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

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!

## 1 Answer

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 $\lambda$.

• what i obtain is $zv=\lambda v$ for $v$ in $V$.can u help in completing this? – Arpit Kansal Feb 19 '15 at 13:28
• Use the fact that a representation is a homomorphism $\rho:G\to{\rm GL}(V)$. What does that mean for $\rho(zg)$? – whacka Feb 19 '15 at 13:30
• $\rho(z)$ commutes with every $\rho(g)$.Right? – Arpit Kansal Feb 19 '15 at 13:35
• Yes that is true - that is how you know $\rho(z)$ acts as a scalar $\lambda$ - although you still need to show $\lambda$ is a root of unity (use the fact $G$ is finite). But what I am getting at in my comment is how to get to the point where you can use the Hint in my answer. – whacka Feb 19 '15 at 13:37
• My hint talks about taking the trace of a scalar times a linear map. Using the fact that $\rho$ is a group homomorphism, how can $\rho(zg)$ be thought of? – whacka Feb 19 '15 at 18:00