Index of Intersection of Subgroups

Theorem
Let $G$ be a group.

Let $H, K$ be subgroups of finite index of $G$.

Then:


 * $\left[{G : H \cap K}\right] \le \left[{G : H}\right] \left[{G : K}\right]$

where $\left[{G : H}\right]$ is the index of $H$ in $G$.

Note that here the symbol $\le$ is being used with its meaning less than or equal to.

Equality holds $H K = \left\{{h k: h \in H, k \in K}\right\} = G$.

Proof
Note that $H \cap K$ is a subgroup of $H$.

From Tower Law for Subgroups, we have:


 * $\left[{G : H \cap K}\right] = \left[{G : H}\right] \left[{H : H \cap K}\right]$

From Index in Subgroup, also:
 * $\left[{H : H \cap K}\right] \le \left[{G : K}\right]$

Combining these results yields the desired inequality.

Again from Index in Subgroup, it follows that:


 * $\left[{H : H \cap K}\right] = \left[{G : K}\right]$

$H K = G$.