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Thread: 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 $\displaystyle \sigma$,$\displaystyle \tau$ are permutations.

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

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

    $\displaystyle \sigma = m_1 m_2 ... m_k$

    $\displaystyle \tau = n_1 n_2 ... n_l$

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

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

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

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

    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 $\displaystyle \mbox{sgn} (\tau \sigma) = \mbox{sgn}(\tau)\ \mbox{sgn}(\sigma)$? 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 $\displaystyle \sigma$,$\displaystyle \tau$ are permutations.

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

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

    $\displaystyle \sigma = m_1 m_2 ... m_k$

    $\displaystyle \tau = n_1 n_2 ... n_l$

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

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

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

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

    Is this right?
    Any help would be appreciated,
    Thanks,
    Katy
    One very circuitous argument leads one to remember that $\displaystyle \text{sgn}:S_n\mapsto\mathbb{Z}_2$ is an epimorphism. And so $\displaystyle \text{sgn}\left(\tau^{-1}\sigma\tau\right)=$$\displaystyle \text{sgn}\left(\tau^{-1}\right)\text{sgn}\left(\sigma\right)\text{sgn}\l eft(\tau\right)$$\displaystyle =\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 $\displaystyle \mathbb{Z}_2$ 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 $\displaystyle \text{sgn}:S_n\mapsto\mathbb{Z}_2$ is an epimorphism. And so $\displaystyle \text{sgn}\left(\tau^{-1}\sigma\tau\right)=$$\displaystyle \text{sgn}\left(\tau^{-1}\right)\text{sgn}\left(\sigma\right)\text{sgn}\l eft(\tau\right)$$\displaystyle =\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 $\displaystyle \mathbb{Z}_2$ 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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