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Math Help - Prove G is abelian

  1. #1
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    Prove G is abelian

    When I get time I take the time to study a little group theory. I am a flyspeck compared to PH and NCA, but here is one I would like some input to see if I have the idea. I will then post my workings.

    "Let G be a group which has the following properties:

    1: G has no element of order 2

    2: (xy)^{2}=(yx)^{2}, \forall{(x,y)} \in{G}

    By the way, what are the typings for "for all" and "element of"?.

    I tried the ones on my list but they wouldn't work.
    Last edited by galactus; July 3rd 2008 at 06:54 PM.
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    Quote Originally Posted by galactus View Post
    By the way, what are the typings for "for all" and "element of"?.

    I tried the ones on my list but they wouldn't work.
    For all: \forall = [tex]\forall[/tex]

    Element of: \in = [tex]\in[/tex]

    Is that what you were looking for?

    -Dan
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    Quote Originally Posted by galactus View Post
    When I get time I take the time to study a little group theory. I am a flyspeck compared to PH and NCA, but here is one I would like some input to see if I have the idea. I will then post my workings.

    "Let G be a group which has the following properties:

    1: G has no element of order 2

    2: (xy)^{2}-(yx)^{2}, \text{for all x,y element of G}
    I think you mean (xy)^2 = (yx)^2.
    See this.
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  4. #4
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    Yes, of course that is what I mean. A stupid typo...duh.

    I tried the 'forall and it didn't work. I also tried 'in' and got the same. Let me try this again.

    {\forall}, \;\ {\in}

    Yep now it's OK. I must've have done something wrong.
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  5. #5
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    I tried to give you a thanks PH, but your button disappears when I try.

    What next?.

    Anyway, you had an old one I asked about and had forgotten from almost 2 years ago. Cool.

    If I may, here is what I done. I think it is similar to what is on the other posting, but I have to admit I didn't look at it in depth.

    It is getting late and I do not feel like typing that much at the moment. I will type mine tomorrow. Okey doke.
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  6. #6
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    Here is my workings. I hope it's OK. I will post while I have time this morning. Today is the 4th. Things to do

    For x, \;\ y \;\ {\in} \;\ G,

    x^{2}y=((xy^{-1})y)^{2}y

    =(y(xy^{-1}))^{2}y, by part 2.

    =(yxy^{-1})(yxy^{-1})y

    That's

    x^{2}y=yx^{2}

    Next, we have:

    x^{-1}y^{-1}x=x(x^{-1})^{2}y^{-1}x

    =xy^{-1}(x^{-1})^{2}x, by part 1.

    that's

    x^{-1}y^{-1}x=xy^{-1}x^{-1}

    Similarly, we get:

    y^{-1}x^{-1}y=yx^{-1}y^{-1}

    Then we get:

    (xyx^{-1}y^{-1})^{2}=xy(x^{-1}y^{-1}x)yx^{-1}y^{-1}

    =xy(xy^{-1}x^{-1})yx^{-1}y^{-1}

    =xyx(y^{-1}x^{-1}y)x^{-1}y^{-1}

    =xyx(yx^{-1}y^{-1})x^{-1}y^{-1}

    =(xy)^{2}(x^{-1}y^{-1})^{2}

    =(xy)^{2}(yx)^{-2}

    =(yx)^{2}(yx)^{-2}

    =1

    {\therefore}, since G has no elements of order 2, we get

    xyx^{-1}y^{1}=1

    And xy=yx, proving that G is abelian.

    WHEW!!. There could easily be a typo in all those x's and y's.
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    That looks good except:
    Quote Originally Posted by galactus View Post
    Next, we have:

    x^{-1}y^{-1}x=x(x^{-1})^{2}y^{-1}x

    =xy^{-1}(x^{-1})^{2}x, by part 1.
    It has nothing to do with part 1, it rather has to do with the line you previously proved.
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