Mutually Permutable Products of Finite Groups

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=𝐺₁𝐺₂=𝐺₁𝐺₃=𝐺₂𝐺₃ is the product of three supersolvable subgroups G₁, G₂, and G₃, where G_i and G_j are mutually permutable for all i and j with 𝑖≠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 G_i and G_j are mutually permutable for all i and j with 𝑖≠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=𝐺₁𝐺₂=𝐺₁𝐺₃=𝐺₂𝐺₃. In [2], Kegel proved that if 𝐺=𝐺₁𝐺₂=𝐺₁𝐺₃=𝐺₂𝐺₃, where 𝐺₁ and 𝐺₂ are nilpotent subgroups of 𝐺, and 𝐺₃ is a supersolvable subgroup of 𝐺, then 𝐺 is supersolvable. In [3], Asaad and Shaalan proved the following result: assume that 𝐺₁ and 𝐺₂ are supersolvable subgroups of 𝐺, 𝐺^ is nilpotent and 𝐺=𝐺₁𝐺₂. Assume further that 𝐺₁ is permutable with every subgroup of 𝐺₂ and 𝐺₂ is permutable with every subgroup of 𝐺₁ (𝐺₁ and 𝐺₂ are mutually permutable). Then 𝐺 is supersolvable. Further, they proved the following result: Assume that 𝐺₁ and 𝐺₂ are supersolvable subgroups of 𝐺 and 𝐺=𝐺₁𝐺₂ and that every subgroup of 𝐺₁ is permutable with every subgroup of 𝐺₂ (𝐺₁ and 𝐺₂ are totally permutable). Then, 𝐺 is supersolvable.

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

Assume that 𝐺=𝐺₁𝐺₂=𝐺₁𝐺₃=𝐺₂𝐺₃, where 𝐺₁, 𝐺₂, and 𝐺₃ are supersolvable subgroups and that 𝐺_𝑖 and 𝐺_𝑗 are mutually permutable for all 𝑖 and 𝑗 with 𝑖≠𝑗. Is 𝐺 supersolvable? The answer is negative as the following example shows.

Example 1 (see [8, pages 8-9]). Let 𝐻=⟨𝑥⟩×⟨𝑦⟩, where |𝑥|=|𝑦|=5. The maps 𝛼∶𝑥→𝑥², 𝑦→𝑦⁻², and 𝛽∶𝑥→𝑦⁻¹, 𝑦→𝑥 are automorphisms of 𝐻 and generate a subgroup 𝐴≤Aut(𝐻) of order 8 (𝐴 is isomorphic with a quaternion group). Take 𝐺=[𝐻]𝐴. Then 𝐺₁=𝐻⟨𝛼⟩, 𝐺₂=𝐻⟨𝛽⟩, and 𝐺₃=𝐻⟨𝛼𝛽⟩ are normal supersolvable subgroups of 𝐺 and 𝐺=𝐺₁𝐺₂=𝐺₁𝐺₃=𝐺₂𝐺₃, but 𝐺 is not supersolvable.

We prove the following result.

Theorem 1.1. If 𝐺=𝐺₁𝐺₂=𝐺₁𝐺₃=𝐺₂𝐺₃  is the product of three supersolvable subgroups 𝐺₁, 𝐺₂, and 𝐺₃ such that 𝐺_𝑖 and 𝐺_𝑗 are mutually permutable for all 𝑖 and 𝑗 with 𝑖≠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≠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=𝐺₁𝐺₂=𝐺₁𝐺₃=𝐺₂𝐺₃  be the product of three subgroups 𝐺₁, 𝐺₂, and 𝐺₃. If 𝐺₁, 𝐺₂, and 𝐺₃ 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=𝐺₁𝐺₂ be a group such that 𝐺₁and 𝐺₂ are mutually permutable. (a)If 𝐺₁∩𝐺₂=1, then 𝐺₁and 𝐺₂ are totally permutable.(b)𝐺₁∩𝐺₂ is a quasinormal subgroup of 𝐺₁ and of 𝐺₂.

Lemma 2.3 (see [3]). Let 𝐺=𝐺₁𝐺₂  be a group such that 𝐺₁and 𝐺₂ are totally permutable subgroups. If 𝐺₁ and 𝐺₂ 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=𝐺₁𝐺₂ be a group such that 𝐺₁ and 𝐺₂ are subgroups of 𝐺. If 𝐻 is a subnormal subgroup of 𝐺₁ and of 𝐺₂, 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/Cor_𝐺(𝐻) is nilpotent, where Cor_𝐺(𝐻)=∩_𝑔∈𝐺𝐻^𝑔.

Lemma 2.7 (see [7, page 29, Theorem  8.8(a)]). 𝐹(𝐺)=⟨𝐻∶𝐻is subnormal subgroup of𝐺and𝐻is nilpotent⟩.

Lemma 2.8 (see [2]). Let the group 𝐺=𝐺₁𝐺₂=𝐺₁𝐺₃=𝐺₂𝐺₃  be the product of three subgroups 𝐺₁, 𝐺₂, and 𝐺₃. If 𝐺₁, 𝐺₂, and 𝐺₃ 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′,𝑝(𝐺) is abelian of exponent dividing 𝑝−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−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 Φ(𝐺)=1, 𝑃=𝐹(𝐺), and 𝐶_𝐺(𝑃)=𝑃. We argue that 𝐺₁∩𝐺₂≠1. If not, 𝐺₁∩𝐺₂=1. Then, 𝐺=𝐺₁𝐺₂ is a totally permutable product of 𝐺₁ and 𝐺₂ by Lemma 2.2(a). Then, by Lemma 2.3, 𝐺 is supersolvable, a contradiction.
Thus, 𝐺₁∩𝐺₂≠1. Analogously, 𝐺₁∩𝐺₃≠1 and 𝐺₂∩𝐺₃≠1. Since 𝐺=𝐺₁𝐺₂ is a mutually permutable product of 𝐺₁ and 𝐺₂, it follows that 𝐺₁∩𝐺₂ is a quasinormal subgroup of 𝐺₁ and of 𝐺₂ by Lemma 2.2(b). Then, by Lemma 2.4, 𝐺₁∩𝐺₂ is a subnormal subgroup of 𝐺₁ and of 𝐺₂. Hence 𝐺₁∩𝐺₂ is a subnormal subgroup of 𝐺 by Lemma 2.5.
If Cor_𝐺1(𝐺₁∩𝐺₂)=1, then, by Lemma 2.6, 𝐺₁∩𝐺₂ is nilpotent. Hence, 𝐺₁∩𝐺₂ is a subnormal nilpotent subgroup of 𝐺. So 𝐺₁∩𝐺₂≤𝑃 by Lemma 2.7. Taken into consideration that 𝐺₁∩𝐺₂ is quasinormal in 𝐺₁, 𝐺₁∩𝐺₂≤𝑃 and 𝑃 is abelian, it follows that 𝐺₁∩𝐺₂ is normal in 𝐺₁ and so 𝐺₁∩𝐺₂=Cor_𝐺1(𝐺₁∩𝐺₂)=1, a contradiction. Thus Cor_𝐺1(𝐺₁∩𝐺₂)≠1. Analogously, Cor_𝐺2(𝐺₂∩𝐺₃)≠1 and Cor_𝐺3(𝐺₃∩𝐺₁)≠1. Set 𝐿₁=Cor_𝐺1(𝐺₁∩𝐺₂), 𝐿₂=Cor_𝐺2(𝐺₂∩𝐺₃) and 𝐿₃=Cor_𝐺3(𝐺₃∩𝐺₁). Then 𝐿^𝐺1=𝐿^𝐺1𝐺2₁≤𝐺₂, 𝐿^𝐺2=𝐿^𝐺2𝐺3₂≤𝐺₃, and 𝐿^𝐺3=𝐿^𝐺3𝐺1₃≤𝐺₁. Now, 𝑃≤𝐿^𝐺𝑖 (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, 𝐶_𝐺(𝑃)=𝑃=𝐹(𝐺) and 𝑃≤𝐺_𝑖 (𝑖=1,2,3), it follows that 𝐶_𝐺𝑖(𝑃)=𝐶_𝐺𝑖(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−1. Hence, 𝐺/𝑃 is abelian of exponent dividing 𝑝−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 Φ(𝐺)=1, 𝑃=𝐹(𝐺), and 𝐶_𝐺(𝑃)=𝑃. Since 𝐺₁, 𝐺₂, and 𝐺₃ have coprime indices, we can assume that 𝑝 does not divide |𝐺∶𝐺₁| and 𝑝 does not divide |𝐺∶𝐺₂|. Then, 𝑃≤𝐺₁ and 𝑃≤𝐺₂ and so 𝑃=𝐶_𝐺1(𝑃)=𝐶_𝐺2(𝑃)=𝐹(𝐺₂)=𝐹(𝐺₁) as 𝑃=𝐹(𝐺)=𝐶_𝐺(𝑃). Then, 𝐺₁/𝑃 and 𝐺₂/𝑃 are abelian subgroups of 𝐺/𝑃 by Lemma 2.11. This together with (|𝐺∶𝐺₁|,|𝐺∶𝐺₂|)=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.