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Symmetry Groups and Group Actions

  1. Apr 7, 2009 #1
    1. The problem statement, all variables and given/known data
    I would like to find the number of distinct elements in S17 that are made up of two 4-cycles and three 3-cycles.


    2. Relevant equations



    3. The attempt at a solution
    This seems like a very simple question but since the group is so huge it's hard to figure out. I have been trying to look at the problem from the point of view of group actions but that has not gotten me anywhere yet.
     
  2. jcsd
  3. Apr 7, 2009 #2

    Dick

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    Isn't it just a combinatorics question? The symmetric groups are very, uh, symmetric. How many ways to pick two four cycles and one three cycle?
     
  4. Apr 7, 2009 #3
    Well, I tried looking at it that way, but it seems really complicated.
    But here is what I have come up with, please let me know if I am missing something since I haven't done this type of thing in a while.

    17 choose 4 = 2380
    13 choose 4 = 715
    9 choose 3 = 84
    6 choose 3 = 20
    3 choose 3 = 1

    then within in each set of 4 elements there are 6 distinct permutations and within in each set of 3 there are 2 distinct permutations.

    So to come up with the total would you just multiply
    (2380)(715)(84)(20)(1)(62)(23)?

    Thanks.
     
  5. Apr 7, 2009 #4

    Dick

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    I thought it should be just C(17,4)*C(13,4)*C(9,3). Shouldn't the rest be fixed? So you don't have any more cycles? I think you still have to account for permutations in those cycles though.
     
  6. Apr 7, 2009 #5
    Would the last two 3-cycles really be fixed? If there were six elements remaining to put into cycles there are still 20 different ways to put them into two separate 3-cycles, correct?
     
  7. Apr 8, 2009 #6

    Dick

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    Sure. I was reading the problem wrong. But you still have to do some more counting, don't you? Once you've chosen the set, say {1,2,3}, (123) and (132) are still different 3 cycles.
     
  8. Apr 9, 2009 #7
    Well, here is a different way of thinking about so I am not sure which way is correct..

    There are 17! ways to line up the 17 elements in a row. Then we group the first 4 in a 4-cycle, the second 4 in a 4-cycle, the next 3 into a 3-cycle and so on until all the elements have been grouped together.

    But there are multiple ways to get get a single element:
    Switch the 2 4-cycles: 2! ways
    Shift the elements inside a 4-cycle while retaining the same permutation: 4 ways
    Switch the 3 cycles: 3!
    Shift the elements inside a 4-cycle while retaining the same permutation: 3 ways
    So this means that for a particular element of S17 there are 2!(42)(3!)(32)

    So our final count for the number of distinct elements with this structure is
    17!/[(2!)(42)(3!)(32)]

    Does this seem right? It seems like the logic is correct but it gives me a different answer than my other method.
     
  9. Apr 9, 2009 #8

    Dick

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    That sounds right. Unless I'm missing something too.
     
  10. Apr 9, 2009 #9
    [itex]\binom {17}{4} 3! \binom {13}{4} 3! \binom {9}{3} 2! \binom {6}{3} 2! \binom {3}{3} 2![/itex] as there are [itex](k-1)![/itex] ways to cycle k elements

    then i think you have to divide by 2 to take into account the repeated 4-cycles and then divide by 3 because of the repeated 3 cycles.
     
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