1. ## question about isomorphic groups

One more question i got this proof that seems trivial but im not sure if there was another way to do it. The question is let G and G' be isomorphic groups and G is abelian, prove G' is abelian.

Can you just say that since G and G' are isomorphic they have the same structure so if G is abelian then G' is abelian?
I know that there is a one to one, onto, homomorphism between them so if G is abelian then phi(ab) should = phi(ba) but how would that apply to G'?

2. The statement that $\displaystyle G' \cong G$ means that there exists an isomorphism $\displaystyle \phi : G' \rightarrow G$. For $\displaystyle a,b \in G'$ we have $\displaystyle \phi(ab)=\phi(a)\phi(b)=\phi(b)\phi(a)=\phi(ba)$. Since $\displaystyle \phi$ is a bijection this implies we must have $\displaystyle ab=ba$.

3. Originally Posted by ChrisBickle
One more question i got this proof that seems trivial but im not sure if there was another way to do it. The question is let G and G' be isomorphic groups and G is abelian, prove G' is abelian.

Can you just say that since G and G' are isomorphic they have the same structure so if G is abelian then G' is abelian?
I know that there is a one to one, onto, homomorphism between them so if G is abelian then phi(ab) should = phi(ba) but how would that apply to G'?
Technically, we only need to have $\displaystyle \phi:G\mapsto G'$ is an epimorphism. Isomorphism is overkill.

4. what is an epimorphism? we only covered homomorphisms and isomorphisms.

5. Originally Posted by ChrisBickle
what is an epimorphism? we only covered homomorphisms and isomorphisms.
It's just a complicated word for a surjective homomorphism. Realize that while Bruno's approach one one-hundred percent valid, it could have been done differently so as to not need bijectivity, but only surjectivity.

Let $\displaystyle \phi:G\mapsto G'$ and suppose that $\displaystyle G$ was abelian. Let $\displaystyle g,g'\in G'$. Since $\displaystyle \phi\left(G\right)=G'$ there exists $\displaystyle a,b\in G$ such that $\displaystyle \phi(a)=g,\phi(b)=g'$. Therefore, $\displaystyle gg'=\phi(a)\phi(b)=\phi(ab)=\phi(ba)=\phi(b)\phi(a )=g'g$. The difference is that in Bruno's example we assumed his $\displaystyle a,b$ were the only elements that mapped to $\displaystyle g,g'$ whereas here we only needed that there needed to only be at least one $\displaystyle a$ and one $\displaystyle b$.