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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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To be honest, you didn't actually tell us why $G/N$ has to be abelian. You just said it is. Try to prove this more general fact, if $A$ is abelian and $f:A\to H$ is a surjective homomorphism then $H$ is abelian. Why does that help us?

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Note that $$(xN) \cdot (yN) = (xy) \cdot N = (yx) \cdot N = (yN) \cdot (xN)$$

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    Thanks, is this enough to show that G/N is abelian?2011-11-03
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    @Lily: Well, I have taken $2$ elements $xN,yN$ from $G/N$ and show that they $\text{commute}$. Is that not enough?2011-11-03
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    It does seem to be enough. I always feel like I need more work when the solutions are on the shorter side. Thank you!2011-11-03
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    Almost every solution to a problem asked in an introductory algebra class (that does not involve calculations in a specific group/ring/...) can be made "on the shorter side." Not Twitter length maybe, but short--- at most two or three short paragraphs. Somebody told me this when I was taking algebra for the first time, and I learned a lot by trying to rewrite my arguments to make them short (but, obviously, not so short that details were left out). Chandrasekhar's calculation, plus about one sentence worth of English to set it up, and summarize what it shows, would be enough for anybody.2011-11-03