Subgroups of $G/N$ are exactly the subgroups of $G$ that contain $N$.Suggest a slogan
Let \(G\) be a group, \(N \normal G\) a normal subgroup, and \(\pi\colon G\to G/N\) the quotient map. Then \[ \theset{H \leq G \suchthat N \subseteq H} \mapstofrom \theset{\overline{H} \suchthat \overline{H} \leq G/N}, \qquad H \mapsto H/N, \qquad \pi^{-1}(\overline{H}) \mapsfrom \overline{H}, \] are mutually inverse bijections between the subgroups of \(G\) containing \(N\) and the subgroups of the quotient group \(G/N\).