Characters of a faithful irreducible Module for an element in the centre

Question

Basically here is my questions. We have a character $\chi$ which is faithful and irreducible of a group $G$. we have an element $g$ which i needs to show belongs to the centre $Z(G)$, i.e. $gh=hg$ for all $g$ in $G$, If and only If $|\chi(g)|=\chi(e)$.

So working on the first half of the proof i.e. if g belongs to the centre then $|\chi(g)|=\chi(e)$. so far i have shown that if $\chi$ is a faithful irreducible character of $G$ then there exists a $CG$-module $V$ which is also faithful and irreducible and as such $Z(G)$ is cyclic (i have proved this, so we are ok here) i have then shown that if $Z(G)$ is cyclic then there is a basis $B=\{u_1,...,u_k\}$ of $V$ such that for $g$ belonging to our centre

$\{g\}b$ is a diagonal matrix with nth roots of unit as the entries on the diagonal where $n=dim(V)$ So the character of $\chi(g)=$ the sum of these nth roots of unity. I also know e belongs to the centre so $[e]b$ can also be written as a diagonal matrix with entries which are nth roots of unity. This is as far as i have got and i dont know if i have made any progress in the right or even wrong direction. Any help would be appreciated.

Answer 1

Assume $G$ is finite and $\chi$ is faithful and irreducible. Our claim: $g\in Z(G) \iff |\chi(g)|=\chi(e)$.

Show both are $\equiv$ to $g$ acting as a scalar: $\Rightarrow$ via Schur, $\Leftarrow$ via triangle inequality on eigenvalues.

Answer 2

In general, if $\chi \in Irr(G)$, then $Z(G/ker(\chi))=Z(\chi)/ker(\chi)$. Here $Z(\chi)=\{g \in G: |\chi(g)|=\chi(1)\}$ and $ker(\chi)=\{g \in G: \chi(g)=\chi(1)\}$. Faithulness of an irreducible character means $ker(\chi)=1$. See also I.M. Isaacs, Character Theory of Finite Groups Lemma (2.27).