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  "body": "\n## Statement\n\nLet $G$ be a [[group|G5S1M9]] and let $H$ and $K$ be finite [[subgroups|0X1JXT47GA]] of $G$.\nThen the [[product|0PT52XQZD7]] $HK$ has exactly $\\dfrac{|H| \\, |K|}{|H \\cap K|}$ elements.\n\n## Remarks\n\nThe formula holds also when $HK$ is not a subgroup.\nThe proof counts the pairs $(h, k)$: two pairs $(h, k)$ and $(h', k')$ give the same product exactly when $h' = hx$ and $k' = x^{-1}k$ for some $x \\in H \\cap K$.\nFor example, two distinct subgroups of order $2$ in $S_3$ meet in $\\{e\\}$, and their product has $2 \\cdot 2 / 1 = 4$ elements.\n\n## Explanations\n\n### L1 human\n\nEach product $hk$ can be written in several ways, because you can move an element of $H \\cap K$ from one factor to the other.\nEach element of $HK$ has exactly $|H \\cap K|$ such ways.\nSo you divide the number of pairs, $|H| \\, |K|$, by $|H \\cap K|$.\n",
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  "name": "Product formula for the order of HK",
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