Supremum of Subgroups in Lattice
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Theorem
Let $\struct {G, \circ}$ be a group.
Let $\mathbb G$ be the set of all subgroups of $G$.
Let $\struct {\mathbb G, \subseteq}$ 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:
- $\sup \set {H, K} = H \circ K$
where $H \circ K$ denotes subset product.
Proof
Recall that Set of Subgroups forms Complete Lattice.
Let $L = \sup \set {H, K}$.
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:
- $\gen {H, K}$
the subgroup generated by $H$ and $K$.
From Subset Product with Normal Subgroup as Generator:
- $\gen {H, K} = H \circ K$
when either $H$ or $K$ is normal.
The result follows.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.6$