilpotent group is solvable.
Thus E(X)
(F*(X))
=:
00
;, (F(X)CX(F(X)))
00
;,
CX(F(X)). (c) Suppose CX(F*(X)) ~ F*(X). Then C (F*(X))F(X)/F(X);1 X = Z(F*(X)) ;, F(X). Thus
CX(F(X))F(X)/F(X) and CX(F*(X)) n F*(X) Cx(F*(X))F(X)/F(X) n F*(X)/F(X)
=:
1.
We see therefore that there exists
a minimal normal subgroup of F(X)Cx(F(X))/'F(X) which avoids F*(X) /F(X) and this is clearly impossible since F*(X)
= socle
This completes the proof. /I
(F(X)CX(F(X)) mod F(X)).
Since by 7.1(b), [F(X), E(X)]
=:
1, we have F(X) n E(X) ;, Z(E(X)).
Also E(X)/F(X) n E(X) '" F*(X)/F(X), a direct product of non-abelian simple groups.
Thus Z(E(X))
= F(X)
n E(X).
Now E(X)/Z(E(X)) is a
direct product of non-abelian simple groups S.IZ(E(X)), 1 Define E.
1
=:
S~oo). 1
1
n.
The groups E. are quasi-simple - thatls E.1Z(E.l 1
is a non-abelian simple group.
1
They are called the components of X.
We will frequently write E. for E./Z(E.). 1
34
< i< -
1
1
1
Lemma 10. 2.
(a) [E., E.]:::: 1 if i J
1
oF
j.
(b) E(X):::: El ... En' Proof.
(a) [E., E.]
C
J::::
1
Z(E(X)) since E(X)/Z(E(X)) is a direct
product of groups E.Z(E(X))/Z(E(X)). 1
[E., E., E.] :::: 1 if i J
1
oF
Thus
j.
1
It follows by the Three Subgroups Lemma that [E., E., E.]:::: 1. 1
J
1
Since
E. is perfect we have [E., E.] :::: 1. 1
J
1
(b) Clearly E(X):::: E ... E Z(E(X)). 1 n Hence E(X):::: E(X)' :::: (E ... E )' :::: E ... E. 1 n 1 n the proof. 1/
This completes
The most remarkable property of the groups E. is contained in 1
the next Lemma.
We see there that any E.-invariant solvable subgroups 1
of X(!) must be actually centralized by E .. 1
Lemma 10. 3.
(a) CX(E mod Z(E(X))) :::: CX(E ) for any comi i
ponent E. of X.
---
1
(b) Any E.-invariant solvable subgroup S of X is centralized --
by
1
-
Er Proof.
(a) Let x
E
Then [E , x] ~ Z(E(X)) i
CX(E mod Z(E(X))). i
and so [E., x, E.] :::: 1. 1
1
The Three Subgroups Lemma gives [E., E., x]:::: 1 and the 1
perfectness of E. shows that [E., x] :::: 1. 1
CX(E/
1
1
Thus CX(E mod Z(E(X))) ;, i
The other containment is obvious. (b) Let S be a solvable E.-invariant subgroup of X. 1
Then
[E , S];, E(X) n S. Now E(X) n S is a solvable E(invariant subgroup i of E(X). Consider the image of E(X) n S in E(X)/Z(E(X)), a direct product of non-abelian simple groups Ex ... xE. Consider the pro1 n jection maps 11. : E(X)/Z(E(X)) -E .. Then 11.(E(X) nS)Z(E(X))/Z(E(X))) J
certainly commutes with E., if j 1
J
oF
i.
J
On the other hand 1T.(E(X) n S)Z(E(X))/Z(E(X))) is a solvable 1
normal subgroup of E., a non-abelian simple group. 1
(E(X) n S)Z(E(X))/Z(E(X)) centralizes with E. modulo Z(E(X)). 1
Thus
E.1 and so E(X) n S commutes
By 7. 3(a), E(X) n S actually centralizes E .• 1
35
Thus [E., S, E.] 1
1
that [E., S]
1.
=:
1
=:
1.
The Three Subgroups Lemma with E;
1
=:
E. shows 1
This completes the proof. ;/
Considerable interest will be attached to the subnormal subgroups of F*(X).
The following Lemma indicates part of their structure.
Lemma 10.4. (a) F(S) (b) S
=:
Suppose that S <J <J F*(X).
F(X) n S
(F(X) n S)(E(X) n S)
=:
Proof.
First F(X) r S
Then by induction F(S) ~
F(S)
F(X) n S.
Then
C
=:
~
F(S)E(S).
=:
F(S).
Let S
F(S.) for all i 1
=:
S <J ... <J S
=:
o
n
=:
F*(X).
1, 2, ... , n and so
This proves (a).
Hence S/F(S)
S/F(X) n S
=:
SF(X) /F(X) is a subnormal subgroup
=:
of F*(X)/F(X) and so is a direct product of non-abelian simple groups or the identity group. so S
=:
S(oo)F(S).
Hence S/F(S)
=:
(S/F(S)(oo)
denote the natural homomorphism modulo F(S).
S(oo)F(S)/F(S) and
=:
Then E(S) ~ S(oo) ~ F*(X)(oo) n S
= E(X)
n S.
Let-
Then we have
E(X) n S ~ C~ and E(X) n S is a normal subgroup of S, a direct product of simple groups. E(S) ;;
s(oo)~ E(X)
Therefore E(X) n S
n S;; E(S)F(S).
(E (X) n S)(F(X) n S).
It follows
C
tha~
E(S) and we have S
=:
E(S)F(S)
=:
This completes the proof. ;/
Lemma 10. 5.
Suppose S <J <J F*(X).
Then
(a) NF*(X)(S) <J<J F*(X) (b) If CF*(X)(S) ~ S, then E(S)
=:
Proof.
E(S)F(S)
Apply 10.4 and get S
=:
E(X).
Clearly E(X) ~ NX(E(X) n S) n CX(F(X) n S).
=:
(E(X) n S)(F(X) n S).
Thus E(X) ~ NX(S) and
(a) follows because F(X) is nilpotent. To verify (b) we show that E(X) 10.4.
Suppose that E
Then if E
1
=:
S and the result follows from
is a component of E(X) not contained in S.
E F(X) /F(X) etc., we have [E , S] 1
[El' S]
~
F(X) n E(X)
[E l' S]
=:
1.
36
1
~
1
=:
Z(E(X)).
=:
1.
Hence
We may apply Lemma 10. 3(a) and get
This contradicts CF*(X) (S) ~ S. ;/
Lemma 10. 6. Proof.
Suppose that A
Suppose that A: A
o
n <J A n +1 : X. If
n: 0 the result is clearly true since [A, F(X)j ~ F(X) n A ~ Z(A) and so [A, F(X)]: 1 by a familiar argument. Suppose A ~ E(A ) ~ X. Then [E(A ), F(X)] ~ F(X) n E(A ) ~ n n n Z(E(A )), Thus [E(A ), F(X)]: 1 and E(A ) C E(A +1)' /I n n n: n Lemma 10. 7.
Suppose A, B ~ X and F*(A) n B <J<J F*(A),
Then F*(A) n B ~ F*(A n B),
Further if B ~ A then F*(A) n B : F*(B).
Clearly F(A) n B ~ F(A n B).
Proof.
By 10.4,
F*(A) n B: (F*(A) n B n F(A))(F*(A) n B n E(A))
= (F(A)
n B)(E(A) n B).
Now E(A) n B ~ F*(A) n B
Thus (E(A) n B)F(A) IF (A)
is a product of components of F*(A) IF(A). Let E Then E
be some component of F*(A) contained in (E(A) n B)F(A). (00) ~ ((E(A) n B)F(A) ~ E(A) n B and El normalizes F(A n B),
I
I
a solvable subgroup of A, that E
C
=
I
Thus E
1
centralizes F(A n B).
Now if B ~ A, then F*(A) n B ~ F*(B). F(B)
It follows
F*(A n B).
= F(A)
n B ~ F*(A) n B.
Hence F*(A) n B
= F*(B).
Lemma 10. 8. E(X)
= (CG(D)
Proof.
Conversely
Since E(B) <J
/I
Let D be any subgroup of X.
Then
n E(X))[E(X), D].
First [E(X), D] ~ E(X).
E(X) not contained in [E(X), D].
Let El be a component of
We show that El ~ CG(D).
[El' D] ~ [E(X), D], we have [El' D] ~ CG(E/ and, by the Three Subgroups Lemma, [E , D] I
= 1.
Since
Thus [El' D, E I ] : l This completes the
proof. /I
37
GROUPS WITH ABELIAN SYLOW 2-SUBGROUPS
H.
J. H. Waiter obtained a characterization of finite groups with abelian Sylow 2-subgroups in [22], [23].
Bender offers a novel approach
in [5]. In this section we commence this classification by Bender of groups with abelian Sylow 2-subgroups.
This depends on the character-
ization also due to Bender of groups which have a strongly embedded subgroup. ,
There are many equivalent formulations of this concept.
The
following definition suffices for our purposes here. Definition H. 1.
A subgroup H of even order of a group G is X
strongly embedded in G if H *- G and H n H
has odd order for all
xEG-H.
In [4], Bender characterized all finite groups which have a strongly embedded subgroup. If G is a non-abelian simple group and has a n 2n+l n strongly embedded subgroup, then G ~ SL(2, 2 ), Sz(2 ), U (2 ) for 3 n suitable n. Here SL(2, 2 ) denotes the group of all 2 x 2 matrices n of determinant 1 with coefficients from the field of 2 elements. The groups SZ(22n+l), U (2 n) denote the Suzuki simple groups, see [20], 3
and the projective special unitary 3 dimensional group over a field of 2n 2 elements, respectively. We are here interested in groups with abelian Sylow 2-subgroups. n
Of the above three classes of groups, only the groups SL(2, 2 ) have abelian Sylow 2-subgroups.
= L 2 (2 n )
The known groups which have abelian
Sylow 2-subgroups in fact include just three more classes of groups.
The
projective special linear 2-dimensional groups L (q), where q is an 2
odd prime power such that q;: ±3(mod 8), all have Sylow 2-subgroups The simple group Ji of 2n+l Janko [16] and the Ree SImple groups G (q), where q = 3 ,
which are elementary abelian of order 4. .
1
2
n
~
1, [19] all have elementary abelian Sylow 2-subgroups of order 8. It is still unresolved whether this completes the list of groups
with abelian Sylow 2-subgroups.
The groups of Janko and Ree are clearly
the most intriguing groups on the above list.
They have a single class
of involutions and if t is one such, C (t) is isomorphic to x E where E
~
L (q) and q is odd. Of course, it follows that q;: ±3(mod 8) since' 2
38
otherwise a Sylow 2-subgroup would be non-abelian.
We proceed to
define a JR (Janko-Ree) group.
n. 2.
Definition
A simple group G with abelian Sylow 2-sub-
groups is called a JR-group if G contains an involution t such that CG(t)
= (0
x E, where L (q) ;; E ;; PI'L(2, q) and q is odd. 2
It follows that [E : L (q)] is odd since otherwise a Sylow 2-sub2
group is non-abelian. It is easy to see that a JR-group has a single 2n class of involutions and so the groups J , G1(q), q = 3 n> 1, are 1
+\
2
groups of type JR.
=
It was shown by Walter [22] that in a group of type JR, the group
E which occurs as the business end of the centralizer of an involution in fact must be isomorphic to L (q). 2
volution t such that CG(t) first has q n
~
=5
= (0
in which case G
Now a simple group G with an in-
x E where E
=J1
= L 2 (q),
q == ±3(mod 8) 2n 1 by [16], or has q = 3 +
1, by [17]. In an early paper, H. N. Ward [24] showed that the
character table of G is determined and Janko and Thompson [17], showed that the 3-Sylow normalizer of G has a uniquely determined structure.
Further results have been obtained by Thompson [21], but
the final determination of the multiplication table of G still eludes us. 1 Thus either a simple group of type JR is J or G (q) or a new simple 1
2
1
group with the same character table (and so order) as G (q) and with 2
very similar structure. Definition
n. 3.
A group G with an abelian Sylow 2-subgroup is
said to be an A*-group, if G has a normal series 1;; N;; M ;; G where N and GM are of odd order and M;N is a direct product of a 2-group and simple groups of type L (q) or JR. 2
Remark.
that of Bender [5].
Our definition of a JR-group differs only slightly from His definition requires (and his proof uses crucially)
the fact that the centralizer of an involution of a simple group G of type JR is a maximal subgroup of G.
This follows from the Definition
n. 2,
as will be proved in 12. 2 below. We state the main theorem of Bender [5] here.
39
Theorem A. 2-subgroup.
Let G be a finite group with an abelian Sylow
Then G is an A*-group.
Remark.
If a group G has an abelian Sylow 2-subgroup T of
rank 1, then G is solvable and 2-nilpotent and clearly an A*-group. If a a T has rank 2, then G is 2-nilpotent unless T is of type (2 , 2 ). But then if a> 1, G is solvable by [7] and clearly an A*-group since it has 2-length L
Thus we may assume that
the results of [13].
IT I =
4 and then we can apply
Again G is an A*-group.
Hence the main thrust of Theorem A is directed at the case of a finite group G with an abelian Sylow 2-subgroup of rank at least 3. We will follow here closely the direction of Bender's proof, expanding the abbreviation portions where necessary.
Minor changes in
the presentation will be made, but the proof itself is so delicately woven and intricate that its beauty would suffer if the changes were too great. For generalizations of the techniques of the following sections, the reader should consult Goldschmidt's 'strongly closed abelian' paper [11].
12.
PRELIMINARY LEMMAS Lemma 12. 1 (Thompson).
Sylow 2-subgroup S. r(R) = r(S) - L
Let G be a group with an abelian
Let R be a subgroup of S such that
If G = 02(G), then any involution t E S is conjugate
in NG(S) to an element of R. Proof.
Of course, NG(S) controls fusion of elements of S by
Burnside's lemma. Consider the transfer V: G -+ S. Let Xl' ... , Xk ' ~+ 1 t , ... , x n ' X t be a system of coset n -1 representatives of S in G chosen so that x.tx. E S for i = 1, ... , k, -1
xitX i
1
(S for i) k.
1
Since [G: S] is odd, k is odd. k
Clearly V(t) = IT x.t£ 1. i=l 1 1 -1 1 h -1 -1 t tx If Xi Xi ' Xi+1 l+1 (R, t en xitX i Xi+1tXi+1 ER, since 1 r(R) = r(S) - L Thus V(t) '" X tx- (mod R). Since 02(G) = 1, V(t) = 1 -1
and so X tx 1
40
1
1
ER.;f
1
Lemma 12. 2. 2-subgroup S.
Let G be a simple group with an abelian Sylow
=
Suppose t E S is an involution such that CG(t)
where L (q) ;; E ;; prL(2, q).
Then CG(t) is a maximal subgroup of G.
2
Suppose CG(t) ~ M
Proof. 2
= M,
If 0 (M)
x E
C
G.
then by 12. 1, t is conjugate in M to an element
Since CG(S) ~ CG(t) ~ M, and NM(S)/CM(S) acts transitively
of E.
on the non-trivial elements of M, we have NM(S) = NG(S) ;; M. Also = t m , m E M and so CG(s) = CG(t m ) ~ M.
for all involutions s EM, s X
Suppose M n M
contains an involution s, where x E G - M.
-1
-1
E M and so there exists y E M such that
SX
x- y E CG(s) ~ M, x E M, a contradiction. 1
y
SX
= s.
Then s,
But then
Hence M is strongly em-
bedded in G. n But by [4], G ~ L (2 ) since the Sylow 2-subgroups of G are 2 n abelian. But the centralizer of any involution in L (2 ) is an abelian 2
2-group.
E
This contradicts our assumption that C(t)
= L 2 (q).
= (t)
::J
2
If 0 (M) C M then let K
= 0 2 (M).
1
2
have odd order and K /K
course
E';; K 2 •
1
2
~
1
2
K
=5
or r
(s) /C 1
K
= (t)
I
then r If K
2
"*
=q
k
for some k.
Thus r -
(s) is a dihedral group of Since s is conjugate to t
q + 1. If q
E ~
=5
= 5, while
then r
= q. 1, then for some involution s
since there exists a four-subgroup of K
1
E
K
1
we have C
normalizing K. 2
K
with a dihedral factor group of order q -
=
x E.
E.
2
=1
1
But then K
(s) 2
This situation does not
Note that any field automorphism of PSL(2, q)
acts non-trivially on a dihedral subgroup of order q Thus K
"*
(s) 2
CG(s) contains a section which has a normal odd order subgroup C
arise in CG(s)
Of
Let s
x E, we see that any dihedral section of CG(s) has
order at most q + 1. if q r
1, K/K1'
2
order r - E where r "" E(mod 4), E = ±1. and CG(t)
;;
2
Since
L (r) for some prime power r.
Hence either q
be an involution in K . Then C
= 2.
Clearly [M: K]
a Sylow 2-subgroup of K;; E is of order 4, K;; K ;; K K
x E where
and K
~
aut L (q). 2
E.
Since clearly t
E
C
M
(K ) ~ M 1
41
we must have t
E
This completes the proof. 11
Z(M), M = CG(t).
The following lemma is not required for some time in the proof of Theorem A. It is of some independent interest, affording as it does the first glimpse of the importance of the F*-subgroup.
Lemma 12. 3.
Let G be a group with abelian Sylow 2-subgroups,
p a prime, P some p-subgroup of G. --
Assume that r(O (G)) p
Then P ~ 0p(G).
that rp, OP(F*(G))) = 1.
Let C = CG(oP(F*(G))) ~ P.
Proof.
by 10.6, F*(C) = F*(G) n C. rp, OP(F*(C))] = 1.
If C
C
< =
2 and
-
G then since C:::J G,
Because rp, oP(F*(G))] = 1,
For F*(G) n C = (F(G) n C) (E(G) n C) and so
OP(F*(G) n C) = OP(F(G) n C) OP(E(G) n C).
By induction P ;;Op(C);;Op(G).
Thus C = G, E(G) = 1, F*(G) = F(G). Now let Z =
Q
1
(Z(F(G) )) for some q"* p, and let G = G/Z. q
First F(G) = F(G) because Z
° (G) = o::TGl p
p
C Z(G). =
and Z is a p f - group.
Let E{G) =E.
Since [E
(00)
= F(G).
= °p (G).
by induction pc
p
<2
because
=
00
,F(G), E ]=1 we have [E
and because F*(G) = F(G), E(oo) = 1. F*(G) = F(G) = F*(G)
Also r(O (G))
Thus E = 1.
(00)
,F(G)]=l
Hence
It follows that rp, OP(F*(G))] = 1 and
Since Z
C
~
Z(G), P cO (G).
=
P
Thus Z = 1 and F(G) is a p-group containing CG(F(G)).
Let
Q be a Thompson critical subgroup of F(G). If p = 2, then P
~
C(F(G))
~
F(G) since the Sylow 2-subgroups of G are abelian. If
p "* 2, we may choose R = Since r(R)::; 2, IR I ::; r
3
Q
1
(Q) and then CG(R) is a p-group by 2.4.
and if R = R/(R), IR I ~ p2.
Still CG(R)
is a p-group and G/CG(R);; GL(2, p), with abelian Sylow 2-subgroups. Every such subgroup has a normal Sylow p-subgroup. PCG(R)/CG(R) ~ 0p(G/CG(R).
Thus
Hence P ~ 0p(G). 11
The following Theorem is, it seems to me, unique in finite group theory.
A theorem about the structure of a general finite simple group
with only a very minor restriction on its subgroups! It is followed by an extension very reminiscent of the uniqueness theorems of Chapter 5. It is true to say that these next two Theorems are the very cornerstone of the whole proof.
42
Theorem B (Bender).
Let A and B be a distinct maximal sub-
group of a simple group G such that F*(A)
~
B and F*(B)
~
A.
Then
F*(A) and F*(B) are p-groups for the same prime p. Proof.
Since F(B) is an E(A)-invariant solvable subgroup of A,
F(B) ~ CA (E(A)) by 10.3. By 10.8, E(A) = CE(A)(E(B))[E(A), E(B)]. Now CE(A)(F*(B)) = CE(A)(E(B)) ~ F*(B). Thus E(A) ~ F*(B) and so E(A) ~ F*(B)(oo) = E(B). By symmetry E(B) ~ E(A), Since A '" B, E(A) = E(B) = 1. Clearly subgroups of coprime orders of F(A) and F(B) centralize each other. p
E:
It follows that 1T(F(A)) = 1T(F(B)).
7T(F(A)) - 7T(F(B)), then [F(B), 0p(A)] = 1.
we have a contradiction. Let p
7T(F(A)) = 1T(F(B)).
E:
Let P
For otherwise if Since F(B);; CG(F(B)),
= °p (A),
Q =
First [E(R), F(R)] = 1 and [Op(A) , E(R)] = 1. centralizes F(R)P = F(A) = F*(A) ;; CG(F*(A)). F*(R) = F(R).
Now [Q, F*(R)] = [Q, F(R)]
= [Q,
°p (B), R=Op ,(A). Thus E(R)
Hence E(R) = 1 and F(A) n R] ~ Q n R = 1.
Since Q centralizes F(R);; CR(F(R)), by 2.2, [Q, R] = 1.
rp,
°
p
,(B)]
Similarly
= 1.
Now R ~ CG(Q)
P ;; CB(oP(F(B))) ~ B. CB(F(B)) ;; F(B).
;!
B.
Thus
Also F(B)
rp,
= W(F(B))
and
CB(Q)] ~ CB(oP(F(B)) n CB(Q) =
Because PF(B) = PQ x oP(F(B)), since subgroups of
coprime orders of F(A) and F(B) commute, we have PQ is normalized by CB(Q).
But now
[PQ, CB(Q)] ~
rp,
CB(Q)] ~ F(B) n PQ = Q.
Thus [PQ, CB(Q), CB(Q)] = 1.
By 0.2 we have [PQ,Op(C (Q))]=1. B Hence 0 ,(A) cO ,(Op(CB(Q))) cO ,(B). P = p = P By symmetry, 0 ,(B) cO ,(A).
°P,(B)== 1 p
Hence 0 ,(A) = p
p
and 7T(F(A)) = 1T(F(B)) = {p L
This completes the proof of Theorem B. 1/ Theorem 12.4.
Let A be a maximal subgroup of a simple group
G, S a subnormal subgroup of F*(A) such that CF*(A)(S);; S.
Let
43
B
~
G be a subgroup of G containing 8. (a) 0 (B) n A = 1 for q
q
-
E
Then
1T(F(A))'.
(b) [0 (B), OP(F*(A))] = 1, if P E 1T(F(A)). P Moreover, if B is a maximal subgroup of G and if either !1T(F*(A))!
> 2 ==
11T(F*(B)) I > 2, then each of the following ensures
or
=
-
that A = B.
(c) A contains a subnormal subgroup S of F*(B) such that CF*(B)(S) ~ S. (d) B is an A*=group, IE(B)
--
q
E
1<= IE(A) I
---
and 0 (B) = 1 for all -q --
1T(F(A))'. (e) A and B are conjugate A*-groups. Remark.
in (d) and (e).
The structure of A*-groups is only minimally required
In fact the relevant fact required of the subgroups A and
Y ~ Z ~ 1 where X/Y, Z are solvable and Y /Z is a direct product of non-abelian simple groups. B is that they have the following structure: X
~
Thus, roughly speaking, this Theorem holds provided the relevant subgroups do not involve groups of type A Proof.
By 10. 5(b), 8
~
E(A).
and since C *(A)(8) ~ 8, Z(F(A)) ~ 8. F (i)
5
wr A . 5
Also by 10.4, 8 = (F(A) n 8)E(A), Thus if P
E
1T(F(A)), P E1T(F(8)).
[0 (B) n A, OP(8)] = 1 for all primes p. p
For 0 (B) n A is an E(A)-invariant solvable subgroup of A. P
Thus
[0 (B) n A, E(A)] = 1 P and [0 (B) n A, OP(8)] = [0 (B) P P (H)
CG(O (8)) q
noq (A)
r,
A, F(8) ,] C 0 (B) n A n F(A) , = l. P = P P
cO (8) for q = q
E
1T(F(A)), q
7'
p.
For CG(O (8)) no (A) centralizes F(8) , C F(A) , and E(A). q q q = q Thus C (Oq(8)) n 0q(A) ~ C(8) n F*(A) ~ 8. It follows that G CG(O (8)) n 0 (A) C 0 (8) = 0 (A) n 8. q q = q q (Hi)
44
[Op(B) n A, 0q(A)] = 1 for q
E
1T(F(A)) - p.
This follows from (i) and (ii) using 2. 2. Thus if P
~
17{F{A)),
°P (B) n A
centralizes F*{A) since by (iii)
0p{B) n A ~ CG{F{A)) and by (ii) 0p{B) n A ~ CG{E{A)). CG{F*{A))
~
F*{A), it follows that 0p{B) n A = 1.
Since
This completes the
proof of (a). Suppose now p E 17{F{A)). (iv)
CG{O (S)) n p
° (B) = ° (B) n A. C
P
P
For CG(Op{S)) ~ CG{Z{F{A)p)) since CG{S) n F*{A) ~ S.
Thus
CG{Op{S)) ~ A. (v)
°P(B)
oPtS) centralizes
°
For [oPtS),
p
(8))
= 1.
By (iv) and (i),
We may now apply 2.2 to the group (vi)
C(F{S) ,) n F{A) , P P
For if x
and so [E{A),
[oPtS), C{O {s))no {B)]=1.
P
° (B)O (S) p
°P(B)] = 1. P
to get the result.
p
F{S) " = P
C
CG{F{S)p') n F{A)p' then [x, optS)] ~ [x, 0p{A)]
E
=1
Clearly [x, E (A)]
and so [x, S]
= 1.
= 1.
Since CG(S) n F* (A) ~ S, we
have (vi). (vii) We may now apply 2. 2 to F(A) , using (v) and (vi). It follows R that [Op{B), F{A)p'] = 1. Hence [Op(B), OP{F*(A))] = 1. We are done with (b). Continuing, let B be a maximal subgroup of G containing S.
We
show that under condition (c), F*(A) is a p-group if and only if F*{B) is a p-group.
For if (c) holds, we are assured of a symmetrical relation-
ship between A and B. p
CG{Op(A)) ~ 0p(A)
Thus OP{F*{B)) ~ CG{Op(A)).
= F*{A),
[0 (B), OP(F*(A))]
P
F*(B) ~ A.
=1
~
it
Since
we see that F*(B) is a p-group also.
/17{F*{A)) I 2: 2 if and only if Now- OP{F*(B))
p
p f. rr(F{B)).
Now if F*(A) is a p-group, then since that p E rr{F(B)).
° (A)] = 1 for all se ° (A) n B follows = p
Thus we have [OP{F*{B)),
° (A) n B = 1 for all
p E 17{F{B)) and
Irr{F*(B)) I 2: 2.
CG{Op{A))
and so
Thus
~A
for all p Err (F(B)).
°P(B) = A C
because OP(F*{A))
Also
"*
1.
Thus
Symmetry gives F*(B);; A and we may apply Theorem B
to get AS B.
This verifies (c).
45
q
E
If B is an A*-group, by hypothesis 0 (B) == 1 for all primes q 1T(F(A)). By (b), it follows that F(B) ~ A. By 10. 6(a), since
CG(S) n F*(A)
~
S <J<J F*(A), E(A)
[F(B), E(A)] == 1.
~
S;; B.
Part (b) also shows that
Since B is an A*-group, it follows that E(A);;E(B).
But jE(A) 12: /E(B)
I.
Thus E(A) == E(B) == 1. Otherwise A == B.
may now apply (c) with S == F*(B) == F(B)
We
S A and get A == B.
(e) follows immediately from (d). 1/ The following two lemmas are easily proved and interesting in themselves.
They are also of interest because they seem not to have
been noticed in even the solvable case. Lemma 12.5.
If G is a group such that F*(G) is a p-group.
If U is a p-subgroup of G, then F*(CG(U)) and F*(NG(U)) are p-
gr oups als o. Let N == NG(U), C == CG(U).
Proof.
F(N)p' and E(N) ~ C because U ~ F(N).
Clearly F(N)p' ~ F(C)p' ;;
By 10.6 E(N) == E(C). It
follows that OP(F*(N)) == oP(F*(C)). Let x
oP(F*(N)) be a pt-element.
E
Then
[cG(u)rF*(G), x];; OP(F*(N)) n F*(G) ~ Z(F*(N)). Hence [CG(U)nF*(G), x, x) == 1.
By 0.2, [cG(u)nF*(G), x)
= 1.
Now
apply 2.2 to <x> acting on F*(G)U. It follows that [x, F*(G)) == 1 and so x
E
F*(G).
Hence x == 1.
Lemma 12. 6. that V
~
U.
Thus oP(F*(N)) == OP(F*(C)) == 1. /I
Let U, V be p-subgroups of a group G sucl!
Then if F*(CG(V)) is a p-group, F*(CG(U)) is a p-group
also. Proof.
If V ~ U, let N
on IGI gives the assume
N
= NG(V).
Clearly if Ne G, induction
res~lt since CG(U) ~ CG(V) ~ N. Hence we may
= G.
Now since F*(CG(V)) is a p-group, F*(NG(V)) is a p-group also. Apply 12.5 to NG(V) == G to get the result. Since V
46
13.
PROPERTIES OF A*-GROUPS Let X be an A*-group, t an involution in X. Lemma 13. 1.
If a 2-subgroup T of X centralizes O(X), then
it is contained in F*(X). Proof.
Since [T, O(X)]
= 1,
= [T,
[T, O(F(X))]
= 1. = 1.
F(O(X))]
But because the Sylow 2-subgroups of X are abelian, [T, F(X)]
But clearly normal subgroups and factor groups of A*-groups are A*groups. Hence F(X)CX(F(X))/F(X) is an A*-group which has no nontrivial solvable normal subgroups.
Thus [F(X)CX(F(X)): F*(X)] is odd.
Hence T;; F*(X). 11 Lemma 13.2.
If X is a simple A*-group, then X has one
class of involutions. Proof.
This is well known if X
= L 2 (q).
If X is of type JR,
it follows from 12. 1. // The next two lemmas are concerned with CX(t)-invariant subgroups of X.
We are able to locate at least part of them within F*(X). Lemma 13. 3.
Let E be a CX(t)-invariant semi-simple subgroup
of E(X).
Then any component K of E is contained in a component L
of E(X).
Moreover (i) or (ii) holds.
= L.
(i)
K
(ii)
K is of type L (q), q odd, L is of type JR and t 2
centralizes K. Proof.
Let T be a Sylow 2-subgroup of X containing t.
Since
K does not centralize a Sylow 2-subgroup T n E(X) of E(X), there exists a component L of E(X) such that [K, T n L] *- 1.
Now T permutes
the components of E(X) and centralizes a Sylow 2-subgroup of each component.
Hence T normalizes both K and L.
and if [K, T n L] ;; Z(K), [K, T n L, K]
= 1,
Thus [K, T n L]
whence [K, T n L]
~
K
= 1.
This is not the case.
47
It follows that [K, T n L]
=K~
L.
If K"* L, then CL (t) normalizes L n E ~ K.
n
If L ~ L (2 ), then t must induce an inner automorphism on L
z
since a Sylow 2-subgroup of prL(2, 2n) is non-abelian if n is even. If L ~ L (pn), p
z
f.
2, then n is odd because otherwise a Sylow 2-subgroup
of L is dihedral of order
~
8. If then t were to induce the 'transpose
inverse' automorphism on L, a Sylow 2-subgroup of L(t) would be non-abelian.
Thus t must induce an inner automorphism T on L if
L ~ L (q).
z
Since CL (T) is dihedral if L ~ L (q), q odd, and an elementary z n abelian 2-group if L ~ L (2 ), CL (T) normalizes no non-solvable sub-
z
group of L except L itself.
Here T is an involution in L.
Hence
K = L.
If L is of type JR, then t centralizes a Sylow 2-subgroup
T n L of L and so t normalizes CL (T) for T E T n L.
Now
CL(T) = (T> x E and so t normalizes E(oo) = L , a group of type l L (q), q odd. As we saw above, t must induce an inner automorphism
z
on L
and so there exists an involution u E L
1
1
such that [tu, L ]
1
= l.
If tELl ~ L, then CL (t) is a maximal subgroup of L and CL (t) ~ K.
This is the result.
Thus L (t> 1
rtu El c C (L ) n C (T)
,
=
L
1
I
= (T>.
= L1
x (tu> and
It follows that [tu, E]
=1
because
rE : I 1 is odd. 1
Now tu centralizes a Sylow 2-subgroup S of L and so tu normalizes N (S), which is modulo CL (S) a non-cyclic group of order 2l. L Let Z /Cl(S) have order 7. Let M = N (S)(tu>. Since L tu E CM(S) ~ M, [tu, Z] ~ CM(S) r Z ~ CL (S). Thus tu centralizes Z modulo CL (S).
It follows that either tu centralizes Z or tu acts on
Z like the unique involution T E S which is centralized by an element of order 3 in N (S) n CL (T). Thus either [tu, Z] = 1 or [tu T , Z] = 1 and L so either tu or tUT centralizes NL(S) CL(T), a maximal subgroup of
1
L by 12. 2.
Since UT is an involution, we lose no generality by assuming
that [tu, L]
=1
subgroup El'
and so CL (t)
= CL (u) = (u>
El for some suitable
By 12.2, CL(u) is a maximal subgroup of L,
CL (t) = CL (u) normalizes L nE;; K. K is of type L (q). I!
z
48
X
Thus K
~
But
Cl(u) = CL (t) and so
Lemma 13.4. that U = F*(U).
Let U be a CX(t)-invariant subgroup of X such
Then [t, U] = V <] <J F*(X).
Proof.
We have already seen that subnormal subgroups V of
F*(X) satisfy V = F*(V) = F(V)E(V).
If we could show that
[t, E(V)] <]<] F*(X), [t, F(V)] <] <J F*(X), then it would follow that [t, V] = [t, F(V)][t, E(V)] <]<] F*(X).
For [t, E(V)]~ F*(X) if
[t, E(V)] <]<] F*(X). Thus we assume first that E(U) = U = [t, U]. [U, F(X)] = 1.
Let D = CX(U) n F(X).
We show first that
Then [CF(X)(t), U] ~ U n F(X),
a nilpotent normal subgroup of U and so [U, Cx(t) n F(X)) ~ Z(U). usual, [U, Cx(t) n F(X)] = 1 and so Cx(t) r, F(X) ~ D. (NX(D)nF(X));D.
As
Thus t inverts
Note that a Sylow 2-subgroup of F(X) lies in
CX(t) n F(X) ~ D. Now U = [t, U] centralizes NX(D) n F(X);D.
Arguing for each
Sylow p-subgroup of the nilpotent group NX(D) n F(X) separately, it follows that U centralizes NX(D) n F(X).
[U,
F(X)]
Thus D = F(X) and
= 1.
Now any Sylow 2-subgroup of U centralizes F(O(X)) ~ F(X). Since CO(X)(F(O(X))) ~ F(O(X)), by 2.2 it follows that any Sylow 2-subgroup of U centralizes O(X).
By 12.1 it follows that any Sylow 2-sub-
group of U lies in F*(X) and so U;; F*(X). 13.3(a), U = [t, U] <]<] E(X).
Thus U ~ E(X).
By
Note that 13.3(b) is not applicable here
since if a component K of U is not a component of E (X), then [K, t] = 1. But U = [t, U] is just the product of components not centralized by t. Now assume that U = [t, U] is a p-group where p is an odd prime and that X is a minimal counter example. First if 0p(X)
"*
1, let X = X/Op(X).
By induction
[t,
U]<]
But any subnormal p-subgroup of F*(G) is contained in 0 (G) for any p group G. Thus [t,U] ~ Op(X) = 1 and so [t, U] ~ 0p(X). Thus
°
p (X)
= 1. Again U centralizes F(X) for otherwise, Cx(t) n F(X) ~
CX(U) n F(X) = D and U centralizes Nx(D) n F(X);D.
Thus
[U, F(O(X))] = 1 and so [U, O(X)] = 1 since (IUI, IF(O(X))I)= 1. For [F(O(X)), U, O(X)) = 1, [O(X), F(O(X)), U] = 1 and by the Three Sub-
49
= 1.
groups Lemma, we have [V, O(X), F(O(X))] Hence [V, O(X)] Now if w w) u 1= w
O(X),
E
= wf IV I,
If O(X)
* 2.
* 1,
Thus [t, V]
diction to 0 (X) p
CX(F(O(X)) ;; F(O(X)).
U E
V, w
U
= wf,
f
E
F(O(X)), and so
a contradiction. let X
and so [f, D] ~ F(X). p
~
= X/O(X).
Then by induction [t, D]<J<J F*(X)
Then [t, V]O(X) ~ 0p(X mod O(X))
=V
~
= O(X)
because
Since V ~ Z(O(X)) we have a contra-
O(X).
= 1.
Now X is an A*-group such that O(X) = 1. It follows that t
E
F*(X).
If F*(X)
X
= F*(X).
C
V]
<::
F*(X).
X, then V <J<J F*(F*(X))
= F*(X),
by induction.
Since X
= F*(X),
N is a simple group.
By induction
But the only subnormal subgroups of X are simple groups
or 2-groups (or products of them) and since V is a p-group, V X is not simple there exists N <J X such that N n N Vc N
=
1
nN
Hence
If X is not simple, let N:sJ X be a minimal normal sub-
group of X. VN <J<J X.
= [t,
Hence V
1
= 1.
=
1
=1
~
N.
If
and then
Thus X is a simple A*-group. n Thus either X ~ 1)2 ) or CX(t) is a maximal subgroup of X.
In the first case, the only odd order group normalized by CX(t) is 1, while in the second case V ~ Cx(t).
Since V
= [V,
t], V
= 1.
This
completes the proof. 1/ Lemma 13. 5. V = OF(CX(t)).
Let T be a Sylow 2-subgroup of CX(t) and
Then V ~ O*(X) = F*(X mod O(X)) and CV(T) ~ O(X).
Further for any component E of E(X), the following hold. (a)
V normalizes E;
(b)
V /CV(E) is cyclic;
(c)
If E is of type L (2 ) or JR, then V centralizes E.
n
-
Proof. (i)
2
Induction on
O(X)
IX I.
= 1.
Suppose O(X)
* 1.
Let
X = X/O(X), C = Cx(t). Then cx(f) = C,
= OF(C) ~ OF(C) = W. D ~ w ~ O*(X) = O*(X) and so
F(X) ~ F(X) and so D By induction
50
V ~ O*(X).
Also by
induction, CW
Hence Cu(T);; O(X).
Moreover, if E is a component of E(X), E is a component of E(X).
For E normalizes the solvable subgroup F(X mod O(X)) and so
E centralizes F(X).
Since
E <J<J X,
and so E is a component of E(X). and so W normalizes EO(X).
Thus W normalizes E by induction
Of course [E, O(X)] ::::: 1 because O(X)
is solvable and E-invariant and so E Now put D::::: Cw(1~).
~
W normalizes (EO(X»)(oo)::::: E.
By induction W ID is cyclic and
U/CU(E) ~ UCW(E)/CW(E);; W/CW(E). [D, E);; O(X).
E;; F*(X),
it easily follows that
[D, E)::::: 1 and so
But
Hence [E, D, E)::::: 1 and [E, D)::::: 1.
Thus D:::::CW(E).
Therefore U/CU(E) is cyclic. Again (c) holds since by induction [W, E]::::: 1 and then [W, E) ::::: 1 as usuaL (ii)
Thus [V, E]::::: 1.
F*(X)::::: O*(X)
~
T where T is a Sylow 2-subgroup of X.
This is clear because X is an A*- group and O(X)::::: 1. (iii)
U normalizes E and so (a) holds.
For T normalizes each component of E(X) and centralizes a Sylow 2-subgroup T n E(X) of E(X).
Also CUrT) permutes the com-
ponents of E(X) and centralizes T n E(X). But U::::: CU(T)[T, U] and by (ii) T [T, V] ;; F*(X) (iv)
'2;
Thus CUrT) normalizes E. ~
F*(X).
Thus
E. It follows that U normalizes E.
X::::: F*(X)U.
Let Y::::: F*(X)U and suppose Y and so U::::: OF(CX(t))
~
Y.
Thus U
~
C
X.
Then T;; Cy(t);; Cx(t)
OF(Cy(t».
By induction
V ;; O*(Y) and CU(T);; O(y). Now [O(Y), F*(X)]::::: [O(X), F(X)] since [E(X), O(Y)] ::::: 1 as usual.
Also [O(y), F(X)];; O(Y) n F(X) 2
group.
Thus O(Y) ~ CX(F*(X) ;; F*(X) and O(Y);1 F*(X).
absurd.
:::::
1 because F(X) is a 2-
Hence O(Y)::::: 1, U ;; F*(Y) ::::: F*(X), CUrT) ::::: 1.
(c) clearly hold in Y and so in X. (v)
This is
Parts (b),
Thus Y::::: X.
X::::: EU, where E is a non-abelian simple group.
First if F*(X)::::: F(X) is a 2-group, then X::::: F(X)U ::::: F(X) x V
51
since [U, F(X)];; F(X) nu == 1. Let F*(X) == EE Y == EUT.
1
This contradicts O(X) == 1.
where E is a component of E(X), and let
Note U normalizes E by (a).
ft, O(Y)] == 1 because ft, U] == 1. nilpotent.
First O(Y) n E == 1 and so
Since Y;ET is nilpotent, O(Y) is
Finally Cy(t) == UT C (t) ;; Cx(t). E x O(Y).
Thus U;; OF(Cy(t)).
Since [E, O(Y)] == 1, EO(Y) == E
By induction if ye X, U;; O*(Y), CU(T) == 1 and (b), (c) hold. O*(Y) == ETO(Y) clearly and so U;; ECU(E). Arguing similarly on Y
1
== E UT we have U 1
U;;EICU(El)nECU(E)==EElCU(EE/
E CU(E). Thus == 1 1 Hence U;;F*(X) andthe C
theorem is true. We may assume that X == EUT. E.
Let T be the projection of t on
If T == 1, then It, E] == 1 and so t E Z(X).
U ;; O(Cx(t)). Thus Y == EU
T"* 1:
But O(X) == 1 and
Since [U, T] == 1, Cx(t) == CX(T).
If
X, we may induct to Y, T in place of X, t and get
C
U;; O*(Y) == E;; F*(X) and CU(T nE) == 1.
Since CU(T nE)==CU(T),
we have (v). (vi)
E is of type L (q), q odd. 2
n
If E is of type L (2 ), then CE(t) == T and [U, CE(t)];;U nE 2
and so [U, CE(t)] ;; O(CE(t)) == 1. Let
U
= U/CU(E) ;; PI'L(2,
n 2 ) and
[D,
T] == 1.
At least one of
the elements of T is moved by any field automorphism and so induce inner automorphisms on E. U == CU(E).
Since CE(T)
= T,
U ==
L (q) ;; H;; PIL(2, q). 2
1 and
x
H where
Therefore O(CE(t)) == 1.
Since
[U, C (t)] ~ CE(t) nu we must have [U, CE(t)] == [U, T] == 1. s "* t.
where L CX(s)
C
is a conjugate of H.
Clearly CX(s) is an A*-group and
X.
= 1,
U == CU(T).
by induction U == CU(T) ;; O(CX(s)). Thus U centralizes (CE(t), CE(s»
== 1.
52
Let
Note U;; CX(s) and so CX(s) == CE(s)U == (s > x L)U
Since [T, U]
O(X)
must
This contradicts O(X) == 1 since EU == X == E x U.
If E is of type JR, then CE(t) == (t>
1 sET ,
U
Now U;; OF(CX(t) n CX(s)) and Then [U, CE(s)];; O(CE(s))
==
E by 12.2.
This completes the proof of (vi).
= 1.
Again this contradicts
(vii)
The final contradiction.
By the above results X = EU, E CE(t) is dihedral of order q -
E,
~
1 (q), q odd.
Therefore
Z
where q '" (mod 4),
E
= ±1, T is of
type (2, 2). If U;: E we are done because E = O*(X), CU(T) = U n CE(T) = T nu = 1 and U = O(CE(t)) is cyclic. As U = CU(T)[U, T] and [U, T];: E, it follows that CU(T)"* 1. Let u E CU(T) have prime order p.
Let q = rn, where r is a prime.
First u must induce a field automorphism on E, since CE(T) = T. CE(u)
~
Thus
m
L (r ) where m = nip.
z
Let CE(t) = rXx> where D is cyclic and x inverts D.
Then
[u, 0 ,(D)] c 0 (CX(t)) no ,(D) = 1. Thus u centralizes 0 ,(D), T. p = P P P On the other hand, u normalizes 0 (D), a cyclic p-group. Thus p
pIC(u) n 0p(D) I ~ IOp(D)!. Thus [CE(t) : CE(t) n CE(u)] ~ p. Now m m ICE(t) n CE(u) I = r ± 1 and so rmp ± 1 $, p(r ± 1). This inequality is not solvable with r
~
3, P ~ 3.
Lemma 13. 5 is completely proved. 11
Lemma 13.6. If U is an abelian subgroup of X of type (2, 2) n and E is a component of E(X) not of type L (2 ), then
-
#
z
--
E = (CE(u) : u EU >. Proof.
Each u E U
inner automorphism on E.
#
normalizes E by 13. 5 and induces an
Further if u ,u 1
Z
E U#, u "* u , then u , 1
Z
1
induce different inner automorphisms on E. But if E is not of type z n L (2 ), then the centralizers of any two distinct involutions of E are
U
z
distinct maximal subgroups of E.
Hence the result. /1
14. PROOF OF THE THEOREM A, PART I Let G be henceforth a minimal counter example to Theorem A. We show as an initial reduction that G is a non-abelian simple group all of whose proper subgroups are A*-groups. Thus let N be a minimal normal subgroup of G.
Clearly O(G)=l
and OZ'(G) = G because otherwise G is immediately an A*-group.
Thus
IN I is even.
53
If N is a 2-group, then G
order. O(G/N)
Thus 1Nl
= 2.
=1
2'
and 0 (G)
= CG(N)
because G/CG(N) has odd
Now GIN is an A*-group by induction.
= G,
it follows that G/N
= SIN
Since
xl /NX ... xLkIN,
where SIN is a 2-group and L' IN are simple A*-groups for 1 i
= 1,
2, ... , k.
Because the Sylow 2-subgroups of G are abelian, it follows by an elementary transfer argument, see [12], that G' n N = 1 and so G' n S = 1. G'
i~
Clearly G' is a perfect A*-group with O(G')
a direct product of simple A*-groups.
=1
Since S ~ G, G
and so
= G'
x S
and G is an A*-group, a contradiction. Thus we may assume that N is a direct product of isomorphic simple A*-groups L., N = LX ... x L . 1 1 k Clearly CG(N) ~ G, CG(N) n N = 1 and so CG(N) is an A*-group which has no non-trivial solvable normal subgroups. NCG(N) of N.
= NXCG(N)
Thus
and GINCG(N) acts as a group of automorphisms
Since a Sylow 2-subgroup T of G is abelian and T permutes
the components L 1 , ... , 1.. of N centralizing T n L., a Sylow 2--k 1 subgroup of L. for all i = 1, ... , k, it follows that T normalizes each 1
component. We show that a simple A*-group has no non-trivial outer automorphism x of order 2 which centralizes a Sylow 2-subgroup. For simplicity write L L (pn).
= L1
and suppose that L is of type
Clearly x must induce a field automorphism on L since a
2
Sylow 2-subgroup of PGL(2, q) is dihedral and non-abelian if q is odd, and so n
= 2m
is even.
But then a Sylow 2-subgroup of L (p2m) is 2
non-abelian. Thus we may assume that L is of type JR.
Let S be a Sylow
2-subgroup of L, x an involution which induces an automorphism on L trivial on S. where F
=L
Let t
x E
As above it follows
that x must induce an inner automorphism on F.
Hence there exists
E
2
F such that y
C
E
= (t)
prL(2, q) and q is odd.
f
(q)
S be an involution and suppose CL (t)
E
C
= = = xf
centralizes F and t.
It is easy to see that y must act trivially on E.
For
[y, E];; CJ(F) n CL (t) = (t) and so [y, E] = 1 because [E : F] is odd. Thus y centralizes CL (t).
54
Now y centralizes S and so normalizes N (S). By Burns idf' 5 L Transfer Theorem, N (S) ;; CL (t). It follows that N (S) /C (S) is ar. L L L odd order subgroup of GL(3, 2) of order> 3. Hence NL(SLCL(Si :5 a non-cyclic group of order 21. Let Z /C (S) ~ N (S) /C (S) have order 7. Either y centralizf5 L L L Z or y acts on N (S) like the unique involution in S which is centralL ized by an element of order 3 in N (S) n CL (t), viz. t. Thus either L [y, NL(S)] = 1 or [yt, NL(S)] = 1. In both cases, since [yt, CL(tl 1 = we have L<x>
=L
x
>,
where z 2
= 1,
because CL (t) is a maximal
subgroup of L by 12. 2. This shows that TL ;; LiCG(L ) for all i and so TN;; NCer"S" . i i 2' Thus G = NCG(N) = N x CG(N) because 0 (G) = G, and then G is an A*-group if G is not simple. Thus G is a simple group all of whose proper subgroups are A'groups. 3.
We remark that a Sylow 2-subgroup T of G has rank at least
For otherwise r(T)
=2
and
IT I = 4
by [7].
Then G ~ L (q) by 2
[13] The proof of Theorem A proceeds by showing that either G is itself an A*-group or G has a strongly embedded subgroup. But in that n 2n+1 n case, by [4], G ~ L (2 ), Sz(2 ) or U (2 ). Of these only the groups 2 3 n L (2 ) have abelian Sylow 2-subgroups and so G is an A*-group every 2
time.
The proof of Theorem A will then be completed. We study the structure of the minimal counter example G to
Theorem A by considering a collection of maximal subgroups containing CG(t), where t is an involution of G.
The maximal subgroups studied
are ingeniously chosen in order to enable the exploitation of the uniqueness Theorem B of Chapter 12 and its immediate consequence Theorem 12. 4. As Bender points out in [5], the explicit definition of the class of maximal subgroups used so frequently in the proof is required for only one resu It, 14. 1. In that result 14. 1, we identify a collection of subnormal subgroups of F*(H), where H is a certain maximal subgroup containing CG(t). These subnormal subgroups are ones over whose G normalizers we ha\'e some control.
They assume crucial importance in the results which
follow.
55
Thus let G be a simple group with abelian Sylow 2-subgroups all of whose proper subgroups are A*-groups.
Assume that G contains a
3-generated abelian 2-subgroup and that G does not contain a strongly embedded subgroup. Let t be an involution in G. G containing CG(t).
o p (H)
*1
Let H be a maximal subgroup of
Choose H in such a way that for some prime p,
but CG(t} no (H) p
maximal subgroup H
~
= 1.
If, for no prime p, can we find a
CG(t) with a subgroup 0p(H) inverted by t,
choose H such that I E(H) I is maximal.
Let M(t) be the set of all
maximal subgroups of G containing CG(t) satisfying the above conditions.
Note that if for some H
then every subgroup M
*1
q, 0 (M)
q
E
E
M(t), 0 (H)
P
*1
and CG(t) no (H) = 1,
P
M(t) has for some (possibly different) prime
and CG(t) r, 0 (M)
q
= 1.
Let T be a fixed Sylow 2-subgroup of G containing t and let 1T
= 1T(F(H)). We now define a set
=
Ci(H)
of subnormal subgroups of
F* (H) satisfying a kind of uniqueness condition. 11T(F*(H)) I ~ 2, then V [1T(F*(H))
I=
1, then V
Let V
<1 <]
F* (H). If
E
E
F* (H) and N (V) ~ H, then V E
that if V
<] <J
F*(CG(V))
= F*(CH(V))
Note
For if
is a p-group also.
Here in 14. 1, we get criteria which allow us to recognize elements of
Let V
*1
be a t-invariant subnormal subgroup of
Each of the following conditions implies that V
E
(a) V is CG(t)-invariant. (b) Some subgroup V
E
et centralizes V and satisfies [V, t]
= V.
(c) There is a non-cyclic abelian subgroup V of odd order, where V ~ CG(t) r, CG(V) such that
Proof.
Assume that BCG, B
Since E(H)
~
Also CG(S) n F*(H)
56
E
S
~
= NG(V)
~
E
# V .
NG(V).
n F*(H} for all X
CG(V) n F*(H)
~
S.
<1<]
F*(H), S
Apply 12. 4(a) and get
<]<]
F*(H).
0q(B)nH==l for all qE1T(F(H))'.
Thus since CG(t);;H and t
normalizes F(B) " t must invert F(B) "
1T
1T
Consider case (b).
Let W == B n F*(H) ~ NG(V) n F*(H) ~ E(H).
Now we may apply 13.4 with X == B and get [t, W]
If
11T(F*(H))
I ==
Now
Hence U
1, put B == NG(V),
{p} == 1T(F*(H)).
Then
E(B) ;; CG(V) ;; CG(U), since U
(Remember
any subnormal p-subgroup of F* lies in F.) Now F(CG(U) n CG(V)) is a solvable E(B)-invariant subgroup of B and so [E(B), F(cG(u)ncG(v))]==l. Since CG(U) n CG(V) is an A*-group, E(B);; E(CG(U) n CG(V)). Now [F(B)p" V];; V n F(B)p' == 1, since
V;; F*(H)
is a p-group.
Also [F(B) " U] C F(B) , r; F(B) == 1, since U is a subnormal p-subp == p p group of F*(B). Thus OP(F*(B));; OP(F*(CG(U) n CG(V)). But by 12.5, if F*(CG(U)) is a p-group, then F*(CG(U) ncG(v) is a p-group also.
Thus F*(B) is a p-group also.
By 12.5 again,
F*(CG(V)) is a p-group and so V E Ci.
If
11T(F*(H)) I ;; 2, let B;; NG(V) be a maximal subgroup of G.
We know U
Thus H n F*(B) ~ E(B)
It follows that U
because NG(U);; H in this case.
1
== HnF*(B)
Now CG(U ) n F*(B);; CG(U) n F*(B);; H n F*(B) == U ' 1
1
Taking S == NG(V) n F*(H) ;; E(H), we have first S
Apply directly 12. 4(c) and get This verifies (b).
Suppose now (b) does not apply. G containing NG(V).
Let B be a maximal subgroup of
By 12. 4(a), 0q(B) n H == 1 for all q E 1T(F(H))'.
Thus since CG(t);; Hand t normalizes 0q(B), if F(B)1T'"* 1, t must invert some non-trivial
°q °
(B) for some q.
But then t must have
inverted some non-trivial
(H) for some prime r! But then U==O (H) r . r is abelian and contained in Z(F*(H)). Clearly U:::: [U, t] and NG(U) ==H. Hence U E Ci.
Since [V, u] == 1, (b) shows that V E Ci(H).
assume that F(B) , == l.
1T
Moreover, t does not invert any subgroup p.
For then
°
p
(H) E Ci(H),
°
p
(H)
C
==
Thus we may
°
(H) for any prime p Z(F*(H)) and by (b) V E Ci(H). Since
57
H E M(t), E(H) must then be of maximal possible order. maximal subgroup containing CG(t) if (a) applies.
IE(H) I > IE(B) I.
2:
H = B ~ NG(V) and V E a(H) if 17T(F*(B))!
Thus we have 2 since otherwise 12.4
However, if 7T(F*(B)) = lpl, then F*(CG(V)) is a
p-group also by 12. 5.
Thus in any case V E a(H).
In case (c), (u>
<J<j
#
(u> ~ F(H) for all u E U. -
Since H E M(t),
It is clear that we have directly reproduced the hypo-
theses of 12. 4(d) since H, Bare A*-groups. does not apply.
But B is a
Suppose 17T(F*(H)) I
containing NG(V).
F*(H), u E u#.
Since (u> is cyclic,
Thus U:= F(H).
2: 2.
Let-B be a maximal subgroup of G
Since u;, OF(CB(t)).
Apply 13. 5(b) and get that every
component of E(B) is centralized by some non-trivial element of U. Thus if N = 07T(F*(B)), then N = (CN(u) : u E U#>. and 17T(F*(H)) I
2: 2,
Hence F*(B);, H.
CG(u) :; H.
Thus N:; H.
Since (u> E a(H)
By 12. 4(b), F(B)1T;' H.
We may apply 12. 4(c) since N(V) n F*(H) ;, B to
get V E a(H). If F*(H) is a p-group and V ,a(H), then F*(CG(V)) is not a p-group.
As remarked above, every component of E(B), where
B = CG(V), is centralized by a non-trivial element u E U. (u> E a(H), F*(CG(u)) is a p-group.
Since
But V;, F*(H) is a p-group and
by 12.5, F*(CG(u>V)) = F*(CB(u)) is a p-group.
But clearly
CG(u) n oP(F*(CG(V))) contains non-trivial pt-elements for suitable u E U#.
Since CG(u) n oP(F*(CG(V)) :; F*(CB(u)) by 10. 7, we have the
required contradiction. Lemma 14. 2. Proof. t
g
This completes the proof of 14. 1. /I H has at least 2 classes of involutions.
g If g E G, t E H n H , where exists h E H such that
-1
h=t, if H has one class of involutions.
and so g E H.
Butthen g-l hECG (t):;H Thus, if H has one class of involutions, H n Hg has odd
order, if g E G - H.
Hence H is strongly embedded in G, a contra-
diction. /l Lemma 14. 3. r(R) = r(T) - 1.
58
Let R be a subgroup of T such that
Then M(s)
* IH)
for some involution s ER.
Proof. M(s)
Suppose M(s)
= M(sg) = M(s)g =
=
#
{H) for all s E ~\ (R).
{H), then clearly g E NG(H)
First if
= H.
Since r(T) ~ 3, R n Rn "* 1 and so there exis::'
Let n E NG(T).
-1
an involution s ERn Rn. M(s)
= M(s n
Then s, sn
ER and by assumption
-1
Thus n EH, NG(T) ~ H.
).
But every involution of G is conjugate by an element of NG(T to an element of R. Hence M(s) = {H) for all involutions s EH. Kc·,;, g if s EH n H , where s is an involution, then g E NG(H) = Hand H is strongly embedded in G. ;/
Lemma 14.4. such that It, DJ "* 1. r(R)
= r(T)
Let D be a T-invariant subgroup of odd order Then T contains a subgroup R such that
- 1 and It, CD(R)J "* 1.
Proof.
Without loss of generality, D is a p- group and C (t);:; [) n and T acts irreducibly and non-trivially on D/CD(tL Since T is abelian, T induces a cyclic group of automorphisms on D CD(t). t acts non-trivially on D/CD(t). r(R)
= r(T)
- 1.
Also CD(R). CD(t)
Cn(R)Cn(t)/Cn(t).
= 0p(H),
CD(t)).
Then clearly
since CD/CD(tl(R)
=
Let p be an odd prime such that [t, -
° (H) J "* 1. P
then P has a Cp(t) invariant subgroup 15 such that
It, PJ "* 1 and V
Proof.
=D
= CT(D
Since It, DJ "* 1, [r, CD(R)J "* 1.
Lemma 14. 5. If P
Let R
Also
E
Cl(H) for any t-invariant subgroup V of
P, V"* 1.
Suppose the result is false.
= Cp(Cp(t)). First if It, V] = 1,
Let V
then It, P]
=1
by 2.2.
This is not the case.
Thus It, V] is a Cp(t)-invariant subnormal subgroup of P(H).
V
= It,
[W, V]
V] E Ci(H) by 14. l(a).
= 1.
Thus
Now for any subgroup W of Cp(t) we have
Since V E (t(H), W
E
Cl(H) by 14. l(b).
If Cp(t) is non-cycliC, there exists a non-cyclic normal 2'-sub-
group VI of type (p, p) of Cp(t), which is, as seen above, an element of Cl(H).
So by 14. l(c), since VI
~
Cp(t) n Cp(V), V E Cl(H).
Finally
if Y is any t-invariant subgroup of V, Y E Cl(H) by the same argument.
59
Letting 15 = V we have the result. If Cp(t) is cyclic, first every Cp(t)-invariant abelian subgroup
A of P is centralized by t.
For otherwise, It, AJ E C1(H) by 14. l(a)
and then any t-invariant subgroup of It, AJ, being centralized by It, AJ E ct(H), is itself an element of ct(H) by 14. l(b). Therefore every Cp(t)-invariant abelian subgroup of P is contained in Cp(t) and so is cyclic.
It follows that Z(P') is cyclic.
Now
any normal subgroup Q of P of type (p, p) lying in P' is clearly contained in Z(P'). P' ;; Cp(Q).
For P /Cp(Q)
~
~
GL(2, p) and so has order
It follows that P' is cyclic since p is odd.
p.
Further
p' ;; Cp(t).
-1
Now if x EP, Thus [x
2
,
X
t
=x
-1
,thenfor y EP' we have y
yJ = 1 and [x, yJ = 1.
xt
=y
Let Q
= ~\ (P).
=y
x
Since P has class
Q has exponent p and Z(Q), being cyclic, has order p.
CQ(t)
X
Since P = Cp(t)I where
-1 ) . t I = {X EP : x = x ,P' ~ Z(P).
:s 2,
Thus
Also
= Z(Q). Every non-abelian subgroup M of Q is an element of ct(H).
For M' = Z(Q) and so NG(M);; NG(M') = H. L
"*
On the other hand, if
1 is an abelian t-invariant subgroup then either L
and L E C1(H), or It, LJ
"*
~
t-
CQ(t) = Z (Q)
1. If x EL is such that x = x-I, then
L E ct(H) if <x> E C1(H) by 14. l(b).
Thus either every t-invariant subt group of Q lies in C1(H) or there exists x E Q such that x = x-I, <x> f C1(H). Consider It, CQ(x)J = a contradiction.
v.
If VEC1(H) then so is (x) by 14. l(b),
Hence V is abelian.
But
Thus CQ(x) is abelian and of index p in Q.
For the centralizer
of any non-central element of Q is of index p in Q. not cyclic, it is not
Cp(t)- invariant
mal subgroups and so
IQ I < P 3•
Now let R
~
T submodule of Q/(Q).
60
Thus Q has two abelian maxi-
T, r(R)= r(T) - 1 and CQ(R)
exists since we could take R t:: CT(Q
1
Since CQ(x) is
/~(Q))
1Z(Q).
where Q
1
Such an R
is an irreducible
g g G such that t ER. Then CG(t ) ~ Z(Q) and so g g g g H "* H. Put K = [t, CQ(t )] ~ H n H since C(t g) ~ H . Let K be 1 g the normal closure of K under CG(t) n H . Since K ~ Q, K ~ Q and 1 g is a p-group. By 12.4, [t, K ]
K:= Qg.
E
1
If K = Qg, then K = Q since K ~ Q. Then g E N(Q) = H, a g contradiction since H "* H. Thus K C Qg, IK S p2. But K = [K, t 1 g and'so CK(t) = 1. Thus IKI = p. But K ~ CG(t ) and so K = Z(QgL g Therefore N (K) ~ H . G g g Now H, H are conjugate A*-groups and OP(F*(H)) ~CG(K) ~ H . -
I
Thus if S
= NG(K)
n F*(H), S
~
E(H) and S
<]
Also
CG(S) n F*(H) ~ S. If 17T(F*(H)) I
"*
1, then H
= Hg
by 12. 4(e), a contradiction.
Therefore F*(H)
=P
also on Q/(Q).
Since then T ~ GL(2, p) and r(T);; 3, we have a
and T acts faithfully on P and also on Q and
contradiction. JI
!!. F*(H) is a p-group, where p is an odd prime.
Lemma 14. 6.
then for any involution s
E
G and any M
E
M(s), F*(M) is also a p-
group.
Proof. of
° (H)
such that [t, P]
p
-
group V of P. r(R)
= r(T)
p-
From 14.5, there is a Co (H)(t)-invariant subgroup
"*
p
1 and V
E
ct(H) for every t-invariant sub-
Now by 14. 4, there exists a subgroup R
- 1 and W
E
[t, Cp(R)]
"*
L
Of course W
E
~
T such that
ct(H).
By 12. L
we may assume that our involutions s ER. Let M E M(s). Let CM(t) , the subgroup generated by the CM(t)-conjugates of W. W1 = W Since W ~ 0p(H), CG(t)~H, W ~ 0p(H) and so W 1
12.4, [t, W
1
]
group of F*(M). If P
(W
E
Thus W
Hence Wc
= 0p(M),
=
U
= [t,
° (M).
W] ~ W
1
since U
~
is a p-group.
By
is a subnormal sub-
p
= Np(W),
since F*(NG(W)) is a p-group
ct(H)!), F*(CG(U)) is a p-group also by 12.5.
CG(U) ~ CG(W) ;; NG(W).
1
For
Now by 12.6, F*(CG(P)) is a p-group also
P. It follows that F*(NG(P))
= F*(M)
is a p-group also.
61
This completes the proof. /I The following major Theorem is the cornerstone on which the whole proof rests.
Its proof is at least staggering in its power and
originality. Of course, it is due entirely to Bender. Theorem 14. 7.
Let p be an odd prime such that [t, 0 (H)] = 1. p
Let V be an elementary abelian p-subgroup of 0 (H) such that
-
r(V)
~ 3.
.
If [t, F*(H)] *- 1, then either W
Proof.
some prime q *- p, or W 14. l(a).
P
#
Then CG(V);; H for all v E V .
2
= [t,
1
= [t,
0
q
(H)] *- 1 for
E(H)] *- 1 is an element of Ci(H) by
For both W ,Ware CH(t)-invariant and subnormal in F*(H). 1
2
But then eve 0 (H) centralizes W
= =
P
implies that E Ci(H) for v E V - 1.
1
or Wand then 14.1(b) 2
Then CG(v);; NG«v» ;; H
since F*(H) is not a p-group. Thus t E CH(F*(H))
= Z(F*(H)).
Let 'U(X) , where X
~
groups K of X such that K
Let W
= CH(V)
n F*(H).
G, be the set of all W-invariant A*-sub-
= F(K) 7i ,E(K)
and no component of E(K)
is contained in H. If 'U(G)
=
{I}, then for any maximal subgroup B of G, B ~ W,
it follows that F(B) is a 7i-group and E(B) is a product of components which all lie in H.
For if a component of E(B) does not lie in H, then
neither does any of its W-conjugates.
Then EWE 'U(G) *- {I).
Thus E(B);; Hand F(B);; H by 12. 4(b). since t E F*(H).
Thus F*(B);; H.
By 12. 4(c), we have B containing W.
= Hand
Note that OP(F*(H)) *- 1
But CG(W) n F*(H) ;; Wand W;; B.
H is the only maximal subgroup of G #
Clearly if v E V , CG(v):; Wand so CG(v);; H.
Thus
we may assume that 'U(G) *- {I}. We derive a contradiction to the simplicity of G from 'U(G) *- { 1 } by a series of steps. (i)
If W;; X c G, then [t, F*(X)] E 'U(X).
For if W;; X;; G, then t inverts F(X) 7i' elementwise by 12. 4(a) since t E W.
Also t centralizes F(X) , C H by 12. 4(b) since if 7i = q E 7i, [0 (X), Oq(F*(H))] = 1 and t E Oq(F*(H)) *- 1. Now [t, E(X)] is q
the product of those components of E (X) not centralized by t, and since
62
H/C (02 (H» H
has odd order, [t, E(X)] is the product of those com-
ponents of E(X) which do not lie in H.
For any component of E(X),
which lies in H, lies in C (02(H» ~ CG(t). Note t E Z(F*(H)). G Consider now [t, F*(X)]. This group is a W-invariant normal subgroup of F*(X) and by the argument of the previous paragraph and the structure of normal subgroups of F*, [t, F*(X)] E 'U(X). (H)
('U(X»
E 'U(X) if Xc G.
Let K E 'U(X) where Xc G.
= [F(X) 11 , t] = 1 [F(X) , t, K] = L 11
[K, F(X) , t] 11
since t centralizes F(X)
C
Hence [K, t, F(X) ] 11
Then K centralizes F(X).
=1
and K
= [K,
11
For
by 12. 4(b).
t] centralizes F(X) . 11
Also t inverts F(X) , and so commutes with K in its action 11
on F(X)1I"
= [K,
Thus K
t] centralizes F(X)1I"
Hence [K, F(X)]
=L
Now if Xc G, X is an A*-group, t E O*(X) and so K
= [K,
t];; O*(X) n CG(F(X». It follows that K ~ F*(X).
= [K,
Thus K ('U(X»
t] ;; [F*(X), t] E 'U(X) by (i).
Hence
~ [F*(X), t] E 'U(X).
There exists R;; V such that V /R is cyclic and
(Hi)
'U(CG(R) *- {11.
vn 1 (Z(Op (H)))
First replacing V by that CG(V)
~
H.
Choose K E 'U(G). If F(K)
For otherwise F(K);; CG(V)
~
H.
if necessary we may assume
n, then [V, F(K)] *- L
But F(K) n H
=1
by 12. 4(a).
Let
Z be a minimal W-invariant subgroup of F(K) such that [V, Z] *- L Let Z
=Z
C
1
be an irreducible V-module.
Clifford Theory, Z irreducible V
~
= Z1
@ •••
W modules.
@
Z
r
Define R
= CV(Z). 1
By
and W permutes transitively the
Since R
~
Z(W), R
~
CW(Z).
Thus
Z E'U(CG(R)
= 1,
If F(K)
of K.
let E be a minimal W-invariant normal subgroup
Then [V, E] *- 1 because otherwise E;; CG(V) ;; Hand E ~ H
since K E 'U(G).
Let Y
every component E on each such E Since V
~
1
= EV(t).
by 13. 5. 1
Then V ~ OF(Cy(t») normalizes
of E and induces a cyclic group of automorphisms Let R
C
=
CV(E ), be such that V /R is cyclic. 1
Z(W) and W permutes the components of E W R;; CV(E-W). Thus E E 'U(CG(R»). Thus (Hi) is done.
W
transitively,
63
Let M = ('U(CG(R))) E 'U(G) by (H). Y
V
Then M ~ Y
= ('U(CG(v))).
v
Let v ER and put
E 'U(G).
We show (iv)
Y= Y
normalizes M.
v
For since Y E 'U(G), F(Y) = Z(Y) normalizes M any component of E (y).
If E ~ M, fine.
-
before. If Cy(E) ~ R, then E ~ E and so Y = RCy(E).
W
Y.
E ('U(CG(R)))
= M.
Let E be
:s- p
as
Thus Cy(E) tR
It follows that [M, Cy(E)] = [M, Cy(E)R] = [M, V].
Now M = F(M)E(M) E 'U(G). F(M) = (CG(y) n F(M))[F(M), V]. 12.4(a).
~
If not, [Y : Cy(E)]
Thus F(M) = (F(M),
As V-group,
But CG(y) n F(M) ~ H n F(M) = 1 by
Yl.
Let L be a component of E(M); y normalizes L as before and [L, y];! L.
If [L, y] ~ Z(L), then [L, Y] = 1 as usual.
Thus
[L, Y] = L and [M, Y] = M = [M, Cy(El]. Now (Cy(E), E, M] = 1; [E, M, Cy(E)] = 1, since E
We have shown that
either a component of E(Y) lies in M or it centralizes M.
Thus
M
=
(v)
v
('U(G)
E 'U(G).
Let S E 'U(G). Remember reV)
~
First S = (CS(v) : v ER #) since R is non-cyclic.
3 and y;R is cyclic and apply 13. 5.
Now if v ER,
Cs(v) n F(S) is a W-invariant nilpotent 1T'-subgroup of S and so CS(v) n F(S) = ('U(CS(v) n F(S)). #
Thus F(S) = ('U(CS(v) n F(S)) : v ER). On the other hand, if E is a component of E (S), choose vER W W such that CS(v) ~ E and then E ~ CS(v) and E ~ 'U(CS(v)). Thus
#
#
S = F(S)E(S) = ('U(CS(v)) : v ER). But then S ~ NG(M) v E Y.
G since ('U(CS(v))) = Y
Now W
64
C
Let W
<J W
n
= H.
Then W
1
per-
= A,
mutes the elements of 'U(G) and hence normalizes ('U(G) Arguing inductively we get H normalizes A is maximal in G we have first A
C
~
NG(M)
H and then A
C
=1
say.
G and since H since A E 'U(GI.
This contradiction completes the proof of 14. 7. ;f Let p E 11, P *- 2, V ~ 0p(H).
Lemma 14. 8.
some involution sET, M *- H. (a) CG(V) ~ H, V
Assume 1T(F*(H));;; 2, V
= [V,
t]:
(c) t E CG(V) ~ H and
l
~
M and either
#
(?) t E CG(v) ~ H for all v E V
or
Let M E M(s) for
and V is abelian of type (P. p :
~ W, where W is of type
(2, 2) such
that CG(w) ~ H for all w E W . Then OP(F*(M)) ~ H, 0p(H) ~ M. r(O (H)) p
= 1<
If [s, V]
= 1,
then
r(O (M)). p
°
°
Proof. Let P = (H), Q = (M). In case (a), let C (t) P P V = V M cO (H). Thus applying 13.4 to V we get It, V ]
=
1
1
V;; 0p(M) = Q and OP(F*(M)) ;; CG(Q) ;; CG(V) ;; H. In case (b) by 13. 5, V;; OF(CM(t)) normalizes each component of E(M) and induces a cyclic group of automorphisms in each such component.
Thus OP(F*(M));; (CG(v) : v E V#) In case (c), V
~
H.
~
OF(CM(t)) normalizes each component of E(M) n and centralizes any component of type L (2 ) or JR. Thus every com2 n ponent of E(M) of type L (2 ) or JR lies in H. Clearly 2
F(M)p' ~ (CG(w) : w E W) ~ H.
If a component of E(M) is of type L (ql. 2
q odd, it is normalized by T and Wand by 13. 6 is contained in . ( CG(w) : w E W # ) ~ H. Thus 0 P (F*(M)) ~ H In every case. If r(Q) ~ 2, then since [P n M, OP(F*(M))] ;; OP(F*(M)) n P, a
solvable normal subgroup of F(M)p,E(M). a familiar argument. [P n M, oP(F*(M))]
=
Thus [P n M, E(M)]
Clearly [P n M, F(M) ,] P 1.
By 12.4, P n M ~ 0p(M)
= Q.
= 1.
=1
by
Thus
In particular CG(CZ(Q)(P n M)) =
CG(Z(Q)) ~ M. If r(Q)
> 3, =
then 1r,(F*(M)) I > 2 because otherwise F*(M) is a
=
p-group and the F*(H) would be a p-group by 14.6.
This is not the case
65
by hypothesis. normal
"* 1, then [s, Q] is a CG{s)-invariant sub-
If [s, Q]
sub~oup
of F*{M) and so [s, Q]
E
Ct{M) by 14. l{a).
Then
C Z (Q) (P n M) centralizes [s, Q] and is clearly s invariant. 14.1{b), CZ{Q){P n M)
C1{M).
E
By
Hence CG{CZ{Q){P n M)) ~ M.
If [s, Q] == 1, then for all v
E
CZ{Q){P n M), CG{v) ~ M by
14. 7 since r{Q) 2:.. 3. Thus in every case we have shown that CG{CZ{Q){P n M)) ~ M. Now V ~ P n M and so CG{P r M) ~ CG{V) ~ H and so CZ{Q){P n M) ~ Q n H.
Thus CG{Q n H) ~ M.
It follows that [oP{F*{M)), Cp{Q r H)] ~ P n oP{F*{M)) and; by
a now very familiar argument, [oP{F*{M)), Cp{Q n H)] == 1. We now have OP(:F*{M)) (Q n H) acting on P and oP{F*{M)) centralizes Cp{Q n H). OP{F*{M))
Thus OP{F*{M)) centralizes P.
Since
"* 1, F ~ M.
We are left only with the last assertion. Suppose therefore that [s, V] == 1. r{Q)
~
2, then P n M == P
Suppose that r{Q) == r{P). ~
~
Q.
We have already seen that if
Thus if r{Q)
~
2, then r{Q)
~
r{P).
Then [2 (Z{Q))P has a subgroup of rank 1
It follows that [21 (Z{Q))P == P and so CG{P) ~ C {[21 (Z{Q)))~M.
r{Q).
Therefore F*{H) ~ M since P ~ M.
G Now OP{F*{M)) ~ Hand P ~ Q.
We have P CQ{P)OP{F*{M))(== S)~ H.
By 12. 4{c), H == M, a contradiction.
Thus if r{Q)
~
2 then r{P)
<
r{Q) and so r{P) == 1, r{Q) == 2.
Assume therefore that r{Q)
2: 3
and [V, s] == 1. If [s, Q]
"* 1,
apply 14. 5, replacing t by s, H by M, to get a subgroup Q is
CG{s)- invariant such that [s, Q1]
s-invariant subgroups Q Q1'
Let V == [C
Q
o (V), s].
of Q1'
"* 1 and Q o E
Now V
Since V, Q
1
1
~
of Q which 1 a{M) for all
CG{s) and so V normalizes
are p-groups, C
Q
(V)
"* 1.
1
V == 1, then by 2. 2, [s, Q 1 ] == 1, a contradiction. Thus V"* 1, V == [V, s] ~ CG{V) ~ H and all s-invariant subgroups of Q 1 are elements
If
of a{M).
!1T{F*{M)) I
Since
2: 2,
NG(V) ~ M.
the hypotheses of (a) apply to H, s, V. and also that Q
66
~
H.
Thus F*{H)
~
We have thus verified that
It follows that OP{F*{H)) ~ M
M, F*{M)
~
Hand H == M by
Theorem A.
This is not true and so [s, Q]
= 1.
Now we find an n (T)P-invariant normal abelian subgroup V of I
Q of type (p, p).
Let A be a maximal abelian normal subgroup of
QPn (T) contained in Q. If A
CQ(A), we can find A
C
l
~
B
C
CQ(A)
such that B/A is an irreducible QPn (T)-module and B/A C Z(QP /A).
=
I
Then n (T) acts irreducibly on B/A and since T is abelian, n (T) I
I
of exponent 2, B/A is cyclic.
Thus B is abelian, a contradiction, and
= CQ(A) E S~~(Q). If A is cyclic, then Q/A is also cyclic and r(Q) < 2, not the case. Thus A is non-cyclic and n (A) has a chief = I
A
QPn (T) series with cyclic factors.
The group V has been located
I
successfully. Any such group V lies in an abelian subgroup of type (p, p, p). For since r(Q)
~
3, there exists X
~
Q of type (P, p, p).
X/CX(V) ~ GL(2, p), [X : Cx(V)] ~ p. abelian and ~ 3-generated. If CX(V) and get CG(v) ~ M for all v
E
If CX(V)
= V,
* V,
Since
then VCX(V) is
then V ~ X.
Apply 14.7
Vi.
If V ~ H, we may apply either conditions (a), (b) to M, V, s.
if [s,
V]
* 1~ then let
and [V s] = V. I
I
V
I
= (v I )
= v-I. I
Hence F*(M) ~ H, F*(H) ~ M, a contra-
diction, using Theorem A and M
* H.
VV and for some v
Hence V ~ H. E
V, CG(v) ~ V.
applies V ~ H, which we have already ruled out.
V ~ (CG(w)-: w
E
W#) because W
must be in case (a), CG(V) If Cp(V)
Then CG(V ) cM I = Thus
If [s, V] = 1, then (b) applies directly.
oP(F*(H)) ~ M and Q ~ H. Now V ~
where vS I
E
~
H, V
~
In case (c),
n I (T) and again
= [V,
Thus if (b)
V ~ H. Thus we
t].
Ci(H), then NG(Cp(V)) ~ H and V ~ H.
Now P' ~ Cp(V) and P'
E
(j{H) if P'
* 1.
So Cp(V)~Ci(H).
Thus if pI
* 1,
V ~ NG(P') ~ H, a contradiction. It follows that P is abelian, It, P] because V
~
P.
For
*1
Apply 14. l(b) and find that every non-trivial t-invariant
subgroup of P is an element of Ci(H). acts faithfully on V.
In particular Cp(V)
=1
and P
Since V is of type (p, p), P is cyclic. ;/
15. PROOF OF THEOREM A, PART IT
Lemma 15. 1.
t
E
F*(H).
67
We show that It, O(H)] == 1 and the result follows from
Proof. 13.1.
Let p E 7f, P *- 2, P == 0 (H). If F*(H) == 0 (H), then O(H) == 1 p 2 and we are done. Assume therefore that It, P] *- 1. By 14. 5, there exists l'
P] *-
It,
~
P which is CH(t)-invariant such that
1 and V E
P. By
14.4, there exists R ~ T such that r(R) == r(T) - 1 and It, Cp(R)] *- 1.
Let V == It, Cp(R)] E Ci(H). By 14.3, {H) *- M(s) for some involution s ER. Suppose M E M(s), M *- H, s ER. applies to V;; CG(R) ~ CG(s) ~ M. since r(M)
>
If P *- F*(H), then 14. 8(a)
Since [s, V] == 1, P is cyclic and,
1, M is not conjugate to H.
Now [R, V] == 1 and so R
centralizes P, a cyclic group. g g Choose g E G such that t ER. Since [t , P] == 1, It, P] *- 1, g g g g t H. Thus H E M(t ) and H *- H. This contradicts the assertion of 14. 8. Thus P == F*(H) and so by 14.6, F*(M) is a p-group for all M E M(s) and for all involutions s E G.
In particular F*(M) is a
p-group for s ER, M E M(s). By the ZJ-Theorem of Glauberman, H == NG(Z(J(S))), where S It follows that S is a Sylow p-subgroup of
is a Sylow p-subgroup of H.
Similarly M == NG(Z(J(sg))) and so H, M are conjugate. g Now V C C (R) C C (s) C M == H .
G.
==
==
P
G
::::
Choose U::J V maximal such that U is a p-group and U g
t H.
=
C
=
g H nH ,
Clearly U is not a Sylow p-subgroup of H by the ZJ-Theorem.
Thus NG(U) ~ H. We break the remainder of the proof into steps. Every p-subgroup P
(0
1
of G containing a Bylow p-subgroup
U 1 of NH(U) lies in H. For if P X
lies in a Sylow p-subgroup ~ of G, then
1
H n H ::J U ::J U and by the choice of U, x E H.
==
1
CG(U) ~ U.
(ii)
o
op (NG(U))U 1 p
(NG(U))
68
C
==
X
H nH
•
C
==
Thus P
1
cH.
==
H by step (i) and if x E NG(U) - H, then Maximality of U forces U == 0 (NG(U)). P
Now
since V
E
(i(H), F*(CG(V)) is a p-group and by 12.6, F*(CG(U)) is a
p-group also.
Thus F*(NG(U))
= 0p(NG(U)) = U
and so CG(U) ;; U.
This verifies (ii). Among all groups NG(Y) ~ NG(U), where Y"* 1 is a p-group,
IN Ip
choose N so that first and then
IN I is maximal.
°
=
°
is maximal, and then
Let
°
P
X
Since NX(U);) NG(U), NX(U);; 0p(NG(U)) some complement A to
is maximal,
(N) = Z, 0, (N) = X. p,p
(N) x p ,(N). p For 0p,(NX(U));; CX(U) ;; U by (ii). (iii)
1op (N) I
°
Hence NX(U) is a p-group.
= U.
Now U normalizes
,(N) in X. Then UA is a p-group. Since p NG(U) n UA = U, UA = U and A;; U. Thus U n X is a Sylow p-subgroup of X and so NG(U n X) ~ NG(U) and ~
Since U n X
ING(U n X) I ~ INl ' p p Z, by the choice of N, U n X = Z. But U n X is a Sylow
°
p-subgroup of X.
°
Hence X = ,(N) x (N). p p N is a p-constrained group.
(iv)
We show that CG(Z) has odd order and so CN(Z) is solvable. Hence CN(Z) cO,
=
p ,p
CG(Z) and take M If [s,
°
E
(N) as is well known.
Let s be an involution in
M(s).
=
°
=
=
(M)] 1, then p 2, since (M) F*(M) and then p p F*(H) is a 2-group by 14.6. This is not the case. Thus by 2.2, [s, CG(Z) n 0p(M)] "* 1. Then [s, CG(Z) n 0p(M)] is a CN(s)-invariant p-subgroup of N. By 13.4, [s, CG(Z) n 0p(M)] <J<J F*(N).
°
(N)
=Z
and so [CG(Z) r,
p contradiction.
°
(M), s, s]
Hence [s, CG(Z) n 0p(M)];;
= 1.
Apply 0.2 and get a p Therefore CG(Z) has odd order and (iv) holds.
We may now apply the ZJ-Theorem to N.
Some Sylow p-sub-
group S of N contains a Sylow p-subgroup of NH(U) and so lies in H by step (i).
By the ZJ-Theorem, N
Z(J(S)) cO, N
=
=
°
°
But then
,(N) x (N) and so Z(J(S)) cO (N). Thus P P P NN(Z(J(S))) and since IN I is maximal, Z(J(S));! N. But
=
p,p
(N)
= 0p,(N) NN(Z(J(S))).
Z(J(S)) ~ H since F*(H) is a p-group. N ~ NG(U) and NG(U) ~ H. Lemma 15.2.
=
This shows that N ~ H.
But
This completes the proof of 15. 1. /1
O(F(H)) has rank at most 2.
69
Proof.
Assume r(O (H)) > 3 for some odd prime p. Let p == P == Q (Z (0 (H))). Since p is odd, P has exponent p. Every normal 1
2
P
subgroup of 0 (H) of type (p, p) lies in P and so P is non-cyclic. p If P/(p)
= P 1 /{P)
x ... x Pk/(P), where P./(P) is an 1
irreducible T-module, define R ~ CT(P 1 /(P)) of rank r(T) - 1.
This
is possible since T is represented on P /(P) as a cyclic group.
If
1
P /(P) is not cyclic, choose Wc R with r(W) == r(R) - 1.
=
1
clearly r(cp(W));; 2.
Then
If PI /4>(P) is cyclic, then k> 2 because
P, (Pl and P are not cyclic.
Choose
WC ker (T - aut P /(P)) n ker (T - aut P /(P))
=
1
such that r(W)
= r(T)
- 2.
2
Again r(Cp(W)) ~ 2 since r(Cp/(P)(W)~ 2.
Now if v E P, (v)Z(O (H)) 3, in which case X lies in an element of S8::J1. (0 (H)), == 3 P or r(X) < 2 in which case X still lies in an element of S8::J1. (0 (H)). == 3 P For by [8] 1. 8.4, S8::J1. (0 (H)) *- i5, and if Y E S8::J1. (0 (H)), 3 P 3 p YCy(X) ~ p. If Cy(X) X then XC y (X);1 0p(H) and r(XCy(x))~ 3. If Cy(X) ~ X, X ~ Y and we are done already.
t
I
-
-
We can thus apply 14. 7 to find that CG(v) By 15.1, 2 E 7T(F*(H)). Since It, 0p(H)] c
5,
V] == 1.
= 1,
~
#
H for all v E P .
Now take V ~ Cp(f) of type (p, p).
t E CG(v) ~ H for all v E V.
Also if SEW, then
Apply 14.8 and get that if M E M(s), M *- H, then r(O (H))==1. p
This is not true by assumption.
Thus
{H) == M(s) for all SEW.
If W is non-cyclic, there exists a fours-group Wo;: Wand t
E CG(V) ~ H, where V is a subgroup of Cp(R).
. 5.
V]
= 1.
Moreover CG(w) ~ H for all w E Wo'
get that if M E M(s), M *- H, then r(O (H)) -
p
\1(s)
=
Also s ER and so
{H) for all s ER.
= 1.
Apply 14. 8(c) to
This shows that
This contradicts 14.3.
Thus we have W is cyclic and r(T)
= 3.
Thus all involutions of
G are conjugate. Consider W == T n F*(H). 1
If V
2
is any subgroup of P of type
P. p), V C Cp(W ) because [W , P] == 1. Now t E CG(v) ~ H for all # 2= 1 1 ': E: V and if SEW, [s, V ] = 1. By 14. 8(b), r(O (H)) = 1, a contra2 1 2 P
70
diction unless M(t)
=
{H 1 = M(s) for all SEW. Of course t E Wand so 1
1
{Hl. If W
is non-cyclic, then since r(T) = 3, we have an immediate
1
contradiction to 14.3.
Thus W = (t) = T n F*(H);: Z(H) and H=CG(t). 1
Now suppose M E M(s), s ER, M t- H.
Then M is conjugate to H
because all involutions are conjugate and M(t) = {H But Cp(S~~Cp(R)t-1 and so pnMt-1. v E P n M.
l. So {M 1 = M(s).
Thus cG(pnM);:CG(V);:H,
It follows that It, CG(P n M)] = 1.
Now we have
(t) x P n M acts on 0 (Z(M)) and by 2. 2, It, 0 (M)] = 1. P P
o
P
(M)
C
=
Thus
H.
Now choose
V;: 0p(M)
jugate to H and so r(Op(M)) M = CG(s).
of type (p, p).
~
3.
Remember M is con-
Then V;: H, [s, V] = 1 since
Interchanging s, t, H, M in 14.8, s E CG(v) ;: M for all
v E V, V is of type p, p; It, V] = 1. r(O (H)) p
> =
Since M t- Hand r(O (M)) = P 3, we have a contradiction. /I
Lemma 15. 3. Proof.
H;E(H) is solvable.
We show that H/CH(F(H)) is solvable and so
H/F(H)CH(F(H)) is solvable. it is solvable.
Hence the result.
Let K = H(oo). for all p, then
Since F(H)CH(F(H))/F*(H) has odd order,
Then K acts on F(H) and if [K, F(H) ] = 1,
CH(F(H))~H(OO).
Thus choose p such that [K,PF(H)p]t-1.
Since K = OP(K) there exists a p'-subgroup X ~ K such that [X, F(H) ] t- 1. p
C =
n1 (D).
Let D be a Thompson critical subgroup of F(H)
and let
p
Of course p t- 2 because H/C(F(H) ) is odd order and 2
solvable and so K;: CH(F(H)/ Now by 15.2, r(O(H)):S 2 and so
Ic I
:s p3.
Letting C=C 4>(C).
we have first CK(C) is a p-group, and K/CK(C) is a subgroup of GL (2, p) and an A*-group.
The only such subgroups are solvable.
Thus
K is solvable, a contradiction. Lemma 15.4.
Let K be a component of E(H). If {Hl =M(s)
for every involution s E CT(K), then NG(T);: H.
71
Proof, S En
1
Let W = CT(K). Assume NG(T)
1 H = M(s)
for all
(W).
(i)
T is elementary abelian.
-1
*
For let g -1 E N ( T ) - H. If W n Wg 1, then there exists G g g w, w E W involutions and {H) = M(w) = M(w ). It follows that g EH, a contradiction.
Since T = (K n T)
Wand K n T is elementary, W
(El
is elementary. Let X, Y be defined as follows:
(ii)
K <J H.
Since CG(T) ~ CG(t) ~ H, X ~ Y.
By assumption X
* Y.
If
y E Y - X, W n wy = 1 and sol W I ~ [T : W] = IQ I, where K n T = Q. In particular there are at most two components of E(H) with Sylow 2subgroups as large as Q.
Moreover if ITI = IQI
2
and E(H) is a
central product of two groups isomorphic to K, then H cannot permute the two groups K. It follows that K ~ H. Let IQI = q.
Choose a subgroup R of K of order q - I which
centralizes Wand is regular on Q.
We show that
*
X = Ny(W) ~ Ny(R 0) for all subgroups R 0 1 of R. n For if n E Ny(R ), then W is centralized by Rn = R. Since o n o 0 R acts regularly on T lW, W = Wand so Ny(R ) ~ Ny(W) because X o y normalizes K. But if y E Y - X, then W n W = 1. Therefore (iii)
X = Ny(W). (iv)
If 1T(F(Y))
If 1T(F (Y))
* 1T(F(Y)
* 1T(F (Y) n X),
n X), then let N
~
Iwl = 2, IQI = 4.
Z (F(Y)) be a minimal normal
subgroup of Y such that N n X * 1. Then every element of R acts fixed-point freely on N since CR(R ) C X for all subgroups R of R, 0 = 0
RoiL o
Put C
= CT(N).
n First C n W ~ W nw, n E N and so C n W = 1.
C=[C, R] since R is regular on TIW. C = [C, R] ~ K n T = Q.
72
R~K,
But R is regular on Q and so C = Q or
C=1. Suppose C = Q.
Because
Hence
Then
= [N,
T
T] EB CT(N)
= [N,
T] EB Q.
Here [N, T], Q are R-modules and so
Since W n C
=W
r Q
= 1,
it follows that W
~
[N, T].
But then
W = [N, T] and this is impossible because N normalizes [N, T] and
W
n
n W = 1 for n E N.
Thus C = 1.
Now T is a direct sum of faithful and irreducible
RN-modules and by [12] 3.4.3, we have only has solutions if
2, q = 4.
This
This verifies (ivl.
is cyclic, then O(H) = 1.
If w
(v)
Iw I =
ITI =q!wl = !w[q-l.
S F*(H) since s E F*(H) for all by assumption and of course K n T S F*(H). By 12.1, G
For T is elementary and so T s E CT(K)
#
has a single class of involutions if W is cyclic.
-
#
Let sET.
H = CG(w), w = (w>, M = CG(s) is conjugate to H and so T
S CG(t)
Thus O(M)
Since
S F*(M).
because t E F*(M) centralizes
F(O(M» ~ CO(M)(F(O(M»).
Thus O(M)
S H.
Similarly O(H);; M.
Now K centralizes O(H), a solvable K-invariant subgroup of H and so F(O(H»;; OF(CM(t»
~
O*(M) and
F(O(H» ;; efT) r OF(CM(t» ;; O(M) by 13. 5. and so F(O(H»;; F(O(M». H = M if O(H)
* 1.
Thus F(O(M»;; O(M) ~ H
By symmetry, F(O(M»;; F(O(H»
But then H == CG(s) for all sET.
and
This is not
the case because H is a maximal subgroup of G and NG(T) ~ H. (vi)
1T(F(Y» = 1T(F(Y) n X).
If 1T(F(Y»
O(H) = 1 by (v). r is odd.
* ll(F(Y)
"X), then
Iwl = 2, IQ! = 4 by (iv) and
Thus F*(H) = W x K and K is of type L (r) where 2
Then G is of type JR, a contradiction.
(This is the only time the exact structure of a group of type JR is used. ) (vii)
Put F = R(F(Y) n X).
For R
=
R acts regularly on Q
#
=
Then NF(R) = F = CF(R). Thus F is a nilpotent group.
But
and so the normalizer of R in the full linear
group on Q is, modulo CF(Q), a subgroup o£-the multiplicative group of the field GF(q) extended by an automorphism
0'.
If 0'
* 1,
such a
73
group is non-nilpotent and so F /CF(Q) = R. Hence R
~
It follows that F =R x CF(Q).
Z(F).
Let p E 1T(F), P =
n1 (Z(Op(F))).
Ny(P) ~ Z.
(ix)
Let x E CF(Q), [x, T] ~ W since T=WfEiQ.
Let Y E Ny(P) - X. Also if z ER #, CT(Z) = W.
-1
First Cp(Q) n Cp(Q)Y = 1.
For if a, aY
E Cp(Q), then
[a, T] ~ W n WY = 1 and so a = 1.
-1
Also (P n R) n (P n R)Y = 1.
For if a, aY
E P n R, then
- 1
cT(a) = W = CT(aY ) and W = WY, a contradiction. Since R is cyclic and P is elementary, Ip n RI
:s:
p.
But
P = P n R fEi Cp(Q). It follows that Ipi = p2. If S is an; subgroup of p such that p = (P n R)S = Cp(Q)S then CQ(S) ~ CQ(P) = 1, CW(S) ~ Cw(P) = 1. Thus S acts without fixed points on T. fixed points on T and are moved by y. changed.
But P n R, Cp(Q) have
Therefore they must be inter-
But then y has even order, a contradiction.
We have now shown that Ny(P) primes p.
C
=
X for P =
n (Z(O (F))) for all 1
P
Consider now R
acting on F(Y). If F(Y) ::> (F(Y) n X) p P p P choose M such that (F(Y) n X) C M C F(Y) and M is the smallest
=
p
such R -invariant group. P
M
C
=
~
p
p
(F(Y) n X) .
C
=
p
Hence M normalizes R (F(Y) n X) = F and so p p P Ny(P) =: X. Thus (F(Y) n X) =: F(Y). Since 1T(F(Y) n X) =: 1T(F(Y)),
we have F(Y) n X R
Then [R , M]
F(Y).
= F(Y).
P
Now [R, F(Y)]
Hence F =: F(Y) and P
~
Y.
P
= [R,
F(Y) n X] =: 1 and so
Thus X=: Y.
This completes
Let K be a component of E(H).
Then CT(K) is
the proof. 11 Lemma 15. 5. not cyclic. Proof. transfer
Let W = CT (K) = <w >.
IW I=:2
Since K n T is elementary, by
and NG(T) is transitive on involutions of T by 12.1. g For K n T has a single class of involutions. If g E G and t E W, g then C(w) ~ H . But K ~ CG(w) and since H/E(H) is solvable by 15.3,
74
K = Kg.
Then g f G and t f W. But then t f CT(K) n F*(H) by 15.1 and so t f Z(H).
Thus
{H) = M(t) and so by 15. 4, N (T) ;; H. But now transf er gives a contraG diction to the simplicity of G. This completes the proof. /I
Lemma 15. 6. E(H).
Assume E(H)"* 1.
Let K be a component of
Then (a) CG(K);; H.
!!
(b)
K;; M f M(s) for some involution s fT, then M = H.
(c) NG(T);; H. n (d) K is of type L (2 ). 2
Proof.
If possible choose, H, t, K such that K is of type JR.
The proof proceeds by verifying (a), (b), (c) with this restriction on K. When (d) is proved, it follows that this restriction on K is vacuous and the Lemma is completely proved. Choose an involution k f T n K, M f M(k).
By 15. 3, H/E(H)
and M/E(M) are solvable.
If K is of type JR, let N be the product of all such components of E(H).
Otherwise put N = E(H).
Note that N is characteristic in
any subgroup S of E(H) which contains it. Let K k
1
"* K be a component of N.
Then K
1
;;
CG(k) ;; M since
Because M/E(M) is solvable, K C E(M). Let C (t) 1 = E = K M . Then E = E(E) lies in E(M) and by 13. 3 any component E
K.
1
of E, for example K , is either a component of E(M) or is of type 1
L (q) contained in a component of E(M) of type JR. 2
if K
1
By choice of H,
is of type L (q), no component of E(M) can be of type JR. 2
Thus K
is a component of E(M) and so K
1 1 =
Let N = KK ... K. Then K ... K <J E(M) n CG(K) <J 1 r 1 r = CG(K) (;; CG(k) .;; M). It follows that N=KK ••• Kr~ KE(CG(K)) = 1
E(KCG(K)).
Hence E(KCG(K));; NG(N) = H and so E(KCG(K));; E(H).
Now N char E(KCG(K)) and so NG(E(KCG(K)));; NG(N) = H.
Therefore
CG(K) ;; H and (a) holds. (b)
If K;; M
component of E(M).
E
M(s), then K ~ E(M) and as above, K is a
Thus by (a) we have CG(K);; M.
Therefore
75
~
E(M)
H and so E(M)
~
E(H).
But also E(H)
~
M.
Hence E(H) = E(M)
and H = M. (c) s E U.
Let U
= CT(K).
By (b), M(s)
=
{H) for all involutions
Then by 15.4, NG(T) ~ H and by 15.5, U is non-cyclic. (d)
Let E be the product of all those components of E(H) not n of type L (2 ). We show that E = 1 and then (a), (b), (c), (d) hold 2
without restriction. Assume E
"*
l.
Since U is non- cyclic and CG(u) {H)
= M(u)
~ H for all u E U #, using
for all u E u#, every T::: U- invariant 2'-subgroup D of
G lies in H.
Also D
~
E for all T-invariant subgroups D of G such n that E(D) = D and no c~mponent of D is of type L (2 ) by 13. 6. 2
Remember T normalizes each component of D. Let M E M(s), M that M
"* H,
= O(M)E(M)NM(T).
sET.
Solvability of M/E(M) implies
For TO(M)E(M)
~
group, and then the Frattini argument applies.
t
O(M) ~ H, E(M) H. H. Let V = CT(L).
M, since M is an A*Since NM(T)
~
H,
Let L be a component of E(M) not contained in
By 15.5 applied to M, V is non-cyclic. # If CG(v) ~ M for all v E V , then E which is V-invariant,
n Also any component of E(M) not of type L (2 ) is 2 n T =:l U invariant. Thus every component of E (M) not of type L (2 ) = 2 n lies in H. Conversely every component of E(H) not of type L (2 ) lies would lie in E(M).
2
in M.
Thus E is the product of all components of E(H) and also n E (M) not of type L (2 ). Thus H = M, a contradiction. Thus there 2
exists v E V such that CG(v) ~ M, where v is an involution. Let R
E
M(v).
Then R
"* M
and of course L
~
R.
Every com-
ponent of E(R) is T-invariant and every component of E(R) of type JR lies in H by 13. 6.
Thus every component of E(Rj of type JR lies in
E(H) and so lies in E. By 14.3 applied to LCR(s C E(R), where CR(s) CR(s) = clearly L is semi-simple, since L ~ E(M), we see that L is a component of E(R) since otherwise L is of type L (q) and lies in 2
a component of E(R) of type JR.
Since L ~ H, this last possibility
does not arise. But now if K is not of type JR, our restriction on K, t, H is
76
vacuous and so (a), (b), (c) hold all the time.
Apply (b) to M, L in place
of H, K and get M
= R,
and so [T : U] = 8.
But U n V = 1 for if x E U n V, CG(x) ~ Hand
CG(x) ;; L
1H.
But
a contradiction.
Iv I ;;;
Thus
Ivl = [T : T
Therefore K is of type JR
8. Thus E(M) ~ L has at most two com-
n L].
ponents and E(R);; L has at most two components.
But then L
~
M, R,
a contradiction. /I Lemma 15. 7. Proof.
H is solvable.
If E(H)
1, let K be a component of E(H), U = CT(K).
=1=
=
By 15.6, NG(T) ;; Hand M(s) by 15.5, U is non-cyclic.
{H} for every involution s E U.
Also
Hence any U-invariant odd order subgroup
of G is contained in H. Let M E M(s), M
=1=
H, sET.
group, M = O(M)E(M)NM(T).
Since M;E(M) is a solvable A*-
Thus E(M)
1H.
Let L be a component of E(M) not contained in H and let #
V = CT(L). By 15. 4, NG(T) ;; M and M(v) = {M} for all v E V . Also U n V
=1
because if x is an involution in U n V,
L ~ CG(x) ;; H. (i) We may assume that L
~
M.
If neither K ~ H nor L ~ M, then both E(H) and E(M) contain
at least two other components isomorphic to K, L respectively.
IU I)
ITn K
I = [T
IV I >
: U],
ITn L
I = [T
Then
: V] and then U n V
=1=
L
Thus we have (i). n Since K is of type L (2 ), K has a cyclic subgroup R which is inverted by some involution Then U = CT(R) and R and CT(L).
~
K and which acts regularly on T n K#.
NG(T) ;; M.
Thus R normalizes L, L n T
Moreover R acts irreducibly on [R, T]
V n [R, T] or V n [R, T] (ii)
i~
=T
n K and so
= V.
U = T n Land RL = R x L.
For if V n [R, T] T modulo [R, T].
=1
then [V, R]
=1
But V n CT(R) = V n U = 1.
if V::> [R, T], then CV(R)
=1=
because R centralizes Thus V ~ [R, T] and
1 and CV(R) ~ U n V.
77
Thus V = [R, T]. It follows that R centralizes L n T. T = V U
=L
q:)
(L n T).
Since CT(R) = U
~
But
L n T and U n V = 1, we have
n T. n Moreover R normalizes L, of type L (2 ) and centralizes a 2
Sylow 2-subgroup L n T of L.
Thus R cannot induce field automor-
phisms on L and must induce inner automorphisms on L.
Since
CL(LnT)=LnT, we have [R, L]=1 and RL=RxL. (iii)
U is a Sylow 2-subgroup of CG(R).
For let U
::J I
=
U be a Sylow 2-subgroup of CG(R). h
Then
h
;; CG(u);; H for u E U and so U ~ T n CG(R ) for sor:;e h EH. I I n Then Rh ~ Kh , a component of E(H) of type L (2 ) and R is a sub2 h n group of order 2 - 1 acting regularly on T n K . But h h h h h . T = (Thn K ) x CT(K ) and CT(K ) ~ CT(R ) ~ U . Smce
U
ICT(K)I = ICT(K)[ =
lul,
[uII
we have
I
=
iul, UI =U
and U is
a Sylow 2-subgroup of CG(R). (iv)
U ~ F*(CH(R)).
Since T n L is elementary by (ii), U is elementary abelian. By 15.1, U ~ F*(H) because M(u) = {H) for all u 15.6. Thus U
~
E
U by
CG(R) n F*(H).
Let F*(H) = KK ... K F(H): r
I
CG(R) r F*(H) = RK
I
.•.
KrF(H);
RF(H) ~ F(CH(R)). Now K ... K
r
I
H.
Hence K
normalizes F(CH(R)), a solvable subgroup of
K centralizes F(CH(R)). Since CH(R) is an A*r group, it follows that K ••• K ;; F*(CH(R)). Hence I r U ;; F*(H) n CG(R) ;; F*(CH(R)). (v)
...
I
U
~
F*(CG(R)) and so L
~
F*(CG(R)).
For O(CG(R)) is a U-invariant subgroup of G. O(CG(R));; H.
Thus
Hence [O(CG(R)), UJ ~ O(CG(R)) n F*(CH(R))~F(CH(R)).
Thus [O(CG(R)), D, D] = [O(CG(R)), D] = 1.
By 13.1, U ~ F*(CG(R)).
Since U is a Sylow 2-subgroup of CG(R) and L ~ CG(R), it follows that L;; F*(CG(R)).
Thus L = E(CG(R)).
NG(R) ~ NG(L) = M.
Hence K =
78
<
It follows that
T n K, NK(R)
>
~ M.
This contra-
diets 14. 6(b) because M *- H.
NG(T) ~ H.
Lemma 15. 8. Proof. O(H)
= 1,
This completes the proof. /I
Since H is solvable and t E F*(H), t E 02 (H). If
NG(T)
~
Let pEn, p *- 2.
H clearly.
Let P be a maximal
T-invariant p-subgroup of G containing 0p(H). pc H.
(i)
For CG(P n H) ~ H and so [t, Cp(P n H] ~ P n 02 (H) [P n H, t] ~ P n 02 (H)
= 1,
= 1.
Since
2.2 implies that P ~ CG(t) ~ H.
(ii) (lI1 (T, p') ~ H. G The Transitivity Theorem 4. 1 obviously applies here and so ~
H acts transitively on the maximal elements of II1 (T, p'). G Since P ~ H, (II1 (T, p') ~ H. G g g Now let g E N (T). Since H ~ T, F(H )2 is centralized by T. G g g g Because F(H )2' E II1 (T, p'), F(H )2' ~ H by (ii). Thus F(H ) ~ H. G g g g It follows that [t, F(H )] ~ [t, F(H )2'] ~ F(H )2' n 02(H) = 1. Because g g [t, F(O(H ))] = 1, [t, O(Hgn = 1 and so O(H ) ~ H. Since H has g g 2-length 1, being a solvable A*-group, O(H ) ~ O(H). Thus H = H CG(T)
and NG(T) ~ H. /I
Lemma 15.9. is non-cyclic.
Proof.
°
°
2
(H) is clearly non-cyclic because, by 15. 8, if
°
2
(H)
(H) ~ Z (NG(T)) n T and transfer then contradicts the 2 simplicity of G.
were cyclic,
Let x E 02(H) be an involution, M E M(x). and so H
=
°(H)NG(T)
~ M.
~
By 15.8, NG(T)
For [x, O(H)] ~ 02 (H) n O(H)
=1
M
and so
O(H) ~ CG(x) ~ M. /I
Lemma 15. la. Proof.
M(s)
=
for all involutions s E T#.
For let M E M(s), sET.
15.9, O(M) ~ H because O(M) NG(T) ~ H.
{H}
Thus M
= H.
°
= (M)NM(T). By # : x E 02 (H) ). By 15.8
Then M
= (cG(x)nO(M)
/I
This contradicts 13.3 and completes the proof of Theorem A.
79
APPENDIX: p-CONSTRAINT AND p-STABILITY These concepts are rather natural generalizations of aspects of the theory of p-solvable groups - see [14].
The definition of p-constraint
is taken from a crucial property of p-solvable groups noticed in the famous Lemma 1. 2. 3 of [14]'
The definition of p-stability is taken from
the famous Theorem B of the same paper.
The reader should be familiar
with both that paper and also the exposition of these concepts in [12].
A
very little discussion of these topics is included to overcome an error in the Gorenstein treatment and also an omission - it is important to know how much induction one has with these concepts. Definition.
Let p be any prime.
A group G is said to be p-
constrained if, when P is a Sylow p-subgroup of 0, (G), then p,p CG(P) cO, (G). = p ,p Definition.
o p (G)
"* 1.
Let p be an odd prime, G a group in which
Then G is said to be p-stable when, for any p-subgroup
A c G and any A-invariant p-subgroup P cO. (G) such that = = p ,p 0p,(G)P;1 G and A, A] = 1 it follows that
rp,
First, it is easy to see that if P is a Sylow p-subgroup of 0, (G) p,p and CG(P) is p- solvable, then G is p- constrained. For by Lemma 0.3, C G /0 (G)(P) = CG(P)O ,(G)/O ,(G) and there is no loss of gener, p' p p ality in assuming 0 ,(G) = 1. Thus CG(P) <J G and we can find
=
p
K
~
G, K
~
P such that K
~
PCG(P) and KIP is a chief factor of
if CG(P):t P. Since P = o. (G), K/P is a p'-group and then p,p a q-group for some prime q"* p. Since K ~ PCG(P), K = PCK(P). Let
GIP,
Q be a Sylow q-subgroup of K contained in CK(P). K = PQ = P x Q and so QC 0 ,(K) cO ,(G) = 1.
=
P
=
P
Then This shows that G
is p-constrained if CG(P) is p-solvable where 0 ,(G)P p a p-group.
= 0, (G), p,p
P
Notice however that the property of p-constraint does not necessarily pass to either subgroups or to factor groups.
80
In order to see this
consider A
x C cA.
5
3= 8
Clearly A
8
5
x C
3
is not 3-constrained. We
can make a group 7 (A xC) by extending an elementary abelian group of order 7
5
8
by A x C 5
3
3
with the natural action.
Now let this new
group act faithfully on any elementary abelian 3-group. The group k G = 3 7 8 (A xC) so constructed is clearly 3-constrained since the 5 3 k elementary abelian group 3 = 0 , 3(G) is self centralizing. Since 3, A x C is both a subgroup and a factor group of G, we see that p5
3
constraint does not induct in either direction. The following result shows that some induction to both factor groups and subgroups is possible under certain circumstances. Lemma 1.
(a) A group G is p-constrained if and only if
G/O ,(G) is p-constrained p
-
(b)
!!
G is p-constrained, then NG(P) and CG(P) are p-
constrained for every p-subgroup P of G. Proof.
(a) This follows immediately from O. 3.
(b) Using O. 3 and (a) we may assume that 0 ,(G) p
= 1.
The
result follows from 12. 5. We turn now to p-stability. It is worth stating the celebrated Theorem B here so that the original genesis of p- stability can be discussed. Theorem B (P. Hall and G. Higman).
Let G be a p-solvable
group of linear transformations in which 0 (G)
= 1,
space V over a field F of characteristic p.
Let x be an element of
p
order pn.
acting on a vector
Then the minimal polynomial of x on V is (X -
= pn
(i)
r
or
(ii)
there exists an integer n
Il
where
n
<
0=
n such that p
0 -
1 is a
power of a prime q for which the Sylow q-subgroups of G are nonabelian. In this case, if n n-n p
n o(p
0 _
°
is the least such integer, then
1) ~ r ~ pn.
Of course, it is no surprise that the minimal polynomial is of the form (X - l)r where r ~ pn,
Clearly the minimum polynomial divides
81
xP
n
n - 1 = (X - l)P.
bound.
The interesting part of this Theorem is the lower
For our purposes we will enquire when the constant r can be
2, i. e., when will x have quadratic minimum polynomial? Clearly it always will if pn = 2 and this is the reason for excluding the prime p = 2 from the definition of p-stability.
n-n
Thus r = 2 occurs only when p
n 0(p
°- 1) = 2
and this holds
only when n = n = 1, P = 3. A Sylow 2-subgroup of G will be non-
°I.
abelian and 3/1 G that r
>
A careful reading of the proof of Theorem B shows
2 unless SL(2, 3) is involved in G.
To get to the hypothesis of Theorem B in an abstract group G, suppose that P is a p-subgroup of G and that 0p(NG(P)/CG(P))=l. A be an abelian p-subgroup of NG(P) and let V = P /
Let
Then
2
[v, a, aJ = v(-1+a) Since [P, A, AJ = 1, (-1 + a)2 is the zero endomorphism of V and so a acts on V with at worst quadratic minimum polynomial. If for example SL(2, 3) is not involved in G, we know that a must act trivially on V. It will follow that A
~
P and so P contains
every abelian subgroup normal in a Sylow p-subgroup of N(P).
Further
information can be found in [12J. Once again, p-stability does not go over to proper sections. For a careful reading of the proof of 3. 8. 3 of Gorenstein [12J shows that, when x and y are conjugate p-elements in a group X such that (x, y) is not a p-subgroup while x acts on a G-vector space of characteristic p with quadratic minimum polynomial, it follows that (x, y) involves SL(2, p). Now consider, A , the alternating group of degree 8. 8
an elementary abelian 3-group of order 3 of V by A X
of A
8
with the natural action of A
8
8
and G the split extension on V. Given any 3-element
there is always a conjugate y of x in A
8
Let V be
8
such that (x, y) ~A .
4
Hence (x, y) does not involve SL(2, 3) and so x cannot act on V with quadratic minimum polynomial.
Thus G is 3-stable.
Since however G contains a section isomorphic to Qd(3), a split
82
extension of a group of type (3, 3) by SL(2, 3) and a classical non3-stable group, we have exhibited a 3-stable group with a non-3-stable proper section. In view of this, the following Lemma is useful. Lemma 2.
° (G) "* 1
A group G in which
p
is p-stable if and only
-
if G/O ,(G) is p-stable. p -
Proof.
X ;;
If G is p-stable, write G = G/O ,(G), etc. Suppose that p
~(P)=G is such that
of G.
Let K =
° ,(G). p
[P, X, X] = 1 where X, Pare p-subgroups
Now A normalizes PK and since the number
of Sylow p-subgroups of PK is prime to p, there exists a k that Ak normalizes P. k k Also rp, A , A ];; K. k
k
[P, A , A ] ~ P n K
E
K such
Hence
= 1.
Therefore since G is p-stable, AkCG(p)/CG(P);;Op(NG(P)/CG(P)), By 0.3, CG(P) = CG(P)K/K, and so G/CG{P) ~ G/CG(P)K = NG(P)C
G
(P)K/CG(P)
~ NG(P)/(NG(P) n CG(P)K) = NG(P)/CG(P).
~
Since AkCG(P)/CG(P);;
Op(NG~)/CG(P)) ~ 0p(G/Ca(P))
and
A = X, CG{P) = CG(P), we have that G is p-stable. Conversely suppose that G = G/O ,(G) is p-stable. p
Let P be a
p-subgroup of G such that 0p,(G)P;; G and suppose that A;; NG(P) is a p-subgroup such that [P, A, A] = 1.
Since [P, X, X] = 1, we
have that
Now
and
83
NG(P). CG(P). 0p,(G)/CG(P). 0p,(G) ~ NG(P)/(NG(P) n (CG(P)Op,(G))
= G/CG(P)Op,(G)
= NG(P)/CG(P).
Thus under the above isomorphism
and G is p-stable.
84
Lemma 2 is completely proved.
References
[1]
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[2]
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Proc. Am. Mat
LONDON MATHEMATICALSOCIETY L E C T U R EN O T E S E R I E S Editor: PRoFEssoR G. C. SI{EPHARD, Universityof East Anglia This series publishes the records of lertures and seminars on advaDc€d topics in mathematics held at universitiss tbroughoBt the world, For the most part, theseare at postgmduatelevel, either presentingnew matqrial or describing older material in a new way. Exceptionally, topics at tlrc uadergraduate level may be published if the treatment is sufficiently original.
Topicsin FiniteGroups TERENCE M. GAGEN Senior Lecturer itt Purc Mathetatics, University of Sydney Thesenotes derive from a course of lectures delivered at the University of Florida in Gainesville d:unt!.g197112.Dr Gagen presents a simplified treatment of recent work by H. Bender on the classification of non-soluble groups with abeliaa Sylow 2-subgroups,together with some background material of wide interest. The book is for researchstudents and specialists in goup theory and allied subjectssuch as finite geometries. For a list of bookspreviouslypublished in this seriesseepsge i
c.u.P. litl ttt u.(.
rL
o 521 210f]/2X
v\.)
