Theorem. let G be a group, then the subset of Sub(G) consisting of all normal subgroups forms a sublattice.
Proof. Let N be the subset of Sub(G) consisting of all normal subgroups. In order to prove that normal subgroups form a sublattice, we will prove that it is a meet subsemilattice and a join subsemilattice separately. Combined these make the normal subgroups a sublattice.
Meet closure:
Let A,B be normal subgroups. Then the meet of A and B is the intersection A \cap B. Let c be an element of A \cap B and let g be an element of G then since A is normal gcg^{-1} is in A and since B is normal gcg^{-1} is in B. Therefore, gcg^{-1} is in a \cap b, which implies it is normal.
Join closure:
Let A,B be normal subgroups, and let A \vee B by their join in the lattice of subgroups. Let c_1 ... c_n be the product of elements in A and B, by closure this is in A \vee B. Now, the conjugate gc_1 ... c_ng^{-1} is equal to (gc_1g^{-1})...(gcg^{-1}) which is the product of conjugates of elements in the normal subgroups, so it is in each of them. So by subgroup closure, this conjugate is in the subgroup A \vee B. By conjugate closure, the subgroup A \vee B is normal.
Remarks: if Sub(G) is the lattice of subgroups, then normal subgroups are a member of Sub(Sub(G)), the lattice of sublattices of the lattice of subgroups. This shows how the subalgebra lattice construction can be nested.
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