Set of Subgroups forms Complete Lattice

Theorem
Let $\left({G, \circ}\right)$ be a group, and let $\mathbb G$ be the set of all subgroups of $G$.

Then $\left({\mathbb G, \subseteq}\right)$ is a complete lattice.

Proof
Let $\varnothing \subset \mathbb H \subseteq \mathbb G$.

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 $\bigcap \mathbb H$ 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 $\left({\mathbb G, \subseteq}\right)$ is a complete lattice.