surjective homomorphism

galactus · #1 $Old$ November 6th, 2009, 01:14 PM

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.

Jose27 · #2 $Old$ November 6th, 2009, 02:27 PM

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