Set of Subgroups forms Complete Lattice

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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 $\mathbb H$.

$\blacksquare$