Closed subgroup of finite index implies open

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Statement

Statement for left-topological, right-topological, or semitopological groups

In a left-topological group or right-topological group, any closed subgroup of finite index (i.e., a closed subgroup that is also a subgroup of finite index) must be an open subgroup.

Note that a semitopological group is both a left-topological group and a right-topological group, so the result applies to semitopological groups.

Statement for topological groups

In a topological group, any closed subgroup of finite index (i.e., a closed subgroup that is also a subgroup of finite index) must be an open subgroup.

Note that topological groups are semitopological groups, so the result applies to these.

Related facts

Proof

Proof for left-topological groups

Given: A right-topological group G, a closed subgroup H of finite index in G.

To prove: H is an open 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 left-topological group G is a left-topological group. [SHOW MORE]
2 Every left coset of H in G is a closed subset of G. Homeomorphisms take closed subsets to closed subsets Step (1) By Step (1), x \mapsto gx is a self-homeomorphism of G, so it takes the closed subset H to the closed subset gH. Thus, for any g \in G, gH is closed in G.
3 The union of all the left cosets of H other than H itself is closed in G Union of finitely many closed subsets is closed H has finite index in G Step (2) Step-fact combination direct.
4 H is open in G A subset is open iff its set-theoretic complement is closed. Step (3) 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 closed. Hence, H is open.

Proof for right-topological groups

The proof is analogous to the proof for left-topological groups, except that we use right cosets instead of left cosets.