Some linear actions of finite groups with q′-orbits

Abstract

Abstract. Let G be a finite group acting faithfully on a finite vector space M in such a way that the centralizer of every element of M contains a Sylow q-subgroup of G as a central subgroup (for a fixed prime divisor q of jGj with ðq; jMjÞ ¼ 1). Then G is isomorphic to a subgroup of the semi-linear group on M.

Groups whose degree graph has three independent vertices

“…Given a prime number q, we say that the pair (H, V ) satisfies N q if (a) q divides |H : C H (V )| and (b) for every non-trivial v ∈ V , there exists a Sylow q-subgroup Q of H such that Q C H (v). We refer to [2] for a thorough analysis of this and related module actions. We recall that, as shown for instance by Lemma 4 of [22], if (H, V ) satisfies N q , then V is an elementary abelian r-group for a suitable prime r and that V is an irreducible GF(r)[H]-module.…”

An Analogue of Huppert’s Conjecture for Character Codegrees

“…Let G be a group acting on a module M over a finite field and let r be a prime divisor of |G/C G (M)|. If, for every ν ∈ M \ {0}, C G (ν) contains a Sylow r-subgroup of G as a normal subgroup, then we say that the pair (G, M) satisfies N r (for more details, see [5]). We use this definition in the proof of the main result.…”

On the character degree graph of finite groups

“…For a proof by contradiction, assume that M is an H-module and q is an odd prime such that (H, M ) satisfies N q . Then, by Proposition 13 of [6], we have u = 2. Now, let Q be a Sylow q-subgroup of H, and let r a = |M |, r b = |C M (Q)|.…”

“…So, for every prime divisor p of |F |, there is a T -stable Sylow p-subgroup P of F . By Lemma 6 of [6], if p = 2, then we have P ≤ C F (T ); recalling that in GL a (r) the centralizer of an element of order t is cyclic, we have that P is cyclic. Now, write E = [O 2 (F ), T ] and observe that, by coprimality, [E, T ] = E. If E is cyclic, then E = [E, T ] = 1 (because the automorphism group of E is a 2-group), so F centralizes T and it is therefore cyclic.…”

“…By [8, Lemma 2.1], we have that s does not divide |L 0 |, whereas d = |S| divides n, and (p n − 1)/(p n/d − 1) divides |L 0 |. Note that p n − 1 has a primitive prime divisor t. In fact, otherwise, either n = 2 or p n = 2 6 . In both cases, as s and p are distinct primes, s divides (p n − 1)/(p n/d − 1), so s divides |L 0 |, a contradiction.…”