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Thread: G/N homomorphic image

  1. #1
    Junior Member
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    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!
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  2. #2
    Super Member Gamma's Avatar
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    Natural 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
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