SOME THiORE M5 ON

EXTENMNG AUTQMOWHISMS ‘

THOSE; {Or the M of DEL. D. ' 

* f mcmm 5mm mmm
Franklin: D. Demana j ‘
1966

 

THiSi;

This is to certify that the

thesis entitled

"Some Theorems on
Extending Automorphisms"

presented by

Franklin D. Demana

has been accepted towards fulfillment
of the requirements for

Ph.D. degree in Mathematics

Qf/Hw

0 Major professo

[Mm March 1H, 1966

 

0-169

 

 
 

  
 

LIBRARY

Michigan Stan:
University

ABSTRACT
SOME THEOREMS ON EXTENDING AUTOMORPHISMS
by Franklin D. Demana

One method to gain some information about the automorphism
group, A(G), of a group G is to consider classes of subgroups of G
on which the automorphisms act as permutations. Two basic problems
must be contended with in any investigation of this kind. First,
given a subgroup A of G and an automorphism a of A , does
there exist an automorphism Y of G such that Y A = a? Secondly,
given a normal subgroup A of G , d an automorphism of A , fl
an automorphism of G/A, does there exist an automorphism Y of G
such that Y A = a and Y induces fi on G/A ? In chapter one we
consider these problems and now list some of the results we have
obtained, This list is not meant to be complete but to give ex—
amples of the type of results obtained.

4U La G=AB,A<G,CIEMM,am fiEAmL Tmn
necessary and sufficient conditions that there exist Y 6 A(G) such
that Y|A = a and Y‘B = e are that

(i) a\A(”\B = e|A(”\B ;

\

-ii; n a = an on A.
( b 'bfi

(2A7 Let G = AB, A Q G, A()B = (l), and a E A(A). Then neces—
sary and sufficient conditions that there exists Y E A(G) such that
YIA = a are that there exist 9 e A(B) and a function r from B

to A satisfying

Franklin D. Demana
b-p
1
(i) f(blb2) = f(bl)f(b2) for all b1, b2 6 B;

(ii) n u = on on A.

b f(b)bB
(3) Let G be the relative holomorph of A by K with
Z(A) = (1). Then a necessary and sufficient condition that a E A(A)
can be extended to G is that there exists an automorphism G of
K such that o‘lso a 99 (mod I(A)) for all e e K.
(A) Let Aid 6, c = x?) Ahi , c/A = {51, Ba, ..., En},

a €.A(A), and B E A(G/A). Then necessary and sufficient conditions

 

that there exists Y e A(G) such that Y‘A = o and Yin/A = p are
that
(i) there exists a function f from G/A to A such that

th
.... = ‘0. _ 13(- = _ B = _
f(bibj) aij f(Ei)f(bj) ai*j* where bibj aijbk and bi bi*

(11) fibia = anf(Ei)bin on A .

(5) Let c = AB, Aid G, A(“3B = (i), and Y e A(G) such that
y|A = o e A(A). If BY = x‘lax for some x e A then the function
f , see (2) above, from B to A induced by Y is f(b) = xhlxbm£3
where hY = r(b)h£3 and e e A(B).

It would be nice to have a result like (2) above in case A is
supplemented in G i.e., there exists a subgroup B of G such
that c = AB and A(“\B ¥ (1). Although this is not possible we
are able to obtain something in this direction by looking at a re-
stricted class of supplements. Next we turn our attention to lift-

ing power and central automorphisms. If A is a normal subgroup of

G = AB we define

Franklin D. Demana

Him (A,Z(A))

ll

{f E H0m(A,Z(A))lf(a) fi’a-l for a ¥ 1}

Ag; (A,2(A)) {r e H’émmemmrm'lao) = slushy}.

Then we show that

(6) Let c = AB with A‘Q c . If r e Hgm(A,Z(A)) and

“r

AK“)B f ker f then the central automorphism of, a = f(a)a , of

A associated with f can be extended to G.
(7) Let c = AB with A‘Q c . Then Y e B(A,G) = {Y e A(B)|AY=A}
is central if and only if there exist f and 9 such that
(i)

(ah)"l for all ab ; l;

e H§m(A,Z(G)(“)A), g e H3m(B,z(c)), and f(a)g(b) a

H)

(ii) f(b‘lab) = f(a) for all a e A and b e B;

(iii) f(b) = g(b) for all b e A(‘\B .
Other results of this type are obtained for power automorphisms.

Now let B(A,G) denote those automorphisms of G which leave
A invariant and c(A,o) those automorphisms which fix A element-
wise. Then we obtain the following results:

(8) Let c = AB, Aid G, and A/‘iB = (1). If |A| # 1 or 2
then A has a nontrivial automorphism that can be lifted to G.
Let c = AB with Aid c . We define B* = {nb e A(A)lb e B}. Then

(9) If u = AB , A‘Q c , CB(A) = (l) , B*ig_A(A) , and A(”)B
is a characteristic subgroup of A then every automorphism of A can
be extended to c i.e., B(A,G)/b(A,G) 3—.ACA).

(10) Let A be a normal abelian subgroup of G . Then
[C(A,c)| = 1 if and only if c is abelian, [G:A] = 2 and (Al.

is odd.

Franklin D. Demana

(10) tells us when an automorphism of A has a unique extension
to G since ‘C(A,G)‘ is the number of ways an automorphism of A
can be extended to G .

In chapter two we turn our attention to the following situa—
tion. Let A be a complemented subgroup of G and denote by S?
the set of all complements of A in G . If D(A,G) denotes those
elements of B(A,G) which fix every complement of A then
B(A,c)/b(A,c) = X1 is a permutation group on Q . Our goal was to
characterize when all complements of A are conjugate in terms of
a permutation condition on (l(,§3). Although we were unable to do
this we did obtain some results in this direction.

(11) If (X ,iZ) is a primitive permutation group then all
complements of A are either conjugate or normal.

(12) If (X ,3?) is a primitive permutation group then either

A is characteristically simple or if H is a characteristic sub-

NA<B>
A

group of A then H NA(B) = <'
(13) If (x:,n ) is 3/2—fold transitive then all complements
of A are either conjugate or normal provided any one of the follow-
ing conditions hold:
(i) (|A|, |B/B'|) = l where B' is the derived group of B;
(ii) if there exists a normal subgroup H of G such that
HfA mdiMMm®)=Uh
(iii) if NA(B) is a Hall subgroup of A with a normal comple-
ment;

(iv) if 2(A)(‘)NA(B) = 2(NAB)) and there exists at least one

x E A — NA(B) such that (‘xl , \NA(B)\) = l;

Franklin D. Demana
(v) if A/NA(B) is not nilpotent.

We also obtained some results in case (X ’§2) is a sharply

doubly transitive permutation group.

SOME THEOREMS ON EXTENDING AUTOMORPHISMS

By

Franklin D. Demana

A THESIS

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY
Department of Mathematics

1966

 

PREFACE

The author is deeply indebted to his major professor, Dr. J.
E. Adney for his patience and helpful guidance in the preparation
of this thesis. The many discussions we had together during the

research were of immeasurable value to me.

ii

TABLE OF CONTENTS

INTRODUCTION .
CHAPTER I . . . . . . . . . . . . . . . . .
1.1 Introduction . . . . .

1.2 Necessary and Sufficient Conditions for Extend-
ing Automorphisms

1.3 Power and Central Automorphisms
l.h The Group B(A,G)
CHAPTER II .
2.1 Introduction . . . . . . . . . . . .
2.2 The Group (X ,KZ) . . .
INDEX OF NOTATION

BIBLIOGRAPHY . . .

Page

27
35
'h3
b3
A6
S9
61

INTRODUCTION

When studying automorphisms of a group G some very natural
questions arise. Can we start with a subgroup A and by "extend—
ing” its automorphisms obtain every automorphism of G ? Of course
the answer to this question is no if A is not characteristic in
G . If A is characteristic then the answer is yes but how to do
this is another problem. One might be tempted to think that if A
is characteristic then every automorphism of A ought to be extend—
able to G . The example in 1.1 shows this conjecture to be false.
Since not every automorphism of a subgroup can be "lifted" to G we
begin by finding necessary and sufficient conditions under which an
automorphism can be ”extended". First we clear up the notion of "ex—
tending” or ”lifting” automorphisms.

If A is a subgroup of G and a an automorphism of A then
u is ”extendable” or ”liftable” if and only if there is an automor—
phism Y of G such that Y A = o. Also if G = AB and T e Hom(B,G)
then the pair a, Y is ”extendable” or ”liftable” if and only if
there is an automorphism Y of G such that Y|A = o and Y|B = r.
Finally, if a is an automorphism of the normal subgroup A and
B an automorphism of G/A then the pair a, B is "extendable" or
"liftable" if and only if there is an automorphism Y of G such
that Y|A = a and Y induces B on G/A.

In chapter one we obtain necessary and sufficient conditions in
each of the cases mentioned above. We are also concerned with the
problem of lifting certain types of automorphisms, namely, power auto—

morphisms and central automorphisms. Let B(A,G) denote the set of

fl

automorphisms of G which leave A invariant and C(A,G) ‘the set
of automorphisms of G which fix A elementwise. Then C(A,G) is
normal in B(A,G) and B(A,G)/C(A,G) is isomorphic to a subgroup
of the automorphism group of A. In fact, lB(A,G)/C(A,G)l is the
number of automorphisms of A that can be extended to G . We try
to find out something about the structure of B(A,G)/C(A,G) and
also find conditions when every automorphism or no automorphism can
be extended to G. Finally, under very special hypothesis on G,
we count the number of ways a particular automorphism of A can be
lifted to G and when an automorphism of A can be paired with a
unique automorphism of a complement of A and lifted to G.

In Chapter Two we restrict our attention to a complemented sub—
group A of a group G. Then we define S2 to be the set of all
complements of A in G. The set B(A,G) of all automorphisms of
G which fix every complement forms a normal subgroup of the group
B(A,G). We then see that the factor group X = B(A,G)/B(A,G) is

a permutation group acting on the set D . Naturally one would like

all complements are conjugate. Although we are not able to do this
at this time we do obtain several interesting results in this di—
rection. What we do is impose the conditions primitive, 3/2—fold
transitive, and sharply 2—fold transitive on X and investigate the
complements in these cases. If (X ,9 ) is primitive then we show

that all complements are either normal or conjugate. By placing

 

3
certain additional restrictions on G we get the same result in
case (}{,$7) is 3/2-fold transitive. We also try to find out
something about G , A , and )( under these hypotheses. The re-
sults are not complete and many interesting questions remain un—
answered.

Our main interest was in finite groups and, in fact, all groups
considered in Chapter Two and 1.h of Chapter One are assumed finite.
However, in 1.2 and 1.3 of Chapter One the results hold as stated
for infinite groups unless otherwise stated.

The reader is asked to consult the index of notation for

identification of symbolic notations of groups, sets and relations.

CHAPTER I

1.1 Introduction

Let A be a normal subgroup of the group G. The two main prob-
lems considered in this chapter are lifting automorphisms of A to G
and pairing an automorphism of‘A with an automorphism of G/A and
lifting to G. In 1.2 we give sets of necessary and sufficient con-
ditions solving these problems under various hypotheses on G. The
question of lifting certain types of automorphisms, namely power
automorphisms and central automorphisms, is considered in 1.3.
Finally, in l.h, we investigate the group B(A,G) to some extent.

It is difficult to make very general statements since the prob—
lem considered is quite complex. To illustrate this we first show
that given any odd prime p and any positive integer n greater than 1

there exists a group G with characteristic subgroup A such that

I‘l-Z IT
c = AB, AflB = (l), |B| = 2, and 12(2) = B—pf’i—‘Q. The latter

implies that for each automorphism of A that can be extended to G

n-z n
there are at least p pr ‘1) — 1 automorphisms of A that cannot

 

be extended to G.

Let A be an elementary abelian p-group of order pn generated

. -l
by a1, ..., an. Let B = <b> where b2 = 1, balb = a1 , and baib = a1

for i = 2,3, ...,n. Now A is a normal p—Sylow subgroup of G = AB

so clearly characteristic. The center of G is generated by a2,...,an

so any automorphism of G must leave < a2,...,an> invariant. Let a be
any automorphism of < a2,...,an> and r and 3 any two integers such

that o _<, r _< p—l and o < s f p-l. Then the mapping Y of G defined

i = aIi aY = 3a (1 = 2, ...,n), and bY = aib is an automorphism

by a 1 i

A

of G and one can easily check that these are all the automorphisms

of G. It is well known [3, pg. 86] that the order of the automorphism

—l —2
(pH -pn ).

group of <a2, ...,a > is (pm—l-l)(pn_l-p) Therefore,

n

(pn— l _pn- 2

we have \A(G)) = p(p—1)(pn_l-1)(pn-l—p) ) so that

(AB); = (amen-p) (tn-pn'li = an‘zon-i)
\A(G)\ p(p-l)(pn'l-l)(pn-1-p) ... (pm—l-pn'z) p-1

1.2 Necessary and Sufficient Conditions for Extending.Automorphisms

Definition 1.1: Let A be a subgroup of G. If B is a proper

 

subgroup of G such that G = AB then B is said to be a supplement of

 

A in G. Moreover if AKIXB = (1) then we say that B is a complement

 

of A in G. We refer to A as being supplemented or complemented in G.

 

 

Let A be a normal subgroup of G which is not contained in the
Frattini subgroup of G. It has been shown [2] that A possesses a
supplement, say B, in this case. The first two results we obtain

are generalizations of those appearing in [5].

Theorem 1.2: Let c = AB, A<flG, a e A(A), and B e A(B). Then

 

necessary and sufficient conditions that there exist Y E.A(G) such
that Y A = o and Y|B = B are that
(l) u‘AY‘wB = B\A(”)B ;
(2) who 2 dan on A.
The proof of this theorem follows immediately from the more

general result:

Theorem 1.3: Let A, A1, B, B1 be subgroups of G with A and Al

 

normal in G, A and B isomorphic to Al and B1 respectively under 0 and

T, and G = AB = A B

l 1. Then necessary and sufficient conditions for

6
the existence of an automorphism Y of G such that Y A = o and Y18 = T
are that
(l) OlA[A\B = T|A(‘\B 3

(2) rho = oan on A.

Proof: (a) Necessity

Let Y E A(G) such that YlA = o and Y‘B = T. Then clearly

ojAr”\B = Y\A{”\B. If a e.A and b e B then (ab)Y = aYhY = aObT

— T .—
bTb TaObT = bTadb. We can also write ab = bb 1at so that (ab)Y

bYao‘lah)Y = bT(b_lab)O = bTabo. Hence we have

(bh‘lab)Y

T (TbT _ T 130 013T bc a _
b a — b a or a a for all a E A. Therefore, nbg _ 0n .
bT

(b) Sufficiency
o

T
Now 6 = AB = A181 with A :' A1, B :—B1, oiA(‘\B =

T‘Af”\B, and who = on T . Let ab (a E A, b E B) be any element of
D

G and define Y by (ab)Y O T

a b . II albl = azbz wlth ai E.A, bi E B

H

- -l
for i = 1,2 then azlal = bzbl e.A(”\B and since O\AflB = T|A(T\B
we have
—1 -l T
(8.2 a1)0 2 (bzbl )

—l T T -1

(albl)lf== (azbz)Y

Hence Y is well defined.
-1
If ab 6 ker Y i.e., if (ah)Y = aObT = 1 then a0 = (bl) e Al(”\B1
_l _ -1 _ . -1
and since 0 ‘Alf’NBl = Y l‘A1{"\B1 we have (a0)0 = [(bT) le or

a = h‘l. Thus ab = 1 and ker Y = (1). So Yjs 1—1 and clearly Y is

onto since G = AlBl.

7

Now let gl = albl and 92 = azbz be any two elements of G where

a1 6 A and b1 6 B for i = 1,2. Then 9192 = 31(blazbillblb2 =
-1 -1
bl Y O' IOTT .
alaz blb2 so that (9192) = alaz blbz. But ”b0 = on T so we have
b

I
Q)
l—fi
Q)
N
U“
..a
D“
N

(9192)Y ‘

l H
in S13
r— i—a
0‘0 £13
H—q N
NO +---
D‘ -\
i—al D"
—i »--
0" O"
H—l N
0"
N-\

I
$13
p...
O”
t-' —l
Q)
N

H

(albl)Y(a2b2)Y

9i 9%”

Therefore Y is an automorphism of G such that YlA = o and Y|B = 7.

Using this result we can obtain a set of necessary and sufficient
conditions for lifting an automorphism of a normal subgroup which is

not contained in the Frattini subgroup.

Corollary l.A: Let G = AB, A <1G, and o e.A(A). Then necessary

 

and sufficient conditions that there exist Y e.A(G) such that Y A = o
are that

(1) there exists a subgroup B1 of G isomorphic, say under T, to
B and G = AB,;

(2) o|Ar')B = T|A(”\B ;

(3) who = on on A.
bT

Prggf: (a) Necessity
Let Y e.A(G) such that Y‘A = o. (l) is clear if we set B1 = BY
and T = Y‘B. (2) and (3) follow immediately from Theorem 1.3.
(b) Sufficiency
From the sufficiency of Theorem 1.3 with A1 = A and

g = a there exists Y E A(G) such that Y‘A = d.

Now if we impose the stronger condition that A be complemented

in G then we can refine the conditions somewhat.

Theorem 1.5: Let G = AB, A <]G, A(AWB = (1), and d E A(A). Then

 

necessary and sufficient conditions that there exists Y E A(G) such
that YlA = d are that there exist B g.A(B) and a function f from B

to A satisfying
b~B
l
(1) f(blbz) = f(bl)f(b2) for all bl,b2 e B;

(2) = A.
nbd Gflf(b)bfi on

3322:: (a) Necessity
Let Y e A(G) such that Y‘A = o. Since A(“\B = (1) each g e c
has a unique representation as g = ab with a E A and b E B. Hence
we can write bY = f(h)b* with f(b) e A and h* e B where f(b) and h*
are uniquely determined by b. Thus f is a function from B to A.
Define B by b‘3 = b*. B is well defined since b* is uniquely determined
by b. If bl, b2 6 B then (b1b2)Y = f(b1b2)(blb2)B. But Y e A(G) so

we also have.

(131132)Y bibg

f(bl)bE f(b2)bE

ll

f(b1)b§ f(b2)(bE)—lebE

at)”
= f(bl) f(bz) :1 bEbE
(bifl
Hence f(b1b2)(blb2)B = f(bl) f(bz) berg and sineeiirrya = (l)
B -l
bl)
it follows that f(blbz) = f(bl) f(bz) and (blbz)fl = bEbE.

Thus f has the desired property and B is an endomorphism of B. Let

‘1
bfi = 1. Then we have [f(b)‘Y b]Y = f(b)‘l f(b)bB =
—1
bi3 = l and since Y E A(G) we must have f(b)—Y b = 1. Now
-1
b = f(b)Y e A since Y‘A e A(A) so b e A/‘NB = (1). So ker B = (l)

b E ker B i.e.

3

and B is 1—1. Clearly B is onto so B e A(B). Define T by b“r = bY =
f(b)bB. Then, by the necessary part of Theorem 1.3, we have

TTCIZGTT :CITT .
b bT f(b)bI3

(b) Sufficiency

Let f be a function from B to A and B E A(B) satisfying
(1) and (2) of the theorem. Define T by bT = f(b)b§. Since f is
a function and B an automorphism T is well defined. Let bl, b2

be any two elements of B and we have

(blb2)T = f(blb2)(blb2)B
bZB
f(bl)i(b2) bEbE

f(bl)b§ f(b2)b15b§bg

f(b,)b§’ f(b2)b§

H

or 'T
bl b2 .

10
Thus T is a homomorphkfinfrom B into G. Let b E ker T then b'r =
f(b)b£3 = 1 so that bB = f(b)-l E A. Since bB E B and A(”\B = (1) we
must have b‘3 = 1. Hence b = 1 since B E A(B). Therefore ker T = (1)
and T is an isomorphism. Set B1 = BT then clearly G = ABl and since
n

a 2 an an we have by thesafifkienqy part of Corllary 1.h

f(b)b‘3 bT
that there exists a Y e A(G) such that Y|A = o.

b

The function f in the above theorem has two other properties

which will be useful later so we will establish them now.

Lemma 1.6: For the function f in Theorem 1.5 we have
(i) f(l) = i
(ii) f(b‘l) = b“3 f(b)‘lbB for all b e B.

bgfi
Proof: In f(blbz) = f(bl) f(bz) set bl = b2 = l and we get

ll

f(l) = f(1)f(l) or f(l) 1. Now set bl = b.1 and b2 = b and we get

f(b—lb) = f(b‘l) f(b)bfi

f(l) = f(b‘l)b“3 f(b)bfi.

Since f(l) = 1 we have f(b—l) = b_‘3 f(b)_le.

We notice that this function f need not be a homomorphism as will

be illustrated in the following example.

Example 1: G=<ab>,A=<a>,B=<b>,an=b2=l,andba=
aIlb. Define the mapping Y by aY = a1, bY = an where (i,n) = 1
and 0.5 j'f n—l. Y e A(G) and the function f from B to A induced by

Y maps b to aJ. Since ‘aJ\ need not be 2,f need not be a homomorphism.

11
If we add to the hypothesis of Theorem 1.5 the condition that
each automorphism of A induced by an element of B is an inner auto—
morphism of A then we can replace condition (2) of the theorem by
one of different form. Under this hypothesis we can write ”b = ”g(b)

with g(b) E A for each b E B and obtain:

Theorem 1.7: Let G = AB, A <|G, A(ANB = (l) and assume for each

 

b E B there is a g(b) E A such that W = ”g(b) on A. Then necessary

b

and sufficient conditions that d E A(A) can be extended to G is that

there exists a function f from B to A and a B E A(B) such that

g(bl-B)
(l) f(blbz) = f(bl) f(bz) for all bl, b2 6 B;

(2) f(b) - g(b)%(b?’)‘l (mod Z(A)) for all b e B.

Proof: (a) Necessity
Let Y E A(G) such that YiA = d. Then, by Theorem 1.5, there
exists a B e A(B) and a function f from B to A such that f(blbz) =

-B
f(bl) f(b2)b1 and who = afif(b)bB. Since ”b = “g(b) on A we can write

f(blbz) = f(bl) f(b2)g(bl ). If x 6 A then it is easy to show that

nxc 2 an a so we have n (b)d = an 6. Thus
x g g(b)

Tig(b)0. = (lTTf(b) bfi

CITT :afif(b)fi

g(b)“ g(bh

g(b)“ : f(b) g(bh ’

Hence n,(b) = n on A so that f(b) E g(b)ag(bB)—l (mod Z(A)>.

g(b)(lg(bBVl

12
(b) Sufficiency

Let a, B, and f be given satisfying (1) and (2).

Since n = n we can rewrite (l) as f(blbz) = f(bl) f(b2)bl

big g(blfi)

for all bl, b2 6 B. Now on = an and since

f(b)bB : anf(b)flbfi f(b)ng(b5)

f(b) a g(b)ag(bB)—l (mod Z(A)) we have nf( on A. Thus

b) = ”can“g<bi>‘l

= on and since

s<b>ag<bi>‘l g(bfi> g(b)“

nf(b>b‘3 : anf(b)fig(b5> z a

Hence an

9(b) 6 A we have Gflg<b)a = ”g(b)a' f(b)bB = ”g(b

CI=1TCI..
)

b

Therefore, by thegaffhgkxny part of Theorem 1.5, there exists Y E A(G)

such that YiA = d.

Definition 1.8: Let c = AK, A(‘\K = (1), K.f A(A) with defining

 

relations 0—136 = acI for all a e A and d E K. Then G is called the

relative holomorph of A by K. If K = A(A) then G is called the

 

holomorph of A.

Next we consider some of the previous theorem if we restrict G
to be a relative holomorph of A. If we try to apply Theorem 1.2 to

G i.e. try to extend a E A(A) O E A(K) to G then the condition of

) 3

the theorem becomes Ba = dB8 for all B E K Since n = B on A. There—

B
fore if Y E A(G) such that YTA = d and Y‘K = G then Y = Na the inner

D
automorphism of the holomorph of A induced by the element a, where
a E NA(A)(K>‘ However, if we apply Theorem 1.5 to G with the added

condition Z(A) = (1) we have a little more success for we are able

to obtain the following result:

13

Theorem 1.9: Let G be the relative holomorph of A by K with

 

Z(A) = (1). Then a necessary and sufficient condition that a E A(A)
can be extended to G is that there exists an automorphism O of K

such that o‘lBo 2 Be (mod I(A)) for all B e K.

Proof: (a) Necessity
Let Y E A(G) such that Y‘A = d. Then,by the necessary part of
Theorem 1.5, there exists a O E A(K) and a function f from K to A

—e
such that f(Ple) = f(Bl) f(B2)[31 for all Bl,B2 E K and n d = dnf

A name”

Now fig = B on A so the latter condition becomes Bu = dnf(p)Be or

d—lBu = ni<$>Be. Hence d—le 5 Be (mod I(A)) for all B E K.

(b) Sufficiency

Let a g A(A), e e A(K) such that o‘le 5 Be (mod I(A))
for all B e K. Then we can write a-le = iTNmB-e where f(B) E A.
First we Show that f is a function from K to A. If f(B) is not
unique then we could have f(B) = a1 or f(B) = a2 where G-lBG =
n e and d-le = n B8 Hence n = n on A Thus a a.1 E Z’A)
a1B a2 ' al a2 ' 1 2 \
and since Z(A) = (1) we have a1 = a2. Therefore, f is a function
from K to A. Let BUB2 be any two elements of K and we have

. e —l
n = d d
f(BlBZ)(PiP2) Ple

_l -1
a firm Bea

e 8

‘ ”f(B.>Bl”f(Bz>fl2
Ge
”f(Bl)Wf(B2) :9 B152

e Now 9 E A(K) so we have n

n - .
we)“e Bl
fl”(Bi)nf(Be)BTe and since Z(A) = (i) it follows that

since B?Wf(fi2> =

fume) 2

I

1A

f(Ble) = f(Bl) f(B2)fi;e. Since fig = B on A we can rewrite d_le =

II

Summarizing we

nf(B)Be as G‘lflBa = ”f(B)WB9' Thus an

f(fi),o'

have a E A(A), G E A(K), and a function f from K to A such that

—e
f(BlBZ) = f(Bl) f(BZ)Bl for all B1,B2 E K and nBa = d on A.

f(B>Be

Therefore, by the sufficiency part of Theorem 1.5, there exists Y E A(G)

such that Y A = a.

Now we consider a relative holomorph of A by a cyclic subgroup
of A(A). However, we do not assume that Z(A) = (1) in the result

that follows.

Theorem 1.10: Let G = A <B>, B e A(A), and B—laB = a‘3 for all

 

a E A. Then necessary and sufficient conditions that d E A(A) can
be extended to G are that

-1
(1) there exists a E A such that d Bd = naBk where (k,‘Bl) = 1;

(2) al+Pk+P2k+"'+B(r-l)k = 1 where r = (B).

Proof: (a) Necessity

 

Let Y E A(G) such that Y‘A = d. Then, by the necessary part of
Theorem 1.5, there exists a 9 € A(<B>) and a function f from <B> to

A such that f(BlBZ) = f(Bl) f(BZ)B18 for all BUB2 E <B> and n ta =

on . . on A. Since O E A(<B>) we have B6 = Bk where (k,(B\) = 1.
imhmhe
Let r = ‘B‘ and we have
1 = 1Y = (Br)Y = (BY)r
= (“w“)r

1+Bk+B2k+...+B(r-l>k

f(B) Brk

15

= f(B)l+§k+§2k+°'-+é(r‘l)k

Thus we have (2) with a = f(B). Now set t = l in n .c =

it f(Bt)(Bt)e

. e
d = on . But NB = B on A so we have Bu = unaB

i f(B)Be

_1 —
or c Ba — WaB .

and we get n

(b) Sufficiency

Let a E A(A), a e A, and k be given satisfying (1) and

k

(2) of the theorem. Define O by B6 = B then clearly O E A(<B>) since

° —k+,,.+ -(i-i)k
(k,\B\) = 1. Let f(B) = a and define f(Bl) = f(g)1+$ B
' - 1. *k ...l -(3+t-1)k
for all i 3 2. Now f(BSBC) = f(fi5+t) s f(g) TB + 1B and
. - k —k _( -l)k -k -(t-l)k -SK
f(BS) f(BE)E S = f(fi)l+B +'°'+B S (f(B)l+B +"'+B )3
-(s+t—1)k

-k
f(B)l+fi +...+B

H

f(BSBt).
—e
Hence we have shown that f(Ple) = f(Bl) f(B2)[31 for all BljBZ E

<B>. Since d_le = nf(B)Bk we have

-1 i _ k i
(a PG) _ (”f(B)B )

a P a = 1k

1Tf(B)1+13_k+~°+B-(i-l)k B

d-lBid = n . ik
f(Bl)

.. . .. —1
Now BB = B on A so we can rewrite the above equation as d n , d r

1
B
OI‘U.C1=O.

g1 ”f(gi)(gi)e . Thus we have a E A(A),

f(Bi> fiik

16
e e.A(<B>), and a function f from <B> to A satisfying f(BlBZ) =
—e
f(Bl) f(BZ)Bl for all B1,B2 e <B> and n .d = an . i e o
B1 f(B1)(B )

Therefore, by the sufficiency part of Theorem 1.5, there exists

n A.

Y e A(G) such that Y A = d.

Now we turn our attention to the following question. Let A be
a normal subgroup of c, d e A(A), and B e A(c/A). Under what condi—
tions can we put a and B together to get an automorphism of G i.e.,
under what conditions does there exist a Y E A(G) such that Yhi = a
and Y induces the automorphism B on G/A? We will use the notation
Y‘s/A = B to mean that Y induces the automorphism B on c/A. First
we consider the Special case when A is supplemented in G and notice
that we have essentially answered this question in Corollary 1.b.
In the following theorem G = AB and we must insist that coset repre—
sentatives be chosen from B so that all the statements made are

meaningful.

Theorem 1.11: Let c = AB, A 4 c, o e A(A), and B e A(G/A).

 

Then necessary and sufficient conditions that there exists Y E A(G)
such that Y‘A = u and Y‘G/A = B are that

(1) there exists a subgroupiBlof G such that c-aABl, B :‘B1 under
., and T induces B on c/A;

(2) u‘A/TAB = T‘A/RAB;

(3) ”ha = an on A.
bT

Proof: (a) Necessity

Let Y E A(G) such that Y‘A = d and Y‘s/A = B. Set B1 = BY and

 

 

17
T = Y‘B. Clearly Y‘G/A = y‘G/A = B where we mean of course that coset
representatives are chosen from B. Now (1), (2), and (3) follow from

Corollary 1.h.

(b) Sufficiency

Let d, B, Y, and B1 be given satisfying (1), (2), and
(3) of the theorem. Then, by Corollary l.b, there exists Y E A(G)
such that Y\A = o and Y‘B = r. Thus Y‘s/A = T‘G/A = B and we have

the result.

In the next result we only assume that A is a normal subgroup
of G. Let a set of coset representatives of A in G be fixed so that
we can write G = Abl(v)Ab2(vj--o(vjAbn where b1 = 1. We denote the

elements of G/A by El,EZ,"'

,55, If b1 and bj are any two coset repre~

sentatives then we can write bibj = aijbk where aij E A and the set

of aij constitute a factor set. For more information concerning

factor sets the reader can consult [3, pg. 218]. Now if B E A(G/A)

we denote the image of Bi under B by E.* where 1* is some positive
1

integer from 1 to n. Then with this notation we have the following

result.

 

n
Theorem 1.12: Let A q<3,<3 = (“)1un, c/A = {El,bz,oo-,bn

1.
1:1
o e A(A), and B e A(c/A). Then necessary and sufficient conditions

that there exists Y e A(G) such that Y|A = d and Y‘s/A = B are that

\

(1) there exists a function f from G/A to A such that f(EiEj) =

b-l

—u — — .*
aij f(bi) f(bj) l ai*j* 3

i

18
(2) nbi Q ~ anf(Ei)b M on A.
in

Proof: (a) Necessity

Let Y e A(G) . a and YlG/A = B.
bl? : f(El.)b where f(El.) e A and SE = b
1,.

such that YlA

We can write
since YlG/A = B. Thus
i
f is a function from G/A to A since e
sentation of the form g

ach g e G has a unique re

pre-
abi where a E A.
coset representatives then bib

Let bi and bj be any two
j:

we have (b.b.)Y = bTbT
1 J 1 J

aijbk and bibj = bk‘ Since Y 6463)
. so that
Y = Y Y
(aijbk> bibj
a?.f(E )b ., = f(E.)b ,f(B.)b .,
lJ k 7“ l .‘n‘ J .m-
k 1 j
a l — — fl — fl — —1
a. .I(b.b.)b ., : f(b.)b \,I(b.)b Vb Vb
lJ 1 J 7: 1 ,7: J _-,r ,W ,7.
k 1 1 1 j
1).].
a f(E.E.)b l = f(E.)f(E.) i“~ a L lb M
lj 1 j k" 1 j inj. k"
b_%
Hence f(EiBj) = a;? f(Ei)f(Ej) l a which establishes (1). Let
1 J'
a be any element of A and bi
~1
-1 b1
b.a = b.ab. b. = a b
1 l 1 1

any coset representative and we can write
1' Hence

b.1
(bia)Y = (a 1 b.)Y

_ a lea
f(bi)b Ma ~

— a l f(E.)b M
1\ l i”
_ a —1 bile
f(bl.)b _,_a b Mb , = a l f(El.)b ,
in in 17V

 

—

19

15%, bfla

O.
f(Ei)a i“ = a 1 f(b.)
cb-:f(Ei)—l b. o
a i" = a

b.

Therefore, n _ld = afi(f(Ei)b x)—1 or flb‘a = aflf(Ei)b V.
1 i”

i/\

(b) Sufficiency
Let c e A(A), B e A(c/A), and f be a function from c/A
to A which satisfies (1) and (2) of the theorem. Define Y by (ab,)Y =

a- . . l . . . . .
a I(Bi)b M. Since each element or G is uniquely represented in the
i/\

form abi with a g A and 1.5 1.5 n Y is well defined. Let 91 = albi
and g2 = azbj be any two elements of G and then we have glg2 =
le

albiazbj = albiazbglbibj = alaz 1 ai.b

where 3.5. = E . Now using
J k 1 J

k

(l) and (2) at the appropriate steps we have

u
(glgz)Y = ala2 a9. I(Ek)b *

s a%f(Ei)b ,egb-j f(E.)'lf(B.)b ,f(b.)b .

1 1 i” j”

. afirmim ,,_a%f(5,)b
i” J”

20

Thus Y is an endomorphism of G. Now let abi E ker Y i.e., (abi)Y =

aaf(bi)b , = 1. Then b 1 a b1 = 1 and since B e A(c/A) and E? = b x
in in i”

we must have bi = bl = 1. If in (1) we set E, = E, = B, we get

f(El) = f(El)f(El) or f(El) = 1 since all = 1. Hence a‘1L = 1 and

since a E A(A) we have a =1" 'Hmnefore ker Y =(1) and we have shown

that Y is 1-1. Finally let 9 = abi be any element of G. Since

B E A(G/A) there exists a bj such that E? = Ei' Hence E? = E M = E,
J
or b M = bi' Since a g A(A) there exists al 6 A such that a? =
J
A -1 n Y a n - A -l A —
aICE.) . Therefore (alb.) = a1 I(b.)b V = a 1(5.) l(b.)b. = ab.
J J J j“ J J 1 1

and Y is onto. We have shown that Y e A(G) and clearly Y‘A = a.

Further (Abi)Y = Ab? =11 f(Ei)b M = Ab V = (Abi)B so that Y‘G/A = B.
i“ i“

We can observe that nowhere in the proof of the above theorem
did we need the fact that G/A was finite. It is easily verified
that this theorem is true even if G/A is infinite.

The conditions obtained in the above theorem leave something
to be desired. However, we must remember that the hypothesis includes
the case that A is contained in the Frattini subgroup of G. From
[3, pg. 156] we have that the Frattini subgroup consists of the non—
generators of G. Thus it is not surprising that the conditions should

be as they are.

21

We can obtain a corollary to the above result in the following
_way. Suppose A14 G and a e A(A). Then necessary and sufficient
conditions that there exists Y e A(G) such that Y\A = o are that
there exists B e A(c/A) and a function f from c/A to A which satisfy
(1) and (2) of Theorem 1.12.

Now we will illustrate Theorem 1.12 with two examples. In the
first f will turn out to be a homomorphism. In general f need not be
a homomorphism, in fact, it may be less well behaved than the func—

tion which appears in Theorem 1.5. This will be pointed out by the

second example given below.

Example 2: G = <a,c>, a8 = c2 = 1 cac = a5. Let A = Z(G) =

,
B(G) = < a2> and as coset representatives we choose bl = 1, b2 = a,
b3 = c, and b4 = ac. The mapping Y defined by aY = ac, cY = a4c is
an automorphism of G. If a = YlA and B = Y‘G/A then we have

<a2>a —> < a2>ac

d : a2 —> a6 and B :
< a2>c —> < a2>c

Direct computation gives f(El) = 1, fCEZ) l, f(Bg) = a4, f(B4) =

a4, and
a11 : 1 a21 = 1 a31 = 1 841 2 1
a12 I 1 a22 = 32 a32 = a4 a42 2 a6
a13 = 1 a23 = 1 a33 = 1 a43 = 1
314 Z 1 a24 = 32 334 = a4 344 = 36

Since A = Z(G) condition (1) of Theorem 1.12 reduces to f(EiEj> r

— I" — -0’ a n a I rs . u o
f(bi)l(bj)aija m V . One can verlfy from the 1nlormatlon given
1.- j"
- ~ (l —X— 9(- —Ti- 9:-
above that ai. : a M M where 1 = 1, 2 = h, 3 = 3, and b = 2.
i /\ jl\

Thus f is a homomorphism.

22

-1
Example 3: G = < a,c >, a27 = c3 = l, cac = a19. Let

 

A = Q(G) = < a3> and as coset representatives we choose bl = 1,
b2 = a, b3 = a2, b4 = c, b5 = c2, b6 = ac, b7 = azc, h8 = acZ, and
b9 = a2c2. The mapping Y defined by

a —> a4c

c —> a9c

is an automorphism of G. If a = Y‘A and B Y‘G/A then we have

< a3> a —> < a3> ac
a : a3 —> a12 and B :
< a3> c —> < a3> c

Now a22 = 1 so agz = 1. Since (I52)[3 = 56 we have 2* = 6. But a66 =

l a . . Gr _ a . . o
a so that a22 + a M M — a66 contrary to the preV1ous example.
2‘“— 2'“-

Now f(EZEZ) = f(ES) = a24 and f(Ez) = a3 so that f(EZEZ) + f(EZ)f(EZ)
and f is not a homomorphism.
We observe from example 2 that if A ::Z(G) then condition (1)

a _ - — _a .
of Theorem 1.12 reduces to f(bib.) = f(b,)f(E.)a.. a M M. Then f is
J J 1J i/‘j ‘
d
a homomorphism iff a.. = a M --
IJ iwj'n'

If, in Theorem 1.12, we impose the further restriction that G =

AB and A/P\B = (1) then we can choose the coset representatives as
elements of B. From [3, pg. 221] we know that the factor set is

trivial in this case i.e., aij = l for all i and j so that (1) reduces

b_i
to f(EiBj) = f(Ei)f(Ej) i” . Since G/A :‘B we can consider f as aT

function from B to A with the property that f(blbz) = f(b1)f(b2)b]
for all bl,b2 g B where TB is the automorphism of B induced by B.

So (2) reduces to n Thus this function is the same

G=CL

b nf(b)bTB

as the function which occurs in Theorem 1.5. We nOtice that f is

23
almost a crossed homomorphism [b, pg. 105] of B to A and, in fact,
would be if we insisted that A be abelian. In the next two results
we try to find out when f behaves like a principal crossed homo-

morphism [h, pg. 106].

Theorem 1.13: Let c = AB, Aid c, AK”)B = (l), u a A(A), B e A(G/A),

 

and TB the automorphism of B induced by B. If there exists an x E A

such that n d = on

b then the pair a, B can be extended to an

-1 T

x b Bx
- n I‘ -l - n 00 n

automorphism Y or G. Moreover, BY 2 x Bx and the lunCtlon I from

B to A induced by Y acts like a principal crossed homomorphism.
- -. . - —1 b B
Procl: For each b e B we define I(b) E A by I(b) = x x .

Clearly f is a function from B to A and if bl,b2 are any two elements

of B we have

~‘T
f(blbz) = x‘1x(blb2) B

x_lxb2 Bbl B

-T -T -T -T
_ A _ A B A
= x lxbl (x l)bl xbz b1

'T

B bgli

I!

—l b— -1 b_T[3

(X X 1 )(X X 2 )
‘T

f<b1>fib2>bl

‘Tg _
T _ T _ T T T _1 T
Since f(b)b B = x lxb b 5 = x 1b Bxb 5b B = x b Bx we have

1T =TT =0. TTCIOTTTCL=CITT

Summarizing we have
T - T ~
f(b)b B x 1b Bx b b

T

f(b)b B

d g.A(A), TB 6 A(B), and a function f from B to A such that f(blbz) =
‘T

f(bl)f(b2)bl for all bljbz E B and ”bu = GW T . Therefore, by

f(b)b B

 

2A

the sufficiency part of Theorem 1.5, there exists Y E A(G) such that
a T = TB - Y a T3 =
Y‘A d and b f(b)b . Now, as was shown preV1ous1y, b f(b)b
_ T _ T T
x 1b Bx so that BY = x le. Finally, (Ab)Y = AbY = Af(b)b B = Ab B = (Ab)B

so that Y|o/A = B-

Theorem 1.1A: Let G = AB, A‘Q G, A(“NB = (1), and Y e A(G) such

 

that Y‘A = d E A(A). If BY = x-le for some x e.A then the function

_ ‘5
f from B to A induced by Y is f(b) = x lxb where bY = f(b)b£3 with

A e A(B).

Proof: By the necessary part of Theorem 1.5 bY = f(b)bB where

B E A(B). Since BY = x_le we can write bY = x_lbx for each b e B

—- __1
1— —1 b l

where b e B. Now bY = x_ bx = x x b so we have f(b)bB = x—lxb b
——1

and since A(T\B = (1) it follows that f(b) = x-lxb and b = bfi.

_~ ‘B
Hence f(b) = x lxb .

If A is complemented in G and we also assume that all complements
of A are conjugate (this could be achieved by insisting (!A|,!Bi) = 1

for example) then we can obtain the following corollary to Theorem 1.1h.

Corollary 1.15: Let G = AB, A <16, A(”\B = (l), and assume all

 

complements of A are conjugate. If Y e B(A,G) then oY = f(b)bB

_ '3
where B E A(B) and f(b) = x lxb for some x E A.

Proof: By the necessary part of Theorem 1.5 bY = f(b)bB where
p e A(B). Since Y e B(A,G) we have cY = (AB)Y = AYBY = ABY so that
BY is a complement of A. Thus there exists an x g A such that BY =

x-le. The result now follows from Theorem 1.1L.

25
One might be tempted to try to obtain a result like Theorem 1.5
in case A is supplemented in G. However, it is not possible to relax

the condition A(AWB = (1) as will be seen by the following example.

Example A: G = < a,b >, a4 a b4 = 1 a2 = b2, and ba = a-lb.

)

Let A = < a > and B = < b > then A(TWB = < a2>. Let Y be any auto—

morphism of G which leaves A invariant. We can write yY = f(y)yF3

for any y E B where f(y) E A and yB E B. However, 1 = 1Y = be2 =

aZb2 so we could choose f(l) = a2 so that f violates (1) of Lemma 1.6.
If G = AB, Aid G, and Y E B(A,G) then we can write bY = f(b)bL3

with f(b) g A and b3 g B. Since ab : bh‘lab we have (ab)Y = (bh‘lah)Y

and exactly as in the proof of Theorem 1.3 we can show that n d =

b

on B on A for all b e B where o = Y‘A. If b g A(‘TB then we have
f(b)b ‘

bCI = f(b)bB. However, in general, we cannot decide if B E A(B) or
if f behaves like the function in Theorem 1.5. If we examine the
first part of the proof of Theorem 1.5 we find that the key step in
concluding that f and B satisfy (1) and (2) was the application of

B -1
Ax’wB = (1) to the equation f(blh2)(hlb2)£3 = f(bl)f(b2)(bl) oEoé.
—1
But if we know that either B e A(B) or f(blbz) = f(bl)f(b2)(bl)

then both must hold. Hence we have:

Theorem 1.16: Let G = AB, A‘Q o, and Y E B(A,G). Then f(blbz) =

 

3 .
f(bl)f(b2)bl for all bl,b2 e B iff B e A(B) where hY = f(b)b£3 with

f(b) g A and bB e B.

26
Throughout the remainder of this section G will be a finite group.

If B1 is a supplement of A in G such that \B1\ = \B\ and B1 =

A
{f(b)bB\B e A(B) or equivalently f(blbz) = f(bl)f(b2)bl for all
bl,b2 E B} then we say that B1 is related to B and write Bl ~B.

We call f an associating function and B an associating automorphism.

 

 

With this notation we have:

Theorem 1.17: Let c = AB with A‘Q G. Then x‘le ~B for all

 

x E A. Moreover we can choose an associating function f which behaves

like a principal crossed homomorphism and an associating automorphism

. -l -1
Proof: We can write x bx = x bxb b = x x b. For each

—1 —l
_l _ _
b e B define f(b) = x xb . Then x le = {f(b)b\f(b) = x lxb for

all b e B}. Hence x'le ~B with the desired properties.
Summarizing the remarks preceeding Theorem 1.16 we have:

Theorem 1.18: Let c : AB with A <16. If Y e B(A,G), hY = f(b)b£3

 

with f(b) e A and h£3 e B, and Y‘A = c then
(1) hCI = f(b)bB for all b g.A(‘\B;

(2) who 2 an on A.

rhino£3

Next we obtain a partial converse of the above theorem.

Theorem 1.19: Let G = AB, A:Q G, and B1 -B with associating

 

function f and associating automorphism B. Then a E A(A) can be

extended to G if

27
(1) bCI = f(b)bL3 for a11 b e A(‘NB;

(2) who = on on A.

f(b)bB

Proof: Define T by bT = f(b)bg. Since f is a function and B
an automorphism T is well defined. Let b1,b2 be any two elements of

B and we have

(blbz)T f(blb2)(blb2)£3

= f(b1)f(b2)b1Bb§bg

= f(bl)b§r(h2)b;@bEbE
= mops? f(o2)hE

= b1 b2 . ‘

Thus Y is a homomorphism from B to B1. Clearly T is onto and since

G is finite T is an isomorphism. From (1) we have bCL = bT for all

b E A(TNB. From (2) we have nbd = an T on A. Therefore, from the
b

sufficiency part of Theorem 1.3, there exists Y e A(G) such that

Y‘A = a.

1.3 Power and Central Automorphisms

In the first three results of this section we consider the prob—
lem of extending a power automorphism of A to G. A power automorphism

. k . .. ... . .
maps each a E A onto a where k is some fixed p051t1ve integer.

Theorem 1.20: Let G = AB, A <lG, and A(AWB = (1). Then every

 

power automorphism of A can be extended to G.

28

a . . . - . a k
Proox: Let a be any power automorphism 01 A i.e., a = a

for all a e A where k is some fixed positive integer. We show that
the pair a 1 can be lifted to G. By Theorem 1.2, we need only show

’ B
. . .1 . . 1 bd
that who dnb on A. Now 11 a 18 any element or A we have a =

H

-lab)(l - (b-lab)k = b-lakb = b—laab = a(lb so that n a = on on A.

(b b b

The above result tells us that when A is complemented then every
power automorphism can be lifted. Now if Y E B(A,G) such that Y‘A
is a power automorphism then what is its effect on the rest of G?
This question is answered by the next theorem when A is supplemented

in G.

Theorem 1.21: Let G = AB with A‘Q G. If Y e B(A,G) such that

 

Y A = d is a power automorphism of A then bY = g(b)b where g is a
-1
function from B to CG(A) and g(blbz) = g(bl)g(b2)bl for all bl,b2 e B.

Proof: By Theorem 1.3 we know that nbd = on Y on A. Let aCI = ak
b

where k is some fixed positive integer. Now

bd de
a = a

(b_lab)a= b-YaabY

(h‘lab)ka b_Yaka

b_lakb = b_Yaka
so that hYb'l g CG(A). Set th‘l = g(b) with g(b) e CG(A) then
clearly g is a function since Y e A(G). Since Y E A(G), (blbz)Y =

bybg for all bl,b2 E B. Hence

g(blb2)blb2 = g(bl)blg(b2)b2
—l
g(blb2)b1b2 = g(bl)blg(b2)bl blbz
. b—
9031132) : g(b1)g(b2) l .

29
Under certain conditions we can take such a function as described
in the above theorem, pair it with a power automorphism of A and ex-

tend to G.

Theorem 1.22: Let G = AB, A <1G, and A/“\B = (1). Let a E A(A)

 

be a power automorphism and g a function from B to CG(A) such that
-1
b _
g(blbz) = g(bl)g(b2) l for all b1,b2 e B and g(b) i b 1 (mod CA))

for b A 1. Then the pair c,T, where bT = g(b)b, can be extended to

Y e A(G) provided 6 is finite.

Proof: We will show that we have all the conditions of Theorem 1.3

satisiied. (I) 18 clear. Since g(b) E CG(A) we have an = fig(b)b =

”b on A. a is a power automorphism of A so that who = Gflb. Hence

an : Gfib 2 who so we have (2). Now T is well defined since g is
bT

a function and if bl,b2 are any two elements of B we have

(bibz)T 9(b1b2)bib2

-l
b
9(b1)9(b2) l blbg

H

, -1
glb1>b19(b2)b1 blbz

9(b1)b19(b2)b2

of cg.

Thus T is a homomorphism. Let b E ker T i.e., bT = g(b)b = 1. Then.
g(b) = b.1 so clearly g(b) 3 b_1 (mod (A)). Thus b = l and ker r = (1).
Since G is finite Y is an isomorphism. Finally we must show that

G = ABl where Bil: BT. Assume otherwise. Then we must have A(ANB1 # (13

since G is finite. So there exists a b e B and an a E A such that

3O
g(b)b = a where a T 1. Then g(b) = ab_l so that g(b) E b_1 (mod (A)).
—1
Hence b = 1 and from g(blbz) = g(bl)g(b2)bl one can easily show that

g(l) = 1. Therefore, a = 1 which is a contradiction. Hence we must

have 6 = ABl.

The remaining results in this section have to do with lifting

central automorphisms of A to G. Y is said to be a central automorphism

 

of the group H if hY = Zhh where 2h 6 2(H) for all h E H. One can

show that the mapping h to z is a homomorphism of H into 2(a). Thus

h
.1 . . . . a .. d a a
11 d is a central automorphism o: A we can write a = 1(a)a for all

a E A where f E Hom (A,Z(A)). Clearly f(a) + a—1 if a T 1. We define

H6R(A,Z(A)) {f e Hom(A,Z(A)Mf(a) + a.1 for a % 1}. It was shown
in [1] that there is a l-l correspondence between the central auto-
morphisms of A and the elements of Hg; (A,Z(A)). Let A be supple—
mented by B in o and define H§E(A,Z(A)) = {f g H§m(A,Z(A))|f(b‘lah) =

-1
b f(a)b for all a E A and b E B}. Then we have

Theorem 1.23: Let c = AB with A <10. If f e Hgm(A,Z(A)) and
an
A(AWB f ker f then the central automorphism of, a I = f(a)a, of A

 

associated with f can be extended to Y e A(G) such that Y|B = 18.
Conversely, if Y e B(A,G) is such that Y|A = of is central and Y B 2 1R

then f6 Hgm(A,Z(A)) and A(‘TB f ker f.

Proof: (a) Let f e Hgm(A,Z(A)) such that A/’\B f ker f. Then
a.
defined by a I = f(a)a, is a central automorphism of A. We will

of,
show that df\A{”\B = lB|A(fi\B and nbdf = dfnb so that, by Thegrem 1.2,
the pair a 1 can be extended to G. Now if b E A{A\B then b I =

f> B
f(b)b = b since b e ker f. Hence dfiA(“\B = iB|A(”\B. Let a be any

31

baf —1 0Li" -1 -1
element of A and we have a = (b ab) = f(b ab)b ab =
-1 —i —i -i CLf afb . .
b f(a)bb ab = b f(a)ab = b a b = a so that n a- = a n

b I f b'

(b) Let Y e B(A,G) such that Y|A = of is central and Y|B = 18.

Then, by Theorem 1.2, we must have n a- on A and df\A("\B =

b I 2 Clffib
C1f
blB = b = f(b)b so that

lB\A(~\B. Let b e A(TTB then we have b

f(b) = l and A(TNB f ker f. Let a be any element of A and we have

a“ b
(b‘lam I = (f(a)a)

f(b—lab)b_lab = h'1f(a)ah

l

f(b~lab)b—lab = h" f(a)bb—lab

f(b-lab) = b—lf(a)b.

Therefore, f E H68 (A,Z(A)).

If we add to the above theorem the further restriction A(‘NB = (1)
then the condition A(TXB < ker f is trivally satisfied so we can re—

phrase the theorem as follows:

Theorem 1.2h: Let c = AB, A‘Q c, and A(“WB = (1). If f e

 

Hom (A,Z(A)) then the central automorphism of of A associated with f

can be extended to Y E A(G) such that Y‘B = l and conversely.

B
What conditions must we put on d E A(A) so that d can be extended

to a central automorphism of G? This question is answered in the

next result under the hypothesis that A is supplemented in G.

Theorem 1.25: Let G = AB with A <1Gy Then Y E B(A,G) is central

 

iff there exist homomorphisms f and g such that

32

(1) f g H3R,(A,z(o)r‘\A), g g wéw (B,Z(G)), and f(a)g(b) A (ab)'l
for all ab t 1;

(2) f(b‘lab) = f(a) for all a e A and b e B;

(3) f(b) = g(b) for all b g A(“TB.

Moreover aa = f(a)a, bT = g(b)b where d = Y|A and Y = Y B.

3322:: (a) Necessity

Let Y e B(A,G) such that Y is a central automorphism of G. Then
we can write xY = h(x)x for all x E G where h E Hdm (G,Z(G)). Define
f and g by f = h|A and g = hlB. Clearly g e Adm (B,Z(G)). Since Y e

B(A,G) aY = f(a)a E A so we must have f(a) E A. Thus f(a) E Z(G)(T)A

and f e Hdm (A,A(”\Z(c)). Since Y e A(G), (ab)Y h(ab)ab i lwif ab A 1

so h(ab)

f(a)g(b) % (ab)—l if ab % 1. Since f = h|A we have

f(b‘lab) h(b’lab) = h(b)‘1h(a)h(b) = h(a) = f(a) since h(a) e 203).

(3) is obvious in view of the way f and g are defined.

(b) Sufficiency

Now suppose f and g are given satisfying (1), (2), and
(3) of the theorem. Define h by h(ab) = f(a)g(b) and Y by (ab)Y a
aabT where aa = f(a)a and bT = g(b)b. Then we have (ab)Y =

f(a)g(b)ab = h(ab)ab where h(ab) E Z(G). If we show

f(a)ag(b)b
that h E Hdm (G,Z(G)) we will have the desired result. Now if a1b1=
.. . . -l *1
azbz with a1 6 A and bi E B for l = 1,2 then a2 a1 = bzb1 E A(TNB
so by (3) we have
a -1 —1
—1- -i
f(az) 1(31) = 9(b2>g(b1)

f(al)g(b1) = f(a2)g(b2)

33
h(albl) = h(azbz).
Thus h is well defined. Let albl and azb2 be any two elements of G
where ai E A and b1 6 B for i = 1,2. Then alblazbz = al(b1a2b11)blb2
so that using (2) at the appropriate step we have
h(alblazbz) = h(al(bla2b;l)blb2)
‘ = r(al<bla2b;l))g<blb2)
= f(al)f<b1a2b11)g(bi)g(b2)
= f(al)f(a2)g(b1)9(b2)
= f(a1)g&n)ifa2)g(b2)
= h(albl)h(a2b2).
Hence h 6 Hon (o,z(o)). Finally if ab + 1 then, by (1), we have

h(ab) = f(a)g(b) + my1 so that h e Hbm (o,2(o)>.

The above result tells us little of what happens to B. If we
further hypothesize A abelian and A(T\B = (1) then the result can
be sharpened. Under these conditions it is easy to show that Z(G) =
AlBl where A1 = CA(B) and B1 = cB(A)(“\2(B). With this notation we

have:

Theorem 1.26: Let G = AB, Aid G, A abelian, and A(“\B = (1).

 

Then necessary and sufficient conditions that Y E B(A,G) is central
are that there exists a homomorphism f from B to Al and a B E A(B)
such that

(1) d(=Y|A) and B are multiplier automorphisms with multipliers
from Al and B1 respectively;

(2) who = on on A.

b

3A

3322:: (a) Necessity

Let Y E B(A,G) be central. Then we can write gY = h(g)g where
h(g) E Z(G) for all g E G. By Theorem 1.5 there exists a function f
from B to A and a B E A(B) such that f(blbz) = f(bl)f(b2)b1B for all
Now bY = f(b)bB so we have bY = f(b)b£3 =

bl,b2 EBandnd =(1TT

b f(b)bB.
h(b)b and by the remarks preceding this theorem we can write h(b) =
albl with al €,Al and b1 6 B, since h(b) e Z(G). Since A("NB = (i)

it follows that f(b) = a1 and bB = blb. Therefore f is a function
from B to Al and B is a multiplier automorphism with multipliers

from Bl. Since A1 = CA(B) we have f(blbz) = f(bl)f(b2)b1fl = f(bl)f(b2)
so that f is a homomorphism from B to Al. Let a E A then since

Y e B(A,G), aY'= h(a)a e A so that h(a) g A. Thus h(a) e.A(‘)z(o)§Al

so that a = Y|A is a multiplier automorphism with multipliers from

A1. Now WCL= on

b on since f(b) E A and

f(b)bp = 0erf(b)b,b = b

b1 6 CB(A). Thus bd = dub.

(b) Sufficiency
Let c,B, and f be given satisfying (1), (2), and (3) of
the theorem. By (2) we can write b;3 = blb with b1 6 Bl. Thus

CL CITT,‘ = CITT

”-(b)bB = I(b)blb Since f(b) E A and b: E 81.5 CB(A). But
I

D

b

by (3), cub = who so that on = nbd. Hence, by the sufficiency

f(b)bB

part of Theorem 1.5, there exists Y E A(G) such that Y A = c and

bY = f(b)bfi. By (2) we can write aCL = ala with al 6 Al. Therefore
(ab)Y = aabY = alaf(b)bB = alaf(b)blb = alf(b)b1ab and since
a1f(b)bl E AlBl = Z(G) we have that Y is a central automorphism of

G.

35

l.h The Group B(A,G)

Becaii that B(A,G) = {Y e.A(6)|Y|A = o e A(A)}. We define

C(A,G) = {Y e B(A,6)\Y|A = 1,}. Then we have

Theorem 1.27: C(A,G) Q B(A,G) and [B(A,G):C(A,G)] = the number

 

of automorphisms of A that can be lifted to G.

Proof: Let Y e B(A,6) and Y e c(A,6). If a is any element of

"J.— _l — ‘1 l "l
A then we have aY YY = (aY )YY: (aY )Y = a A since aY E A. Thus

Y—IYY|A = 1A and Y‘1?Y e c(A,6). Hence c(A,6) q B(A,G).
. - l . .
If Y1,Y2 6 B(A,G) and Y1|A = Y2‘A then Y2 Y1|A = 1A so that
Ylel e c(A,6). Thus Y1C(A,G) = Y2C(A,G) and we have the second

assertion of the theorem.

The next result tells us when no automorphism of A different
from lA can be lifted to G under the hypothesis that A is comple—
mented in G. In view of the above theorem this also tells us when

B(A,G) = c(A,6).

Theorem 1.28: Let 6 = AB, A <16, and A(‘iB = (1). If \A\ A 2

 

then A has a nontrivial automorphism that can be lfited to G i.e.,

lB<A.G>/c(A.G)\ i 1-

Proof: Suppose that B(A,G) = C(A,G). Then since Hg 6 B(A,G)
for all g E G we must have a9 = a. Thus A f Z(G) so B <16 and G = A x B.

Let a be any element of A(A). Since Hb = 1A for all b E B we have

nbd = Gfib. Hence, by Theorem 1.2, the pair 6,1B can be extended to

G. Therefore, we must have a = 1A i.e., \A(A)‘ = 1 and since A is

36
abelian it follows that \A‘ = 2. This is a contradiction so the

theorem is true.

The next theorem gives us a set of sufficient conditions under
which every automorphism of A can be extended to G. But first we
prove a lemma which is of some interest itself. Let B be’a supple—
ment of A in 6 and define B* to be the subgroup of A(A) consisting
of those automorphisms of A induced by the elements of B i.e.,

B* = {nb\A\b e B}. Then we have:

 

Lemma 1.29: Let 6 = AB, A<] 6, and CB(A) = 1. If a g NA(A,(B*)
and A(ANB is c—invariant then there exists a unique B E A(B) such

that the pair 6,B can be lifted to an automorphism of G.

Proof: Let b1 and b2 be any two elements of B. If n = n

b1 b2

on A then bglbl g CB(A) and since CB(A) = i it follows that bl = b2.
We will use this fact several times in the proof. From the above

remarks it follows that B : Bi". Since o e N )(B*) and B : B"" we

A(A

_ _1
have for each b E B a unique b E B such that a n a = n—. Define B

b b

by bB = 5. Clearly B is well defined. Let b e ker B i.e. bB = l.

)

—1 . .
Then a n d = HI = 1 so that n

b A b:A

(l). B is onto since 6 induces an automorphism of BW and B“ :—B.

Now let b1 and b2 be any two elements of B and we have

—1 -1 -1 -1
W = CI 1T 0. = (1 TT 1T (1 = 0. TT C10 TT 0. = TT H = TT .
out .1 b2 b1 b2 bi bi rib‘i

Therefore, since (blbz)B E B and bebg E B we have that (blbz)B =

bEbE. So B e A(B).

l . Hence b E CB(A) = l and ker B :

37

If x e A then it is easy to show that o‘lwxc = nxd. Thus if

-1
= d w a = n since b E A. Since.A(T)B is 6—
bfi b bCL
invariant, bCL 6.A(T)B f B. Therefore w B = n a implies that bB = ba.
b b

Hence G\A(A\B = B\A(A\B. Summarizing welrne d E A(A) and B E A(B)

b 6 A(AXB then H

such that d|A(‘\B = B|A(’\B and who

on Therefore, by Theorem 1.2,

b5'
u and Y‘B = B.

there exists Y E A(6) such that Y|A
Finally suppose that the pairs 6,Bl and 6,B2 can both be extended

a = on and n a = on
b bBl b

Hence W = a W a = W so that bBl = bBZ. Since b is an arbitrary
bgl bfiz

to G. Then, by Theorem 1.2, we have n B .
b2

element of B we must have B1 = B2.

If we add to the hypothesis of the above lemma B* :3 A(A) and
A(TNB characteristic in A then we can say that each a E A(A) can be
paired with a unique B e A(B) and extended to 6 since NA(A,(B*) = A(A)
and A(TNB is 6—invariant for each a e A(A). The next result gives
us a set of sufficient conditions under which every automorphism of

A can be extended to G that was referred to above Lemma 1.29.

Theorem 1.3o: Let 6 = AB, A‘Q 6, CB(A) = (1), B* :3 A(A), and

 

assume that.A(~\B is a characteristic subgroup of A. Then

Bmfiptmfi):AML

Proof: Clearly B(A,6)/t(A,6) :—to a subgroup of A(A) under the
correspondence YC(A,G) <+> Y‘A. By the remarks following Lemma 1.29

each automorphism of A can be extended to G so, by Theorem 1.27,

[B(A,6):c(A,6)] = \A(A)‘. Thus B(A,6)/t(A,6) :'A(A).

38
If an automorphism of a subgroup can be extended to the whole
group then it can usually be done in many ways. In fact Theorem 1.27
tells us that it can be extended \C(A,G)‘ ways. In our next result
we try to find out when an automorphism of a normal abelian subgroup

has a unique extension to G. This, of course, is equivalent to ask-

ing when is \C(A,G)\ 1. First we prove the following:

Lemma 1.31: If A <1G and C(A,G) = (1

 

G) then CG(A) = 2(6).

 

Proof: Clearly Z(G) 5 CG(A). Let g e CG(A) then mg g C(A,G) =
(1G) so x9 = x for all x E G. Hence g E Z(G) and we have CG(A) f

Z(G). Thus CG(A) = Z(G).

Theorem 1.32: Let A be a normal abelian subgroup of G. Then

 

C(A,G) = (16) iff 6 is abelian, [6 A] = 2, and 1A| is odd.

Proof: (a) Necessity

 

Suppose C(A,G) = (lG>° Then, by Lemma 1.31, CG(A) = Z(G). Since
A is abelian we have A.f CG(A) = Z(G). Hence if g e G then Hg 5 C(A,G)
so xg = x for all x E G. Thus 9 E Z(G) and since g was arbitrary we
have that G is abelian. Now we will show that G must split over A.

We can write G = S x °" x S , A = A x ... x A , and G/A =
p1 pr P1 Pr

S x ... x S where S is a .—S low sub rou of G A = A S .

pl Pr Pi P1 y g P . pi f“) pi,

and S :-S /A . We will assume that 6 does not split over A and
pi pi pi

arrive at a contradiction by constructing a nontrivial automorphism

Y of G such that Y|A = 1A and Y‘G/A = 1 By Theorem 1.12 we need

6/A'

fl

only construct a nontrivial homomorphism 1 from G/A to A to accomplish

this. If (‘A\,[G:A]) = 1 then G splits over A so we may assume that

39
for some i, Api # (1) and Spi % (I). If Api is a direct factor of
Sp. for each i then G splits over A so we may assume for some i that
AP: is not a direct factor of Spi. There is no loss in generality

if we assume i = 1. Define f‘Sfi = 1 for i > 1. Now S is a direct
' 1
1

product of cyclic subgroups, say Sp = Cl x ... x CS where C1 is
I . 1

generated by Xi' Then we can write Ap = Al x -°- x AS where Ai =
1

Ap (”\Ci. Since Ap is not a direct factor of Sp there exists at
1 i 1

least one Ai such that l ,<_:Ai F Ci’ :ay Al. Define f‘Ci = l for

i > 1. Let ‘x1\ = p? and A1 = < x1p1> where we may assume 0 < u < t.
The only cosets of G/A on which f has not been defined are of the

k k u k
form xlA. Define f(xlA) = (xlpl) . Clearly f maps G/A into A1. Now
any two elements 51,52 of G/A can be written in the form 5, =

xlklyA, g2 = xlkzzA where y and z belong to C2 x ... x CS x S x "'x 5

P2 Pr.

_ U k _ u k — — u (kl+k2)
Hence f(gi) = (le1) 1. f(gz) = (xipl> 2. and f(gig:> = (x p1)

since glgz = xgkl+k2)yzA. Therefore f is a homomorphism from G/A to

u _
A and f(xlA) = x1pl # 1 so f is nontrivial and we have arrived at a
contradiction. Thus we may assume that G splits over A.

Let G = AB, A(A3B = (l), and B e A(B). Since ”b = 1A for all

b E B we have nblA = 1AnbB
can be extended to Y E A(G). Now Y e C(A,G)= (1

so that, by Theorem 1.3, the pair 1A5 B

G) so we must have

B = 1B i.e., \A(B)\ = 1. Since B is abelian we must have ‘B‘ 2.
Let B = < b > where b2 = 1. If there exists an x g A such that x2 r 1
then, using Theorem 1.3 as before, we can show that the pair 1A, T

where bT = xb can be extended to Y e A(G). But Y E C(A,G) = (10} so we

must have Y = 1G' Hence bT = b i.e., x = 1 and TA! is odd.

(b) Sufficiency

Let G be an abelian group such that G = AB, A(A)B =

\Bl = 2, and ‘A‘ odd. Let B =

then bY is an element of the coset Ab since G

exists an x e A such that bY a
Hence (Xb)2 = X2132 = X2 :

odd unless X = l.

l which contradicts

Therefore, Y = 1

AG

(1).

< b > where b2 = 1. If Y e C(A,G)

A(vjAb. Hence there

xb. But \b\ = 2 so that 1xb\ = 2.

the fact that \A\ is

and we have shown that C(A,G) = (1 ‘.

6 G)

Now we try to find out something about the structure of B(A,G)/C(A,G).

We know that B(A,G)/C(A,G) is isomorphic to a subgroup of A(A) and

also that NG(A)/CG(A) is isomorphic to a subgroup of A(A). The

next result tells us that B(A,G)/C(A,G) contains a subgroup isomorphic

to NG(A)/CG(A) and we make no assumptions about A.

Theorem 1.33:

 

NG(A)/CG(A).

B(A,G)/cm.

G) contains a subgroup isomorphic to

Proof: We define Q: NG(A)/CG(A) ——> B(A,G)/C(A,G) by (gCG(A))e :

ngC(A,G). Since g E NG(A), fig

92 91

gglgl E CG(A) so that H 1 = w

n C(A,G) =

H C(A,G) and 9 is
91 92

be any two elements of NG(A)/CCT

(91CG(A)92CG(A) )8 =

e B(A,6). If gch(A) = gZCG(A) then

-lfigl E C(A,G).
92

Therefore

well defined. Let glCG(A) and gZCG(A}
(A). Then we have

(91(3ch (A) )9

H C A G
9192 ( ’ >

‘IT IT

C A G
91 92 ( ’ >

W C A G H C A G
gl<,>g,<,>

b1
6 e
= (gch(A>) (92CG(A))
. . . . . . 6
so that G is a homomorphism. Let gCG(A) e ker G i.e., (gCG(A)) =

C(A,G). Then mg g c(A,6) so a9 = a for all a e A. Hence g g CG(A)
and ker o = (1). Thus 9 is an isomorphism.

In the remarks preceeding Lemma 1.31 we mentioned that the number
of ways an automorphism of A can be lifted to 6 equals \c(A,6)\. 1n

the next result we calculate |C(A,G)\ under very Special hypothesis

on G.

Theorem 1.3h: Let 6 = AB, A <lG, A(‘)B = (1), CB(A) = (l), A abelian

 

and assume that all complements of A are conjugate. Then |C(A,G)\ =

[A:CA(B)].

Proof: Let 6 E NA(A)(B*)' By Lemma 1.29 there exists a unique
B E A(B) such that the pair 6,B can be extended to G. Then, by

T -1
Theorem 1.2, n 6 = 6n 5 on A. Let x g A and define TB by b g = x be.
b

b
Since A is abelian we have n = n = 6—16 6 on A. Therefore
-1 B P b ’
x b x b
nb6 = an and, by Theorem 1.3, the pair 6,Tfi can be extended to

bTfi
Y E A(G) such that YiA = 6 and Y‘B = YE.

Now let Y E B(A,G) such that Y‘A = 6. Then,by Corollary 1.15,
-o
bY = f(b)be where B e A(B) and f(b) = x-lxb for some x g A. Thus

bY = lexb b8 = x_lbexb_ebe = x-lbex so that, by Theorem 1.3, we
must have n 6 = an = OH . Since x E A and A is abelian this
b bY x-lbex

reduces to Hbd = GW 6' Therefore, by the sufficiency part of Theorem 1.2,

14.2
the pair 6,9 can be extended to 6. But fl is unique so we must have
6 = B- We have shown that if Y e B(A,G) such that y\A = c then
bY = x-lex for some x e A. Finally if x,y e A and x_lex = y—ley
then yx-l E CA(B)’ Hence the number of ways 6 can be extended to G
is the number of distinct cosets of CA(B) in A i.e., [Ach(B)].

Therefore, ‘C(A,G)[ = [A:CA(B)].

If we add 8* :fl A(A) to the hypothesis of the above theorem
then, by Theorem 1.3o, B(A,G)/C(A,G) :,A(A) and it follows that

Bwflfl|=VX%®\LMM\=[MQJEJMMH'

CHAPTER II
2.1. Introduction

Throughout this chapter A will be a complemented subgroup of
G and at first we do not insist that A be normal in G . We will
construct a permutation group on the complements of A in G and
investigate what happens to G under various hypotheses on this per—
mutation group. Por the sake of completeness several definitions
relevant to the theory of permutation groups will be included. One
question we have in mind is what condition on this permutation group
will insure that all complements of A in G are conjugate.

Denote by {2 the set of all complements of A in G . If
Y E B(A,G) and B is any complement of A in G then it follows
that G = GY = (AB)Y = AYBY = ABY so that BY is a complement of
A in G . Thus the elements of B(A,G) can be considered as per—
mutations of the set Q . B(A,G) need not be a permutation group on
0 since different elements of B(A,G) may induce the same permuta—
tion on 52. However, we can obtain a permutation group on Q in the
following way. Let B(A,G) denote those elements of B(A,G) which
fix every element of S? i.e., Y E B(A,G) if and only if BY = B

for all B 6 $2. Then we have:

Lemma 2.1: c(A,6) d B(A,G)

Proof: Let Y E B(A,G) and Q E B(A,G). If B is any comple—
-1 -l -l
ment of A in G then we have BYeY = (BY)eY = (BY)Y = B
. —1 . .
since BY 6 {2. Hence Yoy e B(A,G) and B(A,G) 18 normal in

B(A,G).
A3

Ah
Now )( = B(A,G)/B(A,G) can be considered as permutations on Q
by defining B? = BY where Y = YD(A,G), Y E B(A,G) . If Y1, $2
6 X: induce the same permutation on Q i.e., if BY1 = BY.2 for
all B E S) then BYlYZ-l = B for all B E S) so that Y1Y2_l E B(A,G).
Thus $1 = $2 . For Y e B(A,G) we denote by V the coset of X

which contains Y . At this point we give an example that will be

referred to later.

W: G = < ajbgc >3 a5 = b2 = C4 = l, 1331’) "-'- a..-l 9

bc = cb c_lac = a2 . Let A = < a,b > then the complements of

A are:
B0 = <c> = {l,c,c2,c3} B5 = <bc> = {l,bc,c2,bc3}
B1 = <ac> = {l,ac,a4c2,a3C3} B6 = <abc> = {l,abc,a3c2,a2bc3}
B2 = <a2c> = {1,a2c,a3c2,ac3} B7 = <aZbc> = {1,a2bc,ac2,a4bc3}

B3 = <a3c> {1,a3c,a2c2,a4c3} B8 = <a3bc> {1,a3bc,a4c2,abc3}

B4 = <a4c> [1,a4c,ac2,a2c3} B9 = <a4bc> {1,a4bc,a2c2,a3bc3}

Now the automorphisms of G are:

a —> a1 a —> a1 a —> a1 a —> a1 a —> a1

b —> b b —> a4b b —> a3b b —> a2b b —> ab

c —> c c —> ac c —> a2c c —> a3c c —> a4c

a —> a1 a —> a1 a —> a1 a —> a1 a _> a1

b -> b b —> a3b b —> a4b b —> ab b —> a2b

c —> bc3 c —> aZbc3 c —> abc3 c —> a4bc3 c —> a3bc3
where i = l,2,3,h. From the above it is clear that A is charac—

teristic in 6 so B(A,G) = A(6). The only automorphisms which fix

' i
a—>a

B0 are b —> b for i = l,2,3,h. Of these only the identity fixes
c —> c

b5
—i i _ 2i
131 so clearly D(A,6) = (1). Thus X : A(G). Now a Boa — <a c >
for i = l, 2, 3, b so B0, B1, B2, B3, B4 form a conjugate class.
Similarly a‘1135a1 = < azlbc > for i = l,2,3,A and B5, Be, B7, B8,
B9 form a conjugate class. To show that ( X,§2) is a transitive

permutation group we need only show the existence of an automorphism

a—>a
which sends B0 to B5. Now b —> b sends B0 to 135 so (X, {2)
c —> bc3

is transitive.
If all complements of A in G are conjugate then (X ,9 )

is a transitive permutation group; in fact, the subgroup

I(G)D(A,G)/D(A,G) is transitive on 8?. However, from the above ex—

ample we can see that transitivity of ()<,S?) is not sufficient

to insure that‘all complements of A are conjugate. In the next

example we point out the fact that (}(,§2) need not be transitive.
Example 2: G = < x,y,z > , x5 = y2 = 22 = l, yxy = X.1 ,

yz = zy , zxz = x-l. Let A = < x,y > then the complements of A

are: B0 = < z >, B1 = < xz >, B2 = < x22 > , B3 = < x32 >, B4 =

< x42 >, B5 = <;yz >. Now x-iBOxi = < x312 > for i = 1,2,3,b so

that B0, B1, B2, B3, B4 form a conjugate class. It is easily

verified that Z(G) = B5 so no automorphism of G can map B5 to

any other complement. Consequently (){,§2) is not transitive. We

may observe that this group is a subgroup of the group in example

one where x = a y = b and z = c2.

3 3

Our objective in this study is to find a permutation condition
that will characterize when all complements of A in G are con-
jugate but at present we have not found such a condition. However.

we do have several interesting results in this direction.

A6

2.2 The Group (XL,§Z)

If A = {81, B2, Br} is a subset of f2 and' Y = YD(A,6) e x

O
\O

{BIYJ BZYy "') B Y} :

Then by AY we mean AY r

Definition 2.2: If (X ,9 ) is transitive then a subset A

 

of Q is called a block if and only if

(1) l < lAl < \§2|
(2) AVFWA = {3? for each Y E X.

Definition 2.3: The transitive permutation group (X ,9.) is

 

said to be primitive if and only if it has no blocks.

Now we can obtain the following result:

Theorem 2.b: If ()(,52) is a primitive permutation group

 

then either
(1) a11 complements of A are normal

or (2) all complements of A are conjugate.

Proof: Let B E {2. If Bid G then every complement of A

 

is normal since (X ,0 ) is transitive. Suppose B is not normal

in G and consider the set A = {BHX = X-1BX‘X E A}. Suppose

A;(A\A %’¢’ for some Y €3< i.e., there exist x and y in A such
that (x-le)? = y-lBy. Now (x_le); = (x-le)Y = (xY)—1BYxY where

Y = YD(A,6). Since Y e B(A,G) we have XY 6 A so that BY =
ny‘layx‘Y e A . Thus AT = A . If 1 < tA| < |£l| then A is a
block of (X ,8?) so since (){,§2) is primitive we must have either

\A‘ = 1 or \A\ = \ST‘. ‘A‘ = 1 implies that B‘Q G so we must

A?
have \A‘ = ‘fl‘. Therefore, all complements of A in G appear in

A i.e. all complements of A in G are conjugate.

)
To this point we have not hypothesized A normal in G . How—
ever, if A is normal in G then from the above result we see that
all complements of A in G are conjugate if (){,§2) is a primi—
tive permutation group and A is not a direct factor of G . That

these conditions are not necessary is pointed out by the following,

example.

Example 3: G = < a,b > , a15 = b2 = 1 , bab = a_ . Let A = < a>
then the complements of A are: B. = < an > , j = O,l,2,oo-,1A.

Now A is normal in 6 and (|A|, [GzA]) = 1 so that A is charac—
teristic. Hence B(A,G) = A(G). The automorphisms of G are:
i
a —> a '
Yi .: . where (i,15) = 1 and j = O,1,2,-°',1A. The only
DJ b _> an
automorphisms which fix B0 are of the form Yi o . Of these only
)
the identity fixes 131 so clearly D(A,6) = (1). Hence )1 :- A(6).
Now aSBOa—S :.< aZSb > so all complements are conjugate and (X ,0)

is transitive. However, (X.,§2) is not primitive for we will show

that A = {B0, B5, B10} is a block of order 3. Let Br = < arB > e A

i.e., r

Y. . . .
O,S,1O then Brl’J = < ar1+Jb > . If 5 does not divide

‘Y‘. . Y. .
j then 5 does not divide ri + j and Brl’J E A. Hence A 1’J(~3A =(Z
in this case. If j = O,5,lO then ri + j will take on the values
Y. .
0,5,10 mod 15 as r takes on the values 0,5,lo. Hence [A 1’J(T\A = A.

Therefore, A is a block and (X ,fZ) is imprimitive.
In the next result we obtain some information about the structure of

A in case (){,§2) is primitive.

A8

Theorem 2.5: If (X%,Sl) is primitive then either

 

(1) A is characteristically simple
NA(B>

or (2) if H is a characteristic subgroup of A then HNA(B) ={
A

where B is any element of Q

Proof: Suppose A is not characteristically simple and let H
be a characteristic subgroup of A . By Theorem 2.A all complements
of A are normal or all complements of A are conjugate. If B ES)
and Bid G then H NA(B) = H A = A . Thus we can assume that B is

not normal in 6 . Form the set A = {BAX = x'lbxlx e H} and let

Y 6 it . If AYK‘iA %'¢' i.e., if there exist x and y in H such

that (x_le)Y = y_lBy then BY = ny-lByx-Y where Y = YD(A,G).

Since Y|A e A(A) and H is characteristic in A we must have

xY e H . Thus BY 6 A . If 1 < |A| < |g2\ then A is a block so

since (X ,fZ) is primitive we must have either ‘A‘ = l or
\A\ = \§2| . If \AI = 1 then H j NA(B) . If |A\ = |§2| then
all complements of A appear in A and it follows that 1 6| =

[H : H (T3NA(B)] . Since all complements of A are conjugate we also

 

have |§2‘ = [A : NA(B)] . Thus
|N (B)H| ‘NA(B)A AHA |N (B)\ l l \N (B)|[A N (B)] \A\
A 1H(”\NA(B)\ A A A

so that NA(B)H = A .
We can observe that we did not use the full power of H being
characteristic in A but just invariant under )( . So we have the

following as a corollary to the proof of Theorem 2.5.

Corollary 2.6: If (X ,gz) is primitive and H a subgroup of A

 

A9
NA(B)
invariant under X then HNA(B) =
A
From this point on we will assume that A is a normal comple-
mented subgroup of G . The fixed group of an element B E Q , de-
noted by IX B, is the set of Y in X such that BY = B . It is
well known that [7, pg 15] in a primitive permutation group the
fixed group of a point is maximal. Thus X B is maximal in 3X and
we can obtain the following result where IZAS = I(A)D(A,G)/B(A,G).

Theorem 2.7: If (X ,SZ) is a primitive permutation group and

 

A is not a direct factor of G then X = X BIZAS.

Proof: If Ea e :XB for all a e A i.e., if a‘lBa = B for

 

all a E.A then B‘Q G and A is a direct factor of G since
Aid G. Hence there exists x E A such that fix K'X B' Thus the

group generated byAX B and fix is IX since )AB is maximal in )(

Let Bl, 52 eiX B where B.

1 = OiD(A,G) for i = 1,2. One can easily

Show that elfixez = 9192” so that we have Bl? é @lézfi

X82 x 2 x82

9
Since x e A , x 2 e A and it follows that the group generated by
X B and fix is X BIZAS .

We can replace the hypothesis A not a direct factor of G by
the condition (|A|,|B\) = 1 in the above theorem for if B‘Q 6 then
B is characteristic since (|A|,‘Bl) = 1. But (){,§2) is transi—

tive so that \SZ‘ = 1. Thus we have:

Theorem 2.8: If (X ,S)) is a non—trivial primitive permutation

 

group and (|A\,|B|) = 1 then X = X Ble).

SO .

Definition 2.9: The permutation group (X ,3?) is said to be

 

3/e—fold transitive if and only if (x:,§2) is transitive and the

 

orbits of (3CB,§2—{B}) have equal length.

Definition 2.10: A subgroup N of the permutation group ()(,§2)

 

 

is said to be semiregular on S) if and only if NB = (1) for all
B e O .

Example 1 shows that transitivity of (){,§2) will not insure
that all complements of A are conjugate. Also, from example 3, we
see that primitivity of (:X,§2) is too strong. So the condition
seems to rest somewhere between transitivity and primitivity. Now
we will impose the condition 3/2-fold transitivity on ( X,§2) which
is weaker than primitivity. We will obtain a theorem like Theorem

2.A under additional hypothesis on G . But first we need.

Lemma 2.11: If (X ,g)) is transitive and IZGS is semiregular

 

on {2 then
(a) NG(Bl) = NG(B) = NA(B)B for any two complements Bl, B
in f2 3
(b) Every complement of A is contained in NG(B) for any
B ES) 3 I
(c) NA(B) and NG(B) are normal subgroups of G ;
(d) NA(B) and NG(B) are invariant under B(A,G);
(e) If Y e B(A,G) and bY = f(b)b£3 with f and p as given
in Theorem 1.5 then f E Hom (B,ZCNA(B)));
(f) If B E {2 then B(WB1 is normal in B for any B1 in Q ;

(9) For any B E O we have B',f CB(A) 5 {KY} x—le where B'
ng

is the derived group of B.

51

Proof: (a) Let g be any element of 'NG(B1). Then

Eg E IZGSB = (1) so that fig E D(A,G). Thus 9 E NG(B) and we
1
have NG(Bl) j NG(B). Similarly NG(B) EING(B1) so NG(B) = NG(B1).

(b) From (a) we have NG(B1) = NG(B) and since B1 EING(B1)
we have B1 5 NG(B) for any B1 6 S).

(c) For g e G we have 9-1NG(B)g = NG(g-lBg) EUd, by (a),
NG(g‘lBg) = NG(B) so g‘lNG(B)g = NG(B). Now g’lNA(B)g jiA since
Aid 6 and g‘lNA(B)g jING(B) since N3(B) Q 6. Thus g-lNA(B)g 5
A(TxNG(B) = NA(B).

(d) Let Y e B(A,6) then (NG(B))Y = NG(BY) and NG(BY) = NG(B)
by (a). Thus (NG(B))Y = NG(B). Now (NA(B))Y :iA since AY = A

Y< : Y: Y<
and (NA(B)) _ NG(B) Since (NG(B)) NG(B). Hence (NA(B)) _
A (“\NG(B) = NA(B).

(e) Let B Q Q. and Y E B(A,G). Then for any b E B we have
bY = f(b)bB where B E A(B) and f is a function from B to A
satisfying f(blbz) = f(b1)f(b2)b1 for all bl, b2 6 B. Also, from

Theorem 1.5, we know that N 6 = an where 6 = YlA. By (b)

b f(b)b£3

BY : NG(B) = NA(B)B so f(b) e NA(B) for all b e B. Since 6 = AB,
A.f‘sB = (l), and Aid 6 we know that NA(B) = CA(B). Thus

-A
f(blbz) = f(bl)f(b2)bl reduces to f(blbz) = f(b1)f(b2). By (d),

Y E B(NA(B), NG(B)). Therefore, by Theorem 1.5, nbd = 6nf(b)bI3 on

NA(B). But on NA(B) = CA(B) we have ”b = 1NA(B) so this equation

reduces to Wf(b) = 1NA(B) or f(b) E Z(NA(B)). Hence f E Hom(B,Z(NA(B))).

(f) Let B, B1 be any two elements of S? . Since (X ,§2) is

transitive there exists Y e B(A,G) such that BY = B1. Therefore, by

 

52

Theorem 1.5, we can write B1 = BY = {f(b)bfilb E B} with f and B
as given in the theorem. By (e) f is a homomorphism and we will
show that BI’WBl = (ker f)B . Let b E B (“sBl then there exists
b1 6 B such that b = f(bl)b1A. Since .A(’\ B = (l) and f(bl) e A
we must have f(bl) = 1. Thus b1 6 ker f and b1£3 = b. This
implies EBKDNB1,E (ker f)I3 and it is easily shown that (ker f)A.S
B (“\B1. Thus B (T)B1 = (ker f)fi. Since ker f <3 B we have
B (“)B1 <1 B.

(g) Let x e A and b e B . Then x‘lbx = (x‘lbxb‘1)b. Set
f(b) = x‘lbxb‘l then x‘le = {f(b)blb e B} and, by (e), f is a
homomorphism from B to Z(NA(B)). Therefore, for any bl,b2 E B
we must have f(blbz) = f(bzbl). Thus

x’lblbszglb;l x_1b2blxb;lbg1

l

blbsz'z‘lb;1 — bzblxb11b2_

— - — -l
(bllb21b1b2)x x(bllb2 blbz)

So bllbglblb2 E CB(A) and B' E CB(A). Clearly CB(A) fo—le
XE

so we have the desired result.
We may observe that the transitivity of ( X,§2) was used only
in part (f).

If A is a subset of S? then by XI we mean the set of all

A
Y 6 IX such thatrBY = B for all B E A .

Definition 2.12: The transitive permutation group (IX, Q) is
said to be Probenius if and only if x18 % (l) and x A = (1) when-

ever ‘A‘ = 2.

 

53
If ()(,ST) is a Frobenius group then from [7, pg. 11] we know
that the elements of ){ which fix no element of Q together with
the identity form a nilpotent characteristic subgroup of IX . For
the sake of reference we will state two results which appear on pages

25 and 32 respectively of [7]:

Theorem 2.13: If (IX,ST) is 3/2-fold transitive then either

 

(){,ST) is primitive or ( X, 0) is a Frobenius group.

Theorem 2.1A: If (:X, D) is 3/2—fold transitive and N <3 X

 

then either N is transitive on Q or N is semiregular on Q .

Now we are in a position to obtain the theorem referred to earlier.

Theorem 2.15: If (X ,0 ) is 3/2—fold transitive then all comple—

 

ments of A in G are either normal or conjugate provided any one
of the following conditions hold:
(a) (|A\, ‘B/B") =_l where B' is the derived group of B;
(b) If there exists a normal subgroup H of G contained in
A such that H f‘\NA(B) = (1);
(c) If NA(B) is a Hall subgroup of A with a normal complement;
(d)

H
H)

2(A)(’\:NA(B) = Z(NA(B)) and there exists at least one
element of A — NA(B) whose order is relatively prime to \NA(B)\;

(e) If A/NA(B) is not nilpotent.

Proof: Assume (it, Q) is 3/2-fold transitive and the comple—
ments of A are not all normal or conjugate. By Theorem 2.13 either
(){,ST) is primitive or (X ,8?) is a Probenius group. If (X ,9 )

is primitive then, by Theorem 2.A, a11 complements are normal or

5A

conjugate. So we may assume that (X ,9 ) is a Frobenius group.
Now iTGT <J'X so, by Theorem 2.1A, either ITG7 is transitive on
Q or semiregular on $2 . If $757 is transitive on Q then all
complements are conjugate so we may assume that ITGT is semiregular
on Q . Next we show that if B e Q then (i) 71NA(B) ;!A . If
NA(B) = A Then 13 G G and every complement is normal as (X ,0 )
is transitive. Thus NA(B) ;!A. If NA(B) = (1) then NG(B) =
NA(B)B = B. ITGT is semiregular so, by Lemma 2.ll(6), NG(B) <1 G.
This is a contradiction since B = NG(B) is not normal in G. Hence
NA(B> :1 (1).

Thus (X ,0 ) is a Frobenius group, ITGT is semiregular,
(1) f/NA(B) % A, and all complements of A are not normal or con—

B are the elements of {2 then,

jugate. If B1 = B, B2, -, n

applying Lemma 2.11, we have the following structure on G:

 

(1)

Now we will impose the conditions (a) through (e).

 

55

(a) Let B 6 S? . By Lemma 2.1l(e) if B1 = {f(b)bfiib E B} is
any complement of A we know that f E Hom (B,Z(NA(B)). Since the
image of B under f is abelian we may consider f as an element of
Hom (B/B', Z(NA(B))) in the obvious way. But \f(b)| must divide
‘b‘ and since (‘A‘, lB/B") = 1 we must have f(b) = l for all
b e B. Thus B1 = B so that o = {B}. Hence B is the only comple—
ment so must be normal.

(b) Let H <]G, HfA, and HflNA(B) = (1). If h {H and

16H since H<IG. Now

b is any element of B then h—lbhb_
b e NG(B) and, by Lemma 2.1l(e), NG(B) <1 6 so that h“1bhb‘l e NG(B).
Hence h‘lbhb‘l e HflNG(B) = HmNA(B) = (1) so bh = hb for all
b 6 B . Thus h E NA(B), a contradiction, and the result holds in
this case.

(G) Let A = NA(B)D where DKANNA(B) = (l) and D <3 A. Since
NA(B) is a Hall subgroup of A we have (‘NA(B)\,[A : NA(B)])= 1.
Thus (‘D\, [A : D]) = l and since D <3 A we must have D character—
istic in A. Since A <1 G and D is a characteristic subgroup of
A we have that D <1 6. Then D <1 6, D_<_A, and DflNA(B) =(1)
and we can apply (b).

(d) Let x E A — NA(B) such that (‘x|, (NA(B)|) = 1. Since
x E'NA(B) = CA(B) there is at least one element of B , say b ,
such that x_lbxb—l % 1. But, as in the proof of Lemma 2.11(g), the

1

function f defined by f(y) = x‘lyxy‘ for all y e B is a homo—

l

morphism firm. B to Z(NA(B)). Thus x_lbxb— E Z(NA(B)), say

x‘lbxb‘l = n, and n ,1 1. Since Z(A) fl NA(B) = Z(NA(B)) we have

n E Z(A). Let \x‘ = d then from bxb_l = xn we have

(bxb‘1>d = (mud
bxdb-l = and
1 = nd.

However, (d, (NA(B)‘) = 1 so ndl¥ l and we have a contradiction.
Thus the theorem is true in this case.

(e) Since (){,§2) is a Erobenius group we have, from the re—
marks preceeding Theorem 2.13, that the elements of X which fix no
element of S? together with the identity form a nilpotent character—
istic subgroup, say Y', of X . By Lemma 2.1l(c), NA(B) and
NG(B) are normal in 6. We show that each g e 6 — NG'(B) fixes
no element of {2 so that G/NG(B) is isomorphic to a Subgroup of
Y . Now if mg fixes B1 6 n then Fig emBl = (1) so that
fig 6 D(A,G). Hence fig fixes B i.e., g E NG(B>’ a contradiction.
Therefore, A/NA(B) _~_G/NG(B) and since G/NG(B) is isomorphic to
a subgroup of Y’ [we have A/NA(B) is isomorphic to a subgroup of'y .
But Y is nilpotent so A/NA(B) is nilpotent which is a contra-
diction. Thus the theorem is true in this case.

Whether the cOnditions in the above theorem are necessary is an
open question at this time. We do not have an example of a group G
in which (X ,fZ) is 3/2—fold transitive and the complements of A
are neither conjugate nor normal.

Finally we conclude this chapter by imposing a condition on

(X ,6) which is considerably stronger than primitivity.

Definition 2.16: The permutation group (](,§2) is said to be

 

2-fold transitive if and only if X B is transitive on the set 6 ..{B}.

 

57

From [7, pg 20] every 2—fold transitive group is primitive so

this condition is stronger than primitivity.

Definition 2.17: (}(,§2) is said to be sharply 2—fold transi—

 

 

tive if and only if

 

(l) (X ,fl) is 2—fold transitive;

(2) | X == 1 whenever [AI = 2.

Al
Now we will assume that (){,S)) is sharply 2—fold transitive.

This is a stronger condition than primitivity so all previous re—

sults hold true in this case.

Theorem 2.18: If (X ,fl) is sharply 2—fold transitive and

 

(\Al , |B\) = 1 then XB is fixed—point-free on the set A — NA(B).

Moreover [XBl > 1.

Proof: From the remarks preceeding Theorem 2.8 we note that B
cannot be normal in G . Hence A —NA(B) a'Ot Let a E A — NA(B)
and Y E ){B such that Y E D(A,G). This can be done since (3(,§?)

lS 2-fold tranSltlve. Since [X [B1,a_lBa}[ = 1 and X is not the

identity we must have (a_lBa)? f’a-lBa. Let Y = YD(A,G), Y E B(A,G)
and we have (a-lBa); = (a-lBa)Y = (aY)_lBYaY = (aY)-1BaY so that
aYa—l E NG(B). In particular, aYa_l % 1 so Y is fixed-point—free
on A — NA(B).

Now we will show that [){B[ > 1 by showing that there exists

a b e B such that nb g D(A,G). If n e D(A,G) for all b e B

b
then for any a E A — NA(B) and all b E B we must have b-1(a-1Ba)b =

a_lBa. Thus aba-l

ENG(B). Since A <JG and AmB=1
we have NG(B) = NA(B)B and NA(B) = CA(B). Since

([A| ,[B[) = l , [aba_l| = [b[ , and NA(B> = CA(B> we must have

58
aba‘l e B. Therefore, a e NA(B) which is a contradiction.

In the proof of the above theorem we did not use the condition
(\A], |B[) = 1 to Show that ){B was fixed—point—free on A —NA(B).
However, without this hypothesis it is possible that B <JG so
that the set A — NA(B) is empty. As a consequence of the above

theorem we have:

Theorem 2.19: If ( X,ST) is sharply 2-fold transitive,

 

([Al, [B]) = 1, and A is not an elementary abelian p—group then

(l)._$ NA(B) i A.

Proof: As in the proof of Theorem 2.10 we must have NA(B) $A.

 

By Theorem 2.10 |)<B| > 1 and X18 is fixed—point-free on A — NA(B).
If NA(B) = (1) then ){B contains a fixed—point—free automorphism

of A of prime order. Thompson [6] has shown that if a group has a
fixed-point—free automorphism of prime order then the group is nil—

potent. So A is nilpotent. Now (XIX? ) is primitive and

NA(B) = 1 so, by Theorem 2.5, A must be characteristically simple.
But A nilpotent and characteristically simple implies that .A

must be an elementary abelian p-group which is a contradiction.

Therefore, (1) $ NA(B) $A.

INDEX OF NOTATION

I. Relations:

|/\

Is a subgroup of

Is a proper subgroup of

AH:

Isaiwnmlsmmmmpof

N

Is isomorphic to
= x a y mod A means xy_l E A
Is an element of

E Is not an element of

II. Operations:

G6 The image of the group G under the mapping O
96 The image of the element g under the mapping O
gX x—lgx

Y A The restriction of the mapping Y to the set A
"x The automorphism sending g to x-lgx

991+92 gelgez

1A The identity automorphism of A

G/H Factor Group

x Direct product of groups

[6:H] Index of H in 6

< > Subgroup generated by

{ ] Set whose members are

{x‘P} Set of all X such that P is true

lGl Number of elements in G

59

 

III.

[9] Order of the element g

Groups and Sets:

Hom(6,H)
ker Y
A(6)
B(A,6)
Z(G)
c661)
NG(H)
A(6)
I(G)

$5

A — B

The

The

Th

N

The

Th

(V)

Th

(V

The

The

The

The

The

group of all homomorphisms from G

set of all 9

such that gY = l

automorphism group of G

set of all Y E A(G) such that

center of G

centralizer of H in G

normalizer of H in G

Frattini subgroup of G

inner automorphisms of G

empty set

set of all x

in A and not in

AY

B

to

A

H

 

to

W E’

BIBLIOGRAPHY

J. E. Adney and Ti Yen, Automorphisms of a p-Group, Illinois
Journal of Mathematics, vol. 9 (1965), pp. l37-1A3.

 

W. Gaschutz, fiber die m—Untergruppe endlicher Gruppen,
Mathematische Zeitschrift, vol. 58 (1953), pp. 160—170.

 

M. Hall, The Theory of Groups (New York: Macmillan, 1959).

 

S. Maclane, Homology (New York: Academic Press Inc., 1963).

Cesarina Tibiletti Marchionna, Ampliamenti di automorfismi in
prodotti di gruppi permutabili, Bollettino Unione Matematica

Italiana, (3) vol. 16 (1961), pp. AA9-A6A.

 

J. Thompson, Finite Groups With Fixed—Point—Free Automorphisms

of Prime Order, Proceedings National Academy of Sciences U. S.,

 

vol. 59 (1959). pp. 578—581.

H. Wielandt, Finite Permutation Groups (New York: Academic
Press Inc., 196E).

 

61

 

""llifiNlllNu[NiliNlfiilllll“

192