Compact implies every open subgroup has finite index
Statement for semitopological groups
Statement for topological groups
- Connected implies no proper open subgroup
- Open subgroup implies closed
- Closed subgroup of finite index implies open
The idea is to consider the left cosets of any open subgroup as forming an open cover of the whole group by disjoint subsets. Since the cover uses disjoint subsets, it is minimal, i.e., has no proper subcover. So by compactness, it must be finite, which means that the original subgroup has finite index.
Given: A compact group , an open subgroup of .
To prove: has finite index in , i.e., it has finitely many left cosets in .
|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 topological group||is a topological group.||[SHOW MORE]|
|2||Every left coset of in is an open subset of .||Homeomorphisms take open subsets to open subsets||is open in||Step (1)||[SHOW MORE]|
|3||The left cosets of in form an open cover of by pairwise disjoint subsets.||Fact (1)||Step (2)||By Fact (1), the left cosets form a cover of by disjoint subsets. By Step (2), each of them is open, so we have an open cover.|
|4||has finite index in .||is compact||Step (3)||[SHOW MORE]|