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text: 'The [-497] $S_5$ on five generators has size $5! = 120$ (where the exclamation mark denotes the [-factorial] function).\nBy the result that [4bh in a symmetric group, conjugacy classes coincide with cycle types], we can list the conjugacy classes of $S_5$ easily.\n\n# The table\n\n$$\\begin{array}{|c|c|c|c|}\n\\hline\n\\text{Representative}& \\text{Size of class} & \\text{Cycle type} & \\text{Order of element} \\\\ \\hline\n(12345) & 24 & 5 & 5 \\\\ \\hline\n(1234) & 30 & 4,1 & 4 \\\\ \\hline\n(123) & 20 & 3,1,1 & 3 \\\\ \\hline\n(123)(45) & 20 & 3,2 & 6 \\\\ \\hline\n(12)(34) & 15 & 2,2,1 & 2 \\\\ \\hline\n(12) & 10 & 2,1,1,1 & 2 \\\\ \\hline\ne & 1 & 1,1,1,1,1 & 1 \\\\ \\hline\n\\end{array}$$\n\n# Determining the list of cycle types and sizes\n\nThere are five elements to permute; we need to find all the possible ways of grouping them up into disjoint cycles.\nWe will go through this systematically.\nNote first that since there are only five elements to permute, there cannot be any element with a $6$-cycle or higher.\n\n## Those with largest cycle of length $5$\n\nIf there is a $5$-cycle, then it permutes everything, so its [4cg cycle type] is $5$.\nThat is, we can take a representative $(12345)$, and this is the only conjugacy class with a $5$-cycle.\n\nRecall that every cycle of length $5$ may be written in five different ways: $(12345)$ or $(23451)$ or $(34512)$, and so on.\nWithout loss of generality, we may assume $1$ comes at the start (by cycling round the permutation if necessary).\n\nThen there are $4!$ ways to fill in the remaining slots of the cycle (where the exclamation mark denotes the [-factorial] function).\n\nHence there are $24$ elements of this conjugacy class.\n\n## Those with largest cycle of length $4$\n\nIf there is a $4$-cycle, then it permutes everything except one element, so its cycle type must be $4,1$ (abbreviated as $4$).\nThat is, we can take a representative $(1234)$, and this is the only conjugacy class with a $4$-cycle.\n\nEither the element $1$ is permuted by the $4$-cycle, or it is not.\n\n- In the first case, we have $4$ possible ways to pick the image $a$ of $1$; then $3$ possible ways to pick the image $b$ of $a$; then $2$ possible ways to pick the image $c$ of $b$; then $c$ must be sent back to $1$.\nThat is, there are $4 \\times 3 \\times 2 = 24$ possible $4$-cycles containing $1$.\n- In the second case, the cycle does not contain $1$, so there are $3$ possible ways to pick the image $a$ of $2$; then $2$ possible ways to pick the image $b$ of $a$; then $1$ possible way to pick the image $c$ of $b$; then $c$ must be sent back to $2$.\nSo there are $3 \\times 2 \\times 1 = 6$ possible $4$-cycles not containing $1$.\n\nThis comes to a total of $30$ possible $4$-cycles in this conjugacy class.\n\n## Those with largest cycle of length $3$\n\nNow we have two possible conjugacy classes: $3,1,1$ and $3,2$.\n\n### The $3,1,1$ class\n\nAn example representative for this class is $(123)$.\n\nWe proceed with a slightly different approach to the $4,1$ case.\nUsing the notation for the [-binomial_coefficient], we have $\\binom{5}{3} = 10$ possible ways to select the numbers which appear in the $3$-cycle.\nEach selection has two distinct ways it could appear as a $3$-cycle: the selection $\\{1,2,3\\}$ can appear as $(123)$ (or the duplicate cycles $(231)$ and $(312)$), or as $(132)$ (or the duplicate cycles $(321)$ or $(213)$).\n\nThat is, we have $2 \\times 10 = 20$ elements of this conjugacy class.\n\n### The $3,2$ class\n\nAn example representative for this class is $(123)(45)$.\n\nAgain, there are $\\binom{5}{3} = 10$ possible ways to select the numbers which appear in the $3$-cycle; having made this selection, we have no further choice about the $2$-cycle.\n\nGiven a selection of the elements of the $3$-cycle, as before we have two possible ways to turn it into a $3$-cycle.\n\nWe are also given the selection of the elements of the $2$-cycle, but there is no choice about how to turn this into a $2$-cycle because, for instance, $(12)$ is equal to $(21)$ as cycles.\nSo this time the selection of the elements of the $3$-cycle has automatically determined the corresponding $2$-cycle.\n\nHence there are again $2 \\times 10 = 20$ elements of this conjugacy class.\n\n## Those with largest cycle of length $2$\n\nThere are two possible cycle types: $2,2,1$ and $2,1,1,1$.\n\n### The $2,2,1$ class\n\nAn example representative for this class is $(12)(34)$.\n\nThere are $\\binom{5}{2}$ ways to select the first two elements; then once we have done so, there are $\\binom{3}{2}$ ways to select the second two.\nHaving selected the elements moved by the first $2$-cycle, there is only one distinct way to make them into a $2$-cycle, since (for example) $(12)$ is equal to $(21)$ as permutations; similarly the selection of the elements determines the second $2$-cycle.\n\nHowever, this time we have double-counted each element, since (for example) the permutation $(12)(34)$ is equal to $(34)(12)$ by the result that [49g disjoint cycles commute].\n\nTherefore there are $\\binom{5}{2} \\times \\binom{3}{2} / 2 = 15$ elements of this conjugacy class.\n\n### The $2,1,1,1$ class\n\nAn example representative for this class is $(12)$.\n\nThere are $\\binom{5}{2}$ ways to select the two elements which this cycle permutes.\nOnce we have selected the elements, there is only one distinct way to put them into a $2$-cycle, since (for example) $(12)$ is equal to $(21)$ as permutations.\n\nTherefore there are $\\binom{5}{2} = 10$ elements of this conjugacy class.\n\n## Those with largest cycle of length $1$\n\nThere is only the identity in this class, so it is of size $1$.',
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