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  "body": "\n## Proof\n\nWe apply the first isomorphism theorem to the map $S \\to SN/N$, $s \\mapsto sN$.\n\n1. Let $G$ be a group with identity $e$, let $S$ be a subgroup of $G$, and let $N$ be a [[normal subgroup|0W77CVBCHQ]] of $G$.\n2. By the [[product of a subgroup and a normal subgroup is a subgroup|0TXWV1JGG1]], $SN$ is a subgroup of $G$.\n3. Each $n \\in N$ equals $en$ with $e \\in S$, so $N \\subseteq SN$. Thus $N$ is a [[subgroup|0N0PXCTYA7]] of $SN$.\n4. For $x \\in SN$ and $n \\in N$, $xnx^{-1} \\in N$ because $N$ is normal in $G$. So $N$ is normal in $SN$, and the [[quotient group|0YYWCNTPWR]] $SN/N$ exists.\n5. By [[the intersection of subgroups is a subgroup|0ZTGMDNVTB]], $S \\cap N$ is a subgroup of $G$. It lies in $S$, so it is a subgroup of $S$.\n6. Let $s \\in S$ and $x \\in S \\cap N$. Then $sxs^{-1} \\in S$, since $S$ is a subgroup, and $sxs^{-1} \\in N$, since $N$ is normal. So $S \\cap N$ is normal in $S$.\n7. We record a fact on [[left cosets|05FD317V2W]]: for $a \\in SN$, $aN = N$ holds exactly when $a \\in N$.\n8. Proof of step 7. If $aN = N$, then $a = ae \\in aN = N$. If $a \\in N$, then $aN \\subseteq N$ by closure, and each $m \\in N$ equals $a(a^{-1}m) \\in aN$.\n9. Define $f \\colon S \\to SN/N$ by $f(s) = sN$. This makes sense because $s = se \\in SN$.\n10. For $s, t \\in S$, $f(st) = (st)N = (sN)(tN) = f(s) f(t)$ by the operation of $SN/N$. So $f$ is a [[group homomorphism|0HV9Y7WG5R]].\n11. Let $xN \\in SN/N$, with $x = sn$, $s \\in S$ and $n \\in N$. By step 8, $nN = N$, the identity of $SN/N$. So $xN = (sN)(nN) = sN = f(s)$, and $f$ is surjective.\n12. For $s \\in S$, $f(s)$ is the identity $N$ exactly when $sN = N$, that is, when $s \\in N$ by step 7. So the [[kernel|0Z424HK8CQ]] of $f$ is $S \\cap N$.\n13. By the [[first isomorphism theorem|0CXXJNG5B6]] and steps 11 and 12, the map $s(S \\cap N) \\mapsto f(s) = sN$ is an isomorphism from $S/(S \\cap N)$ onto $\\operatorname{im} f = SN/N$.\n14. Steps 2, 4, 6 and 13 give all the claims of the statement.\n\n## Idea\n\nEvery coset of $N$ inside $SN$ already has a representative in $S$, because the factor from $N$ can be absorbed into the coset. So $s \\mapsto sN$ maps $S$ onto $SN/N$. It forgets exactly the elements of $S$ that lie in $N$, and the first isomorphism theorem turns this into the isomorphism.\n",
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