Finite Groups with ss-embedded Subgroups

International Journal of Algebra, Vol. 11, 2017, no. 2, 93-101 HIKARI Ltd, www.m-hikari.com https://doi.org/10.12988/ija.2017.7311 Finite Groups with ss-embedded Subgroups Xinjian Zhang School of Mathematical Sciences Huaiyin Normal University Huaian, Jiangsu, 223300, China Yong Xu 1 School of Mathematics and Statistics Henan University of Science and Technology, Luoyang Henan, 471003, China Copyright c 2017 Xinjian Zhang and Yong Xu. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract Let H be a subgroup of a group G. Then H is called an ss-embedded subgroup of G if G has an s-permutable subgroup T such that HT is s- permutable in G and H T H ssg, where H ssg is an s-semipermutable subgroup of G contained in H. In this paper we investigate the structure of G under the assumption that some subgroups of P are ss-embedded in G, and some new criteria are obtained. Mathematics Subject Classification: 20D10, 20D15 Keywords: s-permutable subgroup, n-embedded subgroup, ss-embedded subgroup, p-nilpotent group 1 Introduction Throughout this paper, all groups are finite. Let H be a subgroup of a group G. Recall that H is said to be s-permutable (or s-quasinormal, π-quasinormal) 1 Corresponding author

94 Xinjian Zhang and Yong Xu in G if H permutes with every Sylow subgroup of G; H is called to be c- normal in G if there is a normal subgroup T of G such that G = HT and H T H G. Recently, Guo et al. introduced in [1] the following concepts s-embedded subgroups and n-embedded subgroups, which cover s-permutable subgroups and c-normal subgroups. Definition 1.1. Let H be a subgroup of a group G, H SG the subgroup of H generated by all those subgroups of H which are s-permutable in G and H SG the intersection of all such s-permutable subgroups which contain H. Then (1) H is s-embedded in G if G has an s-permutable subgroup T such that T H H SG and HT = H SG. (2) H is n-embedded in G if G has a normal subgroup T such that T H H SG and HT = H G. Recall that a subgroup H is called s-semipermutable in G if H permutes with every Sylow p-subgroup of G with ( H, p) = 1. Denote H ssg be an s- semipermutable subgroup of G contained in H. Here we give a new concept which covers properly both s-semipermutability property and s-embedding property. Definition 1.2. Let H be a subgroup of a group G. Then H is ss-embedded in G if G has an s-permutable subgroup T such that HT is s-permutable in G and H T H ssg. It is clear that every s-permutable subgroup, every c-normal subgroup subgroup, every s-semipermutable subgroup, every s-embedded subgroup and every n-embedded subgroup are ss-embedded subgroups. But the converse does not hold in general. Example 1.3. Let G, A be the groups defined in [1, Example 1.2]. Then A is ss-embedded in G, but neither s-permutable nor c-normal in G. Example 1.4. Let G = C 7 C 3 = [K]c 3 be the regular wreath product, where K is the base group of G and C i = i. Then K = F (G) is the Sylow 7-subgroup of G. Let H = {(a 1, a 2, 1) a 1, a 2 C 7 }. Then H is maximal in K. It is clear that H SG = H G = 1. By the following Lemma 2.2(3) in Section 2, H is not s-semipermutable in G. Let C 7 = x and D = {(x, x, x) x C 7 }. Then D is normal in G and D H=1. Hence DH = K is normal in G and so H is ss-embedded in G. Example 1.5. Let G = A 5, the alternative group of degree 5. Then the Sylow 2-subgroups of G are ss-embedded in G, but neither s-embedded nor n-embedded in G.

Finite groups with ss-embedded subgroups 95 Some properties of the ss-embedded subgroups are given in Section 2 and we shall investigate the influence of ss-embedded subgroups on the structure of finite groups. Our main results are the following theorems. Theorem A Let p be the smallest prime dividing the order of a group G and P a Sylow p-subgroup of G. If every maximal subgroup of P or every cyclic subgroup H of P with prime prime order and order 4 (if P is a non-abelian 2-group and H Z (G)) is ss-embedded in G, then G is p-nilpotent. Theorem B Let F be a saturated formation containing all supersolvable groups and G a group with a normal subgroup E such that G/E F. If for every non-cyclic Sylow subgroup P of E every maximal subgroup of P or every cyclic subgroup of P with prime prime order and order 4 (if P is a non-abelian 2-group and H Z (G)) is ss-embedded in G, then G F. The main results will be proved in Section 3. As applications of these results, many corollaries are given in Section 4. All groups in this paper are finite. The other notations and terminology are standard, as in [2] and [3]. 2 Preliminary Notes The following lemmas about s-permutable subgroups and s-semipermutable subgroups will be used in the paper several times. Lemma 2.1. ([4, Lemma 2.1]) (1) An s-permutable subgroup of G is subnormal in G. (2) If H K G and H is s-permutable in G, then H is s-permutable in K. (3) Let K G. If H is s-permutable in G, then HK/K is s-permutable in G/K. (4) If P is an s-permutable p-subgroup of G for some prime p, then N G (P ) O p (G). Lemma 2.2. ([5, 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 HN/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. Lemma 2.3. ([2, VI, 4.10])Assume that A and B are two subgroups of a group G and G AB. If AB g = B g A holds for any g G, then either A or B is contained in a proper normal subgroup of G.

96 Xinjian Zhang and Yong Xu Now we give some basic properties of the ss-embedded subgroups, which can be proved by direct calculation. Lemma 2.4. Let G be a group and H K G. (1) Suppose that H is normal in G. Then K/H is ss-embedded in G/H if and only if K is ss-embedded in G. (2) If H is ss-embedded in G, then H is ss-embedded in K. (3) Suppose that N is normal in G. Then NH/N is ss-embedded in G/N for every ss-embedded subgroup H of G satisfying ( H, N ) = 1. (4) If H is ss-embedded in G and K is normal in G, then G has an s- permutable subgroup T such that HT K is s-permutable in G and T H H ssg. Lemma 2.5. Let G be a group and H is an ss-embedded subgroup of G. If O p (G) = G and H O p (G), then H is n-embedded in G. Proof Since O p (G) = G, every s-permutable p-subgroup of G is normal in G follows from Lemma 2.1 (d). By hypothesis, G has an s-permutable subgroup T such that HT is s-permutable in G and H T H ssg. By Lemmas 2.4(4), 2.2(3) and 2.1(4), T, HT and H ssg are normal in G. Hence H is n-embedded in G. Lemma 2.6. Let p be the smallest prime dividing the order of a group G. Assume that G = [P ]Q where P is the Sylow p-subgroup of G and Q = q where q p is a prime. If all maximal subgroups of P or all cyclic subgroups H of P with prime order and order 4 (if P is non-abelian and H Z (G)) are ss-embedded in G, then G is nilpotent. Proof Suppose that this lemma is false and let G be a counterexample of minimal order. Then P > p. We first prove that some maximal subgroup of P is not ss-embedded in G. Indeed, suppose that every maximal subgroup is ss-embedded in G. Then by Lemma 2.4(2), the hypothesis is still true for G/N for any minimal normal subgroup N of G and G/N is nilpotent. It follows that N is the unique minimal normal subgroup of G and N Φ(G). But then N = C G (N) and hence N = P is abelian. It follows from Lemma 2.1 every s-permutable subgroup T of P is normal in G. Consequently, either T = 1 or T = P. Let V be a maximal subgroup of P. Then G has an s- permutable subgroup T such that V T is s-permutable in G and V T V ssg. By Lemma 2.2(3), V ssg V SG = 1. It is clear that 1 T P, which contradicts the minimality of N = P. Hence some maximal subgroup of P is not ss-embedded in G and thereby by hypothesis every cyclic subgroup H of P with prime order and order 4 (if P is non-abelian and H Z (G)) are ss-embedded in G. Let A = [A p ]A q be a Schmidt subgroup of G, that is, a non-nilpotent subgroup all proper subgroup of which are nilpotent. Then by

Finite groups with ss-embedded subgroups 97 Lemma 2.4 (2), every cyclic subgroups V of A p with prime order and order 4 (if P is non-abelian and V Z (G)) are ss-embedded in H and so by Lemma 2.6, V is n-embedded in A since evidently O p (A) = A and V A p = O p (A). But this, in view of P > p, contradicts [1, Lemma 2.10]. 3 Proofs of Theorems A and B Proof of Theorem A. Assume that the assertion is false and let G be a minimal counterexample. Then (1) every maximal subgroup of P is ss-embedded in G. Suppose that some maximal subgroup of P is not ss-embedded in G. Then by hypothesis every minimal subgroup of P is ss-embedded in G. By [2, IV, Theorem 5.4], G has a p-closed Schmidt group A = [A p ]A q, where p < q. It is clear that A p > p. Without loss of generality, we may assume that the Sylow p-subgroup A p of A is contained in P. Obviously O p (A) = A. Now by Lemma 2.5, every cyclic subgroup H of A p with prime order and order 4 (if P is a non-abelian 2-group and H Z (G)) is n-embedded in A, then by [1, Lemma 2.10], we have that p is not the smallest prime dividing G, which contradicts the hypothesis. Hence (1) holds. (2) G is not a non-abelian simple group. Let V be a maximal subgroup of P. By the hypothesis, G has an s- permutable subgroup T such that V T is s-permutable in G and V T V ssg. If G is a non-abelian simple group, then by Lemma 2.1 (a), T = G and V T = G, which implies that V = V T = V ssg is an s-semipermutable subgroup of G. Let A be a Sylow q-subgroup of G. Then V A g = A g V for any g G. Now by Lemma 2.3, either V or A is contained in a proper normal subgroup of G. a contradiction. So G is not a non-abelian simple group. (3) G has a unique minimal normal subgroup N, G/N is p-nilpotent and Φ(G) = 1. Let N be a minimal normal subgroup of G. If P N/N is cyclic, then G/N is p-nilpotent by Burnside s theorem. Hence we may assume that P N/N is not cyclic. Then P is not cyclic. Let M/N be a maximal subgroup of P N/N. Then M = V N for some maximal subgroup V of P. By (1) and the hypothesis, G has an s-permutable subgroup T such that V T is s-permutable in G and V T V se. Then T N/N and V T N/N are s-permutable in G/N. Since P V N = V (P N) and p = P N/V N = P (V N)/V N = P/P V N = P/V (P N), P N V and so V N = P N is a Sylow p-subgroup of N. Thus N p = V N = (V N)(T N) p = N V T p. Since (V N)(T N) q = N V T q with q p and (V N)(T N) V T N, we have (V N)(T N) = V T N,. Now by [4, Lemma 2.7], V N T N = (V T )N. Consequently, V N/N T N/N = (V T )N/N V ssg N/N and V ssg N/N is an s-semipermutable subgroup of G/N contained in V N/N by Lemma 2.2(2).

98 Xinjian Zhang and Yong Xu Hence M/N is ss-embedded in G/N and so G/N satisfies the hypothesis. By induction, G/N is p-nilpotent. Obviously N is the unique minimal normal subgroup of G and Φ(G) = 1. (4)O p (G) = 1. If O p (G) 1, then by (3), G/O p (G) is p-nilpotent and so G is p-nilpotent, a contradiction. (5) O p (G) = 1. Assume that O p (G) 1. Then by (3), N O p (G) and G has a maximal subgroup M such that G = [N]M. Obviously N Φ(P ) and so P has a maximal subgroup P 1 not containing N. It follows that P = NP 1. By the hypothesis, G has an s-permutable subgroup T such that P 1 T is s-permutable in G and P 1 T (P 1 ) ssg. First we assume that T G = 1. For T/T G is nilpotent, T is nilpotent. By (4), we have that T is a p-subgroup. Then either P 1 = P 1 T or P = P 1 T. If P 1 = P 1 T, then P 1 is s-permutable in G and so O p (G) N G (P 1 ). It follows that G = P O p (G) N G (P 1 ), that is, P 1 is normal in G. So the unique minimal normality of N implies that N P 1, a contradiction. Hence P 1 T = P is s- permutable in G and so P is normal in G. It follows that N = P. Evidently G is p-solvable. By (4) and [6, Theorem 9.3.1] C G (P ) P. Consider [P ]Q, where Q is a subgroup of order q and q p is a prime. Now by Lemma 2.6, Q C G (P ) P, a contradiction. Now we assume that T G 1. Then N T G T follows from the unique minimality of N. It follows that P 1 N P 1 T (P 1 ) ssg and so P 1 N = N (P 1 ) ssg is s-semipertable in G by lemma 2.2(4). From Lemma 2.2(3), we have P 1 N is s-permutable in G. This implies that P 1 N is normal in G by Lemma 2.1(d) and the fact that P 1 N is normal in P. By the unique minimality of N, we have that N = p. Since G/C G (N) Aut(N) and p is the smallest prime dividing the order of G, G/C G (N) = 1, which implies that N Z(G). Now by (3) G is p-nilpotent, a contradiction. Hence O p (G) = 1. (6) N is not p-nilpotent. This follows from (4) and (5). (7) P N = G. By Lemma 2.4, P N satisfies the hypothesis. Then P N is p-nilpotent if P N is a proper subgroup of G, which implies that N is p-nilpotent, contradicts (6). So P N = G. (8) The final contradiction. By (2) and (7), P N < P. We can choose a maximal subgroup P 1 of P such that N P P 1. Since P 1 is ss-embedded in G, by the hypothesis, G has an s-permutable subgroup T such that P 1 T is s-permutable in G and P 1 T (P 1 ) ssg. If T G = 1, then T is a nilpotent s-permutable subgroup of G, which contradicts (4) and (5). So T G 1. It follows from the unique minimality of N that N T and then N P 1 T P 1 (P 1 ) ssg. By Lemma

Finite groups with ss-embedded subgroups 99 2.2(4), we have that N P = N P 1 = N (P 1 ) ssg is s-semipermutable in G. So P 1 N is s-semipermutable in N by Lemma 2.2(1). By (2) and (7), we have that N is a character simple group and N = N 1 N 2 N s, where N i are conjugated non-abelian simple groups. Without losing generality, we can assume that P 1 N 1 is a non-trivial Sylow p-subgroup of N 1. So P 1 N 1 is s-semipermutable in N 1 by Lemma 2.2. To be similar to (2), we have the final contradiction. Proof of Theorem B. Suppose the theorem is false and consider a counterexample (G, E) for which G E is minimal. Let P be a Sylow p- subgroup of E where p is the smallest prime dividing E. (1) The hypothesis holds on every Hall subgroup of E and every quotient G/X where X is a Hall subgroup of E which is normal in G. (2) If X is a non-identity normal Hall subgroup of E, then X = E. Since X is a characteristic subgroup of E, it is normal in G and by (1) the hypothesis is still true for (G/X, E/X). Hence G/X F by the choice of G. Thus the hypothesis is still true for (G, X) and so X = E by the choice of (G, E). (3) E = P. By (2) and Theorem A, it is easy to see that E = P. (4) O p (G) = 1. If O p (G) 1, then by Lemma 2.4, we have that (G/O p (G), P O p (G)/O p (G)) satisfies the hypothesis and by induction G/O p (G) F. Then G G/P G/O p (G) F, a contradiction. Hence O p (G) = 1. (5) every cyclic subgroup of P with prime prime order and order 4 (if P is a non-abelian 2-group and H Z (G)) is ss-embedded in G. Suppose that some maximal subgroup of P is s-embedded in G. Let N be a minimal normal subgroup of G contained in P. Then by Lemma 2.4, the hypothesis holds on (G/N, P/N) and so G/N F by the choice of (G, E). This implies that N is the only minimal normal subgroup of G contained in P and N Φ(G). Let M be a maximal subgroup of G such that G = [N]M. Then P = P NM = N(P M). Since P F (G) C G (N), P M is normal in G and hence P M = 1. It follows that N = P = G F. If O p (G) < G, then G/O p (G) F and so P = G F O p (G). It follows from (3) and the minimality of G E that O p (G) F. Let q be the largest prime dividing O p (G). Then Q is a character subgroup of O p (G) and so it is normal in G, which contradict to (4). So O p (G) = G. Now by Lemma 2.5, every maximal subgroup of P is n-embedded in G. So by [1, Theorem D], G F, a contradiction. Hence every cyclic subgroup of P with prime prime order and order 4 (if P is a non-abelian 2-group and H Z (G)) is ss-embedded in G. (6) O p (G) = G. Assume O p (G) < G. Let M be a normal maximal subgroup of G containing

100 Xinjian Zhang and Yong Xu O p (G). Then either P M or G = P M. If P M, then obviously (M, P ) satisfies the hypothesis of the theorem and by induction M F. Now assume G = P M. Then P M is maximal in P and P M is a Sylow p-subgroup of M. Notice that M/P M = G/P F. By Lemma 2.4 and (5), the hypothesis holds on (M, P M) and by the choice of (G, E), we have M F. Let q be the largest prime divisor of M and Q be a Sylow q-subgroup of M. The Q is is a character subgroup of M and so Q is normal in G, which contradicts (4). So O p (G) = G. (7) G F. By (3), (5) and Lemma 2.5, every cyclic subgroup of P with prime prime order and order 4 (if P is a non-abelian 2-group and H Z (G)) is n- embedded ing. Now by [1, Theorem D], G F. 4 Some applications of the results Theorems A and B have many corollaries. In particular, in the literature one can find the following special cases of these theorems. Corollary 4.1. ([7]) Let G be a group of odd order. If all subgroups of G of prime order are normal in G, then G is supersolvable. Corollary 4.2. ([8]) Let G be a finite group. Suppose P 1 is c-normal in G for every Sylow subgroup P of G and every maximal subgroups P 1 of P. Then G is supersolvable. Corollary 4.3. ([9]) If every subgroup of G of prime order and cyclic subgroup of order 4 are s-permutable in G, then G is supersolvable. Corollary 4.4. ([10]) If every maximal subgroup of every Sylow subgroup of G is s-permutable in G, then G is supersolvable. Corollary 4.5. ([11]) Let F be a saturated formation containing all supersovable groups. If all minimal subgroups and all cyclic subgroups with order 4 of G F are c-normal in G, then G F. Corollary 4.6. ([12]) Let F be a saturated formation containing all supersovable groups and G a group with normal subgroup E such that G/E F. Assume that a Sylow 2-subgroup of G is abelian. If all minimal subgroups of E are permutable in G, then G F. Acknowledgements. This work was supported by National Natural Science Foundation of China (Grant No. 11501235, 11601225, 11171243) and Natural Science Foundation of Jiangsu Province(No. BK20140451).

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