Completely Reducible Linear Groups

Q: Why not simply say the group algebra is semisimple?

Because for an infinite group kG is never semisimple (6.3), and every interesting linear group in §9 is infinite. The definition therefore refers to the specific module V. The finite-dimensional shadow of kG that does carry the information is Span_k(G) = im(kG → End_k V), and (9.11) says its semisimplicity is exactly what is wanted.

Q: Why must the subgroup in (9.15) be subnormal?

Because Clifford's theorem needs normality, and subnormality is what one gets by iterating it. An arbitrary subgroup can be much worse behaved than the whole group: SL_2(Z) is completely reducible on Q^2 but contains a unipotent copy of Z that is not. The failure is not an artefact of the proof.

Q: Does complete reducibility survive field extension?

Semisimplicity of Span_k(G) can fail after extension when the base field is imperfect: a semisimple algebra S over k may have S ⊗_k K with nonzero radical if the centre of S involves an inseparable extension. Over a perfect field — in particular in characteristic 0 or over a finite field — the property is stable under all extensions of k.

Q: How do I actually compute Span_k(G)?

Spin it up: start with the identity and the generators, repeatedly multiply current basis elements by generators, and echelonise. The process terminates in at most n^2 steps because the span is a subspace of End_k V. Then compute the radical — in characteristic 0 as the kernel of the trace form on S, in characteristic p with the Friedl–Rónyai algorithm.

Q: What is the relationship with the unipotent radical?

By (9.24) the unipotent radical of G is g ∈ G : g - 1 ∈ rad Span_k(G). If G is completely reducible the radical is zero, so the unipotent radical is trivial. The converse fails: the group algebra of a finite group with no normal p-subgroup, acting on itself in characteristic p dividing the order, has trivial unipotent radical and is not completely reducible.

Q: Is an irreducible linear group completely reducible?

Yes, trivially: a simple module is a one-term direct sum of simple modules. The converse fails as soon as V decomposes — the diagonal group in GL_2(k) is completely reducible and reducible at once. When k is algebraically closed, irreducibility is the stronger statement that Span_k(G) is all of M_n(k), by Burnside's theorem.

Executive Summary

Call $G \subseteq G L (V)$ completely reducible if $V$ is a semisimple $k G$ -module. For finite $G$ with $char k ∤ | G |$ this is Maschke's theorem and nothing more need be said. For infinite $G$ the group algebra $k G$ is never semisimple, so the property has to be read as a statement about the single module $V$ .

The translation that makes the notion tractable is $(9.11)$ : $G$ is completely reducible exactly when the finite-dimensional algebra $S = {Span}_{k} (G) \subseteq {End}_{k} V$ is semisimple. A question about an infinite group becomes a question about the radical of an algebra of dimension at most $n^{2}$ .

$rad S = 0$ The criterion

$\leq n^{2}$

\dim_{k} S

subnormalInheritance, via Clifford

localTested on f.g. subgroups

Overview

Complete reducibility is the good behaviour one wants from a representation: it means $V$ breaks into simple pieces with no gluing, so that every invariant subspace has an invariant complement and the module is determined by its composition factors.

For finite groups over a field of coprime characteristic, Maschke's theorem hands this to us for free. The point of §9 is that a lot survives when $G$ is infinite. Three results do the work: complete reducibility is detected by an algebra $(9.11)$ ; it is a local property $(9.13)$ ; and it passes down to subnormal subgroups $(9.15)$ and up along subgroups of finite index prime to the characteristic $(9.12)$ .

What is left over — the obstruction to complete reducibility, packaged as a normal subgroup of $G$ — is the subject of The Unipotent Radical of a Linear Group.

Learning Objectives

State Definition $(9.10)$ and explain why $k G$ semisimple is a strictly stronger condition.
Prove $(9.11)$ : complete reducibility $\Leftrightarrow$ $S = {Span}_{k} (G)$ semisimple.
Prove $(9.12)$ : if $[G : H] = m$ with $char k ∤ m$ and $H$ is completely reducible, so is $G$ .
Prove $(9.13)$ : complete reducibility is detected on finitely generated subgroups.
Prove $(9.14)$ : if every element of $G$ has finite order prime to $char k$ then $G$ is completely reducible.
Prove $(9.15)$ : subnormal subgroups inherit complete reducibility.

Definitions

Definition(9.10)Completely reducible linear group

Let $k$ be a field, $V$ a finite-dimensional $k$ -vector space and $G \subseteq G L (V)$ a linear group. Then $G$ is completely reducible if $V$ is a completely reducible — that is, semisimple — module over the group algebra $k G$ : a finite direct sum of simple $k G$ -submodules.

Equivalently: every $k G$ -submodule of $V$ has a $k G$ -complement in $V$ . The finiteness of the direct sum is automatic, since $\dim_{k} V < \infty$ .

${Span}_{k} (G)$: The $k$ -span of $G$ inside ${End}_{k} V$ . It is a subalgebra, being closed under multiplication, and equals the image of the algebra map $k G \to {End}_{k} V$ ; so $\dim_{k} {Span}_{k} (G) \leq n^{2}$ whatever the size of $G$ .
Irreducible linear group: $V$ is a simple $k G$ -module. Irreducible $\Rightarrow$ completely reducible; the converse fails whenever $V$ splits.
Subnormal: $H ⊴ G_{m - 1} ⊴ \dots ⊴ G_{0} = G$ for some finite chain. Normal subgroups are the case $m = 1$ .
Unipotent element: $g \in G L (V)$ with $(g - 1)$ nilpotent. A nontrivial unipotent element never acts semisimply, so it obstructs complete reducibility as soon as it is normal.
$rad S$: The Jacobson radical of the finite-dimensional algebra $S$ ; here it is nilpotent, and it vanishes exactly when $S$ is semisimple.

Complete reducibility of G is always relative to the given embedding G inside GL(V). The same abstract group can be completely reducible in one representation and not in another.

Core Concepts

Why the group algebra is the wrong object

If $G$ is infinite then $k G$ is never a semisimple ring $(6.3)$ — for instance the augmentation ideal of an infinite group is never a direct summand. So the naive reading of completely reducible as the group algebra is semisimple would make the class empty for infinite $G$ , which is not what is wanted.

The finite-dimensional image $S = {Span}_{k} (G)$ is the right substitute. It forgets everything about $G$ except its action on $V$ , and it is a finite-dimensional algebra, so Wedderburn–Artin theory applies verbatim.

Complete reducibility is a local property

Because $\dim_{k} S \leq n^{2}$ , some finite subset of $G$ already spans $S$ . The subgroup generated by that subset spans the same algebra, so it is completely reducible exactly when $G$ is — this is $(9.13)$ . Consequently, results proved for finitely generated groups, notably Schur's theorem, transfer immediately to arbitrary linear groups.

Which direction inheritance runs

Downward along subnormal subgroups, by Clifford's theorem $(9.15)$ . Upward along subgroups of finite index prime to the characteristic, by relative Maschke $(9.12)$ . Neither holds for arbitrary subgroups or arbitrary overgroups: $S L_{2} (ℤ)$ is completely reducible on $ℚ^{2}$ but its unipotent subgroup $⟨ (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}) ⟩$ is not — and that subgroup is not subnormal.

Key Results

Proposition(9.11)Algebraic criterion

Let $G \subseteq G L (V)$ with $\dim_{k} V < \infty$ and put $S = {Span}_{k} (G) \subseteq {End}_{k} V$ . Then $G$ is completely reducible iff $S$ is a semisimple $k$ -algebra.

Proof

The $k G$ -submodules of $V$ and the $S$ -submodules of $V$ are the same subspaces, since $S$ is spanned by $G$ . So $V$ is semisimple over $k G$ iff it is semisimple over $S$ .

( $\Leftarrow$ ) If $S$ is semisimple then every $S$ -module is semisimple; in particular $V$ is.

( $\Rightarrow$ ) Suppose $V$ is a semisimple $S$ -module. The radical $rad S$ annihilates every simple $S$ -module, hence annihilates $V$ . But $S \subseteq {End}_{k} V$ , so $V$ is a faithful $S$ -module and $rad S = 0$ . As $S$ is a finite-dimensional algebra, it is artinian, and an artinian ring with zero radical is semisimple.

Counterexample(9.11a)The standard non-example

Over any field $k$ , the group $G = {(\begin{matrix} 1 & h \\ 0 & 1 \end{matrix}) : h \in k}$ acting on $V = k^{2}$ is not completely reducible: the line $k e_{1}$ is the unique invariant line, so it has no invariant complement. Here $S = {Span}_{k} (G) = k \cdot I \oplus k \cdot E_{12}$ with $E_{12}^{2} = 0$ , so $rad S = k E_{12} \neq 0$ , in agreement with $(9.11)$ .

Theorem(9.12)Subgroups of finite index

Let $G \subseteq G L (V)$ with $\dim_{k} V < \infty$ , and let $H \leq G$ be a subgroup of finite index $m = [G : H]$ with $char k ∤ m$ . If $H$ is completely reducible on $V$ , then so is $G$ .

Proof

This is Maschke's averaging argument relative to $H$ . Let $W \subseteq V$ be a $k G$ -submodule. Since $V$ is semisimple over $k H$ , there is a $k H$ -linear projection $π : V \to W$ with $π |_{W} = {id}_{W}$ . Choose coset representatives $g_{1}, \dots, g_{m}$ for $H$ in $G$ and set

π' (v) = \frac{1}{m} \sum_{i = 1}^{m} g_{i} π (g_{i}^{- 1} v), v \in V .

The factor $1 / m$ exists in $k$ because $char k ∤ m$ . The definition does not depend on the representatives: replacing $g_{i}$ by $g_{i} h$ with $h \in H$ gives $g_{i} h π (h^{- 1} g_{i}^{- 1} v) = g_{i} π (g_{i}^{- 1} v)$ by $k H$ -linearity of $π$ .

$π'$ maps into $W$ because $π$ does and $W$ is $G$ -stable; $π' |_{W} = id$ because $π |_{W} = id$ ; and $π'$ is $k G$ -linear because left translating the coset representatives by $g$ permutes them. So $W$ is a direct summand of $V$ as a $k G$ -module. Since every submodule of the finite-dimensional module $V$ is a direct summand, $V$ is semisimple.

Lemma(9.13)Complete reducibility is local

Let $G \subseteq G L (V)$ with $\dim_{k} V < \infty$ . If every finitely generated subgroup of $G$ is completely reducible, then $G$ is completely reducible.

Proof

$S = {Span}_{k} (G)$ has dimension at most $n^{2}$ , so we may choose $g_{1}, \dots, g_{m} \in G$ forming a $k$ -basis of $S$ . Let $H = ⟨ g_{1}, \dots, g_{m} ⟩$ , a finitely generated subgroup. Then ${Span}_{k} (H) \subseteq S$ contains a basis of $S$ , so ${Span}_{k} (H) = S$ .

By hypothesis $H$ is completely reducible, so ${Span}_{k} (H) = S$ is semisimple by $(9.11)$ ; applying $(9.11)$ in the other direction, $G$ is completely reducible.

Proposition(9.14)Torsion with orders prime to the characteristic

Let $G \subseteq G L_{n} (k)$ be a group in which every element has finite order prime to $char k$ . (When $char k = 0$ this just says $G$ is torsion.) Then $G$ is completely reducible.

Proof

Let $H \subseteq G$ be finitely generated. It is a finitely generated torsion linear group, hence finite by Schur's Theorem $(9.9)$ . Write $p = char k$ . If $p > 0$ divided $| H |$ , Cauchy's theorem would produce an element of order $p$ in $H$ , contrary to hypothesis; so $p ∤ | H |$ and Maschke's Theorem makes $V$ a semisimple $k H$ -module. Thus every finitely generated subgroup of $G$ is completely reducible, and $(9.13)$ finishes the proof.

Proposition(9.15)Subnormal subgroups

Let $G \subseteq G L (V)$ be a completely reducible linear group with $\dim_{k} V < \infty$ , and let $H$ be a subnormal subgroup of $G$ . Then $H$ is completely reducible.

Proof

By induction along the subnormal chain it suffices to treat $H ⊴ G$ . Write $V = V_{1} \oplus \dots \oplus V_{t}$ with each $V_{i}$ a simple $k G$ -module. Each $V_{i}$ is finite-dimensional, so Clifford's Theorem $(8.5)$ applies: the restriction of $V_{i}$ to $k H$ is a semisimple module. A direct sum of semisimple modules is semisimple, so $V$ is semisimple over $k H$ .

Corollary(9.12a)Finite index in characteristic zero

Let $char k = 0$ , $G \subseteq G L (V)$ and $H \leq G$ of finite index. Then $G$ is completely reducible iff $H$ is.

One direction is $(9.12)$ . For the other, a subgroup of finite index contains a normal subgroup $H_{0} ⊴ G$ of finite index (the core of $H$ ); if $G$ is completely reducible then so is $H_{0}$ by $(9.15)$ , and then so is $H$ by $(9.12)$ applied to $H_{0} \leq H$ , the index being finite and invertible in $k$ .

Proposition—Abelian completely reducible groups over an algebraically closed field

Let $k$ be algebraically closed and $G \subseteq G L (V)$ completely reducible with $\dim_{k} V < \infty$ . Then $G$ is abelian iff $G$ is conjugate in $G L (V)$ to a group of diagonal matrices.

Indeed, $S = {Span}_{k} (G)$ is then a commutative semisimple finite-dimensional algebra over an algebraically closed field, so $S ≅ k \times \dots \times k$ ; the corresponding decomposition of $V$ into simple $S$ -modules is a decomposition into common eigenlines. The converse is clear.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Replace the group by its span

Every proof in this section starts by passing from $G$ to $S = {Span}_{k} (G)$ . The submodule lattices agree, so nothing is lost, and $S$ is finite-dimensional even when $G$ is not.

Move 2

Average with $1 / m$

Maschke's averaging works over any subgroup of finite index, provided the index is invertible in $k$ . The projection is repaired coset by coset; nothing about finiteness of $G$ is needed.

Move 3

Clifford for restriction

Restricting a simple module to a normal subgroup gives a semisimple module. This is the only tool that moves complete reducibility downwards, and it is why subnormal and not arbitrary appears in $(9.15)$ .

A fourth move is worth naming: faithfulness kills the radical. In $(9.11)$ we know $rad S$ annihilates $V$ and $V$ is faithful, so $rad S = 0$ . The same two-line argument identifies the radical in $(9.24)$ and is the standard way to compute $rad S$ for a concrete matrix algebra.

Worked Example

Two groups in $G L_{2} (ℚ)$ , one reducible and one not

Let $A = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ and $B = (\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix})$ in $G L_{2} (ℚ)$ .

** $G_{1} = ⟨ A ⟩ ≅ ℤ$ .** Then $A^{m} = (\begin{matrix} 1 & m \\ 0 & 1 \end{matrix})$ , so ${Span}_{ℚ} (G_{1}) = ℚ I + ℚ E_{12}$ , of dimension $2$ . Since $E_{12}^{2} = 0$ , this algebra is local with $rad = ℚ E_{12} \neq 0$ . By $(9.11)$ , $G_{1}$ is not completely reducible — as is visible directly: $ℚ e_{1}$ is the only invariant line.

** $G_{2} = S L_{2} (ℤ) = ⟨ A, B ⟩$ .** Now $A - I = E_{12}$ and $B - I = E_{21}$ lie in the span, and

B A = (\begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix}), B A - I - E_{12} - E_{21} = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}) = E_{22} .

(E.1)

So ${Span}_{ℚ} (G_{2})$ contains $I, E_{12}, E_{21}, E_{22}$ and therefore equals $M_{2} (ℚ)$ , which is simple, hence semisimple. By $(9.11)$ , $S L_{2} (ℤ)$ is completely reducible on $ℚ^{2}$ ; in fact $ℚ^{2}$ is a simple module.

An infinite completely reducible group by way of $(9.12)$

Let $k = ℚ$ , $V = ℚ^{2}$ , and let

H = {(\begin{matrix} a & 0 \\ 0 & a^{- 1} \end{matrix}) : a \in ℚ^{\times}}, w = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}), G = ⟨ H, w ⟩ .

(E.2)

$H$ is infinite abelian and $V = ℚ e_{1} \oplus ℚ e_{2}$ is a sum of two simple $k H$ -modules, so $H$ is completely reducible. Since $w^{2} = I$ and $w H w^{- 1} = H$ , we have $[G : H] = 2$ , and $char ℚ = 0 ∤ 2$ ; so $(9.12)$ gives that $G$ is completely reducible.

Direct check: the only $H$ -invariant lines are $ℚ e_{1}$ and $ℚ e_{2}$ — a line through $(x, y)$ with $x, y \neq 0$ would need $a = a^{- 1}$ for every $a \in ℚ^{\times}$ — and $w$ interchanges them. So $V$ has no $G$ -invariant line and is a simple $k G$ -module, which is stronger than what $(9.12)$ promised.

Where $(9.14)$ has teeth

Take $k = 𝔽_{5}$ and let $G \subseteq G L_{2} (𝔽_{5})$ be any subgroup all of whose elements have order prime to $5$ — for instance the group of order $8$ generated by $(\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ and $(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ , of exponent $4$ . Proposition $(9.14)$ gives complete reducibility with no need to know $| G |$ in advance; here $5 ∤ 8$ , so Maschke would also do.

Process and Workflow

Is your linear group $G \subseteq G L (V)$ completely reducible?

$G$ finite, $char k ∤ | G |$ Yes, by Maschke. No computation needed.

$G$ finite, $char k = p$ divides $| G |$ Not automatic. Necessary condition: the largest normal

p

-subgroup

O_{p} (G)

must act trivially on

V

. Even that is not sufficient — see the unipotent radical page.

Every element of finite order prime to $char k$ Yes, by

(9.14)

, whatever the cardinality of

G

None of the aboveCompute

S = {Span}_{k} (G)

by spinning up products of generators, then test

rad S = 0

. This is a finite linear algebra problem of size at most

n^{2}

Generate the algebraFrom matrix generators, build an echelonised basis of

S = {Span}_{k} (G)

; it has at most

n^{2}

elements.

Compute the radicalIn characteristic

0

rad S

is the kernel of the trace form

(a, b) \mapsto tr (a b)

S

; in characteristic

p

use the Friedl–Rónyai algorithm.

Decide

rad S = 0

iff

G

is completely reducible, by

(9.11)

If not zero, extract the obstruction

{g \in G : g - 1 \in rad S}

is the unipotent radical of

G

, by

(9.24)

Comparison and Classification

Completely reducible or not: a working list
Linear group $G \subseteq G L (V)$	Completely reducible?	Reason
$G$ finite, $char k ∤ \| G \|$	yes	Maschke
$G$ torsion, all orders prime to $char k$	yes	$(9.14)$
$G L_{n} (k)$ or $S L_{n} (k)$ on $k^{n}$	yes — indeed irreducible	no invariant subspace
Diagonal matrices on $k^{n}$	yes	sum of $n$ eigenlines
$U T_{n} (k)$ , $n \geq 2$	no	unique invariant flag, radical nonzero
$k^{\times} \cdot U T_{n} (k)$ , $n \geq 2$	no	same flag
$k G$ regular module, $G$ finite, $p ∣ \| G \|$	no	$k G$ not semisimple $(6.1)$
$S_{3}$ on the $2$ -dimensional module over $𝔽_{3}$	no	$(1, 1, 1)$ becomes an invariant line

Which operations preserve complete reducibility
	preserved in general	preserved with a hypothesis	reference
Pass to a normal subgroup	yes	no hypothesis needed	$(9.15)$
Pass to a subnormal subgroup	yes	no hypothesis needed	$(9.15)$
Pass to an arbitrary subgroup	no	—	$⟨ A ⟩ \subseteq S L_{2} (ℤ)$
Pass to an overgroup of finite index	no	index prime to $char k$	$(9.12)$
Pass to a directed union	yes	spans stabilise	$(9.13)$
Extend the base field	partial	separability issues	—

Which operations preserve complete reducibility

Relationship Map

The property sits between two much better known ones, and the containments are strict.

All linear groups

G \subseteq G L (V)

\dim_{k} V < \infty

Trivial unipotent radicalno nontrivial normal unipotent subgroup — necessary for complete reducibility, by

(9.24)

Completely reducible

V

semisimple over

k G

; equivalently

rad {Span}_{k} (G) = 0

Irreducible

V

simple over

k G

; equivalently

{Span}_{k} (G)

is simple, and equal to

M_{n} (k)

when

k = \bar{k}

$k G$ semisimpleonly possible for

G

finite with

char k ∤ | G |

k G

semisimple

V

semisimple over

k G

rad {Span}_{k} (G) = 0

unipotent radical trivial

None of the implications in the chain reverses. The last one fails already for a finite group of order divisible by the characteristic with no normal p-subgroup, acting on its own group algebra.

Design Considerations

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

Model the pair, not the group. Complete reducibility belongs to $(G, V)$ . Record the module whenever you assert it; the same $G$ may be completely reducible on one module and unipotent on another.
**Work with $S$ , not $k G$ .** For infinite $G$ the group algebra is intractable and never semisimple. Everything decidable about complete reducibility is visible in the at most $n^{2}$ -dimensional algebra ${Span}_{k} (G)$ .
Choose the base field deliberately. Extending $k$ can only refine the decomposition, and over $\bar{k}$ the theory is cleanest: irreducible becomes absolutely irreducible and abelian completely reducible groups become diagonal. If arithmetic information matters, stay over $k$ and carry the division algebra ${End}_{k G} (V)$ along.
Decide early whether you need semisimplicity of the algebra or of the module. If you need $k G$ itself semisimple you are back to finite groups of order prime to $char k$ ; do not assume the stronger statement when the weaker one is what your argument requires.
Use locality. By $(9.13)$ you may always assume $G$ is finitely generated when proving complete reducibility, which brings Schur's theorem and finite group theory into range.

Failure Modes and Common Mistakes

Do not conclude complete reducibility from trivial unipotent radical; that condition is necessary and not sufficient.
Do not assume the simple summands are unique as subspaces — only the isotypic components are canonical.
Do not read $(9.14)$ as torsion implies completely reducible: in characteristic $p$ , elements of order $p$ are precisely the ones excluded, and a unipotent group in characteristic $p$ is torsion and never completely reducible unless trivial.
Do not forget that a semisimple ${Span}_{k} (G)$ may still be a proper subalgebra of ${End}_{k} V$ — semisimple does not mean everything.

Best Practices

State the module together with the group whenever complete reducibility is asserted.
To prove complete reducibility, exhibit a semisimple ${Span}_{k} (G)$ ; to disprove it, exhibit an invariant subspace with no invariant complement, or a nonzero nilpotent element of the span that is $g - 1$ for some $g \in G$ .
When the group is infinite, reduce to a finitely generated subgroup with the same span before invoking anything from finite group theory.
Check the characteristic against both $| G |$ and $[G : H]$ before applying Maschke or $(9.12)$ ; those are different numbers and both matter.
When working over a non-closed field, record ${End}_{k G} (V)$ — it is what distinguishes irreducible from absolutely irreducible and controls what happens on extending scalars.

Quick Reference

Definition $(9.10)$

V

is a semisimple

k G

-module

Criterion $(9.11)$

{Span}_{k} (G)

semisimple, i.e.

rad {Span}_{k} (G) = 0

Finite index $(9.12)$

H

c.r.,

[G : H] = m

char k ∤ m

\Rightarrow

G

c.r.

Local $(9.13)$ all f.g. subgroups c.r.

\Rightarrow

G

c.r.

Torsion $(9.14)$ all element orders prime to

char k

\Rightarrow

c.r.

Subnormal $(9.15)$

G

c.r.,

H

subnormal

\Rightarrow

H

c.r.

Never

k G

semisimple for infinite

G

(6.3)

Necessary conditionunipotent radical

= {1}

, by

(9.24)

Numbered results of §9 on this page
Reference	Statement	Main hypothesis
$(9.10)$	definition of completely reducible	$\dim_{k} V < \infty$
$(9.11)$	c.r. $\Leftrightarrow$ span is semisimple	none
$(9.12)$	lift along finite index	$char k ∤ [G : H]$
$(9.13)$	local character	none
$(9.14)$	torsion of coprime orders	orders prime to $char k$
$(9.15)$	descend to subnormal subgroups	subnormality

Frequently Asked Questions

Why not simply say the group algebra is semisimple?

Because for an infinite group $k G$ is never semisimple $(6.3)$ , and every interesting linear group in §9 is infinite. The definition therefore refers to the specific module $V$ . The finite-dimensional shadow of $k G$ that does carry the information is ${Span}_{k} (G) = im (k G \to {End}_{k} V)$ , and $(9.11)$ says its semisimplicity is exactly what is wanted.

Why must the subgroup in

(9.15)

be subnormal?

Because Clifford's theorem needs normality, and subnormality is what one gets by iterating it. An arbitrary subgroup can be much worse behaved than the whole group: $S L_{2} (ℤ)$ is completely reducible on $ℚ^{2}$ but contains a unipotent copy of $ℤ$ that is not. The failure is not an artefact of the proof.

Does complete reducibility survive field extension?

Semisimplicity of ${Span}_{k} (G)$ can fail after extension when the base field is imperfect: a semisimple algebra $S$ over $k$ may have $S \otimes_{k} K$ with nonzero radical if the centre of $S$ involves an inseparable extension. Over a perfect field — in particular in characteristic $0$ or over a finite field — the property is stable under all extensions of $k$ .

How do I actually compute

{Span}_{k} (G)

Spin it up: start with the identity and the generators, repeatedly multiply current basis elements by generators, and echelonise. The process terminates in at most $n^{2}$ steps because the span is a subspace of ${End}_{k} V$ . Then compute the radical — in characteristic $0$ as the kernel of the trace form on $S$ , in characteristic $p$ with the Friedl–Rónyai algorithm.

What is the relationship with the unipotent radical?

By $(9.24)$ the unipotent radical of $G$ is ${g \in G : g - 1 \in rad {Span}_{k} (G)}$ . If $G$ is completely reducible the radical is zero, so the unipotent radical is trivial. The converse fails: the group algebra of a finite group with no normal $p$ -subgroup, acting on itself in characteristic $p$ dividing the order, has trivial unipotent radical and is not completely reducible.

Is an irreducible linear group completely reducible?

Yes, trivially: a simple module is a one-term direct sum of simple modules. The converse fails as soon as $V$ decomposes — the diagonal group in $G L_{2} (k)$ is completely reducible and reducible at once. When $k$ is algebraically closed, irreducibility is the stronger statement that ${Span}_{k} (G)$ is all of $M_{n} (k)$ , by Burnside's theorem.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §9 (pp. 154–158), with §6 and §8 for Maschke's and Clifford's theorems.
C. W. Curtis and I. Reiner, Representation Theory of Finite Groups and Associative Algebras, Wiley-Interscience, 1962, §49 (Clifford theory).
B. A. F. Wehrfritz, Infinite Linear Groups, Ergebnisse der Mathematik 76, Springer-Verlag, 1973, Chapter 1.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §9 and §13.
J.-P. Serre, “Complète réductibilité”, Séminaire Bourbaki, Exposé 932, Astérisque 299 (2005), 195–217.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.

AI Suggested Questions

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01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why the group algebra is the wrong object

Complete reducibility is a local property

Which direction inheritance runs

06Key Results

07Proof Techniques and Method

Replace the group by its span

Average with 1/m

Clifford for restriction

08Worked Example

Two groups in GL2(ℚ), one reducible and one not

An infinite completely reducible group by way of (9.12)

Where (9.14) has teeth

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Design Considerations

13Failure Modes and Common Mistakes

14Best Practices

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

Average with $1 / m$

Worked Example

Two groups in $G L_{2} (ℚ)$ , one reducible and one not

An infinite completely reducible group by way of $(9.12)$

Where $(9.14)$ has teeth

Process and Workflow

Comparison and Classification

Relationship Map

Design Considerations

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions