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Math Help - permutation help ,

  1. #1
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    permutation help ,

    Hello,

    I am stuck on part three of this question.

    i)  \sigma = (1 6) (1 4) (1 2) (3 7) (3 5) (8 9) = even permutation

    ii)  \sigma^{2} = 1 2 3 4 5 6 7 8 9
    4 6 7 1 3 2 5 8 9

    III) how do I find T??
    Attached Thumbnails Attached Thumbnails permutation help ,-screen-shot-2014-01-27-21.59.57.png  
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  2. #2
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    Re: permutation help ,

    You do it digit by digit.

    You want: \tau \sigma(1) = 2. You have \tau \sigma(1) = \tau(\sigma(1)) = \tau(2), so \tau(2) = 2.
    You want: \tau \sigma(2) = 1. You have \tau \sigma(2) = \tau(\sigma(2)) = \tau(4), so \tau(4) = 1.
    Etc.
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  3. #3
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    Re: permutation help ,

    oh okay, so the rest just stays the smame?
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  4. #4
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    Re: permutation help ,

    Turning two-line notation into three-line notation:

    \tau \sigma = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 4 & 5 & 6 & 7 & 1 & 3 & 9 & 8 \\ 2 & 1 & 3 & 4 & 5 & 6 & 7 & 8 & 9\end{pmatrix}

    Essentially, the first two lines show where \sigma sends numbers while the second two lines show where \tau sends numbers. So, the first and last line show where the composition \tau \sigma sends numbers. To write \tau, just write the last two lines, but order the columns so that the top row is in order:

    \tau = \begin{pmatrix}2 & 4 & 5 & 6 & 7 & 1 & 3 & 9 & 8 \\ 2 & 1 & 3 & 4 & 5 & 6 & 7 & 8 & 9\end{pmatrix} = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 6 & 2 & 7 & 1 & 3 & 4 & 5 & 9 & 8\end{pmatrix}
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  5. #5
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    Re: permutation help ,

    Quote Originally Posted by Tweety View Post
    III) how do I find T??
    Here is how I would have done it.

    First find \sigma^{-1} then \tau  = \left( {\begin{array}{*{20}{c}}1&2&3&4&5&6&7&8&9\\2&1&3&4  &5&6&7&8&9\end{array}} \right) \circ {\sigma ^{ - 1}}

    That gives the same as above.
    Last edited by Plato; January 27th 2014 at 02:56 PM.
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  6. #6
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    Re: permutation help ,

    Quote Originally Posted by Plato View Post
    Here is how I would have done it.

    First find \sigma^{-1} then \tau  = \left( {\begin{array}{*{20}{c}}1&2&3&4&5&6&7&8&9\\2&1&3&4  &5&6&7&8&9\end{array}} \right) \circ {\sigma ^{ - 1}}

    That gives the same as above.
    In case it is not obvious to the OP, \sigma^{-1} is extremely easy to calculate. You are given that \sigma = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 4 & 5 & 6 & 7 & 1 & 3 & 9 & 8\end{pmatrix}, so its inverse is just flipping the two rows:

    \sigma^{-1} = \begin{pmatrix}2 & 4 & 5 & 6 & 7 & 1 & 3 & 9 & 8 \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \end{pmatrix} = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 6 & 1 & 7 & 2 & 3 & 4 & 5 & 9 & 8\end{pmatrix}

    So, now, the product (composition) of the two permutations: \tau = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 1 & 3 & 4 & 5 & 6 & 7 & 8 & 9\end{pmatrix}\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 6 & 1 & 7 & 2 & 3 & 4 & 5 & 9 & 8\end{pmatrix} is easy to calculate (and gives the same answer as above). Generally, as you continue in mathematics, Plato's solution is going to be easier to apply.
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  7. #7
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    Re: permutation help ,

    Quote Originally Posted by SlipEternal View Post
    In case it is not obvious to the OP, \sigma^{-1} is extremely easy to calculate. You are given that \sigma = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 4 & 5 & 6 & 7 & 1 & 3 & 9 & 8\end{pmatrix}, so its inverse is just flipping the two rows:

    \sigma^{-1} = \begin{pmatrix}2 & 4 & 5 & 6 & 7 & 1 & 3 & 9 & 8 \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \end{pmatrix} = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 6 & 1 & 7 & 2 & 3 & 4 & 5 & 9 & 8\end{pmatrix}

    So, now, the product (composition) of the two permutations: \tau = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 1 & 3 & 4 & 5 & 6 & 7 & 8 & 9\end{pmatrix}\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 6 & 1 & 7 & 2 & 3 & 4 & 5 & 9 & 8\end{pmatrix} is easy to calculate (and gives the same answer as above). Generally, as you continue in mathematics, Plato's solution is going to be easier to apply.

    Thank you,

    the other solution is easier to understand but now I dont understand why  \sigma \sigma^{-1} = \tau ?
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  8. #8
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    Re: permutation help ,

    \tau\sigma = (1 2), so multiplying both sides by \sigma^{-1} on the right, you get \tau \sigma \sigma^{-1} = (1 2) \sigma^{-1}. On the LHS, \sigma \sigma^{-1} is the identity permutation, so the LHS simplifies to \tau. So, it is not that \sigma \sigma^{-1} = \tau, but that you are multiplying \tau by the identity permutation, so you are not changing it. If you send 1 to 2, then send 2 back to 1, it is the same as if you never sent it anywhere at all. That is how multiplying a permutation by its inverse gives the identity permutation.
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  9. #9
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    Re: permutation help ,

    thanks I understand now!
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