Show that a Sylow $q$-subgroup is normal in G, where $|G| = p^2q$.

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.

It's difficult to prove things that aren't true. $$A_4$$ is a counterexample. It has order $$2^2\cdot 3$$, and it does not have a normal 3-Sylow.
Order 12 is the only failing case, though. To get a counterexample, you need $$p^2\equiv1\pmod q$$, since $$p precludes $$p\equiv 1 \pmod q$$. But that means $$q$$ divides $$p^2-1=(p-1)(p+1)$$, so $$q=p+1$$ and since both are primes we have p=2, q=3.
You can also check that $$A_4$$ is the only order 12 counterexample. Since any counterexample has 4 3-Sylows, it has 8 elements of order 3, so only one 2-Sylow, which is normal. So it is a semidirect product of a group of order 4 with a group of order 3 with a nontrivial action (since the 3-Sylow is not normal it can't be a direct product). The cyclic group of order 4 has no order 3 automorphism, so the 2-Sylow must be the Klein 4 group. That has only one conjugacy class of order 3 automorphisms, so there is only one such semidirect product, up to isomorphism.