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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.

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  • $\begingroup$ 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 $\chi(g) = 0$. $\endgroup$ – Derek Holt Mar 4 at 22:05
  • $\begingroup$ @derek Holt , excuse me I left out a key fact that it is irreducible. Changed $\endgroup$ – Matthew Mar 4 at 22:12
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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 $|\lambda|=1$.

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