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H. Wei (Guangxi Teacher’s College, Zhongshan, China),
Y. Wang (Zhongshan Univ., China)
c∗-SUPPLEMENTED SUBGROUPS
AND p
-NILPOTENCY OF FINITE GROUPS*

c -DOPOVNENI PIDHRUPY
TA p-NIL\POTENTNIST\ SKINÇENNYX HRUP
A subgroup H of a finite group G is said to be c -supplemented in G if there exists a subgroup K suchthat G = HK and H ∩ K is permutable in G. It is proved that a finite group G which is S4-free is p-nilpotent if NG(P ) is p-nilpotent and, for all x ∈ G\NG(P ), every minimal subgroup of P ∩ P x ∩ GNpis c -supplemented in P and, if p = 2, one of the following conditions holds: (a) Every cyclic subgroup ofP ∩ P x ∩ GNp of order 4 is c -supplemented in P ; (b) [Ω2(P ∩ P x ∩ GNp ), P ] ≤ Z(P ∩ GNp ); (c) Pis quaternion-free, where P a Sylow p-subgroup of G and GNp the p-nilpotent residual of G. That will extendand improve some known results.
Pidhrupa H skinçenno] hrupy G nazyva[t\sq c -dopovnenog v G, qkwo isnu[ pidhrupa K taka, woG = HK ta H ∩ K [ perestanovoçnog v G. Dovedeno, wo skinçenna hrupa G, qka [ S4-vil\nog, [p-nil\potentnog, qkwo NG(P ) p-nil\potentna i dlq vsix x ∈ G\NG(P ) koΩna minimal\na pidhrupaiz P ∩ P x ∩ GNp [ c -dopovnenog v P ta, qkwo p = 2, vykonu[t\sq odna z nastupnyx umov: a) koΩna cykliçna pidhrupa porqdku 4 iz P ∩ P x ∩ GNp [ c -dopovnenog v P ; b) [Ω2(P ∩ P x ∩ GNp ), P ] ≤ Z(P ∩ ∩ GNp); c) P [ bezkvaternionnog, de P — sylovs\ka p-pidhrupa hrupy G ta GNp p-nil\potentnyjzalyßok hrupy G. Tym samym poßyreno ta pokraweno deqki vidomi rezul\taty.
1. Introduction. All groups considered will be finite. For a formation F and a group
G, there exists a smallest normal subgroup of G, called the F-residual of G and denoted
by GF , such that G/GF ∈ F (refer [1]). Throughout this paper, N and Np will denote
the classes of nilpotent groups and p-nilpotent groups, respectively. A 2-group is called
quaternion-free if it has no section isomorphic to the quaternion group of order 8.
General speaking, a group with a p-nilpotent normalizer of the Sylow p-subgroup need not be a p-nilpotent group. However, if one adds some embedded properties onthe Sylow p-subgroup, he may obtain his desired result. For example, Wielandt provedthat a group G is p-nilpotent if it has a regular Sylow p-subgroup whose G-normalizeris p-nilpotent [2]. Ballester-Bolinches and Esteban-Romero showed that a group G isp-nilpotent if it has a modular Sylow p-subgroup whose G-normalizer is p-nilpotent [3].
Moreover, Guo and Shum obtained a similar result by use of the permutability of someminimal subgroups of Sylow p-subgroups [4].
In the present paper, we will push further the studies. First, we introduce the c - supplementation of subgroups which is a unify and generalization of the permutability andthe c-supplementation [5, 6] of subgroups. Then, we give several sufficient conditions fora group to be p-nilpotent by using the c -supplementation of some minimal p-subgroups.
In detail, we obtain the following main theorem: Theorem 1.1.
Let G be a group such that G is S4-free and let P be a Sylow p- subgroup of G. Then G is p-nilpotent if NG(P ) is p-nilpotent and, for all x ∈ G\NG(P ),one of the following conditions holds: (a) Every cyclic subgroup of P ∩ P x ∩ GNp of order p or 4 (if p = 2) is c - * Project supported by NSFC (10571181), NSF of Guangxi (0447038) and Guangxi Education Department.
ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8 (b) Every minimal subgroup of P ∩ P x ∩ GNp is c -supplemented in P and, if p = 2, [Ω2(P ∩ P x ∩ GNp), P ] ≤ Z(P ∩ GNp); (c) Every minimal subgroup of P ∩ P x ∩ GNp is c -supplemented in P and P is Following the proof of Theorem 1.1, we can prove the Theorem 1.2. It can be consid- ered as an extension of the above-mentioned result of Ballester-Bolinches and Esteban-Romero.
Theorem 1.2.
Let P be a Sylow p-subgroup of a group G. Then G is p-nilpotent if NG(P ) is p-nilpotent and, for all x ∈ G\NG(P ), one of the followings holds: (a) Every cyclic subgroup of P ∩ P x ∩ GNp of order p or 4 (if p = 2) is permutable (b) Every minimal subgroup of P ∩ P x ∩ GNp is permutable in P and, if p = 2, [Ω2(P ∩ P x ∩ GNp), P ] ≤ Z(P ∩ GNp); (c) Every minimal subgroup of P ∩ P x ∩ GNp is permutable in P and, if p = 2, P is As an application of Theorem 1.1, we get the following theorem:
Theorem 1.3.
Let G be a group such that G is S4-free and let P be a Sylow p- subgroup of G, where p is a prime divisor of |G| with (|G|, p − 1) = 1. Then G isp-nilpotent if one of the following conditions holds: (a) Every cyclic subgroup of P ∩ GNp of order p or 4 (if p = 2) is c -supplemented (b) Every minimal subgroup of P ∩ GNp is c -supplemented in NG(P ) and, if p = 2, Our results improve and extend the following theorems of Guo and Shum [7, 8].
Theorem 1.4 ([7], Main theorem). Let G be a group such that G is S4-free and let P
be a Sylow p-subgroup of G, where p is the smallest prime divisor of |G|. If every minimalsubgroup of P ∩ GN is c-supplemented in NG(P ) and, when p = 2, P is quaternion-free,then G is p-nilpotent. Theorem 1.5 ([8], Main theorem). 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 minimal subgroup of P ∩ GN ispermutable in NG(P ) and, when p = 2, either every cyclic subgroup of P ∩ GN of order4 is permutable in NG(P ) or P is quaternion-free, then G is p-nilpotent. 2. Preliminaries. Recall that a subgroup H of a group G is permutable (or quasi-
normal) in G if H permutes with every subgroup of G. H is c-supplemented in G if thereexists a subgroup K1 of G such that G = HK1 and H ∩ K1 ≤ HG = CoreG(H) [5,6]; in this case, if we denote K = HGK1, then G = HK and H ∩ K = HG; of course,H ∩ K is permutable in G. Based on this observation, we introduce: Definition 2.1.
A subgroup H of a group G is said to be c -supplemented in G if there exists a subgroup K of G such that G = HK and H ∩ K is a permutable subgroupof G. We say that K is a c -supplement of H in G. It is clear from Definition 2.1 that a permutable or c-supplemented subgroup must be a c -supplemented subgroup. But the converses are not true. For example, let G = A4,the alternating group of degree 4. Then any Sylow 3-subgroup of G is c-supplementedbut not permutable in G. If we take G = a, b|a16 = b4 = 1, ba = a3b , then b2(aibj) == (aibj)9+2((1)j−1)b2. Hence b2 is permutable in G. However, b2 is not c-supple-mented in G as b2 is in Φ(G) and not normal in G. ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8 c -SUPPLEMENTED SUBGROUPS AND p-NILPOTENCY OF FINITE GROUPS The following lemma on c -supplemented subgroups is crucial in the sequel. The proof is a routine check, we omit its detail.
Lemma 2.1.
Let H be a subgroup of a group G. Then: (1) If H is c -supplemented in G, H ≤ M ≤ G, then H is c -supplemented in M ;(2) Let N ✁ G and N ≤ H. Then H is c -supplemented in G if and only if H/N is (3) Let π be a set of primes, H a π-subgroup and N a normal π -subgroup of G. If H is c -supplemented in G, then HN/N is c -supplemented in G/N ; (4) Let L ≤ G and H ≤ Φ(L). If H is c -supplemented in G, then H is permutable Lemma 2.2.
Let c be an element of a group G of order p, where p is a prime divisor of |G|. If c is permutable in G, then c is centralized by every element of G of order p or4 (if p = 2). Proof. Let x be an element of G with order p or 4 (if p = 2). By the hypotheses,
x c = c x . Clearly, if x is of order p, then c is centralized by c. Now assume that p = 2 and x is of order 4. If [c, x] = 1, then c−1xc = x−1 and (xc)2 = 1. Furthermore,| x c | ≤ 4, of course, [c, x] = 1, a contradiction. We are done.
Lemma 2.3 ([9], Lemma 2). Let F be a saturated formation. Assume that G is a
non-F-group and there exists a maximal subgroup M of G such that M ∈ F and G == F (G)M, where F (G) is the Fitting subgroup of G. Then: (1) GF /(GF ) is a chief factor of G;(2) GF is a p-group for some prime p;(3) GF has exponent p if p > 2 and exponent at most 4 if p = 2;(4) GF is either an elementary abelian group or (GF ) = Z(GF ) = Φ(GF ) is an Lemma 2.4 ([10], Lemma 2.8(1)). Let M be a maximal subgroup of a group G and
let P be a normal p-subgroup of G such that G = P M, where p a prime. Then P ∩ M isa normal subgroup of G. Lemma 2.5 ([11], Theorem 2.8). If a solvable group G has a Sylow 2-subgroup P
which is quaternion-free, then P ∩ Z(G) ∩ GN = 1. Lemma 2.6.
Let G be a group and let p be a prime number dividing |G| with (1) If N is normal in G of order p, then N lies in Z(G);
(2) If G has cyclic Sylow p-subgroups, then G is p-nilpotent;
(3) If M is a subgroup of G of index p, then M is normal in G.
Proof.
(1) Since |Aut(N )| = p − 1 and G/CG(N ) is isomorphic to a subgroup of
Aut(N ), |G/CG(N )| must divide (|G|, p − 1) = 1. It follows that G = CG(N ) andN ≤ Z(G). (2) Let P ∈ Sylp(G) and |P | = pn. Since P is cyclic, |Aut(P )| = pn−1(p − 1). Again, NG(P )/CG(P ) is isomorphic to a subgroup of Aut(P ), so |NG(P )/CG(P )| mustdivide (|G|, p − 1) = 1. Thus NG(P ) = CG(P ), and statement (2) follows by the well-known Burnside theorem.
(3) We may assume that MG = 1 by induction. As everyone knows the result is true in the case where p = 2. So assume that p > 2 and consequently G is of odd order as(|G|, p − 1) = 1. Now we know that G is solvable by the Odd Order Theorem. Let Nbe a minimal normal subgroup of G. Then N is an elementary abelian q-group for someprime q. It is obvious that G = M N and M ∩ N is normal in G. Therefore M ∩ N = 1 ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8 and |N | = |G : M | = p. Now N ≤ Z(G) by statement (1) and, of course, M is normalin G as desired.
3. Proofs of theorems.
Proof of Theorem
1.1. Let G be a minimal counterexample. Then we have the
(1) M is p-nilpotent whenever P ≤ M < G.
Since NM (P ) ≤ NG(P ), NM (P ) is p-nilpotent. Let x be an element of M \NM (P ). Then, since P ∩ P x ∩ M Np ≤ P ∩ P x ∩ GNp, every minimal subgroup of P ∩ P x ∩ M Npis c -supplemented in P by Lemma 2.1. If G satisfies (a), then every cyclic subgroup ofP ∩ P x ∩ M Np with order 4 is c -supplemented in P. If G satisfies (b), then [Ω2(P ∩ P x ∩ M Np), P ] ≤ Z(P ∩ GNp) (P ∩ M Np) ≤ Z(P ∩ M Np). Now we see that M satisfies the hypotheses of the theorem. The minimality of G impliesthat M is p-nilpotent.
(2) Op (G) = 1.
If not, we consider G = G/N, where N = Op (G). Clearly N (P ) = N is p-nilpotent, where P = P N/N. For any xN ∈ G\N (P ), since G p = GNpN/N and P ∩ P xN = P xn for some n ∈ N, we have ∩ G p = (P ∩ P xn ∩ GNpN)N/N = (P ∩ P xn ∩ GNp)N/N. Because xN ∈ G\N (P ), xn ∈ G\N G(P ). Now let P 0 = P0N/N be a minimal ∩ G p. We may assume that P0 = y , where y is an element of P ∩ P xn ∩ GNp of order p. By the hypotheses, there exists a subgroup K0 of P suchthat P = P0K0 and P0 ∩ K0 is a permutable subgroup of P. It follows that P N/N == (P0N/N )(K0N/N ) and (P0N/N ) (K0N/N ) = (P0 ∩ K0N )N/N. If P0 ∩ K0N == P0 then P0 ≤ P ∩ K0N = K0 and consequently P0 = P0 ∩ K0 is permutable inP. In this case, P 0 is permutable in P . If P0 ∩ K0N = 1 then P 0 is complemented inP . Thus P 0 is c -supplemented in P . Assume that G satisfies (a). Let P 1 = P1N/N ∩ G p of order 4. We may assume that P1 = z , where z is an element of P ∩ P xn ∩ GNp of order 4. Since P1 is c -supplemented in P,P = P1K1 and P1 ∩ K1 is permutable in P. We have P N/N = (P1N/N )(K1N/N ) and(P1N/N ) (K1N/N ) = (P1 ∩ K1N )N/N. If P1 ∩ K1N = 1 then P 1 is complementedin P . If P1 ∩ K1N = z2 , since z2 Φ(P ) and z2 is c -supplemented in P, z2is permutable in P by Lemma 2.1. Furthermore, z2 N/N is permutable in P N/N andP 1 is c -supplemented in P . If P1 ∩ K1N = P1 then P1 = P1 ∩ K1 is permutable in Pand P 1 is permutable in P . In a ward, P 1 is c -supplemented in P . Now assume that Gsatisfies (b), then ∩ G ), P = Ω2(P ∩ P xn ∩ GNp), P N/N ≤ Z(P ∩ GNp)N/N, ∩ G ), P ≤ Z(P ∩ G ). If G satisfies (c) then P ∼ = P is quaternion-free. Therefore G = G/N satisfies the hypotheses of the theorem. The choice of G implies that G is p-nilpotent and so is G,a contradiction.
ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8 c -SUPPLEMENTED SUBGROUPS AND p-NILPOTENCY OF FINITE GROUPS (3) G/Op(G) is p-nilpotent and CG(Op(G)) ≤ Op(G).
Suppose that G/Op(G) is not p-nilpotent. Then, by Frobenius’ theorem (refer [12], Theorem 10.3.2), there exists a subgroup of P properly containing Op(G) such that itsG-normalizer is not p-nilpotent. Since NG(P ) is p-nilpotent, we may choice a subgroupP1 of P such that Op(G) < P1 < P and NG(P1) is not p-nilpotent but NG(P2) is p-nilpotent whenever P1 < P2 ≤ P. Denote H = NG(P1). It is obvious that P1 < P0 ≤ Pfor some Sylow p-subgroup P0 of H. The choice of P1 implies that NG(P0) is p-nilpotent,hence NH (P0) is also p-nilpotent. Take x ∈ H\NH(P0). Since P0 = P ∩ H, we havex ∈ G\NG(P ). Again, so every minimal subgroup of P0 ∩P x ∩ HNp is c -supplemented in P0 by Lemma 2.1. If (a) is satisfied then every cyclic subgroup of P0 ∩P x ∩ HNp of order 4 is c -supplemented in P0. If (b) is satisfied then HNp ), P0 ≤ Z(P ∩ GNp) (P0 ∩ HNp) ≤ Z(P0 ∩ HNp). If (c) is satisfied then P0 is quaternion-free. Therefore H satisfies the hypotheses of thetheorem. The choice of G yields that H is p-nilpotent, which is contrary to the choiceof P1. Thereby G/Op(G) is p-nilpotent and G is p-solvable with Op (G) = 1. Conse-quently, we obtain CG(Op(G)) ≤ Op(G) (refer [13], Theorem 6.3.2).
(4) G = P Q, where Q is an elementary abelian Sylow q-subgroup of G for a prime q = p. Moreover, P is maximal in G and QOp(G)/Op(G) is minimal normalin G/Op(G). For any prime divisor q of |G| with q = p, since G is p-solvable, there exists a Sylow q-subgroup Q of G such that G0 = P Q is a subgroup of G [13] (Theorem 6.3.5). IfG0 < G, then, by (1), G0 is p-nilpotent. This leads to Q ≤ CG(Op(G)) ≤ Op(G), acontradiction. Thus G = P Q and so G is solvable. Now let T /Op(G) be a minimalnormal subgroup of G/Op(G) contained in Opp (G)/Op(G). Then T = Op(G)(T ∩ Q).
If T ∩ Q < Q, then P T < G and therefore P T is p-nilpotent by (1). It follows that 1 < T ∩ Q ≤ CG(Op(G)) ≤ Op(G), which is impossible. Hence T = Opp (G) and QOp(G)/Op(G) is an elementary abelianq-group complementing P/Op(G). This yields that P is maximal in G. (5) |P : Op(G)| = p.
Clearly, Op(G) < P. Let P0 be a maximal subgroup of P containing Op(G) and let G0 = P0Opp (G). Then P0 is a Sylow p-subgroup of G0. The maximality of P in Gimplies that either NG(P0) = G or NG(P0) = P. If the latter holds, then NG (P On the other hand, in view of (3), we have GNp ≤ Op(G), hence P ∩ P x ∩ GNp == GNp for every x ∈ G. Now it is easy to see that G0 satisfies the hypotheses of thetheorem. Thereby G0 is p-nilpotent and Q ≤ CG(Op(G)) ≤ Op(G), a contradiction.
Thus NG(P0) = G and P0 = Op(G). This proves (5).
(6) G = GNp L, where L = a [Q] is a non-abelian split extension of Q by a cyclic p-subgroup a , ap ∈ Z(L) and the action of a (by conjugate) on Q is irreducible.
From (3) we see that GNp ≤ Op(G). Clearly, T = GNpQ ✁ G. Let P0 be a maximal subgroup of P containing GNp . Then, by the maximality of P, either NG(P0) = Por NG(P0) = G. If NG(P0) = P, then NM (P0) = P0, where M = P0T = P0Q. ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8 M Np ≤ GNp for all x ∈ M \NM (P0), hence M satisfies the hypotheses of the theorem. By the minimality of G, M is p-nilpotent. It follows thatT = GNp Q = GNp × Q and so Q ✁ G, a contradiction. Thereby NG(P0) = G andP0 ≤ Op(G). This infers from (5) that Op(G) = P0 and hence P/GNp is a cyclicgroup. Now applying the Frattini argument we have G = GNp NG(Q). Therefore wemay assume that G = GNp L, where L = a [Q] is a non-abelian split extension of anormal Sylow q-subgroup Q by a cyclic p-group a . Noticing that |P : Op(G)| = pand Op(G) ∩ NG(Q) ✁ NG(Q), we have ap ∈ Z(L). Also since P is maximal in G,GNp Q/GNp is minimal normal in G/GNp and consequently a acts irreducibly on Q. (7) GNp has exponent p if p > 2 and exponent at most 4 if p = 2.
By Lemma 2.3 it will suffice to show that there exists a p-nilpotent maximal subgroup M of G such that G = GNp M. In fact, let M be a maximal subgroup of G containingL. Then M = L(M ∩ GNp) and G = GNpM. By Lemma 2.4, M ∩ GNp ✁ G, henceM = ( a (M ∩ GNp))Q. Write P0 = a (M ∩ GNp) and let M0 be a maximal subgroupof M containing P0. Then M0 = P0(M0 ∩ Q) and GNpM0 < G. By applying (1) weknow that GNp M0 is p-nilpotent, therefore M0 ∩ Q ≤ CG(Op(G)) ≤ Op(G). It follows that M0 ∩ Q = 1 and so P0 is maximal in M. In this case, if P0 ✁ M, then a = P0 ∩ L ✁ L, which is contrary to (6). Hence NM (P0) = P0 and M satisfies the hypotheses of the theorem. The choice of G implies that M is p-nilpotent, as desired.
Without losing generality, we assume in the following that P = GNp a .
(8) If GNp has exponent p, then GNp ∩ a = 1.
Assume on the contrary that GNp ∩ a = 1 if GNp has exponent p. Then we can take an element c in GNp ∩ a such that c is of order p. Since P is not normal in G, GNp∩ a << a . Consequently c ∈ ap ≤ Φ(P ) and c is permutable in P. By (6), (7) andLemma 2.2, we see that c is centralized by both GNp and L, hence c ∈ Z(G). If G satisfies(c) then, since GNp ≤ GN , c = 1 by Lemma 2.5, a contradiction. If G satisfies (a) or(b), we consider the factor group G = G/ c . It is obvious that N (P ) = N p-nilpotent, where P = P/ c . Now let y c / c be a minimal subgroup of GNp / c ,where y ∈ GNp. Since y is of order p, by the hypotheses, y has a c -supplement Kin P. If y ∩ K = 1 then K is a maximal subgroup of P and c ≤ K. It follows thatP/ c = ( y c / c )(K/ c ) with y c / c ∩ K/ c = 1. If y ∩ K = y then yis permutable in P and hence y c / c is permutable in P/ c . That is y c / c isc -supplemented in P/ c , therefore G satisfies (a) or (b). The choice of G implies thatG/ c is p-nilpotent and so G is p-nilpotent, a contradiction.
(9) The exponent of GNp is not p.
If not, GNp has exponent p. Let P1 be a minimal subgroup of GNp not permutable in P. Then, by the hypotheses, there is a subgroup K1 of P such that P = P1K1 andP1 ∩ K1 = 1. In general, we may find minimal subgroups P1, P2, . . . , Pm of GNp andalso subgroups K1, K2, . . . , Km of P such that P = PiKi and Pi ∩ Ki = 1 for each iand every minimal subgroup of GNp ∩K1 ∩. . .∩Km is permutable in P. Furthermore, wemay assume that Pi ≤ K1 ∩ . . . ∩ Ki−1, i = 2, 3, . . . , m . Henceforth K1 ∩ . . . ∩ Ki−1 == Pi(K1 ∩. . .∩Ki) for i = 2, 3, . . . , m . It is easy to see that GNp ∩Ki is normal in P and(GNp ∩ Ki) a is a complement of Pi in P, so we may replace Ki by (GNp ∩ Ki) a andfurther assume that a ≤ Ki for each i. Now, K1∩. . .∩Km = (GNp ∩K1∩. . .∩Km) a .
Since, for any x ∈ GNp ∩ K1 ∩ . . . ∩ Km, x a = a x , we have ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8 c -SUPPLEMENTED SUBGROUPS AND p-NILPOTENCY OF FINITE GROUPS xa ∈ (GNp ∩ K1 ∩ . . . ∩ Km) ∩ x a = x . This means that a induces a power automorphism of p-power order in the elementaryabelian p-group GNp ∩ K1 ∩ . . . ∩ Km. Hence [GNp ∩ K1 ∩ . . . ∩ Km, a] = 1 andK1 ∩ . . . ∩ Km is abelian.
Now we claim that p is even. If it is not the case, then, by [13] (Theorem 6.5.2), K1 ∩ . . . ∩ Km ≤ Op(G). Consequently, P = GNp(K1 ∩ . . . ∩ Km) ≤ Op(G), a con-tradiction. We proceed now to consider the following two cases: Case 1. | a | = 2n, n > 1.
Since K1 ∩ . . . ∩ Km is an abelian normal subgroup of P and a ∈ K1 ∩ . . . ∩ Km,
Φ(K1 ∩ . . . ∩ Km) = a2 P and so Ω1( a2 ) = c ≤ Z(P ), where c = a2n−1. Again, c ∈ Z(L) by (6), so c ∈ Z(G). If G satisfies (c) then we obtain c = 1 by Lemma 2.5,which is absurd. If G satisfies (a) or (b), then, with the same arguments to those used in(8), we may prove that G/ c satisfies the hypotheses of the theorem. The minimality ofG implies that G/ c is 2-nilpotent and therefore G is also 2-nilpotent, a contradiction.
Case 2. | a | = 2.
Since a acts irreducibly on Q, a is an involutive automorphism of Q; consequently,
Q is a cyclic subgroup of order q and ba = b−1, where Q = b . In this case, GN2is minimal normal in G. In fact, let N be a minimal normal subgroup of G containedin GN2 and let H = N L. Since N a is maximal but not normal in H, we see thatNH (N a ) = N a . Noticing that N a ∩HN2 ≤ N, every minimal subgroup of N a ∩ ∩ HN2 is c -supplemented in NH(N a ) = N a by Lemma 2.1. If further H < G,then the choice of G implies that H is 2-nilpotent. Consequently, N Q = N × Q andso 1 = N ∩ Z(P ) ≤ Z(G). The choice of N implies that N = N ∩ Z(P ) is of order2. This is contrary to Lemma 2.5 if G satisfies (c). Now assume that G satisfies (a) or(b). In this case, if N ≤ Φ(P ), then N has a complement to P. By applying Gasch¨utzTheorem [12] (I, 17.4), N also has a complement to G, say M. It follows that M is anormal subgroup of G. Furthermore, G/M is cyclic of order 2 and so N ≤ GN2 ≤ M,a contradiction. Hence N ≤ Φ(P ). Now we go to consider the factor group G/N. Forany minimal subgroup y N/N of (G/N )N2 = GN2 /N, by the hypotheses, P = y Kand y ∩ K is permutable in P, where y ∈ GN2. Since N ≤ K, we have P/N == ( y N/N )(K/N ) and ( y N/N ) (K/N ) = ( y ∩ K)N/N is permutable in P/N,so y N/N is c -supplemented in P/N. This yields at once that G/N is 2-nilpotent andso is G. Hence H = G and GN2 must be a minimal normal subgroup of G; of course,GN2 is an elementary abelian 2-group. Since GN2 ∩ NG(Q) ✁ NG(Q), we know thatGN2 ∩ NG(Q) = 1 and so b acts fixed-point-freely on GN2. We may assume that N1 == {1, c1, c2, . . . , cq} is a subgroup of GN2 with c1 ∈ Z(P ) and b = (c1, c2, . . . , cq) is apermutation of the set {c1, c2, . . . , cq}. Noticing that ba = b−1 and (c1)a−1ba = (c1)b−1,(c2)a = cq. By using (bi)a = b−i and (c1)a−1bia = (c1)b−i, we see that (ci+1)a == cq−i+1 for i = 1, 2, . . . , (q + 1)/2. Hence N1 is normalized by both GN2 and L andso N1 is normal in G. The minimal normality of GN2 implies that GN2 = N1, thus wehave Z(P ) = {1, c1}. Since GN2 ∩ K1 ∩ . . . ∩ Km is centralized by both GN2 and a , we have 1 < GN2 ∩ K1 ∩ . . . ∩ Km ≤ Z(P ). In view of P is not abelian, we get Φ(P ) = P = Z(P ), namely P is an extra-special 2-group. By applying Theorem 5.3.8of [12], there exists some positive integer h such that |P | = 22h+1. Hence |GN2| = 22h.
However, 22h − 1 = (2h + 1)(2h − 1) and q = 22h − 1, hence h = 1, q = 3 and |P | = 23.
Now we see that L ∼ = A4, therefore G ∼ = S4, which is contrary to the ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8 (10) The final contradiction.
From (7) and (9) we see that p = 2 and the exponent of GN2 is 4. By applying Lemma 2.3, Z(GN2 ) = Φ(GN2 ) is an elementary abelian 2-group. If Φ(GN2 ) ∩ a = 1then there exists an element c in Φ(GN2 ) ∩ a such that c is of order 2. Since Φ(GN2) a < a , we have c ∈ a2 ≤ Z(L). But c is also centralized by GN2 by Lemma 2.2, so c ∈ Z(G). If Φ(GN2) ∩ a = 1 then a induces a power automorphism of 2-powerorder in the elementary abelian 2-group Φ(GN2 ), hence [Φ(GN2 ), a] = 1. In view ofLemma 2.2, Φ(GN2 ) is also centralized by GN2 , hence Φ(GN2 ) ≤ Z(P ). Furthermore,by the Frattini argument, G = NG(Φ(GN2)) = CG(Φ(GN2))NG(P ). Noticing that NG(P ) = P and P ≤ CG(Φ(GN2)), we get CG(Φ(GN2)) = G, namelyΦ(GN2 ) ≤ Z(G). Thus we can also take an element c in Φ(GN2) such that c is of order2 and c ∈ Z(G). This is contrary to Lemma 2.5 if G satisfies (c). Now assume thatG satisfies (a). Denote N = c and consider G = G/N. It is clear that N (P ) = = NG(P )/N is 2-nilpotent because NG(P ) is, where P = P/N. For any y ∈ GN2,since y is c -supplemented in P, there exists a subgroup T of P such that P = y Tand y ∩ T is permutable in P. However, y2 Φ(GN2), hence y2 is permutable inP and y2 T forms a group. Because |P : y2 T | ≤ 2, N ≤ y2 T. It follows thatP/N = ( y N/N )( y2 T /N ) and y N/N ∩ y2 T /N = y2 ( y ∩ T )N/N is permutable in P/N. This shows that G satisfies (a). Thereby G is 2-nilpotent andso is G, a contradiction. Finally we assume that G satisfies (b). Let M be a max-imal subgroup of G containing L. Then M is 2-nilpotent by the proof of (7), henceΦ(GN2 )Q is 2-nilpotent and [Φ(GN2 ), Q] = 1. Write K = CG(GN2/Φ(GN2)). Then,by the hypotheses, P ≤ K ✁ G. The maximality of P yields that P = K or K = G.
If the former holds, then G = NG(P ) is 2-nilpotent, a contradiction. If the latterholds, then [GN2 , Q] Φ(GN2). This means that Q stabilizes the chain of subgroups1 Φ(GN2) ≤ GN2. It follows from [13] (Theorem 5.3.2) that [GN2, Q] = 1 and Q isnormal in G, which is impossible. This completes our proof.
Proof of Theorem 1.3. By applying Theorem 1.1, we only need to prove that NG(P )
If NG(P ) is not p-nilpotent, then NG(P ) has a minimal non-p-nilpotent subgroup (that is, every proper subgroup of a group is p-nilpotent but itself is not p-nilpotent) H. Byresults of Itˆo [2] (IV, 5.4) and Schmidt [2] (III, 5.2), H has a normal Sylow p-subgroupHp and a cyclic Sylow q-subgroup Hq such that H = [Hp]Hq. Moreover, Hp is ofexponent p if p > 2 and of exponent at most 4 if p = 2. On the other hand, the minimalityof H implies that HNp = Hp. Let P0 be a minimal subgroup of Hp and let K0 bea c -supplement of P0 in H. Then H = P0K0 and P0 ∩ K0 is permutable in H. IfP0 ∩ K0 = 1 then K0 is maximal in H of index p. By applying Lemma 2.6 we see thatK0 is normal in H. It follows from K0 is nilpotent that Hq is normal in H, a contradiction.
If P0 ∩ K0 = P0 then P0 is permutable in H. In this case, if P0Hq = H, then Hp = P0is cyclic and H is p-nilpotent by Lemma 2.6, a contradiction. Hence P0Hq < H andP0Hq = P0 × Hq. Thus Ω1(Hp) is centralized by Hq. If further CH(Ω1(Hp)) < H thenCH (Ω1(Hp)) is nilpotent normal in H. This leads to Hq ✁ H, a contradiction. ThereforeΩ1(Hp) ≤ Z(H). If Hp has exponent p, then Hp = Ω1(Hp) and H = Hp × Hq, ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8 c -SUPPLEMENTED SUBGROUPS AND p-NILPOTENCY OF FINITE GROUPS again a contradiction. Thus p = 2 and H2 has exponent 4. If G satisfies (b) then H2 isquaternion-free and, by Lemma 2.5, Hq acts trivially on H2, thus Hq is normal in H, acontradiction. Now assume that G satisfies (a). Let P1 = x be a cyclic subgroup of H2of order 4. Since P1 is c -supplemented in H, H = P1K1 with P1 ∩ K1 is permutablein H. If |H : K1| = 4 then |H : K1 x2 | = 2, hence K1 x2 ✁ H and so Hq ✁ H,a contradiction. If |H : K1| = 2 then K1 ✁ H. We still get a contradiction. ThereforeK1 = H and P1 is permutable in H. Now Lemma 2.6 implies that P1Hq is 2-nilpotentand consequently Hq is normalized by H2. This final contradiction completes our proof.
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ISSN 1027-3190. Ukr. mat. Ωurn., 2007, t. 59, # 8

Source: http://www.imath.kiev.ua/~umzh/Archiv/2007/08/UMZh_2007_08_1011.pdf

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