# Separable profinite group which is not metrisable?

Is there a topological group which is profinite (i.e. compact and totally disconnected) and separable, but is not metrisable (equivalently is not first-countable)?

I know there are such topological space: see here. But among those listed there only the Stone-Cech compactification of $\mathbb{Z}$ has a topology which in theory can be that of a topological group (one of the other two is not Hausdorff and the second is first-countable but not metrisable, so neither can be a topological group). I don't know of any structure of a topological group on the Stone-Cech compactification of $\mathbb{Z}$, though.

If you replace "totally disconnected" with "connected" I have an example: $(\mathbb{R}/\mathbb{Z})^{2^{\aleph_0}}$. I was just wondering if the totally disconnected case had something similar. If $D$ is a finite group, is $D^{2^{\aleph_0}}$ separable?

In fact, I am even more interested in groups which are not just separable but in fact topologically finitely generated (i.e. admit a finitely generated dense subgroup).

I will appreciate all thoughts and comments on this.

• I'm puzzled that you know how to prove that $(\mathbb{R}/\mathbb{Z})^{2^{\aleph_0}}$ is separable but not how to prove that $D^{2^{\aleph_0}}$ is separable if $D$ is finite. – Eric Wofsey Aug 18 '16 at 22:39
• My proof is very twisted: it's by Kronecker's theorem. Using it it's clear this group contains a dense cyclic group, and hence it's separable. I'm sure there are more sensible ways to prove it, though. – Cronus Aug 18 '16 at 23:09

Let $F$ be any finite group and take the product $F^\mathbb{R}$. This is separable: for instance, the set of piecewise constant functions $\mathbb{R}\to F$ where the pieces are intervals with rational endpoints (and there are only finitely many pieces) is dense. As long as $F$ is nontrivial, $F^\mathbb{R}$ will not be first-countable.
On the other hand, there is no topologically finitely generated example. Indeed, a profinite group $G$ is topologically finitely generated iff there is a continuous surjective homomorphism $\widehat{F_n}\to G$, where $\widehat{F_n}$ is the profinite completion of the free group $F_n$ on $n$ generators. But $\widehat{F_n}$ is metrizable (proof: $F_n$ has only countably many different finite quotients up to isomorphism since a finite quotient is just a finite group with a collection of $n$ generators, so $\widehat{F_n}$ embeds in a countable product of finite groups). Any Hausdorff quotient of a compact metrizable space is metrizable, and hence $G$ is also metrizable.
(Note that while there is no natural group structure on $\beta\mathbb{Z}$ as you are hoping to construct, there is a natural universal way to compactify $\mathbb{Z}$ as a group, called the Bohr compactification, and this compactification is not metrizable. However, it turns out that the Bohr compactification of $\mathbb{Z}$ is not profinite! And the universal way to turn $\mathbb{Z}$ into a profinite group is just the profinite completion $\hat{\mathbb{Z}}$, which is metrizable.)
• Very neat! You completely solved all my problems. Yes, now I remember the reason I disqualified $F^{2^{\aleph_0}}$ was because even if it is separable it's clearly not topologically finitely generated. – Cronus Aug 18 '16 at 23:12
• The Bohr compactification is very cool, although at times the map into it is not injective, which is unfortunate. I'm not sure if this is the case with $\mathbb{Z}$; the definition of the Bohr compactification is not very constructive. – Cronus Aug 18 '16 at 23:15
• The map into the Bohr compactification is injective for any locally compact abelian group, since by Pontryagin duality a locally compact abelian group has enough continuous homomorphisms to the compact group $\mathbb{R}/\mathbb{Z}$. – Eric Wofsey Aug 18 '16 at 23:19