Isomorphism proof

  • Thread starter kathrynag
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  • #1
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Homework Statement


[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].



Homework Equations





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.
 

Answers and Replies

  • #2
283
3
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?
 
  • #3
598
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[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
 
  • #4
283
3
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.
 
  • #5
598
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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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