
surjective homomorphism
Hi All:
I study a little abstravt algebra when I have time, which isn't very often. Here is one I was looking over and trying to show.
I know that a homomorphism preserves operations. Example
$\displaystyle e^{x}$ is homorphic because $\displaystyle e^{a}\cdot e^{b}=e^{a+b}$. It has all the criteria of a homomorphism. It maps multiplication to addition.
But, in general how can we show that if G is an abelian group (That is, ab=ba), Let $\displaystyle f: \;\ G\rightarrow H$ be a surjective homomorphism of groups. Prove H is abelian.

Take $\displaystyle a,b \in H$ such that $\displaystyle a=f(x)$ and $\displaystyle b=f(y)$ with $\displaystyle x,y \in G$ then $\displaystyle ab=f(x)f(y)=f(xy)=f(yx)=f(y)f(x)=ba$ And since $\displaystyle f$ is surjective we're finished