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Math Help - Algebra, Problems For Fun (7)

  1. #1
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    Algebra, Problems For Fun (7)

    Definition: A group G is called hopfian if every surjective homomorphism G \longrightarrow G is injective. Clearly every finite group is hopfian.


    Problem: Prove that G=<x,y: \ \ y^{-1}x^2y=x^3> is not hopfian.


    Suggestion:
    Spoiler:
    You may define the homomorphism f: G \longrightarrow G by f(x)=x^2, \ f(y)=y.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    Definition: A group G is called hopfian if every surjective homomorphism G \longrightarrow G is injective. Clearly every finite group is hopfian.


    Problem: Prove that G=<x,y: \ \ y^{-1}x^2y=x^3> is not hopfian.


    Suggestion:
    Spoiler:
    You may define the homomorphism f: G \longrightarrow G by f(x)=x^2, \ f(y)=y.

    In the homomorphism you give what is mapped to x, or to odd powers of x? If you hadn't given it a name I would be sceptical such groups existed: in a homomorphic image you add a relation into the presentation, xy=yx or whatever. However, that makes xy and yx equivalent and so you don't map to one of them.

    Just a thought...
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  3. #3
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    Quote Originally Posted by Swlabr View Post
    In the homomorphism you give what is mapped to x, or to odd powers of x? If you hadn't given it a name I would be sceptical such groups existed: in a homomorphic image you add a relation into the presentation, xy=yx or whatever. However, that makes xy and yx equivalent and so you don't map to one of them.

    Just a thought...
    f is well-defined because f(y^{-1}x^2y)=y^{-1}x^4y=(y^{-1}x^2y)^2=x^6=f(x^3). we also have f(y^{-1}xyx^{-1})=x, which proves that f is surjective.

    so the only thing left is to prove that f is not injective!
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Interesting. I stand corrected. By rights I should probably stare at this group for the next hour or two, but I actually have to study group rings (this site is wonderful - I can procrastinate by studying maths...). Anyway - to complete the solution:

    (x[y,x^{-1}](x^{-1})^y)\phi = x\phi [y,x^{-1}]\phi (x^{-1})^y\phi = x^2 x (x^{-2})^y = x^2 x x^{-3} = 1 and as x[y,x^{-1}](x^{-1})^y \neq 1 is not trivial we have a non-trivial kernel, and so no injection.
    Last edited by Swlabr; May 18th 2009 at 05:24 AM.
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  5. #5
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    Quote Originally Posted by Swlabr View Post

    x^2 x (x^{-1})^y = x^2 x x^{-3} = 1
    (x^{-2})^y not (x^{-1})^y. apart from that your answer is correct.
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Edited.
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