Is it true that cohomological dimension of a group $G$ over a ring $R$ is zero, then group ring $RG$ is semisimple? Let $R$ be a ring and $G$ a group and let $RG$ be the group ring. 
 It is known that  $R$ into a left 
$RG$-module with trivial $G$-action, and define
$cd_R G$ to be projective dimension of  the left module $_{RG}R$.  $cd_R G$ is 
also called the cohomological dimension of $G$ over $R$. 
By this introduction, I want to know that the following statement 
is true or false. 

A group $G$ has cohomological dimension $0$ if and only if its group ring $RG$ is semisimple.

I have seen the above statement in Wikipedia without any reference!
You can see below link:
https://en.wikipedia.org/wiki/Cohomological_dimension
I guess that the above statement is not true.  But I cannot find 
any counter example.     
 A: This is false.  For instance, note that if $G$ is trivial, then $R=RG$ so $R$ is a free module of rank $1$ over $RG$ and so $G$ has cohomological dimension $0$ for any $R$.  But $RG=R$ need not be semisimple, since $R$ could be anything.  More generally, $R$ is always a quotient of $RG$, so $RG$ cannot be semisimple unless $R$ is semisimple.  But if $G$ is finite and $|G|$ is invertible in $R$, then $R$ is a projective $RG$-module, since the quotient map $RG\to R$ is split ($RG$-linearly) by multiplication by the average of all of the elements of $G$.
A: Eric's answer already gives a counterexample, but let me add a few words about when that statement is true. 
It is not hard to show that a group $G$ has zero cohomological dimension over $R$ if and only if $G$ is finite and $|G|$ is invertible in $R$. 
On the other hand, the Corollary in page 660 of the article "On the group ring" by Ian G. Connell, shows that $RG$ is semisimple if and only if $R$ is semisimple, $G$ is finite and $|G|$ is invertible in $R$. So if $R$ is a semisimple ring, then $RG$ is semisimple if and only if $G$ has zero cohomological dimension over $R$.
