Counting Irreducible Representations in Characteristic p

Q: Why do elements of p-power order not contribute?

Because g minus its p-prime part is nilpotent in characteristic p: if g = ab with a the p-prime part and b^p^n = 1, then (g - a)^p^n = a^p^nb^p^n - a^p^n = 0. Over a splitting field (7.17) puts every nilpotent element into T(R), so g and a become equal in R/T(R), and only the p-prime parts survive.

Q: Is the bound in (8.13) ever strict?

Yes. Over F_2 the group A_4 has 3 two-regular classes but only 2 simple modules, because F_2 lacks the cube roots of unity needed to split A_4/V_4 ≅ C_3. Over F_4 the count becomes exact.

Q: Does T(R) being non-ideal cause problems?

Only for careless quotienting. R/T(R) exists as a k-vector space and that is all the argument needs — (7.17) is a statement about its dimension. Attempting to give R/T(R) a ring structure and reason with it will fail.

Q: Does (8.10) really need no hypotheses on G or k?

Correct: G may be infinite and k any commutative ring. Elements of kG have finite support, so each class coefficient sum is a finite sum. Finiteness of G and the field hypothesis are needed only from (8.12) onward, where the radical and the splitting condition enter.

Q: What replaces Brauer's theorem for infinite groups?

Nothing of the same form. The proof depends on kG being finite-dimensional so that rad kG is nilpotent and (7.17) applies. For infinite G the number of simple kG-modules can be infinite, and the notion of a modular character table does not exist.

Executive Summary

In characteristic $0$ over a splitting field, the number of irreducible representations of a finite group equals the number of conjugacy classes. In characteristic $p$ that count drops, and Brauer identified exactly which classes survive: the ** $p$ -regular** ones, those whose elements have order prime to $p$ .

The proof is entirely ring-theoretic. Result $(7.17)$ expresses the number of simple modules over a splitting field as $\dim_{k} R / T (R)$ with $T (R) = rad R + [R, R]$ . For $R = k G$ that codimension is computed directly: the conjugacy class representatives give a basis modulo $[R, R]$ , and the $p$ -regular ones give a basis modulo $T (R)$ .

$\dim_{k} R / T (R)$ Counts the simples

$p$ -regularSurviving classes

$r \leq | J |$ General field

1935Brauer

Overview

Fix a finite group $G$ and a field $k$ of characteristic $p \geq 0$ , and set $R = k G$ . Write $[R, R]$ for the additive span of the commutators $a b - b a$ and

T (R) = rad R + [R, R] .

(7.17a)

A $k$ -subspace of $R$ , generally neither a left nor a right ideal. Its codimension is the object of interest.

Theorem $(7.17)$ says that if $R$ splits over $k$ , then the number of isomorphism classes of simple left $R$ -modules is $\dim_{k} R / T (R)$ , and moreover $T (R)$ contains every nilpotent element of $R$ . Both halves are used below: the first turns a counting problem into a dimension count, the second lets group elements be replaced by their $p$ -regular parts.

This is the counting counterpart of the results on Normal p-Subgroups and the Radical of kG, which showed that a normal $p$ -subgroup acts trivially on every simple module. There the $p$ -elements disappeared from the group; here they disappear from the count.

Learning Objectives

Define $p$ -regular elements and classes, and state the $p = 0$ convention.
Prove $(8.10)$ : membership of $[R, R]$ is the vanishing of every class coefficient sum.
Deduce $(8.11)$ : $R / [R, R]$ is free on the conjugacy class representatives.
Prove $(8.12)$ : over a splitting field of characteristic $p$ , the $p$ -regular representatives are a basis of $R / T (R)$ .
Assemble Brauer's theorem $(8.9)$ from $(7.17)$ and $(8.12)$ .
Derive the inequality $(8.13)$ valid over any field of characteristic $p$ , and exhibit a case where it is strict.

Definitions

Definition—p-regularity

Let $p$ be a prime and $G$ a finite group. An element $g \in G$ is ** $p$ -regular** if $p$ does not divide the order of $g$ . Since conjugate elements have the same order, a conjugacy class is $p$ -regular if one — equivalently every — element of it is. By convention every element and every class is ** $0$ -regular**, so the statements below cover characteristic $0$ uniformly.

$[a, b]$: The additive commutator $a b - b a$ , sometimes called the Lie product.
$[R, R]$: The additive subgroup generated by all $[a, b]$ . For a $k$ -algebra it is a $k$ -submodule, but it is generally not an ideal on either side.
$T (R)$: $rad R + [R, R]$ , for $R$ a finite-dimensional $k$ -algebra.
${a_{i} : i \in I}$: A complete set of representatives of the conjugacy classes of $G$ .
${a_{j} : j \in J}$: The subfamily indexed by $J \subseteq I$ consisting of representatives of the $p$ -regular classes.
Class sum $\hat{C}$: For a conjugacy class $C$ , the element $\sum_{g \in C} g$ of $k G$ . The class sums form a $k$ -basis of the centre $Z (k G)$ , for any commutative $k$ .

In (8.10) and (8.11) G may be arbitrary and k any commutative ring. From (8.12) onward G is finite and k is a field.

Core Concepts

The two congruences that do all the work

Everything reduces to two statements about when group elements become equal in a quotient of $k G$ .

g_{1} \sim g_{2} in G ⟹ g_{1} \equiv g_{2} (\mod [R, R]), g \equiv g_{p'} (\mod T (R)),

(8.12a)

Conjugate elements are congruent modulo commutators; every element is congruent to its $p$ -regular part modulo $T (R)$ . Here $g_{p'}$ denotes the $p$ -prime part of $g$ .

The first congruence is elementary: if $g_{2} = x^{- 1} g_{1} x$ , put $g = x^{- 1}$ and $h = g_{1} x$ ; then $g h = g_{2}$ and $h g = g_{1}$ , so $g_{1} - g_{2} = h g - g h \in [R, R]$ .

The second uses characteristic $p$ twice. Factor $g = a b = b a$ with $a$ the $p$ -prime part and $b$ the $p$ -part; both are powers of $g$ , so they commute. If $b^{p^{n}} = 1$ then

(a b - a)^{p^{n}} = (a b)^{p^{n}} - a^{p^{n}} = a^{p^{n}} b^{p^{n}} - a^{p^{n}} = 0,

(8.12b)

So $g - a$ is nilpotent. When $k$ splits $R$ , $(7.17)$ places every nilpotent element in $T (R)$ , giving $g \equiv a$ .

Why the count cannot be larger

Spanning is therefore easy; the substance is linear independence of the $p$ -regular representatives modulo $T (R)$ . That is where $(7.15)$ enters: raising to a power $q = p^{N}$ chosen with $q \equiv 1$ modulo the exponent of the $p$ -regular part is an additive map modulo $[R, R]$ , fixes each $a_{j}$ , preserves $[R, R]$ , and kills the radical part. The result is an identity in $R / [R, R]$ , where $(8.11)$ already supplies a basis.

(8.11)

: basis of

R / [R, R]

(8.12)

: basis of

R / T (R)

(7.17)

r = \dim_{k} R / T (R)

Brauer

(8.9)

Key Results

Lemma(8.10)Membership in the commutator subspace

Let $G$ be any group, $k$ any commutative ring, and $R = k G$ . An element $a = \sum_{g} a_{g} g \in R$ lies in $[R, R]$ if and only if, for every conjugacy class $C$ of $G$ , $\sum_{g \in C} a_{g} = 0$ .

Proof

Necessity. $[R, R]$ is generated additively by elements $α β - β α$ . Writing $α = \sum_{g} α_{g} g$ and $β = \sum_{h} β_{h} h$ gives $α β - β α = \sum_{g, h} α_{g} β_{h} (g h - h g)$ , so it suffices to treat $g h - h g$ . Now $h g = g^{- 1} (g h) g$ , so $g h$ and $h g$ are conjugate: the element $g h - h g$ has coefficient sum $1 - 1 = 0$ on the class containing them, and $0$ on every other class. Coefficient sums over a class are additive, so they vanish on all of $[R, R]$ .

Sufficiency. First, conjugate elements are congruent modulo $[R, R]$ : if $g_{2} = x^{- 1} g_{1} x$ , set $g = x^{- 1}$ and $h = g_{1} x$ , so that $g h = g_{2}$ and $h g = g_{1}$ , whence $g_{1} - g_{2} = h g - g h \in [R, R]$ .

Now let $C = {g_{1}, \dots, g_{n}}$ be a conjugacy class and $a = c_{1} g_{1} + \dots + c_{n} g_{n}$ supported on $C$ . By the previous paragraph $a \equiv (c_{1} + \dots + c_{n}) g_{1} (\mod [R, R])$ , so if the coefficient sum vanishes then $a \in [R, R]$ . A general element of $R$ is a finite sum of such class-supported pieces, and if every class sum vanishes then each piece lies in $[R, R]$ .

Corollary(8.11)A basis modulo commutators

Let $G$ be any group, $k$ any commutative ring, $R = k G$ , and let ${a_{i} : i \in I}$ be a complete set of representatives of the conjugacy classes of $G$ . Then $R / [R, R]$ is a free $k$ -module with basis ${a_{i} + [R, R] : i \in I}$ .

Proof

Spanning: by the sufficiency argument of $(8.10)$ , every group element is congruent modulo $[R, R]$ to the chosen representative of its class, and the $g \in G$ span $R$ over $k$ .

Independence: if $\sum_{i} c_{i} a_{i} \in [R, R]$ then by $(8.10)$ the coefficient sum over the class of $a_{i}$ , namely $c_{i}$ , is zero for every $i$ .

Lemma(8.12)A basis modulo T(R)

Let $G$ be a finite group and $k$ a **splitting field for $G$ ** of characteristic $p > 0$ , and set $R = k G$ . Let ${a_{j} : j \in J}$ be a complete set of representatives of the $p$ -regular conjugacy classes of $G$ . Then $R / T (R)$ is a $k$ -vector space with basis ${a_{j} + T (R) : j \in J}$ .

Proof

Spanning. Let $g \in G$ and write $g = a b = b a$ with $a$ the $p$ -prime part and $b$ the $p$ -part of $g$ ; both are powers of $g$ , hence commute, and $b^{p^{n}} = 1$ for some $n$ . By $(8.12 b)$ , $g - a = a b - a$ is nilpotent, and since $k$ splits $R$ , $(7.17)$ gives that $T (R)$ contains all nilpotent elements of $R$ ; hence $g \equiv a (\mod T (R))$ . Conjugate elements are congruent modulo $[R, R] \subseteq T (R)$ by $(8.10)$ . So every $g \in G$ is congruent modulo $T (R)$ to the representative $a_{j}$ of the $p$ -regular class of its $p$ -prime part, and these span $R / T (R)$ .

Independence. Suppose $\sum_{j \in J} c_{j} a_{j} \in T (R)$ and write $\sum_{j} c_{j} a_{j} = c + d$ with $c \in rad R$ and $d \in [R, R]$ . Let $m$ be the least common multiple of the orders of the $a_{j}$ ; each is prime to $p$ , so $p$ is a unit modulo $m$ and there is $N_{0}$ with $p^{N_{0}} \equiv 1 (\mod m)$ . Since $p^{t N_{0}} \equiv 1 (\mod m)$ for every $t \geq 1$ , we may choose $q = p^{N}$ with $q \equiv 1 (\mod m)$ and $q$ at least the nilpotency index of $rad R$ — the radical is nilpotent because $R$ is finite-dimensional. Then $c^{q} = 0$ and $a_{j}^{q} = a_{j}$ for every $j$ .

Apply $(7.15) (1)$ to the sum $(\sum_{j} c_{j} a_{j}) + (- d)$ , which equals $c$ , and $(7.15) (2)$ to $d$ :

0 = c^{q} \equiv \sum_{j \in J} c_{j}^{q} a_{j}^{q} + (- d)^{q} \equiv \sum_{j \in J} c_{j}^{q} a_{j} (\mod [R, R]),

since $(- d)^{q} = \pm d^{q} \in [R, R]$ and $[R, R]$ is a $k$ -subspace. By $(8.11)$ the $a_{j}$ are part of a $k$ -basis of $R / [R, R]$ , so $c_{j}^{q} = 0$ and therefore $c_{j} = 0$ for every $j$ .

Theorem(8.9)Brauer

Let $G$ be a finite group and let $k$ be a splitting field for $G$ of characteristic $p \geq 0$ . Then the number of irreducible $k G$ -representations, up to equivalence, equals the number of $p$ -regular conjugacy classes of $G$ .

For $p = 0$ every class is $p$ -regular, so this recovers the classical count *number of irreducibles $=$ number of conjugacy classes*.

Proof

Put $R = k G$ , a finite-dimensional $k$ -algebra split by $k$ . By $(7.17)$ the number of simple left $R$ -modules is $\dim_{k} R / T (R)$ . If $p > 0$ , $(8.12)$ evaluates that dimension as $| J |$ , the number of $p$ -regular classes. If $p = 0$ then $char k ∤ | G |$ , so $rad R = 0$ by Maschke, $T (R) = [R, R]$ , and $(8.11)$ gives $\dim_{k} R / T (R) = | I |$ , the total number of classes — which is also the number of $0$ -regular classes by convention.

Corollary(8.13)The bound over an arbitrary field

Let $G$ be a finite group and $k$ any field of characteristic $p > 0$ . Then the number of irreducible $k G$ -representations is at most the number of $p$ -regular conjugacy classes of $G$ .

Proof

By $(8.3)$ there is a finite extension $K \supseteq k$ that is a splitting field for $G$ . Let $r$ be the number of simple left $k G$ -modules and $r'$ the number of simple left $K G$ -modules. By $(7.18)$ , $r \leq r'$ . By Brauer's theorem applied over $K$ , which has the same characteristic $p$ , $r'$ equals the number of $p$ -regular classes of $G$ . Hence $r \leq$ that number.

Remark(8.14)The dual argument in the semisimple case

Suppose $char k ∤ | G |$ , so that all $s$ conjugacy classes are $p$ -regular and $k G$ is semisimple. The class sums ${\hat{C}}_{1}, \dots, {\hat{C}}_{s}$ form a $k$ -basis of $Z (k G)$ . Writing $k G ≅ M_{n_{1}} (D_{1}) \times \dots \times M_{n_{r}} (D_{r})$ and taking centres, $Z (M_{n} (D)) = Z (D)$ gives

Z (k G) ≅ Z (D_{1}) \times \dots \times Z (D_{r}), s = \sum_{i = 1}^{r} \dim_{k} Z (D_{i}) \geq r,

with equality precisely when every $\dim_{k} Z (D_{i}) = 1$ , in particular when $k$ is a splitting field. This recovers $(8.9)$ and $(8.13)$ in the semisimple case by a shorter route, but says nothing when $p$ divides $| G |$ .

Remark—The exact count over a finite field

For $k = 𝔽_{q}$ with $q$ a power of $p$ , the inequality of $(8.13)$ can be replaced by an equality: the number of simple $𝔽_{q} G$ -modules equals the number of orbits of the $p$ -regular classes under the map $g \mapsto g^{q}$ . This refinement, due to Berman and to Witt, lies outside Lam's §8 and is recorded here only to explain how the deficit in $(8.13)$ arises.

Proof Techniques and Method

How these arguments work, and which move is worth reusing.

Move 1

Count by codimension

Replace how many simple modules? by *what is $\dim_{k} R / T (R)$ ?*. A counting problem becomes linear algebra, and linear algebra over a group ring becomes combinatorics of conjugacy classes.

Move 2

Raise to a $p$ -power

Modulo $[R, R]$ in characteristic $p$ , the map $x \mapsto x^{p}$ is additive. That converts a linear relation into another linear relation with the coefficients raised to a power, and kills any nilpotent term.

Move 3

Split off the $p$ -part

Every $g$ factors as a commuting product of its $p$ -part and its $p$ -prime part. Subtracting produces a nilpotent element, which the splitting hypothesis places inside $T (R)$ .

Move 2 is the least familiar and the most transferable. The Frobenius map being additive modulo commutators — $(7.15) (1)$ — is what makes it possible to clear the radical from an equation without knowing anything about the radical beyond its nilpotence.

Worked Example

$G = S_{3}$ across three characteristics

$S_{3}$ has three conjugacy classes: the identity, the three transpositions (order $2$ ), and the two $3$ -cycles (order $3$ ).

Brauer's count for S_3
$p$	$p$ -regular classes	Count	Splitting field	Irreducibles and their degrees
$0$	${1}$ , transpositions, $3$ -cycles	$3$	$ℚ$	$1, 1, 2$
$2$	${1}$ , $3$ -cycles	$2$	$𝔽_{2}$	$1, 2$
$3$	${1}$ , transpositions	$2$	$𝔽_{3}$	$1, 1$

Each row is consistent with the dimension formula $(8.1) (4)$ : $6 = 0 + 1 + 1 + 4$ over $ℚ$ ; $6 = 1 + 1 + 4$ over $𝔽_{2}$ ; $6 = 4 + 1 + 1$ over $𝔽_{3}$ .

Verifying $(8.12)$ by hand for $R = 𝔽_{2} S_{3}$

By $(8.11)$ , $\dim_{k} R / [R, R] = 3$ , so $\dim_{k} [R, R] = 3$ . The radical is $k σ$ with $σ = \sum_{g \in G} g$ , computed on The Structure of kG modulo Its Radical. Is $σ \in [R, R]$ ? Its class coefficient sums are $1$ , $3$ and $2$ , which in $𝔽_{2}$ are $1$ , $1$ and $0$ — not all zero, so by $(8.10)$ it is not. Hence

\dim_{k} T (R) = \dim_{k} [R, R] + 1 = 4, \dim_{k} R / T (R) = 6 - 4 = 2,

(E.1)

Exactly the number of $2$ -regular classes, and exactly the number of simple $𝔽_{2} S_{3}$ -modules.

$G = A_{4}$ at $p = 2$ : where the inequality is strict

$A_{4}$ has four conjugacy classes: ${1}$ ; the three double transpositions (order $2$ ); and two classes of $3$ -cycles, of four elements each. The $2$ -regular classes are ${1}$ and the two $3$ -cycle classes — three in total.

Over a splitting field of characteristic $2$ , say $𝔽_{4}$ , Brauer's theorem predicts $3$ irreducibles. Check it independently: $O_{2} (A_{4}) = V_{4}$ , so by $(8.4)$ the simple modules are those of $𝔽_{4} [A_{4} / V_{4}] = 𝔽_{4} C_{3}$ ; since $𝔽_{4}^{\times}$ has order $3$ it contains the cube roots of unity, so $𝔽_{4} C_{3} ≅ 𝔽_{4} \times 𝔽_{4} \times 𝔽_{4}$ and there are indeed $3$ simple modules, all $1$ -dimensional.

Now take $k = 𝔽_{2}$ , which is not a splitting field for $A_{4}$ . The same reduction gives $𝔽_{2} C_{3} ≅ 𝔽_{2} \times 𝔽_{4}$ , so there are only $2$ simple $𝔽_{2} A_{4}$ -modules, of $𝔽_{2}$ -dimensions $1$ and $2$ .

r = 2 < 3 = # {2 -regular classes of A_{4}} .

(E.2)

The inequality of $(8.13)$ is strict over a non-splitting field. The deficit is explained by the map $g \mapsto g^{2}$ fusing the two classes of $3$ -cycles.

Process and Workflow

List the conjugacy classesWith the order of a representative in each. This is pure group theory and is usually already known.

Discard the non- $p$ -regular onesKeep exactly the classes whose element order is prime to

p

Check that $k$ splits $G$ If it does, the count of surviving classes is the exact number of irreducibles. If not, it is only an upper bound.

Cross-check with the dimension formulaConfirm

| G | = \dim_{k} rad k G + \sum_{i} n_{i}^{2} \dim_{k} D_{i}

with the number of terms equal to the class count.

If short, look for fusionOver a finite field

𝔽_{q}

, the deficit is accounted for by orbits of

p

-regular classes under

g \mapsto g^{q}

You know the number of $p$ -regular classes. What can you conclude?

$k$ is a splitting fieldYou have the exact number of irreducible

k G

-representations. Combine with

(8.1) (4)

to constrain their degrees.

$k$ is not splittingYou have only an upper bound

(8.13)

. Extend

k

to a splitting field to make it exact, then descend.

$char k = 0$ Every class is

0

-regular, so the bound is the full class count — the classical statement, and an equality over a splitting field.

$G$ is a $p$ -groupOnly the identity class is

p

-regular, so there is exactly one irreducible: the trivial module. No further computation is needed.

Comparison and Classification

Class counts versus irreducible counts
$G$	Classes	$p$	$p$ -regular classes	Irreducibles over a splitting field
$C_{p}$	$p$	$p$	$1$	$1$
$S_{3}$	$3$	$2$	$2$	$2$
$S_{3}$	$3$	$3$	$2$	$2$
$A_{4}$	$4$	$2$	$3$	$3$
$A_{4}$	$4$	$3$	$2$	$2$
$S_{4}$	$5$	$2$	$2$	$2$
$Q_{8}$	$5$	$2$	$1$	$1$

The last row is the extreme case: a $p$ -group has only the identity as a $p$ -regular class, and correspondingly exactly one irreducible in characteristic $p$ — the conclusion already reached by $(8.4)$ .

Which hypotheses each result needs
	$G$ finite	$k$ a field	$k$ splitting	$char k = p > 0$
$(8.10)$ commutator criterion	no	no	no	no
$(8.11)$ basis of $R / [R, R]$	no	no	no	no
$(8.12)$ basis of $R / T (R)$	yes	yes	yes	yes
$(8.9)$ Brauer's count	yes	yes	yes	no
$(8.13)$ upper bound	yes	yes	no	yes
$(8.14)$ centre argument	yes	yes	no	no

Which hypotheses each result needs

Relationship Map

The logical structure is a two-stage reduction: from module counting to a codimension, and from the codimension to conjugacy classes.

Number of simple $k G$ -modules
- $(7.17)$ , needs $k$ splitting
  - $= \dim_{k} R / T (R)$ where $T (R) = rad R + [R, R]$
  - $T (R)$ contains every nilpotent element of $R$
- $(8.10)$ – $(8.11)$ , no hypotheses
  - $a \in [R, R] iff$ every class coefficient sum vanishes
  - $R / [R, R]$ free on the class representatives
- $(8.12)$ , needs $char k = p$ and splitting
  - $p$ -elements are absorbed: $g \equiv g_{p'} (\mod T (R))$
  - $R / T (R)$ has the $p$ -regular representatives as a basis
- $(7.18)$ + $(8.3)$ , arbitrary $k$
  - $r \leq r'$ for a splitting extension, giving the bound $(8.13)$

conjugacy classes

p

-regular classes

\dim_{k} R / T (R)

simple

k G

-modules

Applications and Industry Use

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

Modular representation theory

Sizing the Brauer character table

The number of $p$ -regular classes fixes the shape of the $p$ -modular character table before any module is computed: it is a square table indexed by the $p$ -regular classes, and the decomposition matrix has that many columns.

Computational algebra

A termination criterion

Algorithms that enumerate irreducible modules over $𝔽_{q}$ need to know when to stop. Counting $p$ -regular classes gives the target in advance, so the search can be halted with a certificate of completeness.

Finite group theory

Structural consequences

Bounds relating the number of irreducibles to group structure — for instance that a group with few $p$ -regular classes is close to a $p$ -group — feed into classification arguments and into the study of blocks of defect zero.

Coding and cryptography

Counting components

For codes built from $𝔽_{q} G$ , the number of simple components governs how far the algebra decomposes and hence how many minimal ideals are available as codes. Over a non-splitting field the count drops, and the components are larger.

Honestly framed: this is a counting theorem, and its use is as a bookkeeping constraint. It tells you how many objects to look for, not what they are.

Design Considerations

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

Split before counting. Brauer's theorem is an equality only over a splitting field. Working over $𝔽_{p}$ when $𝔽_{p^{d}}$ is the splitting field will silently undercount, and the undercount is not an error but a genuine difference in the module category.
Which conjugacy classes to store. For modular work only the $p$ -regular classes matter, and there are usually far fewer of them. Restricting the class list to those is a real saving for large groups.
Ordinary or modular first? The decomposition matrix relates the two counts. If ordinary characters are already known, computing the modular count first tells you the shape of the matrix and how much work remains.
** $T (R)$ is not an ideal.** Treating $rad R + [R, R]$ as an ideal is a natural slip and breaks every argument that quotients by it as a ring. It is only a $k$ -subspace, and $R / T (R)$ is only a vector space.

Standards and Notation

Standards here covers notation, symbol and markup standards, and reference implementations, rather than material or design codes.

Number of irreducibles

r

here;

l (G)

l (B)

in the block-theoretic literature

Ordinary count

k (G)

for the number of conjugacy classes; note the clash with the field name

k

Brauer characters

IBr (G)

{IBr}_{p} (G)

for the irreducible

p

-modular Brauer characters

Commutator subspace

[R, R]

here;

[R, R]

K (R)

elsewhere — not to be confused with the commutator subgroup

G'

$p$ -regularalso called *

p

-singular complement*, or

p'

-elements;

G_{p'}

for the set of them

GAPIBr(CharacterTable(G), p), OrdersClassRepresentatives, BrauerTable

MagmaBrauerCharacterTable, AbsolutelyIrreducibleModules(G, GF(q))

MarkupPresentation MathML per ISO/IEC 40314; symbols per ISO 80000-2

Computational Notes

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

Counting $p$ -regular classes is cheap; producing the corresponding modules is not.

Given a conjugacy class list with representative orders, the $p$ -regular count is a single pass — negligible cost, and available from a character table without any module computation.
Constructing the irreducible $𝔽_{q} G$ -modules is far more expensive: the standard route is to build a faithful permutation or matrix module and split it with the MeatAxe, at $O (d^{3})$ per split attempt for degree $d$ .
The Brauer character table is not computable by a uniform algorithm for large groups; the Modular Atlas records known cases, and computing a new one is a research-scale task.
The refinement over $𝔽_{q}$ — orbits of $p$ -regular classes under $g \mapsto g^{q}$ — is also a cheap combinatorial computation, and it is what a system uses to predict the count over a non-splitting field.

Failure Modes and Common Mistakes

Do not assume the number of irreducibles in characteristic $p$ is at most the number in characteristic $0$ by some functorial argument on the modules themselves; the comparison is genuinely through class counts, and the decomposition matrix is what links the two.
Do not forget that the identity class is always $p$ -regular, so the count is at least one — consistent with the trivial module always existing.
Do not apply $(8.12)$ when $char k = 0$ ; the proof uses the Frobenius map, and the $p = 0$ case is handled separately by $(8.11)$ together with Maschke's theorem.

Historical Notes and Lessons Learned

1896Frobenius counts in characteristic zeroThe number of irreducible complex representations of a finite group is shown to equal the number of conjugacy classes, via the group determinant and the centre of the group algebra.
1902–07Dickson observes the dropWorking modulo a prime dividing $| G |$ , Dickson finds fewer irreducibles than classes, without a general formula for how many.
1935Brauer's theoremBrauer proves that the deficit is exactly accounted for by the classes of elements of order divisible by $p$ : the count is the number of $p$ -regular classes over a splitting field.
1940s–50sBrauer characters and decomposition matricesThe $p$ -regular classes become the index set of the modular character table, and the decomposition matrix formalises the relation to the ordinary characters.
1955 onwardCounts over non-splitting fieldsBerman and Witt refine the picture for finite ground fields: the count is the number of orbits of $p$ -regular classes under the $q$ -power map, explaining exactly when $(8.13)$ is strict.

The methodological point is that the proof is not representation-theoretic at all. It computes the codimension of a subspace of a ring, and the group theory enters only through the combinatorics of conjugacy classes. That is why the argument generalises to other algebras with a distinguished basis, and why $(8.10)$ and $(8.11)$ hold with no finiteness hypothesis whatever.

Quick Reference

$p$ -regularorder of

g

not divisible by

p

; all elements are

0

-regular

$(8.10)$

a \in [R, R] iff

every class coefficient sum of

a

0

$(8.11)$

R / [R, R]

free on the conjugacy class representatives

$T (R)$

rad R + [R, R]

; a

k

-subspace, not an ideal

$(7.17)$

k

splitting

\Rightarrow

number of simples

= \dim_{k} R / T (R)

$(8.12)$ basis of

R / T (R)

: the

p

-regular class representatives

Brauer $(8.9)$

k

splitting,

char k = p

: number of irreducibles

=

number of

p

-regular classes

$(8.13)$ any

k

of characteristic

p

: number of irreducibles

\leq

that count

The two key congruences and what they need
Congruence	Modulo	Hypotheses
$g_{1} \equiv g_{2}$ for conjugate $g_{1}, g_{2}$	$[R, R]$	none — any group, any commutative $k$
$g \equiv g_{p'}$	$T (R)$	$char k = p$ , $k$ splits $R$ , $R$ finite-dimensional
$(\sum x_{i})^{p} \equiv \sum x_{i}^{p}$	$[R, R]$	$char R = p$ — $(7.15) (1)$
$d \in [R, R] \Rightarrow d^{p} \in [R, R]$	—	$char R = p$ — $(7.15) (2)$

Frequently Asked Questions

Why do elements of

p

-power order not contribute?

Because $g$ minus its $p$ -prime part is nilpotent in characteristic $p$ : if $g = a b$ with $a$ the $p$ -prime part and $b^{p^{n}} = 1$ , then $(g - a)^{p^{n}} = a^{p^{n}} b^{p^{n}} - a^{p^{n}} = 0$ . Over a splitting field $(7.17)$ puts every nilpotent element into $T (R)$ , so $g$ and $a$ become equal in $R / T (R)$ , and only the $p$ -prime parts survive.

Is the bound in

(8.13)

ever strict?

Yes. Over $𝔽_{2}$ the group $A_{4}$ has $3$ two-regular classes but only $2$ simple modules, because $𝔽_{2}$ lacks the cube roots of unity needed to split $A_{4} / V_{4} ≅ C_{3}$ . Over $𝔽_{4}$ the count becomes exact.

Does

T (R)

being non-ideal cause problems?

Only for careless quotienting. $R / T (R)$ exists as a $k$ -vector space and that is all the argument needs — $(7.17)$ is a statement about its dimension. Attempting to give $R / T (R)$ a ring structure and reason with it will fail.

How does this compare with the count of ordinary irreducibles?

The number of ordinary irreducibles is the total number of classes; the modular number is the number of $p$ -regular classes. The precise relationship between the two sets of representations is the decomposition matrix, whose rows are indexed by ordinary characters and whose columns are indexed by the $p$ -regular classes.

Does

(8.10)

really need no hypotheses on

G

k

Correct: $G$ may be infinite and $k$ any commutative ring. Elements of $k G$ have finite support, so each class coefficient sum is a finite sum. Finiteness of $G$ and the field hypothesis are needed only from $(8.12)$ onward, where the radical and the splitting condition enter.

What replaces Brauer's theorem for infinite groups?

Nothing of the same form. The proof depends on $k G$ being finite-dimensional so that $rad k G$ is nilpotent and $(7.17)$ applies. For infinite $G$ the number of simple $k G$ -modules can be infinite, and the notion of a modular character table does not exist.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §8, (8.9)–(8.14) (pp. 132–136); §7, (7.15)–(7.18).
R. Brauer, “Über die Darstellung von Gruppen in Galoisschen Feldern”, Actualités Scientifiques et Industrielles 195, Hermann, Paris, 1935.
C. W. Curtis and I. Reiner, Representation Theory of Finite Groups and Associative Algebras, Wiley-Interscience, 1962, Chapter XII.
H. Nagao and Y. Tsushima, Representations of Finite Groups, Academic Press, 1989, Chapter 3.
J. L. Alperin, Local Representation Theory, Cambridge Studies in Advanced Mathematics 11, Cambridge University Press, 1986.
D. S. Passman, The Algebraic Structure of Group Rings, Wiley-Interscience, 1977.

AI Suggested Questions

Construct the decomposition matrix of $S_{4}$ at $p = 2$ and check that it has as many columns as there are $2$ -regular classes.
Prove the Berman–Witt count of simple $𝔽_{q} G$ -modules as orbits of $p$ -regular classes under the $q$ -power map.
Give a group and a prime for which the number of $p$ -regular classes is much smaller than the number of classes, and interpret the collapse structurally.
Work through $(7.15)$ in detail: why is the Frobenius map additive modulo the commutator subspace?
How does the count change when one restricts to a single $p$ -block, and what is the analogue of $(8.9)$ there?
For which finite groups is the number of $p$ -regular classes equal to the number of conjugacy classes of $G / O_{p} (G)$ ?
Explain why $(8.10)$ and $(8.11)$ hold over an arbitrary commutative ring while $(8.12)$ needs a splitting field.

Counting Irreducibles in Characteristic p

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

The two congruences that do all the work

Why the count cannot be larger

Key Results

Proof Techniques and Method

Count by codimension

Raise to a $p$ -power

Split off the $p$ -part

Worked Example

$G = S_{3}$ across three characteristics

Verifying $(8.12)$ by hand for $R = 𝔽_{2} S_{3}$

$G = A_{4}$ at $p = 2$ : where the inequality is strict

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Sizing the Brauer character table

A termination criterion

Structural consequences

Counting components

Design Considerations

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

The two congruences that do all the work

Why the count cannot be larger

06Key Results

07Proof Techniques and Method

Count by codimension

Raise to a p-power

Split off the p-part

08Worked Example

G=S3 across three characteristics

Verifying (8.12) by hand for R=𝔽2S3

G=A4 at p=2: where the inequality is strict

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Sizing the Brauer character table

A termination criterion

Structural consequences

Counting components

13Design Considerations

14Standards and Notation

15Computational Notes

16Failure Modes and Common Mistakes

17Historical Notes and Lessons Learned

18Quick Reference

19Frequently Asked Questions

20Related KEVOS Topics

21References

22AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Raise to a $p$ -power

Split off the $p$ -part

Worked Example

$G = S_{3}$ across three characteristics

Verifying $(8.12)$ by hand for $R = 𝔽_{2} S_{3}$

$G = A_{4}$ at $p = 2$ : where the inequality is strict

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Design Considerations

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions