question about an abelian quotient group The question: 
If G is an abelian group and N is a subgroup of G, show that G/N is abelian.
I'm confused as to what it means for a quotient group to be abelian. Isn't G/N a set of sets? How can a set of sets be abelian? Sorry if this is a really stupid question.
 A: For all x,y in G
$(xN)(yN)=xyN=yxN=(yN)(xN)$
A: The elements of $G/N$ happen to be sets, but $G/N$ is a group: there is a group operation defined on that collection of sets. That operation can be commutative, in which case $G/N$ is Abelian, or it can fail to be commutative, in which case $G/N$ is not Abelian. You’re asked to show that if $G$ is Abelian, the group operation on $G/N$ is commutative (and therefore $G/N$ is Abelian).
Specifically, if $+$ is the operation in $G$, the operation $\oplus$ in $G/N$ is defined by $$(x+N)\oplus(y+N)=(x+y)+N\;.$$
A: The quotient set $G/N$ has a group structure, that is, you can add two cosets $(a+N)+(b+N)=(a+b)+N$. If $G$ is abelian, than $a+b=b+a$ so that $(a+N)+(b+N)=(b+N)+(a+N)$.
A: You are right that $G/N$ is a set of sets. Each element of $G/N$ is a set having the form $x+N$ for some $x \in G$. The key now is that we can define an operation between these sets by
$(x+N) + (y+N)=(x+y)+N$. This makes our set of sets $G/N$ into a group. And $G/N$ is an abelian group because $(x+N) + (y+N)=(x+y)+N=(y+x) + N= (y+N) + (x+N)$, where we used the fact that $G$ itself is abelian.
A: More generally, the homomorphic image of an abelian group is abelian:
$$
\phi(x) \phi(y) = \phi(xy) = \phi(yx) = \phi(y) \phi(x)
$$
The result you quote is the special case of the canonical quotient homomorphism. (But it's actually as general as it gets since every homomorphic image is a quotient.)
