# $\chi$ is a character and $\chi(g) \in 2\mathbb{Z}$ then $\frac{1}{2}\chi$ is also a character.

Suppose $$G$$ is a finite group. Let $$\chi$$ be the character of some $$\mathbb{C}G$$-module with the property that $$\chi(g)$$ is an even integer for every $$g \in G$$. Is it true that $$\chi/2$$ defined by $$\chi/2 (g):=\chi (g) /2$$ is the character of some $$\mathbb{C} G$$-module?

Let $$V$$ be a $$\mathbb{C}G$$-module and $$\mathcal{B}$$ be a basis. Then the map $$\chi:G \rightarrow \mathbb{C}$$ defined by $$\chi(g)=trace([g]_{\mathcal{B}}$$) is the character of $$V$$.

I think it is equivalent to ask: Does there exist another $$\mathbb{C}G$$-module $$W$$ and a basis $$B'$$ of $$W$$ such that $$\frac{1}{2}trace([g]_{\mathcal{B}})=trace([g]_{\mathcal{B'}})$$ for every $$g \in G$$ if $$trace([g]_{\mathcal{B}})$$ is an even integer?

No. Let $$G$$ have even order $$n$$ and let $$\chi$$ be the character of the regular representation. Then $$\chi(e)=n$$ and $$\chi(g)=0$$ for $$g\ne e$$. So $$\chi/2$$ is integer valued. But $$\chi/2$$ is not a character. The regular representation includes the trivial representation with multiplicity $$1$$. A representation with character $$\chi/2$$ would have to include the trivial representation with multiplicity $$1/2$$.