# Kaplansky's idempotent conjecture for Thompson's group F

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 $F$?

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I think F is orderable and hence satisfies this conjecture. –  Mustafa Gokhan Benli Feb 24 at 17:16
Thanks Mustafa! –  Mahdi Teymuri Garakani Feb 25 at 15:02
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?.
A reference for properties of Thompson's group $F$ can be found here.