# Difference between revisions of "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).

## 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.