On the conjugacy problem for carter subgroups

Volta, Francesca Dalla; Lucchini, Andrea; Tamburini, M. C.

doi:10.1080/00927879808826135

Cited by 8 publications

(13 citation statements)

References 4 publications

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“…8. Therefore we may suppose that L J = Z(L J ), that is, P J is not a Borel subgroup of G. Consequently L J = H(G 1 * .…”

Section: Carter Subgroups Of Order Divisible By the Characteristicmentioning

confidence: 93%

“…7,8]). With this in mind, we look into the structure of T and N (G, T ), of which we gain an idea by using matrices.…”

Section: Lemma 27mentioning

confidence: 96%

“…P LEMMA 3. 8. Let G, ζg be a finite semilinear group of Lie type over a field of characteristic p (we do not exclude the case G, ζg = G), and G be of adjoint type (recall that g ∈ G σ , but not necessarily g ∈ G).…”

Section: Now Assume Thatmentioning

confidence: 99%

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Carter subgroups of finite almost simple groups

Vdovin

2007

Algebr Logic

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In the paper we complete the classification of Carter subgroups in finite almost simple groups. In particular, we prove that Carter subgroups of every finite almost simple group are conjugate. Togeather with previous results by author and F. Dalla Volta, A. Lucchini, and M. C. Tamburini, as a corollary, it follows that Carter subgroups of every finite group are conjugate.Lemma 2.3. Let G be a finite group and S be a Sylow 2-subgroup of G. Then G contains a Carter subgroup K with S ≤ K if and only if N G (S) = SC G (S).Proof. Assume that G contains a Carter subgroup K with S ≤ K. Since K is nilpotent, it follows that S is normal in K and K ≤ SC G (S) ✂ N G (S). By Feit-Thompson Theorem (see [11]) we obtain that N G (S) is solvable. Thus, by Lemma 2.2(1) we have that SC) is of odd order, it is solvable. Hence it contains a Carter subgroup K 1 . Consider a nilpotent subgroup K = S × K 1 of G. Assume that x ∈ N G (K), then x ∈ N G (S). But K is a Carter subgroup of N G (S), hence x ∈ K and K is a Carter subgroup of G.

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“…8. Therefore we may suppose that L J = Z(L J ), that is, P J is not a Borel subgroup of G. Consequently L J = H(G 1 * .…”

Section: Carter Subgroups Of Order Divisible By the Characteristicmentioning

confidence: 93%

“…7,8]). With this in mind, we look into the structure of T and N (G, T ), of which we gain an idea by using matrices.…”

Section: Lemma 27mentioning

confidence: 96%

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Carter subgroups of finite almost simple groups

Vdovin

2007

Algebr Logic

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“…We intend to improve the results of [1] (see the theorem below). Actually, we will use the ideas of [1] in order to prove a stronger theorem.…”

Section: Introductionmentioning

confidence: 97%

On the Conjugacy problem for Carter subgroups

Vdovin

2006

Sib Math J

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“…We are not aware of any group-theoretical proof of this conjecture. But it was proved in [7] that, if the conjecture is false, a counterexample of minimal order should be an almost simple group. So…”

Section: Introductionmentioning

confidence: 99%

The Conjugacy Conjecture for Carter Subgroups

Tamburini

2007

Milan j. math.

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This paper concerns the problem of determining the nilpotent self-normalizing (Carter) subgroups of a finite group, in relation to the conjecture that they are conjugate, when they exist. We outline the origin of this conjecture and the methods by which it has been studied. For the most substantial contribution, we refer to the recent paper of E.P. Vdovin [27]. As an illustration of the methods, we determine the Carter subgroups of the almost simple groups with socle PSL n (q), in characteristic r 0 ≥ 5.

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On the conjugacy problem for carter subgroups

Cited by 8 publications

References 4 publications

Carter subgroups of finite almost simple groups

Carter subgroups of finite almost simple groups

On the Conjugacy problem for Carter subgroups

The Conjugacy Conjecture for Carter Subgroups

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