Groups with all subgroups subnormal or nilpotent-by-Chernikov

Groups with all Subgroups Subnormal or Nilpotent-by-Chernikov

HOWARDSMITH

It was proved by Asar in [1] that a locally nilpotent group in which every proper subgroup is nilpotent-by-Chernikovis itself nilpotent-by-Cherni- kov. It is easy to see that the main problem here lies with establishing solubility, since a non-trivial soluble group G has a homomorphic image that is either cyclic or a PruÈferp-group, and ifM/N/G;withMnilpotent and bothG=NandN=MChernikov, thenM^Gis easily shown to be nilpo- tent, and of courseG=M^Gis Chernikov. In view of earlier results from [7], Asar's result completes the proof that a locally graded group with all proper subgroups nilpotent-by-Chernikovis nilpotent-by-Chernikov; in particular, a locally graded group with all proper subgroups nilpotent is soluble-by-finite.

Now, by a well-known result due to MoÈhres [6], a group Gwith all subgroups subnormal is soluble, and in [9] and [10] the question was considered as to what might be said about a group G in which every subgroup is either subnormal or nilpotent. With regard to establishing solubility, the main two results in [10] were that a locally soluble-by- finite group with all subgroups subnormal or nilpotent is itself soluble (Theorem 1), and a locally graded group in which every subgroup is either nilpotent or subnormal of defect at most n is also soluble (The- orem 2). (Let us note here that there is an error, but one that is easily corrected, in the proof of Lemma 2:2 of [10], in that the appeal to Lemma 1:2 at the beginning of the second paragraph really requires an amended version of Lemma 1:2; where solubility replaces nilpotency throughout; the details are almost identical. A similar remark applies to the proof of the Proposition in [9], which invokes Lemma 3 of that pa- per, but here an alternative correction is to include in the statement of

Indirizzo dell'A.: Department of Mathematics, Bucknell University, Lewisburg, PA 17837, U.S.A.

E-mail: [email protected]

Lemma 4 of [9] the hypothesis that every soluble subgroup of G be nilpotent.)

Further results in [9] and [10] are concerned with establishing rather more than solubility, and we do not concern ourselves in the present paper with attempting to generalize those results but postpone treatment to a couple of forthcoming articles. Our purpose here is to investigate whether a group with all subgroups either subnormal or nilpotent-by-Chernikovis soluble-by-finite, and as companion results to Theorems 1 and 2 of [10] we have the following.

THEOREM1: Let G be a locally soluble-by-finite group in which every subgroup is either subnormal or nilpotent-by-Chernikov. Then G is so- luble-by-finite and is either soluble or nilpotent-by-Chernikov.

THEOREM2: Let G be a locally graded group in which every subgroup is either nilpotent-by-Chernikov or subnormal of defect at most n;where n is some fixed positive integer. Then G is soluble-by-finite and is either soluble or nilpotent-by-Chernikov.

As is the case for Theorem 1 of [10], Theorem 1 above generalizes quite easily to the case of groupsGthat arelocally generalized radical, that is, every finitely generated subgroup ofGis the union of an ascending series of subgroups whose factors are either locally nilpotent or locally finite. To prove this, it is enough, in view of Theorem 1;to show that such a groupG is locally soluble-by-finite, and to this end we may assume thatGis gen- eralized radical. By (possibly transfinite) induction on the length of an appropriate series ofG;we may assume that there is a normal subgroupK ofGwith G=K locally nilpotent or locally finite andK locally soluble-by- finite and hence, by Theorem 1; soluble-by-finite. Since G=K is also so- luble-by-finite it follows thatGis soluble-by-finite.

We recall from [10] that it is a most difficult problem to decide whether alocally gradedgroup with all subgroups subnormal or nilpotent need be soluble, since it is not known whether there exists an infinite finitely generated residually finitep-group with all subgroups either finite (and hence nilpotent) or of finite index (and hence subnormal). This is our jus- tification for introducing the bound on subnormal defects in the statement of Theorem 2: From now on we shall denote the class of nilpotent-by- Chernikovgroups byNC:

Our first result is easy to prove and is used several times during the course of our discussion.

LEMMA 1: Let G be a group in which every subgroup is either sub- normal or inNC;and let K be a non-NC-subgroup of G:Then the dth term G^(d)of the derived series of G is contained in K, for some integer d;and if G is not soluble-by-finite then every subgroup of G=G^(d) is subnormal.

PROOF. There is a finite subnormal series fromKtoG;and each factor has all subgroups subnormal and is therefore soluble [6]; it follows that G^(d) K for somed: IfGis not soluble-by-finite then neither is G^(d), so G^(d)2=NCand the result follows.

LEMMA2: Let G be a locally nilpotent group that is also locally finite, and suppose that every subgroup of G is either subnormal or inNC(re- spectively, subnormal or soluble). Then either G is inNC(respectively, G is soluble) or there is a primary component Gpof G such that Gpis not inNC (respectively, is not soluble) and GˆGpK;for someNC-subgroup K:

PROOF. Firstly we note that everyNC-subgroup ofGis soluble. Now suppose that every subgroup ofGis subnormal or inNCand that someG_p is not in NC: Then every subgroup of G=G_p is subnormal and so G=Gp2NC[3] andGhas the structure indicated. Suppose, on the other hand, that every primary component is inNCbut thatGis not. Either there are infinitely manyp-componentsGpi that are all nilpotent and are such that the class ofG_pi increases withi, or there are infinitely manyp-com- ponentsG_pi that are not nilpotent. In either case we writeGˆH₁H₂; where eachHjcontains infinitely many of theGpi:It is clear thatH1is not in NC (since a Chernikov group involves just finitely many primes), but H1G=H2;which has all subgroups subnormal and is therefore inNC;

again by [3]. This contradiction concludes the proof in the case where all subgroups are subnormal or inNC:

If every subgroup of Gis subnormal or soluble then much the same argument works: ifGpis insoluble thenG=Gphas all subgroups subnormal and is therefore inNC[3], and ifGpis soluble for all primespbutGis not soluble then we may writeGˆH₁H₂where this time eachH_jcontains primary components of unbounded derived length. Again H₁G=H₂; which is soluble [6], and a contradiction ensues.

Here is another result that is used on more than one occasion.

LEMMA3: Let G be a locally graded group in which every subgroup is either subnormal or in NC; let R denote the soluble residual of G;and suppose that G=R is soluble. Then G is soluble-by-finite.

PROOF. IfGis not soluble-by-finite then neither isR;and soR2=NC:

By results from [1] and [7] (as referred to at the beginning of our discus- sion),Rhas a proper non-NC-subgroupS;and sinceSis subnormal inRwe haveS^R5Rand every subgroup ofR=S^Rsubnormal, soR=S^Ris soluble, contradicting the fact thatRis perfect.

Our next result is a generalization of Lemma 1:2 of [10], referred to above. We have seen that Lemma 2:2 of [10] really required an adaptation of Lemma 1:2 (i.e. a ``soluble version''), and we now state and prove a result that covers both this amended version and other cases. The following may be of use in further discussions of the kind presented here. Notice, for example, that the classNCis an example of a classXsatisfying the given criterion. Notice also that ifeverysubgroup of the groupGis subnormal butG2=Xthen the existence of such a triple (K;F;r) as described in the proposition below follows from a result that is primarily due to Brookes [2]

- see, for example, Lemma 3 of [3].

PROPOSITION1: LetXˆ S¹

iˆ1Xibe a class of groups, where each classXi

ishs;Li-closed and XiXi‡1 for all i: Let G be a group in which every subgroup is either subnormal or inX;and suppose that G2=X:Then there is a positive integer r;a non-Xsubgroup K of G, and a finitely generated subgroup F of K such that, for every non-Xsubgroup L of K that contains F;L is subnormal of defect at most r in K:

PROOF. Suppose the result false and setK₀ˆG;H₀ˆ1:Assume that for somei1 we have subgroupsKi 1;Hi 1ofG;withHi 1 finitely gen- erated and contained inK_{i 1};which is not inX_i:By hypothesis there is a non-X-subgroupK_iofK_{i 1}withH_{i 1}K_iandK_iof defect greater thaniin K_{i 1}: Choose x_i2[K_{i 1};_iK_i]nK_i; then x_i2[K_{i 1};_iL_i] for some finitely generated subgroupL_iofK_i;and (byL-closure) we may chooseL_iso that Li2=Xi:SetHiˆ hHi 1;Lii:Thus we have definedKiandHiinductively for alli1:WriteHˆS¹

iˆ1Hi;thenH2=X;and soHis subnormal inG;of defect d; say. Now x_d2[K_{d 1};_dH_d][G;_dH]H; and sox_d2H_i for some i;

which we may choose to be greater than d: Thus xd2HiKiKd; a contradiction. The result follows.

PROPOSITION 2: Let G be a locally nilpotent group in which every subgroup is either subnormal or nilpotent-by-Chernikov.Then G is soluble.

PROOF. IfTdenotes the torsion subgroup ofGthen every subgroup of the torsion-free group G=T is subnormal or nilpotent; it follows from Theorem 2 of [9] that G=T is nilpotent. Thus we may assume that Gis periodic, and by Lemma 2 we may further suppose thatGis ap-group for some primep:Assume the result false and letRdenote the soluble residual ofG:By Lemma 3;G=Ris insoluble, and we may factor and hence assume thatGis residually soluble.

By Proposition 1, withXthe class of soluble groups, we may also assume that there is a positive integerrand a finite subgroupFofGsuch that every insoluble subgroup ofGthat containsFis subnormal of defect at mostrinG:

LetUbe an arbitrary finite subgroup ofGcontainingF:We claim thatUis subnormal inG:First note that there is a normal subgroupNofGsuch that G=N is soluble and U\Nˆ1; and since N is not soluble we have UN subnormal inG:We may suppose thatGˆUN;so thatG=Nis finite and hence nilpotent, sayg_k(G)N:There is a chainN>N₁>N₂> :::of nor- mal subgroups ofGwith eachG=Nisoluble andT¹

iˆ1Niˆ1;and for eachiwe haveFUN_iandUN_i insoluble, and so [G;rUN_i]UN_i for alli: Thus [G;_r‡kU]T¹

iˆ1UN_i\NˆT¹

iˆ1N_i(U\N)ˆ1; soU is subnormal in G;as claimed. Since every finite subgroup ofGis contained in such a subgroupU;

we see thatGis a Baer group.

Now, as in the proof of Lemma 5 of [5], ifG contains a hyperabelian subgroup Hthat is not soluble then it contains an insoluble subgroupH that has hyperabelian lengthv:By means of an application toHof Pro- position 1, if necessary, we may (by re-labelling) assume thatGˆH:But nowFis contained in some normal soluble subgroupH0ofG;and ifH0has derived lengthnthen we may chooseH₀maximal normal inGwith respect to having derived lengthn;in which caseG=H₀ is also residually soluble.

LetV be an arbitrary normal subgroup ofGthat containsH0and is such thatG=Vis soluble. ThenVis not soluble and so every subgroup ofG=Vis subnormal of defect at mostrinG=V:By Roseblade's Theorem [8],G=Vis nilpotent of bounded class, and so G=H₀ is nilpotent and we have the contradiction thatGis soluble. It follows that every hyperabelian subgroup ofGis soluble.

Again we argue as in [5]: letxbe an arbitrary element ofG:Thenhxiis subnormal and ifhxi^Gis not soluble then (by looking at the normal closure series of hxi inG) we see that there are subgroupsA;B;C ofG, with A soluble,A/B/C;BˆA^CandBinsoluble. SinceBis a product of normal subgroups, each of which is a conjugate ofA, we haveBhyperabelian and

hence soluble, and by this contradictionhxi^Gis soluble. Sincexwas arbi- trary we now have G hyperabelian and hence soluble, and this contra- diction completes the proof of Proposition 2:

LEMMA4: Let G be a group in which every subgroup is either sub- normal or nilpotent-by-Chernikov. Then the product V of all normalNC- subgroups of G is soluble-by-finite, and if G is not soluble then V is nil- potent-by-Chernikov.

PROOF. LetV_lbe a normalNC-subgroup ofG:Then there are char- acteristic subgroupsL_l;D_lofV_lsuch thatL_lD_l;L_lis locally nilpotent, Dl=Llis (divisible) abelian andVl=Dlis finite. The productLof all theLlis locally nilpotent and normal in G;and the product D=L of all D_lL=L is likewise locally nilpotent andG-invariant. By two applications of Proposi- tion 2,Dis soluble. NowV=Dis the product ofG-invariant finite subgroups V_lD=D; so V=D is a periodic FC-group and its finite residual R=D is therefore central.

CertainlyV=Ris residually finite and locally finite, and if it is not so- luble-by-finite then, for every normal subgroup N=R of finite index in V=R;we haveN2=NCand hence every subgroup ofV=Nsubnormal, which gives V=N nilpotent. Thus V=R is residually nilpotent and hence locally nilpotent, and Proposition 2 gives the contradiction thatV=Ris soluble.

ThusV=Ris soluble-by-finite and henceVis soluble-by-finite, so there is a solubleG-invariant subgroupUof finite index inV:IfV2=NCthen all subgroups ofG=Uare subnormal andG=Uis soluble [6]. ThusGis soluble, as required.

PROPOSITION3: Let G be a locally finite group in which every subgroup is either subnormal or nilpotent-by-Chernikov. Then G is soluble-by-fi- nite, and is either soluble or nilpotent-by-Chernikov.

PROOF. If G2NCthenGis soluble-by-finite, so let us assume that G2=NCand proceed to show thatGis soluble-by-finite. Notice that this will imply thatGis soluble, for every finite image ofGhas all subgroups subnormal and is therefore nilpotent (and hence soluble). LetV be as in Lemma 4 and suppose for a contradiction thatGis not soluble-by-finite.

ThenV 2NC;and if N=V is a nontrivial normal subgroup ofG=V then N2=NC;so every subgroup ofG=Nis subnormal andG=Nis soluble and locally nilpotent. Let M=V denote the intersection of all such normal subgroups N=V: Then G=M is residually (locally nilpotent) and locally

finite, soG=M is locally nilpotent and hence soluble, by Proposition 2:If MˆV then G=V is soluble, while if M6ˆV then M=V is the soluble residual ofG=V and Lemma 3 givesG=V soluble-by-finite. In either case we haveGsoluble-by-finite-by-soluble-by-finite and therefore soluble-by- finite.

PROOF OFTHEOREM1: Assume the result false and letRdenote the soluble residual ofG;by Lemma 3G=Ris not soluble. Now a soluble-by- finite group F that is residually soluble is soluble, sinceF modulo its soluble radical is finite and residually soluble and hence soluble. Thus G=Ris locally soluble, and there is no loss in supposing thatGis locally soluble.

LetH1 denote the locally nilpotent radical of Gand, for each i1;

let H_i‡1=H_i be the locally nilpotent radical of G=H_i;set Hˆ [¹_iˆ1(H_i):

Now eachH_iis soluble, by Proposition 2;and ifH_i2=NCfor someithen every subgroup ofG=H_iis subnormal and soG=H_iis soluble, giving the contradiction thatGis soluble. ThusHis contained in the subgroupV of G that is defined in Lemma 4, and since V 2NC so is H: But then HˆH_d for somed;factoring, we may assume thatGhas trivial locally nilpotent radical. ThusGhas no nontrivial subnormal soluble subgroups and hence no subnormal NC-subgroups; so, among the nontrivial sub- groups ofG;the subnormal subgroups, the insoluble subgroups and the non-NC subgroups coincide. If 16ˆN/G then G=N has all subgroups subnormal and is therefore soluble and locally nilpotent. Note that ev- ery finitely generated subgroup of Gis in NC (and is therefore nilpo- tent-by-finite).

Now letG0be the intersection of all nontrivial normal subgroupsSofG withG=Storsion-free and locally nilpotent. Suppose thatG0ˆ1;and letX be an arbitrary finitely generated subgroup ofG:There is a normal nil- potent subgroup Y of X of class c; say, such that X=Y is finite. Every torsion-free nilpotent image ofXis nil-cand henceXis nil-c:But thenGis locally nilpotent and Proposition 2 gives the contradiction thatGis soluble.

It follows thatG06ˆ1 and hence thatG=G0 is locally nilpotent. IfNis an arbitrary nontrivial normal subgroup ofG0thenG^(d)Nfor some integer d;by Lemma 1;and sinceG=G^(d)is locally nilpotent we see thatG₀=G^(d)is the torsion subgroup ofG=G^(d);and so every proper image ofG₀is soluble, periodic and locally nilpotent. We may assume thatGˆG₀.

Next, letG1be the intersection of all nontrivial normal subgroups ofG:

ThenG1is the soluble residual ofGandG=G1is insoluble, by Lemma 3:It follows thatG₁ˆ1:

By Proposition 1 there is an insoluble subgroup K of G; a positive integer r and a finitely generated subgroup F of K such that every insoluble subgroup ofK containingF has defect at mostrinK:LetN be an arbitrary nontrivial normal subgroup ofK:Then N is subnormal inGand soN2=NCand, again by Lemma 1;we haveG^(d) N for some integer d; and G=G^(d) is periodic and locally nilpotent, as therefore is K=N:Every subgroup ofK=Nthat containsFN=Nhas defect at mostr inK;and we see too thatFN=Nis finite. By Theorem 0:2 of [4], there is an integer bdepending only on r such that L:ˆg_b(K) is finite modulo N:Similarly, ifMis an arbitrary nontrivial normal subgroup ofLthen, since K and hence M is subnormal in G there is a term G^(e) of the derived series of G contained in M; and since FG^(e)=G^(e) is a finite subgroup of the locally nilpotent group K=G^(e) it follows from [4] that L=Mis finite and nilpotent. In particular,L=L⁰is finite and so there is a finitely generated subgroup U of L such that LˆUL⁰: But then LˆUg_i(L) for every positive integer i; and if U has derived length at mostdthen so hasL=g_i(L) for everyi:SinceGis residually soluble so is L;and as L=M is finite nilpotent for every nontrivial normal subgroup M of L it follows that L^(d)ˆ1 and L is soluble. This gives the final contradiction that G is soluble, and the proof of the theorem is com- plete.

PROOF OFTHEOREM2: Supposing the result false, we may assume by Lemma 3 thatGis residually soluble, and hence that everyNC-subgroup ofGis soluble. IfNis an arbitrary normal subgroup ofGwithG=Nsoluble thenN is not soluble and so it is not inNC:But then every subgroup of G=N is subnormal of defect at most n and so G=N is nilpotent of class depending only onn[8]. Since the intersection of all suchN is trivial we haveGnilpotent and hence soluble, a contradiction.

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Manoscritto pervenuto in redazione il 26 maggio 2011.