Supremum of Subgroups in Lattice

Theorem
Let $\left({G, \circ}\right)$ be a group.

Let $\mathbb G$ be the set of all subgroups of $G$.

Let $\left({\mathbb G, \subseteq}\right)$ be the complete lattice formed by $\mathbb G$ and $\subseteq$.

Let $H, K \in \mathbb G$. Let either $H$ or $K$ be normal in $G$.

Then:
 * $\sup \left\{{H, K}\right\} = H \circ K$

where $H \circ K$ denotes subset product.

Proof
Recall that Set of Subgroups forms Complete Lattice.

Let $L = \sup \left\{{H, K}\right\}$.

Let either $H$ or $K$ be normal in $G$.

Since $L$ contains $H$ and $K$, then $L$ contains $H \circ K$.

The smallest subgroup of $G$ containing $H$ and $K$ is:
 * $\left \langle {H, K} \right \rangle$

the subgroup generated by $H$ and $K$.

From Subset Product with Normal Subgroup as Generator:
 * $\left \langle {H, K} \right \rangle = H \circ K$

when either $H$ or $K$ is normal.

The result follows.