A G-covering subgroup system of a finite group for some classes of σ-soluble groups
aa r X i v : . [ m a t h . G R ] J a n A G -covering subgroup system of a finite groupfor some classes of σ -soluble groups ∗ A-Ming Liu, W. Guo,
School of Science, Hainan University, Haikou, 570228, P.R. ChinaE-mail: [email protected], [email protected]
Inna N. Safonova
Department of Applied Mathematics and Computer Science,Belarusian State University, Minsk 220030, BelarusE-mail: [email protected]
Alexander N. Skiba
Department of Mathematics and Technologies of Programming,Francisk Skorina Gomel State University,Gomel 246019, BelarusE-mail: [email protected]
Abstract
Let F be a class of group and G a finite group. Then a set Σ of subgroups of G is called a G -covering subgroup system for the class F if G ∈ F whenever Σ ⊆ F .We prove that: If a set of subgroups Σ of G contains at least one supplement to each maximalsubgroup of every Sylow subgroup of G , then Σ is a G -covering subgroup system for the classesof all σ -soluble and all σ -nilpotent groups, and for the class of all σ -soluble P σT -groups.
This result gives positive answers to questions 19.87 and 19.88 from the Kourovka notebook.
Throughout this paper, all groups are finite and G always denotes a finite group. Moreover, P isthe set of all primes and σ is some partition of P , that is, σ = { σ i | i ∈ I } , where P = S i ∈ I σ i and σ i ∩ σ j = ∅ for all i = j .Before continuing, recall some concepts of the papers [1, 2, 3, 4] which play fundamental role inthe theory of σ -properties of groups. ∗ Research was supported by the NNSF of China (No. 11771409.) Keywords: finite group, G -covering subgroup system, σ -soluble group, σ -nilpotent group, P σT -group. Mathematics Subject Classification (2010): 20D10 G is said to be: σ -primary if G is a σ i -group for some i = i ( G ); σ -decomposable or σ -nilpotent if G = G × · · · × G n for some σ -primary groups G , . . . , G n ; σ -soluble if every chief factorof G is σ -primary.A set H of subgroups of G is a complete Hall σ -set of G if every member = 1 of H is a Hall σ i -subgroup of G for some σ i ∈ σ and H contains exactly one Hall σ i -subgroup of G for every i .Recall that a subgroup A of G is said to be σ -permutable in G if G possesses a complete Hall σ -set H such that AH x = H x A for all H ∈ H and all x ∈ G .We say that G is a P σT -group if σ -permutability is a transitive relation in G , that is, if K isa σ -permutable subgroup of H and H is a σ -permutable subgroup of G , then K is a σ -permutablesubgroup of G .Let F be a class of group and G a finite group. Then a set Σ of subgroups of G is called a G -covering subgroup system [5] for the class F if G ∈ F whenever Σ ⊆ F .In this paper, we prove the following Theorem A.
Suppose that a set of subgroups Σ of G contains at least one supplement to eachmaximal subgroup of every Sylow subgroup of G . Then Σ is a G -covering subgroup system for anyclass F in the following list: (i) F is the class of all σ -soluble groups. (ii) F is the class of all σ -nilpotent groups. (iii) F is the class of all σ -soluble P σT -groups.
The theory of
P σT -groups was built in the works [1, 2, 3]. Theorem A gives positive answers toquestions 19.87 and 19.88 from the Kourovka notebook [6] and, also, allows us to give the followingnew characterization of σ -soluble P σT -groups.
Corollary 1.1. G is a σ -soluble P σT -group if and only if each maximal subgroup of every Sylowsubgroup of G has a supplement T in G such that T is a σ -soluble P σT -group.
In the classical case when σ = σ = {{ } , { } , . . . } : G is σ -soluble (respectively σ -nilpotent) ifand only if G is soluble (respectively nilpotent); σ -permutable subgroups are also called S -permutable [7]; in this case a P σT -group is also called a
P ST -group [7].A significant place to the theory of
P ST -groups is given in the book [6]. From Theorem A weget also the following result in this line researches.
Corollary 1.2. G is a soluble P ST -group if and only if each maximal subgroup of every Sylowsubgroup of G has a supplement T in G such that T is a soluble P ST -group. Basic lemmas If n is an integer, the symbol π ( n ) denotes the set of all primes dividing n ; as usual, π ( G ) = π ( | G | ),the set of all primes dividing the order of G . G is said to be a D π -group if G possesses a Hall π -subgroup E and every π -subgroup of G is contained in some conjugate of E .By the analogy with the notation π ( n ), we write σ ( n ) to denote the set { σ i | σ i ∩ π ( n ) = ∅} ; σ ( G ) = σ ( | G | ). G is said to be: a σ -full group of Sylow type [1] if every subgroup E of G is a D σ i -group for every σ i ∈ σ ( E ). Lemma 2.1 (See Theorem A [8]).
Every σ -soluble group is a σ -full group of Sylow type. Lemma 2.2 (Theorem 1 in [9]). G is π -separable if and only if (i) G has a Hall π -subgroup and a Hall π ′ -subgroup; (ii) G has a Hall π ∪ { p } -subgroup and a Hall π ′ ∪ { q } -subgroup for all p ∈ π ′ and q ∈ π . Lemma 2.3 (See Corollary 2.4 and Lemma 2.5 in [1]).
The class of all σ -nilpotent groups N σ isclosed under taking products of normal subgroups, homomorphic images and subgroups. Moreover,if E is a normal subgroup of G and E/ ( E ∩ Φ( G )) is σ -nilpotent, then E is σ -nilpotent. In view of Lemma 2.3, the class N σ , of all σ -nilpotent groups, is a hereditary saturated formationand so from Proposition 2.2.8 in [10] we get the following Lemma 2.4 (See Proposition 2.2.8 in [10]). If N is a normal subgroup of G , then ( G/N ) N σ = G N σ N/N.
In this lemma, G N σ denotes the σ -nilpotent residual of G , that is, the intersection of all normalsubgroups N of G with σ -nilpotent quotient G/N . Lemma 2.5 (See Theorem A in [2]). If G is a σ -soluble P σT -group and D = G N σ , then thefollowing conditions hold: (i) G = D ⋊ M , where D is an abelian Hall subgroup of G of odd order, M is σ -nilpotent andevery element of G induces a power automorphism in D ; (ii) O σ i ( D ) has a normal complement in a Hall σ i -subgroup of G for all i .Conversely, if Conditions (i) and (ii) hold for some subgroups D and M of G , then G is a P σT -group.
Proof of Theorem A.
Assume that this theorem is false. We can assume without loss ofgenerality that σ ( G ) = { σ , σ , . . . , σ t } . 3I) G is not σ -nilpotent. Hence t > and D := G N σ = 1 . Indeed, assume that G is σ -nilpotent. Then G is σ -soluble. Hence Statements (i) and (ii) holdfor G . Moreover, in this case for every i the product H i , of all normal σ i -subgroups of G , is theunique normal Hall σ i -subgroup of G and G = H × H × · · · × H t . Hence every subgroup of G is σ -permutable in G . Thus Statement (iii) also holds for G , contrary to our assumption on G . Hence(I) hods.(i) Assume that this assertion is false and let G be a counterexample of minimal order.(*) G has no non-identity normal σ -primary subgroups. Assume that G has a minimal normal σ -primary subgroup, R say.Let P/R be any non-identity Sylow subgroup of
G/R . Then for some prime p and for a Sylow p -subgroup G p of G we have G p R/R = P/R , so G p is non-identity.Now let V /R be any maximal subgroup of
P/R , that is, | P : V | = | P/R : V /R | = p . Then V = R ( G p ∩ V ), so p = | G p R : R ( G p ∩ V ) | = ( | G p || R | : | G p ∩ R | ) : ( | R || ( G p ∩ V | : | R ∩ G p ∩ V | ) = | G p : G p ∩ V | , so G p ∩ V is a maximal subgroup of G P . Hence G has a σ -soluble subgroup T such that ( G p ∩ V ) T = G .But then RT /R ≃ T / ( T ∩ R ) is a σ -soluble subgroup of G/R such that(
V /R )( RT /R ) = ( R ( G p ∩ V ) /R )( T R/R ) =
G/R.
Therefore the hypothesis holds for
G/R , so
G/R is σ -soluble by the choice of G . But then G is σ -soluble, a contradiction. Hence we have (*).(**) t = 2 , that is, σ ( G ) = { σ , σ } . Assume that t > P i be a Sylow p i -subgroup of G for some p ∈ σ ∩ π ( G ), p ∈ σ ∩ π ( G )and p ∈ σ ∩ π ( G ). Let V i be a maximal subgroup of P i . Then, by hypothesis, G has σ -solublesubgroups T , T and T such that G = V i T i for i = 1 , , . Let R be a minimal normal subgroup of T . Then R is σ -primary, R is a σ k -group say. Since | G : T | = | T T : T | = | T : T ∩ T | is a p -number and | T : T ∩ T | is a p -number, where p ∈ σ and p ∈ σ , we have either R ≤ T ∩ T or R ≤ T ∩ T , R ≤ T ∩ T say. Hence R G = R T T = R T ≤ T , so G has a non-identity normal σ -primary subgroup, contrary to Claim(*). Thus (**) holds. The final contradiction for (i).
Let π = σ ∩ π ( G ). Since T i is σ -soluble, T i has a Hall σ k -subgroup for all k by Lemma 2.1. Then a Hall σ -subgroup of T is a Hall π ′ subgroup of G and aHall σ -subgroup of T is a Hall π − subgroup of G .Now we show that G has a Hall π ∪ { p } -subgroup for every p ∈ σ ∩ π ( G ). If | σ ∩ π ( G ) = 1it is evident. Now assume that | σ ∩ π ( G ) | > q ∈ ( σ ∩ π ( G )) \ { p } . Let V be a maximal4ubgroup of a Sylow q -subgroup Q of G . And let T be a σ -soluble supplement to V in G . Then T is π -separable by Claim (**). Hence T has a Hall π ∪ { p } -subgroup H by Lemma 2.2. But | G : T | is a { q } -number, where p = q σ , so H is a Hall π ∪ { p } -subgroup of G .Similarly it can be proved that G has a Hall π ′ ∪ { p } -subgroup for all p ∈ π . Therefore G is π -separable by Lemma 2.2 and so G is σ -soluble by Claim (**), contrary to the choice of G . HenceStatement (i) holds.(iii) Assume that this assertion is false and let G be a counterexample of minimal order. Then G is σ -soluble by Part (i).(1) If R is a non-identity normal subgroup of G , then the hypothesis holds for G/R . Hence
G/R is a σ -soluble P σT -group (See the proof of Claim (*)).(2) If R is an abelian minimal normal subgroup of G , then R is not a Sylow subgroup of G . Indeed, assume that R is Sylow subgroup of G and let V be an y maximal subgroup of R . Thenfor every supplement T to V in G we have that T ∩ R is normal in G , the minimality of R impliesthat T = G . Hence G is a σ -soluble P σT -group, a contradiction. Hence (2) holds.(3) D is σ -nilpotent. Assume that this is falls. Then D is not σ -primary. Let R be a minimal normal subgroup of G ,so R is a σ i -group for some i since G is σ -soluble. Moreover, from Lemmas 2.3 and 2.4 we get that( G/R ) N σ = G N σ R/R = DR/R ≃ D/ ( D ∩ R )is a Hall σ -nilpotent subgroup of G/R by Claim (1). Hence R is the unique minimal normal subgroupof G , R < D and R (cid:2) Φ( G ) since D is not σ -nilpotent. Therefore C G ( R ) ≤ R and D/R is a Hallsubgroup of
G/R . Moreover,
D/R is not a σ i -group since D is not σ -primary. Let p be a primedividing | D/R | such that p σ i . And let P be a Sylow p -subgroup of D . Then P ∩ R = 1 and P isa Sylow p -subgroup of G since D/R is a Hall subgroup of
G/R .Let V be a maximal subgroup of P and T a supplement to V in G such that T is a P σT -group.Then T N σ ≤ D and T N σ is a Hall abelian subgroup of T such that every subgroup of T N σ is normalin T by Lemma 2.5. Moreover, R ≤ T since | G : T | is a { p } -number. Hence T N σ ∩ R is a normalabelian Hall subgroup of R . Hence either T N σ ∩ R = 1 or T N σ ∩ R = R and so R ≤ T N σ .First assume that T N σ ∩ R = 1. Then T N σ ≤ C G ( R ), so T N σ = 1 and hence T is σ -nilpotent.From P = P ∩ V T = V ( P ∩ T ) it follow that T is not a σ i -group. Hence for a Hall σ ′ i -subgroup E of T we have E = 1 and E ≤ C G ( R ) ≤ R , a contradiction. Therefore R ≤ T N σ , so R = T N σ is a q -group for some prime q = p since C G ( R ) ≤ R and T N σ is abelian. Let Q be a Sylow q -subgroupof T . Then R = Q since R = T N σ is a Hall subgroup of T . Moreover, R is a Sylow q -subgroup of G since p = q and | G : T | is a { p } -number, contrary to Claim (2). Hence we have (3).(4) D is nilpotent. Assume that this false and let R be a minimal normal subgroup of G . Then R ≤ D and5 G ( R ) ≤ R and D/R is a Hall subgroup of
G/R (see the proof of Claim (3)). Hence D ≤ O σ i ( G ) forsome i by Claim (3). Let P be a Sylow p -subgroup of G , where p ∈ π ( G ) \ σ i . Let V be a maximalsubgroup of P and T a supplement to V in G such that T is a P σT -group. Then R ≤ D ≤ T , so T N σ ∩ R is a normal abelian Hall subgroup of R . Hence R ≤ T N σ (see the proof of Claim (*)). Onthe other hand, T N σ ≤ D . Therefore R = T N σ is a Sylow q -subgroup of G for some q = p , contraryto Claim (2).(5) D is a Hall subgroup of G . Suppose that this is false and let P be a Sylow p -subgroup of D such that 1 < P < G p , where G p ∈ Syl p ( G ). We can assume without loss of generality that G p ≤ H .(a) D = P is a minimal normal subgroup of G . Let R be a minimal normal subgroup of G contained in D . Since D is nilpotent by Claim (4), R is a q -group for some prime q . Moreover, D/R = (
G/R ) N σ is a Hall subgroup of G/R by Claim (1)and Lemma 2.3. Suppose that
P R/R = 1. Then P R/R ∈ Syl p ( G/R ). If q = p , then P ∈ Syl p ( G ).This contradicts the fact that P < G p . Hence q = p , so R ≤ P and therefore P/R ∈ Syl p ( G/R ) andwe again get that P ∈ Syl p ( G ). This contradiction shows that P R/R = 1, which implies that R = P is the unique minimal normal subgroup of G contained in D . Since D is nilpotent by Claim (4), a p ′ -complement E of D is characteristic in D and so it is normal in G . Hence E = 1, which impliesthat R = D = P .(b) D (cid:2) Φ( G ) . Hence for some maximal subgroup M of G we have G = D ⋊ M . (c) If G has a minimal normal subgroup L = D , then G p = D × ( L ∩ G p ) . Hence O p ′ ( G ) = 1 . Indeed,
DL/L ≃ D is a Hall subgroup of G/L by Claim (1). Hence G p L/L = RL/L , so G p = D × ( L ∩ G p ). Thus O p ′ ( G ) = 1 since D < G p by Claim (a).(d) V = C G ( D ) ∩ M is a normal subgroup of G and C G ( D ) = D × V ≤ H . In view of Claim (b), C G ( D ) = D × V , where V = C G ( D ) ∩ M is a normal subgroup of G . ByClaim (a), V ∩ D = 1 and hence V ≃ DV /D is σ -nilpotent by Lemma 2.2. Let W be a σ -complementof V . Then W is characteristic in V and so it is normal in G . Therefore we have (d) by Claim (c). The final contradiction for (5).
Let Q be a Sylow q -subgroup of G , where q ∈ π ( G ) \ π ( H ). Let V be a maximal subgroup of P and T a supplement to V in G such that T is a P σT -group. Then T N σ ≤ D and T N σ is a Hall abelian subgroup of T . Then D is not a Sylow q -subgroup of T andso T N σ = 1, which implies that T is σ -nilpotent. But then for a Sylow q -subgroup T q of T we have1 < T q ≤ C G ( D ) ≤ H , a contradiction.(6) Every subgroup H of D is normal in G . Hence every element of G induces a power automor-phism in D . Since D is nilpotent by Claim (4), it is enough to consider the case when H ≤ O p ( D ) for some pπ ( D ). 6et R be any Sylow r -subgroup of G , where r π ( D ). Let V , V , . . . , V t be the set of all maximalsubgroups of R . Let T i be a supplement to V i in G such that T i is a P σT -group with D i = T N .Since G = V i T i , R = V i ( T i ∩ R ). Hence for some a i ∈ T i ∩ R we have we have a i V i . We showthat a i ∈ N G ( H ).First observe that D ≤ T i since | G : T | is a q -number, where r π ( D ). Moreover, D i ≤ D . But D i is a Hall subgroup of T i and every subgroup of D i is normal in T i , so D i is a Hall subgroup of D . So either H ≤ O p ( D ) ≤ D i or O p ( D ) ∩ D i = 1. In the former case we have a i ∈ N G ( H ) sinceevery subgroup of D i is normal in T I . Now assume that O p ( D ) ∩ D i = 1, so D i ∩ O p ( D ) h a i i = 1since D i ≤ D and r π ( D ), so O p ( D ) h a i i ≃ D i O p ( D ) h a i i /D i is σ -nilpotent. Hence [ O p ( D ) , a i ] = 1,so a i ∈ N G ( H ).Let V = h a , a , . . . , a i i . Then V ≤ N G ( H ). Moreover, if V < R , then for some i we have V ≤ V i . But then a i V i and a i ∈ V ≤ V i ≤ V i , a contradiction. Therefore V = R ≤ N G ( H ).Hence R G ≤ N G ( H ). Therefore E G ≤ N G ( H ), where E is a Hall π ( D ) ′ -subgroup of G . But then E G D/E G ≃ D/ ( D ∩ E G ) is nilpotent, so D ≤ E G and hence G = GE = E G . Hence we have (6).(7) If p is a prime such that ( p − , | G | ) = 1 , then p does not divide | D | . In particular, | D | isodd. Assume that this is false. Then, by Claim (4), D has a maximal subgroup E such that | D : E | = p and E is normal in G . It follows that C G ( D/E ) = G since ( p − , | G | ) = 1. Since D is a Hall subgroupof G , it has a complement M in G . Hence G/E = (
D/E ) × ( M E/E ), where
M E/E ≃ M ≃ G/D is σ -nilpotent. Therefore G/E is σ -nilpotent. But then D ≤ E , a contradiction. Hence p does notdivide | D | . In particular, | D | is odd.(8) D is abelian. In view of Claim (5), D is a Dedekind group. Hence D is abelian since | D | is odd by Claim (7).From Claims (5)–(8) we get that G is σ -soluble P σT -group, contrary to the choice of G . HenceStatement (iii) holds.(ii) Assume that this assertion is false and let G be a counterexample of minimal order. Then G is a σ -soluble P σT -group by Part (iii) since every σ -nilpotent group is a σ -soluble P σT -group. Then G N σ is a Hall subgroup of G of odd order and every subgroup of G N σ is normal in G by lemma 2.5.Moreover, the hypothesis holds on G/R for every minimal normal subgroup R of G and hence G/R is σ -nilpotent by the choice of G , so R = G N σ is a group of prime order p for some prime p and R isa Sylow p -subgroup of G . But then the maximal V subgroup of R is identity and so G is the uniquesupplement to V in G , so H is σ -nilpotent, a contradiction. Therefore Statement (ii) holds.The theorem is proved. 7 eferences [1] A.N. Skiba, On σ -subnormal and σ -permutable subgroups of finite groups, J. Algebra , (2015), 1–16.[2] A.N. Skiba, Some characterizations of finite σ -soluble P σT -groups,
J. Algebra , (2018),114–129.[3] A.N. Skiba, On sublattices of the subgroup lattice defined by formation Fitting sets, J. Algebra , , (2020), 69–85.[4] W. Guo, A.N Skiba, On σ -supersoluble groups and one generalization of CLT -groups,
J. Algebra , (2018), 92–108.[5] W. Guo, K.P. Shum, A.N. Skiba, G -covering subgroup systems for the classes of supersolubleand nilpotent groups, Israel J. Math. 138 (2003) 125–138.[6] V.D. Mazurov, E.I. Khukhro, The Kourovka Notebook, Novosibirsk, No 19, 2018.[7] A. Ballester-Bolinches, R. Esteban-Romero, M. Asaad, Products of Finite Groups , Walter deGruyter, Berlin-New York, 2010.[8] A.N. Skiba, A generalization of a Hall theorem,