the sum of the values of a character of a finite group is $0$

The following is an argument that occurs in many proofs involving characters of groups. Here we use additive notation for the group $G$, however this group is not assumed to be abelian.

Lemma 1.

Let $G$ be a finite group, and let $K$ be a field. Let $\chi \mathrm{:}G\mathrm{\to }{K}^{\mathrm{\times }}$ be a character, where $K^{\mathrm{\times }}$ denotes the multiplicative group of $K$. Then:

$$\sum _{g\in G}\chi (g)=\{\begin{array}{cc}\mid G\mid ,\mathit{\text{ if }}\chi \mathit{\text{ is trivial,}}\hfill & \\ 0_K,\mathit{\text{ otherwise}}\hfill & \end{array}$$

where ${\mathrm{0}}_K$ is the zero element in $K$, and $\mathrm{\mid }G\mathrm{\mid }$ is the order of the group $G$.

Proof.

First assume that $\chi$ is trivial, i.e. for all $g\in G$ we have $\chi (g)=1\in K$. Then the result is clear.

Thus, let us assume that there exists $g_1$ in $G$ such that $\chi (g_1)=h\ne 1\in K$. Notice that for any element $g_1\in G$ the map:

$$G\to G,g\mapsto g_1+g$$

is clearly a bijection. Define $𝒮={\sum }_{g\in G}\chi (g)\in K$. Then:

$h\cdot 𝒮$$=$$\chi (g_1)\cdot 𝒮$

$=$$\chi (g_1)\cdot {\displaystyle \sum _{g\in G}}\chi (g)$

$=$${\displaystyle \sum _{g\in G}}\chi (g_1)\cdot \chi (g)$

$=$${\displaystyle \sum _{g\in G}}\chi (g_1+g),(1)$

$=$${\displaystyle \sum _{j\in G}}\chi (j),(2)$

$=$$𝒮$

By the remark above, sums $(1)$ and $(2)$ are equal, since both run over all possible values of $\chi$ over elements of $G$. Thus, we have proved that:

$$h\cdot 𝒮=𝒮$$

and $h\ne 1\in K$. Since $K$ is a field, it follows that $𝒮=0\in K$, as desired.

∎