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

Verbal statement

Any open subgroup of a topological group is closed.

Proof

Let $H$ be an open subgroup of a topological group $G$. Let $g\in G$. Consider the map $G\to G$ given by $x\mapsto g*x$. This is a continuous map. Further, its inverse is the map $x\mapsto g^{-1}*x$. Thus, $x\mapsto g*x$ is a homeomorphism for any $g\in G$.

Now, a homeomorphism takes open subsets to open subsets. Thus, since $H$ is open, every left coset $gH$ is also open. Now, take the union of all the left cosets of $H$, other than $H$ itself. This union is the set-theoretic complement of $H$. Further, since each member is open, and an arbitrary union of open sets is open, the union is an open subset. Thus, the complement of $H$ is open, hence $H$ is closed.

Corollaries

• Connected implies no proper open subgroup

Related results

• Closed and finite index implies open