# Kaplansky Idempotent conjecture and Extension theory

We consider the Idempotent Kaplansky conjecture with $\mathbb{C}$- coefficients, that is the problem of nontrivial idempotents for group algebra $\mathbb{C}\Gamma$ where $\Gamma$ is a torsion free group.

Assume that $G_{1}$ and $G_{2}$ are torsion free groups which satisfies this conjecture. Assume that we have a short exact sequence of groups as $0\to G_{1}\to G_{3} \to G_{2} \to 0$. Does this implies that $G_{3}$ satisfies the Kaplansky conjecture? What about the particular case "semidirect product?(One can consider the same question for Kadison conjecture)

A possible negative answer is difficult as the original Kaplansky conjecture. But this is a reasonable problem if it would have a possible affirmative answer.

There are some indirectly related obstructions and question:

1.A group extension does not give a group algebra extension. But we have a complex of group algebras:$$0\to\mathbb{C} G_{1}\to \mathbb{C}G_{3} \to \mathbb{C}G_{2} \to 0$$ so the cohomology gives us a new algebra associated to a given group extension.

2.There are some known properties about torsion free groups which guarantees the Kaplansky conjecture. We collect such properties in a set $S$. A group which satisfies at least one of these properties, is called an $S$ group. . Is the collection $S$ closed under group extension?(Or at least closed under semidirect product?) Namely: Is an extension of an $S$ group by an $S$ group, an $S$-group?

• Regarding point 2: it's not hard to show that the class of unique product groups is closed under extensions (see Lemma 13.1.8 in Passman's book "The Algebraic Structure of Group Rings"). Commented Dec 31, 2021 at 11:19