Jordan Journal of Mathematics and Statisticscs (JJMS) l (2), 2008, 23-32 WEAKLY C-NORMAL AND Cs-NORMAL SUBROUPS OF FINITE ROUPS * MOAMMAD TASTOUS ABSTRACT A subgrou of a finite grou is weakly c normal subgrou of if there exists a subnormal subgrou N of such that N N core, where contained in If N core ( ) =, and core denotes the core of in, which is the largest normal subgrou of core, then is S c normal subgrou of, where denotes the higher core of in, which is the largest subnormal subgrou of contained in In this aer, we investigate some roerties of weakly c normal and c S normal subgrous of finite grous, and using the weakly c normality and c S normality of some Sylow and maximal subgrous to determine the structure of finite grous INTRODUCTION It is interesting to use some information on the subgrous of a finite grou to determine the structure of the grou The normality of subgrous of a finite grou lays an imortant role in the study of finite grous Wang, 996 initiated the concets of c normal subgrous and used the c normality of maximal subgrous to give some conditions for solvability and suersolvability of a finite grou Lujin hu and et al, 2002 have introduced the concets of weakly c normal subgrous and they have used the weakly c normality of some maximal and Sylow subgrous to determine the structure of a finite grou Definition [0]: Let there is a series from to We say that is a subnormal subgrou of if A subgrou of a grou is called c normal subgrou of if there exists a normal subgrou N of such that = N and N, where = core is the largest normal subgrou of contained in 2000 Mathematics Subject Classification: Primary: 20D35, Secondary : 20E5; 20F22 Keywords: Normality, Weakly c normal, cs normal, Sylow and maximal subgrous Coyright Deanshi of Research and raduate Studies, Yarmouk University, Irbid, Jordan Received on: June 2, 2008 Acceted on: Dec 28, 2008 23
24 MOAMMAD TASTOUS Definition 2 [6]: Let be a grou We call a subgrou weakly c normal subgrou of if there exists a subnormal subgrou N of such that = N and, where core ( ) N = is the largest normal subgrou of contained in Examle 3: Let be a sylow 2 subgrou of the symmetric grou S 3 Then is weakly c normal subgrou of S 3 It is easy to see that, every c normal subgrou of a grou is weakly c normal in, however, the converse is not true, see [6] 2 PRELIMINARIES In this section, we give some definitions and basic results which are essential in the sequel Let π be a nonemty set of rimes, π the comlement set of π in the set of all rime numbers Let be a grou, we denote the set of all rime divisors of the order the grou by π ( ) O ( ) and the Fitting subgrou of by F ( ) ; the maximal normal subgrou of by We introduce the following concet which is closely related with the subnormal subgrous of a grou Definition 2: Let be a grou We call a subgrou cs normal subgrou of if there exists a subnormal subgrou N of such that = N and N, where denotes of the higher core of in which is the maximal subnormal subgrou of contained in Examle 22: Let be a sylow 2 subgrou of the alternating grou A 4 Then is cs normal subgrou of A 4 It is easy to see that, every normal (subnormal) subgrou of a grou, is cs normal subgrou of, but the converse is not true To see this, let S3 =, then the subgrou = ( 2) of is cs normal in, but is not normal subgrou of Clearly, every c normal subgrou of a grou is cs normal subgrou of, also, every weakly c normal subgrou of a grou is also subgrou of cs normal
WEAKLY C-NORMAL AND Cs-NORMAL SUBROUPS OF FINITE ROUPS 25 Lemma 23 [0]: Let be a grou with subgrous A and B such that A is an Abelian subgrou and = AB Then one of the following two conditions is satisfied: (i) A contains a normal subgrou C of such that for all x C, or (ii) B ( x Ax ) =, The following examle shows that the roerty of c S normality cannot imly weakly c normality Examle 24: Let = S3 S3 be the direct roduct of S 3 by itself, and { ( )} K = A3 A3 =, 23, 23, 32, 32 be the diagonal subgrou of = A3 A3 with K = and K = K Then K << since K < < and hence K is cs normal subgrou of But there is no subnormal subgrou, say N of, such that = NK and N K K Suose not, then there exists a subnormal subgrou N of with order having three cases: Case (i): If N < 2, then NK < 36, and therefore there is no subnormal subgrou N of such that = NK Case (ii): If N > 2, then by Lagrange theorem there are two situations; either, (a) N = 8, then NK > 36, or (b) N = 36, then we have = N, and thus NK = K =, but N K = K = K K = Therefore from the two situations there is no subnormal subgrou N of such that = NK Case (iii): Assume that N = 2, and suose that = NK Since K is an Abelian subgrou, then either N is a normal subgrou of, or one cannot find any normal subgrou K such that = NK In this case N is normal in if it is the maximal subgrou of Assume that N is normal in, then (ii) of lemma 23 cannot be satisfied here, and hence we have K x Nx = K N, for all x, hence K N Therefore NK If N is not normal of then K cannot found any subnormal subgrou such that = NK Thus K is not weakly c normal subgrou of
26 MOAMMAD TASTOUS 3 ELEMENTARY PROPERTIES Lemma 3: Let be a grou with subgrou Then (i) <<, and (ii) If L L2 where L and L 2 are two maximal subgrous of with L L2 <, then Proof: (i) We know that ( ) << Thus and << (by definition), then << (ii) If is subnormal of, then the result is obvious, so it is enough to show the case when is not subnormal of By the definition of the higher core and (i), there is a series of minimal length n > that has the form = M n < < M < M 0 =, where M i is not subnormal of for all i =,2,, n Then there exists a unique maximal subgrou of, say M, such that M for all i,2,, n L L This M i imels that n = ence < = which is a contradiction with 2 Lemma 32: Let be a grou with a subgrou Then Lemma 33 [6]: Let be a grou Then the following statements hold (i) If is weakly c normal subgrou of with K, then is weakly c normal subgrou of K (ii) Let K be a normal subgrou of with K Then is weakly c normal of iff K is weakly c normal of K Lemma 34: Let be a grou Then the following statements hold (i) If is cs normal subgrou of with K, then is cs normal subgrou of K (ii) Let K be a normal subgrou of with K subgrou of iff is c K S normal subgrou of K Then is cs normal Proof: (i) Suose that is cs normal in, then there exists a subnormal subgrou N of such that = N and N Then
WEAKLY C-NORMAL AND Cs-NORMAL SUBROUPS OF FINITE ROUPS 27 K = K = K N = ( K N ), and hence ( K N ) and K is subnormal of K, N K = N K K K Thus is cs normal in (ii) Suose that is c K S normal subgrou in K, then there exists a subnormal subgrou N of K K such that ( ) ( N ) ( ) ( N ) ( K K K ) Then we have N K ence is cs normal in =, and K K K = and N Conversely, assume that is cs normal subgrou in, then there exists a subnormal subgrou N of such that = N and N Then we have that ( )( NK ) =, and then NK is a subnormal of, and K K K ( ) ( NK NK K N K ) ( K K K K K K ) = = K ence is c K S normal subgrou of K Definition 35 [0]: For any set π of rime numbers, we denote by π the set of all rimes which do not belong to π If, then is said to be a all : is a π number π subgrou of if is a π number and [ ] Definition 36 [2]: A grou is called π solvable if it has a subnormal series whose factors are π grous or π grous and the π factors are solvable Lemma 37 [9]: Let be a π solvable grou Then has at least one solvable all π subgrou π, and for any π subgrou A of, there is an element x such that x A π In articular, any two all π subgrous are conjugate in For the roof of this lemma, the reader can see [2] and [9]
28 MOAMMAD TASTOUS Definition 38: Let be a grou We call a grou weakly nilotent if has a subnormal comlement in ie, if is a subnormal subgrou of and P is a sylow subgrou of such that weakly nilotent = P and P = then is called Clearly, if is nilotent, then is weakly nilotent, however, the converse is not true The following examle shows that the roerty of weakly nilotent cannot imly nilotent 4 TEOREMS Theorem 4: If is weakly c normal subgrou of a grou, then has a subnormal comlement in, ie, there exists a subnormal subgrou K of such that is the semidirect roduct of K and Conversely, if is a subgrou of such that then is weakly c normal of has a subnormal comlement in, Proof: Let be a weakly c normal subgrou of, then there exists a subnormal subgrou K of such that K K If =, then = and K = ence K is a subnormal comlement of in Assume that, then we can construct the factor grous K and and by Dedekind's Identity, we have K ( K ) = ( ) K = = ence K is a subnormal comlement of in Conversely, if F is a subgrou of such that F is a subnormal comlement of in, then we have that
WEAKLY C-NORMAL AND Cs-NORMAL SUBROUPS OF FINITE ROUPS 29 = F, and F = Then = F, where F is a subnormal subgrou in, and F Therefore is weakly c normal subgrou in Corollary 42: Let be a subgrou of a grou such that has a subnormal comlement in Then is cs normal in The following examle shows that the converse of the above corollary is not necessarily true Examle 43: Let = be the semidirect roduct of a subgrou and the cyclic grou with with; (i), (iii) ( ) is a maximal subgrou in, (ii) is not normal in, = Then < since dose not divide (by Sylow theorem), and hence is cs normal in, also is cs normal in We claim that has no subnormal comlement in, suose not, ie, has a subnormal comlement in, say K Then = K, and ( K ) K = = Then = K, and K = But we know that K K = = K Since = = = K = K = If K =, we get a contradiction with < K By using Cauchy Theorem we have K contains a subgrou of order, say L, since K, then L
30 MOAMMAD TASTOUS will be a sylow subgrou of since ( ), = Since is the unique sylow subgrou of, and its normal in, then This contradicts the maximality of in L = Then < K Theorem 44: A grou has a weakly c normal sylow subgrou if and only if the factor grou O is weakly nilotent Proof: Assume that has a weakly c normal sylow subgrou, say P Then by lemma 33 we have P P has a subnormal comlement in P = O ) ence is weakly nilotent O P (notice that Conversely, assume that O is weakly nilotent Then there exists a subnormal subgrou K O of O such that K P = O O O, and O ( ) O ( ) K P = Then we have = KP such that K is a subnormal subgrou in, and K P = O = P Therefore P is weakly c normal subgrou in a grou Corollary 45: If a factor grou O is weakly nilotent, then has a cs normal sylow subgrou This corollary is obvious and we omit the roof The converse of the Corollary 45 is not necessarily true; regarded with Examle 43, we have seen that is c normal sylow subgrou of such that S O = has no subnormal comlement O = ence
WEAKLY C-NORMAL AND Cs-NORMAL SUBROUPS OF FINITE ROUPS 3 Theorem 46: A grou is metanilotent if and only if every sylow subgrou of is weakly c normal in Proof: Suose that is metanilotent Let π ( ) and P be a sylow subgrou of Since and F F ( ) ( ) are solvable Then is solvable Moreover is solvable By lemma 35, we can relace π by K, and O O Since ( F( ) ) x A by to conclude that has a solvable all subgrou K Since ( F( ) ) is normal in O ( F ( ) ) K and hence KF ( ) is nilotent, then we have F ( ) subgrou of O K = F K F F ( ) ence O K KF is a normal all = = is normal subgrou in, and consequently, we deduce that = P and P = P O K = O P K = O = P Therefore P is weakly c normal in Conversely, Suose that every sylow subgrou of is weakly c normal in Then, by the third isomorhism theorem and Theorem 44 and the fact that F = O, O ( ) O ( ) F ( ) F O ( ) is nilotent for all π ( ) metanilotent Therefore F ( ) is nilotent, and hence is Corollary 47: A grou is metanilotent if and only if every sylow subgrou of is cs normal in This corollary is obvious and we omit the roof
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