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text: 'In the [-497] on a finite set, the [-4bj] of an element is determined exactly by its [4cg cycle type].\n\nMore precisely, two elements of $S_n$ are [group_conjugate conjugate] in $S_n$ if and only if they have the same cycle type.\n\n# Proof\n\n## Same conjugacy class implies same cycle type\nSuppose $\\sigma$ has the cycle type $n_1, \\dots, n_k$; write $$\\sigma = (a_{11} a_{12} \\dots a_{1 n_1})(a_{21} \\dots a_{2 n_2}) \\dots (a_{k 1} a_{k 2} \\dots a_{k n_k})$$\nLet $\\tau \\in S_n$.\n\nThen $\\tau \\sigma \\tau^{-1}(\\tau(a_{ij})) = \\tau \\sigma(a_{ij}) = a_{i (j+1)}$, where $a_{i (n_i+1)}$ is taken to be $a_{i 1}$.\n\nTherefore $$\\tau \\sigma \\tau^{-1} = (\\tau(a_{11}) \\tau(a_{12}) \\dots \\tau(a_{1 n_1}))(\\tau(a_{21}) \\dots \\tau(a_{2 n_2})) \\dots (\\tau(a_{k 1}) \\tau(a_{k 2}) \\dots \\tau(a_{k n_k}))$$\nwhich has the same cycle type as $\\sigma$ did.\n\n## Same cycle type implies same conjugacy class\n\nSuppose $$\\pi = (b_{11} b_{12} \\dots b_{1 n_1})(b_{21} \\dots b_{2 n_2}) \\dots (b_{k 1} b_{k 2} \\dots b_{k n_k})$$\nso that $\\pi$ has the same cycle type as the $\\sigma$ from the previous direction of the proof.\n\nThen define $\\tau(a_{ij}) = b_{ij}$, and insist that $\\tau$ does not move any other elements.\n\nNow $\\tau \\sigma \\tau^{-1} = \\pi$ by the final displayed equation of the previous direction of the proof, so $\\sigma$ and $\\pi$ are conjugate.\n\n# Example\nThis result makes it rather easy to [4bk list the conjugacy classes] of $S_5$.',
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