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Math Help - non abelian group problem

  1. #1
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    non abelian group problem

    Suppose that G = { e, x, x2, y, yx, yx2 } is a non-Abelian group with |x| = 3 and |y|=2. Show xy=yx2. I tried looking at the orders of the elements but i got nowhere doing this, any suggestions?
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  2. #2
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    Quote Originally Posted by wutang View Post
    Suppose that G = { e, x, x2, y, yx, yx2 } is a non-Abelian group with |x| = 3 and |y|=2. Show xy=yx2. I tried looking at the orders of the elements but i got nowhere doing this, any suggestions?

    Hint: show that xy\neq e,\,x,\,x^2,\,y,\,yx

    Tonio
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  3. #3
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    how does this look:
    if xy=e, then x=y^-1, so |x|= 2 contradiction.
    if xy=e, then y=x^-1, so |y|=3 contradiction.
    if xy=x, then y=e contradiction, if x=x(y^-1), then (y^-1)=e contradicition.
    suppose xy=x^2, y=x^3=e contradicition,
    xy=y, then x=e (by right hand cancellation) contradicition.
    I am not sure, what to do if xy=yx, is that a contradicion due to the fact that G is a non-abelian group?
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  4. #4
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    Quote Originally Posted by wutang View Post
    I am not sure, what to do if xy=yx, is that a contradicion due to the fact that G is a non-abelian group?
    Yes. x,y are generators. If they commute, then everything in the subgroup generated by {x,y} (here it's just G itself) commutes with each other. #
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  5. #5
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    would I then have to show that xy=y(x^2), or is it enough to show that I have eliminated all other possibilities ( so that xy=y(x^2) by closure?).
    Thanks for your help again
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  6. #6
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    Quote Originally Posted by wutang View Post
    would I then have to show that xy=y(x^2), or is it enough to show that I have eliminated all other possibilities ( so that xy=y(x^2) by closure?).
    Thanks for your help again

    It is enough since you're given a group and thus xy must be something in that group...

    Tonio
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