Does every group of Order $p^2 q$ have a normal Sylow $p$-Subgroup? This page  talks about a result which says that any group of $p^2 q$ has a normal Sylow  $p$ subgroup.
But then this page proves that any group of $2^2 3$ either has a normal Sylow $2$ subgroup or a normal Sylow $3$ subgroup.
I am extremely confused. Shouldn't the first theorem guarantee a Sylow $2$ subgroup ?
Are groups of order $12$ just an exception ? If yes, are there any other exceptions ?
Please clarify.
 A: The proof wiki statement suffers from very bad notation. It says that every Group of order $p^2 q$ has a normal $p$-Sylow subgroup (where the first $p$ is a fixed number, but the second $p$ in $p$-Sylow subgroup is generic). If you look at the proof given, what they actually show is that a group of order $p^2 q$ has a normal $p$-Sylow subgroup or a normal $q$-Sylow subgroup.
The Sylow theorem does not guarantee a $p$-Sylowsubgroup for a group of order $p^2 q$ if
$$p \equiv 1 \mod q.$$
For example pick $p^2 q = 20$. We have $\mathrm{Aut} \mathbb{Z}/5\mathbb{Z} \cong \mathbb{Z}/4\mathbb{Z}$. Hence, pick $\theta = \mathrm{id}_{\mathbb{Z}/4\mathbb{Z}}$ and let $G$ be the semidirect product of $\mathbb{Z}/4\mathbb{Z}$ and $\mathbb{Z}/5\mathbb{Z}$ with respect to $\theta$. This group $G$ is not abelian, but has a $5$-Sylow subgroup. If it also had a $4$-Sylow subgroup it would be abelian.
The case $p^2 q = 12$ is just some kind of special case, because here the Sylow alone don't give the existence of a normal $p$-Sylow subgroup or a normal $q$-Sylow subgroup without further arguments. I recommend reading the proof given in your first link (there is also a proof of this special case included).
