= Solution
Interchange the two primes if necessary so that $p<q$. By the <Sylow theorems>, the number $n_q$ of Sylow $q$-subgroups satisfies
$$
n_q\mid p^2,
\qquad
n_q\equiv1\pmod q.
$$
If $n_q\ne1$, then $n_q$ is $p$ or $p^2$. The first is impossible because $1<p<q$. The second would imply
$$
q\mid p^2-1=(p-1)(p+1).
$$
Since $q>p$ is prime, it cannot divide $p-1$, so it must divide $p+1$. But $q>p$ then forces $q=p+1$, impossible because $p$ and $q$ are both odd. Therefore $n_q=1$. The unique Sylow $q$-subgroup is a nontrivial proper <normal subgroup>, and hence
$$
\boxed{G\text{ is not simple}}.
$$
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