# Index satisfies intersection inequality

## Contents

## Statement

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

.

(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

- Index satisfies transfer inequality: This states that if , then .
- Index is multiplicative: This states that , then .

## Proof

**Given**: A group with subgroups and .

**To prove**: .

**Proof**: By fact (1), we have:

.

Setting in fact (2) yields:

.

Combining these yields:

as desired.