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Math Help - Groups - presentations

  1. #1
    Super Member Bernhard's Avatar
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    Groups - presentations

    Dummit and Foote exercise 17 Section 1.2 Dihedral Groups reads as follows:

    Let X_{2n} be the group whose presentation is

    X_{2n} = <x,y | x^n = y^2 = 1, xy = y x^2>

    Show that if n = 3k, then X_{2n} has order 6, and it has the same generators and relations as D_6 when x is replaced by r and y is replaced by s.

    Can anyone help with this problem?

    Can anyone with some insight into this problem guess what D&F meant by n = 3k - would it mean n is a multiple of 3 - that is k is a positive integer? I presume they mean this.
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  2. #2
    MHF Contributor

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    Re: Groups - presentations

    yes, that is what is meant. the trick is to prove that if n = 3k, then k = 1.

    note that x = xy^2 = (xy)y = yx^2y . so

    yx = y(yx^2y) = x^2y ,

    and yx^2 = (yx)x = (x^2y)x = x^2(yx) = (x^2)(x^2y) = x^4y .

    but yx^2 = xy, so x^4y = xy \implies x^3 = 1.

    it should be downhill from here.
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