On a graph related to permutability in finite groups

This paper has been published in Annali di Matematica Pura ed Applicata. Series IV, 189(4):567–570 (2010). Copyright 2010 by Springer-Verlag. The final publication is available at www.springerlink.com. http://link.springer.com/article/10.1007%2Fs10231-009-0124-7 http://dx.doi.org/10.1007/s10231-009-0124-7 On a graph related to permutability in finite groups A. Ballester-Bolinches∗John Cossey†R. Esteban-Romero‡ Abstract For a finite group Gwe define the graph Γ(G)to be the graph whose vertices are the conjugacy classes of cyclic subgroups of Gand two conjugacy classes A,Bare joined by an edge if for some A∈ A, B∈ B Aand Bpermute. We characterise those groups Gfor which Γ(G)is complete. Mathematics Subject Classification (2000): 5C25, 20D15, 20D20, Keywords: Finite groups, graphs, soluble groups, permutability 1 Introduction There are many ways in which a graph has been associated with a finite group. Herzog, Longobardi and Maj [8] have defined a graph whose vertices are the conjugacy classes of a group, with two vertices joined by and edge if an element of one vertex commutes with some element of the other vertex. This is a generalisation of the commuting graph of a group, which has the elements of a group as vertices, joined by an edge if they commute (see [9]). In this paper we will consider a generalisation of the graph of Herzog, Longobardi and Maj. For a finite group Gwe define the graph Γ(G)to be the graph whose vertices are the conjugacy classes of cyclic subgroups of G and two conjugacy classes A,Bare joined by an edge if for some A∈ A, B∈ B Aand Bpermute. We characterise those groups Gfor which Γ(G)is complete. ∗Departament d’Àlgebra, Universitat de València; Dr. Moliner, 50; 46100, Burjassot, València, Spain; email: [email protected] †Mathematics Department, Mathematical Sciences Institute, Australian National University; Canberra, ACT 0200, Australia; email: [email protected] ‡Institut Universitari de Matemàtica Pura i Aplicada, Universitat Politècnica de València; Camí de Vera, s/n; 46022 València, Spain, email: [email protected] 1 Recall that a subgroup Hof a group Gis said to be permutable in Gif HL is a subgroup of Gfor every subgroup Lof G. Permutability, like normality, is not a transitive relation in general. We say that a group Gis a PT-group if the permutability is transitive in G, that is, if His permutable in Kand Kis permutable in G, then His permutable in G. According to a classical result of Ore [10] permutable subgroups of finite groups are subnormal. Hence a finite group is a PT-group if and only if every subnormal subgroup is permutable. We prove: Theorem 1. A finite group Gis a soluble PT-group if and only if the graph Γ(G)is complete. 2 Proof of Theorem 1 Suppose that Gis a soluble PT-group. By a result of Zacher [12] G=AH where Ais an abelian normal subgroup of G,His a nilpotent modular subgroup of G,Aand Hhave coprime orders and every subgroup of Ais normal in G. If Xand Yare two cyclic subgroups of G, we can write X=X0X1and Y=Y0Y1, where X0and Y0are subgroups of Aand X1 and Y1have orders dividing |H|. Since His a Hall subgroup of Gwe can, by replacing Xand Yby conjugates if necessary, assume that X1and Y1 are subgroups of H. Since every subgroup of Ais normal in Gwe have Y0X1=X1Y0and since His modular we have X1Y1is a subgroup of G. It now follows that X0X1Y0Y1=Y0Y1X0X1is a subgroup, that is, Xand Y permute. In the other direction, we argue by induction on the order of G. We begin by showing that Gis a soluble PT-group if Ghas at least two minimal normal subgroups. If Mand Nare two minimal normal subgroups of G, then G/M and G/M clearly satisfy the hypothesis of the theorem. Hence G/M and G/N are soluble PT-groups. It follows that NM/M is a minimal normal subgroup of a soluble PT-group, and so is cyclic of prime order because every soluble PT-group is supersoluble. Similarly MN/N is cyclic of prime order and hence Mand Nhave prime orders, pand qsay (and Gis soluble). Then, for any prime r6=p, all r-chief factors of G/M are G-isomorphic. Further, by Zacher’s Theorem [12], Sylow r-subgroups of G/M are abelian if r-chief factors are noncentral and modular if all r-chief factors are central. If p6=qby considering G/N we have all p-chief factors G-isomorphic and Sylow p-subgroups abelian if p-chief factors are noncentral and modular if p-chief factors are central. In this case Gis a PT-group by [3, Corollary 3] and [4, Theorem 2] (note that Gis supersoluble). 2 Thus we suppose that all minimal normal subgroups have the same prime order p. If Mand Nare minimal normal subgroups and pdivides |G/MN|, then both Mand Nare G-isomorphic to a (fixed) p-chief factor of G/MN and so are G-isomorphic. Thus all p-chief factors are G-isomorphic. Therefore G is supersoluble and all chief factors of the same order are G-isomorphic. By [3, Corollary 3] Gis a group in which every subnormal subgroup permutes with all Sylow subgroups (Gis a PST-group). If the p-chief factors are central, then Gis a p-group with all proper quotients modular and so is itself modular, since by Theorem of Longobardi [7] such a group must have a unique minimal normal subgroup. Applying a result of Agrawal [2], G has an abelian Sylow p-subgroup and it then follows that Gis a PT-group by [4, Theorem 2]. Assume now that MN is a Sylow p-subgroup of G. Let M=hmi,N=hni. By hypothesis, given a p0-element y∈G,hmni permutes with a conjugate hygiof hyi. Hence hmnihygi ∩ MN =hmniis normalised by yg. Call mg=ma1,ng=ma2and my=mb1,ny=nb2and (mn)g−1yg = (mn)c. Hence (mn)g−1yg =mb1nb2= (mn)c, which implies that b1≡b2≡c(mod p). Consequently Mand Nare G-isomorphic. Since Ghas all Sylow subgroups modular, it follows that Gis a PT-group by [3, Corollary 3] and [4, Theorem 2]. We now suppose that Ghas a unique minimal normal subgroup N. If Nis not soluble, then N=S1×· · ·×Sr, where the Siare isomorphic (nonabelian) simple groups. Let pand qbe different primes dividing the order of S1and let x1and y1be elements of S1of orders pand q, respectively. For 2≤i≤r, let xi,yibe the images of x1,y1under the isomorphism between S1and Si. Then hx1· · · xripermutes with a conjugate h(y1· · · yr)giof hy1· · · yri. The projection of hx1· · · xrih(y1· · · yr)gionto S1is then a subgroup of S1 of order pq and so S1has subgroups of order pq for every pair of primes dividing its order. A result of Abe and Iiyori [1] shows that this is impossible. Consequently Nis a p-group for some prime p. If Nis not contained in the Frattini subgroup of G, then Gis a primitive soluble group and G=NM, where Mis a maximal subgroup of G,N∩M= 1 and Nis self-centralising. Since Mis isomorphic to the soluble PT-group G/N,Mis the product of its nilpotent residual F=MN, which is an abelian normal Hall subgroup of odd order, and a complement Cwhich acts on Fas power automorphisms ([12]). Let Qbe a cyclic normal subgroup of M. Suppose that QP 6=PQ for some cyclic subgroup Pof N. We have QgP=PQgfor some g∈Gand we can assume that g∈M. Since Qis normal in Mwe have Qg=Q, giving a contradiction. Thus Pis normalised by Qsince P=N∩PQ. It now follows that every element of Facts as a power automorphism on Nand hence Facts as a power automorphism group on N. Since power automorphisms are central in the power automorphism 3 group of N([5, Theorem 2.2.1]), Fis central in Mand so F= 1. Thus M is a nilpotent modular group. In particular Mis a p0-group. Let Qbe a non-abelian Sylow q-subgroup of M. By Iwasawa’s Theorem ([11, Theorem 2.4.14] Qhas an abelian normal subgroup Q0with cyclic supplement Swith Sacting as a power automorphism group on Q0or Qis Hamiltonian. In both cases every cyclic subgroup of Q0is normal in Qand hence in M. Let Ube a cyclic subgroup of Nand let Rcyclic subgroup of Q0. By hypothesis, there exits an element a∈Msuch that RUais subgroup. Since Ra=R, we have that RU is also a subgroup and Uis normalised by R. Further, there exists an element m∈Msuch that Spermutes with Umand so Sand Rnormalise Um. This implies that Qnormalises Uand Qacts as power automorphisms on N. It now follows that Macts as power automorphisms on Nand so N is a cyclic group of order pand Mis cyclic of order dividing p−1and Gis clearly a PT-group. Now suppose that Nis contained in the Frattini subgroup of G. If Gis nilpotent, then Gis a p-group since it has a unique minimal normal subgroup and G/N is an modular group. Assume that Gis not modular. Let M(p) denote the nonabelian group of order p3and exponent pfor podd and the dihedral group of order 8for p= 2. By the Theorem of Longobardi [7] either Gis the central product of a subgroup Pisomorphic to M(p)and another subgroup or Gis isomorphic to G0=ha, b, w :apn=wp= 1, ab=a1+ps, bpj=apn−s, aw=a1+pn−1, bw=bi, where 0< s < n,s≥2if p= 2 and j≥n−s. In the first case it is clear that if Pis generated by aand bof order pno conjugate of awill commute with band hence will not permute with b. Now consider G0and let C=haαbβibe a cyclic subgroup of H=ha, bi. Then Cpermutes with hwgifor some g∈G0and so, being of index pin Chwgi, is normalised by wg. Since [w, g]∈ hapn−1i,Cis also normalised by w. Thus wacts as a power automorphism on H. If pis odd, His regular ([6, III, Satz 11.4]) and so it acts as a universal power automorphism on Hby [5, Theorem 5.3.1], a contradiction. Hence we suppose p= 2. Since wcentralises band bhas order at least 4,wacts as universal power automorphism on Hby a theorem of Napolitani [11, Theorem 2.3.24], again a contradiction. Thus Gcannot be nilpotent. If Gis not nilpotent, then E=GN6= 1 and so N≤E. Since G/N is a PT-group, Gsupersoluble, Eis nilpotent and so it is a p-group. Furthermore E/N is abelian and complemented in G/N by a p0-subgroup B/N say which acts on E/N as power automorphisms. Then there exists a p0-subgroup D of Gcomplementing Ein G. If CD(E/N)6= 1 then CD(E/N)is a nontrivial 4 normal subgroup of G, a contradiction. It follows that Dis cyclic of order dividing p−1. We have that N≤Z(E). Consequently [ap, b] = [a, b]p= 1 for every a,b∈Eby [6, III, Hilfssatz 1.3]. This implies that Ep≤Z(E). Assume that Epis not trivial. Hence Epis cyclic because Epis an abelian normal subgroup of G. Let Ep=hai, where ahas order pn. Suppose that Nis a proper subgroup of Ep. Given x∈G, we have that ax=aiand so all chief factors of Gbelow Ppare G-isomorphic. Appying again [3, Corollary 3] and [4, Theorem 2], Gis a PT-group. Assume now that Ep=N. Let xbe an element of Enot in N. If xhas order greater than p, then N≤ hxpi ≤ Ep and hxi/N is a normal subgroup of G/N . Since all chief factors of Gbetween Nand Eare G-isomorphic and the same happens with all chief factors of Gbelow hxi, we conclude again that all chief factors of Gare G-isomorphic and Gis a PT-group. Thus all elements of Eare of order p. If Eis abelian then Eis cyclic and so E=N≤Φ(G), a contradiction. Suppose that E is non-abelian and x, y ∈Edo not commute, so that N=h[x, y]i. 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