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  "body": "\n## Statement\n\nLet $G$ be a finite [[solvable group|0FXJ7YFMMX]], and let $\\pi$ be a set of [[prime numbers|01826KMG2H]].\nA $\\pi$-subgroup is a subgroup whose order has all its prime factors in $\\pi$.\n\n1. $G$ has a [[Hall $\\pi$-subgroup|KPWW9X14V2]].\n2. Any two Hall $\\pi$-subgroups of $G$ are [[conjugate|0H11ZPX4ND]] in $G$.\n3. Every $\\pi$-subgroup of $G$ is contained in some Hall $\\pi$-subgroup of $G$.\n\n## Remarks\n\nThe proof is by induction on $\\abs{G}$.\nA minimal normal subgroup $N$ of a solvable group is an elementary abelian $p$-group.\nOne applies induction to $G/N$ and takes the preimage $K$ of a Hall $\\pi$-subgroup of $G/N$.\nIf $p \\in \\pi$, then $K$ is a Hall $\\pi$-subgroup of $G$.\nIf $p \\notin \\pi$, then a complement of $N$ in $K$, given by the [[Schur–Zassenhaus theorem|64K1A69C80]], is one.\n\nExample: $G = S_3 \\times \\Z/5\\Z$ of order $30$ is solvable, and $S_3 \\times \\set{0}$ is a Hall $\\set{2, 3}$-subgroup, of order $6$ and index $5$.\n\nSolvability is necessary.\nThe [[simple group|0YMYMNQ0QZ]] $A_5$ of order $60$ has no subgroup of order $15$, so it has no Hall $\\set{3, 5}$-subgroup.\nIn the simple group $\\mathrm{PSL}_2(\\F_7)$ of order $168$ there are two conjugacy classes of subgroups isomorphic to $S_4$; these are Hall $\\set{2, 3}$-subgroups, of index $7$, that are not conjugate.\n\nPhilip Hall also proved a converse.\nLet $p'$ denote the set of all primes other than $p$.\nIf a finite group has a Hall $p'$-subgroup for every prime $p$, then it is solvable.\n\n## Explanations\n\n### L1 human\n\nIn any finite group, the Sylow theorems give subgroups that collect all factors of one prime in the order.\nHall's theorem says that in a solvable group one can do the same for any set of primes at once.\nFor example, a solvable group of order $60$ has a subgroup of order $12$, one of order $15$, and one of order $20$.\nSuch subgroups exist for every choice of primes, and they are unique up to conjugation.\n\n### L2 human\n\nThe theorem extends the [[Sylow theorems|FESTJ0K9TX]] from one prime to sets of primes, but only for solvable groups.\nTogether with the converse, it characterises solvable groups by the existence of Hall subgroups.\nHall $p'$-subgroups are also called Sylow complements, since a Hall $p'$-subgroup and a Sylow $p$-subgroup together make up the whole order of the group.\n",
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