On finite groups generated by strongly cosubnormal subgroups

Document downloaded from: This paper must be cited as: The final publication is available at Copyright Additional Information http://hdl.handle.net/10251/19004 Ballester Bolinches, A.; Cossey, J.; Esteban Romero, R. (2003). On finite groups generated by strongly cosubnormal subgroups. Journal of Algebra. 1(259):226-234. doi:10.1016/S0021-8693(02)00535-5 http://dx.doi.org/10.1016/S0021-8693(02)00535-5 Elsevier This paper has been published in Journal of Algebra, 259(1):226-234 (2003). Copyright 2003 by Elsevier. http://dx.doi.org/10.1016/S0021-8693(02)00535-5 This paper has been published in Journal of Algebra, 259(1):226–234 (2003). Copyright 2003 by Elsevier. The final publication is available at www.sciencedirect.com. http://dx.doi.org/10.1016/S0021-8693(02)00535-5 http://www.sciencedirect.com/science/article/pii/S0021869302005355 On finite groups generated by strongly cosubnormal subgroups A. Ballester-Bolinches Departament d’` Algebra Universitat de Val`encia Dr. Moliner, 50 E-46100 Burjassot (Val`encia) Spain email: [email protected] John Cossey Mathematics Department School of Mathematical Sciences The Australian National University Canberra ACT 0200 Australia email: [email protected]u.edu.au R. Esteban-Romero 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]v.es 23rd January 2002 Abstract Two subgroups Aand Bof a group Gare cosubnormal if Aand Bare subnormal in their join hA, Biand are strongly cosubnormal if 1 every subgroup of Ais cosubnormal with every subgroup of B. We find necessary and sufficient conditions for Aand Bto be strongly cosubnormal in hA, Biand, if Zis the hypercentre of G=hA, Bi, we show that Aand Bare strongly cosubnormal if and only if G/Z is the direct product of AZ/Z and BZ/Z. We also show that projectors and residuals for certain formations can easily be constructed in such a group. Two subgroups Aand Bof a group Gare N-connected if every cyclic subgroup of Ais cosubnormal with every cyclic subgroup of B. Though the concepts of strong cosubnormality and N-connectedness are clearly closely related, we give an example to show that they are not equivalent. We note however that if Gis the product of the Nconnected subgroups Aand B, then Aand Bare strongly cosubnormal. 1 Introduction and statements of results In the sequel it is understood that all groups are finite. Following Wielandt [6], we say that two subgroups Aand Bof a group Gare cosubnormal in Gif Aand Bare subnormal subgroups of their join hA, Bi. More recently, Knapp [5] introduces the notion of strong cosubnormality: two subgroups Aand Bof a group are called strongly cosubnormal if every subgroup of Ais cosubnormal with every subgroup of B. We write Acs Bif Aand Bare cosubnormal and Ascs Bif Aand Bare strongly cosubnormal. Notice that if Aand Bare N-connected, then every cyclic subgroup of A is cosubnormal with every cyclic subgroup of B. Knapp proves in [5] the following characterisation of strong cosubnormality in terms of the hypercentre: Theorem 1 ([5, Theorem 3.3]). Let A,Bbe subgroups of a group G. Then the following are equivalent: 1. Aand Bare strongly cosubnormal. 2. [A, B]≤Z∞(hA, Bi). Here Z∞(G) denotes the hypercentre of a group G. A natural sequel of Knapp’s work would be the study of groups generated by strongly cosubnormal subgroups. On the other hand, Carocca [3] introduces the concept of N-connected subgroups: two subgroups Aand Bof a group Gare N-connected when for 2 every a∈Aand b∈B, the subgroup ha, biis nilpotent (Ndenotes the class of nilpotent groups). It is very easy to show that if Aand Bare two strongly cosubnormal subgroups of a group G, then they are N-connected: if a∈Aand b∈ B, then haiand hbiare nilpotent subnormal subgroups of ha, bi, and so ha, biis nilpotent. However, N-connection and strong cosubnormality are not equivalent in general, as we will show in the Example at the end of Section 2. We prove the following characterisation theorem: Theorem 2. Let Aand Bbe two subgroups of Gsuch that G=hA, Biand let Z=Z∞(G). The following statements are equivalent: 1. Ascs B. 2. Acs Band Aand Bare N-connected. 3. Acs Band if pand qare two different primes, xis a p-element of A and yis a q-element of B, then [x, y] = 1. 4. [A, B]≤Z. We observe from that cosubnormality and N-connection are closely related concepts. In the important case of products, they are indeed equivalent. Theorem 3. If a group Gis the N-connected product of its subgroups Aand B, then Aand Bare strongly cosubnormal. Our next result describes the groups generated by strongly cosubnormal subgroups. Theorem 4. Let G=hA, Biand Z=Z∞(G). Then the following statements are equivalent: 1. Ascs B. 2. G/Z =AZ/Z ×BZ/Z. In [1], Ballester-Bolinches and Pedraza-Aguilera proved that soluble Nconnected products behave well with respect to saturated formations containing N. Following this idea, we study the behaviour of strongly cosubnormal subgroups in the finite (not necessarily soluble) universe with respect to formations. Recall that a formation Fis a class of groups which is closed under taking epimorphic images and subdirect products. Every group Ghas a smallest 3 normal subgroup GF(called the F-residual of G) such that G/GF∈F(see [4, II.2] for details). If Xis a class of groups, a subgroup Eof Gis an Xprojector of Gif EN/N is X-maximal in G/N for all normal subgroups N of G. If Fis a formation, then every group Ghas F-projectors if and only if Fis saturated, that is, if G/Φ(G)∈F, then G∈F(see [4, Chapter 4] for further details). Note that Nis a saturated formation. The following results show that finite (not necessarily soluble) groups generated by strongly cosubnormal subgroups behave well with respect to (not necessarily saturated) formations containing N. Theorem 5. Let Fbe a formation containing Nsuch that either Fis saturated, or Fis contained in the class of soluble groups. Suppose that G= hA, Biand Ascs B. Then GF=hAF, BFi. Theorem 6. Let Fbe a saturated formation containing N. Suppose that G=hA, Biwith Ascs B. Let A1be an F-projector of Aand let B1be an Fprojector of B. Then hA1, B1iis an F-projector of G. Moreover, Apermutes with Bif and only if A1permutes with B1. 2 Proofs of the results We begin with the following Lemma, whose proof is already contained in Knapp’s paper. Lemma 1. Suppose that Aand Bare subgroups of a group Gsuch that the following conditions hold: 1. G=hA, Biand 2. if pand qare two different primes, xis a p-element of Aand yis a q-element of B, then [x, y] = 1. Then: 1. if pis a prime, then Op0(B)≤CGOp(A)and Op0(A)≤CGOp(B) and 2. BA≤CG(AN)and AB≤CG(BN). In particular, ANand BNare normal subgroups of G. Proof. Let pand qbe two different prime numbers. Let Apbe a Sylow psubgroup of Aand let Bqbe a Sylow q-subgroup of B. Then [Ap, Bq] = 1 by hypothesis. 4 Since Bq≤CG(Ap) for every q6=p, we have that Ap≤CGOp(B). Analogously, Bp≤CGOp(A). This proves the first claim. Since AN=Tpprime Op(A), we obtain that Bp≤CG(AN) for all primes p, and hence B≤CG(AN). Bearing in mind that ANis a normal subgroup of A, we get BA≤CG(AN). Analogously, we have that AB≤CG(BN). Proof of Theorem 2. 1 implies 2 has been already noted in the introduction, whereas 4 implies 1 is just one of the implications of Knapp’s result. 2 implies 3. Let pand qbe two different prime numbers. Let xbe a p-element of Aand let ybe a q-element of B. Since hx, yiis nilpotent, it follows that [x, y] = 1. 3 implies 4. We argue by induction on |G|. We have that [A, B] is a normal subgroup of hA, Bi=G. Suppose that [A, B]6= 1, and let Nbe a minimal normal subgroup of Gcontained in [A, B]. If N∩GN= 1, then Nis central in G. Hence, by induction, [A, B]/N ≤Z∞(G/N), which is equal to Z/N because Nis central in G. Consequently [A, B] is contained in Zand the theorem is proved. Therefore we may assume that every minimal normal subgroup of Gcontained in [A, B] is also contained in GN. Since [A, B] centralises ANand BNby Lemma 1, it follows that [A, B] centralises hAN, BNi, which is equal to GNby [5, Theorem W]. This implies that Nis central in [A, B]. Now [A, B]/N ≤Z∞(G/N) by induction. Hence [A, B]/N is nilpotent and so is [A, B]. Suppose that there exists a minimal normal subgroup Cof G,C6=N, and C≤[A, B]. Then, by induction, CN/N ≤Z∞(G/N). Thus Cis central in G. We can argue as in the previous case to conclude [A, B]≤Z. Consequently, [A, B] contains a unique minimal normal subgroup of G. Since [A, B] is nilpotent, we have that [A, B] is a p-group for some prime p. Assume that there exists a minimal normal subgroup N1of G,N16=N. By induction, [A, B]N1/N1≤Z∞(G/N1), and so NN1/N1is centralised by every p0-subgroup of G/N1. In particular, [N, Op(A)] ≤N1and [N, Op(B)] ≤ N1. Since [N, Op(A)] and [N, Op(B)] are both contained in N, it follows that [N, Op(A)] = [N, Op(B)] = 1. This means that N≤CGhOp(A), Op(B)i= CGOp(G), because Op(G) = hOp(A), Op(B)i([5, Theorem W]). This implies that N≤Z. Since [A, B]/N ≤Z∞(G/N) and Z∞(G/N) = Z/N, we have that [A, B]≤Zand so [A, B]≤Z. Consequently we may assume that Ghas a unique minimal normal subgroup, Nsay, and N≤[A, B]. Note that AB=A[A, B] is a normal subgroup of Gand Op(AB) = Op(A) because [A, B] is a p-group. Analogously Op(BA) = Op(B). In particular, Op(A) and Op(B) are normal in G. Suppose that Op(A)6= 1. Then N≤Op(A) and so Op(B)≤CG(N) by Lemma 1. If Op(B)6= 1, we also have Op(A)≤CG(N). This means that Op(G)≤CG(N) 5 and N≤Z. Therefore we may suppose that Op(B) = 1 and Bis a p-group. Then N≤BAand BA≤CGOp(A)by Lemma 1. Since Op(A) = Op(G), it follows that N≤CGOp(G)and then N≤Z. Arguing as above, we have that [A, B]≤Zand the theorem is proved. Proof of Theorem 3. By Theorem 2, we need only prove that Acs Bprovided that Aand Bare N-connected and G=AB. Assume that this is not true and let Gbe a counterexample of minimal order. Note that the hypotheses of Lemma 1 hold for N-connected subgroups. Consequently, ANand BNare normal subgroups of G. Suppose that Ais not subnormal in G. It is clear that G/BNis the N-connected product of ABN/BNand B/BN. Hence, if BN6= 1, we have that ABNis subnormal in Gby the minimality of G. Since A≤CG(BN) by Lemma 1, it follows that Ais normal in ABN. Therefore A is subnormal in G, a contradiction. Consequently, Bis nilpotent. If AN6= 1, we have that A/ANis subnormal in G/ANby the minimal choice of G. Hence Ais subnormal in G, a contradiction. Therefore Aand Bare nilpotent. By [3], Gis nilpotent, a contradiction. Proof of Theorem 4. 1 implies 2. Suppose that Ascs B. Since A∩Bscs B1 for every B1≤B, we have that A∩B≤Z∞(B) by [5, Theorem 2.6]. Since A1scs A∩Bfor every A1≤A, we have that A∩B≤Z∞(A) by [5, Theorem 2.6]. Consequently A∩B≤Z∞(A)∩Z∞(B), which is contained in Z by [5, Proposition 3.2]. On the other hand, [AZ/Z, BZ/Z]≤[A, B]Z/Z = 1, by Theorem 2, whence G/Z =AZ/Z ×BZ/Z. 2 implies 1. Suppose that G/Z =AZ/Z×BZ/Z. Let A1be a subgroup of Aand let B1be a subgroup of B. Since A1is subnormal in A1Zand A1Z/Z is centralised by B1Z/Z, it follows that A1is subnormal in T=hA1Z, B1Zi. Analogously, B1is subnormal in T. Hence A1cs B1, as desired. The proofs of Theorem 5 and 6 depend on the following Lemmas: Lemma 2. Let Fbe a formation containing N. Suppose that G=hA, Bi and Ascs B. If Aand Bbelong to F, then G∈F. Proof. Suppose that the theorem is false. Let G=hA, Bibe a counterexample with |A|+|B|minimal. We can assume without loss of generality that Ais not nilpotent. Then we can write A=ANC, where Cis an N-projector of A. On the other hand, ANis a normal subgroup of Gby Lemma 1 and Theorem 2 and B≤CG(AN). This implies that D=BhB,Ci≤CG(AN). By [2, Lemma 1], bearing in mind that G=ANhC, Bi, there exists an epimorphism θ:X= [AN]hC, Bi −→ G. Let us prove that X∈F. We have that 6 X/AN∈F, because hC, Bi ∈ Fby minimality of G. Now Dis a normal subgroup of X, because Dis centralised by AN. Moreover X/D ∼ =[AN](CD/D)∼ =[AN](C/D ∩C). We see that Y= [AN]C∈F. By [2, Lemma 1], there exists an epimorphism α:Y−→ ANC=Asuch that Ker α∩AN= 1. Now, Y/ Ker α∈F and Y/AN∈F. Since Fis a formation, it follows that Y∈F. It is clear that X/D is isomorphic to a quotient of Y. Therefore X/D ∈F. Since Fis a formation, we have that X/AN∩D=X∈F. This implies that G∈F, because Gis an epimorphic image of X. Lemma 3. Let Fbe a formation containing N. Assume that either Fis saturated or Fconsists only of soluble groups. If Aand Bare strongly cosubnormal subgroups of G,G=hA, Biand Gbelongs to F, then Aand B belong to F. Proof. Assume that Fis a saturated formation. Let Gbe a counterexample of minimal order to the theorem. If Z=Z∞(G) = 1, then A∩B= 1 by Lemma 4 and G=A×B. In particular, Aand Bbelong to F. Hence Z6= 1. Let Nbe a minimal normal subgroup of G. Since G/N satisfies the hypotheses of the theorem, it follows that AN/N ∈Fand BN/N ∈F. In particular, A/A ∩Nand B/B ∩Nbelong to F. If Ghas more than one minimal normal subgroup, we have that Aand Bbelong to F. Hence Ghas a unique minimal normal subgroup. Thus N≤Z, whence N≤Z(G). In particular, A∩N≤Z(A) and B∩N≤Z(B). This implies that Aand B belong to F, as desired. Assume now that Fis a formation of soluble groups. Let G=hA, Bi be a minimal counterexample with |A|+|B|minimal. If, for example, Bis nilpotent, then G=AF(G). By Bryant, Bryce and Hartley’s Theorem ([4, IV.1.14]), it follows that A∈F. Hence we can assume that AN6= 1 and BN6= 1. Since Gis soluble, it follows that there exist a maximal subgroup A0of Asuch that AF(G) = A0F(G) and a maximal subgroup B0of Bsuch that BF(G) = B0F(G). Note that G=hA, B0iF(G) = hA0, BiF(G). From Bryant, Bryce and Hartley’s Theorem ([4, IV.1.14]), we have that hA, B0iand hA0, Bibelong to F. On the other hand, bearing in mind that Ascs B0and A0scs B, the minimality of |A|+|B|implies that A∈Fand B∈F, a contradiction. Proof of Theorem 5. Since N⊆F, we have that GF≤GN,AF≤ANand BF≤BN. Hence BA≤CG(AN) implies that B≤CG(AF). Thus AFand, analogously, BFare normal subgroups of G. Since G/GF=hAGF/GF, BGF/GFi 7