Commun. Math. Stat. DOI 10.1007/s40304-017-0115-8

Some Criteria for p-Supersolvability of a Finite Group Liyun Miao1 · Yangming Li2

Received: 18 February 2017 / Revised: 6 July 2017 / Accepted: 10 July 2017 © School of Mathematical Sciences, University of Science and Technology of China and Springer-Verlag GmbH Germany 2017

Abstract A subgroup H of a finite group G is said to be S-semipermutable in G if H permutes with all Sylow q-subgroups of G for the primes q not dividing the order of H . Some criteria for p-supersolvability of a finite group are given, which are the generalizations of many recent results. Keywords p-Supersolvability · S-semipermutable subgroup · S-permutable subgroup Mathematics Subject Classification 20D10 · 20D15

1 Introduction All groups considered in this paper are finite. We use conventional notions and notation, as in Huppert [7]. Throughout this article, G stands for a finite group and p stands for a prime and π(G) represents the set of distinct primes dividing |G|, the order of G. Suppose that p ∈ π(G). G p denotes a Sylow p-subgroup of G and O p (G) denotes the unique smallest normal subgroup of G for which the corresponding factor group

The project of NSFC (11271085) and NSF of Guangdong Province (CHINA) (2015A030313791) and The Innovative Team Project of Guangdong Province (CHINA) (2014KTSCX196).

B

Yangming Li [email protected] Liyun Miao [email protected]

1

Department of Mathematics, Shanghai University, Shanghai 200444, China

2

Department of Mathematics, Guangdong University of Education, Guangzhou 510310, China

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is a p-group. The class of p-supersolvable group was denoted by U p . Denote the p-supersolvable hypercenter of G by Z U p (G), the largest normal subgroup of G such that every pd-G-chief factor below Z U p (G) is of the order p. Berkovich and Isaacs [3] by G ∗ denote the unique smallest normal subgroup of G for which the corresponding factor group is abelian of exponent dividing p − 1 for a fixed prime p. For a class F of finite groups, F-residual G F of G is the unique smallest normal subgroup of G for which the corresponding factor group is in F [5]. Since people usually use A p−1 to denote the class of all abelian groups whose exponent divide p − 1, we replace G ∗ by G A p−1 to indicate that the subgroup is the A p−1 –residual of G [5]. G A p−1 is an important subgroup of G; it is known that G is p-supersolvable if and only if G A p−1 is p-nilpotent (ref. Proposition 2.3). The recent Berkovich and Isaacs’s following nice result is related to G A p−1 . Theorem 1.1 [3, Theorem A] Fix an integer e ≥ 3, and let P be a p-group with |P| > p e . Let H act on P, and assume that every noncyclic subgroup of P with order p e is stabilized by O p (H ). Then P is centralized by O p (H A p−1 ). Recall that two subgroups H and K of a group G are said to permute if H K = K H , i.e., H K is a subgroup of G. A subgroup H of a group G is said to be S-permutable [8] in G if H permutes with all Sylow subgroups of G, H is said to be S-semipermutable [4] in G if H permutes with all Sylow q-subgroups of G for the primes q not dividing |H |. As indicated in [10], Theorem 1.1 is equivalent to the following: Theorem 1.1* Fix an integer e ≥ 3. Let P be a normal p-subgroup of G with |P| > p e . Suppose that every noncyclic subgroup of P with order p e is S-permutable in G. Then P ≤ Z U p (G). The main result of [6] is the following. Theorem 1.2 [6, 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) is normal in O p (G) for all subgroups H  P with |H | = d. Then G is p-supersolvable, or else |P ∩ O p (G)| > d. In [2], Theorem 1.2 is extended as follows: Theorem 1.3 [2, 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 G is p-supersolvable, or else |P ∩ O p (G)| > d. Stimulated by [3], we give conditions to all noncyclic subgroups of P with order d in [10] and obtain Theorem 1.4 [10, Theorem 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 noncyclic subgroups H of P with |H | = d. Then |P ∩ O p (G)| > d, P ∩ O p (G) is cyclic, or else G is p-supersolvable.

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In [10], we also extend Theorem 1.1∗ as follows: Theorem 1.5 [10, Theorem 5] Fix an integer e ≥ 3, and let P be a normal p-subgroup of a group G with |P| > p e = d. Assume that H ∩ O p (G) is S-semipermutable in G for all noncyclic subgroups H of P with |H | = d. Then P is contained in Z U p (G). For the case that e = 2, we should give conditions to all subgroups of order p 2 . Theorem 1.6 [10, Theorem 6] Let P be a normal p-subgroup of a group G with |P| > p 2 . Assume that H ∩ O p (G) is S-semipermutable in G for all subgroups H of P with |H | = p 2 . Then P is contained in Z U p (G). The goal of this paper is to apply G A p−1 to extend the above results. By the property of S-semipermutability, we know that H ∩ N is S-semipermutable in G if H is a S-semipermutable p-subgroup of G and N is a normal subgroup of G. Hence our following results, Theorems 1.7–1.10, are generalizations of Theorems 1.3–1.6, respectively. Theorem 1.7 Let P ∈ Syl p (G) and let d be a power of p such that 1 ≤ d < |P|. Assume that H ∩ O p (G A p−1 ) is S-semipermutable in G for all subgroups H  P with |H | = d. Then G is p-supersolvable, or else |P ∩ O p (G A p−1 )| > d. Theorem 1.8 Let P ∈ Syl p (G) and let d be a power of p such that 1 ≤ d < |P|. Assume that H ∩ O p (G A p−1 ) is S-semipermutable in G for all noncyclic subgroups H of P with |H | = d. Then |P ∩ O p (G A p−1 )| > d, P ∩ O p (G A p−1 ) is cyclic, or else G is p-supersolvable. Theorem 1.9 Fix an integer e ≥ 3, and let P be a normal p-subgroup of a group G with |P| > p e = d. Assume that H ∩ O p (G A p−1 ) is S-semipermutable in G for all noncyclic subgroups H of P with |H | = d. Then P is contained in Z U p (G). Theorem 1.10 Let P be a normal p-subgroup of a group G with |P| > p 2 . Assume that H ∩ O p (G A p−1 ) is S-semipermutable in G for all subgroups H of P with |H | = p 2 . Then P is contained in Z U p (G). Based on above results, we have the following corollaries which are the generalizations of many recent results, especially those in [9]. Corollary 1.11 Let P ∈ Syl p (G) and let d be a power of p such that 1 < d < |P|. Assume that H ∩ O p (G A p−1 ) is S-semipermutable in G for all subgroups H of P with |H | = d and all cyclic subgroups H of order 4 (if p = 2 and P is not abelian). Then G is p-supersolvable. Corollary 1.12 Let P be a normal p-subgroup of a group G and d be a power of p such that 1 < d < |P|. Assume that H ∩ O p (G A p−1 ) is S-semipermutable in G for all subgroups H of P with |H | = d and all cyclic subgroups H of P of order 4 (if P is a nonabelian 2-group and d = 2). Then P ≤ Z U p (G).

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2 Preliminaries We first collect some properties of S-permutable subgroup of a group. Lemma 2.1 [9, Lemma 2.1] Let G be a group. (1) (2) (3) (4)

An S-permutable subgroup of G is subnormal in G. If H ≤ K ≤ G and H is S-permutable in G, then H is S-permutable in K . If H is S-permutable Hall subgroup of G, then H  G. Let K  G and K ≤ H . Then H is S-permutable in G if and only if H/K is S-permutable in G/K . (5) If H , K are S-permutable in G, then H ∩ K is also S-permutable in G. (6) Suppose that P is a p-subgroup of G for some prime p. Then P is S-permutable  in G if and only if N G (P) ≥ O p (G). The following are some properties of S-semipermutable subgroup of a group. Lemma 2.2 [9, Lemma 2.2] Let G be a group. Suppose that H is an S-semipermutable subgroup of G. Then (1) If H ≤ K ≤ G, then H is S-semipermutable in K . (2) Let N be a normal subgroup of G. If H is a p-group for some prime p ∈ π(G), then H N /N is S-semipermutable in G/N ; (3) If H ≤ O p (G), then H is S-permutable in G. (4) Suppose that H is a p-subgroup of G for some prime p ∈ π(G) and N is normal in G. Then H ∩ N is also an S-semipermutable subgroup of G.  The following is critical to our results, which tells some elementary facts about the influence of the properties of G A p−1 on the structure of G. By definition, G/G A p−1 is abelian of exponent dividing p − 1. Proposition 2.3 Let G be a finite group and p a prime in π(G). Then (1) Suppose that N is a normal subgroup of G. Then (G/N )A p−1 = G A p−1 N /N , O p (G/N ) = O p (G)N /N and O p ((G/N )A p−1 ) = O p (G A p−1 )N /N . (2) G is p-supersolvable if and only if G A p−1 is p-nilpotent. (3) G/O p (G A p−1 ) is p-supersolvable. (4) G is p-supersolvable if and only if O p (G A p−1 ) ≤ Z U p (G). (5) Let P be a normal p-subgroup of a group G. Then P ≤ Z U p (G) if and only if P ∩ O p (G A p−1 ) ≤ Z U p (G). Proof Let A = G A p−1 and U = O p (A). (1) It follows from the definitions. (2) Assume that A is p-nilpotent. By induction, we can assume that A is a p-group. For any G-chief factor C/D below A. Then A ≤ C G (C/D). Since G/A is abelian of exponent dividing p − 1, G/C G (C/D) is abelian of exponent dividing p − 1. Since C/D is an irreducible faithful G/C G (C/D)-module, C/D is of order p [5, Chapter B]. This implies that G is p-supersolvable.

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Conversely, assume that G is p-supersolvable. Then G/C G (C/D) is abelian of exponent dividing p−1 for any p-G-chief factor C/D. Since F p (G) = O p , p (G) is the intersection of centralizers in G of all p-G-chief factors, we have G/F p (G) is abelian of exponent dividing p − 1. Hence A ≤ F p (G). Therefore, A is pnilpotent, as desired. (3) By definition, we know that G/A is abelian of exponent dividing p − 1. By (1), we know that (G/U )A p−1 = A/U is a p-group. Hence G/U is psupersolvable by (2). (4) If U ≤ Z U p (G), then G/U is p-supersolvable by (3). Hence we have G/U = Z U p (G/U ) = Z U p (G)/U. Then G = Z U p (G) and so G is p-supersolvable. The converse is obvious. (5) Since PU/U ≤ G/U and G/U is p-supersolvable by (3), using the Gisomorphism P/P ∩U ∼ = PU/U , we conclude that P/P ∩U ≤ Z U p (G/P ∩U ). If P ∩ U ≤ Z U p (G), then P ≤ Z U p (G). The converse is obvious.  Lemma 2.4 [11, Lemma 2.8] Let P be a normal p-subgroup of G and let d be a power of p such that 1 < d < |P|. Suppose all subgroups H of P with order d and all cyclic subgroups of P of order 4 (if P is a nonabelian 2-group and d = 2) are S-permutable in G. Then P is contained in Z U p (G).

3 Proofs of Theorems 1.7 and 1.8 Proof of Theorem 1.7 Assume that the result is false and G is a counterexample with minimal order. Put A = G A p−1 and U = O p (A). We have that G is not psupersolvable and |P ∩ U | ≤ d. We also mention that P is a Sylow p-subgroup of A by the definition of A. (1) O p (G) = 1. Write V = O p (G) and assume that V = 1. Consider the factor group G = G/V . Let H be a normal subgroup of P of order d. Then there exists a normal subgroup H of P with order d such that H = H V /V . Since H ∩ U is S-semipermutable in G, we have H ∩ U = (H ∩ U )V /V is S-semipermutable in G. Since |U ∩ P| ≤ d, it follows that G is p-supersoluble by the minimal choice of G. Hence G is p-supersoluble and this is a contradiction. Thus V = 1, as required. (2) P is normal in G and P ∩ U = 1. For any subgroups H  P with |H | = d, we have H ∩ U is S-semipermutable in G by hypothesis. Hence H ∩ U is S-semipermutable in A by Lemma 2.2. Applying Theorem 1.3, we have A is p-supersolvable. By (1) and p-length of a p-supersolvable group is less than 2, we have that P is normal in A, then P is normal in G. If P ∩ U = 1, then P ≤ Z U p (G) by Proposition 2.3(5). Therefore, G is p-supersolvable, a contradiction. Hence (2) holds.

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(3) For any subgroup H  P with |H | = d, we have H ∩ U is normal in G. Furthermore, Y ∩ U is normal in G for any maximal subgroup Y of P. By hypothesis, we know that H ∩ U is S-semipermutable in G for any subgroups H  P with |H | = d. Since P is normal in G by (2), we have H ∩ U is S-permutable in G by Lemma 2.2. Hence H ∩ U is normalized by O p (G) by Lemma 2.1. Since H ∩ U is normal in P. Therefore, H ∩ U is normal in G. Since |Y ∩ U | ≤ |P ∩ U | ≤ d and Y ∩ U is normal in P for any maximal subgroup Y of P, we can pick a subgroup H  P with |H | = d such that Y ∩ U ≤ H ≤ Y . Then Y ∩ U = H ∩ U . Hence Y ∩ U is normal in G. (4) d > p. If d = p, then |P ∩ U | = 1 or p. Hence P ∩ U ≤ Z U p (G). Then P ≤ Z U p (G) by Proposition 2.3 (5). Therefore, G is p-supersolvable, a contradiction. (5) Pick a minimal normal subgroup T of G contained in P ∩ U . If |T | < d, then G/T is p-supersolvable. We first argue that G/T satisfies the hypotheses of the theorem. Clearly, 1 ≤ d/|T | < |P/T |. Let H/T be a normal subgroup of P/T of order d/|T |. Then H is a normal subgroup of P with order d. It follows that H/T ∩O p (A/T ) = H/T ∩U/T = (H ∩U )/T , which is S-semipermutable in G/T by Lemma 2.2. This shows that G/T satisfies the hypotheses of the theorem, as claimed. Since |P/T ∩ O p (A/T )| = |(P ∩ U )/T | ≤ d/|T |, it follows that G/T is psupersolvable by the minimality of G. (6) T ≤ (P). If T  (P), then there exists a maximal subgroup Y of P such that P = T Y . Then T ∩ Y is a maximal subgroup of T . Since T ∩ Y = T ∩ Y ∩U , T ∩ Y is normal in G by (3). Hence T ∩ Y = 1 or T ∩ Y = T by the minimality of T . If T ∩ Y = T , then T ≤ Y and P = T Y = Y , a contradiction. Hence T ∩ Y = 1, and then, T is of order p. So |T | < d by (4). By (5), G/T is p-supersolvable, and then, G is p-supersolvable, a contradiction. (7) The final contradiction. By (2) we know that |T | ≤ d. If |T | = d, then T = P ∩ U . Noticing that A/T is p-nilpotent, we have A is p-nilpotent by [7, VI, Beispiele 7.6] and (6). Then G is p-supersolvable by Proposition 2.3(2), a contradiction. If |T | < d, then G/T is p-supersolvable by (5). By [7, VI, Satz 8.6] and (6), we have G is p-supersolvable, the final contradiction. These complete the proof of the theorem.  Proof of Theorem 1.8 Assume that the result is false and let G be a counterexample of least order. Put A = G A p−1 and U = O p (A). Then |P ∩ U | ≤ d, P ∩ U is not cyclic and G is not p-supersolvable. Assume that P ∩ U = 1. Then A is p-nilpotent, and it follows that G is p-supersolvable by Proposition 2.3, a contradiction. Hence P ∩ U = 1. Write H = {H is a noncyclic subgroup of P | |H | = d}. By hypothesis, H ∩ U is S-semipermutable in G for each H ∈ H. (1) O p (G) = 1. Write V = O p (G) and assume that V = 1. Consider the factor group G = G/V . Let H be a noncyclic subgroup of P of order d. Then there exists a H ∈ H such that H = H V /V . Since H ∩U is S-semipermutable in G, we have H ∩U = (H ∩U )V /V

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is S-semipermutable in G. Since |U ∩ P| ≤ d, it follows that G is p-supersolvable by the minimal choice of G. Hence G is p-supersolvable and this is a contradiction. Thus V = 1, as required. (2) P is normal in G; then, G is p-solvable. By the hypothesis of the theorem, for every H ∈ H, H ∩ U is S-semipermutable in G. Therefore, we have that H ∩ U is S-semipermutable in A by Lemma 2.2. By Theorem 1.4, we obtain that A is p-supersolvable. By (1) and [1, Lemma 2.1.6], we get that P is normal in A, and thus, P is normal in G. Hence G is p-solvable. (3) Let T be a minimal normal subgroup of G contained in U . Then |T | ≤ d. By (1) and (2), T ≤ P ∩ U . Hence (3) holds. (4) G/T is not p-supersolvable. Suppose that G/T is p-supersolvable. Since the class of all p-supersolvable groups is a saturated formation by [7, Kapitel VI, Satz 8.6], we may suppose that T  (G). Let M be a maximal subgroup of G such that T  M. Then we get G = T M and T ∩ M = 1. and P = T (P ∩ M). We take a normal subgroup C of P, which is also a maximal subgroup of T . Then we have a subgroup H = C D of P such that |H | = d, where D is a subgroup of P ∩ M. Obviously, H is not cyclic; then, H ∈ H and H ∩ U is S-semipermutable in G by the hypothesis of the theorem. By Lemma 2.2(3), we have that H ∩ T = H ∩ U ∩ T is S-permutable in G; hence, it was normalized by O p (G) by Lemma 2.1. Then we have that C = C(D ∩ T ) = H ∩ T is normalized by O p (G). As C is normal in P, we have that C is normal in G and C = 1 by the minimal normality of T . Then T is of order p. Then G is p-supersolvable, a contradiction. (5) |T | < d. Suppose that |T | = d. Then T = P ∩ U . Noticing that A/T is p-nilpotent, G/T is p-supersolvable by Proposition 2.3(2), contrary to (4). (6) Final contradiction. Clearly, 1 < d/|T | < |P/T |. We argue that G/T satisfies the hypothesis of the theorem. Let H/T be a noncyclic subgroup of P/T of order d/|T |, then H ∈ H. It follows that H/T ∩ O p ((G/T )A p−1 ) = H/T ∩ U/T = (H ∩ U )/T , which is Ssemipermutable in G/T by Lemma 2.2. This shows that G/T satisfies the hypothesis of the theorem, as claimed. If (P ∩ U )/T is cyclic, then P/T ∩ O p ((G/T )A p−1 ) ≤ Z U p (G/T ). Then we have that P/T ≤ Z U p (G/T ) by Proposition 2.3(5). It follows that G/T is p-supersolvable, contrary to (4). So we have that (P ∩ U )/T is not cyclic and since |P/T ∩ U/T | = |(P ∩ U )/T | ≤ d/|T |, it follows that G/T is p-supersolvable by the minimality of G, the final contradiction. This final contradiction completes the proof. 

4 Proofs of Theorems 1.9 and 1.10 Proof of Theorem 1.9 Assume that the result is false and take a counterexample (G, P) with the least order of |G| + |P|. Put P0 ∈ Syl p (G), A = G A p−1 , U = O p (A) and N = P ∩ U . In fact, since PU/U ≤ G/U and G/U is p-supersolvable by Proposition 2.3. Then by using the G-isomorphism P/N ∼ = PU/U , we conclude that P/N ≤ Z U p (G/N ). If |N | > d, we can obtain that N ≤ Z U p (G) by Theorem 1.1∗ . Then we get that P ≤ Z U p (G) by Proposition 2.3, a contradiction. So we

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have that |N | ≤ d, N is not cyclic and N  Z U p (G). In particular, N = 1. Write H = {H ≤ P | H is noncyclic in P and |H | = d}. Since P is not cyclic, we have that H is not empty by [3, Lemma 2.3]. By hypothesis, H ∩ U is S-semipermutable in G for each H ∈ H. (1) Let T be a minimal normal subgroup of G contained in N . Then P/T ≤ Z U p (G/T ). By the choice of T , we have |T | ≤ |N | = d. If N /T is cyclic, then N /T ≤ Z U p (G/T ) and P/T ≤ Z U p (G/T ) by Proposition 2.3(5), as desired. Therefore, we may assume that N /T is noncyclic. If d/|T | ≤ p, then N /T will be cyclic, a contradiction. Hence we have d/|T | ≥ p 2 . First assume that d/|T | = p 2 . Then we have that |N /T | = p 2 and N /T  Z U p (G/T ). Hence we obtain that N /T is a minimal normal subgroup of G/T . As P/N ≤ Z U p (G/N ), we pick a normal subgroup S of G contained in P such that |S/N | = p. Therefore, |S/T | = p 3 . Since S/T is neither elementary abelian of order p 2 nor isomorphic to Q 8 , if S/T has a cyclic maximal subgroup, then by [3, Lemma 2.1], we can have that N /T contains a normal subgroup of G/T of order p, so |N /T | = p, a contradiction. So every maximal subgroup of S/T is noncyclic. Then we get that there exists another noncyclic maximal subgroup M/T = N /T of S/T with |M/T | = p 2 . Obviously, N /T ∩ Z (P0 /T ) = 1. We take a subgroup C/T = aT  of N /T with order p, where aT ∈ N /T ∩ Z (P0 /T ). Then we have a subgroup H/T = C/T · D/T of P/T with order p 2 , where D/T is a subgroup of M/T with order p and D/T  N /T . Obviously, H/T is not cyclic, H ∈ H and (H ∩ U )/T = H/T ∩ U/T is S-semipermutable in G/T by the hypothesis of the theorem. Since N /T ∩ D/T = 1, we have that C/T = H/T ∩ N /T = H/T ∩ N /T ∩ U/T = H/T ∩ U/T = (H ∩ U )/T is S-semipermutable in G/T . As C/T ≤ N /T ≤ O p (G/T ), we have that C/T is S-permutable in G/T by Lemma 2.2. Then C/T is normalized by O p (G/T ) and we get that C/T  G/T since C/T ≤ Z (P0 /T ), contradicting the minimal normality of N /T . So our assumption is incorrect, that is, P/T ≤ Z U p (G/T ). Then we have that p 3 ≤ d/|T | < |P/T |. We argue that (G/T, P/T ) satisfies the hypothesis of the theorem. Let H/T be a noncyclic subgroup of P/T of order d/|T |, then H ∈ H. It follows that H/T ∩ O p (G A p−1 /T ) = H/T ∩ U/T = (H ∩U )/T , which is S-semipermutable in G/T . This shows that (G/T, P/T ) satisfies the hypothesis of the theorem, as claimed. It follows that P/T ≤ Z U p (G/T ) by the choice of (G, P). (2) N ∩ (P) = 1. Suppose that N ∩ (P) = 1. Let T be a minimal normal subgroup of G contained in N . Obviously, T is elementary abelian and T ∩ (P) = 1. By [5, Chapter A, Theorem 9.2(f)], we have that P = T M and T ∩ M = 1. We take a normal subgroup T1 of P0 , which is also a maximal subgroup of T . Then we have a subgroup H = T1 B of P such that |H | = d, where B is a subgroup of M. Obviously, H is not cyclic, then H ∈ H and H ∩ U is S-semipermutable in G by the hypothesis of the theorem. By Lemma 2.1, we have that H ∩ T = H ∩ U ∩ T is normalized by O p (G). Since T ∩ M = 1, we obtain that T1 = T1 (B ∩ T ) = H ∩ T is normalized by O p (G). As

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Some Criteria for p-Supersolvability of a Finite Group

T1 is normal in P0 , we get that T1 is normal in G. Then T1 = 1 and T is of order p. So P ≤ Z U p (G) by (1), this contradiction shows that N ∩ (P) = 1. (3) The final contradiction. (1) and (2) yield P/(P) ≤ Z U p (G/(P)); then, by [5, IV, Theorem 6.7], we have that P ≤ Z U p (G), a contradiction. The proof of the theorem is complete.  Proofs of Theorem 1.10 Assume that the result is false and take a counterexample (G, P) with the least order of |G| + |P|. Put A = G A p−1 and U = O p (A). Since PU/U ≤ G/U and G/U is p-supersolvable, then by using the Gisomorphism P/P ∩ U ∼ = PU/U . Then we have that P/P ∩ U ≤ Z U p (G/P ∩ U ). If |P ∩ U | > p 2 , we can obtain that P ∩ U ≤ Z U p (G) by Lemma 2.4. Then we get that P ≤ Z U p (G), a contradiction. So we have that |P ∩ U | ≤ p 2 . Assume that P ∩ U  Z U p (G); then, we have that |P ∩ U | = p 2 , P ∩ U is not cyclic and P ∩ U is a minimal normal subgroup of G. As P/P ∩ U ≤ Z U p (G/P ∩ U ), we pick a normal subgroup S of a Sylow p-subgroup P0 of G contained in P such that |S/P ∩ U | = p. We can write S = (P ∩ U ) a, where a ∈ / P ∩ U . If (S) = P ∩ U , then S = a is cyclic. So is P ∩ U . Therefore, |P ∩ U | = p, a contradiction. So we assume that (S) < P ∩ U . Since S is normal in P0 , then (S) is normal in P0 . We can pick a maximal subgroup M of P ∩ U containing (S) such that M is normal in P0 . Denote K = M a. Since a p ∈ (S) ≤ M, we have that |K | = p 2 . By hypothesis, K ∩ U is S-semipermutable in G. Then K ∩ U is S-permutable in G by Lemma 2.2. Applying Lemma 2.1, M = K ∩ U ∩ P is normalized by O p (G). Therefore, we get that M is normal in G, contradicting the minimal normality of P ∩ U . So our assumption is  incorrect, that is, P ≤ Z U p (G). This contradiction completes the proof.

5 Proofs of Corollaries 1.11 and 1.12 Lemma 5.1 Let P ∈ Syl p (G). Assume that H ∩ O p (G A p−1 ) is S-semipermutable in G for all subgroups H of P with H cyclic of order p or 4 (if p = 2 and P is not abelian). Then G is p-supersolvable. Proofs of Theorem 1.10 Assume that G is a counterexample of least order. We can assume that O p (G) = 1. Put A = G A p−1 and U = O p (A). By the hypotheses, we have H is S-semipermutable in G for all cyclic subgroups H of U of order p or 4 (if p = 2 and P is not abelian). Applying [9, Theorem 3.3], we have that U is p-supersolvable. Therefore, P ∩ U is normal in U and thus is normal in G. Hence by Lemma 2.2, H is S-permutable in G for all cyclic subgroups H of P ∩ U of order p or 4 (if p = 2 and P is not abelian). By Lemma 2.4, P ∩ U ≤ Z U p (G). Then P ≤ Z U p (G) by Proposition 2.3. This implies that G is p-supersolvable, as desired.  Proof of Corollary 1.11 By induction, we can assume that O p (G) = 1. Put A = G A p−1 and U = O p (A). By Lemma 5.1, we can assume that d > p. Since we can regard P as a Sylow p-subgroup of A, we have A is p-supersolvable by [2, Theorem 5]. Since O p (A) ≤ O p (G) = 1, we have P is normal in A, and then P is normal

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in G. Applying Theorems 1.9 and 1.10, we conclude that P ≤ Z U p (G). Then G is p-supersolvable, as desired.  Proof of Corollary 1.12 By Theorems 1.9 and 1.10, we only need to consider the case that d = p. Then every cyclic subgroup of P ∩ O p (G A p−1 ) of order p and order 4 (if P is a nonabelian 2-group) is S-semipermutable in G. Then every cyclic subgroup of P ∩ O p (G A p−1 ) of order p and order 4 (if P is a nonabelian 2-group) is S-permutable in G by Lemma 2.2. Now applying Lemma 2.4, we have P ∩ O p (G A p−1 ) ≤ Z U p (G).  Hence P ≤ Z U p (G) by Proposition 2.3.

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