# 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

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?

• @usersujo Let $\chi_k$ be the character of a one-dimensional representation then $n_k = 1$. By your Argument above $\langle \chi, \chi_k\rangle = \chi(1)/|G|\cdot n_k = \chi(1)/|G|$ is an integer. Jun 8, 2013 at 17:35
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.
I think it's worth pointing out that we don't need to know that $\chi$ can be decomposed as above, which is something that Serre proves in the next section after this exercise. By direct computation, the (integer!) number of times that the representation that has $\chi$ as its character contains the trivial representation $1$ is $\langle \chi,1\rangle$: $$\langle \chi,1\rangle = \frac{1}{|G|}\sum_{s\in G}\chi(s^{-1})1(s) = \frac{1}{|G|}\chi(1) \ne 0.$$ Moving $|G|$ to the other side, we get $\chi(1) = \langle\chi,1\rangle\cdot|G| = \langle\chi,1\rangle\cdot r_G(1)$. Hence $\chi = \langle\chi,1\rangle\cdot r_G$ since $\chi(s) = r_G(s) = 0$ if $s\ne 1$.