On totally permutable products of finite groups

Document downloaded from: This paper must be cited as: The final publication is available at Copyright Additional Information http://hdl.handle.net/10251/19003 Ballester Bolinches, A.; Cossey, J.; Esteban Romero, R. (2005). On totally permutable products of finite groups. Journal of Algebra. 1(293):269-278. doi:10.1016/j.jalgebra.2005.01.033 http://dx.doi.org/10.1016/j.jalgebra.2005.01.033 Elsevier This paper has been published in Journal of Algebra, 293(1):269-278 (2005). Copyright 2005 by Elsevier. http://dx.doi.org/10.1016/j.jalgebra.2005.01.033 This paper has been published in Journal of Algebra, 293(1):269–278 (2005). Copyright 2005 by Elsevier. The final publication is available at www.sciencedirect.com. http://dx.doi.org/10.1016/j.jalgebra.2005.01.033 http://www.sciencedirect.com/science/article/pii/S0021869305000906 On totally permutable products of finite groups A. Ballester-Bolinches∗John Cossey† R. Esteban-Romero‡ Abstract The behaviour of totally permutable products of finite groups with respect to certain classes of groups is studied in the paper. The results are applied to obtain information about totally permutable products of T,PT , and PST -groups. 1 Introduction If a group Gcan be written as a product of two subgroups Aand B, then somehow the structure of Gis restricted by that of Aand B. Can one transform this general statement into concrete results at least in special situations? In this paper we are concerned with finite groups Gwhich are factorised by their subgroups Aand Bin such way that every subgroup of Apermutes with every subgroup of B. In this case we say that Gis a totally permutable product of Aand B. This sort of products arises when finite products of supersoluble groups are considered and they have been extensively studied even in the non-finite case. More precisely, Asaad and Shaalan [2] first introduced these products and proved that totally permutable products of finite supersoluble groups are supersoluble (Theorem 3.1). Maier [11] generalised Asaad and Shaalan’s result to saturated formations containing all supersoluble groups, and the first author and P´erez-Ramos [4] were able to remove the restriction “saturated” from Maier’s theorem and proved its converse. ∗Departament d’` Algebra; Universitat de Val`encia; Dr. Moliner, 50; E-46100 Burjassot (Val`encia); Spain; email: [email protected] †Mathematics Department; Mathematical Sciences Institute; The Australian National University; Canberra ACT 0200; Australia; email: [email protected] ‡Departament de Matem`atica Aplicada; Universitat Polit`ecnica de Val`encia; Cam´ı de Vera, s/n; E-46022 Val`encia; Spain; email: [email protected] 1 We also refer to Beidleman and Heineken [6] for other interesting facts on totally permutable products of infinite groups. A key point behind the results about finite totally permutable products is a theorem of Huppert stating that the product of two cyclic groups is supersoluble. It also holds in the general non-finite case ([1; 7.4.6]). This theorem shows, in particular, that totally permutable products of nilpotent groups are not in general nilpotent. Therefore a natural question arises: Suppose that G=AB is a totally permutable product of two nilpotent groups Aand B. What can we say about G? Applying Asaad and Shaalan’s result, if Gis finite, then Gis supersoluble. We prove in the paper that in this case Gis abelian-by-nilpotent, that is, its nilpotent residual is abelian. Therefore in the sequel, all groups considered are finite and soluble. Theorem 1. Let Gbe the totally permutable product of the nilpotent groups Aand B. Then Gis abelian-by-nilpotent. This result allows us to think that the nilpotent residual of a group which is a totally permutable product of nilpotent groups plays an important role. Recall that if His a formation, the H-residual GHof a group Gis the smallest normal subgroup of Gsuch that G/GH∈H([7; II, 2.3]). For each normal subgroup Nof G, we have (G/N)H=GHN/N ([7; II, 2.4]). Our next result describes completely the Sylow subgroups of the nilpotent residual of a group Gwhich is a totally permutable product of the nilpotent subgroups Aand B. Theorem 2. Let Gbe as in Theorem 1 and let Kbe its nilpotent residual. If pdivides |K|, then a Sylow p-subgroup of Kis either Ap, or Bp, or Ap×Bp, where Apand Bpare the Sylow p-subgroups of Aand B, respectively. We apply these results to obtain some information of the behaviour of finite totally permutable products with respect to formations Fof the form F=X◦N, where Xis a formation of finite groups containing all abelian groups and Nis the class of all nilpotent groups. It is clear that Fis composed of all finite groups whose nilpotent residual is in X. More precisely, we have: Theorem 3. Let Xbe a formation containing all abelian groups. Let G= AB be a totally permutable product of groups in X◦N. Then G∈X◦N. As the symmetric group of degree 3 shows, Theorem 3 is not true if X does not contain the formation of all abelian groups. In fact, if pis an odd 2 prime and Cpdoes not belong to X, the dihedral group of order 2pis a totally permutable product of its Sylow subgroups, both in X◦N, but the group is not in X◦N. For the converse, we do not require that Xcontains all abelian groups. Theorem 4. Let Xbe a formation. Let G=AB be a totally permutable product of the subgroups Aand Bsuch that Gbelongs to X◦N. Then Aand Bbelong to X◦N. Theorem 4 is not true for non-soluble groups (see for instance [7; X, Exercise 1.12]). Theorem 5. If F=X◦N, where Xis a formation containing all abelian groups, and Gis a finite totally permutable product of Aand B, then: 1. [AF, B] = [A, BF] = 1; in particular, AFand BFare normal subgroups of G, and 2. GF=AFBF. The methods applied in the proofs of the above results allow us to prove the following general theorem about totally permutable products of groups. Theorem 6. If G=AB is a totally permutable product of the subgroups A and B, and ap(respectively, bp) is the number of non-isomorphic non-central p-chief factors in A(respectively, B) for a prime p, then the number cpof non-central non-isomorphic p-chief factors in Gis bounded by a0p+b0p, where a0p= max{1, ap}and b0p= max{1, bp}. The above theorems allow us to derive information about totally permutable products of finite groups which have some group theoretical properties different from those described by formations. They are the ones described by the classes of T,PT,PST, and PSTc-groups. These classes are defined through permutability properties of subnormal subgroups. Let us give a short description of these classes before stating the corresponding results. A subgroup of a group Gis called permutable if it permutes with every subgroup of G. A result of Ore [12] shows that permutable subgroups of a finite group are subnormal in the group, but the converse need not hold. A group is called a PT-group (T-group) if permutability (normality) is a transitive relation. By Ore’s result, PT-groups are exactly those groups whose subnormal subgroups are permutable. In particular, every T-group is a PTgroup. PST-groups are defined in terms of Sylow permutability. A subgroup of a group Gis called S-permutable if it permutes with all the Sylow subgroups of G. A result of Kegel [10] shows that every S-permutable subgroup 3 is subnormal and hence PST-groups are exactly those groups in which all subnormal subgroups are S-permutable. In particular, PT-groups are PSTgroups. Another class containing the beforementioned ones is the class of PSTc-groups, introduced and studied by Robinson in [13], and composed by all groups in which every cyclic subnormal subgroup is S-permutable. One notable fact of the class of all soluble PST-groups is that it is subgroup-closed. Moreover it is closed under taking totally permutable products in which the factors have coprime indices [3] (see also [5] for other results in this direction). The following example shows, in particular, that the subgroup-closed character is absent from PSTc-groups, even if they are factors of coprime indices of a totally permutable product. Example 7. Let W,X,Y, and Zbe, respectively, non-abelian groups of orders 6, 21, 55, and 253. Let G=W×X×Y×Z. For a group H and a prime p,Hpdenotes a Sylow p-subgroup of H. Then Gis a totally permutable product of A=W×X3×Z23 and B=X7×Y×Z11, but none of them is a PSTc-group, because neither all cyclic subgroups of the Sylow 3-subgroup of Apermute with all Sylow 2-subgroups of A, nor all cyclic subgroups of the Sylow 11-subgroup of Bpermute with all Sylow 5subgroups of B. Nevertheless, the group Gis clearly a PSTc-group. However, the extension of the “only if” part of [3; Theorem C] holds. Theorem 8. Assume that the group G=AB is a totally permutable product of the soluble PSTc-groups Aand Band that gcd(|G:A|,|G:B|) = 1. Then Gis a soluble PSTc-group. Corollary 9 ([3]).Assume that G=AB is a totally permutable product of the soluble PST-groups Aand Bsuch that gcd(|G:A|,|G:B|) = 1. Then Gis a soluble PST-group. The classes of PTc-groups and Tc-groups are defined in a similar fashion, by requiring the cyclic subnormals to be permutable or normal, respectively. Theorem 8 also holds for these classes. 2 Proofs Proof of Theorem 1. Assume that the theorem is false, and let G=AB be a counterexample of minimal order. Then, since the class of all abelian-bynilpotent groups is a formation, Ghas a unique minimal normal subgroup M. Since Gis supersoluble by [2], Mis a p-group for some prime p, the Fitting subgroup P= F(G) is a Sylow p-subgroup G,pis the largest prime dividing 4 |G|, and G/ F(G) is abelian of exponent dividing p−1. Moreover P=ApBp, where Apis the Sylow p-subgroup of Aand Bpis the Sylow p-subgroup of B by [1; 1.3.3]. If Aand Bwere abelian, we would have Gmetabelian by Itˆo’s theorem ([1; 2.1.1]), a contradiction. Therefore we may assume that Ais not abelian. As Ap0, the Hall p0-subgroup of A, is abelian, Apis non-abelian. Let Tbe a subgroup of Ap. Then the product TBp0, where Bp0is a Hall p0-subgroup of B, is a supersoluble subgroup of G. Therefore Bp0normalises T, that is, p0-elements of Binduce power automorphisms in Ap. As Apis non-abelian, all p0-power automorphisms are trivial ([8; Hilfssatz 5]), and hence Bp0centralises Ap. Therefore Bp0centralises Pand so Bp0= 1. This means that Bis a p-group. On the other hand, all p0-elements of Ainduce power p0-automorphisms in B. Since Acannot be a p-group, it follows that Ap06= 1 and Ap0cannot centralise B. By [8; Hilfssatz 5], Bis abelian. Then CB(Ap0) = 1 and B= [B, Ap0]. By the minimality of the order of G,G/M is abelian-by-nilpotent. Therefore the nilpotent residual L/M of G/M is abelian. This implies that L/M is complemented in G/M and L/M contains no central chief factors of G/M by [7; IV, 5.18, V, 4.2, and V, 3.2]. Note that Ap0L/L centralises BL/L. In particular, Bis contained in L. Suppose that T/M =L/M ∩AM/M is non-trivial. Then T/M is a normal subgroup of G/M and contains a central minimal normal subgroup of G/M, a contradiction. Consequently, L∩Ais contained in Mand L=BM. Then Lis a p-group and Z(L) is a non-trivial normal subgroup of G. As Mis the unique minimal normal subgroup of G, it follows that Mis contained in Z(L) and either L=Bor L=B×M. In both cases Lis abelian and so Gis abelian-by-nilpotent, final contradiction. The following lemma turns out to be crucial in the proof of our results. Lemma 10. Let G=AB be a totally permutable product of the nilpotent subgroups Aand B. Let Kbe the nilpotent residual of G. For a prime p, denote Apand Bpthe Sylow p-subgroup of Aand B, respectively. The following statements hold: 1. If pis a prime dividing the order of K, then either Ap∩K6= 1 or Bp∩K6= 1. 2. If Ap∩K6= 1, then Apis contained in Kand the Hall p0-subgroup of Bdoes not centralise Ap. 3. If Bp∩K6= 1, then Bpis contained in Kand the Hall p0-subgroup of Adoes not centralise Bp. 5 Proof. By Theorem 1, Kis abelian and, by Asaad and Shaalan’s result [2], Gis supersoluble. We prove the statements by induction on |G|. Let pbe a prime dividing |K|and let qbe the largest prime dividing |G|. Then a Sylow q-subgroup Qof Gis normal in G. Suppose p6=q. By induction, G/Q satisfies 1 and either 2 or 3 because G/Q is a totally permutable product of the nilpotent subgroups ApQ/Q and BpQ/Q. Moreover KQ/Q is the nilpotent residual of G/Q. Assume that ApQ/Q ∩KQ/Q 6= 1, then it is clear that Ap∩K6= 1 because p6=q. Moreover ApQ/Q is contained in KQ/Q and then Apis contained in a Sylow p-subgroup of K. The same argument applies in the case BpQ/Q ∩KQ/Q 6= 1. Consequently we may assume that pis the largest prime dividing |G|. Then P=ApBpis a normal Sylow p-subgroup of Gand a Hall p0-subgroup of Gis abelian of exponent dividing p−1. As Kdoes not contain central p-chief factors by [7; IV, 5.18, V, 4.2, and V, 3.2], Kis not centralised by any Hall p0-subgroup of G. Hence there exist an element z∈Kof p-power order and a p0-element z1of Gin Ap0Bp0, where Ap0is the Hall p0-subgroup of Aand Bp0is the Hall p0-subgroup of B, such that zz16=z. Since z∈ApBp, we can find an element a∈Aand an element b∈Bsuch that z=ab. Moreover z1=a1b1for a1∈Ap0and b1∈Bp0. Suppose that b1does not centralise z. Then (ab)b1=ab1b∈K, and so k=ab1b(ab)−1=ab1a−1is a non-trivial element of K. Since haihb1i is a supersoluble subgroup of G, it follows that haiis normal in haihb1iand so k∈Ap. Consequently Ap∩K6= 1. If a1does not centralise z, the above argument shows that Bp∩K6= 1. Hence 1 holds. Assume now that Ap∩K6= 1. Then K∩Pis a non-trivial normal subgroup of G. Let Mbe a minimal normal subgroup of Gcontained in K∩P. By induction, the lemma holds in G/M because G/M is a totally permutable product of the nilpotent subgroups AM/M and BM/M. Moreover K/M is the nilpotent residual of G/M. Assume that pdivides |K/M|. If ApM/M ∩ K/M 6= 1, then ApM/M is contained in K/M by induction. Hence Ap≤K. Therefore we may assume that ApM/M ∩K/M = 1 and Ap∩K≤M. Since Ap∩K6= 1 and Mis of prime order, we have M=Ap∩K. On the other hand, Bp0acts as a power automorphism group on Apbecause CBp0 is a supersoluble subgroup of Gfor each subgroup Cof Ap. Consequently either Apis abelian or Bp0centralises Apby [8; Hilfssatz 5]. Suppose that Bp0 centralises Ap. Then Mis central in G. This contradicts [7; IV, 5.18, V, 4.2, and V, 3.2]. Thus Bp0cannot centralise Ap. This implies that Apis abelian and Bp0acts as a non-trivial universal power automorphism group on Apby [8; Hilfssatz 5] and [14; 13.4.3]. It follows that [Ap, Bp0] = Apby [7; A, 12.5]. Now ApK/K is centralised by Bp0K/K because G/K is nilpotent. Therefore Ap= [Ap, Bp0]≤Kas desired. Similar arguments to those used above yield Bp≤Kif Bp∩K6= 1. 6 Finally, suppose that Apis contained in Kand Bp0centralises Ap. Let 16=x∈Ap. Then there exists a chief factor E/F of Gbelow Ksuch that E/F =hxFi. It is clear that E/F is central in G, a contradiction. Consequently Bp0does not centralise Apand 2 holds. Analogously Ap0does not centralise Bpif Bp∩K6= 1. The proof of the lemma is now complete. Proof of Theorem 2. Let pbe a prime dividing |K|. Then, by Lemma 10, the Sylow p-subgroup Kpof Kmust contain Apor Bp. Assume that Bpis a proper subgroup of Kp. Then, since Kpis normal in G,Kpis contained in the Sylow p-subgroup ApBpof G. Hence there exists an element ab ∈K with a∈Apand b∈Bpand a6= 1. Since Bpis contained in K, it follows that Ap∩K6= 1. By Lemma 10, Apis contained in Kand Kp=ApBp. Assume that Z=Ap∩Bp. Then Zis centralised by a Hall p0-subgroup of G and by a Sylow p-subgroup of G(note that Kis abelian). Since Kcontains no central chief factors of Gby [7; IV, 5.18, V, 4.2, and V, 3.2], it follows that Z= 1 and Kp=Ap×Bp. It is known that if G=AB is a totally permutable product of Aand B, then [A, BN] = [B, AN] = 1 ([6; Theorem 1]), but, in general, GN6=ANBN. Our next lemma analyses this case. Lemma 11. Let G=AB be a totally permutable product of two subgroups Aand B. Denote M,N, and Kthe nilpotent residuals of A,B, and G, respectively. Suppose that K6=MN, a Sylow p-subgroup Apof Ais contained in Kfor some prime p, and [Ap, Bp0]is not contained in MN for a Hall p0subgroup Bp0of B. Then Bp0acts as a group of power automorphisms on ApM/M,Mis a p0-group, and Apis subnormal in G. Proof. First of all we will prove that Bp0normalises ApM. Denote by Tthe nilpotent residual of Bp0. Note that MT is a normal subgroup of ABp0by [6; Theorem 1]. Let H/MT be the nilpotent residual of ABp0/MT. Since [Ap, Bp0] is not contained in MN, it follows that [Ap, Bp0] is not contained in MT. Hence ApMT/MT ∩H/MT 6= 1. Now ABp0/MT is a totally permutable product of the nilpotent subgroups AMT/MT and Bp0MT/MT. By Lemma 10, ApMT/MT is contained in H/MT and since ApMT/MT is the Sylow p-subgroup of the abelian subgroup H/MT, it follows that ApMT is normal in ABp0. On the other hand, ABp0/M is a totally permutable product of the subgroups AM/M and Bp0M/M. By Beidleman and Heineken’s result [6; Theorem 1], [ApM/M, TM/M] = 1. Hence ApMT/M has ApM/M as a unique Sylow p-subgroup. Therefore ApMis normal in ABp0. This implies that if Xis a subgroup of ApM/M, then X(Bp0M/M) is a subgroup of G/M and Bp0M/M normalises X. This means that Bp0acts as 7