• Jan 26th 2010, 03:54 PM
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'?
• Jan 26th 2010, 04:17 PM
Bruno J.
The statement that $G' \cong G$ means that there exists an isomorphism $\phi : G' \rightarrow G$. For $a,b \in G'$ we have $\phi(ab)=\phi(a)\phi(b)=\phi(b)\phi(a)=\phi(ba)$. Since $\phi$ is a bijection this implies we must have $ab=ba$.
• Jan 26th 2010, 10:09 PM
Drexel28
Quote:

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 $\phi:G\mapsto G'$ is an epimorphism. Isomorphism is overkill.
• Jan 27th 2010, 05:51 AM
ChrisBickle
what is an epimorphism? we only covered homomorphisms and isomorphisms.
• Jan 27th 2010, 09:01 AM
Drexel28
Quote:

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