Closed subgroup implies open?

Question

If $H$ is a closed subgroup of a topological group, $H$ is also open?, I know that an open subgroup of a topological group is also closed, but the converse is true? if isn't, wich could be a counterexample?.

Answer 1

The question as you asked it is wrong, and there are plenty of counter examples like some comment. However it is true that whenever the group is compact the open subgroups are exactly the closed subgroups of finite index.

First I show that if $H$ is closed and of finite index then it is open (this part does not require compactness). H is of finite index and therefore there exists $g_0=e,g_1,...,g_n$ so that $G$ is the disjoin union of $g_iH$ for every $i$. for every $i$ the set $g_iH$ is closed (it is the pre-image of $H$ under the continuous map $x\mapsto g_i^{-1}x$) and we conclude that the finite union $K=\bigcup_{i=1}^n g_i H$ is closed and H=G\K it follows that $H$ is open.

Now if $G$ is compact, Let $H$ be an open subgroup and consider the union $\bigcup_{g\in G}gH$ This is an open cover for $G$, by compactness there exists a finite sub-cover $G=\bigcup_{n=1}^N g_i H$ it follows that $[G:H]\leq N$