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If x is a cycle of length n, x^n is the identity.

  1. May 27, 2009 #1
    Is it true that if [itex]\sigma \in S_n[/itex] is a cycle of length [itex]k \leq n[/itex], then [itex]\sigma^k = \varepsilon[/itex], where [itex]\varepsilon[/itex] is the identity permutation, and that [itex]k[/itex] is the least nonzero integer having this property?
     
  2. jcsd
  3. May 27, 2009 #2

    matt grime

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    That is surely obvious, isn't it?
     
  4. May 27, 2009 #3
    Not to me. :frown: Maybe I'm missing something small...if you can get me started on why it's the case, I can probably finish it out.
     
  5. May 27, 2009 #4

    Hurkyl

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    Special cases are always a good way to get started. Try k=1,2,3.

    P.S. you meant "least positive integer"
     
  6. May 27, 2009 #5

    matt grime

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    A k-cycle has order k - it really is trivial. You only need to consider the case of

    (123..k)

    which just rotates the elements 1,..,k cyclically.
     
  7. May 28, 2009 #6

    matt grime

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    More geometrically, label the vertices of a k-gon with 1,..,k, then (1...k) rotates it by 2pi/k.

    If you don't like that then just think what (1...k) does to the set 1,..,k it sends i to i+1 (wrapping k round to 1). So what happens if apply it r times?
     
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