The Isomorphism Problem for Algebraically Closed Groups

THE ISOMORPHISM PROBLEM FOR ALGEBRAICALLY CLOSED GROUPS B.H.NEUMANN Vanderbilt University, The University of Cambridge, and The Australian National University

1. Known facts and a conjecture

Let @ be a class of algebraic systems (briefly: algebras), all of the same species; for example the class of groups; and let A be an algebra in @.Consider a set C of sentences made from elements of A , element variables xl, x2, ..., y , z, ..., the algebraic operations of the species of 6 , equality, and the operations of the lower predicate calculus. We call Z consistent over A if A can be embedded in an algebra A’ in 6 such that Z is satisfied, that is to say, all sentences of Z: are valid in A ‘ . Next let G be a class of such sets C of sentences. Then A is called algebraically closed (or, if necessary, 6 -Galgebraically closed) if every set Z E E that is consistent over A is satisfied in A . Take, for example, to be the class of (commutative) fields, and G the class of singletons C = { u }, where each sentence u is a polynomial equation in a single variable prefaced by an existential quantifier binding the variable: then we get the usual notion of an algebraically closed field. Again, let (5 consist of singletons C = { u } , where now each sentence u is a finite conjunction of equations

and negations of such equations, 553

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prefaced by existential quantifiers binding all the variables; here L f',g, g' are words formed with the relevant algebraic operations from variables and constants, that is elements of the algebra A under consideration. When 6 is the class of all groups, the 6-Ealgebraically closed algebras are the algebraically closed groups introduced by Scott [71. Other classes of algebraic systems will also serve instead of groups, and in particular the facts that will be stated and the conjecture t o be discussed about algebraically closed groups apply also, mutatis mutandis, t o algebraically closed semigroups [ 61. The following theorem and corollary were proved by Scott [ 71 : Theorem 1 . l . Every group can be embedded in an algebraically closed group. Corollary 1.2. Every countable group can be embedded in a countable algebraically closed group. This latter fact is a corollary of the proof method rather than of the theorem; and the proof applies equally t o other classes of algebras, provided that they admit direct limits and their species is defined by finitely many finitary operations: it is in essence the proof devised by Steinitz [ 81 to prove the corresponding statemen t for fields. We deduce an easy consequence: Theorem 1.3. There are 2 * O mutually non-isomorphic countable algebraically closed groups.

To see this we need only remember that there are 2*0 mutually non-isomorphic 2-generator groups [ 41 , and each can, by Corollary 1.2, be embedded in a countable algebraically closed group; but each countable group contains only countably many pairs of elements and thus only countably many 2-generator groups. Hence countable algebraically closed groups are needed to accom-

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modate all the 2-generator groups. On the other hand there can not be more than 2N0 isomorphism classes of countable algebraically closed groups, because there are no more than 2 N o isomorphism classes of countable groups altogether. We note in passing that the same argument will show the same fact for semigroups instead of groups; but not for fields: there are only countably many countable algebraically closed fields, one for each combination of characteristic (a positive prime number or zero) and transcendence degree (a cardinal number, in this case

< No).

However, no algebraically closed group is explicitly known, the existence proof being highly non-constructive. This stems in part from the fact that there is no useful criterion known that tells one what sentences are or are not consistent over a given group. But assume now that we are given the knowledge that two countable groups, say A and B , are algebraically closed. We then ask how one can tell whether or not they are isomorphic. Now I conjecture that this problem, the isomorphism problem for countable algebraically closed groups, is algorithmically unsolvable. The aim of this note is to give some reasons for entertaining this conjecture. 2. Absolute presentations of groups

A proof of the conjecture would first of all require a more precise formulation of it; in particular we would need to know just how the groups A and B are given, and what algorithms are admitted. But as I am not proving the conjecture, I shall allow myself much imprecision. We may think of A and B as given, for example, in some presentation, that is by defining relations in some set of generators; and we note a fact that is not really relevant:

Theorem 2.1. A n algebraically closed group can not be finitely generated, nor finitely related. Proof. If gl,g z , ..., gn are finitely many elements in an arbitrary group G, then the sentence

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32 : g1z = zg1 .&. g 2 z = zg2 .&. ... .&.gnz = zg, .&. z # 1

is consistent over G, as it is satisfied in any direct product of G and a non-trivial group. It follows that in an algebraically closed group every finitely generated subgroup has a non-trivial centralizer; but it is known [ 51 that every algebraically closed group is simple, and thus has trivial centre: hence it can not be finitely generated. Now in a finitely related group in infinitely many generators infinitely many of these generators occur in none of the defining relations and thus generate freely a free group that is a free factor of the whole group: take any two of these free generators, and observe that the centralizer of the subgroup they generate is trivial: but in an algebraically closed group this can not happen, as we have seen, and the theorem follows. How could we hope to prove that B is not isomorphic to A? We know that B can not contain (isomorphic copies of) all possible 2-generator groups as subgroups; so we might try to spot one that is not in B , but is in A . Thus we pick two elements, say a l and a 2 , of A and try to show that the subgroup H, say, they generate can not be matched in B. To this end we need to know what this subgroup His. A set of defining relations of H in terms of a1 and a 2 , that is a presentation of H , suffices if we are given that it is a presentation, that is to say, if we know that the only relations between a l and a2 are the consequences of the given relations. We observe that to be able to decide whether an arbitrary word in a l and a2 equals the unit element in consequence of the given defining relations, we should be able to solve the word problem for this presentation of H ; and to be able t o decide whether H is isomorphic t o the subgroup K , say, of B generated by a pair b,, b2 of elements picked from B , we should be able to solve the isomorphism problem for our presentation of H . If H is finitely presented, say by

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then the sentence

3 x 1 . 3 x 2 : r 1 ( x 1 , x 2=) 1 .&. r 2 ( x 1 , x 2=) 1 .&. ... .&. rm ( x 1 , x 2 )= 1 .&. x1 # 1 is consistent over every group G, as it is valid in G X H , where we have assumed, without loss of generality as we shall presently see (Lemma 2.5), that H is not cyclic; hence this sentence is satisfied in B. This does not quite mean that H is matched in B and thus useless for telling B from A : if b, , b, in B satisfy this sentence, the group K they generate might still satisfy further relations that are not satisfied by a l and a, in H. This leads us to think of H as specified more precisely by what we shall call an absolute presentation, that is a set of relations

(2.2)

r l ( a l , a 2 ) = 1 , r 2 ( a l , a 2 ) =1 , ...

together with a set of “irrelations”

that jointly ensure that if b, , b , are two group elements that satisfy them in place of a,, a,, then mapping al on b , and a2 on b , necessarily generates an isomorphism of H onto the group K generated by b , and b,. If a presentation by generators and defining relations only is to be distinguished from an absolute presentation, it will be called a relative presentution. We note in passing that all the classical problems on relative group presentations, such as the word problem, the conjugacy problem, the isomorphism problem, also make sense for absolute presentations, and are still in need of solving, or unsolving. It is easy t o see that if H is finitely absolutely presented, then it is matched in B and thus useless for distinguishing B from A :

Lemma 2.4. Every algebraically closed group contains un isomorphic copy of every finitely absolutely presented group.

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Wc need only observe that the sentence obtained by conjunction of the defining relations and irrelations, written in variables xl,x2 that are then bound by existential quantifiers, is consistent over an arbitrary group, hence satisfied in every algebraically closed group. We have already, in claiming that n o generality is lost by assuming u1 # 1 , made use of a fact that is almost a corollary of this lemma:

Lemma 2.5. Every algebraically closed group contains cyclic groups oj' every order. Cyclic groups of finite order are, like all finite groups, finitely absolutely presented, hence contained in every algebraically closed group by Lemma 2.4. For an infinite cyclic group we use the sentence 3x, .3x,.3x3.3x4 : x;2 = x; .&. x;3

= x22 .&. x';4 = x; .&.

whicli is satisfied in a group introduced by Graham Higman [ 1 ] , and which has the property that in every non-trivial epimorph the generators have infinite orders: this sentence must be satisfied in every algebraically closed group, and the lemma follows.

Corollary 2.6. A n ulgebraicully closed group cun not be periodic. 3. Infinite presentations

It follows from what has been said that our hope of picking two elements u l , u2 from the algebraically closed group A such that the subgroup H they generate is not matched in the algebraically closed group B , so as to distinguish B from A , now rests with an H that can not be finitely absolutely presented, and is not infinite cyclic either. However, as we seek an effective procedure that distinguishes our groups, it seems to me that we need to assume a t least

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that H is recursively absolutely presented, in the sense that in some effective enumeration of the elements of the free group on u l , a2 the relators r , (a1,a2 ), r2 (al ,a2 ), r3 (al ,a2 ), ... are recursively enumerated, and so are the irrelators s1 (a1,a2 ), s2 (ul ,u2 ), s3 (al ,u2 ), ..., where the absolute presentation is given by (2.2), (2.3). Note that one and the same group can have absolute presentations that are recursive and others that are not: for example the infinite cyclic group generated by an element a can be given by an arbitrary infinite sequence of irrelations of the form

and the recursiveness or otherwise of this absolute presentation depends on the sequence of numbers n ( i ) chosen. We need a procedure to reduce a countable set of irrelations to a single irrelation; the method is adapted from [ 51 . For notational convenience we describe it here only for a 2-generator group: it is clear that it applies in the same way to groups without this restriction. Lemma 3.1. Let the group H be generated b y a l , u 2 with the defining relations

(3.1.1)

r i ( a l , a 2 ) = l , i = 1 , 2,...,

and assume that H also satisjies the irrelutiotzs (3.1.2)

al f 1 ,

(3.1.3)

s j ( a l , a 2 ) # 1 , j = 1 , 2, ... ;

here i and j range over some or all positive integers. Then H can be embedded in the group H , generated b y H and three elemeuts c, c’, d with the defining relations of H und (3.1.4)

c2 = 1, ( c c ’ ) ~= 1, d 2 = u l ,

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(3.1.5)

( d - - ~ c d ~ . ~ ~ ( u=, 1 , u. j~=) )1,~2, ...,

with j runging over the sume integers us in (3.1.3). Corollary 3.2. With the same notation, i j H is absolutely presented by (3.1.1 ), (3.1.2), (3.1.3), then every homomorphism o f H , under which the image oj'u, is not 1 is u monomorphism on H. Corollary 3.3. ? f ' H is recursively absolutely presented by (3.1. l), (3.1.21, (3.1.31, then H , is recursively relutively presented b y (3.1.1), (3.1.4), (3.1.5).

Proof of the lemma. Denote the order of a, by a ,the order of s,(ul .u2 1 by au). Form the group C generated by c, c' and al with the defining relations c2 =

1,(cc')2 = 1,c'2 = a , , u y = 1 ;

and for each relevant j the group Cj generated by cj and sj(al , a 2 ) with the defining relations

It is known that the order of ul in C i s a, and the order of

sj(ul ,u2 ) in Cj is (YO): hence we can form the generalized free product of H and C, C,, C,, ..., amalgamating u1 in H and C, then s1( u l .u2 ) in the product so obtained and C, , and so on. In the re-

sulting group the elements c, c1 , c 2 , ... all have order 2 and are otherwise not related among one another: that is to say, they generate the free product of their cyclic groups of order 2. The partial automorphism of this free product generated by mapping c on e l , c1 on c 2 , ... can be obtained by conjugation [3] by an element d that we now adjoin. the resulting group is the H , of the lemma. Note that now c, = d--JcdJ, so that the relations c2 = 1 follow from J c2 = 1 and may be omitted. This completes the proof of the lemma. The corollaries are obvious.

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Note that if H is embedded in a simple group, this would serve equally well for Corollary 3.2; but we need an embedding that is constructive, so as t o retain recursiveness; and it is convenient to have H, finitely generated We now use Graham Higman’s famous embedding theorem [ 2 1 t o make our final embedding.

Theorem 3.4.If H is recursively absolutely presented by (3.1.1 ), (3.1.2), (3.1.3), then H can be so embedded in a finite1.y relatively presented group H * that every homomorphism of H* under which the image of al is not 1 is a monomorphism on H. We need only embed H I , which is recursively relatively presented, in a finitely relatively presented group: the rest follows from the lemma and its corollaries.

Corollary 3.5. Every algebraically closed group contains isomorphic copies of every recursively absolutely presented 2-generator group. This is hardly scrprising, as there are only a countable infinity of such groups; but it seems t o me t o leave no hope of distinguishing two algebraically closed groups by picking out pairs of elements and looking a t the subgroups they generate. References [ 11 G . Nigman, A finitely generated infinite simple group, J. London Math. SOC.26 (1951) 61-64. [2] GHigman, Subgroups of finitely presented groups, Proc. Roy. Soc. London (A) 262 (1961) 455-475. [ 31 G. Higman, B.H. Neumann and H. Neumann, Embedding theorems for groups, J . London Math. Soc. 24 (1949) 247-254. [4] B.H.Neumann, Some remarks on infinite groups, J . London Math. Soc. 12 (1937) 122-127. (51 B.H.Neumann, A note on algebraically closed groups, J. London Math. Soc. 27 (1952) 247-249. 161 B.11. Neumann, Algebraically closed semigroups, Studies in Pure Mathematics (ed. L. Mirsky) (Academic Press, New York, London, 1971) 185-194.

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B.H.Neumann, The isomorphism problem f o r algebraically closed groups

[7] W.R.Scott, Algebraically closed groups, Proc. Amer. Math. SOC. 2 (1951) 118-121. [ 8 ] E. Steinitz, Alcebraische Theorie der KBrper, J . reine angew. Math. 137 (1910) 167-309, especially 6 2 1 .