# Isomorphism proof

1. Mar 30, 2009

### kathrynag

1. The problem statement, all variables and given/known data
$$\phi$$: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$$\in$$G, $$\phi$$($$a^{-1}$$)=$$^\phi(a){-1}$$.

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 $$\phi$$(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.

2. Mar 30, 2009

### 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?

3. Mar 30, 2009

### kathrynag

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

4. Mar 30, 2009

### 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.

5. Mar 30, 2009

### kathrynag

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