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Math Help - group homomorphism

  1. #1
    Member Greengoblin's Avatar
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    group homomorphism

    If a:G\to H is a group homomorphism between G and H, then we have the following definition:

    for all g,k\in G we have a(g*k)=a(g)*a(k)

    According to wikipedia, this means that a(1_G)=1_H, where 1_G,1_H, are the identity elements of G and H respectively, and also that a(g^{-1})=a(g)^{-1} for all g\in G.

    Can someone show me how these conditions are true?

    Thanks
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  2. #2
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    Hi

    One way is to write a(1_G)=a(1_G1_G)=a(1_G)a(1_G) thus a(1_G)=1_H.

    Now that we know that, a(g^{-1})a(g)=a(g^{-1}g)=a(1_G)=1_H implies that a(g^{-1})=(a(g))^{-1}.
    Last edited by clic-clac; February 21st 2009 at 01:45 PM. Reason: cor
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  3. #3
    Member Greengoblin's Avatar
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    Thanks, but I don't understand how your first argument shows that a(1_G)=1_H
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  4. #4
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    Ok, that's because a group law is strong enough: let A be a group, e its identity element and a\in A, then a^2=a\Rightarrow a=e
    Indeed, a^2=a\Rightarrow a=a^{-1}a^2=a^{-1}a=e


    That may be a particular case of: \forall a,b\in A,\ \exists !x\ \text{s.t.}\ ax=b

    Indeed if b=a, since e is a solution, by unicity, aa=a\Rightarrow a=e
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  5. #5
    Member Greengoblin's Avatar
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    Thanks, I got it now.
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