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clickbait: 'The conjugacy classes in the alternating group are usually the same as those in the symmetric group; there is a surprisingly simple condition for when this does not hold.',
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text: 'Recall that in the [-497] $S_n$, the notion of "[-4bj]" coincides with that of "has the same [4cg cycle type]" ([4bh proof]).\nIt turns out to be the case that when we descend to the [-4hf] $A_n$, the conjugacy classes nearly all remain the same; the exception is those classes in $S_n$ which have cycle type all odd and all distinct.\nThose classes split into exactly two classes in $A_n$.\n\n# Proof\n\n## Conjugate implies same cycle type\n\nThis direction of the proof is identical to [4bh the same direction] in the proof of the corresponding result on symmetric groups.\n\n## Converse: splitting condition\n\nRecall before we start that an even-length cycle can only be written as the product of an *odd* number of transpositions, and vice versa.\n\nThe question is: is every $\\tau \\in A_n$ with the same cycle type as $\\sigma \\in A_n$ conjugate (in $A_n)$ to $\\sigma$?\nIf so, then the conjugacy class in $A_n$ of $\\sigma$ is just the same as that of $\\sigma$ in $S_n$; if not, then the conjugacy class in $S_n$ must break into two pieces in $A_n$, namely $\\{ \\rho \\sigma \\rho^{-1} : \\rho \\ \\text{even} \\}$ and $\\{ \\rho \\sigma \\rho^{-1} : \\rho \\ \\text{odd} \\}$.\n(Certainly these are conjugacy classes: they only contain even permutations so they are subsets of $A_n$, while everything in either class is conjugate in $A_n$ because the definition only depends on the [even_odd_parity parity] of $\\rho$.)\n\nLet $\\sigma = c_1 \\dots c_k$, and $\\tau = c_1' \\dots c_k'$ of the same cycle type: $c_i = (a_{i1} \\dots a_{i r_i})$, $c_i' = (b_{i1} \\dots b_{i r_i})$.\n\nDefine $\\rho$ to be the permutation which takes $a_{ij}$ to $b_{ij}$, but otherwise does not move any other letter.\nThen $\\rho \\sigma \\rho^{-1} = \\tau$, so if $\\rho$ lies in $A_n$ then we are done: $\\sigma$ and $\\tau$ are conjugate in $A_n$.\n\nThat is, we may assume without loss of generality that $\\rho$ is odd.\n\n### If any of the cycles is even in length\n\nSuppose without loss of generality that $r_1$, the length of the first cycle, is even.\nThen we can rewrite $c_1' = (b_{11} \\dots b_{1r_1})$ as $(b_{12} b_{13} \\dots b_{1 r_1} b_{11})$, which is the same cycle expressed slightly differently.\n\nNow $c_1'$ is even in length, so it is an odd permutation (being a product of an odd number of transpositions, $(b_{1 r_1} b_{11}) (b_{1 (r_1-1)} b_{11}) \\dots (b_{13} b_{11})(b_{12} b_{11})$).\nHence $\\rho c_1'$ is an even permutation.\n\nBut conjugating $\\tau$ by $\\rho c_1'$ yields $\\sigma$:\n$$\\sigma = \\rho \\tau \\rho^{-1} = \\rho c_1' (c_1'^{-1} \\tau c_1') c_1'^{-1} \\rho^{-1}$$\nwhich is the result of conjugating $c_1'^{-1} \\tau c_1' = \\tau$ by $\\rho c_1'$.\n\n(It is the case that $c_1'^{-1} \\tau c_1' = \\tau$, because $c_1'$ commutes with all of $\\tau$ except with the first cycle $c_1'$, so the expression is $c_1'^{-1} c_1' c_1' c_2' \\dots c_k'$, where we have pulled the final $c_1'$ through to the beginning of the expression $\\tau = c_1' c_2' \\dots c_k'$.)\n\n%%hidden(Example):\nSuppose $\\sigma = (12)(3456), \\tau = (23)(1467)$.\n\nWe have that $\\tau$ is taken to $\\sigma$ by conjugating with $\\rho = (67)(56)(31)(23)(12)$, which is an odd permutation so isn't in $A_n$.\nBut we can rewrite $\\tau = (32)(1467)$, and the new permutation we'll conjugate with is $\\rho c_1' = (67)(56)(31)(23)(12)(32)$, where we have appended $c_1' = (23) = (32)$.\nIt is the case that $\\rho$ is an even permutation and hence is in $A_n$, because it is the result of multiplying the odd permutation $\\rho$ by the odd permutation $c_1'$.\n\nNow the conjugation is the composition of two conjugations: first by $(32)$, to yield $(32)\\tau(32)^{-1} = (23)(1467)$ (which is $\\tau$ still!), and then by $\\rho$.\nBut we constructed $\\rho$ so as to take $\\tau$ to $\\sigma$ on conjugation, so this works just as we needed.\n%%\n\n## If all the cycles are of odd length, but some length is repeated\nWithout loss of generality, suppose $r_1 = r_2$ (and label them both $r$), so the first two cycles are of the same length: say $\\sigma$'s version is $(a_1 a_2 \\dots a_r)(c_1 c_2 \\dots c_r)$, and $\\tau$'s version is $(b_1 b_2 \\dots b_r)(d_1 d_2 \\dots d_r)$.\n\nThen define $\\rho' = \\rho (b_1 d_1)(b_2 d_2) \\dots (b_r d_r)$.\nSince $r$ is odd and $\\rho$ is an odd permutation, $\\rho'$ is an even permutation.\n\nNow conjugating by $\\rho'$ is the same as first conjugating by $(b_1 d_1)(b_2 d_2) \\dots (b_r d_r)$ and then by $\\rho$.\n\nBut conjugating by $(b_1 d_1)(b_2 d_2) \\dots (b_r d_r)$ takes $\\tau$ to $\\tau$, because it has the effect of replacing all the $b_i$ by $d_i$ and all the $d_i$ by $b_i$, so it simply swaps the two cycles round.\n\nHence the conjugation of $\\tau$ by $\\rho'$ yields $\\sigma$, and $\\rho'$ is in $A_n$. \n\n%%hidden(Example):\nSuppose $\\sigma = (123)(456), \\tau = (154)(237)$.\n\nThen conjugation of $\\tau$ by $\\rho = (67)(35)(42)(34)(25)$, an odd permutation, yields $\\sigma$.\n\nNow, if we first perform the conjugation $(12)(53)(47)$, we take $\\tau$ to itself, and then performing $\\rho$ yields $\\sigma$.\nThe combination of $\\rho$ and $(12)(53)(47)$ is an even permutation, so it does line in $A_n$.\n%%\n\n## If all the cycles are of odd length, and they are all distinct\n\nIn this case, we are required to check that the conjugacy class *does* split.\nRemember, we started out by supposing $\\sigma$ and $\\tau$ have the same cycle type, and they are conjugate in $S_n$ by an odd permutation (so they are not obviously conjugate in $A_n$); we need to show that indeed they are *not* conjugate in $A_n$.\n\nIndeed, the only ways to rewrite $\\tau$ into $\\sigma$ (that is, by conjugation) involve taking each individual cycle and conjugating it into the corresponding cycle in $\\sigma$. There is no choice about which $\\tau$-cycle we take to which $\\sigma$-cycle, because all the cycle lengths are distinct.\nBut the permutation $\\rho$ which takes the $\\tau$-cycles to the $\\sigma$-cycles is odd, so is not in $A_n$.\n\nMoreover, since each cycle is odd, we can't get past the problem by just cycling round the cycle (for instance, by taking the cycle $(123)$ to the cycle $(231)$), because that involves conjugating by the cycle itself: an even permutation, since the cycle length is odd.\nComposing with the even permutation can't take $\\rho$ from being odd to being even.\n\nTherefore $\\tau$ and $\\sigma$ genuinely are not conjugate in $A_n$, so the conjugacy class splits.\n\n%%hidden(Example):\nSuppose $\\sigma = (12345)(678), \\tau = (12345)(687)$ in $A_8$.\n\nThen conjugation of $\\tau$ by $\\rho = (87)$, an odd permutation, yields $\\sigma$.\nCan we do this with an *even* permutation instead?\n\nConjugating $\\tau$ by anything at all must keep the cycle type the same, so the thing we conjugate by must take $(12345)$ to $(12345)$ and $(687)$ to $(678)$.\n\nThe only ways of $(12345)$ to $(12345)$ are by conjugating by some power of $(12345)$ itself; that is even.\nThe only ways of taking $(687)$ to $(678)$ are by conjugating by $(87)$, or by $(87)$ and then some power of $(678)$; all of these are odd.\n\nTherefore the only possible ways of taking $\\tau$ to $\\sigma$ involve conjugating by an odd permutation $(678)^m(87)$, possibly alongside some powers of an even permutation $(12345)$; therefore to get from $\\tau$ to $\\sigma$ requires an odd permutation, so they are in fact not conjugate in $A_8$.\n%%\n\n# Example\n\nIn $A_7$, the cycle types are $(7)$, $(5, 1, 1)$, $(4,2,1)$, $(3,2,2)$, $(3,3,1)$, $(3,1,1,1,1)$, $(1,1,1,1,1,1,1)$, and $(2,2,1,1,1)$.\nThe only class which splits is the $7$-cycles, of cycle type $(7)$; it splits into a pair of half-sized classes with representatives $(1234567)$ and $(12)(1234567)(12)^{-1} = (2134567)$.',
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