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  "body": "\n## Proof\n\nWe show that two cosets commute exactly when the commutator of their representatives lies in $N$.\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 $(aN)(bN) = (ab)N$.\n2. We record a fact on [[left cosets|05FD317V2W]]: for $a, b \\in G$, $aN = bN$ holds exactly when $a^{-1}b \\in N$.\n3. Proof of step 2. If $aN = bN$, then $b = be \\in aN$, so $b = an$ with $n \\in N$, and $a^{-1}b = n \\in N$.\n4. Conversely, let $n = a^{-1}b \\in N$. Closure gives $nN \\subseteq N$, and each $m \\in N$ is $n(n^{-1}m) \\in nN$. So $nN = N$, and $bN = (an)N = (aN)(nN) = aN$.\n5. For $g, h \\in G$, $(hg)(g^{-1}h^{-1}) = e$. By the [[uniqueness of inverses|0D9NDGE9KP]], $(hg)^{-1} = g^{-1}h^{-1}$.\n6. Let $g, h \\in G$. By step 1, $(gN)(hN) = (hN)(gN)$ holds exactly when $(hg)N = (gh)N$.\n7. By steps 2 and 5, this holds exactly when $g^{-1}h^{-1}gh \\in N$, that is, when the [[commutator|06RGT4V6Y6]] $[g, h]$ lies in $N$.\n8. By steps 6 and 7, $G/N$ is [[abelian|071DDFVG4J]] exactly when $[g, h] \\in N$ for all $g, h \\in G$.\n9. The [[commutator subgroup|08SJ5K2CBG]] $[G, G]$ is the [[subgroup generated by|0RG9SPQSWM]] the set $C$ of all commutators. So it is the intersection of all subgroups of $G$ that contain $C$.\n10. If $C \\subseteq N$, then $N$ is one of the subgroups in this intersection. So $[G, G] \\subseteq N$.\n11. If $[G, G] \\subseteq N$, then $C \\subseteq [G, G] \\subseteq N$.\n12. By steps 10 and 11, $C \\subseteq N$ exactly when $[G, G] \\subseteq N$. With step 8, $G/N$ is abelian if and only if $[G, G] \\subseteq N$.\n\n## Idea\n\nThe cosets $gN$ and $hN$ commute exactly when $gh$ and $hg$ differ by an element of $N$, and the difference is the commutator $[g, h]$. So the quotient is abelian when $N$ contains all commutators. Since $N$ is a subgroup, this is the same as containing the subgroup they generate.\n",
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