Closed subgroup of finite index implies open

Statement

In a topological group, any closed subgroup of finite index (i.e., a closed subgroup that is also a subgroup of finite index) must be an open subgroup.

Related facts

• Open subgroup implies closed
• Connected implies no proper open subgroup
• Compact implies every open subgroup has finite index

Proof

Proof details

Given: A topological group $G$, a closed subgroup $H$ of finite index in $G$.

To prove: $H$ is an open subgroup of $G$

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 a closed subset of $G$.	Homeomorphisms take closed subsets to closed subsets		Step (1)	By Step (1), $x\mapsto gx$ is a self-homeomorphism of $G$, so it takes the closed subset $H$ to the closed subset $gH$. Thus, for any $g\in G$, $gH$ is closed in $G$.
3	The union of all the left cosets of $H$ other than $H$ itself is closed in $G$	Union of finitely many closed subsets is closed	$H$ has finite index in $G$	Step (2)	Step-fact combination direct.
4	$H$ is open in $G$	A subset is open iff its set-theoretic complement is closed.		Step (3)	The set-theoretic complement of $H$ in $G$ is precisely the union of all the left cosets other than $H$ itself, and by Step (3), this is closed. Hence, $H$ is open.