On $s$-semipermutable or ss-quasinormal subgroups of finite groups

Abstract

Suppose that $G$ is a finite group and $H$ is a subgroup of $G$. $H$ is said to be $s$-semipermutable in $G$ if ${ HG}_{p}=G_{p}H$ for any Sylow $p$-subgroup $G_{p}$ of $G$ with $(p,|H|)=1$; $H$ is said to be an $ss$-quasinormal subgroup of $G$ if there is a subgroup $B$ of $G$ such that $G={ HB}$ and $H$ permutes with every Sylow subgroup of $B$. We will study finite groups $G$ saisfying the following: for each noncyclic Sylow subgroup $P$ of $G$, there exists a subgroup $D$ of $P$ such that $1<|D|<|P|$ and every subgroup $H$ of $P$ with order $|D|$ is $s$-semipermutable or $ss$-quasinormal in $G$. Some recent results are generalized and unified.

1 Introduction

All groups considered in this paper are finite. $G$ always means a group, $|G|$ is the order of $G$, $\pi (G)$ denotes the set of all primes dividing $|G|$ and $G_{p}$ is a Sylow $p$-subgroup of $G$ for some $p\in \pi (G)$.

Let $\fancyscript{F}$ be a class of groups. We call $\fancyscript{F}$ a formation, provided that (1) if $G\in \fancyscript{F}$ and $H\unlhd G$, then $G/H\in \fancyscript{F}$, and (2) if $G/M$ and $G/N$ are in $\fancyscript{F}$, then $G/(M\cap N)$ is in $\fancyscript{F}$ for any normal subgroups $M$, $N$ of $G$. A formation $\fancyscript{F}$ is said to be saturated if $G/\Phi (G)\in \fancyscript{F}$ implies that $G \in \fancyscript{F}$. $\fancyscript{U}$ will denote the class of all supersolvable groups. Clearly, $\fancyscript{U}$ is a saturated formation.

A subgroup $H$ of $G$ is called $s$-quasinormal (or $s$-permutable, $\pi$-quasinormal) in $G$ provided $H$ permutes with all Sylow subgroups of $G$, i.e, HP$=$PH for any Sylow subgroup $P$ of $G$. This concept was introduced by Kegel in [5] and has been studied extensively by Deskins [1] and Schmidt [11]. More recently, Zhang and Wang [14] generalized $s$-quasinormal subgroups to $s$-semipermutable subgroups. A subgroup $H$ is said to be $s$-semipermutable in $G$ if $\textit{HG}_{p}=G_{p}H$ for any Sylow $p$-subgroup $G_{p}$ of $G$ with $(p,|H|)=1$. Clearly, every $s$-quasinormal subgroup of $G$ is an $s$-semipermutable subgroup of $G$, but the converse does not hold. Many authors consider minimal or maximal subgroups of a Sylow subgroup of a group when investigating the structure of $G$, such as in [1] and [4–14], etc. For example, Han in [4] proves the following result.

Theorem 1.1

Let $p$ be a prime dividing the order of a group $G$ satisfying $(|G|,p-1)=1$ and $P$ a Sylow $p$-subgroup of $G$. Suppose there exists a nontrivial subgroup $D$ such that $1<|D|<|P|$ and every subgroup $H$ of $P$ with order $|H|=|D|$ or with order $2|D|$ (if $P$ is a nonabelian 2-group and $|P:D|>2$) is $s$-semipermutable in $G$, then $G$ is $p$-nilpotent.

As another generalization of the $s$-quasinormality, Li et al. [7] introduce the following concept: A subgroup $H$ of $G$ is called an $ss$-quasinormal subgroup of $G$ if there is a subgroup $B$ of $G$ such that $G=HB$ and $H$ permutes with every Sylow subgroup of $B$. Many authors consider minimal or maximal subgroups of a Sylow subgroup of a group when investigating the structure of $G$, such as in [6–9, 12], etc. In [13], Wei and Guo provide a result as follows.

Theorem 1.2

Let $\fancyscript{F}$ be a saturated formation containing $\fancyscript{U}$ and $E$ be a normal subgroup of a group $G$ such that $G/E\in \fancyscript{F}$. Then $G\in \fancyscript{F}$ if only if for every non-cyclic Sylow subgroup $P$ of $F^{*}(E)$, there is a subgroup $D$ of $P$ with $1<|D|<|P|$ such that every subgroup $H$ of $P$ with order $|D|$ or $2|D|$ whenever $|D|=2$ is $ss$-quasinormal in $G$.

The aim of this article is to unify and improve above Theorems using $s$-semipermutable and $ss$-quasinormal subgroups. Our main theorems are as follows:

Theorem 1.3

(i.e., Theorem 3.5) Let $\fancyscript{F}$ be a saturated formation containing $\fancyscript{U}$, the class of all supersolvable groups and $G$ a group with $E$ as a normal subgroup of $G$ such that $G/E\in \fancyscript{F}$. Suppose that every non-cyclic Sylow subgroup $P$ of $F^{*}(E)$ has a subgroup $D$ such that $1<|D|<|P|$ and every subgroup $H$ of $P$ with order $|H|=|D|$ or with order $2|D|$ (if $P$ is a nonabelian 2-group and $|P:D|>2$) is either $s$-semipermutable or $ss$-quasinormal in $G$, where $F^{*}(E)$ is the generalized Fitting subgroup of $E$. Then $G\in \fancyscript{F}$.

2 Basic definitions and preliminary results

In this section, we collect some known results that are useful later.

Lemma 2.1

Suppose that $H$ is an $s$-semipermutable subgroup of $G$. Then the following assertions hold.

• (i) If $H\le K\le G$, then $H$ is $s$-semipermutable in $K$;

• (ii) Let $N$ be a normal subgroup of $G$. If H is a p-group for some prime $p\in \pi (G)$, then $HN/N$ is $s$-semipermutable in $G/N$;

• (iii) If $H\le O_{p}(G)$, then $H$ is $s$-permutable in $G$

Proof

(i) is [14, Property 1], (ii) is [14, Property 2], and (iii) is [14, Lemma 3]. $\square$

Lemma 2.2

Let $H$ be an $ss$-quasinormal subgroup of a group $G$.

• (i) If $H\le L\le G$, then $H$ is $ss$-quasinormal in $L$;

• (ii) If $N$ is normal in $G$, then $HN/N$ is $ss$-quasinormal in $G/N$;

• (iii) If $H\le F(G)$, then $H$ is $s$-quasinormal in $G$;

• (iv) If $H$ is a $p$-subgroup($p$ a prime), then $H$ permutes with every Sylow $q$-subgroup of $G$ with $q\ne p$.

Proof

(i) and (ii) are [7, Lemma 2.1], (iii) is [7, Lemma 2.2], and (iv) is [7, Lemma 2.5]. $\square$

Lemma 2.3

([12]) Let $G$ be a group and $P$ a Sylow $p$-subgroup of $G$, where $p$ is the smallest prime dividing $|G|$. If every maximal subgroup of $P$ is $s$-semipermutable in $G$, then $G$ is $p$-nilpotent.

Lemma 2.4

([2] III, 5.2 and IV, 5.4). Suppose that $p$ is a prime and $G$ is a minimal non-$p$-nilpotent group, i.e., $G$ is not a $p$-nilpotent group but whose proper subgroups are all $p$-nilpotent.

• (i) $G$ has a normal Sylow $p$-subgroup $P$ for some prime $p$ and $G=PQ$, where $Q$ is a non-normal cyclic $q$-subgroup for some prime $q\ne p$.

• (ii) $P/\Phi (P)$ is a minimal normal subgroup of $G/\Phi (P)$.

• (iii) The exponent of $P$ is $p$ or 4.

Lemma 2.5

([4]) Let $N$ be an elementary abelian normal $p$-subgroup of a group $G$. If there exists a subgroup $D$ in $N$ such that $1<|D|<|N|$ and every subgroup $H$ of $N$ with $|H|=|D|$ is $s$-semipermutable in $G$, then there exists a maximal subgroup $M$ of $N$ such that $M$ is normal in $G$.

Lemma 2.6

([2], VI, 4.10) Assume that $A$ and $B$ are two subgroups of a group $G$ and $G\ne AB$. If $AB^{g}=B^{g}A$ holds for any $g\in G$, then either $A$ or $B$ is contained in a nontrivial normal subgroup of $G$.

The generalized Fitting subgroup $F^{*}(G)$ of $G$ is the unique maximal normal quasinilpotent subgroup of $G$. Its definition and important properties can be found in [3, X, 13]. We would like to give the following basic facts we will use in our proof.

Lemma 2.7

([3], X, [13]) Let $G$ be a group and $M$ a subgroup of $G$.

• (i) If $M$ is normal in $G$, then $F^{*}(M)\le F^{*}(G)$;

• (ii) $F^{*}(G)\ne 1$ if $G\ne 1$; in fact, $F^{*}(G)/F(G)=Soc(F(G)C_{G}(F(G))/F(G))$;

• (iii) $F^{*}(F^{*}(G))=F^{*}(G)\ge F(G)$; if $F^{*}(G)$ is solvable, then $F^{*}(G)=F(G)$.

Lemma 2.8

([10]) Let $\fancyscript{F}$ be a saturated formation containing $\fancyscript{U}$, the class of all supersolvable groups and $G$ a group with $E$ as a normal subgroup of $G$ such that $G/E\in \fancyscript{F}$. Suppose that every non-cyclic Sylow subgroup $P$ of $F^{*}(E)$ has a subgroup $D$ such that $1<|D|<|P|$ and every subgroup $H$ of $P$ with order $|H|=|D|$ or with order $2|D|$ (if $P$ is a nonabelian 2-group and $|P:D|>2$) is weakly $s$-permutable in $G$, where $F^{*}(E)$ is the generalized Fitting subgroup of $E$. Then $G\in \fancyscript{F}$.

3 Main results

In this section, we will prove our main results.

Theorem 3.1

Let $p$ be the smallest prime dividing the order of a group $G$ and $P$ be a Sylow $p$-subgroup of $G$. If every maximal subgroup of $P$ is either $s$-semipermutable or $ss$-quasinormal in $G$. Then $G$ is $p$-nilpotent.

Proof

Assume that the theorem is not true and let $G$ be a counterexample of minimal order. We derive a contradiction in several steps. $\square$

By Lemmas 2.1 and 2.2, the following two steps are obvious.

Step 1. $O_{p^{\prime }}(G)=1$.

Step 2. $G$ has a unique minimal normal subgroup $N$ and $G/N$ is $p$-nilpotent. Moreover, $\Phi (G)=1$.

Step 3. $O_{p}(G)=1$.

If $O_{p}(G)\ne 1$, then step 2 yields $N\le O_{p}(G)$ and $\Phi (O_{p}(G))\le \Phi (G)=1$. Therefore, $G$ has a maximal subgroup $M$ such that $G=MN$ and $G/N\cong M$ is $p$-nilpotent. Since $O_{p}(G)\cap M$ is normalized by $N$ and $M$, we conclude that $O_{p}(G)\cap M$ is normal in $G$. The uniqueness of $N$ yields $N=O_{p}(G)$. Clearly, $P=N(P\cap M)$. Furthermore, $P\cap M<P$, and, thus there exists a maximal subgroup $P_{1}$ of $P$ such that $P\cap M\le P_{1}$. Hence, $P=NP_{1}$. By hypothesis, $P_{1}$ is $s$-semipermutable or $ss$-quasinormal in $G$. In either case, we have that $P_{1}M_{q}$ is a group for $q\ne p$. Hence

$$\begin{aligned} P_{1}\langle M_{p}, M_{q}|q\in \pi (M), q\ne p\rangle =P_{1}M \end{aligned}$$

is a group. Then $P_{1}M=M$ or $G$ by maximality of $M$. If $P_{1}M=G$, then

$$\begin{aligned} P=P\cap P_{1}M=P_{1}(P\cap M)=P_{1}, \end{aligned}$$

a contradiction. If $P_{1}M=M$, then $P_{1}\le M$. Therefore, $P_{1}\cap N=1$ and $N$ is of prime order. Then the $p$-nilpotency of $G/N$ implies the $p$-nilpotency of $G$, a contradiction. Therefore, $O_{p}(G)=1$.

Step 4. The final contradiction.

Assume that $P$ has a maximal subgroup $P_{1}$ which is $ss$-quasinormal in $G$, then there exists a subgroup $B$ of $G$ such that $G=P_{1}B$ and $P_{1}$ permutes with every Sylow subgroup of $B$. Noticing that $H\cap B$ is $s$-quasinormal in $B$, we have $P_{1}\cap B\le O_{p}(B)$. Since $p$ is the smallest prime dividing the order of $G$ and $H\cap B$ is a maximal subgroup of a Sylow $p$-subgroup of $B$, we see that $B/O_{p}(B)$ is $p$-nilpotent and so $B$ is $p$-solvable. By [2, Chapter VI, Hauptsatz 1.7], we may assume that $K$ is a Hall $p^{\prime }$-subgroup of $B$, $\pi (K)= \{p_{2},\ldots , p_{s}\}$ and $P_{i}\in$ Syl$_{p_{i}}(K)$. Since $P_{1}$ permutes with every $P_{i}$, $P_{1}K=KP_{1}$ and therefore $P_{1}K$ is normal in $G$. Now let $\{H_{j}: j=1,\ldots , t\}$ be the set of all maximal subgroup of $P$. Then, by the above, there is $K_{j}$ such that $H_{j}K_{j}$ is normal in $G$ for $j=1,\ldots , t$, respectively. Let $N=\bigcap \{H_{j}K_{j}: j=1,\ldots , t\}$. Then $N\unlhd G$ and $G/N$ is a $p$-group. Thus we have $P\cap N=\bigcap _{j=1}^{t}((H_{j}K_{j})\cap P)=\bigcap _{j=1}^{t}((P\cap K_{j})H_{j})=\bigcap _{j=1}^{t}H_{j}\le \Phi (P)$. By Tate$^{\prime }$s theorem [2, Chapter IV, Satz 4.7], $N$ is $p$-nilpotent and hence $G$ is $p$-nilpotent, a contradiction. Now we may assume that all maximal subgroups of $P$ are $s$-semipermutable in $G$. Then $G$ is $p$-nilpotent by Lemma 2.3, a contradiction.

Theorem 3.2

Let $p$ be the smallest prime dividing the order of a group $G$ and $P$ be a Sylow $p$-subgroup of $G$. If $P$ has a subgroup $D$ such that $1<|D|<|P|$ and every subgroup $H$ of $P$ with order $|H|=|D|$ or with order $2|D|$ (if $P$ is a nonabelian 2-group and $|P:D|>2$) is either $s$-semipermutable or $ss$-quasinormal in $G$. Then $G$ is $p$-nilpotent.

Proof

Suppose that the theorem is false and let $G$ be a counterexample of minimal order. We will derive a contradiction in several steps. $\square$

Step 1. $O_{p^{\prime }}(G)=1$.

If $O_{p^{\prime }}(G)\ne 1$, Lemma 2.1 (ii) and Lemma 2.2 (iii) guarantee that $G/O_{p^{\prime }}(G)$ satisfies the hypotheses of the theorem. Thus $G/O_{p^{\prime }}(G)$ is $p$-nilpotent by the choice of $G$. Then $G$ is $p$-nilpotent, a contradiction.

Step 2. $|D|>p$.

Suppose that $|D|=p$. Since $G$ is not $p$-nilpotent, $G$ has a minimal non-$p$-nilpotent subgroup $G_{1}$. By Lemma 2.4 (i), $G_{1}=[P_{1}]Q$, where $P_{1}\in$ Syl$_{p}(G_{1})$ and $Q\in$ Syl$_{q}(G_{1})$, $p\ne q$. Let $X/\Phi (P_{1})$ be a subgroup of $P_{1}/\Phi (P_{1})$ of order $p$, $x\in X\backslash \Phi (P_{1})$ and $L=\langle x\rangle$. Then $L$ is of order $p$ or 4 by Lemma 2.4 (iii). By the hypotheses, $L$ is either $s$-semipermutable or $ss$-quasinormal in $G$, thus in $G_{1}$ by Lemmas 2.1 (i) and 2.2 (i). First, suppose that $L$ is $ss$-quasinormal in $G_{1}$, then $L$ is $s$-quasinormal in $G_{1}$ by Lemma 2.2 (iii). Thus $LQ\le G_{1}$. Therefore, $LQ=L\times Q$. Then $G_{1}=P_{1}\times Q$. Hence $G_{1}$ is $p$-nilpotent, contrary to the choice of $G_{1}$. Therefore, $L=\langle x\rangle$ is $s$-semipermutable in $G_{1}$ for every element $x\in P_{1}$. Thus $LQ\le G_{1}$. Therefore, $LQ=L\times Q$. Then $G_{1}=P_{1}\times Q$, a contradiction.

Step 3. $|P:D|>p$.

By Theorem 3.1.

Step 4. $P$ has a subgroup $D$ such that $1<|D|<|P|$ and every subgroup $H$ of $P$ with order $|H|=|D|$ or with order $2|D|$ (if $P$ is a nonabelian 2-group and $|P:D|>2$) is $s$-semipermutable in $G$. Assume that $H\le P$ such that $|H|=|D|$ and $H$ is $ss$-quasinormal in $G$. Then we may assume $G$ has a normal subgroup $M$ such that $|G:M|=p$ and $G=HM$. Since $|P:D|>p$ by Step 3, $M$ satisfies the hypotheses of the theorem. The choice of $G$ yields that $M$ is $p$-nilpotent. It is easy to see that $G$ is $p$-nilpotent, contrary to the choice of $G$.

Step 5. If $N\le P$ and $N$ is minimal normal in $G$, then $|N|\le |D|$.

Suppose that $|N|>|D|$. Since $N\le O_{p}(G)$, $N$ is elementary abelian. By Lemma 2.5, $N$ has a maximal subgroup which is normal in $G$, contrary to the minimality of $N$.

Step 6. Suppose that $N\le P$ and $N$ is minimal normal in $G$. Then $G/N$ is $p$-nilpotent.

If $|N|<|D|$, $G/N$ satisfies the hypotheses of the theorem by Lemma 2.1 (ii). Thus $G/N$ is $p$-nilpotent by the minimal choice of $G$. So we may suppose that $|N|=|D|$ by Step 5. We will show that every cyclic subgroup of $P/N$ of order $p$ or order 4 (when $P/N$ is a non-abelian 2-group) is $s$-semipermutable in $G/N$. Let $K\le P$ and $|K/N|=p$. By Step 2, $N$ is non-cyclic, so are all subgroups containing $N$. Hence there is a maximal subgroup $L\ne N$ of $K$ such that $K=NL$. Of course, $|N|=|D|=|L|$. Since $L$ is $s$-semipermutable in $G$ by the hypotheses, $K/N=LN/N$ is $s$-semipermutable in $G/N$ by Lemma 2.1 (ii). If $p=2$ and $P/N$ is non-abelian, take a cyclic subgroup $X/N$ of $P/N$ of order 4. Let $K/N$ be maximal in $X/N$. Then $K$ is maximal in $X$ and $|K/N|$=2. Since $X$ is non-cyclic and $X/N$ is cyclic, there is a maximal subgroup $L$ of $X$ such that $N$ is not contained in $L$. Thus $X=LN$ and $|L|=|K|=2|D|$. By the hypotheses, $L$ is $s$-semipermutable in $G$. By Lemma 2.1 (ii), $X/N=LN/N$ is $s$-semipermutable in $G/N$. Hence $G/N$ satisfies the hypotheses. By the minimal choice of $G$, $G/N$ is $p$-nilpotent.

Step 7. $O_{p}(G)=1$.

Suppose that $O_{p}(G)\ne 1$. Take a minimal normal subgroup $N$ of $G$ contained in $O_{p}(G)$. By Step 6, $G/N$ is $p$-nilpotent. It is easy to see that $N$ is the unique minimal normal subgroup of $G$ contained in $O_{p}(G)$. Furthermore, $O_{p}(G)\cap \Phi (G)=1$. Hence $O_{p}(G)$ is an elementary abelian $p$-group. On the other hand, $G$ has a maximal subgroup $M$ such that $G=MN$ and $M\cap N=1$. It is easy to deduce that $O_{p}(G)\cap M=1$, $N=O_{p}(G)$ and $M\cong G/N$ is $p$-nilpotent. Then $G$ can be written as $G=N(M\cap P)M_{p^{\prime }}$, where $M_{p^{\prime }}$ is the normal $p$-complement of $M$. Pick a maximal subgroup $S$ of $M_{p}=P\cap M$. Then $NSM_{p^{\prime }}$ is a subgroup of $G$ with index $p$. Since $p$ is the minimal prime in $\pi (G)$, we know that $NSM_{p^{\prime }}$ is normal in $G$. Now by Step 3 and the induction, we have $NSM_{p^{\prime }}$ is $p$-nilpotent. Therefore, $G$ is $p$-nilpotent, a contradiction.

Step 8. The minimal normal subgroup $L$ of $G$ is not $p$-nilpotent.

If $L$ is $p$-nilpotent, then it follows from the fact that $L_{p^{\prime }}$ char $L\lhd G$ that $L_{p^{\prime }}\le O_{p^{\prime }}(G)=1$. Thus $L$ is a $p$-group. Whence $L\le O_{p}(G)=1$ by Step 7, a contradiction.

Step 9. $G$ is a non-abelian simple group.

Suppose that $G$ is not a simple group. Take a minimal normal subgroup $L$ of $G$. Then $L<G$. If $|L|_{_{p}}>|D|$, then $L$ is $p$-nilpotent by the minimal choice of $G$, contrary to Step 8. If $|L|_{p}\le |D|$. Take $P_{*}\ge L\cap P$ such that $|P_{*}|=p|D|$. Hence $P_{*}$ is a Sylow $p$-subgroup of $P_{*}L$. Since every maximal subgroup of $P_{*}$ is of order $|D|$, every maximal subgroup of $P_{*}$ is $s$-semipermutable in $G$ by hypotheses, thus in $P_{*}L$ by Lemma 2.1 (i). Now applying Theorem 3.1, we get $P_{*}L$ is $p$-nilpotent. Therefore, $L$ is $p$-nilpotent, contrary to Step 8.

Step 10. The final contradiction.

Suppose that $H$ is a subgroup of $P$ with $|H|=|D|$ and $Q$ is a Sylow $q$-subgroup with $q\ne p$. Then $HQ^{g}=Q^{g}H$ for any $g\in G$ by the hypotheses that $H$ is $s$-semipermutable in $G$. Since $G$ is simple by Step 9, $G=HQ$ from Lemma 2.6, the final contradiction.

The following corollary is immediate from Theorem 3.2.

Corollary 3.3

Suppose that $G$ is a group. If every non-cyclic Sylow subgroup of $G$ has a subgroup $D$ such that $1<|D|<|P|$ and every subgroup $H$ of $P$ with order $|H|=|D|$ or with order $2|D|$ (if $P$ is a nonabelian 2-group and $|P:D|>2$) is either $s$-semipermutable or $ss$-quasinormal in $G$, then $G$ has a Sylow tower of supersolvable type.

Theorem 3.4

Let $\fancyscript{F}$ be a saturated formation containing $\fancyscript{U}$, the class of all supersolvable groups and $G$ a group with $E$ as a normal subgroup of $G$ such that $G/E\in \fancyscript{F}$. Suppose that every non-cyclic Sylow subgroup of $E$ has a subgroup $D$ such that $1<|D|<|P|$ and every subgroup $H$ of $P$ with order $|H|=|D|$ or with order $2|D|$ (if $P$ is a nonabelian 2-group and $|P:D|>2$) is either $s$-semipermutable or $ss$-quasinormal in $G$. Then $G\in \fancyscript{F}$.

Proof

Suppose that $P$ is a non-cyclic Sylow $p$-subgroup of $E$, $\forall p\in \pi (E)$. Since $P$ has a subgroup $D$ such that $1<|D|<|P|$ and every subgroup $H$ of $P$ with order $|H|=|D|$ or with order $2|D|$ (if $P$ is a nonabelian 2-group and $|P:D|>2$) is either $s$-semipermutable or $ss$-quasinormal in $G$ by hypotheses, thus in $E$ by Lemma 2.1 (i) and Lemma 2.2 (i). Applying Corollary 3.3, we conclude that $E$ has a Sylow tower of supersolvable type. Let $q$ be the maximal prime divisor of $|E|$ and $Q\in$ Syl$_{q}(E)$. Then $Q\unlhd G$. Since $(G/Q, E/Q)$ satisfies the hypotheses of the theorem, by induction, $G/Q\in \fancyscript{F}$. For any subgroup $H$ of $Q$ with $|H|=|D|$, since $Q\le O_{q}(G)$, $H$ is $s$-permutable in $G$ by Lemmas 2.1 (iii) and 2.2 (iii). Since $s$-permutable implies weakly $s$-permutable and $F^{*}(Q)=Q$ by Lemma 2.7, we get $G\in \fancyscript{F}$ by applying Lemma 2.8. $\square$

Theorem 3.5

Let $\fancyscript{F}$ be a saturated formation containing $\fancyscript{U}$, the class of all supersolvable groups and $G$ a group with $E$ as a normal subgroup of $G$ such that $G/E\in \fancyscript{F}$. Suppose that every non-cyclic Sylow subgroup of $F^{*}(E)$ has a subgroup $D$ such that $1<|D|<|P|$ and every subgroup $H$ of $P$ with order $|H|=|D|$ or with order $2|D|$ (if $P$ is a nonabelian 2-group and $|P:D|>2$) is either $s$-semipermutable or $ss$-quasinormal in $G$. Then $G\in \fancyscript{F}$.

Proof

We distinguish two cases:

Case 1. $\fancyscript{F}=\fancyscript{U}$.

Let $G$ be a minimal counter-example.

Step 1. Every proper normal subgroup $N$ of $G$ containing $F^{*}(E)$ (if it exists) is supersolvable. $\square$

If $N$ is a proper normal subgroup of $G$ containing $F^{*}(E)$, then $N/N\cap E\cong NE/E$ is supersolvable. By Lemma 2.7 (iii), $F^{*}(E)=F^{*}(F^{*}(E))\le F^{*}(E\cap N)\le F^{*}(E)$, so $F^{*}(E\cap N)=F^{*}(E)$. For any Sylow subgroup $P$ of $F^{*}(E\cap N)=F^{*}(E)$, $P$ has a subgroup $D$ such that $1<|D|<|P|$ and every subgroup $H$ of $P$ with order $|H|=|D|$ or with order $2|D|$ (if $P$ is a nonabelian 2-group and $|P:D|>2$) is either $s$-semipermutable or $ss$-quasinormal in $G$ by hypotheses, thus in $N$ by Lemma 2.1 (i) and Lemma 2.2 (i). So $N$ and $N\cap H$ satisfy the hypotheses of the theorem, the minimal choice of $G$ implies that $N$ is supersolvable.

Step 2. $E=G$.

If $E<G$, then $E\in \fancyscript{U}$ by Step 1. Hence $F^{*}(E)=F(E)$ by Lemma 2.7. It follows that every Sylow subgroup of $F^{*}(E)$ is normal in $G$. By Lemmas 2.1 (iii) and 2.2 (iii), every non-cyclic Sylow subgroup of $F^{*}(E)$ has a subgroup $D$ such that $1<|D|<|P|$ and every subgroup $H$ of $P$ with order $|H|=|D|$ or with order $2|D|$ (if $P$ is a nonabelian 2-group and $|P:D|>2$) is $s$-permutable in $G$. Applying Lemma 2.8 for the special case $\fancyscript{F}=\fancyscript{U}$, $G\in \fancyscript{U}$, a contradiction.

Step 3. $F^{*}(G)=F(G)<G$.

If $F^{*}(G)=G$, then $G\in \fancyscript{F}$ by Theorem 3.4, contrary to the choice of $G$. So $F^{*}(G)<G$. By Step 1, $F^{*}(G)\in \fancyscript{U}$ and $F^{*}(G)=F(G)$ by Lemma 2.7.

Step 4. The final contradiction.

Since $F^{*}(G)=F(G)$, each non-cyclic Sylow subgroup of $F^{*}(G)$ has a subgroup $D$ such that $1<|D|<|P|$ and every subgroup $H$ of $P$ with order $|H|=|D|$ or with order $2|D|$ (if $P$ is a nonabelian 2-group and $|P:D|>2$) is $s$-permutable in $G$ by Lemmas 2.1 (iii) and 2.2 (iii). Applying Lemma 2.8, $G\in \fancyscript{U}$, a contradiction.

Case 2. $\fancyscript{F}\ne \fancyscript{U}$.

By hypotheses, every non-cyclic Sylow subgroup of $F^{*}(E)$ has a subgroup $D$ such that $1<|D|<|P|$ and every subgroup $H$ of $P$ with order $|H|=|D|$ or with order $2|D|$ (if $P$ is a nonabelian 2-group and $|P:D|>2$) is either $s$-semipermutable or $ss$-quasinormal in $G$, thus in $E$ Lemma 2.1 (i) and Lemma 2.2 (i). Applying Case 1, $E\in \fancyscript{U}$. Then $F^{*}(E)=F(E)$ by Lemma 2.7. It follows that each Sylow subgroup of $F^{*}(E)$ is normal in $G$. By Lemmas 2.1 (iii) and 2.2 (iii), each non-cyclic Sylow subgroup of $F^{*}(E)$ has a subgroup $D$ such that $1<|D|<|P|$ and every subgroup $H$ of $P$ with order $|H|=|D|$ or with order $2|D|$ (if $P$ is a nonabelian 2-group and $|P:D|>2$) is $s$-permutable in $G$. Applying Lemma 2.8, $G\in \fancyscript{F}$. These complete the proof of the theorem.

The following corollaries are immediate from Theorem 3.5.

Corollary 3.6

Let $\fancyscript{F}$ be a saturated formation containing $\fancyscript{U}$. Suppose that $G$ is a group with a normal subgroup $E$ such that $G/E\in \fancyscript{F}$. Then $G\in \fancyscript{F}$ if and only if every maximal subgroup of any Sylow subgroup of $F^{*}(E)$ is $s$-semipermutable in $G$.

Corollary 3.7

Let $\fancyscript{F}$ be a saturated formation containing $\fancyscript{U}$. Suppose that $G$ is a group with a normal subgroup $E$ such that $G/E\in \fancyscript{F}$. Then $G\in \fancyscript{F}$ if and only if every cyclic subgroup of any Sylow subgroup of $F^{*}(E)$ of prime order or order 4 is $s$-semipermutable in $G$.

Corollary 3.8

([6], Theorem 3.3) Let $\fancyscript{F}$ be a saturated formation containing $\fancyscript{U}$. Suppose that $G$ is a group with a normal subgroup $H$ such that $G/H\in \fancyscript{F}$. Then $G\in \fancyscript{F}$ if and only if every maximal subgroup of any Sylow subgroup of $F^{*}(H)$ is $ss$-quasinormal in $G$.

Corollary 3.9

([6], Theorem 3.7) Let $\fancyscript{F}$ be a saturated formation containing $\fancyscript{U}$. Suppose that $G$ is a group with a normal subgroup $H$ such that $G/H\in \fancyscript{F}$. Then $G\in \fancyscript{F}$ if and only if every cyclic subgroup of any Sylow subgroup of $F^{*}(H)$ of prime order or order 4 is $ss$-quasinormal in $G$.

Corollary 3.10

([8], Theorem 3.4) Let $\fancyscript{F}$ be a saturated formation containing $\fancyscript{U}$. Suppose that $G$ is a group with a normal subgroup $E$ such that $G/E\in \fancyscript{F}$. Then $G\in \fancyscript{F}$ if and only if every maximal subgroup of any Sylow subgroup of $F^{*}(E)$ is $s$-quasinormal in $G$.

Corollary 3.11

([9], Theorem 3.3) Let $\fancyscript{F}$ be a saturated formation containing $\fancyscript{U}$. Suppose that $G$ is a group with a normal subgroup $E$ such that $G/E\in \fancyscript{F}$. Then $G\in \fancyscript{F}$ if and only if every cyclic subgroup of any Sylow subgroup of $F^{*}(E)$ of prime order or order 4 is $s$-quasinormal in $G$.