= Coset multiplication is well-defined exactly for normal subgroups
{title2=$(gN)(hN)=ghN$}
If $N$ is a <normal subgroup>, changing representatives to $gn$ and $hm$, where $n,m\in N$, gives
$$
(gn)(hm)=gh(h^{-1}nh)m\in ghN.
$$
The 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.
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