ABSTRACT
A CHARACTERIZATION 0? THE GROUPS Sp 6(2”)

By
John L. Hayden

Some of the most recent research in finite simple groups
has been concerned with the problem of characterizing finite simple
groups by giving the structure of the centralizer of an element of
order 2. The purpose of this paper is to characterize the finite
simple groups Sp6(2n) by giving the structure of the centralizer
of an element of order 2. This paper is a continuation of the work
of'M. Suzuki and H. Yamaki.

The element 7’ = 1 of order 2 is contained

 

 

L1 1..
in the center of a Sylow 2-group of Sp6(2n). Let Hq denote the

centralizer of 1'in Sp6(q) where q = Zn. The object of this paper

is to prove the following two theorems.

Theorem 1. Let G be a finite simple group. Suppose that
G contains a subgroup H of odd index which satisfies the following

two conditions:

Ill

(1) H Hq fer some q = 23:32.

(ii) H CG(z) for any element 2 of order 2 in the center

of H.

Then G is isomorphic to Sp6(Q).

John L. Hayden

Thegggm 2. If G is a finite group which contains a
subgroup H of odd index which satisfies the conditions (i) and
(ii) of theorem 1, then one of the following cases holds:

(1) o E‘ Sp6(q).

(ii) H is normal in G and IG : HJdivides q-l.

(iii) q = 2 and H has a normal complement which is an

abelian group of odd order.

Theorem 1 is an immediate consequence of theorem 2.
This paper assumes qj>2. The result fer q = 2 is exceptional and

has been obtained by H. Yamaki.

A CHARACTERIZATION OF THE GROUPS Sp6(2n)

By

John Leslie Hayden

A THESIS

Suhnitted to
Michigan ‘State University
in partial fulfillment of the requirements

for the degree of
DOCTOR OF PHILOSOPHY
Department of Mathematics

1970

To

Catherine Kennedy

PREFACE

I would like to thank Professor D. L. Winter fer suggesting
this problem and fer his quidance throughout the preparation and
writing of the dissertation. Professor Winter's understanding and
encouragement of'my work has been the source of inspiration for the
completion of this dissertation.

I should also like to thank Professors J. Adney and
J. S. Frame fer stimulating discussions concerning this problem.

A special thanks should be given to Professors M. Suzuki and
H. Yamaki who have sent to me their unpublished papers relating to

this problem.
John L. Hayden

iii

INTRODUCTION . . .

SECTION
1.
2.
3.
4.
5.
6.

THE LOCAL STRUCTURE OF C

THE PARABOLIC SUBGROUPS OF G

TABLE OF CONTENTS

THE NON-SIMPLE CASE

THE WEYL GROUP
THE SUBGROUP Go

IDENTITY G = Go

INDEX OF NOTATIONS .

BIBLIOGRAPHY . . 0

iv

18
28

47
56
7O
72

Introduction

The determination of all finite simple groups has always
been a stimulating problem to group theorists. It was quickly
realized that the cyclic groups of prime order are the only Abelian
simple groups and Galois was the first to show that the alternating
group An is simple if n)5. The group A5 of order 60 is the
smallest non-Abelian simple group.

Further examples of finite simple groups are the classical
matrix groups over finite fields. These groups were introduced by
Jordan [9] in 1870 and were studied in detail by Dickson in the
early years of this century. The classical groups include the
projective special linear grOups PSLn(q), the projective symplectic
groups PSp2n(q), the projective orthogonal groups RI12n+1(q),
Rfién(q), RJ‘;n(q) and the projective unitary groups PSUn(q2).

For a detailed study of the classical groups refer to Artin [l]
and Huppert l:8j] .

The groups we have mentioned were the only finite simple
groups known to Dickson apart from the five simple groups M24,
M23, M22, M12 and M11 described by Mathieu in 1861 and 1873
E 10, 11:], and the simple Lie groups of type G2 discovered by
Dickson in 1901 and 1905 C5, 6] .

No new simple groups were discovered after Dickson's

papers for a period of 50 years, until in 1955 Chevalley [3]
published his fundamental paper showing how simple groups could
be obtained as groups of automorphisms of simple Lie algebras.
This paper of Chevalley included the classical matrix groups as
groups of Lie type and described several other new families of
simple groups.

In 1954 Richard Brauer proposed a program of studying
the known simple groups, particularly various linear groups, by
giving the structure of the centralizers of elements of order 2.
This program was undertaken and lead, surprisingly, to the discovery
of new finite simple groups by Suzuki and Janko. This program
of characterizing known simple groups continues to be a central
problem in group theory today primarily because it has lead in
the past to new knowledge of the known simple groups and has
inspired the discovery of new simple groups. For a survey of the
results in this direction, see [16] .

This paper is devoted to characterizing the finite simple
groups PSp6(2n) by giving the structure of the centralizer of an
element of order 2 and is a continuation of the work of'Suzuki [l5]
and Yamaki [ET] . The groups PSp6(q) are classical matrix groups
and are among the groups of Lie type described by Chevalley.

It is well known that if G = G(q) is a group of Lie type with
q = pr, p a prime and P is a Sylow p-group of G with B = NG(P)
and H a complement of P in B, then there exists a subgroup N of

G with the following properties:

3
(1) Bf‘N = H‘QN and.W = N/H is a group generated by elements of

order 2.

(2) G = U BuB ; equivalently G = BNB.
uJIN

(3) Let "it li‘ifigt, be a defining set for W and let ui be a
representative of "i in N. Then for each u in N and all i,
IQ‘t, we have

BuBuiBCBuBUBuuiB.
(A) Bui 5! B, l‘i‘t.

The integer t is called the rank of G and is the number
of distinguished generators of the group N called the Weyl group.
The group B is called a Borel subgroup of G and any subgroup of G
containing B is called a parabolic subgroup. Finally, H is called
a Cartan subgroup of G and is always abelian. If G = G(q) is not
a twisted matrix group of rank t, then H is the direct product
of t cyclic groups of order q-l ( each thus being isomorphic to
the multiplicative group of GF(q) ). A group G with subgroups B
and N which satisfy the above conditions is said to have a (BN) -
pair. Although the direct knowledge of the theory of groups with
(BN) - pair is unnecessary for this paper, it should be evident
that this is the focal point of this paper. The paper is divided
into six sections which are now briefly described.

Section 1 is primarily concerned with presenting the
centralizer H of an element of order 2 in PSp6(q) where q = 2n
is fixed. A Sylow 2-group S of H is described and is found to

contain three normal subgroups X, Y and D of order q8. The conjugacy

A
of involutions (elements of order 2) in H is discussed and a funda-
mental result (l.16) is obtained. Such information is called
local information since it involves only the properties of the
subgroup H.

Section 2 is devoted to describing the structure of
NG(S), NG(Y), NG(D) and NG(X) where S, Y, D and X denote the groups
discussed in section 1 and G is a group containing H. The
essential tool used in this discussion is Suzuki's theorem (2.2).
The groups NG(X), NG(D) and NG(Y) are called parabolic subgroups
since they contain B = NG(S).

Section 3 shows the structure of a group G containing H
in which NG(X) = NG(S). It is shown that Z = Z(H) is a normal
subgroup of G and IG : HI divides q-l. The remainder of the paper
assumes the hypothesis that NG(S) is a proper subgroup of NG(X).

Section 4 discusses the distinguished generators of the
Weyl group for G. Letting K be a complement for S in NG(S), three
involutions "1’ w2, and w3 are found to belong to NG(K) which
satisfy the relations (w1w2)4'=,(w1w3)2 = (w2'w3)3 = 1. The group
<71, wz, w> generated by wl, "2' and w3 is isomorphic to the
group of all symmetries of the cube and has order 233! = 48.

In section 5 we set G0 = BNB where B = NG(S) and
N = <:H, wl, w2, wi), . Go is found to possess a (BN) - pair
in the sense of our definition above. Two theorems of J. Tits
for groups with (BN) - pair structure are used to conclude that
Go is isomorphic to PSp6(q).

In section 6 we finally resolve the fact that G has

exactly four conjugacy classes of involutions. The involutions
2, v, vz and j are found to be representatives of these classes
and the centralizers in G of z, vz and j are found to be subgroups
of H = CG(Z)' The centralizer CG(v) is found to be a subgroup of

Go and a now standard argument is used to conclude G = Go.

l. The Local Structure of G.

In this section we define the projective symplectic groups
PSp2n(q) over a finite field F of order q. We show that while these
groups have a very geometrical origin, they also admit a completely
analytical description. We will use the latter description to
present the centralizer H of an involution in the group PSp6(2n).
The rest of the section is devoted to describing this centralizer

and some of the local properties of a group G containing H.

Definition. Let V be a vector space over the finite
field F. V is called a symplectic gpggg if there exists a scalar
product (.,.) with the following properties:

(1) (.,.) is bilinear

(alvl + a2v2, w) = a1(v1, w) + a2(v2, w) ,
(w, “1'1 + a2v2) = a1(w, v1) + a2(w, v2)
for all a1 in F and all v, w in V.

O for all v in V.

(ii) (v, v)

Notice that (v+w, v+w) 0 implies (v, w) = -(w, v) using both

properties (i) and (ii).
Definition. Let V be a symplectic vector space. If

U is a subspace of V, then U‘L= {v'v QV, (u, v) = 0 for all 1101]} .
Definition. Let V be a symplectic space. V = R(V) is

called the radical of V. V is app-degenerate if R(V) = 0.

‘Qgfiigigigg. Let V be a symplectic space. We call V
the grthogongI _8_1_J_JTI_ of the subspaces Hi’ i = l, 2, ... . n denoted
v = 11111121....Lhn if v = H1 O H2 9 e Hn and for i # 3, "Hi,
wGHJ, implies (v, w) = 0.

Theorem. If V is a non—degenerate symplectic space, then
dim v = 2n and v = H11H2L..._1Hn where each H1 is a plane with
Hi =<vi’ v5> , (v1, v3) = l ([8] , p. 217).

Definition. Let T be a non-singular linear transformation
of the symplectic space V. T is called an isometg of V if (Tvl, Tvg)
= (v1, v2) for all v1, v2 in V.

Definition. If V is a non-degenerate symplectic space
of dimension 2n over the finite field F, the group of all isometries
of V is called the sup. lectic mg of dimension 2n and is denoted
Sp2n(q) where [Fl = q. Let Z = ‘ TlTéSp2n(q) and T = a1, eel").
The projective symplectic gm of dimension 2n over F is defined
to be the group Sp2n(q)/Z and is denoted PSp2n(q).

Note that if T = a1 is an isometry, (VT, wT) = a2(v, w) =
(v, w) ,5 O for suitable v and w. Thus a2 = l and Z = 11. If
char F = 2, then PSp2n(q) = Sp2n(q).

Let V be a non-degenerate symplectic space and v1, v2,
..., V2n a symplectic basis of V chosen in such a way that H1 =
<v1, V211), H2 =<v2, v2n_1>, ..., Hn =61], vn+1>. We form
the matrix A = (am) where an = (v1, v3), l 5 i, j 5 211. By the

way the basis was chosen, V = H1_LH2_L...J.H .

The matrix A has the form I: I]where I is an n x 11 identity
-I
matrix. Let T be an isometry of V and let Tvi = Eg’ tk,ivk
k=1

relative to the above basis of V. We have, (TV1, TV3) =

( 121 tkeivk' 21 than) = tl,it2n,j(vl’ V211) + +

iéksn n+1£k£2n
aij‘ Further, it is not difficult to see that the matrix of T
t _
satisfies TAT — A if and only if'ZEik itm - Rm a
1:416 = 2114-1 11.5151: 234:1
iékén n+15k52n

aij where tT denotes the transpose of the matrix of T. We conclude
that T is an isometry of V if and only if the matrix of T relative
to v1, ..., V2n satisfies tTAT = T.

Let V be a 6-dimensional vector space over the field F
of q = 2n elements. From our above remarks we can identify the
projective symplectic group PSp6(2n) with the set of all nonpsingular
linear transformations T of V which satisfy tTA6T = A6 where A6

,
is given by, A6 = 1 ,

 

 

Let The the matrix, 1‘ = ’1

 

 

The matrix 7’ is an involution and satisfies t"A6": A6. Let H

q

be the centralizer of'T’in Sp6(q). The object of this paper is

to prove the following two theorems.

Theorem 1. Let G be a finite simple group. Suppose that

G contains a subgroup H of odd index which satisfies the following

two conditions:

Ill

(1) H Hq for some q = 2"}2.
(ii) H = 06(2) for any involution z of the center

Then G is isomorphic to Sp6(q).

Thegrem.2. If G is a finite group which contains
group H of odd index which satisfies the conditions (i) and
of theorem 1, then one of the following cases holds:

(1) G 2’ Sp6(q).

(ii) H is normal in G‘and IG : HI divides q-l.
(iii) q = 2 and H has a normal complement which is

abelian group of odd order.

This paper will assume that q:>2. The result for
is exceptional and has been obtained by Yamaki [:22] . We
investigate the structure of Hq. In the rest of this paper

denote by H a subgroup isomorphic to Hq.

of H.

a sub-

(11)

q = 2
first

we will

10

(1.1) H is the totality of all matrices of Sp6(q) of the

 

form, -1
(1.2) A M
J B 1 i

 

where A is a column vector, B a row vector, and M is a I. x 1. matrix
belonging to Sp4(q), which satisfy tAAAM = B for A4 = 1 .
1

1
1

. C o = 0
£0.02 Lat 8 belong to Sp6(q) (7') Lat 8 (“113) BY
comparing entries of gr: fg, we have from the (i, 1) positions
that ‘16 = O for i = l, 2, ..., 5. From the (6, 1) entry we obtain
mu = “166' From the (6, 1) positions we obtain mli = O for i = 2,
3, ..., 6. All elements of the symplectic groups Sp2n(q) have
determinant 1 ( [a] , p. 22/.) so that det g = 1 implies an? = 1.
We have "11 == 1 and from tgAég = 6 we must have B = tAALM where

g has the form (1.2). This implies g belongs to H and it is not
difficult to see that all matrices of the form (1.2) satisfying

tAAAM = B centralize 7' . This completes the proof.

(1.3) Let U be the kernel of the homomorphism of H onto
Sp4(q) defined by sending the matrix of (1.2) onto M. Then U is
an elementary abelian group of order q5 consisting of all matrices

of the form, —1 _)

 

 

11

(1.1.) Define S to be the totality of matrices of H with

 

 

‘M = Fl "
a 1
b d 1
Lo b+ad a 1.1}

where all entries of M belong to F. Then S is a Sylow 2-group of
H of order q9.
e291. H/U‘=’ 3P4(Q)- l3p4<q)|= q’+<q’+-1)(q2-1> ([8] ,
p. 220) . This implies that IHI = (19014-1) ((12-1) so that the
result follows from Isl = q9.
(1.5) The center of S is elementary abelian of order q2.
Z(S) = 1

 

 

where a4, 36F.

For the remainder of this paper we will use the notation
C for the center of S. A prooflwill not be given for (1.5) and
several other similar calculations.

(1.6) The center of H is elementary abelian of order q.

2(a) = 71 '1

1
1
1
1

LI 1..

where Yt- F. We will use 2 to denote Z(H) in the rest of this paper.

 

 

M. H/U 14' Sp4(q) is a simple group ([8] , p. 227) .

12

Therefore Z(H) Q U. Since U is a subgroup of S, Z(H) CC. The
matrix diag“ 1, m, m'l, m, m’l, 1} belongs to H for any m of 1“".
This implies Z(H)C_;Z. Every involution of Z is centralized by H

so that Z(H) = Z.

(1.7) Define X to be the totality of all matrices of S

with (2, 1) and (6, 5) entries zero. X is a normal subgroup of

S of order q8. The center of X is elementary abelian of order (13.
20:) = “'1 “
l
l
1
a4 2 l l
L'r 4 .J

 

 

(1.8) If t is an involution of 20:) - C, then Cs(t) = x.
2:99;. From (1.7) c f O in the expression for t. If
s 605(t), consider the (5, 1) entry of at = ts. We see that the
(2, 1) entry of s must be zero. All elements of 8 have (2, 1)

and (6, 5) entries the same. This implies x = Cs(t).

(1.9) Define Y to be the totality of all matrices of S

with (3, 2) and (5, 4) entries zero. Then Y is a normal subgroup

of S of order q8. The center of I is elementary abelian of order q3.
2(2) = (.1
1
l
a,z 1
a’ 1
4
a a 1
bx 4 3 ..H

 

 

13

(1.10) If t is an involution of Z(Y) - C, then Cs(t) = Y.

Prgof. The argument is very similar to (1.8).

(1.11) Define D to be the totality of all matrices of
S with (1., 3) entry zero. Then D is a normal subgroup of S of order

Q8. The center of D and the center of S coincide.

(1.12) NH(S) = s . K1 x K2, leIK2 = l where K1 and K2
are cyclic groups of order q-l. Let 4 be a generator of F“, then

K1=¢P 3 K2=¢?o
5=W1 U k -1
1 it
“ n
(1.13) ‘7 "
l -<f
L. 1 a) _ 1 .1
Pro f. A Sylow 2-group of Sp 4(q) has a normalizer of
order q"(q--1)2 in Sp4(q) D5] . Since H/U Sp4(q), NH(S)/U must

have order q’*(q-l)2. This implies INH(S)I = q9(q-1)2 so that

 

 

 

 

NH(S) = S . H1 1: K2 where K1 and K are given in (1.13).

2

(1.11.) No involution-of Z is a square in S.
PM. Let 398 and suppose that s2 belongs to Z. In

2 we must have M2 = I where M is the matrix given

the expression for s
in (1.1.). This implies one of a or d must be zero. Let a = 0 and
let tA = ( a1, a2, a3, a) in the expression (1.2) for s. From the
(4, 1) and (5, 1) entries of s2 we obtain,

cal + be2 = O

alb + da2 = 0.

1.4

.. t ..
This implies B — AA4M — (a4, a3, a2, a1). Thus the (6, 1) entry
of 32 is B ' A = 0. Therefore s2 is the identity of Z. If d = 0,
we may assume a i5 0. From the (3, 1) and (5, 1) entries of 32
we obtain a1 = O and baz + aa3 = 0. Again B 2 (a4, a3, a2, al)
which implies the (6, 1) entry of 82 is zero and s2 is the identity

 

 

 

 

of Z.

(1.15) Every involution of C - Z is a square in S.

1122:. Consider the elements 31 and s2,

81 = Fl 1 32 = '1 -
a1 1 l
1 a2 1
1 b l
c 1 b 1

L cal a1 1; L bq 9.2 ]._J .
For appropriate choices of 3.1, c, a2, b, every involution of C - Z
can be written as either a; or 322 .

(1.16) Every involution of U - Z is conjugate in H to
an involution of C - Z.

M. Let u be an involution of U - Z as given in (1.3).
Then one of a1, a2, a3 is different from zero. Let heH and H be
the A x 1. matrix of Sp4(q) in the given expression (1.2) of h.
Since ‘h A6h = A6, 11-1 = A6thA6. It is not difficult to check

 

 

that — ~—
1 (

huh'lr. HA I
LJ t (m) ‘1. 1 a

where A is a column vector with 13A = ( a1, a2, a3, a4). Therefore

15

u is conjugate in H to an involution of C - z if'MA = t( O, O, O, r )
for some f'# O of F. By hypothesis we may view A and t( 0, O, O, r )
as linearly independent vectors in a 4-dimensional vector space

on which Sp 4(q) acts. Since Sp 4(q) acts transitively on its distinct

one-dimensional subspaces, MA = t( 0, O, O, f ) for some M belonging

to SP4(Q) ( [8] 9 P. 221 )o

(1.17) If u is an involution of U, then 'Cs(u)l‘q7
unless u‘Z(Y).
£33.93..- Let u be an involution of U given by (1.3).
Let s (.Cs(u) and M be the matrix (1.4) which must appear in the
expression (1.2) of s. From an = us we obtain MA = A where A is
a column vector given by 1”A = ( a1, a2, a3, all+ ). This implies,
aal = O
be1 + da2 = 0
cal + (b+ad)a2 + aa3 = 0.
If a1 if 0, a2 = 0, then a = O, c = 0 so that ICS(u)[‘g7. If a1 = O,
O, b = aa3/a2, so that ICS(u)l‘q7. If a1 # 0,
O, b = daz/al, so that [Cs(u)l§17. If a1 = a2 = 0,

then uGZ(Y). In this case u has centralizer of order q8 or q9.

a2 # 0, then d

:3.2 y! 0, then a

(1.18) Define so to be the totality of all matrices of
S with zero column vector A and f: 0. (Refer to (1.2) and (1.4))

Then So is a subgroup of S of order q[‘ and for any involution t
of So, ICS(t)léq7 unless t OZ(X).

16

mg. Let t be an involution of so. From (1.1.), the

(3, 2) or the (1., 3) entry of t must be zero. Let t have (3, 2)

entry zero so that t is given by,
t = r1

1
1
1

where r, g,h belong to P. Let secs(t) be given by (1.2) with
t

:5‘1-5 H
HJUQH

 

 

A = ( a1, a2, a3, a4) and M as given in (1.4). From ts = st

we obtain,
ga = O
rel + ga2 = 0
ha + £32 = 0 e
If g = 0, then r y! 0 since t(z(x). This implies a1 = O, a2 = 0,

so that '03(t)|‘q7. If g i‘ 0, then a = 0, a2 = fol/g so that
|Cs(t)|‘q7’. The case in which the (A, 3) entry of t is zero is

811111181“

(1.19) If t is an involution of S, then 'Cs(t)|‘q7
unless t belongs to Z(X) or Z(Y).

moi. If t belongs to U or So’ the result follows from
(1.17) and (1.18). Since S has order q9 and U is a normal subgroup
of S of order q5, the definition of 80 implies S = U . So, UnSo = 1.
This implies t = uso where we may assume that u is an involution
of U and by the construction of S that so is an involution of So.

If u CC, then sofzu) since we may assume tIZOI).

17

This implies Cs(t) = 63(30) so that '03“! §q7 by (1.18).
If ue-Z(Y) - C, the (1., 1) entry of u is non-zero.

Since t = us t2 = 1, t is given by,

 

 

o,
t {l 7'
l
1
a3 b d 1
a]+ c b 1
LI' at. a‘3 1.1

where a3 34 O and one of b,’ d, c 15 different from zero. It is
not difficult to calculate directly that ICS(t‘ ‘q7.

If ulZ(Y), (1.17) implies that u has at least q2 distinct
conjugates in 8. Since U is elementary abelian and S = U030,
all distinct conjugates of u may be expressed as uxi where xitso.
Let xi, i = 1,2, ..., (12 be elements of 80 such that uxi is a
distinct conjugate of u for i = 1, 2, ..., q2. Suppose txi = txj
for i 3‘ j. Then (use):Ki = (us°)xj which implies uxi = uxJ since U
is normal in S and x1, xii 930. This is not the case and t has

at least q2 distinct conjugates in 3. Therefore 'Cs(t)| ‘q7.

2. The Parabolic Subgroups of G.

In this section we will assume that G is any group satis-
fying the hypotheses of theorem 1. We will determine the structure
of NG(Y), 116(1)) and 116(2) where Y, D, and z retain the same notation-
al meanings given to them in section 1. The first two of these
groups will be shown to involve the simple group L2(q) and NG(Z)
will be shown to involve the simple group Sp 4(q) . Under the
assumption that NG(X) is not 2-closed we will determine that NGOC)
involves L2(q) and from this information determine the structure
of NG(S).

We first need some preliminary lemmas that will be used

throughout the rest of the paper.

(2.1) Let u and v be two involutions in a finite group
G such that u is not conjugate in G to v. Then there is a third
involution w in G such that:

(i) ,w commutes with both 11 and v, and

(ii) wu is conjugate in G to either u or v.

M. This is a well-known result, generally attributed
to Brauer and Fowler [2] . It is a consequence of the fact that
the subgroup (u, v) generated by u and v must be dihedral of order
4k for some integer k. The element w can be taken to be (uv)k,

the involution in the center of (u, b, and the assertion (ii)

19

follows from the fact that (u, b has exactly three conjugate classes

of involutions, one of which consists of w alone.

A group is called pfiglgggg fer a prime number p, if its
Sylow p-group is a normal subgroup. A group G is called a (TI) -
g£23p_if the intersection of any two distinct Sylow 2-groups is
trivial. The structure of (TI) - groups has been determined by
Suzuki. From the main theorem of his paper [12] , we have the

following corollary. A proof may be found in [13] .

(2.2) Let G be a finite (TI) - group with noncyclic
abelian Sylow 2-groups of order r. If G is not 2-closed, then G

contains a normal subgroup of odd index which is isomorphic to L2(r) .

(2.3) let G be the group L2(r) where r = 2m>2. Then
G has the fbllowing properties:

(i) G is a simple group of order r(r2 -l).

(ii) There is a cyclic group R of order 13-1 in G which
normalizes exactly two Sylow 2-subgroups $1 and 32 of G. In fact,
NG(81) = 31R, NG(32) = 323 and NG(Sl)nNG(SZ) = a.

(111) It acts regularly on 31* and 32*.

t=S

(iv) There is an involution t in NG(R) such that $1 2

and t inverts every element of R.
(v) Any two distinct Sylow 2-groups of G generate G.
G is generated by R, 31 and t. In fact,G = SIRwIRtSl.
Pmof.’ Identify G with the group of 2 x 2 matrices

of determinant 1 over the field F of r elements.

20

3:”? 50] : per} 31:) 1313A.”
Sz={[s 21*“ ”H: a] I

Properties (ii) - (iv) can be readily verified. A proof of (i) can
be found in [8]; p. 182. A proof of (v) can be found in ['3]; Lemma

2, Po 34-

(2.4) (Suzuki) Let G be a (TI) - group and S a Sylow 2-
group of G. Assume that S is a noncyclic abelian group of order
r and that G is not 2-closed. Let K be a complement of S in NG(S).
Then K normalizes exactly one more Sylow 2-group. There exists
an involution t of G which normalizes K such that St and S are the
two Sylow 2-groups normalized by K.

2392;. Each element of G induces a permutation of Sylow
2-groups by conjugation. This representation of G is doubly transitive
by (2.2). A subgroup of G consisting of elements which leave two
Sylow 2-groups invariant is conjugate to K. If N denotes the
normal subgroup of G, which is isomorphic to L2(r),ANf)K is a
cyclic group, which satisfies the properties (ii)-(iv) of (2.3).

Since the involution t exchanges S1 = S and 32, t normalizes K.

Let G be a finite group satisfying the hypotheses of

theorem 1.

(2.5) s is a Sylow Z-group of G.

21

2:39;. The index of H in G is odd and S is a Sylow

2-group of H.

(2.6) No involution of U - Z is conjugate to an involution
of Z.

M. Let u be an involution of U - Z and suppose u
is conjugate to an involution z of Z. By (1.16), u is conjugate
in H to an involution of C - Z. Thus we may assume that u belongs

to C - Z. From (1.15), u = s2

for some 3 e8. By a theorem of
Burnside, there exists geNG(S) such that u8 = z. This implies

that z is a square in S , a contradiction to (1.11.).

(2.7) NG(U) = NG(Z), H4NG(Z), NG(s)C:NG(z).

Proof. Let x GNG(Z). Then Zx = Z is the center of Hx
which implies H = H". This implies that H is a normal subgroup
of NG(Z). Further, if: is a normal subgroup of H so that UxU/U
is a normal subgroup of H/U. Since H/U 3' Sp4(q) is a simple group,
11" = u and we have NG(zX;NG(U).

Conversely, let x GNG(U) . Then 2:: is a subgroup of U
so that 2x = Z by (2.6). This implies NG(UQG(Z) and we have
No(U) = No(2>-

Let geNG(S) so that Z8 is a subgroup of C = 2(3). By
(2.6), 28 = Z and we have NG(S)CNG(Z).

(2.8) (i) NG<sI;;NG(x), (ii) uG<sXZZNG(n).
Proof. By (1.9), Z(Y) is a subgroup of U of order q3

containing G. Let gGNG(S) so that Y8 is a subgroup of S of order

22

q8 with center (2(1))5. Let t be an involution of (Z(Y))g - C.

Then 2‘ centralizes t so Cs(t) has order at least q8. By (1.19),
t belongs to 20:) or Z(Y). h'om (2.7), (2(1))3 is a subgroup of
u so we must have that t belongs to 2(1) - o. By (1.10), Cs(t) =
Y = 13. This implies NG(sfi:NG(l).

Let g. 116(5). By (1.11), D contains 11 which implies D8
contains U by (2.7). By the structure of D, D = U 0 U0, Unlo = 1

where Us is an elementary abelian group of order q3.

U = '1 '7

° 1

a1

(2.9) b 1

cbal
1

b -

This implies D3 = U ‘ U08 and Dg/U 3' U0 is an elementary abelian

 

 

subgroup of S/U of order (13. Since S/U g 80(1.18) , it is not
difficult to see that S/U has exactly two elementary abelian sub-
groups of order Q3. This implies Dg/U = Y/U or Dg/U = D/U. In
the first case I)8 = I which is impossible since Z(Y) has order

q3 and Z(D) has order Q2. Thus .113 = D and we have NG(s)_C_NG(D).

(2.10) (1) NH(D)/D is a (TI) - group.
(ii) NH(Y)/Y is a (TI) - group.

Proof. By (2.9), D = u - no, Unlo = 1. Further, a = u . no,

 

Unfio = 1 where Ho 5' Sp4(q) is given by,

H = 1 MESP (q).
(2.11) ° l: M ' 4
1

23

Let heNHw). Then h = uho, ueu, h0 one. Since D = U - U0,
hot-Nam). By the structure of D, Uoho = U0 which implies hoe-
NHO(U°). Let S and 31 be two Sylow 2-groups of NH(D). Then
s = U - so, uflso = l and s1 = u - so, Uf‘lso = 1. This implies
So and So are Sylow 2-groups of NH°(U°). A result of Suzuki [15]
shows NH0(Uo)/UO is a (TI) - group. This implies sense = U 0'
Therefore, SflSl = D and the result follows.

By the structure of Y, I = U 0 To, UflTo = 1 where To
is an elementary abelian group of order q3.

To: 1.

l
1

u 1.1)

O‘QH

l
(2.12) b
c

 

 

Let he-NH(Y). Then h = uho, ul-U, hotfio. By the structure of Y,
Tobe = To so that hoe-NHO(TO). A result of Suzuki [15] shows
NH°(T°)/T° is a (TI) - group and the proof is completed by an argu-
ment similar to that for NH(D) .

(2.13) (i) NG(Y) and N6(D) are subgroups of NG(Z).

(ii) NG(I)/Y and NG(D)/D are (TI) - groups

which are not 2-clossd and have noncyclic abelian Sylow 2-groups
3/2 and S/D respectfully.

M. By (1.9) and (1.11), Z(Y) and Z(D) are subgroups
of U containing Z. Since no involution of U - Z is conjugate to
an involution of Z, N60) and NG(Z) are contained in NG(Z).

All Sylow 2-groups of NG(Z) are contained in H since

24

by (2.7) H is a normal subgroup of NG(Z). It follows that all

Sylow 2-groups of NG(Y) and NG(D) are contained in H. By (2.10) ,

NG(Y)/Y and NG(D)/D are (TI) - groups with noncyclic abelian

Sylow 2-groups 3/! and S/D respectfully. Consider the involutions,
:1 = fl 1 :2 = l

(2.11.) 1 1

 

 

 

 

It is not difficult to see that 31 normalizes U0 (2.9). Thus
:1 normalizes D = U - U0 and NH(D) is not 2-c1osed. Similarly,
32 normalizes I = U - To (2.12) so that NH(Y) is not 2-closed.
(2.15) NG(SCNG(X).
22921:. Let g ¢NG(S) so that Xg is a subgroup of S with
center (Z(X))g. Let t ¢(Z(}{))g - C so that CS(t) has order at
least q8 and t must belong to 20:) or 2(1) by (1.19). If t 22(1) - c,
then 03(t) = Y = Is by (1.10). Thus X = Yg-l contains U by (2.7),
a contradiction. Therefore, t ¢Z(X) - G and X = CS(t) = 13 by
(1.8). This implies NG(s)CNG(x).

(2.16) NG(X)/X is a (TI) - gimp.

M. Let 51 be a Sylow 2-group of NG(X) such that
sflsl ,5 X. Then 31 contains an element of S - X and 31 = S8 for
some geNG(X). This implies 2(31) = CE!(:(z(x))g = 20:). If C“
contains an involution t of Z(X) - C, then t commutes with an
element of S - X. This is impossible since Cs(t) = X by (1.8).
Thus 03 = 0 so that z8 = z by (2.6). By (2.7), this implies

31 = S8 is a subgroup of H.

25

Therefore 31 contains both U and X and we have 31 = U)! = S. This
implies NG(I)/X is a (TI) - group.

In the remainder of this section we will assume that
NG(X) is not 2-closed. Under this condition we will determine the
structure of NG(S) and 1116(1). In section 3 it will be shown that
the non-simple case of theorem 2 holds whenever NG(X) is 2-closed.
(2.17) If NG(X) is not 2-closed, then N60!) contains
a normal subgroup L of odd index such that L/X 9L2“). Further,
Z(L) = V where V is given by, I

V: 11 'aLEF'
(2.18) _ l
> 1
a 1
kal‘ l
L... ..

 

 

2299:. By (2.16), NG(X)/X is a (TI) - group with Sylow
2-group S/X. By hypothesis NG(X) is not 2—closed and 3/}! is a
noncyclic abelian group. Suzuki's (TI) - theorem (2.2) implies
NG(X) contains a normal subgroup L of odd index such that L/X 3’ L2(Q).
Let 81 be a Sylow 2—group of N00!) distinct from S.
By (2.3, v), L =<s, 31> . Clearly z(s)nz(slgz(t). If ace-Z(L),
then M: 2! L2(q) implies xex = 3103. Hence, Z(L) = z(s)nz(s1 ).
Since S.1 belongs to NG(X),Z(31) and 2(8) are subgroups of Z(X).
This implies |2(S)nz(sl)|> q. 21% z(s)fiz(s1 ) )=1 since s1
is not a subgroup of H. This implies Z(S) = Z x ( Z(Sflfil) )

and therefore 'Z(L)I= lZ(SflZ(Sl)I = q.

26

Let X1 and K2 be the groups defined in (1.13). Since
K1 and K2 nomalize 8, they normalize X by (2.15). This implies
that K1 and K2 normalize L and hence Z(L).

Suppose that YnZ(L) = 1. Then Z(S) '2 V x Z(L). For
any involution zOZ, z = vu for some veV, lie-Z(L). We have zk2 =
z = vu for any kzéxz. This implies vk2 = v since K2 normalizes
V and Z(L). By direct calculation, K2 acts regularly on V“ so
that we must have v = 1. This implies z = u, a contradiction to
the fact that zf'tzu.) = 1. we conclude that vnzu.) # l and
finally, Z(L) = V since K2 acts regularly on V” and normalizes Z(L).

(2.19) 216(3) = s - x, snx = 1. If NG(X) is not 2-closed,
K is abelian of order (q-1)3. K = K1 x K2 x K3 where K1 is cyclic
of order q-l for i = l, 2, 3. Further, Knay be chosen such that
KnH = K1 x K2 and K3 = 1:01. where L is the normal subgroup of
116(1) given in (2.17).

M. By the Schur—Zassenhaus theorem, NG(S) = S 0 K,
301: = 1. By (1.12), 113(5) = s. . K1 x K2 = s . mu. Since 143(5)
is a solvable group, (Krill)h = K1 x K2 for some hélH(S). This
implies Kh contains K1 x K2. We conclude that we may asstme that
K has been chosen to contain K1 x K2 and thus m3 = K1 x K2.

If NG(X) is not 2-closed, (2.17) implies the existence
of a normal subgroup L or odd index such that m: a! L2(q). By
(2.3) p L contains a cyclic group R of order q-l which nomelizee
S and acts regularly on 3/1. This implies x01. has order q-l
and acts regularly on s/x. let K3 = xflL.

27

By (2.7), x normalizes z. This implies x/Kfln acts
semi-regularly-cn 2*. Hence, x/Kflh divides q-l and the order
of x is at most (q-l)3. Suppose K3n(K1xK2) # 1. Then k3 = klkz
for some 1:3 “3, kzi-KZ, kleKl. Since K3 is a subgroup of L, x3
centralizes V by (2.17). Let v be an involution of V. We have

v = kglkilv k1k2 = kglv k since K1 also centralizes V. This

2

implies k2 = 1 since K acts regularly on V“. We conclude that

2
k3 = k1, a contradiction since K1 acts trivially on S/X. Therefore,

K3{1(K1 x K2) = 1 and K = K1 x K2 x K3 has order (q-l)3.

(2.20) If'NG(X) is not 2-closed, then NS(K) = 1.

533;. Since xns = 1, NS(K) centralizes x. By direct
calculation, CS(K1 x K2) : Z. Since K3r]H = 1, K3 centralizes no
involution of Z. Therefore, CS(K) = 1 and we conclude NS(K) = 1.

3. The Non-Simple Case.

In this section we investigate the case in which NG(X)
is 2-closed. It will be shown that if NG(X) is 2-closed, then
Z is a normal subgroup of G and [G : H] divides q-l. We will

assume throughout this section that NG(X) is 2-closed.

(3.1) Let so and Ho be the subgroups of H defined in
(1.18) and (2.11) respectfully. Every involution of So is conju-

gate in Ho to an involution of 2(30).

z(s)= '1 7’
° 1
1
b 1
ch 1
u. 1)

 

 

where b, c OF.
2:92;. By the definition of Ho, Ho 2 Sp4(q)e From the

structure of So, it is evident that every involution of So belongs

to either Uo or To. ( Refer to (2.9) and (2.12) ) Every involution

of U0 is conjugate to an involution of Z(So) by an argument analogous

to (1.16) when applied to Sp 4(q). Let t be an involution of To’
p
t = 1 .1

1
l

L 1.1.

We may assume that d # o. If e = 0, 3252 62(80) where :2 is the

00' H
O‘CLH

 

 

29

involution of'H given in (2.14). If c f 0 in the expression for
t, then t is conjugate by an appropriate element of K1 (1.13) to
an involution of T0 with the (5, 2) and (4, 3) entries equal.

Consider the following involution of H,

h:

i'

 

FHHHH

l
1

l
1

1

T

l

 

Since we may assume that t has both the (5, 2) and (4, 3) entries
equal, wzhthwzi-Z(So) by direct calculation. Thus every involution

of S0 is conjugate in no to an involution of Z(So).

(3.2) Every involution of S is conjugate in H to an

involution of T1 or T2 where T1 and T2 are given by,

T1: "'1 ‘1 '12: ('1
l
l
b l a
a c b l a3
4 a l A
L’, ‘4 ._ L."

 

23:22:. Since
of H, (3.1) implies that every involution of S is conjugate in H

 

to an involution of the form,

(3.4)

 

h'I'

l
31 1

a2
a3 b

84:1

where all entries belong to F.

O‘O‘I-I’

l

1
2"‘2“1l

1

 

-l

 

= USD and U is a normal subgroup

 

30

Let e be an involution of s of the form (3.1.). If b y! o in the
expression for s, then a1 = a2 = 0 since a is an involution. By
direct calculation, it is not difficult to see that s is conjugate
to an involution of T1 - Z(X) by an appropriate element of U.

Let s be an involution of S with b = O in the expression
(3.4). If the expression (3.1.) of s has c = 0, then 3 belongs
to U and the result follows from (1.16). we may assume c y! o in

the expression for s. Consider the element hGH given by,

 

 

h= "i 1
1
CM
fDl
(... 1...
where M£L2(q), C = t(c1, c2), D = tCAzM for A2 = O 1 and all

1 0
entries of h belong to F. For apprOpriate choice of M6L2(q),

-l
hsh 0 T2.

(3.5) If 31 is a Sylow 2-group of G containing X or I,
then 51 normalizes X or I respectfully.

2mg. Let SI contain X since the argument is similar
for 1. For any involution tez(x) - 2(31), 031(t) = x by (1.19).

Since 81 and S are isomorphic, X is normal in 81.

(3.6) No involution of Z(X) - Z is conjugate to an
involution of Z.

M. Suppose tl' Z(X) - Z is conjugate to an involution
of Z. Then CG(t) contains a Sylow 2-group 81 of G such that S

1
contains X. By (3.5), $1 normalizes X and by hypothesis S = 31.

31
This implies that tGZ(S) - Z, a contradiction to (2.6).

(3.7) Let T1 be the group defined in (3.2). If t is
an involution of T1 - Z(X), then 03(t) = Z(I)S°. The group Z(Y)So

is a subgroup of S of order q7

with center T1. Moreover,
NS(Z(I)SO) = X and NH(Z(X)SO) is 2—closed.

2:29;. The first two statements are not difficult to
verify directly. Let P be a Sylow 2-group of NH(Z(Y)SO). Then
UP is a 2-group containing USo = S. This implies P = NS(Z(Y)SO).

By direct calculation, NS(Z(Y)S°) = X.

(3.8) NG(Z(Y)SO) is 2-closed.

§£g_£. By (3.7), X normalizes Z(Y)So. Let P be a Sylow
2-group of NG(Z(Y)S°) which contains X. By (3.5), P normalizes
x and by hypothesis P = NS(Z(Y)SO) = x. Suppose that NG(Z(I)SO)
is not 2-closed and X1 is another Sylow 2-group of N5(Z(Y)So) distinct
from X. Since NH(Z(Y)SO) is 2-closed (3.7), X1 is not contained
in B. This implies zf12(x1) = 1. Let 21 be the conjugate of z
in Z(Xl). Since no involution of Z(X) - Z is conjugate to an
involution of_Z (3.6), 21(]z(x) = 1. By the definition of T1,
Z(XM::T1. This implies Z(X1£:;Tl since T1 is the center of Z(Y)S°.
By (3.7), T1 is elementary abelian of order q4 and we have T1 =

Z(X) x 21. Therefore, all involutions of Z belong to T1 - Z(X).

1
By (3.7), Cx(Zl) = Z(Y)S . Let I = XnXl and suppose x ('1- Then,
x OCX(Zl) = Z(Y)S°. We conclude that I = Z(Y)So and that NG(Z(Y)SO)/

Z(Y)So is a (TI) - group which is not 2-closed and which has

32

an abelian noncyclic Sylow 2-group X/Z(Y)So.

By (2.2), NG(Z(Y)S°) contains a normal subgroup L of
odd index such that L/Z(Y)So g L2(q). If X and X1 are two distinct
Sylow 2-groups of L, L =4, X1>by (2.3, v). Since L2(q) is a
simple group, Z(L) = Z(X)nZ(X1). From the fact that Z(X) and
z(xl) are subgroups of T1 of order q3, [z(x)nz(x1)|2q2. Since
zmflzul) intersects z triviely, Z(X) = z x Z(L). we conclude
that IZ(L)| = q2.

The group K2 (1.13) normalizes Z(Y)So and L and thus
Z(L). If VONL) = 1, then Z(X) = V x Z(L) which implies every
involution of Z may be expressed as z = vu for some v 0V, ue»Z(L).
Since K2 centralizes Z, acts regularly on V“, and normalizes Z(L) ,
we have ka = v for any k2 6K2. This implies v = 1 and hence z = u,
a contradiction to Z(X) = Z x Z(L). we conclude that VCZ(L).

Let G = NG(Z(Y)S°) and choose X to be a complement for
X in N_(X) which contains X1 1: K2. By (2.1.) , K normalizes exactly
one megs Sylow 2-group of G. Let X and X:L be the Sylow 2-groups
normalized by K with Z and 21 the unique conjugates of Z in Z(X)
and Z(Xl) respectfully. This implies that K normalizes Z and 21
where 21 = Z8 for some g eL.

By the above remarks, 21 is a subgroup of T1 - Z(X) so
contains an involution t with b 7! o. ( Refer to (3.2) ) Since
K1 normalizes Zl, ZI consists of the q-l involutions kiit 1:11 ,

IsiSq-l. These q-l involutions are given by,

kiit 1:11 == Fl- 1, lgi<q-1.
l
. 1
hi ,1
a4 c ha. 1
a
J 4 ._

 

 

Since Z1 is a subgroup of order q, the above q-l involutions of
Zl must have a4 = c = l’= O. This implies that Zl contains the

involution t given by,

1 e

 

 

Let u be the involution of U with (3, 1) entry equal to l and all
other entries zero. Then utu = tv where v is the involution of
V with (5, 1) entry equal to 1. By construction, t = z8 for some
involution le and some g! L. This implies (utu)8"1 = zv since
Vc:Z(L). This implies that t is conjugate to an involution of

c .- z, a contradiction to (2.6). we conclude that NG(Z(Y)S° is

2-closed.

(3.9) No involution of T1'- Z is conjugate to an in-
volution of Z.

(2:92;. Let t be an involution of T1 - Z such that t is
conjugate to an involution of z. By (3.6), taT1 — 20:). Then
CG(t) contains a Sylow 2-group 81 such that 81 is a Sylow 2-group
of 0 containing Cs(t) = Z(Y)S . By (3.8), N31(z(l)s°)§: so that

N31(Z(Y)So) = Z(Y)S°. This implies 81 = Z(Y)So, a contradiction.

31.

(3.10) Let X and Y be the subgroups of 3 defined in
(1.7) and (1.9). Then X“! is a normal subgroup of S of order q7‘.
The center of X”! is the subgroup T2 defined in (3.2). Moreover,
NH(xf11) is 2-closed with Sylow 2-group 8.

mpg. Since X and Y are normal subgroups of S, X”!
is normal in S. By direct calculation, T2 is the center of XnY e
Let h 6%“ “1). Then h normalizes Y = U(XnY). By (2.11), h = uho
for some u GU, ho 6H0. Since h normalizes T2, ho normalizes the
subgroup of Z(X) containing all involutions with (5, 2) entry non-
zero and all other entries zero. By [15] , ho normalizes U0 and

hence h normalizes D = UUO. Therefore, h normalizes S = YD.

(3.11) NG(xlll) is 2-closed.

Me By (3.10), S is a Sylow 2-group of NG(XnY).
Suppose that 81 is another Sylow 2-group of NG(XnY) distinct from
S. Then 31 is not a subgroup of H since NH(XnY) is 2-closed and
thus Z(Sl) contains a conjugate 21 of Z such that ZflZl = 1. By
(3.6) and (2.6), no involution of z(x) - z or 2(1) - z is conjugate
to an involution of Z. This implies that every involution of 21
belongs to T2”- (z(x)LJz(¥)). From the structure of T2, 03(21) =
XflY. If I = S081 and x 61, then x eCS(Zl) = X”! and we conclude
that I=XnY. Therefore, NG(XnY)/(XnY) is a (TI) - group with
noncyclic abelian Sylow 2—group S/(XnY) of order q2 and by (2.2)

2

NG(X ()1) contains a cyclic subgroup R of order q - 1 which acts

regularly on S/(XnY) and normalizes 8. By (2.15), R nomalizes X.

35

This is a contradiction since X/XnY is a proper subgroup of S/Xflt.

We conclude that NG(Xr]!) is 2-closed.

(3.12) no involution of T2 - z is conjugate to an
involution of Z.

2322:. Let t be an involution of T2 - Z such that t
is conjugate to an involution of Z. By the structure of'T2,
t belongs to T2 - (Z(X)LJZ(I)). Let 31 be a Sylow 2-group of
CG(t) such that 31 is a Sylow 2-group of G containing Cs(t) = Xf‘f.
By (3.11), N51(anCs. This implies mslunr) = X“! since t
commutes with no element of S - XfWY. 'He conclude that 81 = XfWY,
a contradiction to the choice of 31' Therefore, no involution of

T2 - Z is conjugate to an involution of Z.

(3.13) No involution of S - Z is conjugate to an involution
of Z.

{Exggfie By (3.2), every involution of’S - Z is conjugate
in H to an involution of T1 - Z or T2 - Z. The result follows from
(3.9) and (3.12).

(3.14) 2C§(l)

m2. Let u be an involution of Y with (2, 1) and (6, 5)
entries nonpzere and all other entries zero. Let v be an involution
of Y with (5, 2) entry nonpzerc and all other entries zero. Then
(uv)2 belongs to 2'. If w is an involution of Y with (3, l) and

(6, 4) entries non-zero and all other entries zero and t is an

36

involution of I with (4, 2) and (5, 3) entries non-zero and all
other entries zero, then (wt)2 belongs to I'. It is easy to
verify that every involution of Z may be expressed as (uv)2(wt)2
for appropriate involutions u, v, w, and t. Since I/I(Y) is
elementary abelian, CY'C1(Y).

(3.15) Define a to be the totality of all matrices of s

of the form,

 

 

l .7
1
1
8.4 c b 1
Lt “4 “3 1..) °

Then E is a normal elementary abelian subgroup of S of order Q6.

In the remainder of this section, we will use (3.13) ,
(3.14) and (3.15) to show that Z is normal in G. It follows from
(3.13) that no involution of H - z is conjugate to an involution
of Z. Further, if P is a 2-group with PnZ ;é 1 and 31 is a Sylow
2-group of G containing P, then Pr1Zc:Z(Sl) which implies sfiih

and we have that P is a subgroup of H.

(3.17) Let t1! T1 - z(x) be an involution with (5, 2)

entry non-zero. Let t2(»T2 - Z(X). Then CH(t1)ncH(1-,2) = E.

M. Let hGCH(t1). Then h = uho for some uQ-U, hoe-Ho.
Since h centralizes t1, it is not difficult to see that ho centralizes
the involution j given below.

b l b#0,c#0

 

 

Since Ho 9' Sp 4(q), a result of Suzuki [15] implies hoeSo.
Therefore CH(t1) = Cs(t1) = Z(Y)S°. Since Cs(t2£1, we have
campnafluz) = c3(t1)()cg(t2) = E.

(3.18) Let t1 be an involution of T1 - 20:) with (5, 2)
entry non-zero and t2 an involution of T2 .- 200. Then 06(t1)fl0(t2)
is 2-closed with Sylow 2-group E.

M. If P is a Sylow 2-group of CG(H)flG(t2) containing
B, then P is a subgroup of H by the remarks preceding (3.17).

This implies P = E by (3.17). Suppose that E1 is another Sylow
2-group of Cg(tl)nCG(t2) such that I = 301:1 has maximal order.

By (3.17), his not contained in H so that InZ = 1. Since E is
elementary abelian, IE : I|)q. Therefore, Cg(I)/I is a (TI) - group
which is not 2-closed and has Sylow 2-group E/ I which is noncyclic
and abelian. By a theorem of Suzuki, 06(1) contains a cyclic sub-
group R of order |E : II - 1 which normalizes E and acts regularly
on the non-identity elements of E/ I. Since 21/ I is a subgroup of
E/I and no involution of E - Z is conjugate to an involution of Z,

we must have [E : I|§q. Therefore, E = I x Z.

We have 5., E] = [3, Z-EZ since R normalizes E and hence
Z. Since R normalizes Z, R normalizes U by (2.7)and we conclude
that a normalizes I = EU. From [2, a, YIEZ, I]: 1 and Er, E, RJC

38

[:E, RIZ, it follows from the Three Subgroups Lemma that LR, Y, Ep
2 . Since R normalizes Y, El, TI! and by the structure of Y we

have 51, IKE. Therefore, fit, I] =EY, R, er so that R centralizes
I/I(Y) by (3.14). This implies that R centralizes Y, a contradiction.

we conclude that cG(t1)flcG(t2) is 2-closed with Sylow 2-group E.

(3.19) Let t1 be an involution of T1 - z(x) with (5, 2)
entry non-zero. Then CG(t1) is 2-clesed with Sylow 2-group Z(X)S°.

2292:. By the proof of (3.17), cH(t1) = Z(Y)So. If P
is a Sylow 2-group of CG(t1) containing Z(Y)So, then P contains
Z so belongs to H by the remarks preceding (3.17). We conclude
that P = Z(Y)S°.

Suppose that CG(t1) contains a Sylow 2-group Q distinct

from Z(Y)S°. Then Q contains a conjugate z of 2 such that zf'lz1 = 1

l
and 21 lies in the center of Q. Let t2 be any involution of T2 - Z(X)
and let 21 be any involution of 21. By (3.12), t2 and zl are not
conjugate. By (2.1), there is an involution w belonging to <z1, t2),
such that zl, a = [by w] = l and wzl is conjugate to 31 or t2.
The subgroup <21 , t2) is contained in CG(t1) so that w centralizes
both t1 and t2. By (3.18), we E.

If wzl is conjugate to zl, then wzlezl since wz:l 506(21)
and no involution of CG(21) - Z1 is conjugate to an involution of 21.
This implies that w 021, a contradiction to ZInE = 1.

Therefore wz1 is conjugate to t2 so that wzl = tzg for

some gQCG(t1)e This implies that ‘1‘ CG(t1)rF6(t23) =

39

(UG(t1)ncG(t2))g so that zlcsg. This implies 21 = zg. Since

wzl = tzg, w = tzgzl = (t2z)g for some involution 2 OZ. From

the fact that w is an involution of E and théTz - Z(X), we have
ZCCG(w)nCG(t1) = (C(;(tzz)r.)(3c'(t]_))g which'implies ZCEg. Therefore,
Z = Z8 and we have Z = 21, a contradiction to the choice of Zl.

We conclude that CG(t1) is P-closed with Sylow 2-group Z(Y)So.

(3.20) G contains no conjugate of Z different from Z.
w. Let 21 be a conjugate of Z different from Z and
let t1 (T1 - Z(X) be an involution with (5, 2) entry non-zero.
If 251 is an involution of 21, then 21 and t1 are not conjugate
and there exists an involution w such that El, w] = El, zfl = l
with wzl conjugate to zl or t1. Since w centralizes t1, weZ(Y)S°
by (3.19).
Suppose wzl is conjugate to 21. Then wzlé 03(21) and
we have w 621 since no involution of CG(zl) - 21 is conjugate to
an involution of 21' This is impossible since ZInZ(Y)SO = 1.
Therefore, wzl = tlg for some géG. This implies that
zlécG(t18) which is 2-closed with Sylow 2-group (Z(Y)So)g. Hence,
Z1 = Z8 and w ~= tlgzl = (tlz)g for some involution 202. Since
tlz is an involution of T1 - Z(X) with (5, 2) entry non-zero,
ZQCGW) = (CG(tlz))g which implies Z = Zg. We have Z = 2’1’ a contra-

diction to the choice of 21. We conclude that Z is normal in G.

(3.21) H is normal in c and lo : HI divides q-l.
21391. By (2.7) and (3.20), H is normal in G. Since

G/H acts semi-regularly on 2*, IG/HI divides q-l.

4. The Weyl Group.

In this section we assume the hypotheses of theorem 1.
From the results of section 3 we may assume NG(X) is not 2-closed
so that the structure theorems of section 2 may be used. It is
shown underthese hypotheses that G contains three involutions
w1, w2 and w3 which belong to NG(X). The relations among these
involutions are found and the group generated by "1 , "29 w3 is
determined to be the group 0ct3 which is the group of all symmetries
of the regular cube. It will be shown in section 5 and 6 that

W = Oct3 is the Weyl group associated with a (BN) - pair for G.

(4.1) For suitable 25 Z, the involution "1 = wlz belongs
to NG(K).

2229.:- By (2.13) and (2.14), NG(D)/D is a (TI) - group
with noncyclic abelian Sylow 2-group S/D and is not 2-closed. By
(2.8), NG(S£NG(D) and (2.4) implies K normalizes exactly two

Sylow 2-groups S and S of NG(D). From (2.2) and (2.3,11), NG(D)

1
contains a normal subgroup L of odd index such that L/D?! L2(q)
and KnL is a cyclic group of order q-l which normalizes exactly
two Sylow 2-groups of NG(D). Furthermore, NG(DCNG(Z) by (2.13)
so that all Sylow 2-groups belong to H and L =¢, $1) is a sub-
group of H. We conclude by (2.19) that KnICKnH = K1 x K2.

If :1 is the involution given in (2.14) 9 then 3101130)

41

and 31 belongs to L since L is nomal in 110(1)) of odd index. It

is not difficult to calculate directly that 31 normalizes K1 x K2

so that we have :1 normalizes Lml x K9 = LnK. By the structure

of L2(q) , KnL normalizes exactly the two-Sylow 2-groups S and SE1

of NG(D). This implies that S and 81—: Swl are the two Sylow 2-groups
of N G(D) which are normalized by K and (2. 3.11) implies NG(S)nNG(S"1)=
KD- - x :10 so that led = K for some d en. Therefore, wld belongs

to NH(KCHH(K1 x K2) and we have d 6113(K1 3: K2) = Z. We conclude

that "1 = Glz QNG(K) for some ze Z.

(4.2) For suitable z sz, the involution w2 = fizz belongs
to NG(K)e

mg. Consider the involution 32 (2.11.). Since NG(Y)/Y
is a (TI) - group satisfying the hypotheses of (2.2), the proof
is almost identical to the proof of (4.1).

(4.3) NG(X) contains an involution w3 such that wBéNG(K)
and w3 does not belong to the subgroup generated by the involutions
"l and 1:2 given in (4.1) and (4.2).

232;. If NG(K)nNG(X) has odd order, then (2.20) implies
(NG(K)nNG(X))X/X has odd order. By (2.16) and (2.1.). NG(X)/X contains
an involution tX which normalizes KX/X so that K = Ktx for some xeX.
The involution tX belongs to (NG(K)nNG(X) )X/X and we conclude that
NG(K)nNG(X) has even order. Therefore NG(K) nNG(X) contains an
involution w3. If w3 belongs to the subgroup generated by "1 and wz,
then w3 6H and we have that w3 normalizes S = UK, a contradiction to (2.20).

(4e4) (V1W2)4 9- 1
mg. By (4.1) and (4.2), (w1w2)4 = (31:2)4 = 1.

2 ..
(4.5) (vlvp - 1.
21391. By (4.3), w3 normalizes X and hence Z(X). Since
wl = :1: centralizes Z(X), wawlw3 centralizes Z(X). This implies
w3w1w3 ¢0H(Z(X)) since ZCZ(X). If h = uho, u s-U, hog-Ho belongs
to CH(Z(X) , h° must centralize any involution of Z(X) with all entries

zero except the (5, 2) entry. By the structure of Sp4(q), ho
belongs to NH°(U°) [15] . This implies that ho normalizes D = UUo
and hence, cH(z(x)X::NH(D). Therefore, S"3"1"3 is a Sylow 2-group
of NH(D) which is nomalized by K. By the structure of NG(D) ,
Sw3w1w3 must be S or 8"1. If S = Sw3w1w3’ then w3w1w3¢NS(K) = 1’
a contradiction. We conclude that Sw3v1"3 = 3"]- and hence (w3w1) 2
belongs to NG(s)nNG(x) = it.

Since w3w1w3t-H and wleH, (w3w1)2(-CK(Z) = K1 1: K2.
Since (w3w1)2‘ K, (w3w1)2¢. NG(X) so that the involution w1w3w1C-NG(X).
By the structure of NG(X) (2.17) , wlw3w1 belongs to the subgroup
L of odd index in NG(X). This implies (w3w1)26 KnL = K3 (2.19).

We conclude (w3w1)26(K1 x KMK3 = 1.

(4.6) Let E be the subgroup of S defined in (3.15).

Then E is the unique elementary abelian subgroup of S of order q6.

M. By (3.15), E is a normal subgroup of S which is
elementary abelian of order (16. It is not difficult to see that E
has a complement '1' in S such that S = E 9 T, EnT = 1.

43

 

r = E
81 1
(4e?) - 8.2 a 1
l
a 1

 

where b2 = a2 + aal.

Every involution of T must have the (2, l) or (3, 2) entry zero

so that by direct calculation ICE(t)L§q4 for all involutions t in T.
Let 31 be an elementary abelian subgroup of S of order q6

such that E # El‘ Then El contains an involution e1 Such that

s1 = et where e GE and t is an involution of T. This implies

that 1301-31 centralizes t and by the above commentlsnallsq4.

Since every involution of T (4.7) has a1 = 0 or a = 0, a maximal

elementary abelian subgroup of T has order q2. This implies

IEEl/E|$q2 and hence uni-42¢. Therefore, (3033] = q" and the

orders of Z(X) and Z(Y) imply that EnEl must contain an involution e

which does not belong to Z(X) or Z(Y). By (1.19), (Cs(e)l$q7 which

implies that [3031(qu since a and E1 belong to Cs(e). This is

a contradiction and we must have E1 = E.

(4.8) (W2V3)3 = 1.
This is the most difficult relation and is proved through a series
of lemmas.

(a) ZVBV2V3CE.

M. By (4.2) and (4.3), the involutions w2 and w3
normalize I and X respectfully. Since both X and Y contain 3,

w3 and w2 normalize E by (4.6). The result follows since E contains Z.

44

V w2w3

(b) no involution of Z belongs to Z(X).

Progf. 'By (4.3), w3 normalizes K and belongs to the sub—

group L ef odd index in NG(X). This implies that w3 normalizes KnL =
w .
K3 and hence K3 normalizes Z 3. Since K3 acts regularly on Z*, K3

w
acts regularly on the non-identity elements of Z 3. We conclude
w w
that zflz 3 = 1 since w3 does not normalize Z. Therefore, Z 3 is
w
contained in Z(X) - C and every involution of Z 3 has (5.2) entry

nonpzero. The involution'wz interchanges the (5, 2) and (4, 3) entry
w w
of any involution of’E so that every involution of Z 3 2 has (4, 3)

entry nonszero and does not belong to Z(X). If an involution of

Z belongs to Z(X), then Z(X) contains an involution of Z

since w3 normalizes Z(X). This is impossible and the result follows.

w3w2w3
(c) No involution of Z has both its (4, 1) and

(4, 3) entry zero.

"392113 V3V2W3 w

nggf. Suppose z is an involution of Z ith

WBV2V3
both the (4, 1) and (4, 3) entries zero. By (a) and (b), z
w w w
belongs to T1 - Z(X) so that Cx(z 3 2 3) = Z(Y)S°. By the proof
w’w
of (b), 2 3 2 has (4, 3) entry nonszero. By direct calculation,

w w
2 .
Cx(z 3 ) = E. This is a contradiction and (c) follows.

(w3‘w2)2
(d) 2 C200 - c.

ngof. By (e), Z contains an involution of E given by,
H T
1
(4.9) l
83 b d 1
I." c b 1
Ljr a4 a3 1‘ with a3 # 0 or d # 0.

 

 

45

w w w
Since K1 (1.13) normalizes Z 3, by direct calculation we see that
V V2" .
Z 3 3 must contain exactly the q-l involutions belonging to the orbit

"3‘12"

This implies that Z contains

of the involution given in (4.9).

q-l involutions of the form (4.9) with a4 -

 

 

c=t= o. SinceK2 (1.13)

 

 

Z""3 2'3 "3" 2"3
normalizesZ , a similar argument implies that Z contains
the q-l involutions,

- -1
1 i
1 <1
e 1 1‘1 -Q'le
- l
L a3:{ 1
WBWZVB
By (b), no involution of Z ( gen belong to U. Therefore, d # 0
w‘w
in (4.10) and we have that Z 3 contains the q-l involutions,
1 Kigq-l.
1
(‘e11) . ‘
.131 at l
L 53" 1
This completes the result (d).
(w w )2w
(e) Z 3 2 3 = Z. 2
(w W2)
Pzgof. Let t be an involution of Z of the foul

(4.11) .

zero except the (5, 2) entry and veV.

w
involution of Z

w
of z 3

involution z G2 and 3v1

implies that zl"lz(3

acts transitively on Z“, the result (e)'follows.

have the same (5, 2) entry.
01’.

1:2)2w

Then t = jv where j is the involution of Z(X) with all entries

By the proof of (b), every

Therefore, t w3=

This implies z"3 =

3 belongs to Z(X) - C so that no two involutions

jvl for some

jw3v = zvlvi'C This
$13

(93‘42)

3 # 1. Since K normalizes Z 3and

46

(f) (w2w3)36 x.
2.2221:- 'By (a), (w3w2)2u3 eNG(z). This implies that

(w3w2)2w3 3normalize U and E and hence I = UE. By the structure

(w w )
of NG(Y) , S 32 3 is one of the two groups 8 or 3V2 which belong
w )
to Ng(Y) and are normalized by K. If S 3 3 = s, then (w3w2)2w3
WV V2
belongs to IVG(SENG(X). This implies w2w3w26lc (X) so that Z "2 3
G“ 2% VBVZ
is contained in Z(X). This is impossible since 2 = ( )2”
w w
. 2
is not a subgroup of Z(X)by the proof of (b). Therefore, 8 3 3 -
V2
S so that (w2w3)3¢NG(S)nNG(K) = K.
3 ..
(8) (“2"3) - l
"3W2

2:99;. By (f), (w2w3)3(-K so that (w2w3)2w2 = w;
belongs to NG(X). Since (w2w3)2w2 and w3 are involutions of NG(X),
they belong to the subgroup L of odd index in NG(X). This implies
(w2w3)3e KflL- = K3. Similarly, (w2w3)3 belongs to 116(2) so that
(w3w2) 2w3 = "3w2w3 is an involution of H. This implies (w2w3)3
belongs to “111 = K1 x K2. Therefore, (w2w3)3((K1 x 1901(3 = l.

47

5 . The Subgroup Go .

In this section we continue the proof'of theorem 1 by
constructing a subgroup Go of G which is isomorphic to Sp6(q).
We will show that G0 has a (BN) - pair in the sense of Tits [18]
with Weyl group the octahedral group 0ct3. By a theorem of Tits
[19] , Go is shown to be a simple group. Finally, Go is identified
with Sp6(q) by a theorem of Tits [20]. This last theorem of Tits
is part of a large body of semi-published.material which identifies
all finite simple groups with (BN) - pair and Weyl group of rank at
least three with an appropriate known simple group.

First, we list nine elementary abelian subgroups of S.
Each of these subgroups has order q and together they generate S.

The non-zero entries of these groups are arbitrary elements of F.

 

 

 

 

U1 = .1 U2 =1. 1 U3 3t —1[
1 l 1 1
1 32 1 a 1
1 l l
1 l a 1
a1 1 a2 14 L l_l
U - U = 1 - U = E w
l‘ l 5 1 6 l
l 1 1
3 l l l
1 84 1 1
a3 1. _ IL 1- L! l-J

 

 

 

 

 

U -T - U =r q’ U =fi" .-
7 1 8 1 9 1
1 1 l
b 1 l d 1
b 1 c 1 l

 

 

 

 

 

 

Let "1 and w2 be the involutions given in (4.1) and (4.2).
These involutions belong to H and the action of these involutions
on the above subgroups is known. In the following table we list
the action by conjugation of wl and "2 on the groups U1, 1 = l, ...,

9. For example, in the row corresponding to U3 and the column contain-
"1

 

ing “1 we list the group 0-; which means U3 = U7.
"1 m2
U1 U1 02
u2 UL U1
U3 U7 not in S
(5.1) U" 02 Us
U5 U5 U4
U6 U6 U6
U7 03 U7
08 U8 U9
% mus %

 

 

(5.2) Let U4 be the subgroup of S given above. Let w3

W
mmummmgmmum.mwfi=%.

Proof. By (2.17) , w3 centralizes “5 = V. Since (w2w3)3 = 1

V21! w2w3w2
(4.8), t 3 = t for all involutions tGUS. This implies

A9

w2 centralizes U5 . Since w2 and w3 normalize E, U5 is a
w w
subgroup of I. Let t be an involution of U5 2 3given by,

 

 

 

t = "'1 '1
1
l
(5.3) a3 b d 1
a4 c b l
a 8 1 e
. J z. 3 _
Pram the relation wzth = t, we must have a3 = e.4 and d = c in (5.3).
w w -i 1
Since K normalizes U5 = V, K normalizes U5 2 3 and we see k1 t k1

V2“
belongs to US 3 for lgi‘q-l. This implies a3 = at. = c = d = O

in (5.3). If b = O in (5.3), then t is an involution of Z, a contra-

diction to the fact that no involution of C - Z is conjugate to an

w w
involution of Z. Therefore b # 0 in (5.3), so that U5 2 3 consists
-i i w w
of the q-l involutions k1 t k1 . It follows that US 2 3 = U7 so

"3
that U4 = U7 by (5.1).

'3
'3

(SeS) 03 3 U2

» 2 "3 "1"3
mg. By (4.5), (w1w3) = 1 so that 03 = U7 =
‘ijl "1
U7 = U4 = U2 by (5.1.) and (5.1).
Let “’19 i = l, 2, 3 t be the set of involutions of N60!)

which satisfy (4.1.), (4.5) and (4.8). Set,

N=4,w1,i=i, 2, 3> ,B=NG(S).

50

(5.6) Set 0,, = BNB. Then a, is a subgroup of G which
has a (BN) - pair' (B, N) with Weyl group the octahedral group 0ct3.
M. We will prove that Go is a subgroup of G by showing,
tlBtZCBNB for t1, téew = 61, w2, $13).
Since B = SK and t1 GNG(K) we need to consider only tlStZ. Since
t1 is a product of w's, we may assume that t1 is one of the generators
w1, w2, or w3. By (2.7), H is a normal subgroup of NG(Z) of odd index.
The Frattini argument implies that NG(Z) = HB so that the subgroups
B and Q1, w1, w2)form a (BN) - pair for NG(Z). This last statement
may be verified from the observation that H = u . so where Ho 9.! Sp4(q).
This implies that

(5.7) rch; BwSU BrwS
for any we<w1, w2> and r = "1 or r = w2 [18] . Since the group H
generated by w1, "2 and w3 is the octahedral group of order 1.8, the
dihedral group generated by "l and w2 has index 6 in W and elements of
U have the fem

w, W3, wwawz, wwawzwl, w3w2w1w2, ww3w2w1w2w3
where we <w1, w» .

Letr = 1.11 or r = "2' By (5.7) we have

rSquBwSu UBrwSu,
where u = w3, w3w2, w3w2w1, w3w2w1w2, or w3w2w1w2w3. Hence it suffices
to show that

wSuC Bqu

for w Q 61, w> and 1.1 one of the above coset representatives.

51

(i) u = w3
waB = «1ow = wamfi since wBONGOK) , w 9NG(Z) = NG(U) .

(11) u = V3U2
waBwZCwaBSwZ by (i)
= waBU3Iw2

wa3U3wZY since w2 GNG(Y)

UwU2w3w2! by (5.5)

CJJW3w2Y since U2CU.

(111) u = w3w2w1
waBwaICwaBwZSwl by (ii)
= wa3w2Ung1
= wa3w2U9wID since w1 C-NG(D)
= UW3U8w2wID (5.1)
CUwOwazwID since UBVCC by the argument (4.8, e)
Cwa3w2w1D since CCU.

(iv) u = wawzwlw2
wa3w2w1w2Qwa3w2wISw2 = UW3w2w1U 3Iw2
- = UW3w2w1U3w21 since wZQNGU)
= UW3U7w2w1w2Y (5 .1)
= UwULw3w2w1w2! (5.1.)
CUW3w2w1w21 since UFU .

(v) u = w3w2w1w2w3

wa3w2w1w2w3 CwaszwleSwB = wa3w2w1w2U11w3

52

= wa3w2w1w2U1w3X since w3 5110(1)

= M3U5'2'1'2'31 (5.1)

= UwU5w3w2w1w2w32 since w3 centralizes U5 = V
CUW3w2w1w2wBX since 05 CU.

We conclude that
(5.8) :swCsuB U BrwB

for r = '1 or wz and weéll, w2, w3>. It remains to show (5.8)
for r = w3. Let we <w1, w2> , and let u be one of w3, w3w2, w3w2w1,
w3w2w1w2 or w3w2w1w2w3.

(i) u = w3

w33w3 = w3XU1ww3

Clw3w1w3 for w ,5 l and w # "l
= Xw3ww3X since w3 eNG(X) .

By the structure of N60!) , NG(X) = LB where L is a normal subgroup
of odd index such that m ‘=—‘.' L2(q). By (2.3, v), new) = BUBwas.
This implies w3Sw3CBw3S and w33w1w3 = wBSwaICBUBSwl = Dd3U1U1 =
ngwlUl. '

(ii) u = w3w2
w38ww3w2CBw3ww3Sw2U wafiwz by (1)
Consider the coast Bw3ww33w2 since wa38w2 is similar.
B"3"W33W2 = la"3"""3“33“'2
= B“"AWBUBWfi
= ar3wUzw3w2Y (5.5)
CBw§wayt2¥ for w# w2, w 5.! w1w2

53

= BwawaZI
The cases for w = u, and w = w1w2 are Just a little different.
w3Sw2w3w2 = w38w3w2w3cw35w2w3
= Bw3U1w2w3

w3Sw1w2w3w2 = w33w1w3w2w3

= "33'3"1"2"3

C313U1w1w2w3

O'l"2"3"2x
This shows that w38w3w2Cwa3wzsqu3ww3w23 for all w e 611, w2> .
The arguments for u = w3w2w1, w3w2w1w2 and w3w2w1w2w3 can be proved
by using only the relations among w1, w2, and w3 and the information
found in (5.1). (5.2). (5.4), and (5.5). This completes the proof
that Go is a subgroup of G and in fact we have shown that Go has

a (BN) - pair in the sense of Tits [18] .

(5.9) Go = BNB is simple.

2m. Let Bo be the intersection of all conjugates of
B in so. If ant = 1, then 130 centralizes c and belongs to H.
Since 3/0 ‘=" Sp 4(q) , so is a normal subgroup of s and must intersect
C non-trivially. Therefore, Bone 3i 1 and Bo contains an involution
t of C.

mu

Suppose that t belongs to Us = V. Then t st

belong to 8 since no is a normal subgroup of Go and 30:13.

54

w w w2 wgwlw w w
33' (5.1), t 2 1 {-U1 so that t 2 36-8 3 - X. This implies
wzw w w w
t l 2 3fs since SnS 3 = X by the structure of NG(X). This is
a contradiction and t can not belong to Us = V.

Suppose that t = vz for some véV“, zéZ". Then t =

vw2w1WZW3 2“3 does not belong to S since vw2w1w2w3 S. Therefore t
w w2w
must be an involution of Z. However, t does not belong to
w3 w3w2wl
S since t éZ(X) - C and (5.1) implies t lies outside S.
We conclude that B0 = 1.

Let Gl be the subgroup of Go generated by all conjugates
of S in Go. By the structure of NG(X) , <3, Sw3‘> = L is a normal
subgroup of odd index. This implies that w3 and K3 = KnL belong
to 6-1. Similarly w2 and K2 belong to <3, Sw2> and wl, K1 belong
to <3, Sw1>. Therefore, Gl contains B and N so that Gl = 60'

Let U1, 1 = 1, 2, ..., 9 be the nine elementary abelian
subgroups of 8 listed at the beginning of this section. If t is an
involution of U4, it is not difficult to verify that t = u1u7u1u7
for apprOpriate involutions uleUl and my (’U7. This implies that
U4CGi. If t is an involution of U5, then t = u2u7u2u7 for appropriate
involutions u2 6U2 and u7 GU7 so that UgG]... The proof of (3.11.)
shows Z = U6CGi.

If t is an involution of U2, t = (u1u3)2 for appropriate
ulflUl and u3eU3. Hence UQCG-i. If t is an involution of U3,

t = w3u2w3 for some involution u2 eUZ. This implies tug 06-1 so that
tl-Gi since Ugci. Hence U331.

"2 t t
By (5.1), U2 = U1 so that Uzcol implies that Ugcl.

55

w
By (5.1.), u4 3 = U7 so that {14:51 implies U7CC-i. Every involution
w w

of Z 3 = U6 3 belongs to Z(X) — C so that if t is an involution of
w

U8, 2 3 = tu5 for some involution zeZ and u5 e-US. This implies

"3
z

I . '
z tusz (Ci and hence t 601. Hence U8C:Dl. Since (5.1) implies

Uswz = U9, Uchj'L- We conclude that each of the U1, 1 = l, 2, ..., 9
belongs to Gi and hence SQ;61. Therefbre, Gi = Cl = Go. Since the
set of distinguished generators + wl, w2, w3 } is not the union

of two commuting proper subsets, we may apply the result of Tits
[19] . By this theorem, every normal subgroup of Go is contained

in B0 or contains 01. Therefore, Go is simple.

"I

(5.10) so Sp6(q).
ngof. By a theorem of J. Tits announced in his note
[20] , a finite simple group with (BN) -pair of the type presented

in (5.6) is isomorphic to Sp6(q).

56

6. Identity G = Go,

In the preceeding section we established the existence
of a subgroup G0 which is isomorphic to Sp6(q) and is generated by
B, w1, w2, and w3. Although it is known that Sp6(q) contains four
classes of involutions, this fact is not difficult to prove with
our present information about Go' A proof of this fact will be
given in (6.4) and will establish a technical lemma (6.5). Lemma
(6.5) will be used to calculate the centralizers in G of the four
involutions which represent the feur distinct involutionary classes
of G. A now standard argument credited to Suzuki and Thompson will

be used to conclude that G = Go.

Defigition. Let G be any group with X a subgroup. NG(X, 2')
is the set of 2'-subgroups of G which X normalizes and which intersect

X in the identity only.

(6.1) (Iamaki) NH(E, 2') is trivial.

ProOf. Let t1, t2, t3 be the involutions of E defined by,

1 1
1 l l

L} l) L‘l 1' L}
Then <tlt2, t2t3> is a 4-group belonging to E. The involution

(.1
...:

F'P‘ F4

PJFJFJ

 

 

 

 

 

 

“-4

tlt belongs to T - Z(X) with (5, 2) entry non-zero so that the

2 1

57

proof of (3.17) implies CH(t1t2) = Z(Y)So.

If Bought, 2'), then <t1t2, t2t3> normalizes R and
acts fixed - point - free. The Brauer - Wielandt theorem,[2l:]
implies R = CR(t1t2)CR(t1t3)CR(t2t3). The involutions t1t2,ht2t3

2 _ _
and t1t3 are conjugate in H since (tltg) - t2t3 and (t2t3) — tlt3
where w2 is given in (4.2) and h is the involution given in the
proof of (3.1). Since CH(t1t2) is a 2-group, CR(tlt2) = CR(t1t3) =
03(t2t3) = l and we conclude that R = 1. This implies NH(E, 2')

is trivial.

(6.2) If two involutions of E are conjugate in G, they
are conjugate in NG(E).

2199;. Let u, v be two involutions of E conjugate in G.
Then u = g-lv g for some g eG. This implies E and 1".)8 belong to CG(u).
If P and Q are Sylow 2-groups of CG(u) containing E and EE respectfully,
then P = Qy for some yQCG(u). By (4.6), E is the unique elementary
abelian subgroup of P of order q6 so that E = Egy. Therefore,
gy ONG(E) and y'lg'lv gy = y'luy = u so that u and v are conjugate

in NG(E).

(6.3) (Yamaki) NG(E, 2') is trivial.
Prat. Define s1, 82, SB, 84 to be the involutions of E,

F' fl -
81: 1 82: F1 j 33: q
1 l 1
1 l 1
l l 1 1
1 l l 1 l
1 1 111 1 ll l-J

 

 

 

 

 

 

58

 

 

s = _1 '—
4 1
l
l l
1
L} l l_‘ .

Consider t1, t2, t3 presented in (6.1) and the six four-groups

4319 slt1> 9 <31, 81t2>, <31, 311:? , él, tlssz> ,
<1. 8413283), <31, 8432t>. If RGNG(E, 2'), these six four-
groups act fixed - point - free on R since 3162 and NH(E, 2')

is trivial. The Brauer - Wielandt theorem yields,
ca(sl)cR(slt1)cR<t1)
CR(sl)CR(slt2)CR(t2)

CR(SI)CR(31t3)CR(t3)

R

ca(sl)ca(t13233)CR(84t1)

= CR(91)CR(34t233)CR(°2t2)
The group Go is isomorphic to Sp6(q) and therefore contains a sub-
group isomorphic to Sp6(2). This implies 31’ sltl, 81t2’ slt3,
1.13sz, sktzsa and s432t3 are conjugate in Go (Refer to Yamaki [22] ,
Lemma 13). By (6.2), these six involutions are conjugate in NG(E).
Let gGNG(E) such that sltl = s11g and hence CR(slt1) = Cable)-
Since CR(slg) = Ran, CR(slg) QNHgUE, 2') = l by (6.1). Therefore,

1. A similar argument yields CR(sl) = CR(slt1) = CR(slt2) =

CR(81t1)
“R‘slt3)
have R = CR(t1) = CR(t2) = CR(t3) = CR(84t1) = CR(82t2) = 03(83133).

This implies that R centralizes the subgroup of 3 generated by
6

03(t18283) = CR(34t2s3) = CR(8492t3) = 1 so that we

t1, t2, t3, s4t1, sztz and 33t3. This group has order 2 and

59

contains all involutions of E with entries in the field of two
elements. We conclude that R centralizes an involution of Z so that

R‘NH(E, 2') = 10

(6.4) G0 has exactly four classes of involutions.
Every involution of Go is conjugate to one of the four involutions

2, v, vz, or j listed below.

 

 

 

 

z = _1 -1 v = Fl .-
1 l
l 1
l 1
l 1 1
h} 1‘ __ l l_
vz = l .1 j = F1 _1
l 1
l 1
l l 1
1 1 l l
L} 1 1. e 1..

 

 

 

 

2222;. By the structure of Go, B = NG(S) is a subgroup
of Go° By the structure theorems (2.7), (2.15), (2.17) and (2.19),
NG(S) = SK where K normalizes the groups Z and V belonging to the
center of S. The group K acts transitively on both 2* and V“ so
that all involutions of these groups are conjugate to one of z or
v above. Let t be an involution of C of the form t = vz where
2 (2* and v .V'.

kékgf kzkin: k3 are the generators of K2 and K3
respectfully, t = v 22 ranges over all involutions of C - (ZLJV).
This implies that every involution of C is conjugate in Go to

2, v or vz above. By Burnside's theorem, 2, v, vz are representatives

of three distinct classes of involutions in G as well as Go'

60

By (3.2), every involution of S is conjugate to an invo-
lution of T1 or T2} Since T1 = [172(1) and T2 = 0420!) , 1'2"} = T1
by (5.1.). It remains to show that every involution of T1 is
conjugate in Go to one of the above four involutions.

The structure theorem (2.17) shows that N00!) contains
a normal subgroup L of odd index such that L/X 3" L2(q). By (2.3) ,
L contains exactly q+l Sylow 2-groups of N600 and the argument
(2.17) implies that z(sl)nz(sz) = v for any two Sylow 2-groups
81 and $2. This implies that each Sylow 2-group 81 distinct from
s contributes set theoretically q2 - q involutions to z(x) - G.
Since lZ(X)| -|C I = c13-q2 = q(q2-q), every involution of Z(X) - C
is conjugate to an involution of C. Therefore, every involution
of Z(X) is conjugate in Go to 2, v or vz above.

It remains to show that every involution of T1 - Z(X)
is conjugate to one of the above representatives. Let t be an
involution of T1 - Z(X) with (6, 1) entry zero. Then t = u7u3v
where 1175117", “8 ('08 and 70V. This implies t“3 e U - Z by (5.1.)
and the fact that “8'3 is a subgroup of C. By (1.16) , t is conju-
gate to an involution of C - Z so that t is conjugate to v or v2
above.

We may assume that t is an involution of T1 - Z(X) with
(6, 1) entry non-zero. For appropriate “2902, “2W2 has (5’ 1)
entry zero . The involution t may be given by,

61

 

 

t = '1 T
l
l
b 1
c b 1
I 14
where b a»! o and x ;£ 0. Since Go 3' Sp6(q), let K3 = (1:3) be
given by f- "1
’ k3 = 4
cl"
at
.C'

 

 

where A is a generator of F". It is not difficult to see that
K3 is a subgroup of Sp6(q) and may be identified with a subgroup
of Go° Conjugation of t by an appropriate element of K3 implies
that t is conjugate to an involution with (6, 1) entry 1. Conju-
gation of t by appropriate elements of K1 and then by K2 imply

that t is conjugate to one of j or 31 where j is the representative
given above and jl is the involution of T1 - Z(X) presented in

(3.3) with b = I = l and a4 = c = 0. Consider the involution a- ,

6":

El

HHHH

 

b

1
l

1
1

l

1

1
l

 

The involution 6' belongs to Sp6(q) and may be identified with an
involution of :3. If u2 is the involution of U2 with (3 , 1) entry
2
1, then j = jl . We conclude that every involution of T1 - Z(X)
with (6 , 1) entry non-zero is conjugate to j in Go' This proves

that every involution of S is conjugate in Go to one of 2, v, vz or 1.

((((( {itifillfl
1((((((((( I

 

 

 

62

By Burnside's theorem, z, v and v2 are not conjugate in G and
hence are not conjugate in Go. In Go, the involution j is not
conjugate to 2, v, or vz since j - I is a matrix of rank 3 and

z - I, v - I and v2 - I are matrices of rank 1 or 2. This proves

(6.4) .

(6.5) Let 2, v, vz and j be the representatives given
in (6.4). Then all involutions of Z(X) are conjugate in Go to
v, vz or 2. All involutions of T1 - Z(X) with (6, 1) entry non-
zero are conjugate in Go to j. All involutions of T1 - Z(X) with
(6, 1) entry zero are conjugate in Go to v or v2.

Prggf. This follows directly from the proof of (6.4).

(6.6) G has exactly feur classes of involutions with
representatives of each class given in (6.4).

2392;. we have established in (6.4) by the use of Burnside's
theorem that the involutions in the center of a Sylow 2-group of
G are partitioned into three conjugacy classes with representatives
z, v, and vz. It remains to show that the involution j is not
conjugate in G to one of 2, v or vz.

Suppose that j is conjugate to v or vz. Let xeNG(Z(I)S°)
and consider zx where z is the involution given as a representative
(6.4). Since T1 is the center of Z(Y)S°, zx is a conjugate of z
in T1. By (6.5), 2x must belong to Z(X). By the structure of
NG(X), zx lies in the center of a Sylow 2-group of’NG(X) and hence
there exists g QNG(X) such that 2“ = z. This implies xgc-H so

63

that xéHg-l. Since NG(X) = <B, wflGo, we have X€Go and hence
NG(Z(Y)SOCGO.
Suppose that j is conjugate to 2. Let xeNG(Z(Y)So)
and consider the involution) zx which belongs to T1. First assume
that zxéTl - Z(X). By assumption zx is conjugate to j so that
(6.5) implies that the (6, 1) entry of 2x is non—zero. If v is
any involution of V, (vz)x = vxzx is an involution of T1. If
vx¢Z(X), then va V since V is a normal subgroup of NG(X) and
all involutions of Z(X) are conjugate in NG(X) to one of the three
involutions 2, v or vz. This implies that vxzxe T1 - Z(X) and
has (6, 1) entry non—zero. By (6.5), we have that j and vz are
conjugate, a contradiction to our assumption. Therefore, Vx must
belong to T1 - Z(X). ChooSing vx in Vx such that vx and zx have
different (4, 2) entries, vxe T1 - Z(X) and has (6, 1) entry zero
by (6.5) and hence (vz)x = vxzxé T1 - Z(X) with (6, 1) entry non-
zero. By (6.5), j and vz are conjugate, again a contradiction.
'Je conclude that 21‘ can not belong to T1 - Z(X) and hence zxéZ(X).
By the above argument, this implies x600 and hence NG(Z(Y)S°CG°.
Therefore,if j is conjugate to any of 2, v, or vz in G,
NG(Z(Y)SO)CG°. Let P be a Sylow 2-group of CG( j) containing
CS(j) = Z(Y)So. Since Z(Y) is a subgroup of U, (1.16) and (6.4)
imply that j is not conjugate in Go to an involution of Z(Y) or
Z(X). By (1.19), Z(Y)So is a Sylow 2-group of CG°(j) so that
PnGo = Z(Y)So. If j is conjugate to 2, v2 or v in G, then P is
a Sylow 2-group of G and P contains Z(Y)So properly. This implies

NG(Z(Y)SO)¢G0 , a contradiction. The result (6.6) follows.

64

(6.7) CG(szH.

2292;. Let xeCG(vz) and suppose zxeS. By (3.2),
and) belongs to T1 or T2 for appropriate he H. By (6.5), no invo-
lution of T1 - Z(X) or T2 - Z(X) is conjugate to an involution
of 2 so s‘he z(x). This implies a“ = (vvz)xh = thvhz belongs

to Z(X) and hence v"h belongs to Z(X)U.

Suppose ach‘ Z(X). Then v11 = zthP‘hz belongs to Z(X) so

that both v"h and vh must belong to V since V contains all conjugates

of v in Z(X). This implies z:3th = vie for some v1 eV. Since no

xh

involution of C - Z is conjugate to z, z £- 2 and hence v1 = l.

x=

Therefore, th = z and we have 2 2.

Suppose that W‘s U. Then zXh = thvhz 6U so that the Z
by (2.6). This implies M = (vxv)h belongs to Z and hence 195762.
If vxv # 1, then vx = v21 for some involution 21 in Z and we contra-
dict the fact that v and v2 are not conjugate. We conclude that
vxv=lso that am: zandwe have zx=z.

If thc-zuw - (z(x)uU), then vKb is an involution
of the form,
.1 -

1

a2 1

83 l
(6.8) ‘4 o 1 0 fl 0.

sea 1
L__rz.32 _

 

 

By assulnption vxh does not belong to Z(X) so one of a2 or 0.3 is

non-zero in the eXpression (6.8). Let be be the element of HD

81793 by,

 

 

 

 

h. = '1 1
l
C M
D l
L. 1..
where M¢L2(q), ct = (c1, c2) and D = stag»: for A2 = [o ] .
1 0
For appropriate MOL2(q) , thhO is given by
pan, = '3 -
1
l
f l
f; c 1
f f 1
_¥ 2 1 .4)
where f1 # O and c 75 0. Consider the involution,
"3 = F 1 fl
1
1
l
l
__ l _ .

 

 

Since Go g Sp6(q) , w3 may be identified with an involution of Go
and by direct calculation vxhh‘w3 is an involution of T1 - Z(X)
with (6, 1) entry c ,4 0. By (6.5), thh°w3 and j are conjugate,
a contradiction to (6.6).

We conclude that VXhG Z(X) or‘vxheU and in both cases
2x = 2. By Glaubennan's theorem [7] , 262*(CG(vz)) so that
(3)02'(CG(vz))40G(vz) and the Frattini argument yields CG(vz) =
cam, z)02.(CG(vz)). Since mafia), 02.(CG(vz))6NG(E, 2').

By (603), CG(VZ) = (36("9 2010

(6.9) couch.
Proof. Every involution of Z(X) or Z(I) is conjugate

to one of 2, vs or v. By (1.19) and (6.6), a Sylow 2-group of

66

06(3) is 2005,. Let ”06(3) and suppose exezmso. If 2"
belongs to Z(Y)S° - B, then z’c has (4, 3) entry zero and is conju-
gate in NH°(Uo)to an involution of E with (4, 2) entry non-zero.
This implies that 2" is conjugate to an involution of T1 - Z(X)
which is impossible by (6.5) and (6.6). We conclude 2x6- E.

By the structure of NHO(To) [is] , there exists h in

NHO(TO) such that 2“ is an involution of E with (1., 3) entry

 

 

zero. y—u
23th: 1 .1
1
1
840 b 1
a a 1
U 4 3 .1

If b v! o in (6.10), then 2"" is conjugate by an involution of
U to an involution of T1 - Z(X). By (6.5), this is impossible
so that b = O in (6.10). Since every involution of T2 - Z(X)
is conjugate by w3 to an involution of '11 — z(x), a3 = o in (6.10).
This implies s‘hezu).

Let 117 be the involution of U7 with (4, 2) entry 1.
Then 21m = (jwfixh = jh'uqxh belongs to Z(X). By the structure
of NHO(To) [153 , h normalizes U7 and we have jh == (zu.7)h = nil,
where 37 is some involution of U7. This implies that an is an
involution of T1 - Z(X) and from a,“ = 2"th we so. that u,1th
belongs to T1 - Z(X) also. If the (6, 1) entry of n.7,Ch is non-
zero, then uqnth and j are conjugate by (6.5). This is impossible

"2‘13

since v = u7 and v and j are not conjugate. Thus the (6, 1)

entry of a,“ is zero. Furthermore, the (5, 2) entry of u71h is

67

zero since otherwise u7mé U anihcan easily be seen to be conju-
gate to vs. This implies that ugh = i thhere v (N and 3.7

is an involution of U7.XhSince z = j u7 = zu7v'u7 belongs

to Z(X), we must have 2 = 2;. Since no involution of c - z

is conjugate to an involution of Z, zXhe Z and we have ; = 1.
Therefore, th = 2 so that 2x = 2.

By Glaubermanis theorem, <z>02'(CG(j))QCG(j). Thus

00(3) = 06(5, z)0 (06(3)) by the Frattini argument. Since E
pt

centralizes j, 06(3) = CG(j, zgfl by (6.3).

V

w
2
(6.11) CH(v) and CH(v 3) are not isomorphic.

P3002. Since v is an involution contained in the center
of S, S is a Sylow 2-group of CH(v). The center of S is a sub-

group of U so that all Sylow 2-groups of B have centers contained
w w
in U. The involution v 3e U7 so can not lie in the center of
w w
2

a Sylow 2-group of B. This implies that CH(v) and CH(v ) are

not isomorphic since they have Sylow 2-groups of different orders.

VH2

(6.12) The involutions z andz 3 belong to E and

centralize v but are not conjugate in C (v).
"3'23 G
300:. Suppose z
w w g w wzg
(CG(v)nCG(z 3 2)) = CG(v)nCG(z 3 ) = cG(v)ncG(z) = cum.
However,

w w V V w w w
agency. 3 2) = (can 2 may.» 3 2 = (cHe'Z'b) 3”.

w2w
This implies Oak) and CH(v 3) are isomorphic, a contradiction

= z for some g eCG(v). This gives,

to (6.11). The result (6.12) follows.

(6.13) COGCGO.

2mg. Let g eCG(v) and consider the involutions z
and 21 = zw3w2. Both 2 and 21 belong to CG(v) and by (6.12) are
not conjugate in CG(v). This implies the group <2, zlg> is a
dihedral group of CG(v) and there exists an involution e in GG(v)
such that E, 2;] = Ce, 215 = 1 with ez conjugate to z or 215.

Suppose e and v are conjugate in G. The involution
ez centralizes 2 so belongs to H and hence (ez)h belongs to T1
or T2 for appropriate h EH (3.2). By (6.5), no involution of
T1 - Z(X) or T2 - Z(X) is conjugate to an involution of Z so
we must have (ez)h = ehz GZ(X). This implies ehGZ(X) and hence
aha-V since V contains all conjugates of v in Z(X). Therefore,
(ez)h z abs is an involution of C and is conjugate to an involution
of Z. This implies ehz &Z and hence ehOZnV a l, a contradiction.

Thus we may assume that e and v are not conjugate in G.
By (6.6) and (6.4), e is an involution of H and is conjugate in
Go to j, z or vz. For some yeGo,
can) easy) = (canntnycco by (6.9), or

y .

cG(e) Cc‘" ) = (CG(vz))t'Hy:G° by (6.7), or
can.) = age") = (06(2))y = HQ}.-

g
This implies zl QCG(eCG°. Since 21g can not be conju-

gate in Go to v, vz or j by (6.6), there exists gt-Go such that

8? _ w3w
21 = 21. This implies gg “C(21) = H too. Therefore,

g bGo and we have CG(v)G}°.

69

(6.14) G = Go

M. Suppose t is an involution belonging to G - Go.
By (6.6), Go contains an involution u such that t and u are not
conjugate in G. This implies the group <t, u) is a dihedral
group and there exists an involution w such that w, t] = E, E] = 1.
By (6.7), (6.9) and (6.13), Cg(u)CGo so that weGo since w.
centralizes u. Therefore, tQCG(wCGO, a contradiction to the
choice of t. We conclude that Go contains all involutions of G.
Since Go is generated by S, “'19 w2 and w3, Go is generated by
involutions and is thus normal in C. By the Frattini argument,

This completes the proof of theorem 1.

Index of Notations

Is a subgroup of

Is not a subgroup of

Is a normal subgroup of
Is isomorphic to

Is an element of

Is not an element of
Set union

Set intersection
Summation

Set difference

Subgroup generated by
Direct product of groups
The number of elements in G
The centralizer of t
Index of H in G

Factor group

xrlGx

xrlax

1.1“!”

Center of G
Frattini subgroup of G

Derived group of G

71
4-group Non-cyclic abelian group of 4 elements
NG(H) Normalizer of H in G
02. (G) Maximal normal subgroup of G of odd order
2*(0) Subgroup of 6 containing 02,(0) such that z*(e)/ 02, (c)

is equal to 2(0/02,(c))

BIBLIOGRAPHY

1. E. Artin, W M, Interscience, 1957.

2. R. Brauer and K. A. Fowler, 93 mg 2: mp 225122. Ann. of Math.
62(1955). 565-583. '

3. C. Chevalley, _Sg certains groupes sjgaoleg, Tohoku Math. J.
7(1955), 14-66.

4. L. E. Dickson, Ma: gm 319; g}; egositiog 2: 3.313 9.31212
M M, Leipzig, 1901. Dover edition, 1958.

 

5- . Ihee:z.2£ 112222 222222 is as 2:213:22: field.
Trans. American Math. Soc. 2(1901), 363-394.
6. , A as; gzgtgm 9; gm gzoups, Math. Annalen

 

60(1905), 137-150.
7. G. Glauberman, 9233331 W in mm gm, J. Algebra
4(1966). 403-420.
8. B. Huppert, m m 1, Springer-Verlag, New York, 1967.
9. 0. Jordan, gggité figs Substitutions, Paris, 1870.
10. E. Mathieu. uémsims as: liéisee see.isusiiese.ssnnlneisuze
ggggtitég, J. Math. Puree Appl. 6(1861), 241-243. .
u.___&mmmmmmmmnmmmfinm
gyggtitég, J. Math. Puree App1. 18(1873), 25-46.

m inde endent, Ann. Math. 80(1964) , 58-77.

13. , 93 characterization _o_§ linear groups. _LV, J. Algebra
8(1968). 223-247.

14. _____, _Qg characterizations .9; linear ggoups. _Y, (to appear).

 

15.

16.

17.

18.

19.

20.

21.

22.

73

“- Suzuki.QaWafliamwo ill.
(to appear).

, Charactgzizationg g; M gm, Bulletin
American Math. Soc. 75(1969) , 1043-1091.
G. Thomas. A shamanism 2f the assess 02(2").
J. Algebra 13(1969). 87-118.
J. Tits. Me as Brunei 21 sells-22911222 W.
0. a. Acad. Sci. Paris 254(1962), 2910-2912.
, W gag gbstract 2.121212 w, Ann. Math.
80(1964), 313-329.
, W W m, Rand. Math. Pure Appl.
°3(1964), 156-165.
H. Wielandt. W We see W 1211
W ems: 2931121129 91322:. Math. Z-
73(1960), 146-158.

H. Imaki. e We 23311.9. simple mm 3M6. 2).
J. Math. Soc. of Japan 21(1969), 334-356.