Suppose that $G$ is a profinite group with the property that every open compact subgroup is topologically finitely generated and just infinite. Suppose that $H$ is a commensurated subgroup of $G$ with countably infinite index. I am interested in whether such subgroups can exist and what is known about them, in particular whether there are any sufficient conditions on $G$ to rule out their existence. Any useful reading material would be much appreciated.

$\begingroup$ I'm not sure about the status of the question in the normal (existence of normal subgroups of infinite countable index). They exist when $G/[G,G]$ is infinite, I don't know if it's the only case (for f.g. profinite groups). The case of commensurated subgroups sounds like a natural generalization; I don't know if strong condition such as being hereditarily just infinite is a relevant condition here. $\endgroup$– YCorOct 29, 2014 at 13:36
1 Answer
Let me make two comment: If $G$ is finitely generated, than every open subgroup is finitely generated (Edit: I removed the reference). If every open subgroup is just infinite, then the group is called hereditarily just infinite.
A subgroup of countable index cannot be closed in a compact group, e.g. profinite group, because it cannot be measurable.
More information might be found in http://arxiv.org/pdf/1108.5130.pdf and in http://arxiv.org/pdf/1310.3359.pdf.

$\begingroup$ That in an f.g. profinite group every open subgroup is f.g. is not due to NikolovSegal, it's a very basic fact, with the same proof as in the discrete case. Concerning the question about $H$, it is not assumed that $H$ is closed. $\endgroup$– YCorOct 29, 2014 at 13:34

$\begingroup$ Of course you are right about Nikolov and Segal. But Rupert asked what can be said about them. Then you can say that they are not closed. $\endgroup$ Oct 29, 2014 at 15:21