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Let $\chi$ be a non-trivial irreducible character of a finite group $G$. Show that $$\sum_{g \in G} \chi(g)=0.$$ Here $\chi:G \to \Bbb C$ just a function such that $\chi_{\phi}(g)=Tr(\phi_g)$

I am not getting any clue. Please give some hint.

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    Notice that this property is a very general fact : if $G$ is a locally compact group (e.g. a finite group) and $m$ the right Haar measure on $G$, then any non-trivial character $\chi :G \to S^1 \subset \Bbb C^*$ satisfies $\displaystyle \int_G \chi dm = 0$. – 2017-02-08
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    $\sum_{g \in G} \chi(g)=\sum_{g \in G} \chi(g).1=\sum_{g \in G} \chi(g)\bar{\chi_e}(g)=|G|\langle \chi,\chi_e \rangle=0$. Note that $e$ is the trivial representation. – 2017-02-09

2 Answers 2

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Relate $\sum_{g\in G}\chi(g) $ to $\sum_{g\in G}\chi(ag)$ for suitable $a\in G$.

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    So that $\chi(a) \neq 1$ or $\chi(a)=0$. But why this kind of $\chi$ exists? – 2017-02-08
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    @use How do **you** define a character of a finite group? For me it is a homomorphism $\;G\to\Bbb C^*\;$ , so $\;\chi(a)=0\;$ is impossible. What solves your problem is thyat $\;\chi(a)\neq1\;$ as $\;\chi\;$ is a given as *non-trivial character* – 2017-02-08
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    Just a minute for me character is just a function from $G$ to $\Bbb C$ not $\Bbb C^*$ and for the standard representation of $S_3$ I know $\chi ((123))=0$ – 2017-02-08
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    See my edited comment. – 2017-02-08
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    The answer should be for non-abelian groups $\chi(e)$ will solve my purpose and for linear character #Donantonio http://math.stackexchange.com/users/31254/donantonio is correct – 2017-02-08
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    See here I have a vital question that $\chi$ is not in general multiplicative. The definition used here is $\chi_{\phi}(g)=Tr(\phi_g)$ – 2017-02-16
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    I think Swapnil's comment under my question is the right answer. – 2017-02-16
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For any $a \in G$, $\sum_{g \in G} \chi(G) = \sum_{g \in G} \chi(ag) = \chi(a) \sum_{g \in G} \chi(g)$

Since $\chi$ is a nontrivial character, there exists $a$ such that $\chi(a) \neq 1$.

Hence $\sum_{g \in G} \chi(G) = 0$.

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    See my comment in the previous answer. – 2017-02-08
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    What about $0$? – 2017-02-08