1
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.