Let \(p\) and \(q\) be primes with \(q<p\), and let \(G\) be a group of order \(pq\).
If \(q\notdivides p-1\), then \(G\cong C_{pq}\).
If \(q\divides p-1\), then either \(G\cong C_{pq}\), or \(G\) is nonabelian and \[ G\cong C_p\semidirect_\psi C_q \cong \gens{a, b \suchthat a^p=b^q=1,\ bab^{-1} = a^\ell}, \] where \(\psi\colon C_q\to\Aut(C_p)\) is nontrivial and \(\ell\) is any integer with \(\ell \not\equiv 1 \pmod p\) and \(\ell^q \equiv 1 \pmod p\). Up to isomorphism, the nonabelian group does not depend on the choice of \(\psi\) or \(\ell\).
Let \(P\) and \(Q\) be Sylow $p$- and $q$-subgroups of \(G\), so \(P\cong C_p\) and \(Q\cong C_q\). By the Sylow theorems \(n_p \equiv 1 \pmod p\) and \(n_p \divides q\); since \(q < p\), \(n_p = 1\) and \(P\normal G\). As \(P\cap Q=1\), \(\abs{PQ}=pq\), so \(G=PQ\cong P\semidirect_\psi Q\) for the conjugation action \(\psi\colon Q\to\Aut(P)\cong(\ZZ/p\ZZ)^\times\cong C_{p-1}\).
If \(q\notdivides p-1\), then \(\psi\) is trivial, so \(G\cong C_p\times C_q\cong C_{pq}\).
If \(q\divides p-1\) and \(\psi\) is trivial, again \(G\cong C_{pq}\). If \(\psi\) is nontrivial, it is injective because \(\abs{Q}=q\) is prime, and its image is the unique subgroup of order \(q\) of the cyclic group \(\Aut(P)\). Let \(a\) generate \(P\). Two nontrivial homomorphisms \(Q\to\Aut(P)\) with the same image differ by precomposition with an automorphism of \(Q\), which gives isomorphic semidirect products. Choosing a generator \(b\) of \(Q\) with \(\psi(b)(a)=a^\ell\) gives the presentation, and the conditions on \(\ell\) say exactly that \(a\mapsto a^\ell\) has order \(q\) in \(\Aut(P)\). This group is nonabelian because \(bab^{-1}=a^\ell\neq a\).