New criteria of supersolubility of finite groups

1.   Introduction

Throughout this paper, all groups are finite, and $G$ always denotes a group. Moreover, $n$ is an integer, and $\mathbb{P}$ is the set of all primes. The symbol $\pi(n)$ denotes the set of all primes dividing $n$ and $\pi(G)=\pi(|G|)$, the set of all primes dividing the order of $G$.

In what follows, $\sigma=\{\sigma_{i}|i\in I\}$ is some partition of $\mathbb{P}$, that is, $\mathbb{P}=\bigcup_{i\in I}\sigma_{i}$ and $\sigma_{i}\cap \sigma_{j}=\emptyset$ for all $i\neq j$. $\Pi$ is always supposed to be a nonempty subset of the set $\sigma$ and $\Pi^{'}=\sigma\backslash \Pi$. We write $\sigma(n)=\{\sigma_{i}|\sigma_{i}\cap \pi(n)\neq\emptyset\}$ and $\sigma(G)=\sigma(|G|)$.

Following Refs. [1–4], $G$ is said to be $\sigma$-primary if $|\sigma(G)|\leqslant 1$. An integer $n$ is a $\Pi$-number if $\pi(n)\subseteq\bigcup_{\sigma_{i}\in \Pi}\sigma_{i}$. A subgroup $H$ of $G$ is called a $\Pi'$-subgroup of $G$ if $|H|$ is a $\Pi'$-number. A set ${\cal {H}}$ of subgroups of $G$ is said to be a complete Hall $\sigma$-set of $G$ if every nonidentity member of ${\cal {H}}$ is a Hall $\sigma_{i}$-subgroup of $G$ for some ${i}\in I$ and ${\cal {H}}$ contains exactly one Hall $\sigma_{i}$-subgroup of $G$ for every $\sigma_{i}\in \sigma(G)$. $G$ is said to be $\sigma$-full if $G$ possesses a complete Hall $\sigma$-set; a $\sigma$-full group of Sylow type if every subgroup of $G$ is a $D_{\sigma_{i}}$-group for all $\sigma_{i}\in \sigma(G)$. A subgroup $H$ of $G$ is called $\sigma$-subnormal in $G$ if there exists a subgroup chain $H=H_0\leqslant H_1\leqslant\cdots \leqslant H_t=G$ such that either $H_{i-1}$ is normal in $H_{i}$ or $H_{i}/(H_{i-1})_{H_{i}}$ is $\sigma$-primary for all $i=1,\cdots,t$.

It is well known that embedded subgroups and supplemented subgroups play an important role in the finite group theory. For example, a subgroup $H$ of $G$ is said to be $c$-normal^[5] in $G$ if $G$ has a normal subgroup $T$ such that $G=HT$ and $H\cap T\leqslant H_{G}$, where $H_{G}$ denotes the maximal normal subgroup of $G$ contained in $H$. A subgroup $H$ of $G$ is called $n$-embedded^[6] in $G$ if $G$ has a normal subgroup $T$ such that $HT=H^{G}$ and $H\cap T\leqslant H_{sG}$, where $H^{G}$ is the largest normal subgroup of $G$ containing $H$ and $H_{sG}$ is the subgroup of $H$ generated by all those subgroups of $H$ which are $s$-permutable in $G$ $($note that subgroup $A$ of $G$ is said to be $s$-permutable in $G$ if $AP=PA$ for any Sylow subgroup $P$ of $G$$)$. A subgroup $H$ of $G$ is called $s$-embedded^[6] in $G$ if $G$ has an $s$-permutable subgroup $T$ such that $HT=H^{sG}$ and $H\cap T\leqslant H_{sG}$, where $H^{sG}$ is the intersection of all $s$-permutable subgroups of $G$ containing $H$. A subgroup $H$ of $G$ is called $\sigma$-$n$-embedded^[7] in $G$ if there exists a normal subgroup $T$ of $G$ such that $HT =H^{G}$ and $H\cap T\leqslant H_{\sigma G}$, where $H_{\sigma G}$ is the subgroup of $H$ generated by all those subgroups of $H$ that are $\sigma$-permutable in $G$ $($note that a subgroup $H$ of $G$ is called $\sigma$-permutable^[2] in $G$ if $G$ possesses a complete Hall $\sigma$-set ${\cal {H}}$ such that $HA^{x}=A^{x}H$ for all $A\in {\cal {H}}$ and all $x\in G$$)$. A subgroup $H$ of $G$ is called $\sigma$-embedded^[7] in $G$ if there exists a $\sigma$-permutable subgroup $T$ of $G$ such that $HT= H^{\sigma G}$ and $H\cap T\leqslant H_{\sigma G}$, where $H^{\sigma G}$ is the $\sigma$-permutable closure of $H$, that is, the intersection of all $\sigma$-permutable subgroups of $G$ containing $H$. By using the above embedded subgroups and supplemented subgroups, people have obtained a series of interesting results (see, for example, Refs. [5–13]).

Some properties of $\sigma$-embedded subgroups were analyzed in Refs. [7, 8]. In this paper, we continue the study of $\sigma$-embedded subgroups and use them to determine the structure of finite groups. In particular, we obtain some new criteria for the supersolubility of finite groups.

Theorem 1.1. Let $G$ be a $\sigma$-full group of Sylow type and ${\cal {H}}=\{H_1,\cdots,H_t\}$ a complete Hall $\sigma$-set of $G$ such that $H_i$ is a nilpotent $\sigma_i$-subgroup for all $i=1,\cdots,t$. If every cyclic subgroup of any non-cyclic $H_i$ of prime order and order $4$ $($if $2\in \pi(H_i)$ and the Sylow $2$-subgroup of $H_i$ is nonabelian$)$ is $\sigma$-embedded in $G$, then $G$ is supersoluble.

Theorem 1.2. Let $G$ be a $\sigma$-full group of Sylow type and ${\cal {H}}=\{H_1,\cdots,H_t\}$ a complete Hall $\sigma$-set of $G$ such that $H_i$ is a nilpotent $\sigma_i$-subgroup for all $i=1,\cdots,t$. Suppose that $E$ is a normal subgroup of $G$ with $G/E$ supersoluble. If every cyclic subgroup of any noncyclic $H_i\cap E$ of prime order and order $4$ $($if $2\in \pi(H_i\cap E)$ and the Sylow $2$-subgroup of $H_i\cap E$ is nonabelian$)$ is $\sigma$-embedded in $G$, then $G$ is supersoluble.

The rest of this paper is organized as follows. In Section 2, we give some preliminaries which will be used in this paper. In Section 3, we give the proofs of Theorems 1.1 and 1.2. In Section 4, we give some applications of our results.

All unexplained terminologies and notations are standard. The reader is referred to Refs. [14–16] if necessary.

2.   Preliminaries

Let ${\cal {L}}$ be some nonempty set of subgroups of $G$ and $E\leqslant G$. Following Ref. [2], a subgroup $A$ of $G$ is called ${\cal {L}}$-permutable if $AH=HA$ for all $H\in{\cal {L}}$; ${\cal {L}}^{E}$-permutable if $AH^{x}=H^{x}A$ for all $H\in{\cal {L}}$ and all $x\in E$. In particular, a subgroup $H$ of $G$ is $\sigma$-permutable in $G$ if $G$ possesses a complete Hall $\sigma$-set ${\cal {H}}$ such that $H$ is ${\cal {H}}^{G}$-permutable.

Lemma 2.1. ^{[2, Lemma 2.8]} Let $H,K$ and $N$ be subgroups of a $\sigma$-full group $G$. Let ${\cal {H}}=\{H_{1},\cdots ,H_{t}\}$ be a complete Hall $\sigma$-set of $G$ and ${\cal {L}}={\cal {H}}^{K}$. Suppose that $H$ is ${\cal {L}}$-permutable and $N$ is normal in $G$. Then, $HN/N$ is ${\cal {L}}^{\ast}$-permutable, where ${\cal {L}}^{\ast}=\{H_{1}N/N,\cdots ,H_{t}N/N\}^{KN/N}$. In particular, if $H$ is $\sigma$-permutable in $G$, then $HN/N$ is $\sigma$-permutable in $G/N$.

Following Refs. [1, 2], we use $O^{\Pi}(G)$ to denote the subgroup of $G$ generated by all its $\Pi'$-subgroups of $G$. Instead of $O^{\{\sigma_i\}}(G)$, we write $O^{\sigma_i}(G)$.

Lemma 2.2. ^{[2, Lemma 3.1]} Let $H$ be a $\sigma_{1}$-subgroup of a $\sigma$-full group $G$. Then, $H$ is $\sigma$-permutable in $G$ if and only if $O^{\sigma_{1}}(G)\leqslant N_{G}(H)$.

Lemma 2.3. ^{[2, Theorem C]} Let $G$ be a $\sigma$-full group of Sylow type. Then, the set of all $\sigma$-permutable subgroups of $G$ forms a sublattice of the lattice of all $\sigma$-subnormal subgroups of $G$. In particular, for any subgroup $A$ of $G$, $A^{\sigma G}$ and $A_{\sigma G}$ are both $\sigma$-permutable in $G$.

Lemma 2.4. Let $G$ be a $\sigma$-full group of Sylow type and $H$ a subgroup of $G$. Then, $H$ is $\sigma$-embedded in $G$ if and only if there exists a $\sigma$-permutable subgroup $T$ of $G$ such that $HT$ is $\sigma$-permutable in $G$ and $H\cap T\leqslant H_{\sigma G}$.

Proof. The necessity is evident. Now, we prove the sufficiency. Let $T_1=T\cap H^{\sigma G}$. Then, by Lemma 2.3, $T_1$ is $\sigma$-permutable in $G$. Clearly, $HT_1=H(T\cap H^{\sigma G})= HT\cap H^{\sigma G}\leqslant H^{\sigma G}$. Then, by the hypothesis and Lemma 2.3, $HT_1$ is $\sigma$-permutable in $G$. Moreover, since $H^{\sigma G}\leqslant H\leqslant HT_1$, we have that $HT_1=H^{\sigma G}$. It is clear that $H\cap T_1\leqslant H\cap T\leqslant H_{\sigma G}$. This shows that $H$ is $\sigma$-embedded in $G$. The lemma is proved.

Lemma 2.5. ^{[6, Lemma 3.1]} Let $G=P\rtimes Q$, where $P$ is the Sylow $2$-subgroup of $G$ and $|Q|=q$ for some prime number $q\neq 2$. If every cyclic subgroup of $P$ of order $2$ and order $4$ $($if $P$ is a nonabelian group$)$ is $s$-embedded in $G$, then $G$ is $2$-nilpotent.

Lemma 2.6. ^{[7, Lemma 2.7]} Let $G$ be a $\sigma$-full group of Sylow type, $N$ a normal subgroup of $G$ and $H\leqslant K\leqslant G$.

① If $H$ is $\sigma$-embedded in $G$, then $H$ is $\sigma$-embedded in $K$.

② Suppose that $H$ is $\sigma$-embedded in $G$ and $N\leqslant H$ or $(|H|,|N|)=1$. Then, $HN/N$ is $\sigma$-embedded in $G/N$.

Lemma 2.7. ^[17] Let $G$ be a nonsupersoluble group such that all its proper subgroups are supersoluble. Then:

① $G$ has a unique normal Sylow subgroup, says $G_p$.

② $G_p/\Phi(G_p)$ is a minimal normal subgroup of $G/\Phi(G_p)$.

③ If $p>2$, then $G_p$ is of exponent $p$.

3.   Proofs of Theorems 1.1 and 1.2

To prove Theorems 1.1 and 1.2, we first prove the following results.

Proposition 3.1. Let $G$ be a $\sigma$-full group of Sylow type and $P$ a Sylow $p$-subgroup of $G$, where $p$ is the smallest prime dividing $|G|$. Suppose that $p\in \sigma_i$ for some $\sigma_i\in \sigma(G)$ and a Hall $\sigma_i$-subgroup $H_i$ of $G$ is nilpotent. Then, $G$ is $p$-nilpotent if and only if all subgroups of $P$ that have order $p$ and order $4$ $($if $P$ is a nonabelian $2$-group$)$ is $\sigma$-embedded in $G$.

Proof. ($\Longrightarrow$) Since $G$ is $p$-nilpotent, there exists a normal Hall $p'$-subgroup $K$ of $G$. Let $H$ be a subgroup of $P$ with order $p$ or order $4$ (if $P$ is a nonabelian $2$-group). Then, clearly, $HK$ is $\sigma$-permutable in $G$ and $H\cap K=1$. By Lemma 2.4, we have that $H$ is $\sigma$-embedded in $G$.

($\Longleftarrow$) Suppose that this is false and let $G$ be a counterexample of minimal order. We now proceed via the following steps:

(1) Let $E$ be a proper subgroup of $G$. Then, $E$ is $p$-nilpotent.

If $p\notin \pi(E)$, then, of course, $E$ is $p$-nilpotent. If $p\in \pi(E)$, then $E$ satisfies the hypothesis of the theorem by Lemma 2.6①. Hence, $E$ is $p$-nilpotent by the choice of $G$.

(2) $G=P\rtimes Q$ is a minimal nonnilpotent group, where $Q$ is a Sylow $q$-subgroup of $G$ with $p<q$, $P/\Phi(P)$ is a chief factor of $G$, and if $p\neq 2$, the exponent of $P$ is $p$.

This directly follows from Claim (1) and Ref. [18, Chapter Ⅳ, Theorem 5.4].

$(3)$ $|Q|=q$.

Assume that $|Q|>q$. Let $K$ be a maximal subgroup of $Q$. Then, $K\neq 1$ and $K$ is normal in $G$ by Claim (2). Now, we consider the quotient group $G/K$. Clearly, the hypothesis holds for $G/K$ by Lemma 2.6②. Hence, $G/K$ is $p$-nilpotent by the choice of $G$, and so $G$ is $p$-nilpotent, a contradiction. Hence, (3) holds.

(4) There exists some $\sigma_j\in \sigma(G)$ such that $\sigma_j\cap \pi(G)=\{q\}$, where $j\neq i$.

Assume that this is false. Then, $\sigma_i\cap \pi(G)=\{p,q\}$, and thus, $G$ is a $\sigma_i$-group. It follows from the hypothesis that $G$ is nilpotent, a contradiction. Hence, we have $(4)$.

$(5)$ $p\neq 2$.

Assume that $p=2$. In view of Lemma 2.5, $G$ is $2$-nilpotent by Claims (2)–(4), a contradiction. Hence, (5) holds.

(6) Final contradiction.

We claim that $|P/\Phi(P)|=p$. Assume that there exists a minimal subgroup $X/\Phi(P)$ of $P/\Phi(P)$ such that $X/\Phi(P)$ is not $\sigma$-permutable in $G/\Phi(P)$. Let $x\in X\backslash \Phi(P)$ and $L=\langle x\rangle$. Then, $X=L\Phi(P)$ and $|L|=p$ by Claims $(2)$ and $(5)$. If $L=L_{\sigma G}$, then by Lemmas 2.1 and 2.3, $X/\Phi(P)=L\Phi(P)/\Phi(P)$ is $\sigma$-permutable in $G/\Phi(P)$, contrary to the choice of $X/\Phi(P)$. Hence, $1=L_{\sigma G}<L$. Then, by the hypothesis and Lemma 2.4, there exists a $\sigma$-permutable subgroup $T$ of $G$ such that $LT$ is $\sigma$-permutable in $G$ and $L\cap T\leqslant L_{\sigma G}=1$. Let $T_1=T\cap P$. Then, $T_1$ and $LT_1=L(T\cap P)=LT\cap P$ are both $\sigma$-permutable in $G$ by Lemma 2.3. Hence, $O^{\sigma_i}(G)\leqslant N_G(T_1)$ by Lemma 2.2 since $T_1$ is a $\sigma_i$-subgroup. Note that $Q^G\leqslant O^{\sigma_i}(G)$. If $Q^G\neq G$, then $Q$ $char$ $Q^G\unlhd G$ by Claim (2), so $Q\unlhd G$, a contradiction. Hence, $Q^G=G$, and so $T_1\unlhd G$. This shows that $T_1\Phi(P)/\Phi(P)\unlhd G/\Phi(P)$, which implies that $T_1\Phi(P)/\Phi(P)=1$ or $P/\Phi(P)$ by Claim $(2)$. If $T_1\Phi(P)/\Phi(P)=P/\Phi(P)$, then $T_1=P$, and so $L=L\cap T_1\leqslant L\cap T\leqslant L_{\sigma G}=1$, a contradiction. Hence, $T_1\leqslant\Phi(P)$. In addition, since $LT_1$ is $\sigma$-permutable in $G$, we have that $X/\Phi(P)=L\Phi(P)/\Phi(P)=LT_1\Phi(P)/\Phi(P)$ is $\sigma$-permutable in $G/\Phi(P)$ by Lemma 2.1, a contradiction. The contradiction shows that every minimal subgroup of $P/\Phi(P)$ is $\sigma$-permutable in $G/\Phi(P)$. It follows that every minimal subgroup of $P/\Phi(P)$ is $s$-permutable in $G/\Phi(P)$ since $q\notin \sigma_i$ and $\pi(G)=\{p,q\}$ by Claim (2). Hence, $|P/\Phi(P)|=p$ by Ref. [19, Lemma 2.11]. It follows that $P$ is cyclic of exponent $p$, which implies that $Q\unlhd G$ since $p$ is the smallest prime dividing $|G|$, a contradiction. This completes the proof.

Proposition 3.2. Let $G$ be a ${\sigma}$-full group of Sylow type and ${\cal {H}}$$=\{H_1,\cdots,H_t\}$ a complete Hall ${\sigma}$-set of $G$ such that $H_i$ is a nilpotent ${\sigma}_i$-subgroup for all $i=1,\cdots,t$. Let $P$ be a normal $p$-group of $G$ with $G/P$ supersoluble. If every cyclic subgroup $H$ of $P$ of order $p$ and order $4$ $($if $P$ is a nonabelian $2$-group$)$ is $\sigma$-embedded in $G$, then $G$ is supersoluble.

Proof. Assume that this is false and let $(G,P)$ be a counterexample with $|G|+|P|$ minimal. Without loss of generality, we may assume that $P\leqslant H_1$.

$(1)$ Let $E$ be a proper subgroup of $G$. Then, $E$ is supersoluble.

It is clear that $(E,E\cap P)$ satisfies the hypothesis by Lemma 2.6①. Hence, $E$ is supersoluble by the choice of $(G,P)$.

$(2)$ $p$ is the largest prime number dividing $|G|$ and $p>2$.

Assume that this is false and let $q$ be the largest prime number dividing $|G|$. Let $Q$ be a Sylow $q$-subgroup of $G$. Then, $QP/P\unlhd G/P$ since $G/P$ is supersoluble. It follows that $QP\unlhd G$. Hence, by Lemma 2.6①, every cyclic subgroup $H$ of $P$ of order $p$ and order $4$ $($if $P$ is a nonabelian $2$-group$)$ is $\sigma$-embedded in $QP$. In view of Proposition 3.1, we have that $QP$ is $p$-nilpotent, and so $Q\unlhd G$. Next, we consider the quotient group $G/Q$. It is easy to see that $(G/Q,PQ/Q)$ satisfies the hypothesis by Lemma 2.6②. Hence, $G/Q$ is supersoluble by the choice of $(G,P)$. Moreover, since $G/P$ is supersoluble, we obtain that $G$ is supersoluble, a contradiction. Hence, $p$ is the largest prime number dividing $|G|$. It is also clear that $p>2$.

$(3)$ Let $P_1$ be a Sylow $p$-subgroup of $G$. Then, $P_1\unlhd G$.

Since $G/P$ is supersoluble and $p$ is the largest prime number dividing $|G|$ by Claim (2), we have that $P_1/P\unlhd G/P$. This implies that $P_1\unlhd G$.

$(4)$ $P_1$ is the unique normal Sylow subgroup of $G$, $P_1/\Phi(P_1)$ is a minimal normal subgroup of $G/\Phi(P_1)$, and the exponent of $P_1$ is $p$.

This directly follows from Claims (1)–(3) and Lemma 2.7.

$(5)$ $\Phi(P_1)\neq 1$.

Assume that $\Phi(P_1)= 1$. Then, $P_1$ is a minimal normal subgroup of $G$ by Claim (4), and so $P=P_1$. Let $H$ be a subgroup of $P$ with order $p$. Then, by the hypothesis and Lemma 2.4, there exists a $\sigma$-permutable subgroup $T$ of $G$ such that $HT$ is $\sigma$-permutable in $G$ and $H\cap T\leqslant H_{\sigma G}$. Let $T_1=T\cap P$. Then, $T_1$ and $HT_1=H(T\cap P)=HT\cap P$ are both $\sigma$-permutable in $G$ by Lemma 2.3. It follows from Lemma 2.2 that $O^{\sigma_1}(G)\leqslant N_G(T_1)$ since $T_1$ is a $\sigma_1$-subgroup of $G$. In addition, since $T_1\unlhd P$ and the Hall $\sigma_1$-subgroup $H_1$ of $G$ is nilpotent by the hypothesis, we obtain that $T_1\unlhd H_1$, and thus $T_1\unlhd H_1O^{\sigma_1}(G)=G$. Hence, $T_1=1$ or $P$ by Claim $(4)$. In the former case, we have that $H=HT_1$ is $\sigma$-permutable in $G$. In the later case, we obtain that $H=H\cap T_1\leqslant H\cap T\leqslant H_{\sigma G}$, and thus, $H=H_{\sigma G}$ is $\sigma$-permutable in $G$ by Lemma 2.3. It follows that $O^{\sigma_1}(G)\leq N_G(H)$ for $H$ is a subgroup of order $p$. Then, by a similar argument as above, we have that $H\unlhd G$. Therefore, $H=P$ by Claim $(4)$ since $P=P_1$. This shows that $P$ is a group of order $p$. It follows from $G/P$ is supersoluble that $G$ is supersoluble, a contradiction. Hence, we have $(5)$.

$(6)$ $P=P_1$.

It is clear that $P\Phi(P_1)/\Phi(P_1)\unlhd G/\Phi(P_1)$. By Claim $(4)$, we have that $P\Phi(P_1)/\Phi(P_1)=1$ or $P_1/\Phi(P_1)$. The former case shows that $P\leqslant \Phi(P_1)$, so $G/\Phi(P_1)\cong (G/P)/(\Phi(P_1)/P)$ is supersoluble. Ref. [18, Chapter Ⅵ, Theorem 8.6(a)] and Claim (5) show that $G$ is supersoluble, a contradiction. Hence, $P\Phi(P_1)/\Phi(P_1)=P_1/\Phi(P_1)$, and thus, $P_1=P\Phi(P_1)$. This shows that $P_1=P$, as desired.

$(7)$ Final contradiction.

Claims $(6)$ and $(4)$ show that the exponent of $P$ is $p$. Hence, there exists a subgroup $H$ of $P$ with order $p$ such that $H\nleq \Phi(P)$ by Claims (5) and (6). By the hypothesis and Lemma 2.4, there exists a $\sigma$-permutable subgroup $T$ of $G$ such that $HT$ is $\sigma$-permutable in $G$ and $H\cap T\leqslant H_{\sigma G}$.

We claim that $H_{\sigma G}=1$. If not, then $H=H_{\sigma G}$ is $\sigma$-permutable in $G$ by Lemma 2.3. It follows from Lemma 2.1 that $H\Phi(P)/\Phi(P)$ is $\sigma$-permutable in $G/\Phi(P)$. Hence, by Lemma 2.2, we have $O^{\sigma_1}(G/\Phi(P))\leqslant N_{G/\Phi(P)}(H\Phi(P)/\Phi(P))$. Moreover, since $H\Phi(P)/\Phi(P)\unlhd P/\Phi(P)$ and the Hall $\sigma_1$-subgroup of $G/\Phi(P)$ is nilpotent by the hypothesis, we obtain that $H\Phi(P)/\Phi(P)\unlhd G/\Phi(P)$. In view of Claims (4) and (6), we have that $P=H\Phi (P)=H$, and so $P$ is a cyclic group of order $p$. This shows that $G$ is supersoluble since $G/P$ is supersoluble, a contradiction. Hence, $H_{\sigma G}=1$.

Let $T_1=T\cap P$. Then, $T_1$ is $\sigma$-permutable in $G$ by Lemma 2.3. Note that $T_1$ is a $\sigma_1$-subgroup of $G$. Hence, $O^{\sigma_1}(G/\Phi(P))\leqslant N_{G/\Phi(P)}(T_1\Phi(P)/\Phi(P))$ by Lemmas 2.1 and 2.2. With a similar argument as above, we have that $T_1\Phi(P)/\Phi(P)\unlhd G/\Phi(P)$. Hence, $T_1\Phi(P)=\Phi(P)$ or $T_1\Phi(P)=P$ by Claims (4) and (6). If $T_1\Phi(P)=P$, then $P=T_1$, and thus, $H=H\cap T_1\leqslant H\cap T\leqslant H_{\sigma G}=1$, a contradiction. Hence, $T_1\Phi(P)=\Phi(P)$. It is easy to see that $HT_1=H(T\cap P)=HT\cap P$ is also $\sigma$-permutable in $G$ and it is a $\sigma_1$-subgroup of $G$. As arguments above, we obtain that $HT_1\Phi(P)=\Phi(P)$ or $HT_1\Phi(P)=P$. It follows from $T_1\Phi(P)=\Phi(P)$ that $HT_1\Phi(P)=H\Phi(P)=\Phi(P)$ or $HT_1\Phi(P)=H\Phi(P)=P$. However, as $H\nleq \Phi(P)$, we have that $H\Phi(P)=P$, and so $P=H$ is a cyclic group of order $p$. In addition, since $G/P$ is supersoluble, we obtain that $G$ is supersoluble, a contradiction. This contradiction completes the proof.

Proof of Theorem 1.1. We prove this theorem by induction on $|G|$. By Proposition 3.1 and Ref. [18, Chapter Ⅳ, Theorem 2.8], we know that $G$ possesses an ordered Sylow tower of the supersoluble type. Let $p$ be the largest prime dividing $|G|$ and $P$ a Sylow $p$-subgroup of $G$. Then, $P$ is normal in $G$. If $G=P$, then, clearly, $G$ is supersoluble. If $P<G$, then we consider the quotient group $G/P$. It follows from Lemma 2.6② that $G/P$ satisfies the hypothesis, and thus, $G/P$ is supersoluble by induction on $|G|$. In view of Proposition 3.2, we have that $G$ is supersoluble. This completes the proof.

Proof of Theorem 1.2. Suppose that this theorem is false and let $(G,E)$ be a counterexample with $|G|+|E|$ minimal. Then, $E\neq 1$.

(1) $E$ is supersoluble.

By Lemma 2.6①, $E$ satisfies the hypothesis of Theorem 1.1. Hence, $E$ is supersoluble by Theorem 1.1.

(2) Let $p$ be the largest prime number dividing $|E|$ and $P$ a Sylow $p$-subgroup of $E$. Then, $P\unlhd G$ and $P<E$.

Since $E$ is supersoluble, we have that $P$ $char$ $E\unlhd G$, and so $P\unlhd G$. If $P=E$, then $G$ is supersoluble by Proposition 3.2, a contradiction. Hence, $P<E$.

(3) Final contradiction.

We now consider the quotient group $G/P$. It follows from Lemma 2.6② that the hypothesis holds for $(G/P,E/P)$ since $G/E\cong (G/P)/(E/P)$ is supersoluble. Hence, $G/P$ is supersoluble by the choice of $(G,E)$. Therefore, $G$ is supersoluble by Proposition 3.2. The final contradiction completes the proof.

4.   Some applications of our results

It is clear that every $\sigma$-$n$-embedded subgroup and every $\sigma$-permutable subgroup of $G$ are $\sigma$-embedded in $G$. Note that if $\sigma$ is the smallest partition of $\mathbb{P}$, that is, $\sigma_{i}$ is a one-element set for any $i\in I$, then every normal subgroup, every permutable subgroup, every $c$-normal subgroup, every $s$-permutable subgroup, every $s$-embedded subgroup and every $n$-embedded subgroup of $G$ are also $\sigma$-embedded in $G$. However, the converse is not true in general $($see Refs. [6, Example 1.2] and [7, Example 1.2]$)$. Hence, we may directly obtain the following results from Theorems $1.1$ and $1.2$ and Propositions 3.1 and 3.2.

Corollary 4.1. ^{[20, Theorem 3]} Let $G$ be a group of odd order. If all subgroups of $G$ of prime order are normal in $G$, then $G$ is supersoluble.

Corollary 4.2. ^{[9, Theorem 3.1]} If every subgroup of prime order and every cyclic subgroup of order $4$ are permutable in $G$, then $G$ is supersoluble.

Corollary 4.3. If every subgroup of prime order and every cyclic subgroup of order $4$ are $s$-permutable in $G$, then $G$ is supersoluble.

Corollary 4.4. ^{[13, Theorem 3.2]} Let $P$ be a Sylow $p$-subgroup of $G$, where $p$ is the smallest prime belonging to $\pi(G)$. If every cyclic subgroup of $P$ of order $p$ and order $4$ $($if $P$ is a nonabelian $2$-group$)$ is $s$-permutable in $G$, then $G$ is $p$-nilpotent.

Corollary 4.5. ^{[13, Theorem 3.1]} Suppose that $P$ is a normal $p$-subgroup of $G$ with $G/P$ is supersoluble. Suppose further that every subgroup of $P$ of order $p$ and order $4$ $($if $P$ is a nonabelian $2$-group$)$ is $s$-permutable in $G$, then $G$ is supersoluble.

Corollary 4.6. ^{[13, Theorem 3.4]} Let $N$ be a proper normal subgroup of $G$ with $G/N$ is supersoluble. Suppose that every subgroup of $N$ of prime order and order $4$ $($if $2\in \pi(N)$ and the Sylow $2$-subgroup of $N$ is nonabelian$)$ is $s$-permutable in $G$, then $G$ is supersoluble.

Corollary 4.7. ^{[10, Lemma 3.1]} Let $P$ be a Sylow $p$-subgroup of $G$, where $p$ is the smallest prime belonging to $\pi(G)$. If every cyclic subgroup of $P$ of order $p$ and order $4$ is $c$-normal in $G$, then $G$ is $p$-nilpotent.

Corollary 4.8. ^{[5, Theorem 4.2]} Let $G$ be a group. If every cyclic subgroup of $G$ of prime order and order $4$ is $c$-normal in $G$, then $G$ is supersoluble.

Corollary 4.9. ^{[12, Theorem 3.4]} Let $N$ be a normal subgroup of $G$ with $G/N$ being supersoluble. Suppose that every subgroup of $N$ of prime order and order $4$ is $c$-normal in $G$, then $G$ is supersoluble.

Corollary 4.10. ^{[13, Corollary 3.9]} Let $H$ be a subgroup of $G$ with $G'\leq H$. Suppose that every subgroup of $H$ of prime order and order $4$ $($if $2\in \pi(H)$ and the Sylow $2$-subgroup of $H$ is nonabelian$)$ is $s$-permutable in $G$, then $G$ is supersoluble.