Discrete subgroup implies closed

From Groupprops
Jump to: navigation, search
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


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


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 \ne 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 \ne z, y^{-1}z \ne e. Thus, we have found a non-identity element in U \cap H, a contradiction.