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!

Permutations, cycles

  1. Oct 12, 2010 #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. Oct 12, 2010 #2

    tiny-tim

    User Avatar
    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. Oct 12, 2010 #3
    Does it go to 2?
     
  5. Oct 12, 2010 #4
    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. Oct 12, 2010 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper


    a-1
    sends it to 2 …

    so what does ata-1 send it to?
     
  7. Oct 12, 2010 #6
    a(t(2))
    t(2)=2
    a(2)
     
  8. Oct 12, 2010 #7

    tiny-tim

    User Avatar
    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. Oct 12, 2010 #8
    3
    so we are left with a(3)=4
     
  10. Oct 12, 2010 #9

    tiny-tim

    User Avatar
    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. Oct 12, 2010 #10
    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. Oct 12, 2010 #11
    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. Oct 13, 2010 #12

    tiny-tim

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

Have something to add?



Similar Discussions: Permutations, cycles
  1. Permutation and cycles (Replies: 2)

  2. Permutation and cycles (Replies: 1)

  3. Cycle permutation (Replies: 2)

Loading...