Influence of partially τ-embedded subgroups of prime power order in supersolubility and p-nilpotency of finite groups

Abstract

In this paper, we introduced a new concept of $p\tau$ embedded subgroups which belongs to an embedded class of subgroups of finite groups. A subgroup H of a group G is said to be a partially τ-embedded subgroup in G if there exists a normal subgroup K of G such that HK is normal in G and $H\cap K\le H_{p\tau G}$ where $H_{p\tau G}$ generated by all those subgroups of H which are partially τ-quasinormal in G. We investigate the influence of some $p\tau$-embedded subgroups with prime power order on the structure of a finite group G. Some new criteria about the p-nilpotency and supersolubility of a finite group were obtained. Our results also generalized some earlier ones about formations.

Keywords:

• Partially τ-quasinormal subgroup
• τ-quasinormal subgroup
• supersolvable group

1. Introduction

During the twentieth century, mathematicians investigated some aspects of the theory of finite groups in great depth, especially the local theory of finite groups and the theory of solvable and nilpotent groups. As a consequence, the complete classification of finite simple groups was achieved, meaning that all those simple groups from which all finite groups can be built are now known. During the second half of the twentieth century, mathematicians such as Chevalley and Steinberg also increased our understanding of finite analogs of classical groups, and other related groups. One such family of groups is the family of general linear groups over finite fields. Finite group theory fund many applications in chemistry, physics, engineering and other areas of sciences.

In this article, G denotes the finite groups. All the notations are standard, as in [1,2]. Order of G denoted by $\left|G\right|$, Sylow q-subgroup of G denoted by G${}_q$ and Sylow subgroup of G denoted by Syl(G) or simply Syl. partially τ-quasinormal subgroups denoted by pτ-quasinormal and partially τ-embedded subgroups simply denoted by pτ-embedded.

Many authors worked on quasinormal subgroups and gave generalizations of normal subgroups. For example, Kegel [3] gave the extension of S-quasinormal. τ-quasinormality is the generalization of the S-quasinormal subgroups of finite groups. Recently, Li et al. [2] generalized τ-quasinormal subgroups to pτ-quasinormal. H is pτ-quasinormal in G if every Syl(H) is a Sylow subgroup of some τ-quasinormal subgroup of G. Many authors discussed the concept of permutable groups and gave many new concepts and generalizations of permutable groups. Now we present another extension of permutable groups and by the use of the following definitions we extend the permutable groups.

Definition 1.1

H is called partially τ-embedded subgruop in G if K $⊴$ G such that HK is normal in G and H ∩ K ≤ $H_{p\tau G}$ where $H_{p\tau G}$ generated by all those subgroups of H which are pτ-quasinormal in G. We call $H_{p\tau G}$ the pτ-core of H in G.

Clearly, all τ-quasinormal and pτ-quasinormal subgroups are pτ-embedded subgroups.

Remark 1.2

τ-quasinormality and pτ-quasinormality implies pτ-embeddedity.

Example 1.3

Let G=S${}_4$ and H=$⟨\left(14\right)⟩$. Then H is pτ-embedded but H is not τ-quasinormal. Since H and every subgroup of G with order 6 containing H cannot permute with every Syl(G). As T=$A_4$, HT $⊴$ G and H ∩ $A_4$ ≤ $H_{\tau G}=1\le H_{p\tau G}=\mathrm{H}$. Clearly pτ-embedded is an extension of pτ-quasinormal. So pτ-embedded is an extended form of τ-quasinormal and pτ-quasinormal.

Our contribution in this direction are following theorems about pτ-embedded.

Theorem 1.4

Let H $⊴$ G such that G/H is supersolvable. If every maximal subgroup of any Syl of $\mathrm{\Phi }\left(H\right)$ is pτ-embedded in G, then G is supersolvable.

Theorem 1.5

Let Q be a $G_q$, where q is a prime divisor of $|G|$ with ($|G|$, $(q-1)(q^2-1)\cdots (q^n-1))=1$ (n≥1). If every maximal subgroup of Q and every n-maximal subgroup of Q is pτ-embedded in G, then G is q-nilpotent.

Theorem 1.6

Suppose M $⊴$ G and Q be a Sylow q-subgroup of M such that $(|M|,q-1)=1$ provided q is a prime divisor of $\left|M\right|$. If every largest subgroup Q${}_1$ of Q is pτ-embedded in G such that Q${}_1$ does not have a q-supersolvable supplement in G, then each chief factor of G between M and O${}_{q^{\prime }}$(M) is cyclic.

2. Preliminaries

This section contains some basic results that help us in proving our main results.

Lemma 2.1

Suppose that A and B are two subgroups of a finite group G, then the following statements hold:

1. If A is S-quasinormal in G, then $A\cap B$ is S-quasinormal in B [4].

2. If A and B are S-quasinormal in G, then $A\cap B$ is S-quasinormal in G [3].

3. If A is S-quasinormal in G, then $A/A_G$ is nilpotent [4].

Lemma 2.2

[3,4]

Suppose that H be an S-quasinormal subgroup of the finite group G. Then

1. If $H\le K\le G$, then H is S-quasinormal in K.

2. If N is a normal subgroup of G, then HN is S-quasinormal in G and HN/N is S-quasinormal in $G/N.$

3. If $K\le G,$ then $H\cap K$ is S-quasinormal in K.

4. H is subnormal in G.

5. If $K\le G$ and K is S-quasinormal in G, then $H\cap K$ is S-quasinormal in G.

Lemma 2.3

[5]

Let Q is S-quasinormal q-subgroup of the finite group G for some prime q, then we have $N_G\left(Q\right)\ge O^q\left(G\right)$.

Lemma 2.4

Let $M⊴G$ and $H\le T\le G$. Then

1. $H_{p\tau G}⊴H.$

2. $H_{p\tau G}\le H_{p\tau K}$.

3. $H_{p\tau G}M/M\le (HM/M{)}_{p\tau \left(G/M\right)}$.

Proof.

Suppose that L be a partially τ-quasinormal subgroup of G contained in H and q be a prime dividing $\left|L\right|$. Further suppose that Q a Sylow q-subgroup of L and E an S-quasinormal subgroup of G such that $Q\in y{l}_q\left(E\right)$. Then the proof is given as follows:

1. Let $x\in H,$ then $L^x\le H.$ If R is a Sylow q-subgroup of $L^x$, then $R=Q_1^x$ for some Sylow q-subgroup $Q_1=Q.$ Obviously, $Q^x\in Sy{l}_q\left(E^x\right)$ and $E^x$ is an S-quasinormal subgroup of G. Hence $L^x$ is a partially τ-quasinormal subgroup of G. This implies that (1) holds.

2. By Lemma 2.1(1), $E\cap K$ is an s-quasinormal subgroup of K. Since $Q\le E\cap K$, Q is a Sylow q-subgroup of $E\cap K$. Hence $L\le H_{p\tau K}$ and so $H_{p\tau G}\le H_{p\tau K}$.

3. Clearly, $LM/M\le HM/M$ and LM/M is a partially τ-quasinormal subgroup of $G/M.$ Hence $H_{p\tau G}M/M\le (HM/M{)}_{p\tau \left(G/M\right)}$.

Lemma 2.5

Let $H\le G.$ Then we have the following:

1. H is $p\tau$-embedded in G and $H\le N\le G,$ then H is $p\tau$-embedded in N.

2. Suppose that $M⊴G$ and $M\le H.$ If H is a q-group and $p\tau$-embedded in G, then H/M is $p\tau$-embedded in $G/M.$

3. Suppose that H is a q-subgroup of G and M is a normal $q^{\mathrm{\prime }}$-subgroup of G. If H is $p\tau$-embedded in G, then HM/M is $p\tau$-embedded in $G/M.$

4. If H is $p\tau$-embedded in G and $H\le T⊴G,$ then there exists $K⊴G$ such that $HK⊴G,H\cap K\le H_{p\tau G}$ and $HK\le T.$

Proof.

By hypothesis, $T\le G$ such that $HT⊴G,H\cap T\le H_{p\tau G}$.

1. Then by Lemma 2.4 $H(N\cap T)=N\cap HT⊴N$and $H\cap T\le H_{p\tau G}\le H_{p\tau N}.$Hence, H be $p\tau$-embedded in N.

2. We know that $\left(H/M\right)\left(TM/M\right)$ is normal in $G/M,$ so by Lemma 2.4, we have $\begin{array}{rl}H/M\cap TM/M& =(H\cap T)M/M\le H_{p\tau G}M/M\\ & \le (HM/M{)}_{p\tau \left(G/M\right)}\\ & =(H/M{)}_{p\tau \left(G/M\right)}.\end{array}$So H/M is $p\tau$-embedded in $G/M.$

3. Let H be $p\tau$-embedded in G, then $T⊴G$ such that HT be S-quasinormal, $H\cap T\le H_{p\tau G}$.

   Clearly $TM/M⊴G$and $\left(H/M\right)\left(TM/M\right)=HTM/M$is S-quasinormal in $G/M$ by Lemma 2.1(2) of [2].

   Now $\left(\right|HM:H|,|HM:M\left|\right)=1,$also $\begin{array}{rl}HM/M\cap TM/M& =(HM\cap T)M/M\\ & =(H\cap T)(M\cap T)M/M\end{array}$and $=(H\cap T)M/M\le H_{p\tau G}M/M.$From 2.5(3) $H_{p\tau G}M/M\le (HM/M{)}_{H_{p\tau \left(G/M\right)}}.$Hence HM/M is $p\tau$-embedded in $G/M.$

4. Let H be $p\tau$-embedded in G. Then M is normal in G such that HM is S-quasinormal in G and $H\cap M\le H_{p\tau G}.$Let $K=M\cap T,$ then K is normal in G, and $HK=H(M\cap T)=HM\cap T$is S-quasinormal by Lemma 2.1(5) of [6], $HK\le T$and $H\cap K=H\cap M\cap T=H\cap M\le H_{p\tau G}.$

Lemma 2.6

[7]

Consider a group G and prime number q such that $q^{n+1}$ does not divides $\left|G\right|$ for integers $n\ge 1$. If $(|G|,(q-1\left)\right(q^2-1)\dots (q^n-1\left)\right)=1,$ then G is q-nilpotent.

Lemma 2.7

[8]

Consider a group G. If A is subnormal in G and A is a q-subgroup, then $A\le O_q\left(G\right).$

Lemma 2.8

[9]

Suppose that $P,R,S\le G.$ Then the following statements are equivalent:

• $P\cap RS=(P\cap R)(P\cap S).$

• $PR\cap PS=P(R\cap S).$

Lemma 2.9

Let $M⊴G$ such that G = PM for some $Q\le G,$ take $N\le G$ a maximal subgroup with M contained in N, then $Q\cap N$ is maximal in Q.

Lemma 2.10

Lemma 2.8 of [10]

Let q be a prime number which divides $\left|G\right|$, $\left(\right|G|,q-1)=1.$ Then

1. If $M⊴G$ having q order, then $M\le Z\left(G\right).$

2. If G has $G_q$ a cyclic subgroup, then G is q-nilpotent.

3. If $N\le G,|G:N|=q,$ then $N⊴G.$

Lemma 2.11

Let q be a prime divisor of the order of G in such a way $\left(\right|G|,q-1)=1,$ then

1. G is q-nilpotent provided G is q-supersoluble.

2. G is q-nilpotent provided G has cyclic Sylow q-subgroup.

3. X is normal in G provided $|G:X|=q$ and X ≤ G.

4. N lies in Z(G) provided $\left|N\right|=q$ and N is normal in G.

Proof.

Suppose C/D is any random chief factor. If G is q-supersolvable, then there are two possibilities:

1. $\left|C/D\right|=q$ is cyclic.

2. $C/D=q^{\mathrm{\prime }}$-group.

If $\left|C/D\right|=q,$ then $\left|Aut\right(C/D\left)\right|=q-1.$ As $G/C_G\left(C/D\right)$ is isomorphic to a subgroup of Aut$\left(C/D\right),$ then $\left|G/C_G\right(C/D\left)\right|$ will divide $\left(\right|G|,q-1)=1.$ This implies $C_G\left(C/D\right)=G.$ Hence G is q-nilpotent. Proof of (2), (3) and (4) can be seen in [11, Theorem 2.8].

Lemma 2.12

[12], 7.19

Let $Y⊴G,$ then $Y/\mathrm{\Phi }\left(Y\right)\le Z_{\mathcal{U}}\left(G/\mathrm{\Phi }\right(Y\left)\right)$ if and only if $Y\le Z_{\mathcal{U}}\left(G\right).$

Lemma 2.13

Suppose $X⊴G$ is q-subgroup. Then there exists $A⊴G$, such that A is the largest subgroup of X and is $p\tau$-embedded.

Proof.

If the order of X is q, then the theorem holds. Let $Y\le X$ be a normal q-subgroup which is smallest without identity. Let $X\ne Y.$ Using Lemma 2.6(2) of [13], the theorem is still satisfied by $G/Y.$ So with the help of induction some largest subgroup L/Y of $X/Y⊴G/Y.$

Obviously, $L\le X$ and $L⊴G.$ Therefore, the hypothesis is satisfied again. Now let $X=Y.$ Assume that L be any largest subgroup of X. So there will be $E⊴G$ in such a way that LE is S-quasinormal $L\cap E\le L_{p\tau G}.$ Let $L\ne L_{p\tau G}.$ Then $LE\ne L$ and $Y\ne 1.$ If $X\le LE,$ then $X=X\cap LE=L(X\cap E).$ Hence $X\le E,$ which shows that $L=L\cap E=L_{p\tau G},$ a contradiction. Now, if $X\nleqq LE,$ then $L=L(E\cap X)$ So using 2.3 of [13], $LE\cap X$ is S-quasinormal, which is again a contradiction. Thus, $L=L_{p\tau G}$. So using 2.3 of [13], L is S-quasinormal. Consequently, $X⊴G$ by Lemma 2.11 of [14]. Hence the lemma is proved.

3. Proofs of main results

In this section, we give proofs of our main theorems.

Proof of Theorem 1.4

The proof of this follows in the following steps:

1. In this step, we show that $\mathrm{\Phi }\left(G\right)$ ≤ G is supersolvable.

   Since N=N∩G and $\mathrm{\Phi }\left(N\right)=\mathrm{\Phi }\left(N\right)\cap N\le \mathrm{\Phi }\left(G\right)\cap N$, so N, $\mathrm{\Phi }\left(N\right)$ satisfies the theorem, then N is supersolvable.

2. Here, we prove that $\mathrm{\Phi }\left(G\right)$ $⋬$ G.

   Then $Sy{l}_q\left(Q\right)\le \mathrm{\Phi }\left(G\right)$ s.t Q $⊴$ G. Suppose $Q_1$ is maximal in Syl of $\mathrm{\Phi }\left(G\right)$. By hypotheses $Q_1$ is pτ-embedded, then there is T $⊴$ G s.t. G = $Q_1$T and K = $Q_1$ ∩ T is a pτ-quasinormal subgroup. So $Q_1$ ∩ T = K $⊴$ G, by (1) G/T, G/$Q_1$ supersolvable, $\mathrm{G}/Q_1\cap \mathrm{T}\cong \mathrm{G}/Q_1\times \mathrm{G}/\mathrm{T}$then G/$Q_1$ ∩ T be supersolvable. Now ($\left|Q_1\right|$, $\left|T\right|$) = 1, so ($\left|Q_1\right|$, $\left|T\right|$) = 1 and | G | = |$Q_1$T | = $\left|Q_1\right|$ $\left|T\right|=\left|QT\right|=\left|Q\right|$ $\left|T\right|$ / $|Q\cap T|$ and $\left|Q_1\right|=\left|Q\right|/|Q\cap T|$. So $Q_1=Q$, which contradicts the hypothesis. Thus ($\left|Q_1\right|$, $\left|T\right|$) ≠ 1. Since K is pτ-quasinormal in G, then K ≤ N where N is quasinormal.

   Since K ≤ Q ≤$O^q\left(G\right)$, using Lemma 2.7, K is s-quasinormal. Then ∃ a $G_p$ and p ≠ q so KP = PK ≤ G and K ≤$sy{l}_q\left(PK\right)$. By 2.2 of [15], K is τ-quasinormal in PK. Thus by 2.1 and 2.3 of [15] K $⊴$ PK, $O^q\left(G\right)$ ≤$M_G\left(K\right)$. So K = $Q_1\cap T⊴$G. As $Q_1⊴$ G, PK = P×K and then for every $G_p$ also $T_p$, P$⊴$T so T is nilpotent by 5.1.4 of [16]. As P char T, then P $⊴$ G and so nilpotent, which contradict.

3. Here we show that Φ(G) = 1.

   Let Φ(G) ≠ 1, so by (2) Φ(G/Φ(G))=Φ(G)/Φ(G), obviously each Syl of Φ(G/Φ(G)) is pτ-embedded in G/Φ(G) by our supposition, so G/Φ(G) is supersolvable, which implies G is supersolvable, which contradicts the hypothesis.

4. Now we prove that $\mathcal{F}$(G) = Φ(G)

   $\mathcal{F}$(G) is the direct product of minimal normal in G by (3) and Lemma 2.5 of [17]. Then $\mathcal{F}$(G) ≤Φ(G), particularly $\mathcal{F}$(G) < G. Consider $\mathcal{F}$(G) < Φ(G). We have to show (G/$\mathcal{F}$(G))/(H/$\mathcal{F}$(G)) ≅ G/H is supersolvable. From 2.2 (b) of [18], every maximal of $Syl(\mathrm{\Phi }$(G)/F (G)) is pτ-embedded in G/$\mathcal{F}$(G) by lemma 2.2(b) of [18]. Then G/$\mathcal{F}$(G), Φ(G)/F (G) satisfied and G/$\mathcal{F}$(G) is supersolvable. By (2), $\mathcal{F}$(G) is a proper subgroup of a $Sy{l}_p(\mathrm{\Phi }$(G)), particularly $\mathcal{F}$(G) is maximal in Syl(G), then $\mathcal{F}$(G) is abelian or cyclic and their order q, 4, respectively. Hence supersolvable, which contradicts the hypothesis.

5. Finally, we complete the prove with following lines.

   From (4) each maximal of Syl of $\mathcal{F}$(G) is pτ-embedded, so from Theorem 4.3 of [10]. G is supersolvable.

This completes the proof of Theorem 1.4.

Proof of Theorem 1.5

The proof of this theorem is given in the following steps

1. Using Lemma 2.6, | Q | ≥ $q^{n+1}$; thus every n-maximal subgroup $Q_n$ of Q satisfies $Q_n$ ≠1.

2. Now we prove that G is not simple.

   By hypothesis, $Q_n$ is pτ-embedded. Using the definition of the pτ-embedded subgroup and K $⊴$ G s.t. $Q_n$K $⊴$ G, $Q_n$ ∩ K ≤ ($Q_n{)}_{p\tau G}$. Let G be simple. If K = 1, then 1 ≠ $Q_n$K = $Q_n$ $⊴$ G, which contradicts the hypothesis. If K = G, then 1 < $Q_n$ ∩ K = $Q_n$ ≤ ($Q_n{)}_{p\tau G}$. We can write $\begin{array}{rl}(Q_n{)}_{p\tau G}& =<\mathrm{V}|\mathrm{V}\mathrm{i}\mathrm{s}\mathrm{a}\mathrm{n}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{p}\tau \\ & -\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{o}\mathrm{f}\mathrm{G}\mathrm{i}\mathrm{n}{Q}_n>\end{array}$Let V be an arbitrary nontrivial pτ-quasinormal subgroup ≤$Q_n$. Then T ≤ G be S-quasinormal subgroup such that V be $Sy{l}_q\left(T\right)$. As G be a simple group, we have $T_G=1$, By Lemma 2.1, V is S-quasinormal. From the arbitrariness of V and Lemma 2.1 of [2], $Q_n$ is S-quasinormal, so $Q_n=1$, in contrary to (1).

3. Now we prove that M $⊴$ G, where M is unique and minimal.

   Φ(G) is equal to 1. Since G/M satisfies which shows that QM/M is a $Sy{l}_q\left(G/M\right)$. By Lemma 2.6, we may take $\left|QM/M\right|$ ≥$q^{n+1}$. Let $N_n$/M be n-maximal of QM/M. So $N_n=N_n$ ∩ QM = ($N_n$ ∩Q)M = $Q_n$M. Obviously, $Q_n$ is an n-maximal subgroup of Q. According to supposition, $Q_n$ is pτ-embedded. Therefore, there is K $⊴$ G such that $Q_n$K $⊴$ G and $Q_n$ ∩ K ≤$(Q_M{)}_{p\tau G}$. Furthermore, we can see that KM/M is normal in G/M, $M_n$/M. KM/M = $Q_n$M/M. KM/M = $Q_n$KM/M $⊴$ G/M. If M ∩$Q_n$K = 1, then M ∩$Q_n=\mathrm{M}\cap \mathrm{K}=1$, $M\cap Q_n\mathrm{K}=(\mathrm{M}\cap Q_n$)(M ∩ K). If M ∩$Q_n$K ≠ 1, then M ≤$Q_n$K. Since $Q_n$ ∩ M = Q ∩$N_n$∩ M = Q ∩ M is a $M_q$ and $|M:M\cap K|$ = $|MK:K|$ ≤$|Q_{n}K:K|$, $(|M:M\cap Q_n|$, $|M:M\cap K|)$ = 1, (M ∩$Q_n$)(M ∩ K) = M = M ∩$Q_n$K. By Lemma 2.8, $Q_n$M ∩KM = ($Q_n$ ∩ K)M, and thus $Q_n$M/M ∩ KM/M = ($Q_{n}M\cap KM)/M=(Q_n\cap K)\mathrm{M}/\mathrm{M}$. Hence $N_n$/M ∩ KM/M = $Q_n$M/M ∩ KM/M = ($Q_n$ ∩ K)M/M ≤$(Q_n{)}_{p\tau G}$M/M ≤$(Q_{n}M/M{)}_{p\tau \left(G/M\right)}$ by Lemma 2.4. Thus $N_n$/M is pτ-embedded in G/M. Then, the factor group G/M satisfies the hypothesis. It yields that G/M is q-nilpotent. As a consequence, the uniqueness of M and Φ(G) = 1 are clear.

4. Now we prove that $O_{q^{\mathrm{\prime }}}$(G) = 1.

   If L = $O_{q^{\mathrm{\prime }}}$(G) is not equal to 1, then QL/L is a $Sy{l}_q\left(G/L\right)$. Let K/L $⊴$ of QL/L. Then K = $Q_1$L for some $Q_1$ ≤ Q. Then from (1) and Lemma 2.5 that $Q_1$L/L is pτ-embedded in G/L. Besides, $M_G/L\left(QL/L\right)=M_G\left(Q\right)L/L$ (see [11, Lemma 3.6.10]) is q-nilpotent. As a result, G/$O_{q^{\mathrm{\prime }}}$(G) satisfied. It follows that G/L is q-nilpotent and so is G, which contradicts the hypothesis.

5. Now we prove that $O_q$(G) = 1.

   If $O_q$(G) is not equal to 1, according to step (3) M ≤$O_q$(G), there is N ≤ G s.t. G = MN, M ∩ N = 1. Since $O_q$(G) ∩ N is normalized by M and N, thus M yields M = $O_q$(G). Q = Q ∩ MN = M(Q ∩ N). As Q ∩ N < Q, then $Q_1$ ≤ Q is a maximal subgroup which contains Q ∩ N, and hence Q = M$Q_1$. Pick a $Q_n$ maximal of Q contained in $Q_1$. From hypothesis K $⊴$ G s.t $Q_n$K $⊴$ G and $Q_n$ ∩ K ≤$(Q_n{)}_{p\tau \left(G\right)}$. Let V be a nontrivial pτ-quasinormal ≤$Q_n$. Then τ-quasinormal subgroup T ≤ G s.t. V is $T_q$. If $T_G$ ≠ 1, then M ≤$T_G$≤ T, so M ≤ V ≤$(Q_n{)}_{p\tau \left(G\right)}$≤$Q_n$≤$Q_1$. Consequently, Q = M$Q_1=Q_1$, which contradicts the hypothesis. Thus we have $T_G=1$. Furthermore, using Lemma 2.2, V is τ-quasinormal. From the arbitrariness of V and Lemma 2.1 of [2], $(Q_n{)}_{p\tau \left(G\right)}$ is τ-quasinormal. By Lemmas 2.3 of this paper and Lemma 2.1 of [2], $O_q$(G) ≤$M_G\left(\right(Q_n{)}_{p\tau \left(G\right)}$) and $(Q_n{)}_{p\tau \left(G\right)}$ is subnormal in G. By Lemma 2.7, we have $Q_n$ ∩ K ≤$(Q_n{)}_{p\tau \left(G\right)}$≤$O_q$(G) = M, so $Q_n$ ∩ K ≤$(Q_n{)}_{p\tau \left(G\right)}$≤$Q_1\cap$M. Furthermore, $Q_n\cap$K ≤ $(Q_n{)}_{p\tau \left(G\right)}^G=Q_n\cap \mathrm{K}\le (Q_n{)}_{p\tau \left(G\right)}^{O_q\left(G\right)Q}=(Q_n{)}_{p\tau \left(G\right)}^Q\le (Q_1\cap M{)}^Q=Q_1\cap M\le M$. It follows that $(Q_n{)}_{p\tau \left(G\right)}^G=Q_1\cap \mathrm{M}=\mathrm{M}$ or $(Q_n{)}_{p\tau \left(G\right)}^G=1$. If $(Q_n{)}_{p\tau \left(G\right)}^G=Q_1\cap \mathrm{M}=\mathrm{M}$, then $\mathrm{M}\le Q_1$, which contradicts the hypothesis. If $(Q_n{)}_{p\tau \left(G\right)}^G=1$, then $Q_1$ ∩ K = 1 and so $K_q$ ≤$q^n$. So K is q-nilpotent from Lemma 2.6. Suppose $K_{q\mathrm{\prime }}$ is normal q-complement of K, then $K_{q\mathrm{\prime }}$ $⊴$ G, we get $K_{q\mathrm{\prime }}=1$ by step (4), and thus there is q-subgroup K $⊴$ G and K ≤$Q_n$K ≤$O_q$(G) = M. If K ≠ 1, we get K = $Q_n$K = M, so $Q_n$ ≤ K, namely, $Q_n\cap$ K = $Q_n=1$, which contradicts the hypothesis. If K = 1, then $Q_n⊴$ G, so M $\le Q_n\le Q_1$, which contradicts the hypothesis. Now it is clear that (5) holds.

6. Now we complete the proof with the following lines:

   If M ∩ Q ≤Φ(Q), then M is q-nilpotent by $Tat{e}^{\mathrm{\prime }}s$ Theorem [19, IV,4.7]. Therefore, $M_{q^{\mathrm{\prime }}}$ $⊴$ G. So $M_{q^{\mathrm{\prime }}}$ $\le O_{q^{\mathrm{\prime }}}$(G) = 1. Moreover, M be q-group, then M $\le O_q$(G) = 1, which contradicts the hypothesis. As a result, there is a maximal $Q_1$ ≤ Q, s.t. Q = (Q ∩ M)$Q_1$. Take $Q_n$ ≤ Q contained in $Q_1$. By the hypothesis, K is normal in G s.t. $Q_n$K $⊴$ G, $Q_n\cap$K ≤$(Q_M{)}_{p\tau G}$. Let V be a nontrivial pτ-quasinormal contained in $Q_n$. So τ-quasinormal T ≤ G, then V be $Sy{l}_q\left(T\right)$. If $T_G\ne$ 1, then M $\le T_G\le$ T, so V ∩ M is a $Sy{l}_q\left(M\right)$.

   We know V ∩ M ≤$Q_1\cap$ M ≤ Q ∩ M, Q ∩ M be $Sy{l}_q\left(M\right)$, so V ∩ M = $Q_1\cap$ M = Q ∩ M. Consequently, Q = (Q ∩ M)$Q_1=(Q_1\cap \mathrm{M})Q_1=Q_1$, which contradicts the hypothesis. Hence $T_G=1$, V is τ-quasinormal from Lemma 2.1. By Lemma 2.1 of [2] and arbitrariness of V, $(Q_M{)}_{p\tau G}$ is S-quasinormal, and so $(Q_M{)}_{p\tau G}$ is subnormal using Lemma 2.1 of [2]. By Lemma 2.7 that $(Q_M{)}_{p\tau G}\le$ $O_q$(G) = 1, so $\left|K_q\right|\le$ $q_n$, therefore K is q-nilpotent. Similarly, we have $K_{q^{\mathrm{\prime }}}=1$ and so K = 1. It deduces that $Q_n⊴$ G, M ≤$Q_n\le Q_1$.

This completes the proof of Theorem 1.5.

Proof of Theorem 1.6

Here we will prove the theorem by obtaining a contradiction. The proof follows in the following steps:

1. First, we prove that K is q-nilpotent.

   Let Q${}_1$ be the largest subgroup of Q. Q${}_1$ has a q-supersolvable supplement X ∩ K in K provided Q${}_1$ has q-supersolvable supplement X. Because ($\left|K\right|$, q − 1) = 1, this implies X ∩ K is q-nilpotent from Lemma 2.11(1). If Q${}_1$ is pτ-embedded in G, so Q${}_1$ is also pτ-embedded in K from 2.6(1) of [13]. Also, Q${}_1$ does not have any q-nilpotent supplement in K. So by theorem 1.5 of [13], K is q-nilpotent.

2. Now we prove that Q = K.

   Using step (1), O${}_{q^{\prime }}$(K) is the normal Hall q${}_{{}^{\prime }}$-subgroup of K.

   Let O${}_{q^{\prime }}$(K) ≠ 1. We can check it easily that our theorem is true for (G/O${}_{q^{\prime }}$(K, K/O${}_{q^{\prime }}$(K)). Using mathematical induction we can see G/O${}_{q^{\prime }}$(K) be the chief factor, between 1 and K/O${}_{q^{\prime }}$(K)) is cyclic. Following each factor between K and O${}_{q^{\prime }}$(K) is cyclic, which implies O${}_{q^{\prime }}$(K) = 1. Hence Q = K.

3. In this step, we prove that Φ(Q) = 1.

   First, we let Φ(Q) ≠ 1. Then in the light of Lemma 2.3(2), we can check easily that our theorem holds for (G/Φ(Q), Q/Φ(Q)). Every chief factor of G/Φ(Q) under Q/Φ(Q) is cyclic by our selection of (G,K). Hence cyclic by Lemma 2.12, which contradicts the hypothesis.

4. Here we prove that every largest subgroup of Q is pτ-embedded.

   Let us have some largest Q${}_1$subgroup contained in Q in such a way that T is the q-supersolvable supplement of Q${}_1$ in G, thus QT = G with Q ∩ T ≠ 1. Because Q ∩ T $⊴$ T, we may suppose that Q ∩ T contains a smallest normal subgroup L of T. Here Obviously $\left|L\right|=\mathrm{q}$. Since Q is elementary abelian and G = QT, this implies L $⊴$ G. Here we can check that our theorem holds yet for (G/L, Q/L). By our selection of (G,K) we can see every chief factor of G/L under Q/L is cyclic. As a consequence, every chief factor of G under Q is cyclic, which is a contradiction, hence (4) holds.

5. Now we find the smallest normal subgroup.

   Let Q $⊴$ G, so using Lemma 2.13, G contains some largest normal subgroup of Q, which can't be true because Q is of smallest order.

6. Let L $⊴$ Q of G, then Q/L ≤ N${}_{\mathcal{V}}$(G/L), and $\left|Y\right|$ > q. Moreover, using Lemma 2.3(2) of [13], our theorem satisfies (G/L, Q/L). Thus from our selection of (G, K) = (G, Q), every chief factor of G/L under Q/L is cyclic.

   If $\left|L\right|=\mathrm{q}$, then cyclic, which contradiction of our supposition. Now if Q contains two smallest normal subgroups R and L of G, then LR/R ≤ Q/R and from the isomorphism LR/R ≅ L, it follows that $\left|L\right|=\mathrm{q}$, a contradiction again. Thus, step (6) is true.

7. Finally, we prove the contradiction.

   Suppose that L $⊴$ Q of G and L${}_1$ the largest subgroup of L. To show L${}_1$ is S-quasinormal. So we may suppose that B is a complement of L in Q, as Q is an elementary abelian q-group. Also take W = L${}_1$B. Clearly, W is a largest subgroup of Q. Using step (4), W is pτ-embedded in G. So using Lemma 2.6(4) of [13], there will be R $⊴$ G satisfying the condition, W ∩ R ≤ W${}_{p\tau G}$, WR ≤ Q and WR is S-quasinormal. From Lemma 2.3 of [13], W${}_{p\tau G}$ is S-quasinormal. Now if R = Q, so W = W${}_{p\tau G}$ is S-quasinormal. By 2.1(5) of [2], $W\cap L=L_{1}C\cap L=L_1(K\cap L)=Y_1$is S-permutable. If R = 1, this gives W = WR is S-quasinormal. As a result, L${}_1$ is S-quasinormal. Consider 1 < R < Q. Implies L ≤ R by step (6). So by using Lemma 2.1(5) of [2], $L_1=W\cap L=W_{p\tau G}\cap L$is S-quasinormal. This implies $\left|L\right|=\mathrm{q}$, which contradicts step (6).

4. Conclusions

In this paper, we check the supersolvability and nilpotency of $p\tau$-embedded subgroups. We proved that if H $⊴$ G such that G/H is supersolvable and every maximal subgroup of any Syl of $\mathrm{\Phi }\left(H\right)$ is pτ-embedded in G, then G is supersolvable. Further we proved that if Q be a $G_q$, where q is a prime divisor of $|G|$ with ( $|G|$, $(q-1)(q^2-1)\cdots (q^n-1))=1$ (n≥1) and every maximal subgroup of Q and every n-maximal subgroup of Q is pτ-embedded in G, then G is q-nilpotent. At last, we prove that if M $⊴$ G and Q be a Sylow q-subgroup of M such that ( $|M|,q-1$) = 1 provided q is a prime divisor of $\left|M\right|$ and every largest subgroup Q${}_1$ of Q is pτ-embedded in G such that Q${}_1$ does not have a q-supersolvable supplement in G, then each chief factor of G between M and O${}_{q^{\prime }}$(M) is cyclic. Our results are the extension of existing results.

Disclosure statement

No potential conflict of interest was reported by the authors.

ORCID

Lian Chen http://orcid.org/0000-0002-1398-1245

Abid Mahboob http://orcid.org/0000-0003-1576-8517