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Proper π-Hall subgroups in sporadic groups, 2, 3 ∈ π
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In this paper we find the number of conjugate $\pi$-Hall subgroups in all finite almost simple groups. We also complete the classification of $\pi$-Hall subgroups in finite simple groups and correct some mistakes from our previous paper. Comment: article in press in "Journal of algebra"
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Citations
... (ii) Let G/N ∈ X. Then a necessary and sufficient condition that, for H ∈ Hall X (N ), there exist a subgroup K ∈ Hall X (G) such that H = K ∩ N is that H N = H G (that is, when the class H N ∈ Hall X (N )/N is invariant under the action of the group G on the set Hall X (N )/N ; see [23], Lemma 2.1, (e)). ...
... The key role in the proof of Theorem 1 is played by the theorem on the number of classes of conjugate π-Hall subgroups in simple groups (see [23]). We will use the following refined version of this result. ...
... Lemma 6 (see [23], Theorem 1.1). Let S be a simple finite group possessing a π-Hall subgroup for some set π of primes. ...
Let X be a class finite groups closed under taking subgroups, homomorphic images, and extensions, and let k X (G) be the number of con-jugacy classes X-maximal subgroups of a finite group G. The natural problem calling for a description, up to conjugacy, of the X-maximal subgroups of a given finite group is not inductive. In particular, generally speaking , the image of an X-maximal subgroup is not X-maximal in the image of a homomorphism. Nevertheless, there exist group homomorphisms that preserve the number of conjugacy classes of maximal X-subgroups (for example, the homomorphisms whose kernels are X-groups). Under such homomorphisms, the image of an X-maximal subgroup is always X-maximal , and, moreover, there is a natural bijection between the conjugacy classes of X-maximal subgroups of the image and preimage. In the present paper, all such homomorphisms are completely described. More precisely, it is shown that, for a homomorphism φ from a group G, the equality k X (G) = k X (im φ) holds if and only if k X (ker φ) = 1, which in turn is equivalent to the fact that the composition factors of the kernel of φ lie in an explicitly given list.
... Ключевую роль в доказательстве теоремы 1 играет теорема о числе классов сопряженных π-холловых подгрупп в простых группах, доказанная в [23]. Нам она понадобится в следующем уточненном виде. ...
Пусть $\mathfrak{X}$ - класс конечных групп, замкнутый относительно подгрупп, гомоморфных образов и расширений, и $\mathrm{k}_{\mathfrak{X}}(G)$ - число классов сопряженности $\mathfrak{X}$-максимальных подгрупп конечной группы $G$. Естественная задача - описать с точностью до сопряженности $\mathfrak{X}$-максимальные подгруппы данной конечной группы - не индуктивна. В частности, в образе гомоморфизма образ $\mathfrak{X}$-максимальной подгруппы, вообще говоря, не $\mathfrak{X}$-максимален. Существуют гомоморфизмы, сохраняющие число классов сопряженности максимальных $\mathfrak{X}$-подгрупп (например, гомоморфизмы, ядра которых - $\mathfrak{X}$-группы). Относительно таких гомоморфизмов образ $\mathfrak{X}$-максимальной подгруппы всегда $\mathfrak{X}$-максимален и существует естественная биекция между классами сопряженности $\mathfrak{X}$-максимальных подгрупп образа и прообраза. В работе такие гомоморфизмы полностью описаны. Доказано, что для гомоморфизма $\phi$ из группы $G$ равенство $\mathrm{k}_{\mathfrak{X}}(G)=\mathrm{k}_{\mathfrak{X}}(\operatorname{im} \phi)$ выполнено, если и только если $\mathrm{k}_{\mathfrak{X}}(\ker \phi)=1$, а это, в свою очередь, равносильно тому, что композиционные факторы ядра $\phi$ принадлежат известному списку. Библиография: 25 наименований.
... Оказывается, то обстоятельство, что для данных группы G и ее нормальной подгруппы N мощность X-схемы не меняется при переходе от G к G/N, является внутренним свойством группы N. Оно не зависит не только от особенностей вложения N в G, но и от самой G, поскольку влечет справедливость редукционной X-теоремы для N. Именно этот факт является новым в теореме 1 по сравнению с [4, теорема 1]. При этом если [4, теорема 1] была доказана путем сведения общей ситуации частному случаю X = G π , изученному ранее в [23][24][25][26][27][28][29] (см. также обзоры [19,30] и монографию [31]), то утверждение "только если" в теореме 1 является новым даже для этого случая. ...
... Ключевую роль в доказательстве теоремы 1 играет теорема о числе классов сопряженных π-холловых подгрупп в простых группах, доказанная в [29]. Нам она понадобится в следующем уточненном виде. ...
Let X be a class of finite groups closed under taking subgroups, homomorphic images, and extensions. Denote by k_X(G) the number of conjugacy classes X-maximal subgroups of a finite group G. The natural problem to describe up to conjugacy X-maximal subgroups of a given finite group is complicated by the fact that it is not inductive. In particular, generally speaking, the image of an X-maximal subgroup is not X-maximal in the image of a homomorphism. Nevertheless, there are group homomorphisms which preserve the number of conjugacy classes of X-maximal subgroups (for example, the homomorphisms whose kernels are X-groups). Under these homomorphisms, the image of an X-maximal subgroup is always X-maximal and, moreover, there is a natural bijection between the conjugacy classes of X-maximal subgroups of the image and preimage. All such homomorphisms are completely described in the paper. More precisely, it is proved that, for a homomorphism ϕ from a group G, the equality k_X(G)=k_X(imϕ) holds if and only if k_X(kerϕ)=1, which in turn is equivalent to the fact that the composition factors of the kernel of ϕ belong to an explicitly given list.
... Assume thatĜ is not orthogonal. By [11,Lemma 4.1], H stabilises a decomposition ...
... , k}. IfĜ = GL ε n (q), then, by the proof of [11,Lemma 4.3], we can assume that either dim V i = 1 for all i or dim V i = 2 for i < k and dim V k ∈ {1, 2}. IfĜ = GSp n (q), then dim V i = 2 for all i since all one-dimensional subspaces are singular in this case. ...
... Assume nowĜ = GO ε n (q). Since H is solvable, one of (a)-(e) holds in [11,Lemma 6.7]. In cases (a)-(c), H stabilises a decomposition of V as in Lemma 3.1 and the proof as in Lemma 3.1 works. ...
In this paper we show that if S is a simple classical group, a group G is contained in inner-diagonal automorphisms of S and contains S, and H is a solvable Hall subgroup of G, then there exists five conjugates of H, whose intersection is trivial.
... Lemma 2.20 [42,Lemma 4.3], [51,Theorem 8.12] Assume G = SL η n (q) is a special linear or unitary group with the base field F q of characteristic p and n 2. Let π be a set of primes such that 2, 3 ∈ π and p ∈ π. Then the following statements hold. ...
... Lemma 2.21 [42,Lemma 4.4], [51,Theorem 8.13] Let G = Sp 2n (q) be a symplectic group over a field F q of characteristic p. Assume that π is a set of primes such that 2, 3 ∈ π and p ∈ π. Then the following statements hold. ...
... Accepted manuscript to appear in BMS Lemma 2.18[42, Lemma 3.2],[51, Corollary 8.11 ...
Let $\mathfrak{X}$ be a class of finite groups closed under taking the subgroups, homomorphic images and extensions. It is known that if $A$ is a normal subgroup of a finite group $G$ then the image of an $\mathfrak{X}$-maximal subgroup $H$ of $G$ in $G/A$ is not $\mathfrak{X}$-maximal in $G/A$ in general. We say that the reduction $\mathfrak{X}$-theorem holds for a finite group $A$ if, for every finite group $G$ being an extension of $A$ (i.e. containing $A$ as a normal subgroup), the number of conjugacy classes of $\mathfrak{X}$-maximal subgroups in $G$ and $G/A$ is the same. The reduction $\mathfrak{X}$-theorem for $A$ implies that $HA/A$ is $\mathfrak{X}$-maximal in $G/A$ for every extension $G$ of $A$ and every $\mathfrak{X}$-maximal subgroup $H$ of $G$. In the paper, we prove that the reduction $\mathfrak{X}$-theorem holds for $A$ if and only if all $\mathfrak{X}$-maximal subgroups are conjugate in $A$ and classify the finite groups with this property in terms of composition factors.
... Lemma 2.20 [42,Lemma 4.3], [51,Theorem 8.12] Assume G = SL η n (q) is a special linear or unitary group with the base field F q of characteristic p and n 2. Let π be a set of primes such that 2, 3 ∈ π and p ∈ π. Then the following statements hold. ...
... where L = GL η 2 (q) ≀ Sym m ×Z GL n (q) and Z is a cyclic group of order q − η for k = 1 and Z is trivial for k = 0. Lemma 2.21 [42,Lemma 4.4], [51,Theorem 8.13] Let G = Sp 2n (q) be a symplectic group over a field F q of characteristic p. Assume that π is a set of primes such that 2, 3 ∈ π and p ∈ π. Then the following statements hold. ...
... Observe that Theorem 1 cannot be strengthened in the spirit of Tyutyanov's result, that is, the existence of {2, p}-Hall subgroups for every p in π does not guarantee the existence of a solvable π-Hall subgroup. Indeed, the projective special linear group P SL 2 (41) contains {2, 3}-and {2, 5}-Hall subgroups, but it does not contain a {2, 3, 5}-Hall subgroup (see [7,Theorem 5.2] and [8,Lemma 3.1]). ...
... Since G ∈ E {2,s} , it follows from Lemma 2.10 that s is in π(q − ε). If 3 divides q − ε, then G has a solvable π-Hall subgroup and all such subgroups are conjugate by [8,Lemma 3.11]. Suppose that 3 does not divide q − ε and G has a {3, s}-Hall subgroup. ...
... Let G = P SL η n (q) for n > 2. By [8,Lemma 4.3], to prove that G ∈ E π it is sufficient to show that one of the following conditions holds: ...
Let G be a finite group and let π be a set of primes. We prove that G has a solvable π-Hall subgroup if and only if G has a {p, q}-Hall subgroup for every pair p, q ∈ π.
... Observe that this statement cannot be strengthened in a fashion of Tyutyanov's result, that is the existence of {2, r}-Hall subgroups for every r in π does not guarantee the existence of a solvable π-Hall subgroup. Indeed, the projective special linear group P SL 2 (41) contains {2, 3}-and {2, 5}-Hall subgroups (see, [8,Lemma 3.11]), but does not contain {2, 3, 5}-Hall subgroup. ...
... Since G ∈ E {2,s} , it follows from Lemma 2.11 that s is in π(q − ε). If 3 divides q − ε then by [8,Lemma 3.11] the group G has a solvable π-Hall subgroup and all such subgroups are conjugate. Suppose 3 does not divide q − ε and G has a {3, s}-Hall subgroup. ...
... Let n = 2. By [8,Lemma 4.3], to prove that G ∈ E s π it is sufficient to show that one of these two conditions holds: ...
Let $G$ be a finite group and let $\pi$ be a set of primes. In this paper, we prove a criterion for the existence of a solvable $\pi$-Hall subgroup of $G$, precisely, the group $G$ has a solvable $\pi$-Hall subgroup if, and only if, $G$ has a $\{p,q\}$-Hall subgroup for any pair $p$, $q\in\pi$.
... 6.9, 8.1]. P In accordance with [12], for a finite group G, by k π (G) we denote the number of conjugacy classes into which the set Hall π (G) splits. (i) if 2 / ∈ π then k π (S) = 1; ]. ...
... Then every π-subgroup of G is solvable and is an X-group. Hence G ∈ D π , which is a contradiction with Lemma 9 since G / ∈ X and n ≥ 5. Finally, if H is a point stabilizer and π ∩ π(G) = π((n − 1)!) for a prime number n ≥ 5, then G contains a solvable π-subgroup (and hence an X-subgroup) (12) if n = 5. ...
Let $\mathfrak{X}$ be a class of finite groups closed under taking subgroups, homomorphic images, and extensions. Following H. Wielandt, we call a subgroup $H$ of a finite group $G$ a submaximal $\mathfrak{X}$-subgroup if there exists an isomorpic embedding ϕ: G ↪ G* of the group $G$ into some finite group $G*$ under which Gϕ is subnormal in G* and Hϕ = K ∩Gϕ for some maximal $\mathfrak{X}$ -subgroup $K$ of $G*$. We discuss the following question formulated by Wielandt: Is it always the case that all submaximal $\mathfrak{X}$ -subgroups are conjugate in a finite group G in which all maximal $\mathfrak{X}$ -subgroups are conjugate? This question strengthens Wielandt’s known problem of closedness for the class of $D_\pi$-groups under extensions, which was solved some time ago. We prove that it is sufficient to answer the question mentioned for the case where G is a simple group.
... Lemma 2.21 [39,Lemma 4.4], [46,Theorem 8.13] Let G = Sp 2n (q) be a symplectic group over a field F q of characteristic p. Assume that π is a set of primes such that 2, 3 ∈ π and p ∈ π. Then the following statements hold. ...
Let $\mathfrak{X}$ be a class of finite groups closed under taking the subgroups, homomorphic images and extensions. By $D_\mathfrak{X}$ denote the class of finite groups $G$ in which every two $\mathfrak{X}$-maximal subgroups are conjugate. In the paper, the following statement is proven. Let $A$ be a normal subgroup of a finite group $G$. Then $G\in D_\mathfrak{X}$ if and only if $A\in D_\mathfrak{X}$ and $G/A\in D_\mathfrak{X}$. This statement implies that the $\mathfrak{X}$-maximal subgroups are conjugate if and only if the so called $\mathfrak{X}$-submaximal subgroups are conjugate. Thus we obtain an affirmative solution to a problem posed by H.Wielandt in 1964.
