If $|G|=pq^2$ with $p,q$ primes and if $p<q$, with $q\not\equiv\pm1\mod p$, why is $G$ abelian ?
The $3^{rd}$ Sylow theorem implies that $n_p|q^2$ and $n_p\equiv 1 \mod p$,
By hypothesis, $n_p$ should be $1$, thus the $p$-Sylow subgroup is unique and therfore normal by $2^{nd}$ Sylow theorem.
If we consider $q$-sylow subgroups, say one of them is $Q$, then it is normal (because its index is the smallest prime factor of $G$ etc, I know this).
How do I continue?
If two subgroups of a group are normal, have trivial intersection, and generate the group via elements of the form $xy$ with $x$ in the first group and $y$ in the second, then the group is the direct product of the two groups. The group of order $q^2$, is abelian, so the result follows.