Open subgroup implies closed

This article gives the statement and possibly, proof, of an implication relation between two topological subgroup properties. That is, it states that every subgroup of a topological group satisfying the first subgroup property must also satisfy the second
View a complete list of topological subgroup property implications

Statement

For topological groups

Any open subgroup of a topological group is closed.

For other types of groups

The statement is true for algebraic groups as well as for Lie groups, where open and closed are interpreted in terms of the corresponding topologies. This is because algebraic groups and Lie groups are in particular topological groups with the corresponding topologies.

Related facts

Corollaries

• Connected implies no proper open subgroup

Similar facts

• Closed and finite index implies open

Proof

Proof outline

The idea behind the proof is to show that if the subgroup is open, i.e., all its points are well inside it, then each of its left cosets is open, i.e., all points outside it are well outside it. This shows that the subgroup is closed.

Note that instead of left cosets in the proof below, we could use right cosets throughout.

Proof details

Given: A topological group $G$, an open subgroup $H$ of $G$.

To prove: $H$ is a closed 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 an open subset of $G$.	[SHOW MORE] Homeomorphisms take open subsets to open subsets || $H$ is open in $G$ || Step (1) || 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 union of all the left cosets of $H$ other than $H$ itself is open in $G$	Union of open subsets is open		Step (2)	Step-fact combination direct.
4	$H$ is closed in $G$	A subset is closed iff its set-theoretic complement is open.		Step (3)	[SHOW MORE] 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 open. Hence, $H$ is closed.