A. ABDOLLAHI, A. REJALI AND A. YOUSOFZADEH

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1 CONFIURATION OF NILPOTENT ROUPS AND ISOMORPHISM arxiv: v1 [math.r] 14 Nov 2008 A. ABDOLLAHI, A. REJALI AND A. YOUSOFZADEH Abstract. The concept of configuration was first introduced by Rosenblatt and Willis to give a condition for amenability of groups. We show that if 1 and 2 have the same configuration sets and H 1 is a normal subgroup of 1 with abelian quotient, then there is a normal subgroup H 2 of 2 such that 1 H 1 = 2. Also configuration of FC-groups and isomorphism is studied. H 2 1. Introduction The notion of configuration for a finitely generated group was introduced in [4]. It was shown that amenability of is characterized by its configuration equations. In [1], the authors investigated some properties of groups which can be characterized by configurations and studied the question of whether is determined up to isomorphism by its configurations. The configurations of are defined in terms of finite generating sets and finite partitions of. A configuration corresponding to a generating sequence g = (g 1,..., g n ) and a partition E = {E 1,..., E m } of is an (n + 1) tuple C = (c 0,...,c n ), where 1 c i m for each i, such that there is x in with x E c0 and g i x E ci for each 1 i n. The set of all configurations corresponding to the pair (g, E) will be denoted by Con(g, E). Con() = {Con(g, E) g a finite generating sequence of, E a finite partition} The set of all configuration sets of is denoted by Con(). We write 1 2 and 1 is called configuration equivalent with 2, if Con( 1 ) = Con( 2 ). Clearly the relation is an equivalence relation on the class of finitely generated groups. The configuration C = (c 0,..., c n ) may be described equivalently as a labeled tree. The tree has one vertex of degree n, labeled by c 0. Emanating from this vertex are edges labeled 1,...,n, and the other vertex of the i-th edge is labeled c i. When the generators are distinct, this tree is a subgraph of the Cayley graph of the finitely generated group = g 1,...,g n. The edges labels indicate which generator gives rise to the edge and the vertices labels show which set of the partition E the vertex belong to. From this perspective the configuration set Con(g, E) is a set of rooted trees having height 1. This finite set carries information about and the present paper addresses some properties of which can be recovered from such information. We prove that if the group has the same configuration set with the direct product of groups Z n F, where F is a finite group, then is isomorphic with Z n F. This generalizes the result of Abdollahi, Rejali and Willis [1] saying that if 1 is an abelian group with the same configuration sets as 2, then they are isomorphic. 1

2 2 A. ABDOLLAHI, A. REJALI AND A. YOUSOFZADEH In [1] the authors tried to generalize the latter result for nilpotent groups. It was shown that if 1 is a finitely generated nilpotent group of class c having the same configuration sets with 2, then 2 is a nilpotent group of class c. In this paper we show that if 1 is a torsion free nilpotent group of Hirsch length h, then so is 2. We do not know whether the configuration equivalent and isomorphism relations are the same for the class of nilpotent groups. We will prove it for certain nilpotent groups. Let be a finitely generated nilpotent group. Then is isomorphic with a subgroup of direct product of T r (n, Z), and a finite nilpotent group, for some positive integer n, where T r (n, Z) denotes the group of all upper triangular matrices with integer entries and all diagonal entries equal 1. It is well-known that T r (n, Z) is a finitely generated torsion free nilpotent group of class n 1 and Hirsch length n(n 1) 2 (see [3]). We study finitely generated torsion free nilpotent groups of class 2 such that T = Z n and T = Z m, where T = τ() is the isolator of the commutator subgroup of, and this class of groups will be denoted by I(n, m) (see [5]). It is shown that if I(n, m) has the same configuration set with H, then H I(n, m). Therefore τ() = H τ(h) and τ() = τ(h). For n = 3 and m = 2, groups and H will be isomorphic. It is shown in [1] that if a finitely generated group 1 has the same configuration set with 2, and 1 has a normal subgroup H 1 with finite index, then 2 has a normal subgroup H 2 so that 1 H 1 = 2 H 2. For a group, we denote by F() the set of all isomorphism types of finite quotients of. Thus the latter result says that the set of isomorphism types of finite quotients of two configuration equivalent groups 1 and 2 coincide i.e., F( 1 ) = F( 2 ). We show that one can extend finite index property to abelian quotient property. In fact A( 1 ) = A( 2 ), where A() is the set of all isomorphism types of abelian quotients of. Let be a finitely generated infinite simple group. Then F(Z 2 ) = F( Z 2 ), however Z 2 and Z 2 are not configuration equivalent (see [1]). This example shows that the relation of having the same configuration is strictly stronger than the relation of having the same finite quotients. Several authors characterized groups with the same finite quotient sets, under certain conditions (see e.g., [8] and [9]). In [5] it was shown that two polycyclic-by-finite groups 1 and 2 have the same finite quotient sets if and only if Ĝ1 = Ĝ2, where Ĝ is the profinite completion of. Also in [9] cancellation of groups is studied and under certain conditions it is shown that F( 1 ) = F( 2 ) if and only if Z 1 = Z 2. By using this result one can show that if 1 and 2 are finitely generated FC-groups with the same configuration sets, then 1 is isomorphic with a subgroup of 2 of finite index, where by an FC-group we mean a group in which each element has finitely many conjugates. In particular if 1 is a finitely generated nilpotent FC-group, then so is 2. We do not know whether it is true without the nilpotence condition. It is to be noted that a finitely generated group is an FC-group if and only if Z() is finite, where Z() is the center of (see [6]). We prove that if 1 and 2 are two finitely generated nilpotent FC-groups having the same configuration sets, then 1 Z( 1) = 2 Z( and Z( 2) 1) = Z( 2 ). So configuration equivalent nilpotent FC-group of nilpotency class 2 have equivalent upper central series. We do not know if this result is true for general nilpotent groups. If is a finitely generated

3 3 FC-group, then is isomorphic with a subgroup of Z n F, for some finite group F, (see [6]). We show that configuration equivalence and isomorphism are the same for Z n F. Let T r (n, Z) = g i,j : 1 i < j n, in which g i,j = I + E i,j, where E i,j is the matrix with entry 1 in the (i, j) th position and other entries are zero. Also I is the n n identity matrix. One can show that each matrix A = (a i,j ) T r (n, Z) has a unique representation as A = (g a1,n 1,n ga2,n 2,n gan 1,n n 1,n )(ga1,n 1 1,n 1 ga2,n 1 2,n 1 gan 2,n 1 n 2,n 1 ) ga1,2 1,2. Furthermore g i,j g k,t = g k,t g i,j and g k,t g i,k = g i,k g 1 i,t g k,t for all i t and j k. 2. Configuration equivalence of nilpotent groups and isomorphism In this section we are looking for group theoretical properties P satisfying the condition of the question posed in [1]: Which group properties are translated by configuration equivalence? For a group, we denote by F() and A() the set of isomorphism types of finite quotients and abelian quotients of, respectively. We first mention a result in this direction which was proved in [1]. Theorem 2.1. (See Lemma 6.3 of [1] ) Let 1 and 2 be finitely generated groups such that 1 2. Then F( 1 ) = F( 2 ). We actually have further for two configuration equivalent groups, indeed the sets of their abelian quotients are the same. We need the following lemma to prove the latter. Lemma 2.2. Let be a finitely generated abelian group and L be a finite abelian group. If for every prime p, there exists a subgroup N p of such that N p = Z n p L, then has a subgroup N such that N = Z n L. Proof. Suppose that = Z m Z p1 α 1... Z pk α k, where p 1,..., p k are prime numbers. Then every finite factor of is isomorphic to Z t1... Z tm Z p1 β 1... Z pk β k, for some nonnegative integers t 1,..., t m, β 1,...,β k, where β i α i. For, if M is a subgroup of, then there exist a generating set {g 1,...,g m, h 1,..., h k } of and positive integers d 1,...,d m+k such that g i = Z, h i = Z pi α, = g i g m h 1 h k and M = g d1 1,..., gdm m, hdm+1 1,...,h d m+k. Therefore M = k g 1 g m h 1 h k g d 1 1 gdm m h d m+1 1 h d m+k k. This gives the result. Now let p be a prime number greater than L and p α1 1 pα k k. Then by assumption and the first part of the proof, there exist nonnegative integers t 1,...,t m and β 1,...,β k with β i α i such that Z t1 Z tm Z p β Z p β k k = Z n p L.

4 4 A. ABDOLLAHI, A. REJALI AND A. YOUSOFZADEH Now by the uniqueness of the elementary factors of a finitely generated abelian group, we must have n m, Z l1... Z ln = Z n p and Z r1... Z rm n Z p β 1 1 where {t 1,..., t m } = {l 1,...,l n, r 1,...,r m n }. Therefore... Z p β k k = L, Z m Z α p 1... Z α 1 p k k {0} n r 1Z... r m nz Z α p 1 β 1... Z α 1 p k β k k = Z n L. This completes the proof. Proposition 2.3. Let and H be finitely generated groups such that H. Then A() = A(H). Proof. By Lemma 2.2, it is enough to prove that if N is an (infinite) abelian factor of, then N = H M for some normal subgroup M of H. Since is finitely generated, N = Z n L, where n is a positive integer and L is a finite abelian group. Since Z n L has finite factors Z n p L for any prime number p, by Theorem 3.1, there exists a normal subgroup M p of H such that H M p = Z n p L. Now let B := p prime M p. Then H B is an abelian group having Zn p L for all prime number p as a factor group. It follows from Lemma 2.2 that H B has a factor group isomorphic to Zn L, as required. This completes the proof. Proposition 2.4. Let and H be two polycyclic groups such that H. If admits a normal series with infinite cyclic factors, then H is torsion free. Proof. By Theorem 2.1 F() = F(H). Now by using Lemma 1.5 of [7] (see also chapter 10 of [5]), H is torsion free. In a polycyclic group the number of infinite factors in a cyclic series, which is known as the Hirsch length, is independent of the series and is an invariant of. Denote by h() the Hirsch length of a polycyclic group. Proposition 2.5. Let and H be two finitely generated groups such that H. If is torsion free nilpotent of class c with the Hirsch length h, then so is H. Proof. By Corollary 5.2 of [1], H is also nilpotent of class c. Since every finitely generated torsion-free nilpotent group admits a normal (central) series with infinite cyclic series, Proposition 2.4 implies that H is torsion-free. Now one can prove by induction on c, that the Hirsch lengths of and H coincide. Let be a finitely generated torsion-free nilpotent group of class 2. We denote by T = τ() the isolator of in, i.e. the set of all elements x such that x s for some non zero integer s. The set T is a central subgroup of and T/ is the torsion subgroup of /. Then there exist positive integers m and n such that T = Z n and T = Z m. The class of all such groups shall be denoted by I(n, m) (see [5], P. 260). Lemma 2.6. Let I(n, m) and H for some finitely generated group H. Then H I(n, m).

5 5 Proof. By Proposition 2.5, H is a finitely generated torsion-free nilpotent group of class 2. Suppose that H I(n, m ). We have τ() = Z n and H τ(h) = Z n. By Proposition 3.3, there exists a normal subgroup K of H such that H K = Z n. If x τ(h), then x s H for some non zero integer s. Since H/K is abelian, H K and so x s K. As H/K is torsion-free, we have that x K. Thus τ(h) K. It follows that n n. Similarly n n and so n = n. By Proposition 2.5, h() = h(h). On the other hand, h() = h(τ())+h(/τ()) and h(h) = h(τ(h)) + h(h/τ(h)). Therefore h(τ()) = h(τ(h)). It follows that m = m, as required. Theorem 2.7. Let I(n, m) and H for some finitely generated group H. If (n, m) {(3, 2), (3, 3), (n, 1)}, then = H. Proof. By Lemma 2.6, H I(n, m). It follows from Theorem 2.1, F() = F(H). Now results stated in p. 265 of [5] complete the proof. Corollary 2.8. Let be a finitely generated group such that Con() = Con(Tr(3, Z)). Then = Tr(3, Z). Proof. It is easy to see that Tr(3, Z) I(2, 1). Now Theorem 2.7 completes the proof. The following question is natural: Question 2.9. Let be a finitely generated group. Does Con() = Con(Tr(n, Z)) imply = Tr(n, Z)? Remark In [2] the authors gave a positive answer to the above question with another type of configuration, say two-sided configuration. 3. Configuration equivalent for FC-groups It is interesting to know if being FC-group is conserved by equivalence of configuration. We answer this question in some special cases. It is shown that for each finitely generated FC-group, the torsion subgroup Tor() Tor(), and commutator subgroup of are finite and the factor group is an abelian torsion free group [6]. It has been shown that if two groups are configuration equivalent, then the isomorphism classes of their finite quotients are the same [1]. For groups with finite commutator there is an other interesting result: Theorem 3.1. Let and H are two finitely generated groups with finite commutator subgroups. Then the following conditions are equivalent: (a) F() = F(H), (b) Z = H Z. Proof. See [10, Theorem 2.1]. Definition 3.2. For any group, the non cancellation set, χ() is the set of isomorphism classes of groups H such that Z = H Z. It is easy to see that for a finite or abelian group, χ() is trivial. The following theorem is proved in [10]. Theorem 3.3. Let and H be any groups with finite commutator subgroups. If is infinite and χ() is trivial than χ( H) is trivial.

6 6 A. ABDOLLAHI, A. REJALI AND A. YOUSOFZADEH So for example if is an infinite abelian group and H is arbitrary, then χ( H) is trivial. Theorem 3.4. Let Con() = Con(Z n F), where F is a finite group and is an FC-group. Then = Z n F. Proof. By Theorem (3.1) Z = Z Z n F. But χ(z n F) is trivial by Theorem (3.3). Therefore = Z n F. By a direct proof we show in the next theorem that the condition of being FCgroup for is superfluous. This states the main result of this section. For a non empty set A, we denote by χ B the characteristic function of a subset B of A, which is defined as { 1 if a B χ B (a) = 0 if a A\B. Theorem 3.5. Let Con() = Con(Z n F), where F is a finite group. Then = Z n F. Proof. Suppose that F = {x 1,..., x l } where x 1 = e. Let Σ = {σ 1,..., σ 3 n} be the set of all functions from {1,..., n} to {0, 1, 1}. Put E σ,j = σ(1)n... σ(n)n {x j }, for any σ Σ. Then E = {E σ,j : σ Σ, j {1,..., l}} is a partition of Z n F. Let g = (g 1,..., g l+n ) where g 1 = (1, 0,..., 0, e),...,g n = (0, 0,...,1, e), for 1 i n and g n+j = (0, 0,..., 0, x j ) for 1 j l. Then Z n F = g 1,..., g l+n. Let also π : {1,...,l} {1,..., l} {1,..., l} be defined by π(i, j) = k whenever x i x j = x k. We refine the partition E to a new partition E such that if σ = χ {i}, then E σ,1 = A i Bi where A i = {g i } and B i = E σ,1 \A i. if σ = χ {i,j} then E σ,1 = A i,j Bi,j, in which A i,j = {g i g j } and B i,j = E σ,1 \A i,j. Let (f, K) be the configuration pair of, in which K = {K σ,j : σ Σ, j {1,..., l}, (σ, j) (χ {i}, 1), (σ, j) (χ {i,k}, 1)} {C i, D i, C i,k, D i,k : 1 i, k n} is a partition and f = (f 1,...,f l+n ) is a generating set of, such that Con(f, K) = Con(g, E). After writing the members of Con(f, K), we have (I) f i K σ,j K σ i,j where σ(i) = 0 and σ i (i) = 1 and σ(t) = σ i (t) for all t i, (II) f i K σ,j K σ,j where σ(i) = 1, (III) f i K σ,j K σ,j Kσ i,j where σ i (i) = 0 and σ(i) = 1 and σ(t) = σ i (t) for all t i, (IV) f n+i K σ,j = K σ,π(i,j), for 1 i l and 1 j l. Now let σ(i) = 0 then for each j {1,...,l} and each m N we have fi m K σ,j K σ i,j and so m=1 fm i K σ,j K σ i,j. On the other hand f i (K σ,j Kσ i,j) = K σ i,j. Consider the case σ(i) = 1. For each j {1,...,l}, K σ,j = fi 1 (K σ,j Kσ i,j ). It is routine to compute that m=1 f i m K σ i,j K σ,j Finally if K 1 = K σ0,1 where σ 0 0 and 1 K 1 and σ Σ, then f i f j K 1 = f i C j = C i,j = f j C i = f j f i K 1 and, (3.1) m 1=1 m n=1 f m1σ(1) 1 f mnσ(n) n f n+j K 1 K σ,j.

7 We claim that each element of belongs to f m1 1 fn mnf n+jk 1, for some integers m 1,...,m n and some j {1,..., l}. Suppose that K 1 {1} and set L σ0,1 = {1} and L 2 = K σ0,1\{1}. Then if L a = K a, for each a, L = {L a : a σ 0,1 } L 1 L2 is a partition of. So by hypotheses there exist a configuration pair (T, h) of Z n F such that Con(T, h) = Con(L, f). Let h s = (a 1 s,..., a n s,x js ) for some a i s Z and x j s F, where 1 s n+l. We claim that a t n+i = 0 for each 1 i l and 1 t n. If they are not, since the vectors (a 1 1,...,a n 1),..., (a 1 n,...,a n n) and (a 1 n+i,..., an n+i ) are Q linearly dependent in Qn and so in Z n, there exist some integers k 1,...,k n and k i not together all zero such that h k1 1 hkn n h ki n+i = (0,...,0, x l ) for some x l F and so h k1 1 hkn n hki n+i must have a finite order. But it is possible only if k 1 = k 2 = = k n = 0, by (3.1). Therefore h n+i = (0,...,0, x jn+i ). Since h n+1,...,h n+l are l distinct members, (h n+i K σ0,i) we have {x jn+1,..., x jn+l } = F. Now consider that Z n (a 1 1,..., an 1 ),...,(a1 n,...,an n ). Consequently each element of Z n F has the form h k1 1 hkn n h n+j. Also 1 T 1 and h k1 1 hkn n h n+j T σ,j, in which σ(i) = sgn(k i ), is defined 1, 1 and 0 if k i is positive, negative and zero respectively. So Z n F ( T T T)\T 2. Thus T 2 =. This contradiction causes that K 1 = {1} i.e. each element of is of the form f k1 1 fkn n f n+j, and K σ,j = {f k1σ(1) 1 f knσ(n) n f n+j, k 1,...,k n N} Note that the expression of elements are unique. For if x = f k1 1 fkn n f n+j = f s1 1 fsn n f n+t, then there exists a σ such that x K σ,j Kσ,t, and so t = j. Therefore f k1 1 fkn n f k1 s1 1 fn kn = f s1 1 fsn n, = fs2 2 fsn n. So by (3.1), σ(1) = 0, which implies k 1 = s 1. By induction k i = s i for every i {1,..., n}. Define φ : Z n F, φ(f m1 1 fn mn f n+j ) = (m 1,..., m k, x j ), Evidently it is a one to one function onto Z n F. It is also a group isomorphism since 7

8 8 A. ABDOLLAHI, A. REJALI AND A. YOUSOFZADEH φ(f k1 1 fkn n f n+jf k 1 1 fk n n f n+t ) = φ(f k1 1 fkn n fk 1 1 fk n n f n+π(j,t) ) = φ(f k1+k 1 1 f kn+k n n f n+π(j,t) ) = (k 1 + k 1,..., k n + k n, x π(i,j) ). This completes the proof. In the following n = n() = n 1 n 2 n 3, in which n 1 is the exponent of the torsion subgroup of, n 2 is the exponent of Aut(Tor()) and n 3 is the exponent of torsion group of Z(). (The exponent of a group is the least n N such that x n = 1 for all x or if such n does not exist.) Let have finite commutator subgroup and n = n(). If H, then by [11, Theorem 4.2], H is isomorphic to a subgroup L of of finite index in that [ : L] is relatively prime to n. Now let [ : L] ±1 mod(n). Then H = by [11, Theorem 4.3]. Theorem 3.6. Let be a finitely generated nilpotent FC-group and H. Then H is a nilpotent FC-group too. Furthermore Z() = Z(H) and Z() = H Z(H). Also and H have the same Hircsh length. Proof. Since is an FC-group, Z() is finite and Z() = n say. So xyn = y n x for each x, y in. Therefore by [1] this semigroup law is satisfied by the group H H. Thus Z(H) is a periodic finitely generated group. It is also nilpotent. On the other it is known that every periodic finitely generated nilpotent group is finite. H Therefore Z(H) will be finite and so H is an FC-group by [6]. Now using Theorem 3.1, Z = H Z. Therefore Z() Z = Z(H) Z. But Z() and Z(H) are abelian and so Z() = Z(H), by [9, Theorem 7]. Also we can easily see that Z() = H Z(H). On the other hand since Z() and H Z(H) are finite and Z() = Z(H), by [6, p.16] and H have the same Hircsh lengths. The following example shows that configuration equivalence and isomorphism is the same for a group which is neither an FC-group, nor a nilpotent one. Example 3.7. Let be a finitely generated group such that D, where D = x, y x 2 = y 2 = 1 is the infinite dihedral group. Then = D. Proof. Consider the following configuration pair (g, E) of : g = (x, y) and E = {E 1,...,E 5 }, where E 1 = {1}, E 2 = {x}, E 3 = {y}, E 4 = {g 1 g 2 g n g i {x, y}, n N, n > 1, i {1,...,n}, g 1 = x, g i g i+1 }, and E 5 = {g 1 g 2 g n g i {x, y}, n N, n > 1, i {1,...,n}, g 1 = y, g i g i+1 }. Evidently E is a partition of D. We have Con(g, E) = {c 1,...,c 7 }, where c 1 = (1, 2, 3), c 2 = (2, 1, 5), c 3 = (3, 4, 1), c 4 = (4, 5, 5), c 5 = (4, 3, 5), c 6 = (5, 4, 2), c 7 = (5, 4, 4). Since Con() = Con(D ), there exists an ordered generating set h = (h 1, h 2 ) and a partition F = {F 1,...,F 5 } of such that Con(g, E) = Con(h, F) and 1 F 1. Thus h 1 F 1 = F 2, h 1 F 2 = F 1, h 2 F 1 = F 3, h 2 F 3 = F 1, h 1 F 4 = F 3 F 5, h 2 F 5 = F 2 F 4, h 2 (F 2 F 4 ) = F 5, h 1 (F 3 F 5 ) = F 4.

9 9 Therefore h 2 1 F 1 = h 2 2 F 1 = F 1 and for n > 1 and t 1,..., t n {h 1, h 2 } with t i t i+1 we have n 1, Now let t 1 t n F 1 { F 4 if t 1 = h 1 F 5 if t 1 = h 2, (t i {h 1, h 2 }). L 1 = {1}, L 2 = F 1 \{1}, L 3 = F 2, L 4 = F 3, L 5 = F 4, L 6 = F 5. and L = {L 1,..., L 6 }. If L 2, then L is a partition of. Therefore we have Con(h, L) = {d 1,...,d 10 }, where d 1 = (1, 3, 4), d 2 = (2, 3, 4), d 3 = (3, 1, 6), d 4 = (3, 2, 6), d 5 = (4, 5, 1), d 6 = (4, 5, 2), d 7 = (5, 4, 6), d 8 = (5, 6, 6), d 9 = (6, 5, 3), d 10 = (6, 5, 5). Thus there exist an ordered generating set g = (g 1, g 2 ) and a partition K = {K 1,...,K 6 } of D such that Con(g, K) = {d 1,...,d 10 } and 1 K 1. Thus (3.2) g 1 (K 1 K 2 ) = K 3, g 1 K 3 = K 1 K 2, g 1 (K 6 K 4 ) = K 5, g 1 K 5 = K 4 K 6, g 2 (K 1 K 2 ) = K 4, g 2 K 4 = K 1 K 2, g 2 (K 3 K 5 ) = K 6, g 2 K 6 = K 3 K 5. Note that an element of D has finite order if and only if its length as product of a and b is odd. Three different cases are possible: Case I) g 1 (respectively g 2 ) has even order and g 2 (respectively g 1 ) has odd order. In this case g 1 g 2 must have an odd length and consequently finite order, which is impossible since it is easy to see that (g 1 g 2 ) n K 5 K 6 for each non zero integer n. Case II) Both g 1 and g 2 have even lengths. In this case the group g 1, g 2 can not have any non identity element of finite order and it is a contradiction too, since D = g 1, g 2. Case III) Both g 1 and g 2 have odd lengths. Thus they must be of order 2. Also by (3.2) each element of D has a unique form of products of g 1 and g 2. On the other hand 1 K 1, g 1 K 3 and g 2 K 4. We have also for n 1, t 1 t n K 1 { K 5 if t 1 = g 1 K 6 if t 1 = g 2, (t i {g 1, g 2 }). In other word, D K 1 K 3 K 4 K 5 K 6. So K 2 should be empty. This implies that F 1 = {1} and so h 2 1 = h 2 2 = 1. Since h 1 and h 2 are not in F 1, o(h 1 ) = o(h 2 ) = 2. By Proposition 4.3 of [1], is infinite and so is an infinite group generated by two involutions. Hence = D, as required. In [1] the authors proved that if a finitely generated group satisfies a semigroup law and H, then so does H. In the previous example D is a finitely presented group and we show their result for generators in this case. Now the following question is natural.

10 10 A. ABDOLLAHI, A. REJALI AND A. YOUSOFZADEH Question 3.8. Let H and the generating sequence g = (g 1,...,g n ) of satisfies the semigroup law w(g) = 1. Does there exist a generating sequence h of H satisfying w(h) = 1? Acknowledgements. The first two authors s research was partially supported by the Center of Excellence for Mathematics at University of Isfahan. The first author s research was in part supported by a grant from IPM (No ). References [1] A. Abdollahi, A. Rejali and. A. Willis, roup properties characterised by configurations, Illinois J. Mathematics, 48 (2004) No [2] A. Rejali and A. Yousofzadeh, Properties characterized by two-sided configurations, (to appear in Algebra Colloquium). [3] D. J. S. Robinson, A course in the theory of groups, 2nd. edition, Springer-Verlag, New York, [4] J. M. Rosenblatt and. A. Willis, Weak convergence is not strong for amenable groups, Canad. Math. Bull, 44 (2001), [5] D. Segal, Polycyclic groups, Cambridge University Press, London, [6] M. J. Tomkinsin, FC-groups, 2nd. edition, Pitman publishing limited, London, [7] N. Q. Thǎńg, A note on finitely generated nilpotent groups, Expo. Math. 19 (2001), No. 1, [8] E. A. Walker, Cancellation in direct sums of groups, Proc.Amer. Math. Soc. 7 5 (1956), [9] R. B. Warfield, enus and cancellation for groups with finite commutator subgroups, J. Pure Appl. Algebra. 6 (1975), [10] P. J. Witbooi, eneralizing the Hilton-Mislin genus group, J. Algebra. 239 (2001), Ali Rejali, Department of Mathematics, University of Isfahan, Isfahan 81744, Iran Alireza Abdollahi, Department of Mathematics, University of Isfahan, Isfahan , Iran and School of Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O.Box: , Tehran, Iran. Akram Yousofzadeh, Department of Mathematics, University of Isfahan, Isfahan 81744, Iran address: a.abdollahi@math.ui.ac.ir address: rejali@sci.ui.ac.ir address: a.yousofzade@sci.ui.ac.ir