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: Generalized 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.