Ann Univ Ferrara https://doi.org/10.1007/s11565-018-0299-1

On SSH-subgroups of finite groups T. M. Al-Gafri1

· S. K. Nauman1

Received: 7 December 2017 / Accepted: 25 January 2018 © Università degli Studi di Ferrara 2018

Abstract Let G be a finite group. A subgroup H of G is s-permutable in G if H permutes with every Sylow subgroup of G. A subgroup H of G is called an SSHsubgroup in G if G has an s-permutable subgroup K such that H sG = H K and H g ∩ N K (H )  H , for all g ∈ G, where H sG is the intersection of all s-permutable subgroups of G containing H . We study the structure of finite groups under the assumption that the maximal or the minimal subgroups of Sylow subgroups of some normal subgroups of G are SSH-subgroups in G. Several recent results from the literature are improved and generalized. Keywords s-permutable subgroups · H-subgroups · Weakly HC-embedded subgroups · SSH-subgroups · p-nilpotent groups · Supersolvable groups Mathematics Subject Classification 20D15 · 20D20 · 20F16

1 Introduction In this article, all the groups are assumed to be finite. The terminology and notation employed agree with standard usage as in Doerk and Hawkes [14]. Let H be a subgroup of a group G. We write HG for the normal core of H in G and H G for the normal closure

B

T. M. Al-Gafri [email protected] S. K. Nauman [email protected]

1

Department of Mathematics, Faculty of Science 80203, King Abdulaziz University, Jeddah 21589, Saudi Arabia

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of H in G. We say that H permutes with another subgroup K of G if H K = K H , that is, H K  G. Let H be a subgroup of a group G. Recall that H is said to be permutable in G if H permutes with every subgroup of G. We say that H is s-permutable in G if H permutes with every Sylow subgroup of G. This concept was introduced by Kegel [19] in 1962. Wang [26], in 1996, developed the concept of c-normality. We say that H is c-normal in G if there exists K  G such that G = H K and H ∩ K  HG . Another subgroup embedding property was defined by Bianchi et al. [9] in 2000. We say that H is an H-subgroup in G if H g ∩ N G (H )  H , for all g ∈ G. Asaad et al., in 2012, unified the concepts of c-normality and H-subgroup within [4]. Following them, we say that H is a weakly H-subgroup in G if G = H K for some K  G and H ∩ K is an H-subgroup in G. The previous concept was extended by Asaad and Ramadan [6] in 2016 by the notion of weakly H-embedded subgroups. The subgroup H is said to be weakly H-embedded in G if there exists K  G such that H G = H K and H ∩ K is an H-subgroup in G. In 2012, Wei and Guo [27] introduced the concept of HC-subgroups. The subgroup H is said to be an HC-subgroup in G if G = H K for some K  G and H g ∩ N K (H )  H , for all g ∈ G. Recently, in 2016, Asaad and Ramadan [7] generalized the concept of HC-subgroups by introducing the concept of weakly HC-embedded subgroups. We say that H is weakly HC-embedded in G if there exists K  G such that H G = H K and H g ∩ N K (H )  H , for all g ∈ G. We like to stress the fact that a weakly H-embedded subgroup is a weakly HC-embedded subgroup. To see this, assume that the former holds for H . Then H G = H K and (H ∩ K )g ∩ N G (H ∩ K )  H ∩ K , for all g ∈ G, where K  G. As H and N G (H ) ∩ K are normal subgroups of N G (H ), it follows that H ∩ K  N G (H ), hence N G (H )  N G (H ∩ K ). We observe that H g ∩ N K (H ) = H g ∩ K ∩ N G (H ) = (H ∩ K )g ∩ N G (H )  (H ∩ K )g ∩ N G (H ∩ K )  H ∩ K  H , for all g ∈ G. Therefore H is weakly HC-embedded in G. In this article, we introduce a new subgroup embedding property that extends all the previously mentioned concepts as follows: Definition 1.1 A subgroup H of a group G is called an SSH-subgroup in G if G has an s-permutable subgroup K such that H sG = H K and H g ∩ N K (H )  H , for all g ∈ G, where H sG is the intersection of all s-permutable subgroups of G containing H. Let G be a group and H  G. If H is s-permutable in G, then H is an SSHsubgroup in G because H sG = H 1 and H g ∩ N1 (H ) = 1  H , for all g ∈ G. In what follows, some lemmas from Sect. 2 will be used. Assume that H is weakly HC-embedded in G. Then there exists T  G such that H G = H T and H g ∩ N T (H )  H , for all g ∈ G. Note that H sG is s-permutable in G and H sG  H G by Lemma 2.2(a). So, H sG = H sG ∩ H T = H (H sG ∩ T ) = H K , where K := H sG ∩ T . Moreover, K is s-permutable in G by Lemma 2.1(d). Clearly, H g ∩ N K (H ) = H g ∩ N G (H ) ∩ T ∩ H sG = H g ∩ N T (H ) ∩ H sG  H ∩ H sG = H , for all g ∈ G. Thus H is an SSH-subgroup in G. The next three examples show that there is no relationship between the concepts of s-permutability and H-subgroups.

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Example 1.2 Hall subgroups of any group G are H-subgroups by [9, Proposition 3(1)]. We know that s-permutable subgroups are subnormal by Lemma 2.1(c). Since subnormal Hall subgroups are normal, then there exist some Hall subgroups which are not s-permutable subgroups. Example 1.3 We observe that self-normalizing subgroups are H-subgroups. Let M be a maximal subgroup of a group G such that M  G. Then N G (M) = M and so M is an H-subgroup in G. If M is s-permutable in G, then, by Lemma 2.1(c), M  G which is not true. Thus M is not s-permutable in G. Example 1.4 Let G be a nilpotent group. Then every subgroup of G is s-permutable in G. Let H be any subgroup of G such that H  G. Certainly, H is subnormal in G. If H is an H-subgroup in G, then H  G by Lemma 2.7 which contradicts our assumption. Therefore H is not an H-subgroup in G. The subgroups in Examples 1.2 and 1.3 are H-subgroups and so they are SSHsubgroups. Hence there are SSH-subgroups which are not s-permutable subgroups. The following example shows that there exist SSH-subgroups which are not weakly HC-embedded subgroups. Example 1.5 Let G = ((C8  C2 )  C2 )  C2 , where G is the SmallGroup(64,32) in GAP. Then G = a, b, c, d, e, f : a 2 = d, [b, a] = c, [c, a] = e, [e, a] = f, [d, b] = e, [d, c] = f, b2 = c2 = d 2 = e2 = f 2 = [d, a] = [ f, a] = [c, b] = [e, b] = [ f, b] = [e, c] = [ f, c] = [e, d] = [ f, d] = [ f, e] = 1. We set H = b, e. It is obvious that |G| = 64 and |H | = 4. Since H is s-permutable in G, then H is an SSH-subgroup in G. Now we show that H is not weakly HC-embedded in G. There are two distinct conjugate subgroups of H in G which are H itself and C := bc, e f . We know that H ∩ C = 1 and H G = b, c, e, f  is an abelian group of order 16. The normal subgroups of G which are contained in H G are L 1 = H G , L 2 = c, e, f , L 3 = e, f , L 4 =  f  and L 5 = 1. Because H L i  H G , for i = 3, 4, 5, as |H L i | < |H G |, we can exclude L 3 , L 4 and L 5 . For L 1 , we have that H G = H L 1 , but C ∩ N L 1 (H ) = C ∩ L 1 = C  H . Similarly, for L 2 , H G = H L 2 , however, C ∩ N L 2 (H ) = C ∩ L 2 = e f   H . Thus H is not weakly HC-embedded in G. Many authors have investigated the structure of finite groups by using the concepts that were mentioned earlier in the introduction. See Sect. 4 for some examples of results that were obtained by applying them. The aim of this article is to study the structure of finite groups under the assumption that certain subgroups of prime power orders are SSH-subgroups. Several results from the literature are improved and generalized.

2 Preliminaries Lemma 2.1 Let H, K and L be subgroups of a group G such that H is s-permutable in G and L  G. Then the following statements hold: (a) H L/L is s-permutable in G/L. (b) If L  K , then K is s-permutable in G if and only if K /L is s-permutable in G/L.

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(c) (d) (e) (f) (g) (h) (i)

H is subnormal in G. If K is also s-permutable in G, then H ∩ K is s-permutable in G. If K is also s-permutable in G, then H, K  is s-permutable in G. H ∩ K is s-permutable in K . H/HG is nilpotent. If H is a p-subgroup of G, for some prime p, then O p (G)  N G (H ). If x ∈ G, then H x is s-permutable in G.

Proof For (a)–(d), (f)–(g) and (h), see [11,19] and [20, Lemma 2.2], respectively. (e) Let P be a Sylow p-subgroup of G. Then H P = P H and K P = P K . By [14, Lemma 1.6(a), p. 3], H, K P = PH, K , hence H, K  is s-permutable in G. (i) Let Syl(G) be the set of all Sylow subgroups of G. Then H P = P H , for all P ∈ Syl(G). So, H x P x = P x H x , for all P x ∈ (Syl(G))x . Because (Syl(G))x =   Syl(G), we deduce that H x is s-permutable in G. Lemma 2.2 Let G be a group and H  K  G. Then the following hold: (a) (b) (c) (d) (e)

H sG is s-permutable in G and H sG  H G . H s K  H sG . If H  G, then (K /H )s(G/H ) = K sG /H . H sG  K sG . If x ∈ G, then (H sG )x = (H x )sG .

Proof (a)–(c) See [17, Lemma 2.5(1), (2) and (3)]. (d) By part (a), K sG is s-permutable in G. Since H  K  K sG , it follows, by the definition of H sG , that H sG  K sG . (e) Certainly, H x  (H sG )x as H  H sG . Since H sG is s-permutable in G by part (a), then (H sG )x is s-permutable in G by Lemma 2.1(i). Therefore (H x )sG  −1 −1 (H sG )x . Conversely, H x  (H x )sG , hence H = (H x )x  ((H x )sG )x . Note −1 that ((H x )sG )x is s-permutable in G by part (a) and Lemma 2.1(i). Consequently, −1   H sG  ((H x )sG )x and so (H sG )x  (H x )sG . Thus (H sG )x = (H x )sG . Lemma 2.3 Let H and K be subgroups of a group G. Then the following hold: (a) If H is a p-subgroup of G and K is a normal p  -subgroup of G, for some prime p, then N G (H K ) = N G (H )K . (b) If x ∈ G, then (N K (H ))x = N K x (H x ). Proof (a) If x ∈ N G (H )K , then x = yk, where y ∈ N G (H ) and k ∈ K . Hence x H K = y H K = H y K = H x K = H K x and it follows that x ∈ N G (H K ). Conversely, let x ∈ N G (H K ). Then H K = H x K . Clearly, H and H x are Sylow psubgroups of H K , so there exists y ∈ H K such that H x y = (H x ) y = H . We can write y = kh ∈ K H = H K , where k ∈ K and h ∈ H . We see that xkh = x y ∈ N G (H ) which implies that xk ∈ N G (H ). Therefore x ∈ N G (H )k −1 ⊆ N G (H )K . −1 −1 (b) Consider y ∈ (N K (H ))x . Then y x ∈ N K (H ) and hence y x ∈ K with −1 −1 −1 −1 y x H = H y x . This implies that y ∈ K x and y H x = (y x H )x = (H y x )x = H x y. Therefore y ∈ N K x (H x ). Similar arguments hold for the other inclusion. Con  sequently, we have that (N K (H ))x = N K x (H x ).

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Lemma 2.4 Let H , M and L be subgroups of a group G such that H is an SSHsubgroup in G and L  G. Then the following statements hold: (a) If H  M, then H is an SSH-subgroup in M. (b) Assume that L  M. Then M is an SSH-subgroup in G if and only if M/L is an SSH-subgroup in G/L. (c) Assume that H is a p-subgroup of G and L is a p  -subgroup of G, for some prime p. Then H L and H L/L are SSH-subgroups in G and G/L, respectively. (d) If x ∈ G, then H x is an SSH-subgroup in G. Proof (a) Since H is an SSH-subgroup in G, then G has an s-permutable subgroup K such that H sG = H K and H g ∩ N K (H )  H , for all g ∈ G. From Lemma 2.2(b), H s M  H sG . Hence H s M = H s M ∩ H K = H (H s M ∩ K ) = H T , where T := H s M ∩ K . Note that H s M and K ∩ M are s-permutable in M by Lemmas 2.2(a) and 2.1(f). So, T = H s M ∩(K ∩M) is s-permutable in M by Lemma 2.1(d). For all m ∈ M, we have that H m ∩ N T (H ) = H m ∩ N G (H ) ∩ H s M ∩ K = H m ∩ N K (H ) ∩ H s M  H ∩ H s M = H . Thus H is an SSH-subgroup in M. (b) Suppose that M is an SSH-subgroup in G. Then there exists K  G such that K is s-permutable in G, M sG = M K and M g ∩ N K (M)  M, for all g ∈ G. In view of Lemma 2.2(c), (M/L)s(G/L) = M sG /L = (M/L)(K L/L). By Lemma 2.1(a), K L/L is s-permutable in G/L. We realize that (M/L)gL ∩ N K L/L (M/L) = (M g /L) ∩ (N G (M)/L) ∩ (K L/L) = (M g /L) ∩ (N G (M) ∩ K L)/L = (M g /L) ∩ (N G (M) ∩ K )L/L = (M g ∩ N K (M)L)/L = (M g ∩ N K (M))L/L  M/L, for all gL ∈ G/L. Hence M/L is an SSH-subgroup in G/L. Conversely, there exists, by hypothesis, an s-permutable subgroup K /L of G/L such that (M/L)s(G/L) = (M/L)(K /L) and (M/L)gL ∩ N K /L (M/L)  M/L, for all gL ∈ G/L. By Lemma 2.2(c), (M/L)s(G/L) = M sG /L and accordingly M sG = M K . Also, K is s-permutable in G by Lemma 2.1(b). Obviously, we have that (M/L)gL ∩ N K /L (M/L) = (M g /L) ∩ (N G (M)/L) ∩ (K /L) = (M g ∩ N K (M))/L, for all gL ∈ G/L. This implies that (M g ∩ N K (M))/L  M/L, for all g ∈ G. Hence M g ∩ N K (M)  M, for all g ∈ G. Thus M is an SSH-subgroup in G. (c) We first prove that H L is an SSH-subgroup in G. By hypothesis, G has an spermutable subgroup K such that H sG = H K and H g ∩ N K (H )  H , for all g ∈ G. Let P be a Sylow p-subgroup of G with H  P. Since K is, by Lemma 2.1(c), subnormal in G, then K L is subnormal in G, therefore P x ∩ K and P x ∩ K L are Sylow p-subgroups of K and K L, respectively, for all x ∈ G. As [K L : P x ∩ K ] is a p  -number, then P x ∩K is a Sylow p-subgroup of K L and thereby P x ∩K = P x ∩K L. This implies that H x ∩ K = H x ∩ K L because H x  P x , for all x ∈ G. Note that H sG  (H L)sG by Lemma 2.2(d) and so H sG L  (H L)sG . By Lemmas 2.2(a) and 2.1(e), H sG L is s-permutable in G, hence (H L)sG = H sG L. Clearly, (H L)sG = (H L)(K L) and, by Lemma 2.1(e), K L is s-permutable in G. By Lemma 2.3(a), N G (H L) = N G (H )L. Let g ∈ G. Then (H L)g ∩ N K L (H L) = H g L ∩ N G (H )L ∩ K L = H g L ∩ (N G (H ) ∩ K L)L = H g L ∩ N K L (H )L = (H g ∩ N K L (H )L)L. Let P ∗ be a Sylow p-subgroup of N K L (H ). Then P ∗ is a Sylow p-subgroup of N K L (H )L. Because H g ∩ N K L (H )L is a p-subgroup of N K L (H )L, there exists y ∈ N K L (H )L such that (H g ∩ N K L (H )L) y  P ∗  N K L (H ). From the previous arguments, H gy ∩ K = H gy ∩ K L. We know that (H g ∩ N K L (H )L) y = H gy ∩

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N K L (H )L  H gy ∩ N K L (H ) = H gy ∩ K L ∩ N G (H ) = H gy ∩ K ∩ N G (H ) = H gy ∩ N K (H )  H . Since y ∈ L N K L (H ), then y = uv, where u ∈ L and v ∈ N K L (H ). −1 −1 −1 As (H g ∩ N K L (H )L) y  H , we get that H g ∩ N K L (H )L  H v u = H u . −1 This yields that (H L)g ∩ N K L (H L) = (H g ∩ N K L (H )L)L  H u L = H L. Thus (H L)sG = (H L)(K L), K L is s-permutable in G and (H L)g ∩ N K L (H L)  H L, for all g ∈ G. So, H L is an SSH-subgroup in G. Because L  G and L  H L, H L/L is an SSH-subgroup in G/L by part (b). (d) By hypothesis, there exists an s-permutable subgroup K of G such that H sG = H K and H g ∩ N K (H )  H , for all g ∈ G. Using Lemma 2.2(e), we see that (H x )sG = (H sG )x = H x K x and (H g ∩ N K (H ))x  H x , for all g ∈ G. Also, K x is s-permutable in G by Lemma 2.1(i). Now (H g ∩ N K (H ))x = H gx ∩ (N K (H ))x = H gx ∩ N K x (H x ) by Lemma 2.3(b), therefore H gx ∩ N K x (H x )  H x , for all g ∈ G. −1

x −1

−1

Substituting g = y x , where y ∈ G, we attain that H y x ∩ N K x (H x ) = H x yx x ∩ N K x (H x ) = H x y ∩N K x (H x ) = (H x ) y ∩N K x (H x ). Since H gx ∩N K x (H x )  H x , for all g ∈ G, then (H x ) y ∩N K x (H x )  H x , for all y ∈ G. Thus H x is an SSH-subgroup in G.   Lemma 2.5 Let P be a Sylow p-subgroup of a group G, for some prime p. If G is p-nilpotent, then every subgroup of P is an SSH-subgroup in G. Proof Since G is p-nilpotent, then G = P T , where T is the normal Hall p  -subgroup of G. Let H  P and Q be any Sylow q-subgroup of G. If q = p, then Q  T  H T , and if q = p, then G = QT = Q(H T ). Hence H T is s-permutable in G and so H sG  H T . Clearly, H sG = H sG ∩ H T = H K , where K := H sG ∩ T . By Lemmas 2.2(a) and 2.1(d), K is s-permutable in G. We remark that H g ∩ N K (H )   H g ∩ K  H g ∩ T = 1  H , for all g ∈ G. Thus H is an SSH-subgroup in G.  Lemma 2.6 [20, Lemma 2.6] Let N be a nontrivial normal subgroup of a group G. If N ∩ Φ(G) = 1, then F(N ), the Fitting subgroup of N , is the direct product of the minimal normal subgroups of G which are contained in F(N ). Lemma 2.7 [9, Theorem 6(2)] Let G be a group and let H be an H-subgroup in G. If H is subnormal in G, then H is normal in G. Lemma 2.8 Let P be a Sylow p-subgroup of a group G for some prime p. Assume that L is a minimal normal subgroup of G such that L  Z (P). Let H be either a maximal or a minimal subgroup of L. If H is an SSH-subgroup in G, then |L| = p. Proof Let U  L such that U is s-permutable in G. Then O p (G)  N G (U ) by Lemma 2.1(h) and U  P. Thus G = P O p (G)  N G (U ) and hence U  G. Now let H be as above. By hypothesis, there exists an s-permutable subgroup K of G such that H sG = H K and H g ∩ N K (H )  H , for all g ∈ G. From Lemma 2.2(a), H sG is spermutable in G and H sG  H G  L. Assume that K = 1. Then H is s-permutable in G and it follows, from the previous arguments, that H G. If H is a maximal subgroup of L, then H = 1, and if H is a minimal subgroup of L, then L = H . Therefore we may assume that K = 1. Since K  L, then K  G; hence K = L. We remark that H g  L g = L and thereby H g = H g ∩ L = H g ∩ N L (H ) = H g ∩ N K (H )  H ,

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for all g ∈ G, so H  G. Once more, H = 1 or L = H . Thus |L| = p in all cases.   Lemma 2.9 Let G be a supersolvable group. Then the following statements hold: (a) The maximal subgroups of the Sylow subgroups of G  are SSH-subgroups in G. (b) Every cyclic subgroup of G  of order p or 4 is an SSH-subgroup in G. Proof (a) Since G is supersolvable, then G  is nilpotent by [23, Theorem 7.2.13, p. 157]. Let P be a Sylow p-subgroup of G  , for some prime p, and let H be a maximal subgroup of P. Note that P is characteristic in G   G, hence P  G. There exists L  P such that L G and |L| = p. We may assume that H is not s-permutable in G. First consider the case when L  H . Clearly, G/L is supersolvable and L  H . Thus, by induction on |G|, the maximal subgroups of the Sylow subgroups of (G/L) are SSH-subgroups in G/L. We observe that H/L is a maximal subgroup of the Sylow p-subgroup P/L of (G/L) = G  /L. Therefore H/L is an SSH-subgroup in G/L. By Lemma 2.4(b), H is an SSH-subgroup in G. Now consider the case when L  H . Obviously, P = H L, H ∩ L = 1 and, by Lemma 2.2(a), H sG is s-permutable in G with H sG  H G  P. Since H sG = H , then H sG = H L. Also, H g ∩ N L (H )  H g ∩ L = 1  H , for all g ∈ G. Thus, in all cases, H is an SSH-subgroup in G. (b) As G is 2-nilpotent, then every cyclic subgroup of G of order 2 or 4 is an SSHsubgroup in G by Lemma 2.5. We know that G  is nilpotent by [23, Theorem 7.2.13, p. 157]. Let P be a Sylow p-subgroup of G  , for some prime p = 2, and let H  P with |H | = p. We may assume that H is not s-permutable in G. Evidently, H  H G  P G = P. There exists V  H G such that H G /V is a chief factor of G. Since G is supersolvable, |H G /V | = p, that is, V is a maximal subgroup of H G . We observe that H  V , otherwise H  V  H G , a contradiction to the fact that H G is the smallest normal subgroup of G that contains H . Therefore H G = H V and H ∩ V = 1. By Lemma 2.2(a), H sG is s-permutable in G and H sG  H G . So, H sG = H sG ∩ H V = H K , where K := H sG ∩ V . By Lemma 2.1(d), K is s-permutable in G. It is easy to see that H g ∩ N K (H )  H g ∩ V = 1  H , for all g ∈ G. Thus H is an SSH-subgroup in G.   Lemma 2.10 Let G be a minimal non- p-nilpotent group for some prime p (all proper subgroups of G are p-nilpotent, but G is not p-nilpotent). Then: (a) G is a minimal non-nilpotent group (all proper subgroups of G are nilpotent, but G is not nilpotent). (b) G = P Q, where P is a normal Sylow p-subgroup of G and Q is a cyclic nonnormal Sylow q-subgroup of G, for some prime q. (c) If p > 2, then the exponent of P is p and, when p = 2, the exponent of P is at most 4. (d) P/Φ(P) is a minimal normal subgroup of G/Φ(P). Proof For (a), (b) and (c), see [18, Satz 5.4, p. 434 and Satz 5.2, p. 281]. For (d), see [18, Aufgaben 14(b), pp. 285–286].  

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Lemma 2.11 [13] Let G be a minimal non-supersolvable group (all proper subgroups of G are supersolvable but G is not supersolvable). Then: (a) G has a unique normal Sylow subgroup for some prime p, say P. (b) P/Φ(P) is a minimal normal subgroup of G/Φ(P). (c) If p > 2, then P is of exponent p.

3 Main results Theorem 3.1 Let P be a Sylow p-subgroup of a group G, for some prime p. Then G is p-nilpotent if and only if NG (P) is p-nilpotent and every maximal subgroup of P is an SSH-subgroup in G. Proof If G is p-nilpotent, then N G (P) is p-nilpotent and the maximal subgroups of P are, by Lemma 2.5, SSH-subgroups in G. For the converse, we assume that the result is false and let G be a counterexample of minimal order. Then: (1) P is not cyclic. Suppose that P is cyclic. As N G (P) is p-nilpotent, then N G (P) ∼ = P × S, where S is the normal Hall p  -subgroup of N G (P). Note that P, S  C G (P); hence C G (P) = N G (P). Therefore G is p-nilpotent by [18, Hauptsatz 2.6, p. 419], a contradiction. (2) If R  P such that 1 = R  G, then G/R is p-nilpotent. If R = P, then G/R is a p  -group and hence p-nilpotent. So, assume that R  P. Clearly, P/R is a Sylow p-subgroup of G/R and N G/R (P/R) = N G (P)/R is pnilpotent. Let M/R be a maximal subgroup of P/R. Then M is a maximal subgroup of P. By hypothesis, M is an SSH-subgroup in G. Lemma 2.4(b) implies that M/R is an SSH-subgroup in G/R. Therefore the maximal subgroups of P/R are SSHsubgroups in G/R. Thus G/R is p-nilpotent by the minimal choice of G. (3) O p (G) = 1. Suppose that U := O p (G) = 1. Note that P and PU/U are Sylow p-subgroups of PU and G/U , respectively. By using [16, Lemma 3.6.10, p. 134], N G/U (PU/U ) = N G (P)U/U ∼ = N G (P)/(N G (P)∩U ) is p-nilpotent. Let M/U be a maximal subgroup of PU/U . Then M = (M ∩ P)U , where M ∩ P is a maximal subgroup of P. Since M ∩ P is an SSH-subgroup in G and U is a normal p  -subgroup of G, then M/U = (M ∩ P)U/U is an SSH-subgroup in G/U by Lemma 2.4(c). So, G/U is p-nilpotent by the minimal choice of G, thus G is p-nilpotent, a contradiction. (4) If P  T  G, then T is p-nilpotent. Clearly, P is a Sylow p-subgroup of T and N T (P) = N G (P)∩T is p-nilpotent. By Lemma 2.4(a), every maximal subgroup of P is an SSH-subgroup in T . Therefore T is p-nilpotent by the minimal choice of G. For the rest of the proof, we set D = O p (G). (5) Let V be a maximal subgroup of P and let W be the s-permutable subgroup of G that is associated with V from being an SSH-subgroup in G. If V sG is a p-subgroup of G, then V  D. Also, if W is not a p-subgroup of G, then G = P W . Suppose that V sG is a p-subgroup of G. By Lemma 2.2(a), V sG is s-permutable in G. Since V sG is subnormal in G by Lemma 2.1(c), we see that V  V sG  D. Assume that W is not a p-subgroup of G. Note that P W  G. If P W  G, then

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P W is p-nilpotent by (4). Hence W = W p N , where W p is a Sylow p-subgroup of W and N = 1 is the normal Hall p  -subgroup of W . As W is subnormal in G by Lemma 2.1(c), N  O p (G), a contradiction to (3). Thus G = P W . Now we consider the following two cases: Case 1. D = 1. We have the following substeps: (i) D is a minimal normal subgroup of G. In addition, D  Φ(P). If D ∩ Φ(G) = 1, then G/(D ∩ Φ(G)) is p-nilpotent by (2). This implies that G/Φ(G) is p-nilpotent and hence G is p-nilpotent by [18, Hilfssatz 6.3, p. 689], a contradiction. Therefore D ∩ Φ(G) = 1. By Lemma 2.6, D is the direct product of the minimal normal subgroups of G which are contained in D. If N1 and N2 are minimal normal subgroups of G such that N1 = N2 and N1 , N2  D, then G/N1 and G/N2 are p-nilpotent by (2). This yields that G ∼ = G/1 = G/(N1 ∩ N2 ) is p-nilpotent by [14, Theorem 13.4(b), p. 44], a contradiction. So, there exists exactly one minimal normal subgroup of G that is contained in D, thus D must be a minimal normal subgroup of G. If D  Φ(P), then, by [14, Theorem 9.2(d), p. 30], D  Φ(G), a contradiction. Hence D  Φ(P). (ii) G is p-solvable and D  P. Moreover, D is the unique minimal normal subgroup of G. By (2), G/D is p-nilpotent, hence G is p-solvable. If D = P, then G = NG (P) is p-nilpotent, a contradiction; thus D  P. Let N be any minimal normal subgroup of G. Then N is p-solvable and since O p (G) = 1 by (3), it follows that N  D. But D is a minimal normal subgroup of G by (i), therefore N = D. Thus D is the unique minimal normal subgroup of G. (iii) D = C G (D) and P is a maximal subgroup of G. As G is p-solvable by (ii) and O p (G) = 1 by (3), then C G (D)  D by [15, Theorem 3.2, p. 228]. Clearly, D is abelian by (i), so D = C G (D). Since P = G, there exists a maximal subgroup M of G with P  M. By (4), M = P L, where L is the normal Hall p  -subgroup of M. Because D, L  M, then [D, L]  D ∩ L = 1. Hence L  C G (D) = D, that is, L = 1. Thus P = M is a maximal subgroup of G. (iv) The conclusion to the proof of case 1. Since D  Φ(P) by (i), there exists a maximal subgroup H of P such that P = H D. We know that H ∩ D = D, because if not, then H = P, a contradiction. If H ∩ D = 1, then [H, D] = 1 and thereby H  C G (D). This means that P  C G (D) = D by (iii), a contradiction to (ii). Thus we perceive that 1 = H ∩ D  D. If H ∩ D is an H-subgroup in G, then as H ∩ D is subnormal in G, H ∩ D  G by Lemma 2.7, a contradiction to (i). Therefore H ∩ D is not an H-subgroup in G. By hypothesis, there exists an s-permutable subgroup K of G such that H sG = H K and H g ∩ N K (H )  H , for all g ∈ G. If H sG is a p-subgroup of G, then H  D by (5) and so D = P, a contradiction to (ii). Thus H sG is not a p-subgroup of G. By (5), G = P K as K is not a p-subgroup of G. Suppose that K = G. Then H g ∩ N G (H )  H , for all g ∈ G, and accordingly H is an H-subgroup in G. Clearly, D  P  NG (H ), in particular, D ∩ N G (H ) = D. Now (H ∩ D)g ∩ N G (H ∩ D) = H g ∩ D ∩ N G (H ∩ D) = H g ∩ N G (H ) ∩ D ∩ N G (H ∩ D)  H ∩ D ∩ N G (H ∩ D) = H ∩ D, for all g ∈ G. Thus H ∩ D is an H-subgroup in G, a contradiction. This shows that K = G. By Lemma 2.1(c), K is subnormal in G and, by Lemma 2.1(g), K /K G is nilpotent. As

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K is not a p-group, then K has a nontrivial Sylow q-subgroup Q for some prime q that divides |K | with q = p. If K G = 1, then K is a subnormal nilpotent subgroup of G, hence Q  O p (G), a contradiction to (3). Thus K G = 1 and because D is the unique minimal normal subgroup of G by (ii), D  K G  K . If H  G, then H  D and so D = P, a contradiction (ii). Therefore H  G and hence N G (H )  G. Since P  N G (H ) and P is a maximal subgroup of G by (iii), N G (H ) = P. As H ∩ D  P and H ∩ D  G, then N G (H ∩ D) = P. For all g ∈ G, we remark that (H ∩ D)g ∩ N G (H ∩ D) = H g ∩ D ∩ N G (H ) = H g ∩ D ∩ K ∩ N G (H ) = H g ∩ N K (H )∩ D  H ∩ D. Once more, H ∩ D is an H-subgroup in G, a contradiction that completes the proof of this case. Case 2. D = 1. From (1), the maximal subgroups of P are nontrivial. Let {H1 , H2 , . . . , Hn } be the set of all maximal subgroups of P. By hypothesis, Hi is an SSH-subgroup in G, hence g G has an s-permutable subgroup K i such that HisG = Hi K i and Hi ∩ N K i (Hi )  Hi , sG for all g ∈ G, for i = 1, 2, . . . , n. If Hi is a p-subgroup of G for some i, then Hi  D by (5), a contradiction. Thus HisG is not a p-subgroup of G and so does K i , for all i = 1, 2, . . . , n. From (5), G = P K i , for all i = 1, 2, . . . , n. There exists a prime q that divides |G| with q = p. Let Q be a Sylow q-subgroup of G. By is a Sylow q-subgroup Lemma 2.1(c), K i is subnormal in G and it follows easily that Q  n K i . By repeated of K i , for i = 1, 2, . . . , n. We remark that 1 = Q  N := i=1 applications of Lemma 2.1(d), N is s-permutable in G. Moreover, N /N G is nilpotent by Lemma 2.1(g). If N G = 1, then N is nilpotent and since N is subnormal in G by Lemma 2.1(c), Q  O p (G), a contradiction to (3). Thus L := N G = 1. Clearly, L G, P ∩ L is a Sylow p-subgroup of L, and, by (3), P ∩ L = 1. Using Grün Theorem [15, Theorem 4.2, p. 252], we attain that (P∩L)∩L  = (P∩L)∩N L (P∩L) , (P∩L)∩ ((P ∩ L) )x : x ∈ L. Certainly, P ∩ L  G because D = 1. As P  N G (P ∩ L), then N G (P ∩ L) is p-nilpotent by (4), hence N L (P ∩ L) = N G (P ∩ L) ∩ L is p-nilpotent. Evidently, P ∩ L is the unique Sylow p-subgroup of N L (P ∩ L), so N L (P ∩ L) ∼ = (P ∩ L) × S, where S is the normal Hall p  -subgroup of N L (P ∩ L). This implies that (P ∩ L) ∩ N L (P ∩ L) = (P ∩ L) ((P ∩ L) ∩ S  ) = (P ∩ L) . Obviously, for x = 1, (P ∩ L) = (P ∩ L)∩((P ∩ L) )1 . Thus (P ∩ L)∩ L  = (P ∩ L)∩((P ∩ L) )x : x ∈ L. Let x ∈ L. Since Hi  P, we obtain that (P ∩ L) ∩ ((P ∩ L) )x  P N G (Hi ), n Hi . We for i = 1, 2, . . . , n. By [16, Theorem 1.8.8, p. 38], P   Φ(P) = i=1    observe that (P ∩ L)  P  Hi , for i = 1, 2, . . . , n, and (P ∩ L)  L. So, (P ∩ L)  Hi ∩ L and it follows that ((P ∩ L) )x  (Hi ∩ L)x , in particular, (P ∩ L) ∩ ((P ∩ L) )x  (Hi ∩ L)x , for i = 1, 2, . . . , n. Note that L  K i , for i = 1, 2, . . . n. We see that (P ∩ L) ∩ ((P ∩ L) )x  (Hi ∩ L)x ∩ N G (Hi ) = ) ∩ L  Hi ∩ L  Hi , Hix ∩ L ∩ N G (Hi ) = Hix ∩ L ∩ K i ∩ N G (Hi ) = Hix ∩ N K i (Hi  n Hi = Φ(P), for for i = 1, 2, . . . , n. This yields that (P ∩ L) ∩ ((P ∩ L) )x  i=1    x all x ∈ L. Thus P ∩ L = (P ∩ L) ∩ L = (P ∩ L) ∩ ((P ∩ L) ) : x ∈ L  Φ(P). If L  = 1, then L is abelian and since p divides |L|, O p (L)  D, a contradiction. Hence L  = 1. We perceive that L  is characteristic in L  G, therefore L   G. By [18, Satz 4.7, p. 431], L  is p-nilpotent. But O p (G) = 1 by (3) and we conclude that L   D, a final contradiction completing the proof.  

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Corollary 3.2 Let p be the smallest prime that divides the order of a group G and let P be a Sylow p-subgroup of G. Then G is p-nilpotent if and only if every maximal subgroup of P is an SSH-subgroup in G. Proof If G is p-nilpotent, then, by Lemma 2.5, every maximal subgroup of P is an SSH-subgroup in G. We prove the converse by induction on |G|. Suppose that P  G. Then p is the smallest prime dividing |N G (P)| and P is the Sylow p-subgroup of N G (P)  G. By Lemma 2.4(a), every maximal subgroup of P is an SSH-subgroup in N G (P). Therefore N G (P) is p-nilpotent by induction on |G|. Thus G is p-nilpotent by Theorem 3.1, a contradiction. Hence P  G. Let 1 = L  P such that L  G. If L = P, then G/L is p-nilpotent being a p  -group. Assume that L  P. Obviously, p is the smallest prime dividing |G/L| and P/L is the Sylow p-subgroup of G/L. From Lemma 2.4(b), every maximal subgroup of P/L is an SSH-subgroup in G/L. By induction, G/L is p-nilpotent. Now by using the same arguments as in Case 1(i) within the proof of Theorem 3.1, P is a minimal normal subgroup of G and so P is abelian. Let H be a maximal subgroup of P. Then H is an SSH-subgroup in G, hence |P| = p by Lemma 2.8. Thus G is p-nilpotent by [18, Satz 2.8, p. 420], a contradiction completing the proof.   Theorem 3.3 Let G be a group and let N be a normal subgroup of G such that G/N is supersolvable. If the maximal subgroups of the non-cyclic Sylow subgroups of N are SSH-subgroups in G, then G is supersolvable. Proof Assume that G is a counterexample of minimal order. Then: (1) Let p be the largest prime that divides |N | and let P be a Sylow p-subgroup of N . Then P  G. We observe that the maximal subgroups of the non-cyclic Sylow subgroups of N are SSH-subgroups in N by Lemma 2.4(a). In view of [18, Satz 2.8, p. 420] and Corollary 3.2, N has a Sylow tower of supersolvable type. Let P be as above. Then P is characteristic in N  G, therefore P  G. (2) If 1 = L  P such that L  G, then G/L is supersolvable. Clearly, (G/L)/(N /L) ∼ = G/N is supersolvable. Let Q be a non-cyclic Sylow q-subgroup of N /L, for some prime q. Then Q = Nq L/L, where Nq is a non-cyclic Sylow q-subgroup of N . Let M/L be a maximal subgroup of Nq L/L. If q = p, then N p = P by (1) and M/L is an SSH-subgroup in G/L by Lemma 2.4(b). Assume that q = p. Then M = (M ∩ Nq )L, where M ∩ Nq is a maximal subgroup of Nq . Since M ∩ Nq a q-subgroup which is an SSH-subgroup in G and L is a q  -subgroup of G, then M/L is an SSH-subgroup in G/L by Lemma 2.4(c). Therefore the maximal subgroups of the non-cyclic Sylow subgroups of N /L are SSH-subgroups in G/L. Thus G/L is supersolvable by the minimal choice of G. (3) P is a non-cyclic minimal normal subgroup of G. By using (1), (2) and applying similar arguments as in Case 1(i) within the proof of Theorem 3.1, P is a minimal normal subgroup of G. We just need to replace [18, Hilfssatz 6.3, p. 689] by [16, Corollary 1.9.7, p. 49] and [14, Theorem 13.4(b), p. 44] by [23, Exercise 7.2.22, p. 159]. If P is cyclic, then as G/P is supersolvable by (2), G is supersolvable by [23, Theorem 7.2.14, p. 158], a contradiction. (4) p is the largest prime that divides |G|.

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Suppose not and let q be the largest prime that divides |G|. Assume that Q is a Sylow q-subgroup of G. By (3), P is a non-cyclic normal subgroup of G. Since G/P is supersolvable by (2), then Q P/P  G/P, so P Q  G. By Lemma 2.4(a), the maximal subgroups of P are SSH-subgroups in P Q. Note that p < q and P is the Sylow psubgroup of P Q. Corollary 3.2 implies that P Q is p-nilpotent. Now Q is characteristic in P Q  G, hence Q  G. Clearly, (G/Q)/(N Q/Q) ∼ = G/N Q ∼ = (G/N )/(Q N /N ) is supersolvable. By (1), q does not divide |N |. Let R be a non-cyclic Sylow r -subgroup of N Q/Q. We may assume that r = q as the Sylow q-subgroups of N Q/Q are trivial. We observe that R = Nr Q/Q, where Nr is a non-cyclic Sylow r -subgroup of N . Let M/Q be a maximal subgroup of Nr Q/Q. Then M = (M ∩ Nr )Q, where M ∩ Nr is a maximal subgroup of Nr . By Lemma 2.4(c), M/Q is an SSH-subgroup in G/Q. Therefore the maximal subgroups of the non-cyclic Sylow subgroups of N Q/Q are SSH-subgroups in G/Q. Hence G/Q is supersolvable by the minimal choice of G. Consequently, G ∼ = G/(P ∩ Q) is supersolvable by [23, Exercise 7.2.22, p. 159], a contradiction. Thus our supposition is wrong and p is the largest prime that divides |G|. (5) The final contradiction. Let P ∗ be a Sylow p-subgroup of G. It is obvious that P  P ∗ by (1). Because G/P is supersolvable by (2) and p is the largest prime dividing |G| by (4), P ∗ /P  G/P, hence P ∗  G. Note that Z (P ∗ ) is characteristic in P ∗  G, therefore Z (P ∗ )  G. By [15, Theorem 6.4, p. 31], P ∩ Z (P ∗ ) = 1. Since P is a minimal normal subgroup of G by (3) and P ∩ Z (P ∗ )  G, then P  Z (P ∗ ). By Lemma 2.8, |P| = p, a contradiction to (3) completing the proof of the theorem.   As an immediate consequence of Lemma 2.9(a) and Theorem 3.3, we have: Corollary 3.4 A group G is supersolvable if and only if the maximal subgroups of the non-cyclic Sylow subgroups of G  are SSH-subgroups in G. Theorem 3.5 Let p be the smallest prime that divides the order of a group G and let P be a Sylow p-subgroup of G. Then G is p-nilpotent if and only if every cyclic subgroup of P of order p or 4 (if P is a non-abelian 2-group) is an SSH-subgroup in G. Proof If G is p-nilpotent, then every cyclic subgroup of P of order p or 4 (when p = 2) is an SSH-subgroup in G by Lemma 2.5. For the other direction, we proceed by induction on |G|. We have the following steps: (1) If M  G, then M is p-nilpotent. If M is a p  -group, then M is p-nilpotent. So, assume that M p = 1 is a Sylow p-subgroup of M. Surely, p is the smallest prime dividing |M|. Let H be a cyclic subgroup of M p of order p or 4 (if M p is a non-abelian 2-group). There exists x ∈ G −1 such that H x  M px  P. Since H x is an SSH-subgroup in G, then H = (H x )x is an SSH-subgroup in G by Lemma 2.4(d). Hence H is an SSH-subgroup in M by Lemma 2.4(a). Thus, by induction on |G|, M is p-nilpotent. (2) G is a minimal non-nilpotent group. Moreover, G = P Q, where P is a normal Sylow p-subgroup of G and Q is a cyclic non-normal Sylow q-subgroup of G for some prime q > p. Also, if p > 2, then the exponent of P is p and, when p = 2, the

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exponent of P is at most 4. In addition, P/N is a minimal normal subgroup of G/N , where N := Φ(P). By (1), G is a minimal non- p-nilpotent group. Hence (2) holds by Lemma 2.10. (3) There exists a cyclic subgroup H of P of order p or 4 (if P is a non-abelian 2-group) such that H  G. Suppose the contrary of the above. From (2), P/N is a minimal normal subgroup of G/N . Let L/N  P/N with |L/N | = p. Then L = xN for some x ∈ L\N . By (2), |x| = p or 4, so x  G. Therefore L/N  G/N and it follows that P/N = L/N . Since P/N is the Sylow p-subgroup of G/N and p < q, then G/N is p-nilpotent by [18, Satz 2.8, p. 420]. Clearly, N = 1, otherwise G is p-nilpotent, a contradiction. By (2) and [14, Theorem 9.2(e), p. 30], N  Φ(G). Consequently, G/Φ(G) is pnilpotent, hence G is p-nilpotent by [18, Hilfssatz 6.3, p. 689], a contradiction. Thus (3) holds. (4) The final contradiction. Let H be as in (3). By hypothesis, G has an s-permutable subgroup K such that H sG = H K and H g ∩ N K (H )  H , for all g ∈ G. If O p (G)  G, then O p (G) is nilpotent by (2). Therefore Q is characteristic in O p (G)  G, hence Q  G, a contradiction. Thus O p (G) = G. By (2) and Lemma 2.2(a), H sG  H G  P and H sG is s-permutable in G. Suppose that H sG  P. Then H sG Q  G and, by (2), H sG Q is nilpotent; so Q  C G (H sG ), for every Sylow q-subgroup Q of G. Consequently, O p (G)  C G (H sG ) and it follows that H  G, a contradiction to (3). This means that P = H K . Suppose that K = P. Note that H g ∩ N G (H ) = H g ∩ P ∩ N G (H ) = H g ∩ N P (H ) = H g ∩ N K (H )  H , for all g ∈ G, that is, H is an H-subgroup in G. By Lemma 2.7, H  G, a contradiction to (3). This yields that K  P. From Lemma 2.1(h), O p (G)  N G (K ), therefore K  G. If K N = P, then K = P, a contradiction. Hence K N  P. Obviously, K N /N  G/N and since P/N is a minimal normal subgroup of G/N by (2), K N /N = 1. This implies that K  N . Hence P = H N = H is cyclic, so G is p-nilpotent by [18, Satz 2.8, p. 420], a contradiction completing the proof of the theorem.   Theorem 3.6 Let G be a group and let N be a normal subgroup of G such that G/N is supersolvable. If for every non-cyclic Sylow p-subgroup P of N , every cyclic subgroup of P of order p or 4 (if P is a non-abelian 2-group) is an SSH-subgroup in G, then G is supersolvable. Proof Assume that G is a counterexample of minimal order. Then: (1) For every M  G, M is supersolvable. Obviously, M/(M ∩ N ) ∼ = M N /N  G/N is supersolvable. Let Q be any noncyclic Sylow q-subgroup of M ∩ N , where q is a prime number. By Lemma 2.4(a), every cyclic subgroup of Q of order q or 4 (if Q is a non-abelian 2-group) is an SSH-subgroup in M. Hence M is supersolvable by the minimal choice of G. (2) Let R be a Sylow r -subgroup of N , where r is the largest prime that divides |N |. Then R  G. Moreover, G/R is supersolvable and R is not cyclic. From Lemma 2.4(a), for every non-cyclic Sylow p-subgroup P of N , every cyclic subgroup of P of order p or 4 (if P is a non-abelian 2-group) is an SSH-subgroup in N . Thus, by [18, Satz 2.8, p. 420] and Theorem 3.5, N has a Sylow tower of supersolvable type. Let R be as mentioned above. We see that R is characteristic in

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N  G, hence R  G. Also, (G/R)/(N /R) ∼ = G/N is supersolvable. Let P R/R be a non-cyclic Sylow p-subgroup of N /R. Then P is non-cyclic. We may assume that p = r . Let H/R be a cyclic subgroup of P R/R of order p or 4 (if P R/R is a nonabelian 2-group). Note that H = (H ∩ P)R. This implies that |H ∩ P| = [H : R] = p or 4. By Lemma 2.4(c), H/R is an SSH-subgroup in G/R. So, G/R is supersolvable by the minimal choice of G. If R is cyclic, then G is supersolvable by [23, Theorem 7.2.14, p. 158], a contradiction. Thus R is not cyclic. (3) r is the largest prime that divides |G| and r > 2. Suppose that is not true and let q be the largest prime dividing |G|. Since G/R is supersolvable by (2), then Q R/R  G/R, where Q is a Sylow q-subgroup of G, therefore Q R  G. Clearly, r < q and R is a non-cyclic Sylow r -subgroup of Q R. Lemma 2.4(a) tells us that every cyclic subgroup of R of order r or 4 (if R is a non-abelian 2-group) is an SSH-subgroup in Q R. Thus Q R is r -nilpotent by Theorem 3.5 and it follows that Q  G. It is not hard to see that (G/Q)/(N Q/Q) ∼ = (G/N )/(Q N /N ) is supersolvable. Consider a non-cyclic Sylow p-subgroup P Q/Q of N Q/Q, where P is obviously a non-cyclic Sylow p-subgroup of N . We may assume that p = q as q does not divide |N Q/Q| = |N | by (2). Let T /Q be a cyclic subgroup of P Q/Q of order p or 4 (if P Q/Q is a non-abelian 2-group). Then T /Q = (T ∩ P)Q/Q is an SSH-subgroup in G/Q by Lemma 2.4(c). Hence G/Q is supersolvable by the minimal choice of G. Therefore G ∼ = G/(Q ∩ R) is supersolvable by [23, Exercise 7.2.22, p. 159], a contradiction. Thus r is the largest prime dividing |G|. Of course, r > 2, otherwise G is a 2-group, a contradiction. (4) Let R ∗ be a Sylow r -subgroup of G. Then R ∗  G and the exponent of R ∗ is r . In addition, R ∗ /V is a minimal normal subgroup of G/V , where V := Φ(R ∗ ). Since r is the largest prime that divides |G| and G/R is supersolvable by (2) and (3), then R ∗ /R  G/R and hence R ∗  G. By (1), G is a minimal non-supersolvable group. The rest of (4) holds by Lemma 2.11. (5) V = 1. If U  R ∗ such that U is s-permutable in G and U  V , then R ∗ = U . Suppose that V = 1. By (4), R ∗ is an elementary abelian minimal normal subgroup of G. Because 1 = R  G and R  R ∗ , then R ∗ = R. This means that R is a Sylow r -subgroup of G. By Lemma 2.8, |R| = r , a contradiction to (2). So, we see that V = 1. Let U be as above. If O r (G)  G, then O r (G) is supersolvable by (1) and thereby G = R ∗ O r (G) is supersolvable by [23, Exercise 7.2.23, p. 159], a contradiction. Hence O r (G) = G. By Lemma 2.1(h), O r (G)  N G (U ), therefore U  G. We observe that 1 = U V /V  G/V . Since R ∗ /V is a minimal normal subgroup of G/V by (4), R ∗ /V = U V /V . This yields that R ∗ = U V = U . (6) The final contradiction. By (5) and (2), V = 1 and G/R is supersolvable. If R  V , then R  Φ(G) by [14, Theorem 9.2(d), p. 30]. This implies that G/Φ(G) is supersolvable and so does G by [16, Corollary 1.9.7, p. 49], a contradiction. Consequently, R  V ; hence there exists x ∈ R\V and, by (5), R ∗ = R. We set H = x. Since |H | = r by (4), there exists an s-permutable subgroup K of G such that H sG = H K and H g ∩ N K (H )  H , for all g ∈ G. By Lemma 2.2(a), H sG is s-permutable in G and H sG  H G  R. Note that H sG  V , therefore R = H K by (5). If K  V , then R = H V = H is cyclic, a contradiction to (2). So, we may assume that K  V . Hence K = R by (5). We observe that H g ∩ N G (H ) = H g ∩ R ∩ N G (H ) = H g ∩ N R (H ) = H g ∩ N K (H )  H , for all

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g ∈ G. Thus H is an H-subgroup in G and, by Lemma 2.7, H  G . As x ∈ / V , then H  V . Therefore R = H by (5) which is cyclic, a contradiction to (2) completing the proof of the theorem.   As an immediate consequence of Lemma 2.9(b) and Theorem 3.6, we have: Corollary 3.7 A group G is supersolvable if and only if for every non-cyclic Sylow p-subgroup P of G  , every cyclic subgroup of P of order p or 4 (if P is a non-abelian 2-group) is an SSH-subgroup in G.

4 Some applications Let K be a subgroup of a group G. We have the following notations: 1(K ) = {H  K : H is a maximal subgroup of a Sylow subgroup of K }, 2(K ) = {H  K : H is of prime order}, 3(K ) = 2(K ) ∪ {H  K : H is cyclic of order 4},  2(K ) if the Sylow 2-subgroups of K are abelian, 4(K ) = 3(K ) if the Sylow 2-subgroups of K are non-abelian. In what follows, we assume that G is a group, N G and P, Q are Sylow subgroups of G for some primes p, q, respectively, where q is the smallest prime dividing |G|. Also, whenever N is mentioned, G/N is supersolvable. Now we list some results from the literature which may be considered as special cases of our main results. Corollary 4.1 [21, Theorem 3.5] If every H ∈ 1 (N ) is normal in G, then G is supersolvable. Corollary 4.2 [10, Theorem 3] If G is of odd order and every H ∈ 2(G) is normal in G, then G is supersolvable. Corollary 4.3 [2, Theorem 3.1] If every H ∈ 3 (G) is permutable in G, then G is supersolvable. Corollary 4.4 [8, Proposition 2] If the Sylow 2-subgroups of G are abelian and every H ∈ 2(N ) is permutable in G, then G is supersolvable. Corollary 4.5 [25, Theorem 2] If every H ∈ 1(G) is s-permutable in G, then G is supersolvable. Corollary 4.6 [24, Theorem 3.2] If every H ∈ 4 (Q) is s-permutable in G, then G is q-nilpotent. Corollary 4.7 [24, Theorem 3.4] If N is a proper subgroup of G such that every H ∈ 4(N ) is s-permutable in G, then G is supersolvable. Corollary 4.8 [22, Lemma 3.2] If every H ∈ 1 (Q) is c-normal in G, then G is q-nilpotent.

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Corollary 4.9 ([5, Lemma 3.1] and [22, Lemma 3.8]) If every H ∈ 3(Q) is c-normal in G, then G is q-nilpotent. Corollary 4.10 [12, Theorem 3.4] If every H ∈ 4 (N ) is c-normal in G, then G is supersolvable. Corollary 4.11 [3, Theorem 1.1] The group G is p-nilpotent if and only if NG (P) is p-nilpotent and every H ∈ 1(P) is an H-subgroup in G. Corollary 4.12 [3, Corollary 1.2] The group G is q-nilpotent if and only if every H ∈ 1(Q) is an H-subgroup in G. Corollary 4.13 [3, Theorem 1.5] If every H ∈ 1 (N ) is an H-subgroup in G, then G is supersolvable. Corollary 4.14 [4, Theorem 3.2] If every H ∈ 1(G) is a weakly H-subgroup in G, then G is supersolvable. Corollary 4.15 [1, Theorem 3.4] If every H ∈ 3(Q) is a weakly H-subgroup in G, then G is q-nilpotent. Corollary 4.16 [1, Corollary 3.6] If every H ∈ 3(N ) is a weakly H-subgroup in G, then G is supersolvable. Corollary 4.17 [27, Theorem 3.7] If every H ∈ 1(Q) is an HC-subgroup in G, then G is q-nilpotent. Corollary 4.18 [27, Theorem 4.6] If every H ∈ 1(N ) is an HC-subgroup in G, then G is supersolvable. Corollary 4.19 [27, Theorem 3.2] If every H ∈ 3(Q) is an HC-subgroup in G, then G is q-nilpotent. Corollary 4.20 [6, Theorem 1.6] The group G is p-nilpotent if and only if NG (P) is p-nilpotent and every H ∈ 1(P) is weakly H-embedded in G. Corollary 4.21 [6, Corollary 1.7] The group G is q-nilpotent if and only if every H ∈ 1(Q) is weakly H-embedded in G. Corollary 4.22 [6, Theorem 1.8] If every H ∈ 4 (Q) is weakly H-embedded in G, then G is q-nilpotent. Corollary 4.23 [7, Theorem 1.6] The group G is p-nilpotent if and only if NG (P) is p-nilpotent and every H ∈ 1(P) is weakly HC-embedded in G. Corollary 4.24 [7, Corollary 1.7] The group G is q-nilpotent if and only if every H ∈ 1(Q) is weakly HC-embedded in G. Corollary 4.25 [7, Theorem 1.8] If every H ∈ 4(Q) is weakly HC-embedded in G, then G is q-nilpotent. Corollary 4.26 [7, Corollary 3.3] If every H ∈ 1(N ) or every H ∈ 4(N ) is weakly HC-embedded in G, then G is supersolvable.

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