Finite Groups with Given Systems of m-S-Complemented Subgroups

Abstract

Let G be a finite group and H a subgroup of G. We say that H: is generalized S-quasinormal in G if $H=\left\langle A,B\right\rangle$ for some modular subgroup A and S-quasinormal subgroup B of G; m-S-complemented in G if there are a generalized S-quasinormal subgroup S and a subgroup T of G such that $G=HT$ and $H\cap T\le S\le H$. In this paper, we study finite groups with given systems of m-S-complemented subgroups. In particular, we prove that if ${\mathfrak {F}}$ is a saturated formation containing all supersoluble groups and E is a normal subgroup of a finite group G such that $G/E\in {\mathfrak {F}}$ and for every non-cyclic Sylow subgroup P of E every maximal subgroup of P not having a nilpotent supplement in G is m-S-complemented in G, then $G\in {\mathfrak {F}}$ .

1 Introduction

Throughout this paper, all groups are finite and G always denotes a finite group. Moreover, $\pi (G)$ is the set of all primes dividing the order |G| of G; $C_{n}$ denotes a cyclic group of order n.

A subgroup M of G is called modular in G [1, p. 43] if (1) $\langle X,M \cap Z \rangle =\langle X, M \rangle \cap Z$ for all $X \le G, Z \le G$ such that $X \le Z$, and (2) $\langle M, Y \cap Z \rangle =\langle M, Y \rangle \cap Z$ for all $Y \le G, Z \le G$ such that $M \le Z$.

A subgroup H of G is said to be S-permutable [2, 3] or S-quasinormal [4] in G if H permutes with every Sylow subgroup P of G, that is, $HP=PH$. The subgroup H of G is said to be generalized S-quasinormal in G [5] if there are a modular subgroup A and an S-quasinormal subgroup B of G such that $H=\langle A, B\rangle$.

Interesting applications of generalized S-quasinormal subgroups were discussed in the paper [5]. In this paper, we consider the following generalization of such subgroups.

Definition 1.1

We say that a subgroup H of G is m-S-complemented in G if there are a generalized S-quasinormal subgroup S and a subgroup T of G such that $G=HT$ and $H\cap T\le S\le H$.

It is clear that every generalized S-quasinormal subgroup is m-S -complemented. Every modular subgroup and every S-quasinormal subgroup are generalized S-quasinormal. Now consider the following

Example 1.2

(1) Let $C_{3}\wr A_{4}=P\rtimes A_{4}$, where $A_{4}$ is the alternating group of degree 4 and P is the base group of the regular wreath product $C_{3}\wr A_{4}$. Let $G=(P\rtimes A_{4})\times (C_{11}\rtimes C_{5})$, where $C_{11}\rtimes C_{5}$ is a non-abelian group of order 55. Let Q be the Sylow 2-subgroup of $A_{4}$ and R a Sylow 3-subgroup of $A_{4}$. Then, PQ is supersoluble, so some subgroup B of P with $|B|=3$ is normal in PQ. Then, for every Sylow 3-subgroup $G_{3}$ of G we have $B\le P\le G_{3}$, so $BG_{3}=G_{3}=G_{3}B$. On the other hand, for every Sylow 2-subgroup $Q^{x}$ of G we have $Q^{x}\le PQ$, so $BQ^{x}=Q^{x}B$. Hence, B is S-quasinormal in G. In view of [1, Theorem 5.1.9], $A=C_{5}$ is modular in G. Then, $S=\langle A, B\rangle =AB$ is generalized S-quasinormal in G.

Now let $H=(AB)Q=A\times BQ$ and $T=PRC_{11}$. Then, $G= HT$ and $H\cap T=(AB)Q\cap PRC_{11}=B(AQ\cap PRC_{11})=B\le H$. Hence, H is m-S-complemented in G.

Next, we show that H is not generalized S-quasinormal in G. First note that $H_{G}=1$, so for every modular subgroup V of H we have $V^{G}\le C_{11}\rtimes C_{5}$ by Lemma 2.4 below. Therefore, A is the largest modular subgroup of H. Assume that H is generalized S-quasinormal in G and let W be an S-quasinormal subgroup of G such that $H= \langle A, W\rangle =AW$. Then, $W_{G}=1$, so W is a nilpotent subnormal subgroup of G by [2, Theorem 1.2.17]. Hence for a Sylow 2-subgroup $Q_{1}$ of W, we have $1 < Q_{1}\le O_{2}(G)\le P\rtimes (Q\rtimes C_{p})$ and so $Q_{1} \le C_{G}(P)$, a contradiction. Therefore, H is not generalized S-quasinormal in G.

(2) A subgroup H of G is said to be complemented (respectively, c-supplemented [6]) in G, if there is a subgroup T of G such that $G=HT$ and $H\cap T=1$ (respectively, $G=HT$ and $H\cap T\le H_{G}$). It is clear that every complemented subgroup and every c-supplemented subgroup are m-S-complemented.

(3) A subgroup H of G is said to be S-supplemented [7] (respectively, m-supplemented [8]) in G, if there are an S-quasinormal subgroup (respectively, a modular subgroup) S and a subgroup T of G such that $G=HT$ and $H\cap T\le S\le H$. Every S-supplemented subgroup and every m-supplemented subgroup are m-S-complemented.

Let $K\le H$ be normal subgroups of G. Then we say, following [1] that H / K is hypercyclically embedded in G if every chief factor of G between H and K is cyclic. We say also that H is hypercyclically embedded in G if H / 1 is hypercyclically embedded in G.

Hypercyclically embedded subgroups play an important role in the theory of soluble groups (see the books [1,2,3]) and the conditions under which a normal subgroup is hypercyclically embedded were found by many authors (see, for example, the recent papers [9,10,11,12,13,14,15,16,17]).

In this paper, we prove the following results in this line research.

Theorem 1.3

Let E be a normal subgroup of G and let P be a Sylow p-subgroup of E, where p is the smallest prime dividing |E| . If every maximal subgroup of P not having a p-nilpotent supplement in G is m-S-complemented in G, then $E/O_{p^{\prime }}(E)$ is hypercyclically embedded in G.

Theorem 1.4

Let E be a normal subgroup of G. Suppose that for any Sylow subgroup P of E every maximal subgroup of P not having a nilpotent supplement in G is m-S-complemented in G. Then, E is hypercyclically embedded in G.

Recall that the formation ${\mathfrak {F}}$ is a homomorph of groups such that each group G has the smallest normal subgroup (denoted by $G^{{\mathfrak {F}}}$) whose quotient is still in ${\mathfrak {F}}$. A formation ${\mathfrak {F}}$ is said to be saturated if $G\in {\mathfrak {F}}$ for any group G with $G/\Phi (G)\in {\mathfrak {F}}$.

As a first application of Theorem 1.4, we prove also the following theorem which covers many known results (see Sect. 4 below).

Theorem 1.5

Let ${\mathfrak {F}}$ be a saturated formation containing all supersoluble groups, and let $X\le E$ be normal subgroups of G with $G/E\in {\mathfrak {F}}$. Suppose that for any Sylow subgroup P of X every maximal subgroup of P not having a nilpotent supplement in G is m-S-complemented in G. If $X=E$ or $X=F^{*}(E)$, then $G\in {\mathfrak {F}}$.

In this theorem, $X=F^{*}(E)$ denotes the generalized Fitting subgroup of E [18, Ch. X], that is, the product of all normal quasinilpotent subgroups of E.

2 Preliminaries

The first lemma collects the properties of S-quasinormal subgroups used in our proofs.

Lemma 2.1

(See Chapter 1 in [2]). Let A, B and N be subgroups of G, where A is S-quasinormal in G and N is normal in G.

• (1) AN / N is S-quasinormal in G / N.

• (2) If $A \le B$, then A is S-quasinormal in B.

• (3) If $N\le B$ and B / N is S-quasinormal in G / N, then B is S-quasinormal in G.

• (4) A is subnormal in G and $A^{G}/A_{G}$ is nilpotent.

• (5) If B is S-quasinormal in G, then $A\cap B$ and $\langle A, B\rangle$ are S-quasinormal in G.

Lemma 2.2

Let A, B and N be subgroups of G, where A is generalized S-quasinormal in G and N is normal in G. Then

• (1) AN / N is generalized S-quasinormal in G / N.

• (2) If $A \le B$, then A is generalized S-quasinormal in B.

• (3) If $N \le B$ and B / N is generalized S-quasinormal in G / N, then B is generalized S-quasinormal in G.

• (4) If B is generalized S-quasinormal in G, then $\langle A, B \rangle$ is generalized S-quasinormal in G.

Proof

Let $A=\langle L, T \rangle$, where L is modular and T is S-quasinormal subgroups of G.

• (1) $AN/N=\langle LN/N, TN/N\rangle$, where LN / N is modular in G / N by Property (3) in [1, p. 201] and TN / N is S-quasinormal in G / N by Lemma 2.1(1). Hence, AN / N is generalized S-quasinormal in G / N.

• (2) This follows from Property (2) in [1, p. 201] and Lemma 2.1(2).

• (3) Let $B/N=\langle V/N, W/N\rangle$, where V / N is modular in G / N and W / N is S-quasinormal in G / N. Then, $B=\langle V, W\rangle$, where V is modular in G by Property (4) in [1, p. 201] and W is S-quasinormal in G by Lemma 2.1(3). Hence, B is generalized S-quasinormal in G.

• (4) This follows from Property (5) in [1, p. 201] and Lemma 2.1(5).$\square$

The lemma is proved.

Lemma 2.3

Let A, B and N be subgroups of G, where A is m-S-complemented in G and N is normal in G.

• (1) If either $N\le A$ or $(|A|, |N|)=1$, then AN / N is m-S -complemented in G / N.

• (2) If $A \le B$, then A is m-S-complemented in B.

• (3) If $N \le B$ and B / N is m-S-complemented in G / N, then B is m-S-complemented in G.

Proof

Let T be a subgroup of G such that $AT=G$ and $A \cap T \le S\le A$ for some generalized S-quasinormal subgroup S of G. Then, $S=\langle L, M \rangle$, where L is a modular and M is an S-quasinormal subgroups of G.

• (1) Note that $NT \cap NA=(T\cap A)N$. Indeed, if $N\le A$, then $NT \cap NA=NT \cap A=N(T \cap A)$. On the other hand, if $(|A|, |N|)=1$, then from $AT=G$ we get that $N\le T$ and so $NT \cap NA=T\cap AN =N(T\cap A)$. Therefore, $G/N=(AN/N)(TN/N)$ and

  $$\begin{aligned} (AN/N)\cap (TN/N)=(AN\cap TN/N)=(A\cap T)N/N\le SN/N, \end{aligned}$$

  where SN/N is a generalized S-quasinormal subgroup of G / N by Lemma 2.2(1). Hence, AN / N is m-S-supplemented in G / N.

• (2) $B=A(B \cap T)$ and $(B \cap T) \cap A = T \cap A \le S \le A$, where S is m-S-permutable in B by Lemma 2.2(2). Hence, A is m-S-complemented in B.

• (3) See the proof of (1) and use Lemma 2.2(3).$\square$

The lemma is proved.

Lemma 2.4

(See Theorem 5.2.5 in [1]). If H is a modular subgroup of G, then $H^{G}/H_{G}$ is hypercyclically embedded in G.

Lemma 2.5

(See Theorem 1.2 in [12]). If E is a normal subgroup of G and $F^{*}(E)$ is hypercyclically embedded in G, then E is hypercyclically embedded in G.

Lemma 2.6

(See Lemma 2.16 in [7]). Suppose that $G/N\in {\mathfrak {F}}$, where ${\mathfrak {F}}$ is a saturated formation containing all supersoluble groups. If N is hypercyclically embedded in G, then $G\in {\mathfrak {F}}$.

Lemma 2.7

(See Lemma 2.10 in [9]). Let P be a Sylow p-subgroup of G, where p is the smallest prime dividing |G|. If every maximal subgroup of P has a p-nilpotent supplement in G, then G is p-nilpotent.

Lemma 2.8

(See Lemma 2.12 in [19]). Let P be a normal p-subgroup of G. If $P/\Phi (P)$ is hypercyclically embedded in G, then P is hypercyclically embedded in G.

3 Proofs of Theorems 1.3, 1.4 and 1.5

The product of all hypercyclically embedded subgroups of G is denoted by $Z_{{\mathfrak {U}}}(G)$ and it is called the supersoluble hypercentre of G. Note that if A and B are normal hypercyclically embedded subgroups of G, then (in view of the G-isomorphism $AB/A\simeq B/(B\cap A))$ the product AB is also hypercyclically embedded in G.

Proof of Theorem 1.3

Suppose that this theorem is false and consider a counterexample (G, E) for which $|G|+|E|$ is minimal. Then, G is not supersoluble. Let $Z=Z_{{\mathfrak {U}}}(G)$.

• (1) IfRis a minimal normal subgroup ofGandRis either a$p^{\prime }$-group or ap-subgroup contained inEsuch that$R\ne P$, then the hypothesis holds for (G / R, ER / R).

First, we show that PR / R is a Sylow p-subgroup of ER / R. Indeed, $P\cap R$ is a Sylow p-subgroup of R and $PR/R\simeq P/(P\cap R)$ is a p-subgroup of ER / R. On the other hand, from

$$\begin{aligned} |(ER/R):(PR/R)|= & {} |ER:PR|=|E||R||P\cap R|:|E\cap R||P||R|\\= & {} |E||P\cap R|:|E\cap R||P| \end{aligned}$$

we get that |ER / R : PR / R| is a $p^{\prime }$- number since the minimality of R implies that we have either $R\cap E=1$ or $E\cap R=R$. Therefore PR / R is a Sylow p-subgroup of ER / R.

Now let V / R be a maximal subgroup of PR / R. Then, $V=(V\cap P)R$ and

$$\begin{aligned} p= & {} |(PR/R):(V/R)|= |PR:(V\cap P)R|\\= & {} |P||R||(V\cap P)\cap R|:|P\cap R||V\cap P||R| \\= & {} |P||V\cap R|:|P\cap R||V\cap P|. \end{aligned}$$

First, suppose that R is a $p^{\prime }$-group. Then, $p=|P||V\cap R|:|P\cap R||V\cap P|=|P:V\cap P|$, so $V\cap P$ is a maximal subgroup of P. Then, by hypothesis, either $V\cap P$ has a p-nilpotent supplement S in G or $V\cap P$ is m-S-complemented in G. In the first case, $SR/R\simeq S/(S\cap R)$ is a p-nilpotent supplement of $V/R=(V\cap P)R/R$ in G / R. In the second case, V / R is m-S-complemented in G / R by Lemma 2.3(1). Now suppose that R is a p-subgroup contained in E. Then, $R\le P$ and so $p=|(PR/R):(V/R)|= |P:V|$. Then, by hypothesis, either V has a p-nilpotent supplement S in G or V is m-S-complemented in G. Therefore, V / R has a p-nilpotent supplement SR / R in G / R or V / R is m-S-complemented in G / R by Lemma 2.3(1). Hence, the hypothesis folds for (G / R, ER / R).

• (2) IfH / Kis a chief factor ofEbelowEand$|H/K|=p$, then$C_{E}(H/K)=E$.

  Since p is the smallest prime dividing |E| by hypothesis, this follows from the fact that $E/C_{E}(H/K)\simeq V\le \text {Aut}(H/K)$ and from the fact that $\text {Aut}(A)$ is a cyclic group of order $p-1$ for any group A of order p.

• (3) IfRis a minimal normal subgroup ofGandRis either a$p^{\prime }$-group or ap-subgroup contained inEsuch that$R\ne P$, thenER / Risp-nilpotent and$(ER/R)/O_{p^{\prime }}(ER/R)$is hypercyclically embedded inG / R.

  The hypothesis holds for (G / R, ER / R) by Claim (1), so $(ER/R)/O_{p^{\prime }}(ER/R)$ is hypercyclically embedded in G / R by the choice of G. Therefore, ER / R is p-nilpotent by Claim (2).

• (4) $O_{p^{\prime }}(G)=1.$

  Assume that $O_{p^{\prime }}(G)\ne 1$ and let R be a minimal normal subgroup of G contained in $O_{p^{\prime }}(G)$. Then, $(ER/R)/O_{p^{\prime }}(ER/R)$ is hypercyclically embedded in G / R and $ER/R\simeq E/E\cap R$ is p-nilpotent by Claim (3). Hence, E is p-nilpotent and from

  $$\begin{aligned} (ER/R)/O_{p^{\prime }}(ER/R)=(ER/R)/(O_{p^{\prime }}(ER)/R)=(ER/R)/(O_{p^{\prime }}(E)R/R) \end{aligned}$$

  and from the G-isomorphisms

  $$\begin{aligned}&(ER/R)/(O_{p^{\prime }}(E)R/R)\simeq ER/O_{p^{\prime }}(E)R\simeq E/E\cap O_{p^{\prime }}(E)R \\&\quad =E/O_{p^{\prime }}(E)(E\cap R)=E/O_{p^{\prime }}(E) \end{aligned}$$

  we get that $E/O_{p^{\prime }}(E)$ is hypercyclically embedded in G, contrary to the choice of (G, E). Hence, we have (4).

• (5) $Z\cap E\le Z_{\infty }(E)$.

  Since Z is clearly supersoluble, a Sylow q-subgroup Q of Z, where q is the largest prime dividing |Z|, is normal and so characteristic in Z. Then, Q is normal in G, which implies that $Z=Q$ and $q=p$ by Claim (4), so $Z\cap E\le Z_{\infty }(E)\le P$ since p is the smallest prime dividing E by hypothesis.

• (6) $P\ne R$for each minimal normal subgroupRofG.

  Assume that $P=R$ and let V be any maximal subgroup of P. Then, by hypothesis, either V has a p-nilpotent supplement S in G or V is m-S-complemented in G. In the former case, we have $S\ne G$ since G is not p-nilpotent. On the other hand, in this case, we have $P=V(P\cap T)$ , where $P\cap T$ is clearly normal in G and so the minimality of $R=P$ implies that $P\cap T=1$. But then $V=P$. This contradiction shows that V is m-S-complemented in G, so there are an m-S-permutable subgroup S and a subgroup T of G such that $G=VT$ and $V\cap T\le S\le V$. Let A be a modular subgroup and B an S-quasinormal subgroup of G such that $S=\langle A, B\rangle$. Then, $A_{G}=1$, so $A^{G}\le Z$ by Lemma 2.4. Therefore, $A=1$ and so $S=B$ is S-quasinormal in G. But then S is normal in G by Lemma 1.2.16 in [2]. Hence, $S=1$ and so $T\cap V=1$ . But then $1< T\cap R < R$, where $T\cap R$ is normal in G. This contradiction shows that we have (6).

• (7) IfMis a proper subgroup ofGcontainingE, then$E/O_{p^{\prime }}(E)$is hypercyclically embedded inM. Hence$E=P$.

  Let V be a maximal subgroup of P. Then, either V has a p-nilpotent supplement S in G or V is m-S-complemented in G. In the former case, we have $M=V(M\cap S)$, so $M\cap S$ is a p-nilpotent supplement to V in G. In the second case, the subgroup V is m-S-complemented in M by Lemma 2.3(2). Hence, the hypothesis holds for (M, E), so $E/O_{p^{\prime }}(E)$ is hypercyclically embedded in M by the choice of G . Claim (2) implies that E is p-nilpotent. On the other hand, $O_{p^{\prime }}(E)$ is characteristic in E and so it is normal in G. Then, $O_{p^{\prime }}(E)\le O_{p^{\prime }}(G)=1$ by Claim (4). Therefore, E is supersoluble, which implies that a Sylow q-subgroup Q of E, where q is the largest prime dividing |E|, is normal and hence characteristic in E. Hence, $q=p$ and $E=P=Q$ by Claim (4).

• (8) Eisp-nilpotent.

  Assume that this is false. Then, $E\ne P$, so $E=G$ by Claim (7).

(a) $O_{p}(G)\ne 1$.

Assume that $O_{p}(G)=1$. Lemma 2.7 implies that some maximal subgroup V of P has no p-nilpotent supplement in G, so V is m-S-complemented in G. Then, there are a generalized S-quasinormal subgroup S and a subgroup T of G such that $G=VT$ and $V\cap T\le S\le V$. Let A be a modular subgroup and B an S-quasinormal subgroup of G such that $S=\langle A,B\rangle$. Then, $BP^{x}=P^{x}B=P^{x}$ for all $x\in G$, so $B\le P_{G}=O_{p}(G)=1$. Hence, $S=A$ and $A_{G}=1$; therefore, $S\le Z\le Z_{\infty }(G)$ by Lemma 2.4 and Claim (5) since $E=G$.

Since $Z_{\infty }(G)$ is nilpotent, a Sylow p-subgroup of $Z_{\infty }(G)$ is normal in G, so $A=S=1$ since $V_{G}=1$. Therefore, T is a complement to V in G, so for a Sylow p-subgroup $T_{p}$ of T we have $|T_{p}|=p$ . Therefore, T is p-nilpotent by [20, IV, 2.8]. Hence, every maximal subgroup V of P has a p-nilpotent complement in G, so G is p-nilpotent by Lemma 2.7. This contradiction shows that we have (a).

(b) $O_{p}(G)=C_{G}(O_{p}(G))$is a minimal normal subgroup ofGand$O_{p}(G)\nleq \Phi (G)$.

By Claim (a), $O_{p}(G)\ne 1$. Let R be a minimal normal subgroup of G contained in $O_{p}(G)$. Then, G / R is p-nilpotent by Claims (3) and (6). Hence, G is p-soluble. Therefore, every minimal normal subgroup R of G is a p-group by Claim (2). Hence, R is the unique minimal normal subgroup of G and $R\nleq \Phi (G)$, so $R=C_{G}(R)=O_{p}(G)$ by [33, Ch. A, 15.6]. It is clear also that $|R|>p$, so $Z=1$.

Final contradiction for (8).

Let V be any maximal subgroup of P. We show that V has a p-nilpotent supplement in G. Assume that this is false. Then, the subgroup V is m-S-complemented in G by hypothesis.

First suppose that $R\nleq V$. Then, $W=V\cap R$ is normal in P, $|R:W|=p$ and, by Claim (b), $V_{G}=1$. There are an generalized S-quasinormal subgroup S and a subgroup T of G such that $G=VT$ and $V\cap T\le S\le V$. Then, $V\cap T=S\cap T$. Arguing as above, we can show that S is S-quasinormal in G. Hence, S is subnormal in G by Lemma 2.1(4). It follows that $S\le O_{p}(G)=R$ by Claim (b). Hence, $S\le R\cap V=W$ and so $S^{G}=S^{PO^{p}(G)}=S^{W}\le W$ by [2, Lemma 1.2.16], which implies that $S=1$. Then, T is a complement to V in G, so T is p-nilpotent.

Now let V be any maximal subgroup of P containing R, and let M be a maximal subgroup of G such that $G=R\rtimes M$. Then, $M\simeq G/R$ is p-nilpotent, so M is a p-nilpotent supplement to V in G. Thus, every maximal subgroup of P has a p-nilpotent supplement in G. Therefore, G is p-nilpotent by Lemma 2.7. This contradiction shows that we have (8).

The final contradiction. Claims (2) and (8) imply that $E=P$ is a normal p-subgroup of G. Let R be a minimal normal subgroup of G contained in P. Then, P / R is hypercyclically embedded in G by Claims (3) and (6). Therefore, $R\nleq \Phi (P)$ by Lemma 2.8 and [20, III, Hilfsatz 3.3(a)]. Hence, $\Phi (P)=1$, so P is an elementary abelian p-group. If $|R|=p$, then P is hypercyclically embedded in G by the Jordan–Hölder theorem for the chief series. Hence, R is not cyclic. Moreover, R is the unique minimal normal subgroup of G contained in P. Indeed, suppose that for some minimal normal subgroup $N\ne R$ of G we also have $N\le P$. Then, P / N is hypercyclically embedded in G and so from the G-isomorphism $RN/N\simeq R$ we get that $|R|=p$, a contradiction.

Let W be a maximal subgroup of N such that W is normal in a Sylow p-subgroup $G_{p}$ of G. Then, $W\ne 1$. We show that W is S-quasinormal in G. Let B be a complement to N in P and $H=WB$. Then, H is a maximal subgroup of P and $W=H\cap R$. Therefore, W is S-quasinormal in G in the case when H is S-quasinormal in G by Lemma 2.1(5). From now on, we suppose that H is not S-quasinormal in G.

Assume that H has a p-nilpotent supplement U in G and let S be the normal p-complement in U. Then, $P=P\cap HU=H(P\cap U)$, where $P\cap U$ is normal in G since P is abelian. Moreover, $1< P\cap U < P$ since G is not p-nilpotent. Therefore, $R\le P\cap U$. Then, $[R, S]=1$, so $G/C_{G}(R)$ is a p-group and so $C_{G}(R)=G$ since R is a p-group. But then $|R|=p$. This contradiction shows that H has no p-nilpotent supplements in G and hence H is m-S-complemented in G by hypothesis.

Let S and T be subgroups of G such that S is generalized S -quasinormal in G and we have $G=HT$ and $H\cap T\le S\le H$. And let $S=AB$, where A is modular and B is S-quasinormal in G. Then, $N\nleq H$ and so $A_{G}=1$, which implies that $A^{G}$ is hypercyclically embedded in G by Lemma 2.4. But then $A=1$ since otherwise $N\le A^{G}\cap P$ and so $|N|=p$. Therefore, $S=B$ is S-quasinormal in G. Since $T\cap H\le S\le H$ and H is not S-quasinormal in G, it follows that $T<G$ and for the normal subgroup $T\cap P$ of G we have $1<T\cap P$. Then, $N\le T$ and so $N\cap H=N\cap S=W$, which implies that W is S-quasinormal in G by Lemma 2.1(5). But then W is normal in G since $G=G_{p}O^{p}(G)\le N_{G}(W)$ by [2, Lemma 1.2.16] and so $W=1$. Therefore, N is cyclic. This contradiction completes the proof of the result. $\square$

Proof of Theorem 1.4

Suppose that this theorem is false and consider a counterexample (G, E) for which $|G| + |E|$ is minimal. Let p be the smallest prime dividing |E| and let P be a Sylow p-subgroup of E.

Then, E is p-supersoluble by Theorem 1.3 and so E is p-nilpotent since p is the smallest prime dividing |E| (see Claim (2) in the proof of Theorem 1.3). Note also that if X is a non-identity Hall subgroup of E, then $X=E$. Indeed, the hypothesis holds for (G / X, E / X) and for (G, X) by Lemma 2.3(1). Hence in the case $X\ne E$, the choice of G implies that E / X and X are hypercyclically embedded in G. Hence, E is hypercyclically embedded in G by the Jordan–Hölder theorem for the chief series. This contradiction shows that $E=P$, so E is hypercyclically embedded in G by Theorem 1.3. The theorem is proved. $\square$

Proof of Theorem 1.5

This theorem is a corollary of Theorem 1.4 and Lemmas 2.5 and 2.6. $\square$

4 Some Applications of the Results

Theorems 1.3, 1.4 and 1.5 cover many known results. In particular, from Theorem 1.5, we get the following known results.

Corollary 4.1

(Srinivasan [21]). If the maximal subgroups of the Sylow subgroups of G are S-quasinormal in G, then G is supersoluble.

Corollary 4.2

(Asaad [22]). Let ${\mathfrak {F}}$ be a saturated formation containing all supersoluble groups and G a group with a normal subgroup E such that $G/E\in$${\mathfrak {F}}$ . If $G/E\in$${\mathfrak {F}}$ and every maximal subgroup of every Sylow subgroup of E is S-quasinormal in G, then $G\in$${\mathfrak {F}}$ .

A subgroup H of G is said to be c-normal in G [23], if there is a normal subgroup T of G such that $G=HT$ and $H\cap T\le H_{G}$. It is clear that every c-normal subgroup of G is also m-S-complemented in G. Hence, we get from Theorem 1.5 the following known results.

Corollary 4.3

(Wang [23]). If the maximal subgroups of the Sylow subgroups of G are c-normal in G, then G is supersoluble.

Corollary 4.4

(Alsheik Ahmad [24]). If the maximal subgroups of the Sylow subgroups of G not having supersoluble supplement in G are c-normal in G, then G is supersoluble.

Corollary 4.5

(Ramadan [25]). Let E be a normal subgroup of G with supersoluble quotient G / E. If all maximal subgroups of the Sylow subgroups of E are normal in G, then G is supersoluble.

Corollary 4.6

(Li, Guo [26]). Let E be a soluble normal subgroup of G with supersoluble quotient G / E. If all maximal subgroups of the Sylow subgroups of F(E) are c-normal in G, then G is supersoluble.

Corollary 4.7

(Wey [27]). Let ${\mathfrak {F}}$ be a saturated formation containing all supersoluble groups and G a group with a soluble normal subgroup E such that $G/E\in$${\mathfrak {F}}$ . If all maximal subgroups of the Sylow subgroups of F(E) are c-normal in G, then $G\in$${\mathfrak {F}}$.

Corollary 4.8

(Wei, Wang, Li [28]). Let ${\mathfrak {F}}$ be a saturated formation containing all supersoluble groups and G a group with a normal subgroup E such that $G/E\in$${\mathfrak {F}}$ . If all maximal subgroups of the Sylow subgroups of $F^{*}(E)$ are c -normal in G, then $G\in$${\mathfrak {F}}$.

Corollary 4.9

(Asaad, Ramadan, Shaalan [29]). Let E be a soluble normal subgroup of G with supersoluble quotient G / E. Suppose that all maximal subgroups of any Sylow subgroup of F(E) are S-quasinormal in G. Then, G is supersoluble.

Corollary 4.10

(Li, Wang [30]). Let ${\mathfrak {F}}$ be a saturated formation containing all supersoluble groups and G a group with a normal subgroup E such that $G/E\in {\mathfrak {F}}$. If all maximal subgroups of any Sylow subgroup of $F^{*}(E)$ are S-quasinormal in G, then $G\in {\mathfrak {F}}$.

Corollary 4.11

(Li, Wang [30]). Let ${\mathfrak {F}}$ be a saturated formation containing all supersoluble groups and G a group with a normal subgroup E such that $G/E\in {\mathfrak {F}}$. If every maximal subgroup of every Sylow subgroup of $F^{*}(E)$ is S-quasinormal in G, then $G\in {\mathfrak {F}}$.

Corollary 4.12

(Wei [28]). Let ${\mathfrak {F}}$ be a saturated formation containing all supersoluble groups and G a group with a normal subgroup E such that $G/E\in {\mathfrak {F}}$. If every maximal subgroup of every Sylow subgroup of E is c-normal in G, then $G\in {\mathfrak {F}}$.

In view of Example 1.2(ii), we get also from Theorem 1.5 the following known results.

Corollary 4.13

(Wei, Wang and Li [31]). Let ${\mathfrak {F}}$ be a saturated formation containing all supersoluble groups and G a group with a normal subgroup E such that $G/E\in {\mathfrak {F}}$. If every maximal subgroup of every Sylow subgroup of $F^{*}(E)$ is c-supplemented in G, then $G\in {\mathfrak {F}}$.

Corollary 4.14

(Ballester-Bolinches and Guo [32]). Let ${\mathfrak {F}}$ be a saturated formation containing all supersoluble groups and G a group with a normal subgroup E such that $G/E\in {\mathfrak {F}}$. If every maximal subgroup of every Sylow subgroup of E is c-supplemented in G, then $G\in {\mathfrak {F}}$.