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Permutations, cycles

  1. 1. The problem statement, all variables and given/known data
    Let t be an element of S be the cycle (1,2....k) of length k with k<=n.
    a) prove that if a is an element of S then ata^-1=(a(1),a(2),...,a(k)). Thus ata^-1 is a cycle of length k.
    b)let b be any cycle of length k. Prove there exists a permutation a an element of S such that ata^-1=b.




    2. Relevant equations



    3. The attempt at a solution
    We assume t is an element of S and a is an element S.
    By definition of elements of S if t is in S, we have a determined by t(1), t(2),...,t(n)
    Furthermore if a is in S, we have a(1), a(2)....a(n).
    That's as far as I get.
     
  2. jcsd
  3. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    hi kathrynag! :smile:

    just read out that equation in English …
    … so what, for example, does ata^-1 do to a(2) ? :wink:
     
  4. Does it go to 2?
     
  5. Ok if a is defined by 1--->a(1), 2-----a(2),k--->a(k), then a^-1 is defined as a(1)-->1,a(2)--->2, a(k)--->k
    Then ata^-1 for a(2) is a(t(2))=a(2)
     
  6. tiny-tim

    tiny-tim 26,054
    Science Advisor
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    a-1
    sends it to 2 …

    so what does ata-1 send it to?
     
  7. a(t(2))
    t(2)=2
    a(2)
     
  8. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    ah! no, you're misunderstanding the notation for a cycle …

    (1,2....k) means that it sends 1 to 2, 2 to 3, … and k to 1. :wink:

    so t(2) = … ?​
     
  9. 3
    so we are left with a(3)=4
     
  10. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    no!!

    you're not told anything about a, are you?

    a(3) is just a(3) ! :bigginr:

    (and I'm off to bed :zzz: … see you tomorrow!)
     
  11. I thought I could do this:
    By definition of elements of S if t is in S, we have a determined by t(1), t(2),...,t(n)
    Furthermore if a is in S, we have a(1), a(2)....a(n).
     
  12. So I have ata^-1=at(n)
    because a^-1 sends a(1)--->1,a(2)--->2,.....a(n)--->n
    By defininition of t, we have a(1), a(2), a(3)...a(k). We go to k because our definition of t says k<=n.


    For b,
    Let b be any cycle of length k.
    We have (1,2.....k). I'm not sure how to show the rest.
     
  13. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    hi kathrynag! :smile:

    (just got up :zzz: …)
    do you mean we have (a(1), a(2), a(3),...a(k))?

    anyway, before we go any further, i need to know: what exactly is S? :confused:
     
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