This is a question from J.P.Serre's book 'Linear representation of finite groups',section 2.4
The question: Let $G$ be a finite group. Show that each character of $G$ which is zero for all $g \ne 1$ is an integral multiple of the character $r_G$ of the regular representation.
What I have done so far:
$r_G$ satisfies $r_G(g) = 0$ for all $g \ne 1$, and $r_G(1) = |G|$, the order of $G$. If $\chi$ denotes the character, then $\chi(g) = r_G(g) = 0$ for all $g \ne 1$, so it is enough to show that $|G|$ divides $\chi(1)$. If $\chi_1,…,\chi_k$ denotes all the irreducible characters of $G$, with dimension of the representations $n_1,…,n_k$ respectively, then we can write
$\chi = \sum_{i=1}^k \langle \chi,\chi_i\rangle \chi_i$, where $\langle \chi,\chi_i\rangle$ is the inner product. And it is easy to calculate $\langle \chi,\chi_i\rangle = (\chi(1)/|G|)\,n_i$. So each of these values must be integers for all $i$.
But how does one conclude that in fact $\chi(1)/|G|$ is an integer?
thanks in advance.
Best Answer
One of the irreducible representations is the trivial, one-dimensional one, say $\chi_i$ is its character. Then $n_i = 1$, and your argument above gives $\def\<#1>{\left\langle#1\right\rangle}$that $$\<\chi, \chi_i> = \chi(1)/|G| \cdot n_i = \chi(1)/|G| $$ is an integer.