Let $G$ be a finite group and $R(G) = \Bbb Z\chi_1\oplus \cdots \oplus\Bbb Z\chi_h$ be the ring of virtual characters of $G$, so $\{\chi_i\}_{i=1}^h$ is the set of all irreducible characters of $G$. Let $p$ be prime, and let $V_p$ denote the subgroup of $R(G)$ generated by characters induced from $p$-elementary subgroup of $G$.
In Linear Representations of Finite Groups, Chapter 10, Serre shows that if the order of $G$ is $p^n\ell$ with $(p,\ell)=1$, then $\ell\in V_p$.
He then says that this is enough to show that the index $[R(G):V_p]$ is finite and relatively prime to $p$.
That the index is finite is clear: Since $V_p$ is an ideal of $R(G)$, then $\ell R(G)\leq V_p$, and since $$\infty > \ell h = [R(G): \ell R(G)] = [R(G): V_p][V_p:\ell R(G)],$$ is follows that $[R(G): V_p]<\infty$. (Recall that we said $h$ is the number of irreducible characters of $G$).
However, the equality $$[R(G): V_p] = \frac{\ell h}{[V_p:\ell R(G)]} $$ does not in itself guarantee that $[R(G): V_p]$ is prime to $p$, as far as I see. So, why does $\ell\in V_p$ imply $[R(G): V_p]$ is prime to $p$?
Posted this problem on math overflow and it seems the answer lies in an easy mistake: $[R(G):\ell R(G)] = \ell^h$, not $\ell h$, which is prime to $p$.