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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?

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The first one, as the set $\,\{an_1bn_2\;:\; n_i\in N\}\,$ is NOT a left (right) coset of $\,N\,$ in $\,G\,$. – DonAntonio Jan 9 '13 at 21:21
@DonAntonio Are you sure? By normality $aNbN = abNN = abN$? – user56728 Jan 9 '13 at 21:32
In a nutshell, $(aN) * (bN) = abN$ is a candidate for a group operation, and normality is enough to ensure that this operation is well defined. – rschwieb Jan 9 '13 at 21:33
It's just that the "cosets" are equivalence classes, @user56728, and in a quotient group we work with a special representative of each such class, denoted by $\,aN\,$ ,and not with the whole set $\,aN:=\{an\;;\;n\in N\}\,$ – DonAntonio Jan 9 '13 at 21:51
@DonAntonio I'm saying $\{an_1bn_2 : n_i \in N\}$ is a left coset by the normality of $N$, contrary to what you just said. I may be wrong; please correct me if I am. – user56728 Jan 9 '13 at 21:58

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.

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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 $N$.

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