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Math Help - need help in this permutation proof

  1. #1
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    need help in this permutation proof

    Can anyone help me in this proof ?

    If A and B are in S_n, then ABA^{-1} is the permutation that has the same cycle structure as B and that is obtained by applying A to the symbols in B.
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  2. #2
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    Hello,

    Prove it when the length of A is 2 and the B is cyclic:
    A=(i j), B=(m_1 m_2... m_k).
    Consider several cases depending on whether i, j are included in B.
    For example, if i=m_n and j doesn't occur in B,
    ABA^{-1}=(m_1 m_2... m_{n-1} j m_{n+1} ... m_k).

    Bye.
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  3. #3
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    Quote Originally Posted by deniselim17 View Post
    Can anyone help me in this proof ?

    If A and B are in S_n, then ABA^{-1} is the permutation that has the same cycle structure as B and that is obtained by applying A to the symbols in B.
    Let \sigma = (a_1, ... , a_k) (where k\geq 1). Let \tau be a permutation. Prove that \tau \sigma \tau^{-1} = (\tau(a_1), ... , \tau(a_k)). Now let \theta be any permuation. Then we can write \theta = \sigma_1\sigma_2 ... \sigma_m where \sigma_i are disjoint cycles. Therefore, \tau \theta \tau^{-1} = (\tau \sigma_1 \tau^{-1})(\tau \sigma_2 \tau^{-1}) ... (\tau \sigma_m \tau^{-1}). But each \tau \sigma_i \tau has the same structure as \sigma_i by above. Therefore, \theta and \tau \theta \tau^{-1} both have the same number of disjoint cycles in their cyclic decompositition. Therefore \theta and \tau \theta \tau^{-1} have the same structrure.
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