Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question
3  
Dear Vahid, The answer is no in general. E.g. consider an infinite product of non-trivial copies of $\mathbb F_p$ (the finite field with $p$ elements). This will be compact, and the closed subgroups of index $p$ will be precisely the kernels of the continuous projections onto $\mathbb F_p$. But there are many linear functionals on this space that are not continuous, and there kernels will be index $p$ subgroups that are not closed. Regards, –  Emerton Jun 5 '13 at 22:33
6  
P.S. The key feature of this example is that it is profinite, but not topologically finitely generated. For topologically finitely generated profinite groups, finite index subgroups are closed (equivalently, open, since we are in a profinite group); see this paper of Nikolov and Segal: annals.math.princeton.edu/wp-content/uploads/… –  Emerton Jun 5 '13 at 22:36
    
@Emerton: Thank you very much for your answer and also for the link to the paper. –  Vahid Shirbisheh Jun 5 '13 at 22:46
add comment

1 Answer

up vote 1 down vote accepted

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 $C$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.