Get 20M+ Full-Text Papers For Less Than $1.50/day. Start a 14-Day Trial for You or Your Team.

Learn More →

Finite Groups with Some Primary Subgroups Weakly τ-Quasinormal

Finite Groups with Some Primary Subgroups Weakly τ-Quasinormal A subgroup H of a group G is said to be in G if G has a subnormal subgroup T of G such that G = HT and H T H G , where H G is the subgroup generated by all those subgroups of H which are -quasinormal in G. In this paper we investigate the influence of subgroups on the p-nilpotency and p-supersolvability of finite groups. Mathematics Subject Classification 2010: 20D10, 20D15, 20D20. Key words: , p-nilpotent, p-supersolvable. 1. Introduction Throughout this paper, all groups considered are finite. The notation and terminology are standard, as in [3] and [9]. Two subgroups H and K of G are said to be permutable if HK = KH. A subgroup H of G is said to be S-permutable (or S-quasinormal) in G if H permutes with all Sylow subgroups of G (see [4]). In recent years, it has been of interest to use some supplemented properties of subgroups to determine the structure of a group. For example, Wang in 1996 introduced the c-normality of a group and gave some new criteria for the solvability and supersolvability of groups (see [13]). In the paper [11], Skiba introduced the concept of weakly S-permutable subgroup which covers both S-permutability and c-normality. Within the framework of formation theory, Skiba provided a unified viewpoint for The project is supported by the Natural Science Foundation of China (No:11101369) and the Priority Academic Program Development of Jiangsu Higher Education Institutions. a series of similar problems about supersolvable groups. More recently, Lukyanenko and Skiba (see [8]) generalized weakly S-permutability to the following ity. Definition 1. Let H be a subgroup of a group G. Then we say that: (1) H is -quasinormal in G if H permutes with every Sylow subgroup Q of G such that (|H|, |Q|) = 1 and (|H|, |QG |) = 1; (2) H is in G if G has a subnormal subgroup T of G such that G = HT and H T H G , where H G is the subgroup generated by all those subgroups of H which are -quasinormal in G. Let G be a group. A primary subgroup H is a subgroup of prime power order of G. The property of prime subgroups has been studied extensively by many scholars in determining the structure of finite groups; see, for example, [1], [5], [8], [11], [12], etc. In the present paper, we continue this work and characterize p-nilpotency and p-supersolvability of finite groups with the assumption that some primary subgroups are . 2. Preliminaries Lemma 1 ([8, Lemma 2.4]). Let G be a group, H K G and p be a prime. (1) If H is -quasinormal in G, then H is in G. (2) Suppose that K is a p-group and H is normal in G. If K is in G, then K/H is in G/H. (3) If H is in G, then H is in K. (4) Suppose that H is normal in G. Then EH/H is in G/H for every p-subgroup E in G satisfying (|H|, |E|) = 1. (5) Suppose H is a p-group and H is not -quasinormal in G. Assume that H is in G. Then G has a normal subgroup M such that |G : M | = p and G = HM . Lemma 2. Let G be a group, H K G and p be a prime. (1) If H is -quasinormal in G and H Op (G), then H is S-permutable in G. (2) If H is in G and H Op (G), then H is weakly S-permutable in G. Proof. (1) is [8, Theorem 2.2(4)]. In view of (1), we get (2). Lemma 3 ([6, Lemma 2.6]). Let H be a solvable normal subgroup of a group G (H = 1). If every minimal normal subgroup of G which is contained in H is not contained in (G), then the Fitting subgroup F (H) of H is the direct product of minimal normal subgroups of G which are contained in H. Lemma 4. Suppose that G = P Q, where P is a normal Sylow p-subgroup and Q a Sylow q-subgroup of G. If (|G|, p - 1) = 1 and every maximal subgroup of P is in G, then G is p-nilpotent. Proof. By Lemma 2(2), every maximal subgroup of P is weakly Spermutable in G. By [7, Theorem 3.1], we have that G is p-nilpotent. The following lemma is well known (see, for example, [10]). Lemma 5. Suppose that U is S-permutable in G and N is a normal subgroup of G. Then: (1) U is S-permutable in H whenever U H G. (2) U N is S-permutable in G and U N/N is S-permutable in G/N . (3) U is subnormal in G. (4) If U is a p-subgroup for some prime p, then NG (U ) O p (G). Lemma 6 ([3, IV, 4.7]). If P is a Sylow p-subgroup of a group G and N G such that P N (P ), then N is p-nilpotent. 3. Main results Theorem 1. Let p be a prime dividing the order of a group G and H a p-solvable normal subgroup of G such that G/H is p-supersolvable. If all maximal subgroups of Fp (H) containing Op (H) are in G, then G is p-supersolvable. Proof. Suppose that the theorem is false and let G be a counterexample of minimal order. We will derive a contradiction in several steps. (1) Op (H) = 1. If T = Op (H) = 1, we consider G = G/T . First, G/H G/H is p= supersolvable, where H = H/T . Now Op (H) = 1 and Fp (H) = Fp (H)/T . Let M/T be a maximal subgroup of Fp (H). Then M is a maximal subgroup of Fp (H) containing T . Since M is in G, by Lemma 1, M/T is in G/T . Thus G satisfies the hypotheses of the theorem. The minimality of G implies that G is p-supersolvable and so is G, a contradiction. (2) Op (G) = 1. If not, consider G = G/R, where R = Op (G). It is easy to see that G/H G/HR is is p-supersolvable, where H = HR/R. Using Step (1) = we have Fp (H) = Op (H) and hence Fp (H) = Fp (H)R/R. Let P1 R/R be a maximal subgroup of Fp (H). We may assume that P1 is a maximal subgroup of Fp (H). By hypothesis, P1 is in G, so that P1 R/R is in G/R by Lemma 1. The choice of G implies that G is p-supersolvable and so G is also p-supersolvable, a contradiction. (3) H (G) = 1. Write L = H (G). If L = 1, we consider G = G/L. By [3, III, 3.5], we have F (H/L) = F (H)/L and therefore F (H/L) = Op (H)/L. On the other hand, writing K/L = Op (H/L) and letting S be a Hall p -subgroup of K we have K = SL, and by the Frattini argument G = KNG (S) = LNG (S) = NG (S) and S G. Therefore S = 1 and Op (H/L) = 1. This implies that Fp (H/L) = Op (H/L) = Op (H)/L = Fp (H)/L. If P1 /L is a maximal subgroup of Fp (H/L), then P1 is maximal in Fp (H) and, by our hypothesis, it is an subgroup of G. Hence P1 /L is in G/L by Lemma 1. Now the minimality of G implies that G is p-supersolvable and then so is G, contrary to the choice of G. (4) All minimal normal subgroups of G contained in Op (H) are cyclic. Since H is p-solvable and Op (H) = 1, we have CH (Op (H)) Op (H) (see [2, Theorem 6.3.2]). Now (H) = 1 implies that F (H) = Op (H) is a non-trivial elementary abelian p-group by [3, III, 4.5]. Thus CH (F (H)) = F (H). Let N be a minimal normal subgroup of G contained in Op (H). Since H (G) = 1, there exists a maximal subgroup M of G such that G = N M and N M = 1. Let Gp be a Sylow p-subgroup of G. Then Gp = Gp N M = N (Gp M ) and Gp M < Gp . Choose a maximal subgroup G1 of Gp containing Gp M and let P1 = G1 Op (H). Obviously, P1 Gp and P1 N = (G1 Op (H))N = G1 N Op (H) = Gp Op (H) = Op (H). 5 Since |Op (H) : P1 | = |Op (H) : G1 Op (H)| = |Op (H)G1 : G1 | = |Gp : G1 | = p, we have P1 is a maximal subgroup of Op (H). By hypothesis, P1 is in G. Hence there exists a subnormal subgroup T of G such that G = P1 T and P1 T (P1 ) G . Since |G : T | is a power of p and T G, we have Op (G) T . If N Op (G), then N Op (G)/O p (G) is minimal p (G) which is a p-group, so that N N O p (G)/O p (G) is normal in G/O = cyclic of order p. Hence we may assume that N Op (G) T . It follows that N P1 = N (P1 ) G . Let Gq be any Sylow q-subgroup of G with p = q. If (|(P1 ) G |, |(Gq )G |) = 1, then (Gq )G Op (G) = 1, a contradiction. Thus (P1 ) G Gq = Gq (P1 ) G . Since N P1 = N (P1 ) G = N (P1 ) G Gq , we have N P1 is normalized by Gq . On the other hand, N P1 Gp . Therefore, N P1 is normal in G. The minimality of N implies that N P1 = 1 or N P1 = N . If N P1 = N , then N P1 and so Op (H) = N P1 = P1 , a contradiction. Hence we have N P1 = 1. Since |N : P1 N | = |N P1 : P1 | = |Op (H) : P1 | = p, we have that P1 N is a maximal subgroup of N and so |N | = p. (5) Finial contradiction. By Lemma 3, we have F (H) = N1 × N2 × · · · × Nr , where Ni is minimal normal in G of order p. For each i the quotient group G/CG (Ni ) is a subgroup of Aut(Ni ) and so is abelian. Since G/H is p-supersolvable, it follows that G/(H CG (Ni )) = G/CH (Ni ) is p-supersolvable. Therefore G/ r CH (Ni ) is p-supersolvable, and thus G/F (H) is p-supersolvable i=1 because CH (Ni ) = CH (F (H)) = F (H). i=1 But all chief factors of G below F (H) are cyclic of order p and hence G is p-supersolvable, a contradiction. Theorem 2. Let p be a prime, G a p-solvable group and H a normal subgroup of G such that G/H is p-supersolvable. If there exists a Sylow p-subgroup P of H such that every maximal subgroup of P is weakly quasinormal in G, then G is p-supersolvable. Proof. Suppose that the theorem is false and let G be a counterexample of minimal order. (1) G has a unique minimal normal subgroup N contained in H such that G/N is p-supersolvable. Let N be a minimal normal subgroup of G contained in H. Since P is the Sylow p-subgroup of H, P N/N is the Sylow p-subgroup of H/N . Let M/N be a maximal subgroup of P N/N , then M = (M P )N . Let P1 = M P . Obviously, P1 is the maximal subgroup of P . Since G is psolvable, N is elementary abelian p-group or p -group. If N is p -group, then M/N = P1 N/N . If N is p-group, then M/N = P1 /N . By hypothesis, P1 is in G and so M/N is in G/N by Lemma 1. Since (G/N )/(H/N ) G/H is p-supersolvable, G/N satisfies = all the hypotheses of our theorem. It follows that G/N is p-supersolvable by the minimality of G. Clearly, N is the unique minimal normal subgroup of G contained in H as the class of p-supersolvable group is a saturated formation. (2) Op (G) = 1. If T = Op (G) = 1, we consider G = G/T . Clearly, G/H G/HT = = (G/H)/(HT /H) is p-supersolvable by the p-supersolvablilty of G/H, where H = HT /T . Let P1 = P1 T /T be a maximal subgroup of P T /T . We may assume that P1 is a maximal subgroup of P . Since P1 is weakly quasinormal in G, the subgroup P1 T /T is in G/T by Lemma 1. The minimality of G yields that G is p-supersolvable, and so G is also p-supersolvable, a contradiction. (3) Final contradiction. Since G is p-solvable, N is elementary abelian p-group by Step (2). If N is contained in all maximal subgroups of G, then N (G) and so G is p-supersolvable, a contradiction. Hence there exists a maximal subgroup M of G such that G = N M and N M = 1. Let Gp be a Sylow p-subgroup of G containing P . Then Gp = N (Gp M ) and Gp M < Gp . Take a maximal subgroup G1 of G containing Gp M and set P1 = G1 P . Then |P : P1 | = |P : G1 P | = |P G1 : G1 | = |Gp : G1 | = p and so P1 is a maximal subgroup of P . By hypothesis, P1 is weakly quasinormal in G. Then there is a subnormal subgroup T of G such that G = P1 T and P1 T (P1 ) G . Since |G : T | is a power of p and T G, O p (G) T . We know G/O p (G) is p-subgroup, so G/O p (G) is psupersolvable and G/(N Op (G)) G/N × G/O p (G) is p-supersolvable. Then N O p (G) = 1. Since N is the minimal subgroup of G, N O p (G) = N and so N O p (G). It follows that N P1 = N (P1 ) G . Let Gq be any Sylow q-subgroup of G with p = q. By Step (1), (P1 ) G Gq = Gq (P1 ) G . Since N P1 = N (P1 ) G = N (P1 ) G Gq , we have N P1 is normalized by Gq . Since P = Gp H, we have P Gp and so P1 Gp . It follows that N P1 is normalized by Gp . Therefore, N P1 is normal in G. The minimality of N implies that N P1 = 1 or N P1 = N . If N P1 = N , then N P1 and so P = P Gp = P N G1 = N (G1 P ) = N P1 = P1 , a contradiction. Hence we have N P1 = 1. Since |N : P1 N | = |N P1 : P1 | = |P : P1 | = p, P1 N is a maximal of N . Therefore |N | = p, and so G is p-supersolvable by Step (1), a contradiction. Theorem 3. Let P be a Sylow p-subgroup of a group G, where p is a prime divisor of |G| with (|G|, p - 1) = 1. If every maximal subgroup of P is in NG (P ) and P is S-permutable in G, then G is p-nilpotent. Proof. Suppose that the theorem is false and let G be a counterexample of minimal order. Then we have: (1) If H is a proper subgroup of G with P H < G, then H is pnilpotent. Obviously, (|H|, p - 1) = 1 and P is a Sylow p-subgroup of H. Since P is S-permutable in G and P H, we have P is S-permutable in H by Lemma 5. By hypothesis and Lemma 1, every maximal subgroup of P is in NH (P ) since NH (P ) = H NG (P ) NG (P ). Hence H satisfies the hypotheses of our theorem. Now, by the minimality of G, H is p-nilpotent. (2) 1 = P Op (G). Let Q be a Sylow q-subgroup of NG (P ), where q is a prime dividing |NG (P )| with p = q. Obviously, P P Q. If P Q = G, then G is p-nilpotent by Lemma 4, a contradiction. Thus we may assume that P Q < G. By Step (1), P Q is p-nilpotent, and so Q P Q. It follows that P Q = P × Q. Hence Q CG (P ) and so all p -elements of NG (P ) are contained in CG (P ). If P is abelian, then NG (P ) = CG (P ), which implies that G is p-nilpotent by Burnside Theorem (see [9, Theorem 10.1.8]), a contradiction. So we may assume that P = 1. Since P is S-permutable in G, we have P G and so P Op (G) = 1. (3) G is solvable. Let L be a minimal normal subgroup of G contained in Op (G). Obviously, P/L is a Sylow p-subgroup of G/L. Let P1 /L be a maximal subgroup of P/L. Then P1 is a maximal subgroup of P . Since P1 is in NG (P ), we have P1 /L is in NG (P )/L = NG/L (P/L) by Lemma 1. By hypothesis and Lemma 5, (P/L) = P L/L is S-permutable in G/L. It is clear that (|G/L|, p - 1) = 1. Therefore G/L satisfies the hypotheses of the theorem. The minimal choice of G implies that G/L is p-nilpotent. If p = 2, G/L is solvable and so G is solvable. If p > 2, then G is odd and so G is also solvable by the Feit-Thompson Theorem. (4) |G| = pa q b for some prime q = p. Since G is solvable by Step (3), there exists a Sylow system {P1 = P, P2 , · · ·, Ps } of G with Gi = P Pi for 2 i s. By Lemmas 2.1 and 2.5, the hypothesis still hold for each Gi . If |(G)| > 2, then Gi < G and so Gi is p-nilpotent by the minimal choice of G. It follows that Pi Gi and P normalizes Pi for each 2 i s. Hence G is p-nilpotent, a contradiction. Thus we may assume that |G| = pa q b . (5) Oq (G) = 1. Assume that Oq (G) = 1. Now write Oq (G) = N and consider the factor group G/N . Obviously, P N/N is a Sylow p-subgroup of G/N . Let M/L be a maximal subgroup of P N/N . Then M = P1 N , where P1 is a maximal subgroup of P . By hypothesis, P1 is in NG (P ). Thus there is a subnormal subgroup T of NG (P ) such that NG (P ) = P1 T and P1 T (P1 ) NG (P ) . Since P is a Sylow p-subgroup of G, we have NG/N (P N/N ) = NG (P )N/N = (P1 N/N )(T N/N ). Since (|P1 |, |N |) = 1, we have P1 T N = P1 T and so (P1 N/N ) (T N/N ) = (P1 N T N )/N = (P1 T N )N/N = (P1 T )N/N (P1 ) NG (P ) N/N. By [8, Lemma 2.2(3)], we know that (P1 ) NG (P ) N/N (P1 N/N ) NG/N (P N/N ) . Hence every maximal subgroup of P N/N is in NG/N (P N/N ). It follows that G/N satisfies the hypotheses of the theorem. The minimal choice of G implies that G/N is p-nilpotent, and so G is p-nilpotent, a contradiction. (6) Final contradiction. Let N be a minimal normal subgroup of G. By Steps (3) and (5), N is a p-group and G/N is p-nilpotent. Since the class of all p-nilpotent groups is a saturated formation, we have that N is the unique minimal normal subgroup of G and (G) = 1. It follows that Op (G) = F (G) = N . Since P is S-permutable in G, we have NG (P ) Op (G) by Lemma 5(4). Since P normalizes P , we have P G. Then P = Op (G) = N by Step (2) and the minimality of N . Since G/Op (G) is p-nilpotent, we have Op (G)Q G, where Q is a Sylow q-subgroup of G. Since Op (G)Q P = Op (G) = P (P ), Op (G)Q is p-nilpotent by Lemma 6. Thus Q char Op (G)Q G which implies that Q G, i.e., G is p-nilpotent, a contradiction. Theorem 4. Let p be an odd prime dividing |G| and P a Sylow psubgroup of G. If every maximal subgroup of P is in G and NG (P ) is p-nilpotent, then G is p-nilpotent. Proof. Suppose that the theorem is false and let G be a counterexample of minimal order. We will derive a contradiction in several steps. (1) If H is a proper subgroup of G with P H < G, then H is pnilpotent. It is clear to see NH (P ) NG (P ) and hence NH (P ) is p-nilpotent. Applying Lemma 1, we immediately see that H satisfies the hypotheses of our theorem. The minimal choice of G implies that H is p-nilpotent. (2) Op (G) = 1. If Op (G) = 1, we consider G/Op (G). By Lemma 1, it is easy to see that every maximal subgroup of P Op (G)/Op (G) is in G/Op (G). Since NG/O p (G) (P Op (G)/Op (G)) = NG (P )Op (G)/Op (G) is p-nilpotent, G/Op (G) satisfies all the hypotheses of our theorem. The minimality of G yields that G/Op (G) is p-nilpotent, and so G is p-nilpotent, a contradiction. (3) Op (G) = 1 and G = P Q, where Q is a Sylow q-subgroup of G with p = q. Let J(P ) be the Thompson subgroup of P . Then NG (P ) NG (Z(J(P ))) G and every maximal subgroup of P is in NG (Z(J(P ))) by Lemma 1. If NG (Z(J(P ))) < G, then, by Step (1), NG (Z(J(P ))) is p-nilpotent and so G is p-nilpotent by [2, Theorem 8.3.1], a contradiction. Therefore Z(J(P )) G, and this leads to 1 = Op (G) < P . Consider G = G/Op (G) and let G1 be the inverse imagine of NG (Z(J(P ))) in G. Since Op (G) is the largest normal subgroup of G contained in P , we have NG (P ) G1 < G. By Step (1), G1 is p-nilpotent and by [11, Theorem 8.3.1] again, G is p-nilpotent. Then there exists a Sylow q-subgroup Q of G such that P Q is a subgroup of G for any q (G) with q = p by [11, Theorem 6.3.5]. If P Q < G, then P Q is p-nilpotent by Step (1). Hence Q CG (Op (G)) Op (G) by [11, Theorem 6.3.2], a contradiction. Thus P Q = G. (4) Final contradiction. Let N be a minimal normal subgroup of G. From Steps (2) and (3), we have that N Op (G). It is easy to see that G/N satisfies the hypotheses. Hence G/N is p-nilpotent by the choice of G. Since the class of all pnilpotent groups is a saturated formation, N is the unique minimal normal subgroup of G and N (G). Then there exists a maximal subgroup M of G such that G = M N and M N = 1. Since N is elementary abelian p-group, N CG (N ) and CG (N ) M G. By the uniqueness of N , we have CG (N ) M = 1 and N = CG (N ). But N Op (G) F (G) CG (N ), hence N = Op (G) = CG (N ). Let P be a Sylow p-subgroup of M . Then P = N P . If P = N , then NG (P ) = NG (N ) = G is p- nilpotent, a contradiction. Thus P = N . Let P1 is a maximal subgroup of P such that P P1 and P = N P1 . By our hypotheses, P1 is in G. Then there is a subnormal subgroup T of G such that G = P1 T and P1 T (P1 ) G . Since |G : T | is a power of p and T G, we have Op (G) T . Since N is the unique minimal normal subgroup of G, N Op (G). It follows that P1 N = (P1 ) G N . For any Sylow q-subgroup Gq of G (p = q), (P1 ) G Gq = Gq (P1 ) G by Step (2). Since N P1 = N (P1 ) G = N (P1 ) G Gq , we have N P1 is normalized by Gq . Obviously, P1 N P . Therefore P1 N is normal in G. By the minimality of N , we have P1 N = N or P1 N = 1. If P1 N = N , then N P1 and P = N P1 = P1 , a contradiction. Thus P1 N = 1. Since P1 N is a maximal subgroup of N , we have that N is of order p, and so Aut(N ) is a cyclic group of order p - 1. If p < q, then N Q is p-nilpotent by [9, Theorem 10.1.9] and so Q CG (N ) = CG (Op (G)), which contradicts CG (Op (G)) Op (G). Thus we may assume that q < p. Since N = CG (N ), we see that M G/N = NG (N )/CG (N ) is isomorphic = with a subgroup of Aut(N ). Then M and Q are cyclic groups. By using [9, Theorem 10.1.9] again, G is q-nilpotent and so P is normal in G. This implies NG (P ) = G is p-nilpotent, a contradiction. Acknowledgement. The authors are grateful to the referee for his valuable comments on the paper. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Annals of the Alexandru Ioan Cuza University - Mathematics de Gruyter

Finite Groups with Some Primary Subgroups Weakly τ-Quasinormal

Download PDF
12 pages

Loading next page...
 
/lp/de-gruyter/finite-groups-with-some-primary-subgroups-weakly-quasinormal-bttnT63HZ0
Publisher
de Gruyter
Copyright
Copyright © 2014 by the
ISSN
1221-8421
eISSN
1221-8421
DOI
10.2478/aicu-2013-0044
Publisher site
See Article on Publisher Site

Abstract

A subgroup H of a group G is said to be in G if G has a subnormal subgroup T of G such that G = HT and H T H G , where H G is the subgroup generated by all those subgroups of H which are -quasinormal in G. In this paper we investigate the influence of subgroups on the p-nilpotency and p-supersolvability of finite groups. Mathematics Subject Classification 2010: 20D10, 20D15, 20D20. Key words: , p-nilpotent, p-supersolvable. 1. Introduction Throughout this paper, all groups considered are finite. The notation and terminology are standard, as in [3] and [9]. Two subgroups H and K of G are said to be permutable if HK = KH. A subgroup H of G is said to be S-permutable (or S-quasinormal) in G if H permutes with all Sylow subgroups of G (see [4]). In recent years, it has been of interest to use some supplemented properties of subgroups to determine the structure of a group. For example, Wang in 1996 introduced the c-normality of a group and gave some new criteria for the solvability and supersolvability of groups (see [13]). In the paper [11], Skiba introduced the concept of weakly S-permutable subgroup which covers both S-permutability and c-normality. Within the framework of formation theory, Skiba provided a unified viewpoint for The project is supported by the Natural Science Foundation of China (No:11101369) and the Priority Academic Program Development of Jiangsu Higher Education Institutions. a series of similar problems about supersolvable groups. More recently, Lukyanenko and Skiba (see [8]) generalized weakly S-permutability to the following ity. Definition 1. Let H be a subgroup of a group G. Then we say that: (1) H is -quasinormal in G if H permutes with every Sylow subgroup Q of G such that (|H|, |Q|) = 1 and (|H|, |QG |) = 1; (2) H is in G if G has a subnormal subgroup T of G such that G = HT and H T H G , where H G is the subgroup generated by all those subgroups of H which are -quasinormal in G. Let G be a group. A primary subgroup H is a subgroup of prime power order of G. The property of prime subgroups has been studied extensively by many scholars in determining the structure of finite groups; see, for example, [1], [5], [8], [11], [12], etc. In the present paper, we continue this work and characterize p-nilpotency and p-supersolvability of finite groups with the assumption that some primary subgroups are . 2. Preliminaries Lemma 1 ([8, Lemma 2.4]). Let G be a group, H K G and p be a prime. (1) If H is -quasinormal in G, then H is in G. (2) Suppose that K is a p-group and H is normal in G. If K is in G, then K/H is in G/H. (3) If H is in G, then H is in K. (4) Suppose that H is normal in G. Then EH/H is in G/H for every p-subgroup E in G satisfying (|H|, |E|) = 1. (5) Suppose H is a p-group and H is not -quasinormal in G. Assume that H is in G. Then G has a normal subgroup M such that |G : M | = p and G = HM . Lemma 2. Let G be a group, H K G and p be a prime. (1) If H is -quasinormal in G and H Op (G), then H is S-permutable in G. (2) If H is in G and H Op (G), then H is weakly S-permutable in G. Proof. (1) is [8, Theorem 2.2(4)]. In view of (1), we get (2). Lemma 3 ([6, Lemma 2.6]). Let H be a solvable normal subgroup of a group G (H = 1). If every minimal normal subgroup of G which is contained in H is not contained in (G), then the Fitting subgroup F (H) of H is the direct product of minimal normal subgroups of G which are contained in H. Lemma 4. Suppose that G = P Q, where P is a normal Sylow p-subgroup and Q a Sylow q-subgroup of G. If (|G|, p - 1) = 1 and every maximal subgroup of P is in G, then G is p-nilpotent. Proof. By Lemma 2(2), every maximal subgroup of P is weakly Spermutable in G. By [7, Theorem 3.1], we have that G is p-nilpotent. The following lemma is well known (see, for example, [10]). Lemma 5. Suppose that U is S-permutable in G and N is a normal subgroup of G. Then: (1) U is S-permutable in H whenever U H G. (2) U N is S-permutable in G and U N/N is S-permutable in G/N . (3) U is subnormal in G. (4) If U is a p-subgroup for some prime p, then NG (U ) O p (G). Lemma 6 ([3, IV, 4.7]). If P is a Sylow p-subgroup of a group G and N G such that P N (P ), then N is p-nilpotent. 3. Main results Theorem 1. Let p be a prime dividing the order of a group G and H a p-solvable normal subgroup of G such that G/H is p-supersolvable. If all maximal subgroups of Fp (H) containing Op (H) are in G, then G is p-supersolvable. Proof. Suppose that the theorem is false and let G be a counterexample of minimal order. We will derive a contradiction in several steps. (1) Op (H) = 1. If T = Op (H) = 1, we consider G = G/T . First, G/H G/H is p= supersolvable, where H = H/T . Now Op (H) = 1 and Fp (H) = Fp (H)/T . Let M/T be a maximal subgroup of Fp (H). Then M is a maximal subgroup of Fp (H) containing T . Since M is in G, by Lemma 1, M/T is in G/T . Thus G satisfies the hypotheses of the theorem. The minimality of G implies that G is p-supersolvable and so is G, a contradiction. (2) Op (G) = 1. If not, consider G = G/R, where R = Op (G). It is easy to see that G/H G/HR is is p-supersolvable, where H = HR/R. Using Step (1) = we have Fp (H) = Op (H) and hence Fp (H) = Fp (H)R/R. Let P1 R/R be a maximal subgroup of Fp (H). We may assume that P1 is a maximal subgroup of Fp (H). By hypothesis, P1 is in G, so that P1 R/R is in G/R by Lemma 1. The choice of G implies that G is p-supersolvable and so G is also p-supersolvable, a contradiction. (3) H (G) = 1. Write L = H (G). If L = 1, we consider G = G/L. By [3, III, 3.5], we have F (H/L) = F (H)/L and therefore F (H/L) = Op (H)/L. On the other hand, writing K/L = Op (H/L) and letting S be a Hall p -subgroup of K we have K = SL, and by the Frattini argument G = KNG (S) = LNG (S) = NG (S) and S G. Therefore S = 1 and Op (H/L) = 1. This implies that Fp (H/L) = Op (H/L) = Op (H)/L = Fp (H)/L. If P1 /L is a maximal subgroup of Fp (H/L), then P1 is maximal in Fp (H) and, by our hypothesis, it is an subgroup of G. Hence P1 /L is in G/L by Lemma 1. Now the minimality of G implies that G is p-supersolvable and then so is G, contrary to the choice of G. (4) All minimal normal subgroups of G contained in Op (H) are cyclic. Since H is p-solvable and Op (H) = 1, we have CH (Op (H)) Op (H) (see [2, Theorem 6.3.2]). Now (H) = 1 implies that F (H) = Op (H) is a non-trivial elementary abelian p-group by [3, III, 4.5]. Thus CH (F (H)) = F (H). Let N be a minimal normal subgroup of G contained in Op (H). Since H (G) = 1, there exists a maximal subgroup M of G such that G = N M and N M = 1. Let Gp be a Sylow p-subgroup of G. Then Gp = Gp N M = N (Gp M ) and Gp M < Gp . Choose a maximal subgroup G1 of Gp containing Gp M and let P1 = G1 Op (H). Obviously, P1 Gp and P1 N = (G1 Op (H))N = G1 N Op (H) = Gp Op (H) = Op (H). 5 Since |Op (H) : P1 | = |Op (H) : G1 Op (H)| = |Op (H)G1 : G1 | = |Gp : G1 | = p, we have P1 is a maximal subgroup of Op (H). By hypothesis, P1 is in G. Hence there exists a subnormal subgroup T of G such that G = P1 T and P1 T (P1 ) G . Since |G : T | is a power of p and T G, we have Op (G) T . If N Op (G), then N Op (G)/O p (G) is minimal p (G) which is a p-group, so that N N O p (G)/O p (G) is normal in G/O = cyclic of order p. Hence we may assume that N Op (G) T . It follows that N P1 = N (P1 ) G . Let Gq be any Sylow q-subgroup of G with p = q. If (|(P1 ) G |, |(Gq )G |) = 1, then (Gq )G Op (G) = 1, a contradiction. Thus (P1 ) G Gq = Gq (P1 ) G . Since N P1 = N (P1 ) G = N (P1 ) G Gq , we have N P1 is normalized by Gq . On the other hand, N P1 Gp . Therefore, N P1 is normal in G. The minimality of N implies that N P1 = 1 or N P1 = N . If N P1 = N , then N P1 and so Op (H) = N P1 = P1 , a contradiction. Hence we have N P1 = 1. Since |N : P1 N | = |N P1 : P1 | = |Op (H) : P1 | = p, we have that P1 N is a maximal subgroup of N and so |N | = p. (5) Finial contradiction. By Lemma 3, we have F (H) = N1 × N2 × · · · × Nr , where Ni is minimal normal in G of order p. For each i the quotient group G/CG (Ni ) is a subgroup of Aut(Ni ) and so is abelian. Since G/H is p-supersolvable, it follows that G/(H CG (Ni )) = G/CH (Ni ) is p-supersolvable. Therefore G/ r CH (Ni ) is p-supersolvable, and thus G/F (H) is p-supersolvable i=1 because CH (Ni ) = CH (F (H)) = F (H). i=1 But all chief factors of G below F (H) are cyclic of order p and hence G is p-supersolvable, a contradiction. Theorem 2. Let p be a prime, G a p-solvable group and H a normal subgroup of G such that G/H is p-supersolvable. If there exists a Sylow p-subgroup P of H such that every maximal subgroup of P is weakly quasinormal in G, then G is p-supersolvable. Proof. Suppose that the theorem is false and let G be a counterexample of minimal order. (1) G has a unique minimal normal subgroup N contained in H such that G/N is p-supersolvable. Let N be a minimal normal subgroup of G contained in H. Since P is the Sylow p-subgroup of H, P N/N is the Sylow p-subgroup of H/N . Let M/N be a maximal subgroup of P N/N , then M = (M P )N . Let P1 = M P . Obviously, P1 is the maximal subgroup of P . Since G is psolvable, N is elementary abelian p-group or p -group. If N is p -group, then M/N = P1 N/N . If N is p-group, then M/N = P1 /N . By hypothesis, P1 is in G and so M/N is in G/N by Lemma 1. Since (G/N )/(H/N ) G/H is p-supersolvable, G/N satisfies = all the hypotheses of our theorem. It follows that G/N is p-supersolvable by the minimality of G. Clearly, N is the unique minimal normal subgroup of G contained in H as the class of p-supersolvable group is a saturated formation. (2) Op (G) = 1. If T = Op (G) = 1, we consider G = G/T . Clearly, G/H G/HT = = (G/H)/(HT /H) is p-supersolvable by the p-supersolvablilty of G/H, where H = HT /T . Let P1 = P1 T /T be a maximal subgroup of P T /T . We may assume that P1 is a maximal subgroup of P . Since P1 is weakly quasinormal in G, the subgroup P1 T /T is in G/T by Lemma 1. The minimality of G yields that G is p-supersolvable, and so G is also p-supersolvable, a contradiction. (3) Final contradiction. Since G is p-solvable, N is elementary abelian p-group by Step (2). If N is contained in all maximal subgroups of G, then N (G) and so G is p-supersolvable, a contradiction. Hence there exists a maximal subgroup M of G such that G = N M and N M = 1. Let Gp be a Sylow p-subgroup of G containing P . Then Gp = N (Gp M ) and Gp M < Gp . Take a maximal subgroup G1 of G containing Gp M and set P1 = G1 P . Then |P : P1 | = |P : G1 P | = |P G1 : G1 | = |Gp : G1 | = p and so P1 is a maximal subgroup of P . By hypothesis, P1 is weakly quasinormal in G. Then there is a subnormal subgroup T of G such that G = P1 T and P1 T (P1 ) G . Since |G : T | is a power of p and T G, O p (G) T . We know G/O p (G) is p-subgroup, so G/O p (G) is psupersolvable and G/(N Op (G)) G/N × G/O p (G) is p-supersolvable. Then N O p (G) = 1. Since N is the minimal subgroup of G, N O p (G) = N and so N O p (G). It follows that N P1 = N (P1 ) G . Let Gq be any Sylow q-subgroup of G with p = q. By Step (1), (P1 ) G Gq = Gq (P1 ) G . Since N P1 = N (P1 ) G = N (P1 ) G Gq , we have N P1 is normalized by Gq . Since P = Gp H, we have P Gp and so P1 Gp . It follows that N P1 is normalized by Gp . Therefore, N P1 is normal in G. The minimality of N implies that N P1 = 1 or N P1 = N . If N P1 = N , then N P1 and so P = P Gp = P N G1 = N (G1 P ) = N P1 = P1 , a contradiction. Hence we have N P1 = 1. Since |N : P1 N | = |N P1 : P1 | = |P : P1 | = p, P1 N is a maximal of N . Therefore |N | = p, and so G is p-supersolvable by Step (1), a contradiction. Theorem 3. Let P be a Sylow p-subgroup of a group G, where p is a prime divisor of |G| with (|G|, p - 1) = 1. If every maximal subgroup of P is in NG (P ) and P is S-permutable in G, then G is p-nilpotent. Proof. Suppose that the theorem is false and let G be a counterexample of minimal order. Then we have: (1) If H is a proper subgroup of G with P H < G, then H is pnilpotent. Obviously, (|H|, p - 1) = 1 and P is a Sylow p-subgroup of H. Since P is S-permutable in G and P H, we have P is S-permutable in H by Lemma 5. By hypothesis and Lemma 1, every maximal subgroup of P is in NH (P ) since NH (P ) = H NG (P ) NG (P ). Hence H satisfies the hypotheses of our theorem. Now, by the minimality of G, H is p-nilpotent. (2) 1 = P Op (G). Let Q be a Sylow q-subgroup of NG (P ), where q is a prime dividing |NG (P )| with p = q. Obviously, P P Q. If P Q = G, then G is p-nilpotent by Lemma 4, a contradiction. Thus we may assume that P Q < G. By Step (1), P Q is p-nilpotent, and so Q P Q. It follows that P Q = P × Q. Hence Q CG (P ) and so all p -elements of NG (P ) are contained in CG (P ). If P is abelian, then NG (P ) = CG (P ), which implies that G is p-nilpotent by Burnside Theorem (see [9, Theorem 10.1.8]), a contradiction. So we may assume that P = 1. Since P is S-permutable in G, we have P G and so P Op (G) = 1. (3) G is solvable. Let L be a minimal normal subgroup of G contained in Op (G). Obviously, P/L is a Sylow p-subgroup of G/L. Let P1 /L be a maximal subgroup of P/L. Then P1 is a maximal subgroup of P . Since P1 is in NG (P ), we have P1 /L is in NG (P )/L = NG/L (P/L) by Lemma 1. By hypothesis and Lemma 5, (P/L) = P L/L is S-permutable in G/L. It is clear that (|G/L|, p - 1) = 1. Therefore G/L satisfies the hypotheses of the theorem. The minimal choice of G implies that G/L is p-nilpotent. If p = 2, G/L is solvable and so G is solvable. If p > 2, then G is odd and so G is also solvable by the Feit-Thompson Theorem. (4) |G| = pa q b for some prime q = p. Since G is solvable by Step (3), there exists a Sylow system {P1 = P, P2 , · · ·, Ps } of G with Gi = P Pi for 2 i s. By Lemmas 2.1 and 2.5, the hypothesis still hold for each Gi . If |(G)| > 2, then Gi < G and so Gi is p-nilpotent by the minimal choice of G. It follows that Pi Gi and P normalizes Pi for each 2 i s. Hence G is p-nilpotent, a contradiction. Thus we may assume that |G| = pa q b . (5) Oq (G) = 1. Assume that Oq (G) = 1. Now write Oq (G) = N and consider the factor group G/N . Obviously, P N/N is a Sylow p-subgroup of G/N . Let M/L be a maximal subgroup of P N/N . Then M = P1 N , where P1 is a maximal subgroup of P . By hypothesis, P1 is in NG (P ). Thus there is a subnormal subgroup T of NG (P ) such that NG (P ) = P1 T and P1 T (P1 ) NG (P ) . Since P is a Sylow p-subgroup of G, we have NG/N (P N/N ) = NG (P )N/N = (P1 N/N )(T N/N ). Since (|P1 |, |N |) = 1, we have P1 T N = P1 T and so (P1 N/N ) (T N/N ) = (P1 N T N )/N = (P1 T N )N/N = (P1 T )N/N (P1 ) NG (P ) N/N. By [8, Lemma 2.2(3)], we know that (P1 ) NG (P ) N/N (P1 N/N ) NG/N (P N/N ) . Hence every maximal subgroup of P N/N is in NG/N (P N/N ). It follows that G/N satisfies the hypotheses of the theorem. The minimal choice of G implies that G/N is p-nilpotent, and so G is p-nilpotent, a contradiction. (6) Final contradiction. Let N be a minimal normal subgroup of G. By Steps (3) and (5), N is a p-group and G/N is p-nilpotent. Since the class of all p-nilpotent groups is a saturated formation, we have that N is the unique minimal normal subgroup of G and (G) = 1. It follows that Op (G) = F (G) = N . Since P is S-permutable in G, we have NG (P ) Op (G) by Lemma 5(4). Since P normalizes P , we have P G. Then P = Op (G) = N by Step (2) and the minimality of N . Since G/Op (G) is p-nilpotent, we have Op (G)Q G, where Q is a Sylow q-subgroup of G. Since Op (G)Q P = Op (G) = P (P ), Op (G)Q is p-nilpotent by Lemma 6. Thus Q char Op (G)Q G which implies that Q G, i.e., G is p-nilpotent, a contradiction. Theorem 4. Let p be an odd prime dividing |G| and P a Sylow psubgroup of G. If every maximal subgroup of P is in G and NG (P ) is p-nilpotent, then G is p-nilpotent. Proof. Suppose that the theorem is false and let G be a counterexample of minimal order. We will derive a contradiction in several steps. (1) If H is a proper subgroup of G with P H < G, then H is pnilpotent. It is clear to see NH (P ) NG (P ) and hence NH (P ) is p-nilpotent. Applying Lemma 1, we immediately see that H satisfies the hypotheses of our theorem. The minimal choice of G implies that H is p-nilpotent. (2) Op (G) = 1. If Op (G) = 1, we consider G/Op (G). By Lemma 1, it is easy to see that every maximal subgroup of P Op (G)/Op (G) is in G/Op (G). Since NG/O p (G) (P Op (G)/Op (G)) = NG (P )Op (G)/Op (G) is p-nilpotent, G/Op (G) satisfies all the hypotheses of our theorem. The minimality of G yields that G/Op (G) is p-nilpotent, and so G is p-nilpotent, a contradiction. (3) Op (G) = 1 and G = P Q, where Q is a Sylow q-subgroup of G with p = q. Let J(P ) be the Thompson subgroup of P . Then NG (P ) NG (Z(J(P ))) G and every maximal subgroup of P is in NG (Z(J(P ))) by Lemma 1. If NG (Z(J(P ))) < G, then, by Step (1), NG (Z(J(P ))) is p-nilpotent and so G is p-nilpotent by [2, Theorem 8.3.1], a contradiction. Therefore Z(J(P )) G, and this leads to 1 = Op (G) < P . Consider G = G/Op (G) and let G1 be the inverse imagine of NG (Z(J(P ))) in G. Since Op (G) is the largest normal subgroup of G contained in P , we have NG (P ) G1 < G. By Step (1), G1 is p-nilpotent and by [11, Theorem 8.3.1] again, G is p-nilpotent. Then there exists a Sylow q-subgroup Q of G such that P Q is a subgroup of G for any q (G) with q = p by [11, Theorem 6.3.5]. If P Q < G, then P Q is p-nilpotent by Step (1). Hence Q CG (Op (G)) Op (G) by [11, Theorem 6.3.2], a contradiction. Thus P Q = G. (4) Final contradiction. Let N be a minimal normal subgroup of G. From Steps (2) and (3), we have that N Op (G). It is easy to see that G/N satisfies the hypotheses. Hence G/N is p-nilpotent by the choice of G. Since the class of all pnilpotent groups is a saturated formation, N is the unique minimal normal subgroup of G and N (G). Then there exists a maximal subgroup M of G such that G = M N and M N = 1. Since N is elementary abelian p-group, N CG (N ) and CG (N ) M G. By the uniqueness of N , we have CG (N ) M = 1 and N = CG (N ). But N Op (G) F (G) CG (N ), hence N = Op (G) = CG (N ). Let P be a Sylow p-subgroup of M . Then P = N P . If P = N , then NG (P ) = NG (N ) = G is p- nilpotent, a contradiction. Thus P = N . Let P1 is a maximal subgroup of P such that P P1 and P = N P1 . By our hypotheses, P1 is in G. Then there is a subnormal subgroup T of G such that G = P1 T and P1 T (P1 ) G . Since |G : T | is a power of p and T G, we have Op (G) T . Since N is the unique minimal normal subgroup of G, N Op (G). It follows that P1 N = (P1 ) G N . For any Sylow q-subgroup Gq of G (p = q), (P1 ) G Gq = Gq (P1 ) G by Step (2). Since N P1 = N (P1 ) G = N (P1 ) G Gq , we have N P1 is normalized by Gq . Obviously, P1 N P . Therefore P1 N is normal in G. By the minimality of N , we have P1 N = N or P1 N = 1. If P1 N = N , then N P1 and P = N P1 = P1 , a contradiction. Thus P1 N = 1. Since P1 N is a maximal subgroup of N , we have that N is of order p, and so Aut(N ) is a cyclic group of order p - 1. If p < q, then N Q is p-nilpotent by [9, Theorem 10.1.9] and so Q CG (N ) = CG (Op (G)), which contradicts CG (Op (G)) Op (G). Thus we may assume that q < p. Since N = CG (N ), we see that M G/N = NG (N )/CG (N ) is isomorphic = with a subgroup of Aut(N ). Then M and Q are cyclic groups. By using [9, Theorem 10.1.9] again, G is q-nilpotent and so P is normal in G. This implies NG (P ) = G is p-nilpotent, a contradiction. Acknowledgement. The authors are grateful to the referee for his valuable comments on the paper.

Journal

Annals of the Alexandru Ioan Cuza University - Mathematics – de Gruyter

Published: Nov 24, 2014

There are no references for this article.