 Coset multiplication is well-defined exactly for normal subgroups

If $N$ is a normal subgroup, changing representatives to $gn$ and $hm$, where $n,m\in N$, givesThe chosen representatives therefore give the same coset. Conversely, suppose the rule defines an operation on left cosets. Since $nN=N$ for $n\in N$, multiplying by $gN$ gives $ngN=gN$, hence $g^{-1}ng\in N$. Applying this inclusion for every $g$ and $g^{-1}$ proves $g^{-1}Ng=N$, so $N$ is a normal subgroup. This makes normality exactly the condition needed to carry group operation to these classes.