On some classes of supersoluble groups

This paper has been published in Journal of Algebra, 312(1):445–454 (2007). Copyright 2007 by Elsevier. The final publication is available at www.sciencedirect.com. http://dx.doi.org/10.1016/j.jalgebra.2006.07.035 http://www.sciencedirect.com/science/article/pii/S0021869306008118 On some classes of supersoluble groups A. Ballester-Bolinches∗J.C. Beidleman† R. Esteban-Romero‡ 15th August 2006 Dedicated to Professor Hermann Heineken on the occasion of his seventieth birthday Abstract Finite groups Gfor which for every subgroup Hand for all primes qdividing the index |G:H|there exists a subgroup Kof Gsuch that His contained in Kand |K:H|=qare called Y-groups. Groups in which subnormal subgroups permute with all Sylow subgroups are called PST-groups. In this paper a local version of the Y-property leading to a local characterisation of Y-groups, from which the classical characterisation emerges, is introduced. The relationship between PST-groups and Y-groups is also analysed. 1 Introduction and statement of results In this paper, only finite groups will be taken into account. A well-known theorem of Lagrange (see [11, I, 2.7]) states that given a subgroup Hof a group G, the order of Gis the product of the order |H| of Hand the index |G:H|of Hin G. In particular, the order of any subgroup divides the order of the group. The converse, namely, if ddivides the order of a group G, then Ghas a subgroup of order d, is not true in general. Groups satisfying this condition are often called CLT-groups. The ∗Departament d’Àlgebra, Universitat de València, Dr. Moliner, 50, E-46100 Burjassot, València, Spain, e-mail: [email protected] †Department of Mathematics, University of Kentucky, Lexington, Kentucky 405060027, USA, e-mail: [email protected] ‡Departament de Matemàtica Aplicada-IMPA, Universitat Politècnica de València, Camí de Vera, s/n, E-46022 València, Spain, e-mail: [email protected] 1 alternating group of order 12, having no subgroups of order 6, is an example of a non-CLT-group. On the other hand, abelian groups contain subgroups of every possible order, and it is not difficult to prove that a group is nilpotent if and only if it contains a normal subgroup of each possible order [10]. Ore [13] and Zappa [16] obtained a similar characterisation for supersoluble groups: Theorem 1. A group Gis supersoluble if and only if each subgroup H≤G contains a subgroup of order dfor each divisor dof |H|. Of course, we can state Theorem 1 in the following equivalent way, more easily treated: Theorem 2. A group Gis supersoluble if and only if each subgroup H≤G contains a subgroup of index pfor each prime divisor pof |H|. A proof of this theorem can be found in [14, Chapter 1, 4.3]. It must be noted that CLT-groups are not necessarily supersoluble, as the symmetric group of order 4shows. The condition on a group Ggiven in Theorem 2, namely for all H≤Gand for all primes qdividing |H|, there exists a subgroup Kof Gsuch that K≤Hand |H:K|=q, has a dual formulation: for all H≤Gand for all primes qdividing |G:H|, there exists a subgroup Kof Gsuch that H≤Kand |K:H|=q. Groups satisfying the latter condition have been studied by some authors. Following [14, Chapter 1, 4], we will call them Y-groups. Definition 3. A group Gis said to be a Y-group if for all subgroups Hof G and all primes qdividing the index |G:H|of Hin G, there exists a subgroup Kof Gwith H≤Kand |K:H|=q. Note that a group Gis a Y-group if and only if for every subgroup Hof Gand for every natural number ddividing |G:H|there exists a subgroup Kof Gsuch that H≤Kand |K:H|=d. The following characterisation of Y-groups appears in [14, Chapter 1, 4.3]. Theorem 4. Let L=GNbe the nilpotent residual of the group G. Then G is a Y-group if and only if Lis a nilpotent Hall subgroup of Gsuch that for all subgroups Hof L,G=LNG(H). 2 From Theorem 4, we see that if G∈ Y and Xis a normal subgroup of L, then Xis normal in G. In particular, Y-groups are supersoluble. Moreover, if G∈ Y, then Lmust have odd order. Further results on Y-groups can be found in [14, Chapter 6, 6.1]. For example, a soluble group Gis a Y-group if and only if every subgroup of G can be written as an intersection of subgroups of Gof coprime prime-power indices. On the other hand, we say that a subgroup Hof a group Gis S-permutable in Gwhen it permutes with every Sylow subgroup of G. According to [12], S-permutable subgroups are subnormal and the set of all S-permutable subgroups of a group Gis a sublattice of the lattice of all subnormal subgroups of G. A group Gis said to be a PST-group when every subnormal subgroup of Gis S-permutable, that is, when S-permutability is a transitive relation. Some interesting subclasses of the class of all PST-groups are the class of PT-groups or groups in which permutability is a transitive relation and the class of T-groups or groups in which normality is a transitive relation. Soluble PST-groups were studied by Agrawal [1], and, more recently, by Alejandre, the first author, and Pedraza-Aguilera in [2], by the first and the last author [3, 4], and by the second author and Heineken [6], among others. The approach followed in these papers began with a paper of Bryce and Cossey [7] in which a local version of some of the results on T-groups was presented. Let us recall the classical theorem of Agrawal: Theorem 5. A group Gis a soluble PST-group if and only if Ghas an abelian normal Hall subgroup Nof odd order such that G/N is nilpotent and the elements of Ginduce power automorphisms in N. If we add in this result “G/N nilpotent modular group,” we obtain the characterisation of soluble PT-groups given by Zacher [15], and if we put “G/N Dedekind,” we get Gaschütz’s characterisation of soluble T-groups [9]. Agrawal’s theorem has the virtue of showing that the class of soluble PST-groups is closed under taking subgroups. A consequence of Theorems 4 and 5, Gaschütz’s characterisation, and Dedekind theorem [11, III, 7.12] is: Corollary 6. Let Gbe a group. 1. If Gis a soluble PST-group, then Gis a Y-group. 2. Assume that G∈ Y. Then Gis a soluble PST-group if and only if the nilpotent residual of Gis abelian. 3 3. Assume that G∈ Y. Then Gis a soluble T-group if and only if all Sylow subgroups of Gare Dedekind. For a prime p, Bryce and Cossey [7] defined the class Tpof all soluble groups Gfor which every subnormal p0-perfect subgroup of Gis normal. They proved: Theorem 7. A soluble group is a T-group if and only if it is a Tp-group for all primes p. In [2], Alejandre, the first author, and Pedraza-Aguilera introduced in the soluble universe the class PSTpof all soluble groups Gin which every p0-perfect subnormal subgroup in Gpermutes with every Hall p0-subgroup of G. This condition is equivalent to Gbeing p-supersoluble and having all its p-chief factors isomorphic when regarded as modules over G(see [2]). This result not only holds in the soluble universe, but also in the p-soluble one. Theorem 8. A soluble group Gis a PST-group if and only if Gis a PSTpgroup for all primes p. The second author and Heineken defined in [6] the class T00 p, for a prime p, of all soluble groups Gin which every p0-perfect subnormal subgroup of G is S-permutable in Gand proved: Theorem 9. A soluble group Gis a PST-group if and only if it is a T00 p-group for all primes p. A similar result holds for PT-groups replacing S-permutability by permutability. In [3, Theorem A], the following local version of Agrawal’s result was obtained. For each group Xand every prime p,X(p)denotes the p-nilpotent residual of X, that is, the smallest normal subgroup Nof Xsuch that X/N is p-nilpotent, while Op0(X)denotes the largest normal p0-subgroup of X. Theorem 10. Ap-soluble group Gis a PSTp-group if, and only if, one of the following two conditions holds: 1. Gis p-nilpotent, or 2. the subgroup G(p)/Op0G(p)is an abelian normal Sylow p-subgroup of G/ Op0G(p)in which the elements of G/ Op0G(p)induce power automorphisms. 4 Theorem 5 follows from Theorems 8 and 10, as shown in [3]. These local results, together with Corollary 6, encourages us for the search of local versions of the Y-property, leading to a local characterisation of Ygroups, running parallel to the characterisations for PSTp-groups, pa prime. This is the aim of the present paper. In the sequel, pwill denote a fixed prime number. Definition 11. We say that Gsatisfies Zpwhen for every p-subgroup Xof Gand for every power of a prime q,qm, dividing |G:XOp0(G)|, there exists a subgroup Kof Gcontaining XOp0(G)such that |K:XOp0(G)|=qm. Note that if q=p, the condition is obviously satisfied in every group. Definition 12. Let Gbe a group. We say that Gsatisfies Z0 pif Gsatisfies either of the following conditions: 1. Gis p-nilpotent, or 2. G(p)/Op0G(p)is a Sylow p-subgroup of G/ Op0G(p)and for every p-subgroup Hof G(p), we have that G=G(p) NG(H). Our first main result can be regarded as the analogue of Theorem 10: Theorem 13. Let Gbe a p-soluble group. Then Gsatisfies Zpif and only if Gsatisfies Z0 p. Combining Theorems 10 and 13, we have: Theorem 14. Ap-soluble group Gsatisfies PSTpif and only if Gsatisfies Zpand Ghas an abelian Sylow p-subgroup. Our second main result is the analogue of Theorem 8. Theorem 15. Let Gbe a soluble group. Gsatisfies Yif and only if G satisfies Zpfor every prime p. Recall that class of groups which is closed under taking epimorphic images and subdirect products is called a formation. In [5] it has been proved that the largest formation contained in the class PSTpis the class Ep0Spof all p-nilpotent groups. As a consequence, the class of all nilpotent groups is the largest formation contained in the class of all PST-groups. A similar result can be obtained for the class of all groups satisfying Zp: Theorem 16. The class of all p-nilpotent groups is the largest formation contained in the class Zp. 5 As a consequence: Corollary 17. The class of all nilpotent groups is the largest formation contained in the class Y. We bring the paper to a close by giving an alternative proof of Theorem 4 which is based on our local approach. We also show that the class Yis a proper subclass of the class of all supersoluble groups and that the classes Zpare not subgroup-closed in general. The notation used in this paper is standard. For notation not explained, we address the reader to the book [8]. 2 Proofs The following lemmas turn out to be crucial in the proofs of our main results. The first and second ones are very useful in induction arguments. The proof of the first lemma is a routine check. Lemma 18. Let Gbe a group and let Nbe a normal subgroup of G. Then: 1. If Nis a p0-subgroup, then Gsatisfies Zpif and only if G/N satisfies Zp. 2. If Nis a p-group and Gsatisfies Zp, then G/N satisfies Zp. 3. If Gis a group, Gis not p-nilpotent, and Nis a normal subgroup of Gcontained in Op0G(p), then Gis satisfies Z0 pif and only if G/N satisfies Z0 p. 4. If Gis a group satisfying Z0 pand Nis a normal p-subgroup of G, then G/N satisfies Z0 p. Lemma 19. If Gis a group and Nis a normal p0-subgroup of G, then G satisfies Z0 pif and only if G/N satisfies Z0 p. Proof. Assume that Gsatisfies Z0 p. If Gis p-nilpotent, then G/N is also p-nilpotent and so G/N satisfies Z0 p. Suppose that Gis not p-nilpotent. According to Lemma 18, we can suppose that Op0G(p)= 1 by changing Nby NOp0G(p)Op0G(p)if needed. Then (G/N)(p) = G(p)N/N is a Sylow p-subgroup of G/N. It is clear that G/N = NG(H)G(p)/N = NG/N (HN/N)(G/N)(p)for every p-subgroup HN/N of G/N. Conversely, assume that there exists a group Ghaving a normal p0subgroup Nsuch that G/N satisfies Z0 p, but Gdoes not satisfy Z0 p. We 6 choose Gof minimal order. We can suppose that Nis a minimal normal subgroup of Gand Nis not contained in G(p). In this case, (G/N)(p) = G(p)N/N. Moreover, Op0(G/N)(p)=Op0G(p)N/N. It follows that G(p)/Op0G(p)is a Sylow p-subgroup of G/ Op0G(p). Let Hbe a psubgroup of G. Then G/N = NG/N (HN/N)(G/N)(p)=NG(H)N/NG(p)N/N. This implies that G= NG(H)G(p)N. But Ncentralises G(p)and so H. It follows that G= NG(H)G(p), and Gsatisfies condition Z0 p. Lemma 20. Let Gbe a p-soluble group satisfying Zp. Then Gis p-supersoluble. Proof. Assume that the result is false. Consider a p-soluble group Gof minimal order such that Gsatisfies Zp, but Gis not p-supersoluble. From the p-solubility of G, we can assume that Ghas a unique minimal normal subgroup, Nsay, and that Nis a non-cyclic p-group. Moreover Op0(G) = 1. Let Zbe a subgroup of Nof order pcontained in the centre of a Sylow p-subgroup Pof G. Let qbe a prime different from pdividing n=|G:Z| and let qmbe the largest power of qdividing n. Since Ghas property Zp, there exists a subgroup Kof Gsuch that |K:Z|=qm. Moreover Zis a Sylow p-subgroup of Kand is subnormal in G. This implies that Zis normal in K. In particular, Zis normalised by a Sylow q-subgroup of G. Since this happens for all primes q6=pand Zis normal in P, we have that Zis a normal subgroup of G. Hence N=Zbecause Nis a minimal normal subgroup of Gand so Nhas order p. This contradiction shows that no such counterexample exists and the result is proved. Lemma 21. If Gis p-soluble and satisfies Z0 p, then Gis p-supersoluble. Proof. Assume that there exist p-soluble groups satisfying Z0 pwhich is not psupersoluble, and among them we choose a group Gof minimal order. Clearly Gis not p-nilpotent. By Lemma 19, we have that Op0(G)=1. Thus G(p) is a normal Sylow p-subgroup of G. Let Nbe a minimal normal subgroup of G, then Nis a p-group and Ncannot be cyclic, by Lemma 18 and the minimality of G. Let Mbe a minimal normal subgroup of G(p)contained in N. We have that G= NG(M)G(p)=NG(M). Therefore Mis normal in G. This contradiction shows that no such counterexample can exist and the lemma is proved. The following lemma is fundamental to understand the class Yand its local versions. 7 Lemma 22. Let Gbe a p-soluble group satisfying Zpand let Hbe a subgroup of Op(G). Then His normalised by a Hall p0-subgroup of G. Proof. Consider a prime qdifferent from p. There exists a subgroup Kof G such that HOp0(G)≤Kand |K:HOp0(G)|=qm, where qmis the largest power of qdividing |K:HOp0(G)|. A Sylow q-subgroup Kqof Kis a Sylow q-subgroup of Gand His a subnormal Sylow p-subgroup of K. Therefore H is normal in Kand, in particular, Kqnormalises H. Since this happens for every q6=p, we have that hKq|q6=pinormalises H. In particular, there exists a Hall p0-subgroup of Gnormalising H. Proof of Theorem 13. Assume that Gsatisfies Zp. We show that Galso satisfies Z0 p. Since p-nilpotent groups satisfy Z0 p, there is no loss of generality in supposing that Gis not p-nilpotent. By Lemma 20, we have that Gis p-supersoluble. In particular, the derived subgroup G0of Gis p-nilpotent by [11, VI, 9.1(a)]. Since G(p)≤G0, we have that G(p)is p-nilpotent, too. Moreover, by Lemma 19, we can assume that Op0(G) = 1. It is clear then that G(p)is a p-group. We prove that G(p)is a Sylow p-subgroup of G. Assume that this is false and derive a contradiction. In this case, Ghas both central and non-central p-chief factors. Bearing in mind that G0is p-nilpotent and Op0(G0) = 1, we obtain that G0is a p-group. Hence Ghas a normal Sylow p-subgroup, L say. Consider a chief series of Gpassing through Land G(p), and consider two chief factors of that series, K/G(p)and G(p)/M, say. Then K/G(p)is central in Gand G(p)/M is not central in G. Applying [8, IV, 6.7], K/G(p) is not cyclic. Therefore K/M is a p-elementary abelian group of order p2, because Gis p-supersoluble. Let Hbe a Hall p0-subgroup of G. It is clear that K/M is an H-module over GF(p). By Maschke’s theorem [8, A, 11.4], K/M =G(p)/M ×C/M, where C/M is normalised by H. Let G(p)/M =haMiand C/M =hbMi. Note that C/M is H-isomorphic to K/G(p), and so C/M is in fact centralised by H. Consider D=Mhabi. By Lemma 18, we have that G/M satisfies Zp. By Lemma 22, there exists a Hall p0-subgroup H1of Gnormalising D. Since all Hall p0-subgroups are conjugate by [11, VI, 1.7], there exists an element g∈Gsuch that H1=Hg. Moreover, G=HL and so g=hx with h∈Hand x∈L. Hence H1=Hxwith x∈L. Let ybe an element of Hsuch that ayM6=aM. Then ayM=aiM, for some natural number i≥2. Note that K=ha, biMand Kx−1=K. Consequently ha, biM=hax−1, bx−1iMand, bearing in mind that K/M ∼ = Cp×Cp, there exist two natural numbers mand nwith 1≤m≤p−1and 1≤n≤p−1and (ab)x−1M=ambnM. Since h(ab)x−1iMis normalised by 8