= Solution
<Cauchy's theorem for finite groups> states that if a <prime number> $p$ divides the order of a finite group $G$, then $G$ contains an element of order $p$.
Consider
$$
X=\{(x_1,\ldots,x_p)\in G^p:x_1x_2\cdots x_p=e\}.
$$
The first $p-1$ entries determine the last, so $|X|=|G|^{p-1}$ is divisible by $p$. The <cyclic group> $C_p$ acts on $X$ by cyclically shifting the coordinates; the shifted tuple remains in $X$ because $x_2\cdots x_px_1=x_1^{-1}(x_1\cdots x_p)x_1=e$. Every <orbit> has size one or $p$. The fixed points are exactly $(x,\ldots,x)$ with $x^p=e$. Their number is therefore divisible by $p$. Since the identity is one fixed point, another exists; its order is $p$.
Now let the <abelian group> $G$ have order $pq$ for distinct primes $p,q$. Cauchy's theorem supplies $x,y$ of orders $p,q$. Their cyclic subgroups intersect trivially, and commutativity shows that $xy$ has order $pq$. Thus $G$ is cyclic and
$$
\boxed{G\cong C_{pq}\cong C_p\times C_q}.
$$
The corresponding assertion for order $p^2$ is false: $C_{p^2}$ is not isomorphic to $C_p\times C_p$, since the former has an element of order $p^2$ while every nonidentity element of the latter has order $p$.
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