We can use Sylow's theorems to examine a group $G$ of order $pq$ where $p$ and $q$ are primes and $p<q$

Let $n_p$ and $n_q$ denote, respectively, the number of Sylow $p$ subgroups and Sylow $q$ subgroups of $G$

Sylow's theorems tell us that $n_q=1+kq$ for some integer $k$ and $n_q$ divides $pq$ But $p$ and $q$ are prime and $p<q$ so this implies that $n_q=1$ So there is exactly one Sylow $q$ subgroup, which is therefore normal (indeed, fully invariant) in $G$

Denoting the Sylow $q$ subgroup by $Q$ and letting $P$ be a Sylow $p$ subgroup, then $Q\cap P=\{1\}$ and $QP=G$ so $G$ is a semidirect product of $Q$ and $P$ In particular, if there is only one Sylow $p$ subgroup, then $G$ is a direct product of $Q$ and $P$ and is therefore cyclic.

Given $G=Q \rtimes P$ it remains to determine the action of $P$ on $Q$ by conjugation. There are two cases:

Case 1: If $p$ does not divide $q-1$ then since $n_p=1+mp$ cannot equal $q$ we must have $n_p=1$ and so $P$ is a normal subgroup of $G$ This gives $G=C_p \times C_q$ a direct product, which is isomorphic to the cyclic group $C_{pq}$

Case 2: If $p$ divides $q-1$ then $\Aut(Q) \cong C_{q-1}$ has a unique subgroup $P'$ of order $p$ where $P'=\set{x \mapsto x^i \mid i \in \Integer/q \Integer, i^p=1}$ Let $a$ and $b$ be generators for $P$ and $Q$ respectively, and suppose the action of $a$ on $Q$ by conjugation is $x \mapsto x^{i_0}$ where $i_0 \neq 1$ in $\Integer/q \Integer$ Then $G=\inn{a,b \mid a^p=b^q=1, aba^{-1}=b^{i_0}}$ Choosing a different $i_0$ amounts to choosing a different generator $a$ for $P$ and hence does not result in a new isomorphism class. So there are exactly two isomorphism classes of groups of order $pq$