# Difference between revisions of "Index satisfies intersection inequality"

From Groupprops

(New page: ==Statement== Suppose <math>G</math> is a group and <math>H, K</math> are subgroups of finite index in <math>G</math>. Then, we have: <math>[G:H \cap K] \le [G:H][G:K]</math>. (An analo...) |
(→Facts used) |
||

Line 14: | Line 14: | ||

# [[uses::Index satisfies transfer inequality]]: This states that if <math>H, K \le G</math>, then <math>[K: H \cap K] \le [G:H]</math>. | # [[uses::Index satisfies transfer inequality]]: This states that if <math>H, K \le G</math>, then <math>[K: H \cap K] \le [G:H]</math>. | ||

− | # [[uses::Index is multiplicative]]: This states that <math>L \le K \le G</math>, then <math>[G:L] | + | # [[uses::Index is multiplicative]]: This states that <math>L \le K \le G</math>, then <math>[G:L] = [G:K][K:L]</math>. |

==Proof== | ==Proof== |

## Revision as of 23:40, 12 May 2010

## 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.