Set of Subgroups forms Complete Lattice/Proof 2

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

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 $\mathbb H$.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.11 \ \text{(c) (2)}$