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.
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.)