On $S$-propermutable Subgroups of Finite Groups

Abstract

Let $H$ be a subgroup of a finite group $G$. Then we say that $H$ is $S$-propermutable in $G$ provided $G$ has a subgroup $B$ such that $G=N_{G}(H)B$ and $H$ permutes with all Sylow subgroups of $B$. In this paper we analyze the influence of $S$-propermutable subgroups on the structure of $G$.

1 Introduction

Throughout this paper, all groups are finite and $G$ always denotes a finite group. Moreover, $p$ is always supposed to be a prime and $\pi$ is a non-empty subset of the set $\mathbb {P}$ of all primes. We use $\mathcal{M}_{\phi }(G)$ to denote a set of maximal subgroups of $G$ such that ${\Phi }(G)$ coincides with the intersection of all subgroups in $\mathcal{M}_{\phi }(G)$. If for subgroups $A$ and $B$ of $G$ we have $AB=BA$, then $A$ is said to permute with $B$. If $G=AB$, then $B$ is said to be a supplement of $A$ to $G$.

Recall that a subgroup $H$ of $G$ is said to be S-permutable, S-quasinormal, or $\pi$ -quasinormal Kegel [11] in $G$ provided $HP=PH$ for all Sylow subgroups $P$ of $G$. The $S$-permutable subgroups possess many interesting properties (see [3, 11, 15] or Chap. 1 in [1]), and such subgroups are used for the analysis of many questions of the group theory (see Sect. 5 in [20]). This circumstance was the main motivation for the introduction and study of various generalizations of the $S$-permutability. One of the most interesting generalizations of $S$-permutability was found by Shirong Li, Zhencai Shen, Jianjun Liu, and Xiaochun Liu: A subgroup $H$ of $G$ is called SS-quasinormal [18] in $G$ if $H$ permutes with all Sylow subgroups of some supplement of $H$ to $G$. Nice results obtained in the papers [18, 19, 22] were based on applications of this concept.

In this paper we consider another generalization of $S$-permutable subgroups.

Definition 1.1

Let $H$ be a subgroup of $G$. Then we say that $H$ is S-propermutable in $G$ provided there is a subgroup $B$ of $G$ such that $G=N_{G}(H)B$ and $H$ permutes with all Sylow subgroups of $B$.

In fact, we meet $S$-propermutable subgroups quite often.

Example 1.1

• (1) Every maximal subgroup of a soluble group $G$ and every its Hall subgroup $E$ with $|G:N_{G}(E)|=p^{a}$ are $S$-propermutable in $G$. Indeed, since $G$ is soluble, there is a Sylow $p$-subgroup $P$ of $G$ such that $EP=PE$. On the other hand, since $|G:N_{G}(E)|=p^{a}|$ we have $G= N_{G}(E)P$. Hence $E$ is $S$-propermutable in $G$.

• (2) If $|H|=p^{a}$ and $H\le Z_{\infty }(G)$, then $H\le P$, where $P$ is the Sylow $p$-subgroup of $Z_{\infty }(G)$. Therefore, since $G/C_{G}(P)$ is a $p$-group (see Lemma 2.9 below), $G=N_{G}(H)G_{p}$ and $H\le P\le G_{p}$, where $G_{p}$ is a Sylow $p$-subgroup of $G$. Hence $H$ is $S$-propermutable in $G$.

• (3) If $G$ is metanilpotent, that is $G/F(G)$ is nilpotent, then for every Sylow subgroup $P$ of $G$ we have $G=N_{G}(P)F(G)$. Therefore, in this case, every characteristic subgroup of every Sylow subgroup of $G$ is $S$-propermutable in $G$. In particular, every Sylow subgroup of a supersoluble group is $S$-propermutable.

It is clear that every $SS$-quasinormal subgroup is $S$-propermutable. The following elementary example shows that in general the set of all $S$-propermutable subgroups of $G$ is wider than the set of all its $SS$-quasinormal subgroups.

Example 1.2

Let $p > q > r$ be primes such that $qr$ divides $p-1$. Let $P$ be a group of order $p$ and $QR\le Aut (P)$, where $Q$ and $R$ are groups with order $q$ and $r$, respectively. Let $G=P\rtimes (QR)$. Then $R$ is $S$-propermutable in $G$. Suppose that $R$ is $SS$-quasinormal in $G$. Then $Q^{x}R=RQ^{x}$ for all $x\in G$ (see Lemma 1.4 below). But $Q^{x}R\simeq G/P$ is cyclic, so $Q^{G}= PQ\le N_{G}(R)$. Hence $R$ is normal in $G$, which implies that $R\le C_{G}(P)=P$. This contradiction shows that $R$ is not $SS$-quasinormal in $G$.

The results of the above-mentioned papers [18, 19, 22] are motivations for the following our theorem.

Theorem A

Let $E$ be a normal subgroup of $G$ and $P$ a Sylow $p$-subgroup of $E$. Suppose that $|P| > p$.

• (I) If every number $V$ of some fixed $\mathcal{M}_{\phi }(P)$ is $S$-propermutable in $G$, then $E$ is $p$-supersoluble.

• (II) If every maximal subgroup of $P$ is $S$-propermutable in $G$, then every chief factor of $G$ between $E$ and $O_{p'}(E)$ is cyclic.

As a first application of Theorem A, we prove also the following result.

Theorem B

Let $X\le E$ be normal subgroups of $G$. Suppose that every maximal subgroup of every non-cyclic Sylow subgroup of $X$ is $S$-propermutable in $G$. If either $X=E$ or $X=F^{*}(E)$, then every chief factor of $G$ below $E$ is cyclic.

Let $\mathcal F$ be a class of groups. If $1\in \mathcal{F}$, then we write $G^\mathcal{F}$ to denote the intersection of all normal subgroups $N$ of $G$ with $G/N\in \mathcal{F}$. The class $\mathcal F$ is said to be a formation if either $\mathcal{F}= \varnothing$ or $1\in \mathcal{F}$ and every homomorphic image of $G/G^\mathcal{F}$ belongs to $\mathcal{F}$ for any group $G$. The formation $\mathcal{F}$ is said to be solubly saturated if $G\in \mathcal{F}$ whenever $G/\Phi (N) \in \mathcal{F}$ for some soluble normal subgroup $N$ of $G$.

Note that if $\mathcal{F}$ is a solubly saturated formation and $G/E \in \mathcal{F}$, where every chief factor of $G$ below $E$ is cyclic, then $G\in \mathcal{F}$ (see Lemma 2.13 below). Therefore, from Theorem B we get

Corollary 1.1

Let $\mathcal F$ be a solubly saturated formation containing all supersoluble groups and $X\le E$ normal subgroups of $G$ such that $G/E \in \mathcal{F}$. Suppose that every maximal subgroup of every non-cyclic Sylow subgroup of $X$ is $S$-propermutable in $G$. If either $X=E$ or $X=F^{*}(E)$, then $G \in \mathcal{F}$.

Note Theorem A and Corollary 1.4 cover results of many papers and, in particular, some main results in [14, 18, 19] (see Sect. 4).

The proof of Theorem A consists of many steps, and the following useful result is one of them.

Theorem C

Let $E$ be a normal subgroup of $G$ and $P$ is a Sylow $p$-subgroups of $E$. If $P$ is $S$-propermutable in $G$, then $E$ is $p$-soluble.

All unexplained notation and terminology are standard. The reader is referred to [4, 6, 17] or [2] if necessary.

2 Preliminaries

Lemma 2.1

(See [9]) Let $A$ and $B$ be subgroups of $G$ with $G=AB$.

• (1) If $G$ is $\pi$-soluble, then there are Hall $\pi$-subgroups $A_{\pi }$, $B_{\pi }$, and $G_{\pi }$ of $A$, $B$, and $G$, respectively, such that $G_{\pi }=A_{\pi }B_{\pi }$

• (2) For any prime $p$ dividing $|G|$, there are Sylow $p$-subgroups $A_{p}$, $B_{p}$, and $G_{p}$ of $A$, $B$, and $G$, respectively, such that $G_{p}=A_{p}B_{p}$.

Lemma 2.2

(See Lemma 1.6 in [4]) Let $H$, $K$, and $N$ be subgroups of $G$. If $HK=KH$ and $HN=NH$, then $H\langle K, N \rangle =\langle K, N \rangle H$.

We say that $H$ is propermutable in $G$ provided there is a subgroup $B$ of $G$ such that $G=N_{G}(H)B$ and $H$ permutes with all subgroups of $B$.

Lemma 2.3

Let $H\le G$ and $N$ be a normal subgroup of $G$. Suppose that $H$ is $S$-propermutable (propermutable) in $G$.

• (1) $HN/N$ is $S$-propermutable (propermutable, respectively) in $G/N$.

• (2) $H$ permutes with some Sylow $p$-subgroup of $G$ for any prime $p$ dividing $|G|$.

• (3) If $G$ is $\pi$-soluble, then $H$ permutes with some Hall $\pi$-subgroup of $G$.

• (4) $|G:N_{G}(H\cap N)|$ is a $\pi$-number, where $\pi = \pi (N)\cup \pi (H)$.

Proof

• (1) First suppose that $H$ is $S$-propermutable in $G$. By hypothesis there is a subgroup $B$ of $G$ such that $G=N_{G}(H)B$ and $H$ permutes with all Sylow $p$-subgroups of $B$ for all primes $p$ dividing $|B|$. Then

  $$\begin{aligned} G/N=(N_{G}(H)N/N)(BN/N)=N_{G/N}(HN/N)(BN/N). \end{aligned}$$

  Suppose that $p$ divides $|BN/N|$ and let $K/N$ be any Sylow $p$-subgroup of $BN/N$. Then $K=(K\cap B)N$, so by Lemma 2.1, there are Sylow $p$-subgroups $K_{p}$, $P$, and $N_{p}$ of $K$, $K\cap B$, and $N$, respectively, such that $K_{p}=PN_{p}$. Let $P\le B_{p}$, where $B_{p}$ is a Sylow $p$-subgroup of $B$. Then $K/N\le B_{p}N/N$, which implies that $K/N= B_{p}N/N$. But $H$ permutes with $B_{p}$, so that $HN/N$ permutes with $K/N$. Therefore, $HN/N$ is $S$-propermutable in $G/N$. The second assertion of (1) is proved similarly.

• (2) By Lemma 2.1 there are Sylow $p$-subgroups $P_{1}$, $P_{2}$, and $P$ of $N_{G}(H)$, $B$, and $G$, respectively, such that $P=P_{1}P_{2}$. Then

  $$\begin{aligned}&HP=H(P_{1}P_{2})=(HP_{1})P_{2}=(P_{1}H)P_{2}=\\&P_{1}(HP_{2})=P_{1}(P_{2}H)=(P_{1}P_{2})H=PH. \end{aligned}$$

• (3) See the proof of (2) and use Lemma 2.2.

• (4) Let $p$ be a prime such that $p\not \in \pi$. Then by (3) there is a Sylow $p$-subgroup $P$ of $G$ such that $HP=PH$ is a subgroup of $G$. Hence $HP\cap N=H\cap N$ is a normal subgroup of $HP$. Thus, $p$ does not divide $|G:N_{G}(H\cap N)|$.$\square$

Lemma 2.4

Let $H$ and $B$ be subgroups of $G$. If $G=N_{G}(H)B$ and $HV^{b}=V^{b}H$ for some subgroup $V$ of $B$ and for all $b\in B$, then $HV^{x}=V^{x}H$ for all $x\in G$.

Proof

Since $G=N_{G}(H)B$ we have $x=bn$ for some $b\in B$ and $n\in N_{G}(H)$. Hence $HV^{x}= HV^{bn}=Hn(V^{b})n^{-1}= n(V^{b})n^{-1}H=V^{x}H$.$\square$

Lemma 2.5

Suppose that for subgroups $A$ an $B$ of $G$ we have $AB=BA$ and $G=N_{G}(A)B$. Then

• (1) $A^{G}=A(A^{G}\cap B)$.

• (2) If $A$ permutes with all Sylow $p$-subgroups of $B$, then $A$ permutes with all Sylow $p$-subgroups of $A^{G}\cap B$.

Proof

• (1) Since $AB=BA$, $AB$ is a subgroup of $G$ and so $A^{G}=A^{N_{G}(A)B}=A^{B}\le \langle A, B \rangle = AB$. Hence $A^{G}=A^{G}\cap AB=A(A^{G}\cap B)$.

• (2) By (1) we have $A^{G}=A(A^{G}\cap B)$. Let $P$ be any Sylow $p$-subgroup of $A^{G}\cap B$ and $P\le B_{p}$, where $B_{p}$ is a Sylow of $B$. Then $AB_{p}=B_{p}A$ and $P=A^{G}\cap B \cap B_{p}=A^{G} \cap B_{p}$. Hence $AB_{p}\cap A^{G}=A(B_{p}\cap A^{G})=AP=PA$.$\square$

Lemma 2.6

(See Kegel [12]) Let $A$ and $B$ be subgroups of $G$ such that $G\ne AB$ and $AB^{x}=B^{x}A$, for all $x\in G$. Then $G$ has a proper normal subgroup $N$ such that either $A\le N$ or $B\le N$.

In our proofs we shall need the following well-known properties of supersoluble and $p$-supersoluble groups.

Lemma 2.7

Let $N$ and $R$ be normal subgroups of $G$.

• (1) If $N\le \Phi (G)\cap R$ and $R/N$ is $p$-supersoluble, then $R$ is $p$-supersoluble.

• (2) If $G$ is $p$-supersoluble and $O_{p'}(G)=1$, then $p$ is the largest prime dividing $|G|$, $G$ is supersoluble and $F(G)=O_{p}(G)$ is a normal Sylow $p$-subgroup of $G$.

• (3) If $G$ is supersoluble, then $G'\le F(G)$.

Lemma 2.8

(See Knyagina and Monakhov [13]) Let $H$, $K$, and $N$ be subgroups of $G$. If $N$ is normal in $G$, $H$ permutes with $K$ and $H$ is a Hall subgroup of $G$, then

$$\begin{aligned} N\cap HK=(N\cap H)(N\cap K). \end{aligned}$$

We use $\mathcal{A}(p-1)$ to denote the class of all abelian groups of exponent dividing $p-1$. The symbol $Z_\mathcal{U}(G)$ denotes the product of all normal subgroups $N$ of $G$ such that every chief factor of $G$ below $N$ is cyclic.

Lemma 2.9

(See Lemma 2.2 in [21]) Let $E$ be a normal $p$-subgroup of a group $G$. If $E\le Z_\mathcal{U}(G)$ (if $E\le Z_{\infty }(G)$), then

$$\begin{aligned} (G/C_{G}(E))^{\mathcal{A}(p-1)}\le O_{p}(G/C_{G}(E)) \end{aligned}$$

($G/C_{G}(E)$ is a $p$-group, respectively).

Proof

See the proof of Lemma 2.2 in [21].$\square$

Lemma 2.10

Suppose that $G$ is $p$-soluble and $O_{p'}(G)=1$. Then $F^{*}(G)=O_{p}(G).$

Proof

It is clear that $F(G)=O_{p}(G)\le F^{*}(G)$. Suppose that $O_{p}(G)\ne F^{*}(G)$ and let $H/O_{p}(G)$ be a chief factor of $G$ below $F^{*}(G)$. Then, since $G$ is $p$-soluble, $H/O_{p}(G)$ is a non-abelian $p'$-group and $O_{p}(G)\le Z_{\infty }(H)$ by [10], Chap. X, Theorems 13.6 and 13.7]. Hence $H/C_{H}(O_{p}(G))$ is a $p$-group by Lemma 2.9. On the other hand, by the Schur-Zassenhaus theorem, $O_{p}(G)$ has a complement $E$ in $H$. Then $E\le C_{H}(O_{p}(G))$, which implies that $E$ is normal in $H$. Thus, $E$ is a characteristic subgroup of $E$, so $E\le O_{p'}(G)=1$, a contradiction.$\square$

Lemma 2.11

(See Lemma 2.15 in [7]) Let $E$ be a normal non-identity quasinilpotent subgroup of $G$. If $\Phi (G)\cap E=1$, then $E$ is the direct product of some minimal normal subgroups of $G$.

Let $\mathcal F$ be a class of groups. A chief factor $H/K$ of $G$ is called $\mathcal{F}$ -central in $G$ provided $(H/K)\rtimes (G/C_{G}(H/K))\in \mathcal F$.

Lemma 2.12

(See Theorem B in [21]) Let $\mathcal F$ be any formation and $E$ a normal subgroup of $G$. If each chief factor of $G$ below $F ^{*} (E)$ is $\mathcal F$-central in $G$, then each chief factor of $G$ below $E$ is $\mathcal F$-central in $G$ as well.

Lemma 2.13

(See Lemma 3.3 in [7]) Let $\mathcal F$ be a solubly saturated formation containing all supersoluble groups and $E$ a normal subgroups of $G$ with $G/E\in \mathcal{F}$. If every chief factor of $G$ below $E$ is cyclic, then $G\in \mathcal F$.

Recall that $G$ is called a Schmidt group provided $G$ is not nilpotent but every proper subgroup of $G$ is nilpotent. We shall need in our proofs the following facts on Schmidt groups.

Lemma 2.14

(See Theorem 25.4 in [16]) Let $G$ be a Schmidt group. Then

• (a) $G=P\rtimes Q$, where $P$ is a Sylow $p$-subgroup of $G$ of exponent $p$ or exponent $4$ (if $P$ is a non-abelian $2$-group), $Q$ is a Sylow $q$-subgroup of $G$ for some primes $p\ne q$.

• (b) $P/\Phi (P)$ is a chief factor of $G$ and $C_{G}(P/\Phi (P))\ne G$.

Lemma 2.15

Let $E$ be a normal subgroup of $G$ and $P$ a Sylow $p$-subgroup of $E$ such that $(p-1, |G| )=1$. If either $P$ is cyclic or $G$ is $p$-supersoluble, then $E$ is $p$-nilpotent and $E/O_{p'}(E)\le Z_{\infty }(G/O_{p'}(E))$.

Proof

Let $H/K$ be any chief factor of $G$ such that $O_{p'}(E)\le K < H\le E$. Then $|H/K|=p$, so $G/C_{G}(H/K)$ divides $p-1$. But by hypothesis, $(p-1, |G| )=1$. Hence $C_{G}(H/K)=G.$ Thus, $E/O_{p'}(E)\le Z_{\infty }(G/O_{p'}(E))$.$\square$

Lemma 2.16

Let $P$ be a normal $p$-subgroup of $G$. If $P/\Phi (P)\le Z_{\mathcal{U}}(G/\Phi (P))$, then $P \le Z_{\mathcal{U}}(G)$.

Proof

Let $C\!=\!C_{G}(P)$, $H/K$ any chief factor of $G$ below $P$. Then $O_{p}(G/C_{G}(H/K))$ $=1$ by [23], Appendix C, Corollary 6.4]. Suppose that $P/\Phi (P)\le Z_{\mathcal{U}}(G/\Phi (P))$. Then by Lemma 2.9, $(G/C_{G}(P/\Phi (P)))^{\mathcal{A}(p-1)}$ is a $p$-group. Hence $(G/C)^ {\mathcal{A}(p-1)}$ is a $p$-group by [5], Chap. 5, Theorem 1.4 ]. Thus, $G/C_{G}(H/K)\in \mathcal{A}(p-1)$ and so $|H/K|=p$ by [23], Chap. 1, Theorem 1.4]. This implies that $P \le Z_{\mathcal{U}}(G)$.$\square$

Lemma 2.17

(See Corollary 1.11 in [7]) Let $N$ be a normal soluble subgroup of $G$. Then $F^{*}(G/\Phi (N))=F^{*}(G)/\Phi (N)$.

Lemma 2.18

(See Theorem A* in [8]) Let $H$ be a Hall $\pi$-subgroup of $G$. Let $G=HT$ for some subgroup $T$ of $G$, and $q$ a prime. If $H$ permutes with every Sylow $p$-subgroup of $T$ for all primes $p\ne q$, then $T$ contains a complement of $H$ in $G$ and any two complements of $H$ in $G$ are conjugate.

Lemma 2.19

Let $A$ and $B$ be subgroups of $G$. If $A^{x}B=BA^{x}$ for all $x\in G$, then $AB^{x}=B^{x}A$ for all $x\in G$.

Proof

Indeed, from $A^{x^{-1}}B=BA^{x^{-1}}$ we get $AB^{x}= (A^{x^{-1}}B)^{x}=(BA^{x^{-1}})^{x}= B^{x}A$.$\square$

A group $G$ is said to be $\pi$-closed ($p$-closed) provided $G$ has a normal Hall $\pi$-subgroup (a normal Sylow $p$-subgroup, respectively).

Lemma 2.20

(See Corollary 1.7 in [7]) Let $N$ and $R$ be normal subgroups of $G$. If $N\le \Phi (G)\cap R$ and $R/N$ is $\pi$-closed, then $R$ is $\pi$-closed

3 Proofs of Theorems A, B and C

Proof of Theorem C

Suppose that this theorem is false and let $G$ be a counterexample with $|G|+ |E|$ minimal. Suppose that there is a non-identity $p$-soluble normal subgroup $N$ of $G$ such that $N\le E$. If $P\le N$, then $G/N$ is a $p'$-group and so the $p$-solubility of $N$ implies the $p$-solubility of $E$. On the other hand, if $P\nleq N$, then the hypothesis holds for $G/N$ by Lemma 2.3 (1). Hence $E/N$ is $p$-soluble by the choice of $(G, E)$ since $|G/N| < |G|$. Therefore, $E$ is $p$-soluble. But this contradicts the choice of $(G, E)$. Hence every non-identity normal subgroup $N$ of $G$ contained in $E$ is not $p$-soluble.

By hypothesis there is a subgroup $B$ of $G$ such that $G=N_{G}(P)B$ and $P$ permutes with all Sylow subgroups of $B$. We shall show that $E=P^{G}=G=PB$. Indeed, by Lemma 2.5, $P^{G}=P(P^{G}\cap B)$ and $P$ permutes with all Sylow subgroups of $P^{G}\cap B$. Hence $P$ is $S$-propermutable in $P^{G}$. If $P^{G}\ne G$, then $P^{G}$ is $p$-soluble by the choice of $(G, E)$ since $P^{G}\le E$. Therefore, $G$ has a non-identity $p$-soluble normal subgroup, a contradiction. Thus, $E=P^{G}=G=PB$.

Let $Q$ be any Sylow $q$-subgroup of $B$ such that $q\ne p$. Then $p$ divides $|Q^{G}|$ and $P_{0}=P\cap Q^{G}$ is a Sylow $p$-subgroup of $Q^{G}$. We show that the hypothesis holds for $(Q^{G}, P_{0})$. Indeed, let $R$ be a Sylow $r$-subgroup of $Q^{G}\cap B$, where $r\ne p$. Then for some Sylow $r$-subgroup $B_{r}$ of $B$ we have

$$\begin{aligned} R=B_{r}\cap (Q^{G}\cap B)=B_{r}\cap Q^{G}. \end{aligned}$$

By Lemma 2.8 we also know that

$$\begin{aligned} PB_{r}\cap Q^{G}= (P\cap Q^{G})(B_{r}\cap Q^{G})=P_{0}R=RP_{0}. \end{aligned}$$

Therefore, $P_{0}$ is $S$-propermutable in $Q^{G}$. But since $G$ has no non-identity $p$-soluble normal subgroups, the choice of $(G, E)$ implies that $Q^{G}=G$. Note that by Burnside’s $p^{a}q^{b}$-theorem we have $PQ\ne G$. On the other hand, by Lemma 2.4, $PQ^{x}=Q^{x}P$ for all $x\in G$ and so by Lemma 2.6, $P^{G}\ne G$. This contradiction completes the proof of the result.$\square$

Proof of Theorem A

(I) Suppose that this assertion is false and let $G$ be a counterexample with $|G|+ |E|$ minimal. Let $V\in \mathcal{M}_{\phi }(P)$. By hypothesis there is a subgroup $B$ of $G$ that $G=N_{G}(V)B$ and $V$ permutes with all Sylow $q$-subgroups of $B$.

• (1) $V^{G}= V(V^{G}\cap B)$ and $V$ permutes with every Sylow $q$ -subgroup of $V^{G}\cap B$ for all primes $q$ dividing $|V^{G}\cap B|$ (This directly follows from Lemma 2.5).

• (2) $O_{p'}(N)=1$ for every normal subgroup $N$ of $G$ contained in $E$.

Suppose that for some normal subgroup $N$ of $G$ contained in $E$ we have $O_{p'}(N)\ne 1$. Since $O_{p'}(N)$ is a characteristic subgroup of $N$, it is normal in $G$. On the other hand, by Lemma 2.3 (1), the hypothesis holds for $(G/O_{p'}(N), E /O_{p'}(N))$. Hence $E/O_{p'}(N)$ is $p$-supersoluble by the choice of $(G, E)$. Thus, $E$ is $p$-supersoluble, a contradiction.

• (3) If $L$ is a minimal normal subgroup of $G$, then $L\nleq \Phi (P)$.

Indeed, in the case, where $L\le \Phi (P)$, we have $L\le \Phi (E)$ and the hypothesis holds for $(G/L, E/L)$ by Lemma 2.3 (1). Hence $E/L$ is $p$-supersoluble by the choice of $(G, E)$. Therefore, $E$ is $p$-supersoluble by Lemma 2.7 (1), which contradicts to our assumption on $E$.

• (4) If $D$ is a normal $p$ -soluble subgroup of $G$ contained in $E$, then $D$ is supersoluble and $p$ -closed.

By (2), $O_{p'}(D)=1$. Therefore, $O_{p}=O_{p}(D)\ne 1$. Let $N$ be a minimal normal subgroup of $G$ contained in $O_{p}$. In view of (3) we have $N\nleq \Phi (P)$. Hence for some subgroup $W\in \mathcal{M}_{\phi }(P)$ we have $P=NW$. Let $S=N\cap W$. Then $S$ is normal in $P$. On the other hand, by Lemma 2.3 (4), $|G:N_{G}(S)|$ is a power of $p$. Hence $|E:N_{E}(S)|=|E:N_{G}(S)\cap E|=|EN_{G}(S):N_{G}(S)|$ is a power of $p$. Thus, $S$ is normal in $E$. By Proposition 4.13 (c) in [4], Chap. A], $N=N_{1}\times \ldots \times N_{t}$, where $N_{1}, \ldots , N_{t}$ are minimal normal subgroups of $E$, and from the proof of this proposition we know also that $|N_{i}|=|N_{j}|$ for all $i, j$. Therefore, there is a minimal normal subgroup $L$ of $E$ such that $N=SL$ and $S\cap L=1$. Hence $P=L\rtimes W$, which implies by Gaschütz’s theorem [9], Chap. I, Satz 17.4] that $L$ has a complement $M$ in $E$. Thus, $N\nleq \Phi (E)$ and $N_{1}, \ldots , N_{t}$ are groups of order $p$. It is clear that $\Phi (E)\cap O_{p}$ is normal in $G$. Therefore, $\Phi (E)\cap O_{p}= 1$. Hence $O_{p}=L_{1}\times \ldots \times L_{t}$, where $L_{1}, \ldots , L_{t}$ are minimal normal subgroups of $E$ by Lemma 2.11. If for some $i$ we have $L_{i}\nleq \Phi (P)$, then, as above, one can show that $|L_{i}|=p$. Therefore, there are normal subgroups $F$ and $M$ of $E$ such that $O_{p}=FM$, every chief factor of $E$ below $M$ is cyclic and $F\le \Phi (P)\le \Phi (E)$. Now consider $D/F$. It is clear $O_{p}(D/F)=O_{p}/F=MF/F$. On the other hand, by Lemma 2.20, $O_{p'}(D/F)=1$ since $O_{p'}(D)=1$. Therefore, by Lemma 2.10, $F^{*}(D/F)=O_{p}/F$, where every chief factor of $D/F$ below $F^{*}(D/F)$ is cyclic. Hence $D/F$ is supersoluble, so $D$ is supersoluble by Lemma 2.7 (1). But $O_{p'}(D)=1$, so $O_{p}$ is a Sylow $p$-subgroup of $D$ by Lemma 2.7 (2).

• (5) $E$ is $p$ -soluble.

Assume that $E$ is not $p$-soluble.

• (a) If $O_{p}(E)\ne 1$, then $P$ is not cyclic.

Suppose that $P$ is cyclic. Let $L$ be a minimal normal subgroup of $G$ contained in $O_{p}(E)\le P$. Suppose that $C_{E}(L)=E$, so $L\le Z(E)$. Let $N=N_{E}(P)$. If $P\le Z(N)$, then $E$ is $p$-nilpotent by Burnside’s theorem [9], Chap. IV, Satz 2.6], which contradicts the choice of $(G, E)$. Hence $N\ne C_{E}(P)$. Let $x\in N\backslash C_{E}(P)$ with $(|x|, |P|)=1$ and $K=P\rtimes \langle x \rangle$. By [9], Chap. III, Satz 13.4], $P=[K, P]\times (P\cap Z(K))$. Since $L\le P\cap Z(K)$ and $P$ is cyclic, it follows that $P= P\cap Z(K)$ and so $x\in C_{K}(P)$. This contradiction shows that $C_{E}(L)\ne E$.

Since $P$ is cyclic, $|L|=p$. Hence $G/C_{G}(L)$ is a cyclic group of order dividing $p-1$. If $|P/L| >p$, then the hypothesis holds for $(G/L, E/L)$ by Lemma 2.3 (1). Hence $E/L$ is $p$-supersoluble by the choice of $(G, E)$ and so $E$ is $p$-soluble, a contradiction. Thus, $|P/L|=p$ and hence $V=L$ is normal in $G$. Therefore, the hypothesis holds for $(G, C_{E}(L))$, so $C_{E}(L)$ is $p$-supersoluble since $C_{E}(L)\ne E$. But then $E$ is $p$-soluble since $E/C_{E}(L)=E/ E\cap C_{G}(L)\simeq EC_{G}(L)/C_{G}(L)$ is cyclic. This contradiction shows that we have (a).

• (b) If $P\nleq V^{G}$, then $V$ is normal in $G$.

Indeed, since $P\nleq V^{G}\le E$, $V$ is a Sylow $p$-subgroup of $V^{G}$. On the other hand, by (1) we have $V^{G}=V(V^{G}\cap B)$ and $V$ is $S$-propermutable in $V^{G}$. Therefore, $V^{G}$ is $p$-soluble by Theorem C. Thus, $V$ is normal in $V^{G}$ by (4). Since $V$ is a Sylow $p$-subgroup of $V^{G}$, $V$ is characteristic in $V^{G}$. Hence $V=V^{G}$ is normal in $G$.

• (c) $P$ is not cyclic.

Suppose that $P$ is cyclic. Then $\mathcal{M}_{\phi }(P)=\{V \}$, and by (1), (a) and (b) we have $P\le V^{G}=V(V^{G}\cap B)$ and $V$ permutes with every Sylow $q$-subgroup of $V^{G}\cap B$ for all primes $q$ dividing $|V^{G}\cap B|$. Hence the hypothesis holds for $(V^{G}, V^{G})$. Assume that $V^{G}\ne G$. Then $V^{G}$ is $p$-supersoluble by the choice of $(G, E)$. Hence by (4), $P$ is normal in $G$, which contradicts (a). Therefore, $V^{G}=G$, which implies that $G=VB$ by (1). Hence $P=P\cap VB=V(P\cap B)$, so $P\le B$ since $P$ is cyclic. Therefore, $B=G$, so $V$ is $S$-permutable in $G$. Hence $V\le P_{E}\le O_{p}(E)$, which contradicts (a). Hence $P$ is not cyclic.

• (d) $P$ permutes with every Sylow $q$ -subgroup $Q$ of $P^{G}$ for all primes $q\ne p$ dividing $|P^{G}|$.

Let $D=P^{G}$. In view (c), there is a subgroup $W\in \mathcal{M}_{\phi }(P)$ such that $V\ne W$. Then $P=VW$. Hence in view of Lemma 2.2 we have only to show that $V$ and $W$ permute with $Q$. In view of (b) we may suppose that $P\le V^{G}$ and $P\le W^{G}$. Then $D=P^{G}\le V^{G}$ and so by (1), $D=V(D\cap B)$ and $V$ permutes with every Sylow $q$-subgroup $Q_{1}$ of $D\cap B$. It is also clear that $Q_{1}$ is a Sylow $q$-subgroup of $D$. Therefore, for some $x\in D$ we have $Q_{1}= Q^{x}$. Hence $V$ permutes with $Q$ by Lemma 2.4. Similarly, it may be proved that $W$ permutes with $Q$.

Final contradiction for (5). By (d) and Lemma 2.18, $P^{G}$ has a Hall $p'$-subgroup. Hence by (d), $P$ is $S$-propermutable in $P^{G}$. Therefore, by Theorem C, $P^{G}$ is $p$-soluble. Hence by (4), $P$ is normal in $G$. Therefore, $E$ is $p$-soluble. This contradiction completes the proof of (5).

By (5), $E$ is $p$-soluble. Hence $E$ is supersoluble by (4). This contradiction completes the proof of (I).

(II) Suppose that this assertion is false and let $G$ be a counterexample with $|G|+ |E|$ minimal. Let $Z=Z_\mathcal{U}(G)$. First we show that $O_{p'}(E) =1$. Indeed, suppose that $O_{p'}(E)\ne 1.$ It is clear that $O_{p'}(E)$ is normal in $G$. Moreover, the hypothesis holds for $(G/O_{p'}(E), E/O_{p'}(E))$ by Lemma 2.3 (1). Therefore, every chief factor of $G/O_{p'}(E)$ below $E/O_{p'}(E)$ is cyclic by the choice of $(G, E)$. Hence every chief factor of $G$ between $E$ and $O_{p'}(E)$ is cyclic, a contradiction. Thus, $O_{p'}(E) =1$.

By (I), $E$ is $p$-supersoluble. Hence by Lemma 2.7 (2), $E$ is supersoluble and $P=F(E)$. Hence the hypothesis is true for $(G, P)$. If $P\ne E$, then every chief factor of $G$ below $P$ is cyclic by the choice of $(G, E)$. Hence every chief factor of $G$ below $E$ is cyclic by Lemma 2.12, contrary to the choice of $(G, E)$. Hence $P=E$.

Let $N$ be any minimal normal subgroup of $G$ contained in $P$. Then the hypothesis holds for $(G/N, P/N)$, so every chief factor of $G/N$ below $P/N$ is cyclic by the choice of $(G, E)$. Thus, $|N| > p$. Moreover, $N\nleq \Phi (P)$, otherwise every chief factor of $G$ below $P$ is cyclic by Lemma 2.16. Thus, $\Phi (P)=1$ and so $P$ is elementary abelian $p$-group. Let $W$ be a maximal subgroup of $N$ such that $W$ is normal in a Sylow $p$-subgroup $G_{p}$ of $G$. Let $V=WS$, where $S$ is a complement of $N$ in $P$. Then $W= V\cap N$ and $V$ is $S$-propermutable in $G$ by hypothesis. Hence by Lemma 2.3 (4), $G=G_{p}N_{G}(W)$. Therefore, $W$ is normal in $G$, so $W=1$. This contradiction completes the proof of Assertion (II).

Theorem is proved.$\square$

Proof of Theorem B

First we assume that $X=E$. Suppose that in this case the theorem is false and consider a counterexample $(G,E)$ for which $|G| +|E|$ is minimal. Let $p$ be the smallest prime dividing $|E|$ and $P$ a Sylow $p$-subgroup of $E$. Then $E$ is $p$-nilpotent by Lemma 2.15 and Theorems A. Let $V$ be the normal Hall $p'$-subgroup of $E$. Since $V char E\lhd G$, $V$ is normal in $G$. Moreover, the hypothesis holds for $(G, V)$ and for $(G/V, E/V)$ by Lemma 2.3 (1). Hence in the case when $V\ne 1$ we have $V\le Z_\mathcal{U}(G)$ and $E/V\le Z_\mathcal{U}(G/V)$ by the choice of $(G, E)$. This induces that $E\le Z_\mathcal{U}(G)$, a contradiction. Therefore, $E=P$ and consequently $E\le Z_\mathcal{U}(G)$ by Theorem A.

Finally, if $X=F^{*}(E)$, then as above we have $F^{*}(E)\le Z_\mathcal{U}(G)$. Therefore, $E\le Z_\mathcal{U}(G)$ by Lemma 2.12.$\square$

4 Some Applications of Theorem A and Corollary 1.4

In the literature one can find many special cases of Theorem A and Corollary 1.4. Here we discuss only some of them.

From Theorem A and Lemma 2.15 we get

Corollary 4.1

(See Theorem 1.1 in [18]) Let $P$ be a Sylow subgroup of $G$, where $p$ is the smallest prime dividing $|G|$. If every number $V$ of some fixed $\mathcal{M}_{\phi }(P)$ is $SS$-quasinormal in $G$, then $G$ is $p$-nilpotent.

Corollary 4.2

Let $P$ be a Sylow subgroup of $G$. If $N_{G}(P)$ is $p$-nilpotent and every number $V$ of some fixed $\mathcal{M}_{\phi }(P)$ is $S$-propermutable in $G$, then $G$ is $p$-nilpotent.

Proof

If $|P|=p$, then $G$ is $p$-nilpotent by Burnside’s theorem [9], IV, 2.6]. Otherwise, $G$ is $p$-supersoluble by Theorem A. The hypothesis holds for $G/O_{p'}(G)$ by Lemma 2.3(1), so in the case, where $O_{p'}(G)\ne 1$, $G/O_{p'}(G)$ is $p$-nilpotent by induction. Hence $G$ is $p$-nilpotent. Therefore, we may assume that $O_{p'}(G)=1$. But then, by Lemma 2.7(2), $P$ is normal in $G$. Hence $G$ is $p$-nilpotent by hypothesis.$\square$

From Corollary 4.2 we get

Corollary 4.3

(See Theorem 1.2 in [18]) Let $P$ be a Sylow subgroup of $G$. If $N_{G}(P)$ is $p$-nilpotent and every number $V$ of some fixed $\mathcal{M}_{\phi }(P)$ is $SS$-quasinormal in $G$, then $G$ is $p$-nilpotent.

Corollary 4.4

Let $P$ be a Sylow subgroup of $G$. If $G$ is $p$-soluble and every number $V$ of some fixed $\mathcal{M}_{d}(P)$ is $S$-propermutable in $G$, then $G$ is $p$-supersoluble.

Proof

In the case, when $|P|=p$, this directly follows from the $p$-solubility of $G$. If $|P| > p$, this corollary follows from Theorem A.$\square$

The next fact follows from Corollary 4.4.

Corollary 4.5

(See Theorem 1.3 in [18]) Let $P$ be a Sylow subgroup of $G$. If $G$ is $p$-soluble and every number $V$ of some fixed $\mathcal{M}_{\phi }(P)$ is $SS$-quasinormal in $G$, then $G$ is $p$-supersoluble.

Corollary 4.6

If, for every prime $p$ dividing $|G|$ and $P\in Syl _{p}(G)$, every number $V$ of some fixed $\mathcal{M}_{\phi }(P)$ is $S$-propermutable in $G$, then $G$ is supersoluble.

Proof

Let $p$ be the smallest prime dividing $|G|$. Then $G$ is $p$-nilpotent by Corollary 4.1, so $G$ is soluble by Fait-Thompson’s theorem. Hence $G$ is supersoluble by Corollary 4.4.$\square$

From Corollary 4.6 we get

Corollary 4.7

(See Theorem 1.4 in [18]) If, for every prime $p$ dividing $|G|$ and $P\in Syl _{p}(G)$, every number $V$ of some fixed $\mathcal{M}_{\phi }(P)$ is $SS$-quasinormal in $G$, then $G$ is supersoluble.

The formation $\mathcal{F}$ is said to be saturated if $G\in \mathcal{F}$ whenever $G/\Phi (G) \in \mathcal{F}$. It is clear that every saturated formation is soluble saturated. Hence from Corollary 1.4 we get

Corollary 4.8

Let $\mathcal F$ be a saturated formation containing all supersoluble groups and $X\le E$ normal subgroups of $G$ such that $G/E \in \mathcal{F}$. Suppose that every maximal subgroup of any non-cyclic Sylow subgroup of $X$ is $S$-propermutable in $G$. If either $X=E$ or $X=F^{*}(E)$, then $G \in \mathcal{F}$.

The following results are special cases of Corollary 4.8.

Corollary 4.9

(See Theorem 1.5 in [18]) Let $\mathcal F$ be a saturated formation containing all supersoluble groups and $E$ a normal subgroup of $G$ such that $G/E \in \mathcal{F}$. Suppose that for every maximal subgroup of every non-cyclic Sylow subgroup of $E$ is $SS$-quasinormal in $G$. Then $G \in \mathcal{F}$.

Corollary 4.10

(See Theorem 3.2 in [19]) Let $E$ a normal subgroup of $G$ such that $G/E$ is supersoluble. Suppose that for every maximal subgroup of every Sylow subgroup of $F^{*}(E)$ is $SS$-quasinormal in $G$. Then $G$ is supersoluble.

Corollary 4.11

(See Theorem 3.3 in [19]) Let $\mathcal F$ be a saturated formation containing all supersoluble groups and $E$ a normal subgroup of $G$ such that $G/E \in \mathcal{F}$. Suppose that for every maximal subgroup of every Sylow subgroup of $F^{*}(E)$ is $SS$-quasinormal in $G$. Then $G \in \mathcal{F}$.

Corollary 4.12

(See Theorem 3.2 in [14]) Let $\mathcal F$ be a saturated formation containing all supersoluble groups and $E$ a normal subgroup of $G$ such that $G/E \in \mathcal{F}$. If all maximal subgroups of $F^{*}(E)$ are $S$-permutable in $G$, then $G\in \mathcal F$.