TITLE: Show that a group of order $p^2q^2$ is solvable
QUESTION [10 upvotes]: I am trying to prove that a group of order $p^2q^2$ where $p$ and $q$ are primes is solvable, without using Burnside's theorem. Here's what I have for the moment:

If $p = q$, then $G$ is a $p$-group and therefore it is solvable.
If $p \neq q$, we shall look at the Sylow $p$-subgroups of $G$. We know from Sylow's theorems that $n_p \equiv 1 \pmod p$ and $n_p \mid q^2$, therefore $n_p \in \{1, q, q^2\}$.
If $n_p = 1$, it is over, because the Sylow $p$-Subgroup $P$ is normal in $G$ of order $p^2$, and $G/P$ has order $q^2$. Thus both are solvable and $G$ is solvable.
If $n_p = q^2$, we have $q^2(p^2-1)$ elements of order $p$ or $p^2$ in $G$, and we have $q^2$ elements left to form a unique Sylow $q$-subgroup. By the same argument as before, $G$ is solvable.
That's where I'm in trouble. I don't know what to do with $n_p = q$. It seems to lead nowhere. 

Thanks in advance for any help!
Laurent

REPLY [3 votes]: Assuming that you know that groups of order $p^2q$, $pq$ and $p^k$ are solvable, it is enough to prove that a group of order $p^2q^2$ is not simple.
Suppose that $G$ is a simple group of order $p^2q^2$. By symmetry (and since $p$-groups are solvable) we may assume $p > q$. Steps to reach a contradiction:

Prove the following: if $G$ is a finite group with a subgroup $H$ of index $r$, where $r$ is the smallest prime divisor of $|G|$, then $H$ is a normal subgroup.
By 1. $n_p = q$ cannot happen since $G$ is simple. Therefore $n_p = q^2$.
If there exist distinct Sylow $p$-subgroups $P_1$ and $P_2$ such that their intersection $D = P_1 \cap P_2$ is nontrivial, then $D$ has order $p$. Now $D$ is normal in both $P_1$ and $P_2$, but not normal in all of $G$, so $N_G(D)$ has order $p^2q$. This is a contradiction by 1.
Therefore distinct Sylow $p$-subgroups of $G$ have pairwise trivial intersection. By 2. this means that there are $q^2(p^2-1)$ elements of order $p$ or $p^2$. But then $G$ has a normal Sylow $q$-subgroup.