# surjective homomorphism

• November 6th 2009, 11:14 AM
galactus
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

$e^{x}$ is homorphic because $e^{a}\cdot e^{b}=e^{a+b}$. 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 $f: \;\ G\rightarrow H$ be a surjective homomorphism of groups. Prove H is abelian.
• November 6th 2009, 12:27 PM
Jose27
Take $a,b \in H$ such that $a=f(x)$ and $b=f(y)$ with $x,y \in G$ then $ab=f(x)f(y)=f(xy)=f(yx)=f(y)f(x)=ba$ And since $f$ is surjective we're finished