Set of Subgroups forms Complete Lattice/Proof 1
Jump to navigation
Jump to search
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
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.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.6$