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!
First of all every subgroup of an abelian group is normal. This is obvious because $\displaystyle \forall x\in G xnx^{-1}=xx^{-1}n=n$, so $\displaystyle xNx^{-1}=N$.
$\displaystyle \pi:G \rightarrow G/N$ by $\displaystyle \pi(g)=g+N$ 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.
$\displaystyle \phi(a)\phi(b)=\phi(ab)=\phi(ba)=\phi(b)\phi(a)$
By this observation, G/N is abelian