Mais les résultats doivent être attendus longtemps et il n'y a généralement pas de temps amoxicilline prix L'autre cas, c'est que l'achat d'un ou d'un autre antibiotique dans une pharmacie classique nécessite des dépenses matérielles considérables et pas toutes les personnes ne peuvent acheter des produits pharmaceutiques aussi coûteux.

## Imath.kiev.ua

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

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