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If G is an abelian group and N is a subgroup of G, show that G/N is an abelian group.

What I have so far: N is abelian since N is a subgroup of the abelian group G. N is also normal to G because of this reason. Since G and N are abelian, GmodN (or G/N) is abelian.

Is this sufficient or am I missing some details?

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    I would probably want some details of the last sentence, showing explicitly that $ab=ba$ for all $a,b\in G/N$, working from the definition of $G/N$.2011-11-03

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