Group ring

Remarks

With these operations $R[G]$ is a ring with identity $1_R \, e$, where $e$ is the identity of $G$. The maps $r \mapsto r e$ and $g \mapsto 1_R \, g$ embed $R$ as a subring and, when $R \ne 0$, $G$ into the group of units of $R[G]$. If $R$ is commutative, $R[G]$ is an algebra over $R$, and if $R$ is a field, it is the group algebra.

Example: let $G = \langle t \rangle$ be infinite cyclic. A finite sum $\sum_n a_n t^n$ is a Laurent polynomial, and the product formula is the usual product of Laurent polynomials, so $R[G] \cong R[t, t^{-1}]$.

Non-example: for infinite $G$, the set of all functions $G \to R$ with the same product formula is not a ring in general. For $G = \Z$ and $a = b = \sum_{n \in \Z} t^n$, the coefficient of $t^0$ in $ab$ would be the infinite sum $\sum_{n} 1$, which is not defined in $R = \Z$.

If $G$ has an element $g \ne e$ of finite order $n$, then $(1 - g)(1 + g + \cdots + g^{n-1}) = 1 - g^n = 0$, so $R[G]$ has zero divisors when $R \ne 0$. The group ring is the monoid ring of $G$ over $R$.