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.

  • $\begingroup$ it might be nice to mention what are $A$ and $B$ (although I'll admit it's a fairly common shortcut) $\endgroup$
    – ARG
    Commented Mar 9, 2021 at 10:04
  • $\begingroup$ I realized that some time after posting but I figured it's not worth bumping the post over. Inverses included in presentation now. $\endgroup$
    – Ville Salo
    Commented Mar 9, 2021 at 12:52
  • 1
    $\begingroup$ Maybe I'm missing a point in your last question, but the zero divisor conjecture (which holds for the group $P$ since it is solvable) implies the idempotent conjecture. If $p$ had finite order, that would contradict the zero divisor conjecture, since $0 = p^n-1 = (p-1)(p^{n-1}+p^{n-2} + \ldots + p^2 + p + 1)$. (Also the convention that a capital letter is an inverse is very misleading there, since $p$ is an element and $P$ is either the inverse of $p$ or the group $P$; but again it's just me being pointy) $\endgroup$
    – ARG
    Commented Mar 9, 2021 at 14:44
  • $\begingroup$ Ok I did not connect the dots there. So on $P$, we can immediately say that the group of units of the group ring is torsion-free. $\endgroup$
    – Ville Salo
    Commented Mar 9, 2021 at 15:09
  • 1
    $\begingroup$ FWIW, I computed 5000 elements of the Gardam group (I mean the one generated by $a,b,p$), always multiplying previous ones by $a,b,p$ and always keeping only the 5000 with minimal ($> 1$) length and secondarily minimal total length (with some arbitrary convention), until this converged. I found $5000$ elements of support $21$, nothing with less. (This is about the exchange between Abdollahi and Gardam, maybe Gardam went further already?) $\endgroup$
    – Ville Salo
    Commented Mar 23, 2021 at 8:04

4 Answers 4


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).

  • $\begingroup$ Very nice! Just to correct my previous comment: assume $x^n=1$ in $\mathbb{F}_p[G]$. WLOG assume $n$ is the first integer to do so. Then if $p|n$, $0=x^n-1=(x^{n/p}-1)^p$ so $x^{n/p}-1$ and some power of $x^{n/p}-1$ form a zero divisor. If $n=2$ then $x^2-1=(x-1)(x+1) =0$ contradicts zero divisor again. If $n$ is $>2$ and coprime to $p$, $0=x^n-1=(x-1)(x^{n-1}+\cdots+x+1)$. Now sending $y \in \mathbb{F}_p[G]$ to the sum of its coefficients $\epsilon(y)$ (augmentation map), one has that $\epsilon(x)^n=1$. Since $\epsilon(x)^p=\epsilon(x)$, then $\epsilon(x)^{n \text{ mod}(p-1)}=1$. So ... $\endgroup$
    – ARG
    Commented Mar 12, 2021 at 13:37
  • $\begingroup$ are there $\epsilon \in \mathbb{F}_p$ with $\epsilon^n=1$, $2<n <p-1$ and $ \epsilon^{n-1} + \cdots + \epsilon + 1 =0$? It's not too hard to check that this is not the case for $p=3$ or 5. But then this fails with $p=7$: one has $n=4$ and $\epsilon(x) = -1$. However one can then exclude 4-torsion using $x^4-1 = (x^2-1)(x^2+1)$, since $\epsilon(x)^2 = -1$ has no solution with $p=7$. So is there a general argument for any $p$? $\endgroup$
    – ARG
    Commented Mar 12, 2021 at 13:46
  • $\begingroup$ An element of a field that is a root of $x^{n-1}+\cdots+x+1-0$ is an $n$th root of 1. So for $n$ a prime that does not divide $p-1$ there won't be any such elements. $\endgroup$
    – IJL
    Commented Mar 12, 2021 at 14:59
  • $\begingroup$ @IJL many thanks, but I found the "correct" argument (which is I believer much simpler); see next comment... $\endgroup$
    – ARG
    Commented Mar 12, 2021 at 22:35
  • $\begingroup$ Assume $x^n=1$ in $K[G]$ for $K$ a field of characteristic $p$. WLOG assume $n$ is minimal. Then if $p|n$, $0=x^n-1=(x^{n/p}-1)^p$ so $x^{n/p}-1$ and some power of $x^{n/p}-1$ form a zero divisor [or contradicts minimality]. Otherwise one has $\epsilon(x)^n = 1$. So look at $x' = x/\epsilon(x)$ (we are in a field after all so $1/\epsilon(x)$ exists). Then $\epsilon(x')=1$ and $x'^n=1$.Then $0=x'^n-1=(x'-1)(x'^{n-1}+\cdots+x'+1)$. Since $\epsilon(x')=1$, one has that $\epsilon(x'^{n-1}+\cdots+x'+1)=n \neq 0$ as $(n,p)=1$. The factorisation of $x'^n -1$ contradicts zero divisor. $\endgroup$
    – ARG
    Commented Mar 12, 2021 at 22:35

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 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.

  • $\begingroup$ nice! I also wrote up a "clean" version of my comments here; the arguments work for any commutative ring (but without the algebraic closure). Which raises two questions: 1- is it easier to get the zero divisor conjecture on the algebraic closure than on the base field? 2- did you mean "Since the degree is positive, $P$ has a root $s$ in $\overline{K}$"? $\endgroup$
    – ARG
    Commented Mar 21, 2021 at 21:04
  • $\begingroup$ @ARG For 2: I need algebraically closed in Lemma 1, it disappeared by some cut-paste accident (it's clearly false otherwise, just considering a nontrivial finite extension). I corrected $\endgroup$
    – YCor
    Commented Mar 21, 2021 at 21:15
  • $\begingroup$ @ARG As regards your question 1: of course the zero divisor conjecture is stronger when the algebra is larger. So I'm just assuming something a bit stronger than it for $KG$. In the known instances, the proof might be the same for all fields. Also, for given $G$, the zero divisor (resp. idempotent) conjecture for every field is equivalent to the zero divisor (resp. idempotent) conjecture for every finite field, by a simple classical approximation argument. $\endgroup$
    – YCor
    Commented Mar 21, 2021 at 21:20
  • $\begingroup$ For those who are interested in the details of "finite fields $\implies$ all fields" see this other post $\endgroup$
    – ARG
    Commented Mar 22, 2021 at 9:57
  • $\begingroup$ @ARG here are the details in a more concise way. Suppose the result known for all finite fields. Let $K$ be a domain and suppose that $KG$ fails one of the conjectures (has a idempotent, zero divisor, unit, nilpotent...). Passing to a finitely generated subdomain we can suppose $K$ finitely generated. Since every f.g. domain is residually a finite field, taking a suitable quotient we get a finite field $F$ with a counterexample of the same conjecture for $FG$. $\endgroup$
    – YCor
    Commented Mar 22, 2021 at 10:01

Non-trivial torsion in the group of units of $K[G]$ contradicts the zero-divisor conjecture for $K[G]$. Here non-trivial means "which does not come from $K$ itself".

This answer rather answers a question of Giles in the comments and is essentially an expansion of the comment to Giles' answer; the aim is to show that if a commutative ring $K$ (of any characteristic!) is so that $K[G]$ satisfies the zero divisor conjecture, then there is no element $x \in K[G]$ such that $x^n = 1$ for $n>1$, unless $x \in K e_G$ (an element supported solely at the identity element of $G$; i.e. the ring $K$ itself has an element of $n$-torsion).

So assume $x^n=1$ and $x \notin K e_G$, and take $n$ to be the smallest integer to do so. There are two cases:

Case 1: if $K$ is of characteristic $p \neq 0$ and $p$ divides $n$, $0=x^n-1=(x^{n/p}-1)^p$ (Frobenius map; i.e. the binomial coefficients of a prime $p$ are all divisible by $p$). It cannot happen that $x^{n/p}-1 =0$ (since it would contradict the minimality of $n$). This means that $x^{n/p}-1$ and some power of $x^{n/p}-1$ form a zero divisor.

Case 2: if $p$ does not divide $n$ or $K$ is of characteristic 0, consider the augmentation map $\epsilon:K[G] \to K$... if $x = \sum_{g \in G} x_g g$ then $\epsilon(x) = \sum_{g \in G} x_g$. Since $\epsilon$ is a ring homomorphism, $\epsilon(x)^n = 1$ (so $\epsilon(x)^{n-1}$ is a multiplicative inverse of $\epsilon(x)$). Consider the element $x' = x \epsilon(x)^{n-1}$ Note that $x' \neq 1$, since that would mean that $x \in K e_G$. Then $\epsilon(x') = \epsilon(x) \epsilon(x)^{n-1}=1$ and $x'^n= x^n \epsilon(x)^{n(n-1)}=1$ (this uses commutativity of the ring $K$). Then $0=x'^n-1=(x'-1)(x'^{n-1}+\cdots+x'+1)$. Since $\epsilon(x')=1$, one has that $\epsilon(x'^{n-1}+\cdots+x'+1) = \epsilon(x')^{n-1} + \cdots + \epsilon(x') + 1 =n \neq 0$ as either $n$ is not divisible by $p$ or $K$ has characteristic 0. In particular $x'^{n-1}+\cdots+x'+1 \neq 0$ and $x'-1 \neq 0$. But $(x'-1)(x'^{n-1}+\cdots+x'+1) =0$ form a zero divisor.

Remark 1: Note that if $K[G]$ has no zero divisors, then any element $x$ such that $x^n = x^k$ for $n \neq k$, is a torsion element. Indeed, $0 = x^n-x^k = x^k (x^{n-k} -1)$. Since $x^k \neq 0$, this means that $x^{n-k}=1$.

Remark 2: I guess this is all well-known, but for some reason I could only find references to the case $n=2$ (which is much easier since $x^2-1 = (x-1)(x+1)$ and $x \pm 1$ is never 0 [unless $x = \pm e_G$]).

  • $\begingroup$ This looks correct. So it seems your original argument only needed a little bit of augmentation to deal with all commutative rings. $\endgroup$
    – Ville Salo
    Commented Mar 15, 2021 at 6:11
  • $\begingroup$ @VilleSalo a bit of augmentation and a bit of Frobenius :D $\endgroup$
    – ARG
    Commented Mar 15, 2021 at 8:47
  • $\begingroup$ @VilleSalo it's not really "all commutative rings" but rather "all domains". Indeed the zero divisor conjecture for fields implies it for domains, and conversely $RG$ has zero divisors when $R$ is not a domain. $\endgroup$
    – YCor
    Commented Mar 22, 2021 at 9:20
  • $\begingroup$ So we agree that ARG shows that "no zero divisors in $RG$ implies no torsion in $RG$" holds in all commutative rings (as I assume I meant with my comment), and you're saying it's silly to include non-domains, since the implication is trivial for them? $\endgroup$
    – Ville Salo
    Commented Mar 22, 2021 at 9:26
  • $\begingroup$ Actually the main point of my comment was probably to make the augmentation pun. $\endgroup$
    – Ville Salo
    Commented Mar 22, 2021 at 9:28

Let me describe how I would try amenability. I'll make it in two steps (the first step is fine; the second is more of a work in progress) by going through the [possible] amenability of the monoid $R[G] \setminus \{0\}$. I discuss below why I think this strategy seems fine.

The first step is relatively straightforward. Let me call a partially-absorbing element of a monoid $M$ an element $x$ such that for some $m \in M$ with $m \neq 1$, $xm = x$. Note that if the ring $R[\Gamma]$ has no zero divisors, there cannot be such an element except 0 (since $xm = x$ implies $x(m-1) =0$.

Lemma 1: let $M$ be an [right-]amenable monoid without partially-absorbing elements and $G$ be a subgroup of $M$. Then $G$ is amenable.

Proof: Let $\mu$ be a right-invariant mean on $\ell^\infty(M)$, that is $\forall m \in M, \forall f \in \ell^\infty(M)$ one has $\mu(\rho_mf) = \mu(f)$ (here $\rho_mf(x) := f(xm)$). Now consider the right-$G$-cosets of $M$, and choose representatives $m_i$ so that $\{m_iG\}$ covers $M$. (Note that being in the same $G$-coset is an equivalence relation because $G$ is a group; also note that the cosets are "naturally" identified with $G$ since there are no partially-absorbing elements).

Define a mean $\mu_G$ on $G$ as follows. For $f\in \ell^\infty(G)$ consider $f_M \in \ell^\infty(M)$ by $f_M(m) = f(g)$ where $m \in Gm_i$ so that $m = m_ig$. (Basically $f_M$ is a copy of $f$ on each $G$-coset.) Define $\mu_G(f) = \mu(f_M)$. Note that $\rho_gf_M(x) = f_M(xg)$. Then $\mu_G(\rho_gf) = \mu(\rho_gf_M) = \mu(f_M) = \mu_G(f)$. From there $G$ has an invariant mean and is consequently amenable (on the right, but then also on the left)$\qquad \square$

Remark: It follows from Lemma 1 that if $R[\Gamma] \setminus \{0\}$ is amenable [and satisfies the zero divisor conjecture] then $\Gamma$ is amenable (because $\Gamma$ is embedded in $R[\Gamma] \setminus \{0\}$ by sending $\gamma$ to the element solely supported on $\gamma$ and with coefficient 1.

Now the next "step" (the amenability of $R[G] \setminus \{0\}$) should be well-known (or known to be open? or perhaps well-known to be false? ). I posted it as a separate question.

Before moving on, let me just point out why this strategy does not seem so bad. Start with $\Sigma_0 =\Gamma$ (embedded as Dirac masses), then consider the monoid $\Sigma_1$ which is generated by elemetns of the form $1+\gamma$, then consider the monoid $\Sigma_2$ which is generated by elements with support of size $\leq 3$, etc. This gives a step by step process which seems easier to tackle (either in postive [amenable] or negative [non-amenable]). One might think that this brings too many elements, but the conjugate of the unit element $p$ under $\Gamma$ has an infinite orbit. So it seems likely that any positive [or negative] argument for the monoid will transfer to the group of units. The obvious argument against this strategy, is that there are many more tools to conclude of the amenability of groups, than of monoids.

Problem 2: Assume $\Gamma$ is a countable amenable group, $R$ is a finite field and that the group ring over $R[\Gamma]$ satisfies the zero divisor conjecture. Show that the multiplicative monoid $M = R[\Gamma] \setminus \{0\}$ is a [right-]amenable monoid.

Applying a positive answer to Problem 2 would show that $R[\Gamma] \setminus \{0\}$ is an amenable monoid, and then applying Lemma 1 to the group of units of $R[\Gamma]$ shows it is an amenable group.

[EDIT: as updated in Gilles' answer above, the group of unit is not amenable. So the interesting problem would either be: 1- is there any non-abelian group for which the multiplicative monoid (without 0) is amenable? 2- which $\Sigma_i$ is the first to be non-amenable?]

In support of the amenability of $\Sigma_1$ (the monoid generated by the Dirac masses and the elements of the from $1+\gamma$). Each element of $\Sigma_1$ can be written in the form $\gamma \prod_{\eta \in S} (1+ \eta)$ where $S \in (\Gamma)^N$ (is an ordered $N$-tuple). The element $\prod_{\eta \in S} (1+ \eta)$ is closely related to $\prod_{\eta \in S} \eta$. Indeed it will be (at most) supported on all the possible partial products of $\prod_{\eta \in S} \eta$. So this second part of the semigroup is very close to $\Gamma$ itself. The difference with $\Gamma$ can be "measured" by how the different ways of writing an element $h$ as a product so that the partial products are distinct. In some sense, this makes it sound that groups which are far from being Abelian should have a non-amenable $\Sigma_1$. The question is then whether virtually-Abelian groups are far enough from being Abelian.

Note: Bartholdi has a paper about amenable $K$-algebras, but the notion of amenability he uses does not match with the amenability of the multiplicative monoid $K[G] \setminus \{0\}$. He uses the sequences $F_i$ (which clearly don't work here) in his paper.

  • $\begingroup$ I don't understand Lemma 2. For instance if $F_{i,k}$ happens to be just the set of all elements with support contained in $B_i$ (is that possible?) then it seems the "boundary effect" is quite big: almost all elements will have something on the support (even if only a small proportion of the "entropy" is generated there). $\endgroup$
    – Ville Salo
    Commented Mar 14, 2021 at 19:01
  • $\begingroup$ Indeed the estimates are quite hard to go through ... (and I do not claim to have done them). I'd be really surprised if $F_{i,k}$ would coincide with $F_i$ (especially since $i >> k$). But if the boundary effect is what bothers you here is another variant. Take $F_{i,j,k}$ to be the all the translates (in the sense of the usual action/embedding of $\Gamma$ in $R[\Gamma]$) of the elements of $F_{j,k}$ so that the support still lies in $B_i$. If $i >> j$ and $j$ is much larger than the support of any of the first $k$ elements in your enumeration, it "should" be fine. $\endgroup$
    – ARG
    Commented Mar 14, 2021 at 21:32
  • $\begingroup$ @VilleSalo Come to think of it, there might be another variant which is easier to compute. Let $\Pi_{k,n}$ to be the elements which can be written as products of the $k$ first in your enumeration, with at most $n$ uses of each of these elements. Then let $\Phi_{i,k,n}$ to be the translates of these elements so that the support lies in $B_i$. Then $|\Phi_{i,k,n} \cap \Phi_{i,k,n} m| \geq |$all those products with at most $n-1$ appearances of $m|$ and the boundary effect is also negligible. $\endgroup$
    – ARG
    Commented Mar 14, 2021 at 22:15
  • $\begingroup$ Quite possibly I just have a point in every open set, but I'm not really seeing these. One overarching problem is that everything you say seems to work just as well for the monoid of $2$-by-$2$-matrices over this group ring, and I know that one is not amenable. $\endgroup$
    – Ville Salo
    Commented Mar 15, 2021 at 6:01
  • $\begingroup$ @VilleSalo So there is clearly one thing which does not go through for the monoid of $2\times2$ matrices. Namely look at $\big( \begin{smallmatrix} m_{11} & m_{12} \\ m_{21} & m_{22} \end{smallmatrix} \big)$ then, even if the support of these elements is contained in $B_i$, by multiplying with a simple matrix like $m=\big(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \big) $ one gets that the resulting elements have support in $(B_i)^2$ and not $B_i S$ for a set $S$ which only depends on the element $m$. So I'm absolutely sure my arguments fails dramatically for this monoid. $\endgroup$
    – ARG
    Commented Mar 15, 2021 at 8:46

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