Set of Normal subgroups is a sublattice of a set of subgroups

Question

I need to show that if $ G$ is a group then $\mathcal N(G)$ is a sublattice of $S(G)$. Obviously $N(G) \subseteq S(G) $. How do I show that operations join and meet agree with those of the original lattice?

Answer 1

It suffices to show that the join and meet of two normal subgroups $H,K$ are also normal.

The meet of $H,K$ is $H \cap K$, and it is easy to verify that the intersection of normal subgroups is normal.

The join of $H,K$ is $HK$, which we can show is also normal: $$ x hk x^{-1} = x h x^{-1} x k x^{-1} = h'k' $$ (Also, because one of the factors is normal, $HK =KH$)

Answer 2

Take two normal subgroups $H$ and $K$ of $G$. The meet of those is simply the intersection of them, which you should have no problem in proving to be also a normal subgroup (simply use the definition). The joij of them is the set of elements of the form $a_1\cdots a_n$, with $a_i\in H\cup K$, and this is also a normal subgroup of $G$ (associativity is crucial here).