Let \(G\) be a group and \(N\normal G\) a normal subgroup.
If \(N\leq K\leq G\), then \(K/N\) is a subgroup of \(G/N\).
For \(N\leq K\leq G\), \(K\normal G\) if and only if \(K/N\normal G/N\).
The map \(K\mapsto K/N\) is a bijection from the subgroups \(K\) of \(G\) containing \(N\) to the subgroups of \(G/N\), and it restricts to a bijection between normal subgroups.
If \(N\leq K\) and \(K\normal G\), then \[ \frac{G/N}{K/N} \cong \frac{G}{K}. \]
Let \(\pi\colon G\to G/N\) be the quotient map. (a) \(K/N=\pi(K)\) is the image of a subgroup. (c) The inverse of \(K\mapsto\pi(K)\) is \(\bar K\mapsto\pi^{-1}(\bar K)\): \(\pi^{-1}(\bar K)\) is a subgroup containing \(N\), \(\pi(\pi^{-1}(\bar K))=\bar K\) since \(\pi\) is surjective, and \(\pi^{-1}(\pi(K))=KN=K\) since \(N\leq K\). (b) Since \(\pi(gKg^{-1})=\pi(g)\pi(K)\pi(g)^{-1}\) and \(\pi\) is a bijection on subgroups containing \(N\), \(gKg^{-1}=K\) for all \(g\in G\) if and only if \(\bar g(K/N)\bar g^{-1}=K/N\) for all \(\bar g\in G/N\); this also gives the second half of (c). (d) The map \(G/N\to G/K\), \(gN\mapsto gK\), is well defined because \(N\leq K\), and it is a surjective homomorphism with kernel \(K/N\); apply the first isomorphism theorem.