Prove that a group of order $p^2q$ has a proper normal subgroup and is solvable.

For the contradiction: Since $p^2-1=qj$ then $p^2-1$ is a multiple of $q$. Hence, $q$ divides $p-1$ or $p+1$. Since $q>p$ then $q=p+1$, which implies $q=3$ and $p=2$.

This paper lists all possible groups of order $12$ (and includes a proof that none of them are simple).

You can use a counting argument. Note that if there are $p^2$ subgroups of order $q$, each of them has $q-1$ distinct elements of order $q$ generating that subgroup. Thus, only $p^2 = p^2q - p^2(q-1)$ elements can have orders $1, p, p^2$. This implies that there can be only one subgroup of order $p^2$, as desired.

Hint: If there are $q$ Sylow $p$-subgroups, then either two of them intersect non-trivially, or between them the Sylow $p$-subgroups cover $1+q(p^2-1)=p^2q-(q-1)$ elements.

• In the latter case there is room for only a single Sylow $q$-subgroups.
• In the former case let $H$ be a non-trivial intersection of two Sylow $p$-subgroups: $H=P_1\cap P_2$. Show that the normalizer of $H$ has more than $p^2$ elements.