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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


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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

  5. Apr 7, 2009 #4


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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


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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

    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


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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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