Giles Gardam recently found that Kaplansky’s unit conjecture fails on a virtually abelian torsion-free group: Consider the fours group $P = langle a, b ;|; (a^2)^b = a^{-2}, (b^2)^a = b^{-2} rangle$ (Gardam calls it the Promislow group). This group is a join of two Klein bottle groups and fits in a non-split exact sequence $1 rightarrow mathbb{Z}^3 rightarrow P rightarrow mathbb{Z}_2^2 rightarrow 1$. Now in the group ring $mathbb{F}_2(G)$ we have $pq = 1$ where

$$ p = 1 + aa + aaa + aab + AABA + aaBA + aaBABA + aabABA + aabb + ab + aba + abaBa + ababa + aBAbb + Abb + b + bA + BABA + bABA + BAbb + bb $$

and

$$ q = 1 + a + AA + aaab + aaababb + AAAbb + aaB + AABB + Ab + aB + AbaB + Abab + abaB + abab + AbaBa + Ababa + abaBB + abbb + B + BA + BB, $$

so Kaplansky’s unit conjecture is false.

So an the obvious question is:

What can we now say about the group of units in $mathbb{F}_2(G)$?

I claim no originality in observing that this question can be asked (I saw Avinoam Mann discuss this question on twitter). I gave this five minutes of thought myself and I see that the group is residually finite and contains a copy of $P$ (:P). Two random questions that are particularly interesting to me (but any information is welcome).

Is this group finitely-generated? Is it amenable?

Such questions do not seem obviously impossible on a quick look (for the first, some kind of Gaussian elimination? for the latter, use the fact the group has very slow growth somehow?), but they don’t seem obvious either. I assume there is no information in the literature explicitly about the units in group rings of torsion-free groups, because we didn’t know they can be nontrivial. But people have studied group rings a lot and I don’t have much background on them, so maybe more is known than I was able to see. One fun thing to look at is the following:

What is the isomorphism type of the subgroup $langle p, P rangle$ of the group of units?

I checked that p has order at least $10$, and I suppose it must have infinite order. I checked also that the $3$-ball generated by $p$ and $a$ has no identities, but $p$ and $b$ satisfy some identities.