Some New Applications of Weakly ℋ-Embedded Subgroups of Finite Groups

Abstract

A subgroup H of a finite group G is said to be weakly $\mathcal{H}$-embedded in G if there exists a normal subgroup T of G such that $H^G=HT$ and $H\cap T\in \mathcal{H}\left(G\right)$, where $H^G$ is the normal closure of H in G, and $\mathcal{H}\left(G\right)$ is the set of all $\mathcal{H}$-subgroups of G. In the recent research, Asaad, Ramadan and Heliel gave new characterization of p-nilpotent: Let p be the smallest prime dividing $\left|G\right|$, and P a non-cyclic Sylow p-subgroup of G. Then G is p-nilpotent if and only if there exists a p-power d with $1<d<\left|P\right|$ such that all subgroups of P of order d and $pd$ are weakly $\mathcal{H}$-embedded in G. As new applications of weakly $\mathcal{H}$-embedded subgroups, in this paper, (1) we generalize this result for general prime p and get a new criterion for p-supersolubility; (2) adding the condition “$N_G\left(P\right)$ is p-nilpotent”, here $N_G\left(P\right)=\{g\in G|P^g=P\}$ is the normalizer of P in G, we obtain p-nilpotence for general prime p. Moreover, our tool is the weakly $\mathcal{H}$-embedded subgroup. However, instead of the normality of $H^G=HT$, we just need $HT$ is S-quasinormal in G, which means that $HT$ permutes with every Sylow subgroup of G.

1. Introduction

Throughout this paper, “G is a group” always means that “G is a finite group”. For convenience, one can refer to [1,2,3,4] for the definitions and notions in the paper.

The T-groups are defined as the groups G in which normality is a transitive relation, that is, if $H⊴K⊴G$, then $H⊴G$. In 2000, Bianchi Gillio Berta Mauri, Herzog and Verardi [5] proved a characterization of soluble T-groups by means of $\mathcal{H}$-subgroup: a subgroup H of a group G is called an $\mathcal{H}$-subgroup in G if $N_G\left(H\right)\cap H^g\le H$, for every element $g\in G$, where $N_G\left(H\right)=\{x\in G|H^x=H\}$ is the normalizer of H in G. They proved that a group G is a supersolvable T-group if and only if every subgroup of G is an $\mathcal{H}$-subgroup of G. Later, except for the exploration of T-groups, $\mathcal{H}$-subgroups were widely used to character finite groups. Cs$\ddot{o}$rg$\ddot{o}$ and Herzog [6] obtained that a group G is supersolvable if every cyclic subgroup of G of prime order or order 4 is an $\mathcal{H}$-subgroup. Asaad [7] proved that a group G is supersolvable if every maximal subgroup of every Sylow subgroup of G is an $\mathcal{H}$-subgroup. The set of all $\mathcal{H}$-subgroups of a group G is denoted by $\mathcal{H}\left(G\right)$. Moreover, Guo and Wei [8] gave new characterization of p-nilpotent or supersolvable by assuming some subgroups of G of the same order all belong to $\mathcal{H}\left(G\right)$, which provide a unified version of the results mentioned above if the order of G is odd. Moreover, Li, Zhao and Xu [9] considered the case when G is of even order.

Recently, Asaad et al. [10] introduced a new subgroup embedding property called weakly $\mathcal{H}$-subgroup, which generalizes both c-normality and $\mathcal{H}$-subgroup, called weakly $\mathcal{H}$-subgroup. Soon after, Asaad and Ramadan [11] gave the definition of weakly $\mathcal{H}$-embedded subgroup. Please note that a subgroup H of G is said to be a weakly $\mathcal{H}$-embedded subgroup (weakly $\mathcal{H}$-subgroup) of G if there exists a normal subgroup T of G such that $H^G=HT$$(G=HT)$ and $H\cap T\in \mathcal{H}\left(G\right)$, where $H^G$ is the normal closure of H in G. Clearly, c-normal subgroups, $\mathcal{H}$-subgroups and weakly $\mathcal{H}$-subgroups imply weakly $\mathcal{H}$-embedded subgroups. However, the converse does not hold in general, see [11] (Examples 1.3, 1.4 and 1.5).

In fact, these subgroups were widely used to investigate the structure of finite groups. As a result, many interesting results have been subsequently obtained, such as [7,10,11,12,13].

In the recent research about $\mathcal{H}$-subgroups, Asaad, Ramadan, and Heliel gave a new characterization of p-nilpotency.

Theorem 1.

([12] Theorem A) Let p be the smallest prime dividing $\left|G\right|$, and P a non-cyclic Sylow p-subgroup of G. Then G is p-nilpotent if and only if there exists a p-power d with $1<d<\left|P\right|$ such that all subgroups of P of order d and $pd$ are weakly $\mathcal{H}$-embedded in G.

However, according to this result, some natural questions arise:

Problem 1.

(1)

If delete the condition “p is the smallest prime dividing $\left|G\right|$”, can we claim that G is p-supersoluble?

(2)

Does there exist another condition to obtain p-nilpotence rather than “p is the smallest prime dividing $\left|G\right|$”?

(3)

As we know, the condition that $HT$ is the smallest normal subgroup of G containing H, is too strict. Can we replace it by a weaker embedding subgroup property?

In this paper, we further explore weakly $\mathcal{H}$-embedded subgroups and pay attention to Problem 1. However, instead of the normality of $HT$, we just consider $HT$ is S-quasinormal in G. As we know, a subgroup K is S-quasinormal in G, means that K permutes with every Sylow subgroup P of G, that is $KP=PK$. However, for convenience, we also called it a weakly $\mathcal{H}$-embedded subgroup, that is:

Definition 1.

A subgroup H of a group G is said to be weakly $\mathcal{H}$-embedded in G if there exists a normal subgroup T of G such that $HT$ is S-quasinormal in G and $H\cap T\in \mathcal{H}\left(G\right)$.

As an application of these subgroups, we give a positive answer to Problem 1 in the class of p-soluble groups, for detail:

Theorem 2.

Let E be a p-soluble normal subgroup of a group G such that $G/E$ is p-supersoluble, where p is a prime divisor of $\left|E\right|$. Let P be a Sylow p-subgroup of E. Suppose that P has a subgroup D with $1⩽\left|D\right|<\left|P\right|$ such that all subgroups of P of order $\left|D\right|$ and $p\left|D\right|$ are weakly $\mathcal{H}$-embedded in G. When $\left|D\right|=1$ and P is a non-abelian 2-group, we further assume that all cyclic subgroups of P of order 4 are weakly $\mathcal{H}$-embedded in G. Then G is p-supersoluble.

Moreover, to avoid the condition “p is the smallest prime dividing $\left|G\right|$” of Theorem 1, we further prove that the conclusion holds if this condition is replaced by “$N_G\left(P\right)$ is p-nilpotent”. Consequently, we give an answer to Problem 1.

Theorem 3.

Let E be a normal subgroup of G such that $G/E$ is p-nilpotent, and P be a non-cyclic Sylow p-subgroup of E, where p is a prime dividing $\left|E\right|$. Assume that $N_G\left(P\right)$ is p-nilpotent and P has a subgroup D with order $1<\left|D\right|<\left|P\right|$ such that all subgroups of P of order $\left|D\right|$ and order $p\left|D\right|$ are weakly $\mathcal{H}$-embedded in G. Then G is p-nilpotent.

In the second section, we list some lemmas which will be useful for the proofs of the above results. The proofs of Theorems 2 and 3 are put in the third section. Some previously known results are generalized by our theorems, and we list some in the fourth section.

2. Preliminaries

Lemma 1.

(see ([1], Chapter 1) or ([3], Chapter 1, Lemmas 5.34 and 5.35)) Assume that H, E are subgroups of G and $N⊴G$.

$\left(1\right)$If H is S-quasinormal in G, then $H\cap E$ is S-quasinormal in E, and $HN/N$ is S-quasinormal in $G/N$.

$\left(2\right)$Assume that H is a p-group. Then H is S-quasinormal in G if and only if $O^p\left(G\right)\le N_G\left(H\right)$.

$\left(3\right)$The set of S-quasinormal subgroups of G is a sublattice of the subnormal subgroup lattice of G.

$\left(4\right)$If H is a p-group and H is subnormal in G, then $H\le O_p\left(G\right)$.

Lemma 2.

([11] Lemma 2.1) Let H, N be subgroups of G satisfying $H\in \mathcal{H}\left(G\right)$ and $N⊴G$. Then:

$\left(1\right)$If E is a subgroup of G containing H, then $H\in \mathcal{H}\left(E\right)$;

$\left(2\right)$If H is subnormal in G, then H is normal in G;

$\left(3\right)$Assume that $N\le N_G\left(H\right)$. Then $NH\in \mathcal{H}\left(G\right)$;

$\left(4\right)$If E is a subgroup of G satisfying $N\le E$, then $E\in \mathcal{H}\left(G\right)$ if and only if $E/N\in \mathcal{H}(G/N)$;

$\left(5\right)$If H is a p-group and $p\nmid \left|N\right|$, then $NH\in \mathcal{H}\left(G\right)$ and $HN/N\in \mathcal{H}(G/N)$.

Lemma 3.

Let H be a weakly $\mathcal{H}$-embedded subgroup of a group G.

$\left(1\right)$Assume that E is a subgroup of G containing H. Then H is weakly $\mathcal{H}$-embedded in E.

$\left(2\right)$If N is a normal subgroup of G satisfying $N\le H$, then $H/N$ is weakly $\mathcal{H}$-embedded in $G/N$.

$\left(3\right)$Assume that H is a p-group and N a normal $p^{\prime }$-subgroup of G. Then $HN/N$ is weakly $\mathcal{H}$-embedded in $G/N$.

Proof. 

By the hypothesis, G has a normal subgroup T such that $HT$ is S-quasinormal in G and $H\cap T\in \mathcal{H}\left(G\right)$.

(1) Clearly, $T\cap E$ is a normal subgroup of E such that $H(T\cap E)=HT\cap E$ is S-quasinormal in E and $H\cap (T\cap E)=H\cap T\in \mathcal{H}\left(E\right)$ (see Lemmas 1(1) and 2(1)). This shows that H is weakly $\mathcal{H}$-embedded in E.

(2) Consider the normal subgroup $TN/N$ of $G/N$. Please note that $N\le N_G(H\cap T)$, so $(H\cap T)N\in \mathcal{H}\left(G\right)$ by Lemma 2(3). Furthermore, we have that $(H/N)(TN/N)=HT/N$ is S-quasinormal in $G/N$ and

$$(H/N)\cap (TN/N)=(H\cap T)N/N\in \mathcal{H}(G/N)$$

(see Lemmas 1(1) and 2(4)). By the definition, $H/N$ is weakly $\mathcal{H}$-embedded in $G/N$.

(3) By Lemma 1(1), the normal subgroup $TN/N$ of $G/N$ such that $(HN/N)(TN/N)=HTN/N$ is S-quasinormal in $G/N$. Please note that

$$\left(\right|HN\cap T:H\cap T|,|HN\cap T:N\cap T\left|\right)=\left(\right|N\cap HT|,|H\cap NT\left|\right)=1,$$

so $HN\cap T=(H\cap T)(N\cap T)$. Combining with Lemma 2(5),

$$(HN/N)\cap (TN/N)=(HN\cap T)N/N=(H\cap T)N/N\in \mathcal{H}(G/N).$$

Hence $HN/N$ is weakly $\mathcal{H}$-embedded in $G/N$. ☐

Recall that a class of groups $\mathfrak{F}$ is called a formation if for every group G, every homomorphic image of $G/G^{\mathfrak{F}}$ belongs to $\mathfrak{F}$, where $G^{\mathfrak{F}}=\bigcap \{N⊴G|G/N\in \mathfrak{F}\}$. Furthermore, a formation $\mathfrak{F}$ is said to be saturated if $G\in \mathfrak{F}$ whenever $G/\mathsf{\Phi }\left(G\right)\in \mathfrak{F}$. The intersection of all formations containing the set $\{G/O_{p^{\prime },p}\left(G\right)|G\in \mathfrak{F}\}$ is denoted by $\mathfrak{F}\left(p\right)$, and $F\left(p\right)$ denotes the class of all groups G such that $G^{\mathfrak{F}\left(p\right)}$ is a p-group. Associated with a saturated formation $\mathfrak{F}$, there is a function f of the form $f:\mathbb{P}\to${group formations}, where $f\left(p\right)=F\left(p\right)$ for any prime p, which divides $\left|G\right|$ for some $G\in \mathfrak{F}$, and $f\left(p\right)=\varnothing$ otherwise. The function f is called the canonical local satellite of $\mathfrak{F}$. For more detail, please turn to ([3] P. 3) or ([2] Chap. IV, Theorem 3.7 and Definitions 3.9). Now we recall the subgroup $Z_{\mathfrak{F}}\left(G\right)$ of G, which is called the $\mathfrak{F}$-hypercenter of G. In fact, $Z_{\mathfrak{F}}\left(G\right)$ the product of all such normal subgroups N of G whose G-chief factors $H/K$ satisfying $(H/K)⋊(G/C_G(H/K))\in \mathfrak{F}$.

Lemma 4.

Let $\mathfrak{F}$ be a saturated formation and f the canonical local satellite of $\mathfrak{F}$. Let P be a normal p-subgroup of G. Then $P\le Z_{\mathfrak{F}}\left(G\right)$ if and only if one of the following holds:

$\left(1\right)$$G/C_G\left(P\right)\in f\left(p\right)$ (([3] Chap. 1, Lemma 2.26) or ([14] Lemma 2.14));

$\left(2\right)$$P/\mathsf{\Phi }\left(P\right)\le Z_{\mathfrak{F}}(G/\mathsf{\Phi }\left(P\right))$ ([15] Lemma 2.8).

Lemma 5.

([1] Lemma 2.1.6) If G is p-supersoluble and $O_{p^{\prime }}\left(G\right)=1$, then G has the unique Sylow p-subgroup.

Lemma 6.

([2] Chap. A, Lemma 8.4) Let N be a nilpotent normal subgroup of G and M a maximal subgroup of G such that $N\nleqq M$. Then $N\cap M$ is a normal subgroup of G.

3. Proofs of Main Results

The following proposition plays an important role in the proof of Theorem 2.

Proposition 1.

Let P be a normal p-subgroup of a group G. Assume that P has a subgroup D satisfying $1⩽\left|D\right|<\left|P\right|$, such that all subgroups of P of order $\left|D\right|$ and $p\left|D\right|$ are weakly $\mathcal{H}$-embedded in G. When $\left|D\right|=1$ and P is a non-abelian 2-group, we further assume that all cyclic subgroups of P of order 4 are weakly $\mathcal{H}$-embedded in G. Then $P\le Z_{\mathfrak{U}}\left(G\right)$.

Proof. 

Assume by contradiction that $(G,P)$ is a counterexample of minimal order $\left|G\right|+\left|P\right|$. We proceed via the following steps.

(1) P is not a minimal normal subgroup of G.

Assume that P is minimal normal in G. Let H be a subgroup of P of order $\left|D\right|$ or $p\left|D\right|$, which is normal in some Sylow subgroup of G. By the hypothesis, H is weakly $\mathcal{H}$-embedded in G. So G has a normal subgroup T such that $HT$ is S-quasinormal in G and $H\cap T\in \mathcal{H}\left(G\right)$. Please note that $P\cap T$ is normal in G, so $P\cap T=1$ or $P\cap T=P$ by the minimality of P. If $P\cap T=1$, then $H=H(P\cap T)=P\cap HT$ is S-quasinormal in G. However, by the choice of H and Lemma 1(2), $H⊴G$, a contradiction. So $P\le T$. In this case, $H=H\cap T\in \mathcal{H}\left(G\right)$ and then $H⊴G$ by the relationship $H⊴P⊴G$ and Lemma 2(2), which is impossible. Thus, P is not a minimal normal of G.

(2) If every maximal subgroup of P is weakly $\mathcal{H}$-embedded in G, then $P\le Z_{\mathfrak{U}}\left(G\right)$.

Let N be a minimal normal subgroup of G contained in P. By Lemma 3(2), $(G/N,P/N)$ satisfies the hypothesis. So, the choice of $(G,P)$ implies that: (i) $P/N\le Z_{\mathfrak{U}}(G/N)$; (ii) N is non-cyclic; (iii) N is the unique minimal normal subgroup of G contained in P. Now assume that $\mathsf{\Phi }\left(P\right)=1$. In this case, P is elementary abelian and $P=N\times B$, where B is a complement of N. Let $N_1$ be a maximal subgroup of N such that $N_1$ is normal in some Sylow p-subgroup $G_p$ of G. Then $P_1=N_{1}B$ is a maximal subgroup of P. By the hypothesis, G has a normal subgroup T such that $P_{1}T$ is S-quasinormal in G and $P_1\cap T\in \mathcal{H}\left(G\right)$. Please note that $P\cap T$ is a normal subgroup of G contained in P, so $N\le P\cap T$ or $P\cap T=1$ by (iii). First, assume that $N\le T$. Then $1<N_1\le P_1\cap T$. However, $P_1\cap T⊴G$ by the relationship $P_1\cap T⊴P⊴G$ and Lemma 2(2). Thus, the uniqueness of N deduces that $N\le P_1\cap T\le P_1$, a contradiction. Secondly, if $P\cap T=1$, then $P_1=P_1(P\cap T)=P\cap P_{1}T$ is S-quasinormal in G, moreover $P_1\cap N=N_1$ is S-quasinormal in G by Lemma 1(3). Hence Lemma 1(2) and the choice of $N_1$ imply that $N_1⊴G$, a contradiction. The above shows that $\mathsf{\Phi }\left(P\right)\ne 1$ and consequently, $N\le \mathsf{\Phi }\left(P\right)$. Furthermore, $P/\mathsf{\Phi }\left(P\right)\le Z_{\mathfrak{U}}(G/\mathsf{\Phi }\left(P\right))$. However, we have $P\le Z_{\mathfrak{U}}\left(G\right)$ by Lemma 4. This contradiction shows that (2) holds.

(3) If every cyclic subgroup of P of order p or 4 (when P is a non-abelian 2-group) is weakly $\mathcal{H}$-embedded in G, then $P\le Z_{\mathfrak{U}}\left(G\right)$.

If P is not a non-abelian 2-group, then we use $\mathsf{\Omega }$ to denote the subgroup ${\mathsf{\Omega }}_1\left(P\right)$ of P. Otherwise, $\mathsf{\Omega }={\mathsf{\Omega }}_2\left(P\right)$.

Let R be a normal subgroup of G such that $P/R$ is a G-chief factor. Obviously, R satisfies the hypothesis. So $R\le Z_{\mathfrak{U}}\left(G\right)$ and $P/R$ is non-cyclic by the choice of $(G,P)$. Moreover, for any normal subgroup L of G satisfying $L<P$, we have $L\le R$. In fact, if $L\nleqq R$, then similarly $L\le Z_{\mathfrak{U}}\left(G\right)$, and $P=RL\le Z_{\mathfrak{U}}\left(G\right)$, a contradiction. Now, assume that $\mathsf{\Omega }\le R$. Then $\mathsf{\Omega }\le Z_{\mathfrak{U}}\left(G\right)$. From Lemma 4 and ([16] Lemma 2.4), it follows that $G/C_G\left(\mathsf{\Omega }\right)\in F\left(p\right)$ and $C_G\left(\mathsf{\Omega }\right)/C_G\left(P\right)\in {\mathfrak{N}}_p$, where F is the canonical local satellite of $\mathfrak{U}$ and ${\mathfrak{N}}_p$ is the class of p-groups. Consequently, $G/C_G\left(P\right)\in {\mathfrak{N}}_{p}F\left(p\right)=F\left(p\right)$, and thereby $P\le Z_{\mathfrak{U}}\left(G\right)$ by Lemma 4 again. This contradiction shows that $\mathsf{\Omega }=P$.

Let $L/R$ be a minimal subgroup of $Z(G_p/R)\cap P/R$ and $x\in L\setminus R$, where $G_p$ is a Sylow p-subgroup of G. Then $H=\langle x\rangle$ has order p or 4 and $L=HR$. By the hypothesis, H is weakly $\mathcal{H}$-embedded in G, so G has a normal subgroup T such that $HT$ is S-quasinormal in G and $H\cap T\in \mathcal{H}\left(G\right)$. Please note that $P\cap T⊴G$. Combining with the above result, we have $P\cap T=P$ or $P\cap T\le R$. If $P\cap T=P$, that is, $P\le T$, then $H=H\cap T\in \mathcal{H}\left(G\right)$. Moreover, the relationship $H⊴⊴P⊴G$ and Lemma 2(2) deduce $H⊴G$. By the choice of H, we have $P/R=L/R$ is cyclic, which is a contradiction. Now assume that $P\cap T\le R$. Then

$$L/R=HR/R=H(P\cap T)R/R=P/R\cap HTR/R$$

is S-quasinormal in $G/R$ by Lemma 1(3). From Lemma 1(2) and the choice of $L/R$, it follows that $L/R⊴G/R$, which also shows that $P/R=L/R$, a contradiction. This completes the proof of (3).

(4) $p<\left|D\right|<\frac{\left|P\right|}{p^2}$ (it follows directly from (2) and (3)).

(5) $\mathsf{\Phi }\left(P\right)=1$.

Suppose that $\mathsf{\Phi }\left(P\right)>1$. We compare the order of $\mathsf{\Phi }\left(P\right)$ with $\left|D\right|$. First, assume that $\left|\mathsf{\Phi }\right(P\left)\right|>\left|D\right|$. In this case, we have $\mathsf{\Phi }\left(P\right)\le Z_{\mathfrak{U}}\left(G\right)$ by the hypothesis and the choice of P. Let N be a minimal normal subgroup of G contained in $\mathsf{\Phi }\left(P\right)$. Clearly, $\left|N\right|=p$ and by (4), $P/N$ satisfies the hypothesis. Thus, $P/N\le Z_{\mathfrak{U}}(G/N)$ and consequently $P\le Z_{\mathfrak{U}}\left(G\right)$, a contradiction. So $\left|\mathsf{\Phi }\right(P\left)\right|⩽\left|D\right|$. Please note that $P/\mathsf{\Phi }\left(P\right)$ is elementary abelian, so we can easily prove that $P/\mathsf{\Phi }\left(P\right)$ satisfies the hypothesis. Therefore, $P/\mathsf{\Phi }\left(P\right)\le Z_{\mathfrak{U}}(G/\mathsf{\Phi }\left(P\right))$ and by Lemma 4, we further have $P\le Z_{\mathfrak{U}}\left(G\right)$. This contradiction shows that $\mathsf{\Phi }\left(P\right)=1$.

(6) Final contradiction.

Let N be a minimal normal subgroup of G contained in P. Clearly, $N<P$. Compare the order of N with $\left|D\right|$. If $\left|D\right|<\left|N\right|$, then N satisfies the hypothesis and the choice of P implies that $N\le Z_{\mathfrak{U}}\left(G\right)$. Consequently, $\left|N\right|=p$ and then $\left|D\right|=1$, which contradicts (4). Thus, $\left|D\right|⩾\left|N\right|$. By (5), P is elementary abelian, and all subgroups of $P/N$ of order $\left|D\right|/\left|N\right|$ and $p\left|D\right|/\left|N\right|$ are weakly $\mathcal{H}$-embedded in G (see Lemma 3(2)). Therefore $P/N\le Z_{\mathfrak{U}}(G/N)$ by the choice of P. Please note that $|P/N|⩾\left|P\right|/\left|D\right|>p^2$. So there exists a normal subgroup E of G contained in P satisfying $N\le E\le P$ and $|P/E|=p$. Consider the subgroup E. Then $E\le Z_{\mathfrak{U}}\left(G\right)$ by the hypothesis and the choice of P, which implies $\left|N\right|=p$. Combining with $P/N\le Z_{\mathfrak{U}}(G/N)$, we finally obtain $P\le Z_{\mathfrak{U}}\left(G\right)$, which is a contradiction. The final contradiction completes the proof of the proposition. ☐

Now we give the proof of Theorem 2:

Proof. 

Suppose that the assertion is false and consider a counterexample $(G,E)$ with minimal $\left|G\right|+\left|E\right|$. We proceed via the following steps.

(1) $O_{p^{\prime }}\left(E\right)=1$.

Clearly, $(G/O_{p^{\prime }}\left(E\right),E/O_{p^{\prime }}\left(E\right))$ satisfies the hypothesis by Lemma 3(3). If $O_{p^{\prime }}\left(E\right)>1$, then the choice of G implies that $G/O_{p^{\prime }}\left(E\right)$ is p-supersoluble. Furthermore, G is p-supersoluble, which is a contradiction. Thus, $O_{p^{\prime }}\left(E\right)=1$.

(2) $E=G$.

Suppose that $E<G$. Please note that Lemma 3(1) shows that $(E,E)$ satisfies the hypothesis, so E is p-supersoluble. Combining (1) with Lemma 5, we have $P⊴E$ and consequently, $P⊴G$. From the hypothesis and Proposition 1, it follows that $P\le Z_{\mathfrak{U}}\left(G\right)$. This result implies $E\le Z_{p\mathfrak{U}}\left(G\right)$ and then G is p-supersoluble, which is a contradiction. Thus, $E=G$.

(3) If every maximal subgroup of P is weakly $\mathcal{H}$-embedded in G, then G is p-supersoluble.

Let N be a minimal normal subgroup of G. Since G is p-soluble and $O_{p^{\prime }}\left(G\right)=1$, $N\le O_p\left(G\right)$. By Lemma 3(2), $G/N$ satisfies the hypothesis, so: (i) $G/N$ is p-supersoluble; (ii) $\left|N\right|>p$; (iii) N is the unique minimal normal subgroup of G. Obviously, $N\nleqq \mathsf{\Phi }\left(G\right)$, so there exists a maximal subgroup M of G such that $G=N⋊M$. By Lemma 6, $O_p\left(G\right)\cap M⊴G$. So $O_p\left(G\right)\cap M=1$ by the uniqueness of N, and then

$$O_p\left(G\right)=N(O_p\left(G\right)\cap M)=N.$$

On one hand, $O_p\left(G\right)\le C_G\left(O_p\left(G\right)\right)$ by the minimality of $O_p\left(G\right)$. On the other hand, since G is p-soluble and $O_{p^{\prime }}\left(G\right)=1$,

$$C_G\left(O_p\left(G\right)\right)=C_G\left(F\left(G\right)\right)\le F\left(G\right)=O_p\left(G\right).$$

In general, $C_G\left(O_p\left(G\right)\right)=O_p\left(G\right)$. Now we show that $O_p\left(G\right)<P$. In fact, if $P⊴G$, then $P\le Z_{\mathfrak{U}}\left(G\right)$ by Proposition 1. Similar to step (2), it is impossible.

Using the above symbol, $G=O_p\left(G\right)⋊M$ and then $P=O_p\left(G\right)⋊(P\cap M)$. Let $P_1$ be a maximal subgroup of P containing $P\cap M$. Then $P_1\cap O_p\left(G\right)>1$ and it is not normal in G. In fact, if $P_1\cap O_p\left(G\right)⊴G$, then $O_p\left(G\right)\le P_1\cap O_p\left(G\right)\le P_1$ by the minimality of $O_p\left(G\right)$ and consequently, $P=P_1$, a contradiction. By the hypothesis, $P_1$ is weakly $\mathcal{H}$-embedded in G. So G has a normal subgroup T such that $P_{1}T$ is S-quasinormal in G and $P_1\cap T\in \mathcal{H}\left(G\right)$. If $T=1$, then $P_1$ is S-quasinormal in G, which implies that $P_1\le O_p\left(G\right)$ by Lemma 1(3)(4) and then $O_p\left(G\right)=P$. However, it contradicts the above result. So, the uniqueness of $O_p\left(G\right)$ implies that $O_p\left(G\right)\le T$. Next, we prove that

$$P_1\cap O_p\left(G\right)\in \mathcal{H}\left(G\right).$$

First, we show that $N_G(P_1\cap O_p\left(G\right))=N_G(P_1\cap T)$. On one hand, note that

$$P_1\cap O_p\left(G\right)=(P_1\cap T)\cap O_p\left(G\right),$$

so

$$P\le N_G(P_1\cap T)\le N_G(P_1\cap O_p\left(G\right))<G.$$

On the other hand, $N_G(P_1\cap O_p\left(G\right))$ is p-supersoluble by Lemma 3(1) and the relation

$$N_G(P_1\cap O_p\left(G\right))<G.$$

Please note that $C_G\left(O_p\left(G\right)\right)=O_p\left(G\right)$, so it is rather clear that $O_{p^{\prime }}\left(N_G(P_1\cap O_p\left(G\right))\right)=1$. Thus, P is normal in $N_G(P_1\cap O_p\left(G\right))$ by Lemma 5. At this moment, we have

$$P_1\cap T⊴P⊴N_G(P_1\cap O_p\left(G\right)),$$

and by Lemma 2(1),

$$P_1\cap T\in \mathcal{H}\left(N_G(P_1\cap O_p\left(G\right))\right).$$

Consequently, $P_1\cap T⊴N_G(P_1\cap O_p\left(G\right))$ by Lemma 2(2), that is, $N_G(P_1\cap O_p\left(G\right))\le N_G(P_1\cap T)$. Together with the above proof, we finally obtain $N_G(P_1\cap O_p\left(G\right))=N_G(P_1\cap T)$. Please note that $P_1\cap T\in \mathcal{H}\left(G\right)$. So, for any element $g\in G$,

$${(P_1\cap O_p\left(G\right))}^g\cap N_G(P_1\cap O_p\left(G\right))={(P_1\cap T)}^g\cap O_p\left(G\right)\cap N_G(P_1\cap T)\le P_1\cap T\cap O_p\left(G\right)=P_1\cap O_p\left(G\right).$$

This shows that $P_1\cap O_p\left(G\right)\in \mathcal{H}\left(G\right)$. By Lemma 2(2), we further have $P_1\cap O_p\left(G\right)⊴G$, a contradiction. This completes the proof of (3).

(4) If every cyclic subgroup of P of order p or 4 (when P is a non-abelian 2-group) is weakly $\mathcal{H}$-embedded in G, then G is p-supersoluble.

Let M be any proper subgroup of G and $M_p$ a Sylow p-subgroup of M. Clearly, ${\left(M_p\right)}^g\le P$ for some element $g\in G$. Then consider $M^g$, which has a Sylow p-subgroup ${\left(M_p\right)}^g$ contained in P. So, without loss of generality, assume that $M_p\le P$. By Lemma 3(1), M satisfies the hypothesis, so the choice of G implies that M is p-supersoluble. As a result, G is a minimal non-p-supersoluble group.

By ([17] Theorem 1), $G^{{\mathfrak{U}}^p}\mathsf{\Phi }\left(G\right)/\mathsf{\Phi }\left(G\right)$ is the unique minimal normal subgroup of $G/\mathsf{\Phi }\left(G\right)$, where ${\mathfrak{U}}^p$ is the class of all p-supersoluble groups. Clearly, $p\mid |G^{{\mathfrak{U}}^p}\mathsf{\Phi }\left(G\right)/\mathsf{\Phi }\left(G\right)|$, so $G^{{\mathfrak{U}}^p}\mathsf{\Phi }\left(G\right)/\mathsf{\Phi }\left(G\right)$ is a p-group and $G^{{\mathfrak{U}}^p}$ is solvable. From ([18] Theorem 3.4.2), it follows that $G^{{\mathfrak{U}}^p}$ is a p-group of exponent p or 4 (when $G^{{\mathfrak{U}}^p}$ is a non-abelian 2-group). By the hypothesis, every cyclic subgroup of $G^{{\mathfrak{U}}^p}$ of order p is weakly $\mathcal{H}$-embedded in G. When $G^{{\mathfrak{U}}^p}$ is a non-abelian 2-group, clearly, P is also a non-abelian 2-group, so every cyclic subgroup of $G^{{\mathfrak{U}}^p}$ of order 4 is also weakly $\mathcal{H}$-embedded in G in this case. Hence, we have $G^{{\mathfrak{U}}^p}\le Z_{\mathfrak{U}}\left(G\right)$ by Proposition 1, and then G is p-supersoluble, a contradiction. So (4) holds.

(5) $p<\left|D\right|<\frac{\left|P\right|}{p^2}$ (It follows directly from (3) and (4)).

(6) $p>2$ (It follows directly from (2), (5) and Theorem 1).

(7) $O_p\left(G\right)$ is the unique minimal normal subgroup of G and $G/O_p\left(G\right)$ is p-supersoluble.

Let N be a minimal normal subgroup of G. Clearly, $N\le O_p\left(G\right)$. If $\left|N\right|>\left|D\right|$, then $N\le Z_{\mathfrak{U}}\left(G\right)$ by Proposition 1, which shows that $\left|N\right|=p$ and $\left|D\right|=1$, a contradiction. So, we have $\left|N\right|⩽\left|D\right|$. Please note that $p>2$, so it is easy to show that $(G/N,P/N)$ satisfies the hypothesis. Thus, the choice of G implies that: $G/N$ is p-supersoluble; $\left|N\right|>p$; N is the unique minimal normal subgroup of G. Since $N\nleqq \mathsf{\Phi }\left(G\right)$, there exists a maximal subgroup M of G such that $G=N⋊M$. By Lemma 6, $O_p\left(G\right)\cap M⊴G$, so $O_p\left(G\right)\cap M=1$ and $O_p\left(G\right)=N(O_p\left(G\right)\cap M)=N$. Thus, (7) holds.

(8) Final contradiction.

Let R be a normal subgroup of G such that $O_p\left(G\right)\le R\le G$ and $G/R$ is a G-chief factor. Please note that $G/O_p\left(G\right)$ is p-supersoluble. So $|G/R|=p$ or $p\nmid |G/R|$. First, assume that $|G/R|=p$. Then $|P:P\cap R|=p$ and by (6), R satisfies the hypothesis of the theorem. So R is p-supersoluble. Please note that $O_{p^{\prime }}\left(R\right)\le O_{p^{\prime }}\left(G\right)=1$. Together with Lemma 5, R has the unique Sylow p-subgroup $P\cap R$, and furthermore, $P\cap R⊴G$. By (6), $P\cap R$ satisfies the hypothesis of Proposition 1. Thus, $P\cap R\le Z_{\mathfrak{U}}\left(G\right)$, that is, $R\le Z_{p\mathfrak{U}}\left(G\right)$, which deduces that G is p-supersoluble, a contradiction. Then assume that $p\nmid |G/R|$, that is $P\le R$. In this case, R satisfies the hypothesis and so R is p-supersoluble by the choice of G. Similarly, we have $O_{p^{\prime }}\left(R\right)=1$ and by Lemma 5, $P⊴R$, which implies that $P⊴G$. By Proposition 1, $P\le Z_{\mathfrak{U}}\left(G\right)$ and consequently, G is p-supersoluble, a contradiction. The final contradiction completes the proof of the theorem. ☐

Next we give the proof of Theorem 3:

Proof. 

Suppose that the assertion is false and consider a counterexample G of minimal order. According to Theorem 1, we only need to consider that p is odd. We proceed via the following steps.

(1) $O_{p^{\prime }}\left(E\right)=1$.

If $O_{p^{\prime }}\left(E\right)>1$, then it is normal in G. Consider $\overline{G}=G/O_{p^{\prime }}\left(E\right)$. Please note that $\overline{P}$ a Sylow p-subgroup of $\overline{E}$ and $N_{\overline{G}}\left(\overline{P}\right)=\overline{N_G\left(P\right)}$ is p-nilpotent. Moreover, by hypothesis and Lemma 3(3), all subgroups of $\overline{P}$ of order $\left|D\right|$ and order $p\left|D\right|$ are weakly $\mathcal{H}$-embedded in $\overline{G}$, that is $\overline{G}$ satisfies the hypothesis for G. Thus, the choice of G implies that $\overline{G}$ is p-nilpotent. Consequently, G is p-nilpotent, a contradiction. So $O_{p^{\prime }}\left(E\right)=1$.

(2) $E=G$.

By Lemma 3(1), all subgroups of P of order $\left|D\right|$ and order $p\left|D\right|$ are weakly $\mathcal{H}$-embedded in E. Since $N_E\left(P\right)=N_G\left(P\right)\cap E$, $N_E\left(P\right)$ is p-nilpotent. Then E satisfies the hypothesis. If $E<G$, then E is p-nilpotent by the choice of G. Let $E_{p^{\prime }}$ be the normal $p^{\prime }$-Hall subgroup of E. Clearly, $E_{p^{\prime }}⊴G$. So, by (1), $E_{p^{\prime }}=1$, that is, $E=P$. In this case, $G=N_G\left(P\right)$ is p-nilpotent. This contradiction shows that $E=G$.

(3) $O_p\left(G\right)>1$.

Let $J\left(P\right)$ be the Thompson subgroup of P. Then clearly, $Z\left(J\right(P\left)\right)>1$, $P\le N_G\left(Z\left(J\left(P\right)\right)\right)$ and $N_{N_G\left(Z\left(J\left(P\right)\right)\right)}\left(P\right)$ is p-nilpotent. Assume that $N_G\left(Z\left(J\left(P\right)\right)\right)<G$. Please note that $N_G\left(Z\left(J\left(P\right)\right)\right)$ satisfies the hypothesis by Lemma 3(1). So, the choice of G implies that $N_G\left(Z\left(J\left(P\right)\right)\right)$ is p-nilpotent. However, it contradicts ([19] Theorem 8.3.1). Thus, $N_G\left(Z\left(J\left(P\right)\right)\right)=G$, that is $Z\left(J\right(P\left)\right)⊴G$, which shows that (3) holds.

(4) G is not p-soluble.

Suppose that G is p-soluble. Then G is p-supersoluble by the Theorem 2. Please note that $O_{p^{\prime }}\left(G\right)=1$. So $P⊴G$ by Lemma 5, which shows that $N_G\left(P\right)=G$ is p-nilpotent, a contradiction. Thus, (4) holds.

(5) Let N be a minimal normal subgroup of G contained in $O_p\left(G\right)$. Then $\left|N\right|>\left|D\right|$.

If $\left|N\right|=\left|D\right|$, then every subgroup of $P/N$ of order p is weakly $\mathcal{H}$-embedded in $G/N$ by Lemma 3(2). Denote $\overline{G}=G/N$. Let $\overline{M}$ be a proper subgroup of $\overline{G}$ and $\overline{M_p}$ a Sylow p-subgroup of $\overline{M}$. Clearly, ${\overline{M_p}}^{\overline{g}}\le \overline{P}$ for some $\overline{g}\in \overline{G}$. Now consider $\overline{M^g}$, which has a Sylow p-subgroup ${\overline{M_p}}^{\overline{g}}$ contained in $\overline{P}$. Without loss of generality, we can assume that the Sylow p-subgroup $\overline{M_p}$ of $\overline{M}$ contains in $\overline{P}$. By Lemma 3(1), every cyclic subgroup of $\overline{M_p}$ of order p is weakly $\mathcal{H}$-embedded in $\overline{M}$. Moreover, $N_{\overline{M}}\left(\overline{P}\right)=\overline{N_M\left(P\right)}$ is p-nilpotent. So $\overline{M}$ satisfies the hypothesis, and the choice of G implies that $\overline{M}$ is p-nilpotent. Consequently, G is a minimal non-p-nilpotent group. However, in this case, G is soluble, which contradicts (4). Suppose that $\left|N\right|<\left|D\right|$. Then all subgroups of $P/N$ of order $\left|D\right|/\left|N\right|$ and $p\left|D\right|/\left|N\right|$ are weakly $\mathcal{H}$-embedded in $G/N$ by Lemma 3(2), that is $G/N$ satisfies the hypothesis for G. SoSo, from the choice of G, we deduce that $G/N$ is p-nilpotent. Similarly, G is p-soluble in this case, a contradiction. Thus, $\left|N\right|>\left|D\right|$.

(6) Final contradiction.

By (5), all subgroups of N of order $\left|D\right|$ and $p\left|D\right|$ are weakly $\mathcal{H}$-embedded in G. Then $N\le Z_{\mathfrak{U}}\left(G\right)$ by Proposition 1. From this result, we deduce that $\left|N\right|=p$ and $\left|D\right|=1$, that is, every subgroup of P of order p is weakly $\mathcal{H}$-embedded in G. Similarly, as the proof of (5), we can prove that in this case G is soluble, a contradiction. The final contradiction completes the proof. ☐

4. Some Applications

In this section, we list some applications of our results.

Corollary 1.

Let E be a normal subgroup of G. For every non-cyclic Sylow subgroup P of E, assume that P has a subgroup D such that $1<\left|D\right|<\left|P\right|$ and all subgroups of P of order $\left|D\right|$ and $p\left|D\right|$ are weakly $\mathcal{H}$-embedded in G. Then $E\le Z_{\mathfrak{U}}\left(G\right)$.

Proof. 

Assume that p is the smallest prime divisor of $\left|E\right|$ and P is a Sylow p-subgroup of E. If P is cyclic, then E is p-nilpotent by the famous Burnside Theorem. Otherwise, by Lemma 3(1) and the hypothesis, all subgroups of P of order $\left|D\right|$ and $p\left|D\right|$ are weakly $\mathcal{H}$-embedded in E. So E is p-nilpotent by Theorem 1, and then E is soluble. By Lemma 3(1) again, we have that for any prime p dividing $\left|E\right|$, E satisfies the hypothesis of Theorem 2. So E is supersoluble. Let q be the maximal prime dividing $\left|E\right|$ and Q the unique Sylow q-subgroup of E. Clearly, $Q⊴G$. Note that Q satisfies the hypothesis of Proposition 1, so $Q\le Z_{\mathfrak{U}}\left(G\right)$. Now consider $E/Q$. By Lemma 3(3), $E/Q$ satisfies the hypothesis of corollary. So $E/Q\le Z_{\mathfrak{U}}(E/G)$ by induction. Therefore, $E\le Z_{\mathfrak{U}}\left(G\right)$. ☐

Corollary 2.

([12]) Assume that the Sylow subgroups of G are non-cyclic for all primes p dividing $\left|G\right|$. Assume further that for each such p there is a p-power d with ${1<d<\left|G\right|}_p$ such that all subgroups of P of order d and $pd$ are weakly $\mathcal{H}$-embedded in G, then G is supersoluble.

Proof. 

Let p be the smallest prime dividing $\left|G\right|$. By Theorem 1, G is p-nilpotent. Consequently, G is soluble. From the Theorem 2, it follows that G is q-supersoluble, for any prime divisor q of $\left|G\right|$, that is, G is supersoluble. ☐

Corollary 3.

([10]) Let P be a normal p-subgroup of a group G. If all maximal subgroups of P are weakly $\mathcal{H}$-subgroups in G, then $P\le Z_{\mathfrak{U}}\left(G\right)$.

Corollary 4.

([10]) Let $\mathfrak{F}$ be a saturated formation containing the class of supersolvable groups $\mathfrak{U}$. A group G lies in $\mathfrak{F}$ if and only if it has a normal subgroup H such that $G/H\in \mathfrak{F}$ and all maximal subgroups of every Sylow subgroup of H(or $F^{\ast }\left(H\right)$) are weakly $\mathcal{H}$-subgroups in G.

Corollary 5.

G is supersolvable, if one of the following holds:

$\left(1\right)$G has a normal subgroup H such that $G/H$ is supersolvable and all maximal subgroups of every Sylow subgroup of H belong to $\mathcal{H}\left(G\right)$ [7];

$\left(2\right)$all maximal subgroups of every Sylow subgroup of $F^{\ast }\left(G\right)$ belong to $\mathcal{H}\left(G\right)$ [7];

$\left(3\right)$all maximal subgroups of every Sylow subgroup of a group G are weakly $\mathcal{H}$-subgroups in G [10].

5. Conclusions

In this paper, we further explore weakly $\mathcal{H}$-embedded subgroups. As new applications, we generalize the characterization of p-nilpotent given by Asaad, Ramadan and Heliel and get a new criterion for p-supersolubility for general prime p. Moreover, adding condition “$N_G\left(P\right)$ is p-nilpotent”, we obtain p-nilpotence for general prime p.