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Let $\rho$ be a representation of $G$ on $V$. Why are its eigenvalues roots of unity?

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up vote 8 down vote accepted

I assume $G$ is finite. In that case any $g \in G$ has some finite order $n$, hence $\rho(g)^n = 1$. It follows that the characteristic polynomial of $\rho(g)$ divides $x^n - 1$.

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