Index satisfies intersection inequality
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).
- 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 , then .
- Index is multiplicative: This states that , then .
Given: A group with subgroups and .
To prove: .
Proof: By fact (1), we have:
Setting in fact (2) yields:
Combining these yields: