Order of Subgroup Product/Corollary

Theorem

Let $G$ be a group.

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

Then:

$\size {H \vee K} \ge \dfrac {\order H \order K} {\order {H \cap K} }$

or

$\dfrac {\size {H \vee K} } {\order H} \ge \dfrac {\order K} {\order {H \cap K} }$

where $H \vee K$ denotes join and $\order H$ denotes the order of $H$.

Proof

From Order of Subgroup Product:

$(1): \quad \size {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$

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:

$(2): \quad \size {H \vee K} \ge \size {H K}$

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

$\blacksquare$

Sources

• 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.9$: Theorem $21$: Corollary