Prove that: Z(G) = { g∈G ∣ |χ(g)| = χ(1) for all χ ∈ Irr(G)} (Representation Theory)

189 Views Asked by At

I'm working on representation theory and I came across a statement that Z(G) = { g ∈ G ∣ |χ(g)| = χ(1) for all χ ∈ Irr(G)}.

I'm trying to see why this is the case.

(1) For the backwards direction, I've worked out that if χ the character of an irreducible representation p, then if |χ(g)|=χ(1), as χ(g) is the sum of eigenvalues of p(g) which are roots of unity, we must have all the eigenvalues are the same and so p(g) = kI, where k is some root of unity. Clearly this p(g) commutes with every element in the image of p. From this I want to conclude that as this is the case for all irreducible representations p, g therefore commutes with every element in G - but I'm not sure why that is the case.

(2) Not sure where to start with the forward direction.

Many thanks for your help! :)