# The sum of character of finite groups

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.

• I don't understand, what happened if $\chi$ is the trivial character, $\chi(g) = 1$ $\forall g \in G$ ?
– user171326
Nov 28 '16 at 13:42
• @ N.H. Thank you very much. I require that $\chi$ is a non-trivial character.
– bing
Nov 28 '16 at 13:45
• Ok. I did answer to your question, but I don't know if this is easy to deduce without orthogonality relations.
– 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 $\sum_g \chi(g) = 0$.
• 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 $\displaystyle \sum_g \chi(g)=0$. Nov 28 '16 at 13:50
• Why is $\chi(g_0)\sum_g\chi(g)=\sum_g \chi(g_0 g)$? Also, how about $\chi(g_0)=0$? Aug 31 '20 at 19:17