 Solution

Let $G$ act on its $n$ left cosets by $g\cdot (xH)=gxH$. This is well-defined and gives a group homomorphism $\rho :G\to S_n$. Its kernel $K$ is a normal subgroup. An element lies in this kernel exactly when it fixes every coset, soThis is the normal core of a subgroup. By the first isomorphism theorem, $G/K\cong \rho (G)$, a subgroup of $S_n$. The Lagrange theorem therefore givesIn particular $K$ has finite index even when $G$ is infinite.

Now let $H$ be generated by an element of order $q$ in a group of order $pq$. Then $|H|=q$ and $[G:H]=p$. The preceding result gives a normal $K\subset H$ with $[G:K]\mid p!$. Since $H$ has prime order, its subgroup $K$ is either $\{e\}$ or all of $H$. The first choice would imply $pq\mid p!$, which is impossible because the prime $q>p$ divides none of the factors of $p!$. Therefore $K=H$, proving that every subgroup generated by an element of order $q$ is normal. No classification of groups of order $pq$ is needed.