Order of Subgroup Product

Theorem

Let $G$ be a group.

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

Then:

$\left|{H K}\right| = \dfrac {\left|{H}\right| \left|{K}\right|} {\left|{H \cap K}\right|}$

Proof

From Intersection of Subgroups, we have that $H \cap K \le H$.

Let the number of left cosets of $H \cap K$ in $H$ be $r$.

Then the left coset space of $H \cap K$ in $H$ is $\left\{{x_1 \left({H \cap K}\right), x_2 \left({H \cap K}\right), \ldots, x_r \left({H \cap K}\right)}\right\}$.

So each element of $H$ is in $x_i \left({H \cap K}\right)$ for some $1 \le i \le r$.

Also, if $i \ne j$, we have $x_j x_i^{-1} \notin H \cap K$.

Let $h k \in H K$.

We can write $h = x_i g$ for some $1 \le i \le r$ and some $g \in K$.

Thus $h k = x_i \left({g k}\right)$.

Since $g, k \in K$, this shows $h k \in x_i K$.

• Now suppose the left cosets $x_i K$ are not all disjoint.

Then $x_i K = x_j K$ for some $i, j$ by Coset Spaces form Partition.

So $x_j^{-1} x_i \in K$ by Congruence Class Modulo Subgroup is Coset.

Since $x_i, x_j \in H$, we have $x_j^{-1} x_i \in H \cap K$, which is contrary to the definition.

Therefore, the cosets $x_i K$ are disjoint for $1 \le i \le r$.

This leads us to $\left|{H}\right| / \left|{H \cap K}\right| = \left|{H K}\right| / \left|{K}\right|$

whence the result.

$\blacksquare$

Sources

• Richard A. Dean: Elements of Abstract Algebra (1966): $\S 1.9$: Theorem $21$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 38 \alpha$
• John F. Humphreys: A Course in Group Theory (1996): $\S 5$: Proposition $5.18$