Question about quotient group - is the operation defined or is it a consequence? Given a normal subgroup $N \le G$.  Do we define the operation $*$ on $G/N$ to be
$$(aN) * (bN) = abN$$
or is the group operation the usual product, 
$$aNbN = \{an_1bn_2 : n_1, n_2 \in N \}$$
with the above being $abN$ due to normality?
 A: We do have to define some sort of operation on $G/N$ to get a group structure, and so we can either write $aN\cdot bN=aNbN$, using the operation in $G$, or we can formally set $aN\cdot bN=abN$. You are right that either definition follows from the other via normality.
If we define $aN\cdot bN=aNbN$ using multiplication in $G$, then we ought to make sure that $aNbN$ is a left coset of $N$. You can check that all such products $aNbN$ will be left cosets of $N$ if and only if $N$ is normal.
If we formally define $aN\cdot bN=abN$, we should make sure that our multiplication is well-defined, which means our multiplication rule does not depend on the choice of representative. In other words, if $aN=a^\prime N$, and $bN=b^\prime N$, then we need $abN=a^\prime b^\prime N$ for our multiplication rule to be well-defined. You can check that normality is precisely the condition that makes this true.
A: You can take both as a definition of $aN \cdot bN$, since
$$(aN)(bN) = a(Nb)N = a(bN)N = ab(NN) = abN$$
by normality of $N$.
