Bulletin of the Iranian Mathematical Society https://doi.org/10.1007/s41980-018-0081-2 ORIGINAL PAPER

p-Supersoluble Hypercenter and s-Semipermutability of Subgroups of a Finite Group J. X. Shen1 · S. H. Qiao1 Received: 14 March 2017 / Accepted: 18 October 2017 © Iranian Mathematical Society 2018

Abstract Let G be a finite group and H a subgroup of G. We say that H is s-semipermutable in G if H G p = G p H for any Sylow p-subgroup G p of G with ( p, |H |) = 1. In this paper, we consider the s-semipermutability of prime-power order subgroups and prove the following result which generalizes some known results concerning s-semipermutable subgroups. Theorem: Suppose that p is a prime dividing the order of a finite group G and E is a normal subgroup of G. Then, E ≤ Z U p (G), if there exists a normal subgroup X of G, such that F p∗ (E) ≤ X ≤ E, and a Sylow p-subgroup P of X satisfies: 1. H ∩ O p (G ∗ ) is s-semipermutable in G for all subgroups H ≤ P with |H | = d, where d is a power of p with 1 < d < |P|. 2. If p = d = 2 and P is non-abelian, we further suppose H ∩ O p (G ∗ ) is ssemipermutable in G for H ≤ P cyclic of order 4. Finally, we also prove a partially generalized version of this theorem, which extends some results in [Y. Berkovich and I.M. Isaacs, p-supersolvability, and actions on pgroups stabilizing certain subgroups, J. Algebra 414 (2014) 82–94.] and [L. Miao, A. Ballester-Bolinches, R. Esteban-Romero, and Y. Li, On the supersoluble hypercentre of a finite group, Monatsh.Math. 184 (2017), no. 4, 641–648]. Keywords Finite group · s-semipermutable subgroup · p-supersoluble hypercenter Mathematics Subject Classification Primary 20D10; Secondary 20D20

Communicated by Hamid Mousavi.

B

S. H. Qiao [email protected] J. X. Shen [email protected]

1

School of Applied Mathematics, Guangdong University of Technology, Guangzhou 510006, People’s Republic of China

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1 Introduction In this paper, only finite groups are considered and G always denotes a finite group. Recall that two subgroups H and K of G are said to be permutable if H K = K H . A subgroup H of G is said to be s-permutable in G if H permutes with all Sylow subgroups of G [12]. A. N. Skiba introduced weakly s-permutability in the paper [14]. Many nice results on the structure of finite groups are presented by assuming that some subgroups of fixed order are weakly s-permutable. Recently, in the beautiful paper [5], Y. Guo and I.M. Isaacs considered the condition H ∩ O p (G)  O p (G) for a subgroup H of p-power order. This condition is less restrictive than weakly s-permutability, and the following result is proved. Theorem 1.1 [5, Theorem B] Let P ∈ Syl p (G) and let d be a power of p, such that 1 ≤ d < |P|. Assume that H ∩ O p (G)  O p (G) for all subgroups H  P with |H | = d. Then, either G is p-supersoluble, or else |P ∩ O p (G)| > d. Ballester-Bolinches, Esteban-Romero, and the second author of the present paper worked along this line, instead of assuming that all subgroups H ∩ O p (G) are normal in O p (G), they assumed that all of them are s-semipermutable in O p (G). This is based on the following observation. Let A ≤ G be a p-subgroup, A ∩ O p (G)  O p (G) if and only if A ∩ O p (G) is s-permutable in O p (G) [1, Lemma 1.2.16]. Clearly, O p (G) contains all Sylow q-subgroups of G for any prime q = p. Hence, A is ssemipermutable in G if and only if A is s-semipermutable in O p (G). We also note that, for a p-subgroup H ≤ O p (G), H is a s-semipermutable p-subgroup of G which implies that H is s-permutable in G [10, Lemma 2.2]. Theorem 1.2 [3, Theorem 2] Let P ∈ Syl p (G) and let d be a power of p, such that 1 ≤ d < |P|. Assume that H ∩ O p (G) is s-semipermutable in G for all subgroups H  P with |H | = d. Then, either G is p-supersoluble, or else |P ∩ O p (G)| > d. More recently, Miao, Ballester-Bolinches, Esteban-Romero, and Li in [11] presented some interesting results on supersoluble hypercenter of p-subgroup P  G by assuming H ∩ O p (G) is s-semipermutable in G for some fixed p-power order non-cyclic subgroup H with p 3 ≤ |H | < |P|. By [10, Lemma 2.2], it is equivalent to the condition H ∩ O p (G) is s-permutable in G; furthermore, it is equivalent to the condition H ∩ O p (G)  O p (G). Following [4], we use G ∗ to denote the smallest normal subgroup of G, such that G/G ∗ is abelian of exponent dividing p − 1, where p is a prime divisor of |G| that we hold fixed in this paper. Clearly, G ∗ contains all Sylow p-subgroups of G. It is not hard 2 : Z : Z to see that (G ∗ )∗ < G ∗ is possible. To see this, just see the example G = Z 19 3 2 for p = 19. Lemma 1.3 Let P ∈ Syl p (G) and let d be a power of p, such that 1 ≤ d < |P|. Assume that H ∩ O p (G ∗ ) is s-semipermutable in G for all subgroups H  P with |H | = d. Then, either G is p-supersoluble, or else |P ∩ O p (G ∗ )| > d. Based on Lemma 1.3, we could prove our main theorem:

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Theorem 1.4 Suppose that E is a normal subgroup of G, p is a prime dividing |G|. Then, E ≤ Z U p (G) if there exists a normal subgroup X of G, such that F p∗ (E) ≤ X ≤ E, and a Sylow p-subgroup P of X satisfies the following: 1. H ∩ O p (G ∗ ) is s-semipermutable in G for all subgroups H ≤ P with |H | = d, where d is a power of p with 1 < d < |P|. 2. If p = d = 2 and P is non-abelian, we further suppose that H ∩ O p (G ∗ ) is s-semipermutable in G for H ≤ P cyclic of order 4. Here, we remark the notations used in this paper. U denotes the class of all supersoluble groups. The supersoluble hypercenter Z U (G) of G is the product of all normal subgroups H of G, such that all G-chief factors under H have prime order, the psupersoluble hypercenter Z U p (G) is the product of all normal subgroups H of G, such that all p-G-chief factors under H have order p for some fixed prime p, and Z U p φ (G) denotes the product of all normal subgroups H of G, such that all non-Frattini p-Gchief factors of H have order p. The generalized p-Fitting subgroup F p∗ (G) follows the definition in [2]. In [10], Li, Su, Wang, and the second author of the present paper defined weakly s-semipermutability of subgroups: a subgroup H of G is said to be weakly ssemipermutable in G if there is a subnormal subgroup T of G, such that G = H T and H ∩ T ≤ HssG , where HssG is the subgroup of H generated by all those subgroups of H which are s-semipermutable in G. They proved the following result which extends L. A. Shemetkov and A. N. Skiba’s result [13, Theorem 1.4]. Theorem 1.5 [10, Main Theorem] Assume that p is a fixed prime in π(G) and E is a normal subgroup of G. Then, E ≤ Z U p φ (G) if there exists a normal subgroup X of G, such that F p∗ (E) ≤ X ≤ E, where F p∗ (E) is the generalized p-fitting subgroup of E, and X satisfies the following: for any Sylow p-subgroup P of X , P has a subgroup D, such that 1 < |D| < |P|, and all subgroups H of P with order |H | = |D| and all cyclic subgroups of P with order 4 (if P is a non-abelian 2-group and |D| = 2) are weakly s-semipermutable in G. Actually, by [10, Lemma 2.3(6)], our theorem (Theorem 1.4) extends Theorem 1.5.

2 Proofs We prove our main theorem (Theorem 1.4) via a series of results. The first lemma is easy but important which enables us to make a change on the condition H ∩ O p (G) by condition H ∩ O p (G ∗ ). It is frequently used in our proofs. Lemma 2.1 Suppose a group G has a decomposition G = P : M, such that M is abelian of exponent dividing p − 1, P ∈ Syl p (G). Then, P ≤ Z U (G). In particular, G is supersoluble. Proof Let T be a chief factor of G under P. Then, G acts (in a natural way) on T irreducibly. By [6, Chapter 3, Theorem 1.3], P acts trivially on T , and so, M acts on T irreducibly. By [6, Chapter 3, Theorem 2.4], T has order p. The lemma is proved.  

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The proof of Lemma 1.3 basically follows the idea of the proof of [3, Theorem 2]. Proof of Lemma 1.3 For the case G ∗ = G, this is [3, Theorem 2]. Suppose that G ∗ < G. The hypotheses are inherited by G ∗ . By induction, either G ∗ is p-supersoluble, or else |P ∩ O p ((G ∗ )∗ )| > d. If |P ∩ O p ((G ∗ )∗ )| > d, then |P ∩ O p (G ∗ )| ≥ |P ∩ O p ((G ∗ )∗ )| > d, as desired. Suppose that G ∗ is p-supersoluble and |P ∩ O p (G ∗ )| ≤ d. Clearly, by induction, we may suppose that O p (G) = 1. Then, P  G, in particular, G is p-soluble. If |P ∩ O p (G ∗ )| ≤ p, then P ∩ O p (G ∗ ) ≤ Z U (G). By Lemma 2.1, G is p-supersoluble. Thus, d > p. Let N be a minimal normal subgroup of G contained in P ∩ O p (G ∗ ). Then, |N | ≤ |P ∩ O p (G ∗ )| ≤ d. Let |N | < d. We show that G/N satisfies the hypotheses. Clearly, 1 ≤ d/|N | < |P/N |. Let H /N be a normal subgroup of P/N of order d/|N |. It follows that H /N ∩ O p ((G/N )∗ ) = H /N ∩O p (G ∗ )/N = (H ∩O p (G ∗ ))/N , which is s-semipermutable in G/N . This shows that G/N satisfies the hypotheses of the theorem. Since |P/N ∩ O p (G ∗ )/N | = |(P ∩ O p (G ∗ ))/N | ≤ d/|N |, it follows that G/N is p-supersoluble by induction. We may suppose that N  (P). Let Y be a maximal subgroup of P, such that N  Y . There holds Y ∩ N ≤ Y ∩ (P ∩ O p (G ∗ )) ≤ H ≤ Y for some H  P, since Y ∩ (P ∩ O p (G ∗ )) is normal in P of order at most d. Then, Y ∩ N = H ∩ N = N ∩ (H ∩ O p (G ∗ )) which is s-semipermutable in G, since H ∩ O p (G ∗ ) is s-semipermutable in G. Since Y ∩ N ≤ P = O p (G), we have Y ∩ N is s-permutable in G, and so N G (Y ∩ N ) ≥ O p (G)P. Since N is minimal normal in G, it follows that Y ∩ N = 1. Thus, N has order p, and G is p-supersoluble. Let |N | = d. Then, N = P ∩ O p (G ∗ ). By Lemma 2.1, G/N is p-supersoluble. Since the class of all p-supersoluble groups is a saturated formation, N is a proper subgroup of P which is not contained in (P). Let Y be a maximal subgroup of P not containing N . Arguing as in the previous paragraph, we could obtain that Y ∩ N is normal in G. This implies that Y ∩ N = 1 and |N | = p, against our assumption |N | = d > p. This final contradiction completes the proof.   Lemma 2.2 Let G p ∈ Syl p (G), P a normal subgroup of G p , and N is minimal normal in G contained in P. Suppose that H ∩ O p (G ∗ ) is s-semipermutable in G for all subgroups H ≤ P with |H | = d, where d is a power of p and 1 < d < |P|. Then, |N | ≤ d. Furthermore, N has order p if one of the following holds: 1. |N | = d. 2. |N | < d and N  (P). Proof By Lemma 2.1, we may suppose that N ≤ O p (G ∗ ). Suppose that |N | > d. Choose H  G p , such that H ≤ N with |H | = d. Then, H = H ∩ O p (G ∗ )  G. Since N is minimal, it follows that H = 1, contrary to 1 < d < |P|. Thus, |N | ≤ d. Pick S  G p , such that N ≤ S ≤ P and |S| = pd. Assume that |N | = d. Then, (S) ≤ N . If (S) = N , then S and N are cyclic, we are done. Suppose that (S) < N . Clearly, (S)  G p . Then, there is a normal subgroup chain (S)  N1  N  S  G p of G p , such that |N : N1 | = p. Since S/N1 is elementary abelian, there is another maximal subgroup K /N1 , such that K = N . Then, K ∩ N = N1 . Since |K | = d, by hypotheses, K ∩ O p (G ∗ ) is s-semipermutable in G. Then, N1 = N ∩ K = N ∩ K ∩ O p (G ∗ ) is s-semipermutable in G by [10,

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Lemma 2.2]. Since N1  G p , we have N1  G. This implies that N has order p, since N is minimal normal in G. Assume that |N | < d and N  (P). Clearly, N  (S), N ∩ (S) < N . Then, there is a normal subgroup chain N ∩ (S)  N1  N  S  G p of G p , such that |N : N1 | = p. In elementary abelian factor group S = S/(S)N1 , N has order p and has a complement K = K /(S)N1 , and K has order d. Since N = K and N1 ≤ K ∩ N , we have N1 = N ∩ K . Arguing as in the previous paragraph, we could obtain |N | = p, as desired.   Lemma 2.3 Let P be a normal p-subgroup of G. Suppose that P/(P) ≤ Z U (G/(P)). Then, P ≤ Z U (G). Proof Let R be the smallest normal subgroup of G, such that G/R is abelian and has exponent dividing p − 1. Then, G acts supersoluble hypercentrally on P/(P) by conjugation. Write P = P/(P). Since G acts supersoluble hypercentrally on P, assume that 1 = P0  P1  · · ·  Pt = P is a chief series of P, such that G acts on Pi for each i. Since Pi /Pi−1 has order p, it follows that G/C G (Pi /Pi−1 ) is cyclic of order dividing p − 1. Thus, R ≤ C G (Pi /Pi−1 ) for each i. Then, R stabilizes a normal subgroup chain of P/(P). Let Q ∈ Sylq (R) with p = q. Then, Q stabilizes a normal subgroup chain of P/(P). Thus, Q centralizes P/(P), since the action is co-prime. It follows that Q centralizes P, and so, R/C R (P) is a p-group. Let M be a minimal supplement of R in G, such that G = R M. Then, R ∩ M ≤ (M). Since G/R is abelian, it follows that M is nilpotent, and so, M is p -subgroup of G, since M is minimal. Consider the action of G = G/C G (P) by conjugation on P. Clearly, G = R : M, M is abelian and has exponent dividing p − 1. Let X be any G-chief factor under (P). By [6, Chapter 3, Theorem 1.3], R acts on X trivially, and so, G induces an abelian automorphism group AutG (X ) of exponent dividing p − 1. By [6, Chapter 3, Theorem 2.4], X has order p, as desired.   Lemma 2.4 Let P be a normal p-subgroup of G and p ∈ π(G). Suppose that H ∩ O p (G ∗ ) is s-semipermutable in G for every maximal subgroup H of P. Then, P ≤ Z U (G). Proof Let T be a minimal normal subgroup of G contained in P. Then, |T | ≤ |H | by Lemma 2.2. Clearly, (G/T , P/T ) satisfies the hypotheses of the lemma, and so P/T ≤ Z U (G/T ). Then, by Lemmas 2.2, 2.3, P ≤ Z U (G), as desired.   Lemma 2.5 Let P be a normal p-subgroup of G and p ∈ π(G). Suppose that d is a power of p, such that 1 < d < |P| and H ∩ O p (G ∗ ) is s-semipermutable in G for every subgroup H of P of order d and every cyclic subgroup of P with order 4 (if P is a non-abelian 2-subgroup and d = 2). Then, P ≤ Z U (G). Proof Suppose the lemma is false and let (G, P) be a counterexample with |G||P| minimal. If P ∩ O p (G ∗ ) = 1, by Lemma 2.1, then we are done. Thus, we may suppose that P ∩ O p (G ∗ ) = 1. Step 1. |P| > pd. This follows from Lemma 2.4. Step 2. d > p.

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Suppose that d = p. If P  O p (G ∗ ), then P ∩ O p (G ∗ )  G is properly contained in P. Let P/V be a chief factor of G, such that P ∩ O p (G ∗ ) ≤ V . Then, O p (G ∗ )P/O p (G ∗ )V is G-isomorphic to P/V ; by Lemma 2.1, P/V has order p. By Step 1, |V | > d. The hypotheses are also true for the pair (G, V ), which implies that V ≤ Z U (G) by the choice of (G, P). Hence, P ≤ Z U (G), a contradiction. Thus, P ≤ O p (G ∗ ) ≤ O p (G), and each subgroup H of P of order d and every cyclic subgroup of P with order 4 (if P is a non-abelian 2-subgroup and d = 2) is normal in O p (G). First, assume that p is the smallest prime dividing |G|. Then, G has no p-closed Schmidt subgroup of the form S = S p :Sq with S p ≤ P. Indeed, H  O p (G) for every subgroup H of S p of order p and every cyclic subgroup of S p with order 4 (if S p is a non-abelian 2-group). By [7, Chapter III, Theorem 5.2-Chapter IV, Theorem 5.4], S p /(S p ) is minimal normal in S/(S p ), so we have S p /(S p ) ≤ O p (S)(S p )/(S p ), since S p /(S p ) ∩ O p (S)(S p )/(S p ) = 1. This yields that S = O p (S), each subgroup of S of order p and every cyclic subgroup of S with order 4 (if S p is a non-abelian 2-group) are normal in S, a contradiction. By [7, Chapter IV, Theorem 5.4], every non- p-nilpotent group has a p-closed Schmidt subgroup. This implies that, for any p -element x of G, P:x = P × x . Thus, x centralizes P, G/C G (P) is a p-group, so P ≤ Z ∞ (G) ≤ Z U (G), a contradiction. Now, suppose that p is not the smallest prime dividing |G|. Then, p > 2. We know that every minimal subgroup of P is normal in O p (G). Consider the chain 1 ≤ 1 (P) ≤ · · · ≤ t (P) = P. By [14, Lemma 2.4], every subgroup of i (P)/i−1 (P) is normalized by O p (G)/i−1 (P). Clearly, choose suitable Pi , such that the chain can be refined to be a G p -chief series 1 = P0 ≤ P1 ≤ P2 ≤ · · · ≤ Pr = P, where P ≤ G p ∈ Syl p (G). Now, it is easy to see P ≤ Z U (G), a contradiction. Thus, d > p. Step 3 . If T is a minimal normal subgroup of G contained in P, then |T | < d. By Lemma 2.2, |T | ≤ d. If |T | = d, again by Lemma 2.2, we have d = |T | = p, contrary to Step 2. Then, |T | < d, and Step 3 holds. Step 4. A final contradiction. Let T be any minimal normal subgroup of G contained in P. By Step 3, the hypotheses hold for (G/T , P/T ). Hence, P/T ≤ Z U (G/T ). By Lemma 2.2, we   may suppose that T ≤ (P). By Lemma 2.3, P ≤ Z U (G). This finishes the proof. Theorem 2.6 Let p ∈ π(G) and P ∈ Syl p (G). Then, G is p-supersoluble if P satisfies: 1. H ∩ O p (G ∗ ) is s-semipermutable in G for all subgroups H ≤ P with |H | = d, where d is a power of p with 1 < d < |P|. 2. If p = d = 2 and P is non-abelian, we further suppose that H ∩ O p (G ∗ ) is s-semipermutable in G for H ≤ P cyclic of order 4. Proof If G ∗ = G, by [3, Theorems 4-5], G is p-supersoluble. Suppose that G ∗ < G. Then, the group pair (G ∗ , P) satisfies the hypotheses of the theorem. By induction, G ∗ is p-supersoluble. Since the hypotheses are inherited by (G/O p (G), P O p (G)/O p (G)), we may suppose that O p (G) = 1. Then,   O p (G ∗ ) = 1, and P is normal in G. By Lemma 2.5, G is p-supersoluble.

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Now, we are ready to prove our main theorem. Proof of Theorem 1.4 Consider the subgroup X . If H ∩ O p (G ∗ ) is s-semipermutable in G, then H ∩ O p (X ∗ ) = (H ∩ O p (G ∗ )) ∩ O p (X ∗ ) is s-semipermutable in G, so H ∩ O p (X ∗ ) is s-semipermutable in X by [10, Lemma 2.2]. By Theorem 2.6, we have that X is p-supersolvable. Therefore, F p∗ (X ) = F p (X ) = F p∗ (E) = F p (E) by [2, Lemma 2.10]. Write G = G/O p (G) and E = E O p (G)/O p (G) ∼ = E/O p (E). Then, F p∗ (E) = F p∗ (E)O p (G)/O p (G). Since F p∗ (X ) = F p (X ) = F p∗ (E) = F p (E), we have F p (E)O p (G)/O p (G) = F p (X )O p (G)/O p (G). It follows that F p∗ (E) ≤ X ≤ E. Suppose that O p (G) = 1. Let H be a subgroup of X of order d. Then, H = H O p (G)/O p (G) for some H ≤ X of order d. By the hypotheses, H ∩ O p (G ∗ ) is s-semipermutable in G. Hence, H ∩ O p (G ∗ ) = (H ∩ O p (G ∗ ))O p (G)/O p (G) is s-semipermutable in G. Then, G satisfies the hypotheses of the theorem, and so E ≤ Z U p (G) = Z U p (G). It follows that E ≤ Z U p (G), as desired. Let O p (G) = 1. Therefore, O p (X ) = O p (E) = 1. This yields F ∗ (X ) = F p∗ (X ) = F ∗ (E) = F p∗ (E) = F p (E) = F(E) = O p (E) = O p (X ) = P. For any H ≤ P of order d, by hypotheses, H ∩ O p (G ∗ ) is s-semipermutable in G. By Lemma 2.5, P ≤ Z U (G) ≤ Z U p (G). By [13, Lemma 2.4], E/C E (P) is supersoluble. Since F ∗ (E) = P, C E (P) ≤ P. Then, E/P is supersoluble, so is E. It follows that p-length of E is 1. Since O p (E) ≤ O p (G) = 1, P = F p (E) is a Sylow subgroup of E, this means that E ≤ Z U p (G), as desired. This finishes the proof of the theorem.  

3 A Generalization In this section, we give a partial generalization of our main result which also extends the results in [4] and [11]. First, we prove two lemmas. Lemma 3.1 Fix an integer e ≥ 3 and let P be a non-cyclic Sylow subgroup of G with |P| > p e . Suppose that H ∩ O p (G) is s-semipermutable in G for each non-cyclic subgroup H of order p e . Then, G is p-supersoluble. Proof Write U = O p (G). By [4, Theorem D], we may suppose that U < G. By [11, Theorem 3], we may suppose that either |P ∩ U | > p e or P ∩ U is cyclic. First, we suppose that |P ∩ U | > p e . By induction, we have U is p-supersoluble, since U < G. Thus, the p-length l p (U ) = 1. The hypotheses are inherited by G/O p (G), and thus, we may suppose that O p (G) = 1. Then, U has a normal Sylow p-subgroup. According to [11, Theorem 5], we have that G is p-supersoluble. Suppose below P ∩ U is cyclic. In addition, we may suppose that |P ∩ U | ≤ p e , and G is not soluble (Otherwise, U is soluble and so p-supersoluble), p = 2. If P ∩U ≤ H ≤ P with H being non-cyclic and |H | = p e , by [8, Theorem A], (H ∩U )G is soluble. Then, P ∩U ≤ (H ∩U )G , and we are done. If there is a non-cyclic subgroup H of order p e , such that U H < G, then we are done. Suppose that P = (P ∩ U )H for any non-cyclic subgroup H ≤ P of order p e , that is, U H = G. Let U ∩ H = 1.

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By hypotheses, H ∩ U = 1 is s-semipermutable in G, so in O p (G). By [4, Lemma 3.8], U is p-supersoluble, and so is G. Let U ∩ H = 1. Then, P = (P ∩U ) : H . Since H is subnormal in P, pick P ≥ P1 > H , such that |P1 : H | = p. Then, P1 = H : P0 for some subgroup P0 ≤ P1 . Then, (P ∩ U ) ∩ P1 = P0  P1 , so P1 = H × P0 . Let H0 be a maximal subgroup of P1 containing P0 . Then, H0 is non-cyclic of order p e . By hypotheses, P0 = H0 ∩ U = 1 is s-semipermutable in G, so in O p (G). By [4, Lemma 3.8], U is p-supersoluble, and so is G. This finishes the proof.   Lemma 3.2 Let P ∈ Syl p (G) and |P| > p e ≥ p 3 . Assume that H ∩ O p (G ∗ ) is s-semipermutable in G for all non-cyclic subgroups H of P with |H | = p e . Then, G is p-supersoluble. Proof For the case G ∗ = G, this is Lemma 3.1. Suppose that G ∗ < G. The hypotheses are inherited by G ∗ . By induction, G ∗ is p-supersoluble. Clearly, by induction, we  may suppose that O p (G) = 1, and then, P  G. By [11, Lemma 5], we are done.  By Lemma 3.2, following the idea of proof of Theorem 1.4, we can prove the following theorem which extends the results in [4] and [11]. Theorem 3.3 Suppose that p ∈ π(G) and E  G. Then, E ≤ Z U p (G) if there exists a normal subgroup X of G, such that F p∗ (E) ≤ X ≤ E, and a non-cyclic Sylow p-subgroup P of X satisfies: 1. |P| > p e ≥ p 3 . 2. H ∩ O p (G ∗ ) is s-semipermutable in G for all non-cyclic subgroups H ≤ P with |H | = p e . Acknowledgements The authors are very grateful to the referees who read the paper carefully and provided helpful comments. This work is supported by Cultivation Program for Outstanding Young College Teachers (Yq2013061) and Pei Ying Yu Cai Project of GDUT. This work forms part of the Master’s thesis of the first author JiaXin Shen.

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