Confirmation for Wielandt's conjecture
aa r X i v : . [ m a t h . G R ] J u l Confirmation for Wielandt’s conjecture ∗ Wenbin Guo
Department of Mathematics, University of Science and Technology of China,Hefei 230026, P. R. ChinaE-mail: [email protected]
D. O. Revin, E. P. Vdovin
Sobolev Institute of Mathematics and Novosibirsk State University,Novosibirsk 630090, RussiaE-mail: [email protected], [email protected]
Abstract
Let π be a set of primes. By H.Wielandt definition, Sylow π -theorem holds fora finite group G if all maximal π -subgroups of G are conjugate. In the paper, thefollowing statement is proven. Assume that π is a union of disjoint subsets σ and τ and a finite group G possesses a π -Hall subgroup which is a direct product of a σ -subgroup and a τ -subgroup. Furthermore, assume that both the Sylow σ -theoremand τ -theorem hold for G . Then the Sylow π -theorem holds for G . This resultconfirms a conjecture posed by H. Wielandt in 1959. Key words: finite group, Hall subgroup, Sylow π -theorem, condition D π , Wie-landt’s conjecture. MSC2010:
Introduction
In the paper, the term ‘group’ always means ‘finite group’, p is always supposed to bea prime, π is a subset of the set of all primes, and π ′ is its complement in the set of allprimes.Lagrange’s theorem states that the order | G | of a group G is divisible by the order ofevery subgroup of G . This simple statement has extraordinary significance and largelydetermines the problems of finite group theory. Lagrange’s theorem shows the extent towhich the order of a group determines its subgroup structure. For example, it turns outthat every group of prime order is cyclic and contains no non-trivial proper subgroups.It is well-known that the converse of Lagrange theorem is false. However, Sylow’stheorem states that if p is a prime then • every group G contains a so-called Sylow p -subgroup, i. e. a subgroup H such that | H | is a power of p and the index | G : H | is not divisible by p , and ∗ Research is supported by a NNSF grant of China (Grant every p -subgroup of G (i. e. a subgroup whose order is a power of p ) is conjugatewith a subgroup of H .Thus, the converse of Lagrange’s theorem holds for certain divisors of the group order.Moreover, it turns out that, in a finite group, the structure and properties of every p -subgroup are determined in many respects by the structure and properties of any Sylow p -subgroup. In fact, Sylow’s theorem is considered by specialists as a cornerstone of finitegroup theory.A natural generalization of the concept of a Sylow p -subgroup is that of a π -Hallsubgroup.Recall that a subgroup H of G is called a π -Hall subgroup if • all prime divisors of | H | lie in π (i.e. H is a π -subgroup ) and • all prime divisors of the index | G : H | of H lie in π ′ .According to P.Hall [15], we say that a finite group G satisfies D π (or is a D π -group ,or briefly G ∈ D π ) if G contains a π -Hall subgroup H and every π -subgroup of G isconjugate in G with some subgroup of H . Thus, for a group G the condition D π meansthat the complete analog of Sylow’s theorem holds for π -subgroups of G . That is why,together with Hall’s notation, the terminology introduced by H. Wielandt in [47, 48] isused. According to Wielandt, the Sylow π -theorem holds (der π -Sylow-Satz gilt) for agroup G if G satisfies D π . Sylow’s theorem implies that the Sylow π -theorem for G isequivalent to the conjugacy of all maximal π -subgroups of G .Recall that a group is nilpotent if and only if it is a direct product of its Sylowsubgroups. In [46], H.Wielandt proved the following theorem. Theorem 1. [46, Satz]
Assume that a group G possesses a nilpotent π -Hall subgroup fora set π of primes. Then G satisfies D π . This result is regarded to be classical. It can be found in well-known textbooks [5,13, 18, 29, 37, 38]. Wielandt mentioned the result is his talk at the XIII-th InternationalMathematical Congress in Edinburgh [48].There are a lot of generalizations and analogs of Wielandt’s theorem which was provedby many specialists (see, for example, [2, 6, 7, 15, 16, 30–34, 39, 41, 45, 50]). Wielandt’stheorem plays an important role in the study of D π -groups (cf. [8–12, 14, 19, 20, 22–28, 42–44]).One of the earliest generalizations of Wielandt’s theorem was obtained by Wielandthimself in [47]: Theorem 2. [47, Satz]
Suppose that π is a union of disjoint subsets σ and τ . Assume thata group G possesses a π -Hall subgroup H = H σ × H τ , where H σ is a nilpotent σ -subgroupand H τ is a τ -subgroup of H , and let G satisfy D τ . Then G satisfies D π . This result also is included in the textbooks [29, 37, 38] and in the talk [49] and hasmany generalizations and analogs (see [21, 31, 33, 35, 36]).In the same paper [47], Wielandt asked whether, instead of the nilpotency of H σ ,it is sufficient to assume that G satisfies D σ ? Thus, Wielandt formulated the followingconjecture. Conjecture. (Wielandt, [47]) Suppose that a set of primes π is a union of disjoint subsets σ and τ , and a finite group G possesses a π -Hall subgroup H = H σ × H τ , where H σ and2 τ are σ - and τ -subgroups of H , respectively. If G satisfies both D σ and D τ , then G satisfies D π .This conjecture was mentioned in [17, 33, 35, 36] and was investigated in [33, 35, 36].The main goal of the present paper is to prove the following theorem which, in particular,completely confirms Wielandt’s conjecture. Theorem 3. (Main Theorem)
Let a set π of primes be a union of disjoint subsets σ and τ . Assume that a finite group G possesses a π -Hall subgroup H = H σ × H τ , where H σ and H τ are σ - and τ -subgroups, respectively. Then G satisfies D π if and only if G satisfies both D σ and D τ . In view of Theorem 1, one can regard Theorem 3 as generalization of Theorem 2.Moreover, by induction, it is easy to show that Theorem 3 is equivalent to the followingstatement.
Corollary 1.
Suppose a set π of primes is a union of pairwise disjoint subsets π i , i =1 , . . . , n . Assume a finite group G possesses a π -Hall subgroup H = H × · · · × H n where H i is a π i -subgroup for every i = 1 , . . . , n . Then G satisfies D π if and only if G satisfies D π i for every i = 1 , . . . , n . In view of Sylow’s Theorem, Corollary 1 is a generalization of Theorem 1 since thestatement of the theorem can be obtained if we assume that π i = { p i } for i = 1 , . . . , n ,where { p , . . . , p n } = π ∩ π ( G ) . Our notation is standard. Given a group G we denote the set of π -Hall subgroups of G by Hall π ( G ). According to [15], the class of groups G with Hall π ( G ) = ∅ is denotedby E π . We denote by Sym n and Alt n the symmetric and the alternating groups of degree n , respectively. For sporadic groups we use the notation from [4]. The notation for groupsof Lie type agrees with that in [3]. A finite field of cardinality q is denoted by F q .Let r be an odd prime and q be an integer coprime to r . We denote by e ( q, r ) theminimal natural e such that q e ≡ r ) , that is, e ( q, r ) is the multiplicative order of q modulo r . Lemma 1. [15, Lemma 1]
Let A be a normal subgroup and let H be a π -Hall subgroup ofa group G . Then H ∩ A ∈ Hall π ( A ) and HA/A ∈ Hall π ( G/A ) . Lemma 2. [22, Theorem 7.7]
Let A be a normal subgroup of G . Then G ∈ D π if and only if A ∈ D π and G/A ∈ D π .
3e need the criterion obtained in [25] for a simple group satisfying D π . In order toformulate it, we need, according with [25], to define Conditions I–VII for a pair ( G, π ),where G is a finite simple group and π is a set of primes. Condition I.
We say that (
G, π ) satisfies Condition I ifeither π ( G ) ⊆ π or | π ∩ π ( G ) | . Condition II.
We say that (
G, π ) satisfies Condition II if one of the following cases holds.(1) G ≃ M and π ∩ π ( G ) = { , } ;(2) G ≃ M and π ∩ π ( G ) = { , } ;(3) G ≃ M and π ∩ π ( G ) = { , } ;(4) G ≃ M and π ∩ π ( G ) coincide with one of the following sets { , } and { , } ;(5) G ≃ M and π ∩ π ( G ) coincide with one of the following sets { , } and { , } ;(6) G ≃ J and π ∩ π ( G ) coincide with one of the following sets { , } , { , } , { , } , and { , } ;(7) G ≃ J and π ∩ π ( G ) coincide with one of the following sets { , } , { , } , { , } , { , } , and { , } ;(8) G ≃ O ′ N and π ∩ π ( G ) coincide with one of the following sets { , } and { , } ;(9) G ≃ Ly and π ∩ π ( G ) = { , } ;(10) G ≃ Ru and π ∩ π ( G ) = { , } ;(11) G ≃ Co and π ∩ π ( G ) = { , } ;(12) G ≃ Co and π ∩ π ( G ) = { , } ;(13) G ≃ Co and π ∩ π ( G ) = { , } ;(14) G ≃ M (23) and π ∩ π ( G ) = { , } ;(15) G ≃ M (24) ′ and π ∩ π ( G ) = { , } ;(16) G ≃ B and π ∩ π ( G ) coincide with one of the following sets { , } and { , } ;(17) G ≃ M and π ∩ π ( G ) coincide with one of the following sets { , } and { , } . Condition III.
Let G be isomorphic to a group of Lie type over the field F q of charac-teristic p ∈ π and let τ = ( π ∩ π ( S )) \ { p } . We say that (
G, π ) satisfies Condition III if τ ⊆ π ( q −
1) and every prime in π does notdivide the order of the Weyl group of G . 4 ondition IV. Let G be isomorphic to a group of Lie type with the base field F q ofcharacteristic p but not isomorphic to B ( q ), F ( q ) and G ( q ). Let 2 , p π . Denoteby r the minimum in π ∩ π ( G ) and let τ = ( π ∩ π ( G )) \ { r } and a = e ( q, r ) . We say that (
G, π ) satisfies Condition IV if there exists t ∈ τ with b = e ( q, t ) = a andone of the following statements holds.(1) G ≃ A n − ( q ), a = r − b = r , ( q r − − r = r , (cid:20) nr − (cid:21) = h nr i , and e ( q, s ) = b forevery s ∈ τ ;(2) G ≃ A n − ( q ), a = r − b = r , ( q r − − r = r , (cid:20) nr − (cid:21) = h nr i + 1, n ≡ − r ),and e ( q, s ) = b for every s ∈ τ ;(3) G ≃ A n − ( q ), r ≡ a = r − b = 2 r , ( q r − − r = r , (cid:20) nr − (cid:21) = h nr i and e ( q, s ) = b for every s ∈ τ ;(4) G ≃ A n − ( q ), r ≡ a = r −
12 , b = 2 r , ( q r − − r = r , (cid:20) nr − (cid:21) = h nr i and e ( q, s ) = b for every s ∈ τ ;(5) G ≃ A n − ( q ), r ≡ a = r − b = 2 r , ( q r − − r = r , (cid:20) nr − (cid:21) = h nr i +1, n ≡ − r ) and e ( q, s ) = b for every s ∈ τ ;(6) G ≃ A n − ( q ), r ≡ a = r −
12 , b = 2 r , ( q r − − r = r , (cid:20) nr − (cid:21) = h nr i +1, n ≡ − r ) and e ( q, s ) = b for every s ∈ τ ;(7) G ≃ D n ( q ), a ≡ n = b = 2 a and for every s ∈ τ either e ( q, s ) = a or e ( q, s ) = b ;(8) G ≃ D n ( q ), b ≡ n = a = 2 b and for every s ∈ τ either e ( q, s ) = a or e ( q, s ) = b . Condition V.
Let G be isomorphic to a group of Lie type with the base field F q ofcharacteristic p , and not isomorphic to B ( q ), F ( q ) and G ( q ). Suppose, 2 , p π . Let r be the minimum in π ∩ π ( G ), let τ = ( π ∩ π ( G )) \ { r } and c = e ( q, r ) . We say that (
G, π ) satisfies Condition V if e ( q, t ) = c for every t ∈ τ and one of thefollowing statements holds.(1) G ≃ A n − ( q ) and n < cs for every s ∈ τ ;(2) G ≃ A n − ( q ), c ≡ n < cs for every s ∈ τ ;(3) G ≃ A n − ( q ), c ≡ n < cs for every s ∈ τ ;54) G ≃ A n − ( q ), c ≡ n < cs for every s ∈ τ ;(5) G is isomorphic to one of the groups B n ( q ), C n ( q ), or D n ( q ), c is odd and 2 n < cs for every s ∈ τ ;(6) G is isomorphic to one of the groups B n ( q ), C n ( q ), or D n ( q ), c is even and n < cs for every s ∈ τ ;(7) G ≃ D n ( q ), c is even and 2 n cs for every s ∈ τ ;(8) G ≃ D n ( q ), c is odd and n cs for every s ∈ τ ;(9) G ≃ D ( q );(10) G ≃ E ( q ), and if r = 3 and c = 1 then 5 , τ ;(11) G ≃ E ( q ), and if r = 3 and c = 2 then 5 , τ ;(12) G ≃ E ( q ), if r = 3 and c ∈ { , } then 5 , , τ , and if r = 5 and c ∈ { , } then7 τ ;(13) G ≃ E ( q ), if r = 3 and c ∈ { , } then 5 , , τ , and if r = 5 and c ∈ { , } then7 , τ ;(14) G ≃ G ( q );(15) G ≃ F ( q ), and if r = 3 and c = 1 then 13 τ . Condition VI.
We say that (
G, π ) satisfies Condition VI if one of the following statementsholds.(1) G is isomorphic to B (2 m +1 ) and π ∩ π ( G ) is contained in one of the sets π (2 m +1 − , π (2 m +1 ± m +1 + 1);(2) G is isomorphic to G (3 m +1 ) and π ∩ π ( G ) is contained in one of the sets π (3 m +1 − \ { } , π (3 m +1 ± m +1 + 1) \ { } ;(3) G is isomorphic to F (2 m +1 ) and π ∩ π ( S ) is contained in one of the sets π (2 m +1) ± , π (2 m +1 ± m +1 + 1) ,π (2 m +1) ± m +2 ∓ m +1 − , π (2 m +1) ± m +2 + 2 m +1 ± m +1 − . Condition VII.
Let G be isomorphic to a group of Lie type with the base field F q ofcharacteristic p . Suppose that 2 ∈ π and 3 , p π , and let τ = ( π ∩ π ( G )) \ { } and ϕ = { t ∈ τ | t is a Fermat number } . We say that (
G, π ) satisfies Condition VII if τ ⊆ π ( q − ε ) , where the number ε = ± q − ε , and one of the following statementsholds. 61) G is isomorphic to either A n − ( q ) or A n − ( q ), s > n for every s ∈ τ , and t > n + 1for every t ∈ ϕ ;(2) G ≃ B n ( q ), and s > n + 1 for every s ∈ τ ;(3) G ≃ C n ( q ), s > n for every s ∈ τ , and t > n + 1 for every t ∈ ϕ ;(4) G is isomorphic to either D n ( q ) or D n ( q ), and s > n for every s ∈ τ ;(5) G is isomorphic to either G ( q ) or G ( q ), and 7 τ ;(6) G ≃ F ( q ) and 5 , τ ;(7) G is isomorphic to either E ( q ) or E ( q ), and 5 , τ ;(8) G ≃ E ( q ) and 5 , , τ ;(9) G ≃ E ( q ) and 5 , , , τ ;(10) G ≃ D ( q ) and 7 τ . Lemma 3. [25, Theorem 3]
Let π be a set of primes and G be a simple group. Then G ∈ D π if and only if ( G, π ) satisfies one of Conditions I–VII . Lemma 4.
Let G be a simple group and π be such that , ∈ π ∩ π ( G ) . Then G ∈ D π if and only if π ( G ) ⊆ π. Proof. If π ( G ) ⊆ π , then evidently G ∈ D π . Conversely, suppose G ∈ D π . Then Lemma 3implies that ( G, π ) satisfies one of Conditions I–VII. Without loss of generality, we mayassume that π ∩ π ( G ) = π .If Condition I holds, then either | π | π = π ( G ) . Since 2 , ∈ π , we have that | π | > π = π ( G ).Clearly, ( G, π ) cannot satisfy one of Conditions II and IV–VII since otherwise either2 / ∈ π (Conditions II and IV–VI) or 3 / ∈ π (Condition VII).Finally, ( G, π ) also cannot satisfy Condition III. Indeed, if Condition III holds, then G is a group of Lie type over a field of characteristic p ∈ π and every prime in π (inparticular 2) does not divide the order of the Weyl group of G . But this is impossiblesince the Weyl group is nontrivial and is generated by involutions (see [3, page 13 andProposition 13.1.2]). Lemma 5.
Suppose that n > and π is a set of primes with | π ∩ π ( n !) | > and π ( n !) * π. Then (1)
The complete list of possibilities for
Sym n containing a π -Hall subgroup H is givenin Table . (2) K ∈ Hall π (Alt n ) if and only if K = H ∩ Alt n for some H ∈ Hall π (Sym n ) . n π H ∈ Hall π (Sym n )prime π (( n − n − { , } Sym × Sym { , } Sym ≀ Sym Table 2:
G π ∩ π ( G ) M { , } M { , }{ , } Ru { , } F i { , } J { , }{ , }{ , }{ , } Co { , } G π ∩ π ( G ) M { , } M { , }{ , } F i ′ { , } J { , }{ , }{ , }{ , }{ , } Co { , } G π ∩ π ( G ) M { , } Ly { , } O ′ N { , }{ , }{ , } B { , }{ , } M { , }{ , } Co { , } In particular, if either / ∈ π or / ∈ π , then Alt n / ∈ E π .Proof. See [15, Theorem A4 and the notices after it], [40, Main result], and [22, Theo-rem 4.3 and Corollary 4.4].
Lemma 6. [8, Corollary 6.13]
Let G be either one of sporadic groups or the Tits group.Assume that / ∈ π . Then G ∈ E π if and only if either | π ∩ π ( G ) | or G and π ∩ π ( G ) are given in Table . In particular, | π ∩ π ( G ) | . Lemma 7. [23, Theorem 4.1]
Let G be either a simple sporadic group or the Tits groupand π be such that ∈ π, π ( G ) * π, and | π ∩ π ( G ) | > . Then the complete list for G containing a π -Hall subgroup H is given in Table . Inparticular, if / ∈ π , then G = J and π ∩ π ( G ) = { , } . Table 3:
G π ∩ π ( G ) Structure of HM { , } : Q . { , , } Alt . M { , , } : Alt { , } : (3 × A ) : 2 { , , } : Alt { , , } : (3 × Alt ) : 2 { , , , } L (4) : 2 { , , , } : Alt { , , , , } M M { , , } : 3 · Sym J { , } × Alt { , } : 7 { , , } × A { , , } : 7 : 3 J { , , } : (2 : 3 · Sym ) Lemma 8.
Let G be a group of Lie type with base field F q of characteristic p . Assumethat π is such that p ∈ π , and either / ∈ π or / ∈ π . Suppose G ∈ E π and H ∈ Hall π ( G ) .Then one of the following statements holds. (1) π ∩ π ( G ) ⊆ π ( q − ∪ { p } , a Sylow p -subgroup P of H is normal in H and H/P is abelian. (2) p = 2 , G ≃ B (2 n +1 ) and π ( G ) ⊆ π .Proof. See [8, Theorem 3.2] and [11, Theorem 3.1].
Lemma 9.
Let G be a group of Lie type over a field of characteristic p . Assume that π is such that p / ∈ π , and either / ∈ π or / ∈ π . Denote by r the minimum in π ∩ π ( G ) .Suppose G ∈ E π and H ∈ Hall π ( G ) . Then H possesses a normal abelian r ′ -Hall subgroup.Proof. See [9, Theorems 4.6 and 4.8, and Corollary 4.7], [43, Theorem 1], and [42, Lem-ma 5.1 and Theorem 5.2].
Lemma 10.
Let G be a simple nonabelian group. Assume that π is such that π ( G ) π and either / ∈ π or / ∈ π . Suppose G ∈ E π and H ∈ Hall π ( G ) . Then H is solvable and,for any partition π ∩ π ( G ) = σ ∪ τ, where σ and τ are disjoint nonempty sets, either σ -Hall or τ -Hall subgroup of H is nilpo-tent.Proof. Consider all possibilities, according to the classification of finite simple groups(see [1, Theorem 0.1.1]).C a s e 1: G ≃ Alt n , n >
5. By Lemma 5 it follows that | π ∩ π ( G ) | = 1 and apartition π ∩ π ( G ) = σ ∪ τ onto nontrivial disjoint subsets is impossible.C a s e 2: G is either a sporadic group or the Tits group. By Lemmas 6 and 7 itfollows that either | π ∩ π ( G ) | = 1 , or 2 / ∈ π and | π ∩ π ( G ) | = 2 , or 3 / ∈ π, G ≃ J and π ∩ π ( G ) = { , } . | π ∩ π ( G ) | = 1, then a partition π ∩ π ( G ) = σ ∪ τ onto nonempty disjoint subsets is impossible. If | π ∩ π ( G ) | = 2, then H is solvable byBurnside’s p a q b -theorem [5, Ch. I, 2], the orders of its σ -Hall and τ -Hall subgroups arepowers of primes, and thus σ -Hall and τ -Hall subgroups of G are nilpotent.C a s e 3: G is a group of Lie type over a field of characteristic p ∈ π . We mayassume, without loss of generality, that | π ∩ π ( G ) | > p ∈ σ . By Lemma 8, H issolvable and its τ -Hall subgroup T is isomorphic to a subgroup of the abelian group H/P ,where P is the (normal) Sylow p -subgroup of H . Whence T is abelian and, in particular,is nilpotent.C a s e 4: G is a group of Lie type over a field of characteristic p / ∈ π . We mayassume, without loss of generality, that | π ∩ π ( G ) | >
1. Denote by r the minimum in π ∩ π ( G ), and assume that r ∈ σ . By Lemma 9, it follows that H is solvable and its τ -Hall subgroup T is included in the normal abelian r ′ -Hall subgroup of H . Thus weagain obtain that T is abelian.Thus in the all cases, Lemma 10 holds. Assume that the hypothesis of Theorem 3 holds, i.e. we have a partition π = σ ∪ τ of π onto disjoint subsets σ and τ , and a group G satisfying condition:(1) G possesses a π -Hall subgroup H such that H = H σ × H τ , where H σ and H τ are σ - and τ -subgroups, respectively.It is easy to see that H σ and H τ are, respectively, σ -Hall and τ -Hall subgroups of both H and G . Moreover, H σ coincides with the set of all σ -elements of H , while H τ is the set ofall τ -elements of H .We prove first that G ∈ D π implies G ∈ D σ ∩ D τ . We need to prove that a σ -subgroup K of G is conjugate to a subgroup of H σ in order to prove that G ∈ D σ . Since K is,in particular, a π -subgroup and G ∈ D π , there exists g ∈ G such that K g H . Hence K g H σ , since H σ is the set of all σ -elements of H . Thus we obtain G ∈ D σ . The samearguments show that G ∈ D τ .Now we prove the converse statement: if G ∈ D σ ∩ D τ , then G ∈ D π . Assume that itfails. Without loss of generality we may assume that G satisfies the following conditions:(2) G ∈ D σ ∩ D τ ;(3) G / ∈ D π ;(4) G has the smallest possible order in the class of groups satisfying conditions (1)–(3).10ow we show that the assumption of existence of such group leads us to a contradiction.In view of (3) we have π ( G ) * π and G is nonabelian.Note that G must be simple. Indeed, assume that G possesses a nontrivial propernormal subgroup A . Then Lemma 1 impies that A and G/A satisfy (1), and Lemma 2implies that they both satisfy (2). In view of (4), neither A nor G/A satisfies (3), andthus A ∈ D π and G/A ∈ D π . Hence by Lemma 2, we obtain G ∈ D π , which contradictsthe assumption (3).Assume that either 2 or 3 does not lie in π . Then by Lemma 10 either H σ or H τ isnilpotent. Hence by Theorem 2 we obtain G ∈ D π , which contradicts (3). Hence 2 , ∈ π .By Lemma 4 and the condition (2), the numbers 2 and 3 cannot simultaneously lie inthe same subset σ or τ . We may, therefore, assume that2 ∈ σ and 3 ∈ τ. Let S be a Sylow 2-subgroup of H σ (hence of both H and G ), and T be a Sylow3-subgroup of H τ (hence of both H and G ). Since[ S, T ] [ H σ , H τ ] = 1 , we see that G possesses a nilpotent { , } -Hall subgroup h S, T i ≃ S × T. It follows from Theorem 1 that G ∈ D { , } . Now by Lemma 4 we have π ( G ) = { , } ⊆ π, which implies that G is solvable by Burnside’s p a q b -theorem [5, Ch. I, 2]. This contradic-tion completes the proof.Notice that the proof of Theorem 3 implies the following statement. Corollary 2.