A group of order $p^2q^r , p\lt q$ is not simple

Question

Prove: A group of order $p^2q^r , p\lt q$ is not simple

I only see proofs for $r=1$, how can I deduce for $r\in\mathbb{N}$?

Any help would be appreciated.

Answer 1

Assume the group is simple. Then the number of $q$-Sylows is $\ne1$, hence is $p$ or $p^2$ and also $\equiv 1\pmod q$. As $p<q$, $p\equiv 1\pmod q$ is impossible. Hence $p^2\equiv 1\pmod q$, i.e., $p\equiv \pm1\pmod q$, but then only the option $p=q-1$ remains. That makes $p=2$ and $q=3$ and there are four $3$-Sylow groups. The action by conjugation on the set of 3-Sylows gived us a homomorphism $G\to S_4$, where the image must be transitive, henve non-trivial. Thus we can view $G$ as subgroup of $S_4$. From $|S_4|=24$, we conclude that $|G|=12$. You may already have enumerated all subgroups of $S_4$ and know which of them are simple ...