# Compact implies every open subgroup has finite index

## 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.

## Facts used

1. Left cosets partition a group

## 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 $G$, an open subgroup $H$ of $G$.

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

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 For all $g \in G$, the map $G \to G$ given by $x \mapsto gx$ is a self-homeomorphism of $G$. Definition of topological group $G$ is a topological group. [SHOW MORE]
2 Every left coset of $H$ in $G$ is an open subset of $G$. Homeomorphisms take open subsets to open subsets $H$ is open in $G$ Step (1) [SHOW MORE]
3 The left cosets of $H$ in $G$ form an open cover of $G$ by pairwise disjoint subsets. Fact (1) Step (2) By Fact (1), the left cosets form a cover of $G$ by disjoint subsets. By Step (2), each of them is open, so we have an open cover.
4 $H$ has finite index in $G$. $G$ is compact Step (3) [SHOW MORE]