Index of Intersection of Subgroups

Theorem
Let $G$ be a group.

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

Then:


 * $\index G {H \cap K} \le \index G H \index G K$

where $\index G H$ denotes 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 $H K = \set {h k: h \in H, k \in K} = G$.

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

From Tower Law for Subgroups, we have:


 * $\index G {H \cap K} = \index G H \index H {H \cap K}$

From Index in Subgroup, also:
 * $\index G {H \cap K} \le \index G K$

Combining these results yields the desired inequality.

Again from Index in Subgroup, it follows that:


 * $\index H {H \cap K} = \index G K$

$H K = G$.