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  "body": "\n## Statement\n\nLet $G$ and $H$ be finite [[cyclic groups|0KVGH0J5B3]] of orders $m$ and $n$.\nLet $G \\times H$ be the group of pairs $(g, h)$ with $g \\in G$ and $h \\in H$, with the operation $(g, h)(g', h') = (gg', hh')$.\nThen $G \\times H$ is cyclic if and only if $\\gcd(m, n) = 1$.\n\n## Remarks\n\nWhen $\\gcd(m, n) = 1$, 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.\nWhen $\\gcd(m, n) \u003e 1$, each element of $G \\times H$ has order at most $\\operatorname{lcm}(m, n) \u003c mn$.\nFor 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.\n\n## Explanations\n\n### L1 human\n\nThe order of a pair $(g, h)$ is the least common multiple of the orders of $g$ and $h$.\nA pair of generators has order $mn$ exactly when $m$ and $n$ have no common factor.\nFor example, $(1, 1)$ in $\\mathbb{Z}/2\\mathbb{Z} \\times \\mathbb{Z}/3\\mathbb{Z}$ has order $6$, so this group is cyclic.\n",
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