Let $K$ be a field and $G$ be a torsion-free group. Kaplansky's idempotent conjecture states that the group ring $K[G]$ does not contain any non-trivial idempotent, i.e. if $x^2=x$ then $x=0$ or $x=1$.


Is Kaplansky's idempotent conjecture known for Thompson's group $F$?



Kaplansky's idempotent conjecture for Thompson's group F
Thompson's group $F$ satisfies the idempotent conjecture, because it is torsion-free and 
orderable. For torsion-free groups it is known that 
the zero-divisor conjecture for group rings implies the idempotent conjecture. Malcev has proved in $1948$ that orderable groups satisfy the zero-divisor conjecture. Hence the claim follows for Thopson's group. For more details see [What is the current status of the Kaplansky zero-divisor conjecture for group rings?](https://mathoverflow.net/questions/79559/what-is-the-current-status-of-the-kaplansky-zero-divisor-conjecture-for-group-ri).   

A reference for properties of Thompson's group $F$ can be found [*here*](http://www.cs.lafayette.edu/~meierj/talkspdfs/ThompsonI.pdf).


