JOURNAL OF ALGEBRA 200, 571᎐605Ž. 1998 ARTICLE NO. JA977205

A Classification of Fully Residually Free Groups of Rank Three or Less

Benjamin Fine

Department of Mathematics, Fairfield Uni¨ersity, Fairfield, Connecticut 06430 View metadata, citation and similar papers at core.ac.uk brought to you by CORE Anthony M. Gaglione* provided by Elsevier - Publisher Connector

Department of Mathematics, United States Na¨al Academy, Annapolis, Maryland 21402

Alexei Myasnikov

Department of Mathematics, City College of CUNY, New York, New York 10031

Gerhard Rosenberger

Fachbereich Mathematik, Uni¨ersitat¨ Dortmund, 44227, Dortmund, Germany

and

Dennis Spellman

Department of Mathematics, St. Joseph Uni¨ersity, Philadelphia, Pennsyl¨ania 19131

Communicated by Alexander Lubotzky

Received February 5, 1996

A group G is fully residually free provided to every finite set S ; G _ Ä41of non-trivial elements of G there is a free group FSSand an epimorphism h : GªFSSsuch that hgŽ./1 for all g g S.If nis a positive integer, then a group Gis n-free provided every subgroup of G generated by n or fewer distinct

* The research of this author was partially supported by the Naval Academy Research Council.

571

0021-8693r98 $25.00 Copyright ᮊ 1998 by Academic Press All rights of reproduction in any form reserved. 572 FINE ET AL.

elements is free. Our main result shows that a fully residually free group of rank at most 3 is either abelian, free, or a free rank one extension of centralizers of a rank two free group. To prove this we prove that every 2-free, fully residually free group is actually 3-free. There are fully residually free groups which are not 2-free and there are 3-free, fully residually free groups which are not 4-free. ᮊ 1998 Academic Press

1. INTRODUCTION

A group G is residually free if for each non-trivial g g G there is a free group Fggand an epimorphism h : G ª Fgsuch that hggŽ./1 and is fully residually free provided to every finite set S ; G _ Ä41 of non-trivial ele- ments of G there is a free group FSSand an epimorphism h : G ª FSsuch that hgSŽ./1 for all g g S. Clearly fully residually free implies residually free. Both properties have been of great interest recently in terms of the study of questions in logic pertaining to certain other freeness properties in groups and in connection with free tree actionsŽ seewx F-G-R-S , wxG-S1᎐GS6, Re and the references there. . In particular a group G is uni¨ersally free if it has the same universal theory as the class of non-abelian free groupsŽ seewx F-G-R-S for a precise formulation. ; n-free for a positive integer n, provided every subgroup of G generated by n or fewer distinct elements is free; commutati¨e transiti¨e if commutativity is transitive on the non-identity elements in G; and tree-free if G acts freely on some ⌳-tree in the sense of Basswx B where ⌳ is some ordered abelian group. Gaglione and Spellmanwx G-S6 and independently Remeslennikovwx Re , extending a theorem of B. Baumslag w Ba1 x , proved that if G is a non-abelian residually free group then the following are equivalent:Ž. 1 G is fully residually free; Ž. 2 G is commutative transitive; Ž.3 Gis universally free. Further they showed that universally free groups are tree freewx G-S5, Re . Fine, Gaglione, Rosenberger, and Spellman wxF-G-R-S and again independently Remeslennikovwx Re showed that the converse is not true; that is, there exist tree-free groups which are not universally free. In this paper we prove the result that every 2-free residually free group is 3-freeŽ. Theorem 4 . Since 2-free groups are commutative transitiveŽ. see Section 2 it follows from the Baumslag result that 2-free residually free groups are fully residually free. There are fully residually free groups which are not 2-free and there are 3-free, fully residually free groups which are not 4-freewx F-G-R-S . Applying the techniques used in proving the 2-free implies the 3-free result, together with the result itself, we can prove that all fully residually free groups of rank 3 or less are of one of three easily describable types. FULLY RESIDUALLY FREE GROUPS 573

This is a classification in the same vein as the classification of finite simple groups into finitely many types in that our result does not contain an algorithm to recognize isomorphic members of our list of 3-generator fully residually free groups. The principal result of this paper is that every fully residually free group of rank 3 or less may be obtained in a particularly simple way. Namely, starting with infinite cyclic groups one applies free rank one extensions of centralizers, free products, and the taking of subgroups. These construc- tions preserve full residual freeness. Moreover all known, finitely gen- erated, fully residually free groups may be obtained in this fashion. Myasnikov and Remeslennikov have conjectured that there are no other finitely generated examples. We note that since this paper has been under review this conjecture has been proved. Residual properties can be applied to rings as well as groupsŽ see Section 4. and there is a parallelism between the theories of fully residu- ally free groups and fully residually ޚ ringsŽ. or ␻-residually ޚ rings . We prove a resultŽ. Theorem 1 on fully residually ޚ rings analogous to the Baumslag result on fully residually free groups. We then exploit this parallelism to shed further light on fully residually free groups by embed- ding such groups in a hybrid object, a group admitting exponents from a fully residually ޚ ring. The prototypical example is Lyndon’s free ޚwxx group admitting exponents from the polynomial ring ޚwxx . Myasnikov and Remeslennikov define a ring of Lyndon type as a commutative ring A whose additive group Aq is torsion-free and contains the additive group ޚ of its minimum subring as a pure subgroup. In particular any fully residually ޚ ring is of Lyndon type. We show that if A is of Lyndon type then any finitely generated group embeddable in a free A groupŽ see Section 4. is already embeddable in a free ޚwxx group. If G is a residually free group, then one may embed G into an unrestricted direct power of the free group F2 of rank 2. If G is fully residually free, then one does not lose injectivity when one projects to a suitable ultrapower of F2 . Furthermore, if G is 3-generator and fully residually free, then using the theory of equations over free groups and a formal power series representation of a free exponential groupŽ see wxG-M-Re-S. , one deduces that G is embedded in a free exponential subgroup of an ultrapower of F2 . From this, after several technical lemmas, one deduces that every 2-free, residually free group is 3-free since 2-free, residually free groups are necessarily fully residually free. The outline of this paper is as follows. In Section 2 we give some preliminary results and definitions followed in Section 3 by some necessary material on equations in free groups. In Section 4 we discuss free-A groups where A is a ring and prove a resultŽ. Theorem 1 analogous to the B. Baumslag theorem on fully residually free groups. We also prove that if 574 FINE ET AL.

A is an ␻-residually ޚ ring then every free A-group is fully residually freeŽ. Theorem 2 . In Section 5 we have a discussion of ultrapowers which are necessary for our characterization of fully residually free groups. In Section 6 we prove two of our main results.

THEOREM 3. E¨ery 3-generator, fully residually free group lies in F, where F is the smallest class of groups containing the infinite cyclic groups and closed under the following four ‘‘operators’’: Ž.1 Isomorphism Ž.2 Finitely Generated Subgroups Ž.3 Free Products of Finitely Many Factors Ž.4 Free Rank One Extensions of Centralizers.

THEOREM 4. E¨ery 2-free, residually free group is 3-free. Using these results we present in Section 7 our main classification result.

THEOREM 5. Let G be a fully residually free group. Then

Ž.1 if rank ŽG .s 1 then G is infinite cyclic. Ž.2 if rank ŽG .s 2 then either G is free of rank 2 or free abelian of rank 2. Ž.3 if rank ŽG .s 3 then either G is free of rank 3, free abelian of rank 3, or a free rank one extension of centralizers of a free group of rank 2. That is, G has a one-relator presentation

y1 G s ² x1233, x , x ; x ¨x 3s ¨ :, where ¨ s ¨Ž.x12, x is a non-tri¨ial element of the free group on x12, x which is not a proper power. In Section 8 we give some additional applications and finally in Section 9 we give proofs of two important auxiliary results.

2. PRELIMINARIES

By a ring we shall always mean an associative ring with multiplicative identity 1. All subrings are assumed to contain the multiplicative identity of the overring in question and every ring homomorphism is assumed to map the multiplicative identity to the multiplicative identity. We use the symbol ޚ to represent variously the set of integers or an additively written infinite cyclic group with generator 1ᎏin particular the additive group of the minimum subring of a ring of characteristic zero or the minimum FULLY RESIDUALLY FREE GROUPS 575 subring itself. The usage will depend on the context. The symbol ␻ shall represent the non-negative integers as well as the first limit ordinal. ގ shall be the set of positive integers.

DEFINITION 1Ž B. Baumslagwx Ba1. . Let n g ގ. A group G is n-residu- n ally free provided to every ordered n-tuple Ž.Žg1,..., gng G_Ä41of non-identity elements of G there is a free group F and an epimorphism h:GªGsuch that hgŽ.i /1 for all i s 1,...,n. G is fully residually free or ␻-residually free provided it is n-residually free for every n g ގ. It is an easy consequence of the Nielsen᎐Schreier subgroup theorem that full residual freeness is inherited by subgroups. B. Baumslag proved in wxBa1 that if A / 1 and B / 1 are residually free groups, then A) B is residually free if and only if each of A and B is fully residually free.

DEFINITION 2. A group G is commutati¨e transiti¨e provided the rela- tion of commutativity is transitive on the non-identity elements.

LEMMA 1Ž Harrisonwx H. . Let G be a group. The following three statements are pairwise equivalent.

Ž.i G is commutati¨e transiti¨e. Ž.ii The centralizer of e¨ery non-tri¨ial element in G is abelian. Ž.iii E¨ery pair of distinct maximal abelian subgroups in G has tri¨ial intersection. B. Baumslag proved inwx Ba1 that a residually free group is fully residually free if and only if it is commutative transitive. Remeslennikov wxRe has deduced from that result that fully residual freeness actually follows from 2-residual freeness. Suppose that G is 2-residually free. To show that G is fully residually free it suffices to show that G is commuta- tive transitive. Suppose then there are elements a, b, c g G with b / 1 such that ab s ba and bc s cb but ac / ca. Then b / 1 and wxa, c / 1. There is a free group F and an epimorphism h: G ª F, x ª x such that b/1 and wxa, c / 1. But this is impossible since ab s ba and bc s cb and b / 1 in the free group F. This contradiction shows that G is commutative transitiveᎏhence fully residually free as claimed.

DEFINITION 3. Let G be a group and H a subgroup of G. H is 1 malnormal in G or conjugately separated in G provided gy Hg l H s 1 unless g g H. Now suppose G is a fully residually free group with more than one element. Let u g G _ Ä41 and let M be its centralizer which we will denote by ZuGŽ.. Then M is maximal abelian in G. We claim that M is malnormal in G.IfGis abelian, then M s G and the conclusion follows 576 FINE ET AL.

1 trivially. Suppose that G is non-abelian. Suppose that w s gy zg / 1 lies 1 in gMgy lM.If gfMthen wxg, u / 1. Thus, there is a free group F and an epimorphism h: G ª F, x ª x such that w / 1 and wxg, u / 1. Let y1 CsZuF Ž.. Then w g gCglC. However, the maximal abelian sub- groups in a free group are malnormal. This implies g g Cᎏcontradicting 1 wxg,u/1. This contradiction shows that gMgy lM/1 implies g g M and hence the maximal abelian subgroups in G are malnormal.

DEFINITION 4Ž Myasnikov and Remeslennikovwx M-Re2. . A group G is a CSA-group or conjugately separated abelian group provided the maximal abelian subgroups are malnormal. Every CSA-group is commutative transitive. Let G be a group in which maximal abelian subgroups are malnormal and suppose that M12and M are maximal abelian subgroups in G with z / 1 lying in M12l M . Could y1 we have M12/ M ? Suppose that w g M12_ M . Then w zw s z is a y1 non-trivial element of w Mw22lM so that w g M2. This is impossible and therefore M12; M . By maximality we then get M12s M . Hence, G is commutative transitive whenever all maximal abelian subgroups are mal- normal. In any non-abelian CSA-group the only abelian normal subgroup is the trivial subgroup 1. This follows from the following argument. If N is any normal abelian subgroup of the non-abelian CSA-group G then N is contained in a maximal abelian subgroup M. Let g f M. Then N s 1 1 gy Ng l N ; gy Mg l M. N / 1 would imply that g g Mᎏa contradic- tion. This observation allows us to deduceŽ as did Myasnikov and Remeslennikov inwx M-Re2. that the class of CSA-groups is properly smaller than the class of commutative transitive groups. For example, let p and q be distinct primes with p a divisor of q y 1. Let G be the non-abelian group of order pq. Then it is not difficult to prove that the centralizer of every non-trivial element of G is cyclic or order either p or q. Thus G is commutative transitive. However, theŽ. necessarily unique Sylow q-subgroup of G is normal in G.

DEFINITION 5Ž Myasnikov and Remeslennikovwx M-Re. . Let G / 1bea commutative transitive group. Let u g G _ Ä41 and let M s ZuGŽ.. Let B / 1 be a torsion free abelian group. Then

GuŽ.,B s²:G,B; rel Ž.Ž.G , rel B , wxB, M s 1 is the B-extension of the centralizer M of u in G.If Bs²:t; is infinite cyclic, then

1 GuŽ.,B s²G,t; rel Ž.G , tzty sz, for all z g M : is the free rank one extension of the centralizer M of u in G. FULLY RESIDUALLY FREE GROUPS 577

Observe that if B / 1 is an arbitrary torsion free abelian group, then GuŽ.,B is a direct union of free rank one extensions of centralizers. One can prove that if G / 1 is fully residually free, then so is every free rank one extension of a centralizer of G. From this it follows that if B / 1 is any torsion free abelian group and u g G _ Ä41 , then GuŽ.,B is locally fully residually free. In the special case where B is residually ޚ,in particular if B is free abelian, then GuŽ.,B is fully residually free.

DEFINITION 6. Let n g ގ. A group G is n-free provided every sub- group generated by n or fewer distinct elements is free. Observe that if G is an n-free group then G is also an m-free group for all 1 F m - n. The 1-free groups are precisely the torsion free groups. 2-free groups are commutative transitive. To see this let G / 1 be 2-free and let u g G. Let M s ZuGŽ.. We claim that M is locally cyclic and therefore abelian. Suppose a, b g M. Since G is 2-free, a, u generate a free group. Since a, u commute this must be cyclic and hence they are ␣ ␤ both powers of a single element g. Thus a s g , u s g for some ␣, ␤. Similarly since b and u commute and b, u generate a free group we have ␦ ␥ an element h with b s h , u s h for some ␦, ␥. Now consider ²:h, g ; G. ␤ ␥ This is free because G is 2-free; however, this has the relation g s h . Therefore ²:h, g must be cyclic and h and g are both powers of a single element. Hence a and b are both powers of this element and any 2-generator subgroup of M is cyclic. A straightforward induction then shows that any n-generator subgroup ²:a1,...,an ;M is also cyclic. Thus, centralizers of non-trivial elements are locally cyclic and 2-free groups are commutative transitive as claimed. In addition 2-free groups are also CSA-groups. Assume as above that y1 G/1 is 2-free, u g G _ Ä41 , and M s ZuGŽ.. Suppose gMglM/1 1 1 and suppose w s gzgy /1ggMgy lM. Certainly z g M _ Ä41. If g commutes with z, then g would commute with u by commutative transitiv- ity and we would have g g M. Suppose that g f M. Then g cannot commute with z. Consider F s ²:g, z which is free since G is 2-free. Since g and z do not commute F is free of rank 2 with basis Ä4g, z . Thus, y1 CsZzF Ž.sMlFs ²:zis malnormal in F. But w g gCglCand w / 1. This implies that g commutes with z contrary to the assumption 1 g f M. Hence, gy Mg l M / 1 implies that g g M and therefore M is malnormal in G. Thus, 2-free groups are CSA-groups as claimed. For each integer n g ގ the non-orientable surface group

² 22иии : a1 ,...,anq11;a anq1s 1 of genus n q 1is n-free but not Ž.n q 1 -free. 578 FINE ET AL.

A fully residually free group is 1-freeŽ. i.e., torsion free but not necessar- ily 2-free. Indeed, every residually free group is torsion free. Suppose G is residually free and g g G _ Ä41 . Then there is a free group F and an epimorphism h: G ª F, x ª x such that g / 1. If n ) 1 is an integer, then g nn/ 1. It follows that g / 1 and G is torsion free as claimed. ޚ = ޚ is an example of a fully residually free group which is not 2-free. Of course, every free group is both fully residually free and n-free for every positive integer n. We shall prove in this paper that every 2-free, fully residually free group must also be 3-free. This result is best possible since, 22 22 for example, the non-orientable surface group ²a12341234, a , a , a ; aaaa s 1: is a 3-free, fully residually free group which is not 4-free. We remark that it was shown in Fine, Gaglione, Rosenberger, and Spellmanwx F-G-R-S that

² иии 2 n: a12,...,a ny212,b ,b ;wxwa 12,a a 2ny32, a ny2121 xwxs b , bb is an n-free group which is not residually free.

3. EQUATIONS IN FREE GROUPS

Let X s Ä4xn: n g ގ be a free basis for a free group of countably infinite rank. For each n g ގ, let Xn s Ä4x1,..., xn be the initial segment of X consisting of the first n elements. Let r G 2 be a cardinalŽ finite or infinite. and let Fr be a group free of rank r.Byan n-¨ariable word ଙ equation with coefficients in Frrwe mean an element w g FX²:n.A ଙ²: < retraction h: FXrnªF r,hFrsid is a solution to the equation w g F r provided hwŽ.s1. The equation w is consistent over Fr provided it has at least one solution in Frr. Otherwise, it is inconsistent over F .If wg²:Xn, then w is an n-¨ariable word equation without coefficients. Henceforth, we shall write our equation w as a formal expression w s 1 and say that Ž.Ž.x1,..., xn s b1,...,bn is a solutionᎏor simply that Ž.b1,...,bn is a solution provided the retraction given by f ª f for all f g Fr , x11ª b , ..., xnnªb is a solution. Every word equation without coefficients is consistent since, for example, Ž.Ž.x1,..., xn s 1, . . . , 1 is always a solution. However, if there are coefficients an equation can be inconsistent. For y2 example, let F212s ²:a , a ; and consider the equation xa112wx,as1or 2 wxa12,asx 1. It was shown by Schutzenbergerwx Sc that no non-trivial commutator can be a proper power in a free group. Hence, this equation is inconsistent over F2 . Let us briefly consider, for the sake of illustration, the problem of solving a 2-variable word equation without coefficients wxŽ.12,x s1 over a free group Fr . Here wxŽ.12,x is a freely reduced, non-empty word on FULLY RESIDUALLY FREE GROUPS 579

y1 X22jX . For i s 1, 2, let ewiiŽ.be the exponent sum on x in w. Suppose Ž.Ž.x12,xsb 12,bis a solution. Then b1and b 2must satisfy the non-trivial relation wbŽ.12,b s1in Fr. Clearly then we can conclude that b12and b have to commute. They must therefore lie in a maximal cyclic subgroup ni ²:c of Frr, c / 1, c not a proper power in F . Thus, bis c , i s 1, 2 for some integers n12, n . It follows that the solutions to wxŽ.12,x s1in Fr ␯ i are exhausted by the values in Friof the parametric words x s u in a single group variable u and the integral parameters ␯ 12and ␯ . Although all solutions are found by substituting a group element for u and integers for ␯ 12and ␯ in general not every substitution will yield a solution. Thus, n1 n2 if b12s c and b s c is a solution in Frto wxŽ.12,x s1, where as before c / 1 is not a proper power in Fr , then ne11Ž. w qne 22 Ž. ws0. Hence, in order for the assignment ␯iiª n , i s 1, 2, to yield a solution it is necessary and sufficient that y1122s n , y s n be a solution to the linear Diophantine equation ewy1122Ž.qewy Ž. s0 whenever an element c / 1, not a proper power in Fr , is substituted for the group variable u. Now let us consider a 3-variable word equation without coefficients wxŽ.123,x,x s1 where wx Ž.123,x,x is a freely reduced, non-empty word y1 3 on the alphabet X33j X . Call a solution Ž.b123, b , b g Frtrivial pro- vided ²:b123, b , b is cyclic. Hmelevskiiwx Hm showed that the non-trivial solutions in Fr to wxŽ.123,x,x s1 are exhausted by the values in Frof a finite set of triples Ž.W1, j, W2, j, W3, j , j s 1,...,N, of parametric words in two group variables u and ¨ and a finite set ␯ 1,...,␯m of integral parameters. Not every substitution will yield a solution; however, one can effectively determine those substitutions which do yield solutions. In order to solve consistent one variable equations over Fr Lyndonwx L1, L2 intro- A duced the free object Ž.FrAin the category L of groups admitting exponents from the polynomial ring A s Zwx␯ 1,...,␯m . More generally, he defined for an arbitrary ring A with identity 1 / 0an A-group to be a ␣ group admitting operators from A, G = A ª G, Ž.g, ␣ ª g subject to the conditions

␣ ␤␣␤ Ž.igg sgq ␣ ␤ ␣␤ Ž.ii Žg . s g 1 Ž.iii g s g 1 ␣ 1 ␣ Ž.iv Žhghy .shghy for all g, h g G, ␣, ␤ g A.

A It follows from abstract nonsense that the rank r free A-group, Ž.Fr , exists and is unique up to an A-isomorphism. One can show that if the q A additive group A of A is torsion free, then Ž.Fr is commutative A transitive and the centralizer of any non-trivial element u / 1inŽ.Fr is a 580 FINE ET AL. right A-moduleᎏin fact a free rank one right A-module. However, Myasnikov and Remeslennikovwx M-Re have given an exampleŽ necessarily non-free. of an A-group G containing a maximal abelian subgroup which is not a right A-module. They study the subclass MAAof L consisting of those A-groups G satisfying the additional condition

␣ ␣␣ gh s hg implies Ž.gh s gh

A for all g, h g G, ␣ g A. We remark that Ž.Fr can fail to be commutative transitive if A is not torsion free. Indeed, if A s ޚ3, then the free A Burnside group Ž.F2 is a non-abelian, nilpotent group and no such group can be commutative transitive.

4. FREE A-GROUPS

DEFINITION 7. Let A be a ring with identity 1 / 0. For a natural number n, A is n-residually ޚ provided to every ordered n-tuple n Ž.Ž␣1,...,␣n g A_Ä40. there is a ring homomorphism ␳: A ª ޚ such that ␳Ž.ai / 0 for i s 1,...,n. A is fully residually ޚ or ␻-residually ޚ provided it is n-residually ޚ for every natural number N. A is residually ޚ if it is 1-residually ޚ.

LEMMA 2. Let A be a ring with identity 1 / 0. Suppose that A is residually ޚ. Then A is commutati¨e and has characteristic zero. Proof. Let ␣, ␤ g A.If ␣␤ / ␤␣ then ␣␤ y ␤␣ / 0 and there is an epimorphism ␳: A ª ޚ such that ␳␣␳␤Ž.Ž.y␳␤␳␣ Ž .Ž.s␳␣␤ Ž y ␤␣.Ž.Ž.Ž.Ž./ 0. Then ␳␣␳␤ /␳␤␳␣, which is impossible in ޚ. There- fore A is commutative. Suppose n .1s0in A. Since A is residually ޚ and 1 / 0in Athere is a retraction ␳: A ª ޚ such that ␳Ž.1 s 1 / 0. Then ␳ Žn .1 .sn.␳ Ž.1 s n.1s0inޚ. Hence n s 0 and therefore A has characteristic zero. If a ring A has characteristic zero we shall identify its minimum subring with ޚ and consider an epimorphism ␳: A ª ޚ to be a retraction. The next result is analogous to the Baumsalg resultwx Ba1 on residually free groups.

THEOREM 1. Let A be a ring with identity 1 / 0. Suppose that A is residually ޚ. Then the following three statements are pairwise equi¨alent. Ž.i Ais2-residually ޚ. Ž.ii A is an integral domain. Ž.iii Ais␻-residually ޚ. FULLY RESIDUALLY FREE GROUPS 581

Proof. Ž.i ± Žii . . Suppose ␣, ␤ / 0 and ␣␤ s 0in A. Then there is a retraction ␳: Aªޚ such that ␳␣Ž./0, ␳␤ Ž ./0, ␳␣␤ Ž .s␳␣␳␤ Ž.Ž.s 0. Since this is impossible in ޚ, A must be an integral domain. Ž.ii ± Žiii . . Suppose ␣i / 0 for i s 1,...,n. Let ␣ s ␣1 иии ␣n. Then ␣/0 and hence there exists a retraction ␳: A ª ޚ such that ␳␣Ž.s ␳␣Ž.Ž.Ž.1 иии ␣n s ␳␣1 иии ␳␣ni/0. It follows that ␳␣ Ž./0 for all i s 1,...,n. Ž.iii ± Ž.i . This is obvious. Examples of ␻-residually ޚ rings are the integral polynomial rings

ޚwx␯1,...,␯n . Another example is the following. Let d ) 1 be an integer which is not a perfect square. Then the residue class ring A s 22 ޚwx␯12,␯rŽ␯ 1yd␯ 2y1is.␻-residually ޚ. Note that if A is any locally ␻-residually ޚ ring, then the additive group Aq of A is torsion free and ޚ is a pure subgroup of Aq Žsee M. Hallwx Ha, p. 198 for a definition of pure. .

DEFINITION 8. A is a ring of Lyndon type if its additive group Aq is torsion free and ޚ is a pure subgroup of Aq.

Now let A be the integral polynomial ring ޚwx␯ 1,...,␯n . Let r G 1bea A cardinal. Lyndonwx L2 showed that if Ž.Fr is A-freely generated by X of cardinal r, then the subgroup ²:X generated as a group by X is free with basis X. We write Fr s ²:X . It follows easily from the universal property A of the A-tensor completion Ž.Frrof F that every ring retraction ␳: A ␣ Aªޚinduces a group retraction ␳: Ž.Frrª F such that ␳Ž.g s ␳Ž␣. A ␳Ž.g for all g g Ž.Fr , ␣ g A. Furthermore, Lyndon showed that to A every element g g Ž.Fr there corresponds an element ␣ g A satisfying the following property.

Ž.) If ␳: A ª ޚ is any ring retraction, then ␳Ž.g s 1 if and only if ␳␣Ž.s0. Moreover, he showed that if g / 1, then any element ␣ g A corre- sponding to g in the sense of Ž.) is non-zero. Thus, let g1,..., gn be A finitely many non-trivial elements of Ž.Fri. Let ␣ g A correspond to giin the sense of Ž.) for i s 1,...,n. Then since A is ␻-residually ޚ there is a retraction ␳: A ª ޚ such that ␳␣Ž.i /0 for i s 1,...,n. It follows that A ␳Ž.gir/1 for i s 1,...,n. Thus, Ž.F is fully residually free. Since full A residual freeness is inherited by subgroups, every subgroup of Ž.Fr is fully residually free. A careful analysis of Lyndon’s paperwx L2 reveals that his arguments go through if A is any ␻-residually ޚ ring. The full residual A A freeness of Ž.Frralso follows from residual freeness since Ž.F is commu- tative transitiveᎏthe centralizers of non-trivial elements being free rank one right A-modules. This result can be extended to further rings of 582 FINE ET AL.

Lyndon type. Specifically:

THEOREM 2. If A is any ␻-residually ޚ ring, or more generally a ring of Lyndon type, where Aq, the additi¨e subgroup of A, is fully residually ޚ where ޚ is the additi¨e subgroup of the minimum subring of A, then e¨ery free-A group is fully residually free. Consider the additively written abelian group ޚ N which is the unre- stricted direct power of countably infinite copies of the infinite cyclic group ޚ. Every countable subgroup of ޚގ is free abelianwx Sp . ޚގis residually ޚ as is easily seen by taking projections and hence it is residually free. Clearly it is commutative transitive and therefore it is fully residually free. ޚގ is uncountable so it cannot be embedded in the countable group Zw␯x Ž.F␻ . Along the way to proving that every 2-free, fully residually free group is 3-free we shall prove that every 3-generator, fully residually free group ޚw␯x is embeddable in Ž.F␻ . In order to make progress in that direction we need a construction of the free A-group F A when A is a ring of Lyndon type. We follow the treatment of Myasnikov and Remeslennikovw M-Re1, M-Re2x . We define an increasing chain F s FŽ.0 F F Ž.1 F иии F Fn Ž .F A Ž. A иии of subgroups of F whose union D n- ␻ Fn is F as follows. FŽ.0sFand if FnŽ.has already been defined, then FnŽ.q1 is the subgroup of F A generated by all elements g ␣ as g varies over FnŽ.and ␣ varies over A. Call an element of F A irreducible if it generates its own centralizer as a right A-module. Thus u / 1 is irreducible provided ␣ A ZuF AŽ.is Äu : ␣ g A4. Choose a set V of irreducible elements of F with A the property that if u is any irreducible element of F then ZuF AŽ.is A conjugate in F to ZF AŽ.¨ for exactly one ¨ g V. For each n g ␻ let VnŽ.sVl ŽFn Ž._Fn Žy1 .. where F Žy1 . is empty. Fn Žq1 . may be constructed as a tree product as follows. There is a root with vertex group FnŽ.. Emanating from this root there is one edge for each element q ¨gVnŽ.. The terminal vertex of that edge has vertex group ZF A Ž.¨ ( A and the edge group is the infinite cyclic group ²:¨ ,

FnŽ.lAq. ²:¨

Note that if there were just a single edge then the group so constructed would be ²Ž.ŽFn ¨,Aq.²; ¨ :sޚ :. For this reason we say that FnŽ.q1is atree extension of centralizers of FnŽ.and that F is an iterated tree extension of centralizers Ž.up to the first limit ordinal ␻ . Suppose ¨ g VnŽ.. Let C be a system of coset representatives for Aq Ž.mod ޚ containing 0 and let R be a system of right coset representatives m for FnŽ.Žmod ²:.¨ containing 1. Any element g of FnŽ.is uniquely ¨ g FULLY RESIDUALLY FREE GROUPS 583 for some m g ޚ, g g R. Every element g of ²Ž.ŽFn ¨,Aq .²:; ¨ sZ :has cc1иии k ²: a unique canonical form g s gg01¨ gkk¨ g q10where g g ¨ , gig R,gii/1, i s 2,...,k,c /0, c g C Žsee Myasnikov and Remeslennikov . Ac1иии ck wxM-Re2 . Every element of F _ F has the form gg011¨ gkk¨ g kq1 where for some n g ␻, giig FnŽ.and ¨ g Vn Ž.for all i while cig C _ Ä40 for all i. Neither the giinor the ¨ need be distinct. In view of the ޚw␯x construction of Ž.F␻ as an iterated tree extension of centralizers, we may characterize the class F of finitely generated groups embeddable in ޚw␯x Ž.F␻ Ž.see Section 9 as the smallest class of groups containing the infinite cyclic groups and closed under the following four ‘‘operators’’: Ž.1 Isomorphism Ž.2 Finitely Generated Subgroups Ž.3 Free Products of Finitely Many Factors Ž.4 Free Rank One Extensions of Centralizers.

A Moreover, since if A is any ring of Lyndon type Ž.F␻ is similarly A constructed, we see that any finitely generated subgroup of Ž.F␻ must lie in F for such a ringŽ. see Section 9 . Remeslennikovwx Re showed that if G is any k-generator, fully residually free groupŽ. where k is finite , then G is ⌳-free where ⌳ is free abelian of rank at most 3k. The question of whether or not an arbitrary finitely generated fully residually free group is ⌳-free for finitely generated ⌳ remains open. However, it follows from Basswx B that in the special case of members of F the question has a positive answer. Suppose we write I for the isomorphism operator, Sf for the finitely generated subgroup operator, Ff for the free product of finitely many factors operator, and EZ for the free rank one centralizer extension operator. Suppose G is constructed from an infinite cyclic group via a finite sequence of the above operators. In view of Corollary 4.4 and Proposition 4.15 of Basswx B , G acts freely without inversions on a ⌳-tree X where ⌳ is free abelian of rank n q 1if n is the number of occurrences of the EZ operator in the sequence and moreover, the order on ⌳ is lexicographic. From this and Proposition 1.7Ž. d and Corollary 1.10 of Basswx B it follows that members of F and in particular 3-generator finitely generated fully residually free groups satisfy the maximal condition for abelian subgroups. A We need one other fact about Ž.Fr . Suppose that A is locally ␻-residu- ally ޚ. Let CAŽ.,ގ be the smallest ring containing A and closed under the formation of ‘‘binomial coefficients’’ CŽ.␣, n s ␣␣ Žy1 .Žиии ␣ y nq1.Ž.Žrn!, ␣ g CA,ގ,ngގ. Recall that if A is any ␻-residually ޚ ring, then A is an integral domain of characteristic zero. It easily follows that any locally ␻-residually ޚ ring also enjoys those properties.. One can prove that CAŽ.,ގ may be achieved by adjoining to A the binomial 584 FINE ET AL.

A coefficients CŽ.␣, n where ␣ g A. Let Ž.Fr have A-basis x1,..., xr . Let y1,..., yr be a set of r non-commuting indeterminates over CAŽ.,ގ. Extend the free associative CAŽ.,ގ-algebra CA Ž.,ގwxy1,..., yr to a ring A of formal infinite sums CAŽ.,ގwwy1,..., yrr xx. Then Ž.F may be faithfully represented in the group of units of CAŽ.,ގwwy1,..., yri xx via x ª 1 q y i, i ␣ 2 s1,...,r. Moreover,Ž. 1 q u is defined as 1 q CŽ.␣,1 uqC Ž.␣,2 u n q иии qCŽ.␣, nu qиии if u is an element of CAŽ.,ގwwy1,..., yr xx with zero constant termŽ see G-M-Re-Sx. .

5. ULTRAPOWERS

Let I be a non-empty set. Let PIŽ.be the power set of I.If JgPI Ž. let JЈ be its complement I _ J. A subset Mo ; PIŽ.will be an ideal in the ring PIŽ.if л g Moo, M is closed under finite unions and Mois closed under the formation of subsets. Mo will be a proper ideal in PIŽ.if it is an ideal in PIŽ.and I f Mo. The dual of a proper ideal is a filter. Specifi- cally, a subset Do ; PIŽ.is a filter on I provided

Ž.1 IgDo

Ž.2 A,BgD0±AlBgDo

Ž.3 AgDooand A ; B ; I ± B g D

Ž.4 лfDo. A subset SIŽ.satisfies the finite intersection property provided whenever

A1,..., An are finitely many members of S the intersection A1 l иии l An is non-empty. In view of propertiesŽ. 2 and Ž. 4 above no subset S ; PIŽ. failing to satisfy the finite intersection property can lie in any filter on I. However, every subset SIŽ.which satisfies the finite intersection property is contained in a least filter Dooon I. Specifically D is the family of all elements BgPIŽ.such that B contains some finite intersection A1 l иии l Anoof elements of S. A filter D is principal if it is generated by a single element A g PIŽ.. A principal filter is the dual of aŽ. proper principal ideal. A filter D on I is an ultrafilter on I provided it is maximal in the class of filters on I. An ultrafilter D on I is the dual of a maximal ideal M in PIŽ.. One can show that a filter D on I is an ultrafilter on I if and only if for every element A g PIŽ.exactly one of A or AЈ belongs to D. One can also show that for every element i og I the set D s Ä A g PIŽ.: io gA4is a principal ultrafilter on I and that these exhaust the principal ultrafilters on I. It is then easy to show that if I is finite then every ultrafilter on I is principal. If I is infinite then we may appeal to Zorn’s

Lemma to prove that every filter Do on I extends to an ultrafilter D on I. FULLY RESIDUALLY FREE GROUPS 585

The result just asserted is known as The Prime Ideal Theorem. Let I be an infinite set. Call an element A g PIŽ.cofinite if AЈ is finite. Then the set S of cofinite subsets of the infinite set I satisfies the finite intersection property so by the Prime Ideal Theorem extends to an ultrafilter on I. Such an ultrafilter is never principal. Assuming the consistency of ZF, the Prime Ideal Theorem is provably weaker than the Axiom of Choicewx Hp . Now let ⍀ be an operator domain and let A be an ⍀-algebra and I a non-empty index set. Form the unrestricted direct power AI consisting of all functions I ª A with operations defined componentwise. Let ␦: A ª I ␦ A be the diagonal map a ª ␦Ž.a , ␦ Ž.Ž.aisafor all i g I. Now let D be an ultrafilter on I and let M s Ä4JЈ: J g D be the maximal ideal dual to D. Then, up to isomorphism, the residue class ring PIŽ.rMis the finite field of two elements ޚ2 and hence there is a ring homomorphism ␮: PIŽ.ªޚ2 with kernel M. Specifically the two residue classes of M in PIŽ.are M and I q M s D. Suppose J1,..., Jn are finitely many pairwise disjoint subsets of I. Then their symmetric difference J1 q иии qJn reduces to their union J1 j иии j Jn. Thus, ␮Ž.Ž.J1 j иии j Jn s ␮ J1 q иии qJn s ␮Ž.J1 qиии q␮ Ž.Jn . Hence, ␮ is a finitely additive ޚ2-valued measure I defined on all subsets of I. We may define for two elements f, g g A equality almost everywhere in the sense of D by f s g a.e. provided Äi g I: fiŽ.sgi Ž.4gD. This is equivalent to f and g differing only on a set of measure zero. Equality almost everywhere is a congruence on AI.We I I write frD for the equivalence class of f g A . The quotient algebra A rD is the ultrapower of A with respect to the ultrafilter D on the index set I. I I The map d: A ª A rD, a ª ␦Ž.a rD embeds A in A rD. Further d is also an elementary embedding. That means if ␺ Ž.x1,..., xn is any formula of the first-order language with equality L⍀ appropriate for ⍀-algebras where ␺ Ž.x1,..., xn contains free at most the distinct variables x1,..., xn n and Ž.a1,...,an gA then ␺ Ž.a1,...,an is true in A if and only if I ␺ŽŽda1 .,...,da Žn ..is true in A rD. By observing that a sentence of L⍀ is by definition a formula of L⍀ containing no free variables we see that A I and A rD must satisfy precisely the same sentences of L⍀ . Two such ⍀-algebras are called elementarily equi¨alent. In the case where D is a principal ultrafilter the construction is not interesting because in that I event A rD is isomorphic to A. Followingwx J-Le we call a ring elementarily equivalent to ޚ a Peano ring. Thus, if I is an infinite index set and D is a non-principal ultrafilter on I ଙ I then the ultrapower ޚ s ޚ rD is an example of a Peano ring. LEMMA 3. E¨ery Peano ring is locally ␻-residually ޚ. Proof. Let R be any Peano ring. Then R is easily seen to be commuta- tive. Hence, every subring is commutative. Let A be a finitely generated subring of R. A is a homomorphic image of an integral polynomial ring 586 FINE ET AL.

ޚwx␯1,...,␯n . Let ␪: ޚ wx␯ 1,...,␯n ªA be an epimorphism. By the Hilbert Basis Theorem, ޚwx␯ 1,...,␯n is Noetherian.Ž See, e.g., Langwx La .. Thus, if K is the kernel of ␪ then K is finitely generated as an ideal. Let f1,..., fm generate K. Suppose ␣1,...,␣k are finitely many non-zero elements of A. Choose preimages g j g ޚwx␯ 1,...,␯njj, such that ␪Ž.g s ␣ for j s 1,...,k. Let ␪␯Ž.be the ordered n-tuple ŽŽ␪␯1 .,...,␪␯ Žn ..of elements of A. Then

mk f␪␯ 0 g ␪␯ 0 HHŽ.ijŽ.Ž.s n Ž.Ž.Ž./ Ž.) ž/i1 j1 s žs is true in R. Therefore the existential sentence

mk ᭚xfx0 gx 0 HHijŽ.sn Ž./ Ž.)) ž/i1 j1 s ž/s

is true in R where x s Ž.x1,..., xn is a tuple of distinct variables. We must have that Ž.)) is true in ޚ also. Hence there is an ordered n-tuple n usŽ.u1,...,un gޚ such that

mk fu 0 gu 0 HHijŽ.s n Ž./ Ž.))) ž/i1 j1 s ž/s

is true in ޚ. We may then define a retraction h: ޚwx␯ 1,...,␯n ªޚ determined by ␯iiª u for i s 1,...,n. Since K is contained in the kernel of h there is a retraction ␳ such that the triangle

␪ 6 ޚwx␯1n,...,␯ A

␳ 6 6 ޚ

commutes. But then ␳␣Ž.jjjjs␳␪ Ž.g s hg Ž.sgu Ž./0 for j s 1,...,k. Thus A is ␻-residually ޚ. Hence R is locally ␻-residually ޚ completing the proof. In the preceding lemma the only property of Peano rings that was utilized is that every Peano ring has the same universal theory as ޚ. Hence every ring having the same universal theory as ޚ is locally ␻-residually ޚ. The converse is also true. That is, a ring A with identity 1 / 0 has the same universal theory as ޚ if and only if it is locally ␻-residually ޚ. Now suppose that G is a non-abelian, fully residually free group. Let

a and b be a fixed pair of non-commuting elements of G and let Ž g1,..., n gn.ŽgG_Ä41. be an arbitrary finite tuple of non-trivial elements of G. Then there is a free group F and an epimorphism ␺ : G ª F such that FULLY RESIDUALLY FREE GROUPS 587 w␺Ž.a,␺ Ž.bxs␺Žwxa,b.Ž./1 and ␺ gi / 1 for i s 1,...,n. However, ev- ery non-abelian free group is fully residually-Ž. free of rank 2 . F is non- abelian since w␺ Ž.a , ␺ Ž.b x/ 1 and therefore, there is an epimorphism ␾: FªF2 such that ␾␺ŽŽ..gi /1 for i s 1,...,n. It follows that every non-abelian, fully residually free group is actually fully residually F2 . Let I I be the set of all epimorphisms h: G ª F22. G may be embedded in F via I ␾:GªF2 ,␾Ž.Ž.ghshg Ž.for all h g I. Now define for each non-trivial element g g G _ Ä41 its support by SuppŽ.g s Äh g I: hgŽ./14 . The fact that G is fully residually F2 is equivalent to the assertion that the family ÄSuppŽ.g : g g G _ Ä441 satisfies the finite intersection property. Hence this family extends to an ultrafilter D on I. G may be embedded in the ଙ I ଙ ଙ ଙ ultrapower F22s F rD via ␾: G ª F2, ␾Ž.g s ␾ Ž.g rD.

6. 2-FREE RESIDUALLY FREE GROUPS

In this section we prove our first two main results: that every 3-genera- tor fully residually free group lies in the class F and that every 2-free residually free group must be 3-free. ଙ I ଙ I Let F22s F rD be an ultrapower of F2. Let ޚ s ޚ rD be the corresponding ultrapower of the ring ޚ. Suppose A is a subring of ଙޚ. ଙ Then exponentiation from A can be shown to be well-defined on F2 by aD aŽi. Ž.grD rsfrDwhere fiŽ.sgi Ž. for all i g I. Moreover, this expo- nentiation satisfies the axioms of an A-group.

LEMMA 4. Let urD and ¨rD be a pair of non-commuting elements of ଙ ଙ F22.Let²: urD, ¨rDA be the A-subgroup of F generated as an A-group by urD and ¨rD. Then urD and ¨rD are A-free generators of²: urD, ¨rD A . A Proof. Let Ž.F21have A-basis Ä4x , x2. Then there is an A-epimor- A phism ␺ : Ž.F212ª ²:urD, ¨rD A determined by x ª urD, x ª ¨rD. A View Ž.F22and F as contained in the group of units of CAŽ.,ގwwy12,y xx and ޚwwy12, y xx, respectively, where y1and y 2are non-commuting indeter- minates. Then one inducts on the recursive construction of an element A WgŽ.F2 to show that if W s 1 q ÝdegŽt.)0 ␣ttrn . tyŽ.12,y where ␣ts atttrDgA,ngގ,␣rntgCAŽ.,ގthen ␺ Ž.W s frD where

fiŽ.s1q ÝaittŽ.rntŽ. U Ž. i ,Vi Ž. degŽ. t )0 for all i g I and for which uiŽ.s1qUi Ž.and ¨ Ž.i s 1 q Vi Ž.for all i g I. Since urD and ¨rD do not commute then UV / VU a.e. Suppose that W / 1 lies in kerŽ.␺ . Now we may order the monomials on y12and y by increasing degree and those of the same degree we order lexicographi- 588 FINE ET AL.

cally where we take y12to precede y . Then there is a first non-vanishing coefficient ␣tt/ 0. Hence, a / 0 a.e. for that t. Call the earliest such monomial s. But ␺ Ž.W s frD s 1. Gilbert Baumslag showed inwx GB that if U and V are any two non-commuting elements each with zero constant term in ޚwwy12, y xx, then they freely generate an associative algebra so that 1qUand 1 q V are free generators of a group. Hence, frD s 1 implies that 1 q ÝdegŽt.)0 aittŽ.rntUi Ž Ž.,Vi Ž..s1 for almost all values of i and that asss 0 a.e. That contradicts a / 0 a.e. Therefore, the assumption that kerŽ.␺ contains a non-trivial element is untenable and ␺ is an A isomorphism from Ž.F2 onto ²:urD, ¨rD A . This completes the proof of Lemma 4.

THEOREM 3. E¨ery 3-generator, fully residually free group lies in F.

Proof. It suffices to assume that the 3-generator, fully residually free group G is neither free nor abelian and has rank 3. The other cases are easy to handle. Under the above assumptions G s ²:g123, g , g where the generators g123, g , g satisfy a non-trivial relation RgŽ.123,g,g s1 with y1 RxŽ.123,x,x being a non-empty, freely reduced word on X33j X .We ଙ ଙ have an isomorphic copy ␾Ž.G embedded in F2 Žsee the discussion at the end of the previous section. . Thus it suffices to show that ଙ␾Ž.G lies in F. ଙ The generators ␾Ž.gi , i s 1, 2, 3, must satisfy the relation ଙ ଙ ଙ RŽ ␾Ž.g123,␾Ž.g,␾Ž..g and therefore, RhgŽŽ.Ž.Ž..123,hg ,hg s1in F 2 for almost all epimorphisms h: G ª F2 . Moreover, since G is non-abelian ²Žhg123 .,hg Ž .,hg Ž .:is almost everywhere non-abelian. Thus, Ž.x123, x , x sŽŽhg123 .,hg Ž .,hg Ž ..is almost everywhere a non-trivial solution in F2to the 3-variable coefficientless word equation RxŽ.123,x,x s1. By Hmelevskiiwx Hm the solutions are exhausted by a finite set Ž.W1, j,W2, j,W3, j ,js1,...,N, of triples of parametric words in two group variables u and ¨ and a finite set of integral parameters ␯ 1,...,␯m. For each 1 F j F N let Tj be the set of all epimorphisms h: G ª F2 such that Žhg Ž123 .,hg Ž .,hg Ž .. is of the form ŽW1,j, W2, j, W3, j.Ž. Conceivably some h can lie in more than one Tj..ŽŽ.Ž.Ž.. Since hg123,hg ,hg is a.e. of at least one of the forms Ž.W1, j, W2, j, W3, j we must have T1 j иии j TN g D.We claim that Tk must lie in D for at least one 1 F k F N. Suppose not. Then XXX Tj gDfor all js1,...,N. Therefore, T1 lиииlTN sŽ.T1 jиииjTN Јg D. But that contradicts T1 j иии j TNkg D. Therefore T g D for at least one 1 F k F N. Fix one such k. Then ŽŽ.Ž.Ž..hg123,hg ,hg is a.e. of the form Ž.W1, k , W2, k, W3, k . Having fixed k we henceforth drop the sub- script and write Ž.Ž.W123, W , W for W 1,k, W2, k, W3, k . Now for i s 1, 2, 3 we have that hgŽ.i is a.e. the result of the substitution u ª uhŽ.,¨ª¨ Ž.h, ␯11ªnhŽ.,...,␯mmªnhŽ.into the parametric word W i. Let A be the FULLY RESIDUALLY FREE GROUPS 589

ଙ I subring of ޚ s ޚ rD generated by ␣11s n rD,...,␣mmsn rD. Then ଙ ଙ ଙ A ଙ ²␾Ž.g123,␾Ž.g,␾Ž.:g lies in ²urD, ¨rDA :( Ž.F2. Thus ␾Ž.G is a A 3-generator subgroup of Ž.F2 where A is ␻-residually ޚ. It follows that ଙ␾Ž.Gand therefore G lies in F. The result that F contains the 3-generator fully residually free groups allows us to begin the proof of our main result that all 2-free residually free groups are 3-free.

THEOREM 4. E¨ery 2-free, residually free group is 3-free. The proof depends on the following three results: Lemma 5, Lemma 6, and Lemma 7, which are of interest in their own right. After putting together these results to give a proof of Theorem 4, we will then return to the proofs of the lemmas, which are quite technical.

DEFINITION 9. A group G satisfies property P if every 2-free subgroup of G must be 3-free.

LEMMA 5. LetGbea3-free, residually free group and let u g G, u / 1 and M s ZuGŽ..Then the HNN extension

1 K s² t, G; relŽ.G , tzty sz,zgM:

satisfies property P.

LEMMA 6. Let G be a finitely generated 2-free residually free group and let M be a maximal 2-generator subgroup of G. Then M is malnormal in G. Using Lemma 6 we can remove the restriction that the base be 3-free in preserving property P under rank one extensions of centralizers.

LEMMA 7. Let G be a residually free CSA group satisfying property P.

Further let u g G, u / 1, and M s ZuGŽ..Then the HNN extension

1 K s² t, G; relŽ.G , tzty sz,zgM: also satisfies property P. We now use these three results to prove Theorem 4. Proof of Theorem 4. Since being 2-free implies commutative transitiv- ity, it follows from the theorem of B. Baumslagwx Ba1 that a 2-free residually free group is actually fully residually free. Now first we claim that every member of F satisfies property P. This is clear since property P is easily seen to be preserved by subgroups and free products. Further from Lemma 8 it is preserved under rank one extensions of centralizers. 590 FINE ET AL.

Now suppose G is a 2-free, residually free group. From the remarks above G is fully residually free. If H is a 3-generator subgroup of G then H is also a 2-free fully residually free group. But every 3-generator fully residually free group lies in F. Therefore H lies in F. But since every member of F satisfies property P every 2-free subgroup of a member of F must be 3-free. Since H is a 2-free subgroup of itself and lies in F,it follows that H is 3-free. Since H is a 3-generator H is then free. Therefore, every 3-generator subgroup of G is free and consequently every 2-free, residually free group is 3-free.

We now prove the three required lemmas.

Proof of Lemma 5. Since G is a CSA group the centralizer M is malnormal in G. K is then an HNN group with malnormal associated subgroups and therefore the Nielsen reduction arguments found inw F-R- R2xwx and based on the work of Peczynski and Reiwer P-R can be adapted analogously to this situation. It suffices to prove that a 3-generator, 2-free subgroup H of K is actually free. Suppose then that H is a 2-free, 3-generator subgroup of K. We may assume that H actually has rank 3, that is, H cannot be generated by two elements. Using the Nielsen reduction method described inwx F-R-R2 we may assume that we have a minimal generating system Ä4x123, x , x for Hand we wish to show that Ä4x123, x , x is actually a free generating system for H. This is certainly the case if H l M s Ä41 by the theorem of H. Neumann on subgroups in HNN groups. Hence let H l M / Ä41 and let H11s G l H. Since H ; H it is 2-free and therefore 3-free since H1; G and G is 3-free. Analogously as in the proof of Proposition 2 and Theorem 1 ofwx F-R-R2 , Ä4x123,x,xwill freely generate a free groupŽ either they are in the base G and so generate a free group by property P or some t symbols remain uncancelled in forming a reduced word in Ä4x123, x , x .except, after a suitable conjugation, possibly in the following cases:

rr Ž.1 x12,xgG,x 3sht or x3s th with r g ޚ _ Ä40,hgG_M, hf²:²:²:x12,x,x 12,xlMsy,y/1. rr ryr Ž.2 x12,xgG,x 3stkt or x3s tkt with r g ޚ _ Ä40, kg G_M,²:x12,xlMs ²:y,y/1.

r Consider first caseŽ. 1 with x3 s ht . The argument is identical with r y1 yr r y1 y1 x33s t h. Notice that hyh s xt yt x3s xyx33gH1sHlG. y1y1 ÄSince x33, y g H, xyx3gHand since h g G, hyh g G.4 Since H is y1 2-free we have h f M for otherwise y s xyx33g²:x 12,x gives a non- trivial relation. Therefore ²y, hyhy1:is free of rank two and y, hyhy1 constitute a free basis because M is malnormal in G. FULLY RESIDUALLY FREE GROUPS 591

Since x12, x are in the base G of the HNN group K and generate a free group and x3 f G the only cancellations can occur via conjugates of the associated subgroup relations. Therefore a non-trivial relation between x1, y1 x23, and x implies a non-trivial relation between x12, x , and hyh . Now we regard the subgroups N s ²:x12, x , h ; G and Q s y1 ²x12,x,hyh :; G. N and Q are both free groups because G is 2-free and satisfies property P and Q is a subgroup of N l H1. We have h f Q r because H is 2-free. If h were in Q then h g H and hence t g H and then t r ytyr y would define a non-trivial relation in the two generator r y1 subgroup ² y, t : of H.If x12,x ,hyh freely generate Q then there is no y1 non-trivial relation between x12, x , and hyh and therefore in this case His freely generated by x123, x , x . Now let Q be free of rank two ŽQ cannot be cyclic since x12, x g Q.. Let Ä4z12, z be a free generating system of Q. Since Q is a subgroup of the free group N we may now apply ␣ Lemma 2.1 ofwx Ro . Let y s y11, ␣ G 1 where y g N is not a proper power 1 in N. Since h f Q but y g Q and hyhy g Q there is a Nielsen transfor- mation from Ä4z12, z to a system in which an element is conjugateŽ. in N to ␣ a power of y11. Hence without loss of generality let y s y be a primitive element in Q. Since y g ²:x12, x ; Q a conjugate of y is also a primitive element in ²:x12, x . Without loss of generality we may then assume that ␥ x11szsy. We cannot have h s hy11for some h1g Q s ²:z 12, z ; H ␥ r ␥ r yr y␥ and ␥ g ޚ _ Ä40 for if so then yt11gHand then y t yt y1s y defines ␥r a non-trivial relation in the two-generator subgroup ²yt1 ,y:of H. Therefore we get a Nielsen transformation from Ä4Ä4z12, z s y, z 2to a ␤y1 system Äy, gy11 g 4for some g g N, g f ²:y and some ␤ G 1. Since 1 ygQ,hfQbut hyhy g Q we then also get a Nielsen transformation ␤y1 from Ä4Äz12, z to y, hy1 h 4. Also we have necessarily ␣ s ␤␦ for some y1 ␤y1␦ ␦G1, that is, hyh s Žhy1 h .. Now we have

␤y1r ␤y1 r Hs²:²:²x123,x,xsz 123,z,xsy,hy1 h , ht :²s hy1 h , ht : r 1 1 r because y s ty hy hyhy ht . This implies that H is a 2-generator group contradicting the fact that H has rank three. Therefore Q must be free of rank 3 and hence H is freely generated by x12, x , and x 3. r yr In caseŽ. 2 suppose first that x3 s tkt . Then a non-trivial relation between x12, x , and x 3would imply such a relation between y and k. Let ryr H2 stHt . This is also 2-free being a conjugate of H and contains y and k. Therefore there is no such relation and x123, x , x generate a free group. r r yr r r yr In the final situation where x334s t kt let H s t Ht , H s t Ht . Both H34and H are 2-free. A non-trivial relation between x123, x , and x implies a non-trivial relation between y and ky1 yk or between y and y1 kyk . But these pairs lie in either H34or H so no such relation is possible completing the proof. 592 FINE ET AL.

Proof of Lemma 6. The 2-free residually free groups are commutative transitive and torsion-free. It follows that the centralizers of non-trivial n elements are isolated. This means that for all n g ގ, b g ޚGŽ.a implies that b g ޚGŽ.a . Further any such root must necessarily be unique since 2-free residually free groups are universally free and universally free groups have unique roots. Now suppose that G is a finitely generated 2-free residually free group and let M be a maximal 2-generator subgroup of G. If rank G F 2 then M s G and clearly then M is malnormal. Thus we assume that rank G G 3. Suppose that M is not malnormal in G. Then there exist non-trivial 1 u, w g M and g g G _ M with w s gy ug. It follows from the work of Gaglione and Spellmanwx G-S2 that maximal 2-generator subgroups in finitely generated 2-free residually free groups are isolated. Since M is isolated we may assume without loss of generality that neither u nor w is a proper power in G. First of all u and w cannot be conjugate within M for if so there would y1 y1 y1 exist an h g M such that gugshuhand so gh g ZuGŽ.s ²:u. n But then g s hu for some n g ޚ and so g g M. Next M is non-abelian. If it were abelian, it would have to be cyclic since G is 2-free. Since G has rank at least 3, G is non-abelian and therefore a non-abelian commutative transitive group. It follows that G must have a trivial center. Suppose M s ²:m . Since ZG Ž.sÄ41 there exists a g g G with gm / mg. Then M is properly contained in the rank two free subgroup ²:g, m contradicting the maximality of M. Since M is non-abelian and G is 2-free it follows that M s ²:a, b is free of rank two. Since G is fully residually free there exists an epimorphism ␾: G ª F with F free such that ␾Ž.a / 1, ␾ Ž.b / 1, ␾ Žwxa, b ./ 1. It follows that the image of M is also free of rank two and ␾ < M is an isomorphism of M onto F0 s²:␣,␤with ␣ s ␾ Ž.Ž.a , ␤ s ␾ b . It follows that uŽ.Ž.␣, ␤ , w ␣, ␤ are non-trivial conjugate elements in F and hence wŽ.␣, ␤ is not conju- y1 gate to uŽ.␣, ␤ in F and therefore also not conjugate in F0. Moreover since ␾ < M is an isomorphism we have that uŽ.Ž.␣, ␤ , w ␣, ␤ are not conjugate in F00and neither is a proper power in F .Ä However, since they are conjugate in F we say that they are ¨irtually conjugate in F0 Žsee Section 8. .4 y1 y1 Next we have that the element xux312312Ž.Ž.,xxwx,x cannot be primitive in the rank three free group on x123, x , x . If it were consider 1 1 ²:²M,gsa,b,g :. The element guayŽ.Ž.,bgwa,byis a relator in ²:M,gso we may rewrite a presentation for ²:M, g with generators y1 y1 ¨123,¨,¨where ¨ 3s guaŽ.Ž.,bgwa,b and delete ¨ 3via a Tietze transformation. Then M ; ²:²:M, g s ¨ 12, ¨ . Then by the maximality of Mwe have M s ²:M, g and hence g g M contradicting g f M. FULLY RESIDUALLY FREE GROUPS 593

Now since G is finitely generated so is ␾Ž.G s F. Consider the group

1 K s² F, s; relŽ.Ž.F , suy ␣,␤ssu Ž.␣,␤: . Since F is a finitely generated free group it is certainly 3-free, so by Lemma 6, K satisfies property P.In K we have that

1 1 1 y1 wŽ.␣ , ␤ s ␥y u Ž.␣ , ␤␥s␥y Ž.suy Ž.␣,␤s␥s Ž.Ž.Ž.s␥ u␣,␤ s␥, where ␥ s ␾Ž.g . Letting t s s␥ we have

1 H s² ␣ , ␤ , t; tuy Ž.␣,␤tsw Ž.␣,␤: . Now uŽ.Ž.␣, ␤ , w ␣, ␤ are non-trivial elements of the free group on ␣, ␤ and further wŽ.␣, ␤ is not conjugate within ²:␣, ␤ to u Ž.␣, ␤ "1. It follows from the work of Fine, Rohl,¨ and Rosenbergerwx F-R-R1 that H is 2-free. Since H ; K and K satisfies property P, H is also 3-free and hence a free group. However, the only way that a group with the presentation

1 ² ␣ , ␤ , t; tuy Ž.␣,␤tsw Ž.␣,␤:

y1 y1 could be free is if xux3Ž.Ž. 123,xxwx 12,x is primitive in ²:x123, x , x ; which we have showed to be impossible. This contradiction shows that it is impossible for M not to be malnormal in G. Proof of Lemma 7. The beginning of this proof exactly mirrors that of the proof of Lemma 6. Since G is a CSA group satisfying the maximal condition on abelian subgroups, the centralizer M is malnormal in G. K is then an HNN group with malnormal associated subgroups and there- fore the Nielsen reduction arguments found inwx F-R-R2 and based on the work of Peczynski and Reiwerwx P-R can be adapted analogously to this situation. As in Lemma 5 it suffices to prove that a 3-generator, 2-free subgroup H of K is actually free. Suppose then that H is a 2-free 3-generator subgroup of K. We may assume that H actually has rank 3, that is, H cannot be generated by two elements. Using the Nielsen reduction method described inwx F-R-R2 we may assume that we have a minimal generating system Ä4x123, x , x for H and we wish to show that Ä4x123, x , x is actually a free generating system for H. This is certainly the case if H l M s Ä41by the theorem of H. Neumann on subgroups in HNN groups. Hence let

HlM/Ä41 and let H11s G l H. Since H ; H it is 2-free and there- fore 3-free since H1 ; G and G satisfies property P. Analogously as in the proof of Proposition 2 and Theorem 1 ofwx F-R-R2 , Ä4x123,x,xwill generate a free groupÄ either they are in the base G and so 594 FINE ET AL. generate a free group by property P or some t symbols remain uncan- celled in forming a word in Ä4x123, x , x except possibly in the following cases4 :

rr Ž.1 x12,xgG,x 3sht or x3s th with r g ޚ _ Ä40, hgG_M, hf²:²:²:x12,x,x 12,xlMsy,y/1. Note that by the conditions on G,Ml²:x12,xis either cyclic or trivial. If it were trivial there could be no cancellation between x12, x , and x 3and since x12, x generate a free group so do Ä4x123, x , x . rr ryr Ž.2 x12,xgG,x 3stkt or x3s tkt with r g ޚ _ Ä40, kg G_M,²:x12,xlMs ²:y,y/1. The proof of caseŽ. 2 in Lemma 5 did not use that the base was 3-free only the residual freeness and 2-free property of the subgroup H. There- fore caseŽ. 2 follows exactly as in Lemma 6. r Now consider caseŽ. 1 with x3 s ht and we suppose as before that H has rank three. As in the proof of Lemma 6, H will be free unless there is y1 y1 a non-trivial relation between x12, x , and hyh s xyx33. Consider the subgroup H1 s H l G. Since G is 2-free with property P and H is 2-free y1 it follows that H11is 3-free. Now x , x212, hyh g H and therefore H s y1 ²x12,x,hyh :is a non-abelian free group. If the rank of H2is three we are done since then no non-trivial relation is possible. Rank one is impossible since H12is non-abelian so we suppose that H has rank two, y1 that is, ²x12, x , hyh :²s z12, z :; H. From work of Gaglione and Spell- manwx G-S2 , 2-free residually free groups satisfy 2-A.C.C., that is, any two generator subgroup is contained in a maximal two-generator subgroup.

Therefore since H is a 2-free residually free group we have ²:z12, z ; ²:X12,Xwhere ²:X12, X is maximal in H. Now ²:²:x12, x ; X 12, X and y1 y1 hence y g ²:X12, X . Further hyh g ²:X12, X so therefore xyx33g ²:X12,X. The subgroup ²:X12, X is a maximal 2-generator subgroup of the 2-free residually free group H and therefore from Lemma 7 it is malnormal in H. It follows that x312g ²:X , X . This implies that ²x123,x,x :sH; ²:X1,X 2and hence H s ²:X1, X 2contradicting the fact that H has rank three. Therefore rank two for H2 is impossible and hence H also is free of rank three. We note that there is a converse result to Lemma 6 which gives another characterization of 2-free fully residually free groups.

LEMMA 8. Let G be a non-abelian finitely generated fully residually free group. If e¨ery maximal two-generator subgroup of G is malnormal in G then Gis2-free. Proof. Suppose G were a non-abelian finitely generated fully residually free group with the property that every maximal two generator subgroup is FULLY RESIDUALLY FREE GROUPS 595 malnormal. Since G satisfies the maximal condition for abelian subgroups it would follow that if G were not 2-free there would exist a non-trivial agGsuch that M s ZaGŽ.is free abelian of finite rank k ) 1. Suppose b1,...,bk is a basis for M. Since G is a non-abelian commutative transitive group it has a trivial center. Thus there exists a non-trivial g g G with gb11/ bg. Consider ²:g, b1and extend to a maximal 2-generator sub- group H s ²:²:c12, c > g, b 1. Since gb1/ bg 1,H is non-abelian and therefore cc12/cc 21. In a fully residually free group every pair of non- commuting elements freely generates a subgroup. Therefore H is free with y1 basis c12, c . But 1 / b1s b 2bb 12gHand hence it follows that b2g H since H is malnormal. But then the free abelian group of rank two

²:b12,bis contained in a non-abelian free group which is impossible. This contradiction shows that G must be 2-free.

QUESTION 1. Can one gi¨e an alternati¨e characterization of the elements of F as fundamental groups of graphs of group?

QUESTION 2. Which elements of F are indecomposable with respect to free products?

Benjamin Baumslag inwx Ba1 defined for each n g ގ the condition n-A.C.C. for a group G to mean that every non-decreasing chain H0 F H1 F иии F Hk F иии of subgroups of G generated by n or fewer distinct elements must stabilize after finitely many steps. Observe that if 1 F m F n then n-A.C.C. implies m-A.C.C.

QUESTION 3. Let n G 3 be an integer. Must the elements of F satisfy n-A.C.C.?

Recall that Remeslennikovwx Re showed that if k g ގ and G is a k-generator, fully residually free group then G is ⌳-free where ⌳ is free abelian of rank not exceeding 3k. From that it follows that finitely generated, fully residually free groups satisfy the maximal condition for Abelian subgroups and in particular satisfy 1-A.C.C. It was shown in Gaglione and Spellmanwx G-S2 that finitely generated, fully residually free groups satisfy 2-A.C.C. It follows easily from commutative transitivity that 2-free residually free groups have the property that abelian subgroups are locally cyclic. How- ever, it can be shown that they are actually cyclic. To see this suppose that

G is a non-trivial 2-free residually free group and M0 is an abelian subgroup of G. Extend M0 to a maximal abelian subgroup M of G. Since G is residually free so is M and thus M is residually an infinite cyclic group. Hence M is, up to isomorphism, a subdirect power of ޚ and is embeddable in ޚ I for some non-empty index set I. Since G is 2-free, M is 596 FINE ET AL. locally cyclic and torsion-free and hence M is also embeddable in ޑ.In particular M is countable. But every countable subgroup of ޚ I is free abelian. Since M is locally cyclic it cannot now fail to be cyclic. Therefore

M0 ; M is also cyclic.

7. CLASSIFICATION THEOREM

We can now give a complete classification of the fully residually free groups of rank at most three.

THEOREM 5. Let G be a fully residually free group. Then

Ž.1 if rank ŽG .s 1 then G is infinite cyclic; Ž.2 if rank ŽG .s 2 then either G is free of rank 2 or free abelian of rank 2; Ž.3 if rank ŽG .s 3 then either G is free of rank 3, free abelian of rank 3, or a free rank one extension of centralizers of a free group of rank 2. That is, G has a one-relator presentation

y1 G s ² x1233, x , x ; x ¨x 3s ¨ :,

where ¨ s ¨Ž.x12, x is a non-tri¨ial element of the free group on x12, x which is not a proper power.

Proof. If rankŽ.G s 1 then clearly G must be infinite cyclic since residually free groups are torsion-free. If rankŽ.G s 2 then suppose G s ²a, b :.IfGis non-abelian then wxa,b/1 and since G is fully residually free there exists a free group F and an epimorphism ␾: G ª F with ␾Ž.a / 1, ␾ Ž.b / 1, ␾wxa, b / 1. Let ␣s␾Ž.a,␤s␾ Ž.b. Since ␾wxwa, b s ␣, ␤ x/ 1 it follows that ²:␣, ␤ freely generate a free subgroup of rank 2. Hence ␾ is a monomorphism and G is isomorphic to ²:␣, ␤ and hence is also free of rank 2. We now concentrate on the case where rankŽ.G s 3. We first extend property P.

DEFINITION 10. A group G has property-Q if every non-abelian 3- generator subgroup of G is either free or has a one-relator presentation 1 ² x,y,t;ty ¨ts¨:Žwhere 1 / ¨ s ¨ x, y.is not a proper power in ²x, y;.: In the previous section it was shown that all finitely generated fully residually free groups of rank F 3 are contained in the class F character- ized as the smallest class of groups containing the infinite cyclic groups FULLY RESIDUALLY FREE GROUPS 597 and closed under the following four operators: Ž.1 Isomorphism Ž.2 Finitely Generated Subgroups Ž.3 Free Products of Finitely Many Factors Ž.4 Free Rank One Extensions of Centralizers. Clearly infinite cyclic groups satisfy property-Q as do finitely generated subgroups of groups with property-Q. Further property-Q is preserved under taking free products. The proof of Theorem 5 then follows directly from the following crucial lemma which says that property-Q is preserved under free rank one extensions of centralizers of fully residually free groups. This is analogous to our proof of Theorem 4.

LEMMA 9. Let G be a fully residually free group of rank at least 2 which satisfies property-Q. Let 1 / u g G and M s ZuGŽ..Let

1 K s² t, G; relŽ.G , tzty sz,for all z g M: .

Then K satisfies property-Q. Proof of Lemma 9. Since G is fully residually free it follows as before that M is malnormal in G and that K is also fully residually free. Let

Hs²:x123,x,xbe a 3-generator subgroup of K. Since K is fully residu- ally free we may without loss of generality assume that the rank of H is 3 since if the rank is less than three the result follows as above. Since K is an HNN group with both associated subgroups malnormal and equal to M we can get analogous results as inwx F-R-R2 and hence use the Nielsen reduction method for HNN groups with malnormal associated subgroups as described inwx F-R-R2 and based on the work of Peczynski and Reiwer wxP-R . In particular each x g K may be written in the reduced form rry1 rn xshgGor x s ht h1 t иии htn k with n G 0, h, k g G, r g ޚ _ Ä40 and hiig G _ M, r g ޚ _ Ä40,is1,...,n if n G 1. If H is 2-free then since H is fully residually free it follows from Theorem 4 that H is free and hence free of rank 3. Therefore suppose that H is not 2-free. Then there is a 2-generator subgroup N of H which is free abelian of rank 2. By the Nielsen reduction method we may assume ᎏpossibly after a Nielsen transformation and a suitable conjugationᎏthat one of the following cases holds:

Ž.A x12,xgGand xx12sxx 21.

Ž.Bx12,xgG_M,²:x12,xfree of rank 2, ²:x12, x l M s ²:y s and x3 s t g with s g ގ, g g M. s Ž.C x12gMand x s tgwith s g ގ, g g M. 598 FINE ET AL.

We first consider CaseŽ. A . If x3 g G then the result follows since G has property-Q. Therefore we suppose that x3 f G. If x12, x f M then there can be no non-trivial relations between x12, x , and x3. Therefore

y1 ଙ H s ² x123121, x , x ; xxxsx 2: sޚޚŽ.=ޚ which is a free rank one extension of centralizers of the free group

²:x23,x; . If either x1or x 2is M they both are, since M is malnormal in r try1 trn G. Now suppose x12, x g M. Let x3s ht h 1 иии hnk with n G 0, h, kgG,rgޚ_Ä40 and hiig G _ M, r g ޚ _ Ä40,is1,...,n if n G 1. If nG2orns1 and h f M or n s 1 and k f M then again we have

y1 H s ² x123121, x , x ; xxxsx 2: sޚ)Ž.ޚ=ޚ.

So we may assume that these don’t hold. Suppose first that n s 1 and h, k g M. Then we may assume that r r1 h s k s 1, that is, x31s t ht . In forming a cyclically reduced word in x123,x,xwhich contains x3some t-symbols will remain uncancelled and hence H will form a free group except possibly ifŽ. 1 r syr11or Ž. 2 r s r . If r syr11then there is a non-trivial relation between x , x2, x3which is not a consequence of wxx12, x s 1 if and only if there is a non-trivial relation of the form

␣1 ␤1 ␥1 ␣ m ␤m ␥ m x121x h иии x 1x 2h 1s 1, mG1, all ␣iii, ␤ , ␥ g ޚ which is not a consequence of the relation wxx12,xs1. However, this is impossible since h1f M and G has prop- erty-Q. Hence if r syr1 again we have that

y1 H s ² x123121, x , x ; xxxsx 2: sޚ)Ž.ޚ=ޚ.

If r s r11then there is a non-trivial relation between x , x2, x3which is not a consequence of wxx12, x s 1 if and only if there is a non-trivial relation of the form

␣1 ␤1 ⑀ ␥1 ␦1 ␣ m ␤m ϱ ␥ m ␦ m y⑀ x12112x h x x иии x 1x 2h 112x x h 1s 1, mG1, ⑀ s 1or⑀sy1, all ␣iii, ␤ , ␥ g ޚ which is not a consequence of the relation wxx12, x s 1. But this is also impossible since h1f M and M is malnormal in G. Therefore in this case again we have that

y1 H s ² x123121, x , x ; xxxsx 2: sޚ)Ž.ޚ=ޚ. FULLY RESIDUALLY FREE GROUPS 599

r Now let n s 0 so that x3 s ht k with r / 0, h, k g G.If h,kgMthen H is abelian. If both h, k f M then as before we have that

y1 H s ² x123121, x , x ; xxxsx 2: sޚ)Ž.ޚ=ޚ.

Hence without loss of generality we may assume that k g M, h f M.We r may assume that k s 1 so that x3 s ht . Then there is a non-trivial relation between x123, x , x which is not a consequence of wxx12, x s 1if and only if there is a non-trivial relation of the form

␣1 ␤1 ␥1 ␦1 y1 ␣ m ␤m ␥ m ␦ m y1 x12x hx 12x h иии x 1x 2hx 1x 2h s 1, mG1, all ␣iii, ␤ , ␥ g ޚ which is not a consequence of the relation wxx12,xs1. But this is also impossible since h f M and M is malnormal in G. Therefore in this case again we have that

y1 H s ² x123121, x , x ; xxxsx 2: sޚ)Ž.ޚ=ޚ which completes CaseŽ. A . We now consider CaseŽ. B . Here we have immediately that

y1 H s ² x1233, x , x ; xyx 3sy:. Further, M must be isolated in G Ž.see Section 6 since G is fully residually free so that y is not a proper power and hence H is a free rank one extension of centralizers of the desired form. Finally we consider CaseŽ. C . We may assume that g s 1, that is, s x23s t . The case g / 1, g g M is analogous. If x g M then H is abelian. If x31g G _ M then both ²:²:x , x3and x2, x3are necessarily of rank 2 and therefore

y1 H s ² x123121, x , x ; xxxsx 2:.

Therefore we assume that x3 f G and let

r r1 rn x31s ht ht иии htnk with n G 0, h, k g G, r g ޚ _ Ä40 and hiig G _ M, r g ޚ _ Ä40,is 1,...,n if n G 1. If h, k f M then as in CaseŽ. A and as above we have

y1 H s ² x123121, x , x ; xxxsx 2:.

If n G 1 and h, k g M then we may assume that h s k s 1 and using yr r yr r the facts that t xt1122sx,t xtsx we may replace x3by

Xrr r1 rnqr x331stxtsht иии htn. 600 FINE ET AL.

X If in this case x3 g G then we are done since it reverts to the previous X rr X argument. If x3f G then we continue with HЈ s tHt s²x123,x,x:. Since each hi g G _ M and n G 1 there can be no non-trivial relation X between x123, x , x which is not a consequence of wxx12, x s 1. Hence

X y1 HЈs²x123121,x,x;xxxsx 2: is a free rank one extension of centralizers and therefore so is its isomorphic conjugate H. Hence suppose h f M and k g M or h g M and k f M if n G 1. Without loss of generality we may assume that h f M and k g M and thus k s 1if nG1. If sr< nn, that is, r s ␣ s for some ␣ g ޚ then we may y␣ replace x332by xx to get a shorter system and we continue with this new one. Hence let n G 1 and s ¦ rn. Then in forming a cyclically reduced word in x123, x , x which contains x3some t-symbols remain uncancelled and again we get that

y1 H s ² x123121, x , x ; xxxsx 2:. r Therefore we now suppose that n s 0 so that x3 s ht k. Since the case h, k f M has already been done, we may assume as above that either r h f M and k s 1 so that x3s ht with h f M or h, k g M.If h,kgM r then H is abelian. Hence let x3 s ht with h f M. As before if sr< n we may pass to a shorter system and so we may assume that s ¦ rn. Then there is a non-trivial relation between x123, x , x which is not a conse- s quence of the relation wx1, t xs 1 if and only if there is a non-trivial relation of the form

␣1 s␤1 ␥1 s␦1 y1 ␣ m ␤m ␥ m ␦ m y1 x11t hx t h иии x 1212x hx x h s 1, mG1, all ␣iiii, ␤ , ␥ , ␦ g ޚ which is not a consequence of the relation s wx1,txs1. From cancellation arguments this is possible only if ␤iis ␦ s 0 for all i s 1,...,m; that is, we would have a non-trivial relation of the form

␣1 ␥1 y1 ␣ m ␥ m y1 x11hx h иии x 1hx 1h s 1 which is impossible since necessarily ²:x1, h is free of rank 2 since h f M. Therefore again we get that H must have the presentation

y1 H s ² x123121, x , x ; xxxsx 2: completing CaseŽ. C and the proof of the lemma. The verification of this lemma then completes the proof of Theorem 5. The cancellation arguments used in the proof of the above Lemma together with Lemma 2.1 ofwx Ro also give the following result which is a corollary of the lemma. FULLY RESIDUALLY FREE GROUPS 601

1 COROLLARY 1. Let K s ²a, b, t; ty ¨t s ¨ :Žwhere 1 / ¨ s ¨ a, bis. not a proper power in the free group²: a, b;.LetÄ4 x123, x , x be a generating system of K. ThenÄ4 x123, x , x is Nielsen equi¨alent toÄ4 a, b, t .

8. SOME APPLICATIONS TO VIRTUALLY CONJUGATE ELEMENTS IN FREE GROUPS

We now give some applications to determining when two elements of a free group are virtually conjugate.

DEFINITION 11Ž. Remeslennikov . Let F0 be a free group and let u, w be two elements of F00. Then u and w are ¨irtually conjugate in F if there exists a free overgroup F > F0 in which u and w are conjugate. We have the following two observations.

Ž.1IfF0 is a countable free group and u and w are virtually conjugate elements in F0 then there is a free group F of rank two, F > F0 in which u and w are conjugate.

Ž.2IfF0 is a free group and u, w g F _ Ä41 are virtually conjugate in y1 F00, then w cannot be conjugate to u in F .

Now let F212s ²:b , b ; be free of rank two and suppose that ubŽ.Ž.12,b ,wb 12,b gF 2_Ä41 and suppose that neither u nor w is a proper power in F2 . Then the following theorem is a consequence of the proof of Lemma 6. THEOREM 6. Under the abo¨e assumptions, if u and w are ¨irtually conjugate in F2 , then either u and w are already conjugate in F2 or y1 y1 bub312312Ž.Ž.,bbwb,b is primiti¨einF3s²:b 123,b ,b;.

Now for each ubŽ.Ž.12,b ,wb 12,b gF 2_Ä41 such that neither is a proper y1 power in F22and w is not conjugate to u in F and bub3Ž.1,bbw23 y1 Ž.b12,bis not primitive in F3s ²:Ž.b 123, b , b ; set Gu,w to be the HNN group y1 GuŽ.,w s²b123,b,b;wbŽ.Ž. 12,b sbub 3 123,bb:.

COROLLARY 2. Under the abo¨e assumptions the images of b12 and b commute in any free homomorphic image of GŽ. u, w .

Let euiiŽ.respectively ew Ž .be the exponent sum in u Žrespectively in w . of bi, i s 1, 2. Let ⌬Ž.u, w be the determinant ⌬Ž.u, w s euew12 Ž.Ž.y eweu12Ž.Ž.. COROLLARY 3. If, in addition to the hypotheses of the pre¨ious corollary we assume that ⌬Ž.u, w / 0, then G Ž. u, w has only abelian free homomor- phic images. 602 FINE ET AL.

Proof. Suppose h: GuŽ.,w ªFis an epimorphism from gu Ž.,w onto the free group F. Then f is generated by hbŽ.i ,is1, 2, 3. Moreover hbŽ.Ž.12,hb commute. If hbŽ.12shb Ž.s1, then F s ²Ž.:hb3is abelian. Hence assume that ␤i ²Žhb12 .,hb Ž .:s ²:c with c / 1. Let hbŽ.isc ,is1, 2. Applying h to y1 wbŽ.12,b sbub 3Ž. 123,bbwe get

1 e1122Žw.␤qeŽw.␤ ye1122Žu.␤qeŽu.␤ c shbŽ.33 c hbŽ..

If hbŽ.3 commutes with c then F is abelian. Suppose not. Since the centralizer ZcF Ž.is malnormal in F we infer the following system of two linear equations in the two unknowns ␤12, ␤ :

eu1122Ž.␤qeu Ž.␤s0

ew1122Ž.␤qew Ž.␤s0.

Since ⌬Ž.u, w / 0 this implies that ␤12s ␤ s 0 giving that hbŽ.1s hbŽ.2 s1 and contradicting that c / 1. From this contradiction we have that hbŽ.3 commutes with c and hence F is abelian.

9. ALTERNATIVE DESCRIPTION OF THE CLASS F

Suppose that F is the class of groups defined as in Section 4 as the smallest class of groups containing the infinite cyclic groups and closed under the following four ‘‘operators’’: Ž.1 Isomorphism Ž.2 Finitely Generated Subgroups Ž.3 Free Products of Finitely Many Factors Ž.4 Free Rank One Extensions of Centralizers.

ޚw␯x We first show that every element of F is embeddable in Ž.F␻ . Subsequently we will show that if A is any ring of Lyndon type, then every A finitely generated subgroup of Ž.F␻ lies in F. Since ޚwx␯ is of Lyndon type this will establish that F is precisely the class of all finitely generated ޚw␯x groups embeddable in Ž.F␻ .

ޚw␯x THEOREM 7. E¨ery element of F is embeddable inŽ. F␻ .

Proof. Certainly every infinite cyclic group is embeddable in F␻ which ޚw␯x in turn is embeddable in Ž.F␻ . Any group isomorphic to one embed- ޚw␯x ޚw␯x dable in Ž.F␻ is clearly also embeddable in Ž.F␻ . Now suppose the FULLY RESIDUALLY FREE GROUPS 603

ޚw␯x finitely generated groups G and H are each embeddable in Ž.F␻ . Each ޚw␯x of the images of G and H in Ž.F␻ involve parametric words on a finite set of generators of a free basis X for F␻. By applying the isomorphism operator, if necessary, we may assume that the images of G and H in ޚw␯x Ž.F␻ , are such that the subsets Y and Z of X containing the generators actually appearing in the parametric words in the images of G and H, respectively, are disjoint. Then GHଙ is embeddable in ŽŽ..FY ޚw␯xଙ ޚw␯x ޚw␯x ޚw␯x ŽŽ..FZ which in turn is a subgroup of ŽŽFX .. sŽ.F␻ . ޚw␯x Now suppose G / 1 is embeddable in Ž.F␻ and u g G _ Ä41 . Let Bs²:t be infinite cyclic and let K s GuŽ.,B be the free rank one extension of ZuGŽ.. Now if F is any free group and A is any ring of A Ž. Ž. Ž . Lyndon type then F s D n-␻Fn where F 0 s F and Fnq1s ␣ A ²g:ggFnŽ.,␣gA :. Define for g g F the le¨el of g by levelŽ.g s minÄn g ␻: g g FnŽ.4. Now every level n ) 0 element g has a unique canonical form

c1 иии ck gg011¨ gkk¨ g kq1,

Ž.Ž. ޚ where g0 ,..., gkq1 gFny1 , level ¨iis n y 1, and c g A _ for i s 1,...,k. Suppose a1,...,ar generate G, which we now identify with its ޚw␯x image in Ž.F␻ . Then at each level in the construction of each aj only finitely many coset representatives cŽ.␯ g ޚwx␯ _ ޚ are involved so there is a finite bound N g ގ such that degŽŽc ␯ ..- N for all such c Ž␯ .. ޚw␯x Suppose that the centralizer of u in Ž.F␻ is generated as a ޚwx␯ -module ␯ N by u00. Then ²G, u : ( K. This completes the proof. A THEOREM 8. E¨ery finitely generated subgroup ofŽ. F␻ lies in F when- e¨er A is of Lyndon type. A Suppose G s ²:Ž.a1,...,ar ; F␻ where r is finite. We use induction on n s maxÄ levelŽ.a1 , . . . , level Ž.ar 4.If ns0 then G ; FŽ.0 s F␻ so G is a finitely generated free group and certainly lies in F. Now suppose the result is true for all 0 F m - n. At least one generator then ai s a must c1 иии ck ޚ have level n. Suppose a s gg011¨ gkk¨ g kq1has level n. Now is a q pure subgroup of A so if c1,...,cm are all the coset representatives Ž.perhaps with repetitions involved in all the generators of level n, then by Theorem 13.4.1 of M. Hallwx Ha , ޚ s ²:1 is a direct summand in the q subgroup of A generated by 1, c1,...,cm. Let 1, b1,...,bq be a basis for that subgroup. Then ciis n ,0 q nbi,1 1 qиии qnbi,qq,is1,...,m. The el- ²: ement a may be reached from g011, g , ¨ ,..., gkkk,¨ , g q1by repeated c1 centralizer extensions. For example, ¨ 1 lies in the extension of the centralizer of ¨ 1 adjoining one step at a time commuting elements c1 n1, 0 n1, 1 n1, q t1, 1,...,t 1, q to commute with ¨ 1111,11,. Then ¨ s ¨ t иии t q. The same may be done for each distinct ¨ in a and for each generator of level n 604 FINE ET AL. taking care to adjoin new stable letters only for ¨ which have not already appeared. Thus G may be obtained as a finitely generated subgroup of repeated centralizer extensions of a group generated by finitely many elements of levels less than n. This completes the induction and proves the result.

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