Finite groups with specific $S_*$-embedded subgroups

Abstract

We examine the p-nilpotency and supersolvablitiy of a finite group under the assumption that certain subgroups of prime power order are $S_*$- embedded in the group itself. In particular, we extend and generalize the recent results of Li (Commun. Algebra 50(4):1585–1594, https://doi.org/10.1080/00927872.2021.1986056, 2022) and the related results in the literature.

1 Introduction

All the groups are considered to be finite and G always stands for a finite group. $G_{p}$ denotes a Sylow p-subgroup of G where p is a prime number. We use conventional notions and notation as in [1]. Recall that a subgroup A of G is said to be S-permutable in G if $AG_{p}=G_{p}A$ for every Sylow p-subgroup $G_{p}$ of G [8]. Zhang and Wang [18] called a subgroup A of G S-semipermutable in G if A permutes with every Sylow p-subgroup $G_{p}$ of G such that $(p,|A|)=1$. Ballester-Bolinches and Pedraza-Aguilera [2] called a subgroup A of G S-permutably embedded in G if all Sylow p-subgroups of A are also Sylow p-subgroups of some S-permutable subgroup F of G. Obviously, the class of all S-permutably embedded subgroups is wider than the class of all S-permutable subgroups. More recently, Li [9] introduced a new embedding property which covers both of S-permutably embedded and S-semipermutable concepts as follows: A subgroup A of G is said to be $S_*$-embedded in G if G has an S-permutable subgroup F such that AF is S-permutable in G and $A \cap F \le A_*$, where $A_*$ is a subgroup contained in A which is either S-permutably embedded or S-semipermutable in G.

By using this new concept, Li [9] studied the group structure when certain subgroups are $S_*$-embedded and obtained new results generalised many classical and recent results in the literature. More precisely, he proved:

Theorem A

([8, Main Theorem]) Assume for each non-cyclic Sylow p-subgroup $G_{p}$ of G, either of the following two conditions is held:

• 1. All maximal subgroups of $G_{p}$ not having a supersolvable supplement in G are $S_*$-embedded in G.

• 2. All cyclic subgroups of $G_{p}$ of prime order or of order 4, that are without a supersolvable supplement in G, are $S_*$-embedded in G.

Then, G is supersolvable.

Our main object in this paper is to go further in studying the influence of $S_*$- embedded subgroups on the group structure. In fact, we prove:

Theorem B

Suppose $A\mathrel {\unlhd } G$ such that G/A is p-nilpotent for some prime divisor p of |G|. If A has a Sylow p-subgroup $A_{p}$ such that $N_G(A_{p})$ is p-nilpotent and all maximal subgroups of $A_{p}$, that are without a p-nilpotent supplement in G, are $S_*$-embedded in G, then G is p-nilpotent.

Theorem C

Suppose $A \mathrel {\unlhd }G$ such that G/A is supersolvable. If all maximal subgroups of each non-cyclic Sylow subgroup of A, that are without a supersolvable supplement in G, are $S_*$-embedded in G, then G is supersolvable.

Theorem D

Suppose $A\mathrel {\unlhd }G$ such that G/A is supersolvable. If all cyclic subgroups of each non-cyclic Sylow subgroup of A of prime order or of order 4, that are without a supersolvable supplement in G, are $S_*$-embedded in G, then G is supersolvable.

2 Preliminaries

Lemma 1

([9, Theorem 3.2]) Assume p is some prime divisor of |G| satisfying $(|G|,p-1)=1$, and $G_{p}$ is a Sylow p-subgroup of G. If all maximal subgroups of $G_{p}$, that are without a p-nilpotent supplement in G, are $S_*$-embedded in G, then G is p-nilpotent.

Lemma 2

([9, Lemma 2.5]) Let A be an $S_*$-embedded p-subgroup of G.

• 1. If $A \le B \le G$, then A is $S_*$-embedded in B.

• 2. If $L \mathrel {\unlhd } G$ and $L \le A$, then A/L is $S_*$-embedded in G/L.

• 3. If $L \mathrel {\unlhd } G$ and $(|L|,|A|)=1$, then AL/L is $S_*$-embedded in G/L.

• 4. If $L \mathrel {\unlhd } G$ and $A \le L$, then G has an S-permutable subgroup F contained in L such that AF is S-permutable in G and $A \cap F \le A_*$.

Lemma 3

([12, Theorem A]) Suppose that A is an S-permutable p-subgroup of G. Then $O^p(G)$ is contained in $N_G(A)$.

Lemma 4

([17, Lemma 2.1(d)]) Suppose that A is a p-subgroup of G and B is a normal p-subgroup of G. If A is either S-permutably embedded or S-semipermutable in G, then $A \cap B$ is also S-permutable in G.

Lemma 5

Let A be a subgroup of G.

• 1. [3] If A is S-permutable in G, then $A/A_G$ is nilpotent.

• 2. [8] If A is S-permutable in G, then A is subnormal in G.

• 3. [16] If A is a subnormal p-subgroup of G, then A is contained in $O_p(G)$.

• 4. [13] $A_{SG}$ is an S-permutable subgroup of G, where $A_{SG}$ is a subgroup of A generated by all S-permutable subgroups of G that contained in A.

• 5. [12] If A and B are S-permutable subgroups of G, then $A \cap B$ is also S-permutable in G.

Lemma 6

([10, 18]) Suppose that A is either S-permutably embedded or S-semipermutable in G. If A is a p-subgroup contained in $O_p(G)$, then A is S-permutable in G.

Lemma 7

([9, Lemma 2.1(3)]) Suppose that p is some prime divisor of |G| satisfying $(|G|,p-1)=1$. If G is p-supersolvable, then G is p-nilpotent.

Lemma 8

([15, Lemma 2.4]) If A is a maximal subgroup of G and B is a normal p-subgroup of G such that $G=AB$, then $A\cap B \mathrel {\unlhd } G$.

Lemma 9

([11, Theorem 3.5]) If $A \mathrel {\unlhd } G$ such that G/A is supersolvable and all maximal subgroups of any Sylow subgroup of A are normal in G, then G is supersolvable.

Lemma 10

([8]) Let A be an S-permutable subgroup of G.

• 1. If $A \le B \le G$, then A is S-permutable in B.

• 2. If $L \mathrel {\unlhd } G$, then AL/L is S-permutable in G/L.

Lemma 11

([5]) Suppose that A and B are normal supersolvable subgroups of G with $G=AB$. If the indices |G : A| and |G : B| are relatively prime, then G is supresolvable.

Lemma 12

([16]) If A is a subnormal subgroup of G such that the number |G : A| is not divisible by p, then every Sylow p-subgroup $G_{p}$ of G is contained in A.

3 Proofs

Theorem 1

Suppose $G_{p}$ is a Sylow p-subgroup of G for some prime divisor p of |G|. If $N_{G}(G_{p})$ is p-nilpotent and all maximal subgroups of $G_{p}$, that are without a p-nilpotent supplement in G, are $S_*$-embedded in G, then G is p-nilpotent.

Proof

If $p=2$, then, by Lemma 1, G is p-nilpotent. So, it can be assumed that $p>2$. Suppose the result is not true providing G as a counterexample of minimal order. Following the method of the first part of the proof of [6, Theorem 2.3], we conclude that all maximal subgroups of $G_{p}$ are $S_*$-embedded in G. Now, we build up the proof by the following steps:

• (1) $O_{p'}(G)=1$.

  Assume that $O_{p'}(G) \ne 1$. Let $X/O_{p'}(G)$ be a maximal subgroup of the Sylow p-subgroup $G_{p}O_{p'}(G)/O_{p'}(G)$ of $G/O_{p'}(G)$. Then $G_{p}$ has a maximal subgroup $X_1$ such that $X=X_1O_{p'}(G)$. Since $X_1$ is $S_*$-embedded in G, it follows that $X_1O_{p'}(G)/O_{p'}(G)$ is $S_*$-embedded in $G/O_{p'}(G)$ by Lemma 2(3). Also, we have $N_{G/O_{p'}(G)} (G_{p}O_{p'}(G)/O_{p'}(G)) = N_{G}(G_{p})O_{p'}(G)/O_{p'}(G)$ is p-nilpotent. Our choice of G implies that $G/O_{p'}(G)$ is p-nilpotent. Consequently, G is p-nilpotent which is a contradiction.

• (2) If $G_{p} \le A < G$, then A is p-nilpotent.

  Clearly, $N_A(G_{p}) \le N_G(G_{p})$ is p-nilpotent, and by Lemma 2(1) all maximal subgroups of $G_{p}$ are $S_*$-embedded in A. The minimal choice of G implies that A is p-nilpotent.

• (3) G is p-solvable.

  By Thompson’s result [14, Corollary], $G_{p}$ has a non-trivial characteristic subgroup R such that $N_{G}(R)$ is not p-nilpotent. Let L be any characteristic subgroup of $G_{p}$ such that $O_p(G)< L\le G_{p}$. Since L char $G_{p} \mathrel {\unlhd } N_G(G_{p})$, we have $L \mathrel {\unlhd }N_G(G_{p})$. Hence $G_{p} \le N_G(G_{p}) \le N_G(L)<G$ which implies that $N_G(L)$ is p-nilpotent by step (2). If $O_p(G)=1$, then $1<R\le G_{p}$ and hence $N_G(R)$ is p-nilpotent; a contradiction. Thus, we may assume that $O_p(G) \ne 1$. Clearly, $N_{G/O_p(G)}(G_{p}/O_p(G)) = N_G(G_{p})/O_p(G)$ is p-nilpotent. By using Lemma 2(2), it is easy to see that $G/O_p(G)$ satisfies the hypothesis of the theorem. Our choice of G implies that $G/O_p(G)$ is p-nilpotent, and thereby G is p-solvable.

• (4) There exists a unique minimal normal subgroup L of G such that G/L is p-nilpotent and $G=L \rtimes X$, where X is a maximal subgroup of G. Moreover, $L= C_G(L)=F(G)=O_p(G)$.

  Let L be a minimal normal subgroup of G. Then L is an elementary abelian p-group by steps (1) and (3). This implies $L \le O_p(G) \le G_{p}$. If $L=G_{p}$, then clearly, G/L is p-nilpotent. Thus, we may assume that $L<G_{p}$. In view of Lemma 2(2), we can see that the hypothesis still holds for G/L. Our choice of G implies that G/L is also p-nilpotent. Since the class of all p-nilpotent groups is a saturated formation, it follows that L is the unique minimal normal subgroup of G and $L \nleq \Phi (G)$. Hence, there exists a maximal subgroup X of G such that $G= L \rtimes X$, $L= C_G(L)=F(G)=O_p(G)$.

• (5) Finishing the proof.

  Let $X_p$ be a Sylow p-subgroup of X such that $G_{p}=LX_p$, and $P_{1}$ be a maximal subgroup of $G_{p}$ containing $X_p$. Then $G_{p}=LX_p=LP_1$. We may assume that $L \ne G_{p}$ (Otherwise, $G=N_{G}(L)=N_{G}(G_{p})$ is p-nilpotent; a contradiction). Since $P_1$ is $S_*$-embedded in G, then there exists an S-permutable subgroup F of G such that $P_1F$ is S-permutable in G and $(P_1 \cap F)\le (P_1)_*$. If $F=1$, then $P_1$ is S-permutable in G which implies that $P_1 \mathrel {\unlhd } G_{p}O^p(G)=G$ by using Lemma 3 and since $P_{1} \ne 1$, we have $L \le P_{1}$ by step (4) which means $P_1=G_{p}$; a contradiction. Thus, we may assume that $F \ne 1$. Assume that $F_G \ne 1$. Then $L\le F_G\le F$ which implies that $P_1\cap L \le P_1\cap F\le (P_1)_*\cap L$ and so $P_1\cap L=(P_1)_*\cap L$. Lemma 4 yields $P_1\cap L$ is S-permutable in G. If $P_1\cap L \ne 1$, then $L \le (P_1 \cap L)^{O^p(G)G_{p}} \le (P_1)^{G_{p}}=P_1$ by Lemma 3; a contradiction. So, assume $P_1\cap L=1$. This implies that $|L|=p$. Since $X \cong G/L=G/C_G(L) \lesssim Aut(L)$, we have $|X|\ |\ (p-1)$. It follows that L is a Sylow p-subgroup of G. Hence $L=G_{p}$; a contradiction. Thus, $F_G=1$. In view of Lemma 5(1), we have F is a nilpotent group. Then from step (1), F is a p-subgroup. By Lemma 5(2) and Lemma 5(3), we have $(P_1)_*\le P_1\le P_1F\le O_p(G)$ which implies that $(P_1)_*\le (P_1)_{sG}$ by using Lemma 6 and Lemma 5(4). Now, we have $P_1 \cap F\le (P_1)_*\le (P_1)_{sG}$ which implies $P_1\cap F \le (P_1)_{sG}\cap F$. Hence, $P_1\cap F= (P_1)_{sG}\cap F$. In view of Lemma 5(5), we have $P_1 \cap F$ is S-permutable in G. If $P_1 \cap F=1$, then $|F|=p$. It follows that $P_1F$ is a Sylow p-subgroup of G. Hence, $P_1F=G_{p}=O_p(G)=L$; a contradiction. Thus, it can be assumed that $P_1\cap F\ne 1$. Then, by using Lemma 3 and step (4), we have $L\le (P_1\cap F)^G =(P_1\cap F)^{O^p(G)G_{p}} \le (P_1)^{G_{p}}=P_1$. Hence $G_{p}=LP_1=P_1$; a final contradiction.$\square$

Proof of Theorem B

Assume the result is not true providing G as a counterexample of minimal order. Lemma 2(1) with Theorem 1, imply A is p-nilpotent. Let $A_{p'}$ be the normal p-complement of A. Since $A_{p'}$ char $A \mathrel {\unlhd } G$, then $A_{p'} \mathrel {\unlhd } G$. Assume that $|A_{p'}|>1$. Clearly, $A/A_{p'} \mathrel {\unlhd } G/A_{p'}$ and $(G/A_{p'})/(A/A_{p'}) \cong G/A$ is p- nilpotent. Let $X/A_{p'}$ be a maximal subgroup of the Sylow p-subgroup $A_{p}A_{p'}/A_{p'}$ of $A/A_{p'}$. Then $A_{p}$ has a maximal subgroup $X_1$ such that $X=X_1A_{p'}$. If $X_1$ possesses a p-nilpotent supplement D in G, then $DA_{p'}/A_{p'}$ is a p-nilpotent supplement of $X_1A_{p'}/A_{p'}$ in $G/A_{p'}$. If $X_1$ is $S_*$-embedded in G, it follows that $X_1A_{p'}/A_{p'}$ is $S_*$-embedded in $G/A_{p'}$ by using Lemma 2(3). Our choice of G implies that $G/A_{p'}$ is p-nilpotent. Consequently, G is p-nilpotent which is a contradiction. Thus, it can be assumed $A_{p'}=1$. Then $A=A_{p}$, which implies $G=N_G(A)=N_G(A_{p})$ is p-nilpotent; a final contradiction. $\square$

Remark 1

The condition $N_G(G_{p})$ is p-nilpotent in Theorem 1 and Theorem B is necessary. For example, consider $G=A_5$ and $p=3$. Then all maximal subgroups of any Sylow 3-subgroup $G_{3}$ of G are $S_*$-embedded in G, but G is not 3-nilpotent.

We work toward the proof of Theorem C:

Theorem 2

Suppose that p is some prime divisor of |G| satisfies $(|G|,p-1)=1$, and $A \mathrel {\unlhd } G$ such that G/A is p-nilpotent. If A has a Sylow p-subgroup $A_{p}$ such that all maximal subgroups of $A_{p}$, that are without a p-nilpotent supplement in G, are $S_*$-embedded in G, then G is p-nilpotent.

Proof

Assume the result is not true providing G as a counterexample of minimal order. By Lemma 2(1) and Lemma 1, we have that A is p-nilpotent. Let $A_{p^{'}}$ be the normal p-complement of A. Since $A_{p^{'}}$ char $A \mathrel {\unlhd } G$, then ${A_{p^{'}}} \mathrel {\unlhd } G$. Assume that $|{A_{p^{'}}}|>1$. Clearly, ${A/A_{p^{'}}} \mathrel {\unlhd } {G/A_{p^{'}}}$ and $({G/A_{p^{'}}})/({A/A_{p^{'}}}) \cong G/A$ is p-nilpotent. Let ${X/A_{p^{'}}}$ be a maximal subgroup of the Sylow p-subgroup ${A_{p}A_{p^{'}}}/{A_{p^{'}}}$ of ${A/A_{p^{'}}}$. Then $A_{p}$ has a maximal subgroup $X_1$ such that $X=X_1{A_{p^{'}}}$. If $X_1$ possesses a p-nilpotent supplement D in G, then $D{A_{p^{'}}}/{A_{p^{'}}}$ is a p-nilpotent supplement of $X_1{A_{p^{'}}}/{A_{p^{'}}}$ in $G/{A_{p^{'}}}$. If $X_1$ is $S_*$-embedded in G, then $X_1{A_{p^{'}}}/{A_{p^{'}}}$ is $S_*$-embedded in $G/{A_{p^{'}}}$ by Lemma 2(3). So $G/{A_{p^{'}}}$ is p-nilpotent due to the minimal choice of G. It follows, G is p-nilpotent; a contradiction. Thus, it can be assumed ${A_{p^{'}}}=1$ which yields $A=A_{p}$ is a p-group. Let $L/A_{p}$ be the normal p-complement of $G/A_{p}$. Schur-Zassenhaus Theorem implies that L has a Hall $p^{'}$-subgroup $L_{p^{'}}$ such that $L=A_{p} \rtimes {L_{p^{'}}}$. Since L is p-nilpotent by Lemma 2(1) and Lemma 1, it follows that $L=A_{p} \times {L_{p^{'}}}$. Hence, we have $L_{p^{'}}$ is the normal p-complement of G, and thereby G is p-nilpotent; a contradiction. $\square$

Lemma 7 and Theorem 2 lead to the following corollary:

Corollary 1

Suppose $A \mathrel {\unlhd } G$ provided G/A is p-nilpotent, where p is the smallest prime divisor of |G|. If A has a Sylow p-subgroup $A_{p}$ such that all maximal subgroups of $A_{p}$ that are without a p-supersolvable supplement in G are $S_*$-embedded in G, then G is p-nilpotent.

Proof

Clearly, $(|G|,p-1)=1$ as p is the smallest prime divisor of |G|. Lemma 7 suggests all maximal subgroups of $A_{p}$ that are without a p-nilpotent supplement in G are $S_*$-embedded in G. Using Theorem 2, gives G is p-nilpotent. $\square$

Now we can prove Theorem C:

Proof of Theorem C

Assume the result is not true providing G as a counterexample of minimal order. In view of Theorem A(1), we have A is supersolvable. Let $G_{p}$ be a Sylow p-subgroup of G, where p is the largest prime divisor of |G|. We distinguish two cases.

• Case 1. $G_{p} \le A$.

  Then $G_{p} \mathrel {\unlhd } A$ as A is supersolvable. Since $G_{p}$ char $A \mathrel {\unlhd }G$, it follows that $G_{p} \mathrel {\unlhd } G$. Now, we show that $G/G_{p}$ is supersolvable. Clearly, $(A/G_{p}) \mathrel {\unlhd } (G/G_{p})$ and $(G/G_{p})/(A/G_{p}) \cong (G/A)$ is supersolvable. Let $X/G_{p}$ be a maximal subgroup of the Sylow q-subgroup $A_{q}G_{p}/G_{p}$ of $A/G_{p}$. Then $A_{q}$ has a maximal subgroup $X_1$ such that $X=X_1G_{p}$. If $X_1$ has a supersolvable supplement B in G, then $BG_{p}/G_{p}$ is a supersolvable supplement of $X_1G_{p}/G_{p}$ in $G/G_{p}$. If $X_1$ is $S_*$-embedded in G, then $X_1G_{p}/G_{p}$ is $S_*$-embedded in $G/G_{p}$ by using Lemma 2(3). The minimal choice of G yields $G/G_{p}$ is supersolvable and $G_{p}$ is not cyclic. Let N be a minimal normal subgroup of G contained in $G_{p}$. It is also easy to see that G/N is supersolvable. Further, since the class of all supersolvable groups is a saturated formation, it follows that N is the unique minimal normal subgroup of G contained in $G_{p}$ and $N \nleq \Phi (G)$. Hence G possesses a maximal subgroup X such that $G=NX$ and $N \cap X=1$. Since $G_{p} \cap X$ is normalized by G by using Lemma 8, it follows that $N=G_{p}$ which yields $G_{p}$ is an elementray abelian p-group. Now, Let $N_1$ be a maximal subgroup of N. If $N_1$ has a supersolvable supplement B in G, then $G=N_1B=NB$ and $N=N \cap N_1B=N_1(N\cap B)$, which implies that $N \cap B \ne 1$. Since $N \cap B \mathrel {\unlhd } G$ and N is a minimal normal subgroup of G, we have $N \cap B=N$. Consequently, $N \le B$ which implies that $G=B$ is supersolvable which is a contradiction. Thus, we can assume that $N_1$ is $S_*$-embedded in G. In view of Lemma 2(4), G possesses an S-permutable subgroup F contained in $G_{p}$ such that $N_1F$ is S-permutable in G and $N_1 \cap F \le (N_1)_*$. If $F=1$, then $N_1$ is S-permutable in G and $N_1 \mathrel {\unlhd } G_{p}O^p(G)=G$ by using Lemma 3 and so $|G_{p}|=p$; a contradiction. Thus, $F \ne 1$. Since $G_{p}$ is an elemntary abelian p-group, then $F \mathrel {\unlhd } G_{p}$. Applying Lemma 3 again, we get $F \mathrel {\unlhd } G_{p}O^p(G)=G$. This implies $F=G_{p}=N$. Hence $N_1 \cap F=N_1=(N_1)_*$, is S-permutable in G by Lemma 6; again a contradiction as above.

• Case 2. $G_{p} \nleq A$.

  In this case we distinguish the following two subcases.

  • Subcase (i). $G_{p}A<G$.

    Clearly, $A \mathrel {\unlhd } G_{p}A$ and $G_{p}A/A \cong G_{p}/G_{p}\cap A$ is supersolvable. By Lemma 2(1), all maximal subgroups of any non-cyclic Sylow subgroup of A not having a supersolvable supplement in G are $S_*$-embedded in $G_{p}A$. Therefore, $G_{p}A$ is supersolvable due to the minimal choice of G. Since $G_{p}A/A$ is a Sylow p-subgroup of G/A, where p is the largest prime divisor of |G| and G/A is supersolvable, it follows that $G_{p}A/A \mathrel {\unlhd } G/A$ and so $G_{p}A \mathrel {\unlhd } G$. Since $G_{p}$ char $G_{p}A \mathrel {\unlhd } G$, then $G_{p} \mathrel {\unlhd } G$. Hence $G_{p} \cap A \mathrel {\unlhd } G$, where $G_{p} \cap A$ is a Sylow p-subgroup of A. By using the same arguments as in Case 1, we have $G/G_{p}\cap A$ is supersolvable and $G_{p} \cap A$ is a minimal normal subgroup of G. Set $N=G_{p} \cap A$. By our choice of G together with Lemma 9, N has a maximal subgroup $N_1$ such that $N_1 \ntrianglelefteq G$. If $N_1$ possesses a supersolvable supplement B in G, we have G is supersolvable as in Case 1; a contradiction. Thus, we can assume that $N_1$ is $S_*$-embedded in G. In view of Lemma 2(4), there exists an S-permutable subgroup F of G contained in N such that $N_1F$ is S-permutable in G and $N_1 \cap F \le (N_1)_*$. By the maximality of $N_1$ in N, we have either $N_1F=N_1$ or $N_1F=N$. If the former holds, then $N_1$ is S-permutable in G. So, $N_1 \mathrel {\unlhd } NO^p(A)=A$ and hence $A \le N_G(N_1)$ by using Lemma 10(1) and Lemma 3. Applying Lemma 3 again, we get $O^p(G) \le N_G(N_1)<G$. Hence G possesses a maximal subgroup X such that $O^p(G) \le N_G(N_1) \le X<G$ with $|G:X|=p$. Since $X/O^p(G) \mathrel {\unlhd } G/O^p(G)$, then $X \mathrel {\unlhd } G$. Due to the minimal choice of G, we have X is supersolvable. Therefore, $G=G_{p}X$ and consequently G is supersolvable by using Lemma 11; a contradiction. Thus, we can assume that $N_1F=N$. If $F \mathrel {\unlhd } G$, then $F=N$ which implies $N_1 \cap F=N_1=(N_1)_*$ is S-permutable in G by Lemma 6; a contradiction as above. Thus, $F \ntrianglelefteq G$. Since N is an elementary abelian p-group, then $F \mathrel {\unlhd } N$. Hence $F \mathrel {\unlhd } NO^p(A)=A$ by Lemma 10(1) and Lemma 3 and so $A \le N_G(F)$. Applying Lemma 3 again, there exists a maximal normal subgroup X of G such that $O^p(G) \le N_G(F) \le X <G$ with $|G:X|=p$. It follows that X is supersolvable by our choice of G. So, $G=G_{p}X$ and consequently G is supersolvable again by Lemma 11; a contradiction.

  • Subcase (ii). $G=G_{p}A$.

    Lemma 12 and Corollary 1 yield A contains all Sylow q-subgroups of G with $q \ne p$ and G is a $q_r$-nilpotent, where $q_r$ is the smallest prime divisor of |G|. This implies that G has a Sylow tower group of supersolvable type. Therefore, $G_{p}\mathrel {\unlhd } G$ as p is the largest prime divisor of |G|. Applying Lemma 11, G is supersolvable; a final contradiction.$\square$

In order to show Theorem D, we need the following useful theorem:

Theorem 3

Suppose A is a normal p-subgroup of G such that G/A is supersolvable. If all cyclic subgroups of A of order p or of order 4 not having a supersolvable supplement in G are $S_*$-embedded in G, then G is supersolvable.

Proof

Assume the result is not true providing G as a counterexample of minimal order. It is easy to see that all proper subgroups of G are supersolvable by using Lemma 2(1). Hence G is a minimal non-supersolvable group. By Doerk’s result [4] G has a Sylow p-subgroup $G_{p}$ such that $G_{p} \mathrel {\unlhd } G$ for a prime divisor p of |G|, $G_{p}/\Phi (G_{p})$ is a minimal normal subgroup of $G/\Phi (G_{p})$, and the exponent of $G_{p}$ is either p or 4. Clearly, $A\le G_{p}$ (Otherwise, $G\cong G/G_{p}\cap A \lesssim G/G_{p} \times G/A$ is supersolvable; a contradiction). We build up the proof by the following two steps.

• (1) $A=G_{p}$.

  Since $A\Phi (G_{p})/\Phi (G_{p}) \mathrel {\unlhd } G/\Phi (G_{p})$, we have either $A\Phi (G_{p})=G_{p}$ or $A\le \Phi (G_{p})$. If the latter holds, then $G/\Phi (G_{p})$ is supersolvable. It follows that, from $\Phi (G_{p}) \le \Phi (G)$, $G/\Phi (G)$ is supersolvable and so G is also supersolvable by a well-known result of Huppert [7, p.713]; a contradiction. Thus, $A\Phi (G_{p})=G_{p}$ and thereby $A=G_{p}$ as required.

• (2) Finishing the proof.

  Assume that $|A/\Phi (A)|=p$. Then there exists x in A such that $A/ \Phi (A) = <x\Phi (A)>$ which implies that A is cyclic and consequently G is supersolvable which is a contradiction. So, $|A/\Phi (A)|=p^n$, $n>1$ and $A/\Phi (A)=<x_1\Phi (A),x_2\Phi (A),...,x_n\Phi (A)>$ as $A/ \Phi (A)$ is an elementary abelian p-group. Hence, we have $A=<x_1,x_2,...,x_n>$. Set $A_i=<x_i>$ for all $i=1,2,...,n$. So, we have $|A_i|=p$ or 4. Now, the hypothesis of the theorem assures that $A_i$ either has a supersolvable supplement in G say B or $A_i$ is $S_*$-embedded in G. If $A_i$ is not $S_*$-embedded in G, then $G=A_iB$ and so $A=A\cap G=A\cap A_iB=A_i(A\cap B)$. Obviously, ${(A\cap B)\Phi (A)}/\Phi (A) \mathrel {\unlhd } G/\Phi (A)$ as $A/\Phi (A)$ is abelian. In view of step (1), $A/\Phi (A)$ is a minimal normal subgroup of $G/\Phi (A)$ which implies that either $(A\cap B)\Phi (A)=A$ or $(A\cap B)\le \Phi (A)$. If the latter holds, then $A=A_i$ is cyclic and so G is supersolvable; a contradiction. Hence $(A\cap B)\Phi (A)=A$ and so $A\cap B=A$ which implies that $G=B$ is supersolvable; contradicts our choice of G. Thus, we can assume that $A_i$ is $S_*$-embedded in G. In view of Lemma 2(4), then G possesses an S-permutable subgroup F contained in A such that $A_iF$ is also S-permutable in G and $A_i\cap F\le (A_i)_*$. By Lemma 10(2) and the fact that $A/\Phi (A)$ is abelian, it is easy to see that ${F\Phi (A)}/\Phi (A)$ is S-permutable in $G/\Phi (A)$ and ${F\Phi (A)}/\Phi (A) \mathrel {\unlhd } A/\Phi (A)$. Applying Lemma 3, we have ${F\Phi (A)}/\Phi (A) \mathrel {\unlhd } (A/\Phi (A))(O^p(G/\Phi (A)) = G/\Phi (A)$. Again, the minimal normality of $A/\Phi (A)$ in $G/\Phi (A)$ implies that either $F\Phi (A)=A$ or $F\Phi (A)\le \Phi (A)$. If the latter holds, ${A_i\Phi (A)}/\Phi (A)= {A_iF\Phi (A)}/\Phi (A)$ is S-permutable in $G/\Phi (A)$ by using Lemma 10(2). If $F\Phi (A)=A$, then we have $F=A$. Therefore, $A_i\cap F=A_i=(A_i)_*$ is S-permutable in G by Lemma 6, and so ${A_i\Phi (A)}/\Phi (A)$ is S-permutable in $G/\Phi (A)$ by Lemma 10(2). By [13, Lemma 2.11], there exists a maximal subgroup ${X\Phi (A)}/\Phi (A)$ of $A/\Phi (A)$ such that ${X\Phi (A)}/\Phi (A) \mathrel {\unlhd } G/\Phi (A)$; a final contradiction.$\square$

Proof of Theorem D

Assume the result is not true providing G as a counterexample of minimal order. If the order of A is of prime power, then G is supersolvable by Theorem 3; a contradiction. Thus, we can assume that the order of A is divisible by at least two distinct primes. By Lemma 2(1) and Theorem A(2), we have A is supersolvable. Hence A possesses a normal Sylow p-subgroup $A_{p}$, where p is the largest prime divisor of |A|. Since $A_{p}$ char $A \mathrel {\unlhd } G$, we have $A_{p} \mathrel {\unlhd } G$. Let $U/A_{p}$ be a cyclic subgroup of the Sylow q-subgroup $A_{q}A_{p}/A_{p}$ of $A/A_{p}$ such that $|U/A_{p}|=q$ or 4. Then $A_{q}$ has a cyclic subgroup R such that $U=RA_{p}$ and $|R|=q$ or 4. If R has a supersolvable supplement B in G, then $BA_{p}/A_{p}$ is a supersolvable supplement of $RA_{p}/ A_{p}$ in $G/A_{p}$. If R is $S_*$-embedded in G, then $RA_{p}/A_{p}$ is $S_*$-embedded in $G/A_{p}$ by using Lemma 2(3). Our choice of G yields $G/A_{p}$ is supersolvable. Applying Theorem 3, we get G is supersolvable; a contradiction. $\square$