Order of Subgroup Product/Corollary

Theorem
Let $G$ be a group.

Let $H$ and $K$ be finite subgroups of $G$.

Then:
 * $\left|{H \vee K}\right| \ge \dfrac {\left|{H}\right| \left|{K}\right|} {\left|{H \cap K}\right|}$

or
 * $\dfrac {\left|{H \vee K}\right|}{\left|{H}\right|} \ge \dfrac {\left|{K}\right|} {\left|{H \cap K}\right|}$

where $H \vee K$ denotes join and $\left|{H}\right|$ denotes the order of $H$.

Proof
From Order of Subgroup Product:
 * $(1): \quad \left|{H K}\right| = \dfrac {\left|{H}\right| \left|{K}\right|} {\left|{H \cap K}\right|}$

From Subset Product is Subset of Generator, we have that:
 * $H K \subseteq H \vee K$

where $H K$ is the subset product of $H$ and $K$.

Thus:
 * $(2): \quad \left|{H \vee K}\right| \ge \left|{H K}\right|$

The result follows by substituting for $\left|{H K}\right|$ from $(1)$ into $(2)$.