A note on the supersolvability of a finite group with prime index of some subgroups

TL;DR

This paper establishes a characterization of supersoluble finite groups by showing that such a group is supersoluble if and only if it contains a supersoluble subgroup of prime index for every prime dividing its order.

Contribution

It provides a new necessary and sufficient condition for supersolvability based on the existence of specific prime index subgroups.

Findings

A finite group is supersoluble iff it has a supersoluble subgroup of prime index for each prime dividing its order.

The result offers a practical criterion for verifying supersolvability in finite groups.

The characterization simplifies the analysis of group structure through subgroup properties.

Abstract

In this paper, we proved that a group $G$ is supersoluble if and only if for any prime $p\in \pi (G)$ there exists a supersoluble subgroup of index $p$.

Equations

$$\Phi (P)\le \Phi (G)\le \bigcap _{\begin{array}{c}i=1,2,\dots ,n\end{array}}{H}_i$$

$$X^{h^{-1}}(Y^{h^{-1}}{)}^u=X^{h^{-1}}{Y}^{v{h}^{-1}}=(X{Y}^v{)}^{h^{-1}}.$$

Full text

All groups considered in this paper will be finite. Denote by $\pi(G)$ the set of all prime divisors of order of $G$. The notation $H\leq G$ means that $H$ is a subgroup of $G$.

By Huppert [1, Theorem 6], a group $G$ is supersoluble if and only if every maximal subgroup of $G$ has prime index. The following observation is generated by Huppert’s result.

Theorem 1. Let $G$ be a group. Then $G$ is supersoluble if and only if for any prime $p\in\pi(G)$ there exists a supersoluble subgroup of index $p$.

Proof. If $G$ is supersoluble, then for every $p\in\pi(G)$ maximal subgroup of $G$ that contains a Hall $p^{\prime}$-subgroup $G_{p^{\prime}}$ is supersoluble and has index $p$.

Conversely, let $\pi(G)=\{p_{1},p_{2},\ldots,p_{n}\}$, $p_{1}>p_{2}>\ldots>p_{n}$ and $H_{i}$ be a supersoluble subgroup such that $|G:H_{i}|=p_{i}$, where $i=1,2,\ldots,n$. It is clear that $H_{n}$ is normal in $G$. Let $P$ be a Sylow $p_{1}$-subgroup of $H_{n}$. Then $P$ is normal in $G$. We use induction on the order of $G$. Suppose that $\Phi(P)\neq 1$. Then

[TABLE]

and $G/\Phi(G)$ is supersoluble by induction. So $G$ is supersoluble. Therefore, we consider that $\Phi(P)=1$ and $P$ is elementary abelian. By Mashke’s theorem [2, Theorem I.17.6], $P=N_{1}\times N_{2}\times\ldots\times N_{m}$, where $N_{j}$ is a minimal normal subgroup of $G$, where $i=1,2,\ldots,m$. Since $|G:H_{1}|=p_{1}$, it follows that $P$ is not contained in $H_{1}$ and there is a subgroup $N_{j}$ for some $j$ such that it is not contained in $H_{1}$. Hence $G=N_{j}H_{1}$, $N_{j}\cap H_{1}=1$ and $|N_{j}|=p_{1}$. Now $G$ is supersoluble by [3, Lemma 4.46]. The theorem is proved.

Corollary 1. ([4, Theorem 3.1]) Let $H$ and $K$ be supersoluble subgroups of $G$ and $G=HK$. If each subgroup of $H$ permutes with every subgroup of $K$, then $G$ is supersoluble.

Proof. We use induction on the order of $G$. We can assume that $H_{1}K\neq G\neq HK_{1}$ for all proper subgroups $H_{1}$ of $H$ and $K_{1}$ of $K$. Let $p\in\pi(H)$, $A\leq H$ and $|H:A|=p$. By induction, $AK$ is supersoluble and $|G:AK|=p$. Similarly, if $q\in\pi(K)$, $B\leq K$ and $|K:B|=q$, then $HB$ is supersoluble and $|G:HB|=q$. Since $\pi(H)\cup\pi(K)=\pi(G)$, it follows that $G$ is supersoluble by Theorem 1.

Asaad and Shaalan’s theorem [4, Theorem 3.1] was developed by Guo W., Shum  K. P., Skiba A. N., see [5, Theorem A]. This result can also be obtained from Theorem 1. We introduce the following

Definition. The subgroups $H$ and $K$ are said to be $tcc$-permutable if for any $X\leq H$ and $Y\leq K$ there exists an element $u\in\langle X,Y\rangle$ such that $XY^{u}=Y^{u}X$. Note that the equality $XY^{u}=Y^{u}X$ is equivalent to $XY^{u}$ is a subgroup.

Lemma 1. Let $H$ and $K$  be subgroups of $G$. If $H$ and $K$ are $tcc$-permutable, then the subgroups $H$ and $K^{h}$ are $tcc$-permutable for any $h\in H$.

Proof. Let $X\leq H$ and $Y\leq K^{h}$. Then $X^{h^{-1}}\leq H$, $Y^{h^{-1}}\leq K$ and there exists an element $u\in\langle X^{h^{-1}},Y^{h^{-1}}\rangle$ such that $X^{h^{-1}}(Y^{h^{-1}})^{u}$ is a subgroup. Since $\langle X^{h^{-1}},Y^{h^{-1}}\rangle=\langle X,Y\rangle^{h^{-1}}$, it follows that there exists an element $v\in\langle X,Y\rangle$ such that $u=v^{h^{-1}}$. Because $h^{-1}u=vh^{-1}$, we have

[TABLE]

Hence $XY^{v}$ is a subgroup. Consequently, $H$ and $K^{h}$ are $tcc$-permutable.

Corollary 2. ([5, Theorem A]) Let $H$ and $K$ be supersoluble subgroups of $G$ and $G=HK$. If for any $H_{1}\leq H$ and $K_{1}\leq K$ there exists an element $x\in\langle H_{1},K_{1}\rangle$ such that $H_{1}K_{1}^{x}=K_{1}^{x}H_{1}$, then $G$ is supersoluble.

Proof. It is obvious that $H$ and $K$ are $tcc$-permutable. We use induction on the order of $G$. We can assume that $H_{1}K\neq G\neq HK_{1}$ for any proper subgroups $H_{1}$ of $H$ and $K_{1}$ of $K$.

Let $p\in\pi(H)$, $H_{1}\leq H$ and $|H:H_{1}|=p$. By hypothesis, there exists an element $g\in\langle H_{1},K\rangle$ such that $H_{1}K^{g}=K^{g}H_{1}$. Since $g=hk$ for some $k\in K$ and $h\in H$, we have $H_{1}K^{g}=H_{1}K^{h}=K^{h}H_{1}$ and $h\in\langle H_{1},K\rangle$. Because $G=HK^{h}$, it follows that $|G:H_{1}K^{h}|=p$. By Lemma 1, $H$ and $K^{h}$ are $tcc$-permutable. Hence the subgroups $H_{1}$ and $K^{g}$ satisfy the hypothesis of the corollary. By induction, $H_{1}K^{g}$ is supersoluble.

Similarly, if $q\in\pi(K)$, $K_{1}\leq K$ and $|K:K_{1}|=q$, then there exists an element $k\in K$ such that $H^{k}K_{1}$ is supersoluble in $G$ and has index $q$. Now $G$ is supersoluble by Theorem 1.