Units and Idempotents in Integral Group Rings

Q: Why is the integral group ring so much poorer in idempotents than the rational one?

Because the idempotents of QG carry denominators. By the central idempotent formula, the coefficient of the identity in e_i is n_i^2/|G|, which is an integer only when n_i^2 = |G|, that is when r = 1 and G is trivial. Theorem (8.26) says this obstruction is complete: no cleverer integral idempotent exists either.

Q: Does the theorem on idempotents extend to infinite groups?

Not by this proof, and not as a theorem at all. For an infinite torsion-free group the assertion that CG has only trivial idempotents is Kaplansky's conjecture, still open in general; it is known for large classes of groups, for example those satisfying the Baum–Connes conjecture. The finite case here is genuinely arithmetic and does not generalise by analogy.

Q: What is known about non-central torsion units?

Much less. The first Zassenhaus conjecture predicted that every torsion unit of augmentation 1 in ZG is conjugate in QG to a group element; it holds for many families, including nilpotent groups, but Eisele and Margolis produced a metabelian counterexample in 2018. Weaker statements about orders of torsion units remain active.

Q: Where exactly does the proof of the unit theorem use that the field is Galois over the rationals?

In the norm step. The estimate |a_g| ≤ 1 holds for each Galois conjugate because conjugating the coefficients of a central torsion unit produces another central torsion unit. Multiplying the conjugates gives |N(a_g)| ≤ 1, and only a Galois extension makes that product a rational integer. Enlarging k to a Galois closure at the start is what buys this.

Q: Is the corollary that isomorphic integral group rings have isomorphic centres the best possible?

It is best possible in the sense that the full isomorphism problem has a negative answer: Hertweck exhibited non-isomorphic finite groups G and H with ZG ≅ ZH. Any such pair must have isomorphic centres, and indeed the same character table and the same order, so the invariants that do survive are quite strong without being complete.

Q: How do I actually find a nontrivial unit in an integral group ring?

The standard constructions are the bicyclic units 1 + (1-g)hg, where g is the sum of the powers of g, and the Bass cyclic units built from the identity _j x^j raised to suitable powers. For ZC_5 the element 1 - x^2 - x^3 works and is easier to check by hand; its inverse is 1 - x - x^4.

Executive Summary

Over a field the group algebra of a finite group is rich in idempotents and units. Over a ring of algebraic integers almost all of that richness disappears. Lam closes §8 with two theorems that make this precise, both proved by the same arithmetic technique: expand in the idempotent basis, bound each coefficient by the unit-modulus lemma, and use that an algebraic integer of small norm is forced.

The results are $(8.21)$ — central torsion units of $A G$ are trivial up to a root of unity — and $(8.26)$ — $A G$ has only the idempotents $0$ and $1$ . The corollary $(8.25)$ of Higman and Berman is the first serious positive result on the isomorphism problem for integral group rings.

$ω g$ Central torsion units

${0, 1}$ All idempotents

$Z (G) ≅ Z (H)$ Forced by

A G ≅ A H

$\leq 1$ Every coefficient modulus

Overview

Fix a field $k$ of algebraic numbers with ring of algebraic integers $A$ , and a finite group $G$ . The group ring $A G$ sits inside $k G$ as an order: it spans $k G$ over $k$ and is finitely generated over $A$ . Passing from $k G$ to $A G$ costs semisimplicity — $A G$ is not a product of matrix rings — and the two theorems here measure exactly what is lost.

A G \subseteq k G ≅ \prod_{i = 1}^{r} M_{n_{i}} (k) after enlarging k to a splitting field,

(8.21a)

The idempotents $e_{i}$ of the right-hand side have denominators dividing $| G |$ , so they do not lie in $A G$ — and $(8.26)$ says nothing else does either.

Note the contrast with the Group Ring Problems page. There the questions concern infinite torsion-free groups over a field, where the absence of idempotents is conjectural and reflects the group. Here the group is finite and the field has plenty of idempotents; the obstruction is purely arithmetic, and it is a theorem.

Both arguments are due, in outline, to Higman in the abelian case; Lam presents the general-character version, with the idempotent theorem in the form given by Takahashi after Swan's projective-module proof.

Learning Objectives

State $(8.21)$ with its hypotheses on $k$ , $A$ and $G$ .
Justify the reduction to a finite Galois splitting field.
Carry out the modulus estimate $| a_{g} | \leq 1$ and the norm argument that makes it an equality.
Deduce $(8.24)$ and the Higman–Berman corollary $(8.25)$ on centres.
Prove $(8.26)$ : $A G$ has no nontrivial idempotents.
Exhibit a unit of infinite order in $ℤ C_{5}$ and confirm the idempotent theorem there.

Definitions

Standing hypotheses: k is a field of algebraic numbers, A is its ring of algebraic integers, and G is a finite group. In the proofs k is enlarged to a finite extension that is Galois over the rationals and a splitting field for G; that enlargement is always harmless.

$U (A G)$: The group of units of $A G$ . It always contains the trivial units $ω g$ with $ω$ a root of unity in $A$ and $g \in G$ .
Central unit: A unit lying in $Z (A G)$ . For $G$ abelian every unit is central; for $G$ with trivial centre the central torsion units are as small as possible.
$N_{k / ℚ}$: The field norm of a finite Galois extension $k / ℚ$ , $N (x) = \prod_{σ \in Gal (k / ℚ)} σ (x)$ . It maps $A$ into $ℤ$ .
Augmentation $ε$: The ring map $A G \to A$ , $\sum a_{g} g \mapsto \sum a_{g}$ . A torsion unit has $ε (u)$ a root of unity in $A$ .
Order: A subring of a finite-dimensional $ℚ$ -algebra that is a finitely generated $ℤ$ -module spanning the algebra. $ℤ G$ is an order in $ℚ G$ .

Core Concepts

Two coordinate systems, again

A central element $α \in Z (A G)$ can be written in the group basis, $α = \sum_{g} a_{g} g$ with $a_{g} \in A$ , or in the idempotent basis, $α = \sum_{i} b_{i} e_{i}$ with $b_{i} \in k$ . Multiplicative conditions are transparent in the second: $α^{m} = 1$ says exactly $b_{i}^{m} = 1$ for every $i$ , so every $b_{i}$ is a root of unity and in particular $| b_{i} | = 1$ in any embedding of $k$ into $ℂ$ .

Integrality conditions are transparent in the first. The proof of $(8.21)$ consists in translating between them with the formula of the Central Idempotents and Characters page.

a_{g} = \frac{1}{| G |} \sum_{i = 1}^{r} n_{i} b_{i} χ_{i} (g^{- 1}) = \frac{1}{| G |} \sum_{i = 1}^{r} \sum_{j = 1}^{n_{i}} n_{i} b_{i} w_{i j},

(8.22a)

Here $w_{i 1}, \dots, w_{i n_{i}}$ are the eigenvalues of the action of $g^{- 1}$ on $M_{i}$ , all roots of unity.

The counting that makes the bound tight

Read the right-hand side as $\sum_{i} n_{i}^{2} = | G |$ complex numbers, each of modulus $1$ : the term $b_{i} w_{i j}$ repeated $n_{i}$ times, for each $i$ and each $j$ . The unit-modulus lemma $(8.20)$ gives $| a_{g} | \leq | G | / | G | = 1$ , with equality only when all $| G |$ of those numbers coincide.

From rigidity to the group

Once all the numbers $b_{i} w_{i j}$ equal a single $ω$ , the character values are determined: $χ_{i} (g^{- 1}) = n_{i} ω / b_{i}$ and $χ_{i} (g) = n_{i} b_{i} / ω$ . Feeding these into the second orthogonality relation kills every coefficient of $α$ off the conjugacy class of $g$ , and then computes $| C_{G} (g) | = \sum_{i} n_{i}^{2} = | G |$ . A centraliser of full order means $g$ is central.

α^{m} = 1

| b_{i} | = 1

| a_{g} | \leq 1

norm forces

| a_{g} | = 1

all

b_{i} w_{i j}

equal

α = ω C_{g}

g \in Z (G)

Key Results

Theorem(8.21)Central torsion units of an integral group ring

Let $k$ be a field of algebraic numbers with ring of algebraic integers $A$ , and let $G$ be a finite group. If $α$ is a central unit of finite order in $A G$ , then $α = ω g$ for some $g \in Z (G)$ and some root of unity $ω \in A$ .

For $G$ abelian this is Higman's theorem; the general statement is obtained by running Higman's method through the character theory of $(8.15)$ – $(8.18)$ .

Proof

Reduction. As noted above we may assume $[k : ℚ] < \infty$ , that $k / ℚ$ is Galois, and that $k$ is a splitting field for $G$ ; enlarging $k$ only has to be undone at the end, and $A' \cap k = A$ does that.

Setting up coordinates. Since $α \in Z (k G) \cap A G$ we may write $α = \sum_{i} b_{i} e_{i} = \sum_{g} a_{g} g$ with $b_{i} \in k$ and $a_{g} \in A$ . Fix $m$ with $α^{m} = 1$ . Computing in $Z (k G) ≅ \prod_{i} k e_{i}$ gives $b_{i}^{m} = 1$ , so each $b_{i}$ is a root of unity and $| b_{i} | = 1$ for any fixed embedding $k ↪ ℂ$ .

The modulus estimate. Expressing the $e_{i}$ in the group basis by $(8.15) (1)$ gives $a_{g} = | G |^{- 1} \sum_{i} n_{i} b_{i} χ_{i} (g^{- 1})$ . Fix $g$ with $a_{g} \neq 0$ and write $χ_{i} (g^{- 1}) = w_{i 1} + \dots + w_{i n_{i}}$ , the eigenvalues of the $g^{- 1}$ -action on $M_{i}$ ; these are roots of unity because $g$ has finite order. The resulting expression is an average of $\sum_{i} n_{i}^{2} = | G |$ complex numbers $b_{i} w_{i j}$ of modulus $1$ , so by $(8.20)$

| a_{g} | = \frac{1}{| G |} | \sum_{i = 1}^{r} \sum_{j = 1}^{n_{i}} n_{i} b_{i} w_{i j} | \leq \frac{1}{| G |} \sum_{i = 1}^{r} n_{i}^{2} = 1 .

The norm argument. For $σ \in Gal (k / ℚ)$ , applying $σ$ to the coefficients of $α$ produces another central unit of $A G$ of order dividing $m$ , so the same estimate gives $| σ (a_{g}) | \leq 1$ . Hence $| N_{k / ℚ} (a_{g}) | = \prod_{σ} | σ (a_{g}) | \leq 1$ . But $N_{k / ℚ} (a_{g}) \in ℤ$ and $a_{g} \neq 0$ , so the norm is a nonzero integer of absolute value at most $1$ ; therefore it has absolute value exactly $1$ , and each factor has modulus exactly $1$ . In particular $| a_{g} | = 1$ and the inequality above is an equality.

Rigidity. By the equality case of $(8.20)$ all the numbers $b_{i} w_{i j}$ are equal to a common $ω$ with $| ω | = 1$ . Then $χ_{i} (g^{- 1}) = n_{i} ω / b_{i}$ , so

a_{g} = \frac{1}{| G |} \sum_{i = 1}^{r} n_{i} b_{i} \cdot \frac{n_{i} ω}{b_{i}} = \frac{ω}{| G |} \sum_{i = 1}^{r} n_{i}^{2} = ω .

The $g$ -action on $M_{i}$ has eigenvalues $w_{i j}^{- 1} = b_{i} / ω$ , whence $χ_{i} (g) = n_{i} b_{i} / ω$ , that is $n_{i} b_{i} = ω χ_{i} (g)$ .

Locating the support. For $h \in G$ not conjugate to $g$ ,

a_{h} = \frac{1}{| G |} \sum_{i = 1}^{r} n_{i} b_{i} χ_{i} (h^{- 1}) = \frac{ω}{| G |} \sum_{i = 1}^{r} χ_{i} (g) χ_{i} (h^{- 1}) = 0

by the Second Orthogonality Relation $(8.16)$ (B). Since $α$ is central its coefficients are constant on classes, so $α = ω C_{g}$ .

**Centrality of $g$ .** Applying $(8.16)$ (B) with $h = g$ ,

| C_{G} (g) | = \sum_{i = 1}^{r} χ_{i} (g) χ_{i} (g^{- 1}) = \sum_{i = 1}^{r} \frac{n_{i} b_{i}}{ω} \cdot \frac{n_{i} ω}{b_{i}} = \sum_{i = 1}^{r} n_{i}^{2} = | G | .

So $C_{G} (g) = G$ , that is $g \in Z (G)$ , and then $m_{g} = 1$ and $C_{g} = g$ , giving $α = ω g$ . Finally $α^{m} = 1$ and $g$ of finite order force $ω$ to be a root of unity, and $ω = α g^{- 1} \in A G$ has all coefficients in $A$ , so $ω \in A$ .

Remark(8.23)The hypothesis can be weakened

The proof never used that all coefficients of $α$ lie in $A$ — only that the one coefficient $a_{g}$ singled out is a nonzero element of $A$ . So a central unit of finite order in $k G$ with at least one nonzero coefficient in $A$ is already of the form $ω g$ . If every coefficient lies outside $A$ , the conclusion can fail: $α^{m} = 1$ need not make $α$ a $k$ -multiple of a group element.

Corollary(8.24)Two special cases

If $G$ is a finite group with $Z (G) = 1$ , then every central unit of finite order in $A G$ is a root of unity in $A$ .
If $G$ is a finite abelian group and $A$ is the ring of algebraic integers of a number field $k$ that is not totally imaginary — that is, $k$ has at least one real embedding — then every unit of finite order in $A G$ has the form $\pm g$ with $g \in G$ .

Proof

(1) is $(8.21)$ with $Z (G) = {1}$ . For (2), $G$ abelian makes every unit central, so $(8.21)$ gives $α = ω g$ with $ω$ a root of unity in $A$ . A real embedding sends the finite group of roots of unity of $k$ injectively into the roots of unity of $ℝ$ , which are $\pm 1$ ; hence $ω = \pm 1$ .

Corollary(8.25)Higman–Berman: integral group rings detect the centre

Let $A$ be as in $(8.21)$ and let $G$ and $H$ be finite groups. If $A G$ and $A H$ are isomorphic as rings, then $Z (G) ≅ Z (H)$ as groups.

Proof

A ring isomorphism carries central units of finite order to central units of finite order, so those two groups are isomorphic. Let $U$ denote the (finite, cyclic) group of roots of unity in $A$ . By $(8.21)$ the central torsion units of $A G$ are exactly the elements $ω g$ with $ω \in U$ and $g \in Z (G)$ , and the map $U \times Z (G) \to U (A G)$ is injective because $G$ is an $A$ -basis of $A G$ . Hence $U \times Z (G) ≅ U \times Z (H)$ as finite abelian groups, and cancellation in the Fundamental Theorem of Finite Abelian Groups gives $Z (G) ≅ Z (H)$ .

Theorem(8.26)Only trivial idempotents

Let $k$ be a field of algebraic numbers with ring of algebraic integers $A$ , and let $G$ be any finite group. Then the only idempotents of $A G$ are $0$ and $1$ .

Proof

Enlarge $k$ as before so that it is a finite splitting field for $G$ of characteristic $0$ ; an idempotent of $A G$ remains one in $A' G$ , and $0$ and $1$ are unaffected. Let $e = \sum_{g} a_{g} g$ be an idempotent and $e' = 1 - e = \sum_{g} a'_{g} g$ its complement.

Let $θ_{i 1}, \dots, θ_{i n_{i}}$ be the eigenvalues of the action of $e$ on $M_{i}$ . Since $e^{2} = e$ that action is an idempotent operator, so each $θ_{i j} \in {0, 1}$ and $χ_{i} (e) = θ_{i 1} + \dots + θ_{i n_{i}} \in {0, 1, \dots, n_{i}}$ .

Apply the regular character $χ_{reg} = \sum_{i} n_{i} χ_{i}$ , using $χ_{reg} (e) = a_{1} | G |$ :

a_{1} = \frac{1}{| G |} χ_{reg} (e) = \frac{1}{| G |} \sum_{i = 1}^{r} n_{i} χ_{i} (e) \in ℚ,

(8.27)

and since $0 \leq χ_{i} (e) \leq n_{i}$ we get $0 \leq a_{1} \leq | G |^{- 1} \sum_{i} n_{i}^{2} = 1$ . But $a_{1} \in A \cap ℚ = ℤ$ , so $a_{1} \in {0, 1}$ ; the same applies to $a'_{1}$ , and $a_{1} + a'_{1} = 1$ . Replacing $e$ by $e'$ if necessary, assume $a_{1} = 1$ and $a'_{1} = 0$ .

Then $(8.27)$ gives $\sum_{i} n_{i} χ_{i} (e) = | G | = \sum_{i} n_{i}^{2}$ , and since $χ_{i} (e) \leq n_{i}$ term by term, equality forces $χ_{i} (e) = n_{i}$ for every $i$ . Hence every $θ_{i j} = 1$ and $e$ acts as the identity on each $M_{i}$ . As $char k = 0$ , $k G$ is semisimple and $k G ≅ ⨁_{i} n_{i} M_{i}$ as a left module, so $e$ acts as the identity on $k G$ ; applying this to $1$ gives $e = 1$ .

Remark—Two other proofs

Swan proved that a finitely generated projective left $A G$ -module $P$ satisfies $k \otimes_{A} P$ free over $k G$ , so the $A$ -rank of $P$ is a multiple of $| G |$ . A nontrivial idempotent $e$ would make $P = A G e$ projective of $A$ -rank strictly between $0$ and $| G |$ , which is impossible; $(8.26)$ follows. Coleman's argument is shorter still: $\dim_{k} (e k G) = χ_{reg} (e) = a_{1} | G |$ shows directly that $a_{1}$ is a non-negative rational integer at most $| G |$ divided by $| G |$ , hence $0$ or $1$ .

Proof Techniques and Method

How these proofs work, and which move to reuse.

Enlarge the coefficient fieldPass to a finite Galois splitting extension. Check that the conclusion descends — usually because the object produced is visibly defined over the smaller field.

Change to the idempotent basisMultiplicative hypotheses (

α^{m} = 1

e^{2} = e

) become coordinatewise and trivial to solve.

Return to the group basis and estimateUse

(8.15) (1)

to write a coefficient as an average of

| G |

unit-modulus numbers; the triangle inequality gives modulus at most

1

Invoke arithmeticA nonzero algebraic integer whose conjugates all have modulus at most

1

has norm of absolute value

1

; the estimate is then an equality and the equality case bites.

Apply orthogonalityThe rigidity produced by the equality case is fed into the second orthogonality relation to locate the support and to compute a centraliser order.

Note where the two proofs diverge. For units the constraint is multiplicative and the estimate is genuinely complex-analytic. For idempotents the constraint is on eigenvalues in ${0, 1}$ , so the estimate is a finite count and no absolute values are needed beyond the trivial bound.

Worked Example

Take $A = ℤ$ , $k = ℚ$ , and $G = ⟨ x ⟩$ cyclic of order $5$ . Then $ℚ G ≅ ℚ \times ℚ (ζ_{5})$ , where $ζ_{5}$ is a primitive fifth root of unity.

Idempotents

$ℚ G$ has exactly four idempotents: $0$ , $1$ , $e_{1} = \frac{1}{5} (1 + x + x^{2} + x^{3} + x^{4})$ and $1 - e_{1}$ . Both nontrivial ones have denominator $5$ , so neither lies in $ℤ G$ — and $(8.26)$ says that no other idempotent could have appeared either. The integral group ring has only $0$ and $1$ .

Torsion units

$G$ is abelian and $ℚ$ has a real embedding, so $(8.24) (2)$ applies: the units of finite order in $ℤ G$ are exactly the ten trivial units $\pm 1, \pm x, \pm x^{2}, \pm x^{3}, \pm x^{4}$ .

A unit of infinite order

Torsion units are trivial, but units are not. Put $u = 1 - x^{2} - x^{3}$ and $v = 1 - x - x^{4}$ . Expanding with $x^{5} = 1$ :

\begin{aligned} u v & = (1 - x^{2} - x^{3}) (1 - x - x^{4}) \\ = (1 - x - x^{4}) + (- x^{2} + x^{3} + x) + (- x^{3} + x^{4} + x^{2}) = 1 . \end{aligned}

(E.1)

The middle line uses $x^{6} = x$ and $x^{7} = x^{2}$ ; every non-identity term cancels.

So $u \in U (ℤ G)$ . It has infinite order: under the ring map $ℤ G \to ℤ [ζ_{5}]$ , $x \mapsto ζ_{5}$ , one computes $x u \mapsto ζ - ζ^{3} - ζ^{4} = (1 + ζ)^{2}$ , using $1 + ζ + ζ^{2} + ζ^{3} + ζ^{4} = 0$ . Since $1 + ζ$ is a fundamental unit of $ℤ [ζ_{5}]$ , its square has infinite order, and so does $u$ .

Units and idempotents of $ℤ C_{5}$ against $ℚ C_{5}$
	$ℤ C_{5}$	$ℚ C_{5}$
Idempotents	$0, 1$ only, by $(8.26)$	four: $0, 1, e_{1}, 1 - e_{1}$
Torsion units	$\pm x^{j}$ , ten of them, by $(8.24)$	infinitely many
Full unit group	$⟨ u ⟩ \times (\pm G) ≅ ℤ \oplus ℤ / 2 \oplus ℤ / 5$	$ℚ^{\times} \times ℚ (ζ_{5})^{\times}$
Ring type	an order, not semisimple	semisimple, a product of fields

Comparison and Classification

Idempotents and torsion units of $R G$ for $G$ finite
Coefficient ring $R$	Idempotents	Torsion units	Reason
$k$ of characteristic $0$	many, unless $G = 1$	many	$k G$ semisimple; $e_{i}$ available
$A$ , algebraic integers	$0$ and $1$ only	central ones are $ω g$	$(8.26)$ , $(8.21)$
$ℤ$	$0$ and $1$ only	central ones are $\pm g$	$(8.26)$ , $(8.24)$
$ℤ_{(p)}$ or $ℤ_{p}$	nontrivial ones exist when $p ∤ \| G \|$	more than trivial	idempotents lift modulo $p$
$𝔽_{p}$ , $p ∤ \| G \|$	many	many	semisimple again
$𝔽_{p}$ , $G$ a $p$ -group	$0$ and $1$ only	$1 +$ augmentation ideal	$𝔽_{p} G$ is local

Which theorem applies to which kind of unit
	$G$ finite, $R = A$	$G$ finite, $R$ a field	$G$ infinite torsion-free
Central torsion units classified	yes	no	no
All torsion units classified	partial	no	no
Idempotents classified	yes	no	no
Statement is a theorem, not a conjecture	yes	yes	no

Which theorem applies to which kind of unit

Row two is deliberately partial. $(8.21)$ says nothing about non-central torsion units, and the sharpest general statement in that direction — the first Zassenhaus conjecture, that every torsion unit of augmentation $1$ in $ℤ G$ is conjugate in $ℚ G$ to a group element — is now known to be false in general.

Relationship Map

$U (A G)$ the full unit group; infinite as soon as

G

is large enough

Torsion unitsfinite order; augmentation a root of unity

Central torsion unitsclassified by

(8.21)

: exactly

{ω g : ω \in U (A), g \in Z (G)}

Trivial units

\pm g

when

A = ℤ

(8.15)

idempotent formula

(8.16)

B orthogonality

(8.20)

modulus lemma

(8.21)

(8.25)

$(8.26)$ branches off earlier: it needs the regular character and $A \cap ℚ = ℤ$ , but neither the orthogonality relations nor the modulus lemma. That is why it holds for any finite group with no centrality hypothesis anywhere.

Applications and Industry Use

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

Integral representation theory

The isomorphism problem

$(8.25)$ was an early positive result: $ℤ G ≅ ℤ H$ forces $Z (G) ≅ Z (H)$ . Hertweck's 2001 counterexample shows $ℤ G$ does not determine $G$ , so results of this shape mark exactly how much information survives.

Algebraic K-theory

Projective modules over orders

Swan's route to $(8.26)$ goes through projective $A G$ -modules; the same circle of ideas produces the class group of an order and the finiteness obstruction in surgery theory.

Topology

Whitehead groups and torsion

The Whitehead group $W h (G)$ is built from $U (ℤ G)$ modulo trivial units. Knowing that the trivial units are exactly $\pm g$ for central torsion is what makes the quotient well behaved.

Computational algebra

Unit group computations

GAP and Magma compute unit groups of $ℤ G$ for small $G$ using Bass and Bass–Milnor cyclic units; the classification of the torsion part supplied by $(8.21)$ is what makes the computation terminate with a proof.

The honest summary: these are internal results about the arithmetic of orders. Their downstream users are algebraic topology and K-theory, where $ℤ G$ is the natural coefficient ring for a space with fundamental group $G$ .

Design Considerations

Design considerations here means the choices made when modelling a problem with these algebraic structures.

You need a group ring. Which coefficient ring should you use?

A field of characteristic $0$ Choose this when you want semisimplicity and the full Wedderburn decomposition. You gain idempotents and lose all arithmetic information.

$ℤ$ or a ring of integersChoose this when arithmetic invariants matter — class groups, Whitehead groups, isomorphism problems. Expect no idempotents at all and a unit group that is hard but finitely generated.

A local ring or $p$ -adic integersChoose this to isolate one prime. Idempotents lift modulo the maximal ideal, so the ring is between the two extremes and is the natural home for block theory.

A field of characteristic $p$ dividing $| G |$ Choose this for modular representation theory. The radical is nonzero and the interesting structure is in the blocks and the defect groups.

Which units to study. Central torsion units are completely understood; general torsion units are not, and non-torsion units require Bass's cyclic unit constructions. Decide early which class the problem needs.
Whether to enlarge the field. Enlarging $k$ to a splitting field is free for the theorems here but changes $A$ , hence changes which roots of unity are available. If the conclusion mentions $A$ , check that it descends.
Which side. $A G$ is noncommutative when $G$ is, but every result on this page concerns the centre or two-sided data, so no left–right choice arises. That is unusual in this collection and worth noticing.
What replaces idempotents. With no nontrivial idempotents, $A G$ has no ring direct decomposition; the working substitute is the block decomposition of $A_{𝔭} G$ after localising at a prime.

Failure Modes and Common Mistakes

$(8.26)$ requires $G$ finite. For infinite groups the idempotent question is Kaplansky's conjecture, open in general even over $ℂ$ — see the Group Ring Problems page.
$(8.26)$ requires $A$ to be a ring of algebraic integers. Over $ℤ_{(p)}$ with $p ∤ | G |$ , nontrivial idempotents do exist, because $| G |$ becomes invertible.
In $(8.24) (2)$ the hypothesis is that $k$ has a real embedding. Over $ℚ (i)$ the roots of unity include $\pm i$ and the torsion units of $ℤ [i] G$ are correspondingly larger.
$(8.25)$ concludes only about centres. It does not say $ℤ G$ determines $G$ — that statement is false, by Hertweck's counterexample.
Do not expect the estimates to work over an arbitrary integral domain of characteristic $0$ . The proof needs a finite Galois group acting on the coefficients and a norm landing in $ℤ$ .

Historical Notes and Lessons Learned

1940Higman's thesisGraham Higman studies the units of group rings, proves the abelian case of the theorem on central torsion units, and determines which finite groups have only trivial units in their integral group rings.
1955BermanS. D. Berman obtains related results on torsion units and on what the integral group ring remembers about the group, in particular its centre.
1960Swan on projective modulesSwan proves that finitely generated projective modules over an integral group ring become free after extension to the rational group algebra, which yields the triviality of idempotents as a corollary.
1960s–70sCharacter-theoretic proofsTakahashi gives the direct character proof of the idempotent theorem reproduced here, and Coleman finds a still shorter dimension-counting argument.
2001Hertweck's counterexampleThe isomorphism problem for integral group rings is settled negatively: there exist non-isomorphic finite groups with isomorphic integral group rings. Their centres are necessarily isomorphic, by the Higman–Berman corollary.
2018The first Zassenhaus conjecture failsEisele and Margolis produce a metabelian group whose integral group ring has a torsion unit of augmentation one not conjugate in the rational group algebra to any group element.

The lesson is that positive results here are narrow and negative results took sixty years. What survives is exactly what the arithmetic forces: the idempotents, and the central part of the torsion. Everything beyond that turned out to be false.

Quick Reference

Hypotheses

k

a field of algebraic numbers,

A

its integers,

G

finite

$(8.21)$ central torsion unit

\Rightarrow α = ω g

g \in Z (G)

ω \in U (A)

a root of unity

$(8.23)$ one nonzero coefficient in

A

already suffices

$(8.24) (1)$

Z (G) = 1 \Rightarrow

central torsion units are roots of unity

$(8.24) (2)$

G

abelian,

k

with a real embedding

\Rightarrow

torsion units are

\pm g

$(8.25)$

A G ≅ A H \Rightarrow Z (G) ≅ Z (H)

$(8.26)$ idempotents of

A G

are

0

and

1

only

Key estimate

| a_{g} | \leq | G |^{- 1} \sum_{i} n_{i}^{2} = 1

Key arithmetic

A \cap ℚ = ℤ

;

N_{k / ℚ} (A) \subseteq ℤ

Where each ingredient is used
Ingredient	Used in	Role
$(8.15) (1)$	$(8.21)$ , $(8.26)$	converts idempotent coordinates to group coordinates
$(8.16)$ (B)	$(8.21)$	kills coefficients off the class of $g$ ; computes $\| C_{G} (g) \|$
$(8.20)$	$(8.21)$	modulus bound and its equality case
Galois norm	$(8.21)$	turns an inequality into an equality
$χ_{reg}$	$(8.26)$	extracts the coefficient $a_{1}$
$A \cap ℚ = ℤ$	both	converts an estimate into a finite list of possibilities

Frequently Asked Questions

Why is the integral group ring so much poorer in idempotents than the rational one?

Because the idempotents of $ℚ G$ carry denominators. By the central idempotent formula, the coefficient of the identity in $e_{i}$ is $n_{i}^{2} / | G |$ , which is an integer only when $n_{i}^{2} = | G |$ , that is when $r = 1$ and $G$ is trivial. Theorem $(8.26)$ says this obstruction is complete: no cleverer integral idempotent exists either.

Does the theorem on idempotents extend to infinite groups?

Not by this proof, and not as a theorem at all. For an infinite torsion-free group the assertion that $ℂ G$ has only trivial idempotents is Kaplansky's conjecture, still open in general; it is known for large classes of groups, for example those satisfying the Baum–Connes conjecture. The finite case here is genuinely arithmetic and does not generalise by analogy.

What is known about non-central torsion units?

Much less. The first Zassenhaus conjecture predicted that every torsion unit of augmentation $1$ in $ℤ G$ is conjugate in $ℚ G$ to a group element; it holds for many families, including nilpotent groups, but Eisele and Margolis produced a metabelian counterexample in 2018. Weaker statements about orders of torsion units remain active.

Where exactly does the proof of the unit theorem use that the field is Galois over the rationals?

In the norm step. The estimate $| a_{g} | \leq 1$ holds for each Galois conjugate because conjugating the coefficients of a central torsion unit produces another central torsion unit. Multiplying the conjugates gives $| N (a_{g}) | \leq 1$ , and only a Galois extension makes that product a rational integer. Enlarging $k$ to a Galois closure at the start is what buys this.

Is the corollary that isomorphic integral group rings have isomorphic centres the best possible?

It is best possible in the sense that the full isomorphism problem has a negative answer: Hertweck exhibited non-isomorphic finite groups $G$ and $H$ with $ℤ G ≅ ℤ H$ . Any such pair must have isomorphic centres, and indeed the same character table and the same order, so the invariants that do survive are quite strong without being complete.

How do I actually find a nontrivial unit in an integral group ring?

The standard constructions are the bicyclic units $1 + (1 - g) h \hat{g}$ , where $\hat{g}$ is the sum of the powers of $g$ , and the Bass cyclic units built from the identity $\sum_{j} x^{j}$ raised to suitable powers. For $ℤ C_{5}$ the element $1 - x^{2} - x^{3}$ works and is easier to check by hand; its inverse is $1 - x - x^{4}$ .

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §8, (8.20)–(8.26).
D. S. Passman, The Algebraic Structure of Group Rings, Wiley-Interscience, 1977.
S. K. Sehgal, Topics in Group Rings, Monographs and Textbooks in Pure and Applied Mathematics 50, Marcel Dekker, 1978.
G. Higman, “The units of group-rings”, Proceedings of the London Mathematical Society (2) 46 (1940).
R. G. Swan, “Induced representations and projective modules”, Annals of Mathematics 71 (1960).
C. W. Curtis and I. Reiner, Methods of Representation Theory, Volume I, Wiley-Interscience, 1981, Chapters 3 and 5.

AI Suggested Questions

Verify by hand that 1 minus x squared minus x cubed is a unit in the integral group ring of a cyclic group of order five and identify its image in the cyclotomic ring.
How does Swan's theorem on projective modules over an integral group ring imply that its only idempotents are trivial?
What is the current status of the first and third Zassenhaus conjectures for integral group rings?
Describe Hertweck's counterexample to the isomorphism problem and explain why its centres must agree.
Construct the Bass cyclic units and explain when they generate a subgroup of finite index in the unit group.
How is the Whitehead group of a finite group related to the unit group of its integral group ring?
Why does the idempotent theorem fail if the coefficient ring is localised at a prime not dividing the group order?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Two coordinate systems, again

The counting that makes the bound tight

From rigidity to the group

06Key Results

07Proof Techniques and Method

08Worked Example

Idempotents

Torsion units

A unit of infinite order

09Comparison and Classification

10Relationship Map

11Applications and Industry Use

The isomorphism problem

Projective modules over orders

Whitehead groups and torsion

Unit group computations

12Design Considerations

13Failure Modes and Common Mistakes

14Historical Notes and Lessons Learned

15Quick Reference

16Frequently Asked Questions

17Related KEVOS Topics

18References

19AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

Comparison and Classification

Relationship Map

Applications and Industry Use

Design Considerations

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions