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Math Help - Group presentation question

  1. #1
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    Group presentation question

    I have the group presentations

    $\left\langle x,y|x^2=1,y^2=1,(xy)^n=1\right\rangle$

    and

    $\left\langle x,y|x^2=1,y^2=1\right\rangle$

    and am told to say what well-known groups they define. I know the second is the infinite dihedral group and so am guessing the first is $D_n$, but don't know how to show why they are these groups.

    Thanks

    P.S. in an unrelated question, given a non-abelian group of order 60, how could you show that it has no odd permutations (without using the fact that it is $A_5$)?
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Re: Group presentation question

    Quote Originally Posted by alsn View Post
    I have the group presentations

    $\left\langle x,y|x^2=1,y^2=1,(xy)^n=1\right\rangle$

    and

    $\left\langle x,y|x^2=1,y^2=1\right\rangle$

    and am told to say what well-known groups they define. I know the second is the infinite dihedral group and so am guessing the first is $D_n$, but don't know how to show why they are these groups.

    Thanks

    P.S. in an unrelated question, given a non-abelian group of order 60, how could you show that it has no odd permutations (without using the fact that it is $A_5$)?
    Are you familiar with Tietze transformations? If so, try and get this presentation to look like the presentation of the Dihedral group which you should know,

    \langle a, b; a^2, aba^{-1}=b^{-1}, b^n\rangle.

    Otherwise, re-write the presentation you know so well so it "looks like" the presentation you are given. What should your isomorphism be? Then, prove it is an isomorphism!

    For your unrelated question, ask it in a new post!
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  3. #3
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    Re: Group presentation question

    i would suggest adding the generator u = xy, deriving from yu = yxy that yu = u^-1y (if x^2 = y^2 = 1, then (xy)^-1 = yx ,since (xy)(yx) = x(y^2)x = x^2 = 1),
    and then re-writing u = xy as x = uy, so that we can eliminate x (because (uy)^2 = uyuy = uu^-1yy = 1).

    all of that works for both groups.
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  4. #4
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    Re: Group presentation question

    Thanks Deveno, I did something similar but replacing x with zy^-1, then got that (zy^-1)^2=1 implies yzy^-1=z^-1 because y=y^-1, which gives the familiar presentation for D_n. i think this works too?
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