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title: 'Conjugacy classes of the alternating group on five elements: Simpler proof',
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text: 'The [-4hf] $A_5$ on five elements has [4bj conjugacy classes] very similar to those of the [-497] $S_5$, but where one of the classes has split in two.\n\nNote that $A_5$ has $60$ elements, since it is precisely half of the elements of $S_5$ which has $5! = 120$ elements (where the exclamation mark is the [-factorial] function).\n\n# Table\n\n$$\\begin{array}{|c|c|c|c|}\n\\hline\n\\text{Representative}& \\text{Size of class} & \\text{Cycle type} & \\text{Order of element} \\\\ \\hline\n(12345) & 12 & 5 & 5 \\\\ \\hline\n(21345) & 12 & 5 & 5 \\\\ \\hline\n(123) & 20 & 3,1,1 & 3 \\\\ \\hline\n(12)(34) & 15 & 2,2,1 & 2 \\\\ \\hline\ne & 1 & 1,1,1,1,1 & 1 \\\\ \\hline\n\\end{array}$$\n\n# Working\n\nFirstly, the identity is in a class of its own, because $\\tau e \\tau^{-1} = \\tau \\tau^{-1} = e$ for every $\\tau$. \n\nNow, by the same reasoning as in [4bh the proof] that conjugate elements must have the same cycle type in $S_n$, that result also holds in $A_n$.\n\nHence we just need to see whether any of the cycle types comprise more than one conjugacy class.\n\nRecall that the available cycle types are $(5)$, $(3,1,1)$, $(2,2,1)$, $(1,1,1,1,1)$ (the last of which is the identity and we have already considered it).\n\n## Double-transpositions\n\nAll the double-transpositions are conjugate (so the $(2,2,1)$ cycle type does not split):\n\n- $(ab)(cd)$ is conjugate to $(ab)(ce)$ if we conjugate by $(ab)(de)$; symmetrically this covers all the cases where one of the two transpositions remains the same.\n- $(ab)(cd)$ is conjugate to $(ac)(bd)$ by $(cba)$; this covers the case that $e$ is not introduced.\n- $(ab)(cd)$ is conjugate to $(ac)(be)$ by $(bc)(de)$; this covers the remaining cases that $e$ is introduced and neither of the two transpositions remains the same.\n\n## Three-cycles\n\nAll the three-cycles are conjugate (so the $(3,1,1)$ cycle type does not split): \n\n- $(abc)$ is conjugate to $(acb)$ by $(bc)(de)$, so three-cycles are conjugate to their permutations.\n- $(abc)$ is conjugate to $(abd)$ by $(cde)$; this covers the case of introducing a single new element to the cycle.\n- $(abc)$ is conjugate to $(ade)$ by $(bd)(ce)$; this covers the case of introducing two new elements to the cycle.\n\n## Five-cycles\n\nThis class does split: I claim that $(12345)$ and $(21345)$ are not conjugate.\n(Once we have this, then the class must split into two chunks, since $\\{ \\rho (12345) \\rho^{-1}: \\rho \\ \\text{even} \\}$ is closed under conjugation in $A_5$, and $\\{ \\rho (12345) \\rho^{-1}: \\rho \\ \\text{odd} \\}$ is closed under conjugation in $A_5$.\nThe first is the conjugacy class of $(12345)$ in $A_5$; the second is the conjugacy class of $(21345) = (12)(12345)(12)^{-1}$.\nThe only question here was whether they were separate conjugacy classes or whether their union was the conjugacy class.)\n\nRecall that $\\tau (12345) \\tau^{-1} = (\\tau(1), \\tau(2), \\tau(3), \\tau(4), \\tau(5))$, so we would need a permutation $\\tau$ such that $\\tau$ sends $1$ to $2$, $2$ to $1$, $3$ to $3$, $4$ to $4$, and $5$ to $5$.\nThe only such permutation is $(12)$, the transposition, but that is not actually a member of $A_5$.\n\nHence in fact $(12345)$ and $(21345)$ are not conjugate.',
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