l LIBRARY
200% Michigan State
University

This is to certify that the
dissertation entitled

CENTRALIZERS OF ELEMENTS OF PRIME ORDER IN
LOCALLY FINITE SIMPLE GROUPS

presented by

Elif Seckin

has been accepted towards fulfillment
of the requirements for the

PhD. degree in Mathematics

 

 

t

11:1,,
MajogPfi’J/fessor’s Signature
(-11 (73/2007

Date

 

MSU is an Affirmative Action/Equal Opportunity Institution

- —--.-.-n---.—.- -

 

PLACE IN RETURN BOX to remove this checkout from your record.
TO AVOID FINES return on or before date due.
MAY BE RECALLED with earlier due date if requested.

 

DATE DUE DATE DUE DATE DUE

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5/08 K:lProi/Acc&Pres/ClRC/DateDueindd

 

CENTRALIZERS OF ELEMENTS OF PRIME ORDER IN
LOCALLY FIN ITE SIMPLE GROUPS

By

Elif Segkz'n

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements

for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

2008

ABSTRACT

CENTRALIZERS OF ELEMENTS OF PRIME ORDER IN LOCALLY FINITE
SIMPLE GROUPS

By

Elif Seckin

This thesis is mainly focused on the centralizer of elements of prime order. We prove
a result which gives all the cases where Ca(gl) 2: 00(92) holds for G’ = PGLK(V)
or G = PSLK(V) if V is a finite dimensional K—space and 91, 92 E G of prime order
7" such that r 75 charlK. A similar result is obtained for finite alternating groups.
We prove that a simple locally finite group containing an element of prime order
p whose centralizer is abelian is either linear or a group of p-type. Another result
presented is that any non-linear simple locally finite group contains a p-subgroup
which is not Cernikov. This in turn proves that such a group contains an infinite

abelian p-subgroup.

T0 the memory of Dr. Richard E. Phillips

iii

ACKNOWLEDGMENTS

I would like to express my deepest and sincerest gratitude to my advisor, Ulrich
Meierfrankenfeld, for his exceptional support and constant encouragement, without
which this thesis would not have been possible. I am very thankful to him for his
understanding, boundless patience with me, and not giving up on me. It has been
an honor to work with him and the time I have spent as his apprentice will forever
be appreciated.

I would also like to express my gratitude to Jonathan 1. Hall, Jeanne Wald, Ronald
Fintushel, and Abdul Sami for agreeing to be on my committee, for their kindness,
and support. I am thankful to Dr. F intushel for introducing me to Beamer and
for his joyful chats. I would like to thank to Abdul Sami for his friendship and for
valuable conversations.

I would like to thank to my dear friend Jui-Ling Yu for supporting me when I
needed help most and for encouraging me to continue. Her friendship have helped
me immensely. I am also very grateful to my friend Onur Aglrseven for his continual
help. Finally, I want to thank my brothers H. Murat Seckin and Sahin Seckin for
their love and support.

I would like to once again dedicate this work to the memory of Dr. Richard E. Phillips,

my thesis advisor at the beginning of my doctoral studies.

iv

TABLE OF CONTENTS

Introduction ................................

1 Preliminaries ..............................
1.1 Structure of Centralizer in GLK(V) ...................
1.2 Field Extensions .............................
1.3 Exceptional Cases of Theorems 2.1.1 and 2.2.3 .............

2 Centralizers in PGLK(V) and PSLK(V) ...............
2.1 The PGLK(V) Case ...........................
2.2 The PSLK(V) Case ............................

3 Centralizers in Alt(Q) .........................
3.1 Centralizers in Alt(n) ..........................

4 On Abelian Centralizer in Locally Finite Simple Groups
4.1 The Non-regular Alternating Case ...................
4.2 The Finitary Case ............................

5 On Infinite Abelian Subgroups in Locally Finite Simple Groups

BIBLIOGRAPHY .............................

Introduction

Centralizers have long played a very important role in the theory of finite as well as
locally finite groups. These subgroups are one of the key tools that can be used to

obtain detailed information about the structure of the group itself.

In this dissertation, we focus on the centralizer of elements of prime order. The
thesis is organized as follows. In Chapter 2 and 3, we investigate the cases when two
such centralizers are equal in certain groups. In other words, the following question
is considered in these chapters:

Question 1. Let G be a group and a,b E G be elements of prime order r such

that Cg(a) = Cc(b). Then what can be said about (G,a, b) .9

In Chapter 2, we deal with the case when G is a projective linear or a special
linear group. More precisely, in Section 2.1 we prove a result (Theorem 2.1.1) that
characterizes exactly when 00(91) = 03(92) holds for G = PGLK(V) if V is a
finite dimensional IK-space and 91,92 6 G of prime order r such that r # charK.
A similar result is obtained for the group PSLK(V) in Section 2.2 (Theorem 2.2.3).
The key lemmas giving the structure and properties of Cg(g,-) that are needed for the
proofs of theorems in Chapter 2 are provided in the first two sections of Chapter 1.
Some small cases (i.e., when dim V g 3 and [Kl S 5) appearing in these results are

investigated in Section 1.3.

In Chapter 3, we consider the finite alternating group Alt(n) and determine all

possibilities for (n,a:,y) such that :r,y E Alt(n) of prime order, (as) 74 (y) and

CA1t(n)(-’L‘) = CAlt(n)(y)-

Chapter 4 is, roughly speaking, about the centralizers of elements of prime order in a
simple locally finite group. In order to give a more precise description of each section
we shall need some definitions as well as the classification theorems that are used in

our proofs.

e Let G be a group. G is called locally finite if every finite subset of G generates

a finite subgroup of G. G is a LFS—group if G is a simple, locally finite group.

a A set of pairs {(Hi, M,) | i E I} is called a Kegel cover for G if, for all i E I,
H, is a finite subgroup of G, M, is a maximal normal subgroup of H,- and for
each finite subgroup F S G there exists i E I with F S H,- and F n M,- = 1.

The groups Hi/A/Ii, i E I , are called the factors of the Kegel cover.

It has been shown in [13, p.113] that every LFS-group has a Kegel cover. Kegel
covers are one of the important tools in the study of locally finite groups because
using Kegel covers many question about LFS—groups can be transferred to questions
about finite simple groups, which in turn may be answered using the classification

of finite simple groups.

0 A group G is called finitary if there exists a field K and a faithful KG-module

V such that dimK[V,g] < 00 for all g E G where [V,g] := {v(g —1)|v E V}.

The classification of infinite finitary LFS-groups has been completed:

(a) those that are linear have been classified independently by several authors:
Theorem 1 ([1, 2, 10, 17]) Each LFS-group which is not finite but has a faithful
representation as a linear group in finite dimension over a field is isomorphic to a

group of Lie type over an infinite locally finite field.

(b) those that are non-linear have been classified by J. 1. Hall:
Theorem 2 ([8]) Each LFS—group which is not linear in finite dimension but has
a faithful representation as a finitary linear group over a field is isomorphic to one
the following holds:

(1) an alternating group Alt(Q) with Q infinite;

(2) afinitary classical group: FS’pK(V,s), FSUK(V, u), FQK(V, q);

(3) a special transvection group TK(W, V).
Here K is a (possibly finite) subfield of Fp, for some prime p; the forms
3, u, and q are non-degenerate on the infinite dimensional K —space V; and W

is a subspace of the dual V* whose annihilator in V is trivial: 0 = {u E V] uW = 0}.

Let G be a non-finitary LFS-group. Then

0 G is of alternating type if G has a Kegel cover all of whose factors are alter-

nating groups.

0 G is of p-type for some prime p if every Kegel cover for G contains at least
one factor which is isomorphic to a classical group defined over a field in char-

acteristic p.

In [14, Theorem A], U. Meierfrankenfeld proved an important result on the structure
of an arbitrary LFS-group showing that any LFS-group can be sorted into one of

the following classes:

Theorem 3 ([14]) Let G be a LFS-group. Then one of the following holds:

(a) G is finitary.

(b) G is of alternating type.

(c) There exists a prime p and a Kegel cover {(Hi, Ali) | i E I} for G such that
G is of p-type and for all i E I, Hi/Op(H,-) is the central product of perfectcentral
extension of classical groups defined over a field in characteristic p and Hi/Mi is

a projective special linear group.

It is possible to split the groups of alternating type into two categories. First, let
us give more notation and terminology. Let G be an infinite LFS-group, H a finite
subgroup of G and {I an H -set such that [Q] 2 7 and H /CH(Q) E“ Alt(Q). Let A
be the class of such pairs (H, (I). Note that G is of alternating type if and only if
for each finite subgroup F of G there exists (H, (2) E A such that F _<_ H and F

acts faithfully on 9. Let G be of alternating type, F S G be finite, and define

Areg(F) :2 {(H,Q) E A I F S H and F has a regular orbit on 9}.

Then
0 F is called regular if Areg(F) is a Kegel cover for G.

o G is of regular alternating type if G is locally regular, that is, every finite

subgroup of G is regular.

0 F is called non-regular if there exists a finite subgroup F * of G with F g F*

such that for all (H, D) E A with F“ S H, F has no regular orbit on Q.

o G is called non-regular alternating type if G is of alternating type and G has

a non—regular finite subgroup.

There is another characterization of regular and non-regular alternating type groups

given in [4, Theorem 1.4], for which the following definitions are needed.

0 G is of 1-type if every Kegel cover has a factor which is an alternating group.

0 We say that G is of 00 -type if the following property holds:
Let S be any class of finite simple groups such that every finite group can be
embedded into a member of 5. Then there exists a Kegel cover for G all of

whose factors are isomorphic to a member of 8.

Theorem 4 ([4]) Let G be a LFS-group of alternating type.
(a) G is of non-regular type if and only if G is of 1-type.

(b) G is of regular alternating type if and only if G is of oo-type.

In [6], it is proven that if G is a locally finite simple group of alternating type, p is a
prime, and Z s G is an elementary abelian subgroup of order p2, then there exists
1 7E z E Z with C005) 5f Cc(Z). One might ask whether a stronger result is true;
namely,

Question 2. Let G be a locally finite simple group of alternating type and p be
aprime. Does Cg(a) # 00(b) holdfor all a, b E G with [a] = [b] = p and (a) 7£ (b).

Note that this is a special case of Question 1. In Section 4.1, we observe that Question
2 is true when G is a regular alternating group (see Theorem 4.1.2), and we also
prove that if G is a non-regular alternating group, 00(23) is non-abelian for m E G
with [11:] = p, p a prime. One of the corollaries we derive from these results is the
following: If G is a LFS-group of alternating type, then 00(23) is non-abelian for
any a: E G with [r] = p (see Theorem 4.2.5). In addition to this, Section 4.2 answers

the following question.

Question 3. What can be said about the structure of a simple locally finite group

containing an element of prime order p whose centralizer is abelian.

This question is stated by Hartley in [9, page 39] and it is mentioned that “We have
not been able to say anything about the structure of such a group even with the
assumption that the centralizer is elementary abelian.” We shall show that the group
under consideration must be either linear or a group of p-type. For the proof of

this result, we shall be using the classifications mentioned above in Theorems 2 and 3.

Finally, in the last chapter, we prove that a non-linear LFS-group contains a p-
subgroup which is not Cernikov where p is a prime. This enables us to show that

such a group contains an infinite abelian p-subgroup.

Chapter 1

Preliminaries

This chapter provides the definitions, notation and lemmas that will be used in
Chapter 2. Some well-known material is included in order to make the presentation

self-contained. Throughout this chapter we assume the following:

K is a field of characteristic p, p is a prime or zero, V is a finite dimen-
sional K-space, G = GLK(V), G = G/Z(G) = PGLK(V), S = SLK(V),
S = PSLK(V) and r is a prime.

Also, we regard K as a subring of EndK(V), that is, we identify k with kidV for

kEK.

1.1 Structure of Centralizer in GLK(V)

Lemma 1.1.1 Assume that r 75 p. Let x" — 1 = H20 t,(:r) where t,(:c) E IK[:r:] is
irreducible and t0 = a: — 1. Let IE be a splitting field for 51:" — 1 over K. Let 5,- be a

root of t,- in IE. Then
(a) :13?" — 1 has r distinct roots in E.

(b) ti(:c) 7E tj(:c) for all 0 g i <j _<_ m.

(c) IE = K[§,-] for all 1 S i S m.
(d) Let d = dimK IE. Then deg t,- = d for all 1 S i S m.

Proof: (a) r33“1 3% 0 since charK # r. So gcd(:r" — 1,r:rr_1) = 1 and hence
:cr — 1 has no multiple roots in IE.

(b) This is immediate from part (a).

(c) Any root of t,(:z:), 1 S i S m, is a primitive r-th root of unity and 52' is algebraic
over K[:1;]. Therefore K[€,-] = K(§,~) and it contains all the roots of 237' — 1, that is,
it contains IE. The other inclusion, K(£,-) C_Z IE, is obvious. Thus K(§,-) 2 IE 2 K[§,~].
(d) For all 1 S i S m, we have d = dimKIE 2 [IE : K] = [K(§,j) : K] 2 degree of the

minimal polynomial of 5,; over K[.L‘] 2 deg t,-.

Definition 1.1.2 Let K be a subfield of a field IF and V be an IF —space. Then
I‘K GLIF(V) is the set of all IF -semilinear isomorphisms of V which are K-linear.
So if h E GLK(V), then h E PK GLIF(V) if and only if there exists a E AutKOF)
with h(fv) = o(f)(hv) for all v E V and f E IF.

Lemma 1.1.3 Let IF be a subfield 0f EndK(V) containing K. If h E I‘K GLIF(V)
acts o-semilinearly, then hfh‘1 = o(f) for all f E IF. Moreover, I‘K GLIF(V) =

NGLK(V)(IF)-
Proof: Let h E P K GLIF(V) be o-semilinear. Then h(fv) = o(f)(hv) for all

v E V and f E IF by definition. Thus hf = o(f)h and hence hfh‘1 = 0(f) for all

f E IF, proving the first statement.

For the proof of the second statement, let h E F K GLIF(V) with o E AutK(IF) as

the corresponding automorphism. Then h fh‘1 = o( f) E IF for all f E IF by the

first part. Hence h normalizes IF, giving us I‘ 1K GL1F(V) Q NGLK(V)(IF). For the
converse inclusion, take h E NGLK(V)(]F)- Then h f h"1 E IF for all f E IF. Defining
0(f) :22 hfh'l, we see that o E AutKUF) and h(fv) = o(f)(hv) for u E V, f E IF.
Hence h E PK GLIF(V).

Lemma 1.1.4 Let g E G and g = gZ(G) E G. Assume that IE := K[g] S EndK(V)
is a field. Then

GLIE(V) = 00(9) 9 Cd?) Q Pk GLIE(V)- (1-1)

Proof: Let us first prove GLE(V) = Cg(g):

If a: E Cc(g), then arg" = gna: for all n E Z+. Therefore, :cp(g) :2 p(g):r for
any p(:r) E K[sr] and hence me 2 ea: for e E IE, that is, a: E GLE(V) which gives
00(9) 9 GLE(V). For the converse inclusion, observe that any element in GLE(V)
commutes with 9 since IE = K[g]. Hence GLE(V) Q 00(9), giving the first equality
in (1.1).

Now we shall show the last inclusion as the other one is obvious. Let h E 00(g).
Then 9" = 779 E IE for some n E K. Hence hgnh—1 E IE for all n E Z+ and so
heh‘l 6 IE for all e 6 IE. That is, h e N0(1E) and hence h. 6 PK GLE(V) by Lemma
1.1.3.

Lemma 1.1.5 Assume that r 75 p and let 9 E G with gr = 1. Let f(:r) be the

minimal polynomial of g. Then the following holds:
(a) f(:1:) divides :r'r — 1 in K[ar].

(b) f (:13) = 1‘]le f,(:1:) where f, ’s are pairwise distinct and f2: = ti’ for some

OSi’Sm.

(c) For 1 S i S 3, let V, = Ann(f,-(g)) := {v E V I f,(g)v = 0}. Then V,- 31$ 0 for

all i and

3
V = G9 V,.
i=1
(d) Let g,- be the restriction of g to V,. Then f,(g,-) = 0.

(e) Let E,- = K[g,] be the K-subalgebra of EndK(V,~) generated by g,. Then
Er '5 IMEI/fr:(IL‘IIKIIICI-

(f) E,- = K if f,(:c) = a: — 1 and E,- E’ E if f,(:c) 7E :13 — 1 where E is a splitting
field for at" — 1 over K.

(g) V, is a vector space over E,- and

>51 0143,04): 00(9)

where we identify Xi=1GLIE,(Vi) with its image in GLK(V).

Proof: (a) This is obvious as 9 satisfies the polynomial JCT — 1 E K[1r].

(b) Follows from Lemma 1.1.1 since f (:r) | at? —— 1.

(c) Each V, is a g-invariant subspace and V is a direct sum of these subspaces are
from a theorem about decomposition of a vector space via a linear transformation
(see [11, Theorem 12, p.220]). Also note that, for each 1 S i S s, V; 7E 0 because
otherwise f (III) / f,-(:c) would be the minimal polynomial of g, contradiction.

((1) By definition of V}, we have f,(g)v, = 0 for all u,- E V,. Hence f,(g,-) = 0.

(e) Let m(:c) be a polynomial in K[ar] such that m(g,-) = 0. Then m(g)q,(g) = 0
where q,(:r) = f (.v) / f,(:z:) Since f is the minimal polynomial of g, f divides mg,
and hence f, divides m. This shows that f,(:r:) is the minimal polynomial of g,.

Consider now the map 29 : K[x] —> EndK(V,-) given by 29(nt(x)) = m(g,-). We

10

observe that the kernel of i9 is f,(.r)K[;r] and the image is K[g,], giving the required
isomorphism of part (e).

(f) If f,(:1:) = :1: — 1, then f,(g,-) = 0 implies that g, = 1 and hence IE,- = K[g,] = K.
If f,(:1;) sé 17— 1 then g, # 1. Part (e) and Lemma 1.1.1 easily imply that E, =
K[zr]/f,(:r)K[.r] 9: K[g]] = E where 6,] is a root of the irreducible polynomial f,(:z:).

(g) For any v E V, and e E E, S EndK(V,), let us define e - v z: e(v). It is straight
forward to check that this multiplication gives an E, -module structure on V,. Note
also that E,- is a field because it is isomorphic to either K or E by part (f), giving
us the first part.

For the second part, let 1 S i S 3. If h E Cg(g), then hg” = gnh for all
n E 2+ and hence hp(g) = p(g)h for any polynomial p(x) E K[ar]. In particular,
h f,(g) = f,(g)h. Applying v, E V, to both sides of this equation and using
part (d), we get 0 = f,(g)h(v,-). So h(v,) E V, by definition of V,. Define h,
as the restriction of h on V,. Then h, is a K-linear map on V,. In fact it is
IE, -linear since h,g, = g,h, and E, = K[g,]. Hence, h, E GLEi(V,-) implies that
C0(g) S Xf=1GLlE,-(Vi) with the correspondence h r—> (h1,h2, - -- ,hs). For the
converse inclusion, note that since E, is a field for each 1 S i S s, we can apply

Lemma 1.1.4 and conclude that GLEi(l/,) = GGLK([/i)(g,) Q CGLK(V)(9)- Thus
Xi=1 GLEz-(Vi) S CG(9)-

Remark 1.1.6 Let H be a group and assume that V = VIEBVQEB- - cEBVs where V, ’s
are simple KH -submodules of V for 1 S i S 3. Then {V,- | 1 S i S s} is the set of

all simple KH —subm0dules of V if and only if the V, ’s are pairwise non-isomorphic.

Proof: (:=>) Without loss of generality, assume that V1 g V2 as KH-modules.
Then there exists a K-linear map f : V1 ———> V2 such that f(hv1) = h f (v1) for

all h E H and U} E V1. Let us denote the image of v1 under f by 51. Then

11

W := {(v1,v'1,0,...,0) I v1 E V1} is a nonzero simple KH-submodule of V and

W 75 V, for any i, a contradiction. Hence V, ’s are pairwise non-isomorphic.

(4:) Let W be a nonzero simple KH-submodule of V. Then the projection 1r, :
W —~> V, is nontrivial for some 1 S i S 3. Since both W and V, are simple

submodules, W C‘—_’ V, by Schur’s Lemma. So W $5 V]- and thus it, = 0 for all j # i.

Therefore W = V, for some i, as desired.

Lemma 1.1.7 Let E, be a field containing K for 1 S i S s. Assume that V 2
V1 GB V2 69 - - - 69 V, where V, is an E, -space and dimIE, V, 7é 1 for all i. Let H =
Xi=1SLIE,-(Vr)- The“

(a) V, ’s are pairwise non-isomorphic.

(b) {V, | 1 S i S s} is the set of all simple KH-submodules of V.

Proof: (a) Since for any j E {1.2, . . .,s}

S
CHM) = X SLE,(v.->.
2%]

we observe that V]- and V, have different centralizer if i ,i j. Hence V, 9E V,.
(b) The assumption dimEi V, > 1 implies that V,- is a simple SLEI.(V,)-submodule

and in particular a simple KH-submodule. Remark 1.1.6 finishes the proof.

Notation: We will denote the set of nonzero elements of a set S by S”.

Lemma 1.1.8 Assume that r 7% p. Let g E G with lg] = r and f, s, f,,IE,,V, be

defined as in Lemma 1.1.5. Then the following holds:

(a) V, ’s are pairwise non—isomorphic.

12

(b) {V, | 1 S i S s} is the set of all simple KGg(g) -submodules of V.
(C) 132' = CEndK(V,-)(GLIE,(Vi)) = CEndK(r/,)(CG(9)) for 1 S 2' S s.

Proof: (a) Observe that

S
CCG(g)(Vr) = 5 GLEk(Vk)-
7.

Suppose to the contrary that CCG(g)(V,-) = CCg(g)(le for some distinct i and j .
Then GLIE,(V2‘) = GLEJ. (V3) 2 {1} and this implies dimlEk Vk = 1 and E, = K = IF2
for k = i,j. By Lemma 1.1.5(f), E}, = K = E where E is a splitting field of :13" — 1
over K and hence K contains a primitive r-th root of unity, a contradiction to
[K] = 2. Thus CCG(g)(V,) 7e CCG(g)(VJ-) for any i,j. Since isomorphic KC'G(g)-
submodules must have the same centralizer in Cg(g), we conclude that V,’s are
pairwise non-isomorphic.

(b) By Lemma 1.1.5 (g),
3
00(9) = >_<1 (ELEM.

Since GLEi(l/,) acts transitively on V}, so does Cg(g). Thus, V,’s are simple

KCG(g)—modules. Now, Remark 1.1.6 completes the proof of part (b).

(c) Note that the second equality is trivial and we only need to verify the first one:
(Q) Denote C :2 CEndK(V,-)(GLIE,(V2'I) and let e E E, S EndK(V,). Then for any
h E GLIE,(V23) we have h(ev,) = eh(v,) for v, E V,, that is, e E C.

(2) Let 0 75 h E G. Then h commutes with every element in GLlE,(Vi) and,
in particular, it commutes with every element in E, since E? (_i GLIE,(V2')- Thus

h E Z(GL1Ei(I/,)) which implies that h = Eidvz. for some 6 E E,. So h E E,.

13

Lemma 1.1.9 Assume that r 7E p and forj = 1,2, let 9,- E G with [ng = r.
Let f,, 3,, f,,, E,, and V,, be defined as in Lemma 1.1.5. Then the following are

equivalent:

(a) 00(91) = CC(92)-

(b) s := 81 = 32 and (possibly after permuting f12,f22,...,f32) V,1 = V,2 and

IE,"1=IE,'2 f07‘ all 137:3 8.

Proof: (b): (a): Trivial by Lemma 1.1.5(g).

(a): (b): Assume that C5191) = Gg(g2). By Lemma 1.1.8(b), {V,, | 1 S i S 8,}
is the set of all simple KCg(gj)-submodules of V for j = 1, 2. Then the assumption
00(91) = 03(92) implies that s := 31 = 32 and possibly after permuting the f,, ’s

V,1 = V,2 for all 1 S i S 3. By Lemma 1.1.8(c), we have

[En 1‘ CEndK(V,-1)(CG(91)) = CEndK(v:,,)(CG(92)) = Era-

Lemma 1.1.10 Assume that r 7E p and let 9 E G with [g] = r. Let the notation be

as in Lemma 1.1.5. Then exactly one of the following holds:

1' Cd?) = 00(9)-
2. (a) E = K.
(b) fa.) : xr —- 1.

(c) s = r and there exists 1 aé E E K’1 with {T = 1 such that for all 1 S i S r
f,(x) = x — {i‘l (possibly after reordering the f, ’s).

(d) dimKV, =dimKVj for all 1 S i <j S r.

(e) Cal?) = ma) where h E G with hr = 1, hi, 2 HI for all

1Si<ranthr=V1.

14

Proof: Assume that (1) does not hold. Then 05(g) ; 00(9) and hence there
exists y E 05(g) such that y E Cg(g). So 1 75 [g,y] E Z(G). Let 1 76 [g, y] =5 for
some 5E Kl]. As y‘lgy = 59 and |y‘1gy| = lg] = r, we have [5] = r. That is, 5 is
a primitive r-th root of unity in K’1 and hence E = K. This proves 2(a).

We shall now prove the parts 2(b)—(d) together: Since f (x) = 1]le f,(x) divides
xr — 1 and xr — 1 splits in K[x], we may let f,(x) = x — 5,, 1 S i S s, where 5, is

an r-th root of unity in K. Note that, for each 1 S i S 5,

Vi = Ann(fr(9)) = {’U E V | 9’0 = {iv} = Ker(g - Er)-

That is to say, V, is the eigenspace of 9 corresponding to the eigenvalue 5,. For any

A E Kit and v E V, we have
gyv = 5A1) 41) 59v 2 5/\v <:> gv = Av,

which means that

Ker(g — A) = Ker(gy — 5).). (1.2)

Using (1.2) with A = 5,, it follows that V,- = Ker (g — 5,) = Ker(gy — 55,). Let us

now consider Ker(gy — 5,) and observe that

Ker(gy — it) = {v E V l 9"?) = érv} = {v E V | 69m = {iv} = {v E V | 9v = (lit-v}

is the eigenspace of 9 corresponding to the eigenvalue 5 ‘15,. Thus, Ker(gy — 5,)

gives another eigenspace V,, where j 7é i. Also, note here that

dimK V, = dim}, Ker(g — 5,) = dimK hI-IIKerfgy — 51'»

= dimK Ker(gy — 5,) = dimK V-,

15

which yields 2(d). By (1.2), we have Ker(gy —— 595,) :2 Ker(g — 5j‘15,). Since
I5] = r, for each j E {1, 2, . . . , r}, Ker(g-y —— 535,) gives a different eigenspace of 9.
Therefore, 3 = r as claimed in part 2(c) and hence f (x) = xr — 1. Without loss of

generality, let f,(x) = x — 5"."1 for 1 S i S r.

2(e) First we shall show that an arbitrary h E G with the conditions hr = 1,

 

hV, = V,+1 and hV, 2 V1, for all 1 S i < r, must satisfy C§(g) = Cg(g)(h). Note

that, for any i, we have
gh’vt = (ll—lgthi) = h_19(h(vi)) = h—1(5ihu,) = 6% for all vi 6 Vi- (1-3)

On the other hand, 5 gv, = 55‘i’1v, for all u, E V,. Combining this with (1.3), we
conclude that gh 2 5g on each V, and hence on V. Now gh = 59 implies h E Ody)
and hence Cg(§) Q Cg(g)(h). Conversely, take an element d E C§(§). We need to
show that dh'k E Cg(g) for some I: E Z. Let [g,d] = A E Z(G) for some A E K”.

Since Ar = 1, we have A = 5k for some k. Then dh’k E Gg(g) easily follows.

Next we shall show the existence of such an h. For this, we let h, : V, ——> V,+1
be arbitrary K-linear maps for all 1 S i < r and define hr : V, ——+ V1 as
h, = (h,._1h.,.._2 - . -h1)_1. Now let it e G with 15],, := h,‘. Then obviously h" =1

and above observation implies that Cd?) = Gg(g)(h).

Let us mention some further observations that will be needed later.

Remark 1.1.11 Assume that r 7e p and let 9 E G with [g] = r be as in Lemma
1.110(2). If y E Cg(g) with yV1 2 V2, then yV, = V,+1 for all 1 S i < r and

16

Proof: y E Cg(g) implies y"1gy = Ag for some A E K. Then, for any v1 E V1,
y‘lgyvl = AQ’UI- AS 9’01 E V2 and gvl = v1, we get y—15yv1 = Avl and hence

5 =2 A. Now take v, E V, and observe that
rm 6 Vi+1 41* 9(yvt) = ézyvt er 9—199'Ui = E’vt 4: Egvt = 62222--

The last equality above does hold by the definition of V,. Hence yv, E V,.”. Also

we know that V, ’s have the same dimension.

Proposition 1.1.12 Assume that r 74 p and let g E G with [g] = r be as in Lemma
1110(2). Further assume that dimKV, = 1 for 1 S i S 7'. Then either 00(9) is

the unique abelian subgroup of index r in CG(g) or r = 2, [K] = 3, dimKV = 2.

Proof: Let A _—_ Cc(g) and B = Cg(g). Note that A is an abelian normal

subgroup of B and lB/Al = r. Suppose that there is a subgroup D of B such that
D is abelian, |B/D| = r and D 74 A. Then AD = B. Now

A n D 3 CAD) =. 0MB) = CAB) = CA(A<h>) = Cm
D abelian A abelian

and lA/A n D| = |AD/D| = |B/D| = r imply that lA/CA(h)| _<. r. Since
dimKV, =1, A 9-: XL, Kn. Let y = (k1,k2,...,k,~) E A where k, E KII for all i.
Since h permutes the k, ’s, y E CA(h) if and only if y is of the form y = (k, k, . . . , k)
for some k E K“. Therefore CA(h) E’ Kll and hence [A/CA(h)] = [Killr—1 S r.
Since K contains an r—th root of unity, [Ki] 2 r. Thus r7".1 S r = r1 which gives
r = 2. Now 7' S [Ki] S r implies that [K] = 3. Furthermore, dileV = 2 follows

from r==2 and dimKV,=1 for lSiSr.

Now we will state a similar result where GLK( V) is replaced by SLK( V).

17

Proposition 1.1.13 Assume that r 7é p. Let g E G with |g| = r be as in Lemma
1110(2). Assume that dimK V,- = 1 for l S i S r. Then either C5(g) is the unique

abelian subgroup of index r in C3(g) or we have one of the following cases:
(1) r =3, [K] = 4, and dimK V = 3.
(2) r = 2, [K] = 3, and dimK V = 2.
(3) r = 2, [K] = 5, and dimKV = 2.

Proof: Let h E Cg(g) \ Cg(g) and d := det(h). Consider the element x E G
which acts as d"1 on V1 and as identity on the remaining V, ’s. Trivially x E 06(9)
and hx E Cg(‘g’) has determinant 1, that is, hx E 05(g). It is also obvious that
hx E CS(g). Thus, we have C3(g) 7t 03(9).

Since 00(9) S CG(QICs(§) S Cal?) and ICG(§)/CG(9)| = 7", W8 have
00(9) = CGIQICS(§) 0r 00(9)Cs(§) = Coda). The first case implies

that 03(g) = 03(9) which is not possible. Therefore, by the latter case
7“ = ICG(ZI')/Co(9)| = lColgle(§)/Co(9)l = ICs(§)/Cs(g)l-

The rest of the proof is essentially the same as the proof of Proposition 1.1.12: Let
A 2 05(9) and B = C5(g) and suppose that there exists an abelian subgroup D of
index 7" in B and D 7‘- A. Then AD 2 B and [B/D] = r implies IA/A 0 D] = r.
Since B = 5' fl (Cg(g)(h)), we have B Q A(3 (1 (h)) and

I‘ll-ID < CA(D)

D alTelian A abelian

CA(AD) = CA(B) S CA(5 0 (h))-

Thus [A/CA(S H (h))] S r. We also have A =2 5' (1 X521 K” and |A| = [KIWI—1.
The elements in CA(S n (h)) are of the form (k,k,...,k) with k? = 1 where

k E Ki. Since K contains an r-th root of unity, [CA(S D (h))] = r and hence

18

[A/CA(S n (h))] = [Kim—1h“ S r. Therefore
IKW"1 S r2 and [Ki] 2 r.

From these inequalities we get r’“1 S r2. Thus r = 2 or r = 3. The first case

yields [K] = 3 or 5 ([K] 75 4 since r % p) while the second case gives [K] = 4.

Remark 1.1.14 Assume that r 75 p. Let g E G with [gl = r. Put 5 = gr E K“.

Then the following are equivalent.
(a) [gkl 75 r for any k E K”.
(b) 5 ¢ KT-
(c) xr - 5 is irreducible over K.

Proof: (a) 4:) (b):
ngl = r for some I: E K1 (Er (gk)r =1 for some k E K” 4:) 9’ = k‘r for some
k E K3 4:) 5: k"" for some k E K“ <=> 5E KT for some k E K”.

(b) 4:) (c): See [12, Lemma 16.3].

Lemma 1.1.15 Assume that r 75 p and let 9 E G with |g| = r. Suppose |gk| ,5 r
for any ls: E Ka and let 5 E K[1 be such that gr = 5. Put E := K[g] S EndK(V).
Then the following holds.

(a) f(x) :2 xr — 5 is irreducible.
(b) E E K[x]/f(x)K[-x] is a field with dimKE = r.
(e) V is a vector space over E and Cg(§) 2 PK GLE(V).

(o) |K| > 2.

19

Proof: (a) This is from Remark 1.1.14.

(b) Obvious since f (x) is the minimal polynomial of g and dining E 2 deg f (x) = r.
(c) Define e - v = e(v) for all e E E and v E V. This defines an E-module structure
on V. Note that by Lemma 1.1.4 we have Cg(g) 9 PK GLE(V). Thus, it remains
to show the converse inclusion. Let h E PK GLE(V) with a E AutK(E) being the
corresponding automorphism. We need to prove that h E Ody), which is equivalent
to hgh"1g—1 E K. By Lemma 1.1.3, heh"1 = 0(6) for all e E E. Letting e = g E E,

-—1

we get hgh—1 = 0(9) and hence hgh‘lg—1 = o(g)g . Therefore, we are done if

we show that cr(g)g”1 E K.

—1

Since gr =5 = o(g)r, we have (o(g)g"1)’" = 1. Hence o(g)g is a root of xr — 1.

’1 is strictly less than r because

The degree of the minimal polynomial of o(g)g
x7" — 1 is reducible. Then [E : K] = r and [K(o(g)g‘1) : K] divides [E : K] imply
that o(g)g_1 E K.

(d) If [K] = 2, then gr 2 0 or 1, contradiction.

Remark 1.1.16 Let 37 E S with |y| = r where r 7t p. Then C§(y) is not solvable

if n > 2r(r — 1) where n = dim}, V.

Proof: Assume first that [yk] # r for any k E K. Put yr 2 5 E K. Then, by
Lemma 1.1.15. dimKE = r where E = K[y] and moreover C—G—(y) = I‘K GL(V).
Hence C§(y) 2 SH PK GL(V) 2 SLE(V)Z(G)/Z(G). Choosing n > 2r implies
that dimE V > 2. Then SLE(V)Z(G)/Z(G) and hence C—S—(y) is non-solvable,

giving the desired result in this case.

Now suppose that |yk| = r for some I: E K. Without loss of generality, we may
assume [y] z: r and then use Lemma 1.1.10. In Case 1.1.10(1), we have C-S-(y) 2
IXi=1 SLIE2.(V,)]Z(G)/Z(G), where E, = E or K and [E : K] S r-l. Thus choosing

20

n > 2r(r — 1) implies that dimEz. V, > 2 for some i and hence C3-(y) is not solvable.
In Case 1.1.10(2), we observe that C—S—(y) 2 IXizl SLK(V,)]Z(G)/Z(G). Similarly,

if n > 2r then dimK V, > 2 for some i, which leads to desired result.

1.2 Field Extensions

Lemma 1.2.1 Assume that r at p and K contains a primitive r-th root of unity.
Suppose a is an element of an extension field of K such that or E K. If b E K(a)

with bf E K, then b = all: for some j E Z and some k E K.

Proof: It is trivial if a E K, so we assume a E K. Since K has a primitive r-th root
of unity, we have a" E K“. Put c := a’" E K and g(x) := x" —— c E K[x]. By Remark
1.1.14, g(x) is irreducible and it is also separable since gcd(g(x), g'(x)) = 1. Thus
K(a) is a splitting field of g(x) and K(a) / K is a Galois extension with [K(a) : K] = r.
Let 1 74 o E AutK(K(a)). Then 0(a) = 5a where 5 is a primitive r-th root of unity.
Hence 0(ai) = 5iai for all 0 S i < r, which means 5i is an eigenvalue of o with the
corresponding eigenvector ai for O S i < r. Obviously Kai Q Ann(o — 5'). Since
K (a) is a vector space of dimension r over K, each eigenspace of a has dimension 1
and thus Kai = Ann(o -— 57:). By assumption d :2 br E K. So both b and 0(b) are
the roots of the polynomial x’" —- d E K[x]. Hence 0(b) = b5j for some j. Therefore,

b is in the eigenspace of a corresponding to the eigenvalue 57 and thus b E Kaj .

Lemma 1.2.2 Assume that r # p and let a be an element of an extension field of
K such that or e K \ KT. If b e K(a) with or e K, then b = ajk for some j e z
and k E K.

Proof: Let 5 denote a primitive r-th root of unity in an extension field of
K(a). Then [K(5) : K] S r — 1. Since a’" E KT, we have [K(a) : K] = r. Hence
[K(a) : K] and [K(5) : K] are relatively prime and so K(a) fl K(5) = K. We

21

are now in a position to apply Lemma 1.2.1 to the field extension K(5)(a)/ K(5)
and conclude that b = kaj for some j E Z and k E K(5). On the other hand,
it = bo—J' e K(5) n K(a) = K.

The previous lemma will in fact be needed and used in Chapter 2 only in the

following set up and, for convenience, we would like to mention it here.

Lemma 1.2.3 Let g]- E G be such that l'g‘j] = r where r 79 p. Assume that E]- :=
K[gj] is a field for j = 1,2. Assume further that for at least one of 9, we have
[Itng 7b r for any I: E KI]. If E1: E2 then ('91) 2: (g2).

Proof: For j = 1,2 we have g; E K and, say, gg E K". Using Lemma 1.2.2, we

get 92 = gilt for some j E Z and k E K. Then g, E (m) and the lemma follows.

Lemma 1.2.4 Let E be an extension field of IF. Then

[EIi : IF t1] = number of 1-dimensional IF-subspaces of E.

> dimIF E unless IF = E.

Proof: Let IFe be a 1—dimensional IF -subspace of E, where e E E”. It is easily
seen that StabE(IFe) = IF. Let Q be the set of all 1-dimensional IF-subspaces of E.
Since ElI acts transitively on El], and hence on 9, we have [IEu : StabIEII (lFe)| = IEII :
IF [I] = [Q]. The last part is from the fact that each element in an IF-basis of E gives

a 1-dimensional IF-subspace. But there are 1-dimensional subspaces other than this

type.

Lemma 1.2.5 Let V be an E-space and K be a subfield of E. If dimE V at 1, then

CEndK(V)(SLIE(V)) = E-

22

Proof: Let us denote GEndK(V)(SL1E(V)) by D. Note that D = EIIdKSLIE(V)(V)
and V is a simple KSLIE(V)-module, so D is a division ring by Schur’s Lemma. If
e E IE and h E SLE(V), then e commutes with h since h is IE-Iinear. Hence e E D,
giving the inclusion E Q D. Now, let t E SLIE(V) be a transvection. Then [V, t] is
a 1-dimensional space over E. Let (1 E D and 0 7t v E [V, t]. Since D centralizes t,
[V, t] is invariant under D. Hence dv E [V, t], and dv = ev for some 6 E E. Then

d = e since D is a division ring. Therefore D Q E, completing the proof.

Lemma 1.2.6 Let E1, E2 be subfields of EndK(V) containing K with SLE2(V) S
PK GLE1(V). Then one of the following holds:

(1) dimIE2 V =1.
(2) 1E1§ E2.
(3) K 2 E2 '5 IF2, E12” IF4, and dimKV = 2.

Proof: Note that since V is finite dimensional, dimK E]- < 00 for j = 1,2. We

may assume that dimIE2 V > 1. Let $2 = SLE2(V).

Case (a) Assume that (dimly,2 V, [E2|) 75 (2,2), (2,3).

Then SQ is quasisimple. Since 32 S Ng(E1), C32(Et]) S1 82. Suppose for
a contradiction that [Sg,Er[] # 1. Then 52/052(EI[) # 1. Furthermore, .32
being quasisimple implies that C32(E[]) S Z(Sg). Since Sg/CS2(El]) is iso-
morphic to a subgroup of AutK(lE1), AutK(E1) has a section isomorphic to
Sg/Z(5'2) = PSLIE2(V). Also since AutK(E1) is finite, PSLIE2(V) is finite. Thus
IE2, and so K and E1, are finite. Hence AutK(E1) is cyclic, but Sg/Z(Sg) is not, a
contradiction. Therefore, [Sg,El[] = 1 and hence E] E E2 by Lemma 1.2.5. Thus

(2) holds in this case.

23

Case (b) Assume that (dimly,2 V, [E2]) = (2,2) or (2,3).
Since K E E2, K 2 E2. If E1 = E2, then (2) holds. So we may assume that
E2 = K 9; E1. Since 2 = dimIE2 V = dim],2 El'dImIEl V, it follows that dimIE2 E1 = 2

and dim],1 V = 1. Therefore,
K=E2=le and E1=IFp2 2V (1.4)

where p = 2 or 3. Since SLIE2(V) S I‘K GLE1(V), we have

(192 --1)(p2 — p)

ISLE,(V)I = p_,

 

s [PKGLIE1(VII s 2(p2 — 1).

This implies p = 2 and hence (3) holds by (1.4).

Proposition 1.2.7 Assume that r # p and let E1 and E2 be subfields of EndK(V)

containing K.
(a) If It] -_<_ N0(IE2) and leg 3 Ng(IE1), then [shag] = 1.

(b) Assume that GLE1(V) S PK GLE2(V) and GLE2(V) S I‘K GLE1(V). Then
E1: E2 or the following holds: K 2 F2, {E1,E2} = {ng,IF4}, and V = IF4.

Proof: (a) For i = 1,2, let us define IL, by IL, :2 CEZ.(E]I3_1). Obviously, IL, is a
subfield of E,. Since EndK(V) is finite dimensional over K, we have [E, : K] < 00
and hence [E, : IL,] < oo.

1e2e1 for

Now consider the map 61 : IE1] ——+ AutL2(E2) defined as 6(e1)(e2) = e,—
all e1 E El] and e2 E E2. The first isomorphism theorem implies that Eg/ ILI] is
isomorphic to a subgroup of AutlL2(E2) and thus [Ea/Lg] S [AutL2(E2)|. Let

a E E2 be in the fixed field of AutIL2(E2). Then, in particular, a is fixed by the

24

automorphisms 61(e1) for each 81 E El]. So a commutes with every element in E”,
hence a E L2. This shows that the fixed field of AutlL2 (IE2) is equal to L2 , that is,

Eg/ I142 is a Galois extension and ] AutL2(IE2)| = dimIL2 E2. Therefore,
|lE[/ll.[| g dim,2 1E2. (1.5)
We define the map 62 : IEn2 —-> AutL1(E1) in a similar manner and obtain
log/ 11%| 3 dim,1 E,. (1.6)
Combining (1.5) and (1.6), along with Lemma 1.2.4, gives
lEl/Lll s dint, IE2 s list/Lil s aim, at s ital/ill s dimlL21P32-

Thus [Eg/ Lg] = dimLz. E, and hence E, 2 IL, by Lemma 1.2.4 for i = 1, 2. In other

words, E1 and E2 do commute, proving part (a).

(b) Observe that the assumption GLEJ.(V) S PK GLE3_J.(V) implies that E]; S
N0(IE3_,) for j = 1, 2. Then by the previous part e1(e2v) = e2(e1v) for all e, E E,
and v E V. That is, El] S GLE2(V) and E; S GLE1(V). Without loss, assume

CllmlEl V S dimIE2 V. (1.7)

If dimE2 V = 1, then IE1] S GLE2(V) = Eg. Moreover, since dimIE1 V = 1 we have
E; S GLIE1 (V) = Et]. Hence E] 2 Eg. We are done in this case, so assume that
dimng2 V > 1. By Lemma 1.2.6, either E1 Q IE2 or K = IE2 E’ IF2, E1 g IF4, and

dimK V = 2. In the latter case we are done and in the former case E1 = IE2 by (1.7).

25

Definition 1.2.8 Let K be a subfield of IE with [E : K] < 00. Let 6 E E. Then the

norm on E over K is defined by
N,E§(o) := detlflre)

where re is the left multiplication by e; namely, r,3 : E —r E such that re(a) = ea

for all a E IE. Note that re is a K-linear map. Also, we define E by
~ ._ K _
E.— {eEEINlE(e)—1}.

Remark 1.2.9 (a) N11,}? : IEII ————> KIi is multiplicative.
(b) N150.) = k" for all r; e K where [IE : K] = n

(c) If V is a 1-dimensional vector space over E and g E GLE(V), then 9 is
multiplication by an element of E. 50 g 2 el = re for some 6 E E”. It easily

follows that det[§(g) = det%(re) = ngfie).

(d) If E/ K is a finite Galois extension with Galois group A, then

NEW) = H 0(6)

aEA

for all e E E. See [15, Corollary 8.13].

Lemma 1.2.10 Let V be an E-space and K a subfield of E with [E : K] < 00.
Let g E GLE(V). Then

(a) SLE(V) S SleU/I-

(b) detii‘tv) = N§(detfx(g))-

26

Proof: (a) Note that SLE(V) is generated by transvections. Let t = I + aE,j be
a transvection in SLE(V), where O 74 a E E and i % j. Clearly, detfiflt) = 1, that
is, t E SLK(V).

(b) Let g E GL1E(V) and d := detIF/(g). We can write g in the form 9 = kh where
k = diag(d, 1, 1, . . . , 1) and h is a product of transvections in SLlE(V). By part (a),
det[§(h) = 1. Note that we can view It as a linear transformation on a 1-dimensional

vector space, and hence Remark 1.2.9(c) gives detl5(k) = N,]§(d). Thus
dot“,§(g) = detléa) det][§(h) = det[§(k) = Ngw) = N§(detl[3(g)).
Lemma 1.2.11 Let K S IF S E be a chain offields with [E : K] < 00. Then
ngfie) = Ntulvga» for all e 6 till.

Proof: Let V = E. We can view V as a vector space over both K and IF. Let

c E E”. Using Lemma 1.2.10(b) for the field extensions IF / K and E/IF, we get

dot",§(e) = N§(det£(e)) and (1.8a)

IF I IF IE
detv(e) = NIE(det-V(e)) (1.8b)
respectively. Combining these equations and using det1f}(e) = e, we obtain

N1K =NKdtlE _—. dtKr = NIKtltIF
112(8) lei e V<e>)1.2.10(b) 9 Me) (1.8a) 1F( 9 vie»

(1gb) N§(Ng(detl[3(e))) = Nlii<(N1E(€II-

27

Lemma 1.2.12 Let E be a separable extension field of K of degree n > 1. Assume
that K is maximal in E and let N = NlflEi. Then e"/N(e) E K for some 6 E E\K.

In particular, there are elements in E \ K whose norm is 1.

Proof: First observe that e"/N(e) has norm 1 by Remark 1.2.9(a)-(b).
Suppose for a contradiction that for any 6 E E \ K we have 6” /N (e) E K. Then

e" E K. Let q be a prime dividing it. We shall show that n = q and charK 74$ q.

Since 6" = (eq)"/q E K, we have [K(eq) : K] S n/q < it. Moreover, [E : K] = n
implies that K(eq) 75 E. Therefore eq E K by maximality of K. Now since
e E E \ K, E = K(e) again by maximality of K. Hence it = [K(e) : K] S q and so
n = q. Now xq -— 69 E K[x] is the minimal polynomial of e and so it is irreducible.

Then E/ K is a separable extension implies that charK yé q.

Since xq—eq is irreducible, eq E Kq by Remark 1.1.14. By our assumption bq /N (b) E
K, and so bq E K for any b E E \ K. Now we observe that the hypothesis of Lemma
1.2.2 are satisfied. Therefore, for any b E E, b E (e)Ku and hence E”- = (e)KII. As
Ell/Kti = (eKlI) and 6‘1 E K, we have [Ell/Kl] : q. On the other hand, [Ell/Kl] >

dimK E = g by Lemma 1.2.4. This contradiction completes the proof.

Corollary 1.2.13 Let E be a finite separable extension field of K. Then
lE=K(oelE|N[I§(o)=1).

Proof: Suppose that E # K(e E E | N,E{(e) = 1). Let IF be a maximal field in E
containing K(e E E | NIHE<(e) = 1 ) and note that E/IF is separable. By Lemma 1.2.12,
there exists an x E E\IF with Ng(x) = 1. Then NII§(x) = NHE<(NIEIF(x)) = NI],5(1) = 1

by Lemma 1.2.11. Hence x E IF by definition of IF, contradiction.

28

Lemma 1.2.14 Assume that r # p and let E be an extension field of K of degree

r. Then there exists e E E such that er E K.

Proof: Assume that eT E K for all e E E. Let 6 E E \ K. By as-
sumption both (1 + e)r and er are in K, so is their difference. That is,
(1+e)"—e" = [237:0 (;)ej]—e" = k for some k E K. Put f(x) = rxr—1+- - -+1—-k.
Then f (x) is a polynomial of degree r — 1 in K[x] with the root 6 E E \ K which

implies that [K(e) : K] S r — 1, a contradiction since K(e) = E and [E : K] = r.

Lemma 1.2.15 Let E/K be a Galois extension of degree r where r E p. Assume
that K contains a primitive r-th root of unity. If (r, IKI) aé (2,3) then there exists

an element e E E such that Nllgfle) = 1 and er E K.

Proof: Assume to the contrary that whenever e E E with N (e) = 1 we have
er E K”. By Corollary 1.2.13, there are elements in E \ K whose norm is 1. Let B
be such an element. By assumption fir E K, so let [3" = d E K. Since [E : K] = r,
we have E = K(5). Let 1 ,5 o E AutKE. Note that 0(5) = 513 where 5 is a
primitive r-th root of unity. We observe that ao(a)‘1 has norm 1 for a E E. Thus
a’" = ko(a)’" for some k E K. Choosing a = 1+)3 gives (1 +fi)r = k(1+5fl)r. Note
that {1, fi, 52, . . . ”Br-1} is a basis for E/K. We expand both sides to get

(1+ d) + rfi + (2)52 + . - - + rflT—l = k(1+ d) + k5rfi + - - - + kr5r_lflr_1.

If (1 aé —1, then I; = 1 = 5, a contradiction Thus, d = —1. Comparing the coeffi-
cients of B and of fir-1, we get k5 = 1 and l~c5"‘1 = 1 correspondingly. Hence
5"-2 = 1, which implies r = 2. Replacing a by 1+ cf} where c E K11 gives
1+ do2 = 0. Thus c2 = 1 for all c E K’1 and hence [K] = 3.

29

Proposition 1.2.16 Assume that r 31$ p. For j = 1,2, let E, be a subfield of

EndK(V) containing K with dimK E]- = r.

(a) Suppose that E1 S NC;(E2) and E2 S N0(E1). Then [Eu,Eu2] = 1 or r = 2,
[K] = 3, [E,] = 9.

(b) If GLE1(V)HSLK(V) _<_ PK (311132 (V) and GLE2(V)OSLK(V) S PKGLE1(V),

then one of the following holds:

(t) IE1: IE2.

(ii) r=2, [KI =3, dimKV=2, and [E,] =9 forj=1,2.

Proof: (a) Certainly [AutK(Eg)| S dimK E2 = r. But in fact [AutK(E2)] = 1 or
r. To see this, let E0 = Fix(AutK(E2)), the fixed field of AutK(E2). Then Eg/Eo
is a Galois extension. If E0 = K, then [AutK(E2)] = dimK E2 = r. If E0 = E2,
then AutK(E2) = {id} by definition of E0. These arguments remain true if IE2 is
replaced by E1. Hence, we have [AutK(E1)| = 1 or r.

Obviously, E,- is a multiplicative group for j = 1,2. Now we consider the map
19 : E1 ———> AutK(E2) defined by t9(81)(€2) = el—legel for e1 E E1,e2 E E2. The
assumption E1 S N0(E2) implies that el—legel E E2. It is easy to check that 19 is
well-defined and Ker(d) = CEIUEZ)‘ Therefore, [E1 /CIE1(1E2)| divides [AutK(E2)|.
By symmetry, [EQ/CIE2(E1)] divides [AutK(E1)].

Case(1) Suppose that [E,/CE(E3_J')| = 1 for j = 1 or j = 2.
.7
Without loss, assume that j = 1. Then E1 = CE1(E2) and hence [E1,E2] = 1.

Since r 75 p, E1 / K is separable, thus there exists an element a E E1 \K such that

E1 = K(a) by Corollary 1.2.13. So [E1,E2] = 1, and we are done in this case.

30

Case(2) Suppose that
IIEj/CffijUE3—jll = 7‘ for j=1,2. (1.9)

Then Cfij(E3_j) g E, and hence CE]. (E3_j) S E,. Note that certainly K S

CEJ.(E3_,). Therefore, K = CIEJ-(IEB—j) é E, by dimK E]- = r. We now have
[1131.152] E 1E1“ 1E2 9 CEJ-(IE3—j) = K- (1-10)

Using (1.9), together with

it follows that [EjKIl/KIII = r. Thus there exist elements 8, E E, \K for j = 1, 2.
Then by (1.10) we have [61,632] E [E1,E2] Q K, which implies 85-18182 = elk for
some k E KI]. Note that since CE1(E2) = K and E2 = K(eg), we have [e1,e2] 75 1
and hence k 75 1.

Denote ngi, by N,. Since e2 E E2 S NC;(E1), we define 6 : E1 —+ E1 by 9(8) =
82—1682 where e E E1. Then 6 E AutK(E1). Note that |Aut1K(E1)] = r by (1.9)
and thus E1 / K is a Galois extension. Since 9 is a nontrivial automorphism of E1,
AutK(E1) = (6). Using 82—18162 = elk, we obtain 9"(e1) = knel for all n. Then

n = r gives hr = 1. Hence I: is an r-th root. of unity. By Remark 1.2.9(d),
1= N1(e1) = e1 .kel - - JCT-161 = e'l'ltrV—ll/2 = 61.196) (1.11)

If r # 2, then (S) is divisible by r. Since is is an r-th root of unity, we have

h(gl = 1. Hence, (1.11) simplifies to e7, = 1. Since K contains a primitive r-th

31

root of unity, namely [9, we have e1 E K, which contradicts to the choice of e1.
Therefore r = 2.

It remains to show [K] = 3 and [E,-l = 9. By (1.11), e? = k‘1 = -1 as k2 = 1
and k 7E 1. Then 61': ii where i2 = —1. If k E K with N1(k) = 1, then k2 =1
and hence k = :l:1. We thus conclude that E1 = {:l:1,:l:i}. For any a E E1, the
N1 -norm of aT/Nl(a) is equal to 1, that is to say, aT/N1(a) E E1. Therefore
a2 = f or a2 = fi for some f E K. Also E1 = K(i). Let k E KI] and consider the
element It +i E E1. If (k + i)2 = k2 — 1 + 2ki = f E K then k = 0, contradiction.
Hence we may assume (k + i)2 = fi. It implies that k2 — 1 = 0. Thus k = 21:1
and [K] = 3 since k E KII is arbitrary. Finally, IE3] = 9 follows from dimK E, = 2,

completing the proof of part (a).

(b) The assumption GLE1(V) flSLK(V) S I‘K GLE2(V) implies that E1 S Ng(E2).
Similarly, E2 S N0(E1). Hence [ELEQ] = 1 or we have r = 2, |K] = 3, and
[E,] = 9 by part (a).

Assume first that [Efi,Eg] = 1. Then E]; S GLE3_J.(V) for j = 1,2. Without loss

of generality, suppose
dim“;1 V S dimIE2 V. (1.12)

If dimlEZV = 1, then El] S GLE2(V) = E2. Moreover since dimly;1 V = 1 we
have E% S GLE1(V) = El]. Thus E] = Eg and part (i) holds. So assume that
dimly;2 V > 1. Since SLIE].(V) S PK GLIE3_j (V), by Lemma 1.2.6 either E1 Q IE2
or K = E2 E’ IF 2, E1 ’2-‘1 IF4 and dimK V = 2. The latter case is a contradiction to

the assumption dimK E,- = r. Thus E1 Q E2. Now it clearly follows from (1.12)
that IE1 = IE2.

32

Next assume that r = 2, [K] = 3, [E,] = 9. If E1 = IE2, then again part (i) holds.
So assume E1 yé E2, and without loss, say E1 Q IE2. Then dimly;2 V = 1 by Lemma

1.2.6. Hence dimK E2 = 2 implies that dimK V = 2 and so (ii) holds in this case.

1.3 Exceptional Cases of Theorems 2.1.1 and 2.2.3

In this section we investigate the existence of the exceptional cases that will

show up in the results of the next Chapter. As in the previous sections, we let

G = GLKW), a = G/Z(G), s = SLK(V) and is" = SZ(G)/Z(G) g PSLK(V).

Lemma 1.3.1 Let G = GL2(3) and g,- E G such that |ng = [9,] = 2 for j = 1,2.
If (g1) # (g2) and C5(g1) = (75(272), then the following holds:

(a) det g]- = —1 and C—G—(gj) = (g1, g2) is an elementary abelian group of order 4.

(b) CC(91) 75 CG(92)-
(c) g, satisfies Lemma 1.110(2), j = 1,2. In particular, (70(gj) 75 C0(gj).
Moreover, with respect to some suitable basis, g1 = ((1) _01) and 92 = (9 (1)).

Proof: We shall prove (a)-(c) together. Since 91 75 i1 and it has order 2, its
minimal polynomial is x2 — 1 and we may assume that g1 = ((1) _01) with respect

to a basis {v1,v2}. Then

CGigll=iiIiigli and Coai)={i1,igi.i(95).i(_‘l5)}.

The assumptions Cc('g’1) = Cdgg) and (g1) # (g2) with 92 has order 2 imply
that i((1) (1)) are the only options for gg. If necessary we may change the basis
to {——vl,vg} so that g2 = ((1) (1)). Note that the matrix of 91 remains unchanged.

Then Cg(g2) = { i I , :l: 92} and the lemma follows easily.

33

Lemma 1.3.2 Let G = GL2(3) and g, E G such that |gj, = |ng = 2 for j = 1,2
071d (S1) 75 (§2I- If C§(§1)= C§(§2), then

(a) 05(9'1) = 05(52)-

(b) 9, satisfies Lemma 1110(2), j = 1, 2. In particular, Cc(g,) 75 Gg(gj).

(C) 05(9j) = i1 and Cowl) it Co(92)~

Moreover, g1 = ((1) _01) and g2 = ((1) 6) for some suitable basis.

Proof: As ill the previous lemmas, we may assume that 91 = ((1) _1). Then

)} and 05(31) = {i1,:t(_01 5)}.

(Di—-

Coal) = {iI,:l:g1,i((1) (1,) ,1(_0,

Let h := (91,13). Then h E 03(g1) and hence [92,h] E Z(G). Let g2 = (g 3).
Using 9% = 1, if 92h = hgg, we get 92 = :I:1, a contradiction. Thus 92h = —hg2

and so 92 = :I:((1) (1)). As in Lemma 1.3.1 by changing the basis we may assume

g2 = ((1) 5). Then Cg(g2) = { i1, 21:92} and all parts of the lemma follow.

Lemma 1.3.3 Let G = GL2(5) and g, E G such that |ng = |ng = 2 forj = 1,2.
Assume that C§(g,) = C§(g2) and Cay-(g1) # 05(g2). Then the following holds:

(a) C§(gj) = (mm) is an elementary abelian group of order 4.
(b) Lemma 1.1.10/2) holds for gj, j = 1,2. In particular Cg(gj) aé Gg(gj).

(C) C3(91) 7‘5 03(92).

Furthermore, there is a basis so that g1 = ((1) _01) and 92 = ((1) (1)).

34

Proof: Let K = {0,:I:1,:l:i}. Since g1 7£ :l:1, we may assume 91 = ((1) _01), say

with respect to the basis {v1, v2}. Then

camera ,o_,,,(_g_.,)...o}.t.

Also C§(§1) is elementary abelian of order 4. Now ig1 E S and g, E C§(g1)
imply that glgg = ggglz, where z E Z(S). Let 92 = (‘0‘ 3). Having z = I would
imply 92 equals igl or 2121, either of which is not possible. Hence 2 = —I and

92 = i (9 (1)) or 92 = :I: (_01 (1)). The second case does not hold as [g2] = 2. Thus

92 = :I: ((1) (1)). If necessary changing the basis to {—vl,v2} gives 92 = ((1) (1))
_ O '
Then CS(gg)—{i1,:l:(i 6)} and

05(y2)={:t1,i(6_0,),i(?8)ii(_016)}

Now observe that (9 6) E CS(g2)\CS(gl), giving part (c). Also (% i) E 05(g2)\

C5(g,). Now the lemma follows.

Lemma 1.3.4 Let G = GL3(4) and g, E G with |ng = |ng = 3 forj = 1,2.
Assume that g]- has three difierent eigenvalues. If C§(g,) = C§(g2) and C§(g1) #
05(g2), then the following holds:

(a) C§(gj) = (mfg-2) is an elementary abelian group of order 9.
(b) Lemma 1110(2) holds for 9,, j = 1,2. In particular Cg(gj) sé Cg(gj).

(CI 05(91) # 08(92)-

Moreover, there exists a suitable basis and 5 E KI] with [5] = 3 so that

g1: 0’5 0 and g2=(001).
0052 100

35

Also the pair (91,92) is unique up to conjugation by an element of FGL3(4).

Proof: Let K = {0,1,5,52} where 53 = 1. Since g1 has three distinct eigenvalues,

100
00g2

with respect to some basis {v}, v2, v3}. Hence

*00 00 0*0
C _ = 0 O, 0 ,00 *EKfl’
C(91) {(061) (ll * ) (* 0 6)] }

|G5(g1)] = 27, and C§('g'1) is elementary abelian of order 9. Since gl E CS(§1)

we may assume

CDC-X-

we have 9291 = Q1922: for some 2 E Z(S). Let g2 = (a,,) and use [92] = 3. Then
if z = 1, we get g2 E Z (S) or g2 = legit1 for some k E K, a contradiction. Hence
2 75 1 and so 2 E K” with [z] = 3. Without loss of generality (by changing v2 and

0 0
v3 if necessary), we assume that z = 5. Then g2 = (0 (81 8) where abc = 1. In
c

A
#00

1 0
8 ). Note that

this case changing the basis to {v1,av2.abv3}, we get 92 = (1)

the matrix of g1 is unchanged with respect to this basis.

Now a trivial calculation gives parts (a) and (b). We also note that

92 E C5(92) \ 05(91) and (

OJHO
«m CO

1
8) E (1069”,) \ CG('£72)-

Lemma 1.3.5 Let G = GL2(3), g]- E G with lgj| = 2, |ng 75 2 for j = 1,2 and
(E1) ?9 (Eal- Assume that 05951) 7'5 05(32) and C§(§1) = C§(g2). The” |ng = 4
and C§(gj) = (g,,g2) is an elementary abelian group of order 4. Moreover, with

respect to some basis,

Proof: Note that ng-l = 4, since g?- = —1. Furthermore, vgl 74 :I:v, for any
0 75 v E V. Hence, with respect to the basis {v,vg1}, we have g1 = (_01 (1)). Also,
with respect to the same basis, the only elements of G of order 4 other than :l:gl

are the followings:

(i 31), (_11 i) (—11 :i) (:i_11)-

If we have the first case above, then we are done. If not, we can change the basis
to {v + v91, —v + v91}. {’0 - 1191.22 + v91}. and {—vg1.v}, reSPectively, to get

92 = (] _11 ) while the matrix representation of g1 remains unchanged. Elementary

calculations give ((1) ]) E Gg(§2) \ Cg('g'1) and C§(gj) = (g1,g2).

37

Chapter 2

Centralizers in PGLK(V) and
PSLK(V)

In this chapter, we prove two main results (Theorems 2.1.1 and 2.2.3) which
describe when two distinct elements of prime order in the finite dimensional projective
and special linear group shall have the same centralizer. We adopt the notation and

set up of Chapter 1. Furthermore, throughout this chapter we will assume that r E p.

2.1 The PGLK(V) Case

Theorem 2.1.1 For j = 1,2, let g,- E G with |ng = r and choose 9, so that
|kg,| 2 |ng for all k E K”. Then 05(g1) = 05(g2) if and only if one of the

following holds.

(a) (S1) = (to)-

(b) CG(91I = 00(92), |ng = 7‘, and CG(9jI = 05%) f07‘ .7' = 1,2-

(c) r = 2, [K] = 3, and dimKV = 2. Moreover, there exists a basis of V with

respect to which g1: ((1) _(i) and g2 = ((1) (1)).

38

Proof: ( <2) See the proof of Lemma 1.3.1.

(———>) Assume that 05(g1) = 05(9)) and hence

Co(§1) = Co(§2)- (2-1)

Due to lemmas 1.1.10 and 1.1.15, it is reasonable to split the proof into the following

five cases:
Case 1. [9,] = r and CG(gj) = Cg(g,) for j = 1,2.
Case 2- |ng = 7“ for j = 1,2, and 00(21'1) = 00(91), C(;(§2) 75 00(92)
Case 3. |ng = r and Cg(g,) 75 Cg(g,) for j = 1,2.
Case 4. |ng # r for j = 1,2.
Case 5- l91| = 7‘, I92] 7t 7‘.

Case 1. By the hypothesis of this case and (2.1), clearly Theorem 2.1.1(b) is

attained.

Case 2. Note that the assumptions of this case imply that 91 and 92 satisfy parts

(1) and (2) of Lemma 1.1.10. respectively. Then, using Lemma 1.1.5, we have

Cool =CG()91 =XGLIE,(V, o) (22)

00(9 2:) C(1(92)h(2)-[)_1:<Gle(Vt2)(]h2) (2-3)

We observe that V is a simple Gg(g2)-module. Then (2.1), together with (2.2),
gives 31 = 1, and hence V11 = V. Moreover, by Lemma 1.1.10 (2a), E = K.

Recall that E was the splitting field of xT — 1 over K and E,1 = K or E by

39

Lemma 1.1.5(f). Hence E11 = K. Therefore (2.2) turns into 00(91) = GLK(V),

that is, g1 E Z (G), a contradiction to |g1| = r. Hence, this case does not occur.

Case 3. Note that we are in the situation of Lemma 1.1.10(2) for both g1 and 92.

In particular, we have (2.3) and

Coat) = Colgtxhi) = Ii>=<1 GLa<ni)]<ht>. (24)

Furthermore, the V,j’s have the same dimension over K for all i and j, namely

dimK V,j = dim], V/T‘.

Step 1. If Cg(gl) = Cg(gg), then (g,) = (g2) (and hence Theo-
rem 2.1.1(a) holds):

We use the notation and results of Lemma 1.1.5 and Lemma 1.1.10(2) and write
V = V1, 89 V2, EB - . - 69 V,., where V3,]- : Ann(f,j(gj)) = {v E V I go =5;‘1v} for
all 1 S i S r and j = 1, 2 where 5]- E KI1 is a primitive r-th root of unity. Now, gj

acts as (1,5,,52,...,5}"—1) on (V1,,V2,,--- ,V,,). Put g: 51. By Lemma 1.1.9,

{th 1$iSTI=lVi2I 15257:}-

Hence 92 is of the form (5i1,5i2,...,5ir) on V where the exponents i], are in Z.

 

Note that g, = 54192. Replacing 92 by 5":1g2, we may assume that g2 acts as
an identity on V11 and hence g2 = (1,512,. . .,5”). On V21, we have 93v = 571in
for any integer n. Choose an n so that 57”? = 5. Note that (g2) = (3). Then

replacing g2 by g? gives g2 = (1,5,5l2, . . . ,5”). Thus,

V11 = V12 and V21 = V22. (2.5)

40

Take y E Cg(g,) with yV11 = V21. Then, by Remark 1.1.11, yV,1 = V,+1,1 for all
i. Now (2.1), together with (2.5), implies that y E Gc(§2) and yVlg = V22. Hence

yV,2 = V,+1,2 for all i. It follows that V,1 = V,-2 for all i and thus g1 = g2 and

(31) = (g2).
Step 2. If dimK V,, > 1, then Cc;(g1) = Cg(gg) and part(a) holds:

First assume [K] > 3. Since Cg(gj)/C(;(gj)= (h ) is abelian, we have CG(g,~)” S
00(9)) = i=1(GLK(V2jII = i=1 SLKWij) Conversely. we have SLIK(Vij) =
(SLK(th))” S 00%)" for 93311 i- Thus 00%)” = i=1Sle(Vt'j)- If IKI S 3.
let us consider the group OP'(C(;(g,-)). Since the quotient Cg(gj)/Cg(g,) is a p’ -
group. 0P’1Co1s,» = 0P’1Co1gjl). Besides

T
0 ’11Cog,) )=X0P’1 (GLa1V o) =>___<0P’1 1SL111Voll= -X13Lk(Vij)-
2:

For the last equality we used the fact that SLK(V,J-) is generated by transvections
and if t E SLK(V,,-) is a transvection then it has order p, which implies that t E
Op,(SLK(V,J-)). Thus, the assumption (2.1) yields

7' 7'
><1 SLK(Vi1) = ><1 SLKU/fll-
i: 2:

regardless of the order of K. Therefore, {V,l I 1 S i S r} = {V,g I 1 S i S r} by
Lemma 1.1.7. Since E,, = K, we deduce from Lemma 1.1.9 that 00(91) = Gg(g2).
By step (1), we conclude that whenever dimK V, > 1 part (a) of the theorem holds.

It remains to treat the case dimK IQ]- = 1.

41

Step 3. If dimK V,,- = 1 for all i, then (70(91) = (50(92) or r = 2,
[K] = 3, dimKV = 2:

This statement follows from Proposition 1.1.12 and by mentioning that both 00(91)
and 00(92) are contained in Cg(gj).

The case Cc(gl) = Cg(gg) implies that part (a) holds again by step (1) Therefore,
we may assume r = 2, [K] = 3 and dimKV =- 2. Lemma 1.3.1 shows that this
exceptional case does really occur and the properties stated in part (c) are satisfied.
Moreover, since (g1) # (g2) and Gglgl) # Gg(g2), we get a new case as claimed

in the theorem.

Case 4. Note that in this case we have Ikgjl 71$ r for all k E K”, j = 1,2.
Put E]- := K[ng S Eridi'). Then Lemma 1.1.15 implies that E,- is a field
and [K] > 2. By Lemma 1.1.4, Cg(gj) = GLEJIV) Q Cg(§j) Q PKGLEj(V)
which implies, together with the assumption (2.1), that GL1};1 (V) S I‘K GLE2 (V)
and GLE2(V) S PK GLE1(V). Therefore, E1 = E2 or K = IF2 by Proposition
1.2 7. The latter case does not hold. Thus E1 = IE2 and by Lemma 1 2.3, we get

(a1) = (a2), giving part (a).

Case 5. Recall that by the hypothesis of this case, g1 satisfies Lemma 1.1.10 and

hence one of the following holds:

31
CG(§1) = C(3(91) = X GL11.,,(I21) OF (2561)
2:1
,_ , . , I r q , ~l - 1
Cool) -- Cocoa]; 1X ‘JLK()1.1)](hII- 12.6b)

Liz-z]

By the assumption of this case gg satisfies Lemma 1.1.15 and in particular, E2 :=

K[gg] is a field with dimK IE2 = 7‘. Thus Lemma 1.1.4 implies

00(92) = GLIE2(V) Q CG(§2) Q PK GLE2(V)-

We observe that Cg(§2) acts transitively on V and so is primitive. Thus, CG(§1)
is primitive. Hence (2.6b) does not hold and 31 = 1, which means that the minimal
polynomial of g1 is irreducible. Then V11 = V and E11 2 K[gl]. By Lemma 1.1.5(f),
E11 2 K or IE. If IE“ 2 K, then (2.6a) gives 91 E Z(G), a contradiction to
|§1| = 7‘. Thus IE” 2 1E. Note that since IE is the splitting field of mr — 1 over 1K,
dimKlEll < 7‘. By Lemma 1.1.4, C0(gl) = GLIE11(V) g Cg("g’1) 2 PK GLE11(V).
We now can imitate the proof given in Case (4) and apply Proposition 1.2.7 where IE1
is replaced by E11 and conclude that E11 2 IE2. But, this is a contraction because

these fields have different dimensions over 1K. Thus, Case (5) does not hold.

2.2 The PSLKW) Case

In the PSLK(V) case, we have an analogue to Theorem 2.1.1. For its proof, we

will use the following lemmas.

Lemma 2.2.1 Let g E G with |g| = 7‘. Let the notation be as in Lemma 1.1.5.
Then the following holds:

(a) V,- is a simple KCS(g)—submodule of V for all i.

(b) Assume that g 75 —I. Then {V,- I 1 S i S s} is the set of all simple KCS(g)-

submodules of V if and only if the following does not hold:
7' = 2, IKI = 3, and dimK V = 2. (T)

43

(c) IE,- 2 EndKCS(g)(Vil for all 1 S i S s.

Proof: Note that we have

C 3(9) =(VSLKV)r1>:<1 GLE.(V >XSL (2.7)

(a) Fix an i E {1,2...,s}.

If dimIE, V, > 1, then V,- is a simple SLEi(V,-)-submodule and, in particular, it
is a. simple KCS(g)-submodule of V. Now assume dimIEiV, = 1. If K = lEi,
then V,- is a 1-dimensional K-space and hence it is simple. Therefore, we may
suppose that K 7é lE,. Note that for all e E El, we have detlléi(e) = NE, (e) by
definition. Moreover, by (2.7) and by the assumption dimlE, V,- = 1, we have
05(9) 2 {e 6 IE? | det§i(e) = 1} = {e E E? | NlnEi(e) = 1} = E,. By Lemma
1.1.5(f), E,- = IE (recall that IE is the splitting field of :I:" —— 1 ove K). Hence, IE,-
is a Galois extension over K and, in particular, it is separable. Observe that V,- is
simple if IE, 2 K(e 6 IE? I ngi(e) = 1). But, this is immediate by Corollary 1.2.13,

proving part (a).

(b) We observe, by Remark 1.1.6 and part (a), that the V, ’s are the set of all simple
submodules if and only if they are pairwise non-isomorphic. Thus, it is enough to
show that V,- ’s are non-isomorphic if and only if (1‘) does not hold.

(<=) Suppose that there exist j and k such that 1 S j 74 k S s and V- E’ Vk as

KC S(g)-submodules. We shall show that (T) holds.
Step 1. dimlEj V-— — dimlEk Vk— — 1. In particular, |IEJ- |— — W I- — lel= IlEkI:

Suppose false and without loss of generality let dimEj V]- 7-‘ 1. Then 1 ¢ SLEJ,(V]-)

acts nontrivially on Vj and trivially on Vk, contradiction.

44

Step 2. K = lEj = IE)c and |K| S 3:
Since VJ- 1%” Vk, there exists a KCS(g)-linear isomorphism a : Vj ——-> Vk. Define the
map 5 : EndK(g)(Vj) ———> EndK(g)(Vk) by 52(0)(vk) = a(o(of‘1(vk))) where o E
EndK< g)(Vj) and U}, E Vk. It is straight forward to check that (32 is an isomorphism of
K-algebras. Since dimlEj V]- : 1, we have EndIEj (Vj) = lEj, that is, Endm g)(Vj) =
lEj and, by symmetry, we also have EndK(g)(Vk) = lEk. Therefore 5? : E, ——-+ lEk.

Take an element 6 E IE3 and define E E G by

ev if v E Vj
ev = 5(e)‘1v if v E Vk

v ifvEVblaéch

Since by definition IEJ- = K[g]-l where gj = glvj, and similarly for k, we see that
e commutes with g. Also 5(e) = ozeoz"1 by definition. Taking the determinant
of both sides gives det(5(e)) = det(ozea—l) -—- det(e). Thus, det(é’) = 1 and

hence E 6 03(9). By KCS(g)-linearity of a, we have a(§vj) = Ea(vj) for all

'0} E Vj. Expanding this equality gives a(evj) = (1(Evj) = 'e'a(vj) = 5(e)_loz(vj) =

-1

5(e"1)o(vj) = a(e_1a“1(owj)) = a(e 22]), that is, a(evj) = a(e“1vj) for all

vj E V]- and e 6 Eg. Now by injectivity of a, evj = e_1vj. Hence e = e‘l, i.e.,

2 = 1. As e is an arbitrary element of E‘g, we conclude that IlEJ-l S 3. Hence

6
lEj = K as K g lEj. Then lEj = K = IE)C because IIEJI = IlEkI and K g IEk. Also

IKI S 3.

Step 3. [Kl = 3, r = 2, and dimKV = 2:
By Lemma 1.1.5(f), 1E,- # E for at most one i, lS‘i S 3. So lEj = K = IE1, = E and
since |K| S 3, 'r # p, and IE contains a primitive r—th root of unity, we conclude

that [KI = 3 and 7' = 2. Since 3 S r = 2, we also have dimKV = 2, completing

45

the proof of this direction.

(=>) Suppose that (T) holds. Since 9 5A —1 and g has order 2, its minimal
polynomial is .732 — 1, and V = V1 EB V2. Without loss of generality, we may
1 O

assume g = (0 1). Then V1 and V2 are isomorphic as KCS(g) -modules because

03(9) = 1U-

(c) The proof splits into two cases depending on the dimension of dimlE, Vi.

First assume that dimIE, V,- = 1. Then IE,- _C_ 05(9). Since IE,- = K or lEi/K is
Galois, we have E,- = K(lE,) g p(KC'3(g)) by Corollary 1.2.13 where d is the canon-
ical homomorphism d) : KC'S(g) -—> EndK(V). Therefore, IE,- Q EndlKCS(g)(Vi) Q
EndEz.(V,-) = EndEi(lE,-) = IE,- which gives IEZ- = EndKCS(g)(V,-), as claimed.

If dimEi V, # 1, then we have IE,- = CEndK(V,-)(SLIE,-(Vi)) by Lemma 1.2.5. Now
let «p E EndKCS(g)(Vi)- Then 90 commutes with every element in 05(9). Because
of (2.7), in particular, to commutes with every element in SLIEz-(Vz') which gives

CEndK(V,-)(SLIE,-(Vi)) Q EndKCS(g)(Vi)- Combining these two, we obtain

1E1 = CEndK(V,-)(SLIE,-(Vi)) 2 EndKCS(g)(Vl) 2 El,
which completes the proof of part (c).

Proposition 2.2.2 Let gj E G with Igjl = 7‘ and gj 31$ —I forj = 1,2. Then
C's-(91) = 05(92) if and only if one of the following holds:

(a) CC(91) = 00(92)-

(b) |ng = 2, detgj = —1, |K| = 3, dimKV = 2, and 05(91): 321.
Proof: (<=) Obvious.

(=>) Suppose that 03(91) -= Cs(.(12)-

Assume first that (b) does not hold. Then for j = 1, 2, {V,j | 1 S i S 3]} is the set

46

of all simple KCS(gj)-submodules of V by Lemma 2.2.1. Since C5(gl) = C5(gg),
we conclude that s := 31 = 32 and Vil = V12 for all 1 S i S 3. Furthermore,
IE“ = EndKCS(gl)(Vi1) = EIidKC-S(92)(V,-2) = Eiz by part (c) of previous lemma.
Hence Cg(gl) = 00(92) by Lemma 1.1.9.

It remains to treat the case r = 2, |K| = 3 and dimK V = 2. The following claim
implies that part (b) does occur.

Claim: Let G = GL2(3) and g E G be such that [g] = 2 and g 75 —1. Then
detg = —1 and CS(gj) = 21:]. In particular, if 91,92 6 G with |ng = 2 and

gj 79 i1. then Cs(91) = C5(92)-

Proof of claim: The minimal polynomial of g is 1:2 —— 1 and. without loss, we put

g1 = ((1) _1). Take 1? E C3(g) and let :I: = ((613). Then [.r,g] = 1 and :1: E 5'

imply that .r = i1, and so (75(g) = {i1}.

Theorem 2.2.3 Let g,- E G with |ng = r for j = 1,2, Choose 93‘ so that lkgjl _>_
I gjl for all k E K“. Then C§(§1) = C§(g2) if and only if one of the following holds:

(a) C§(§1) = 05(g2).

(b) |ng = r, r = 3, |K| = 4, dimK V = 3, and there exists a basis of V and some

6 E Kt1 with [5| 2 3 such that

91: 050 and 92=(001).
0 0 g2 1 0 0
(c) |gj| = r, r = 2, |K| 2: , dimKV = 2, and there exists a basis ofV such that
(a. .0.) (95>.

(d) |ng 2 4, r = 2, |K| = 3, dimKV = 2 and g1: (91(1)) and 92 = (T _11)

for some suitable basis of V.

47

Proof: (<=) See the proof of Lemmas 1.3.3, 1.3.4, and 1.3.5.

(=>) Assume that

05871) = Cs(§2)- (2-8)

As before, the proof splits into five cases and those are exactly the same cases as in

the proof of Theorem 2.1.1 (see page 39).

Case 1. If (gl) 2 (g2), then (a) holds. So assume (g1) yé (352). By the
hypothesis of this case, we have Cg('g'j) = Cg(gj) and hence CS(§]-) = 03(gj).
Then by (2.8), we obtain 03(91) 2 05(92). By Proposition 2.2.2, we have
either the case r = 2, |K| = 3, dimKV = 2 or 00(91) = 00(92). But,
the first case does not satisfy the assumption Cg(§j) = Cg(gj) by Lemma 1.3.2.

Thus Cg(gl) = Cg(g2). By the assumptions of Case (1), we get 0— C(gl)— — G—(gz)

Case 2. In this case, 91 and g2 satisfy part (1) and (2) of Lemma 1.1.10, respectively.

Then, using Lemma 1.1.5, we have

31
Cow 1= Co()91 =>__<1GL22,<V.-11 (2.91

ng< 21: SnCa<g21<h2>= Sn([>_< Gnu v.21<]h 21) (2.101

and, by Lemma 1.1.10 (2a) we have IE = K. Furthermore, IE“ 2 K 2 IE by Lemma
1.1.5(f).

48

Claim: V is a simple KCS(§2)-module.
Observe that g2 # —I. Assume first that r = 2, |K| = 3, and dimK V = 2. Without

loss of generality, we put 92 = ( (1) _1). Then

(1))» 2(5), (11)} =<$‘= I—Ol (ill

Note that :1: has order 4 and its eigenvalues are 4-th roots of unity. Suppose that

Ot—I

Csa21= {2(

W := Kw = (w) is a 1-dimensional KCS(g2)-invariant subspace of V. Then for an
arbitrary h E 03(g2), we have hw = kw for some k E Kl. Thus the only eigenvalues
of h are :I:1, a contradiction. Hence such a IV does not exist and V is a simple
module.

Now assume that we are not in the above case, let 0 # W S V be a KCS(§2)-
submodule and 0 75 U be a simple KGS(gg)-submodule of W By Lemma 2.2.1,
U = ng for some 1 S k S 7'. Let :1: :2 (t,1,1,...,1) E XLI GLK(V,-2) be
chosen such that det(t) = det(h2)—1. Then :I:/22 E CS('§2) and (:rhg) permutes the
subspaces V22 for all i. So V S Vk<2xh2> = U (Th?) E W, that is V = W, proving
the claim.

The above claim, along with (2.8), implies that V is a simple CS(§1)-module. This
implies 81 = 1 and V = V11. Consequently, (2.9) simplifies to 00(91) = GLK(V),

that is, 91 E Z (G), a contradiction to |§1| = r. Hence Case (2) does not hold.

Case 3. In this case, Lemma 1.1.10(2) holds for both 91 and 92. In particular we
have (2.10) and a similar formula for g1 holds. The proof of this case is essentially

the same as the proof of Case (3) of Theorem 2.1.1:

Suppose first that dimTK Vi]: > 1 for all i,j.
If TKT > 3, then CS(§]-)” = Xf=1SLK(Vij) and if |K| S 3, then Op,(CS(§J-)) =

49

Xlilzl SLK(V,-j). Hence in both cases we get

7‘ 7'
_X1 SLIM-21: X1 SLK(V221.
7.: 1,:

Therefore, by Lemma 1.1.7, {Vz’l | 1 S i S r} = {Viz II S i S 7'}. By Lemma 1.1.9,
(30(91) = 03(92). Hence, by Step(1) of Case (3) in the proof of Theorem 2.1.1 (see

page 40), we conclude that (gl) 2 (9‘2) and so Cg('g‘1) = 00(g2).

Suppose next that dimK Vij = 1 for all i,j.
Then Proposition 1.1.13 implies that one of the following holds:
(a) C5(91) = C3(92) or (b) (7‘2 IKladimK V) = (22322)2(22522) or (31423)-
Suppose first that Case (a) holds. Then by Proposition 2.2.2, we get either
00(91) 2 00(92) which as above implies 00(31) = 00(32), or we have
r = 2,|K| = 3,dimKV = 2 which again implies that 00(g1) = Cg(g2) by
Lemma 1.3.2.
Now suppose that 05(g1) # 05(92) and let us look at the cases listed in (b). By
Lemma 1.3.2 r = 2,|K| = 3 is not possible. By Lemma 1.3.3, r = 2, |K| = 5
gives (c). Finally note that 93- ’s have three different eigenvalues in the case r = 3,

|K| = 4, dimK V = 3 and Lemma 1.3.4 gives (d).
Case 4. For j = 1,2, put IEJ- = K[g]-T Q EndK(V). Since lkgjl 51$ r for all k E K,

IEJ- is a field, V is a vector space over IEj, and dimK IEJ- = r by Lemma 1.1.15. Note

that

05(5),) .—_ SflGLEjW) g 03%) g san GLij). (2.11)

50

dj

Write V :,$Ej where dj 2: dimlEj V. Use the notation Nj := NIHE: and let

i=1
ej E IEj. Then by Lemma 1.2.10(b) we have

dImIEj V dimlEj V imIE , V

detT§(ej) = [det.Téj(eJ-)] = [NJ-(detEflejDT : [NJ-(ejfld J

(2.12)
Thus whenever [VJ-(e3) = 1, ej E SLK(V) by (2.12). Also note that ej commutes
with gj by definition of IEj, hence IEj Q 05(gj) where IEJ- := {ej E IEJ- I Nj(ej) = 1}.
Moreover, we have GLE].(V) F1 SLK(V) Q PIK GLE3_J.(V) for j = 1,2. So by
Proposition 1.2.16 one of the following two situations holds:

(1) IE1 = IE2. This gives (g1) = (g2) by Lemma 1.2.3 and thus (a) holds.
(2) r = 2, |K| = 3, dimK V = 2. In this case part (d) holds, see Lemma 1.3.5.

Case 5. In this case, |91| = r and |g2| 75 r. Then IE2 = K[gg] is a field and

dimK IE2 = r by Lemma 1.1.15. Moreover, we have
SLEQWI E 05(92) = GLEQWI OSLKW) Q Cs(§2) Q Pk GLE2(V)-

As for gl, we have either

31
03a.) = SLK(V1 r1 >_<1 GL2,,(V.-21 or (213221
03(21): snav1r1([>=<1 minimum». (21%)

by Lemma 1.1.10.

51

Claim( 1) V is an irreducible KC5(gg) -module:
If dimIE2 V # 1, then V is a simple KSLE2(V)-module, and the claim follows
easily. If dimIE2 V = 1, then 03(92) 2 {e E IE2 I N(e) = 1} where N := NIHE(2' Let
IF := K[e E IE2 I N (e) = 1]. In order to prove the claim it suffices to show that
IF = IE2. By Lemma 1.2.14, there is e E IE2 such that er E K. Let a := er/N(e).
Since [IE2 : K] = r, we have IE2 = K(a). On the other hand, a E IF, thus K(a) S IF,

and II" = IE2, as required.

We now deduce, using both (2.8) and the above claim, that V is an irreducible

KCS(§1)-module. Hence 31 = 1 or (2.13b) must hold.

Claim(2) 31 74 1 and hence (2.13b) holds:
Suppose to the contrary that 31 = 1. Then V = V11 and IE11 = K[glI is a field.
Therefore SLIE11 (V) Q C5(§1) Q PK GL1};11 (V) by Lemma 1.1.4 and so the hypoth-
esis of Proposition 1.2.16 are satisfied. Hence one of the following holds:
(1) IE11 2 E2. Then (gl) 2 (g2) by Lemma 1.2.3. However, this implies ggk has
order 7‘ for some k E K“, a contradiction to the choice of 92.

(2) r = 2, |K| = 3. Then Ig1I = 2 implies 91 = —1 E K”, contradiction.

Claim(S’) We have dimflg2 V = 1 and dimK V“ = 1 for all 1 S i S r:
If dimlE2 V 79 1, then 1 75 SLE2(V) Q CS(§2). So C5(§g) is transitive (and primi-
tive) on V. But C 5(‘g‘1) is not primitive, a contradiction. Thus dim1E2 V = 1. Now

dimK IE2 = r implies that dimK V = r and hence dimK Vil = 1 for all i. Therefore,

,.
Csol1=SLKW1ni><1KIJ<h21 and Cs<§21=SLKW1r1EI2AutKIE2
z:

52

Claim(.l) K is finite:
Let us take y = (/\,/\‘1,1,...,1) E C5(§1) = 03(g2) where A is any nonzero
element in K. Since ICS(g2)/ lEgI divides IAutK(IE2)I and IAutK(IE2)I = 1 or
r, we see that |C5(g2)/IE2II7~ and hence yr E IE2. Note that if a E IE2 has an
eigenvalue in K, then an 2 ICU for some 0 # v E V = IE2 and k E K, giving us
a = 1:. Therefore. since yr has an eigenvalue in K, we get yr 2 (Ar,/\T,...,Ar).

That is, if r = 2 then X” = A” and if r > 2 then X” = 1. In any case, K is finite.

As our final step let q :2 |K|. The determinant map det : Ca(g1) —> Kti is onto
with Ker(det) :2 05(g1). Also ICC;(§1)I = rIKIIT = r(q — 1)". Hence |C'3(g1)| =
r(q — 1)r/q — 1. On the other hand, IIE2| = qr and |C6(§2)| = IIEQIr. Furthermore,
the norm map N : IEg ——+ K:1 is onto since both fields are finite and image of N is a
cyclic group of order q — 1. Thus ICS(‘g‘2)| = (qr —1)r/q — 1. But (q — 1)’ < qr — 1

implies that 03(g1) # 05(g2), a contradiction. Hence Case (5) does not occur.

53

Chapter 3

Centralizers in Alt(fl)

3.1 Centralizers in Alt(n)

Throughout this section assume the following: G = Alt(Q) where Q is a finite
set of size n and x and y are elements of G of prime order p such that (3:) 7E (y).
Our aim is to prove a theorem which lists all possibilities for x, y, and n so that x

and y have the same centralizer in G.
Let

x=x1x2--'$r and y=y1y2---y3

be the decompositions of x and y into the product of disjoint p-cycles where

1132' = (0212ai2w-202p) and yj:(bj11bj21-~abjp)

foralllSiSrandlSsz.

Theorem 3.1.1 Let x, y, and G be as above. Then the cases where Cg(x) = Gg(y)

are exactly the following:
(a) p is odd, n 2 2p or 2p+1, and x = x1232, y = .23,fo with k, l E Z,
1 S k 74 l < p.

54

k

(b) p23, n=6 andxlexg. yzxf oruiceuersawherelSk<p

(c) p = 3, n = 6 and x = x1, y = y1 with supp(x) fl supp(y) = (b.
(d) p = 2, n = 4 or 5 and supp(x) = supp(y) has size 4.

For the proof of this theorem, the following two lemmas will be needed.

Lemma 3.1.2 Cg(x) does not act transitively on supp(x) if and only if p is odd,

Isupp(x)| = 2p, and n 2 2p or 2p + 1.

Proof: Note that 00(27) acts on supp(x).
Case 1. Suppose r = 1.
Then x = x1 and p is necessarily odd since :1: E Alt(n). Clearly, Cg(x) contains

the subgroup (x1) and hence acts transitively on supp(x).

We may now assume that r _>_ 2. Let a and B be in supp(x). If they belong to the
same orbit of x, then there exists an element a E (x) which moves a to S. Hence
it is enough to consider only the case where a and S are in two different orbits of
x. Without loss of generality, assume that a = all and fl = 0.21.

Case 2. Suppose r _>_ 3.

Let a = (an,a21,a31)(a12,a22,a32)...(a1p,a2p,a3p). It is evident that o is an
even permutation, commutes with x, and 0(a11) = am. This proves that Ca(x)
acts transitively on supp(x) when r 2 3.

Case 3. Suppose r = 2.

Let u 2:: (a11,a21)(a12,a22) - - -(a1p,a2p). If n 75 2p, 2p+ 1, then define a by

u if p is even
0' :

an if p is odd

55

where 7r is a transposition whose support is contained in Q \ supp(x). Such a trans-
position exists since n > 2p + 1. Then a E C0(x) and 0 sends all to am, which
proves the transitivity of 00(1) on supp(x) in this case.

Finally, if n = 2p or 2p + 1, then we have CSym(n)(~T) = (x1, x2, 7') where
r 2: (a11,a21) - 2 -(a1p,a2p). If p is even (so n :- 4 or 5), then r E Alt(n) and thus
Cg(x) is transitive on supp(x). On the other hand, if p is odd then 00(23) 2 (.151, x2).

Hence Cg(x) does not act transitively on supp(x).

Lemma 3.1.3 Let 1 S i S r and 1 S j S s and let p be an odd prime.
Put :1: = x1 x2---.’L‘7~ and y = y1 yg-«-y3 be as above with Cg(x) = Cg(y).
Then either supp(xi) = supp(yj) or supp(xi) fl supp(yj) = 0. Furthermore, if

supp(x,) = supp(yj) then yj 2 xi” for some 0 < k < p.

Proof: Since p is odd, :I:,- is an even permutation and xi E Gg(x) = Cg(y). Thus
supp(xi) is y-invariant and so it is a union of orbits of y. Now the first part of the
lemma follows from the facts that |supp(x,-)I = p and orbits of y have length 1 or
p. For the second part, note that [yj, y] = 1 and hence yj E Cg(y) = Cg(x). Then
III/ja'l‘il = 1 as SUPP($i) = 5111310ij Since 21) E CSym(supp(xz-))(xi) = (932'). we get

yjzxf forsome 0<k<p.
Proof of Theorem 3.1.1. We split the proof into two cases:

Case 1. Cg(z) does not act transitively on supp(z) for z = a: or z = y.
Without loss of generality, assume that Cg(x) does not act transitively on supp(x).
Then by Lemma 3.1.2, Isupp(x)| 2 2p where p is odd and n 2 2p or 2p+ 1. Thus,
supp(x) flsupp(y) 7t 0. By Lemma 3.1.3, y = xfl x52 for some 0 S k,- < p, i = 1, 2.
k

First assume that both k1 and k2 are nonzero. If k 2: k1 = k2, then y = x a

contradiction to (x) 75 (y). Thus k1 % k2 which gives part (a) of the theorem. If

56

one of the k,- is zero, say k2, then y 2 xi” where 1 S In < p. Note that p Z 5 is
not possible since otherwise we can construct an element 1 75 o E Alt(n) such that
supp(o) Q supp(xg). Then 0 E Cg(y) \ Cc(x), a contradiction. Therefore p = 3
and hence it = 6 or 7. When n = 7, we can define a := (a21,a22)(a23,b) where
b E Q \ supp(x). Then 0 commutes with y but not with x, a contradiction. Hence

n = 6 and (b) is attained.

Case 2. 00(2) acts transitively on supp(z) where z E {x, y}.

Since supp(x) is an orbit of (70(1):), supp(y) is an orbit of Cg(y), and

00(1) = Cow). we have Stipptr) fl supp(y) = 0 or supp(x) = supp(y)-

Case 2a. supp(x) fl supp(y) = (b:
In this case, x E Alt(Q \ supp(y)). Moreover, Alt(Q \ supp(y)) Q Cg(y) = Gg(x).
So 1 7E 51: E Z(Alt(Q \ supp(y))) and hence IQ \ supp(y)] = 3. This implies

Isupp(x)| = 3. We use the same argument for y instead of x and get Isupp(y)] = 3,

giving us (c).

Case 2b. supp(x) = supp(y):

Suppose to the contrary that p is odd. We write y as y = £13,161 (tyne? for

some 0 < k,- < p by Lemma 3.1.3. If all the kj’s are equal then y = 113’“, a
contradiction. Thus, if necessary by replacing y with some power of y and re-

. , . k .
ordering :13,- s we choose the notation as y = $111322 ”.5135? With k2 7e 1. If r 2 3,

the permutation (011,021,031Xa12,(12226132) - - - (a1p1a2p1a3p) is in CC(13) \ 00(9).
a contradiction. Therefore r = 2. By Lemma 3.1.2, n > 2p + 1. Now the element
a := (a11,a21)(a12,a22) - - -(a1p,a2p)7r, where 7r is a transposition whose support is

in Q \ supp(x) satisfies Io, x] = 1 and [0, y] 75 1, a. final contradiction. So p = 2.

57

Next we will show that r 2 2. Assume for a contradiction that r 2 3. Let 1 S i S r
be arbitrary and pick 3' and k such that I{i,j,k}] = 3 where 1 S j, k S r.
Since xixj E Cg(x) = Cg(y), supp(xixj) is y-invariant. Similarly, supp(xixk)
is y-invariant. Then their intersection, supp(xixj) fl supp(xixk) = supp(xi), is
y—invariant as well. That is, supp(x,) is an orbit of y. As i is arbitrary, we get

x = y which is a contradiction. Thus r = 2.

Now p = 2 and r = 2 imply that supp(x) = supp(y) has size 4. Finally, the
assumption 00(2)) = Cg(y) forces n S 5. To see this, without loss of generality let
x = (a,b)(c,d) and y = (a, c)(b, d). If n 2 6, take e,f E Q \ supp(x) and consider
0 = (b, d)(e, f) Clearly, a commutes with y but not with x, a contradiction. This

gives (d) and completes the proof of the Theorem.

58

Chapter 4

On Abelian Centralizer in Locally

Finite Simple Groups

In this chapter, we show that the centralizer of an element of prime order in a

group of alternating type as well as in a nonlinear finitary group is non-abelian.

4.1 The Non-regular Alternating Case

Recall the definition of regular and non-regular alternating groups from the In-

troduction.

Lemma 4.1.1 Let G be a LFS-group of alternating type and 0;, x 0’1, ’—-_‘-’ Z S G a
regular subgroup of G. Then 00(2) 75 Cc(Z) for all 1 75 z E Z.

Proof: Since Z is regular, there is an element (H, II) E [C such that Z has at least
t regular orbits on Q for all Kegel covers IC and for all non-negative integers t by
[4, Theorem 1.2]. Choosing t _>_ 53113 implies 011(2) 919 CH(Z) for some 1 75 z E Z
by [6, Theorem 6.1]. In fact, the proof of [6, Theorem 6.1] gives a stronger result; .
namely, CH(z) 75 (111(2) for all 1 # z E Z. In that proof the assumption “for all

1 # z E Z” is used only in one place, the forth line before the end of the proof.

59

Instead, one could have said “for some 1 75 z E Z” because the lemma referred '
to only requires the existence of some such element. Thus Cg(z) aé 00(2) for all
1 75 z E Z.

Proposition 4.1.2 Let G be a regular alternating group. Then Cg(a) 75 Cg(b) for

all a,b E G with la] 2 lb] 2 p and (a) 75 (b) where p is a prime.

Proof: Assume it is false. Let a,b E G be of order p such that (a) 75 (b) and
Cg(a) = Cg(b). Put Z := (a,b). Since G is regular, Z ’=“ 0,, x Cp is regular

and Cg(Z) = Cg(a)flCg(b) = Cg(a) = Cg(b), a contradiction to the above lemma.

For the remaining of the section more definitions and terminology will be needed. Let
(H, 9) E A. Then, by [14, Lemma 2.8], there exists a unique minimal (sub)normal
supplement R to CHM) in H. That is, R is a normal subgroup of H and min-
imal with respect to H = RC'H(Q). For w E 9, we denote the minimal normal

supplement to CHIQ) in CH(w) by Rw.

Definition 4.1.3 Let A be an H-set and 2 be an orbit for H on A. Then
(a) Z is called Q-essential if GHQ?) S 011(9).
(b) E is called Q-natural if 23 and Q are isomorphic as H-sets.

(c) E is called Q-bloch-natural if for some H-invariant partition A of E, A is
Q—natural and NH(D) = CH(D)CH(Q) for all D E A.

(d) If all the Q-essential orbits on A are Q-block-natural, then A is said to be

9 block-diagonal.

Remark: (a) The condition N H(D) = CH(D)CH(Q) in the above definition is
equivalent to CH(D) S 011(9).

60

(b) E is an Q-essential orbit for H on A <=> R acts non-trivially on 2.

Proof: (a) Obvious since NH(D)/C'H(Q) 2’ Alt(IQI — 1) is simple.

(b) Assume that R acts non-trivially on 2. Then R S 011(2) and hence
CH(E)CH(Q) 7t H. Since H/C’H(Q) is simple, 011(2) S CH(Q). For the converse,
assume that Z is an Q-essential orbit. If R S 011(2) S GH(Q), then H = GH(Q),
a contradiction. So R S GHQ).

Let G be a group of alternating type. For A E .A, we define H A and (I A by
A = (H A,Q A)- Let D be a subset of A. Then D is called a Kegel cover for G if
{(H, GH(Q)) I (H, O) E D} is a Kegel cover for G. For any finite subgroup F of G,
we define D(F) :2 {(H,Q) E ’D I F S H and CF(Q)=1}.

Remark: Let A and E be finite groups with E perfect and acting transitively on a
finite set 9. Denote the base group of AZQE by A9 and put (A9)o := Afln(AlQE)’ .
Then (A 29 E)’ = [A9E,A9E] = (A9)'[A9,E] E. Furthermore,

(A912=<A“1’ IAQ1EI={(aw)weQ€AQ I Hate/4'}. (4.11
wEQ

The first equality in (4.1) is clear. For the second equality, we will first assume that
A is abelian and Show that [A9, E] = {(awlwen E AQ I HwEQ aw = 1} :

(Q) is obvious since E permutes the coordinates of the elements of the base group
and A is abelian. For the converse inclusion, without loss of generality, we put
52 = {1,2, . . .,n} and let a = (a,),-EQ E AQ such that l—IILI a,- = 1. We need show
that a E [A9,E]. Since E is tran51tive, there exists 8, E E such that lei = i for
iE e. For each 1% k e 51 define h(k) e AQ by h(k) = (ak,1,1,...,1). Then

Ih(k),ek] = (a;1,1,...,1,ak,1,...,1) where ak is in the kth position.

61

Hence

n n
IIIhIk1eII = I I1 a;1,a2,...,a..1
k=2 k=2

which is equal to a since {121:1 a, = 1. Thus, a E [A9, E].

For the general case, since AQ/(AQY = (A/A')Q is abelian, we get
n
<A'1QIA9, EI/M’)” = [(A/A’W. E1 = {<a.A’1.ee | Hoe/1’: A’}.
i=1

This implies (A’WIAQ. E1 ={(a.-1.-ee I Item a2- 6 A’}. as claimed-

We now quote a theorem proven in [4, Theorem 4.3].

Theorem 4.1.4 ([4]) Let (H, D) E .A and suppose that H is faithful and Q-block-
diagonal on some set. Let R be the minimal normal supplement to CHM) in H.

Let u) E Q and put K = GR(w)/Rw. Then R E’ (K In Alt(Q))'.

Let us denote the isomorphism defined in the proof of the above theorem by

(p : R i (K 29 Alt(Q))’ . We will show that it can be extended to H as follows:

Lemma 4.1.5 Let H, R,w,Rw,Q and K be as above. Put L := Lw = GH(w)/Rw
and D := {(dwlwefl E L9 I dwK = dw/K for allw,w' E 0}. Then there exists a

monomorphism 6 : H ——+ L 29 Alt(Q) such that 6] R = ()5. Moreover,
(K9)o s K9 r1 9(H) s K9 :1 (9(o,.,(a))xQ = D g 9mm“ 2: DAlt(Q)

with KQ/(KQ)O a: K/K’ and D/KQ a L/K.

Proof: Let us first show that R8 = Rwh for all h E H and w E 0.
By definition of Ru), GH(w) = RwCHIQ) and conjugating it by h gives
CHM) = agonh) = egon). Thus aw, g 1255. by definition of Rwh. In

62

. a similar way, we have R“, S Rf);1 and so Rf, = Rwh' Next we will Show that
Lw 9: Lw; for any w,w' E Q. Since H/GH(Q) g Alt(fl) and H = RGH(Q), H and
hence R acts transitively on Q. So choose r E R such that w" = w’ and define
19 : GH(w) -—> CH(w')/Rw/ by 19(h) = r-lherl. One can easily check that it is an

epimorphism with kernel Ru), giving the isomorphism Lu T—1 Lwr.

Definition of 6 and the proof that it is a monomorphism will be similar to that of the
one given in [4]. But they are included below as well because the explicit definition
of 0 and some further observations will be needed later on.

Without loss of generality, assume 9 = {1,2, . . . ,n} and w = 1. For i E SI, pick
2,- e R such that 1% = 2'. Since Whig-7.1 e CH(1) for all h e H and 2' e a, we

obtain a map
9 : H —2 L IQ Alt(Q) : h —2 ((r,h225,131),€9, 7T(h))

where 7r : H -—> Alt(fl) is the onto homomorphism arising from the action of H on

Q. 0 is a homomorphism since for any h,t E H we have

901W” = ((TihTi—thlliena”(h))(ITItrglRiliena”(ill
= ((TihrfilTMTfiRihen, ”(hlfltll
= ((TihtfiItheQ» WWI)

= 6(ht).

Now let h E H such that 6(h) = 1. Then 7r(h) = 1, that is, 7r(h) acts trivially on
Q. Hence 6(h) :2 ((r,~hr27’1R1),'EQ, 1) and so rihrfl‘ E R1 and h E R? = R,- for all
i. By assumption H acts faithfully and Q—block-diagonally on some set, say A, and

nieQR’i acts trivially on A by [4, 4.1(b)]. Thus h = 1 and 6 is one-to—one.

63

Note that (KQ)O = K9 {16(R) S KQ 06(H). To show that KQ/(Kfl)o T—i K/K’,
define If) : KQ —> K/KI by 1b : (kjljefl -—+ (Hjefl kj)K’ and observe that it is an
epimorphism and the kernel consists of elements (19]) jg) such that Hjefl kj E K’.

Then Ker(w) = (K9)o by (4.1).

Since rj,r;C E R and R S H, we have (rjhrjfhl)_1(rkhr;hl) E 03(1) for
any j,k E 9. This implies rjhrj—th = rkhrgth for j,k E Q and from
this it is immediate that 6(03(f2)) Q D. Obviously, K0 S D and hence
6(oH(o))Kn g D. In fact, we will show that e(c'H(r2))KQ = D. For this, let
(1 = (dz-R1369 E D. So d,- E 03(1) and di03(1) = dj03(1) for all i,j E 9.
Since 03(1) = R103(Q), there exist t E R1 and h E 03(9) such that d1 = th.
Recall that 0(h) = (rihri— 1R1),EQ by definition. Putting s = (siRl) = d0(h)"1,
we have s,- = d,(r,-h_1ri—1) = d,r,(dl_1t)rg'l = (dirid{1)(d,d1_1)(tri—l). Note that
di—ldl E 03(1) and since R _<_] H and d,- E H, we also have dirz-dg'l E R, thus
3,- E R. In fact, 3,- E 03(1) because 5, = d,(r,~h_1ri"l) and both d,- and rih‘lrfl
fixes 1. Therefore, 3 E KS2 and hence d = 36(h) E K90(03(Q)) which gives
D g 0(CH(Q))K9.

Next let us consider the map ib : D —> L/K defined by it: : (dj)jeg ——> dlK. It
can be easily checked that w is an onto homomorphism and if (dj) E Ker(w), then
d1K = K, that is, d1 E K. Thus dj E K for all j E {2, giving Ker(I/i) = K9 and
D/KQ ’2-‘1 L/K. Finally, 0(R) = (K 29 Alt(Q))’ 2: (Kn)o Alt(n) and H = 03(Q)R
imply that 6(H) = 6(CH(Q)I9(R)) = 0(CH(o))(K9)oAlt(o). Multiplying this by
K0 and using (KQ)O S K”, we obtain 6(H)KQ = DAlt(Q), completing the proof.

64

Put B := ¢_1((KQ)O) and note that g E 03(9) if and only if 7r(g) = 1 if
and only if (p(g) = ((rigri—glRl),Eg, 1) E (KQ)O. In other words, B = 03(0) and
hence B = R 0 03(9) S H. Without loss of generality, we let (I = {1, 2, . . . ,n} for
the remaining of the section. Moreover, whenever convenient, we shall identity the

group with its image under the isomorphism (15. In particular, for the next lemma

we identify R with (K In Alt(n))' and B with (K9)...
Lemma 4.1.6 Let R* :2 {(g,)7r E R I 91 =1 and 17r =1}. Then R1: R*.

Proof: It is straight forward to check that R* is a group.

Step 1. R1 S R*:

It is enough to show R* S 03(1) and 03(1) = R*03(Q), since then R1 S R* fol-
lows from the minimality of R1. For R*S03(1); let u E 03(1) and y = (yi)7r E 12*.
Since y1 = 1, 17r 2: 1, and (p(u) = (u,)o where u,- = riurfile with 1" = 1, we
get ((yilfll‘l’m = 0_1(Ui)"1(yi)7rluz')0 = [(Uil—l(yi)(Ui)"_1l00—17W- The first
coordinate of this element is (rlur1—1)_1 - 1 -r1ur1_1R1 = 1 K and 0‘171'0 fixes 1.
Thus, ((yi)7r)¢(“) E R* and hence R* S 03(1). For the second part, notice that
Alt(Q \ {1}) Q R* and hence 03(1) = 03(9) Alt(Q \ {1}) Q 03(Q)R* Q 03(1).

Step 2. (K’)Q* S R1 where 9* = Q\{1}:

Let k E K and r = (r,)7r E R1 be such that 27r = 3. Let a :2 (k,h"1,1,...,1) E B.
Then of = (k,1,k‘f2, 1,. . .,1) and a‘lar = (1,k,k"r2,1. ..,1). Since R1 S03(1)
and B 2 03(9) S 03(1), B normalizes R1 and hence a—lar = [a,r] E R1 r1 B.
Let s = (37;)0 E R1 such that 2‘7 = 4 and consider 0 :2 (l,l‘1, 1,. . . , 1) E B where
l E K. Similarly, c‘lcs = (1,l,1,l'32,1,...,1) E R1 08 and the commutator
Ia—lar,c‘1c3] gives (1, [k,l], 1,1,. . .,1) E R10 B. In fact, for 2 Sj S n, if we put
k‘1 and 1‘1 into the jth position in the definition of a and b respectively, and

choose r and s so that j" 76 1,j and js 75 1,j,j" we get the commutator [k,l] in

65

the jth position. This proves Step 2 as Ik, l] ’s generate K ' .

Step 3. R108: R*flB:

Obviously, R1 (1 B S R* I) B by Step 1. For the converse, let 9 E R* (I B. Then
9 = (9,),69 where gl 2 1 and 1],”:19, E K’. Observe that 9 can be written as
9 = (12922951212---i1)(121292932(9293I'1212m,1) "'(12---t12y2y_1)(12---112$)
where y = :I:-219.,- and x = ng 97;. Since x E K’, the last factor of g is in R1
by Step 2. In order to conclude g E R1, we shall show that the other factors are
in R1 as well. Without loss, take (1,/c, k’l, 1, . . . , 1) for some I: E K. Recall that
d :2 (1,k,k"r2, 1,. . ., 1) E R1 from the previous step. Moreover, since a—lar E B,
[ck—7‘2 E K’ and so does its inverse and hence e :2 (1,1,k—1kr2,1,...,1) E R1.

Thus de= (1,k,t-1,1,...,1) ER1.

Step 4. R1 = R*:
Since R1S03(1) and R* S 03(1), we have RlSR*. Also R*/R*flB E Alt(Q\{1})
is simple. Then R1 S B, together with Step (3), implies that R* H B é R1 S R*

and hence R1 = R*, completing the proof of the lemma.

Let (H, II), (HA,QA) E .A such that H S HA, 03(QA) = 1 and H is Q-block-
diagonal on HA. Since R aé 1, R S 03(QA) and hence there exists an orbit 23
for H on Q A such that R S 03(2). Thus, 23 is an Q-essential orbit. Let A* be
the union of all Q-essential orbits for H on 9,4. Then there exists an H -invariant
partition A of A” such that A 92’ Q as H-sets. Thus, set A 2 (A,- I i E {2} Define
31 = {(gilten E BI91=1}- Then:

Lemma 4.1.7 03(A1) S BI.

66

Proof: Put J1 := {91 I g = (gm-69 E 03(A1)}. In other words, J1 is the
projection of 03(A1) onto the first coordinate and hence J1 S K. We define
J 2: {g E B I g,- E J1 for all i E Q} and observe that it is a group. Since
B = {(9,359 E KQ I Hie!) g,- E K’}, we can choose any element of K as gl and
choose rest of the coordinates so that the product of these coordinates is in K’.
This shows that if we take the projection of the groups 03(A1) and B onto the
first coordinate and use 03(A1) S B, we get J1 S K. We now claim that J S R.
To see this, take 9 E J and r E R. Since R = BAlt(Q), we put r = (r,)7r where
(r,) E B. Then g7" 2: (iii—1933359 and each of these coordinates is in J1 since

92- E J1 and r,- E K. Hence J S R, as claimed.

Next we shall show that R1 S 03(A1). Let \P be an Q-essential orbit for H on
QA- Then 03(\II 0 A1) S 03(9) by Remark (a) on page 60. Put H1 := 03(1).
Then 031(\II 0 A1) SI H1 and so 03101! n A1)03(Q) S H1. Since H1/031(Q)
is simple, we obtain H1 = 031011 I”) A1)03(.Q). Minimality of R1 implies that
R1 S 031(\II HA1) and since ‘11 is arbitrary, we get R1 S 03(A1). Then R1 S R

gives R1 S 03(A1).

Now we claim that 03(A1) = {g E B | 91 E Jl}. Observe that (Q) is trivial. For
the converse, let 9 E B with 91 E J1. Then g1 = hl for some h E 03(A1) and
hence h‘lg E B 0 R* = B flR1 S B 003(A1). Now h"lg E 03(A1) implies
that g E 03(A1), proving the converse. As our final step, we shall show that
J = 1. Note that trivially J E {9 E B I gl E J1} = 03(A1). Let w E 52. Then
to = 19 for some 9 E R and since J S R, J = J9 Q 03(A1)9 = 03(Aw). Since
w is arbitrary, we get J S 03(A*). On the other hand, R and in particular B
acts trivially on 9A \ A*. Hence 03(A*) S 03((QA \ A*) U A*) = 03(QA) = 1.

Thus J = 1 which implies that J1 = 1 and thus 0 3(A1) S Bl, completing the proof.

67

Proposition 4.1.8 Let G be a non-regular alternating group and x E G be of prime
order p. Then there exists 9 E G such that Z := (x, x9) E’ 0p x 0,, and Z is a

regular subgroup of G.

Proof: By I4, Theorem 3.4], there exists an alternating Kegel cover D for G such
that, for any A, B E D with H A S H B, H A is Q A -block-diagonal on 93. Without
loss of generality, x E H A for all A E D. Since the set {degQA(x) | A E D}
is unbounded by [8, Corollary 3.13], we pick an element (H, D) E D such that

deg9(x) 2 2p2, that is, x has at least 2p non-trivial orbits on 0. Thus, we write

P P
71’ = IH(a,;1, a152, . . . ,a,p)] THU)“, big, . . . , biplla (4.2)
i=1 i=1

where n is the image of x in Alt(Q), {oz-j, bij} Q Q, and 0 denotes the action of x
on the remaining elements of (2.
Without loss of generality, we may assume that D = D(H) and hence H acts
faithfully on Q A for all A E D by definition of D(H). We continue the notation
used above and recall the definition of gb : R i (K 29 Alt(Q))’ . In particular, recall
that B = Cale) = e‘1((K“)o) 21H and (K912 = {(giIien e K“ ! Hana- e K’}.
We now consider the cases pI IK I and p I IK I separately.
Case (a) Assume that pI IK I
Let H = H/03(Q) and note that Alt(Q) 9—“ H = R03(Q)/03(Q) = R and (KQ)O
is a p’ -gr0up. Observe that there exists an element g E R so that
P P
E? = IH(a1,-, agi, . . . , ap,)] IH(b1,-, bgi, . . . , bpi)]o. (4.3)
i=1

i=1

68

Indeed, we can let

91=I H (aijaajillI H (bijabjill'

1Si<jSp 1Si<jSp

Note that g is a product of even number of transpositions and hence g E Alt(Q).
Clearly (313) S (ES), [fig] = 1, and IE] = ITS] = p. Thus (Eff?) ’5 0,, x 0p. Note
that {a,j I 1 S i,j S p} (or {by I 1 S i,j S p}) is a regular orbit for (5,316?) on 52.
Since (35,5?) is abelian, [x,x9] E 03(9). Moreover, since 9 E R and RSH, we have
xR = ng. Thus Ix,x9] E 03(0) 2 B and so x3 = x9B. On the other hand, ob-
serve that (x)B = (x9)B would imply that (x)03(§2) = (x)BCH(Q) = (x9)03(§2),
a contradiction. Therefore (x)B S (x9)B and hence (x,x9)B/ B E 0,, x 0p.
Let T be a Sylow p-subgroup of (x,x9)B containing x. Since B is a p’ -group,
TF1 B = 1. Then (x,x9)B 2 TB and 0;, x 0p ’5 TB/B '5 T. Let y be such
that Tflng = {y}. Since (y) and (x9) are Sylow p-subgroups of (x9)B, there
exists an element h = (x9)ib E (x9)B with (x9)h = (y). Hence (xgb) = (y), which
implies y-lxgb E (y). Moreover, xgbB = ng = yB since 9 E R and B S H. Thus,
y"1x9b E (y) (I B = 1 and y = xgb. By the definition of g, y = xgb ¢ (2:) and hence
T = (x,y) = (x,x9b) ":14 0p x 0p. Since b E B acts trivially on Q, T has a regular

orbit on I) as well. So let w E 9 such that 0T(w) = 1.

Let A E D be arbitrary and let )3 be an Q-essential orbit for H on DA. Let A be
an H -invariant partition of 2 such that Q E“ A as H -sets and let D be the element
in A corresponding to w. Then 0T({D}) = 0T(w) = 1, that is, Dt S D for any
1S t E T. Hence dt S d for any 1 S t E T and d E D. So 0T(d) = 1, in other

words, T has a regular orbit on CA. Therefore, we have shown that

Dreg(T) z: {(HA, 9A) E D I T S HA and T has a regular orbit on 9,4} = D.

69

Since Dreg(T) Q Areg(T), we conclude that Areg(T) is a Kegel cover for G and

hence T is a regular subgroup of G. This proves Case (a) with ‘Z ’=‘T’ and ‘ g ’=‘ gb’.

Case (b) Assume that p I IKI.

Since :I: E H S L 29 Alt(Q), write x = (x,),-Egrr where (x1369 E La. Recall that
by (4.2) we already defined the action of x on 9. However, for the rest of the proof
we shall change this notation and put 7r 2 (1, 2, . . . ,p)(p + 1,}? + 2,. . .,2p)o where
0 denotes the remaining orbits of x on 9. This is done to simplify the notation.
Besides the proof of Case (b) requires only that x has at least two non-trivial orbits.
Let k E K be of order p and define h E KQ by h := (k,1,...,1,k-1,1,...,1)

where k"1 is on the (p+ 1)St coordinate. Put

9 := fi hixi—l = h - 1121‘ . h3x2 - 2 - h(P—1)$p_2 -1 (4.4)
i=1

and observe that since h E B and B S H, we have 9 E B. For any 1 S n S p,
hm = (1,km1, 1,. . . , Lit—MPH, 1,. . ., 1) where Ic—nmp‘l’l appearing in the (p+2)nd
coordinate. In a similar way, hmj = (1, . . . , 1, *, 1, . . . , 1, *, 1, . . . , 1) where the non-
trivial elements * are on the (j+1)St and (p+j+1)St coordinate for any 1 S j < p.
This shows that [hmi, hmxj] = 1 for all n,m and for all 1 S i, j S p.
Let us now consider the product [1211 h(i'l'llmi = hzx - h3g32 - - - (hp)""p_1 - (hp+1)$zf
Since hp = x3D = 1, the last two factors of this product are equal to 1 and h,

respectively. Since the factors do commute, we get 9 = H121 h(l‘l'llxz.

Then g 121—11121 h (”In and

p . . p .
= H h-—(2+1):irz .311 = H h—l‘z, (4.5)
i=1 2'21

70

Since the first coordinate of g’lgx is k‘1 S 1, 9—19”: E B1 by definition of Bl.
In particular, g‘lg‘” S 1 which means x S x9. Next assume to the contrary
that (x) = (x9). Then x”1x9 E (x). Since 9 E B and B S H, x8 = ng
and hence x—lxg E B and x—lxg E B (1 (x). Note that B D (x) is a group of
order 1 or p and since x does not act trivially on 9, x E B. Thus B r1 (x) = 1
which implies x‘lxg = 1, a contradiction. Hence (x) S (x9). Notice that
(g'lgg’):lc = g_1g$ by (4.5) and so x9 and x do commute and (x,x9) E’ 0,, x 0p.

Put Z := (x, 2) 2 0,, x 0,, where z := [g,x]. Note that Z = (x, x9).

Let A E D. We shall show that Z has a regular obit on 9,4. Since :5 = g"1g“c E B1,
2 E 03(A1) by Lemma 4.1.7. Therefore, there exists A E A1 such that A2 S A. We
claim that AZ is a regular orbit for Z. Since 2 E B = 03(9) acts trivially on A,
we have 2 E NZ(A1). On the other hand, x E NZ(A1). Thus, (2) S NZ(A1) S Z
and since I3] = IZI = p2 we get NZ(A1) = (2). For any y E 020‘), we have
A9 = /\ E A1. Since A is an H -invariant partition, A? = A1, that is, y E NZ(A1).
Thus, 020.) S NZ(A1) = (2). This implies 020‘) = 0<z>(/\) = 1 and hence Z
has a regular orbit on A1 and so on DA. By an argument similar to the one in the

previous case, we deduce that Z is a regular subgroup of G, completing the proof.

Theorem 4.1.9 Let G be a non-regular alternating group and x E G with Ix] = p

where p is a prime. Then 00(x) is non-abelian.

Proof: Let x E G be of order p. There exists an element 9 E G such that
Z := (x,x9) '5 0,, x 0;, and Z is a regular subgroup of G by Proposition 4.1.8.
Then by Lemma 4.1.1 00(2) S 00(Z) for all 1 S 2 E Z and, in particular, 00(x) S
00(x9). Suppose to the contrary that 06(x) is abelian. Then x9 E 00(x) implies
that 0g(x) S 00(x9) and similarly 0g(x9) S 00(x). Thus 0g(x) = 00(x5'), a

contradiction. Therefore, 00(x) is non-abelian.

71

4.2 The Finitary Case

Lemma 4.2.1 Let P be a finite p—group, p a prime, and x E P with Ix] = p. If
X = (x) S Z(P), then there is an element 9 E P with (x) S (x9) and Ix, x9] = 1.

Proof: X S Z (P) implies that 0p(X) is a proper subgroup of P. By the normal-
izer condition, 03(X) S N p(Cp(X )) Since the quotient N p(X ) / 03(X ) is isomor-
phic to a subgroup of Aut(X) whose order is p — 1, we also have 03(X) = N p(X )
Let g be an element in NP(CP(X))\NP(X). Then 9 E Np(X) implies (x9) S (x)

and g E Np(Cp(X)) gives us x9 E 03(X) and hence [x,x9] = 1.

Lemma 4.2.2 Let G be a locally finite group and x E G such that Ix] = p, p a
prime. If Ix,x9] S 1 for any g E G with (x) S (x9), then every finite p-subgroup is

conjugate to a subgroup of 0g(x).

Proof: Let P be a finite p-subgroup of G. Since G is locally finite, (P, x) is a
finite subgroup. Choose a Sylow p-subgroup S of (P, x) with x E S. Then P9 S S
for some 9 E (P, x) If x E Z (S), we get a contradiction by Lemma 4.2.1. Thus
x E Z(S) and hence P9 S S S 00(x).

Theorem 4.2.3 Let G be a non-linear LFS—group and n E 2+. Then there exist
A S B S G with B finite and B/A 9:“ Sym(n).

Proof: It is a well known result. See, for instance, [9, Theorem 2.6].

Corollary 4.2.4 Let G be a LES-group and p a prime. If every finite p-subgroup

of G is abelian, then G is linear.

Proof: Suppose that G is non-linear. Then, by Theorem 4.2.3, for any n E Z+
there exist A S B S G such that B/A 21-“ Sym(n). We can choose n large enough

so that Sym(n) has non—abelian Sylow p-subgroups. So let PA /A be a non-abelian

72

Sylow p-subgroup of B/A where P E Sylp(B). As PA/A "2" P/P I) A, P is non-

abelian. Hence G contains a non—abelian p-subgroup, a contradiction.

Theorem 4.2.5 Let G be a LFS-group of alternating type. Then 00(x) is not

abelian for any x E G with Ix] = p where p is a prime.

Proof: Assume that 00(x) is abelian for some x E G with Ix] = p. Suppose
for a contradiction that Ix,x9] S 1 for all g E G with (x) S (x9). Then, by
Lemma 4.2.2, every finite p-subgroup is conjugate to a subgroup of 00(x). So every
finite p-subgroup is abelian. Then G is linear by Corollary 4.2.4, a contradiction.
Hence there exists t E G such that (x) S (xt) and Ix, xt] = 1. Since 00(x) is
abelian and xt E 00(1), we have 00(x) S 00(xt). By a similar argument we
obtain 06(xt) S 00(x) and thus 03(xt) = 00(x). By Proposition 4.1.2, G is

non-regular. Now by Theorem 4.1.9, we get a final contradiction.

Lemma 4.2.6 Let K be a field, V a vector space over K, X a finite dimensional
subspace of V and s a nondegenerate bilinear form on V. Then there exists a finite

dimensional subspace U of V containing X such that SIUxU is nondegenerate.

Proof: Write V = (X + X J.) EB Y for some K-space Y. Note that dimY =
dim V/X+XJ- S dim V/Xi = dimX < 00. Let U := X+Y. We need to show that
UnUi=0.ameUnULgvnxicxx+yhux+xih=X+onuX+xin=
X, we have UflUi Q X. Moreover, XflUi = XflXiflY‘L Q (X+X‘L)‘LflYJ- =

vi=aihnUnUigxnUi=a

Lemma 4.2.7 Let s : V x W ——2 K be a nondegenerate bilinear map and X1
and Y1 be finite dimensional subspaces of V and W, respectively. Then there are
finite dimensional subspaces X and Y satisfying the following: X1 S X S V,
nngucV=XemedW=Xiex

73

Proof: See [8, Lemma 3.5].

Lemma 4.2.8 Let G be a group, K a field, and A and B be KG-modules. Let
s : A x B ——> K be a G-invariant bilinear map, that is, s(ag,b9) = s(a, b) for all

aEA, bEB andgEG. IfAi =0, then IA,g]J-=03(g) for allgEG.

Proof:

b E IA,g]‘L <=> s(Ia,g],b) = O for all a E A.
4: s(ag—a, b) = O for all a E A.
(it s(a9,b) — s(a,b) = O for all a E A.
4:) s(a, b9_1) — s(a, b) = 0 since 3 is G-invariant.
e» s(a,bg—l — b) = 0 for all a e A.
©b9_1—bEAi@b9—l =b.

<$ b E CB(9)-

Corollary 4.2.9 Let G be a LES-group and x E G such that Ix] = r where r
is a prime. If 03(x) is abelian, then G is either a group of r-type or linear. In
particular, if there are two elements in G with distinct prime orders and with abelian

centralizers. then G is linear.

Proof: Let x E G with Ix] = r and 0c(x) abelian. By Theorem 4.2.5, G can
not be of alternating type. Hence, using Theorem 3 (page 4). we will assume that
G is either a group of p—type for some prime p S r or a non-linear finitary group

and obtain a contradiction in both cases.

Case 1. Assume that G is a group of p-type where p S r.
Then there exists a Kegel cover IC for G such that. if (H, N) E [C then H / Op(H ) is

74

the central product of perfect central extensions of classical groups defined over a field
in characteristic p and H /N is a projective special linear group, again by Theorem
3. So H /N 9.’ PSLn(pk) for some n and k and, without loss of generality, we
may assume n >> r. Note that H /N is simple, so Op(H/N) = 1 and Op(H) S N
for any (H, N) E IC. Now choose (H, N) E K such that x E H \ N and write
H :2 H/OP(H) = G1 - G2 - - -Gl where [Gi,G]-] = 1 for all i S j, Gi’s are perfect,
and G,/Z(G,°) is isomorphic to a classical group. Put T = 171172°~5171 where x,- E 0,.
Since N is a maximal normal subgroup of H, there exists 1 S i S I, say i = 1,
such that G1 S N. Then H = GIN. Also IGj,G1] = 1 implies that GjN/N S
Z(H/N). Then since H/N is simple, Gj Q N for j = 2,3...,l. Thus x1 E N
and, in particular, x1 S 1. Note that Ix] = r implies x’l' E G1 (1 H552 Gj = Z (H),

that is, x1Z(Gl) has order r in Gl/Z(G1). Moreover,
PSLn(pk) 11—: H/N a H/N = GW/N = (111/01 o ‘N a GI /Z(Gl ).

Denote D/Z(G1) :2 CGl/Z(G1)(17ll- Then D/Z(G1) and hence D’Z(Gl)/Z(G1)
is not solvable by Remark 1.1.16. So D' is non-abelian. Since [D,x1] S Z (G1),
by Three Subgroup Lemma we have D' S 001(x1). Therefore 061(x) = 001(x1)
is non-abelian. Moreover 077(x) = C3—C5 by [16, Theorem 8.13, p.238]. Since
001(17) S 03(x), we conclude that 03(1) and hence 00(1‘) is not abelian. This

contradiction completes the proof of this case.

Case 2. Assume that G is a non-linear finitary group.

Then by Theorem 2 (p. 3), G is isomorphic to one of the following:
(a) an alternating group Alt(Q) where Q is infinite,
(b) a finitary classical group, or

(c) a finitary special transvection group.

75

(a) This is not possible: Clearly, Alt(Q \ supp(x)) Q CA1t(Q)(5’3) implies that

0A1t(m(x) is not abelian, a contradiction to the assumption.

(b) Assume G is a finitary classical group and let G S FGLK(V) and s be the cor-
responding bilinear form on V. Since G is finitary, X 2: [V,x] is finite dimensional.
By Lemma 4.2.6, there exists a finite dimensional subspace U of V containing X
and V = UEBU where U 2: UL. Now we get [U,x] S UflU = 0 and note that
U is infinite dimensional and induces a full classical group K :2 0lK(U, 3). Hence
Ix, K ] centralizes both U and U, which implies that Ix, K ] = 1. Thus K Q 00(x).

Hence 00(x) is not abelian, contradiction.

(c) Suppose now that G is a finitary special transvection group, that is,
G = TK(VV, V) :2 (t(go,v) I (,9 E W,v E V, v<p = 0) S GLK(V)

where W is a subspace of the dual V* and AnnVW = 0. Observe that
AnnWV = 0. We note here that t(<p,v) is defined by U.t((,0,’U) := u + (ucp)v for
all u E V. Observe that W is a G-submodule of V* with the action given by

u.Ag 2: (ug‘1)A where u E V, A E IV and g E G.

Let x E G. Since G is finitary, dimIV, x] < oo. Define s : V x W' ——> K by
s(v, A) = v.A. Obviously, s is a G-invariant, non-degenerate bilinear map. Since
Vl = O and WT = 0, we have [I/V, x]i = 0V(x) and [V,.xI-L = 0W(x) by Lemma
4.2.8. Then dimIW, x] = codim CW(x) = codimIV,x]J- = dimIV, x] < 00. Therefore,
there are finite dimensional subspaces X and Y such that [V,:r] S X S V and

[W,x] g Y _<_ W with V = X 69 vi and W 2 xi e Y by Lemma 4.2.7.

76

We now claim that H := T(Xi,Yi) s 00(2). Clearly, [V,x,H] g [X,H] = 0.
Let u E V and t(<,9,v) E H where (p E Xi and v E YJ“ such that v99 = 0. Then
Iu,t(tp,v)] = (unp)v E Kv S Yi implies that [V, H] S YT. Since [IV,x] S Y, we
have YJ- S [W,x]l = 0V(x). Thus [V, H, x] S IYJ-,x] = 0. Using Three Subgroup
Lemma, we conclude that [V, Ix,H]] = 0. Hence Ix, H] = 1, that is, H S 00(33)-
Therefore 00(x) is not abelian, a contradiction. This completes the proof of the

first statement of the theorem.

The second statement follows from the fact that a LFS—group cannot be a p-type

group for two different primes.

77

Chapter 5

On Infinite Abelian Subgroups in

Locally Finite Simple Groups

Recall that a group is said to have the minimum condition, or min, if every
descending chain of subgroups terminates in finitely many steps and G is called a

Cernikov group if it is abelian-by-finite and satisfies min.

Notation: (0 oo)k 2: 0

p 00 X 01,00 x - -- x 01,00 where 01,00 stands for Priifer

\ J
v

k—times
groups.

Lemma 5.1.1 Let Y S X S R where X/Y is a finite elementary abelian p-group
and R a (opee)k. Then IX/YI g pk.

Proof: First consider the special case when X is finite. Then X ”S
0pr1 x 0pr2 x x Cprk for some r,- E Z+ where 1 S i S k and the Frat-
tini subgroup of X is @(X) ’3—1 Cpr1_1 x Cpr2_1 x x Cprk—l' Since X/Y is
elementary abelian, <I>(X/Y) = 1. Then <1>(X)Y/Y S (D(X/Y) gives <I>(X) S Y
and hence IX/YI g |X/<I>(X)| = pk.

78

For the general case, write X = Ute 1X,- where X,- is of finite order with X,- < Xi+1
for all i E I. Note here that X is a locally finite group. By letting Y,- := Y H Xi,
we get Y = UYZ- and XiY S Xi+1Y for all i E I. Assume to the contrary that

IX/YI > pk. Then we can choose i E I such that IXiY/YI > pk. But
Xi/Yi 3 Xi/YnXi gXiY/Y

implies that IX,/Y,-| > pk for some i and Xi/Yz- is elementary abelian since Xi/Yz-
is isomorphic to a subgroup of X / Y. As X,- is finite, we get a contradiction to the

special case. Thus IX / YI < pk.

Lemma 5.1.2 Let G be a Cernikov p-group. Then there exists an integer n such

that IA/BI S p" for all B S A S G with A/B elementary abelian.

Proof: Let A/ B be an arbitrary elementary abelian section of G. Since A/B
satisfies min, the order of A/ B is finite. Recall that G is a Cernikov p-group
implies that there exists an abelian normal subgroup R of G such that G/ R is
finite and R ’5 (Cpoo)k for some k E Z+, see [5, Theorem 1.5.5 I. Let IG/RI = p1
and A/B a (opp. We shall show that t g n := k +1.
R S RB S RA S G implies that RA / RB, which is isomorphic to A/A (1 RB, has
order divisible by pl. Hence

IA/AflRBI 5 pl. (5.1)

We also have AflBR/B E” (AflR)B/B ”E AflR/AflBflR := X/Y. Since X/Y is
isomorphic to a subgroup of A/ B, it is elementary abelian. Note that X S R. By
Lemma 5.1.1, we get

IA r1 BR/BI = X/YI _<_ pk. (5.2)

 

Combining (5.1) and (5.2), we obtain IA/BI S pk“.

79

Theorem 5.1.3 Let G be a non-linear LES—group and p a prime. Then there exists

a p-subgroup of G which is not Cernikov.

Proof: For any integer n, there exist An S Bn S G with 3,, finite and
Bn/An 2’ Sym(n) by Theorem 4.2.3. Let 3,, :2 In/p]. Sym(n), and thus Bn/An,
has elementary abelian p-subgroups of order p3". Let Cn/An be such a group and

let Pn, E Sylp(0n). Then 0n = PnAn and

C'n_PnAn,_\_, Pn

— __ _ —— is an elementar ' abelian rou of order 3".
An An An r1 Pn y g p p

 

This means Pn has arbitrarily large elementary abelian sections as n gets arbitrarily
large. Let Q1 2 P1. Let Q2 be a Sylow p-subgroup of (P1,P2) containing the
p—subgroup P1. Continuing like this, we choose Qn+1 as Qn+1 E SYlp(<Qn, Pn+1))
containing Qn. Then we get a chain Q1 S Q2 S -~Qn S Qn+1 S Define
Q := UQn and note that Q is a p-group. Since Pn is a p-subgroup and Qn is a
Sylow p-subgroup of (Qn_1,Pn), we have Qn 2 P7517 for some 9 E G. As P", has
elementary abelian sections of order p3”, so does Pg. So Q has elementary abelian
sections of unbounded order as n gets arbitrarily large. Therefore, Lemma 5.1.2

implies that Q can not be Cernikov.

Corollary 5.1.4 Let G be a non-linear LES-group and p a prime. Then there exists

an infinite elementary abelian p-subgroup of G.

Proof: By Theorem 5.1.3, there exists a p-subgroup Q of G which is not Cernikov.

Then, by [13, 1.G.6], Q contains an infinite elementary abelian p-subgroup and so
does G.

80

Corollary 5.1.5 Let G be a LFS—group. Then the following are equivalent.
(a) G is non-linear.
(b) For all prime p, there exists an infinite elementary abelian p-subgroup of G.

(c) There exist distinct primes p1 and p2 such that G has an infinite elementary

abelian p, subgroup for i = 1, 2.

Proof: (a) => (b) This is Corollary 5.1.4.

(b) :> (c) Trivial.

(c) => (a) Suppose for a contradiction that G is linear and let G S GLIF(V) where V
is a finite dimensional vector space over a field IF. Let p E {p1, p2} with p S charIF
and let H be an infinite elementary abelian p-subgroup of G. Clearly, H E“ 0,, x
0,; x - ~ - and does not satisfy min. On the other hand, this gives a contradiction to
the fact that a linear p-group over a field of characteristic different than p satisfies

min condition, see I13, 1.L.3I.

Corollary 5.1.6 Let G be a LFS—group. Then G is infinite if and only if G has

an infinite elementary abelian p-subgroup for some prime p.

Proof: (<=) Obvious.

( =>) If G is not linear then we are done by Corollary 5.1.5.

If G is linear, then by Theorem 1 (page 3) G is a group of Lie type defined over an
infinite locally finite field IF. Let A be a long root subgroup in G. Then A ’—‘_-’ (IF, +)

and so A is an infinite elementary abelian p-subgroup of G.

81

BIBLIOGRAPHY

82

BIBLIOGRAPHY

[1] V. V. Belyaev, Locally finite Chevalley groups, in Studies in Group Theory,
Urals Scientific Center of Academy of Sciences of USSR, Sverdlovsk (1984),
pp. 39—50.

[2] A. V. Borovik, Periodic linear groups of odd characteristic, Dokl. Akad. Nauk.
SSSR 266 (1982), pp. 1289-1291.

[3] R. W. Carter, Simple Groups of Lie Type, Wiley-Interscience, 1972.

[4] S. Delcroix and U. Meierfrankenfeld, Locally Finite Simple Groups of I-type,
J. Algebra 247 (2002) pp. 728-746.

[5] M. R. Dixon, Sylow Theory, Formations and Fitting Classes in Locally Finite
Groups, World Scientific Publishing, 1994.

[6] N. Flowers and U. Meierfrankenfeld, 0n the center of maximal subgroups in lo-
cally finite simple groups of alternating type, J. Group Theory 5 (2002), pp. 429-
439.

I7] J. I. Hall, Locally Finite Simple Finitary Groups, in: Finite and Locally Fi-
nite Groups (Istanbul, 1994), eds. B. Hartley, G. M. Seitz, A. V. Borovik,
R. M. Bryant, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci. 471 (1995), pp. 147-
188.

I8] J. I. Hall, Periodic simple groups of finitary linear transformations, Ann. of
Math. (2) 163 (2006), no. 2, pp. 445—498.

9 B. Hartle , Sim le Locall Finite Grou s, in: Finite and Locall Finite Grou s

I I y r y p y p
(Istanbul, 1994), eds. B. Hartley, G. M. Seitz, A. V. Borovik, R. M. Bryant,
NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci. 471 (1995), pp. 1-44.

[10] B. Hartley and G. Shute, Monomorphisms and direct limits of finite groups of
Lie type, Quart. J. Math. (2) 35 (1984), pp. 49-71.

83

[11] K. Hoffman and R. Kunze, Linear Algebra, 211d. Edition, Prentice-Hall Inc.,
1971.

I12] G. Karpilovsky, Field Theory: Classical Foundations and Multiplicative Groups,
Marcel Dekker, 1988.

[13] O. H. Kegel and B. A. F. Wehrfritz, Locally Finite Groups, North-Holland,
Amsterdam, 1973.

[14] U. Meierfrankenfeld, Non-finitary simple locally finite groups, in: Finite and Lo-
cally Finite Groups (Istanbul, 1994), eds. B. Hartley, G. M. Seitz, A. V. Borovik,
R. M. Bryant, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci. 471 (1995), pp. 189-
212.

I15I P. Morandi, Field and Galois Theory, Springer-Verlag, New York, 1996.
[16] M. Suzuki, Group Theory I, Springer-Verlag, 1982.

[17] S. Thomas, The classification of the simple periodic linear groups, Arch. Math.
41 (1983), pp. 103-116.

84

 

M G VEHS R

IIIIIIIIIIIIIIIIIIIIIIIIIIII

3 1 95 7884