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After Gardam had found a counterexample to the Kaplansky unit conjecture for the group ring $K[G]$ with $K = \mathbf F_2$ and $G = P$, the Promislow group, Murray extended this to $\mathbf{F}_p[P]$ for all primes. Of course, it would be great to refute the conjecture in the complex case too but the proof methods are susceptible to positive characteristic.

However, I can prove using standard model-theoretic trickery that if for all (or at least infinitely many) primes we could get examples of non-trivial units in $\mathbf{F}_p[P]$ whose supports have uniformly bounded cardinalities, then one can cook up a complex counterexample.

The problem is that the counterexamples found by Murray lack this property as they contain the factor $(1-(ab)^2)^{p-2}$ where $a, b$ are the standard generators of $P$. So I would like to get my head around it:

Are there any obvious obstacles preventing the existence of non-trivial units in $\mathbf{F}_p[G]$ whose supports have uniformly bounded cardinality regardless of $p$?

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    $\begingroup$ Isn’t it the case that any non-trivial unit in the group ring over $\mathbb{Z}$ would reduce to such a unit for all but finitely many $p$? If so, then it seems like a positive answer to your question would be close to a proof of the unit conjecture for $\mathbb{Z}[P]$. $\endgroup$
    – HJRW
    Commented Jul 24, 2021 at 18:56
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    $\begingroup$ The short answer is no, there isn't. You are indeed right that bounded support as $p$ grows is exactly what is needed to solve this problem, at least over something like $\mathbb{Q}$, if not over $\mathbb{Z}$. $\endgroup$
    – David A. Craven
    Commented Jul 27, 2021 at 20:47

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There are no such obstacles a posteriori since the complex unit conjecture is false.

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