Decision Problems in Algebra: Word Problems: Working Conference Proceedings

WORD PROBLEMS I1 The Oxford Book Edited by S. I. ADIAN W. W. BOONE 7he Steklou Mathematical Institute University of...

Author: S. I. Adian | W. W. Boone | G. Higman

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WORD PROBLEMS I1 The Oxford Book

Edited by

S. I. ADIAN

W. W. BOONE

7he Steklou Mathematical Institute

University of Illinois Urbana, USA

Moscow, USSR

G. HIGMAN University of Oxford Oxford. England

1980

NORTH-HOLLAND PUBLISHING COMPANY A M S T E R D A M . N E W YORK . O X F O R D

@ NORTH-HOLLAND PUBLISHING COMPANY - 1980

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the copyright owner.

ISBN: 0 444 85343 X

Published by: North-Holland Publishing Company - Amsterdam . New York . Oxford Sole distributors for the U.S.Aa n d Canada: Elsevier North-Holland, Inc. 52 Vanderbilt Avenue New York, N.Y. 10017

Library of Congress Cataloging in Publication Data

Main entry under t i t ) e : Word problems 11.

(Studies i n l o g i c and t h e foundations o f mathematics V-

,851.

This book grew o u t o f t h e working conference 'Decis i o n problems i n algebra' held i n Oxford t h e summer o f 196.'' Bibliography: p. 1. Groups, Theory of--Congresses. 2. G6del's I. Adian, S. I. 11. Boone, theorem--Congresses. W i l l i a m W. 111. Higman, Graham. IT. Series. ~ 7 1 . ~ 8 6 5I2l.22 79-1P76 ISBN 0-444-85343-X

PRINTED IN THE NETHERLANDS

;

Dedicated to the memory of Kurt Godel (1906-1978) in awe and affection

In angusto vivebamus, si quicquam esset cogitationibus clausum. Seneca, Letters 55.11

INTRODUCTION This book grew out of the working conference “Decision Problems in Algebra”, held in Oxford the summer of 1976, under the auspices of the Science Research Council of the United Kingdom. This work is a sequel to the volume “Word problems: decision problems and t h e Burnside problem in group theory”, which itself was the result of a similar working conference held in Irvine the summer of 1969. The Oxford conference was organized by t h e editors of this volume. The secretary of the conference was Donald J. Collins, without whom the entire endeavour could not have been successfully carried out. What is to be said about t h e present book? Like t h e Irvine book, a major intention is that it serves as a means of entry for the reader into the field of word problem. For this reason we have included various surveys; but, moreover, it is hoped that various articles which, while on the very borderline of advancing ideas, have still been presented in such a way as to be highly accessible to the working mathematician. The field of “Word problems” would seem to have flourished between Irvine and Oxford. Unsolvability results have been sharpened in various ways. Thus, as explained in an expository article herein, various classical decision problems about groups have been attacked, but with the class of groups considered now restricted to some familiar variety; and in certain cases this has led to new unsolvability results e.g., in the case of solvable groups. Now, too, the full story is known about the existence of word problems of the various finer degrees of Post. But positive results have not been lacking either. Indeed, we include an article solving t h e conjugacy problem for matrices with integer entries. (This was problem 22 of t h e Irvine book.) A short article on algebraically closed groups appeared in t h e Irvine book, but in subsequent years, this has become a large area of inquiry both with regard to decision problems and with regard to purely algebraic questions. An in-depth study and a shorter paper are included. Much the same thing could be said about a focus of interest on simple groups within the word problem field, and we include an article on this matter also. Small cancellation theory and generalizations (surveyed in the Irvine book) have truly come of age. We include two papers in the general area, one of which solves a well-known problem of Kurosch and Bjarni J6nsson. vii

viii

Introduction

A brief list of open questions is given as well. We call t h e reader’s attention to the note by J.L. Britton, “Erratum: The existence of infinite Burnside groups” in which he says that a mistake occurs in his article in the Irvine volume. This mistake was first noted in the book “The Burnside problem and identities in groups”, by S.I. Adian (Nauka, Moscow, 1975), page 4. The existence of infinite Burnside groups of large odd exponent was proved in a joint article by P.S. Novikov and Adian in 1968. The much stronger proposition of the announcement of Novikov in 1959 must be regarded as not proved. We wish to thank all the mathematicians who have helped us by refereeing individual articles, but to single out here, by name, only Gerhard Hesse for his especially valuable help in this regard. Originally, this volume was intended as a Festschrift to mark the seventieth birthday of Kurt Godel. Now, sadly, we can only dedicate it to his memory. Sergei I. Adian William W. Boone Graham Higman

S.I. Adian, W.W. Boone, G . Higman, eds., Word Problems I1 @ North-Holland Publishing Company (1980) 1-16

MODULAR MACHINES, THE WORD PROBLEM FOR FINITELY PRESENTED GROUPS AND COLLINS’ THEOREM StAl AANDERAA and Daniel E . COHEN University of Oslo and Queen Mary College, London

We shall define a class of machines which we call modular machines, related to Minsky machines [15] (these were called r-machines in the lecture by the first author on which this paper is based). These machines act on NZ,the set of pairs of natural numbers, with a very simple transition function. It will then be almost immediate that any function computable by a modular machine is partial recursive. On the other hand, modular machines are defined in such a way that Turingcomputable functions are computable by modular machines (which can also be regarded as a new way of Godelising Turing machines). This provides a new proof that Turing-computable functions are partial recursive. It also provides an easy proof of the normal form theorem for partial recursive functions, since t h e data for a modular machine, being numerical in nature, can easily be encoded by a natural number with t h e decodings being primitive recursive. There will be a modular machine M,, whose halting problem is unsolvable. Readers are invited to construct their own proof of t h e undecidability of elementary number theory using this machine M,, and to see how it compares in difficulty with standard proofs. (Various results on the degrees of halting, word and confluence problems for modular machines, similar to known results for Turing machines, are proved in [9] but are not needed in this paper.) However the main aim of this paper is not to apply modular machines in computability theory and logic, but to give some of their applications in group theory. In this and t h e following paper t h e grouptheoretic results are all known, but the new proofs are very much easier than any previous proofs. We prove the following theorems.

Theorem A (Novikov-Boone). There is a finitely presented group whose word problem is unsolvable. Theorem B (Collins). For any recursively enumerable unbounded truth table degree, there is a finitely presented group whose word problem has that degree. 1

S.Aanderaa, D.E. Cohen

2

Theorem A was first proved by Novikov [16] and Boone [2] independently. The original proofs were somewhat complicated and lengthy, but were later simplified. In particular, Britton [5] made very important simplifications by proving results about HNN extensions. Lemma 4 of his paper is now referred to as Britton’s Lemma and is a very important tool in combinatorial group theory, which will be used in our papers also. Our proof of Theorem A has t h e same basic idea as Britton’s, but is very much easier because of the use of modular machines rather than Turing machines. Other proofs of Theorem A are known. It can be proved using Higman’s embedding theorem [0, 131. A very different proof from any of the others has been given by McKenzie and Thomson [14]. A weaker form of Theorem B, with “unbounded truth-table degree” replaced by “Turing degree” was proved independently by Boone [3, 41, Clapham [6], and Fridman [ l l , 121. Collins [lo] was able to improve Boone’s proof to obtain Theorem B itself. In $1 we develop the theory of modular machines. Theorem A is proved in 02. In 54 we prove Theorem B, having obtained the relevant degree results for modular machines in 03. The paper following this used similar techniques to prove the Higman embedding theorem. 01. Modular machines

We shall follow Turing and consider Turing machines as defined by quintuples q,a,a,q,D (where a,, a, are letters, q,, qs are states, and D is one of the two symbols L, R ) instead of the more common definition by quadruples due to Post. The two definitions are equivalent, a non-moving quadruple q,aa ’qs being replaced by a quintuple q,aa’q,s,R (where q,s. is a new auxiliary state) together with quintuples qrsaxxqsLfor all letters x. We shall use the word configuration rather than “instantaneous description”. We write uqau for the configuration with uau on the tape (u , u being words, a being a letter), t h e machine being in state q scanning letter a. Let T be a Turing machine. We regard its alphabet as consisting of the natural numbers 0 , 1 , . . . ,n (where 0 is the blank) and its states as consisting of n i- 1 , . . . , m - 1 and possibly 0. Take a configuration . . . b l b o q a c , , c I. . Define u, u by u = C. b m ’ , u = C c,rn Minsky [ 151 represents this configuration by (u, u, a, q ) E N4. It can also be represented by either of the pairs (urn + a, urn + q ) or (urn + q, um + a ) . We shall use whichever pair is most convenient at the time, sometimes using one pair and sometimes the other for t h e I.

3

Modular machines I

same configuration. Notice that both mappings from configurations to pairs are recursive. Also P ( T ) = {(a,p ) E N Z ;(a, p ) represents a configuration} is recursive and the map sending (a,p ) to t h e corresponding configuration is partial recursive with domain P ( T ) . Let qaa’q‘R be a quintuple of T. It is easy to see that (with u, u as above) one element of N Zcorresponding to the next configuration is (urn’+ a’m + q ’ , u ) . Similarly if qaa‘q’l is a quintuple of T then (u, urnz + a ’ m + 9’)is a pair corresponding to t h e next configuration. This motivates the definition of a modular machine. Definition. A modular machine M consists of an integer m > 1, an integer n with 0 < n < m, and quadruples (a, b, c, R ) and (a, b, c, L ) such that 0 S a, b < h,0 S c < m ’, and, for each a, b, at most one quadruple begins with the pair a, b.

p

A configuration of M is an element (a,p ) of N 2 . Write a = um

+ b, where

+ a,

0 S a, b < m. If n o quadruple begins with a, b we call (a,p ) terminal. If (a, p ) is not terminal we say (a, p ) yields ( a ’ , p ’ ) ,written ( a , p ) 3 (a‘,p‘), if either M has a quadruple ( a , b, c, R ) and a ’ = urn2+ c, p’ = u or M has a quadruple (a, b, c, L ) and a ’ = u, p’ = urn2 + c. If ( a I ,PI) 3 (a2, p2) 3 * . * 3 (at, pk) we write ( a , ,p l ) + ( a k , p k ) . We define a (partial) function gM : N --+ N 2 by g M ( a ,p ) = (a’, p’) iff (a, p ) + (a’,p ’ ) and (a‘,p ‘ ) is terminal. The class of functions { g M ; all M } is rather strange. For if gM is somewhere defined, there is a pair a, b such that n o quadruple begins with a, b. T h e n g M ( a , p ) = (a,P) if a = a (mod m), p = b (mod m). We shall use t h e integer n (which has played no part as yet) to define input and output functions. The name “modular machine” is given because t h e action of M on a pair depends on its class modulo m. For any r E N we can write r .uniquely as r = C i b,n’, 1 S b, S n (for r = 0, take the empty sum). In a Turing machine with alphabet 0,1,. . . , n, this means that b, . . b,, is the tape description corresponding to r. Let iM : N 4N 2 be given by iMr = (C b,mI, n + 1). Write a E N as C c,m’, 0 6 c, < m , and take y with c, = 0 but c, # 0 for j < y. Define uM : N 2 + N by u M ( a p, ) = X:a,n’-I. Then uMgMMiM is a partial function from N to N , which we call the function computed by M. We shall associate with a Turing machine T a modular machine M such that M and T compute the same function. We shall use a slightly unusud definition of the function computed by T. The class of Turing computable functions is (without using their characterisation as partial =

urn

4

S. Aanderaa, D.E. &hen

recursive functions) easily seen to be the same for both definitions. The reader who prefers a more usual definition should change the definition of the input function iM and the output function uM accordingly. Let T be a Turing machine with alphabet 0, 1,.. . , n, and states n + 1,. . .,m - 1 and (perhaps) 0. Each r E N has a tape description as above. The output of a configuration will be the integer corresponding to the portion of the tape lying strictly between the scanned square and the first blank to its left. The function fT : N + N computed by T is defined by fTr = s if T when started in state n + 1 on the rightmost square of the description of r ultimately halts in a configuration with output s. Note that, if convenient, we may modify T so that whenever it halts the scanned square is blank. We now define a modular machine M associated with T. M will have the integers m, n previouslyAe6ned for T.M will have two quadruples (a, q,a’rn + q’,R (or L))and (4,a, a‘rn + q’,R (or L)) corresponding to each quintuple qaa’q’R(or L ) of T. The account preceding the definition of modular machines explaining how to associate members of N 2with configurations of T makes it clear that M simulates T. Precisely, if (a,p ) corresponds to C then (a,p ) is terminal iff C is terminal, while if C 3 C’ then (a, p ) 3 (a’,p ‘ ) where (a’,p ’ ) is a pair corresponding to C’. The definition of the functions computed by T and M now makes it clear that T and M compute the same function. The numerical nature of M makes it obvious that the function computed by M is partial recursive, so we have a proof that Turing computable functions are partial recursive. We could also obtain the normal form theorem fairly easily by a Godel numbering of modular machines. ) Define H f l ( M )to be 0 if (0,O) is not terminal for M and H f l ( M = {(a,p ) ; (a, p ) + (0,O)) if (0,O) is terminal. It is obvious that H o ( M ) is r.e. For any r.e. set S there is a Turing machine T such that fT is the (partial) characteristic function of S. Further we may assume that if f T r is defined then T halts on a blank tape (see [18], where T is constructed to simulate the action of a single-register machine computing the function). If M is t h e modular machine associated with T,plainly H o ( M ) is not recursive if S is not recursive. A stronger result is proved in 03. 02. Unsolvability

We begin with an account of HNN extensions and Britton’s Lemma

Modular machines I

5

for the reader who is unfamiliar with these topics. A knowledge of free groups and free products will be assumed. Let A , , A _ , (for i in some index set Z which need not be countable) be subgroups of a group H. Let cp, : A, + A - , be isomorphisms. Let G be the free product H * * ( p , ) (i E I). Let N be the normal closure in G of the set of elements p;la,pi(cp,a,)-lfor all i and all aiE A,. The quotient group GIN is called t h e HNN extension of H with stable letters p , and associated pairs of subgroups A, and A - , ; it is usually written as ( H , p , ; p ; ' A , p , = A - , ) .Any element of G can be written as

. hn-lp::hn, where n 2 0 , h ,,..., hn E H and hop::hI

* .

,..., E , = + I . If for some r we have i, = i,,', E , + = 0 and either E , = - 1 and h, E A,, or E , = 1 and h, E A _ , ,we say this expression has a pinch (more precisely, that it has a p,,-pinch). In such a case t h e same element of GIN is represented by an expression with fewer p-symbols (since we can replace p ; ' a p , with a E A, by cp,a without altering the image) and we call this process pinching out. We may now state Britton's Lemma (for a proof see [7] or [S]). Britton's Lemma. A n y non-trivial element of G lying in N has a pinch. The following results are immediate consequences (and are all we need in this section; the full force of Britton's lemma is used in 54). (I) H embeds in the HNN extension (by t h e map H -,G + G I N ; we regard this as inclusion). (11) If K is a subgroup of H such that cp, (K n A , ) = K r l A _ , for all i then H n ( K ,p a(all i ) ) = K. For it is easy to check that if a pinch in G can be applied to an element of t h e subgroup ( K , p , (all i ) ) of G the hypotheses ensure that the new element still lies in this subgroup. Hence any element of t h e subgroup ( K , p , )of GIN can be represented in a form without pinches. By Britton's Lemma it cannot lie in H if any p , occurs, i.e. if it is in H it is in K. (It also follows that the map from the HNN extension of K with associated pairs K n A, and K fl A _ , to the group ( K , p , )is an isomorphism. This part is used in [O], not in this paper.) (111) In t h e group ( H , p ; p - ' A p = A ) , the isomorphism being t h e identity, p - l h p = h (with h E H) o n l y if h E A . Let A be the group ( t , x, y ; x y = y x ) = ( t ) * ( x , y ; x y = y x ) , which is also an HNN extension of the free group ( r , x ) with stable letter y . Let T = ( t ) ^ (the normal subgroup of A generated by t ) and let t ( r , s ) = y -'x-'tx'y '. We easily obtain the following properties.

S. Aanderaa. D.E. Cohen

6

(i) T is free with basis { t ( r , s)}. This follows from the KuroS subgroup theorem, but is more easily proved directly by expanding a product of elements t ( r , s). (ii) T f l ( t ( i , j ) ,x"', y " ) has basis { t ( r , s); r = i mod rn, s = j mod n } . These elements are plainly contained in the subgroup, and any element of ( t ( i , j ) , x " , y " ) is ux'"'yp" for integers a, p and some element u of ( t ( r , s); r = i mod m , s = j mod n ) , whence the result. (iii) The subgroups ( t ( i , j ) , x " ' , y " )for an m , n # O and any i , j are all isomorphic. For this subgroup is a conjugate of ( t , x " , y " ) = ( t ) * (x"', y " ) which is isomorphic to A. (iv) Let rn, n, rn,, n l # 0 and i,j, i l , j l arbitrary. Then the map r ( i , j ) H t ( i l , j l ) , x"' H y" H induces isomorphisms of ( t ( i , j ) , x " ' , y " ) with ( t ( i l , j l ) , x ' " ~ , yand " ~ ) of T n ( t ( i , j ) , x " , y " ) with T f l (t(il,jl), ~ ' " y1",~ ) This . is immediate from (iii) and (ii), t h e latter isomorphism sending t ( u m + i, urn + j ) to ((urnl+ i l , unl + jl). Now let M be a modular machine with quadruples (a,,b , , c , , R ) for i E Z and (a,, b,, c,, L ) for j E J. Let T ( M )= ( t ( a ,p ) ; (a,p ) E Ho(M)). Let B ( M ) be t h e group (A, r,, 1,; r;'t(a,, b,)r, = t(c,,O), r;'x"r, = x m Z , r;'y"r, = y, l;'t(u,, b,)l, = t ( 0 , c,), l~'x"'1,= x, lJly"1, = y m 2 for all i E I, j E J). By (iii) this is an H N N extension of A with stable letters r,, 1,. (v) Let (a, b, c, R ) be a quadruple of M with corresponding stable letter r. Then under t h e isomorphism of the associated subgroups their subgroups T ( M ) n ( t ( a b, ) , x " , y " ' ) and T ( M ) f l ( f ( c , O ) , x m 2 , are y) isomorphic. For by (iv) t h e isomorphism sends t(a,p ) to t ( a l ,PI) where a = a, p = b mod rn and (a,p ) 3 ( a I PI). , Then (a,p ) E H o ( M ) iff ( a l ,PI)E H o ( M ) .The similar result holds for L. (vi) A fl ( T ( M ) ,r,, 1, ; all i, j ) = T ( M ) . This follows from (11), using (v). (vii) ( T ( M ) ,r,, 1, ; all i, j ) = ( t , r,, 1, ; all i, j ) . We show r(a, p ) E ( t , r,, 1,) for (a,p ) E H o ( M ) by induction on the length of the computation. If (a,p ) is terminal we have a = /3 = 0 as (a,p ) E H o ( M ) .Let (a,p ) j ( a I PI), , and let a = a, p = b mod m where (a, 6, c, R ) is quadruple with corresponding stable letter r (similarly for L and I ) . Then t ( a l ,PI)E ( t , r,, 1,) by induction and r - ' t ( a , P ) r = t ( a l ,PI), giving the result. ~"'1,

y"l

Theorem 1. Let G ( M )= ( B ( M ) ,k ; kt = tk, kr, = r,k, kl, = ljk, all i,j). Then G ( M ) is finitely presented and H o ( M ) is Turing reducible to the word problem for G ( M ) . Theorem A follows immediately, since we can choose M with H o ( M ) unsolvable .

Modular machines 1

7

Proof. G ( M ) is plainly finitely presented. Also G ( M ) is an HNN extension of B ( M ) with stable letter k. By (111), k f ( a ,p ) = t(a,P ) k iff t ( a ,p ) E ( t , r,, 1,). By (vii) and (vi) this holds iff (a,p ) E H o ( M ) , proving the theorem.

T h e most natural way of showing the existence of finitely presented groups whose word problem has arbitrary r.e. Turing degree would be to reduce the word problem for G ( M )to H o ( M ) .G(M)is nice enough for this to be possible, using the Bokut normal form; this is done in an unpublished paper by K. A. Kallorkoti. W e shall prove a stronger result in $4. 93. Modular machines revisited It is frequently inconvenient to have two configurations of M corresponding to one of T. Suppose q is a state such that no quintuple ends with qL.Then if a configuration with state q occurs in a computation the corresponding element of N Zis (urn + q, urn + a ) , with the previous notation. Consequently we need only associate this pair (and not the pair (urn + a, vrn + 4 ) ) with the configuration. Also corresponding to a quintuple qaa’q’R (or L ) we need take only one quadruple in M , namely (4, a, a’m + q ’ ,R (or L)). Similarly if no quintuple ends with qR. In particular if T is a directed-state machine, i.e. n o state q occurs both in a quintuple ending with qR and in one ending with qL, then we need only associate one pair with each configuration. Any Turing machine can be replaced by a directed-state machine computing the same function. If qo is a state of the Turing machine T, and there is a quintuple beginning qoaodefine H o ( T )to be 0; if there is no such quintuple then the configuration Co,of a blank tape in state qO,is terminal, and we define H o ( T ) as {C;C + Co}. We call the state qo special if the following conditions hold: no quintuple ends with qoR, the only quintuples ending with qoL are q*aaqoL for all a and some fixed state q*, and n o quintuple ends with q*L. A s above, in this case we only take one pair corresponding to configurations in states qo or q * , and only take one quadruple of M corresponding t o each quintuple starting with qo or q * . For convenience when regarding states as integers, we take q* as rn - 1. Many complications occur because 0 may stand for the letter a. or the state qo; requiring qo t o b e special is one way of managing the complications.

8


Theorem 2. (i) Let T be a Turing machine with special state qo and let

M be the associated modular machine. Then H o ( T ) and H , ( M ) have the same many-one degree. (ii) Let d be an r.e. many-one degree, not that of N . Then there is a modular machine Md with Ho(Md)of degree d.

Proof. (i) (0,O) is the unique pair corresponding to Co. Since M simulates T, if (a,p ) corresponds to C we have C E H o ( T )iff (a, p ) E Ho(M).Hence Ho(T) is many-one reducible to Ho(M), the function showing the reduction being C (a, p ) where (a,p ) is chosen to be (um + a, um + q ) if there are two possibilities. The set P ( T ) of pairs corresponding to configurations of T is recursive. Let P * ( T )= {(a,p ) ; (a, p)+(a’, p ‘ ) for some (a’, p’)E P ( T ) } . Plainly H o ( M )C P * ( T ) . If P * ( T ) is recursive we can define a total recursive function 7c from N 2to configurations of T as follows. If (a,p ) $Z P * ( T ) let ~ ( ap ), be qoaoa,(which is not in Ho(T),since it is terminal if Ho(T)#O). If ( a l p ) €P * ( T ) ,let ( a ’ , p ’ ) be the first pair in the computation starting from ( a , @ which ) lies in P ( T ) and let ~ ( ap ), = C‘, the configuration corresponding to (a’, p’). Then C’E H o ( T )iff (a’, p’) E H o ( M ) iff (a,p ) E Ho(M).Hence (a,p ) E H o ( M ) iff ~ ( ap ), E Ho(T), as required. Since P ( T ) is recursive, to show P * ( T ) is recursive it is enough to show the following. (a,p ) E P * ( T )- P ( T ) iff p = b modm where 0 s b s n, a = um i + ‘ + qm + m ‘ - 1 for some q with n < q < m and some i >O, and, if i is the largest integer such that a can be written as above, then (um + q , p m ’ ) E P ( T ) . We prove this by induction on the number of steps needed to obtain an element of P ( T ) starting from (a, p ) . Let (a,p ) 3 (a’, p ’ ) where (a’,p ’ ) satisfies conditions similar to the above. The quadruple which applies to (a,p ) cannot correspond to a quintuple ending with q * R (recall that q * is represented by t h e integer m - 1, and no quintuple ends with q * L ) . For such a quadruple leads to a pair whose first entry is congruent mod m * to am + m - 1 where 0 S a S n, but a ’ is not of this form. The only other possibility is that in (a’,p ‘ ) the state-symbol is not m - 1 (occurring in a’) but is 0 (occurring in p’). This means that the quadruple applied to (a, p ) was ( m - 1 , b, bm, L ) for some b with 0 s b s m. Hence p = b mod m and a = a ’ m + m - 1, so (a, p ) has the required form. We must start the induction by showing that if ((a,@) (a’,@’))€ P ( T ) then either (a,p ) E P ( T ) or (a,0) has the stated form.

+

Modular machines I

9

Suppose the quadruple that applies is (9, a, a ‘ m + 9’, L ) (other cases are similar). If 9 ’ # 0, and C’ is t h e configuration corresponding to ( a ’ , p ‘ )it is easy to see that C ’ =W,q’ba‘W, for some letter b and tape descriptions WI, W,. Then (a,p ) E P ( T ) , corresponding to the configuration WlbqaW2.If, however, 9’ = 0, when t h e quadruple can only be (m - 1, a, am, L ) for some a, in addition to this case there is another one. We may regard 9 ’ as being the letter a. rather than a state. Thus we may have a‘ = um + 91 for some state 91, and p‘ = um ; as ( a ’ , p ’ ) EP ( T ) neither u nor u can be congruent modm to an integer strictly between n and m. In this case a = a r m + m - 1 and p = u, giving t h e stated form of an element in P * ( T ) - P ( T ) . (ii) H f l ( M )is not many-one equivalent to N for any machine M, since if Ho(M)‘#0 then (m,O) fZ H,,(M) as n o quadruple begins with 0,O. Also we can find M with H f l ( M = ) 0. By (i) it is enough to find a Turing machine Td with a special state such that H[)(Td)has many-one degree d for any other d. The work of Borger [ l ] and Overbeek [I71 show that H o ( T )can be chosen to have any r.e. many-one degree (except that of N) for a Turing machine T defined by quadruples. This is explicit in [ I ] for a programmed Turing machine. Replacing program lines by the appropriate quadruples we get a Turing machine by quadruples whose configurations are bijective with those of t h e programmed machine. The work in [17] is already expressed in terms of quadruples; Lemma 5 of [17] shows that the halting problem may have any degree. If we add to the routine given there further subroutines clearing all the arguments it is easy to see that the resulting machine T has H f l ( T )of the required degree. From t h e Turing machine T defined by quadruples we get, as explained earlier, a Turing machine T’ defined by quintuples by adding auxiliary states 9,s, (indeed, a simpler set of auxiliary states may be used here; the full force of this labelling is only needed for other problems). Any configuration C of T can be regarded as a configuration of T’ and C E H,(T) iff C E Ho(T’).T‘ has other configurations corresponding to t h e auxiliary states. If C’ is such a configuration then C’ C, where C can be regarded as a configuration of T, and C‘E Ho(T’)iff C E Ho(T’)iff C E Ho(T). Hence Ho(T’)and H o ( T ) are many-one equivalent. Finally we obtain Td by adding states to T’. Let H,,(T’)be defined in terms of the state 96. Obtain Td from T‘ by adding states 9 * , 9,) and quintuples 9*aaqoL for all letters a and the quintuple 9,koao9*R. Plainly a configuration C of T‘ is in Ho(T’)iff it is in Ho(Td). A configuration of Td in states 9 * or 90 is in Ho(Td) iff it consists of a

+

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S. Aanderaa, D.E. Cohen

blank tape. Thus H o ( T d )and H,,(T') are many-one equivalent, as required. Remarks. 1. H,,(T) and H , , ( M ) are special halting problems. By contrast to Theorem 1, the general halting problem for M can have any r.e. Turing degree greater than that of T. A full account of these results is given elsewhere [9]. 2. Instead of choosing t h e state n + 1 as initial state and t h e state 0 as halting state we may, once having ensured that the machine only halts scanning 0, make 0 into t h e initial and final state (so no quintuple begins 00 but there is a quintuple beginning O i for 1 c i c n). This is useful in t h e proof of Higman's embedding theorem, and the details are given in t h e paper following [O].

84. Truth-table degrees

Let M be a modular machine. We shall associate with M a finitely generated group whose word problem has the same truth-table degree as H o ( M ) .There are several ways of doing this; indeed t h e group G ( M ) of 02 is suitable. We shall use a different group, partly as an illustration of t h e techniques that can be used. It is convenient to regard M as acting on Z 2 instead of on N 2 ; H o ( M ) is unaltered by this change, as one may prove by induction on the length of the computation that if (a, p ) E H,,(M) with a,p E Z , then a, /3 3 0. Let G , ( M )= (r,, k, for a,p E 2 ; r,k, = k,ra for (a,p ) E H,(M)). We sometimes write r, k for to, k,, and r ( a ) ,k ( p ) for r,, k , to avoid complicated subscripts. GI(M) may be regarded as an HNN extension with stable letters { k , } of t h e free group with basis {r,} or as an HNN extension with stable letters { r a } of the free group with basis { k , } . It follows immediately that r,k, = kpra in GI(M) (if and) only if (a,p ) E H o ( M ) . y - ' k , y = k p e l for all Let G 2 ( M )= (GI(M),x, y ; x - ' r a x = a, p E Z). Since {r,} and { k , } are bases for free subgroups of GI(M), we see that G 2 ( M )is an HNN extension of GI(M)with stable letters x, y . As before let M have quadruples (an,b,, c,, R ) for i E I and (a,,b,, c,, L ) for j E J. We obtain G,(M) by adding new generators ri(i E I ) and I, (j E J) to G 2 ( M )and corresponding relations r;'x"r, = xm2, r;'y"r, = y , r;'r,r, = r,,, r;'ker, = k,, for all pairs ((u, p ) to which (a,,b,, c,, R ) applies and yields (a,, PI) (more simply we have

Modular machines I

11

r;'t(um + a,)r, = t(urn2+ c,), r;'k(um + b,)r#= f ( u ) ) and the similar relations for l,, interchanging x and y. W e shall write G I , G2, G3 instead of GI(M), G t ( M ) , G 3 ( M ) .W e shall prove shortly that G3 is an HNN extension of G2. From the two ways of regarding GI as an HNN extension of a free group we obtain both t,-pinches and k,-pinches. As G2 is an HNN extension of G I we also have x - and y-pinches. Notice that if a word in the generators of Gz has an x-pinch this means that it contains a ~ E = ? 1 and u is a word in { f a , k,} which equals subword X - ~ U Xwhere in G I a word in { t m } ;this does not mean that u freely equals a word in it-). (i) If a word in the generators of G2 has an x-pinch then either it has an x-pinch of the form X - ~ U X ' with E = 2 1 where u is a word in { t < , }or it has a k,-pinch for some p (and similarly for y). For, as already remarked, it must have a subword X - ~ U X where ~ u is a word in { f a , k,} which equals in G I a word u in { f a } . Either uu-' freely equals 1 or uu-l has a k,-pinch for some p in which case SO does u (as no k, occurs in u ) . (ii) G3is an HNN extension of Gz. To check the isomorphism conditions we have four cases to consider, one of which is the following (the others are similar). Let W be a word in the symbols X, Y, T,, K,. Let (a, b, c, R ) be a quadruple of M. If W(x"', y"', t @ m + a ) , k (qm + b)) = 1 in G2 then W(x"', y, t @ m 2+ c), k (4)) = 1 in Gz. We use induction on the length of W, which we may assume freely reduced. If W is non-trivial, W(x "',y "',f (prn + a ) , k (qm + b)) must contain a pinch. By (i) W freely equals either WIX-'UXpW2 where U is a word in {T,} and E = 2 1 or W freely equals WlK,'UK,W2 where U is a word in those T, for which @m + a, qm + b ) E H o ( M ) (or a similar case with the roles of X and Y and of Tp and K , interchanged). In the first case let W' = WIU' Wz where U' is obtained from U by replacing each T, by T,,,; in the second case let W' = WlUW2. I n both cases we find that W(x", y"', t @ m + a ) , k(qm + b)) = W'(x"',y",t(pm + a ) , k ( q m + b ) ) and that W(x"",y,t(pm2+c),k(q))= W'(x"', y, t @ m 2+ c), k ( q ) ) , so the result follows by induction. The process of repeated HNN extension is called a Britton tower. (iii) G , is finitely presented. It is finitely generated, with generators t, k, x, y, r,, I,, since f a = x-"tx" and k, = [email protected] infinitely many relations involving r, (and similarly 1,) are consequences of four of them, namely r;'x"'r, = X"'', r , 'y"'r, = y, r , Ix "~tx"~r, = x-'atxc,, r;ly-bikyber, = k. The remaining relations include tk = kt. However they are all consequences of this

12

S.Aanderaa, D.E. Cohen

relation and the relations involving r, and f,. This follows by induction on the number of steps in the computation from (a, to (0,O)since if (a,p ) 3 (a,, p,) using the quadruple (a,,b,, c,, R ) then the relations involving r, show that r, conjugates the commutator of x-"tx" and y - B k y Pinto the commutator of x-"1txU1 and ~ - ~ i k y @Hence 1. the former is trivial if the latter is. (From this presentation we find a much easier expression for G,as a Britton tower, G , 2 (t, x , y, r,, 1,) 2 ( t , x , y ), the latter being free. This is easily seen to be a Britton tower, G,being obviously finitely presented, and the construction being analogous to that of 02. The construction given, however, makes questions of truth-table degrees easier to look at .) Let W be a word in the generators of G,.By a reduction of W we mean one of the following: (I) Free reduction; also replacing x p . x q by x p + q if p + q # O and deleting x " . x-,; also similar reductions for y . (11) If W freely equals UxPt:V where U does not end in x e , replacing W by U t : - g p V ,and similarly for y and kp. (111) If W freely equals Ur-'WorV where r is a stable letter corresponding to (a, b, c, R ) and Wofreely equals a word in x m , y m , t(pm + a ) , k(qm + b ) (for all p , q E Z ) , replacing W by V W ,V where W 1is obtained from Wo by replacing x",y",t(pm + a ) , k(qm + b ) by xm2,y, t(pm * + c), k (4);also similar replacements corresponding to r - l , 1 and I-'. (IV) Replacing UkitZV by UtZkEV or vice versa for all integers a, p. A reduction is called correct if it is of types (I), (11) or (111) or if it is of type (IV) with (a,p ) E Ho(M).A reduction sequence starting from W is a sequence W = W , , W2,..., W, of distinct words such that W,+, is a reduction of W, for all p ; a correct reduction sequence is one in which all reductions are correct. Define the weight of W to be the sum of the following: the number of occurrences of r,, kp, r,, I, and their inverses, the number of occurrences of powers of x and y , and the number of times a power of x precedes (not necessarily immediately) some t : (for E = +1) and similarly for y and k,. Plainly, reductions of types (I), (11), and (111) reduce the weight while type (IV) reductions preserve it. Since there is a bound on the number of type (IV) reductions which can be applied consecutively to W without getting a repetition, we see that only finitely many reduction sequences can start with W (indeed, there is a recursive function of W bounding their number). Because we apply type (111) reductions only when Wo freely equals a

6)

Modular machines I

13

word in the relevant generators, and apply type (IV) reductions for all a, p, it is clear that the finite set { W ' ;W' is a reduction of W } depends recursively on W. The finite set {W'; W' occurs in a reduction sequence starting from W } will also depend recursively on W, as will {(a, p ) E 2'; a type (IV) reduction involving fa, k, occurs in some reduction sequence starting from W } .This set will be used to construct a truth-table depending recursively on W. (iv) If a word in the generators of G , has a pinch it has a correct reduction sequence ending in a word of less weight. Such a sequence will be called a weighr -reducing sequence or w.-r. sequence. By (i) if there is an x- or y-pinch either a type (11) reduction applies or there is a f a - or k,-pinch. Suppose there is a k,-pinch for some Po (the other case is similar). Then there is a subword k;;Uk;, where U is a word in those r, with (a,Po)E H,(M). Type (IV) correct reductions replace this subword by Uk;,'kE, to which a type (I) reduction applies. Let r be a stable letter corresponding to (a, b, c, R). Suppose the pinch is the subword r-'Wr (the other cases being similar). Then W equals in G 2a word in x'", y"', r,, k, for a = a mod m, p =_ 6 mod m. If W freely equals such a word we may apply a type (111) reduction to the original word, while the original word has a w.-r. sequence if W has. We use induction on the length of W to show that if W equals such a word and has no w.-r. sequence then it freely equals such a word. Suppose W has no w.-r. sequence. Then W has no pinch since it only involves x, y, r,, k,, and the result has been proved for such words. Let W equal W , in G2, where W , is a word in x"', y'", fa, k, for a = a, p = b mod m. We may assume W , has no pinch, since pinching out produces a shorter element in the same subgroup. Let U and U , be the largest final segments of W and W 1involving only r, and k,. Suppose there is cancellation in U I U - ' .Then, for instance, W = W'r, where a. = a mod m and W' is shorter than W. Then W' has no w.-r. sequence since W has none, and W' = W,t;i. Hence, inductively, W' freely equals a word in x"', y"', t ,k, (a = a, p = b). Then W will also freely equal such a word. Suppose there is a k,-pinch in U I U - ' but no cancellation. Since U and U , have no pinches, W and W , must freely equal W'kiV and WlkEV, where V, V, are words in { a } and p = b (since k, occurs in W J . The pinch now tells us correct type (IV) reductions take us from W to W'Vk;. This must be freely reduced, since it has no w.-r. sequence as W has none (and free reduction reduces weight). But then W'V has no w.-r. sequence, equals in Gz a word of the required form, and is shorter than W. Hence it freely equals such a word (by induction on length). Since W'V is freely reduced and V involves only

14

S.Aanderaa. D.E. Cohen

both W' and V, and hence W itself, must be as required. Similar aruments apply to a t,-pinch. As W ,W - ' = 1 in Gz it must have a pinch. By (i) the only possibility left is that W and W , freely equal W'xqU and W:xp"U1where U, UI are words in { t a } (or similarly for y and k p ) . U must be empty, else W has a type (11) reduction; we may also assume UI empty since a type (11) reduction applied to W, gives a word of the same form. If q = p m , as before W ' , and hence W, is as required. If q # pm then WIX~~-~W must - ' have a pinch which can only be a pinch of a term x * I in x ~ m - q with a similar term in Wl or W - ' . In the first case (by (i) as W, has no pinches) a further type (11) reduction could have been made in W,; we may choose W, so that this is impossible. In the second case, for the same reason, a type (11) reduction could be made in W, contrary to hypothesis.

{t,},

Lemma 3. Let W be a word in the generators of G,. If W = 1 in G , there is a correct reduction sequence starting at W and ending at 1. Conversely, i f there is such a sequence then W = 1 in G,.

Proof. Each correct reduction is a consequence of the defining relations so W = 1 if there is such a sequence. If W = 1 in G , and is non-trivial it must have a pinch. By (iv) there is a correct reduction sequence from W to W , where W , has smaller weight than W. This sequence shows that W , = W in G3,and the lemma follows by induction on weight. The next lemma is used only for the Higman embedding theorem [O] but its proof fits in naturally here. Lemma 4. Let S be a subset of Z . If a word W equals in G , a word in ( k , t, ( a E S ) ) there is a correct reduction sequence from W to W , where W , freely equals a word in ( k , t, ( a E S ) ) .

Proof. It is enough to show that if W does not freely equal such a word then it has a w.-r. sequence. For if it has a w.-r. sequence ending in W' then W' = W in G , and by induction on the weight there is a correct reduction sequence from W' to such a W , , which gives a correct reduction sequence from W to W,. If W involves x,y,r,, or 1, it must have a pinch since it equals an element of GI. By (iv) it then has a w.-r. sequence. Suppose W is a word in (t,, k p (all a,p ) ) which equals in G I a word W , in (k,t, (a E S)). If W has a pinch it has a w.-r. sequence so we

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may assume it has no pinch. We may also assume that W , has n o pinch, since pinching o u t gives an element of t h e same subgroup. Since W ,W - ’= 1 in G , it must contain a cancellation or a pinch. We now proceed as in (iv). Theorem B is an immediate consequence of Theorem 2 and Theorem 5 below. Theorem 5 . G , ( M ) is a finitely presented group whose word problem has the same unbounded truth-table degree as H,(M).

Proof. We have already shown that G , ( M ) is finitely presented. H o ( M ) is one-one reducible to t h e word problem for G , ( M ) since ( a ,p ) E H o ( M ) iff x - ” t x ” and y - P k y P commute in G,. We show how the word problem for G3 is truth-table reducible to H o ( M ) .Let W be a word in t, k , x, y , r,, 1,. Our truth-table will have question columns corresponding to all pairs ( a ,p ) occurring in type (IV) reductions in some reduction sequence starting at W. A line in the answer column will be “Yes” iff there is a reduction sequence ending in 1 all of whose relevant questions are answered by “Yes”. This is a recursive construction, by the remarks on reduction sequences. This construction gives a “Yes” somewhere in the answer column iff there is a correct reduction sequence starting at W and ending at 1. By Lemma 3 this holds iff W = 1 in G,. This proves Theorem 5. The crucial point that leads to truth-table reducibility rather than Turing reducibility is that we make reductions in a general situation, and only decide after all reductions are made whether or not they have been made correctly. If w e had partial recursive functions which allowed us to make correct reductions, but (for instance) required us to determine if ( a ,p ) E H , ( M ) or not before determining if kp’t,kp could be replaced by t, t h e procedure would only give Turing reducibility.

References [l] E. Borger, A new general approach to the theory of the many-one equivalence of decision problems for algorithmic systems, Z . Math. Logik Grundlagen Math. (to appear). [2] W.W. Boone, The word problem, Annals of Math. 70 (1959) 207-265. [3] W. W. Boone, Word problems and recursively enumerable degree of unsolvability A sequel on finitely presented groups, Annals of Math. 84 (1966) 49-84.

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[4] W.W. Boone, Word problems and recursively enumerable degrees of unsolvability. An emendation, Annals of Math. 94 (1971) 389-391. IS] J.L. Britton, The word problem, Annals of Math. 77 (1963) 16-32. [6] C.R.J. Clapham, Finitely presented groups with word problems of arbitrary degrees of unsolvability, Proc. London Math. SOC.(3) 14 (1964) 633-676. [7] D.E. Cohen, Residual finiteness and Britton’s Lemma, J. London Math. Soc. (2) 16 (1977) 232-234. [8] D.E. Cohen, Combinatorial group theory; a topological approach, Queen Mary College Lecture Notes, 1978. [9] D.E. Cohen, Degree problems for modular machines (to appear). [lo] D.J. Collins, Truth-table degrees and the Boone groups, Annals of Math. 94 (1971) 392-396. [ 111 A.A. Fridman, Degrees of unsolvability of identity in finitely presented groups, Soviet Math. 3 (1962) 1733-1737. [I21 A.A. Fridman, Degrees of unsolvability of the word problem for finitely defined groups (Izdatel’stvo Nauka, Moscow, 1967). [13] G. Higman, Subgroups of finitely presented groups, Proc. Roy. SOC.A 262 (1961) 455-475. [14] R. Mckenzie and R. J. Thompson, An elementary construction of unsolvable word problems in group theory, in Word Problems, Studies in Logic and the Foundations of Mathematics (North-Holland Publishing Co., Amsterdam, 1973). [15] M.L. Minsky, Recursive unsolvability of Post’ problem of ‘Tag’ and other topics in the theory of Turing machines, Annals of Math. 74 (1961) 437-455. [16] P.S. Novikov, O n the algorithmic unsolvability of the word problem in groups, Trudy Mat. lnst. Steklov 44 (1955). [ 171 R. Overbeek, The representation of many-one degrees’by decision problems of Turing machines, Proc. London Math. Soc. (3) 26 (1973) 167-183. [ 181 J.C. Shepherdson and H.E. Sturgis, Computability of recursive functions, J. Ass. Computing Machinery 10 (1963) 217-256. And also the immediately following paper in this volume. 0,s. Aanderaa and D.E. Cohen, Modular machines and the Higman-Clapham-Valiev embedding theorem, in Word Problems 11: The Oxford Book, Studies in Logic and the Foundations of Mathematics (North-Holland Publishing Co., Amsterdam, 1979).

S.I. Adian, W.W. Boone, G. Higman, eds., Word Problems I1 @ North-Holland Publishing Company (1980) 17-28

MODULAR MACHINES AND THE HIGMAN-CLAPHAMVALIEV EMBEDDING THEOREM Stal AANDERAA and Daniel E. COHEN University of Oslo and Queen Mary College, London

In this paper we use the modular machines introduced in [2] to give simple proofs of the following theorems. Theorem A (Higman [5]). A n y finitely generated recursively presented group can be embedded in a finitely presented group. Theorem B (Valiev [lo]). Let C be a finitely generated recursively presented group. Then there is a finitely presented group containing C whose word problem. has the same unbounded truth-table degree as the word problem for C. Since Higman’s original proof of Theorem A, several other proofs have appeared [ l , 3, 4, 6, 91. Our proof is closest in spirit to that of Aanderaa [l]. As in [2], the use of modular machines instead of Turing machines greatly simplifies the proof. Higman also showed (which is not difficult) that any finitely generated subgroup of a finitely presented group is recursively presented. He also observed that the existence of a finitely presented group with unsolvable word problem follows, by Theorem A, from the existence of a recursively presented group with unsolvable word problem, and that such a group is easy to find. Theorem B was first proved by Valiev [lo]. A weaker result, with Turing degree replacing truth-table degree, was proved earlier by Clapham [4]. Readers interested only in Theorem A should omit 01. They may omit 02 if they give their own proof of the following fact, shown in the middle of 92. For any r.e. set S of words in the alphabet {a,,. . . , a,} there is a Turing machine T on ao,a,, . . . ,a, such that if W is not empty W E S iff the configuration with tape description W in state qo scanning the last square of W is in Ho(T).(Note the unusual features: qo is both the initial and the final state, and the last square of W is scanned.) The proof of Theorem A is contained in parts (i)-(iv) of 03. 1 1

18


91. Some reduction lemmas

Let X be a (finite or countable) set of generators of a group H, and the natural map from the free group F = F ( X ) to H. A map f from H to H is called recursive if there is a recursive cp : F + F with n-cp = fn-. If X is infinite, changing the ordering of X could change recursiveness; in the applications there will always be a natural ordering of X . If X is finite, any homomorphism cp from F to F is recursive; if X is infinite, cp is recursive iff its restriction to X is recursive. n-

Lemma 1. Let A be a subgroup of H, f : H + A a recursive map which is the identity on A. Then the membership problem for A is many-one reducible to the word problem for H. Proof. Take recursive cp : F + F with n-cp = fn-. Let w E F. If T W E A, n-cpw = fn-w = T W , while if n-w = n-cpw we have n-w E fH C A. Hence n-w E A iff n- (w-'cpw) = 1, which reduces the membership problem for A to the word problem for H. Lemma 2. Let A, ( i = ? I ,..., k n ) be subgroups of H, fi : H + A - , recursive maps which restrict to isomorphisms A, -+ A _ , (the restrictions of fz and f-l being inverse). Let H * be the HNN extension ( H , p , ; ~ ; ' A Q= , A-#, i = 1,. . . , n ) . Then the word problem for H * is truth-table reducible to the join of the word probiem for H and the membership problems for A,, i = ? 1,. . . , k n. Proof. Denote p;' by p-#. Let cp# : F + F be maps with ~ c p ,= fin-. (Note that fi and cp, need not be homomorphisms.) W e use the methods of 93 of [2]. Let w be a word in X and the p , ( i = 1,. . .). A reduction of w is one of the following: (I) delete a subword p - # p # ; (11) delete a subword p-,up,, where u E F; delete the whole of w if wEF; (111) replace a subword p - p p , by cp,u, where u E F. A reduction is correct if it is of type (I), of type (11) with the deleted subword equalling 1 in H, or of type (111) with TU E A,. A (correct) reduction sequence is a sequence w,, w2,.. . . ,wk where w,+' comes fOm w, by a (correct) reduction, j = 1,. . . , k - 1. Plainly there is a bound to the length of a reduction sequence starting at w, and there are only finitely many reduction sequences starting at w. W e associate a truth-

*

Modular machines I1

19

table recursively with w as follows. There will be a question column m = l?” to each (sub)word u which is deleted in a type (11) reduction in a reduction sequence starting from w ; there will be a question column “ I T U E A,?” for each u occurring in a type (111) reduction in some reduction sequence starting with w. In t h e answer column a line receives the answer “Yes” if the questions answered by “Yes” o n this line include all the questions in some reduction sequence ending in 1. This means that there is a “Yes” in the answer column iff there is a correct reduction sequence from w to 1. If there is such a correct reduction sequence then w = 1 in H*, since each correct reduction corresponds to a relation of H*. If w = 1 in H * either w E F and ITW = 1 or w contains a subword p-,pl or w contains a pinch p-#up, with u E F and ITU E A,. In each case a correct reduction can be performed, yielding a word w ’ with w’ equalling 1 in H * and w ‘ either trivial or with fewer p-symbols than w. Inductively, w’ and hence w itself, has a correct reduction sequence ending in 1. Hence w = 1 iff the truth-table for w has a “Yes” in t h e answer column, giving the result. “

Note that Lemmas 1 and 2 may be generalised slightly. In Lemma 1 we only require a recursive map cp with ~ c p wE A for all w and ncpw = TW if T W E A ; then TQ is a map from F to A which need not induce a map from H to A. In Lemma 2 we may take maps ‘pi such that ITQ,FC A-, and pi induces the isomorphism from A, to A _ , (but need not induce a map from H to A - , ) .

Lemma 3. Let A, ( i = 2 1,..., * n ) be subgroups of H . Let cp, : F + F be partial recursive functions such that cpa induces an isomorphism from A, to A _ , (and cp-, induces the inverse isomorphism). Let 0, be a , that ITW E r 0 , w if recursive map from F to finite subsets of I T - ’ Asuch r w E A,, and 0,F dom cp,. Then rhe word problems for H and H * = ( H , p , ; p;’A,p, = A-,, i = 1 , . . . , n ) are truth-table equiualent.

Proof. The proof combines and slightly generalises Lemmas 1 and 2. Replace in the proof of Lemma 2 t h e reductions of type (111) by reductions of type (111’): replace P - ~ U P , by cpiu for some u E Oiu. As Oi is recursive and cpiu is defined for u E Oiu it is clear that { w ’ ; w ’ is a reduction of w } is a finite set depending recursively on w. The set of all reduction sequences starting from w is finite and depends recursively on w as do the finite sets { u ; u is used in some reduction of type (11) in a reduction sequence starting from w } and (u, u ) ; the pair (u, u ) is used in a type (111’) reduction in a sequence starting from

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S. Aanderaa, D.E. a h e n

w. A reduction of type (111’) is correct if m = PU. Build a truth-table as before with columns corresponding to “mu = l?” for u in a type (11) ) for reduction in some sequence starting from w and ‘ ‘ ~ ( u u -=~ l?” pairs (u, v ) in such a type (111’) reduction. As before this provides a truth-table reduction of the word problem for H * t o the word problem for H. The word problem for H is trivially one-one reducible to that for H*. In Lemma 2 the existence of partial recursive isomorphisms from Ai to A-i would enable us to perform pinches and so Turing reduce the word problem for H * to the join of the word problems for H and the membership problems for Aki in H. The crucial feature that gives truth-table reducibility is that we have functions from H to A-i. These enable us to perform reductions first and check correctness later. Many similar results can be proved, in particular versions of Lemmas 2 and 3 for amalgamated free products. We will simply prove some results on solvable problems.

Lemma 4. Let H, A, and H * be as before. Suppose the isomorphisms from Ai to A-i are partial recursive. Zf the membership problems for Aei in H are solvable, then the membership problem for H in H * is solvable. Zf, in addition, the word problem for H is solvable, so is the word problem for H * . Remark. Frequently we are given partial recursive maps from Yi to A _ , which will always extend to partial recursive maps from (Y,) to A+. If P ( Yi) = Ai, and ker P is r.e. (i.e. H is recursively presented) this defines a partial recursive map from .rr-’Ai to A - , which is what we need.

Proof. Any element of H * can be written as the product of elements of H and the stable letters (and their inverses). We can see if there is a subword p-,up,. If so, as the membership problem for A, is solvable we can determine whether or not u E A,. If u E A, we can replace p-,up, by the corresponding element of A _ , (as the isomorphisms are partial recursive). This enables us to obtain recursively a reduced form for any element of H*. An element of H * equals 1 in H * iff its reduced form lies in H and equals 1 in H. Lemma 5. Let A1, A2 be subgroups of HI, H2.Suppose there is an isomorphism from A , to A2 which is partial recursive with partial recursive inverse. Let the membership problems for A t in H,and for A 2

Modular machines I1

21

in H , be solvable. Let K , be a subgroup of H I with A 1C K I and the membership problem for K I in HI solvable. Then the membership problems for HI, K , , and ( K l ,H,) in the amalgamated free product ( H I ,H,: A , = A,) is solvable. Proof. As in Lemma 4 we recursively find reduced forms for elements of the amalgamated free product. An element lies in HI iff its reduced form lies in H I (or in A,). Because K, 2 A , an element lies in (K1,H,) iff its reduced form consists of elements alternately from K1 and H,. 02. Turing machines

Let S be an r.e. subset of N. It is proved by Overbeek (Lemma 5 of

[7]) that there is a Turing machine (defined by quadruples) on the alphabet {0,1} whose halting problem is many-one equivalent to S. A

glance at the construction shows that the machine halts on qlOl' iff r E S ; also there is a halting state (i.e. a configuration is terminal iff its state is the specified one). It is easy to check that if S # N and we add further subroutines clearing all registers and a final subroutine moving to the next blank square to the right (this latter is convenient but not necessary) we obtain a machine T I with halting state qo such that H,,(TI)(the problem of halting with a blank tape) has many-one degree that of S, while r E S iff qlO1' E H o ( T I )(and T I does not halt on qlO1' if r & S). The final subroutine (move to next blank right) ensures that there is only one quadruple involving qo, which is qOOqo for some state

4.

Routine technical arguments (e.g. [8]) extend this result to an arbitrary alphabet. Le., if S is a proper r.e. set of word in the alphabet { a , , .. . , a,,} then there is a Turing machine T I on the alphabet {ao,a , , . . . ,a.} with the following properties: TI has a halting state qo; only one quadruple involves qo and this is qaoaoqofor some state q ; H o ( T I )is many-one equivalent to S; if TI halts on q l a o W (where W is , q l a o W E H o ( T Iiff ) a word in u , , ..., a,,) then q l a o W E H o ( T I )and W E S. For more details look at the discussion in [S] of machines on an arbitrary alphabet and translate the subroutines in Overbeek's machine (and his con,struction of machines computing total recursive functions) accordingly. We wish to consider Turing machines defined by quintuples. These are obtained by adding auxiliary states corresponding to non-moving quadruples. We shall also need another modification. Let T; be as above. Let T be defined by adding auxiliary states q :

22


for each state q. and two further states q’, q”. T h e quintuples consist of q,a,a,q,R (or L ) corresponding to the quadruple q,a,q,R (or L ) , q,a,a,qTR corresponding to q,alalqs,and for all s and i quadruples q fa,a,q,l, and finally quintuples qoa,a,q’l for i # 0, q’a,a,q’L for i # 0, q’aoaoq’’R and q”a,a,qIL for all i. Note that qo is special in the sense of Theorem 2 of [2] and the remarks preceding it. If T is started in state q,, scanning the last square of W, and W is non-empty (if W is empty the tape is blank and the configuration is terminal) it will reach the configuration q,aoW; from this it will not halt if W tif S but will halt on a blank tape if W E S. A configuration in state q, (for r # O ) will lie in H n ( T ) iff it is in H,(T,) (note that TI only halts on a non-blank square if the configuration we start with is in state q,,). If the configuration C is in state q f then C j CI with C , in state 4.; for s # O we have C E H n ( T ) iff C, E Ho(TI)(again because TI halts on blank squares the computation for T does not continue after it has simulated the computation for TI). For configurations with state qn scanning a blank square or in state q : immediately to the right of a blank we can tell whether or not they are in Hn(T).For any other configuration C we can recursiv,ely find a configuration CI with C + C , and C , in state q l (all configurations between C and C, being in states qo, q’ or 4”). Then C E H o ( T )iff C , E Hll(Tl).These remarks show that H n ( T )is many-one equivalent to Hn(Tl). So we have found t o any proper r.e. set S in the alphabet a l ,. . . ,a, a Turing machine T defined by quintuples such that (i) H,,(T) is manyone equivalent to S and (ii) W E H,,(T)iff the configuration in state qn scanning the last square of W lies in H,(T) (the holding also for W empty). This machine is used in the next section, S being the set of relators of a recursively presented group. 03. The embedding theorems Let C be a non-trivial finitely generated recursively presented group. Let C be generated by c , , . . .,c,. W e regard a word in the letters cf’, . . . , c:’ as a positive word in the letters c,,. . . , c2,, where c,+~ denotes c;’. Let W be the r.e. set of all relators of c. By 92 we may take a Turing machine T whose alphabet is cl,. . . , cz. and co (blank) such that H o ( T )is unbounded truth-table equivalent to W and such that a word w is in W iff the configuration with tape description w in state q,, scanning the last square of w lies in H , ( T ) (note that w = 1 corresponds to the blank tape). As remarked in the introduction for

Modular machines II

23

the proof of Theorem A only the second of these two properties is needed, and readers may supply the proof of this property themselves if they do not wish to read § 2 . Let M be the modular machine associated with T. Each word w defines an integer a, by regarding w as the representation in m-ary notation of a with cl,. . ,cZn being taken as the integers 1,. . ., 2 n . The set I of integers represented by a word is recursive. For a E Z let W, denote the corresponding word. In particular wo = 1 and, for 1 C i C 2n, we have wi = ci and w,,+, = w,ci. If a E Z then (a,O)E H o ( M ) iff w, E W by the construction of M from T. It was shown in 94 of [ 2 ] that the group G 3 = G 3 ( M )constructed there is finitely presented. We never refer to the quadruples of M explicitly so we can use letters a, b, c for generators of groups without fear of confusion. The notation used is chosen to be as close as is convenient to the notation in the proof of Theorem A by Aanderaa [l]. Let H3=C*G,. Let H4= ( H 3 ,bi ( i = 1,. . . ,n ) ; br'qb, = cj ( i , j = 1,. . . , n ) , b;'kbi = kc;' (i = 1,. . . , n ) ) . Observe that if bn+i denotes b;' then we have b;'cjbi = c, and bi'kb, = kc;' for i , j = 1,. . . , 2 n . (i) H4 is an HNN extension of H,. For the subgroups associated with b, are both C*(k), and the relevant map is a homomorphism which plainly has an inverse. LetLH,=(H4,d;d-'kd = k ) and let Hs=(H,,p;p-'kp = k, p-'r,p = r,w,(b)d for all a E I ) . Here if w is a word in the ci, w(b) denotes the corresponding word in the b , ; other similar notations will be used. (ii) H , is (obviously) an HNN extension of H4. Also H6 is an HNN extension of H,. We must show that the subgroups (ra, k (a E I ) ) and (rawa(b)d, k (a E I ) ) are isomorphic, with the named generators corresponding. The latter plainly maps to the former by the homomorphism of H,which maps G , identically and kills all the other generators. From the construction of G3 as a Britton tower we know that the former has defining relations r,k = kr, for all a such that a E Z and (a,0) E H o ( M ) .Thus we have a homomorphism of the former to the latter provided r,w,(b)dk = kr,w,(b)d for all such a. If a E Z we have w, (b)dk = kw, (bc)d since k-'bik = bici. Also w, (bc) = w, (b)wa (c) since bici = cibi for all i, j. Since w, ( c ) = 1 iff w, E W iff (a,0) E H o ( M ) the required relations hold. The last group needed is H7= (H6, a ] ,. . .,a z n ;a;'rai = ti, a;'xai =


24

x"', a;'pa, = p , a;'da, = b,d, a;'b,a, = b, (i = 1,. . . , 2 n ; j = 1,. . . , n)).

Notice that a;'b,a,

=

b, will also hold for j

=

n

+ 1,. . . ,2n.

(iii) H7is an HNN extension of H,. The corresponding associated subgroups which we must prove isomorphic are (for given i) K = (r, x, b,, d , p (j = 1,. . . ,n)) and L = (r,, x m , b,, b d , p ) . Recall that if B is a subgroup of a group A, and A * is any HNN extension of A, then the group generated by B and the stable letters will be an HNN extension of B provided B meets the associated subgroups suitably (see, e.g., the remarks following Britton's Lemma in [2]). Also an HNN extension with trivial associated subgroups is just a free product. It follows from these remarks that, since ( t , x ) n ( k ) is trivial (and hence so is ( r , x ) r l ( C * ( k ) ) ,the subgroup (r, x, b, 0' = 1,. . . ,n ) ) is free on these generators. In turn we find that (r, x, b,, d (j = 1,. . . ., n)) is free on these generators. We next show that ( t , x, b,, d

) n ( k , taw, ( b ) d (a E I)) = ( a w , ( b ) d (a E I)).

For the former clearly contains the latter. Let h be in the former. In the homomorphism from Hs to G 3 which is the identity on G,and kills the other generators h maps to an element g E (r, x ) n ( k , r, (aE I)) which equals (r, (a E I)) from the construction of G, (indeed this follows from the expression of G,as an HNN extension of G1).Since h is the image of g under the map sending k to k and C, to r,w,(b)d we find that h is in the required subgroup. From this and other similar properties we can identify K as the HNN extension with stable letter p of the group freely generated by {r, x, b,, d } , the defining relations being (where r, = x - " t x " ) p-'r,p = raw,(b)d for all a E I. Similarly L is an HNN extension with stable letter p of the group freely generated by r,, x m , b,, d, the relations being p-'r,p = t,w,(b)d for all a E Z with a = i mod m. Since w U m + , ( b=) w, (b)b,,the isomorphism condition is now easily checked. (iv) H7is finitely presented. Since G , is finitely presented it is clear that H7is finitely generated and is presented by: (I) the relations of C, i.e. w , ( c ) = 1 for all a with a E Z and ( a , O ) E H,(M), (11) p-'t,p = r,w,(b)d for a E I, and (111) a further finite set of relations, namely, the relations of G,, all the relations not in (11) which have been explicitly given in the cogstruction of H,, and the relation p-'rp = rd (which also occurs in (11)). We first show that the set of relations (I) follows from (11) and (111).

Modular machines I1

25

From (111) we have, for all a E I, both k-'p-'t,pk = p-'k-'t,kp and k-'t,w, (b)dk = k-'t,kw, ( b ) w , ( c ) d . Hence, using (11), we find that w , ( c ) = 1 if k-'t,k = t,. But by definition of G,, this holds as a consequence of the defining relations of G , provided (a,0) E H,(M). This gives the result. Next we show that (11) follows from (111), proving H , is finitely presented. We first show inductively that, for (Y E I, w i ' ( a ) t w , ( a ) = t, and w,'(a)dw, (a) = w, ( b ) d are consequences of (111). For these are true by definition if 0 < a < 2n. Since /3 > 0 is in I iff /3 = am + i for some a E I and 1S i S 2 n , it is enough to show that if these are true (a) = for a then they are true for am + i (1 s i =s2n). Now warn+, w,(a)a,, SO w 2 , + , ( a ) t w , , + , ( a = ) a ; ' w 2 ( a ) t w a ( a ) a ,= a;'r,a, (inductively) = a;'x-"tx"a, = x-Pma;'ra,x"m = tam+,. Similarly w 2 + , ( a ) d w a m + , (=a a) ; ' w i ' ( a ) d w , ( a ) a , = a;'w,(b)da, = w,(b)b,d = warn+,( b ) d Hence, as a consequence of (111) we have, for all a E I, p-'r,p = p-'w;'(a)rw,(a)p = w:'(a)p-'tpw, ( a )= wi'(a)tdw, ( a )= raw, ( b ) d , i.e. (11) follows from (111). This completes the proof of Theorem A, since H,, being obtained by a Britton tower (i.e. sequence of HNN extensions) from C*C,, will contain C. We now proceed to the proof of Theorem B. We use the same construction as above, and show that at each stage the degree of the word problem is preserved (by using the lemmas of 01). If H, H' are any groups we write W P ( H ) = W P ( H ' ) if H and H' have word problems which are unbounded truth-table equivalent. (v) WP (H3)=W P ( C ) . For by 04 of [2] the word problem for G , is truth-table equivalent to H , ( M ) which is many-one equivalent to H o ( T ) by Theorem 2 of [2]. Hence W P ( G 3 ) =WP(.C)by the construction of T. The result now follows from the analogue of Lemma 2 for free products (the details of this are left to the reader). (vi) WP(H,) = WP(H,).

As already remarked, the associated subgroups are both C * ( k ) .The isomorphisms plainly lift to partial recursive functions. We may now apply Lemma 3 (in the weak form where the maps 8 map to elements rather than to subsets). The relevant maps send c, to c, (j= 1 , . . . , n),

map all the finite set of generators of G , except k to the identity, and send k either to k or to kc;'.

26

S. Aanderaa, D.E. &hen

(vii) WP(H,)= WP(H,). This is a very easy application of Lemma 3, the map 8 being the homomorphism which kills all the generators except k. (viii) WP(H,) = WP(H,). For the only time we use the full force of Lemma 3. Let F,, F3 be the free groups on the generators of H , and G3, IT : F , + H , the projection and p : F5+ F3 the retraction (i.e. maps the generators of F. to themselves and maps the generators of F, not in F3 trivially). Let U and V be the subgroups (k,t, (a E I)) and (k,taw, ( b ) d (a E I)) of H,. Let cp be the homomorphism from the subgroup (k,t, (a E I)) of F, into F5 defined by cpk = k, cpt, = t,w,(b)d (here t, is the element x - " t x a of F,, and the homomorphism cp exists because the subgroup is free on the stated generators). As I is recursive, cp is partial recursive and pcpw = w if w E dom cp. Plainly cp and p induce the isomorphisms between the associated subgroups U and V. Refer to the discussion of reduction sequences in §4 of [2]. It is shown in Lemma 4 of that paper that if w E F3 with ITW E U then there is a correct reduction sequence (in the sense of that section) from w to w ' where w ' is in the subgroup (k,t, (a E I))of F3; also ITW = ITW' since the reductions are correct. For w E F3 let Bw be the subset consisting of all words w ' E (k,t, (a E I)) which occur in some reduction sequence starting at w. By the above ITW E d w if ITW E U. Also B is plainly recursive. Extend 8 t o F5 by defining Ow t o mean Bpw for any w E F,. Then 6 has the required properties provided ITW = ITPW whenever ITW E U. This certainly holds if w E F3, since then pw = w. If w @ F3 but ITW E U there must be a pinch in w . Let w ' be obtained from w by pinching out the relevant subword. It is easy t o check that pw' = pw. Since I T W ' = ITW it follows by induction on the number of generators not in F3 which occur in w that ITW = ITPW,as required. Finally we must define B ! with ITW E d ' w if ITW E V. W e notice that Bw Cdomcp and define B'w to be pew. If ITW E V then ITPW E U. Hence we can find w ' E Ow = Bpw with ITW' = rrpw. W e have rrcpw'E ITB'W so we need only show I T P W ' = ITW. Since m p w ' E V and p induces an isomorphism from V to U we need only show rpcpw' = ITPW,which is true since pcpw' = w'. (ix) WP(H,)= wP(H6). The isomorphisms between the.associated subgroups K and L lift t o partial recursive maps. By Lemma 3 it will be enough to find recursive maps from H6 to K and L which are the identity on K and L. It will

Modular machines II

27

be enough to find a subgroup HL of H6 containing K (and hence L ) such that there is a retraction from H6 to H i and to show that the membership problems for L in K and K in HL are solvable. For then a map (which is not a homomorphism) from HL to K can be defined to be the identity on K and trivial outside K,and this gives a map from H6 to K by composing with the retraction. Let Hk be the subgroup of H6 generated by r, x, d, p , b, (1 S j S n) and all the r - and I-symbols (which are among the generators of G3). We obtain a retraction from H6 to HA by killing the remaining generators. This retraction sends those defining relators of H6 involving only the generators of HL to themselves and sends all other relators to the trivial relator. It follows that HL has as defining relators those (for HL is defining relators of H6 involving only the generators of certainly a quotient of the group with the same generators and only these relators, but in addition H6 and so HL maps to this latter group). In particular HA is an HNN extension of K with stable letters the r and I-symbols; each associated subgroup is (ru,xy)for some integers u, U.

The required results have been proved by Kalorkoti (unpublished) using the Bokut normal form. The messy proof given here does not seem to admit much simplification. It follows from Lemma 4 that the membership problem for K in H L is solvable if t h e membership problems for ( t u , x " ) in K are solvable for all u, u. The membership problem for (tu,x u ) in (r, x, b,, d ) is clearly solvable, since the latter is free on t h e named generators. Hence it is enough to show that the membership problem for ( t , x, b,, d ) (which we denote by K') in K is solvable. Now K is in turn an HNN extension of K' with associated subgroups (r,(a E I)) and (r,w,(b)d (a E I ) ) . Using Lemma 4 again it is enough-to show the membership problems for these in K' are solvable. Now an element of K ' is in the first of these iff (i) it is a word in t and x only, (ii) its exponent sum in x is zero, so that it can be written as a word in r, (all a ) , and (iii) when so written it involves only those t,with a E I. Since I is recursive, this is solvable. With the notation of part (viii) an element w of K ' is in the second subgroup iff pw is in the first subgroup and w = cppw, showing that this problem is also solvable. We still have to show that the membership problem for L in K is solvable. Now K is t h e amalgamated free product of the free group (I, x ) with (t,, (all a ) , d , p, b, (1 S j s n)), the amalgamated subgroup being (r"). Since (t,, d, b , , p ) is an HNN extension of the free group (a,d, b,), Lemma 5 and the previously proved results show that the

28


membership problem for the subgroup L ' = (r, (all a),x'", b, (1C j C n), d , p ) in K is solvable. Since ( x ' " , r, (all a)) is free on x'" and r,, (1 =su =sm ) we may , d , p ; r;lxmpx-mr, = x " p d - ' w 2 + , ( b ) x - present L' as ( r , , , ~ "b,, (1 c j C n, 1 d u m, all u ) ) . Thus L' is an HNN extension of the free group (x'", bj, d , p ) with several stable letters, and so is the amalgamated free product of HNN extensions with one stable letter. Since L is the subgroup of L' with the stable letter rj, it follows from Lemma 5 that the membership problem for L in L' is solvable if the membership problems for the base group in each extension are solvable. A further application of Lemma 4 tells us that this holds if the membership problems for ~"'"px-"'" (all u ) and ~ " ' " p d - ~ w ; ~ , , ( b ) x - "(' u" fixed, all u ) in the free group x'",b,,d,p are solvable. These problems are seen to be solvable in the same way as for r, and r,w,(b)d in K'. These results complete the proof of (ix) and so prove Theorem B.

References [l] S. Aanderaa, A proof of Higman's embedding theorem using Britton extensions of groups, in Word Problems. Studies in Logic and the Foundations of Mathematics (North-Holland Publishing Co., Amsterdam, 1973). [2] S. Aanderaa and D.E. Cohen, Modular machines, the word problem for finitely presentcd groups and Collins' theorem, in Word Problems 11: The Oxford Book, Studies in Logic and the Foundations of Mathematics (North-Holland Publishing Co., Amsterdam, 1979 (i.e. this volume)). [3] C.R.J. Clapham, Finitely presented groups with word problem of arbitrary degrees of unsolvability, Proc. London Math. SOC. (3) 14 (1964) 637-676. [4] C.R.J. Clapham, An embedding theorem for finitely generated groups, Proc. Londoi Math. SOC. (3) 17 (1967) 419430. [5] G. Higman, Subgroups of finitely presented groups, Proc. Roy. SOC. A 262 (1961) 455475. [6] R.C. Lyndon and P.E. Schupp, Combinatorial Group Theory, Ergebnisse der Mathematik, Bd. 89 (Springer, Berlin-Heidelberg-New York, 1977). [7] R. Overbeek, The representation of many-one degrees by decision problems of Turing machines, Proc. London Math. SOC. (3) 26 (1973) 167-183. (81 J.C. Shepherdson and H.E. Sturgis, Computability of recursive functions, J. Ass. Computing Machinery 10 (1963) 217-256. [9] M.K. Valiev, On a theorem of G. Higman, Algebra i logika, 8 (1969) 93-128. [lo] M.K. Valiev, On polynomial reducibility of the word problem under embedding of recursively presented groups in finitely generated groups, Lecture Notes in Computer Science, vol, 32 (Springer, Berlin, 1975), pp. 432438.


MALCEV’S PROBLEM AND GROUPS WITH A NORMAL FORM L.A. BOKUT’* Institute of Mathematics, Novosibirsk

Introduction

The following problem was known as Malcev’s problem.

Does there exist an associative ring R, without zero divisors, which is not embeddable in a skew field whose multiplicative semigroup R * of non-zero elements is embeddable in a group. This problem occurred to A.I. Malcev in connection with his wellknown work o n embedding rings in skew fields and semigroups in groups (see (191). In 1966 at the International Congress in Moscow it emerged that three authors had, independently, constructed examples of rings which solve Malcev’s problem (see [ 5 ] , (131, [IS]). The solution given in Bokut’ (51 to this problem leads to the notion of a group with standard basis, a notion which provides a uniform approach to certain well-known constructions in group theory. This paper begins with a brief account of the solution in [5] to Malcev’s problem and then goes on to provide the detailed definition of a group with standard basis, together with some examples. 61. Malcev’s problem

Let k = GF(2) be the prime field of characteristic 2. Let 0 be the semigroup with generators

a,, b,, c,, s,, t,, v, where 1 d i

S

4, 0 d j d 1, and with defining relations

(1)

a,s, = c,vo, b,s,+l= c,uI, b,t,+,= a,t,

where 1 s i s 4 and ss = sl,

tr = t l .

With D.J. Collins, Queen Mary College, London. 29

30

L.A. Bokut'

It is easy to show that the ring kQ contains no zero divisors and is not embeddable in a skew field. (The latter follows from the fact that if kQ C D, where D is a skew field, then the equality ( u ; ' u , ) = ~ 1 holds and hence u o = u I , which is impossible.)

Theorem 1. The semigroup ( k Q ) * is embeddable in a group.

The proof of this theorem appears in Bokut' [S], [9]. The idea of the proof is as follows. Since Q is given byhomogeneous relations, the ring kQ can be embedded in the ring kQ of infinite series. Then, to prove Theorem 1, it suffices to prove that t h e semigroup (kQ)*is embeddable in a group. Now it follows from the results of [2] that the semigroup has a presentation of the form

(m)*

(2)

(kQ)*= (r,s ;PIX#,

= p k x k p , xtpt = pix:)

where r is the group of units of the ring kQ (i.e. t h e set of all series with non-zero constant term), 92 is a set of atoms of kQ consisting of one atom from each class of associated atoms (see [14] for terminology), t h e elements p with subscripts lie in \n and the elements X with subscripts belong to the group r. We consider the corresponding group of quotients

given by the same generators and defining relations as the semigroup (2). To establish Theorem 1 it is necessary to prove that two words in the alphabet of t h e semigroup (2) are equal in the group (3) if and only if they are equal in the semigroup (2). The proof of this last assertion is carried out in the following manner. Firstly we enlarge the initial system of defining relations of the group (3) by adding relations of the form X,~q,X,~klp;'X~~IPLX:!;/ p I' = pk y;k/p L' y;/ pi y$/p y'

yt,kl

where p E 2 , X , Y E r. These additional relations are consequences of the initial defining relations and so we may write (4)

G((kQ)*)= g p K 9 ;@)

where Qi is the enlarged system of relations. Then we prove that every group word in the generators f U S has a normal form, which is defined in terms of the presentation (4). It turns out that the process

Malceo's problem and groups with a normal form

31

whereby a positive word in f U % is reduced to normal form is carried out by means of transformations in the semigroup (2). This enables one to prove Theorem 1. 92. Groups with standard basis

The process of constructing normal forms in the group G ( ( k Q ) * ) , given by (4), can be formalized and, in particular, the definition of a group with standard basis can be formulated. This was done in t h e papers [3], [4], and [6] of Bokut'. These papers also contain proofs that Novikov's groups ?I,,,, (see [20]) and Boone's groups G ( T , q )(see [lo]) are groups with standard basis. This makes it possible to obtain many well-known properties of these groups (see [ l ] and [12]) in a uniform and relatively simple way. Furthermore, with the help of the same method, it was possible (Bokut' [7]) to prove that for any recursively enumerable Turing degree of unsolvability there exists a finitely presented group (of the kind ?I,,a) whose conjugacy problem is of t h e given degree. This last result was obtained, independently, by Collins 1151. We begin with the definition and some properties of groups with stable letters, following, by and large, P.S. Novikov's work [21]. If 2 is an alphabet, then by a 2-word or a word in C (on 2, from 2 ) we shall always understand a group word constructed from this alphabet. For a fixed presentation G = ( 2 ;@) of some group we write X E G if X is a 2-word. Graphic equality of words will be denoted by the sign E and definition or denotation by the sign e. Let d = (2;@) be the group with generators 2 and defining relations @. The group

G = (2,'$; @, A p m , = pn,B,, i E I) (5) where 2 r l '$ is empty, p,,, pn, E '$ and A,, B, are 2-words is the group with stable letters '$ and base group (?. Groups with stable letters were first considered by G . Higman, B.H. Neumann and H. Neumann in their well-known paper [22]. Thus the group G is often called an HNN extension of the group d. (We do not at present assume any conditions which ensure, for example, that d is naturally embedded in G although such conditions are often incorporated in the definition of HNN extension.) The relations (6)

A ;'pn, = pm,B;',

iEI

are, of course, valid in G. We say that the letters pm and p . are related

32

L.A. Bokut'

if the relations (5) and (6) contain, among them, a sequence of equalities A :,"pn,* = p n,* l B2, . . . A :,'p n,, = p n,{)B

r,'

7

-

where p m 2 p n , *and p. K p " , ? . When pm and p,, are related via t h e above sequence, we use the notation

Bpdm B :,I . . . B :,* In other words we use ?Ipmpnand Bpmp, as variables for pairs of words of the form (7). Obviously we always have \ ) I p _ p ~=mp n 8 p n P n . In t h e book [21], t h e words ?Ipmpnand Bpsmin (7) are referred to as occurring in the correspondence L,. We shall call the words (7) connected. For brevity we write ?Ipmpne A f; . . . A f;,

(7)

?IP"z ?IPd", BP" e BPd", Y p ; p ; '

e BPGm,BP"lPrn1 e ?IP&.

By the extension system of relations of t h e group G we mean the system of relations in (5) enlarged by the equalities

B ; ' p , '= pLfA;',

iE

I.

We now turn to the notion of individuality of letters. In using the extension system of relations of G, we perform transformations

XUY + X V Y where either U = V or V = U is a member of the extension system of relations (or else a trivial relation of the kind v f ' = 1, pp-' = 1, u E Z,p E Is). We say that the individuality of the letters in X and Y is preserved by the transformation. Further, if, for example, U K A,pm, and V K p , B , or U K B l ' p i , ' and V K p G l A ; ' we also say that pm, and pn,, or pi,' and p;f have the same individuality. We define individuality for letters occurring in a sequence of transformations of the extension system of relations by transitivity. (For brevity we call such sequences extension sequences.) We give three lemmas, which are explicitly or implicitly in Novikov [21], and which we shall use later. Lemma 1. Ler there be given an extension sequence of the group G

w p : v + w l p : , v I ~ * " - * Wkp:,vk where p : , p ; , , . . . . ,p z kF p L have the same individuality. Then for some connected words ?Ipk; and Bpk; the equalities w k

hold in G.

=

w?Ip;p;,

v, = 82 .v pq-

Malcev's problem and groups wirh a normal form

33

Lemma 2. If in any extension sequence of G which starts with the word Wp;'Up;V the letters p i a and p ; (strictly, letters of the same individuality as p i e and p 2) cancel one another, then U = ?lpap:in G, for some '?Ipap:.. For the third lemma we need a definition. The system @ of stable letters of the group G is called regular if for any p E @ and any two connected words ?I, and '23, the condition

?I,

= 1 in

G

if and only if '23,

=

1 in

G,

is satisfied. (This is often called the isomorphism condition for G and is assumed as part of the definition of HNN extension.)

Lemma 3 (Britton's lemma). Let @ be a regular system of stable letters of the group G, with base group G and let W be a ( 2 U P)-word. I f or W contains W = 1 in G then either W is a 2-word and W = 1 in a subword of the form

pieup:,

(8)

where U is a 2-word and for some !I',;

u = ?I,;,,

(8')

in

G.

A (2 Ug)-word is called $3-reduced if it has no subword (8) for which (8') is satisfied. We now embark on the definition of a group with standard basis. (Our account is a little different from that in [4].) Let G = ( 2 ;@) be a group. Let us suppose that the given presentation of G satisfies the conditions: SB 1) 2 = UZa,0 S a < 7,where 7 is an ordinal and t h e sets S,, are non-empty and mutually disjoint. SB 2) @ = 0 S a < 7 and, for every a, all the relations in @a are of t h e form

u@=,

A p . = p,B

(9)

where p,,p, E 2- and A, B are non-empty words in Up O and E , = S I , Ro = So%,,:lq;, . From SB 6 it follows that p l Eq, and Ra = So in i.e. R o 9 S o .The proof of this assertion we need is completed by induction o n 1.

em,

We make some remarks that we shall use later on. Remark 1. Let (Y > 0 and suppose that for each /3 < a t h e set C, is a basis of Gp.We consider a.reduced word W of the form (12) in which every R, is normal. If W is not normal, then by condition C 2 it contains, for example, the subword x 'Epn,

c,

where E E and E = Bx.,A2in C?, (for some Bxsx). Since E is ) . condition C 2 is equivalent to the normal and so E E C ( B ~ . , A ~Thus condition : C 2') a reduced word W of the form (12) belongs to C, if and only if every R, is a normal word of G, and W contains n o subword of the form x 'C (B.,A2)pn,

(13)

y 'C(B,,,B,)p; I ,

x r - l C ( B x , - i x;')pm, -~A

y

C(By .~ y - IB ;' ) pm', 1

corresponding to a relation (10) from @., Remark 2. In Go, each word is transformed to its normal form by means of cancellations. Let us suppose that for some a > O , it is already proved, for every /3 < a , that C is a basis of G and that rules for transforming words R E G, into normal form have been formulated. Then the following constitute rules whereby a word

W K RopL:.. . p;R,


31

of the group G,, where s S O , p E Z,, R, E G,, is transformed into normal form: (a) transform each R, into normal form, 0 s i c s ; (b) make all possible cancellations of letters p., in the word obtained from (a); (c) exclude from the resulting word the leftmost occurrence of a word (13); (d) return to (a). We understand condition (c) to mean that one has to apply to the word under consideration a transformation of, for example, the kind x ' C ( B ~ ~ , A-+ ~ %,,,A ) P ~ ;'p,B.

Similar transformations are to be used for the other words from (13). We call the application of transformations (a)-(d) the process of reducing W to normal form in G,. In general this reduction process may not terminate in a finite number of steps. To realise the application of the given process in a finite number of steps it is necessary and sufficient that in reducing each word Rp'., where R E G,, p. E Z,,only finitely many transformations (a) and (c) are applied. Remark 3. The process described for reducing a word W of G, to its normal form is some extension sequence of G, beginning with W. Therefore we can refer to t h e individuality of letters p F , p E 2, in this sequence of transformations. We shall say that p F is cancelled (uncancelled) according as a letter of t h e same individuality as p F is cancelled (remains uncancelled) when W is reduced to normal form. We note that Lemmas 1 and 2 can both be applied to the extension sequence of transformations whereby a word is reduced to normal form. 03. Novikov groups

In this section we shall prove that a Novikov group ?Ip,, is a group with standard basis. We give the definition of such a group ?Ip,,, following [20]. Let I

. . , a,,

5 = (a],.

q l , . . . ,q,, r j , . . . , rA, l I , .. . , l A )

where n and A are positive integers. We also consider the alphabet 5' = ( a ; , q:, r:, I : , 1 sj s n, 1 s i S A}. Further we fix a system (A,,B,),1 s i c A, of pairs of non-empty positive words in the alphabet

38

L.A. Bokut'

{ a l , .. .,a n } .Then we denote by ?lp,m = G the group given in t h e alphabet 2=BU3'U{p,,p,}

by the following system of defining relations (the bold face italic letters are those which we distinguish in the relations)

(14)

1. q,a, = aiq,q,

1'. a;q: = q:q:a;

2. r,r,al = a,r,

2'.

3. l,a, = all,

3'. 1:a;

r:a; = u;r:r: = a;l:

4. A : p l A , = q:l:pllt4,

5. r:plrz = p l

6 . B,p,B := r,l,pzl:r: 7. q,p24:

= p2

where i = 1,2, .. . , A , j = 1 , 2 , . . ., n and ( a , ; . . a,,)'= a t . . . a:!. We partition 2 and @ into disjoint subsets: z o = { q , , r , , +4 # ,+r ,l,S i S A } ,

-Zz= {la,I:, 1 i Qo = 4,

S

A 1,

21={uJ,u;,lSjSn}

Z3= { p l ,p 2 ) ;

@, = {(14.1),(14.2), (14.17, (14.2+)}

= {(14.3),(14.3+)},

@d

= ((14.4) - (14.7)).

We view all the relations of the corresponding group G,, 0 S cr as being in the form (15)

AixAzp,

6 3,

= pmBiyB2

where p.,p, E 2, and A l x A r , B I y B z are 2,-l-words. In each case x and y in (15) are the bold italic letters in (14) making the obvious modifications to (14.4)-(14.7) so that they have t h e form (15). We define the sets C,, 0 < cr < 3, as described in the previous section, and prove that C, is a basis of G,, 0 S a S 3. We shall use, in turn, the process of reducing to normal form given in Remark 2. The set Co consists of all reduced words in &. It is clear that Co is a basis of Go. The set Cl consists of all reduced words of t h e form

-

w x Uox f ' UlX ;*

. . x ;"u, *

where k 3 0 (if k = 0 then WK U,,), x, E S1, U, E C,, and W contains no subword from the following list:

Malcev's problem and groups with a normal form -2

(16)

39

-1

q:a,,q, a , ,q,a;', rfa,, r:'a,, rra;'

q +2 i a +, ,qT-ia;,q?a;-',

r?a;, rTa;-', r:-2a;-'.

Let us verify conditions SB 4-SB 6 for LY = 1. We begin with SB 5 . For the connected words 21a, and Bal we have =

V ( q t ,rf),

8 , = V(q:,r r )

where V(q,,r:) is a word in the arguments ql,rf, 1 s i s A and V ( q f ,r,) is the same word but in the arguments q:, r,. If V(q,,rf) is reduced then, clearly, V ( q : , r , ) is reduced. Hence SB 5 follows. Next we show that every word of G I is equal to some normal word. By Remark 2 it suffices to prove that the process of eliminating subwords of the form (16) from a word Rp; R E C,, p. E 2' is finite. Suppose, for example, that p i n a,. In the transformations q:a, + a,qf' and rfa, + a,r,, the length of the word R is decreased. In the transformations r;'a, + r,a,r;l, the length of R does not increase and the number of negative letters in R is decreased. Therefore after finitely many such transformations Ra, will be reduced to normal form. The argument is similar when p',Zu;', a:, u;-'. Let us now verify SB 6. Let, for example, Ra, and Ta, be normal words and let R = TV(q,,r f ) , where V = V(q,,r f ) is a reduced word. Suppose that V i l . If T V is reduced then R ends in q : or rf' which is impossible since Ra, is normal. Otherwise if V begins with q : then T ends with q;' and thus Ta, is not normal. Similarly if V begins with rf', and rf' can be cancelled in TV then Ta, is not normal. So suppose only r : can be cancelled between T and V. If V $ r f Ethen Ra, is not normal while if V I r f ' then either Ra, or Tu, is not normal according as E = - 1 or E = 1. This condition is verified in an analogous way in the other cases. Now let us consider the group G2.From the equalities

?Il,= V ( a , ) , B,, = V ( a , ) ,

?XC =

V ( a T ) , BC

=

V(a;),

it follows that {l,, l:} is a regular system of stable letters. Then, by C 2' (in Remark 1) t h e reduced word

WnUox;'UI * * . x : u , where k 2 0 , x, E Z2, U, E CI,is normal if and only if it contains no subword from the list: (17)

a , V ( q f ,r,)Z:, a:V(q:, rT2)l?,

a;'V(q,, rf)Z: aT-'V(q:', r:)l:',

L.A. Bokut'

40

where the V's are reduced words. To verify SB 4 in this situation it suffices, by Remark 2, t o note that, for instance, after the transformation

R l f sR'ajV(qf,ri)lf+ R'V(qi,rT)r;a, has been performed, R'V(qi,r f ) has fewer occurrences of the letter a, than R. Let us verify condition SB 6. Let Rli and TI, be normal words and is *a - non-empty *a~ reduced let R = TV(aj),where V ( a , ) K u , ~ , ' a ~ ~ word. We consider the process of reducing TV to normal form. If a;,' is not cancelled in this reduction process then none of a;,',. . .,a;: will be cancelled. By Lemma 1 it follows, writing E = E, and j = js, that R must end in a;E where E = Ba;. We see immediately that Rli is not normal. If u f : is cancelled in the reduction process, then, by Lemma 2, T ends in a,"E, where E = Ba;=l,and this contradicts the normality of TIi.Thus V-1 and SB 6 is verified. Finally we consider G,. By C 2', the reduced word

w zu0X;'u'- x ?Uk, * *

where k 3 0 , xi E Z3and Ui E C2,is normal if and only if it contains no subword from the list:

rtEppt,

r:pi',

I ;V( a ;)PI, (18)

qfpz,

q?pi',

1;' V( a, ) p i',

I+-lC(V(af)q'-'At)pi, IiC(V(aj)qiA;')p;', IV(a,)p2,

l+-'V(at)p;',

1 ;' C ( V(a,)r ;'Bi)pz,

I + C ( v ( a f ) r + B+-')p;I,

where 1 s i s A, 1 s j s n and the V's are reduced words. The validity of SB 4 for the group G, follows from Remark 2 and the fact that eliminating subwords (18) from the word Rp:, where R E G2,either decreases the number of letters I f , I + " in R or leaves fixed the number of these letters but decreases the number of letters r:, r:', qf, 97' in R. Now we verify SB 5 . The relations of G, which involve p , and p 2 can be presented in the form

E'piE = pi,

F'zF'

=PZ

x p * * . x:,". It remains where E, F are words in P and (x y,' . . x::)' to note that the mapping U + U' defines an antiisomorphism of the group with generating set 1" and defining relations (14.1)-(14.3) and the and defining relations (14.1+)-(14.3+). group with generating set

a+

Malcev's problem and groups with a normal form

Before verifying SB 6 we consider the elements more detail these are

am= V(BY1r,f8, q,),

'BPI, ?Irn, Bm.In

BPI= V(A,q;'l;',r ; ' )

?Ip,= V ( A : - ' q : l : , r:), (19)

41

'Brn= V(B:rT-'f:-',q:-')

where we may assume that in the words V there are no adjacent inverse pairs ( A:-'q;l:)-'(A ;-'q,f;)',r:-'r;', . . . ,(q;-')-'(q:-')'.We q,). Obviously this is normal if it consider, for example, am= V(B;'r,f#, contains no occurrences of I,. On the other hand, if it does contain occurrences of f, then it is, in fact, {l}-reduced. Suppose, for example, that V contains the subword

f ;'r;IB, W(q,)B and r;'B,W(qf)B;'r,= V , ( a f ) where , W and V , are reduced. The normal forms (in G I ) of r;'B,WB;'r, and V , must coincide graphically and so C(r;lB,WB;'r,)contains no occurrences of letters 4,. But this happens only when W 3 1 and then V contains an inverse pair, which we ruled out above. Now, to verify SB 6, suppose that Rp2 and Tp2 are normal and that R = TV(B;'rJ,,q,),where V contains n o adjacent inverse pairs (B;'r,l,)-'(B;'r,l,)~, q;'qf. If V K V ( q , ) ,then, since neither R and S can end in q f , it follows that V K l . Let V contain an occurrence of a letter I : . We take the leftmost of such occurrences and suppose that this I f is cancelled in reducing TV to normal form. Two cases arise: 1) E = 1. Then, by Lemma 2, we have

T3T,I;'T2,

T2€ GI,

V K Vi(qJ)B;'r,l,V2

where T2V,(qJ)B;'r, = V3(aJ) in G I . It follows that T2KC(V3(aJ)r;'.B,V4(q,)), where V3 and V 4are reduced words. The last equality shows that Tp2 is not normal. (If V4Z1,then Tp2 contains q;pz while if V4E1,then Tpt contains l;'C(V3(aJ)r;'B,)p2.) 2) E = - 1. Here we have Tz€ GI,

T K TIl,T2,

V K Vi(q,)l;'Vz

where T 2 V l ( q , ) =V3(aJ). Thus T , = V3(aJ)V4(qJ), with V,, V 4 reduced and again Tp2 is not in normal form. We conclude that the leftmost letter 1: of V(B;'r,l,,q,)is not cancelled in reducing TV to normal form. But then, by Remark 3 and Lemma 1, the word C ( T V ) H R ends in one of the following words: 18

v3(aJ) v4( qJ

)?

I;'

c(v,(

a,)rrlBt

v4(q,1)

42

L.A. Bokut'

Arguing as above we show in both cases that Rpz is not in normal form. The other cases occurring in the verification of SB 6 (pEKpG',pl,p71) are dealt with in an analogous way. This completes the proof of the following theorem. Theorem 2. A Nouikou group ?Ip,, is a group with standard basis. Corollary (See [l]). The word problem in any Nouikou group ?Ip,, is algorithmically soluable.

To establish the corollary it suffices to inspect the list (16)-(19) of words which we eliminate in reducing to normal form and be convinced that the question of recognising whether an arbitrarily given word belongs to the indicated list is algorithmically solvable. In fact the only words in this list whose explicit form is not indicated are those words, for example, of the type C ( V ( a , ) r ; ' B 0 )However, . a normal word T is of this indicated form if and only if C ( T B ; ' r , ) K V ( a , ) which fact one can always determine. W e derive from Theorem 2 the result of Novikov [20] concerning the conjugacy problem which may be unsolvable in a group '%p,h. W e recall the definition of a certain calculus introduced by Post. A Post system K is a pair K. = [Zn; GO]where 20 is a finite alphabet and @n = {(A,,B,); 1 s i s A } is a finite set of pairs of words from the alphabet Zo.By elementary transformations of K we mean the following transformations of words in the alphabet Zo: (20)

X-X,

A,X-XB,,

XB, + A , X

where (A8, B , )E Go. Two words U and V (from Zo) are called equivalent in K, written U W, if there exists a sequence of elementary transformations which transforms U into W. It is well known (see the references to Post and Markov in [20]) that there exists a Post system with unsolvable problem of equivalence. With each Novikov group '%, one can associate a Post system

-

K ( 2 ) = [ { a l , . . ,a"};(A,,B#), 1S i

S

A]

Novikov's first theorem. Let X and Y be positive words from { a , 1 n } . If BP,X2lIh = Y in 21P,,, then X Y in the corresponding Post system K(2l).

-

Proof. From the given condition we have

ic

Malcev's problem and p u p s with a normal form

(21)

43

V(A , q : 'l: ',r,)X(a,)W(B;'r,l,,q,) = Y(a,)

where V and W are suitably reduced (see (19)). If V 4 V (r , ),then W I W ( q , )- otherwise the normal form of the left-hand side of (21) will contain letters I:, which is obviously impossible. In this case V Z 1 and W K l because Y is in normal and contains no letters r;, qf. Then X K Y and so X Y in K(?I). Now let V and W contain letters I:. Then the last letter 1: of V has to be cancelled (in reducing the left-hand side of (21) to normal form) with the first letter I;' of W . There are two alternative cases. 1) E = - 1. We have

-

(22)

V K V ,A,q;I 1;I V2(r, ),

W K W,(q,) B ;'r,l,W 2

and, by Remark 3 and Lemma 2, we obtain (23)

V2(rJ)XWl(q,)B;'rt = U(a,).

As the normal form of the left side of (23) cannot contain q, or r, we have W , S l , X n X I B , and V2(rl)Xr,= X I in G I . Substituting in (22) and (23) we obtain

V , ( A , q ; ' K 'r,)A,Xl , W'(B;'r,l,,q , ) = Y where V , and W'Eq;'IX~'W2 contain fewer occurrences of letters 1: then do V and W respectively. (Here IXII is the length of X I . ) 2) E = 1 . In the same way as above, we show that X 4 A , X I ,and from (22) there follows the equality V I X I BW, , = Y where V , and W , contain fewer occurrences of letters I:. The proof is completed by an obvious induction on the number of occurrences of letters 1: in V.

Novikov's second theorem. Let X and Y be positive words from {ai,1 S i G n } . Then the words p I X p 2 X +and p l Y p 2 Y +are conjugate in the group ?I,,, if and only if X Y in the corresponding Post system K (%).

-

Proof. Write E E p lX p 2 X+and D 4 p l Y p 2 Y +and assume that E is conjugate to D in %plm. Let {&, 1 6 k S N } be the set of all words obtained from E by cyclic permutation of letters. Among all normal words U satisfying (24)

D = UEkU-',

for some k we choose a word Uo in which the number of occurrences of p ; , p ; is least. We claim that this chosen Uo contains no occurrences , U2 is pf-free, then of p ; , p ; . Suppose otherwise; if UoK U l p f U 2where we have

L.A. Bokut'

44

D = Uip :UzE,U;'p yEUY1.

(25)

Take E = - 1; as D contains only positive occurrences of p l and p2, Britton's lemma applied to (25) yields (26)

Ek KE{p,E:,

U2E

= 21p8.

Substituting in (25), this yields

D

=

UIBpiE;E{p,Bj,lUil

and C(UIBp,)contains fewer occurrences of p f than Uo.This contradicts our choice of Uo. Now we consider (24) with U K Uo.So we have (27)

I/oEku~'=piYPzY+

Since Uo has no occurrences of p ; , p ; we can, without loss of generality assume that Ek KplXp2X'. Then Britton's lemma applied to (27) yields

Bp,X21,= Y

-

in 21plh. It follows from Novikov's first theorem that X Y in K(2l). Y in K(?I). Inductively it suffices to Conversely suppose X consider the two cases:

-

XKA,X,,

Y K X , B , and X F X I B , , Y K A , X , .

Let, for example, the first of these occur. Then in 21pl, we have

UplA,XipzX;A; U-' = piXiB,pzB TX: where U K(r:)-21xfi11:-1q;-1A :. The second case is dealt with similarly and so the theorem is proved. Novikov's second theorem shows that the equivalence problem for the system K(clI) is reducible to the conjugacy problem for the group ?IPlh. The converse reduction was established in [7] and the following result proved.

Theorem 3. For any finitely presented semigroup IZ one can construct a Post system K ( l 7 ) such that 1 ) the word problem for Il is equivalent (in the sense of Turing reducibility) to the equivalence problem for K ( l 7 ) , 2) the equivalence problem for K ( l 7 ) is equivalent (in the sense of Turing reducibility) to the conjugacy problem in the Novikov group ?Iplh for which K ( l 7 ) is the corresponding Post system.

Malceu's problem and groups wirh a nonnal form

45

From this theorem it follows that there exist finitely presented groups with conjugacy problem of arbitrary recursively enumerable Turing degree of unsolvability. As noted earlier, this last result was obtained independently by Collins [15]. 94. Boone groups

In this section, we shall prove that a Boone group G(T,q), as defined in [lo], is a group with standard basis and, using this fact, we shall derive some properties of such groups. By a special semigroup, we understand a semigroup T with generators Sb, 9.

( b E B,

A)

and defining relations

~ , = r , ,i s i c ~ where 2,and r, are special words, i.e. words of t h e form Sq.S', s and s' are (positive) {Sb}-WOTdS.

where

Let T be a special semigroup with a distinguished letter q1= q. We construct, for the pair ( T , q ) ,the following group G ( T , q ) .This group is generated by t h e set of elements I

=;

=(sb,qo,k,t,x,y,I,,r,,b E B , a € A , l S i S N }

and has the following defining relations (the bold face italic letters are to be distinguished): 1.

Sby

= yysb,

XSb =

2. Sbla = yltysb,

risb

sbxx

= sbxr,x

3. 2,= l,T,r,, (28)

4. ti' = f i t ,

5 . r,k = kr,,

ry

= yt

xk

=

kx

6. (q-'tq)k = k(q-'tq),

where b E B, 1 s i c N and Z,, r, are t h e words from the defining system of equalities of the semigroup T. We add to this system of relations t h e equalities

4'. l,V(y)r = rl,V(y), (29)

6'. q-'tqV(r,, x)k

=

kq-'tqV(r,, x),

LA. Bokut'

46

where V ( y ) and V ( r i , x )are reduced words. The system of relations (28) and (29) we shall denote by @. Since each relation in (29) is easily seen to be a consequence of the relations (28), (8;@) is still a presentation of G ( T , q ) ,though now one with infinitely many relations. In our present approach the adjunction of these extra relations is essential to the definition of normal words. Let us consider the following partitions of 8 and @:

E2= {Ii, r,, 1 s i s N }

3 0

= {x, y l ,

81= { S b , b E B},

8 3

= {qa,u E A } ,

5 4

= {t},

8 5

={k},

@I

= {(28.1)},

@2

= {(28.2)},

@5

= {(28.5), (29.6')).

@o= @3

4,

43, = {(28.4),(29.4')},

= {(28.3)},

These partitions define the groups G,, 0 c a c 5, as explained in Section 2. (For reasons of convenience we change notation somewhat; instead of 2 we write 8,and so on.) We take all relations from @,, 0 < a s 5 to be given in the form A 1 Z A2pn = pmB1 U B Z

where p n , p mE E,, AIzA2and BluB2 are 8,-l-words, and z, u are distinguished letters in (28) and (29). Let us construct the sets C,, 0 s a 6 5 , and prove, for each a, that C, is a basis of G,. The group Go is free and Co is defined to be the set of reduced words in { x , y } . The group G I was essentially considered in the previous section. We note only that C1is the set of reduced words W 4 U o x f ' U l . - - x ; ' Uwhere k, k 2 0 , U,ECo, x , E ~ l = { ~ b } ~ which contain no subwords

y2sb,y-lsb, Y's;', X E S b , X-'sb1,Xsbl. (30) Now we consider G2. By virtue of C 2', the set C2 consists of all the reduced words W E Uox UI . . . x ;"U,, where k 3 0, U, E CI, x , E 9,= {Z,, r,}, containing no subwords sbv(y,X 2 ) I : , (31)

sbV(y,xZ)xE1r,,

sbl

v(y',X)y'f!;' s i l v ( y 2 ,x ) r f

where V, Vy', Vx'1 and V are reduced words, b E B and 1 s i N . Condition SB 4 is verified in a routine way - elimination of subwords (31) from Rp:, where R E C1,p,, E Z2,leads to a decrease in the number of letters s;' in the word R . Condition SB 5 follows from the fact that

Malccv's problem and groups with a normal form %Il

=

V(y-'sb)= y s y - ' ,

a,l = V(sbX)= xsx-',

47

81, = v(ysb)= y-'sy

B,, = V(sbX-')= x-'Sx

where S is a reduced {sb}-Wordand the mapping sb + sb x +x - ' , y + y - l induces an automorphism of the group GI. Condition SB 6 is verified in a routine way. For example, let 7'li and Rl, be normal with R = TB,,, i.e. R = TySy-' where S is a reduced word. Let S nsi: ,:s k 3 1. We consider two cases. 1) In reducing TySy-' to normal form, the letter s 2 K s i is cancelled. Then by Lemma 2 and Remark 3, we obtain

---

TZT1s;'Tz, Tz€ Go, TzKC(Bs,*y-')

This contradicts the fact that 7'h is normal. 2) In reducing TySy-I, the letter sb: is not cancelled. Then in reducing this word to normal form no letter of the word S is cancelled, and thus, by Lemma 1, the word R EC(TySy-') ends in s;C(Bay-'), where b = bk, E = &k. This contradicts the fact that RI, is normal. Other possibilities (p;K l;', ri, r ; ' ) are dealt with in an analogous way. Let us consider the group G,.Writing 2,and r, in the forms

2,I 2,I q ~ i z ,

r,Irt,qrn1rh

respectively, we rewrite the relations (28.3) as

(28.3') Then, if

(32)

X,'Wilqm, = q,,,S2r;1r;i'. and B,, '%,m,m

are connected words, we have

n v(2;,'1,r,J,

B,_

V(Z,dr;:)

where we take a reduced word V ( x , )in some new variables x,, 1 S i C N,and '21qmqeand a,&_are obtained from V ( x , )by substituting 2;,'l,I',l and 2,2r;1r;:, respectively, for x,. We shall call V ( x , )the diagram of the words (32). Let 21qmqm have diagram V(x,),where V ( x , )is reduced and nonempty. Then the projection of 21qmqm on the alphabet { I , } is reduced and non-empty. This projection is a homomorphism from Gzto the free 1 in Gz.By a similar argument the group on {I,} and hence 1 in G2and SB 5 holds for G,. connected word B,# We note that the words aqmqn, Bqmqm are uniquely defined by the and sequence of letters lf,r: which appear in them. Specifically if B, have diagram V ( x , )I x * x 2, then we have

r,' - -

(33) where

'21qmqm~ ? l ( I f ; ., . .,l;;),

BqAm FB(r;ei,. . .,r;..),

L.A. Bokuf'

48

and

For Gs, the verification of condition SB 4 is straightforward. We note only that C, consists of all reduced words w K uoq 2 U1* q kuk, where k 3 0 , U, E C,, which contain no subwords

ItC(y-'syrJqm,, (34)

r ;1 ~ ( x s -1r;;)q x

;:,

I;'C(ysy -'Z 1)qn,

r , c ( x +sxz ;;)q

associated with the relations of (28.3'). The verification of SB 6 is also straightforward. For example, let Rq, and Tq. be in normal form, where R, T E Gzand R = T'uqmqm = T'u(I:;,. . . ,I:;). Suppose 'uqmqn f 1; if 1:; is cancelled in reducing T21qmqn to normal form, then, by Lemma 2, the word T ends in I;'C(ySy-'s,ll) or in I,kC(y-'SyI',ll), according as E' = 1 or E' = - 1. In either case Tq. is not normal. If 1:; is not cancelled, then R ends in I;'C(ySy-'2,J or I,kC(y-'SyI'ikl)and Rq, is not normal. For the group G4we have 'ur = V(Ly),

'8, = v(Ly)

where V(l,,y ) is a reduced and hence normal word. The set C4consists of all reduced words W K Uot'lUl * * * tekUk,where k a 0 and U, E C,, which contain no subwords

(35)

y v ,

lC(y-'SyV(y))t', I;'C(ySy-'V(y))t'

where S = k 1 and S, as usual, is a reduced {sb}-word. Condition SB 5 is clear because 'u, = 23, while condition SB 4 follows from the fact that elimination of subwords (35) from Rt', R E C,, either decreases the number of letters I : in R or decreases the number of letters y' in R. To verify SB 6, consider, for example, Rt and St where R = S a , = SV(I,,y ) , and V(l,,y ) is reduced. Suppose V(I,,y ) f 1; if V(I,,y ) K V ( y ) then either S or R ends in y 6 which is impossible. If V ( l , , y )involves some If, then considering, as usual, the first and last of the occurrences of these letters, we deduce that Rt or St contains a subword (35), which is impossible when Rt and St are normal. Finally we consider Gs. Bearing in mind that we added relations (29.6'), strictly we have

%k = w ( q - ' t q v ( r #x), , r,, x ) = 8

k

;


49

of course we can write W(q-'rqV(ri,x ) , ri, x ) = W'(q-'rq,r,, x). The set C5consists of all reduced words W K UokelUl. . . k'SU, where s 2 0 and Ui E C4,which contain no subwords

rfk', X'k',

r"C(W(li,y)qV(ri,x))ke

where 6 = & 1. The verification of SB 4, SB 5 and SB 6 for G5is similar to that for G4 and we omit these straightforward arguments. We do, though, note precisely why the additional relations (29) are needed. Let 2 be a special word of T and suppose that 2 = q in T, Then Z'rZk = k 2 - ' r 2 in G (see [lo] or Lemma 6 below). If we had used only the relations (28) in our definition of normal form, then we would be forced to regard 2-'r2k as normal since we are not able, in general, to show that Z = V(li,y)q in G. Since kZ-'r2 would also be normal, SB 6 would fail. The relations (29.6') are added since, when 2 = q in T, one can prove that 2 = V(l,,y)qW(ri,x ) in G.,. Our conclusions above amount to the following result.

Theorem 4. Any Boone group G ( T , q ) is a group wirh standard basis. We shall now apply Theorem 4 to show that the word problem for G ( T , q ) is equivalent to the problem of equality with q, for special words, in the semigroup T. For this we determine a number of properties of G ( T , q ) (see [12], [17]).

Lemma 5. The word problem for rhe group G4 is solvable.

Proof. We consider, in turn, the groups Go, 0 s a s 4 and verify that, for each a, the basis C, is effective in the sense that the process of reducing words to normal form is effective. Since Go is free, its basis Co is clearly effective. The effectiveness of CI and C2 follows from the fact that the words (30) and the words (31) are explicitly defined. By contrast the words (34) are not given in an explicit way. Suppose though, that we are given a normal word T. To decide, for example, whether or not there exists a word S such that TKC(y-'SyTiI),it suffices to decide whether or not C(yTT;;'y-')is an {sb}-wordand this can be done effectively. So C3 is effective. In a similar way, we can decide of a normal word T of C , whether or not TKC(y-'Sy"),for some reduced {sb}-wordS. For this occurs if and only if S is the projection of T on the alphabet { s b } and C(S-'yT)Ey".This shows that C, is effective and the lemma is proved.

50

L.A. Bokut'

Lemma 6. Let 2 and 2'be two special words of T. Then 2 = 2' in T if and only if there exist words V(lz,y ) and W(ri,x ) such that (36)

2

=

V(li,y)2'W(r0,x )

in the group G,. Proof. Let 2 = 2'in T. Inductively, it suffices to consider t h e case when 2 is transformed into 2' with the aid of one transformation of the type 2,+ or r,+ 2,. Suppose, for example, that 2 KS12,Szand 2'KSlf,Sz.Then, applying relations (28.1)-(28.3) we see that 2 = V(l,,y)X'W(r,,x)in G3,for some words V(l,,y ) , W(r,,x). We now prove the converse. Let the equality (36) hold. We suppose that

r,

2ns1qns2, 2 ' n S ; q m S ; , v ( l ' , y ) ~ y " . l ~ ~ . . . l ~ y " h , and that V is reduced. By our assumption, the normal form of

(37)

y " . 1 : ; . * *1:;s;q,

is of the form SlqnU,where U E GI. However, by Remark 2, the normal form of (37) is obtained by eliminating words (34) which correspond to relations (28.3) of G ( T , q ) .Since each such elimination represents the application of a relation (28.3'), accompanied by use of the relations of G I except insertions of letters I,, we obtain

. 1 2 Y " k S I = SI?l(Q,.. . . ,If;) in GI, where 2l(l:;, . . .,l:;)Z!!lqmqmand the diagram of ?lqmqm is reduced. (38)

y".Q

*.

From (36) and (38), it follows that

(39)

B(r;EI,.. . , r ~ ) s ; w ( r ' = , xsz. )

From (39) and Britton's lemma (or the algorithm for reducing to normal form) we have

W(r,,x)Kx"V:;..* r:;xm~. We prove that 2 = 2' in T by induction o n k. If k = 0, then 2 K2'. Let k > O and suppose, for example, that &k = 1. Then from (38) we have

I,y"'SIqm = I,y-'Syftklqm where qmZqmfk and S is a positive {s,,}-word. From a first reduction of the right-hand side of (36), we obtain

(W

2 = y v : ; . - - i ~ , l y " k - l s 2 , , , q n , k ~ r ~ r , 1 r , : s : x m-k.r ~i . p

.

Malcev's problem and groups with a normal form

From (39), with

El,

51

= 1, it follows that = xs~x-i

r,:s:xmk

where S' is a positive word, and thus S:KT,,,S'. In this way we obtain, from (40),

2 = y % l : l . . l'k-ly"k-ISZ 4 ~ B k 2 X m ~ - l Irk -eI k - l* * . r , : # X m O (41) k-l I d nnk and Z'ZST,,,q ,,kI',kzSr.The inductive assumption applied to (41) yields 2 = SZ,klqn,kZ,kzS' in T and so Z = 2' in T as required. Lemma 7. The problem of equality of an arbitrary word W of G , with a word of the form V ( Ly)ZW(ri,x),

where

Z is special,

is solvable.

Proof. If suffices to consider a normal word 0 4 Q1qmQzwhere Q,, Q2E Cz. Suppose that (42)

Q = V(l,,y)ZW(r,,x )

where Z KZ'q,Z'' is special and V, W are reduced. Among all equalities (42), for our given Q, we select one for which the (1,)-length of V is least. Then we have (43)

Q IC ( V Z W ) IQiqmC(gqmqmZ"W(rt,x))

where Q 1K Ql(l,, y , s b ) contains no letters sbl, 8q,qm 9 U(Z,zr;lK') and the diagram of this word is reduced. It follows from (42) that

Q1Vlq&"

= VZ'

where '?Iqnqm I LJ(Z;:U',l). We claim that BqmqnZ"W(r,, x ) is (1,)-reduced. Let us suppose that this is not so and that, for example, Bqmqn ends with Z,zr;lr;;. Then W ( r , , x ) begins with x'r,, say W K x ' r , W l ( r , , x ) ,8t'Kr,zSand Sx' = xSX-'. It follows from (43) that, when reducing VZ'q. to normal form, the first transformation will be

i,y-lslyr,lqn+ y ~ ~ y - ~ ~ , ~ q ~ , ~ , ~ r ; ~ r ; : , that is, V KVl(l,,y ) 4 y k , Z'FSlr,l and y k S 1= y-'S,y. Then (42) gives

Q = V I l ykSJ',~qnI"2Sx% Wl = vlyk'sIZ,Iqn,Z,zsx'wl

52

L.A. Bokut'

and V l y k 'has smaller (1,)-length than V . This contradicts our choice and so our claim is proved. W(ri,x ) is {r,}-reduced, the characterisation of Bqmq., in Since BqmqmZ" the form of the sequence of letters rT occurring therein, remains in the word C(B,P"W(r,, x ) ) K Q2 which we know. Thus Q2KRRorT; * rzRk, k P 0 and for some 1, 0 d 1 d k, B(r:;, .. .,r?) has the form Bqmq,, for some qm. Furthermore the words

--

1,

C(Ql~lqmqm

c(aAQ2)

contain no letters s i ' . These last conditions can be easily checked having found the normal forms of the respective words. Thus the lemma is proved.

Corollary 1 ([ll], [17]). The Turing degree of unsolvability of the word problem for the group G (T,q ) coincides with the degree of unsolvability of the problem of equality with q for special words in the semigroup T. Proof. The word problem for Gs is reduced to the process for transforming words of Gs into normal form. Since the word problem for G4 is solvable the process is reduced to the problem of whether or not a word 0 of the group G3 is equal to a word of the form V ( Qy)qW(ri,x). By Lemmas 6 and 7, this is reducible to the problem of equality with q for special words in T. To prove the converse reduction, it suffices to note that 2 = q in T if and only if Z - ' t Z k = k Z - ' t 2 in Gs. One part of this assertion follows from Lemma 6 and the other part from applications of Britton's lemma. Since there exist special semigroups in which the problem of equality with q for special words has arbitrary Turing degree of unsolvability we have the following result.

Corollary 2 ([ll], [16], [17)). For an arbitrary recursively enumerable Turing degree of unsolvability, there exists a finitely presented group (namely a Boone group G ( T , q )for some semigroup T ) with word problem of the given degree of unsolvability. In conclusion we should like to express our thanks to Professor W.W. Boone at whose invitation this article was written.

Malceo’s problem and groups with a normal form

53

References [I] S.I. Adjan, On the work of Novikov and his pupils concerning algorithmic questions in algebra, Trudy Matem. Inst. imeni V.A. Steklova AN SSSR 133 (1973), 23-32. [2] L.A. Bokut’, Factorisation theorems for some classes of rings without zero divisors, 1. Algebra and Logic 4, No. 4 (1965), 25-52; 11, Ibid. 5 , No. 1 (1965), 5-30. [3] L.A. Bokut’, On a property of Boone’s groups, Algebra and Logic 5 , No. 5 (1966), 5-23; 11, Ibid. 6, No. 1 (1967), 15-24. [4] L.A. Bokut’, On Novikov’s groups, Algebra and Logic, 6, No. 1 (1967), 25-38. [5] L.A. Bokut’, On the embedding of rings in skew fields, Doklady AN SSSR 175, No. 4 (1967), 755-758. [6] L.A. Bokut’, Groups with a relative standard basis, Sib. Mat. Z. 9, No. 3 (1968) 499-521. [7] L.A. Bokut’, The degrees of unsolvability for the conjugacy problem for finitely presented groups, Algebra and Logic 7, No. 5 (1968) 4-70; 7, No. 6 (1968) 4-52. [8] L.A. Bokut’, Groups of fractions of multiplicative semigroups of certain rings, I, Sib. Mat. Z. 10, No. 2 (1969) 246-286; 11, Ibid. 10, No. 4 (1969) 744-799; 111, Ibid. 10, No. 4 (1969) 800-819. [9] L.A. Bokut’, On Malcev’s problem, Sib. Mat. Z. 10, No. 5 (1969) 965-1005. [lo] W.W. Boone, The word problem, Ann. of Math. 70 (1959) 207-265. [I I ] W.W. Boone, Word problems and recursively enumerable degrees of unsolvability. A first paper on Thue systems, Ann. of Math. 83 (1966) 520-571. [ 121 W.W. Boone, Word problems and recursively enumerable degrees of unsolvability. A sequel on finitely presented groups, Ann. of Math. 84 (1966) 49-84. [13] A.J. Bowtell, On a question of Malcev, J. Algebra 7 (1967) 126-139. [ 141 P.M. Cohn, Free rings and their relations (Academic Press, London, New York, 1971). [ 151 D.J. Collins, Recursively enumerable degrees and the conjugacy problem, Acta Math. 122 (1969) 115-160. [16] C.R.J. Clapham, Finitely presented groups with word problems of arbitrary degree of unsolvability, Proc. Lond. Math. Soc. Series 3, 14 (1964) 633-676. [I71 A.A. Fridman, Degrees of unsolvability of the word problem for finitely presented groups (Nauka, Moscow, 1967). [ 181 A.A. Klein, Rings nonembeddable in fields with multiplicative semigroups embeddable in groups, J. Algebra 7 (1967) 100-125. (191 A.1. Malcev, Selected works, Vol. I (Nauka, Moscow, 1976). [20] P.S. Novikov, Unsolvability of the conjugacy problem in the theory of groups, Izv. Akad. Nauk SSSR, Ser. Mat. 18 (1954) 485-524. [21] P.S. Novikov, On the algorithmic unsolvability of the word problem in group theory, Trudy Mat. Inst. Steklov No. 44 (1955). [22] G. Higman, B.H. Neumann and H. Neumann, Embedding theorems for groups, J. Lond. Math. Soc. 24 (1949) 247-254.

S.I. Adian, W.W. Boone, G. Higman, eds., Word Problems 11 @ North-Holland Publishing Company (1980) 55-69

DECISION PROBLEMS FOR RING THEORY L.A. BOKUT’ Institute of Mathematics, Novosibirsk

In my talk I shall discuss the following questions concerning the theme of the title: 1. The word problem for Lie algebras. 2. Embeddings in simple associative algebras. 3. Associative algebras with one relation. 4. Finite approximation of algebras. 5. Free Lie algebras. 6. Alternative, Jordan, and Malcev rings. I have not included the results on free associative algebras and associative division rings in my report, but have presented here some well-known theorems by P. Cohn, J. Bergman, A. Macintyre and others worth analysing in some detail. I do not touch upon constructive rings, because the theory of these rings is part of the general theory of constructive models (A.I. Malcev, Yu. L. Ershov and others). 91. The word problem for Lie algebras

The first results on the word problem for finitely presented (f.p.) algebras ( a , ..., a n ;f l = 0,. . . . ,fm = 0)s

in the given variety of linear algebras X belongs to A.I. h k o v (1950). He proved that if X is the class of all (non-associative) algebras over a field k, then the word problem in X is algorithmically solvable. A.I. SirSov published two papers in 1962. In the first he pointed out one simple algorithm “exclusions of the leading words of the defining relations” which solves the word problem in the class of all the algebras and in the class of commutative ( x y = y x ) or anti-commutative ( x y = - y x ) algebras. In the second paper SirSov obtained the first result on the word problem for Lie algebras - he proved the solvability of this problem for Lie algebras with one relation

(at, . ., an ; f = 55

56

L.A. Bokut'

In the same paper SirSov proposed a composition method, which I will say more about below. The status of the word problem for arbitrary f.p. Lie algebras was settled in 1972. I was able to prove

Theorem 1. Let k be an arbitrary field. There exists an f.p. Lie algebra over field k with unsolvable word problem. I shall speak briefly about the idea of the proof. Let M be a recursively enumerable non-recursive set of natural numbers. Then the Lie algebra

( a , b, c, a l ,c l ;ab"c.= a l b ; c l ,n E M ) , where ab"c = ( . * . ( a b )* * b ) c has unsolvable word problem. In this way, it suffices to prove that this (or some similar) Lie algebra is embeddable in a finitely presented Lie algebra. This is proved by means of Matijasevich's theorem on the existence of a Diophantine representation of the set M and uses the following simple analogy of one of Higman's lemmas.

Lemma. Let A be an f.g. Lie algebra and let B be a Higman subalgebra A. Then the algebra ( A ,A'"; b = b"), b E B),,,

is embeddable in an f.p. Lie algebra. Definition. B A the Burnside group B ( m , n ) on m generators and with exponent n is infinite. (This was proved originally by Adian and Novikov [ l ] and subsequently by Adian [ 2 ] . )Conceivably, such a proof might contain ideas from [I], [ 2 ] and [3]. It is possible that such a proof would also throw light on the outstanding unsolved problem in the field: is B ( m , 2 ' ) infinite for any k 2 I? However I am inclined to pessimism since I believe that the answer is negative. Reference [I] P.S. Novikov and S.1. Adjan, On infinite periodic groups, Izv. Akad. Nauk SSSR, Ser. Mat. 32 (1968) 212-214, 224-251, 709-731. [2] S.I. Adian, The Burnside problem and identities in groups, Nauka (1975). [3] J.L. Britton, The existence of infinite Burnside groups, in Word Problems: Decision problems and the Burnside problem in group theory, Studies in Logic and Foundations of Mathematics (North-Holland Publishing Co., Amsterdam, 1973).

71

S.I. Adian, W.W. Boone, G. Higman, eds., Word Problems I1 North-uolland Publishing Company (1980) 73-80

ON SEMIFIR CONSTRUCTIONS P.M. COHN Bedford College, London

$1. Introduction

Free ideal rings or firs (the definition is recalled in §2) form an analogue of principal ideal domains, to which they reduce in the commutative case, and to obtain examples of firs one can use the weak algorithm and its variants. Semifirs form a wider class and it is less easy to construct semifirs (that are not also firs) directly. A possibility is to form direct limits or ultraproducts of firs, but a more interesting way is to start from n-firs, for the examples so obtained show that semifirs do not form an elementary class. This is in sharp contrast to the commutative case, where semifirs correspond to Bezout domains (an elementary class). After describing the necessary background in $52-3 (including a 'small cancellation' theorem needed to construct n-firs) we examine the ultraproduct construction more closely. It turns out that a countable ultraproduct of n-firs for increasing n (with a non-principal ultrafilter) is always a semifir but never a fir, except in trivial cases. On the other hand, an uncountable ultraproduct may well be a genuine fir. $2. Firs, semifirs, n-firs Throughout, all rings are associative, with a unit-element 1 which is inherited by subrings and preserved by homomorphisms. Mostly our rings have a commutative field k as coefficient domain, i.e. they are k -algebras. A free k-algebra on a set X, k ( X ) can be defined by a universal property, although often a more pragmatic definition (in terms of a normal form for its elements) is preferred. If we examine the sense in which relations in k ( X ) can be described as trivial, we are led to the following Definition. Let R be a ring; a relation (1)

x.y =xlyl+***+x"y,=o 13

(Xi,YiER)

74

P.M. Cohn

is said to be trivial if for each i = 1,. . ., n either x , = 0 or y , = 0 . If there is an invertible n x n matrix P over R such that the transformed relation xP-’ . Py = 0 is trivial, t h e relation (1) is said to be triuialized by P. It turns out that every relation in k(X)is trivializable in this way; this is proved by first showing that k(X)possesses a ‘weak algorithm’, an analogue of the Euclidean algorithm to which it reduces for commutative rings (cf. [S], Chapter 2). Next it is shown that in a ring with a weak algorithm all relations are trivializable. We shall not repeat the proofs, but to elucidate the connexion we quote the following result from [4] (cf. [S], Chapter 1):

Theorem 1. Let R be a non-zero ring and n 2 1, then the following conditions are equivalent : (a) every relation (1) with at most n terms is trivializable, (b) any right ideal of R generated by m d n elements which are right linearly dependent ouer R can be generated by fewer than m elements, (c) any right ideal on at most n generators is free, of unique rank. A non-zero ring satisfying these equivalent conditions is called an n-fir. By (a) this condition is left-right symmetric while (b) shows that the class of n-firs is elementary, i.e. it can be defined by a single elementary sentence. Thus for any n 2 1, both the class of n-firs and its complement are closed under ultraproducts (cf. [ 5 ] , Chapter 5 ) . It is easily checked that I-firs are just (non-commutative) integral domains; for increasing n the n-firs form smaller and smaller classes until we reach semifirs, defined as n-firs for all n. From Theorem I we see that a ring R is a semifir if and only if every finitely generated right ideal is free, of unique rank. Thus semifirs can be defined by an infinite set of elementary sentences, so t h e class of semifirs is closed under ultraproducts. As we shall see later, its complement is not closed under ultraproducts, so the class of semifirs is not elementary. By contrast, in t h e commutative case a 2-fir is already a semifir (a commutative 2-fir is a Bezout domain, i.e. a domain in which every finitely generated ideal is principal), so the class of commutative semifirs is elementary. Now a right fir ( = free ideal ring) is defined as a ring in which every right ideal is free, of unique rank. Unlike semifirs right firs are n o longer left-right symmetric (cf. [S], Chapter 2), so we must also define left firs, and by a fir we shall understand a left and right fir. In the

O n semifir consmctwm

75

commutative case firs reduce to principal ideal domains, while semifirs reduce to Bezout domains, as noted earlier. This remark already shows that t h e class of firs is not closed under ultraproducts, and so cannot be defined by a set of elementary sentences. Examples of firs exist in profusion; they include (a) principal ideal domains (even non-commutative), (b) free algebras over a field, (c) free products ( = coproducts) of skew fields. Semifirs are a little harder to come by; as in the commutative case we can obtain them as direct limits of firs, or also as ultraproducts of firs. But now we can also form ultraproducts (or direct limits) of n-firs, for increasing n. For this method to be of use we shall need a source of n-firs; this is provided by a result to which we now turn. 83. A ‘small cancellation’ theorem for rings Our aim in this section is to describe a result which enables one to recognize from certain ring presentations that t h e ring defined is an n-fir. By definition this just means that every relation (1) of at most n terms is trivializable, and one would expect a condition for this to be expressed by saying that all the defining relations can be expressed as relations of N > n terms with little interference, so that no nontrivializable relations of S n terms can be deduced by cancellation. Such a theorem was proved in [7] (see also [6] for a special case). To state it we need some definitions. Let k be any commutative field and let R be a k-algebra, generated by a set U ; then R can be expressed as a homomorphic image of the free k -algebra F = k( U) : R = F / N . On F we define a filtration u by assigning arbitrary positive integer values u ( u ) to each u E U, putting u ( u I * u,) = u(ul) + . . . + u ( u , ) and for any f = C a,m E F writing

where m runs over all products of u ’ s and a, E k. If the homomorphism F + R is denoted by a H a * , then the filtration u can be defined on R by writing for r E R, We now turn to t h e defining relations. If one of them is linear, C a,ui + /3 = 0 (ai,/3 E k), we can use it to eliminate one of t h e generators, because k is a field. So we may assume that there are no linear relations. Now relations of higher degree may be expressed as

P.M. Cohn

76

relations of degree two, by introducing further generators. Thus the defining relations for R may all be taken to be of the form

C N

(2)

I

~ i yi

b = 0,

where xi, yi E U and b is an expression in the generators with no term having xi as left factor. We shall assume that all defining relations are of the form (2) with the same value of N ; this can always be achieved by incorporating supernumerary terms in b. Now each element f of R can be expressed as a linear combination of products of the generators. Such an expression is said to be in reduced form for the suffix 1 if no term contains a factor x l y l for any of the defining relations (2). Any f E R can be expressed in reduced form by writing down any expression for f and then applying the move

arising from (2), for all the defining relations, wherever possible. Under suitable conditions a reduced form for the suffix 1 for f (not necessarily unique) is reached after a finite number of moves. If for each f E R there is just one reduced form we shall call it the normal form of f and denote it by [ f l l ; we also say then that a normal form for the suffix 1 exists. In a similar way we can define a reduced and a normal form for the suffix v ( v = 2 , . . . , N ) by using in place of (3) the move

Such a normal form will be denoted by [f].. Now we can state the result:

Theorem 2. Let R be a k-algebra generated by a set U containing a family of distinct elements (xi,) ( i E I, v = 1 , . . . ,N ) , another family of distinct elements (y,) (jE J, v = 1,. . . ,N ) and possibly other elements z h ( h E H ) . These generators are assigned positive integer values in any way such that v(xi.) = 1 and v(y,) is independent of v. Assume that R has a complete set of defining relations indexed by some subset of I x J :

where bij is an expression of value v(bii)d 1+ v ( y p j )in U and it has no term of this value with any xi,, as left factor. Moreover, assume that a normal form exists for each v = 1,. . . , N, satisfying

O n semi@ consmcctions

n

N , If f g is in normal form for v and v ( g ) > 0, then #or any h E R, the terms of highest value in f [ g h I yare in normal form for v. Then R is an n-fir, for all n C N The proof consists in verifying that R satisfies a truncated form of the weak algorithm ((N - 1)-term weak algorithm, cf. [7]). In spite of its complexity the condition is a very natural one, easy to check in most concrete cases. 04. Direct limits and ultraproducts

It is clear that any direct limit (over a directed system) of firs is again a semifir; more generally, a routine argument shows the truth of

be a direct system of rings. Suppose that for Theorem 3. Let {R,, each n 2 1 there exists A, E I such that R, is an n-fir for all a > An, then + lim R, is a semifir. This result enables one to construct semifirs from n-firs, but there is no guarantee that the limit will be non-trivial; it may well reduce to 0 or to the ground field k unless precautions are taken. For ultraproducts the situation is rather better. We recall that in a fir each element not zero or a unit can be written as a product of a finite number of atoms ( = unfactorable elements) and any two complete factorizations of c have the same number of factors; this is called the length of c, written l(c). Moreover, the factors can be paired off into pairwise similar ones; the precise condition need not concern us here, it is expressed by saying that the ring is a unique factorization domain, or UFD for short. In the commutative case this reduces to the usual definition (cf. [8], Chapter 3). We also recall that a semifir is a UFD if and only if each non-unit # 0 can be written as a product of a finite number of atoms. Theorem 4. Let {R,}I be any family of rings and R = ll R,/D their ultraproduct with respect to a non-principal ultrafilter D. If for each n 2 1, the set {a E I R, is an n-fir} is in D, then R is a semifir. If moreover, I is countable, then R is either a skew field or not a fir.

I

Proof. That R is a semifir is clear from the definition of ultraproduct, because n-firs form an elementary class. Now let I = N say; if the set { n E N R, is a field} is in D, then R is a field. Otherwise there exists

1

P.M. Cohn

78

N o € D such that R. is not field for all n E No. Let a. E R. be neither 0 nor a unit for n E No and a. = 1 for n 6Z No, and consider a = (a:) E nR.. The image ti of a in R is a non-unit, and it can be written as a product of n non-units, for each n. Hence R is not a UFD and a fortiori not a fir. For uncountable index sets the situation can be very different, especially if we have an wl-complete ultrafilter. Here w I denotes the first uncountable cardinal, and for any infinite cardinal d an ultrafilter D on a set I is called d-complete if the intersection of any set of fewer than d members of D belongs to D. This is the case if and only if, for any partition of I into fewer than d parts, one of the parts belongs to D (cf. [3] p. 180). Theorem 5 . Let { R , } , be an uncountable family of f i r s and let D be an wl-complete non-principal ultrafilter on I, then the ultraproduct R = Il R,lD is a unique factorization domain. Proof. As noted earlier, in a fir each non-zero element a has a finite length l ( a ) . Given a = ( a , ) E n R,, put P. = {a E I1 l ( a , ) = n } , P, = {a E I a, = 0 } , then the P's form a partition of I into w < w1 sets, hence one of them, say P,, lies in D, and a is a product of nl atoms (or a = O ) . Thus R is a semifir in which each non-zero element has finite length, i.e. R is a UFD. In the commutative case this result shows that an ultraproduct of an uncountable family of principal ideal domains with an wl-complete non-principal ultrafilter is again principal. The same method of proof will show that under the conditions of the theorem R has ACC. (i.e. the ascending chain condition on ngenerator right ideals) for any n, but it is not known whether R is necessarily a fir. If however, each R, is a ring with weak algorithm, then R will again have weak algorithm and hence be a fir.

I

95. Examples

With the help of Theorem 2 it is easy to give examples of (n - 1)-firs that are not n-firs. We simply take 2n generators x , , y , with the defining relation xly,+ * *

*

+ x.y.

= 0.

This is not trivializable (otherwise every n-term relation would be

O n semifir constructions

79

trivializable and every ring would be an n-fir!). But the conditions of Theorem 2 are satisfied, so we have an (n - 1)-fir Llsay. If we take the ultraproduct of LI,Lz, .. . with a non-principal ultrafilter, we obtain a ring which is an n-fir for all n, i.e. a semifir. Thus semifirs do not form an elementary class. A slight modification shows that the class of rings embeddable in skew fields is not elementary. This problem, raised by Mal’c,ev [lo] was answered in [9] using the compactness theorem. Here is a more direct proof: Let R,-, be generated by 2(nZ+ n ) generators aiA,bAi ( i = 1 , . . . , n, A = 1,. . . , n,+ 1) with the defining relations (4)

In matrix terms we have an n x ( n + 1) matrix A and an (n + 1 ) x n matrix B such that AB = I, BA = I. Thus R.-l does not have invariant basis number (cf. [6]) and so is not embeddable in a skew field. But the relations (4) satisfy the conditions of Theorem 2 (cf. [6, 7]), hence R.-l is an ( n - 1)-fir. Now the ultraproduct of all the R. (with a nonprincipal ultrafilter) is a semifir, and by the results of [8], Chapter 7, is embeddable in a skew field. Therefore the class of rings embeddable in skew fields is not elementary, although, by general results of universal algebra (cf. [5, lo]) it can be defined by an infinite set of sentences. This can also be seen from the explicit form of the embeddability conditions given in [8]. In the last example we saw that when a relation such as ab = 1 is interpreted as a relation between n x n matrices, an (n - 1)-fir results. This idea has been made precise and proved generally by G.M. Bergman [2]. His result (referring to square matrices only) may be stated as follows: Let R be any k-algebra; if we take any presentation of R and interpret all relations as relations between n x n matrices, we obtain an n x n matrix ring over an (n - 1)-fir. Bergman gives two proofs, one based on results in [l], the other based on [7]. In this way one obtains a family of n-firs (for all n ) from any given k-algebra; in general there is no way of making this family into a direct system, but by taking ultraproducts one can again obtain semifirs. References [l] G.M. Bergman, Modules over coproducts of rings, Trans. Amer. Math. SOC. 200 (1974) 1-32.

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P.M. Cohn

[2] G.M. Bergman, Coproducts and some universal ring constructions, Trans. Amer. Math. Soc. 200 (1974) 33-88. [3] C.C. Chang and H.J. Keisler, Model theory, Vol. 73, Studies in logic (North-Holland Publishing Co., Amsterdam, 1973). [4] P.M. Cohn, Free ideal rings, J. Algebra 1 (1964) 47-69. (51 P.M. Cohn, Universal algebra (Harper and Row, New York, London, Tokyo, 1965). [6] P.M. Cohn, Some remarks on the invariant basis property, Topology 5 (1966) 2 15-228. [7] P.M. Cohn, Dependence in rings 11. The dependence number, Trans. Amer. Math. SOC. 135 (1969) 267-279. [8] P.M. Cohn, Free rings and their relations, LMS monographs No. 2 (Academic Press, London and New York, 1971). [9] P.M. Cohn, The class of rings embeddable in skew fields, Bull. London Math. Soc. 6 (1974) 147-148. [lo] A.I. Mal’cev, Algebraic systems (Springer, Berlin, 1973).


CONJUGACY AND THE HIGMAN EMBEDDING THEOREM Donald J. COLLINS Queen Mary College, London

The Higman embedding theorem asserts that if a finitely generated group has a recursively enumerable set of defining relations, then it can be embedded in a finitely presented group. Subsequently, Clapham showed that t h e Higman construction preserves the solvability of the word problem in the sense that if the original finitely generated group has solvable word problem, then so does t h e finitely presented group in which it is embedded. Our purpose here is to show that the analogue of Clapham's theorem for the conjugacy problem is false. Let C = (cl,c 2 , .. . ,c , ; r = 1, r E R ) be a finitely generated group with a recursively enumerable set R of defining relations. It is convenient to assume, as we may, that R = R ( c , ) is the normal subgroup of the free group F ( c , )= F(cl,c2,.. . , c , ) consisting of every relation of C. As proved by Higman [4], C can be embedded in a finitely presented group G in the following manner. Let F ( a , ) and F ( b , ) be free groups isomorphic to F ( c , ) . From these form the amalgamated free product

H = ( F ( a , ) * F ( b , ) ; R ( a , )R=( b , ) ) . Then H can be embedded in a finitely presented group L. Next form the direct product L x C and finally the HNN extension

G

=

(L x C,r ; t-'a,r = a,, r-'b,t = b,, i

=

1,2,.

. . ,n ) .

Then G is, in fact, finitely presented. We fix t h e above notation and are ambiguous, leaving the context to clarify, in our use of notation for an element of a free group and its image in a factor group.

Proposition 1. Let x ( a i )E F ( a i ) and suppose (a) x ( a i ) is cyclically reduced in F ( a , ) , (b) x ( a i ) is nor a proper power in F ( a i ) , (c) x ( a i ) @ R ( a i ) . 81

82

D.J. Collins

Then for any u(ci), the subgroup of C generated by x ( c i ) contains u ( c i ) if and only if x(a,)u(ci)t-'and x(a,)t-' are conjugate in G. Proof. Suppose x(a,)u(c,)t-' and x(a,)t-' are conjugate in G. Then by Collins' Lemma (see [ 5 ] ) , they must be conjugate by an element w(a,, b,). The conjugacy is then valid in the group

( H x C, t ; t-'a,t

=

a,c,, t - ' b , t

=

b,, i

=

1,2,.

. . ., n ) .

From the equation w(a,,b,)-'x(a,)u(c,)t-'w(a,, b,) = x(a,)t-' we obtain, via the relations t-'a,t = ax,,

(1)

t-'b,t =

b,,

w(a,,b,)-'x(a,)w(a,,b,) = x ( a , )

in H,

(2)

U(C,)Z(Ci)=

1

in C, where w (a,c,,b,) = w(a,,b,)z(c,)in H X C. As x ( a i ) R ( a i ) it follows from (1) and the structure of H as amalgamated free product that w(a,,b i )E F(a,). Then from (a) and (b) in the hypotheses, we see that, for some integer m, w(ai,6 , )= x(a,)"' in F(a,). This means that z ( c # = ) X ( C , ) " ' in C as required. Conversely suppose that u(c,)x(ci)"'= 1 in C. A simple calculation verifies that x (a,)-? (a,) u (c, ) t - ' x (a,)" = x (a,) t - ' .

Corollary 2. Let C have unsolvable power problem and solvable order problem. Let C be embedded in G via the Higman embedding. Then G has unsolvable conjugacy problem. Proof. Let X be t h e set of all cyclically reduced words of F ( c , ) that are not proper powers in F ( c , ) and do not lie in R ( c i ) . Since C has solvable order problem this is a recursive set. On the other hand, since C has unsolvable power problem there is no algorithm to decide of an arbitrary element z of F(c,) and an arbitrary element x of X whether or not z lies in the cyclic subgroup of C generated by x. By Proposition 1 G must have unsolvable conjugacy problem. We give another demonstration of the way in which the Higman embedding fails to preserve the solvability of the conjugacy problem.

Conjugacy and the Higman embedding theorem

83

Proposition 3. Let xk, y k , k = 1,2,3,4 be elements of F(c,), none of which lies in R ( c , ) . Let w = w(a,,b,) be an element of the free product F ( u ~*)F(bi). If w - 1 ~ l t ~ z t - I ~ 3wt ~=4yltyzt-ly3ry4r-1 t-l

in G then, in C, x I = y l , z-'xzz = y 2 , x3 = y 3 and

z-'x4z = y 4 ,

where w(aici,b i )= w ( a , b i ) z ( c i )in H x C. Proof. The postulated equality must hold in the HNN extension (H x C, t ; t-'ait = sic,, t-'bit = bi, i

=

1 , 2 , . ..,n ) .

Using t as stable letter, we see that x4w(aici,bi)y;' lies in the subgroup (sic,, bi). But then, clearly, x4w(aic,bi)y;' = w(aic,bi). This gives Z - I X q Z = y 4 and

w (ai,b i ) - 1 ~ l t ~ Z (ai, r - Ib ~i )= 3 ~yltyzt- l y 3 . From this we obtain x 3 w ( a ,bi)y;' = w ( a , b i ) and hence x3 = y 3 and w-lxltxzt-lw = y l f y 2 f - l . A similar argument gives z-'x2z = y z and X I = y1.

Corollary 4. Let C have solvable conjugacy problem but suppose there is no algorithm to decide of two arbitrary pairs u I , u1 and u2,u2 of elements of F ( c i ) whether or nor there exists z in F ( c i ) such that z-'ulz = u1 and z-'u2z = u2 in C. If C is embeddable in G via the Higman embedding then G has unsolvable conjugacy problem.

Proof. Let x and y be elements of F ( c i ) such that x # y in C and x # 1, y # 1 in C (i.e. x f y mod R ( c i ) and x, y 6Z R ( c i ) ) .For any two pairs ul, uI and uz, uz of elements of F ( c i ) , none of which lies in R ( c i ) , it follows from Proposition 3 that if ~ t u ~ f - ~ and y f ~ xtulf-1ytu2f-' ~ f - ~ are conjugate in G then there exists z in F ( c i ) such that z-'uIz = u1 and z-'u2z = uz in C. Conversely, if such a z = z ( c i ) exists then the given words ~ t u ~ t - ~ y t and u ~ txtult-'ytuzt-' -~ are easily seen t o be conjugate in G by z ( a i ) . Since C has solvable conjugacy problem it follows that even when the assumption that none of ul, u2, ul, u2 lies in R ( c i ) is added there is still no algorithm to decide whether or not there exists z such that z-'ulz = ul and z-'u2z = u2 in C. Thus G must have unsolvable conjugacy problem.

D.J. Collins

84

Example. We specify a group in which t h e hypotheses of Corollary 4 are satisfied. Let C have presentation (abandoning our systematic notation):

a l , a2, b l , b2, p l , p2, 41, 92,

C I , CZ,

dl, d2,

T I , TZ, S I , SZ;

p;’b’,alb’,p’, = r;’d’,cld{r’, p;’bia2bipi = r;’dic2dM q;’q;’bf(l)a I bf(” I q;q;= s~’s~’d~“’cldf“’s;s; q ;‘q ;‘b{“’a2b{(“q;q ; = s 7’s;‘d {“’c2d{(‘)s;s ; where i and j range through the positive integers and f is a one-to-one recursive function of t h e positive integers with non-recursive range. Then C has solvable conjugacy problem and there exists z such that

z-‘b’,alb:z= d’,cld’, and

z - ’ b i a 2 b := dic2di

if and only if j lies in t h e range of f. We shall not give the arguments for this since they are standard and rather similar to t h e kind of argument given by Miller on pp. 46-54 of [ 5 ] . We conclude with some general remarks. There seems at present to be no hope of establishing t h e analogue of Clapham’s theorem. T o prove Higman’s theorem one firstly achieves a kind of logical embedding in a finitely presented group (in our notation C is logically embedded in L since w ( a , )= w ( b , ) in L if and only if w ( c , )= 1 in C). Then, in Paul Schupp’s graphic phrase, one performs the Higman “rope-trick’’ to achieve the actual group embedding. It follows from the author’s paper [3] that at least for Aanderaa’s version [l] of the Higman theorem, solvability of the conjugacy problem can be achieved for t h e logical embedding section of the construction. But, as we have seen, there are substantial difficulties in dealing with t h e “rope-trick’’ part. Furthermore, these difficulties seem to be more or less inevitable given t h e structure of t h e proof and probably a wholly new strategy will be needed to avoid them. For the present the most that can be hoped for is t h e isolation of conditions on C that are necessary and sufficient for the preservation of the solvability of the conjugacy problem in t h e Higman embedding.

References [I]

s. Aanderaa, A

proof of Higman’s embedding theorem using Britton extensions Of groups, in: W.W. Boone, F.B. Cannonito and R.C. Lyndon (eds.), Word problems (North-Holland, Amsterdam, 1973).

Conjugacy and the Higman embedding theorem

85

[2] C.R.J. Clapham, An embedding theorem for finitely generated groups, Proc. Lond. Math. Soc. 3, 17 (1967) 419-430. [3] D.J. Collins, Representation of Turing reducibility by word and conjugacy problems in finitely presented groups, Acta Math. 128 (1972) 73-90. [4] G. Higman, Subgroups of finitely presented groups, Proc. Roy. Soc. Series A, 262 (1%1) 455475. [S] C.F. Miller, On group-theoretic decision problems and their classification (Princeton, 1971).


SOME SOLVABLE WORD PROBLEMS Trevor EVANS* Emory University, Atlanta, Georgia

Introduction

This is a survey of some algorithms which have been used to solve decision problems (mainly word problems) in various varieties of algebras, e.g., lattices, commutative semigroups, quasigroups. O u r interest is in algebraic properties which imply the existence of such algorithms. There is n o attempt to be encyclopaedic but instead the emphasis is on examples of algorithms which have a universal algebra flavor. In Section 1 we discuss the connection between embedding of partial algebras in a variety and the solvability of the word problem for finitely presented (f.p.) algebras in t h e variety. In Section 2 we consider algorithms based on finite separability properties. Finally, we look at some aspects of normal form theorems. We will work within a finitely presented variety V, i.e., a variety defined by a finite number of finitary operations and a finite set of identities, and by an algebra, we will always mean an f.p. algebra in such a variety.

81. Partial algebras Let V be an f.p. variety and let V* be t h e variety with the same operation type as V but defined by the empty set of identities. For example, if V is t h e variety of commutative semigroups, V* is t h e variety of groupoids. If R is t h e set of operations in a V-algebra, by a partial V*-algebra 9 = (P,R ) we mean a set P of elements and, for each n-ary operation f in 0,a mapping f : S + P where S C P". By a partial V-algebra, we mean a partial V*-algebra 9 = ( P , R ) such that the following conditions are satisfied. (i) Insofar as the identities of V apply to 9, they are satisfied. That * This research was supported in part by U.S. NSF Grant MCS76-06986.

88

T. Evans

is, if u ( x l ,x 2 , x 3 , . . . ) = v ( x I x, 2 , x 3 , . . . ) is one of the defining identities of V and if for any elefnents a r , a 2 , a ,.... , of P substituted for the variables in u = v, the partial 0-operations of 9’ allow both u(al, a2,a , , . . . ) and v ( a , ,a2,a 3 , .. . ) to be computed, then the identity u = v holds for these values. (ii) The partial operations defined in 9’ and the defining identities of V do not allow us to extend the domain of the partial operations of P. In other words, if a substitution of elements of P for the variables in an identity u = v of V enables us to compute one side as a, and the other as f ( a l ,a2,a 3 , .. .), for some f E 0 and a l ,a 2 , a 3 ,... in P, then f ( a l ,a2,a 3 , .. . ) = a, is already given as a value of the partial operation f . We say that the embeddability problem is solvable for V if there is an algorithm for deciding whether any finite partial V-algebra (actually, we could just as well say “any finite partial V*-algebra”) can be embedded isomorphically in a V-algebra. The following theorem is proved in [4]. Theorem. The embedding problem is solvable for the variety V if and only if the word problem is solvable for V.

This theorem is useful in solving the word problem for two reasons. (i) In the most favorable case, any finite partial V-algebra can be embedded in a finite V-algebra. This occurs more frequently than one would expect. (ii) We may be able to give reasonable conditions on V or on partial V-algebras which imply a test for embeddability. For example, V may have the property that if a finite partial V-algebra is embeddable at all, then it can be embedded in a finite V-algebra. If V has the property that any finite partial V-algebra can be embedded, then the algorithm for solving the word problem for an f.p. V-algebra goes as follows. Let d be a V-algebra given by generators a l , a2,a 3 , .. . , and relations r, (ar,a2,a s , .. . ) = r : ( a r a2, , a 3 , .. .), i = 1,2,3,.. . . Let w , , w2 be two words in the generators of d. We wish to decide whether w I = w2 in d. We begin by introducing new generators b l , bz, b 3 , .. . for every word in the a, which occurs as a subword of r,, r:, w 1 or w2. This enables us to rewrite the defining relations of d so that each relation is either of the form b, = b, or f(b,,b,, bk,.. . ) = b, where f is an operation of d. Now, by direct applications of the identities of V and the defining relations of d, we alternately remove redundant generators and introduce new relations of the form f(b,,b,, bk,. . . ) = b,. We arrive at a

h

e solvable word problems

89

presentation for d which has the form of a partial V-algebra 9. In doing this, the b,, b, corresponding to wl,w 2 may have been indentified, in which case we know that w1= w 2 in d.On the other hand, b,, b, may remain distinct in 8. Since 9 can be isomorphically embedded in a V-algebra, bi# b, in the V-algebra freely generated by 8 and so, isomorphically, w1# w 2 in d.For more details see [3], [4]. The above procedure solves the word problem for the varieties of (i) lattices (ii) quasigroups, loops and many subvarieties of these which arise in the study of combinatorial designs [17]. In a number of papers (see, for example, [13]) Gluhov and Gvaramija have given conditions on varieties of quasigroups (and of other algebras) which imply that any finite partial algebra in the variety can be finitely embedded. We can differentiate between the different kinds of embeddability of finite partial algebras in a variety V by the following conditions. El: Any finite partial V-algebra can be finitely embedded. E2:Any finite partial V-algebra can be embedded (although not necessarily in a finite V-algebra). E,: If a finite partial V-algebra can be embedded, then it can be embedded in a finite V-algebra. Clearly, El @ E2A E,. One of the simplest examples of a variety V with property E2but neither El nor E , is given by binary operations x y , x\y and the identity x . ( x \ y ) = y . For details, see [17]. The situation is clarified by the following theorem [17]. (See also [2], [7].)

Theorem. A variety V has the property E , if and only if every f.p. V-algebra is residually finite. An f.p. residually finite algebra has a solvable word problem. (See [17], although the result is apparently first due to Malcev.) We repeat briefly the simple argument. Let d be an f.p. residually finite Valgebra and let w l , w z be words in the generators of d.We enumerate all consequences of the defining relations of d.If w 1= w 2 in d,it will appear in this enumeration. We also enumerate the finite V-algebras which are homomorphic images of d.If w , # w 2 in d,then in one of these homomorphic images of d w 1 and w 2 will map onto distinct elements. Combining these enumerations, we have a procedure which, in a finite number of steps, will decide whether w , = w 2 in d. In the table below we list these embeddability properties for f.p. algebras in some familiar varieties. We note that the free algebras in all of the listed varieties other than inverse property loops and modular lattices are residually finite and hence have solvable word problem. It is not known whether free modular lattices are residually finite, i.e. have property E,, nor is this known for inverse property loops.

T. Evans

90

W.P.

Variety

E,

E,

E, solvable

Groups, rings semigroups, modular lattices

no

no

no

no

no

yes

Quasigroups, loops, lattices

Yes

yes

yes

Inverse property loops

?

yes

?

Non-associative rings

no

no

?

Abelian groups, commutative rings and semigroups

For commutative rings and semigroups, residual finiteness (or E,) at present provides the only procedure for solving the word problem for f.p. algebras. Problems. 1. Prove that free modular lattices are not residually finite. 2. Prove that f.p. non-associative rings are residually finite. 3. Prove that finite partial inverse property loops can be finitely embedded. (Inverse property loops are one of the few varieties of loops which have been studied in detail but for which the finite embeddability property has not been verified.) 82. Finite separability properties

An algebra d is residually finite if, for any two disjoint finite subsets SI, S 2 of d , there is a homomorphism of d onto a finite algebra such that the image of SI and Sz are disjoint. There are numerous generalizations of this concept. If x is an element and Y a subalgebra of d , then d has the finite separability property (abbreviated to f.s.p.) if, for any x fZ Y there is a homomorphism a onto a finite algebra in which x a fiiYa. If Y,, Y2are subalgebras, then d has the subalgebra property if for any such disjoint Y,, Y2,there is a finite separability Y2are disjoint. homomorphic image of d such that the images of Y,, If d is a group, then it has the conjugacy separability property if for

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any elements x, y not conjugate in d, there is a finite homomorphic image of d in which the images of x and y are not conjugate. The relationship of such finite separability properties to algorithms for solving corresponding decision problems is obvious. We have given one example earlier - residual finiteness implies solvability of the word problem - a similar proof shows that finite separability of an element and an f.g. subalgebra implies solvability of the generalized word problem. Free groups have the finite separability property for f.g. subgroups (M. Hall [14]), hence have solvable generalized word problem. So do free semigroups and free commutative semigroups (these are easy to show). In fact, free algebras in any variety defined by balanced identities (each variable in a defining identity u = u occurs exactly the same number of times in u as in u ) . Surprisingly, however, free rings do not have the finite separability property. Let 9 be the free ring on one generator x and let Y be the subring generated by 2x,2x2+ x. Then Y# 9, i.e. x E Y.However, in any homomorphism a of 9 onto a finite ring, x a E Ya.This result is due to K. Mandelberg [17]. The generalized word problem for free rings is apparently an open question. However, one can show, using methods similar to those of Mihailova [19] for groups, that the generalized word problem is unsolvable for F @ F, the direct sum of two copies of a free ring. Let S = (gl,gz, . . . ,g, ; u I = u,, u z = u2, u, = u,) be a semigroup with an unsolvable word problem and let F be the free ring generated by g,, gz,. . . , g,. In the direct sum F @ F, let S * be t h e subring generated by (u,, u,), i = 1,2,. . . , n and (g,,g#), i = 1,2,. . . , m. It is easy to prove that if u, u are monomials in F with coefficient one, then (u, u ) E S * if and only if u = u in S. It follows that t h e G.W.P. is unsolvable for F@E For the problem of deciding for a V-algebra d whether two f.g. sub-algebras have a non-empty intersection, we appeal to t h e subalgebra separability property. For quasigroups (and various subvarieties) we can use the fact that any finite partial quasigroup can be finitely embedded to show that t h e variety of quasigroups has the subalgebra separability property [17]. There is an interesting contrast for this property between free semigroups and free commutative semigroups. Let 9 be an f.g. free semigroup and YI, Y2be disjoint f.g. subsemigroups of 9.By a standard result in the theory of finite automata, Y,and Y2can each be expressed as the union of congruence classes of some congruence of finite index o n 9.(See, for example, [ll].) The intersection of these two congruences is a congruence 8 of finite index. In the quotient

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semigroup S / O , the images of 9,and YZwill be disjoint. Now let sd be the free commutative semigroup generated by g , , gz and let 58,, 58, be the subsemigroups generated by {glgz,g , g : } and {gl,g:gz} respectively. Although 9 ,n 58z = 0 in d,in any homomorphism of .d onto a finite semigroup the images of 58, and BZintersect (Garrett [121). We conclude this section with a decision problem which is known to be solvable [16] but for which it would be interesting to know whether there is an algorithm based on a finite separability property. Let 3 be a group. If, for any x, y in % such that x and y are not conjugate in 3, there is a homomorphism a of 3 onto a finite group such that x a , ya are not conjugate in 3-, then 3 is conjugacy separable or residually finite with respect to conjugacy. Stebe [23] has shown that free groups are conjugacy separable. If we describe conjugacy separability in groups as finitely separating elements and conjugacy classes, it suggests the following generalization. Let JB be an algebra and 3 a group of automorphisms of d. Does d have the finite separability property with respect to orbits of 3? A particularly interesting case of this concerns free algebras. Let 9,b e an f.g. free algebra in'a variety V. Consider all homomorphisms of 9, onto n-generator finite algebras which are free in some subvariety of V. If an element x E 9,maps onto a primitive element in each such finite relatively free image, is x primitive in S,,? (Here, primitive means an element of some free generating set.) We are actually asking here whether 9,has the finite separability property with respect to the orbits of its automorphism group. No examples of varieties V appear to be known in which F . ( V ) does not have this property but on the other hand only a few rather trivial examples, such as semigroups and groupoids, are known of varieties in which the free algebras d o have this property. Motivation for t h e above questions comes from the following theorem [17]. Theorem. Let 9,be a residually finite free V-algebra which also has the f.s.p. If wl,w z , .. . ,w, in Snmap onto a free generating set for every homomorphism of 9,onto a finite relatively free V-algebra, then w I ,w z , .. . , w, is a free generating set of 9,. From this, we obtain by a familiar argument Theorem. If 9"is a free V-algebra on n generators which is residually finite and has the f.s.p., then there is an algorithm for deciding whether a set of n elements of 9,is a free generating set.

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This theorem applies to groups, semigroups, groupoids, quasigroups, loops, lattices. It is tempting to conjecture that if each f.p. algebra in a variety V is residually finite (so that it has a large number of finite homomorphic images) then any two f.p. V-algebras having the same homomorphic images are isomorphic. This is the case for abelian groups unfortunately no other non-trivial example of this situation is known. A result of Pickel [21] states that if H ( d ) is the set of finite homomorphic images of an f.g. nilpotent group d,then there are only a finite number of f.g. nilpotent groups 93 such that H(46)= H ( d ) . However, examples are known (Remeslennikov [22], Dyer [2]) of nonisomorphic f.g. nilpotent groups d,93 such that H ( d ) = H(93). Let us call an f.p. algebra in an f.p. variety Vfinirely derermined if the set of its finite homomorphic images determines it uniquely (to within isomorphism). A similar argument to that used to show residual finiteness implies solvability of the word problem shows that the isomorphism problem is solvable for f.p. V-algebras if every f.p. Valgebra is finitely determined. However, as we remarked above, few examples of this situation are known. We conjecture, o n rather flimsy evidence, that the varieties V of quasigroups with the property that every finite partial V-quasigroup can be finitely embedded d o have every f.p. algebra finitely determined. The variety of lattices is also another good candidate for this property. A curious connection between t h e properties we have been discussing is that if the isomorphism problem is solvable for f.p. algebras in a variety V and if every such algebra is hopfian, then the word problem is solvable for f.p. V-algebras. Let d be an f.p. Valgebra and u, u words in the generators of d. Let d,be the algebra obtained from d by adding the additional defining relation u = u. Since d is hopfian, u = u in d if and only if d and I, are isomorphic. Problems. 1. Solve the generalized word problem for free rings. 2. Prove that free lattices (or f.p. lattices) have the f.s.p. and the subalgebra separability properties. 3. Prove that w is a primitive element in an f.g. free group 9" if and only if w maps onto a primitive element in every homomorphism of 9" onto a relatively free finite group. Prove corresponding theorems for free loops and free quasigroups. 4. Let d,93 be two f.p. groupoids (or f.p. algebras in a variety V defined by the empty set of identities). Prove that if d, 46 have the same sets of finite homomorphic images, then d, 46 are isomorphic.

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5. The same as Problem 4, for lattices. 6. It is known that residual finiteness of f.p. algebras in a variety V is not sufficient to prove t h e property described in 4 above for f.p. V-algebras. Find finite separability properties on V which d o imply this property.

93. Normal form theorems We recall the usual procedure for solving the word problem in a free group. Consider the set of all words in the generators and their inverses. We introduce contractions and expansions of words (deleting and inserting subwords g g - ’ , g ’ g ) and define two words to be equivalent if one can be transformed into t h e other by a finite sequence of expansions and contractions. It is easy to prove that every for which no equivalence class [ w ] contains a unique word contractions are possible, and that starting from any word in [ w ] and applying contractions, we eventually arrive at This word EJ is called the normal form (or reduced form) of any word in [ w ] .Clearly, when we have such a theorem we have a procedure for solving the word problem for F. There are numerous varieties to which we can apply this method or some variation of it. The general setting is as follows. Let d be an algebra (gl, g,, g,, . . . ; rl = r ; , r2 = r:, . . . ) in a variety V defined by identities u, ( x l , x z , x 3 , . . . ) = u:(xl,x2, x,, . . .), i = 1 , 2 , 3 , . . . . The elements of d are represented by words w ( g , ,g,, g,, . . . ) built up from t h e generators and operations of d.Two words u, u‘ are connected by an elementary transformation u W- u’ if one can be transformed into the other by an application of either a defining relation r, = r: of d or a defining identity u, = u : of V. Words u, u’ are equivalent u = u‘ if one can be transformed into the other by a finite sequence of elementary transformations. The equivalence classes [ w ] are the elements of d and the word problem for d consists of finding a method (or showing that one does not exist) which will decide when two words lie in the same equivalence class. We may represent the situation by a graph. The points correspond to words and connecting segments to elementary transformations. The maximal connected components of the graph are the classes of equivalent words. We turn the graph into a directed graph by classifying the elementary transformations as either contractions or expansions. We write u + u’ in the directed graph if we can get u’ from u by a contraction. We attach a “weight” to each point, usually

* *.


95

called the length l ( u ) of the corresponding word. These weights can come from any well-ordered set although most frequently one uses the non-negative integers and we require that if u + u’ then I ( u ) > l(u’). Our aim in doing this is to have in each connecting component exactly one point of minimum weight such that any directed path from any other point in the component ends at this “sink”. This point will correspond to the normal form of every word in its equivalence class. There is a general theorem of M.H.A. Newman [20] which enables us to achieve this desired end once we have assigned directions and weights in the graph. We require the resulting graph to satisfy the following confluence property (sometimes called the Diamond Lemma). If u + u’, u + u” are any two contractions of u, then there is a point u”’ + u’”. such that there are directed paths u ’+ * * + u”’, up’+

-

I

J

If this condition is satisfied it is an easy matter, using induction on the length l ( u ) of u, to show that any sequence of contractions starting at a word u ends a unique normal form fi of u. A simple induction on the number of elementary transformations connecting two words w, w’ also enables us to show that words w, w‘ are equivalent iff they have the same normal form. Thus, to use this approach to solve the word problem for algebras in a variety V we have to find an appropriate length function on words and we have to characterize the contractions of words so that we can prove that the confluence condition is satisfied. Once this is done, we may decide whether two words in an algebra are equivalent by comparing their normal forms. Actually, what we have described is the bare bones of just one normal form approach t o solving the word problem. In putting flesh on the bones, we may have to (even in the most direct application of this method)

T. Evans

%

(i) rewrite the defining relations so that they are in some special form (or consider only algebras whose relations can be put in this special form) (ii) add redundant identities to the defining identities of V. Such adjustments (and others) may be necessary in order to obtain a confluence condition for the contractions. However, the above approach, with the refinements described, is sufficient to give a solution of the word problem for f.p. algebras in many different varieties. Some examples are given below. Example 1. If V is a variety not satisfying any laws, then the word problem for free V-algebras is trivial. For f.p. algebras in such a variety we rewrite the defining relations in the form f(gl, g,, g,, . . . ) = g k (see [3]). Now, with the obvious definitions of length and contraction, the confluence condition holds. Example 2. Let V be the variety of groupoids defined by the identity

x y . y z = y, for all x, y , z. Let F be a free V-algebra. If, using only the reduction x y . y z + y, we try to prove that the confluence condition holds, then we fail but we do discover that the identities x ( x y . z ) = x y and ( x . y z ) z = y z are consequences of x y .y z = y .

We now try to verify the confluence condition using the contractions

x y . y z + z, x ( x y . z)+ x y , ( x . y z ) z + y z where x , y, z are any words in

the generators. This time, we are successful. This example illustrates a situation which occurs quite often - the confluence condition (if it holds) “generates” its own proof. Example 3. Let V be the variety of quasigroups defined by three binary operations x y (multiplication), x \ y (left-division), x / y (rightdivision) and the identities x . ( x \ y ) = y , (XlY1.Y = x x \ ( x y ) = y, (XY )/Y = x .

Let d be an f.p. V-algebra generated by g,, g, g, . . . . In order to obtain the confluence property and hence, a normal form theorem, we first rewrite d in terms of generators a , , az,a s , .. . and defining relations of the form aiai = a k . Then we add the redundant identities x l ( y \ x ) = y, (YlX)\Y = x

to the defining identities of V and the redundant relations ai\ak = a,, ak/ai= ai for every defining relation aiai = ak.


97

Now we use as contractions aia, + at, ai\ak+ ah at / a ,+ a , where aiaj = at is a defining relation and x ( x \ y ) + y, ( x / y ) y + x, x \ ( x y ) + y, ( x y ) / y + x , x / ( y \ x ) + y , ( y / x ) \ y + x , where x, y are any words in the

generators. The confluence condition holds and a normal form theorem results. See [ 5 ] for the use of this in studying the structure of f.p. algebras in various varieties of quasigroups and loops. Actually, our original description of how we solve the word problem for free groups is not an example of the normal form approach we have described - it illustrates a generalization. When, in a free group, we define a word to be a string of generators and their inverses such as glg;'g3g3g2', the associative law (and the identities ( x y ) - ' = y - l x - ' , (x-')-' = x ) are sneaking in through the back door. Formally, we should define a word to be, for example, ((glg;')g3)(g2g;')-'. In writing glg;'g3g3g;' as a "word" we are identifying all words which are equivalent by using group identities which preserve length. In other words, the confluence condition using xx-'+ 1, x - ' x + 1 as reductions, is being applied to equivalence classes of words, rather than individual words, each equivalence class consisting of all words connected by the elementary transformations ( x y ) z f;*x(yz), (x-')-'-x, (xy)-'f* y - ' x - ' . The general set-up is this. We have a variety V and a V-algebra d. With what seems to be the appropriate definition of length of a word, it may happen that some of the V-identities preserve length and so cannot be used to shorten a word. We proceed by first considering an equivalence relation on words using only these length-preserving identities. We define contractions so that if u + u' and u u, then there is a contraction u + u' where u ' - u ' . If this can be done, we have essentially defined contractions on ( )-equivalent classes of words. If we can now prove a confluence theorem for these contractions applied to the ( )-equivalence classes, then providing that we can recognize when words are ( - )-equivalent, we have a solution of the word problem. An example of this procedure is used in [9] where the length-preserving identities are essentially generalized commutative laws for the operations. We conclude this section with an example from group theory. The result is familiar - the normal form theorem for free products with an amalgamated subgroup - but the elementary transformations and length function used are not the familiar ones. Let A, B be groups with subgroups A', B1 which are isomorphic. We will identify A', Bl according to the isomorphism and regard A, B as groups sharing a common subgroup C = A n B. We construct a group A *=B, the generalized free product, as follows. The generators

-

-

-

-

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of A *cB are the elements A U B and the defining relations are the multiplication tables of A and B. By a word, we will mean a finite sequence x I x 2 * . x , of elements of (A U B )- { e } ( e is the common neutral element of A and B). 1 denotes the empty word. By an elementary transformation of a word we mean either a contraction or its inverse, where contractivn is defined as follows. We begin by choosing fixed left coset representatives for the subgroup C in the groups A and B. W e take 1 as the coset representative of C and denote by x * the left coset representative of C in A or B to which x belongs. Hence, x = x * c for some c in C. Let w = x 1 x 2* x, where x , E ( A U B )- { e } . (i) If x, belongs to A - C or to B - C, then replacing x, by x t c where x , = x : c in A or B, is a contraction of w. (ii) If adjacent elements x,, x , + ~in w belong t o the same group A or B, and x,x,+l = x = x * c in that group, then replacing X , X , + ~ by x * c is a contraction of w. Thus, contractions have the effect of replacing elements of A - C and B - C by coset representatives and of shifting elements of C to the right. A reduced word in the free product of A and B with x I c where the x t are amalgamated subgroup C has the form x : x f coset representatives from A and B, and c is either empty or in C. We now introduce a length function with the right property. We define the empty word to have length (0, 0). If w = x I x 2 . . . x,, we assign as its length an ordered pair of non-negative integers (p,q). Each x , in C is given p-weight 1, each x , which is a coset representative is given p-weight 3, all other x, are given p-weight 5 . W e define p to be the sum of the p-weights. To each x, which is in C we assign a q-weight the number of x,’s in w which are coset representatives and lie to the right of x , in w. W e define w to be the sum of the q-weights of w. We order lengths lexicographically. It is a routine matter to check that the confluence condition holds and hence we obtain the usual normal form theorem that every element in the generalized free product can be written uniquely in the form x : x f - . . x t c . Despite the simplicity of the procedure described in this section, it is enormously versatile. There is an extensive survey for other applications of the Diamond Lemma in a forthcoming expository paper by G. Bergman. P. Hall’s expository paper [15] is a detailed study of an application of this technique. A number of examples are given in Knuth and Bendix [18].

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Some solvable word ptvblems

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Problems. 1. In the paper by Knuth and Bendix referred to above, there is a universal algebra version of the algorithm we have described for free algebras in a variety satisfying certain conditions. Can one extend the results of Knuth and Bendix to apply to varieties where some of the identities preserve length? 2. Is there a normal form theorem of the type we have described for free lattices? The usual solution of the word problem (Whitman [24]) involves explicit use of the order relation. The same question applies to f.p. lattices. 3. A version of the algorithm we have described for solving the word problem is used by the author (Trans. Amer. Math. SOC.1963) to solve the isomorphism problem for some varieties of loops and quasigroups. Is there a universal algebra version of this? References

RADCLIFFE

[l] B. Banaschewski and E. Nelson, On residual finiteness and finite embeddability, Algebra Universalis 2 (1972) 361-364. [2] J.L. Dyer, On the isomorphism problem for polycyclic groups, Math. Z. 112 (1%9) 145-153. [3] T. Evans, The word problem for abstract algebras, J. London Math. Soc. 26 (1951) 64-71. [4] T. Evans, Embeddability and the word problem, J. London Math. Soc. 28 (1953)

76-80.

[5] T. Evans, On multiplicative systems defined by generators and relations, I. Normal form theorems, Proc. Cambr. Phil. Soc. 47 (1951) 637-649. [6] T. Evans, The isomorphism problem for some classes of multiplicative systems, Trans. Amer. Math. Soc. 109 (1963) 303-312. [7] T. Evans, Residual finiteness and finite embeddability. A remark on a paper by Banaschewski and Nelson, Algebra Universalis 2 (1972) 397. [8] T. Evans, Identities and relations in commutative Moufang loops, J. of Algebra 31 (1974) 508-513. [9] T.Evans, A decision problem for transformations of trees, Can. J. Math. 15 (1%3) 584-590. [lo] T. Evans, Some connections between residual finiteness, finite embeddability and the word problem, J. London Math. Soc. (2). 1 (1969) 399-403. [ I l l E. Engeler, Introduction to the Theory of Computation (Academic Press, New York, 1973). [12] R. Garrett, Some problems concerning residual finiteness in algebras, Ph.D. Thesis, Emory University (in preparation). [I31 A.A. Gvaramija and M.M. Gluhov, A solution of the fundamental algorithmic problems in certain classes of quasigroups with identities, Sibirsk. Mat. Z. 10 (1969) 297-317 (Russian). English translation; Siberian Math. J. 10 (1969) 211-224. [I41 M. Hall, Jr., Coset representations in free groups, Trans. Amer. Math. Soc. 67 (1949) 421432. [IS] P. Hall, Some word-problems, J. London Math. Soc. 33 (1958) 4824%.

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P.J. Higgins and R.C. Lyndon, Equivalence of elements under automorphisms of a free group, Mimeographed Notes. Queen Mary College, London. C. Lindner and T. Evans, Finite embedding theorems for partial designs and algebras (Les Presses de L’Universite de Montreal, 1977). 1181 D. Knuth and P. Bendix, Simple word problems in universal algebras, Proc. Conf. o n Computational Problems in Abstract Algebra, Oxford. 1967. 1191 K.A. Mihailova, The occurrence problem for direct products of groups, Dokl. Akad. Nauk SSSR 119 (1958) 1103-1105 (Russian). [20] M.H.A. Newman, On theories with a combinational definition of “equivalence”, Ann. Math. 43 (1942) 223-243. [21] P.F. Pickel, Finitely generated nilpotent groups with isomorphic finite quotients, Trans. Amer. Math. Soc. 160 (1971) 327-341. [22] V.N. Remeslennikov, Finite approximability of groups with respect to conjugacy, Sibirsk. Mat. Z. 12 (1971) 1085-1099. [23] P.F. Stebe, A residual property of certain groups, Proc. Amer. Math. SOC.26 (1970) 37-42. [24] P. Whitman, Free lattices, Ann. Math. 42 (1941) 325-330.


SOLUTION OF THE CONJUGACY PROBLEM IN CERTAIN ARITHMETIC GROUPS Fritz J. GRUNEWALD University of Bielefeld (FRG)

The purpose of this paper is to give an algorithm which decides for two matrices T, f E GL, (Q) whether there exists a matrix X E GL, (Z) such that: XTX-'=

f'

This is problem 22 of the problem section of [l]. The algorithms we give can all be transformed into quite practical computer programs. They have as a general feature that one can always easily compute a bound for the number of steps necessary to complete the computation. To explain our results we make the definitions: Definition. Let KI C K 2 be two commutative rings with 1 . For T E GL, ( K z ) define:

1

%,,(T)= { X E GL, (KJ XTX-'

=

T}

GL,(K,) being the group of invertible n x n matrices with entries in the ring K , . n is a natural number. We of course consider GL,(KI) as being contained in GL,(Kz). For T, f € GL,(K2) we write T

--K,

if and only if there is a X E GL, ( K , ) with XTX-'=

f.

We write SL.(Kz) for the group of n x n matrices with entries in K Z of determinant 1. We prove: Theorem A. Given two matrices, T, f E GL, (Q), there is an algorithm for deciding whether : 101

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EJ. Grunewald

T-zf

If the answer is “yes” the algorithm constructs a conjugating matrix. Another algorithm does the following:

Theorem B. Given two matrices T, f E GL. (Q), there is an algorithm for deciding whether there is a matrix X E SL. (2) such that

XTX-’= If the answer is ‘‘yes’’ the algorithm produces such an X . We get Theorem B from Theorem A together with the following:

Theorem C. Given T E GL.(Q), there is an algorithm for computing a finite set of generators of Cez(T). q Z ( T )is an arithmetic group in the sense of [3]; a main theorem of this paper says that q Z ( T )will be finitely generated. As generalizations we mention some results on general and special linear groups over rings of algebraic integers. Let K be an algebraic number field, that is a finite extension of the field of rational numbers. We assume that K is given by a field generator 8 together with its minimal polynomial over Q. Let OK be the ring of algebraic integers in K. By algorithm A6 listed in the last section of the paper it is possible to compute a Z-basis for OK together with a multiplication table for it. Almost a word for word translation (using OK instead of Z) gives now: Theorem A’. Given T, f E GL. (K), there is an algorithm for deciding whether: T-oKf If the answer is ‘‘yes’’ the algorithm computes a conjugating matrix. A little more involved is:

Theorem B’. Given T,f E GI, (K), there is an algorithm for deciding whether there is a matrix X E SL. (OK)such that: XTX-’ =

f.

If the answer is ‘‘yes’’ such an X may be computed.

n t e conjugacy problem in arithmetic groups

103

Theorem B’ follows from: Theorem C‘. Given T E GL,(K), there is an algorithm for computing a (T). finite set of generators of gOK The groups GL, (OK),SL, (OK),goK(T) are again arithmetic groups in the sense of [3] as may be seen by “reduction to the ground field” in [24] Chapter I. Hence they are all finitely presented by [3], [3a]. Our algorithm therefore has some group-theoretic meaning. Presentations of some of them are computed in [21], [13], [14]. A more thorough look into the methods used here even leads to an effective algorithm for constructing a finite presentation for the g o K ( T ) for a matrix T E GL, (K). As an application to another type of arithmetic group we mention: Corollary A. Given T, -i. E PGL, (K), there is an algorithm for deciding whether there is an X E PGL, (OK)with

XTX-’ =

f

in PGL, ( K )

and similarly : Corollary B. Given T, -i. E PGL,(K), there is an algorithm for deciding whether there is an X E PSL. ( O K ) with :

XTX-’=

-i.

in PGL,(K).

Our method allows one to deduce Theorems A-C for Z replaced by an order in a (explicitly given) ring of algebraic integers or for an S-arithmetic ring if S is an explicitly given finite set of primes of the algebraic number field in question. The proof again is an almost word for word translation. It is also clear that Theorems A’-C‘ decide the conjugacy problem in every subgroup of finite index in GL.(0) or SL”( 0 ) . As a further generalization we mention the possibility of applying a quite similar procedure to Sp2,(Z) or to O,U, n ) where f is an ndimensional integral quadratic form, or to the group of matrices having entries in an order in a (explicitly given) finite dimensional skew field over Q, whose reduced norms lie in the unit group of an order in the center of the skew field. Details will appear elsewhere. The particularly nice group-theoretic structure of GL2(Z) and SL(Z) as amalgamated products of finite groups gives a more group-theoretic procedure for the decision problems in Theorems A and B (for 7’’ f € GL(Z)!). For this see for example [12], [13], [27], and for generalizations [ 111.

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We have not tried to write out very effective algorithms, a lot of them depend highly exponentially on the data. But for dimensions 2 and 3 it is possible to modify the procedures (because there are not SO many unipotent Jordan matrices) to actually obtain not too inefficient computer programs. In the first part of the paper we give a number of effective reduction steps which will reduce our algorithm to a couple of algorithms known from algebraic number theory. These will be listed at the end of the paper. I wrote this paper while doing a project on quite similar questions with Daniel Segal. I thank him very much for his advice. I also thank Professor J. Mennicke for help with an argument.* A. Reductions Before beginning, I want to mention that we often use a form of Gauss’ algorithm: Given a finitely generated free abelian group N with Z-basis n,,. . . ,n,. Let { f l , . . . ,fs} and { g , , . . . , g l } be two sets of elements in N , given in terms of t h e basis n l , . . ., n,. Let:

. ,fs}> Nz = ({gi,. . . ,g o ) Ni = ( { f i r . .

be the subgroups generated by these two sets. It is then effectively possible to compute the index:

1 NI: (NIn N2)I and if it is finite to give a set of coset representatives for: N , I N ~n N ,

For this see e.g. [ 5 ) . We will not always mention the use of this algorithm. We add some further definitions: If R is a ring and E a subset of R we put ( E ) ~for the ideal generated by E. We often leave the subscript R.

* (Added September 1977): After this paper was submitted in May 1977, Professor Adian informed me in September 1977 that similar results have been independently obtained by A. Sarkisjan.

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77ae conjugacy problem in arithmetic groups

Step I Let n be a natural number. We are given two n x n matrices T, f with integer entries and both of determinant # 0. We hand these to algorithm A, which returns:

f=g+o

T=S+U

where S, S are rational semisimple matrices and U, nilpotent matrices. Furthermore we have:

S U = US

and

0

are rational

S f i = I%.

These decompositions are clearly unique. We now find a scalar matrix k . E n where

En =

(1

:)

* a .

is the n-dimensional unit matrix and where 0 # k E Z, such that:

kE, . S, kE. . U E M,, (Z) kE,,S, kEnU E M, (Z). Put now:

TI = kE,, T

fl= kE, f

S , = kE. * S

S,=

UI= kEnU

U ,= kE,o

1

*

kE,

*S

We find then:

Lemma 1. (1) fa ~ , - ~ f , . ( 2 ) S , , 3, are semisimple with integer entries. ( 3 ) U , , ir, are nilpotent with integer entries. (4) TI= S1+ U , ; 7 3, + 0,. ( 5 ) S , U , = U , S , ; s,u,=U$,.

TI

These statements are obvious. Because of the uniqueness of t h e decomposition T = S + U we get:

Lemma 2. TI- z f,e (1) 3X E GL. (Z) with XSIX-'= 3, and ( 2 ) 3Y E %'z(S,) with Y X U I X - ' Y - '= 0,. So, to decide the conjugacy problem we are essentially left with the two tasks:

106

F.J. Grunewald

I. Given two semisimple matrices S , , 3, with integer entries and nonzero determinant, decide whether S I and if t h e answer is "yes" to produce a conjugating matrix X . 11. Given a semisimple matrix 3, with integer entries and two nilpotent matrices U , , U z with integer entries and:

-zsl

s,u,= u,s,; s,u2= uzs, decide whether there is X E %,(s,) with XUIX-' =

u2.

We now define for any n x n matrix T with integer entries a Z[x,,]module, which as a Z-module will be isomorphic to Z". The operation of xo o n Z" is defined by: XoU

=

Tv'

for any vector v E Z". Here u' is t h e transpose of v. We call this module Z [ T ] . We obviously have:

Lemma 3. Let T I ,TIE Mn(Z), then TI modules.

--z

Z[ TI] Z[ TI] as Z [ X O ] -

In Step I1 we shall reduce our problems I and I1 and the problem in Theorem C to t h e following two questions: Let K be an algebraic number field and O C OK an order in t h e ring of integers of K (i.e. 0 is a subring of finite index in OK). Ia. Let M, M be two explicitly given Z-torsion free, finitely generated

O[~ ] / ( ~ t , - m o d u ldecide e s , whether: M=M

as O[y]/cy~-rnodules. IIa. Given a Z-basis for M --. Z", compute a Cfinite) set of generators for the group of O[y]/,,~,-module automophisms, viewed as a subgroup of GLn (Z). Step 11.

This step gives reductions of our problems to more number-theoretic questions, which are then dealt with in Step I11 and Step IIIa. We first compute the minimal polynomials min(SI) and min(3,) by algorithm A*. Both will be monic and will have integer coefficients. We assume that:

m e conjugacy problem in arirhmeric groups

107

P ( X )= min (sI) = min (3,). Otherwise t h e answer to our original question would be “no”. We now split P ( x ) by algorithm A,,: P ( x )= P l ( X ) . * .P , ( x ) ,

(1)

P , ( x ) being Q-irreducible, monic with integer coefficients for i = 1,. . . ,r. O u r matrices being semisimple, the P, will all be distinct. Put:

n

Pi@)= P / ( x ) jZi

for i = 1,. . . , r. The polynomials p,(x) are coprime, so algorithm A1 will give us polynomials Ql, . . . ,Q, with integer coefficients and a natural number A such that:

QIPl+ . . . + Q,P, = A

(2)

We now define:

R = Z [x ]/(rw)) where ( P ( x ) ) denotes t h e ideal generated by P ( x ) . The ring R decomposes: R

(3)

Z [ X ] / ( P , , ,X, , * . . X Z[x]/(rAx))

and we put 0, = Z [ x ] / , , , , ) for i = 1,. . ., r. The polynomials P I , .. .,P, being Q-irreducible, monic with integer coefficients, the rings 0, are rings of algebraic integers. We then find a natural number 1 such that:

ul = 0:= 0. We define on Z [ T , ] resp. Z [ f , ] now R[y]/(,~)-modulestructures by: xv = S l V ’ yv

=

u l v ‘ for all v E z [ P , ]

with t h e corresponding definition for Z [ f,]. By Lemma 1 this definition , Z[T,]. leads to a ring operation of R [ y ] / ( , i on Putting everything so far together we get:

Lemma 4. The following two statements are equivalent: (i) Z[ T I ]= Z [ PI] as Z[xo]-modules (ii) Z[ T I ]= Z [ PI] as R [y]/(,?-modules. We now decompose ZITl] and Z [ F , ] according to the decomposition (2). For this put:

F.J. Grunewald

108

M, = p, ( x ) . Z[ TI]

M,= F,(X).Z[Ti]. The M, and 9, are a variant of Gauss' and M,.Equally we on these new bases.

clearly R[y]/(,~,-submodulesof ZITI] and Z['fl]. By algorithm we are able to find Z-bases in both M, find a representation for the operations of x and y Put then:

+ M, C Z [ T i ] M = M I+ * +M,C Z [ T 1 ] M = Mi i - .. * * .

Lemma 5. (i) M, 6 f are R [y]/,,~,-submodulesof Z[T,] resp. Z[ TI]. (ii) M,, M,are annihilated by P,(x) for i = 1,. . ., r. (iii) 1 Z[ TI]/M 1 S A " and 1 Z[ 'f.,]/~, 1 S A ". (iv) M, n (C,,,M,) = (O), n (C,,, = (0). So M and A? are direct sums of their submodules M, and d,. (v) M, is a characteristic submodule of M for i = 1,. . . , r (i.e. normalized by any automorphism of M ) .

a

G)

Proof. Lemma 5 follows from our decomposition (2) and t h e fact that the polynomials P, (x). are distinct and Q-irreducible. We may consider then M,, &Ij as explicitly given O,[y]/(,~,-modules. Furthermore M and M are given explicitly as R [y]/(,~)-modules.

Lemma 6. Let cp : Z[ TI]+ Z[ TI] be a n R [y]/(,l)-module isomorphism then : cp(M)=

M.

Proof. For any u E ZITI] we have: cp

( R (x)

*

0 )=

Po( X I . cp (u ),

hence cp even induces R[y]/(,~)-rnoduleisomorphisms of the M, and

A?,.

We now first decide whether the nicer R[y]/(,~)-modulesM and $2 are isomorphic. The following is clear:

Lemma 7. The following two statements are equivalent: (i) M = h a s R [y]l(y~)-modules. (ii) Mi = hia s Oi [y]/,,~,-modulesfor all i = 1,. . . ,r.

7he conjugacy problem in arithmetic groups

109

We now go with our modules M,, together with their rings O , [ Y ] / ( ~toI , Step I11 of this paper. We are supplied there with explicit 0, module isomorphisms: f o r i = 1 , . . . , r.

~,:M,+M,

By our lemmata so far we are able to construct an R[y]/(,~,-module isomorphism: y:M+M by putting the information about t h e y, together. If t h e application of Step 111 to a pair of modules M,, M, yields that these are not isomorphic as 0,(y]/,y~,-modules,then our matrices T, f will certainly not be integrally conjugate. We put now:

fi = A . Z [ f l ]

N = y(AZ[Ti])

We note that: and

A Z [ f l ]= k

&f

Y(AZ[TI])Cy ( M ) = A?.

N and N may be viewed as given by Z-bases in terms of a fixed Z-basis of Id. Lemma 8. The following statements are equivalent: (i) Z[ T I ]= Z[ f,]a s R [y]/(yymodules. (ii) 3 a n R [y]/(,l,-module automorphism K : M + M such that K(N) = N. Proof. Assume we have an R [y]/(,~,-moduleisomorphism cp

:z[Tl]+Z[fl]

then q ( M ) = M has already been observed. Furthermore we have: cp ( A

Z[ TI])= A Z[ ?I].

We take K = cpy-' Then K is a R[y]/,,+-module automorphism of M with K ( N )= fi. To prove the reverse direction we put cp = KY.Then cp : M + A is a R [ y ] / ( y ~ ~ - m ~isomorphism d~le with cpy-I(N) = fi this means: cp(AZITI])= AZ[fl].

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F.J. Grunewald

We define:

(4)

cp0(u)= h - l c p ( W

for all LJ E ZIT1]. cp,, is evidently an R[y]/(,~,-moduleisomorphism of Z[T,] and Z [ f , ] . The formula (4) even gives a recipe of how to compute cptr from an explicitly given K. We are now left with our nice R[y]/,,l,-module and two explicitly given abelian subgroups N and both of finite index in and we have to decide whether there is a R[y]/(,~,-module automorphism: ( 5 ) with

K

:M-*M

K(N)= N.

To do this and to prove Theorem C, C' we first construct a set of generators for the group of R[y]/(,l,-module automorphisms f of A?. Our module has a chosen Z-basis, this enables us to view f as a subgroup of GL,(Z). We will give finitely many matrices for t h e generators of f. Let f, be the group of 0,[y]/(,~,-module automorphisms of 0,. By our previous lemmata we know:

M=@M, i=l

and each M gis a characteristic R[y]/~,~,-submodule of M. So:

f =f

x

' ' '

x

f,

But from Step IIIa we know a finite set of generators for each individual f,. We then compute the index:

IM:(Nn

= hz

Our Z-basis of M gives us an explicit isomorphism: M/AzM

(z/A*Z)n

and a homomorphism:

f

+

GLn (Z/A2z)

Let f ( h J be the image and f[Az] be the kernel of this homomorphism. We know a set of generators of f ( A z ) in GL,,(Z/,,=) and we want to list all elements of f ( h z ) . We may assume to be in

111

7'he conjugacy problem in arithmetic groups

possession of the order ( S (A;)!) GL, (Z/A2z). We then employ:

and a multiplication table of

Lemma 9. Let G be a finite group of order m , and H c G generated by yl, . . . ys. One can find all elements of H amongst the products of length c m formed from the y l , . . . , y s .

.

Proof. To compute t h e length one adds up t h e number of occurrences of t h e generators y,. We claim that each element of H has an expression of length s m in terms of t h e generators y l , . . . , y.. This is so because any word of length > m in the y, may be shortened in G ; for among the first m + 1 initial segments of such a word, at I'east two must take the same value in G. To finish off o u r decision problem (5) put Obviously we get: of N resp. A in (Z/A2Z)n.

fl and A

for t h e images

Lemma 10. The following statements are equivalent: (i) There is an R [ ~ ] / ( ~ t , - m o dauromorphism ule

with

K1M-M

K(N)= A

(ii) There is an element

with

y Ef(Az)

y ( N ) = N.

Being in possession of all elements of f ( A z ) we may decide (ii) of this lemma. Any preimage of a y satisfying (ii) is then suitable as a K in (i). The preimage is found as a word in the generators of f. To prove theorems C, C' we still have to find a set of generators for the group of R[y]/(,+module automorphisms of Z [ f , ] which comes by our lemmata to the same as giving a set of generators for Vz(Fl) = Vz(F). By our formula (4) in Lemma 8 this comes to the same as giving a finite set of generators for the group of R[y]/(,)I module automorphisms of Id which leave N invariant. We call this group f,,CP. Again we want matrices in GL,(Z)Z P as generators for Po = {& E P &(& = lir}. By our previous argument we find all elements E l , . . .,E, E P ( h J which satisfy:

I

F.J. Grunewald

112

E(N) = N. . . . , E , is a set of preimages in f for them. The ei can be found as words in the generators of r'. Secondly we can effectively solve the word problem in f ( A 2 ) by reduction modA2. This enables us to define a set of coset representatives in terms of the generators of r' for the quotient: together with an effective coset representative function. Algorithm A, will then construct a finite set of generators vl,. . ., v. of f[A2] as words in the generators of r'. We then note that: f[Az]

C fo

and find: Lemma 11.

Po is generated

by

el,

.: . , E ~ 7, 7 1 , . . . vS.

We have proved now theorems A, C modulo the following Steps I11 and IIIa.

Step I11 Let P ( x ) be a monic, Q-irreducible polynomial with integer coefficient?, of degree m. Put:

0 = Z[X]/(P(l)) Let 1 be a natural number. Furthermore let MI, fit be two 0 [ y ] / ( , 1 , modules which are Z-free of rank n. This step decides whether MI= fi,as B[y]/~,~,-modules. If the answer is "yes" an isomorphism with respect to a given pair of Z-bases in MI, fit is exhibited. is a ring of algebraic integers in an In this situation 0 = Z[X]/(~(,), algebraic number field K = Q [ X ] / ( ~ (which ~ ) ) is the quotient field of 6. Our fixed Z-bases in MI and n"ll enable us to define K-vector spaces QMI and Q d 1 . Both QM1 and Q f i l carry an induced K[y]/(,,structure. We are going to investigate certain nice subgroups of MI resp. fit;for this we mak,e the following definition.

Definition. Let R I be a ring (commutative with 1). An R,[y]/(,l,-module where the di are M is called standard of rype { d o , .. . , nonnegative integers; if:

The conjugacy problem in arithmetic groups

113

as R,[y]/,,,-modules. We put in short M = R l { d o , . . , d I - , } .For an Rl[y]/(,~-moduleM, we put M ' = {kernel of the endomorphism y' of M}, with the convention M" = (0). We mention now the following easy characterization of standard modules:

Lemma 12. Let M be an Rl[y]/(,~-module.The following two conditions are equivalent : (i) M is standard. (ii) is finitely generated free as R,-module for 1 Q i Q 1. M'IyM1+l+~l-l

Proof. (i) j (ii) is obvious. (ii) j (i). For each i there is a free R1-submodule Fn of M, of rank r ( i ) say, such that M ' = (yM'+'+ MI-') @ F,.

In particular, M' for each i,

=

F', @ MI-', and an inductive argument shows that

M'= E@yE+,~...@y'-'F',@M'-'. Therefore by a trivial induction:

@j y'E.

M' = &I ,=I

But

y'E = 0.

,=o

y'E = (Rl[y]/(yL))'('), as Rl[y]/(,~)-modulesince E C M ' implies

Thus M is standard of type { r ( l ) ,. . . ,r(1)) as claimed. It is also clear that explicit bases for the free R1-modules M ' / ( y M ' " @ MI-') can effectively be lifted to R1-bases of the-F,'s, and thus yield an explicit isomorphism of M onto the standard module of the correct type. We now go back to our original situation. Let OK be the full ring of algebraic integers in K. Then 6 C OK and algorithm As computes a Z-basis { w l , . . . ,on}for OK together with the index: f =(OK

:61.

F.1. Grune wald

114

We have MI C Q M , and Q M , carries a K-vector space structure. Hence it makes sense to say of an abelian subgroup N of MI that it is OK-stable, i.e., OKNC N. We now make the following definition. Definition. An abelian subgroup N of MI is said to be good of level A, A being a natural number, iff: (i) ( M I : N I S A . (ii) N is OK-stable. (iii) N is y-stable. (iv) N is a standard OK[y]/(yl,-m~dule. Lemma 13. It is effectively possible to compute a natural number A, such that the set of good subgroups of level A, in MI is not empty. Proof. From linear algebra we know that any K[y]/,,~,-moduleis of standard type. By a simple algorithm based on Lemma 12 we are able to pick a K-basis inside MI for t h e standard K[y]/,,ymodule Q M , . So we have: (6)

O{dn,. . . ,dr-i} C Mi C QMi = K { d o , .. . , dr-il

if {d,,, . . . ,d r - , } is t h e type of t h e K[y]/(,~-moduleQ M , . By Gauss' method we compute the index A., of the copy of O{do,. . . ,dr-,} in MI. We then multiply our previous K-basis by A;' an.d get:

(7)

MI C B{do, . . . , dr-I} C O K { d o , . . . ,dr-I} C QMi

=

K { d o , . . ., d1-11

: MI1" . t". Here n is the Z-rank of MI Put now: A, = I O{do,.. . . , and the index 1 O{do, . . . . , d r - , }: M I ]is supposed to be computed from

(7)-

It would have been possible to define t h e notion of good submodules and to find a As without referring to t h e K[y]/(,i-module structure of QMI and an embedding (7). For this one must use Steinitz's theory in [18], [19]. This would have been more natural, but more complicated. Lemma 14. There are no more than 2(A3)"good subgroups of level A s in MI (or MI). It is effectively possible to make a list of all of them: {NI,.. . ,Ns},

each one together with an explicit OK[y]/(,~,-module isomorphism :

. . .,d f - l } .

yi : Ni + OK{do,

{ d o , .. .,d r W lis} here the type of the K[y)/(,i-moduleQMI.

77ae conjugacy problem in arithmetic groups

115

Proof. The first claim follows from the fact that for any good subgroup N of level A, in MI we have: AsM1 S N.

There order of 1 MI : A,M,( is A;. We then list all abelian subgroups of index S A,. These will be only finitely many, again less than 2(*3)".Let No be one of them, given by a Z-basis in terms of our Z-basis of MI. To test whether N o is OK stable it is enough to check whether:

wA"C No for our Z-basis of OK. The same method decides whether No is y-stable. We end up with a finite list of OK[y]-submodulesof MI of index A,. Each one is (say) given by a Z-basis and the operation of OK[y]on it is given in terms of this Z-basis: Let N be one of them. By algorithm A we compute Z-bases for t h e groups N ' , i = 0 , . . .,1. So we get a set of generators and relations for the groups

N' l y N '

+

1 +N'- 1,

The OK-module structure on them is given by a multiplication table for the w,, i = 1 , . . .,m, our Z-basis of OK. We hand these objects to algorithm Als. It decides whether the

N'/yN'+l+N' 1

are free OK-modules. If they are free we get an OK-basis for each of them. The second part of the proof of Lemma 12 shows how to put this information together to obtain an explicit OK module isomorphism y : N + O K {do,. . .,&1}. Let now { N I , .. . , Nsl}and {fil,.. . ,fi%} be the sets of good subgroups of level A, of MI and fhI.y,, are t h e OK[y]/(y~,-module isomorphisms +#

and

y : Nt

oK{dO, *

9, : fi,

OK

*.

df-I}

{do, . . . , df-l}.

We may clearly assume that the types of the K[y]/(y~)-m~dules QM1 and Q 2 , are the same, otherwise t h e answer to our question would be "no". For each i = 1, ..., s1 let: At

=

A3MI C N ,

and correspondingly for i

=

1,. . . , sz put:

116

F.J. Gnrnewald

di = A 3 M l C fii. The A, resp. di are subgroups of

Ni resp.

N,of finite (known) index.

Lemma 15. The following two statements are equivalent: (i) There is an O[y]/,,,-module isomorphism : y : M1-,MI.

(ii) The set of good subgroups of M Iof level A3 is not empty. There are i , j with 1 6 i 6 sI and 1 6 j 6 s2 and an OK[y]l(,~-module isomorphism : ~:N,+fi

with K

(Ai) = d,.

Proof. (i) j (ii). Let y : MI+ dlbe an 0-module isomorphism. Let N C MI be an OK[y]/,,~-submoduleof MI. We claim that Y :N+Y(N) is an OK-module isomorphism. For this let w E OK and u E N. We put:

oK.MI GQM,.

w = w .y(u)- y ( w u ) ~

Because

toKC 0 we get: tw = ( t w ) y ( u ) - y ( ( t w ) u ) = 0.

Hence w = O . Put now: fii

= y(Ni).

By the above remark that:

fi, is a good

subgroup of

MI. It

is then clear

~ ( A M I =) Asy(Mi) = A M I . (ii) j(i). Let define:

K

: Ni -P

fi, be the OK-module isomorphism.

We

Y ( V )= A ; ' K ( A ~ U )

for u E MI. By our assumption we have: K ( A ~ u= ) A3w

for a w E MI.Hence y defines an isomorphism of abelian groups between MI and MI. y is clearly compatible with the B[y]/,,~,-module structures on both sides.

m e conjugacy pmblem in arithmetic groups

117

Having our lists of good subgroups

{N, ,..., N,,} and {Al,..., &J in MI and dltogether with the isomorphism yi, f i we end up with the following decision problem: (8) Let A, d^ be two abelian subgroups of OK{do,. .., & I } of finite index, given in terms of a Z-basis of &{do,. . ., decide whether there is an OK[y]/(,~-moduleautomorphism: with

Y : oK{do,. . . ,di-i}+ o ~ { d o ., .. ,d1-J y ( A ) = d.

The decision procedure we shall describe will certainly exhibit such an automorphism, if there is any. We write r = ToK{do,. . . , d r - l }for the group of OK[y]/(,~-module By extending our Z-basis of O K automorphisms of M = OK{&, . . ., we think of r as a subgroup of GLn(Z). We are first going to construct a finite set of generators for r by giving certain matrices in GLn (Z). For each i between 1 and 1 we find an BK-freesubmodule F, of M of OK rank di such that: Mi = (,Mi+'

+ M i - ' )@ F,

1 s i s 1.

We put: I

n;i=@ECM i=l

We obviously have:

&!@.

n;i

..@

&-I

as OK-modules. Now let a E M. Then there is a unique set of elements Po,.. . ., PI-]E such that: (9)

Q!

=

2 Y'(P")

"=-0

under the convention y o = id,. Let f = GL,(OK) x * x GL4-,(OK).l= operates naturally on &f. Let y € f and a E M . Put then:

118

EJ. Grunewald

using expression (9). Elementary arguments show that now y is an OK[y]/cyi-m~dule automorphism of M. In fact we have defined a group monomorphism: Q1:f-r

W e shall identify f with its image in r. Any OK[y]/cyg-moduleautomorphism y of M induces an OK-module automorphism of each: Mi/(yM'+l+M1-l)

Let C be the subgroup of on all of these quotients. Furthermore put:

r which

induces the identity automorphism

I

fi = @ Homo, (E,(yM"' + MI-')). ,=I

We now define a map rpz of fi t o C. For this let f E HomoK(E,(,MI+' + MI-')) and a E M, we put: rpz(f)(a)

=

3 Y '(P.)+ 3 Y

"-0

"-0

' ( f ( P Y ) )

using the expansion (9) for a. By f ( & ) we ith projection of Py E lii. This means rpz(f) O n e easily sees that the so defined rpzcf) is automorphism of M. If 1 > 1 then rpz is not But we have:

understand f applied to the centralizes all F, with i # j. an OK[y]/cy~-module a group homomorphism.

Lemma 16. (i) rpz is a surjective map from H to C. (ii) Let h , , . . . , h, be a set of generators for the abelian group H, then the rp2(hl),. . . ,rpz(hS)generate C. (iii) is the semidirect product of C by f i f f is assumed to be contained in r via rp,.

r

Proof. (i) Any OK[y]/cyi)-moduleautomorphism of M is determined by its values on A? C M ; if y E C then for P E E ,

v(P)=P+At(P) with A, ( P ) E (yM'+' + MI-'). A, is clearly in Hom, (F, (yM'+' + M'-')), and y = rp2(AI,. . . , A , ) . of

(ii) Let AUto,(M'/Me-k-f)be the group of OK-modules automorphisms M ' / M ~ - kfor - ~ i = 2,. . . , I ; k = 1,. . . , i - 1. Each element of C


119

induces an element of this group. One easily sees that the composition map: C + AutoK( M i / , w - k - l ) Homo,(E, M i - k )

4

Homo, (E,(yMi+'+ Mi-')) for each k = 1,.. . i - 1 is a group homomorphism. (i) together with an

obvious induction under the assumption that the automorphism in question already centralizes all F, with j > i, finishes then (ii). (iii) is obvious.

Both E and (yMi+'+ M i - ' ) are free OK-modules. It is then easy to describe a finite set of generators for the group fi effectively. By algorithm AI9we may construct a finite set of generators for f. The maps cpl, cp2 are then explicit enough to deduce together with Lemma 16 a set of generating matrices for f in GL,(Z). We now go back to our decision problem (8): The first thing is to determine the index: As=(OK{do7...,dr-1}:(dnd)l.

Our Z-basis of OK{ d o , .. . ,dl-l}gives us an explicit reduction isomorphism: OK{& .. .

9

df-l}/AlO,( ~ , . . . d. ,.- , )

2,(z/AsZ)"

of abelian groups. Let T(A,) the image of f in GL,(Z/,,,) via this homomorphism. Because we know a finite set of generators for f we know a set of generators for f ( A , ) . We may assume to be in possession of a multiplication table for the finite group GL, (Z/+). The order of this group is certainly less than (A:)! So by an_ argument used resp. d be the images before we may list all elements of f ( h s ) . Let We then obviously have: of A or d^ in (Z/AsZ)n.

a

Lemma 17. The following statements are equivalent: (i) There is an OK[ y ] / , , ~ - m o d u automorphism fe : ~ ~ K ~ d O 7 ~ ~ ~ 7 ~ f - l } ~ ~ K { d O ~ ~ ~ ~ ~ d f - l }

with

y ( A ) = d^

(ii) There is an element K

E f(A5)

-

with K

(a)= A.

C GL. (UA+)

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F.J. Grunewald

By the above remarks we know all the elements of f ( A , ) . There are not many things which are then easier than to decide the second condition of Lemma 17. Having found a K satisfying this condition, any preimage of K in f will then serve as y. y may be found as some expression in t h e generators of r. Step IIIa Let P ( x ) be a monic, Q-irreducible polynomial of degree m with integer coefficients. Put:

6 = Z[X]/,P,,)) Let 1 be a natural number. Let M i be an O[y]/,,l,-module which is Z-free of rank n, with a fixed Z-basis. Let f l be the group of B[y]/,,l,-module automorphisms of M I . The Z-basis in MI allows us to view f I as a subgroup of GL. (Z). Step IIIa gives a finite set of generators for f lC GL. (Z). We use t h e terminology of Step 111. We first compute a natural number A ? such that t h e set of good subgroups of level A 3 in M2 is not empty. We then compute a list: {Nl, . . . ,Ns,}

o f all of them, together with explicit OK[y ]/,,l,-isomorphisms: i = 1,. . . ,S I

y, : N, +-6K{d,,,. . . , d1-l)

where {d,,,. . . ,d , - i } ,is the K-standard type of Q M I . Put again: A, = AXMi C N , . By the method described in Step 111 we decide for each pair: (Ni,N , )

2 s i =ssI

whether there is an B[y]/,,~)-moduleautomorphism: with

E,

: M i + Mi

E,

: (NI) = N , .

Step I11 then gives us a finite list of { E ,,..., E , } , to get:

Lemma 18. {E

fl

E fl

is generated by the

E,’S

.. . , E , }

{ E ~ ,

I E ( N I ) = Nil.

Proof. This follows from Lemma 15.

which we renumber to

together with :

f i e conjugacy problem in nrirhmeric groups

121

Using our isomorphism: 71:

NI4

. . df}

OK{&,.

7

We end up with the problem: Give a set of generators for t h e group:

1

Tz = { E E r & ( A l )= A l }

(10)

where r is the group of OK module automorphisms of and A , is a subgroup of finite index in OK{&, . . . , O K { & , . . ., At t h e end of Step I11 we showed how to give explicitly a finite set of generators for r. Knowing a set of generators for r2on a Z-basis of we could rewrite the corresponding O[y]/,,~,-module OK{ d o , .. . , automorphisms of M I by the method described in the proof of Lemma 15. and To deal with problem (10) we again put M = OK{&, . . . , compute:

I M : A l l = h6

-

We put r[h6] = k e r { r GL, (Z/+)}. This homomorphism is defined by use of our Z-basis in M.The was called r(h6). Knowing all elements of image of r in GL,(Z/,) r(h6) we give a set of coset representatives for:

rlmi

in r. Because we are able to solve the word problem in r/r[b] effectively by reduction modh6 we have an effective coset representative function required by algorithm As. This algorithm then turns our set of generators for r into a set of generators { T ~. ,. . ,q v } for r[h6]. Next let Al be the image of A l in (Z/-)". Knowing all elements of r(h6) we make a list of: {E

E r(A6)

1

E

(51) = 31)

This is a finite list with no more than (A:)! members. Let l, . ., . ,5. be a set of preimages in of all elements of this set. We are able to find them as expressions in terms of the generators of r. Any element y of r [ & ] now satisfies:

r

r(Ai)=Ai and we obviously get:

Lemma 19.

r2is generated

by

771,.

. . ,vU, 51,. . .-5..

EJ. Grunewald

122

This finishes the problem of Step IIIa.

Proof of Theorems A‘, C’ A glance through Steps I and I1 shows that everything stays correct if one replaces Z by OK.One has to keep in mind that we consider OK to be given by a Z-basis together with a multiplication table for this Z-basis. Steps I11 and IIIa may then be used as stated to obtain a proof of Theorems A’ and C’. Proof of Theorems B, B’ We give our matrices T, f to the algorithm described in Theorem A and decide whether there is an X E GL. (Z) such that:

XTX-’

=

?

If there is none the answer to our question is “no” too. If the answer is “yes” we construct such an X by the same algorithm and compute the determinant of X . If det X = 1 then we are in good shape. Otherwise we find the following two equivalent statements: (i) 3Y E SL.(Z) with Y T Y - ’ = f. (ii) 3Y E Cez(T)with det Y = - 1. Then we construct a set of generators ll,. . . , Ls for the group Wz(T) by Theorem C and look whether:

det (6,) = - 1 for a generator 5;. The answer to our original question is then “yes” if and only if det (&) = - 1 for at least one i. The proof of Theorem B’ is slightly more involved. We again assume to have found an X E GL. (OK)with:

XTX-‘=

f

by Theorem A’. Put u = det X .

We then construct a set of generators and put u, = det

6,

{ 3 ] ,. . . ,& } of the group W,(T),

for i = 1,. . . ,s

and V = ( u l , . . . , u s )C 0:. We then get the following two equivalent statements: (i) Y E SL.(OK) with Y T Y - ’ = f. (ii) u E V. To decide the second condition we express u and ul,. . . , us in terms of

f i e conjugacy problem in arithmetic groups

123

our set of generators for 0: by the second part of algorithm Ae. Having done this we apply Gauss' method. Proof of Corollaries A, B Take two preimages To,foof T and

a

= det

To;

ci

= det

f

in GL,(K) and put:

fo

We factorize then the polynomial:

over K. By this method we find all zeroes b,, . . . , b, of this polynomial in K. For each zero put:

6, = and

(; ;#)

T,=T6,

...

i

=

1, ..., r

i = l , ..., r.

Then we find the following equivalent statements: (i) 3 X E P GL. ( O K )with XTX-' = f in P GL, (K). (ii) 31 S i S r and X E GL, (OK)with XT,X-' = f0. Similarly: (i) 3 X E P SL, ( O K )with XTX-' = f in P GL, (K). (ii) 31 S i S r and X E SL, (OK)with X T X - ' = fo. The second conditions may then be decided by applications of the algorithm from Theorems A', B'. B. Some algorithms

Let K be a finite extension field over the rational numbers Q. We consider K to be known if we are given a Q-irreducible polynomial P ( x ) of degree m with rational coefficients and are told that K is generated as a field by a zero of P ( x ) , i.e.,

K = C?[x]l(P(x)) as fields. By a suitable transformation x+Ax

AEZ

we may assume that P ( x ) has integer coefficients and highest coefficient 1. Hence all zeroes of P ( x ) are algebraic integers. We write

EJ. Grunewald

124

OK for the full ring of algebraic integers in K . We consider OK to be } 0, in terms of the field known if we have a Z-basis { w , , . . . , u r nfor generators I , x , . . . , x ' " - ' , i.e., w , = COl+ w, = Corn

CllX

+

+

* * *

+ c,,x +

*.

a

G - I . 1 .

+ Crn-l.,

Xm-I

* Xm-l

with c,- E Q. One of the following algorithms will compute such a 2-basis for OK. A lot of t h e following algorithms are taken from Steinitz's papers [19], [20]. We often have only written out why t h e reasoning of [19], [20] is effective. For the algorithms given in section A of this paper we need only A,,-A6 and AI7, Aln,A,, of the algorithms listed below. Algorithm A,, Let K be an algebraic number field and let P ( x ) E K [ x ] be a polynomial with coefficients in K. Algorithm A, splits P ( x ) into irreducible factors:

P ( x )= Pl(X)

-

* * *

*

P,(x).

P, ( x ) are K-irreducible for i = 1,. . . ,r. If P ( x ) is monic with coefficients in t h e ring of algebraic integers OK of K , then each P i ( x ) will be monic with coefficients in OK. For this w e use Kronecker's algorithm as described in [ 9 ] . Algorithm A , Let K be an algebraic number field. Given 1 polynomials PI,.. . , PI E K [ x ] , this algorithm decides whether P I , .. . , P I are coprime and if the answer is "yes" it produces polynomials Q,, . . . , QI E K [ x ] such that:

OIP, + . . . + QIPI = 1. If t h e polynomials P I , . . , PI have coefficients in t h e ring of algebraic integers OK of K and if they are coprime, algorithm A l will compute t h e polynomials Q,, . , . , Ql with algebraic integer coefficients together with an element A E OK such that

OlP,+ . . . + QIPI = A. A repeated application of Euclid's algorithm in K [ x ] will do t h e job of A,.

7 h e conjugacy problem in arithmetic groups

125

Algorithm A2 Let K be an algebraic number field and T E GL, (K) for a natural number K. Algorithm A z computes the characteristic polynomial char(T) of T together with the minimal polynomial min(T). Knowing the characteristic polynomial char(T) we factorize it by algorithm Ao. In the list of all divisors of char(T) we then find the polynomial of minimal degree which is satisfied by T. Algorithm A, Let K be an algebraic number field. Let T E GL, (K). Algorithm A, computes a semisimple matrix S and a nilpotent matrix U both with coefficients in K such that:

and

T=S+U

su= us

We first compute t h e minimal polynomial min(T) of T by algorithm A2. By successive factorisation of min(T) we construct a multiplication table for the normal extension field K O of K which is generated by all zeroes of min(T). By an algorithm of linear algebra we find y E GL, ( K O )such that yTy-' = J, where J E GL,(Ko) is in Jordan normal form. How to decompose: with

J = S o + Uo

UoSo= UOSO

where Uo is nilpotent and SO is semisimple is then evident. We put:

S

= y-'S0y,

u = y-'u,y.

A Galois-type argument then shows that both

S, U E M. (K). For all this see [4] Chapter 7, 05. Algorithm A, Let M = Z" be a free Z-module with basis. Let A be an n x n matrix with integer entries representing an endomorphism with respect to this basis. Algorithm Aq computes a Z-basis for the kernel of A.

126

EJ. Grunewald

By elementary row and column operations (over the euclidian ring Z!) we find element P, Q E SL,(Z) such that

PAQ = D where D is a diagonal matrix. Algorithm A, (Reidemeister-Schreier) Given a finite set of generators g,, . . . ,g, for a group G and a subgroup H S G of finite index together with a set of coset representatives Y and an effective coset representative function

@:G+Y As computes a set of generators for H. We use the method described in [6] Lemma 7.2.2. Algorithm A6 Let K be an algebraic number field with integeral generator 19. Let 6 be the subring of K generated by 4, this is the same as the Zsubmodule of K generated by 1,6,6',. ..,I?,-'where rn is the degree of the minimal polynomial of 6. This algorithm computes a Z-basis WI

=

Cll6

COl+

w, = Corn

+

* .

+ CI,6 +

.+ C m 1 6 m - l * * *

+ Cmm6m-l

Ci.1

EQ

for the ring of algebraic integers OK of K. Furthermore the algorithm gives the index Let

1 6 K : 6I = t.

Tr:K+Q

be the trace of the field K. The trace of an element a of K is the trace of the linear automorphism, which is induced by multiplication with a. It ,can therefore be computed, for example, with our field basis 1,6,.. ., a"-'. Let

BK( a ,p ) = Tr(a * p ) be the trace form induced on K. If we first compute the matrix of BK with respect to 1,6,.. . ,a"-', we are able to give a complementary basis d l , .. . , 4, for 1,6,. . . , $?"-I with respect to BK. Let d be the Z-module generated by d,,. . .,4,. Then by [ 5 ] Chapter 11, 02, Theorem 6:

OK

d


127

Next we list a set of coset representatives a I ,. . .,a,for 6/O. Let a,,. . . ,a, be those which are integral. This can.be checked by computing the characteristic polynomials for the linear maps corresponding to the 0's on K. Because t h e existence of an algebraic integer in a coset b/O would imply that any element of the coset is integral, we find OK

=(o,al,...,a,)

We then compute by Gauss' method a basis for OK and the number t. Algorithm A, Let K be an algebraic number field with ring of algebraic integers OK. Let n be a natural number. This algorithm lists a set of elements AI,. . .,A, E OK with such that for any A E OK with

1 OK : A O K 1 c ~t we can find a unit u E OK such that

A =A~u for some i E (1,. . . ,r } . Here we use the algorithm described in the book of Borevich-Shafarevich [ 5 ] Chapter 11, 55.4. Algorithm As Let K be an algebraic number field with ring of algebraic integers OK and let OE be the unit group of OK. (i) The first part of this algorithm gives a set of generators:

5,u1,. . . , us for the unit group of OK. ul,... , u s are a set of fundamental units and 4' is a root of unity. (ii) Given a unit u E 0: in terms of our Z-basis of O K : u = alwl

+ - + a,w, * *

ai E Z

the second part finds f , e l , .. . ,e, E Z with: u = 5fu;1.. . u >

(i): Let 9 be a generator of K over Q ; we first compute t h e discriminant d of K , for example by obtaining a Z-basis for OK. By [7]

128

EJ. Grunewald

928.1 the order of 5 is a divisor of 2 )d 1. By algorithm A. we then decompose the polynomial: X2ldl - 1

in K into irreducible factors. By this we find generators for the torsion group of 0:. T o find a set of fundamental units { u l , . . . , u s }we employ the algorithm described in [ 5 ] Chapter 11, 05. 3. (ii): For the second part of this algorithm we have to explain Dirichlet's method in more detail. For a description of this method see [ 5 ] Chapter 11, 05. We want to prove: Let u E 0: be a unit given in terms of t h e chosen Z-basis of OK. There is an effectively computable number c E N such that

(*I

u E { 5 c w ; ~ *. .

. . u > Jle, I s c, Vi = 0 , . . . , s }

where 5, u , , . . . ,us is t h e set of generators computed in (i). For this let al, . . .,a,, be the distinct embeddings of K into R and T ~. ., ., T , ~ , f l , .. . , f,zbe the embeddings of k into C which do not map K entirely into R. Put:

L : K,{O)-, R'I+'~ L : a +(In l a l ( a ) l , .. . ,In lurl(a)l,lnl ~ ~ ( a ). l. .,In ~ , Irr2(a)I2) L is a homomorphism of the multiplicative group K,{O} into t h e additive group of R'l+'z. The image of 0: under L is a discrete lattice of dimension rl + r2 - 1 in R'l+'z. The kernel of L is exactly the group of roots of unity contained in 0:. By the proofs of Lemma 1 and Theorem 2 of (51 Chapter 11, 05 it is clear how to find effectively a rational number c I# 0 such that:

where:

B ( C J ~~ ( 0 ; =) (0) B ( c ) = {x E R I " .

IIlx I1 = +d/xxIcc }

is the (rI + r2)-dimensional euclidian ball of radius c. By approximating all zeroes of a minimal polynomial of a field generator of K in C up to the first decimal of their real and imaginary parts, we find for our explicitly given unit u E 0: a rational number c2 such that u €5 B ( C 2 ) .

In (*) we may then clearly put:

(

o =max 2 ) d ) ,c2+ 1) CI

m e conjugacy problem in arithmetic groups

129

To finish this algorithm we spread out each element of the finite set mentioned in (*) in terms of our Z-basis of OK and look which element is equal to u.

Algorithm As Let K be an algebraic number field, with ring of algebraic integers OK. Given r ideals al, . . . ,a, by Z-bases, which are pairwise coprime 1.e.: (ai,ai) = OK

for i # j

and elements e l , .. . ,8, E OK. We give here an algorithm for determining an x E OK such that x=8,modal x

= 8, moda,.

We first determine a Z-basis for the ideal a = a1fl .

- . fl a,.

This has as a consequence, that we explicitly know the order of OK/a and a multiplication table for the ring: OK//8.

In this finite ring we check for one element after the other whether it satisfies the corresponding conditions. The Chinese Remainder Theorem makes sure that we find an f in 6 / a satisfying our congruences. As x we may take any preimage of f.

Algorithm Alo Let K be an algebraic number field, with ring of algebraic integers OK. Let a be an ideal of OK given by a set of ideal generators or, which comes to the same (because we know a Z-basis of O K ) , by a Z-basis. Here we compute Z-bases for prime ideals pl, . . . ,p, and natural numbers e l , . . .,e, such that: a = p71. . . . pfr

.

By Gauss’ method we know the order of the ring 4 / a . We are also , .We are given a multiplication table by reduction of our table for 6 then able to decompose:

F.J. Grunewald

130

O K I a = R1 X . . . X R,

effectively into a direct product of Artin local rings R,. Let p, be t h e maximal ideal of R, and e, be t h e least natural number such that p? = (0). Let: ~ p ,: O K

+

R,

be the projection homomorphism, which we may consider to be known explicitly. Put pr = Cp;l(pt). By algorithm A5 we are able to compute a set of generators for the abelian groups p,. By Gauss’ method we reduce these sets to Z-bases. Algorithm A l l Let K be an algebraic number field, with ring of algebraic integers OK. Let a, b be two ideals given by Z-bases. We decide here whether there is a c E K (cZ0) with c a = b.

If the answer is “yes” we determine such a c. In particular we decide whether ideals are principal, and in case they are, we give generators for them. We determine first a A E Z with: AbCa. A c satisfying c a = b would then satisfy:

Aca= AbCa.

Hence we get Ac E OK. On t h e other hand putting: A o = la:Ab/

we get:

I

OK

I

: ACOK = A ~ .

We now take the list of nonassociate elements yl, . . . ,ys with )OK:Y,OKI=AO

computed by algorithm A, and check whether one of the numbers A-ly,, i = 1,. . . , s, satisfies: A-’y,a = b.


131

Algorithm A12 Let K be an algebraic number field, with ring of algebraic integers OK.Let K be the class number of OK and JK the ideal class group of OK. This algorithm computes K together with ideals a,, . . . ,a, which lie in distinct classes. Furthermore a multiplication table for JK is given. Let m be the degree of K and D be the absolute discriminant of OK. From the proof of [ 5 ] Theorem 3 of Chapter 11, 96 it follows that each ideal class of OK contains an ideal a with

We then list all subgroups of OK of index s c . There are not more than 2(Cd)of them. For each such subgroup we check whether w,NCN

i = l , ..., m

for our Z-basis of OK. This leaves U S with a list of all ideals in OK with index at most c. This gives a first upper bound for K . By algorithm A l l we deduce from our list a set of inequivalent ideals together with a multiplication table for JK. Algorithm A1, Let K be an algebraic number field, with ring of algebraic integers OK. Let a, b be two ideals of OK. This algorithm finds a number c E K such that c b C OK and c b is coprime to a. In other words, given an ideal class of OK and an ideal a we want to find an element of 6 which is prime to a. We first split a by algorithm Alo into prime factors: a = p;i.

...

- pf,

Then we determine by algorithm A,, an ideal bo in the ideal class inverse to the class of b. Again by application of algorithm A,,, we find: bo = p{i . . p{rq;i . . . q3 Here t h e f, might be zero. We then find elements: 0, E OK for i and with

8, E

$8;

8,

pfs+l

1,.

for i = 1 , . . . , r

vg€OK

f o r i = I , ..., s

7,Eq:

for i

=

=

1,..., s.

. . ,r with:

132

E l . Grunewald

Then we determine by algorithm As a solution of the congruences: z

= O1 mod

z

= 8, modp!." = v 1modq?

z

z = 9, modq:.

The ideal z OK is clearly divisible by bo hence we find an ideal bl such that:

z * OK

= bob1

bl may be determined by: b i = { v E OK IvboCZ ' O K ) .

But bl then lies in the ideal class of b and by the choice of our congruences it is prime to a. Algorithm AI4 Let K be an algebraic number field, with ring of algebraic integers OK. Let A = (a,,)be an n x m matrix with coefficients from O K such that:

and

I

(a,, 1 d i d n ; 1 c j s m ) = O K

rgKA 3 2

Algorithm A14determines a vector x = (xI,.. .,x.) with entries in OK such that: with

XA = y = (yi,. . ., y m ) (yli.. ., y m > = OK.

The rank condition for A allows us to assume that: d = a l l a 2 2 -uZ1u12

is not zero. We first compute a Z-basis for the ideal d . OK and decompose it by algorithm A; dOK = p;' ' . ' p:',

For each ideal p. we find an entry a , , , , , , , ,such that: a I(v).l(u)

P..


133

By algorithm As we solve the congruences: if p = i ( v )

else

mod p.

for p = 1,. . ., n and v = 1 , . . . , r . NOWput: x ’ = ( x : ,..., x:)

and y ‘ = x ’ A = ( y I ,..., y A ) .

a22and azl not being simultaneously 0 (because of (11)) we may furthermore assume that:

y;zo. We apply algorithm Alo again to find:

y;OK = pi,. . . p!rq;, . . . q:: where the fi might be zero, but where none of the q, is a p.. We now find an a E OK with: (12)

v = 1, ..., s v = l , ..., r

a =lmodq, a=Omodp,

It then follows that: ad$Omodq, a d =Omodp,

v = 1, ..., s v = 1,..., r.

Then we find a number b E OK such that:

y { + abd

+ 0 mod q.

v = 1,. . . , s

This is possible because of (12). It is then obvious that the ideal

( y : + abd, y 4)

has only prime divisors from the set {pl,. . . ,p,}. Now put: c r = ba2* and

r = ba2,.

We make the further definitions: x,=x~+aa, for v

= 3,. . . , n

x2=x;+7a,

x,=x:

and

y = xA = ( ~ 1 , . . . , y m ) .

Because of our choice of a the congruences (11) are still valid for the Xi.

EJ. Gruncwold

134

Hence n o p. for v = 1 , . . . ,r divides (contains) the ideals ( y l , . . .,ym). On the other hand an easy computation shows: yl = y l + abd

and

y2 = y;,

so by the above every prime division of (yl, . . . , y m ) is amongst the PI,. . . ,p.. Thus ( y i , . . . ,ym)= OK. Algorithm A1, Let K be an algebraic number field, with ring of algebraic integers OK. Let A = (ali) be an n x rn matrix with coefficients in OK such that r g K A G n - 2.

:. ,x,)

Here we find a vector x and

(xi,. . .,x n )

=

= (xl,.

with entries in OK such that

OK

X A = 0.

We first find two linearly independent solutions with entries in OK: Put:

a = (a,. . .,a")

p r = (Pi,.. . ,PA)

( p ' )= (pi,. . ., P 3 By algorithm AI3we find in the ideal class of ( p ' ) an ideal (a) = ( a l , .. . ,a.)

and

coprime to (a).In other words we find a c E K such that: CP

= : P I , . . . ,C P :=:

P n

a which is

EO K

and (a)is coprime to ( p ) where:

P

= (PI,. . ., P n ) .

By algorithm AI4we then find yl, y2 E OK such that:

satisfies: (x,, . . . ,x,)= OK. Algorithm Al, Let K be an algebraic number field with ring of algebraic integers OK. Let x = (xI1,x I 2 , .. . ,xln) be a vector with entries in OK such that: (xll,.

. ., X I " )

=

OK,


135

Algorithm AI6 finds a matrix

with det X = 1. We assume we have constructed the matrix X up to the rth row

with r < n - 2 under the hypotheses that: (619

* . ., 6s) = O K

where 61,.. . ,6, are the r x r subdeterminants. Here s = (:). There is a n x (,:J matrix A ( r ) whose entries are the 6, with appropriate signs attached, such that the column:

( r l , .. . ,z.). A ( r )

has the ( r + 1)X ( r + 1) subdeterminants of

as entries. The rank of A is n - r. As long as r S (n - 2) we may therefore employ algorithm A I 4to find a ( r + 1)th row such that the ( r + 1)x ( r + 1) subdeterminants still generate t h e unit ideal. For the linear algebra involved see [4] Chapter 3. By this method we find X up to the last row. Let S , , . . . , be t h e (n - 1) x (n - 1) subdeterminants of the so found ( n - 1)X n matrix. We know that:

(61,. . .

9

&+I)

=

OK-

By writing out a basis for the abelian subgroup generated by oi6, ( i = 1,. . . ,m ; j = 1,. . . , n ) inside OK we find a solution of the equation: xn161 + * ' * + XnnS. = 1 i = 1,. . . ,fl xni E O K Algorithm AI7 Let K be an algebraic number field with ring of integers OK.Let -~

136

F.J. Grunewald

ul,. . .,anE 62 for natural numbers n, m. This algorithm decides . is free. If the whether the.OK-module M generated by the u l r..,a,, answer is “yes” an OK-module basis is computed in terms of the (TI,.

. .,an.

Let r be the K-dimension of the vector space generated by the uI,. . . , a nBy . successive application of t h e algorithms Al, and AI6 we effectively find a set of generators: for M

rl,.. . ,T , + ~

in terms of the u I , . .,a,,. We then form the matrix: all,.

x=(

i

. . ,a1.n i

)

. . ,a r + 1 . m ~. ., . ,T , + ~ X . has rank r. Let * X be the left

a+1,1,.

whose rows are the T adjuncted matrix to X. Its entries are certain r x r subdeterminants of X , with appropriate signs attached. * X has ( r + 1) columns and many rows and satisfies:

*XX = 0 X having rank r. X has nonzero entries. For each row (XI,.. .,x,!+~)of * X we test by algorithm All whether is a principal ideal.

(x I, . . . ,

Let the answer be “yes” for the row (xi,. . .,x;+,) and put: cxi = x :

for i = 1 , ..., r + l

for a generator c found by algorithm A l l . Then: and

(XI, * * .

9

-&+I)

=

OK

x,71+ . . . + xr+17;+1= 0

This follows from the corresponding properties of the xi. By algorithm Alh we may then effectively find a set of generators for M consisting of r vectors. Then M is free and we have found an OK-basisin terms of the ul,. . . ,an. Suppose now that n o row of * X generates a principal ideal. We claim that then M will not be free. For this see [19] Theorem 27. Algorithm Al, Let K be an algebraic number field with ring of algebraic integers OK. Let B be an explicitly given finitely generated abelian group


137

equipped with an OK-modulestructure, which is given on a set of generators of B. Algorithm Ale decides whether B is a free OKmodule. If the answer is “yes” an OK-modulebasis is computed. We first test whether B is a free Z-module by Gauss’ method. This condition is clearly necessary for B to be a free OK-module. B being a free Z-module we pick a Z-basis. Then we have: B = Z” C Q“ = Q B and Q B carries a K vector space structure. We pick then in Q B a K-basis such that: B C OZC K‘ = Q B . Then we express our Z-basis for B as vectors in OZ and use algorithm An.

Algorithm AI9 Let K be an algebraic number field with ring of algebraic integers OK.Let n be a natural number. This algorithm lists a finite set of generators for GL. (OK). If for a given n we would have computed a finite set of generators y l , . . . ,y, for SL. (OK) we would get a finite set of generators for G L . ( b ) by adding matrices

where u runs through a set of generators for the unit group of OK which is computed by algorithm As. It suffices then to compute a set of generators for SL.(OK): (1) n = 2, OK has infinitely many units. The main result of Wasserstein’s paper [25] says that SL(OK)is then generated by

G

-;) (; y),.-.,(; ”;) 9

where wI,. . .,wm is a Z-basis of OK. (2) n = 2, OK is imaginary quadratic. In this case Swan’s paper [21] even gives a presentation for !%(OK). It can be checked that the constructions of [21] are all effective. See the remarks in section 8 of this paper.

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F.J. Grunewald

These two cases exhaust SL2 because of Dirichlet’s theorem o n units (see [51). (3) n 2 3 . By [O] and [I51 Corollary 16.3 it follows that SLn(O;O is generated by elementary matrices. It is quite easy to extract a system of generators for SLn(OK)n 3 2 from the paper [lo] of A. Hurwitz. The method used there can also be seen to be effective. A feature of all these methods is that one can give an a priori bound for the number of generators needed for SLn(OK)in terms of n, the discriminant of OK,the degree of K, and the number of complex valuations of OK. References [O] H. Bass, J. Milnor and J.-P. Serre, Solution of the congruence subgroup problem for

SL, ( n 3 3) and Sp,. (n 2 2), Publ. Math., IHES 33 (1%7). (11 W.W. Boone, F.B. Cannonito and R.C. Lyndon, Word problems (North-Holland Publishing Comp., Amsterdam, London, 1973). [2] A. Borel, et al., Seminar on algebraic groups and related finite groups, Lecture Notes in Mathematics 131 (Springer Verlag, Berlin-Heidelberg-New York, 1%9). [3] A. Borel and Harish-Chandra, Arithmetic subgroups of algebraic groups, Annals of Mathematics Vol. 75, No. 3 (1962). [3a] A. Borel, Arithmetic properties of algebraic groups, Proc. Int. Congr. Math., Stockholm (1962). [4] N. Bourbaki, Algbbre, Chap. 3 and Chap. 7 (Hermann, Paris, Deuxibme Edition, 1958). [5] Z.I. Borevich and I.R. Shafarevich, Zahlentheorie (Birkhauser Verlag, Base1 u. Stuttgart, 1966). [6] M. Hall, The theory of finite groups (Macmillan Comp., 1973). [7] H. Hasse, Zahlentheorie (Akademie-Verlag, Berlin, 1%9). (81 E. Hecke, Vorlesungen iiber die Theorie der algebraischen Zahlen (Akademische Verlagsgesellschaft, Leipzig, 1923). [9] G. Hermann, Die Frage der endlich vielen Schritte in der Theorie der Polynomideale, Math. Annalen 95 (1926) 736788. [lo] A. Hunvitz, Die unimodularen Substitutionen in einem algebraischen Zahlkorper. Nachrichten von der k. Gesellsc.haft der Wissenschaften zu Gottingen (1895). [ll) R.D. Hurwitz, On the conjugacy problem in a free product with commuting subgroups, Math. Annalen, Vol. 221 (1976). [ 121 W. Magnus, Noneuclidian tessalations and their groups (Academic Press, 1974). [13] W. Magnus, A. Karass and D. Solitar, Combinatorial group theory (Interscience Publishers, 1966). [14] Charles Miller, 111, On group-theoretic decision problems and their classification, Annals of Math. Studies, No. 68, Princeton University Press (1971). [15] J. Milnor, Introduction to algebraic K-theory, Annals of Math. Studies, No. 72, Princeton University Press (1971). [16] M. Newman, Integral matrices (Academic Press, 1972). [17] I. Schur, Uber Gruppen linearer Substitutionen mit Koeffizienten aus einem algebraischen Zahlkorper, Math. Annalen, Vol. 71 (1912).

fie conjugacy problem in arithmetic groups

139

[ 1x1 A. Seidenberg, Constructions in algebra, Transactions of the Am. Math. Soc.,Vol. 197 (1974).

[ 19) E. Steinitz, Rechteckige Systeme und Moduln in algebraischen Zahlkorpern I, Math. Annalen, Vol. 71 (1912). 1201 E. Steinitz, Rechteckige Systeme und Moduln in algebraischen Zahlkorpern 11, Math. Annalen, Vol. 72 (1912). 1211 R.G. Swan, Generators and relations for certain special linear groups, Advances in Math. 6 (1971) 1-77. 1221 0. Tausky, Matrices of rational integers, Bull. Am. Math. Soc. 66 (1960) 327-345. 1231 B.L. van der Waerden, Eine Bemerkung uber die Unzerlegbarkeit von Polynomen, Math. Annalen, Vol. 102 (1930). 1241 B.L. van der Waerden, Moderne algebra I + 11, 2nd edition (Springer Verlag, Berlin, 1937). [2S] L.N. Wasserstein. On the group SL, over Dedekind domains of arithmetic type (Russian), Mat. Sbornik, Vol. 89 (131) No. 2(10) (1972). 1261 A. Weil. Adkles and algebraic groups, Inst. Adv. Study (1961) (notes polycopies). 127) M.J. Wicks, Presentations of some classical groups, Bull. Aust. Math. Soc., Vol. 13 (1975).


ALGEBRAICALLY CLOSED GROUPS: EMBEDDINGS AND CENTRALIZERS Kenneth HICKIN* and Angus MACINTYRE** Michigan State University, East Lansing, MI 48824, USA Yale University, New Haven,

W W ,USA

90. Introduction

Here we present several results on t h e algebraic structure of a.c. groups. We refer t h e reader to [2: Chapter IV, §8], [3], and [6] for discussions of t h e basic properties of a.c. groups. Theorem 1. No a.c. group can be embedded into a finitely generated subgroup of itself. This suggests the following problem: Is there an o-homogeneous group (in the sense of Jonsson) which can be embedded into a f.g. subgroup of itself? If A and B are groups, A C B means that A is a subgroup of B. Theorem 2. Suppose G is a countable a.c. group. There is a set % of 2-1 a.c. groups, each of power w I and m-o-equivalent to G, with the property: each U E % has a subgroup U,,= G such that, for all U # V E % and for all countable B and D with U,,C B C U and V,C D c V, neither of the centralizers CU(B)and Cv(D) is embeddablp in the other. Hence distinct members of 91 are not isomorphic. If the C.H.is assumed, then we can arrange that distinct members of % are mutually non-embeddable. In [ 5 ] Shelah and Ziegler give a complete solution to the spectrum problem for a.c. grdups using stability theory. Our proof of Theorem 2 The first author gratefully acknowledges the support and splendid hospitality of the Wanvick Symposium on Infinite Groups and Group Rings during the preparation of this paper in May of 1978 at the University of Wanvick. ** Both authors were aided in this research by NSF grant MCS77-07731. 141

142

K. Hickin, A.

Maclntyre

shows that, at least in the case of wl, extremely elementary techniques involving centralizers suffice to construct the maximum number of CQ-wequivalent a.c. groups which are not isomorphic for obvious reasons. In the remainder of this section G denotes an arbitrary a.c. group. MIN denotes the minimum condition on subgroups.

Theorem 3. Suppose A C G satisfies MIN and A is the intersection of finitely many f g . subgroups of G. Then, for every M with CG( A )C M C G , there exists H C A such that

(the normalizer of H in G ) .

CG( H )C M C NG ( H )

Zn particular, if A is finite, then there are only finitely many subgroups of G containing the centralizer of A. Corollary. If A C G is finite and characteristically simple, then NG ( A ) is a maximal proper subgroup of G.

Proof. Suppose NG ( A )C M C G. By Theorem 3, CG (H)C M C NG (H) for some H C A. Since G is a.c., every automorphism of A is induced by an element of N G ( A ) . Hence, H is characteristic in A and either H = l , M = G o r H = A , M=NG(A). A more complete characterization of the maximal subgroups of countable a.c. groups will be given in a future paper by the first author. If P C G and 9 acts on P, we define N Z ( P ) = { x E NG ( P ) x induces a 9-automorphism on P}. Aut (P) is the automorphism group of P. Further information on the subgroup lattice between C G ( A )and G is given by

I

Theorem 4. Suppose B,C G , 1 C i c n, are fig. and B,f l . . r l B, has MIN. Suppose 23,C Aut ( B , ) , 1 C i d n, and let K = the largest subgroup * n B, which is inoariant under 9 ,U . . . U 93“. Then, of B ,fl.

I

N ~ ~ . . . . . ” J= ( K( )N ~ ( B , 1) s i s n ) . In the special case of centralizers Theorem 4 becomes

Lemma 1. Z f B , C G , l d i d n , aref.g ., t h e n C G ( B I n . . . n B , ) = ( C G ( B i )1(s i C n ) . Proof of Theorem 3. Induct on the subgroups of A. Suppose C G ( A ) C

Embeddings and centralizers

143

M C G. If M C No (A), there is nothing to prove. So, suppose x E M - NG(A).We have Ao=A n A " < A since A f A' and A has MIN. Now, M 3 (CG(A), X ) 3 (CG(A),CG(A')). Put A = Dl n * * n D, (DiC G f.g.). Thus, M 3 CG (Di)and M 3 CG(DI) for all 1 < i < m. Since A. = DIf DYn l * fl D, n D i , we have by Lemma 1, M 3 CG(Ao). By induction we now conclude that CG(H)C M C NG ( H ) for some H C A".

-

Next we consider the normal structure of normalizers.

Theorem 5 . Suppose A C G is f.g. and x E NG ( A ) - A. Then CG(A) C X ~ G ' ~ Thus ) . CG(A)/CA(A) is simple. ( x c is the subgroup generated by all the c-lxc, c E C ) . It is well known that if A C G is centerless and f.g., then CG(A) is a.c. and m-o-equivalent to G. If CA(A)# 1, it is in many cases easy to see that CG(A)/CA(A)has a pair of isomorphic f.g. subgroups which are not conjugate-in it. For example, suppose CA(A) is not divisible and let x E CA(A) have no pth root. Choose z, y E CG(A) such that y p = x and z p = 1, z & CA(A). It is easily checked that (yCA(A)) and (zCA(A))are not conjugate in CG(A)/CA(A) since the former 'cosets have no elements of order p . Next we note that Theorem 3 fails if A is infinite cyclic.

Proposition 1. Suppose Z C G is infinite cyclic. Put c = U{C,(Z")In 2 11, N = U { N ~ ( Zn 3~ I), )) and M = {x E G IZn Z x # 1). Then C a N a M, M is maximal in G, C is simple, N / C has order 2, and M / N is free abelian of rank w. Proof. Clearly M # G and its maximality follows from Lemma 1 since x $Z M implies Z n Z" = 1 whence ( M , x ) 3 (CG( Z ) , CG(Z')) = G. The simplicity of C follows from Theorem 5 : if 1 # x E C, there exists m 3 1 such that x E CG( Z " ) - Z" for all multiples n of m . Thus, CG( Z " ) C X ~ C ( ~ for " ) such n which gives C C x c . Clearly N / C has order 2 and N a M . It remains to consider M / N . For each pair ( m , n ) of positive integers, there exists T = T ( m , n ) E M such that T-'Z"'T = Z". We will view T as an element of M / N . Thus, 7 ( n ,n ) = 1 , T(n, m ) . T ( m , p )= ~ ( n , p )and , T(nm, n p ) = T(m,p)(mod N ) for all n, rn, and p . Thus,

K. Hickin,A. Maclntyre

144

I

M / N is generated by { T ( a ,b ) a, b are relatively prime} and multiplication in M / N is given by T(a,

( y $)

b ) . ~ ( cd ,) = 7 -

3

where A is t h e least common multiple of b and c. We check easily that M / N is also generated by { T ( l , p ) ( pis prime} and that these generators have infinite order, commute, and are independent. Finally, we will show that a basic construction of group theory can be performed in every a.c. group. If A is a f.g. group, W(A) denotes the words in some finite generating set of A which are relations of A ; W-(A) denotes the corresponding set of words non-trivial in A. If S and T are (coded) sets of natural numbers, S S eT is the relation “S is enumeration reducible to T” [4: p. 1451. Theorem 6 . Suppose A, B C G are f g . and there exists an f . g . F C G such that W-(B) SeW(F). Then, the restricted wreath product A wrB is embeddable in G. In particular, this is true i f B has a solvable word problem.

Corollary. If A C G is f.g., there exists a f.g., perfect, centerless group D such that A C D C G. Proof. Let P # 1 be a finite perfect group. Since P * P is SO-universal [2: Th. 11.31, there is a group H = ( P I ,P J with P, = P such that A C H. Since G is a.c., we can assume WLOG that A C H C G. Now P I *Pf has a solvable word problem; so, by Theorem 6, D = H w r ( P , * P&is embeddable in G. D is f.g., perfect, and centerless.

Notice that the group D of the corollary cannot always be chosen simple or even subdirectly irreducible. For example, if every f.g. subgroup of G is recursively presented [3: Th. 81 and A has an unsolvable word problem, D cannot be so chosen; and also, if S is a finite non-abelian simple group, then S wr A is not embeddable in G because S wrA is subdirectly irreducible. This same reasoning shows that for arbitrary G, the converse to Theorem 6 is also true: namely, if S wr B is embeddable in G, this group is an adequate choice for F since W-(B) c > = d for all n E P.

From (l), Lemma 2, and t h e fact that P and R are r.e. in W(A), it follows that w ( r ) seW(A 1.

So, t h e General Higman Theorem implies that r is a subgroup of an f.g. group H with relations W ( A ) U F, F finite, and that t h e generators (2) form a recursive subset of H. Since G is a x . there is a homomorphism p : H + G such that p ( d )# 1 . From the inclusions p ( r ) C p ( H )C G 6 A , the first of which is recursive, we have as in (1) that W-(p seW(A). So,

(r))

K. Hickin, A. MacIntyre

146

C = { n E w 1 p ( c , , )# l } s eW(A), and, trivially, C is r.e. in W(A). Likewise, W(p (f)) 6,W(A) and w - C is r.e. in W(A). Hence C is recursive in W(A). Since p ( d )# 1, we have R C C and P C w - C contrary to the W(A)-recursive inseparability of P and R.

62. Proof of Theorem 2 From Ziegler [6] we need Lemma 3. If G is a.c. and A C G is f . g . , then A in G.

X

A is embeddable

Lemma 4. Suppose G is a countable a.c. group, H C G, and X,, C G, n 3 1, are isomorphic to G. Then, there exist embeddings cp : G -+ G and J, : H + G such that (1) c p ( G ) @ J , ( H )exists in G and ( 2 ) for all n 3 1, Cc(cp(Xn)) = cp(G(X))@ J,(H).

1

Proof. Let H = u { H , i 3 1) where H, C H , + ,is f.g. and possibly H = H , for some n. Let G = U{G, i 2 1) where G, C G , + ,is f.g. We construct q and J, inductively. Suppose we have defined cp(G,) and + ( H a )so that c p ( G , ) @ J , ( H , )exists in G. At the next step it is possible (A) to extend J, to H,,, so that S = c p ( G , ) @ + ( H , + ,exists ) in G and (B) given elements g l , . . . , g, E G - S and x,, . . . ,x , E ( G 8 +-, G , )r l ( u n a , X n )to, extend cp to G , + ,so that (a) q ( G , + , ) @ J,(H,+Jexists in G, and (b) no cp (x,), 1 s j s 9,commutes with any g,, 1 S j S p. (A) and (a) can be done because G , + , @H , + , is embeddable in G and G is homogeneous, while (b) can be done because we need only force finitely many inequations, which hold in the free amalgamated product ( S , x l , .. . ,x , ) * ~ ( S , g l , . . . . ,g,), to hold in G.

I

We claim that (A) and (B) suffice to build the desired properties of the embeddings cp and J,. Note that, by Theorem 1 , no X,, is contained , Gi for in a f.g. subgroup of G, so that every X,,intersects G i +infinitely many values of i. Hence, by choosing the sets { g , , . . .,g , ) to eventually exhaust G - (cp(G)@J , ( H ) ) and the sets { x , , . . . , x , } to intersect every X,, for infinitely many values of i, we will guarantee that, for all n 3 1,

147


yielding conclusion (2) of the lemma. Proposition 2. Let G be any countable a.c. group and let H ( ( Y ) ,(Y < w l , be groups embeddable in G. There exists an a.c. group G * of power w1 which is m-w-equivalent to G and there exists GoC G * with Go= G such that, for all countable S with GoC S C G *, there exists p < w1 such that

CG.(S)=A x

(n

H((Y)

/3GO Cfd-f(Uf,))=A , x

(n

@ ( B ~I )1 s i s n ) .

Lemma 5. Assume the hypotheses of Theorem 4. There exists a group J 3 H such that, for all a E %, there exists y E N, such that 1 y I = W, ( y ) n H = 1, H n H Y= K, and y induces a on K (by conjugation). Proof. Let J have the presentation (H, r ( i , a ) where a E 9, and 1 S i s n : W ( H ) and “ t ( i ,a ) induces a and B, by conjugation”). J is a (multiple) HNN extension of H and we will use Britton’s Lemma (B.L.). Put Q, = ( t ( i ,a ) a E 9 , )C J and Q = (Q, 1 S i S n ) . Note that Q C N,. Q is free on the generators { r ( i , a)}and Q n H = 1 by B.L. Also,

I

1

K = n { A q1q E Q}. (1) Suppose w = u1* uk is a product such that adjacent factors belong to different Q,’s. Using B.L. in J, we check easily that (2)

[

for all h E H, h E H iff h E H for all initial segments s = ~ l . . . u, ( l s i s k ) of w.

Suppose a E % is given as in the hypothesis. We will construct y as ..., u k + l of Q to be a product of a sequence of elements u,, determined inductively. We define u1as follows. Let 1 # q, E Q,, 1 d i d n and put 1 , = t ( i , L,).Define s1 = q1 and inductively s,+~ = s,lr+ls;lq,+l, 1 s i d n, and define u I= s,,. We claim

(3)

H nHUT’C

A.

To prove this, assume inductively

if h E H - ( B l n . - . n B , ) , then hsa6fH, which is obvious if i = 1 from B.L. Suppose

K. Hickin, A. MacZntyn

150

h E H - B1n

- - - f l Bi+land h'i+lE H; this implies h &

or else (4), is contradicted using (2).

Thus, h'l E H by (2). If h'l then h'll~+ltir H by B.L. and hence h'i.1 fZ H by (2). So, we have h'd E Bi+land hence h'i'i+l= h'i, whence h =h H contradicting ( 5 ) and proving (4)i+1.Thus (4)" holds and this proves (3). Suppose ul,. . . ,ui have been selected such that the last letter of a, 1 S j S i, belong to distinct Qk's.Put 7ri = u1 * * * ui. and the first of Note that '~';'~i+l

(6)

e

Li = H

n H "L1C A

by (3) and (2).

Suppose there exists a E Li - K . Thus a-4 fZ K and by (1) there exists q E Q such that a"@6Z A, and we can arrange that a"dq bZ H (by replacing q by qr where 1 # r E C?, if u"fl E H - B,).Putting m+l= q,

we have a fZ Li+l(see (6)). So, by (6) and'(2) we have A 3 L 1 > * * * > L i > L i + l > . * * >K.

Since A has MIN, we must have for some k, L, = K. Finally we Q so that y = ? T k u k + + I induces Q on K (we can begin u k + l choose with some l i to prevent unwanted cancellation). is the set-theoretic union of two groups, An amalgam d = = B U C,which intersect in a common subgroup A = B f l C. gp,(d) is the free product of €3 and C with amalgamated subgroup A.

d

Lemma 6. Suppose K C H C J, y E J with y 2 fZ H, y normalizes K , and H f H l Y= K. Let SP = 'Y' be a symmetric amalgam ( - : J + is an isomorphism which fixes H elementwise). Put 3 = gp*(SP). (i) There exists t E X , such that 1 t 1 = m, t normalizes K, ( t )n H = 1, and =

(t)KYJ generates gp,(SP')

in 91, and (ii) Suppose X C Aut (K), y induces a X-automorphism on K and for all a E X, there exists ye E J - H which induces a on K. Then, t can be chosen in ( i ) so that it induces any a E X on K. Proof. To prove (ii), put t = yyy,yg-' where p E X is chosen so that will induce a on K. (To prove (i) we would use y in place of yp.)

t


151

Clearly It I = m since y 2 @ H and ( t ) r l H = 1. We have ( r ) K r lJ = K because every z = t:k E r'K with i f 0 involves y implying z $Z J. Hence the amalgam d'does exist in g p , ( d ) . Let w = po(t"k1)pi(t"kz)p2.. . (r'"k.)p,, where p, E J - K, i, # 0, and k, E K, 1 =sk S n (and possibly P o or p . do not occur) be a reduced product of members of d'.We must show that if n 2 1 (i.e., some t occurs), then w f 1 in ?I. This will be so provided not too much reduction in ?I occurs in segments of the form t 'ik,p,t'1

+

1.

To check this we can ignore k, E K by consolidating it with p, and consider subsegments of the possible forms tpr, rpr-', r-'pt, and

t-'pr-'.

All cases being similar, we consider only t h e first. We have

YY%YF-'PYYYPYY-'. If p E J - H , this segment is already an &?-reduced product. If p E H - K , then u = y - ' p y E J - H since by hypothesis H r l HF-' =K; and, making the replacement u = y - ' p y in rpt yields an &-reduced product. Thus, the middle letter of every r occurring in w survives in'the &?-reduced form of w, which implies w # 1 in ?l and finishes t h e proof of Lemma 6. tPt

=

Lemma 7. Suppose K C B C G where G is a x . and B is f.g. If Y l C Aut (K), then N Z ( K ) is generated by elements of infinite order.

Proof. Suppose x E N E ( K ) . Since B is f.g., there exists c E G with lc I = co such that (c)@ (x, B ) exists in G (e.g., by Lemma 3). Now x and c + x induce the same automorphism on K and c + x has infinite order; so we only need to show that C G ( K )is generated by elements of infinite order. Suppose x E CG(K). With c as above, we have c and c + x centralize K, have infinite order, and generate x, so we are done. Lemma 8. Suppose & = "YR, M C Q, and dl= M Xgenerates p g p , ( d , ) in Q. Then, dz= '"wR generates gp,(dZ) in gp,(&).

Proof. A reduced product in gp,(&) can be written as r l r I* r n r n where ri E R - P and each ri is a reduced product in gp,(dl) (with

K. Hickin, A.

152

Maclnryre

n,, r, optional). We have 7ri 6Z P for all i, otherwise the product is not reduced. Hence, this equals a reduced product in g p , ( d ) .

Proof of Theorem 4. We will use t h e notation of Lemma 5. By Lemma 7, N g ( K ) is generated by elements of infinite order. Let x be such an element. We must show that x ENG. Let F = ( a,,a z ,a,) be free on {aI, az,a 7 }and let FK be t h e semidirect product such that a z and x induce on K the same automorphism and aI, a? centralize K . Put

with x = ( Y I ( Y Z ( Y 7 . First, we claim (*)

[

the amalgams d, = (L.,)Ky(x.H) ( i = 1,2,3) generate g p , ( d , ) in g p , ( d ) .

To prove this, use Lemma 8 with M = ( a , ) K . Since ( a , ) K f l ( x ) K = K in FK, we must show that

generates gp,(d',) in FK. Since K is normal, this is so iff a, and x generate ( a , ) * ( x )in F, which is easily checked. Second, we claim there is a group P 3 g p , ( d ) such that aiE N p ( i

=

1,2,3)

and hence x E N p :

To obtain P we first use Lemma 5 to get J 3 H and y E N , such that ly I = w, (y) f l H = 1, H n H Y= 4, and y = ya can be chosen to induce any a E X o n K. Now apply Lemma 6 to obtain

( f , ) fl H = 1, 1, and a, induce t h e same automorphism such that I f, 1 = on K, t, EN,,, and (r,)K3?' generates its free product in J,. Thus, ( ' j K Y H generates its free product in J , , which is, by (*), isomorphic to the subgroup of g p , ( d ) generated by (a>KY under the map H = H and t, * a , . So, we can amalgamate g p , ( d ) with the J, via these isomorRhisms to obtain P = ( J , ,Jz, J,) 3 g p , ( d ) = ( H , F). Finally, by making direct use of the algebraic closedness of G and QJ,


153

the existence of P in (**), we can obtain P with ( H , x ) C P C G such that x E Np C NG, as required. To obtain P, use (**) to express x = z1. . * z, where z , E N%~)(Bn(,)). So, the relations {x = f l. . . fe and “2, induces the same automorphism on B,(,)as does z a ” (1 s i s E ) } (which can be taken to be a finite set because each B, is f.g.) are consistent over (H,x). So, there are elements fl,. . . , f, E G which satisfy them and we take

P

= ( H , x, 51,.. . , fc).

Proof of Lemma 1. In the case of centralizers, in the proof of Theorem 4, we have Bi= 1 (1 s i S n), X = 1, and hence K = Bl fl... f l B, and the hypothesis of MIN is not needed in the proof of Lemma 5 in this case, and we obtain Lemma 1. 04. Proof of Theorem 5

In the notation of Theorem 5, suppose x E NG ( A )- A. Put C = CG ( A). In the group with presentation (A,x, (Y : (Y centralizes A ) let f = ( Y - I X ( Y * X - I ( YThen -I. If 1 = 03, 5 E x C p ) and , ( Z ) @ A exists. Since G is a.c., there exists z E G such that ) z1 = CQ, z E x‘, and ( z ) @ A exists in G. We will show that (*)

{

C is generated by elements w such that 1 w I = 03 and ( ( z ) * ( w ) ) @ Aexists in G.

First note that C is generated by elements c such that I c I = Q), (c)@ A exists in G, ( z ) n ( A , c ) =1. To see this, let u E C and choose u so that I u 1 = 00, and ( u ) @ ( A ,z, u ) exists in G. The elements u and v + u satisfy the requirements on c above and generate u. To prove (*), let c be a generator of C mentioned above, let F = (5, be a free group, and put

p)

J = gP*

( ; @ A Q % c J ) ) c =@)@A

(p),

Since ( E ) , and ( z ) all intersect the amalgamated subgroup trivially, we see that ( G ) * ( z ) @ Aand ( p ) * ( z ) $ A exist in J. Since G is a.c., there exist a,p E G such that c = ap and ( ( a ) * ( z ) ) @ Aand ( ( P ) * ( z ) ) @ Aexist in G (because these groups are embeddable in

K . Hickin, A. Maclntyre

154

G). This proves (*) by taking

(Y

and p as values of w, thus generating

c and hence all of C. Now, for every w E C satisfying (*), there exists T E C such that T-’z= ~ w, 17 = 2, and T E C. Thus C is generated by elements of t h e form z T E xc. This concludes t h e proof.

I

85. Proof of Theorem 6 We shall deduce Theorem 6 from the following result due essentially to Belegradek [ l ] who proves it for recursively presented groups. Proposition 3. Suppose A and B are f . g . groups with generating sets X and Y. Let ( X ) and ( Y ) be the free groups on X and Y . B is embeddable in every a.c. group which contains A iff W ( B ) % W ( A ) and there is a function 9 S cW ( A ) whose domain is ( Y ) and whose range consists of certain sequences in ( X ) (which we denote 9 ( y ) = (xb ,..., x; ,... )) such that, for all y E ( Y ) , y E W ( B ) i f f x ; E W ( A )for all n 2 1.

I

By 9 S c W ( A )we mean that the set of pairs { ( y , x;) y E ( Y ) , n 3 1) is enumeration reducible to W ( A ) . The “if” part of Proposition 3 - which we will use - is proved in a manner similar to the proof of B.H. Neumann that every group with a solvable word problem can be embedded into every a.c. group [2: Chapter IV, §8] and the method can also be gleaned from Belegradek’s proof for r.p. A and B ; but one must use the General Higman Embedding Theorem ($1). Let us assume the hypothesis of Theorem 6. We can also assume that A , B C F and hence that W ( B )ScW ( F ) as well as W - ( B )S cW ( F ) because of obvious properties of enumeration reducibility. Let X , P, and Q be finite generating sets of F, A , and B respectively and put Y = P U Q. So, Y generates A wrB. Since A , B C F, we can assume Y C X. Using Proposition 3 we need only find a function 9 s cW ( F ) whose domain is ( Y ) and such that, for all y E (Y), 9 ( y ) = ( x i , . . . ,x;‘’)) (with x ; E (X)) and y E W ( A w r B ) iff x : E W ( F )for all 1 S i S m ( y ) . We will now describe an algorithm for computing 9 given an enumeration of W ( F ) . Suppose y E (Y). Using the fact that A w r B = A B B is a semidirect product we can collect P’s and Q’s in y to obtain


y =p;‘.

. . p>w

where

155

pgE P ; w , w , E (0).

As we enumerate W ( F ) we can also enumerate W ( B ) and W - ( B ) since these sets are seW ( F ) . Thus, w will be enumerated either in W ( B )or in W - ( B ) in finitely many steps. In the latter case, we have y FZ W ( A w r B) and we put 9 ( y ) = ( w ) . In the former case, we likewise determine for all 1 c i, j s n that either w,w;I E W ( B ) or E W - ( B ) in finitely many steps. In A wrB, p 7 and p 7 will commute provided w , # w,. So, the above information allows us to rearrange y as follows: y = p ; ’ . . . p > = u y , . . . UZrn where u, E (P), u, E (Q), and, if 1 s i f j s m, then u , # u, (each u, is one of the original w , ) .

Since the factors up,,1 s i =Sm, belong to distinct direct factors of A B in A wrB, we have y E W ( A w r B ) if and only if { u , , ..., u,,,}C W ( A ) , so we put 9 ( y ) = ( u I , .. . , u,,,).

Note added in proof

The paper “Maximal subgroups of a.c. groups” mentioned in 00 has been subrmtted for publication. Theorem 3 is generalized in it to effectively embedded subgroups, and applied to study properties of Ziegler’s construction [ 5 ] .

References [ I ] O.V. Belegradek, On algebraically closed groups, Algebra and Logic 13 (1974) 135-143. [2] R. Lyndon and P. Schupp, Combinatorial group theory (Springer-Verlag, 1977). [31 A. Macintyre, On algebraically closed groups, Annals of Mathematics 96 (1972) 53-97. [4] H. Rogers, Theory of recursive functions and effective computability (McGraw Hill, 1967). [ 5 ] S. Shelah and M. Ziegler, Algebraically closed groups of large cardinality, to appear. [6] M. Ziegler, Algebraisch abgeschlossene Gruppen, to appear.


SMALL CANCELLATION THEORY OVER GROUPS EQUIPPED WITH AN INTEGER-VALUEDLENGTH FUNCTION Bernard M. HURLEY University of Zam bia, Lusaka, Zambia

01. Introduction

In this paper, small cancellation theory is developed over a class of groups equipped with an integer-valued length function satisfying certain “natural” axioms. Such groups can also be characterised by the fact that they are equipped with a normal form (for elements) satisfying certain other axioms. Since free groups, free products with amalgamation, and HNN groups all possess a normal form of the required type, our theory includes the “classical” small cancellation theories of Lyndon [13], Schupp [19] and Sacerdote and Schupp [18] However, the class of over the above mentioned classes of groups.groups that we shall consider (we call them NFS groups) is much wider than this and is closed under a fairly large number of constructions, as will be shown in 93. As with the “classical” theories, there are three “cases” of our theory. The observant reader will notice that we only apply one of the “cases” in 08, whereas we develop the theory of all three in 05. We make no apology for this. We consider the mathematics of 95 to be interesting mathematics in itself. The applicatiqns of 08 have been selected according to two criteria. The first being that the result in question can be proved without too much extra work. The second being that the result is of interest. The applications include various embedding theorems, a characterization of the solubility of the word problem for groups, and proofs of the residual infinite simplicity and the infinite height (in the sense of Pride [16]) of various groups. It is obvious, from the results of 03 and 97, that many of the theorems of 08 could be followed by a corollary( giving a vast list of groups satisfying the theorem (e.g. we could follow Theorem 8.10 with a corollary to the effect that non-abelian free groups, certain HNN groups, etc.. . are residually infinite simple). Many writers would do this, but we do not see the neccessity for it. 157

B.M. Htcrley

158

Many interesting applications have been left out of 98, in view of their failure to satisfy our first criterion. Perhaps the most interesting being the use of the theory to prove a version of Higman’s Embedding Theorem. In 99 we have given a list of such applications. These, perhaps, should be regarded as “signposts” to things to come. Of course, whether or not one needs signposts depends on who one is. (Anglo-Romany has no word for signpost.) T o t h e reader unfamiliar with small cancellation theory, we recommend Schupp [20] for an incomplete, but clear, introduction to the “classical” theories. Having read this paper, he/she may be surprised that we d o not impose any length restrictions in our Cconditions. Our use of special sequences, in 95, makes this unneccessary . We have tried to strike a balance between formality and informality. If the reader finds t h e construction of the diagram, in $ 5 , too informal, he/she should bear in mind that, written out formally, it would involve a four-fold induction. Notation : We believe that most of our undefined notation is standard. However, we draw to the reader’s attention that, according to context,

n=

[

proof ends here, proof omitted, proof to occur later in text.

It is natural to think of elements of a group that possesses one of our axiomatic normal form structures as “words”, by analogy with the words of a free group. We shall often use the word “word” in this loose sense. When we use it in the “strict” sense of a reduced word on a set S of letters, we shall speak of a “word o n S”, unless the context makes it obvious what we mean. -

82. NFS groups, NFS groups and length functions

In general terms a length function o n a group G is a function 1 g 1 satisfying a “reasonable” set of axioms. Length functions on groups were first studied axiomatically by Lyndon [12]. He imposed five axioms, of which t h e following two seem to capture the “essence” of the notion of a length function. (The numbering is Lyndon’s.):

G + R, g

Small cancellation theory

A4.

159

d ( x , y ) < d ( x , z ) implies d ( y , z ) = d ( x , y ) ,

where d ( x , y ) = f { J x J + l y I - J x y - ’ ( } .Chiswell [4] and Hoare [8] have studied length functions satisfying these two axioms. A length function is normalised if it satisfies: Al’.

IlI=O.

If g I+ 1 g I is a length function G + R satisfying A2 and A4, then the function g llg (1, defined by ((g(1 = ( g1 - 1, is a normalised length function that still satisfies A2 and A4 (see the comment after [8] Proposition 5 ) . Thus there is no loss of generality in considering only normalised length functions. Geometric small cancellation theory was first developed by Lyndon [13] over free groups and free products. Later Schupp [19] developed the theory over free products with amalgamation, and Sacerdote and Schupp [I81 developed it over HNN groups. It is a striking fact that in every case the group under investigation possesses a “natural” length function satisfying Al’, A2 and A4. Moreover this length function is always well-based in the sense of Hurley [lo]. That is, in addition, it is integer-valued and satisfies the extra axiom: N1*.

G

= gp{x E G : I x

I s 1).

Intuitively, the function d ( x , y ) measures the amount of cancellation in the product x y - ’ . Thus, for instance, we say that no cancellation occurs in this product if d ( x , y ) = 0, and that almost no cancellation occurs if d ( x , y ) S 1. However in order to develop a small cancellation theory we need to do more than merely give a numerical value to the cancellation in a product, and, in each of the above mentioned cases, a central role is played by the appropriate Normal Form Theorem, whereas the length function only appears as a way of simplifying some of the “small cancellation” hypotheses. Abstract normal form structures (NF structures) on groups were first studied by Hurley [9], however we shall use the axiomatisation of Hurley [lo]. We can associate a length function in a natural way with each NF structure on a group. This leads us to the Main Theorem of [lo].

Theorem 1.1. A function G -+Z, g H ( g1 on a group is a well-based length function (i.e. it satisfies N1*, Al’, A2 and A4) if and only if it is the associated length function of some NF structure on G. 0

B.M. Hurley

160

We shall now briefly explain our axiomatisation of the normal form. First we need some definitions in which we shall suppose G is a fixed group, X is a fixed subset of G,satisfying 1 bZX, and B is a fixed subgroup of G called the base group. Definition. Let S = S(G,B, X) be the set of all sequences u = (ui);ET', where n 3 0, such that u21+1 E B, for j = 0,. . . , n, and uziE X U X-', for j = 1 , . . . , n. We call n the length of such a sequence, and denote it by u 1. We define the value of the sequence to be 6 = u l u 2 .. un.

I

If u = (0;)E S and for some 1 s j s n - 1, u21uzl+1u21+2 E B, then the sequence r of length n - 2 defined by: 7, =

[

if l s i s 2 j - 2 ~ ~ , - ~ u ~ , u ~ , + 1ifu i2= , 2j + 2-u1Z I + 3 if 2j s i < 2n - 4, u,+4 u#

clearly lies in S and satisfies 7 = 6.We say that r is obtained from u by means of a pinch between u2,and u ~ , + ~ . If u = (a,)E S and for some 1 s j s n, uzlu21+lu21+2 E B(X U X-')B, = bgc for some g E X U X-' and some 6, c E B and then u21u21+1u21+2 the sequence p of length n - 1 defined by:

PI =

I

if u21-lb if g if cu21+3if u , + ~ if u,

lsis2j-2 i = 2j - 1 i = 2j i = 2j + 1 2j+2sis2n-2,

clearly lies in S and satisfies p = 6. We say that p is obtained from u by means of an amalgamation between u2,and u ~ , + ~ . A (G, B,X)-sequence u is reduced if n o pinch or amalgamation can be performed between any two elements of u.The set of all such sequences shall be denoted by RS(G,B , X ) or just RS. Intuitively, RS consists of normal forms for elements of G. If u E RS and g = 5, then we say that u represents g . Clearly if u E S then there exists r E RS such that 7 = 5.It follows that to ensure that every element g E G has a normal form (i.e. that every element of G is represented by some member of RS),we need merely impose the axiom:

N1.

G = g p W , B).

IT

We say that a, 7 E S are parallel, written u 1) T , if, either I u 1 = 1 = 0 and 6 = 7, or 1 u 1 = I T 1 = n > 0 and there exist h, E B for i = 1 , 2 , . . . ,, 2 n such that ul= T 1 h l , u,= h ;!'T,h, for 2 S i S 2n, and


161

IT^,,+^ = h ; , ! ~ ~It~ is + easy ~ . to see that if u 11 T then 5 = .T. The right sort of uniqueness for normal forms turns out to be uniqueness “up to parallelism”. This can be obtained by imposing the axiom:

N2.

If a,T E RS(G, B, X) and 5 = ?, then u I(T.

Definition. A triple ( G , B , X ) ,consisting of a group G, a subgroup B, and a subset X , satisfying 1 ex,that satisfies axioms N1 and N2 is called an NFS group (NFS = normal form structure). Alternatively we say (GI B, X) is an NF structure o n G. Given an NFS group (G, B, X), we define the associated length function in the “obvious” way. It easily follows from N1 and N2 that any two normal forms for (i.e. sequences in RS representing) an where element g E G have the same length, and so we define Ig = ( u ) u is such a normal form. An obvious question to ask is “Can distinct N F structures on a group G give rise to the same length function?”. The answer is “Yes”. Following [lo], we call such structures isometric. If (G, B, X) is an NFS group, then it is trivial to check that (G, B, XI).where Xl = B(X U X - ’ ) B , is also an NFS group and that these structures are isometric. Clearly these structures are the same if and only if .(G,B, X ) satisfies:

I

F.

X

= B ( X UX - ’ ) .

An NFS group satisfying F shall be called a full NFS group or an group, and, correspondingly, we speak of full NF structures or F F structures. We easily see that XI = {x E G : I x I = l}, and it follows that (G,B,X& always full. Moreover, by [lo] lemma 2.3, this is the unique NFS group isometric to (G, B, X ) . Thus F is a sort of normalisation axiom. We shall work entirely with NFS groups; by the usual abuse of language we shall speak of the NFS group G. The reader who wishes to reformulate our results in terms of NFS groups will find it a routine, if somewhat messy, matter. It is easy to see that any u E RS, in an NFS group, is parallel to one of the form (1, xl, 1, xz,. . ., x,, 1) if n 2 1, and is equal to one of the form ( h ) if n = 0. If u has value g E G, then we say that, according to case, g = xlxz * * * x, or g = h, in normal form, and we abbreviate the sequence t o (x,, xz, .. . ,x,) or ( h ) . Finally, we note that there are alternative axiomatisations of the NFS groups. Consider the following set of axioms:

N3.

If u E RS and 5 = 1, then u = (1).

162

B.M. Hurley

N3*.

If u E S and 6 = 1, then either T E S - RS such that u 11 7.

N4.

If g, h, gh E XI but hk 6Z B U XI, then ghk 65 B U XI.

N4*.

If g, h, hk E XI but gh

EB

(T

=

(1) or there exists

U XI, then

ghk

EB

U XI,

where, as before, X1= B(X U X - ' ) B , and so, when F is in force, we may drop the suffix 1 in N4 and N4*. By [lo] Lemma 2.2, N3 N3*, N4 ~3 N4* and N2 e N3 & N4, giving us various axiomatisations of the NFS groups.

-

63. Examples of NFS groups

-

The purpose of this section is to provide examples of NFS groups. It is to be hoped that these examples will help motivate the remainder of the paper. We shall also show that, together with the appropriate homomorphisms, the NFS groups form a category JWW,in which constructions analogous to free product, free product with amalgamation and HNN extension are possible. We 'first show that certain familiar constructions yield NFS groups.

-

Lemma 3.1. In the following cases, (G, B,X ) is an NFS group: (i) G is any group, B is any subgroup of G and X = G - B. (ii) G is the free group freely generated by the set Y,B = 1 and = Y u Y-I. (iii) G = (H * K : B1= B2), a free product with amalgamation, B = BI and X = ( H - B ) U ( K - B). (iv) G = (t, B : t-'At = C ) an HNN extension, and X = B{t, t-'}B.

x

Note on Proof. We feel that the proof of this result may safely be left to the reader. The following remarks may be useful. Parts (ii), (iii) and (iv) are restatements of well known normal form theorems (Britton's Lemma, see [3], in part (iv)). However an alternative way of proving the lemma would be to note that the natural length function satisfies N1*, Al', A2 and A4, in each case, and then to apply Theorem 1.1.

-

Definition. If (G1,B1, XI) and (G,, B2,X , ) are NFS groups, then by an F S homomorphism (p : (G1,B1, XI)+ (G2,B2,X2), we mean a group homomorphism (p : G1+ G2 satisfying Bl = (p-'(B2)and XI = (p-l(X2).

Our next result, whose proof isroutine and therefore omitted, sums up the more trivial properties of NFS homomorphisms.

163


-

-

Lemma 3.2. (i) The NFS groups together with the NFS homomorphisms form a category, composition being composition of mappings. (ii) Let (G2,B2,X2) be an NFS group and let cp : G I+ G2 be a - Let B1= cp-'(B2) and XI = cp-'(Xz). Then (group) homomorphism. (GI,BI,XI) is an N a r o u p if and only if G I = gp(Xl, B1). When this is the case, cp is an NFS homomorphism. (iii) Let (GI,B I ,XI) and (G2,B G 2 ) be NFS groups. Then a homomorphism cp : GI-+ G2 is an NFS homomorphism if and only if Igcp I = l g 1 for all g E G (i.e. i f n d only if cp is length preserving). (iv) Let (GI,BI,XI) be an NFS group and let cp : G I + G2 be a (group-imorphism. Let B2= cp(B,) and X2= q ( X , ) . Then (G2,Bz,X 2 ) is an NFS group if and only if ker(cp)S B1. In this case cp is an NFS homomorphism. 0

-

Problem 1. Let (GI,B I ,XI) be an NFS group and let (o : G I+ Gz be a (group) homomorphism that is not an epimorphism. Under what conditions isitpossible to construct an NFS structure on Gz that makes cp an NFS homomorphism? A necessary condition is ker(cp) 6 B,.

-

-

Notation. We denote the category of NFS groups and NFS , homomorphisms by N 9 Y . We often use the suffix N to indicate that we are working in this category. Thus, for example, to indicate that two NFS groups are isomorphic in N99,we often write G I= N G2 instead of (GI,B,, XI)= (G2,B2,X2).

-

Definition. Suppose (GI,BI,XI) a n d x 2 ,Bz,XZ) are NFS groups and G I sG2. Then we say GI is a sub-NFS-group (a subgroup,) of G2, and write (GI,B1,X s (G2,B2,X2) (or G1SNGz), if the natural injection G I + G2 is an NFS homomorphism (i.e. if B1= G Ifl B2 and XI = G In Xz or, by Lemma 3.2(ii), if the length function on Gz extends that on GI). This relation is clearly transitive. Warning. sNis not the same relation as

2. By symmetry we may suppose ( x 1 C I y 1. Again the lemma is trivial if 1 x 1 = 0. If 1 x I = 1, then, by our assumptions, we can write y = c I c 2 .. * c, (s 2 2 ) in normal form. If (xcll= 0, then Ixy 1 = 1 y I - 2 so by induction (i) or (ii) holds for ( x ,x y ) and thus by (*), (i) or (ii) holds for ( x , y ) . If 1 xcI 1 = 1, then ( x c I ) c 2 . c, and c1c2* ( G X ) are both normal forms for x y so by N3, c-'xcI E B and hence (ii) holds for ( x , y ) . If I xcI 1 = 2, then xclcz * * c, and L1c2 . CJ are both normal forms for x y and so by N3, ( x - ' c I1 = 0. But then (i) or (ii) holds for ( x - ' , y ) and so by (*), (i) or (ii) holds for ( x , y ) . If 1 x 1 2 2 , then we can write x = c 1 c 2 * * - c( ,r 2 2 ) and y = d l d 2 . . . d , (s 2 r ) both in normal form. Since xyx-'y-' = 1, either (c,d;'( = 0 or else 1 c;'d;'J = 0. By (*), we may assume 1 c,d;'( = 0. If 1 c,cl(< 2, then I c j c ; ' ) < x and I c,yc;'( C y, thus by induction (i) or (ii) holds for the pair ( c j c ; ' , c,yc;') and thus (i) or (ii) holds for (x, y ) . If 1 c,cl 1 = 2, then no cancellation occurs in either product xy or yx. Comparing normal forms (i.e. using N2) in the equation x y = yx, we see ( x - l y 1 = s - r. By induction (i) or (ii) holds for the pair ( x , x - ' y ) and so by (*) holds for the pair ( x , y ) . 0

I

-

- -

B.M. Hurley

176

We define a pseudo-proper power (p.p.p.) to be an element of the form: x - ' h x w k where h E B ; x, w E G ; k 3 2 and x-'hx and w commute.

or 1 x " I = I n I for all n E 2. (ii) Suppose I x 1, 1 y 1, I x y 1 3 1 , but are not all equal to 1, xy = yx and x is s.c.r., then y and x y are s.c.r. and there exist w E G, h, h ' E B and j , k E Z such that: Lemma 4.9. (i) If 1 x 1 = 1 , then either ( x )C B U X

x = hw',

y = h'w',

w is s.c.r. with 1 w z J= 2 )w 1, and h, h' and w commute in pairs. (iii) Let x E X be s.c.r., then either or :

CG(X)CBu x , CG( x ) = ( x ) x ( B n CG( x )) and

1 x 2I = 2.

Proof. (i) If ( x ) B U X, then there exists r 3 1 such that x'+' 6Z B U X. Let r be chosen to be as small as possible. If E B, then write b = x'-'. N o cancellation occurs in (bx)x, and so no cancellation can occur in xx. Thus r = 1. If x'-' E X, then writing g = x-'+', h = x and k = X ' and applying N4, we see that xx = ghk 6Z B U X, and again r = 1. By the minimality of r, there are n o other cases and it is now easy to see that no cancellation occurs in any product of the form xxx . . . x. Thus 1 X " 1 = 1 n 1 1 x 1 for all n E Z. (ii) We first prove: (**) Suppose I u 1, 1 u 1, 1 uu 1 3 1, but are not all equal to 1, and uu = uu, then u is s.c.r. (j u is s.c.r. @ uu is s.c.r. We first note that, by symmetry, we need only prove that if u is s.c.r. then so is uu. Suppose then that u, u and uu satisfy the hypotheses of (**), and that u is s.c.r. If 1 uu 122, then the first and last letters of uu are the same as, respectively, the first and last letters of u. This implies that uu is s.c.r., so suppose 1 uu 1 = 1. From [lo] lemma 2.4 and our assumptions on u and u, the following cases can occur. Cpse I. u = u-'uI, where 1 u l ( = 1, and where no cancellation occurs in this product. Since uu = uu, we have that u-'u-'ul = u - ' u l u - ' , and no cancellation occurs on either side of this equation because u is s.c.r. Comparing normal forms for u-'ul and ulu-', we easily see that there exists b E B and r 3 1 such that u = ui'b, u = u;+'b-', where ulb = bul. We easily see that uu = uI, and that 1 u : l = 2. If uu is not s.c.r., then, XI-'


177

by Lemma 4.l(iii), there exists t E X such that t - ' u l t E B. It follows that t-'u:f E B, from which we conclude that there exist c, d E B satisfying u1 = fc = d t - ' . But now u: = dc E B, contradicting u:l = 2. Hence uu is s.c.r. Case ZZ. u = uIu-', where lull = 1, and where no cancellation occurs in this product. This case is similar to Case I. Case 111. u = u l z - ' , u = zul, where no cancellation occurs in these products, and where I ul( = lull = I u l u l \= 1. Since we cannot have I u I = I u I = ( u u l = 1, we conclude that I z ( 3 l . Now uu = uu, and so Cases I, I1 and I11 can occur with the roles of u and u reversed. We may assume Case I11 obtains, for otherwise we could argue as before to obtain that uv = uu is s.c.r. Now we easily see that there must exist f, g E G such that u = ulfu;', u = u;'gul, with no cancellation occurring in these products. We now see that u - ' u = uu-', with no cancellation occurring because u is s.c.r. W e can also see that u I = I z 1 + 1 = I u 1, and so, comparing normal forms, u = u-'b for some b E B f l CG( u ) . But now uu = b E B, contradicting 1 uu 1 = 1. Thus this final case cannot occur. Thus (**) is proved. Now suppose that x and y satisfy the hypotheses of (ii). By.(**), x, y and x y are s.c.r., and so, by symmetry, we may suppose that ( x 13 2, and that either 1 y 1 3 2, or 1 y 1 = I xy 1 = 1. By Lemma 6.2, there exist g, w E G,'h,h ' E B and j , k E Z, satisfying x = ghg-lw' and y = gh'g-'w', where ghg-', gh'g-' and w commute in pairs. We assert that w' 6Z B, for, if not, then we see that x = g ( h g - ' w ' ) = (w'gh)g-', both products being normal forms, and hence x = gh"g-' for some h " E B, contradicting x being s.c.r. We also assert ghg-'E B, for, if not, then, since I x 132, we can apply (**) to the pair ghg-' and w', to obtain the contradiction that ghg-' is s.c.r., but does not have length 0. It now easily follows from our assumptions on x and (when 1 w I = 1) (i) of the lemma, that w is s.c.r. and satisfies 1 w21 = 2 )w 1. Applying a similar argument to y, we see that w' 6Z B, and that, if ( y 1 = 2, then gh'g-'€ B, so that, in this case, we may assume g = 1 and (ii) is proved. It remains to prove (ii) when y I = 1 x y 1 = 1, and it is clear that it is sufficient to show that gh'g-'E B. Suppose this is not so, then we have gh'g-', w' E X , for, if not, we could apply (**) to this pair to obtain the contradiction that gh'g-' is s.c.r. But now we can obtain the same contradiction by observing that )gh'g-'x 12 1, and then applying (**) to gh'g-I and x. Thus gh'g-'€ B, and the proof of (ii) is completed. (iii) Suppose x E X is s.c.r. If CG( x ) B U X, then x commutes with some element y with ( y 1 2 2 . Since ( x y I = 1, we can write x = hw', y = h'w', as in the conclusion of (ii). Since I w21= 2 )w I and I x 1 = 1, by

I

I

I

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B.M. Hurley

(i), j = + - 1. We may clearly suppose j = 1. Writing h" = h'h-*, we see that h " E CG(x) f l B and that y = x k h " . Hence we see CG(x) = ( X ) X ( B n C G ( X ) ) . Since l x Z l = ( h z w Z ( = 2 (iii) , is proved. 0 Our next result is an immediate corollary of part (ii) of the lemma. Corollary 4.10. If ghg-'w' is S . C . ~ .pseudo proper power of length at least 2, then ghg-' E B, so that we may suppose that g = 1, and that no cancellation occurs in the product hw'.

Proof of Theorem 4.7. It is easy to see that cases (i), (ii), and (iii) of the theorem are mutually exclusive, so that at most one case obtains. We shall assume that we have an NFS group (G, B, X), for which case (i) does not hold, i.e. for which Z ( G ) & B , and shall prove that one of the remaining two cases obtains. Let x be an element of Z ( G ) - B of smallest possible length. Clearly every element of Z ( G ) is s.c.r. and so x is s.c.r. If ( x ( = 1, then, by Lemma 4.9(iii), either G = CG(x) C B U X , when clearly X = G - B and (ii) holds, or G = C G ( x ) = ( x ) X ( B fl c G ( x ) ) = ( x ) X B, and we easily see that (x) is infinite cyclic and that X = B{x,x-'}B. In this case, x induces the identity automorphism on B, which is clearly root inner, and since the centre of G obviously is (x) x Z ( B ) , case (iii) obtains. Now suppose 1 x I 3 2. Since G # B, X is nonempty, and so we can choose t E X. Clearly ( x t 1 z= 1, so, by Lemma 4.9(ii) and (iii), t is s.c.r., x E CG( t ) = ( t ) x ( B n CG( t ) ) and ( t ) is infinite cyclic. Thus there exists b E B such that x = t"6, where ( n ( = ( x I. Replacing t by r-', if necessary, we may assume n is positive. If s E X , then a similar argument shows that x = sene,for some c E B and E = & 1, in normal form. Comparing these two normal €orms for x yields s = (td)', for some d E B. So G = gp(t, B). Now suppose e E B, then te E X , and so x = (te)'"f, for some f E B and E = & 1. If E = - 1, then te = ht-', for some h E B, and so ( x I = I(te)-"fl = I he(te)-'"-Z'f)d n - 2, contradicting ( x ( = n. Thus E = 1, and, comparing the two normal forms (et)"-'ef and t"-'b for t - ' ~ ,we see that et = th, for some h E B. But e E B was chosen arbitrarily, and so G is a normal extension of B by (x), and, since (x) is free, it is a split extension. It is now easy to see that X = B { t , t-'}B. Now, x = t"b E Z ( G ) , and so for y C B, we have t-"yt" = byb-'. Since b E B n CG(t),we have t-'bt = b, and so t induces an .nth root of y H byb-'. To see that it induces a primitive nth root of this automorphism, suppose t-"yt" = cyc-' and t-'ct = c, for some c E B, some m 3 1, and all y E B, then we see

Small cancellation rheory

179

t"c E Z ( G ) - B. By the minimality of the length of x = t"b, n S m, as required. It is now easy to see that the centre of G is of the required form, and so case (iii) obtains. 0 85. Small Cancellation Theory over

NFS groups

In this section, we shall assume familiarity with Lyndon (131, and we shall concentrate on the points of difference between Lyndon's theory and ours. For the whole of this section, (G, B,X) shall be a fixed NFS group. Definitions. A set R of c.r. elements of G is symmetrised if whenever r E R so is r-' and so is any c.r. conjugate of r. (From now on R is a fixed symmetrised set.) An element p E G is a piece (or an R-piece) if there exists rI # r2 E R such that rl = p s l , r2 = ps2 with almost no cancellation (see §2) occurring in both products. (Note that we do not require all the elements of R to be s.c.r. and that if R # 0, then any element of B is a piece.) We now state our small cancellation conditions, of which all except

T: are restatements, for NFS groups, of corresponding conditions, from [13], for free products. C @ ) : No element of R is a product of fewer than p pieces. C'(A): If r E R has the form r = ab with almost no cancellation occurring, where a is a piece, then 1 a 1 < A 1 r 1. It is easy to see that C'(A) with A s l/(p - 1) implies C@). Definition. If w E G and w = uxu, with no cancellation occurring in this product, and x EX, then we say that x is a letter occurring in w. If rl, r2 and 13 are s.c.r. elements of R, then, in at least one of the products rlrz, rzr3, 1311, no cancellation occurs. If a l , az and a , are letters occurring in s.c.r. elements of R, then T: alazas# 1. Ti: If r l , . . . , r5 are s.c.r. elements of R , with each r,ri+' # 1, then there exists a G such that each r l = aria-' is in R, and in at least one of the products r : ~ i l +no ~ , cancellation otcurs. Ti: If a ] , az, a,, a4 are letters occurring in s.c.r. elements of R and a l a z , aza3,a3a4& B, then alaza3a4# 1. We now take conditions T: and T: to be the obvious analogs of T: and T:.

Ti:

B.M. Hurley

180

We now state the main result of this section. It is an analog, for NFS groups, of Theorems I and IV of [13].

Theorem 5.1. Let (G,B , X ) be an NFS group, R be a symmetrised subset of G, N be the normal closure of R in G, and w be a nontrivial element of N. Under the additional hypothesis (9

C(6),

it follows that w contains some r from R with three pieces missing: w and r can be written w = bac and r = axIx2x3with almost no cancellation occurring in these products, where x1,x2and x3 are pieces. Under the hypotheses (ii)

C(4), T3 and T:,

or (iii)

C(3),

Tf,Tf, i = 3,4,5,

it follows that w contains some r from R with two pieces missing. 0

It is pointed out by Lyndon [13] p. 225 that, in the free product case, the next result is not quite an immediate corollary of the last. 1 C 1 x 11 + Ix21+ Ix3I. This is because, for example, we may have 1 xIx2x3 However corresponding considerations strengthen the hypothesis C’(A), for free products as opposed to free groups, in a way that more than compensates for this. A similar argument clearly works for general NFS groups.

Corollary 5.2. Let (G,B, X ) , R, N and w be as before. If we have (i)

C‘(A)

for some A d 1/5,

I

then we contains a part a of some r in R, with I a > ( 1 - 3A)l r I. If we have either (ii)

C(A),for some A

S f ,Ti

C‘(h), for some A

d 4,

and

T:,

or

(iii)

and

T:, Tf, i

= 3,4,5.

then w contains a part a of some r in R, with 1 a 1 > ( 1 - 2A)l r I. 0

-

To prove Theorem 5.1, we must first construct diagrams over NFS groups, and then show that these diagrams have all the “usual” properties. We turn to this construction now. It is essentially the same


181

as that used by Lyndon [I31 for diagrams over ordinary free products, and that used by Schupp [19] for diagrams over free products with amalgamation. Let R be a symmetrised subset of the NFS group (G, B, X), satisfying (i), (ii) or (iii) of Theorem 5.1. With certain sequences pl,p 2 , . . . ,p , of conjugates of elements of R we shall associate a diagram M(pl,p 2 , . . . ,p , ) which will be a connected, simply-connected, oriented planar map with a distinguished vertex 0 E M’, the boundary of M. M will be labelled by a function cp into G satisfying: (1) If sl, . . . ,s, are, in order, the edges in the boundary cycle of M beginning at 0, then, writing w = pl * * p,, we have w = cp(sl).. * cp (s,) with almost no cancellation occurring in this product of labels. (2) If D is any region of M and e l , .. . ,ei are the edges in a boundary cycle S of D, then cp ( e l )* . . cp ( e i ) is equal to a c.r. conjugate of one of the pi, and almost no cancellation occurs in this product. By a weak diagram for a sequence pl, . . .,p,, we mean a diagram M satisfying (2), but only a weaker form of (l), namely that the product of the boundary labels, taken in their “correct” order, is equal to w = pl * * p,,. Suppose the (weak) diagram M‘ can be obtained .from the (weak) diagram M by: (a) choosing a new origin 0’, not necessarily lying on the boundary M ’ of M, for M’, (b) deleting boundary edges ending up with a, possibly disconnected, diagram MI’, of which 0’ is a boundary vertex, and (c) deleting the connected components that d o not contain 0‘, then we say M‘ is a subdiagram of M. It is important to notice that, in (b), we not only allow deletion of edges on the boundary of the original diagram M, but we also allow deletion of edges that become boundary edges in virtue of other edges having been deleted. We say a (weak) diagram is reduced if the product of the boundary labels (taken in their “correct” order) of any subdiagram is equal to 1 if and only if that subdiagram is a tree. We say a sequence p , , p2,.. .,p . of conjugates of elements of R is minimal if no shorter such sequence has the same product. 9

Lemma 5.3. A ( w e a k ) diagram for a minimal sequence is reduced. Here we impose no cancellation hypotheses on R .

Proof. Let M be a non-reduced (weak) diagram for a minimal sequence. Then M contains a subdiagram M’ the product of whose boundary labels is 1, and which is not a tree. We can now divide M into diagmms A?’ and M” as shown in Figure 1. The product of the boundary labels of M“ is clearly the same as the product w of the boundary labels of M. Since f i r is not a tree, M” contains less regions

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B.M. Hurley

than M, and so must be a diagram for a shorter sequence. This contradicts M being a diagram for a minimal sequence. 0 Exactly as in [13], we may prove the following lemma. Lemma 5.4. If M is a reduced (weak) diagram, then the label on an interior edge is a piece, provided the boundary path of every path of every region is a simply connected closed path. 0 We recall the initial construction of the diagram M, over a free product F as described by Lyndon [13]. In this case vertices are divided into two classes, primary and secondary. The labels on every edge belong to a free factor X i of F, with the labels on edges meeting while the labels on at primary vertices belonging to different factors Xi, edges meeting at secondary vertices belong to the same factor. Moreover in traversing any path through the diagram, one comes across, alternately, primary and secondary vertices. We shall use an analogous initial construction; however, following Schupp [ 191, we allow 1 to be a label. The precise relationship between the two classes of vertices and the labels on the edges in our initial construction shall be as follows: (a) The label on any edge has length 0 or 1. (b) If vl, v2, v3 are vertices of M joined by edges El = ( v , , v2) and E2 = ( v 2 ,v,) and v I is primary, then 1q(E1)cp(E2) Cl1. Moreover if v l , v2, v3 are consecutive vertices of either the boundary path of some region or the boundary path of the whole diagram (starting at 0), then Q(EI)Q(E2)I= 1. (c) Suppose vi ( i = 1 , . . ., 5 ) are consecutive vertices of the boundary path of some region, v 1 is primary and Ei= (vi,v ~ + ~ for) ,i = 1 , . . . ,4, then 1 Q (El)cp (E,)Q(E,)Q(&)I = 2. Unless we are dealing with a weak diagram, we require an analogous condition to obtain when the vi are consecutive vertices of the boundary path of the whole diagram (starting at 0).

I


183

We introduce two ways of adjusting the labels on the edges of a diagram. (i) Suppose the vertices u, (i = 1,. . . , r) are joined to t h e vertex u by edges E, = (uz,u ) (i = 1,. . . , r ) with labels cp(E,) (i = 1,. . . , r). Then we perform an adjustment of the first kind if, for each i, we change the label on E, to cp(E,)h,where h is a fixed element of B. (ii) Suppose the primary vertices u, (i = l , . . . , r ) are joined to the secondary vertex u by edges E, = (ua,u ) ( i = 1,. . . ,r ) with labels cp(E,) (i = 1,. . . , r). Then we perform an adjustment of the second kind if, for each i, we change t h e label on E, to cp(E,)(cp(E1))-'. Figure 2 illustrates an adjustment of t h e second kind. It is easy to see that adjustments of both kinds preserve properties (a), (b) and (c) of the labeling. Moreover, the precise set of quintuples of boundary points for which the analog of (c) fails remains unchanged by these adjustments. The product w of t h e labels on the boundary arc is unchanged by any adjustment with u # O . If u = 0, then an adjustment of the first kind changes w to K ' w h , for some h E B, and, since we shall take care to make 0 a primary vertex, adjustments of the second kind will have no effect on w. l B = 0, for if R # 0 then any element of B is a piece, and Now R r so, by Lemmas 4.3 and 4.5, any conjugate p of an element of R can be written as p = wrw-' with no cancellation occurring and r E R. Note that, since every element of R is c.r., if p = wfrfwf-Iwith no cancellation and r f E R , then ( w I = I w f l . W e d e f i n e s t ( p ) = I w / . We say a sequence p l , p 2 , . . . ,p . of conjugates of elements of R is special if it is minimal, and, for i = 1,. . . , n, p i p ; . . p : = p I p 2* * p , , with p : a conjugate of something in R for 1 9 j 9 i, implies st @,) =zst @:). Our initial construction of a diagram M ( p l , p 2 , .. . , p n ) for a special sequence is as follows. If n = 0, then the diagram consists in the

x.

P?,rn.."

01

S.arm.rr "art."

".,I..

Figure 2.

184

B.M. Hurley

Tq'f 0

0

Figure 3.

primary vertex 0. If n 3 1, then ,...,pn-l is special sequence, and so we may assume, inductively, that we have constructed a diagram M(pl,.. . . , P , , - ~ )for it. We also assume, inductively, that t h e origin is primary, and that, when all vertices of degree 2 have been deleted, the diagram is, according as R satisfies (i), (ii) or (iii) of Theorem 5.1, a (3, 6), (4, 4) or (6, 3) map. Here, by a ( p , q ) map, we mean one in which each interior vertex has degree at least p and each interior region has degree at least q. We now form the usual 'lollypop' diagram for p,,, making the origin of this diagram primary, and form a weak diagram for pl, p 2 , . . . ,p,, by indentifying t h e origins of the two diagrams. We now form the diagram M ( p l , p 2 ,.. . . , p . ) from this diagram by a process of 'cancellation' o n the boundary, which we shall describe fully below. Before doing so, we note that, in such a process, only edges on the boundary of the region of the 'lollypop' diagram can become identified with an edge on the boundary of some region of M(pl,.. . . ,P , - ~ ) ,for otherwise we could clearly construct a sequence p i , p i , . . . . , pL of conjugates of elements of R, with plp;...pA= pIp2* * .p . and st (p:) < st (p"),contradicting pl,p 2 , . . . . ,p,, being special. Informally, this says that the diagram M(pl, . . .,p , ) will 'look like' one of t h e three diagrams of Figure 3, and t h e same is true of all the intermediate diagrams formed during the cancellation process. In Figure 3, D is the region from t h e 'lollypop' diagram for p,, and u is the only vertex, if any, at which our cancellation process can take place. To see how the cancellation process works, let M * be an intermediate diagram formed during the process. Then there is precisely one quintuple of consecutive vertices ui (i = 1 , . . . ,5) of the ) , i = 1,. . . ,4, for which boundary path joined by edges Ei = (ui, u ~ + ~for


185

1, with u1 primary. Since M * 'looks like' Icp(El)cp(E2)cp(E3)cp(E4)1 one of the diagrams of Figure 3, with u3 = u, we see that u2 f u4. (Our sole reason for using only special sequences in our construction is that it enables us to force this inequality.) We now transform the diagram as follows. (i) If 1 cp (El)cp(Ez)cp(E,)cp(E4)I= 1, then because u2 # u4, it is possible to force cp(E2)= cp(E,)-', by making adjustments of the first kind. Having made such adjustments, we identify u2 with u4 and E2 with E;', and then we make another adjustment of the first kind, if necessary, to restore the value of the product of the labels on the boundary. Now no further cancellations are possible on the boundary, and M ( p , , . . . , p , ) has been constructed. (ii) If 1 cp(El)cp(Ez)cp (E,)cp(E4)I= 0, then we assert ul f u5. For, if not, then we could construct a reduced diagram whose boundary consisted of a single edge with a label of length 0, and which would be (for reasons which will become clear below), according to case, a (3,6), (4,4)or (6,3) map. Corollary 2.4 of [13] now implies that some element of R is a product of two pieces, a contradiction. Since we also have uz # u4, it is possible to force cp(E,)= cp(E4)-'and cp(Ez)= cp(E3)= 1, by means of adjustments of both kinds. Having made such adjustments, we identify uz with u4, E2with Eq', u1 with us and El with ES', and again we make a final adjustment of the first kind, if neccessary. We now see that the diagram, so obtained, 'looks like' one of the diagrams of Figure 3, and, if further cancellation is possible on the boundary path, then it takes place at u I = us, which 'looks like' the vertex u of Figure 3. It is clear that, when we identify two primary vertices in this cancellation process, we obtain a primary vertex. To see that identifying two secondary vertices yields a secondary vertex, consider the situation of Figure 4, where we wish to show, for instance, that ( c d - ' (s 1. We know ( c a - ' ( ,(ad-'ls 1, and if either of these is zero, then we are done. Suppose then that ( c a - ' (= lad-'l= 1, but ( c d - ' ( = 2. Then, putting g = ca-', h = ad-' and k = d, we see that g, h, k E X but gh B U X. So, by N4*,c = ghk $Z B U X, which contradicts our assumption that all labels have length 0 or 1. Thus Icd-ll S 1, as required. We now wish to show that the cancellation process preserves the property of being, according to case, a (3,6), (4,4)or a (6,3) map, when all vertices of degree 2 have been deleted. We easily see that the T-conditions have the effect of excluding interior vertices of the 'wrong' degrees, except that T,' does not exclude that the diagram should contain an interior vertex P of degree d(P) = 4. However

186

B.M. Hurley

Figure 4.

suppose that in the passage from M * to M * * , the diagram obtained by ‘cancelling’ at u3, we have ‘created’ such a vertex. We wish to use a similar construction to that of Lyndon [13] pp. 224-225 to eliminate this vertex. However, with Lyndon’s notation, in our case, Ti only allows us to suppose azalE B, which implies x1 = hx,, for some h E B. However, the vertices PI and P , must be distinct, for otherwise we should have an impossible subdiagram, which is both a (6,3) map and has label of length 0 on its boundary. Hence it is possible to force x , = x 3 , by means of an adjustment of the first kind, and then to eliminate P as in (131. We now show that the boundary of any region of our final diagram M(pl,.. . . , p . ) is a simple closed path. Suppose not, then there exists a region D such that its boundary D’ is not a simple closed curve, but that the submap L obtained by taking 0 ’ and all of M interior to D contains no other such region. Let 1 be a loop contained in D ’ , but not the whole of D , such that the subdiagram K obtained by taking 1 and the whole of M interior to 1 contains as few regions as possible. We see that at most one vertex uo, say, of 1 lies in D ’ . It follows from our T- and C-conditions that every region of K has degree at least p and every interior vertex of K has degree at least q (i.e. K is a [ p , g ] map), where, according to case, (p, q ) = (3,6), (4,4) gr (6,3). K contains more than one region, for otherwise C ( p ) is contradicted, and by [13] corollary 2.4, must therefore, contain more than one vertex. Then, by [13] corollary 2.2, Z.’(p/g + 2 - d ( u ) ) s p , where this sum is taken over the boundary vertices of K. But the T-conditions imply that only uo can make a positive contribution to this sum, which gives us the contradiction that Z:’(p/4+ 2 - d ( u ) ) < p. All the hard work in the proof of Theorem 5.1 is now over. We merely now need to remark that it is obvious that every w in N, the normal closure of R, can be written as a product of a special sequence,


187

and the theorem now follows in the same manner as Theorem I of Lyndon [13]. It is also possible to use corollary 2.4 of [13], and our construction of a diagram, in a similar way to prove our next result.

Lemma 5.5. Let (G,B, X ) , R, N and w be as before. Suppose further that w is not conjugate to an element of R. If we have 6)

C’(A), for some A

S

1/5,

then some s.c.r. conjugate of w contains two disjoint subwords, a l and az, such that, for i = 1,2, each ai is a part of some ri in R, with 1 ai 1 > (1 - 3A ) 1 ri I . If we have either (ii) or (iii)

C’(A), for some A

S ! , Ti

C’(A), for some A

S;,

and Ti,

T:,Tf,

i

= 3,4,5,

then some s.c.r. conjugate of w contains two disjoint subwords, a l and a2, such that, for j = 1,2, each a, is a part of some rj in R, with I a, I > (1 - 2A)(rj I. There is, of course, a non-metric version of this last lemma, which the interested reader might like to formulate for himself/herself. We end this section with a few remarks on our results. Firstly it is clear that, by making appropriate adjustments of the second kind to our final diagram, the subwords a of w, in Theorem 5.1 and Corollary 5.2, and the two subwords ai ( i = 1,2) of an s.c.r. conjugate or w, in Lemma 5.5, can be taken to consist of a consecutive sequence of ‘letters’ from the normal form of w, or, respectively, the s.c.r. conjugate of w. However, we cannot take a (respectively ai, for i = 1,2) to consist of a consecutive sequence of ‘letters’ from a normal form for r (respectively ri, for i = 1,2), without allowing for amalgamations to take place when we express it as a subword of w (respectively an s.c.r. conjugate of w ) . Our second remark concerns the strength of the C-conditions. Suppose R satisfies C’(A), and r E R has the form r = ab with almost no cancellation occurring, where a is a piece. By definition, there exists r ’ E R, which can be written r’ = ab’ with almost no cancellation occurring, with r# r’. This fact can be illustrated by the two region diagram of Figure 5. If r is not s.c.r., then, provided I a 1 # 0 , we can make an adjustment of the second kind at the vertex v, which, obviously, is secondary, to make r s.c.r., while leaving r‘ c.r. and

B.M. Hurley

188

Figure 5 .

leaving the length of a constant. We thus see that C‘(A) implies that if r E R and r is not s.c.r. and r = ab, almost n o cancellation occurring, and a is a piece, then 1 a < r 1 - 1). Our third remark concerns the formulation of Corollary 5.2 and Lemma 5.5. For example, in Corollary 5.2(i), to conform with the ‘usual’ way of formulating such results, we have written l a 1 > (1 - 3A)(r 1. But it is clear that a stronger result holds, namely, that we can write r = ab, almost no cancellation occurring, with I b < 3 A I r 1. We shall have occasion to use this stronger form of our results in the next section. The easiest way if doing this, while avoiding undue circumlocutions, is simply to agree that, when we say that a is more than a of r, what we mean is that r = ab, for some b, almost no cancellation occurring, where b 1 < (1 - a ) [r 1. This convention has the advantage that, if a is more than half of r, then I b < 1 a Our fourth remark concerns the applications of the results of this section. We shall only apply Case (i) of the various results; we have already given our reasons for including the other cases. The results are so fundamental to the subject, and shall be used so often, that it would become tiresome to refer to them by name at each application. Accordingly, we shall not do so, but we d o not anticipate that this will cause the reader undue difficulty. We shall, in fact, be mainly concerned with proving embedding theorems, for which purpose, we shall, for the most part, need Corollary 5.2, or, to be precise, a trivial corollary of it. We leave the reader to judge whether or not, given Corollary 5.2, the next result is worth stating at all. However, if it is to be stated, it must clearly be called a theorem.

I 4

I

I

I

I.

-

Theorem 5.6. Let (G,B, X) be an NFS group, R be a symmetrised subset of G, N be the normal closure of R in G, and w be a non-trivial element of N . If we have C‘(A), for some A S 115, and R contains no element of ( B UX)’, then G I N embeds B U X . Proof. N contains no element of (B U X)’.(Note that if r E R and ( r (S 5 , then every R-piece contained in r has length 1 . ) 0

189


06. Control over torsion, conjugacy and the centre in 9th and 10th groups

-

In this section, ( G , B , X ) is a fixed NFS group, R is a fixed symmetrised subset of G, N is the normal closure of R in G, and F is the factor group GIN. Lemma 6.1. Suppose R contains no p.p.p. Zf x m a E R, where m, ( x m aI 3 2, x is s.c.~.with Ix 1 3 1, and almost no cancellation occurs in the product ( x " ) a , then ax # xa, and x and x " - I are pieces. Proof. Suppose ax = xu. Since R contains no p.p.p., 1 a 15 1. Applying Lemma 4.9(ii), x = hw' and a = h w ', for some h, h ' E B, w E G and j , k E Z - { 0 } , where w iss.c.r., ( w 2 1 = 2 ( w (and , h , h ' a n d w commute in pairs. Now x m a = (hm'h'k)wm'+kis not a p.p.p., and so mj + k = & 1. Obviously, ~ j ~ , ~ k ~ and , / w so ~ ~ l ,

d ( x " , a - ' ) = l { m l j l l w ( +Ik IIw I- Imj

+ k 1 1 w /}a1,

and so cancellation occurs in ( x m ) a ,contrary to hypothesis. Thus a x # x a , and so x " - ' ( x a ) # x " - ' ( a x ) E R and x ( x " - ' a ) # x ( x " - * a x ) E R, with almost no cancellation occurring in any of these products. Hence x and x"-' are pieces. 0 Corollary 6.2. Suppose R satisfies Cf(1/5), and contains no element of ( B U X)' and no p.p.p. Suppose x E X , b E B, m 2 1 and 1 x ' J = 2, then x"b# 1 in F. Proof. Since F embeds B U X , we may assume m 3 2. Since b E B, it is a piece. If x"b E R, then, by the lemma, it is a product of three pieces, contradicting Cf(1/5). Thus, if x"b = 1 in F, then some s.c.r. conjugate of x"b contains two disjoint subwords, each of which is more than 2/5 of some element of R. From this it easily follows that some element of R has the form ~ " awith , almost no cancellation occurring, and a being a product of at most three pieces. If 1 a = 0, then, as above, this element of R is a product of three pieces, contradicting C'(1/5). Thus I a 3 1, and so, by C'(1/5), lxma I 3 6 , and so Ix" 13 2. From the lemma, x and x"-' are pieces, and so x m a is a product of 5 pieces, contradicting C'(1/5). Hence x"b# 1 in F, as required. 0

I

I

Corollary 6.3. Suppose Z ( G )P B, then, if R satisfies C'(1/5) and contains no element of ( B U X)' and no p.p.p., then R = 0.

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Proof. By Theorem 4.7, if Z ( G ) e B, then either X = G - B, when there is nothing to prove, or every element of G - B has the form x"b, where x E X , b E B, m 3 1 and ( x ' ) = 21x I. Thus, in the latter case, by Corollary 6.2, F embeds G, and so R = 0. 0 Theorem 6.4 (Torsion theorem for 9th groups). Let (G, B, X ) , R, N and F be as above. Suppose R satisfies Cr(1/9), and contains no element of ( B U X)' and no p.p.p. Then every element of finite order in F is conjugate in F to an element of B U X . Hence F contains an element of finite order n i f and only if G does. Proof. Let w have finite order n in F, and let C be the set of all elements of G whose images in F are conjugate, in F, to w. Let z E C have minimum possible length. If z 13 2, then z is s.c.r. and no c.r. conjugate of z can have the form ab, almost no cancellation occurring, with a more than half an element of R. Since R contains no p.p.p., z " @ R, and so contains more than 5/9 of some element of R. Replacing z by an s.c.r. conjugate, if necessary, we may suppose Z" = ub, no cancellation occurring, where u is more than 6/9 of some element of R. Since u is not a subword of z, by [lo] lemma 2.4, u = zmt and z = ts for some t, s E G, no cancellation occurring in either product. Thus some r E R has the form r = uu = (ts)"rv, with almost no cancellation occurring in the product uu and no cancellation anywhere else. If m 2 2 , then, by Lemma 6.1, (ts)"-' and ts are pieces, and we easily see that t is also a piece, which implies that u is the product of three pieces contradicting its being more than 6/9 of an element of R. If m = 1, then r = tstu and t is a piece unless tstu = tuts. But, in this case, either u = suI,no cancellation occurring, when, by Lemma 6.1, ts, and hence t, is a piece, or s = us1, almost no cancellation occurring, when again t is a piece. Thus t is less than 1/9 of r, and, since tst is more than 6/9 of r, we see that u is less than 3/9 of r. Hence:

I

But z = ts = (tu)-' in F, which contradicts the minimality of 1 z Hence 1 z 13 1, and the theorem is proved. 0

1.

Lemma 6.5. Suppose R satisfies Cr(l/lO), contains no element of ( B U X ) ' , and contains no element that is conjugate to its own inverse. Then for each g E G, there exists g E G such that g = g in F, and, for all s, t E B U X , if g-'sg = t in F then g-'sg = t in G.


191

Proof. We prove the stronger result (***). (***) Suppose R satisfies the hypotheses of Lemma 6.5. If g E G, then any element g E gN of shortest possible length satisfies the conclusion of the lemma. Suppose (***) to be false. Let g E G be an element of smallest possible length such that there exists g, of smallest possible length in gN, that falsifies (***). Clearly g has smallest possible length in gN = gN, g-'sg = t in F if and only if g-'sg = t in F, and, by the minimality of the length of g , 1 g 1 = I g 1. Thus we may assume g = g. Note that here we are taking a new choice for g ; we d o not claim that, by taking g to be our original g , the conclusion of the lemma would not hold. Now there exist s, t E B U X such that g-lsg = t in F, but not in G. It easily follows that ( g 1 s 2 . Write g = g l * . . g , ( r 3 2 ) in normal form. We assert I g;'sgll, I g,tg;'( s 2 . Suppose, for instance, that g;'sgl = sIE B U X. Then h = g 2 * g , must be a shortest element of hN, and, by the minimality of g , since h-'s,h = t in F, this equation also holds in G. But this implies the contradiction that g-'sg = f in G. Hence 1 g;'sgl 1 3 2, and, similarly, 1 g,tg;') 2 2 . The word g-'sgt-' must have an s.c.r. conjugate that can be written, in normal form, as g ; ' . . . ( g ; ' s ) g l * * (g,t-') or as one of the eight similar forms obtained by allowing s to amalgamate with g l instead of g ; ' , or by allowing t - ' to amalgamate with g;' instead of g , or by allowing both s to amalgamate with g l and t - ' to amalgamate with g ; ' , or by allowing less amalgamations. Call this s.c.r. conjugate of g-lsgt-', w. If w E R, then some c.r. conjugate of w can be written ga, with almost no cancellation occurring, and some c.r. conjugate of w - ' can be written gb, with almost no cancellation, but 1 g I S w 1, and so, by C'(l/lO), w w - ' , contrary to hypothesis. Thus w $Z R, and since w = 1 in F, some s.c.r. conjugate of w contains two disjoint subwords each of which is more than 7/10 of some element of R. At least one of these subwords does not "involve" the letter t - ' , and so there exists an element u E R, 7/10 of which has the form:

-

-

u = g;'

* * *

g;'sgl

*.

*

gk,

where, at most, an amalgamation occurs between g;' and s or between s and g , , but not at both places. From the minimality of g, neither g l * gj nor g l . . . g k is more than half an element of R, and it easily follows that both have length > 1/10]u I. By C'(l/lO), u u-I, contrary to hypothesis. This contradiction proves (***). 0

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Corollary 6.6. If R satisfies the hypotheses of Lemma 6.5, and contains

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no p.p.p., then F embeds the centraliser of every s.c.r. element of length 1, and, if x is such an element, then C F ( x )= CG( x ) , with the “obvious” identification. Proof. It is immediate from t h e lemma that if x E X is s.c.r., then C,(x) is the image of C, (x). If C, (x) C B U X, then it is embedded in F, and the result follows. If CG(x) B U X , then, by Lemma 4.9(iii), every element of CG( x ) has t h e form x”b, where m E Z and b E B. Since 1 x z (= 2, it follows from Corollary 6.2, that F embeds C, (x), and the result holds. 0 The next result is an immediate corollary of Lemma 6.5, but, in view of its content, we label it a theorem. Theorem 6.7. Let (G,B, X ) , R, N and F be as above. Suppose R satisfies C’(l/lO), and contains no element of ( B U X ) * and no element that is conjugate in G to its own inverse. Then any two elements of B U X are conjugate in F i f and only if they are conjugate in G. 0 Theorem 6.8. Let (G,B, X), R, N and F be as above. Suppose R satisfies C’(l/lO), and contains no element of (B U X)’, no p.p.p., and no element that is conjugate in G to its own inverse. Then Z ( G ) is embedded in F, and, with the “obvious” identification, Z ( G )= Z ( F ) . Proof. If R = 0, then the result is trivial, so suppose R # 0. By Corollary 6.3, Z ( G ) cB, and so is embedded in F. But z E Z ( F ) if and only if z-’sz = s, for all s E B U X , and so, by Lemma 6.5, Z ( F ) must be the image in F of Z ( G ) , and the theorem is proved. 0 97. SSC groups

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In this section, we consider a class of NFS groups, each member of which has many “nice” small cancellation quotients. In the next section, we shall be concerned mainly, but not exclusively, with quotients of groups from this class. We start with some definitions.

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Definitions. Let R be a symmetrised subset of the NFS group ( G , B , X ) . By the basis number b ( R ) of R we mean the number of equivalence classes in R under the equivalence relation “ r is conjugate to s or s-”’. A basis for R is a set of representatives for these classes. If S is a basis for R, then Card(S) = b ( R ) S Card(R), but, in general, we need not have b ( R ) = Card(R).

193


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Definition. Let K be an infinite cardinal. Then an NFS group ( G , B , X ) is an SSC-K group, if G I = K and it contains a symmetrised set R satisfying C’(1/6),b ( R ) = K , and the centraliser condition (C.C.) below. C.C. Every element r E R has an s.c.r. conjugate r‘ satisfying

I

CG(r’) n B

= Z(G).

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An SSC group is defined to be an NFS group that is an SSC-K group for some K . Before we give examples of SSC groups, we need a concept introduced by Schupp [ 191. Definition. Let A be a subgroup of the group H. Let {xl,x2}be a pair of distinct elements of H, neither of which are in A. We say that {x,,x2}is a blocking pair for A in H if the following two conditions are satisfied: (i) x f x p & A , l s i , j G 2 , E = k l , S = ? l , u n l e s s x f x ~ = l , (ii) if a E A , a # 1 , then x f a x p @ A ,l s i , j s 2 , E = k l , S = * I . Theorem 7.1. (i) A non-abelian free group of rank r, with the “usual” NFS structure, is an SSC-K group, where K = max (r, No). (ii) A free product of two or more non-trivial groups is an SSC-No group, with the single exception of the group C2* C2, provided it has order No. (iii) A free product with amalgamation ( A * B : H = K ) , where H A, K B, H = K , A # H and B contains a blocking pair over K , is an SSC-No group, provided it has order No. (iv) A n H N N group ( H ,t : t-’At = A+), where : A B is an isomorphism between subgroups of H and I H 1 = No, is an SSC-No group under any of the following conditions: (a) There exists x E H, x bZ A, x $Z B such that x-’Bx n A = 1. (b) There exist x, y E H such that x - ’ A x f l A = y-’By n B = 1. (c) There exists m 2 1 such that C,,,= D,,, = 1, where Co= A, Do = B, and, for n 2 0 , Cn+,= A n +(Cn) and Dn+, =B f +-‘(Dn). l (v) A group that can be presented with at least three generators and only one relator can be equipped with an NFS structure in such a way that it becomes an SSC-No group. In all the above cases, the SSC groups have a trivial centre.

+

-

Proof. These are the “classical” cases. The existence of the appropriate symmetrised sets follows from the results of [13],[18] and [19].The fact that these groups have trivial centre follows easily from the

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construction used in each case. Moreover C.C., which now has the form: “Every element r E R has an s.c.r. conjugate r‘ satisfying CG( r ’ ) r l B = l”, is easy to check. For example, in case (iii), if k E K, k # 1, then k cannot commute with either element of the blocking pair { x I , x z }in B over K, and so t h e group has trivial centre. If R is t h e symmetrised set constructed in [19] (to be precise, R is the symmetrised set of [19] 04 with C = l), then for r E R, let r’ be an s.c.r. conjugate of r whose first letter in its normal form is xT’. If r’ commutes with k E K, then cancellation takes place in r’kr’-’, which implies that x ; k x : E K, for some E = 1, 7 = & 1. From the definition of a blocking pair, we now see k = 1, and so C . C . holds. We remark that Murasugi [15] first noticed that the one relator groups of (v) have trivial centre, however, Sacerdote and Schupp [18] show that they are all HNN groups of type (iv)(a), and, in the present context, this is t h e easiest way of seeing that they have trivial centre. 0

*

Lemma 7.2. Let (G,B, X ) be an SSC-K group. Let real numbers A, n > 0 be given. Then G contains a symmetrised set R satisfying C’(A), b ( R ) = K , C.C., I r > n for all r E R, and containing no p.p.p. and no element conjugate to its own inverse.

I

Proof. By definition, G contains a symmetrised set R I satisfying C’(1/6), b ( R ) = K , and C.C. Partition a basis SI for R 1into pairwise disjoint triples. To each such triple U = (s,,s2,SJ, say, we associate a word W ( U ) , which we define to be an s.c.r. conjugate of:

n M

,=I

{SIS2(SIS3)’),

where M is some large number. We now take the set of all W ( U )to be a basis for a symmetrised set R. Now in each W ( U ) ,less than 1/6 of each s, is lost through cancellation either at the beginning or-at the end of this s,, which leaves at least 2 as a subword of W ( U ) . But since each s, can be identified by as much as 1/6 of it, it is clear that we can force R to satisfy all the required conditions, except, possibly, C.C., merely by making a large enough choice for M. To see that we can force R to satisfy C.C., consider how cancellation can occur on the left-hand side of the equation W ( U ) - ’ h W ( U ) =h, which we shall assume holds for some h E B. Lemma 2.l(i) of [lo] tells us that, as soon as an amalgamation takes place, the left-hand side becomes reduced, which implies that no amalgamations take place. But this implies that some a E G exists such that when we write W ( U )= uav


195

and r = ab, n o cancellation occurring in either product, with a more than 1/6 of si E U, and r an s.c.r. conjugate of si,then an equation of the form k , a = ak2, with k , , k Z E B, and k, a conjugate of h, must hold. Replacing si by an s.c.r. conjugate, if necessary, in the definition of W ( U ) , we can write W ( U )in t h e above form where r satisfies C G ( r )n B = Z ( G ) . Since a is not a piece and kla = akz, we see that k, must commute with r. Hence k l E Z ( G ) , and, since h is conjugate to k,, h E Z ( G ) . 0

Remark. By a similar argument to that used above, we see that if R 1 satisfies C.C. and contains no element of X , then any s.c.r. element r E R I satisfies C G ( r )n B = Z ( G ) , and so no special choice of a c.r. conjugate for si in the definition of W ( U )is needed, in this case, for the conclusion of the lemma to hold. The next result is easy to see and its proof shall be omitted.

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Lemma 7.3. Let R be a symmetrised set, in the NFS group (G,B,X ) , satisfying C'(1/6), C.C., 1 r 1 2 2 for all r E R, and containing no p.p.p. and no element conjugate to its own inverse. Then, for all r E R, CG ( r ) f l ( B U X ) = Z ( G ) . 0 Corollary 7.4. If M is a normal subgroup of the SSC group (G,B, X ) and M d B, then M d Z ( G ) . Proof. By Lemma 7.2, G contains a symmetrised set R satisfying the hypotheses of Lemma 7.3. Let r E R and rn E M. Then r-lrnr = m , E M. Since r cannot be an R-piece and mrm = rm ,rn E R, we see m = m , , and so m E C,(r) f l B = Z ( G ) . 0 It is, perhaps, worth remarking that the last result can be strengthened to the statement that any bounded normal subgroup of an SSC group lies inside the centre, where we say S C G is bounded if S C (B U X)", for some n.

Lemma 7.5. If (G,B, X ) is an SSC group, then for any subgroup M Z ( G ) , G / M is an SSC group and Z ( G / M )= Z ( G ) / M . In particular, G / Z ( G ) is an SSC group with trivial centre. Proof. Let us write a bar over symbols to denote image in = G/M. It is obvious, from Lemma 7.2, that X # G - B, and that we d o not have X = B { t , t-'}B, with G a split extension of B by an infinite cycle

B.M. Hurle y

1%

x)

(t). It follows that (G, B, does not have either of these two forms, and so, by Theorem 4.7, Z ( G )S B. Let R be the symmetrised subset of G constructed in the proof of Lemma 7.2. Let p E G, and suppose that is an I?-piece. (Note that, since the map G + G is length preserving, R is a symmetrised subset of G.) Then there exist a, b E G and rl # rz E R, such that rI = p a (mod Z ( G ) )and r, = pb (mod Z ( G ) ) , and it follows that p is an R-piece. Thus R satisfies C'(1/6) if R does. Now suppose r E R and h E B. If h E CG(T) n B, then r-lhr = hz, for some z E Z ( G ) . If R satisfies C'(1/6), then z = 1. Since R was constructed as in the proof of Lemma 7.2, we may assume it satisfies the hypotheses of Lemma 7.3. Thu? h E Z ( G ) and R satisfies C.C. Thus G/M is an SSC group. Since Z ( G / M )C B, the inverse image, in G, of Z ( G / M ) lies in B, and hence, by Corollary 7.4, lies in Z ( G ) . Hence Z ( G / M )= Z ( G ) / M . 0

Lemma 7.6. Let (G,B , X ) , A, n be as in Lemma 7.2, let R1 be a symmetrised subset of G satisfying C'(1/6) and C.C., and suppose T is a finite subset of G - 1. Then there exist symmetrised subsets R, and R, of G, both satisfying the conclusion of Lemma 7.2, such that (i) every subword a or r E R,, satisfying 1 a I S A 1 r 1, contains every element of T as a subword, (ii) R, C gp(R,, T ) and (iii) R, C gp(R,, T) fl (t)").

(n,,

Proof. Let T = {tl, t,, . . . ,t s } . By Lemma 7.2, there exists a symmetrised subset R * of G satisfying C'(1/20), b ( R * )= K , C.C., J r (>20(ti 1 for all r E R * and all ti E T, and containing no p.p.p. and n o element conjugate to its own inverse. From the form of the construction of R * , we see R * E gp(RI). It follows from Lemma 6.1, that there exists, for each i = 1,. . . , s and each E = ? 1, rn ( i , E ) 3 1, such that, for all r E R * , r(t:)m(i.')rcontains t ; as a subword, but

Now, we construct R, from R *, by first partitioning a basis S * for R * into pairwise disjoint triples. If U = (sl, s,, s3) is such a triple, we first form the modified triple U' = (si, s2, ss), where s; = S l t ; " ( l . l ) S , t ~ ( Z 1s1 )

. ..

t~(s.l)slt;m(l.-l)

st..

.

t ; m ( s . - 1 ) s1.

We now construct R,, by a similar construction to that of Lemma 7.2, but using modified triples U' instead of U. It is clear that we can make R, satisfy (i) and (ii). We illustrate the construction of R, in the case T = {t}. The general case follows from an obvious induction argument. The construction is


197

similar to the construction of R2, but we take our modified triples to be U ”= ( s y , s;, s‘& where s’i = s;’tsi. Since R * contains no element conjugate to its own inverse, we see that this works. Obviously R 3 C gp(Rl, t ) n 0)“. 0 We now come to our most important “New symmetrised sets for old” construction.

Lemma 7.7. Let (G,B, X ) be an NFS group, and suppose A, p are real numbers satisfying 0 < 3A < p < f. Let R , be a symmetrised subset of G, which satisfies C‘(A), and which contains no element of length less than 2 + 2/(p - 3A), no element conjugate to its own inverse, and no p.p.p. Suppose S , is a basis for R,. Let cp : S , + B U X be a function such that, for all s E S , , (scp)s is c.r. and CG(s) n ( B U X ) C CG(scp). Let S = {(scp)s : s E S , } , and let R be the symmetrised closure of S. Then R satisfies C ’ ( p ) , and contains no element of length less than 2 / ( p - 3A), no element conjugate to its own inverse, and no p.p.p. Proof. Since the length of an element of G is decreased, at most, by 2 when it is premultiplied by an element of B U X , it is clear that R contains no element of length less than v = 2 / ( p - 3 A ) . Now suppose we can write r, = pa, r2 = pb, almost n o cancellation occurring, with r , # r2 E R. By the comments following Corollary 5.2, we may suppose r, to be s.c.r. Suppose 1 p 1 3 3A I r , I 2, then it is clear from the construction of R that p = qlflq2f2q3, with no cancellation occurring, with f , , f 2 E B U X, and with q,, q2 and q3 being subwords of r , and r2 that were not “disturbed” during the construction of R from R,. Clearly at least one of the qi is not an R,-piece. This qi identifies an element s E S , , from which r, and r2 were both constructed, and it is clear from the fact that R, contains n o element that is conjugate to its own inverse that scp occurs in r, and r2 t o the same power. Thus taking inverses of both elements, if necessary, r, = s1(scp)s2 and r2 = s3(scp)s4, where sIs2= s3s4 and s2sI= s4s3= s. It follows from these equations that s;IsI commutes with s. Write h = s;IsI. There exists x E X such that x-lsx is s.c.r., and so, from Lemma 4.9(ii) and the fact that R , contains no p.p.p., we see that x-lhx E B. Thus 1 h 13 2. If 1 h 1 = 2, then from hs = sh, we see that s = xux-’, with n o cancellation, and is not c.r., contradicting R , being symmetrised. Thus h E CG (s) n ( B U X ) C CG ( ~ c p ) . Thus r , = s , ( s ~ ) s=~ ~ 3 (h~ c p ) s Z = s3(scp)hs2= s3(scp)s4 = r2. Thus if r , # r2, then we must have:

+

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B.M. Hurley

Thus R satisfies C'(p). Suppose R contains some element conjugate to its own inverse, then we easily see that this element has two subwords that are mutually inverse, have not been disturbed in the construction of R from R l , and are too long to be R1-pieces. Thus R , contains an element that is conjugate to its own inverse, contrary to hypothesis. Thus no element of R is conjugate t o its own inverse. Suppose R contained a p.p.p. Then S I would contain an element of t h e f o r m bxm,where b E B U X , m a 2 , I b x " I a 3 , x iss.c.r. and ( x 13 1. If almost no cancellation occurs in b(x"), then, by Lemma 6.1, x m - l is an R1-piece and C'(1/6) fails for R,. If cancellation occurs, then either m 3 3, x " - ~is an R1-piece and C'(1/6) fails for R , , or m = 2, and b ( x 2 )= x i * ' * x,xlx2 . x, in normal form, where x = x1* x, and x i x i ' E B, and again C'(1/6) fails for R,. But this contradicts A < 1/6. Thus R contains no p.p.p., and the lemma is proved. 0

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Remarks. When applying Lemma 7.7, it is usually easiest not to compute exact values for A and p, but to think of the lemma as showing that if we wish to construct R, satisfying "strong enough" cancellation conditions, then it is merely necessary to impose "strong enough" conditions. However, we have also imposed conditions on the mapping 9. The condition that (scp)s be c.r., for all s E S I , is nearly redundant, and one could d o without it altogether by insisting that p and A be small enough. In any case it does not seem to lead to any difficulties in practice. The condition that CG(s) r l (B U X ) C CG(scp), is, by Lemma 7.3, automatically satisfied if R satisfies C.C. and G is an SSC group. However, if G is not an SSC group, then it requires careful checking. 88. Applications

Application I. Simultaneous quotient groups Definition. Let 2 = { L j } j Ebe J a family of groups, then, by a simultaneous quotient group of 2, we mean a group that is a homomorphic image of each Lj, if J # 0, while, if J = 0, we allow any group to be a simultaneous quotient group of 2. Theorem 8.1(A). Let K be an infinite cardinal. Let 2 = { L j } j Ebe J a family of SSC-K groups, with IJ 1 S K . Suppose G is any group satisfying 1 G 1 d K . Then G can be embedded in a simultaneous quotient group Q of 2. 0


199

Before we prove Theorem 8.1(A), we prove a technical lemma. Before stating the lemma, we will define some notation that will be of use throughout this section.

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Definition. We say that a symmetrised set R, in an NFS group satisfies [A, n ] ' , for A, n > 0 , if it satisfies C'(A), and contains no p.p.p., no element conjugate to its own inverse, and no element of length less than n. We say it satisfies [A, n ] , if it satisfies [A, n]' and C.C., and we say it satisfies [A, n, K ] , if it satisfies [A, n ] and b ( R ) = K .

Lemma 8.2. Let J be a non-empty set, { K , } , E J be a family of infinite cardinals, let L, = (L,,B,, X , ) be an SSC-K,group, for each j E J, and let R, be a symmetrised subset of L,, with basis S,, and satisfying [ A , n, K , ] , for each j E J, where A 6 1/6 and n > 12. Let

F

=

@ IEJ

(N)

Li,

-

the free product with centralised centres of the L,, with the NFS' structure induced on it by the L,. Let R be the symmetrised closure of U , R , in F, and let S = u , S , . Then S is a basis for R, and R satisfies [ A , n, E, K , ] in F. Proof. We first prove the lemma in the case Z ( L j ) =1, for all j E J. In this case, F is simply the free productN of the SSC groups Lj. The base group or F is simply B = * Bp It is clear from Corollary 4.3 that any element of R can be obtained from one in u R j by conjugating by an element of B. Let p be an R-piece, then there exist p u f p u E R, with almost no cancellation occurring in these products. Let pu = b;'rlbl and p u = b;'r2b2 where b l , b2E B, rl E Ljcl,and r2 E Lj(*),for some j(l),j(2)E J. It is clear that we may assume that when bi is written in free product normal form, it does not end in a letter from Ljci,.Now let $aibe the first letter in the (Lj(i),Bjci),Xj(i,)-normal form for ri. If ( p 1 2 2 , in F, then a;'b1b;'a2E B, and from the free product normal form for this element, and the fact that ai E Bjci,,we see that bl = b2 and j(1) = j(2)= j , say. Thus premultiplying p by an element of B, we may assume, if ' ( p1 3 2, that pu = rl and p u = r2 where rl, r2 E Rj. A similar argument applied to the products up and up, shows that, postmultiplying by an element of B, we may assume that p E L p It is now clear that R satisfies C'(A), at least as far as pieces of length at least 2 are concerned. But our assumption n > 12 ensures that shorter pieces cannot give us any trouble. It follows from the above argument that p E F, with J p1 3 2, is an

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R-piece if and only if there exist b, c E B, u,u E F, u',u ' E L,, for some j E J, and an R,-piece p ' such that p ' u ' f p ' u ' E R,, almost no cancellation occurring, and p = b-lp'c, u = c-'u'b and u = c-'u'b. Using this, it is easy to see that R contains n o p.p.p., and no element conjugate to its own inverse. Clearly S is a basis for R , and the remainder of [A, n, 2 K , ] is easy to check. The general case is proved as follows. We write L, = L , / Z ( L , ) ,and F = *,€, (N)L,. The homomorphisms L, + L,, induce a unique homomorphism F F, which is clearly length preserving, and hence, an N9.Y-homomorphism. We write for the image of Y under this homomorphism, for any Y. As in the proof of Lemma 7.5, R, satisfies [A, n, K , ] , and clearly has basis $. From the Z ( L , )= 1 case, it is clear that R satisfies [A, n, C K , ] , and has basis Since R is the inverse image, in F, of R, it clearly must have all the desired properties, except, perhaps, C.C. But C.C. follows from the fact that every element of R is conjugate to an element of R,, for some j E J. 0

-

s.

Proof of Theorem 8.1(A). If J = 0, then t h e result is trivial, so suppose J f 0. Let H = (G *N F : [G,Z ( F ) ]= l), where G has the NFS structure (G, 1, G - l), and F is as in Lemma 8.2. We pick a sufficiently small A > O , and a sufficiently large n > O , and we let S, be a basis for a symmetrised set R, satisfying [A, n, K ] in each L,. By Lemma 8.2, s = U S , is a basis for a symmetrised set R in F satisfying [A, n, K ] . Clearly R is also symmetrised in H and satisfies the same condition. Now, for all j E J, let T, = uk+,€J (&. U xk U G), where & xk are as in Lemma 8.2. Then 1 T, I S K = 1 S, 1, for all j E J, and so there is an onto map (p, : S, T,. Let (p : S + F be the map whose restriction to S, is cp,, for all j . Since R satisfies C.C., and (scp)s is clearly c.r. for all s E S, we see from Lemma 7.7, that the set S * = {(s(p)s : s E S } , is a basis for a symmetrised set R * in H, satisfying [ p , m,K ] , for some convenient p, m, provided only we choose A, n "correctly". If we choose them so that p s 1/6, and m 3 6 , then letting N * be the normal closure of R * in H, H / N * embeds everything in H of length 0 or 1. Hence H / N * embeds G. However, it is clear, from the definition of S*,that given the obvious presentation for H / N * , we can remove from it, by means of Tietze transformations, all generators the generators of L,, for some fixed, but arbitrary, j E J. It follows that H / N * is a simultaneous quotient group of 2, and so the theorem is proved with Q = H / N * . 0

-

Before we turn to other versions of Theorem 8.1, we note the following corollary.

20 1


Corollary 8.3. A n y countable group G can be embedded in a countable group Q with the property that, for any word w(a, 6, c ) in ( a ,b, c ) , the free group on {a,b, c } , there exists a generating set {x,, y,, 2,) of Q such that w(x,, y,, 2,) = 1, in Q. Proof. By Theorem 7.1 (v), the three generator one relator groups are SSC-N,, groups. There are clearly No of them, and so we can apply the theorem to embed G in a simultaneous quotient Q of them. This group clearly satisfies the conclusion of the corollary. 0 It is well known that any group of order K , where K is an infinite cardinal, can be embedded in a simple group of order K . We vary t h e proof of Theorem 8.1(A) as follows. (a) Replace the family K , of SSC groups, by 9 = 9 U F,, where F. is the free group of rank K . a simple group of order K (b) Replace the group G by containing G. (c) Add to S * all elements of the form:

c,

where g is some fixed element of G, M is some “large” number, and t is any element of some basis of F,, to form the set S * * . (d) Let R * * be the symmetrised closure of S** in H, and let N** be the normal closure of R * * in H, and put Q**= H / N * * . (e) Let N be a normal subgroup of Q** that is maximal with respect to the property that g 6Z N, where g is t h e element used in (c), and let Q = Q * * / N . Then we see that, by choosing M large enough, we can make R * * satisfy any condition [ p , m, K ] that R * satisfies. By requiring p S 1/6 and 1 m I > 6, we can ensure that Q** embeds and since N** 2 N*, where N * is constructed as in the proof of Theorem 8.l(A), Q** is a simultaneous quotient of the family 9. Now Q embeds for, if not, then N n C?# 1, and since is simple g E n N, contradicting t h e definition of N. Finally, Q is simple. For, let P be a non-trivial normal subgroup of Q, then the inverse image P** of P in Q** contains g, and since N must contain all the elements of H introduced in (c), we see that t E P**,for every generator t of F,, and since P** is a quotient of F., P**= Q * * , and so P = 0.Thus we have proved the following version of Theorem 8.1.

c,

c

c,

Theorem 8.1(B). In Theorem 8.1(A), we may assume Q to be simple. 0

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If we let K = No,and we suppose a two generator group to occur among the L,, we obtain the following result, which has also been proved by Schupp [21], Goryushkin [6], and Hurley [9]. Corollary 8.4. Every countable group can be embedded in a 2 generator simple group. 0 In the proof of Theorem 8.1(A), we can replace G by any generating set for G in the definition of the sets T,, and the conclusion still holds. But note that, if this is done, then no use has been made of the relations between these generators. It follows that, if M a G, then M' n G = M, and so G / M is embedded in Q / ( M o ) . We express this property by saying that G is an E-subgroup of Q. If we ensure p s 1/10, and m 123, then since the embedding of G in H is Frattini, the embedding of G in Q is also Frattini, by Lemma 6.5. From the constructions of 03, we see that the group H c o n c c t e d in the proof of Theorem 8.1(A), can be given the additional NFS structure (H,2, Y),where Y = Z ( ( G - 1) U (Li - Z(L,))))Zand Z is the (restricted) direct product Z = x i e , Z ( L i ) .It follows from Theorem 4.7 that, if J # 0, then Z ( H ) = Z. From Theorem 4.6, it follows that every element of H of finite order is conjugate to one of the form gz, where g E L,, for some j E J, and z E Z, and both have finite order. If we ensure that p < 1/10, and I m I > 3 , we can apply Theorems 6.4 and 6.8 to the construction of Q from H to obtain the following version of Theorem 8.1.

I

(u,,,

Theorem 8.1(C). In Theorem S.l(A), we can assume that G is an E subgroup of Q, and that the embedding of G in Q is Frattini. Moreover, i f J # 0, we may assume that Z ( Q )= X j e , Z ( L i ) the , restricted direct product of the Z ( L j ) , and that any element of Q of finite order is conjugate, in Q, to an element of the form gz, where g E Li, for some j EJ, and z E Z ( Q ) . 0 If, in the proof of Theorem 8.l(A), we assume Z ( L i )= 1, for all j E J, and we replace H by H' = H x,A, where A is an abelian group satisfying / A 1 < K , and T, by Ti = T, U A, then Z ( H ' )= A, and Q is still a simultaneous quotient of 2 that embeds G. If we ensure that p < 1/10, and 1 m I < 3, we see, from Theorem 6.8, that Z ( Q ) = A. If we do not have Z ( L , )= 1, for all j E J, then, by Lemma 7.5, L,/Z(Li) is an SSC group with trivial centre, and since any simultaneous quotient of these groups is also a simultaneous quotient of the Li, we obtain the following version of Theorem 8.1.


203

Theorem 8.1(D). In Theorem 8.1(A), we can assume that Z ( Q )= A, where A is any abelian group satisfying / A =sK . 0

I

The following corollary is immediate.

Corollary 8.5. Every countable abelian group is the centre of a two generator group. 0 Application 11. Embeddings into the join of given groups. For this application, it is convenient to work with countable groups. However, the results can clearly be generalised so that they work for groups of higher cardinality.

Theorem 8.7. Let {G,},eM be a family of at least two groups, and let G be any group. Suppose 1 M /,I G 1 s No, and 1 G, I S No,for all p E M. Suppose each G, contains a proper subgroup B,, such that, for all, except perhaps one, p E M , G, contains a blocking pair over B,. Finally, suppose there exists a group B, generated by a family of subgroups { C , } , , , satisfying C, = B,, for all p E M, and let SB denote the amalgam, obtained from the G, and B, by setting B, = C,, for all p E M. Then there exists a group Q, that embeds the amalgam SB and the group G, and is generated by any pair of groups (G,, G w ) where , p#vEM.

Proof. The proof is very similar to the proof of Theorem 8.l(A), and so we confine ourselves to a few remarks. In the structure product STR (G,, B), with its natural NFS structure, for every pair p, v E M, we can construct a symmetrised set R,, satisfying [A, n,No],for convenient A, n > 0 , with a basis S,," that lies entirely inside G, U G , This is possible because at least one of the groups G, or G, contains a blocking pair over B, or B , and the construction is similar to that used in a free product with amalgamation. We now proceed as in the proof of Theorem 8.1(A), but we replace the use of Lemma 8.2 by the construction described above. 0 v)

Variants of this theorem can be proved, as in the last application. In fact analogs of Theorems 8.1(B), (C) and (D) hold, where, in case (C), we replace Z ( Q ) = X Z ( L j )by Z ( Q )= 1. We leave their formulation and proof to the reader. (Note that, in case (B), a little extra work is needed to ensure that d is embedded in the simple,group Q.) It is also possible to combine Applications I and 11.

B.M. Hurley

204

Application 111. Decision problems in group theory.

Theorem 8.8. A necessary and sufficient condition for a finitely generated, recursively presented, group G to have soluble word problem is that there exists a non-trivial finitely presented group H, such that G can be embedded in every simple factor group of H . Proof. Suppose we have a finitely generated group G with soluble word problem. By Theorem I of Boone and Higman [2], we can embed G in a simple group S, which, in turn, can be embedded in a finitely presented group K. As i s p e l l known, we may assume K to be generated by two elements, a and b, say. Let ( u , u ) be the free group of rank 2, and form the free product L = K * ( u , v ) . Then the set S consisting of the four words:

n 80

a

,=I

( u v i ) ,b

n 80

i=l

n ( s u ~ ) s u -n~ ~( s~u~~, ) s u - ~ ~ ~ ~ , 80

80

( U U ~ + ~ ) , i=l

i=l

is a basis f o r g m m e t r i s e d subset R of L, equipped with an appropriate NFS structure, where s is some fixed element of S - 1. We let N be the normal closure of R, and put H = L / N . H is visibly finitely presented, and, since R satisfies C’(1/6), embeds K and hence embeds G. Moreover, we easily see that every non-trivial quotient group of H embeds G, and so, in particular, every simple quotient group of H contains a copy of G. Now suppose we have a finitely generated group G, with presentation (gl, g 2 , . . . ,g, : K ) , which can be embedded in every simple quotient group of the non-trivial finitely presented group H = ( x l , . . . ,x, : R). We let u I ,u 2 , .. . , v, be variables. If L # 1 is a group, then we say p # 1 E L is consistent in L if ( p ) = # L, and otherwise, we say it is inconsistent in L. We define a, possibly finite, ascending sequence N o d N , d N 2 of finitely generated normal subgroups of H as follows, where we suppose we are given an enumeration U of all s-tuples of words on { x l , .. .,x,}. We define No = 1. For i 3 1, suppose N,-] has been defined. Let ( w , , w2,..., w.) be the ith element in the enumeration U. There are three cases. (a) If there exists a word u ( v , ) such that u ( g , ) = 1, in G, but u ( w , ) is inconsistent in H / N , - , , then define N, = N,-,. (b) If (a) does not obtain, but there exists a word u ( v , ) such that u ( g , ) # 1, in G, but u ( w , ) is consistent in HIN,-,, then define N, to be the normal closure, ,in H, of N,-, and u ( w , ) . (c) If neither (a) nor (b) obtains, then Nk is undefined for k z = i .

205


Suppose this sequence were infinite, then, since H is finitely generated, and the N, are proper subgroups of H, there is a maximal normal subgroup M of H containing all the N,. But H cannot be embedded in the simple quotient group H I M of H, because, for every s-tuple ( w l , .. . ,w s ) , of words on {x,, . . . ,x,}, either there exists a word u ( u , ) such that u ( g , ) = 1, in G, but u ( w , ) # 1 in HIM, or there exists a word u ( u , ) such that u ( g , ) # 1, in G, but u ( w , ) = 1 , in HIM. This contradicts our assumption that G can be embedded in every simple quotient group of H, and shows that the ascending sequence of normal subgroups of H is finite. Let Nk be the last member of the sequence { N , } .And write = H / N k . From the definition of the sequence, if ( w ~wz, , . . . , w , ) is the (k + 1)th element of V , then: { u ( u , ) : u ( g , ) # l in G } = { u ( u , ) : u ( w ,is) inconsistent in

R}.

Since R is finitely presented, the set on t h e right-hand side is r.e., by an obvious diagonal enumeration, and hence the set on the left-hand side is r.e. Since G is recursively presented, { u ( u , ) : u(g,) = 1 in G} is also r.e., and so must be recursive. Hence G has soluble word problem. 0 We remark that, using the techniques of Application I, we may assume that H is a simultaneous quotient group of any finite family {Lj},EI of finitely presented SSC groups. Problem 2. Can the hypothesis that G be recursively presented be removed from Theorem 8.8? The proof that G has soluble word problem in Theorem 8.8 is nonconstructive, and there seems to be n o obvious way of replacing it with a constructive proof. Thus there seems to be no reason why an analog of Theorem I11 of Boone-Higman [2] should hold, in this case. Indeed, it seems rather unlikely that there should be such an analog. This situation prompts us to ask the next question. Problem 3. Does there exist a non-trivial finitely presented group H with the property that every finitely generated group G with soluble word problem is embeddable in every simple quotient group of H?

Our Theorem 8.1(B) is a generalisation of Theorem I of Schupp [21]. It is possible to use our theory to obtain a generalisation of his Theorem 11. The proof is a simple adaption of Schupp’s proof, using

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methods from the proof of Theorem 8.1. We therefore content ourselves with a statement of this theorem. For the definitions of the terms used in the theorem, we refer the reader to Schupp’s paper.

Theorem 8.9. Let A be a finitely presentable group which has a finitely presentable quotient group L which is an SSC group. Let rr be any finite presentation of A, and let P be a Markov property which is possessed by the trivial group. Then the problem of determining P restricted to quotients of rr is insoluble. Also the generating problem for A x A is insoluble. 0 Application IV. Residual simplicity. Theorem 8.10. A n SSC group with trivial centre is residually infinite simple. Proof. We actually prove the following stronger statement.

Let K be an infinite cardinal. Suppose L is an SSC-K group, with trivial centre, and let G be a group. Suppose I G I,[ L I K . Then, if P is any finite set of non-trivial elements of L, there exists a simple quotient group Q of L, containing a copy of G, such that p f 1, in Q, for all p E P. Proof of above statement. By Lemma 7.6, L contains a symmetrised set R, satisfying [A, n, K ] , for some convenient A, n > 0 , which is contained in We embed G in a simple group G, of order K , and we let H = d * L * F., where F, is the free group of rank K . We proceed, as in the proof of Theorem 8.1(B), to construct a symmetrised set R * , in H, satisfying [ p , r n , ~ ]for , suitable p , m . We may suppose R * contains (i) words of the form gr, where r E R and g is a generator of d or F., (ii) words that express the generators of L in terms of those of F,, and (iii) words that ensure that all non-trivial quotient groups of Q * = H / N * , where N * is the normal closure of R * in H, contain a copy of G. We chose p , m so that G is embedded in Q*, and, as before, construct a simple quotient group Q of Q*. Now Q is a simple quotient group of L, containing G. If p = 1, in 0, for some p € P, then r = 1, in Q, for all r E R, and hence g = 1, in Q, for all generators of F.. But Q is a quotient group of F,, and so this would imply Q = 1, a contradiction. Hence p # 1, in Q, for all p€P. 0

npEp(p)L.


207

Application V. The height of SSC groups. We assume familiarity with Pride [16], and we shall freely use the terminology of Pride’s paper. Theorem 8.11. Countable SSC groups have infinite height.

Proof. Suppose G is a countable SSC group. Let So,S,,. . . be an infinite sequence of countable groups. By Theorem 8.9(A), we can construct an infinite group 0 that is a simultaneous quotient of the family {G}U {Si * Si)T=,,. For n = 0,1,2,. . ., define Q, to be the group whose presentation is obtained from the obvious presentation (i.e., the presentation constructed in the proof of Theorem 8.9(A)) for Q by deleting the generators of Si * Si, for i 2 n, and deleting all relators that involve these generators. It is clear that Q, is a simultaneous quotient of the family {G} U { S , * Si}yZd, and that Qn+lis a homomorphic image of Q,, for all n. It is immediate from [16] Theorem 2.1, and the definition of S , which follows [16] corollary 2.1.1, that G 3 Qo t QI3 Q2 ?=. . .?=Q, ?= . . . . The theorem will be established if we can show # On, for all n. We note that, for some choice of the groups Si, On+, that the construction of Q can be done in such a way that it does not depend on the defining relations of the Si, and so, for all i, we can choose Si to be a countable simple group that is not contained in Qi. (Note that by Theorem 8.1(B) there are 2”. two generator infinite simple groups, and Qi can only contain Nosuch groups.) Since Q,+, is a homomorphic image of S, * S,, it is clear that any non-trivial homomorphic image of a subgroup of finite index in must contain a copy of S,, and so cannot be contained in 0,. By [16], Theorem 2.1, we have Q n + l # On,for all n, as required. 0 If G is a countable SQ-universal group, then this can usually be established by arguments that use small cancellation theory and the fact that SQ-universality is a large property, in the sense of [16]. This prompts us to ask the following question. Problem 4. Do all finitely generated SQ-universal groups have infinite height? We remark that Hurley (unpublished) has constructed a countable SQ-universal group that is minimal, in the sense of Pride [16]. However, a finitely generated SQ-universal group cannot be minimal, because it must have a just infinite quotient group.

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208

Application VI. The rank of certain direct products. Definition. If G is a finitely generated group, then by t h e rank r ( G ) , of G , we mean the minimum size of a generating set for G. Theorem 8.12. If G i s a countable SSC group, and C is any countable group, then C can be embedded in a quotient group 0,of G , of rank 2, such that, for all non-trivial finitely generated groups B,

r ( B )+ 1, if B is free or cyclic, B ) = ,.(B), otherwise.

[

r(Q

Remark. The theorem says that usually r(Q x B ) = r ( B ) . It is clear that an exception to this rule must occur when B is cyclic. If B is free, then r(Q x B ) = r ( B ) would imply that Q x B was free, which contradicts the fact that a free group cannot be decomposed into a non-trivial direct product. Hence an exception must also occur when B is .free. Thus the value of r ( Q x B ) given by the theorem is always as small as possible. Proof. Let uI, ur, u 3 , . . . be an infinite list of variables, and let U be an enumeration of all pairs (r, w ( u , ) ) ,where r 3 2 is an integer, and w ( v , ) is a reduced word on {ul, u 2 , . . . , u,} that both begins and ends with vT'. Note that we do not require w ( u , ) to involve all the variables in u I , u 2 , . . . , u,. Let R be a symmetrised subset of G satisfying [1/30,30, No].Let S be a basis for R. Since Card(S) = No,we can write it as a disjoint union

s = I3 s,, =o a

where Card(So)= N o , and, for i = 1,2,. . . ,Card(S,) = r,, where (r,,w , ( v , ) )is the ith element in the enumeration U. We suppose that each S, is given a standard ordering (s,,s2,. . . ,s,,). For each i = 1,2,. . . , we define the set T,(u,)of words on { u , , u 2 , . . ., u,,} to be the set of all words of the form:

for all k larger than 1 w,(u,)I. Note that the r,,k are reduced as written. We now write T,(u,) as a disjoint union T:(u,)U T';(u,),where Card(T:) = r,, Card(T',')= No, and we equip both sets with a standard ordering.


209

We now suppose that cl, czrc,, . . . is a generating set for C, and that . . . is a generating set for G consisting of elements of length 0 and 1. We set g,, gz,g,

H = C * G * ( i g= l K ( i ) ) , where K ( i ) is a free group with basis { u , , ~u ,, . ~. ., . u, ,,}. We now let R * be t h e symmetrised closure, in H, of all words of the forms: (a) cqlzqr where z, is t h e 4th word in the standard ordering of So, (b) u;:z,, where z , is the 4th word in the standard ordering of S,, (c) u;:t,, where t, is the 4th word in the standard ordering of T ( u , . , ) , and (d) g i l t , , where r, is the 4th word in the standard ordering of T':(u,.,). By an argument similar to that used in the proof of Theorem 8.1, we see that R * satisfies [1/10, 10,No], in H. Thus, if N * is the normal closure of R * in H, then Q = H / N * embeds C. As before, Q is a quotient group of G. We must now show that Q satisfies the conclusion of the theorem. We first note that, from the construction, Q is a simultaneous quotient group of the family {P,}:=l, of groups, presented as follows:

P, = ( v , , u z , . . ., u,, : vi'r, (t, is the 4th word in the standard ordering of T:(v,),1 c 4

r,)).

Let us fix some i 2 1, and write (r, w ( v , ) )= (r,, w,(u,)). Let E be the one relator group with presentation

E = (xi, ~

..,X, ;~ ( x , ) ) .

2 , .

Let Y be t h e split extension of E by t h e free group with basis {vl, v 2 , . . ., u,}, with the v, acting as follows: v;'x,u, = x,'x,x,, for all i , j = 1 ,. . . , r. Let J = E { v ? , . . . ,u"}E. Then, by Lemma 3.2(ii), ( Y , E , J )is an NFS group. Note, however, that it need not be an SSC group. Let R * * be the symmetrised closure, in Y, of all words of t h e form v;'?,x,, where, for the fixed i, and for q = 1 , . . . ,r, the q th word in the standard ordering of T:(u,)is r,. We leave it to the reader to check that R * * satisfies C'(l/lO), in Y . Note that, as well as doing the usual calculations, one must check that the elements r E R ** are not, themselves, R **-pieces. This amounts to showing that, if b, c E E and r E R **, then br = rc implies b = c. This is very easy t o do, but since, for instance, the symmetrised closure of {fixl} does no? satisfy this condition, it is non-trivial.

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Let N * * be the normal closure of R * * in Y , and set D = Y / N * * . . Since the r, are words in the v,, we can eliminate the x, from the obvious presentation for D, by means of Tietze transformations, and hence, we see, r ( D )s r. Let P' be the normal closure, in D, of w ( u , ) , and let D' = D/P'. From their definitions, we see, r, E P', in D, for q = 1,. . . , r. It follows that v, = x,, in D', that D ' = E, and that P' fl E = 0. Where, since R ** satisfies C'(l/lO), E is a subgroup of D. But note that E a D, and, since r, E P', for all q, D = gp(P', E ) . It follows that D = P'x E. Setting x, = 1, in the presentation for D, and deleting the x,, we obtain the presentation which we used to define P,, for our fixed i. Thus D = P, x E. We now let i vary, and accordingly add the suffix i to E defined above. We have shown r(P, X E , ) s r,, for all i = 1,2,. . . . Now let B be any non-free group of finite rank r 2 2 . For some i, B is a quotient group of E,, and r, = r. Since Q is a quotient group of P,, we see that Q x B is a quotient group of P, x E,. Hence r ( Q x B ) s r(P, X E,)< r, = r ( B ) . Now suppose B is cyclic or free, and let it have rank r. For some i, B is a quotient group of E,, with w ( v , )= u l , and r, = r + 1, and we conclude r ( Q X B ) < r ( B )+ 1. Since the value of r ( Q x B), given in the statement of the theorem is as small as possible, the above inequalities can be strengthened to equalities. To see that r ( Q ) = 2, we merely need to note that it is a homomorphic image of the two generator group Q x (x, y : x 2 ) , that Q f 1, for it contains a copy of C, and that Q is not cyclic, for cyclic groups do not satisfy the conclusion of the theorem. 0 We remark that, as with Theorem 8.1, one can prove variants of Theorem 8.12. In particular, a version of the theorem holds in which Q is simple. The next result is a corollary of this version of the theorem. Proposition 8.13. There exists a family 9 of simple groups satisfying : (i) Every countable group C can be embedded in some member of 9, and (ii) if S , , S 2 , .. . ,S, E 9,where n 3 1 is finite, then S , x S2 x . . x S, is a two generator group.

69. Concluding remarks The theory bas many applications which, for reasons of space, we have been unable to include in 08. We shall mention a few of these here, without going into details.


21 1

One construction we have hardly used is the HNN construction in the category N9.9'. Indeed, the only reason we included it in $3 is that it provides the quickest way we know of proving the results we need about structure products. One of the difficulties with this construction is that it does not, in general, preserve the property of being an SSC group. However, it can be used, for instance, to prove a version of Theorem 8.1, in which all elements of Q of the same finite order k are conjugate to one another. Moreover, it is possible to exploit the pathologies that can occur when using the HNN construction in N9.9'. For instance, observe that, if R is a symmetrised set in an SSC group G, and R * is its symmetrised closure in an N 9 Y HNN extension G *, then there is no obvious reason why R and R * should not satisfy the same C- and T-conditions, and yet also satisfy b ( R *) < b ( R ) . This observation has lead us to a new proof of the well-known Embedding Theorem of Higman [7). The construction is rather nice and allows us to prove results such as the following.

Theorem 9.1. There exists a finitely presented group G, with soluble word, power and order problems, with a subgroup F, which is free of rank 2, and a recursive functional 0 from the natural numbers to words on the generators of G, such that: (i) if Re is the r.e. subset of F with r.e. index e, then, in G, F f l @(e)" = RP, and so ( G : @(e)) embeds (F: Re),and (ii) the embedding of ( F : Re) in ( G : @(e)) is Frattini, and preserves the Turing degrees of the word, power and order problems. 0 Theorem 9.2. Every recursively enumerable c -degree contains the word problem of some finitely presented group. 0 Theorem 9.3. A necessary and sufficient condition for an abelian group A to be the centre of some finitely presented group is that it should be recursively presentable. 0 Remarks. We refer the reader to Rogers [17] for the definitions of r.e. index and c-reducibility, and to Collins [ 5 ] for the definitions of the power and order problems. For the definition of r.e. index of a subset of the free group F, of rank 2, to make sense, we, of course, assume some Godel numbering of F into N to be given. In Theorem 9.3, by recursively presentable we mean, of course, possessing a presentation with an infinite, but recursive, set of generators a l , a 2 , .. . , and an r.e. set rl, rz, . . . of relators on these generators.

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A topic we have not mentioned is t h e study of symmetrised sets, - that are fixed, setwise, by some satisfying small cancellation conditions, group of automorphisms of an NFS group. A typical example of a result obtainable by such study is the following.

Theorem 9.4. Let H, K be countable groups, neither of which can be decomposed into a non-trivial free product. Suppose further that 1 H 1 3 2, 1 K I 2 3 , and that K is not an infinite cyclic group. Then all normal extensions of H * K are SQ-universal. 0 We remark that P. Hall (unpublished) has constructed an example of a countable SQ-universal group G, that has a normal extension (actually a split extension by an infinite cyclic group) that is not SQuniversal. The reader will notice that Theorem 9.4 gives us no information about what happens when both H and K are infinite cyclic groups, and so we ask the following question.

Problem 5. Are all normal extensions of a free group of rank 2 SQuniversal? Another topic we have not mentioned, is that of small cancellation towers. That is, possibly infinite, towers G IS G2C . * . of groups, where Gi, for i 3 2 , embeds Gi-l, by a small cancellation theory argument. We remark that a group with the same property as the group of P. Hall, mentioned above, can be constructed by means of such a tower, that the countable, minimal (in the sense of Pride [16]), SQ-universal group, mentioned after the proof of Theorem 8.11, was constructed by such a tower, and that Problem 7 of Pride [16] has been settled, in the negative, by a small cancellation tower argument. Finally, a small cancellation tower construction can be used to prove the following.

Theorem 9.5. Every countable group C can be embedded in a nonsimple countable group G that is isomorphic to all its non-trivial quotient groups. 0 We shall say a word about the future of small cancellation theory. It is widely believed that it should be possible to iterate small cancellation arguments. That is to “do” small cancellation theory over small cancellation groups. We believe that this should be possible, in the near future. It is possible that such a development would have some relevance to the question of whether or not every group with soluble word problem can be embedded in a finitely presented simple


213

group, the answer to which we believe to be “No”. The reader who thinks of using Theorem 8.8, in some way, to obtain the answer “Yes” to this question had better be warned that one can prove a version of this theorem, in which every non-trivial quotient group of H has insoluble word problem, in which case, it can have n o recursively presented simple homomorphic image. Returning to t h e non-iterated theory, we have a feeling that constructions of t h e type used in the proof of Theorem 8.12 might have some relevance to disproving a famous conjecture in Topology. And now a word about the history of t h e results in this paper. The idea of doing the “classical” cases of small cancellation theory in a uniform way has its origins in my (for the remainder of this section, for the sake of naturalness, the pronoun “I” shall be used instead of the conventional mathematical “we”) M.Sc. dissertation [9]. This was conceived, written and typed within the space of three weeks, and so contains many errors. However, it does contain a correct proof (The first, I believe, but then, what does it matter who proves things first?) of the fact that every countable group can be embedded in a two generator simple group. The theory remained dormant, for some time, mainly because I got bored with it. My interest revived, when, at the Oxford meeting, I learned of the existence of the axiomatic lengthfunction theory, and it seemed to me that there might be some connection between this theory and my theory of normal forms in groups. As to the value of this work, I am unsure. Dividing things, roughly, into “reality” and “waxworks”, most Mathematics falls within the category of “waxworks”, and if one thinks that then, presumably (apologies to Lewis Carroll) one ought to pay. Perhaps one can do no better than t o quote Wittengenstein “Wenn es einen Wert gibt, der Wert hat, so muss e r ausserhalb alles Geschehens und So-Seins liegen”. 110. Acknowledgements

Knowledge is not gained in isolation, and my ideas on this subject have benefited from discussion with many people, and I would find it impossible to compile a complete list. However certain people need special mention. The idea of doing small cancellation theory over NFS groups originated in my M.Sc. dissertation, and I must thank my (long suffering) supervisor Professor G . Higman. Incidentally, it was h e who started me thinking about small cancellation theory by showing me a paper of Schupp. I thank Professor R. Lyndon, who read an earlier draft of this paper and made many detailed suggestions as to how it

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might be improved. I must thank Professor W.W. Boone, without whose encouragement, this paper might never have been written. Finally, I must apologise to the other contributors to this volume for the late appearance of the final version of this paper.

References [I] B. Baumslag, Residually free groups, Proc. London Math. Soc. 17 (1967) 402-418. [2] W.W. Boone and G. Higman, An algebraic characterization of groups with soluble word problem, J. Austral. Math. Soc. 18 (1974) 41-53. [3] J.L. Britton, The word problem, Ann. of Math. 77 (1963) 16-32. [4] I.M. Chiswell, Abstract length functions in groups, Math. Proc. Camb. Phil. Soc. 80 (1976) 451-463. [5] D.J. Collins, The word, power, and order problems in finitely presented groups, In: Word problems (North-Holland Publishing Co., Amsterdam, 1973) pp. 401-420. [6] A.P. Goryushkin, Imbedding of countable groups in 2-generator groups, Mat. Zametki. 16 (1974) 231-275. (71 G. Higman, Subgroups of finitely presented groups, Proc. Royal Soc. London, Ser. A 262 (1961) 455-475. [8) A.H.M. Hoare, On length functions and Nielsen methods in free groups, J. London Math. Soc. 14 (1976) 88-92. [9] B.M. Hurley, Embedding theorems for groups, M.Sc. dissertation (Oxford, 1972). [lo] B.M. Hurley, On length functions and normal forms in groups, Math. Proc. Camb. Phil. Soc. (to appear). [ I l l R.D. Hunvitz, On the conjugacy problem in a free product with commuting subgroups, Math. Ann. 221 (1976) 1-8. [12] R.C. Lyndon, Length functions in groups, Math. Scand. 12 (1963) 209-234. [13] R.C. Lyndon, On Dehn’s algorithm, Math. Ann. 166 (1966) 208-228. [I41 W. Magnus, A. Karrass and D. Solitar, Combinatorial group theory (Wiley, New York, 1966). [IS] K. Murasugi, The center of a group with a single defining relation, Math. Ann. 155 (1964) 246-251. [16] S.J. Pride, The concept of “largeness” in group theory, in this volume, pp. 299-335. [17] H. Rogers, Jr., Theory of recursive functions and effective computability (McGrawHill, New York, 1967). [I81 G.S. Sacerdote and P.E. Schupp, SQ-universality of HNN and I-relator groups, J. London Math. Soc. 7 (1974) 733-740. [19] P.E. Schupp, Small cancellation theory over free products with amalgamation, Math. Ann. 193 (1971) 244264. [20] P.E. Schupp, A survey of small cancellation theory, In: Word Problems (NorthHolland Publishing Co., Amsterdam, 1973) pp. 569-589. [21) P.E. Schupp, Embeddings into simple groups, J. London Math. Soc. 13 (1976) 90-94.


ON ISOMORPHISMS OF DIRECT POWERS J.M. Tyrer JONES Department of Pure Mathematics and Mathematical Statistics, Cambridge University

81. Introduction

In [2] Corner shows that, given any positive integer n, there exists a countable abelian group H such that H' = H sif and only if r is congruent to s modulo n. (Where H' denotes the direct sum of r isomorphic copies of H.) In this paper, we show that similar results can be obtained for finitely generated non-abelian groups. For the case n = 1 it is sufficient to construct a non-trivial finitely generated group isomorphic to its own direct square. This was done in [3], and the method of construction used there can easily be adapted to the needs of the general case. For given n > O we wish to construct a group G such that (i) G G"" (ii) G g G ' for l < r < n + l . The fact that G' = G " if and only if r is congruent to s modulo n follows easily from (i) and (ii). For ease of notation, we shall take the given integer to be n - 1, so that we need to construct a group G satisfying (i)' G = G" (ii)' G g G' for 1 < r < n. The 'building blocks' of our construction are t h e simple groups constructed by Camm in [l]. We give a brief description of these groups at t h e beginning of $2, but will assume a general familiarity with the notation and methods of calculation used by Camm. We will make frequent use of the structure theorems for amalgamated free products and HNN extensions. For the terminology of amalgamated free products we will follow [4], and for that of HNN extensions [ 5 ] . General structural results to be found in either of these two sources will be utilised without further comment.

82. The construction A = ( a , p ) and B = ( b , q ) are two-generator free groups. Z is t h e set 71 C

J.M.T. Jones

216

{ 2 1, 2 2, 2 3 , . . . } and p, a, T are permutations of I. Let U be the subgroup of A (freely) generated by the set {a'pP'":i E I } and V be the subgroup of B generated by the set {b'q"'":i E I}. Set g, = alpp(') and h, = b'q"'", and form the free product R of A and B amalgamating U with V under the isomorphism g, h,(,),i = 21, 2 2 , 2 3)... . In [l], Camm shows that if p, a and 7 are suitably chosen, then the group R is simple. Her definitions specify the effect of p, a and 7 on some of the positive integers. The definitions elsewhere can be made arbitrarily, subject to one technical condition, which states essentially that I p ( i ) l , J a ( i ) (and I ~ ( i ) lmust not be very much bigger than l i l . It is easy to see that if, for a given integer t (however large), the negative integers are divided into 'blocks' B 1= { - 1, - 2, - 3 , . . . - t } , B 2= { - ( t + l), - (t + 2), . . . - 2t}, B , = { - (2t + l), - (2t + 2), . . . - 3 t } etc., and p, a and T are so defined that for each r = 1,2,3,. . . p ( - rt) = a( - rt) = T( - rt) = - rt and p, a and T fix each B, setwise, then Camm's conditions are satisfied, and a simple group R results. We now proceed to the details of our construction. W e are given an integer n 3 3 , and wish to construct a group G satisfying conditions (i)' and (ii)' of $1. W e divide the negative integers into blocks of size 9n, and define permutations p, a and T of the negative integers by p ( - 9nt + r ) = - 9nt

+ pl(r) a( - 9nt + r ) = - 9nt + a l ( r ) 7(-9nt+r)= -9nt+7,(r)

O c r c 9 n - 1 , r = l , 2 , 3 ,...

where p l , ul, T~ are those permutations of the set {0,1,. . . 9 n - 1) defined by pl = (13,14), u1= ( 5 , 6 ) , = (9n - 10,9n - 9). By using Camm's definitions on the positive integers, we obtain a simple group R = A *"="B. The subgroup A = ( a , p ) of R is free, and so we may form an H N N extension H of R by adding a generator c and relations c-'ac = a, c-lpc = p", viz. H = ( a ,p , b, q, c : a ' p = bT(')qm('), i = * 1 , 2 2 , * 3 ,..., c - ' a c = a , c - ' p c = p " ) . We next construct a group T = ( d , e,f : d-'fd = f', e-lde = d ' ) as follows: Let (f) be an infinite cyclic group, and form an H N N extension (d,f : d-lfd = f') of (f). The element d in this group has infinite order, and so we may make a further H N N extension by adding a generator e and the relation e-'de = d 2 . The resulting group we call T.

217

On isomorphisms of direct powers

Britton's Lemma shows that the subgroups ( b - ' a , c ) of H and (e,f) of T are both free groups of rank two, and so we may form the free product of H and T amalgamating ( b - ' a , c ) with (e,f) according to the relations

b-'a = e

and

c = f

The resulting group S has presentation

S = ( a , p , b , q , c , d : a ' p P " ' = b T c ' ) q ~ (*1,*2,*3 '),i= ,..., c - ' a c = a , c-lpc

= p " , d-lcd = c2,

( b - ' a ) - ' d ( b - ' a )= d 2 ) .

We now form the Cartesian product C = II:,,S, of copies of S indexed by N = {1,2,3,. . .}. Elements of C will be written as infinite vectors. The corresponding direct product D = II,,,S, is the subgroup of C consisting of all those elements whose projection on S, is the identity for all but finitely many i E N . We shall denote by H,, T,, A,, etc. the subgroups of S, corresponding to the subgroups H, T, A, etc. of s. G is defined to be the subgroup of C generated by the elements x = (b-'a, b-'a, b - ' a , . . .),

Y = (P, P, P , . * * ), z = (P9ap-', p "ap-", ~ ~ ' a p. . -. ),~ ~ ,

,... ), u = ( d , d , d ,... ). u =(c,c,c

We shall show that G satisfies conditions (i)' and (ii)' of 91.

63. The isomorphism In this section we prove that the group G defined above satisfies condition (i)' of $1. Theorem 3.1. G is isomorphic to G". Proof. It is easy to check (by using the definitions p(1)= a(1) = ~ ( 1 ) =1, specified by Camm, and the definitions of p, u and T on the negative integers given in $1) that the following equations are consequences of the amalgamation relations of R, and hence hold in S. (1) b - I a ( P 9 n l - 9 k a p -9n1+9k+I )a-lb = P 9 n r - 9 k a p - 9 n r + 9 k + I for all k = 0 , 1 , ... n - 2 , a n d a l l t s l .

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J.M.T. Jones

(2) b - l a (pgnl-Y(n-l)ap-'hr+%n-I)+l )a-Ib = p % T - % n - I ) + I 131. (3) b - 1 a (PYnl-14aP-9nf

2

a p

-%tt+%n-l)+l

for all

for all t 3 1. -yn'+4qfor all t 3 1. Using equations (1) and (2) we deduce that )a-lb

= P%-13ap-ynf+15

(4) b-la(pYnf-xap-Ynt+4)a - lb = p'"l-3ap

[ x - i , z - l=~@lnap-', i , i,... 1 , ~ ~ ~ + ~ ~ ~ ~ i,p1Hn+1nap-18n-9,i - ~ ~ - ~ , 1 , 1 ,...I , . . . =

z I say.

Equation (3) gives [X-l,yYn-24 - I Yn+24 ]=@,1,1, ... l , p , l , l , ... l , p , l , ... ) = y l say zIy and equation (4) gives [yYn-llz

;ly9n+13

, x-'1

=

(q,I,l,. . . 1,q, 1, 1, . . . l , q, 1,.. .).

Since b-'a = qp-l, we deduce that G also contains

xI= (b-'a, 1,1,. . . 1, b-la, 1,1,. . . 1, b-'a, 1,. . .). The relations [ d , b-'a] = d and [c, d ] = c of S show that the elements u I =

and

[u,xI]=

(d, 1,1, ... 1,d, 1,1, ... l , d , l , . . . )

uI= [ u , u l ] = (c, 1 , l )... l , c , l , l , ... l , c , l , ...)

also lie in G. Let G I be t h e subgroup of G generated by xI,y , , z I , u1 and u I . Then G I is t h e projection of G on the subgroup C , of C consisting of all those elements of C whose projection on S, is trivial for all i f 1 mod n. By considering the commutator [x-I, Y ' " - ~ zi'y-"'+'] and proceeding as above, we deduce that G also contains its projection G2 on C2 (the set of all elements of C whose projection on S, is trivial for all i f 2 mod n ) , and similarly for eath i, 1 i n. Clearly, G = G I X G2 x . * * X G,. The groups G,, 1 s i s n, are all isomorphic, and so it only remains to prove that G I s G. Now

GI = (

~

1

~

z~I , 1 u ,I , v I )

= (XI,YI, y?-10Zly;9n+9, U I ,

But

= (Xl,Yl, Uly~-~oZly;yn+9U;'yl,it,, Ul).

y~~-10fly;9n+Y

so

V')

- (p9.up-9-, 1,1,. . . l,plH"ap-IH", 1, 1,. . . 1,p27nap-27n, 1,. . . )


lyy -loz ,y ;Yn

and

+Yu

;I

= (pyap-y,1,1,. Yn-lOZly;Yn+Y

uly 1

219

-1

u1

y1

= (p9ap-8,1,1,.

. . l , p'8ap-'8,1,1,. . . 1, pZ7ap-Z7,1,. . .) . .1,p'8ap-'7,1,1,. . .1,p2'ap-26,1,. ..).

The isomorphism betwen G and G I is now clear. This completes the proof of Theorem 3.1. 04. Some preliminary lemmas

We wish to show that G 8 G' for 1 < r < n. Since G = G", it is clear that each of G and G' is both a subgroup and a homomorphic image of the other. Consequently many of the standard methods of demonstrating that two groups are not isomorphic are not applicable in our case. For this reason we must proceed by direct calculation. We first prove Lemma 4.1. G contains the direct product D = II,,,Si. Proof. As in 43, we see that t h e following relations hold in S. ( 5 ) b - I a (p9nr-(9k -8)ap -Ynr+(Yk -8)+ I)a -1b = p 9 n r - ( Y k - 8 ) a p - Y n ~ + ( 9 k - 8 ) + 1 for all 121 andQSkSn. (6) b - 1 a ( p - 1 a p 2 ) a - 1=b ~ - ~ a ~ p ~ q - ' . (To prove (6), we have to use some of Camm's definitions for p, u and T on t h e positive integers, namely, p ( l ) = U ( l ) = T(1)= 1,

p(2)=7(2)=2, a ( 2 ) = 3

p ( 3 ) = 3 , U(3)=7(3)=4,

p(4)=5, V(4)=2, T(4)=3

p ( 5 ) = 4 5 ) = 4 5 ) = 5,

p ( 6 ) = 4 6 ) = ~ ( 6=) 6.)

From (5) and (6) it follows that [x-l,y-loz-ly'o] = @-3a3p6q-5p-2a-1p,1,1,. . . l , . . .).

The first entry is a non-trivial element of the subgroup R 1of SI. But the projection of G on Sl is the whole of SI, and R , is simple, so G contains elements (g, 1,1,. . . l , 1,. . . ) for each g E R. In particular, G contains the element h = (b-la, 1,1,. . .1,1,. . .), and hence t h e elements [u,h ] = (d, 1,1,. . . 1,1,. . . ) and [u, [u,h ] ]= (c, 1,1,. . .1; 1,. . . ) also lie in G. Thus G contains t h e whole of Sl. Similarly, by considering [ ~ - l , y - ' ~ z - l y we ~ ~ ]see , that G contains S2. Repetition of the argument shows that G contains Si for all i 3 1, and hence that G contains D. 0

220

J.M. T. Jones

Lemma 4.2. S is centreless. Proof. By t h e amalgamated free product structure of S, we have

Z ( S )c Z ( ( b - ' a ,c ) ) . But (!.-'a,c ) is a two-generator free group, and so has trivial centre. 0 Lemma 4.3. Zf N , and N 2 are normal subgroups of S which intersect trivially, then one of NI and N 2 is trivial.

Proof. Suppose that neither NI nor N 2 is trivial. We first show that one of them lies in t h e amalgamated subgroup K = ( b - ' a ,c ) of S. Suppose N , p K. Then N , must contain a cyclically reduced element of S of normal form length greater than one. For suppose not. If NI contains a cyclically reduced element g, say, of length one, then g lies in one of the factors H, T of S, but not in K. Let x be any element of the other factor not lying in K. Then the commutator [x,g] is a cyclically reduced element of length four, lying in N , . So suppose that every cyclically reduced element of N , lies in K. NI K, so there is an element ghg-'(g E S, h E K ) , of N , which does not lie in K. By taking a suitable conjugate of this, we find that NI contains an element which lies in a factor, but not in K , and t h e result then follows as above. Let sIs2---sm be a cyclically reduced element of N , , written in normal form with m > 1. Without loss of generality, we may assume s1 E H and s, E T. Since IH : K I # 2, we may find an element g E H \ K such that gsl $Z K. Thus g~,s~---s,g-~ E N1 and is not cyclically reduced. Consequently N , contains both cyclically reduced and non-cyclically reduced elements. Let x be any element of N,. Since N , and N 2 intersect trivially they generate their direct product, so x commutes with both a cyclically reduced and a non-cyclically reduced element, and thus, by a length argument, x E K. Thus N 2 C K, as require$. Restricting our attention to H, we see that H has a non-trivial normal subgroup N 2 lying entirely within K = ( b - ' a ,c ) . We shall show that this cannot happen. Let g = (b-'a)"lcaI(b-'a)".c~---(b-'a)",c Br E N2.

Then bg6-I also lies in N2, and we have an equation of the form

6 (b-'a)"ic6 ___ (b-'a)"rcarb-'

= (b-'a)'ic *z(b-la)%c* 2 - - -

There are, without loss of generality, two cases to consider:

(b-'a)'sc **.


22 1

(i) p,# 0 and (ii) g = (b-la)”]. In case (i), an application of Britton’s lemma to the HNN extension H of R shows that one of b-’ and b-’(b-la)-ys lies in A, and the normal form theorem for R shows that neither case is possible. In case (ii), Britton’s lemma shows that b(b-’a)”lb-’ = (b-’a)?I, which again is impossible by the normal form theorem in R. Thus N 2 = 1 , as required. 0 We are now ready to prove the important Proposition 4.4. If G = A x B, then there is a partition {PI,P2}of N such that A = GPI,B = Gp2,where Gpdenotes the set of all g E G such that the projection of g on S, is the identity for all i P. Proof. Suppose there exists an i E N such that A and B both have non-trivial projection on S,. Then A 3 a = ( a l ,a2,.. .a,, . . . ) with a, # 1 . By Lemmas 4.1 and 4.2, there exists an element g, E S such that [ g , , a , ] # I ,a n d g = ( l , l , ... l , g , , l , l ,. . . ) E G . Now l # [ g , a ] = (1,1,. . . l , [g,, a , ] ,1,. . . ) E A since A is normal in G. Thus A has a non-trivial intersection with S,, A fl S, = NI, say. Similarly, B has a non-trivial intersection with S,, B nS,= N 2 say. But then NI and N 2 are non-trivial normal subgroups of S,, which intersect trivially, and this contradicts Lemma 4.3. Thus there are disjoint subsets PI and P2 of N such that A has non-trivial projection on S, if and only if i E PI and B has non-trivial projection on S, if and only if i E P2.Since S, c G for each i, and since G = A x B, we see that {PI,P2} is a partition of N. Now A C GPlr B c Gp2and every element g of G can be written in the form g = ab with a E A and b E B. Hence A = GPlrB = Gp2as required. 0

85. The admissible partitions {PI,P2} Since G 2 0,several decompositions of G as a direct product of the form G = S‘ x G suggest themselves. Other direct decompositions were obtained in 03. In this section, we prove that the only decompositions of G as Gplx Gp2can be obtained by making repeated use of these two methods of decomposition. In particular, we prove Proposition 5.1. Let P be a subset of N. Then G = Gpx GNiP if and only if either (i) P is finite or

J.M.T. Jones

222

(ii) There exist integers k , m and t, t c n', and congruence classes CI,Cz,. . . C, mod n' such that P = X U Cl U CzU . . . U C,, where X is a subset of {1,2,. . . m } . The methods used are those of direct calculation, using the normal form theorems for amalgamated free products and HNN extensions. We begin with a trivial observation. Lemma 5.2. The subgroup of G generated by x, u and v is isomorphic to T under the map x -+ e, u +f , v -+ d. We now attempt to derive similar descriptions of other subgroups of G, in particular, of t h e subgroup generated by x, y , z and u. T o this end, we introduce new elements of C as follows: For any g E C, let g, denote the projection of g on S,. For each positive integer k , and each integer I, 1 S I S n', define new elements x ( I , k ) , y ( I , k ) , u ( l , k ) and v(I, k ) of C by x(I, k ) , = y ( l , k ) , = u(I, k ) , = v(I,k ) , = 1 if i f I mod n'. x ( I , k ) , = b-'a, y ( l , k ) , = p , u(I, k ) , = c, u ( l , k ) , = d if i = I mod n'.

For each k , I as above, and each integer s 3 1, define an element

z ( l , k , s ) of C by

z(I, k , s ) , = 1

if i f I mod n'

z(I, k , s), = p'n'fap-9n1f+1if i = n * ( t - 1)+ 1.

The proof of Theorem 3.1 shows that G contains the elements x ( l , k ) , y ( l , k ) , u(I, k ) , v(I, k ) and z(I, k , s ) for all I, k , s as above. Let C 1 . k be t h e subgroup of c consisting of all those elements of c whose projection on s, is trivial for all i f 1 mod n', and let D { , k be the corresponding direct product. The proof of Lemma 4.1 can easily be adapted to show that for each I, k , s t h e group L I . k . s = ( x ( I , k ) , y ( I , k ) , z(I, k, s ) ) contains the direct product DI.'. We will use bars to denote cosets of D1.k in L I . k . s . Lemma 5.3. The group ~ , . k . ~ l D ~is, ka free product of free groups A l . k . s = ( j ( I ,k ) , f ( I , k, s ) ) and B 1 , k . s = ( % ( I , k ) j ( I , k ) , f ( I , k, s ) ) with the subgroups ~ I . k . ~ = ( j ( I , k ) - i f ( I , k , s ) j ( I , kif12,13,14mod9n, )i, j(1, k)-IZ+Ynfz(I, k , s ) j ( l , k)13-9"f,y(I, k)-I3+'"'-z ( I , k , s ) j (I, k )14-9nr, j (I, k )- 14+9nr z(I, - k , s)f(l, k)I2-'"', I E Z ) , and v f . k . s = ((f(1, k ) j ( 1 , k ) ) - i f ( I ,k , s ) ( f ( I , k ) j ( l , k ) ) ' , i f 4 , 5 , 6 m o d 9 n , (a(/, k)ji(I, k))-4+9"f2(I,k , s)(f(l,k ) j ( I , k))s-9"f,


223

@ ( I , k)f(I, k))-'+''''f(I, k , s)(f(I,k ) j ( I , k))6-'n', ( X ( I , k ) y ( I , k))-"""'f(I, k , s)(X(I,k ) j ( I , k))4-'n', t E Z ) amalgamated under the isomorphism j ( I , k ) - ' f ( l , k , s)f(I, k ) ' = (%(I, k ) j ( l , k ) ) - ' f ( I , k, s)(X(l, k)f(l, k ) ) '

for all i f 4 , 5 , 6 , 1 2 , 1 3 , 1 4 , 9 n- 11,9n - 10,9n - 9 mod 9n j(1, k ) - 4 f Y n t - z(I, k , s ) j ( I ,k)4-'n' = (f (I,

j ( I , k )-S+YW

z ( I , k, s )f ( I , k )5-'n'

k

~ ( lk , s ) j ( l ,k)'-'"'

=

j (I,

)-6+Ynl-

=

k )f ( I , k ))-4+'n'f ( I , k , s)(i I, ( k ) j (I, k ))5-'n'

(3( I , k ) j (I, k ))-'+'"'f

( I , k , s)(X ( I , k ) j ( I , k ))4-'n'

(3(I,k ) j ( I , k))-'+yn'f(I,k , s)(.?(I, k)f(l, k))6-'n'

j ( I , k ) - l 2 + Y n l - z(I, k , s ) j ( I ,k)13-'n1 = (X ( I ,

k )?(I, k ))-12+'"'Z ( I , k, s )(X ( I , k ) j (I, k ))"-'"'

jj ( I , k )- 13+0nl-z(I, k , s ) j ( I ,k)I4-'"' = (%(I,

k ) j ( I , k))-'4+'"'2(I, k , s ) ( f ( l , k ) j ( I k,))'@'"

j ( I ,k ) - l 4 + Y n l -z(I, k , s)f(I, k)'*-'"' = ( X ( I , k ) j ( I , k))-13+'n'f(I,k , s)(i(l, k ) j ( I , k))I3-'"' j

( I , k )-(Yn-

I I )+Ynl-

z(I, k , s ) j ( l , k)'n-ll-9n' = ( X ( l , k ) j ( I , k))-('n-ll)+'n'~ ( 1 k, , s) *

(X ( I , k )? ( I , k ))-If ( I , k , s )(f( I , k ) j ( I, k j(1, k

)-(Ym-lll)+Ynl-

~ ( lk , s ) j ( I ,k)yn-l"-yn'

= (a(/,k)f(I, k))-('"-l")+'n'z ( I , k, s)-'.

(3( I , k ) j ( I , k ))'" j ( I , k ) - ( Y n - Y ) + Y m l - z ( I , k , s ) j (I, k )'n-'-9nr

- I"-'n'

- (%(I, k ) j i ( / , k))-(9n-10)+ynfz(l,k,s)*

(X ( I , k ) j (I, k ))-'Z( I , k , s )(X ( I , k )f ( I , k ))'n-'-'n' Proof. An equation w ( 3 ( I ,k ) , j ( I , k ) , f ( I , k , s)) = 1 holds in L 1 , k . s if and only if t h e projection of the corresponding word w(x(I, k ) , y ( I , k ) , z(I, k , s ) ) o n S, is trivial for all but finitely many i E N. Thus it is easy to check that the listed equations hold in L 1 . k . s . To show that the subgroups k i 1 . k . s and B 1 . k . s are free, we need only note that for any i = I mod n' the projections of Al.k.s and B,.k.son S, are the free subgroups A, and B, of S,. Hence i j , . k . s and are free on the listed generators, and the given equations represent an isomorphism between U,.k.sand Thus L,.k..is a homomorphic image of t h e rgquired amalgamated free product.

vl,k.s

v,.k.s.

224

J.M.T. Jones

Let g = K s ~ s ~ - - - s , be an element of L 1 , k . s written in normal form with respect to the relations of the amalgamated free product, so that is a word o n the generators of U 1 . k . s and the S, are powers of j(1, k ) and Z(1, k ) y ( l , k ) taken alternately. If we can prove that g # 1 unless r = 0 and K = 1, then the result will be established. Let J ( 1 , k)'"IZ(I, k , s)J(l, k)'"2 be any generator of 0 l . k . s and consider t h e corresponding word w = y(1, k)'"lz(I, k , s)y(l, k)". of L1.k.$.Now w, lies in t h e subgroup R, of S,, and it is easy to check (by taking each generator in turn) that w, also lies in t h e amalgamated subgroup U, of R, for all sufficiently large i = 1 mod n k . Now K involves only finitely many generators of 0 l . k . . and consequently the projection h, of h on S, lies in U, for all sufficiently large i E 1 mod n k . It follows that, for all sufficiently large i = 1 mod n k , g, is a word of R, of normal form length r. Thus g = 1 in L 1 . k . s if and only if r = 0 and 6 = 1, as required. We next consider the subgroup of G generated by the elements

x(1, k ) , y(1, k ) and z ( l , k , s ) when s is allowed to vary through a finite

set. We prove

Lemma 5.4. Let CY and p be fixed positive integers. Then the groups Ll,k . 0 0 S a S s S p generate their free product amalgamating the subgroup (Z(1, k ) , j ( 1 , k ) ) . Proof. The group fi,,,,,,,, generated by t h e L 1 , k . s is clearly a homomorphic image of the required amalgamated free product. We note that any element g of f i , , k , - . O may be written in the form = w(X(1, k ) , J ( 1 , k ) ) f i i i i z - - - i i r

where each 6, is a non-trivial word on t h e generators of U , , k . s , for some s = s ( i ) and where s(i + 1)# s(i) for any i = 1 , 2 , . . . r - 1. We need only show that g = 1 if and only if r = 0 and w ( 3 ( f ,k ) , J ( f ,k ) ) is the trivial word. As in the proof of Lemma 5.3, the projection of u I u 2 - - - u , on Si lies in the amalgamated subgroup U, of R, for all sufficiently large i = 1 mod n k . A length argument in R, then shows that if g = 1, w must be the trivial word. Thus it is sufficient to consider words of the form fi,iiz---iir. Suppose z.i,iiz-r-ii, = 1 with r > 1, and set i = n k ( t- 1)+ 1. The projection of u, on S, is a word in finitely many elements from among p - m @9n~rap-9nsr+l )p", m # 12, 13, 14 mod 9n,

P -12+9nm

@9nslap

-9n

+

%I I

lP

13-9nm,

P - 1 3 + 9 n m @9"'lap

-9n'l+l

>P14-9nm,

225


- I4+Ynm (p 9nslap -Yn*r+ 1 )p12-ynm, m E Z. Now u I u 2 - - - u , E D, so there P must exist some j , 1 s j s r - 1, such that for infinitely many values of i = 1 mod n k , all adjacent p-symbols in the projection of U , U , + ~on S, cancel. Suppose that ii,, written as a word on the generators of u , , k , s ( , ) ends in f(l, k)'Z(l, k, s)=f(I,k)", y, 6 E Z , E = f 1, and s = so'), and ii,+l begins in f(l, k)"Z(l, k, s')''f(l, k)"', y', S'E 2, E ' = f 1, and s ' = s ( j + 1)# s. The projections of these segments on S,, i = n k ( t - 1)+ I, are p'(py"rap-yn"+1)ep6 and v'(pYn.'rap -9n*',+ I ) 'p"'. Thus one of the following must hold for infinitely many values of t : (i) - 9n't + 1+ 6 + y ' + 9n"t = 0 (ii) - 9n't + 1 + 6 + y ' + 9n"'t- I = O (iii) - 9n"t + 6 + y ' + 9n"'t = O (iv) - 9n't + 6 + y + 9n"'t- 1= 0. Rewriting these, we see that one of the following must hold for infinitely many values of t : (i)' 9 t ( n s ' - n ' ) + 1+ 6 + y ' = O (ii)' 9 t ( n s ' - ns)+ 6 + y' = 0 (iii)' 9 t ( n S ' - ns)+ 6 + y ' - 1 = 0. But 6, y ' , s and s' are fixed with s f s', so each of these equations can hold for at most one value of t. Thus ti,&---&# 1. Since u , . k . s is free, the case r = 1 is easily dealt with, and so the structure of & & , k . m , @ is established. 0 E

We are now in a position to investigate the structure of the subgroup

(x, y, z, u ) of G. We prove

Lemma 5.5. Suppose w is an element of (x, y, z, u ) written in the form w = u"Ig,u"2g*---U('rg, where each gi E (x, y, z ) . Set k = C:=, 1 ai1 + 1. Then there exists an integer m, and congruence classes C , ,C2,.. . C, mod n k such that the projection of w on Si is trivial if i > m and i 61 C1U C2U * . U C,, and non-trivial if i > m and i € C , U C2U . . . U Cr.

-

Proof. The projection w, of w on S, lies in H,, and H, is an HNN extension of R,. It follows that if Z:=, a,#0, then w , # 1 for all i E N, and the result is proved. Thus we may assume C : = , a ,= O . It is clear that there exists a word

G

= G ( x ( l , k ) , y ( l , k ) , z ( I ,k , k ) , u ( l , k ) )

such that $, = w, for each i = l mod n k . Thus it is sufficient to prove

226

J.M.T. Jones

that for each I = 1,2, ..., n k , there is a positive integer mf such that if S,# 1 for some i > mf , then S,# 1 for every i > rnf such that i 3 I mod n k . We may write S in the form

GJ= ~ ( lk)"lglu(l, , k)U2gz---U(I,

k)"rg,

where each g, is a word in x ( I , k ) , y ( l , k ) and z(I, k , k ) . Now the following relations clearly hold in G: (1) u ( I , k ) - ' z ( I , k , s ) y ( I , k ) - ' u ( I , k ) ' =z ( l , k , s + r ) y ( l , k ) - ' for alJ r S O . (2) u ( l , k)'z(I, k , s ) y ( I , k ) - ' u ( I , k)-' = z ( l , k , s - r ) y ( I , k ) - ' for all O < r < s - 1. (3) u ( I , k 1-'Y (1, k )u (1, k = Y ( I , k 1". We will call a word in n ; i , k . , , Z k reducible if it contains a subword of the t Y Pe (i) ti(/, k ) - ' g P ( I , k ) with g E r ; i , , k , l , Z k - l or (ii) f i ( l , k ) g l ( l , k ) - ' with g E N J . k . 2 . Z k = ( f ( l , k , s ) j ( l ,k ) - ' , j ( f ,k ) " 2sss2k). Otherwise the word is called irreducible. Lemma 5.4 shows that we can always tell of- a given word of n ; i J , k , I . Z k whether or not it lies in -

M L I C . IOr . ~N - I, k , , x .

By using the images of relations (l), (2) and (3) in G / D we see that any reducible word can be replaced by a word equal to it, and involving fewer appearances of f i ( 1 , k ) . Thus any word can be replaced in a finite number of steps, by an equal irreducible word. We apply this process to 6.The choice of k ensures that the resulting reduced word contains no subword of the form ti(/, k)-'gP(f, k ) with g E f l f . k . l . 2 k or of the form # ( I , k ) g P ( l , k ) - ' with g E N f , k : l . Z k . Let V be the reduced word thus obtained. Suppose that V still involves an appearance of i i ( l , k ) , but that the projection of u o n S, is trivial for infinitely many i = 1 mod n k . It follows from the structure theorems of H , that u contains a subword of the type (i) u ( l , k ) - ' g u ( l ,k ) with g, lying in A, for infinitely many i = l m o d n k or (ii) u ( I , k ) g u ( I , k ) - ' with g, lying in ( a , p " ) for infinitely many i = 1 mod n k . By Lemma 5.4, g may be written in reduced from as w ( i ( l , k ) , j ( l ,k))ii1ii2---&, and for sufficiently large i = 1 mod n k , the projection of g o n S, is equal to the projection of w ( x ( I , k ) , y ( I , k ) ) u l u 2 - - - u r on S,. The projection of each u, on S, lies in the amalgamated subgroup U, of R, and so it follows from the normal form theorem for R, that w ( x ( l , k ) , y ( l , k ) ) = y ( l , k ) " for some integer s. Thus g E n ; i J , k . ] , , k - l and so case (i) leads to a contradiction.

O n isomorphisms of direct powers

227

Case (ii) therefore holds, and the projection of y ( l , k ) s u l ~ 2 - - - ~on, S, lies in (a, p " ) for infinitely many i = f mod n '. Now the projection of z(f, k, s)y(f, k)-' (1 C s s 2 k ) on S, lies in ( a , ~ "for ) all i, and so we see that, by writing g as a word in j ( I , k ) and ? ( I , k , s ) y ( l , k)-', 1 C s s 2 k , we may conclude that g lies in N r , k . 2 . 2 k contradicting the fact that V is irreducible. Thus if 0 contains an appearance of U ( I , k), then u, is non-trivial for all but finitely many i = f modnk. If, on the other hand, V contains no appearance of U ( I , k ) , we may use Lemma 5.4 to write V in normal form, and hence we may deduce that there is an integer mr such that, if u,# 1 for some i > mr then u, # 1 for all i >' mr such that i = f mod n k . This completes the proof of Lemma 5.5. 0

Lemma 5.6. Let w E ( x , y , z, u ) be written in the form w = u u ~ g I u u * g z - - - u a ~where g, each g , E ( x , y , z ) , and let k = C : = , l c r , ) + I . Let G be the corresponding word of ( x ( I , k ), y ( I , k ), z (f, k, k ), u ( I , C)), and let 77 = k"ol(I,k ) v % ( f , k)'z---u -(f, k)S"'h;, (&, = 2 1, h E n;il,k.1,2k)be the reduced form of 6.Then the projection u, of u on S, lies in (b-la,c ) for infinitely many i = 1 mod n k i f and only if there is an element 6 E D, and a word g = g ( u ( f ,k ) , x ( f ,k ) ) such that u = (g(u(I,k ) , x ( I ,k ) ) . Proof. We proceed by induction on m. If m = 0, then V E n;il,k.l,Zkand the result follows from Lemma 5.4, and t h e structure theorems of H,. Suppose m > O . A slight adaptation of Lemma 5.4 shows that we may write 6, in t h e form k,,, = ii,i&----ii,w(f(f,k ) , j ( f ,k)) where each U, is a word on the generators of u l , k , $ for some s = so'), and where s(j + 1) # s (j ) for each j = 1,2,. . . r - 1. Now the projection of u on S, lies in (b-la, c ) for some suitably large i = f mod n k . An application of Britton's lemma in H, th en shows that k, = iiliiz---ii,j(f, k)@f(I,k)' for some p, y, and that, if E, = f 1, the projection of u I u z - - - u r y ( I , k)@ on S, lies in ( a , ~ " As ) . in the proof of Lemma 5.5, we see that filGz---Li,j(l, k ) P E g i . k , i , z k - i if &, = - 1, and ~ ~ i i ~ - - - i ik)@ , j (E ~ ,N l , k . 2 , 2 k if E, = + 1. Hence E(f, k)'-h;, may be rewritten in t h e form ~ u ( kI) ,F m f ( f , k)', where g E n;if.k.l~Zk. We may, therefore, rewrite B in t h e form

-

-

6 = k,,u(f, k)'lk,E(f,

k)'.---k

-

,-rE(I, k ) € m f ( f ,

k)'.

The projection of u on S, then lies in ( b - ' a , c ) if and only if the same is true of 0 = L o u ( / ,k)'nk,u(I,k ) E 2 - - - k , - l , and the result follows by induction. 0

228

J.M.T. Jones

We now take a general word in (x, y, z, u, u ) . G is clearly a homomorphic image of the free product of the groups (x,y, z, u) and (x, u, u ) with the subgroup (x, u) amalgamated. Any element of G can be written in the form w = g l h , g 2 h 2 - - - g l , where each giE (x, y, z, u) and each hi E (x, u, u ) \ ( x , u). By Lemma 5.2, the projection of hi o n Si does not lie in (b-la, c ) for any j = 1 , 2 , . . . r. If r = 1, and w = g,, then the result of Lemma 5.5 applies. Otherwise the normal form theorem in S, shows that if wi = 1 for infinitely many values of i, then there is some j such that t h e projection of g, o n Si lies in ( b - ' a , c ) for infinitely many values of i, and so the result of Lemma 5.6 applies. We are now ready to prove Proposition 5.1. Let P be a subset of N . Then G = G p X GN\p if and only if either (i) P is finite or (ii) there exist integers k , m and t C n', and congruence classes CI,C 2 , .. . C, mod n k such that P = X U CIU C2U.. . U C, where X is a subset of { 1,2, . . . m }. Proof. Suppose G = G px GN\P. Then G p is a homomorphic image of G, and, as such, is finitely generated. Each of t h e generators of G p lies in G, and thus may be written as a word on t h e generators x, y, z, u, u of G. We have seen that any such word may be written in the form g l h l g n h 2 - - - g l , where each g, E (x,y, z, u) and each h, E (x, u, u ) \ ( x , u). Each g, can be written in the form U ~ ~ W ~ U " ~ W ~ - - - U ' "with ~ W , each w, E (x,y, z ) . We set k, = C:=, la, 1 + 1 and for each of t h e finitely many generators g of G , we set k, = max k,. Let k = max k,, where t h e maximum is taken over our set of generators for Gp.We now split each g l h , g 2 h z - - - g l , into a product of n' commuting words on the generators x ( l , k ) , y ( l , k ) , z(1, k , k ) , u ( l , k ) , u ( l , k ) , 1 S 1s nk...Let w = iI6,g2&---g,k, be any such word, and note that 6, E ( x ( l , k ) , u ( l , k ) , u(l, k ) ) \ ( x ( l , k ) , u(l, k)). Lemma 5.6 shows that either t h e projection w, of w on S, is non-trivial for all sufficiently large i = I mod n k , or some g, can be written in the form g, = &(x(l, k ) , u(l, k ) ) where 6 E D. But if -this be t h e case, may be rewritten in the form giKig;h;---gkh A with conditions as above, and with m < r. Repeated application of the above enables u s to conclude that there is an integer m such that, if w, # 1 for some i > m, then w,# 1 for all i > m such that i = 1 mod n k . We deal thus with each generator of Gp.Since P is equal to the set

.


229

of all those i E N such that the projection of g on Si is non-trivial for some generator g of G , t h e required result follows. 0 96. Equations in S Suppose G = G'. Proposition 4.4 shows that there is a partition {PI,Pz, .. . P,} of N such that each Gplis isomorphic to G, and G = Gp,x . * * x Gp,. Let cp : G + Gpl be an isomorphism. Then cp(x), cp(y), cp(z), cp(u), cp(u) generate G , and satisfy any relations satisfied by x, y, z, u, u. If i E P, then t h e projections cp(x),, cp(y),, cp(z),, cp(u),, cp(v), of cp(x), cp(y), cp(z), cp(u), cp(u) on S, must generate S, and must satisfy any relations satisfied by x, y, z , u and u. We will use t h e structure theorems of S to establish the possible values of cp(x),, cp(y),, cp(z),, cp(u), and cp(u), and will then use the ) results of 95 to reconstruct the elements cp(x), cp(y), p(z), ~ ( u and cp(u) cf G,. We will then know which maps from G to G , can be isomorphisms, and can use this information to deduce that G = G , if and only if the number t of congruence classes mod n k given in Proposition 5.1 is congruent to 1 mod n - 1. This shows that G $ G' if 1 0, in which case g-'wg = f'"" = w2'. In case (ii), g -' wg = d -'f -'d -Of "d @f'd'. Since p 3 0, this is d -'f ""d '. Now 6 S O , so t h e result is a proper power of w if and only if p > ( 61. In this case d-'f'"d6 = f'"-'a' ". This completes t h e proof. 0

-

Lemma 6.3. Suppose 1 # w E TI,, and w + T f " for any a. Suppose further that w - T ~ m with m # 51. Then w LITd" for some a#O. Moreover, if g-'d"g = dam,g E T, then either g = e p for some p > O , or g = ePd'e' for integers p, y, 6 satisfying p > 0, 6 < 0, and either 6 = 0 or y is odd and /3 > 16 1. Moreover, m = 2' in the first case, and 2s-'6' in the second. Proof. As in Lemma 6.2, w may be written in the form d"fPd' with w"'. But a + y # O . A consideration of d-length shows that w w - I ~ w " ' and , SO we must have w -T,d" for some a. Suppose g E T satisfies g-'d"g = d"", m # 5 1, and g is written in e-reduced form as woe"lwle"~w2---e"~w,where each wi E TI,.Clearly r # 0, and so w i l d " w o E (d). Suppose wo = dPf'd' with y # 0. We must have f-'d"f' E (d), and so it follows by a consideration of d-length in To that f-'d"f' = d", that is, d"f'd-" = f', and this relation clearly does not hold in To. Hence w o E (d), and repetition of the argument shows that w, E ( d ) for each i = 1,2,. . .r. Hence g E (e, d ) and, since the relation e-'de = d 2 holds in (e, d), we may use the method of Lemma 6.2 to write g in one of the forms (i) es, p # 0, or (ii) ePd'e'


231

with p 3 0, 6 < 0, and either 6 = 0 or y odd. The remainder of the proof is identical to the corresponding part of Lemma 6.2. 0

Lemma 6.4. Let w = d"f @d'with T to an element of ( e ,f).

CY

+ y # 0.

Then w is not conjugate in

Proof. Suppose w is conjugate in T to a cyclically reduced element g of (e, f). A consideration of e-length shows that g = f 6 for some 8 f 0 . Let K ' w h = f 6 and suppose that h is not e-free. By considering the last e-reduction in the word h-'wh, we see that f " - % , d Afor some AZO.But a consideration of d-length in To shows that this is impossible. Hence h is e-free, and d"fadT-%,f6, and a consideration of d-length again yields a contradiction. 0 Lemma 6.5. Suppose w E ( e ,f ) and re( w ) # 0. Let g be an element of T such that g-'wg E (e,f). Then g E (e,f).

Proof. Without loss of generality, we may take w to be a cyclically reduced element of the free group (e,f). We first show that g cannot be a power of d. We may write w in e-reduced form as w = e'~f"~eC~fo2---e'~f"~, where each E, = -+1. (Here the a, are permitted to take the value zero, but if a, = 0, then E , + ' # - E , . ) If d-@wdPE (e,f), an application of Collins' Lemma shows that we have an equation of the form d-@e ---e'rf "rd = e 'I f 71---e ',f 1, 'if"i

that is, d

'1f"1

_ _ _ e ef%d pf-yre

-er

___ f-71e

-cI

= 1.

Now f"dPf' is equal in To to a power of d if and only if it is equal to d P (by d-length in To).Moreover, e-'dPe' is equal to a power of d if and only if. it is equal to d2'@(where, if E = - 1, this should be interpreted to mean that p is even). Consequently our equation is possible if and only if it results in d-Pd2"P= 1 where Y = El-, e i . But this is dP""-l) = 1, which is possible only if 2"*("'= 1. Since a,( w ) # 0, this cannot occur. An application of Collins' Lemma now shows that the only cyclically reduced conjugates of w lying in (e,f) are those achieved by cyclically permuting w. Thus if g-'wg E (e,f) there is an element h E (e,f) such that g-'wg = h-'wh. It remains to show that if g commutes with w then g E (e,f). Suppose g is written in e-reduced form as g = goe"1glee2g2---ee,g, where each E, = & 1 and each gi E To. We will assume that as large as

J.M.T. Jones

232

possible an initial segment of g of the form goe'lgleez--- g,e'm lies in (e,f), viz. go,gl,. . .g, are powers of f, g,+l is not, and if g can also be then at least one of written in reduced form as &eel~leez---ee~& is not equal in To to a power of f. If g E (e,f), we are go,gl,.. finished, so assume not. We have g = hd"f@d'k where

h = goe'lgle"2---g,eEm+l E&,f) gm+i

=

d"f@d' (f)

k = eEm+2gm+2eEm+3--e Erg,. Case (i). k E ( e , f ) . Let w , and w 2 be the reduced forms of h-lwh and kwk-' in the free group (e,f). Then ( d " f @ d T ) - I w l ( d " f @=d w2. Y ) We have already established that conjugation by a power of d cannot transform a nontrivial element w of (e,f) with u , ( w ) # O to another element of (e,f), and so, without loss of generality, we may take (Y > O , y < O and p odd. Let w , = f"oe@lf"I---e@-f"" and w 2 -- fyoeS,fYI---esfYiwhere each p, and each 6, = 2 1. We have (d"f@d')-'(f"oe @if"i---e (d"f@d7 ) (f-y'e-9 ___ e-*if-Yu)= 1. @.fun)

An application of Britton's Lemma yields that p. = 6, and that fad"fedrf-'~lies in ( d ) if Pn = - I, or in ( d ' ) if Pn = + I. Since a > O and y 2 and k,, and B;(,+,) = p9'"a'ps for some integers r, s, r. In case (i) an easy calculation shows that y-4zy4 = ( ~ y ) - ~ z ( x y and )~, since x y = q # 1, relation (ii) fails to hold. Thus m is odd, and we have case (ii). Again we consider the equation ~ y - ' ~ " - ~ ' z y ~ " -=~ x - ' -(Yn-IO)zy -Izy 9"-9 . The right-hand side of this equation is 1

1


p -(vn-v)B i B z - - - B t(,,, - I )B,(:

+I

) p-'B!(,

+

24 1

-I - I))-'pYn-* l)p (BIBZ---Bj(,,,

and this must be a product of m blocks. It follows that k . B;(,+l,p-lB+(,,,+I,p IS a block for some integer k , that is pVnra rp s- I + Y w a r p + k is a block. This is possible only if s = - 9nt + 1, and this can occur only if r = 1. Thus Bi(,,,+l) = p9"'ap-'n''', as required. The case m = 1 is easily dealt with, by similar methods, and results in = p'"' ap-'"'+I for some t > 2 . I

97. The possible isomorphisms between G and Gp = @Vnap-Vn+I,plXnap-~~n+I , . . . ,p '"lap-yn'+ I , . . . ). L~~ = u - l z y - I u y Then x, y , 2, u, u generate G, and, by projecting onto each S, (i E N), it is easy to check that the following relations hold in G : (i) u - l y u = y " (ii) u-luu = u 2 (iii) x - I u x = u 2 (iv) x - l i x = i (v) x ( y - ' i y ) x - l = y - ' i y (vi) x(y-"iy")X-' = y - ' 3 i y l 4 (vii) y-'iy'= ( ~ y ) - ~ i ( x y ) ' . Suppose P N is such that G = Gpx GN,P,and cp : G --* Gp is an isomorphism. cp(x), q ( y ) , cp(i),cp(u), cp(u) generate Gp and satisfy the , ~(i),, above equations. It follows that for each i E P, ~ ( x ) , v(J),, cp(u), and cp(u), generate S, and satisfy t h e above equations. Their values are, therefore established, within an inner automorphism of S,, by Lemma 6.10. Let CI,Cz,. . . C, be the congruence classes of n k found in Proposition 5.1, and suppose t > O . Let C, be the congruence class of I, mod n k . For each of the listed generators g of G, q ( g ) is equal modulo D to a product of commuting words cpl,(g) on the elements x(I,, k ) , y(I,, k ) , z(I,,k , k ) , u(I,,k ) and u(L, k ) . We restrict our attention to one of these words, with 1, = 1, say. Now t h e results of $5 show that, provided k is chosen sufficiently large, a necessary and sufficient condition for cpI(g), to lie in T, for all but finitely many i = I mod n k is that there is an element 5 E D, and a word w on the generators x(I, k ) , u(I, k ) and u(I, k ) such that q c r ( g= ) (w. The results of 86 then show that there is a word gr on x(I, k ) , y(I, k ) , r(l,k , k ) , u(I, k ) and u(I, k ) such that VI(X)=

5ig;'x(I, k)gr

C P , ( U )= S z g ; ' U ( l

k)gr

1.M. T. Jones

242 (pi ( u ) =

t3g ; ' u ( I k )gr

where t,,t2,t 3 ED. We n e x t consider the word g l ( p l ( y ) g ; ' . By the results of $6, t h e projection of this element o n S, is equal to a-"pP aa (some a , P ) for all sufficiently large i = 1 mod nk. The results of §5 force gl(pl(y)g;' = t4y(1, k ) @ for some t 4 E D. It then follows by Lemma 6.11 that /3 = 1. Now consider g l ( p l ( i ) g ; ' . Lemma 6.10 shows that t h e projection of this element on S, must lie in U, for all sufficiently large i = 1 mod n k . We conclude, from the results of $5 that g , ( p , ( f ) g ; ' = t5uIu2---ur where t5E D and each y is a word on finitely many of the generators of u l , k . . , , and where s,+' # s, for each ] = 1 , 2 , . .. r - 1. We now project 5 5 ~ 1 ~ 2 - -onto - ~ , S, where i is a very large integer congruent to 1 mod nk, to obtain a word of the form p a ~ a P ~ ~ u ~ a P ~ - - - p o ~ a P - p n ~ + ~ where t h e a, and P, satisfy the conditions of Lemma 6.12. It can easily be checked (by projection o n each S,, i E N ) that the elements x, y, z of G satisfy equations (i) to (x) of Lemma 6:12, and hence so also do (p,(x),, (p, ( y ) , , (p, (i),for each i = 1 mod n '. We may, therefore, use Lemma 6.12 to obtain information about the projection of ~ 5 u I u ~ - - - u , o n S, for all sufficiently large i = 1 mod nk. We see that g l ( p l ( i ) g ; ' can be written in t h e form g,(p1(i )g

=

= ~ ~ ( W ~ W ~ - - - W , , , )kY) '( "I ,' ~ ( lk, , s ) Y ( ~k)-'"'y(I, ,

k ) - ' ( w i ~ *. .. w , , , ) - ' Y ( ~k, )

where t h E D, and each w, is a word of the form y(1, k)9n'i(z(1,k, s,)y(I,k)-')9"kny(1,k)-y"'~+y"k~ for some integers s, t, k,, s,, t,, with, for each i = 1 , 2 , . . . rn - 1, either s,+, # s, or s,+' = s, and t,+i # t, - kt. Set glcpr(i)g-' = h. Now x(1, k ) , y ( l , k ) , u(1, k ) , u(1, k ) and h generate G,, and so there is a word w on these generators such that z(1, k , k ) = w(x(1, k ) , y ( l , k ) , u(l, k ) , u(1, k ) , h ) . For all suitable t, let h ( ' )be the element obtained from h by replacing each occurrence of z(1, k , r ) (any r ) in h by z(1, k, r + t ) . The analysis of $5 can now be applied to the group ( x ( l , k ) , y ( l , k ) , u(1, k ) , u(l, k), h ) . Analogues of Lemmas 5.3 to 5.6 with h taking the place of ~ ( 1 k, , k ) and h") taking the place of z(1, k , k + t ) can be obtained (a certain amount of care being necessary about t h e range of values that t is allowed to take). It thus becomes clear that the equation z ( l k , k ) = w(x(1, k ) , y (1, k ) , u ( l k ) , 4 1 ,

k), h )

is possible only if h = t 6 y ( I ,/ ~ ) ~ " 'k, z (sl),y ( l , k)-"" for some t 6 E D, and some integers s, t.


243

We now put together the results we have achieved so far to obtain

Proposition 7.1. Suppose P is an infinite subset of N such that G = Gpx GN,Pand suppose cp : G + Gpis a n isomorphism. Then there exist integers k , m , and t n k and congruence classes C I ,C2,. . . Cl mod n k such t h a t P = X U C I U C 2 U . . . U C ,w h e r e X C { l , 2 , ... m } , and, if C, is the congruence class of 1, mod n', there exist elements g, of Get, integers s,, t, and elements &, &, e4, e5 of D such that

el,

Moreover, each s, is greater than or equal to two. Proof. Only the inequality for s, remains to be proved. Let s, = min s, and suppose that s, = 1. The equation 9nt, - 9n'v = 9nt, - 9nS:rlmay be rewritten as t, - t, = ns,-'r - r l , and so it has a solution for the pair (r, rl). Moreover, if (r, r l ) is one solution, then ( r + i f rrI + r'n'd-'!) is another. Now G , C G, for each i = 1,2,. . . t, and consequently (ignoring a few terms of the direct product) we have Gc, Gp.There is therefore a word w on cp(x), cp(y), c p ( f ) , cp(u) and cp(u) whose projection on S, is trivial for all sufficiently large i fiZ C,, but non-trivial for infinitely many i E C,. But a consideration of the equation 9nt, - 9 n ' ~= 9nt, - 9nSlrlshows that the projection of w on the r t h term of Cl is a conjugate of the projection of w on the r,th term of C,, and this pattern is repeated for all sufficiently large r. Thus the projection of w on S, is trivial for all but finitely many i E C,, and we have our contradiction. 0 This establishes the form of all possible isomorphisms between G and Gp.We will use this information in the next section to show that G Z G' for any 1 < r < n.

J.M.T. Jones

244

88. The non-isomorphisms We are now ready to prove the final theorem. Theorem 8.1. G $ G' for any r such that 1 < r < n. Proof. Suppose that G = G px GNiP,and that G = Gp. We first show that P is infinite. Suppose not. An easy adaptation of Proposition 4.4 shows that given any finite integer k , the only possible non-trivial decomposition of S' as a direct product is of the form S' = S' x S'-' for some r, 1 zs r < k. It follows that S' can be decomposed into a direct product in only finitely many different ways. But Lemma 4.1 shows that G can be factored as a direct product with an arbitrarily large number of factors. Thus G $ S' for any k. Hence P is infinite, and so the result of Proposition 7.1 holds. We will prove that the number t of congruence classes mod n k given in Proposition 7.1 is congruent to 1 mod n - 1. We proceed by induction on t. If t = 1 the result is proved, so assume t > 1. Now the proof of Theorem 3.1 shows that the normal closure in G of the commutator [ x - ' , z - ' ] is a subgroup of G isomorphic to G. q induces an isomorphism between this subgroup and some subgroup of G, We will consider the form of this induced isomorphism, and hence establish the basis of an inductive proof. Now z = u i y - ' u - ' y , and so we have

q ( r )= 5 6

fl gll(u(l,,k ) ~ ( 1 , , k ) ~ " ' ~k,s,)y(I,, z ( 1 , , k)-9"7(I,,k)-'u(1,k)-'~(l,k))g,' ,=I

Since s, 3 2 for each i, this is 56n:=,gI,(y(1,,k)"iz(I,, k , s , - J y ( l , , k)"i)g,'. We now consider the projection of cp([x-', z - ' ] ) on S, for large i = 1, mod nk. If t, is not congruent to one modulo n, cp([x-', z-']), = 1. Consequently, if each of the t, satisfies t,f 1 mod n, then ~ ( [ x - ' 5, ' 1 ) lies in D, and so cp induces an isomorphism between G and S k for some k . We have already seen that no such isomorphism exists, and so there are some integers i, 1 c i I, such that t, = 1 mod n. Set R = uI,-,(,,)C,. Then, if we ignore a few terms of the direct product, cp induces an isomorphism between G and GR. We investigate the form of this isomorphism. The calculations made in the proof of Theorem 3.1 show that if 4 is the induced isomorphism, then there exist elements vl, qt, v3, v4, q 5 fo D such that


245

Similarly, (as in Theorem 3.1) t h e normal closure of (y9n-9Zy-(9n-9) ) -I ] in G is another subgroup isomorphic to G. If we apply t h e above arguments to this subgroup, we obtain similar results for those congruence classes C,for which t, = 2 mod n, and, continuing in this way, we derive similar results for those congruence classes C, for which t, = r mod n, for each r, 1 < r < n. Our inductive hypothesis allows us to conclude that the classes C, can be divided into n sets, each containing a number of classes congruent to 1 mod n - 1. Thus t is congruent to ( n x 1) mod n - 1, that is, t = 1 mod n - 1. Now suppose that G = G ' with 1 < r < n. Then there is a partition {PI,P2,.. . P,} of N such that G = GP,x GP2 x . . . x Gp,,and each Gp,= G. By the above reasoning, each set P, consists of a subset of D, and a number t, of congruence classes modulo n k (where k is taken sufficiently large to be the same for each GP,),such that t, = 1 mod n - 1. Thus the total number of congruence classes i n N = PI U P2U... UP, is congruent to r x 1 = r mod n - 1. But the total number is n k ,which is congruent to 1 mod n - 1. Thus r = 1 mod n - 1, contradicting 1 < r < n. Thus, G f G' for any r such that 1 < r < n, and t h e result is established. [x-l,

References [l] R. Camm, Simple free products, J. London Math. SOC.28 (1953) 66-76. [2] A.L.S. Corner, On a conjecture of Pierce concerning direct decompositions of Abelian groups, Proc. Colloq. Abelian Groups (Tihany, 1963) (Akademiai Kiado, Budapest, 1964) pp. 43-48. [3] J.M. Tyrer Jones, Direct products and the Hopf property, J. Australian Math. SOC. Vol. 17, Part 2 (1974) 174-196. [4] W. Magnus, A. Karrass and D. Solitar, Combinatorial group theory, Pure and Applied Mathematics, Vol. 13 (Interscience Publishers, 1966). [5] C.F. Miller, 111, On group-theoretic decision problems and their classification, Annals of Maths. Studies, No. 68 (Princeton University Press, 1971).

S.I. Adian, W.W. Boone, G . Higman, Word Problems I1 @ North-Holland Publishing Company (1980) 247-254

CANCELLATION THEORY IN FREE PRODUCTS WITH AMALGAMATION Roger C. LYNDON* University of Michigan, Ann Arbor, Michigan

The technical details of the original talk appear elsewhere [27], and need not be repeated here. Instead, we first discuss the general context of the problem under consideration, and then give a brief summary of the specific results obtained. 01. Cancellation theory in group theory

There are two principal methods, admittedly related, in the study of free groups, free products, free products with amalgamation, HNN extensions, and similar group constructions. One is the method of cancellation arguments, already used extensively by Nielsen [31, 32, 331. The second, equally old but less systematically used until recently, derives from topology, and is often described as using covering space arguments. One could say that cancellation arguments are based on nothing more than the sheer group axioms, while covering space arguments employ elaborations of Cayley’s observation that every abstract group can be realized as a group of transformations of a suitable object. For general subgroup theorems, covering space arguments give by far the shortest and most transparent proofs. The most elegant theory at present seems to be that of graphs of groups, or groups acting on trees, introduced by Serre and Bass [43]. The fundamental theorem of this theory can be developed without topology, as is done by Serre and Bass, although a very simple proof can be given, using only a minimum of the most elementary topology, following the ideas of Tretkoff [MI. Once the fundamental theorem is known, no further topology is needed to obtain immediately a subgroup theorem for graph products of groups, which contains all the main classical subgroup theorems. * The author gratefully acknowledges support of the National Science Foundation 247

248

R.C. Lyndon

Chiswell [4] has used this theory to obtain a fairly direct and transparent proof of the Grushko-Neumann Theorem, which is comparable in its simplicity (and presumably related) to the topological proof by Stallings [45]. A number of the more technically specialized theorems concerning subgroups of groups of the sorts under consideration have been given improved statements and proofs by Cohen [7, 81, exploiting t h e implicit interpretation of the Serre-Bass theory in terms of graphs of cosets. The theory has other applications, as noted by Serre, notably to the structure of linear groups, especially PSL2’s, where it is closely related to a fruitful method of Behr [ l , 21. There remains, nonetheless, an area where cancellation arguments appear indispensable. Cancellation arguments themselves can be divided into two kinds. There are those, exemplified historically by the work of Dehn [9], that are concerned typically with the question of membership in the normal subgroup defined by a given set of elements; of this theory, which contains most of small cancellation theory, we shall say no more here. The remaining area is exemplified by the problem of characterizing the elements of a subgroup generated by given elements. A first example is Nielsen’s proof of his subgroup theorem. For this limited purpose his method is hardly preferable to that of Schreier [a] (which can be interpreted topologically), and is certainly inferior to the standard simple covering space argument; but it is very fruitful for other purposes. First, it gives immediately a set of generators for the automorphism group of a finitely generated free group (Nielsen [31, 32, 33]), and so also for G L ( n , Z ) . A most important direct outgrowth of this method is the theorem of Whitehead ([46, 471; see also Rapaport [36], Higgins and Lyndon [17]), giving an effective criterion for two finite sequences of elements from a free group to be equivalent under automorphism. From this has grown in turn McCool’s [28, 291 recovery, in simple form, of Nielsen’s finite presentation of the automorphism group of a finitely generated free group (and with it of GL(n, Z)), as well as McCool’s [30] finite presentations for stabilizers in this group, and, consequently, for the classical mapping-class groups. In another direction, Nielsen transformations and their generalizations play an important role in the study of low-dimensional topology. In the original setting of free groups, the main role of cancellation theory lies in transforming a given finite sequence of elements into one that is Nielsen reduced. A Nielsen reduced sequence is then a basis for the subgroup it generates. (An elegant refinement of t h e Nielsen reduction process was presented at this conference by Hoare [18].) This cancellation argument was formulated axiomatically by Lyndon [22],

Cancellation theory in free products

249

using length functions, to obtain a uniform and sharper form of the Nielsen-Schreier and Kurosh subgroup theorems, as well as a very down-to-earth proof of the Grushko-Neumann Theorem ([23]; see also Cohen [6]). Chiswell [5] has recently shown a close connection between the theory of length functions and the Serre-Bass theory. Cancellation theory has long been applied to free products, free products with amalgamation, and HNN extensions, to the extent of providing normal form theorems for such groups (Britton’s Lemma, in the case of HNN groups), and relative solutions of the word and conjugacy problems. But the role of Nielsen transformations and the status of Nielsen reduced sets had received only small, though important, consideration. In 1970 Zieschang [54] initiated an extension of the full theory, with the necessary modifications, to free products with amalgamation. The essential difficulty in such an extension is that a Nielsen reduced set, even with any reasonable modification of the definition, cannot be expected to be a basis for a free group..However, Zieschang managed to pinpoint the special configurations that must occur in the exceptional cases. By these means he, later together with Peczynski and Rosenberger [35], managed to determine the ranks of all Fuchsian groups. This method was used also by Rosenberger [37] to analyze the situation in which an analog of the Grushko-Neumann Theorem fails for free products with amalgamation. The work to be described here uses Zieschang’s formalism to obtain a proof of a version of the theorem of Karrass and Solitar [19] on subgroups of a free product with amalgamation. Although the proof is admittedly lengthy, we believe that the detailed information obtained is instructive and should be of use in other applications. Indeed, the result as obtained contains the main content of Zieschang’s theorem. We mention only one application that we have immediately in mind. Recent work of Edmunds and others [3, 10, 11, 12, 13, 14, 151 refines earlier work on cancellation theory in free products to extend significantly work of Lyndon, McDonough, Newman, Rosenberger, and Zieschang [21, 24, 25, 26, 38, 39, 51, 521 on solutions of quadratic equations in free products, as well as work of Schupp [41, 421 and Wicks [48, 49, 501 on this and on the closely related endomorphism problem: given two elements w and w ’ in a free group F, when is there a map from F into itself carrying w into w’? A suitable analogous result for free products with amalgamation should put in a more general setting the result of Nielsen ([34]; see also Zieschang [53]) on lifting automorphisms of surface groups. The theory under discussion already gives some promise of revealing to what extent Edmunds’ work carries over to free products with amalgamation.

R.C. Lyndon

250

42. Cancellation arguments

We sketch below, omitting details, the derivation of a form of the Karrass-Solitar Subgroup Theorem for free products with amalgamation, using the method of Zieschang. Let G = * A {G, : A E A}, the free product of groups G A with a common subgroup A amalgamated. Every element of G either belongs to A or can be written in the form w = u1 u,, n 3 1, where each ui is in some GA, no ui is in A, and no successive ui, ui+,are in the same GA.We define the length of an element w in A to be 1 w 1 = 0; otherwise w determines the number n uniquely, and we define I w l = n. We write u , . . u, = 11, * * u, to mean that equality holds and that the right member is reduced in the sense that 1 u1 * * u, I =

---

-

I

u11+ * * *

-

+ 1 u. I .

Following Zieschang, we make constant use of the fact that every element of G can be written in the form

U'phq-',

IpI=lql, J h l S l .

Here u uniquely determines pA, qA, and A h A . If 1 h 1 = 0 we may choose h = 1. If p A = q A we may refactor, writing u =ph1p-', (h,l = 1. Paralleling the usual development of the Nielsen theory for free groups, we define a well-ordering of the cosets A p such that (1) if l P l < l 4 l t h e n p A < q A , (2) ifp =plp2, q = q l q 2 , Ipll = lqll,Ipzl= /q21, and p l A < q A , then pA 2 or for A,. We conjecture that they d o not exist, but we have not been able to prove this. However, we shall contribute to the problem and we shall try to explain why it may deserve some interest. We shall show first that it would suffice to prove our conjecture for Az. We have Theorem 1. A 2 is isomorphic with a subgroup of A *, for all n > 2.

Since A :+, 3 A z , it suffices to prove this for n = 3. Let a, 6, c be free generators of F3. Let a be an automorphism of F2 with free generators a, 6. Then a also acts o n A , as an automorphism by defining a ( c )= c. But unless a is the identity automorphism, its action on A, is never that of an inner automorphism. Next, we observe that t h e obvious methods for proving our conjecture d o not apply. The groups A. are residually finite according to a theorem of Baumslag. See [l]. The groups Al: are residually finite according to E. Grossman [2]. Therefore, Malcev’s Theorem (see [4]) according to which finitely generated linear groups over a field are residually finite, cannot be used to prove our conjecture. The same remark applies to Selberg’s Theorem [5] according to which a finitely generated finite-dimensional linear group over C has a torsion free subgroup of finite index. Indeed, this is most easily seen to be satisfied by Az which has a subgroup S of finite index with t h e following 255

256

W.Magnus, C.'Iketkoff

property: S is the extension of a free group Zz of rank two (the group of inner automorphisms of Fz) by another free group of rank 2, namely the group generated by the elements represented by the matrices

c 3'(; 3

of GL(2,Z). This remark also shows that Az cannot contain ''large'' abelian subgroups the existence of which might contradict the existence of a linear representation. We now prove

Theorem 2. Z f Az has a finite-dimensional faithful linear representation R over C , then at least one of the irreducible components of R represents Az faithfully. Proof. Let RA, A = 1,. . ., I , be the irreducible components of R. (Of course, R need not be fully reducible, i.e., R need not be the direct sum of the RA.) Assume first that none of the R , represents Zz faithfully, and let K , be the normal subgroup of ZZ represented by the identity in R,. Let (KA, K , ) denote the normal subgroup generated by commutators of elements of K , with elements of K,. Then (K,, K , ) C KA f l K,. Since Zz is free of rank 2, it follows that the intersection K of all of the K, is free of rank 3 2 . On the other hand, K would be represented in R by supertriangular matrices which would form a solvable group. Therefore R itself could not represent ZZ faithfully. Suppose now that, in RA,an automorphism a # 1 would be represented by the identity. Let I ( w ) denote the inner automorphism which maps any element f in Fz onto w-'fw, where w € Fz.Since a-'Z(w)a = I ( a w ) where aw is the image of w in Fz under the action of a, we would have that, in R,, the inner automorphism

I ( a w ) Z - ' ( w )= Z ( ( a w ) w - ' ) would be represented by the identity. This can happen only if ZZ is not represented faithfully in R, since there does not exist any automorphism a # 1 which maps every w E Fzonto itself, and since the center of F2 is trivial. We shall now analyze the possibilities for unitary representations over C of A2.We note that there exist such representations of finite quotient groups in which both the image of ZZ and of its quotient group are non-solvable. See Stork [6].

A presentation of A, can be given as follows: Let x, y be free generators of the free group F,. We present the group A2 of its automorphisms as follows: Generators. PI, uI,uz,U,V, U*,V*,defined as:

-

P: x + y , y + x ,

u1:

u: x +xy, u*:x + x ,

v:x + yx, y y. v*:x - x , y - x y .

y + y, y+yx,

x+x-',

y+y,

u2:x - x ,

y-y-'.

If w E F2 is any word in x, y, the inner automorphism Z ( w ) is defined by Z ( w ) : x +w-'xw, y +w-'yw.

We denote Z ( x ) , Z ( y ) respectively by X, Y and we have (1)

x = v*-1u*, y = v-'u.

For any a E Az, we have (2)

a-'Z(w)a =Z(a(w))

where a ( w ) is the image of w under the action of a . The defining relations for Az can be put into the form [3, p. 1621

(5)

P2= a:= a:= 1, UlU2 = U2U1, PU'P = uz PUP = u*,PVP = v* UIUUI = v-',U,U*U, = v*-'

(6)

u2uuz=

(7)

u,vu,= v-I,u1 V*UI =

(3) (4)

(8) (9)

U-I, UIU*UI =

u*-'

v*-' U-'u*V-'=U'P,uv*-'v = uv = vu,U * V *= V*U*.

u2P

As an easy consequence of (l), (4), (8), we find

(10)

u-lxu = x u ,

a relation which also would be derived from (2) and from the definition of u. To build up a unitary representation of A,, we start with the maximal abelain subgroup generated by U and V. We can diagonalize U and V simultaneously and we may assume that U (which, according to (lo), is conjugate with U-I) has in its main diagonal successively in certain multiplicities the distinct eigenvalues

+ 1, - 1,

Al,

A;’,

. . . , A r , A;’

We call any matrix commuting with U an w-matrix. An w-matrix consists of blocks whose sizes equal the multiplicities of the respective eigenvalues. We call blocks belonging to eigenvalues AZ and A;’ ( 3 = 1 , . . . ,r ) “conjugate blocks.” By using the proof of Clifford’s theorem on page 5 in [7], we find Lemma 1. The subgroup of AZ generated by U, V,ul,uzhas unitary representations which can be put into monomial form, with U, V, in diagonal form. The permutation associated with the monomial matrix u2 exchanges conjugate blocks in any w-matrix. Relations (lo), (6), (7) show that (11)

x-’ux= v =

u 1

U-Iul = u1u2Uu;’u;’.

Therefore xu1u2

=

R

is an w-matrix, and by using (l),(9,(6) we find

(Xu,)’= (OU2)’ = O 2= 1. (13) These remarks prove Lemma 2. If U has simple eigenvalues in a unitary representation of A2, then X’ commutes with U (since R is diagonal) and therefore (10) shows that Y X Y = X and that the representation cannot be faithful. We can sharpen Lemma 2. Let el and 5’ be the permutations associated with the monomial matrices representing u1and uz.Then we have Theorem 3. Let Il be the permutation group generated by Sl, Sz and by the permutations rr which exchange only eigenvalues of U which have the same value. If Il is imprimitive with the eigenspaces of U as a system of imprimitivity, then the unitary representation of A2 is not faithful. It should be noted that the permutations 7~ together with 6’ certainly generate an imprimitive group which satisfies the conditions of Theorem 3 and that el commutes with E2.That is, 6’is given and Sl is not entirely arbitrary.

Representations of automorphism groups

259

Proof. We show that a system of imprimitivity of II is also a system of imprimitivity for the group generated by X and Y. To see this, we observe that R is of order 2. Therefore the group ring generated by the unitary matrices v,,vz,R can also be generated by diagonal matrices and the matrices in Z7. Therefore X is imprimitive if and only if II has this property. But if a power of X commutes with U, we have, as in Lemma 2, a relation

X"

=

(XYy

which cannot hold in a faithful representation since, in Az, X and Y freely generate a free group. Acknowledgment Work on this paper was supported in part by a grant of the U.S. National Science Foundation. References [ l ] G. Baumslag, Automorphism groups of residually finite groups, Proc. London Math. Soc. 38 (1963) 117-118. [2] E. Grossman, On the residual finiteness of certain mapping class groups, J. London Math. Soc. (2) 9 (1974) 160-164. [3] W. Magnus, A. Karrass and D. Solitar, Combinatorial group theory (Dover Publications, 1976) pp. 162-165. [4] A.1. Malcev, On the faithful representation of infinite groups by matrices, Amer. Math. Soc. Translations, Ser. 2, Vol. 45 (1965) 1-18. [5] A. Selberg, O n discontinuous groups in higher dimensional symmetric spaces, Int. Colloq. Function Theory, Tata Inst. Fundamental Res., Bombay (1964) pp. 147-164. [6] D. Stork, Structure and applications of Schreier coset graphs, Comm. Pure Appl. Mat. 24 (1971) 797-805. [7] B.A.F. Wehrfritz, Infinite linear groups (Springer Verlag, New York, 1973).

S.I. Adian, W.W. Boone, G. Higman, eds., Word Problems I1

@ North-Holland Publishing Company (1980) 261-295

ON REDUCIBLE BRAIDS James McCOOL* a p t . of Mathematics, Uniwrsity of Toronto, Toronto, Canada

11. Introduction The Braid Group B. ( n 3 1 ) may be defined as t h e abstract group o n the generators vl, uz,, . . , C T , - ~ with defining relations cr,q = up,

(1.1)

if I i

-

j

1 3 2, I s i, j

sn

-I

and ~r,a,+~rr, = U,+~U,U,+~,1 s i

(1 - 2 )

S

n -2.

We denote by 93 the set of all ordered pairs ( p , r ) , where r is a and positive integer and p is a word on the generators ul,a 2 .... , their inverses, and we call 93 t h e set of (algebraic) braids. In [ I ] J. Birman has defined a braid (p, r ) to be reducible if p is conjugate in B, to an element of the form ya;ll, where y is a word on u *I I , . . . , a?!2,and has asked how one can recognise, given a braid (p, r ) , whether or not it is reducible. The interest of this question stems from a result of Markov, which says that in order to provide an algorithm to determine, given any two links, whether or not they are combinatorially equivalent, it is sufficient to describe an algorithm to determine, given any two algebraic braids, whether or not they are combinatorially equivalent; here braids (p, r ) and ( P ’ , r ’ ) are said to be combinatorially equiualenr if there is a sequence of braids (p, r ) = (PI,rl), (&, r2),. . . ,(&, r.) = (p’, r’), such that each braid in the sequence can be obtained from its predecessor by applying one of t h e following operations: A,

replace ( P , r ) by ( y , r ) if p and y represent conjugate elements of

B“ A,

replace (p, r ) by (pa:’, r + l), or, if p = pa:!l in B, where p is a word on a?, . . . ,a,f12,replace (p, r ) by ( p , r - 1). * Research supported by a grant from the National Research Council of Canada. 261

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A full account of Markov’s result is given in [l]. The restricted question of finding an algorithm to determine when two given braids (p, r ) and (7, r ) are equivalent under A, operations, i.e. the conjugacy problem for B, has been solved by Garside [3]. The investigation of reducible braids would seem to be a natural next step in the study of the general problem of combinatorial equivalence of braids, since to say a braid ( p , r ) is reducible is to say that it is joined to a braid ( y , r - 1) by a single application of operation A2 (preceded by an application of A,). The aim of this paper is to describe an algorithm to determine, given a braid (p, r ) , whether or not (p, r ) is reducible. In addition, we shall show that if (p, r ) is reducible to (y, r - l), i.e. if p’ = yu:!, in B, for some conjugate p’ of p, then y must belong to a determinable, finite set of conjugacy classes of B,-,. 811. Statement of results

In order to state our results precisely, we begin with a brief description of some of the results obtained by Garside [3]. A word W on u?,. . . ,u:!, is said to be positive if it is a word on u,,. . . , u,-~only, i.e. no inverse of any generator occurs in W. Of particular interest are the positive words and

A,

~

A,

As-IAs-2*

1

~ + 2U, ’

* *

A,,

(where = denotes equality of words). When we are considering B,, the words An-, and A. will usually be denoted by A and A, respectively. A positive word P is said to be prime to A if there does not exist a positive word Q such that P = A Q in B,. Garside showed that each element y of B. has an expression of the form y = A ‘ P for some integer k and positive word P prime to A. Moreover, if A ‘ l P I is another such expression for y, then k = kl,and so P = PI.The integer k is called the power of y, and such an expression is called a standard form for y. For y E B. we denote by (e, ( y ) the set of conjugates of y of power i . Garside proved that %,(y) is finite for each i, that there is a maximum value of i, called the summit power of y, such that % , ( y ) is non-empty, and that elements p, y of B. are conjugate if, and only if, they have the same summit power t and (e, ( y ) = %?, ( p ) . Garside also provided an algorithm to determine, given y E B,, the summit set (e, ( y ) of y.

On reducible braids

263

We now observe that if r, s are integers such that 1 S r S s S n - 1 and B,,,is the subgroup of B, generated by the set {u,, . .,us}, then it follows easily from the usual geometric definition of B , that the map uI+ u,,uz+ u,+~, . . . ,us-,+l+ usextends to an isomorphism from B,-,+2 to B,,,(this may also be established algebraically by rioting that the case for was proved by Chow [2], and then using Garside’s automorphism 22 of B,, defined in I11 below, to derive the general case from this). This observation enables us to identify the groups B z , .. . ,B,-z with the subgroups B1.l,. . . , B l , n - of 2 B,, respectively. We say that an element y of B, is positively reducible to y ’ ~ ” if- ~ Our main result is there is a conjugate y ’ of y such that y ’ ~ , , -E~ Theorem A. Let y E B, be positively reducible to Y ’ U , , - ~ ,and let y have ) C,(y) summit power t. Then t C - 1 and there is y l E V,-z(y) U % , - ~ ( yU such that ylun-lE B.-l, and ylun-I is conjugate in to y’~,-~. Now given y E B, it is easy to see, by using Garside’s results and Theorem 2-7 of [l], that the finite set Vf-z(y) U V,-,(y) U V,(y) can be computed, and that it can be determined if any element y l of this set is such that Y ~ U , ,E - ~Bn-l.Thus, given y E B,, it can be determined if y is positively reducible; moreover, there is only a finite set of conjugacy classes of containing an element of the form Y ’ U , - ~ , where y ’ is a conjugate of y, and this set can be determined. for some conjugate y ’= P y P - ’ If y E B. is such that y’u;!,E of y, we say that y is negatively reducible to y ’ u ; ! , . Now PyP-’a;ll E if, and only if, (u,-l~)y-’(u,-l~)-’u.-lEi.e. y is negatively reducible to y ’ u ; ! , if, and only if, y-’ is positively reducible to (y’a;ll)-l. Now obviously y is reducible if, and only if, y is either positively reducible or negatively reducible. We thus have Corollary B. Let y E B,. Then it can be determined i f y is reducible. If y is reducible and V is the set of conjugacy classes in B.-I which contain elements to which y is reducible, then V is finite, and can be determined.

OIII. Notation and preliminary results Let U, V be words on u?,.. .,u:!,.We write U = V if U and V are equal as words, U = V if U and V represent the same element of B., U V if U and V represent conjugate elements of B.. If P = XIX2 X, is any word, where each Xi is a generator or the

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inverse of a generator, then by the reverse of P, revP, is meant the word X,Xl-l* * * XI. We have rev QP = rev P rev Q. (1 Q i d n - l), extends to an We note that the mapping u,+ automorphism 52 of B. of order two. We shall write X for 9 X when this notation is more convenient. with defining Let S . be the abstract semigroup on uI, u2,.. . , relations (1.1) and (1.2). We write X = Y to mean X and Y are .. We note that if positive words which represent the same element of S X = Y then rev X = rev Y . If X = U V we say that X begins with the positive word U,and ends with the positive word V . The diagram D ( W ) of the positive word W is defined to be the set of positive words which can be obtained from W by a succession of applications of the operations replace a subword u,u,by up,, if 1 i - j

(3.1)

1 3 2,

replace a subword u,u,u,by u,u,u,,if 1 i - j

(3.2)

I = 1.

We note that D ( W )is finite, and consists of all positive words which are equal to W in S .. We have the following results, due to Garside [3].

Theorem 1. S. is embedded in B. under the identity map on the generating set ul,. .. ,

x

Thus, for positive words X and Y, we have X = Y if, and only if, =

Y.

Lemma 2. Let X, Y be positive words in B,. If uix= U k Y , then (a) X = Y i f k = i. (b) x = u k z , Y = uiZ, for some 2,if I k - i 1 zz 2. (c) x = UkqZ, Y = uiukz, for some Z , if I k - i 1 = 1. If Xui = Y u k , then (d) X = Y if k = i. = Zo,, for some Z , i f 1 k - i 1 3 2. (e) X = zuk, (f) x = z u i v k , Y = zukui, for some Z , if I k - i I = 1. Lemma 3. In B., u,A, = Atu,-l

if 1 < s

a&, = A,U,+~

if 1 Q s < t c n - 1..

ss t d

n - 1,

Lemma 4. Let XI,Xz,. . . , X I be generators of B, which commute with ui, and such that no Xiis ui. We have

On reducible braids

265

(a) Zf uiP = X l X 2 . XfQ, then Q begins with ui. (b) Zf Pai = Q X , X , * X f , then Q ends with a.

--

Lemma 5. (a) Zf ui+lP= AiQ in B., then Q begins with U ~ + ~ U , . (b) Zf Pai+l= Q(rev Ai), then Q ends with uiui+l. Lemma 6. In B., (a) XA = Ag, AX = X A , for any word X . (b) A = = rev A . (c) A begins and ends with ui,for each i E {1,2,. . . ,n - 1). (d) Zf r s n - 1 and W is a positive word which begins (ends) with ui, for each i E {1,2,. . . ,r } , then W begins (ends) with A,+l. (e) Zf W = A V and W = PQ, then, for each i E {1,2, .-..,n - l}, either P ends with ui or Q begins with ui. (f) Zf A = IF, then for each i E {1,2, . . .,n - l}, either Z ends with ui or F begins with a,but not both.

a

The next result, due to Birman [l], is of great importance for our work.

Theorem 7. Let /3 = A “ P be in standard form in B.. Zf /3 is not in summit form (i.e. m is less than the summit power of p ) then there are positive words Z, F, R with A = ZF, such that (a) P = FRZ if m is even, (b) P = FRf if m is odd. We shall also need the following result from [l].

Lemma 8. Let A = ZF. Then A = Ff = fp = FZ. We now introduce two notational conventions which facilitate the statements of the results to follow. Let i, j be non-negative integers. We define a ( i ;j ) to be the product aiui+l - .ujif 1 d i d j d n - 1, and to be the identity element of B. otherwise; p ( j ; i ) is defined to be the product U , U ~ - ~ * al if 1d i d j d n - 1, and to be the identity element otherwise. We shall write Q = Q ( r , s) to indicate that the word Q involves (at most) the generators a, .. ,a,,where 1 d r d s d n - 1. In a context where we have established that Q = Q ( r , s), and i is a positive integer such that i + s - r d n - 1, we shall denote by Q(i, i + s - r ) the ..,a, by ui, . . ,ai+,-, word obtained from Q by replacing a, a,+l,.

-

--

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respectively, and by Q(i + s - r, i ) the word obtained from 0 by replacing u,,u,+~, .. . , us by a,+,-, . . ,ui respectively. For example, if Q = Q(r, s) then 6 may be written as Q ( n - r, n - s), and if s < n - 1 then we may write AQ(r, s) = Q ( r + 1, s + l ) A . Finally, in this section, we note that, in view of the isomorphism the results listed above may be exhibited earlier from Bs-,+2to B,.,, applied to the subgroup B,.s. To give just one example, we have An-2(1, n - 3), and from (a) of Lemma 6 we have u i A , - 3 ( 2 , n - 3 ) = A , - 3 ( 2 , n - 3 ) u , - i - 1 , if 2 S i S n - 3 .

OIV. Further preliminary results We now obtain a number of results which are required for the proof of Theorem A. Lemma 9. Let X , Y be positive words in B, with X = Y. Then a generator ui occurs in X if, and only if, it occurs in Y.

Proof. This follows immediately from the definition of D ( X ) . Lemma 10. Let i, r be integers with 1 s i s r s n - 2. We have (a) If P a ( i ; r ) = Qu,,, in B,, then P = Z a ( i ; r + l), for some Z. (b) If P ( r ; i ) P = C,+~Q in B,, then P = P ( r + 1 ; i ) Z , for some Z. (c) I f P P ( r + l ; i + l ) = Q Q ai,n & , t h e n P = Z p ( r + l ; i ) , f o r s o m e Z . (d) If a ( i + 1; r + l ) P = uiQ in B., then P = a(i, r + 1)Z, for some Z.

Proof. We note that (b), (d) follows from (a), (c) respectively using the fact that X = Y implies rev X = rev Y, while (c) follows from (a) by applying the automorphism 92. Thus it is only necessary to prove (a). We note that (a) is true if r = i, by part (f) of Lemma 2. We suppose that (a) holds for i, r as stated, with r < n - 3, and that P l a ( i ;r + 1 ) = Qlu,+z. Applying (f) of Lemma 2 again, we see that P l a ( i ;r) = Qzu,+lu,+z and QI= QZu,+2u,+l, for some Q2. Hence, by (b) of Lemma 4, we have PI= P2ur+2for some Pz, and we then see that P z a ( i ;r) = Qzu,+l.It now follows, from our assumption that t h e result holds for i, r, that P2 = Z a ( i , r + 1) for some Z, so that PI= Za ( i ;r + 2), as required. Lemma 11. Let i,j,r be integers with 1 s i sj s r s n - 1. (a) If P a ( i ; r ) = Qui, and j < r, then P ends in uj+,. (b) If P p ( r ; i ) = Qui, and j > i, then P ends in uj-l.

On reducible braids

267

(c) If a ( i ;r ) P = a,Q, and j > i, then P begins with a,-1. (d) If P ( r ; i ) P = a@, and j < r, then P begins with q + , . Proof. As in Lemma 10, we need only prove part (a). By Lemma 4, we have Q = Q , a ( j + 2; r ) for some Q,, so that P a ( i ;j + 1 ) = Qlaj.Hence, by Lemma 2, P a ( i ;j ) = PIu,+,a, for some P I , and so P a ( i ;j - 1 ) = as Plaj+l. Applying Lemma 4 again, we see that P ends in required.

Corollary 12. Let W be a positive word and r a non-negative integer. (a) If W A { a A } ' ends in a, and 1 s i < n - 1 , then W ends in a,+,. (b) If W{AA}' ends in a, and 1 S i < n - 1, then W ends in mi. Proof. These are immediate consequences of Lemma 11.

We note, comparing with Lemma 10, that each part of Corollary 12 has three variants obtained using t h e operations rev and 2.We leave it to the reader to state these variants. Lemma 13. Let U, V , W be positive words with W = UV. Suppose that U ends only in an-,,V = V ( l ,n - 2) and W ends in a,, for some i E {1,2,. . . , n - 2). Then V ends in a,. Proof. We must clearly have L ( V )3 1, where L ( V ) is t h e length of the word V. Now if each generator in V is distinct from, and commutes with t h e generator a,,then by Lemma 4, U ends in a,. Since this is not the case, we must have either V ends in a,, or V = V,a,+rV2, where E = 5 1 and each generator in V 2is distinct from and commutes with a$. In the latter case, we have W = UV,a,+.V2=Ka, say, and by Lemma 4 we'see that K = K1V2for some K , , so that UVlal+r= Kla,. We then have, by Lemma 2, that UV, = K 2 a l + r a gfor , some K 2 . Now if L ( V )= 1, then V , = 1 and U must end in a,, which is a contradiction. Hence L ( V )> 1 and we have, inductively, that V , ends in a,, V ,= V3a,say. We now have UV3= K2u,+.,and i + E f n - 1 , since a,+,occurs in V. Hence, inductively, V 3must end in a,+E, V3= V4u,+,say. We then have V = V , C T # +=~V4u,+,a,u,+.V2 V~ = V4a,u,+,V2u,, as required.

We leave it to the reader to state t h e three variants of Lemma 13 obtained using rev and 2.

J. McGwl

268

Lemma 14. Let W be a subword of {AA}", m > 0. Then D ( W )= W. Proof. This follows immediately from the definition of D ( W ) . As a consequence of Lemma 14, we note that A{AA}' (r PO) begins only with uland ends only with

Lemma 15. Let X , Y be positive words in B.. (a) Z f X Y = a ( 2 ; n - l ) a ( l ; n - 2 ) , then forsome k,j with Ocjs n - 2 , l c k c n - 1 and k - j a l , wehaue Y = a ( k + 1 ; n - l ) a ( j+ 1 ; n -2). X r a ( 2 ;k ) a ( l ; j ) , (b) If X Y = P(n - 2; 1)P(n - 1 ; 2), then for some k , j with 2 s k d n, l s j c n - 1 and k - j P 1 , wehaue

X = P(n - 2 ; j ) p ( n - 1 ; k ) ,

Y = P(j

- 1 ; 1)P(k - 1;2).

Proof. W e need only prove part (a). W e let 9 be the set of words of the form UIV 1U 2 V 2 . . UsVs,where s d n - 2 and

-

U 1 U 2 . . U,= ( ~ ( 2n; - l ) ,

-

V l V 2 . . V,E ( ~ ( 1n; - 2),

V, = a(pj ;pj+i - l ) , for l s i d s , with 2 = A 1 < A 2 < - . . < A , + l = n, 1 = p l < p 2 < - . . < p s + I = n - 1, and Ui

(Ai

;A i + l - l ) ,

( i = 1,2,. . . ,s). - (pi+I - 1)5 2 (4.1) W e claim that Ed is the diagram of W = a ( 2 ; n - l ) a ( l ;n - 2). W e note firstly, from (4.1), that UiV,= V,Ui whenever 1 s j d i - 1. Thus, if UIVl UsVsis in 9, then

-

u,v,*.* u,vs= u,u2..*usv,v2**. v,= w, so that Ed C D ( W ) . Now if 2 E 9 it clear that no operation of type

(3.1) can be applied t o 2,while any application of (3.2) t o 2 will result in an element of 9. It follows that Ed = D ( W ) . Hence if W = XU, then there is an element U 1V, U,V, of Ed such that X Y = U 1VI * . U,V,, and the result follows easily from this.

--

Lemma 16. Let P be a positive word in B., with P prime to A. Then Pun-lhas power zero or one, and the latter occurs if, and only if, P ends in a ( 2 ; n - l)Am-l. Proof. Suppose Pun-lhas power at least one, Pun-1= SA say. Then, by

On reducible braids

269

(e) of Lemma 6, P ends in each of a], u2,.. . , so that by (d) of Lemma 6, P ends in Am-l, P = QA.-, say. Thus QA.-lun-l = SA, and using Lemma 8, we have QA.-lan-l = Sa(2; n - l)Am-la,,-l, so that Q = S a ( 2 ; n - 1). Hence P ends in a ( 2 ; n - l)A,,-], as required. = ZA. Now Z cannot Conversly, if P = Zcu(2; n - l)An-l then Pan-1 contain A, since P is prime to A, and therefore Pa,,-1 has power one. This proves the Lemma. Lemma 17. Let Q be a positive word, and r a non-negative integer. We have : (a) {Aa}'AQ contains A if, and only if, Q begins with (b) {aA}'aQ contains A if, and only if, Q begins with Proof. Part (b) follows easily from (a) and the observation that a positive word P contains A if, and only if, P contains A. To prove (a), we observe firstly that if Q begins with then so that {Aa}'AQ contains A. Q = A,-IQI say, and A = Conversely, suppose that W = { A a } ' A Q contains A, so that W begins with each of a2, ma,.. . , Using (a variant of) Corollary 12, we see that Q must begin with each of u I , a,..., 2 and so Q begins with A n - I . Lemma 18. Let V be a positive word such that some generator a,does not occur in V. Then A 'V has summit power r, for each integer r. Proof. We know, by Lemma 9, that V cannot contain A, so that A ' V is in standard form. Suppose that A ' V has summit power greater than r. Then, by Theorem 7 there exist positive words Z, F, Y with A = IF and V = FYI*, where I*= Z if r is even, I*= 1 if r is odd. Since ai does not occur in V, it does not occur in F or I*. However, we have A = Z*F if r is even, and, by Lemma 8, A = FZ* if r is odd. Since a, occurs in A, we obtain a contradiction from Lemma 9. Lemma 19. Suppose that WV(2, n - 1) = U p ( s ;1) in B., where 0 S s S n - 1, and let r be the greatest integer such that W ends in P ( r ; l), W = W l p ( r ; 1) say. Then there exists I 3 0 , with s + I s r, and positive words Hi = Hi (2, s + i - l), Ki = Ki (s + i + 1, n - l), 1 G i s 1 + 1, such that ,

and U

=

V 7 H i + i K i + i ~ ~ + r H i K i ~* ~ * *+ Hr -~iK N , + I H ~ K I

Wlp(r; s + 1 + I ) H ; + I K i + i a ~ + i - i H ; K i ~ * + i - ~* *(HX2aSa,+iHIKi, ~~+i-i.

where H : = Hi (1, s + i - 2), and Hi= 1 if 2 > s + i - 1.

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Proof. If s = 0 then we may take 1 = 0, HI= 1 and K , = V. Thus we may assume that s 3 1. We consider firstly the case that some word in D ( V ) is of t h e form Hl(2,s ) K , ( s + 2, n - 1). It then follows from Lemma 4 that U = UIKl for some U , , so that WH, = UIP(s;1). Now applying Lemma ll(b), we see that U , = TH,(l, s - l), for some T. Hence W = TP(s;l), and the result holds with 1 = 0. We now suppose that n o element of D ( V ) is of the form H I K I ,with H , K , as above. It follows that there must be a word in D ( V ) of the form Vl(2, n - 1)uS+,Hl(2, s)Kl(s + 2, n - 1). As in the previous case, we then see that U = UIHIKI,for some Ul, so that w V ~ u , +=, UIP(s;1). From Lemma 5(b) we now obtain UI= U2u,us+l, for some U2, SO that U , p ( s ;1) = Uzusus+lp(s; 1) = U 2 p ( s+ 1; 1)us+,, and therefore WV, = U , p ( s+ 1; 1). Since L ( V I )< L ( V ) an inductive argument now completes the proof of the Lemma.

Corollary 20. Suppose that /3 ( r ;1) V(2, n - 1) = U p(s ; l), where 1 s r, s s n - 1. Then s c r and A-'U has summit power - 1. Proof. From t h e Lemma we have s

+ 1S r

and

U = P ( r ; s + 1 + l)H~+lKl+lus+~-lus+l . . . H;K2usus+lHIK1

+ 1 + l)H~+lus+l-lus+lHi' - . H ~ ~ , u , + , H ~ K l +. IKK,l = H;+lus+l-lH H~ ; uu, H s$+( r,; ~s 2 + ~l)Kl+lKl ~ K,. =P(r;s

*

*

* *

Now writing Z, = H ~ + l u ~ + l - l * .H H;u,H:, ~ u s +and l-2~ Z, = P ( r ; s + l)Kl+,K1 K1, we have U = ZIZ2.Noting that ZI= Zl(l,n - 2), Z2= Z2(2,n - l), since 1 c r, s < n - 1 and s + 1 r, we see that Z22, is a positive word on u,,. . . ,u,,-,. Since A-'Z2gIis conjugate to A-IU, it follows from Lemma 18 that A-'U has summit power - 1.

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We pause to observe that B. is the trivial group if n = 1, and is infinite cyclic if n = 2. From this, it is easy to see directly that Theorem A holds if n = 2. We shall therefore assume, in what follows, that n 3 3.

Lemma 21. Let r be a positive integer. Then (a) A:-, is in standard form in B.. (b) A -*r+'{AA}r-'Ais the standard form of A ;?',+I in B.. (c) A -,'{AA}' is the standard form of A i?', in B..

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271

Proof. Part (a) follows immediately from Lemma 9. Now if k is a positive integer, it is easy to prove, by induction on k, that i-k-1

W'(A)]

and therefore

Now IIP-k-192i(A)is prime to A, by Lemma 17, and this prove? the result. Lemma 22. Let P be a positive word in B. which is prime to A, and let r be a positive integer. Then: does not occur in P. (a) P E Bn-l if, and only if, (b) A'P does not belong to B,-l. (c) A-"P E B,-l if, and only if, P = {AA}'Q, where Q is a positive word in B,-l which is prime to Moreover, if P has this form, then A,?lQ is a standard form of A-"P in B,,-l. (d) A-"+'P E B,-, if,. and only if, P = {AA}'-'AQ, where Q is as in (c). Moreover, if P has this form then A,?,+;"Ois a standard form of A-''+'P in Bn-l. Proof. Suppose that, for some integers k and s, we have A *P = AL1Q, where Q is as in (c). If not both k < O and s < O hold, then it is easy to obtain a contradiction, using Lemma 9 and the fact that P is prime to A. In particular, this establishes (a) and (b). Now suppose that A 'P = A i-lQ, with k,s < 0. Then A *P = A'{IIPe-,-, Bi(A)}Q, and from Lemma 17 it follows that the right-side of this equality is in standard form. Hence we must have P = {l-IP=-s-l wi(A)}Q, as required. Conversely, if P has this form then A 'P = A i-IQ. This proves the Lemma.

#V. Proof of Theorem A The theorem is an immediate consequence of the following two Lemmas. Lemma 23. Suppose that A-"-'M is in standard form in B. ( r b O ) , and A-2'-1Mun-l E B.-l. Let t be the summit power of A-"-'M. We have: (a) t = - 1 if r = 0.

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J. McCool

(b) Zf r 2 1 then either t s - 2r + 1 , or there is a conjugate A -2rM1of A-"-'M such that A-"M, is in standard form, A ~2'Mlu,-l E En-,and A -Z'Mlun-l is conjugate to A-2'-1Mun-lin En-'. Lemma 24. Suppose that A-"M is in standard form in B, ( r 3 l), and A MU^-^ E En-,. Let t be the summit power of A-"M. W e have: (a) t s - 1 i f r = l . (b) If r 3 2 then either t s - 2r + 2, or there is a conjugate A -'"'MI of A-"M such that A-2"'MI is in standard form, A-2'+'MIun-l E En-' and A ~ 2 r + 1 M l uisn nconjugate -l to A -2rMun-lin

In order to prove Lemma 23, we shall require the following two results. Lemma 25. Let p = A -2rA{AA}'-' V ( 2 ,n - 1 ) be in standard form in B. ( r 3 l ) , and suppose that p has summit power t > - 2r. Then (a) t = - 1 i f r = l . (b) Zf r 3 2 we have V ( 2 ,n - 1 ) = X ( 3 , n - 1)V1(2,n - 1 ) Y ( 2 ,n - 2 ) for some X , Y , V, with Y ( 2 , n - 2 ) X ( 2 , n - 2) = An-,(2, n - 2). Also, p A -2r+1A{AA}'-2ul V , ( 2 ,n - l ) , and the latter word has power -2r+1.

-

Lemma 26. Let 4 = A -2r+1A {AA}'-'ul V l ( 2 ,n - 1 ) be in standard form in B. ( r 3 2 ) , and suppose that 4 has summit vower t > - 2r + 1. Then V 1 ( 2 n, - 1 ) ends with a ( 2 ; n - 1).

The proofs of Lemmas 25 and 26 will be given after that of Lemma 23. Similarly, the following two results will be required for the proof of Lemma 24, and their proofs will be given after that of Lemma 24. Lemma 27. Let p = A -2r+1{AA }'-I V ( 2 ,n - 1 ) be in standard form in B, ( r 3 2), and suppose that p has summit power t > - 2r + 1. Then V ( 2 ,n - 1 ) = X ( 2 , n - 2 ) V l ( 2 ,n - 1 ) Y ( 2 ,n - 2 ) for some X , Y,V , with V l ( l ,n - 2), and the P X = A,-,(2, n - 2). Also, p A-2r+2{AA}r-2un-l latter word has power - 2r + 2.

-

Lemma 28. Let 4 = A -2r"{~A}r-2un-I V l ( l ,n - 2) be in standard form in B. ( r 3 2), and suppose that 4 has summit power t > - 2r + 2. Then V , ( l ,n - 2) ends with ( ~ ( 1n; - 2), and t = - 1 if r = 2.

We now give the proofs of these Lemmas.

On reducible braids

273

Proof of Lemma 23. Suppose that A-"-'M is in standard form in B. (r 3 0), and that A-2'-1Mun-iE Z3,-,. We suppose, moreover, that A-2'-1M has summit power t > - 2r - 1. The plan of t h e proof is as follows. We begin by establishing that A-2'-'Man-, has power - 2r - 1. We then have, by Lemma 22, Since Man-,= {Aa}'AQ(l, n - 2), for some 0 which is prime to t > - 2r - 1 we also have, by Theorem 7, M = FRf for some F, R, Z with IF = A, so that FRfan-, = {Aa}'AQ(l, n - 2). We are then able, using the results of t h e previous sections, to analyse the contributions that Q(1, n - 2) makes to F and f respectively, and this in turn gives us a description of R. Now A -2r-1M = A -"-'FRf A -2rR, and the description obtained for R enables us to complete t h e proof. As was noted above, since t > - 2r - 1 we have M = FRf, where ZF = A. Also, A -"-'FRf A -"R, and A -"R has power - 2r, since R is prime to A. We note, by Lemma 16, that A-"-'Man-, has power - 2r - 1 or - 2r. We suppose that its power is - 2r, and consider firstly E Bn-,, we then have, by Lemma 22, the case r > 0. Since A -2'-1Mun-l that FRfu,-, = A{aA}'Q(I,n - 2), where Q is prime to A,-,. Hence, by Lemma 3, FRfu,-, = AQ(1, n - 2){AA}'. However, since A ends in u,-,, this would imply that FRf contains A, which is not the case. Thus A-2'-1Mu,-, has power - 2r - 1 if r > O . Now suppose r = 0 and A-'FRfun-, has power zero. From Lemma 16 it follows that F R f = Sa(2; n - l)A,-,, so that, in particular, FRf ends with each of u,, uz,. . ., It follows that f cannot end with u,-,, since otherwise, by Lemma 6, FRf would contain A. Now A = IF, so that, by Lemma 6, F must begin with uI. We have A-'FRfan-, = Q(1, n - 2), where Q is a positive word, and therefore

-

-

FRfun-,= AQ(1, n - 2)- a(1; n - l)An-lQ(l; n - 2) = A.-,a(l;

n - l)Q(l,n - 2)

= A,-,Q(2,

fl - l)Cy(l; n - l),

so that FRf = A,-,Q(2, n - 1)a(1; n - 2). Now An-l begins with each of a 2 ,u3,. . . , un-,, so that FRf begins with each of a l , u2,. . . , un-'and hence contains A. This contradicts t h e fact that M is prime to A. Thus we have established that A-2'-'Man-i has power - 2r - 1. Since A - 2 ' - 1 F R f ~ nE - , B,-,and has power - 2r - 1, we have, by Lemma 22, that FRfu,-, = {Aa}'AQ(l, n - 2), where Q is prime to An-,. Hence FRfun-, = {Aa}'Q(2, n - 1)A, so that (5.1)

FRf

= {AA}'Q(2,

)t

- l ) a ( l ; n - 2).

We now observe that if FRf begins with some a, (i > I ) , then, by (a

1. McCool

274

variant of) Corollary 12, O(2,n - l)a(l; n - 2) must begin with ui, and hence, by (a variant of) Lemma 13, O(2, n - 1) must begin with ui. Also, if FRf ends with some a,(i C n - 2) then, by Lemma 11, {AA}' must end with ui+l,so that, by Lemma 13, O(2, n - 1) must end with ui+t.

We now write F = FIF2,where FI= F1(2,n - 1) and F2 does not begin with any of u2,u3,. . .,u.-~.We also write Z = Ill2,where Z2 = Z2(3,n - 1) and I, does not end with any of u3, u4,. . . ,u.-~.From the observations of the preceding paragraph we see, firstly, that Q(2, n - 1) = F101(2, n - 1) for some O1= Q42, n - l), and secondly, since f2= Z2(n - 3, l), that O1= Q2(2,n - 1)Z2(n- 2,2) for some O2= Q2(2, n - 1). Thus we have

O(2, n - 1) = F1(2,n - 1)02(2,n - l)Z2(n- 2,2). We can now write (5.1) as

F2Rfl= {AA}'Q2(2, n - 1)a(1; n - 2).

(5.2)

In order to 'solve' this equation for R, we seek descriptions of F2 and I,. We begin with F2. We have

fp1p2= UA,-lp2 U,since fpl ends with each of

A for some ends with

A

= ZFIF2 =

Hence

ul,u2,.. . , c,,-~, and therefore

An-IF2U7 A.-IP(n - 1; l), . . A

7

so that p20= P ( n - 1; l), and therefore we have F2 = a(1; I) and U = a(I + 1; n - l), for some I with 0 d 1 d n - 1. We now consider ZI. Since Z, does not end with any of u3, u4,.. .,u , , - ~we , know that either (a) f, ends with both and u,,-~, or (b) 1, ends only with c,,-~, or (c) ZI = 1, or (d) f, ends only with u.-~.We shall consider each of these four cases separately, and show that the Lemma holds in each case.

f, ends with both and u,,-~. We have A = Z,Z2F= f,f# = flAn-zX

Case ( a ) .

for some X, since

CT,,-~

f2pbegins with each

of uI,uz,. . .,

Hence

A = XflA,,-2=a(1; n - 1)a(1; n - 2)A.-2, so that XI, = a(1;n - l)a(l; n - 2). Since f, ends in both we have 1, = Vu,,-2u.-lu,,-zfor some V, so that

and

On reducible braids

275

Xfl = X ~ u , - ~ u . - ~ =u ,a(1; - ~ n - l)a(1; n - 2) = a(2; n - 2)a(1;

n -2)~.-~,

and therefore X V = a(2; n - 2)a(1; n - 3). Applying Lemma 15(a) (with n - 1 in place of n ) we obtain R = a ( 2 ; k ) a ( l ; j ) and V = a ( k + l ; n - 2 ) a ( j + l ; n - 3 ) f o r s o m e k,j with O S j s n - 3 , 1s k n - 2 and k - j 3 1. Hence i,= a ( k + 1; n - 2)aG + 1; n - l)un-2. Before using the descriptions obtained for F2 and I, to study (5.2), we obtain a description of 12(3,n - 1)F1(2,n - 1). This will be used firstly to show that 1 = j , and then will be used again later in the proof of the Lemma for case (a). We have

A = Z2FIF2f, = X f l A , - 2 = d.-2a(2; k ) a ( 1 ;j)!,, so that

(5.3)

143, n - 1)F1(2,n - l ) a ( l ; I ) = A,-2(3, n - l)a(2; k ) a ( l ; j ) ,

= A,-2(n - 1,3) = A,-,(3, n - 1). We wish to show that 1 = j since and that

(5.4)

Z2(3,n - 1)F1(2,n - 1) = A,-2(3, n - 1)a(2; k).

Suppose, firstly, that 1 = 0. Then u Idoes not occur in the left side of (5.3), and hence cannot occur on the right, so that j = 0. Now suppose that 12 1. Then u,occurs on the left side of (5.3), so it must occur on the right, and therefore j 3 1. We now apply (a variant of) Lemma 13 to (5.3) as follows. The word a(1; I ) on the left begins only with ul, while the word A,-,(3, n - l)a(2; k ) on the right does not involve ul; it now follows, from Lemma 13, that An-,(3,n - 1)a(2; k ) is a beginning subword of L(3, n - 1)F1(2,n - l), Z2(3,n - 1)F1(2,n - 1)= A,-2(3, n - l)a(2; k ) X say. Similarly, since a(1;j ) on the right begins only with ulrit follows that A.-,(3, n - l)a(2; k ) = Z2(3,n - 1)F1(2,n - 1)Y, for some Y. Hence YX = 1, and since YX is a positive word we have Y = 1, X = 1. This establishes (5.4), and 1 = j follows. We may now rewrite (5.2) as (5.5)

a ( l ; j ) R a ( k+ 1; n - 2)aO + 1; n - 1) = {A,4}'Q2(2,n

- l ) a ( l ;n - 3).

Now, since j c n - 3, the right side of (5.5) must end in u,-~, so that, by Lemma 4, {A,4}rQ2(2,n - 1) must end in un-l. Since AA ends only in ul, it follows, using Lemma 13, that Q2(2, n - 1) ends in gn-1, Q2(2,n - 1)= Q42, n - l)un-, say. We then have

. I McCwl .

276

a ( l ; j ) R a ( k+ 1 ; n - 2 ) a G

(5.6)

+ 1 ; n - 2)

= { A A}'Q3(2, n - 1)a( 1 ; n

- 3).

Since the right side of (5.6) must end in u ~ . .we ~ ,see, using Lemma 10, that { A A } ' Q 3 ( 2 ,n - 1 ) must end i n a(1;n - 2 ) . Now if r = 0 this is impossible, so we must have r 3 1 . Since { A A } ' Q , ( 2 , n - 1 ) must end in a(2; n - 2) we see, using Lemma 13, that Q42, n - 1) ends in a ( 2 ; n - 2), 0 4 2 , n - 1) = Q4(2, n - l)(r(2; n - 2) say. We then have { A A } ' Q 4 ( 2 ,n - l ) a ( 2 ;n - 2 ) = { A A } ' ~ ' A Q 4 ( 1 , f l - 2 ) p ( f l1-; 2 ) a ( l ; n - 2 ) ,

and (5.6) yields Ra(k+l;n-2) = a(j

+ 1; n - 1){,&4}r-'04(1, n - 2 ) P ( n - 2 ; 2 ) a ( 2 ;n - 2 ) a ( l ; j ) .

Since j s k - 1 we have a ( 2 ; n - 2 ) a ( 1 ;j ) = n ( 2 ; k ) a ( k + 1; n - 2 ) a ( l ; j ) = a ( 2 ;k ) a ( l ;j ) a ( k

+ 1 ; n - 2),

so that

R

=

00' + 1 ; n - J){AA}r-'Q4(l,n - 2 ) P ( n - 1 ; 2 ) a ( 2 ;k ) a ( l ;j ) .

We conclude the proof of the Lemma for case (a) by finding a conjugate A-''M, of A-"R which has t h e required properties. We have A-"R - A - 2 ' A { A A } ' - ' Q 4 ( 1 , n - 2 ) p ( n - 1 ; 2 ) a ( 2 ; k )

- A-"A{AA}'-'P(n - 1; n - k + l)Q4(n - 1 , 2 ) a ( l ;n - 2 ) 5

A -'.MI

say. NOW ~ - ~ ' ~ ~ c r , - ~ = ~ - ~ ~ { A ~ ) ~ p ( n - 2 ; n - k ) ~ , ( n - 2 , 1 ) = A i Y l p ( n - 2 ; n - k ) Q 4 ( n - 2, l), so that A-Z'Mlan-l E Bn-l. to A - 2 ' M l ~ m - l Next, we check that A - 2 ' - ' M 5 n -is ~ conjugate in We have

A--2'-'M~m = A,2'L1Q(1, -l n - 2) =

A iCL'Fl(1, n - 2)Q41, n - 2 ) a ( ;n - 2)Z2(n- 3 , l )

- A ,?';'p(n =

(using (5.4))

- 2; 1)Z2(2,n - 2 ) F l ( l

n - 2)Q4(1, n - 2 )

A i?L'p(n - 2 ; 1)A,-42, n - 2 ) . 1; k - 1)Q4(1, n - 2 )

277

On reducible braids

= A ,?,cy(l;

- A ;?,/3(n =

k - l)Q4(l, n - 2) - 2; n - k)Q4(1, n - 2)

A -Z'Mlu.-,,

and the two conjugacy relations above hold in J3-,. Finally we must show that A-"M has power -2r. Suppose not. Then MI must contain A, so that, by Lemma 17, P ( n - 1; n - k + l)Q,(n - 1,2)a(1; n - 2) must begin with An-,(n - 1,2). Applying Lemma 13, we then have that P ( n - 1; n - k + 1)Q4(n - 1,2) begins with A,-,(n - 1,2), a n d so P ( n - 2; n - k)Q4(n - 2 , l ) must contain An-,. This implies that a(1; k - 1)Q4(1,n - 2) contains A,-,. However, we have Q(2, n - 1) = F,(2, n - 1)Q4(2,n - 1)a(2; n - l)Z,(n - 2,2), and applying Lemma 13 to (5.4), we see that F,(2, n - 1) = Xa(2; k ) for s o m e X = X(2, n.- l), so that

Q(l, n - 2) = X(1, n - 2 ) a ( l ; k - 1)Q4(1, n - 2 ) a ( l ; n - 2)Z2(n - 3,l). Since Q(1, n - 2) is prime t o A m - , , so therefore must be a(1; k - l)Q4(l, n - 2). This contradiction proves t h e L e m m a for case (4. . Case (b). f, ends only with a,-1. W e have A = ZIZzF= I1An-, Y , for = a(1; n - l)Ao-l, and therefore 9 = some Y , so that A = PfIAn-, cy(l;s), f I = a ( s + l ; n - l ) , f o r s o m e s w i t h O s s s n - 2 . Since A = ZzFIFrf,= d , - , a ( l ; n - l), we now have -

12(3, n - 1)Fl(2,n

-

1

l ) a ( l ; I ) = A.-,(2, n - l ) a ( l ; s).

As in t h e previous case we now see, using L e m m a 13, that 1 = s and

(5.7)

Z2(3, n - 1)F1(2,n - 1) = An-,(2, n - 1) = A,-2(3, n

- l)a(2; n - l),

and applying Lemma 13 again, we have (5.8)

F,(2, n - 1) = W a ( 2 ;n - 1 )

for some W = W(3, n - 1). W e now consider (5.2), which may b e written c y ( I ; s ) R a ( s +1 ; n - 1 ) = { A ~ } r Q 2 ( 2 , n - l ) a ( l ; n - 2 ) . Now s G n - 2, so the left side above ends with urn-,, and from Lemma 10 it follows that {Aa}'Q2(2, n - 1) ends with a(1; n - 1). This is

J. McGwl

278

impossible if r = 0, so we must have I 3 1. We then see that Q2(2,n - 1) = Q3(2,n - l)a(2; n - l), and then we have R =a(s

+ I; n - I ) { , ~ A } ~ - ' Q ~n( I-, 2 ) ~ ( -n I; 2)a(2; n - 1)a(1; s).

We now find a conjugate A-"M1 of A-"R which has the required properties. We have A-"R

- A-"a(2;

n - l)A{AA}r-1Q3(l,n - 2)P(n - 1;2)

=A-2'A{aA}r-'a(l;n-2)Q,(1,n-2)p(n-1;2)

- d-2rA{AA}r-'p(n- 1;2)Q,(fl - 1,2)a(1;

?I2)

-

= A -''MI say, and A-2'Mla.-1 = A ,?'$(n We have, using (5.7),

- 2; l)Q3(n - 2,l).

A-2'-'M~n-I= Ai?'L'Q(l, ~t- 2) =A,?';-'Fl(l,n-2)Q3(1,n-2)a(l;n-2)Z2(n-3,1)

- A ;?;;-'P(n- 2; 1)Z2(2,n - 2)Fl(l,n - 2)Q3(l, n - 2) = A,?la(l; n

-A

,?'l@(n

- 2)03(1, n - 2)

- 2; l)Q,(n - 2, l),

Hence A-2'-1Man-l is where the conjugation relations hold in conjugate to A-2'Mla.-1 in To complete the proof of the Lemma for this case, we need to show that A-2'MI has power -2r. If this is not so, then MI must contain A, so that P ( n - 1; 2)Q3(n - 1,2)a(1; n - 2) must begin with A.-42, n - l), and therefore P ( n - 1;2)Q3(n - 1,2) must begin with A.-,(2, n - 1). This implies that a(1; n - 2)Q3(1,n - 2) contains A,,-'. However, Q(1, n - 2) = Fl(l, n - 2)Qs(l, n - 2)a(l; n - 2)Z2(n - 3 , l ) =

W(2, n - 2)a(l; n - 2)Q3(1,-.n

2)a(l; n - 2)Z2(n- 3, l),

and Q(1, n - 2) is prime to An-l, so that a(1; n - 2)Q3(l, n - 2) must be prime to A,-I. This proves the Lemma for case (b). Case ( c ) . ZI = 1. We have

A = Z2(3,n - 1)F1(2,n - 1)F2= & - l a ( l ; n - 1) P ( n - 1 - 1; 1)An-1(2,n - l ) a ( l ; I ) , so that

On reducible braids

Z2(3,n - 1)Fl(2,n

279

1)- P ( n - 1 - 1; 1)An-42,n - 1).

-

Since uIdoes not appear on the left side of t h e above, we must have 1 = n - 1, and so Z2(3,n - 1)F1(2,n - 1) = An-1(2,n - 1). We now have, from (5.2),

a(1; n - l)R = {A,4}'Q2(2, n - l ) a ( l ; n - 2). If r = 0 then, by using 'rev' of Corollary 20, we obtain a contradiction. Hence we must have r t 1, so that

R

= ,4{A,4}'-lQ2(2,

n - l ) a ( l ; n - 2),

and A -2'Run-1= A i?'lQz(l,n - 2). Now A -2'-IMun-I= A ;?;'Q(l, n - 2) =

A i?'ilFl(l,n

- A,?'i1Z2(2,n =

A ;!'lQ2(1,

-

2)Q2(1, n - 2)Z2(n - 3 , l )

2)F1(1,n - 2)Q2(1, n - 2)

n - 2).

Since A -"R has power - 2r and is conjugate to A -*'-'M, this proves the Lemma for case (c). Finally we consider Case ( d ) . 1, ends only with of ZI.We have A

We obtain, as before, a description

= ZiZzF = Ild.-z~,-lH, A , . *

A

for some H, so that A

= An-zan-lHI1 = An-zP(n -

1; l)P(n - 1;2),

and therefore HII= P ( n - 2; l ) P ( n - 1; 2). Hence by Lemma 15, H = P ( n - 2; a ) P ( n - 1; b), ZI= P ( a - 1; I)P(b - 1;2) for some a, b with lSaSn-l,3GbSn andb-asl. Using A = An-$3(n - 1; a ) p ( n - 1; b)Zl = 1$,F2Z1,we obtain

12(3,n- 1 ) F 1 ( 2 , n - I ) a ( l ; 1 ) = A n - 2 ( 3 , n - 1 ) ~ ( 1 ; n - a ) a ( l ; n - b ) = An-2(3, n -

l)a(2; n - b + l ) a ( l ; n - a ) ,

since b - a 5 1. As in the previous cases, we see that 1 = n - a and that, for some W = W(3, n - l), (5.9)

Z2(3,n - l ) W = An-2(3,n - l), F1(2,n - 1) = Wa(2; n - b + 1).

We may now rewrite (5.2) as

280

J. McCool

a(1; n - a ) R a ( n- a 7 {AA}rQ2(2, n

+ 1; n - l ) a ( n - b + 1;2)

- l ) a ( l ; n - 2),

so that a(1;n - a ) R a ( n - a

+ 1; n - 1) = {Aa}'Q2(2, n - l ) a ( l ; n - b).

If now follows, since (n - a + 1)- ( n - b) 3 2, that {Aa}rQ2(2rn - 1) ends with a ( n - a + 1; n - l), and this implies that Q42, n - 1) = 0 4 2 , n - l ) a ( n - a + 1; n - l), for some Q3. We then have a(1; n

- a ) R = {Aa}'Q3(2, n - l ) a ( l ; n

- b).

Now if r = 0 we conclude from Lemma 19 that n - b 2 n - a, which contradicts the fact that b - a 3 1. Hence we must have r 3 1, and

R

=a(n-a

+ 1; n - 1)a{AA}'-'Q3(2, n - l ) a ( l ; n - b).

We now observe that

A-"R

- A-''a{Aa}'-'a(2;

n - b + 1)Q3(2, n - l ) a ( n - a

+ 1; n - 1)

A -"A{AA}'-'V(2, n - 1) say. We show that this latter word has power -2r. To see this, note that, by (5.9),

Q(2, n - 1) = F1(2,n - 1)Q42, n - l)12(n - 2,2) =

W(3,n - 1)a(2; n - b

+ 1)Q3(2,n - I ) a ( n - a + 1; n - 1)12(n- 2,2),

and the desired conclusion follows easily from t h e fact that Q(1, n - 2) is prime to An-'. Thus we have shown that A -"-'A4 is conjugate to A-"A{Aa}'-'V(2, n - l), and that t h e latter word has power - 2r. Now from Lemma 25 we see that if r = 1 then A-"-'M has summit power t = - 2 or t = - 1. Thus we may suppose that r 3 2. Now if t s - 2r + 1 there is nothing to prove, so we suppose that t > - 2r + 1. Applying Lemmas 25 and 26 we obtain

V(2, n - 1) = X ( 3 , n - 1 ) V,(2, n - 1) Y(2, n - 2) = X(3,

n - 1)V2(2,n - l)a(2; n - 1)Y(2, n - 2)

(for some V,) 7

X(3, n - 1) V2(2,n - 1)Y(3, n - 1)a(2; n - 1)

=

v42, n - 1)a(2; n - 1)

say, where Y(2, n - 2)X(2, n - 2) = A.-2(2, n - 2).

On reducible braids

281

We now obtain the desired A-2'M1.W e have

-

A-~'-'M A - ~ ~ A { A A } ~ - -~ 1) v(~,~ = A - ~ ~ { A A } ~ - ' A v ,n( ~ -,1)a(2; n

= A-2r{AA}r-'V3(1, n -/2)&(2;

- 1)

n - 1)

-A-*'A(AA}'-IV~(~,~I -2)P(n-1;2)

-A

{AA l r - l v3(n - 1,2)a (I, n - 2)

= A -"M1 say, and A -2'M1un-l= A ;!" V3(n - 2,l). We check that A-2'M1 has power -21. If not, then V3(n - 1,2)a(l, n - 2) must begin with A,-'(2, n - l), so that V3(n - 1,2) must begin with A,-'(2, n - l), and so therefore must v3(2, n - 1). This {AA}'-' V(2, n - 1) has powef - 21. contradicts the fact that A Hence A-"M1 has power - 21. To complete the proof of this case, we show that A -2'MIu.-lis conjugate to A-2'-'Ma,-l in E L l . We have

-2ra

A-2'-'Mun-l = A;f;-'Q(l, n - 2) =A

;?;'Fl(l,

n - 2)Q3(l, n - 2)a(n - a ; n - 2)Z2(n - 3 , l )

- A ;?i'Z2(2, n - 2)Fl(1, n - 2)Q3(1, n - 2)a(n - a ; n - 2) = A ; 2 _ ; - ' A , - 2 ( 2 , n - 2 ) a ( 1 ; n - b)Q3(1,n-2)a(n - a ; n l - 2 ) = A ;?;'An-2(2,

n - 2)V(1, n - 2)

= A,?~'A,-2(2, n - 2)V3(1, n - 2 ) a ( l ; n - 2)

-A

I'!;

-A

v3(1, n

-

2)

V3(n- 2, l),

as required. This concludes the treatment of case (d). In the consideration of the above cases, we have shown that if r = 0 then t = - 1, and that if I 3 1 then (b) of the Lemma holds. Thus we have established that the Lemma follows from Lemmas 25 and 26. Proof of Lemma 25. We have p = A -"A{AA}'-' V(2, n - 1) is in standard form, r 3 1 and p has summit power t > - 21. We thus have A{Aa}'-'V(2, n - 1)= FUZ for some F, U, Z with ZF = A. We write F = FlF2where FI3 Fl(l, n - 2) and Fz does not begin with any of W e also write Z = Z1Z2, where Z2 = Z2(2,n - 1) and Zl does ul, .. .,

J. McCool

282

not end with any of u2,.. . It is then easy t o see that V(2, n - 1)= F1(2, n - 1)V2(2,n - 1)Z2(2,n - 1) for some V2. We now obtain descriptions of F2 and ZI. We have A = Z , Z 2 F = Z1A.-1(2, n - 1)W for some W, so that A

= A,-,(2,

n - l)Wfl = An-42, n - l ) a ( l ; n - l),

sothat f , = a ( j ; n - l ) f o r s o m e j WlAn-IF2for some W,, so that A

= An-,F2W, =

with l G j C n . A l s o , A = Z F , F 2 =

A,-IP(n - 1; l),

and therefore F2 = P ( n - 1; k ) for some k with 1 c k c n. We note that not both F2 = 1 and ZI = 1 can hold, since otherwise we would have A = Z2(2, n - l)Fl(l, n - 2) = F , ( l , n - 2)Z2(n - 2, l), which would imply that does not occur in A. We consider separately the cases (a) F2# 1 and Z, # 1, (b) F2 = 1, ZI # 1, (c) F2# 1, Z, = 1.

Case ( a ) . F2 # 1 and ZI # 1. We have j A

1..

n - 1, k C n - 1 and

,. = P ( n - 1; k ) a ( j ;n - 1)f2PI. C

7 F2Z1Z2FI

From Lemma 6 it follows that k # j . We consider separately the subcases j > k, j < k.

Subcase (1). j > k. We note that A

= A,-2(2,

n - 2)&(2; n - 1)

= A,-2(2,

n - 2)P(n - 1; k ) P ( k - 1; l)a(2; j

'-

1)a(j;n - 1)

= A,-2(2, n

- 2 ) a ( l ; j - 2)P(n - 1; k ) a ( j ;nl- 1)P(k

=a(n -k

+ 1; n - 1)A,-2(2, n - 2 ) a ( l ; j

- 1; 1)

-2)P(n - 1; k ) a ( j ;n - l),

so that

12(2,n - l)Fl(l, n - 2) = a ( n - k

+ 1; n - 1)A,-2(2, n - 2)a(1; j - 2).

Using Lemma 13, we now see that

Z42, n - I ) = a ( n - k

+ 1; n - 1)Y,

Fl(l,n - 2) = Xa(1; j - 2), for some X(2, n - 2) and Y(2, n - 2) with YX We now observe that

V(2, n - 1) = X(3, n - 1)a(2; j - 1)V2(2,n - l ) a ( n - k

= A,-2(2,

n - 2).

+ 1; n - 1)Y[2, nl--2)

283


is a description of V(2, n - 1) of the required form, with V42, n a(2; j - 1)V2(2,n - l ) a ( n - k + 1; n - 1). Now if r 3 2 we have A -"A{AA }'-I V(2, n

-

=

n

-

-

1)V,(2,n

-

1)Y(2, n - 2)

2)X(2, n - 2)A{AA}r-' V'(2, n - 1)

A-2'A,-2(2, n - 2)AV1(2,n

-

l){AA}r-'

- A -"A

{AA }'-'AAn-2(2, n - 2)A V'(2, n - 1)

=A

{AA } r - 2 ~ cv,(2, + I n - 1)

=A

1) =

1)

= A-2rA{AA}r-'X(3,n

- A -"Y(2,

-

-"+'A{AA}r-2alVI(2, n - 1).

Thus to establish the Lemma in subcase (l), when r 3 2 , we need only check that this last word has power - 2r + 1. If this is not so, then, since A{Aa}'-'a, ends only in ulr V1(2,n - 1) must begin with A.-42, n - 1). This implies that V(2, n - 1) contains An-'(2, n - l), which contradicts the fact that p has power - 2 r . We now suppose that r = 1. We then have V(2, n - 1) = FUZ, so that AV2(2, n - 1) = a ( n - 1; k)UP(n - j ; l), and therefore P(k - 1; 1)V2(2,n - 1 ) = U @ ( n- j ; l), where 1 5 k < n - 2 - 2 5 j 5 n - 1 and k < j . Now if k = 1 we obtain a contradiction, since u 1would then occur in U p ( n - j ; 1) but not in P ( k - 1 ; 1)V2(2,n - 2). Hence k 3 2 , and it now follows from Corollary 20 that A-'U has summit power - 1. Since 1 A-'U, this concludes the proof of the Lemma for Subcase (1).

A

-

Subcase (ZZ). k > j . We have A

= AA,-r = a(1; j - 1)a(j;

n

- 1)P(n - 2; k - 1)P(k -2;

= P ( n - 1; k ) a ( l ; j = a ( 1 ; j - l)P(n -

l)a(j; n - 1)P(k

- 2;

l)A.-2(2, n - 2)

1)6.-2(2, n - 2)

1; k)a(j; n - 1)P(k - 2; 1)An-2(2,n - 2)

= P ( n - 1; k ) c y ( j ; n -

1)p(k -2; 1)~¶,-~(2, n - 2)P(n - 1; n - j

+ l),

so that

i2F1= P ( k - 2; 1)A.-2(2, n - 2)P(n - 1; n - j + 1). We then obtain F , ( l , n - 1 ) = Xa(1; j - l), 142, n - 1) = a ( n - k + 2; n - l ) Y for some X,Y with YX = An-2(2,n - 2). If r

32

J. McCool

284

the result then follows as in Subcase (1). If r = 1 then we have P ( k - 1; 1)V2(2,n - 1)= U p ( n - j ; l), as in Subcase (I), where 1 j n - 2 and 2 k n - 1; an application of Corollary 20 now shows that r = - 1.

Case ( b ) . Fz = 1, ZI# 1. We have A

=

ZIZzFl = P ( n - j ; 1)Z2Fl= P ( n - j ; 1)An-,(2,n - l ) a ( l ; j - l),

so that 142, n - l)FI(l, n - 2)= An-,(2, n - l ) a ( l ; j - 1) = ( ~ ( 2n; -

1)An-2(2,n - 2)a(l; j - l),

and if r 2 2 this gives the desired result, as in t h e previous case. If r = 1 then we have AV2(2,n - 1) = U p ( n - j ; l), so that Vz(l,n -2)P(n - l ; n - j

-

+ 1 ) = U.

Hence A - ' U A - ' a ( l ; j - l)Vz(l, n - 2), and j d n - 1 since I , # 1, so that, by Lemma 18, A - ' U has summit power - 1, as required.

Case ( c ) . F2# 1, ZI= 1. We have

A

= Z2FIF2 7 I2F,P(n -2;

= &(n

-k

k)

+ 1; n - l)A.-IP(n

- 2; k ) ,

so that Z2(2,n - l)Fl(l, n - 2) = a ( n - k =a(n-

k

+ 1; n - 1)An-,

+ 1; n - 1)A.-2(2, n - 2)a(l; n - 2).

As before, this establishes the result if r 3 2. If r = 1 we have AV2(2, n - 1) = P ( n - 1; k)U, so that U = P ( k - 1; 1)V2(2,n - 1) and A-'U A-'V2(2, n - l ) a ( n - k + 1; n - 1).

-

Now k d n - 1 since F2# 1, and the result follows from Lemma 18. This concludes the proof of Lemma 25.

Proof of Lemma 26. We have 4 = A - 2 ' + ' A { A a } r -V42, 2 ~ ,n - 1) is in standard form, r B 2 and 4 has summit power t > - 2r + 1 . We therefore have

A { A A } r - 2VI(2, ~ l n - 1)

FWf,

for some F, .W, Z with A = IF. We write F = FIF2,where F , = F,(2, n - 2) and F2 does not begin

On reducible braids

285

with any of u l ,u2,. . . , un-z.Now alVl(2, n - 1) must begin with F1(3, n - l), V,(2, n - 1) = F1(3,n - 1)V2(2,n - 1) say, so that A{AA}'-2alV2(2, n - 1) = F2Wf.Now if F2 begins with u,,then alV2(2,n - 1) must begin with u2.However, this would imply that V2(2,n - 1) begins with a2a1, which is impossible. Hence either F2 = 1, or F2 begins only with am-1. We also write Z = Z1Z2, where Z2 = Z2(1, n - 2) and ZI does not end u 2 , .. . , un-z. We then have V2(2, n - 1) = V42, n - l)i2, with any of ul, for some V,. We now find descriptions of F2 and ZI.W e have A = ZFlF2= UA,-IF2for some U, so that A.-,A = An-IF2fi, and F2 = P ( n - 1;j ) for some j with 1 s j s n. Also, A = ZIZ2F= ZIA,-IH for some H, so that H i , = A and f, = P ( k ; 1) for some k with 0 k n - 1. We note that not both ZI= 1 and Fz = 1 can hold, since otherwise we would have A = Z2(1, n - 2)Fl(2, n - 2), which is impossible. We consider separately the cases (a) I , # 1 and F2 # 1, (b) F2# 1, ZI= 1, (c) Fz= 1, ZI # 1.

Case ( a ) . ZI# 1, F2# 1. We have 1 S j , k

A

= Z2FlFzf, = Z2Flp(n-

S

n - 1. Now

1;j ) p ( k ; l),

so that j # k . If j < k then A has a subword P(n

1; j ) P ( k ;j + 1)Po'; 1)- P ( k - l ; j ) P ( n - 1;j)Po'; I),

which is impossible. Hence k < j. Now if j > k A

+ 1 then

j > 1 and

+ 1)P(k; l)P(n - 1; j ) P G - 1; 1) = a ( n - j + 1; n - l)A...2P(n - 2; k + l)P(n - l ; j ) P ( k ; 1) = a ( n - j + 1; n - l)A.-2P(n - 2; k + 1)F2fl, = A,-$(n

- 2; k

so that Z2(l,n - 2)Fl(2, n - 2) = a ( n - j

+ 1; n - 1)An-$(n

- 2; k

+ 1).

However, occurs on the right side above, but not the left, which is impossible. Hence we must have j = k + 1. Since j = k + 1 we have A = Z2F,A,and therefore

Iz(1, n - 2)Fl(2, ~t- 2) = An-I

=

P ( n - 2; 1)Am-2(2,n

- 2).

Applying Lemma 13, we see that Z2(l,n - 2) = P ( n - 2; 1)Z, for some Z = Z(2, n - 2), so that 1, = ( ~ ( 2n; - l ) Z ( n - 2,2) = Z ( n - l,3)a(2; n - 1). Since Vl(2, n - 1) ends with f2, this proves the required result.

286

3. McCool

Case ( b ) . 1 1 = 1 ,Fz#l. We have k = 0 , l s j s n - 1 and

Iz(l,n - 2)Fl(2, n - 2 ) = UA.-,

= a ( n -j

+ 1; n - l)An-l.

Now a,,-, does not occur in IzFl, and so we must have j IzFl = An-l, which yields the result as in Case (a).

Case ( c ) . II# 1, F2= 1. We have 1 < k

Iz(1, n - 2)F1(2, n - 2) = A,-IH

=

1 and hence

n - 1, j = n and A,-iP(n - 1; k

+ l),

which is impossible unless k = n - 1. Thus LF, = Am-l,and the result follows. This concludes the proof of Lemma 26.

Proof of Lemma 24. Suppose that A-"M is in standard form, r a 1, and A -"M has summit power t > - 2r. We then have M = FRI for some F, R, I with A = IF. We note that A-"FRI A-"+IR, and that A-'"'R has power - 2r + 1. We show, firstly, that A-2'Mu,,-I has power -2r. If this is not the case then, by Lemma 16, it has power - 2r + 1, and by Lemma 22 we have FRIU,,-~= A{Aa}'-'AQ, where 0 = Q(1, n - 2) and Q is prime to A,,-,. We then have FRIu.-, = A{Aa}'-'Q(2, n - 1)A, so that FRI must have power contains A. This is a contradiction, and so A - 2r. We now have, by Lemma 22, that FRIun-, = {aA}'Q(l, n - 2), where Q is prime to Hence

-

FRI = a{Aa}'-lQ(2, n - l ) a ( l ; n - 2). We write F = Fl(l, n - 2)F2, where Fz does not begin with any of u,,. . . , v,,-~, and I = 1112(1,n - 3), where II does not end with any of ul,. . . ,c.-~.We then have Q(2, n - 1)= F1(2,n - 1)Q1(2,n - 1)Iz(2,n - 2) for some Q1, so that

(5.10)

FzRIl= A{AA}'-IQl(2, n - l ) a ( l ; ~t- 2).

We now obtain a description of Fz.We have A = IFIFz=UAn-IF2 for some V,so that A = A.-1F20= An-lfl(n - 1; 1). Hence F2= P ( n - 1; I) for some 1 with 1 S 1 d n. We now consider separately the four possibilities for I,: (a) II ends in both and (b) ZI ends only in (c) ZI = 1, (d) I I ends only in

Case ( a ) . I , ends in both and u n - 2 . We have A = Z l I 2 F = IIA.-zH, for some H, so that A = f i I I A n - z = a(1;n - l)a(1;n - 2)A,-*, and

287


~ n - 2)a(l; t~ - l)un-2. NOW therefore f i Z l a(1; n - l ) a ( l ; n - 2 ) a(2; z1= WU,-2un-lu,-2 say, so that fiW = a(2; n - 2)a(l; n - 3). Applying Lemma 15 we obtain fi = a(2;k ) a ( l ; j ) and W

a(k

+ 1; n - 2)a(j + 1; t~ - 3),

forsome k, j w i t h O s j s n - 3 , l ~ k ~ n - 2 a n d k - j * 1 , s o t h a t Z I = a ( k + 1; n - 2)a(j+ 1; n - 1)3,-2. We now obtain a description of Zz(l,n - 3)Fl(l, n - 2). We have A

7ZIZ2FiF2

=a(k

+ 1; n - 2)a(j + 1; n - I)U,-~A.-,P(~- 2; n - k ) P ( n - 1; n - j),

so that Z2(1,n - 3)Fl(l, n - 2)P(n - 1; 1 ) = A.-2P(n - 2; n - k ) P ( n - 1; n - j). Using Lemma 13 we see tht 1 = n - j, so that Z2(1,n - 3)Fl(l,n - 2) = A,-$(n

- 2; n

- k),

and, using Lemma 13 again, we have

Fi(1, n - 2) 7 Z(1, II - 3)P(n - 2; n - k ) , for some Z. We now have, from (5.10), that

P ( n - 1; n

- j)Ra(k

+ 1; n - 2)a(j + 1; n - 1)

= A{AA}'-IQ1(2, n

- l ) a ( l ; n - 3).

Since the left side of this ends in u , - ~we , have Ql(2, n - 1) = Q2(2,n - l ) ~ "say, - ~and P ( n - 1; n - j ) R a ( k = A{AA}'-IQ2(2,

+ 1; n - 2)aG + 1; n - 2) n - l ) a ( l ; n - 3).

Since the left side of this ends with (242, n - l)a(2; n - 2) say, and

we have Q2(2,n - 1)=

P ( n - 1; n - j)Ra(k + 1; n - 2) ={AA}~-*Q~(I, n - 2 ) ~ ( -n 1;2)a(2; n - 2)a(1; j ) = {AA}r-1Q3(1, n - 2)P(n - 1;2)a(2;k ) a ( l ; j ) a ( k

+ 1; n - 2),

so that

P ( n - 1; n - j ) R = {AA.)-'Q3(1,n - 2 ) P ( n - 1;2)a(2; k ) a ( l ; j ) . This equation may be solved directly for R if r a2, while if r = 1 a solution may be obtained using Lemma 19. However, in order to avoid

J. McCool

288

separate consideration of these two cases, we proceed somewhat differently. We have A -"M = A -"FRI = A -"FIP(n - 1; n - j)RZIZ2 = A-"FI{~A}'-'Q3(1, n

- 2)P(n - 1;2)a(2; k ) a ( l ; j ) a ( k + 1; n - 2)

(Yo' + 1; n - l)an-2Z2

- A-"Z2Fl{~A}'-'Q3(l,n - 2)P(n - 1;2)0(2; n - 2)a(l; n - l ) ~ . - ~

- A-"a(l;

t~ -

l ) ~ ( t~ l ;- 2)ZzFl{AA}'-'Q3(1, n - 2)P(n - 1;2),

and substituting the expression obtained for Z2F1, we have

A-"M

- A-"a(l; =A

n - l ) a ( l ; n - 2)A.-z/3(n - 2; n - k){aA}'-' Qd1, n - 2)P(n - 1; 2)

- ~ ~ + ' { A A }-~2;- n~ -~ k( )~a 3 ( 1 ,n - 2 ) ~ ( n 1;2)

- A - " + ' { A a } ' - ' a ( 2 ; k ) Q 3 ( n - 1 , 2 ) a ( l ; n -2) A -Z'+IMI say, and A -"+lMlan-l= A i?,+'a(l;k - 1)Q3(n - 2,l). We now show that A-"MU~-~and A-2'+'MIan-Iare conjugate in B,-]. We have

A-"Man-I

= Ai?'lQ(l,

n - 2)

= A i?IFl(l, t~ - 2)Q3(1, n

- 2 ) a ( l ; n - 2)Z2(1, n - 3)

- A if;a(l;n - 2)Z2(1,n - 3)Fl(l, n - 2)Q3(l, n - 2) = A i?',+IP(n- 2; n - k)Q3(1,n - 2)

-

A i?';'a(l; k - l)Q,(n - 2, l), as required. Next we check that A-"+'MI has power - 2r + 1. If this is not the case, then MI contains A, so that a(2; k ) Q 3 ( n - 1,2)(1; n - 2) begins with An-'(2, n - l), and therefore a(2; k ) Q 3 ( n - 1,2) also begins with An-1(2,n - 1). However, this would imply that Ql(l, n - 2) contains An-,, since

01(l,n-2)=F~(l,n-2)03(1,n-2)a(l;n-2)Zz(l,n-3) = Z(1, n - 3 ) P ( n - 2; n - k)Q3(I, n

- 2)a(1; n - 2)I2(1, n - 3).

This contradiction shows that A-"+'MI has power - 2r + 1. It remains to show, in Case (a), that r = - 1 if r = 1 . Now when r = 1 we have

On reducible braids

289

A-'M 1 = A - ' a (2; k ) Q 3 ( n - 1 , 2 ) ~ ~ (n1-2) ; -A-'p(n-

1;2)a(2;k)Q3(n- 1,2),

so that A-'MI has summit power - 1, as required. Case (b). ZI ends only in un-'.We then have A = Z l Z 2 F = ZIAn-lH for some H, so that f i Z l = A, and therefore ZI = a ( s ; n - 1) for some s with l ~ s ~ n - 1 . Now A = ZlZ2FlF2 = ZIA,-lP(n - 1; n - s + l), so that

Z 2 ( 1 , n - 3 ) F ~ ( l , n - 2 ) ~ ( n - 1 ; 1 ) = A n - ~ P ( n - 1 ; n - ~1). + From this we conclude that 1 = n - s + 1, so that

141,n - 3)F1(1,n - 2)

A,-'

An-2/3(n- 2; l),

and therefore FI(l,n - 2) = J(1, n - 3 ) P ( n - 2; l), for some J. We now see, from (5.10), that

P ( n - 1; n

-s

+ l)Ra(s; n - 1 ) = a{Aa}'-'Q1(2, n - l ) a ( l ; n - 2).

Since the left side of the above ends in Q2(2,n - l)a(2; n - 1) say, and then

we have 0 4 2 , n - 1) =

P ( n - 1; n - s + l)R ={(aA}'-'Q,(l,n-2)P(n-l;2)a(2;n - l ) a ( l ; ~ ' - l ) . We now observe that

A-"M = A-"FRZ

= A-"Fip(n

- 1; n - s + 1)RZIZZ

= A-2rFl{~A}'-'Q2(1, n - 2)P(n

,

- 1;2)a(2;n - l ) a ( l ; s - l ) a ( s ; n

- A-2'a(1; n - 1)Z2FI{~A)'-'Q2(1,n - 2)P(n - 1;2)a(2; n - 1) = A-2r+'{AA}r-'Q2(1,n - 2)P(n - 1;2)a(2; n

- A-2"'{Aa}'-'a(2;

- 1)

n - 1)Q2(n - 1,2)0(1;n - 2)

A -2r+1M1

say, and A-2'+'Mlun-l= A ;!',+'a(l;n - 2)Q2(n - 2,l). Now A 2'lQ(l,n - 2) = A 2'lFIQ2(l, n - 2)a(1; n - 2)Z2

- A,?'T.;''p(n - 2; l)Q2(1, n - 2) - A ;!',+'a(l; n - 2)Q2(n - 2, l),

so that A-2'Mun-I is conjugate in to A-2'+'MIun-I. Now it is easy to see that A-2'+'M1 has power - 2r + 1, so to

-

1)Z2

J. McCool

290

complete the proof of the Lemma in Case (b) we only have to examine what happens when r = 1. In that situation, we have A-'M1 A-'P(n - 1 ; 2 ) a ( 2 ;n - l)Q2(n - 1 , 2 ) , so that t = - 1, as required.

-

Case (c). Z,= 1. We then have

Z2(l, n so that

- 3)Fl(1, n - 2)/3(n

- 1; I ) = A

=a(n- 1

+ 1; n - l)A.-lP(n

- 1; l ) ,

Z2(1, n - 3)Fl(1,n - 2 ) = a ( n - 1 + 1; n - 1)An-2,

and therefore 1 = 1. We may now rewrite (5.10) as

R

= {AA}'-'Q1(2,

n - 1)a(1;II - 2).

Hence A - Z ' + I R ~ n=- lA ;?,+'QI(l,n - 2 ) . Now

A-2'M~n-I = A,?'Q(l, n - 2 ) = A ;!',F1QI(l, n

- 2)ZZ- A ,!',+'Q,(l,n - 2).

Noting that if r = 1 then A - ' R has summit power result is established for Case (c).

- 1, we see that

the

Case (d). ZI ends only in We then have A = Z1Z2F= ZlA.-2~n-,H, for some H, so that A = fiZlAn-2cn-l = a ( 2 ; n - 1)a(1;n - 2)A.-zun-l, and therefore f i Z l = a ( 2 ; n - l ) a ( l ; n - 2). From Lemma 15 it follows that I, = a ( a + 1; n - l ) a ( b + 1; n - 2 ) and fi = a ( 2 ; a ) a ( l ;b), for some a , b with l S a S n - l , O S b S n - 3 a n d a - b a l . = Z,ZzFlF2, so that Hence A = Z l A n - 2 ~ n - l H

Z2(l, n - 3)Fl(1,n - 2 ) P ( n - 1; I )

= A . - , p ( n - 1 ; n - a ) P ( n - 1; n - 6 ) = An-$(n - 2; n -

since a - b

2

b - 1)P(n - 1; n - a ) ,

1. It then follows that 1 = n - a, so that

Z2(1, n - 3)FI(1,n - 2 ) = An-2P(n- 2 ; n - b - l),

and therefore FI(l,n - 2 ) = K ( 1 , n - 3 ) P ( n - 2 ; n - b - 1) and Z2(l, n - 3 ) K ( 1 ,n - 3) = for some K. We may now write (5.10) as

p ( n - 1; n - a ) R a ( a + 1; n - l ) a ( b + 1; n - 2 ) F A{AA}'-'QI(2,

n - l ) a ( l ; n - 2),

On reducible braids

291

so that

/3(n-l;n-~)Ra(~+l;n-l)~A{AA}'-'Q,(2,n-l)~(l;b). Since a - b

1, it then follows that

3

Ql(2, n - 1) = Q2(2, n - l ) a ( a

+ 1; n - 1)

say, and R

/3(n - u - 1; l){AA}'-'Q2(2, n - l ) a ( l ; b).

We then have A-"+'R

- A-"+'P(n

+ 1; n - 1) = A-2r+'{AA}'-'/3(n - 1; n - b)Q2(2, n - l ) a ( a + 1; n - 1)

say. We note that

- 1; n - b){AA}'-'Q2(2, n - l ) a ( a

A -''+'{AA}'-' V(2, II - 1)

Q(1, n - 2) = K(1,

n - 3 ) P ( n - 2; n

- b - l)Qz(l, n - 2 ) a ( a ; n - 2)12(1,n - 3)

and that Q(1, n - 2) is prime to Am-'. From this it follows easily that A -'"'{AA}'-'V(2, n - 1) has power - 2r + 1. Now if r = 1 then A-'V(Z, n - 1) has summit power - 1. Hence we may assume that r 2 2, and t > - 2r + 2. Applying Lemmas 27 and 28, we obtain V(2, n - 1) = X(2, n - 2) V1(2,n - 1)Y(2, n - 2) = X(2,

n - 2)V2(2, n - l ) a( 2 ; n - 1)Y(2, n - 2)

= X(2, n - 2) V42, n - 1)Y(3, n - 1)a (2; n - 1)

v3(2, n - 1)a (2; ? -I1) say, and ?X = A,-42, n - 2). W e now obtain the desired A -"+'MI. We have A - z r + 1 { ~ A } r - l v (~t 2, 1) = A - 2 r + 1 { ~ A } r - 1 ~n3 (-2i)a(2; , n - 1) =

A - 2 r + 1 ~ { A ~ } r - 2 ~-32)Aa(2; (i, n - 1)

- A-2r+'{AA}r-'V3(l,n - 2)P(n - 1;2) - A - ~ ~ + ' { A A } ~ -- ' i~, 2, () ~a ( i ; n - 2) A -"+'MI say, and A-Z'+'Mla.-I= A 2',+'V3(n - 2 , l ) .

I. McGwl

292

Now

A-"Mun-l

= Ai?'lQ(l, ~t- 2)

= A ;FIK(l, n

- 3)V(1, n - 2)12(1,n - 3)

- A ;!'lZz(l, n - 3)K(1, n - 3)V(1, n - 2) = A ,!'lAn-zV3(1,

-A

I;?'

n

- 2)(w(l;

- 2)

~t

V3(l, n - 2),

so that A-z'Mun-l is conjugate to A-Z'+lM1u,,-l in &-,. It remains to establish that A-"+'M1 has power - 2r + 1. If this is not the case, then V3(n - 1 ,2 )a(l; n - 2) must begin with A,,-,(2, n - l), and so therefore must V3(n - 1,2). However, this would imply that A-2'Mu,,-I has power greater than - 2r in Eln-,. Hence A -"+'MI has power - 2r + 1. The concludes the proof of Case (d) and thereby establishes the Lemma. Proof of Lemma 27. We have p = A-zr+1{AA}'-1V(2,n - 1) is in standard form, r 3 2 and p has summit power t > - 2r + 1. We then have {AA}'-'V(2, n - 1 ) = FWf for some F, W, I with A = IF. We write F = F1(2,n - l)Fz, Z = ZIZz(l,n - 2), where Fz does not begin with It is any of u 2 , .. . , C T - ~and ZI does not end with any of u l , .. ., then easy to see that V(2, n - 1) = F1(2,n - 1)V2(2,n - 1)12(n- 1,2), for some Vz. We obtain descriptions of II and F2. We have A = 1112F=IIAn-lU for some U,so that A,-IUfl = A, and therefore f, = P ( k ; l), for some k with 0 S k S n - 1. Also, A = ZFIFz= HAn-1(2,n - l)Fz for some H, so that F 2 H = A and F 2 = a ( 1 ; j ) , forsome j with O s j s n - 1 . We note that not both Z, = 1 and Fz = 1 can hold, since otherwise A = F,(2, n - l)Zz(n - 1,2), which is impossible. We consider separately the cases (a) F2 # 1 and ZI # 1, (b) ZI # 1, F2 = 1, (c) ZI = 1, F2# 1.

Case ( a ) . We have A = 12FIFzfl = ZzFIa(l;j ) p ( k ; l), so that j # k . We consider separately the subcases (I) j > k , (11) j < k. Subcase (I). j > k . Then A

= P(n -j

- 1; 1)A,-2(2, n - 2)P(n - 1; k

+ 2 ) a ( l; j ) p ( k ;l),

so that Zz(l, n - 2)Fl(2, n - 1) = P ( n - j - 1; 1)An-z(2,n - 2)P(n - 1; k

+ 2).

293

On reducible braids

It now follows that there exist X(2, n - 2), Y(2, n - 2) with Z2(l,n-2)=p(n-j-1;1)Y(n-2,2), F1(2,n-l)= X(2, n - 2)P(n - 1; k + 2) and Y(n - 2,2)X(2, n - 2) = A,-42, n - 2), and therefore

V(2, n - 1) = X(2, n - 2)V1(2,n - 1)Y(2, n - 2), where V1(2,n - 1) = P ( n - 1; k + 2)V2(2,n - 1)a(j + 1; n - 1). To establish the Lemma for the subcase under consideration, we Vl(l, n - 2), and that this need to show that p A -2"2{~A}r-2an-, latter word has power - 21 + 2. We have

-

p = A -2'"{AA}'-1X(2,

n - 2)V1(2,n - 1)Y(2, n - 2)

= A-"+'XA{AA}'-'VI(l,

- A-"+'AY(n =

n - 2)AY(2, n - 2)

- 2,2)XA{AA}'-ZVl(1,n

- 2)

A -2'c1Aan-1 V1(l,n - 2){dA}'-'

- A -2r+z{AA}'-2an-IVl(l, n - 2).

Now if this last word has power greater then - 2r + 2, then VI(l, n - 2) must contain An-l, which would imply that p does not have power - 2r + 1. Hence A -ZrfZ{~A}'-2an-I V1(l,n - 2) has power - 2r + 2, as required.

Subcase (IZ). j < k. Then A

=P(n

- j -2; 1)An-2(2,n -2)P(n - 1; k

+ l ) a ( l ; j ) P ( k ; l),

so that

IzFl= P ( n - j - 2; 1)An-42, n - 2)P(n - 1; k

+ l),

and the result follows as in the previous subcase. Case (b). II# 1, F2 = 1. We have IzFi = An-lU = An-iP(n - 1 ; k

+ 1)

=a(l;n-2)An-2(2,n-2)P(n-l;k and the result follows as above.

+l),

Case (c). ZI= 1, F z # 1. Then IzF1 = HAn-l(2, n - 1) = P ( n - j - 1; 1)An-42, n

and the result follows.

-'2)P(n - 1;2),

1. McCool

294

This concludes the proof of Lemma 27. Finally, we have Proof of Lemma 28. We have + = A --2"2{AA}r-2a.-l Vl(l, n - 2) is in standard form, r 3 2 and has summit power t > - 2r + 2. W e Vl(l, n - 2) = FWI, for some F, W, I with therefore have {AA}'-2an-l A = IF. We write F = F l ( l , n - 3)F2, where F2 does not begin with any of uI,. . . , an-3.We then have Vl(l, n - 2) = F I ( l , n - 3)v2(l, n - 2) for V2(1,n - 2) = F2WI. Now if F2 begins with some V2, so that {AA}r-2an-l then an-l V2(1,n - 2) begins with so that V2 begins with ~ , - ~ a . -which ,, is impossible. Hence either F z = 1 or Fz begins only with We also write I = 1,12(1,n - 2), where 1, does not end with any of aI, . . . ,an-2. We then have V2(1,n - 2) = V3(l, n - 2)12(1,n - 2), for some V,. We now describe I, and F2. W e have A = II12F= l l A n - , U, for some U , so that Uf,= A and f, = P ( j , 1) for some j with O S j S n - 1. Also, A = l F l F 2 = HAn-,F2,for some H, so that F2H = and F 2 = P ( n - 1; k ) for some k with 1 S k S n. We note that not both I,= 1 and Fz = 1 can hold. We consider separately the cases (a) I,# 1, F2 # 1, (b) II= 1, F2# 1, (c) II# 1, F2 = 1.

+

a

Case (a). 1 1 # 1 , F2#1. We have 1 S j S n - 1 , 1 S k S n - 1 and A

= I2FIFzfl = I z F , P ( n -

1; k)p(j; l),

so that k # j . If j > k then we obtain a contradiction as in Lemma 26. If k > j + l then

A

= A,-2P(n

- 2; j

+ l)P(n - 1; k ) P ( j ; 1)P(k - 1; l),

so that

+ 1; n - l)A.-2/3(n - 2; j + 1). k = 1, but since k > j + 1 we cannot

I2(1, n - 2)Fl(l, n - 3)= a ( n - k This, however, is impossible unless have k = 1. Hence k = j + 1. We now have A = I2Fla,so that

12(1,n - 2)F1(1,n - 3) = a ( 1 ; n - 2)A.-2, and therefore Z2(1,n - 2) = a ( 1 ; n - 2)X(1, n - 3) = X(2, n - 2 ) a ( l ; n - 2), for some X. This shows that VI(l, n - 2) ends with a ( 1 ; n - 2).

On reducible braids

295

To complete the proof of this case, we note that when r = 2 we have u , - , V 3 ( l , n - 2 ) = F 2 W Z 1 = ~ ( n -l ; k ) W a ( n - j ; n - l ) , so that V3(l, n - 2) = P ( n - 2; k)Wa(n - j; n - l), which is impossible. Hence Case (a) cannot occur when r = 2.

Case ( b ) . ZI= 1, F2# 1. We have j

=0

and 1 s k < n - 1. Now

Z2(l, n - 2)Fi(l, n - 3) = HAn-]7 c*(n - k

+ 1; n - l)An-i,

so that we must have k = 1, and Z2Fl= A,-I. If r z 3 then this proves the result, as in the previous case. If r = 2 we have a,,- V3(1,n - 2) = F2 W =

a W,

so that V3(1,n - 2) = P ( n - 2; 1) W, and therefore W = W(1, n - 2). Now A -2a,-l Vl(l, n - 2) A - I W, and A - I W has summit power - 1, as required.

-

Case (c). ZI # 1, F2 = 1. We have 1 s j < n - 1 and k

= 0.

Now

12(1,n - 2)Fl(1, n - 3) = An-IU= A , - ] P ( n - 1; j + l),

so that we must have j = n - 1 and Z2Fl= A,-]. If r 2 3 this establishes the result. If r = 2 then we have an-I V2(1,n - 2) = WA, so that u1V2(n - 1,2) = Wa.However, from Corollary 20 we see that this is impossible. This concludes the proof of the Lemma. References [ l ] J.S. Birman, Braids, links and mapping class groups (Princeton University Press, 1975). [2] W.L. Chow, On the algebraic braid group, Ann. of Math. 49 (1948) 654-658. [3] F.A. Garside, The braid group and other groups, Quart. J. Math. Oxford 20 (1969) 235-254.

S.I. A d i ~W.W. , Boone, G. Higman, eds., Word Problems I1 @ North-Holland Publishing Company (1980) 297-298.

ON SOME GROUP PRESENTATIONS* B.H. NEUMANN Instintte of Advanced Studies, me Australian National University, Canberra

Let p, q, r, s, r, u be integers, and denote by G(p,q, r, s, r, u ) , or G for short, the group with three generators a, b, c and defining relations

a P b= bas, bqc = cb‘, c’a = ac”. It is known that G(1,1,1, - 1, - 1, - 1) is infinite [2, 021 and that is trivial (Higman [I], see also [2, 923)). If

G(1,1,1,2,2,2)

(*)

2spslsJ,

2sqsJr1,

2 ~ r ~ l u J ,

then it can be shown that G is infinite. The method consists of constructing a “normal form” for the elements of G. Though the idea is simple, the technical details are complicated. The groups G(p, q, r, p + 1, q -+ 1, r + 1) have no proper subgroups of finite index, and thus no non-trivial finite epimorphic images. The proof follows one of Higman [ l ] (see also [2], 023). On the other hand, if one of 1 s - p 1, I t - q 1, 1 u - r I is different from 1, the group is easily seen to have non-trivial finite factor groups. All the presentations considered have solvable word problem; this is an immediate consequence of the fact that there is a normal form, which is readily computable, for the elements of the groups presented. The question with which the investigation started is, whether G(2,2,2,3,3,3) is finite (and then necessarily trivial) or infinite; and the author is indebted to Dr A.M. Biunner for reminding him of it. This group is infinite. By contrast it follows from recent, more general results of Post 13) that G(2,2,1,3,3,3) is finite, and G(2,2,1,3,3,2) is * Most of this work was carried out at the Conference, but not formally presented there; it was only shown privately to a few interested participants. I t was reported to the Conference on finire groups and permutarion groups at the Mathematisches Forschungsinstitut Oberwolfach, Germany, in August 1976. 297

298

B.H. Neumann

trivial. The author is grateful to Dr Michael J. Post for permitting him to quote his unpublished results. A more detailed account is to be published in the Canadian Journal of Mathematics. References (11 Graham Higman, A finitely generated infinite simple group, J. London Math. SOC.26 '

(1951) 61-64.

[2] B.H. Neumann, An essay on free products of groups with amalgamations, Phil. Trans. Roy. SOC.London, Ser. A 246 (1954), 503-554. [3] Michael J . Post, A note on finite three-generator groups with zero deficiency, to be published.


RADCLIFFE

THE CONCEPT OF “LARGENESS” IN GROUP THEORY Stephen J. PRIDE* Open University, Milton Keynes

91. Introduction 91.1. Summary of results

There are many group-theoretic properties which are traditionally thought of as measuring the “largeness” of a group - for example, having cardinality greater than or equal to K for some infinite cardinal K , being SQ-universal, having a free subgroup of rank 2, having a properly ascending chain of normal subgroups. There are also many groups which are traditionally thought of as being “large” - nonabelian free groups and Fuchsian groups for example. At the other end of the scale there are groups, which although infinite, are still considered to be rather “small” - infinite groups in which every nontrivial normal subgroup has finite index for instance. The aim of this paper is to develop a theory which gives precision to these vague notions of “largeness”. This theory provides a framework in which to discuss several well-known, but until now, unrelated ideas. It also gives a different viewpoint, which leads to new results. By examining several properties 9 which are traditionally thought of as indicating the “largeness” of a group, one finds that they all satisfy three basic conditions, and moreover, it is these conditions which account for the fact that the property is a measure of “largeness”. (LP 1) If G has 9 and K maps onto G then K has 9. (LP 2) If H s G with H of finite index in G then G has 9 if and only if H has 9. (LP 3) The trivial group 1 does not have 9. A group-theoretic property satisfying (LP 1)-(LP 3) will be called a large property. In 02.1, partly for illustration, and partly for use in later discussion, several well-known large properties are listed. In 02.2 it is shown how to associate with any infinite group A a large property 2 ( A ) called the large property generated by A . This notion enables one to compare two infinite groups G, H. Thus G is said to be larger than H (written * Present address: University of Glasgow. 299

S.J. Pride

300

H < G ) if G has 6P(H); G and H are said to be equally large if G has 9 ( H ) and H has 9 ( G ) ;and G is said to be strictly larger than H if G is larger than H, but G and H are not equally large. If G is larger than H then any large property enjoyed by H is also enjoyed by G ; and if G and H are equally large then they have the same large properties. The relation of larger than is a quasi-order, and one would hope that groups traditionally thought of as being “large” would be “high up” in this quasi-order, and groups traditionally thought of as being “small” would be “low down”. These two extremes are examined in Sections 3 and 4 respectively. If G is a finitely generated group then it is easily shown that F2, the free group of rank 2, is larger than G. Thus G is “as large as it can possibly be” if it is equally as large as F2. In §3.2 it is shown that many finitely generated groups which are traditionally thought of as being “large” are actually equally as large as F2. In particular, this is verified for: (i) Fuchsian groups (Theorem 3.2); (ii) one-relator groups with torsion on at least two generators (Theorem 3.3); (iii) certain arithmetically defined linear groups PSL(2,

1, SL(2, 1, GL(2, 1.

Some remarks are also made concerning small cancellation groups. The verification that one-relator groups with torsion on at least two generators have 6P(F2)provides a new and straightforward proof that these groups are SQ-universal. Also in 83.2, an example is given of a finitely generated one-relator group B which, although having several well-known large properties, nevertheless is not equally as large as F2. To be precise, it is shown that if B = ( a ,t ; t-’a2t = a3) then no subgroup of finite index in B maps onto a free group of rank greater than 1 (Theorem 3.4). In 03.3 the question of which free products have 9 ( F 2 )is discussed. Let G = H * K with H , K # l . If ( H I = I K ( =then 2 G iscyclic-byfinite and is “small”. On the other hand, if max{ 1 H I , IK I} > 2 then G has many well-known large properties. However, it is often quite a complicated affair to establish that G has such properties, and this leads one to think that perhaps G need not necessarily be “naturally large” in the sense of having 9 ( F 2 ) .This turns out to be the situation. It is shown in 83.3 that G has 8 ( F 2 ) if and only if either: ( i ) one of H, K has 9(F2);or ( i i ) both H and K have non-trivial finite homomorphic images fi, E,with max { I fi 1, I R I} > 2 (Theorem 3.7).

Largeness in group theory

301

Section 4 turns to “small” groups. Firstly, in 54.1, several concepts of “smallness” are discussed. Thus, an infinite group G is said to have height m if whenever one has groups G I , . . .,G, with G, = G and Gi+l strictly larger than G, for i = 1,. . . , r - 1, then r m. Also, G is said to satisfy Max- < (Min- < ) if there is no infinite sequence of groups G l , G Z ,... with G larger than G, and Gi+lstrictly larger than Gi (Gi strictly larger than Gi+l)for i = 1,2,. . . . Related to these two latter properties are chain conditions on certain subgroups of G (see Theorem 4.1). In $54.2, 4.3, 4.4 consideration is given to a class of infinite groups which are traditionally thought of as being ‘‘small’’ - namely, infinite groups in which every non-trivial normal subgroup has finite index (the so-called just-infinite groups). Let A denote the class of minimal groups with respect to the quasi-order relation of larger than. It is easy to show that any infinite group in which every non-trivial subnormal subgroup has finite index lies in A. However, just-infinite groups which have non-trivial subnormal subgroups of infinite index need not be minimal. Nevertheless, if G is a just-infinite group satisfying the maximal condition on subnormal subgroups then G has finite height (see Theorem 4.2). For just-infinite groups not satisfying the maximal condition on subnormal subgroups, one finds that such a group is either minimal or does not satisfy Min- < (see Theorem 4.3). An example is given in 54.4 of a just-infinite group lying in A but not satisfying the maximal concition on subnormal subgroups. In 04.5 questions concerning extensions are raised. In particular, are the properties “having finite height”, Max- < , Min- =S extension closed? The main result (Theorem 4.4) is that the answer is negative for the first two properties (it remains unresolved for the third). In fact, it is shown that there is a split extension of one minimal group by another which does not satisfy Max- < . In contrast to this result it is also shown that i f MI,.. . ,M. are minimal groups then MI x * . * X M. has height less than or equal to n (see Theorem 4.5). Although poly-minimal groups d o not in genela1 have finite height, it can be shown that for certain classes 8 of minimal groups, poly-9/ groups do have finite height. In particular, poly-cyclic groups, and groups with a composition series, have finite height. These results will be obtained in a ganeral setting in a sequel to the present paper.* The work presented in this paper raises many questions, some of which are stated in the text below. *

“Groups of finite height”, J. Austral. Math. Soc. (Series A) 28 (1979) 87-99.

302

S.J. Aide

61.2. Notes to the reader The reader who wishes to get the main results and ideas of the paper, but is not interested in details should read as follows: 02.1, 02.2, 03.2, 43.3 (up to the end of the statement of Theorem 3.7), 04.1, 04.2 (up to Problem 6), 04.5 (up to Problem 7). The casual reader should note that = does not signify isomorphism. 61.3. Acknowledgements This paper has benefited from conversations with D. Goldrei, A.J. Wilkie, and most especially with I.M.S. Dey. Visits to New York University and the University of Illinois at Urbana in Spring 1976 were also very beneficial. I thank these universities for their hospitality, and the Overseas Research Committee of the Open University for financial support. Some of the work in this paper was carried out while I was attending the Conference on Word and Decision Problems in Group Theory and Algebra, Oxford, 1976. The work of the organizers and the financial support of the SRC is gratefully acknowledged. 61.4. Notation Throughout the paper the text book [14] by Magnus, Karrass and Solitar will be used for matters relating to combinatorial group theory. In particular, the techniques of the Reidemeister-Schreier method as described in [14, 02.31 will be used in several places, and in such computations the notation of [14, 02.31 will be employed. Additional notation and background material will be needed as follows. Let G be a group. If A is a subset of G then sgp A will denote the subgroup of G generated by A. If {H, : i E I } is a set of subgroups of G then sgp{Hi : i E I } will denote the subgroup of G generated by the elements of the Hi. Let H be a subgroup of G and let g E G. The normalizer of H in G will be denoted by N G ( H ) .The conjugate subgroup {g-lhg : h E H } of H by g will be denoted by H E .The intersection of all the conjugates of H in G will be called the core of H (in G ) and will be denoted by Cor(H). Clearly Cor(H)G G. The index of H in G will be denoted by J G : H ( If. J G : H l < w then, as is well-known (and easily established), H has only finitely many distinct conjugates under G and so J G:Cor(H)I 0, p > 1 such that y r ( r ) 3 a p ‘ for r = 0 , 1 , 2 , .. . (i.e. yr grows exponentially). Suppose G is a group with a subgroup H of finite index. Obviously if H has exponential growth then so does G. Conversely, suppose of G such that yr grows exponentially. Let there is a finite subset g I+ g be a right coset representative function for G modH, and let A be the subset of H consisting of all elements

r

--I

ua. ua ,

where U is an arbitrary representative and a E f‘. Then it follows from [14, pp. 89-90] that if g is an element of G expressible as a word of length less than or equal to r in the elements of r, then g = hg, where h is an element of H expressible as a word of length less than or equal to r in the elements of A. This implies that y r ( r ) S I G : H 1 y,, ( r ) , so that H has exponential growth. Example 2.9. Let 9 be a group-theoretic property which is closed under taking homomorphic images; then the property of not being locally 9-by-finite is a large property.

To verify that (LP 2) holds, suppose H is a subgroup of finite index in a group G. Obviously if G is locally 9-by-finite then so is H. Conversely, suppose H is locally 9-by-finite, and let r be a finite subset of G. Let A be as constructed in Example 2.8. Then sgpA has finite index in sgpr. Since sgpA is L?-by-finite, so it sgpr. Thus G is locally 9-by-finite. Example 2.10. The properties i Max-sn, i Min-sn.

i

Max,

iMin, i Max-n, 1Min-n,

The properties Max, Min satisfy the conditions (**). (See [ l l , Lemma 1.E.11 and [19, Lemma 1.48, Corollary] in this connection.) The properties Max-n, Min-n, Max-sn, Min-sn satisfy the conditions (*). (See [19, Lemma 1.48, Corollary] and [26] in this connection.) There are several group-theoretic properties which are not known to be large properties, although they “ought to be”. Two of these are mentioned here. In [25] Wiegold discusses the class 8!? of groups G with the property

S.J. Pride

306

that whenever G acts transitively on a set R, there is some element of G displacing every element of R. Problem 1. Is the property of not lying in

I a large property?*

In [2] Chiswell discusses the concept of length functions on groups. O n e can define a length function on any group by setting the length of every element equal t o 0, or by setting the length of the identity equal t o zero and the length of every other element equal t o 1. However, there are various meanings one can attach t o a “non-trivial” length function. For example, one can require the existence of at least one element g such that the length of g 2 is greater than the length of g . Problem 2. With some interpretation of “non-trivial”, is the property of having a “non-trivial” length function a large property?

82.2. Large properties generated by groups Let A b e an infinite group. Then one can associate with A a large property 9 ( A ) , called the large property generated by A, defined as follows. A group G has 9 ( A ) if and only if there is a chain of groups Go,G I , .. . , G, with Go= A, G. = G, and where for i = 1,. . . , n either

Gi is a subgroup of finite index in GiF1,

(2.1)

or Gi-l is a subgroup of finite index in Gi,

(2.2)

or

(2.3)

G, maps onto Gi-,.

It is clear that Z ( A ) actually is a large property. Theorem 2.1. Let A be an infinite group. (i) A group G has 9 ( A ) if and only if there are groups B, C with B of finite index in A, C of finite index in G , and C mapping onto B. (ii) It can be assumed in ( i ) that either B S A or C a G. finite

G-2F e\:if index

I B

Added in proof. The answer is “No” (P.M. Neumann).


307

Proof. First consider (i). Obviously if the conditions are satisfied then G has 8 ( A ) . Conversely, if G has 2 ( A ) then there is a chain of groups Go,G I , .. . ,G, with Go= A, G. = G, and where for i = 1,. . . , n one of (2.1)-(2.3) holds. It will be shown by induction on n that the conditions stated in (i) are satisfied. The result is trivial if n = 0. Suppose n > O . Since G,-, has 2 ( A ) the inductive hypothesis guarantees that there are groups B,,C , with B, of finite index in A, C , of finite index in G,-, and C , mapping onto B,by a homomorphism 8 say. Case ( i ) . G, is of finite index in G,-,. Let C = C , r l G,. Then C has finite index in both C , and G.. Let B = 8 ( C ) . The fact that C has finite index in C , implies that B has finite index in B,.Thus B also has finite index in A. Case ( i i ) . G,-, is of finite index in G,. Simply take B = B I and C = C , . Case ( i i i ) . G, maps onto G,-,. Suppose JI : G, + G.-,is the homomorphism. Let C = JI-'(C,).Then C is of finite index in G.. Let B = B,.Then 84 maps C onto B. Now consider (ii). If C is not normal in G one can replace C by Cor(C) and replace B by the image of Cor(C) under the homomorphism C + B. On the other hand, if B is not normal in A one can replace B by Cor(B) and replace C by the inverse image of Cor(B) under the homomorphism C + B. This completes the proof of the theorem. Considerable'use will be made of Theorem 2.1 in the sequel. For the present, the following simple corollary is noted.

Corollary 2.1.1. Let G and A be infinite groups and suppose G has no proper subgroups of finite index. If G has 8 ( A ) then A has a (necessarily normal) subgroup G of finite index, where G is n homomorphic image of G. In particular A has a subgroup of finite index which has no proper subgroups of finite index. The first part of this result is an immediate consequence of Theorem 2.1 (i). The second part follows from the fact that no homomorphic image of G can have a proper subgroup of finite index.

308

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The concept of the large property generated by a group is useful in that it gives a way of comparing two groups. Let G and H be infinite groups. It will be said that G is larger than H (written H G ) if G has Y ( H ) ; G and H will be said to be equally large (written G = H) if H < G and G < H; and it will be said that G is strictly larger than H (written H < G ) if H < G but H is not larger than G. It is easy to see that H < G if and only if 9 ( G ) implies 9 ( H ) . Obviously the relation of larger than is a quasi-order and the relation of equally large is the induced equivalence relation. Another trivial but important fact is that if 9 is any large property, and if H has 9,then every group larger than H also has 9.Thus two groups which are equally large have the same large properties. The referee has pointed out that an alternative way of viewing the relation < might be profitable for future work. Define two groups to be equivalent if one is a subgroup of finite index in the other, and let fi denote the equivalence class of G. Then G =S H if and only if some Ho in is a pre-image of some Go in &.

93. “Large” groups 93.1. Large properties of non-abelian free groups

P and F be free groups of ranks r, s respectively with 2~r<s. (i) If s is infinite then P < F. (ii) If s is finite then P = F.

Theorem 3.1. Let

Proof. First consider (i). Since there is a homomorphism of F onto F it follows that P =S F. To show that P < F it suffices to find a large property enjoyed by F but not by I? Clearly %(s) (Example 2.2) is such a property. To prove (ii), let F2 denote the free group of rank 2 freely generated by x, y . The normal closure in Fz of {x”, y } ( n 2 1) has finite index and is free of rank n + 1 by Schreier’s formula [14,Theorem 2.101.Thus Fz has subgroups of finite index isomorphic to P and F, so F2= P 1:F, as required. This proves the theorem. It is now easy to describe the large properties of non-abelian free groups.


309

Corollary 3.1.1. Let F be a free group of rank r 3 2, and let 9 be a large property. (i) If r is infinite then F has 9 if and only if there is an r-generator group with 9. (ii) If r is finite then F has 9 if and only if there is a finitely generated group with 9. Proof. (i) is straightforward. For (ii), note that if there is a finitely generated group G with 9 then there is a free group P of rank greater than or equal to 2 mapping onto G and therefore having 9. Thus if r is finite then F has 9 since F = ?I Taking 9 to be Y ( A ) in the above result one obtains: Corollary 3.1.2. Let F be u free group of rank r 2 2. (i) If r is infinite then F is larger than euery r-generator group. (ii) Zf r is finite then F is larger than every finitely generated group. 93.2. Groups which are equally as large as F2 It was seen in 03.1 that F2, the free group of rank 2, is larger than every finitely generated group. Consequently, a finitely generated group is as large as it can possibly be if it is equally as large as F2. The aim of this section is to show that many groups traditionally thought of as being “very large” are actually equally as large as F2. It is clear that any group which has a non-abelian free subgroup of finite index has Y ( F 2 ) .The structure of such groups has been determined [3], [lo], [22]. For future reference it should be noted that the class includes all free products A * B with A, B finite and [ A1 3 2 , 1 B 13 3. It also includes all HNN groups, where the base H is finite and the associated subgroups K-l, K ,are properly contained in H. 1. Fuchsian groups

Theorem 3.2. All finitely generated Fuchsian groups are equally large as F2. Proof. The following argument is adapted from [17]. A finitely generated Fuchsian group G has a presentation with generators

a l , . . .,a,,, b,,. . ., b,, c I , .. .,c,,, e l , . . .,e,,

S.J. Pride

310

and defining relations i = 1, ..., r,

e:= 1 [ a , ,bl]

*

-

[a,, b P ] c *l . cqel . * e, = 1,

where ni 3 2 for i = 1,.. . , r, and where

By a theorem of Nielsen, Bundgaard and Fox [8, p. 681, G has a subgroup H of finite index which can be presented with generators UI,.*.

9

urn, V I , . . ., Urn, X I , .

9

XI,

and defining relator [UI,v , ] *

* *

[u,,

V,]XI

*

XI =

1.

Here (m,I ) # (O,O), (0, I), (0,2), (1,O). If 1 > O then H is free of rank greater than 1, while if 1 = 0 then H maps onto F2 (put u , = * = urn= v3 = . . . = v, = 1). Thus in either case H, and therefore G, has 2’(F2). 2 . One-relator groups There are two well-known heuristic principles in the theory of onerelator groups: (i) one-relator groups which do not satisfy a non-trivial law behave “like” free groups; (ii) one-relator groups with torsion which do not satisfy a non-trivial law behave “more like” free groups (G. Baumslag). The following two theorems further illustrate, and in fact give a certain substance, to these two principles.

Theorem 3.3. Every one-relator group with torsion which does not satisfy a non-trivial law has 2‘(F2). Theorem 3.4. There is a two -generator torsion -free one -relator group B which does not satisfy a non-trivial law but which is such that B < F2. It follows immediately from Theorem 3.3 that if G is a one-relator group with torsion which does not satisfy a non-trivial law then G has every large property enjoyed by F2. In particular, G is SQ-universal. The SQ-universality of one-relator groups has been investigated by Sacerdote and Schupp in (201. They obtain results for torsion-free groups as well as groups with torsion, but their methods (small cancellation theory over HNN groups) are much more complicated than those needed to prove Theorem 3.3.

Largeness in group rheory

311

As well as the two heuristic principles mentioned above, there is a third principle, namely that one-relator groups on at least three generators behave "more like" free groups. One is led to ask, Or to state the therefore, whether such groups actually have 9(F2). question in an alternative form: Problem 3. Does every one-relator group on at least three generators have a subgroup of finite index mapping onto Fz? In this connection it is worth remarking that Stallings [23] has shown that for any positive integer n there is an n-generator one-relator group which does not map onto F2. A question related to the above has been raised by A.M. Macbeath (private communication): is there an integer d > 0 such that whenever a group G can be presented with r + d generators and r relations, then G has 2(F2)?*(See [17,p. 5 ) for a similar question concerning SQ-universality.) Proof of Theorem 3.3. Let G = (t, a, b, . . . ; R " ) ( n > 1). Then G does not satisfy a non-trivial law provided G has at least two generators [9]. Now by [20, Lemma 4.11 it can be assumed that u,(R) = 0. Let a, = t-'at', b, = t-'bt', . . . ( i E Z). Then R can be rewritten as a word P in the a,, b,, . . . . There is an integer m 3 0 such that if a, occurs in P then 1 i 1 C m. Now it is easy to'see that G has presentation with generators

t , a - , , . . . , a,,,, b, ( i E Z), . . ., and defining relations

P", t-'a,t

=

a,,,

( i = - m , . . .,m - l),

t-'b,t = b,,, ( i E Z), . . . . The result will therefore follow from the Lemma 3.1. Let H be a group which has a presentation with generators t, a-,,

. . . , a , , bi ( i E Z), . . .

and defining relations This question has now been answered affirmatively (with d

= 2).

See "Groups with

two more generators than relators" by Benjamin Baumslag and Stephen J. Pride, J.

London Math. Soc. 17(2) (1978) 425426.

S.J. Pride

312

V, W,. . ., t-'a,t = a,,'

t-'b,r

=

( i = - m,. . . , m - l),

b,,, ( i E Z ) , . . . ,

where m 3 0 , and where V, W, . . . are t-free words. Suppose that there is an integer n > 1 such that for i m each of aa, ( V ) ,aa, ( W ) ,. . . is divisible by n. Then H has 2 ( F 2 ) .

I I

To prove this lemma consider the factor group K of H obtained by imposing the additional relations: a:= 1

(lilsm),

b, = 1 ( i E Z ) ,..., [a,,a,]=1

(lil,ljl~m).

Then V, W, ... are superfluous as defining relators in K , and so K is an H N N group with base the direct product of the finite cyclic groups sgp{a-,}, . . . ,sgp{a,}, associated subgroups sgp{a-,, . . . , a m - J , sgp{a-,+l,. . . ,a,}, and stable letter t. Thus K has a non-abelian free subgroup of finite index (see above) and therefore has 2 ( F z ) . Consequently, H has 2 ( F 2 ) , as required. This proves the lemma, and Theorem 3.3 follows. Before going on to prove Theorem 3.4 it is worth remarking that Lemma 3.1 can be used to show many groups with more than one relator have 2 ( F z ) . For example:

Theorem 3.5.* Let G be a group with presentation of the form

(t, a, b,. . . ;R",S", .. .), where n > 1. Suppose that a, ( R )= a,( S ) = * . = 0 , and furthermore there is an integer K 2 0 such that the number of t-symbols in each of R", S", . . . does not exceed K . Then G has 2 ( F 2 ) . The requirement in this theorem that there be a bound on the number of t-symbols in the relators R", S",. . . cannot be dispensed with in general. For consider the group Go with generators a, t and * 1 thank A.H.M. Hoare for pointing out a mistake in my original formulation%f this theorem.


313

defining relators [ WI, Wz]", where WI, W2 range over all words in a, t. Then all the conditions of Theorem 3.5 are satisfied except that there is no bound on the number of r-symbols in the relators. Moreover, it is clear that Go does not have the large property 9 ( 2 ) , and therefore certainly cannot have Y ( F z ) . Lemma 3.1 can also be used to show that many torsion-free onerelator groups have Y ( F z ) .For example, it is easily shown that if G = ( a , t ; a " ~ t ' ~ . . . a "where ~ t ' ~ )E T ,= O and where the a, have a common factor greater than 1, then G has Y ( F z ) . (On the other hand, if the a, do not have a common factor greater than 1 then G need not have Y ( F z ) ,as the group B in Theorem 3.4 shows.) Proof of Theorem 3.4. Let B = (a, t ; t - ' a 2 t = a'). It will be shown that B < Fz. By Theorem 2.1 it must be shown that n o normal subgroup of finite index in B can map onto a subgroup of finite index in Fz. Suppose N a B, 1 B : N 1 finite. Since t and a have infinite order in B, some positive power of t and some positive power of a must belong to N , otherwise 1 B : N 1 would be infinite. Suppose I " , u p ( n , p >0) belong to N , and assume p is the smallest integer i such that a ' E N. Let M be the normal closure of {t",a P } ,so that M N. Let B = B / M . Then

B = ( a , t ; t - l a z t= a 3 , f n , . p ) . Now the minimality of p guarantees that ( 2 , p ) = (3,p) = 1. For example, if p = 24 then a 4 = a-Zqt-lazqt E N , and 0 < q < p. It now follows that sgp{a} is normal in B. Moreover, the minimality of p guarantees that a has order exactly p in B. Thus every element of can be expressed uniquely in the form

(3.1)

t 'a'

B

i < n, 0 c j < p ) .

(0

In particular, M has finite index in B. Thus in order to show N does not map onto a subgroup of finite index in Fz it suffices to show that M does not map onto a subgroup of finite index in Fz.Now a subgroup of finite index in Fz is free of rank greater than 1 and therefore maps onto F2. Consequently, it is enough to show M does not map onto Fz. The elements in (3.1) constitute a Schreier system for M in B, and so M is generated by the elements --I

(3.2)

t'a'a. t'a'a ,

(3.3)

t'a't. t'a't

--1

,

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S.J. Ride

where 0 S i < n, 0 S j < p. An element (3.2) is freely equal to 1 unless --1

j = p - 1. Let'a, denote the element t'aP-la. t'aP-la ( = t'aPt-'), and let y denote the element t". Then it is easily shown that the following relations hold:

a:, . . . ,

(3.4)

a;

(3.5)

y - l a r y = a:.

=

=

a;-l,

Suppose M maps onto a free group F. Then the relation (3.5) holds in F and so, in F, y and a . commute. Thus, in F, a. has finite order and must therefore define the identity. Consequently, by (3.4), all the a, define the identity in E It follows that the factor group M of M obtained by putting all the a, equal to 1 still maps onto F. Let s(i, j ) denote the generator (3.3) of M. Now the defining relators of M are obtained by rewriting the relators 'a"ta y

t a -2a

-at

-7

(0 S T < n, 0 S a < p ) of B in terms of the s(i, j ) and the a,. Thus the defining relations of &f on the generators s(i. j ) are S(T, ~ ) s - ' ( T a ,

+ 2) = 1,

where 0 d 7 < n, 0 =scy < p, and where LY + 2 is reduced modulo n to lie between 0 and n - 1. Since (2,p) = 1 it follows easily that for T = 0, ..., n - 1 S (7,o) = S ( T , 1 ) = * * * = S (7,p - 1).

But t't.t't-I is freely equal to 1 unless T = n - 1. Consequently, M is infinite cyclic generated by s ( n - 1 , O ) ( = y ) , and so F is of rank less than 2. This compietes the proof of Theorem 3.4. 3. Arithmetically defined linear groups Let d be a positive square-free integer, and let I d be the ring of integers in the quadratic field Q(d-d). Fine and Tretkoff [ 5 ] have investigated the SQ-universality of the groups PSL(2, I d ) , SL(2, I d ) , GL(2, I d ) in case I d is a Euclidean domain - that is, when d = 1,2,3,7,11. Their work carries over to show that these groups have 2(Fz). Since SL(2, I d ) maps onto PSL(2, I d ) , and since, for the values of d under consideration, SL(2, I d ) has finite index in GL(2, I d ) , it suffices to show PSL(2, I d ) has ~ ( F Z ) . For d = 1,2,3,11 the arguments of Lemmas 1 , 2, 4 of [5] can easily


315

be modified. Consider the case d = 7. It is shown in Lemma 3 of [5] that PSL(2,17) has a subgroup G of finite index with presentation

(x,r,s, w ; x 3 = r 3 = s 3 = 1, w-’[s-’,r]w = [s-’,r], [r-l,x-’] = [s-I, r ] , w-’sw Setting s = x

=

= x-’).

1 in G gives the group

(r, w ; r’

=

G

with presentation

1).

Since has 6p(F2)it follows that PSL(2,17) has 2?(F2). Fine and Tretkoff also show that for all d, t h e subgroup E(2, Id) of SL(2, Id) generated by the transvections is SQ-universal [5, Theorem C]. Their argument is easily modified to show that E(2, Id) has 6p(F2). R. Zimmert [28] has shown that for many values of d, SL(2,Id) has Fz as a homomorphic image. For such values of d it is obvious that sL(2, I d ) has 6p(F2). It follows from [l],[16] that for n 3 3, SL(n, Id) does not have 6p(Fz).

4. Small cancellation groups Let F be a free group and let r be a symmetrized subset of F which satisfies t h e small cancellation conditions C ( p ) , T ( q ) for ( p , q ) one of (6,3), (4,4), (3,6). Let G be t h e quotient of F by t h e normal closure of r. Then G is called a small cancellation group. The reader unfamiliar with small cancellation theory should consult [21]. Small cancellation groups tend to be “quite large”. For example, it has been shown by Collins [4] that if r satisfies C(4) and T(4) then, with a few trivial exceptions, G has the large property 9(2). Recently it has been shown by M. al-Janabi in his PhD thesis at London University that if r is finite and satifies C(6), then again, apart from a few trivial exceptions, G has %(2). Al-Janabi has also established that if r is finite and satisfies C(7) then G is actually SQ-universal whenever it has 9 (2 ). One is led to ask the following question. Problem 4. D o “most” finitely presented small cancellation groups have 2?(Fz)? I suspect that the answer is negative (see the comments in t h e second paragraph of 03.3 in this regard). It should be remarked that many small cancellation groups can be shown to have 6p(F2) using Lemma 3.1. The following lemma is also useful.

316

S.J. Pride

Lemma 3.2. Let G = ( a ,b, x, y , . . . ;R, S, . . . ). Suppose there are integers p,q greater than 1 and not both 2 such that all the a-exponents in R, s,. . . are divisible by p and all the numbers c b ( R ) ,u b ( S ) , . . . are divisible by q. Then G has 2?(Fz). Proof. Let d be the image of G obtained by putting a p = b q = 1, and = (a, b ; a p ,b q ) , so that d has setting x,y,. . . equal to 1. Then 2?(F2).Consequently G has 6p(F2). Finitely generated infinitely related small cancellation groups need not have 2?(F2),even though they can be "quite large". Roughly speaking, this is because there is considerable scope to "impose relations which destroy largeness".

Theorem 3.6. Let u be a number greater than or equal to 5, and let F be the free group on a, b. There is an infinite symmetrized subset r of F satisfying C'(l/u) such that the factor group G of F by the normal closure N of r is such that G < F2. The group G is SQ-universal. Proof. The construction of G is similar to the construction given in [18,§2]. Let n > 2 v and let a I ,. . . , a, be numbers satisfying 1 a,1 > 1, C a,= 1, I a , J # I a , Ifor i # j . Let a = m a x { ) a , ( : i= 1, ..., n } , and suppose 2a/XIa,I < l/v. Let r be the symmetrized closure of the following elements of F: ab"1ab"2.. . ab"., (3.6)

+

a"lbPa"2bP* . . a"nbP,p= a + 3, a 4,.. . . (3.7) Then it is easily shown that pieces relative to r are subwords of (b"lab"~+1)*', (a"lbPao~+l)*', (bPa"nbP)"for i = 1,. . .,n (where subscripts are calculated modulo n ) , and it thus follows that r satisfies C'(l/v). Now G has no non-trivial finite homomorphic image. For if G maps onto with C? finite, then in d,b P = 1 for some p 2 (Y + 3, and so it follows from the relators (3,6), (3,7) (which of course hold in that a = b = 1. Thus is trivial. Now since F2 does not have a subgroup of finite index which has no proper subgroups of finite index, it follows from Corollary 2.1.1 that G 0. Moreover H = [C,( i E I ) , s] where C, = A f. In particular, the rank of H is the same as the rank of G. Proof. Now there is a normal subgroup N of finite index in G containing H. In particular, t P E N for some p # 0, and so it follows from the previous lemma that B s N s H. Thus H = sgp{B, t'} for some q > O . Now let s = t", and for (i, n ) E I x Z let C(i,n ) = A ( i , n q ) x A ( i , nq + 1 ) X - . . X A ( i , nq + q - 1). T h e n for all n E Z , C(i,n ) is isomorphic to A f , and it is readily seen that H = [C,( i E I ) , s ] .

Lemma 4.7. Suppose A, = n?S, ( i ) ( i E Z), where each S, ( i ) is a nonabelian simple group. Let G be an infinite homomorphic image of G. Then G = [fly S, (j) (j E J ) , f], where J C I and where for each j E J, f, is a non-empty subset of A,. In particular, the rank of is less than or equal to the rank of G. Proof. In the copy A(i,n ) of A,, the copy of S, ( i ) will be denoted by S,(i,n), so that A ( i , n ) = n ? , S , ( i , n ) . Some of the following calculations are similar to those in Example 4.3 but are repeated in detail here for convenience. Let N be the kernel of the homomorphism G - G . It follows from Lemma 4.5 that N s B. Consider the subgroups N n A ( i , n ) . It will be shown that these generate N . Let b be an element of B belonging to N , and suppose that the (i, n ) coordinate, a say, of b is non-trivial. Now A (i, n ) = n?,S, (i, n ) . Suppose the A th coordinate, s say, of a is non-trivial. Let z E S, (i, n ) . Then [b,z ] E N and [b, I ] = [s, 21. Since [s, I ]# 1 for at least one z, it follows that SA(i,n ) < N. Consequently, a E N, and so b E sgp{N n A (i, n ) : (i, n ) E I x Z},as required. Now similar calculations to those in the previous paragraph show that N fl A (i, n ) is the subgroup generated by S, (i, n ) with A ranging over some subset ei of A. This subset ei depends only on i, and not on n. For if S A ( i ; n S ) N n A ( i , n ) , then conjugating by r p shows that S,(i, n + p ) s N n A ( i , n + p ) .

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S.J. Ride

It is now not difficult to see that one can take J to be the subset of I consisting of those elements i for which Si# A , and then put & = Ai'\6Ji for j E J. This proves the lemma. Proof of Theorem 4.4. In the above construction take Z = {1,2,. . .}, and take all the Ai to be isomorphic copies of some finite non-abelian simple group. Then B is minimal (see Example 4.3), and G is a split extension of B by the minimal group ( t ) . It will be shown that G does not have Max- < . For p = 1,2,.. . let G, = [Ai(i = 1,. . . , p ) , t ] . Then G I + Gz< < G. It will be shown that G, < G,+,. Suppose, by way of contradiction, that G,+,< G,,. Then by Theorem 2.1, G, would have a subgroup H, of finite index mapping onto a subgroup H,+] of finite index in GP+,.Now the rank of H, would be p, by Lemma 4.6. Moreover, H, = [Af(i = 1,. . . , p ) , t ] for some q. Consequently, by Lemma 4.7, H,+, would have rank less than or equal to p. On the other hand, H,+, would have rank p + 1, by Lemma 4.6. This contradiction shows G, < GP+,,as required. Proof of Theorem 4.5. The proof is by induction on n. If n = 1 the result is trivial. Suppose n > 1, and let H =S G. By Theorem 2.1 there is a subgroup A of finite index in G mapping, by a homomorphism 8 say, onto a subgroup of finite index in H. Let Ai = A n Mifor i = 1,. ..,n. Then I M i : A i ) < w a n d s oI G : A , x . - . x A , ( < ~Notice . that 8(Al X . * . X A,) has finite index in H. Let N be the kernel of the restriction of 8 to A , x * x A,. For i = 1,. . . ,n let

Ni = { a : a E Ai, there exists u E A, x

. . . x-Ai-,x A,+,x

-

*

x A. such that au E N}.

Then Ni Ai. Case 1. For some i, 1 Ai : Ni 1 < w.

Assume for definiteness that 1 A, : N , 1 < 03. Then (A2x . . X A,)N/N has finite index in (A, x * * x A,)/N. For let z,, . . . ,z, be coset representatives of N , in A,, and let au E A , X . X A, with a E A,, u E Az x * * x A,. Then a = zia' for some zj and some a ' € N , . Moreover, since a ' € N , , there exists u E A 2x * X A, such that a ' u E N . Thus au = ~ ~ ( a ' u ) ( u - and ~ u )so auN = z j ( u - ' u ) N . Consequently (A, x * x A,)/N = U:(zjN)((A2 X * X A.)N/N).

-

-

- -

-

- -


333

It now follows that Since

(A1 X

* * .

X

A,,)/N 2: (A2X .. . X A.)/N n(A2X

* * .

(A2X

* * *

X

A,,)/Nn (A2X

X

* * *

X

A,,) 6 M2X

0

.

*

X

A").

M,,,

the inductive hypothesis now applies to give the required result.

Case 2. For all i, / A i: Ni I = w. NOW N sgp{NI,N2,. . . ,N,,}.Thus ( A lx * * * x A,,)/sgp{NI,.. .,N.} < ( A l x - . . x A . ) / N But . ( A I x ~ ~ ~ x A n ) / s ,..., g p {N.} N l is isomorphic to A I / N Ix . . . x A,,/N,,,and since for each i, A i / N i Mi, it follows that AJN1X * * * X A,/N,, = G. Thus H G. This completes the proof. 2 :

2 :

BS. A final question Let 9 be a large property and let % be a class of groups. Say that 9 is large for 9 if every %-group with 9 has infinite height.

With the possible exception of SQ-universality, none of the properties listed in $2.1 is large for finitely presented groups. In fact, for all the properties except SQ-universality there are finitely presented minimal groups with the property. On the other hand, finitely generated SQ-universal groups cannot be minimal (since finitely generated groups have just-infinite quotients). Now SQ-universality is generally agreed to be quite a good measure of largeness for countable groups. Can this be made more precise? Problem 8. Is SQ-universality large for finitely presented, finitely generated, or even countable groups?"

Of course, SQ-universality is not large for all groups since there are simple (and hence minimal) SQ-universal groups. References [ l ] H. Bass, J. Milnor and J.-P. Serre, Solution of the congruence subgroup problem for SL, ( n Z 3 ) and Sp,. ( n Z 2 ) . Inst. Haute Etudes Sci. Publ. Math. 33 (1%7) 59-137. [2] 1.M. Chiswell, Abstract length functions on groups, Math. Proc. Camb. Phil. Soc. 80 (1976) 451463. Added in proof. B. Hurley has constructed a countable SQ-universal minimal group.

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[3] D.E. Cohen, Groups with free subgroups of finite irldex, Conference on Group Theory, Springer Lecture Notes No. 319, 26-44. [4] D.J. Collins, Free subgroups of small cancellation groups, Proc. London Math. SOC. ( 2 ) 26 (1973) 193-206. [5] B. Fine and M. Tretkoff, The SQ-universality of certain arithmetically defined linear groups, J. London Math. SOC. (2) 13 (1976) 65-68. [6] F.P. Greenleaf, Invariant means on topological groups, Van Nostrand Mathematical Studies No. 16 (Van Nostrand-Reinhold Company, New York, Toronto, London, Melbourne, 1969). [7] P. Hall, On embedding a group i n t o the join of given groups, J. Austral. Math. SOC. 17 (1974) 434-495. [8] A.H.M. Hoare, A. Karrass and D. Solitar, Subgroups of infinite index in Fuchsian groups, Math. Z. 125 (1972) 59-69. [9] A. Karrass and D. Solitar, Subgroups of HNN groups and groups with one defining relation, Can. J. Math. 23 (1971) 627-643. [lo] A. Karrass, A. Pietrowski and D. Solitar, Finite and infinite cyclic extensions of free groups, J. Austral. Math. SOC.26 (1973) 458-466. [ 1 I] O.H. Kegel and B.A.F. Wehrfritz, Locally finite groups, North-Holland Mathematical Library, Vol. 3 (North-Holland, Amsterdam, London, New York, 1973). [I21 D. McCarthy, Infinite groups whose proper quotient groups are finite I, Comm. Pure Appl. Math. 21 (1968) 545-562. [13] D. McCarthy, Infinite groups whose proper quotient groups are finite 11, Comm. Pure Appl. Math. 23 (1970) 767-790. [14] W. Magnus, A. Karrass and D. Solitar, Combinatorial group theory: Presentations of groups in terms of generators and defining relations (Interscience Publishers, New York, London, Sydney, 1966). [I51 J. Milnor, A note on curvature and fundamental group, J. Differential Geometry 2 (1968) 1-7. [16] J.L. Mennicke, Finite factor groups of the unimodular group, Ann. of Math. (2) 81 (1965) 31-37. [17] P. Neumann, The SQ-universality of some finitely presented groups, J. Austral. Math. SOC.16 (1973) 1-6. [I81 S.J. Pride, On quotients of certain countable groups, Bull Austral. Math. SOC.,16 (1977) 225-228. [ 191 D.J.S. Robinson, Finiteness conditions and generalized soluble groups, Parts 1 and 2, Ergebnisse der Mathematik und ihrer Grenzgebiete, Bands 62 and 63 (SpringerVerlag, New York, Berlin, 1972). [20] G. Sacerdote and P.E. Schupp, SQ-universality in HNN groups and one relator groups, J. London Math. SOC.(2) 7 (1974) 733-740. [21] P.E. Schupp, A survey of small cancellation theory, in Word Problems (NorthHolland Publishing Company, Amsterdam, London, 1973). [22] G.P. Scott, An embedding theorem for groups with a free subgroup of finite index, Bull. London Math. SOC.6 (1974) 30&306. [23] J.R. Stallings, Quotients of the powers of the augmentation ideal in a group ring, in Knots, Groups and 3-Manifolds, Annals of Mathematics Studies No. 84 (Princeton University Press, Princeton, 1975). [24] C. Tretkoff, Some remarks on just-infinite groups, Commun. Algebra 4 (1976) 43-49.


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(251 J . Wiegold, Transitive groups with fixed-point free permutations, Arch. Math. 27 (1976) 473-475. [26] J.S. Wilson, Some properties of groups inherited by normal subgroups of finite index, Math. Z . 114 (1970) 19-21. [27] J.S. Wilson, Groups with every proper quotient finite, Proc. Cambridge Philos. SOC. 69 (1971) 373-391. [28] R. Zimmert, Zur SL, der ganzen Zahlen eines imaginar-quadratischen Zahlkorpers, Inventidnes Math. 19 (1973) 73-81.


ALGORITHMIC PROBLEMS FOR SOLVABLE GROUPS V.N. REMESLENNIKOV and N.S. ROMANOVSKII Institute of Mathematics, Nooosibirsk

$1. Introduction

Classical algorithmic problems - t h e word problem, conjugacy problem, and isomorphism problem - arose in the theory of groups from topolocal considerations and were formulated for t h e first time at the beginning of this century by Dehn. In the middle fifties it was shown that these problems have negative solutions in the class of all groups. However, for important classes of groups such as nilpotent and solvable groups, these problems remained unsolved. At t h e present moment a number of interesting results have been obtained in this field, and this paper is intended to survey them. Let us recall the main definitions and t h e formulation of t h e classical algorithmic problems for t h e class of all groups. We call the pair ( X , R ) t h e presentation of group G, where X is a non-empty set and R is a set of t h e words in t h e alphabet X U X - ' . The group G is a factor group of the free group with the base X by the normal subgroup, generated by the set R . With this understanding we shall write G = (X, R ) ; further, X is called the set of the generators of group G, and R - the set of defining relators. If t h e sets X and R are finite, we say that group G is finitely presented. We shall also consider the case when X is finite and R is recursively enumerable. In this case G is called a recursively presented group. Let G be a given finitely presented (recursively presented) group. The word problem, conjugacy problem and membership problem are the problems of establishing the existence of algorithms to decide respectively (i) whether or not an arbitrary element (word) of group G equals 1; (ii) whether or not two arbitrary elements from G are conjugate in G ; (iii) for a given arbitrary finite set { u , u , , . . . ., u,} of elements of G 337

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whether or not the element u belongs to the subgroup generated by UI,. . . , V".

Finally, let = {GiI i E N } be a recursive class of finitely presented groups. The isomorphism problem for the class C#I consists in establishing the existence of an algorithm to decide for arbitrary i, j whether or not the groups G, and G, are isomorphic. P.S. Novikov [lo] and W.W. Boone [27] constructed examples of finitely presented groups, for which the word problem was solved negatively. This implied the negative solution of the conjugacy and membership problems. Aided by this result, S.I. Adian [l] proved that the isomorphism problem in the class of all groups also has a negative solution. In addition, algorithmic unsolvability was proved for a number of other problems, such as the recognition of the following properties: nilpotency, finiteness, simplicity or unity of a given group presentation. Let H be a subgroup of the finitely presented group G, generated by the elements u I , . .. , u r nThe . set of all the words from the u, is recursively enumerable, as is the set of all the words from G equal to 1. The intersection of these two sets forms the recursively enumerable set of the presenting relations of group H in the generators u I , .. . ,u,. Therefore, the finitely generated subgroup of the finitely presented group is recursively presented. H igman (301 showed that the converse statement is also correct; i.e. every recursively presented group is embeddable in some finitely presented group.

92. The setting of the problems

Thus we see that the majority of algorithmic problems has a negative solution in the class of all groups. Therefore, it is natural to study algorithmical problems with additional restrictions on t h e groups considered. In a number of papers, such restrictions were on the form of the defining relations, e.g., Magnus' theorem [32], proved in 1932, and which deals with the solvability of the word problem for groups with one defining relation. However, we shall consider restrictions having a natural group-theoretic character, such as nilpotency, polycyclicity and solvability. Here, two approaches to such problems are possible. The first approach consists in the algorithmic problems being considered for a finitely presented group (or a set of groups) under the assumption that such group turns out to be, say, solvable of given steps of solvability;

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i.e. it is assumed that t h e condition of solvability results from the defining relators. In t h e second approach, groups which are finitely presented in a given variety are studied. This means t h e following: Let V be a variety given by a finite set of identities. We say that a group G is given in the variety V by generators X and relations R, if it is a factor group of t h e free group of the variety V with the base X by the normal subgroup, generated by the set R. Finitely presented group and recursively presented group in the variety V are defined in the obvious way. Let us introduce the corresponding designations FP, RP, FPV, RPV for the classes: finitely presented, recursively presented, finitely presented in the variety V, recursively presented in the variety V. In t h e language of these classes t h e first of the formulated approaches leads to the study of groups from F P n V, if the additional condition consists of membership in t h e variety V. In the second approach groups from FPV are studied. Evident inclusions take place:

F P n V c F P V C R P n V=RPV. For some varieties these relations may be made more exact. If V is t h e variety of groups of nilpotency class c, then it is well known that FP f l V = FPV = RPV. where 1 3 2 For the variety of 1-step solvable groups, denoted by d’, strict inclusion takes place: F P n V 3 the inclusion is strict FPV < RPV, as it was shown by A.L. Shmelkin [23], who proved that the free metabelian group is not finitely presented in d’. Recalling Higman’s theorem about the embedding of recursively presented groups into finitely presented ones, we may formulate a few which acquire an important meaning, problems for the varieties d’, taking into account V.N. Remeslennikov’s theorem, which we shall deal with below.

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Problem 1. Can any RPd'-group, where I >3, be embedded into an FPdm-groupfor suitable m ? Problem 2. Can any FPdI-group, where I >3, be embedded into an FP r l d"-group for suitable m ? Problem 3. Can any RPd-group, where I > 3 , be embedded into an FP rl A -group for suitable m ? For I = 2 these problems have been solved positively, since G. Baumslag [25] and V.N. Remeslennikov [13, 151 have shown that any finitely generated metabelian group can be embedded into a finitely presented metabelian group. 63. Classical algorithmic problems for solvable groups Before supplying the formulations of the main results obtained, we shall take note of the connection of the algorithmic problems with the finite approximatability of the groups. Let T be some group-theoretic predicate. We say that the group G is finitely approximated with respect to the predicate T , if for any collection of elements which are not related under T, there exists a homomorphism of group G to a finite group H such that the images of the elements considered are not related under T in H. Let us consider, for instance, the predicate: to be equal to the element 1. Groups which are finitely approximated with respect to this predicate are called residually finite. Let G be a group finitely presented in the variety V. Let the variety V be given by a finite set of identities. If G is residually finite, then the word problem can be positively solved in G. I.e., the set of words representing 1 is recursive. It is evident that this set is recursively enumerable. Therefore, it is enough to show that its complement is also recursively enumerable. Actually, because of the residual finiteness the complement may be enumerated by searching through all possible homomorphisms of group G to finite groups. Similarly, if the group G is residually finite with respect to conjugacy or membership, then in it the problems of conjugacy and membership are solved positively. Let us consider now the word problem for groups in d'.By Hall's theorem [29] finitely generated metabelian groups are residually finite, therefore, the word problem for groups finitely presented in the variety d 2 ,is solved positively. On the other hand, V.N. Remeslennikov's


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theorem [14] states that for 1 > 5 there exists a group finitely presented irl the variety d’, for which the word problem is solved negatively. To be more precise, for each I a specific representation of this group can be obtained. Since FP n d’< F P d ’ for 1 > 3, we are led to the following question:

Problem 4. Does there exist a solvable finitely presented group with unsolvable word problem? Let us note that a similar question was formulated by S.I. Adian, who asked if there exists a group with unsolvable word problem which satisfies some nontrivial identity. It is clear that an affirmative answer to problem 2 and 3 gives an affirmative answer to problem 4. The cases 1 = 3, 4 are not covered by V.N. Remeslennikov’s theorem. Hence, we have the following question:

Problem 5. Is the word problem solvable for groups from F P d 3 or FPd4? As for the conjugacy problem, we know still less about it. It follows from V.N. Remeslennikov’s theorem that for 1 > 5 there exist examples of groups finitely presented in the variety d’, for which the conjugacy problem has negative solution. However, this problem remains uninvestigated for solvable groups of 2, 3 and 4 steps. It should be noted that it cannot be solved by using the finite approximation method, because M.I. Kargapolov and E.I. Timoshenko [5], and Wehrfritz [35],have constructed examples of finitely generated metabelian groups which are not residually finite with respect to conjugacy. As for the membership problem, there exist two main results. On the one hand, N.S. Romanovskii [18] proved that it is solved positively for metabelian groups. On the other hand V.N. Remeslennikov [14], for 1 3 4 , has constructed an example of a group which is finitely presented in the variety d‘,and with a finitely generated subgroup for which the membership problem is solved negatively. Hence, we have the following question:

Problem 6. Is the membership problem solvable for groups in F P d 3 ? By using V.N. Remeslennikov’s theorem, for each I2 7, there was for which there constructed in [6] a group G finitely presented in d’,

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exists no algorithm, deciding for any group H finitely presented in d' whether or not it is isomorphic to G. By this example the isomorphism problem for F P d ' for 1 3 7 is solved negatively. Hence, we have the following

Problem 7. Is the isomorphism problem solvable for metabelian groups? We recall that in the class of all groups the isomorphism problem for a unit group is solved negatively. The analogous statement is wrong for the variety d', because there G # 1 if and only if G / G ' # 1. Therefore, the problem is reduced to the same problem for abelian groups, where, as is well known, it has a positive solution. We may also remark that for groups from F P d ' there exists an algorithm to decide whether a given group is finite or not, and in the positive case to find its multiplication table. This can be proved by induction on the solvability step of the group. It should be noted that, for solvable groups, there does not exist a general theorem on the algorithmic undecidability of group-theoretic properties, similar to the theorem of Adian-Rabin. Perhaps, we can prove such a theorem if we can solve the following:

Problem 8. Is there an algorithm which decides if groups in d', 1 > 2, are nilpotent? Now a few words about the methods of proof of the majority of theorems of this kind. They use the so-called Magnus embedding of a free solvable group, the essence of which lies in the following. Let 5 be a free solvable group of step 1 and rank n and T the left free module of rank n over the ring Z[F,]. Then the free solvable group Fl+Iof the step 1 + 1 and rank n can be embedded into the group of matrices M of the type (6 :), where f E R and t E T.We may consider the group M as the wreath product of the free abelian group of rank n with R. The Magnus embedding turns out to be convenient, for instance, in the case that N is a normal subgroup of E + , ;for in this case the embedding may be extended to the normal of group M,which represents the semidirect product of subgroup the subgroups consisting of unitriangular and diagonal matrices such n F,+,= "181. that Thanks to the Magnus embedding many questions concerning solvable groups can be replaced by corresponding questions about modules over group rings of lesser solvability step. In particular, a


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number of algebraic problems for two-step solvable groups become problems for free modules over the ring Z[%',,. . . ,Zn]. 94. Free solvable groups

As has been mentioned above, free solvable groups of finite rank are recursively defined in the class of all groups, and evidently finitely defined in the corresponding variety YI'. By the use of the Magnus embedding for a free solvable group, the positive solution of the word problem is easily established. Matteus [33] showed that in the wreath product of groups A and B the conjugacy problem is solved positively, if it is solved positively in A and B, and additionally, the membership problem for cyclic subgroups is positively solvable in B. Making use of this fact and of the Magnus embedding, M.I. Kargapolov and V.N. Remeslennikov [3] positively solved the conjugacy problem for free solvable groups by induction on the solvability step. Later on, V.N. Remeslennikov and V.S. Sokolov [16] established the finite approximatability of those groups with respect to conjugacy. R.A. Sarkisjan [19] proved that the conjugacy problem is positively solved for free polynilpotent groups. The next problem is still open. Problem 9. Is the membership problem solvable for free solvable groups? This problem is closed connected with the following. Problem 10. Is there an algorithm for solving finite systems of linear equations over Z[F,]? Here, F. is a free solvable group with solvability length 1. Let us also note that in an absolutely free group there exist algorithms (for instance, Nilsen's method), to decide if a given finite set is a basis. A.F. Krasnikov (7) has found a similar algorithm for free solvable groups. The class of groups with one defining relation in the variety a' is the closest to free solvable groups. We know little about these groups so far. Let us still note that N.S. Romanovskii [17] proved the analog of Magnus' Freiheitssatz for a group with one relation in the variety a', and the word problem with some restrictions on the single relation has been solved [18]. However, the general case remains open:

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Problem 11. Is the word problem solvable for groups with one defining relation in the variety ‘ill’1,3 3? 05. Algorithmic problems for nilpotent groups

In 1955 A.I. Mal’cev [8] showed that the word problem can be solved positively for nilpotent groups. He also showed [9], that finitely generated nilpotent groups are residually finite with respect to membership, hence the membership problem is solvable. Blackburn [26] has positively solved the conjugacy problem for nilpotent groups by the same method. M.I. Kargapolov and his students [4] raised and solved the following algorithmic problems for $3-power groups over a those of word, conjugacy, membership, separating of “good” ring ‘0-periodical part, intersection, the description of subgroups in the terms of relations. Thus, we see that in nilpotent groups the picture is diametrically opposed to that in the class of all groups we have the following:

a:

Problem 12. Is the isomorphism problem solvable for nilpotent groups? This problem cannot be solved by the method of finite approximation, for at present we know many examples of nonisomorphic finitely generated nilpotent groups with the same families of finite homomorphic images. Concerning this we should note Pickel’s theorem [34]. H e showed that there can only exist a finite number of groups with the same family of finite homomorphic images in the class of finitely generated nilpotent groups. It is highly probable that problem 12 will be answered negatively, hence, we have another interesting problem: Problem 13. Is the isomorphism problem solvable for groups with one defining relation in the variety of nilpotent groups with given nilpotency class? Such an algorithm exists for class 2-nilpotent groups [20]. 06. Algorithmic problems for polycyclic groups

It should be noted at once that any polycyclic group is finitely presented, so it is possible to discuss algorithmic problems for

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polycyclic groups. All results known to us here were obtained by means of finite approximatability. Initially, Hirsch [31] showed that polycyclic groups are residually finite. Then A.I. Mal’cev [9] showed that they are finitely approximated with respect to membership. Finally, V.N. Remeslennikov [12] and Formanek [28] proved the finite approximatability of polycyclic groups with respect to conjugacy. The isomorphism problem for polycyclic groups is still unsolved.

References [ I ] S.I. Adjan, Insolvability of some algorithmic problems of group theory, Trudy Moskovoskogo Matem. ob-va 6 (1957) 213-298 (Russian). [2] M.I. Kargapolov, Finite approximatability of supersolvable groups with respect to conjugacy, Algebra i logika 6, No. 1 (1967) 63-68 (Russian). [3] M.I. Kargapolov and V.N. Remeslennikov, Conjugacy in free solvable groups, Algebra i logika 5, No. 6 (1966) 15-25 (Russian). [4] M.I. Kargapolov, V.N. Remeslennikov, N.S. Romanovskii V.A. Roman’kov and V.A. Churkin, Algorithmic problems for 0 power groups, Algebra i logika 8, No. 6 (1969) 643-659 (Russian). [5] M.I. Kargapolov and E.I. Timoshenko, O n the problem of finite approximatability with respect to conjugacy of metabelian groups, 4 Vsesoyuznij simpozium PO teorii grupp (Tezisy dokladov, Novosibirsk 1973) (Russian). [6] A S . Kirkinskii and V.N. Remeslennikov, The isomorphism problem for solvable groups, Matematiceskie zametki 18, 3 (1975) 437-443 (Russian). [7] A.F. Krasnikov, On the bases of free solvable groups, to appear (Russian). [8] A.I. Mal’cev, Two remarks on nilpotent groups, Matematiceskij sbornik 37 (1955) 567-572 (Russian). [9] A.I. Mal’cev, On homomorphisms to finite groups, Ucenye zapiski lranovskogo ped. instituta 18 (1958) 49-60 (Russian). [lo] P.S. Novikdv, On the algorithmic unsolvability of the word problem in the theory of groups, Trudy matematiceskogo instituta AN SSSR 44 (1955) 1-144 (Russian). [ I l l V.N. Remeslennikov, Conjugacy of subgroups in nilpotent groups, Algebra i logika 6, No. 2 (1967) 61-76 (Russian). [I21 V.N. Remeslennikov, Conjugacy in polycyclic groups, Algebra i logika 8, No. 6 (1969) 712-725 (Russian). [ 131 V.N. Remeslennikov, O n finitely presented groups, 4 Usesoyuznyj simpozium PO teorii grupp, Novosibirsk 1973, pp. 164-169 (Russian). (141 V.N. Remeslennikov, An example of a group, finitely presented in the variety ?I5, with the unsolvable word problem, Algebra i logika 12, No. 5 (1973) 577-602 (Russian). [IS] V.N. Remeslennikov, Investigations of infinite solvable and finitely-approximated groups, Doctoral dissertation, Novosibirsk, 1974 (Russian). [ 161 V.N. Remeslennikov and V.G. Sokolov, Some properties of Magnus’ imbedding, Algebra i logika 9, No. 5 (1970) 566-578 (Russian). [I71 N.S. Romanovskii, The freedom theorem for groups with one defining relation in

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the varieties of solvable and nilpotent groups of given steps, Matematiceskij sbornik 89, No. 1 (1972) 93-99 (Russian). [18] N.S. Romanovskii, On some algorithmic problems for solvable groups, Algebra i logika 13, No. 1 (1974) 2 6 3 4 (Russian). [19] R.A. Sarkisjan, Conjugacy in free polynilpotent groups, Algebra i logika 11, No. 6 (1972) 694-710 (Russian). [20] N.F. Sesekin, On the classification of metabelian (two-step nilpotent) torsion free groups, Ucenye zapiski Ural’skogo universiteta 1 9 (1965) 27-41 (Russian). [21] V.G. Sokolov, An algorithm for the word problem for a class of solvable groups, Sibirskij matematiceskij iurnal 12, No. 6 (1971) 140.5-1410 (Russian). (221 E.I. Timoshenko, Some algorithmic problems for metabelian groups, Algebra i logika 12, No. 2 (1973) 232-240 (Russian). [23] A.L. Shmel’kin, Wreaih products and varieties of groups, Izv. AN SSSR, ser. mat. 29 (1965) 149-170 (Russian). [24] G. Baumslag, Wreath products and finitely presented groups, Math. Z. 7.5 (1961) 22-28. [2.5] G. Baumslag, Subgroups of finitely presented metabelian groups, J. Austral. Math. Soc. 16 (1973) 98-110. [26] N. Blackburn, Conjugacy in nilpotent groups, Proc. Amer. Math. Soc. 16 (1965) 143- 148. [27] W. Boone, The word problem, Ann. Math. 70, No. 2 (1959) 207-265. [28] E. Formanek, Conjugate separability in polycyclic groups, J. Algebra 42, No. 1 (1976) 1-10, [29] P. Hall, On the finiteness of certain solvable groups, Proc. London Math. Soc. 9 (1959) 595-622. [30] G. Higman, Subgroups o n finitely presented groups, Proc. Roy. Soc. London A262 (1961) 455-475. (311 K.A. Hirsch, On infinite solvable groups, IV, J. London Math. Soc. 27 (1952) 81-85. 1321 W. Magnus, Das Identitats problem fur Gruppen mit einer definierenden Relation, Math. Ann. 106 (1932) 295-307. [33] J. Matteus, The conjugacy problem in wreath products and free metabelian groups, Trans. Amer. SOC. 121 (1966) 329-339. [34] P.F. Pickel, Finitely generated nilpotent groups with isomorphic finite quotients, Trans. Amer. Math. Soc. 160 (1971) 327-341. [35] Wehrfritz, Two examples of solvable groups that are not conjugacy separable, J. London Math. Soc. (2). 7 (1973) 312-316.


QUADRATIC EQUATIONS IN GROUPS, CANCELLATION DIAGRAMS ON COMPACT SURFACES, AND AUTOMORPHISMS OF SURFACE GROUPS Paul E. SCHUPP* The University of Illinois at Urbana -Champaign

91. Introduction

In recent years cancellation diagrams have become a fundamental tool of combinatorial group theory, and we assume familiarity with the basic facts about such diagrams. (See, for example, [9], [lo], or [15].) The basic inspiration for this paper is a remark of Roger Lynaon who once stated that “There is an obvious intuitive connection between the question of commuting elements and cancellation diagrams o n t h e torus, which has yet to be made precise.” It turns out that the connection can indeed be made precise, not only for the basic commutator, but for any quadratic word. In terms to be defined precisely below, we show that a “non-trivial” solution of a “quadratic equation” W = 1 in a group G = (X: R ) induces a “non-trivial” cancellation diagram on a compact surface S defined by an endomorphic image of W. If we assume that R satisfies a suitable small cancellation hypothesis, such a diagram cannot exist, and we can thus conclude that all solutions of W = 1 in G are “trivial”. Consequences of this result are discussed below. It turns o u t that the present methods also yield a very simple proof of the important theorem of Nielsen [I41 that if G is the fundamental group of a compact surface then all t h e automorphisms of G are induced by automorphisms of the corresponding free group. It is a pleasure to acknowledge many helpful conversations with C.C. Edmunds during the research leading to this paper. We now turn to definitions and precise statements of results. Let @ = (a,,a z ,. . . ) be a free group of countably infinite rank. We shall use the letters a, p, y, etc. to denote generators or inverses of generators of @. A word W E @ is quadratic if every generator of @ * This research was supported by the U.S. National Science Foundation and the National Research Council of Canada. 347

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which occurs in W occurs exactly twice. Each quadratic word W defines a closed compact surface as the quotient space of a polygon whose sides are labelled, in counterclockwise order, by W. (We assume that the reader is familiar with the proof of the classification theorem for compact 2-manifolds; see, for example, Massey [12].) It is wellknown that if W is a quadratic word, then there is an automorphism of @ sending W to one of the canonical forms [ a I PI] , * * * [a8, p,] or a:. . * a;. (The proof of t h e classification theorem tells one how to construct the automorphism.) We say that W is orientable (non-orientable)if the surface defined by W is orientable (non-orientable). It is well-known that W is orientable if and only if for every generator ai which occurs in W, the two occurrences of a, have opposite signs. We define the Euler characteristic of W, written x( W ) , to be the Euler characteristic of the surface defined by W. To calculate the Euler characteristic of W, we write W counterclockwise around the boundary of a polygon in the plane, and then count the numbers V, E, and F of vertices, edges, and faces after identifying each pair of edges labelled by the same generator, and set x( W )= V - E + F. If W contains n generators, then there are n edges after identification. If W is a canonical form it is easy to verify that there is only one vertex. Since there is exactly one face, we have x( W )= 2 - n for W a canonical form containing n generators. We shall often use this observation. Let F = (X) be a free group, and let G = (X; R ) be a quotient of F. We shall always assume that the set R of defining relators for G is symmetrized, that is, all elements of R are cyclically reduced and R is closed under taking cyclic permutations and inverses. Let N be the normal closure of R in F, so that G = F / N , and let n- : F+ G be the natural map. We want to investigate solutions of quadratic equations in G. Within the set of all solutions we distinguish those solutions which arise in an obvious manner from solutions in the free group F. Definition. Let W ( a l ,. . .,a,) E @. A solution of the equation W = 1 in G is an n-tuple of words ( a l , . . , a , ) on X" such that W ( a l , . . , a , ) = 1 in G. We say that the equation is quadratic if W is quadratic. We say that the solution ( a l , .. . , a,) is free if there exist words b l , . . . ,b, in F such that n-(b,)= a, for i = 1 , . . . ,n, and W ( b l , . .,b,) = 1 in the free group F. (Note in which groups the equations hold.) For example, since two elements of a free group commute if and only if they are powers of a common element, a non-free solution of ( Y ~ ( Y ~ ( Y ; I ~= ; ~ 1 in G is a pair ( a l ,a z ) representing elements of G which commute but which are not powers of a common element.

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For complicated quadratic equations, exact descriptions of all possible solutions are not known even in the free group F. There is, however, one general result which we shall find very useful. This is the “rank formula” of Lyndon [ l o ] which states that if W ( a , ,. . . , a,) is a canonical form involving n generators and ( a , ,. . . , a,) is a solution of W = 1 in F, then the rank of the subgroup H = Gp{al,. . .,a,} generated by the entries in the solution must satisfy the bound rank(H) S [ n / 2 ] (where, as usual, [ m ] denotes the greatest integer not exceeding m ) . W e shall give a simple proof of the rank formula in the next section. In studying solutions of a particular equation W = 1 it turns out that we must consider a larger class of equations. W e shall denote by % ( W ) the set of all quadratic words of @ which are endomorphic images of W. (If W is orientable then %( W ) consists of all orientable quadratic words which have Euler characteristic greater than or equal to x ( W ) . If W is non-orientable, %( W ) consists of all non-orientable quadratic words of characteristic greater than or equal to x ( W ) , and all orientable quadratic words with Euler characteristic strictly greater than

XW.1

We now turn to the definition of a cancellation diagram M on a compact surface S. First of all, M is a tessellation of S and there is a function assigning to each oriented edge e of M a label + ( e ) in the free group F such that + ( e ) # 1 and + ( e - ’ ) = + ( e ) - ’ . Let e be an unoriented or “geometric” edge of M. If e is on the boundary of two distinct regions D , and D z , we say that e occurs with multiplicity one on the boundaries of DI and D,. If e occurs only on the boundary of a single region D, we say that e occurs with multiplicity two in the boundary of D. If D is any region of M , the degree of D, written d ( D ) , is the number of unoriented edges, counted with multiplicity, in the boundary of D. Let R be a symmetrized.subset of F. W e say that M is an R diagram if for every region D of M there is a closed path 6 with the following properties. All the edges in 6 lie in dD, and for each pair { e , e - ’ }of oriented edges lying in the boundary of D, the total number of occurrences of either e or e-’ in S is equal to the multiplicity with en say, which the geometric edge e occurs in aD. Further, if S = e l and +(el) = c, E F, then the product r = c I . c, is cyclically reduced without cancellation and is an element of R. If we are given one such S then, since R is symmetrized, any path which is a cyclic permutation of S or S-I also has the same properties. W e shall assume then that, when we have an R-diagram M, there is specified for each region D of M a class of paths, called distinguished boundary cycles of D, such

+

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that each path has t h e properties of S above and the class is closed under cyclic permutations and inverses. We now recall the basic “non-triviality” condition on R -diagrams. If there are elements rl = cbl and r2 = cb2 of R with r2# rl then c is called a piece. In the plane, we say that an R-diagram is reduced if the label on every interior edge is a piece. If M is an R-diagram which tessellates a compact surface S, we say that M is reduced if the label on every edge is a piece. Geometrically, this means that if e is an edge on the boundaries of regions DI and D 2(not necessarily distinct) and if Si is the boundary cycle of Di beginning with e, then +(Sl)# +(S2). (In the case that D 2= D1, S1 and S2 are boundary cycles of D1beginning with distinct occurrences of e.)

Figure 1.

The basic connection between non-free solutions of quadratic equations and reduced diagrams is given by the following.

Theorem 1. Let W E @ be quadratic, and let G = ( X ;R). If W is nonorientable, assume that R contains no proper powers. If G admits a non-free solution of the equation W = 1, then there is a reduced R diagram on a surface S defined by some U E %( W). The above theorem requires n o hypothesis on R except the exclusion of proper powers in the case where W is non-orientable. In this generality, we cannot hope to say much about whether or not all solutions of specific equations are free. In recent years, small cancellation theory has become an important part of combinatorial group theory, and it turns out to be precisely the tool needed here in order to obtain sharp results. We now assume that the reader is familiar with the basic notation and hypotheses of small cancellation theory. (See, for example, [9], [lo], or [16].) We need here mainly an understanding of the hypotheses, for we shall actually use very few results.


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Theorem 2. Let W E @ be a quadratic word with x = x ( W) S 0. Let G = ( X ;R ), and assume that if W is non-orientable then R contains no proper powers. Suppose that R satisfies any one of the small cancellation hypotheses C(7- 6x), or C(5 - 4x) and T(4), or C(4- 3x) and T(6). Then all solutions of W = 1 in G are free. We turn to a consideration of some of the consequences of the theorem. Solutions of quadratic equations W = 1 where the Euler characteristic of W is non-negative have previously been investigated by a variety of ad hoc methods.

Remark 1. Let W be a 2 ,which defines the projective plane. Since, of course, x ( W ) = 1, the statement of the theorem does not apply. We shall see from the proof however, that if we assume that R contains no proper powers and satisfies C(6), or C(4) and T(4), we can conclude that G has no elements of even order. It is fairly well known that if R contains no proper powers and satisfies one of the stronger metric hypotheses, either C’(b) or C’(t) and T(4), then G is, in fact, torsionfree. These results were established by Greendlinger [5] and Soldatova [18]. It has been strongly conjectured that the torsion result is still true assuming only C(6), or C(4) and T(4), but the methods here get only “half way” towards this goal. Seymour [17] has shown that if R is finite, contains no proper powers, satisfies C(4) and T(4), and no piece has length greater than one, then G is torsion-free. Chris Chalk [l] has recently proven that if R contains no proper powers and satisfies C(7), or C(5) and T(4), then G is torsion-free. Remark 2. Let W be aPa-’P-’, which defines the torus. If we assume that R satisfies C(7), or C(5) and T(4), we conclude that two elements of G commute only if they are powers of a common element. This result was established by Greendlinger for R satisfying C(!)[6], and subsequently [7] strengthened to C’(i). Comerford has extended these results to small cancellation quotients of free products. Remark 3. Let W be either aPaP-’ or a2P2,which define the Klein bottle. We call an element of G real if it is conjugate to its own inverse. Since the only real element in a free group is the identity, if we assume that R contains no proper powers and satisfies C(7), or C(5) and T(4), then we conclude that G has no real elements except the identity. Equivalently, we conclude that whenever square roots exist in G, they are unique. Comerford [2] established that if R contains n o even powers and satisfies C’(a), or C’(a) and T(4), then G

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has n o non-trivial real elements. Earlier, Gowdy [4] had proved that if R contains no proper powers and satisfies C'(k), then G has n o real elements. Remark 4. Let W be a 2 p Z y z which , defines the non-orientable surface with Euler characteristic - 1. It is a well-known result of Lyndon [8] (and an immediate consequence of his rank formula), that if a 2 b Z c = 21 in the free group F then a, b, and c are all powers of a common element. Thus if we assume that R contains n o proper powers and satisfies either C(13), or C(9) and T(4), the same result holds for G = ( X ;R ) .

For more complicated quadratic equations W = 1 we lack explicit descriptions of the solutions even in the free group F. We do, however, transfer the rank formula to all those groups G = (X; R ) for which we can prove that all solutions of W = 1 are free. Remark 5. Another interesting consequence of the theory is the following result. Let W ( a l , . .,a,) be a canonical form involving n generators, and let S be the surface defined by W. The standard presentation of the fundamental group G of S is G = ( x l , . . . ,x, ; W ( x , ,. . ., x.)). Let V be any quadratic word involving k generators where k < n. Then all solutions of V = 1 in G are free. It is tempting, in view of the known results on torsion elements and real elements, to attempt to weaken the hypothesis that R contains no proper powers to assuming only that R contains no even powers. The following example, due to Comerford, shows that this cannot be done in general, regardless of how good a small cancellation condition is assumed. Let G = ( x,..., ~ x ~ - ~ ; ( x : . - . x : - ~2k+1 ) )

where g > 2 . Let W be a ; . - - a i Then . (xl,..., X ~ - ~ , ( X ; - - * X ~ - , ) is ~ )a solution of W = 1 in G. The subgroup generated by the entries in the solution is all of G, and has rank g - 1. Thus the solution is non-free by the rank formula. The above presentation satisfies an arbitrarily prescribed cancellation condition C ( m ) if k is chosen large enough. It is an important theorem of Nielsen [14] that if G is the r) where r is fundamental group of a compact surface S, say G = ( X ; one the canonical relators, then every automorphism of G is induced by an automorphism of the free group F = (X). (See also Zieschang, Coldewey, and Vogt [19].) Let r = W(xl,. . . ,xm). If 8 : G + G is any


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endomorphism of G, then, of course, t h e n-tuple (8(xl), . . . , 8(x.)) is a solution of W ( a I ,... , a,)= 1 in G. Assuming that 8 is one-to-one, the rank of the subgroup generated by (8(x,), . . . , 8(x,)) is n, the rank of G, and we thus have a non-free solution. Let R be the symmetrized subset of F generated by r. It turns out that if x ( S ) < O , then the possible tessellations of S by reduced R -diagrams are severely limited; indeed, such a tessellation must have only one region. We shall use this fact to deduce that the n-tuple (8(xl), . . . , 8(x,)) is obtained from (xl, . . . ,x,) by Nielsen transformations, and thus 8 is induced by an automorphism of F. 02. Quadratic words and equations

In this section we collect the results which we need about quadratic words and equations in free groups. We assume that the reader is familiar with Nielsen transformations and the fact that they generate the automorphism group of a free group of finite rank. (See [lo] or [ll].)Let @ = ( a l ,az,. . .), and let W ( a l , . . , a,)€ @ be quadratic. Let F = (X) be a free group, and let u = ( a l , .. . , a,) be an n-tuple of elements of F. If we substitute ai for a, in W, we obtain the element W(al, . . . , a , ) = w € F. Thus ( a l , .. . , a,) is a solution of the equation W ( a I ,...,a,)= w in F. We need to consider the process of performing certain attached transformations, which will be defined by the list below, to the triple W, w, and u.The transformations will, in each case, consist of applying an endomorphism of @ to W to obtain a quadratic word U(pl,. . .,p k ) , applying an inner automorphism of F to w to obtain u, and replacing the n-tuple u = (al,.. .,a,) by a k-tuple 7 = ( b l , .. . ,b k )such that U ( b l , .. . ,b,) = u. For ease of notation we adopt the following convention. We use a,p, y,. . . to denote a letter of @, that is, a generator of @ or its inverse. If we are considering a particular transformation, we shall write W(. . . ,a,., . ,p, . . . ) displaying only the generators which are involved in the transformation. In defining an endormorphism of @, t h e endomorphism is understood to fix all generators not explicitly mentioned in the definition. Given a tuple ( . . . , a , . . . ,b , . . . ) it is understood that the element of F denoted by a Roman letter is substituted for the corresponding Greek letter. If we replace the tuple (. ..,a , . , . ,b, . . .) by a new tuple (2,.. .,a , . . ., b, . . . ) it is understood that all entries not explicitly mentioned are the same in the new tuple as in the old, and that new letters denote new entries in the tuple. We write a = bc to mean that a is freely reduced as written, that is, there is no cancellation between b and c.

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Given a quadratic W ( a l , ..,a,) . and a tuple u = (a,,...,an) with W ( a , ,. . . ,a,) = w, a transformation of any of t h e following types is said to be attached to the triple (W, w, u). (1) Suppose that ap occurs in W ( . . .,a,.. . ,p,. . . ) and that a = alz, b = z - ' b , where z # 1. Obtain U ( [ ,. . . , a , . . . , p,. . . ) from W by applying the automorphism of @ defined by a + ap, p + p - ' p where [ does not occur in W. Replace the n-tuple ( . . . , a , . . . , b,. . . ) by t h e (n + 1)-tuple (2,. . . , a , , . . , ,b , , . . .). It is clear that U is quadratic and that U ( z ,. . . , a , , . . . , b , , . . .) = w. Leave w unchanged. (2) Suppose that ap or p - l a - ' occurs in W. Obtain U ( . . .,a,. . . , p,. . . ) from W by applying t h e automorphism of @ defined by a + ap-', and replace the tuple ( . . . , a , . . . , b,. . . ) by the tuple ( . . ., a , , .. . , b,. . . ) where a , = ab. Note that t h e new tuple is obtained from the old by a Nielsen transformation. (3) Suppose that a occurs in W and that a = alz. Obtain U ( [ , .. . , a,. . . ) from W by applying the automorphism of @ defined by a + a[ where [ does not o c p r in W. Replace the n-tuple ( . . . , a , . . . ) by the (n + 1)-tuple (2,. . . , a l ,. . .). Leave w unchanged. (4) Suppose that W begins with a. Obtain U from W by applying the conjugation W + a-'Wa, and do not change the tuple ( . . . ,a , . . . ). Replace w by a-lwa. ( 5 ) Suppose that a occurs in W ( . . . , a,.. . ,p, . . . ) and that a = 1. Obtain U from W by the endomorphism of @ defined by a + 1. Replace the n-tuple ( . . . , a , . . . , b,. . .) by the (n - 1)-tuple (. . . , b,. . .). Leave w unchanged. We call a transformation of type ( 5 ) a singular transformation. Suppose that we are given a quadratic word W and a solution u = (a,,. . . ,a,) of W ( a , ,. ..,an) = w in F. We say that the equation V ( p , ,. . . , p k ) = u and t h e solution 7 = ( b , ,. . . , b k ) are derived from W, w and a by a sequence of attached transformation if there is a sequence

(W, W ,u)= (Woi ~

0 ~ ,o

) +

*+

(Wn, w n , a n ) =

(U, u, 7 )

where each (W,,,,w , + , ,a,+l) is obtained from (W,,w,, a,)by a transformation attached to (W,, w,,a,). If a E F, then la I will denote the length of a relative to t h e fixed basis X of F. We define the length of an n-tuple by adding the lengths of the entries, I ( a , , .. . , a , ) [ = Z:=, 1 a, 1. Note that if we derive the tuple ( b , , .. . , b,) from the tuple (a,,. . .,a,) by an attached transformation of ~ ,an)[. type (l), then I(bl ,..., b , ) / < ( ( a,... Let W E @, say W = a,,* * * a,m.We say that the n-tuple (a,,. . . , a , ) (of elements of the free group F) is cancellation-free in W if no a, = 1, and if w = W ( a , ,. . . ,a,) = a,, . . . a," is cyclically reduced as

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written. In other words, no cancellation occurs, even cyclically, between any of t h e a,. Thus ( x y x , y x y ) is cancellation-free in azpz,while (xy-I,y x ) is not cancellation-free in a’@’. Lemma 1. Let W ( a ,..., I a , ) € @, let ( a ,, . . . , a n ) be an n-tuple of elements of F, and let W ( a , ,. . . , a n )= w. Then there is a triple ( U ( p ,..., , pk),(bl, . . . , b k ) , u )derioed from ( W , ( a l ,..., a n ) , w )by a sequence of attached transformations such that u is a cyclically reduced conjugate of w and ( b , , . . . , bk) is cancellation-free in u. Proof. Suppose that a,a,occurs in W while a, = a : r and a, = z - l a ; with z # 1. Then apply the transformation of type (1) where U is obtained by applying t h e automorphism a, -+a,( and a,+ ( - l a , and the tuple ( . . . , a,, . . . , a,, . . . ) is replaced by the tuple (2,. . . , a : , . . ., a ; , . . .). In this case the length of the new tuple is less than the length of the old tuple. If any of the entries ak = 1, apply the appropriate singular transformation sending a k + 1. If W ( a , ,. . . , a,) is not cyclically reduced, a conjugation of type (4) brings us to t h e case considered above. Since each transformation of type (1) produces a tuple of shorter length, a finite number of transformations must lead to a U and ( b , , .. . , b k ) which is cancellation-free in u.

Lemma 2. Let W be a quadratic word of @, and let V E @ be any quadratic word which is an endomorphic image of W. Then a cyclically reduced conjugate of V is obtainable from W by a sequence of attached transformations. Proof. By assumption, there are words Y,, . . . , Y,,of @, not necessarily By Lemma 1, some cyclically quadratic, such that V = W (Y , ,. . ., Yn). reduced conjugate V* of V is obtaina,ble by a cancellation-free substitution ( Z , , .. . , Z k )into tr(pI,.. . , p k ) where U is derived from w by a sequence of attached transformations. Since both U and V are quadratic and the substitution is cancellation-free, a generator which occurs in some Z , can occur only in that Z , . By transformations of type (3) we can “subdivide” each pi into a product of (Zi1 generators. If the result is U * ( y , ,..., y m ) then V* is obtainable from U * by a letter for letter substitution and thus differs from U * only by a permutation of the generators and their inverses. Such a permutation is obtainable by a sequence of attached transformations. 0

The following lemma is well-known.

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Lemma 3. Let W E 4j be a quadratic word. If U is obtained from W by an attached transformation which is an automorphism of 0,then ,Y( U ) = x ( W ) . If U is obtained from W by applying one singular transformation then x ( W )s y, ( U )S y, ( W )+ 2. In view of Lemmas 1 through 3 we have the

Corollary. I f W is quadratic and U is a quadratic endomorphic image of W , then x ( U ) a , y ( W ) . Let F be a free group. Recall that a sequence y l , . . . , y , of elements of F is said to be Nielsen-reduced if the following three conditions hold. (Nl) Each y,# 1. (N2) More than half of y , does not cancel in either of the products y,-,y, or Y , Y , + ~ . (N3) No y , cancels exactly half in both products y,-,y, and y , ~ , + ~ . A set S of elements of F is said to be Nielsen-reduced if every sequence y I , . . . , y , of elements y, E S" such that n o y,+, = y;' is Nielsen reduced. If T is a finite subset of F, then a finite number of Nielsen transformations lead from T to a set S which is Nielsen reduced. (For a fuller discussion see [lo].) We shall need the following relativization of the concept of being Nielsen reduced. Let W ( a l , .. . , a,)E @, say W = a :,'* . * a : I where each E , = k l . Let ( a l , . .. , a , ) be an n-tuple of elements of F. We say that (a,,. . . ,a,) is Nielsen-reduced relative to W if the particular sequence a:,', . . . ,a:; satisfies conditions (Nl) through (N3). In short, we consider only the products which result from substituting the a, into W. Transformations of types (2) and ( 5 ) effect Nielsen transformations on the tuples involved. It is easy to see that given any W ( a , , a , ) E @ and any n-tuple ( a l , .. . , a n ) we can, by a sequence of attached transformations, of types (2) and ( 5 ) , obtain u(&,. . . ,p k ) and tuple (b,, . . . , b k )which is Nielsen-reduced relative to U. Lemma 4 (Lyndon's Rank Formula). Let W E @ be a quadratic word which involves n generators and which is of minimal length among its images under automorphisms of @. Let F be a free group, and let (a,, . . . , a n ) be any n-tuple of elements of F such rhar W ( a , ,. . . ,a,) = 1. Then the rank of the subgroup H generated by { a l , .. . , a.} is less than or equal to [ n / 2 ] .

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Proof. It is well-known that any canonical form has minimal length among its automorphic images. Let W * be the canonical form which is an image of W under an automorphism of @. Since W is of minimal length under automorphism, 1 W*l = I W I , and thus W * involves n generators. Hence, x ( W )= x ( W * )= 2 - n. There is a sequence of attached transformations of types ( 2 ) and ( 5 ) leading from W and ( a l , .. . ,a.) to U ( p l , .. . , p k ) and a tuple ( b , ,. . ., b k )which is Nielsen reduced relative to U and with U ( b l , .. . , b,) = 1 . Now the product of elements in a Nielsen reduced sequence y,, . . .,ym with m 3 1 cannot be equal to the identity. We conclude that U = 1. Let d be the number of singular transformations applied in obtaining U from W. Since a singular transformation is applied only when an entry in the tuple is the identity, it is clear that rank(H) z s ( n - d ) . Since x ( W )= 2 - n, and x(1) = + 2, in going from W to U the Euler characteristic is raised by n. According to Lemma 3, each automorphism preserves the Euler characteristic, while one singular transformation can increase the Euler characteristic by at most two. Thus the number d of singular transformations must be at least n / 2 if n is even, and at least ( n + 1)/2 if n is odd. Hence, rank(H)S(n - d ) S [ n / 2 ] . 0 As usual, let F = ( X ) be a free group, and let G = (X; R ) be a quotient group of E Let W ( a , ,. . . , a , ) € @, and let ( a l , .. . , a , ) be a non-free solution of W = 1 in G. In the next section we shall transform given equations and solutions, and we shall need to know that we preserve the non-freeness of the solutions. All the situations which will be encountered are covered by the following lemma.

Lemma 5. Suppose thar ( a l , .. . , a , ) is a non-free solution of W ( a l , . ., a , ) = 1 in G. Suppose thar there are words Z, ( P I , .. ., p k ) of @, and a k-tuple of ( b , ,. . . , b k )of words of F such that a, = Zi(bl,.. . , b k ) in G for i = 1,. . . , n, and U ( p , ,. . . , P k ) = W ( Z l ,. . . , Z n ) in @. Then ( b l ,. . . ,b , ) is a non-free solution of U = 1 in G .

Proof. If the solution ( b ] , . . , b , ) of U = 1 in G were free, there would be words b : , . . . , b : such that each b f = b, in G and U ( b : , .. . , b : ) = 1 in F. But Z , ( b : , . .., b : ) = a, in G and

W ( Z l ( b : ,..., b : ),..., Z . ( b : , .. ., b : ) ) = U ( b : , . . . , b : ) = 1 in F. This contradicts ( a l , .. , , a n ) being a non-free solution of W=l. 0

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%3.Cancellation diagrams on compact surfaces

We now assume that the reader is familiar with the basic facts about cancellation diagrams in the plane. (See Chapter V of [lo].) Let F = (X) be a free group, let R be a symmetrized subset of F, and let G = (X; R). Let N be the normal closure of R in F, so that G = F / N , and let T : F + G be the natural map. Let W ( a , ,. . . , a , ) E 4j be a quadratic word, and suppose that there is a non-free solution ( a l , .. ., a , ) of W = 1 in G. Note that W ( a l , .. ., a , ) is a non-trivial element of N. We consider the following

Choice. Among all U E 8( W ) and all tuples of words (bl, . . . , b k ) which are non-free solutions of I/ = 1 in G, pick a pair U and ( b l , . ..,b k ) such that (1) The number, rn, of regions in t h e minimal connected simply connected R-diagram M in t h e plane with boundary label U ( b l , . . , b k ) is as small as possible. (2) Subject to (l), the length of the free group element U ( b l , .. . ,bk) is as small as possible. (3) Subject to (1) and (2), t h e length l(bl, ..., b k ) l of t h e solution tuple is as small as possible. . and ( b l , . . . ,b k ) refer to t h e pair In the next sequence of lemmas, U chosen above, and M is a minimal connected simply connected R diagram in the plane with boundary label U ( b l , .. . ,b k ) . Lemma 6. The tuple ( b , , . . . , b k ) is cancellation-free in U. Proof. If cancellation occurs we can, by an attached transformation as in Lemma 1, obtain V ( y l , .. . , y m ) and a shorter tuple ( c l , .. . , c”). Since the free group element u = V(cl, . . . , c,) is a conjugate of u = U(b1,.. . , b k ) , the minimal R-diagram with boundary label u still has only rn regions. This contradicts our minimal choice. 0 By using transformations of type (3) we can “subdivide” the variables so U ( p , ,. . . ,& ) is such that each entry bi in ( h , . . , b k ) is a generator or the inverse of a generator of F. We now assume that this has been done.

Lemma 7. The boundary cycle of M is a simple closed path. Proof. Suppose that M is not bounded by a simple closed path and let K be an extremal disk of M. (See [lo] or [13].) Both K and M \ K


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contain regions. Since each letter of U is replaced by a letter from XI',the label o n t h e boundary of K is a substitution instance of V ( p , , , ..,&) . where some cyclic permutation of U has the form VZ. First suppose that V is quadratic, so that for each p, which occurs in V, both of t h e occurrences of p, are in V . Then both V and Z are quadratic. Substituting the appropriate b, into V and Z yields solutions of the quadratic equations V = 1 and Z = 1 in G. If both of these solutions were free, then ( b , ,..., b k )would be a free solution of U = 1 in G, which is not the case. On t h e other hand, if at least one solution is non-free, say that of V = 1, then there is an R-diagram, namely K, with fewer than m regions verifying that V (b , ,., . .,b,")= 1 in G. This contradicts t h e minimality of our choice. We are left with the possibility that some generator p occurs exactly once in V. Write V = V l p V 2 .Since V(b,,,. . . , b,[)is t h e boundary label of an R-diagram, it follows that V( b , ,., . . , b,,)= 1 in G. Let u1= V2(b,,, . . . , b,,),and u 2 = V 2 ( b , , ., . ,b,[).Solving for 6 we have b = u ; ' u ; ' in G. Obtain U * from U by replacing p by V ; ' V ; ' and replace ( . . . , b , . . . ) by ( . . . , u ; ' v ; I , . . . ). Obtain the R-diagram MI from M by removing t h e extremal disk K and reattaching K by sewing, in the appropriate orientation, the edge labelled by p in K to t h e edge labelled by p in M \ K . Now M 2 has the same number of regions as MI, and a boundary label of MI is u * = U * ( .. . , u;Iu;', . . . ). But u * is shorter than u. This contradicts the minimality of our choice, so we must conclude that M is bounded by a simple closed path. 0 Since t h e label on the boundary of M is a cancellation-free substitution instance of U, we can regard aM as a polygon whose boundary "has t h e form U". We can then identify the edges in aM to

M

Figure 2.

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obtain an R-diagram, which we also call M , on the surface S defined by U. Let D be a region of the diagram M in the plane, and let 6 = e l * * e, be a boundary cycle of D. Considering D as a region of the diagram M o n the surface S, a distinguished boundary cycle of D will be the sequence e l * et after identification.

Lemma 8. The diagram M is still reduced on the surface S provided that, if W is non-orientable, we make the additional assumption that R contains no proper powers. Proof. Recall that M is reduced if, for any edge e on the common boundary of regions D1and D 2(which may coincide), the labels +(al) and +(az) are not the same, where Si is the boundary cycle of D i which begins with e. Since the diagram M was reduced in the plane, we need only consider the edges identified according to U. Suppose that the identification of two edges e , and e2 o n the boundaries of two distinct regions makes the diagram not reduced. (See Figure 3 which pictures the orientable case.)

M

Figure 3.

Obtain the diagram M' in the plane by removing the two edges e l and e2. Now M' has boundary label equal in G to U ( . . . , b,. . .), for if b is the label on the edge e, and the boundary cycle S of D has label +(a) = bh E R, then we have replaced b by h-I. The diagram M' has fewer regions than M, contradicting the minimal choice. It is conceivable that two edges of the same region D are identified in such a way that the diagram becomes not reduced on the surface S. If the identification of the two edges is an orientable identification (in the obvious sense that the edges are labelled by b and b - ' ) then the diagram being non-reduced means that the label r on D is conjugate to r-' in the free group F, which is impossible.


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Figure 4.

There remains the possibility of a non-orientable identification making the diagram not reduced. (See Figure 5 . )

Figure 5 .

Let D have boundary cycle e l v l e 2 q 2where c#J(e,)=b = c$(ez), and c#J(v2) = z. The diagram being non-reduced on S means that bybz = bzby in the free group E Thus y b z y - ' b - ' z - ' , which is equal to the commutator ( y b ) ( z b ) ( y b ) - ' ( z b ) - 'is, equal to 1 in E Since this implies that yb and z b are powers of a common element, the label r = bybz must be a proper power, but we have assumed that this is not the case.

c#J(vl)= y ,

We have completed the proof of Theorem 1 and now consider small cancellation theory. In our constructions so far, it has been convenient to assume that edges are labelled by generators or their inverses. We now want to d o t h e opposite, eliminating as many unnecessary vertices as possible. If u is a vertex of a diagram M, either in the plane or on a surface, we use d ( u ) to denote the degree of 11, that is, the number of edges incident at u, where, if both endpoints of an edge are a? u, we count t h e edge twice.

Lemma 9. Let M be an R-diagram on a compact surface S, where M is obtained by {dentifying the boundary of an R-diagram in the plane as in the proof of Theorem 1. (Thus S is not the sphere.) W e can modify M so

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P.E. Schupp

that M has no vertices of degree less than three unless S is the projective plane. If S is the projective plane, we may assume that M has at most one vertex of degree two. Proof. Consider t h e diagram M in the plane before identification. Since M is an R-diagram with cyclically reduced boundary label, M has no vertices of degree one. If u is an interior vertex of degree two, say u separates edges e l and e2, we can simply delete u and combine e l and e2 into a single edge e with label 4 ( e ) = 4(el)4(e2). Suppose that there are successive edges e l and e2 in dM separated by a vertex u which has degree two in the plane, and where the labels on e l and ez correspond to successive occurrences of a and p in U with p # a?'.Now u will have degree at lest three after identifying the boundary edges of M unless the edges e3 and e4 corresponding to the two other occurrences of a and p in U are situated so that the end of e , is the beginning of e4. This means that in U (or a cyclic permutation of U ) the occurrences of a and p are in two subwords of the form (ap)%'.We can replace these occurrences of ( Q P ) ~ by ' y" where y is a new variable not occurring in U . Delete the vertex u, combining e l and ez into a single edge e with label 4 ( e ) = +(el)4(e2). Similarly, combine the two edges corresponding to the other occurrence of ( a p ) = l into a single edge. (In short, all that we have done is to eliminate an unnecessary subdivision.)

Figure 6.

Iterate the above process, eliminating as many vertices of degree two as possible. The only situation not covered is that of a vertex u separating edges e l and e2 corresponding to successive occurrences of a letter a in U. Now u will certainly have degree at least three after the identification of dM unless e l and ez are all of dM. In this case, U is a2,the surface S is the projective plane, and u is the only exceptional vertex. 0 From now on, when we consider a reduced R-diagram M tessellating a compact surface S, we assume that M has no vertices of

Quadratic equatwm in groups

363

degree less than three; except for the possibility of one exceptional vertex in the case of the projective plane. Since M is reduced, the label on every edge is a piece and t h e small cancellation hypotheses have their usual geometric meanings. (See [9] or [ l o ] . )If R satisfies the condition C ( k ) , then no element of R is a product of fewer than k pieces, and thus d ( D ) a k for every region D of M. If R satisfies the condition TO') and u is a vertex of degree at least three, then, in fact, d ( u ) 3 j . Thus every vertex of M has degree at least j , except possibly the one exceptional vertex in the case that S is the projective plane. The basic tool of small cancellation theory is the summation formula given below. We use the following convention on summation signs: if M is a fixed diagram, CM [ p - d ( D ) ] denotes the sum of [ p - d ( D ) ] for all regions D of M, while E M [q - d ( u ) ] denotes [q - d ( u ) ] summed over all vertices u of M. We shall usually omit the subscript naming the diagram.

Lemma 10 (The summation formula). Let M be a tessellation of a compact surface S. Let p and q be positive integers with l l p + l/q = 112. Then

Proof. Let M have V vertices, E edges, and F regions. (This is our only departure from using F to denote a free group.) Euler's formula is

(1)

x ( S )= V - E

+ F.

Since we have counted degrees with the appropriate multiplicity,

(2)

2E

=

2 d ( u )= c d ( D ) .

Let n be a positive real number. We have

(3)

2(n + l)X(S)= 2(n + l)V - 2(n + l)E + 2(n + 1)F.

Using (2), we can eliminate E from (3), (4)

2(n + l ) X ( S )= 2(n + l ) F -

2 d ( D ) + 2(n + l)V - n c d ( D ) .

Since V is the number of vertices and F is t h e number of regions,

P.E. Schupp

364

Let p = 2(n + 1) and q = 2(n Substitution in (5) yields

+ l)/n. Then

n = p / q and l/p

+ l/q = 1/2.

which is the desired formula. 0 We can now prove Theorem 11. Suppose that we start with a quadratic equation W = 1 where x = x ( W ) is not positive; and a nonfree solution of W = 1 in the group G = (X; R). If W is nonorientable we assume that R contains no proper powers. Then there is a reduced R-diagram M on a surface S defined by an endomorphic image U of W. Thus x ( S ) = x ( U )a x ( W ) . Consider the summation formula with @, q ) = (6,3).We have

6x(W)"6x(S)=C [6- d(D)]+2 C [3- d ( ~ ) ] . M

M

The basic idea is simply to assume a good enough small cancellation condition to force the right-hand side to be less than p times x ( W ) . Supposing that S is not the projective plane, all vertices of M have degree at least three. Since M has at least one region, the equation (*) is impossible if R satisfies the small cancellation condition C(7- 6 x ) . If S is the projective plane, x(S)= + 1, and we have

6=

(**)

c [6 M

+ 2 C [ 3 - d (U )].

- d(D)]

M

Since there is at most one vertex of degree two, this equation is impossible as long as R satisfies C(6). In summary, if R satisfies C(7- 6 x ) then the required diagram M cannot exist. We conclude that there cannot be a non-free solution of W = 1 in G. By letting @, q ) = (4,4), we see that it is sufficient to assume that R satisfies C(5 - 4 x ) and T(4). If we set (p, q ) = (3,6),we see also that it suffices to assume that R satisfies C(4- 3 x ) and T(6).

64. Automorphisms of surface groups Let W ( a I,..., a,) b e o n e o f thecanonicalforms[al,az]...[a.-l,a,] or a:. . at. Let F be the free group on X = { x l , .. . ,x,,}, let r = W ( x , ,. . . ,x,,), and let R be the symmetrized closure of r in F. We say that (X; R ) is the standard presentation of the fundamental group G of t h e surface S defined by W. We first observe some facts about the standard presentations.

-

Quadraiic equaiions in groups

365

Lemma 11. Let ( X ;R ) be the standard presentation of the group of the surface S defined by W(crI,.. . , a n ) .Then R satisfies both of the cancellation conditions C ( 2 n ) and T ( 2 n ) . Proof. We prove t h e lemma in t h e case where R is the symmetrized closure of x i . . . x'.; t h e orientable use is analogous. T o verify C ( 2 n ) it suffices to observe that any two-letter word, of the form x: or X , X , + ~ , has at most one ocurrence in any element of R. Thus a piece consists of only one letter. To verify T ( 2 n ) our assumption is that we have elements T I , . . .,r, of R with each r,+' # r ; ' , r h # r;', and that there is cancellation in each of the products r l r z , r2r3,. . . , rh-'rh,rhrl. Now h > 2 and one of r I or rz must be of t h e form (xf . . . x'.xX:. ~ f - ~ ) and * ' the other of the form X : - ~ X , ) ~ ~By . symmetry, we may suppose that rl = x i * * x'.. (X,X:+~ Then we must have rz = X;'X,!~ * * * x;'x;'. Consequently, r3 = XZ.xl2 . .. x i - , . Continuing thus, we see that it takes at least 2n steps to obtain an r h which can cancel against r l . Thus h 2 2 n . 0

Lemma 12. Let G = ( X ;R ) be given by the standard presentation of the fundamental group of the surface S defined by a canonical form W ( a I ,...,a,)where W inuolues n > 2 generators. Let V ( P l , . . , P k ) be a quadratic word inuoluing k generators where k < n. Then all solutions of V = 1 in G are free. Proof. The canonical form V* to which V is equivalent under an automorphism of @ involves at most ( n - 1) generators. Thus x ( V ) s 2 - ( n - 1) = 3 - n. If G admits a non-free solution of V = 1, then there is a reduced R-diagram M on a surface S * defined by an endomorphic image of V: Using t h e summation formula with (p, q ) = (4,4) we have

4(3-n)includes Mu.In fact, every element of Ma is a product of length no elements from S, U { a } (where no is the length of the r in Fact 2.4). (iii) For limit S, M~= Uac8Ma. We do the induction step later, and now let us prove that M = u,,,+M, is as desired. Suppose by contradiction that M has a proper sub-semigroup, M*, of cardinality A'. Clearly, for some a(0)< A + , S = M *r l Ma(o)has cardinality A, and let a E M - M * ,for some a ( l ) , a E Ma('). Now S is a subset of A' of cardinality A, hence for some a(3), S = &). Let a = max{a(O), a(1), ( ~ ( 3 ) )Since . M * has cardinality A + , it is not included in Mu and hence for some b E M *- Ma.So necessarily for some p 2 a, b E MB+'- MB. Thus by (ii) (p, a(3) here includes stands for a,y there) the subgroup generated by b and

384

S. Shelah

and hence a belongs to it. But b E M*, S,(9 C M*, hence a E M*, contradicting its choice. This finishes the proof of 2.1(1). As for part 2 of Theorem 2.1, M is a Jonsson semi-group as just proved; moreover checking how the group generated by b and S,(3) includes M, (by (ii)) we see that we have proved that if S is any subset of M of cardinality A', any element of M is equal to the product of no elements of S, here no is the length of the word r from Fact 2.4 and is a fixed natural number. From 2.1(2) only the simplicity remains, but if a E M, a not t h e unit, for some a , a E Ma,so for every /3 a'cy a has a conjugate in M,+l-M, (as M,+I is a malnormal extension of M,, for each b E MScl- M, we have bab-'E M,,, - M,). Hence the set of conjugates of a has power A + , so the subgroup they generate is the whole group M. In other words t h e normal subgroup a generates is M, so M is simple.

Ma.Therefore it includes

The induction step 2.6. So M a is given and we should define Ma+,. Let < A } be a list of t h e elements of Ma.Let {(a,, y,, b,): /3 < A } be a - Ma ( M a + ,is not list of all triples (a, y, b), b E Ma,y S a, a E yet defined, but - Ma is A, = (5 : A(1 + a )S 5 < A(l + a + 1))) and each triple appearing A times. (This can be done by cardinality considerations.) Now we shall define groups Lo, H, (/3 < A ) (more = uPc,L,. accurately L;, H ; ) and later let Ma+, We define them such that: (a) IL,I n w eg e t y p ' . * . y ; ' ~ ; i - ~ y ~ P. . 2- =y ke = y p ' . . . Y ; ' z " - " Y ~ .* . yP-I(ypk2). As before, zT-"#e. As y l . - . z l . . . are in K U L - H , if k 2 E H , z"-"tZ H then the word is in canonical form. If k 2 E H, z"-" E H , y l E L - H, by t h e madness condition w = y,'...y;'(y;'z"-"yl)y2...y p-l(ypk2) is in canonical form ( z m - " # e as K is torsion free) when p > 1, and (y;lzrn-"ylh2)EL - H is in canonical form when p = 1. When k2 $Z H we have a similar situation. So in all cases in t h e word w is not e in L * . As it is e it should be long, SO we can make it weakly cyclically reduced by small changes and we get easy contradiction by t h e strong 4000-randomness. The case k l , k 2 # e is similar. Case 11. q > 1 hence q is even. We get that

e

-

k l x m k 2 x - "= k l y , ' . . y ; ' z l . . . z , . . . z I z,yl * . . ypk2 y p ' . . . y -I ' z'l- 1 . . . z i t . . . . z;'y,. . . yp = d f w * *

=

The word is not weakly cyclically reduced only, possibly in k l , k2,y;'zI,z4yI,y;'zq', z;'y,. But if kl,k z # e, the needed changes involve few letters (much less than a hundred) so we ignore them. Let w * be the weakly cyclically reduced form. If k , = k 2 = e we get e = x"-" = z I . z, . z l . z,. If k l = e # k 2 -

e

= Xm-"k2=

a

y;' .

y;'z,

. . -2, . .

e z , .

. .z q y l .

*

- y,k2.

Both cases are easier, and we leave them to the reader. So by 1.9 there is t = tl t, a part of w * and of a word w from R I t 1 a 997 1 w I/ 1000.

---

O n a problem of Kurosh

389

If for some natural number j > 1 w 1 / 1000, 1 < i < i < i + j < 9, t contains 2, . . . z,+, from two copies of z, we get a contradiction to t h e strong 4000-randomness of r. Similarly if t contains z1 * z,+,, 1 z -,+, . . . z ;'. Also if for some j > 1 w 1 /4000, 1< i < i + j < p, t contains y ;+!,* . y ;', and y , * . . y , + , (or two copies from one of them) we get a contradiction. But t is a part of w * , so looking at it we can see that if it intersects two among the four copies of y, y - ' in w * with length > I w 1/4000, we get a contradiction to the above. With the one left its intersection has length =G I w ) / 2 (by A). Necessarily the sum of its intersection with z", z - " is Z= (w1(99/100-3/4000- 1 / 2 ) > 4 9 6 ( w ( / l O O O so with one of them, e.g. z'", it is a 2 1 w ( / 1 0 . If the length of z is < I w I/lOO, then for some i, 1 < i < 1, for every (or t,+,-J contradicting the strong j , i S j < i + 1 w 1/10 t, is equal to 4000-randomness of r. So the length of z is > 1 w ) / l o o , hence t cannot contain two copies of z (see above). So summing our observations we have the following possibilities only : a) t is contained in y - ' z z , and is not disjoint to y - l . So its intersection with the second z has length < I w ( / 1 0 0 0 , SO y - ' z contains a subword of w of length > (91/100 - 1/100))w I = 91 w 1/10, contradiction to (B). b) r is contained in zzy and is not disjoint to y . The same contradiction. c) r is contained in zz, so it is z:z,+1* . . Z , Z l . . . z,-1z:.

We get contradiction to (C). d) t is contained in z z y k 2 y - ' z z (or similarly with k l ) . We can get similar contradictions. (iv) Left to the reader. Proof of Theorem 2.9. The construction is as in 2.5, but the list {S, : y < A'} is n o longer necessary; and we make the changes mentioned in 2.7 and assume all groups are torsion-free. The main point is that in the induction step 2.6 we use Fact 2.11 rather than Fact 2.4. So in 2.5, condition (ii) is replaced by (ii)' For every a E Ma+, - Ma,for an S C Ma which is not included in Mas (a, S ) . Let us prove that t h e a finitely generated subgroup of Ma, definition of Ma+l(see 2.6) satisfies this. So let b E Ma,and we should prove b E (a, S ) . Clearly for big enough n, a E L., b E H.; and we can find m 3 n such that

S. Shrlah

390

Sf (H,,,+, l - H,)# 0 (as Ma= u,,H,, but for no m is S C H,,,). Now - If,,,).As we have used Fact 2.11 in La+], choose c E S n (H,,,+l b E (a, c ) ; so we finish.

Proof of Theorem C. We assume 2*0 = N,, and let M be the group is a non-trivial topology which makes M a from 2.7. Suppose topological group. Choose a neighborhood Uo# M of the unit, and define inductively neighbourhoods U, of the unit such that Un+lC U,, and x , y E 3 xy E U,. If U,,, is uncountable, every element of M is the product of no elements of it and hence belongs to Uo, so U o = M. Contradiction. If U,,, is countable, it is a subset of Ma for some a < ol. Choose x E M - M a ;xU,,,x-’ is necessarily open, so U,,, n xU,,,x-’ is a neighbourhood of the unit; but by the malnormality condition (Ma SmM,see 2.5(i)) this intersection contains the unit alone. So any singleton is an open set, so any subset of M is open, a contradiction. is either {M,0} or the discrete topology. We can conclude that

v

Proof of Theorem D. Let M be the group of Theorem 2.9. Clearly we cannot find a strictly increasing sequence of sub-semi-groups M,, M = u.,,M, (as some M , is necessarily uncountable, so M , = M). Hence if K is a field, K ( M ) the group-ring, K ( M ) = U.,,A., then M = (A,, n M). Now A,, fl M is a sub-semi-group of M. Clearly M E A,, so A, n M is strictly increasing and get a contradiction to the previous observation.

u,,,

Additional information 2.15. Simplicity of Jonsson groups Macintyre has shown that in fact “almost” any Jonsson group is simple. More exactly, there is no Jonsson abelian group, hence M has centre Z ( M ) of power < [ M I . For any a E M - Z ( M ) , its centralizer again (is by its choice # M, hence) has cardinality < ( M I ; so the number of conjugates of a is 1 M 1, so the normal subgroup it generates is M. So M / Z ( M ) is a Jonsson group, and is simple. 2.16. Jonsson groups with centre This naturally raises the question whether there are Jonsson groups with a non-trivial centre. We can repeat the proofs and constructions of 2.1, 7, 9 by starting with an abelian group Z , IZ 1 S A, and change the definitions and requirements accordingly. So all groups will extend

O n a pmblem of Kumh

391

Z, and Z will be in their centre; H =%L means a E L - H, b E H Z j aba-' 62 H, etc. S o the generalization will be easy. 2.17. On torsion Jonsson groups The groups we constructed are torsion free, and moreover satisfy x" = y" x = y for n# 0. We may like to build torsion Jonsson groups. Now free products with "good" amalgamation for some torsion groups (i.e. the given groups and the result are torsion) exists (by Adian [l]) but not with small cancellation, so we can only hope.

+

2.18. Jonsson group in Nz The proof of Theorem 2.9 works also for A = Nz without any CH but for any N,, we need more complicated amalgamations, and the situation is not clear.

Appendix Another proof of Theorem 2.9 avoiding 2.11. This proof is due to G. Hesse. Lemma. Let H , K , L be groups such that K n L = H and H S,,, L, and let 0 be a subset of H x (K\ H ) X ( L \ H)* satisfying : (i) If (h, a, b, b') E 0,then b, b' are good fellows over H. (ii) If ( h l ,a', bl, bl) and (h2,az, bz,b;) are different elements of 0, then at least one of the pairs bl, bz and bl, b; is a pair of good fellows. Suppose L * = K * H L , w ( x l , x z )is the word xlxzxlx:...xlx?, R C L * is the symmetrized closure of {h-'w(ba, b ' a ) ((h,a, b, b')€ a},N is the normal closure of R in L * and L** = L*/N. Then: (1) R satisfies C'(l/lO), and the natural map 7r : L*+ L** embeds K and L, so that their intersection does not increase. (2) K s , L * * . (3) I f a E K \ H , b E L \ H a n d O S m < n € o , r h e n (ab)",(ab)"are good fellows over K in L * . (4) If b,, bz E L \ H are good fellows over H, then b l , bS are good fellows over K in L**. Proof. (1) Let g * = g" * * * g', g,. = gl * . . g, be canonical representations of elements of R such that g*g, # e, and let g' * g 'gl * * gr € H for some 1 S min{m, n } =: p. An easy computation shows m, n E { k , k + l}, where k = 6640. We may assume that there are

-

-

392

S. Shelah

elements (hl, a , , b l ,bl) and (hz, az,bz,b;) of 0,canonical representations h;'w(blal,bIal)= u l * * * Uk and hi'w(bzaz,b h z ) = V, * V,, p , q E (1,. ..,k } and 6, E E { - 1,1} such that (a) g ' = uf-&, g, = VG+e,(1 c i < p ) , (b) g'g" * * * g' = U ; - ~ U : - & - * u:-+ g , * * * g,,gl= V;++ * * * VG+akVG+L, where u, = u, and V, = V, when i = j mod K. Case 1. p # q or E = 6. Then by (i) and the choice of w we can find i E (2,. . .,664) such that g' = u;-& and g, = VG+e,are good fellows over H. Hence g'hg, H for every h E H, and we have 1 C i d 664 C p /lo. Case 2. p = q and E = - 6. We find i , j E (2,. . . ,664) such that w.l.0.g. g ' = u& = bye, g, = V;+.,= b;, g' = u&= b!-' and g, = V;+,= bl-e. If we assume 12664, then none of the pairs b,, bz and bI, b; is a pair of good fellows. By (ii), ( h , , a , , b,, b : )= (hz,az,bz,b;) and w.1.0.g. u, = V,( i = 1,. . . , k). Choose u E {2,3} and b E L \ H with g' = b-', g . = be. Since by assumption b-'(g"-' * g'g, . . . g,-,)b" E H, by the malnormality of b over H we have g "-I * g'g, . . g V - ,= e. Therefore by (a) and (b):

--

-

-

g * g * = g" = g"

.. .g'g'-'. . . g " g y . ..g,-,g,. . . g , . . . g'g' . . . g, . . .g . . . gigm . g'g, . . - g,g, . g')-'(g,. . . gV-,)-' = e,

= (g v-I

')-Ig

"-1

* *

*

-

*

g"-l(gl . . . gJ'

-(gu-l..

a contradiction. The embedding property follows now from 1.9. The proof of (2) is exactly the same as the proof of Fact 2.4(ii). (3) Suppose n, n E w, rn # n, E E { - 1, I}, a , , az E K and (ab)"a,(ab)""az= e in L**.Let w = g l . * . gr be a canonical representation of the element (ab)"al(ab)''"azof L *. Obviously there is at most one i E { l , . . . , l } such that gi, b are good fellows over H in L. Since m # n, we have 1 2 1. Thus by (1) and the Main Theorem 1.9, g , . . . gr contains a long subword W o which is also a piece of some w , E R. It follows from (i) that there are at least two i E (1,. . .,I } such that gi, b are good fellows over H in L, a contradiction. The proof of (4) is left to the reader.

Proof of Theorem 2.9. As before M is constructed as the union of an increasing continuous chain ( M , a < N,) of countable groups, where no M, is finitely generated If M , has already been constructed, we choose a strictly increasing chain ( H , n E w ) of finitely generated

I

I

O n a problem of Kurosh

393

I

subgroups such that H o = {e} and M, = u { H , n E w } . By induction, we now define a strictly increasing chain (L, n < w ) of groups with the following properties: (a) I L n ) = N o . (b) L, n M, = H k > m , and @ ( A t+,)= A This can be shown for E and D by refining the tables which were used to define them, and for W, with m 2 2 by induction on m, as EW,,,E-' is easily computed. For p E P * , p belongs to the subgroup of P generated by E and D. This can be shown by induction: First, there is only one element of P * , t h e identity, which corresponds to a finite table whose left partition set is To. Now let us suppose that n 1 and that p corresponds to a finite table Zl, while every element of P* corresponding to a finite table whose left partition set is for some j < n belongs to the group generated by E and D. Let us choose (+ E 2 , such .that (+ is of maximum length. As n 2 1, (+ is either P O or p l for some p E "2 of length 2 1. Further, both p0 and p l belong to S , (since 2 , is a partition set and (+ is of maximum length). Case 1: p l is + ( A Z + , ) . Then, p 0 must be +(A,,), and if we define the function with domain rn-,by setting $ ' ( & ) = + ( A k ) for k < n and +'(A t) = p, and put 2 ; equal to the range of +', p will correspond to the finite table Zi, 4' (for replacing p0 and p l in ZIby p still leaves a finite partition set), and thus belongs to the subgroup generated by E and D. Case 2: p l is $(A,,,+l) for some m < n. Then, p 0 must be $(A,,,), and we define the function with domain by setting + ' ( A k ) = + ( A k ) for k < m, +'(A,) = p, $ ' ( A k ) = + ( A k + l ) for n > k > m, and +'(At) = +(At+,), and put 2;equal to the range of +', so that 2;is a finite partition set. If p * is the element of P* corresponding to the finite table r,,-,, S ; , +', then p * must belong to the subgroup generated by E and D. But it is now easy to compute that p = W,,,* p * , so that p also belongs to the subgroup generated by E and D. (Replacing the references above to the group generated by E and D by references to the semigroup generated by the W,,,, m 30, we see that this semigroup includes P* and so coincides with it.) Now, every r E P* corresponds to a finite table So,ZI, in which So and 2 , both contain n 2 elements for some n E w . Let p be t h e

r,,

r.

+

r,

r,,

z.

r.,

+,

r,

+'

r,,-,,

+'

r.-,

+

+

409

Embeddings which preserve rhe word problem

unique element of P which corresponds to a finite table f,,211, 5 and let 9 be the unique element of P which corresponds to a finite table f., ,XI, v. From p , 9 E P it follows that p , 9 E P*. Further, the element p - ' q corresponds to t h e finite table XI,, X I , 5 - l ~(as p - ' corresponds to the finite table 2,),f,,t-'), and thus must be equal to r, as both p - ' q and r belong to P (so that 5 - l ~is equal to +). As p and q belong to the group generated by E and D, so does p - ' q = r, and we conclude that P is generated by E and D. The definition of TI,, r I ,and r: has been given in the statement of the theorem. For m > 0 , let r,+,= Er,,,E-' and r:+, = E7r:E-I. Then for n > m, r,,,corresponds to the finite table f,,,f",8' where for k # m, m + 1 we have e ' ( h k ) = Ak with O'(A,) = A,+, and O'(A,+l) = A,, while r t corresponds to a finite table f,,f,, 0" where e " ( A k ) = Ak for k < n with O"(A,) = A t + I and e"(A *,+ I ) = A. For r ,this is true as inspection shows: t h e rest follows by computation and induction. Now, every g E Ft ( K ) corresponds to a finite table -XI,, Sl, in which -XI, and 2, both contain n + 2 elements and n > 0. As before, let p and 9 be the unique elements of P corresponding to finite tables f., -XI,, 6 and f., XI, v. Then ,$+v-' is a permutation of f, so that f,, f., (+v-' is a finite table corresponding to an element g ' E Ft (K). Consequently, as the tables involved can be directly multiplied to show that g = p-'g'q, to show that g belongs to the group generated by E, D, rlj,r I ,and r f it is enough to show that g' belongs to this group. But ,$+q-' is a permutation of f,,and is thus equal to a product of transpositions, in fact of transpositions which transpose A, and A,,, for some i < n or transpose A,, and A This means, in turn, that g' is a product of the elements r,,i < n, and r t . So g' and consequently g belong to t h e group generated by E, D, rlj,r l ,and r f ,which thus coincides with Ft (K).

+

Proposition 1.5. For K E Fr, Ft (K) is generated by two elements. Two such elements are r : r l D and r o r l m : . F t ( K ) is also generated by the four elements E, D, ro,and r f ,aud by the three elements E, D, and ror fro.

Proof. As E, D, rll,r i ,and r f are a set of generators by Theorem 1.4, and r l .= m O r f D - ' r l , D r f r oas , a computation will show, Ft(K) is generated by E, D, ro,and r f .In fact, it is generated by E, D, and r , l r f r las j , rlr = D - ' E ( r l , m : r n ) D ( m n m f r o ) D - ' and E r f= D ( r , , r f r , , ) D - ' E ( r o r f m 1 ~The ) D .last reduction is more lengthy. Let g l be r t r l D and let gz be rllrlr?. Put g 3 equal to (g:gl)3 and g4 equal to g:glg;glg2. and then set g5 equal to g3g4g3g:g3.(So g s = rt.)

R.J. 7kompson

410

Computation then shows that T ? = g2gsg:gl and D = g,T:gl and E = g:Dg2D.

TI)=

g : g s g : , while

Now we turn to results connected with simple subgroups of Pa(K). Proposition 1.6. For K E Fr there is for every non-identity element g E Pa(K) some h E Ft(K) such that h-'gh corresponds to a table 2 : . 2 ? , $ * such that $*((0)) = (10).

Proof. Let g be a non-identity element of Pa(K). Then there will be some table 2,),,El, $ corresponding to g ; we can assume that (0) 6Z 2,) and ( I ) 6Z Co,since a line 0- p could be replaced by lines 00- p 0 and 01 + p 1, and 1 p' could be replaced by 10- p'0 and 1 1 + p ' l without changing t h e corresponding element of Pa(K). By Proposition 0.6 there will then be some finite partition set 2 with a, T E 2, where a E ,El) and T E ,El have been chosen according to Proposition 0.7 so that $(a)= T and [a]f l [ T ]= 0. As a # T , 2 has cardinality 3 2 . In fact, 2 must have cardinality 3 3 , as if 2 had cardinality equal to 2 then we would have 2 = {@),(I)}, and so a = (0) or a = (I). Thus 2 has the same cardinality as some f., n 2 I (where f, is as in the proof of Theorem 1.4). Because n 2 1, (0) = A,, E f. and (10) = A I E f,. We can take $' to be some one-one mapping of 2 onto f, such that $ ' ( a )= (0) and $ ' ( T ) = (10). Then there is h E Ft ( K ) corresponding to t h e finite table 2, f., $'. Now, for p E K , ( h - ' g h ) ( ( O ) " p=) ( g h ) (h-'((O)"P))= ( g h ) ( a " P ) = h ( P p ) = (lo)"& so by Theorem 0.3 there will be a table 2 $ , 2 : , $ * corresponding to h - ' g h such that for some p C (O), p E 2 : . Thus (as h-'gh, like g, is not t h e identity) (0) = p E 2 : and we must have $*((0)) = (10).

-

Proposition 1.7. For K E Fr, if H is a group juch that Ft ( K )C H C Pa(K) then every non-trivial normal subgroup of H contains rhe elemenr of F t ( K ) corresponding to th.e finite table 00-000, 010-001, 011 -01, 1000- 100, 1001 1010, 101 1011. 11 1 I .

-

- -

Proof. Suppose H is as in the proposition, and N is a normal subgroup of H with g E N and g not equal to t h e identity. By Proposition 1.6 there will be some j E N corresponding to a table 2 : , 2 ? , $ * such that (0)E 2 : , (10)E 27, and $*((0))= (10). Let f E F t ( K ) correspond to the finite table 0-0, 100- 1000, 1010- 1001, 1011- 101, 11 11. Then, jfi-' is t h e element corresponding to t h e finite table 00-000, 010+001, 011 +01, 1- 1. For, given any p E K we have j((l)"p) 6f [lo] so that ( j f j - ' ) ( ( l ) " p= ) j - ' ( f ( j ( ( l ) " p ) )=)

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411

j-'(j((l)"P)) = (I)"@. On the other hand, if (T is (0), (lo), or (11) and (T* is (OO), (Ol), or (1) respectively, then (ifi-')((O)"a"p) = Cf,-')((lO>"a"P)= j-'((lO>"a*"P) = (O)"a*"P. AS f E F t ( K ) c H, [ j , f ] E N, and it is easily computed that [j,f] = (ifi-')f-' corresponds to the finite table given in the proposition. Proposition 1.8. For K E Fr, if H is a group such that Ft ( K )C H C Pa(K) then Ft (K) is a subgroup of every non-trivial normal subgroup of

H.

Proof. Let H be as in the proposition. It is enough, in view of Proposition 1.7, to show that Ft (K) coincides with the normal subgroup N of Ft (K) generated by the element k of Ft (K) which is mentioned in Proposition 1.7. Now, let g , be D - ' E D - ' E - ' D . Then, as k E N , [k,gl]= k g l k - ' g ; ' E N. Computation shows that k g l k - ' g ; ' = D -'ED - I EDE -'D * = ( D-2E ) ( D- I EDE - ') ( E - I D ') ; D - I EDE - I is conjugate to this and so belongs to N. Next we observe that D - ' E N as D - ' = (D.rrT.rr"E)-'(D-'EDE-')(D.rrTik,,E), and E - ' D = (.rra.rrl.rrT)-'(D~'EDE-')(.lrl,.rrl.rrT) so that E - ' D E N. So D E N and E = ( ( E - ' D ) D - ' ) -E' N. Thus, since (.rr,,.lrTE)-'E(.lr0r~E)E N there N . So, as .rri'D.rro~ N, follows .rro.rrT = E(no.lr:E)-'E(.rr,,.rrTE)~ .rro.rrT.rr,, = E-'(.rr,,.rrT)D-'(.rrn'D.lro)E-'(.rr[,.rrT)D-' E N. Since, by ~ , lgenerators of Ft(K), N must Proposition 1.5, E, D, and . r r ~ l ~ Tare coincide with Ft (K).

Corollary 1.9. For K E Fr, Et ( K ) is a simple group. Proof. Take H

=

Ft(K) in Proposition 1.8.

Theorem 1.10. For K E Fr, is H is a group such that Ft ( K )C H C Pa(K), then Ft (K ) is a subgroup of the commutator group of H , and this group is a subgroup of every non-trivial normal subgroup of H. Proof. Let H be as in the theorem. As Ft (K) is a non-Abelian group, so is H. Thus the commutator group of H is a non-trivial normal subgroup, and so has Ft(K) as a subgroup by Proposition 1.8. Now let N be a non-trivial normal subgroup of H, so that Ft(K) C N by Proposition 1.8. Given f , g E H, by Proposition 1.6 there will be h,,,hl E Ft(K) such that hi'fh,, and h ; ' g h , correspond to tables Zb, 2 { , and Z:, z:, respectively with (0)E Zb, (10)E (0)E z f , (10)E Z?,and +'((O)) = (10) = +*((O)). Let x = hi'fhll.rroand y * = h;'gh,.rr,,. Then, for every /3 E K, the restriction of x to [(O)] is the identity mapping, as x ((0)"P) = .rro((hi'fh,J((O)"P1) = . r r ~ ( ( l O ) ~=P(O)"P, )

+'

+*

zi,

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while x permutes t h e elements of [(l)]; similarly, t h e restriction of y * to [(O)] is t h e identity map. Note that r o r : E corresponds to t h e finite we then have, for every table 0 + 1, 1-0. If we set 7~ = rI,r?:E p E K, ( r - ' y * r ) ( ( l ) " p )= ( y *r)((O)"p) = r ( ( 0 ) " p )= (l)"p, so that the restriction of r - ' y * r to [(l)] is t h e identity mapping, while r - ' y * r permutes t h e elements of [(O)]. If we let y = r - ' y * r , we then have by Lemma 0.10 (i) that x y = yx. So [f,g ] = fgf-'g-' = f g y - ' x - ' y x f - ' g - ' = (fx-')(x(gy-')x-')(y(xf-')y-')(yg-') E N, N being a normal subgroup of H,provided that ( f x - I ) , ( g y - ' ) E N (so that (xf-I), ( y g - I ) E N ) . But f x - ' = fr;'h;'f-'h,, = ( f ( r ; ' h i ' ) f - ' ) h o E N (since .rri'hi', h o E Ft(K)G N) and gy-' = g.rr-'.rri'h;'g-'h,a = (g(.rr-'ni'h;')g-')hl.rr E N (since .rr-'.rri'h;', h l a E F t ( K ) c N). As f and g were arbitrary elements of H,we conclude that t h e commutator group of H is a subgroup of N, and the last part of the theorem follows. As a matter of act, for H as in t h e last theorem the commutator group of H is simple; this result is not needed later however, the following corollary being sufficient.

Corollary 1.11. For K E Fr, if H is a subgroup of Pa ( K) a n d H is equal to its commutator group, then the subgroup of Pa(K) generated b y H U Ft (K) is simple. Proof. Let H be as in the proposition, and let B be the subgroup of P a ( K ) generated by H U F t ( K ) , so that F t ( K ) c B C Pa(K). Then t h e commutator group of H is a subgroup of t h e commutator group of B, so that H is a subgroup of the commutator group of B, and Ft (K) is a subgroup of the commutator group of B by Theorem 1.10. Thus B is equal to t h e commutator group of B, and from Theorem 1.10 it now follows that B is simple. 92. Splinter groups Definition 2.1. Suppose K E Fr and Y is a group of permutations of a set L. Then S p ( Y , K), the splinter group corresponding to Y and K, is t h e group of permutations of L x K generated by the set Y = {g : g E Y } , where g ( ( x , a))= ( g ( x ) ,a ) for g E Y, x E L, and a E [Ol] and where g ( ( x , a))= ( x , a ) for g E Y, x E L, and a 6Z [Ol], and t h e set Ft ( K ; L ) = {F : F E Ft (K)}, where F ( ( x , a))= ( x , F ( a ) ) for all x E L and a E K.


413

Note that both Y and Ft ( K ;L ) are subgroups of Sp(Y, K ) , and that Y = Y and Ft ( K ;L ) = Ft ( K ) . The name "splinter group" refers to the way in which t h e action of an element of Sp(Y, K ) on a subset of L x K whose projection on K is contained in a sufficiently small interval in K is determined by a single element of Y; see also Proposition 2.4. Throughout the rest of t h e section we assume that Y is a group of permutations of a set L and that K E Fr. Proposition 2.2. For every g E Y, if F,, FzE Ft ( K ) are given by the finite tables 00-000, 010+001, 011-01, 1 + 1 and 00-00, 01-010, 10+011, 11+ 1 respectively, then [g,F,]= Proof. We can show that ( [ g ,F , ] ) ( ( x ,a ) )= ( F ; ' g F 2 ) ( ( x ,a ) ) for arbitrary x E L and a E K by checking that this holds in the four separate cases of a E [OO], a E [OlO], a E [Oll], and a E [l ]. Theorem 2.3. Sp( Y, K ) is equal to its commutator group. Proof. This follows from Proposition 2.2, which implies that Y is contained in the commutator group of Sp(Y, K ) , and the fact that Ft ( K ;L ) Ft ( K ) , which is simple (and so equal to its commutator group, as i t is non-Abelian) by Corollary 1.9.

Let us now define a splinter table to be a quadruple 20,XI, $, 5 such that Z:,,, Z,, $ is a finite table and 5 is a mapping of Z:,, into Y. By the permutation of L x K corresponding to the splinter table XI,, Z I , $, $, is meant that permutation Q of L x K such that for cr E Z,,, a E K , and x E L, Q((x, a " a ) )= ( ( 5 ( a ) ) ( x ) ,$ ( a ) " a ) . Proposition 2.4. Sp( Y, K ) is equal to the set of permutations of L x K corresponding to splinter tables. Proof. First of all, every splinter table corresponds to an element of Sp(Y, K ) . For, given a splinter table Z,,, Zl, 4, 6, t h e corresponding permutation of L x K is the product of the permutations corresponding to the splinter tables Zll, Z,,, $*, 6 and XI,, El, $, (*, where $ * is the identity mapping on Zll and 5* is the mapping whose domain is Z l , and whose range consists of a single element, the identity of Y. The permutation corresponding to the second table evidently belongs to Ft ( K ;L ) . As for the permutation corresponding to the first table, if Z,,= { u l , .. . , u"},take for 1 S i s n the mapping ti to have domain Z:,,

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with < , ( a , )[((T,) = but with & ( a , being ) t h e identity of Y for j # i. Then this permutation will be equal t o t h e product o f permutations corresponding to splinter tables C,,, XI,, Q*. 5, for 1 S i S n (of course, by Lemma 0.10 (i) these later permutations will commute), s o it will be enough to remark that such p e r m u t a t e a r e , when ,Vl, has cardinality greater than 1, equal to elements FT'{(a,)E of Sp(Y,K) where t ( a i ) E and E is an element of Ft(K;L) which maps L x [Ol] onto L x [a,]. When C,,has cardinality 1, the permutation corresponding to C,,, Z,,, Q*,5, will also corespond to {(O), (l)}, {(0),(1)},Q', [i, where Q ' is the identity on { ( O ) , (l)}, and [{((O)) and [{((l)) are both equal t o that element of Y which is the range of tl; this reduces t o the previous case. On the other hand, the elements of Ft(K; L ) correspond t o those C,,Q, [ where Z,,, C,,4 is a finite table and t h e splinter tables C,,, range of 5 is the identity element o f Y, while g E corresponds t o the splinter table {(OO), (01). (I)}, {(OO), (Ol), (I)}, Q*,[ * where Q * is the identity map and [*((()I)) = g with [ * ( ( O O ) ) and [*((I)) being equal to the identity of Y. So to show that every element of Sp( Y , K ) and F t ( K ; L ) corresponds t o a splinter table it is enough (with being closed under inverses) to show that t h e product of permutations corresponding to splinter tables also corresponds to a splinter table. Now, if a E Z,, the splinter tables C,,, C,,Q. [ and Z,',Xi, , Q', 5' correspond to the same permutation when Z'b, Xi, Q' is obtained from Z,,, C,,II, by replacing the line (T Q(a)by its direct extracts, and [' is obtained by setting 5 ' ( p ) = [ ( p ) for p # (TO, ( T I , with [ ' ( a ( )=) ['(al) = [(a). So splinter tables can be refined, just like finite tables. Thus, by refining the right partition set of one splinter tableand the left partition set of another splinter table until they are equal, we can multiply the two splinter tables: for the product of the permutations corresponding to the splinter tables C,,, C,,Q, [ and Z,, Zz,Q', 6' is Z2,$4'. (*, the permutation corresponding to the splinter table Z,,, where ( * ( p ) is, for p E Z,,, [ ( p ) - ( Q [ ' ) ( p ) . (The inverse of the permutation corresponding to Z,,, Z l , Q, (,, is the permutation Z,,, I,!-', where ( , ( p ) = ([ll($-l(p)))-' for p E corresponding to Z,, 21.)

v

-

el,

Definition 2.5. An embedding of a group G into a group H is a Frattini embedding just in case, whenever g, g ' E G are conjugate in H . they are conjugate in G. Theorem 2.6. Suppose that only the identity element of Y has any fixed points. Then the embedding of Y irito Sp(Y, K ) is a Frattini embedding.


415

Proof. Suppose a, b E Y and g E Sp( Y, K ) are such that g-'iig = b. If a = b, obviously ci and 6 are conjugate in Y. If a # b, then both d and b are distinct from the identity permutation of K x L and map K x [OI] onto itself but have no fixed points on this set (because, for x E L, a ( x ) , b ( x ) # x ) , although they are pointwise fixed o n t h e rest of K x L. Thus Lemma 0.10 (ii) applies and g must map K x [Ol] onto K x [OI]. By Proposition 2.4 there will be some splinter table Z(,, Z l , $, 6 corresponding to g. In fact (because we can refine a splinter table if necessary) we can assume that there is some a E Zl, such that (01) C m. As g maps K x [Ol] onto K x [Ol] it follows that (01)C +(aj.If we put h = 6(a)we have for all x E L and p E K that (g-'iig)((x,$ ( a ) " P ) ) = ( a g ) ( ( h - ' ( x ) on@>) , = g((a(h-'(x))a , " P ) ) = ( h ( a ( h - I ( x ) h+ ( a ) " P ) , while b ( ( x ,+(a)"p)) = ( b ( x ) ,+ ( m ) " p ) , so that for all x E L, b ( x ) = h ( a ( h - ' ( x ) ) ) =( h - ' a h ) ( x )and consequently b = h - ' a h . As h E Y, a and b are thus conjugate in Y, so ii and 6 are conjugate in Y. Proposition 2.7. If the word problem for Y is solvable, so is the word problem for Sp(Y, K ) . Proof. We can appeal to Proposition 2.4 to see that the word problem for Sp(Y, K ) is solvable if we have an effective means of multiplying and inverting splinter tables and determining whether or not a splinter table corresponds to the identity mapping of K x L onto K x L. This last determination can certainly be made effectively if the word problem for Y is solvable, for a splinter table Zo, Z l , $, 6 corresponds to the identity mapping of K x L onto K x L just in case XI = Zll, $ is the identity mapping of Z(, onto Z(,, and 6 is the mapping whose range consists of the identity element of Y . The last half of t h e proof of Proposition 2.4 shows that splinter tables can be multiplied and inverted effectively in terms of Y (the word problem being solvable for Ft(K; L ) = F t ( K ) ) .

Although no proofs will be given here, it is worth remarking that Sp(Y, K ) is such a well-behaved extension of (an isomorphic image of) Y that it has a finite presentation if Y has a finite presentation. On the other hand, Sp(Y, K ) has the interesting property (which is not hard to prove) that the countable weak direct product of a collection of subgroups of Sp(Y, K ) is always isomorphic to a subgroup of Sp(Y, K ) . This property is also possessed by F t ( K ) ; in the former case Proposition 2.4 can be used, while finite tables are used in the latter case.

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03. The main construction Throughout this section J will denote the set of all positive integers, and K O will denote the subset of " 2 consisting of every (Y E "2 such that there is, for every n E w, some m > II with cr(m)# a ( n ) . (Thus K,, consists of every sequence belonging to " 2 which is not "eventually" either 0 or 1.) It is easy to see that K u EFr. Note that every P E Ko is equal to (I)'"(O>'"(I>"P' for some unique i E w, j E J, and P ' E KIJ;by t h e index of p will be meant j , and by the parity, the integer j - i. For t h e remainder of this section A will be a fixed group of permutations of J, possibly satisfying these conditions: C1: The word problem for A is solvable. C2: Given g E A and j , k E J it is always effectively decidable whether or not g o ' ) = k. C3: If g E A is such that { j E J : g o ' ) 2'j } is finite, then g is t h e identity of A. C4: If g E A is not the identity of A, then go') # j for every j E J. For g E A let g * be t h e function with domain K,, such that, for P E K,, with parity not equal to 0, g * ( p ) = p, while for (Y E KOand m EJ, g*((1)""(0)""(l)"a)= (l)~cm)n(0)R(m)n(l)"(Y. As g is a permutation of J , g * is a permutation of K,, which maps the set of elements of parity 0 onto itself. Further, in view of Definition 0.2, g * E Pa(&). If we set A * = {g* : g E A } it is evident that (taking the group operation for A * to be functional composition) we have A = A *, with A * a subgroup of Pa(K,,). In order to deal with t h e subgroup of Pa(&) generated by A * U Ft (KO) using objects finitely expressible in terms of A we have devised A-tables. Thus t h e reader should be careful to note that t h e constructions to be given in connection with A-tables are effectively obtained, granted that C1 and C2 hold. Let F i ( A ) be the set of all functions whose domain is a finite, non-empty subset of J and whose range is a subset of A. For b E F i ( A ) let ( ( b ) = { ( m + l , g ) : ( r n , g ) E b}; then ( ( b ) E F i ( A ) . For i E w , by ("'(b) will be meant ((('(b)), ("(b) being b.

Definition 3.1. An element b of F i ( A ) will be called n-bolsrered, for n E w, just in case for every (m, g ) E b and every j E J such that j 3 m we have g ( j ) + n 3 m.


417

Lemma 3.2. Suppose that b E Fi ( A ) is n - bolstered. Then : (i) For n' 3 n, b is n'-bolstered. (ii) For i E w, ( ' ( b ) is ( n + i)-bolstered. (iii) For every s E J there is some r E J , r 3 s, such that ( ' ( b ) is ( n + r - s)-bolstered.

Proof. W e obtain (i) a n d (ii) directly from Definition 3.1. For (iii), given s E J let us first set m * equal t o the largest element in the domain of b. As the range of b is finite, and every g E A is a permutation of J, there must b e some r 3 s such that for every x 3 r we have g ( x ) 3 m * + s for each g in the range of b. T h u s if ( m ,g') E ( ' ( b ) we will have r < m s m * + r so that, for every j 3 m , g ' ( j ) + ( n + r - s ) 3 ( m * + s ) + ( n + r - s) 2 m * + r 3 m. This yields (iii). Note that, given s in Lemma 3.2 (iii), r can b e obtained effectively when C2 holds: for each g belonging to t h e range of b we compute in succession g(l), g ( 2 ) ,. . . until every positive integer less than m * + s has appeared (which must happen as we are dealing with permutations); then r can b e taken to b e t h e smallest integer which has not been used a s an argument in o u r computations. By a regular line is meant what was called a line in $1; in this section a line will b e either a regular line or an A-line,-defined as follows: Definition 3.3. An A -line is a quadruple a, b, T , n such that a, T E "2 and b E F i ( A ) , while n E w is such that b is n-bolstered; it will b e written as a, b + T , n. Definition 3.4. (i) T h e line function corresponding to a regular line a+ T will b e that function x with domain [a]such that x ( a " P ) = ~ " p for p E K,,. (ii) T h e line function corresponding to an A-line a, b + T , n is that function x with domain [a]such chat for p E K O with parity not p , for p E KO belonging to the domain of b, x ( a " P ) = ~ ~ ( l ) " "while such that p has parity m (where m belongs to the domain of b ) SO that p = (I>""'"(0)"(1)"p' for some p ' E KO and k a m , x ( a " P ) = n( 1 ) ( R ( k P - m )n(0)8(k"'(1)"p', where g = b ( m ) . (Observe that from k 3 m and b being n-bolstered it follows that g ( k ) + n 3 m.) T h e left sequence of a line a + T or a, b + T,n will b e a,a n d t h e righr sequence, T. Definition 3.5. A partial ruble is a finite set of regular lines a n d A -

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lines such that all of the left sequences of these lines are distinct and the intervals in K o they determine are all disjoint. Definition 3.6. The left sequence set of a partial table is the set consisting of all the left sequences of lines belonging to it. Definition 3.7. An A-table is a partial table whose left sequence set is a finite partition set for K,,. By the function defined by a partial table or A-table will be meant the union of the line functions (which is again a function, as the left sequences of t h e lines determine disjoint intervals in Kll). For an A table this corresponding function will have a domain equal to K,, (the left sequence set being a finite partition set for K,,); as a line function is always a one-one function an A-table will correspond to a one-one function just in case the ranges of the line functions (for the lines belonging to t h e A-table) are disjoint. Note that, if a regular line appears in a finite table corresponding to an element of Ft ( K , ) ,this element of Ft (Kl,) will be an extension of t h e line function corresponding to the regular line, which gives: Proposition 3.8. If an A-table and a finite table have the same lines belonging to them (which will thus all be regular lines) then they correspond to the same element of Ft ( K O ) . Proposition 3.9. For every A-table having only regular lines and corresponding to an element of Pa(K,,), this element belongs to Ft (Kll). Proof. An A-table has a left sequence set which is a finite partition set; if all its lines are regular it can only define an element of Pa(Ko) if the set of t h e right sequences of its lines is a finite partition set of the same cardinality as its left sequence set and the set of its lines. In this case there is a finite table with the same lines as the A-table. From t h e relevant definitions we directly obtain: Proposition 3.10. For g E A, i f b = ((1,g)) then b E Fi(A), b is 0bolstered, and the A -table with the lines (0) +(0) and ( I ) , b + (I), 0 corresponds to g * E A *. Proposition 3.11. For k E J and u,T E e2, if b E Fi( A) is n-bolstered for n E w (and so ( n + k)-bolstered) then a,b + 7, n + k and a,b + ~ " ( 1 ) ~n. determine the same line function.

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The direct extracts of a regular line have already been defined, in Definition 1.1. For A-lines we make the following definition: Definition 3.12. The direct extracts of the A -line u,b + 7,n are, where m * is the largest element in the domain of b, the regular line u"(0)"'+l-+7n(l)nn(0)m*+I, and for 0 < m S m * the regular line u"(O)""(l)-+ 7n(l)nn(0)mn(l)(when m does not belong to the domain O ) ~ ( m~ ) ~ ( l ) of b ) or the regular line u"(O>""(l)+ ~ ~ ( l ) ( ~ ( ~ ) + ~ - ~ ) ~ ((when belongs to the domain of b and b ( m )= g ) together with the A-line a"(l),((b)-+ 7,n + 1. (Noting that ( ( b ) is ( n + 1)-bolstered by (ii) of Lemma 3.2.) Definition 3.13. An extract of a line is either the line itself or a line obtained from it as the last in a series of direct extracts, starting from the given line. From Definition 1.1 or Definition 3.12, using Defin'ition 3.5, etc., we obtain: Proposition 3.14. The function determined by the partial table consisting of the direct extracts of a line is the line function corresponding to the line.

Corollary 3.15. The line function corresponding to a line is an extension of every line function corresponding to an extract of the line. Definition 3.16. Suppose n E w and b, c E Fi(A). Then t h e n-alignment of b and c, or al(n, b, c), is the function whose domain is the union of the domains of b and ["(c), while (al(n, b , c ) ) ( r n ) = ( b ( m ) ) .( ( l " ( c ) ) ( m )for ) m belonging to both domains, (al(n, b , c ) ) ( m ) = ( l " ( c ) ) ( m )for m belonging to t h e domain of [ " ( c ) only, and (al(n, b, c ) ) ( m )= b ( m ) for m belonging to the domain of b only. (Thus al(n, b , c ) € F i ( A ) . ) Further consideration of the separate cases of the definition shows: Proposition 3.17. Zf b, c E Fi(A), n, n' E w, b is n-bolstered, and c is n'-bolstered, then al (n, b, c ) is (n + n')-bolstered. Definition 3.18. Suppose p, a, 7 E "2, a'nd b, c E Fi(A), with b being n bolstered and c being n'-bolstered. Then the (symbolic) composition of lines is stipulated to be such that:

R.J. Thompson

420

(1) ( a + ~ ) . ( ~ + p ) = ( a - + p ) . (2) (a+ 7 ). (7,b + p, n ) = (a,b + p, n ). (3) (a,b + T , n ) * ( T p ) = (a,b + p, n ) . ( 4 ) (a,b + T , n ) . (7,c p, n ' ) = (a,al(n, b, c)+ p, n -+

-+

+ n').

Proposition 3.19. The line function corresponding to the composition of two lines is the same function as the composition of the two line functions which correspond to the lines. Proof. This is clear enough for the cases (l)-(3) of Definition 3.18, using the definition of line function. For (4) we have to consider the effect of the two compositions on a"P, where /3 E K O ;there are four subcases depending on whether or not the parity m of /3 belongs to the domains of b or l " ( c ) , and each is easily checked. Just as in §1, an A-table (or partial table) obtained from another A-table (or partial table) by an iterated process of replacing a line by its direct extracts will be called a refinement of the other table; if, in this process, we also allow the replacement of a line a, b -+ T , n + k (where k E J ) by t h e line a, b -+ ~ " ( l ) n~ ,when b is n-bolstered, we will call the result a semi-refinement. Appealing to Proposition 3.14 and Proposition 3.11 we now see that: Proposition 3.20. If an A-table (or partial table) is obtained from another A -table (or partial table) as a refinement or semi-refinement, it corresponds to the same function. Definition 3.21. T a ( A ) is the subset of Pa(&) consisting of all elements of Pa(&) which are defined by A -tables. Definition 3.22. If a E " 2 is equal to p "(1) for some p E " 2 it will be said to be (Z)-ending;a line whose right sequence is (1)-ending will also be said to be ( 2 )-ending. Proposition 3.23. Every extract of an A-line with left sequence a is either a regular line or has a left sequence equal to ~ " ( 1 for ) ~ some k E w. Proof. This follows because, given an A-line with left sequence a, the only direct extract which is an A-line has ~ " ( 1 ) as its left sequence while all extracts of regular lines are regular lines.


42 1

Lemma 3.24. Suppose a,u'E " 2 with u'C u while a', b --* T, n is an A line. If u is either (1)-ending or else, for some p E " 2 of length t, is equal to p"(O)"'+' for some m E J which is greater than every element of the domain of b, and u'C p in the second case, then there exists exactly one line which is a n extract of this line and has u as a left sequence.

Proof. The intervals in K,, determined by the left sequences of the direct extracts of a regular line or A-line are all disjoint; as these left sequences are distinct from the left sequence of the line itself (but are extensions of it), it follows that there can be at most one line which is an extract of a given line and has a given left sequence. It remains to show that whenever u, u', b, T, and n are as in the lemma there is some line with u as its left sequence which is an extract of the A-line u',b + T, n. Let u" be the element of " 2 of maximum length such that u"C u and u" is either u' or is equal to ~ ' " ( 1 ) ' for some s E J. Then a''is the left sequence of an A-line which is an extract of the A-line u',b + T, n, being either this line itself or, for some s E J , the line u'"(l)', l ' ( b ) - , T,n + s. So if a"= a, the lemma holds. Otherwise, ~ " " ( 0C ) u..Then, if u is (1)-ending, there will be a regular line which is a direct extract of the A-line with left sequence d" and which has u as an extension of its left sequence; this regular line will then have an extract which has u as its left sequence, and t h e lemma will again hold. Finally, if u is equal to pn(0)"'+' for some p E " 2 of length t and m E J which is greater than every element of the domain of b, while u'c p, then u"C p (since u"C a ) so'that if u" is ~ ' " ( 1 ) ' for s E J then s < t and thus m + s m + t so that m + t is greater than every element of the domain of l ' ( b ) , just gs m is greater than every element of the domain of b ; consequently, there will be (as u " C p and ) u )a direct extract of the A -line w i t h left sequence a'' also ~ " " ( 0C which is a regular line and has u as an extension of its left sequence, and so has an extract whose left sequence is u, so that the lemma holds in this case also. Lemma 3.25. Suppose f I is an A -table or partial table. Then, for every k , m E J there is a n A -tGble or partial table, respectively, which is a semi-refinement of r ,'and in which every line has a right sequence of length 2 . k and is either (1)-ending or else is a regular line with a right sequence equal to P ~ ( O ) ~ +for ' some t E J and p E "2 with length t. Proof. First we show that there is a semi-refinement f 7 of r , in which every A-line has length 3 k and is (1)-ending. Suppose that, for n E J , there are exactly n A-lines in f l which are either not (1)-ending or are

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of length less than k. Let a, b + 7 , j be one such A-line. By (iii) of Lemma 3.2 there is some r E J with r z=k such that l ' ( b ) is (j+ r - k)-bolstered. For 1 c m C r let f,,,,, be obtained from f,,,by replacing the A -line belonging to f,,,which is an extract of a, b + 7, j by its direct extracts. Then r,+l has ~"(1)'.l ' ( b ) + ~ ,j + r as one of its A-lines (with this being an extract of a, b + 7 , j ) and has exactly n - 1 other A-lines which are either not (1)-ending or are of length less than k. Now, f,+, is a refinement of fl,and if f;+l is the A-table resulting from replacing an(l)', l ' ( b ) + 7 , j + r by ~ " ( 1 ) ' . l ' ( b ) + ~ ~ ( lj + ) r~ -, k then f;+,will be a semi-refinement of f,,,and f,.Note that f;+Ihas only n - 1 A-lines which are either not (1)-ending or are of length less than k. Proceeding from f;+,we can continue to refine t h e table in the same way until we obtain a semi-refinement f f of f l in which all A-lines are (1)-ending and of length 3 k. (Of course, if there are n o A-lines in f,which are not (1)-ending or are of length less than k, we Now f';can be obtained from f T by replacing each take f T = f,.) regular line a + 7 of f f by t h e 2' extracts consisting of all the lines a n y + T ~ such V that v E "2 is of length k. T h e n f';is a semirefinement of f ? and so of f I, while every line of ft;is of length 3 k and is (1)-ending if it is an A-line. Finally fl is obtained from f; by replacing each regular line ar--,7'by the regular lines (where t is t h e length of 7') a r n ( l ) + ~ ' " ( l ) ,a'"(0)"""(1)-+ ~~"(0)'"'"(1) (for each m r such that 0 < m' < m + t ) , and arn(0)"'+'-+ ~'"(0)'"+~; f will be a semirefinement of f: (and so of f,)with all t h e required properties.

Theorem 3.26. The set T a ( A ) is (with functional composition as the operation) a subsemigroup of Pa(K,,).

Proof. Suppose that F and G are elements of T a ( A) , with f l and f2 being A-tables which define F and G respectively. Let k be t h e least integer greater than the lengths of all left sequences of f2,and let m be the least integer greater than every element belonging to t h e range of some b E Fi(A) occurring in an A-line of f2.By Lemma 3.25 there will be an A-table f ;which is a semi-refinement of f l and has only lines, with right sequences of length a k, that are (1)-ending or else are regular lines with a right sequence equal to p ,(0)'"+' for some t E J and p E ?2 with length s t. By Proposition 3.20, r I also defines F. Now we will define an A-table f3 (as t h e "composition" of the two A-tables f I and f2)as follows: we replace each line a 7 or a, b + T, n of f I by the composition of this line with an extract of a line of fzto obtain a line with a as its left sequence. Given a line of f I, say with left sequence a and right sequence T, it follows that, as . -+

Embeddings which preserw rhe word problem

423

the left sequences of f2are a partition set for K,,, and all of them are of length < k, while T is of length 3 k, that T is an extension of some left sequence T ' of r2. Case 1: T ' is the left sequence of a regular line ~ ' + pof f2.Then let p E "2 be such that T = ~ ' ~and p ,replace the line of f I which has u as its left sequence and T as its right sequence with t h e composition (according to (1) or (3) of Definition 3.18) of this line and t h e extract ~'"+ p p " p of the line ~ ' + p . Case 2: T' is t h e left sequence of an A -line T', c + p , n' of f2.By Lemma 3.24 there will exist exactly one line with T as its left sequence which is an extract of T ' , c + p, n'. Let the line of f I which has u as its left sequence and T as its right sequence be replaced by t h e composition (according to (2) or (4) of Definition 3.18) of this line of f I with the extract of T',C + p,n' which has T as its left sequence. Now, we obtain in this way an A-table r3with t h e same number of lines as f I and the same set of left sequences. If H is t h e function defined by f3 then t h e restriction of H to an interval [u], where u is a left sequence of f3 and so of f I, is evidently a line function equal by Proposition 3.19 to t h e composition of t h e restriction of F to [u]and a function which is, by Corollary 3.15, a restriction of G (because it is the line function corresponding to a line which is an extract of a line of f2).Thus t h e restriction of H to an interval [u], u being a left sequence of f3,is a function of which FG is an extension. It follows that H itself has FG as an extension; as they have t h e same domain, H is equal to FG, and thus the theorem holds.

Theorem 3.27. If A satisfies condition C3, then an A-table f will define the idenity element of Pa(K,) (that is, the identity permutation of K,,) if and only if every right sequence of a regular line belonging to f is the same as its left sequence and, for every A-line a,b -+ T,n of f, u = ~"(1)"and the range of b consists of only one element, the identity of A.

Proof. The sufficiency of t h e requirement concerning f is clear from the definitions of t h e line functions of regular nnes and A-lines. The necessity is clear enough, as regards regular lines, since a regular line U - + T corresponds to a line function which maps [a] onto [TI. For an A-line u,b + T, n of an A-table defining the identity element of Pa(&) we must have u = ~"(1)"as, for m E J sufficiently large, [a"(O)'"] will be mapped onto [ ~ ~ ( 1 ) " ~ ( 0by ) "t'h]e line function corresponding to u,b + T,n. Then, in view of Definition 3.4, this line function will be the identity mapping on [u]only if every g E A belonging to t h e range of b is equal to the identity of A, as if m E J

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belonging to the domain of b is such that g = b ( m ) is not equal to the identity of A then by C3 there will exist k 3 m such that g ( k ) # k and thus [~"(1)"-""(0)"(1)] will be mapped onto [.n( 1y k ) + n - m n(0)p(k)n(l)], which must be distinct as g ( k ) # k. From the next result t h e significance of Theorem 3.26 and Theorem 3.27 becomes apparent.

Theorem 3.28. The subgroup of Pa (KII)generated by A * U Ft ( K O )is a subsemigroup of T a ( A ) . Proof. As every inverse of an element of A * U Ft(K,,) belongs to this set, the subgroup of Pa(&) generated by A * U Ft(K,,) is t h e same as the subsemigroup of Pa(Kll) generated by A * U Ft (KJ,and thus, because of Proposition 3.8 and Proposition 3.10, is by Theorem 3.26 a subsemigroup of T a( A ) . Theorem 3.29. Suppose that A satisfies conditions C1, C2, and C3. Then the word problem for the subgroup of Pa(K,,) generared by A * U Ft (I&) is solvable (for a presentation appropriately related to the presentation for which A satisfies C1 and C2). Proof. Given that A satisfies condition C3, Theorem 3.27 will apply, and we will be able to effectively determine whether or not an element of the subgroup of Pa(&) generated by A * U Ft (&) which is given in terms of an A-table is equal to the identity element (of Pa(&) and so of the subgroup) using condition C1. In view of Proposition 3.10 and Proposition 3.8 we can effectively find A-tables which correspond to given elements of A * U Ft ( K J ; as this set is closed under inverses, and t h e proof of Theorem 3.26 shows that there is an effective procedure (relative to a solution of the word problem for A, and an effective method of determining for j , k E J and g E A whether or not g o ) = k ) which yields an A-table corresponding to the composition of two elements of T a ( A ) given by A-tables, we have a procedure that effectively determines whether or not an element of the subgroup of Pa(&) generated by A * U Ft ( K , , )is equal to t h e identity element, so that the word problem is solvable.

Theorem 3.30. Suppose that A satisfies condition C4. Then the embedding of A * into the subroup of Pa(&) generated by A * U Ft ( K O ) is a Frattinis embedding.


425

Proof. Let B be the subgroup of Pa(Ko) generated by A * U Ft ( K O ) , and let M be the set of all p E K,, with parity equal to 0. Suppose g, h E A and f E B are such that f - ' g * f = h *. If g = h, obviously g * and h * are conjugate in A *. If g # h then both g * and h * are distinct from the identity permutation of K,, so that they both map M onto itself but have no fixed points on this set (by C4 and t h e definition of g * and h *) and are pointwise fixed on t h e rest of K,,. Thus Lemma 0.10 (ii) applies and f must map M onto M. Now, by Theorem 3.28, f is defined by some A-table f(,. Then there is some semi-refinement rl of f,,(also defining f ) which has at least two lines, thus possessing a unique line with (for some r E J ) a left sequence a = (l)', this unique line being (1)-ending. (After first obtaining, if necessary, a refinement of f,,with more than one line, t h e unique line with a left sequence of the form (1)" for some s E J is either replaced by its direct extracts if it is a regular line, or if it is an A-line an appeal is first made to Lemma 3.25.) Whether the line with a as its left sequence is a regular line or an A-line, the corresponding line function will map [ a ]into [ T I , where T is the right sequence of the line. If T is not (1)' for some s E J , the parity of every element of [TI will be the same (as T is (1)-ending), so that f either maps [ a ]= [(l)']into M or maps it into the complement of M in K,,. But this is impossible, as [ ( l ) ' ]r l M # 0 and [ ( l ) ' ] is not contained in M, so that the image of [ a ]under f cannot lie wholly within either M or its complement. Thus T is (1)' for some s E J. If T is the right sequence of a regular line, then s = r, as we can choose (Y E K,, such that (Y has parity equal t,o r, so that (+"a has parity 0 and belongs to M and thus the value of f for una,which is T ~ ( Ywith parity r - s, also belongs to M. Further, if T is the right sequence of an A-line a, b + T,n then s = r - n, as we can choose a E K,, such that a has parity equal to r, so that aria has parity 0 and belongs to M , and thus the value of f for v n a , which has parity equal to r - (s + n ) by Definition 3.4, belongs to M. Now we remark that there is some u E A such that for every (Y E K,,, and for every k 3 r , f ( ( l)k"(o)kn(1)"a)= ( l)u(k)n(0)u(k)n( 1)"CY. If the line with a as its left sequence is a regular line (i.e., if it is ( l ) ' + ( l ) ' ) we can take u to t h e identity element of A (so that u(k)= k for all k E J), and if v is the left sequence of an A-line a, b + T , n we can take u (in view of Definition 3.4) to be the identity element of A when r does not belong to the domain of b, and b ( r ) when r

belongs to the domain of b. Then, u - l g u = h. For otherwise u-lguh-' will not be the identity permutation and so, by C4 (in fact, by C3

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alone), as u, g, and h are all permutations there will be some m E J such that ( u - l g u h - ' ) ( m ) # m while for f = u - ' ( m ) (thus ( g u ) ( t ) # ( u h ) ( t ) )we have f 3 r and g ( t ) 3 r. For such an m and corresponding t we would then have, for every cr E KO,

( f - ' g *f)((l)"~""(o)U't'n(l)na) = ( g *f)((l)'"(O>l"(l>"a) = f(( I ) g ( t ) n ( 0 ) s ( r ) n ( 1)"a)

0u ( m ) l n a # 1h(u(t))Oh(u(t))l"cr -

lu(g(r))

=

h *(( l)u(t)n(0)u(r)n( l)"a),

so that f-'g *f# h *, contrary to hypothesis. Thus if g * is conjugate to h * in B, g is conjugate to h in A and so, as A = A *, g * is conjugate to h * in A * . The results that are really needed for 04 have now been obtained, but the question naturally arises as to whether or not T a ( A ) is actually a group - in particular, the subgroup of Pa(&) generated by A * U Ft(K,,). It actually is the case that T a ( A ) is a subgroup of Pa(K,,), the one generated by A * U Ft(K,,). This makes both T a ( A ) and A * appear less arbitrarily chosen, t h e exact form of t h e definitions not being so important.

Theorem 3.31. The subgroup of Pa(Ko) generared by A * U Ft ( K , , )is equal to T a ( A ) .

Proof. Let us say that the A-multiplicity of a regular line is 0, the A-multiplicity of an A-line a,b + T,n is equal to t h e cardinality of the domain of b, and the A-multiplicity of an A-table is t h e sum of t h e A-multiplicities of its lines. Let B be t h e subgroup of Pa(K,,) generated by A * U Ft (K,,); t h e theorem follows (in view of Theorem 3.28) once it is shown that. every element of T a ( A ) belongs to B. Note that if p E $2 is non-empty there will be a finite table f,,with the line p + (1); if-FIE Ft (K,,)is defined by this table then, for g E A, G = F1g* E B will be defined by t h e A -table obtained from f,,by replacing t h e line p + ( l ) by the line p , { ( I , g ) } + ( l ) , O = (p+(l)).((l),{(I,g)}~(l),o). Now suppose F E Pa(KI,) is defined by an A -table f I with A multiplicity d 1. Then, if f,has A -multiplicity 0, F E Ft (KO) C B, by Proposition 3.9; otherwise f I will have t h e single A -line: (1)

a,{ ( m , g

+ 7,

n.


427

If necessary, f l can be refined to ensure that T E "2 is a non-empty sequence. Case 1: n 3 m - 1. As { ( m , g ) } is ( m - 1)-bolstered, if n is ( m - 1)+ j for j E J then line (1) can be replaced by the line a , { ( m , g ) } + ~ ~ ( l y , m - 1 which will define the same line function. Thus we can assume n = m - 1. As noted above there will be G ' E B defined by an A-table f2whose only A-line is (2)

7,((1, g-%+

(I), 0.

Since n = m - 1, the composition of lines (1) and (2) is, by (4) of Definition 3.18: (3)

a , { ( m 9 g g - ' ) } - - + ( 1 ) , m- 1.

By refining f,we can get an A-table f I of which (1) is still the only A-line while every other line has a right sequence equal to the left sequence of some extract of a line of f2,this extract being a regular line in view of Proposition 3.23 and the fact that (1) is a line of f I while F E Pa(&). Composing the A -tables f I and f2we obtain an A-table f3which defines FG' and has (3) as its only A-line. As the same line function corresponds to the regular line a+(l)"' as to (3) (because gg-' is the identity permutation on J), by replacing (3) with this regular line in f, we obtain an A-table f; also defining FG', but with only regular lines. So FG'E B, and F = (FG')(G')-'€ B as G' E B, as required. Case 2: n < m - 1. Let f4 be a finite table with one of its lines being T - , ~ ~ ( l ) ( " - ' ) - Then, ". where G"E Ft(K,,) corresponds to f4,FG" will be defined by an A-table whose only A-line is cr,{(m,g)}-,~n(l)(m-')-n, n (the composition of ( 1 ) and T + ~ ~ ( l ) ( " ' - ' ) - " according to (3) of Definition 3.18), which defines the same line function as u,{ ( m , g ) } - * ~ ,m - 1. So there is an A-table f, which has this line as its only A-line and which defines FG"€ Pa(K,,); by case 1 we have FG"€ B so that, as G"EB, F = (FG")(G")-'E B. Thus every F E Pa(&) defined by an A-table with A-multiplicity C 1 belongs to B. Now let us assume that p 2 1 and every element of Pa(&) defined by an A-table of A-multiplicity S p belongs to B. Let F E Pa(&) be defined by an A-table f6 of A-multiplicity p + 1. Then f6 will have an A-line; let (4)

a,b + 7, n

be such a line, and choose m to belong to the domain of b. As g = b ( m ) is a permutation of $ there will be some w E J such that

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for k 3 w, g - ' ( k ) 3 m. Then X = { j E J : j < rn + w and g o ' ) 3 rn + w } = { j E J : rn j < rn + w and g ( j ) z rn + w } and the set Y = { j E J :j 3 rn + w and g o ' ) < rn + w } have the same finite cardinality and there is a one-one mapping 6 of X onto Y.For j E X, let IiG) be u"(1)'-""(0)'"( 1) + u"(l)eo)-mn(0)eo)n( l),

which is a regular line. Then there is a partial table r* whose lines consist of the lines lie) for j E X, together with t h e A-line (5)

u"(l)", {(rn

+ w, g - ' ) } - +

u,w.

Let 2 be the set of all left sequences of r6 except a, let 2 * be the left sequence set of r*,and let 2" = 2 U 2 *; as r6 is an A -table and the left sequence of (5) and of every line li(j) for j E X is an extension of a, Proposition 0.6 and Proposition 0.5 apply and there is a finite partition set which is an extension of 2(,.Note that we can arrange that the added sequences of this partition set 2' are all left sequences of some extract of line (4) (which will be, by Proposition 3.23, a regular line, as it cannot have a"(1)' for a left sequence for any j E w, since ~ " ( 1 ) " belongs to t h e parition set Z,). Let r7be the A-table with left partition set 2' which i s a n extension of 2 * by regular lines with the same left and right sequences. The A-multiplicity of r7 is 1, with (5) the sole A-line; thus the function H defined by r7 belongs to B as it belongs to Pa(&) (as inspection shows that the partial table 2 * determines a one-one function whose range equals its domain). 6 yields the A-table Tx:First, each line of r 7 Composing r7 with r whose left sequence belongs to 2 is replaced by the line of r6 with the same left sequence. Second, every regular line of r7 with a right sequence u such that u C u' is replaced by the composition of this line, according to (1) of Definition 3.18, with the unique regular line which is an extract of line (4) and has u' as its left sequence (this extract exists as, for j E X , the line liu) has a (1)-ending right sequence, and the other regular lines have right sequences equal to their left sequences, chosen to be left sequences of extracts). Third, letting b' be the function with the same domain as l " ( b ) , and agreeing with it except that b'(m + w ) = g-lg, the line (5) of r7 is replaced by ~ " ( l ) ' " ,b'+ T , n + w. (6) The A-table r, defines the function HF and has exactly the same A -lines as r6,except that the A -line (5) is replaced by the A -line (6). If m + w is t h e only element belonging to the domain of b', (6) can be replaced in rxby the regular line u"(1)" -+ ~ " ( l ) "and + ~ otherwise , b' can be replaced in (6) by the b " E F i ( A ) which is like b' except for not


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having m + w as an element of its domain. In this way we obtain an A -table r9,also defining HF, which has A -multiplicity less than that of r, and r6.So HF E B and F = ( H - ’ ) ( H F )E B as H E B. Hence every element of Pa(&) defined by an A-table of A-multiplicity p + 1 belongs to B. Thus, by induction, every element of T a ( A ) belongs to B, and so B is equal to Ta(A). 14. Conclusions

In this section, as in $3, J will be t h e set of all positive integers. It was shown by Rabin (in [8]) that if a finitely generated group has a word problem which is solvable with respect to one presentation o n a finite set of generators then the word problem is solvable with respect to every presentation o n a finite set of generators. Thus we can say that the word problem for a finitely generated group is solvable or unsolvable without making reference to a specific presentation (on a finite set of generators). As only countable groups will be discussed in this section, it will be tacitly assumed that all the presentations referred to are on a finite or countably infinite set of generators. In the case where a presentation is given on a countably infinite set of gernerators of a group G there will be a function A whose domain is J and whose range is a set of generators of G, with A 0‘) ,being, for j E J, the element of G correspodning to the j-th element of an enumeration of t h e generators of the presentation. Note that A need not be a one-one function. From the proof of Proposition 4.1 it will be apparent that the word problem for t h e presentation depends in fact only on G and A. The ordering of the generators of t h e presentation (when there are more than finitely many of these) is important, however, so it will be assumed that with every presentation on a countably infinite set of generators there is also “given” an ordering, of the. same order type as J, of the generators; when the presentation has only finitely many generators we will assume that a simple ordering of these is also provided, but here we rely on the fact that the specific finite simple ordering assigned does not affect the Turing degree of the word problem (as this is explicated in the proof of Proposition 4.1). Proposition 4.1. Given a presentation of a countable group C there exists an isomorphic group C‘ which is a group of permutations of J such that only the identity element has a fixed point. Further, C’ can be construed as given by the same presentation as C in such a way that for j , k E J

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and a word w on the generators of the presentation, with g E C’ corresponding to w, it can be effectively determined whether or not g ( j ) = k relative to a solution of the word problem for the given presentation (of C and C’).

Proof. It will be convenient to make the obvious adjustments enabling us to speak of recursive functions on J instead of w. Let F be a free group with the same number of independent elements required to generate it as t h e number of generators in the presentation of the countable group C.The words on the generators of F (which will include the empty word) form a countably infinite set (even if F is finitely generated), so there is a one-one mapping 6 of this set onto J. The words of length one on F correspond either to the generators or their inverses; let K ,be the subset of J which is the image under 6 of the set of words corresponding in this way to the generators. Now, 6 can easily be chosen in such a way that K , is a recursive subset of J, while 4 and 8 are recursive functions, where 4 and 8 are the unique binary and unary functions from J into J such that for words u and v on the generators of F we have + ( t ( u ) , 6 ( v ) )= [ ( u u ) and 8 ( [ ( u ) )= [ ( u ’ ) where uv is the concatenation of the words u and v and u ’ is the inverse of t h e word u (obtained by reversing the order of u and then replacing generators by their inverses and vice-versa - the empty word being its own inverse, of course). The generators of the presentation of C can now be identified with the elements of K,,with the first generator of t h e presentation identified with the smallest element of K,,the second generator identified with the second-smallest element of K,,etc. Words on the generators of the presentation are thus implicitly identified with the image under 6 of corresponding words on the generators of F. Let K 2 be the set of all elements j E J such that the word on the generators of the preseitation with which j is identified corresponds to the identity element of C. Note that K 2 depends only on the choice of generators of C and the ordering given to the generators. We can take the Turing degree of the word problem for the presentation to be the Turing degree of K,.A recursive permutation of the ordering given the generators will not result in a change in the Turing degree of K2;in particular, the Turing degree will be independent of the ordering if the presentation has only finitely many generators. Using K2,the group C’ of permutations of J can be defined. First of all, let us define t h e equivalence relation = on J by setting j = k for j , k E J just in case the two words (on the generators of the presentation) with which they are identified correspond to the same

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element of C. Then we have + ( e ( j ) , k ) € K Z if and only if j = k . As 9 and 8 are recursive functions, we can determine effectively (relative to KZ)whether or not j = k for j , k E J. Note that for every element of C there are infinitely many distinct words on t h e generators of the presentation, and so elements of J, which correspond to it; thus we can now effectively determine (relative to K 2 ) the smallest, second-smallest, third-smallest . . . members of J corresponding to a given element of C (given by specifying a word on the generators of the presentation and so an element of J - to which it corresponds). We now define the function r by letting r ( j ) be, for j E J, the unique n E J such that j is the nth smallest element of the equivalence class it determines. For g E C we let 6 ( g ) E C’ be t h e function with domain J such that, for j E J, ( S ( g ) ) ( j ) is the unique element i E J such that r ( i ) = r ( j ) and i belongs to the equivalence class determined by + ( j , k ) , where k is the smallest element of J corresponding to g. Note that is a congruence relation relative to = , while for i, j , k E J we have + ( + ( i , j ) , k ) = + ( i , +(j,k ) ) . For g, h E C we have for every j E J (where k l , k z , and k 3 are respectively the smallest elements of J corresponding to gh, g , and h ) ( S ( g h ) ) ( j ) - + ( j , k l ) - + ( j , + ( k z , k 3 ) ) (as k l = + ( k z , k 3 ) ) = +(+kkz), k 3 ) = +((S(g))W7 k 3 ) - ( 6 ( h ) ) ( ( 6 ( g ) N hwhile r((S(gh))O’))= T o ’ ) = r ( ( 6 ( g ) N )) = r ( ( ~ ( h ) ) ( ( ~ ( g ) K Thus, i ) ) ) . ( 6 k h ) ) U )= ( S ( h ) ) ( ( a ( g ) ) ( j )for ) every j € J and so C‘ is a homomorphic image of C. Note that S ( g ) only has a fixed point if g is the identity element of C, as for j E 7 (with k E J being t h e smallest element of J correspomding to g) (6(g))(j) = j if and only if j = +(j, k ) , which holds if and only if k E Kz. Thus C’ is isomorphic to C and can be given by the same presentation - and as it can be effectively determined, relative to K z , whether or not for i, j , k E J we have both r ( i ) = r ( j ) and i a member of the equivalence class determined by +(j, k ) , the same applies to t h e determination of whether or not g ( j ) = i for g E C’.

+

Lemma 4.2. Given a presentation Q of a countable group G there exists a group V with a presentation Q’ such that V is equal to its commutator group, G is a subgroup of V, and the embedding of G in V is a Frattini embedding, while Q’ is obtained from Q by adjoining 2 generators and further relations, with the word problems for them Turing equivalent. Proof. Given Q and G, we see by Proposition 4.1 that G is isomorphic to a group Y of permutations of J such that only t h e identity element of Y has any fixed points; by Theorem 2.6 and Definition 2.1 we have Y = t where P is a subgroup of Sp(Y,”2)

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whose embedding into Sp(Y,"'2) is a Frattini embedding. Further, it appears from the proof of Proposition 2.7 that the word problems for the presentation Q of Y (and G and Y) and the presentation Q' of Sp( Y, "'2) obtained by adjoining to Q two generators corresponding to of Ft(Y;"'2) (these will the elements ( r : r l D ) " and (rorlrTTT)" generate Ft ( Y ; "2), in view of Proposition 1.5 and t h e fact that Ft( Y; "'2) = Ft("'2)) and all the relations valid in Sp( Y, "'2) will have the same Turing degree. We now just choose V so that there is an isomorphism of Sp(Y,"2) onto V whose restriction to Y is an isomorphism of Y onto G. AsV=Sp(Y,"'2), V is equal to its commutator group by Theorem 2.3.

Lemma 4.3. Given a presentation S of a countable group V equal to its commutator group there exists a simple group W with a presentation S' such that V is a subgroup of W, the embedding of V in W is a Frattini embedding, and S' is obtained from S by adjoining 2 generators and further relations, with the word problems for S and S' being Turing equivalent. Proof. Given S and V, we see by Proposition 4.1 that V is isomorphic to a group A of permutations of J satisfying condition C4 of $3 (and so C3) and conditions C1 and C2 relative to a solution of the word problem for the presentation S of V. Let A * be defined as in 43, and . let B be the subgroup of Pa(Ko) generated by A * U Ft ( K O )Then A = A * and the embedding of A * into B is a Frattini embedding by Theorem 3.30. Further, it appears from the proof of Theorem 3.29 that the word problems for the presentation S of A (and V and A * ) and the presentation S' of B obtained by adjoining to S two generators corresponding to the elements r : r l D and rorlrTTTof Ft(Ko) (these will generate Ft ( K O ) ,in view of Proposition 1.5) and all the relations valid in B have the same Turing degree. We choose W so that there is an isomorphism of B onto W whose restriction to A * ,is an isomorphism of A * onto V. As W = B, W is simple by Corollary 1.11, taking H there to be A *, equal to its commutator group as A * = V. Theorem 4.4. Given a presentation Q, of a countable group G there exists a simple group H with a presentation Qz such that G is a subgroup of H , the embedding of G in H is a Frattini embedding, and Q2 is obtained from Q1 by adjoining 4 generators and further relations, with the word problems for Q1 and Q2 being Turing equivalent. Proof. Given Q1 and G we obtain Q' and V by Lemma 4.2, adjoining


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two generators; as V is equal to its commutator group we can apply Lemma 4.3 (with Q‘ 9s S) to obtain Q2 (that is, S’) and the simple group H (that is, W), adjoining two more generators. The word problem for Q, is Turing equivalent to that for Q’, which is Turing equivalent to that for Qz, so the word problems for Q, and Q2 are Turing equivalent. As t h e embedding of t h e subgroup G of V into V is a Frattini embedding, and t h e embedding of the subgroup V of H into H is also a Frattini embedding, G is a subgroup of H and its embedding into H is a Frattini embedding.

Lemma 4.5. Given a finitely generated simple group H , the word problem for H will be solvable if and only if there is a presentation of H on a finite set of generators with a recursively enumerable set of defining relations. Proof. If the word problem for H is solvable, there will evidently be a recursively enumerable set of defining relations for H: after choosing a finite set of generators for H we can take the set of defining relations to consist of those words on the generators which correspond to t h e identity element of H ; this set is recursive. But o n the other hand, if there is a presentation of H on a finite set of generators with a recursively enumerable set of defining relations there must be a recursive enumeration of those words on the generators of the presentation which correspond to the identity element of H. Let u s choose a fixed word wo on the generators of the presentation which does not correspond t o the identity element of H. Then (because the defining relations of the presentation can be recursively enumerated) there is a recursive enumeration of those words on t h e generators of the presentation such that their (individual) adjunction to the defining relations determines a group in which wo corresponds to the identity element. As H is simple, every word on the generators of the presentation will occur in one of these two enumerations, but not both; thus t h e word problem for H is solvable.

If we start with a countable group which is not finitely generated, it is necessary to supplement Theorem 4.4 in order to obtain an embedding of the group in a finitely generated simple group. Using the Higman-Neumann-Neumann embedding technique we can embed a countable group in a finitely generated group (Rotman’s book [9] serves as a reference for t h e particular proof used here for Proposition 4.6).

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Proposition 4.6. Given a presentation Q of a countable group C on an infinite set of generators, there is a group C' generated by 3 elements such that C is a subgroup of C', the embedding of C in C' is a Frattini embedding, and the word problem for C' is Turing equivalent to the word problem for the presentation Q.

Proof. Let the generators of the presentation Q be, in order, g l , g 2 , .. . ,g,, . . . ; call the set of these generators f. Let F be the free group on the two (independent) generators a and b, and let G I be the free product C * F of C and F. A subgroup U of F with infinitely many free generators can be chosen so that there is a recursive set of words on a and b corresponding to t h e (independent) generators of U :let these words be u I ,u 2 , .. . , u., . . . . For i E J let wi be the word giui. Let GI and G',' be, respectively, t h e factor groups of G I obtained by putting all t h e elements of C and F, respectively, equal to t h e identity. Let W be the subgroup of G I generated by the wi, i E J ; then W is isomorphic to U via the mapping q defined by setting q ( w i ) = ui for i E J. (For in GI the images of t h e generators of W are independent free generators of t h e image of U.) As F is a free group and GI is a free product, G I has a presentation QI whose generators are, in order, a, b, g l ,g, . . . , g n , . . . and whose defining relations are exactly the defining relations of Q. We now let C' be t h e group given by the presentation Q' obtained by adjoining the generator t and the defining relations t-'wit = ui for i E J to the presentation Q1.Note that C' is generated by { a , b, t } , for in C' we have t-'giuit = ui and thus gi = tu,t-lu;' for i E J. As q is an isomorphism we can apply Britton's Lemma to conclude that G I is a subgroup of C' (that is, words on r U { a ,b } corresponding to distinct elements of GI remain so for C') and that a word w on f U { t , a, b } in which t or t - I occurs can only be equal to the identity element of C' if it contains (where u and u' are words on f U { a ,b } ) either a subword t-lut or t u f t - ' , where u is equal in G I to an element of W, and u' to an element of U (so the subword is equal to an element of U or W in C'). Now, given words p and q on f not corresponding to the identity element of C, if p and q are conjugate in C' there will be a word s with a minimum number of occurrences of t and t-I such that s-lps = q in C', and thus s-lpsq-' is t h e identity element of C'. By Britton's Lemma there will be a subword of s-lpsq-' of the form t - ' u t or t u r f - ' (with u and u' as above) if s is not a word on r U { a , b } . But the subword cannot be a subword of s-' or s (as it could be replaced in s-I or s by a word on f U {a,b } equal to it in C', and s would not be minimal in terms of number of occurrences


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of t and t - ' ) , nor can it contain p as a subword and thus be conjugate to p (because an element of U or W distinct from the identity will have, in GI, an image distinct from the identity, no element of C except the identity element can be conjugate in G I to an element of V or W). As these are the only possibilities for a subword, s must be a word on r U {a, b } so that elements of C conjugate in C' are conjugate in G I . And if two elements of C are conjugate in G I they are conjugate in C, since if they were conjugate in G I they would be conjugate in GY. Thus the embedding of C in C' is a Frattini embedding. By using the normal form for free products (and the fact that F is a free group) we observe that a freely reduced non-empty word on U {a,b } will be equal to the identity in G I only if there is some nonempty subword on r which is equal to the identity in C. This allows us to reduce the word problem for Q, t o that for Q, so their Turing degrees are the same. Similarly, Britton's Lemma allows us, relative to a solution of the word problem for Q1,t o either effectively determine that a word on r U {r, a, b } is not equal to any word on r U {a, b } or else obtain (by replacing subwords t-lvt or tv't-' by words on r U {a,b } ) a word on r U { a ,b } t o which it is equal in €' and thus in G I (because we can effectively determine of a word on r U { a ,b } what element of V,or W, it could be equal to in GI by examining the factor group GI). So the word problems for Q, and Q' have the same Turing degree. Finally, C' has a presentation Q" on the generators t, a, and b whose defining relations consist of the defining relations of Q with each gi,for i E J, replaced by the word tuit-Iu;I. That Q" is a presentation of C' can be seen from the fact that by defining gi as tuit-'u;' for i E J we obtain the relations t - ' w i t = ui for i E J as well Because we can effectively eliminate as the relations of Q (and 0,). the generators belonging t o r from words o n r U {r, a, b } by using the equality gi = tuit-Iu;I holding in C', it follows that the word problems for Q' and Q " have the same Turing degree. Thus, so d o the word as Q" is a presentation of C f on a finite problems for Q and 0"; number of generators, the word problem for C' is Turing equivalent to the word problem for the presentation Q.

r

The construction in the proof of Proposition 4.6 could be continued so as to reduce the number of generators to two, but (for finitely generated groups) it is more convenient here to rely on a result which can, for example, be extracted directly from the main theorem of [7]:

Proposition 4.7. Every countable group G has a Frattini embedding into

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a two-generator group H, with the Turing degree of the word problem for H being the same as that for G, if G is finitely generated. Proposition 4.8. Given a presentation Q of a countable group C there is a group C' generated by 2 elements such that C is a subgroup of C', the embedding of C into C' is a Frattini embedding, and the word problem for C' is Turing equivalent to the word problem for the presentation Q. Proof. If Q is a presentation of C on finitely many generators, we can use Proposition 4.7. If Q is a presentation of c on infinitely many generators, we can put Proposition 4.6 and Proposition 4.7 together, much as was done with Lemma 4.2 and Lemma 4.3 to obtain Theorem 4.4. The main theorem of Higman's paper [5] gives a characterization of finitely generated groups with recursively enumerable sets of defining relations. We state it in the following form: Proposition 4.9. A finitely generated group can be embedded in a finitely presented group if and only if it has a presentation on a finite set of generators with a recursively enumerable set of defining relations. Applying Proposition 4.9 to Lemma 4.5 yields: Proposition 4.10. A finitely generated simple group has a solvable word problem if and only if tt can be embedded in a finitely presented group. From Theorem 4.4 it follows that every finitely generated group can be embedded in a finitely generated simple group (by adjoining at most four extra generators) as was proved originally by Philip Hall; Theorem 4.11 below shows that, as a crude estimate (the best result will be given later), there will always be a Frattini embedding of a countable group into a simple group generated by no more than six elements. (In [4], Hall shows that every countable group can be embedded in a simple group with three generators; A.P. Goryushkin reduced the number to two in [3], and P.E. Schupp in [ I l l showed that these two generators could be chosen to be elements of order two and order three.) Theorem 4.11. Given a presentation 0 of a countable group C there is a Frattini embedding of C into a simple group H which is generated by 6 elements and has a word problem with the same Turing degree as the

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word problem for the presentation Q. Further, if the word problem for Q is solvable, there will exist a finitely presented group K such that H is a subgroup of K and K, like H, has a solvable word problem. Proof. In Clapham’s paper [2] it is shown that when a finitely generated group possesses a presentation o n a finite set of generators with a recursively enumerable set of defining relations it can be embedded in a finitely presented group in a way that preserves the Turing degree. So, in view of Lemma 4.5, the last part of t h e theorem follows from the rest. As for the rest, we put Proposition 4.8 and Theorem 4.4 together (in the same way that Lemma 4.2 and Lemma 4.3 were put together to obtain Theorem 4.4). Corollary 4.12. For any Turing degree of unsolvability, not necessarily recursively enumerable, there exists a simple group, generated by 6 elements, whose word problem has this degree. Proof. For any Turing degree of unsolvability there is a presentation Q of a countable group C which has this degree. Then by Theorem 4.11 the appropriate simple group H exists. At this point it is easy to show that all t h e results obtained by Boone and Higman in [ l ] concerning groups still remain valid when “simple group” is replaced’by “finitely generated simple group.” For instance, when this replacement is made in Theorem I‘ and Theorem IV of [ l ] we obtain stronger results which are weakened versions of Theorem 4.11 (weakened by failing to specify the number of elements required to generate it and omitting the requirement that t h e embedding be Frattini). When the “Corollary to Theorem IV” is strengthened by this replacement it becomes a weaker version of Corollary 4.12. The same replacement applied to Theorem I11 of (11 yields t h e first sentence of t h e next theorem: Theorem 4.13. A necessary and sufficient condition that any recursively enumerable class r of finitely presented groups have a uniformly solvable word problem is that there exist a finitely generated simple group H, and a finitely presented,group K , such that for each member G of r, G is a subgroup of H, and H is a subgroup of K. Further, it can be required that the embedding of G in H be a Frattini embedding for each member G of r, while H is generated by 6 elements. Proof. As the condition is stronger than that of the condition of

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Theorem 111 of [l], the proof of sufficiency given there (of a constructive form, Theorem III’, in which t h e replacement of “simple” by “finitely generated and simple” also produces a weaker result) applies here also. As regards necessity, we note as in [ I ] that if has a uniformly solvable word problem then the obvious presentation of the free product GI- of the members of r has a solvable word problem, and Gr contains each member of r as a subgroup. Further, t h e embedding of each member G of r i n t o GI-is a Frattini embedding (we examine, as in t h e proof of Proposition 4.6, the factor groups obtained by putting all the elements of G I .not belonging to the given G equal to t h e identity). Applying Theorem 4.11 to Gr and its presentation, we obtain t h e required H.

r

Without the use of Proposition 4.8 or Proposition 4.6 we can obtain Theorem 4.14 and (using exactly t h e same proof) Theorem 4.15. (When “simple group” is replaced by “finitely generated simple group” in Theorem I of [ I ] the result is Theorem 4.15.) Theorem 4.14. A finitely generated group G has a solvable word problem if and only if there is a Frattini embedding of G into a finitely generated simple subgroup of a finitely presented group. Proof. When there is an embedding of G into a finitely generated simple subgroup H of a finitely presented group K, H will have a solvable word problem by Proposition 4.10, and as G is a finitely generated subgroup of H, so will G. Since G has a presentation on a finite set of generators, the other part follows from Theorem 4.4 and Proposition 4.10. Theorem 4.15. A necessary and sufficient condition that a finitely generated grodp G have a solvable word problem is that there exist a finitely generated simple group H, and a finitely presented group K , such that G is a subgroup of H, and H is a subgroup of K . An algebraic characterization of finitely generated groups with solvable word problems is provided by Theorem 4.15, just as it is provided by Theorem I af [l]. Using only the construction of §3 as a basis we could prove a version of Lemma 4.3 in which V is not required to be equal to its commutator group and W (instead of being required to be simple) is only required to be a group such that every proper factor group is Abelian. This would lead to a proof of the variant of Theorem 4.15 obtained by replacing “finitely generated


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simple group H” by “finitely generated group H whose proper factor groups are Abelian.” Boone and Higman ask in [l] whether or not every finitely generated group with a solvable word problem is embeddable in a finitely presented simple group. If so, t h e finitely generated subgroups of finitely presented simple groups would consist precisely of those finitely generated goups which have a solvable word problem. The construction of 93 (unlike that of §2) does not preserve finite presentability (because applied to finitely presented groups with unsolvable word problems it yields a finitely generated simple group); something like the piecemeal approach (depending on a genetic characterization of the recursive function which solves t h e word problem for the group to be embedded) which Higman (in [ 5 ] ) and Clapham (in [ 2 ] )employ is probably needed for a positive answer. The question also arises as to whether or not a finitely generated simple group which is sufficiently well-behaved to have a solvable word problem must have a solvable conjugacy problem. The answer is given by Proposition 4.16, and follows from the fact that there exist finitely generated groups with solvable word problems and unsolvable conjugacy problems. Proposition 4.16. There exists a simple group generated by 6 elements which has a solvable word problem but an unsolvable conjugacy problem. Proof..Starting with a finitely generated group G with a solvable word problem but unsolvable conjugacy problem we obtain by Theorem 4.1 1 a Frattini embedding of G into a simple group H which is generated by 6 elements and has a solvable word problem. Since G is a finitely generated subgroup of H, H cannot have a solvable conjugacy problem, as then we would be able to determine effectively whether or not two elements of G were conjugate in H - and so in G. Although the proofs cannot be given here, there are some significant observations that can be made regarding the number of generators required for the simple groups refcrred to in the theorems above. First of all: Remark 4.17. For K E Fr and Y a finitely generated permutation group, Sp( Y,K) is generated by 2 elements of finite order.

The proof that Sp( Y,K ) can be generated by three elements relies on Proposition 2.4 (further reduction depends on a careful examination

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of the generators of Ft ( K ) ) . Note that Proposition 4.7 follows from Remark 4.17 (and the proof of Lemma 4.2), and Proposition 4.6 is only required when we deal with infinitely generated groups. So the construction of $2 can serve as a partial substitute for the Higman-Neumann-Neumann embedding technique. Consideration of Theorem 3.31 eventually leads to: Remark 4.18. When A is a group of permutations of J generated by k 3 2 elements, two of which are of finite order, then T a ( A ) can be generated by k elements, two of them of finite order. It is now a straightforward matter to show that, using Remark 4.17 and Remark 4.18, we can replace 6 by 2 in Theorem 4.11, Corollary 4.12, Theorem 4.13, and Proposition 4.16 (with these two generators both being elements of finite order); in addition, we can require that the simple group generated by two elements be such that the countable weak direct product of a collection of subgroups is always isomorphic to a subgroup. This last part depends on the following remark: Remark 4.19. When A is a group of permutations of J , the countable weak direct product of a collection of subgroups of T a ( A ) is always isomorphic to a subgroup of T a ( A ) . George Sacerdote’s paper [ 101 has been included in the references below because it uses Theorem 4.14 to give (among other characterizations) an algebraic characterization of groups with a solvable conjugacy problem. References [ I ] W.W. Boone and G. Higman. An algebraic characterization of groups with solvable word problem, J . Aust. Math. Soc. 18 (1974) 41-53. [2] C.R.J. Clapham. An embedding theorem for finitely generated groups. Proc. London Math. Soc. (3) 17 (1967) 419430. 1-71 A.P. Goryushkin, Imbedding of countable groups in 2-generated simple groups (in Russian), Mat. Zametki 16 (1974) 231-235. The English translation appears in Math. Notes 16 No. 2 (1975; for 1974) 725-727. 141 P. Hall. On the embedding of a group in a join of given groups. J. AUSI.Math. Soc. 17 (1974) 434495. [S] G. Higman, Subgroups of finitely presented groups, Proc. Roy. Soc. London, Ser. A. 262 (1961) 455475. 161 Ralph McKenzie and Richard J . Thompson. An elementary construction of unsolvable word problems in group theory. in: Word problems. ed. W.W. Boone.


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F.B. Cannonito, R.C. Lyndon, (North-Holland Publishing Co., Amsterdam, 1973) pp. 457478. [7] C.F. Miller and P.E. Schupp, Embeddings into Hopfian groups, J. Algebra 17 (1971) 17 1- 176. [S] M.O. Rabin, Computable algebra, general theory and theory o f computable fields. Trans. Amer. Math. Soc. 95 (19M) 341-360. [Y] J.J. Rotman, The theory of groups, an introduction (second edition, Allyn and Bacon, Boston, 1973). [lo] George S. Sacerdote, The Boone-Higman theorem and the conjugacy problem. Preprint version; for a later version see J. Algebra 49 (1977) 212-221. [ I 1 1 P.E. Schupp, Embeddings i n t o simple groups. J. London Math. Soc. (2) 13 (1976) 0-94.

S.I. Adian, W.W. Boone, G. Higman, eds., Word Problems I1 @ North-Holland Publishing Company (1980) 4 4 3 4 8

LAWS IN FINITE ALGEBRAS M.R. VAUGHAN-LEE Mnthematical Institute, Oxford

In the first part of this talk I shall describe a method which has been used with great success by several authors to show that certain classes of finite algebras have finite bases for their laws. (Here I am using the word ‘algebra’ in the sense of Universal Algebra.) In t h e second part of the talk I shall give some examples of finite algebras which do not have finite bases for their laws. If A is a finite algebra then t h e variety generated by A, VarA, is a locally finite variety. Any locally finite variety is generated by its critical algebras, a d so the critical algebras in Va r A play an important role. (A finite algebra is critical if it is not contained in the variety generated by its proper subalgebras and proper homomorphic images.) A locally finite variety V is Cross if (a) V has a finite basis for its laws, (b) V has only finitely many critical algebras. Finite groups, rings, Lie rings, and finite algebras in varieties all of whose algebras have distributive congruence lattices, all generate Cross varieties. The basic method of proving that a finite algebra A genergtes a Cross variety is to find a finite set r of first order sentences with the following properties. (1) Every algebra in V a r A satisfies r. (2) Every finitely genetated algebra satisfying r is finite. (3) There-are only finitely many critical algebras satisfying r. Then because V a r A is axiomatized by its laws, it follows from (1) that each sentence in r is a consequence of t h e laws of A. Since r is finite this implies that the sentences in r are all consequences of a finite set V of laws of A. V determines a finitely based variety V of algebras containing VarA, and (2) and (3) imply that V is Cross. The fact that VarA is Cross follows from the result proved below that a subvariety of a Cross variety is Cross. In most applications the sentences in r are laws, but this is not necessary, and is not the case in the proof that a finite algebra generates a Cross variety if it lies in a variety all of whose algebras have distributive congruence lattices. The power of t h e

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method is that it is not necessary to find an explicit basis for the laws of A, but only necessary to find a finite set of first order sentences which imply t h e existence of a Cross variety containing A. The following theorem shows the strength of property (b) of a Cross variety. Theorem. I f V is a locally finite variety then the following conditions are equivalent. (b) V has only finitely many critical algebras. (c) V has only finitely many subvarieties. (d) V satisfies the maximal and minimal conditions on subvarieties. The equivalence of (b) and (c) is well known (see L’vov [ l ] ) . The maximal condition o n subvarieties is equivalent to the condition that every subvariety (including V) is generated by a finite algebra. The minimal condition on subvarieties is equivalent to t h e condition that every subvariety is finitely based as a subvariety, that is every subvariety is determined by the laws of V together with a finite set of additional laws. Thus the result that subvarieties of Cross varieties are Cross follows from the fact that (b) implies the minimal condition on subvarieties. Proof of Theorem. Every subvariety of V is generated by its critical algebras, that is by some subset of the critical algebras of V. Thus (c) follows from (b). Trivially (c) implies (d). To show that (d) implies (b) we will assume that V is a locally finite variety which satisfies the minimal condition on subvarieties, but which has infinitely many critical algebras. We will then prove that V fails to satisfy t h e maximal condition o n subvarieties. First some notation: if A is an algebra then ( 0 s - l ) A is t h e set of proper factors (or sections) of A. A is critical if A @ V a r ( 0 S - l ) A . We will show that if V has a subvariety U such that V has infinitely many critical algebras A with Var(QS - l ) A 3 U, then V has a subvariety W with this property which properly contains U. By assumption V has infinitely many critical algebras, and if A is any algebra then V a r ( 0 S - l ) A contains t h e trivial variety determined by the law x = y. It follows by induction that V has a strictly ascending sequence of subvarieties. Now let U, W be subvarieties of V, and let U be a proper subvariety of W . Since the subvarieties of V satisfy the minimal condition, U is determined by t h e laws of V together with a finite set of additional laws. Let this additional set consist of n-variable laws. Then W must

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fail to satisfy some n-variable law of U . So W contains an n-generator algebra B,say, which does not lie in U. Then V < U v V a r B s W. Now there are only finitely many n-generator algebras in V, and so this implies that there are only finitely many subvarieties of V which are minimal with respect to strictly containing U. Let these be W , ,W2,. .., W,,,. Now suppose that A is a critical algebra in V and that Var(QS - l )A 3 U, but that Var(QS - I)A$ W, for any i = 1,2,. . . , in. Then Var(QS - l )A = V , which implies that A is n generator. But V has o n l y finitely many n-generator algebras, and so if V has infinitely many critical algebras A such that Var(QS - l ) A 3 U then for some i, 1 s i < rn, V has infinitely many critical algebras A such that Var(QS - l ) A 3 W,. This completes the proof of the theorem. Note that the proof actually gives that (d) implies that V has only finitely many S-critical algebras. (A finite algebra is S-critical if i t is not contained in t h e variety generated by its proper subalgebras.) This theorem, and the success of the method I have just described, suggest that the two properties (a) and (b) of a Cross variety may be closely related. However the examples below show that this is not the case. The first is a finite algebra which generates a variety satisfying (a) but not (b), and the second is a finite algebra which generates a variety satisfying (b) but not (a). The examples are non-associative algebras over an arbitrary finite field F. (A non-associative algebra over F is a vector space over F with a bilinear product.) From the point of view of Universal Algebra, non-associative algebras are very well behaved. They are unital, that is every algebra contains the unique one element algebra { O } . They have regular congruences, that is two congruences of an algebra are equal if and only if the corresponding congruence classes of 0 are equal. Most important of all, they have permutable congruences and their congruence lattices are moduJar. Finite associative algebras do generate Cross varieties, but Polin (21 has given examples of finite n o n associative algebras which do not have finite bases for their laws. The first of the examples below is a finite algebra A such that Var A does not satisfy the minimum condition on subvarieties. Thus Polin's examples and this example show that the variety generated by a finite non-associative algebra need not be finitely based, and may have subvarieties which are not finitely based even us subvarieties. First we establish some notation. A left normed convention is used: thus abc denotes ( a b ) c . We define ab" inductively by a b " = a, ab" = (ab"-')b. Thus ab' = ( a b ) b and a b 3= ((ab)b)b. F is an arbitrary finite field, and we let IF1 = q.

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The first example is an algebra A with basis a, b, c, d, ab, abc, abd, abcb, abcd, abdb, abdc, abcdb over F. The products of basis vectors are all zero, except for those of the form xy, x E {a, ab, abc, abd, abcb, abcd, abdb, abdc, abcdb}, y E {b, c, d } . These products are given by the following table.

a ab abc abd abcb abcd abdb abdc abcdb

b

C

d

ab ab abcb abdb abcb abcdb abdb abcdb abcdb

0 abc abc abdc abc abdc abdc abdc abdc

0 abd abcd abd abcd abcd abd abcd abcd

It is straightforward to check that the following three laws are a basis for the laws of A (as a non-associative algebra over F). XI(X2X3)

= 0,

XIX2X3X4X5 = XIX2X4~X3XSr XIX2X1X4

= XIX2X3X4.

We show that V a r A does not satisfy the minimum condition o n subvarieties by finding a finite algebra B E Var A which does not have a finite basis for its laws. Furthermore it can be shown that V a r B has only finitely many subvarieties. This implies the existence of a finite algebra C E Var B with the following properties. (1) C does not have a finite basis for its laws. (2) Every proper subvariety of V a r C is Cross. (3) V a r C has only finitely many subvarieties and only finitely many critical algebras. The algebra B has a basis a, b, c, ab, ac, abc, acb, abcb over E All products of these basis vectors are zero except for those of the form xy, x E {a, ab, ac, abc, acb, abcb}, y E {b, c}. Products of this form are given by t h e following table.

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b

C

a ab ac abc acb

ab ab acb abcb acb

abcb

abcb

ac abc ac abc abc + acb - abcb abc

The following laws a r e a basis for the laws of B. XI(X2X3)

= 0, XIX2X4X3XSr

xIx2x3x4xs=

x1x2xqx4 = XIX2X3X4,

u,=O,

i=1,2,3,4,5,

w, = 0,

i = 6 , 7 , . . . , where

UI

:= x l x : + ' -

u2:= x l x f x : u3:=

x1x:x3x4-

XIX2X2,

x1x2xq-

x1x;x.1+

xIxsIx4x.1-

u4:= x I x 2 x . 1 x 4 + x I x . 1 x 4 x 2 +

- xIx3x2x4-

XIX2X3,

XIXZX.1X4+ XIX2X4X.1, XIX4X2X3

x1x2x4x1-

XIX4X3X2,

u5:= xIx2x3x4xs - xIx3x2x4xs-xIx2x3x5x4+xIx3x2xsx4,and

w, := XIX2X3XOX7 *

* * x,x4x5

- XIX2X3XhX7

' '

- XIXlX2XhX7

* * * x,x4x5

. x,x5x4 + x I x 3 x 2 x h x 7 '

* '

xgX5x4,

for i = 6 , 7 , . . . .

Note that the first three laws of this basis form a basis for the laws of A, a n d so B E Var A. It is straightforward to check that B satisfies these laws, but it is extremely tedious to check that they d o form a basis f o r t h e laws of B. However there a r e no real difficulties. T h e main step is to notice that it is sufficient to consider laws of the form u = 0 , where u is a linear combination of terms of t h e form XlX,X, * . * Xk with i,j,. . . ,k 3 2. To show that B does not hbve a finite basis for its laws we proceed as follows. It is not difficult to show that for i = 1,2,3,4,5 any

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consequence of u, = 0 is of the form u + u = 0, where u = 0 is a law of A, and u is a (non-associative) polynomial whose degree is the same as the degree of u,. Similarly for i = 6,7,. . . any consequence of w, = 0 is of the form u + u = 0, where u = 0 is a law of A and u is a polynomial of degree i. Now if B has a finite basis for its laws, then for some integer n greater than the degree of u, for i = 1 , 2 , 3 , 4 , 5 , B has a basis consisting of the laws of A together with u, = 0 for i = 1 , 2 , 3 , 4 , 5 and w, = 0 for 6 s i < n. But then w. = 0 must be a consequence of these laws, and so, by t h e above remarks, w, can be expressed in the form u + u where u = 0 is a law of A and u is a polynomial of degree at most n - 1. Clearly we may assume that u and u only involve the variables x,, x z , . . . , x,. Now, for i = 1 , 2 , . . . , n, let T, be the endomorphism of the free non-associative algebra over F generated by xl, x2,. . . ,x. which maps x, to 0 and maps the other generators to themselves. Let p be the endomorphism (1 - .rr,)(l- r2).. . (1 - T,).Then if u is any polynomial of degree less than n, up = 0. So w, = wnp = up + up = up. This implies that w, = 0 is a law of A. Howeverit is easy to see that w, = 0 is not a law of A for any i 3 6 , and so B cannot have a finite basis for its laws. Let V be the subvariety of Var A determined by the laws u s= 0, w, = 0 for i = 6,7,. . . . Then the argument given above shows that V does not have a finite basis for its laws, and hence that V a r A does not satisfy the minimum condition o n subvarieties. It is possible to prove that any subvariety of V is determined by 4-variable laws. It follows that V has only finitely many subvarieties, and since B E V this implies that Var B has only finitely many subvarieties. However since V has only finitely many subvarieties it follows from the theorem that V is itself generated by a finite algebra. Thus the example B is, in a sense, redundant, but I have included it as it seems more satisfactory to have a specific example of a finite algebra which does not have a finite basis for its laws, rather than just to have proved the existence of one. References [ I ] I.V. L'vov, Varieties of associative rings, Algebra i logika 12 (1973) 269-297. [2] S.V. Polin, The identities of finite algebras, Sib. Math. J. 17 (1976) 1356-1366.


ALGEBRAISCH ABGESCHLOSSENE GRUPPEN Martin ZIEGLER Mathematical Insitute, Technical University, Berlin

Introduction

We characterize algebraically closed groups up to partial isomorphism by “recursion-theoretic” invariants and determine their properties by looking at these invariants. In t h e course of our investigations we obtain most of the known results* about algebraically closed groups (in [21, 22, 14, 15, 2, 31) from more general theorems. We started from the problems of A. MacIntyre in [14, 15, 171 and solve many of them. We survey some of our results: A non-trivial group M is algebraically closed if every finite system of equations with coefficients from M, which is solvable in a supergroup of M, has a solution in M. By the theorem of Higman, Neumann, and Neumann (“every isomorphism of subgroups is extendable to an inner automorphism of a supergroup”) one can conclude: Algebraically closed groups are w-homogeneous, i.e. every isomorphism of finitely generated subgroups is extendable to an automorphism (I, 1.10). w-homogeneous groups M are determined up to partial isomorphism (u.t.p.i.) by their skeleton Sk(M) (the class of all finitely generated groups which are embeddable in M) (I, 2.10). So in the study of properties which are compatible with partial isomorphism, it is enough to study the skeletons of algebraically closed groups. First we characterize those classes X of finitely generated groups which occur as skeletons of algebraically closed groups. X has to satisfy the following conditions (I, 3.8): (U) If G E X and H is a finitely generated group, which is embeddable in G, then H E X. * This paper - my “Habilitationsschrift” at the Technische Universtitat Berlin - was finished in its original version in June 1976. I here omit the last chapter of that version, which dealt with the construction of uncountable groups, because of the great progress made in this area since then; see e.g. [34] and [35]. 449

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(JEP) If H, G E X,then there is F E X,in which both H and G are embeddable. (AA) Every finite system of equations with coefficients from G E X which is solvable ifi a group containing G has a solution in a supergroup of G which belongs to X. B.H. Neumann showed that we can disregard the finitely generated groups with solvable word problem ([22], cf. [27]): Every finitely generated group with solvable word problem is in the skeleton of every algebraically closed group. MacIntyre proved the converse: If G is in the skeleton of every algebraically closed group, it has solvable word problem. And moreover, if G is in the skeleton of every algebraically closed group which contains the finitely generated group F, the word problem of G is Turing-reducible to the word problem of F ([15]). It was asked in [14] whether this theorem has a converse. The answer is negative ([3], 111, 3.5.3). We define a new notion of reducibility S * , stronger than Turing reducibility, and prove: for all finitely generated groups F, G the following are equivalent (111, 1.8.1): (a) for all algebraically closed M, F E Sk(M) 3 G E Sk(M). (b) W ( G ) s *W ( F )( W ( G )stands for the word problem of G). S * induces an equivalence relation =* (on t h e subsets of the set of natural numbers), which partitions 9 ( w ) into *-degrees. V, the set of *-degrees, is - like the set of Turing degrees - an upper semilattice in a natural way. “(b)+(a)” of 111, 1.8.1 shows that the skeleton of an algebraicall closed group M is already determined by W , ( M )= { W ( G ) / = *[ G E Sk(M)}. The following characterization of these subsets Z of V, which are of the form W , ( M ) for an algebraically closed group M, turns problems about algebraically closed groups into recursion theoretic problems. (Note that there is u.t.p.i. at most one M s.t. I = W , ( M ) (111, 3.12).) There is an algebraically closed group M s.t. I = W , ( M ) iff the following conditions hold: (i) If b E I, a s b, then a E I. (ii) If a, b E I, then sup(a, b ) E Z. (iii) If (A v B)E X for a r.e. horn class X and A E UI,then there is C E U I s.t. (A v C) E X. We call I C V satisfying (i), (ii), (iii) a closed ideal. (A v B is the disjoint union of A and B, A, B C w. R.e. horn classes - our recursion theoretic substitute for finite systems of equations - are defined from r.e. sets {(nb, . . . n:,) i E w } of finite sequences of natural numbers by X = { X C w ( V i ~ : E X A . . . A ~ : , E X ~ ~Cf. ~ E 111, X } . 3.8.) The main result of Chapter I1 is: For every r.e. horn class X and every natural number m, there is a finite system S ( & i ) , which is

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solvable in a supergroup of a group generated by g,, . . .g, if the word problem of ( g l , .. . gm) is contained in X (11, 2.6). As a first application of (111, 3.12) we give in 111, 4.1 a recursion theoretic proof of 111, 1.8.1 (a)+ (b): Let a, b E V. If for every closed ideal Z C V,a E Z j b E Z, then b s a. We prove the following three theorems in 111, 4 using the equivalence of algebraically closed groups and closed ideals: 1. Let the word problem of F be a complete set, then there is a smallest algebraically closed group M which contains F, i.e. M is embeddable over F in every algebraically closed supergroup of G (cf. 111, 2.2). Such an M exists uniquely. 2. M,, the countable algebraically closed group whose skeleton consists of all finitely generated groups with r.e. word problem, is minimal, i.e., isomorphic to each algebraically closed subgroup. 3. There are 2-1 algebraically closed groups of power wl, which are pairwise not partially isomorphic. All algebraically closed groups satisfy the same (elementary) Vzsentences (i.e sentences of the form V.? 39 cp(x, y ) , where cp is a quantifier-free formula). In [14] MacIntyre exhibited two algebraically closed groups and a V,-sentence which holds only in one of them. One of the problems in [14] was the question whether there is a V,-sentence by which one can distinguish algebraically closed groups. What are the elementary properties of algebraically closed groups in terms of their associated closed ideal? We assign to W , ( M ) the omodel o(M)= (w, + , * , W , ( M ) , E). We speak about w ( M ) in the language L z - the elementary language of second-order number theory. It is shown in Chapter IV that the properties of w ( M ) which are expressible by Lz-sentences, correspond to the elementary properties of M: For every Lz-sentence p there is an elementary sentence cp, which holds in an algebraically closed group M iff p holds in o ( M ) , and vice versa (IV, 2.4). We have a good control over the mutual dependence of he complexity of p and the complexity of 9. E.g., for every r.e. set A there is a V,-sentence cp, which is true in an algebraically closed group M iff A W , ( M ) . cp is false in M,,. If A is not recursive, there are also (by [lS]) algebraically closed groups where cp holds. This solves the mentioned problem from [14]. There are in fact 2”-many algebraically closed groups which satisfy pairwise different V,-sentences (IV, 3.5.1). (This was also proved in [MI.) By this method we can answer a further question from [14]. Which properties of algebraically closed groups are definable by elementary formulas? [ 141 contains some examples of “infinitary notions”, which

u

u

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are in algebraically closed groups elementarily expressible. There is, e.g., a V2-formula cp(xo,. . .x n ) , which holds in M for a l , .. . a, iff the group generated by a , , . . . a, is simple (cf. IV, 3.1.1). In all w-homogeneous M the L*-properties of w(M) correspond canonically to t h e “arithmetical” properties of M, which are described in I11 in two ways: As the properties which are obtained from “recursive” properties by quantification, and as the properties which are expressible in L*, a weak second order logic with “in-built number theory”. In algebraically closed groups all arithmetical properties are “elementary”. The examples in [14, 21 are easily seen to be arithmetical properties. This Characterization was independently given in [33]. The algebraically closed groups which we construct in this paper e.g., in the proof of I, 3.8 (characterization of the skeletons of algebraically closed groups) or in the proof of 111, 1.8.1 - are built up using step-by-step methods, not by “forcing in model theory”. On the contrary we used our methods to study generic groups. In V, 1 we give a simple description of the w-generic groups. An algebraically closed group M is a-generic iff w ( M ) is an elementary substructure of (w, + ,* ,9 ( w ) , E ). V, 2 gives a new approach to finite generic groups: We define the forcing relation “game theoretically”, in a way which clarifies the connection with the step-by-step methods. The usual syntactic definition of forcing is now Lemma V, 2.6. The results of IV (“arithmetical properties are elementary”) allow us to prove a simple characterization of the generic groups, in which the forcing relation and elementary sentences are not used. A group M is generic iff every “dense”, arithmetical family of finite systems of equations has a member which has a solution in M (V, 3.8). In Chapter VI we study the propertie’s of W , ( M ) , for generic M. A family 9 of r.e. horn classes is “dense” if 2 n US#0 for all nonempty r.e. horn classes 2. B C w is said to be “subgeneric” if B belongs to the union of every arithmetical dense family of r.e. horn classes. Theorem (VI, 3.3) is: An algebraically closed group M is generic iff W , ( M ) contains only subgeneric sets. We list four theorems, which we prove by this characterization: (VI, 3.3) For every generic group M there is a unique b E V s.t. W , ( M )= { a E VI a S b} (the b’s which occur in this way can be easily characterized). (VI, 3.4.4) Countable generic groups are minimal. (VI, 4.4.2) Every generic group M contains a finitely generated group F s.t. F * F & Sk (M).

u

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(VI, 2.10) A V3-sentence, which holds in some algebraically closed group, holds in all generic groups. The last theorem answers a question from [14]: Is there a V,sentence true in generic groups which does not hold in m-generic groups? It is enough to note that every V,-sentence which holds in w-generic groups, holds in all algebraically closed groups (IV, 3.6.2). C.F. Miller I11 proved many of our theorems (e.g. 11, 3.10, 111, 1.8.1) independently and at about the same time. I thank Susanne Ziegler and Robert Fittler for their help and encouragement. Bezeichnungen Eine Ordinalzahl ist die Menge aller kleineren Ordinalzahlen. w, die kleinste Limesordinalzahl, ist die Menge der naturlichen Zahlen. a + 1 ist der Nachfolger der Ordinalzahl a. Wenn K eine Kardinalzahl ist, ist K + die Nachfolgerkardinalzahl. Statt w + schreiben wir w , . Die Kontinuumshypothese (CH) sagt 2" = w , . ( A 1 ist die Machtigkeit der Menge A. A ist abzahlbar, wenn ( A 1 = w. Die Potenzmenge von A bezeichnen wir mit P ( A ) . 1st K eine Kardinalzahl, bedeutet PK(A) {X C A ( X ( < K } . AB ist die Menge der Funktionen f : A + B von A nach B. Fur C C A ist f[C] = {f(x)( x E C } . A ist der Vorbereich Vb f[A] = N b f der Nachbereich. idA ist die Funktion {(x, x ) xvonf, E A}. f setzt g fort, wenn g C f. f l c bezeichnet die Einschrankung f fl (CXNbf) auf C. Wenn f injektiv ist, ist f-' die Umkehrfunktion von f. 1st der gemeinte Vorbereich klar, schreiben wir Ax x * * . fur die Funktion, die jedem x . * x * * . zuordnet. fg ist die Funktion A x f ( g ( x ) ) . Eine Funktion f konnen wir als die Familie cf(x)( x E V b f ) auffassen. 1st a eine Ordinalzahl, nennen wir (ap /3 < a ) eine Wohlordnung von {aB /3 < a}.9B:= U { @ B/3 < a}.Mit 6 meinen wir eine endliche Folge ( b o , .. . ,bn-') aus yB.n ist die Lange l(6) von b. Wir schreiben haufig kurzer 6 E B. Wenn g E "C, meinen wir mit g(6) die Folge (g(bo),. . . g(bn-')).6"(c) ist die Folge ( b o , .. . ,bn-',c ) . Elemente von "B bezeichnen wir mit 6. 1st = eine Aquivalenzrelation auf A, ist A / = die Menge der Aquivalenzklassen a / =, a E A. C trennt A und B, wenn A C C und

I

r

-

1

1

1

~nc=0.

Wir betrachten L-Strukturen a, 23... fur eine abzahlbare Sprache L. ( L ist eine Menge von Konstanten, Funktionszeichen und Relationszeichen. AuSer im ersten Kapitel wird L immer die Sprache L G der "Gruppentheorie" sein: L G enthalt eine Konstante e fur das neutrale Element, ein zweistelliges Funktionszeichen fur die

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Gruppenmultiplikation und das einstellige Funktionszeichen-I, die Inversenbildung.) Wir unterscheiden vielfach die Struktur nicht von ihren Grundbereich, z. B. in Schreibweisen wie X f l ? ! = 0, x E ?I, ii E 2I, f[2I]. Mit 2I C B meinen wir aber, daB 2I Unterstruktur von B ist. 9f < B heiBt: 2I ist (isomorph) in B einbettbar. Ein Homomorphismus v :2I+ 2 ist “uber C”, wenn v auf C identisch operiert. Wir schreiben dann v : B und ?I < c B, wenn es eine Einbettung von 2I in B uber C gibt. ?I=cB sagt, daB es einen Isomorphismus von 2I und B uber C gibt. Aut(2I) ist die Automorphismengruppe von ?I. id, ist der triviale Automorphismus. Wenn B C a, bezeichnet (B) die von B erzeugte Unterstruktur. Entsprechende Bedeutung haben (B,C), (B, C ) , (ii), (B,ii), (6,ii), (a, i E I ) , . . .usw. Wir erweitern jedes L durch Hinzufugen einer Menge C = {c, i E w } von neuen Konstanten zu L ( C ) . Wenn ii €?I, wird ?I zur L(co,. . .~ , - ~ ) - S t r u k t u(r8 ,ii), in der c, durch a, interpretiert wird. Fur c, schreiben wir dann a,. ((ii), a‘) kurzen wir mit (6)’ab. v :(ii)+ (6) ist also genau dann ein Homomorphismus von (ii)’ nach (6)’,wenn v ein Homomorphismus von (ii) nach (6) ist, und v ( i i ) = 6. Entsprechend meinen wir mit (B,a)’ die L(E)-Struktur ((B, i),(z) usw. a und b haben den gleichen Typ uber C, wenn ( a ) ’ = =(b)’ (kurz fur (C, a)’s C(C, b)‘). Der Sprache L ( L ( C ) )ordnen wir die Menge L,, (L,,(C)) der elementaren L- (L(C)-) Formeln zu. Die Formeln bauen sich aus den atomaren Fomeln tl f 2 und R(rl,. . . t , ) mit 3x, A , i auf ( t , L- ( L ( C ) - ) Terme). Die Formeln von L-., erhalt man mit der zusatzlichen Bildungsregel “ist S eine Menge von Formeln, so sind auch A S und V S Formeln”. Diese Schreibweise benutzen wir, wenn S endlich ist, auch, um L,,-Formeln zu beschreiben. Statt A{cp,( i E I} verwenden wir auch A I E I q , .Wenn S leer ist, ist A S = T und V S = I (“wahr” und “falsch”). Wenn man nur hochstens abzahlbare Konjunktionen und Disjunktionen zulaBt, gelangt man zu den Formeln von L-.,. DaB in der L,,(C)-Formel cp hochstens die freien Variablen xo, . . .x. (und die neuen Konstanten co,. . . c k ) vorkommen, drucken wir durch cp(x0,. ..,x,) (cp(co,.. .ck, xo,. .. x,)) aus. Fur das Resultat der Ersetzung der x, durch Terme r, (und der c, durch t : ) schreiben wir cp (to,.. .) (cp ( t &. . . t o , . . .)). Entsprechend benutzen wir die Schreibweise S(xo, . . .) fur Formelmengen. Wenn 6 E 2I, bedeutet 2I I= cp(a’), daB (a, i)I= q ( 6 ) (oder ( 8 , i i ) C cp(E)). t(ii), fur einen L-Term t(Z), bedeutet genauer f(G)%. 1st f(2) eine Folge von L-Termen, ist die Folge to(Li), tl(ii), . . . .

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Die Menge S ( i ) von Formeln ist in !?1 erfullbar, wenn '?I I= S ( G ) fur geeignete ii aus !)I. Wenn T eine Menge v o n Aussagen ist, ist {cp T t- cp} die Menge der logischen Folgerungen von T. ?I = B heiBt, daB 91 und 23 elementar aquivalent sind, !?I < B,dal3 91 elementare Unterstruktur von % ist. Eine 3,-Formel +b hat in pranexer Normalform die Gestalt 3 i 1 V i 2 3 . . Qi.cp, cp quantorfrei. Dann ist -I t,b eine V,,-Formel. 91 B meint: alle 3,-Aussagen gelten genau dann in 91, wenn sie in B gelten. !?I ist 1-elementare Unterstruktur von %, wenn 91 C 93 und fur alle ii € 1 ' 1 jede 3,-Formel aus L ( t i ) , die in '8 gilt, auch in 9[ gilt, ? I K l B . Es sei n u n L = L G .T G ist die Menge der Gruppenaxiome. Gruppen sind Modelle von T G .Wir schreiben . . . k G * fur * * U T G t- . . . . Eine Gleichung (Ungleichung) ist eine Formel der Form ( i ) t l ( 2 ) t2(2),ti L(C)-Terme. Relativ zu T G ist jede Gleichung aquivalent zu einer Gleichung der Form t ( X ) = e. Weiter k o n n e n wir Terme mit Wortern identifizieren; das sind endliche Folgen von x, und x;'. Wenn W ein Wort in x ,,,... xk ist und gl,,...g, Elemente einer Gruppe G sind, bedeutet W(g,,,. . . g,) - wenn k s m - W ( g o , . . g k ) , und - wenn m < k - W(g,,,. . . g,, e, e,. . .). Gl(G, 2) ist die Menge aller Gleichungen aus L ( g ) ,Ungl(G, g) die Menge aller Ungleichungen aus L ( g ) , die in G gelten. Diag (G, g ) := Gl(G, 2) U Ungl( G, 2). Wenn 90,. . . ,q,, Aussagenvariable sind, nennen wir die Formel A q,, + (das bedeutet q1A q2 A * A q,, + 90 bzw. q 1A q2 A (ql A A q,, + I)eine positive bzw. negative aussagenlogische Implikation). Eine (positive bzw. negative) Implikation bekommt man, wenn man in einer aussagenlogischen Implikation die Aussagenvariablen durch Gleichungen ersetzt. ist die Godelnymer von (9) Vermoge einer Godelnumerierung iibertragen wir rekursionstheoretische Begriffe auf L,, (C). z. B. heil3t eine Menge von Formeln rekursiv aufzahlbar (r.a.), wenn die entsprechende Menge von Godelnummern r.a. ist. Fur Worter nehmen wir eine besondere Godelnumerierung. Wir wahlen eine rekursive Bijektion von yw und w. Dadurch ist jede naturliche Zahl s Godelnummer einer Folge (s(O), s(l), . . . s ( l ( s )- 1)). Dem Wort W = x 2 . . . x t , E~ E (1, - l}, ordnen wir die Folge mo,. .. m,, zu, wobei mi = 2ki ( = 2ki + l), wenn E~ = 1 ( = - 1). Die Godelnummer von W ist die Godelnummer s der Folge m o , .. . m,,.Wir schreiben W = V,. Wenn g' E G, definieren wir W ( G ,g) = {i V, (g) = e}. Wenn G endlich erzeugt ist, und g und 6 zwei Erzeugendensysteme von G sind, sind W(G, 2) und W(G, 6 ) rekursiv isomorph (W(G,2) =, W(G, 6)).Wir schreiben dann einfach W(G) fur W ( G , g ) ,das Wortproblem von G.

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G hat losbares Wortproblem, wenn W ( G )rekursiv ist. G ist rekursiv prasentiert, wenn W ( G )rekursiv aufzahlbar ist. Fur eine Untergruppe U von G definieren wir W(G U, g):= { i V,(g)E U } . Mit einer rekursiven Bijektion n, m (n, m ) ( 2n, m ) identifizieren wir w 2 mit w. Wir konnen so eine Teilmenge von w auch als zweistellige Relation auffassen. (a, b, c , . . . ) bedeutet ( . . ((a, b ) , c ) . * -). Die disjunkte Vereinigung A v B von A, B ist 2 A U (2B + 1) d. h. { 2 x I x E A } U { 2 x + l l x E B } . A v B v C v - - -bedeutet (( . . . ( ( A v B ) v C ) v . * . ). Wir verwenden Godelnumerierungen i w D,, D : , Df aller endlichen Mengen, aller Mengen mit hochstens einem Element, aller Mengen mit hochstens zwei Elementen. W, bezeichnet die z-te r.a. Menge, z ist ein Z?-Index von W,. A ist r.a. in B, A CraB, wenn fur ein z E w A = { x ( ~ u , w ( x , u , w )W € , A D , C B A D , C ~ \ B }A . ist TuringB. A ist m reduzierbar auf B, A S T B , wenn A % B und w \ A Sra reduzierbar (1-reduzierbar) auf B vermoge f, wenn f eine (injektive) rekursive Funktion ist und A = { x ( f ( x ) EB } , Schreibweise: A S m B , A SIB. Eine 2;-Menge A C w k ist eine Menge der Form { ( x l , .. . x k ) I 3 v l v v 2 3 * . QV.R ( v l , . .x t ) } fiir eine rekursive Relation R. A ist n;,,wenn w k \ A 2 ; ist, und A,", wenn A 2; und n: ist. A ist arithmetisch, wenn A 2 ; ist fur ein n E w. 4'"' ist die vollstandige Z$Menge aus [24]. 9"= {(n, k ) k E 4'")).Entsprechend definiert man n; (Z:,arithmetische) Teilmengen von w k X (P(w))'", (in [24] Zt), Z7:)). A ist implizit arithmetisch definierbar in B, wenn { A } = { Y C w ( Y ,B ) E R } fur ein arithmetisches R C (P(w))'. = Zu jeder Teilmenge Y von P ( w ) gehort das w-Modell ( w , 9) (w, + ,. , E ). Wir beschreiben w-Modelle mit L2-Aussagen. L 2 hat Variable v I , . . . w,,. . . fur naturliche Zahlen, X , Y,2,. . . fur Elemente von Y und fur jedes n E w eine Konstante. Aus den atomaren Formeln rl = rz, rl E X (r, Zahlterme) baut man die Formeln von L 2 mit 7 ,n , I u , und 3X, auf. R ist analytisch, wenn R in (0, P ( w ) ) mit einer L2-Formel definiert werden kann. A ist analytisch in B, wenn A = {x (B,x ) E R} fur ein anaIytisches R. Die L2-Formeln ohne gebundene Mengenvariable beschreiben gerade arithmetische Eigenschaften. D a die Gultigkeit einer solchen Formel cp nicht von Y abhangt, iibernehmen wir haufig cp in die Metasprache und schreiben q ( A ) statt (w, * . )I= cp(A). Wir nennen eine solche Formel X;-Formel, wenn sie eine ZZ-Menge definiert. Wir denken uns alle arithmetischen Relationen als Relationszeichen in L z aufgenommen.

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Eine L2-Formel der Form 3G,,XlVG2,X 2 .* . Qtj,,,XnR fur ein rekursives R n e n n e n wir eine 2 :-Formel. Entsprechend definiert man fiL-Formeln. Wenn A eine Teilmenge der Gruppe G ist, bezeichnet ( A ) Nden von A erzeugten Normalteiler. G ist genau dann einfach, wenn fur . ist die v o n (von den Elementen von) S alle a, b f e a E ( b) N Fr(S) erzeugte freie Gruppe. Fr(6) ist Fr({bo,. . . bn-l}). G wird von der Menge von Gleichungen S(g) prasentiert, wenn - falls S(g) = { W ( g )= e W E A } - (G, 6) isomorphes Bild von (Fr(g)/( W(g)l W E A ) Ng) , ist. Insbesondere wird also G von den 2 erzeugt. Wenn H, F Untergruppen von G sind, ist "H .F direkt", wenn H F = H x F (inneres Produkt) ist. Das heiat H n F = E (E ist die triviale Gruppe) and H C Z'( F ) ( Z'( F ) ist der Zentralisator von F in G). G sei Untergruppe von H, E Wir fassen das amalgamierte Produkt H * G F als Obergruppe von H und F auf. Das amalgamierte der Produkt der Obergruppen F; von G, j E J ist * G j E J}. @X ist die direkte Summe aller Gruppen aus 9, wir schreiben auch BG,,G oder X,,,F,. Z ( G ) ist das Zentrum von G, [g, h ] der Kommutator g h g - l h - ' . r, = Axgxg-' ist der zu g gehorende innere Automorphismus von G.

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I. Algebraisch abgeschlossene Gruppen als o-homogene Strukturen

Im ersten Abschnitt zeigen wir, daB sich jede Gruppe in eine algebraisch abgeschlossene (a.a.) Gruppe einbetten laBt (1.3). D a jede 1-elementare Unterstruktur einer a.a. Gruppe wieder a.a. ist (1.5.2), findet man zu jeder abzahlbaren .Gruppe eine abzahlbare a.a. Obergruppe (1.6). Das zeigt, weil es 2" nicht-isomorphe endlich erzeugte Gruppen gibt, daB es auch 2" nicht-isomorphe abzahlbare a.a. Gruppen gibt. Um die Isomorphietypen der abzahlbaren a.a. Gruppen zu bestimmen, beachten wir, daB a.a. Gruppen w-homogen sind (2.1): Jeder partielle Isomorphismus endlich erzeugter (e.e.) Unterstrukturen laBt sich zu einem Automorphismus fortsetzen (1.9). Abzahlbare w-homogene L-Strukturen (fur eine Sprache L ) sind bis auf Isomorphie eindeutig durch ihr Skelett - die Klasse aller einbettbaren e.e. L-Strukturen - bestimmt (2.11.1). Allgemeiner :Zwei w-homogene Strukturen sind genau dann partiell isomorph (oder Lm,uaquivalent), wenn sie das gleiche Skelett haben (2.10.1). Wir konnen

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also (zumindest theoretisch) Eigenschaften w-homogener Strukturen, die nur vom “partiellen Isomorphietyp” abhangen, am Skelett erkennen. So ist z. B. (wenn L die Sprache der Gruppentheorie ist) eine w-homogene L-Struktur genau dann eine a.a. Gruppe, wenn ihr Skelett aus Gruppen besteht und algebraisch abgeschlossen ist, d. h. daB jedes uber einer Gruppe des Skeletts losbare Gleichungssystem in einer Obergruppe aus dem Skelett losbar ist (3.7). Um das Studium der w-homogenen Strukturen auf das Studium der Skelette zuruckzufuhren, mussen wir die Klassen der e.e. Strukturen kennzeichnen, die als Skelett vorkommen. (3.5) sagt: Eine Klasse X endlich erzeugter Strukturen (in der hochstens w , viele verschiedene Isomorphietypen vorkommen) ist genau dann Skelett einer w homogenen Struktur, wenn X “unterstruktur-abgeschlossen” ist (d. h. mit einer Strcktur auch jede in sie einbettbare e.e. Struktur enthalt), die “joint embedding property” und die Amalgamierungseigenschaft besitzt. Dieser Satz liefert uns ein einfaches Verfahren zur Konstruktion a.a. Gruppen. Wir bekommen z. B. eine (dadurch eindeutig bestimmte) abzahlbare a.a. Gruppe deren Skelett gerade aus den e.e. Gruppen besteht, deren Wortproblem rekursiv aufzahlbar ist (3.9.2). Im vierten Abschnitt konstruieren wir mit den Methoden von 3 zu jeder abzahlbaren a.a. Gruppe eine partiell isomorphe Gruppe der Machtigkeit wl. In 4.6 und 4.7 werden einfache Bedingungen fur w homogene Strukturen angegeben zu einer Struktur der Machtigkeit w 1 partiell isomorph zu sein. Wir zeigen auf drei verschiedene Arten, daB a.a. Gruppen diesen Bedingungen genugen: in 4.3, 4.9, 4.10. 1. Einfache Konstruktionen algebraisch abgeschlossener Gruppen Ein Korper heiBt algebraisch abgeschlossen, wenn fur jedes nichtkonstante Polynom p E K [ X ] - ein Polynom also, das in einem Oberkorper eine Nullstelle hat - die Gleichung p ( x ) = 0 eine Losung hat. Hilberts Nullstellensatz zeigt, daB in allen algebraisch abgeschlossenen (a.a.) Korpern sogar jedes endliche System von Gleichungen

pI(xI, . - . x , ) = O , . . * , ~ n ( ~ l ,..* x , ) = O 9

3

@I,.

. . ,pn E K [ X l , .

a])

das in einem Oberkorper von K eine Losung besitzt, schon in K losbar ist. D a man nicht erwarten kann ein ahnliches Resultat fur Gruppen zu erhalten, definiert man:

Delinition 1.1 (Scott, Neumann). Eine nichttriviale Gruppe G ist algebraisch abgeschlossen, wenn fur jede endliche Folge von Wortern


459

Wl(yll,.. . ,y,, xo, . . . ,xs), . . . W,(yo,. . . ,x,) und Koeffizienten go,. . .,g, € G das Gleichungsystem (“mit Koeffizienten aus G”) { Wl(g0,.. . ,g, xll,.. . ,x.)

=

e, . . . , W, (gll,.. . ,x,) = e } (=: S(g,i ) )

in G losbar ist, wenn es in einer Obergruppe von G losbar ist. (E, die triviale Gruppe, hat diese Eigenschaft, gilt aber per definitionem nicht als algebraisch abgeschlossen. Eine Losung von S(g, i )in G ist eine Folge von Elementen f € G, auf die alle Gleichungen von S zutreffen, fur die also G I= A S ( g , j ) . Statt “a.a. Gruppe” sagen wir auch “a.a.Gr.”, und fur “in einer Obergruppe von G losbar” “uber G losbar”.) Wenn F und H Obergruppen von G sind, sind sie uber G in das amalgamierte Produkt F * G H einbettbar. Wir schliefien daraus Bemerkung 1.2. Ein Gleichungssystem mit Koeffizienten aus G C F ist uber G losbar gdw. es uber F losbar ist.

In [26] wurde der erste Satz iiber a.a. Gruppen bewiesen: Satz 1.3. Jede Gruppe ist Untergruppe einer algebraisch abgeschlossenen Gruppe. Beweis. Sei G eine Gruppe und ( S , ] p E a \{O}) eine Wohlordnung aller endlichen Gleichungssysteme mit Koeffizienten aus G, die uber G losbar sind. Wir definieren rekursiv eine aufsteigende Folge von Gruppen Go, p < a : Go:= G ; fur p > 0 sei G, eine Obergruppe von U{G,(y < p } , in der S, eine Losang hat (vgl. 1.2). Wir bilden G + : =U{G,I p < a } und iterieren diese Konstruktion: G o : =G ; G“+’:=(G”)’. Die gesuchte Obergruppe von G ist U{G‘‘I n < w } . 0 Folgerung 1.4. Z u jeder Menge X von Gruppen, gibt es eine a.a.Gr., in die jedes Element von X einbettbar ist. Beweis. Man wende 1.3 auf

BGEtKG an. 0

Wir geben noch zwei andere Konstruktionsverfahren fur a.a. Gruppen an. Lemma 1.5. 1. Die Vereinigung einer aufsteigenden Kette von a.a. Gruppen ist algebraisch abgeschlossen. 2. Jede l-elementare Unterstruktur einer a. a. Gncppe ist algebraisch abgeschlossen.

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Wir werden spater sehen, daB sich 2 umkehren 1aBt: a.a. Untergruppen sind 1-elementare Unterstrukturen. Zum Beweis von 1: Ein endliches Gleichungssystem, das iiber der Vereinigung einer Kette a.a. Gruppen losbar ist, besitzt schon eine Losung in jedem Kettenglied, in dem alle Koeffizienten des Systems liegen. Z u 2: Sei G < I M a.a. Wenn S ( g , i ) iiber G losbar ist, ist S auch iiber M und also in M losbar. Aus MI= 3 i A S(2, i) folgt wegen der Voraussetzung G I= 32 A S(g, i ) :Es gibt eine Losung von S in G. 0

Folgerung 1.6. 1. A sei eine Teilmenge der a.a.Gr. M. Dann gibr es eine a.a.Gr. A C N C M mir 1 NI = 1 A I + o. 2. Zu jeder Gruppe G und jeder Kardinalzahl K 2 1 GI gibt es eine a.a.Gr. M 3 G, K = / M I . (Aus 4.5.2 wird sich ergeben, daB a.a. Gruppen nicht endlich sind.) 1 bekommt man aus 1.5.2 mit Hilfe des Satzes von Lowenheim Skolem, der sogar ein A C N < M, 1 N I = I A I + o liefert (siehe auch 2.11.5). Nach 1.3 ist G in beliebig groBen Gruppen enthalten, also folgt 2 aus 1. 0 In jeder Machtigkeit gibt es also a.a. Gruppen. Uberabzahlbare a.a. Korper fester Charakteristik sind durch ihre Machtigkeit (bis auf Isomorphie) eindeutig bestimmt. Es gibt abzahlbar viele abzahlbare a.a. Korper. Wieviel a.a. Gruppen gibt es in jeder Machtigkeit? Im Laufe dieser Untersuchung werden wir zeigen, daB es in jeder iiberabzahlbaren Machtigkeit 2-1 a.a. Gruppen gibt. (Tatsachlich gibt es in jeder Machtigkeit K genau 2" nichtisomorphe a.a. Gruppen, siehe [341-)

Satz 1.7 [22]. Es gibt 2" abzahlbare a.a. Gruppen. Der Satz folgt aus

Lemma 1.8 [19] (siehe auch 111, 3.5.2). Es gibt 2" uon zwei Elementen erzeugte Gruppen. Denn jede abzahlbare Gruppe enthalt nur abzahlbar viele von zwei Elementen erzeugte Untergruppen. Jede von zwei Elementen erzeugte Gruppe ist aber in einer abzahlbaren a.a.Gr. enthalten Andererseits gibt es iiberhaupt nur 2" abzahlbare Gruppen. 0 Die Losbarkeit eines endlichen Gleichungssystems S mit Koeffizienten g E M in der a.a.Gr. M hangt nach 1.2 nur von ( g ) (bzw. den Obergruppen von ( g ) ) ab. Der nachste Satz zeigt, daB alle


Eigenschaften von 2 in M nur vom Isomorphietyp von M ist “w-homogen”.

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Satz 1.9. In algebraisch abgeschlossenen Gruppen laPt sich jeder Isomorphismus z wischen endlich erzeugten Untergruppen zu einem inneren Automorphismus fortsetzen. Z u m Beweis verwenden wir Lemma 1.10 [ 8 ] . Jeder Isomorphismus zwischen zwei Untergruppen einer Gruppe lafit sich zu einem inneren Automorphismus einer geeigneten Obergruppe fortsetzen.

Sei nun M algebraisch abgeschlossen, g E M und p : (2)- M eine Einbettung. In einer Obergruppe von M gibt es einen inneren Automorphismus 7, 3 p. f ist also eine Losung des Gleichungssystems {xg,x-’ = p ( g , ) i = 0 , . . . . , r}. Wenn f ’ eine Losung in M ist, leistet 7,. das verlangte.

1

Im nachsten Abschnitt untersuchen wir beliebige Strukturen fur eine (abzahlbare) Sprache L, die ein Analogon von 1.10 erfullen. 2. a-homogene Strukturen Definition 2.1 (siehe [7]). Eine Struktur $%I ist w-homogen, wenn sich jeder Isomorphismus zwischen endlich erzeugten (e.e.) Unterstrukturen “schrittweise” fortsetzen lafit; d. h. wenn 91 e.e., ?I Cm, p :?1+m eine Einbettung ist, gibt es fur jedes a E \JJ7 eine Fortsetzung p c v : (?I, a)-+s%I. Bemerkung 2.2. Nach 1.10 sind a.a. Gruppen w-homogen. 1st rol w-homogen und ii E YA, so ist auch (die L(E)-Struktur) (m, ii) w-homogen. Lemma 2.3. 1 . $YR . ist genau dann w-homogen, wenn fur jedes e.e. ?I C ‘%I jede endlich erzeugte Oberstruktur von ?I, die in lm einbettbar ist, iiber ?I in 2J2 einbettbar ist. 2. Die Vereinigung einer aufsteigenden Kette von w-homogenen Strukturen ist w-homogen. 3. A sei Teilmenge der unendlichen w-homogenen Struktur m. Dann gibt es eine w-homogene Struktur A C ?I C lllz mit (?[ ( = ( A( + w.

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Beweis. 1. Sei 9Jl w-homogen, e.e. '? CI W,2l C 23 e.e. und p :23 +9Jl eine Einbettung. Wir setzen den Isomorphismus p &, :p [a]+ '3 fort Dann ist vp : 2 3 - ~2172 die gesuchte Einbettung. Sei zu v : p [23]+92. umgekehrt die Bedingung von 1 erfult und p : l?l+9Jl eine isomorphe Einbettung einer e.e. Unterstruktur, a E 92.Wir wahlen eine Oberstruktur p [a]C 23 mit einem Isomorphismus p C v :(a,a ) + 23. 9' 3 ist vermoge v-' in W einbettbar. Also gibt es eine Einbettung T : 23 +ciurl 9Jl. ~p : (a, a ) + W ist die gesuchte Fortsetzung von p. Wir uberlassen dem Leser den Beweis von 2 und 3. 0 w-homogene Strukturen sind bis auf "partielle Isomorphie" durch ihre endlich erzeugten Unterstrukturen eindeutig bestimmt:

Delinition 2.4 ([ll]). 1. Zwei L-Strukturen W,8 sind partiell isomorph - W = p 8 -, wenn es eine nichtleere Familie I von Isomorphismen zwischen Unterstrukturen von 9Jl und '% gibt, die folgende Eigenschaften hat: (a) fur alle 7r E I, a E 9Jl gibt es v 3 T , v E I, a E Vb v. (b) Fur alle T E I, b E W gibt es v 3 T , v E I, b E Nb v. 2. 9Jl ist partiell isomorphe Substruktur von 8 - 9Jl 2 . D a n n g i b t e s i n F * u F ' f u r a l l e g E F \ U g ' E F ' \ U , h E (F*,F')\ U ein x E F*"F', so dal3 [ h ,xg'gx-'1 6! U. (Wenn zum Beispiel h = kk', k E F\ U, k ' E F'\ U, wahle y E F\(U U U g ) und y ' E F'\(U U ( k ' ) - ' U ) und setze x = y'y.) Fur g E F sei g' das entsprechende Element beim Isomorphismus (f)' (f')'. Dann ist g U gdw. g'g-' # e. Weiter ist go, gl $Z U gdw. ex. x E F *,F' mit [g&;',xg;g;'x-'] # e usw. Also g o , . . . ,g, U gdw. ex. xI, x z . . . x, E F *,F', so dal3

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- . [[g~g;',xlf~f;'x;'], xZf;f;'~;'], AISO ist w \ W ( F 1 w(F*"F'). [ [ e

. . .,I,xnfiffnlxi'] # e.

Satz 2.16. M sei eine algebraisch abgeschlossene Gruppe. Es sind

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(a) Fur alle G C F und G C H aus Sk(M) mit F r l H = G gibt es einKESk(M) mitFUHCK. (b) Wenn H, F E Sk(M) und G eine e.e. gemeinsame Untergruppe von H und F isr, ist H *G F E Sk(M). (c) Fur alle G E Sk(M) und e.e. H ist W ( H ) G .W ( G )3 H E Sk(M). Bemerkung. Fur algebraisch abgeschlossene Gruppen M rnit der Eigenschaft von 2.16 gilt: Jede Gruppe N rnit Sk(N) C Sk(M) ist in einer a.a. Gruppe N' enthalten, die zu M partiell isomorph ist. Insbesondere gibt es beliebig, groi3e zu M partiell isomorphe Gruppen. Beweis der Bemerkung: S(g^,i) sei uber N losbar. G sei von den g erzeugt. Wenn H von S uber G prasentiert wird, ist W(H)=S,W ( G ) und also H E Sk (M). Man sieht leicht, dai3 Sk ( N * G H) C Sk ( M ) . Auf diese Weise kann man N algebraisch abschliei3en. Beweis. (b)+(a) is klar. (c)+ (b). Aus 11, 3.3 folgt W ( H G ) seW ( H )und W(F G )6 =W(F).Wie im Beweis von 2.15 folgt W(H *OF)seW ( H )v W ( F )=* W ( H x F ) . Weil H x F E Sk(M), ist also H*GFESk(M). (a)+ (c). Wenn (a) gilt, ist W(F G )6 *W ( F )fur alle G C F aus dem Skelett von M.Das folgt aus 2.14. Wir brauchen eine Konstruktion, die uns noch spater nutzlich sein wird: c, b seien Erzeugende einer Gruppe 8 . Wir definieren bk = c-'bck fur k E Z und ai = bi+lbob;21b;1 fur i E o.8 sei durch die Relationen caiG ac, aia,A aiai, a;& e, aib= bai, i, j E u,prasentiert. Aus den Relationen fOlgt Uibk = b&, bk+i+:bk= aibkbk+i+lfur alle i E 0, k E 2.Man sieht leicht, dai3 man jedes Element von 3 in der Form c'b;; b;: schreiben kann. Daraus gewinnt man eine Darstellung jedes Elementes * . . ainb;; * b = ; mit von H in der Form c 'aai, i l < i 2 < * * - i nk, l < k Z < . . . k , , k , n i E Z , n , m Eo, n , # 0 . In ([29]) wurde (implizit) gezeigt, dai3 diese Darstellung eindeutig ist (siehe auch ([3])). 23 sei die von den ai erzeugte Untergruppe von 8. '?Bliegt im Zentrum von 8 und ist direkte Summe der zyklischen Gruppen der Ordnung 2 ( a i ) .Jede Untergruppe von 23 ist Normalteiler von 8 . Sei nun G endlicherzeugte Untergruppe von M and W ( H )6,W ( G ) .Wir konnen G f E annehmen. Dann gibt es eine rekursive Funktion f : 0 2 + 9- ( w ) mit

I

I

I

-

i E W ( H ) gdw. ex. k rnit f ( i , k ) C W ( G ) .


499

Wir indizieren die ai doppelt vermoge einer rekursiven Bijektion zwischen w und w 2 . Es gibt also eindeutig j , k rnit = ai, und umgekehrt kann man zu j , k genau ein i mit a,= aj,kfinden. Die Untergruppe N von B sei erzeugt von

{ai,j * a i l l f ( i , j ) C W ( G ) , f ( ik, ) c W ( G ) I . Wir betrachten a / N . Die Erzeugenden cN and bN nennen wir E und & Ebenso bezeichenen wir aiN mit iii und biN rnit 6. Wenn k # O oder n,#O, und k l < k t < * ist C k i i i l * . . iiim6;;* * 6;; ! i i BIN. Wir nehmen nun eine isomorphe Kopie 5%’ von a, die von c’ und b’ erzeugt wird. A bezeichne die von den a: fur i E W ( H )erzeugte Untergruppe von 23’.Wenn a : E A, gibt es ein k E w mit f ( i , k ) C W ( G ) ;wir identifizieren a: mit i i , k (Wenn f ( i , j ) C W ( G ) ,ist di,k = iii.j) und bilden das amalgamierte Produkt BIN *a; B’. Das folgende System von Gleichungen und Ungleichungen in den Die Menge der “Unbekannten” 6, c, b‘, c ’ gilt offenbar in B / N *A 8’. Gleichungen ist aufzahlungsreduzierbar auf W ( G ) ,die Menge der Ungleichungen 1-aufzahlungsreduzierbarauf W ( G ) .Nach Satz 11, 3.15 gibt es eine Losung des Systems in M. Die ai, a:, 6k*,6;. fassen wir als Worter in c, 6, c f , b ’ , auf. ( i , j , k, i l , . . . in E w ; k l < k 2 < * . * < k , E Z)

0. ai,ks a: wenn f ( i , k ) C W ( G ) . 1. a 2 -- e, cai = aic, aiaj A up,, crib =hi. 2. ickai;..ai~6;;...b;;=aa: wenn k # O oder n,#O. 3. i aj,kai,* - * aim=a: wenn i l , . . . in# (i,k ) , und f ( i , k ) l t W ( G ) . 4. i a:, . . = a: wenn i 6i { i l , . . . i n } . Wir andern unsere Bezeichnungen und bezeichnen mit b, c, b’, c f eine Losung des Systems in M. Wir zeigen, dab i E W ( H )gdw. a : € (b, c ) . Daraus folgt dann die Behauptung mit W ( H )6, W((b,c, c’, b‘) (b, c)) und nach unserer Bemerkung am Anfang W((b,c, c’, b’) (b, c ) )6*W((b,c, c’, b’)) aus 1.4. Sei i E W ( H ) .Dann gibt es ein k rnit f(i, k ) C W ( G ) .Wegen 0 ist a : = aik E ( b , c ) . Sei umgekehrt a : € (6, c ) . Weil 1 gilt, ist a : = ckui,- - - a,b;; - - * b;: fur geeignete k, i l . . ., nl .. . ,k , - * * . Aus 2 folgt k = 0 und m = 0, also fur a : = ai,* . aim.Wegen 3 kann man schreiben a : = ajI,,]* * * f ( j , m , ) C W ( G ) ( r = 1,.. . n ) . Aus 0 folgt a : = a/,* - ai’.. Nach 4 mu0 i E Ql,. . .jn} sein, d. h. i E W ( H ) . 0

I

1

3. Ideale von *-Graden und a.a. Gruppen Definition 3.1. Wir ordnen jeder Gruppe M die Menge W , ( M ) C V der *-Grade aller endlich erzeugten Untergruppen von M zu. X C V

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ist durch UX eindeutig bestimmt: X W , ( M ) schreiben wir W * ( M ) .

u

= {a E

I

V aC

UX}. Fur

Aus 1.4 schlieBt man sofort

Lemma 3.2. M sei algebraisch abgeschlossen. Dann ist fur alle e.e. F F E Sk(M) gdw. W ( F )E W * ( M ) . Mit diesem Lemma konnen wir I, 2.10/2.11 fur a.a. Gruppen anders formulieren. Satz 3.3. M, N seien Gruppen, M algebraisch abgeschlossen. 1. N < p M gdw. W , ( N ) C W , ( M ) . 2. Wenn N abzuhlbar ist, N < N gdw. W , ( N ) C W , ( M ) . Sei auch N algebraisch abgeschlossen. 3. N = p M gdw. W , ( N ) = W , ( M ) . 4. Wenn N und M abziihlbar sind, N = M gdw. W , ( N ) = W , ( M ) .

W , ( M ) sol1 an die Stelle des Skeletts w-homogener Strukturen treten. Ebenso, wie wir in I, 3 untersucht haben, welche Klassen als Skelette vorkommen, wollen wir jetzt die Teilmengen von V bestimmen, die von der Form W , ( M ) fur ein algebraisch abgeschlossenes M sind. Zunachst zeigen wir, daB alle *-Grade als Wortproblem einer endlich erzeugten Gruppe vorkommen. Satz 3.4. Fur alle A C w gibt es eine von zwei Elementen erzeugte Gruppe F mit A C ,W ( F )< + A .

Dieser Satz wurde (unabhangig) in ([3]) und [29]) bewiesen. Beweis. % sei wie im Beweis von 2.16. A bezeichne jetzt die von den ai, i E A erzeugte Untergruppe von '23. Setze F = %/AT.Weil c "ai, * * biz E A gdw. { i l , .. . in}C A ist W ( F )6+A. Weiter ist f : A C ,W ( F ) fur f ( i ) = Godelnummer von ai. Wir fassen ai dabei als Wort in b,c auf. 0

-

Folgerungen 3.5. 1. Fur alle a E V gibt es ein e.e. F mit W ( F )=* a. 2. Es gibt 2" von zwei Elementen erzeugte Gruppen. 3. Es gibt W ( F )6, W ( G )und eine a.a.Gr. M mit G < M , F A M. (Wiihle w \ A & A , W ( G )=* A, W ( F )=* w \ A , M nach 1.7.) 4. Es gibt e.e. F < M a.a., so dap fur kein HZ E F * H < M. (Wiihle A mit A* & * A (2.9), W ( F )= * A , wende 2.13.3 an.)


50 1

Bemerkung 3.6. In [29] und [30] wird gezeigt, dal3 man Satz 3.5 nicht verscharfen kann, indem man s +durch scharfere Reduzierbarkeitsbegriffe - zum Beispiel s,,,- ersetzt. Es gibt ein nichtrekursives r.a. A, so daa fur keine endlich erzeugte Gruppe F mit nicht losbarem Wortproblem W ( F )%I1 A, [30]. ( s b r r : beschrankte wahrheitstafel Reduzierbarkeit.) Lemma 3.7. Wenn M algebraisch abgeschlossen ist, ist W , ( M ) ein Ideal in (V, s). Beweis. Sei b E W , ( M ) , also b =* W ( G )fur ein G E S k ( M ) . Wenn a s b fur ein a E V, wahlen wir ein e.e. F mit W ( F )=* a. Dann ist W ( F ) z) : = y I

•

(siehe Beweis V, 1.1)

Z

ip . (x, YI, Y2, z):= 3w (wxw "

= y, II WY2W-1 = z) V(y, = ell z =e)

ipw(x,y):= 3YI'" y43w,··· W4Y= WI' W2' W3' W4" 4

/\ (E1(y"x)

II

ip.(x,y" y" w.)

i=J

ip,(y, WI, W2):= W1(YW2y-l)

=(yW2y-I)W1

Dann gilt fiir aile a.a. Gruppen M und a EM 1. a hat genau dann unendliche Ordnung, wenn M 1= ipo(a). Wenn a unendliche Ordnung hat, ist fiir aile b, c E M, n, mEw. 2. bEl a) gdw. M 1= ipw (a, b) _ 3. c = anEBa m gdw. Ml=ip+(an,am,c) 4. c=anOa m gdw. Ml=ip.(a,an,am,c) 5. anESc.bgdw. Ml=ip,(an,c,b). Beweis. 1. a hat unendliche Ordnung gdw. (a 2) == (a) und (a 2) eine echte Untergruppe von (a) ist d. h. a (a 2). 3. ist trivial. Wenn ani e, gibt es in Meinen inneren Automorphismus T, der a in an iiberfiihrt. Es ist dann T(a m ) = a":", Aus dieser Uberlegung folgt 4. (jl+ und cpo definieren sogar Addition und Multiplikation auf (a) und mach en so (a) isomorph zum Ring der ganzen Zahlen. In diesem Ring sind aber die natiirlichen Zahlen die Summen von vier Quadraten. Das beweist 2. 5 ist trivial. 0

e

Foigerung 4.3 (siehe 2.4.5). Man kann fiir jede L 2-Formel p(v, X) eine Lw.w-Formel p(y, w, w', x) angeben, so daft fiir aile a.a. Gruppen M, ii E w, a, 6, c E' M, wenn a unendliche Ordnung hat,

w(M)1= p(ii, S"".'1" SCI,b" ... ) gdw.

M 1= p(a nll, an" ... ,

c, 6, a).

A lgebraisch abgeschlossene Gruppen

537

Die Aussage p gilt also genau dann in w ( M ) , wenn MI= 3 x ( i ; ( x ) A qn(X)). Beweis. Wir definieren i; induktiv: Wir kiinnen dabei annehmen, dal3 p nur atomare Formeln der Form u I + U? A u3, u I . u Z = u3 und Oil E X o enthalt. Wir setzen dann (Ul

und

+ uz = u3)" = cp+(y,,yz, y 3 ) , ( U l . uz = 4" = cp .(x, Y l . y z 7 y 3 )

(~n€X~i)"=cp,(yi,,wii,w6),(1~)"=1b~ ( p A p')" = b A b', ( 3 x o p ) ' = 3W o , WI!@ (3Vii

p ) " = 3yn(b

A

(x, y o ) ) .

c ~ w

0

Wir wollen eine Ubersetzung von L *-Formeln in elementare Formeln angeben. Dazu ubersetzen wir zunachst L * in eine n e u e Sprache L**. Diese Ubersetzung nutzt aus, dal3 wir in algebraisch abgeschlossenen Gruppen jede endlich Folge bo, . . . , b, in der Form Nl,(c,,cz),. . . Nn(cI,cz) schreiben konnen. L * * hat Variable uo, . . . fur naturliche Zahlen und x, y, . . . fur Gruppenelemente. Zahlterme werden aus 0 und den Zahlvariablen mit + und . gebildet. Individuenterme aus e, den Variable fur Gruppenelemente durch "Gruppenmultiplikation", und wenn g ein Individuenterm und f ein Zahlterm ist, ist g' ein Individuenterm (mit der Interpretation g . . . g -t mal). Die Formeln bauen sich wie ublich aus den atomaren Formeln g l = g z und t l fz ( g , Individuenterme, t, Zahlterme) auf. Offenbar beschreiben die Formeln von L * * arithmetische Eigenschaften von Gruppenelementen. Also ist nach 1.7 (in allen Gruppen) jede L **-Formel durch eine L *-Formel ausdruckbar. In a.a. Gruppen gilt auch die Umkehrung:

- -

Lemma 4.4. Z u jeder L *-Formel cp(uo,.. . X I ) , . . . a,,... ) gibt es eine L * * Formel cp * ( u o , .. . w,,,. . . xi], . . . zo, . . .z6,. . . ), so dap fur alle a.a. Gruppen M, an,. . . € M, Co,. . . E M, ~ 6 ,... E M und n o , .. . , m o , .. . E w M C cp (no, . . . an, . . . ,(Ni(Co, c ; ) i < mo), . . . ) gdw. M I = cp * ( n o , . . m,,, . . . an,. . . en,. . . CC . . . ).

I

Beweis. Wir ersetzen in cp die Folgenvariable a,durch die Variablen z,, z:, w, mit der Idee, dal3 l(a,)= w, und a, = N o ( Z , , 2:). . . Nw,-l(z,,2:). Wir nehmen an, daB die atomaren Formeln von cp nur entweder rein "zahlentheoretisch" sind oder atomare Formeln von L,, oder von der Form a,( v , ) = x, oder l(a,)= u, sind. Fur zahlentheoretische atomare Formeln und atomare Formeln setzen wir cp * = cp. Wir definieren

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M. Ziegler

x,) v (w, G u, A x,= e ) und (u, (u,) = x,)* = (u, < w, A N,,(z,,z:) (l(a,) = u,) = w, A u,. ( N , ( x , y ) ist der Individuenterm [[x, y2"+'],x ] aus L * * . ) Wir gewinnen nun cp * durch die rekursive Definition ( i q ) * = i c p * , (cp A +)* = c p * A +*, (3u,cp)* = 3u,cp*, (3x,cp)* = 3x,cp* und (3a,cp)*=3w,3z,,z:cp*. 0 Satz 4.5. Z u jeder L**-Formel cp(uo,... X I ) , . . . ) kann man eine LW,-Formel 6(yll,.. . x(,, . . . x ) angeben, so dap fur alle n o , .. . E w und

a, ao,a , ,. . . aus einer a.a. Gruppe M, a uon unendlicher Ordnung, M t= cp (no,. . .a(),. . . ) gdw. M != . . a l l , .. . a ) . Die Aussage cp gilt also genau dann in M , wenn MI= 3 x ( @ ( x ) A cpo(x)). $(a"~l,.

Folgerung. Z u jeder L*-Aussage kann man eine in algebraisch abgeschlossenen Gruppen aquiualente elementare Aussage angeben. Beweis. Wir ubersetzen zuerst die Formel x = x,: Nun ist aber, wenn a unendliche Ordnung hat, b" = c gdw. (6,c)' ein homomorphes Bild von (a, a")' ist. In 11, 2.5 wurde explizit ein endliches Gleichungssystem H z ( d l ,d z ,d l , d,, z,, z2) angegeben, das in einer Obergruppe genau dann losbar ist, wenn ( d ? ,d4)' homomorphes Bild von ( d , , d2)' ist. Wir setzen also (x: = x,). = 3 z l ,z2 A Hz(x,y,, x, x,, z , , z2). Wir konnen annehmen, dal3 cp sonst nur noch atomare Formeln der Form u, + u,= Uk, u, * u,= u k und x, .x, = Xk enthalt. Wir setzen (u, + u, = U k ) . = cp+(y,,y,, y k ) , = (x, y,, y,, y k ) und (x, . XI A x k ) " = (x, . XI x k ) . Wir (u, * 0, definieren dann rekursiv uber den Aufbau von cp ( i c p ) ^ = 1 4 , (cp A +)^ = @ A 4, (3x,cp)^ = 3x,@ und (3u,cp)^= 3y, (4 A cpW(x,y,)). Man sieht leicht, dal3 die so definierten @ die gewunschte Eigenschaft haben. 0 V. Generische Gruppen

"Forcing in der Modelltheorie" wurde in ([l, 14, 15, 23)) benutzt, um a.a. Strukturen (a.a. Gruppen) zu konstruieren. Wahrend ein grol3er Teil dieser Konstruktionen durch die einfachen "schrittweisen" Methoden der Kapitel I und I11 ersetzt werden kann, (siehe z. B. I, 3.5, I, 4.7, die Konstruktion von Mo (I, 3.9), 111, 1.7) sind generische Gruppen weiterhin wegen ihrer relativ einfach zu bestimmenden elementaren Eigenschaften von Interesse (siehe z. B. V, 3.4, VI, 2.9, VI, 3.3.7, VI, 4.1(2), 4.4(2), 4.5(2)). Im ersten Abschnitt von V behandelen wir die unendlich-generischen Gruppen. Das bekannte Resultat, dal3 die unendlich-generischen


539

Gruppen gerade die elementaren Substrukturen der a.a. w -universellen Gruppen sind, fuhrt zu einer einfachen Kennzeichnung: (1.5) Beim Ubergang zu w ( M ) , dem zur a.a. Gruppe M gehorenden w-Modell, entsprechen die unendlich-generischen Gruppen gerade den elementaren Unterstrukturen von (w, 9 ( w ) ) . Dieser Satz macht die Untersuchung der m-generischen Gruppen sehr leicht. Wir erhalten z.B., wenn V = L, eine kleinste a-generische Gruppe M“. Das Skelett von M” besteht aus den e.e. Gruppen mit analytischem Wortproblem (1 .S). Die Untersuchung der (endlich-) generischen Gruppen beginnen wir im zweiten Abschnitt. Wir geben eine neue Definition der Erzwingungsbeziehung Il- : Eine Bedingung p (eine endliche Menge von atomaren und negiert atomaren Aussagen aus L G ( C ) ,die mit T G vertraglich ist) erzwingt die Eigenschaft d,wenn im dem Spiel, in dem zwei Spieler I und I1 abwechselnd Elemente einer aufsteigenden Folge von Bedingungen p = p o C p , C p 2 C . . * wahlen, I eine Strategie hat, die gewahrleistet, daB - wie auch I1 spielt - die von U{p, i E w } beschriebene Gruppe die Eigenschaft d hat. Dieser Zugang hat den Vorteil, daB keine syntaktischen Uberlegungen vorkommen. Man kann z. B. formulieren: O!= “ . . . ist generisch” (2.4). Aul3erdem wird der Zusammenhang mit den “schrittweisen” Methoden klar. Aus 2.6 folgt, daB fur elementare Eigenschaften d unsere Definition rnit der ublichen Definiton von IF iibereinstimmt. Im dritten Abschnitt bestimmen wir die durch “Vxo,. . . x, S?( W((xo,. . . x,)))” ausdruckbaren Eigenschaften generischer Gruppen. (Das sind fur r.a. A gerade die durch V,-Aussagen beschriebenen elementaren Eigenschaften.) Als Folgerung bekommen wir den Satz aus ([14]), daB in generischen Gruppen nicht MAX gilt: es gibt keine universelle e.e. Untergruppe. SchlieBlich geben wir noch zwei Charakterisiereungen der generischen Gruppen, o h n e modelltheoretische Begriffe zu benutzen: Die Gruppe M ist generisch, wenn jede “dichte” arithmetische Menge von endlichen GI. & Ungl.systemen ein losbares System enthalt (3.8). Auf ahnliche Weise kann man die endlich erzeugten Untergruppen der generischen Gruppen (die “subgenerischen” Gruppen) charakterisieren. Die generischen Gruppen sind dann die a.a. Gruppen, deren endlich erzeugte Untergruppen subgenerisch sind (3.11, 3.12). Im VI Kapitel werden wir die generischen Gruppen durch rekursionstheoretische Eigenschaften des zugehorigen w-Modells kennzeichnen.

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1. Unendlich generische Gruppen

Algebraisch abgeschlossene Gruppen sind “grol3e” Gruppen. Ein Gleichungssystem, S ( a , i ) , das uber (5)losbar ist, ist schon in der Gruppe selbst losbar. Wir werden einige Verstarkungen dieser Definition behandeln. Die starkste derartige Forderung ist, dal3 jeder Isomorphietyp einer endlich erzeugten Obergruppe von (ii) schon in M realisiert wird. Diese M sind gerade die a.a. w-universellen Gruppen. Denn aus I, 2.3.1 und I, 3.9.1 folgt Lemma 1.1. Eine Gruppe M ist genau dann a.a. und w-universell, wenn sich fur alle ii E M alle endlich erzeugten Obergruppen von (Z) uber ( a ) in M einbetten lassen.

Weil es 2‘” endlich erzeugte Gruppen gibt, hat jede w-universelle Gruppe mindestens die Machtigkeit 2‘”. Aus I, 1.4 folgt, daR sich jede Giuppe in eine a.a. w-universelle Gruppe einbetten 1aBt (man wahlt in I, 1.4 3% so, daR jede endlich erzeugte Gruppe zu einer Gruppe aus X isomorph ist). I, 2.11.5 zeigt, daR man zu jedem G eine a.a. w-universelle Obergruppe M finden kann mit I M 1 = max (2”, 1 G I ). Wir suchen abzahlbare a.a. Gruppen, die ahnlich “groB” sind wie a.a. w-universelle Gruppen. Die endlich erzeugten Untergruppen (a,,,. . . a,) sollen in diesen Gruppen dieselben elementaren Eigenschaften haben wie in w-universellen a.a. Gruppen. Diese elementaren Eigenschaften hangen nur vom Isomorphietyp von (a,,,. . . )’ ab; denn aus I, 2.10.l(a) folgt: Wenn M , N a.a., w-universe11 sind, ii E M , 6 E N, so ist (M,ii) = p (N, 6), also auch ( M , ii) = (N, 6), falls (ii)’ = (6)’. (Wenn M, N beliebige a.a. Gruppen sind, gilt nur ( M , ii) E ~ * ( N6), (IV, 3.6.1).) Fur endliche erzeugte Gruppen F und ii E F konnen wir also sinnvoll definieren: (cp ( 2 ) eine Formel aus Lqw)F IF cp (a) gdw. cp (ti) in einer (allen) a.a. w-universellen Obergruppe(n) von F gilt. Mit folgendem Schema konnen wir durch Rekursion uber den Aufbau von cp “berechnen”, ob F IF cp (a): Lemma 1.2. 1. Wenn cp quantorenfrei ist, ist F It cp ( a ) gdw. F I= cp (a). 2. Fb 3icp(u, 2 ) gdw. es eine e.e. Obergruppe G 3 F und 6 E G gibt rnit G IF cp (ti, 6). 3. F It cp ( a )A +(ti) gdw. F I t cp (6)und F IF +(a). 4. F I t i cp ( t i ) gdw. nicht F It cp (a).


54 1

Folgerung. “(ii) It cp(&)” ist fur jedes cp eine analytische Eigenschaft von W ( ( i i ) ) .Das beweist man leicht durch Induktion iiber den Aufbau von cp, ist aber auch auf folgende Weise einzusehen: Nach IV, 1.2 gibt es eine L’-Formel cp(X), so dap fur alle a.a. w-universellen Gruppen M und alle ii E M MI= cp(Z) gdw. (w, B ( w ) ) C + ( W ( ( 6 ) ) )Die . genaue Angabe von 6 in IV, 2.1.1 zeigt, dap - wenn cp eine 3,-Formel ist “ ( a ) l tcp(&)” eine ZL-z-Eigenschaft von W ( ( i i ) )ist ( n 2 3). Definition 1.3. M heiBt unendlich generisch (m-generisch), wenn fur alle ii E M und alle Formeln cp(2) MI= cp(u) gdw. (ii)Itcp(u). Satz 1.4. 1 . N sei Untergruppe der m-generischen Gruppe M. N ist genau dann m-generisch, wenn N elementare Unterstruktur von M ist. 2. N ist genau dann m-generisch, wenn N elementare Unterstruktur einer a.a. w-universellen Gruppe ist. 3. m-generische Gruppen sind algebraisch abgeschlossen. 4. Jede Gruppe G is in eine m-generische Gruppe der Machtigkeit 1 G 1 + w einbettbar (es gibt also 2” viele abzahlbare m-generische Gruppen ) . 5 . M ist genau dann a-generisch, wenn fur alle ii E M und alle q ( i , j ) , fur die (ii)Il- 39cp(ti, j ) , 6 E M existiert mit (6, 6 )It cp(ti,6). 6. Alle m-generischen Gruppen sind elementar aquivalent. Folgerung aus 6. m-generische Gruppen erfullen nicht max, es gibt also auch keine universellen endlich erzeugten Untergruppen. Jede %Aussage, die in einer a.a. Gruppe gilt, gilt in allen a-generischen Gruppen. (Man vergleiche dazu IV, 2.8.2, die Bemerkung nach dem Beweis von IV, 3.3 und IV, 3.6.2.) Bemerkung zu 5. Wegen IV, 3.6.1 gilt in allen a.a. Gruppen M, ii E M, fur alle g2-Formeln cp ( 2 ) M I= cp (Z) gdw. ( t i )IF cp (ti). Umgekehrt sind die Gruppen mit dieser Eigenschaft algebraisch abgeschlossen. Ahnlich wie in 5 gilt: Die Gruppe M ist genau dann algebraisch abgeschlossen, wenn fur alle ti E M und alle 3,-Formel cp (2, j ) , fur die ( i i ) I t 3 j p ( i i 7 j )6 , E M existiert mit (ti,6)bcp(ti,6). Beweis. 1 folgt unmittelbar aus der Definition. 2 folgt aus 1, weil a.a. w-universelle Gruppen m-generisch sind und jede Gruppe in eine a.a. w-universelle Gruppe einbettbar ist. 3 folgt mit I, 1.5.2 aus 2. 4: Man bettet G zuerst in eine a.a. w-universelle Gruppe ein. Die gesuchte Gruppe ist eine elementare Unterstruktur von geeigneter Machtigkeit, die G enthalt und mit dem Satz von Lowenheim-Skolem gewonnen

542

M. Ziegler

wird. 6 folgt sofort aus der Definition. 5: M sei Untergruppe der m-generischen Gruppe N. M ist unendlich generisch gdw. M < N gdw. (Nach einem Kriterium von Tarski) fur alle q ( i , j ) , ii E M

N I= 3yq(ii,j ) j ex.

EM

N I= ~ ( i i6). ,

Daraus ergibt sich die Behauptung. 0 1.4.2 liefert eine sehr einfache Beschreibung der m-generischen Gruppen. (Man vergleiche auch [28].) Satz 1.5. 1. Eine algebraisch abgeschlossenen Gruppe M ist genau dann

m-generisch, wenn w ( M )< ( w , 9 ( w ) ) . 2. Sei ( w , 9')< ( w , 9 ( w ) ) . D a m gibt es eine m-generische Gruppe M mit w ( M )= (0, 9')und I MI = 1 91.Wenn 9'abziihlbar ist, ist M eindeutig bestimmt.

Beweis. Eine a.a. Gruppe M ist genau dann w-universell, wenn W * ( M ) =9 ( w ) . 1 folgt so aus 1.4.2 und IV, 2.4.6. Sei Y wie in 2. (0,9 ')ist a.a. (vgl. Bem. vor IV, 1.2). Sei (G, i E I ) eine Familie von e.e.Gr. mit Y/=* = { W(G,)/=* i E I } . Setze No = @ (G, i E I). Da Y/=* ein Ideal ist, ist W * ( N o ) =9'.Sei N, schon konstruiert mit W * ( N , )= 9.S J ( i J ) j, E J seien alle endlichen Systeme von Gleichungen uber N,, I J 1 = I 9'1.Wir bilden das amalgamierte Produkt aller freien Losungen N,,, = * N , ( ( N ,* F r ( i J ) ) / ( S J ( i J ) )jNE J). Wegen 1.6.4 ist W * ( N , + J =9 Setze M = u{N, i E w } . 0

1

1

1

1

1

Folgerungen 1.6. 1. Wenn M m-generisch ist, ist

Th(M)=,Th(w,9 ( w ) ) .

2. F und G seien endlich erzeugte Gruppen, W ( F ) analytisch in W ( G ) .Dann ist F in jede m-generische Gruppe einbettbar, in die G einbettbar ist (ugl. 111, 1.4). 3. M a ist in jede m-generische Gruppe einbettbar. 4. U C F, U C G seien e.e. Untergruppen der m-generischen Gruppe M. Dann ist F * " G < M. Beweis. 1. Nach IV, 2.4.5 lassen sich die Aussagen uber M und w ( M ) eff ektiv ineinander ubersetzen. 2. gilt, weil, wenn (w, Y)< ( w , 9 ( w ) ) , jede Menge, die analytisch in einem Element von Y ist, selbst Element von Y ist. 3. D a alle arithmetischen MTngen analytisch in W ( E ) sind, ist fur jede m-generische Gruppe N W * ( M a )C W * ( N ) . Die Behauptung folgt nun aus 111, 3.3.2.

543


4. Wenn W ( F )=zeW ( G ) ,ist W ( F ) analytisch in W ( G ) .4 folgt also aus 2 und 111, 2.16. Der letzte Satz gestattet uns eine weitere Charakterisierung der generischen Gruppen, die zeigt, daR m-Generizitat eine naturliche Verstarkung der Algebraischen Abgeschlossenheit ist.

m-

Satz 1.7. Die Gruppe M ist genau dann a-generisch, wenn es zu jeder analytischen Klasse 9 C P ( w ) und fur alle 6 E M, fur die es 6 in einer Obergruppe uon M gibt mit W ( ( 6 ,6))E 9,Z E M mit W ( ( 6 ,E ) ) E %; gibt.

Bemerkung. Nimmt man fur die 9 in 1.6 nur r.a. Hornklassen, charakterisiert die Bedingung von 1.6 gerade die a.a. Gruppen.

SO

Beweis. Sei die Bedingung fur M erfullt. Wir zeigen, daR M die Eigenschaft von 1.4.5 hat. Sei 6 E M und (6)It 3jjcp(a, 9). (6, d ) It c p ( i , 6 ) ist eine analytische Eigenschaft von W((ii,6)).Da fur eine Obergruppe (6, E ) 3 (2) (ii, E ) It cp(& E) und man wegen der Amalgamierungseigenschaft der Klasse aller Gruppen E aus einer Obergruppe von M wahlen kann, existiert nach Voraussetzung ein 6 E M mit (ii,6)IF cp (i,6 ) . Wir zeigen nun, daR umgekehrt alle m-generischen Gruppen die angegebene Eigenschaft haben. M, 9 C 9 ( w ) und 6 seien wie in 1.7. %; werde durch die L2-Formel cp (X) beschrieben, d.h. A E %; gdw. (o, P(o))!=cp(A). DaB fur eine Obergruppe (ii, 6) von ( 6 ) 6 = ak,. . . a,) W((ii,6))E 9 bedeutet (wenn ii = a o , .. . ( o , P ( w ) ) ! = 3 Y ( Y J k =W((ii))hYEW,hcp(Y)).Nach 1.5ist w ( M ) C 3 Y ( Y J k =W ( ( i i ) ) ~ Y E W , ~ c p ( Y ) ) . S e i a l s o ( M ist ohomogen, man beachte den Beweis von IV, 1.2) E E M mit o(M)!= cp(W((ii,E))). Dann ist wieder wegen 1.5 (o,P(o))l=cp(W((ii,E))),d.h. W ( ( i , E ) ) E9. 0 Satz 1.8 ( V = L). 1.

Es gibt eine kleinste m-generische Gruppe M o :M o ist in jede m-generische Gruppe einbettbar. ( W * ( M o )besteht gerade aus den analytischen Mengen.) 2. Fur e.e. F, G sind iiquiualent: (a) W ( F ) ist analytisch in W ( G ) , (b) F ist in jede m-generische Gruppe einbettbar, in die G einbettbar ist (ugl. 111, 1.8.1). Bemerkungen. M o ist durch die angegebenen Eigenschaften eindeutig bestimmt. Es gibt keine kleinste a.a. Gruppe (111, 4.5 zu 3). ( V = L ) W ( F ) analytisch gdw. F in alle m-generischen Gr. einbettbar.

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Beweis. 1. Wie in 2 folgt aus V = L, daB (w, d )< (w, B ( w ) ) - d die Klasse der analytischen Mengen. Mn sei nach 1.5.2 die abzahlbare agenerische Gr. rnit W * ( M n= ) d. Aus 1.6.2 folgt, daB M” in jede mgenerische Gruppe einbettbar ist. 2. Aus V = L folgt: R sei eine n!,-Relation zwischen Mengen. Wenn fur ein A C w { X R ( X , A ) } # 0, gibt es ein B, das n!,+, in A ist und R ( B , A ) erfullt. Wenn nun W ( F )nicht analytisch in W ( G )ist, ist nach dieser Bemerkung fur X = { X X analytisch in W ( G ) }( w , X ) eine elementare Unterstruktur von ( w , B ( w ) ) . Nach 1.5.2 gibt es ein w-generisches M mit W * ( M )= X. F ist nicht in M einbettbar. Damit ist (b)+ (a) bewiesen. (a)+(b) folgt aus 1.6.2. 0

1

1

2. Forcing in der Gruppentheorie Wir haben fruher (111, 1.7) eine algebraisch abgeschlossene Gruppe M, die eine vorgegebene e.e. Gruppe G mit nicht-losbarem Wortproblem nicht enthalt, auf folgende Weise konstruiert: M war eine Losung der Vereinigung einer aufsteigenden Folge po(E) C p l ( E ) C . . . von endlichen Mengen von Gleichungen und Ungleichungen. Wir muBten bei der Wahl der p , dafiir sorgen, daB alle endlichen Gleichungssysteme S, (t,i ) ,die ev. spater in einer Obergruppe von M losbar sind Losungen bekommen und daB, wenn G = (g), fur jede endliche Folge (E) von Termen (G,g’) nicht isomorph in (M, (2)) eingebettet werden kann. Wir numerieren diese Forderungen E, i = 0,1,. .. und wahlen, wenn p,-l schon vorgegeben ist, p, so, daB, egal wie weiter konstruiert wird, M die Forderung E erfullt. Da wir dabei nicht ausnutzen, daB p,-l “richtig” gewahlt war, ist es moglich, mit dieser Konstruktion noch mehr Forderungen zu erfullen. Die Sicherheit mit der dieses Verfahren zum Ziel fuhrt, kann man auch dadurch beschreiben, daB man sagt, daB I ein Spiel der folgenden Art immer gewinnen kann. B sei die Menge aller Gleichungen und Ungleichungen ohne freie Variable aus L ( C ) (die Menge der basischen Aussagen), P die Menge aller endlichen Teilmengen von B, die mit T G konsistent sind. Die Spieler I und I1 wahlen abwechselnd Elemente einer aufsteigenden Folge p o C pl C * von Elementen aus P. Sei d eine Klasse von Teilmengen von B und eine Folge (ein Spiel) C p l C * . . gewahlt. i E w } E d,ist “d gewonnen”. Eine Gewinnstrategie von Wenn I fur d ist eine Folge (fi i E a),f, : P + P, mit V i , p f, (p) 3 p , so daB in allen Spielen p o C pl C . . . , in denen I nach dieser Strategie spielt, d.h. fur alle i > 0 f, ( p 2 , - l ) = p z , und f o ( 4 ) = Po, d gewonnen ist. Ahnlich definiert man Gewinnstrategien von I1 fur d :eine Folge von

u{p,I

1


545

Funktionen f,],f l , . . . von P nach P, so dal3 in allen Spielen poC p l C . . ., fur die f.a. i P ~ , = . ~fg(pz,),d gewonnen ist. I1 hat eine Gewinnstrategie fur d gdw. I fur jedes p E P eine Gewinnstrategie (f, i E w ) fur d rnit fo(4)= p hat (f “beginnt mit p”). p ist eine (I-) Gewinnposition fur d , wenn I eine Gewinnstrategie fur d hat, die rnit p beginnt. Wir schreiben dann p ltd (p erzwingt d).p E P ist eine 11-Gewinnposition fur d , wenn es ein p’ 3 p mit p ’ t d gibt. .d heiBt determiniert, wenn jedes p E P Gewinnposition fur d oder 11-Gewinnposition fur B(B)\ d ist (p kann nicht gleichzeitig Gewinnposition fur d und 11-Gewinnposition fur B ( B ) \ d sein. Alle d,die wir spater betrachten, werden determiniert sein).

1

Bemerkung. Es gibt zwei andere Begriffe fur “Gewinnstrategie” fur d, eine starkere und eine schwachere. Eine starke Gewinnstrategie von I fur d ist eine Funktion F : st‘ --* P mit p, C F ( ( p o , .. . ,p,)) f.a. i und p(,,. . . ,p, aus P, so dal3 jedes Spiel po C p, C . . rnit pz, = F ( ( p o , .. . , P ~ , - ~f.a. ) ) i d gewinnt. Eine schwache Gewinnstrategie von I fur d ist eine Funktion f : P-, P mit Vp f(p) 3 p, sodal3 cf i E w ) eine Gewinnstrategie von I fur .d ist. Aus jeder Gewinnstrategie von I fur d,($ i E a),erhalt man eine starke ) (pZt-,)setzt. Man kann Gewinnstrategie, indem man F ( ( p o , .. . ,P ~ , - ~=) f, sich n u n uberlegen, dal3 man aus einer starken Gewinnstrategie eine schwache Gewinnstrategie gewinnen kann. Die drei Begriffe sind also aquivalent. +

1

1

Beispiele. 1. Sei G eine e.e. Gruppe mit nicht-losbarem Wortproblem und d die Klasse aller Teilmengen von B, die mit T G konsistent sind und fur die jedes Modell, das von den Interpretationen der ci erzeugt wird, algebraisch abgeschlossen ist und G - bis auf Isomorphie nicht enthalt. Der Beweis von 111, 1.7 zeigt, dal3 4 It d. 2. cp sei eine atomare Aussage von L ( C ) . Dann ist 4 II- {G c B cp E G oder i cp E G}. Denn wenn I1 0 C pl wahlt, so kann I p2 3 p l so wahlen, dal3 cp E p2 oder i cp E p2. (Bei der Gewinnstrategie von I kommt es also nur auf f l an.) Statt p I t { G C B I . . . G . . . } s c h r e i b e n wirauch p I I - “ - - . G . * . ” .

I

Lemma 2.1. d,do, d l ,... seien Teilmengen von s ( B ) , r, p , q E P. 1. Wenn p Ik d,gibt es ein G E d,G U T G konsistent, p E G. 2. p u q P (p,q unvertraglich) gdw. p It “ q f G”. “ q C G ” ist determiniert.

M. Ziegler

546

3. Aus p Il- P ( B ) \ d folgt V q 3 p q Y d.Wenn d determiniert ist, ist auch 9 ( B ) \ d determiniert und p Il- P ( B ) \ d gdw. V q 3 p q Y d. 4. p Ik i E w } gdw. fur alle i p IF d,. Wenn alle d,determiniert sind, ist auch I i E o} determiniert. 5 . A ~ VSq 3 p 3 r 3 q 3i E w r IF d,folgt p IF U{d#i E w } . Wenn i E w } determiniert und alle d,determiniert sind, ist auch

n{d# I n{d,

1

u{d,1

Vq 3 p 3 r 3 q 3 i E w r l t - d ,

gdw. p l l - U { d , l i i E w } .

6. Wenn p C q, p Ik d,so auch q Il- d.

u{p,1

Beweis. 1. Man gewinnt G als i E w } aus einem Spiel po C p , C . . . , bei dem I seine bei p beginnende Gewinnstrategie fur d anwendet. 2. Wenn p und q unvertraglich sind, ist jede Strategie, die bei p beginnt, eine Gewinnstrategie fur “ q JZ G”. Wenn p und q vertraglich sind, kann I1 auf die Wahl po = p von I mit der Wahl p U q antworten. Also ist p J$“ q z G” und p ist 11-Gewinnposition fur “ q C G”. “ q g G” ist also determiniert. Nach 3 ist also auch “ q C G ” determiniert . 3. Sei p Il- P ( B ) \ d und q 3 p. Wenn I eine bei q beginnende Gewinnstrategie fur d hat, und I1 nach dieser Strategie spielt, gewinnt jedes Spiel p C q C p2C * d.Wid. Sei d determiniert, p Y P ( B ) \ d . Wenn also I po = p wahlt, kann I1 p , = q 3 p so wahlen, daB q keine 11-Gewinnposition fur 9 ( B ) \ d ist. q ist dann I-Gewinnposition fur d. 4. Jede Gewinnstrategie fur i E w } ist auch eine Gewinnstrategie fur jedes d,. Sei umgekehrt fur jedes i ( f t l j E w ) eine die mit p beginnt. ( ( i k , j k ) l k E o)sei Gewinnstrategie von I fur d,, eine Durchzahlung von w 2 mit j o = 0 und V k 3 k ’ > k (ill*=j k + 1, i k = ik,). (fa1 k E o)ist dann eine Gewinnstrategie von I fur i E w } , die bei p beginnt. Denn wenn I diese Strategie beim Spiel p = po C p , C . . . angewendet hat, gibt es fur jedes i eine Teilfolge . * C p,, C f ~ , , C) . * . Die Vereinigung p = po C p,, C f;@,() C C der p,, liegt aber in d,. 5 . Sei V q 3 p 3 r 3 q 3i E w r Il- d,. Dann hat I die folgende Strategie fur U{d, i E d}:I beginnt rnit p ; wenn I1 q 3 p wahlt, wahlt I ein r 3 q mit r Il- d,fur ein i und spielt mit einer Strategie fur d,, die bei r beginnt, weiter. Also ist p Il- U{d, i E w } . Der Rest von 5 folgt aus 3, 4 und der Darstellung U{d, i E w } = g ( B ) \ ( n { S ( B ) \ d , i E w } ) . 6. Wenn (f, i E w ) eine Gewinnstrategie von I fur d ist, die bei p beginnt, ist (f: i E w ) rnit f A = q und f: = f i sonst eine Gewinnstrategie von I fur d,die bei q beginnt.

n{d,1

n{d,I

I

I

1

1 I

I


547

Beispiel 2.2. Wenn G C B, bezeichnet MG die von den Gleichungen in G prasentierte Gruppe. 1st G U T G konsistent, gelten alle Aussagen von G in MG. co,... d o , ... seien in MG durch go,... ho,... interpretiert. Es ist dann

0 It “MG ist algebraisch abgeschlossen”. Nach 2.1.4 mussen wir zeigen, dai3 fur jedes endliche Gleichungs- & Ungl.-system S(E,X’j 0It“Wenn S ( g , i ) uber MG losbar ist, hat S(g,X’)eine Losung in MG”. I hat folgende Gewinnstrategie: pl sei von I1 gewahlt; wenn pl U S(E, i ) (mit T G )konsistent ist, wahlt I p 2 = pl U S(E, 6) fur Konstanten 6, die in S und pl noch nicht vorkommen. Auf die weiteren Zuge von I kommt es nicht mehr an. X sei eine Klasse von determinierten Teilmengen von P(B). Wir nennen G C B X-generisch, wenn G Ergebnis eines Spiels ist, bei dem “um jedes Element von X gespielt wurde”. Das sol1 bedeuten:

Definition 2.3. G C B heiBt X-generisch, wenn G U T G konsistent ist und fur alle d E X G E d gdw. 3 p C G p It- d. Lemma 2.4. Wenn X abzahlbar, ist 011 “G ist X-generisch ”. (Also gibt es nach 2.1.1 X-generische Mengen.) Beweis. In Anbetracht von 2.1.4 mussen wir zeigen: Fur alle d E X, 1. 0lt- “ G E d es gibt p C G p It d” und fur alle p mit p It d, 2. 0lt “Wenn p C G, ist G E d”. 1. Wir geben eine Gewinnstrategie von I an: Wenn I1 pl wahlt, und es ein p 3 pl mit p k d gibt, setzt I p z = p (in dem Spiel ist dann offenbar “es gibt p C G, p I!- d” gewonnen). Wenn es solch ein p nicht gibt, ist pl C “G $Z d”.I spielt nun eine Gewinnstrategie von “G $Z d ” , die mit pl beginnt. 2. Sei pl gegeben. Wenn pl und p vertraglich sind, ist pl C p , U p I!- “G E d”.Sonst ist p , It- ‘‘pg G”. 0

+

Lemma 2.5. X sei eine Klasse uon determinierten Teilmengen von ??(B). Wenn d E X, sei auch P ( B ) \ d E X. Dann ist fur alle p E P und d E X p It- d gdw. fur alle X-generischen G p C G 3 G E d. Beweis. Wenn p It- d,G X-generisch, p C G, d E X , ist nach Definition G E 1.Sei andererseits p F SQ. Dann gibt es ein q 3 p mit

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548

q It- $P(B)\d. D a q It- “G ist X-generisch”, gibt es ein X-generisches G, das q enthalt. D a Y ( B ) \ d E X,ist G @ d. 0 Wir interessieren uns fur elementare Eigenschaften von Gruppen, also fur jede Aussage cp E L ( C ) fur die Klasse A, = {G C B MG I= cp}. Wir schreiben p It (o fur p It- A,. In folgendem Lemma geben wir an, wie man p It- cp rekursiv “berechnen” kann.

I

Lemma 2.6. cp, Ic, seien Aussagen aus L ( C ) , p, q, r E P, konstante L ( C ) -Terme. 1. Wenn cp quantorenfrei ist, ist p Ik cp gdw. p kG cp. 2. plt-icp gdw. V q 3 p qIIccp. 3. p It- cp A cp gdw. p It- cp und p Ik Ic,. 4. p It cp v cp gdw. V q 3 p 3 r 3 q r Ik cp oder r It5. p It VXcp ( i )gdw. fur alle I p I t cp ( I )gdw. fur alle E p It- cp ( E ) . 6. 0lk “Es gibt c, MG I= t = c”. 7. Alle A, sind determiniert. 8. W e n n p ( ~ ) t t - c p ( ~i)s, t p ( l ) ~ t - c p ( IC) f a l l s p ( l ) E P ) .

+.

Folgerung. “ p I t cp (E)” ist fur jedes cp eine arithmetische Relation zwischen (den Godelnummern v o n ) p und E. M a n rechnet leicht nach, dap, wenn cp E V., n z=1, ist, diese Relation n”,,ist. Beweis. 1: Wenn p kG cp und das Spiel p o C pl C * * mit p beginnt, ist i E w } E A,. Also ist p It- A,. Sei p )cGcp. Dann gibt es ein q 3 p cp und die mit q kG i cp, also q It- “ G @ A,”. Daraus folgt p Determiniertheit von A,. 2, 3, 4, 7 und der erste Teil von 5 folgen sofort aus 2.1. 6: Sei p , E P. Wenn c weder in pl noch in t vorkommt, ist pl U { t l = c } E P. Aus pl U { t = c)’lt- t = c folgt die Behauptung. Wir zeigen den zweiten Teil von 5: Sei fur alle E p I1 cp(E) und I eine endliche Folge von konstanten Termen. Nach 6 ist dann p It “Es gibt c mit MG I= = ; E A cp (E)”. Daraus folgt p It cp ( I ) .8: Es ist klar, dal3 fur eine Bijektion T :C+Cp(E)lkcp(E) gdw. p ( ~ ( E ) ) l t c p ( ~ ( E ) ) .Sei nun p(E)It- c p ( E ) und p ( I ) gegeben. q E P sei eine beliebige Erweiterung von p ( r ) . Fur neue Konstanten d ist p ‘ = q U {to+ do, t l d l , . . . } U p ( 4 ) aus P. Die Bemerkung am Anfang zeigt p’I1 cp(d) und daher p’lt- cp(t). 0

u{p,I

Wir nennen G endlich generisch, wenn G X-generisch fur cp Aussage aus L(E)}. Das bedeutet also

X = {A,

I

Definition 2.7. 1. G C B heiDt endlich generisch, wenn G U T G

Algebraisch abgeschlossene Gmppen

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konsistent ist und fur jede L(C)-Aussage cp MG C cp gdw. es ein p C G mit p IF cp gibt. 2. Eine Gruppe M heiBt (endlich) generisch, wenn fur alle Formeln cp(E), ii E M MI= c p ( i ) gdw. es 6 E M und ein p(E, d ) E P gibt rnit MI= p ( i , 6 ) und p(E, d ) IF cp(Z). Lemma 2.8. 1. G C B sei mit T G konsistent. Dann ist G genau danil endlich generisch, wenn MG generisch ist und fur jede atomare Aussage cp G t-cp oder G I - i c p . 2. (gi i E w ) sei ein Erzeugendensystem der Gruppe M und G die Menge aller Aussagen aus B, die in ( M ,g’) gelten. Dann is; M genau dann generisch, wenn G endlich generisch ist. 3. p(E, d ) IF cp ( E ) gdw. fur jede generische Gruppe M ii, 6 E M MI=p(i,6)+ MCcp(i).

1

Beweis. 1. G sei generisch, ii E MG, MG C c p ( i ) . D a MG von den Interpretationen der c‘ erzeugt wird, gibt es Terme F(d), Konstanten d aus C,die in MG durch h‘ interpretiert werden, mit ii = F(h’). D a MG I= cp(t(d)),gibt es ein q ( d , d r )C G rnit q IF cp(l(d)). Naturlich ist Ma I= q ( i , i‘). Wenn E neue Konstanten aus C sind, ist p = q U {to A _co, tl = cl,. . . } aus P. Dann ist p IF c p ( E ) und MG I = p ( i , i , h ’ ) . Wenn M G - C p ( i ) , p(E)IFcp(E) folgt MG I= c p ( i ) . Denn cp (2). D a aber p sonst gibt es 6 E Me, q(E, d ) E P, MG I= q(i,6), q IF i und q vertraglich sind, wurde p U q IF (1 cp A cp)(c) folgen. Wid. Sei nun umgekehrt MG generisch und erfulle G die genannte Bedingung. Sei p C G und p IF cp. Dann ist MG C p und somit MG C cp. Sei umgekehrt MG C cp. Dann gibt es ein q(E, d ) IF cp mit MG I= q(E, 6 ) fur 6 E M. Wir konnen annehmen, daB die d von den E verschieden sind. Es sei 6 = f ( j r )fur die Interpretationen j ‘ der Konstanten E’. Aus 2.6.8 folgt q(E, F(E’))IF cp. D a MG C q(E, ?(I?’)),ist G I-q(E,F ( E f ) ) . Also gibt es p C G rnit p t- A q(E, t(E’)). p IF cp. 2. Folgt aus 1, weil M = MG. 3. Die eine Richtung der Behauptung folgt sofort aus der Definition. Wenn p(E, d)JV cp(Z), gibt es ein endlich generisches G mit p C G und MG bi’ c p ( E ) . Es ist aber MG C p. CI

3. Endlich generische Gruppen In diesem Abschnitt behandeln wir zuerst einige einfache Eigenschaften generischer Gruppen. Dann geben wir verschiedene Charakterisierungen der generischen Gruppen. Satz 3.1. 1. Alle generischen Gruppen sind elementar aquivalent.

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2. Generische Gruppen sind algebraisch abgeschlossen. 3. Die Theorie der generischen Gruppen ist rekursiu isomorph zu 0 c. ([28]). Beweis. cp sei eine Aussage aus L,,, p(E) It- cp. Wir zeigen @IFQ ; Es gibt sonst q ( d ) E P, q ( d ) l F i c p . Wenn fur die Permutation .rr von C T ( E ) von d disjunkt ist, sind p(.rr(E)) und q vertraglich. Aber q Up(.rr(E))It- i c p A cp. Wid. Damit ist 1 bewiesen. 2. Das endliche GI. & UngLsystem S ( 6 , i ) sei uber der generischen Gruppe M losbar. Wenn S keine Losung in M hat, gibt es 6 EM, p(E, d ) It- V f i A S(E, i ) ,M C ~(6,s). Da S ( & i )in einer Obergruppe von (5,6) Iosbar ist, sind p(E, d) und S(E, E’) fur neue C’ vertraglich. Es ist aber p C p U S(E, E‘) It- 32 A S(E, if). Wid. 3. Beim Beweis von 1 haben wir gesehen, dal3 eiae elementare Aussage genau dann in allen generischen Gruppen gilt, wenn sie von 0 erzwungen wird. Also ist Q genau dann aus Tf, der Theorie der generischen Gruppen, wenn die arithmetische Aussage “0 IF cp” gilt. Daraus folgt T’). (c) Fiir alle i gibt es eine a.a. Gruppe M, so dap fiir alle an,.. . a,,, E M S$( W((a,,. . . a,,,))) gilt. Zusatz. Wenn alle A, arithmetisch sind, sind ( a ) , ( b ) , ( c ) iiquiualent zu (d). Fiir alle generischen Gruppen M, alle i E w, a(),.. . E M gilt SPa(W((ao,.. . a, ))I. Beweis. Wenn die A, arithmetisch sind, folgt (a)@(d) weil “f.a. ao,. . . Spt( W ( ( a o., . . )))” arithmetische Eigenschaften sind, die sich auf elementar aquivalente Gruppen ubertragen. (a)+ (c) ist klar. Um (c)+ (b) einzusehen, geben wir uns ein i E o und ein endliches G1. & Unglsystem S(X,j) vor. S hat eine Losung ti, 6 in der a.a.Gr. M, wo S$( W ( ( a o , . .))) gilt. . . ) ) € 2”’’ (go,. . . (b)+(a): Wir zeigen 01t“f.a. i ex. n E A, W((go,. sind die Interpretationen der co, . . . in MG). Das bedeutet

V i E w V p 3q 3 p 3 n E A, q IF “W((g0,.. .)) E 2,”’. Sei also p(c0,.. . c,,,, d ) E P und i € w gegeben. Wegen (b) (S(X,9 ) = p(xo,. . . y)) gibt es ao,. . .6 E H mit HI= p(ao,.. . 6 ) une ein n E A, mit W ( ( a o ., . . a,)) E X,,. Es gibt ein endliches System S ( a o , . ..Z) von GI. & Ungl., das uber H genau dann losbar ist, wenn W((ao,. . .)) E 2“ (siehe Beweis von IV, 2.1.2, n = 1). q = p U S(co,.. . d ’ )(fur neues d r ) ist aus P, weil S(ao,. . . ) uber H losbar ist. Andererseits ist q Ik “ W((g,,,. . . )) E %,,”. Wie in 2.6.8 folgt nun 0 t “f.a. ao,. . . aus MG, f.a. i Sgh( W((a,, . . .)))”. Da auch

0 It “G generisch”, gibt es ein generisches G, so da8 MG (a) erfullt. 0

Folgerung 3.3 (Bezeichnungen wie in 3.2). Aus 1 und 2 folgtjeweils, dap (a), (b), (c) des letzten Satzes rutrefen. 1. Fiir alle i E w und Erzeugende ao,. . . a, einer Gruppe rnit r.a. Wortproblem gilt S:h( W ( ( a o ., . . a,,,))). 2. Kein A, ist durch eine r.a. Menge uon T,,,:= { n W, C 2-1 trennbar.

I

Beweis. Wenn 1 gilt, trifft fur jedes i (c) von 3.2 auf MI, zu; denn alle e.e. Untergruppen von MI, haben r.a. Wortproblem.

M. Ziegler

552

2. Wir zeigen (b). Sei also i E w und S ( i , 9 ) gegeben. Zu S gehort eine r.a. Hornklasse X,die genau dann W ( ( a , ,.,. . a , ) ) enthalt, wenn S ( & 9 ) in einer Obergruppe losbar ist. Wir setzen D = { n ,%, n %‘ n W, = 0). D ist r.a. Wenn wir S als losbar annehmen, folgt D n T, = 0. Es muB also A i f D sein. DaB fur ein n E Ai ,%, f l X n W, # 0 bedeutet aber, daB es eine Gruppe ( a o , . . a,) gibt mit W((ao,. . .)) E % n 2, d.h. Sf( W ( ( a o ,...))) und S(ao,.. . ) losbar uber ( a o , .. .). 0

1

Beispiel 3.4 ([14]). Generische Gruppen erfullen MAX nicht (siehe IV, 1.6 Beisp. .2). Beweis. E sei die Menge aller endlichen Systeme S(xo, . . .) von G1. & Ungl., die hochstens eine Ungleichung enthalten. Jedem S E E konnen wir effektiv eine r.a. Hornklasse gnzuordnen, die genau dann das Wortproblem einer Gruppe ( a o , . . a,) enthalt, wenn S in ( a o , .. .) sei die nicht losbar ist; n berechne sich durch g ( S ) rekursiv aus S. in die Menge der losbaren Systeme von E . Da jedes System von in jeder jeder a.a. Gruppe losbar ist, muB auch jedes System von universellen e.e. Untergruppe einer a.a.Gr. losbar sein. In generischen Gruppen gibt es keine e.e. Untergruppen mit dieser Eigenschaft: Wenn a o , .. . a, Elemente einer generischen Gruppe sind, gibt es ein S E mit W ( ( i i ) E ) ,%g(s). Wenn A = g [ ~ , ] heil3t , das Sf(W((ao,. . .)). Um das zu zeigen, zeigen wir, daB A nicht durch eine r.a. Menge von T, getrennt werden kann, und wenden dann 3.3.2 an. Nehmen wir an, D sei r.a. und trenne A von T,. Da g E \ E~ in T, abbildet, ware dann r.a. Das widerspricht der Unlosbarkeit des aber = g-’(D) d.h. Wortproblems fur Gruppen. 0 Die generischen Gruppen sind eine “elementare” Klasse von a.a. Gruppen:

Satz 3.5 (siehe [33]). Es gibt eine rekursiue Menge uon elementaren Aussagen, die genau dann in einer a.a. Gruppe gelten, wenn die Gruppe generisch ist.

Bemerkung. Da alle generischen Gruppen elementar aquivalent sind, ist das - von der Rekursivitat abgesehen - dasselbe wie: Wenn N = M a.a. und N generisch, ist M generisch.

1

Beweis. Wir wahlen eine Godelnumerierung (pi i E w ) von P. Dann ist fur jedes cp{i [ p ikcp} arithmetisch. ‘‘(M,fi)t=p’”ist in M eine


553

arithmetische Eigenschaft der a,,, . . . und i. Also ist V i ( c p ( i ) o 3 u3~"( . . . ; i C, p~" )" ~ p " t c peine ) arithmetische

Eigenschaft von Gruppen. In a.a. Gruppen sind diese arithmetischen Aussagen durch elementare Aussagen beschreibbar. Wir nehmen fur jedes cp die zugehorige elementare Aussage. Wir geben jetzt andere Charakterisierungen fur generische Gruppen. Lemma 3.6. Die Gruppe M ist genau dann generisch, wenn es fur alle GEM, p E P m i t ( M , i i ) / = p u n d a l l e c p , $ a u s L ( C ) e i n q E P u n d 6 E M m i t p C q u n d (M,i,6)Cqgibt, sodaJ?pItcpvq 3 q l t c p o d e r q It $ und p It3xcp(x) j ex. c E C q IF cp(c). Beweis. Die eine Richtung folgt sofort aus der Definition. Wenn die Bedingung erfiillt ist, zeigen wir durch Induktion uber den Aufbau von cp E L ( C ) , daJ? fur alle i E M, p E P ( M , ii) t= p , p It cp 3 (M, ii) C cp. Wir konnen annehmen, daB cp sich aus atomaren und negiert atomaren Formeln mit A , v , Vx, 3 x aufbaut. Fur quantorenfreie cp ist die Behauptung klar (siehe 2.6). Der Beweis ist beim Induktionsschritt fur A , Vx sehr einfach und bei v ahnlich wie bei 3 x . Sei (M, ii)t= p und p IF 3 x Q (x). Nach Voraussetzung gibt es q 3 p , 6 E M, rnit (M, ii, 6 ) C q und c E C rnit q It cp(c). Nach der Induktionsvoraussetzung ist ( M ,Ci, 6)C Q (c). Also gilt 3 x Q (x) in ( M ,ii). Wir mussen noch zeigen, daB es, wenn ( M ,ii) I= cp, ein p E P 6 E M rnit p It cp und ( M , ii, 6)k p gibt. Nun ist aber 0ll- Q v i c p . Also gibt es 6 E M und 0 C p rnit ( M , i i , 6 ) t = p u n d p l t c p o d e r p t i c p . Aus p It i cp wurde aber ( M , ii, 6)F cp folgen, also ist p It cp. 0 Definition 3.7. Eine Familie 9 von endlichen Gleichungs- und Ungleichungssystemen mit Koeffizienten g o , .. . ,g, aus einer Gruppe G heiBt dicht uber G, wenn zu allen . . ,gk E G und jeder endlichen Menge S(go,. . . gk) von Gleichungen und Ungleichungen, die in G gelten, es ein F ( g o , .. . , g n , 2 ) aus 9 gibt, sodaB S U F U T G widerspruchsfrei ist. Bemerkung. 9 ist also dicht uber G, wenn fur kein q E P, g E M (M,g ) t= q und q It i V FESr 3 2 F . Beispiel. Wenn das endl. G1. & UngLsystem S' uber G Iosbar ist, ist {S'} dicht.

In der nachsten Charakterisierung der generischen Gruppen kommt die Erzwingungsrelation nicht mehr vor.

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Satz 3.8. Eine Gruppe M ist genau dann generisch, wenn es fur jede

arithmetische, uber M dichte Familie 9 ein F E 9gibt, das eine Losung in M hat. Beweis. Die Bedingung sei erfullt. Mit 3.6 zeigen wir, daB M generisch ist. Sei ( M , i)t= p ( E ) und p(E)It- 3xcp(x). Die Familie von endlichen Systemen von Gleichungen und Ungleichungen uber M 9 : = { q ( t i , i ) l q ( E , d ) E P ,p C q , pIkcp(d’), d ‘ E C ) ist arithmetisch. Sei a” E M und M I = S(ti, ti‘) fur eine endliche Menge S(ti, ti’) von Gleichungen und Ungleichungen. Dann sind p und S(E, E’) vertraglich und es gibt ein q 3 p U S(E, t’)mit q IF cp(d’) fur ein d ’ E C. 9 ist also das eine Losung 6 E M hat. Das dicht und es gibt ein q ( 6 , i ) aus 9, wurde gerade in 3.6 verlangt. Der Fall cp v I,/I wird ebenso behandelt. Sei M generisch und $ ( g o , . . . g , ) eine arithmetische Familie von endlichen GI. & Unglsystemen. M I = 3 i V { A F(g, i ) F E 9)ist eine arithmetische Eigenschaft von M , g o , . . . ; wenn M, g o * diese Eigenschaft nicht hat, gibt es p ( c o ,. . . ,ck) E P g n + , ,. . . g k E M mit M k p ( g 2 . ..gk) und p IF “MG b v i A {-I A F ( E , i ) / F E 9)”. Also iSt fur alle d E C und alle F E 9p Ik -I A F(Z, 2). Das bedeutet aber, daB p ( g o , . . . gk) U F(go,. . . g,, i )U T G fur alle F E 9 inkonsistent ist. 9 ist also nicht dicht. 0

I

-

Nach 3.5 ist jede zu einer generischen Gruppe partiell isomorphe Gruppe wieder generisch. Die generischen Gruppen miissen sich also unter den a.a. Gruppen durch ihr Skelett auszeichnen lassen. Dazu definieren wir:

Definition 3.9. Ein Gruppe G heiBt subgenerisch, wenn jede arithmetische Familie 9 ( g o , .. . g n ) von endlichen Systemen von Gleichungen und Ungleichungen uber G, die in folgendem Sinn “dicht” ist:

Zu jedem endlichen G1. & Unglsystem S(go,. . . gk, y), das so daB uber G losbar ist, gibt es ein F ( g o , .. .g., 2 ) aus 9, F U S U T G widerspruchsfrei ist. ein Mitglied besitzt, das uber G losbar ist.

Bemerkung 3.10. Nach 3.8. ist eine a.a. Gruppe genau dann generisch, wenn sie subgenerisch ist. Aus I, 1.2 folgt daB jede Untergruppe einer subgenerischen Gruppe wieder subgenerisch ist. Satz 3.11. Eine algebraisch abgeschlossene Gruppe ist genau dann

generisch, wenn jede endlich erzeugte Untergruppe subgenerisch ist.

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A lgebraisch abgesch lossene Gruppen

Beweis. 3.11 folgt aus 3.10. Denn eine Gruppe ist, wie man leicht sieht, genau dann subgenerisch, wenn jede e.e. Untergruppe subgenerisch ist. Folgerung. Jede a. a. Untergruppe einer generischen Gruppe ist generisch. Satz 3.12. Jede abzahlbare subgenerische Gruppe isf Untergruppe einer

generischen Gruppe.

I

Beweis. Es sei G = {g, i E w } subgenerisch. Wir denken uns die Menge der Konstanten C disjunkt zerteilt in {g, i E w } und eine unendliche Menge C’. Wir konstruieren eine aufsteigende Folge p o C p , C . von Elemeden aus P, die mit Diag(G, g ) U T G vertraglich sind. Die generische Obergruppe M von G wird dann eine von den Interpretationen der c E C erzeugte Losung von Diag(G, g ) U u{pi i E w } . $,, i E w sei eine Numerierung aller Aussagen aus L ( C ) . Sei pi schon gewahlt und t+bi = (p1 v cpz (oder = 3xcp(x)). In t+bi mogen nur die Parameter g o , .. . ,g, vorkommen. Wir betrachten die arithmetische Familie 9 = {q(go,. . . g,, i ) q 3 p,, q Il- 14, oder q IF cpl oder q IF cpz (oder q IF cp(c) fur ein c E C)}. 9 ist im Sinn von 3.9 dicht. q(go, . . . g,, i )E 9 sei uber G Iosbar. Wir setzen dann p i + ,= q(go, . . . E ’ ) U pi fur neue Konstanten c“ aus C’. Das so konstruierte M erfullt offenbar die Bedingungen von 3.6.

I

1

I

Bemerkung. Wir werden sehen, daB fur generische M W*(M) abzahlbar ist, G ist also genau dann partiell in eine generische Gruppe einbettbar, wenn G subgenerisch ist und W*(G) hiichstens abzahlbar. Wir werden im nachsten Kapitel die endlich erzeugten subgenerischen Gruppe auf einfache Weise durch ihr Wortproblem charakterisieren. Das direkte Product zweier subgenerischer Gruppen ist i.A. nicht subgenerisch. Problem. 1st W*(G) fur subgenerische G immer hochstens abzahlbar? VI. Generische o-Modelle Wir bestimmen die generischen Gruppen durch die zugehorigen w Modelle. Zuerst geben wir eine einfache rekursionstheoretische Beschreibung der subgenerischen Gruppen: Eine Menge von naturlichen Zahlen heiBt subgenerisch, wenn sie Element der Vereinigung jeder “dichten” Familie von r.a. Hornklassen ist (1.1). Die

M . Ziegler

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e.e. Gruppe F ist subgenerisch gdw. W ( F ) subgenerisch ist (1.11). D a mit der Menge G auch jedes H mit H S * G subgenerisch ist, ist die a.a. Gruppe M genau dann generisch, wenn W * ( M )nur aus subgenerischen Mengen besteht (2.4). Wir nennen ein w-Modell daher generisch, wenn es algebraisch abgeschlossen ist und nur subgenerische Mengen enthalt. Die generischen Gruppen entsprechen dann gerade den generischen w-Modellen. Im zweiten Abschnitt definieren wir eine Erzwingungsrelation fur w Modelle: eine r.a. Hornklasse X erzwingt die L’-Formel cp(X), wenn fur alle generischen ( w , 9’) und alle A E Y n %‘ (w, 9’)t= p ( A ) (2.7). Wenn man n u n ausnutzt, daB r.a. Hornklassen endlichen Gleichungssystemen, L ‘-Formeln Formeln aus L generische w Modelle generischen Gruppen entsprechen, und bemerkt, daB (2.7) dem Lemma V, 2.5 entspricht, sieht man, daB unsere neue Erzwingungsrelation die Ubertragung der Erzwingungsrelation von V auf w-Modelle ist. Aus (der Definition) V, 2.7 kann man also schliessen : In einem generischen w-Modell ( w , 9’) gilt cp(A) genau d a m , wenn eine r.a. Hornklasse X mit A E X cp(X) erzwingt (2.8). Wir beweisen 2.8 aber nicht auf diese Weise, sondern direkt, ohne die Resultate von V uber generische Gruppen zu benutzen. Ebenso zeigen wir “nocheinmal” direkt die Existenz generischer w -Modelle (1.3, 2.5). Das zeigt weitere Moglichkeit generische Gruppen einzufuhren (namlich analog zur Einfuhrung der generischen w Modelle): Man definiert generische Gruppen durch V, 3.8 und die Erzwingungsrelation analog zu 2.7. Die Theorie der generischen Gruppen 1aBt sich dann wie in VI entwickeln. Als erste Anwendung zeigen wir, daB alle in a.a. Gruppen erfullbaren V,-Aussagen in allen generischen Gruppen gelten (2.10). Man kann also generische Gruppen “klein” nennen, denn V,-Aussagen, die in einer a.a.Gr. gelten, gelten auch in jeder a.a. Untergruppe. In diesem Sinn sind w-generische Gruppen “groB”: Eine V,-Aussage, die in einer w-generischen Gruppe gilt, gilt in allen a.a. Gruppen. Durch diese “Kleinheit” werden die generischen Gruppen nicht charakterisiert. Denn eine zu einer generischen Gruppe elementar aquivalente a.a. Gruppe ist zwar generisch (V, 3.5). Fur jedes n gibt es aber eine nicht-generische a.a. Gruppe, in der alle 3,-Aussagen gelten, die in generischen Gruppen gelten (3.5). Der Beweis von 3.5 benutzt schon das Hauptresultat des dritten Abschnitts: Zu jeder generischen Gruppe M gibt es eine (bis auf rekursive Isomorphie eindeutig bestimmte) Menge B C w, so daB W * ( M )= {A C w A

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