I'm studying profinite groups, and I know that every group can be endowed with the profinite topology, whose basis consists of the cosets of subgrous with finite index. Moreover this topology is Hausdorff if and only if the group is residually finite. Up to these things everything is clear in my mind.

Then I wonder what this topology is on the group of real numbers $(\mathbb{R}, +)$. So my question is: what are the subgroups of $\mathbb{R}$ of finite index? Does it admit any? Is $\mathbb{R}$ residually finite?

(I know these are many questions: my aim is to understand the profinite topology on $\mathbb{R}$)

  • $\begingroup$ The profinite topology is trivial on any abelian divisible group, because such a group has no proper subgroup of finite index $>1$. $\endgroup$ – egreg Dec 31 '14 at 11:55

The reals (or any field of characteristic $0$) does not have any subgroups of finite index.

For if $H $ was such a subgroup, say of index $n > 0$, then we would have $n\mathbb {R}\subseteq H $. But when $n\neq 0$ we have $n\mathbb {R} = \mathbb {R} $ contradicting the assumption.


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