normal subgroups form sublattice of a subgroup lattice

Consider $L(G)$, the subgroup lattice of a group $G$. Let $N(G)$ be the subset of $L(G)$, consisting of all normal subgroups of $G$.

First, we show that $N(G)$ is closed under $\wedge$. Suppose $H$ and $K$ are normal subgroups of $G$. If $x\in H\wedge K=H\cap K$, then for any $g\in G$, $gx{g}^{-1}\in H$ since $H$ is normal, and $gx{g}^{-1}\in K$ likewise. So $gx{g}^{-1}\in H\cap K=H\wedge K$, implying that $H\wedge K$ is normal in $G$, or $H\wedge K\in N(G)$.

To see that $N(G)$ is closed under $\vee$, let $H,K$ be normal subgroups of $G$, and consider an element

$$x=x_1{x}_2\mathrm{\cdots }{x}_n\in H\vee K,$$

where $x_i\in H$ or $x_i\in K$. If $g\in G$, then

$$gx{g}^{-1}=g{x}_1{x}_2\mathrm{\cdots }{x}_n{g}^{-1}=(g{x}_1{g}^{-1})(g{x}_2{g}^{-1})\mathrm{\cdots }(g{x}_n{g}^{-1}),$$

where each $g{x}_i{g}^{-1}\in H$ or $K$. Therefore, $gx{g}^{-1}\in H\vee K$, so $H\vee K$ is normal in $G$ and $H\vee K\in N(G)$.

Since $N(G)$ is closed under $\wedge$ and $\vee$, $N(G)$ is a sublattice of $L(G)$.

Remark. If $G$ is finite, it can be shown (Wielandt) that the subnormal subgroups of $G$ form a sublattice of $L(G)$.