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Find the order of a k cycle in group Sn

  1. May 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that a k-cycle in the group Sn has order k.
    2. Relevant equations



    3. The attempt at a solution
    I'm mostly confused on how to write this in math notation. I know it will have order k because a1 → a2 → a3 .... ak-1 → ak → a1 if we do the compositions K times. and so everything will get mapped back to itself and hence we have the identity permutation.

    Let f = (a1,a2,a3...,ak)
    then f(a1) = a2 this is the original permutation
    then if we do the compoisition of f o f we have
    a1 → a3 , a2→ a4. .. etc. so f^2(a1) = a3
    then if we do f o f o f we have a1 → a4 so
    f^3(a1) = f(a3) = a4
    ..
    f^k(a1) = f(ak) = a1
    and this process would hold for any element in f, so we see that f^k =(1) , and so f has order of K. therefore any k cycle group in Sn has order K.

    Any help is appreciated.Thanks
     
  2. jcsd
  3. May 25, 2012 #2
    Looks fine to me. I guess you're looking for a more concise way of stating this fact though? You could show that for any [itex] i \in \{1,\ldots, k \} [/itex] that
    [tex] f^n(a_i) = a_{i + n \mod k}[/tex]
    which is just what you said in more words in your post. This holds for any i, so [itex] f^k(a_i) = a_{i+k \mod k} = a_{i} [/itex], and we conclude that the order of f divides k. All you have to claim is that k is in fact the smallest number and hence is the order. But I think that this is evident.
     
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