Let $G$ be a locally compact (Hausdorff) group and let $H$ be a finite index subgroup of $G$. Can we say that $H$ has to be a closed subgroup of $G$? If it is not correct, do you know any counterexample?


Remember, we can always replace $H$ by a normal finite index subgroup (i.e. $N:=\bigcap\_{g\in G} gHg^{-1}$). This might help to find an answer.


Sorry, if my question looks very elementary!



Finite index subgroups of  locally compact groups
Here's a cw answer so as not to leave this unanswered. If $p$ is prime, $D$ is a countable abstract set, then


(1) the product $C^D$ has exactly $\aleph\_0$ closed subgroup of index $p$ (recall that for a finite index subgroup, closed=open), but


(2) it has $2^{2^{\aleph\_0}}$ abstract subgroups of index $p$.


The argument for (1) is topological: this is because the Cantor set has only countably many clopen subsets.


The argument for (2) is algebraic (use that any vector space over $\mathbf{Z}/p\mathbf{Z}$ has a basis): if $R$ is an abstract set of cardinal $2^{\aleph\_0}$ then $C^D$ is abstractly isomorphic to $C^{(R)}$. The set of subgroups of index $p$ is the projective space of $\text{Hom}(C^{(R)},C)=C^R$, and thus has $2^{|R|}$=$2^{2^{\aleph\_0}}$ elements.


Some of these non-closed subgroups can, in a sense, be made explicit by considering an ultrafilter and consider the kernel of the natural map from the product to the ultraproduct. This is of course not really explicit since the ultrafilter itself cannot be made explicit. On the other hand, this latter argument works for an arbitrary nontrivial finite group $C$.


