# Compact implies every open subgroup has finite index

From Groupprops

## Contents

## Statement

### Statement for semitopological groups

In a compact semitopological group, any open subgroup must be a subgroup of finite index in the whole group.

### Statement for topological groups

In a compact group, any open subgroup must be a subgroup of finite index in the whole group.

## Related facts

### Similar facts

- Connected implies no proper open subgroup
- Open subgroup implies closed
- Closed subgroup of finite index implies open

### Applications

## Facts used

## Proof

### Proof outline

The idea is to consider the left cosets of any open subgroup as forming an open cover of the whole group by *disjoint* subsets. Since the cover uses disjoint subsets, it is minimal, i.e., has no proper subcover. So by compactness, it must be finite, which means that the original subgroup has finite index.

### Proof details

**Given**: A compact group , an open subgroup of .

**To prove**: has finite index in , i.e., it has finitely many left cosets in .

**Proof**:

Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|

1 | For all , the map given by is a self-homeomorphism of . | Definition of topological group | is a topological group. | [SHOW MORE] | |

2 | Every left coset of in is an open subset of . | Homeomorphisms take open subsets to open subsets | is open in | Step (1) | [SHOW MORE] |

3 | The left cosets of in form an open cover of by pairwise disjoint subsets. | Fact (1) | Step (2) | By Fact (1), the left cosets form a cover of by disjoint subsets. By Step (2), each of them is open, so we have an open cover. | |

4 | has finite index in . | is compact | Step (3) | [SHOW MORE] |