Prove that there exists an $m$th root of unity $λ ∈ C$ such that for all $g ∈ G$, $\chi(zg) =λ \chi(g)$

Question

Suppose $\chi$ is an irreducible character of $G$. Suppose $z ∈ Z(G)$ and that $z$ has order $m$. Prove that there exists an $m$th root of unity $λ ∈ C$ such that for all $g ∈ G$, $\chi(zg) =λ \chi(g)$.

I think I need to use Schur's Lemma but I am not getting a way to use it.Please help!

Answer 1

Hint: ${\rm tr}(\lambda A)=\lambda{\rm tr}(A)$ for all scalars $\lambda\in\Bbb C$ and linear transformations $A$.

Use Schur's lemma to relate $z\in Z(G)$ to a scalar $\lambda$.