Normal p-Subgroups and the Radical of kG

Q: Does (8.6) mean rad kG is determined by O_p(G)?

No. It says only which *group elements* h have h - 1 in the radical. The radical usually contains much more, and F_2 S_3 shows it can be nonzero while O_p(G) = 1. The full radical is determined by O_p(G) only in the Wallace situation, where a Sylow p-subgroup is normal.

Q: Is Clifford's theorem true without any finiteness hypothesis?

The proof needs a simple kH-submodule to exist, which is where finiteness enters. That is automatic when V is finite-dimensional over k or when H is finite. For an infinite H and an infinite-dimensional simple V one must assume it. The conclusion, once a simple kH-submodule exists, requires no further hypotheses.

Q: Why does J^|G| = 0 rather than some smaller power?

|G| is simply the number of composition factors of the regular module, and the radical shifts a composition series down by one step each time it is applied. The bound is a termination guarantee, not an estimate: for G elementary abelian of order p^n the exact nilpotency index is n(p-1)+1, far below p^n.

Q: How does this interact with Maschke's theorem?

They are complementary. Maschke says rad kG = 0 when p |G|, so O_p(G) = 1 and the results here are vacuous. When p |G| they supply the part of the radical that group theory can see. The two together explain the whole picture when a Sylow p-subgroup is normal, since then kG/rad kG ≅ k[G/H] is exactly the Maschke case.

Q: Is kG local only for p-groups?

Over a field of characteristic p, kG is local if and only if G is a finite p-group. If G is not a p-group, G/O_p(G) is a nontrivial group of order divisible by some prime other than p, and its group algebra has more than one simple module, so kG has more than one simple module and cannot be local. This is developed on *When Is a Group Algebra a Local Ring?*.

Executive Summary

In the modular case $char k = p$ with $p$ dividing $| G |$ , the radical of $k G$ is nonzero and there is no formula for it. The results of this page supply the part that is computable: everything contributed by normal $p$ -subgroups.

The central statement $(8.4)$ is that a normal $p$ -subgroup $H ⊴ G$ acts trivially on every simple $k G$ -module, so the simple $k G$ -modules coincide with those of $k [G / H]$ . Its sharpest form $(8.6)$ identifies the $p$ -core: $h - 1 \in rad k G$ if and only if $h \in O_{p} (G)$ , a condition on $G$ alone with no reference to $k$ .

$O_{p} (G)$ Detected by the radical

$k [G / H]$ Same simple modules

$[G : H] (| H | - 1)$ Wallace's dimension

$J^{| G |} = 0$

p

-group case

Overview

Fix a field $k$ with $char k = p > 0$ and a finite group $G$ . If $p ∤ | G |$ then Maschke's theorem gives $rad k G = 0$ and there is nothing to say. The interesting case is $p ∣ | G |$ , where $k G$ is not semisimple and the simple modules are fewer and smaller than in characteristic $0$ .

The mechanism is elementary and worth stating first. If $h \in G$ has order $p^{n}$ , then inside the commutative subring $k [h]$ of $k G$ ,

(h - 1)^{p^{n}} = h^{p^{n}} - 1^{p^{n}} = 0 (char k = p),

(8.4a)

The freshman's dream, applied to two commuting elements. Every $p$ -element of $G$ contributes a nilpotent to $k G$ .

So $h - 1$ is always nilpotent when $h$ is a $p$ -element. Whether it lies in the radical is a different question — nilpotent elements need not lie in the radical of a noncommutative ring. The answer, $(8.6)$ , is that $h - 1 \in rad k G$ exactly when $h$ lies in the largest normal $p$ -subgroup.

Learning Objectives

Prove Clifford's theorem for a normal subgroup $H ⊴ G$ and identify the hypothesis it needs.
Prove $(8.4)$ : a normal $p$ -subgroup acts trivially on every simple $k G$ -module in characteristic $p$ .
Deduce that for a $p$ -group $G$ the only simple $k G$ -module is $k$ .
Prove the three-way equivalence of $(8.6)$ characterising $O_{p} (G)$ .
Apply Wallace's formula $(8.7)$ and check it against the dimension count of $(8.1) (4)$ .
Show that $k G$ is local when $G$ is a $p$ -group and $char k = p$ .

Definitions

Definition—The p-core

For a finite group $G$ and a prime $p$ , let $O_{p} (G)$ denote the intersection of all Sylow $p$ -subgroups of $G$ . Since the Sylow $p$ -subgroups form a single conjugacy class, this intersection is normal; it is a $p$ -group, and it contains every normal $p$ -subgroup of $G$ , because a normal $p$ -subgroup lies in every Sylow $p$ -subgroup. Hence $O_{p} (G)$ is the **largest normal $p$ -subgroup** of $G$ .

$p$ -element: An element of $G$ whose order is a power of $p$ .
$h^{g}$: The conjugate $g^{- 1} h g$ . For $H ⊴ G$ and $h \in H$ we have $h^{g} \in H$ , which is what makes the arguments below work.
Augmentation ideal: $\ker ε$ where $ε (\sum a_{g} g) = \sum a_{g}$ ; a $k$ -space with basis ${g - 1 : g \neq 1}$ and $k$ -dimension $| G | - 1$ .
$\sum_{h \in H} k G (h - 1)$: The kernel of the natural surjection $k G ↠ k [G / H]$ when $H ⊴ G$ ; a two-sided ideal of $k$ -dimension $| G | - [G : H]$ .
Semisimple module: A direct sum of simple submodules; equivalently a sum of simple submodules.

Throughout, k is a field of characteristic p > 0 and G is a finite group, except in Clifford's theorem where both hypotheses are relaxed.

Core Concepts

Why normality is essential

The whole argument turns on one computation. If $H ⊴ G$ and $M$ is a $k H$ -submodule of a $k G$ -module $V$ , then for $g \in G$ the translate $g M$ is again a $k H$ -submodule, because

h (g M) = g (g^{- 1} h g) M = g h^{g} M = g M (h \in H, h^{g} \in H) .

(8.5a)

Normality is used exactly once, and it is indispensable: for non-normal $H$ the translate $g M$ is a module over $g H g^{- 1}$ , a different subgroup.

That single line gives Clifford's theorem, and Clifford's theorem reduces $(8.4)$ from a statement about $G$ to a statement about $H$ alone.

Why a $p$ -group acts trivially

For $H$ a $p$ -group in characteristic $p$ , take a central element $h \neq 1$ of $H$ — the centre of a nontrivial $p$ -group is nontrivial. By $(8.4 a)$ , $h - 1$ acts nilpotently on any module, so its kernel is nonzero. Centrality makes that kernel a submodule, and simplicity forces it to be everything. Induction on $| H |$ finishes the job.

H ⊴ G

Clifford:

V |_{k H}

semisimple

H

p

-group acts trivially on simple

k H

-modules

H

acts trivially on

V

Key Results

Theorem(8.5)Clifford

Let $k$ be any field, $G$ any group (not necessarily finite), $H ⊴ G$ a normal subgroup, and $V$ a simple left $k G$ -module. Assume $V$ contains a simple $k H$ -submodule — which is automatic if $\dim_{k} V < \infty$ , and also if $H$ is finite. Then $V$ is a semisimple $k H$ -module.

Proof

Let $M \subseteq V$ be a simple $k H$ -submodule. For $g \in G$ , the subspace $g M$ is a $k H$ -submodule by $(8.5 a)$ . It is simple as a $k H$ -module: if $0 \neq M' \subseteq g M$ were a $k H$ -submodule, then $g^{- 1} M'$ would be a nonzero $k H$ -submodule of $M$ — again by $(8.5 a)$ applied to $g^{- 1}$ — hence equal to $M$ , so $M' = g M$ .

Put $V' = \sum_{g \in G} g M$ . As a sum of simple $k H$ -submodules, $V'$ is a semisimple $k H$ -module. It is also stable under $G$ and under $k$ , hence is a nonzero $k G$ -submodule of $V$ . Since $V$ is simple, $V' = V$ , and $V$ is semisimple over $k H$ .

For the parenthetical hypotheses: if $\dim_{k} V < \infty$ , a nonzero $k H$ -submodule of least $k$ -dimension is simple. If $H$ is finite and $0 \neq v \in V$ , then $k H v$ has $k$ -dimension at most $| H |$ , so the same argument applies inside it.

Theorem(8.4)Normal p-subgroups act trivially

Let $k$ be a field of characteristic $p > 0$ , $G$ a finite group, and $H ⊴ G$ a normal $p$ -subgroup. Then $H$ acts trivially on every simple left $k G$ -module. Consequently the simple left $k G$ -modules are precisely the simple left $k [G / H]$ -modules.

In particular, if $G$ is itself a $p$ -group then the only simple left $k G$ -module is $k$ with trivial $G$ -action.

Proof

Let $V$ be a simple left $k G$ -module; $V$ is finite-dimensional over $k$ , being a quotient of $k G$ . By $(8.5)$ , $V$ is a semisimple $k H$ -module, so it is a sum of simple $k H$ -modules. It therefore suffices to prove that $H$ acts trivially on every simple $k H$ -module $M$ .

Induct on $| H |$ , the case $| H | = 1$ being vacuous. Suppose $| H | > 1$ . Since $H$ is a nontrivial finite $p$ -group its centre is nontrivial; choose $1 \neq h \in Z (H)$ , of order $p^{n}$ say. In the commutative subring $k [h] \subseteq k H$ we have $(h - 1)^{p^{n}} = h^{p^{n}} - 1 = 0$ , so $h - 1$ acts on $M$ as a nilpotent operator and

M_{0} : = {m \in M : h m = m} = \ker ((h - 1) ∣ M) \neq 0 .

Because $h$ is central in $H$ , $M_{0}$ is a $k H$ -submodule: for $x \in H$ and $m \in M_{0}$ , $h (x m) = x (h m) = x m$ . Simplicity of $M$ gives $M_{0} = M$ , so $h$ acts trivially and $M$ is a module over $k [H / ⟨ h ⟩]$ . It is simple as such, and $| H / ⟨ h ⟩ | < | H |$ , so by induction $H / ⟨ h ⟩$ acts trivially. Hence $H$ acts trivially on $M$ .

Finally, a $k G$ -module on which $H$ acts trivially is the same thing as a $k [G / H]$ -module, and simplicity is preserved in both directions because the submodule lattices coincide.

Corollary(8.6)The radical detects the p-core

Let $k$ be a field of characteristic $p > 0$ and $G$ a finite group. For $h \in G$ the following are equivalent:

$h \in O_{p} (G)$ ;
$h$ acts trivially on every simple left $k G$ -module;
$h - 1 \in rad (k G)$ .

Although (2) and (3) refer to the field $k$ , condition (1) does not; so the truth of (2) and (3) is independent of which field of characteristic $p$ is used.

Proof

**(2) $\Leftrightarrow$ (3).** $rad k G$ is the intersection of the annihilators of the simple left $k G$ -modules, and $h$ acts trivially on $M$ precisely when $(h - 1) M = 0$ .

**(1) $\Rightarrow$ (2).** $O_{p} (G)$ is a normal $p$ -subgroup, so $(8.4)$ applies.

**(2) $\Rightarrow$ (1).** Assume (2). Choose a composition series of the left regular module $_{k G} k G$ ; by (2), $h - 1$ annihilates every factor, so $h - 1$ acts nilpotently on $k G$ , and in particular $(h - 1)^{N} = 0$ in $k G$ for some $N$ . Taking $p^{n} \geq N$ and using $(8.4 a)$ in the commutative subring $k [h]$ , $0 = (h - 1)^{p^{n}} = h^{p^{n}} - 1$ , so $h^{p^{n}} = 1$ and the order of $h$ is a power of $p$ .

Let $H = {h \in G : h satisfies (2)}$ . By (2) $\Leftrightarrow$ (3), $H = G \cap (1 + rad k G)$ , which is closed under multiplication and inverses and is stable under conjugation because $rad k G$ is a two-sided ideal invariant under the inner automorphisms of $k G$ . So $H ⊴ G$ , and by the previous paragraph every element of $H$ is a $p$ -element; a finite group all of whose elements are $p$ -elements is a $p$ -group. Hence $H$ is a normal $p$ -subgroup, so $H \subseteq O_{p} (G)$ , which with (1) $\Rightarrow$ (2) gives $H = O_{p} (G)$ .

Corollary(8.7)Wallace

Let $k$ be a field of characteristic $p > 0$ and let $G$ be a finite group possessing a normal Sylow $p$ -subgroup $H$ . Then

rad k G = \sum_{h \in H} k G \cdot (h - 1), \dim_{k} rad k G = [G : H] (| H | - 1) .

Proof

Write $𝔄 = \sum_{h \in H} k G (h - 1)$ . It is a left ideal by construction, and a right ideal because $(h - 1) g = g (h^{g} - 1)$ with $h^{g} \in H$ by normality; so $𝔄$ is two-sided.

$H$ is a normal $p$ -subgroup, so by $(8.4)$ every $h - 1$ with $h \in H$ annihilates every simple left $k G$ -module; hence $𝔄 \subseteq rad k G$ .

The surjection $k G ↠ k [G / H]$ induced by $G \to G / H$ kills each $h - 1$ , so it factors through $k G / 𝔄$ ; conversely $k G / 𝔄$ maps to $k [G / H]$ compatibly and one checks the two maps are mutually inverse, giving $k G / 𝔄 ≅ k [G / H]$ . Since $H$ is a Sylow $p$ -subgroup, $p ∤ [G : H] = | G / H |$ , so $k [G / H]$ is semisimple by Maschke's theorem, i.e. $rad (k G / 𝔄) = 0$ .

Now $𝔄 \subseteq rad k G$ and $rad (k G / 𝔄) = (rad k G) / 𝔄$ by the quotient formula for radicals of ideals inside the radical. Therefore $rad k G = 𝔄$ . Counting dimensions, $\dim_{k} 𝔄 = \dim_{k} k G - \dim_{k} k [G / H] = | G | - [G : H] = [G : H] (| H | - 1)$ .

Corollary(8.8)Group algebras of p-groups

Let $k$ be a field of characteristic $p > 0$ and $G$ a finite $p$ -group. Put $J = rad k G$ . Then:

$J$ equals the augmentation ideal of $k G$ , of $k$ -dimension $| G | - 1$ ;
$J^{| G |} = 0$ ;
if $G$ is generated as a group by $g_{1}, \dots, g_{n}$ , then $J$ is generated as a left ideal by $g_{1} - 1, \dots, g_{n} - 1$ .

In particular $k G / J ≅ k$ is a division ring, so $k G$ is a local ring.

Proof

(1). Apply $(8.7)$ with $H = G$ (a $p$ -group is its own Sylow $p$ -subgroup and is normal in itself): $J = \sum_{g \in G} k G (g - 1)$ . The right-hand side is spanned as a $k$ -space by the elements $g - 1$ , which is exactly the augmentation ideal; and the dimension formula gives $[G : G] (| G | - 1) = | G | - 1$ .

(2). By $(8.4)$ the only simple $k G$ -module is $k$ , so every composition factor of $_{k G} k G$ is $1$ -dimensional and there are exactly $| G |$ of them. The radical annihilates every composition factor, so $J$ carries each term of a composition series into the next; applying $J$ a total of $| G |$ times gives $J^{| G |} k G = 0$ , i.e. $J^{| G |} = 0$ .

(3). Let $L$ be the left ideal generated by the $g_{i} - 1$ . From $g h - 1 = g (h - 1) + (g - 1)$ and $g^{- 1} - 1 = - g^{- 1} (g - 1)$ one sees by induction on word length that $g - 1 \in L$ for every $g \in G$ . Since those elements span $J$ over $k$ by (1), $L = J$ .

Locality. $k G / J ≅ k$ by (1), a division ring, so $k G$ is local.

Proof Techniques and Method

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

Move 1

Translate a submodule around

For $H ⊴ G$ , replace a $k H$ -submodule $M$ by $\sum_{g} g M$ . The sum is $G$ -stable and still semisimple over $k H$ . This is Clifford's whole proof and the standard way to promote local information to global.

Move 2

Use a central element for a fixed-point submodule

If $h$ is central in $H$ and acts nilpotently, $\ker (h - 1)$ is nonzero and is a submodule. Simplicity then forces it to be everything. The nontriviality of $Z (H)$ for a $p$ -group is the input.

Move 3

Squeeze the radical from both sides

Show a candidate ideal $𝔄$ lies in $rad$ , then show $k G / 𝔄$ has zero radical. Together these force equality. Wallace's corollary is a two-line application once Maschke supplies the second half.

Move 3 is the reusable one. Identifying an ideal inside the radical is usually easy — exhibit nilpotence, or annihilation of all simple modules. Bounding the radical from above is the hard half, and semisimplicity of the quotient is almost the only available tool.

Worked Example

$G = S_{3}$ over $k = 𝔽_{3}$

Here $p = 3$ and $| G | = 6 = 2 \cdot 3$ . The Sylow $3$ -subgroup $H = ⟨ (123) ⟩ = A_{3}$ is normal, so $O_{3} (S_{3}) = A_{3}$ and Wallace's corollary applies.

By $(8.4)$ the simple $𝔽_{3} S_{3}$ -modules are the simple $𝔽_{3} [S_{3} / A_{3}] = 𝔽_{3} C_{2}$ -modules. Since $3 ∤ 2$ , Maschke gives $𝔽_{3} C_{2} ≅ 𝔽_{3} \times 𝔽_{3}$ , so there are exactly two, both $1$ -dimensional: the trivial module $M_{1}$ and the sign module $M_{1}'$ .

\dim_{k} rad (𝔽_{3} S_{3}) = [G : H] (| H | - 1) = 2 \cdot 2 = 4 .

(E.1)

Wallace's formula. Cross-check against $(8.1) (4)$ : $6 = 4 + 1^{2} \cdot 1 + 1^{2} \cdot 1$ .

The count closes exactly, so there are no further irreducibles and $𝔽_{3}$ is a splitting field for $S_{3}$ . Note what has happened to the $2$ -dimensional representation $M_{2} = (k e_{1} \oplus k e_{2} \oplus k e_{3}) / k (e_{1} + e_{2} + e_{3})$ : in characteristic $3$ the vector $e_{1} + e_{2} + e_{3}$ lies in the augmentation-zero subspace, so $M_{2}$ acquires a $1$ -dimensional submodule spanned by the image of $e_{1} - e_{2}$ , and is no longer simple. Consistently with $(8.4)$ , the $3$ -cycle acts trivially on that submodule.

One can go further. Writing $I$ for the augmentation ideal of $k H$ , normality gives $𝔄^{j} = k G \cdot I^{j}$ , and $k G$ is free of rank $[G : H] = 2$ over $k H$ , so $\dim_{k} 𝔄^{j} = 2 (3 - j)$ . Hence $rad$ has dimension $4$ , ${rad}^{2}$ has dimension $2$ , and ${rad}^{3} = 0$ .

$G = C_{p}$ over $k = 𝔽_{p}$ : the sharp case

Let $G = ⟨ g ⟩$ of order $p$ . Then $k G ≅ k [x] / (x^{p} - 1) = k [x] / ((x - 1)^{p})$ , and putting $t = x - 1$ ,

𝔽_{p} C_{p} ≅ 𝔽_{p} [t] / (t^{p}), J = (t), \dim_{k} J = p - 1, J^{p} = 0 \neq J^{p - 1} .

(E.2)

A local ring with a single chain of ideals $k G \supset J \supset J^{2} \supset \dots \supset J^{p} = 0$ .

Every conclusion of $(8.8)$ is visible: $J$ is the augmentation ideal, of dimension $p - 1 = | G | - 1$ ; $J^{| G |} = J^{p} = 0$ , and the exponent $| G |$ is attained; and $J = k G \cdot (g - 1)$ is generated by the single element $g - 1$ , matching the single generator of $G$ .

$G = S_{4}$ over $k = 𝔽_{2}$ : reduction by the $p$ -core

Here $O_{2} (S_{4}) = V_{4}$ , the Klein four-group of double transpositions, and $S_{4} / V_{4} ≅ S_{3}$ . So $(8.4)$ gives: the simple $𝔽_{2} S_{4}$ -modules are the simple $𝔽_{2} S_{3}$ -modules, namely the trivial module and the $2$ -dimensional one. Two irreducibles for a group of order $24$ .

Wallace's formula is not available: the Sylow $2$ -subgroups of $S_{4}$ are dihedral of order $8$ and are not normal. But $(8.1) (4)$ still gives $\dim_{k} rad (𝔽_{2} S_{4}) = 24 - (1^{2} + 2^{2}) = 19$ , since $𝔽_{2}$ splits $S_{3}$ and hence $S_{4}$ .

Process and Workflow

Compute $O_{p} (G)$ The largest normal

p

-subgroup. Cheap: intersect the Sylow

p

-subgroups, or read it off from a composition series.

Replace $G$ by $G / O_{p} (G)$ By

(8.4)

the simple modules are unchanged. The new group has trivial

p

-core, so no further reduction of this kind is possible.

Check whether a Sylow $p$ -subgroup is normalIf it is, Wallace's formula gives

rad k G

outright, and

k G / rad k G ≅ k [G / H]

is semisimple.

Otherwise fall back on the countFind the simple modules by other means and recover

\dim_{k} rad k G

as the residual in

(8.1) (4)

VerifyConfirm that the number of simple modules does not exceed the number of

p

-regular conjugacy classes.

What does $O_{p} (G) = 1$ tell you?

About group elementsNo

h \neq 1

has

h - 1 \in rad k G

. The radical contains no element of the form

h - 1

with

h \in G

About the radicalNothing.

rad k G

can still be nonzero —

𝔽_{2} S_{3}

has a

1

-dimensional radical with trivial

2

-core.

About reductionThere is no further quotient to pass to by this method. Determining

rad k G

now requires block theory or explicit computation.

Comparison and Classification

The radical of kG in characteristic p, worked cases
$G$	$k$	$O_{p} (G)$	Sylow normal?	$\dim_{k} rad k G$	$r$
$C_{p}$	$𝔽_{p}$	$C_{p}$	yes	$p - 1$	$1$
$Q_{8}$	$𝔽_{2}$	$Q_{8}$	yes	$7$	$1$
$S_{3}$	$𝔽_{3}$	$A_{3}$	yes	$4$	$2$
$S_{3}$	$𝔽_{2}$	$1$	no	$1$	$2$
$A_{4}$	$𝔽_{2}$	$V_{4}$	yes	$9$	$2$
$S_{4}$	$𝔽_{2}$	$V_{4}$	no	$19$	$2$
$S_{3}$	$ℚ$	—	—	$0$	$3$

The fourth row is the instructive one: $O_{2} (S_{3}) = 1$ , yet $rad (𝔽_{2} S_{3}) = 𝔽_{2} \cdot σ \neq 0$ where $σ = \sum_{g \in G} g$ . A trivial $p$ -core does not force a trivial radical; $(8.6)$ says only that no group element $h \neq 1$ has $h - 1$ in the radical. In the fifth row, $A_{4}$ over $𝔽_{2}$ has $r = 2$ because $𝔽_{2}$ is not a splitting field — the second simple module has endomorphism ring $𝔽_{4}$ .

Which conclusion needs which hypothesis
	$H ⊴ G$	$H$ a $p$ -group	$H$ Sylow	$char k = p$
Clifford $(8.5)$ : $V \|_{k H}$ semisimple	yes	no	no	no
$(8.4)$ : $H$ acts trivially on simples	yes	yes	no	yes
$(8.6)$ : $h - 1 \in rad k G \Leftrightarrow h \in O_{p} (G)$	yes	yes	no	yes
$(8.7)$ : Wallace's formula for $rad k G$	yes	yes	yes	yes
$(8.8)$ : $rad k G$ is the augmentation ideal	yes	yes	yes	yes

Which conclusion needs which hypothesis

Relationship Map

The results form a chain of increasing hypotheses, each buying a sharper conclusion.

$H ⊴ G$ , $k$ any field — Clifford $(8.5)$
- $\dots$ and $H$ a $p$ -group, $char k = p$
  - $(8.4)$ : $H$ acts trivially on every simple $k G$ -module
  - simple $k G$ -modules $=$ simple $k [G / H]$ -modules
  - $(8.6)$ : $O_{p} (G) = G \cap (1 + rad k G)$
- $\dots$ and $H$ is a Sylow $p$ -subgroup
  - $(8.7)$ : $rad k G = \sum_{h \in H} k G (h - 1)$ , of dimension $[G : H] (| H | - 1)$
  - $k G / rad k G ≅ k [G / H]$ , semisimple by Maschke
- $\dots$ and $H = G$ is a $p$ -group
  - $(8.8)$ : $rad k G$ is the augmentation ideal, $J^{| G |} = 0$
  - $k G$ is a local ring with residue field $k$

O_{p} (G)

rad k G

k G / rad k G

simple 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

Reduction to $p$ -core-free groups

Every classification of modular irreducibles begins by quotienting out $O_{p} (G)$ . The remaining group has trivial $p$ -core, which is the standing hypothesis of local representation theory and of the theory of blocks with normal defect groups.

Computational group theory

Shrinking the problem

Computing simple $𝔽_{p} G$ -modules for $| G |$ in the thousands is expensive. Replacing $G$ by $G / O_{p} (G)$ is free and can reduce the order by a large factor — for a group with a big normal $p$ -subgroup, dramatically so.

Commutative-style algebra

Local group algebras

$(8.8)$ makes $k G$ local for a $p$ -group in characteristic $p$ , with a nilpotent maximal ideal. Nakayama's lemma, projective covers and the theory of local rings then transfer wholesale — the basis of the modular theory of $p$ -groups.

Coding theory

Codes over $p$ -group algebras

Ideals of $𝔽_{p} G$ for a $p$ -group $G$ form a chain-like lattice inside a local ring, which is why abelian $p$ -group codes such as generalised Reed–Muller codes have such rigid parameter sets.

The honest framing is that these results are a reduction step. They do not solve the modular problem; they remove the part of it that has a clean answer, leaving the genuinely hard case of groups with trivial $p$ -core.

Design Considerations

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

Quotient early. Passing to $G / O_{p} (G)$ costs nothing in simple modules and can cost a great deal if postponed. Record what was quotiented out, since the non-simple modules do change.
Which normal subgroup? $(8.4)$ needs $H$ normal and a $p$ -group. Applying it to a normal $p'$ -subgroup is wrong, and applying it to a non-normal $p$ -subgroup is wrong; both errors produce plausible-looking false counts.
Left or right ideals. $(8.8) (3)$ generates $J$ as a left ideal by the $g_{i} - 1$ . The same elements generate it as a right ideal by the symmetric argument, but the two generating statements are separate assertions and the proofs, though mirror images, are distinct.
When to bring in blocks. Once $O_{p} (G) = 1$ , the useful next structure is the block decomposition of $k G$ and the defect groups. That is beyond the scope of this section but is where the subject continues.

Standards and Notation

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

$p$ -core

O_{p} (G)

; occasionally

O_{p} (G)

in the finite-group literature

$p'$ -core

O_{p'} (G)

, the largest normal subgroup of order prime to

p

— a different object

Augmentation ideal

ω (k G)

I (G)

Δ (G)

;

Δ

is standard in Passman

Radical

rad k G

here;

J (k G)

in most of the representation-theory literature

Conjugation

h^{g} = g^{- 1} h g

(Lam, and most group theory); some authors use

g h g^{- 1}

GAPPCore(G, p), RadicalOfAlgebra, AugmentationIdeal

MagmapCore(G, p), JacobsonRadical, AugmentationIdeal

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.

The results here are unusually computation-friendly, because they replace a linear-algebra problem of size $| G |$ by a group-theoretic one.

Computing $O_{p} (G)$ for a permutation or matrix group is polynomial time and is a standard library routine; it does not require constructing $k G$ at all.
Wallace's formula gives $rad k G$ with no linear algebra when a Sylow $p$ -subgroup is normal: generators are the $| H | - 1$ elements $h - 1$ , and the dimension is known in advance.
Without a normal Sylow $p$ -subgroup, computing $rad k G$ in characteristic $p$ needs the Friedl–Rónyai algorithm rather than the trace form, since the trace form is degenerate in characteristic $p$ . Cost is polynomial in $| G |$ but the constant is large.
For $p$ -groups the local structure of $(8.8)$ means the Loewy series $k G \supset J \supset J^{2} \supset \dots$ can be computed by repeated multiplication of a single ideal, and the Loewy length is at most $| G |$ — usually far less.

Failure Modes and Common Mistakes

Do not conclude from $(8.4)$ that $H$ acts trivially on every $k G$ -module. Only the simple ones are covered; the regular module is a counterexample as soon as $H \neq 1$ .
Do not use the $p$ -part of $| G |$ in place of $O_{p} (G)$ . A group can have order divisible by a large power of $p$ and still have trivial $p$ -core — $S_{3}$ at $p = 2$ , or any simple group.
Do not assume $rad k G$ is generated by elements $h - 1$ with $h \in G$ in general; that is precisely the Wallace situation and it fails otherwise.

Historical Notes and Lessons Learned

1902–07Dickson opens the modular caseDickson studies representations over fields whose characteristic divides $| G |$ and observes that the count of irreducibles drops. The mechanism is not yet understood.
1935Brauer's countBrauer proves that over a splitting field of characteristic $p$ the number of irreducibles equals the number of $p$ -regular classes, giving the first quantitative grip on what characteristic $p$ destroys.
1937Clifford's theoremClifford analyses the restriction of a simple module to a normal subgroup, proving it is semisimple and that its isotypic components are permuted transitively. This becomes the standard bridge between $G$ and $N ⊴ G$ .
1940s–50sRadicals of group algebrasThe systematic study of $rad k G$ begins, with the $p$ -core emerging as the part of the group visible inside the radical.
1960sWallace's formulaD. A. R. Wallace obtains explicit descriptions of $rad k G$ in favourable cases, including the normal Sylow subgroup formula recorded here as $(8.7)$ .
1970s onwardBlock theoryBrauer's theory of blocks and defect groups becomes the framework for the general case, in which $O_{p} (G) = 1$ is the standing normalisation.

The lesson is about the right level of generality. Clifford's theorem is proved for an arbitrary group and an arbitrary field, and uses normality once. The specialisation to $p$ -groups in characteristic $p$ is where all the arithmetic lives. Separating the two makes both proofs short.

Quick Reference

Clifford $(8.5)$

H ⊴ G

V

simple over

k G

with a simple

k H

-submodule

\Rightarrow

V |_{k H}

semisimple

$(8.4)$

H ⊴ G

p

-group,

char k = p

\Rightarrow

H

acts trivially on simple

k G

-modules

Reductionsimple

k G

-modules

=

simple

k [G / O_{p} (G)]

-modules

$(8.6)$

h \in O_{p} (G) iff h - 1 \in rad k G

Wallace $(8.7)$

H

normal Sylow

p

rad k G = \sum_{h \in H} k G (h - 1)

Dimension

\dim_{k} rad k G = [G : H] (| H | - 1)

$p$ -group $(8.8)$

rad k G

is the augmentation ideal;

J^{| G |} = 0

;

k G

local

Key identity

(h - 1)^{p^{n}} = h^{p^{n}} - 1

in characteristic

p

Hypotheses and conclusions at a glance
Result	Hypotheses	Conclusion
$(8.5)$	$H ⊴ G$ , $V$ simple over $k G$ , simple $k H$ -submodule exists	$V$ semisimple over $k H$
$(8.4)$	$char k = p$ , $G$ finite, $H ⊴ G$ a $p$ -group	$H$ acts trivially on simple $k G$ -modules
$(8.6)$	$char k = p$ , $G$ finite, $h \in G$	$h \in O_{p} (G) \Leftrightarrow h - 1 \in rad k G$
$(8.7)$	$char k = p$ , $G$ finite with normal Sylow $p$ -subgroup $H$	$rad k G = \sum_{h \in H} k G (h - 1)$
$(8.8)$	$char k = p$ , $G$ a finite $p$ -group	$rad k G =$ augmentation ideal, $J^{\| G \|} = 0$

Frequently Asked Questions

Why must the subgroup be normal in

(8.4)

Because the proof runs through Clifford's theorem, whose only use of the hypothesis is the identity $h (g M) = g h^{g} M = g M$ , valid because $h^{g} \in H$ . For non-normal $H$ the translate $g M$ is a module over a conjugate subgroup and the sum $\sum_{g} g M$ is not a $k H$ -module. The conclusion genuinely fails: a transposition in $S_{3}$ acts nontrivially on the $2$ -dimensional simple $𝔽_{2} S_{3}$ -module.

Does

(8.6)

mean

rad k G

is determined by

O_{p} (G)

No. It says only which group elements $h$ have $h - 1$ in the radical. The radical usually contains much more, and $𝔽_{2} S_{3}$ shows it can be nonzero while $O_{p} (G) = 1$ . The full radical is determined by $O_{p} (G)$ only in the Wallace situation, where a Sylow $p$ -subgroup is normal.

Is Clifford's theorem true without any finiteness hypothesis?

The proof needs a simple $k H$ -submodule to exist, which is where finiteness enters. That is automatic when $V$ is finite-dimensional over $k$ or when $H$ is finite. For an infinite $H$ and an infinite-dimensional simple $V$ one must assume it. The conclusion, once a simple $k H$ -submodule exists, requires no further hypotheses.

Why does

J^{| G |} = 0

rather than some smaller power?

$| G |$ is simply the number of composition factors of the regular module, and the radical shifts a composition series down by one step each time it is applied. The bound is a termination guarantee, not an estimate: for $G$ elementary abelian of order $p^{n}$ the exact nilpotency index is $n (p - 1) + 1$ , far below $p^{n}$ .

How does this interact with Maschke's theorem?

They are complementary. Maschke says $rad k G = 0$ when $p ∤ | G |$ , so $O_{p} (G) = 1$ and the results here are vacuous. When $p ∣ | G |$ they supply the part of the radical that group theory can see. The two together explain the whole picture when a Sylow $p$ -subgroup is normal, since then $k G / rad k G ≅ k [G / H]$ is exactly the Maschke case.

k G

local only for

p

-groups?

Over a field of characteristic $p$ , $k G$ is local if and only if $G$ is a finite $p$ -group. If $G$ is not a $p$ -group, $G / O_{p} (G)$ is a nontrivial group of order divisible by some prime other than $p$ , and its group algebra has more than one simple module, so $k G$ has more than one simple module and cannot be local. This is developed on When Is a Group Algebra a Local Ring?.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §8, (8.4)–(8.8) (pp. 129–133).
A. H. Clifford, “Representations induced in an invariant subgroup”, Annals of Mathematics 38 (1937), 533–550.
D. S. Passman, The Algebraic Structure of Group Rings, Wiley-Interscience, 1977 — radicals of group rings in characteristic p.
H. Nagao and Y. Tsushima, Representations of Finite Groups, Academic Press, 1989, Chapters 1 and 5.
J. L. Alperin, Local Representation Theory, Cambridge Studies in Advanced Mathematics 11, Cambridge University Press, 1986.
C. W. Curtis and I. Reiner, Methods of Representation Theory, Volume I, Wiley-Interscience, 1981.

AI Suggested Questions

State the full version of Clifford's theorem, including the transitive permutation of isotypic components, and prove it.
For which finite groups $G$ and primes $p$ is $rad (𝔽_{p} G)$ generated by elements of the form $h - 1$ with $h \in G$ ?
Compute the Loewy series of $𝔽_{2} D_{8}$ and compare its length with the bound $J^{| G |} = 0$ .
How does the defect group of a block generalise the role played by $O_{p} (G)$ here?
Give an example of a group with trivial $p$ -core and a large radical, and explain what controls the radical there.
What is the analogue of $(8.4)$ for a normal $p'$ -subgroup, and why is the conclusion completely different?
Derive the nilpotency index of the augmentation ideal of $𝔽_{p} G$ for $G$ elementary abelian of order $p^{n}$ .

Normal p-Subgroups and rad kG

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Why normality is essential

Why a $p$ -group acts trivially

Key Results

Proof Techniques and Method

Translate a submodule around

Use a central element for a fixed-point submodule

Squeeze the radical from both sides

Worked Example

$G = S_{3}$ over $k = 𝔽_{3}$

$G = C_{p}$ over $k = 𝔽_{p}$ : the sharp case

$G = S_{4}$ over $k = 𝔽_{2}$ : reduction by the $p$ -core

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Reduction to $p$ -core-free groups

Shrinking the problem

Local group algebras

Codes over $p$ -group algebras

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

Why normality is essential

Why a p-group acts trivially

06Key Results

07Proof Techniques and Method

Translate a submodule around

Use a central element for a fixed-point submodule

Squeeze the radical from both sides

08Worked Example

G=S3 over k=𝔽3

G=Cp over k=𝔽p: the sharp case

G=S4 over k=𝔽2: reduction by the p-core

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Reduction to p-core-free groups

Shrinking the problem

Local group algebras

Codes over p-group algebras

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

Why a $p$ -group acts trivially

Key Results

Proof Techniques and Method

Worked Example

$G = S_{3}$ over $k = 𝔽_{3}$

$G = C_{p}$ over $k = 𝔽_{p}$ : the sharp case

$G = S_{4}$ over $k = 𝔽_{2}$ : reduction by the $p$ -core

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Reduction to $p$ -core-free groups

Codes over $p$ -group algebras

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