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If $\chi$ is a nontrivial irreducible character of $G$ (a finite group), define $S_{\chi}:= \sum_{x \in G} \chi(x)$. In terms of conjugacy classes $\mathcal{C}$, this is $\sum_{\mathcal{C}} |\mathcal{C}| \chi(\mathcal{C})$. Is there a nice condition that guarantees $S_{\chi}=0$?

I've noticed that this occurs, for instance, with $S_5$. I'd love a description of this phenomenon and a proof, if possible.

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    $\begingroup$ Notice that $\sum_{g\in G} \chi(g) = |G|\langle \chi, \mathbb 1\rangle$, so $S_{\chi} = 0$ iff $\mathbb 1$ doesn't appear as a direct summand of $\chi$. $\endgroup$ – ah11950 Apr 25 '14 at 0:02
  • $\begingroup$ Of course. Wow, that was an obvious one. Sorry for the time wasted. $\endgroup$ – José Siqueira Apr 25 '14 at 0:06
  • $\begingroup$ No worries! It can be dreadfully easy to miss the most simple things at times; I certainly know it happens to me! (I've just noticed that you've assumed $\chi$ is irreducible, so obviously it's possible to make the stronger statement that $S_{\chi}=0 \iff \chi \neq \mathbb 1$...) $\endgroup$ – ah11950 Apr 25 '14 at 0:09
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As @ah11950 states in the comments, $$\sum_{g\in G} \chi(g) = |G|\langle \chi, \mathbb 1\rangle,$$ so $S_{\chi} = 0$ if and only if $1$ doesn't appear as a direct summand of $\chi$.

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