# 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 $g\in G$, $gxg^{-1}\in H$ since $H$ is normal, and $gxg^{-1}\in K$ likewise. So $gxg^{-1}\in H\cap K=H\wedge K$, 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

 $x=x_{1}x_{2}\cdots x_{n}\in H\vee K,$

where $x_{i}\in H$ or $x_{i}\in K$. If $g\in G$, then

 $gxg^{-1}=gx_{1}x_{2}\cdots x_{n}g^{-1}=(gx_{1}g^{-1})(gx_{2}g^{-1})\cdots(gx_{% n}g^{-1}),$

where each $gx_{i}g^{-1}\in H$ or $K$. Therefore, $gxg^{-1}\in H\vee K$, 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 $L(G)$.

## References

• 1 H. Wielandt Eine Verallgemeinerung der invarianten Untergruppen, Math. Zeit. 45, pp. 209-244 (1939)
Title normal subgroups form sublattice of a subgroup lattice NormalSubgroupsFormSublatticeOfASubgroupLattice 2013-03-22 15:48:24 2013-03-22 15:48:24 CWoo (3771) CWoo (3771) 5 CWoo (3771) Example msc 20E15