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!

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- Apr 23rd 2009, 10:52 AMmpryalG/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! - Apr 23rd 2009, 02:56 PMGammaNatural projection
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