Is the following correct way of showing that there is no simple group of order $pq$ where $p$ and $q$ are distinct primes?
If $|G|=n=pq$ then the only two Sylow subgroups are of order $p$ and $q$.
From Sylow's third theorem we know that $n_p | q$ which means that $n_p=1$ or $n_p=q$.
If $n_p=1$ then we are done (by a corollary of Sylow's theorem)
If $n_p=q$ then we have accounted for $q(p-1)=pq-q$ elements of $G$ and so there is only one group of order $q$ and again we are done.
Is that correct?