On the chief factors of parabolic maximal subgroups in finite simple groups of normal Lie type

On chief factors of parabolic maximal subgroups of the group 2 E 6(q 2)

“…It is known that U = X γ , where the product is taken in some fixed order over all roots γ ∈ Φ + \ Φ + J . Denote the jth term of the lower central series of the group U by U (j) , where j ≥ 1 and U = U (1) .…”

“…The study of unipotent subgroups of a group of Lie type is a key to understanding its structure and properties. In a previous paper [1], the author obtained a refined description of chief factors of parabolic maximal subgroups contained in the unipotent radical for all groups of normal Lie type, except for special groups (defined below). In the present paper, we continue the study in this direction and consider the twisted group 2 E 6 (q 2 ).…”

Finite simple exceptional groups of Lie type in which all the subgroups of odd index are pronormal

“…By [17,Table 3], C acts on V by conjugation and induces on V a faithful irreducible F q Cmodule. The Clifford theorem [3,Theorem 3.4.1] implies that F q Z-module V is completely reducible and is equal to a direct product of irreducible F q Z-modules which are pairwise conjugate in C. By [3,Theorem 3.2.4], since the order of the cyclic group Z divides q − 1, the group Z acts on V with scalar action, i. e. for any element z from Z there exists an element β ∈ F * q such that v z = βv for each v ∈ V .…”

A strong version of the Sims conjecture for primitive parabolic permutation representations of finite simple groups Lie types $G_2$, $F_4$ and $E_6$

“…В работе [4] Соответствующий этой подстановке графовый автоморфизм группы E 6 (p n ) отображает параболическую максимальную подгруппу [4254] Случай p ̸ = 3. Из работ автора [5,6] следует, что для нахождения рядов взаимных ядер для подгрупп M 1 и M 2 в тройке (G, M 1 , M 2 ) из Π, где G = G 2 (p n ), p ̸ = 3 и M 1 , M 2 -различные сопряженные параболические максимальные подгруппы в G, достаточно рассматривать в качестве M 1 только параболическую максимальную подгруппу P 1 . Нам понадобится Лемма 4 ([5], [6]).…”