I am reading a book in representation theory and it says for any group $G$ and a commutative ring $R$ with a $1$. Then the map $$ \epsilon : RG \rightarrow R \\ , g \mapsto 1 \text{ for all } g \in G $$ is called the augmentation map and is a ring homomorphism.
Does $R$ need to be commutative here. I can show it’s a ring homomorphism and as far as I can tell there is no necessity for $R$ to be commutative for this to still be the case. Is this true?
No, the augmentation ideal is defined exactly the same way when $R$ is noncommutative, and it is still very useful.
The most important commutativity a group ring needs is that $rg=gr$ for all $r\in R$, $g\in G$ (that is part of the definition of the group ring) but this stops short of requiring elements of $R$ to commute with each other.