linear character of a finite group

Question

I am reviewing my notes of algebra. It's not a long proposition so I tried to prove it by myself but failed.

We have a finite group $G$ and a linear character $\chi$ of $G$. I need to show $\chi(\sigma)\overline{\chi(\sigma)}=1$ for any $\sigma\in G$.

I know the dimension of linear characters is $1$, i.e., $\chi(e)=1$. I guess this may help to show the proposition above.

Thanks.

Answer 1

A linear character $\chi$ of $G$ is just a homomorphism $\chi:G\to\mathbb{C}^\times$. Because $G$ is a finite group, for any $\sigma\in G$ there is some $n$ such that $\sigma^n=e$. Therefore $\chi(\sigma)^n=1$, so that $\chi(\sigma)$ is a root of unity.

Answer 2

Let $g\in G$ and consider $\chi(g)\in \mathbb{C}^{\ast}$. We know that $G$ is finite so $g^n=e$ for some positive integer $n$, i.e., $\chi(g^n)=\chi(e)=1$ for some positive integer $n$.

I hope this helps!