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  "body": "\n## Proof\n\nWe show that taking the preimage under $g \\mapsto gN$ is an inverse of $K \\mapsto K/N$, and then check inclusions and normality.\n\n1. Let $G$ be a group with identity $e$, and let $N$ be a [[normal subgroup|0W77CVBCHQ]] of $G$. The [[quotient group|0YYWCNTPWR]] $G/N$ has the operation $(aN)(bN) = (ab)N$ and the identity $eN = N$.\n2. For $b \\in G$, $(bN)(b^{-1}N) = N$. By the [[uniqueness of inverses|0D9NDGE9KP]], $(bN)^{-1} = b^{-1}N$.\n3. Let $a \\in G$. If $a \\in N$, then $aN = N$: closure gives $aN \\subseteq N$, and each $m \\in N$ is $a(a^{-1}m) \\in aN$.\n4. Let $\\mathcal{A}$ be the set of [[subgroups|0N0PXCTYA7]] $K$ of $G$ with $N \\subseteq K$, and $\\mathcal{B}$ the set of subgroups of $G/N$.\n5. For $K \\in \\mathcal{A}$, let $K/N = \\{kN : k \\in K\\}$, the set of [[left cosets|05FD317V2W]] of $N$ with a representative in $K$.\n6. Let $K \\in \\mathcal{A}$. The set $K/N$ contains $eN$. For $a, b \\in K$, step 2 gives $(aN)(bN)^{-1} = (ab^{-1})N$, and $ab^{-1} \\in K$. By the [[subgroup criterion|05N4TQKPH0]], $K/N \\in \\mathcal{B}$.\n7. For $Q \\in \\mathcal{B}$, let $\\beta(Q) = \\{g \\in G : gN \\in Q\\}$.\n8. Let $Q \\in \\mathcal{B}$. Then $e \\in \\beta(Q)$, since $eN$ is the identity of $Q$. For $a, b \\in \\beta(Q)$, $(ab^{-1})N = (aN)(bN)^{-1} \\in Q$, so $ab^{-1} \\in \\beta(Q)$.\n9. By step 8 and the subgroup criterion, $\\beta(Q)$ is a subgroup of $G$. For $n \\in N$, step 3 gives $nN = N \\in Q$, so $N \\subseteq \\beta(Q)$. Thus $\\beta(Q) \\in \\mathcal{A}$.\n10. Let $K \\in \\mathcal{A}$. We claim $\\beta(K/N) = K$. Each $k \\in K$ has $kN \\in K/N$, so $K \\subseteq \\beta(K/N)$.\n11. Conversely, let $g \\in \\beta(K/N)$, so $gN = kN$ for some $k \\in K$. Then $g = ge \\in kN$, so $g = kn$ with $n \\in N \\subseteq K$. Hence $g \\in K$.\n12. Let $Q \\in \\mathcal{B}$. We claim $\\beta(Q)/N = Q$. Each element of $\\beta(Q)/N$ is $gN$ with $gN \\in Q$. Each element of $Q$ is a coset $gN$, and then $g \\in \\beta(Q)$.\n13. By steps 10 to 12, the maps $K \\mapsto K/N$ and $Q \\mapsto \\beta(Q)$ are inverse to each other. So $K \\mapsto K/N$ is a bijection from $\\mathcal{A}$ to $\\mathcal{B}$.\n14. Let $K, L \\in \\mathcal{A}$. If $K \\subseteq L$, then each $kN$ with $k \\in K$ lies in $L/N$, so $K/N \\subseteq L/N$.\n15. Let $K/N \\subseteq L/N$. Each $g$ with $gN \\in K/N$ has $gN \\in L/N$, so $\\beta(K/N) \\subseteq \\beta(L/N)$. By step 10, $K \\subseteq L$. So $K \\subseteq L$ exactly when $K/N \\subseteq L/N$.\n16. Let $K \\in \\mathcal{A}$ be normal in $G$. For $g \\in G$ and $k \\in K$, step 2 gives $(gN)(kN)(gN)^{-1} = (gkg^{-1})N$, and $gkg^{-1} \\in K$. So $K/N$ is normal in $G/N$.\n17. Conversely, let $K/N$ be normal in $G/N$, and let $g \\in G$ and $k \\in K$. Then $(gkg^{-1})N = (gN)(kN)(gN)^{-1} \\in K/N$.\n18. By step 17, $gkg^{-1} \\in \\beta(K/N)$, which is $K$ by step 10. So $K$ is normal in $G$.\n19. Steps 13, 15, 16 and 18 prove all the claims of the statement.\n\n## Idea\n\nA subgroup of $G$ that contains $N$ is a union of cosets of $N$, so it can be recovered from the set of these cosets. This is why taking the preimage undoes $K \\mapsto K/N$. Since products and conjugates in $G/N$ are computed on representatives, the inclusions and the normality pass back and forth.\n",
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