Index satisfies intersection inequality

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

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

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

Related facts

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

Facts used

 * 1) uses::Index satisfies transfer inequality: This states that if $$H, K \le G$$, then $$[K: H \cap K] \le [G:H]$$.
 * 2) uses::Index is multiplicative: This states that $$L \le K \le G$$, then $$[G:L] = [G:K][K:L]$$.

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

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

Proof: By fact (1), we have:

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

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

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

Combining these yields:

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

as desired.