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On s*--supplemented subgroups of finite groups (1).

Abstract A subgroup H of a group G is said to be [s.sup.*]--supplemented subgroups in G if there exists a subgroup K of G such that G = HK and H [intersection] K [less than or equal to] [H.sub.SG], where [H.sub.SG] is the largest subnormal subgroup of G contained in H. In this paper we determine the structure of finite groups with [s.sup.*]--supplemented primary subgroups,and obtain some new results about p--nilpotent groups.

Keywords P--nilpotent groups, primary subgroups, [s.sup.*]--supplemented subgroups, solvable group.

[section] 1. Introduction

A subgroup H of a group G is said to be supplemented in G, if there exists a subgroup K of G such that G = HK. Furthermore, a subgroup H of G is said to be complemented in G if there exists a subgroup K of G such that G = HK and H [intersection] K = 1. It is obvious that the existence of supplements for some families of subgroups of a group gives a lot of information about its structure. For instance, Kegel [1-2] showed that a group G is soluble if every maximal subgroup of G either has a cyclic supplement in G or if some nilpotent subgroup of G has a nilpotent supplement in G. Hall[3] proved that a group G is soluble if and only if every Sylow subgroup of G is complemented in G. Arad and Ward [4] proved that a group G is soluble if and only if every Sylow 2-subgroup and every Sylow 3-subgroup are complemented in G. More recently, A. Ballester-Bolinches and Guo Xiuyun[5] proved that the class of all finite supersoluble groups with elementary abelian Sylow subgroups is just the class of all finite groups for which every minimal subgroup is complemented. Yanming Wang[6] defined a new concept, c--supplementation, which is a generalization of c--normality and complement. Applying it, the supersolvablity of G and some related results were got. In 2003, Zhang Xinjian, Guo Wenbin and Shum K. P.[7] defined another concept: s--normal subgroup, which is a generalization of normality and c--normality.

In this paper, we remove the c--supplement condition and replace the c-- normality assumption with s-- normality assumption for the some primary subgroups of G, We obtain a series of new results for the p-- nilpotency of finite groups.

Definition 1.1. We say a subgroup H of a group G is [s.sup.*]--supplemented (in G) if there exists a subgroup K of G such that G = HK and H [intersection] K [less than or equal to] [H.sub.SG], where [H.sub.SG] is the largest subnormal subgroup of G contained in H. We say that K is an [s.sup.*]--supplement of H in G.

Recall that a subgroup H of a group G is called s--normal in a group G if there exists a subnormal subgroup K of G such that G = HK and H [intersection] K [less than or equal to] [H.sub.SG], where [H.sub.SG] is the largest subnormal subgroup of G contained in H. It is easy to see that suppose H is an [s.sup.*]--supplemented subgroup of G, K is an [s.sup.*]--supplement of H in G. If K = G, then H must be a subnormal subgroup of G.

All groups considered in this paper are finite. Our notation is standard and can be found in [8] and [9]. We denote that G is the semi-product of subgroup H and K by G = [H]K, where H is normal in G.

Let F be a class of groups. F is a S--closed if any subgroup K of G is in F when G [member of] F. We call F a formation provide that (1) if G [member of] F and H [??] G, then G/H [member of] F, and (2) if G/M and G/N are in F, then G/M [intersection] N is in F for normal subgroups M and N of G. A formation F is said to be saturated if G/[PHI](G) [member of] F implies that G [member of] F. As we all know, the class of all p--nilpotent groups is a saturated formation.

[section] 2. Preliminaries

For the sake of easy reference, we first give some basic definitions and known Results from the literature.

Lemma 2.1. Let G be a group and N a normal subgroup of G. Then the following statements hold.

(1) If H [less than or equal to] M [less than or equal to] G and H is [s.sup.*]--supplemented in G, then H is [s.sup.*]--supplemented in M.

(2) If N is contained in H and H is [s.sup.*]--supplemented in G if and only if H/N is [s.sup.*]--supplemented in G/N.

(3) Let [pi] be a set of primes. If N is a normal [pi]'--subgroup and let A be a [pi]--subgroup of G, if A is [s.sup.*]--supplemented in G then AN/N is [s.sup.*]--supplemented in G/N. If furthermore N normalizes A, then the converse also hold.

(4) Let H [less than or equal to] G and L [less than or equal to] [PHI](H). If L is [s.sup.*]--supplemented in G, then L [??] [??] G and L [less than or equal to] [PHI](G).

Proof. (1) If HK / G with H [intersection] K [less than or equal to] [H.sub.SG], then M = M [intersection] G = H(M [intersection] K) and H [intersection] (K [intersection] M) [less than or equal to] [H.sub.SG] [intersection] M [less than or equal to] [H.sub.SM]. So H is [s.sup.*]--supplemented in M.

(2) Suppose that H/N is [s.sup.*]--supplemented in G/N. Then there exists a subgroup K/N of G/N such that G/N = (H/N)(K/N) and (H/N) [intersection] (K/N) [less than or equal to] [(H/N).sub.S(G/N)]. It is easy to see that G = HK and H [intersection] K [less than or equal to] [H.sub.SG].

Conversely, if H is [s.sup.*]--supplemented in G, then there exists K [less than or equal to] G such that G = HK and H [intersection] K [less than or equal to] [H.sub.SG]. It is easy to check that KN/N is [s.sup.*]--supplemented of H/N in G/N.

(3) If A is [s.sup.*]--supplemented in G, then there exists K [less than or equal to] G such that G = AK and A [intersection] K [less than or equal to] [A.sub.SG]. Since [|G|.sub.[pi]'] = [|K|.sub.[pi]'] = [|KN|.sub.[pi]'], we have that [|K [intersection] A|.sub.[pi]'] = [|N|.sub.[pi]'] = |N| and hence N [less than or equal to] K. It is clear that (AN/N)(K/N) / G/N and (AN/N) [intersection] (K/N) = (A [intersection] K)N/N [less than or equal to] ([A.sub.SG]N/N) [less than or equal to] [(AN/N).sub.S(G/N)]. Hence AN/N is [s.sup.*]--supplemented in G/N.

Conversely, assume that AN/N is [s.sup.*]--supplemented in G/N. Let K/N be [s.sup.*]--supplement of AN/N. Then AK = ANK = G and (A [intersection] K)N/N [less than or equal to] L/N = [((AN)/N).sub.S(G/N)]. By hypothesis, NA = N x A. This means NA is a both [pi]--nilpotent and [pi]--closed and [A.sub.1] = A[intersection]L hence L = [A.sub.1] x N with [A.sub.1] = A [intersection] L and [A.sub.1] [??] [??] G. Now we have A [intersection] K [less than or equal to] [A.sub.1] [less than or equal to] [A.sub.SG] and A is [s.sup.*]--supplemented in G.

(4) In fact, if L is [s.sup.*]--supplemented in G with supplement K, then LK = G and L [intersection]K [less than or equal to] [L.sub.SG]. Now H = H [intersection] G = L(H [intersection] K) = H [intersection] K since L [less than or equal to] [PHI](H). Therefore L = (H [intersection] K) [intersection] L = (H [intersection]L)[intersection]K = L[intersection]K [less than or equal to] [L.sub.SG] and hence L = [L.sub.SG] and L [??] [??] G. If L [??] [PHI](G), then there exists a maximal subgroup M of G such that LM = G. Now H = H [intersection] G = L(H [intersection]M) = H [intersection]M [less than or equal to] M. Therefore G = LM [less than or equal to] HM [less than or equal to] M < G, a contradiction.

Lemma 2.2.([14]) Let F be an S--closed local formation and H a subgroup of G. Then H [intersection] [Z.sub.F](G) [subset].bar] [Z.sub.F](H).

Lemma 2.3.([11, Lemma 4.1]) Let p be the smallest prime dividing the order of the group H and P a Sylow p--subgroup of H. If |P| [less than or equal to] [p.sup.2] and H is [A.sub.4]--free. then H is p--nilpotent.

Lemma 2.4. Let G be an [A.sub.4]--free group and p be the smallest prime dividing the order of the G. If G/L is p-- nilpotent and [p.sup.3] not dividing the order of the subgroup L, then G is p-- nilpotent.

Proof. By hypothesis and lemma 2.3, We know that L is p--nilpotent and L has a normal p--complement [L.sub.p']. Since [L.sub.p'] char L and L is normal in G, We have that [L.sub.p'] [??] G. Therefore G/L [??] (G/[L.sub.p'])/(L/[L.sub.p']) is p--nilpotent. There exists a Hall p'--subgroup (H/[L.sub.p'])/(L/[L.sub.p']) of (G/[L.sub.p'])/(L/[L.sub.p']) and H/[L.sub.p'] [??] G/[L.sub.p']. By Schur-Zassenhaus Theorem, we have that H/[L.sub.p'] = [L/[L.sub.p']] [H.sub.1]/[L.sub.p'], where [H.sub.1]/[L.sub.p'] is a Hall p'--subgroup of H/[L.sub.p']. Then by lemma 2.3, we have [H.sub.1]/[L.sub.p'] [??] H/[L.sub.p'] and so [H.sub.1]/[L.sub.p'] char H/[L.sub.p'] [??] G/[L.sub.p']. Therefore [H.sub.1]/[L.sub.p'] [??] G/[L.sub.p'] and hence G/[L.sub.p'] is p--nilpotent. Thus G is p--nilpotent.

Lemma 2.5.([13]) Suppose that G is a group which is not p--nilpotent but whose proper subgroups are all p--nilpotent. Then G is a group which is not nilpotent but whose proper subgroups are all nilpotent.

Lemma 2.6.([13]) Suppose that G is a group which is not nilpotent but whose proper subgroups are all nilpotent. Then

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

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

3) If P is non-abelian and p [not equal to] 2, then the exponent of P is p.

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

5) If P is abelian, then P is of exponent p.

Lemma 2.7.([12]) Let K be a subgroup of G. If |G : K| = p, where p is the smallest prime divisor of |G|, then K [??] G.

Lemma 2.8.([15]) Let P be an elementary abelian p--group of order [p.sup.n], where p is a prime. Then |Aut(P)| = [k.sub.n] x [p.sup.n(n-1)/2], here [k.sub.n] = [[PI].sup.n.sub.i=1] ([p.sup.i] - 1).

Lemma 2.9. Let G is a group which is not p--nilpotent but whose proper subgroups are all p--nilpotent. N [??] G and P is a normal Sylow p--subgroup of G. If G/N is p--nilpotent, then P [less than or equal to] N.

Proof. By Lemma 2.6, we have that G = P[Q.sub.0], where [Q.sub.0] is a non-normal cyclic Sylow q--subgroup. If P is not a subgroup of N, by p--nilpotenty of G/N, there exists a normal p--complemente, QN/N say, here Q is a Sylow q--subgroups of G, such that G/N = (PN/N)(QN/N). Now we have QN < G since P is not a subgroup of N, then QN is nilpotent by Lemma 2.5 and Q char QN, furthermore Q E G. a contradiction. thus P [less than or equal to] N.

Lemma 2.10.([9,10.1.9]) Let p be the smallest prime dividing the order of the finite group G. Assume that G is not p--nilpotent. Then the Sylow p--subgroups of G are not cyclic. Moreover |G| is divisible by [p.sup.3] or by 12.

Lemma 2.11.([12]) Let G be a finite group, P is a p--subgroup of G but not a Sylow p--subgroup of G, then P < [N.sub.G](P).

[section] 3. Main results

Theorem 3.1. Let G be an [A.sub.4]--free group and p be the smallest prime dividing the order of the group G. If there exists a normal subgroup N in G such that G/N is p--nilpotent and each subgroups of N of order [p.sup.2] is [s.sup.*]--supplemented in G, then G is p--nilpotent.

Proof. Assume that the claim is false and choose G to be a counterexample of minimal order.

By Lemma 2.4 and hypothesis, we have [|N|.sub.p] > [p.sup.2]. Let L be a proper subgroup of G. We prove that conditions of the theorem 3.1 is inherited by L. Clearly, L/L [intersection] N [??] LN/N [less than or equal to] G/N implies that L/L [intersection] N is p--nilpotent. If [|L [intersection] N|.sub.p] [less than or equal to] [p.sup.2], then L is p--nilpotent by Lemma 2.4. If [|L [intersection] N|.sub.p] > [p.sup.2], then each subgroup of L [intersection] N of order [p.sup.2] is [s.sup.*]--supplemented in G and hence is [s.sup.*]--supplemented in L. Then L is p--nilpotent by induction. Thus G is a minimal non-p--nilpotent group. Now Lemma 2.5 implies that G is a group which is not nilpotent but whose proper subgroups are all nilpotent. Then by Lemma 2.6, we have G = PQ, where P [??] G and Q is a non-normal cyclic Sylow q--subgroup of G. It is clear that P/[PHI](P) is a minimal normal subgroup of G/[PHI](P).

By Lemma 2.9, we have P [less than or equal to] N. Let A [less than or equal to] N and |A| = [p.sup.2]. By hypothesis, there exists a subgroup K of G such that G = AK and A [intersection] K [less than or equal to] [A.sub.SG], where [A.sub.SG] is the largest subnormal subgroup of G contained in A. If K < G, We have K = [K.sub.p] x [K.sub.p'] since that G is a group which is not nilpotent but whose proper subgroups are all nilpotent.

(i) If [K.sub.p] / 1, then P = A and hence G is p--nilpotent, a contradiction.

(ii) If [K.sub.p] [not equal to] 1, then we consider subgroup [N.sub.G]([K.sub.p]). Since K [less than or equal to] [N.sub.G]([K.sub.p]) and [K.sub.p] < [N.sub.G]([K.sub.p]), we have that |G : [N.sub.G]([K.sub.p])| = p or [N.sub.G]([K.sub.p]) = G. If |G : [N.sub.G]([K.sub.p])| = p, then we have [N.sub.G]([K.sub.p]) < G. Let [P.sub.1] = P [intersection] [N.sub.G]([K.sub.p]), Then [P.sub.1] [??] G since G = PQ and Q [less than or equal to] [N.sub.G]([K.sub.p]). If [P.sub.1] [less than or equal to] [PHI](P), then P/P [intersection] A[N.sub.G]([K.sub.p])/A(P [intersection] [N.sub.G]([K.sub.p]))/A, a contradiction. If [P.sub.1] [??] [PHI](P), then [P.sub.1][PHI](P)/[PHI](P)/P/[PHI](P) by Lemma 2.6. In this case, P/[P.sub.1] and hence [N.sub.G]([K.sub.p])/G, a contradiction. If|G : [N.sub.G]([K.sub.p])| = 1, then [K.sub.p] [??] G. We consider the factor group G/[K.sub.p]. Since G = AK and |A| = [p.sup.2], we have [p.sup.3] [??] |G/[K.sub.p]|. By lemma 2.4, G/[K.sub.p] is a p--nilpotent and P [less than or equal to] [K.sub.p], we know P = [K.sub.p]. Furthermore, we have G = AK = K via A [less than or equal to] [K.sub.p] [less than or equal to] K, a contradiction.

Now we let K = G. It is easy to see A [??][??]G. By A [less than or equal to] P, we have A[PHI](P)/[PHI](P) [less than or equal to] P/[PHI](P). Furthermore, A[PHI](P)/[PHI](P) [??] P/[PHI](P), where P/[PHI](P) is either a minimal normal subgroup or characteristic subgroup of G/[PHI](P). So we have A[PHI](P)/[PHI](P) [??] G/[PHI](P) and A [[subset].bar] [PHI](P) or A = P. If A = P, then |A| = |P| = [p.sup.2] and [|N|.sub.p] [less than or equal to] [p.sup.2] here because P [less than or equal to] N, a contradiction to [|N|.sub.p] > [p.sup.2]. If A [[subset].bar] [PHI](P), considering A be a arbitrary abelian group, we have [PHI](P) [greater than or equal to] P, hence [PHI](P) = P, a contradiction.

The final contradiction completes our proof.

Corollary 3.2. Let G be an [A.sub.4]-free group and p be the smallest prime dividing the order of the group G. If each subgroup of G of order [p.sup.2] is [s.sup.*]--supplemented in G, then G is p--nilpotent.

Theorem 3.3. Suppose that G is a group with a normal subgroup H such that G/H is p--nilpotent for some prime divisor p of |G|. If every cyclic subgroup of H of order 4 is [s.sup.*]--supplemented in G and every subgroup H of order p is contained in [Z.sub.F](G), where F is the class of all p--nilpotent groups, then G is p--nilpotent.

Proof. Assume that the claim is false and choose G to be a counterexample of smallest order.

The hypothesis is inherited by all proper subgroups of G. Then K/K [intersection] H [??] KH/H [less than or equal to] G/H implies that K/K [intersection] H is p--nilpotent. Every cyclic subgroups of K [intersection] H of order 4 is [s.sup.*]--supplemented in K by Lemma 2.1. Every subgroup of H [intersection] K of order p is contained in K [intersection] [Z.sub.F](G) [less than or equal to] ZF(K) by Lemma 2.2. Thus G is a minimal non-p-nilpotent group. Now Lemma 2.5 implies that G is a group which is not nilpotent but whose proper subgroups are all nilpotent. Then by Lemma 2.6, G has a normal Sylow p--nilpotent subgroup P and G/P [??] Q, where Q is a non-normal cyclic Sylow q--subgroup of G, P/[PHI](P) is a minimal normal subgroup of G/[PHI](P). We consider the follow cases.

Case 1. P is abelian. By Lemma 2.6, P is an elementary abelian p--group. Since G/H is p--nilpotent, we have P [less than or equal to] H. By hypothesis, every subgroup of H of order p is contained in [Z.sub.F](G), then P [less than or equal to] [Z.sub.F](G) and hence G is p--nilpotent, a contradiction.

Case 2. P is non-abelian and p > 2. By Lemma 2.6, the expotent of P is p and every subgroup of H of order p is contained in [Z.sub.F](G). Therefore P [less than or equal to] [Z.sub.F](G) and we have that G is p--nilpotent, a contradiction.

Case 3. P is non-abelian and p = 2. Let A be a cyclic subgroup of H of order 4. By hypothesis, A is [s.sup.*]--supplemented in G and there exist a subgroup L of G such that G = AL and A[intersection]L [less than or equal to] [A.sub.SG]. Since L < G and L is nilpotent, we have L = [L.sub.p] x [L.sub.p']. If [L.sub.p] = 1, then P = A and P is abelian subgroup. a contradiction. Now we have [L.sub.p] [not equal to] 1 and we consider [N.sub.G]([L.sub.p]). Since L [less than or equal to] [N.sub.G]([L.sub.p]), we have that |G : [N.sub.G]([L.sub.p])| = 2 or |G : [N.sub.G]([L.sub.p])| = 1. If |G : [N.sub.G]([L.sub.p])| = 2, then [N.sub.G]([L.sub.p]) [??] G by Lemma 2.7 and hence G is 2-nilpotent, a contradiction. If |G : [N.sub.G] ([L.sub.p])| = 1, then [L.sub.p] [??] G. Since P/[PHI](P) is the minimal normal subgroup of G/[PHI](P), we have P/[L.sub.p] or Lp [less than or equal to] [PHI](P). If P/[L.sub.p], then L / G since G = AL, a contradiction. If [L.sub.p] [less than or equal to] [PHI](P), then P/A[L.sub.p] = A, a contradiction.

Now we let K = G. It is easy to see A [??] [??] G. By A [less than or equal to] P, We have A[PHI](P)/[PHI](P) [less than or equal to] P/[PHI](P). Furthermore, A[PHI](P)/[PHI](P) [??] P/[PHI](P), where P/[PHI](P) is either a minimal normal subgroup or characteristic subgroup of G/[PHI](P).So we have A[PHI](P)/[PHI](P) [??] G/[PHI](P) and A [[subset].bar] [PHI](P) or A = P. If A = P, then |A| = |P| = [p.sup.2] and [|N|.sub.p] [less than or equal to] [p.sup.2] here because P [less than or equal to] N, a contradiction to [|N|.sub.p] > [p.sup.2]. If A [[subset].bar] [PHI](P), considering A be a arbitrary abelian group, we have [PHI](P) [greater than or equal to] P, hence [PHI](P) = P, a contradiction.

The final contradiction completes our proof.

Corollary 3.4. Let G be a group and p be the prime devisor of |G|. If every cyclic subgroup of order 4 is [s.sup.*]--supplemented in G and every subgroup H of order p is contained in [Z.sub.F](G), where F is the class of all p--nilpotent groups, then G is p--nilpotent.

Theorem 3.5. Let G is a group and (|G|, 21) = 1. If each subgroup of G of order 8 is [s.sup.*]--supplemented in G, then G is 2-nilpotent.

Proof. Assume that the theorem is false and choose G to be a counterexample of smallest order. Let [2.sup.[alpha]] be the order of a Sylow 2-subgroup P of G.

If 2 [??] |G|, then G is 2-nilpotent. If P is cyclic, then G is 2-nilpotent by Lemma2.10. So we can suppose that P is not cyclic. Let L be a proper subgroup of G. We prove that L inherits the condition of the theorem. If 8 [??] |L|, then L is 2-nilpotent by Lemma 2.3. If 8||L|, then each subgroup of L of order 8 is [s.sup.*]--supplemented in G and hence is [s.sup.*]--supplemented in L by Lemma 2.1, so L is 2-nilpotent by induction. Thus we may assume that G is a minimal non-2-nilpotent group. Now Lemma 2.5 implies that G is a group which is not nilpotent but whose proper subgroups are all nilpotent. Then by Lemma 2.6, we have G = PQ, where P is a normal in G and Q is a non-normal cyclic Sylow q--subgroup of G(q [not equal to] p), P/[PHI](P) is a minimal normal subgroup of G/[PHI](P).

Let H be a subgroup of G of order 8. By hypothesis, there exists a subgroup K of G such that G = HK and H[intersection]K [less than or equal to] [H.sub.SG], where [H.sub.SG] is the largest subnormal subgroup of G contained in H.

First, we claim that H [not equal to] P. Otherwise |P| = |H| = 8. If [PHI](P) [not equal to] 1, then G/[PHI](P) is 2-nilpotent by Lemma 2.3 and hence G is 2-nilpotent and P [less than or equal to] [PHI](P) by Lemma 2.9, a contradiction. It follows from [PHI](P) = 1 and Lemma 2.6 we have that P is an elementary abelian 2-group. Next we consider [N.sub.G](P)/[C.sub.G](P). It is clear that [N.sub.G](P)/[C.sub.G](P) is isomorphic to a subgroup of Aut(P). By Lemma 2.8 we have |Aut(P)| = 8 x 7 x 3. Since (|G|, 21) = 1, we see that [N.sub.G](P) = [C.sub.G](P) in this case. Then, by Burnside Theorem(cf,[2]), we have that G is 2-nilpotent, a contradiction.

If K < G and G is a group which is not nilpotent but whose proper subgroups are all nilpotent, we have K is nilpotent and K = [K.sub.2] x [K.sub.2']. We claim that [K.sub.2] [??] G. Otherwise, if [K.sub.2] [??] G, then P = [K.sub.2] or [K.sub.2] [less than or equal to] [PHI](P). If P = [K.sub.2], then H [less than or equal to] [K.sub.2]. We have G = K by G = HK. This shows that P = [K.sub.2] is impossible. If [K.sub.2] [less than or equal to] [PHI](P), then P = H[K.sub.2] = H, a contradiction. Therefore, we note subgroup [N.sub.G]([K.sub.2]). Since K [less than or equal to] [N.sub.G]([K.sub.2]) and Lemma 2.9, we have [K.sub.2] [less than or equal to] [N.sub.P] ([K.sub.2]) and |G : [N.sub.G]([K.sub.2])| = 4 or |G : [N.sub.G]([K.sub.2])| = 2. If |G : [N.sub.G]([K.sub.2])| = 2, then [N.sub.G]([K.sub.2]) is a nilpotent normal subgroup of G by Lemma 2.7 and G is 2-nilpotent, a contradiction. If |G : [N.sub.G]([K.sub.2])| = 4, then we consider [N.sub.G]([N.sub.G][([K.sub.2])).sub.2]). If [([N.sub.G]([K.sub.2])).sub.2] [??] G, then G/[([N.sub.G]([K.sub.2])).sub.2] is 2-nilpotent by Lemma 2.3 and we have that P [less than or equal to] [([N.sub.G]([K.sub.2])).sub.2], a contradiction to |G : [N.sub.G]([K.sub.2])| = 4; Since [([N.sub.G]([K.sub.2])).sub.2] < Np([N.sub.G][([K.sub.2])).sub.2]) by Lemma 2.11, We may assume that |G : [N.sub.G][(([N.sub.G]([K.sub.2])).sub.2])| = 2. By Lemma 2.7 we know that [N.sub.G][(([N.sub.G]([K.sub.2])).sub.2]) [??] G. Since [N.sub.G][(([N.sub.G]([K.sub.2])).sub.2]) is a nilpotent normal subgroup of G, we know that G is 2-nilpotent, a contradiction.

If K = G, we have H = H [intersection] G = H [intersection] K [less than or equal to] [H.sub.SG], that is, H = [H.sub.G] and H [??] [??] G. It is clear that H [less than or equal to] P. So we claim H[PHI](P)/[PHI](P) [less than or equal to] P/[PHI](P), where P/[PHI](P) is either a minimal normal subgroup or a character subgroup of G/[PHI](P). Furthermore, H[PHI](P)/[PHI](P) [??] G/[PHI](P). Because H [not equal to] P, we have H [[subset].bar] [PHI](P) = Z(P). Notice H is arbitrary, we have [PHI](P) [greater than or equal to] P, that is, [PHI](P) = P, a contradiction.

The final contradiction completes the proof.

(1) This research is supported by the Natural Science Foundation of Guangxi autonomous region (No. 0249001).

Received May 5, 2007

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Yingwu Xu ([dagger]) and Shirong Li ([double dagger])

([dagger]). School of Science, Xi'an Polytechnic University Xi'an, Shanxi, China, 710048 Email: ywxu256@sina.com

([double dagger]). Department of Mathematics, Guangxi University, Nanning, Guangxi, China, 530004 Email: shirong@gxu.edu.cn
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Title Annotation:P--nilpotent groups, primary subgroups, supplemented subgroups, solvable group
Author:Xu, Yingwu; Li, Shirong
Publication:Scientia Magna
Geographic Code:9CHIN
Date:Jun 1, 2007
Words:5076
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