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Math Help - Conjugacy Classes

  1. #1
    Junior Member universalsandbox's Avatar
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    Conjugacy Classes

    The symmetry group S_{6} has a total number of elements 6!
    How do you find the number of conjugacy classes and their sizes?

    I have done the following:

    6------------------(123456) ---- SIZE = 6!/6 = 120
    5+1---------------(12345)(6) -- SIZE = 6!/5 = 144
    4+2---------------(1234)(56)
    4+1+1------------(1234)(5)(6)
    3+3---------------(123)(456)
    3+2+1------------(123)(45)(6)
    3+1+1+1---------(123)(4)(5)(6)
    2+2+2------------(12)(34)(56)
    2+2+1+1---------(12)(34)(5)(6)
    2+1+1+1+1------(12)(3)(4)(5)(6)
    1+1+1+1+1+1---(1)(2)(3)(4)(5)(6) = e ---- SIZE = 6!/6! = 1

    It appears there is 11 conjugacy classes in total. The sizes must sum to 720.

    Is this correct what I'm doing? How do you compute the size of the remaining classes? it's 6!/(something) I'm not quite sure.

    Thanks.
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  2. #2
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    4+2---------------(1234)(56)
    Over here once the 4-cycle is determined the 2-cycle immediately follows. The number of 4-element subsets of {1,2,3,4,5,6} is given by {6\choose 2}=15. For each 4-cycle there are 3!=6 ways to write it. Thus, we should get (15)(6) = 90.

    Try the other ones.
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  3. #3
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    Quote Originally Posted by universalsandbox View Post
    The symmetry group S_{6} has a total number of elements 6!
    How do you find the number of conjugacy classes and their sizes?

    I have done the following:

    6------------------(123456) ---- SIZE = 6!/6 = 120
    5+1---------------(12345)(6) -- SIZE = 6!/5 = 144
    4+2---------------(1234)(56) -- SIZE = \color{red}{6\choose2}3! = 90
    4+1+1------------(1234)(5)(6) -- SIZE = \color{red}{6\choose2}3! = 90
    3+3---------------(123)(456) -- SIZE = \color{red}{6\choose3}\times2 = 40
    3+2+1------------(123)(45)(6) -- SIZE = \color{red}{6\choose3}{3\choose1}\times2 = 120
    3+1+1+1---------(123)(4)(5)(6) -- SIZE = \color{red}{6\choose3}\times2 = 40
    2+2+2------------(12)(34)(56) -- SIZE = \color{red}{6\choose2}{4\choose2}\div3! = 15
    2+2+1+1---------(12)(34)(5)(6) -- SIZE = \color{red}{6\choose2}{4\choose2}\div2 = 45
    2+1+1+1+1------(12)(3)(4)(5)(6) -- SIZE = \color{red}{6\choose2} = 15
    1+1+1+1+1+1---(1)(2)(3)(4)(5)(6) = e ---- SIZE = 6!/6! = 1

    It appears there is 11 conjugacy classes in total. The sizes must sum to 720.

    Is this correct what I'm doing? How do you compute the size of the remaining classes? it's 6!/(something) I'm not quite sure.
    Notice that the number of possible k-cycles formed from a given collection of numbers is (k1)!, because you can select one of the numbers to come at the start of the cycle, and then each permutation of the remaining numbers will give a different cycle.

    "3+3" is the trickiest one. There are 6\choose3 ways of selecting a triple of numbers from 1 to 6. That splits the six numbers into two triples, each of which can form 2 cycles ((abc) or (acb) if the numbers are a,b,c), but then we have to divide by 2 because we can interchange the triples. So the correct number is {6\choose3}\times2\times2\div2 = 40.
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