# 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 $\chi\colon G\to K^{\times}$ be a character, where $K^{\times}$ denotes the multiplicative group  of $K$. Then:

 $\sum_{g\in G}\chi(g)=\begin{cases}\mid G\mid,\text{ if }\chi\text{ is trivial,% }\\ 0_{K},\text{ otherwise}\end{cases}$

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:

 $\displaystyle h\cdot\mathcal{S}$ $\displaystyle=$ $\displaystyle\chi(g_{1})\cdot\mathcal{S}$ $\displaystyle=$ $\displaystyle\chi(g_{1})\cdot\sum_{g\in G}\chi(g)$ $\displaystyle=$ $\displaystyle\sum_{g\in G}\chi(g_{1})\cdot\chi(g)$ $\displaystyle=$ $\displaystyle\sum_{g\in G}\chi(g_{1}+g),\quad(1)$ $\displaystyle=$ $\displaystyle\sum_{j\in G}\chi(j),\quad(2)$ $\displaystyle=$ $\displaystyle\mathcal{S}$

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:

 $h\cdot\mathcal{S}=\mathcal{S}$

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

Title the sum of the values of a character of a finite group is $0$ TheSumOfTheValuesOfACharacterOfAFiniteGroupIs0 2013-03-22 14:10:30 2013-03-22 14:10:30 alozano (2414) alozano (2414) 6 alozano (2414) Theorem msc 11A25