Let g ∈ G, where G is an arbitrary finite group. Then there exists χ ∈ Irr(G) such that ker(χ) ∩ g = 1 and every prime divisor of the order o(g) divides the codegree of χ. This improves a recent result of Qian, in which G was assumed to be solvable.
Mathematics Subject Classification (2010). 20C15.In his recent paper [2] in this journal, Guohua Qian proved the following theorem under the additional hypothesis that the group G is solvable. The purpose of this note is to give an easy proof with no solvability assumption. To state the result, we recall Qian's definition: the codegree of a character χ ∈ Irr(G) is the integer |G : ker(χ)|/χ(1).where G is a finite group. Then there exists a character χ ∈ Irr(G) with ker(χ) ∩ g = 1, and such that every prime divisor of the order o(g) divides the codegree of χ.The key to our proof is the following.Lemma. Let n be a square-free positive integer. Then the set X of primitive n th roots of unity in the complex field C is linearly independent over the rational field Q.Proof. Suppose that a δ δ = 0, where a δ ∈ Q and the sum runs over δ ∈ X. We show by induction on the number k of prime divisors of n that a δ = 0 for all δ ∈ X. If k = 0 then n = 1 and X = {1}, and the result clearly holds in this case. We can thus assume that k > 0, and we let p be a prime divisor of n. Since n is square-free, each element δ ∈ X is uniquely of the form δ = μν, where μ is a primitive p th root of unity and ν is a primitive m th root of unity, where m = n/p. Let Y and Z, respectively, be the sets of all primitive p th and m th roots of unity.