The <cosets> of an open <subgroup> $H$ form an open cover of the <compact space> $G$. A finite subcover exists. Since distinct <cosets> are disjoint, that subcover must contain every <coset>. Thus \b[$[G:H]$ is finite]. Neither normality nor Hausdorffness is needed for this argument.

Solution

Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 5 / 1 / ii / Solution

 Solution (source code)