Discrete 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

Any discrete subgroup of a T0 topological group (i.e., a subgroup that is discrete in the subspace topology), is a closed subgroup.

Facts used

• Every open subset of a topological group containing the identity has a symmetric open squareroot: If $U$ is an open subset of a topological group containing the identity element of the topological group, there exists an open subset $V$ such that $V$ is symmetric: it contains the identity element and is closed under the inverse map, and further, such that $V.V\subset U$

Proof

Given: A T0 topological group $G$, a discrete subgroup $H$

To prove: $H$ is a closed subgroup of $G$

Proof: Let $e$ denote the identity element of $G$. Since $H$ is discrete, there exists an open set $U\ni e$ such that $U\cap H=\{e\}$. By the fact stated above, there exists a symmetric open subset $V$ such that $V.V\subset U$.

Now, suppose $H$ is not closed. Then there exists an element $g\in G$ such that every open subset containing $g$ intersects $H$. This yields that every open subset containing the identity intersects $g^{-1}H$. In particular, $V$ intersects $g^{-1}H$. Note that since $V$ does not intersect $H$, $g\notin H$. Hence, $y\neq e$. Thus, $V\setminus \{y\}$ is also an open subset containing the identity, and hence it again intersects $g^{-1}H$. Thus, we can find another point $z\in V\cap g^{-1}H$.

Now consider $y^{-1}z$. This is an element of $H$. Moreover, since $V$ is symmetric, $y^{-1}\in V$, so $y^{-1}z\in V.V\subset U$. Finally, since $y\neq z$, $y^{-1}z\neq e$. Thus, we have found a non-identity element in $U\cap H$, a contradiction.