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Astonishing but true!
Dec. 9th, 2008 09:35 pmEvery transitive abelian subgroup H of S_n has order n!
Proof: For any i, j in 1..n, the stabilizer of i in H is conjugate to the stabilizer of j, since the group is transitive. But the group is abelian, so they're all the same! But if any element h ∈ H leaves all of 1..n fixed, then h is id. So, the stabilizer of everything is trivial, which means #H = #OrbitH(1) #StabH(1) = n.
Proof: For any i, j in 1..n, the stabilizer of i in H is conjugate to the stabilizer of j, since the group is transitive. But the group is abelian, so they're all the same! But if any element h ∈ H leaves all of 1..n fixed, then h is id. So, the stabilizer of everything is trivial, which means #H = #OrbitH(1) #StabH(1) = n.
