Compact implies every open subgroup has finite index

Statement for semitopological groups
In a uses property satisfaction of::compact semitopological group, any  uses property satisfaction of::open subgroup must be a  subgroup of finite index in the whole group.

Statement for topological groups
In a uses property satisfaction of::compact group, any  uses property satisfaction of::open subgroup must be a  subgroup of finite index in the whole group.

Similar facts

 * Connected implies no proper open subgroup
 * Open subgroup implies closed
 * Closed subgroup of finite index implies open

Applications

 * Equivalence of definitions of profinite group

Facts used

 * 1) uses::Left cosets partition a group

Proof outline
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.

Proof details
Given: A compact group $$G$$, an open subgroup $$H$$ of $$G$$.

To prove: $$H$$ has finite index in $$G$$, i.e., it has finitely many left cosets in $$G$$.

Proof: