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?

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    $\begingroup$ 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"). $\endgroup$ Commented Dec 31, 2021 at 11:19

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This paper contains some results on this topic



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