Let $G$ be a finite group. A character of $G$ is a homomorphism $\chi: G \to K^*$, where $K^*$ denotes the multiplicative group of non-zero elements of a field $K$. From the definition of $\chi$, it follows that $\chi(g)$ is a root of unity for any $g \in G$.

I am stuck that $\sum_{g \in G}\chi(g) = 0$ for a non-trivial character $\chi: G \to K^*$. I know $\sum_{g \in \langle a \rangle}\chi(g) = 0$ if $G=\langle a \rangle$ is a cyclic group. How to prove the fact that $\sum_{g \in G}\chi(g) = 0$ for any finite group $G$.

Any help will be appreciated.

  • 1
    $\begingroup$ I don't understand, what happened if $\chi$ is the trivial character, $\chi(g) = 1$ $\forall g \in G$ ? $\endgroup$
    – user171326
    Nov 28 '16 at 13:42
  • $\begingroup$ @ N.H. Thank you very much. I require that $\chi$ is a non-trivial character. $\endgroup$
    – bing
    Nov 28 '16 at 13:45
  • $\begingroup$ Ok. I did answer to your question, but I don't know if this is easy to deduce without orthogonality relations. $\endgroup$
    – user171326
    Nov 28 '16 at 13:49

The characters verify an orthogonality property : if $\chi$ is not isomorphic to $\chi'$ then $\langle \chi, \chi' \rangle = 0$ (assuming they are both irreducible, but here any 1-dimensional character is by definition irreducible). If $\chi$ is not the trivial character, then $\chi$ is not isomorphic to the trivial character $1$. In particular, $\langle \chi, 1 \rangle = 0$ and it follows that $\sum_{g \in G}\chi(g) = 0$.

Edit : as pointed Alex Youcis in the comment, there is a simpler way of seeing it. If $\chi$ is non-trival, there is $g_0$ with $\chi(g_0) \neq 1$. But then $\sum_{g} \chi(g) = \sum_{g} \chi(g_0g) = \chi(g_0) \sum_g \chi(g)$ and therefore $\sum_g \chi(g) = 0$.

  • 3
    $\begingroup$ This is way overkill (although definitely good to recognize it as a basic case of orthogonality)--+1. Since $\chi$ is non-trivial you can choose $g_0$ such that $\chi(g_0)\ne 1$. Then, $\displaystyle \chi(g_0)\sum_g \chi(g)$ equals $\displaystyle \sum_g \chi(g_0g)=\sum_g\chi(g)$ by reindexing. But, since $\chi(g_0)\ne 1$ this implies that $\displaystyle \sum_g \chi(g)=0$. $\endgroup$ Nov 28 '16 at 13:50
  • $\begingroup$ Oh you're totally right ! You should post this as an answer, this is way more elementary than my solution. $\endgroup$
    – user171326
    Nov 28 '16 at 13:55
  • $\begingroup$ No need to clutter the answers--you can just edit yours and add it as an alternative way of seeing it. $\endgroup$ Nov 28 '16 at 13:56
  • 1
    $\begingroup$ Having stumbled across this while doing something else, I don't think the original idea was overkill at all. I think it is more natural than reindexing, which always feels... dirty. $\endgroup$
    – The Count
    May 15 '17 at 18:22
  • 1
    $\begingroup$ Why is $\chi(g_0)\sum_g\chi(g)=\sum_g \chi(g_0 g)$? Also, how about $\chi(g_0)=0$? $\endgroup$ Aug 31 '20 at 19:17

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