garcia de galdeano PRE-PUBLICACIONES del seminario matematico 2003 n. 22 Groups with Polycyclic - by - finite conjugate classes of subgroups

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1 PRE-PUBLICACIONES del seminario matematico 2003 Groups with Polycyclic - by - finite conjugate classes of subgroups garcia de galdeano Leonid A. Kurdachenko Javier Otal Panagiotis Soules n. 22 seminario matemático garcía de galdeano Universidad de Zaragoza

2 Groups with Polycyclic by Finite Conjugate Classes of Subgroups. LEONID A. KURDACHENKO Department of Algebra, National University of Dnepropetrovsk, Vul. Naukova 13. Dnepropetrovsk 50, UKRAINE JAVIER OTAL Department of Mathematics, University of Zaragoza, Pedro Cerbuna Zaragoza, Spain. PANAGIOTIS SOULES Mathematics Department, University of Athens, Athens, Greece. Abstract B.H.Neumann characterized the groups in which every subgroup has finitely many conjugates only as central by finite groups. If X is a class of groups, a group G is said to have X conjugate classes of subgroups if G/Core G(N G(H)) X for every subgroup H of G. In this paper, we generalize Neumann s result by showing that a group has polycyclic by finite classes of conjugate subgroups if and only if it is central by (polycyclic by finite). Key Words: Conjugacy class, Polycyclic by finite group. A.M.S. Classification: 2OC07, 20D10, 20F24. 1 Introduction A group G is said to have finite conjugacy classes or G is an F C group if every element of G has a finite number of conjugates. Equivalently if G : C G (x) is finite or G/C G (x G ) is a finite group for all x G. Nowadays, the theory of F C groups forms a well developed topic and represents one of the main branches of group theory with finiteness conditions. In this area many famous mathematicians have given relevant contributions (see the books [9], [15] and the survey [16]). Let G be a group. If H is a subgroup of G, then we denote the set of all conjugates of H in G by Cl G (H) = {H g g G}. This research was supported by Proyecto BFM of CICYT (Spain), Proyecto 100/2001 of Gobierno de Aragón (Spain) and the University of Athens-R.C., grant No 3403

3 This set is called the conjugacy class of H in G. In the middle of the fifties, B.H. Neumann [11] characterized the groups G such that Cl G (H) is finite for every subgroup H of G. He showed that these groups are central by finite. In passing, we recall that, if X is a class of groups, then a central by X group is a group G such that G/ζ(G) X, where ζ(g) is the centre of G. Later, I.I.Eremin [2] obtained the first generalization of B.H.Neumann s theorem. He proved that the groups G, in which Cl G (A) is finite for every abelian subgroup A of G are likewise central by finite. Further, in the paper [3], I.I.Eremin began to consider groups with finite conjugate classes of infinite subgroups. Actually, locally (soluble by finite) groups with finite conjugate classes of infinite subgroups were described later by N.N.Semko, S.S.Levischenko and L.A.Kurdachenko [14]. In other papers as [1, 4, 5], combinatorial extensions of theorems of B. H. Neumann and I. I. Eremin were considered as they studied groups in which Cl G (H) < λ for all subgroups H (respectively, for all abelian subgroups H), where λ is a cardinal number. In this paper, we follow a different approach to consider another direction to generalize B.H.Neumann s theorem. This approach concerns to classes of groups rather than combinatorial ideas. Recall that Cl G (H) = G : N G (H) so that the size of the conjugacy class of H in G can be ruled out through the index of the normalizer N G (H) of H in G. this can be carried out by means of the core of that subgroup N G (H g ) = N G (H) g, g G which, by definition, we call the normalizer of the conjugacy class of H in G and denote N G (Cl G (H)). Let X be a class of groups. If G is a group, a subgroup H of G is said to have X classes of conjugate subgroups if the factor-group G/N G (Cl G (H)) X. In other words, if G/Core G (N G (H)) X. The group G is said to have X classes of conjugate subgroups if every subgroup H of G has X classes of conjugate subgroups. If I is the class of all identity groups, then it is clear that G has I classes of conjugate subgroups if and only if every subgroup of G is normal, i.e. G is a Dedekind group. If we match X = F, the class of all finite groups, then groups with finite conjugacy classes of subgroups (in the sense of above definition) are exactly the groups considered by B.H.Neumann. In fact, the finiteness of G/N G (Cl G (H)) implies the finiteness of G/N G (H), that is, the finiteness of Cl G (H). The next natural step is to consider classes of infinite groups close to that of finite groups and that have been studied very good from different points of view. The first candidates are the formation C of all Chernikov groups and the formation P of all polycyclic by finite groups. In the paper [12], Ya.D.Polovizky started the study of groups with Chernikov classes of conjugate subgroups. He proved that a periodic group with Chernikov classes of conjugate subgroups is central by Chernikov. The general case was considered recently by L.A.Kurdachenko and J.Otal in [10]. In the present paper, we study the dual case: groups with polycyclic by finite classes of conjugate subgroups and obtain the following description of these groups. Main theorem. A group G has polycyclic by finite classes of conjugate subgroups if and only if G/ζ(G) is polycyclic by finite. Throughout this paper, most of our group-theoretical notation is standard and follows that of [13]. The ideas involved in the paper are closely related to groups with polycyclic by finite conjugacy classes of elements or P C groups. These are the groups G such that G/C G ( g G ) is polycyclic by finite for all g G. We refer to [7] for the basic properties of these groups. g G 2

4 2 Auxiliary results The proof of the next two results is rather immediate and we omit. Lemma 2.1 Let G be a central by (polycyclic by finite) group. If H is a subgroup of G, then H/Core G (H) is polycyclic by finite. Lemma 2.2 Let G be a group with polycyclic by finite classes of conjugate subgroups. (1) If H is a subgroup of G, then H has polycyclic by finite classes of conjugate subgroups. (2) If H is a normal subgroup of G, then G/H has polycyclic by finite classes of conjugate subgroups. (3) If L H G, then H/L has polycyclic by finite classes of conjugate subgroups. Lemma 2.3 Let G be a group with polycyclic by finite classes of conjugate subgroups. If G is periodic, then G is central by finite. PROOF. Indeed, if H is a subgroup og G, then N G (H) includes a normal subgroup U such that G/U is polycyclic by finite. Since G/U is periodic, it has to be finite. In particular, G : N G (H) is finite and by Neumann s theorem [11], G is central by finite. Corollary 2.4 Let G be a group with polycyclic by finite classes of conjugate subgroups. If G is an F C group, then G is central by finite. PROOF. Since G is an F C group, by a result due to Baer (see, for example, [13, Theorem 4.32]), G/ζ(G) is periodic. Let U be a maximal torsion-free subgroup of ζ(g). Then U is normal in G and G/U is periodic. By Lemma 2.3, Z/U = ζ(g/u) has finite index in G/U. Given g G and z Z, we have [g, z] U. On the other hand, by another result of B.H.Neumann (see, for example, [13, Theorem 4.32]), [G, G] is periodic, and so it follows that [g, z] = 1. Thus Z = ζ(g), and hence G is central by finite. Lemma 2.5 Let G be a group with polycyclic by finite classes of conjugate subgroups. Suppose that G includes a normal periodic subgroup P. If g G, then C P (g) has finite index in P. PROOF. Put H = N G (Cl G ( g )) so that G/H is a polycyclic by finite group. First, we suppose that g has infinite order. Since P is periodic, R = H P has finite index in P. Since R N G ( g ), [a, g] g for every element a R. On the other hand, R is normal in G, so that [a, g] R. Hence [a, g] g R = 1. In follows that R C G (g) and, in particular, P : C P (g) is finite. If g has a finite order, then g, P is periodic. By Lemma 2.3, g, P is central by finite. Again P : C P (g) is finite, as required. Corollary 2.6 Let G be a group with polycyclic by finite classes of conjugate subgroups. Suppose that G includes a normal periodic subgroup P such that G/P is polycyclic by finite. Then G/ζ(G) is polycyclic by finite. PROOF. By Lemma 2.3, P is central by finite. Put A = ζ(p ) so that A is normal in G and G/A is polycyclic by finite. Hence there exists a finitely generated subgroup 3

5 X such that G = AX. Suppose that X = x 1,..., x n. By Lemma 2.5, all indexes A : C A (x 1 ),..., A : C A (x n ) are finite. If Z = C A (x 1 )... C A (x n ), then A/Z is finite and Z ζ(g). It follows that G/ζ(G) is polycyclic-by-finite, as required. Corollary 2.7 Let G be a group with polycyclic by finite classes of conjugate subgroups. Then G is a P C group. PROOF. Pick g G and put H = N G (Cl G ( g )). Then G/H is a polycyclic by finite group and H N G ( g ). If g has finite order, then H/C H ( g ) is a finite cyclic group. Suppose that H/C H ( g ) = n. If g has infinite order, then H/C H ( g ) 2. Put H 1 = C H ( g ) and H 2 = Core G (H 1 ). Then H 2 = (C H ( g )) x = C H ( g x ) = C H ( g x ). x G x G Since H/(C H ( g )) x = H/C H ( g ) for every x G, by Remak s theorem, we obtain the embedding H/H 2 x G H/H x 1 and, in particular, we deduce that H/H 2 is a bounded abelian group. By Lemma 2.2, G/H 2 has polycyclic by finite classes of conjugate subgroups and, by Corollary 2.6, G/H 2 is central by (polycyclic by finite). By Lemma 2.1, (H/H 2 )/Core G/H2 (H/H 2 ) is polycyclic by finite. Then H/H 2 is finite since it is bounded. Hence G/H 2 is a polycyclic by finite group and therefore G/C G (g G ) is a polycyclic by finite group for every element g G. This means that G is a P C group, as required. A group G is said to have finite special rank r if every finitely generated subgroup of G can be generated by r elements and r is the least positive integer with this property (see [13] for details). Lemma 2.8 Let G be a P C group and suppose that G includes an abelian normal subgroup A such that G/A has no proper subgroups of finite index. Then G is abelian. PROOF. Since a P C group with no proper subgroups of finite index has to be divisible abelian, G/A is divisible abelian. It follows that G/A is a direct product of copies of the full rational groups and Prüfer p groups (see, for example, [8, Theorem 22.1]). In other words, G has an ascending series of normal subgroups x G A = A 0 A 1... A α A α+1...a γ = G such that A α+1 /A α = Q + or A α+1 /A α is a Prüfer p group, for every α < γ. Since G is a P C group, G/C G ( a G ) is polycyclic for each element a A. In particular, G/C G ( a G ) is residually finite. On the other hand, A is abelian, so that A C G ( a G ) and hence G/C G ( a G ) is also divisible. All this at once gives that G = C G ( a G ) for every a A, that is A ζ(g). Since A 1 /A is an abelian group of special rank 1, A 1 is likewise abelian. Similarly, A 1 ζ(g). Thus it suffices to apply transfinite induction to obtain that G itself is abelian. Corollary 2.9 Let G be a P C group and suppose that G includes a polycyclic by finite normal subgroup H such that G/H has no proper subgroups of finite index. Then G/ζ(G) is polycyclic by finite. 4

6 PROOF. Put C = C G (H). Since a P C group having no proper subgroups of finite index is abelian, G is soluble by finite. Thus G/C is polycyclic by finite (see, for example, [13, Theorem 3.27]). In particular, G/C is residually finite. Since G/H has no proper subgroups of finite index, CH/H = G/H, that is G = CH. Now C/(C H) = CH/H = G/H, so that C/(C H) has no proper subgroups of finite index. By the election of C, C H ζ(h). By Lemma 2.8, C is abelian. It suffices to apply Lemma 3.1. Lemma 2.10 Let G be a P C group and let A be an abelian normal subgroup of G. If A is not periodic, then A includes a G invariant free abelian subgroup C such that A/C is periodic. PROOF. There exists some a 1 A having infinite order. Put A 1 = a 1 G. By [7, Theorem 2.2], A 1 is finitely generated and it follows that there exists some m N such that C 1 = (A 1 ) m is torsion-free. Clearly C 1 is G invariant. If G/C 1 is periodic, it suffices to take C = C 1. Otherwise, G/C 1 has some element a 2 C 1 of infinite order. Put A 2 /C 1 = a 2 G C 1 /C 1. As above, there exists some t N such that C 2 /C 1 = (A 2 /C 1 ) t is torsion-free. Clearly C 2 is a G invariant free abelian subgroup of A. Thus, it suffices to apply transfinite induction t obtain a G invariant free abelian subgroup C of A such that A/C is periodic, as required. Let A be an abelian group and G Aut(A). Then A is said to be rationally irreducible (more precisely, G rationally irreducible) if for every non-identity G invariant subgroup B of A the factor-group A/B is periodic. If G = g, then we say that g induces on A a rationally irreducible automorphism. Lemma 2.11 Let G be a group, g G and A a g invariant finitely generated torsion-free abelian subgroup. Suppose that A includes a g invariant subgroup C satisfying the following conditions: (i) [C, g] = 1 ; (ii) A/C is not periodic; and (iii) [A/C, g] 1 and g induces on A/C an rationally irreducible automorphism. Then A includes a g invariant subgroup B such that [B, g] 1 and g induces on B as a rationally irreducible automorphism. PROOF. Since A is abelian, the mapping γ : a [a, g], a A is a Z g endomorphism of A such that Ker γ = C A (g) and Im γ = [A, g]. Moreover, Ker γ and Im γ are g invariant subgroups of A. Since A is torsion-free abelian, C A (g) is a pure subgroup of A. Let T/C be the periodic part of A/C. Then A T and clearly T = C A (g). Clearly A/T is torsion-free and g induces on A/T a rationally irreducible automorphism. We have B = [A, g] = Im γ = A/Ker γ = A/T, so that B is a g invariant subgroup of A such that [B, g] 1 and g induces on B a rationally irreducible automorphism. Corollary 2.12 Let G be a group, g G and A a g invariant finitely generated torsion-free abelian subgroup. Suppose that A, g is not nilpotent. Then A includes a g invariant subgroup B such that [B, g] 1 and g induces on B a rationally irreducible automorphism. 5

7 PROOF. Since A has finite rank, A has a series of pure g invariant subgroups 1 = A 0 A 1... A m = A such that g induces a rationally irreducible automorphism on each factor of this series. Since A, g is non-nilpotent, there is a minimal number j such that [A j+1 /A j, g] 1. If j = 0, then it suffices to take B = A j+1. If j > 0, then after finitely many steps and applying Lemma 2.11, we construct a subgroup B as required. 3 Particular cases In this intermediate section we collect some particular cases of the Main Theorem of this paper, which are needed for its proof. Lemma 3.1 Let G be a P C group and suppose that G includes a normal abelian subgroup A such that G/A is finitely generated. Then G/ζ(G) is polycyclic by finite. PROOF. Choose a finite subset S such that G = S A. Put L = S G. By [7, Theorem 2.2], L is polycyclic by finite. In particular, there exists a finite subset M such that L = M and it readily follows that C G (L) = C G (M). Note that C G (L) is normal in G as L is normal in G. Since G is a P C group, G/C G (L) is polycyclic by finite and hence A/C A (L) is finitely generated. Since ζ(g) includes Z = C A (L) and both factor-groups G/A and A/Z are polycyclic by finite, we finally deduce that G/Z is likewise polycyclic by finite. Lemma 3.2 Let G be a group with polycyclic by finite classes of conjugate subgroups. Suppose that G includes a normal torsion-free abelian subgroup A such that G/A is a direct product of finite cyclic subgroups. Then G/ζ(G) is polycyclic by finite. PROOF. By Lemma 3.1, we may assume that G/A is a direct product of infinitely many finite cyclic subgroups. Suppose that the result is false. Let H be a normal subgroup of G such that G/H is polycyclic-by-finite. Suppose that H A ζ(h). In particular, H is nilpotent subgroup and so the set T of all elements of H having finite order is a characteristic subgroup of H. Since A is torsion-free, T A = 1 and so T is abelian. On the other hand, H/T is a torsion-free nilpotent group including an abelian normal subgroup (H A)T/T such that (H/T )/((H A)T/T ) is periodic. Thus H/T is also abelian. Since T A = 1, by Remak s Theorem, we obtain an embedding H H/(H A) H/T, which shows that H is abelian. By Lemma 3.1, G/ζ(G) is polycyclic by finite, a contradiction to our assumption. Consequently, if H is a normal subgroup of G such that G/H is polycyclic by finite, then ζ(h) cannot include H A. In particular, ζ(g) does not include A itself. In other words, there are some elements a 1 A and g 1 G \ A such that [a 1, g 1 ] 1. Put A 1 = a 1 g1 and note that this is a finitely generated subgroup. There is no loss if we assume that g p1 1 C G(A 1 ) for some prime p 1. We choose inside A 1 a g 1 invariant subgroup C 1 of minimal nonzero rank. If [C 1, g 1 ] 1, then we define B 1 = C 1. Suppose that [C 1, g 1 ] = 1. By [6, Corollary 2.8], there is a g 1 invariant subgroup D 1 of A 1 such that C 1 D 1 = 1 and A 1 /C 1 D 1 is finite. It follows that [D 1, g 1 ] 1. Proceeding in the same way, by the finiteness of the rank of A 1, we deduce that there exists a g 1 invariant subgroup B 1 of A 1 such that [B 1, g 1 ] 1 and g 1 induces on B 1 a rationally irreducible automorphism. Put now U 1 = C G ( B 1, g 1 G ). By Corollary 2.7, G is a P C group and hence 6

8 G/U 1 is polycyclic by finite. Since A/(A U 1 ) is finitely generated and G/(A U 1 ) is a P C group, E/(A U 1 ) = C G/(A U1)(A/(A U 1 )) determines a polycyclic by finite factor-group. Then A/(A U 1 ) ζ(e/(a U 1 )) and (E/(A U 1 ))/(A/(A U 1 )) is periodic. Similarly as above, E/(A U 1 ) is abelian. By Lemma 3.1, G/(A U 1 ) is central by (polycyclic by finite). It follows that U 1 includes a G invariant subgroup V 1 A U 1 such that U 1 : V 1 is finite and (U 1 /(A U 1 )) (V 1 /(A U 1 )) = 1. We have already noted that ζ(v 1 ) does not include A V 1 = A U 1. Proceeding in the same way, we choose an element g 2 V 1 \ A and a non-identity g 2 invariant subgroup B 2 V 1 such that g p2 2 C G(B 2 ) for some prime p 2, [g 2, B 2 ] 1 and g 2 acts on B 2 as a rationally irreducible automorphism. Put U 2 = U 1 C G ((B 2 g 2 ) G ). By Corollary 2.7, G is a P C group and therefore G/U 2 is polycyclic by finite ([7, Theorem 2.2] again). Proceeding in this way, we construct a set {g n n N} of elements and a family {B n n N} of subgroups of A satisfying the following conditions: (i) B n is a finitely generated g n invariant subgroup; (ii) gn pn C G (B n ) for some prime p n ; (iii) [g n, B n ] 1 and g n acts on B n as a rationally irreducible automorphism; and (iv) [ g n, B n, g m, B m ] = 1 for every pair n m. Define W = g n, B n n N. By (ii) and (iv), gn pn ζ(w ) for each n N. Therefore, we may replace W by W/ gn pn n N to assume gn pn = 1 for each n N. Then W = BX, where B = B n n N and X = Dr n N g n. By (iii) and (iv), N W (X) = X and Core W (X) = 1. In particular, W/Core W (N W (X)) is not polycyclic by finite, which contradicts Lemma 2.2. This contradiction proves that G/ζ(G) has to be polycyclic by finite, as required. Lemma 3.3 Let G be a torsion-free group with polycyclic by finite classes of conjugate subgroups. Suppose that G/ζ(G) is abelian. Then G/ζ(G) is polycyclic by finite. PROOF. We Note that G/ζ(G) is likewise torsion-free. Suppose that the result is false. In this case, by Lemma 3.1, a normal subgroup H of G such that G/H is finitely generated cannot be abelian. There are some g 1, g 2 G such that [g 1, g 2 ] = c 1 1. If Z = ζ(g), we claim that g 1 Z g 2 Z = 1. Otherwise, if g 1 Z, g 2 Z = yz for some y, there exist k, m Z and z 1, z 2 Z such that g 1 = y k z 1 and g 2 = y m z 2. Thus [g 1, g 2 ] = [y k z 1, y m z 2 ] = [y k, y m ] = 1, a contradiction that shows our claim. Put A 1 = g 1, g 2. Since G/ζ(G) is abelian, the mapping x [g, x], x G is an endomorphism for every g G. It follows that [g, G] = G/C G (g) and, in particular, G/C G (g) is a torsion-free group. Put C 1 = C G (A 1 ) = C G (g 1 ) C G (g 2 ) so that G/C 1 is a torsion-free finitely generated group, because G is a P C group by Corollary 2.7. Hence C 1 cannot be abelian. Then there are some g 3, g 4 C 1 such that [g 3, g 4 ] = c 2 1. As above g 3 Z g 4 Z = 1, moreover g 1 Z, g 2 Z g 3 Z, g 4 Z = 1. Put A 2 = g 3, g 4 and C 2 = C G ( A 1, A 2 ) = C G (g 1 ) C G (g 2 ) C G (g 3 ) C G (g 4 ) so that G/C 2 is a torsion-free finitely generated group. Proceeding in the same way, we construct a set {g n n N} of elements and a family {B n n N} of subgroups of A satisfying the following conditions: (i) [g 2n+1, g 2n+2 ] = c n 1; (ii) g 2n+1 Z, g 2n+2 Z g 2k+1 Z, g 2k+2 Z k n = 1 ; and (iii) [ g 2n+1, g 2n+2, g 2k+1, g 2k+2 ] = 1 if k n. Define W = g n n N and U = g 2n+1 n N. Then V = ζ(w ) = c n n N and W/V = Dr n N g n V. Clearly N W (U) = UV is normal in W, so that 7

9 W/Core W (N W (U)) = W/UV = Dr n N g 2n+2 V is not finitely generated, what contradicts Lemma 2.2. This contradiction proves that G/ζ(G) is polycyclic by finite, as required. Corollary 3.4 Let G be a group with polycyclic by finite classes of conjugate subgroups. Suppose that G/ζ(G) is abelian. Then G/ζ(G) is polycyclic-by-finite. PROOF. Let T be the periodic part of G. Then G/T is torsion-free and, by Lemma 3.3, (G/T )/ζ(g/t ) is polycyclic by finite. Put U/T = ζ(g/t ). By Lemma 2.3, T is central by finite. By a theorem due to Schur (see, for example, [13, Theorem 4.12]), V = [T, T ] is finite. Consequently U/V includes an abelian normal subgroup T/V such that (U/V )/(T/V ) is torsion-free abelian. If u U, then (U/V )/C U/V (uv ) = [uv, U/V ] T/V and, in particular, (U/V )/C U/V (uv ) is periodic finitely generated, i.e this factor-group is finite. In other words, U/V is an F C group. Since V is finite, U is likewise an F C group. By Corollary 2.4, U/ζ(U) is finite. Put Z = ζ(u). Since G/U is polycyclic, so is and G/Z. Now it suffices to apply Lemma 3.1. Corollary 3.5 Let G be a group with polycyclic by finite classes of conjugate subgroups. Suppose that [G, G] is polycyclic by finite. Then G/ζ(G) is polycyclic by finite. PROOF. Put K = [G, G] and C = C G (K). Since G is soluble by finite, G/C is polycyclic by finite (see, for example, [13, Theorem 3.27]). We have C K ζ(c) and, by Corollary 3.4, C/ζ(C) is polycyclic. Put Z = ζ(c). Thus Z is an abelian normal subgroup of G such that G/C is polycyclic by finite. Now it suffices to apply Lemma 3.1. Corollary 3.6 Let G be a group with polycyclic by finite classes of conjugate subgroups. Suppose that G includes a torsion-free abelian normal subgroup A such that G/A is a periodic abelian group. Then G/ζ(G) is polycyclic by finite. PROOF. Since G/A is periodic abelian, G/A = Dr q Π(G/A) S q /A, where S q /A is the Sylow q subgroup of G/A. Choose in every subgroup S q /A a q basic subgroup B q /A; the existence of B q /A is ensured by [8, Theorem 32.3]). Put B/A = Dr q Π(G/A) B q /A. Then B/A is a direct product of cyclic subgroups and G/B is divisible. By Lemma 3.2, B/ζ(B) is polycyclic by finite. Then K = [B, B] is polycyclic by finite (see, for example, [13, pp 115]). By Corollary 2.7 and Lemma 2.8 at once, G/K is abelian. It suffices to apply Corollary 3.5. Corollary 3.7 Let G be a group with polycyclic by finite classes of conjugate subgroups. Suppose that G includes an abelian normal subgroup A such that G/A is periodic abelian. Then G/ζ(G) is polycyclic by finite. PROOF. Let T be the periodic part of A. Then, by Corollary 3.6, (G/T )/ζ(g/t ) is polycyclic by finite and, by Lemma 3.3, A includes a G invariant free abelian subgroup C such that A/C is periodic. Since G/A is periodic, G/C is periodic. By Lemma 2.3, (G/C)/ζ(G/C) is finite. Since C T = 1, we may embed G G/C G/T and so G/ζ(G) is polycyclic by finite, as required. Corollary 3.8 Let G be a group with polycyclic by finite classes of conjugate subgroups. Suppose that G includes an abelian normal subgroup A such that G/A is an abelian group of finite special rank. Then G/ζ(G) is polycyclic by finite. PROOF. Pick a subgroup B of G such that B A, B/A is free abelian and G/B is periodic. By Corollary 3.6, (G/T )/ζ(g/t ) is polycyclic by finite. By Lemma 2.10, 8

10 A includes a G invariant free abelian subgroup C such that A/C is periodic. Since G/A is periodic, G/C is also periodic. By Lemma 2.3, (G/C)/ζ(G/C) is finite. Since C T = 1, we have the embedding G G/C G/T, from which we deduce that G/ζ(G) is polycyclic by finite. Corollary 3.9 Let G be a group with polycyclic by finite classes of conjugate subgroups. Suppose that G includes an abelian normal subgroup A such that G/A is a soluble group of finite special rank. Then G/ζ(G) is polycyclic-by-finite. PROOF. G has a series of normal subgroups A = G 0 G 1.. G k = G such that every factor G j+1 /G j is an abelian group of finite special rank. We proceed by induction on k. If k = 1, the assertion follows from Corollary 3.8. Let k > 1 and suppose that we have already proved that G k 1 /ζ(g k 1 ) is polycyclic by finite. Put H = G k 1. Then L = [H, H] is polycyclic by finite (see, for example, [13, pp 115]). Put Z/L = ζ(g/l). By Corollary 3.8, (G/L)/(Z/L) is polycyclic by finite. By Corollary 3.5, Z/ζ(Z) is also polycyclic by finite, so that G/ζ(Z) is likewise polycyclic by finite. Now it suffices to apply Lemma 3.1. Lemma 3.10 Let G be a group with polycyclic by finite classes of conjugate subgroups. Suppose that G includes a torsion-free abelian normal subgroup A such that G/A is torsion-free abelian. Then G/ζ(G) is polycyclic by finite. PROOF. By Corollary 3.8, there is no loss if we assume that G/A has infinite (special) rank. Suppose that the result is false. Let H be a normal subgroup of G such that G/H has finite special rank and H A ζ(h). In particular, H is nilpotent. By Lemma 3.3, H/ζ(H) is polycyclic by finite and then G/ζ(H) is a soluble group of finite special rank. By Corollary 3.9, G/ζ(G) is polycyclic by finite, what contradicts our assumption. Consequently, if H is a normal subgroup of G such that G/H has finite special rank, then ζ(h) does not include H A. In particular, ζ(g) does not include A. In other words, there are some a 1 A, g 1 G \ A such that a 1, g 1 is not nilpotent. Put A 1 = a 1 g1 so that A 1 is finitely generated. Thus, A 1, g 1 is not nilpotent. By Corollary 2.12, A 1 includes a g 1 invariant subgroup B 1 such that [B 1, g 1 ] 1 and g 1 induces on B 1 a rationally irreducible automorphism. Now put U 1 = C G ( B 1, g 1 G ). By Corollary 2.7, G is a P C group and hence G/U 1 is polycyclic by finite. Since A/(A U 1 ) is finitely generated and G/(A U 1 ) is a P C group, E/(A U 1 ) = C G/(A U1)(A/(A U 1 )) determines a polycyclic by finite factor-group. Then A/(A U 1 ) ζ(e/(a U 1 )) and (E/(A U 1 ))/(A/(A U1 ) ) is abelian. By Corollary 3.5, G/(A U 1 ) is central by (polycyclic by finite). It follows that U 1 includes a G invariant subgroup V 1 A U 1 such that U 1 /V 1 has finite special rank and (U 1 /(A U 1 )) (V 1 /(A U 1 )) = 1. Then G/V 1 has finite special rank. We have already noted that in this case ζ(v 1 ) does not include A V 1 = A U 1. Iterating the arguments, there exist g 2 V 1 \ A and a non-identity g 2 invariant subgroup B 2 V 1 such that [g 2, B 2 ] 1 and g 2 acts on B 2 as a rationally irreducible automorphism. Put U 2 = U 1 C G ((B 2 g 2 ) G ). By Corollary 2.7, G is a P C group and therefore G/U 2 is polycyclic by finite ([7, Theorem 2.2]). Proceeding in this way, we construct a set {g n n N} of elements and a family {B n n N} of subgroups of A satisfying the following conditions: (i) B n is a finitely generated g n invariant subgroup; 9

11 (ii) [g n, B n ] 1 and g n acts on B n as a rationally irreducible automorphism; and (iii) [ g n, B n, g j, B j ] = 1 provided n j. Put W = g n, B n n N. Replacing (if necessary) W by W/ z n n N, where z n = g n C G (B n ), we may assume that g n C G (B n ) = 1 for each n N. Then W = BX, where B = B n n N and X = Dr n N g n. By (ii) and (iii), N W (X) = X and Core W (X) = 1. In particular, W/Core W (N W (X)) is not polycyclic by finite, what contradicts Lemma 2.2. This contradiction shows that G/ζ(G) is polycyclic by finite, as required. Corollary 3.11 Let G be a group with polycyclic by finite classes of conjugate subgroups. Suppose that G includes a normal torsion-free abelian subgroup A such that G/A is abelian. Then G/ζ(G) is polycyclic by finite. PROOF. Let B be a normal subgroup of G such that B A, B/A is free abelian and G/B is periodic. By Lemma 3.10, B/ζ(B) is polycyclic by finite. Then K = [B, B] is polycyclic by finite (see, for example, [13, pp 115]). Put Z/K = ζ(g/k). By Corollary 3.7, (G/K)/(Z/K) is polycyclic by finite. By Corollary 3.5, Z/ζ(Z) is also polycyclic by finite, so that G/ζ(Z) is likewise polycyclic by finite. Now it suffices to apply Lemma 3.1. Lemma 3.12 Let G be a group with polycyclic by finite classes of conjugate subgroups. Suppose that G includes a periodic abelian normal subgroup A such that G/A is torsion-free abelian. Then G/ζ(G) is polycyclic by finite. PROOF. By Corollary 3.8, there is no loss if we assume that G/A has infinite (special) rank. Suppose that the result is false. Let H be a normal subgroup of G such that G/H has finite special rank and H A ζ(h). In particular, H is nilpotent. By Corollary 3.4, H/ζ(H) is polycyclic by finite and then G/ζ(H) is a soluble group of finite special rank. By Corollary 3.9, G/ζ(G) is polycyclic by finite, what contradicts our assumption. Consequently, if H is a normal subgroup of G such that G/H has finite special rank, then ζ(h) does not include H A. In particular, ζ(g) does not include A. In other words, there are a 1 A and g 1 G \ A such that a 1, g 1 is not nilpotent. Put A 1 = a 1 g1 so that A 1 is a finite subgroup. Thus A 1, g 1 is not nilpotent. Therefore A 1 = B 1 C 1, where both subgroups B 1 and C 1 are g 1 invariant, C 1, g 1 is nilpotent and [B 1, g 1 ] = B 1 ([17, Theorem 1 ]). Moreover, we may suppose that B 1 is a minimal g 1 invariant subgroup. Now we put U 1 = C G ( B 1, g 1 G ). By Corollary 2.7, G is a P C group and hence G/U 1 is polycyclic by finite. Since A/(A U 1 ) is finitely generated and G/(A U 1 ) is a P C group, E/(A U 1 ) = C G/(A U1)(A/(A U 1 )) determines a polycyclic by finite factor-group. Then A/(A U 1 ) ζ(e/(a U 1 )) and (E/(A U 1 ))/(A/(A U 1 )) is abelian. By Corollary 3.5, G/(A U 1 ) is central by (polycyclic by finite). It follows that U 1 includes a G invariant subgroup V 1 A U 1 such that U 1 /V 1 has finite special rank and (U 1 /(A U 1 )) (V 1 /(A U 1 )) = 1. Then G/V 1 has finite special rank We have already noted that in this case ζ(v 1 ) does not include A V 1 = A U 1. Iterating the arguments, we choose g 2 V 1 \ A and a non-identity g 2 invariant subgroup B 2 V 1 such that B 2 is a minimal g 2 invariant subgroup and [g 2, B 2 ] = B 2. Put U 2 = U 1 C G ((B 2 g 2 ) G ). By Corollary 2.7, G is a P C group and therefore G/U 2 is polycyclic by finite ([7, Theorem 2.2]). Proceeding in the same way, we construct a set {g n n N} of elements and a family {B n n N} of subgroups of A satisfying the following conditions: 10

12 (i) B n is a finite minimal g n invariant subgroup; (ii) [g n, B n ] = B n ; and (iii) [ g n, B n, g j, B j ] = 1 provided n j. Put W = g n, B n n N. Then W is an F C group. By (ii) and (iii), W/ζ(W ) is infinite, what contradicts Lemma 2.2 and Corollary 2.4. This contradiction shows that G/ζ(G) is polycyclic by finite, as required. Corollary 3.13 Let G be a group with polycyclic by finite classes of conjugate subgroups. Suppose that G includes a periodic abelian normal subgroup A such that G/A is abelian. Then G/ζ(G) is polycyclic by finite. PROOF. Let T/A be the periodic part of G/A. Then T is also periodic and by Lemma 2.2 and Lemma 2.3 éat once, T/ζ(T ) is finite. By a theorem due to Schur (see, for example [13, Theorem 4.12]), K = [B, B] is finite. Put Z/K = ζ(g/k). By Lemma 3.12, (G/K)/(Z/K) is polycyclic by finite. By Corollary 3.5, Z/ζ(Z) is also polycyclic by finite, so that G/ζ(Z) is likewise polycyclic by finite. It suffices to apply Lemma 3.1. Corollary 3.14 Let G be a group with polycyclic by finite classes of conjugate subgroups. Suppose that G includes an abelian normal subgroup A such that G/A is abelian. Then G/ζ(G) is polycyclic by finite. PROOF. Let T be the periodic part of A. Then T is G invariant. By Corollary 2.7, G is a P C group. By Lemma 2.10, A includes a G invariant free abelian subgroup U such that A/U is periodic. By Corollary 3.11, (G/T )/ζ(g/t ) is polycyclic by finite. By Corollary 3.13, (G/U)/ζ(G/U) is polycyclic by finite. Since U T = 1, by Remak s Theorem, G G/C G/T, which gives that G/ζ(G) is polycyclic by finite, as required. Corollary 3.15 Let G be a group with polycyclic by finite classes of conjugate subgroups. If G is soluble, then G/ζ(G) is polycyclic by finite. PROOF. Let 1 = G 0 G 1... G m = G be a series of normal subgroups with abelian factors. We proceed by induction on m. If m = 2, the result follows from Corollary Let m > 1 and suppose that we have already proved that G m 1 /ζ(g m 1 ) is polycyclic by finite. Put H = G m 1. Then L = [H, H] is polycyclic by finite (see, for example, [13, pp 115]). Put Z/L = ζ(g/l). By Corollary 3.14, (G/L)/(Z/L) is polycyclic by finite. By Corollary 3.5, Z/ζ(Z) is also polycyclic by finite, so that G/ζ(Z) is likewise polycyclic by finite. Now it suffices to apply Lemma Towards the Proof of Main Theorem Before dealing with the proof of Main Theorem, we need to show a couple of results. Lemma 4.1 Let G be a group with polycyclic by finite classes of conjugate subgroups. Suppose that A is a maximal abelian normal subgroup of G. Then every soluble normal subgroup of G/A is polycyclic. 11

13 PROOF. Let S/A be a normal soluble subgroup of G/A and suppose that S/A is not polycyclic. By Corollary 3.15, S/ζ(S) is polycyclic. Since S is normal in G, ζ(s) is likewise normal, so that Aζ(S) is normal in G. Since Aζ(S) is abelian and includes A, A = Aζ(S), that is ζ(s) A. Since S/ζ(S) is polycyclic, S/A must be polycyclic, and we obtain a contradiction. This contradiction shows that S/A is polycyclic, as required. Lemma 4.2 Let G be a P C group and suppose that T is a periodic locally soluble normal subgroup of G. If T is infinite, then T includes a G invariant infinite abelian or metabelian subgroup. PROOF. Put Z = ζ(t ). If Z is infinite, we are done. Assume that Z is finite. If x T, by [7, Theorem 2.2], x G is polycyclic and hence finite. It follows that T/C T ( x G ) is also finite. Clearly x T C T ( x G ) = Z, and, by Remak s theorem, T/Z x T T/C T ( x G ). In other words, T has a family of G invariant subgroups {T λ λ Λ} such that T/T λ is finite for every λ Λ and λ Λ T λ = Z. Consider now the factor-group Q = G/Z. Put P = T/Z and pick 1 y 1 P. Then Y 1 = y 1 Q is a finite soluble subgroup and therefore it includes a Q invariant finite non-identity abelian subgroup U 1. There exists a Q invariant subgroup V 1 of P such that U 1 V 1 = 1 and P : V 1 is finite. Let 1 y 2 V 1 and put Y 2 = y 2 Q. Then again Y 2 includes a Q invariant finite non-identity abelian subgroup U 2. Since U 2 V 1, U 1 U 2 = 1 and U 1 U 2 = U 1 U 2. In particular U 1 U 2 is abelian. Again there is a Q invariant subgroup V 2 of P such that U 1 U 2 V 2 = 1 and P : V 2 is finite. Proceeding in this way, we construct inside of P an infinite Q invariant a subgroup U. If W the preimage of U in G, then W is an infinite G invariant metabelian subgroup of G. Lemma 4.3 Let G be a P C group and suppose that L is a torsion-free locally nilpotent normal subgroup of G. If L is not polycyclic, then L includes a G invariant abelian or metabelian subgroup, which is not polycyclic. PROOF. Put Z = ζ(t ). If Z is not polycyclic, we are done. Therefore we may suppose that Z is polycyclic. If x L, by [7, Theorem 2.2], x G is polycyclic. Furthermore G/C G ( x G ) is also polycyclic, so that L/C L ( x G ) is likewise polycyclic. We note that the centralizers of arbitrary subsets in a locally nilpotent torsion-free group are pure subgroups. It follows that L/C L ( x G ) is a torsion-free finitely generated nilpotent group. Clearly x L C L( x G ) = Z, and by Remak s theorem L/Z x L L/C L ( x G ). In other words, L has a family of subgroups {L λ λ Λ} satisfying the following conditions: (i) every L λ is G invariant; (ii) every factor-group L/L λ is torsion-free finitely generated; and (iii) λ Λ L λ = Z. We consider the factor-group Q = G/Z. Put K = L/Z and pick 1 y 1 K. By [7, Theorem 2.2], Y 1 = y 1 Q is a polycyclic subgroup. Choose in Y 1 a Q invariant abelian subgroup U 1 of minimal non-zero rank. If H is a normal subgroup of Q, then 12

14 either H U 1 = 1 or U 1 /(H U 1 ) is finite. Let 1 u 1 U 1. By (i)-(ii)-(iii), there exists a Q invariant subgroup V 1 of K such that K/V 1 is torsion-free and u 1 V 1. If U 1 V 1 1, then U 1 /(U 1 V 1 ) is finite. On the other hand, U 1 /(U 1 V 1 ) = U 1 V 1 /V 1, a contradiction since K/V 1 is torsion-free. Hence U 1 V 1 = 1. Pick 1 y 2 V 1 and put Y 2 = y 2 Q. As above, Y 2 includes a Q invariant abelian subgroup U 2 of minimal non-zero rank. Since U 2 V 1, U 1 U 2 = 1 and U 1 U 2 = U 1 U 2. In particular, U 1 U 2 is abelian. One more time, there exists a Q invariant subgroup V 2 of V 1 such that V 1 /V 2 is torsion-free and U 2 V 2 = 1. In particular, U 1 U 2 V 2 = 1. Proceeding in the same way, we construct inside of K an abelian Q invariant subgroup U, which is not polycyclic. If W is the preimage of U in G, then W is a G invariant metabelian subgroup of G, which is not polycyclic. PROOF OF MAIN THEOREM. Let A be a maximal abelian normal subgroup of G. Denote by L/A the locally nilpotent radical of G/A. Let T/A be the periodic part of L/A. By Corollary 2.7, G is a P C group. Suppose that T/A is infinite. By Lemma 4.2, T includes a G invariant subgroup S A such that S/A is abelian or metabelian and infinite. However this contradicts Lemma 4.1. Hence T/A is finite. The factor-group L/T is torsion-free. Suppose that L/T is not polycyclic. By Lemma 4.3, L includes a G invariant subgroup D T such that D/T is abelian or metabelian and not polycyclic. Then D/A is a soluble normal subgroup of G/A, which is not polycyclic. But this contradicts Lemma 4.1. This contradiction shows that L/T is polycyclic and hence L/A is polycyclic. Let R/A be a maximal locally soluble normal subgroup of G/A; the existence of R/A is ensured since G is a P C group. Moreover, R/A is hyperabelian. By a result due to B.I.Plotkin (see, for example, [13, Theorem 3.31]), R/A is also polycyclic. Put Q = G/R so that Q includes no non-identity abelian normal subgroups. Suppose that Q is infinite. Pick 1 x 1 Q and put X 1 = x 1 Q. By [7, Theorem 2.2], X 1 is polycyclic by finite, and, by the election of Q, X 1 is finite. Put C 1 = C Q (X 1 ). Since Q : C 1 is finite, C 1 is infinite. Our assumption on Q yields that X 1 C 1 = 1. Pick 1 x 2 C 1 and put X 2 = x 2 Q. Then X 2 is a finite normal subgroup and X 1, X 2 = X 1 X 2. Proceeding in the same way, we construct an infinite family {X n n N} of finite normal subgroups such that X = X n n N = Dr n N X n. By Lemma 2.2 and Lemma 2.3, X is central by finite. Since X is infinite, ζ(x) is a non-identity abelian normal subgroup of Q, a contradiction. Hence Q is finite and so G is soluble by finite. By Corollary 3.15 and Lemma 3.1 at once, we obtain that G/ζ(G) is polycyclic by finite. References [1] J. Brendle. Set-theoretic aspects of periodic F C groups extraspecial groups and Kuteva trees. J. Algebra 162 (1993), [2] I. I. Eremin. The groups with finite classes of conjugate abelian subgroups. Mat. Sbornik 47 (1959), [3] I. I. Eremin. The groups with finite classes of conjugate infinite subgroups. Uch. Zap. Perm University 17 (1960), n o 2, [4] V. Faber. Large abelian subgroups of some infinite groups. Rocky Mountain J. Math. 1 (1971),

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