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Mutually Permutable Products of Finite Groups, II
Journal of Algebra 218, 563᎐572 Ž1999. Article ID jabr.1999.7858, available online at http:rrwww.idealibrary.com on
Mutually Permutable Products of Finite Groups, II A. Ballester-Bolinches Departament d’Algebra, Uni¨ ersitat de Valencia, CrDoctor Moliner 50, ` 46100 Burjassot (Valencia ` ), Spain E-mail: [email protected]
and M. C. Pedraza-Aguilera Departamento de Matematica de Valencia, ´ Aplicada, E.U.I., Uni¨ ersidad Politecnica ´ Camino de Vera, srn, 46071 Valencia, Spain E-mail: [email protected] Communicated by George Glauberman Received March 6, 1998
This paper is a natural continuation of w4x and represents an attempt to extend and improve the main results of the above paper through the theory of formations. Throughout the present article it is understood that all groups considered are finite. First of all, recall that the subgroups H and K of a group G are mutually Žrespectively totally. permutable if H Žrespectively each subgroup of H . permutes with every subgroup of K and K Žrespectively each subgroup of K . permutes with every subgroup of H. A group G is said to be the mutually Žrespectively totally. permutable product of the subgroups H and K if G s HK and H and K are mutually Žrespectively totally. permutable. In w4x, the following result is proved: THEOREM 1. Let G s HK be the mutually permutable product of the subgroups H and K. Assume that one of the following conditions is satisfied: Ža. Žb.
K is supersoluble and G⬘ is nilpotent. K is nilpotent.
Then G U s H U. 563 0021-8693r99 $30.00 Copyright 䊚 1999 by Academic Press All rights of reproduction in any form reserved.
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Here, G U denotes the supersoluble residual of G, that is, the smallest normal subgroup of G with supersoluble quotient. The main goal here is to prove that if F is a saturated formation containing the class U of supersoluble groups, then the F-residual respects the operation of forming mutually permutable products with nilpotent commutator subgroup. We finish by showing that unfortunately the F-projectors Žand so the F-normalizers. of a mutually permutable product with nilpotent commutator subgroup cannot be obtained from the corresponding projectors of the factor subgroups. This is true for totally permutable products as is shown in w2x. We need the following preliminary results: PROPOSITION 1 w5x. Let the group G s HK be the product of the mutually permutable subgroups H and K. Then: Ža. If H l K F X F H and H l K F Y F K, then X and Y are mutually permutable. In particular, if H l K s 1, then H and K are totally permutable. Žb. H l K is a permutable subgroup of H and K and then H l K is a subnormal subgroup of G. PROPOSITION 2 w3x. Let F be a formation containing U such that F is either saturated or F : S . If G s G1G 2 ⭈⭈⭈ Gr is the product of the pairwise totally permutable subgroups Gi for i g 1, 2, . . . , r 4 , then G F s G1F G 2F ⭈⭈⭈ GrF. Remark w4x. Let G s HK be a group such that H and K are mutually permutable subgroups. For every normal subgroup N of G, we have that HNrN and KNrN are mutually permutable subgroups of the group GrN s Ž HNrN .Ž KNrN . and for every subgroup L of G such that H F L or Ž K F L., we have that L is the product of the mutually permutable subgroups H and L l K Žor H l L and K .. PROPOSITION 3 w7x. If Q is a permutable subgroup of the finite group G, then Q G rCoreG Ž Q . is contained in the hypercenter Z⬁Ž GrCoreG Ž Q .. of GrCoreG Ž Q .. Our objective is to prove the following result. THEOREM A. Let G s HK be the mutually permutable product of the subgroups H and K. Let F be a saturated formation containing the class U of supersoluble groups. If G⬘ is nilpotent, then G F s H F K F. Ž Here G F denotes the F-residual of a group G, that is, the smallest normal subgroup of G such that GrG F belongs to F..
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We refer the reader to w6x for notation, terminology, and results about formations. Theorem A will follow after proving some auxiliary results. It is known that a subgroup U supplementing the Fitting subgroup of a group G lies in the formation generated by G Žsee w6, IV, 1.14x.. Our first result shows that something similar happens with arbitrary products with nilpotent commutator subgroup and saturated formations. LEMMA 1. Let G s HK be the product of the subgroups H and K such that G⬘ is nilpotent. Suppose that F is a saturated formation such that G belongs to F. Then H and K belong to F. Proof. Assume that the result is false and let G be a minimal counterexample. Notice that G is a soluble group. By the minimality of G, we can assume that G has a unique minimal normal subgroup N, say. Let p be the prime dividing the order of N. Since G⬘ is nilpotent, we have that G⬘ is contained in F Ž G . s Op Ž G . and then G⬘ is a p-group. In particular,G has a unique Sylow p-subgroup P and P s F Ž G .. By w1, Lemma 1.3.2x, we can take A a Hall p⬘-subgroup of H and B a Hall p⬘-subgroup of K such that X s AB is a Hall p⬘-subgroup of G. Then G s PX and P l X s 1. Since G g F , it follows that GrOp⬘ p Ž G . lies in F Ž p ., where F is the canonical local definition of F. Since Op⬘Ž G . s 1, we have that Op⬘ p Ž G . s Op Ž G . s P. Therefore X belongs to F Ž p .. Now G⬘ is contained in P. This implies that X is an abelian group. We can now apply w6, IV, 1.14x to conclude that A and B belong to F Ž p .. Since H and K have a normal Sylow p-subgroup, it follows that H and K belong to Sp F Ž p . s F Ž p .. Consequently, H and K are F-groups and the proof of the lemma is complete. The next result shows that the converse of the above lemma holds for mutually permutable products with nilpotent commutator subgroup and saturated formations containing U . LEMMA 2. Let G s HK be the mutually permutable product of the subgroups H and K. Let F be a saturated formation containing U . Suppose that H and K belong to F. If G⬘ is nilpotent, then G belongs to F. Proof. Assume that the result is not true and let G be a minimal counterexample to the lemma. By the above remark and the minimality of G, we know that every proper quotient of G belongs to F. Consequently, G is a soluble primitive group because F is a saturated formation. Hence F Ž G . s SocŽ G . is the unique minimal normal subgroup of G and if p is the prime dividing its order, then F Ž G . s G⬘ is the unique Sylow p-subgroup of G.
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On the other hand, by Proposition 1, D s H l K is a permutable subgroup of H and K. Suppose that Core H Ž D . s 1. Then, applying Proposition 3, we have that D is a nilpotent subnormal subgroup of G. Hence D is contained in F Ž G .. Let A and B be Hall p⬘-subgroups of H and K, respectively, such that AB is a Hall p⬘-subgroup of G. Notice that w D, A x is a normal p-subgroup of the group DA and D is a Sylow p-subgroup of DA. Therefore w D, A x is contained in D and so D is a normal subgroup of DA. Analogously, D is a normal subgroup of DB. In addition, D is contained in F Ž G . which is abelian. So D is normalized by F Ž G .Ž AB . s G. If D s 1, then G is the totally permutable product of the subgroups H and K. So, by Proposition 2, we have G g F , a contradiction. Consequently, D / 1 and then F Ž G . s D. Suppose now that Core H Ž D . / 1 and Core K Ž D . / 1. Then the normal closure Core H Ž D . G is different from 1 and so F Ž G . F Core H Ž D . G . Notice that Core H Ž D . G s Core H Ž D . K which is contained in K. Analogously, F Ž G . F Core K Ž D . G and Core K Ž D . G is contained in H. In particular, F Ž G . is contained in D. Therefore, in both cases, we have that F Ž G . is contained in H l K. Moreover, since G⬘ s F Ž G ., we also have that H and K are normal subgroups of G. Consider F Ž G ., regarded as a H-module over GF Ž p .. By Clifford’s theorem, we know that F Ž G . is a completely reducible H-module, that is, F Ž G . s N1 = ⭈⭈⭈ = Nr , where Ni is a minimal normal subgroup of H for all i. Now H g F. This means that HrCH Ž Ni . g F Ž p . for all i Žhere F denotes the canonical local definition of F .. Since F Ž p . is a formation, we have that HrCH Ž F Ž G .. g F Ž p .. So HrF Ž G . g F Ž p .. Analogously, KrF Ž G . g F Ž p .. Now the group GrF Ž G . is an abelian group which is the product of the subgroups HrF Ž G . and KrF Ž G .. Since formations are closed under central products, it follows that GrF Ž G . g F Ž p .. This implies that G g S p F Ž p . s F Ž p . : F , the final contradiction. As the symmetric group of degree 3 and the saturated formation of nilpotent groups show, the condition U : F is essential in the above lemma. LEMMA 3. Let G s HK be the mutually permutable product of the subgroups H and K. Assume that G⬘ is nilpotent. If F is a saturated formation containing U , then G F s ² H F, K F :. Proof. Suppose that the result is not true and let G be a counterexample of minimal order. If N is a minimal normal subgroup of G, then GrN is the mutually permutable product of the subgroups HNrN and KNrN. Moreover Ž GrN .⬘ is nilpotent. So, by the minimal choice of G, we have
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that G F N s ² H F , K F : N. By Lemma 1 applied to GrG F s Ž HG FrG F .Ž KG FrG F ., ² H F, K F : is contained in G F. Hence G F s ² H F, K F :Ž G F l N .. Therefore, if G F l N s 1, it follows that G F s ² H F, K F :, a contradiction. Consequently, we may assume that SocŽ G . is contained in G F and G F s ² H F, K F : N for every minimal normal subgroup N of G. In particular, Core G Ž² H F, K F :. s 1. On the other hand, G F is contained in G⬘ which is nilpotent. So H F and F K are subnormal subgroups of G. By a result due to Wielandt w6, A, 14.4x, ² H F, K F : is subnormal in G. Hence SocŽ G . normalizes ² H F, K F : by another result of Wielandt w6, A, 14.3x. This means that ² H F, K F : is a normal subgroup of G F. Let p be the prime dividing the order of a minimal normal subgroup N of G. Since G is soluble, it follows that N is an abelian p-group. In particular, G FrŽ² H F, K F :. is an abelian p-group and so O p Ž G F .Ž G F .⬘ is contained in ² H F, K F :. Since O p Ž G F .Ž G F .⬘ is normal in G and Core G Ž² H F, K F :. s 1, it follows that O p Ž G F .Ž G F .⬘ s 1 and G F is an abelian p-group for some prime p. This means that F Ž G . is a p-group and then G⬘ is also a p-group because G⬘ is nilpotent. Consequently, F Ž G . is the unique Sylow p-subgroup of G. Let H p and K p denote the Sylow p-subgroups of H and K, respectively. It is clear that F Ž G . s Hp K p . Let Hp⬘ and K p⬘ be Hall p⬘-subgroups of H and K, respectively. Since H and K are mutually permutable, it follows that HK p⬘ and KHp⬘ are subgroups of G. Suppose that G s HK p⬘. Then F Ž G . is contained in H and so H is a normal subgroup of G. Hence H F is a normal subgroup of G. Since Core G Ž² H F, K F :. s 1, it follows that H F s 1 and so H g F. If K F s 1, then G g F by Lemma 2. Therefore K F / 1 and so KHp⬘ should be a proper subgroup of G Žotherwise F Ž G . F K and K is a normal subgroup of G .. Let M be a maximal subgroup of G containing KHp⬘. Then M s Ž M l H . K. By Proposition 1, M is the mutually permutable product of the subgroups M l H and K. The minimal choice of G yields M F s ²Ž M l H . F, K F :. Moreover H s H p Ž M l H . because G s Hp M s F Ž G . M. By w6, IV, 1.17Žb.x, Ž M l H . F is contained in H F because H p is a nilpotent normal subgroup of H. Therefore M F is contained in ² H F, K F :. Suppose that G F is not contained in M. Then G s MG F. Since G F is abelian and M normalizes M F, we have that M F is a normal subgroup of G. In particular, M F F Core G Ž² H F, K F :. s 1 and so K g F , a contradiction. Consequently, G F is contained in M. Let N be a minimal normal subgroup of G. Then N F G F F M. Since G s MF Ž G . and F Ž G . centralizes N, it follows that N is a minimal normal subgroup of M. If N l M F s N, then N F ² H F, K F : and G F s ² H F, K F :, a contradiction. So N l M F s 1 and NM FrM F is a minimal normal subgroup of MrM F g F. In particular, N is F-central in M and so N is F-central in G. By w6, V, 3.2x, N is contained in every F-normalizer of
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G. Let E be one of them. Then G s G F E and E l G F s 1 by w6, IV, 5.18; V, 4.2x. However N F E l G F, a contradiction. Consequently, HK p⬘ is a proper subgroup of G. Analogously, KHp⬘ is a proper subgroup of G. If H does not belong to F or K does not belong to F , we can use the above argument to get a contradiction. So H g F and K g F. By Lemma 2, G g F , yielding the final contradiction. LEMMA 4. Let G s HK be the mutually permutable product of the subgroups H and K. Suppose that G⬘ is nilpotent. If F is a saturated formation containing U , then Ž H l K . F is contained in G F. Proof. We argue by induction on the order of G. Let N be a minimal normal subgroup of G. Then, by the above remark, the group GrN is the mutually permutable product of the subgroups HNrN and KNrN and Ž GrN .⬘ s G⬘NrN is nilpotent. By induction, ŽŽ HN l KN .rN . F is contained in Ž GrN . F s G F NrN. Now, HN l KN s Ž HN l K . N s Ž H l KN . N and ŽŽ HN l KN .rN . F s Ž HN l K . F NrN s Ž H l KN . F NrN. Hence Ž HN l K . F N s Ž H l KN . F N is contained in G F N for every minimal normal subgroup N of G. We distinguish two cases: Case 1. G does not belong to F. In this case G F / 1 and we can take a minimal normal subgroup N of G such that Ž HN l K . F N s Ž H l KN . F N F G F. Consider the subgroup T s H Ž HN l K . s HN. Then, by Proposition 1, T is the mutually permutable product of the subgroups H and HN l K. So, if T is a proper subgroup of G, we have Ž H l K . F F T F. On the other hand, T F s ² H F, Ž HN l K . F : by Lemma 3. Therefore Ž H l K . F is contained in ² H F, G F : s G F by Lemma 3, and the lemma is true. So we may assume that G s HN. Analogously, G s KN. Consequently, H and K are maximal subgroups of G Žnotice that if H s G or K s G, the result is true by Lemma 3.. It is clear that H and K cannot be conjugate in G. So, by w6, A, 16.1x, Core G Ž H . / Core G Ž K . and either CoreG Ž H . / 1 or Core G Ž K . / 1. Suppose that CoreG Ž H . / 1. Now, if CoreG Ž K . is contained in H, then HrCoreG Ž K . is a maximal subgroup of the primitive group GrCore G Ž K . whose Fitting subgroup is precisely N Core G Ž K .rCoreG Ž K .. Moreover CoreG Ž H .rCoreG Ž K . is a non-trivial normal subgroup of GrCoreG Ž K .. Therefore N CoreG Ž K .rCoreG Ž K . is contained in Core G Ž H .rCoreG Ž K . and N is contained in H, a contradiction. Consequently, Core G Ž K . is not contained in H. Since H is maximal in G and Core G Ž K . is a normal subgroup of G, it follows that G s H Core G Ž K . and then K s Core G Ž K .Ž H l K .. It is clear that K is the mutually permutable product of the subgroups Core G Ž K . and H l K
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because H l K is a permutable subgroup of K by Proposition 1. By Lemma 3, we have K F s ²ŽCore G Ž K .. F, Ž H l K . F : and K F is contained in G F. So Ž H l K . F is contained in G F and the lemma is true in this case. Case 2. G g F. Let D s H l K. By Lemma 1, we have that H and K are F-groups. If either Core H Ž D . s 1 or Core K Ž D . s 1, it follows by Propositions 1Žb. and 3 that D is nilpotent and D g F and the lemma is true. Consequently, we may assume that Core H Ž D . / 1 and Core K Ž D . / 1. This implies that Core H Ž D . G is a non-trivial normal subgroup of G contained in K and Core K Ž D . G is a non-trivial normal subgroup of G contained in H. Let A and B be minimal normal subgroups of G contained in H and K, respectively. Applying induction to GrA and GrB, we conclude that Ž H l KA.rA and Ž K l HB .rB belong to F. Since Ž H l KA.rA is isomorphic to Ž H l K .rŽŽ H l K . l A. and Ž K l HB .rB is isomorphic to Ž H l K .rŽŽ H l K . l B ., we have that Ž H l K .rŽŽ H l K . l A l B . g F. So if A / B, then H l K g F and we have finished. Therefore we may assume that A s B and then A is contained in H l K. In particular, Ž H l K . F is contained in A. Suppose there exists a minimal normal subgroup N of G such that N / A; we know that Ž H l KN . F and Ž K l HN . F are contained in G F N s N. The group T s K Ž H l KN . is the mutually permutable product of K and H l KN. So, if T - G, we have Ž H l K . F F T F s ² K F, Ž H l KN . F : by Lemma 3. Since K F s 1, it follows that Ž H l K . F F N. Consequently, Ž H l K . F F N l A s 1 and D g F. The lemma is true in this case. Consequently, we can assume that G s HN s KN. Arguing as in Case 1, we have D g F. Therefore we may assume that A s SocŽ G . is the unique minimal normal subgroup of G. Since G is soluble, A is an elementary abelian p-group for some prime p. Then F Ž G . s Op Ž G . s Op⬘ p Ž G . and F Ž G . is the unique Sylow p-subgroup of G and G⬘ is contained in F Ž G .. Since G g F , we know that GrF Ž G . g F Ž p ., where F is the canonical local definition of F. This implies that the Hall p⬘-subgroups of G are in F Ž p .. Moreover they are abelian. By w6, IV, 1.14x, every p⬘-subgroup of G is in F Ž p .. In particular, the Hall p⬘-subgroups of H l K are in F Ž p .. Since H l K has a normal Sylow p-subgroup, we have that H l K g Sp F Ž p . s F Ž p . : F and the lemma is proved. Proof of Theorem A. Assume that the result is not true and choose for G a counterexample of smallest order.
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If N is a minimal normal subgroup of G, then G F N s H F K F N by the minimal choice of G. Moreover H F and K F are contained in G F by Lemma 3. Again by the minimality of G, we can deduce easily that SocŽ G . is contained in G F and G F s H F K F N for every minimal normal subgroup N of G. Let p be a prime dividing the order of an arbitrary minimal normal subgroup N of G. We claim that G F is a p-group. Otherwise since G F is nilpotent, it has a unique normal Hall p⬘-subgroup Ž G F . p⬘ / 1. It is clear that Ž G F . p⬘ is normal in G and it is the product of the unique Hall p⬘-subgroup Ž H F . p⬘ of H F and the unique Hall p⬘-subgroup Ž K F N . p⬘ of K F N. So H F permutes with Ž K F N . p⬘ which is equal to Ž K F . p⬘ because N is a p-group. Now, if we take a minimal normal subgroup B of G contained in Ž G F . p⬘ , then G F s H F K F B. Arguing in the same way as above, we have that H F permutes with the unique Sylow p-subgroup Ž K F . p of K F. Consequently, H F permutes with K F and G F s ² H F, K F : s H F K F by Lemma 3, a contradiction. Therefore there exists a prime p such that G F is a p-group. Since SocŽ G . is contained in G F, it follows that Op⬘Ž G . s 1 and F Ž G . s Op Ž G .. Hence G⬘ is a p-group and F Ž G . is the unique Sylow p-subgroup of G. Denote D s H l K. Suppose that D g F. Consider the subgroup T s Ž H F D . K and assume that T is a proper subgroup of G. Then, by Proposition 1, T is the mutually permutable product of the subgroups H F D and K. By the minimal choice of G, T F s Ž H F D . F K F. This implies that H F F NG ŽT F . and H F T F is a subgroup of G. Since H F is normal in H, it is clear that H F D is a mutually permutable product of H F and D. So, by minimality of G, we have Ž H F D . F s Ž H F . F D F s 1. Therefore T F s K F and H F permutes with K F. This is a contradiction as G F s ² H F, K F : by Lemma 3. Consequently, T s G and then G s H F K. Arguing in the same way with the subgroup Ž K F D . H, we have G s H F K s HK F. Hence H s H F D is the mutually permutable product of H F and D. Therefore H F s ²Ž H F . F, D F : s 1 and H g F. Analogously, K g F. From Lemma 2, G g F , a contradiction. Hence we may assume that D does not belong to F , that is, D F / 1. Let Hp and K p be the Sylow p-subgroups of H and K, respectively, such that F Ž G . s H p K p . Since H p is a normal subgroup of H, we have that Hp and D are mutually permutable subgroups of H. Hence Ž H p D . F s D F by the minimal choice of G. Analogously, Ž K p D . F s D F and then F Ž G . normalizes D F. Let now H p⬘ and K p⬘ be Hall p⬘-subgroups of H and K, respectively, such that X s H p⬘ K p⬘ is a Hall p⬘-subgroup of G. Since, by Proposition 1, D is a permutable subgroup of H, it follows that DH p⬘ is a subgroup of G.
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Let Dp⬘ be a Hall p⬘-subgroup of D such that Dp⬘ H p⬘ is a Hall p⬘-subgroup of DH p⬘. Then Dp⬘ H p⬘ s H p⬘ because the maximality of H p⬘ as p⬘-subgroup of H. Since X is abelian, Dp⬘ permutes with K p⬘ and again Dp⬘ is contained in K p⬘. It is clear that Dp is the unique Sylow p-subgroup of the subgroups DH p⬘ and DK p⬘. Hence X normalizes D. Since D F is characteristic in D, we have that X normalizes D F. Consequently, D F is a normal subgroup of G. We distinguish two cases: Case 1. G s HK p⬘ s KHp⬘. In this case F Ž G . is contained in D. So H and K are normal subgroups of G and so are H F and K F. Then, by Lemma 3, G F s H F K F. Case 2. Either HK p⬘ is a proper subgroup of G or KHp⬘ is a proper subgroup of G. Suppose that HK p⬘ - G and let M be a maximal subgroup of G containing HK p⬘. Then K p is not contained in M and G s MK p . Moreover M s H Ž M l K . is the mutually permutable product of H and M l K by Proposition 1. By the minimality of G, M F s H F Ž M l K . F and 1 / D F is contained in M F by Lemma 4. The fact that Ž M l K . F is contained in K F by w6, IV, 1.17x yields M F is contained in H F K F. Consequently, since D F is a non-trivial normal subgroup of G, a minimal normal subgroup N of G is contained in H F K F. This implies that G F s H F NK F s H F K F, the final contradiction. One might wonder whether some F-projector of a mutually permutable product with nilpotent commutator subgroup could be the product of F-projectors of the factor groups. Unfortunately, this is not true in general as the next example shows: Let G be the direct product of a cyclic group ² a: of order 3 with the alternating group A 4 of degree 4. Let V be the Klein 4-group of A 4 . Then G is the mutually permutable product of H s ² a: = V and K s A 4 . Moreover G⬘ s V is nilpotent. Notice that H is the supersoluble projector of H and a Sylow 3-subgroup B of A 4 is a supersoluble projector of K. However HB s G is not supersoluble.
ACKNOWLEDGMENT This research was supported by Proyecto PB 94-0965 of DGICYT, MEC, Spain.
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REFERENCES 1. B. Amberg, S. Franciosi, and F. de Giovanni, ‘‘Products of Groups,’’ Clarendon, Oxford, 1992. 2. A. Ballester-Bolinches, M. C. Pedraza-Aguilera, and M. D. Perez-Ramos, On finite ´ products of totally permutable groups, Bull. Austral. Math. Soc. 53 Ž1996., 441᎐445. 3. A. Ballester-Bolinches, M. C. Pedraza-Aguilera, and M. D. Perez-Ramos, Finite groups ´ which are products of pairwise totally permutable subgroups, Proc. Edinburgh Math. Soc. Ž 2 . 41 Ž1998., 567᎐572. 4. A. Ballester-Bolinches, M. C. Pedraza-Aguilera, and M. D. Perez-Ramos, Mutually per´ mutable products of finite groups, J. Algebra 213 Ž1999., 369᎐377. 5. A. Carocca, p-supersolvability of factorized finite groups, Hokkaido Math. J. 21 Ž1992., 395᎐403. 6. K. Doerk and T. Hawkes, ‘‘Finite Soluble Groups,’’ de Gruyter, BerlinrNew York, 1992. 7. R. Maier and P. Schmid, The embedding of permutable subgroups in finite groups, Math. Z. 131 Ž1973., 269᎐272.
Mutually Permutable Products of Finite Groups, II A. Ballester-Bolinches Departament d’Algebra, Uni¨ ersitat de Valencia, CrDoctor Moliner 50, ` 46100 Burjassot (Valencia ` ), Spain E-mail: [email protected]
and M. C. Pedraza-Aguilera Departamento de Matematica de Valencia, ´ Aplicada, E.U.I., Uni¨ ersidad Politecnica ´ Camino de Vera, srn, 46071 Valencia, Spain E-mail: [email protected] Communicated by George Glauberman Received March 6, 1998
This paper is a natural continuation of w4x and represents an attempt to extend and improve the main results of the above paper through the theory of formations. Throughout the present article it is understood that all groups considered are finite. First of all, recall that the subgroups H and K of a group G are mutually Žrespectively totally. permutable if H Žrespectively each subgroup of H . permutes with every subgroup of K and K Žrespectively each subgroup of K . permutes with every subgroup of H. A group G is said to be the mutually Žrespectively totally. permutable product of the subgroups H and K if G s HK and H and K are mutually Žrespectively totally. permutable. In w4x, the following result is proved: THEOREM 1. Let G s HK be the mutually permutable product of the subgroups H and K. Assume that one of the following conditions is satisfied: Ža. Žb.
K is supersoluble and G⬘ is nilpotent. K is nilpotent.
Then G U s H U. 563 0021-8693r99 $30.00 Copyright 䊚 1999 by Academic Press All rights of reproduction in any form reserved.
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Here, G U denotes the supersoluble residual of G, that is, the smallest normal subgroup of G with supersoluble quotient. The main goal here is to prove that if F is a saturated formation containing the class U of supersoluble groups, then the F-residual respects the operation of forming mutually permutable products with nilpotent commutator subgroup. We finish by showing that unfortunately the F-projectors Žand so the F-normalizers. of a mutually permutable product with nilpotent commutator subgroup cannot be obtained from the corresponding projectors of the factor subgroups. This is true for totally permutable products as is shown in w2x. We need the following preliminary results: PROPOSITION 1 w5x. Let the group G s HK be the product of the mutually permutable subgroups H and K. Then: Ža. If H l K F X F H and H l K F Y F K, then X and Y are mutually permutable. In particular, if H l K s 1, then H and K are totally permutable. Žb. H l K is a permutable subgroup of H and K and then H l K is a subnormal subgroup of G. PROPOSITION 2 w3x. Let F be a formation containing U such that F is either saturated or F : S . If G s G1G 2 ⭈⭈⭈ Gr is the product of the pairwise totally permutable subgroups Gi for i g 1, 2, . . . , r 4 , then G F s G1F G 2F ⭈⭈⭈ GrF. Remark w4x. Let G s HK be a group such that H and K are mutually permutable subgroups. For every normal subgroup N of G, we have that HNrN and KNrN are mutually permutable subgroups of the group GrN s Ž HNrN .Ž KNrN . and for every subgroup L of G such that H F L or Ž K F L., we have that L is the product of the mutually permutable subgroups H and L l K Žor H l L and K .. PROPOSITION 3 w7x. If Q is a permutable subgroup of the finite group G, then Q G rCoreG Ž Q . is contained in the hypercenter Z⬁Ž GrCoreG Ž Q .. of GrCoreG Ž Q .. Our objective is to prove the following result. THEOREM A. Let G s HK be the mutually permutable product of the subgroups H and K. Let F be a saturated formation containing the class U of supersoluble groups. If G⬘ is nilpotent, then G F s H F K F. Ž Here G F denotes the F-residual of a group G, that is, the smallest normal subgroup of G such that GrG F belongs to F..
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We refer the reader to w6x for notation, terminology, and results about formations. Theorem A will follow after proving some auxiliary results. It is known that a subgroup U supplementing the Fitting subgroup of a group G lies in the formation generated by G Žsee w6, IV, 1.14x.. Our first result shows that something similar happens with arbitrary products with nilpotent commutator subgroup and saturated formations. LEMMA 1. Let G s HK be the product of the subgroups H and K such that G⬘ is nilpotent. Suppose that F is a saturated formation such that G belongs to F. Then H and K belong to F. Proof. Assume that the result is false and let G be a minimal counterexample. Notice that G is a soluble group. By the minimality of G, we can assume that G has a unique minimal normal subgroup N, say. Let p be the prime dividing the order of N. Since G⬘ is nilpotent, we have that G⬘ is contained in F Ž G . s Op Ž G . and then G⬘ is a p-group. In particular,G has a unique Sylow p-subgroup P and P s F Ž G .. By w1, Lemma 1.3.2x, we can take A a Hall p⬘-subgroup of H and B a Hall p⬘-subgroup of K such that X s AB is a Hall p⬘-subgroup of G. Then G s PX and P l X s 1. Since G g F , it follows that GrOp⬘ p Ž G . lies in F Ž p ., where F is the canonical local definition of F. Since Op⬘Ž G . s 1, we have that Op⬘ p Ž G . s Op Ž G . s P. Therefore X belongs to F Ž p .. Now G⬘ is contained in P. This implies that X is an abelian group. We can now apply w6, IV, 1.14x to conclude that A and B belong to F Ž p .. Since H and K have a normal Sylow p-subgroup, it follows that H and K belong to Sp F Ž p . s F Ž p .. Consequently, H and K are F-groups and the proof of the lemma is complete. The next result shows that the converse of the above lemma holds for mutually permutable products with nilpotent commutator subgroup and saturated formations containing U . LEMMA 2. Let G s HK be the mutually permutable product of the subgroups H and K. Let F be a saturated formation containing U . Suppose that H and K belong to F. If G⬘ is nilpotent, then G belongs to F. Proof. Assume that the result is not true and let G be a minimal counterexample to the lemma. By the above remark and the minimality of G, we know that every proper quotient of G belongs to F. Consequently, G is a soluble primitive group because F is a saturated formation. Hence F Ž G . s SocŽ G . is the unique minimal normal subgroup of G and if p is the prime dividing its order, then F Ž G . s G⬘ is the unique Sylow p-subgroup of G.
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On the other hand, by Proposition 1, D s H l K is a permutable subgroup of H and K. Suppose that Core H Ž D . s 1. Then, applying Proposition 3, we have that D is a nilpotent subnormal subgroup of G. Hence D is contained in F Ž G .. Let A and B be Hall p⬘-subgroups of H and K, respectively, such that AB is a Hall p⬘-subgroup of G. Notice that w D, A x is a normal p-subgroup of the group DA and D is a Sylow p-subgroup of DA. Therefore w D, A x is contained in D and so D is a normal subgroup of DA. Analogously, D is a normal subgroup of DB. In addition, D is contained in F Ž G . which is abelian. So D is normalized by F Ž G .Ž AB . s G. If D s 1, then G is the totally permutable product of the subgroups H and K. So, by Proposition 2, we have G g F , a contradiction. Consequently, D / 1 and then F Ž G . s D. Suppose now that Core H Ž D . / 1 and Core K Ž D . / 1. Then the normal closure Core H Ž D . G is different from 1 and so F Ž G . F Core H Ž D . G . Notice that Core H Ž D . G s Core H Ž D . K which is contained in K. Analogously, F Ž G . F Core K Ž D . G and Core K Ž D . G is contained in H. In particular, F Ž G . is contained in D. Therefore, in both cases, we have that F Ž G . is contained in H l K. Moreover, since G⬘ s F Ž G ., we also have that H and K are normal subgroups of G. Consider F Ž G ., regarded as a H-module over GF Ž p .. By Clifford’s theorem, we know that F Ž G . is a completely reducible H-module, that is, F Ž G . s N1 = ⭈⭈⭈ = Nr , where Ni is a minimal normal subgroup of H for all i. Now H g F. This means that HrCH Ž Ni . g F Ž p . for all i Žhere F denotes the canonical local definition of F .. Since F Ž p . is a formation, we have that HrCH Ž F Ž G .. g F Ž p .. So HrF Ž G . g F Ž p .. Analogously, KrF Ž G . g F Ž p .. Now the group GrF Ž G . is an abelian group which is the product of the subgroups HrF Ž G . and KrF Ž G .. Since formations are closed under central products, it follows that GrF Ž G . g F Ž p .. This implies that G g S p F Ž p . s F Ž p . : F , the final contradiction. As the symmetric group of degree 3 and the saturated formation of nilpotent groups show, the condition U : F is essential in the above lemma. LEMMA 3. Let G s HK be the mutually permutable product of the subgroups H and K. Assume that G⬘ is nilpotent. If F is a saturated formation containing U , then G F s ² H F, K F :. Proof. Suppose that the result is not true and let G be a counterexample of minimal order. If N is a minimal normal subgroup of G, then GrN is the mutually permutable product of the subgroups HNrN and KNrN. Moreover Ž GrN .⬘ is nilpotent. So, by the minimal choice of G, we have
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that G F N s ² H F , K F : N. By Lemma 1 applied to GrG F s Ž HG FrG F .Ž KG FrG F ., ² H F, K F : is contained in G F. Hence G F s ² H F, K F :Ž G F l N .. Therefore, if G F l N s 1, it follows that G F s ² H F, K F :, a contradiction. Consequently, we may assume that SocŽ G . is contained in G F and G F s ² H F, K F : N for every minimal normal subgroup N of G. In particular, Core G Ž² H F, K F :. s 1. On the other hand, G F is contained in G⬘ which is nilpotent. So H F and F K are subnormal subgroups of G. By a result due to Wielandt w6, A, 14.4x, ² H F, K F : is subnormal in G. Hence SocŽ G . normalizes ² H F, K F : by another result of Wielandt w6, A, 14.3x. This means that ² H F, K F : is a normal subgroup of G F. Let p be the prime dividing the order of a minimal normal subgroup N of G. Since G is soluble, it follows that N is an abelian p-group. In particular, G FrŽ² H F, K F :. is an abelian p-group and so O p Ž G F .Ž G F .⬘ is contained in ² H F, K F :. Since O p Ž G F .Ž G F .⬘ is normal in G and Core G Ž² H F, K F :. s 1, it follows that O p Ž G F .Ž G F .⬘ s 1 and G F is an abelian p-group for some prime p. This means that F Ž G . is a p-group and then G⬘ is also a p-group because G⬘ is nilpotent. Consequently, F Ž G . is the unique Sylow p-subgroup of G. Let H p and K p denote the Sylow p-subgroups of H and K, respectively. It is clear that F Ž G . s Hp K p . Let Hp⬘ and K p⬘ be Hall p⬘-subgroups of H and K, respectively. Since H and K are mutually permutable, it follows that HK p⬘ and KHp⬘ are subgroups of G. Suppose that G s HK p⬘. Then F Ž G . is contained in H and so H is a normal subgroup of G. Hence H F is a normal subgroup of G. Since Core G Ž² H F, K F :. s 1, it follows that H F s 1 and so H g F. If K F s 1, then G g F by Lemma 2. Therefore K F / 1 and so KHp⬘ should be a proper subgroup of G Žotherwise F Ž G . F K and K is a normal subgroup of G .. Let M be a maximal subgroup of G containing KHp⬘. Then M s Ž M l H . K. By Proposition 1, M is the mutually permutable product of the subgroups M l H and K. The minimal choice of G yields M F s ²Ž M l H . F, K F :. Moreover H s H p Ž M l H . because G s Hp M s F Ž G . M. By w6, IV, 1.17Žb.x, Ž M l H . F is contained in H F because H p is a nilpotent normal subgroup of H. Therefore M F is contained in ² H F, K F :. Suppose that G F is not contained in M. Then G s MG F. Since G F is abelian and M normalizes M F, we have that M F is a normal subgroup of G. In particular, M F F Core G Ž² H F, K F :. s 1 and so K g F , a contradiction. Consequently, G F is contained in M. Let N be a minimal normal subgroup of G. Then N F G F F M. Since G s MF Ž G . and F Ž G . centralizes N, it follows that N is a minimal normal subgroup of M. If N l M F s N, then N F ² H F, K F : and G F s ² H F, K F :, a contradiction. So N l M F s 1 and NM FrM F is a minimal normal subgroup of MrM F g F. In particular, N is F-central in M and so N is F-central in G. By w6, V, 3.2x, N is contained in every F-normalizer of
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G. Let E be one of them. Then G s G F E and E l G F s 1 by w6, IV, 5.18; V, 4.2x. However N F E l G F, a contradiction. Consequently, HK p⬘ is a proper subgroup of G. Analogously, KHp⬘ is a proper subgroup of G. If H does not belong to F or K does not belong to F , we can use the above argument to get a contradiction. So H g F and K g F. By Lemma 2, G g F , yielding the final contradiction. LEMMA 4. Let G s HK be the mutually permutable product of the subgroups H and K. Suppose that G⬘ is nilpotent. If F is a saturated formation containing U , then Ž H l K . F is contained in G F. Proof. We argue by induction on the order of G. Let N be a minimal normal subgroup of G. Then, by the above remark, the group GrN is the mutually permutable product of the subgroups HNrN and KNrN and Ž GrN .⬘ s G⬘NrN is nilpotent. By induction, ŽŽ HN l KN .rN . F is contained in Ž GrN . F s G F NrN. Now, HN l KN s Ž HN l K . N s Ž H l KN . N and ŽŽ HN l KN .rN . F s Ž HN l K . F NrN s Ž H l KN . F NrN. Hence Ž HN l K . F N s Ž H l KN . F N is contained in G F N for every minimal normal subgroup N of G. We distinguish two cases: Case 1. G does not belong to F. In this case G F / 1 and we can take a minimal normal subgroup N of G such that Ž HN l K . F N s Ž H l KN . F N F G F. Consider the subgroup T s H Ž HN l K . s HN. Then, by Proposition 1, T is the mutually permutable product of the subgroups H and HN l K. So, if T is a proper subgroup of G, we have Ž H l K . F F T F. On the other hand, T F s ² H F, Ž HN l K . F : by Lemma 3. Therefore Ž H l K . F is contained in ² H F, G F : s G F by Lemma 3, and the lemma is true. So we may assume that G s HN. Analogously, G s KN. Consequently, H and K are maximal subgroups of G Žnotice that if H s G or K s G, the result is true by Lemma 3.. It is clear that H and K cannot be conjugate in G. So, by w6, A, 16.1x, Core G Ž H . / Core G Ž K . and either CoreG Ž H . / 1 or Core G Ž K . / 1. Suppose that CoreG Ž H . / 1. Now, if CoreG Ž K . is contained in H, then HrCoreG Ž K . is a maximal subgroup of the primitive group GrCore G Ž K . whose Fitting subgroup is precisely N Core G Ž K .rCoreG Ž K .. Moreover CoreG Ž H .rCoreG Ž K . is a non-trivial normal subgroup of GrCoreG Ž K .. Therefore N CoreG Ž K .rCoreG Ž K . is contained in Core G Ž H .rCoreG Ž K . and N is contained in H, a contradiction. Consequently, Core G Ž K . is not contained in H. Since H is maximal in G and Core G Ž K . is a normal subgroup of G, it follows that G s H Core G Ž K . and then K s Core G Ž K .Ž H l K .. It is clear that K is the mutually permutable product of the subgroups Core G Ž K . and H l K
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because H l K is a permutable subgroup of K by Proposition 1. By Lemma 3, we have K F s ²ŽCore G Ž K .. F, Ž H l K . F : and K F is contained in G F. So Ž H l K . F is contained in G F and the lemma is true in this case. Case 2. G g F. Let D s H l K. By Lemma 1, we have that H and K are F-groups. If either Core H Ž D . s 1 or Core K Ž D . s 1, it follows by Propositions 1Žb. and 3 that D is nilpotent and D g F and the lemma is true. Consequently, we may assume that Core H Ž D . / 1 and Core K Ž D . / 1. This implies that Core H Ž D . G is a non-trivial normal subgroup of G contained in K and Core K Ž D . G is a non-trivial normal subgroup of G contained in H. Let A and B be minimal normal subgroups of G contained in H and K, respectively. Applying induction to GrA and GrB, we conclude that Ž H l KA.rA and Ž K l HB .rB belong to F. Since Ž H l KA.rA is isomorphic to Ž H l K .rŽŽ H l K . l A. and Ž K l HB .rB is isomorphic to Ž H l K .rŽŽ H l K . l B ., we have that Ž H l K .rŽŽ H l K . l A l B . g F. So if A / B, then H l K g F and we have finished. Therefore we may assume that A s B and then A is contained in H l K. In particular, Ž H l K . F is contained in A. Suppose there exists a minimal normal subgroup N of G such that N / A; we know that Ž H l KN . F and Ž K l HN . F are contained in G F N s N. The group T s K Ž H l KN . is the mutually permutable product of K and H l KN. So, if T - G, we have Ž H l K . F F T F s ² K F, Ž H l KN . F : by Lemma 3. Since K F s 1, it follows that Ž H l K . F F N. Consequently, Ž H l K . F F N l A s 1 and D g F. The lemma is true in this case. Consequently, we can assume that G s HN s KN. Arguing as in Case 1, we have D g F. Therefore we may assume that A s SocŽ G . is the unique minimal normal subgroup of G. Since G is soluble, A is an elementary abelian p-group for some prime p. Then F Ž G . s Op Ž G . s Op⬘ p Ž G . and F Ž G . is the unique Sylow p-subgroup of G and G⬘ is contained in F Ž G .. Since G g F , we know that GrF Ž G . g F Ž p ., where F is the canonical local definition of F. This implies that the Hall p⬘-subgroups of G are in F Ž p .. Moreover they are abelian. By w6, IV, 1.14x, every p⬘-subgroup of G is in F Ž p .. In particular, the Hall p⬘-subgroups of H l K are in F Ž p .. Since H l K has a normal Sylow p-subgroup, we have that H l K g Sp F Ž p . s F Ž p . : F and the lemma is proved. Proof of Theorem A. Assume that the result is not true and choose for G a counterexample of smallest order.
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If N is a minimal normal subgroup of G, then G F N s H F K F N by the minimal choice of G. Moreover H F and K F are contained in G F by Lemma 3. Again by the minimality of G, we can deduce easily that SocŽ G . is contained in G F and G F s H F K F N for every minimal normal subgroup N of G. Let p be a prime dividing the order of an arbitrary minimal normal subgroup N of G. We claim that G F is a p-group. Otherwise since G F is nilpotent, it has a unique normal Hall p⬘-subgroup Ž G F . p⬘ / 1. It is clear that Ž G F . p⬘ is normal in G and it is the product of the unique Hall p⬘-subgroup Ž H F . p⬘ of H F and the unique Hall p⬘-subgroup Ž K F N . p⬘ of K F N. So H F permutes with Ž K F N . p⬘ which is equal to Ž K F . p⬘ because N is a p-group. Now, if we take a minimal normal subgroup B of G contained in Ž G F . p⬘ , then G F s H F K F B. Arguing in the same way as above, we have that H F permutes with the unique Sylow p-subgroup Ž K F . p of K F. Consequently, H F permutes with K F and G F s ² H F, K F : s H F K F by Lemma 3, a contradiction. Therefore there exists a prime p such that G F is a p-group. Since SocŽ G . is contained in G F, it follows that Op⬘Ž G . s 1 and F Ž G . s Op Ž G .. Hence G⬘ is a p-group and F Ž G . is the unique Sylow p-subgroup of G. Denote D s H l K. Suppose that D g F. Consider the subgroup T s Ž H F D . K and assume that T is a proper subgroup of G. Then, by Proposition 1, T is the mutually permutable product of the subgroups H F D and K. By the minimal choice of G, T F s Ž H F D . F K F. This implies that H F F NG ŽT F . and H F T F is a subgroup of G. Since H F is normal in H, it is clear that H F D is a mutually permutable product of H F and D. So, by minimality of G, we have Ž H F D . F s Ž H F . F D F s 1. Therefore T F s K F and H F permutes with K F. This is a contradiction as G F s ² H F, K F : by Lemma 3. Consequently, T s G and then G s H F K. Arguing in the same way with the subgroup Ž K F D . H, we have G s H F K s HK F. Hence H s H F D is the mutually permutable product of H F and D. Therefore H F s ²Ž H F . F, D F : s 1 and H g F. Analogously, K g F. From Lemma 2, G g F , a contradiction. Hence we may assume that D does not belong to F , that is, D F / 1. Let Hp and K p be the Sylow p-subgroups of H and K, respectively, such that F Ž G . s H p K p . Since H p is a normal subgroup of H, we have that Hp and D are mutually permutable subgroups of H. Hence Ž H p D . F s D F by the minimal choice of G. Analogously, Ž K p D . F s D F and then F Ž G . normalizes D F. Let now H p⬘ and K p⬘ be Hall p⬘-subgroups of H and K, respectively, such that X s H p⬘ K p⬘ is a Hall p⬘-subgroup of G. Since, by Proposition 1, D is a permutable subgroup of H, it follows that DH p⬘ is a subgroup of G.
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Let Dp⬘ be a Hall p⬘-subgroup of D such that Dp⬘ H p⬘ is a Hall p⬘-subgroup of DH p⬘. Then Dp⬘ H p⬘ s H p⬘ because the maximality of H p⬘ as p⬘-subgroup of H. Since X is abelian, Dp⬘ permutes with K p⬘ and again Dp⬘ is contained in K p⬘. It is clear that Dp is the unique Sylow p-subgroup of the subgroups DH p⬘ and DK p⬘. Hence X normalizes D. Since D F is characteristic in D, we have that X normalizes D F. Consequently, D F is a normal subgroup of G. We distinguish two cases: Case 1. G s HK p⬘ s KHp⬘. In this case F Ž G . is contained in D. So H and K are normal subgroups of G and so are H F and K F. Then, by Lemma 3, G F s H F K F. Case 2. Either HK p⬘ is a proper subgroup of G or KHp⬘ is a proper subgroup of G. Suppose that HK p⬘ - G and let M be a maximal subgroup of G containing HK p⬘. Then K p is not contained in M and G s MK p . Moreover M s H Ž M l K . is the mutually permutable product of H and M l K by Proposition 1. By the minimality of G, M F s H F Ž M l K . F and 1 / D F is contained in M F by Lemma 4. The fact that Ž M l K . F is contained in K F by w6, IV, 1.17x yields M F is contained in H F K F. Consequently, since D F is a non-trivial normal subgroup of G, a minimal normal subgroup N of G is contained in H F K F. This implies that G F s H F NK F s H F K F, the final contradiction. One might wonder whether some F-projector of a mutually permutable product with nilpotent commutator subgroup could be the product of F-projectors of the factor groups. Unfortunately, this is not true in general as the next example shows: Let G be the direct product of a cyclic group ² a: of order 3 with the alternating group A 4 of degree 4. Let V be the Klein 4-group of A 4 . Then G is the mutually permutable product of H s ² a: = V and K s A 4 . Moreover G⬘ s V is nilpotent. Notice that H is the supersoluble projector of H and a Sylow 3-subgroup B of A 4 is a supersoluble projector of K. However HB s G is not supersoluble.
ACKNOWLEDGMENT This research was supported by Proyecto PB 94-0965 of DGICYT, MEC, Spain.
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REFERENCES 1. B. Amberg, S. Franciosi, and F. de Giovanni, ‘‘Products of Groups,’’ Clarendon, Oxford, 1992. 2. A. Ballester-Bolinches, M. C. Pedraza-Aguilera, and M. D. Perez-Ramos, On finite ´ products of totally permutable groups, Bull. Austral. Math. Soc. 53 Ž1996., 441᎐445. 3. A. Ballester-Bolinches, M. C. Pedraza-Aguilera, and M. D. Perez-Ramos, Finite groups ´ which are products of pairwise totally permutable subgroups, Proc. Edinburgh Math. Soc. Ž 2 . 41 Ž1998., 567᎐572. 4. A. Ballester-Bolinches, M. C. Pedraza-Aguilera, and M. D. Perez-Ramos, Mutually per´ mutable products of finite groups, J. Algebra 213 Ž1999., 369᎐377. 5. A. Carocca, p-supersolvability of factorized finite groups, Hokkaido Math. J. 21 Ž1992., 395᎐403. 6. K. Doerk and T. Hawkes, ‘‘Finite Soluble Groups,’’ de Gruyter, BerlinrNew York, 1992. 7. R. Maier and P. Schmid, The embedding of permutable subgroups in finite groups, Math. Z. 131 Ž1973., 269᎐272.