Let $G$ be a group such that $|G| = p^2q$, where $p, q$ are primes such that $p < q$.
Consider a Sylow $q$-subgroup of $G$. I want to show that such subgroup is normal in $G$.
Now, by Sylow's Theorems, it will suffice to show that there is only one such subgroup. But the usual divisibility and congruence conditions aren't helping me out here (e.g., I can't seem to derive a contradiction by assuming that the number of Sylow $q$-subgroups is $p$ or $p^2$).
Another approach I've attempted is considering the normalizer of such a group: Call such a subgroup $Q$. It will be sufficient to show that $N_G(Q) = G$. I've got it down to an argument regarding the order of $N_G(Q)$, where I've eliminated all possibilities except for $q$... but, again, I see no reason why $Q$ can't be self-normalizing...
Basically, I've hit this problem with everything I've got and am out of ideas.