On the supersolubility of a finite group with NS-supplemented Sylow subgroups

TL;DR

This paper investigates the conditions under which a finite group is supersoluble, focusing on the role of NS-supplemented Sylow subgroups and maximal subgroups, and establishes new criteria for supersolubility.

Contribution

It introduces the concept of NS-supplemented Sylow subgroups and proves their role in ensuring a group's supersolubility, extending previous results in group theory.

Findings

Groups with NS-supplemented non-cyclic Sylow subgroups are supersoluble.

Solubility is established for groups with NS-supplemented maximal subgroups.

Abstract

A subgroup $A$ of a group~$G$ is said to be {\sl NS-supplemented} in $G$, if there exists a subgroup~$B$ of $G$ such that $G=AB$ and whenever $X$~is a normal subgroup of~$A$ and $p\in \pi(B)$, there exists a Sylow $p$-subgroup~$B_p$ of~$B$ such that $XB_p=B_pX$. In this paper, we proved the supersolubility of a group with NS-supplemented non-cyclic Sylow subgroups. The solubility of a group with NS-supplemented maximal subgroups is obtained.

Equations

$$((A\cap X)K)B_p=B_p((A\cap X)K)$$

$$G=⟨a,b,c\mid a^3=b^3=c^8,ab=ba,a^c=b,b^c=ab⟩$$

$$M_1=[(⟨a⟩\times ⟨b⟩)]⟨c^2⟩,M_i=⟨c^x⟩,x\in ⟨a⟩\times ⟨b⟩.$$

Full text

1 Introduction

Throughout this paper, all groups are finite and $G$ always denotes a finite group. We use the standard notations and terminology of [1]. The set of all prime divisors of the order of $G$ is denoted by $\pi(G)$. The notation $Y\leq X$ means that $Y$ is a subgroup of a group $X$. The semidirect product of a normal subgroup $A$ and a subgroup $B$ is denoted by $[A]B$.

By the Zassenhaus Theorem ([1, IV.2.11]), a group $G$ with cyclic Sylow subgroups has a cyclic Hall subgroup $H$ such that the quotient $G/H$ is also cyclic. In particular, $G$ is supersoluble.

A group $G$ with abelian Sylow subgroups may be non-soluble (for example, $PSL(2,5)$) and the compositional factors of $G$ are known [2].

In some papers, the sufficient conditions of solubility and supersolubility of a group in which Sylow subgroups permute with some subgroups are established. For example, the supersolubility of a group $G$ such that every Sylow subgroup $P$ of $G$ permutes with subgroups of some supplement of $P$ in $G$ is obtained in works [3]–[4].

The following concept is introduced in [5].

Definition 1.1**.**

Two subgroups $A$ and $B$ of a group $G$ are said to be NS-permutable, if they satisfy the following conditions:

$(1)$ whenever $X$ is a normal subgroup of $A$ and $p\in\pi(B)$, there exists a Sylow $p$-subgroup $B_{p}$ of $B$ such that $XB_{p}=B_{p}X$;

$(2)$ whenever $Y$ is a normal subgroup of $B$ and $p\in\pi(A)$, there exists a Sylow $p$-subgroup $A_{p}$ of $A$ such that $YA_{p}=A_{p}Y$.

Moreover, if $G=AB$, we say that $G$ is an NS-permutable product of the subgroups $A$ and $B$.

The totally permutable [6] and totally c-permutable [7] subgroups are NS-permutable [5, Lemma 2]. The supersolubility of a group $G=AB$ which is the NS-permutable product of supersoluble subgroups $A$ and $B$ is obtained in [5].

We introduce the following

Definition 1.2**.**

A subgroup $A$ of a group $G$ is said to be NS-supplemented in $G$, if there exists a subgroup $B$ of $G$ such that:

$(1)$ $G=AB$;

$(2)$ whenever $X$ is a normal subgroup of $A$ and $p\in\pi(B)$, there exists a Sylow $p$-subgroup $B_{p}$ of $B$ such that $XB_{p}=B_{p}X$.

In this case we say that $B$ is a NS-supplement of $A$ in $G$.

In this paper, we proved the supersolubility of a group in which every non-cyclic Sylow subgroup is NS-supplemented. The solubility of a group with NS-supplemented maximal subgroups is obtained.

2 Preliminaries

Definition 1.2 implies the following result for $X=A$.

Lemma 2.1**.**

Let $A$ be an NS-supplemented subgroup of $G$ and $B$ is its NS-supplement in $G$. Then for every $p\in\pi(B)$ there exists a Sylow $p$-subgroup $B_{p}$ of $B$ such that $AB_{p}=B_{p}A$.

Lemma 2.2**.**

Let $K$ be a normal subgroup of $G$. If $A$ is NS-supplemented in $G$ and $B$ is its NS-supplement in $G$, then $AK/K$ is NS-supplemented in $G/K$ and $BK/K$  is its NS-supplement in $G/K$.

Proof.

It’s obvious that $G/N=(AK/K)(BK/K)$. Let $X/K$ be a normal subgroup of $AK/K$ and $p\in\pi(BK/K)$. Then $X=(A\cap X)K$ and $A\cap X$ is normal in $A$. By the hypothesis, for every $p\in\pi(B)$ there exists a Sylow $p$-subgroup $B_{p}$ of $B$ such that $(A\cap X)B_{p}=B_{p}(A\cap X)$. Hence

[TABLE]

and $X/K$ permutes with Sylow $p$-subgroup $B_{p}K/K=(BK/K)_{p}$ of $BK/K$. ∎

Lemma 2.3**.**

([8, Theorem 2])* Let $G$ be a group with $p\in\pi(G)$ and $p\neq 3$. If $G$ has a Hall $\{p,r\}$-subgroup for every $r\in\pi(G)$, then $G$ is $p$-soluble.*

Lemma 2.4**.**

([9, Corollary 3])* Let $G$ be a group such that every maximal subgroup has prime power index. Then $G=S(G)$ or $G/S(G)\simeq PSL(2,7)$.*

Here $S(G)$ is the maximal normal soluble subgroup of $G$.

3 Groups with NS-supplemented subgroups

Theorem 3.1**.**

If a Sylow $p$-subgroup $P$ of $G$ is NS-supplemented in $G$, then $G$ is $p$-supersoluble in each of the following cases:

$(1)$* $p\neq 3$;*

$(2)$* $p=3$ and $G$ is 3-soluble.*

Proof.

Let $B$ be an NS-supplement of $P$ in $G$. By Lemma 2.1, for any $q\in\pi(B)\setminus\{p\}$ there exists a Sylow $q$-subgroup $Q$ of $B$ such that $PQ=QP$. The subgroup $PQ$ is a Hall $\{p,q\}$-subgroup of $G$. Since $q$ is an arbitrary prime of $\pi(G)\setminus\{p\}$, it follows that by Lemma 2.3, $G$ is $p$-soluble for $p\neq 3$ and by the hypothesis, $G$ is 3-soluble for $p=3$.

We use induction on the order of $G$. Let $N$ be an arbitrary non-trivial normal subgroup in $G$. Then by Lemma 2.2, a Sylow $p$-subgroup $PN/N$ is NS-supplemented in $G/N$. By induction, $G/N$ is $p$-supersoluble, $O_{p^{\prime}}(G)=1$ and $N=O_{p}(G)\neq 1$.

We choose a subgroup $X$ of $G$ such that $X\leq N\cap Z(P)$ and $|X|=p$. Since $X$ is normal in $P$, it follows that for every $r\in\pi(B)$ there exists a Sylow $r$-subgroup $R$ of $B$ such that $XR=RX$. If $p\neq r$, then the subgroup $N\cap XR=X(N\cap R)=X$ is normal in $XR$. This is true for any prime $r$, hence $X$ is normal in $G$. Since the quotient $G/X$ is $p$-supersoluble by induction, $G$ is $p$-supersoluble. ∎

Corollary 3.2**.**

If all non-cyclic Sylow subgroups of $G$ are NS-supplemented in $G$, then $G$ is supersoluble.

Proof.

Let $p$ be the smallest prime of $\pi(G)$ and $P$ be a Sylow $p$-subgroup of $G$. If $P$ is cyclic, then $G$ is $p$-nilpotent [1, IV.2.8]. If $P$ is non-cyclic, then $P$ is NS-supplemented in $G$ and by Theorem 3.1, $G$ is $p$-nilpotent. In particular, $G$ is soluble and we apply Theorem 3.1 for each $r\in\pi(G)$. Let $R$ be a Sylow $r$-subgroup of $G$. If $R$ is cyclic, then $R$ is $r$-supersoluble. If $R$ is non-cyclic, then $R$ is NS-supplemented in $G$ and $G$ is $r$-supersoluble by Theorem 3.1. Thus, $G$ is $r$-supersoluble for any $r\in\pi(G)$. Consequently, $G$ is supersoluble. ∎

Example*.*

The group $PSL(2,7)$ is an NS-supplement of its Sylow 3-subgroup. Hence we can not omit the condition $\ll$group is 3-soluble$\gg$ in Theorem 3.1.

Theorem 3.3**.**

If all maximal subgroups of $G$ are NS-supplemented in $G$, then $G$ is soluble.

Proof.

We use induction on the order of $G$. By Lemma 2.2, all non-trivial quotients are soluble, hence $S(G)=1$. Let $M$ be a maximal subgroup of $G$ and $B$ is its NS-supplement in $G$. By the hypothesis, for every $p\in\pi(B)$ there exists a Sylow $p$-subgroup $B_{p}$ of $B$ such that $B_{p}M=MB_{p}$. Since $M\neq G$, there exists $r\in\pi(B)$ and a Sylow $r$-subgroup $B_{r}$ such that $MB_{r}=G$. Hence $|G:M|=r^{b}$. Consequently, every maximal subgroup of $G$ has prime power index. By Lemma 2.4, $G$ is either soluble, or $G\simeq PSL(2,7)$. The group $PSL(2,7)$ has a maximal subgroup $H\simeq[Z_{7}]Z_{3}$ and it has not a subgroup of the order $7\cdot 2^{3}$. Hence $H$ is not NS-supplemented in $PSL(2,7)$. Consequently, $G$ is soluble.

∎

The following example shows that the group that satisfies the hypotheses of Theorem 3.3 can be non-supersoluble.

Example*.*

In the group ([10], IdGroup=[72,39])

[TABLE]

all maximal subgroups are the following subgroups:

[TABLE]

Moreover, all maximal subgroups are NS-supplemented in $G$ and the subgroup $M_{1}$ is non-supersoluble.