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Permutation groups

  1. Nov 14, 2004 #1
    what is the number of elements of order 5 in the permutaion group S7??
    so what we're concerned with here is, after decompositon into disjoint cycles the l.c.m of the lengths must be 5. since 5 is a prime, the only possible way we could get 5 as l.c.m would be to fix ANY 2 elements amongst the 7 to themselves....so we end up getting 2 cycles of length 1 each. the remaining five elements can be arranged in 4! ways...
    so, the answer is 7C2 * 4! = 21*24 = 504.
    :smile: but unfortunately, this answer is WRONG!!
    CAN ANYBODY TELL ME WHY?! PLEEEEASE HELP!
     
  2. jcsd
  3. Nov 14, 2004 #2

    Galileo

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    Since 5 is prime, you MUST have cycle of length 5.
    Then there could be a 2-cycle in addition.
     
  4. Nov 15, 2004 #3
    hello

    well....yes, it's possible to have a cycle of length 2 in addition to the 5 cycle...but then the l.c.m becomes 10. so that rules out such a consideration!
     
  5. Nov 15, 2004 #4

    Galileo

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    Yes, so the only permutations in S7 of order 5 are the 5-cycles.
     
  6. Nov 15, 2004 #5

    matt grime

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    Your answer is wrong because you've counted, for example:

    12345, 23451, 34512, 45123, and 51234 as different elements.
     
  7. Nov 15, 2004 #6
    ya....so what further? that's a valid point you've raised...
    so do you divide by 4?
     
  8. Nov 15, 2004 #7

    Galileo

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    You have [itex]{7 \choose 5}[/itex] ways to pick 5 elements from a set of 7.
    There are 5! ways you can order 5 elements in a cycle.
    For a given cycle of length 5, 5 orderings are give the same permutation.
     
  9. Nov 16, 2004 #8
    hey galileo

    please read stuff carefully...
    we left this a long while ago, right astronut?! :smile:
     
  10. Nov 21, 2004 #9
    i think the solution 504 is correct....
    i don't see any fallacy in it.
     
  11. Nov 21, 2004 #10

    matt grime

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    The general solution is:

    suppose t is a permutation of type

    [tex]1^{m_1}2^{m_2}....r^{m_r}[/tex]

    then the order of the centralizer is

    [tex]\prod_i i^{m_i}m_i![/tex]

    in this case it is 1^2.5

    so the centralizer's order is

    2!.5

    hence the conjugacy class has order

    7!/10

    which is indeed 504
     
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