Question on irreducible character. Suppose that $\chi \text{Irr}(G)$, i.e $\chi$ is an irreducible character, and assume that $G/Z(\chi)$ is abelian, where $Z(\chi)=\{g \in G : \mid\chi(g)\mid = \chi(1) \}$. How can I prove thet $\chi(g)=0$ if $g \in G \setminus Z(\chi)$.
The hint was let $\phi:G \to \text{Aut}(V)$ be a representation of $G$ with character $\chi$ and look what happens with $\phi(hgh^{-1}g^{-1})$  where $h$ runs through $G$ and $g$ is picked from $G \setminus Z(\chi)$. I can't see how this hint should be used, any advice?

Since $Z(\chi)=\phi^{-1}(Z(\phi(G)))$, then $G/Z(\chi)\cong \phi(G)/Z(\phi(G))$, hence $\phi(G)/Z(\phi(G))$ is abelian. Let $g\in G\setminus Z(\chi)$, then $\phi(g)\in\phi(G)\setminus Z(\phi(G))$. Since $\phi(G)/Z(\phi(G))$ is abelian, then $[\phi(g),\phi(h)]\in Z(\phi(G))$ for all $h\in G$. From Schur's lemma we deduce, that $[\phi(g),\phi(h)]=\lambda_h 1_V$ for some $0\neq\lambda_h\in\mathbb{C}$. On the other hand, $\phi(g)\in\phi(G)\setminus Z(\phi(G))$ follows that there exists $h\in G$ such that $[\phi(g),\phi(h)]\neq 1_V$. Then $\phi(g)^{-1}\phi(h)^{-1}\phi(g)\phi(h)=\lambda_h 1_V$, where $\lambda_h\neq 1$. Hence $\phi(h)^{-1}\phi(g)\phi(h)=\lambda_h\phi(g)$. Comparing traces of left and right hand sides we conclude, that $\chi(g)=\lambda_h\chi(g)$ and $\chi(g)(1-\lambda_h)=0$. But $\lambda_h\neq 1$, so $\chi(g)=0$.
