1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

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

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Permutation groups
  1. Permutation Group (Replies: 2)

  2. Permutation Group (Replies: 29)

Loading...