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Math Help - Proof on abelian and isomorphic groups

  1. #1
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    Proof on abelian and isomorphic groups

    If a function f, defined f(g)=g^-1 maps a group G to G, is isomorphic, then G is abelian.

    How do I go about proving this one?

    Edit: Sorry, wrong section, delete please.
    Last edited by Tripod87; April 20th 2009 at 07:03 PM.
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  2. #2
    Super Member Gamma's Avatar
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    take a,b\in G and a\not =b. Then f(a)f(b)=f(ab)=(ab)^{-1}=b^{-1}a^{-1}=f(b)f(a). But f is an isomorphism (in particular an automorphism), so it has an inverse which is also an isomorphism from G to G.

    f^{-1}(f(a)f(b))=f^{-1}(f(b)f(a))

    f^{-1}(f(a))f^{-1}(f(b))=f^{-1}(f(b))f^{-1}(f(a))

    ab=ba
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Given  g,h \in G, \; (gh)^{-1} = h^{-1} g^{-1} since  (gh)(h^{-1} g^{-1}) = (h^{-1} g^{-1})(gh) = 1 .

    But we know  f is an isomorphism and hence a homomorphism, so  f(gh) = f(g)f(h) .

    Hence  h^{-1} g^{-1} = (gh)^{-1} = f(gh) = f(g)f(h) = g^{-1} h^{-1} .

     \exists \; a,b \in G s.t.  a = g^{-1}, \; b = h^{-1} .

    Therefore  ab = ba \; \forall a,b \in G which means  G is abelian.
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