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Math Help - Alternating groups

  1. #1
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    Alternating groups

    Hey guys.

    I'm sitting with an exercise I can't quite finish.

    Let \sigma \in S_6 be defined as follows: \sigma = (\frac{ 1 2 3 4 5 6}{4 5 2 1 6 3}) (of course it's not actually a fraction, I just dont know how to write a permutation.)

    I have already answered the first three exercises:

    (a) - Write \sigma as a product of disjoint cycles : \sigma = (14)(2563).
    (b) - Determine sgn(\sigma) and the order of \sigma : sgn(\sigma) = (-1) ^{n(\sigma)} = (-1)^8 = 1 and \sigma^4 = e so the order of \sigma is 4.
    (c) - How many simple transpositions are needed to write \sigma as a product of simple transpositions : From (b) I had that n(\sigma) = 8 and that is the number of simple transitions needed.

    The exercise I can't quite figure out is this:

    (d) - Find \tau \in A_6, the alternating group, so \tau \sigma \tau^{-1} = \sigma^{-1}.

    Could someone give me a tip on how to solve this? Thanks on beforehand.

    /Morten
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  2. #2
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    Re: Alternating groups

    Quote Originally Posted by m112358 View Post
    Hey guys.

    I'm sitting with an exercise I can't quite finish.

    Let \sigma \in S_6 be defined as follows: \sigma = (\frac{ 1 2 3 4 5 6}{4 5 2 1 6 3}) (of course it's not actually a fraction, I just dont know how to write a permutation.)
    Here is some help posting:
    In LaTeX
    [TEX] \binom{123456}{452163} [/TEX] gives  \binom{123456}{452163} .
    Thanks from topsquark and m112358
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  3. #3
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    Re: Alternating groups

    Heh thanks.. Now I just need help with the exercise

    /Morten
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  4. #4
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    Re: Alternating groups

    Hi,
    You did not define the sign of a permutation. Presumably, you are using the definition that \text{sgn}(\sigma) is (-1)^k if \sigma can be written as a product of k transpositions. For my money, this is an awkward definition of the sign; however, it is a workable definition. (The problem comes with showing sgn is well defined.) An important fact about sgn is that \text{sgn}(\sigma\tau)=\text{sgn}(\sigma)\text{sgn  }(\tau)

    Now any cycle of length n can be written as a product of n-1 transpositions; e.g. (2 5 6 3) = (2 3)(2 6)(2 5) (composition is right to left), so sgn(2 5 6 3) = (-1)3 = -1. So the sign of your \sigma is 1; i.e. \sigma\in A_6. I don't know what 8 has to do with this.

    You correctly found the order of \sigma.

    Now if \alpha is any cycle and \tau\in S_n, the conjugate \tau\alpha\tau^{-1} is the cycle where for each letter i in the cycle, i is replaced by \tau(i). Example: \alpha=(2\,5\,6\, 3),\text{ and } \tau=(1\, 2\, 3) so \tau^{-1}=(3\,2\,1) Then \tau\alpha\tau^{-1}=(1\,2\,3)(2\,5\,6\,3)(3\,2\,1)=(3\,5\,6\,1). I leave it to you to verify this fact.

    You should be able to use the above paragraph to find a \tau with \tau\sigma\tau^{-1}=\sigma^{-1}. Just write down \sigma^{-1} and write a \tau in two rowed form; figure out what the second row of \tau should be.
    Thanks from topsquark and m112358
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  5. #5
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    Re: Alternating groups

    First of all thanks. You made me realize how to solve the exercise!

    Second, about the sign and my solution. We have a nifty algorithm for finding n(\sigma). Something about inversion count. This can then be used to find the sign of \sigma by calculating (-1)^{n(\sigma)}.

    Anyway, thanks a million for the help!

    /Morten
    Last edited by m112358; December 4th 2013 at 09:14 AM.
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  6. #6
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    Re: Alternating groups

    Hi again,
    Ah, you are using the "correct" definition of sign. Namely, if \sigma is a permutation, \text{sgn}(\sigma)=\epsilon where \epsilon=\pm1 is defined by

    \prod_{i<j}(j-i)=(-1)^\epsilon\prod_{i<j}\sigma(j)-\sigma(i)

    That is, \epsilon is the number of pairs (i,j) with i < j such that \sigma(i)>\sigma(j) or the number of inversions of \sigma. With this definition, it's "easy" to prove that if the permutation is a product of k transpositions, then the sgn(k)=(-1)k.
    Thanks from topsquark
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