On m-Embedded Primary Subgroups of Finite Groups

Abstract

Suppose that A is a subgroup of a group G. A is called to be m-embedded in G if G has a subnormal subgroup T and a $\{1\le G\}$-embedded subgroup C such that $G=AT$ and $A\cap T\le C\le A$. In this paper, we shall investigate the structure of finite groups by using m-embedded subgroups and obtain some new characterization about p-supersolvability and generalized hypercentre of finite groups. Some results in Guo and Shum (Arch Math 80:561–569, 2003) , Ramadan et al. (Arch Math 85:203–210, 2005), Tang and Miao (Turk J Math 39:501–506, 2015), and Xu and Zhang (Can Math Bull 57(4):884–889, 2014) are generalized.

1 Introduction

Every group considered in this paper is finite. Most of the notation is standard and can be found in [3, 10]. Let $|\,G|$ denote the order of a group G, $|\,G|_{p}$ denote the p-part of $|\,G|$ and $\pi (G)$ denote the set of all prime divisors of $|\,G|$. Let $A\rtimes B$ denote the semidirect product of groups A and B, where B is an operator group of A. Let $\mathcal F$ be a class of groups and I / K be a chief factor of a group G. I / K is called Frattini provided that $I/K \le \Phi (G/K)$. Moreover, I / K is called $\mathcal F$-central if $I/K \rtimes (G/C_G(I/K))\in \mathcal F$. Otherwise, I / K is called $\mathcal F$-eccentric. The symbol $Z_\mathcal{F}(G)$($Z_{p\mathcal F}(G))$ denotes the $\mathcal F$-hypercentre($p\mathcal F$-hypercentre) of a group G, which is the product of all such normal subgroups H of G whose G-chief factors(whose G-chief factors of order divisible by p)are $\mathcal F$-central. In addition, $\mathcal {U}$ and $p\mathcal {U}$ denote the class of all supersolvable groups and all p-supersolvable groups, respectively.

Suppose that A is a subgroup of G, $K\le H\le G$. (1) if $AH=AK$, then A covers the pair (K, H); (2) if $A\cap H=A\cap K$, then A avoids (K, H). In 1939, Ore [11] introduced the notion of quasinormal subgroups. Furthermore, if E is a quasinormal subgroup of G, then for every maximal pair of G, that is, a pair (K, H), where K is a maximal subgroup of H, E either covers or avoids (K, H). In 1992, Doerk and Hawkes [3] gave the definition of CAP-subgroups, that is, a subgroup A of G is called a CAP-subgroup if A either covers or avoids each pair (K, H), where H / K is a chief factor of G. Based on the definitions and observations above, Guo and Skiba introduced new concepts as follows:

Definition 1.1

[6, Definition 1.1] Let $\Sigma =\{G_0\le G_1\le \ldots \le G_n\}$ be some subgroup series of G and A be a subgroup of G. Then A is $\Sigma$-embedded in G if A either covers or avoids every maximal pair (K, H) such that $G_{i-1}\le K<H\le G_i$, for some i.

In [6], quasinormal subgroups, CAP-subgroups, and partial CAP-subgroups [1] (or a semi cover-avoiding subgroups [8]) are $\Sigma$-embedded subgroups.

Definition 1.2

[6, Definition 2.7] Let A be a subgroup of G. Then A is m-embedded in G if G has a subnormal subgroup T and a $\{1\le G\}$-embedded subgroup C in G such that $G=AT$ and $A\cap T\le C\le A$.

In [6, Example 2.8], every c-normal subgroup of G is also m-embedded in G.

On the other hand, In 2014, Xu and Zhang [14] investigated p-nilpotency of a group by using m-embedded property of primary subgroups. In 2015, Tang and Miao [13] obtained some results about p-supersolvability of finite groups by using m-embedded primary subgroups. They proved the following theorem:

Theorem 1.3

[13, Theorem 1.3] Let G be a p-solvable group and P be a Sylow p-subgroup of G where p is an odd prime divisor of |G|. If every maximal subgroup of P is m-embedded in G,  then G is p-supersolvable.

It is clear that the p-solvability in [13, Theorem 1.3] is essential. Naturally, the question is that

What is the structure of a group if the p-solvability is removed in [13, Theorem 1.3]?

Along the clue, we obtained the following result:

Theorem 1.4

Let E be a normal subgroup of G and P be a Sylow p-subgroup of E where p is an odd prime divisor of |E|. If every maximal subgroup of P is m-embedded in G,  then every G-chief factor A / B below E satisfies one of the following conditions : 

$$\begin{aligned} (1) \ A/B\le \Phi (G/B); \quad \quad \quad (2) \ A/B\text { is a } p{'}-\text { group}; \quad \quad \quad (3) \ |A/B|_{p}=p. \end{aligned}$$

With the further consideration, we will study the p-supersolvability and generalized hypercentre of a finite group by using m-embedded subgroups, where p is a prime. Some theorems in [9, 12,13,14] are corollaries of our results.

2 Preliminaries

For the sake of convenience, we first list here some known results which will be useful in the sequel.

Lemma 2.1

[6, Lemma 2.13] Let K and H be subgroups of G. Suppose that K is m-embedded in G and H is normal in G. Then

• (1) If $H\le K$ $,$ then K / H is m-embedded in G / H.

• (2) If $K\le E\le G$ $,$ then K is m-embedded in E.

• (3) If $(|H|, |K|)=1$ $,$ then KH / H is m-embedded in G / H.

Lemma 2.2

[6, Lemma 2.14] Let P be a normal non-identity p-subgroup of G with $|P|=p^{n}$ and $P\cap \Phi (G)=1$. Suppose that there is an integer k such that $1\le k<n$ and the subgroups of P of order $p^{k}$ are m-embedded in $G$ $,$ then some maximal subgroup of P is normal in G.

Lemma 2.3

[6, Lemma 2.5] Every $\{1\le G\}$-embedded subgroup of G is subnormal in G.

Lemma 2.4

[5, Lemma 2.8] Let G be a p-supersolvable group. If $O_{p{'}}(G)=1,$ then G is supersolvable.

3 Main Results

Theorem 3.1

Let p be an odd prime divisor of |G| and P be a normal p-subgroup of G. If every minimal subgroup of P is $\{1\le G\}$-embedded in G,  then $P\le Z_\mathcal{U}(G)$.

Proof

Assume that the assertion is false and choose (G, P) to be a counterexample of minimal order.

• (1) G has a unique normal subgroup N such that P / N is a chief factor of G, $N\le Z_{\mathcal {U}}(G)$ and $|P/N|>p$.

  Let P / N be a chief factor of G. Then, clearly, (G, N) satisfies the hypothesis of the theorem. The choice of (G, P) implies that $N\le Z_{\mathcal {U}}(G)$. If $|P/N|=p$, then $P/N\le Z_{\mathcal {U}}(G/N)$ and so $P\le Z_{\mathcal {U}}(G)$, a contradiction. Hence $|P/N|>p$. Assume that P / L is a chief factor of G with $P/N\ne P/L$. With the same discussion as above, we have that $L\le Z_{\mathcal {U}}(G)$. Then $P/N=NL/N\le NZ_{\mathcal {U}}(G)/N\le Z_{\mathcal {U}}(G/N)$. It follows from $N\le Z_{\mathcal {U}}(G)$ that $P\le Z_{\mathcal {U}}(G)$, a contradiction.

• (2) The exponent of P is p.

  Let C be a Thompson critical subgroup of P. If $\Omega (C)<P$, then $\Omega (C)\le N\le Z_{\mathcal {U}}(G)$ by (1), so $P\le Z_{\mathcal {U}}(G)$ by [7, Lemma 4.4], which is impossible. Hence $P=C=\Omega (C)$. Then by [7, Lemma 4.3], the exponent of P is p.

• (3) P is a minimal normal subgroup of G.

  If not, then $N\ne 1$. Let H / N be a minimal subgroup of P / N. Then there exists an element $x\in H\backslash N$, $H=\langle x\rangle N$ and $|\langle x\rangle |=p$ by (2). By hypothesis and [6, Lemma 2.3], H / N is $\{1\le G/N\}$-embedded in G / N. Then $P/N\le Z_{\mathcal {U}}(G/N)$ by the choice of (G, P). Hence $|P/N|=p$, a contradiction. Hence $N=1$ and (3) holds.

• (4) $P\le \Phi (G)$.

  If not, then $P\nleq \Phi (G)$. By (1), we may choose a minimal subgroup H of P such that $G=HM=PM$ and $P\cap M=1$. Since $|P:P\cap M|=p$, $|P|=p$ and $P\le Z_\mathcal{U}(G)$, a contradiction.

• (5) The final contradiction.

  By [6, Lemma 2.3], every minimal subgroup of P is $\{1\le M\}$-embedded in M. Then $P\le Z_\mathcal{U}(M)$ by the choice of (G, P), for every maximal subgroup M of G. We assert that $C_{G}(P)\le \Phi (G)$. If not, then $C_{G}(P)\nleq \Phi (G)$ and $G=C_{G}(P)M_{1}$ for some maximal subgroup $M_{1}$ of G. Next, we choose a minimal normal subgroup N of $M_{1}$ contained in P. Then $|N|=p$. Further, $N^{G}=N^{C_{G}(P)M_{1}}=N^{M_{1}}=N$ and $N\unlhd G$. Then $|N|=|P|=p$, a contradiction. Set $Z=\cap (C_{M}(X/Y))$, where X / Y is an M-chief factor below P for every maximal subgroup M of G. Then M / Z is an abelian group of exponent dividing $p-1$ and $O^{p}(Z)\le C_{G}(P)\le \Phi (G)$. Hence $M/\Phi (G)$ is a strictly p-closed group and $M/\Phi (G)$ is supersolvable by [15, Theorem 1.9]. Then $G/\Phi (G)$ is minimal non-supersolvable and G is solvable by [15, Theorem 2.3]. Further, we have $F(G)\le C_{G}(P)\le \Phi (G)<F(G)$, a contradiction.

The final contradiction completes our proof.$\square$

Theorem 3.2

Let p be an odd prime divisor of |G| and P be a Sylow p-subgroup of G. If every minimal subgroup of P is m-embedded in G,  then G is p-supersolvable.

Proof

Assume that the assertion is false and choose G to be a counterexample of minimal order. Furthermore, we have that

• (1) $O_{p{'}}(G)=1$.

  Assume that $T=O_{p{'}}(G)\ne 1$. By Lemma 2.1(3), G / T satisfies the conditions of the theorem, and the minimal choice of G implies that G / T is p-supersolvable. Hence G is p-supersolvable, a contradiction.

• (2) $P\cap E\unlhd E$, where E is a proper normal subgroup of G.

  Assume that E is a proper normal subgroup of G. By (1), $P\cap E\ne 1$. By Lemma 2.1 and the choice of G, E is p-supersolvable. Hence E is supersolvable by (1) and Lemma 2.4. Then $P\cap E\unlhd E$.

• (3) There exists a minimal subgroup H of P such that H has a normal complement in G. Otherwise, all minimal subgroups of P are $\{1\le G\}$-embedded in G. Then all minimal subgroups of P are contained in $O_{p}(G)$ by Lemma 2.3. Further, $O_{p}(G)\le Z_\mathcal{U}(G)$. Hence G is p-supersolvable by [2, Theorem 6], a contradiction.

• (4) The final contradiction.

  By (3) and hypothesis, $G=HM$, $M\unlhd G$, $H\cap M=1$. Then $P\cap M\unlhd M$ by (2). If every minimal subgroup of $P\cap M$ is $\{1\le G\}$-embedded in G, then $P\cap M\le Z_\mathcal{U}(G)$ by Theorem 3.1 and $1\unlhd P\cap M\unlhd M\unlhd G$ a normal subgroup series of G such that every G-chief factor either cyclic of order p or $p{'}$-group. Hence G is p-supersolvable, a contradiction.

Now we assume that there exists a minimal subgroup $H_{1}$ of $P\cap M$ such that $H_{1}$ has a normal complement in G. Further, $G=H_{1}M_{1}$, $M_{1}\unlhd G$, $H_{1}\cap M_{1}=1$. Similar to the previous discussion, we consider $P\cap M \cap M_{1}$. Now, we set $G_{0}=G$, $G_{1}=M$, $G_{2}=M\cap M_{1}$. Repeat above discussion, we have $1=G_{s+1}\unlhd G_{s}\unlhd \cdots \unlhd G_{3}\unlhd G_{2}\unlhd G_{1}\unlhd G_{0}=G$ a normal subgroup series of G such that every G-chief factor either cyclic of order p or $p{'}$-group. Then G is p-supersolvable, a contradiction.

The final contradiction completes our proof.$\square$

Theorem 3.3

Let E be a normal subgroup of G and P be a Sylow p-subgroup of E where p is an odd prime divisor of |E|. If every minimal subgroup of P is m-embedded in G,  then $E\le Z_{p\mathcal U}(G)$.

Proof

Assume that the assertion is false and choose (G, E) to be a counterexample of minimal order. Furthermore, we have that

• (1) $O_{p{'}}(E)=1$.

  If $K=O_{p{'}}(E)\ne 1$, then we consider G / K. (G / K, E / K) satisfies the hypothesis of the theorem by Lemma 2.1(3). The minimal choice of (G, E) implies that $E/K\le Z_{p\mathcal U}(G/K)$, and so $E\le Z_{p\mathcal U}(G)$, a contradiction.

• (2) $P\unlhd G$.

  By Theorem 3.2, E is p-supersolvable. By (1) and Lemma 2.4, E is supersolvable. Then $P\unlhd G$.

• (3) There exists a minimal subgroup H of P such that H has a normal complement in G.

  Otherwise, all minimal subgroups of P are $\{1\le G\}$-embedded in G. By Theorem 3.1, $P\le Z_\mathcal{U}(G)$. Hence $E\le Z_{p\mathcal U}(G)$, a contradiction.

• (4) The final contradiction.

  By (3) and hypothesis, $G=HM$, $M\unlhd \unlhd G$, $H\cap M=1$. Then $M\unlhd G$, $|P:P\cap M|=p$ and $P\cap M\unlhd G$. By the choose of (G, E), we have that $P\cap M\le Z_{p\mathcal U}(G)$. Hence $P\le Z_{p\mathcal U}(G)$ and $E \le Z_{p\mathcal U}(G)$, a contradiction.

The final contradiction completes our proof.$\square$

Corollary 3.4

Let E be a normal subgroup of G such that G / E is p-supersolvable and P be a Sylow p-subgroup of E where p is an odd prime divisor of |E|. If every minimal subgroup of P is m-embedded in G,  then G is p-supersolvable.

By Theorem 3.3, it is easy to prove the following corollaries:

Corollary 3.5

[9, Theorem 3.8] Let p be an odd prime number dividing the order of a group G and $\mathcal F$ a saturated formation containing the class $p\mathcal U$ of all p-supersolvable groups. Also let N be a normal subgroup of G such that $G/N \in \mathcal F$. If P is a Sylow p-subgroup of N and every minimal subgroup of P is c-normal in G,  then $G \in \mathcal F$.

Theorem 3.6

Let G be a group and P be a Sylow p-subgroup of G where p is an odd prime divisor of |G|. If every maximal subgroup of P is m-embedded in G,  then every chief factor A / B of G satisfies one of the following conditions : 

$$\begin{aligned} (1) \ A/B\le \Phi (G/B); \quad \quad \quad (2) \ A/B \ \text { is a } p{'}\text {-group} ; \quad \quad \quad (3) \ |A/B|_{p}=p. \end{aligned}$$

Proof

Assume that the theorem is false and let G be a counterexample of minimal order.

• (1) $O_{p{'}}(G)=1$.

  Assume that $O_{p{'}}(G)\ne 1$. The hypothesis also holds for $G/O_{p{'}}(G)$ by Lemma 2.1, and for G, a contradiction.

• (2) If $O_{p}(G)\ne 1$, then $O_{p}(G)\cap \Phi (G)=1$.

  Assume that $O_{p}(G)\cap \Phi (G)\ne 1$. We may choose a minimal normal subgroup L of G such that $L\le O_{p}(G)\cap \Phi (G)$. By induction, G / L holds and so G holds, a contradiction.

• (3) $O_{p}(G)=1$.

Assume that $O_{p}(G)\ne 1$. By (2) and [4, Theorem 1.8.17], $O_{p}(G)=L_1\times L_2\times \cdots \times L_t$ where $L_i$ are the minimal normal subgroups of G, $i=1,2,\ldots ,t$. For every $L\in \{L_i\}$ and we consider G / L. Clearly, G / L holds by the choice of G. Further, we assert that $O_{p}(G)$ is a minimal normal subgroup of G. Otherwise, there exists two different minimal normal subgroups $L_1$ and $L_2$ such that $G/L_{j}$ satisfies the hypothesis of theorem and so every $G/L_{j}$-chief factor holds by the choice of G where $j=1, 2$. If $L_{1}L_{2}/L_{2}\le \Phi (G/L_{2})$, then $L_{1}L_{2}\le \Phi (G)L_{2}$ by [3, A. Lemma 9.11]. Since $L_{1}L_{2}\le O_{p}(G)$, $L_{1}L_{2}\le O_{p}(G)\cap \Phi (G)L_{2}=(O_{p}(G)\cap \Phi (G))L_{2}=L_{2}$ by (2), a contradiction. Hence $L_{1}\cong L_{1}L_{2}/L_{2}$ satisfies the condition (2) or (3), then every G-chief factor holds, a contradiction.

Hence $O_{p}(G)$ is a minimal normal subgroup of G and $\Phi (G)=1$. Then there exists a maximal subgroup M of G such that $G=O_{p}(G)M$. We assert that $O_{p}(G)<P$. If not, by Lemma 2.2, $|\,P|=p$ and so G holds, a contradiction. Hence we may choose a maximal subgroup $P_{1}$ of P such that $M_{p}\le P_{1}$ and $O_{p}(G)\nleq P_{1}$. By hypothesis, $P_{1}$ is m-embedded in G, there exists a subnormal subgroup T in G and a $\{1\le G\}$-embedded subgroup C in G such that $G=P_{1}T$ and $P_{1}\cap T\le C \le P_{1}$. We assert that $C=1$. Otherwise, $C\ne 1$. If $C<O_{p}(G)$ by Lemma 2.3, then we obtain C neither covers nor avoids maximal pair (M, G) since $O_{p}(G)\cap M=1$, a contradiction. Hence we may assume that $C=O_{p}(G)$ by Lemma 2.3, that is, $O_{p}(G)\le P_1$, a contradiction. Then we have $|\,T_{p}|=p$.

If $O_{p}(G)\cap T\ne 1$, then $O_{p}(G)\cap T=T_{p}\unlhd T$ and so T is p-solvable. Furthermore, $T_{p{'}}$ is a Hall ${p{'}}$-subgroup of G. By [4, Theorem 1.8.19] and (1), $C_{T}(T_{p})=T_{p}$. Hence $N_{T}(T_{p})/C_{T}(T_{p})=T/C_{T}(T_{p})\hookrightarrow C_{p-1}$. By Schur–Zassenhaus Theorem, $T_{p{'}}$ is cyclic and T is supersolvable. We assert that p is the largest prime divisor of |G|. Otherwise, if $q\ne p$ is the largest prime divisor of |G|, then $Q\unlhd G$, which contradicts (1). Then we assume that $p_{1}<p_{2}<\cdots <p_{n}=p$, where $\pi (G)=\{p_{1}, p_{2}, \ldots , p_{n}=p\}$. Since $T_{p{'}}$ is cyclic, $G_{p{'}}$ is cyclic and $G_{p_{1}}$ is cyclic where $G_{p_{1}}$ a Sylow $p_{1}$-subgroup of G. By Burnside Theorem, G is $p_{1}$-nilpotent and $G_{p_{1}{'}}\unlhd G$. Next, we consider $G_{p_{1}{'}}$. Similar to the previous discussion, $G_{\{p_{1}, p_{2}\}{'}}\unlhd G_{p_{1}{'}}$. Repeat above discussion, we get a normal subgroup series of G: $1\unlhd P \unlhd \ldots \unlhd G_{\{p_{1}, p_{2}\}{'}}\unlhd G_{p_{1}{'}}\unlhd G$. Hence G has supersolvable type Sylow tower and $P\unlhd G$, a contradiction.

If $O_{p}(G)\cap T=1$, then $O_{p}(G)\le N_{G}(T)$ by [3, Lemma A.14.3] and $O_{p}(G)T=O_{p}(G)\times T$. Since $O_{p}(G)\cap Z(P)\ne 1$, we may pick a minimal subgroup H of $O_{p}(G)\cap Z(P)$ and $H\unlhd G$. Clearly, G / H satisfies the hypothesis of Theorem and so G / H holds by the choice of G. Then every chief factor of G satisfies one of the three conditions in the conclusion of Theorem, a contradiction.

• (4) The final contradiction.

Let $P_{2}$ be a maximal subgroup of P. By hypothesis and (3), $P_{2}$ is m-embedded in G, we may choose a subnormal subgroup $K_{2}$ of G such that $G=P_{2}K_{2}$ and $P_{2}\cap K_{2}=1$. Hence there exists a maximal normal subgroup K such that $|G:K|=p$. Clearly, $K_{p}$ is m-embedded in G, where $K_{p}$ is a Sylow p-subgroup of K, we may choose a subnormal subgroup $K_{3}$ of G such that $G=K_{p}K_{3}=KK_{3}$ and $K_{p}\cap K_{3}=1$. Since $|\,K_{p}K_{3}|=|\,KK_{3}|$, we have $|\,K\cap K_{3}|=\frac{|\,K|}{|\,K_{p}|}$. If $K\cap K_{3}\ne 1$, then $K\cap K_{3}$ is a $p{'}$-group, which contradicts (1). Hence $K\cap K_{3}=1$ and K is a normal p-subgroup, which contradicts (3).

The final contradiction completes our proof.$\square$

From Theorem 3.6, it is easy to prove the following corollaries:

Corollary 3.7

[14, Theorem 3.1] Let G be a group and P be a Sylow p-subgroup of G where p is an odd prime divisor of |G|. If every maximal subgroup $P_{1}$ of P is m-embedded in G and $N_{G}(P_{1})$ is p-nilpotent,  then G is p-nilpotent.

Proof

Clearly, G is not a non-abelian simple group and $O_{p{'}}(G)=1$. Then we may pick a minimal normal subgroup L of G. Further, L satisfies one of the three conditions in Theorem 3.6. We only need to consider the condition that $|L|_{p}=p$. Then we consider the group $N_{G}(L_{p})$ where $L_{p}$ is a Sylow p-subgroup of L. Next, we prove that $|L|=p$. If $N_{G}(L_{p})<G$, then $N_{G}(L_{p})$ is p-nilpotent since $P\le N_{G}(L_{p})$. Further, $N_{L}(L_{p})=C_{L}(L_{p})$ and so L is p-nilpotent by Burnside Theorem. Then $|L|=p$. If $N_{G}(L_{p})=G$, then $|L|=p$.

Since G / L is p-nilpotent, G is p-supersolvable and so G is supersolvable by Lemma 2.4. Hence $P\unlhd G$, p is the largest prime divisor of |G|. Since $G=L\rtimes M$, $P\cap M$ is a maximal subgroup of P and $P\cap M \unlhd G$. Then $G=N_{G}(P\cap M)$ is p-nilpotent.$\square$

Corollary 3.8

[13, Theorem 1.2] Let G be a group and P be a Sylow p-subgroup of G where p is an odd prime divisor of |G|. If every maximal subgroup of P is m-embedded in G and $N_{G}(P)$ is p-nilpotent,  then G is p-nilpotent.

Proof

See the proof of Corollary 3.7.$\square$

Corollary 3.9

[9, Theorem 3.1] Let p be an odd prime dividing the order of a group G and P a Sylow p-subgroup of G. If $N_{G}(P)$ is p-nilpotent and every maximal subgroup of P is c-normal in G,  then G is p-nilpotent.

Corollary 3.10

[13, Theorem 1.3] Let G be a p-solvable group and P be a Sylow p-subgroup of G where p is an odd prime divisor of |G|. If every maximal subgroup of P is m-embedded in G,  then G is p-supersolvable.

Proof

Clearly, G is not a non-abelian simple group. Then we may pick a minimal normal subgroup L of G. Further, L satisfies one of the three conditions in Theorem 3.6. Since G / L satisfies the hypothesis of Theorem 3.6, G / L is p-supersolvable by induction. Then G is p-supersolvable.$\square$

Theorem 3.11

Let E be a normal subgroup of G and P be a Sylow p-subgroup of E where p is an odd prime divisor of |E|. If every maximal subgroup of P is m-embedded in G,  then every G-chief factor A / B below E satisfies one of the following conditions : 

$$\begin{aligned} (1) \ A/B\le \Phi (G/B); \quad \quad \quad (2) \ A/B \text { is a } p{'}\text {-}\,\text {group}; \quad \quad \quad (3) \ |A/B|_{p}=p. \end{aligned}$$

Proof

Assume that the theorem is false and let (G, E) be a counterexample with |G||E| minimal.

• (1) $O_{p{'}}(E)=1$.

  Assume that $O_{p{'}}(E)\ne 1$. The hypothesis also holds for $(G/O_{p{'}}(E), E/O_{p{'}}(E))$ by Lemma 2.1, and for (G, E). Then every G-chief factor below E holds, a contradiction.

• (2) If $O_{p}(E)\ne 1$, then $O_{p}(E)\cap \Phi (G)=1$.

  Assume that $O_{p}(E)\cap \Phi (G)\ne 1$. We may choose a minimal normal subgroup L of G such that $L\le O_{p}(E)\cap \Phi (G)$. By induction, (G / L, E / L) satisfies the hypothesis of theorem and so every G / L-chief factor below E / L holds. Then every G-chief factor below E holds, a contradiction.

• (3) $O_{p}(E)=1$.

Assume that $O_{p}(E)\ne 1$. By (2) and [4, Theorem 1.8.17], $O_{p}(E)=L_1\times L_2\times \cdots \times L_t$ where $L_i$ are the minimal normal subgroups of G, $i=1,2,\ldots ,t$. We assert that $O_{p}(E)$ is a minimal normal subgroup of G. Otherwise, there exists two minimal normal subgroups $L_1$ and $L_2$, then we consider $(G/L_1, E/L_1)$ and $(G/L_2, E/L_2)$. Clearly, $(G/L_{j}, E/L_{j})$ satisfies the hypothesis of theorem and so every $G/L_{j}$-chief factor below $E/L_{j}$ holds by the choice of (G, E) where $j=1, 2$. If $L_{1}L_{2}/L_{2}\le \Phi (G/L_{2})$, then $L_{1}L_{2}\le \Phi (G)L_{2}$ by [3, A. Lemma 9.11]. Since $L_{1}L_{2}\le O_{p}(E)$, $L_{1}L_{2}\le O_{p}(E)\cap \Phi (G)L_{2}=(O_{p}(E)\cap \Phi (G))L_{2}=L_{2}$ by (2), a contradiction. Hence $L_{1}\cong L_{1}L_{2}/L_{2}$ satisfies the condition (2) or (3), then every G-chief factor holds, a contradiction. Clearly, $O_{p}(E)\nleq \Phi (G)=1$. Then there exists a maximal subgroup M of G such that $G=O_{p}(E)M$. Then $P=O_{p}(E)(P\cap M)$. We assert that $O_{p}(E)<P$. If not, by Lemma 2.2, $|\,P|=p$ and so (G, E) holds, a contradiction. Hence we may choose a maximal subgroup $P_{1}$ of P such that $P\cap M\le P_{1}$ and $O_{p}(E)\nleq P_{1}$. By hypothesis, $P_{1}$ is m-embedded in G, there exists a subnormal subgroup T in G and a $\{1\le G\}$-embedded subgroup C in G such that $G=P_{1}T$ and $P_{1}\cap T\le C\le P_{1}$. We assert that $C=1$. Otherwise, $C\ne 1$. If $C<O_{p}(E)$ by Lemma 2.3, then we obtain C neither covers nor avoids maximal pair (M, G) since $O_{p}(E)\cap M=1$, a contradiction. Hence we may assume that $C=O_{p}(E)$ by Lemma 2.3, i.e., $O_{p}(E)\le P_1$ and so $P\le O_{p}(E)(P\cap M)\le P_1<P$, a contradiction. Then we have $|\,T_{p}|=p$.

If $O_{p}(E)\cap T\ne 1$, then $O_{p}(E)\cap T=T_{p}\unlhd T$ and so T is p-solvable. Furthermore, $T_{p{'}}$ is a Hall ${p{'}}$-subgroup of G. By [4, Theorem 1.8.19] and (1), $C_{T}(T_{p})=T_{p}$. Hence $N_{T}(T_{p})/C_{T}(T_{p})=T/C_{T}(T_{p})\hookrightarrow C_{p-1}$. By Schur–Zassenhaus Theorem, $T_{p{'}}$ is cyclic and T is supersolvable. Then $E\cap T$ is supersolvable. We assert that p is the largest prime divisor of |E|. Otherwise, if $q\ne p$ is the largest prime divisor of |E|, then $Q\le O_{q}(E)=1$ where Q is a Sylow q-subgroup of $E\cap T$, which contradicts (1). Then we assume that $p_{1}<p_{2}<\cdots <p_{n}=p$, where $\pi (E)=\{p_{1}, p_{2}, \ldots , p_{n}=p\}$. Since $T_{p{'}}$ is cyclic, $E_{p{'}}=E\cap T_{p{'}}$ is cyclic and $E_{p_{1}}$ is cyclic where $E_{p_{1}}$ a Sylow $p_{1}$-subgroup of E. By Burnside Theorem, E is $p_{1}$-nilpotent and $E_{p_{1}{'}}\unlhd E$. Next, we consider $E_{p_{1}{'}}$. Similar to the previous discussion, $E_{\{p_{1}, p_{2}\}{'}}\unlhd E_{p_{1}{'}}$. Repeat above discussion, we get a normal subgroup series of E: $1\unlhd P \unlhd \cdots \unlhd E_{\{p_{1}, p_{2}\}{'}}\unlhd E_{p_{1}{'}}\unlhd E$. Hence E has supersolvable type Sylow tower and $P\unlhd E$, a contradiction.

If $O_{p}(E)\cap T=1$, then $O_{p}(E)\le N_{G}(T)$ by [3, Lemma A.14.3] and $O_{p}(E)T=O_{p}(E)\times T$. Since $O_{p}(E)\cap Z(G_{p})\ne 1$ where $G_{p}$ is a Sylow p-subgroup of G, we may pick a minimal subgroup H of $O_{p}(E)\cap Z(G_{p})$ and $H\unlhd G$. Clearly, (G / H, E / H) satisfies the hypothesis of Theorem and so (G / H, E / H) holds by the choice of (G, E). Then every G-chief factor below E holds, a contradiction.

• (4) The final contradiction.

Let $P_{2}$ be a maximal subgroup of P. By hypothesis and (3), $P_{2}$ is m-embedded in E, we may choose a subnormal subgroup $K_{2}$ of E such that $E=P_{2}K_{2}$ and $P_{2}\cap K_{2}=1$. Hence there exists a maximal normal subgroup K such that $|E:K|=p$. Clearly, $K_{p}$ is m-embedded in E, where $K_{p}$ is a Sylow p-subgroup of K, we may choose a subnormal subgroup $K_{3}$ of E such that $E=K_{p}K_{3}=KK_{3}$ and $K_{p}\cap K_{3}=1$. Since $|\,K_{p}K_{3}|=|\,KK_{3}|$, we have $|\,K\cap K_{3}|=\frac{|\,K|}{|\,K_{p}|}$. If $K\cap K_{3}\ne 1$, then $K\cap K_{3}$ is a $p{'}$-group, which contradicts (1). Hence $K\cap K_{3}=1$ and K is a normal p-subgroup, which contradicts (3).

The final contradiction completes our proof.$\square$

Corollary 3.12

Let E be a normal subgroup of G and P be a Sylow p-subgroup of E where p is an odd prime divisor of |E|. Suppose that $G/E=\overline{G}$ and every chief factor $\overline{A}/\overline{B}$ of $\overline{G}$ satisfies one of the following conditions : 

$$\begin{aligned} (1) \,\overline{A}/\overline{B}\le \Phi (\overline{G}/\overline{B}); \quad \quad \quad (2)\, \overline{A}/\overline{B} \text { is a} p{'}\text {-}\,\text {group}; \quad \quad \quad (3)\, |\overline{A}/\overline{B}|_{p}=p. \end{aligned}$$

If every maximal subgroup of P is m-embedded in G,  then every chief factor A / B of G satisfies one of the following conditions : 

$$\begin{aligned} (1)\, A/B\le \Phi (G/B); \quad \quad \quad (2)\, A/B \text { is a} p{'}\text {-}\,\text {group}; \quad \quad \quad (3)\, |A/B|_{p}=p. \end{aligned}$$

Corollary 3.13

[12, Theorem 3.1] Let p be a prime, G be a p-solvable group and let H be a normal subgroup of G such that $G/H\in$ $p\mathcal U$ $,$ $p\mathcal U$ is the class of all p-supersolvable groups. If the maximal subgroups of the Sylow p-subgroups of H are c-normal in G,  then $G \in$ $p\mathcal U$.