= 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, so
$$
K=\bigcap_{x\in G}xHx^{-1}\subset H.
$$
This 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 gives
$$
\boxed{[G:K]=|\rho(G)|\text{ divides }n!}.
$$
In 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 \b[every subgroup generated by an element of order $q$ is normal]. No classification of groups of order $pq$ is needed.
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