Access the full text.

Sign up today, get DeepDyve free for 14 days.

Annals of the Alexandru Ioan Cuza University - Mathematics
, Volume 60 (2) – Nov 24, 2014

/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

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.

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

**Published: ** Nov 24, 2014

Loading...

You can share this free article with as many people as you like with the url below! We hope you enjoy this feature!

Read and print from thousands of top scholarly journals.

System error. Please try again!

Already have an account? Log in

Bookmark this article. You can see your Bookmarks on your DeepDyve Library.

To save an article, **log in** first, or **sign up** for a DeepDyve account if you don’t already have one.

Copy and paste the desired citation format or use the link below to download a file formatted for EndNote

Access the full text.

Sign up today, get DeepDyve free for 14 days.

All DeepDyve websites use cookies to improve your online experience. They were placed on your computer when you launched this website. You can change your cookie settings through your browser.