Let $p,q$ be primes and let $G$ be a group of order $p^2q^2$, what's the best way to show $G$ is non-simple?
I know it suffices to show that one of the Sylow-p or Sylow-q subgroup of $G$ is normal, but the counting elements argument doesn't work here since different Sylow subgroups may have non-trivial intersection.
We may as well assume $p < q$. The number of Sylow $q$-subgroups is $1$ mod $q$ and divides $p^2$. So it is $1, p$, or $p^2$. We win if it's $1$ and it can't be $p$, so suppose it's $p^2$. But now $q \mid p^2 - 1$, so $q \mid p+1$ or $q \mid p-1$.
Thus $p = 2$ and $q = 3$. The case of order $36$ has been proved in an earlier question:
Using a little more group theory allows us to prove something stronger (and avoid the reduction to $|G|=36$):
A group of order $p^2q^2$ has either a normal Sylow $p$-group or normal Sylow $q$-group.
For assume that $p<q$, then there are either $1$ or $p^2$ Sylow $q$-groups in $G$.
If there is $1$, it is normal, and we are done.
If there is $p^2$, then the Sylow $q$-groups are self-normalizing. But any group of order $q^2$ is abelian, so Burnside's transfer theorem implies the Sylow $p$ group is normal.
I'll give another proof similar to user29743's answer but avoiding to examine groups of order $36$.
Suppose that $p<q$ and $n_q=p^2$.
If $Q_i\cap Q_j=1$ for every two distinct Sylow $q-$subgroups then by simple counting we find that $n_p=1$ hence $G$ is not simple.
Let $Q_i,Q_j$ be two Sylow $q-$subgroups with $1\not= I= Q_i\cap Q_j$. Since $|Q_i|=|Q_j|=q^2$ these are abelian groups and $I\lhd Q_i,Q_j \Rightarrow 1\not=I\lhd \langle Q_i,Q_j\rangle=M$. It has to be $|M|>|Q_i|=q^2\Rightarrow |M|=p^2q^2$ or $pq^2$.
If $|M|=p^2q^2=|G|$ then clearly $G$ is not simple $\checkmark$
If $|M|=q^2p$ then $|G:M|=p$ so we have a homomorphism $h:G\to S_p$. If $kerh=1$ we have a contradiction since $p^2q^2\not| p!$ so it has to be $1\not= kerh \leq M\lneq G \ \checkmark$
