The nilpotency of some groups with all subgroups subnormal

Publicacions Matem`atiques, Vol 42 (1998), 411–421. THE NILPOTENCY OF SOME GROUPS WITH ALL SUBGROUPS SUBNORMAL Leonid A. Kurdachenko and Howard Smith Abstract Let Gbe a group with all subgroups subnormal. A normal subgroup Nof Gis said to be G-minimax if it has a finite G-invariant series whose factors are abelian and satisfy either max-Gor minG. It is proved that if the normal closure of every element of G is G-minimax then Gis nilpotent and the normal closure of every element is minimax. Further results of this type are also obtained. 1. Introduction Let Gbe a group with all subgroups subnormal. If the normal closure in Gof every element is finitely generated then Gis nilpotent [15, Theorem 1]. We deal in the present paper with the case where every normal closure is minimax. Let us note at the outset that there is no corresponding result for the case where normal closures have finite (Pr¨ufer) rank; indeed, an example in [12] shows that even if Gitself has finite rank then it need not be nilpotent. Using standard notation, we denote by S2the class of soluble minimax groups; thus a group Gbelongs to S2if it has a finite normal series the factors of which are abelian and satisfy either max or min. Since every group with all subgroups subnormal is known to be soluble [9], the hypothesis that the normal closure of every element belong to S2 is less restrictive than it might at first appear. In fact, we shall establish a somewhat stronger result than that hinted at above. A normal subgroup Nof a group Gis said to be G-minimax if it has a finite Ginvariant series of subgroups the factors of which are abelian and satisfy either max-Gor min-G. Our first result is as follows. 412 L. A. Kurdachenko, H. Smith Theorem A. Let Gbe a group with all subgroups subnormal. If hxiG is G-minimax for all xin Gthen (i) Gis nilpotent and (ii) hxiGis minimax for all xin G. A group Gbelongs to the class S1if it has a finite normal abelian series the factors of which are torsion-free of finite rank or Chernikov groups. (Thus S2⊆S1. Additional care needs to be taken when dealing with the class S1since it is not closed under forming quotients, as may be seen by considering the additive group of rationals.) Theorem B. Let Gbe a group with all subgroups subnormal and suppose that hxiG∈S1for all xin G. Then Gis nilpotent. Using Theorem B we are able to deduce a further result. For a class X of groups, a group Gis said to be an XC-group if G/CG(xG) belongs to Xfor all xin G. In the case where X=S2we have the class of groups with “minimax conjugacy classes”. Now, by Theorem 2 of [6], if G∈S2Cthen hxiG∈S2for all x∈G, and we have the following consequence of Theorem B. Corollary. Let Gbe a group with all subgroups subnormal and suppose that Ghas minimax conjugacy classes. Then Gis nilpotent. There are no doubt several possible generalisations of Theorem B; we content ourselves with establishing a result that has both Theorem A(i) and Theorem B as special cases. Theorem C. Let Gbe a group with all subgroups subnormal and suppose that, for each xin G,hxiGhas a finite G-invariant series each of whose factors is either G-minimax or torsion-free abelian of finite rank. Then Gis nilpotent. 2. max-Gand min-Gsubgroups of locally nilpotent groups We begin with a couple of definitions. Let Gbe a group, Han infinite normal subgroup of G. Then His G-quasifinite if every proper G-invariant subgroup Nof His finite and His the join of all such subgroups N, while His G-just infinite if every nontrivial G-invariant subgroup Nof Hhas finite index in Hand the intersection of all such Nis trivial. For a group G, just infinite ZG-modules were first studied in [11] and [5], quasifinite ZG-modules in [16]. For further references the reader is invited to consult the survey [4]. Our first requirement is as follows. Groups with all subgroups subnormal 413 Lemma 2.1. Let Gbe a locally nilpotent group, Han infinite normal subgroup of G.IfHsatisfies min-Gthen Hcontains a G-invariant subgroup Athat is G-quasifinite. Proof: By min-Gwe see that Hcertainly contains a G-invariant subgroup Aminimal with respect to being infinite. If Ais not G-quasifinite then the join Fof all finite G-invariant subgroups of Ais finite; then A/F is a chief factor of Gand therefore finite (of prime order) and we have the contradiction that Ais finite. Now let Gand Abe as above. If Bis a proper G-invariant subgroup of Athen Bis finite and therefore so is A/CA(B), and it follows that B is contained in the centre of A;thusAis abelian. Clearly Aisap-group for some prime p. If the subgroup Cgenerated by all elements of order p in Ais infinite then C=A; otherwise Cis finite and Ais Chernikov [10, 25.1] and therefore divisible. Thus we have the following. Lemma 2.2. Let Gbe a locally nilpotent group, AaG-quasifinite subgroup of G. Then Ais abelian and either of exponent por a divisible Chernikov p-group, where pis a prime. Next we establish a result that rules out the first of these two possibilities in certain circumstances. Recall that a Baer group is a group in which every cyclic subgroup is subnormal. Lemma 2.3. Let Gbe a Baer group, AaG-quasifinite subgroup of G. If G/CG(A)is hypercentral then A≤Z(G); in particular Ais divisible Chernikov. Proof: Supposing the result false, choose z∈Gwith zCG(A) a nontrivial element of Z(G/CG(A)). By Lemma 2.2 Ais abelian; from the choice of zit follows easily that both [A, z] and CA(z) are normal in G. Further, if [A, z]<Athen [A, z] is finite and so CA(z) has finite index in Aand hence equals A, a contradiction; thus A=[A, z]. Write H=Ahzi. Since Gis a Baer group, His nilpotent. But H0=[A, hzi]= A=[A, H]=[H 0 ,H] and it follows that H0= 1. This again contradicts the choice of zand establishes the result. We are now able to prove the following. Lemma 2.4. Let Gbe a Baer group, Aa normal subgroup of Gsuch that G/CG(A)is hypercentral. Suppose that Asatisfies min-G. Then A is Chernikov and A≤Zn(G)for some positive integer n. 414 L. A. Kurdachenko, H. Smith Proof: If Ais finite the result is clear. Otherwise, we may apply Lemma 2.1 to obtain a G-invariant subgroup B1of Athat is G-quasifinite and hence, by Lemma 2.3, divisible Chernikov and central in G. Assuming the statement of the lemma false, repeated application of this argument gives an ascending chain of G-quasifinite factors Bi/Bi−1where, for each i≥1, Bi≤Zi(G) and Bi/Bi−1is divisible Chernikov (interpreting B0as 1). Write B= ∞ [ i=1 Bi. Each Biis divisible Chernikov and hence abelian, so that Bis abelian but not Chernikov. By min-G, some prime component Pof Bhas infinite rank and hence contains an infinite G-invariant subgroup Q(= Ω1(P)) of exponent p[1, 25.1]. But Q satisfies min-Gand so by Lemma 2.1 contains a G-invariant subgroup R that is G-quasifinite. Lemma 2.3 now gives the contradiction that Ris divisible, and the lemma is proved. We turn now to discussion of normal subgroups satisfying max-G, beginning with the counterpart to Lemma 2.1. Lemma 2.5. Let Gbe a locally nilpotent group, Han infinite normal subgroup of G.IfHsatisfies max-Gthen Hcontains a G-invariant subgroup Asuch that H/A is G-just infinite (that is, G/A-just infinite). Proof: Let Abe a G-invariant subgroup of Hmaximal with respect to H/A being infinite and let Bdenote the intersection of all G-invariant subgroups Nof Hthat properly contain A.IfB>Athen H/B is finite and B/A is a chief factor of Gand therefore finite. This gives the contradiction that H/A is finite. Lemma 2.6. Let Gbe a locally nilpotent group, AaG-just infinite subgroup. Then either Ais torsion-free or Ais an elementary abelian p-group for some prime p. Proof: Let Tdenote the torsion subgroup of A; then Tis normal in Gand so T=1orT=Aand we may assume that Ais torsion and therefore a p-group for some prime p. Certainly Ais not minimal normal in Gand so Acontains a proper G-invariant subgroup Cof finite index. By local nilpotency Cmay be chosen so that |A/C|=p. Let Bdenote the intersection of all subgroups of index pin A, so that Bis normal in G.IfB= 1 the result follows, so we shall assume for a contradiction that A/B is finite and hence that A=KB for some finite subgroup K. Now let Fbe some finite subgroup of A. There exists a G-invariant subgroup Eof finite index in Asuch that F∩E= 1. Let L/E be the Frattini subgroup of A/E;thusB≤Land we have A/E =KL/E, which Groups with all subgroups subnormal 415 implies that A/E =KE/E ∼ =K/K ∩E(by the usual property of the Frattini subgroup). But then the rank of Fis at most that of K, and the fact that Fwas arbitrary tells us that Ahas finite rank and is therefore Chernikov (see, for example, Corollary 2 of [10, Theorem 6.36]). But A is residually finite and we obtain the contradiction that Ais finite. With the hypotheses of the above lemma, if Ais torsion-free then it is in fact central in Gand therefore cyclic. This result is not essential for the proofs of the theorems but, apart from the fact that it allows us to establish a “just-infinite version” of Lemma 2.4 without the hypothesis that Gbe a Baer group, it appears to be of interest in its own right. It is no further trouble to establish a somewhat stronger result, namely the following, which may indeed be well known. Theorem 2.7. Let Gbe a locally nilpotent group, Na normal torsionfree subgroup of G, and supose that N/M is periodic for all nontrivial G-invariant subgroups Mof N. Then Nis central in Gand hence of rank (at most) one. Proof: Suppose that Nis not central, let a∈N,g∈Gwith [a, g]6=1 and write c=[a, g]. By hypothesis N/hciGis periodic; in particular ∃n∈Nsuch that an∈hci Gand hence an∈hci Ffor some finitely generated subgroup Fof G. Let H=ha, g, Fi,A=haiH, so that c∈A. Since His nilpotent we have [A, rH] = 1 for some r∈N.Nowc∈[A, H], a normal subgroup of H, and so an∈[A, H]. Since A=hai[A, H]it follows that An≤[A, H]. For each i≥0 write Ai=[A, iH], and suppose that An i≤Ai+1 for some i. Then An i+1 =[A i ,H] nand, modulo Ai+2,we have [Ai,H] central in Hand generated by elements [x, y], where x∈Ai, y∈H, and so (mod Ai+2)[Ai,H] nis generated by the nth powers of such commutators. But [x, y]n≡[xn,y]modA i+2, by centrality, and we deduce that An i+1 ≤[An i,H]A i+2 =Ai+2. By induction, therefore, Anr≤Ar= 1. But A≤Nand Nis torsion-free and so we have A=1 and hence the contradiction [a, g] = 1. Thus Nis central. If zis a nontrivial element of Nthen hziis normal in Gand therefore N/hziis periodic. The result follows. Corollary 2.8. Let Gbe a locally nilpotent group, NaG-just infinite subgroup of G.IfNis torsion-free then Nis central and cyclic. Proof: By Theorem 2.7 Nis central. If zis a nontrivial element of N then hziCGand N/hziis finite, so Nis cyclic. 416 L. A. Kurdachenko, H. Smith The following result may be compared with Lemma 2.3, where the hypothesis that Gbe Baer could not be replaced by that of local nilpotency, as shown by the example G=A]hgiwhere A∼ =Cp∞and g∈Aut Ais determined by ag=ap+1 for all a∈A. Lemma 2.9. Let Gbe a locally nilpotent group, AaG-just infinite subgroup of G, and suppose that G/CG(A)is hypercentral. Then A≤Z(G). In particular, Ais infinite cyclic. Proof: By Lemma 2.6 and Corollary 2.8 we need only dispose of the case where Ais assumed to be an elementary abelian p-group. Let zCG(A) be a nontrivial element of the centre of G/CG(A) and choose a nontrivial element dof Asuch that [d, z] = 1 (such exists by local nilpotency). Let g∈G; by the choice of zwe have 1 = [d, z]g=[d g ,z] and so zcentralises hdiG.NowA/hdiGis finite and the map a→[a, z] for all a∈Ais a homomorphism whose kernel contains hdiG, hence [A, z] is finite (and G-invariant) and therefore trivial. This yields the contradiction that z∈CG(A) and thus establishes the lemma. We have been unable to decide whether the hypothesis of solubility is necessary in the following. Lemma 2.10. Let Gbe a locally nilpotent group, Aa normal soluble subgroup of Gsuch that G/CG(A)is hypercentral. Suppose that A satisfies max-G. Then Ais finitely generated and A≤Zn(G)for some positive integer n. Proof: An easy induction allows us to assume that Ais abelian. Let Tdenote the torsion subgroup of Aand suppose first that Tis infinite. By Lemma 2.5, Tcontains a G-invariant subgroup Usuch that T/U is G-just infinite; Lemma 2.9 now gives a contradiction. Thus Tis finite and, factoring, we may assume that Ais torsion-free. Again by Lemma 2.5, there is a G-invariant subgroup A1of Awith A/A1G-just infinite and hence, by Lemma 2.9, infinite cyclic. If Ais not finitely generated then we obtain easily an infinite descending chain of G-invariant subgroups Aiwith Ai/Ai−1infinite cyclic for each i≥1 (with A0=A). Clearly A/Aiis a free abelian group of rank exactly ifor each i.Nowfix a prime pand consider A/Ap; since this is a torsion group the previous argument shows that A/Apis finite of order pr, say. But A/Ar+1 has a finite image of exponent pand order pr+1, a contradiction that shows that Ais finitely generated. We may now apply Lemma 6.37 of [10]to deduce that A≤Zn(G) for some finite n, thus concluding the proof. We are now ready to establish the final result of this section. Groups with all subgroups subnormal 417 Proposition 2.11. Let Gbe a Baer group, Aa normal subgroup of Gsuch that G/CG(A)is hypercentral. If Ais G-minimax then Ais minimax and A≤Zn(G)for some positive integer n. Proof: By definition Ais soluble and, by induction on the length of an appropriate series, we may assume that Ais abelian and satisfies either max-Gor min-G. Lemmas 2.4 and 2.10 now give the result. 3. Conclusion Our main objective now is to prove Theorem C, since Theorems A and B are easy consequences. The work of the previous section allows us to establish the following key result without difficulty. Proposition 3.1. Let Gbe a soluble Baer group and suppose that, for each element xof G,hxiGhas a finite G-invariant series with abelian factors that are either torsion-free of finite rank or G-minimax. Then G is hypercentral, with hypercentral length at most ω. Proof: Let G/N be a hypercentral image of G,U/V aG-invariant section of N.IfU/V is torsion-free of rank rthen [U, rG]≤Vby Lemma 6.37 of [10], while if U/V is G-minimax then, applying Proposition 2.11 to the group G/N0, we see that there is an integer nsuch that [UN0,nG]≤VN0 . Now let a∈N,D=haiG. The given hypotheses, together with the above argument, imply that ∃m∈Nsuch that [DN0,mG]≤N0, and it follows that N/N0is contained in the hypercentre of G/N0and hence that G/N0is hypercentral. Since G/G0is certainly hypercentral, an easy induction on the derived length shows that G is hypercentral. Now let xbe an arbitrary element of G,X=hxiG,U/V aG-invariant section of X. Again by Lemma 6.37 of [10] and Proposition 2.11 we have [U, rG]≤Vfor some integer r, so that X≤Zm(G) for some integer m. Since xwas arbitrary we have G=Zω(G)asrequired. Our final prerequisite is a result that is probably well known. For the basic properties of isolators in locally nilpotent groups the reader is referred to [3]. Lemma 3.2. Let Gbe a countable locally nilpotent group. Then there exists a torsion-free subgroup Kof Gwhose isolator in Gis G. 418 L. A. Kurdachenko, H. Smith Proof: Write G= ∞ [ i=1 Fiwhere each Fiis finitely generated and Fi≤ Fi+1 for all i. Since the torsion subgroup of F1is finite there is a positive integer n1such that K1=: Fn1 1is torsion-free; clearly |F1:K1|is finite. Suppose that for some i≥1 we have a torsion-free subgroup Kiof finite index in Fiand let Ki+1 be a torsion-free subgroup of Fi+1 maximal with respect to containing Ki. Claim. |Fi+1 :Ki+1|is finite. Supposing this false, write F=Fi+1, H=Ki+1,I=IF(H) (the isolator of Hin F). Then |I:H|is finite and ∃n>0 with In≤H. By hypothesis I≤Fand so (by nilpotency) I<N F (I), hence ∃g∈F\Isuch that IChI,gi. Clearly ghas infinite order module I. Since I/Inis finite ∃k>0 such that [I,hgki]≤In; but then [H,hgki]≤In≤Hand His normal in hH, gki. Since hH,gki/H is torsion-free we have a contradiction that establishes the claim. Inductively, therefore, we may construct a chain K1≤K2≤ ··· such that each Kiis a torsion-free subgroup of finite index in Fi. Set K= ∞ [ i=1 Ki; clearly Ksatisfies the desired condition, and the lemma is proved. Proof of Theorem C: Let Gbe as stated. By a result of M¨ohres [9]G is soluble and, by Proposition 3.1, Gis hypercentral of length at most ω. Let g∈G,D=hgiG. Applying Proposition 2.11 we see that G-minimax sections of Dare in fact minimax and hence that Dhas finite rank. Whereas the original hypothesis on normal closures is not in general inherited by subgroups and quotients of G, the hypothesis that each hxiGhave finite rank certainly is, and we now show that this condition is sufficient to ensure the nilpotency of the ω-hypercentral group G. By induction on the derived length of Gwe may assume that G0is nilpotent. If G/G00 is nilpotent then so is G[2, Theorem 7] and so we may factor and assume that Gis metabelian. We may also assume that Gis countable. Let Abe a normal abelian subgroup of Gwith G/A abelian. By Lemma 3.2 there is a torsion-free subgroup Kof Gwith IG(K)=G. Since Kis hypercentral of length at most ωit is nilpotent [14] and so KA, as a product of a normal and a subnormal nilpotent subgroup, is also nilpotent (see, for example, Proposition 3.3.12 of [7]). If G/(KA)0 is nilpotent then so is G; factoring if necessary we may therefore assume that G/A is periodic. Now let Tbe the torsion subgroup of G.By[8], Tis nilpotent and therefore so is TA. Since (TA)0≤T, the torsion subgroup of G/(TA)0is T/(TA)0; factoring once more we may assume Groups with all subgroups subnormal 419 that T≤A. Next, G/T is torsion-free and abelian-by-periodic and hence abelian [10, Lemma 6.33]. Let Ube the divisible component of T,C=CG(U). Then A≤Cand G/C is periodic and, arguing as in the proof of Lemma 3.13 of [10], we deduce that G/C is trivial. Then U≤Z(G) and we may (finally) assume that U= 1 and hence that Tis reduced. Let x∈Gand write X=hxiG,Y=X∩T. Each p-component of Y is Chernikov and reduced and therefore finite. Let NbeaG-invariant subgroup of finite index in Y,E=CG(Y/N), so that G/E is finite. Since G0≤Twe see that Gcentralises X/Y , and hence that [X,E,E]≤N.By the Three Subgroup Lemma [10, Lemma 2.13], therefore, [X,E0]≤N. Now consider the group G/E0, which is abelian-by-finite. Every abelian group Jis residually of rank 1 —this is well known and may be seen by noting that, for every nontrivial element jof J,ifMis maximal with respect to not containing jthen J/M is locally cyclic. It follows that G/E0is residually of finite rank. Now from the structure of Ywe have that the intersection of all Ndefined as above is trivial and hence that the intersection Vxof all the corresponding E0centralises X. But G/Vxis also residually of finite rank and, further, so is G/W, where Wis the intersection of all Vxobtained as xruns through the set G. By Theorem 2 of [13]G/W is nilpotent. But W≤Z(G) and so Gis nilpotent and the proof of Theorem C is complete. Theorem B is an immediate consequence of Theorem C, as is part (i) of Theorem A. Part (ii) follows from Proposition 2.11, and Theorem A is therefore proved. Finally, we recall that paper [13] was concerned with establishing the nilpotency of ω-hypercentral groups with all subgroups subnormal that have, in addition, certain rank restrictions on their structure. It does not appear to be known whether every ω-hypercentral group with all subgroups subnormal is nilpotent (the examples in [12] having length ω+ 1) and so it is perhaps worth recording the following result, which is what much of the proof of Theorem C was concerned with establishing. Theorem 3.3. Let Gbe a group with all subgroups subnormal and suppose that Gis hypercentral of length at most ω.Ifhxi Ghas finite rank for all xin Gthen Gis nilpotent. References 1. L. Fuchs,“Infinite abelian groups,” vol. 1, Academic Press, New York, 1970.