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ISRN Algebra
VolumeΒ 2011Β (2011), Article IDΒ 867082, 4 pages
http://dx.doi.org/10.5402/2011/867082
Research Article

Mutually Permutable Products of Finite Groups

Department of Mathematics, Faculty of Science 14466, King Abdulaziz University, Jeddah 21424, Saudi Arabia

Received 19 June 2011; Accepted 10 July 2011

Academic Editors: M.Β Asaad, G.Β Buskes, G.Β Mason, and L.Β Vinet

Copyright Β© 2011 Rola A. Hijazi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Let G be a finite group and G1, G2 are two subgroups of G. We say that G1 and G2 are mutually permutable if G1 is permutable with every subgroup of G2 and G2 is permutable with every subgroup of G1. We prove that if 𝐺 = 𝐺 1 𝐺 2 = 𝐺 1 𝐺 3 = 𝐺 2 𝐺 3 is the product of three supersolvable subgroups G1, G2, and G3, where Gi and Gj are mutually permutable for all i and j with 𝑖 β‰  𝑗 and the Sylow subgroups of G are abelian, then G is supersolvable. As a corollary of this result, we also prove that if G possesses three supersolvable subgroups 𝐺 𝑖 ( 𝑖 = 1 , 2 , 3 ) whose indices are pairwise relatively prime, and Gi and Gj are mutually permutable for all i and j with 𝑖 β‰  𝑗 , then G is supersolvable.

1. Introduction

Throughout this paper, 𝐺 will denote a finite group. We write 𝜎 ( 𝐺 ) for the set of prime divisors of the order of 𝐺 and | 𝜎 ( 𝐺 ) | for their number. In [1], Doerk determined the structure of minimal non-supersolvable groups (nonsupersolvable groups and all of whose proper subgroups are supersolvable). He proved that if 𝐺 is a minimal non-supersolvable group, then 𝐺 is solvable and 2 ≀ | 𝜎 ( 𝐺 ) | ≀ 3 . Therefore, if 𝐺 is a minimal non-supersolvable group with | 𝜎 ( 𝐺 ) | = 3 , then 𝐺 possesses three supersolvable subgroups 𝐺 𝑖 ( 𝑖 = 1 , 2 , 3 ) such that 𝐺 = 𝐺 1 𝐺 2 = 𝐺 1 𝐺 3 = 𝐺 2 𝐺 3 . In [2], Kegel proved that if 𝐺 = 𝐺 1 𝐺 2 = 𝐺 1 𝐺 3 = 𝐺 2 𝐺 3 , where 𝐺 1 and 𝐺 2 are nilpotent subgroups of 𝐺 , and 𝐺 3 is a supersolvable subgroup of 𝐺 , then 𝐺 is supersolvable. In [3], Asaad and Shaalan proved the following result: assume that 𝐺 1 and 𝐺 2 are supersolvable subgroups of 𝐺 , 𝐺 ξ…ž is nilpotent and 𝐺 = 𝐺 1 𝐺 2 . Assume further that 𝐺 1 is permutable with every subgroup of 𝐺 2 and 𝐺 2 is permutable with every subgroup of 𝐺 1 ( 𝐺 1 and 𝐺 2 are mutually permutable). Then 𝐺 is supersolvable. Further, they proved the following result: Assume that 𝐺 1 and 𝐺 2 are supersolvable subgroups of 𝐺 and 𝐺 = 𝐺 1 𝐺 2 and that every subgroup of 𝐺 1 is permutable with every subgroup of 𝐺 2 ( 𝐺 1 and 𝐺 2 are totally permutable). Then, 𝐺 is supersolvable.

In this paper, we are interested in the following question.

Assume that 𝐺 = 𝐺 1 𝐺 2 = 𝐺 1 𝐺 3 = 𝐺 2 𝐺 3 , where 𝐺 1 , 𝐺 2 , and 𝐺 3 are supersolvable subgroups and that 𝐺 𝑖 and 𝐺 𝑗 are mutually permutable for all 𝑖 and 𝑗 with 𝑖 β‰  𝑗 . Is 𝐺 supersolvable? The answer is negative as the following example shows.

Example 1 (see [8, pages 8-9]). Let 𝐻 = ⟨ π‘₯ ⟩ Γ— ⟨ 𝑦 ⟩ , where | π‘₯ | = | 𝑦 | = 5 . The maps 𝛼 ∢ π‘₯ β†’ π‘₯ 2 , 𝑦 β†’ 𝑦 βˆ’ 2 , and 𝛽 ∢ π‘₯ β†’ 𝑦 βˆ’ 1 , 𝑦 β†’ π‘₯ are automorphisms of 𝐻 and generate a subgroup 𝐴 ≀ A u t ( 𝐻 ) of order 8 ( 𝐴 is isomorphic with a quaternion group). Take 𝐺 = [ 𝐻 ] 𝐴 . Then 𝐺 1 = 𝐻 ⟨ 𝛼 ⟩ , 𝐺 2 = 𝐻 ⟨ 𝛽 ⟩ , and 𝐺 3 = 𝐻 ⟨ 𝛼 𝛽 ⟩ are normal supersolvable subgroups of 𝐺 and 𝐺 = 𝐺 1 𝐺 2 = 𝐺 1 𝐺 3 = 𝐺 2 𝐺 3 , but 𝐺 is not supersolvable.

We prove the following result.

Theorem 1.1. If 𝐺 = 𝐺 1 𝐺 2 = 𝐺 1 𝐺 3 = 𝐺 2 𝐺 3   is the product of three supersolvable subgroups 𝐺 1 , 𝐺 2 , and 𝐺 3 such that 𝐺 𝑖 and 𝐺 𝑗 are mutually permutable for all 𝑖 and 𝑗 with 𝑖 β‰  𝑗 , and the Sylow subgroups of 𝐺 are abelian, then 𝐺 is supersolvable.

As a corollary of Theorem 1.1, we have the following.

Corollary 1.2. If 𝐺 possesses three supersolvable subgroups 𝐺 𝑖 (i= 1,2,3) whose indices in 𝐺 are pairwise relatively prime, and 𝐺 𝑖 is permutable with every subgroup of 𝐺 𝑗 , for all 𝑖   and 𝑗 with 𝑖 β‰  𝑗 , then 𝐺 is supersolvable.

2. Preliminaries

We list here some basic results which are needed in this paper.

Lemma 2.1 (see [2]). Let 𝐺 = 𝐺 1 𝐺 2 = 𝐺 1 𝐺 3 = 𝐺 2 𝐺 3   be the product of three subgroups 𝐺 1 , 𝐺 2 , and 𝐺 3 . If 𝐺 1 , 𝐺 2 , and 𝐺 3 have normal Sylow p-subgroups for a certain prime p, then also 𝐺 has a normal Sylow p-subgroup.

Lemma 2.2 (see [4]). Let 𝐺 = 𝐺 1 𝐺 2 be a group such that 𝐺 1 and 𝐺 2 are mutually permutable. (a)If 𝐺 1 ∩ 𝐺 2 = 1 , then 𝐺 1 and 𝐺 2 are totally permutable.(b) 𝐺 1 ∩ 𝐺 2 is a quasinormal subgroup of 𝐺 1 and of 𝐺 2 .

Lemma 2.3 (see [3]). Let 𝐺 = 𝐺 1 𝐺 2   be a group such that 𝐺 1 and 𝐺 2 are totally permutable subgroups. If 𝐺 1 and 𝐺 2 are supersolvable subgroups of 𝐺 , then 𝐺 is supersolvable.

Lemma 2.4 (see [5, page 213, Theorem  7.1.2]). If 𝐻 is a quasinormal subgroup of 𝐺 , then 𝐻 is subnormal in 𝐺 .

Lemma 2.5 (see [5, page 239, Theorem  7.7.1]). Let 𝐺 = 𝐺 1 𝐺 2 be a group such that 𝐺 1 and 𝐺 2 are subgroups of 𝐺 . If 𝐻 is a subnormal subgroup of 𝐺 1 and of 𝐺 2 , then 𝐻 is subnormal in 𝐺 .

Lemma 2.6 (see [6]). Let 𝐺 be a group and 𝐻 is a quasinormal subgroup of 𝐺 . Then, 𝐻 / C o r 𝐺 ( 𝐻 ) is nilpotent, where C o r 𝐺 ( 𝐻 ) = ∩ 𝑔 ∈ 𝐺 𝐻 𝑔 .

Lemma 2.7 (see [7, page 29, Theorem  8.8(a)]). 𝐹 ( 𝐺 ) = ⟨ 𝐻 ∢ 𝐻 is subnormal subgroup o f 𝐺 and 𝐻 is nilpotent ⟩ .

Lemma 2.8 (see [2]). Let the group 𝐺 = 𝐺 1 𝐺 2 = 𝐺 1 𝐺 3 = 𝐺 2 𝐺 3   be the product of three subgroups 𝐺 1 , 𝐺 2 , and 𝐺 3 . If 𝐺 1 , 𝐺 2 , and 𝐺 3 are nilpotent subgroups of 𝐺 , then 𝐺 is nilpotent.

Lemma 2.9 (see [8, page 196, Theorem  5.1(15)]). If 𝐺 is a supersolvable group, then 𝐺 / 𝑂 𝑝 β€² , 𝑝 ( 𝐺 ) is abelian of exponent dividing 𝑝 βˆ’ 1 for all primes 𝑝 .

Lemma 2.10 (see [8, page 6, Theorem  1.9]). Let 𝑃 be a normal Sylow 𝑝 -subgroup of 𝐺 . If 𝐺 / 𝑃 is abelian of exponent dividing 𝑝 βˆ’ 1 , then 𝐺 is supersolvable.

Lemma 2.11 (see [8, page 5, Theorem  1.6]). The commutator subgroup of a supersolvable group is nilpotent.

3. Proofs

Proof of Theorem 1.1. Assume that the result is not true, and let 𝐺 be a counterexample of minimal order. Since 𝐺 𝑖 ( 𝑖 = 1 , 2 , 3 ) is supersolvable, it follows that 𝐺 𝑖 has a normal Sylow 𝑝 -subgroup, where 𝑝 is the largest prime dividing | 𝐺 | . Then, by Lemma 2.1, 𝐺 has a normal Sylow 𝑝 -subgroup, say 𝑃 . Certainly, every proper quotient group of 𝐺 satisfies the hypothesis of the theorem. So every proper quotient group of 𝐺 is supersolvable by the minimal choice of 𝐺 . But the class of all supersolvable groups is a saturated formation, so Ξ¦ ( 𝐺 ) = 1 , 𝑃 = 𝐹 ( 𝐺 ) , and 𝐢 𝐺 ( 𝑃 ) = 𝑃 . We argue that 𝐺 1 ∩ 𝐺 2 β‰  1 . If not, 𝐺 1 ∩ 𝐺 2 = 1 . Then, 𝐺 = 𝐺 1 𝐺 2 is a totally permutable product of 𝐺 1 and 𝐺 2 by Lemma 2.2(a). Then, by Lemma 2.3, 𝐺 is supersolvable, a contradiction.
Thus, 𝐺 1 ∩ 𝐺 2 β‰  1 . Analogously, 𝐺 1 ∩ 𝐺 3 β‰  1 and 𝐺 2 ∩ 𝐺 3 β‰  1 . Since 𝐺 = 𝐺 1 𝐺 2 is a mutually permutable product of 𝐺 1 and 𝐺 2 , it follows that 𝐺 1 ∩ 𝐺 2 is a quasinormal subgroup of 𝐺 1 and of 𝐺 2 by Lemma 2.2(b). Then, by Lemma 2.4, 𝐺 1 ∩ 𝐺 2 is a subnormal subgroup of 𝐺 1 and of 𝐺 2 . Hence 𝐺 1 ∩ 𝐺 2 is a subnormal subgroup of 𝐺 by Lemma 2.5.
If C o r 𝐺 1 ( 𝐺 1 ∩ 𝐺 2 ) = 1 , then, by Lemma 2.6, 𝐺 1 ∩ 𝐺 2 is nilpotent. Hence, 𝐺 1 ∩ 𝐺 2 is a subnormal nilpotent subgroup of 𝐺 . So 𝐺 1 ∩ 𝐺 2 ≀ 𝑃 by Lemma 2.7. Taken into consideration that 𝐺 1 ∩ 𝐺 2 is quasinormal in 𝐺 1 , 𝐺 1 ∩ 𝐺 2 ≀ 𝑃 and 𝑃 is abelian, it follows that 𝐺 1 ∩ 𝐺 2 is normal in 𝐺 1 and so 𝐺 1 ∩ 𝐺 2 = C o r 𝐺 1 ( 𝐺 1 ∩ 𝐺 2 ) = 1 , a contradiction. Thus C o r 𝐺 1 ( 𝐺 1 ∩ 𝐺 2 ) β‰  1 . Analogously, C o r 𝐺 2 ( 𝐺 2 ∩ 𝐺 3 ) β‰  1 and C o r 𝐺 3 ( 𝐺 3 ∩ 𝐺 1 ) β‰  1 . Set 𝐿 1 = C o r 𝐺 1 ( 𝐺 1 ∩ 𝐺 2 ) , 𝐿 2 = C o r 𝐺 2 ( 𝐺 2 ∩ 𝐺 3 ) and 𝐿 3 = C o r 𝐺 3 ( 𝐺 3 ∩ 𝐺 1 ) . Then 𝐿 𝐺 1 = 𝐿 𝐺 1 𝐺 2 1 ≀ 𝐺 2 , 𝐿 𝐺 2 = 𝐿 𝐺 2 𝐺 3 2 ≀ 𝐺 3 , and 𝐿 𝐺 3 = 𝐿 𝐺 3 𝐺 1 3 ≀ 𝐺 1 . Now, 𝑃 ≀ 𝐿 𝐺 𝑖 ( 𝑖 = 1 , 2 , 3 ) as 𝑃 is a unique minimal normal subgroup of 𝐺 . Hence, 𝑃 ≀ 𝐺 𝑖 ( 𝑖 = 1 , 2 , 3 ).
Now, we finish the proof of the theorem. Since 𝐺 𝑖 is supersolvable, 𝐢 𝐺 ( 𝑃 ) = 𝑃 = 𝐹 ( 𝐺 ) and 𝑃 ≀ 𝐺 𝑖 ( 𝑖 = 1 , 2 , 3 ), it follows that 𝐢 𝐺 𝑖 ( 𝑃 ) = 𝐢 𝐺 𝑖 ( 𝐹 ( 𝐺 𝑖 ) ) = 𝑃 ( 𝑖 = 1 , 2 , 3 ) and 𝐺 𝑖 / 𝑃 is abelian. Hence, by Lemma 2.8, 𝐺 / 𝑃 is nilpotent, and, since the Sylow subgroups of 𝐺 are abelian, it follows that 𝐺 / 𝑃 is abelian. On the other hand, by Lemma 2.9, 𝐺 𝑖 / 𝑃 ( 𝑖 = 1 , 2 ) is of exponent dividing 𝑝 βˆ’ 1 . Hence, 𝐺 / 𝑃 is abelian of exponent dividing 𝑝 βˆ’ 1 and so 𝐺 is supersolvable, by Lemma 2.10, a final contradiction completing the proof of the theorem.

Proof of Corollary 1.2. Assume that the result is not true and let 𝐺 be a counterexample of minimal order. Let 𝑃 be a Sylow 𝑝 -subgroup of 𝐺 , where 𝑝 is the largest prime dividing | 𝐺 | . Then, 𝑃 is normal in 𝐺 by Lemma 2.1. Certainly, every proper quotient group of 𝐺 satisfies the hypothesis of the corollary. So every proper quotient group of 𝐺 is supersolvable by the minimal choice of 𝐺 . But the class of all supersolvable groups is a saturated formation, so Ξ¦ ( 𝐺 ) = 1 , 𝑃 = 𝐹 ( 𝐺 ) , and 𝐢 𝐺 ( 𝑃 ) = 𝑃 . Since 𝐺 1 , 𝐺 2 , and 𝐺 3 have coprime indices, we can assume that 𝑝 does not divide | 𝐺 ∢ 𝐺 1 | and 𝑝 does not divide | 𝐺 ∢ 𝐺 2 | . Then, 𝑃 ≀ 𝐺 1 and 𝑃 ≀ 𝐺 2 and so 𝑃 = 𝐢 𝐺 1 ( 𝑃 ) = 𝐢 𝐺 2 ( 𝑃 ) = 𝐹 ( 𝐺 2 ) = 𝐹 ( 𝐺 1 ) as 𝑃 = 𝐹 ( 𝐺 ) = 𝐢 𝐺 ( 𝑃 ) . Then, 𝐺 1 / 𝑃 and 𝐺 2 / 𝑃 are abelian subgroups of 𝐺 / 𝑃 by Lemma 2.11. This together with ( | 𝐺 ∢ 𝐺 1 | , | 𝐺 ∢ 𝐺 2 | ) = 1 imply that the Sylow subgroups of 𝐺 / 𝑃 are abelian. Now Theorem 1.1 implies that 𝐺 is supersolvable, a contradiction completing the proof of the corollary.

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