Problem 24 of section 2 of Noncommutative Algebra by Farb & Dennis states:
Let $R$ be an artinian ring and let $G$ be a finite group. Show that $R[G]$ is semisimple if and only if $R$ is semisimple and $|G|$ is invertible in $R$.
I tried to make some progress in assuming that $R$ is semisimple and $G$ is invertible in $R$ and proving that $R[G]$ is semisimple. I'm trying to prove it by showing that in $R[G]$, any submodule is a direct summand. So, given a submodule $M$ of $R[G]$, I'd first like to find a complement for it as a direct summand as $R$-modules (not sure how to do that) and then define a projection into $M$ and turn its kernel into a complement as $R[G]$ module. This is an attempt to immitate techniques learnt in class, but I'm very confused because $R$ is not a field.
This exercise is from the chapter discussing the Jacobson radical so it may be related.
I would appreciate a hint.
EDIT: I managed this direction. Now I'm trying the other one. $R[G]$ is semisimple. I can show that $R$ is then semisimple, but how do I show that $|G|$ is invertible in $R$? I tried to assume that $|G|$ is not invertible and find a nil ideal in $R[G]$ to get a contradiction, but couldn't do it. I considered 2 ideals so far:
- $R\sum_{g \in G}1 \cdot g$
- $\{\sum_{g \in G} a_g g \mid \sum_{g \in G}a_g = 0\}$.