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Math Help - Proof involving permutations

  1. #1
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    Proof involving permutations

    Hi, I have this question and I'm not sure if i'm going about it in the right way.

    Let \sigma, \tau are permutations.

    Prove that sgn( \tau^{-1} \sigma \tau) = sgn( \sigma)

    I know that permutations can be written as a product of transpositions. So is this right?

    \sigma = m_1 m_2 ... m_k

    \tau = n_1 n_2 ... n_l

    \tau^{-1} = p_1 p_2 ... p_l

    where m_i, n_i, p_i are transpositions. (I'm not sure if the inverse of \tau does have the same number of transpositions as \tau?)

    Then sgn( \sigma) = (-1)^k
    sgn( \tau) = (-1)^l
    sgn( \tau^{-1}) = (-1)^l

    So sgn( \tau^{-1} \sigma \tau) = (-1)^{k + 2l} = (-1)^k = sgn( \sigma)

    Is this right?
    Any help would be appreciated,
    Thanks,
    Katy
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Well, are you allowed to use that \mbox{sgn} (\tau \sigma) = \mbox{sgn}(\tau)\ \mbox{sgn}(\sigma)? This makes it quite trivial.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by harkapobi View Post
    Hi, I have this question and I'm not sure if i'm going about it in the right way.

    Let \sigma, \tau are permutations.

    Prove that sgn( \tau^{-1} \sigma \tau) = sgn( \sigma)

    I know that permutations can be written as a product of transpositions. So is this right?

    \sigma = m_1 m_2 ... m_k

    \tau = n_1 n_2 ... n_l

    \tau^{-1} = p_1 p_2 ... p_l

    where m_i, n_i, p_i are transpositions. (I'm not sure if the inverse of \tau does have the same number of transpositions as \tau?)

    Then sgn( \sigma) = (-1)^k
    sgn( \tau) = (-1)^l
    sgn( \tau^{-1}) = (-1)^l

    So sgn( \tau^{-1} \sigma \tau) = (-1)^{k + 2l} = (-1)^k = sgn( \sigma)

    Is this right?
    Any help would be appreciated,
    Thanks,
    Katy
    One very circuitous argument leads one to remember that \text{sgn}:S_n\mapsto\mathbb{Z}_2 is an epimorphism. And so \text{sgn}\left(\tau^{-1}\sigma\tau\right)= \text{sgn}\left(\tau^{-1}\right)\text{sgn}\left(\sigma\right)\text{sgn}\l  eft(\tau\right) =\left(\text{sgn}\left(\tau\right)\right)^{-1}\text{sgn}\left(\tau\right)\text{sgn}\left(\sigm  a\right)=\text{sgn}\left(\sigma\right) using the fact that \mathbb{Z}_2 is abelian and the properties of homomorphisms.
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  4. #4
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    Quote Originally Posted by Drexel28 View Post
    One very circuitous argument leads one to remember that \text{sgn}:S_n\mapsto\mathbb{Z}_2 is an epimorphism. And so \text{sgn}\left(\tau^{-1}\sigma\tau\right)= \text{sgn}\left(\tau^{-1}\right)\text{sgn}\left(\sigma\right)\text{sgn}\l  eft(\tau\right) =\left(\text{sgn}\left(\tau\right)\right)^{-1}\text{sgn}\left(\tau\right)\text{sgn}\left(\sigm  a\right)=\text{sgn}\left(\sigma\right) using the fact that \mathbb{Z}_2 is abelian and the properties of homomorphisms.
    Thank you for your help

    So it this enough? The breaking down into transpositions is unecessary?
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