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 $L(G)$.
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 |
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Canonical name | NormalSubgroupsFormSublatticeOfASubgroupLattice |
Date of creation | 2013-03-22 15:48:24 |
Last modified on | 2013-03-22 15:48:24 |
Owner | CWoo (3771) |
Last modified by | CWoo (3771) |
Numerical id | 5 |
Author | CWoo (3771) |
Entry type | Example |
Classification | msc 20E15 |