The Unipotent Radical of a Linear Group

Q: Why is a trivial unipotent radical not enough for complete reducibility?

Because the unipotent radical is a group-level invariant and complete reducibility is a module-level one. The radical of Span_k(G) records how the composition factors of V are glued together; the subgroup H records only those gluings realised by group elements of the form 1 + (radical element). When G is a finite group with O_p(G) = 1 in characteristic p dividing |G|, the gluing is present and the subgroup is not: H = 1 while radSpan_k(G) ≠ 0.

Q: Does the unipotent radical depend on the representation?

Yes in general, no for finite groups. For finite G in characteristic p, normal unipotent subgroups are exactly normal p-subgroups by (9.17), so H = O_p(G) whatever the faithful module. For infinite groups the module matters: Z has trivial unipotent radical when acting by diag(2^m, 3^m) on Q^2 and is its own unipotent radical when acting unitriangularly.

Q: Is there a Levi decomposition G ≅ (G/H) H?

Not for abstract linear groups. For linear algebraic groups over a field of characteristic zero, Mostow's theorem provides a Levi decomposition, and in characteristic p it can fail even there. The theorem proved here gives only that G/H is completely reducible; the extension of G/H by H need not split.

Q: How does this relate to the Jacobson radical of a group algebra?

For a finite group G in characteristic p, the augmentation ideal of O_p(G) generates a nilpotent ideal of kG contained in rad(kG), which is why O_p(G) acts trivially on all simple modules. Corollary (9.24) is the linear group version: H is what the radical of Span_k(G) contributes to the group. The correspondence is one-way, since the radical carries more information.

Q: What is the unipotent radical of a Borel subgroup?

For B_n(k), the full upper triangular group acting on k^n, the composition factors are the n coordinate lines, and an element acts trivially on all of them exactly when its diagonal entries are all 1. So H = UT_n(k), matching the algebraic groups definition where UT_n is the unipotent radical of the standard Borel and B_n/UT_n is the diagonal torus.

Q: Can I compute H from generators?

Yes. Spin up S = Span_k(G) from the generators — at most n^2 basis elements — compute rad S by a nullspace computation, and then H = G (1 + rad S). Equivalently, compute a composition series of V with the MeatAxe and take the kernel of the induced action on the factors, which is how matrix group recognition software proceeds.

Executive Summary

Let $G \subseteq G L (V)$ be any linear group over any field, with $\dim_{k} V < \infty$ . Among the normal subgroups of $G$ that are unipotent there is a largest one, $H$ — the unipotent radical — and $G / H$ is isomorphic to a completely reducible linear group over $k$ .

Two descriptions make $H$ computable. It is the set of elements acting trivially on every composition factor of $V$ ; equivalently, writing $S = {Span}_{k} (G)$ , it is ${g \in G : g - 1 \in rad S}$ . The obstruction to complete reducibility that lives inside the group is therefore visible in the radical of an algebra of dimension at most $n^{2}$ .

$g - 1 \in rad S$ Membership test

$G / H$ Completely reducible

necessary

H = 1

for complete reducibility

not sufficientThe converse fails

Overview

Complete reducibility is the property one wants and rarely has. It is natural to ask what part of $G$ is responsible for its failure, and to hope for a normal subgroup that carries the blame — a radical in the group-theoretic sense, mirroring $rad R$ for rings.

The unipotent radical is that subgroup, and Kolchin's theorem is what makes it exist. A normal unipotent subgroup must act trivially on every composition factor of $V$ : Clifford's theorem makes the restriction semisimple, and $(9.18) (1)$ makes each constituent a line with trivial action. So all normal unipotent subgroups are contained in one explicit subgroup, which is itself normal and unipotent.

The notion is borrowed by, and named for, the theory of linear algebraic groups, where connected groups with trivial unipotent radical are precisely the reductive groups.

Learning Objectives

Prove $(9.22)$ : a unique maximal normal unipotent subgroup $H$ exists and $G / H$ is completely reducible.
Identify the mechanism: Clifford's theorem plus $(9.18) (1)$ force a normal unipotent subgroup to act trivially on composition factors.
State $(9.23)$ with the correct reference class of modules.
Prove $(9.24)$ : $H = {g : g - 1 \in rad {Span}_{k} (G)}$ .
Exhibit a linear group with $H = {1}$ that is not completely reducible.
Compute $H$ , ${Span}_{k} (G)$ and $rad {Span}_{k} (G)$ for $S_{3}$ on a two-dimensional module over $𝔽_{3}$ .

Definitions

Normal unipotent subgroup: $N ⊴ G$ with every element unipotent. In characteristic $p$ this means a normal $p$ -subgroup, by $(9.17)$ .
Unipotent radical: The unique maximal element of the set of normal unipotent subgroups of $G \subseteq G L (V)$ ; written $H$ here, and $R_{u} (G)$ in the algebraic groups literature.
$rad S$: The Jacobson radical of the finite-dimensional algebra $S = {Span}_{k} (G)$ ; a nilpotent ideal, equal to the set of elements annihilating every composition factor of $V$ .
Composition factors of $V$: The simple subquotients in a $k G$ -composition series of $V$ ; well defined up to isomorphism and order by Jordan–Hölder.
Reductive: Trivial unipotent radical, for a connected linear algebraic group. The analogue for abstract linear groups is the necessary condition $H = {1}$ .

The unipotent radical depends on the module V, not only on the abstract group G — except for finite groups in characteristic p, where it is always the largest normal p-subgroup.

Core Concepts

Why a largest normal unipotent subgroup exists

Define $H$ directly as the set of elements acting trivially on every composition factor of $V$ . This is the kernel of $G \to \prod_{i} G L (V_{i})$ , so it is normal without argument, and it is unipotent because $(g - 1)$ shifts each term of a composition series into the next, giving $(g - 1)^{t} = 0$ where $t$ is the length of $V$ .

The work is in showing $H$ contains every normal unipotent subgroup. That is where Clifford's theorem and Kolchin's theorem combine: restricted to a normal unipotent $N$ , each composition factor becomes semisimple; each of its simple $k N$ -constituents is a subquotient of $V$ , hence carries a $λ$ -potent action of $N$ with $λ \equiv 1$ , hence is a line with trivial action by $(9.18) (1)$ ; and a semisimple module all of whose constituents are trivial is itself trivial.

N ⊴ G

unipotentClifford:

V_{i} |_{N}

semisimpleKolchin: constituents are trivial lines

N

acts trivially on each

V_{i}

N \subseteq H

The radical of the span, in two lines

Let $S = {Span}_{k} (G)$ and let $J$ be the set of elements of $S$ annihilating every composition factor of $V$ . Then $J^{t} V = 0$ where $t$ is the length of $V$ , and $V$ is a faithful $S$ -module, so $J^{t} = 0$ and $J \subseteq rad S$ ; the reverse inclusion is automatic because the composition factors are simple $S$ -modules. Hence $rad S = J$ , and $(9.24)$ follows immediately.

Key Results

Theorem(9.22)Existence of the unipotent radical

Let $k$ be an arbitrary field, $V$ a finite-dimensional $k$ -vector space, and $G \subseteq G L (V)$ a linear group. Then $G$ has a unique maximal normal unipotent subgroup $H$ , called the unipotent radical of $G$ . Moreover $G / H$ is isomorphic to a completely reducible linear group over $k$ .

Proof

Let $V_{1}, \dots, V_{t}$ be the composition factors of $V$ as a $k G$ -module (a composition series exists because $\dim_{k} V < \infty$ ), and put

H = {g \in G : g acts as the identity on every V_{i}} = \ker (G \to \prod_{i = 1}^{t} G L (V_{i})) .

** $H$ is normal**, being a kernel. ** $H$ is unipotent**: fix a composition series $V = W_{0} \supset W_{1} \supset \dots \supset W_{t} = 0$ ; for $g \in H$ the operator $g - 1$ maps $W_{j}$ into $W_{j + 1}$ , so $(g - 1)^{t} = 0$ on $V$ .

** $H$ contains every normal unipotent subgroup.** Let $N ⊴ G$ be unipotent and let $M$ be any composition factor of $V$ . Then $M$ is a finite-dimensional simple $k G$ -module, so by Clifford's Theorem $(8.5)$ the restriction $M |_{N}$ is a semisimple $k N$ -module. Each simple $k N$ -constituent of $M |_{N}$ is a subquotient of $V$ as a $k N$ -module, so $N$ acts on it unipotently, and $(9.18) (1)$ makes it one-dimensional with $N$ acting by the scalar $1$ . A semisimple module whose constituents are all trivial is trivial, so $N$ acts trivially on $M$ . As $M$ was arbitrary, $N \subseteq H$ .

The quotient. Put $\bar{V} = V_{1} \oplus \dots \oplus V_{t}$ . By construction $H$ is the kernel of the representation $G \to G L (\bar{V})$ , so $\bar{G} : = G / H$ embeds in $G L (\bar{V})$ as a linear group. Since $\bar{V}$ is a direct sum of simple $k G$ -modules, it is a semisimple $k \bar{G}$ -module, and $\bar{G}$ is completely reducible.

Corollary(9.23)Characterisation by composition factors

With $H$ as above,

H = {g \in G : g acts trivially on every composition factor of V} .

For a finite group $G$ in characteristic $p$ acting faithfully on $V = k G$ , every simple $k G$ -module occurs among the composition factors, so in that case $H = O_{p} (G)$ is exactly the set of elements acting trivially on all simple $k G$ -modules — the statement to be compared with $(8.6)$ .

Remark(9.23a)The reference class of modules matters

The characterisation is about the composition factors of the given $V$ , not about all finite-dimensional irreducible representations of the abstract group $G$ . The discrete Heisenberg group $G \subseteq U T_{3} (ℂ)$ is unipotent, so it is its own unipotent radical, and the composition factors of $ℂ^{3}$ are three trivial modules; yet $G$ has irreducible $ℂ G$ -modules of every finite dimension $m$ , on which it acts faithfully. Those modules are not unipotent, so $(9.18) (1)$ does not apply to them and they are irrelevant to $H$ .

Corollary(9.24)The unipotent radical inside rad Span

Let $G \subseteq G L (V)$ with $\dim_{k} V < \infty$ and $S = {Span}_{k} (G)$ . Then the unipotent radical of $G$ is

H = {g \in G : g - 1 \in rad S} = G \cap (1 + rad S) .

In particular, if $G$ is completely reducible then $rad S = 0$ by $(9.11)$ and hence $H = {1}$ : triviality of the unipotent radical is a necessary condition for complete reducibility.

Proof

The composition factors of $V$ as a $k G$ -module are exactly its composition factors as an $S$ -module, and they are simple $S$ -modules. Let $J = {s \in S : s V_{i} = 0 for all i}$ , the intersection of their annihilators.

Then $J \subseteq rad S$ is clear, since $rad S$ is contained in the annihilator of every simple module and conversely $J$ annihilates the simple modules occurring. For the reverse inclusion: with a composition series of length $t$ , $J$ maps each term into the next, so $J^{t} V = 0$ ; since $S \subseteq {End}_{k} V$ acts faithfully, $J^{t} = 0$ , so $J$ is a nilpotent ideal and $J \subseteq rad S$ . Hence $rad S = J$ .

Now $g \in H$ iff $g$ acts as the identity on each $V_{i}$ , iff $(g - 1) V_{i} = 0$ for all $i$ , iff $g - 1 \in J = rad S$ .

Counterexample(9.24a)Trivial unipotent radical is not sufficient

Let $k$ have characteristic $p > 0$ and let $G$ be a finite group with $p ∣ | G |$ and with no nontrivial normal $p$ -subgroup. Let $V = k G$ with $G$ acting by left multiplication; the action is faithful, so $G \subseteq G L (V)$ .

By $(9.17)$ the normal unipotent subgroups of $G$ are exactly its normal $p$ -subgroups, so the unipotent radical is $O_{p} (G) = {1}$ . But $p ∣ | G |$ makes $k G$ non-semisimple $(6.1)$ , so $V = k G$ is not a semisimple $k G$ -module and $G$ is not a completely reducible linear group.

Concretely, take $p = 2$ , $G = S_{3}$ , $k = 𝔽_{2}$ : the only proper nontrivial normal subgroup of $S_{3}$ is $A_{3}$ of order $3$ , so $O_{2} (S_{3}) = 1$ , while $𝔽_{2} S_{3}$ has nonzero radical.

Remark—Finite groups: the radical does not see the module

If $G$ is finite and $char k = p$ , then for every faithful finite-dimensional representation $V$ the unipotent radical of $G \subseteq G L (V)$ equals $O_{p} (G)$ , because normal unipotent subgroups are normal $p$ -subgroups by $(9.17)$ and this is a property of the abstract group. For infinite linear groups no such module-independence holds.

Proof Techniques and Method

How these proofs work, and which move to reuse.

1. Define the candidate firstDo not search for a maximal normal unipotent subgroup; write down the kernel of the action on the composition factors and prove afterwards that it is maximal.

2. Normality and unipotency are freeA kernel is normal; and

g - 1

shifting a composition series gives

(g - 1)^{t} = 0

3. Clifford to restrictRestriction of a simple module to a normal subgroup is semisimple. This is the only step needing

N

normal rather than merely unipotent.

4. Kolchin to collapseEach constituent is a line with trivial action, so the whole restriction is trivial and

N

lands in the candidate.

5. Read off the algebra statementFaithfulness plus nilpotence identify

rad S

with the annihilator of the composition factors, giving the membership test

g - 1 \in rad S

The reusable pattern is define by an action, prove maximality by a restriction theorem. It is the same shape as the definition of $rad R$ as the annihilator of all simple modules: an invariant defined by what it does to representations is easier to prove maximal than one defined by an internal property such as nilpotence.

Worked Example

$S_{3}$ on a two-dimensional module over $𝔽_{3}$

Let $k = 𝔽_{3}$ , let $G = S_{3}$ permute $e_{1}, e_{2}, e_{3}$ , and let

V = (k e_{1} \oplus k e_{2} \oplus k e_{3}) / k (e_{1} + e_{2} + e_{3}), \dim_{k} V = 2 .

Write ${\bar{e}}_{i}$ for the images, so ${\bar{e}}_{3} = - {\bar{e}}_{1} - {\bar{e}}_{2}$ . In the basis $({\bar{e}}_{1}, {\bar{e}}_{2})$ , with $σ = (1 2 3)$ and $τ = (1 2)$ ,

σ = (\begin{matrix} 0 & - 1 \\ 1 & - 1 \end{matrix}), τ = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) over 𝔽_{3} .

(E.1)

Composition factors

The line $V_{1} = k ({\bar{e}}_{1} - {\bar{e}}_{2})$ is $G$ -invariant: $τ ({\bar{e}}_{1} - {\bar{e}}_{2}) = - ({\bar{e}}_{1} - {\bar{e}}_{2})$ , while $σ ({\bar{e}}_{1} - {\bar{e}}_{2}) = {\bar{e}}_{2} - {\bar{e}}_{3} = {\bar{e}}_{1} + 2 {\bar{e}}_{2} = {\bar{e}}_{1} - {\bar{e}}_{2}$ in characteristic $3$ . So $V_{1}$ is the sign module. Since $\det σ = 1$ and $\det τ = - 1$ , the quotient $V / V_{1}$ is the trivial module. These are the two composition factors.

The unipotent radical

By $(9.23)$ , $H$ consists of the elements acting trivially on both factors, i.e. of the kernel of the sign character: $H = A_{3} = ⟨ σ ⟩$ , of order $3$ . This agrees with the general principle for finite groups: $H = O_{3} (S_{3}) = A_{3}$ .

The spanned algebra and its radical

From $σ^{2} + σ + I = 0$ one gets $σ^{2} = - σ - I$ , and a direct check gives $σ τ = - I + σ + τ$ . So ${I, σ, τ}$ spans $S = {Span}_{𝔽_{3}} (G)$ , and these three are linearly independent, whence $\dim_{𝔽_{3}} S = 3 < 4$ . Indeed $S$ is the algebra of all matrices preserving the line $V_{1}$ .

(σ - I)^{2} = σ^{2} - 2 σ + I = (- σ - I) - 2 σ + I = - 3 σ = 0, rad S = 𝔽_{3} (σ - I) = 𝔽_{3} (\begin{matrix} - 1 & - 1 \\ 1 & - 2 \end{matrix}) .

(E.2)

A one-dimensional square-zero ideal; $S / rad S ≅ 𝔽_{3} \times 𝔽_{3}$ .

Now check $(9.24)$ element by element. $σ - I$ is nilpotent and lies in $rad S$ , so $σ \in H$ . For $τ$ ,

τ - I = (\begin{matrix} - 1 & 1 \\ 1 & - 1 \end{matrix}), \det (τ - I) = 0, tr (τ - I) = - 2 = 1 \neq 0,

(E.3)

eigenvalues $0$ and $1$ : not nilpotent, so $τ - I \notin rad S$ and $τ \notin H$ .

So $H = ⟨ σ ⟩ = A_{3}$ , exactly as the composition factor computation gave. And since $rad S \neq 0$ , $(9.11)$ says $G$ is not completely reducible on $V$ — consistent with $H \neq {1}$ and $(9.24)$ .

A trivial radical without complete reducibility

Change the field to $𝔽_{2}$ and the module to $V = 𝔽_{2} S_{3}$ , the regular module of dimension $6$ . Then $O_{2} (S_{3}) = {1}$ , so the unipotent radical is trivial, yet $2$ divides $| S_{3} | = 6$ and $𝔽_{2} S_{3}$ is not semisimple. The group is not completely reducible even though its unipotent radical vanishes.

Process and Workflow

Span the algebraFrom matrix generators build a basis of

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

by closing under multiplication; at most

n^{2}

basis elements.

Compute the radicalIn characteristic

0

rad S

is the kernel of the trace form on

S

; in characteristic

p

use the Friedl–Rónyai algorithm. Both are polynomial time in

\dim_{k} S

Test membership

g

lies in the unipotent radical iff

g - 1 \in rad S

, by

(9.24)

. For a group given by generators this identifies

H

as the kernel of

G \to G L (\bar{V})

Decide complete reducibility

rad S = 0

iff

G

is completely reducible

(9.11)

. If

rad S \neq 0

but

H = {1}

, the failure is in the gluing, not in a normal unipotent subgroup.

What does the pair $(rad S, H)$ tell you?

$rad S = 0$

G

is completely reducible, and necessarily

H = {1}

$rad S \neq 0$ , $H \neq {1}$ There is a genuine normal unipotent obstruction. Pass to

G / H

acting on

⨁ V_{i}

, which is completely reducible, and study the extension separately.

$rad S \neq 0$ , $H = {1}$ The failure is invisible to the group: the composition factors are glued nontrivially but no normal unipotent subgroup records it. The regular module of a finite group with

O_{p} (G) = 1

in characteristic

p

is the standard case.

Comparison and Classification

Unipotent radicals of familiar linear groups
$G \subseteq G L (V)$	Unipotent radical $H$	Completely reducible?
$U T_{n} (k)$ on $k^{n}$	all of $U T_{n} (k)$	no
$k^{\times} \cdot U T_{n} (k)$ on $k^{n}$	$U T_{n} (k)$	no
Borel $B_{n} (k)$ on $k^{n}$	$U T_{n} (k)$	no
$G L_{n} (k)$ or $S L_{n} (k)$ on $k^{n}$	${1}$	yes — irreducible
Diagonal group on $k^{n}$	${1}$	yes
$S_{3}$ on the $2$ -dimensional module over $𝔽_{3}$	$A_{3}$	no
$S_{3}$ on $𝔽_{2} S_{3}$	${1}$	no — the converse fails
$G$ finite, $char k ∤ \| G \|$	${1}$	yes, by Maschke

What each invariant detects
	sees normal unipotent part	sees gluing of factors	module independent
Unipotent radical $H$	yes	no	for finite $G$ only
$rad {Span}_{k} (G)$	yes	yes	no
Composition factors of $V$	partial	no	no
Complete reducibility	yes	yes	no

What each invariant detects

Relationship Map

The unipotent radical sits between the trivial subgroup and the whole group, and its two descriptions bracket the algebra.

$H$ , the unipotent radical of $G \subseteq G L (V)$ — largest normal unipotent subgroup
- equals
  - ${g : g trivial on every composition factor of V}$
  - $G \cap (1 + rad {Span}_{k} (G))$
  - $O_{p} (G)$ when $G$ is finite and $char k = p$
- contains
  - every normal unipotent subgroup of $G$
  - every normal $p$ -subgroup, in characteristic $p$
- controls
  - $G / H$ is completely reducible on $⨁_{i} V_{i}$
  - $H = {1}$ is necessary for $G$ to be completely reducible
- does not control
  - complete reducibility of $G$ on $V$ itself
  - the algebra $rad {Span}_{k} (G)$ , which can be nonzero with $H = {1}$

Linear group $G$ acting on

V

\dim_{k} V = n

$H$ = unipotent radicalnormal, unipotent, maximal such

Trivial on all composition factors

H = G \cap (1 + rad S)

${1}$ attained exactly when no nontrivial normal unipotent subgroup exists — necessary, not sufficient, for complete reducibility

Applications and Industry Use

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

Algebraic groups

Reductive groups

A connected linear algebraic group with trivial unipotent radical is reductive; the Levi decomposition writes a general group as a semidirect product of a reductive group by its unipotent radical. The entire classification of reductive groups by root data presupposes that the unipotent part has been quotiented away.

Automorphic forms

Parabolic subgroups and Eisenstein series

Parabolic subgroups $P = M N$ have $N$ as their unipotent radical, and constant terms along $N$ are the basic operation in the theory of automorphic forms and the Langlands programme.

Modular representation theory

$O_{p} (G)$ acts trivially on simple modules

For a finite group in characteristic $p$ , the unipotent radical of the regular representation is $O_{p} (G)$ , and $(9.23)$ recovers the classical statement that $O_{p} (G)$ acts trivially on every simple $k G$ -module — the starting point of block theory and of the Green correspondence.

Computational algebra

Structural decomposition of matrix groups

Algorithms that analyse a matrix group given by generators compute a composition series of the natural module, then the kernel of the action on the factors. That kernel is the unipotent radical, and it is the first splitting in the matrix group recognition project implemented in GAP and Magma.

Inside this collection the notion completes the analogy with ring theory begun by the Jacobson radical: an obstruction that is always defined, always normal, always quotient-able — and, as here, not always a complete answer.

Design Considerations

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

Decide whether the group or the algebra is your object. If you need the gluing data, keep ${Span}_{k} (G)$ and its radical. If you need a normal subgroup to quotient by, take $H$ . They are not interchangeable.
Fix the module before speaking of the radical. Except for finite groups in characteristic $p$ , the unipotent radical depends on $V$ ; the same abstract group can have trivial radical in one representation and be its own radical in another.
Do not expect a Levi complement. For algebraic groups over a perfect field, Mostow's theorem provides a Levi decomposition; for an abstract linear group there is no reason for $G \to G / H$ to split, and the extension must be handled on its own terms.
**Use $H = {1}$ as a filter, not a certificate.** It is cheap to test and eliminates many groups, but a positive answer still requires computing $rad {Span}_{k} (G)$ to decide complete reducibility.
Choose the characteristic deliberately. In characteristic $0$ a nontrivial unipotent radical is torsion-free and infinite; in characteristic $p$ it is a $p$ -group and may be finite. Which of these you want often determines the right field to work over.

Failure Modes and Common Mistakes

Do not assume $rad {Span}_{k} (G)$ is generated by ${g - 1 : g \in H}$ ; the containment can be strict, and that gap is exactly the case $H = {1}$ with nonzero radical.
Do not confuse the unipotent radical with the solvable radical or with the Fitting subgroup; they agree in special cases only.
Do not expect $H$ to be closed under passing to overgroups: a normal unipotent subgroup of $G$ need not be normal in a larger linear group containing $G$ .
Do not conflate this $H$ with the unipotent radical $R_{u} (G)$ of an algebraic group, which is defined as a closed connected normal unipotent subgroup; for abstract subgroups of $G L (V)$ no topology is involved.

Quick Reference

$(9.22)$ unique maximal normal unipotent

H

;

G / H

completely reducible on

⨁ V_{i}

$(9.23)$

H = {g : g trivial on all composition factors of V}

$(9.24)$

H = G \cap (1 + rad {Span}_{k} (G))

Necessary condition

G

completely reducible

\Rightarrow

H = {1}

Not sufficient

S_{3}

𝔽_{2} S_{3}

H = {1}

, not completely reducible

Finite groups

H = O_{p} (G)

in characteristic

p

, for every faithful

V

Tools usedClifford

(8.5)

and Kolchin

(9.18) (1)

Algebraic groupstrivial unipotent radical

+

connected

=

reductive

The three descriptions of $H$ , and what each is good for
Description	Best for	Reference
Largest normal unipotent subgroup	structural statements, quotients	$(9.22)$
Trivial on all composition factors	computing $H$ from a composition series	$(9.23)$
$g - 1 \in rad {Span}_{k} (G)$	membership tests, machine computation	$(9.24)$
$O_{p} (G)$	finite groups in characteristic $p$	$(9.17)$ plus $(9.22)$

Frequently Asked Questions

Why is a trivial unipotent radical not enough for complete reducibility?

Because the unipotent radical is a group-level invariant and complete reducibility is a module-level one. The radical of ${Span}_{k} (G)$ records how the composition factors of $V$ are glued together; the subgroup $H$ records only those gluings realised by group elements of the form $1 + (radical element)$ . When $G$ is a finite group with $O_{p} (G) = 1$ in characteristic $p$ dividing $| G |$ , the gluing is present and the subgroup is not: $H = {1}$ while $rad {Span}_{k} (G) \neq 0$ .

Does the unipotent radical depend on the representation?

Yes in general, no for finite groups. For finite $G$ in characteristic $p$ , normal unipotent subgroups are exactly normal $p$ -subgroups by $(9.17)$ , so $H = O_{p} (G)$ whatever the faithful module. For infinite groups the module matters: $ℤ$ has trivial unipotent radical when acting by $diag (2^{m}, 3^{m})$ on $ℚ^{2}$ and is its own unipotent radical when acting unitriangularly.

Is there a Levi decomposition

G ≅ (G / H) ⋉ H

Not for abstract linear groups. For linear algebraic groups over a field of characteristic zero, Mostow's theorem provides a Levi decomposition, and in characteristic $p$ it can fail even there. The theorem proved here gives only that $G / H$ is completely reducible; the extension of $G / H$ by $H$ need not split.

How does this relate to the Jacobson radical of a group algebra?

For a finite group $G$ in characteristic $p$ , the augmentation ideal of $O_{p} (G)$ generates a nilpotent ideal of $k G$ contained in $rad (k G)$ , which is why $O_{p} (G)$ acts trivially on all simple modules. Corollary $(9.24)$ is the linear group version: $H$ is what the radical of ${Span}_{k} (G)$ contributes to the group. The correspondence is one-way, since the radical carries more information.

What is the unipotent radical of a Borel subgroup?

For $B_{n} (k)$ , the full upper triangular group acting on $k^{n}$ , the composition factors are the $n$ coordinate lines, and an element acts trivially on all of them exactly when its diagonal entries are all $1$ . So $H = U T_{n} (k)$ , matching the algebraic groups definition where $U T_{n}$ is the unipotent radical of the standard Borel and $B_{n} / U T_{n}$ is the diagonal torus.

Can I compute

H

from generators?

Yes. Spin up $S = {Span}_{k} (G)$ from the generators — at most $n^{2}$ basis elements — compute $rad S$ by a nullspace computation, and then $H = G \cap (1 + rad S)$ . Equivalently, compute a composition series of $V$ with the MeatAxe and take the kernel of the induced action on the factors, which is how matrix group recognition software proceeds.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §9 (pp. 161–162).
J. E. Humphreys, Linear Algebraic Groups, Graduate Texts in Mathematics 21, Springer-Verlag, 1975, §19 and §30.
A. Borel, Linear Algebraic Groups, 2nd enlarged edition, Graduate Texts in Mathematics 126, Springer-Verlag, 1991.
C. W. Curtis and I. Reiner, Methods of Representation Theory, Volume I, Wiley-Interscience, 1981, §5 and §8.
B. A. F. Wehrfritz, Infinite Linear Groups, Ergebnisse der Mathematik 76, Springer-Verlag, 1973.
D. F. Holt, B. Eick and E. A. O'Brien, Handbook of Computational Group Theory, Chapman and Hall/CRC, 2005.

AI Suggested Questions

State Mostow's theorem on Levi decompositions and explain what fails in characteristic p.
Prove that the augmentation ideal of O_p(G) generates an ideal contained in the radical of kG for a finite group G.
How do matrix group recognition algorithms in GAP and Magma compute the unipotent radical in practice?
Compare the unipotent radical with the solvable radical and the Fitting subgroup for solvable linear groups.
Give an infinite linear group whose unipotent radical is trivial but whose spanned algebra has nonzero radical.
What is the unipotent radical of a parabolic subgroup of GL(n,k), and how does it appear in the theory of Eisenstein series?
For which finite groups G and fields k of characteristic p is the regular module a faithful module with trivial unipotent radical and nonzero radical of the span?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why a largest normal unipotent subgroup exists

The radical of the span, in two lines

06Key Results

07Proof Techniques and Method

08Worked Example

S3 on a two-dimensional module over 𝔽3

Composition factors

The unipotent radical

The spanned algebra and its radical

A trivial radical without complete reducibility

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Reductive groups

Parabolic subgroups and Eisenstein series

Op(G) acts trivially on simple modules

Structural decomposition of matrix groups

13Design Considerations

14Failure Modes and Common Mistakes

15Quick Reference

16Frequently Asked Questions

17Related KEVOS Topics

18References

19AI Suggested Questions

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