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. Index satisfies transfer inequality: This states that if $H,K\le G$, then $[K:H\cap K]\le [G:H]$.
2. 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:

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

as desired.