Proof of Isomorphism: Proving \phi(a^{-1})=^\phi(a){-1}

[kathrynag]

Homework Statement

\phi:G-->G'
Let \phi be an isomorphism. Prove that \phi maps the e identity of G to e', the identity of G' and for every a\inG, \phi(a^{-1})=^\phi(a){-1}.

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 \phi(x)=\phi(y), implying x=y.
Then \phi(G)=\phi(G') implying e=e'.
Now \phi(a*a^{-1})=\phi(a)*\phi(a^{-1}) is what we want to prove.

Now I get stuck.

 

[Focus]

For the first part try
\phi(e)=\phi(ee)=...

\phi(e)=\phi(a a^{-1})=\phi(a)\phi(a^{-1})=e
Now what do you know about inverses?

 

[kathrynag]

\phi(e)=\phi(ee) implies e=ee
\phi(e)=\phi(aa^{-1})=\phi(a)\phi(^a{-1})=e
I think I understand where all this comes from and I know aa^{-1}=e

 

[Focus]

Firstly no it doesn't, try using the homomorphic property and then cancel.

On the second one you found one inverse for \phi(a), use the uniqueness of the inverse. Sorry my post is a bit confusing, use the second equation for the second part.

 

[kathrynag]

Sorry, I'm blanking on what a homomorphic property is. Is that f(xy)=f(x)f(y)?
so \phi(e')=\phi(aa^{-1})=\phi(a)*\phi(a^{-1})=e