Let $|G|=p^{(k+l)}q$ with $k, l \geq 0$ and $p \nmid q$. Show that G acts on $X:=\{A \subset G : |A| = p^k \}$ by left multiplication?

Question

I have the idea of how to prove this for when $|G|=p^kq$, which is as follows.

Let $|G|=p^kq$ such that $\gcd(p,q)=1$. Let $$ X=\{A \subset G : |A|=p^k \}. $$ $G$ acts on $X$ by left translation, that is, $gA:= \{ga : a \in A \}$. $gA$ is an element of $X$ because $a \mapsto ga: A \to G$ is an injection and so $|gA|=|A|$.

But, how do I manipulate this to be shown for when $|G|=p^{(k+l)}q$? Then how do I show that $v_p(|X|)=l?$