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  "body": "\n## Statement\n\nLet $G$ be a [[group|G5S1M9]], let $H$ be a [[subgroup|0X1JXT47GA]] of $G$, and let $N$ be a [[normal subgroup|0CB0DX5BVV]] of $G$.\nThen the [[product|0PT52XQZD7]] $HN$ is a subgroup of $G$, and $HN = NH$.\n\n## Remarks\n\nIf $H$ is also normal, then $HN$ is normal in $G$.\nThe proof uses $hN = Nh$ for each $h \\in H$, so that $HN = \\bigcup_{h \\in H} hN = \\bigcup_{h \\in H} Nh = NH$.\nThis product appears in the second isomorphism theorem.\n\n## Explanations\n\n### L1 human\n\nThe product of two subgroups can fail to be a subgroup, but not when one of them is normal.\nA normal subgroup lets you move its elements past other elements, at the cost of changing them inside the subgroup.\nSo products of the form $hn$ stay of that form when you multiply them.\n",
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