Let $G$ be a group, $R$ be a ring. One can then define the groupring $RG = \{f: G \to R \mid \sup(g) \ \mathrm{is \ finite}\}$, with pointwise addition and with the following multiplication:
If $\alpha, \beta \in RG$, then the product is defined as the function defined by $$\alpha \beta(z) = \sum_{x,y \in G, xy = z} \alpha(x)\beta (y) = \sum_{x \in G} \alpha(x)\beta(x^{-1}z)$$
Why does the function $\alpha\beta$ have finite support? I can see that for any $z \in G$, the sum above is finite, so the sum is well defined, but can't seem to find an easy reason why $\sup(\alpha\beta)$ is finite.
$supp(\alpha\beta)\subseteq supp(\alpha)supp(\beta)=\{xy\mid x\in supp(\alpha), y\in supp(\beta)\}$.
Therefore $|supp(\alpha\beta)|\leq |supp(\alpha)||supp(\beta)|<\infty$
Suppose $z\in supp(\alpha\beta)$. Since $\alpha\beta(z)$ is nonzero, then at so is $\alpha(x)\beta(x^{-1}z)$ for at least one $x$. For this to be true, both $\alpha(x)$ and $\beta(x^{-1}z)$ have to be nonzero, so that $x\in supp(\alpha)$ and $x^{-1}z\in supp(\beta)$. So finally, $z=xx^{-1}z\in supp(\alpha)supp(\beta)$.