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Math Help - [SOLVED] Expressing as a product of transpositions

  1. #1
    Super Member craig's Avatar
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    [SOLVED] Expressing as a product of transpositions

    Let \beta = (1 \; 7\; 8)(3 \;10)(6 \;11\; 12) in the symmetric group S_{12}

    Express \beta as a product of transpositions in the form (1 \; b), with b \in \{ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 \}

    I can express it as normal transpositions, but not sure how to go about it with the first number only being 1.

    Thanks
    Last edited by craig; May 18th 2010 at 10:05 AM.
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    Quote Originally Posted by craig View Post
    Let \beta = (1 \; 7\; 8)(3 \;10)(6 \;11\; 12) in the symmetric group S_{12}

    Express \beta as a product of transpositions in the form (1 \; b), with b \in \{ 2, 3, 4, 5, 6, 7, 9, 10, 11, 12 \}

    I can express it as normal transpositions, but not sure how to go about it with the first number only being 1.

    Thanks

    Lemma (boring): we have (i_1\,i_2\,\ldots,i_n)=(1\,i_1)(1\,i_n)(1\,i_{n-1})\ldots(1\,i_2)(1\,i_1)

    Tonio
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    Super Member craig's Avatar
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    Quote Originally Posted by tonio View Post
    Lemma (boring): we have (i_1\,i_2\,\ldots,i_n)=(1\,i_1)(1\,i_n)(1\,i_{n-1})\ldots(1\,i_2)(1\,i_1)

    Tonio
    Cheers for the reply.

    How would I apply that to the problem where instead of (i_1\,i_2\,\ldots,i_n), you have = (1 \; 7\; 8)(3 \;10)(6 \;11\; 12)?
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    Quote Originally Posted by craig View Post
    Cheers for the reply.

    How would I apply that to the problem where instead of (i_1\,i_2\,\ldots,i_n), you have = (1 \; 7\; 8)(3 \;10)(6 \;11\; 12)?

    Apply the lemma for each of the cycles...!! Of course, you need 8 (which you didn't include among your possible b's in your OP. This must be a mistake, of course)

    Tonio
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    Super Member craig's Avatar
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    Quote Originally Posted by tonio View Post
    Of course, you need 8 (which you didn't include among your possible b's in your OP. This must be a mistake, of course)

    Tonio
    Yeh must of missed that one out.

    In the answers they have:

    So (1 \; 7\; 8)(3 \;10)(6 \;11\; 12) =

    (1\;8)(1\;7)(1\;3)(1\;10)(1\;3)...

    How is it that they've got 2 (1\;3)s?

    Sorry if this is something stupid, thanks again for the replies.
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    Quote Originally Posted by craig View Post
    Yeh must of missed that one out.

    In the answers they have:

    So (1 \; 7\; 8)(3 \;10)(6 \;11\; 12) =

    (1\;8)(1\;7)(1\;3)(1\;10)(1\;3)...

    How is it that they've got 2 (1\;3)s?

    Sorry if this is something stupid, thanks again for the replies.

    I wouldn't say it is stupid but it definitely seems to be careless...please do look carefully at the lemma in my first post!

    Tonio
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  7. #7
    Super Member craig's Avatar
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    Quote Originally Posted by tonio View Post
    I wouldn't say it is stupid but it definitely seems to be careless...please do look carefully at the lemma in my first post!

    Tonio
    Ahh sorry, just had a look again this morning, not sure how I didn't get it last night.

    Using the fact that (a \; b) = (1 \; a)(1 \; b)(1 \; a), we have:

    (1 \; 7\; 8)(3 \;10)(6 \;11\; 12) =
    (1 \; 8)(1 \; 7)(1 \; 3)(1 \; 10)(1 \; 3)(6 \; 11)(11 \; 12) =
    (1 \; 8)(1 \; 7)(1 \; 3)(1 \; 10)(1 \; 3)(1 \; 6)(1 \; 11)(1 \; 6)(1 \; 11)(1 \; 12)(1 \; 11)

    Thanks again for the help, (and patience )
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