Group Rings and Semigroup Rings: Construction and First Properties

Q: Is kG ever a division ring?

Only when G is trivial and k is a division ring. Any g ≠ 1 of finite order produces zero-divisors by (1.4c), and any g of infinite order generates a copy of k[x,x^-1], which has non-units. So the group ring construction never produces new division rings — the Laurent series construction of the *Polynomial and Laurent Series Rings* page does.

Q: How does kG relate to representations of G?

Exactly: a representation of G on a k-module M is a group homomorphism G → Aut_k(M), and by the universal property that is the same as a k-algebra map kG → End_k(M), that is, a kG-module structure on M. Irreducible representations correspond to simple kG-modules, and Maschke's theorem says all of them are direct summands exactly in the semisimple case.

Q: What was the unit conjecture and how was it disproved?

Kaplansky conjectured that for a field k and a torsion-free group G, every unit of kG is trivial, that is of the form ug. Gardam disproved it in 2021 by exhibiting an explicit nontrivial unit in F_2 P, where P is the torsion-free Promislow group, also known as the Hantzsche–Wendt or Fibonacci group of that name. The zero-divisor conjecture, which is formally weaker in flavour, is still open.

Q: What is the centre of a group ring?

For k commutative, Z(kG) is the free k-module on the class sums of the finite conjugacy classes of G. For finite G this gives _k Z(kG) equal to the number of conjugacy classes, which over a splitting field of characteristic zero is also the number of irreducible representations.

Q: Why does the free ring appear as a semigroup ring?

Because the free monoid on X has the words in X as its elements and concatenation as its product, so the monoid ring k is k-linear combinations of words multiplied by concatenation — which is precisely the definition of k X. The universal properties match as well: monoid maps out of are free choices of images, exactly as for the free ring.

Executive Summary

The group ring $k G$ is the smallest ring in which the group $G$ sits as a subgroup of the units and the coefficients from $k$ commute with everything in $G$ . Representations of $G$ over $k$ are exactly $k G$ -modules, so the entire representation theory of finite groups is the module theory of one ring.

The construction is elementary but the resulting rings are not. Even $ℤ C_{5}$ has units that are not of the form $\pm g$ , and the questions of whether $k G$ can have zero-divisors or nontrivial units for torsion-free $G$ occupied ring theorists for seventy years — one of them was settled only in 2021, and negatively.

$| G |$ Rank of

k G

over

k

$ε$ Augmentation to

k

$\pm g$ The trivial units of

ℤ G

2021Unit conjecture disproved

Overview

Let $k$ be a ring and let $G$ be a group or, more generally, a monoid written multiplicatively. Define

k G = ⨁_{σ \in G} k σ,

(1.4a)

finite formal sums $\sum_{σ} a_{σ} σ$ with $a_{σ} \in k$ , almost all zero.

Multiplication is convolution: extend the monoid law bilinearly, requiring coefficients to commute with group elements. Explicitly,

(\sum_{σ} a_{σ} σ) (\sum_{τ} b_{τ} τ) = \sum_{μ} c_{μ} μ, c_{μ} = \sum_{σ τ = μ} a_{σ} b_{τ} .

(1.4b)

Associativity comes from associativity in $G$ and in $k$ ; the identity is $1_{k} \cdot 1_{G}$ . The ring $k G$ is commutative if and only if both $k$ and $G$ are, which is already enough to manufacture a large supply of noncommutative rings from finite groups.

Two special cases connect this page to its neighbours. If $G$ is the free monoid on a set $X$ , then $k G$ is the free ring $k ⟨ X ⟩$ . If $G$ is infinite cyclic, $k G ≅ k [x, x^{- 1}]$ , the Laurent polynomial ring.

Learning Objectives

Define $k G$ for a monoid $G$ and verify that convolution is associative with identity $1_{k} \cdot 1_{G}$ .
Prove the universal property: $k$ -algebra maps $k G \to A$ correspond to group homomorphisms $G \to U (A)$ for commutative $k$ .
Show that the augmentation ideal is free on ${g - 1 : g \neq 1}$ .
Produce a zero-divisor in $k G$ whenever $G$ has an element of finite order greater than one.
Verify by hand that $1 - z^{2} - z^{3}$ is a unit of $ℤ C_{5}$ .
State Maschke's theorem and identify where it fails.

Definitions

Definition(1.4)Group and semigroup rings

For a ring $k$ and a monoid $G$ , the monoid ring $k G$ is the free left $k$ -module on $G$ with the convolution product $(1.4 b)$ . When $G$ is a group this is the group ring; when $G$ is a semigroup without identity the same formula defines a ring without identity, and this collection assumes a monoid throughout.

Definition—Augmentation

The augmentation map is the $k$ -linear map $ε : k G \to k$ with $ε (g) = 1$ for all $g \in G$ , so $ε (\sum a_{g} g) = \sum a_{g}$ . It is a surjective ring homomorphism, and its kernel $Δ (G) = \ker ε$ is the augmentation ideal.

$supp (α)$: The support of $α = \sum a_{g} g$ : the finite set of $g$ with $a_{g} \neq 0$ .
Trivial unit: An element $u g$ with $u \in U (k)$ and $g \in G$ . These are always units; the question is whether there are others.
Class sum: For $G$ finite and $k$ commutative, the sum of the elements of a conjugacy class. The class sums form a $k$ -basis of $Z (k G)$ .
$Δ (G)$: The augmentation ideal, a two-sided ideal of $k G$ with $k G / Δ (G) ≅ k$ .
Skew group ring $k * G$: Same underlying module, with $(a σ) (b τ) = a σ (b) σ τ$ for an action of $G$ on $k$ by automorphisms. It equals $k G$ exactly when the action is trivial.
Symmetrising idempotent: For $G$ finite and $| G |$ invertible in $k$ , the element that averages over the group; it is an idempotent and generates the trivial-representation summand.

Group elements are identified with their images $1_{k} \cdot g$ , and coefficients with $a \cdot 1_{G}$ . Under this identification $G \subseteq U (k G)$ and $k \subseteq k G$ as a subring.

Core Concepts

Torsion produces zero-divisors immediately

If $g \in G$ has finite order $n > 1$ , then in $k G$

(1 - g) (1 + g + g^{2} + \dots + g^{n - 1}) = 1 - g^{n} = 0,

(1.4c)

with both factors nonzero, since $1, g, \dots, g^{n - 1}$ are distinct basis elements.

So $k G$ has zero-divisors whenever $G$ has torsion, for every nonzero $k$ . Kaplansky's zero-divisor conjecture asks the converse: if $k$ is a field and $G$ is torsion-free, must $k G$ be a domain? It remains open in general, and is known for large classes — orderable groups, and more.

Units beyond the obvious ones

The trivial units $U (k) \cdot G$ always sit inside $U (k G)$ . Whether they exhaust it is delicate. Higman determined the finite abelian answer in 1940: for $G$ finite abelian, $U (ℤ G) = \pm G$ precisely when the exponent of $G$ is $1, 2, 3, 4$ or $6$ . Since $C_{5}$ has exponent $5$ , $ℤ C_{5}$ must have nontrivial units — and the worked example produces one explicitly.

U (k) \cdot G

U (k G)

hard in general

Semisimplicity: Maschke's dividing line

For $k$ a field and $G$ a finite group, $k G$ is semisimple if and only if $char k$ does not divide $| G |$ . The forward direction uses the averaging idempotent $\frac{1}{| G |} \sum_{g} g$ , which exists exactly when $| G |$ is invertible. When the characteristic does divide the order, the radical is nonzero and the whole of modular representation theory is the study of what survives passing to $k G / rad (k G)$ .

Key Results

Theorem(1.4)Universal property of the group ring

Let $k$ be a commutative ring, $G$ a group and $A$ a $k$ -algebra. Then restriction to $G$ is a bijection

{Hom}_{k -alg} (k G, A) \tilde{⟶} {Hom}_{Grp} (G, U (A)) .

(1.4d)

So $k G$ is the universal $k$ -algebra containing $G$ in its unit group.

Proof

If $ϕ : k G \to A$ is a $k$ -algebra homomorphism and $g \in G$ , then $ϕ (g) ϕ (g^{- 1}) = ϕ (1) = 1$ and likewise on the other side, so $ϕ (g) \in U (A)$ ; and $ϕ (g h) = ϕ (g) ϕ (h)$ , so $ϕ |_{G}$ is a group homomorphism $G \to U (A)$ .

Conversely let $ρ : G \to U (A)$ be a group homomorphism. Since $k G$ is free as a $k$ -module on $G$ , there is a unique $k$ -linear map $ϕ$ with $ϕ (g) = ρ (g)$ . It is multiplicative: on basis elements $ϕ (g h) = ρ (g h) = ρ (g) ρ (h) = ϕ (g) ϕ (h)$ , and bilinearity of the product in $A$ together with the commutativity of $k$ — needed so that coefficients may be pulled out of both arguments — extends this to all of $k G$ . It sends $1_{G}$ to $1_{A}$ .

The two constructions are mutually inverse because a $k$ -linear map is determined by its values on the basis $G$ .

Proposition—The augmentation ideal

Let $k$ be a ring and $G$ a monoid. The augmentation $ε : k G \to k$ is a surjective ring homomorphism, and its kernel $Δ (G)$ is free as a left $k$ -module with basis ${g - 1 : g \in G, g \neq 1}$ . In particular $k G / Δ (G) ≅ k$ .

Proof

Multiplicativity: with $α = \sum a_{σ} σ$ and $β = \sum b_{τ} τ$ , the coefficient sum of $α β$ is $\sum_{μ} \sum_{σ τ = μ} a_{σ} b_{τ} = \sum_{σ, τ} a_{σ} b_{τ} = ε (α) ε (β)$ , the middle step being a reindexing of a finite sum. Surjectivity is clear from $ε (a \cdot 1_{G}) = a$ .

Spanning: if $α = \sum_{g} a_{g} g$ lies in $Δ (G)$ then $\sum_{g} a_{g} = 0$ , so

α = \sum_{g} a_{g} g - (\sum_{g} a_{g}) \cdot 1 = \sum_{g \neq 1} a_{g} (g - 1) .

Independence: a relation $\sum_{g \neq 1} a_{g} (g - 1) = 0$ expands to $\sum_{g \neq 1} a_{g} g - (\sum_{g \neq 1} a_{g}) 1 = 0$ , and comparing coefficients of each $g \neq 1$ in the basis $G$ gives $a_{g} = 0$ .

Proposition—Group rings are self-opposite over a commutative base

Let $k$ be commutative and $G$ a group. The $k$ -linear extension of $g \mapsto g^{- 1}$ is an involution of $k G$ ; hence $k G ≅ (k G)^{op}$ , and $k G$ is left noetherian if and only if it is right noetherian, left artinian if and only if right artinian, and so on.

Proof

Write $α^{*} = \sum_{g} a_{g} g^{- 1}$ for $α = \sum_{g} a_{g} g$ . This is additive and involutive. For products,

(α β)^{*} = \sum_{g, h} a_{g} b_{h} (g h)^{- 1} = \sum_{g, h} a_{g} b_{h} h^{- 1} g^{- 1}, β^{*} α^{*} = \sum_{g, h} b_{h} a_{g} h^{- 1} g^{- 1},

and the two agree because $k$ is commutative. It fixes $1$ , so it is an involution and therefore an anti-automorphism, which is what self-oppositeness requires.

TheoremMaschkeSemisimplicity of finite group algebras

Let $k$ be a field and $G$ a finite group. Then $k G$ is semisimple if and only if $char k ∤ | G |$ . Both hypotheses matter: for infinite $G$ the statement is false — $k G$ is then not even artinian — and for $char k$ dividing $| G |$ the element $\sum_{g \in G} g$ spans a nonzero nilpotent ideal.

Corollary—Units of a free monoid ring

Let $k$ be a domain and $Σ$ the free monoid on a set $X$ . Then $U (k Σ) = U (k ⟨ X ⟩) = U (k)$ . The hypothesis that $k$ is a domain cannot be dropped: over $ℤ / 4 ℤ$ , $(1 + 2 x) (1 - 2 x) = 1$ .

Worked Example

A nontrivial unit in $ℤ C_{5}$

Let $G = ⟨ z ⟩$ be cyclic of order $5$ , so $z^{5} = 1$ , and work in $ℤ G$ . Put

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

(E.1)

Expand $α β$ term by term, reducing exponents modulo $5$ :

$1 \cdot (1 - z - z^{4}) = 1 - z - z^{4}$
$- z^{2} \cdot (1 - z - z^{4}) = - z^{2} + z^{3} + z^{6} = - z^{2} + z^{3} + z$
$- z^{3} \cdot (1 - z - z^{4}) = - z^{3} + z^{4} + z^{7} = - z^{3} + z^{4} + z^{2}$

Adding: the constant term is $1$ ; the $z$ terms give $- 1 + 1 = 0$ ; the $z^{2}$ terms give $- 1 + 1 = 0$ ; the $z^{3}$ terms give $1 - 1 = 0$ ; the $z^{4}$ terms give $- 1 + 1 = 0$ . Hence

α β = 1 .

(E.2)

Since $ℤ C_{5}$ is commutative, $β α = 1$ too, so $α, β \in U (ℤ C_{5})$ . Neither has the form $\pm z^{i}$ , so both are nontrivial units. Note $ε (α) = 1 - 1 - 1 = - 1$ and $ε (β) = - 1$ , consistent with $ε (α β) = 1$ : augmentation of a unit must be a unit of $ℤ$ , hence $\pm 1$ .

Maschke on both sides of the line

Characteristic does not divide the order. Take $k = ℚ$ and $G = C_{3} = ⟨ z ⟩$ . Then $ℚ C_{3} ≅ ℚ [x] / (x^{3} - 1) = ℚ [x] / ((x - 1) (x^{2} + x + 1))$ , and since $x^{2} + x + 1$ is irreducible over $ℚ$ the Chinese remainder theorem gives

ℚ C_{3} ≅ ℚ \times ℚ (ω), ω = e^{2 π \sqrt{- 1} / 3},

(E.3)

a product of two fields — semisimple, as Maschke requires.

Characteristic divides the order. Take $k = 𝔽_{2}$ and $G = C_{2} = {1, z}$ . Then $(1 + z)^{2} = 1 + 2 z + z^{2} = 1 + 0 + 1 = 0$ , so $1 + z$ is a nonzero nilpotent. Hence

𝔽_{2} C_{2} ≅ 𝔽_{2} [x] / (x + 1)^{2}, rad (𝔽_{2} C_{2}) = 𝔽_{2} (1 + z),

(E.4)

a local ring with residue field $𝔽_{2}$ — as far from semisimple as a four-element ring can be.

Torsion is fatal for the domain property

In the same $ℤ C_{5}$ , $(1 - z) (1 + z + z^{2} + z^{3} + z^{4}) = 1 - z^{5} = 0$ with both factors nonzero. This is $(1.4 c)$ in the smallest interesting instance, and it shows that no group ring of a group with torsion is a domain.

Comparison and Classification

Group rings of small groups
Ring	Structure	Semisimple?	Reason
$k C_{\infty}$	$≅ k [x, x^{- 1}]$	no	not artinian; $G$ is infinite
$ℚ C_{2}$	$≅ ℚ \times ℚ$	yes	$char = 0$
$ℚ C_{3}$	$≅ ℚ \times ℚ (ω)$	yes	$char = 0$
$𝔽_{2} C_{2}$	local, $≅ 𝔽_{2} [x] / (x + 1)^{2}$	no	$2 ∣ \| C_{2} \|$
$𝔽_{3} S_{3}$	not semisimple	no	$3 ∣ \| S_{3} \| = 6$
$ℂ S_{3}$	$≅ ℂ \times ℂ \times M_{2} (ℂ)$	yes	$char = 0$ ; three irreducibles of degrees $1, 1, 2$
$k Σ$ , $Σ$ free monoid	$≅ k ⟨ X ⟩$	no	not even noetherian for $\| X \| \geq 2$

Which properties of $k G$ follow from which hypotheses
	$G$ finite, $char k ∤ \| G \|$	$G$ finite, $char k ∣ \| G \|$	$G$ infinite torsion-free	$G$ free monoid
Semisimple	yes	no	no	no
Artinian	yes	yes	no	no
Noetherian	yes	yes	partial	no
Has zero-divisors	yes	yes	open in general	no if $k$ is a domain
Only trivial units	no	no	no	yes if $k$ is a domain
Self-opposite ( $k$ commutative)	yes	yes	yes	no

Which properties of $k G$ follow from which hypotheses

Relationship Map

The monoid-ring construction specialises to several rings that look unrelated at first sight.

Monoid rings $k G$ free

k

-module on

G

, convolution product

Free monoid

k ⟨ X ⟩

, the free ring — no relations at all

One generator

k [x]

, the polynomial ring

Groups

G \subseteq U (k G)

; augmentation available

Infinite cyclic

k [x, x^{- 1}]

, Laurent polynomials

Finite cyclic $C_{n}$

k [x] / (x^{n} - 1)

Twisting the construction gives the skew group ring $k * G$ , where $(a σ) (b τ) = a σ (b) σ τ$ for an action of $G$ on $k$ by automorphisms. Two facts tie it back:

For $G = ⟨ σ ⟩$ infinite cyclic acting on $k$ , $k * G ≅ k [x, x^{- 1}; σ]$ , the skew Laurent polynomial ring of Skew Polynomial Rings and Hilbert's Twist.
If $G = T ⋊ H$ is a semidirect product, then $k G ≅ (k T) * H$ , with $H$ acting on $k T$ by conjugation. Skew group rings are therefore not exotic: they appear inside ordinary group rings.

k

k T

(k T) * H = k G

Applications and Industry Use

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

Representation theory

The whole subject, as module theory

Characters, induction, blocks and defect groups are all statements about $k G$ -modules. Modular representation theory is precisely the study of $k G$ when $char k$ divides $| G |$ , that is, when Maschke fails.

Coding theory

Cyclic and group codes

A cyclic code of length $n$ over $𝔽_{q}$ is an ideal of $𝔽_{q} C_{n} = 𝔽_{q} [x] / (x^{n} - 1)$ . Generalising the group to a noncyclic one gives group codes, and the ring-theoretic decomposition of $k G$ is what supplies generator idempotents.

Signal processing

Convolution and the group Fourier transform

Multiplication in $k G$ is convolution, and for abelian $G$ the Wedderburn decomposition of $ℂ G$ into a product of copies of $ℂ$ is the discrete Fourier transform. Fast transforms are fast changes of basis in the group algebra.

Computational chemistry and physics

Symmetry-adapted bases

Projection operators built from the symmetrising idempotents of $ℂ G$ block-diagonalise Hamiltonians by irreducible representation, which is how molecular symmetry reduces the size of an electronic structure calculation.

Topology and geometry

Fundamental group actions

Chains on a universal cover form a module over $ℤ π_{1}$ , so the group ring is the coefficient ring of equivariant topology; Whitehead torsion and surgery obstructions live in its $K$ -theory.

Ring theory internally

A machine for examples

Group rings supply noncommutative rings with prescribed behaviour: torsion produces zero-divisors, free groups produce domains that are far from commutative, and infinite groups break every chain condition on demand.

Computational Notes

Computational notes cover algorithms, cost and library behaviour rather than manufacturing process.

Representation as arrays. For finite $G$ , elements of $k G$ are length- $| G |$ coefficient vectors and multiplication is a convolution, costing $O (| G |^{2})$ naively. For abelian $G$ a fast Fourier transform over the character group reduces this to $O (| G | \log | G |)$ when the characters are available in $k$ .
Wedderburn decomposition. Over a field with $char k ∤ | G |$ , decomposing $k G$ into matrix blocks is equivalent to finding a complete set of primitive central idempotents; the standard method computes the centre — spanned by class sums — and splits it.
The radical in the modular case. When $char k = p$ divides $| G |$ , computing $rad (k G)$ is a finite-dimensional radical computation and uses the Friedl–Rónyai algorithm rather than the trace form, which is degenerate in characteristic $p$ .
Software. GAP's GroupRing and its character-table library, Magma's GroupAlgebra, and Sage's GroupAlgebra all implement the construction; the Meataxe handles the module theory over finite fields.
Infinite groups. For infinite $G$ only finitely supported arithmetic is computable, and questions such as invertibility of a given element are undecidable in general, since the word problem for finitely presented groups is.

Failure Modes and Common Mistakes

Do not assume the units of $ℤ G$ are $\pm G$ . They are only for a short list of finite abelian $G$ , and the general problem is open.
Do not assume $k G$ is a domain when $G$ is torsion-free. It is conjectured, not known; and the analogous unit conjecture turned out to be false.
Do not confuse $k G$ with the skew group ring $k * G$ . They agree exactly when the action of $G$ on $k$ is trivial, and the ideal theory differs sharply otherwise.
Do not treat a semigroup ring as having an identity. Only monoid rings do, and every convention in this collection presumes one.

Historical Notes and Lessons Learned

1854CayleyIntroduces formal linear combinations of group elements while developing the abstract notion of a group, the first appearance of the construction.
1896–97FrobeniusCreates character theory by studying the group determinant, effectively factorising the commutative case of $ℂ G$ .
1899MaschkeProves complete reducibility of representations of a finite group in characteristic zero, the theorem that fixes the semisimple boundary.
1929NoetherRecasts representation theory as the module theory of $k G$ , which is the point of view used throughout this collection.
1940HigmanDetermines the finite abelian groups for which $ℤ G$ has only trivial units — exponents $1, 2, 3, 4, 6$ — and thereby shows nontrivial units are the norm.
1950s–70sKaplansky's problemsThe zero-divisor, idempotent, unit and direct-finiteness conjectures for torsion-free groups are formulated and become a programme; Kaplansky proves direct finiteness over fields of characteristic zero.
1977PassmanPublishes the systematic ring-theoretic treatment of group rings, consolidating the subject.
2021GardamDisproves the unit conjecture: a nontrivial unit exists in $𝔽_{2} P$ for $P$ the torsion-free Promislow group. The zero-divisor conjecture remains open.

The lesson is that a construction can be elementary and its questions still be hard. Nothing in the definition of $k G$ hints that deciding whether $𝔽_{2} G$ has a nontrivial unit for torsion-free $G$ would take seventy years and a computer search.

Quick Reference

Definition

k G = ⨁_{σ \in G} k σ

with convolution product

Rankfree left

k

-module of rank

| G |

Universal property

k

-algebra maps

k G \to A

\leftrightarrow

group maps

G \to U (A)

, for

k

commutative

Augmentation

ε (\sum a_{g} g) = \sum a_{g}

;

Δ (G)

free on

{g - 1}

Torsion

g

of order

n > 1

gives

(1 - g) (1 + \dots + g^{n - 1}) = 0

Maschke

G

finite,

k

a field: semisimple

iff char k ∤ | G |

Self-opposite

g \mapsto g^{- 1}

is an involution when

k

is commutative

Special casesfree monoid

\to k ⟨ X ⟩

;

C_{\infty} \to k [x, x^{- 1}]

;

C_{n} \to k [x] / (x^{n} - 1)

The Kaplansky problems for torsion-free $G$ and a field $k$
Conjecture	Statement	Status
Zero-divisor	$k G$ is a domain	open in general; known for orderable groups
Idempotent	$k G$ has no idempotents other than $0$ and $1$	open in general; known in many cases
Unit	every unit of $k G$ is trivial	false — Gardam, 2021
Direct finiteness	$k G$ is Dedekind-finite	known for fields of characteristic zero

Frequently Asked Questions

Why must the coefficients commute with the group elements?

Because that is what the definition imposes, and it is what makes $k G$ a free module on $G$ with a well-behaved product. If you want the group to act on the coefficients instead, the correct object is the skew group ring $k * G$ , in which moving a coefficient past $σ$ applies the automorphism $σ$ to it.

k G

ever a division ring?

Only when $G$ is trivial and $k$ is a division ring. Any $g \neq 1$ of finite order produces zero-divisors by $(1.4 c)$ , and any $g$ of infinite order generates a copy of $k [x, x^{- 1}]$ , which has non-units. So the group ring construction never produces new division rings — the Laurent series construction of the Polynomial and Laurent Series Rings page does.

How does

k G

relate to representations of

G

Exactly: a representation of $G$ on a $k$ -module $M$ is a group homomorphism $G \to {Aut}_{k} (M)$ , and by the universal property that is the same as a $k$ -algebra map $k G \to {End}_{k} (M)$ , that is, a $k G$ -module structure on $M$ . Irreducible representations correspond to simple $k G$ -modules, and Maschke's theorem says all of them are direct summands exactly in the semisimple case.

What was the unit conjecture and how was it disproved?

Kaplansky conjectured that for a field $k$ and a torsion-free group $G$ , every unit of $k G$ is trivial, that is of the form $u g$ . Gardam disproved it in 2021 by exhibiting an explicit nontrivial unit in $𝔽_{2} P$ , where $P$ is the torsion-free Promislow group, also known as the Hantzsche–Wendt or Fibonacci group of that name. The zero-divisor conjecture, which is formally weaker in flavour, is still open.

What is the centre of a group ring?

For $k$ commutative, $Z (k G)$ is the free $k$ -module on the class sums of the finite conjugacy classes of $G$ . For finite $G$ this gives $\dim_{k} Z (k G)$ equal to the number of conjugacy classes, which over a splitting field of characteristic zero is also the number of irreducible representations.

Why does the free ring appear as a semigroup ring?

Because the free monoid on $X$ has the words in $X$ as its elements and concatenation as its product, so the monoid ring $k Σ$ is $k$ -linear combinations of words multiplied by concatenation — which is precisely the definition of $k ⟨ X ⟩$ . The universal properties match as well: monoid maps out of $Σ$ are free choices of images, exactly as for the free ring.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §1, Examples (1.4) and (1.11), pp. 7–9 and 14–15; Maschke's theorem in §6.
D. S. Passman, The Algebraic Structure of Group Rings, Wiley-Interscience, New York, 1977.
G. Higman, “The units of group-rings”, Proceedings of the London Mathematical Society (2) 46 (1940), 231–248.
G. Gardam, “A counterexample to the unit conjecture for group rings”, Annals of Mathematics 194 (2021).
C. W. Curtis and I. Reiner, Methods of Representation Theory, Volume I, Wiley-Interscience, 1981, Chapters 1–3.
S. K. Sehgal, Units in Integral Group Rings, Pitman Monographs and Surveys in Pure and Applied Mathematics 69, Longman, 1993.

AI Suggested Questions

Prove Maschke's theorem and identify precisely where invertibility of the group order is used.
Describe Gardam's nontrivial unit in the group ring of the Promislow group over the field of two elements.
For which finite groups is the integral group ring determined up to isomorphism by the group?
How is the augmentation ideal used to define group cohomology?
Compute the Wedderburn decomposition of the rational group algebra of the symmetric group on three letters.
What is known about the zero-divisor conjecture for orderable and for right-orderable groups?
How do skew group rings arise from semidirect products, and what does that say about the ideal structure of the ordinary group ring?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Torsion produces zero-divisors immediately

Units beyond the obvious ones

Semisimplicity: Maschke's dividing line

06Key Results

07Worked Example

A nontrivial unit in ℤC5

Maschke on both sides of the line

Torsion is fatal for the domain property

08Comparison and Classification

09Relationship Map

10Applications and Industry Use

The whole subject, as module theory

Cyclic and group codes

Convolution and the group Fourier transform

Symmetry-adapted bases

Fundamental group actions

A machine for examples

11Computational Notes

12Failure Modes and Common Mistakes

13Historical Notes and Lessons Learned

14Quick Reference

15Frequently Asked Questions

16Related KEVOS Topics

17References

18AI Suggested Questions

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Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Worked Example

A nontrivial unit in $ℤ C_{5}$

Comparison and Classification

Relationship Map

Applications and Industry Use

Computational Notes

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions