Just recently I've stumbled across Warren Dicks' book *Groups, trees and projective modules* (1980) and I was pretty stunned. I know nothing of group cohomology, but I gather the "tree" component is a special case of space studied in general group cohomology.


The results that caught my attention were:



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> The augmentation ideal of $R[G]$ is projective iff $G$ has cohomological dimension at most $1$, and it has cohomological dimension exactly $1$ iff it is an infinite group of cohomological dimension $1$.
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> The augmentation ideal of $R[G]$ is projective iff $G$ is the fundamental group of a graph of finite groups having order invertible in $R$.
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For me this resonated with two other results I know well about group rings:



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> (Renault) $R[G]$ is right self-injective iff $R$ is right self-injective and $G$ is finite.
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and



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> (Connell) $R[G]$ is von Neumann regular iff 1) $R$ is von Neumann regular; 2) $G$ is locally finite; and 3) the order of every finite subgroup of $G$ is invertible in $R$.
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Suppose from here on that $R[G]$ is von Neumann regular, $G$ is (at most) countable, and you don't know Connell's proof about regular group rings.


Then the augmentation ideal is countably generated, and by a result of Kaplansky the augmentation ideal must be projective. Apparently now $G$ is "the fundamental group of a graph of finite groups having order invertible in $R$."


**First question:** is it obvious somehow that $G$ is locally finite and that the groups which are vertices of the graph of groups tell us about the finite subgroups of $G$ and the invertibility of their orders? (I.e., can you recover Connell's theorem from Dunwoody's theorem?)


**Second question:** now additionally assume $R[G]$ is right self-injective and you don't know Renault's result. Can right self-injectivity be interpreted in this context to explain why $G$ is finite, i.e. $G$ has cohomological dimension $0$? (I.e., can you recover Renault's theorem in the special case of VNR right self-injective rings from Dunwoody's theorem by explaining why injectivity of $R[G]$ decides that $G$ is finite?)


I'm just probing around here seeing if I can get some connection between the two disciplines.



Using Dunwoody's results on cohomological dimension to learn about a von Neumann regular group ring
I'm adding a new answer since my other answer was the converse.


Question 1:


I can `simplify' Connell's argument using Dunwoody-Dicks. Assume that $RG$ is von Neumann regular. Note that if $g\in G$, then $(1-g)r(1-g)=1-g$ for some $r\in R$. So $(1-g)(1- r(1-g))=0$. As $r(1-g)$ is in the augmentation ideal, it follows that $1-r(1-g)\neq 0$ and so $1-g$ is a left zero divisor. This implies $g$ has finite order by standard group ring arguments (in order for $a=ga$ we must have $g$ permutes the finite support of $a$ and hence has finite order). So $G$ is torsion. 


If $G$ is countable, it has projective augmentation ideal (since countably generated left ideals in a von Neumann regular ring are projective) and so $G$ acts on a tree with finite stabilizers with order invertible $R$ by Dunwoody-Dicks and because the augmentation ideal is projective. Thus each finitely generated subgroup of $H$ also acts on a tree with finite stabilizers of order invertible in $R$ and we prove that $H$ is finite. But any finitely generated torsion group acing on a tree has a global fixed point (see Serre's book, where it is shown any finitely generated group of elliptic automorphisms has a global fixed point) and so $H$ is finite with order invertible in $R$. Alternatively, a finitely generated group acting on a tree with finite stabilizers has a finite index free subgroup by Bass-Serre theory, which must be trivial in our case since $H$ is a torsion group. So $H$ is finite and hence is a subgroup of a vertex stabilizer so its order is invertible in $R$. Thus $G$ is locally finite.


