# Is a topological abelian group with no open subgroup connected?

You may assume that the groups we consider are all Hausdorff for the purpose of this question:

• Clearly, the implication converse to the one in the question holds, as all open subgroups are also closed and therefore if $$G$$ is connected the only open subgroup is $$G$$ itself.
• Is it the case that if the only open subgroup of an abelian group $$G$$ is $$G$$ itself, then $$G$$ is connected ?.
• If not, are there any nice sufficient conditions to ensure this is the case ?.
• Consider $(\Bbb Q,+)$ with Euclidean topology. Commented Jul 13 at 12:43
• Local path-connectedness is another condition under which this implication holds (not requiring abelianness or Hausdorffness) - see here. This includes a lot of nice groups like Lie groups. Commented Jul 13 at 13:14

If you add a compactness condition, then the result is true.

Consider a topological group $$G$$ with no proper open subgroup (I am not assuming $$G$$ to be abelian). The connected component of the identity of $$G$$, denoted $$G^0$$, is a topological subgroup which is normal and closed. Hence $$G/G^0$$ is again a Hausdorff topological group, for which the only connected components are the singletons (this space is totally disconnected). Moreover, $$G/G^0$$ again has no proper open subgroups (otherwise you could pull them back to $$G$$).

$$G$$ is connected iff $$G = G^0$$, so your question (without the abelian condition) amounts to: are there any totally disconnected Hausdorff topological groups who do not have any proper open subgroups ? The answer is yes (e.g. Giulio's comment). However, the answer becomes no if you assume the group to be compact, because compact totally disconnected subgroups are precisely profinite groups, which do admit proper open subgroups when they are not trivial.

• I do not see that "$𝐺/𝐺^0$ again has no nontrivial open subgroups". The discrete topology is Hausdorff, pick a group with a subgroup, give it the discrete topology and we have a $𝐺/𝐺^0 = G$ with a nontrivial subgroup. Commented Jul 13 at 15:39
• @user24142 I should have specified that the $G$ I am considering is assumed to have no proper open subgroup. In particular, under this hypothesis, $G$ cannot have the discrete topology unless it is trivial. Commented Jul 13 at 15:43
• i think that works now Commented Jul 13 at 15:45
• Thank you very much. It is clear to me that profinite groups (seen as inverse limits of finite groups) are compact and totally disconnected, as they are a closed subgroup of a product of finite (and thus compact) groups. Do you have any reference for the opposite implication? Commented Jul 13 at 18:16
• @LucaMarchiori This implication is not easy, you can find it as Proposition 1.1.3 in Neukirch, Cohomology of number fields. Additionnally, this question is discussed in this MSE post. Commented Jul 13 at 21:27