Exercise 4.4.2 (An abelian group of order $pq$ is cyclic.)

Proof. By Cauchy’s Theorem, there exists $x\in G$ and $y\in G$ such that $|x|=p$ and $|y|=q$.

Then ${(\mathit{𝑥𝑦})}^{p}q={(x^p)}^q{(y^q)}^p=1$.

Note that $x^q\ne 1$, otherwise $x^q=1$ and $|x|=p$ imply $p\mid q$, where $p,q$ are prime numbers, thus $p=q$. This contradicts the hypothesis $p\ne q$, so $x^q\ne 1$, and similarly $y^p\ne 1$. Therefore ${(\mathit{𝑥𝑦})}^p=y^p\ne 1$ and ${(\mathit{𝑥𝑦})}^q=x^q\ne 1$, so

If $\delta =|\mathit{𝑥𝑦}|$, then $\delta \mid \mathit{𝑝𝑞}$, but $\delta \nmid p$ and $\delta \nmid q$. Hence $\delta =\mathit{𝑝𝑞}$. Therefore $|⟨\mathit{𝑥𝑦}⟩|=\mathit{𝑝𝑞}=|G|$, where $⟨\mathit{𝑥𝑦}⟩\subseteq G$, therefore

is cyclic. □