1 Introduction

Guo et al. [9–12, 14] introduced the following new concepts of generalized permutable subgroups. Let A and B be subgroups of a group G and X a nonempty subset of G. Then A is said to be X-permutable with B if there exists some element x in X such that \(AB^x=B^xA\) (in particular, if \(X=G\), then, in [10], A is said to be conditionally permutable with B); A is said to be X-semipermutable in G if A is X-permutable with all subgroups of some supplement T of A in G. Based on these generalized permutable subgroups, one has given a series of new and interesting characterizations of the structure of finite groups (see [2, 6, 9–16, 24]).

Later on, as a generalization of X-semipermutability, Hao et al. introduced the concept of X-s-semipermutability in [19]. Let A be a subgroup of a group G and X a non-empty subset of G. Then A is said to be X-s-semipermutable in G if A is X-permutable with every Sylow subgroup of some supplement T of A in G. Obviously, the X-semipermutability and S-permutability imply the X-s-semipermutability. However, the converse does not hold. For example, let \(G=[\langle a, b\rangle ]\langle \alpha \rangle \), where \(a^4=1\), \(a^2=b^2=[a, b]\) and \(a^{\alpha }=b\), \(b^{\alpha }=ab\). Let \(A=\langle \alpha \rangle \) and \(X=1\). Clearly, A is X-s-semipermutable in G. But A is not X-semipermutable in G. On the other hand, let \(G=[C_5]C_4\), where \(C_5\) is a group of order 5 and \(C_4\) is the automorphism group of \(C_5\) of order 4. Let H be a subgroup of \(C_4\) of order 2. Then H is G-s-semipermutable in G but not S-permutable in G.

Note that Li et al. [28], introduced the concept of SS-quasinormality. A subgroup H of a group G is said to be SS-quasinormal in G if H has a supplement T in G such that H is permutable with every Sylow subgroup of T. Clearly, SS-quasinormality implies that X-s-semipermutability, where \(X=1\). But the converse does not hold in general. The group \(G=[C_5]C_4\) mentioned in the foregoing paragraph is a counterexample. Let H be a subgroup of \(C_4\) of order 2. Then H is G-s-semipermutable in G, but not SS-quasinormal in G.

Hao et al. [19, 20] investigated the influence of X-s-semipermutable subgroups on the supersolubility and p-nilpotency of finite groups. Our object in this paper is to study further this kind of generalized permutable subgroups. Moreover, we will present some new characterizations of p-nilpotency and supersolubility of finite groups under the assumption that some subgroups are X-s-semipermutable. One of our results obtained in this paper characterizes the structure of groups G all of whose subgroups are all G-s-semipermutable.

All groups considered in this paper are finite. For notation and terminology not given in this paper, the reader is referred to [8, 18, 22] if necessary. For some related topics, the reader is also referred to [1, 5, 21, 25–27, 29, 33, 35, 36].

2 Preliminaries

We begin by stating some elementary facts about the classes of finite groups.

Let \(\mathcal {F}\) be a class of groups. \(\mathcal {F}\) is said to be a formation if \(\mathcal {F}\) is a homomorph and every group G has a smallest normal subgroup (denoted by \(G^{\mathcal {F}}\)) whose quotient is still in \(\mathcal {F}\). A formation \(\mathcal {F}\) is said to be saturated if \(G/\Phi (G)\in \mathcal {F}\) always implies \(G\in \mathcal {F}\). A chief factor H / K of a group G is said to \(\mathcal {F}\)-central (or \(\mathcal {F}\)-eccentric) in G if \([H/K](G/C_G(H/K)) \in \mathcal {F}\) (or \([H/K](G/C_G(H/K)) \notin \mathcal {F}\) respectively). In this paper, \(Z_{\infty }^{\mathcal {F}}(G)\) denotes the \(\mathcal {F}\)-hypercenter of a group G, that is, the product of all such normal subgroups H of G whose G-chief factors are \(\mathcal {F}\)-central. We use \(\mathcal {N}\) and \(\mathcal {U}\) to denote the class of all nilpotent groups and the class of all supersoluble groups, respectively.

Lemma 2.1

[19, Lemma 2.1] Let A and X be subgroups of a group G and let N be a normal subgroup of G.

  1. (1)

    If A is X-s-semipermutable in G, then AN / N is XN / N-s-semipermutable in G / N.

  2. (2)

    If A is X-s-semipermutable in G, \(A\le D\le G\) and \(X\le D\), then A is X-s-semipermutable in D.

  3. (3)

    If A is X-s-semipermutable in G and \(X\le D\), then A is D-s-semipermutable in G.

Lemma 2.2

[23, Lemma 3.3] Let G be a group and X a normal p-soluble subgroup of G. Then G is p-soluble if and only if a Sylow p-subgroup P of G is X-permutable with all Sylow q-subgroups of G, where \(q\ne p\).

Lemma 2.3

[32, Lemma 2.10] Let G be a group. Suppose that p is the smallest prime dividing the order of G and P is a non-cyclic Sylow p-subgroup of G. If every maximal subgroup of P has a p-nilpotent supplement in G, then G is p-nilpotent.

Lemma 2.4

[31, Corollary 1] Let A be an S-permutable subgroup of a group G. Then A is subnormal in G.

Lemma 2.5

[6, Lemma 2.8] Let G be a group, p a prime and \((|G|, p-1)=1\). If M is a subgroup of G with index p, then M is normal in G.

Lemma 2.6

[17, Lemma 2.6] Let H be a nilpotent normal subgroup of a group G. If \(H\ne 1\) and \(H\cap \Phi (G)=1\), then H has a complement in G and H is a direct product of some minimal normal subgroups of G.

Lemma 2.7

[29, Theorem 1.3] Let p be a prime dividing the order of a group G and P a Sylow p-subgroup of G. If every maximal subgroup of P has a p-nilpotent supplement in G, then G is p-nilpotent.

3 Main Results

Theorem 3.1

Let \(\mathcal {F}\) be a saturated formation containing all supersoluble groups. A group \(G\in \mathcal {F}\) if and only if G has a normal soluble subgroup E such that \(G/E\in \mathcal {F}\) and for every non-cyclic Sylow subgroup P of F(E), every cyclic subgroup of P of order prime or order 4 (if P is a non-abelian 2-group and \(H\nsubseteq Z_{\infty }(G))\) not having a supersoluble supplement in G is G-s-semipermutable in G.

Proof

The necessity is clear and we need only to prove the sufficiency.

First, we claim that any chief factor of G below F(E) is of prime order. Assume that the assertion is not true and let L / K be a counterexample with |K| minimal, that is, L / K is not of prime order but for every chief factor U / V of G below F(E) with \(|V|<|K|\), U / V is of prime order. Since E is soluble, we see that L / K is a p-chief factor for some prime p. Noticing that \(L/K\simeq L\cap O_p(E)/K\cap O_p(E)\), we obtain by the choice of L / K that \(L/K=L\cap O_p(E)/K\cap O_p(E)\) and so \(L\subseteq O_p(E)\). Let P be the Sylow p-subgroup of F(E). If P is cyclic, then L / K is cyclic of order p, a contradiction. Hence we can assume that P is non-cyclic. Let R / K be a chief factor of \(G_p/K\), where \(G_p\) is a Sylow p-subgroup of G and \(R\subseteq L\). Then \(R=\langle x\rangle K\) for any \(x\in R\setminus K\). Now we assume that there is some element \(x\in R\setminus K\) of order p or 4 (if P is non-abelian 2-group and \(\langle x\rangle \nsubseteq Z_{\infty }(G)\)) not having a supersoluble supplement in G is G-s-semipermutable in G and prove that L / K is of order p, reaching a contradiction. If \(x\in Z_{\infty }(G)\), then \(xK/K\in L/K\cap Z_{\infty }(G/K)\) and so \(L/K\subseteq Z_{\infty }(G/K)\), which implies that L / K is of order p, a contradiction. If \(\langle x\rangle \) has a supersoluble supplement T in G, then \(L/K\cap TK/K=1\) or L / K. If \(L/K\cap TK/K=L/K\), then L / K is a chief factor of \(G/K=TK/K\), which is supersoluble. Therefore, L / K is cyclic of order p, a contradiction. If \(L/K\cap TK/K=1\), then \(L/K=L/K\cap (\langle x\rangle K/K) (TK/K)=\langle x\rangle K/K(L/K\cap TK/K)=\langle x\rangle K/K\), a contradiction again. These contradictions together with our hypothesis show that \(\langle x\rangle \) is G-s-semipermutable in G. Therefore, G has a subgroup T such that \(\langle x\rangle \) is G-permutable with every Sylow subgroup of T. Let \(T_q\) be a Sylow q-subgroup of T, where \(q\ne p\). Then \(\langle x\rangle (T_q)^g=(T_q)^g\langle x\rangle \) for some \(g\in G\). Since \(R/K=\langle x\rangle K/K\) is subnormal in G / K, \(\langle x\rangle K/K\) is subnormal in \((\langle x\rangle K/K)((T_q)^gK/K)\) and so \(\langle x\rangle K/K\) is normalized by \((T_q)^gK/K\). Now one can see that \(R/K=\langle x\rangle K/K\) is normal in G / K and, therefore, \(L/K=R/K\) is cyclic. This contradiction means that all elements of \(R\setminus K\) of order p or order 4 (if P is a non-abelian 2-group) are contained in K. Since \(L/K=(R/K)^{G/K}=R^G/K\), we have that all elements of L of order p or 4 (if P is a non-abelian 2-group) are contained in K.

Let U / V be any chief factor of G below K. Then, by the choice of L / K, U / V is of order p and so \(G/C_G(U/V)\) is abelian of exponent dividing \(p-1\). Put \(X=\bigcap _{U\subseteq K} C_G(U/V)\). Then X is normal in G and G / X is abelian of exponent dividing \(p-1\). Let Q be any Sylow q-subgroup of X, where \(q\ne p\). Then Q acts trivially on K by [18, Lemma 3.2.3]. Moreover, since all elements of L of order p or 4 (if P is a non-abelian 2-group) are contained in K, Q acts trivially on L / K by the well-known Blackburn’s theorem, from which we conclude that \(X/C_X(L/K)\) is a p-group. It follows that \(X\subseteq C_G(L/K)\) as \(O_p(G/C_G(L/K))=1\) by [18, Lemma 1.7.11] and thereby \(G/C_G(L/K)\) is abelian of exponent dividing \(p-1\). Now, by [34, I, Lemma 1.3], we have that L / K is of order p, which contradicts our assumption for L / K. Hence our claim holds. Thus \(F(E)\subseteq Z_{\infty }^{\mathcal {U}}(G)\) and thereby \(F(E) \subseteq Z_{\infty }^{\mathcal {F}}(G)\) (see [18, Theorem 3.1.6]).

Let M / N be any chief factor of G below F(E) and put \(C=\bigcap C_E(M/N)\). Then \(F(E)\subseteq C\) since \(F(G)\subseteq C_G(M/N)\). We assert that \(F(E)=C\). Suppose that it is not true and let R / F(E) be a minimal normal subgroup of G / F(E) with \(F(E)<R\le C\). Then \(R\subseteq Z_{\infty }(R)\) and R / F(E) is an elementary group as E is soluble. It follows that R is nilpotent and consequently \(R\subseteq F(E)\), a contradiction. Hence \(F(E)=C\). Since \(G/C_G(M/N)\) is abelian by the preceding argument and \(\mathcal {F}\) is a saturated formation, \(G/F(E)=G/C\in \mathcal {F}\). Since \(F(E) \subseteq Z_{\infty }^{\mathcal {F}}(G)\), we obtain that \(G\in \mathcal {F}\). Thus the proof is complete. \(\square \)

By Theorem 3.1, we have the following corollary.

Corollary 3.2

(Asaad and Cs\(\ddot{o}\)rg\(\ddot{o}\) [3]) Let \(\mathcal {F}\) be a saturated formation containing all supersoluble groups. Then a group \(G\in \mathcal {F}\) if and only if G has a normal soluble subgroup E such that \(G/E\in \mathcal {F}\) and the subgroups of prime order or order 4 of F(E) are S-permutable in G.

Theorem 3.3

Let G be a group and \(\mathcal {F}\) a saturated formation containing all supersoluble groups. Then \(G\in \mathcal {F}\) if and only if G has a normal soluble subgroup E such that \(G/E\in \mathcal {F}\) and every maximal subgroup of each non-cyclic Sylow subgroup of the Fitting subgroup F(E) not having a supersoluble supplement in G is G-s-semipermutable in G.

Proof

The necessity part is obvious. We only need to prove the sufficiency part. Assume that the assertion is false and let G be a counterexample of minimal order. Then

  1. (1)

    \(\Phi (G)\cap E=1\). Suppose that \(\Phi (G)\cap E\ne 1\). Let p be a prime divisor of \(|\Phi (G)\cap E|\) and P a Sylow p-subgroup of \(\Phi (G)\cap E\). Since \(\Phi (G)\cap E\) is a nilpotent normal subgroup of G, P is normal in G and so \(P\le F(E)\). Consider the factor group G / P. It is clear that \(F(E/P)=F(E)/P\) (see [18, Lemma 1.8.1]) and \((G/P)/(E/P)\simeq G/E\) is contained in \(\mathcal {F}\) by the hypothesis. Then by Lemma 2.1(2), we can see that G / P satisfies the hypothesis. Hence \(G/P\in \mathcal {F}\) by the choice of G. It follows that \(G\in \mathcal {F}\) as \(\mathcal {F}\) is a saturated formation, a contradiction.

  2. (2)

    \(F(E)=N_{1}\times N_{2}\times \cdots \times N_{t}\), where \(N_{i}\) is a minimal normal subgroup of G, for \(i=1, 2, \ldots , t\). This follows directly from Lemma 2.6 and (1).

  3. (3)

    \(N_i\) is a cyclic group of prime order, for all \(i\in \{1, 2,\ldots , t\}\). Without loss of generality, we may assume that \(P=N_{1}\times N_{2}\times \cdots \times N_{s}\) is a Sylow p-subgroup of F(E), where \(s\le t\). Let \(L_1\) be a maximal subgroup of \(N_1\) such that \(L_1\) is normal in some Sylow p-subgroup \(G_p\) of G and write \(B=N_{2}\times \cdots \times N_{s}\). Then \(L=L_1B\) is a maximal subgroup of P. If P is cyclic, then clearly \(N_1=P\) is cyclic of order p. Hence we assume that P is not cyclic. Now, by the hypothesis, L has a supersoluble supplement in G or is G-s-semipermutable in G. Suppose that L has a supersoluble supplement T in G. Then \((N_1\cap BT)^G=(N_1\cap BT)^{L_1BT}\subseteq N_1\cap BT\) and so \(N_1\cap BT=1\) or \(N_1\). If \(N_1\cap BT=1\), then \(N_1=N_1\cap L_1BT=L_1(N_1\cap BT)=L_1\), a contradiction. If \(N_1\cap BT=N_1\), then \(G=BT\) and, therefore, G / B is supersoluble. Since \(N_1B/B\) is a chief factor of G / B, \(N_1\simeq N_1B/B\) is of order p, as desired. Now assume that L is G-s-semipermutable in G. Then G has a subgroup T such that L is G-permutable with every Sylow subgroup of T. Let \(T_q\) be a Sylow q-subgroup of T, where \(q\ne p\). Then, for some element g of G, \(L(T_q)^g=(T_q)^gL\). Since L is subnormal in G, L is subnormal in \(L(T_q)^g\) and so L is normalized by \((T_q)^g\). Since L is also normalized by \(G_p\), we conclude that L is normal in G. Consequently \(L_1=L_1(N_1\cap B)=N_1\cap L_1B=N_1\cap L\) is normal in G, which implies that \(N_1\) is cyclic of order p. Similarly we can prove that \(N_i\) is a cyclic group of prime order for \(i=2,\ldots , t\).

  4. (4)

    Final contradiction. By (3), we see that \(G/C_{G}(N_i)\) is abelian, where \(i=1, 2, \ldots , t\). Hence \(G'\le C_G(N_i)\) and so \(G'\le C_G(F(E))\). It follows that \(G'\cap E\le C_H(F(E))=F(E)\). Hence by (2) and (3), every G-chief factor below \(G'\cap E\) is cyclic, from which we have that every chief factor of G below \(G'\cap E\) is \(\mathcal {F}\)-central. On the other hand, since \(\mathcal {F}\) is a saturated formation, \(G/(G'\cap E)\in \mathcal {F}\). This induces that \(G\in \mathcal {F}\). The final contradiction completes the proof. \(\square \)

The corollaries below follow from Theorem 3.3.

Corollary 3.4

(Ramadan [30]) Assume that G is a soluble group and every maximal subgroup of the Sylow subgroups of F(G) is normal in G. Then G is supersoluble.

Corollary 3.5

(Asaad et al. [4]) A soluble group G is supersoluble if and only if G has a normal subgroup E such that G / E is supersoluble and every maximal subgroup of each Sylow subgroup of F(E) is normal in G.

Corollary 3.6

(Asaad et al. [4]) Let G be a group with a normal supersoluble subgroup E such that G / E is supersoluble. If all maximal subgroups of any Sylow subgroup of F(H) is S-permutable in G, then G is supersoluble.

Corollary 3.7

(Chen and Li [6]) A group G is supersoluble if and only if G has a normal soluble subgroup E such that G / E is supersoluble and every maximal subgroup of each Sylow subgroup of F(E) is F(E)-semipermutable in G.

Now, we can characterize the structure of groups G with all subgroups G-s-semipermutable in the light of the preceding results.

Theorem 3.8

Let G be a group. Every subgroup of G is G-s-semipermutable in G if and only if

  1. (1)

    \(G=[H]K\), where \(H=G^{\mathcal {N}}\) is a nilpotent Hall subgroup of G with odd order, and

  2. (2)

    \(G=HN_G(L)\) for every subgroup L of H.

Proof

We first prove the necessity. Suppose that each subgroup of G is G-s-semipermutable in G. Then G has a Hall \(\{p, q\}\)-subgroup for different primes p and q dividing the order of G. By the well-known Arad’s result, we see that G is soluble. Moreover, by Theorem 3.1, G is supersoluble. It follows that \(G^{\mathcal {N}}\) is nilpotent. We claim that \(G^{\mathcal {N}}\) is of odd order. If not, assume that \(2\in \pi (G^{\mathcal {N}})\) and let P be a Sylow 2-subgroup of \(G^{\mathcal {N}}\). Then, P is normal in G and every chief factor of G below P is of order 2. Thus, \(P\le Z_{\infty }(G)\). Let D be a Hall \(p'\)-subgroup of \(G^{\mathcal {N}}\). Then \(G^{\mathcal {N}}\) is contained in D, a contradiction. Hence \(G^{\mathcal {N}}\) is of odd order.

Let \(H=G^{\mathcal {N}}\). We prove H is a Hall subgroup of G by induction. It is trivial if \(H=1\) and so we suppose \(H>1\). Let N be a minimal normal subgroup of G contained in H and \(|N|=p\), where p is a prime. Assume that G has a minimal normal subgroup R of prime order q with \(q\ne p\). Since the hypothesis holds for the factor group G / R, \((G/R)^{\mathcal {N}}\)=\(G^{\mathcal {N}}R/R\)=HR / R is a Hall subgroup of G / R by induction. Then the Sylow p-subgroup of H is also a Sylow p-subgroup of G. If there exists \(r\in \pi (H)\) with \(r\ne p\), then, by considering the factor group G / N, we conclude that the Sylow r-subgroup of H is a Sylow r-subgroup of G. Therefore, H is a Hall subgroup of G. Hence we can suppose that every minimal normal subgroup of G is a p-subgroup. Since G is supersoluble, \(O_p(G)\) is a Sylow p-subgroup of G and consequently \(H\le O_p(G)\). If \(N<H\), then, by induction, we see that H is a Hall subgroup of G. Hence, we now assume that \(H=N\) is a minimal normal subgroup of G. If \(H=O_p(G)\), then the conclusion is obvious. Thus, we suppose H is a proper subgroup of \(O_p(G)\). We assert that \(\Phi =\Phi (O_p(G))=1\). Assume this is not true. Then \((G/\Phi )^{\mathcal {N}}\)=\(H\Phi /\Phi \) is a Sylow p-subgroup of G. Since the class of all nilpotent groups ia a saturated formation, we have that H is not contained in \(\Phi \). Therefore, \(H\Phi \) is a Sylow p-subgroup of G, which implies that H is a Sylow p-subgroup of G, a contradiction. Hence \(\Phi =1\) and so \(O_p(G)\) is elementary abelian. Let L be any subgroup of \(O_p(G)\). We show that L is normal in G. By the hypothesis, L has a supplement T in G and L is G-permutable with the Sylow subgroups of T. Let \(T_q\) be a Sylow q-subgroup of T, where \(q\ne p\). Then, for some \(x\in G\), \(LT_q^x\) is a subgroup. Since L is subnormal in G, L is normal in \(LT_q^x\), which means that \(T_q^x\) normalizes L. In addition, since \(O_p(G)\) is an elementary abelian Sylow p-subgroup, L is normal in G, as wanted. Let \(O_p(G)=\langle a\rangle \times \langle a_2\rangle \times \cdots \times \langle a_t\rangle \) and \(H=\langle a\rangle \), where \(|a|=|a_i|=p\) for all \(i=2, \ldots , t\). Set \(a_1=a a_2\ldots a_t\). Then we have that \(O_p(G)=\langle a_1\rangle \times \langle a_2\rangle \times \cdots \times \langle a_t\rangle \). Since \(1\ne H\le O_p(G)\), \(O_p(G)\) is not contained in Z(G). Hence there exists an index \(i\in \{1,2,\ldots ,t\}\) such that \(a_i\) is not contained in Z(G). Pick a \(p'\)-element \(g\in G\backslash C_G(a_i)\). Then \(y=[a_i, g]\ne 1\). Since G / H is nilpotent, we know that \(y=[a_i, g]\in H\). On the other hand, \(y=[a_i, g]\in \langle a_i\rangle \) as \(\langle a_i\rangle \) is normal in G. Hence \(\langle a_i\rangle =H\), a contradiction. Therefore, \(H=G^{\mathcal {N}}\) is a Hall subgroup of G.

By the well-known Shur-Zassenhaus theorem, we see that H has a complement K in G. Since G / H is nilpotent, K is a Hall nilpotent subgroup in G and \(G=[H]K\), and therefore (1) holds. Finally, let L be any subgroup of H. By the preceding argument, \(N_G(L)\) contains a Hall \(\pi \)-subgroup of G, where \(\pi =\pi (K)\). It follows that \(K^x\le N_G(L)\) for some element x in G. Thus, \(G=HK\)=\(HK^x\)=\(HN_G(L)\), completing the proof of (2).

From now on, we prove the sufficiency. Suppose that G is a group satisfying (1) and (2). We will show that every subgroup of G is G-permutable with all Sylow subgroups of G and so is G-s-semipermutable in G. Let \(\pi =\pi (H)\) and \(\pi '\) the set of all primes not in \(\pi \). Let D be an arbitrary subgroup of G. By the hypothesis, G is soluble and so \(D=D_1D_2\), where \(D_1\) and \(D_2\) are Hall subgroups of D with \(\pi (D_1)\subseteq \pi \) and \(\pi (D_2)\subseteq \pi '\). Let P be any Sylow p-subgroup of G.

Suppose \(D_2=1\). Then \(D=D_1\). If \(p\in \pi \), then P is normal in G by the hypothesis and, therefore, \(DP=PD\). If \(p\in \pi '\), then by condition (2), there exists an element x in G such that \(P^x\le N_G(D)\). It follows that \(DP^x=P^xD\). Hence, in this case, D is G-permutable with all Sylow subgroups of G. Similarly, one can show that D is G-permutable with every Sylow subgroup of G provided \(D=D_2\).

Hence, we suppose that \(D_1\) and \(D_2\) are non-trivial. Note that \(D_1\le H\) by (1). Since \(D_1\) is subnormal in D by condition (1), \(D_1\) is normal in D. This means that \(D\le N_G(D_1)\). Since \(G=HN_G(D_1)\) by (2), \(N_G(D_1)\) contains a nilpotent Hall \(\pi '\)-subgroup of G by the solubility of G, B say. Without loss of generality, we may suppose that \(D_2\le B\). If \(p\in \pi \), then, clearly, \(PD=DP\) as P is normal in G. If \(p\in \pi '\), then G has an element x such that \(P^x\le B\). Since B is nilpotent, \(P^xD_2\) is a subgroup of \(N_G(D_1)\) and consequently \(P^xD_2D_1=P^xD\) is a subgroup of G. Thus, in this case, D is also G-permutable with all Sylow subgroups of G, completing the proof of the sufficiency. \(\square \)

Lemma 3.9

Let p be a prime dividing the order of a group G with \((|G|, p-1)=1\), P a Sylow p-subgroup of G and \(X=O_{p'p}(G)\). Then G is p-nilpotent if and only if every maximal subgroup of P not having a p-nilpotent supplement in G is X-s-semipermutable in G.

Proof

The necessity is obvious and we only need to prove the sufficiency. Suppose that the result is false and let G be a counterexample of minimal order. Then

  1. (1)

    P is not cyclic. Assume that P is cyclic. Then \(N_G(P)/C_G(P)\) is a \(p'\)-group. Since \(N_G(P)/C_G(P)\) is isomorphic to a subgroup of Aut(P) and \((|G|, p-1)=1\), we have \(N_G(P)=C_G(P)\) and, therefore, G is p-nilpotent by [22, IV, Theorem 2.6], a contradiction.

  2. (2)

    \(O_{p'}(G)=1\). Suppose that \(O_{p'}(G)\ne 1\). Then, by Lemma 2.1, it is easy to see that \(G/O_{p'}(G)\) satisfies the hypothesis. The minimal choice of G implies that \(G/O_{p'}(G)\) is p-nilpotent and so G is p-nilpotent, a contradiction.

  3. (3)

    \(O_p(G)\ne 1\). If not, then \(O_p(G)=1\) and so \(X=1\). First, we assume that every maximal subgroup of P has a p-nilpotent supplement in G. If \(p=2\), then by Lemma 2.3, G is p-nilpotent, a contradiction. Hence p is an odd prime and so G is also p-nilpotent by Lemma 2.7. Therefore, by the hypothesis, some maximal subgroup R of P is X-s-semipermutable in G. Then G has a subgroup T such that \(G=RT\) and R is X-permutable with every Sylow subgroup of T. Indeed, one can easily see that R is permutable with every Sylow q-subgroup of G, where \(q\ne p\). We claim that \(R\cap T\) is an S-permutable subgroup of T. In fact, let Q be a Sylow subgroup of T. Then \(RQ=QR\), whence \((R\cap T)Q=Q(R\cap T)\), as claimed. Thus, by Lemma 2.4, \(R\cap P\) is subnormal in T and so \(R\cap T\le O_p(T)\) by [7, A, Lemma 8.6]. Since \(|T: R\cap T|=|RT: R|=|G:R|\), \(|T/O_p(T)|\le p\). Similar to (1), we have that \(T/O_p(T)\) is p-nilpotent. It follows that T is p-soluble. Let K be a Hall \(p'\)-subgroup of T. Then \(RK=KR\) since R is permutable with every Sylow subgroup of T. The fact that \(|G: RK|=p\) and \((|G|, p-1)=1\) imply that RK is normal in G by Lemma 2.5. Since R is permutable with all Sylow q-subgroups of G, where \(q\ne p\), it follows from Lemma 2.2 that RK is p-soluble, which implies that either \(O_{p'}(RK)\ne 1\) or \(O_p(RK)\ne 1\). Consequently \(O_p(G)\ne 1\) by (2), a contradiction. Thus (3) holds.

  4. (4)

    \(O_p(G)\) is a minimal normal subgroup of G. It is easy to verify that \(G/O_p(G)\) satisfies the hypothesis. The minimal choice of G implies that \(G/O_p(G)\) is p-nilpotent. It follows that G is p-soluble. Let N be a minimal normal subgroup of G. Then N is an elementary abelian p-group by (2). Obviously G / N satisfies the hypothesis and so G / N is p-nilpotent. Since the class of all p-nilpotent groups is a saturated formation, N is the unique minimal normal subgroup of G and \(\Phi (G)=1\). Now it is easy to see that \(O_p(G)=F(G)=C_G(N)=N\). Hence \(O_p(G)\) is a minimal normal subgroup of G.

Final contradiction.

Since \(G/O_p(G)\) satisfies the hypothesis, \(G/O_p(G)\) is p-nilpotent and so G is p-soluble. By (3) and (4), we have that \(G=[O_p(G)]M\) for some maximal subgroup of G. In view of Lemmas 2.3 and 2.7, P has a maximal subgroup R not having a p-nilpotent supplement in G. By the hypothesis, R is \(O_p(G)\)-s-semipermutable in G since \(X=O_p(G)\) by (1). Hence G has a subgroup T such that R is \(O_p(G)\)-permutable with every Sylow subgroup of T. Since R is normalized by \(O_p(G)\), we can see that R is permutable with every Sylow subgroup of T. Let K be a Hall \(p'\)-subgroup of T. Then RK is a subgroup of G of index p by the above arguments and so RK is normal in G by Lemma 2.5. Consequently \(RK\cap O_p(G)=1\) or \(O_p(G)\). Note that \(O_p(G)\) is not contained in R (if not, R has a p-nilpotent supplement M in G, a contradiction) and so \(P=O_p(G)R\). Now, if \(O_p(G) \cap RK=O_p(G)\), then \(O_p(G)\) is contained in R, a contradiction. Therefore, \(O_p(G)\cap RK=1\) and so \(O_p(G)\) is of order p. Thus \(O_p(G)\) is contained in Z(G) as \(G/C_G(O_p(G))\) is isomorphic to a subgroup of \(Aut(O_p(G))\) and \((|G|, p-1)=1\). Since \(G/O_p(G)\) is p-nilpotent, it follows that G is p-nilpotent, a final contradiction. \(\square \)

Theorem 3.10

Let p be a prime dividing the order of a group G with \((|G|, p-1)=1\) and \(\mathcal {F}\) a saturated formation containing all p-nilpotent groups. Then \(G\in \mathcal {F}\) if and only if G has a normal subgroup E such that \(G/E\in \mathcal {F}\) and E has a Sylow p-subgroup P with the property that every maximal subgroup of P not having a p-nilpotent supplement in G is X-s-semipermutable in G, where \(X=O_{p'p}(E)\).

Proof

The necessity is clear and it needs only to prove the sufficiency. By Lemma 3.9, we have that E is p-nilpotent. Let K be a normal p-complement of E. If \(K\ne 1\), then G / K satisfies the hypothesis by Lemma 2.1 and so belongs to \(\mathcal {F}\) by induction. Let A / B be a chief factor of G below K. Since K is a \(p'\)-group, \(G/C_G(A/B)\) is \(\mathcal {F}\)-central by [18, §3.1, Example 2] and [18, Corollary 3.1.16]. It follows that \(G\in \mathcal {F}\). Now assume that \(K=1\). Then \(E=P\) is a normal p-subgroup of G. Let Q be a Sylow q-subgroup of G, where \(q\ne p\). Then PQ is p-nilpotent by the hypothesis and Lemma 3.9. Therefore, \(Q\le C_G(N)\). Let L / M be a chief factor of G with \(L\le E\). Then \(QM/M\le C_{G/M}(L/M)\) by above argument. Let \(G_p\) be a Sylow p-subgroup of G. Then \(L/M\cap Z(G_p/M)\ne 1\) (see [8, II, Theorem 6.4]). Let \(L_1/M\) be a subgroup of \(L/M\cap Z(G_p/M)\) of order p. Then \(G/M\le C_{G/M}(L_1/M)\) and so \(L_1/M\le Z(G/M)\). Consequently, \(L/M=L_1/M\le Z(G/M)\) as L / M is a chief factor of G, which implies that \(E\subseteq Z_{\infty }(G)\). Since \(G/E\in \mathcal {F}\) by the hypothesis, we have that \(G\in \mathcal {F}\) by [18, Theorem 3.1.6] and so the theorem follows. \(\square \)

From Theorem 3.10, we have

Corollary 3.11

(Chen and Li [6]) Let p be a prime dividing the order of a group G with \((|G|, p-1)=1\), P a Sylow p-subgroup of G and \(X=O_{p'p}(G)\). Then G is p-nilpotent if and only if every maximal subgroup of P not having a p-nilpotent supplement in G is X-semipermutable in G.