A question on partial CAP-subgroups of finite groups
Abstract
A subgroup H of a finite group G is a partial CAP-subgroup of G if there is a chief series of G such that H either covers or avoids every chief factor of the series. The structural impact of the partial cover and avoidance property of some distinguished subgroups of a group has been studied by many authors. However there are still some open questions which deserve an answer. The purpose of the present paper is to give a complete answer to one of these questions.
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This paper has been published in Science China Mathematics, 55(5):961–966 (2012). Copyright 2012 by Science China Press and Springer-Verlag. The final publication is available at www.springerlink.com. http://link.springer.com/article/10.1007/s11425-011-4356-9 http://dx.doi.org/10.1007/s11425-011-4356-9
A question on partial CAP-subgroups of finite groups A. Ballester-Bolinches∗R. Esteban-Romero† Yangming Li‡ Abstract A subgroup Hof a finite group Gis a partial CAP-subgroup of G if there is a chief series of Gsuch that Heither covers or avoids every chief factor of the series. The structural impact of the partial cover and avoidance property of some distinguished subgroups of a group has been studied by many authors. However there are still some open questions which deserve an answer. The purpose of the present paper is to give a complete answer to one of these questions. Keywords: finite groups, minimal subgroups, partial cover and avoidance property, saturated formations. Mathematics Subject Classification (2000): 20D10, 20D20. 1 Introduction In this paper all groups are assumed to be finite. A subgroup Aof a group Gis said to have the cover-avoidance property in Gand is called a CAP-subgroup of Gif either HA =KA or A∩H=A∩Kfor every chief factor H/K of G. CAP-subgroups have played an important role in the structural study of soluble groups because, in that universe, some important subgroups such as Hall subgroups, maximal subgroups, normalisers associated to saturated formations and prefrattini subgroups enjoy the property. In fact, as it is shown in [1, Chapter 4], the cover and avoidance property and conjugacy of ∗Departament d’Àlgebra, Universitat de València, Dr. Moliner, 50, E-46100 Burjassot, València, Spain, email: [email protected] †Institut Universitari de Matemàtica Pura i Aplicada, Universitat Politècnica de València, Camí de Vera, s/n, E-46022 València, Spain, email: [email protected] ‡Department of Mathematics, Guangdong University of Education, Guangzhou, 510310, People’s Republic of China; email: [email protected] 1
the members of some of the above families characterise solubility. Despite of this, there was a little evidence to suggest a huge proliferation of results in the area. However, Ezquerro [5] began to study the structural impact of the cover and avoidance embedding of some relevant families of subgroups. He observed that it is enough to impose the cover and avoidance property to the maximal subgroups of the Sylow subgroups to get supersolubility. This result establishes the standpoint for a research project consisting in characterising some formations by the cover and avoidance property of some distinguished subgroups such as maximal subgroups of Sylow subgroups, second maximal subgroups of Sylow subgroups or minimal subgroups. For an overview, the reader is referred to [2, 7, 11] and the papers cited therein. Our point of departure is the observation that the cover and avoidance property is not persistent in intermediate subgroups, that is, if His a CAPsubgroup of Gand His contained in K, then His not a CAP-subgroup of K in general (cf. [2, Example 1.3]). Surprisingly, the following weaker property introduced by Y. Fan, X. Guo, and K. P. Shum in [6] has this property: a subgroup Aof a group Gis said to be partial CAP-subgroup of Gif there exists a chief series ΓAof Gsuch that Aeither covers or avoids each factor of ΓA. These type of subgroups are also called semi CAP-subgroups (see [6]) or SCAP-subgroups (see [11]). Of course, every subgroup of a supersoluble group is a partial CAP-subgroup. Hence the natural question arising is whether there are more restricted families of subgroups whose partial cover and avoidance property could guarantee supersolubility. In [2] and [11] it is proved, as a particular case of a more general result, that the partial CAPproperty of a much more restricted family of subgroups, namely the maximal subgroups of the Sylow subgroups of the generalised Fitting subgroup, implies supersolubility. In addition, in [2] a local approach is used. It allows us to discover new situations and see how the global properties can be deduced from the local ones. A further step in this analysis is to consider a dual family, that is, the one of all subgroups of order por order 4, and wonder whether a group is supersoluble provided that all the members of this family are partial CAPsubgroups. In fact, the following more general question was asked by the third author in [10]: Question 1.1. Let Fbe a saturated formation containing U, the class of all supersoluble groups, and Ha normal subgroup of a group Gsuch that G/H ∈F. Suppose that, for every non-cyclic Sylow subgroup Pof F∗(H), Phas a subgroup Dsuch that 1<|D|<|P|and all subgroups Eof Pwith order |E|=|D|and with order |E|= 2|D|(if Pis a non-abelian 2-group and |P:D|>2) are partial CAP-subgroups of G. Does Gbelong to F? 2
The following example gives a negative answer to Question 1.1. Example 1.2. Let H=ha, b |a5=b5= 1, ab =baibe an elementary abelian group of order 52, and let αbe an automorphism of Hof order 3 satisfying that aα=b,bα=a−1b−1. Let H1,H2=ha0, b0ibe two copies of Hand denote by G= [H1×H2]hαithe corresponding semidirect product. If Ais a subgroup of Gof order 25, there exists a minimal normal subgroup Nsuch that A∩N= 1. Hence Acovers or avoids the following chief series of G: 1< N < AN < G. Consequently Ais a partial CAP-subgroup of Gand so Gis a non-supersoluble group whose every second maximal subgroups are partial CAPsubgroups. Note that His an absolutely irreducible C3-module over the finite field of 5-elements. In fact, the main result of [3] characterises when the partial cover and avoidance property of the second maximal subgroups implies supersolubility. The main objective of the present paper is to show that Question 1.1 has an affirmative answer for minimal subgroups. Theorem 1.3. Let Fbe a saturated formation containing U, the class of all supersoluble groups, and let Gbe a group with a normal subgroup Hsuch that G/H ∈F. Then G∈Fif every cyclic subgroup of F∗(H)of prime order or order 4is a partial CAP-subgroup of G. We note that the saturation of Fcannot be removed in Theorem 1.3. Consider the formation Fof all groups whose supersoluble residual is trivial or a direct product of copies of the alternating group S=A5of degree 5. Then Fis a formation which contains the class Uof all supersoluble groups. The group G= SL2(5) has a normal subgroup H= Z(G)of order 2and obviously H, which is the unique subgroup of prime order of F∗(H), is a partial CAP-subgroup of G. However, Gdoes not belong to F. The proof of the above result depends on the following local theorem. Theorem 1.4. Let pbe a prime and let Gbe a group. If every cyclic subgroup of Gof order por order 4is a partial CAP-subgroup of G, then Gis psupersoluble. As a consequence of Theorem 1.3, we get another local result. Corollary 1.5. Assume that pis a prime and Fis a saturated formation containing all p-supersoluble groups such that Ep0F=F. Suppose that Gis a group with a normal subgroup Nsuch that G/N belongs to F. If every cyclic subgroup of F∗ p(N)of order por 4is a partial CAP-subgroup of G, then G belongs to F. 3
Recall that for every group Xand for every prime p, the subgroup F∗ p(X) is defined to be the subgroup of Xsuch that F∗ p(X)/Op0(X) = F∗X/Op0(X). 2 Preliminaries This section contains the results which are needed to prove our main theorems. We begin with a lemma containing the basic properties of the partial CAP-subgroups which turn out very useful in induction arguments. Lemma 2.1 (see [6]).Every CAP-subgroup of Gis a partial CAP-subgroup of G. Furthermore, if Sis a partial CAP-subgroup of a group G, then: 1. If S≤K≤G, then Sis a partial CAP-subgroup of K. 2. If N≤Sand NEG, then S/N is a partial CAP-subgroup of G/N. 3. If NEGand (|S|,|N|)=1, then SN/N is a partial CAP-subgroup of G/N. The next lemma describes a configuration often encountered in the study of partial CAP-subgroups. Although its proof is part of the proof of Lemma 2.2 in [2], we include it here for the sake of completeness. Lemma 2.2. Let Hbe a partial CAP-subgroup of a group G. Suppose that Qis a normal subgroup of Gsuch that His contained in Q. Then there exists a chief series ΩHof Gpassing through Qsuch that Heither covers or avoids each chief factor in ΩH. Proof. Since His a partial CAP-subgroup of the group G, there exists a chief series ΓH:1=G0< G1<· · · < Gn=G of Gsuch that Heither covers or avoids each chief factor in ΓH. Since Qis a normal subgroup of G, then Qis a CAP-subgroup of G. Therefore ΓH∩Q: 1 = G0∩Q < G1∩Q < · · · < Gn∩Q=Q is, avoiding repetitions, part of a chief series of G. Moreover, if Hcovers (respectively, avoids) Gi+1/Gi, then Hcovers (respectively, avoids) (Gi+1 ∩ Q)/(Gi∩Q). We can complete ΓH∩Qto obtain a chief series ΩHof G. Note that H avoids all chief factors above Q. Hence there exits a chief series ΩH: 1 = G∗ 0< G∗ 1<· · · < G∗ r=Q<G∗ r+1 <· · · < G∗ n=G of Gsuch that Hcovers or avoids all chief factors of Gin ΩH. 4
Recall that a class of groups Fis called a formation if it is closed under taking epimorphic images and subdirect products. Fis said to be saturated if G/Φ(G)∈Fimplies G∈F. Let Fbe a non-empty formation. Each group Ghas a smallest normal subgroup whose quotient belongs to F; this is called the F-residual of Gand it is denoted by GF. Clearly GFis a characteristic subgroup of G(cf. [1, 4] for details). Let pbe a prime. A group Gis said to be p-supersoluble if Gis p-soluble and every chief factor of order divisible by pis cyclic. The class Upof all p-supersoluble groups is a saturated formation. Clearly the intersection of all Upis again a saturated formation which is composed of all soluble groups whose chief factors are cyclic. This class is the class of all supersoluble groups and is denoted by U. Let Hbe a non-empty class of groups. According to [1, 1.2.9, 2.3.18], a chief factor H/K of a group Gis said to be H-central in Gif [H/K]∗Gbelongs to H, where [H/K]∗Gis the semidirect product [H/K]G/CG(H/K)if H/K is abelian and G/CG(H/K)if H/K is non-abelian. A normal subgroup N of a group Gis called H-hypercentral in Gif every chief factor of Gbelow Nis H-central in G. By virtue of the generalised Jordan-Hölder theorem [1, 1.2.36], we obtain that the product of H-hypercentral normal subgroups of a group Gis again H-hypercentral in G. Thus every group Gpossesses a unique maximal normal H-hypercentral subgroup called the H-hypercentre of Gand denoted by ZH(G). Applying again the generalised Jordan-Hölder theorem, every chief factor of Gbelow ZH(G)is H-central in G. The following theorem is a good illustration of how the partial cover and avoidance property of the minimal subgroups influences the embedding of a normal p-subgroup and it plays a crucial part in the proof of our main results. Theorem 2.3. Let pbe a prime and let Pbe a normal p-subgroup of a group G. If every cyclic subgroup of Pof order por 4is a partial CAP-subgroup of G, then Pis contained in ZU(G). Proof. Suppose, by way of contradiction, that the theorem is false, and choose a pair (G, P)for which it fails. Then there exists a chief factor of Gbelow P which is not of prime order. Among the non-cyclic chief factors of Gbelow P, we choose one L/K with |L|as small as possible. Assume there exists an element xof Lof prime order or order 4which is not in K. Then hxiis a partial CAP-subgroup of G. Applying Lemma 2.2, there exists a chief series Γ : 1 = L0< L1<· · · < Ls−1< Ls=L < · · · < G 5
of Gpassing through Lsuch that hxieither covers or avoids each chief factor in Γ. By the choice of L/K, we have each chief factor Li/Li−1is of prime order for i= 1,2, . . . , s −1. If hxicovers Ls/Ls−1, then Ls/Ls−1is of prime order. Hence L≤ZU(G). This implies that L/K is of prime order, which is not the case. Therefore hxiavoids Ls/Ls−1. Then x∈Ls−1and so Ls−1K > K. Hence L=Ls−1K. Since L/K =Ls−1K/K is G-isomorphic to Ls−1/(Ls−1∩ K), it follows that Ls−1/(Ls−1∩K)is a chief factor of G. The choice of L/K implies that Ls−1/(Ls−1∩K)is of prime order. Consequently L/K is of prime order. This contradiction proves that every element of Lof order p or order 4is contained in K. Let Xdenote the intersection of the centralisers of the chief factors of Gbelow K. Then Xstabilises a chain of subgroups of K. Applying [4, A, 12.4], Op(X)centralises K. In particular, Op(X)centralises every element of prime order or order 4of L. By [8, IV, 5.12], Op(X)centralises L. Thus X/CX(L/K)is a normal p-subgroup of G/CG(L/K). By [4, B, 3.12], X centralises L/K. This implies that L/K can be regarded as an irreducible G/X-module over the finite field of p-elements. Note that every chief factor U/V of Gbelow Kis of order pand so G/CG(U/V )is cyclic of order dividing p−1. Consequently, G/X is abelian of exponent dividing p−1. Applying [4, B, 9.8], L/K has order p. This final contradiction shows that no such counterexample Gexists. Since every cyclic chief factor of order 2in a given chief series of a group is central, we have: Corollary 2.4. Suppose that Pis a normal 2-subgroup of G. If every cyclic subgroup of Pof order 2or 4is a partial CAP-subgroup of G, then P≤ Z∞(G), the nilpotent hypercentre of G. 3 Proofs of the main theorems We are now ready to prove our main results. Proof of Theorem 1.4. Suppose that the theorem is false, and let Gbe a counterexample of minimal order. The structure of Gis analysed, and eventually a contradiction is reached. For the ease of reading we break the argument into separately-stated steps. Step 1. Every proper subgroup of Gis p-supersoluble, that is, Gis a minimal non-p-supersoluble group. Let Mbe a maximal subgroup of Gand let Lbe a cyclic subgroup of Mof order por order 4. Then Lis a partial CAP-subgroup of G. By 6
Lemma 2.1 (1), Lis a partial CAP-subgroup of M. Hence Msatisfies the hypotheses of the theorem. The minimal choice of Gyields that Mis psupersoluble. Consequently, every proper subgroup of Gis p-supersoluble, since the class of all p-supersoluble groups is subgroup-closed. This is to say that Gis a minimal non-p-supersoluble group. Step 2. Op0(G) = 1. Therefore F(G)=Op(G). Set G=G/Op0(G). Suppose that Lis a cyclic subgroup of Gof order p or 4. Then we can write L=LOp0(G)/Op0(G), where Lis a cyclic subgroup of Gof order por 4. By hypothesis, Lis a partial CAP-subgroup of G. By Lemma 2.1 (3), Lis a partial CAP-subgroup of G. Hence Gsatisfies the hypothesis of the theorem. If Op0(G)were non-trivial, then Gwould be a p-supersoluble group by the minimal choice of G. In this case, Gwould be a p-supersoluble group. This contradicts Step 1. Hence Op0(G) = 1. Applying Theorem 2.3, we have: Step 3. F(G)≤ZU(G). Step 4. G= F∗(G). Assume that F∗(G)is a proper subgroup of G. Then F∗(G)is p-supersoluble and so F∗(G) = F(G)=Op(G). Since F(G)is U-hypercentral in G, we can apply [4, IV, 6.10] to conclude that G/CGF∗(G)∈U. By [9, X, 13.12], CGF∗(G)≤F(G)and so G/F(G)∈U. Since every G-chief factor of F(G) is of order p, it follows that G∈Up, contrary to supposition. Hence we have G= F∗(G). Step 5. G/Z(G)is non-abelian simple and Gis perfect. By Step 4 G= F∗(G) = F(G)E(G), where E(G)is the layer of G, that is, the product of all components of G(cf. [9, X, 13.18]). Since Gis not nilpotent, we have that E(G)is not contained in F(G). Let Hbe a component of G. Then His normal in Gand H/Z(H)is non-abelian simple. Moreover, by Step 2, pdivides the order of H. In particular, His not p-supersoluble. By Step 1, H=G. Step 6. The conclusion of the proof. Assume that Mis a maximal subgroup of Gnot containing Z(G). Then MZ(G) = Gand so Mis a normal subgroup of Gbecause Mis normalised by Mand centralised by Z(G). But then G/M is isomorphic to the abelian group Z(G)/M∩Z(G), which contradicts that Gis perfect. Consequently Z(G)is contained in the Frattini subgroup of G. Let Abe a normal subgroup of Gsuch that G/A is a chief factor of G. Then AZ(G)6=G. Therefore A contains Z(G)and so A= Z(G). Hence Z(G)belongs to all the chief series of G. Let xbe an element of Gof order por 4. We know hxiis a partial CAP-subgroup of G. Hence hxicovers or avoids the G-chief factor G/Z(G). This implies that x∈Z(G). Applying [8, IV, 5.5], Gis p-nilpotent. This final contradiction completes the proof. 7
Proof of Theorem 1.3. Clearly we may assume that H6= 1. Let pbe a prime dividing the order of F∗(H). By hypothesis, every cyclic subgroup of F∗(H)of prime order or order 4is a partial CAP-subgroup of G. By Lemma 2.1 (1), every cyclic subgroup of F∗(H)of prime order or order 4is a partial CAP-subgroup of F∗(H). By Theorem 1.4, we know that F∗(H) is p-supersoluble for all primes p. Therefore F∗(H)is supersoluble and so F∗(H) = F(H). Moreover, applying Theorem 2.3, every Sylow subgroup of F(H)is contained in the supersoluble hypercentre of G. Hence F(H)≤ ZU(G). By virtue of [4, IV, 6.10], G/CGF(H)∈U. Since G/H belongs to Fand Fcontains U, it follows that G/CHF(H)∈F. This implies that G/F(H)∈Fbecause CHF∗(H)≤F(H)(cf. [9, X, 13.12]). Since every chief factor of Gbelow F(H)is of prime order and Fcontains U, we obtain that Gacts F-hypercentrally on F(H). It follows that G∈F. Proof of Corollary 1.5. We argue by induction on the order of G. Applying Lemma 2.1 (1), every cyclic subgroup of F∗ p(N)of order por 4is a partial CAP-subgroup of F∗ p(N). Applying Theorem 1.4, F∗ p(N)is p-supersoluble. If F∗ p(N)is a p0-group, then F∗ p(N) = Op0(N)and so Nis a p0-group. In this case G∈Ep0F=F. Therefore, we may assume that pdivides the order of F∗ p(N). Since every abelian chief factor in a given chief series of F∗ p(N)with order divisible by pis central in F∗ p(N), we conclude that F∗ p(N)is p-nilpotent, that is, F∗ p(N)=Op0,p(N). Set G=G/Op0(G)and N=NOp0(G)/Op0(G). Clearly F∗ p(N) = F∗ p(N) Op0(G)/Op0(G). If Lis a cyclic subgroup of F∗ p(N)of order por 4, we can write L=LOp0(G)/Op0(G), where Lis a cyclic subgroup of F∗ p(N)of order por 4. By hypothesis, Lis a partial CAP-subgroup of G. Hence Lis a partial CAP-subgroup of Gby Lemma 2.1 (3). Thus Gsatisfies the hypothesis of the theorem. If Op0(G)6= 1, then G∈F. Then G∈F, as Ep0F=F. Hence we can assume that Op0(G) = 1. Therefore F∗(N) = F∗ p(N)=Fp(N) = F(N) = Op(N). Applying Theorem 1.3, G∈F. Acknowledgements The first and the second authors have been supported by MEC, Spain, and FEDER, European Union (Grant No. MTM-2007-68010-C03-02) and MICINN, Spain (Grant No. MTM-2010-19938-C03-01). The third author has been supported in part by NSFC (Grant No. 11171353/A010201) and NSF of Guangdong (Grant No. S2011010004447). Part of this research was carried out during a visit of the third author to the Departament d’Àlgebra, Universitat de València, Burjassot, València, Spain, and the Institut Universitari de Matemàtica Pura i Aplicada, Universitat Politècnica de València, 8