Let $p$ and $q$ be prime numbers with $p<q$.
If $q \not\equiv 1 \pmod{p}$, then every group of order $p^2q$ is
abelian.
From the Third Sylow Theorem I found that the number of $p$-Sylow subgroups is 1, and since it's the only one, it is normal, thus is isomorphic to $\mathbb{Z}/p^2\mathbb{Z}$ or to $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$.
But form here I'm stuck. I know that the number of $q$-Sylow subgroups $n_q \in\{1 , p, p^2\}$. Maybe if I could find that $n_q = 1$, which implies that the $q$-Sylow subgroup is normal, I could find an isomorphism between the group of order $p^2q$ and the direct product of the two Sylow subgroups, but I don't know how to proceed.
I looked also at this question (A group of order $p^2q$ will be abelian), but I dont' have the assumption that $p^2\not\equiv 1\pmod{q}$.
Thank you for your help!
Best Answer
This follows from the question you linked to...the assumption that $p^2\not \equiv 1 \pmod q$ comes for free.
To see this note that $p^2-1=(p+1)(q+1)$ so $q\,|\,(p^2-1)\implies q\,|\,p- 1\,\text{or}\, q\,|\,p+1$.
Since $q>p$ we can't have $q\,|\,p-1$ so we must have $q\,|\,p+1$. But that only happens when $p=2,q=3$ and in that particular case we have $q\equiv 1 \pmod p$ so this instance is excluded.