@TomDeMedts Yes of course you are right insofar as 1 goes. What struck me as interesting about it is the simplicity with which it indicates something reminiscent of the monumental classification of FSGs. 2 I think does not immediately follow from Wedderburn, but again it is not difficult.

The groups you list always have torsion when $n\ge2$. The virtual cohomological dimension is defined to be the cohomological dimension of a torsion-free subgroup of finite index (if there is one), or $\infty$ (if not). One always has $$\mathrm{cd}_{\mathbb Q}(G)\le \mathrm{vcd}_{\mathbb Z}(G)\le\mathrm{cd}_{\mathbb Z}(G).$$ These facts may help you extract the information you need from the results about vcd that you have already found.

@BCLC yes there are capital and small of each letter. For example $\mathfrak G$ and $\mathfrak g$. Use $\backslash$mathfrak to get these in latex.

Historically, this answer is correct: this is how Fraktur symbols were written until the mid-nineteen-eighties at least. Of course, times change, and today this form of handwriting is rare even in Germany.

I don't get the closure either. Btw, the underlining convention goes back to the time before computer typesetting when authors needed an efficient way of indicating to typesetters what symbols to use. Another examples, Greek letters were often handwritten and then underlined in red to mean "please note this is a Greek letter".

That is very true

Yes, sorry, what I meant to say was that if a group $G$ is a subgroup of a non-standard free group and is finitely generated (in the standard sense) then $G$ has a faithful 2 dimensional linear representation over $\mathbb C$.

@MoisheKohan I was using the term 'linear' loosely. I would regard the projective general linear group as close enough to linear. The idea of model theory is really inspired by the Wilkie -- van den Dries proof of Gromov's polynomial growth theorem using non-standard analysis. So $\mathbb Z^*$ is a non standard model of the integers and could be used to approximate ordinary real numbers.

@OlaSande The other point worth making is that, given the way the question is stated I think it would be unnatural to ignore the topology on $S^1$ but as YCor pointed out (and I tried to in a different way) there are considerable restrictions on how the topology of a (mathematical) ring can be so rarely do we find any attractive topological forms to have interesting compatible ring structures. In this was, $S^1$ is not a special case but a typical case.

@OlaSande First, to a non-mathematician, the circle $S^1$ looks very like a ring insofar as the words \emph{circle} and \emph{ring} might conjure up geometric shapes. So I can't help momentarily imagining that the person who raises this question also notices, with a wry smile, the non-mathematical rather obvious answer 'yes it is'. That raises serious points about the way we use words from everyday language and employ them with strict new definitions.

@OlaSande I think there are two additional comments I want to make that go a little beyond what has been said. In both cases I am interpreting your question to mean $\mathbb R/\mathbb Z$ should be interpreted as a topological space with compatible structure of additive group.

@BenjaminSteinberg OK, You make me suspect that my above comment is not helpful so I will probably delete it. Before I do, here is a simpler comment which is also a special case of the Question. Suppose that $G$ is a group of finite cohomological dimension $n$ over $\mathbb Z$: does this guarantee that $\textrm{gl.dim}(\mathbb ZG)=n+1$?

@BenjaminSteinberg Yes, I agree that $R$ being a (commutative?) PID means its contribution to the global dimension is relatively minor (like 0 or 1). However if $R=\mathbb Q$ then the cohomological dimension of $G$ over $R$ and the global dimension of $RG$ can still be finite even if $G$ has non-trivial finite subgroups.