A group of order $p^2q^2$ is never simple

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:

No group of order 36 is simple

• How do you arrive at $p=2$ and $q=3$? Commented Apr 11, 2015 at 0:03
• $p<q$, so we must have $q=p+1$ if $q$ divides either $p-1$ or $p+1$. Commented Apr 11, 2015 at 0:49
• "But now $q \mid p^2 - 1$" - why? Commented May 16, 2023 at 19:08
• @Robin By the third Sylow theorem, $p^2 \equiv 1 \pmod q$. Commented Oct 19, 2023 at 0:47

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 LOVE when people prove extra things, especially when they state explicitly that they are doing so. It seems the answerer has vanished, but thanks anyway, ghost person! Commented Sep 29, 2016 at 13:35

I'll give another proof similar to user29743's answer but avoiding to examine groups of order $$36$$.

Suppose that $$p and $$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 $$pq^2$$.

1. If $$|M|=p^2q^2=|G|$$ then clearly $$G$$ is not simple $$\checkmark$$

2. 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 $$1\not= kerh \leq M\lneq G \ \checkmark$$

• Really nice one as well! Commented Dec 28, 2022 at 0:51