Supremum of Elements of Sublattice not necessarily Same as for Lattice

Theorem
Let $\struct {S, \preceq}$ be a lattice.

Let $\struct {T, \preceq_T}$ be a sublattice of $S$.

Let $a, b \in T$.

Then it is not necessarily the case that:
 * $\sup_S \set {a, b}$

is the same as:
 * $\sup_T \set {a, b}$

Proof
Proof by Counterexample:

Let $\struct {G, \circ}$ be a group.

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

Let $\powerset G$ denote the power set of $G$.

Let $\struct {\powerset G, \subseteq}$ be the complete lattice formed by $\powerset G$ and $\subseteq$.

From Power Set is Complete Lattice, $\struct {\powerset G, \subseteq}$ indeed forms a complete lattice.

Let $\struct {\mathbb G, \subseteq}$ be the complete lattice formed by $\mathbb G$ and $\subseteq$.

From Set of Subgroups forms Complete Lattice, $\struct {\mathbb G, \subseteq}$ indeed forms a complete lattice.

By definition, $\struct {\mathbb G, \subseteq}$ is a sublattice of $\struct {\powerset G, \subseteq}$.

Let $H, K \in \mathbb G$.

We have that:
 * $H \cup K$ is the supremum of $H$ and $K$ in $\struct {\powerset G, \subseteq}$.

But from Union of Subgroups, $H \cup K$ is not a subgroup of $G$ element of $\mathbb G$.

However, from Supremum of Subgroups in Lattice, we do have that the subset product $H K$ in $G$ is a subgroup of $G$ such that:
 * $\sup_{\mathbb G} \set {H, K} = H K$