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

Solution 1:

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.

Solution 2:

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 $s\ne 1$.