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Math Help - cycle decomposition of permutations

  1. #1
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    Question cycle decomposition of permutations

    How can I show that if alpha is any permutation then:
    alpha (i1,i2,....ir) alpha -1 = (alpha(i1) alpha(i2)....alpha(ir))
    Thank for any help.
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  2. #2
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    Quote Originally Posted by thomas_donald View Post
    How can I show that if alpha is any permutation then:
    alpha (i1,i2,....ir) alpha -1 = (alpha(i1) alpha(i2)....alpha(ir))
    Thank for any help.
    Let \sigma   = (\alpha(i_1),\alpha(i_2),...,\alpha(i_r)) and \tau = \alpha (i_1,i_2,...,i_r) \alpha^{-1}.
    To show that \sigma = \tau you need to show that \sigma(n) = \tau(n) for all n.

    If n = \alpha (i_j) for j=1,2,...,r then \sigma(n) = \alpha (i_{j(\bmod r)+1}) and \tau (n) is the same thing.

    If n\not \in \{ \alpha (i_j)|j=1,2,...,r\} then \sigma(n) = n while \tau(n)=n, the same value.

    This is Mine 12,6th Post!!!
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  3. #3
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    Thumbs up

    Thanks You! Thank You! Thank You!
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  4. #4
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    Question

    Maybe I am asking a basic question but what is the difference between: and ?
    This is a HM problem and I am not clear what the question is
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  5. #5
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    Quote Originally Posted by thomas_donald View Post
    Maybe I am asking a basic question but what is the difference between: and ?
    This is a HM problem and I am not clear what the question is
    What do you mean the difference? We proved they are the same.
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  6. #6
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    Question

    I understand why but I am not clear why is the same thing.Could you show it in more details? Or, is it possible to get an example?
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  7. #7
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    Quote Originally Posted by thomas_donald View Post
    I understand why but I am not clear why is the same thing.Could you show it in more details? Or, is it possible to get an example?
    If n=\alpha(i_j) then \tau (n) = \alpha (i_1,...,i_r) \alpha^{-1} (\alpha(i_j)) = \alpha(i_1,...,i_r)(i_j) = \alpha (i_{(j\bmod r)+1}).
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  8. #8
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    Smile

    Thanks ! Now I get it slowly and surely.
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