Let $G$ be a nonabelian group of order $p^{2}q$ where $p>q$. As $G$ is nonabelian we have $q\mid p-1$ or $q\mid p+1$. I want to show the two following facts, but I couldn't complete my proofs.
1) If $q\mid p+1$ then $G$ has no subgroup of order $pq$.
In this case $q\mid p+1$ is equvalent to $q\nmid p-1$ and so by the classification of groups of order $pq$ if $G$ has a subgroup of order $pq$ it should be cyclic. Therefore here we should show that $G$ has no element of order $pq$.
I suppose that $x$ be an element of $G$ with order $pq$ but I couldn't find a contradiction.
2) If $q\mid p-1$ then $G$ has a nonabelian subgroup of order $pq$.
In this case as $q\mid p-1$ and $G$ is nonabelian, the number of sylow $q$-subgroups of $G$ is equal to $p$. So by assumption that $Q$ is a sylow $q$-subgroup of $G$, we have $|G:N_{G}(Q)|=p$ and therefore $N_{G}(Q)$ is a subgroup of order $pq$.
I don't know how to show $N_{G}(Q)$ is nonabelian or maybe there is another nonabelian subgroup of order $pq$.