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Math Help - Proof that homomorphism of an inverse equals the inverse of the homomorphism

  1. #1
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    Proof that homomorphism of an inverse equals the inverse of the homomorphism

    g is an element of group G. f(g) is a homomorphism from G to H. Prove that f(g^{-1})=f(g)^{-1}.

    I'm told the proof is by multiplying the two terms, like this (I guess)
    f(g^{-1}) f(g)^{-1} = ??
    but still cannot see what to do.

    I also have that e_G, e_H, are the identities in G and H, respectively.
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  2. #2
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    Re: Proof that homomorphism of an inverse equals the inverse of the homomorphism

    Oops, sorry. It is not "inverse of a homomorphism" but "homomorphism of an inverse".

    That is, if ab= e1, the identity in the first group (ring, field?), then you want to show that f(a)f(b)= e2. Well, the whole point of a homomorphism is that f(ab)= f(a)f(b). What does that tell you if ab= e1?
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  3. #3
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    Re: Proof that homomorphism of an inverse equals the inverse of the homomorphism

    Quote Originally Posted by HallsofIvy View Post
    That is, if ab= e1, the identity in the first group (ring, field?), then you want to show that f(a)f(b)= e2. Well, the whole point of a homomorphism is that f(ab)= f(a)f(b). What does that tell you if ab= e1?
    g in G implies g^{-1} in G.
    f(g)*f(g^{-1})=f(g*g^{-1})=f(e1)=e2

    Then just left multiply the first and last element by f(g)^{-1}.
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    Re: Proof that homomorphism of an inverse equals the inverse of the homomorphism

    How can I post things by using this beautiful notation?
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  5. #5
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    Re: Proof that homomorphism of an inverse equals the inverse of the homomorphism

    Quote Originally Posted by mohammedlabeeb View Post
    How can I post things by using this beautiful notation?
    See this post.
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