Solution
= Solution
Relabel the two primes so that $p<q$. By the <Sylow theorems>, the number $n_q$ of Sylow $q$-subgroups divides $p$, so it is either $1$ or $p$. But $n_q\equiv1\pmod q$ and $1<p<q$, ruling out $n_q=p$. Thus there is a unique <Sylow subgroup> of order $q$, which is a <normal subgroup>. It is nontrivial and proper because $1<q<pq$. Consequently \b[no <group> of order $pq$ is simple]. No additional assumption on divisibility of $q-1$ by $p$ is required.