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ON IHFIIIITE INTEGRAL LINEAR GROUP

BY

I.H. FAROUOI

A thesis submitted in the partial fulfilment of the requirements for the degree of Doctor of Philosophy in The Australian

ITational University.

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STATE-iENT

Unless otlierc'jise s t a t e d t h i s t h e s i s i s my oim '^rork.

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(ii)

CONTMTS

Preface

Chapter 0:

0.1.

0.2.

Chapter 1:

1.1.

1.2.

Chapter

2%

Chapter 3;

3.1.

3.2.

Chapter 4:

Chapter 5:

5.1.

5.2.

Chapter 6;

".eferences<

'age

1X

Introduction

Notation and terminology.

Some preliminary results.

K

Definitio^is of some normal subgroups of

and r(Tn)'\,

•^.elations between the automorphisms of A and those 27

of A/pA.

normal subgroups of

On the normal closure of an element of T.

The definitions of A(m,q), E(f) and ECf^f'').

35

The number of normal subgroups of the type r(f).

The number of normal subgroups of the type

E(fsf ). 100

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PREFACE.

I prepared this thesis as a post-graduate student in the

Department of lathematics, Institute of Advanced Studiesj The

Australian National University. I am grateful to the University

for granting me a Post-graduate Pvesearch Scholarship. During the

time I spent in this Department, I have been on leave from the

Mathematics Department 5 University of Karachi? I am thankful to that

University•for granting me leave for this period to enable me to take

up this Scholarship.

I feel most grateful to Professor Ilanna i-^eumann for her

encouragement and very inspiring supervision and Professor B.H. ileumann

for his advice, and readiness to discuss my problems.

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CEAPTER 0

0.1, Introduction

It is knovm ([8]) that in a symmetric group of countably

infinite degree, the subgroup of all finitary permutations, that isj those permutations which keep almost all the symbols fixeds is the unique maximal normal subgroup. ^lore generally ([1]), the normal subgroups of the group of all permutations of an infinite set of an arbitrary cardinal n, say^ can all be characterized in terms of the cardinals m < n in that a normal subgroup consists of all permutations that move only the symbols in a subset of cardinal lees than m. m - ^ ^ o The situation is similar in the case of the group of all automorphisms, that isjinvertible linear transformations in this casej of a vector space of infinite dimension over a division ring. This was shown by Alex Rosenberg in [7], In particular^ in the case of countably

infinite dimensions the subgroup of all finitary transformations which are the identity on a subspace of finite codimension is again essentially the unique maximal normal subgroup, that is, it is maximal normal

modulo the centre.

T]^ese analogies in different situations suggested the question whether something similar is true for the automorphism groups of free groups or of free abelian groups, of infinite rank.

It turned out that the position is vastly different. Consequently this thesis restricts itself to the case of free abelian group of

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defined in analogy to the finite dimensional case are inmiediately found to be normal subgroups other than and not contained in In one way this subgroup of finitary automorphisms is special: we find that every other normal subgroup intersects it non-trivially.

Ue have used known results, in particular by Brenner [2] and

lennicke [6] on the finite dimensional integral linear groups to find the normal subgroups contained in Since the factor group of the free abelian group A by its subgroup pA of the pth multiples is a vector space over the field of p elements 5 we have also made use of the results of Rosenberg [7] to get some additional information about the normal subgroups of F and the relation between the automorphisms of A and those of A/pA. The first three chapters of this thesis deal with the normal subgroups in $ and contain some results on the relation between the automorphisms of A and those of A/pA. Chapter 2 was to contain the proof that the group of automorphisms of A possesses maximal normal subgroups, one for each prime p. The proof was almost

entirely based on Rosenberg's description of the group of automorphisms of A/pA, Unfortunatelyp the proof of a lemma used there broke doxm at the last moment so that the lemma is now merely a hypothesis. This

a

affects the later parts of Clpter 2, We have nevertheless retained these parts: they may be provable without the lemma, and the questions raised that way, as well as the question whether the lemma is true or not5

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Chapters 4 and 5 deal with normal subgroups which have no

analogue in the group of automorphisms of the vector space A/pA or

in the finite dimensional integral linear groups, because these normal

subgroups arise entirely out of the fact that our linear group is bothy

integral and of infinite dimension.

In the last chapter we have discussed some questions vjhich are

left unansvxered in the course of this thesis and are of some interest.

0.2. Notation and

terminology;-For notation and terminologyj in general, we refer to Ilarshall Hall

[ 4 3 . In particular the follov/ing notation will be used throughout the

text.

A: a free abelian group of countably infinite rank,

rI the group of automorphisms of

A ,

Zp : the prime field x^ith elements {O^l^Z,...,p~l}, p being a prime.

A(p) ; a vector space over Zp of countably infinite dimension

when there is no danger of confusion, we shall denote it by A .

r(A): the group of all automorphisms (invertible linear transformations)

of A(p) or A .

the ring of endomorphisms of A .

Z: the set of all integers.

Z^: the set of all positive integers.

zr

the group ring of T , consisting of all^elGmenta of the form

Y a . Y . , y . e T , a. e Z for all i.

^ 1' 1 i 1

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n o t a t i o n , r(A/H)5 to denote the rank of A/r! when A/H is free abelian. In case A/H is not free abelian x-re shall

replace r(A/H) by |x] where X is a smallest set of generators of A / H ,

Codi!n(H); the rank or dimension of a complement of IIs where H is contained in A , or in A .

I: an identity matrix of infinite dimension, an (n x n)-identity m a t r i x ,

r; the normal closure of a in r , a e r .

a

a

the subgroup generated by x^jo^.^x^, <X>: the subgroup generated by the subset X ,

the general linear group consisting of all (n x n)"matrices with integral coefficients and determinants equal to ±1. SL^[Z]: the special linear group consisting of all those elements of

GL [Z] of determinent equal to + 1 . n

cn{s^}: the Cartesian product of the groups S^,

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He shall al«o write A when tlie dimension of

-•m X n

A is to be mentioned explicitly; then A denotes a

^ ' =m X n

matrix having m rows and n columns.

abbreviation for ''greatest coiTi.mon divisor"o

(m.n)- the g.c.d. of m and n.

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CHAPTER 1

1.1. Some preliminary

results;-In this section r-je have collected results about the structure of

free abelian groups of countably infinite rank. He shall need these

results in the next section and subsequent chanters to define or

characterize certain hinds of normal subgrouns of r.

Theorem 1.1.1. ([3l3§2)°.- H is a direct summand of A if and only

if A/H is free abelian.

P r o o f L e t H be a direct summand of A, and let A = II C+) K.

Let IT : A A be a retractive mapping of A into itself defined by air !->• k vjhere a e A, k e K^ and a = h + k, h e II. tt is an

endomorphism of A, I: is the kernel of it and Air = K. Thus

A/il = IC: and since K as subgroup of A is free abelian A/H is

free abelian. Ito-r let A/II be free abelian. and let

{x. + lili £ Z' C Z, X. e A) be a basis of A/H. Since x. + H

1 — X 1

is non-trivial for all i e Z', x^ ^ 11^ for any i e Z'; and since

A/II is free abelian no linear combination of the form

I a^ ^ Oj can be trivial, so 11 intersects trivially with i

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Corollary 1.1.1.1. If (j) is an endo^orphism of A then ker^ is a direct suimnand of A.

Proof;- A/ker(t> = A(j>-; and A^ s as a subgroup of A. is free abelianj therefore^ A/ker<() is also free abelian. Hence by T'ieorein 1.1.1. ken^) is a direct summand of A.

Theorem 1.1.2. ([ 5] Exercise 51(c));- The intersection of any number of direct summands of A is a direct summand of A,

Proof;- Let {"._.|i e ' tie a family of direct summands of A, and let H = ^ .. -Je define a mapping y,

i e M ^

y : A/'l -v C n {A/I-I.} ie M

by setting (a+H)y '(j) (y) where (|)(y) is the function whose value at i e [V) is the coset a + IL. (We regard the cartesian product

C n {A/E.} as the set of all functions ^ defined on (Vj such that iefyl ^

the value of (j) at i e jV| is an element of the factor group A/II^,) One can easily check that the mapping y is an isomorphism. Thus A/K is isomorphically contained in the Cartesian product of all factor groups A/H^. Since each K^ is a direct summand of A, by

Theorem 1.1.1. each A/F^ is free abelian and so it is a direct sum of the infinite cyclic groups C.,, :i c Mi CZ |V1» Thus

3-:] V "

c .^AA (A/U.) is contained in C n {C..} and so A/I! is

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47.1 3 [ 3 ] , every countable subgroup of a Cartesian product of infinite cyclic groups is free abelian; so A/H is free abelian. Hence by Theorem 1.1.1. H is a direct summand of A. This completes the proof.

The follox-7ing Corollary is immediate.

Corollary 1.1.2.1. Let H £ A. There is a unique least direct summand B of A containing li. B has finite rank if and only if II has finite rank.

Theorem 1.1.3. Let H £ A and let B be tha least direct sunraiand of A containing E. If an automorphism y e T maps H onto itself; then it maps B onto itself.

Proof;- Since Hy = H, if B is a direct summand of A containing H then By ^ Hy = II and By is a direct summand. Let

{B^|i e Z' C- Z} be the set of all direct summands B^ of A, x/nich contain H. Then C^.li e Z' C Z} = {B.y i e Z' C Z}.

1 — i' — Hence

B = i'\ B. contains K. and by Theorem 1.1.>. B is a direct summand ieZ' ^

of A and so

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Theorem 1 . 1 . 4 . Let K^, K^ be tv/o direct summands of A3 both

of f i n i t e codimension. Then ' ' f i n i t e codimension.

Proof t- • Let A = H^ I'.^ = E, (+. K^,

hie

use again the

isomorphism

p t A/(K^M K^) A/K^ a A/K^

defined by {a + (K^-'^ = (a + K^.a+K^), a e A. Thus

A / ( K ^ n K^) i s isoniorphically contained i n A/K^ C+) A/K^- Since

Kj^ and K^ are direct summands i i i A3 both having f i n i t e codimension5

A/K^ and A/K^ are free abelian of f i n i t e ranks^ and therefore

A/K^ (+) A/K^ i s also free abelian and has f i n i t e rank. Hence

A/(K P K,) has f i n i t e rank and so K^ H K, has f i n i t e codimension.

1 ^

J- z

(That Kj^ n K^ i s a direct summandj i s obvious i n this case, and also

follows from Theorem 1 . 1 . , ' . . ) This completes the proof.

Lemma 1 . 1 . 5 . ( [ 3 ] § 2 ) g ~ I f H i s a direct summand i n A and B

contains H, B A, then H i s a direct summand i n B.

Proof;- Since H i s a direct summand i n A, we can xiTrite A = H

Obviously B K intersects t r i v i a l l y with II. He only check that

B = (B n K) + K. Let b e B, then ws can write

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is trivial. "ence B = ® (B 0 K ) and I', is a direct suiranand

in B. This completes the proof.

Theorem 1.1.6. Let A = H (+) K and let B be another direct summand

of A containing H. Then we can vjrite A = B '3 C where C £ K.

Proof;- Since B and Y. are direct summands^ B Ti K is a direct summand of A by Theorem 1,1.S; and by Lemma 1.1. f-. BP, K is a

direct simmiand of IC. Thus we can write K = (B (") K) C-f; C, Wow

it can easily be checked that B = II c-f i (B i !') ^ and so A = B 4-} C;

X'/here G _< K. This completes the proof,

1.2. In this section we introduce certain special normal subgroups

of r. In the first place we define finitary automorphisms and show

that all finitary automorphisms of A form a normal subgrotip of F.

We denote it by confusion with the use of this symbol for the

Frattini subgroup of a group will not arise as the latter concept is

not used in this thesis, rlext we define two kinds of congruence

subgroups" of r ; they are natural generalizations of the congruence

subgroups of the finite dimensional linear groupsj and we therefore use

the notation T (m), T(m) in agreement with the notation that

seems to be most commonly used in the finite dimensional case. The

definition of these congruence subgroups arises naturally from a

general relation between characteristic subgroups of A and normal

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11

We can write every automorphism y of A in the form

Y = 1 -r p 3 V7here 1 is the identity automorphism of A and p = y ~ 1 is an endomorphism of A.

Definition 1,2.1. The automorphism y e T is called finitary if Y = 1 + p where the image Ap has finite rank.

Theorem 1. 2., 2, The following are equivalent (i) Y E r and y is finitarys

(ii) there exists a basis X {x^ji e of A and a positive integer n such that x^Y ~ ^^ i > (iii) Y n'.aps a direct summand of A of finite codimension/

identically onto itself«

Proof:- That (ii) and (iii) are equivalent and (ii) or (iii) implies (i) is obvious. We only prove that (i) implies (iii). Let Y e r and let y be finitary so that y = ^ + P where Ap has

finite rank. Let K = kerp. Gince K is mapped onto the zero element of A by p, it is mapped identically onto itself by Y• ^V

Corollary 1.1.1.1,, kerp is a direct summand of A; since

A/kerp - Ap, kerp = K has finite codimension. This shows that (i) implies (iii) and the proof of the theorem is complete.

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Theorem 1,2.3, If y e T is finitary then x-je can w r i t e A = B C+) C xdiere y is the identity on C , 3 is of finite rank and B y- = B.

Proof;- It follows from Theorem 1.2.2. (iii) that if y is finitary then there exists a decomposition of Aj namely, A = H ® K where H has finite rank and K is mapped identically onto itself by y.

A A ^ A Consider <H (j Hy> = K , Let h e K » We can vjrite h = h + I:,

* *

h e H, k e K so that h y = hy + ky or h y = hy + k, since y is the identity on K. ilow hy e liy H and h + k = h e H and

sV * A * A

h e H so k e H and h y = hy + k e H . Since h is chosen

A *

arbitrarily we get H y = H . Corollary 1.1.2.1. gives that there is a unique least direct summand B of A containing H , and B has

A

finite rank, since H has finite rank. And Theorem 1.1.3. gives that A >•«

By = B, since H y = H . And finally^by Theorem 1.1.6. v;e can write A = B C vrhere C j< K. This is a decomposition of A with the required (properties and the proof of the Theorem is complete.

From now on we shall use the equivalent characterizations of a

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matrix (finite or infinite) with respect to a basis, say

X = {x^ji e Z^} of A. In that case we shall say that X is a

basis of A v/ith respect to y ^^ ^ so chosen that

n

all j > n. This is possible by Theorem 1.2.2,. Thus in this case

:,Y = y a.,x., 1 < i < n, for some n r Z . and x y = x for

^ nil iJ J

the matrix A representing y is the direct sum of an (n x n)-matrix

with coefficients and an infinite identity matrix.

Theorem 1.2.4. Let $ be the set of all finitary automorphisms of

A. Then <$> is a n o m a l subgroup of r =

Proof; - Let a^B e $ and suppose that A = H^ K = H^ + K,

1

L i . i.

are two decompositions of A corresponding to a and g respectively.

It follows from Theorem 1.1.4. that K^ rs ^ direct summand of

A and has finite codimension. Clearly, a3 maps •

identically onto itself. Eence by Theorem 1.2.2. (iii) a3 Is

finitary. Since K^ is mapped identically onto Itself by a, a ^

is the identity on K^ cmd so a"^ is also finitary. Finally,^ let

Y e r 5 then A = H^y K^y where K^y has finite codimension and

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a3 = (l-Hi)(l+v) = 1 + + V + yvo By definition 1.2.1= r(Ay)

and r(Av) are f i n i t e , so r(ACy+v+vv)) < r(Ay) + r(Av) + r(A(yv))

is also finite. Hence aB is finitary. Now let = 1 + n so

that = (l-Hi)(l-fn) = 1 + y + n + yn ^ that is„ y + n + yn = 0

or y ( l + n ) -J- n = 0 or y ( l + n ) = -n or Ay(l4-n) = A ( - n ) = An? but

r ( A y ) is finite so r(Ay(l+ri)) is also finite and hence r(Ari) is

finite; and a ^ is finitary. Finally^let y e T ,, then

Y ^ay = 1 + Y ^yY and now A(Y ^iy) = A y y . Since r(Ay) is finite, r(A(yY)) is also finite. Hence r(A(Y "^yy)) is finite and y ^ay

is also finitary. Since a a n d y are chosen arbitrarily, w e

conclude that is a normal subgroup of r . This completes

the p r o o f .

W e shall; from now onj use the same symbol <[> for botli the set

of all finitary automorphisms of A and the subgroup consisting of

these elementss as is customary w h e n the carrier of the subgroup

generated by the set is the set itself„ W e shall tacitly use this

convention whenever this situation arises.

W e also point out that the definition of finitary automorphisms,

their characterization and the fact that they form a normal subgroup

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15

group of an absolutely free group of infinite r a n k . This is also true

for the relation between characteristic subgroups of A and normal

sub-groups of r to w h i c h w e now turn.

Theorem 1.2.5. L e t 5 be a characteristic subgroup of A , then the

subset 0 of r given b y

G = {Y!Y e r . A(I--Y) £ m

is a n o r m a l subgroup of r .

Proofs- Let e 0 so that w e have A ( l ~ a ) £ H and

A(l~-6) £ that is, for every element x e A there exists an h

and an h ' , h s h ' e 11, such that

x ( l - a ) = h and x ( l - e ) = h ' .

Now x ( l - a 3 ) = x(l+6"B-ae)

or x ( l - a B ) = x { ( l - B ) + (l-a)6}

= x ( l - 3 ) + x(l-a)B

= h ' + h B .

Since H is a characteristic subgroup of A , hB e H , and so

x ( l a B ) e Hj w h i c h gives aB e 0

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= "ha""^ E H.

So a ^ e 0.

Finally, let y t- T ^ and let x e A; then

x(l-Y ^ay) = xy ^(l--a)y

= x' (l--a)Y x' £ A, = h'y h' e H, - h-^ h^' e H,

because x(l-a) e H, for all x e A and H is a characteristic subgroup of A. This gives y ^ay e 0 for all y e r and for all a e 0 since y and a are chosen arbitrarily. Hence 0 is a normal subgroup of r. This completes the proof.

Theorem 1.2,6. Let H be a characteristic subgroup of A and let

*

X be the set of all primitive elements of A. Then the subset 0 of r gi'^^en by

0* = {y y e r, X(1-Y) e H, for all x e X}

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Proof:- One can prove that q is a subgroup of r iii t^e same way as w e proved in Theorem 1.2,5. that e is a subgroup of r io> as defined in Theorem 1.2,5.). ^'e only check that 0 is a normal subset

k „1 of r. Let y z T and 6 e 0 • Thus vre have xCl-y Qy) ==

Xy ^(1~9)y = x'(l~e)Y' Since x is a primitive element of A and Y ^ is an automorphism of k^ x' is also a primitive element of A

*

and so by the property defining 0 ^ x ' ( l - e ) is an element of H, And since II is a characteristic subgroup of A x'(1-0)y also

- 1 * belongs to H, Hence x(l y Qy) e H and so 0 is a normal subset of r. This complet;^ the proof,

•k

Theorem 1,2,7. Let 0 and Q be the normal subgroups of r as defined in Theorems 1.2, '^. and 1.2.0,, Then 0 = 0 ,

j'c ic

Proof;- Clearly, 0 ^ 0 . We only show that 0 £ 0. Let *

6 e 0 , so that for every primitive element x e A there exists an h e H such that x ( l ~ e ) = h. Lat y e A, We can write

y = y n,x. where the n.'s belong to Z and the x.'s are

4- i 1 3. X

i *

primitive elements of A. By the property defining 0 we have x . ( l - e ) = h,, h. e H for all i and therefore y(l-e) = I n x (1-0)

i i i i = ^ n h. e Ho Since y is chosen arbitrarily, it follows that

i

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Theorem 1.2.(3. Let 0 be a normal subgroup of r and let

H = <x(l-9) }x e A, e e 0> .

Then H is a characteristic subgroup of A.

Proof;- It is sufficient to prove that {x(l-e)|x e A, 9 e 0} is

a characteristic subset of A. Thus let y £ T, x e A and 9 e 0.

We show that x(1~9)y is also an element of the form y(l-e')s y c A,

and 9' e 0. ^Tow

x

(l

~e)Y = x ( y Q y )

= X(Y-YY~^9Y)

= x'Y(1"Y~

V)

= y(l~e')

where 9' = y ^©Y ^ © because 0 is a normal subgroup of A;

and y = XY. Thus H is generated by a characteristic subset of A.

Hence H is a characteristic subgroup of A. This completes the

proof.

IJhile it x-zill become clear that different characteristic subgroups

of A, lead, by the process of Theorem 1.2.5., to different normal

subgroups of r, this is .lot true for the reverse process given by

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19

Theorem 1,2.9» Let 9 be a normal subgroup of T and let X be the set of all primitive elements of A. Then

">group

li = <x(l-e)|x e Xj 0 e G> is a characteristic subgroup of k.

The proof is similar to the proof of Theorem 1.2.6. o

ii

Theorem. 1,2.10, Let H and K be as defined in Theorems 1.2,8. and l,2,9o then R = p/.

The proof is similar to the proof of Theorem 1,2.7,,

Theorem 1.2.11, Let 11 be a characteristic subgroup of Aj

0 = {yIy e r, x(l-Y) e for all x e a)

and let K = <y(l-0)ly e A, 6 s G>. Then E = K.

Proof;- Clearly K ^ 1! by the properties defining 0 and K, Ue prove that H K. Let h e H, He construct an automorphism Y e r such that for some x e A^ x(1-y) = h and for rll y e A,

Y D - Y ) £ H; in X/hich case y e 6 and H = X(1-Y) e Thus let n

X = {x. li e Z } be a basis of A such that h = ^ a i i=l ^ ^ a. e Zj for 1 £ i £ n, and let y be the automorphism

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Obviously y i s a required element of p because f o r evrry y e A, yCl-y) = m ( y ) . h , m(y) e Z and so yCl-y) e 11, f o r a l l y ^ A. Hence y e Q^ Mow = h wiiere e y £ 9 »

and so h e K. Since h i s chosen a r b i t r a r i l y , this gives H 11. This completes the p r o o f .

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of A. We claim that n is a multiple of m. By the previous argument5 also mz e E„ Write n = mq + r where 0 ^ r < m, then nz - mqz = rz e H; but m was mimimal with respect to m-th multiples of primitive elements occuring in H. Hence r = and so m

divides n as required. Thus every characteristic subgroup of A is of the form mk for some positive integer m. NOXJ one can

easily see that if K and K are two characteristic subgroups of A, H K, H A, K A and if G = <Y|Y e T, ACl-y) ^ H> , and

* i<

0 = <Y Y C FS A(1~Y) 1. K> 9 then 0 ^ e A^NEAI: L.-XFAR

(CViapter 3) that different normal subgroups of r may give rise to the same characteristic subgroup of A by the process of Theorem 1.2o8.,

In fact it will follo"-/ from t'le results in Ciiar'tcr 3 tb.at the lar.st

normal su' [^rou-n of T w'aich leads to a particular characteristic subgroup mA of A by tb.e process of Theorem loZ.S. is uniquel.y determined as the

intersection F (m) H ? to be defined later (po23 and •d„53)„ We denote by A(m) the factor group A/mA. Then A(p) may be looked upon as a vector space of countably infinite dimension over the field of p-elements. One can see that every basis of A gives a basis of A(m) under the natural map, v ' A A(m) defined by

av a + mA, a e A; thoughj obviously, not every basis of A(m) can be so obtained. We note that every automorphism of A induces an

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follows:-Let Y e To show that the mapping y ; A(m) A(in) defined by ^ where a. denotes the coset a + mA. for all a e A; is an

automorphism of A(m)»

(i) Let e A(m) and a = b. vhen a - b e mA and so ay ~ by e mA which gives ^ = So y is v/ell defined.

(ii) Let ay = Then ay - by e niAj so that (a-b)y e mA, Since y is an automorphism of k, a -- b e mA and so a = ]b. Thus y is o-i>e-to--one.

(iii) Since a + t = a+b, we get (a+^y = (a+b)y = (a+bj^ = (ay+by) = ^ + bj^. Bence y is a homomorphism.

(iv) FinallyJ let £ e A(m). Since y is an automorphism of A there exists an a' e A such that a'y = a. So we have

^'y = a'y = ^ and so y is onto. Hence y is an automorphism of A(m). We denote the group of all automorphisms of A(m) by

rA(m). Clearly3 if the automorphisms a,3 e T induce the automorphisms a,^ e rA(m), then a6 induces ^ ^ induces oi_

Now let R be a characteristic subgroup of A and G the n o m a l subgroup of r corresponding to H as defined in Theorem 1.2.5.. Then every element of G induces the identity automorphism of A/H. For let 6 e Q, so that we have x(l-9) e II, for all x e A (by Theorem 1.2.5.). Mow (x+H)e = xG + H0 = x6 + H (since H is

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some h e H , so that (x+H)e = xe + T = x + Ii + 11 = x + II. On the other hand5 let y e T and suppose that y induces the identity automorphism of A ( m ) . By definition of an induced

automorphism as given a b o v e , we have ^ = a. for all a e Aj which gives a - ay e m A or a(l-y) e m A , for all a e. A , But by

definition of 6 , y e G . Thus w e can also describe the normal subgroup 0 associated with H according to T'l-^oren 1,2,5. as the normal subgroup of T consisting of all those automorphisms of A which induce the identity automorphism^ of A/Ii.

H e have noted above that every characteristic subgroup of A is of t'le form m A , for some positive integer m . We denote t-ie set of all automorphisms of A which induce tie identity automorphism of A(m) = A/TIA by R( m ) . Thus y e F (m) means that w e can write y = 1 + p w h e r e xp e m A , for all x e A , 1 is the identity automorphism of A and p is an endomorphism of A . In other

words if y e r (m) is represented by a matrix v/lth respect to a given basis X of A , then the same matrix with all of its coefficients reduced m o d u l o m represents the automorphism y_ e RA(m) w i t h respect to the basis Xv w h e r e v is the natural ma:, from A

onto A/mA., av a + m A , for all a e A; and y is induced by Y , The normal subgroups T (m) so obtained are analogous to one type of congruence subgroups of the general linear group G L ^ [ Z ] , namely the ones which consist of all matrices "1, w h e r e h I mod (ra), I

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being the identity matrix in GL^[Z] and (m) being the principal ideal of Z generated by the integer m .

In a similar manner w e can also define another type of normal subgroup of r , namely the ones vzhich consist of all those elements of r which induce an element of the centre of rA(m). The centre of rA(m) is known. It consists of those elements of rA(m) for which x ^ = qx for all x e A ( m ) , q being fixed for y s independent of x e ACm), 0 < q < m , and (qjm) = 1. W e show that in fact all such elements y t T which induce a centre element of rA(m) form a normal subgroup of T and that every element of the centre of

rA(m) is induced by an element of r .

Leinma 1.2.12. ([2 ]);-• Let A be an (n X n)-matrix with integral coefficients such that detA 5 1 mod m . Then t'rere exists a matrix A ' with integral coefficients such that a^^ s a^j mod m , for all i,i" detA' = 1 , (^ij'^ij t)eing the (i,j)th coefficients of A' and A respectively).

First w e prove that every element of the centre of rA(ra) can be induced by an automorphism y of A .

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P r o o f S i n c e ^ i s in the centre of rA(m), we have = qx f o r a l l ^ e A(m), 0 < q < m, (qjin) = 1. There e x i s t s an integer n such that q"^^ s 1 mod m. Let X = {x^|i e z"^} be a basis of A and l e t X = {jc^ = x^ + mA[i e Z^} be the corresponding basis of A(in) o By Lemma 1.2.12 there e x i s t s an i n t e g r a l matrix A such that

" =nxn detA = 1 and a^^ = q mod m, a^^ = 0 mod m, f o r a l l i j j s i ^ j .

Thus a required automorphism ^ of A which induces ^ i s the following n

x'C = y a . . = q mod m. a. , E 0 mod m, t 13 J l i i j

t =• 0 , 1 , 2 7 T V T,

r »

, {l-trO^j,

^ ^ = <5/ ly - - • .

This complets the proof of the Theorem.

We n o t i c e that ? can be v/ritten as the sum of two endomorphisms, ? = q „ l + p , where 1 i s the i d e n t i t y automorphism of A and

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check that all automorphisms of A of the form q.l + p. v/here

p e mfi and q is any element of the set of all positive integers xi;hich

are less than m and prime to m, form a normal subgroup of r.

We denote the normal subgroup of r consisting of all such elements •k *

by r(m) . Clearly r(m) consists of all those elements of T

which induce a centre element of rA(m), Thus we have obtained the

following

*

Theorem 1.2.14, T(m) is the normal subgroup of r which consists

of all those elements of T which induce an element of the centre of

rA(m).

Thus if Y e r(m)* then if X = e z"*"} is any basis of A,

and vre write ~ ^ ^ij^j' ^^^ ^ ^ ^

3 J. A a.. = 0 mod m, for all i,j e Z , i j. Clearly F (m) < T (m)

IJ • 9

-+ *

for all m e Z . In fact T(m) is properly contained in T(m) for

•k

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This chapter was to contain the proof that r contains maximal

normal subgroupss one for each prime p. Unfortunately the proof

has brol-en do^m at the last raoment. It is probably still true,

but one lemma that x-jas used, has noH become a mere hypothesis. Since it may be true, we have retained the argument that is based on it,

but Tie indicate clearly what is affected by this uncertainty. As the

arguments were based largely on "Rosenberg s description of the group

of automorphisms of the vector space A(p) of countably infinite

dimension over the prime field of characteristic p., we begin by some

comments on the relation between automorphisms of A and of A(p).

Since A(p) is isomorphic to A/pA, every automorphism of A

induces an automorphism of A(p). >Iow in the finite dimensional

case, one knows that not all automorphisms of A(p) can be induced

by one of A, simply because A(p) has automorphisms of determinant

other than ± 1 mod p. This difficulty disappears in the case of

infinite dimension; the reasons appear in the rjroof of Theorem

2.2.1. and in Lemma 2.2.3. But one also used that to any basis in

ACp), there exists a basis in A, which maps under the natural mappings

on non-zero multinles (modulo p) of the given basis in A(t)). This is

true for finite dimension as is shoT-m by the proof of Lemma 2.1.9.

It is the extension of this proof to the infinite dimensional case that

has broken doTm. ^Te discuss at the end of this chapter the resulting

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2.1. Rosenberg's results and Preliminaries

Throughout this chapter we shall consider the vector space A(r»)

for a fixed prime p and therefore, from viow on, we shall write simply

A for ^(p). ':;osenberg has proved the results in [ 7 ] for

a general case, namely, the case of a vector space over a

division ring where the vector space has infinite dimension

y^Sj 5 being an ordinal, 6 ^ 0. He shall mention the results

of his that

ue

need, and these only In the case 6 = 0 which

concerns us here. One of Rosenberg's main tools is the concept

of an element of class two of the group of automorphisms of A,

and a further generalization of the concept of elements of

class two, namely, the locally algebraic elements of

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Definition 2.1.1. [ 7 ]. ^ p (A) is of class two if (^-1)2 = 0 (i.e. X ~ 1 where ^ is an endomorphism of A and = O).

Definition 2.1.2. ["/']. An endomorphism ^ of A is called locally algebraic if = 03l,2,...> is finite dimensional for all

a. e A.

From the two definitions above we derive the follov/ing simple lemma.

Lemma 2.1.3. Every element of class two of r(A) i s locally algebraic.

Proof -; - Let y_ = (1+p^) e r (A) and let be of class two. Tlien for all a e A, x-re have ^ = = a.(l+2g^) and

inductively ay^ =£(1+1^)3 hence < ay ^[i = 031,2,...> = <a,ap> is of dimension at most tvjo. The lemma follows.

Rosenberg cliaractv^rizes the locally algebraic elements as follows s

Theorem 2.1.4. [ 7]. Let y e T(A) be locally algebraic, then y_ = x^here has the property taat there exists a direct

decomposition of A into finite dimensional subspaces invariant under

T_ and a_ is such that v7ith respect to a basis X = i e z"*"}

which is the union of bases of these subspaces, it has the form X,a = y a. .X.,

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In fact this theorem (Lerama 2,4. [ 7]) describes the action of on the invariant subspaces more precisely and this more precise form is needed to derive his main result that r(A) is generated by the

elements of class two. As every such element is locally algebraic (Lemma 2.1.3)5 we have in particular the weaker forms

Theorem 2.1.5. [ 7]. r(A) is generated by its locally algebraic elements.

This is the result of Rosenberg's that we rely on to show that

every automorphism of A = A/pA can be induced by an automorphism of A-^ assuming the truth of the lemma on bases mentioned above.

The otiier main result of Rosenberg's that we use also is based on the fact that the elements of class two generate r(A). He

succeeds in determining all normal subgroups of r(A). In the case of countably infinite dimension, the result is as follows! If Z is the multiplicative group of non-zero integers modulo p, we denote

A

by Z .1 the automorphisms of A v/hich consist in multiplying every it

element of A by a constant; Z .1 is the centre of r(A). Let $ ( A ) denote the subgroup of finitary automorphisms of A. Then

Theorem 2.1.6. [ 7]. The group ^"(A), which is generated by Z .1 and ^ ( A ) is the unique maximal normal subgroup of r(A)» and

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31

Then every proper normal subgroup of p(A) is contained in

* *

'J' (A). In fact all the normal subgroups of (A) turn out to be normal in T(A).

If now every element of T (A) can be induced by an automorphism of Aj then Theorem 2.1.6„ \:ill enable us to derive v/hat was

the main result of this chapter, namely, that the normal subgroups of r consisting of all automorphisms which induce an element of ^'"(A) is maximal in T.

We need a fev; preliminary results.

Lemma 2.1.7. ([ ] Lemma 10.6,). Let H' be a non-zero subgroup of a free abelian group H where H has finite rank n. Then H has a basis and H' has a basis x' ...jx' such that

X n J. n

x^ = ^^^ non-negative integers and divides (i = 2j3J...,n).

In fact we shall only need that the l-^'s (as in Lemma 2.1.7. above) are non-negative integers.

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P r o o f L e t K be a subgroup of A having finite rank and let A/H

be torsion free. Let (4. be the least direct sutnmand of A containing

*

H,

That

H

exists follows from Corollary 1.2.1. and it also

*

follows froni there that H has finite rank (since H has finite

rank). Since H is free abelian of finite rank, n, say; and H

is a non-zero subgroup of H . there is a basis {x^5...^x } of

i m

ii

H and a basis {y-,s...sy } of II such that y = k.x for all i,

J- iTt i i i

1 i 1 Lemma 2.1.7. "c note that k^ 0 for all i. For,

let k^ = 0 for some j, 1 £

J

f. then H is contained in

<x .. jX ,,,, X > which is a direct suirimand of A having rank

-L 1~i l+i' "m

(m-1) and contains H. This contfa^dicts the assumption that H is the

least direct summand of A containing I!. And since A/H is

torsion free, and k^x^ e H, x^ also belongs to H for all i,

A

1 £ i ^ m, and therefore

II

= H . Kcnce II is a direct summand of A.

It is obvious that if H is a direct summand of A then A/H is

torsion free. This completes the proof.

Corollary 2.1.8.1, Let li be a ctir^eeL-siacaaad of A having finite rank

it

and let H' be the least direct Gtmrnand of A containing H. Then

r(H) = rC/).

I-Je omit the proof of this Corollary. In fact this result

(Corollary 2.1.8.1.) follc-73 directly from Lerana 2.1.7., only using

Vf

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33

From now o n , let v : A A denote the natural epimorphism given by av = ^ = a + pA for all a £ A .

L e m m a 2.1. 9, Let A = K K be a decomposition of A such that ri has finite d i m e n s i o n . W e can find a corresoonding decomposition of As n a m e l y , A = H :+. K such that r ( H ) = dim(H) and II and K are m a p p e d homomorphically onto H and K respectively under the n a t u r a l homomorphism v : A A .

P r o o f ; - L e t A = H K where dimli = n; and let

{x^ = x ^ + p A | x ^ e A \ p A , i = be a basis of H . We first show that these arbitrary inverse images x ^ are linearly independent in A . Assume ^ b x,' = 0 ; as A is free abclianj

j ^

w e m a y assume that the coefficients b^^ have greatest common divisor 1 . T h e n also ^ bj^/ x ; = 0 in A and so each coefficient is congruent to

J J 3

zero m o d u l o p . Since t?ieir greatest common divisor is 1 , they m u s t all b e zero. Thus the group H ' = = lj2 5...5n> has basis { x ' and H'v = H . Then if {yis^.-jy'} is another basis for H ' , then images y W = ^^so form a basis for H . How let H b e the least direct summand of A containing H ' . By Corollary 2 . 1 . 8 , 1 . r(Ii) = r(H') = n and by Lemma 2.1.7. there is

a basis { y , I i = 1 , 2 5 . . . jn} of H ' such that y . = k.x , for all i J

. 1 1 X 1 /\

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is prime to p for all i^ there exist integers n^ such that n^k^ e 1

mod p for all i. Hence the elements also form a basis for

H" but n.y, = n.k.x. = x, for all i and so — i—i 1 X—1 — i

{Xj_ = x^ pA|i = lj2,...,n} is a basis of H. lliis shows that

H = <x^,...5x^> is mapped onto H homomorphically under the natural

homoTOorphism v.

A^ow let K' be any complement of II in A and let A = H (+ K*.

Let k' be an element of K'. Then k'v = h + k' where h e H

and k' e K. Since v maps .1 onto H there is an element h e H

such that hv = h. Then we have (k'-h)v = k'v ~ hv = h + k' - h = k' .

This suggests the process of making a complement of II having the

required properties. VJe construct a complement K of H as

foHoxi7s;-Let {y^|i e z"*"} be a basis of IC and let K = I i e

where y^v = h , + k., h, e K, k. e K and y ' v = h,. Then one has •'i — -' i —i

A = H '-H K and H and K are mapped onto H and K respectively

under the natural map . This completes the proof of the Lemma.

At this point I thouglit I had proved the extension of this result

to the case v/hen A = I Hj is a direct sum of finite dimensional e i

vector spaces: after constructing the direct summands H^ of A

inductively, using Lemma 2.1.9.3 the additional informaticn that each

H^ maps onto an H^ so that ^H^ together with pA generates A

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Tp.e

fact on tlie connectioii of bases in and A = A/pA mentioned

in the introduction to this chapter would have been an iirmiediate

corollaiy. It is now raerely a hypothesis j on v/hich the proof

that every locally algebraic eleraent in F (A) c£in be induced by an

element of

T

is b a s e d .

Proposition 2.1.^0^ Let {x. i e Z"^} be a basis for A . Then there

1 _

is a basis {x. 1 e z"^} of A such that x.v = k.x. x/nere

X 1 i—X

= 1 for all i.

Alt w i ^ - c f w ^

e'^

p A e ^ l W ^ -

^'/.fo-2,2 The automorphisms of

A"-Mow V7e turn to the automorphisms of A . To begin x-rith we

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characterized a finitary automorphism of A . Thus if e $(A) where <Ii(A) denotes the group of all finitary automorphisms of A then vje can assume that there exists a basis X = f x . | i e Z"^} of A and a

n ^

p o s i t i v e integer n such thatj x , ^ = ^ ^ . .X-j 1 < i , i £ n 1 13 3

and ~ X.. for all j > r. or equivalently ^ that there exists a decomposition of A^ n a m e l y , A = K v-t^ K such that II has finite rank and is manped onto itself by ^ and K is mapped identically onto itself by

Theorem 2 . 2 . 1 . Let ^ e $ ( A ) . Then ^ can be induced by an element of r .

Proof:- Let ^ e $ (A) and let A = H (J^ K be a decomposition of A such that H has finite d i m e n s i o n , say n , and H is mapped onto itself by ^ and ^ is the identity on K . By Lemma 2.1.9. there exists a corresponding decomposition of A5 namelyj A = K K such tliat H and H have the same rank n and H and K are mapped onto H and K respectively under the natural map v : A A . Let

X = {x^. I i f: Z^} be a basis of A such that {x^|l £ i 1 n} is a basis of H and {x.li > n} is a basis of K . W e can choose

1

X = = x ^ + p A | i e Z'} as a corresponding basis for A x^here { x . | l ; i i f _ n } is a basis of -IH and {x.| i > n} , is a basis of K .

n ^

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37

for all j > n. Let A be the (n x n)~matrix with its (i5j)th

coefficient being a^^. Since = Hs the matrix A has an

inverse and so detA cannot be zero. Let detA E q mod p;

0 < q < p. There exists an integer m such that q™ 5 1 mod p.

Consider the automorphism ^^ of A defined as

follows:-n

= I 1 £ i £

-L 13 3

We show that ^^ can be induced by an element of r. We represent

^^ in the basis X of Aj by a matrix B where b^^ = a^^ j for

1 £ i, j £ n, b^^ = qj for all i > n, b^^ = O3 for all i,j > n,

i ^ j.

Consider the matrix C with c.. = b..^ for all

— C^L IJ IJ

I j ^ i , j £ t , t = m + n - l . Clearly detC = q"' 5 1 mod p and

therefore by Lemma 1.2.12. there exists a matrix such that

c! , = c,, mod p and detC = 1. Similarly let D be the matrix with

13 xj = HTixm

= q, for all i, and d _ = 0 , i ^ j. Obviously,

detp = q"^ = 1 mod p and so again by Lemma 1.2.12. we can find a matrix

D' such that d'. , ^ d.. mod v and detD' = 1. We note that

13 13 '

B = B^ where B^ = C and B^ = D for all i > 1. Now since each of

the matrices C' and D' has determinant 1 and is integral^ we

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o f A ) . Ue denote t h i s automorphism o f A by w h i c h i s g i v e n by

^ i * ^ ! ~ \ ^ ' i j ^ j ' ^ ^ ^ C l e a r l y , (j)^ i n d u c e s S i n c e

0 < q < p , t h e r e e x i s t s an i n t e g e r 0 < Z < p s u c h t h a t qj, E 1 mod p.

Thus 5 s i m i l a r l y , we d e f i n e a n o t h e r automorphism ^ ^ o f A as f o l l o w s °

-x.d). = X , f o r a l l i < n ,

g

Now t h e r e a l s o e x i s t s an i n t e g e r s s u c h t h a t £ E 1 mod p .

C o n s i d e r the m a t r i x E vahere e . . - fL, e , . ~ f o r a l l i ^ j ? i ^ —S^S X I

We have d e t E = 5 1 mod p and so t h e r e e x i s t s a m a t r i x S ' s u c h

t h a t e : , E e . . mod p and d e t E ' = 1. We can r e p r e s e n t A , by a

m a t r i x P xirhere P = ^ P ^ j ~ ^^^ ^^^ ^ n ) - i d e n t i t y m a t r i x ,

and Pj^ = E f o r a l l i > 1. C l e a r l y , r e p r e s e n t e d by P and t h e n t h e m a t r i x = ^ glj^s = I ^ s l i "" 1 ' ' ^^^ i > 1 ,

+

r e p r e s e n t s an automorphism o f A b e c a u s e a g a i n the m a t r i x P i s a d i r e c t

sum o f f i n i t e i n t e g r a l i n v e r t i b l e m a t r i c e s ( d e t P ^ = 1 , f o r a l l i ) .

We denote t h i s automorphism o f A by ^^ t'^hich i s d e f i n e d by

x.(j)2 = \ p ' ^ . x . , f o r a l l i j j . Then (p.^ i n d u c e s ^^^ F i n a l l y s we

o n l y have t o check t h a t ^ = have

n

X'd)^ = y a. .x, 1 < i < n , 1 3 3

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39

~ •] ^^^ j > rij q£ E 1 uod p. n

Thus X. .d), -- y 1 < i < n and -J jijj X3-J - ~

~ j ~ —3 ^^^ J > Hence ^ =

Since (}) = ^ ~ ^ automorphism of A having the required propertiec. This completes the proof.

It folloxi?s from the last theorem that every element of <|)(A) can be induced by some e l m a n t of r. As we have pointed out before in Chapter 1, if a^B e T induce e T (A) respectively, then a3 induces ^ . We denote by the normal subgroup of r which consists of all those elements of r which induce an element of $(A). Obviously3 i contains $.r(p). We have already seen

(Theoreia 1.2.14.) that r (p) is the nor.nal subgroup of r which

consists of all those elements of r that induce an element of the centre A

of r(A). We recall that y ^ T (p) is of the form y = + P where 0 < q < p, 1 is the identitji ciutciuorphism of A and p is an

endomorphism of A such that xp e pA, for all x £ A. Thus we see that every element of r(p) induces an element of the maximal

i'

J-normal subgroup $ ' (A) = z\l x $(A) of r(A). We now check that •A

if any ele.aent y e r induces and element of $ (A) then y must -A * -:<

be in r(p) . . For brevity we denote r(p) by Avp) o Thus

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can be vrritten as = x''A '^'^here x' 1 arid ^ e $ (A), Now both X - induced

by soitie eXements of p 5 say y' and (j) e r respectively = This gives that also induces But since y induces v/e have that y^y'^') ^ induces the identity of r(A). Now we know that r(p) consists of all those elements of r waica irxduce the identity element of T (A), Hence ^ £ r(p)«

A

Since r(p) ^ r ( p ) < A(p) and (y'tj)) e A(p) we finally get Y £ A(p)« Thus we have proved

Theorem 2.2,2. A(p) = T(p) is the normal subgroup of r consisting of all those elements of r which induce an element of the maximal

normal subgroup $ (A) of T(A),

We now showj using Z.l.^Of^ that every locally algebraic element of r(A) can be induced by an automorphism of A. For this we need a

lemma. The idea behind the lemma is very simple. Let ^^ ^^ infinite sequence of integers which are not congruent to zero modulo

a prime p. Then by rearranging the terms of the sequence we can obtain another sequence which can be divided into segments

of finite length such that the product of the integers in any one segment except possibly the first is congruent to 1 modulo p. We form.ulate this in a Biore precise way in the following lemraa.

Lemma 2.2.3. Let infinite sequence of positive integers

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for all i. There ^^xists a perniutatxori

j

|

- of Z such that the

sequence (a ^.x), = (b,) . has the following property

7i

1 I X

There Is an infinite increasing saquence of positive integers

such that

n (b.) E 1 mod p, for all i.

1

(Note that v/e say nothing at this stage about the product of the first

s^ elements, that their product is in Z folloxjs from the assumptions.)

Proof;" We assume that the sequence (^j^)j does not have the

property required in the statement of the above lemmao First

T-7e consider the following

case;-{*) If z £ Z and a. s z mod n for some

± e 7.

then a. = z mod p

for infinitely many integers j e Z .

Nov; suppose that z^ ^z^ j • •. jZ^^ (and only these) are the integers

such that z. e Z , for all ij z^ ^ z^ for i j and a^ = p' mod p^

0 < p' < p implies that p' ~ ^ y some j, 1 j <_ k. Since

the z.'s are all distinct ^ k £ p - 1 and since 0 < z^ p -- 1

^ n,

for all i, there exist nositive integers n. such that z. e 1 mod p,

for all i. Let ^^^^ infinite sequences of integers.,

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We

wrd.te the xndsxing set Z as a disioint union of its subsets E^ as follows

I-"2 "

k

(n^-t-1) (n^+2) ' • • •

^rr+s +l)'^s,(rn '^s,(r-l-l)n ^

s s s

for all r e Z"^ and s = 1,25,.. 5k. The sets E^ so constructed are disjoint and Z"^ = I'E,, because every term of ev?rv sequence

^ +

(t^^)^ belongs to some E^ and Z = ^ t _ i = l52,...5k5 j = 051,2,...} I-Je notic-^that II (a.) = 1 mod p because |e | = n , for all

T-, 1 ITkT'S S leE , ,

rlc+s

n^

rjS, and a. = z mod p for all r,s and z s 1 ro.od p, as we have

i s s

assumed. Now we start arranging the terms of the sequence

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43

magnitude of the terras^ that is, t _ < t^Z If and only If j < £:

for all i = IsZj...^!'- I^e also assume that the elements of each set

E^ are listed in ascending order of magnitude, How (a^)^ is

rearranged as follows:-- a. preceeds a. iff

(i) i,^ e E and i < i or

q -J

(ii) i e E^i , j e E^;, and q' < q „ By this rule we order the

set {a^ji e z"^} and f-jrite this ordered set as the sequence

This ordering uniquely determines a permutation of the indexing :.set

Z\

The result of it is that the first n^ elements of the sequence

(b^)^ are the elements of E^ in the natural order of their second

subscripts. The next n^ terms of the sequence the elements

of E2 and so on. And nov7 we can write dox-m the sequence

follows:- s^ = ~ ^ ^

s , ^ = 2n + n^ + ... + n, J o.. etc.. The sequences (b,). and (s.).

iv+J- 1 Z R i 1 1 i

are the ones which have all the properties required in the lemma. This

proves the lem-^a in the case (*).

Now we assume that {*) do3s not hold. Then since the sequence

(a^)^ is an infinite sequence there exists an integer z such that

a^ = z mod p, 0 < z < p for infinitely many integers j e Z^.

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b^ = a^ for i £ q and for i > q is obtained from the sequence (a^)^ i > q by the same method that w e used in the case

(*). And this sequence (b!). has tha property that

X X

II (b'.) = 1 mod PJ for all i, s, = q.

This completes the proof.

We need next t>.e substance of Lemma 2.1.4. In order to be able to use it, we outline its proof (for details see Rosenberg [7 ]j Lemma 2.4.).

By the definition of a locally algebraic automorphism the 1 elements where ^ is a generator of A, generate a subspace of A of finite dimension and clearly invariant under Denote it by H^. Then induces an automorphism of A/H^, itself a vector space of dimension „ over Z , and this induced automorphism is

'J p

locally algebraic itself. Hence picking a basis element of A

not contained in H^ defines another subspace, generated by the orbit of this basis element which is invariant modulo We can carry on inductively to obtain H^ a subspace of finite dimension which is invariant modulo K^ ij-) ... (fB under If a basis of A

is chosen to be the union of bases of the H., then in terms of this basis^ —X

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45

C =

* A A,

where the blocks A^ represent the automorphism induced by on

H^. / , and the blank space above the diagonal ' consists of zeros, j=l

All the matrices are invertible modulo p, hence denoting by T the direct sum of the matrices A.„ T is invertible and

=1- =

C = T(I+S)

where I + S is the same as C in the blocks indicated by asterisks above, and has identity matrices of the appropriate size in place of the A^. T and I + S define the automorphism and 1 + £ of Lemma 2.1.4.

The particular basis in which is represented by this matrix C need not be the image of a basis of A under the natural homomorphism.

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a non-zero element of Z^, But this preserves the shape of the A

matrix C representing in the new basis. Thus Proposition 2 . 1. 1 0 c implies

if

Lemma 2.2.4. if ^ is a locally algebraic automorphism of A, there exists a basis of A which is the image of a basis of A under the natural homomorphism v such that is represented by a matrix of

the form

C =

A.

=1

* A =2

•k * *

with respect to this basis. Thus = where T_ is the

automorohism represented by the matrix sum I = J A. and - + £ i^ represented by a matrixi+S with zeros above the diagonal and I's on the diagonal.

The fact that this can be done with respect to a basis {X . IE

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47

crucial. It gives at once that 1 ^ is induced by an automorphism of Aj namely, the one that is represented by the matrix I 4- S

whose elements are nov; Interpreted as integers in Z^ with respect to the basis i e z"^} • To see that t_ can be induced by an automorphism of A is not quite so immediate: the individual

matrices A^ along the diagonal of the matrix T are invertible modulo p whicii only means that their determinants are some non-zero integers

modulo PJ not necessarily one. It is here that we use Lemma 2,2.3.;, applying it to the sequence of integers mod p obtained by taking the determinants of the A^. in order. Then the lemma tells us that we can permute the ^^ (that isj the whole blocks - this also corresponds to a perm.utation of basis elements and no more and the new basis is again the image of a basis of A under v j namely that obtained by the

corresponding permutation from X = i e z"^} ) , so that in the new basis say Y = {y^| i e z"^"} where {y^|i e z"*"} is a basis of A obtained by permuting the elements of X, T_ is rapresented by a matrix C' of the following form:

£i

.;

_

£2

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v7here C^ is a finite block of non-zero determinant modulo p, and each Cj(j > 1 ) , is the direct sum of a finite number of the A^,

say A.. 5 so that detC_. = ji det(A. ,)= 1 mod p. From this reoresentation of deduce that it is induced by an automorphism of A as

folloT-Ts. We v/rite as a product = \vhere n_ is the automorphism of A which can be represented by a matrix with respect to the

basis Y of A; where N = C^^ 4- I and can be represented by a matrix 0 where 0 = 1 , 4 " ) C. where d. is the dimension of

- = a^ V =1 1 1 -j- 1 1

the matrix C^, It is sufficient to prove, now, that loth _n and £ can be induced by some elements of r. That t]_ can be induced by

an element of T follows immediately from Theorem 2.2.1. because

is a finitary automorphism of A. Now since detC^ s 1 mod p, for all i > 1, by Lemma 1.2.12. there exists a matrix say, where

q..i = c.., mod p, q. and c. being the (jk)th coefficients i j tX. ij R xj iv i J ii

of and C^ resprecively, and detO^ = 1. Thus the matrix o'= I,-h 0. represents an automorphism say S of A (with respect to the basis Y of A ) and K induces Thus both r^ and C can be induced and we have, on the basis of Proposition 2.1J6. :

^ T h e o r e m 2.2.5. Let be a locally algebraic automorphism of A. Then ;y can be induced by an automorphism of A,

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49

Tlieoren 2.2.G. Let ^ ^ ^ ^A) • Then can be induced by an

automorphism of A .

Finally x/e prove:

Theoren 2,2.1. If every automorphism of r(A) can be induced by an automorphism

•k *

in r , then A(p) = r(p) is a m a x i m a l normal subgroup of Tj

(p being a prime).

Proof;- Consider the mapping tt ,

-FT : r/A(p) ^ r(A)/<I>*(A),

A

defined by yA(p)tt ^ x i ^A^ where y e T s X ^ ^ ^A) X

induced by y . First w e check that tt is in fact a well-defined

mapping. Let YJ^JY-,^ T and suppose that ir^duce the same

element ^ of r (A). Then Yj^Y^^ induces the identity element of r(A).

Clearly y ^ ^'(p)s where r ( p ) consists of all those elements of r ft

that induce the identity of r(A) and r( p ) is contained in r( p )

which is contained in , ^ ^^^^ ^^^ therefore Y-J^A(p) =Y^A(p).

Now let Y2^>Y2 ^ ^ suppose that Y2^>Y2 induce Xi'X2 ^ ^^A) respectively. Then

*

= (A)

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(A) = (A). Then we have ^ ^ ' Since every element

of r (A) can be induced by an element of r^ xje can assume that

Y e r induces ^^^ have sho\-m earlier (Theorem 2.2.2.) that

A(p) consists of all those elements of r whicri induce an element

A

of $ (A)

5

therefore

y z

A(p). Now let Y

J

^ Y2

arbitrary elements of r such that

YJI

induces X

Q

^ and Y2 induces

so that

Y T Y 2 ^

induces Thus

Y

and

y

^ induce

the same elauent X] ^ therefore ^ induces the

identity of F(A) and so it must be in r(p). We write

3 =

YCYJ^Y^^)

~> where 6 e T (p) and so B also belongs to A(p). This

gives

Y2^Y2^

=

B Y

^ and since both

6

and

Y

^ belong to ACp),

so does

Y-,Y2^.

Thus Y^A(p) =

Y 2 ^ (p ) '

Since

Y^^

and

Y 2

arbitrary elements of

T

which induce X]^sL2 respectively 5 we conclude

that fr is a one-to-one mapping. It only remains to shov; that ir

is onto. But this we have already shotm in Theorem 2.2.6. where we

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H e n c e r / A ( p ) is i s o m o r n h i c to r(A)/<l"(A). By T h e o r e m 2 . 1 , 6 ,

•k it •I" (A) is the m a x i m a l n o r m a l s u b g r o u p of r (A) so that r ( A ) / $ (..)

is s i m p l e . H e n c e r / A ( p ) is a l s o s i m p l e and thus A ( p ) is a

m a x i m a l n o r m a l s u b g r o u p of T . T h i s c o m p l e t e s t h e p r o o f of the t h e o r e m .

W e s m m a r i z e t h e p o s i t i o n w h i c h a r i s e s if P r o p o s i t i o n 2 , 1 , 9 , is

n o t t r u e .

I f a b a s i s o f A = A / p A is g i v e n it is c e r t a i n l y true that t h e r e

is an i n v e r t i b l e t r a n s f o r m a t i o n that is o n e r e p r e s e n t e d b y a r o w - f i n i t e

i n t e g r a l m a t r i x 'J w h i c h is i n v e r t i b l e j n o t o n l y i n v e r t i b l e m o d u l o pj

w h i c h t r a n s f o r m s the g i v e n b a s i s into o n e for w h i c h 2 , 1 . 9 . is t r u e .

T h i s doas^ h o w e v e r , n o t seem to b e s u f f i c i e n t to use R o s e n b e r g ' s

r e s u l t r.s a b o v e to s h o w that e v e r y l o c a l l y a l g e b r a i c e l e m e n t of r ( A )

can b e i n d u c e d b y an e l e m e n t of r .

E v e n if this last thieorem is not t(Q^es the c o x n a c t u r e that the

s u b g r o u p s A ( p ) a r e m a x i m a l n o r m a l s u b g r o u p s of r is likely to b e

t r u e ; and i n fact w e b e l i e v e t h a t they a r e the o n l y m a x i m a l n o r m a l

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He divide this chapter into

trio

sections. The first section

consists of results about the group of all finitary automorphisms

of A. He have determined all normal subgroups of It turns out

t'.at every normal subgroup of

<

i

> is a normal subgroup of T and that

the subgroups $ D F (m) are the only normal subgroups of

with-one exception. ^ has a normal subgroup v/hicli contains $

r

(m)

for all m > 2, is not a congruence subgroup itself and is maximal

normal in The only other maximal normal subgroup of

<

l

> is

$ n r(2). In this sense the normal structure of $ is different

from that of the finite dimensional general linear groups ,

n ^ 2: the latter are kno^.ra to have proper nor.nal subgroups containing

for given m the congruence subgroup mod m properly.

The normal subgroup $ of T leads immediately to another normal

subgroup, generated by

<

J

> and the centre of

V

which is the

2-cycle generated by the inverting automorphism ~1. Part of the

significance of $ for

T

is that every normal lisubgroup of

T

intersects $ non-trivially. In order to nrove this we have

investigated the normal closure in T of a

s i n g l e

element, a

say, which is not contained in $ or iJliile we cannot determine

r

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53

of the normal subgroups of including $ itself.

In this chapter we have used knot-m results on the finite

unimodular groups, naaely, the results by Brenner [-2 ] and Ileninicke [6 Ij which T-7e mention here vzithout proofs.

3.1. The iiormal subgroups of

p;-Let (|) e 4 and let X = {x.li e z"*"} be a basis of A with n

respect to cj) so that x,(|) = Y c x^^ 1 < i j ^ rijX ({> = x. for 1 j ij i " i ^ all i > n where n is some positive integer. lie denote by B

V

the infinite matrix of this representation of (j). Then § = C + I where I is the infinite identity matrix. Me shall often represent an element of 6 by a finite matrix which Biay be of the form

C or C ' I where I is the (m ^ m)-identity matrix. Since = = m m

( ( ) m a p s onto itself and C has integral coefficients, C is invertible and detC = ±1. The same automorphism (j) can also be represented by some other finite matrix, say E, v/ith respect to some ot'ier basis of A. However, we shall show that detC = detE, so that the determinant of a finite matrix C representing an element 4) of 4> is independent of a choice of basis of A to which C

corresponds. Thus we can speak of the determinant of an element

(j> e $, denoted detcj), by vzhich we mean the determinant of any finite matrix representing (j). We denote the set of all those elements

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of index two in <i>. We shall also show that all other normal subgroups of $ except $ r r(2) are contained in

Lemma 3.1.1. Let a,B be two elements of $ and let

A = !' (•© K = H' ^-f) K' Me two decompositions of A corresponding to a and 0 respectively. Then v/e can vjrite A = L (rb) 11 v/here "J _< K K' and La = L6 = L, L has finite rank and contains both

H and H'.

Proof:- By Theorem 1,1,4, K o K' is a direct sumraand in A and has finite codimension; and by Lemma 1.1.5. K O K ' is a direct sumraand in both K and 7J. Thus we can write

A = H ® K^ (J; K O K ' = II' (cB ® I- n K', where < X, and K^^ < K'. Then IC r> K' has finite codimension in both K and E' and so both iq and have finite ranks. Let H ^ - U U H

Let h e H and let h = h ' k ' where h' e H' and k' e K*. Then hB = (h'+k')3 = h'3 + k'B = h'3 + k' where h'B e H' because by the property of the decomposition A = H' @ K'

H'3 = H' and BjK' = ijK'. But K' = h - h' e < H U H ' > ; therefore, h3 = h'B + k' e < H U H'> which gives hB £ H*. Hence HB £ H'^.

Similarly one can show that H'a, K|B, and K^a are also

contained in H*. That H'B, Ha, K|a and K^B are also contained in H* follows from the property defining the decompositions

A = H © K and A = H' (© K'i This implies that H* is

invariant under a and B. We also note that H* has finite rank.

References

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