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
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:
- $\sup_{\mathbb G} \set {H, K} = H K$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings