Giles Gardam recently found (arXiv link) that Kaplansky's unit conjecture fails on a virtually abelian torsion-free group, over the field $\mathbb{F}_2$.
This conjecture asserted that if $\Gamma$ is a torsion-free group and $K$ is a field, then every invertible element in the group ring $K\Gamma$ is a nonzero scalar multiple of a group element.
Consider the fours group
$$ P = \langle a, b, A, B \;|\; a = A^{-1}, b = B^{-1}, (a^2)^b = a^{-2}, (b^2)^a = b^{-2} \rangle $$
(Gardam calls it the Promislow group). This group is a join of two Klein bottle groups and fits in a non-split exact sequence $1 \rightarrow \mathbb{Z}^3 \rightarrow P \rightarrow C_2^2 \rightarrow 1$. 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 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 $3$-sphere generated by $p$ and $a$ has no identities, but $p$ and $b$ satisfy some identities.
Best Answer
Good questions! To bump the discussion of torsion out of the comments: the group of units of $\mathbb{F}_2[P]$ is torsion-free. Suppose we have $q$-torsion and factor $0 = x^q - 1 = (x - 1)(x^{q-1} + \dots + x + 1)$. For $q = 2$ this immediately contradicts the zero divisor conjecture unless $x = 1$. For odd $q$ we use the fact that $x$ must map to $1 \in \mathbb{F}_2$ under the augmentation map to see that $1 + x + \dots + x^{q-1} \neq 0$ and get the contradiction.
Any non-trivial unit in $\mathbb{Z} G$ for $G$ torsion-free (if such a thing exists!) is infinite order by a theorem of Sehgal.
Sehgal, Sudarshan K., Certain algebraic elements in group rings, Arch. Math. 26, 139-143 (1975). ZBL0322.20002.
Overdue update: my paper now has a corollary showing that the group of units has free subgroups and is not finitely generated. Furthermore, Murray has given non-trivial units in all positive characteristics (see arXiv).