Classification of groups of order $p^2q$, when $q|p+1$

Question

When considering non-abelian groups of order $p^2q$, where $p>q$ and $p,q>3$, two cases can be identified, when $q|p^2-1$.

Case 1: $q |p+1, q \nmid p-1$

Case 2: $q |p-1, q \nmid p+1$

Can someone guide me to prove that in Case 1, a unique semidirect product is present.

$p,q$ are distinct primes.

Thanks a lot in advance.

Answer 1

Let your group be $G$.

First of all, the Sylow $p$-subgroup $P$ will be a unique, as it has index $q < p$.

If $P$ is cyclic, then there is no (non-abelian) group $G$, as $q$ does not divide the order $p (p - 1)$ of the automorphism group of $P$.

Let then $P$ be elementary abelian, so that its automorphism group $A$ has order $$a = (p^{2} - 1) (p^{2} - p) = p (p - 1)^{2} (p + 1).$$

As $q > 2$, $q$ is the highest power of $q$ which divides $a$.

Let $Q$ be a Sylow $q$-subgroup. The groups $G$ are parameterised by the non-trivial homomorphisms $\phi : Q \to A$. The image of such a $\phi$ is thus a Sylow $q$-subgroup of $A$, and all these Sylow subgroups are conjugate.

This, and an easy additional argument with powers, show that all the groups arising from these $\phi$ are isomorphic.

Answer 2

Let $G$ be a group of order $p^2q$, with $q|p^2-1$ and $p>q$.
You know that $n_p\equiv_p1$ and $n_p|q\implies n_p\in\{1,q\}$. If $n_p=1\implies\exists !P\in Syl_p(G)$ such that $P\trianglelefteq G$.
If $n_p=q\implies p|q-1\implies p<q$, which is not possible because we are supposing $p>q$, hence $n_p=1$ and we have a unique normal p-group in $G$.
Similarly we know that $n_q\equiv_q 1$ and $n_q\in\{1,p,p^2\}$.
If $n_q=p^2$, then we would have $p^2-1$ elements of order $q$ because every q-Sylow has prime order, hence it's cyclic.
Therefore we would have $|G|-1-|Syl_q(G)|q=p^2q-1-(p^2-1)=p^2(q-1)$ elements of order $p$, but this is not possible because we've shown that there is only one p-Sylow subgroup in $G$.