On the σ-Length of Maximal Subgroups of Finite σ-Soluble Groups

Abstract

Let $\sigma =\{{\sigma }_i:i\in I\}$ be a partition of the set $\mathbb{P}$ of all prime numbers and let G be a finite group. We say that G is $\sigma$-primary if all the prime factors of $\left|G\right|$ belong to the same member of $\sigma$. G is said to be $\sigma$-soluble if every chief factor of G is $\sigma$-primary, and G is $\sigma$-nilpotent if it is a direct product of $\sigma$-primary groups. It is known that G has a largest normal $\sigma$-nilpotent subgroup which is denoted by $F_{\sigma }\left(G\right)$. Let n be a non-negative integer. The n-term of the $\sigma$-Fitting series of G is defined inductively by $F_0\left(G\right)=1$, and $F_{n+1}\left(G\right)/F_n\left(G\right)=F_{\sigma }(G/F_n\left(G\right))$. If G is $\sigma$-soluble, there exists a smallest n such that $F_n\left(G\right)=G$. This number n is called the $\sigma$-nilpotent length of G and it is denoted by $l_{\sigma }\left(G\right)$. If $\mathfrak{F}$ is a subgroup-closed saturated formation, we define the $\sigma$-$\mathfrak{F}$-length $n_{\sigma }(G,\mathfrak{F})$ of G as the $\sigma$-nilpotent length of the $\mathfrak{F}$-residual $G^{\mathfrak{F}}$ of G. The main result of the paper shows that if A is a maximal subgroup of G and G is a $\sigma$-soluble, then $n_{\sigma }(A,\mathfrak{F})=n_{\sigma }(G,\mathfrak{F})-i$ for some $i\in \{0,1,2\}$.

1. Introduction

All groups considered in this paper are finite.

Skiba [1] (see also [2]) generalised the concepts of solubility and nilpotency by introducing $\sigma$-solubility and $\sigma$-nilpotency, in which $\sigma$ is a partition of $\mathbb{P}$, the set of all primes. Hence $\mathbb{P}={\bigcup }_{i\in I}{\sigma }_i$, with ${\sigma }_i\cap {\sigma }_j=\varnothing$ for all $i\ne j$.

In the sequel, $\sigma$ will be a partition of the set of all primes $\mathbb{P}$.

A group G is called $\sigma$-primary if all the prime factors of $\left|G\right|$ belong to the same member of $\sigma$.

Definition 1.

A group G is said to be σ-soluble if every chief factor of G is σ-primary. G is said to be σ-nilpotent if it is a direct product of σ-primary groups.

We note in the special case that $\sigma$ is the partition of $\mathbb{P}$ containing exactly one prime each, the class of $\sigma$-soluble groups is just the class of all soluble groups and the class of $\sigma$-nilpotent groups is just the class of all nilpotent groups.

Many normal and arithmetical properties of soluble groups and nilpotent groups still hold for $\sigma$-soluble and $\sigma$-nilpotent groups (see [2]) and, in fact, the class ${\mathcal{N}}_{\sigma }$ of all $\sigma$-nilpotent groups behaves in $\sigma$-soluble groups as nilpotent groups in soluble groups. In addition, every $\sigma$-soluble group has a conjugacy class of Hall ${\sigma }_i$-subgroups and a conjugacy class of Hall ${\sigma }_i^{\prime }$-subgroups, for every ${\sigma }_i\in \sigma$.

Recall that a class of groups $\mathfrak{F}$ is said to be a formation if $\mathfrak{F}$ is closed under taking epimorphic images and every group G has a smallest normal subgroup with quotient in $\mathfrak{F}$. This subgroup is called the $\mathfrak{F}$-residual of G and it is denoted by $G^{\mathfrak{F}}$. A formation $\mathfrak{F}$ is called subgroup-closed if $X^{\mathfrak{F}}$ is contained in $G^{\mathfrak{F}}$ for all subgroups X of every group G; $\mathfrak{F}$ is saturated if it is closed under taking Frattini extensions.

A class of groups $\mathfrak{F}$ is said to be a Fitting class if $\mathfrak{F}$ is closed under taking normal subgroups and every group G has a largest normal subgroup in $\mathfrak{F}$. This subgroup is called the $\mathfrak{F}$-radical of G.

The following theorem which was proved in [1] (Corollary 2.4 and Lemma 2.5) turns out to be crucial in our study.

Theorem 1.

${\mathcal{N}}_{\sigma }$ is a subgroup-closed saturated Fitting formation.

The ${\mathcal{N}}_{\sigma }$-radical of a group G is called the $\sigma$-Fitting subgroup of G and it is denoted by $F_{\sigma }\left(G\right)$. Clearly, $F_{\sigma }\left(G\right)$ is the product of all normal $\sigma$-nilpotent subgroups of G. If $\sigma$ is the partition of $\mathbb{P}$ containing exactly one prime each, then $F_{\sigma }\left(G\right)$ is just the Fitting subgroup of G.

If G is $\sigma$-soluble, then every minimal normal subgroup N of G is $\sigma$-primary so that N is $\sigma$-nilpotent and it is contained in $F_{\sigma }\left(G\right)$. In particular, $F_{\sigma }\left(G\right)\ne 1$ if $G\ne 1$.

Let n be a non-negative integer. The n-term of the $\sigma$-Fitting series of G is defined inductively by $F_0\left(G\right)=1$, and $F_{n+1}\left(G\right)/F_n\left(G\right)=F_{\sigma }(G/F_n\left(G\right))$. If G is $\sigma$-soluble, there exists a smallest n such that $F_n\left(G\right)=G$. This number n is called the $\sigma$-nilpotent length of G and it is denoted by $l_{\sigma }\left(G\right)$ (see [3,4]). The nilpotent length $l\left(G\right)$ of a group G is just the $\sigma$-nilpotent length of G for $\sigma$ the partition of $\mathbb{P}$ containing exactly one prime each.

The $\sigma$-nilpotent length is quite useful in the structural study of $\sigma$-soluble groups (see [3,4]), and allows us to extend some known results.

The central concept of this paper is the following:

Definition 2.

Let $\mathfrak{F}$ be a saturated formation. The σ-$\mathfrak{F}$-length $n_{\sigma }(G,\mathfrak{F})$ of a group G is defined as the σ-nilpotent length of the $\mathfrak{F}$-residual $G^{\mathfrak{F}}$ of G.

Applying [5] (Chapter IV, Theorem (3.13) and Proposition (3.14)) (see also [3] (Lemma 4.1)), we have the following useful result.

Proposition 1.

The class of all σ-soluble groups of σ-length at most l is a subgroup-closed saturated formation.

It is clear that the $\mathfrak{F}$-length $n_{\mathfrak{F}}\left(G\right)$ of a group G studied in [6] is just the $\sigma$-$\mathfrak{F}$-length of G for $\sigma$ the partition of $\mathbb{P}$ containing exactly one prime each, and the $\sigma$-nilpotent length of G is just the $\sigma$-$\mathfrak{F}$-length of G for $\mathfrak{F}=\left\{1\right\}$.

Ballester-Bolinches and Pérez-Ramos [6] (Theorem 1), extending a result by Doerk [7] (Satz 1), proved the following theorem:

Theorem 2.

Let $\mathfrak{F}$ be a subgroup-closed saturated formation and M be a maximal subgroup of a soluble group G. Then $n_{\mathfrak{F}}\left(M\right)=n_{\mathfrak{F}}\left(G\right)-i$ for some $i\in \{0,1,2\}$.

Our main result shows that Ballester-Bolinches and Pérez-Ramos’ theorem still holds for the $\sigma$-$\mathfrak{F}$-length of maximal subgroups of $\sigma$-soluble groups.

Theorem A.

Let $\mathfrak{F}$ be a saturated formation. If A is a maximal subgroup of a σ-soluble group G, then $n_{\sigma }(A,\mathfrak{F})=n_{\sigma }(G,\mathfrak{F})-i$ for some $i\in \{0,1,2\}$.

2. Proof of Theorem A

Proof. 

Suppose that the result is false. Let G be a counterexample of the smallest possible order. Then G has a maximal subgroup A such that $n_{\sigma }(A,\mathfrak{F})\ne n_{\sigma }(G,\mathfrak{F}))-i$ for every $i\in \{0,1,2\}$. Since $A^{\mathfrak{F}}$ is contained in $G^{\mathfrak{F}}$ because $\mathfrak{F}$ is subgroup-closed, we have that $G^{\mathfrak{F}}\ne 1$. Moreover, $n_{\sigma }(A,\mathfrak{F})\le n_{\sigma }(G,\mathfrak{F})=n$ and $n\ge 1$. We proceed in several steps, the first of which depends heavily on the fact that the $\mathfrak{F}$-residual is epimorphism-invariant.

Step 1.If N is a normal σ-nilpotent subgroup of G, then N is contained in A, $n_{\sigma }(A,\mathfrak{F})=n_{\sigma }(A/N,\mathfrak{F})$ and $n_{\sigma }(G/N,\mathfrak{F})=n-1$.

Let N be a normal $\sigma$-nilpotent subgroup of G. Applying [7] (Chapter II, Lemma (2.4)), we have that $G^{\mathfrak{F}}N/N={(G/N)}^{\mathfrak{F}}$. Consequently, either $n_{\sigma }(G/N,\mathfrak{F})=n$ or $n_{\sigma }(G/N,\mathfrak{F})=n-1$.

Assume that N is not contained in A. Then $G=AN$ and so $G/N\cong A/A\cap N$. Observe that either $n_{\sigma }(A/A\cap N,\mathfrak{F})=n_{\sigma }(G/N,\mathfrak{F})=n$ or $n_{\sigma }(A/A\cap N,\mathfrak{F})=n_{\sigma }(G/N,\mathfrak{F})=n-1$. Therefore $n-1\le n_{\sigma }(A,\mathfrak{F})\le n$. Consequently, either $n_{\sigma }(A,\mathfrak{F})=n$ or $n_{\sigma }(A,\mathfrak{F})=n-1$, contrary to assumption.

Therefore, N is contained in A. The minimal choice of G implies that $n_{\sigma }(A/N,\mathfrak{F})=n_{\sigma }(G/N,\mathfrak{F})-i$ for some $i\in \{0,1,2\}$, and so either $n_{\sigma }(A/N,\mathfrak{F})=n-i$ or $n_{\sigma }(A/N,\mathfrak{F})=n-i-1$. Suppose that $n_{\sigma }(A,\mathfrak{F})\ne n_{\sigma }(A/N,\mathfrak{F})$. Then $n_{\sigma }(A,\mathfrak{F})=n_{\sigma }(A/N,\mathfrak{F})+1$. Hence either $n_{\sigma }(A,\mathfrak{F})=n-i+1$ or $n_{\sigma }(A,\mathfrak{F})=n-i$. In the first case, $i>0$ because $n\ge n_{\sigma }(A,\mathfrak{F})$. Hence $n_{\sigma }(A,\mathfrak{F})=n-j$ for some $j\in \{0,1,2\}$, which contradicts our supposition. Consequently, $n_{\sigma }(A,\mathfrak{F})=n_{\sigma }(A/N,\mathfrak{F})$.

Suppose that $n_{\sigma }(G/N,\mathfrak{F})=n$. The minimality of G yields $n_{\sigma }(A/N,\mathfrak{F})=n-i$ for some $i\in \{0,1,2\}$. Therefore $n_{\sigma }(A,\mathfrak{F})=n_{\sigma }(G,\mathfrak{F})-i$ for some $i\in \{0,1,2\}$. This is a contradiction since we are assuming that G is a counterexample. Consequently, $n_{\sigma }(G/N,\mathfrak{F})=n-1$.

Step 2.$Soc\left(G\right)$ is a minimal normal subgroup of G which is not contained in $\mathsf{\Phi }\left(G\right)$, the Frattini subgroup of G.

Assume that N and L are two distinct minimal normal subgroups of G. Then, by Step 1, $n_{\sigma }(G/L,\mathfrak{F})=n-1$. Since the class of all $\sigma$-soluble groups of $\sigma$-$\mathfrak{F}$-length at most $n-1$ is a saturated formation by Proposition 1 and $N\cap L=1$, it follows that $n_{\sigma }(G,\mathfrak{F})=n-1$. This contradiction proves that $N=Soc\left(G\right)$ is the unique minimal normal subgroup of G.

Assume that N is contained in $\mathsf{\Phi }\left(G\right)$. Since $n_{\sigma }(G/N,\mathfrak{F})=n-1$ and the class of all $\sigma$-soluble groups of $\sigma$-$\mathfrak{F}$-length at most $n-1$ is a saturated formation by Proposition 1, we have that $n_{\sigma }(G,\mathfrak{F})=n-1$, a contradiction. Therefore N is not contained in $\mathsf{\Phi }\left(G\right)$ as desired.

According to Step 2, we have that $N=Soc\left(G\right)$ is a minimal normal subgroup of G which is not contained in $\mathsf{\Phi }\left(G\right)$. Hence G has a core-free maximal subgroup, M say. Then $G=NM$ and, by [5] (Chapter A, (15.2)), either N is abelian and $C_G\left(N\right)=N$ or N is non-abelian and $C_G\left(N\right)=1$. Since G is $\sigma$-soluble, it follows that N is $\sigma$-primary. Thus, N is a ${\sigma }_i$-group for some ${\sigma }_i\in \sigma$.

Step 3.Let H be a subgroup of G such that $N\subseteq H$. Then $F_{\sigma }\left(H\right)=O_{{\sigma }_i}\left(H\right)$.

Since N is contained in $F_{\sigma }\left(H\right)$, it follows that every Hall ${\sigma }_i^{\prime }$-subgroup of $F_{\sigma }\left(H\right)$ centralises N. Since $C_H\left(N\right)=N$ or $C_H\left(N\right)=1$, we conclude that $F_{\sigma }\left(H\right)$ is a ${\sigma }_i$-group, i.e., $F_{\sigma }\left(H\right)=O_{{\sigma }_i}\left(H\right)$.

Step 4.We have a contradiction.

Let $X=F_{\sigma }\left(G\right)$, and $T/X=F_{\sigma }(G/X)$. Suppose that T is not contained in A. Then $G=AT$, $G/T\cong A/A\cap T$, and $n_{\sigma }(G/T,\mathfrak{F})=n_{\sigma }(A/A\cap T,\mathfrak{F})$. By Step 1, $n_{\sigma }(G/X,\mathfrak{F})=n-1$. Hence $n_{\sigma }(G/T,\mathfrak{F})\in \{n-2,n-1\}$. Now, $X\subseteq A$ and $n_{\sigma }(A,\mathfrak{F})=n_{\sigma }(A/X,\mathfrak{F})$ by Step 1. Consequently, $n_{\sigma }(A/A\cap T,\mathfrak{F})\in \{n_{\sigma }(A,\mathfrak{F})-1,n_{\sigma }(A,\mathfrak{F})\}$. This means that $n_{\sigma }(A,\mathfrak{F})=n-j$ for some $j\in \{0,1,2\}$. This contradiction yields $T\subseteq A$.

By Step 3, we have that $X=O_{{\sigma }_i}\left(G\right)$. Assume that $E/X$ and $F/X$ are the Hall ${\sigma }_i$-subgroup and the Hall ${\sigma }_i^{\prime }$-subgroup of $T/X$ respectively. Then $T/X=E/X\times F/X$ and E and F are normal subgroups of G. Since X and $E/X$ are ${\sigma }_i$-groups, it follows that E is a ${\sigma }_i$-group and hence $E\subseteq X$. In particular, $T/X$ is a ${\sigma }_i^{\prime }$-group.

On the other hand, $F_{\sigma }\left(A\right)=O_{{\sigma }_i}\left(A\right)$ by Step 3. Consequently $F_{\sigma }\left(A\right)/X\subseteq C_A(T/X)$. Applying [1] (Corollary 11), we conclude that $C_A(T/X)\subseteq T/X$. Therefore $X=F_{\sigma }\left(A\right)$.

By Step 1, $n_{\sigma }(A,\mathfrak{F})=n_{\sigma }(A/X,\mathfrak{F})$. Now $n_{\sigma }(A/X,\mathfrak{F})=l_{\sigma }(A^{\mathfrak{F}}X/X)$. Since $A^{\mathfrak{F}}/A^{\mathfrak{F}}\cap X=A^{\mathfrak{F}}/F_{\sigma }\left(A^{\mathfrak{F}}\right)$, it follows that $n_{\sigma }(A/X,\mathfrak{F})=n_{\sigma }(A,\mathfrak{F})-1$ which yields the desired contradiction. □

3. Applications

As it was said in the introduction, the $\mathfrak{F}$-length $n_{\mathfrak{F}}\left(G\right)$ of a group G which is defined in [6] is just the $\sigma$-$\mathfrak{F}$-length of G for $\sigma$ the partition of $\mathbb{P}$ containing exactly one prime each, and the $\sigma$-nilpotent length of G is just the $\sigma$-$\mathfrak{F}$-length of G for $\mathfrak{F}=\left\{1\right\}$.

Therefore the following results are direct consequences of our Theorem A.

Corollary 1.

If A is a maximal subgroup of a σ-soluble group G, then $l_{\sigma }\left(A\right)=l_{\sigma }\left(G\right)-i$ for some $i\in \{0,1,2\}$.

Corollary 2.

([6] (Theorem 1)). If A is a maximal subgroup of a soluble group G and $\mathfrak{F}$ is a saturated formation, then $n_{\mathfrak{F}}\left(A\right)=n_{\mathfrak{F}}\left(G\right)-i$ for some $i\in \{0,1,2\}$.

Corollary 3.

([7] (Satz 1)). If A is a maximal subgroup of a soluble group G, then $l\left(A\right)=l\left(G\right)-i$ for some $i\in \{0,1,2\}$.

4. An Example

In [6], some examples showing that each case of Corollary 2 is possible for the partition $\sigma$ of $\mathbb{P}$ containing exactly one prime each. We give an example of slight different nature.

Example 1.

Assume that $\sigma =\{\{2,3,5,7\},\left\{211\right\},{\{2,3,5,7,211\}}^{\prime }\}$. Let X be a cyclic group of order 7 and let Y be an irreducible and faithful X-module over the finite field of 211 elements. Applying [5] (Chapter B, Theorem (9.8)), Y is a cyclic group of order 211. Let $L=\left[Y\right]X$ be the corresponding semidirect product. Consider now $G=A_5\wr L$ the regular wreath product of $A_5$, the alternating group of degree 5, with L. Then $F_{\sigma }\left(G\right)=A_5^{*}$, the base group of G. Then $l_{\sigma }\left(G\right)=3$. Let $A_1=A_5^{*}X$. Then $A_1$ is a maximal subgroup of G and $l_{\sigma }\left(A_1\right)=1$. Let $A_2=A_5^{*}Y$. Then $A_2$ is a maximal subgroup of G and $l_{\sigma }\left(A_2\right)=2$.