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Let $F$ be an arbitrary field, and $(\rho, V)$ be an irreducible representation of $G$. Then $$\sum_{g \in G} \rho(g) = \begin{cases} 0 & \text{ if } \rho \neq 1_G, \\ |G|1_V & \text{ if } \rho = 1_G. \end{cases}$$

The case when $\rho =1_G$ is clear. But why is the sum $0$ for nontrivial irreducible representations?

Thanks very much.

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  • $\begingroup$ You probably want $G$ to be a finite group and $V$ to be a vector space over $F$. $\endgroup$ – tomasz Jun 17 '13 at 14:13
  • $\begingroup$ You mean $g \neq 1_{G}$ (and $g = 1_{G}$) since $\rho : G \rightarrow \text{GL}(V)$. $\endgroup$ – Gil Jun 17 '13 at 14:14
  • $\begingroup$ @GilYoungCheong: I think he means what he wrote in this case. $1_G$ is simply the trivial representation. $\endgroup$ – tomasz Jun 17 '13 at 14:16
  • $\begingroup$ @tomasz Thanks. Then I learn another notation. $\endgroup$ – Gil Jun 17 '13 at 14:17
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Here's a hint:

Suppose that $x = \rho(\sum g) v \neq 0$ for some $v$. What happens if you act $g \in G$ on $x$? What does that mean for the subspace spanned by $x$?

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