In order $pq$ with $p<q$, the Sylow $q$-subgroup is unique and normal.Suggest a slogan
Let \(p<q\) be primes and let \(G\) be a group with \(\abs G=pq\). Then \(G\) has a normal subgroup of order \(q\).
Let \(n_q\) be the number of Sylow \(q\)-subgroups of \(G\). By Sylow's third theorem, \(n_q\equiv1\pmod q\) and \(n_q\divides p\). The divisors of \(p\) are \(1\) and \(p\), and \(1<p<q\) gives \(p\not\equiv1\pmod q\), so \(n_q=1\) and \(G\) has a unique Sylow \(q\)-subgroup \(Q\). For every \(g\in G\), the conjugate \(gQg^{-1}\) is again a Sylow \(q\)-subgroup, so \(gQg^{-1}=Q\). Therefore \(Q\trianglelefteq G\), and \(\abs Q=q\).