Solvable groups with many BFC-subgroups

Publicacions Matem`atiques, Vol. 44 (2000), 491–501 SOLVABLE GROUPS WITH MANY BF C-SUBGROUPS O. D. Artemovych Abstract We characterize the solvable groups without infinite properly ascending chains of non-BF C subgroups and prove that a non-BFC group with a descending chain whose factors are finite or abelian is a ˇ Cernikov group or has an infinite properly descending chain of non-BFC subgroups. 0. Introduction In a series of papers Belyaev-Sesekin [2], Belyaev [3], Bruno-Phillips [4], [5], Kuzucuo˘glu-Phillips [12], Leinen-Puglisi [13], Asar [1], Leinen [14] have obtained the results on mimimal non-FC groups. In particular, in [2] are characterized the minimal non-BFC groups, i.e. the non-BFC groups in which every proper subgroup is BFC. Recall that a group Gis called a BFC-group if there is a positive integer dsuch that no element of Ghas more than dconjugates. Due to the well known result of B. H. Neumann (see e.g. [16, Theorem 4.35]) the BFC-groups are precisely the groups with the finite commutator subgroups. We say that a group Gsatisfies the minimal condition on non-BFC subgroups (for short Min-BFC) if for every properly descending series {Gn|n∈N}of subgroups of Gthere exists a number n0∈Nsuch that Gnis a BFC-group for every integer n≥n0and a group Gsatisfies maximal condition on non-BFC subgroups (for short Max-BFC) if there exists no infinite properly ascending series of non-BFC subgroups in G. Every minimal non-BFC group satisfies Min-BFC and Max-BFC. S. Franciosi, F. de Giovanni and Ya. P. Sysak [11] have investigated the locally graded groups with the minimal condition on non-FC subgroups. In this paper we characterize the solvable groups satisfying Max-BFC and Min-BFC, respectively. Namely, we prove the two following theorems. 2000 Mathematics Subject Classification. 20E15, 20F16, 20F24. Key words. BFC-group, minimal non-BFC group, maximal condition, minimal condition, solvable group. 492 O. D. Artemovych Theorem 1. A solvable group Gsatisfies Max-BFC if and only if it is of one of the following types: (i) Gis a BFC-group; (ii) G=BU is a finitely generated group, where Bis a proper torsion normal subgroup of G,Uits polycyclic subgroup and Bxis either a BFC-subgroup or a finitely generated subgroup for every element xof U; (iii) G=DU is a locally nilpotent-by-finite group with the torsion commutator subgroup G, where Dis a normal divisible abelian p-subgroup, Uis a polycyclic subgroup, and if uacts non-trivially on Dfor an element uof U, then Dis an indecomposable injective Qu-module and Auis a BFC-subgroup for every proper submodule Aof a Zu-module Dwith the action induced by the conjugation of uon D. Theorem 2. Let the group Ghave a descending series whose factors are finite or abelian. If Gsatisfies the minimal condition on non-BFC subgroups, then it is a BFC-group or a ˇ Cernikov group. Throughout this paper pis a prime. For a group G,Z(G) will always denote the centre of G,G,G ,... ,G (n)the terms of derived series of G,τ(G) the set of all torsion elements of G,Gp=gp|g∈G. In the sequel we will use the following notation: Qthe rational number field; Fpthe finite field with pelements; Qpthe additive group of all rational numbers whose denominators are p-numbers; Zthe additive group of all rational integers; Cp∞the quasicyclic p-group; Rxthe group ring of a cyclic group xover a commutative ring R. We will also use other standard terminology from [10] and [16]. 1. Solvable groups with Max-BF C In this section we study the solvable groups with the maximal condition on non-BFC subgroups. Groups with Many BFC-Subgroups 493 Lemma 1.1. Let Gbe a group satisfying Max-BFC and Hits subgroup. Then: (i) Hsatisfies Max-BFC; (ii) if His normal in G, then the quotient group G/H satisfies MaxBFC; (iii) if His a normal non-BFC subgroup of G, then G/H satisfies the maximal condition on subgroups. Proof: Is immediate. Lemma 1.2. Let Gbe a group which satisfies Max-BFC. If Gcontains a normal abelian subgroup Nwith the quasicyclic quotient group G/N, then Gis a nilpotent group. Proof: We prove this lemma by the same arguments as in the proof of Lemma 2.3 from [2]. Since G/N is a quasicyclic p-group for some prime p, G/N =∞ n=1an, where anp=an−1,a0=N. Put An=N,an. Then AnG,AnG and by Lemma 1.1(iii) Anis a BFC-subgroup. Hence An≤Z(G) and consequently G=∞ n=1An≤Z(G), as desired. Lemma 1.3. If Gis a ˇ Cernikov group with Max-BFC, then it is a BFC-group or the quotient group G/Gis finite. Proof: Assume that the quotient group G=G/Gis infinite and Gis not a BFC-group. Then by Theorem 21.3 of [10]G=D×Fis a direct product of the non-trivial divisible part Dand a reducible subgroup F. Let Dand Fbe the inverse images of Dand Fin G, respectively. By Corollary 2.2 of [2]G=DF. Since Gis not a BFC-group, Fis a finite group. It is clear that D∼ =Cp∞for some prime pand Ghas a normal BFC-subgroup Nwith G/N ∼ =Cp∞. By Theorem 1.16 of [7] G=NZ(G) and so G=N, a contradiction with our assumption. The lemma is proved. Proposition 1.4. If a group Gsatisfies Max-BFC, then it is a BFC-group or the quotient group G/Gis finitely generated. 494 O. D. Artemovych Proof: As it is well known G=G/G=D×Sis a direct product of the divisible part D=D/Gand a reducible subgroup S=S/G. (1) First, let Dbe a non-trivial subgroup. Then Sand Gare the BFC-subgroups. It is clear that Gis a BFC-group or Dis a quasicyclic group. We suppose that D∼ =Cp∞. Let F=F/Gbeap-basic subgroup of S.IfF=S, then G/F is a direct product of a quasicyclic p-subgroup and an infinite p-divisible abelian subgroup. By Lemma 2.2 of [2] and Lemma 1.1 Gis a BFC-group. Assume that F=S. Then by Lemma 26.1 and Proposition 27.1 from [10]G/Fp=D∗×F∗is a direct product of a quasicyclic p-subgroup D∗ and a p-subgroup F∗of exponent p. Lemma 2.2 of [2] implies G=DS. If Fis not a finitely generated subgroup, then in view of Lemma 1.1 D and Gare the BFC-groups. Therefore we assume that Fis a finitely generated subgroup. Since Fis a BFC-subgroup, |G:D|<∞.By Lemma 1.2 D/G is a nilpotent group and so D/Dis a ˇ Cernikov group. This yields that Dis a ˇ Cernikov group. By Lemmas 1.1 and 1.3 Dis a BFC-group and as a consequence Gis the ones. (2) Now let the divisible part Dis trivial. If F=S, then the quotient group G/Gis finitely generated or G/Fpis a direct product of infinitely many cyclic subgroups of order pin which case Gis a BFC-group. Therefore we assume that F=S.IfFis not finitely generated, in the same manner as above we can prove that Gis a BFC-group. Let Fbe a finitely generated subgroup. (a) Assume that the quotient group G1=G/Fis non-torsion. Then there exists a subgroup F0such that F≤F0≤Gand G/F0is torsionfree. As noted in [6] (see also [7, Chapter 2, §6]) G/F0contains a subgroup T/F0isomorphic to Qp.IfZ/F0is a subgroup of T/F0isomorphic to Z, then T/Zis a quasicyclic p-group, and it follows that Ghas a normal BFC-subgroup Xwith G/X ∼ =Cp∞. By Lemma 1.2 G0=G/XFp is a nilpotent group and so by Lemma 26.1 and Proposition 27.1 from [10]G0/G 0=F1×K1is a direct product of a finite p-subgroup F1and an infinite p-divisible abelian subgroup K1. Let K0be an inverse image of K1in G0. From what is proved above it follows that K0has a normal subgroup K∗with K0/K∗∼ =Cp∞.IfK∗=(K∗)p, then Theorem 1.16 of [7] yields that K0/(K∗)p=X×Yis a direct product of a quasicyclic p-subgroup Xand some divisible p-subgroup Y. Since Yis a non-trivial subgroup, it is infinite. Consequently Gis a BFC-group. Therefore we suppose that K∗=(K∗)p. As above we can prove that K∗contains a G-invariant subgroup Lwith K∗/L ∼ =Cp∞. Hence K0/L ∼ =Cp∞×Cp∞ and so Gis a BFC-group. Groups with Many BFC-Subgroups 495 (b) Let G1=G/Fbe an infinite torsion p-group. Then without loss of generality we can assume that G1is an infinite q-group for some prime qdifferent from p.ByBwe denote a basic subgroup of G1.If B=G1, then the quotient group G/Gis finitely generated or Bis an infinitely generated subgroup in which case Gis a BFC-group. Let B=G1.IfBis not a finitely generated subgroup, then Lemma 26.1 and Proposition 27.1 of [10] give that G1/Bq∼ =B×Cq∞, where Bis an infinite abelian q-subgroup of exponent q, and this yields that Gis a BFC-group. Therefore we assume that Bis a finitely generated subgroup. Then without loss of generality let B= 1 and G1∼ =Cq∞.We would like to prove that the commutator subgroup Gis torsion. Since the subgroup G is finite, without restricting of generality let G =1. But then ˆ F=F/τ(G) is an abelian subgroup of ˆ G=G/τ(G) and from G1∼ =ˆ G/ ˆ Fit follows that ˆ Gis an abelian group. This means that G is a torsion subgroup. By Lemma 1.2 G/Fis a nilpotent group and it has the torsion commutator subgroup. So Corollary 3.3 of [2] yields that G/Fis a torsion group. Hence Gis a torsion group and G∼ =Cq∞×M, where Mis a finite subgroup, a contradiction with our assumption. (c) Finally, if G1=G/Fis a torsion group and it has a non-trivial p-subgroup, then without loss of generality we can assume that G1is a quasicyclic p-group. As in the line (b) this gives that Gis a BFC-group. The proposition is proved. Lemma 1.5. Let G=Bxbe a product of a normal abelian torsionfree subgroup Band a cyclic subgroup x. If Gsatisfies Max-BFC, then it is either an abelian group or a polycyclic group. Proof: If Fis any finitely generated subgroup of B, then F, xis a polycyclic subgroup in Gand F, x=Axfor some G-invariant subgroup Aof B. Assume that the quotient group G/A is not finitely genereted. Then Axis a BFC-subgroup in view of Lemma 1.1 and consequently it is abelian. Therefore a non-polycyclic group Gis abelian, as desired. Lemma 1.6. If Gis a solvable group satisfying Max-BFC, then one of the following conditions holds: (i) G=BU is a finitely generated group, where Bis a proper torsion normal subgroup of G,Uits polycyclic subgroup and Bxis either a BFC-subgroup or a finitely generated subgroup for every element xof U; (ii) Gis a BFC-group; 496 O. D. Artemovych (iii) G=DV is a product of a normal divisible abelian p-subgroup D and a polycyclic subgroup V. Proof: Suppose that Gis not a BFC-group. Let nbe the derived length of G. Then there exists an integer ksuch that G(k−1) is not a BFC-group, but G(k)is a BFC-group, where 1 ≤k≤n−1 and G(0) = G. Proposition 1.4 implies that G(k−1) =G(k)Ufor some polycyclic subgroup U. By Lemma 1.5 UG(k−1), where G(k−1) =G(k−1)/τ(G(k))= G(k)U, and so U=G(k−1). This means that G(k−1) =τ(G(k))U.We denote τ(G(k))byB. (a) First we assume that Gis not a finitely generated group. Clearly that there is an element uof Usuch that H1=G(k)uis a non-BFC group. We would like to prove that H=Buis the ones. Indeed, if His aBFC-group, then the quotient group H1/HG(k+1) is a nilpotent group and by Theorem 2.26 of [10] and Proposition 1.4 it is finitely generated. But then H1(and consequently G) is also a finitely generated group, a contradiction. Hence His a non-BFC group. (1) Assume that Bis an abelian π-subgroup for some set πof primes. If B=B1×B2is a direct product of an infinite π1-subgroup B1and an infinite π2-subgroup B2, where π1and π2are the disjoint subsets of πsuch that π=π1∪π2, then it is not difficulty to prove that His a BFC-group, a contradiction. Thus πis a finite set and B=P×S, where Pis an infinite p-subgroup for some prime p∈πand Sis a finite p-subgroup. Moreover Puis a non-BFC group. (2) If Bis not necessary an abelian subgroup, then from the line (1) it follows that B/T is a divisible abelian p-group for some finite H-invariant subgroup T. By Theorem 1.16 of [7] there exists a divisible abelian p-subgroup Dof Bsuch that D≤Z(B) and B=DT.ThusG=DV , where Vis a polycyclic subgroup. (b) Now let Gbe a finitely generated group. Then G=BU for some polycyclic subgroup U. Suppose that Bxis not a BFC-group for some x∈U.IfBxis not finitely generated, then, as in the line (1) and (2), we can prove that Bx=D1V1, where D1is a normal divisible p-subgroup, V1is a polycyclic subgroup and D1≤B. By Theorem of [2]Bxcontains a proper non-BFC subgroup K. Since D1K= D1K/(D1∩K)=D1Kand D1is a non-trivial divisible p-subgroup, we conclude that D1K(and consequently G) contains an infinite properly ascending series of type K<K 1<···<K n<··· , Groups with Many BFC-Subgroups 497 a contradiction. This means that Bxis a finitely generated subgroup. The lemma is proved. Example 1.7. If G=At, where tis an infinite cyclic subgroup, A∼ =Cp∞and at=a1+p(a∈A), then Gsatisfies Max-BFC. If Dis a commutative Dedekind domain, Aright D-module, Spec(D) the set of non-trivial prime ideals of Dand P∈Spec(D), then AP={a∈A|aPn={0}for some positive integer n=n(a)∈N} is said to be the P-component of A, and Ais said to be a D-torsion module if A={a∈A|Ann(a)={0}}. Lemma 1.8. Let G=Axbe a semidirect product of a normal abelian subgroup Aof exponent pand an infinite cyclic subgroup x. If Gsatisfies Max-BFC, then it is either a finitely generated group or a BFC-group. Proof: It is clear that Ais a right Fpx-module with the action determined by the conjugation of xon A. Assume that Gis not neither a finitely generated group nor a BFC-group. Then Ais a Fpx-torsion module and by Proposition 2.4 of [8,§8.2] A=⊕ P∈Spec(Fpx)AP is a module direct sum of its P-component AP. Without loss of generality we can suppose that |A:AQ|<∞for some Q∈Spec(Fpx). Let Bbe a basic submodule of AQ. By our hypothesis B=AQ. Since Bcan be written as a direct product of two infinite G-invariant subgroup of infinite index, we obtain that Bx(and consequently G)is aBFC-group, a contradiction. The lemma is proved. Proposition 1.9. If Gis a non-“finitely generated” non-BFC solvable group satisfying Max-BFC, then: (1) Gis a locally nilpotent-by-finite group; (2) G=BU is a product of a normal divisible abelian p-subgroup B and a polycyclic subgroup U; (3) Buis a BFC-subgroup for an element u∈Uif and only if u∈CU(B); (4) if Buis a non-BFC subgroup for some element u∈U, then [B,u]=B; 498 O. D. Artemovych (5) if Buis a non-BFC subgroup for some element u∈U, then B is an indecomposable injective Qu-module; (6) if Buis a non-BFC subgroup for some u∈U, then Auis a BFC-subgroup for every proper Zu-submodule Aof B, where the action is induced by the conjugation of uon B; (7) Gcontains a normal subgroup Hof finite index in which every non-BFC subgroup is subnormal; (8) Gis a torsion subgroup of G. Proof: (1) Is obvious. (2) Follows from Lemma 1.6. (3) Assume that H=Buis a BFC-subgroup for some element u∈ U.Ifuhas a finite order, then H/Huis a divisible group and by Theorem 1.16 of [7]H=Z(H)Hu. Consequently Z(H) is a subgroup of finite index in Hand His an abelian group. Let ube an element of infinite order. Since the subgroup Hus and the quotient group Bus/(Hus)are nilpotent for some integer s, Busis a nilpotent group by Hall theorem [16, Theorem 2.27]. But then Busis an abelian group and therefore as proved above H/(Z(H)∩u) is abelian. This yields that His an abelian group. (4) If Buis a non-BFC subgroup for some element uof Uand [B,u]=B, then T=[B,u]uis a BFC-subgroup. Since Bu/T is a nilpotent group, it is abelian, a contradiction. (5) It is clear that Bis a right Qu-module with the action induced by the conjugation of uon B. Furthermore, Bis a divisible Qu-module and therefore it is injective (see e.g. [11, Theorem 5.28]). By Theorem 2.5 of [15]Bhas a decomposition as a module direct sum of indecomposable injective Qu-submodules. Since Busatisfies Max-BFC,Bis an indecomposable module. (6) Let Bube a non-BFC group and Aa proper submodule of a right Zu-module B, where the action is induced by the conjugation of uon B.ByFwe denote a basic subgroup of A.IfA=F, then Au is either a polycyclic group or a BFC-group in view of Lemmas 1.8 and 1.5. Therefore we assume that F=A. Since Bis an indecomposable Qu-module, we conclude that Fis an infinite group. But then A/Apis also infinite and so Au/Apis a BFC-group by Lemma 1.8. This yields that Auis the ones. Groups with Many BFC-Subgroups 499 (7) If Vis a nilpotent subgroup of finite index in Uand Kis any nonBFC subgroup of DV , then D≤K. Hence Kis a subnormal subgroup of DV . (8) Is obvious. The proposition is proved. Proof of Theorem 1: (⇒) Follows from Proposition 1.9. (⇐) Suppose that Kis a non-BFC subgroup of a non-BFC group G. Let Gbe a group of type (ii) and BK =BK/B(B∩K)=BK. Since BK is a finitely generated subgroup, S=(B∩S)K, where S is a subgroup of BK which contains K, and BK satisfies the maximal condition on normal subgroups by Theorem 5.34 of [10], we conclude that every properly ascending series of type K<K1<···< Kn<··· is finite. This means that BK (and consequently G) satisfies Max-BFC. If Gis a group of type (iii), then it is clear that K=(K∩D)F, where F=u1,... ,u tis some finitely generated subgroup. Assume that Ki=(K∩D)uihas the finite commutator subgroup Kifor all i (1 ≤i≤t). Since the subgroup K1,... ,K t,Fis a finitely generated and K1,... ,K t,F=K0Ffor some finite F-invariant subgroup K0≤ K∩D,(K/K0)=(FK0/K0)is a finite subgroup and therefore K is a BFC-subgroup, a contradiction. Hence (K∩D)uis non-BFC subgroup for some u∈Fand by our hypothesis D=K∩D≤K. The theorem is proved. Corollary 1.10. A solvable group Gsatisfies Max-BFC if and only if it is of one of the following types: (i) Gis a BFC-group; (ii) G=BU is a finitely generated group, where Bis a proper torsion normal subgroup of G,Uits polycyclic subgroup and Bxis either a BFC-subgroup or a finitely generated subgroup for every element xof U; (iii) G=DU is a product of a normal divisible abelian p-subgroup D and a polycyclic subgroup Uwith D≤{H|His a non-BFC subgroup of G}. 2. Groups with Min-BF C In this section we prove that a group which have a descending series with abelian or finite factors and satisfying Min-BFC is either a BFC-group or a ˇ Cernikov group. Lemma 2.1. If Gis a non-perfect group in which every proper normal subgroup is a BFC-subgroup, then Gis a BFC-group or G=Gx, where xpn∈Gfor some prime pand some positive integer n.