TITLE: Units in the group ring over fours group after Gardam
QUESTION [21 upvotes]: 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.

REPLY [4 votes]: This is an extended comment, follow-up to Giles Gardam's answer and the comments therein. Recall that the group of nonzero scalars $K^\times$ is a central subgroup of $(KG)^\times$ (for each group $G$ and field $K$).

Proposition. Let $G$ be a group and $K$ a field, such that $\bar{K}G$ has no zero divisor. Then for every field $K$, the group $(KG)^\times/K^\times$ is torsion-free. In particular, every torsion element in $(KG)^\times$ is a scalar.

Here $K\subset \bar{K}$ is an algebraic closure. For arbitrary fields, the assumption on $G$ is satisfied by torsion-free virtually solvable groups (and hence residually torsion-free solvable groups, such as free groups and many more).
Lemma 1.  Let $K$ be an algebraically closed field. Let $A$ be a $K$-algebra (=unital associative $K$-algebra) with no zero divisors. If $P\in K[t]\smallsetminus\{0\}$ and $P(x)=0$ for $x\in A$ then $x$ is a scalar.
Proof: just consider the unital $K$-subalgebra generated by $x$. It is commutative, finite-dimensional over $K$, and has no nonzero divisor, hence is a field, and hence is 1-dimensional since $K$ is algebraically closed.
In turn this implies:
Lemma 2. Let $G$ be a group and $K$ a field for which $\bar{K}G$ has no zero divisor. Then for every nonzero polynomial $P\in K[t]$, every element $x$ in $KG$ such that $P(x)=0$ for some nonzero polynomial $P$, is a scalar.
Proof: by Lemma 1, $x$ is a scalar in $\bar{K}G$, and hence is a scalar in $KG$.
Lemma 2 in turn, applied to $P(t)=t^n-\lambda$ for any fixed $\lambda\in K$, immediately implies the proposition.

Edit: the second sentence of the proposition can be restated as: if $K$ is a domain and $\overline{\mathrm{Frac}(K)}G$ has no zero divisors, then every torsion element in $(KG)^\times$ is a scalar. This is obtained in ARG's answer just assuming that $KG$ itself has no zero divisors.