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  "body": "\n## Statement\n\nLet $G$ be a finite [[group|G5S1M9]] and let $N$ be a [[normal subgroup|0CB0DX5BVV]] of $G$ such that $\\abs{N}$ and the [[index|0DTXBQPA2S]] $[G : N]$ are [[coprime|0VZ48N083F]].\n\n1. $N$ has a complement in $G$: a subgroup $H$ with $NH = G$ and $N \\cap H = \\set{e}$. Hence $G$ is a [[semidirect product|0H3WZXCZNF]] $N \\rtimes H$.\n2. Any two complements of $N$ in $G$ are [[conjugate|0H11ZPX4ND]] in $G$.\n\n## Remarks\n\nIn other words, every [[extension|MF4N3VGGFV]] of a group of order $m$ by a group of order $n$ with $\\gcd(m, n) = 1$ splits.\nWhen $N$ is abelian, the proof is cohomological.\nThe obstruction to a complement lies in $H^2(G/N, N)$, and the obstruction to conjugacy lies in $H^1(G/N, N)$.\nBoth cohomology groups are zero, because they are killed by both $\\abs{N}$ and $[G : N]$.\nThe general case of part 1 reduces to the abelian case by induction, using a Sylow subgroup of $N$ and the Frattini argument.\nFor part 2 in general one needs $N$ or $G/N$ to be [[solvable|0FXJ7YFMMX]]; one of them has odd order, so this follows from the Feit–Thompson theorem.\n\nExample: $G = S_3$ and $N = A_3$, with $\\abs{N} = 3$ and $[G : N] = 2$.\nEach subgroup of order $2$, such as $\\set{e, (1\\,2)}$, is a complement, and the three of them are conjugate.\n\nThe coprimality is necessary.\nIn $G = \\Z/4\\Z$ with $N = \\set{0, 2}$, both $\\abs{N}$ and $[G : N]$ are $2$, and $N$ has no complement, because the only subgroup of order $2$ is $N$ itself.\n\n## Explanations\n\n### L1 human\n\nSuppose a finite group has a normal subgroup $N$, and the size of $N$ shares no prime factor with the number of its cosets.\nThen the group splits cleanly: there is a second subgroup that meets $N$ only in the identity and fills up the rest.\nSo the group is built from $N$ and the quotient in the simplest possible way, as a semidirect product.\n\n### L2 human\n\nA normal subgroup $N$ with $\\gcd(\\abs{N}, [G : N]) = 1$ is a normal [[Hall subgroup|KPWW9X14V2]], so the theorem says that normal Hall subgroups have complements, unique up to conjugacy.\nIt is a key step in the proof of Hall's theorem on solvable groups and in many arguments with coprime actions.\nWithout coprimality, extensions may fail to split, and then one needs cohomology to describe them.\n",
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