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Permutation and cycles

  1. Dec 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Let P be a permutation of a set. Show that P(i1i2...ir)B-1 = (P(i1)P(i2)...P(ir))


    2. Relevant equations

    N/A

    3. The attempt at a solution

    Since P is a permutation, it can be written as the product of cycles. So I figured that showing that the above equation holds for cycles will be sufficient to show that it holds for all permutations.

    Let C = (im1im2...imk) be a cycle and let D = (i1i2...ir). Then, for mk [tex]\neq[/tex] r,

    imk[tex]\stackrel{C^{-1}}{\rightarrow}[/tex]imk-1[tex]\stackrel{D}{\rightarrow}[/tex]imk-1+1[tex]\stackrel{C}{\rightarrow}[/tex]imk+1

    Let D` = (C(i1)C(i2)...C(ir)), then imk[tex]\stackrel{}{D`\rightarrow}[/tex]imk+1
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 29, 2009 #2
    I accidentally made two threads. Just ignore this one.
     
  4. Dec 29, 2009 #3

    Dick

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    Science Advisor
    Homework Helper

    Well, for example (P(i1)P(i2)...P(ir)) maps P(i1) to P(i2). You want to show P(i1,i2,...ir)P^(-1) does the same thing. Try it. Evaluate P(i1,i2,...ir)P^(-1)P(i1).
     
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