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  "body": "\n## Proof\n\n1. Let $H$ be a subgroup of a group $G$ with [[index|04QEBF8S49]] $2$.\n2. There are two [[left cosets|03DEH43F4H]]. One is $H$, so the other is $G \\setminus H$.\n3. In the same way, there are two right cosets, $H$ and $G \\setminus H$.\n4. Let $g \\in G$. If $g \\in H$, then $gH = H = Hg$.\n5. If $g \\notin H$, then $gH \\ne H$ and $Hg \\ne H$. Thus $gH = G \\setminus H = Hg$.\n6. So $gH = Hg$ for all $g \\in G$, and [[$H$ is normal|01E7SWYJX3]].\n\n## Idea\n\nWith only two cosets, one of them is the subgroup and the other is its complement. This is true on the left and on the right, so the left and right cosets agree.\n",
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