$\chi$ is an irreducible faithful character of $G$ then $Z(G)=\{g \in G: |\chi(g)|=\chi(1)\}$

Question

$\chi$ is an irreducible faithful character of $G$ then prove that $$ Z(G)=\{g \in G: |\chi(g)|=\chi(1)\} \tag{1} $$

I am reading I am reading from James and Liebeck. So there on chapter 13 I have read a theorem: Let $p:G \to GL(n,\Bbb C)$ be a representation of G and let $\chi$ be the character of $p$.

For $g \in G$, $$|\chi(g)|=\chi(1) \iff p(g)=cI_n$$, for some $c \in \Bbb C$.

Now using this I am able to prove one side

$|\chi(g)|=\chi(1) \iff p(g)=cI_n \Rightarrow p(g)p(h)=p(h)p(g) \iff p(gh)=p(hg) \iff gh=hg [\text{as $p$ is faithful for $\Rightarrow$ direction}] \iff g \in Z(G)$

So what we get is in (1) R.H.S $\subseteq Z(G)$. Basically I am stuck in proving $p(g)p(h)=p(h)p(g) \Rightarrow p(g)=cI_n$. How to prove that?

Answer 1

One method of proof is as follows:

Suppose that $g \in Z(G)$, and that $\rho(g) \neq cI$ for any $c \in \Bbb C$. Then there exists a $\lambda$ such that the eigenspace $V = \ker(\rho(g) - \lambda I) \subsetneq \Bbb C^n$ is non-zero. Show that $\rho|_V:GL(V) \to GL(V)$ given by $g \mapsto \rho(g)|_{V}$ is a subrepresentation of $\rho$, contradicting irreducibility.

Answer 2

$g\in Z(G)$ iff $|C_G(g)|=|G|$.

We have that $|G|=\displaystyle\sum_{\chi\in Irr(G)}\chi(1)^2$, and so $|C_G(g)|=\displaystyle\sum_{\chi\in Irr(G)}|\chi(g)|^2$, by a direct consequence of the Second Orthogonality Relation. Therefore, $g\in Z(G)$ iff $$\displaystyle\sum_{\chi\in Irr(G)}\chi(1)^2=\displaystyle\sum_{\chi\in Irr(G)}|\chi(g)|^2.$$ Since $0\leq|\chi(g)|\leq\chi(1)$, we have that the above holds iff $|\chi(g)|=|\chi(1)|$ for all $\chi\in Irr(G)$.