The Group Ring Problems: Units, Zero Divisors, Idempotents and J-Semisimplicity

Q: Where exactly is torsion-freeness used in (6.21)?

It is not, directly. (6.21) assumes only that k is nonzero, that G is nontrivial, and in part (1) that k is reduced with G free of elements of order 2. Torsion-freeness enters when the proposition is applied: it supplies the no-element-of-order-2 hypothesis and rules out the exceptional case.

Q: What is a unique product group and why does it help?

A group in which any two finite nonempty subsets A, B admit an element of AB with exactly one factorisation ab. That element then has coefficient a_gb_g ≠ 0 in the product of two group ring elements with those supports, so no zero divisors arise. Orderable groups are unique product groups; Promislow's group is torsion-free but is not, which is where the counterexample came from.

Q: How does the idempotent question fit in?

A domain has only the idempotents 0 and 1, so Problem D implies the idempotent conjecture. The converse is not known. The idempotent question has an analytic life of its own as the Kadison-Kaplansky conjecture for reduced group C^*-algebras, where it is a consequence of the Baum-Connes conjecture.

Executive Summary

Fix a domain $k$ and a torsion-free group $G$ . Four questions about $A = k G$ have driven the theory of group rings for seventy years: are all units trivial (Problem U), is $A$ reduced (Problem R), is $A$ a domain (Problem D), and is $A$ J-semisimple (Problem J)? Torsion-freeness is not optional: an element of finite order manufactures zero divisors immediately.

The implications known are $U \Rightarrow R$ , $U \Rightarrow J$ and $R \Leftrightarrow D$ . Of these, $D \Rightarrow R$ is a triviality and $R \Rightarrow D$ is a hard theorem resting on Passman's $Δ$ -methods. The first two come from one elementary but genuinely clever proposition, which is given in full below together with its single exceptional case, $| k | = | G | = 2$ .

4Problems

$| G | = 5$ First nontrivial units in

ℤ G

1Exceptional case in (6.21)

2021Unit conjecture refuted

Overview

In any group ring $k G$ the elements $a g$ with $a \in U (k)$ and $g \in G$ are units, with inverse $a^{- 1} g^{- 1}$ . These are the trivial units, and they form a group isomorphic to a semidirect-style extension of $U (k)$ by $G$ . The interesting question is whether there are others.

There usually are. Already in the integral group ring of the cyclic group of order five one finds nontrivial units, whereas for every group with at most four elements $ℤ G$ has only trivial units. Torsion is the culprit, and once torsion is banned the situation changes character completely: no example of a nontrivial unit in $k G$ was known for $k$ a domain and $G$ torsion-free until 2021, and none is known in characteristic zero even now.

(x - 1) (x^{n - 1} + x^{n - 2} + \dots + x + 1) = x^{n} - 1 = 0 (x \in G of order n > 1),

(6.18a)

Why torsion-freeness is a necessary hypothesis for Problem D: an element of finite order gives zero divisors in any $k G$ .

Problem J extends the J-semisimplicity theme of this chapter from fields to domains. Note that the $p'$ -hypothesis needed in Group Rings in Characteristic p disappears here, because torsion-freeness already implies $G$ is a $p'$ -group for every prime $p$ .

Learning Objectives

Distinguish trivial from nontrivial units and produce an explicit nontrivial unit in $ℤ C_{5}$ .
State Problems U, R, D and J with their hypotheses on $k$ and $G$ .
Prove that torsion in $G$ obstructs Problem D.
Prove $(6.21)$ : trivial units imply reduced, and imply J-semisimple outside one exception.
Draw the implication diagram $(6.20)$ and say which arrow is deep.
Name the classes of groups for which the problems are settled, and the counterexample that settled Problem U negatively.

Definitions

Definition(6.16d)Trivial units

Let $k$ be a nonzero ring and $G$ a group. A unit of $k G$ is trivial if it has the form $a g$ with $a \in U (k)$ and $g \in G$ ; equivalently, if its support is a single group element and the coefficient there is a unit of $k$ . All other units are nontrivial. Writing $U (k) \cdot G$ for the group of trivial units, Problem U asks whether $U (k G) = U (k) \cdot G$ .

$supp (α)$: The set of $g \in G$ with $a_{g} \neq 0$ , for $α = \sum a_{g} g$ . It is finite by definition of the group ring, and $supp (α β) \subseteq supp (α) supp (β)$ .
Reduced: No nonzero nilpotents; for a ring it suffices to check $α^{2} = 0 \Rightarrow α = 0$ .
Domain: A nonzero ring with no zero divisors; not assumed commutative.
Torsion-free: Only the identity has finite order. Implies $G$ is a $p'$ -group for every prime $p$ .
Idempotent: $e^{2} = e$ . In a domain the only idempotents are $0$ and $1$ , so Problem D subsumes the idempotent question.

Standing hypotheses for Problems U, R, D and J: $k$ is a domain and $G$ is a torsion-free group. $(6.21)$ is proved under much weaker assumptions and is stated accordingly.

Core Concepts

Where nontrivial units come from

Torsion produces units by a norm trick. If $x$ has order $n$ then $\hat{x} = 1 + x + \dots + x^{n - 1}$ satisfies ${\hat{x}}^{2} = n \hat{x}$ , and combinations of $\hat{x}$ with group elements can be inverted when the arithmetic cooperates. The classical Bass cyclic units and the bicyclic units of $ℤ G$ are built this way, and all of them evaporate when $G$ is torsion-free.

This is the intuition behind Problem U: with no elements of finite order there seems to be no mechanism for producing a unit with support of size greater than one. The intuition survived sixty years of scrutiny and is nevertheless false in characteristic $2$ .

Why reducedness is the flexible hypothesis

Problem R sits between U and D because reduced is a condition on single elements — $α^{2} = 0$ implies $α = 0$ — whereas domain is a condition on pairs. Single-element conditions can be attacked by the unit trick: if $α^{2} = 0$ then

(1 - α) (1 + α) = 1 - α^{2} = 1 = (1 + α) (1 - α),

(6.21a)

A square-zero element yields a unit, and knowledge of the units then constrains $α$ severely.

Upgrading a statement about single elements to a statement about pairs is exactly the content of the implication $R \Rightarrow D$ , and it needs the $Δ$ -subgroup machinery developed on the page *FC Groups, the $Δ$ -Subgroup and Trace Methods*.

The idempotent question

A fifth question rides along: are $0$ and $1$ the only idempotents of $k G$ ? A domain has no nontrivial idempotents, so Problem D implies it. Kaplansky proved unconditionally that for a subfield $k$ of $ℂ$ and any idempotent $e \in k G$ , the trace $tr (e)$ — the coefficient of $1$ — is a real number with $0 \leq tr (e) \leq 1$ , equal to $0$ only for $e = 0$ and to $1$ only for $e = 1$ . Zalesskii sharpened this to rationality of the trace in characteristic zero and membership in the prime field in characteristic $p$ .

Key Results

Throughout $(6.16)$ – $(6.19)$ , $k$ is a domain and $G$ is a torsion-free group.

Problem(6.16)Problem U

Is every unit of $k G$ trivial, that is, of the form $a g$ with $a \in U (k)$ , $g \in G$ ?

Problem(6.17)Problem R

Is $k G$ always reduced?

Problem(6.18)Problem D

Is $k G$ always a domain?

Problem(6.19)Problem J

If $G \neq {1}$ , is $k G$ J-semisimple, that is, $rad k G = 0$ ?

Remark(6.20)Known implications

$U \Rightarrow R$ , $U \Rightarrow J$ , and $R \Leftrightarrow D$ . Here $D \Rightarrow R$ holds because a domain has no nonzero nilpotents; the converse $R \Rightarrow D$ is a substantial theorem. The first two implications follow from $(6.21)$ below, which assumes far less than $k$ a domain and $G$ torsion-free.

Proposition(6.21)Consequences of having only trivial units

Let $k \neq 0$ be a ring and $G \neq {1}$ a group such that $A = k G$ has only trivial units. Then:

if $k$ is reduced and $G$ has no element of order $2$ , then $A$ is reduced;
$A$ is J-semisimple, except when $| k | = | G | = 2$ .

Proof

(1) It suffices to show $α^{2} = 0 \Rightarrow α = 0$ . From $α^{2} = 0$ , $(1 - α) (1 + α) = (1 + α) (1 - α) = 1$ , so $1 - α \in U (A)$ and by hypothesis $1 - α = a g$ for some $a \in U (k)$ , $g \in G$ .

Suppose $g \neq 1$ . Then $α = 1 - a g$ and $0 = α^{2} = 1 - 2 a g + a^{2} g^{2}$ . Since $G$ has no element of order $2$ we have $g^{2} \neq 1$ , and $g \neq 1$ , so the identity element occurs in the support of the right-hand side only through the leading $1$ ; its coefficient is $1 \neq 0$ in $k$ , contradicting $0$ . Hence $g = 1$ and $α = 1 - a \in k$ . Now $α^{2} = 0$ inside the reduced ring $k$ forces $α = 0$ .

(2) First the exception. If $| k | = | G | = 2$ , say $k = 𝔽_{2}$ and $G = ⟨ g ⟩$ of order $2$ , then $U (A) = {1, g} = G$ , so $A$ does have only trivial units, yet $rad A = {0, 1 + g} \neq 0$ .

Now assume we are not in that case, and let $α \in rad A$ . Then $1 - α \in U (A)$ , so $1 - α = a g$ with $a \in U (k)$ , $g \in G$ . Suppose $g \neq 1$ .

If $| k | \geq 3$ , pick $b \in k ∖ {0, 1}$ . Since $rad A$ is an ideal, $α b \in rad A$ and $1 - α b \in U (A)$ . But $α b = (1 - a g) b = b - (a b) g$ , so $1 - α b = (1 - b) + (a b) g$ has support ${1, g}$ with both coefficients nonzero — a nontrivial unit, contradiction.
If $| G | \geq 3$ , pick $h \in G ∖ {1, g^{- 1}}$ . Then $α h \in rad A$ and $1 - α h = 1 - h + a g h$ is a unit whose support contains the distinct elements $1$ and $h$ — again a nontrivial unit, contradiction.

Outside the excluded case one of $| k | \geq 3$ , $| G | \geq 3$ holds, so $g = 1$ and $α = 1 - a \in k$ . Finally pick any $h \neq 1$ in $G$ (possible as $G \neq {1}$ ). Then $1 + α h$ is a unit with support inside ${1, h}$ and coefficient $1$ at the identity; triviality forces the coefficient $α$ at $h$ to vanish. Hence $α = 0$ and $rad A = 0$ . $■$

Corollary(6.21c)U implies everything

Let $k$ be a domain and $G \neq {1}$ a torsion-free group such that $k G$ has only trivial units. Then $k G$ is reduced, is a domain, has no idempotents other than $0$ and $1$ , and is J-semisimple.

Proof

A domain is reduced and torsion-freeness excludes elements of order $2$ , so $(6.21) (1)$ gives reducedness. The exceptional case $| k | = | G | = 2$ cannot occur, since a group of order $2$ has torsion, so $(6.21) (2)$ gives $rad k G = 0$ . Reducedness upgrades to domain by the implication $R \Rightarrow D$ , proved by Passman's $Δ$ -method, and a domain contains no idempotents besides $0$ and $1$ . $■$

Remark(6.21r)Status of the problems today

Problem U has a negative answer: in 2021 Gardam exhibited a nontrivial unit in $𝔽_{2} P$ , where $P$ is the torsion-free crystallographic group of Promislow — a group already known to fail the unique product property. Further counterexamples in positive characteristic have followed. Problems R, D and J remain open in general, and Problem U itself is open over fields of characteristic zero and over $ℤ$ .

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Turn nilpotence into a unit

$α^{2} = 0$ makes $1 - α$ a unit with explicit inverse $1 + α$ . Any hypothesis controlling units then controls nilpotents. The same trick reads $rad$ as a source of units, since $1 + rad A \subseteq U (A)$ .

Move 2

Perturb to break triviality

Given a supposed trivial unit $1 - α = a g$ with $g \neq 1$ , multiply $α$ by a scalar or a group element. The result stays in the radical, so it is still a unit, but its support has size two — contradiction. Both perturbations need room, which is why $| k | = | G | = 2$ escapes.

Move 3

Count the support

Almost every elementary group-ring argument is bookkeeping on supports: an identity in $k G$ is a family of identities in $k$ , one for each group element, and a group element in the support of a product cannot be cancelled unless something else lands on it.

The perturbation move deserves emphasis. It shows why the exceptional case is genuinely exceptional rather than an artefact: over $𝔽_{2}$ with $| G | = 2$ there is no scalar to perturb by and no third group element to translate by, and indeed the conclusion is false there.

Worked Example

A nontrivial unit in $ℤ C_{5}$

Let $G = ⟨ x ⟩$ be cyclic of order $5$ and work in $ℤ G$ . Put

α = 1 - x^{2} - x^{3}, β = 1 - x - x^{4} .

(E.1)

Expanding $α β$ gives nine terms, with exponents reduced modulo $5$ : $1$ from $1 \cdot 1$ ; $- x$ and $+ x$ from $1 \cdot (- x)$ and $(- x^{2}) (- x^{4}) = x^{6} = x$ ; $- x^{2}$ and $+ x^{2}$ from $(- x^{2}) \cdot 1$ and $(- x^{3}) (- x^{4}) = x^{7} = x^{2}$ ; $- x^{3}$ and $+ x^{3}$ from $(- x^{3}) \cdot 1$ and $(- x^{2}) (- x)$ ; $- x^{4}$ and $+ x^{4}$ from $1 \cdot (- x^{4})$ and $(- x^{3}) (- x)$ . Everything cancels except the identity:

α β = β α = 1 .

(E.2)

Both $α$ and $β$ have support of size three, so these are nontrivial units of $ℤ C_{5}$ .

Only trivial units for $| G | \leq 4$

Take $G = ⟨ g ⟩$ of order $2$ . Then $ℤ G ≅ ℤ [t] / (t^{2} - 1)$ , and $a + b g \mapsto (a + b, a - b)$ identifies it with

{(u, v) \in ℤ \times ℤ : u \equiv v (\mod 2)} .

(E.3)

A unit must map to a pair of units of $ℤ$ , so $u, v \in {\pm 1}$ : four possibilities, giving $\pm 1$ and $\pm g$ . All four are trivial. The groups of order $3$ and $4$ are handled the same way, and Higman's theorem explains the pattern: for finite $G$ , $ℤ G$ has only trivial units precisely for the abelian groups of exponent $1, 2, 3, 4$ or $6$ together with the Hamiltonian $2$ -groups.

The torsion-free model case

Let $k$ be a domain and $G = ⟨ x ⟩$ infinite cyclic, so $k G = k [x, x^{- 1}]$ . Writing a nonzero element as $a_{m} x^{m} + \dots + a_{n} x^{n}$ with $m \leq n$ and $a_{m}, a_{n} \neq 0$ , the product of two such elements has lowest term $a_{m} b_{m'} x^{m + m'}$ and highest term $a_{n} b_{n'} x^{n + n'}$ , both with nonzero coefficients because $k$ is a domain. Hence $k G$ is a domain, and $α β = 1$ forces $m = n$ and $m' = n'$ : only trivial units. This is the prototype for the ordered-group argument.

The exception in $(6.21) (2)$

With $k = 𝔽_{2}$ and $G = ⟨ g ⟩$ of order $2$ , $A = 𝔽_{2} G$ has four elements $0, 1, g, 1 + g$ . Since $(1 + g)^{2} = 1 + g^{2} = 0$ , the element $1 + g$ is nilpotent, so $U (A) = {1, g}$ — all trivial — while $rad A = {0, 1 + g} \neq 0$ .

Frameworks and Models

The classes of groups for which the problems are settled nest inside one another. Each inclusion is strict, and the outermost gap is where the counterexample was eventually found.

Torsion-free groupsThe standing hypothesis; Problems U, R, D, J all open in general

Unique product groupsSome product

a b

A B

is uniquely represented; gives U and D at once

Orderable groupsA bi-invariant total order; leading and trailing terms survive multiplication

Torsion-free abelianOrderable by Levi's theorem

Free groupsOrderable by Birkhoff, Iwasawa and Neumann

Beyond orderability, positive results have come from very different directions: Kropholler, Linnell and Moody proved the zero-divisor conjecture for torsion-free elementary amenable groups over fields of characteristic zero, using $K$ -theoretic induction rather than combinatorics of supports.

Comparison and Classification

The four problems compared
Problem	Assertion for $k$ a domain, $G$ torsion-free	Status
U $(6.16)$	$U (k G) = U (k) \cdot G$	False in general — Gardam 2021, char $2$ ; open in char $0$
R $(6.17)$	$k G$ has no nonzero nilpotents	Open
D $(6.18)$	$k G$ has no zero divisors	Open; equivalent to R
J $(6.19)$	$rad k G = 0$ for $G \neq {1}$	Open
I	only idempotents are $0$ and $1$	Open; implied by D

What is known for particular classes of torsion-free groups
	U	R and D	J
Ordered groups	yes	yes	yes
Unique product groups	yes	yes	yes
Torsion-free abelian	yes	yes	yes
Free groups	yes	yes	yes
Torsion-free elementary amenable, char $0$	partial	yes	partial
Promislow's group $P$ over $𝔽_{2}$	no	partial	partial

What is known for particular classes of torsion-free groups

In the last row, U fails by Gardam's counterexample while R, D and J for that group are not settled by it — a nontrivial unit is not a zero divisor.

Relationship Map

The implication diagram $(6.20)$ , with the difficulty of each arrow marked.

U — all units trivial — the strongest of the four
- $\Rightarrow$ R by $(6.21) (1)$
  - needs $k$ reduced and no element of order $2$
- $\Rightarrow$ J by $(6.21) (2)$
  - needs $G \neq {1}$ and excludes $| k | = | G | = 2$
- $\Rightarrow$ D via R
  - through the deep implication R $\Rightarrow$ D
  - hence also no nontrivial idempotents

RDno nontrivial idempotents

The reverse arrow $D \Rightarrow R$ is immediate, so R and D are equivalent; the content is entirely in $R \Rightarrow D$ . That proof runs through the $Δ$ -subgroup of finite conjugacy classes: reducedness forces $γ k G γ' = 0$ for a hypothetical pair of zero divisors, Passman's proposition pushes the vanishing down to $k Δ$ , and $Δ$ is torsion-free FC hence abelian, so $k Δ$ is a domain — contradiction.

No implication is known from J to any of the others, and none from R or D back to U.

Applications and Industry Use

Applications here means where this structure is used — inside mathematics and in the engineering and computing disciplines that consume it.

Topology

Whitehead groups and surgery

Units of $ℤ G$ control the Whitehead group $W h (G)$ , which classifies $h$ -cobordisms. The unit problem for torsion-free $G$ is the algebraic shadow of the vanishing conjectures for $W h (G)$ in geometric topology.

Operator algebras

Kadison-Kaplansky

The idempotent conjecture for the reduced group $C^{*}$ -algebra of a torsion-free group is the analytic counterpart of the algebraic idempotent problem, and follows from the Baum-Connes conjecture — one of the main motivations for that programme.

Computer search

Certified counterexamples

Gardam's unit was found by an SAT-solver-assisted search over $𝔽_{2} P$ with bounded support, then verified by direct multiplication. The episode is a case study in machine-assisted algebra: the object is small, the search space is enormous, and verification is trivial.

Coding and cryptography

Group ring codes and key exchange

Group-ring based schemes need to know their unit groups: whether a candidate key is invertible, and whether inverses are unique, is exactly the unit problem restricted to a concrete finite group.

The honest description is that these problems are load-bearing conjectures inside algebra and topology. Their value is in what assumes them: whole theories in $K$ -theory and geometric topology are stated under the hypothesis that the zero-divisor or idempotent conjecture holds.

Failure Modes and Common Mistakes

Do not assume $R \Rightarrow D$ is formal. It is a theorem requiring Neumann's covering lemma and the $Δ$ -machinery.
Do not conflate no nontrivial units with the unit group is small: even trivial units already form a copy of $U (k) \cdot G$ , which is infinite whenever $G$ is.
Do not assume $k$ must be a field. The problems are stated for a domain, and the passage from a field to a domain is not vacuous — $ℤ G$ is the case of greatest interest in topology.
Do not use $(6.21) (1)$ when $G$ has an element of order $2$ ; the proof needs $g^{2} \neq 1$ to keep three group elements distinct.

Best Practices

State which of U, R, D, J you mean; the literature calls all four the Kaplansky conjecture.
Record the coefficient ring: results over $ℤ$ , over a field of characteristic zero and over $𝔽_{p}$ have genuinely different status.
When claiming a positive result, name the class of groups; almost nothing is known for arbitrary torsion-free groups.
Before believing an alleged nontrivial unit, verify the product directly — support bookkeeping in a group ring is checkable by machine and by hand.

Historical Notes and Lessons Learned

1940Higman's thesisGraham Higman determines the units of $ℤ G$ for large classes of finite $G$ and classifies when only trivial units occur, founding the subject.
1950sKaplansky's problem listsKaplansky circulates the unit, zero-divisor and idempotent questions for torsion-free groups; they become known as the Kaplansky conjectures.
1962-70Passman's $Δ$ -methodsThe finite conjugate subgroup $Δ (G)$ becomes the standard tool, yielding $R \Rightarrow D$ and much of the semiprimitivity theory.
1988Promislow's groupA torsion-free crystallographic group is shown to fail the unique product property, removing the main heuristic support for the conjectures.
1988Kropholler-Linnell-MoodyThe zero-divisor conjecture is proved for torsion-free elementary amenable groups in characteristic zero by $K$ -theoretic methods.
2021Gardam's counterexampleA nontrivial unit in $𝔽_{2} P$ refutes the unit conjecture, eighty years after Higman.

The lesson is about the evidential status of a heuristic. No mechanism produces nontrivial units without torsion was a plausible principle supported by every verified class, and it was wrong. The classes for which the results are theorems — ordered, unique product, elementary amenable — were never a proof of the general case, and the group that failed was one already known to be exceptional.

Quick Reference

Trivial unit

a g

with

a \in U (k)

g \in G

Problem U

U (k G) = U (k) \cdot G

Problem Ris

k G

reduced?

Problem Dis

k G

a domain?

Problem J

rad k G = 0

for

G \neq {1}

ImplicationsU

\Rightarrow

R, U

\Rightarrow

J, R

\Leftrightarrow

Standing hypotheses

k

a domain,

G

torsion-free

Exception in $(6.21) (2)$

| k | = | G | = 2

Small cases to keep in mind
Ring	Units	Radical
$ℤ C_{2}$	$\pm 1, \pm g$ — all trivial	$0$
$ℤ C_{5}$	contains $1 - x^{2} - x^{3}$ — nontrivial	$0$
$𝔽_{2} C_{2}$	${1, g}$ — all trivial	${0, 1 + g}$
$k [x, x^{- 1}]$ , $k$ a domain	$a x^{n}$ — all trivial	$0$
$𝔽_{2} P$ , $P$ Promislow	contains a nontrivial unit	unknown

Frequently Asked Questions

Why is the case

| k | = | G | = 2

excluded from

(6.21) (2)

Because both perturbation arguments in the proof need room. To contradict a trivial unit $1 - α = a g$ with $g \neq 1$ one multiplies either by a scalar $b \notin {0, 1}$ or by a group element $h \notin {1, g^{- 1}}$ . Over $𝔽_{2}$ with a two-element group neither exists, and the conclusion genuinely fails: $𝔽_{2} C_{2}$ has only trivial units and a nonzero radical.

If the unit conjecture is false, why keep Problem U?

Because it remains open in the cases of greatest interest — over $ℤ$ and over fields of characteristic zero — and because it is still the most efficient hypothesis: $(6.21)$ and the implication R implies D turn a positive answer for a class of groups into all four conclusions at once. Ordered groups are proved this way.

Does J-semisimplicity of

k G

say anything about zero divisors?

No. J-semisimplicity is a statement about the radical, and there are many J-semisimple rings with zero divisors — any product of two fields, for example. The implications run from U downwards only.

Where exactly is torsion-freeness used in

(6.21)

It is not, directly. $(6.21)$ assumes only that $k$ is nonzero, that $G$ is nontrivial, and in part (1) that $k$ is reduced with $G$ free of elements of order $2$ . Torsion-freeness enters when the proposition is applied: it supplies the no-element-of-order- $2$ hypothesis and rules out the exceptional case.

What is a unique product group and why does it help?

A group in which any two finite nonempty subsets $A$ , $B$ admit an element of $A B$ with exactly one factorisation $a b$ . That element then has coefficient $a_{g} b_{g} \neq 0$ in the product of two group ring elements with those supports, so no zero divisors arise. Orderable groups are unique product groups; Promislow's group is torsion-free but is not, which is where the counterexample came from.

How does the idempotent question fit in?

A domain has only the idempotents $0$ and $1$ , so Problem D implies the idempotent conjecture. The converse is not known. The idempotent question has an analytic life of its own as the Kadison-Kaplansky conjecture for reduced group $C^{*}$ -algebras, where it is a consequence of the Baum-Connes conjecture.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §6, results (6.16)-(6.21) (pp. 95-99).
D. S. Passman, The Algebraic Structure of Group Rings, Wiley-Interscience, New York, 1977, Chapters 12-13.
S. K. Sehgal, Topics in Group Rings, Monographs and Textbooks in Pure and Applied Mathematics 50, Marcel Dekker, New York, 1978.
G. Higman, “The units of group-rings”, Proceedings of the London Mathematical Society (2) 46 (1940).
G. Gardam, “A counterexample to the unit conjecture for group rings”, Annals of Mathematics 194 (2021).
P. H. Kropholler, P. A. Linnell and J. A. Moody, “Applications of a new K-theoretic theorem to soluble group rings”, Proceedings of the American Mathematical Society 104 (1988).

AI Suggested Questions

Reconstruct Gardam's nontrivial unit in $𝔽_{2} P$ and verify the product by hand or by machine.
What exactly does Promislow's group look like, and why does it fail the unique product property?
Explain how the implication R implies D uses Neumann's covering lemma.
State the Kadison-Kaplansky conjecture and its relation to Baum-Connes.
For which finite groups $G$ is $U (ℤ G) = \pm G$ , and how does Higman's classification go?
Is the zero-divisor conjecture known for torsion-free hyperbolic groups?
Construct the Bass cyclic units of $ℤ G$ and explain why they need torsion.

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Where nontrivial units come from

Why reducedness is the flexible hypothesis

The idempotent question

06Key Results

07Proof Techniques and Method

Turn nilpotence into a unit

Perturb to break triviality

Count the support

08Worked Example

A nontrivial unit in ℤC5

Only trivial units for |G|≤4

The torsion-free model case

The exception in (6.21)(2)

09Frameworks and Models

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Whitehead groups and surgery

Kadison-Kaplansky

Certified counterexamples

Group ring codes and key exchange

13Failure Modes and Common Mistakes

14Best Practices

15Historical Notes and Lessons Learned

16Quick Reference

17Frequently Asked Questions

18Related KEVOS Topics

19References

20AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

A nontrivial unit in $ℤ C_{5}$

Only trivial units for $| G | \leq 4$

The exception in $(6.21) (2)$

Frameworks and Models

Comparison and Classification

Relationship Map

Applications and Industry Use

Failure Modes and Common Mistakes

Best Practices

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions