MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am looking for an example of the following situation:

$G$ is an infinite profinite group, with a dense normal subgroup $N$. However $N$ does not contain any non-trivial closed normal subgroup of $G$.

There are easy abelian examples of this, such as $\mathbb{Z}$ inside $\mathbb{Z}_p$. There are probably also tricks to get examples of 'diagonally embedded' dense normal subgroups inside a direct product in a way that fails to contain any closed normal subgroups. To get something essentially different from these, let's say we want all the open subgroups of $G$ to be directly indecomposable and to have no abelian normal subgroups.

Can it be done with these extra conditions? My motivation is to try to understand topologically simple t.d.l.c. groups and the ways they can fail to be abstractly simple. It's interesting to know what restrictions can be put on dense normal subgroups just at the local level.

Edit: I should add that I am aware of work by Nikolov and Segal on dense subgroups of (topologically) finitely-generated profinite groups. In such groups, under some mild additional conditions every dense normal subgroup contains the derived subgroup, which is itself closed. This severely limits the possibilities for finitely-generated examples. But as far as I know, there is no hope at present of generalising Nikolov-Segal to any other class of profinite group, as their methods are critically dependent on properties of d-generated finite groups for fixed d.

share|cite|improve this question

Your Answer


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

Browse other questions tagged or ask your own question.