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  "body": "\n## Proof\n\nWe first show $HN = NH$ and then apply the subgroup criterion.\n\n1. Let $G$ be a group with identity $e$, let $H$ be a subgroup of $G$, and let $N$ be a [[normal subgroup|0W77CVBCHQ]] of $G$.\n2. By the [[definition of the product|0NS1YTGM19]], $HN = \\{hn : h \\in H, n \\in N\\}$. In the same way, $NH = \\{nh : n \\in N, h \\in H\\}$.\n3. Let $h \\in H$ and $n \\in N$. Then $hn = (hnh^{-1}) h$. Since $N$ is normal, $hnh^{-1} \\in N$. So $hn \\in NH$, and $HN \\subseteq NH$.\n4. Let $n \\in N$ and $h \\in H$. Then $nh = h (h^{-1} n h)$. Since $N$ is normal, $h^{-1} n h = h^{-1} n (h^{-1})^{-1} \\in N$. So $nh \\in HN$, and $NH \\subseteq HN$.\n5. By steps 3 and 4, $HN = NH$.\n6. For $a, b \\in G$, we have $(ab)(b^{-1}a^{-1}) = e$. By the [[uniqueness of inverses|0D9NDGE9KP]], $(ab)^{-1} = b^{-1} a^{-1}$.\n7. Since $e \\in H$ and $e \\in N$ by the [[definition of a subgroup|0N0PXCTYA7]], $e = ee \\in HN$. So $HN$ is not empty.\n8. Let $a, b \\in HN$, say $a = h_1 n_1$ and $b = h_2 n_2$ with $h_1, h_2 \\in H$ and $n_1, n_2 \\in N$.\n9. By step 6, $b^{-1} = n_2^{-1} h_2^{-1}$. So $ab^{-1} = h_1 n h_2^{-1}$, where $n = n_1 n_2^{-1}$ lies in $N$.\n10. The element $n h_2^{-1}$ lies in $NH$. By step 5, it lies in $HN$, so $n h_2^{-1} = h' n'$ with $h' \\in H$ and $n' \\in N$.\n11. Then $ab^{-1} = (h_1 h') n'$. Since $h_1 h' \\in H$, this shows $ab^{-1} \\in HN$.\n12. By steps 7 and 11 and the [[subgroup criterion|05N4TQKPH0]], $HN$ is a subgroup of $G$. Together with step 5, this proves the statement.\n\n## Idea\n\nNormality lets an element of $N$ pass to the other side of an element of $H$, at the price of being replaced by a conjugate in $N$. This gives $HN = NH$. Then in $h_1 n h_2^{-1}$ we can move the middle factor past $h_2^{-1}$, so the product keeps the form \"element of $H$ times element of $N$\".\n",
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