Normal subgroups form a sublattice

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.