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.

Related facts

Similar facts

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

Applications

• Equivalence of definitions of profinite 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] From the definition of topological group, the multiplication map $G\times G\to G$ is continuous. The map $x\mapsto gx$ is a composite of the fiber inclusion $x\mapsto (g,x)$ and the multiplication map, hence it is a continuous map. Further, its inverse is the map $x\mapsto g^{-1}x$. Thus, $x\mapsto gx$ is a homeomorphism for any $g\in G$.
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] By Step (1), $x\mapsto gx$ is a self-homeomorphism of $G$, so it takes the open subset $H$ to an open subset. Thus, for any $g\in G$, $gH$ is open in $G$.
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] Since $G$ is compact, the open cover of Step (3) must have a finite subcover. But the cover uses disjoint subsets, so its only possible subcover is itself. So the cover must be finite to begin with, i.e. $H$ must have finitely many left cosets in $G$.