# G/N homomorphic image

• April 23rd 2009, 11:52 AM
mpryal
G/N homomorphic image
If G is an abelian group and N is a subgroup of G, show that G/N is an abelian group observing that G/N is a homomorphic image of G.

Please show steps, I'm super confused.

Thanks!
• April 23rd 2009, 03:56 PM
Gamma
Natural projection
First of all every subgroup of an abelian group is normal. This is obvious because $\forall x\in G xnx^{-1}=xx^{-1}n=n$, so $xNx^{-1}=N$.

$\pi:G \rightarrow G/N$ by $\pi(g)=g+N$ this is a homomorphism from onto G/N with kernel N. The point is it is a homomorphism from an abelian group, so the image group is abelian. G/N is a group because N is normal.

$\phi(a)\phi(b)=\phi(ab)=\phi(ba)=\phi(b)\phi(a)$

By this observation, G/N is abelian