Closed subgroup of finite index implies open
Statement for left-topological, right-topological, or semitopological groups
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
Note that topological groups are semitopological groups, so the result applies to these.
- Open subgroup implies closed
- Connected implies no proper open subgroup
- Compact implies every open subgroup has finite index
Proof for left-topological groups
Given: A right-topological group , a closed subgroup of finite index in .
To prove: is an open subgroup of
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||For all , the map given by is a self-homeomorphism of .||Definition of left-topological group||is a left-topological group.||[SHOW MORE]|
|2||Every left coset of in is a closed subset of .||Homeomorphisms take closed subsets to closed subsets||Step (1)||By Step (1), is a self-homeomorphism of , so it takes the closed subset to the closed subset . Thus, for any , is closed in .|
|3||The union of all the left cosets of other than itself is closed in||Union of finitely many closed subsets is closed||has finite index in||Step (2)||Step-fact combination direct.|
|4||is open in||A subset is open iff its set-theoretic complement is closed.||Step (3)||The set-theoretic complement of in is precisely the union of all the left cosets other than itself, and by Step (3), this is closed. Hence, 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.