If $R$ is an integral domain (I am having $\mathbb{Z}$ or a field in mind) and $G$ a (not necessarily finite) group then we can form the group ring $R(G)$.
Note that if $g^{n+1} = e$ then $(e-g)(e+g\ldots + g^n) = e - g^{n+1} = 0$. This means if $G$ has torsion then $R(G)$ always has zero-divisors.
What about the inverse? So if $G$ is torsion-free does that imply $R(G)$ having no zero-divisors.