# Character of An Element in the Center of a Group Representation

Let V be an n-dimensional irriducible complex representation of a finite group G, let C be it's center. Show that $$|\chi (s)| = n$$ when $$s \in C$$. Where $$\chi$$ is the character function.

My attempt at proof:

Given that $$\rho_s \rho_g = \rho_g\rho_s$$ when $$g \in G, s \in C$$, by Schur's Lemma we have that $$\rho_s=\lambda \ Id$$, $$\lambda \in \mathbb{C}$$. Therefore $$\chi(s)=Tr(\rho_s)=\lambda n$$

My question is why does $$|\lambda|=1$$?

I have been unable to prove this any further so any hints or help would be greatly appreciated.

• I don't think this is correct. Consider the $2$-dimensional representation of the cyclic group $\langle g \rangle$ of order $2$ that maps the generator $g$ to the diagonal matrix with entries $1$ and $-1$. Then $\chi(g) = 0$. – Derek Holt Mar 4 at 22:05
• @derek Holt , excuse me I left out a key fact that it is irreducible. Changed – Matthew Mar 4 at 22:12

Since $$s$$ is an element of the finite group $$G$$, we have $$s^k=1$$ for some positive integer $$k$$. This means that for your $$\lambda$$ we must have $$\lambda^k=1$$, whence $$\lambda$$ is a root of unity, so $$|\lambda|=1$$.