5‘ L ’2
:hgfl‘ld

:
9

:0)” 53',
"#355

2 N
4!
Z‘S’n?

 

THESlQ

W ”‘llllllllll ill

will I .

"' l
|“H3 ‘1‘ 93 oi555 2361

 

l

 

 

 

 

 

 

 

 

 

 

 

 

This is to certify that the

dissertation entitled

Core-free Maximal Subgroups
of Locally Finite Groups

presented by

Neil Henry Flowers

has been accepted towards fulfillment
of the requirements for

Ph . D . degree in Mathemfl L .I 95

@% zx/

Major professor

Date 7/l 6/?g

MSU i: an Affirmative Action/Equal Opportunity Institution 0-12771

 

LIBRARY

Michigan State
University

 

 

 

PLACE |N RETURN BOX to remove thin chookout from your record.

‘ TO AVOID FINES rotum on or More dd. duo.

DATE DUE DATE DUE DATE DUE

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

usu loAn Affirmative ActiorVEquol Opportunw Institution
Want

 

CORE-FREE MAXIMAL SUBGROUPS OF LOCALLY FINITE
GROUPS

By

Neil Henry Flowers

A DISSERTATION
Submitted to
Michigan State University

in partial fulfillment of the requirements
for the Degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

1996

ABSTRACT

COREFREE MAXIMAL SUBGROUPS OF LOCALLY FINITE GROUPS
By

Neil Henry Flowers

In this dissertation we determine the structure of the centers of certain maximal
subgroups in both finite groups and locally finite simple groups. In Theorem 1.1
we show that in a finite group G the center Z ( M ) of a solvable core—free maximal
subgroup M is cyclic modulo 02(Z (M ))03(Z (M )) In Theorem 3.1 we show that, for
certain primes p, a nonfinitary locally finite simple group cannot contain a subgroup
of type (p, p) in which all nontrivial elements have the same centralizer. As a result
we are able to show that the center of a maximal subgroup of a nonfinitary locally

finite simple group is locally cyclic modulo one of its Sylow subgroups.

DEDICATION

To my mother Nona and my wonderful daughter Felicia

iii

ACKNOWLEDGEMENTS

I would like to thank my Advisor, Ulrich Meierfrankenfeld, for working closely with
me through out the completion of this dissertation. It is from him that I’ve learned
many advanced Group Theory techniques and I feel I’ve grown tremendously as a
result of our interaction. I cannot think of any other way one could have been a
better advisor and I am extremely fortunate to have had the oppourtunity to have

worked with him.

I would also like to thank Richard Phillips and Susan Schuur for entertaining my

various questions concerning my dissertation and other aspects of my personal life.

Finally, I would like to thank Bernd Stellmacher for helping me in the early stages
of my dissertation and for being a great Group Theory teacher. It was through taking

a class from him that I initially became attracted to the subject.

iv

Contents

Introduction 1
1 Core-free maximal subgroups of finite groups 3
2 The Amalgam Method 18
3 Maximal subgroups of locally finite simple groups 29
3.1 Locally finite simple groups of q—type .................. 32
3.2 Locally finite simple groups of alternating type ............. 39
Bibliography 111

Introduction

A common theme throughout Group Theory has been the interplay between a group
and its maximal subgroups. Of particular interest has been the phenomena in which
properties of the maximal subgroup of a group impose a certain structure on the group
itself. In [11], for example, John G. Thompson showed that any finite group which
has a nilpotent, maximal subgroup of odd order must be solvable. Its the influence
that a maximal subgroup can have on the whole group that makes the study of the

structure of maximal subgroups worth while.

A core-free maximal subgroup of a group G is a maximal subgroup in which the

intersection of all its conjugates is trivial, that is, Coreg(M) = H M9 = 1. In

geG
Chapters 1 and 2, we explore the centers of such subgroups and show that in a finite

group any core—free maximal subgroup which is solvable has a cyclic center modulo

02(Z(M))03(Z(M))-

In [6] Liebeck and Saxl proved that any core-free maximal subgroup of a finite
group must have cyclic center. However their proof of this theorem relied on the
classification of finite simple groups, a reasonably powerful tool. Although Theorem
1.1 is weaker than this result, the proof of Theorem 1.1 does not rely on the classifica—
tion of finite simple groups. Rather it depends on The Amalgam Method, a method
developed by Daniel Goldschmidt [2] and Bernd Stellmacher [9]. This method is a

way of analyzing properties of a group through its action on the coset graphs of some

of its subgroups.

After the completion of the classification of finite simple groups a new interest
arose in the structure of locally finite simple groups. In [7] Ulrich Meierfrankenfeld
showed that a locally finite simple group must be one of three types, a finitary group, a
group that is covered by a system of finite subgroups each of which having alternating
quotients, or a group that is covered by a system of finite subgroups each of which
having quotients isomorphic to projective special linear groups defined over finite
fields of prime characteristic. Using this classification, in Chapter 3 we explore the
structure of the centers Z (M) of maximal subgroups M of nonfinitary locally finite
simple groups G'. Indeed, we prove that, for certain primes p, G does not contain
a subgroup Z of type (p,p) in which every nontrivial element of Z has the same.
centralizer in G. Therefore, as a consequence, we show the center Z ( M ) of M,

modulo one of its Sylow-subgroups, is locally cyclic.

Chapter 1

Core-free maximal subgroups of
finite groups

In this section we will start a proof of Theorem 1.1 (below) which does not depend
on the classification of finite simple groups. The proof of Theorem 1.1 is primarily

an application of the Amalgam Method.

Theorem 1.1 Let G be a finite group and M be a core-free maximal subgroup of G.
Suppose further that M is solvable and p a prime with p Z 5. Then Op(Z(]W)) is

cyclic.

Let (G, M, p) be a minimal counter example to Theorem 1.1. Then, since 0,,( Z (M ))

is noncyclic and abelian, there exists Z S 0,,( Z (M )) of type (p, p).

Lemma 1.2 (a) UN S G andN S M, then N: 1.
(b) For anyC S Z(M) andl #- c€ C, Cc;(c) = Cc(C) = M.

(c) IfQ S G is a p’-group and Z S Ng(Q), then [Z,Q] =1 and Q S It].
Proof: For (a), since N S G and N S M, we have

N S n M9 = Corec(M) =1

QEG
3

and therefore N = 1.

Now C S Z(M) implies M S Cc(c) S G. Hence, by the maximality of M in
G, we get M = Ca(c) or Cc;(c) = G. If G = Ca(c), then (c) S G and (c) S M.
Thus, by (a), we get (c) = 1, a contradiction. Therefore M = Ca(C). But then, since
C S Z(M), we get 06(0) S Ca(C) = M S 00(0). Thus M = Ca(c) = Cc(C)
and we have (b). Finally, for (c), since Z is a noncyclic abelian p—group acting on a

p’-gr0up Q, by [3, 6.2.4] we have
Q=(CQ(Z)|1#26 Z) (1)

Now by (b), Cq(z) S Cg(z) = M for each 1 aé z E Z. Thus, (1) implies Q S M.
Since Z S Z(M), we get [Z, Q] = 1.

Lemma 1.3 Suppose T is a finite solvable group and P S T is a p-subgroup. Then
0.4mm» 3 0.4T).

Proof First assume 0,,:(T) = 1. Then we want to show 0pr(NT(P)) = 1. Let
A = 0,,v(NT(P)) and B = 0,,(T). Since A and P are normal subgroups of NT(P)
and (IAI, IPI) = 1, we have [A, P] S Afl P =1. Also, since B S T, A x P acts on B.
Further [A,C3(P)l S Am B = 1, and so by [3, 5.3.4] [A, B] = 1. Thus A S CT(B).
If A 7b 1, then CT(B) is not a p—group and so B < C'T(B)B S T. Let T = T/B
and G171?) be the image of CT(B) in T. Then we can pick N = N/B S T minimal

 

such that N S T and N S CT( B) Now T is solvable implies T is solvable and so N
is a elementary abelian q-group. If g = p, then N is a p-group. But, since N S T,
we get N S 0,,(T) = B and hence N = 1, a contradiction. Therefore q 75 p. Let
Q 6 Squ(N). Since N is a q-group we have Q = N and therefore N = QB. But
Q S BCT(B), and so

 

’z—Z’ is a p—group.

CT(B) CT(B) n B

 

Thus QCT(B)/CT(B) = 1 and so Q S CT(B). But then N = Q x B and Q charN S
T. Therefore Q is a normal p'-subgroup T and hence Q S Op:(T) = 1. But then we
get N = QB = B and N =1, a contradiction. Thus A = OpI(NT(P)) =1.

Now if 0p:(T) ;£ 1, let T = T/OpI(T) and P be the image of P in T. Then

OAT) = 1 and so by the above argument applied to T and P we get 0,,I(NT(P)) = 1.

Next we claim that N-T-(P) = NT(P). Clearly NT(P) S NT(P). Let t 6 NT P). Then
"i = r implies (190,473)t = P‘0,.(T) = P0,..(T). Therefore P and P‘ are p—Sylow
subgroups of P0,,:(T). Therefore there exists x E 0,,:(T) with P" = P. But then
tx e NT(P) and therefore i 6 W797. Thus NfiP) 3 TV???) and the claim holds.

Now we have

0p’(NT(P)) S Op’(NT(P)) = Op’(NT(?)) =1

and so OpI(NT(P)) S Op:(T) and the lemma is proved.

Lemma 1.4 (a) OPI(G)0,,(G) = 1, in particular F(G) = Z(G) =1.

(b) Iffll(Z(M)) S H < G, H H M is maximal in H, and Q :2 CoreH(H n M),
the" 91(Z(M)) .<_ 91(Z(0p(Q)))-

(c) There exists g E G with 91(Z(M)) S M9 andg é M.

(d) 0,.(M) = 1.

Proof: Clearly, since Z(G) S F(G) S 0,,I(G)0,,(G), 0p:(G)OP(G) = 1 implies
F (G) = Z(G) = 1. Now 0pI(G) S G implies Z normalizes the p’-group OpI(G) and
so, by Lemma 1.2 (c), we have [Z,0,,:(G)] = 1. But then, 0p:(G) S Cc(Z) = M, by
Lemma 1.2 (b). Therefore, since OPI(G) S G, by Lemma 1.2 (a), we have 0pr(G) = 1.
Next suppose I 75 0,,(G). Then, since 0,,(G) is a normal p-subgroup of G, we have
1 # C = Cc(Op(G)) S G. Therefore, by Lemma 1.2 (a) and the maximality of
M in G, C S M and G = CM. But then Mfl0p(G) S CM = G and so, by
Lemma 1.2 (a), M n 0,,(G) = 1. Now Z acts on 0p(G) and for each 1 76 z E Z,
C0P(G)(z) S Ca(z)r10p(G) = MflOP(G) = 1. Thus, Cop(G)(Z) = 1 for each 1 75 I: 6 Z
and Z acts fixed—point freely on 0,,(G). But, since Z and 0,,(G) are p—groups, this is
a contradiction to [3, 2.6.3]. Therefore, 0pI(G)0p(G) = 1 and we have (a).

For (b), suppose Z 0 Q = 1. Let H: H/Q and W be the image of H n M
in H. Then m is a core-free maximal subgroup of H. Moreover, M is solvable
implies W is solvable. Thus, by the minimality of G, 0,,(Z (HO—117)) is cyclic.
But Z (1 Q = 1 implies Z 2 Z and Z S 0p(Z(H—FW)) is of type (p,p), yeilding a

contradiction. And so we may assume Z 0 Q at 1.

Since Q S M, [521(Z(M)),Q] = 1 and so 91(Z(M)) S CH(Q). Now, by Lemma
1.2 (b), we have CH(Q) S Cg(Z (1 Q) = 00(2) = M and therefore CH(Q) S
H n M. But then CH(Q) S H implies CH(Q) S CoreH(H n M) = Q. Thus,
01(Z(M)) S CH(Q) S Q. Now, since Q centralizes {21(Z(M)), we have 01(Z(M)) S
n1(Z(0p(Q))-

For (c), if (91(Z(M)),01(Z(M))9) = G for each 9 6 G \ M, then M 0 M9 S
(91(Z(M)),QI(Z(M))9) = G for eachg E G\M. Thus, by Lemma 1.2 (a), MGM? =
1 for each 9 6 G \ M. Now Frobenius’ theorem [3, 2.76] implies G 2 NM, where
N = (G\UgEG M”) U {1} S G and ([Nl, [M]) = 1. But, since p E 7r(|M|), N is a p’-

group. Hence, by Lemma 1.4 (a), N S Opr(G)0p(G) = 1 and we get G = MN = M,

a contradiction to the maximality of M.

So there exists 9 E G \ M with (521(Z(M)),S21(Z(M))9) 75 G. We may assume
(Q1(Z(M)),QI(Z(M))9) S M, otherwise, 91(Z(M)) S M-"-1 and we are done. Thus,
we can choose H < G minimal such that 91(Z(M)) S H and H S M. Then H n M
is a maximal subgroup of H. For if H n M 3 H0 3 H, then o,(z(M)) 3 Ho 3 H.
Thus, by the minimalty of H, Ho S H D M or Ho = H. Now by (b), {21(Z(M)) S
CoreH(HflM) s 0,6,, M. Since H g M, pick ah e H\M. Then n,(Z(M)) s M[1

and we are done.

Next we want to show 0p.(M) = 1. By (c), there exists 9 e G \ M with Z s
o,(Z(M)) 3 M9. Then [2,29] = 1 and so 29 s 00(2) = M, again by Lemma 1.2
(b). Now Z9 acts on the p’-group Op:(M) and so, by Lemma 1.2 (c), [Z9, Op:(M)] = 1
and Opv(M) S M”. Hence, Opr(M) S M (1 M9, and so Op:(M) S OpI(M 0 M9). But
Opr(M 0 M9) = Op:(Ng(Z) 0 M9) = 0,,:(NM9(Z)) S 0pI(M9) by Lemma 1.3. Thus,
0,,:(M) S 0pr(M9). Now by the symmetry of this argument, we get Opr(M) =
OpI(M9). And so if 0pI(M) 7:4 1, then

M = moot» ——- NG(0p'(M9)) = M9

Thus, M = M9 and g E Na(M) = M, a contradiction to the choice of g.

Lemma 1.5 Let T be a finite group, P E Sylp(T), and H S Z(P). Then

}

0p(T/Cr(<HT>)) =1

Proof Let V = (HT), T = T/CT(V) and K = K/CT(V) = 0,,(T). Then K S T
implies K S T. Now P E Sylp(T) so T = NT(P fl K)K by the Frattini Argument.
Since, K is a p-group and K 0 P 6 Sylp(K), we have K = W and so K =
(K n P)CT(V). Thus, T = NT(K fl P)CT(V). By assumption H S Z(P), and so
V = (HT) = (HNTlKnpl) S CT(K n P) Thus, K 0 P S CT(V) and so K = CT(V).

But then, 0,,(T) = K = 1.

Lemma 1.6 Suppose A is a elementary abelian p-group which acts on a nilpotent

p'-group T. Then

[T,A] = (ICT(B)9AIIB _<_,, A) (1)
and if in addition A is noncyclic,
[T,A]=([CT(a),A]|1¢ a E A) (2)

Proof Since A acts on T, A also acts on [T, A]. Since A is an elementary p-group

and [T, A] is a pI-group, we have

[T,A] = (ICIT,A1(B)|B so A)-

Let C = ([CT(B),A]]B S, A). Since A is a p-group and T is a p’-group, by [3, 5.3.6],
we have [T , A] = [T, A, A]. We claim (CIT'AI) = [T, A]. First, by the definition of C,

we have C S [T, A] and so (CIT'AI) S [T, A]. On the other hand, by the commutator

9

laws [3, 2.2.1], (1) implies [T, A] = [T,A,A] S (CIT'Al). Thus [T, A] = (CIT'AI) and
the claim holds.

Now suppose C < [T, A] and let C S D S [T, A], where D is a maximal subgroup
of [T, A]. Then, since T is nilpotent, [T, A] is nilpotent and so D S [T, A]. But
then we get [T, A] = (CIT'Al) S D and hence [T, A] = D. This is a contradiction to
the maximality of D. Therefore C = [T, A] and the first statement of the lemma is
proved.

If A is noncyclic, then [T, A] = (CIT.A](a)|1 75 a 6 A) by [3, 6.2.4]. Now the
second statement in the lemma follows by the same argument used above applied to

C = (C[T,A](a),A]]1;Ié a E A).

Lemma 1.7 Let T be a finite group, H S T be solvable, and V S H be an abelian
p-group with Op(H/CH(V)) = 1. Suppose there exists a noncyclic elementary abelian
p-group P S T such that CT(P) = CT(x) for each 1 924 x E P. Then,

H. = (P‘IP‘ s H) 3 Cum. (1)

Proof: Suppose not. Then there exists t E T with P‘ S H and Pt S CH(V). Let
E = Pt and H = H/CH(V). Then, B 75 1 and OP(H)=1. Now EflCH(V) = 1,
otherwise there exists 1 75 e E E0 CH(V) and V S 01(6) 2 CT(E) would imply B =
1. Thus, E E“ E is a noncyclic abelian p-group. Since 0,,(H) = 1, F(H) is a p’-group.
Hence, by [3, 6.2.4] applied to E and F(H), we get F(H) = (CF(7,-)(E)|1 aé E 6 E).

But then, by Lemma 1.6, we have

[F(H), l=(lCr(H)(E)»Ell19‘é E E E) (2)

 

10

Let W; = [CF(-,7)(E), ] for each 5 e 73'. Since, (Jr/(a) = Cv(e) g CT(e) =
CT(E), we have [CV(E),-—] = 1 and so [CV(E),F,CF(fi)(E)] = 1. Also, V H—invariant
implies [ct/(5), omega] 3 C‘s/(a). Thus, [cm-me), co(e),F] = 1. So by the Three
Subgroup Lemma we get [W;, CV(E)] = 1. Now the p’-group W; acts faithfully on
the abelian p-group V to give V = CV(WE) x [V, W;]. But, [W3,CV(E)] = 1 implies
C[v,w;](§) = CV(E) fl [V,Wg] S Cv(Wg) fl [V,Wg] = 1. Thus, C[V.We](5) = 1. Since
both (E) and [V, W;] are p—groups, we conclude [V, W;] = 1. Thus, as H acts faithfully
on V, we get W; = 1. Now, since E E E was arbitrary, by (2) we get [B, F(H)] = 1.
But H is solvable implies H is solvable and so, by [3, 6.1.3], B S Cfi( F (H)) S F(H).

This is a contradiction, since 1 ¢ B is a p-group and F (H) is a p’-group.

Lemma 1.8 M contains a Sylow p-subgroup of G.

Proof: Suppose not. Let S E Sylp(M) and T E Sylp(G) with S S T, S 7t T. Then,
since S and T are p-groups, there exists x E NT(S) \ S. Now Z S Z(M) a p—group
implies Z S Z(S). Thus, since x E NT(S), we have Z: S Z(S) and [Zx,0p(M)] = 1.
Now M is solvable, so by [3, 6.1.3], we have CM(F(M)) S F(M). But OpI(M) = 1,
by Lemma 1.3 (d), and so we get Z’ S CM(0p(M)) S 0,,(M) and, consequently,
z: s 2(0.(M)).

Let V = ((Z‘”)M). Then V is a normal abelian p-subgroup of M. Moreover,
by Lemma 1.5, Op(M/CM(V)) = 1. Thus, M, Z, and V satisfy the hypothesis of
Lemma 1.5 to give Mo = (Z9IZ9 S M) S CM(V). But then Mo S M 0 M“c and
M5“ = (Z9IZ9 S M‘”) 2 Mo. Therefore M5c = M0 and since 1 75 Mo S M, by Lemma

1.2 (a), we get

11

M = NG(M0) = NG(Mg) = Me

This means x 6 Nc;(M) = M and therefore x E M 0 T = S. This contradicts the

choice of x.

The following proposition follows quite easily from Thompson’s work on quadratic

pairs for p 2 5, but we prefer to give an elementary proof.

Proposition 1.9 Let p Z 5 be a prime, T a finite group, V a finite dimensional
GF(p)T-module, P E Sylp(T), P S H S T, and Q = {t 6 T|[V,t,t] = 0}. Suppose

that

(a) H is the unique maximal subgroup ofT containing P,

(b) H is solvable,

(6}QQH-

Then CV(H) = Cv(T).

Proof Clearly Cv(T) S CV(H), and so we only need to show CV(H) S CV(T).
The proof is by induction on |T| - dimGF(p)V. If CT(V) 95 1, let T = T/C7~(V),
Q = {t E T|[V,i,fl = 0}, and S, H be the images in T of S and H respectively.
Then P 6 Sylp(T), H is solvable, and V is a GF(p)T—module. If H 2 CT( V), then
T = HCT(V) by the maximality of H. Hence, CV(H) S CT(V) and we are done.
So we may assume that CT(V) S H. Then hypothesis (a) and (c) imply that H is
the unique maximal subgroup of T containing S and Q S H. Now by induction we
get Cv(H) = CV(H) 2 CV (T) = CV(T) and we are done. Therefore we may assume
that CT(V) = l and V is a faithful T—module.

12

Let U = (Cv(H)T). If U = 0 then we get CV(H) = 0 S Cv(T) as desired. Hence,
we may assume U 79 0. Suppose U at V. Then, since Q Q QU = {t E T|[U,t,t] = 0},
Q0 S H and so T, P, H, U, and QU fullfill the assumptions of the theorem. By

induction we conclude that CU(H) = CU(T). But then,

Cv(H) S CU(H) = CU(T) S CV(T) =1

and so CV(H) = 1, again we are done. So we may assume that

V = (Cv(H)T). (I)

Then by (1) CoreT(H) S CT(V) = 1 and hence

CoreT(H) = 1 (2)

Let X S H be a p—group maximal with respect to X = (X 0 Q) and NT(X) S H
(we can make this choice since NT(1) S H). Since X is a p-group there exists h E H
with X S P". Now, since V is a p—group, by [3, 5.3.6] all elements of Q are p-
elements. Therefore, since P" E Sylp(T), each element of Q lies in a conjugate of
Ph. Suppose (Ph D Q) S T. Then, since (Ph (1 Q) S H, (Ph 0 Q) S CoreT(H) = 1
by (2). But then, since the set Q is T—invariant, ((P")T 0 Q) = 1. Thus, we get
Q _C_ ((P")T 0 Q) = 1, a contradiction to assumption (c). Therefore (Ph (1 Q) S T.
Now we have P g NT((Ph n Q))"" as T and so by (a) NT((Ph n Q)) g H. Hence,
X are <P" 0 Q)-

Suppose Nph(X) D Q Q X. Then, since X is a group and X 0 Q Q Nph(X) 0 Q,

we have

13

(Nph(X) r1Q) S X = (XOQ) S (Nph(X)flQ)

and, hence X = (Nph(X)flQ). Now, since Q is a T—invariant set, we have Nph (Nph(X)) S
Nph(X) and so Nph(X) = Nph(NPh(X)). But then Ph is ap-group and IVPh(X) #1
implies Ph = Nph(X). Thus, we get X = (Nph(X)flQ) = (Ph 0Q), a contradiction.
Therefore Nph(X) 0 Q S X.

Let a E Nph(X) O Q \ X, R = X(a), and choose K S NT(X) minimal such that
K S H and R S K. Then clearly

K 0 H is the unique maximal subgroup of K containing R (3)

Suppose a 6 0,,(K). Then, since K S NT(X) and X is a p—group, we have
X S 0,,(K) and so R = X(a) S 0p(K). Now (3) implies 0,,(K) S H or K = 0,,(K).
In the first case put Y = (0,,(K) 0 Q) and in the second put Y = (K O H D Q).
Then in both cases X < R S Y and Y is a p—subgroup of H. Moreover, we claim
K S NT(Y) in both cases. In the first case its clear, since 0,,(K) S K. In the second,
K is a p-group with K 0 H < K. Thus R S K H H < NR-(K n H) S K and so
NK(K D H) = K, by (3). Hence, again K S NT(Y). Therefore, in either case, we

get a contradiction to the maximal choice of X. Hence, a é 0,,(K).

Now by Baer’s Theorem [3, 3.8.2] there exist conjugates x, y of a in K such that
D = (x, y) is not a p—group. Since x and y are p-elements, there exists a composition
factor W for D on V such that [W, D] at 0. Let D = D/CD(W). If [W,x] = 0,
then D = ('37) is a p-group and acts faithfully and irreducibly on D. Hence, by [3, ],
D = 1 and we get [W, D] = [W, y] = 0, a contradiction. Therefore [W, x] 31$ 0 aé [W, y]

and D 2 (any). Moreover, E and y are p—elements with [W,E,§] = [W,ygg] = 0.

14

Since [3, 3.81] holds for the field GF(p) as well as its closure, by [3, 3.81] there exists
E = E/CD(W) g E with r e ‘E and E/CD(W) g SL2(p). Now x is conjugate to
a in K and so there exists k E K with a = 2:". Hence a 6 E“. Moreover E" S H,
otherwise E" S H would imply E“ is solvable and therefore E would be solvable.
But then we get E E SL2(p) is solvable, a contradiction as p Z 5. Now we have

Ii’SXE’c S K and XE" S H, and so K=XE" by (3).

Suppose that K = T. Then X S T and so X = 1 by (2). Thus, T = K = Ek.

Also as X = 1 and K = T, from (3) we conclude that

H is the unique maximal subgroup of T containing a (4)

Let N = N/CD(W) = Z(E). Then 1 at N S E and so 1 # N S E. Thus
1% N" S E” = T. Now a E Nk(a) S T and so N"(a) S H or T = Nk(a), by (4). If
Nk(a) S H then by (2) we get Nk S CoreT(H) = 1, a contradiction. On the other

hand, if T = N"(a) then a E P" implies T = NkP" and therefore E" = NkPh. Thus

E" NkPh Ph ,
TV; E T = W 18 ap—group.

and consequently E /N and E/N are p—groups. But then, since TV— : Z (E), E/Z(N)

is a p—group and therefore E E’ SL2(p) is nilpotent, a contradiction.

Therefore K 79 T. Let P; E Sylp(K 0 H) with R S P, and P2 6 Sylp(K) with
P1 S P2. If P2 S H, then K = P2 by the minimal choice of K. But then K is a p—group
and a 6 0,,(K), a contradiction. Thus P2 S H and P1 = P2. Suppose KflQ Q KflH.
Then x, y E KflQ implies x, y 6 KflH and therefore E S D = (x,y) S KflH. But
then we get E S H, a contradiction. Hence K 0 Q S K n H. It follows that K, P1,

K 0 H, K D Q, and V fullfill the assumptions of the theorem. By induction we have

15

Cv(1\" n H) g (It/(K). But then (Jr/(H) g (It/(K n H) g Cv(K) and :r = (tug)

implies Cv( H) S Cv(T) and the lemma is proved.

Recall, in Lemma 1.8 we found S E Sylp(G) with S S M. Next we show that

there is another subgroup of G containing S.

Lemma 1.10 There exists H S G, H 75 G, with S S H and H S M.

Proof By Lemma 1.4 (c) we can choose a proper subgroup H of G with 91(Z(M)) S
H, H S M and, in this order,

(1) [H 0 MIp maximal
(2) |H|JD minimal

(3) |H| minimal

Since 01(Z(M)) is ap-group and Q,(Z(A{)) S HOM there exists T E Sylp(HflM)
with 01(Z(M)) s T. If T e sy1,(M), then T = Sm for some in e M. And so
S = T"‘—1 S Hm"l and we are done. Therefore it is enough to show that T E Sylp(M).
Suppose T ¢ Sylp(M). Then we claim T 6 Sylp(H). For if NH(T) S M, then
T S NH(T) S H 0 M. Hence T 6 Sy1p(NH(T)) as T 6 Sylp(H n M). But then
T 6 Sylp(H). On the other hand, if NH(T) S M, then, N0(T) S M, 91(Z(M)) S
Nc;(T), and Ng(T) 75 G. Moreover, since T e Sylp(M), we have

IM 0 NG(T)lp = INM(T)lp > IT] = ”'1 fl Mlp

16

a contradiction to the choice of H. Thus in any case T 6 Sylp(H). Clearly (1), (2),

and ( 3) imply

H (1 M is the unique maximal subgroup of H containing T

and so, by Lemma 1.4 (b), Z S 91(Z(M)) S 01(Z(OP(Q))) S T, where Q =

CoreH(H 0 M), and V = (Z H ) is a elementary abelian p—subgroup of T.

Suppose [J(T), V] ;£ 1. Then choose A E A(T) such that [V, A] 75 1 and [VflAI is
maximal. If V S Na(A), then by Thompson’s Replacement Theorem [3, 8.25], there
exists A" 6 .A(T) with V O A < V n A" and A“ S Ng(A). Then [V,A‘] = 1 by the
choice of A and therefore VA“ is an abelian subgroup of T containing A‘. Hence,
we have A" = VA‘ as A" 6 .A(T). But then V S A“ S NG(A), a contradiction.
Therefore V S Ng(A). Let Q0 = {h E H|[V,h,h] =: 1}. Then [V,A,A] S [A,A] = 1
and A S Q0. Now Q0 S H n M, otherwise, Q0 Q M and [Q0, Z] = 1 as Z S Z(M).
But, since Q0 is a H-invariant set, we get [Qo, V] = 1 and [A, V] = 1, a contradiction
to the choice of A. Thus Q0 3; H n M. Applying Proposition 1.9 to H, T, H n M,
V, and Q0, we get CV(H 0 M) = Cv(H). But then

Z S CV(H0M) =Cv(H)

implies H centralizes Z and therefore H S M by Lemma 1.2 (b). This is a contra-

diction to the choice of H. Therefore we have shown [J (T), V] = 1.

Now T E Sylp(H) implies CT(V) E Sylp(CH(V)) and so by the Frattini Argument
we have

H = NH(CT(V))CH(V) (1)

17

Since CT(V) is a p—group and [J(T),V] = 1, we have J(T) char S CT(V) and con-
sequently NH(CT(V)) s NH(J(T)). Then H = NH(J(T))CH(V), by (1), and so
NH(J(T)) g M as H g M and 0,,(V) g M. Thus, N = NG(J(T)) g M,
91(Z(M)) s T g N and, |M n N|, > |NM(T)|,, > [Tl = |H n M1,. This is a

contradiction to our choice of H and this proves the lemma.

In the following proposition we summarize the main results of this section.

Proposition 1.11 Let S E Sylp(M). Then S E Sylp(G) and there exists H < G

such that

(a) H S M and S E Sylp(H),

(b) H (1 M is the unique maximal subgroup ofH containing S.

Chapter 2

The Amalgam Method

In this section we continue the proof of Theorem 1.1 by applying the Amalgam
Method. This method uncovers information about a given group by studying its

action on the coset graph of some of its subgroups.

Let M and H be as in Proposition 1.11 and I‘ = {Mx,Hy|x,y E G}. Then we
can define an adjacency relationship on I‘, namely for a, )8 E P, S is adjacent to a
if a 75 B and a 0 fl 75 (A. Then G acts on F by right multiplication and this action
preserves adjacency in I‘. Let A(a) = {fl 6 I‘lfl is adjacent to 01}. Since G = (M, H),
it follows that the graph 1‘ is connected. Therefore given any two points a, [3 E P,
we can find a path from a to )8, by starting with a and jumping from one point of F
to another adjacent point of I‘ until we reach S. We can speak of the length of such
a path as the number of jumps taken to make it. By gathering all such paths from
a to S, we can choose one with the fewest jumps and let d(a,fi) = the number of

jumps in this path. Define

G0, = {g E Glag = a}
Qa = 0,,(G'a)

cl.” = {g e Glfig = e for all a with d(o,e) g 1}

18

19

Za = (Ufilfi E Mal)

It turns out that Z0, G9), and U0, are all normal subgroups of Go, and Go, is
conjugate to M or H for all a E F. Moreover, since cosets of a subgroup are either
the same or disjoint, I‘ has an alternating adjacency structure. That is, if a 6 ALB),
then either Ga ~ M and G5 ~ H or Go ~ H and G3 ~ M. Let a E F and S 6 A(a).
By Proposition 1.11 there exists S E Sylp(G) with S S M n H. It follows from the
definition of adjacency that there exists g E G with 39 S Ga 0 G3. And so all normal

p—subgroups of G0 are in G5 and vice-versa.

From here on let a E I‘ with Go, ~ H. Then, for any 3 6 A(a), by Proposition
1.11, Ga (1 G5 is the unique maximal subgroup of Ga containing 59 for some 59 E
Sylp(G). Also, since Ga ~ H, G3 ~ M and so U3 contains a conjugate of Z. In

particular, Z0 75 1. Hence since
flGAS fl GyzflMx=Coreg(M)=1
AEF GxNM xEG

there exists 7 e r with 2,, g at”. Let b = min{d(a,7)|Za g 0.9)}. Then if n,
n E F with G“ ~ H and Z“ S G9), we have b S d(p,n). Now choose a, E l" with

d(o,o’) = b and z, z GS).

Lemma 2.1 Let a and a, be choosen as above. Then

(a) For any 5 E F, G5 acts transitively on AM).
(6) 1m, 6 e r with e e mi), then of,” = CoreG6(G5 n GB).

(c) Z0, is a elementary p-abelian subgroup of Gan

20
(d) 000w...) = at”.
(e) IfA Sp Za, then 03(14) = Ca(Zo,).
(f) Iffl, 6 E F with ,6 E A(6), then G = (G5,Gg).
PPOOf For (a) let ,5, '7 E A(6) and, without loss of generality, let 6 = Mx, i3 = Hya

and 'y = Hz. Then, by the definition of adjacency, 6 0 fl 74 a and 6 fl '7 ¢ 0. Hence,

there exists elements gl and g; of G with

91 = mtx=h1y
and
92 = m2$=h22

Put g = (mflmg)? Then 9 E MJr = G5 and

flg=(Hy)g = (Hint/)9
= (Hm1)mr‘1m2$
= ngx
= thz

= H227.

Therefore B9 = ‘7 and we are done.

For (b), for any 3 E A(6) we have

CoreG6(GgflGg)= fl(GtoGr)-"= fl (Gaflaé) = fl (GtoGoa)

géGa 966'; 9606

= fl (armor)

Aeaus)

= 0]”

21

where the fourth equality holds by (a).

For (c), suppose Z, S Got. Let (a,a +1,a + 2.6! + 3,...,a' — 1,01') be a path
from a to o’. Then, 2., 5g GEL, as GEL, 3 Gay. And so we get, d(o,o' — 1) <
d(a,a') = b, a contradiction to the minimality of b. Therefore Z, S Gal. Next we
claim that Z, is elementary p—abelian. Since Z, = (Uelfl 6 A(a)) and the U3’s are
elementary p-abelian, its enough to show that Z, is abelian. Let B E A(a). Then
U3 S Go, and, by Proposition 1.11, Ga (1 G3 is a maximal subgroup of Ga. Thus,
by Lemma 1.4 (b), U3 S 01(Z(OP(Q))), where Q = CoreGa(Ga 0 G3). But by (b),
G9) = CoreGa(Ga 0G3) and so we get U3 S Z(Ggl)). But since 3 E A(a) was chosen

arbitrarily, we have Z, S Z (Glyn) is abelian.

For (d), from the proof of (c) we have Z, S Z(GW) and so G9) S CGO(ZO). On
the other hand, for any B E A(a), we have G3 ~ M. Hence, by Lemma 1.2 (b), we
have Cc(U3) = G3. Therefore, U3 S Z0 implies Cc;(Za) S Cc(U3) 2 G3. Thus,
since [i 6 11(0) was chosen arbitrarily, CGO(ZO) S G9) and ((1) holds.

For (e), its clear that Cg(Za) S Cg(A). Since A S, Z0, there exists fl 6 A(a)

with U3 S A and Z0, 2 AU3. Moreover, since G3 ~ M, U3 contains a conjugate of

Z. Thus,

_ IZol lUsAl anl P2

—— _' —— — -—_—

Hence, [U3 0 A] 2 p and so U3 0 A ¢ 1. Therefore, by Lemma 1.2 (b), we get
00(A) S 00(U3 n A) = 00(U3) and

Cg(A) S CG(U3A) = CG(ZC,)

and we have (e).

22

Finally, for (f), without loss of generality, let 5 = M x and fl = Hy for some x,
y 6 G. Then S 6 11(6) implies 506 76 ¢. Thus, there exists 9 = mx = hy E MxflHy.
Now G5 = M’ and G3 = H 3’. Moreover, since M is a maximal subgroup of G, M x is
also a maximal subgroup of G with M 3 S (M x, H y). Therefore either M I = (M I, H 3’ )
or G = (M‘,Hy). Suppose M” = (M’,H”). Then Hy S Mx. But mx = by implies
y’lmx = h” and so there exists mo 6 M such that y"1mx = mg. By tranposing this
equation we get m‘ly = mglx 6 My 0 Mx. Thus, Mx = My and so yx‘l E M.
But then H” S M“c implies ny-l S M. Thus, since yx‘l E M, we get H S M, a
contradiction to the choice of H. Therefore G = (M I, H y) = (G5, G3), which proves

the lemma.

It should be noted that Lemma 2.1 parts (c), (d), and (e) hold for any u, 77 E F
with 0,, ~ H, 2,, g 05,", and d(u,n) = b.

At this point the proof splits into the three cases Gav ~ M or H and b > 1 or

Go: ~ M and b = 1. In each case we get a contradiction.

Case 1 Ga: ~H andb> 1.

Lemma 2-2 (a) 1m e A(a), then Z, = (Ugo),
(b) 2,, 3 G5.
(6) 0r(Ge/Ca.,(Zs))=1.
(d) [Zmzai] st 1.

(e) Z, S Qa+k for allk E {1,2,...,b— 1}.

Proof For (a), since U3 5 Z, and Z, 51 Ga, we have ((150) g 2,. On the other

hand, let '7 E A(a). Then by (a) there exists g 6 Go, such that 7 = 139. But then

23

U7 = U39 = U; S (U?) Therefore Z, = (UBG°).

For (b), suppose Z): S Ga. Then Z): S GS], and as above we have d(a', 0+ 1) <

d(a', a) = b, a contradiction to the minimality of b.

For (c) let S E A(a). Then there exists P E Sylp(Ga) fl Sylp(G3) and, by (a),
z, = ((1%). Hence, since Up 3 Z(P), Lemma 1.2 (b) implies 0405/0042,» = 1

and we have shown (c).

For (d), if [2,20,] = 1 then, by Lemma 2.1 (d), 2,, s 00042,) = GS). This is

a contradiction to the choice of (a, 0').

Finally for (e) we use induction on k. If k = 0, then the result follows as Z,
is a normal p-subgroup of Ga. Now suppose (e) holds for k with k < b — 1. Then
k +1 S b —1 and so a + (k +1) S a +(b-1). Hence, by the minimality of b and

l) S G5,”, and Z, S Qa+k. Therefore,

. . (
induction, we get Z, S Ga+(k+1)

(1) (1)
Zn 5 Ga+(k+l) fl Qa+k S 0P(Ga+(k+l)) S Qa+(k+l)

and we are done.

Now since [ZmZaI] ¢ 1, there exists 5 E A(a') with [Za,U3] # 1. Let P E
Sylp(GaI) fl Sylp(G3), and Q = {g E GaI|[Zar,g,g] = 1}. Then, by Lemma 2.1 (a)
and Proposition 1.11, Z0: is a GF(p)Gat-module and Gov 0 G3 is the unique maximal
subgroup of Ga: containing P with Ga: 0 G3 solvable. Since, by (a) Zar S Ga, and

Z, S Ga we have

24

[Za’a ZaaZa] S [ZonZa] = 1

and so Z, S Q. Moreover, Q S Got 0 G3, otherwise Z, S Q S Ga: 0 G3 and
U3 S Z(G3) would imply [Za,U3] = 1, a contradiction to the choice of S. Thus,
Zar, Gar, P, Ga: {'1 G3 and Q satisfy the hypothesis to Lemma 1.9 which yields,
CZ°I(Ga') = Cza,(Ga: (1 G3). But then Gar S Cc(U3) = G3, as U3 S Z(G3) fl Z01.
Thus, we get Gar S G3. But, since 5 E A(a'), by Lemma 2.1 (f) we get G =

(M, H) = (Gar, G3) = G3, a contradiction to the maximality of M.

Case 2 Ga, ~M andb> 1.

Let G: = Gar/Q0: and Q = Q/Qat = F(G—ar). Then 0,,(‘G—07) = 1 and Q is a

p'-group.

Lemma 2.3 (a) Z 75 1

(b) [Q3 Z0] g CGOI(Za'-1)Qa'

(c) There exists A S,, Z—a with [C§(_),Z;] S CGat(Za’—1)

 

(d) Let H be as in (c). Then there exists 5 E [C§(A),Z—5] with A S Cc;(Z:,_1)

(e) Let (aI — 1)” = a, + 1. Then A S Gag, fl Ga’+l

Proof If Z; = 1, then Z, S Q0: S G2,). Hence we get, Z, S G2,), a contradiction
to the choice of (0,0,). Therefore Z, 75 1 and this shows (a). For (b), again we

proceed by contradiction. Suppose [Q, Z5] S CG r(Zo'—1)Qo’- Then,

25

 

[5,—ZZISCG.(ZO'_1)= fl 6.306s

a
yEA(a'—l)

And so by Lemma 2.2 (e), we have [Q,Zo: Z] S Qar_2 (1 Q = 1 as Q is a p’-group
and Q—a _2 is a p-group.

Therefore [Q, 2;, Z] = 1 and so by the coprime action of the p—group Z: on the p’-

groupQ, wehave[Q,Z—:]=[Q, Za,Za]=1. Butthenby[3, 6..,13] Z5, SCGT. —(Q)S

Q. Since Z; is a p—group and Q 18 a p -group we get Z, = 1, a contradiction to (a).

For (c ), by the coprime action of Z, on Q, we have Q: ( CQ(A)|A Sp Z—a). Now

by applying Lemma 1.6 to Z and Q we get

[Size 1 = ([C—( (A) ZsllZ Sr Z)‘ (1)

 

Now by b(), [Q,Z—a ] S CG (Z a _1) and so by (1) implies there exists A Sp Z; with
[05(1) Zal S Coarl Zn '-_1)

Suppose ((1) does not hold and let K = K/Qal =[C—(A ),Z —]. Then A S CG(ZO)

 

where Z0 = (Z554). Now Z S, Z; implies A S,D Z5. Therefore, by Lemma 2.1 (b),
Cg(Z5,) = Ca(A) and consequently Z, S Cg(Zo). Now by the coprime action of the

p-group Z: on the p’—group C§(A) we get

and so,

K S [K’ ZalQa' S CGOI(ZO)Q0' S CGOI(Za'-1)Qa"

 

Thus, by taking images in G_ar, we get [Cb-(A), Z,.] S CG , (Z ’—1 ), a contradiction to

a O

26

(C)-

Finally, for (e), Lemma 2.3 (e) implies A S Z, S QC];1 S Gag]. Hence, since
QaI S Gal_l, we get AQOI S GaI_1. Now 5 E [C5(Z),Z;] S Cal) implies :1: E
Ng(AQaI). Therefore

AQa' : (AQa')x S G:'_1 : G(a'—l)1 = Gen/+1

and A S Ga’—1 fl Ga’+1-

Lemma 2.4 Let 6 E 1‘. Define V5 = (ZAIA E A(6)). Then

(a) V'o,+1 S Ga, and Va' S Ga+1
(b) Va.” S Ga+1 and Va' fl Ga,
(c) Va.” and VaI are elementary abelian p-groups.

(d) [Va/1 Va+la Var-H] = 1'

Proof For (a), suppose Va“ S GOI. Then there exists 7 6 [3(1) + 1) with Z, S

GOI. Then, since €8.11 S GaI, Z, S GE}: Hence, G, ~ H, Z, S 02.11, and

1'
d('y, a, — 1) < d(a, a') = b. This contradicts the minimality of b. A similar argument

shows VaI S Go,“

For (b), let 9 6 Ga.” and 7 E A(a + 1). Then, since the action of G on the graph

F preserves adjacency, we have 7-" E A((a + 1)9) = A(a + 1) and

Z: = 2’79 S Var-+1

27

Thus, V0,.“ S Go,“ and a similar argument shows VaI S GaI.

For (c), since Va.“ and VaI are generated by elementary abelian p—groups, it is
enough to show that Va.“ and VaI are abelian. Let 7, 6 E A(a + 1). Then 7, 6 71$ (1’,
since G, and G5 are both conjugate to H. Moreover, b > 1, GaI ~ M, and the
alternating adjacency structure of F imply b 2 3. Hence, since d(7,6) = 2 < 3 S b,
by the minimality of b we have Z, S G551). But Ggl) = C0(Z5), by Lemma 2.1 (d) and
so [Z,, Z5] = 1 and V0,.“ is abelian. A similar argument shows VOI is abelian. Finally,
(a), (b). and (c) imply

[Va”VO+13VO+1] S leVal = 1

and the lemma is proved.

Now, since A S Z0, S V0,“ and ZaI+1 S VaI, by Lemma 2.3 (d) and Lemma
2.4 (d), we have [ZOI+1,A,A] = 1 and [ZOI+1,A] 3:4 1. Thus, there 7 E A(a" + 1)
with [A,U,] # 1. Now as before, letting P E Sylp(G,) fl Sylp(GaI+1) and Q = {g E
GaI+l|[ZaI+1,g,g] = 1}, we see that Ga’+1s Za’+1a P, Q, and G, F) Ga'+1 satisfy the

hypothesis to Proposition 1.9. Just as before we get GOI+1 S G,, a contradiction.

Cl

Case3 b=1

Since b = 1 and Z, S G,,, we have, QaI S Ga and [QaI,Za,Za] S [Za,Za] = 1. Let
L = (230'). Then L is not a p—group; otherwise, since L S GOI, we get 20 S L S
QaI S G2,), a contradiction to the choice of (a, 0'). Since L is not a p—group 0”( L) 5:4
1. Thus, since GaI is solvable and OpI(GaI) = 1, we have [0”(L), QaI] 91E 1. Therefore,

28

by [3, 5.3.2], there exists a composition factor V for L on QaI with [V,O”(L)] # 1.
Now <I>(V) char S V implies (b(V) is L—invariant. Since <I>(V) is a proper subgroup
of V, we have <I>(V) = 1, by the irreducibility of L on V. Thus, since V is a p—group,
by [3, 3.1.3], V is elementary p—abelian. Let L = L/GL(V). Then V is a faithful and
irreducible GF(p)L—module and so, by [3, 5.1.3], 0,,(L) = 1. Since [V, L] 75 1, there
exists g E GaI with [V,Zg] 75 1. Thus, Z2,— ;£ 1 and [V,—Zia] = 1. This means L
is not p—stable, and so, by [3, 3.8.3], L involves SL2(p). Since L S GOI and GaI is
solvable, L and L are solvable. But SL2(p) is not solvable for p 2 5. Thus since L

involves SL2(p) we have a contradicition and therefore we have shown Theorem 1.1.

C]

Chapter 3

Maximal subgroups of locally finite
simple groups

In this chapter we will show that Theorem 1.1 extends to locally finite simple groups.
This result is Theorem 3.1 and its proof relies indirectly on the classification of finite

simple groups.

Theorem 3.1 Let G be a nonfinitary locally finite simple group, p a prime with p at q
ifG is of q-type, and Z a subgroup ofG of type (p, p). Then there exist nontrivial
elements 2, z, E Z with Cg(z) 71$ (30(5). In particular, ifM is a maximal subgroup
ofG then 0,,(Z(M)) is locally cyclic.

_Qe_f A group G is called locally finite if |(H)| < 00 for any finite subset H g G.

D_ef G is a LFS-group ifG is locally finite and simple. A set of pairs IC = {(Hg, N,)Ii E

I } is called a Kegel cover for G if,
(1) H.- S G and IHgl < 00 for all i E I.
(2) N,- is a maximal normal subgroup of H,- for all i E I .

(3) For each finite subgroup H S G, there existsi E I with H S H,- and HON,- = 1.
29

30

The groups Hg/N,’ are called the factors of the Kegel cover. It has been shown in
[5, 4.3] that every LFS-group has a Kegel cover. Next we define some terminology
that will be useful for us. Let G be a LFS-group and IC = {(H,, N,)|i 6 I } be a Kegel

cover for G. Also let V be a vector space over a field F and Q be a nonempty set.

Qe_f Let X be a finite subgroup of G. Then [C(X) = {(H,, N.) E ICIX S H,- and X0
N,- = 1}.

Qe_f G is called finitary if there exists a field F and a faithful F G—module V such

that dimp[V,g] < 00 for all g E G.

D_ef G is of alternating type if it possesses a Kegel cover all of whose factors are

isomorphic to alternating groups.

22f If q is a prime, G is of q-type if G is non-finitary and every Kegel cover for
G has a factor which is isomorphic to a classical group defined over a field of

characteristic q.

M Suppose a: is acting on V, then

degv(:1:) = dimp]V, :13]

and

pdegv(a:) = min{degv(z\:c)|0 79 A E F}.

Let V be a vector space over a field F and Q be a nonempty set.

D_e_f Suppose a: is acting on 51. Then Supr(:r) is the set of elements of Q that are

not fixed by :r and

31

(1989(3) = pdesntv) = ISUPP9($)I-

_D_ef Let x E T and T 2’ PSLF(V) or T E’ A1t(fl) and y be the image of at under this

isomorphism. Then

(1987“?) = degv(y) or degT(x) = (1989(9)

respectively.
In [7, 3.3] U.Meierfrankenfeld proved:
Theorem 3.2 Let T be a LFS-group. Then one of the following must hold:

(a) T is finitary
(b) T is of alternating type
or

(c) There exists a prime q and a Kegel cover {(H,,N,-)|i E I} for T, such that T
is of q-type, H,/O,,(H,-) is the central product of perfect central extensions of
classical groups defined over a field of characteristic q, and H,/N,- is isomorphic

to a projective special linear group for all i E I.

Proof of Theorem 3.1 First suppose Theorem 3.1 holds and let M be a maximal
subgroup of G. Indeed, if 0,,(Z(M)) were not locally cyclic for such a prime p,

then, since G is locally finite, we can find a finite noncyclic abelian p-subgroup H of

32

0,,(Z ( M )) Now inside of H, by the Classification of Finite Abelian Groups, we can
find a subgroup K of type (p, p). But then, since G is simple and M is maximal, we

get M = Cc(k) = CG(k') for all nontrivial elements k k, E K, a contradiction.

Suppose that Theorem 3.1 does not hold. Then Cc(z) = Cab-2') for all nontrivial
elements 2, z, E Z. Although G is a locally finite group, Lemma 1.2 (c) still holds
for Z and will be used throughout the proof of Theorem 3.1. Since G is a nonfinitary
LFS-group the proof of Theorem 3.1 splits naturally into the two cases (b) and (c)

referred to in the statement of Theorem 3.2.

3.1 Locally finite simple groups of q-type

In this section we consider the case of Theorem 3.1 in which G is of q-type for some

prime q # p.

Case 1 There exists a prime q 75 p and a Kegel cover IC={(H,-, N,)|i 6 I} for G, such
that G is of q-type, H,/OP(H,) is the central product of perfect central exstensions of
classical groups defined over fields in characteristic q, and H,/N,- is isomorphic to a

projective special linear group for all i E I.

For any (HivNi) 6 K1 let Hi/Ni g PSLn.‘(qti1‘/z)a E : Hi/Oq(Hi)a Ti!- : ({L—Jlj E
J}) with [L—j,L—k] = 1 for allj sé k and L—J- = Lj/0q(H,-) are quasi simple for allj E J.

Lemma 3.3 Let (H,, N.) E K: and F, = H,/N,-. Then

(a) IfT S 11,, then H,- = TN,- or T S N,. In particular, 0q(H,) S N,.

(b) There existsj 6 J such that L, S N,.

33

(C) If LJ' S Ni, then Z(L—j) = N; 0 LJ' and Hi/Ng E L—j/Z(Z-;)

Proof For (a), T S H, implies N,- S TN,- S H5. And so by the maximality of N,,
we get N,- = NgT or N,T = H,. On one hand, N,- = N,T implies T S N,. And on the
other hand, H,- = NgT. In particular, Oq(H,-) S H,, and if H,- = Oq(H,)N,, then

PSLn.(€1")"—"’ H,/N,- = 0,(H,-)N,-/N,- "3 O,(H,-)/O,(H,-) n N,.

Hence we get, PSLn,(q"’) is a q—group, a contradiction. Therefore O,,(H,-) S N,.

Next we show (b) by contradiction. Suppose L, S N,- for allj E J. Then E S W,

for all j E J and

F.- = ({L—J-IJ' 6 J}) s W.»

Hence, F, = 7V7. Since, by (a), Oq(H,) S N,, we get H,- = N,, a contradiction to the
maximality of N,.

For (c), Z(L—j) char S L:- = K,- S Eimplies Z(LD S H_,. Let Z(L—j) = U/0q(H,-).
Then, since Z (L_J) is a abelian normal subgroup of If, we have U S H,- and U I S

0,,(Hg) S N,. Thus, by (a), we have U S N,- or H,- = NgU. Now H, = N,U implies
PSLn,(qt‘) E Hg/N,‘ = NgU/N, 9: U/N, 0 U.
But U/N, n U is abelian as U, S N,- O U. Hence, we get PSLm(q“) is abelian,

a contradiction. Hence, U S N,- 0 Lj and Z(E) = U S N, 0 Lj. On the other

hand, N,- S H,- implies N,- 0 LJ- S Lj and N,- 0 L, S E. Thus, since L—J- is quasi-

 

simple, either N, n L, S Z(L—J) or L:- = (N,- O Lj)Z(-L:). Since Z(L—j) S N,- 0 Li,

34

E = (N,- 0 Lj)Z(L_j) implies L: = N,- H L, and L, S N,, a contradiction to the choice

of Lj. Therefore, N.- 0 L, S Z(L—j).

Finally, since L,- S N, and L,- S H,, (a) implies H,- = N,LJ-. Thus,

 

Hi/Ni = NiLj/Ni '5 Lj/Ni 0 Li 3 L—j/Ni 0 L1 = E/ZFLT)

and we have (c).

Lemma 3.4 Let q be a prime, k > 0 an integer, and V be a n-dimensional vector
space over GF(q"). Also let 0 76 W S V be a subspace ofV, and Qw = CGL,(V)(W)n
CGL..(V)(V/W)' Then,

(a) Qw is a q-subgroup of SLn(V).
(b) CaLn(v,(Qw) = QWZ(GL.(V)).

Proof For (a) let {w,~}f":l be a basis for W. Extend this basis to B = {w,- in=1 U

{v,-}?=m+l, a basis for V. Let a E Qw. Then the matrix MB(a) for a in the basis B is

MB(a)=(/{1 (1))

where I is the m x m identity matrix, 0 is the m x (n — m) zero matrix, and A is

some (n — m) x m matrix. Since the field GF(q") has characteristic q, we have

a,_ 10
— qAI

35

where 1,, denotes the n x n identity matrix. Hence every element of Qw is a q-element
and therefore Qw is a q—group. Also, since M3(a) is a lower triangular matrix with

1’s along the main diagonal, we have det(MB(a)) = 1 and Qw S SLn(V).

Next we claim that Qw is abelian. Let a, b 6 Qw. Then

(23%;?)

ab

I!

NC

(A B l
= b(B+A 3)

Hence, Qw is abelian and Qw S CGL..(V)(QW)° But then we get QwZ(GLn(V)) S
CGL..(V)(QW)' On the other hand, let g E CGL,,(V)(QW) with

MB(9):(§ (S!)

where R, S, T, andUaremxm,m><n—m,n—mxm,andn—mxn—mmatrices

‘3)

SA=0and UA=AR (1)

over GF(q") respectively. Let a 6 Qw with

MEN) = (A

Then ya = ag implies

Now, since (1) holds for any n — m x m matrix A over GF(qk), let A = (e,J-) be
the n — m x m matrix with e,,- = 1 and all other entries zero. Then from (1) we
get S(e,~,~) = 0 and consequently the ith column of S is zero. By letting i vary we
eventually get all the columns of S are zero and thus 5 = 0. Again, by (1) applied

to (e,,-) we get U(e,-J-) = (e,,-)R. Letting i and j vary we get u,,- = r,, = /\ for all i and

36

u,, = 13-,- = 0 for all i 3% j. But this means R = U = AI. Hence,

MBlQ) = ($3 (5,)

_A10
‘ TAI

with x E Qw and y E Z(GLn(V)). Thus 9 E QwZ(GLn(V)) and the lemma is

proven.

Lemma 3.5 Let p sé q be primes, k > 0 an integer, and V a n-dimensional vec-
torspace over GF(q"). Suppose P S GLn(V) such that 1 # P/Z(GL,,(V)) is a
p-group and |P/Z(GL.,(V))| < n, then

(a) There exists a nontrivial, P-invariant, subspace W S V with [Qw, P] S Z(GLn(V)).

(b) Let P denote the image of P in PGLn(V). Then there exists a nontrivial,

q-subgroup Q of PSLn(V) on which P acts nontrivially.

Proof Let 0 at v E V and W = (11”). Then W is a nonzero, P-invariant, subspace
of V. Let IP/Z(GL,,(V))| = t < n and (x,- f=1 be a transversal of Z(GLn(V))
in P. If x 6 P then, x = 22:,- for some 1 S i S t and some z E Z(GLn(V)).
Hence, vJr = vn" = (Av)x" = A(v“) E ({v" i=1) for some A E GF(q"). Therefore
W = (v$"|1 S i S t) and so W is properly contained in V, as W is spanned by less
than n vectors. Since W is P-invariant, it follows from the definition of Qw that Qw

is P-invariant. Now suppose [Qw, P] S Z(GLn(V)). Let Po 6 Sylp(P) and P, Po be

37

the images of P, P0 in PGLn(V). Then, since P is a p-group, we get P = Pfi- NOW
[Qw,P] s Z(GL,(V)) implies [Qw,P, P] = 1 and therefore [QW,P0,P0] = 1. But,
by Lemma 3.4 (a), Qw is a q-group, and so by the coprime action of P0 on Qw we
get [Qw,Po] = 1. Hence, Po S CGL,.(V)(QW)‘ But, CGL,.(V)(QW) '2 QwZ(GLn(V)),
by Lemma 3.4 (b). Therefore, since Qw is a q-group and Z(GLn(V)) S GLn(V),
we get P0 3 Z(GLn(V)). But then r = E, = 1 and P g Z((GLn(V)). This is a
contradiction, since P is noncentral. Therefore [Qw, P] S Z (GLn( V)) and the lemma

is proved.

For (b), let W denote the image of Qw in PGLn(V). Then W is P-invariant
implies Q; is P-invariant. Moreover, by Lemma 3.4 (a), Q_v; is a q-subgroup of
PSLn(V). Finally, by (a), we have [QW,P] S Z(GLn(V)). Thus, as Z(GLn(V)) S
GLn(V), we get Qw S Z(GLn(V)) and [W,P} 75 1. Therefore P acts nontrivially

onQ—fiI-and Q7751

In [7, 3.1] U.Meierfrankenfeld showed

Theorem 3.6 Let T be a nonfinitary LFS—group and T be a Kegel cover for T. Then

if]: is a positive integer and 1 919 X S T with [X] < oo

73 = {(H,N) E lC(X)|pdegH/N(x) Z k for all 1 75 x E X}

is also a Kegel cover for T.

Let (H,, N,) E lC(Z). Then, since [Z] = p2 < 00, by Theorem 3.6 we may assume

that pdegH,/N,(z) 2 p2 +1 for all 1 at z E Z. Set H = H,, N = N,, V = V,, and

38

t = t,. Then

n = dimcnqu Z PdegH/MZ) > P2

for any 1 75 z E Z and so p2 < n. Moreover, by Lemma 3.3 (c), there existsj 6 J
such that L,- S N. Let L = L,.

Then by Lemma 3.3 (c), we have

ml
h
bl
] bl

WET- NnL = Z(L)
Hence, since Z n N = 1, L/N 0 L contains a copy Z" of Z. It follows from (1) that

(1)

 

 

 

Z“ =‘Z‘an/N0L.

Now, since L/N—fi—L- E PSLn(q‘) and [Z‘] = p2 < n, by Lemma 3.5 (b), there
exists a q-group U of 17W on which Z‘ acts nontrivially on U by conjugation.
Let U = Q/W. Then Z" acts nontrivially on Q/l—VTl—L- implies W D L acts
nontrivially on Q. Let Q be the preimage of Q in L. Then, since both N and L are
normal subgroups of H, we have [N , Q] S N n L and therefore [N,—Q] S m S Q.

Thus, N normalizes Q and so (Wm L)N = Z acts nontrivially on Q. But then Z

acts on Q.
Since Q/N O L = Q/Z(L) is a q—group, Q/Z(L) is nilpotent. Hence, since Q S
S _

Q
implies Z acts on Q—o. Moreover, since a 6 Squ(Q) and Q/ N F) L is a q-group, we

L, the latter implies Q is also nilpotent. Let Q7, 6 Squ(Q). Then Q6 char

 

have Q = _Q_0(N n L). Suppose [Z,m] = 1. Then, since [N,Q] S N n L, we have
[N,QE] S N n L and therefore [W0 1%] S N Fl L. Thus we get,

 

7W0?) Q;(NOL)
NnL ’ NnL

 

 

 

 

 

[(W03) 6

_( _
NnL 1"[ 1‘1

39

a contradiction. Hence Z acts nontrivially on a and therefore Z acts nontrivially
on Q0, where Q0 is the preimage of a; in H. But, since H = H/Oq(H), Q0 is a
q-group. Thus Z acts nontrivially on a q—group, a contradiction to Lemma 1.2 (c).

This concludes case 1 of Theorem 3.1.

3.2 Locally finite simple groups of alternating type

In this section we consider those LFS-groups which admit a Kegel cover all of whose

factors are alternating groups, the so—called groups of alternating type.
Case 2 G is a LFS-group of alternating type.

Before we begin this case we start by giving a list of some notations used through-
out. Suppose H is a group acting on a set (I with H/N '5’ Alt(2), for some set E and

some normal subgroup N of H. Let t be a positive integer with t S IEI/‘Z.

_I_)_ef A system of imprimitivity A for H on Q is a set of proper subsets of 9 such
that |D| Z 2for at least one DE A, D" E A for all DE Aand h E H, and Q

is the disjoint union of the members of A.

m H acts t-pseudo naturally on Q with respect to N if H acts transitively on Q
and if there exists a system of imprimitivity A for H on D such that CH( A) = N
and the action of H on A is isomorphic to the action of H on subsets of size t

of 23.

kt: H acts essentially on Q with respect to N if CH(Q) S N.

40

Lemma 3.7 Let H be afinite group and N 51 H with H/N perfect, Then there exists

a unique subnormal subgroup R which is minimal with respect to H = RN.

Proof The proof is by induction on IHI. We first remark that if such an R indeed
exists then R S] H. For Rh would be a subnormal supplement to N in H for each
h 6 H. Hence, by uniqueness of R, we have R S Rh for each h E H, and therefore
R 31 H.

For the proof, suppose R1 and R2 are two minimal subnormal supplements to N
in H. Let K]- be proper normal subgroups of H with Rj Eli] K,- for j = 1,2. Then
H = KIN = KgN and, since H/N is perfect, H = H'N. Hence, since K]- 31 H for

j = 1,2, we get

H =-. H’N = [K1 N, KQN] = [K1, K2]N (1)

Letj 6 {1,2}. Then N 31 H and (1) implies Nfl[K1, Kg] 51 [K1, K2] and NflK, SJ Ki-

We also have

[K1, '2] 2 [K1,K21N : g

 

 

 

[K1,K2]ON_ N N
and
A, gKJ-N_H
AJON- N —N'

Thus, both [K1,K2]/[K1, ’2] n N and Kj/N D K,- are perfect. Since [K1,K2] and
K,- are proper subgroups of H, by induction on the IH I, there exists unique minimal
subnormal supplements U and V for [K 1, K2] 0 N and K,- 0 N in [K1, K2] and K]-

respectively. Hence we have

 

 

41

[K1, K2] = U([K1,K2] n N) and K, = V(K, n N) (2)

Also, since [K1, K2] S K,- and R,- S K], (1) and (2) together imply

K, = K, n H = K, n R,N = (K, n N)R,

and

K, = K, n H = K, n [K1, K2]N = (K, n N)[K1, K2] = (K, n N)U.

Thus, since R,- 3131 K, and U 3g [K1, K2] _<_1 K], by the uniqueness of V we get,

VSRjandVSU (3)

But H = KjN = (K,- 0 N)VN = VN and V 33 K,- 51 H. Hence, by the minimality
of R], we get R,- = V S U. Now (1) and (2) imply

H = [5,1,5th = U(lK1, "2] fl N)N = UN- (4)

Thus, as before, N 31 H implies N n U _<_1 U and U /N 0 U is perfect. Thus, since
U < H, by induction there exists a unique minimal subnormal supplement R for
U0 N in U. Since 12,3 U and H = RjN, we have U = (U0 N)R,-, and therefore R,-
is a supplement for U 0 N in U. Moreover, R,- _<_l£l H and R,- S U implies R,- flfl U.

Thus R S R,- by the uniqueness of R. Now R 5131 U 513 H and by (4),

H=UN=R(UflN)N=RN.

Therefore by the minimality of R], we get R,- = R. Since j 6 {1,2} was chosen

42

arbitrarily, we have R1 = R = R2 and the lemma is proven.

Lemma 3.8 Let H be a finite group acting on a set 0 and N a proper normal
subgroup of H such that H/N is perfect and simple. If R is the unique subnormal
supplement to N in H, then an orbit 0 for H on Q is essential with respect to N if
and only ifO g Fix(R).

Proof First suppose O is an essential orbit for H on (I. Then 011(0) S N. If
O C_: Fix(R), then R S 011(0) S N. Hence we get, R S N and H = RN = N,
a contradiction to the choice of N. Therefore 0 Q F ix(R). Conversely, suppose
(9 Q Fix(R) for some orbit of H on Q. Then, since 0 is H—invariant, CH(0) S H.
Hence, since H/N is simple, we have 03(0) S N or H = CH(O)N. IfH = CH(O)N
then CH(O) would be a subnormal supplement to N in H. Thus, by the uniqueness
of R, we get R S CH(0) and 0 Q Fix(R), a contradiction. Therefore 03(0) S N

and O is essential.

Lemma 3.9 Let T be a LFS-group with Kegel cover lC. UK: is the union offinitely
many subsets, IC = UL, Kg, then at least one of these subsets K, is a Kegel cover for

T.

Proof Suppose the lemma is false. Then none of the K.- are Kegel covers for T.
Thus, for each 1 S i S n there exists a finite subgroup L, of T such that for every
memeber (H,N) of K.- either L.- S H or L.- S H and L.- D N #1. Let L = (L,|1 S

i S n). Then since T is locally finite and L is finitely generated, |L| < 00. Therefore

 

43

there exists (H,N) E 1C, with L S H and Ln N = 1. But, since (H,N) E K 2
UL, Kg, there exists 1 S i S n with (H,N) E A}. Hence, we get Li _<_ L S H and

L,- 0 N S L D N = 1, a contradiction to the choice of Lg.

Lemma 3.10 Let p be a prime, n an integer with n > 4p, and Q a set of order n. If

d), 0 E Alt(fl) = A, such that

(a) pdegn(¢), pdegnW) 2 4p

and

(5) ($90) is 0f type (M?)
Then CAn(¢) # CAn(6).

Proof Suppose the lemma is false. Then CAn(¢) = CA..(9). Let

¢:¢1¢2-~¢r and6=0102 ..... 0,

be the decompositions of d) and 0 into the product of disjoint p—cycles where

4% = (aklaalfla - --9alcp) and 9: = (511.1712. . - ~ ablp)

for all 1 S k S r and 1 S l S s. We claim that Supp(qb) = Supp(0). First, suppose
Supp(¢) fl Supp(9) = (6. Then if p is odd, let 1r 6 3,, be the permutation exchanging
the orbits (b1 and d); of 43 and o = 7r(bu,b12). Then a E An and [0, (15] = 1. But
[0,0] 75 1 as 00(b12) = bu 74 b13 = 0(b12). Hence we get, a E (7,4,,(9‘5) \ 0,4,,(0), a
contradiction. If p = 2, let 7r E 5,, be the permutation exchanging the orbits 451
for (152 and 0' = 7r(b11,521)(b12,b31). Then a E An and [mob] = 1. But [0,0] ¢ 1 as

44

00(b11) = b22 # bl; = 0(bn). Thus, again we get a contradiction to the assumption

that d) and 0 have same centralizer in An. Therefore Supp(cb) fl Supp(0) 75 (6.

Without loss of generality, let an E Supp(dh) fl Supp(01). Suppose there exists
1 S l S s with Supp(01) fl Supp(qS) = e. We may assume I = 2. Let 0 E An be
the permutation exchanging 01 for 02 and 03 for 04. Then [0, 0] = 1 but 45" ¢ 0') as
Supp(02) fl Supp(¢) = o and b22 E Supp(qfi") fl Supp(02) # o and so [0, ct] gé 1. Again

we get a contradiction, therefore Supp(0z) fl Supp(qfi) at o for all 1 S l S s.

Now suppose Supp(0) Q Supp(ct). Then there exists 1 S l S s with Supp(0l) Z
Supp(qb). Assume l = 2 and let b2.- 6 Supp(02) \ Supp(cb). Then from above we know
that Supp(02) meets Supp(qb). Without loss, assume a3,- 6 Supp(02) fl Supp(oa). Let
0 E A, be the permutation exchanging the orbits (#1 for 452 and (153 for 454. Then
[0,43] = 1. But [0, 0] = 1 implies 0" = 0 and so 0; and 02 are two orbits of 0. Since
b2,- 6 Supp(02) fl Supp(0f,’ ) and orbits are either the same or disjoint, we have 02 = 0; .
But 0 does not centralize 02 as 0(b2,-) = b2.- and 0(a3,~) = a4,- 74 03,-. Thus we get,
[0, 0] 75 1, a contradiction. Therefore we have shown Supp(0) Q Supp(cb). Now by

the symmetry of this argument we get Supp(gb) = Supp(0).

Next we claim that two elements of the same orbit of 45 cannot lie in different
orbits of 0. First, assume p is odd and, without loss of generality, suppose a“ 6 01
and an E 02. Then, since Supp(0)) = Supp(0) there exists 1 < k S r with Supp(01) fl
Supp(¢k) 75 e). Let b1.- 6 Supp(01)flSupp(</>k). Sincep is odd, qbl 6 An and [(1)], ab] = 1.
Now 0"” = 0 implies 01 and 0‘," are two orbits of 0 with b1,- 6 Supp(01) fl Supp(0'11’1 ).
Hence, as before, we get 0‘," = 01. But this is impossible as an, b1,- 6 Supp(01) and 4‘),
fixes b1; but moves an. Thus (151 centralizes (,b but not 0. On the other hand, suppose
p = 2 and an E 01, and an E 02. Consider 0 = ¢k¢k' for some k, ¢ {1,k}. Then

0 E An and [0,q5] = 1. But [0, 0] 75 1, as a“, b1; 6 01 and 0 fixes all but moves bu.

45

Thus, in any case we can find an element of A, that centralizes 05 but not 0 yeilding

a contradiction to our original assumption.

Altogether we have shown Supp(qb) = Supp(0), ¢ and 0 have the same orbits, and
consequently, r = 3. Without loss of generality we may assume Supp(cbi) = Supp(0,-)
for all 1 S i S r. Next we claim that 0 is actually a power of (b. If p = 2 then, since
they have the same orbits, we get (1) = 0. Thus, I(¢, 0)] = |(¢)| 2: p, a contradiction
to (b). Therefore from here on we may assume p is odd. Suppose a1,- is adjacent to
01, in the orbit 01 of 0. Then p odd implies qbl 6 An. Moreover, [451,gb] = 1 implies
[451,0] = 1. Now, since 431 only acts on elements of 01, we have [051,01] = [¢1,0] = 1.
Hence, a1,- adjacent to 01,- in 01 implies of? is adjacent to of? in 01 for all 1 S m S

p — 1. Therefore 01 = #4. Similarly we can show 0.- : it)? for all 1 S i S r. That is,

_ 1112 l
a- , ,...¢;

Now let 0 E A, be the permutation exchanging the orbits 451 for $2 and $3 for $4-

Then [0, 9b] = 1 and so, by assumption, [0, 0] = 1. But then we get

l1 l2 l3 l4 lr _ l2 l1 l4 [3 l,-
12 3 4"'¢r —¢l 2 3 4"'¢r

and therefore 11 = [2 and [3 = 14. By the same argument applied several times we
eventually get an integer l with l,- = l for all 1 S i S r. But then 0 = 45' and
[(4), 0)] = [(45,431)] = [(03)] = p. Hence, we get a contradiction to (b) and this proves

the lemma.

46

Now we remark that Lemma 3.3 (a) still holds for Kegel covers that have fac—
tors which are alternating groups and will be used throughout this case. Let IC =
{(H,, N,)|i E I} be a Kegel cover for G with H,/N,- E Alt(Q,-) = Am, where IQ] = n.-.
Then, by Lemmas 3.6 and 3.9, we may assume that K = [C(Z) and pdeg Iii/~13) 2
p2(a + 2) for all (H;,N.°) E [C(Z), where a = max(l(u),5uw(Z)|Z|2,9|ZI4) and

u 2 (210g4/3IZI) + 2 from [7, 2.5 and 2.14]. Also, for any positive integer t let

IC,(Z) = {(H,,N.-) E lCIZ has at least t regular orbits on 51,-}.

Proposition 3.11 For any integer t > 0, lCt(Z) is a Kegel cover for G.

We prove this by contradiction through a series of lemmas. Suppose there exists
a positive integer t for which IC¢(Z) is not a Kegel cover for G. Then, by Lemma
39, IC (Z ) \ IC¢(Z) is a Kegel cover for G. Without loss of generality we may assume
IC = lC(Z) \ [Ct(Z) and Z has less than t regular orbits on Q, for all i E 1.

Lemma 3.12 Z has no regular orbits on Q,- for alli E 1.

Proof Since IC is a Kegel cover, by embedding H, into larger and larger H ,’s, we
can find (H, N) E IC such that |H|/|Z| = r 2 t. Then H has no regular orbits on Q],
for any k 6 I and (2;, on which H acts. For if s 6 Q], with s” a regular orbit for H,
then sh‘z would be a regular orbit for Z on 91,, where {h;}f=1 is a transversal for Z

2’s are distinct since cosets of Z in H are either equal

in H. Moreover, each of the s’“
or distinct. Hence, Z would have r 2 t regular orbits on 91., a contradiction to our

assumption.

 

47

Thus, H has no regular orbits on Q]. for all k 6 1. Therefore, by [7, 3.4], there
exists a Kegel cover T Q [C such that whenever (H1, N1), (Hm, Nm) 6 T with H, S
Hm, H; n Nm = 1, every essential orbit of H; on Kim is pseudo natural with respect

to N1. As before we may assume K: = T.

Let (1 = maxre 1{ number of regular orbits of Z on 51,-} and suppose d > 0. Let
(H,,N.-), (H,,N,-) 6 [C such that Z has (1 regular orbits on Q.- and H,- S H, and
H,- H N, = 1. If 0 is an essential orbit of H,- on Q,- then 0 is pseudo natural with
respect to N,. Thus, there exists a system of imprimitivity A for H,- on 0 such that
CH,(A) = N, and the action of H,- on A is isomorphic to the action of H,- on 0,. Now

Z has (1 regular orbits on Q,- implies Z has (1 regular orbits on A.

Let {Uk}fi=l be d members of A such that U? are regular orbits of Z on A. Let
1 S k S d and pick any uk 6 Uk. Then uf is a regular orbit of Z on 0,, otherwise
there exists 1 # z 6 Z with u; 2 uk and consequently uk 6 Uk n U;. But then, as A
is a system of imprimitivity, we get U}, = Uf, contradicting the assumption that Ukz

is a regular orbit of Z on A.

Thus Z has 22:, [Uh] regular orbits on (2,. Since (1 is the maximum number of
regular orbits of Z on any 9, for any I E I, we get [Ur] = 1 for each 1 S k S d. Thus,

we get

d
o = Um} u U U
lc=1 {uk}¢UEA
where the above union is disjoint. Moreover, since Hi-orbits on Q, form a partition
of Q,- and d is maximal, we get 0 is the unique essential orbit of H,- on (2,. Since

Hi/N, E Alt(fl.-) is perfect, by Lemma 3.7, there exists a unique minimal subnormal

supplement to N,, R;, in Hg. Then, by Lemma 3.8, 0 is the unique orbit of H,- on Q,- on

48

which R,- acts nontrivially. Therefore 9,- \ 0 Q F1X(R,’). Also, since H,/N,- E Alum),
we have 9.- Q Fix(Ni) and therefore A Q F1X(N,'), as the actions of H.- on A and
H,- on {2,- are isomorphic. Since N,- S Hg, H,- leaves Fix(N,) invariant. Thus, for any

I: 6 {1,2, ...,d}, we have uk 6 Fix(Ni) and
0 = {uf‘} g Fix(Ni).

Thus, since R.- acts trivially on all nonessential orbits and N.- acts trivially on the only

essential orbit O, we have 0,- _C_ Fix(R, Fl Ni) and therefore R,- D N,- S N,- H H,- = 1.

Now, since R, and N,- are both normal subgroups of H,, we have [R,, N,] S R, n

N,- = 1. Also, since Z n N,- = 1,

Z ZNg<fi
NgnZ g—N,‘

II?
II?

N
II?

An.

 

and therefore a copy of Z sits inside of A"... Moreover, we claim the centralizer
property of Lemma 1.2 (b) for Z still holds inside An... That is, all elements of Z N, / N.-
have the same centralizer in A"... To see this let 2 E Z, H,- = H,/N,~, and m be
the image of CH,(z) in 717. We claim that G,,—[(3) = m. Clearly DEF) g 07,7(2)
and so to prove the claim its enough to show (IE—XE) S m. Let h 6 C313).
Then, since H,- = RgNi, we have h = rn for some r E R,- and n 6 Ni. Thus, h = F
and [F,'z'] = 1. But then [r,z] E N.- and, since R,- S Hg, we get [r,z] 6 R.- H N,- = 1.
Therefore r E CH,(z) and h = F 6 m. Thus 0717(3) S C—H.(7) and so we have

shown 07-17(3) = CH..(2) . Now if 31’, 72 E Z, by Lemma 1.2 (b), we have

49

 

0717(5) = CH.(ZI) = Gib-(Z2) = GF(E)

Hence, all elements of Z have the same centralizer in 77,-, H.- 9-1 Am , and Z is of type
(p, p). Now, since pdeg H,/N.-(Z) 2 4p for all 1 ¢ 3 E Z, we get a contradiction to

Lemma 3.10 and this proves Lemma 3.12.

Now by Lemma 3.12 we know that Z has no regular orbits on Q.- for all i E I .
Let 1 95 216 Z and (H,,N.-) 6 [C with Z S Hg, Z (1 N,- = l, Hg/N; E’ Alt(Q,-). Since
has ‘no regular orbits on 9,, every element of Q,- is fixed by at least one element of Z.

Hence, we have

Supp(21)= U Supp(zl)flFiX(z') (1)
1¢z'ez

Since [Supp(zi)| Z p2(a+2), (1) implies there exists z; E Z with [Supp(zi)flFix(zg)| 2
a + 2. Now 22 g (21), otherwise we get Fix(zi) = Fix(zz) and Supp(zi) fl Fix(zz) = o,
a contradiction. Therefore Z = (21,22). Similarly we can find 1 34$ 23 E Z with

[Supp(zz) n Fix(23)| 2 a + 2. Let
A1 = Supp(z1)fl F1X(22)
A2 = Supp(zg) n Fix(zg)
A3 = Q.- \ (A1 u A2)

Then by the definition of a we have |Ak[ Z a > 5 for k = 1,2. Let A = {g E Hglg E

NH..(A3) and g is even on A1UA2}. Then A S Hg. Moreover we claim Z S A. Since

50

Z is abelian, Z normalizes A3. If p is odd, then all p-cycles are even. Hence zl,
22 E A and therefore Z S A. If p = 2 and [Ale] E 0( mod 4) for all k 6 {1,2} then
again Z S A. On the other hand, ifp = 2 and IAkI E 2( mod 4) for some I: 6 {1,2}
we must modify the definition of Ak. In this case, we take one zk-orbit out of Ak.
Let Ah 2 CA(A3_k) and B = A1 0 A2 for k = 1,2. Then A1 and A2 normalize

each other, 23 E A], 22 6 A2 and

A/B e Alt(A1U A2) and Ak/B e Alt(Ak) for k = 1,2.

Thus, by Lemma 3.7, there exists a unique subnormal supplement R for B in A.

Finally we let A; = (254“?) and A; = (23%“).

Lemma 3.13 (0) A2 = AEB.
(b) Z 0 B =1.

(c) SuppHi/sz) 2 A1 or A2 for all 1 96 z E Z.

Proof Since A; S1 A; and (42/8 g Alt(A2) is simple, either A; S B or A2 = A58.
If A; S B then 22 E B as 22 6 A3. But B = A1 0 A2 implies 22 6 A1 = CA(A2), a

contradiction as A2 9 Supp(zg). Therefore A2 = A38 and we have (a).

For (b), let z E 208. Since Z = (21,22), we havez = zine“; for some] S m,n S p.
Let s 6 A1. Then z E B 2 A10 A2 implies 2(3) = s and so zine-51s) = 5. Since Z
is abelian and 22(5) 2 s, we have 2}”(3) = 5. But 3 E Supp(zl) implies m = p and
therefore 2 = 23. Let s 6 A2. Then again 2 6 B implies 23(s) = s and n = p. Thus,
2 = :«zf‘z2 = 1.

Finally for (c), let 1 95 z E Z and suppose A1 Q Supp(z). Then 2 = 2:1an for

some 1 S m,n S p and there exists 3 6 (A1 \ Supp(z)). Hence 2(3) = s and, as

51

before, we get 2;"(3) = s and m = p. Therefore 2 = z; and z ¢ 1 implies n # p.

Thus Supp(z) 2 A2 and we are done.

Let (H,,N,-) 6 1c with A g H,, An N, = 1, and H,/N, s Alt(Q,-).

Lemma 3.14 Let (H,,N,-) 6 K and TT, be as above. Then [ALAE] S CAO) for all

essential orbits 0 ofA on 0,.

Proof First, let A = A/CA(O). Then, since 0 is essential for A, we have CAO) S
B. Thus A F) N,- = B 0 [Vi-:1 implies
A ANj/N, A/A H N,- A
==——§———E’—EAltA UA
B BNj/N, B/BnN, B ( 1 2)
Therefore 2 S Sym(0), B S A, A/B E Alt(A1 U A2), |A1U A2] 2 5, Z 0 B = 1,
pdeg-KEG) Z a for all 1 7f 3 E Z and Z has no regular orbits on 0. Thus, by [7,

2.14], 0 is t-pseudo natural for t S [ZI — 2.

Let Pt(A1 U A2) denote the set of subsets of size t of A1 U A2 and F be a set of

imprimitvity for A on 0 such that F and P¢(A1 U A2) are isomorphic as A-sets.

We claim that t = 1. To prove this we proceed by contradiction. Suppose t 2 2.
Let at E A]. and XI. = af for k =1,2. Also, let X Q A1 U A2 such that [X] = t and
XflX;c = {at} for k = 1, 2. We can find such a set X since the order of A1 UA; is large
compared to t. That is, sincet S p2 —2 and [Xi] S p2, we have |A1UA2| 2 3p2 —4 =

p2-2+2p2—2 Z t+2p2-2 and SO [AIUAzl—(IXIUX2l) Z IAIUA2I_2P2 Z t—2

Now Z has no regular orbits on 0 implies Z has no regular orbits on I‘. Hence,

since F and P,(A1 U A2) are isomorphic as H—sets. Z has no regular orbits on P¢(Ai U

52

A2). Therefore there exists 1 74 z E Z with X’ = X. If ai # at for some k E {1.2},
we get at, a: E X n Xk, a contradiction as [X n Xkl = 1. Thus, a: = or for k = 1,2.
But, by Lemma 3.27, Supp(z) 2 A1 or A2 and so 2 cannot fix both a, and 02. Thus,

it must be the case that t = 1.

Since t = 1, we know that F and A; U A2 are isomorphic as Z-sets. Let i E {1, 2}
and F = F1 U F 2 where F,- are the images of A.- under this isomorphism. Also for
i E {1,2}, let Y;- = Uxer,Xa and y E Y.- with y E X for X E F,-. Since Z has no
regular orbits on 9,- there exists 1 ¢ 2 E Z with y2 = y. But then X2 = X, otherwise
Xz ¢ X and we get y = y2 E X2 0 X = o, a contradcition as F is a system of
imprimitivity. But then, since X E F .- and F .- and A.- are isomorphic as Z-sets, we
have Xz‘“ = X but XZ # X. Therefore (2,“) = (z) and so yz‘“ = y. Thus, 2,“
fixes all elements of Y,.

Now, since A; and A2 normalize K, A;_, z (2331’4’) fixes all elements of Y,- and so

[A1, A3] fixes all elements of Y,- for i = 1, 2. Therefore [AL A3] fixes 0 and this proves

the lemma.

If AgnR = 1, then [AgflR,A;] S AEOR = land so [A20R,A;] = 1. Then, since
A2 = (A2 0 R)B = A;B, we get [(A2 0 R)B,AEB] S B and [A2,A2] S B. Therefore
A2/B is abelian, a contradiction as A2/B 9-1 Alt(A2) is nonabelian. Thus, A; F) R at 1.

Now by Lemma 3.14, {A}, A; n R] fixes every essential orbit of A on 51,-. Also R
fixes all nonessential orbits of A on (2,. Hence {A}, A; O R] fixes all orbits of A on Q,-
and therefore {A}, A; D R] fixes all of (2,. But then [AL A; n R] S H,- 0 N,- = 1 and so
[A;', A; n R] = 1. Since 23 E A; and all elements of Z have the same centralizer in H,,
we have [23,A; n R] = 1 and [22,A; F) R] = 1. Hence, [A3,A; 0 R] = 1 as A1 and A2

normalize A; n R. Now [A2 0 R, A3] S A; n R implies [A2 0 R, A}, A3] = 1. But then,

53

as before, we have [(A2 0 R)B,A;B,A;B] S B and [A2,A2,A2] S B. Thus, we get
A;/ B 9.5 Alt(A2) is nilpotent, a contradiction as [A2] 2 5. This proves Proposition

3.11.

Now we may assume K = K4(Z) and so Z has at least four regular orbits on Q,-
for all i E I. Let (Hg,Ng) E K, with Z S Hg, ZflNg = 1, and Hg/Ng '-‘_-’ An, = Alt(Qg).
Also let “reg = UL, 0;. where 0;. is a regular orbit for Z on Qg. Finally let [firegl = r

and put

H“ = {h e H,|h e NH,(Qreg) and h is even on Qreg}

and

N" = 0H,.(oreg)

Then, since Z is a semiregular acting subgroup of Sym(Qi-eg) of type (p, p), we have
Z S H‘. Moreover, Z n N“ = 1 and H‘/N' E Alt(flreg) = A... Thus, A, contains
a copy of Z and, since Z acts semiregularly on Qreg, r = 4p2 > 5 and therefore AT
is simple. Now choose K S H“ minimal with respect to Z S K and H‘ = KN" and

let L: N‘flK.

Lemma 3.15 (a) Z S K and Z O L =1.
(b) K/L ’5 A,.
(c) L is maximal with respect to being normal in K.

(d)IfZSUSKandK=UL,thenKzU.

54

Proof By definition of K we have Z S K. Also, ZflL = Zfl(N"flK) _<_ ZflN‘ = 1

and so we have (a). For (b), we have

K- K threat,
L_N‘flK— N‘ —N"_ r

For (c), suppose L S K0 S K. Since K/L E“ A, is simple, we have either K0 S L or

K = K0 and so we have (c).

Finally for (d), K = UL implies

H‘ = KN" = ULN" = UN".

Hence, H" 2 UN" and, since Z S U S K, we get K = U by the minimality of K.

C]

Lemma 3.16 (a) IfU S K, then either U S L or K = UZ.

(b) OAK) S L-

(c) K = (ZK).

((1) Let K = K/0p(K) and L be the image of L in K. Then L is a pi-group and
I = <I>(K).

(e) K' is the unique subnormal supplement for L in K.

(f) K' = 0”(K')-

Proof Suppose U S K. Then L S UL S K and so L 2: UM or K 2 UL, by
the maximality of L. If L = UL then U S L. On the other hand, K = UL implies

55

K = UZL. Moreover, Z S UZ and UZ S L. Hence, by Lemma 3.15 (d), we have
K = UZ and this proves (a).

Now 0,,(K) S K, and so by (a) either 0,,(K) S L or K = 0,,(K)Z. But
K = 0,,(K)Z implies K is a p—group and consequently A, 3 K/L is a p—group, a
contradiction. Thus Op( K) S L and we have (b).

Since (ZK) S K and Z 0 L = 1, by (a) we have K = Z(ZK) = (ZK), which
implies (c).

For ((1) let Z S To E Sylp(K). Then, since L S K, T = To 0 L E Sylp(L)
and by the Frattini Argument K = Ng,»(T)L. Now Z S To and L S K implies Z
normalizes T = To 0 L and so Z S NK(T). Thus, by Lemma 3.15 (d), K 2' NA-(T)
and therefore T S K. But then T S 0,,(K) and T = 1. Now T E Sylp(L) implies
1 = T E Sylp(L) and therefore L is a p'-group. Suppose L S @(K). Then there
exists a maximal subgroup U = U / Op( K) of K with L S U. Since U is a maximal
subgroup of K, K = U—L. Also, since L is a pI-group, U contains a Sylow p-subgroup
of K and so Z S U: for some T:- E K. Then K = UFL and therefore K = U k L with
Z S U". Hence, by Lemma 3.15 (d), we get K = U". Thus we get K = U; = U, a
contradiction to the maximality of U. Therefore L S (MK). Let (MK) = R/0p(K).
Then L S (MK) S K implies L S R S K. Thus either K = R or L = R by the
maximality of L. If K = R we get, K = R = (b(K), a contradiction. Therefore

L = R and L = (MK) as claimed.

For (e), since K/ L '5 A, is simple, by Lemma 3.7 there exists a unique minimal
subnormal supplement K0 for L in K. Since K’ S K we have L S K'L S K and so
either K = K’L or K, S L by the maximality of L. Now Ki S L implies A, g K/L
is abelian, a contradiction. Therefore it must be the case that K = K'L and so

K0 S K', as K, is a subnormal supplement for L in K. Now K0 S L and so by (a)

56

we have K = KoZ. But then

r I
It 0 2

K0 Bo

is abelian.

 

 

ZflKo

Hence [1" S K0 and so K, 2 K0 is the unique subnormal supplement for L in K.

Finally for (f), its clear that 0”(I\") S K’. Also since 0P(K') char S K, S K
we have own") 51 K. Thus OP(K') g L or K = 0P(K')L by the maximality of L.
Suppose 0”(K') S L and let Q E Squ(K') for some q 75 p. Then Q S O”(K') S L
and therefore Q S L. Let K = K/L, and K7 and Q be the images of K, and Q in K.
Then Q = 1 and so Q E Squ(K') implies l = Q E Sylp(K_'). But K' is a supplement
for L in K by (d) and so P = K. Hence 1 = Q E Sylp(K) and therefore K is a
qi-group for all q ¢ p. But this implies K = K/ L g A, is a p-group, an impossibility.
Therefore K = O”(K')L and so K, = O”(K') by (e).

In Lemmas 3.17 through 3.21, F will be a finite field. T a finite group, and V and

E a finite dimensional FT-modules.

_Dej Let a E Aut(F). Then a map 3 from V x V to E is called F a-sesquilinear if

(a) 3(u + v,w) = s(u,w) + s(v,w)
(b) 3(u, v + w) = s(u, v) + s(u. w)
(c) s(Au,v) = /\s(u,v)

(d) s(u,/\v) = X’s(u,v)

M A F a-sesquilinear map 3 from V x V to E is called T—invariant if s( u, v)‘ =

s(u‘,v‘) for all u,v E V and t E T. for all u, v, w E V and A E F.

57

Lemma 3.17 Let F be afinite field, T afinite group, V and E be finite dimensional
FT—modules, and s be a T-invariant F a-sesquilinear map from V X V to E where
a E Aut(F) with 02 =1. Also let C = {t 6 Tlt is a scalar on V}, ITI/ICI = m, and

dimFE = n. Then for any subspace W ofV there exists a subspace X of W with

4mn—1

dime 2 (dimpW - 4 3

 

)/4"‘”

and s IUxuz 0 where U = (XT).

Proof Let L = {a}; be a transversal for C in T, i.e T = U21 tiC, and {qty-”=1

be a basis for E‘. Define a map s” from V x V to F by

sij(u,v) = ¢js(u,vt') for all u,v E V

Then, since 3 is F a-sesquilinear and Q and t.- are are F -linear, 3'5 is F o-sesquilinear.
Consider s” on W. By [7, 2.1] there exists a subspace X1 of W with s” lexX1= 0

and dimel > fldimpW — 4). Similarly there exists a subspace X2 Q X1 with

12 _
S |X2XX2— 0 and

1 1
dlmFX2 Z E(dime1 - 4) > 4—2(dlmFW — 4 — 42)

By continuing this process we eventually get a subspace Xn ; Xn_1 with 31" anxXn‘;

0 and

. 1 . n-l k
dimpxn 2 47(d1mpw 4:4 ).

k=0

58

Next we look at s21 on X”. Again, by [7, 2.1], there exists a subspace Xn.“ Q Xn
with s”1 an+lxxn+l= 0 and dimen.“ 2 zfiddimFW — 422:0 4"). Eventually we

get X2" Q Xgn-1 with s2" |X2nxx2n= 0 and dimegn Z 4—%,;(dimpW — 4 22:01 4").

Continuing this process for each 1 S i S m we get a subspace an C an_1 with

Smn l/Ymnxxmnz 0 and

mn-l
. 1 . k _ . "m - 1
dlmFan Z 4—m;(d1mFW — 4 E 4 ) — (dimpW — 4 3

=0

)/4mn.

 

Now an Q ng for all 1 S i S m and 1 S j S n and so 3'5 ianmen= 0
for each i and j. Let u,v E an, i E {1,2,...,m} and j 6 {1,2,...,n}. Then
s‘j(u,v) = 0 implies qus(u,v"') = 0. Since i 6 {1,2,.. . ,m} was chosen arbitrarily
and (Xgn) 2 (X531), we get ¢js(u,v) = O for all u E an and v E (Xgn). Now as
{cpj}3‘=l is a basis for E" andj 6 {1,2, . . . ,n} was chosen arbitrarily we get s(u, v) = 0
for all u E an and v E (Xgn). But then s(u,v) = 0 for 21,!) 6 ML) as s is T-

invariant.

Qe_f Suppose s is a F o-sesquilinear map from V X V to E. A subspace U S V is

called isotropic if s lyxyz 0.

Lemma 3.18 Let V, E, F, T, C,m, n, s, and o be as in Lemma 3.17. Then
there exists a increasing function f, defined on the positive integers with the following
property: for every subspace W S V with dimpW Z f(mn) there exists 0 7b w E W

such that (wT) is isotropic.

Proof Let f(zc) = 4’ + 431—311 and let W S V with dimpW Z f(mn). Then by

Lemma 3.17 there exists a subspace X S W with (X T) isotropic and

 

59

 

 

 

 

1 . 4m" — 1
dime 2 F(dlmpW — 4 3 )
1 4m" - 1 4m" -— 1
> 4'“ 4 — 4
_ 4m"( + 3 3 )

Thus, dime 2 1 and in particular X # 0. Let 0 76 w E X _C_ W. Then (XT) is

isotropic implies (wT) is isotropic as desired.

Lemma 3.19 Let V, E, F, n, and s be as in Lemma 3.17. IfW S V is a subspace
ofV and

Wl = {v E V|s(v,w) = s(w,v) = 0 for all w E W}

then dimpV/WJ' S 2n - dimpW.

Proof Define the map:

a : V —) Homp(W,E)

by a(v)(w) = s(v,w) for all 2) E V and w E W. Then, since 3 is linear in its first

coordinate, a is a F-linear map. Moreover, Kern a = {v 6 V|s(v, w) = 0 for all w 6

W} and

dimpV/Kern a S dimpa(V) S dimp Homp(W, E) = n ' dimpW.

60

Similarly define a map:

3 : V —> HomF(W,E)

by fi(v)(w).= s(w,v) for all 2) E V and w 6 W. Then, as above, we get Kern 3 =
{v E V|s(w,v) = 0 for all w 6 W} and dimpV/Kern [3 S n - dimpW. Thus, WJ- =

Kern a H Kern fl and,

dimV/WJ‘ = dimF(Kern a + Kern B) + dimFV — dimpKern a — dimpKern 5
S dimpV — dimpKern a + dimpV — dimpKern fl
= dimpV/Kern a + dimpV/Kern ,8

S 2n - dimpW

and the lemma is proved.

Lemma 3.20 Let V, E, F, T, C, m, n, s, and o be as in Lemma 3.17, f be as in

Lemma 3.18, and S = {W S VIW is a T-invariant, isotropic subspace of V}. Then

(a) IfU is a maximal element ofS then dimFU > —l-(dimpV — f(mn)).

2n+l

(b) If dimpV Z (2n+1)2mn+f(mn) and x E T with pdegv(x) > f(mn) then there

exists U E S such that x does not act as a scalar on U.

Proof Since 0, the zero subspace, is in S we have 5 ;é (6. Let U be a maximal
element of S and Ul = {v E V|s(v,u) = s(u,v) = 0 for all u E U}. Since U is

isotropic, U S U i and U l /U is T—invariant. Moreover, by definition of UL, s on

61

Ui/U X Ul/U is well defined. Hence, as U 6 5 is maximal, Ui/U has no nontrivial

isotropic T-invariant subspaces. Therefore, by lemma 3.18

dimpUi/U < f(mn)

where

it = |T|/l{t 6 T|t is a scalar on U‘L/U}|.

But, since f is an increasing function and T72 S m, we have dimpUL/U < f(mn) S

f(mn). Furthermore, by Lemma 3.19, CodimpUi S 2n - dimpU. Thus,

dimpV = dimpV/Ui + dimpUi/U + dimpU < 2n - dimpU + f(mn) + dimpU

Therefore dimFU > 4(dimpV — f(mn)) and (a) holds.

2n+l

We show (b) by contradiction. Suppose x is a scalar on all elements of 5. Let X
and Y be maximal elements of S, L = {ti}f’;l be a transversal for C in T, 0 94 y E Y,
and W = (yT). Since W = (yT) = ({y“}}';1), W is spanned by the m vectors {y"}f’;1

and so dimpW S m. Also as X E S maximal, (a) implies

 

dime >

 

(dimpV—f(mn)) Z ((2n+1)2mn+f(mn)—f(mn)) = 2mn.

(1)

2n+1 2n+1

and so dimFX > 2mn. But then, by Lemma 3.19, we get

X . X J“ . .
= dlmF—i-W— S dlmp% S 2n - dimpW S 2mn. (2)

dimFx n W-L Wi

62

Thus, (1) and (2) imply X 0 W‘L -+‘ 0. Now X 0 W" + W is a T-invariant isotropic
subspace of V different from X and Y and Therefore (X 0 Wl) + W E S and x acts
as the same scalar on (X n Wi) + W, X, and Y. Since the choice of X, Y E S
was arbitrary, x acts as a scalar on all of (5). Let x(v) = Av for all 2) E (S) and
V = (S) 69 D for some subspace D of V. Then D has no T-invariant isotropic
subspaces and so by Lemma 3.18 dimpD < f(mn). Now consider the map A’lx from

V onto [V,A'lx]. Since (8) Q Kern()\‘1x) we get,

pdegv(x) S dimp[V,)\'1x] = dimpV/Kern(x\‘1x) S dimpV/(S) = dimpD < f(mn)

a contradiction as pdegv(x) > f (mn) by assumption. This proves the lemma.

Lemma 3.21 Let V be a n-dimensinal vector space over a field F with charF = p

for some prime p. Then for any p-subgroup P of GLF(V)

dimpV S [Pl dimFCV(P)

and in particular

dimpV S [Pl dimpV/[V, P]

Proof For the proof we use induction on |P| - dimpV. For |P| - dimpV = 1 then
both |P| and dimpV are equal to 1 and the result only states 1 S 1 which is true.
Assume the result holds for all vector spaces U over F and p—subgroups T of GLF(U)
with |T|~ dimpU < [P] dimpV.

63

Then we can assume that V is indecomposable. For suppose V = U a; W for some
P-invariant subspaces U and W. Let P = P/Cp(U) and P = P/Cp(W). Then the

p—groups P and P act on U and W respectively. Thus, by induction we get

dlmFU S dlmFCU(F)IFI

Now V = U 69 W implies Cv(P) = CU(P) EB CW(P). Therefore we get,

dimpV = dimpU + dimpW g amped?) . |F| + dimpr(P) - lfil
< dimpCU(P) . |P| + dimpr(P) - |P|

= (aimed?) + dimpr(P))lP|

dimpCv(P)|P|

as desired.

Let Q be a maximal subgroup of P and assume Q 75 1. Then Q S P and lP/Ql <
IPI. Now, since Q is a maximal subgroup of P, we have CCV(Q)(P/Q) S CV(P).

Thus, since P/ Q acts on CV(Q), by induction have

dimFCv(Q) S IP/Ql dimFCcv(Q)(P/Q)-

Also, by induction applied to Q on V, we get

dimpV S |Q| dimpCv(Q).

But then

dimpVSIQldimpCv(Q) s lQl(%-:dimpCcv(Q)(P/Q))

64

= IPI dimFCcV(Q)(P/Q)

S IPI dierv(P)

as desired.

Thus we may assume P has no nontrivial maximal subgroups. Then P is cyclic of
order p. Let P = (x) where :1:p = 1. Now x” = 1 and charF = p implies (x — I)” = 0
on V. Hence, since 1 is the only characteristic root of x and V is indecomposable,

there is a basis B of V for which the matrix MB(x) for x has the form

10 0
l ..0\

 

 

 

 

Now J n = 0 and, in fact, 12. is the smallest positive integer with this property. Suppose

dimpV > p. Then n > p and so J” # 0. Hence, we get

(z—1)P=(I+J—1)P=JP¢0

a contradiction. Thus dimpV S p S |P| dimpCv(P). Now P also acts on the dual
space V‘ of V by (¢)9(v) = (b(vg-l) for any 9 6 P, v E V and (b E V‘. Under this
action we have (V/[V, P])"' E Cv-(P) and so

65

dimFV = dimpV‘ S IPI dlmFCvo(P) = IPI dlmF(V/[V,P])' = IPI dlmFV/[l/7,P]

and we are done.

Now let f be as in Lemma 3.18 and I be as in [7, 2.5]. Then by Lemma 3.6 and

Lemma 3.11, the set

r = {<H.. N.) e Ic.(2)mc.(K)IpdegH./N,(z) 21((2p2+1)2p2lK|+f(p2|1\'|))for all 1 ¢ x e K}

is a Kegel cover for G. Let (Hk,Nk) E T, and set H = Hk, N = Nk, 7n, = n, and
0;, = 0. Then since H/N E An is perfect, by Lemma 3.7, there exists a unique

subnormal supplement R for N in H.
Proposition 3.22 CH(OP(H)) S N

We prove Proposition 3.22 by contradiction through a series of lemmas. Let
U = CH(OP(H)) and suppose U S N.
Lemma 3.23 (a) [R, F(H)] = 1.

(b) R, = R.

(c) Z normalizes each component of H.

(d) R is a component of H.

Proof 0,,(H) S H implies U S H. Hence, since U S N, Lemma 3.3 implies

H = UN. Thus U is a subnormal supplement to N in H, and so by uniqueness

66

of R we have R S U. Since Z S N, a similar argument shows R S (2”). Now
[U,O,,(H)] = 1 and R S U implies [R,0,,(H)] = 1. Also, Z normalizes OPI(H) and
so by Lemma 1.2 (c) [Z,OPI(H)] = 1. Since 0pr(H) S H and R S (ZH), we have
[(ZH),OPI(H)] =1 and so [R,0Pr(H)] = 1. Thus, [R,F(H)]=1 and we have (a).

For (b), R, char S R S H implies R, S H. Thus, by the uniqueness of R and

Lemma 3.3, we get R = R' or R, S N. If R’ S N, then R, S RN N and we get,

RN R

2

7v— RnN

 

ll?

am

A, is abelian

a contradiction. Hence, R = R' and R is perfect.

Next we prove (c) by contradiction. If (c) does not hold, then there exists 1 #
u E Z and a component C; of H with Cf‘ 7b C1. Let C, = Cf"! for 1 S r S p
and C = (C1,C2, . . . ,Cp). Since C1 is a component of H and C, are conjugate to C1
for all 1 S r S p, C, is also a component of H for each 1 S r S p. Furthermore,
if C, = C, for some 1 S r,s S p, then CF“ 2 Cf,"1 and therefore Cf”, = C1.
But then u’"3 E NH(C1) and, since la] 2 p, we get u E (u) = (u"’) S NH(C1), a
contradiction. Therefore we have shown C, 75 C, for all r # 3. Hence, as C, and C,

are components of H, we have [C,,C,] =1 for all r 95 3. Thus C = C1 -C2 - . . . - Cp.

Define a map 0 from C1 to Cc(u) by

11.2

9(c) = cc“c ...c“p_l for each c 6 C1.
Since [C,,C,] = 1 for all r 75 s, we have cue“2 ...c“p-l E Cc(u) for all c 6 C1. Thus,
0 does indeed map C1 into Cc(u). Moreover, since conjugation by powers of u is a

homomorphism and [C,, C,] = 1 for distinct r and s, 0 is a homomorphism.

67

Suppose 0(c) = 1 for some c 6 C1. Then cc“c“2 . . . cu'D-l = 1 and so

c=(c“l“2...c"p-1)_1EClflCg-C3-....Cp.

Thus c 6 Z(Cl) and, since c 6 C1 was chosen arbitrarily, we have shown kern 0 S
Z(Cl).

Let X = 0(C1). Then X is a nonabelian, otherwise, by The First Isomorphism

Theorem, we get

 

Hence, Cl/kern 9 is abelian and therefore C1 = C; S Kern 0, as C1 is a component

of H. But then we get C1 = Kern 0 S Z(Cl) and C1 is abelian, a contradiction.

Next we claim that Z normalizes the set {C1,C2, . . . ,Cp}. For otherwise, there
exists '0 E Z and 1 S r S p with C: Q’ {C1,C2,...,Cp}. Now C1 is a component of
H implies C: is a component of H. Thus, Cf ¢ {C1,C2, . . . ,Cp} implies [C,f’,C,] = 1
for all 1 S s S p. But then [C,EfiC] = 1 and so, by conjugating by powers of u, we
get [CLC] = 1 for all 1 S r S p. Thus, [C”,C] = 1 and, since X S C, [X”,X] = 1.
But X S CC(u) = Cc(v), by Lemma 1.2 (a). Thus, X“ = X and we get [X,X]=1,

a contradiction.

Now since Z normalizes the set {C1,Cg,...,Cp}, there exists 1 75 w E Z with
Cf” =2 C1. Let c 6 C1 and k : cucui...c“p—1. Then I: 6 C2 - C3 - - C, and so
k E CC(C1). Also, since 0(c) = ck and 0(c) E Cc(u) = Cc(w), we have

68

ck = 9(c) = 0(c)‘” = cwkw

Also, since w E NH(C1) and k E CC(C1), we have c‘” 6 C1 and kw E CC(C1). But

then

[c, w] = c'lc'” = kk'w E CC(C1)

and so, since c 6 C1 was chosen arbitrarily, we have [C1,w] S Cc(C1). But then we
get [C1,w,C1] = 1 and [w,C1,C1] = 1. Thus, by the Three Subgroup Lemma, we

have [C1,C1,w] =1 and so [C1,w] = 1, as C, is perfect. Hence, we get

C1 S CH(w) = CH(u)

and so [C1,u] = 1, a contradiction.

Finally let E (H ) be the subgroup generated by the components of H and comp( H) =
{01,Cz,...,C,~} be the set of components of H. Let i E {1,2,...,r}. Then by (c),
we know that Z normalizes C,. Thus, since H normalizes the set comp(H), (Z H ) also
normalizes C,. But R S (2”), and so R also normalizes C,. Let N,- = NH(C,-). Then
R S N,- and Z(C,) char S C,- S N,- implies Z(C,) S N,. Let N,- = N,/Z(C,-) and R
and C,- be the images of R and C,- in K. Then R normalizes C,- and C:- is simple. But
R is perfect implies R is perfect. Thus by the Schrier Conjecture, Out(C,-) is solvable
therefore R induces inner automorphisms on C7. But then R S CN—(C:)C,i. Hence,

by taking preimages, we get R S C H(C,-)C,~.

69

Now we claim C~,(€,) = CN,(C,-). Let C = CN,(_C'I). Then clearly C~,(C,-) g C.
On the other hand, [C,C,] = 1 implies [C,C,] S Z(C,) and hence [C,C,, C,] = 1. But
also [C,,C,C,-] = 1 and so by the Three Subgroup Lemma, [C,, C,, C] :2 [C;,C] = 1.
Since C,- 6 comp(H), we have C: = C,- and so we get [C,,C] = 1. Therefore C S

CN,(C,-) and the claim holds.

Now we have R S CN,(—CT,-)C,~ = CN,(C,-)C,- S CH(C,)C,. Since distinct components

of H commute and the choice of i 6 {1,2, . . . ,r} was arbitrary, we get

R 3 fl omen-)0.- = (fl CH(C,))(CI . c2 - . . . . c.) g CH(E(H))E(H).

Thus R S E(H)CH(E(H)). Now E(H) S H implies CH(E(H)) S H. As before
we get CH(E(H)) S N or R S CH(E(H)). If R S CH(E(H)) then [R,E(H)] = 1.
Also, by (a), [R,F(H) = 1, and so [R,F"(H)] = 1, where F'(H) = E(H)F(H)
is called the generalized fitting subgroup of H. But then, by [1, 11.31.13], R S
CH(F"(H)) S F‘(H) and so R S Z(F"(H)). Thus, in this case, we get R is
abelian and so R = R’ = 1, a contradiction. Therefore CH(E(H)) S N. But
R S CH(E(H))E(H) and R S N implies E(H) S N. Hence, there exists a com-
ponent subgroup C,- of H with C,- S N. Now since C,- SS H and C, S N, C,-
supplements N in H. That is, H = C,N and so by the uniqueness of R we have
R S C, and therefore R S 0,. Let C,- = C,/Z(C,-) and R be the image of R in C,.
Since C,- is a component of H, C; is simple. Thus, R = 1 or R = C,. If R = 1 we
get R S Z (C,) and therefore R is abelian, a contradiction. Hence, R = C,. But then
C, = RZ(C,~) and (),/R is abelian. Therefore C, = C; s R and so R = C, and R is a

component of H.

70

Lemma 3.24 (a) Z(R) = N 0 R

(5) [EN] =1

Proof Now Z(R) char S R S H implies Z(R) S H. Thus, since R is nonabelian,
we must have Z(R) S N. Hence, Z(R) S N H R. Also, N S H implies N n R S R.
But, by Lemma 3.23, R is a component of H and therefore N n R S Z(R) or
R = N n R. The latter implies R S N and H = RN = N, a contradiction to the

maximality of N. Therefore N F) R S Z(R), and so Z(R) = N O R.

For (b), since R and N are normal subgroups of H we get [R, N] S NOR = Z(R),
by (a). Hence, [R, N, R] = 1 and [N, R, R] = 1. Thus, and so by the Three Subgroups
Lemma we get [R, R, N] = [R,, N] = 1. But, by Lemma 3.23, R = R, and so we get
[R, N] = 1.

Now since Z(R) = N 0 R, we have

R R 5.1!-
Z(R) NOR N '

II?
II?

 

An

Zlm

Let H = H/N and R and Z be the images of R and Z in H. Now, since ZflN =1, H
contains a copy of Z. From here on we will identify H and A, as they are isomorphic.
Now, by [8], R/Z(R) E An and R' = R implies [Z(R)] = 1 or 2. If Z(R) = 1
then, by the same argument used in Lemma 3.12, the centralizer property of Z in
Lemma 1.2 (b) passes over to H to give a contradiction to Lemma 3.10. Therefore
we may assume |Z(R)| = 2.
Suppose p is odd. Then again the centralizer property of Z in Lemma 1.2 (b) passes

over to H. Since p is odd, Z(R) is a p’-group and Z(R) S H implies Z S NH(Z(R)).

71

Thus, by Lemma 1.2 (c), [Z,Z(R)] = 1 and so Z(R) S CH(Z). Next we claim
C-g(E) = m for all 2' E Z. For ‘2' 6 Z, clearly m S Cy(3) and so we only
need to show CFO?) S m. Let h E Cy('z'). Since H = RN, there exists r E R
and n E N with h = rn. Then [hi] = [if] = 1 implies [r,z] 6 Z(R). Hence, since
|Z(R)| = 2, we get [r,z]2 = 1. But, since [r, z] E Z(R) S CR(Z), [r, z] commutes with
both 2 and r and so, by [3, 2.2.2], [r,.z]2 = [r, 22 = 1. Thus, r E CH(::2) and therefore

Lemma 1.2 (b) implies r E CH(2). By taking images we finally get F 6 CH(::) and so

we have shown Cy(§) = CH(z

V

Now letting Z = (31', '23) we have H 2’ An, Z S H and, by Lemma 1.2 (b),

07(17): CHM) = CHM) = CF95)-

Moreover, Z is of type (p, p) and, by the choice of the (H, N), we have pdegn(?:') 2 4p.

Hence, again we get a contradiction to Lemma 3.10.

Therefore it must be the case that p = 2 = |Z(R)|. Since (H,N) E lC4(Z), we
can find two regular orbits 01 and 02 for Z on 9. Since Z is regular and transitive
on 0,, the action of Z on O,- is isomorphic to the action of Z on Z by right multipli-
cation. Let Z = (5,72). Then E has cycles (1,7?)(72, M) on Z while 27 has cycles
(1,5)(2—1,M) on Z. Let 1, 2, 3, 4, and 5, 6, 7, 8 be the images of 1, El“, '5, and T23
under the above isomorphisms respectively. Then '2'? has cycles (12)(34)(56)(78) on
01 U 02 while 33 has cycles (13)(24)(57)(68) on 01 U 02.

Now let r 6 R with F = (135)(246). Then [772,27] = 1 while [F235] # 1. Thus,
[r2,zl] 6 Z(R) and [r2,22] Q’ Z(R). Also F3 = 1 and F2 at 1 implies r3 E Z(R)
and r2 ¢ Z(R). Let s = r2. Then, since |Z(R)| = 2 and r3 E Z(R), we have

s3 = 1 and [3,21] 6 Z(R). Suppose [3,21] = 1. Then, by Lemma 1.2 (b), we get

 

72

[3,22] = 1 E Z(R) a contradiction. Thus [3,21] 75 1. But then, since |Z(R)| = 2, we
get Z(R) = {1,[s,zl]}. Now Z(R) char S R S H implies Z(R) S H and therefore
[3,21] 6 Z(H). Thus [s,zl] commutes with both 3 and zl. Now [3, 2.2.2] implies
l = [53,21] = [s,zf] = [3, 21]. Hence, we get [3,21] = 1, a contradiction and therefore

Proposition 3.22 holds.

Now U = CH(0,,(H)) S N implies R S U. Hence, [R,0p(H)] 7b 1 and so we can
pick S S 0,,(H) minimal with respect to S S H and [S, R] 75 1. Then alot can be

said about the structure of S.

Lemma 3.25 Let S be chosen as above, then

(a) S = [S, R].

(b) C5(R) is the unique subgroup ofS maximal with respect to being normal in H.
Moreover, either S is abelian or Z(S) = C5(R).

(c) US is nonabelian, then S/Z(S) and S, are elementary abelian p-groups.

Proof Since R and S are normal subgroups of H we have 1 ¢ [5, R] S S and
[S, R] S H. So by minimality of S either S = [S, R] or [S, R, R] = 1. If [S, R, R] = 1
then also [R, S, R] = 1. Thus, by the Three Subgroup Lemma we get [R, R, S] = 1.
But then, since R is perfect, we get [R, R, S] = [R', S] = [R, S] = 1, a contradiction.
Thus, S = [S, R] and this proves (a).

For (b), since S S H and R S H, we have C5(R) S H and C5(R) # S. Now let
So < S with So S H. Then by minimality ofS we have [30, R] = 1 and so So S C5(R).

Thus, C5(R) is the unique subgroup of S maximal with respect to being normal in

 

73

H. In particular, if S is nonabelian, then Z (5') char S S S H implies Z (S) S H
and therefore Z(S) S C5(R). Also, since R S H, (a) implies S = [S, R] S R. Hence,
[C5(R),S] = 1 and we get C5(R) S Z(S). Therefore Z(S) = C5(R) and we have (b).

Finally, since both <I>(S) and S, are characteristic subgroups of S, they are both
normal in H. Hence, by (b), <I>(S) and SI both lie in Z(S). Therefore S/Z(S) is
abelian. Now, since S is a p—group, by [3, 5.1.3], S/<I>(S) is elementary p—abelian.
Therefore S/Z(S) is also elementary p—abelian as <I>(S) S Z(S). Let 2 = [x,y] E S'.
Then S/Z(S) elementary p—abelian implies y” E Z(S). Moreover, since z: 6 S, S

p

Z(S), z commutes with both x and y. Thus, by [3, 2.2.2], 1 = [x,y”] = [x,y]p =

N

and so all commutators have order p. Since S ' S Z (S) is abelian, all elements of S '

must have order p and therefore S I is elementary p—abelian.

Lemma 3.26 We may assume H = KR.

Proof If H # KR, then we can replace (H, N, R) by (KR, N n KR, R). For
Z S KR, Z 0 (Nfl KR) S Zfl N = 1, and N0 KR is maximal with respect to being

normal in KR as

KR KRN _

NflKR N

Ill

’2.”

A, is simple .

ZID:

Moreover, we claim that R is the unique minimal subnormal supplement to N 0 KR
in KR. Clearly R S KR and KR = KRflH = KRfl RN = R(NflKR). Hence R is
a subnormal supplement for N D KR in KR. Now suppose R0 S R with R0 SS KR
and KR = R0(N fl KR). Then

74

H = RN = (KR)N = R0(N n KR)N = RON

and so H = RON. Moreover, R0 SS R S H, and so R0 SS H. Thus R = R0 by
the uniqueness of R and therefore R is a minimal subnormal supplement for N O KR
in KR. Thus, by Lemma 3.7, R is the unique minimal subnormal supplement for

N n KR in KR.

Proposition 3.27 There exists an abelian subgroup A ofG with K S Nc(A) and

CK(A) S L.

We now show Proposition 3.27 through a series of lemmas.

Lemma 3.28 We may assume

(a) [1135'] =1

(b) Z(S) < S

Proof Since Z(S) is abelian and Z(S) char S S S H implies Z(S) S H, we
may assume CK-(Z(S)) S L. Otherwise, we are done by taking A = Z(S). Now
CK(Z(S)) S K and CK(Z(S)) S L and so by Lemma 3.16 K = CK(Z(S))Z. Thus,
K/CK(Z(S)) is abelian and so K’ g CK(Z(S)). Now mm = 1 and K, g G,,-(Z(S))
implies C'K(Z(S)) g N. Hence, we have CH(Z(S)) ,<_ N and CH(Z(S)) 3 H. But
then CH(Z(S)) supplements N in H and therefore R S CH(Z(S)). Since H = KR
and R is perfect, we have H' = K'R and therefore [H', Z(S)] = 1. By Lemma 3.25

75

(b) S" S Z(S) and so [H',S'] = 1. Finally, since [5, R] 54 1 and [Z(S),R] = 1 we

have Z(S) < S and we are done.

Lemma 3.29 (a) H = H'Z.
a)Rgmun

I

(C) 0”(H) S H -

Proof For (a), recall that H = KR by Lemma 3.26 and K = K'Z by Lemma 3.16.
Also H, S H and H, S N implies H = H’N and therefore R S H'. Hence,

HzKaszRgHZRsz

and so H = H'Z.

For (b), since 0"(H) S H, by Lemma 3.3, we have either R S Op(H) or O”(H) S
N. Suppose 0P(H) S N. Then, since H/0P(H) is a p-group, we get

H/O”(H) .
W)- 18 a p-group

II?
”2

H
An —
N

a contradiction. Therefore R S 0” ( H)

Finally, by (a) H = H’Z and therefore H/H' = H'Z/H' 2’ Z/H'flZ is ap—group.
Thus, 09(H) S H' and we are done.

76

Let V = S /Z (S ) then, by Lemma 3.25, both V and S, are elementary p-abelian.
Since S S H, H acts on V and S, by conjugation making V and S, into GF(p)H-
modules. Moreover, by Lemma 3.25 (b) and Lemma 3.28 (b), H acts irreducibly on

V. Define a map 5 from V X V to S, by

flab) = [a,b] for all 'a’,b 6 V.

Lemma 3.30 (a) fl is well defined.
(b) B is GF(p)H-bilinear.

(C) flab) = M55)“-

Proof For (a), let (ab) = (E,d) E V X V. Then 5 = E and b = d. Thus there exists
:1, 22 E Z(T) with a = 21c and b 2 22d. Then, writing the operations of V and TI

multiplicatively, we have
[a,b] = [C21,ng] = [21,22d]C[C, 22d]
= [21,d]°[z,,22]“d[c,d][c,22]°‘
= [c,d] as 21,22 6 Z(T)
Thus Maj) = mad) and e is well defined.
For (b) recall T’ g Z(T) by Lemma 3.25. Thus for any a, B, and e e v we have,
fi(ab,z) = fi(ab,E) = [ab,c]
= [a.c]"[b.cl
= [a,c][b,c]

= Miamfla.

77

A similar argument shows that ,8 is linear in the second component as well. Now
let /\ E GF(p). Then, since [a,b] E T, S Z(T), [a,b] commutes with both a and b.

Therefore,

II
F
.2.

y

where the third equality is due to [3, 2.2.2]. Again, a similar argument will show
—.\

B(‘a’,b ) = B(E,b)*. Thus fl is GF(p)-bilinear and we have (a). Next we show that S

commutes with the action of H. Let h E H, then

WW = [a,bl" - [a’fibhl

For (c), we have

fl(Ea-6) = [a,b] = [bid-1 = Mail—l-

Finally, since V is H-invariant, CH(V) S H and so by Lemma 3.3 (a) either
CH(V) S N or H = NCH(V). IfH = NCH(V) then R S CH(V), as R is the unique
subnormal supplement for N in H. Hence, [R, V] = 1 and therefore [R, S] S Z (S )
But recall S = [R, S], by Lemma 3.25 (a). Thus we get S S Z(S), a contradiction.
Therefore CH(V) S N and we have ((1).

78

Def Let F be a finite field of characteristic p, F0 a subfield of F, T a finite group

and V and W finite dimensional F T—modules. Let C(F, F0) the group of field
automorphisms of F which fix F0. For a E C(F, F0) let V" be the FT module
defined by V" = V as a FoT—module but the following new scalar multiplication

by elements of F:

A-,v=)r"v

for all A E F, v E V. Note that for elements of F0, this is just the old scalar

multiplication.

Lemma 3.31 Let V, W, F0, F, and T be as above.

(a) The map It ®l —> {kal},EG(F,FO) induces a Fo-isomorphism from F @170 F to

FlG(FvFO)l .

(b) Let a E C(F,Fo). Then the map It ®l —> 1:01 induces a F-isomorphism from
F®Fo F” to F”.

(c) The map v (8) w -—) {v (8) UJ}a€G(F,F0) induces a FoT-isomorphism from V ®po W

to ®UEG(F,F0) V ®F W0

Proof (a) Since the map (1:, l) —> {kal},€G(p,po) is Fo-bilinear, there indeed exists a
F0 linear map a from F (81:0 F to FGlF'Fb) with a(k ® I) = {kal}aea(p’po). Moreover,
F (81:, F becomes a F vector space by A - (k (8)1) = k (X) /\l and a is F-linear. Since
dimp(F (81:, F) = dimFo F = |G(F,Fo)| = (limp F'GlF’F°)' it is enough to show that

a is surjective. If a is not surjective, then a(F ®Fo F) aé F'GlF'F0“ and therefore

79

F|G(F»Fo)|/a(F (8)170 F) 91$ 0. Thus, there exists 0 74 0 E (FIG(F’F°)I/Q(F (31% F))'-

Define d) E (FlG(F.Fo)|)* by

¢({7‘a}aeG(F.Fo)) = 9({7‘alaeG(F.Fe) + 04F 891% F))

for [r,} E F'GlF'FOH. Then, for any h, l 6 F, we have d) o a(k (8) l) = 0(a(k 6:) l) +
cr(F®,c~o F)) = 0(0) 2 0 and so (poo =0.

Let e, be the element of FlGlF’FOH which has a 1 in the oth-coordinate and a 0
everywhere else. Then {ea},eg(p,po) is a basis for F'G(F'F°)l. Let {(1),},eqnpo) be

the dual basis corresponding to {ea},€G(p,po). Then (15 = ZOEG(F.F0) A,(,b, for some

A, 6 Fand so

<f9,({7‘o}oeG(FFo))-— Z /\ ore

aEG(FFo)

for all {TaloeG(F.Fo) E FG(F'F°). But then

o=¢(a(ke>1)) meow.) 2 Ask“
oeG(F,Fo)

Now each a E C(F, F0) is of the form a : l —> lpi" for some integer i, Z 0 with
p‘° < |F|. Now 0 96 0 implies ¢ ¢ 0. Thus f(rr) = ZaeGWfo) bear“ is a non-zero
polynomial over F of degree less than IF I with at least [Fl-roots, a contradiction,

which completes the proof of (a).

For (b), define

80

6 : F X F" -—> F" by 6(k,l) = kal. Then, since a E G(F,F0), 6 is additive on

F X F " in both coordinates. Moreover,

6(Ak,l) = (Ak)"l = Aakal = A -, (k0!) = A -, 6(k,l)

and

6(k,)1-, 1) = 6(k,)1"l)= kw: = A”(k"l) = A .., 6(k,l).

Hence 6 is F -bilinear and so there exists a F -linear map o5 from F ®p F a to F "
with (b(k <81) = 6(k,l). Now 6 ab 0, otherwise 5(k,l) = 0 implies k"l = 0 for all k,
l E F. Letting k = 1 we get l = 0 and, as the choice ofl E F was arbitrary, F = 0,
a contradiction. Therefore 5 7t 0 and consequently ab 76 0. Thus, since both F (8);: F a

and F a are 1—dimensonal over F, (b is a F -isomorphism and (b) holds.

(c) Since (v, w) —> EBa€G(F’FO)(v (8 w) is Fo—bilinear there exists a Fo-linear map

fi:V®p,W—> ® V®FW0
oEG(F,Fo)

with

,6(v®w)= $ (v®w).

a€G(F,Fo)

Next we show that B injective. Since V and W are finite dimensional vector spaces
over F, both V and W are isomorphic to the direct sum of copies of F. Therefore,
since 3 clearly commutes with action of T and the tensor product of a direct sum is

the direct sum of the tensor products, we may assume that T = 1 and V = F = W.

81

By (b) there exists a Fo-linear map

7: $ F®FF° —> F'G(F'F°)
U€G(F,Fo)

with

{ka (3) la}oEG(F,Fo) _) {kgld}a€G(F,Fo)

Leta be as in (a). Then, for any k, l E F, yfiUc <81) 2 7(®0€G(F.Fo)(k ® 1)) =

{1201} = a(k ® 1) and so a = 75. Now since a is injective, fl is injective as well.

Finally we claim dimpo(F (8);}, F) = dimpo($a€G(FJ,—o) F ®F F). For dimpo(F (8);},
F) = (dimp0 F)2 = [C(F, F0)|2. On the other hand, dimp(F®pF") 2 land dimpoF =
[C(F, Fo)| implies dimpo(F ®p F") = |C(F, F0)|. Thus, we get

dimFo( ® F®FFO) = 2 dimpo(F®pF°)= Z [G(F,Fo)] = [C(FaFoliz
a€G(F,Fo) OEG(F,F0) OEG(F,F0)

and so dimpo(F (8);, F) = dimpo(€Baec(F,Fo) F (8)}: F) as claimed.

But then S is also surjective and is therefore a FoT-isomorphism and the lemma

is proven.

Lemma 3.32 Let p be a prime, T be a finite group, V a FT-module and F a finite

field of characteristic p. Then

V = [V, T] if and only ifV = [V,0p(T)]

82

Proof Clearly, since [V, 0”(T)] S [V, T], V = [V, OP(T)] implies V = [V, T]. On the
other hand, suppose V = [V, T]. Let V = V/[V, OP(T)]. Since OP(T) S T, [V, 0”(T)]
is T-invariant and so T acts on V. Moreover, since [V, 0"(T)] = 0, T" = T/OP(T) acts
on V. But T" is a p—group and, since IF I is a power of p, V is also a p—group. Hence
semidirect product S = T" a): V is a p—group. Therefore S is nilpotent and there exists
a integer n 2 0 such that [S, n] = 1. Now V = [V, T] implies V = [V, T] = [V,T‘],
and hence V = [V, H ‘]. Also, since V is abelian, we have V = [V, T‘] = [V, T" *V] =
[V, S]. Thus, we get V = [V,S,n — 1] S [S, n] = 1 and V = 1. But this implies
V = [V, 0”(T)] and we are done.

Lemma 3.33 Suppose F0 is a finite field of characteristic p, V is irreducible as FoT-
module on which T acts nontrivially, F = HompoT(V, V) and that [W, 0P(T)] = O for

some FoT-module W. Let 0 # s : V X V —> W be a Fo-bilinear T-invariant map.

(a) There exists a unique 0 E C(F, F0) with [V ®p V",O”(T)] # V <3); V".

(b) Put E = V (8)}: Va/[V (8)1: V",OP(T)]. Then there exist an T-invariant, F-o

sesquilinear map 3‘ : V X V —> E and a Fo-linear map p : E —> W with s = ps“.
(Cl dimFE S lT/OP(T)|-
(d) Ifs(u,v) = —s(v,u) for all 21,12 6 V, then 02 =1.
Proof Note first that 3 defines a Fo—linear T-invariant map a : V (8)170 V —+ W with

a(u®v) = s(u, v) and, since 0”(T) centralizes W, we have [V®FO V, 0”(T)] S Kern a.

Note also that, as FoT-modules,

83

V (8F, V
[V ®F0 Va Tl

 

)‘gCWspoVrlTl '5 Cv'ep0V'(T)

ll?

CHomro(v,v-)(T)
= HompoT(V, V‘).

(3.1)

Since 3 75 O, a is not zero. Hence, V (85, V/[V (8)170 V, 0”(T)] # O and therefore,
by Lemma 3.32, V (81:, V/[V (8);}, V, T] 51$ 0. Thus, (3.1) implies HOI‘DFOT(V, V“) aé 0.
Also, by [1, 4.14.5], V is a irreducible FoT-module implies V“ is also a irreducible
FoT-module. Thus, by [4, 5.1.1], we conclude that V and-V‘ are isomorphic as
FoT-modules. But then, as FoT-modules, Homp,T(V,V') ’5 HompoT(V,V) = F.
Therefore (3.1) implies V 8);}, V/[V (81:, V, T] ’-_‘-’ (V @170 V/[V (8);}, V, T])" E“ F.

Notice that, since V is a irreducible FoT-module, [4, 5.1.1] implies F = HomFoT(V, V)
is a division ring. Hence, since IF | = [HOmFoT(V, V)| is finite, F is a field by Wedde-
burn’s Theorem. Now under the action of elements of F, V becomes a F T—module.
Moreover, since F0 acts on V by scalar multiplication and V is a FoT-module, F
contains a copy of F0 and therefore G(F,Fo) makes sense. For a E C(F,F0) let

E, = V (8);: V". Then by Lemma 3.31 (c),

V ®F0 V '5 e E0 (1)
aEG(F,Fo)

and hence

F

H?

 

V®FOV ~ $ E,

[v or. V. T) = [E.. T]' (2)

oeG(F,Fo)

Since 0 at F, (2) implies there exists a E C(F, F0) such that E, 515 [E,,, T]. Moreover,

since E, is an F T-module, |E,/[E,,T]| 2 IF I Therefore a is the only element of

84

G(F,F0) for which E, 75 [E,,T]. But then E, 34 [E,,T] implies E, # [E,,,OP(T)].
Now suppose p E G(F,Fo) with ,u 76 0. Then E“ = [EMT] and so, by Lemma 3.32.
E“ = [E,,,O”(T)] and (a) holds.

For (b), applying (2) with 0”(T) in place of T, (a) implies there exists a FUT-

isomorphism

_ E0 1, V ®Fo V
— [Ee.0’"(T)l _ [V 80% V»0p(T)l'

 

an

with 77((u (29 v) + [E,,0”(T)]) = (u (8) v) + [V ®Fo V, O”(T)].

Define s" : V X V —> E by s‘(u,v) = (u®v) + [E,,,0”(T)] and 6: V<E§>po V/[V Ebro
V, 090)] —> W by on + [v esp, V,0P(T)] = a(r). Finally, let p = on. Then,
s' is F o-sesquilinear and, since a is Fo-linear, 6 is Fry-linear. Thus, since 77 is a

FoT-isomorphism, p = an is Fo-linear and

ps"('u, v) = p((u (8) v) + [Em 0P(Tll) = 5’1““ ® v) + [Em 0P(T)l)

= E((u e) v) + [v or, V. 0”(T)])

= a(u (8) v)
= s(u,v)
(3.2)
For (c), notice that by (a) and (2), we have
E E, V 0 V
g (8‘7 '5 F. (3)

 

 

[E,T] = [E,,T] [v er, v, T]
Also, since 0”(T) centralizes E, T = T/OP(T) acts on E. Moreover, (3) implies

 

E E

_—
.———-— _——_
_—

[E, E, T]

ll?
"1':

and therefore dimFE/[E,T] = 1. Now since T is a p—group, by Lemma 3.21, we get

E _ _
dimpE S dimF[E,T] - |T| = |T|

 

and therefore (c) holds.

For ((1), after replacing W by a(V (8)170 V), we may assume that a is surjective.
Let s“ and p be as in (b) and i : W —> W/[W,T] and j: E —> E/[E,T] be the
natural maps. By (3) there exists a FoT-isomorphism k : E/[E,T] —) F which is
also F -linear. Now a is T-invariant implies H is T-invariant. Thus, since 17 is a FoT-
isomorphism, p = an is also T-invariant and so [E,T] S Kern ip. Hence, by the
First Isomorphism Theorem, there exists a map 7r : E / [E,T] —-> W/[W, T] such that
7rj = ip. Finally define the maps 6 : V X V -—) F and 7r" : F -> W/[W, T] by 6 = kjs“
and 1r" = wk”. Then, since 3" is F o—sesquilinear and la and j are F -linear, 6 is F

o-sesquilinear. Also 7r"' is Fo-linear as 7r and k are Fo-linear. Moreover,

is = ips“ = 7rjs' = nk—lkjs" = u‘b. (4)

and so is = 7r‘6.
Let D = kern 7r". Since s(u,v) = —s(v,u) for all u,v 6 V, we have s(u,v) +
s(v,u) = 0 and therefore is(u,v) + is(v,u) = 0. Now, since is = «'6, we get

7r*(6(u,v) + 6(v,u)) = 0 and so 6(u,v) + 6(v,u) E D for all u, v E V.

Now W 75 [W, T]; otherwise, by Lemma 3.32, we get W = [W,O”(T)] = 0 and
therefore 3 = 0, a contradiction. Hence, since a is surjective, W 75 [W, T] implies

a(V @170 V) S [W, T]. Therefore there exists u,v E V with a(u <8) v) a [W, T] and

 

86

consequently is(u,v) # 0. But then (4) implies 1r‘6(u, v) 31$ 0 and therefore 6(u,v) ,e

0.

Put a = 6(u,v) and b = 6(v,u) and let A, p E F. Then

5(AU.uv) +5(W~Av) = Au°a+wa€ 0- (5)

Applying (5) to A’ = 1 and p' = pA" we obtain

paAaia + pAab E D. (6)

Subtracting (6) from (5) we obtain

na(,\ — A02)a e D. (7)

Suppose A — A"2 315 0. Then, since a E Aut(F), Add — A” ¢ 0. Applying (7)
to p" = p(A“—l — A”)’1 we get paa E D. Let f E F. Then, as p E F was chosen
arbitrarily and a 75 0, p = (fa'1)°_l is a well-defined element of F. Moreover, f =
((fa‘l)”-l )"a = [fa E D and so F = D. But a surjective implies 5 is surjective and
therefore p = an is surjective. Hence, since i is surjective, 11' and therefore 7r“ = 7rlc"l
is surjective as well. Thus, 1r"‘(F) = 0 implies W/[W, T] = 0 and W = [W, T], a
contradiction. Therefore A — A"2 = 0 and, since A was a arbitrary element of F, we

get 02 = 1 and we have shown ((1).

Let F = Homgp(p)H(V). Now V and S, are GF(p)H-modules with V irreducible
and, by Lemmas 3.28 (a) and 3.29 (d), [S',O"(H)] = 1. Moreover, by Lemma 3.30,
S : V X V —> S, is a GF(p)-bilinear H-invariant map with flab) = F(bfi)”.

 

87

Therefore by Lemma 3.33 (b) there exist an F H -module E, a e Aut( F) with 02 = 1,

a H invariant F 0—sequilinear map 5" : V X V —> E and G'F(p)-linear map p : E —> S I

with B = ps". Moreover, by Lemmas 3.26, 3.16 (a), (e), and (f) and 3.29 (c), we get
R(K fl O”(H))

0P(H) = H n 0P(H) = KR r) O”(H)
R(K’z n 0P(H))
RK'(Z r) 0”(H))

H'(ZflOP(H)).

Thus, OP(H) = H'(Z fl 0”(H)) and so, by Lemma 3.33 (c), we have
IH’Zl/IH’I
IZMZHHTSF-

dimFE s IH/O”(H)| = IH/H’(Zr) O”(H))| s IHI/lH'l

Let H = H/CH(V) and N and K; be the images of N and K' in H. Let 3? E K"

be an element of order q for some q E w(|fi|). Since V is a FH-module and N S H,

H g GLF(V) and N g H. Also, CH(V) g N, by Lemma 3.29 (b), and so

H H
72—3/1”.
N N

Let I be as in [7, 2.5] and f be as in Lemma 3.18. Recall that H was choosen so that

11’ +f(|K|P2)l-

 

 

pdegmnfi) = pdegn/Mx) 2 “(2192 +1)2r2
Thus by [7, 2.5] applied to (H, N, iii, V) we get pdegVF(A) 2 (2p2+1)2p2|K|+f(|1\"|p2).

Since dimFV Z pdegvpfi) we also have
(21?2 +1)2P2|K|+ f(lKlP2)

Z
(2dimpE +1)2dimp(E)(|Kl)+ f(IKI dimp E).

dimpV
2

88

Therefore by lemma 3.20 (b) applied to (V, F, K, 51?, E, 3") there exists a K-invariant
subspace A = A/Z(S) of V such that s" leTi’: 1 and [2,3] 74 1. Since B = p o s" we
get B fix]: 1. By the definition of the map B the latter implies [A, A] = 1 and A is

an abelian subgroup of S.

Also [A,fi] 75 1 implies [A,x] # 1 and so x ¢ CK(A). Since x E KI we have
K’ S CK(A) and CK(A) S K. Therefore CK(A) S L, and we have shown proposition

3.27.

Lemma 3.34 We may assume A is elementary p-abelian.

Proof If its not the case that A is elementary p—abelian, then we can replace A by
Q1(A). Since p | [A], we know 01(A) 75 1. Also (21(A) char S A and so K S NG(A)
implies K g NG(o,(A)). Moreover we claim CK(Q,(A)) s L. For if G,,-(91W) g
L then K = 0,,(a,(A))L and so K’ g CK(o,(A)) by Lemma 3.16 (e). Hence
[K',QI(A)] = 1. Now K, = 0”(K') by Lemma 3.16 (f) and so K, is generated by
p’-elements. Thus by the coprime action of the pi-elements of K, on the p—group A,
[3, 5.2.4] implies [K',A] = 1. But then we get K, S CK(A) S L, a contradiction.

Therefore CK(01(A)) S L and we can replace A with {21(A) if need be.

Lemma 3.35 Let A and B be abelian groups with B acting on A and suppose CA(b) =
CA(B) for all 1 at b E B. For Y Q B define CA(Y,i) inductively by CA(Y,0) =1 and
CA(Yii)/CA(Y9i — 1) = CA/CA(Y,£—l)(Y)' The”

(a) CA(Y,i) is a subgroup ofA maximal with respect to [CA(Y,i), Y, i] = l.

89
(b) CA(b,i) = CA(B,i) for all 1 75 b E B and i _>_’ O.

(c) If both A and B are elementary abelian p-groups, then [A, B,p] = 1.

Proof For (a) its clear that CA(Y,i) is a subgroup of A. Next we claim that
[CA(Y,i),Y,i] = 1. For this we use induction on i. Notice that CA(Y,O) = 1 and
CA(Y, 1) = CA(Y) and so the claim holds for i = 0,1. Now suppose [C,,(Y, k), Y, 1:] =
1 for all k < i. Then, by the definition of CA(Y,i), we have [CA(Y, i), Y] S CA(Y, i— 1)

and so, by induction, we get

[[CA(Y,i),Y],Y,i — 1] _<_ [C,,(YJ — 1),Y,z' — 1] = 1.

But then [CA(Y,i),Y, i] = [[CA(Y,i),Y],Y,i — 1] = 1 and so [CA(Y,i),Y, i] = 1 and
the claim holds for each i Z 0. Now we claim that CA(Y,i) is indeed maximal
with respect to the above property. Again we use induction on i. Let U S A with
[U,Y,1] = 1. Then [U, Y] = 1 and so U S CA(Y) = CA(Y,1). Therefore the claim
holds for i = 1. Assume the claim holds for all k < i and let U S A with W, Y, i] = 1.
Then [[U, Y], Y,i — 1] = 1 and so, by induction, [U, Y] S CA(Y,i — 1). But then

UC'A(Y,2'— 1)
CA(Y,i— 1)

__ CA(Y,2')
' CA(Y,i — 1)‘

 

S CA/CA(Y,i—l)(Y)

 

Thus, U S CA(Y,i), and we have shown (a).

For (b), we use induction on i. For i = 0, both subgroups are equal to 1. If
i = 1 then, by assumption, we get CA(b,1) = CA(b) = CA(B) = CA(B,1). Suppose
CA(b,k) = CA(B,lc) for all k < i+1 and let A = A/CA(B,i — 1). Then, since A is
B-invariant and B is abelian, B acts on A. Moreover, the assumptions on (A, B) still

hold for (A, B). That is,

90

CA(b,i) CABJ)
07(5) = Grimm—1)“): CA(b.z' — 1) = CA(B,2' — 1)

 

= CX(B)

where the first and third equalities hold by induction. Now let X = C A(b, i + 1). Then
[X, b] S CA(b,i) and, by induction, CA(b,i) = CA(B,i). Thus, [X, b] S CA(B,i) and
therefore [[X,b],B] S CA(B,i — 1). Also, since B is abelian, we have [b,B,X] S
CA(B,i— 1). Hence, by The Three Subgroup Lemma, we get [X, B,b] S CA(B,i— 1).
But then [[X,—B], b] = 1 in A and so, since b and B have the same centralizer on A, we
get [W, B] = 1. Thus, [X,B, B] S CA(B,i—— 1) and therefore [X, B,B, B,i— 1] =
[X, B,i+1] =1 by (a). Hence, again by (a), we get X S CA(B,i+ 1). On the other
hand, since [CA(B,i + 1),B,i + 1] = 1 implies [CA(B,i + 1),b,i + 1] = 1, by (a) we
have CA(B,i + 1) S CA(b,i+1). Thus CA(b,i+ 1) = CA(B,i + 1) and we are done.

Finally, suppose A and B are elementary abelian p-groups and let 1 7t b E B.

Then bp = 1 and, since A is a vector space over GF (p), we have [A,b, p] = 1. Thus,
by (a) and (b), we get A = CA(b,p) = CA(B,p) and so [A,B,p] = 1.

Now, since A and Z are elementary p—abelian and Z acts on A, by Lemma 3.35

(c), we have [A, Z, p] = 1. Let

AEAlEAzEAsE---2Ak=1 (1)

be a composition series for K on A where A,- are all K-invariant normal subgroups
of A,-1 for all 1 S i S k and K acts irreducibly on the factors A,/A,-+1 for all

1SiSk—1.

Suppose K ' acts trivially on all the composition factors A,/A,~+1 for K on A.

Then the action of the p’-elements of K ' stabilize the chain (1) of the p—group A and

 

91

therefore centralize A by [3, 5.3.2]. Since K, is generated by p’-elements, the latter

implies K, centralizes A, a contradicition. Therefore there exists a composition factor

B for K on A such that [K',B]7$1. Let K = K/CK(B).

Lemma 3.36 Let B and K be chosen as above. Then

(a) B is a faithful irreducible K-module for which CK(B) S L and [B, Z,p] = 1.
(b) L is a p'-group with L = @(K).
(c) R/ciu?) 95 A,.
Proof For (a), since B is a composition factor for K on A and [A,Z, p] = 1, its
clear that B is a faithful irredducible K—module with [B, Z, p] = 1. Moreover, since

CK(B) S] K and I", S CK(B), we have (33(3) S L. Since 019(11’) S CK(B), (1))

follows immediately by the same argument used in the proof of Lemma 3.16 (d).

Finally for (c), since CK(B) S L, we have

ll?

1~r| >1
II?
E“

 

>
NI >3)

D_ef Let T be a finite group and t E T. Then L, = CT(t)°° is the terminal member

of the derived series of CT(t) defined by

Lt = CT(t)°° = n CT(t)(i)
i=1

where CT(t)°° denotes the (i + 1)th member of the derived series of CT(t).

92
Lemma 3.37 Let 1 7b k E K. Then

(a) L, H 0,05).
(b) L; is generated by p'-elements.
(c) Let 1?‘ = femur?) and L; be the image of L; in 1?. Then L; = Ls. for

anyl¢zEZ.

Proof Clearly, since L; is the intersection of normal subgroups of C R(F), we

have L; S C505). Also, since L; is perfect, L;/0”(L;) is a perfect p—group.
Thus, L;/0’°(L;) = 1 and therefore L; 2: 0”(L;) and L; is generated by p,-

elements. Therefore (a) and (b) hold.

Finally for (c), we claim CR.(’2\‘) = CR(3)'. Let Y‘ = F/<P(K) = CR4?)
Then [?t,?] = 1 implies [17,2] 3 (MR). But then (aw?) g i7 and (s) e
Sylp((?)<I>(K)). Thus by The Frattini Argument we get F = Np((3))<I>(K).

But, since @(K) is a pI-group, we have [N,7((§)), (5)] S (3) fl <I>( ') = 1. Thus
N,((3)) = 0,,(<2)) = 0,,(2) and i7 = cp(2)<1>(1?). But then

 

A Corina?) A. A.
Y I; :C*(Z) SCR(Z)

and so we have shown CR.(?) S CRCE)‘. On the other hand, [CK(§),E] = 1

implies [CR(?)',?] = 1 and so CRC‘E)‘ S CK..(?‘) Hence altogether we have

shown Cit-(’2‘) = CR(?)' as claimed. Now we get

93

Let q 74 p be an odd prime. Then by Fermat’s Little Theorem we have p I

q”'1 — 1. Thus, we can choose t minimal such that p I q‘ — 1.

Lemma 3.38 Let p, q, andt be chosen as above and eq be an elementary

q-abelian group of order q‘. Then

(a) P l [Aut(eq)I.

{b} Ifx E Aut(eq) with it” = 1, x acts irreducibly on Eqr.

Proof For (a) eq is a vectorspace over GF (q) and therefore Aut(eq) ’5
GL.(q). Hen p | q. —1 end lAut(eq)| = IGLt(q)l = new — q").
we get p | [Aut(eq)I.

For (b), suppose H is a x-invariant subgroup of E9: of order q‘ where 0 < s < t.

Then at E Aut(H) and Aut(H) ’5’ GL,(q). Therefore

p = In) I lAut(H)l = II(q* — q‘).

i=0

But since q 75 p we get p I (q‘0 — 1), for some so S s < t. This is a contradiction

to the minimality of t.

Let eq and x be as in Lemma 3.38 and (y) be a cyclic group of order p. Then

we can form the semidirect product eq =1: (x). Consider the group

W = Eq‘ * (1') X (y)

94

Then [W] = quta 0,,(VV) 2 Eq‘» and <33) X (y) E Sylp(W). We can embeed W

into S, by letting W act on
1“ -— W u (W/0.<W))H"”’2q"

by right multiplication. Here (W/Oq(W))1—F(r-p2qt) denotes #(r — p2q‘) distinct
copies of W/ 0,,( W) Moreover, we claim that through this action W becomes a.
subgroup of A,. Since E9: = 0,,(W), eq acts trivially on (W/0q(W));]7(r-p2qtl.
Furthermore, since q is odd, any q-cycle is even. Thus, since eq is elementary
q-abelian and acts semiregularly on W, all elements of E,: are even on W and
are therefore even on F. Now x and y act semiregularly on W. Also, since
Stabw(w0q(W)) = 0,,(W)“’ is a q-group for any w E W, both x and y act
semiregularly on (W/O,(W))F’(r—p2qi). Thus, (at) X (y) is of type (p,p) and acts
semiregularly on I‘. If p is odd then all elements of (x) X (y) are even on P. On
the other hand, ifp = 2, then 4 = p2 = [(x) X (y)I divides IFI. Hence, again
every element of (x) X (y) is even on F. Therefore, through this action, W is a

subgroup of A,.

Now by Lemma 3.36 (b), K" 2’ A,., and so henceforth we will identify K' and

A,.. Finally let W" denote the image of W in A,.

Lemma 3.39 Let Q be a set with IQ] = n for some positive integer n 2 3 and

A and B be subgroups of Sym(0) = 5,, such that

(a) A and B act semiregularly on 0.

(b) there exists an isomorphism d) from A onto B.

Then there exists x E S, such that 45(a) = a“ for all a E A.

95

Proof First we claim that A and B have the same number of orbits on 9.

Suppose

I: l
QzUrfanszUs? (1)
i=1 j=l

are two partitions of 9 into disjoint unions of orbits of A and B. Then we want.

to show k = I. Since A acts semiregularly on Q, from (1), we get,

I: k

A

IQI=IUT§4| = eril
i=1 i=1

i=1

= kIAI.
Hence IQI = kIAI and similarly we get IQI = lIBI. But then, kIAI = lIBI and,
since A E“ B, we have IAI = IBI. Therefore I: = l as claimed.
Now we assume that A and B both have I: orbits on Q and define x E S, by

(r9)” = st“) for all a e A and 1 g i g k

Then x is well defined on (2 since gt is well defined and A is semiregular. More-
over, cb is surjective implies x maps 0 onto 9. Finally, B is semiregular and d)

is injective implies x injective on Q and therefore x is indeed an element of Sn.

Let s E Q and a E A. Then there exists 1 S i S ls: and a1 E A such that

s = 3;?(01). Hence, we have

96

S(a’)"¢(a) ___ (Syphon-we) ___ (r?1)0—11‘¢(a) = (rilla‘l)x¢>(a)
d: a a“ a
: (Si(i ))o()
sifla‘) = 3.
Thus, since 3 E Q was chosen arbitrarily, we have (a’)"q'>(a) = 1 and so

gb(a) = a” for all a E A.

Lemma 3.40 Let 1 ¢ 2 E Z. Then

(a) We may assume Z' E Sylp(W‘) and ’2‘ E Z(W').

(b) 2* g Lg.

Proof Let 3", 23' E Z“ such that Z“ = (3",26'). Then (9,2)”) E (5529').
Moreover both of these subgroups of A, act semiregularly on Greg. Therefore,
by Lemma 3.39, they are conjugate in 5,. Hence there exists 5 E S.- with
’5“:3 = g“ and £5" = ff". But then 2'“ E Z(W“) and (Z‘)’ E Sylp(W‘) and we

have (a).

For (b), since [Greg] = 4p2, r/p is an integer. Define the subgroup U" of A, by

where (f‘) is a cyclic group of order p and acts on ”reg by permuting each of
the p subsets of size r/ p of Qreg cyclicly and A‘ acts as A; on each of these p
subsets. Then, since r = 4p2, r/ p = 4p > 5 and so A’ is perfect and simple.
Again, as 4p = r/ p, we can choose it; E A‘ such that t7). acts semiregularly

on Sly-cg and t?!” = 1. Then as above (Ff?) E (2‘35?) and both of these

97

subgroups act semiregularly on nreg- Therefore, by Lemma 3.39, there exists
3 E S, such that 2’3" = t7). and 3:“ = f“. Now t; E A‘ S CR4?) and, since

A‘ is perfect, we get

{3' e 2* g flame)“ = L,

i=1

1, we get 25' E Lg». Therefore we

Thus t?)“ E L;. and so, by conjugating by s”
have shown 55" E L; for all 23" E (33"). Now 26" 3" E (’2‘) and so 50' 3:“ E L2.
But then we have p < IZ" fl Lg] I p2 and therefore Z" n L; = Z". Thus Z' S L;

and this proves the lemma.

Lemma 3.41 Let T be a finite perfect group and B and C subgroups of T.

Suppose

(a) BST

(b) C is solvable

and
(c) T = BC.
Then T = B.

Proof Since T is perfect T/B is also perfect. But, C is solvable implies

BC
B BflC

ll?

 

is solvable .

ml“)

98

Thus, T/ B is both perfect and solvable. The only way this is possible is if
T/leande-B. C]

Lemma 3.42 For any 1 ¢ 2" E Z’ let C = CR.(?), C, be a cyclic group of
order p, and CE = C, X C, X X C, be the direct product of; distinct copies

of Cp. Then

(a) Up is odd, C E’ C, wr Ag.
(b) Up = 2, C ~ (0,? n A,.)Sg.
(c) Solv(L;) is a p-group, L; = Solv(L;~)A%, and Lg/Solv(L§) E“ Ag.

(d) L; = (2'14)-

(6) L3 = (21‘?)

Proof For (a), let S, = Sym(Qreg) and, since K“ E A,, identify K‘ and A,.
Now Z " acts semiregularly on Greg and 2‘ has order p, thus 3?" is the product
of f)- distinct p-cycles. Therefore, by [10, 3.2.13], we have C3,(?) ’5 C, wr 53'
But then Cs.(2>') = AU where A 2 CE, A g 05,09), U .2: 5%, A n U =1,
and C5,(’z"')/A L.“ U. Now since p is odd, every p-cycle is an even permutation.
Therefore, since every element of A is the product of p—cycles, we have A S A,.
Moreover, if u E U is an even permutation in S,, then since p is odd, u must be

an even permutation on each subset of size % of Qreg and therefore UflA, 2 Ag.

Thus we get,

C = C,,(?) = A, n C,,(2*) = A, n AU = A(A, r) U) = AAg.

Now A S C3,(?') and A n U = 1 implies A S CA,(§') and A 0 A: = 1.
Therefore C = C, wr Ag.

99

For (b), p = 2 implies every permutation of U is the product of two disjoint
permutations of the same type. Thus, every element of U is even and therefore

U S A,. But then
C = A, 0 CS,(?) = A, 0 AU = (A, n A)U.

For (c), first suppose p is odd. Then by (a) we have C = A»: Ag. Now L; = C°°
is the unique maximal perfect subgroup of C. Therefore, since Ag is perfect,

we have Ag S L}. But then we get

gngnCngnAAgzwgnAM (1)

Let S = Solv(L;~). We claim S = AOL}. First A S C5,(?‘) implies AOL; S L}.
Therefore, since A is a p-group, A is solvable and A n L; S S. On the other
hand, SflA: S A: and so SflAr. = 1 or Ag, as A: is simple. But 80A; 2 A:

P P P P P P P

implies Ag S S. Hence, in this case we get, Ag is solvable, a contradiction.

Therefore S 0 Ag = 1 and so

S=SnL§=SnAg(AnL;)=(SflAg)(L;nA)=Lg~nA (2)

and the claim holds. Moreover, since A is a p—group, S is a p—group. Now, by

combining (1) and (2), we get L321: SAg and, since S S L; and S f) A; = 1,
p

we have Lg/S 92’ Ag.

Now if p = 2, then (a) implies C = (A, n A)U. Thus, since L; is perfect, A is

abelian, and S; = Ag, we get
P

L; g ((A, n A)U)’ = [A,. n A,U]Ag. (3)

100

Then, again since Ag S Lg, from (3) we have

;= L;n[A,nA,U]Ag = (Lgn [A,nA,U])Ag. (4)

Let So 2 L; D [A, F) A,U] and S = Solv( g). We claim S0 = S. Since A S
C5,(?), So S A and therefore So is ap-group. Also, [A,flA, U] S (A,flA, U) 2
C513") implies So = Lgr) [A, n A, U] S L; and therefore So S S. But then, as

before, S 0 Ag 2 1 and so from (4) we get

S=SnL§=SnSoAg=(SnAg)Sp=So.

Therefore S = So, L; = SAg, S is a p-group and Lg/S 2’ Ag.

For ((1), let L0 = (Z’LE), S = Solv(Lg), L—g = 'zi/S and L; be the image of L0
in L_;. Then we want to show L; = Lo. Well Lo S L; implies L_o S L—g. But, by

(c), :35 Ag is simple and so we get L_o =1 or L—o = L_;.
IfL_o = 1 then Lo S S. Now, from the proof of Lemma 3.40, there exists 3 E S,
with A"—1 S L}. Thus, since Lo S L-;, LoflA‘rl S A“—l. But 14”“ is simple

and so L0 0 14"": = 1 or 14"“. Now

1 a :26" = {0‘8“ 6 L0 n AH"

and therefore L0 0 21"“ ¢ 1. On the other hand, if A"?! 2 Lo 0 A“?! then
14"“ S Lo. But, from above we have, Lo S S is solvable and therefore we get

A —l . . o
A” 13 solvable, a contradiction.

Therefore it must be the case that L—o = L; But then L; = LOS. Now L; is

perfect, Lo S L}, and S is solvable. Thus, by Lemma 3.41, we get L; = Lo and

101

((1) holds.
Finally, by Lemma 3.37, L; = (2%) implies L; g (2L?)<I>(i{'). Let L0 = (2L2).
Then L, 3 Low?) and L0 5 L, implies

L, = L, 0 Loan?) = L,(L, n <1>(1?)). (5)

Thus, since L; is perfect, L0 S Lg, and L0 (1 <I>(K) is solvable, again by Lemma

3.41, (5) implies L; 2 L0 and therefore (e) holds.

Lemma 3.43 Let Q be a finite set, p a prime, and B a semiregularly acting

subgroup of Alt(Q) of type (p,p). IfIQI Z max{p2 +1,5p} then

Alt(Q) : (La, Lb)

for all a, b E B with B 2 (a,b).

Proof First, let a, b E B with B = (a,b) and consider the set (2' = R X X
where R = GF(p) and X is a set of order IQI/p. For any (r,x) E 9’, define a',
b, E Sym(Q') by a, : (r,x) —) (r+1,x) and b, : (r,x) —) (r, x”) where p E Alt(X)
is the product of IX I / p p—cycles. Then a’p = b'p = 1, [a',b'] = 1 and therefore
B' = (a',b') is of type (p, p). Moreover, since a' and b, are semiregular on 9',
it follows that B, is also semiregular on 9'. Now since IQI = IQ'I, we have
Alt(Q) ’5 Alt(fl'). Let this isomorphism be 45. Suppose we are able to show
that Alt(fl’) is generated in the desired way, that is, Alt(fl’) 2 (L01, L61). Then,

since both (a, b) and (¢(a),qb(b)) are isomorphic semiregular acting subgroups

102

of Sym(fl), by Lemma 3.39 there exists 3 E Sym(Q) such that gb(a)‘ = a and
o5(b)’ = b. Thus, Alt(fl) = (L,,,(,,),L¢(b)) implies Alt(fl)’ = (L;(a)’L;(b)l =
(Lacy, L¢(b)s) = (L,, Lb). Now by the normality of Alt(fl) in Sym(fl), we get
Alt(fl) = (La,Lb). Therefore we may assume it = 0', a, = a, b' = b, and
B’ = B.

Let 7r E Alt(X). Define 7r“ E Sym(Q) by 7r" : (r,x) —> (r, x") for each (r,x) E 9.
Then 71' E Alt(X) implies 7r"‘ E Alt(fl). Let A = {7r‘I7r E Alt(X)}. Then clearly

A ’95 Alt(X) and A S CAlt(o)(a)' Moreover, since IQI 2 5p, we have X

 

 

25.

Hence A is perfect and A S L,,.

For x E X, define a, E Sym(fl) by

I (r+1,x) ifx'=x
(r,x) ifx'yéx

Then a, is the cycle of a containing (0,x) and therefore a, E CSym(n)(a)°
Moreover, a]: = (ll-n. Let x and x, be two distinct elements of X. Then
there exists 7r E Alt(fl) with 7r(x) = x'. But then, since 7r" E A S La and
L, S CSym(o)(a)1 a, E CSym(o)(a) implies [a,,7r‘] E L,,. But

1 l

[ear] = a; a3." = a; as

and so we have shown that for any two cycles c and c, of a we have c‘lc' E L,,.

Moreover, since a and b are conjugate in Sym(fl), the corresponding statement

holds for b and Lb.

For i E R define, X,- = {i} X X, and A,- = {a E Alt(Q)Io E CAlt(n)(Q\
X,) and o is even on X,}. Then, since A,- fixes the first components of elements

of (2, we have A,- E’ Alt(X,) E’ Alt(X) E’ A. Now ID] > p2 implies IXI > p.

103

Thus, since b acts semiregularly on 0, p has at least two different cycles, say d,
and d2. Define c1 and c2 E Sym(f2) by

(r, xd“) ifr = i

c),(r,x) ={ (r,x) if r # i
for k = 1, 2. Then c), is just the cycle ofb containing (i, xk) where x, E Supp(dk)

for k = 1,2. Therefore 6 = cl'lc2 E Lb.

Next we claim A, = [e, A]. First, since it \ X, Q Fix(e) we have 6 E CAlt(o)(Q\
X,) = A,. Also, since A normalizes Q \ X,, we get (2 \ X, Q Fix([e,A]) and
therefore [e,A] S A,. On the other hand, A E Alt(X) g Alt(X,) 2’ A, and
e acts nontrivially on X,. Thus, identifying A and e as subgroups of Alt(X,),
we get 1 7t [e,A] S A. But, since A is simple, we get A = [e,A]. Hence
[e,A] g Alt(X,) and therefore I[e,A]I = IA,I. But then A, = [e,A], and the

claim holds.

Let L = (La,Lb). Then, since 6 E L, and A S L,,, we have A, = [e,A] S L.

But then L contains three cycles, as A, g Alt(X,).

Finally, we claim L acts primitively on 9. Since A,- E Alt(X,), we know A, acts
primitively on X,. Suppose that A is a block for L on Q with 2 S IAI < IQI.
Then, since A, normalizes X,, A 0 X,- is a block for A,- on X,. But then A, is

primitive on X, implies either IA 0 X,I S 1 or X, S A.

Suppose that IA 0 X,I = 1. Then, since IAI Z 2, we have A \ X,- 7E 9. Now
A,- fixes A \ X,- and A is a block implies A,- normalizes A. But then A, also

normalizes A n X,, a contradiction, since A, E’ Alt(X,).

Hence we proved that for all i E R either A H X, = a or X, Q A. Since 9 is the
disjoint union of the X,’s we conclude that A = U je J X ,- for some proper subset

J of R. Pickj E J so that j+ 1 E J ( This is possible since A # 9 implies

104

J 75 R ). Let :r, :r', and x" be distinct elements of X and put u = (11:10:31. Then
u fixes (j, 1:”). Moreover, since u E L, S L and A is a block, (j, :r") E A implies
that u normalizes A. But this is a contradiction as (3,15,) E A, (j + 1,:r') E A

and (j, :L")u = (j + 1,27,). Therefore L acts primitively on Q.

Now since L is primitive on Q and L contains three cycles, by [12, 2.13.3].

L = Alt(fl) and the lemma is proved.

Lemma 3.44 Let I? = R/ou?) and 21, 22 e Z with 2- = (5123;). Then

I? = (1...; , L33).
Proof Since R" 9: A,., 2" is a semiregularly acting subgroup of R" of type
(Pap), and 7‘ = 4P2 2 maX{p2+1,5p}, by Lemma 3.43, we have R" = (Lao, Lao).

But then I? = (L5,L;,)<I>(R'). But then, by [3, 5.1.1], we get I? = (L5, L23).

CI

Lemma 3.45 There exists a 1 75 ’2‘ E 2 and a composition factor C for L; on

[3,3] with [L;, C] 9a 1.

Proof First notice that both L3 and 2 centralize 3 and so both L; and 2
act on [8,3]. Now suppose the lemma is false. Then, for all 1 ¢ 3 E 2, L5
centralizes all the composition factors of L; on [B, 27]. But L; is generated by p,-
elements, by Lemma 3.37 (b). Hence, by the coprime action of these p'-elements
on the p—group [8,3], [3, 5.3.2] implies [B,E, L3] = 1 for all 1 3:5 3 E 2. Let 2"],

32, and 23 e 2 with 2 = (2,22) = (21,23) = (23,23). Then,

105

[08(2).221=Ica(a),<a22>1 = [Cami]
= [Gamma]

and therefore

[Cams] =[CB(2.),231=[CB(21>.21. (1)

Now, by Lemma 3.44, R' = (L5,L5). Hence, since [B,ZLg] = 1 for all 1 #

3E Z, we have

[03(51). 5‘2] = [C3(?1), 23] S CB(L23) 0 CB(Lz3)

|/\

03((LzaaLal)

= 03(2).

Now, since 030?) S B is R-invariant and B is a irreducible R—module, either
CAI?) = 1 or B. Now 03(3)) = B implies [K, B] = [R, B] = 1 and therefore
K = CK(B) S L, a contradiction. Thus 03(1? ) = 1 and so from above we get

[Cigars] =[CB(§1),2‘3] = [03(3021 = 1. But this implies

03(21): 03(23) = 03(2). (2)

Again, since R = (qu , L13), (2) implies CB(2) is a R-invariant subgroup of B.
Hence, by the irreducibility of B, we get C30?) = 1 or B. But, since 2 is a
p—group and B is a vector space over GF(p), by [3, 2.6.3], (33(2) 75 1. Thus
03(2) = B and so [B,Z] = [B,Z] = 1. Hence we get, Z S G,,-(B) S L, a

contradiciton.

106

Let 1 74 3 E 2 and C be a composition factor for L; on [B,E] such that.

[0,14] 7e 1.

Lemma 3.46 p is odd.

Proof Suppose p = 2. Then, by Lemma 3.35, [A, Z, Z] = 1 and so also
[B,3,2] = 1. Hence [0,2] = 1, as C is a composition factor of [BE]. But

then, by Lemma 3.42 (b), we get

[0. L5] = [042112)] :1

a contradiction to Lemma 3.45. Therefore p must be odd.

By Lemma 3.40 we may assume W" has the form

A .

W'=<2‘o*>xE‘.: *(3‘)

where B;- is an elementary abelian q-group of order q‘, q is an odd prime,

2“ = (’2‘, 23"), and 236' acts nontrivially and irreducibly on Bis...

Lemma 3.47 Let W”, B; , ’2", 5.5", and C, be as above. Then
(a) E9‘ = [Eq‘ ’2‘]-
(b) V17" 3 Lg.

(C) 2 S L3.

107
(d) [0,3] = 1.

Proof Since B;- is Z‘-invariant we have IB;¢.,2‘] S B}. and [B21221 is
Eif-invariant. But, since 26' acts nontrivially and irreducibly on B}: we have
51:13.21.

For (b), since B}. = [Fist/Zn] S (Z‘W') and 2" S (A'Wi), we have W" =
(Z‘W'). But, by Lemma 3.40 (b), 2" S L; S C34?) and W" S CR-(?)-
Thus W‘ = (B‘W.) S L} and we have (b).

For (c), 2" S L; implies Z S L;<I>(R’). Hence

2 S L;<I’( ’)nCR(E)

= L5C¢(R,(E)

2L5 L§C¢(R)(E) 2, Came)
L3 — L; — C<I>(I?)(E) n L;

 

 

is a p, — group

Therefore, since 2 is a p—group, we have 2Lg/L; = 1 and so 2 S L5.

Since L; centralizes 3, we have Cc(§) is a Lyinvariant subgroup of C. Thus,
since L; acts irreducibly on C, we have either 00(3) = 1 or C. But since both
C and (E) are p-groups CC(§) 75 1 by [3, 2.6.3]. Therefore CC(?) = C and we
have (d).

[:1
Now 2,3" acts nontrivially on 13:1. and 1 74 Bic. S W" implies E3. is 2'-

invariant and [IE/$223] 791. Thus 1% B; S W and [1521,21 aé 1 where W and

B; denote the preimages of W‘, B}. in I? and Z is the image of Z in R.

108

Lemma 3.48 Let W, 2 and B; be as above. Then

(a) B; is a pI-group.
(b) E} n L; +1 1.
(C) IE; 0 [5,2] 5f 1.

(d) There exists a Z—invariant q-Sylow subgroup B; ofBZflL; with [B3, 2] 96 1.

Proof For (a), since B?“ is a q—group, q 75 p, and (NI?) is a p'—group, we have

A

E,: is a pI-group.
For (b), 177;: S F17" S L; implies B; S L;‘I>(R') and therefore

A

E? = E; n L.<1>(1?)=(E‘,. n L2)<I>( ').

Hence, since 1 # Fig.1., we have B; 75 (DU?) and therefore B; n L; 3f 1.

A

For (c), B; = (ch fl L;)<I>(I?) implies Fit. = (B; O 113).. Therefore, since
[Ii/21‘, 2"]751, we have [(B; n Lg)‘,2.] 95 1 and so [(17:20 Lg),2];£ 1.

Finally, since 2 S L; and B; is B-invariant, we know that 1 75 B; (1 L; is a 2-
invariant pI-group. Therefore by the [3, 6.2.2], there exists B; E Squ(B:¢ 0 L3)
with B; Z-invariant. Since (B; 0 L;)" = 173;: is a q-group, we have B3. = E;

Thus, {B}: 2"] ;£ 1 implies [B32 2"] 31$ 1 and therefore [B3, 2] 3i 1.

Now let B = [233, 2]. Then, by Lemma 3.48 (d), 1 # B is a B-invariant q—group.
Also let )7 = B2 and X = 17).: F/CflC).

Lemma 3.49 Let )7, X, and B be choosen as above. Then

109

(a) E = [E 21.
(b) L; = (3%) and LE: (E14).

(c) 0,,(X) =1.

Proof For (a), by the coprime action of the p—group 2 on the q-group BB, [3,

5.3.6] implies

E = [233,2] = [33,221 = [13:2].

For (b) let L0 = (13%;), s = sch/(Lg), 17;, = Lg/S, and E; be the image of L0

 

in L_§. Then, as in Lemma 3.42, the simplicity of L—; implies LT, = 1 or L—o = L3}.
Now L; = 1 implies L0 S S and therefore B“ S 5'. But, since 5 is a p—group

and B" is a q-group, we get B" = 1 2 [B32 2*] 2 {B}: 2"], a contradiciton.

On the other hand, if L; = L; then L; 2 L05. Thus, and therefore by Lemma
3.41, we get L; = L0. The same argument used in Lemma 3.42 (e) now shows

L = (1%).

N)

Finally, suppose (c) is false. Then 0,,(X) 7f 1 and therefore IX

 

,, 75 1. Now
2 E Sylp(i>) implies E E Sylp(F) = Sylp(X). Therefore [BI = p or p2. Suppose

Ii] = p2. Then, since [2| = p2, we have

2 @2191 = ___1_2_1_ = P2
ICp(C)I ZflCp(C) I200?(C)I.

 

P2=IZI

Thus we get 2 fl C?(C) = 1, a contradiction to Lemma 3.47 ((1). Therefore
[2| = p and so 2 = 0,,(X) S X. Now 2 and B normalize each other. Hence,

since 2 is a p-group and B is a q—group, we have [B,?] S B n E = 1. Thus

 

110

[B32] S C7(C). But B = [3,2] by (a) and so B S C5;(C). Hence, we get

[B, C] = 1 and therefore [L3, C] = [(BLE), C] = 1, a contradiction. D

N ow, since X is a pq—group, X is solvable and therefore p-solvable. Moreover, C
is a vector space over GF( p) and so X acts as a group of linear transformations
on C. Consider E}; E X. By Lemma 3.35 (c) we have [A, Z, p] = 1 and so, since
B is a section of A, [B,Zp] = 1. Hence, [B,E,§B,p — 1] = 1 and therefore
[C,Zm — 1] = 1, as C is a section of [B,?]. But then (.2?) —1)”‘1 = 0 on
C. Now suppose p is a fermat prime. Then, since the sylow-2-subgroups of X
are abelian (in fact, since p and q are odd, there trivial), by [3, 11.1.2], we get
mindg) = (a: - 1)”, a contradiction as S satisfies a polynomial of degree less
than p. On the other hand, suppose p is not a fermat prime. Then, since p aé 2,
again, by [3, 11.1.2], we get mindg) = (ac—1)”, a contradiction. This concludes

the case in which G is of alternating type and therefore we have shown Theorem

3.1.

Bibliography

Bibliography

[1] Micheal Aschbacher, Finite group theory, Cambridge University Press,
1986.

[2] Daniel Goldschmidt, Automorphisms of Trivalent Graphs, Ann. of
Math.,111(1980) 377-404.

[3] D.Gorenstein, Finite Groups, Chelsa Publishing Company, New York 1982.
[4] Larry C. Grove, Algebra, Academic Press, Inc, 1983.
[5] O.Kegel and B.A.F Wehrfritz, Locally Finite Groups, North—Holland, 1973.

[6] Martin W. Liebeck and Jan Saxl, 0n point stabilizers in primitive permu-

tation groups, Communications in Algebra, 19(10), (1991) 2777-2786.
[7] Ulrich Meierfrankenfeld, Non-Finitary Locally Finite Simple Groups.

[8] I.Schur, Uber die Dantellungen der symmetrischen Gruppen and alternieren
der gruppen durch gebrochenen linearen substitutionen, Crelle .1., 139 (1911)
155—250.

[9] Bernd Stellmacher, D.Goldschmidt, A.Delgado, Groups and Graphs: New
results and Methods, Birkhausser, Basel 1985.

[10] Michio Suzuki, Finite Groups, Springer-Verlag, 1982.

[11] John. G Thompson, Finite with normal p-complements, Proceedings of

Symposia in Pure Mathematics, voll (Finite Groups), 1-4 (1959).

111

 

112

[12] Helmut Wielandt, Finite Permutation Groups, Academic Press Inc. New

York and London, 1964.