{
  "aliases": [
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  "body": "\n## Statement\n\nLet $G$ be a [[group|G5S1M9]], let $S$ be a [[subgroup|0X1JXT47GA]] of $G$, and let $N$ be a [[normal subgroup|0CB0DX5BVV]] of $G$.\nThen $SN$ is a subgroup of $G$, $N$ is normal in $SN$, and $S \\cap N$ is normal in $S$.\nMoreover, the map $s(S \\cap N) \\mapsto sN$ is an [[isomorphism|08MG7H8JW5]] from $S/(S \\cap N)$ to $SN/N$.\n\n## Remarks\n\nThe proof applies the [[first isomorphism theorem|0CXXJNG5B6]] to the homomorphism $S \\to SN/N$, $s \\mapsto sN$, which is surjective with kernel $S \\cap N$.\nFor finite groups, it gives $|SN| = |S| \\, |N| / |S \\cap N|$.\nIt is enough that $S$ normalizes $N$, that is, $sNs^{-1} = N$ for each $s \\in S$.\n\n## Explanations\n\n### L1 human\n\nPicture a diamond with $SN$ at the top, $S$ and $N$ in the middle, and $S \\cap N$ at the bottom.\nThe theorem says that the two slanted sides opposite each other give the same quotient group.\nFor integers, with $S = 4\\mathbb{Z}$ and $N = 6\\mathbb{Z}$, it gives $4\\mathbb{Z}/12\\mathbb{Z} \\cong 2\\mathbb{Z}/6\\mathbb{Z}$, both cyclic of order $3$.\n",
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