Lemma (Lattice isomorphism theorem) Let α:G→H be some group homomorphism. There is a natural bijection between the (normal) subgroups of G that contain ker(α) and the subgroups of im(α).
proof: For K≤G let Kα={α(k)∈H|k∈K} (note this is a group) and for M≤im(α) let M∗={g∈G|α(g)∈M} this too is a group. Clearly (M∗)α={α(g)∈G|α(g)∈M}=M since every element of M is in the image of α. Now (Kα)∗≤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⊴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⊲G/N then ∃K,B≃K/N.
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