Set of Subgroups forms Complete Lattice
Theorem
Let $\struct {G, \circ}$ be a group.
Let $\mathbb G$ be the set of all subgroups of $G$.
Then:
- $\struct {\mathbb G, \subseteq}$ is a complete lattice.
where for every set $\mathbb H$ of subgroups of $G$:
- the infimum of $\mathbb H$ necessarily admitted by $\mathbb H$ is $\ds \bigcap \mathbb H$.
Proof 1
Let $\O \subset \mathbb H \subseteq \mathbb G$.
By Intersection of Subgroups: General Result, $\bigcap \mathbb H$ is the largest subgroup of $G$ contained in each of the elements of $\mathbb H$.
Thus, not only is $\ds \bigcap \mathbb H$ a lower bound of $\mathbb H$, but also the largest, and therefore an infimum.
The supremum of $\mathbb H$ is the smallest subgroup of $G$ containing $\bigcup \mathbb H$.
Therefore $\struct {\mathbb G, \subseteq}$ is a complete lattice.
$\blacksquare$
Proof 2
From Group is Subgroup of Itself:
- $\struct {G, \circ} \in \mathbb G$
Let $\mathbb H$ be a non-empty subset of $\mathbb G$.
From Intersection of Subgroups is Subgroup: General Result:
- $\ds \bigcap \mathbb H \in \mathbb G$
Hence, from Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice:
- $\struct {\mathbb G, \subseteq}$ is a complete lattice
where $\ds \bigcap \mathbb H$ is the infimum of $\mathbb H$.
$\blacksquare$