G is a finite group. Let $\chi$ be an irreducible character of degree $n$ (that is $\chi(1)=n$). Show that $|\chi(g)|=n$ for every element in $Z(G)$.
I am basically asking to prove that $Z(G)\subseteq\ker \chi$ which can be shown by using Schur's Lemma but I would like to know if there can be a proof without it. Perhaps it could be fully utilising the orthogonality relations or some other properties of the character $\chi$.
By orthogonality you have $\sum_{i=1}^s|\chi_i(g)|^2=|C(g)|$ summing over irreducible characters where $C(g)=\{x \in G:xg=gx\}$. Since $g\in Z(G)$ you have $C(g)=G$ and thus $\sum_{i=1}^s|\chi_i(g)|^2=|G|=\sum_{i=1}^s\chi_i(1)^2 \implies \sum_{i=1}^s(\chi_i(1)^2-|\chi_i(g)|^2)=0$. Now since $\chi_i(1)\geq{}|\chi_i(g)|$ we have $\chi_i(1)^2-|\chi_i(g)|^2\geq 0$ forcing $\chi_i(1)^2=|\chi_i(g)|^2$ and thus $\chi_i(1)=|\chi_i(g)|$.