On Finite Groups with Three Supersolvable Subgroups of Pairwise Relatively Prime Indices

Abstract

Let G be a finite group. In this paper, we obtain some new conditions for supersolvability of G with three supersolvable subgroups of pairwise relatively prime indices in G. Our main results improve and extend some known results in the literature.

1 Introduction

All groups considered in this paper will be finite. We use conventional notions and notation (see [7, 13]). G always denotes a finite group, |G| is the order of G and $\pi (G)$ denotes the set of all primes dividing |G|.

Recall that a class of groups $\Im$ is a formation if $\Im$ contains all homomorphic images of a group in $\Im ,$ and if G/M and G/N are in $\Im$, then $G/(M\cap N)$ is in $\Im$ for any normal subgroups M and N of G. A formation $\Im$ is said to be saturated if $G/\Phi (G)\in \Im$ implies that $G\in \Im .$ Let $\mathcal {U}$ be the class of all supersolvable groups. Clearly, $\mathcal {U}$ is a saturated formation. The $\mathcal {U}{ -}hypercenter$ of G will be denoted by $Z_{ \mathcal {U}}{ (G)}$, i.e., $Z_{\mathcal {U}}{ (G)}$ is the product of all normal subgroups N of G such that every chief factor of G below N has a prime order (see [10]).

A subgroup H of G is said to be $s-permutable$ in G if H permutes with every Sylow subgroup of G (see [15]); H is said to be $s-semipermutable$ in G if H permutes with every Sylow subgroup L of G with $(|H|,|L|)=1$ (see [16]). We say that two subgroups H and K of a group G are mutually permutable if H permutes with every subgroup of K and K permutes with every subgroup of H (see [7]). It is well known that the product of two normal supersolvable subgroups is not supersolvable in general (see [20, Chapter 1, pages 8/9]). In [5], it is proved that the product of two mutually permutable supersolvable subgroups is supersolvable provided that the commutator subgroup is nilpotent. In [1], it is also proved that if G is a product of two mutually permutable supersolvable subgroups H and K with Core$_{G}(H\cap K)=1,$ then G is supersolvable. These two results have been generalized in [8] and [11].

There exists a nonsupersolvable group G with three supersolvable subgroups $H_{i},$ $i=1,$ 2,  3,  such that $(|G:H_{i}|,|G:H_{j}|)=1$ if $i\ne j$ as the following example shows:

Example 1.1

• (1) There exists a nonsupersolvable group G of order $7^{2}.2.3$ (see [20, Chapter 3, Corollary 4.6]).

• (2) G is solvable and every proper subgroup of G is supersolvable (see [20, Chapter 3, Lemma 4.3(i)]).

• (3) G has three supersolvable subgroups $H_{1},$ $H_{2}$ and $H_{3}$ of orders $7^{2}.3,$ $7^{2}.2,$ and 3.2, respectively (see [20, Chapter3, Lemma 4.3(ii)]). Now this example raises the following question:

Question. Suppose that G is a group with three supersolvable subgroups $H_{i},$ $i=1,2,3,$ such that $(|G:H_{i}|,|G:H_{j}|)=1$ if $1\le i\ne j\le 3.$ What are the conditions that should be imposed on G to gurantee it is supersolvable?

Many researchers showed interest in addresing this question. In [9], Doerk proved the following theorem:

Theorem 1.2

If a group G possesses four supersolvable subgroups whose indices in G are pairwise relatively prime, then G is supersolvable.

In [2, 3], we proved the following two theorems:

Theorem 1.3

Let G be a group. Let $H_{i},$ $i=1,2,3,$ be supersolvable subgroup with $|G:H_{i}|=p_{i},$ $i=1,2,3,$ where $p_{1},$ $p_{2}$ and $p_{3}$ are distinct primes. Then G is supersolvable.

Theorem 1.4

Let G be a group. Let $H_{1}$ be a nilpotent subgroup of G. Let $H_{i},$ $i=2,3,$ be supersolvable subgroups of G. If the indices of $H_{i},$ $i=1,2,3,$ in G are pairwise relatively prime, then G is supersolvable.

In [12], Flowers and Wakefield proved the following theorem:

Theorem 1.5

Let G be a group and let $H_{i},$ $i=1,2,3,$ be supersolvable subgroups of G with pairwise relatively prime indices in G. Suppose that $G^{^{\prime }}$ is nilpotent. Then G is supersolvable.

In [6], Ballester-Bolinches and Ezquerro proved the following theorem:

Theorem 1.6

Let G be a group with three supersolvable subgroups $H_{i},$ $i=1,2,3,$ with pairwise relatively prime indices in G. Suppose that $H_{1}$ and $H_{j}$ are mutually permutable subgroups for every $j\ne 1.$ Then G is supersolvable.

In this paper, we prove the following theorems:

Theorem 1.7

Let $H_{1}$, $H_{2}$, $H_{3}$ be three supersolvable subgroups of a group G. Let $|G:H_{1}|=p_{1}^{e_{1}}p_{2}^{e_{2}}\ldots p_{s}^{es},$ $|G:H_{2}|=p_{s+1}^{e_{s+1}}p_{s+2}^{e_{s+2}}\ldots p_{t}^{e_{t}},$ $|G:H_{3}|=p_{t+1}p_{t+2}\ldots p_{m},$ where $p_{1},p_{2},\ldots ,p_{s},$ $p_{s+1},\ldots ,p_{t},p_{t+1},\ldots ,p_{m}$ are distinct primes such that $p_{1}<p_{2}<\cdots<p_{s}<p_{s+1}<\cdots<p_{t}<p_{t+1}<\cdots <p_{m}$. Then G is supersolvable.

Theorem 1.8

Let $H_{1}$, $H_{2}$, $H_{3}$ be three supersolvable subgroups of a group G. Let $|G:H_{1}|=p_{1}^{e_{1}}p_{2}^{e_{2}}\ldots p_{s}^{es},$ $|G:H_{2}|=p_{s+1}^{e_{s+1}}p_{s+2}^{e_{s+2}}\ldots p_{t}^{e_{t}},$ $|G:H_{3}|=p_{t+1}^{e_{t+1}}p_{t+2}^{e_{t+2}}\ldots p_{m}^{e_{m}},$ where $p_{1},p_{2},\ldots ,p_{s},$ $p_{s+1},\ldots ,p_{t},p_{t+1},\ldots ,p_{m}$ are distinct primes such that $p_{1}<p_{2}<\cdots<p_{s}<p_{s+1}<\cdots<p_{t}<p_{t+1}<\cdots <p_{m}$. Suppose that every subgroup of G of prime order p is s-semipermutable in G for every p dividing $|G:H_{3}|.$ Then G is supersolvable.

Theorem 1.9

Let $H_{1}$, $H_{2}$, $H_{3}$ be three supersolvable subgroups of a group G. Let $|G:H_{1}|=p_{1}^{e_{1}}p_{2}^{e_{2}}\ldots p_{s}^{es},$ $|G:H_{2}|=p_{s+1}^{e_{s+1}}p_{s+2}^{e_{s+2}}\ldots p_{t}^{e_{t}},$ $|G:H_{3}|=p_{t+1}^{e_{t+1}}p_{t+2}^{e_{t+2}}\ldots p_{m}^{e_{m}},$ where $p_{1},p_{2},\ldots ,p_{s},$ $p_{s+1},\ldots ,p_{t},p_{t+1},\ldots ,p_{m}$ are distinct primes such that $p_{1}<p_{2}<\cdots<p_{s}<p_{s+1}<\cdots<p_{t}<p_{t+1}<\cdots <p_{m}$. Suppose that $H_{3}$ and $H_{2}\cap H_{1}$ are mutually permutable subgroups of G. Then G is supersolvable.

Theorem 1.10

Let $\Im$ be a saturated formation containing $\mathcal {U}$. Let G be a group and let $H_{1},$ $H_{2}$ and $H_{3}$ be subgroups of G belong to $\Im$ with pairwise relatively prime indices in G. Suppose that the commutator subgroup $G^{^{\prime }}$ of G is nilpotent. Then $G\in \Im .$

2 Preliminaries

Lemma 2.1

[2, Theorem 11]. Let G be a group. Suppose that $H_{1},$ $H_{2}$ and $H_{3}$ are subgroups of G whose indices in G are pairwise relatively prime. If $H_{1},$ $H_{2}$ and $H_{3}$ have Sylow towers of supersolvable type, then G has a Sylow tower of supersolvable type.

Lemma 2.2

[18, II, Lemma 7.9]. Let P be a nilpotent normal subgroup of a group G. If $P\ne 1$ and $P\cap \Phi (G)=1,$ then P is the direct product of some minimal normal subgroups of G.

Lemma 2.3

[16, Lemma 2.1]. Suppose that H is an s−semipermutable subgroup of G. Then

• (a) If $H\le K\le G$, then H is s−semipermutable in K.

• (b) If $H\le O_{p}(G),$ then H is s−permutable in G.

Lemma 2.4

[17, Theorem 3.3]. Let G be a group. Suppose that P is a normal $p-$subgroup of G,  where p is an odd prime dividing the order of G. If every subgroup of P of order p is s$-$permutable in G,  then $P\le Z_{\mathcal {U}}{ (G).}$

Lemma 2.5

[1, Theorem A]. Let the group $G=HK$ be the product of the mutually permutable subgroups H and K. If H and K are supersolvable subgroups of G with Core$_{G}(H\cap K)=1,$ then G is supersolvable.

Lemma 2.6

[7, Theorem 5.2.21]. Let $G=HK$ be the product of the mutually permutable subgroups H and K. Let $\Im$ be a saturated formation containing $\mathcal {U}$ If H and K belong to $\Im$ and $G^{{\prime }}$ is nilpotent, then $G\in \Im .$

3 Main Results

Theorem 3.1

Let G be a group and let $H_{1},$ $H_{2}$ and $H_{3}$ be supersolvable subgroups of G. Suppose that $|G:H_{1}|=p_{_{1}}^{e_{1}},$ $|G:H_{2}|=p_{2}^{e_{2}},$ $|G:H_{3}|=p_{_{3}},$ where $p_{1},$ $p_{2},$ $p_{3}$ are distinct primes such that $p_{1}<p_{2}<p_{3}.$ Then G is supersolvable.

Proof

Suppose that the theorem is false and let G be a counterexample with minimal order. Let P be a Sylow $p-$subgroup of G,  where p is the largest prime dividing the order of G. Then:

• (1) $P\trianglelefteq G.$ By $H_{1},$ $H_{2}$, $H_{3}$ have Sylow towers of supersolvable type. Then by Lemma 2.1, G has a Sylow tower of supersolvable type. Hence by [14, VI, Satz 9.1(c)], $P\trianglelefteq G.$

• (2) If N is a nontrivial normal $p-$subgroup of G,  then G/N is supersolvable. If $G=H_{3}N,$ then G/N is supersolvable. Suppose now that $N\le H_{3}.$ It is clear that $H_{3}/N,$ $H_{2}/N,$ $H_{1}/N$ are proper supersolvable subgroups of G/N. Then by the minimal chioce of G,  G/N is supersolvable.

• (3) $p=p_{3}.$ Suppose that $p>p_{3}.$ By (1), $P\trianglelefteq G,$ where P is a Sylow $p-$subgroup of G and p is the largest prime dividing the order of G. It is clear that $P\le H_{i},$ $i=1,2,3.$ Then by (2), G/P is supersolvable. By Schur−Zassenhouse Theorem [13, Theorem 6.2.1], G has a subgroup K which is a complement to P in G. It is clear that $(|G:H_{i}|,$ $|G:K|)=1$ for any $i=1,2,3.$ Hence by Theorem 1.2, G is supersolvable, a contradiction.

• (4) The final contradiction. By (1), $P\trianglelefteq G.$ Put $A=P\cap \Phi (G).$ If $A\ne 1,$ then by (2), G/A is supersolvable. Hence by [20, Chapter 1, Corollary 3.2], G is supersolvable, a contradiction. Thus $A=1.$ Then by Lemma 2.2, $P=N_{1}\times N_{2}\times \cdots \times N_{n},$ where $N_{i}$ is a minimal normal subgroup of G,  $i=1,2,3,\ldots ,n.$ By (3), P is a Sylow $p_{3}-$ subgroup of G. Then $G=H_{3}N_{j}$ for some j,  where $1\le j\le n.$ Since $N_{j}$ is a minimal normal subgroup of G and $|G:H_{3}|=p_{3},$ we have $|N_{j}|=p_{3}.$ Hence G is supersolvable, a contradiction. $\square$

Proof of Theorem 1.7

Suppose that the theorem is false and let G be a counterexample with minimal order. By Wielandt’s Theorem [13, Theorem 6.4.4], G is solvable. Let $M_{i}$ be a maximal subgroup of G such that $H_{i}\le M_{i},$ $i=1,2,3.$ Since $|G:M_{i}|||G:H_{i}|,$ we have $|G:M_{1}|=p_{r}^{l_{r}},$ $|G:M_{2}|=p_{j}^{l_{j}},$ $|G:M_{3}|=p_{k},$ where $p_{r},$ $p_{j},$ $p_{k}$ are distinct primes such that $p_{r}<p_{j}<p_{k}.$ It is clear that $G=H_{2}M_{1}=H_{3}M_{1}$ and so $|G:H_{2}|=|M_{1}:M_{1}\cap H_{2}|$ and $|G:H_{3}|=|M_{1}:M_{1}\cap H_{3}|$. Then by the minimal choice of G,  $M_{1}$ is supersolvable. Similarly, $M_{2}$ and $M_{3}$ are supersolvable. Hence by Theorem 3.1, G is supersolvable, a contradiction. $\square$

As a corollary of the proof of Theorem 1.7, we have

Corollary 3.2

[4]. If a group G has three supersolvable subgroups $H_{1},H_{2},H_{3}$ whose indices in G are pairwise relatively prime and $|G:H_{i}|=$ square free order, where $i=1,2,3,$ then G is supersolvable.

The following result follows at once from Theorem 3.1

Corollary 3.3

[19, Theorem 3]. A group G is a supersolvable group whose order has at least three prime divisors if and only if there exist three maximal supersolvable subgroups of G whose indices are three distinct primes.

Theorem 3.4

Let G be a group and, for $i=1,2,3,$ let $H_{i}$ be a supersolvable subgroup of G. Suppose that $|G:H_{1}|=p_{1}^{e_{1}},$ $|G:H_{2}|=p_{2}^{e_{2}},$ $|G:H_{3}|=p_{3}^{e_{3}},$ where $p_{1},$ $p_{2},$ $p_{3}$ are three distinct primes such that $p_{1}<p_{2}<p_{3}.$ Suppose further that every subgroup of G of order $p_{3}$ is s−semipermutable in G. Then G is supersolvable.

Proof

Suppose that the theorem is false and let G be a counterexample with minimal order. Then:

• (1) $P\trianglelefteq G,$ where P is a Sylow $p-$subgroup of G and p is the largest prime dividing the order of G. By a similar argument as in step (1) in the proof of Theorem 3.1, we have $P\trianglelefteq G.$

• (2) $p>p_{3},$ where p is the largest prime dividing the order of G. Suppose that $p=p_{3}.$ By (1), $P\trianglelefteq G.$ Then by Lemma 2.3(b), every subgroup of P of order $p_{3}$ is s−permutable in G. Hence by Lemma 2.4, $P\le Z_{\mathcal {U}}(G).$ It is clear that $G=PH_{3},$ where P is a Sylow $p_{3}-$subgroup of G. Since $G/P\cong H_{3}/P\cap H_{3}$ is supersolvable and $P\le Z_{\mathcal {U}}(G),$ it follows easily that G is supersolvable, a contradiction.

• (3) The final contradiction. By Wielandt’s Theorem [13, Theorem 6.4.4], G is solvable. Then by Hall’s Theorem [13, Theorem 6.4.1], G has a Hall subgroup K such that $\pi (K)=\{p_{1},$ $p_{2},$ $p_{3}\}.$ By (2), $p>p_{3}>p_{2}>p_{1}$ and so $K<G.$ It is clear that $G=H_{i}K$ and so $|G:H_{i}|=|K:K\cap H_{i}|,$ $i=1,2,3.$ By Lemma 2.3(a), every subgroup of K of order $p_{3}$ is s− semipermutable in K. Then K satisfies the hypothesis of the theorem. The minimal choice of G yields that K is supersolvable. It is clear that ($|G:H_{i}|,|G:K|)=1,$ $i=1,2,3.$ Then by Theorem 1.2, G is supersolvable, a contradiction. $\square$

Proof of Theorem 1.8

Suppose that the theorem is false and let G be a counterexample with minimal order. Let $M_{i}$ be a maximal subgroup of G such that $H_{i}\le M_{i},$ $i=1,2,3.$ Then we have $|G:M_{1}|=p_{r}^{l_{r}},$ $|G:M_{2}|=p_{j}^{l_{j}},$ $|G:M_{3}|=p_{k}^{l_{k}},$ where $p_{r},$ $p_{j},$ $p_{k}$ are distinct primes such that $p_{r}<p_{j}<p_{k}$ (see the proof of Theorem 1.7). It is clear that $G=H_{2}M_{1}=H_{3}M_{1}$ and so $|G:H_{2}|=|M_{1}:M_{1}\cap H_{2}|$ and $|G:H_{3}|=|M_{1}:M_{1}\cap H_{3}|.$ By hypothesis, every subgroup of G of order p is s−semipermutable in G for every prime p dividing $|G:M_{3}|.$ By Lemma 2.3(a), every subgroup of $M_{1}$ of order p is s−semipermutable in $M_{1}$ for every prime p dividing $|M_{1}:M_{1}\cap H_{3}|=|G:H_{3}|.$ Then $M_{_{1}}$ satisfies the hypothesis of the theorem. Hence by the minimal choice of G,  $M_{1}$ is supersolvable. Similarly, $M_{2}$ and $M_{3}$ are supersolvable. Hence by Theorem 3.4, G is supersolvable, a contradiction. $\square$

Proof of Theorem 1.9

Suppose that the theorem is false and let G be a counterexample with minimal order. Let p be the largest prime dividing the order of G. Then:

• (1) $p=p_{m}.$ Suppose that $p>p_{m}.$ By Wielandt’s Theorem [13, Theorem 6.4.4], G is solvable. Then by Hall’s Theorem [13, Theorem 6.4.1], there exists a Hall subgroup K of G such that $\pi (K)=\{p_{1},$ $p_{2}$,..., $p_{s},$ $p_{s+1},$..., $p_{t},$ $p_{t+1},$..., $p_{m}\}.$ It is clear that $G=H_{3}(H_{2}\cap H_{1})$ and $H_{3}$ and $H_{2}\cap H_{1}$ are mutually permutable. Since $G=KH_{i},$ we have $|G:H_{i}|=|K:K\cap H_{i}|,$ $i=1,2,3.$ By [7, Lemma 4.1.21(1)], $(K\cap H_{3})(K\cap H_{2}\cap H_{1})$ is a subgroup and $(K\cap H_{3})$ and $(K\cap H_{2}\cap H_{1})$ are mutually permutable. Then K satisfies the hypothesis of the theorem. By the minimal choice of G,  K is supersolvable. It is clear that $(|G:H_{i}|,|G:K|)=1$ for any $i=1,2,3.$ Hence by Theorem 1.2, G is supersolvable, a contradiction.

• (2) $P\trianglelefteq G,$ where P is a Sylow $p_{m}-$subgroup of G. By (1), $p_{m}$ is the largest prime dividing the order of G. By a similar argument as in Step (1) in the proof of Theorem 3.1, we have $P\trianglelefteq G.$

• (3) If N is a nontrivial normal $p_{m}-$subgroup of G,  then G/N is supersolvable. If $G=H_{3}N,$ then $G/N\cong H_{3}/H_{3}\cap N$ is supersolvable. Assume now that $H_{3}N/N<G/N.$ It is clear that $N\le H_{2}$ and $N\le H_{1}.$ By [7, Lemma 4.1.10], $H_{3}N/N$ and $H_{2}\cap H_{1}/N$ are mutually permutable subgroups of G/N. Then G/N satisfies the hypothesis of the theorem. By the minimal choice of G,  G/N is supersolvable.

• (4) $P\cap \Phi (G)=1.$ Suppose that $P\cap \Phi (G)\ne 1.$ Then by (3),  $G/P\cap \Phi (G)$ is supersolvable. Hence by [20, Chapter 1, Corollary 3.2], G is supersolvable, a contradiction.

• (5) $Core_{G}(H_{3}\cap H_{2}\cap H_{1})\ne 1$ and $G/Core_{G}(H_{3}\cap H_{2}\cap H_{1})$ is supersolvable. Suppose that $Core_{G}(H_{3}\cap H_{2}\cap H_{1})=1.$ Then by Lemma 2.5, G is supersolvable, a contradiction. Thus $Core_{G}(H_{3}\cap H_{2}\cap H_{1})\ne 1.$ Put $A=Core_{G}(H_{3}\cap H_{2}\cap H_{1}).$ Now, it is easy to see that G/A satisfies the hypothesis of the theorem. Then by the minimal choice of G,  G/A is supersolvable.

• (6) P is a minimal normal subgroup of G.

  By (4), $P\cap \Phi (G)=1.$

  Then by Lemma 2.2, $P=N_{1}\times N_{2}\times \cdots \times N_{n},$ where $N_{i}$ is a minimal normal subgroup of G,  $i=1,2,\ldots ,n.$ By (3), $G/N_{i}$ is supersolvable, $i=1,2,3,\ldots ,n.$ Since the class of supersolvable groups is a saturated formation, we have P is a minimal normal subgroup of G.

• (7) The final contradiction. By (5), G/A is supersolvable, where $A=Cor_{G}(H_{3}\cap H_{2}\cap H_{1})\ne 1.$ By (2) and (3), G/P is supersolvable. Since G is not supersolvable, we have $P\cap A\ne 1.$ By (6), P is a minimal normal subgroup of G and so $P\le A\le H_{3},$ a contradiction. $\square$

Proof of Theorem 1.10

Suppose that the theorem is false and let G be a counterexample with minimal order. Since $G^{^{\prime }}$ is nilpotent, we have G is solvable. Clearly, the order of G is divided by at least three distinct primes. Let L be a minimal normal subgroup of G. Since G is solvable, it follows that $|L|=p^{n},$ for some prime p. If $G=H_{i}L$ for some $i=1,2,3,$ then $G/L\in \Im .$ Assume now that $H_{i}L<G,$ $i=1,2,3.$ Then $1<H_{i}L/L<G/L,$ $i=1,2,3.$ Clearly, $H_{i}L/L\cong H_{i}/H_{i}\cap L\in \Im$ and $(G/L)^{^{\prime }}=G^{^{\prime }}L/L\cong G^{^{\prime }}/G^{^{\prime }}\cap L$ is nilpotent. By the minimal choice of G,  $G/L\in \Im .$ Since $\Im$ is a saturated formation, L is a unique minimal normal subgroup of G and $\Phi (G)=1.$ By Lemma 2.2, $F(G)=L$ and so $G^{^{\prime }}=F(G)=L.$ Since $|L|=p^{n}$ and G possesses three subgroups $H_{i},$ $i=1,2,3,$ whose indices in G are pairwise relatively primes, we can assume that $p\not \mid |G:H_{i}|$, $i=1,2.$ Now it follows easily that $G^{^{\prime }}=F(G)\le H_{i},$ $i=1,2.$ Then $H_{i}\trianglelefteq G,$ $i=1,2.$ Hence $G=H_{1}H_{2}$ is a mutually permutable product of $H_{1}$ and $H_{2}.$ Applying Lemma 2.6, $G\in \Im ,$ a contradiction. $\square$

As a corollary of Theorem 1.10, we have

Theorem 1.5 ([12, Theorem 1.1]). Let G be a group and let $H_{i},$ $i=1,2,3,$ be supersolvable subgroups of G with pairwise relatively prime indices in G. Suppose that $G^{^{\prime }}$ is nilpotent. Then G is supersolvable.