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  "body": "\n## Statement\n\nLet $G$ be a [[group|G5S1M9]], and let $N$ and $K$ be [[normal subgroups|0CB0DX5BVV]] of $G$ with $N \\subseteq K$.\nThen $K/N$ is a normal subgroup of $G/N$, and the map $(gN)(K/N) \\mapsto gK$ is an [[isomorphism|08MG7H8JW5]] from $(G/N)/(K/N)$ to $G/K$.\n\n## Remarks\n\nThe proof applies the [[first isomorphism theorem|0CXXJNG5B6]] to the surjective homomorphism $G/N \\to G/K$, $gN \\mapsto gK$, whose kernel is $K/N$.\nFor example, $(\\mathbb{Z}/12\\mathbb{Z})/(4\\mathbb{Z}/12\\mathbb{Z}) \\cong \\mathbb{Z}/4\\mathbb{Z}$.\nFor finite groups, it gives $[G : K] = [G/N : K/N]$.\n\n## Explanations\n\n### L1 human\n\nIf you divide by $N$ and then by the image of $K$, you get the same result as dividing by $K$ in one step.\nIt looks like cancelling fractions: $(G/N)/(K/N) = G/K$.\nThe theorem makes this cancellation precise.\n",
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  "name": "Third isomorphism theorem",
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