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Math Help - Permutations and Abelian Group proof.

  1. #1
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    Permutations and Abelian Group proof.

    Where N=10,



    I'm really struggling even starting both questions.

    Thanks to anyone who can help.
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  2. #2
    MHF Contributor Amer's Avatar
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    (N 2) = \left(\begin{array}{cccc}1&2&...&N\\1&N&...&2\end{  array}\right)

    \alpha=(N 2)(12 (N-2) N) = \left(\begin{array}{ccccccccc}1&2&3&...&6&...&N-2&N-1&N\\1&N&3&...&6&...&N-2&N-1&2\\2&1&3&...&6&...&N&N-1&N-2\end{array}\right)

    same to find beta

    (23(N-1)) = \left(\begin{array}{ccccccccc}1&2&3&...&6&...&N-2&N-1&N\\1&3&N-1&...&6&...&N-2&2&N\end{array}\right)

    \beta = (23(N-1))(6(N-1)N) = \left(\begin{array}{ccccccccc}1&2&3&...&6&...&N-2&N-1&N\\1&3&N-1&...&6&...&N-2&2&N\\1&3&N&...&N-1&...&N-2&2&N\end{array}\right)

    want to find n such that

    \beta =  \alpha \cdot n

    let's write alpha and beta to see how we can find n

     \alpha \cdot n=\beta

     \left(\begin{array}{ccccccccc}1&2&3&...&6&...&N-2&N-1&N\\2&1&3&...&6&...&N&N-1&N-2\end{array}\right)n= \left(\begin{array}{ccccccccc}1&2&3&...&6&...&N-2&N-1&N\\1&3&N&...&N-1&...&N-2&2&N\end{array}\right)
    Last edited by Amer; February 17th 2010 at 12:12 PM.
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  3. #3
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    RIGHT, so you can solve it algebraically then sub in N=10 at the end?

    EDIT: To clarify, I was starting off with:

    Alpha = (10,2) (1,2,8,10)
    Beta = (2,3,9) (6,9,10)

    but it was the whole cycle notation/product of disjoint cycles thing that really threw me off. I find it very hard to understand.
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by MickQ View Post
    RIGHT, so you can solve it algebraically then sub in N=10 at the end?
    at least N should equal 26 since you have 23 and N-2, at least N-2=24 , so at least N=26 .
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  5. #5
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    For question 2:

    \forall a, b \in G:

    abba = ab^2a = a(1)a = aa = a^2 = 1

    abab = (ab)^2 = 1

    Let ab = c.

    Then cba = cab

    c^{-1}cba = c^{-1}cab

    ba = ab
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  6. #6
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    Quote Originally Posted by Amer View Post
    at least N should equal 26 since you have 23 and N-2, at least N-2=24 , so at least N=26 .
    The question states that N=10, I wrote it in the initial post. In hindsight, I should've made it a bigger font...really sorry.
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  7. #7
    MHF Contributor Amer's Avatar
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    okay I asked you

    in alpha (12(N-2)N) , 12 is 1 then 2 or it is twelve
    and in beta is 23 twenty three or 2 then 3

    if 12 is twelve and 23 is twenty three this is the solution
    so it is not twelve and twenty three

    I will edit my previous post
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  8. #8
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    Quote Originally Posted by Amer View Post
    okay I asked you



    so it is not twelve and twenty three

    I will edit my previous post

    Right, I understand now. Sorry for not clarifying, here is what it would be:

    Alpha = (10,2) (1,2,8,10)
    Beta = (2,3,9) (6,9,10)
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    Icemanfan, even though we the question doesn't state that b^2=1, we can still assume that to be true?

    Apologies for the dumb questions, this is really new to me!
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  10. #10
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    Quote Originally Posted by MickQ View Post
    Icemanfan, even though we the question doesn't state that b^2=1, we can still assume that to be true?

    Apologies for the dumb questions, this is really new to me!
    a^2 = 1 for all a \in G. This means that if b \in G, then b^2 = 1. It also means (ab)^2 = 1, since ab \in G by the definition of a group.
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  11. #11
    MHF Contributor Amer's Avatar
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    is there anything you did not understand
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    Quote Originally Posted by Amer View Post
    is there anything you did not understand
    I was just trying to work it all out using N=10. I ended up with these for Alpha and Beta, but I don't think they are correct:

    Alpha: (1,2) (10,8)

    Beta: (6,9,3,2,10)

    Could you confirm if those are correct?
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  13. #13
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by MickQ View Post
    I was just trying to work it all out using N=10. I ended up with these for Alpha and Beta, but I don't think they are correct:

    Alpha: (1,2) (8,10)

    Beta: (3,2,10,6,9)

    Could you confirm if those are correct?
    alpha is correct, in beta u just made a mitake in the order of two numbers can you fix it.
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  14. #14
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    Quote Originally Posted by Amer View Post
    alpha is correct, in beta u just made a mitake in the order of two numbers can you fix it.
    I double-checked and I can't seem to work out where I've gone wrong. Any hints?
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  15. #15
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    Actually, is it (6,9,3,2,10)?
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