# Eigenvalues of depictions

Answers: 1

I think $G$ is limited. Because instance any kind of $g \in G$ has some limited order $n$, therefore $\rho(g)^n = 1$. It adheres to that the particular polynomial of $\rho(g)$ separates $x^n - 1$.

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Qiaochu Yuan 2019-12-03 05:08:35

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