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Thread: 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; Apr 20th 2009 at 06:03 PM.
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  2. #2
    Super Member Gamma's Avatar
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    take $\displaystyle a,b\in G$ and $\displaystyle a\not =b$. Then $\displaystyle f(a)f(b)=f(ab)=(ab)^{-1}=b^{-1}a^{-1}=f(b)f(a)$. But f is an isomorphism (in particular an automorphism), so it has an inverse which is also an isomorphism from G to G.

    $\displaystyle f^{-1}(f(a)f(b))=f^{-1}(f(b)f(a))$

    $\displaystyle f^{-1}(f(a))f^{-1}(f(b))=f^{-1}(f(b))f^{-1}(f(a))$

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

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

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

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

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