CENTRE-BY-METABELIAN GROUPS WITH A CONDITION ON INFINITE SUBSETS

In this note, we consider some combinatorial conditions on infinite subsets of groups and we obtain in terms of these conditions some characterizations of the classes L(Nk)F and FL(Nk) for the finitely generated centre-by-metabelian groups, where L(Nk) (respectively, F) denotes the class of groups in which the normal closure of each element is nilpotent of class at most k (respectively,

Let k be a fixed positive integer.Denote by E * k the class of groups such that for every infinite subset X there exist two distinct elements x, y in X, and integers t 0 , t 1 , . . ., t k depending on x, y, and satisfying z t0 0 , z t1 1 , . . ., z t k k = 1, where z i ∈ {x, y} for every i ∈ {0, 1, . . ., k} and z 0 = z 1 .Denote also by E # k the class of groups G ∈ E * k for which the integers t 0 , . . ., t k belong to {−1, 1}.In [3], it is proved that if G is a finitely generated soluble group in the class E * k (respectively E # k ), then there is an integer c, depending only on k, such that G is in N c F (respectively FN c ); where N c and F denote respectively the class of nilpotent groups of class at most c and the class of finite groups.In [3], it is also proved that a finitely generated metabelian group G is in E * k (respectively E # k ) if, and only if, G belongs to N k F (respectively FN k ); and it is observed that these results are not true if the derived length of G is ≥ 3.
Among the examples cited, which are due to Newman [14] (see also [2]), there is a finitely generated torsion-free nilpotent group G of class In [7], it is proved that a metabelian group G is (k + 1)-Engel if, and only if, G belongs to L(N k ).Morse [12] extended this result to a certain class of soluble groups of derived length ≤ 5 which contains the centreby-metabelian groups.So our theorems improve Morse's result for the centre-by-metabelian groups.

Denote by B *
k the class of groups such that every infinite subset contains an element x such that x is subnormal of defect k.It is proved in [8,Corollary 2.5] that a metabelian non-torsion group is a k-Baer group (that is every cyclic subgroup of G is subnormal of defect k) if, and only if, G is a k-Engel group.Here, using Theorem 1.2, we shall improve this result with the following: In particular, a torsion-free centre-by-metabelian group G belongs to B * k if, and only if, G is k-Engel.

Lemma 2.1. Let G be a finitely generated torsion-free nilpotent group of class at most
Proof: Let G be a group in E * k and assume that G is not k-Engel.Therefore there exist x, y in G such that [x, k y] = 1.The group G, being a finitely generated torsion-free nilpotent group, is a residually finite p-group for every prime p.So G has a normal subgroup N such that [x, k y] / ∈ N and |G/N | = p r for some positive integer r.Considering the infinite subset x p r+i y : i integer , there are integers n, m, t 0 , t 1 , . . ., t k since G is torsion-free.Put z 0 = x p r+s 0 y and z 1 = x p r+s 1 y, where Conversely, suppose that G is in the class FL(N k ).Therefore there is a finite normal subgroup H such that G/H is (k + 1)-Engel.Since G is a finitely generated soluble group, G/H is therefore nilpotent.It follows that G is finite-by-nilpotent, so G is residually finite.Consequently, there is a normal subgroup N of finite index such that H ∩ N = 1.Since G/N is finite, if X is an infinite subset of G, then there are x, y ∈ X such that x = y and xN = yN .We have [x, k+1 y] ∈ H and x,y N N is cyclic, since G/H is (k + 1)-Engel and xN = yN .Thus, [x, k+1 y] ∈ H ∩ N .It follows that [x, k+1 y] = 1 and, therefore, G belongs to E # k+1 .
Now we suppose that G is a torsion-free centre-by-metabelian group in the class E # k+1 and let x, y 1 , . . ., y k+1 ∈ G. Then H = x, y 1 , . . ., y k+1 is a torsion-free finitely generated centre-by-metabelian group.It follows, from the first part of the proof, that H belongs to FL(N k ), and consequently H ∈ L(N k ) since it is torsion-free.Hence, [x y1 , . . ., x y k+1 ] = 1 and, therefore, G belongs to L(N k ).
For the proof of Theorem 1.3, we need further lemmas.Note that it is proved in [8,Theorem 2.3] that every non-torsion k-Baer group is a k-Engel group.But the converse is shown only in the metabelian case.As a consequence of Morse's result [12], we will extend this result with the following lemma: Lemma 2.3.Let G be a non-torsion centre-by-metabelian group.Then, G is a k-Baer group if, and only if, G is a k-Engel group.
Proof: Let G be a non-torsion centre-by-metabelian group, and suppose that G is a k-Engel group.From [12,Theorem 2] k , then G is a k-Engel group.Proof: Let x, y in G; since G is torsion-free, the subset x i : i positive integer is infinite.Therefore there is a positive integer i such that x i is k-subnormal in G. Thus, x i , y, k−1 x i ∈ x i , so x i , y, k−1 x i = x r for some integer r.Since G belongs to L(N k ), we have that G is a (k + 1)-Engel group.Hence, 1 = x i , k+1 y, k−1 x i = x r k+1 ; and this gives that r = 0 as G is torsion-free.It follows that . Once again, as G is torsion-free, we obtain that [y, k x] = 1; this means that G is a k-Engel group.
Proof of Theorem 1.3: Let G be a finitely generated centre-by-metabelian group in the class B * k .So every infinite subset of G contains an element x such that x is k-subnormal in G. Hence, for any y in G we have [y, k+1 x] = 1.Thus, G belongs to E # k+1 .It follows, from [11,Theorem 1], that G is finite-by-nilpotent.Therefore there is a finite normal subgroup T such that G/T is a torsion-free centre-by-metabelian group which belongs to E # k+1 .It follows from Theorem 1.2 that G/T is in L(N k ), and by Lemma 2.4, we obtain that G/T is a k-Engel group.Therefore, G is finite-by-(k-Engel); as claimed.Now, assume that G is a torsion-free centre-by-metabelian group in B * k and let x, y in G.Then, from the first part of the proof, H = x, y is finite-by-(k-Engel).Since G is torsion-free we deduce that H is k-Engel.
Hence, [y, k x] = 1, so G is a k-Engel group.Conversely, suppose that G is a torsion-free centre-by-metabelian and a k-Engel group.From Lemma 2.3 we get that G is a k-Baer group, so G is in B * k .
).Note that if a group belongs to N k , then it is in L(N k−1 ), where L(N k−1 ) denotes the class of groups in which the normal closure of each element is nilpotent of class at most k − 1. Considering this weaker condition we are able to prove the following results: Now x p r+n , x p r+m ∈ N , so z i N = yN .It follows that [x, k y] N = N ; this means that [x, k y] ∈ N , a contradiction which completes the proof.It is proved in[12,Theorem 1]that if G is nilpotent of class at most k+2, then G is (k+1)-Engel if and only if G ∈ L(N k ).So combining this result and Lemma 2.1, we have the following consequence: Let G be a finitely generated nilpotent group of class at most k + 2. If G is in E * k+1 , then G belongs to FL(N k ).In particular, a torsion-free nilpotent group G of class at most k + 2 is in E * k+1 if, and only if, G belongs to L(N k ).Let G be a finitely generated nilpotent group of class at most k+2 and suppose that G is in E * k+1 .Then T , the torsion subgroup of G, is finite and G/T is a finitely generated torsion-free group of nilpotency class at most k + 2 which belongs to E * k+1 .It follows, from Lemma 2.1, that G/T is a (k + 1)-Engel group, and by [12, Theorem 1], G/T belongs to L(N k ).Hence, G is in FL(N k ); as claimed.Now, we suppose that G is a torsion-free group of nilpotency class at most k + 2 which belongs to E * k+1 and let x, y 1 , ..., y k+1 ∈ G. Then H = x, y 1 , ..., y k+1 is a finitely generated group of nilpotency class at most k + 2 which belongs to E * k+1 .It follows, from the first part of the proof, that H is in FL(N k ).So H is in L(N k ) since it is torsion-free.Hence, [x y1 , ..., x y k+1 ] = 1, and this means that G belongs to L(N k ).Hence, G belongs to N k+2 F. Since finitely generated nilpotent groups are (torsion-free)-by-finite[15, 5.4.15(i)],G has a normal subgroup H, of finite index such that H is a torsion-free nilpotent group of class at most k + 2 which belongs toE * k+1 .It follows, by Lemma 2.2, that H is in L(N k ); so G belongs to L(N k )F.Conversely, suppose that G is in L(N k )F.Therefore there is a positive integer n and a normal subgroup H such that H ∈ L(N k ) and |G/H| = n.So H is a (k + 1)-Engel group and x n , y n ∈ H for any x, y in G. Hence, [x n , k+1 y n ] = 1 and consequently G belongs to E * k+1 .