When multiplying two terms, G and R commmute past each other, though they may not be commutative themselves. Thus r1g1 * r2g2 = r1r2g1g2. Verify this is associative.
Expressions in the group ring are multiplied term by term, multiplying each term in the first expression by each term in the second, then gathering terms together according to the members of G. By its very definition, multiplication distributes over addition.
Finally, multiplication is associative. This is true per term, and it holds for expressions as well, once everything is expanded. Therefore RG is a ring.
If R contains 1, RG contains 1e, which is the multiplicative identity. Sometimes I simply call it 1 for shorthand, though that is a bit confusing.
Since R acts on the members of RG via multiplication in R, RG is a left R module, or a right R module.
Embed R into RG via R*e. Thus R is a subring of RG. Ideals in RG contract to ideals in R in the usual way.
If R is commutative, then R lies in the center of RG, and RG is an R algebra. In this case RG is called the group algebra over R. Prime ideals in RG contract to prime ideals in R.
The units of RG include xc, where c is any element of G and x is a unit in R. These are called the trivial units of RG. There may be other units as well.
If M is an RG module, then M is also an R module, and a G module. (Since members of G can't be added together, this is more like G acting on a set.) Note that any c in G performs an R endomorphism on M. This works because G and R commute past each other, hence we can apply cx or xc, and the result is the same. Since c has an inverse in G, the endomorphism is actually an automorphism. Therefore G maps into the group of R automorphisms of M. Since M is a left module, group action follows the reverse convention. Thus cd acts on M through d, and then through c. To be completely rigorous, G acts on the right, and R acts on the left. We just understand that cdM really means Mdc, so we can retain our left module notation.
Conversely, let M be a left R module, and let G be a group of R automorphisms of M. Verify that M is an RG module. A term acts on M through the associated automorphism of G and the element of R, and an expression acts on M by taking the sum of the actions of the terms of the expression, applied to M. Fill a page with algebra, and you can prove M is a left RG module. It's pretty intuitive.
Assume M is finitely generated over R. Each automorphism in G is a matrix, taking generators of the R module to linear combinations of generators. If M is free, matrices and automorphisms correspond 1-1. Of course G only comprises invertible matrices. Conversely, if G maps into the n×n matrices over R, then G acts on the free module M = Rn, and M is an RG module.
If R is a division ring, then all modules are free. Members of G are uniquely represented as matrices, mapping basis elements to linear combinations of basis elements.
A subgroup of G, or a subring of R, produces a subring of the original group ring RG.
The direct product of groups, or the direct product of rings, creates a direct product of group rings.