Let $\mathbb{Q}$ be the rationals and $G$ a group. Then we consider the group ring $\mathbb{Q}[G]$.
Since the operation on $\mathbb{Q}[G]$ restricted to $G$ is just the group operation, I know that $G$ is a subgroup of the units of $\mathbb{Q}[G]$.
How can we describe all units of $\mathbb{Q}[G]$?
For a general group $G$ this is a hard problem.
Let me mention the partial progress that I remember.
For one thing, if $G$ isn't torsion-free, there is a theorem in Passman's Algebraic structure of group rings that says that the units of $F[G]$ for a field $F$ are trivial iff $G$ is $F_2[C_2]$, $F_2[C_3]$ or $F_3[C_2]$. ($F_i$ denotes a field of order $i$ and $C_i$ the cyclic group order $i$.)
So attention is usually restricted to torsion-free groups, and the open conjecture is the unit conjecture that asks if $F[G]$ has trivial units for any field $F$ and any torsion-free group $G$.
Two good resources I can recommend are Passman's Algebraic structure of group rings and Lam's First course on noncommutative rings I think has some information on this. I don't have copies of Milies' group ring books, but I remember when I consulted them they were very good too.
Finally, through this MO post I found a very interesting slide set for a talk about the difficulty of the conjecture, which looks very good.