Lemma 1   Let $G$ be a finite group, and let $K$ be a field. Let $\chi\colon G\to K^{\times}$ be a character, where $K^{\times}$ denotes the multiplicative group of $K$ . Then:

where $0_K$ is the zero element in $K$ , and $\mid G \mid$ 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 $\chi(g_1)=h\neq 1\in K$ . Notice that for any element $g_1\in G$ the map: $$ G \to G,\quad g\mapsto g_1+g $$ is clearly a bijection. Define $\mathcal{S}=\sum_{g\in G} \chi(g)\in K$ . Then: \begin{eqnarray*} h\cdot\mathcal{S} &=& \chi(g_1)\cdot\mathcal{S}\\ &=& \chi(g_1)\cdot \sum_{g\in G}\chi(g)\\ &=& \sum_{g\in G}\chi(g_1)\cdot\chi(g)\\ &=& \sum_{g\in G}\chi(g_1 + g),\quad (1)\\ &=& \sum_{j\in G}\chi(j),\quad (2)\\ &=& \mathcal{S} \end{eqnarray*}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: $$h\cdot \mathcal{S}=\mathcal{S}$$ and $h\neq 1\in K$ . Since $K$ is a field, it follows that $\mathcal{S}=0\in K$ , as desired.