Let $G$ be a finite group with order $pq$, where $p < q$ are primes. Let $$[G, G] = \langle xyx^{-1}y^{-1} \mid x, y \in G \rangle$$ denote the commutator subgroup.
Then, the index of $[G, G]$ in $G$ is a multiple of $p$.
I have a solution to the above using representation theory. (Written below.)
Is the above a "standard"/"basic" fact of group theory that one shows using class equations/Sylow theorems? I have not seen it before and a search on this site didn't yield any results. Is there an "elementary" proof of the above?
Usually, with class equations and Sylow theorems, one focuses on the centre of a group or an element. The commutator subgroup does not usually show up, which is why I am curious.
Proof. Let $d_1, \ldots, d_s$ denote the degrees of the irreducible representations of $G$. Then, $$pq = d_1^2 + \cdots + d_s^2$$ and $d_i \mid pq$ for all $i$. The two facts show that for each $i$, $d_i = 1$ or $p$.
Letting $m$ ($n$, resp.) denote the number of degree $1$ ($p$, resp.) representations gives $$pq = m + np^2$$ and thus, $p \mid m$.
However, $m$ is precisely the order of $G/[G, G]$ or the index of $[G, G]$ in $G$. $\Box$
This follows very easily from knowing that the $q$-subgroup is normal (the existence of a $q$-subgroup follows from Cauchy; normality is a trivial consequence of the Sylow theorems, but can be proven by counting arguments on the action of the $p$-subgroup).
If $Q\leq G$ has $|Q|=q$, then $G/Q$ is of order $p$, hence abelian. This means that $G’=[G,G]\leq Q$. Hence $$[G:G’] = [G:Q][Q:G’] = p[Q:G’],$$ and thus a multiple of $p$. (Of course, it is either $p$ if $G$ is nonabelian, or $pq$ if $G$ is abelian).