# The influence of SCAP and S-supplemented subgroups on the p-nilpotency of finite groups.

[section] 1. IntroductionThroughout the paper, all groups are finite. We use conventional notions and notation, as in Huppert [1]. G always denotes a group, [absolute value of G] is the order of G, [O.sub.p](G) is the maximal normal p-subgroup of G, [O.sup.p](G) =< g [member of] G | p [??] o(g) > and [PHI](G) is the Frattini subgroup of G.

Let L/K be a normal factor of a group G. A subgroup H of G is said to cover L/K if HL = HK, and H is said to avoid L/K if H [intersection] L = H [intersection] K. If H covers or avoids every chief factor of G, then H is said to have the cover-avoiding property in G, i.e., H is a CAP-subgroup of G. This conception was first studied by Gaschutz (see [2]) to study the solvable groups, later by Gillam (see [3]) and Tomkinson (see [4]). In Ezquerro (see [5]) gave some characterizations for a group G to be p-supersolvable and supersolvable under the assumption that all maximal subgroups of some Sylow subgroups of G have the cover-avoiding property in G. For example, Ezquerrohas proved: Let G be a group with a normal subgroup H such that G/H is supersolvable. Then G is supersolvable if one of following holds: (1) all maximal subgroups of the Sylow subgroups of H are CAP-subgroups of G; (2) H is solvable and all maximal subgroups of the Sylow subgroups of F(H) are CAP-subgroups of G. Asaad (see [6]) said that it is possible to extend Ezquerro's results with formation theory. Recently, Guo and Shum pushed further this approach and obtained some charaterizations for a solvable group and a p-solvable group based on the assumption that some of its subgroups are CAP-subgroups (see [7]). More recently, in Fan et al. (see [8]) introduced the semi cover-avoiding property, which is the generalization not only of the cover-avoiding property but also of c-normality (see [9]). A subgroup H of a group G is said to have the semi cover-avoiding property in G, i.e., H is an SCAP-subgroup of G, if there exists a chief series of G such that H either covers or avoids every G-chief factor of this series. The results in Guo and Shum (see [7]) and Wang (see [9]) were extended with the requirement that the certain subgroups of G are SCAP-subgroups (see [10, 11]). More recently, many authors invest presented some conditions for a group to be p-nilpotent and supersolvable under the condition that some subgroups of Sylow subgroup are SCAP-subgroups (see [12, 13, 14]).

A subgroup H of a group G is said to be S-quasinormal (or [pi]-quasinormal) in G if H permutes with all Sylow subgroups of G, i.e., HS = SH for any Sylow subgroup S of G. This concept was introduced by Kegel in [15]. As another generalization of S-quasinormal subgroups, A. N. Skiba (see [16]) introduced the following concept: A subgroup H of a group G is called weakly S-supplemented (or S-supplemented) in G if there is a subgroup T of G such that G = HT and H [intersection] T [less than or equal to] [H.sub.sG], where [H.sub.sG] is the subgroup of H generated by all those subgroups of H which are S-quasinormal in G. In fact, this concept is also a generalization of c-supplemented subgroups given in [17]. By using S-supplemented subgroups, many interesting results in finite groups were obtained (see [18, 19, 20]). For example, Skiba proved: Let E be a normal subgroup of a finite group G. Suppose that for every non-cyclic Sylow subgroup P of E, either all maximal subgroups of P or all cyclic subgroups of P of prime order and order 4 are S-supplemented in G. Then each G-chief factor below E is cyclic.

There are examples to show that semi cover-avoiding property and S-supplementation can not imply from one to the other one. In this paper, we will try an attempt to unify the two concepts and establish the structure of groups under the assumption that all maximal subgroups or all minimal subgroups of a Sylow subgroup or are SCAP or S-supplemented subgroups. Our theorems generalize and unify some known results, such as in [11, 13, 26, 27, 28].

[section] 2. Preliminaries

In this section, we list some lemmas which will be useful for the proofs of our main results. Lemma 2.1.[11] (Lemma 2.5 and 2.6) Let H be an SCAP subgroup of a group G.

(1) If H [less than or equal to] L [less than or equal to] G, then H is an SCAP subgroup of L.

(2) If N [??] G and N [less than or equal to] H [less than or equal to] G, then H/N is an SCAP subgroup of G/N.

(3) If H is a [pi]-subgroup and N is a normal [pi]'-subgroup of G, then HN/N is an SCAP subgroup of G/N.

Lemma 2.2. [16] (Lemma 2.10) Let H be an S-supplemented subgroup of a group G.

(1) If H [less than or equal to] L [less than or equal to] G, then H is S-supplemented in L.

(2) If N [??] G and N [less than or equal to] H [less than or equal to] G, then H/N is S-supplemented in G/N.

(3) If H is a [pi]-subgroup and N is a normal [pi]'-subgroup of G, then HN/N is S-supplemented in G/N.

Lemma 2.3. [16] (Lemma 3.1) Let p be a prime dividing the order of the group G with ([absolute value of G], p = 1) = 1 and let P be a p-Sylow subgroup of G. If there is a maximal subgroup [P.sub.1] of P such that [P.sub.1] has the semi cover-avoiding property in G, then G is p-solvable.

Lemma 2.4.[22 (Lemma 2.8) Let M be a maximal subgroup of G and P a normal p-subgroup of G such that G = PM, where p is a prime. Then P [intersection] M is a normal subgroup of G.

Lemma 2.5. [23] (Lemma 2.7) Let G be a group and p a prime dividing [absolute value of G] with ([absolute value of G], p - 1) = 1.

(1) If N is normal in G of order p, then N [less than or equal to] Z(G).

(2) If G has cyclic Sylow p-subgroup, then G is p-nilpotent.

(3) If M [less than or equal to] G and [absolute value of G : M] = p , then M [??] G.

Lemma 2.6. [25] (Lemma 2.6) If P is a S-quasinormal p-subgroup of a group G for some prime p, then [O.sub.p](G) [less than or equal to] [N.sub.G](P).

Lemma 2.7. [24] (Main Theorem) Suppose that G has a Hall [pi]-subgroup where [pi] is a set of odd primes. Then all Hall [pi]-subgroups of G are conjugate.

Lemma 2.8. ([1], IV, 5.4) Suppose that G is a group which is not nilpotent but whose proper subgroups are all nilpotent. Then G is a group which is not nilpotent but whose proper subgroups are all nilpotent.

Lemma 2.9. ([1], III, 5.2) Suppose G is a group which is not p-nilpotent but whose proper subgroups are all p-nilpotent. Then

(a) G has a normal Sylow p-subgroup P for some prime p and G = PQ, where Q is a non-normal cyclic q-subgroup for some prime q [not equal to] p.

(b) P/[PHI](P) is a minimal normal subgroup of G/[PHI](P).

(c) If P is non-abelian and p > 2, then the exponent of P is p; If P is non-abelian and p = 2, then the exponent of P is 4.

(d) If P is abelian, then the exponent of P is p.

(e) Z(G) = [PHI](P) x [PHI](Q).

[section] 3. P-nilpotentcy

Theorem 3.1. Let p be a prime dividing the order of a group G with ([absolute value of G], p - 1) = 1 and H a normal subgroup of G such that G/H is p-nilpotent. If there exists a Sylow p-subgroup P of H such that every maximal subgroup of P is either an SCAP or an S-supplemented subgroup of G, then G is p-nilpotent.

Proof. We distinguish two cases:

Case I. H = G.

Suppose that the theorem is false and let G be a counterexample of minimal order. We will derive a contradiction in several steps.

(1) [O.sub.p'] (G) = 1.

Assume that [O.sub.p'](G) [not equal to] 1. Then P[O.sub.p'](G)/[O.sub.p'](G) is a Sylow p-subgroup of G/[O.sub.p'] (G). Suppose that M/[O.sub.p'](G) is a maximal subgroup of P[O.sub.p'](G)/[O.sub.p'](G). Then there exists a maximal subgroup [P.sub.1] of P such that M = [P.sub.1][O.sub.p'](G). By the hypothesis, [P.sub.1] is either an SCAP or an S-supplemented subgroup of G, then M/[O.sub.p'](G) = [P.sub.1][O.sub.p'](G)/[O.sub.p'](G) is either an SCAP or an S-supplemented subgroup of G/[O.sub.p'](G) by Lemma 2.1 and 2.2. It is clear that ([absolute value of G/[O.sub.p'](G)], p - 1) = 1. The minimal choice of G implies that G/[O.sub.p'](G) is p-nilpotent, and so G is p-nilpotent, a contradiction. Therefore, we have [O.sub.p'](G) = 1.

(2) [O.sub.p](G) [not equal to] 1.

If all maximal subgroups of P are S-supplemented in G, then G is p-nilpotent by [21, Lemma 3.1]. Therefore we may assume that there is a maximal subgroup [P.sub.1] of P which is an SCAP subgroup of G. By Lemma 2.3, G is p-solvable. Since [O.sub.p'](G) = 1 by Step (1), we have [O.sub.p](G) [not equal to] 1.

(3) G is solvable.

If p [not equal to] 2, then G is odd from the assumption that ([absolute value of G], p - 1) = 1. By the famous Odd Order Theorem, G is solvable. If p = 2, then [O.sub.2](G) [not equal to] 1 by Step (2). Suppose that M/[O.sub.2](G) is a maximal subgroup of P/[O.sub.2](G). Then M is a maximal subgroup of P. By the hypothesis, M is either an SCAP or an S-supplemented subgroup of G, then M/[O.sub.2](G) is either an SCAP or an S-supplemented subgroup of G/[O.sub.2](G) by Lemma 2.1 and 2.2. Therefore G/[O.sub.2](G) satisfies the hypothesis of the theorem. The minimal choice of G implies that G/[O.sub.2](G) is 2-nilpotent, and so G/[O.sub.2](G) is solvable. It follows that G is solvable.

(4) [O.sub.p](G) is the unique minimal normal subgroup of G.

Let N be a minimal normal subgroup of G. By Step (3), N is an elementary abelian subgroup. Since [O.sub.p'](G) = 1, we have N is p-subgroup and so N [less than or equal to] [O.sub.p](G). It is easy to see that G/N satisfies the hypothesis of the theorem. The minimal choice of G implies that G/N is p-nilpotent. Since the class of all p-nilpotent groups is a saturated formation, N is a unique minimal normal subgroup of G and N [??] [PHI](G). Choose M to be a maximal subgroup of G such that G = NM. Obviously, G = [O.sub.p](G)M and so [O.sub.p](G) [intersection] M is normal in G by Lemma 2.4. The uniqueness of N yields N = [O.sub.p](G).

(5) The final contradiction.

By the proof in Step (4), G has a maximal subgroup M such that G = M[O.sub.p](G) and G/[O.sub.p](G) [congruent to] M is p-nilpotent. Clearly, P = [O.sub.p](G)(P [intersection] M). Furthermore, P [intersection] M < P. Thus, there exists a maximal subgroup V of P such that P [intersection] M [less than or equal to] V. Hence, P = [O.sub.p](G)V. By the hypothesis, V is either an SCAP or a S-supplemented subgroup of G. First, we assume that V is an SCAP of G. Since [O.sub.p](G) is the unique minimal normal subgroup of G, V covers or avoids [O.sub.p](G)/1. If V covers [O.sub.p](G)/1, then V[O.sub.p](G) = V, i.e., [O.sub.p](G) [less than or equal to] V. It follows that P = [O.sub.p](G)V = V, a contradiction. If V avoids [O.sub.p](G)/1, then V [intersection] [O.sub.p](G) = 1. Since V [intersection] [O.sub.p](G) is a maximal subgroup of [O.sub.p](G), we have that [O.sub.p](G) is of order p and so [O.sub.p](G) lies in Z(G) by Lemma 2.5. By the proof in Step (4), we have G/[O.sub.p](G) is p-nilpotent. Then G/Z(G) is p-nilpotent, and so G is p-nilpotent, a contradiction. Now, we may assume that V is an S-supplemented subgroup of G. Then there is a subgroup T of G such that G = VT and V [intersection] T [less than or equal to] [V.sub.sG]. From Lemma 2.6, we have [O.sub.p](G) [less than or equal to] [N.sub.G]([V.sub.sG]). Since [V.sub.sG] is subnormal in G, we have V [intersection] T [less than or equal to] [V.sub.sG] [less than or equal to] [O.sub.p](G). Thus, [V.sub.sG] [less than or equal to] V [intersection] [O.sub.p](G) and

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It follows that [([V.sub.sG]).sup.G] = 1 or [([V.sub.sG]).sup.G] = V [intersection] [O.sub.p](G) = [O.sub.p](G). If [([V.sub.sG]).sup.G] = V [intersection] [O.sub.p](G) = [O.sub.p](G), then [O.sub.p](G) [less than or equal to] V and P = [O.sub.p](G)V = V, a contradiction. If [([V.sub.sG]).sup.G] = 1, then V [intersection] T = 1 and so [[absolute value of T].sub.p] = p. Hence, T is p-nilpotent by Lemma 2.5. Let [T.sub.p'] be the normal p- complement of T. Since M is p-nilpotent, we may suppose M has a normal Hall p'-subgroup [M.sub.p'] and M [less than or equal to] [N.sub.G]([M.sub.p']) [less than or equal to] G. The maximality of M implies that M = [N.sub.G]([M.sub.p]') or [N.sub.G]([M.sub.p]') = G. If the latter holds, then [M.sub.p]' [??] G and [M.sub.p'] is actually the normal p-complement of G, which is contrary to the choice of G. Hence, we may assume M = [N.sub.G]([M.sub.p']). By applying Lemma 2.7 and Feit-Thompson's theorem, there exists g [member of] G such that [T.sup.g.sub.p'] = [M.sub.p']. Hence, [T.sup.g] [less than or equal to] [N.sub.G]([T.sup.g.sub.p']) = [N.sub.G]([M.sub.p']) = M. However, [T.sub.p'] is normalized by T, so g can be considered as an element of V. Thus, G = [VT.sup.g] = VM and P = V(P [intersection] M) = V, a contradiction.

Case II. H < G.

By Lemma 2.1 and 2.2, every maximal subgroup of P is an SCAP or S-supplemented subgroup of H. By Case I, H is p-nilpotent. Now, let [H.sub.p'] be the normal p-complement of H. Then [H.sub.p'] [??] G. Assume [H.sub.p'] [not equal to] 1 and consider G/[H.sub.p']. Applying Lemma 2.1 and 2.2, it is easy to see that G/[H.sub.p'] satisfies the hypotheses for the normal subgroup H/[H.sub.p']. Therefore, by induction G/[H.sub.p'] is p-nilpotent and so G is p-nilpotent. Hence, we may assume [H.sub.p'] = 1 and so H = P is a p-group. Since G/H is p-nilpotent, we can let K/H be the normal p-complement of G/H. By Schur-Zassenhaus's theorem, there exists a Hall p'-subgroup [K.sub.p'] of K such that K = H[K.sub.p']. A new application of Case II yields K is p'-nilpotent and so K = H x [K.sub.p']. Hence, [K.sub.p'] is a normal p-complement of G, i.e., G is p-nilpotent.

Corollary 3.2. Let P be a Sylow p-subgroup of a group G, where p is the smallest prime divisor of [absolute value of G]. If every maximal subgroup of P is either an SCAP or an S-supplemented subgroup of G, then G is p-nilpotent.

Proof. It is clear that ([absolute value of G], p = 1) = 1 if p is the smallest prime dividing the order of G and so Corollary 3.2 follows immediately from Theorem 3.1.

Corollary 3.3. Suppose that every maximal subgroup of any Sylow subgroup of a group G is either an SCAP or an S-supplemented subgroup of G, then G is a Sylow tower group of supersolvable type.

Proof. Let p be the smallest prime dividing [absolute value of G] and P a Sylow p-subgroup of G. By Corollary 3.2, G is p-nilpotent. Let U be the normal p-complement of G. By Lemma 2.1 and 2.2, every maximal subgroup of any Sylow subgroup of U is either an SCAP or an S-supplemented subgroup of U. Thus U satisfies the hypothesis of the Corollary. It follows by induction that U, and hence G is a Sylow tower group of supersolvable type.

Corollary 3.4. [26] (Theorem 3.1) Let G be a group, p a prime dividing the order of G, and P a Sylow p-subgroup of G. If ([absolute value of G], p - 1) = 1 and every maximal subgroup of P is an SCAP subgroup of G, then G is p-nilpotent.

Corollary 3.5. [11] (Theorem 3.2) Let P be a Sylow p-subgroup of a group G, where p is the smallest prime divisor of [absolute value of G]. If P is cyclic or every maximal subgroup of P is an SCAP subgroup of G, then G is p-nilpotent.

Proof. If P is cyclic, by Lemma 2.5, we have G is p-nilpotent. Thus we may assume that every maximal subgroup of P is an SCAP subgroup of G. By Corollary 3.2, G is p-nilpotent.

Theorem 3.6. Suppose N is a normal subgroup of a group G such that G/N is p-nilpotent, where p is a fixed prime number. Suppose every subgroup of order p of N is contained in the hypercenter [Z.sub.[infinity]](G) of G. If p = 2, in addition, suppose every cyclic subgroup of order 4 of N is either an SCAP or an S-supplemented subgroup of G, then G is p-nilpotent.

Proof. Suppose that the theorem is false, and let G be a counterexample of minimal order.

(1) The hypotheses are inherited by all proper subgroups, thus G is a group which is not p-nilpotent but whose proper subgroups are all p-nilpotent.

In fact, [for all]/K < G, since G/N is p-nilpotent, K/K [intersection] N [congruent to] KN/N is also p- nilpotent. The cyclic subgroup of order p of K [intersection] N is contained in [Z.sub.[infinity]](G) [intersection] K [less than or equal to] [Z.sub.[infinity]](K), the cyclic subgroup of order 4 of K [intersection] N is either an SCAP or an S-supplemented subgroup of G, then is either an SCAP or an S-supplemented subgroup of K by Lemma 2.1 and 2.2. Thus K, K [intersection] N satisfy the hypotheses of the theorem in any case, so K is p-nilpotent, therefore G is a group which is not p-nilpotent but whose proper subgroups are all p-nilpotent. By Lemma 2.8 and 2.9, G = PQ, P [??] G and P/[PHI](P) is a minimal normal subgroup of G/[PHI](P).

(2) G/P [intersection] N is p-nilpotent.

Since G/P [congruent to] Q is nilpotent, G/N is p-nilpotent and G/P [intersection] N [??] G/P x G/N, therefore G/P [intersection] N is p-nilpotent.

(3) P [less than or equal to] N.

If P [not less than or equal to] N, then P [intersection] N < P. So Q(P [intersection] N) < QP = G. Thus Q(P [intersection] N) is nilpotent by (1), Q(P [intersection] N) = Q x (P [intersection] N). Since G/P [intersection] N = P/P [intersection] N x Q(P [intersection] N)/P [intersection] N, it follows that Q(P [intersection] N)/P [intersection] N [??] G/P [intersection] N by Step (2). So Q char Q(P [intersection] N) [??] G. Therefore, G = P x Q, a contradiction.

(4) p = 2.

If p > 2, then exp(P) = p by (a) and Lemma 2.9. Thus P = P n N [less than or equal to] [Z.sub.[infinity]](G). It follows that G/[Z.sub.[infinity]] (G) is nilpotent, and so G is nilpotent, a contradiction.

(5) For every x [member of] P\[PHI] (P), we have o(x) = 4.

If not, there exists x [member of] P\[PHI](P) and o(x) = 2. Denote M =< [x.sup.G] >[less than or equal to] P. Then M[PHI](P)/[PHI](P) [??] G/[PHI](P), we have that P = M[PHI](P) = M [less than or equal to] [Z.sub.[infinity]](G) as P/[PHI](P) is a minimal normal subgroup of G/[PHI](P) by Lemma 2.9, a contradiction.

(6) For every x [member of] P\[PHI](P), < x > is supplemented in G.

Let x [member of] P\[PHI](P). Then x either an SCAP or an S-supplemented subgroup of G by Step (5) and the hypothesis. We assume that x has the semi cover-avoiding property in G. In this case, there exists a chief series of G

1 = [G.sub.0] < [G.sub.1] < ... < [G.sub.t] = G,

such that x covers or avoids every [G.sub.j]/[G.sub.j-1]. Since x [member of] G, for some k, x [not member of] [G.sub.k] but x [member of] [G.sub.k+1]. It follows from [G.sub.k] [intersection] < x > [not equal to] [G.sub.k+1] [intersection] < x > that [G.sub.k] < x > = [G.sub.k+1] < x >= [G.sub.k+1]. Hence [G.sub.k+1]/[G.sub.k] is a cyclic group of order 4. The normality of P [intersection] [G.sub.k] implies that (P [intersection] [G.sub.k])[PHI](P)/[PHI](P) is normal in G/[PHI](P). Since P/[PHI](P) is a minimal normal subgroup of G/[PHI](P), we see that (P [intersection] [G.sub.k])[PHI](P) = [PHI](P) or P. If (P [intersection] [G.sub.k])[PHI](P) = P, then P [intersection] [G.sub.k] = P, contradicting x [not member of] P [intersection] [G.sub.k]. Thus P [intersection] [G.sub.k] [less than or equal to] (P). Since x [not member of] [PHI](P) but x [member of] P [intersection] [G.sub.k+1], P [intersection] [G.sub.k+1] = P, i.e., P [less than or equal to] [G.sub.k+1]. Therefore, P = P [intersection] [G.sub.k] < x > = < x > (P [intersection] [G.sub.k]) = [PHI](P) < x >=< x >. By Lemma 2.9, P is an elementary abelian group and so P does not have an element of order 4, a contradiction.

(7) Final contradiction.

For any x [member of] P\[PHI](P), we may assume that x is supplemented in G by Step (6). Then there is a subgroup T of G such that G = < x > T and < x > [intersection]T [less than or equal to] < x > [sub.sG]. It follows that P = P [intersection] G = P[intersection] < x > T =< x > (P [intersection] T). Since P/[PHI](P) is abelian, we have (P [intersection] T)[PHI](P)/[PHI](P) [??] G/[PHI](P). Since P/[PHI](P) is the minimal normal subgroup of G/[PHI](P), P [intersection] T [less than or equal to](P) or P = (P [intersection] T)[PHI](P) = P [intersection] T. If P [intersection] T [less than or equal to] [PHI](P), then < x >= P [??] G, a contraction. If P = (P [intersection] T)[PHI](P) = P [intersection] T, then T = G and so < x >=< x >[s.sub.G] is s- permutable in G. We have < x > Q is a proper subgroup of G and so < x > Q =< x > x Q, i.e., <x> [less than or equal to] [N.sub.G](Q). By Lemma 2.9, [PHI](P) [subset or equal to] Z(G). Therefore we have P [less than or equal to] [N.sub.G](Q) and so Q [??] G, a contradiction.

Corollary 3.7. Suppose N is a normal subgroup of a group G such that G/N is nilpotent and every minimal subgroup of N is contained in the hypercenter [Z.sub.[infinity]] (G) of G. If p = 2, in addition, suppose every cyclic subgroup of order 4 of N is either an SCAP or an S-supplemented subgroup of G, then G is nilpotent.

Corollary 3.8. [27] (Theorem 4.3) Suppose N is a normal subgroup of a group G such that G/N is nilpotent and every minimal subgroup of N is contained in the hypercenter [Z.sub.[infinity]] (G) of G. If p = 2, in addition, suppose every cyclic subgroup of order 4 of N is c-supplemented subgroup of G, then G is nilpotent.

Corollary 3.9. [28] (Theorem 2.5) Suppose that p is a prime and K = [G.sup.N] be the nilpotent residual of G. Then G is p-nilpotent if every minimal subgroup of K is contained in [Z.sub.[infinity]] (G) and every cyclic < x > of K with order 4 is c-supplemented in G.

Corollary 3.10. [29] (Theorem 2.4) Let G be a finite group and K = [G.sup.N] be the nilpotent residual of G. Then G is nilpotent if and only if every minimal subgroup < x > of K lies in the hypercenter [Z.sub.[infinity]] (G) of G and every cyclic element of P with order 4 is c-normal in G.

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Changwen Li ([dagger]) and Xuemei Zhang ([double dagger])

([dagger]) School of Mathematical Science, Xuzhou Normal University, Xuzhou, 221116, China

([double dagger]) Department of Basic Sciences, Yancheng Institute of Technology

Yancheng, 224051, China

E-mail: lcw2000@126.com zhangxm@ycit.edu.cn

(1) The project is supported by the N. S. F. of China (No: 11071229) and the N. S. F. of the Jiangsu Higher Education Institutions (No: 10KJD110004).

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Author: | Li, Changwen; Zhang, Xuemei |
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Publication: | Scientia Magna |

Article Type: | Report |

Date: | Jun 1, 2011 |

Words: | 5049 |

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