Product of cyclic groups is cyclic iff the orders are coprime

Statement

Let $G$ and $H$ be finite cyclic groups of orders $m$ and $n$. Let $G \times H$ be the group of pairs $(g, h)$ with $g \in G$ and $h \in H$, with the operation $(g, h)(g', h') = (gg', hh')$. Then $G \times H$ is cyclic if and only if $\gcd(m, n) = 1$.

Remarks

When $\gcd(m, n) = 1$, the group $\mathbb{Z}/mn\mathbb{Z}$ is isomorphic to $\mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$, and this is a form of the Chinese remainder theorem. When $\gcd(m, n) > 1$, each element of $G \times H$ has order at most $\operatorname{lcm}(m, n) < mn$. For example, $\mathbb{Z}/12\mathbb{Z}$ is isomorphic to $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}$, but $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z}$ is not cyclic.

Explanations

L1 human

The order of a pair $(g, h)$ is the least common multiple of the orders of $g$ and $h$. A pair of generators has order $mn$ exactly when $m$ and $n$ have no common factor. For example, $(1, 1)$ in $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ has order $6$, so this group is cyclic.