I am trying to show that any group of order $p^2q$ has a normal Sylow subgroup where $p$ and $q$ are distinct primes.
In the case $p>q$ I have no problem.. By Sylow $n_p|q$, so $n_p$ is either $1$ or $q$. But we also have $n_p\equiv1\mod p$ which rules out $q$. So in the case $p>q$, $n_p=1$ and the unique p-Sylow subgroup is normal.
In the case $p Does this mean I need to check groups of order $2^2\cdot3=12$ or did I miss something along the way that lets me conclude $n_q\neq p^2$ and thus that $n_q=1$.