A FINITENESS CONDITION CONCERNING COMMUTATORS IN GROUPS PATRIZIA LONGOBARDI AND MERCEDE MAJ

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1 HOUSTON JOURNAL OF MATHEMATICS Volume 19, No. 4, 1993 A FINITENESS CONDITION CONCERNING COMMUTATORS IN GROUPS PATRIZIA LONGOBARDI AND MERCEDE MAJ 1. Introduction. In [3] P.S. Kim, A. H. Rhemtulla and H. Smith posed the following question: Suppose that 12 is a variety of groups defined by means of a law v(x,...,x, ) = 1, and say that a group G belongs to the class 12' if for any infinite subsets X,...,X, there exist elements xi Xi, for all i {1,...,n}, such that V(Xl,...,x, )= 1. Obviously the class 12 and the class jf of all finite groups are contained in 12'. Is it true that 12' = ]2 U jf? The answer is affirmative for the variety 4 of abeljan groups by a famous theorem of B. U. Neumann (see[6]) and for the variety Ark of nilpotent groups of class at most k (see[4]). Questions of a similar nature for other classes of soluble groups are discussed in [3] and in [7]. In this paper we give a positive result for the variety Q2 defined by the law Ix, y]2 = 1, by proving the following Theorem A. Let G be an infinite group in Q. Then G is in Q2. The variety Q2 has been studied by I.D. MacDonald in [5] and S. Bachmuth and J. Lewin in [1]. If G Q2 then the seconderived subgroup G" of G has exponent at most 2 and is in the centre of G. Q2 is contained in the variety with the law [[x, y], Ix, z]] -- 1, which in turn is exactly the class of groups with the property that every 3-generator subgroup is metabelian. Finally a free group in the variety Q2 is residually a finite 2-group. Work partially supported by M.U.R.S.T. and G.N.S.A.G.A. (C.N.R.) 505

2 506 PATRIZIA LONGOBARDI AND MERCEDE MAJ 2. Preliminary results. We start with the following very easy remark: 2.1. Let G ß Q. Iœ Z(G) is infinite, then G ß Proof. Let x, y ß G and consider the infinite sets xz(g), yz(g). Then, for suitable z, z2 ß Z(G), [xz,yz2] = I since 1. [] The next result will play a key role in the following Let G ß Q be an infinite group. Then CG(x) is infinite, for any x ß G oœ order 2. Proof. Assume that Co(x) is finite, for some x ß G of order 2. First we construct, by induction on n, a sequence (x, ), er of elements of G satisfying the following properties (I)' XiX; 1 CG(I), if i j, (II): xix;lxkx C (/), ifi,j,h,k are pairwise different. For, assume that x,...,x, satisfy (I) and (II). If G = D, with = U C(x)x u U u U i=1 i,j,h= l i,j,h---1 then G is finite, a contradiction. Thus there exists xm+ ß G - D, and it is easy to verify that/,--., x,, x, + satisfy (I) and (II). Now let N = A U B with A and B infinite and disjoint. Then also the sets (xx /i ß A} and {xx /j ß B} are infinite and, since G ß Q, there exist s ß A, t ß B such that [/xs,/x,] = 1. Thus --1 and I x ß C (I). Write A = A-{s} and B = B-(t}. Then, arguing --1 analogously, we find v ß A, w ß B such that x x ß C (x). Hence, by induction, we construct infinite subsets I C_ A, J C_ B such that x x ß C (x), for an infinite number of pairs (i,j) (with all i's different and all j's different). But Cc(x) is finite, so there exists; ß I, ß J such that, for _ infinitely many h I, k ß J, x = x. Then x x' C (x), a finite set, hence there exist h, h" ß I, xx X, -1

3 A FINITENESS CONDITION IN GROUPS 507 k" such that xx X x;, -2 x, XXh"xk". Thus xxh'xk _ ' = x a" -'], and xa, x, x,,x, e C (x), contradicting (II). Therefore C (x) is finite, required. Corollary 2.3. Let G ß Q be an infinite group. Then G has an infinite abelian subgroup. Proof. We show that in any infinite group G ß Q there exists an element x with CG(x) infinite. Then the result will follow, arguing as in Corollary 2.5 of [2]. If G has element of order 2, then the result follows from 2.2. Assume that G has no elements of order 2, then if X, Y are infinite subsets of G there exist x ß X, y ß Y such that Ix, y] - 1. Thus G ß A* and G is abelian (see[6] and [4]). [] Corollary 2.4. Let G ß Q be an infinite FC-group. Then G ß Q2. Proof. Let x,y ß G. Then IG'CG(x)l and IG' C (y)l are finite, hence G- C(x) n is finite. Thus C (I) Cl CG(y) is an infinite group in Q and, by Corollary 2.3, there exists an infinite abeljan subgroup A _< C (I) Cl C (y). Therefore Z ({A, x, y}) is finite, and [/, y]2 = 1, by 2.1. [] We will often make use of the following Lemma. Lemma 2.5. Let G ß Q, x ß G and let A be an infinite abelian subgroup of S. Then there exists an infinite subset B of A such that ([b, x]/b ß B} is an elementary abelian 2-group. Furthermore, if IA' CA(x)l is infinite, there exists C c_ A such that ([c, x]/c ß C} is an infinite elementary abelian 2-group. Proof. Consider the sets Y and Ax, where Y is any infinite subset of A. Since G ß Q there exist b ß Y, v ß A such that I = [b, vx] 2 = [b,x] 2. Therefore A = {a ß A/[a,x] 2-1} is a cofinite subset of A. Now fix a ß A and consider the sets a V - and Ax, where V is any infinite subset of A. Then there exists c V and w ß A such that 1 = [a c- wx] 2 [a c- x] 2 (([a,x][x,c]) - ) 2, =, =, and so [[a,x], [c,x]] = 1. Therefore {a ß Ai/[a,x] ß CG([at,x])} is a cofinite subset of A1. We are now able to construct by induction on n an infinite sequence (a,),er of elements of A such that, for any n N, (II): ([ai, x]/i ß {1,...,n}) is an elementary abelian 2-group.

4 508 PATRIZIA LONGOBARDI AND MERCEDE MAJ For, assume that al,...,a A satisfy I) and II). By the previous remark there exist cofinite subsets Bx,..., B of A such that, for any i {1,..., s) and any a Bi, [a,x] C6 ([ai, x]). Hence also B 1 [-]... [' Bs is a cofinite subset of A and we have, for any y B1 N... Bs, [y, x] C6([al, x], [a2,x],...,[as, X]). If asnc 1 is any element of (Bx... B ) - {al,...,a ), the elements al,...,as, as+l satisfy (I) and (II). Therefore ([ai, x]/i N) is an elementary abeljan 2-group. Now assume that [A'C t(x)[ is infinite. Then there exists an infinite subset D of A such that d C t(x) for any d D and c-xd C t(x) if c, d are different elements of D. Arguing as before we can construct a sequence (b ) e of pairwise different elements of D such that H -- ([bi, x]/i N) is an elementary abeljan 2-group. l rom [bh, x] -- [bk, x] it follows that bhb Ca(x) and h k; therefore the subgroup H is infinite. [] Corollary 2.6. Iœ G Q, G is infinite and G does not have an infinite elementary abeljan 2-subgroup, then G Q.. Proof. Let x, y be elements of G. By 2.3 there exists an infinite abeljan subgroup A _ G, and 2.5 implies that [A'CA(x)[ and A'C (y)[ are both finite. Then A' C t(x) C t(y)[ is finite and, arguing as in 2.4, we get that = 1. [] The following result generalizes 2.2. Corollary 2.7. Let G Q be an infinite group. Then C6(x) is infinite, œor any x G oœ order 2 (œor all n in N). Proof. We argue by induction on n. If n - 1, then the result follows from 2.2. Let n 1 and assume that C6(x 2) is infinite. Then by 2.3 there exists an infinite abelian subgroup A _< Co(x2). If[A'C (x)l is finite, then C (x) is infinite, as required. Thus assume that [A'C t(x)[ is infinite. Then by 2.5 there exists a subset C C_ A such that ([c, x]/c C) is an infinite elementary abelian 2-group. Now for any c C we have 1 - [c, x 2] - = e e C) c_ and C (x) is infinite, as required. [] 3. The soluble case. In this section we will prove Theorem A under the additional hypothesis that G is soluble. We start with the following useful Lemma.

5 A FINITENESS CONDITION IN GROUPS 509 Lemma 3.1. Let G 6 Qt. If there exists an infinite abelinn subgroup A normal in G, then G 6 Q.. Proof. Let x, y be elements of C. We will show that [x,y] 2 = 1. If IA'CA(x)I and IA'CA(y)l are both finite, then Z ({A,x,y)) is infinite and [x,y] 2 = 1 by 2.1. So we can assume that A ß C/t(x)l is infinite, thus by 2.5 there exists an infinite subset B of A such that {[b, x]/b ß B) is an infinite elementary abelinn 2-group. Hence {a ß A/a = 1) is infinite, and we can assume without loss of generality, that A = {a ß A/a 2 = 1} is an elementary abelinn 2-group. Arguing as before we can also suppose that IA'CA(y)I is infinite (replacing y by xy if necessary). First we prove that (*) there exist infinite subsets C ß D A such that for any c C, d D, (Ix, y] [c, x]) = (Ix, y] [d, y])2 =. That is, [c,/]-1 Ix, y] [c, x] = [y,x], [d, y]-l[x, y] [d,y] = [y,x], and [c,/] [d,y] commutes with For, by a famous theorem of B. H. Nedann (see for instance [8] p. 105), A is not the union of finitely many cosets of C (x)uca(y). Then, guing in 2.2, we can construct infinite sequence (ai)ien of elements of A such that: 1) aiaj Co(x) U Co(y), if i j; 2) aiaja a Co(x) U Co(y), for i,j, h, k pairwise different. Then, guing again in the proof of 2.2, we get infinite subsets X = (bi/i N) and Y = (ci/i N) of (ai/i N} such that X Y = and [x b', y ']2 = 1, for any i e N. Hence I = [x b' ',y]2 = [x[x, bici],y]2 = ((Ix y] Ix, bier,y])2) [ ' ' '1, and (Ix, y] [/, b c, y])2 = ; an ogously from = [y ' ',x] 2 we get (Ix, y] [y, c b, x]) =. Then we have (*) with C = {[y,c bd/i e N} d D = {[x,b cd/i e Now consider the infinite sets xd, yc, then there exist d 6 D, c 6 C such that I = [xd, yc] = (Ix, c] [x, y] a[d, y]) = (([x,c] Ix, y] [d,y]) a) 2, and = (Ix, y] [c,x][d,y]) 2 = Ix, y] 2, since [c,x][d,y] centr izes Ix, y] by (.). Hence [x, y]2 = 1, required. Now we can prove Theorem 3.2. Let G ß Q be an infinite soluble group. Then G ß Q2.

6 510 PATRIZIA LONGOBARDI AND MERCEDE MAJ Proof. We argue by induction the derived length d of G. Assume d > 1, and let A be the last non-trivial term of the derived series of G. Then A is abelian. If A is infinite, then G Q2 by Lemma 3.1, so we can assume that A is finite. Thus G/A is an infinite group in Q and, by induction, G/A Q2. First we prove that (*) If b is any element of G of order 2, then [b, g]2 _ I for every g G, so that (b) G is abelian. For, let b e G be of order 2. Then ((b)ga)/a is abelian since G/A Q, and ((b) ) _ A is finite. Thus C (b)f C (b g) is infinite and, by 2.3, therexists em infinite abelian subgroup B C (b) C (b ). If lb'cs(g)[ is finite, then [b,g] = 1 since G Q. Therefore we can sume that [B ß Cs(g)[ is infinite and so, by 2.5, ther exists a subset C of B such that ([c, g]/c C) is an infinitelement y abelian 2-group. Now consider the infinite subsets {bc-1/c C} and go. There exists an infinite sequence (ai)ie in C such that [ba l,g] = 1, hence ([b,g] [g, ai]) = 1 and [ai, g] conjugates lb, g] to its inverse for y i e N. But, if we consider the infinite sets {b(ala;1)-l/i e N} and gc we have, for some k e N, c e C, 1= [b(ala;1)-l,gc]2= (([b,g] [g, ala;1]) 2) (a a; Dd [g, ala; 1] = --1 ([, ][, ]) m s Iv, ] to t o i= o o. Si.ce e C (IV, ]).a [ 1, ] [, ] e C (IV, ]) we IV, ] = 1, req.iroa. Now, if therexists b G of order 2 such that <b) a is infinite, then G Q by Lemma 3.1. Thus we c sume that <b) a is finite for any b G of order 2. By 2.6 we can suppose that G has an i nite elementary abelinn 2-sub oup X. Let D be a maximal normal elementary abelinn 2-subgroup of G. If D is infinite, then G Q by Lemma 3.1. Assume that D is finite. T o= e' C (V)I is nnite,a so IX.Cx(V)I is nnito. T,s t o e e ist b Cx(D) - D, d <b) D is an elementary abelinn norram 2-subgroup of G, contradicting the maxima ty of D. Therefore, G Q. Proof of Theorem A. Let G be an infinite group in Q. First we show that (*) if A is an infinite abelinn subgroup of G and x G has order 2, then In, x] = 1, for any a A. o, i n' c ( )l <, t o. c ( ) is i.n.ite, so t t z (<n, >) is infinite and the result follows. Now sume that [A'Cn(x)[ is infinite. Thus, by Lemma 2.5, there exists a subset C of A such that the sub-

7 A FINITENESS CONDITION IN GROUPS 511 group X = ([c,x]/c e C) is an infinite elementary abelian 2-group. If c e C, from [c,x] 2 = 1 = [c,x 2] we get [c,x,x] = 1. Now, let a e A. By the hypothesis G e Q there exist c, b e C such that 1 = [a[c, x], xb] 2 = ([a, x]b[c'x][c, x, xb]) 2 = ([a, x]o[c'x][c, x, x[x, hi]) 2 = ([a, x]o[c'z]), hence [a, x] = 1, as required. Then we have (**) [a,x] = 1, for any x e G of order œ and any a e G of order In fact, by 2.7, C (a) is infinite and, by 2.3., there exists an infinite abelian subgroup B 5 C (a). Then (**) follows from (.) with A = (a, B). Now put V = (x e G/x = 1). If V is finite, then G/V is an infinite group in A*, hence G/V is abelian (see[6] and [4]), G t is finite, and G is an FC-group. Then 2.4 implies G Q, required. Now assume that V is infinite. Then if x, y G there exist v, w V such that [xv, yw] = 1, thus [x, y] e V and G/V is abeljan. We will prove that V e Q, then V is soluble (see[5]), G is soluble and G e Q by Theorem 3.2. IfXl,X,...,x G d xi[ = 2 for alli, it is easy to prove, by induction on n and using (**), that ]x x... x [ divides 2. Hence V is a 2-group. If x e V h order 2 and y e V, then (**) implies [x, y]" = therefore Ix, xv] = 1 d (x)v is abeli. If (x)v is infinite for some x e V of order 2, then V Q by Lemma 3.1. Otherwise y element of order 2 is an FC-element, then V is an FC-group and, by 2.4, again V Q, as required. REFERENCES 1. Bachmuth, S., and Lewin, J., The Jacobi identity in groups, Math. Z. 85 (1964), Kegel, O.tt. and Wehrfritz, B.A.F., Locally Finite Groups, North-ttolland, Amsterdam, Kim, P.S., Rhemtulla, A.H., and Smith, H., A characterization of infinite metabelian group, Houston J. of Math. 17 n. 3 (1991), Longobaxdi, P., Maj, M., Rhemtulla, A. H., Infinite groups in a given variety and Ramsey's theorem, Comm. Alõebra 20 (1992), MacDonald, I.D., On certain varieties of groups, Math. Z. 76 (1961), Neumann, B.H., A problem of Paul ErdSs on groups, J. Austral. Math. Soc. (Series A) 21 (1976), Rhemtulla, A.H., Smith, H., On infinite solvable groups, to appear.

8 512 PATRIZIA LONGOBARDI AND MERCEDE MA3 8. Robinson, D. J. S., Finitenezz Conditionz and Generalized Soluble Groupz, l, Springer- Verlag, Berlin, Received: February 23, 1993 Revised Version Received: May 18, 1993 DIPARTIMENTO DI MATEMAT,CA --. APPLICAZIONI "R. CACCIOPPOLI", UNIVER- SITA' DEi2LI STUDI DI NAPOLI, VIA CINTIA, NAPOLI - ITALY