Supremum of Subgroups in Lattice

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 $14.6$