# 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 is an open subset of a topological group containing the identity element of the topological group, there exists an open subset such that is symmetric: it contains the identity element and is closed under the inverse map, and further, such that

## Proof

**Given**: A T0 topological group , a discrete subgroup

**To prove**: is a closed subgroup of

**Proof**: Let denote the identity element of . Since is discrete, there exists an open set such that . By the fact stated above, there exists a symmetric open subset such that .

Now, suppose is not closed. Then there exists an element such that every open subset containing intersects . This yields that every open subset containing the identity intersects . In particular, intersects . Note that since does *not* intersect , . Hence, . Thus, is also an open subset containing the identity, and hence it again intersects . Thus, we can find another point .

Now consider . This is an element of . Moreover, since is symmetric, , so . Finally, since , . Thus, we have found a non-identity element in , a contradiction.