Sohei YASUDA gave me a proof of a lemma whose proof is instructive and interesting, using many little facts from group theory and topology. I thought I should blog it.

Lemma: G is a compact topological group and H is an open subgroup. Then H is also closed (!) and of finite index.

Proof. G can be written as a union of cosets, each of which is open in G thanks to continuity of the group product. Thus the complement of H is a union of open (co)sets so is open. H is thus closed.

Consider now the quotient map from G to G/H. (Here, I’m not assuming H is normal in G but rather thinking of G/H as the quotient topological group with it’s topology coming from the quotient map). By definition of the quotient map, H cl-open in G implies it’s image, the identity coset of G/H too is cl-open.

Compactness of G inherits to the topology on G/H so the identity element is discrete on a compact set. The set G/H better be finite now, thus proving the second assertion. $\blacksquare$