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1 ath-net.ru All Russian mathematical ortal I.. Dergacheva, I. P. Shabalina, E. A. Zadorozhnyu, A criterion for a finite grou to belong a saturated formation, PFT, 017, Issue (31), Use of the all-russian mathematical ortal ath-net.ru imlies that you have read and agreed to these terms of use htt:// Download details: IP: July 1, 018, 05:18:33

2 Problems of Physics, athematics and Technics, (31), 017 УДК МАТЕМАТИКА КРИТЕРИЙ ПРИНАДЛЕЖНОСТИ КОНЕЧНОЙ ГРУППЫ НАСЫЩЕННОЙ ФОРМАЦИИ И.М. Дергачева, И.П. Шабалина, Е.А. Задорожнюк Белорусский государственный университет транспорта A CRITERION FOR A FINITE ROUP TO BELON A SATURATED FORATION I.. Dergacheva, I.P. Shabalina, E.A. Zadorozhnyu Belarusian State University of Transort Доказывается следующий результат: пусть такая наследственная насыщенная формация -разрешимых групп, содержащая все -сверхразрешимые группы, что Пусть AT где A холлова -подгруппа из, и T -сверхразрешимая подгруппа из. Предположим, что для силовской -подгруппы P из T мы имеем P Если A перестановочна с холловой -подгруппой из T и со всеми такими максимальными подгруппами V из P, что P V то Ключевые слова: конечная группа, насыщенная формация, -разрешимая группа, -сверхразрешимая группа, холлова подгруппа. We rove the following result: Let be a hereditary saturated formation of -soluble grous containing all -suersoluble grous such that Let AT where A is a Hall -subgrou of, and T is a -suersoluble subgrou of. Suose that for a Sylow -subgrou P of T we have P If A ermutes with a Hall -subgrou of T and with all maximal subgrous V of P such that P V then Keywords: finite grou, saturated formation, -soluble grou, -suersoluble grou, Hall subgrou. Introduction Throughout this aer, all grous are finite and always denotes a finite grou. oreover, is always suosed to be a rime and is a non-emty subset of the set of all rimes; denotes the set of all rimes q A subgrou H of is said to ermute with a subgrou K of if HK KH By the well nown Hall theorem [1], is soluble if every Sylow subgrou P of has a comlement T in, that is, a subgrou of such that PT and PT 1 The examle of the alternating grou A 5 shows that such a result is incorrect in general if we consider only the Sylow -subgrous for some fixed. Nevertheless, B. Huert [] roved that if a Sylow -subgrou P of has a comlement T in P and T ermutes with every maximal subgrou of P then is -soluble. This result was imroved in some directions. V. Sergieno [3] on the base of this result roved that if a Sylow -subgrou P of has a comlement T in, there is a number such that 1 P and T ermutes with all subgrous of P of order and P is abelian in the case then is -so-luble and the -length of is equal to 1. Further, Dergacheva I.., Shabalina I.P., Zadorozhnyu E.A., Boroviov [4] roved that under these conditions, is even -suersoluble. In [5] W. uo, K.P. Shum and A.N. Siba roved that if AT where A is a Hall -subgrou of T is nilotent, and A ermutes with all Sylow subgrous of T and with all maximal subgrous of any Sylow subgrou of T, then is -suersoluble, for each rime such that T for a Sylow -subgrou T of T See also aers [6], [7]. In this aer we rove the following result in this line researches. Theorem. Let be a hereditary saturated formation of -soluble grous containing all - suersoluble grous such that Let AT where A is a Hall -subgrou of, and T is a -suersoluble subgrou of Suose that for a Sylow -subgrou P of T we have P If A ermutes with a Hall -subgrou of T and with all maximal subgrous V of P such that P V then All unexlained notation and terminology are standard. The reader is referred to [8] [10] or [11] if necessary.

3 A criterion for a finite grou to belong a saturated formation 1 Preliminaries Lemma 1.1. Let be a hereditary formation. Let H E and E where E and are Sylow -subgrous of E and resectively. Suose also that H E (1) If N is a normal subgrou of and ( N) ( PN N) HN N then P H () If E E H then H Proof. (1) Assume that P H Then N ( P) NH so ( N) ( PN N) ( N N) ( PN N) N( N P) N N ( P)( NP) N N ( P) N NH N a contradiction. Hence we have P H () Since the formation is hereditary, E E E Hence this assertion directly follows from the inclusion E E Lemma 1.. If is -suersoluble and O ( ) 1 then is suersoluble and F ( ) O ( ) is normal Sylow -subgrou of where is the largest rime dividing Lemma 1.3. Let be a saturated formation containing suersoluble grous and E a minimal normal subgrou of such that E If E is abelian and -central, then Proof. Clearly, we can suose that E ( ) Let be a maximal subgrou of such that E and let C C ( E) Then C ( E ) ( ) since C Thus E Lemma 1.4 (O.H. Kegel [1]). Let A and B be x x subgrous of such that AB and AB B A for all x Then has a roer normal subgrou N such that either A N or B N Lemma 1.5 (V.N. Knyagina and V.S. onahov [13]). Let H K and N be subgrous of If H is a Hall subgrou of and H ermutes with K then N HK ( N H)( N K) Proof of Theorem Assume that this theorem is false and let be a counterexamle of minimal order. Then 1 We roceed our roof by roving the following claims: (1) O ( ) 1 In view of Lemma 1.1(1), the hyothesis still holds for D D by the choice of But then since a contradiction. Thus we have (1). () P 1 Indeed, if P 1 then is a -grou. Hence 1 by Claim (1), a contradiction. (3) T is suersoluble, O ( T) 1 and P is normal in T Since O ( T) is normal in T AT A ( O( T)) ( O) ( O) AT TA where T is a Hall -subgrou of T Hence ( O( T)) O( ) 1 so O( T) 1 Hence, since T is -suersoluble by hyothesis, T is suersoluble and P is normal in T by Lemma 1.. (4) is not -soluble. Hence is not -soluble. Assume that is -soluble. Let L be a minimal normal subgrou of Then by Claim (1), L is a -grou L P Next note that LF Indeed, if PL then the assertion follows from Lemma 1.3. On the other hand, if PL the hyothesis is true for L by Lemma 1.1 (1). Hence L by the choice of Therefore L ( ) Hence L and L ( T) Let be a maximal subgrou of T such that L T Then every Hall -subgrou of is a Hall -subgrou of T. Since T is soluble, any two Hall -subgrous are conjugate in T Hence without loss of generality we may suose that where is a Sylow -subgrou of and is a Hall -subgrou of such that A A Since T is suersoluble, T so is a maximal subgrou of P Note also that L Indeed, if L then from the -isomorhism L LL we deduce that L is -central in and hence by Lemma 1.3, contrary to the choice of Hence A ermutes with Therefore A A A is a subgrou of with A and with L A But then L a contradiction. Thus we have (4). (5) If H is a minimal normal subgrou of and H then P Indeed, if P the hyothesis is still true for H H by the choice of Hence by Lemma 1.3, contrary to the choice of Problems of Physics, athematics and Technics, (31),

4 I.. Dergacheva, I.P. Shabalina, E.A. Zadorozhnyu (6) If H is a normal subgrou of and H A A then H is -soluble. It is clear that H ( AH)( T H) Let E ( H A) T Let V be a maximal subgrou of P Suose that E P V Then, by Lemma 1.1 (), P V Hence AV VA is a subgrou of Therefore AV ( A H ) P ( AH)( AV P) ( AH) V( AT) ( A HV ) V( AH) Thus the hyothesis is still true for E If E then A A( H A) P ( H A)( AT) H A a contradiction. Hence, E E by the choice of Since every grou in is -soluble by hyothesis, we conclude that H E is -soluble. (7) O ( ) Suose that O ( ) Since the hyothesis holds for O ( ) by Lemma 1.1 (), O ( ) by the choice of But then is -soluble, contrary to Claim (4). (8) If H is a -soluble minimal normal subgrou of then H and H Z ( ) First note that if H and C C ( H) then C as a grou of automorhisms of H is a cyclic grou of order dividing 1 Hence in this case we have H Z ( ) by Claim (6). Therefore we need only show that H Clearly, H is either -grou or -grou. But the former case is imossible by Claim (1), so H for some natural a a If either H P or PH then is clearly -soluble, contrary to Claim (4). Hence H is a maximal subgrou of P Suose that a 1 Then P is not cyclic. Therefore for some maximal subgrou V of P we have P HV Suose that P V Then and H Thus, in view of Claims (1) and (6), H Since is -soluble and O ( ) there is a normal maximal subgrou of such that and Since H it follows that H Hence H a contradiction. Then P V which imlies that A ermutes with V Now, as in the roof of Claim (4), it may be roved that there is a subgrou W of such that W and H W. But then H a contradiction. Hence we have (8). (9) P is not cyclic. Suose on the contrary that P is cyclic. First we show that in this case does not have a roer normal subgrou E with EP Indeed, if 48 EP where E is normal in and E then for any Sylow q-subgrou Q of A we have EN ( ) Q by the Frattini argument. Hence P D N for some Sylow -subgrous D of D and N of N ( Q) But P is cyclic P N ( ) Q Now let W be the Hall -subgrou of T such that AW AW Then AWP AW Q Q Q QAW AW where AW WA is a -subgrou of Hence Q O ( ) which contradicts Claim (1). Now suose that and let where is a normal subgrou of with simle quotient In view of Claim (7), divides But then, since consists of -soluble grous, is a -grou and hence P This contradiction shows that so A ermutes with the maximal subgrou Z of P Since T is suersoluble by Claim (3), Z is normal in T Hence AT A D Z Z Z ZA By Lemma 1.4, D ( AD)( T D) Assume that either D AT of T P Then D is -soluble. Indeed, in the former case we have DA A and so, by Claim (6), D is soluble. On the other hand, if A D and T P then the hyothesis still holds on DP Since DP DP is -suersoluble by the choice of Now, let H be a minimal normal subgrou of contained in D Then since D is -soluble, H and H Z ( ) by Claim (8). Let N N ( P) If P Z( N) then is -nilotent by the Burnside theorem [14], which contradicts the choice of Hence N CN ( P) Let x N\C( P) with ( x P) 1 and E P x By [8, III, 13.4], P [ EP] ( PZ( E)) Since H P Z( E) and P is cyclic, it follows that P P Z( E) x C ( ) P This contradiction shows that T P and D Let q be a rime dividing A and Q be any Sylow q-subgrou of A. Let N N ( ) Q Clearly, Q is a Sylow subgrou of D by the Frattini argument we have DN P D N for some Sylow subgrou D of D and Sylow subgrou N of N But P is cyclic P N Hence AP A Q Q Q A which contradicts Claim (1). Hence we have (9). (10) P Suose on the contrary that P By Claim (9), P is not cyclic. If Z is a maximal subgrou of P, then Z AZ so by Claim (3). Therefore T has Проблемы физики, математики и техники, (31), 017

5 A criterion for a finite grou to belong a saturated formation at least three different subgrous Z1Z Z3 of order such that P Zi Let Ni Zi be the normal closure of Z i in Then Ni AZi Ni N j is contained in O ( ) 1 for any different i j{1 3} Hence P Ci C( Ni) for all i Assume that for some i Ci Then C i is -soluble by Claim (6), is -soluble since C i is a -grou. This contradiction shows that Ci for all i It follows N1 N and N 3 are abelian grous Ni Zi for all i P is normal in It follows that is -soluble, which contradicts Claim (4). Thus we have (10). (11) O ( ) 1 Let D O ( ) 1 and H a minimal normal subgrou of contained in D Then H by Claim (8) P by Claim (5), which contradicts Claim (10). Final contradiction. Let V be a maximal subgrou of P and N V be the normal closure of V in Suose that P V Then N A If N A A then N is -soluble by Claim (5) and hence O ( ) 1 which contradicts Claim (11). Therefore N A A Hence N AV and N so N Therefore P N P AV P V Thus P( P) But then is -nilotent by Tate s theorem [8]. It follows that is -soluble, contrary to Claim (4). REFERENCES 1. Hall, P. A characteristic roerty of soluble grous / P. Hall // J. London ath. Soc Vol. 1,. P Huert, B. Zur Sylowstrutur auflosbarer ruen / B. Huert // Arch. ath Vol. 1. P Sergieno, V.I. A criterion for the -solubility of finite grous / V.I. Sergieno // at. Zam Vol. 9. P Boroviov,.T. rous with ermutable subgrous of mutually simle orders /.T. Boroviov // Questions of Alg Vol. 5. P uo, W. Criteria of Existence of Hall Subgrous in Non-soluble Finite rous / W. uo, A.N. Siba // Acta ath. Sinica, English Ser Vol. 6,. P uo, W. Finite grous with some given systems of X m -semiermutable subgrous / W. uo, K.P. Shum, A.N. Siba // ath. Nachr Vol. 83. P Yi, X. On some generalizations of ermutability and S-ermutability / X. Yi, A.N. Siba // Problems of Physics, athematics and Technics (17). P Huert, B. Endliche ruen I / B. Huert. Berlin, Heidelberg, New Yor: Sringer, Doer, K. Finite Soluble rous / K. Doer, T. Hawes. Berlin, New Yor: Walter de ruyter, uo, W. The Theory of Classes of rous / W. uo. Beijing, New Yor, Dordrecht, Boston, London: Science Press, Kluwer Academic Publishers, Ballester-Bolinches, A. Classes of finite grous / A. Ballester-Bolinches, L.. Ezquerro. Dordrecht: Sringer, Kegel, O.H. Produte nilotenter ruen / O.H. Kegel // Arch. ath Vol. 1. P Knyagina, V.N. On -roerties of finite grou having a Hall -subgrou / V.N. Knyagina, V.S. onahov // Siberian ath. J P orenstein, D. Finite rous / D. orenstein. New Yor, Evanston, London: Harer & Row Publishers, Поступила в редакцию Problems of Physics, athematics and Technics, (31),