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Powers of permutations

  1. Nov 19, 2017 #1
    1. The problem statement, all variables and given/known data
    In the following problems, let ##\alpha## be a cycle of length s, say ##\alpha = (a_1a_2 ... a_s)##

    3)Find the inverse of ##\alpha## and show that ##\alpha^{-1} = \alpha^{s-1}##
    2. Relevant equations
    I've observed in the previous problem that there are ##s## distinct powers of ##\alpha##, similar to how ##\mathbb{Z}_n## has n elements.

    3. The attempt at a solution
    The inverse of ##\alpha## is ##\alpha^{-1} = (a_sa_{s-1} ... a_2a_1)##

    I need to show ##\alpha^1 = \alpha^{s-1}##
    I've tried ##\alpha^{-1}\alpha^1 = \alpha^{-1}\alpha^{s-1}##
    => ##e = \alpha^{s-2}##

    and ##\alpha^1\alpha^{-1} = \alpha^{s-1}\alpha^{-1}##
    => ##e = \alpha^{s-2}##

    and ##\alpha^{-1}\alpha^s = \alpha^{s-1}\alpha^s##
    => ##\alpha{s-1} = \alpha^{2s-1}##, and since there are only ##s## distinct permutations, maybe somehow we can conclude ##\alpha^{2s-1} = \alpha^{2s-1-s} = \alpha^{s-1}##.. But if that's the case, I can use that same argument starting with ##\alpha^1 = \alpha^{s-1}##

    I am asking for a small hint.
     
  2. jcsd
  3. Nov 19, 2017 #2

    haruspex

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    Um... No. Reread the question.
     
  4. Nov 19, 2017 #3

    StoneTemplePython

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    hint: raising to what power leaves things intact? i.e. what ##\alpha## to the what power is the identity operation?
    - - - -
    note: it's perhaps a matter of taste (and what course you are in), but I would use permutation matrices to show this. Then again, I really like matrices and matrix groups.
     
  5. Nov 19, 2017 #4
    OK i made a mistake reading the question, sorry about that, but think I can use post 3's hint..

    We observed that raising a permutation, ##\alpha## of length s, to the ##s^{th}## power, we have ##\alpha^s = e##.
    Therefore ##\alpha^{-1} = \alpha^{-1}e = \alpha^{-1}\alpha^s = \alpha^{s-1}## []
     
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