Index satisfies intersection inequality
From Groupprops
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.