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  "body": "\n## Proof\n\nWe apply the first isomorphism theorem to the map $G/N \\to G/K$, $gN \\mapsto gK$.\n\n1. Let $G$ be a group with identity $e$, and let $N \\subseteq K$ be [[normal subgroups|0W77CVBCHQ]] of $G$. Both [[quotient groups|0YYWCNTPWR]] $G/N$ and $G/K$ exist.\n2. Since $N \\subseteq K$ and $N$ is normal in $G$, $N$ is normal in $K$. So $K/N = \\{kN : k \\in K\\}$ is a subset of $G/N$.\n3. We record a fact on [[left cosets|05FD317V2W]]. For $a \\in G$, $aK = K$ holds exactly when $a \\in K$.\n4. Proof of step 3. If $aK = K$, then $a = ae \\in K$. If $a \\in K$, then $aK \\subseteq K$ by closure, and each $m \\in K$ equals $a(a^{-1}m) \\in aK$.\n5. Let $g, g' \\in G$ with $gN = g'N$. Then $g' = g'e \\in gN$, so $g' = gn$ with $n \\in N \\subseteq K$.\n6. By step 4, $nK = K$. So $g'K = (gK)(nK) = (gK)K = gK$, using the operation of $G/K$ and that $K$ is its identity.\n7. By steps 5 and 6, the formula $f(gN) = gK$ defines a map $f \\colon G/N \\to G/K$.\n8. For $a, b \\in G$, $f((aN)(bN)) = f((ab)N) = (ab)K = (aK)(bK) = f(aN) f(bN)$. So $f$ is a [[group homomorphism|0HV9Y7WG5R]].\n9. Each element $gK$ of $G/K$ equals $f(gN)$. So $f$ is surjective.\n10. Let $gN \\in G/N$. By step 3, $f(gN) = K$ exactly when $g \\in K$. So the [[kernel|0Z424HK8CQ]] of $f$ is the set of cosets $gN$ with $g \\in K$.\n11. A coset $gN$ with $g \\in K$ lies in $K/N$. Conversely, each element $kN$ of $K/N$ has $k \\in K$. So $\\ker f = K/N$.\n12. By [[the kernel is a normal subgroup|0C9ZKYSPYR]] and step 11, $K/N$ is a normal subgroup of $G/N$.\n13. By the [[first isomorphism theorem|0CXXJNG5B6]], the map $(gN)(K/N) \\mapsto f(gN) = gK$ is an isomorphism from $(G/N)/(K/N)$ onto $\\operatorname{im} f$.\n14. By step 9, $\\operatorname{im} f = G/K$. Together with step 12, this proves the statement.\n\n## Idea\n\nSince $N$ lies in $K$, each coset of $N$ lies in one coset of $K$, so $gN \\mapsto gK$ is well defined. This map is onto, and it kills exactly the cosets of $N$ that lie inside $K$, namely $K/N$. The first isomorphism theorem does the rest.\n",
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