Some Notes on the Second Maximal Subgroups of Finite Groups

Abstract

Under study are the arithmetic properties of second maximal subgroups of finite groups. Generally speaking, we investigated the problem by Monakhov [1, Problem 19.54] and developed the research of Meng and Guo [2, Theorem B] by weakening the condition of solvability.

1. Introduction

All groups is finite. We will adhere to the notation of [3, 4]. In particular, $|G|$ denotes the order of a group $G$ (or a set $G$), while $\pi(G)$ denotes the set of all prime divisors of $|G|$. Let $H_{G}$ be the core of $H$ in $G$ when $H\leq G$ and let $M<\cdot G$ signify that $M$ is a maximal subgroup of $G$. Put

$$ \operatorname{Max}(G,H)=\{M<\!\cdot\,G\mid H\leq M\},\quad\operatorname{Max}_{1}(G,H)=\{M<\!\cdot\,G\mid H<\!\cdot\,M\}. $$

Let $\operatorname{Irr}(G)$ be the set of all irreducible complex characters of $G$. An element $x$ of $G$ is called nonvanishing if $\chi(x)\neq 0$ for all $\chi\in\operatorname{Irr}(G)$.

Next, we recall some known research that is tightly related to our study.

Firstly, we need to list the following problem by Monakhov [1, Problem 19.54] which motivated our research.

Problem

What are the chief factors of a finite group whose no $2$-maximal subgroup is $m$-maximal for any $m\geq 3$?

Addressing the problem, Meng and Guo [2] considered the properties of the second maximal subgroup of a group and the structure of a $WSM$-group under the universe of solvable groups, where the $WSM$-group is equal to the group satisfying the condition of the above problem.

On the other hand, Isaacs, Navarro, and Wolf conjectured in [5] that every nonvanishing element of a solvable group $G$ is contained in the Fitting subgroup $F(G)$. In [6], Guo, Skiba, and Tang introduced the concept of boundary factors and traces of subgroups in finite groups and investigated the solvability of a group by considering the traces of maximal subgroups.

Continuing to study the Problem and developing the research of Meng and Guo [2], we will investigate the numerical structure of a second maximal subgroup of a group by weakening the condition of solvability. Meanwhile, viewing from the conjecture in [5] and the result in [6], we also consider the relationship between the conjecture and the traces of second maximal subgroups of a group. Here we obtained the following results:

Theorem 1.1

Let $G$ be a $WSM$-group and let $x$ be a nonvanishing element of $G$. If each second maximal subgroup of $G$ has a nilpotent trace, then $G$ is solvable and $x\in F(G)$.

Theorem 1.2

Let $H<\cdot M<\cdot G$ and $|M:H|=p^{\alpha}$, where $p\in\pi(G)$. If $H$ is a $CAP$-subgroup of $G$, then $|\operatorname{Max}(G,H)\backslash\operatorname{Max}_{1}(G,H)|\leq 1$.

2. Preliminaries

For the sake of convenience, we start with listing some known results that will be useful in this paper.

Lemma 2.1 [2, Lemma 1]

Let $G$ be a group and let $H$ be a subgroup of $G$. If there exists $M$, $X\in\operatorname{Max}(G,H)$ such that $H$ is maximal in $M$ and $H$ is not maximal in $X$, then $H_{G}=M_{G}$.

Lemma 2.2 [2, Theorem B]

Let $G$ be a solvable group and let $H$ be a weak second maximal subgroup of $G$. Then there exists at most one $X\in\operatorname{Max}(G,H)$ such that $H$ is not maximal in $X$.

Lemma 2.3

Let $G$ be a group. If every second maximal subgroup of $G$ is nilpotent, then $G$ is either solvable or isomorphic to $PSL(2,5)$ or $SL(2,5)$.

Proof

See [7] and [8].

3. The Main Results

In [6], Guo, Skiba, and Tang introduced the concept of boundary factors and traces of subgroups in finite groups. Next, we will study the construction of a nonvanishing element of a group by the nilpotency of the traces of second maximal subgroups of $G$.

Theorem 3.1

Let $G$ be a $WSM$-group and let $x$ be a nonvanishing element of $G$. If each second maximal subgroup of $G$ has a nilpotent trace, then $G$ is solvable and $x\in F(G)$.

Proof

By [2, Theorem A], we only need to prove that $G$ is solvable.

If there exists a second maximal subgroup $H$ of $G$ such that $H=1$, then there is a maximal subgroup $M$ of $G$ such that $|M|$ is a prime. By [9, Chapter IV, Theorem 7.4], $G$ is solvable. Hence every second maximal subgroup of $G$ is nontrivial. Now we assert that $G$ is not nonabelian simple. Otherwise, $G$ is nonabelian simple. Since each second maximal subgroup of $G$ has a nilpotent trace, every second maximal subgroup of $G$ is nilpotent. By Lemma 2.3$G\cong A_{5}$, where $A_{5}$ is the alternating group of degree $5$. However, $A_{5}$ is not a $WSM$-group by [10].

Further, we may choose a minimal normal subgroup $L$ of $G$ and consider the quotient group $G/L$. If $L$ is maximal in $G$, then $G/L$ is of order $q$, with $q$ a prime. If $L$ is not maximal in $G$, then $G/L$ satisfies the hypothesis and $G/L$ is solvable by induction on $|G|$. Hence $G/L$ is solvable for every minimal normal subgroup $L$ of $G$. Further, $L\nleq\Phi(G)$ and $L$ is the unique minimal normal subgroup of $G$.

Let $L_{p}$ be a Sylow $p$-subgroup of $L$ where $p$ is a largest prime divisor of $|L|$. Clearly, $p>3$. By the Frattini argument, $G=LN_{G}(L_{p})=LM$, where $M$ is maximal in $G$ and $N_{G}(L_{p})\leq M$. Further, $L\cap M\neq 1$ and $M_{G}=1$. Hence there is a maximal subgroup $H$ of $M$ such that $L\cap M\leq H$ and $L\cap H=L\cap M$ is nilpotent by hypothesis. Then $N_{L}(L_{p})=L\cap N_{G}(L_{p})\leq L\cap M=L\cap H$ is nilpotent and $N_{L}(L_{p})/C_{L}(L_{p})$ is a $p$-subgroup. Further, $O^{p}(G)<L$ by [11, Chapter X, Theorem 8.13] and $L$ is a $p$-subgroup of $G$. Hence, $G$ is solvable since $G/L$ is solvable.

In view of [12, Theorem 3.7], we can weaken the condition of solvability in [2, Theorem B] by the following condition and arithmetic description of a second maximal subgroup of a group.

Theorem 3.2

Let $H<\cdot M<\cdot G$ and $|M:H|=p^{\alpha}$, where $p\in\pi(G)$. If $H$ is a $CAP$-subgroup of $G$, then $|\operatorname{Max}(G,H)\backslash\operatorname{Max}_{1}(G,H)|\leq 1$.

Proof

Clearly, we may assume that $H_{G}=1$ and $X_{i}\in\operatorname{Max}(G,H)\backslash\operatorname{Max}_{1}(G,H)$, where $i$=1,2. Since $H<\cdot M$, $H=M\cap X_{1}=M\cap X_{2}$. Also, $H_{G}=M_{G}=1$ by Lemma 2.1.

Since $G$ is primitive, $G$ has one of the following structures by [3, Chapter A, Theorem 15.2]:

$(1)$ $G=LM$, $C_{G}(L)=L$ and $L$ is abelian, where $L$ is the unique minimal normal subgroup of $G$;

$(2)$ $G=LM$ and $L$ is nonabelian, where $L$ is the unique minimal normal subgroup of $G$;

$(3)$ $G$ has exactly two minimal normal subgroups $L$ and $L^{*}$ of $G$, while $G=LM=L^{*}M$ and $L\cap M=L^{*}\cap M=1$. Also, $C_{G}(L)=L^{*}$, $C_{G}(L^{*})=L$ and $L\cong L^{*}\cong LL^{*}\cap M$. Moreover, if $V<G$ and $LV=L^{*}V=G$, then $L\cap V=L^{*}\cap V=1$.

By Lemma 2.2, we may assume that $G$ is not solvable. Since $H$ is a $CAP$-subgroup of $G$, we assert that $G$ is not simple. Otherwise, $H=1$. Then $M$ is a maximal subgroup of $G$ of prime order and $G$ is solvable by [9, Chapter IV, Theorem 7.4]; a contradiction. To prove, we will coincide the following cases:

Case I: $G$ has the structure $(1)$ above.

If $L\nleq X_{i}$ for some $i\in\{1,2\}$, then $G=LX_{i}$ and $L\cap X_{i}=1$. Further, $M\cong ML/L$ and $X_{i}\cong LX_{i}/L$. Since $HL/L<\cdot ML/L=LX_{i}/L$, $H<\cdot X_{i}$; a contradiction. So $L\leq X_{1}$ and $L\leq X_{2}$. Then $LM\cap X_{1}=LM\cap X_{2}=LH$ and $|\operatorname{Max}(G,H)\backslash\operatorname{Max}_{1}(G,H)|\leq 1$.

Case II: $G$ has the structure $(3)$ above.

Since $G=LM=L^{*}M$, $L\cap M=L^{*}\cap M=1$, If there exists some $X_{i}$ such that $(X_{i})_{G}=1$ for some $i\in\{1,2\}$, then $G=LX_{i}=L^{*}X_{i}$ and $L\cap X_{i}=L^{*}\cap X_{i}=1$. Further, $HL/L<\cdot ML/L=LX_{i}/L$, $H<\cdot X_{i}$; a contradiction. So, $(X_{1})_{G}\neq 1$ and $(X_{2})_{G}\neq 1$. Then we assert that $L\leq(X_{1})_{G}\cap(X_{2})_{G}$ and $L^{*}\leq(X_{1})_{G}\cap(X_{2})_{G}$. Otherwise, there exists $R\in\{L,L^{*}\}$ and $X_{i}$ for some $i\in\{1,2\}$ such that $RX_{i}=G$ and $R\cap X_{i}=1$. With the similar discussion of the above, $H<\cdot X_{i}$; a contradiction. Hence $LM\cap X_{1}=LM\cap X_{2}=LH$ and $|\operatorname{Max}(G,H)\backslash\operatorname{Max}_{1}(G,H)|\leq 1$.

Case III: $G$ has the structure $(2)$ above.

Since $H$ is a $CAP$-subgroup of $G$, $H\cap L=1$. Further, we consider the subgroup $HL$.

If $HL=G$, then $M=M\cap G=M\cap HL=H(M\cap L)$ and $M\cap L$ is a $p$-subgroup since $|M:H|=p^{\alpha}$. Also, $M\cap L$ is a minimal normal subgroup of $M$ since $H\cap L=1$ and $H<\cdot M$. Now we assert that $N_{G}(L\cap M)=M$. Otherwise, $N_{G}(L\cap M)=G$ and $L=L\cap M$ by the minimal normality of $L$. Further, $HL=G=M$; a contradiction. Hence $N_{L}(L\cap M)=L\cap N_{G}(L\cap M)=L\cap M$. Clearly, $L\cap M$ is a Sylow $p$-subgroup of $L$. Thus, $N_{L}(L\cap M)=L\cap M=C_{L}(L\cap M)$. By the Burnside Theorem, $L$ is $p$-nilpotent. Since $L\cap M\neq 1$, $L$ is a $p$-subgroup of $G$ and $L=L\cap M\leq M$. Then $HL=G=M$; a contradiction.

If $HL<G$, then $M=M\cap HL=H(M\cap L)$ since $L\cap M\neq 1$ and $H<\cdot M$. Hence $L\leq M$ and $LM=G=M$; a contradiction.

The authors proved in [2, Lemma 1] that $H_{G}=M_{G}$, where $M\in\operatorname{Max}_{1}(G,H)$ and $\operatorname{Max}_{1}(G,H)$ is properly included in $\operatorname{Max}(G,H)$. By dual consideration, we will show the following relationship between $H$ and $X$, where $X\in\operatorname{Max}(G,H)\backslash\operatorname{Max}_{1}(G,H)$:

Theorem 3.3

Let $G$ be a group and let $H$ be a subgroup of $G$. If there exist $M\in\operatorname{Max}_{1}(G,H)$ and $X\in\operatorname{Max}(G,H)\backslash\operatorname{Max}_{1}(G,H)$, then either $H_{G}=X_{G}$ or $HX_{G}=X$.

Proof

Assume that $H_{G}\neq X_{G}$. Prove that $HX_{G}=X$.

If $H_{G}=1$, then $H_{G}=M_{G}=1$ by Lemma 2.1. Further, $X_{G}\neq 1$, $G$ is primitive and $G=LM$, where $L$ is a minimal normal subgroup $L$ of $G$ which lies in $X_{G}$. Since $H$ is maximal in $M$, $H=M\cap X$ and $LH=X$. Hence $HX_{G}=X$.

If $H_{G}\neq 1$, then $1<H_{G}<H$ or $H_{G}=H$. To proceed the proof, we will consider the following cases:

Case I: $1<H_{G}<H$.

We consider the quotient subgroup $G/H_{G}$. So $(H/H_{G})_{G/H_{G}}=(X/H_{G})_{G/H_{G}}$ or $(H/H_{G})(X/H_{G})_{{G/H_{G}}}={X/H_{G}}$ by the induction on $|\,G|$. Further, $H_{G}=X_{G}$ or $HX_{G}=X$. Hence $HX_{G}=X$ by assumption.

Case II: $H_{G}=H$.

Since $H$ is maximal in $M$; therefore, $|M:H|=p$ and $H=M\cap X=M\cap X_{G}$. Then $G=MX_{G}$ and $|G:X_{G}|=|M:M\cap X_{G}|=|M:H|=p$. Hence $X_{G}=X$ and $HX_{G}=X$.

Corollary 3.4

Under the hypothesis of Theorem 3.3, if $H$ is subnormal in $G$, then either $H_{G}=X_{G}$ or $X\trianglelefteq G$.

Proof

If $G$ is simple then $H_{G}=X_{G}=1$. Assume now that $G$ is not simple. Since $H$ is subnormal in $G$; therefore, $\operatorname{Soc}(G)\leq N_{G}(H)$ by [3, Chapter A, Theorem 14.3] and $N_{G}(H)=M$ or $N_{G}(H)=G$. If $N_{G}(H)=G$, then $H$ is normal in $G$ and $H=X_{G}$ or $X\trianglelefteq G$ by Theorem 3.3. If $N_{G}(H)=M$, then $\operatorname{Soc}(G)\leq N_{G}(H)=M$ and $M_{G}=H_{G}\neq 1$ by Lemma 2.1. Hence $1<H_{G}<H$. Next, we consider the quotient group $G/H_{G}$. By induction on $|G|$, we see that $(H/H_{G})_{G/H_{G}}=(X/H_{G})_{G/H_{G}}$ or $X/H_{G}\trianglelefteq G/H_{G}$. Then $H_{G}=X_{G}$ or $X\trianglelefteq G$.