Quotienting $SN$ by $N$ is the same as quotienting $S$ by its overlap with $N$.Suggest a slogan
Let \(G\) be a group, \(S \leq G\) a subgroup, and \(N \normal G\) a normal subgroup. Then \[ \frac{SN}{N} \cong \frac{S}{S\intersect N}. \] If \(S\) and \(N\) are finite, then \[ \abs{SN} = \frac{\abs S \abs N}{\abs{S\intersect N}}. \]
