Index satisfies intersection inequality

Statement

Suppose $G$ is a group and $H, K$ are subgroups of finite index in $G$ . Then, we have:

$[G : H \cap K] \leq [G : H] [G : K]$ .

(An analogous statement holds for subgroups of infinite index, provided we interpret the indices as infinite cardinals).

Conjugate-intersection index theorem: A somewhat stronger bound when the two subgroups are conjugates to each other.

Index satisfies transfer inequality: This states that if $H, K \leq G$ , then $[K : H \cap K] \leq [G : H]$ .
Index is multiplicative: This states that $L \leq K \leq G$ , then $[G : L] = [G : K] [K : L]$ .

Given: A group $G$ with subgroups $H$ and $K$ .

To prove: $[G : H \cap K] \leq [G : H] [G : K]$ .

Proof: By fact (1), we have:

$[K : H \cap K] \leq [G : H]$ .

Setting $L = H \cap K$ in fact (2) yields:

$[G : H \cap K] = [G : K] [K : H \cap K]$ .

Combining these yields:

$[G : H \cap K] \leq [G : H] [G : K]$

as desired.