Lemma (Lattice isomorphism theorem) Let $\alpha : G \to H$ be some group homomorphism. There is a natural bijection between the (normal) subgroups of $G$ that contain $\ker(\alpha)$ and the subgroups of $\text{im}(\alpha)$.
proof: For $K \le G$ let $K^\alpha = \{ \alpha(k) \in H | k \in K \}$ (note this is a group) and for $M \le \text{im}(\alpha)$ let $M^* = \{ g \in G | \alpha(g) \in M \}$ this too is a group. Clearly $(M^*)^\alpha = \{ \alpha(g) \in G | \alpha(g) \in M \}= M$ since every element of $M$ is in the image of $\alpha$. Now $(K^\alpha)^* \le K$ easily by definition and the reverse inclusion comes down to the something about kernel. As for the inclusions unproved.
Theorem (Noether4). Let $G$ be a group and $N \unlhd G$. There is a natural bijection between the lattice of subgroups of $G$ that contain $N$ and the lattice of subgroups of $G/N$. The bijection preserves relations "subgroup", "normal subgroup" and properness.
proof: Let $H$ be a subgroup of $G$ which contains $N$ i.e. $N$ is also a subgroup of it then by Noether3 $N$ is normal in $H$ too so $H/N$ is well defined homomorphism. Now apply lattice isomorphism.
Corollary: If $B \lhd G/N$ then $\exists K, B \simeq K/N$.
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