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Isomorphism proof

  1. Mar 30, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\phi[/tex]:G-->G'
    Let [tex]\phi[/tex] be an isomorphism. Prove that [tex]\phi[/tex] maps the e identity of G to e', the identity of G' and for every a[tex]\in[/tex]G, [tex]\phi[/tex]([tex]a^{-1}[/tex])=[tex]^\phi(a){-1}[/tex].



    2. Relevant equations



    3. The attempt at a solution
    We have an isomorphism, therefore one to one, onto and has a homomorphism.
    Phi is one to one therefore [tex]\phi[/tex](x)=[tex]\phi[/tex](y), implying x=y.
    Then [tex]\phi[/tex](G)=[tex]\phi[/tex](G') implying e=e'.
    Now [tex]\phi[/tex](a*[tex]a^{-1}[/tex])=[tex]\phi[/tex](a)*[tex]\phi[/tex]([tex]a^{-1}[/tex]) is what we want to prove.

    Now I get stuck.
     
  2. jcsd
  3. Mar 30, 2009 #2
    For the first part try
    [tex]\phi(e)=\phi(ee)=...[/tex]

    [tex]\phi(e)=\phi(a a^{-1})=\phi(a)\phi(a^{-1})=e[/tex]
    Now what do you know about inverses?
     
  4. Mar 30, 2009 #3
    [tex]\phi[/tex](e)=[tex]\phi[/tex](ee) implies e=ee
    [tex]\phi[/tex](e)=[tex]\phi[/tex](a[tex]a^{-1}[/tex])=[tex]\phi[/tex](a)[tex]\phi[/tex]([tex]^a{-1}[/tex])=e
    I think I understand where all this comes from and I know a[tex]a^{-1}[/tex]=e
     
  5. Mar 30, 2009 #4
    Firstly no it doesn't, try using the homomorphic property and then cancel.

    On the second one you found one inverse for [itex]\phi(a)[/itex], use the uniqueness of the inverse. Sorry my post is a bit confusing, use the second equation for the second part.
     
  6. Mar 30, 2009 #5
    Sorry, I'm blanking on what a homomorphic property is. Is that f(xy)=f(x)f(y)?
    so [tex]\phi[/tex](e')=[tex]\phi[/tex](a[tex]a^{-1}[/tex])=[tex]\phi[/tex](a)*[tex]\phi[/tex]([tex]a^{-1}[/tex])=e
     
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