Schur’s Theorem on Torsion Linear Groups

Q: Why can finite generation not be dropped?

Because infinite torsion linear groups exist. The group of all complex roots of unity is torsion and infinite inside GL_1(C); the additive group of F_p, realised as unitriangular 2 × 2 matrices, is torsion of exponent p and infinite. Both are locally finite, in accordance with (9.9'), and neither is finitely generated.

Q: Where does the proof use the characteristic?

Only in the reducible step. The irreducible step counts eigenvalues among the roots of t^N-1, and that set is finite in every characteristic. In the reducible step Burnside's argument would kill the kernel using Nh = 0 and N invertible; when char k N that fails, and Schur substitutes the observation that the kernel is abelian of finite index and appeals to (9.7).

Q: Does Schur's theorem bound the order of G?

No. For fixed n and a fixed number of generators the order is unbounded: _m ⊆ GL_1(C) is cyclic on one generator and has order m. A bound requires extra data — the exponent, as in Burnside's N^n^3, or arithmetic constraints on the field, as in Minkowski's bound for subgroups of GL_n(Q).

Q: Are torsion subgroups of GL_n(Z) finite?

Yes, and uniformly so. Every finitely generated subgroup is finite by (9.9), and by Minkowski's theorem the order of a finite subgroup of GL_n(Q) divides an explicit function of n. A locally finite group whose finitely generated subgroups have uniformly bounded order is itself of that bounded order, so torsion subgroups of GL_n(Z) are finite with a bound depending only on n.

Q: How is this used to prove a group is not linear?

Contrapositively. Exhibit the group as finitely generated, torsion and infinite — Golod's p-groups, the Grigorchuk group, the free Burnside groups of large exponent — and (9.9) forbids any faithful finite-dimensional representation over any field. This is usually easier than showing failure of residual finiteness, and it applies where the Tits alternative gives nothing new.

Q: Is the analogue true for algebras?

The corresponding question is the Kurosh problem: must a finitely generated algebraic algebra over a field be finite-dimensional? The answer is no, by the same Golod–Shafarevich construction that answers the General Burnside Problem — the group example is manufactured from the algebra example. Within a fixed finite-dimensional matrix algebra, however, the answer is yes for trivial reasons of dimension.

Executive Summary

**Schur's Theorem $(9.9)$ .** Let $k$ be a field and $G \subseteq G L_{n} (k)$ a finitely generated torsion subgroup. Then $G$ is finite.

Schur proved this in 1911 for $k = ℂ$ ; Kaplansky supplied the modifications for characteristic $p$ . It settles the General Burnside Problem affirmatively for linear groups, and therefore tells us that every negative answer to that problem — Golod's infinite finitely generated $p$ -groups, the large-exponent free Burnside groups — must be a group with no faithful finite-dimensional representation over any field at all.

1911Schur, characteristic 0

$(9.8)$ torsion → bounded exponent

$(9.7)$ abelian-by-finite repair

anyCharacteristic allowed

Overview

Burnside's First Theorem already settles bounded exponent linear groups when $char k ∤ N$ , with an explicit bound $N^{n^{3}}$ and no finite generation needed. Two things are missing from that result: torsion groups need not have bounded exponent, and the characteristic may divide the exponent. Schur's theorem removes both gaps at the price of assuming finite generation.

The repairs are exactly two lemmas from Linear Groups and Burnside's Problem. Lemma $(9.8)$ says a finitely generated torsion linear group automatically has bounded exponent — this is where finite generation is spent. Lemma $(9.7)$ says a finitely generated torsion group with an abelian subgroup of finite index is finite — this covers the case where the characteristic divides the exponent and the block kernel refuses to vanish.

The theorem is best read as a linearity obstruction. If you can exhibit an infinite finitely generated torsion group, you have exhibited a group that embeds in no $G L_{n} (k)$ — a conclusion that is otherwise hard to reach.

Learning Objectives

State $(9.9)$ and $(9.9')$ precisely, including the finite generation hypothesis.
Reconstruct the proof: bounded exponent by $(9.8)$ , then the irreducible/reducible dichotomy, then $(9.7)$ .
Explain why the irreducible case needs no hypothesis on $char k$ .
Exhibit infinite torsion linear groups and verify they are not finitely generated.
Deduce that Golod's groups and the large-exponent free Burnside groups are not linear.
Compare Schur's theorem with Selberg's lemma and the Tits alternative as routes to the same conclusion.

Definitions

Torsion: Every element has finite order. Orders may be unbounded.
Locally finite: Every finitely generated subgroup is finite. Always implies torsion, since $⟨ g ⟩$ is finitely generated.
Exponent: The least $N \geq 1$ with $g^{N} = 1$ for all $g$ , when such $N$ exists.
Virtually P: Has a subgroup of finite index with property $P$ . Schur's proof produces a virtually abelian group at its last step.
Linear over $k$: Admits an injective homomorphism into $G L_{n} (k)$ for some $n < \infty$ .
Residually finite: For every $g \neq 1$ there is a homomorphism to a finite group not killing $g$ . By Mal'cev, every finitely generated linear group is residually finite.

Finite generation is a hypothesis on the group, not on the field. The field k is arbitrary and may be of any characteristic and any cardinality.

Core Concepts

Where finite generation is actually used

Exactly once, in Lemma $(9.8)$ , and once more in Lemma $(9.7)$ . In $(9.8)$ it lets us shrink $k$ to a field finitely generated over its prime field, so that the minimal polynomials of the elements of $G$ range over a finite set; that finite set bounds the orders and hence the exponent.

Everything after that point uses only bounded exponent — until the very last step, where $(9.7)$ needs finite generation again to conclude that a finitely generated torsion group with abelian subgroup of finite index is finite.

Why the irreducible case is characteristic-free

Suppose $G$ has exponent $N$ and acts irreducibly on ${\bar{k}}^{n}$ . Whether or not $p ∣ N$ , the polynomial $t^{N} - 1$ has at most $N$ roots in $\bar{k}$ , so eigenvalues of elements of $G$ range over a finite set and $| tr (G) | \leq N^{n}$ . Burnside's theorem makes the action absolutely irreducible, the Trace Lemma applies, and $G$ is finite.

Torsion versus locally finite

For abstract groups these are genuinely different: Golod's $p$ -groups are torsion and not locally finite. For linear groups the difference vanishes, because local finiteness is tested on finitely generated subgroups and each such subgroup is again linear of the same degree $n$ — the degree does not grow as one passes to subgroups. That stability of $n$ is the reason the induction closes.

Key Results

Lemma(9.8)Bounded exponent

Let $k$ be a field and $G \subseteq G L_{n} (k)$ a finitely generated torsion subgroup. Then $G$ has finite exponent.

Sketch: reduce to $k$ finitely generated over its prime field $P$ , then to $G \subseteq G L_{r n} (k_{0})$ with $k_{0}$ purely transcendental over $P$ and $r = [k : k_{0}] < \infty$ . Minimal polynomials of torsion elements have roots of unity as their roots, so their coefficients are algebraic over $P$ and therefore lie in $ℤ$ (when $P = ℚ$ , with bounded absolute value) or in $𝔽_{p}$ . Only finitely many such polynomials of degree $\leq r n$ exist, and the minimal polynomial determines the order. Full proof in Linear Groups and Burnside's Problem.

Lemma(9.7)Abelian-by-finite torsion groups are finite

Let $G$ be a finitely generated torsion group with an abelian subgroup $H$ of finite index. Then $G$ is finite. Note that no linearity is assumed; this is a statement about abstract groups.

Theorem(9.9)Schur

Let $k$ be a field of arbitrary characteristic and let $G \subseteq G L_{n} (k)$ be a finitely generated torsion subgroup. Then $G$ is finite.

Proof

By $(9.8)$ , $G$ has finite exponent $N$ . Replace $k$ by its algebraic closure — this changes neither $G$ nor $N$ nor finite generation — and induct on $n$ .

**Base $n = 1$ .** $G \subseteq k^{\times}$ consists of roots of $t^{N} - 1$ , so $| G | \leq N$ .

** $G$ acts irreducibly.** Every $g \in G$ satisfies $g^{N} = 1$ , so its eigenvalues lie in the finite set of roots of $t^{N} - 1$ in $k$ and $tr (G)$ is finite, of cardinality $r \leq N^{n}$ . Since $k$ is algebraically closed, irreducible means absolutely irreducible $(7.3)$ , so the Trace Lemma $(9.3)$ gives $| G | \leq r^{n^{2}} < \infty$ . No hypothesis on $char k$ enters.

** $G$ acts reducibly.** Choose a basis adapted to a proper nonzero $k G$ -submodule, so that every $g \in G$ takes the block form

g = (\begin{matrix} g_{1} & h \\ 0 & g_{2} \end{matrix}), n_{1} + n_{2} = n, n_{i} \geq 1,

and let $G_{i} \subseteq G L_{n_{i}} (k)$ be the group of blocks $g_{i}$ occurring. Each $G_{i}$ is a homomorphic image of $G$ , hence finitely generated and torsion, so by the inductive hypothesis $G_{1}$ and $G_{2}$ are finite.

Therefore $H : = \ker (G \to G_{1} \times G_{2})$ has index at most $| G_{1} | \cdot | G_{2} | < \infty$ in $G$ . By $(9.6)$ , $H$ — consisting of the matrices $(\begin{matrix} I & h \\ 0 & I \end{matrix})$ lying in $G$ — is abelian.

So $G$ is a finitely generated torsion group with an abelian subgroup of finite index, and $(9.7)$ gives $| G | < \infty$ . $■$

Theorem(9.9')Torsion equals locally finite

A linear group $G \subseteq G L_{n} (k)$ over a field $k$ is torsion iff it is locally finite.

Proof

If $G$ is locally finite then every cyclic subgroup $⟨ g ⟩$ is finite, so $G$ is torsion — this direction holds for all groups. Conversely, let $G$ be torsion and let $H \subseteq G$ be finitely generated. Then $H$ is a finitely generated torsion subgroup of the same $G L_{n} (k)$ , so $H$ is finite by $(9.9)$ .

Corollary(9.9a)A linearity obstruction

An infinite finitely generated torsion group admits no injective homomorphism into $G L_{n} (k)$ , for any $n$ and any field $k$ . In particular Golod's infinite finitely generated $p$ -groups and the free Burnside groups $B (d, N)$ for the exponents $N$ where they are known to be infinite are not linear over any field.

Remark—Jordan–Schur

Schur proved more in characteristic zero. Jordan's theorem provides a function $j (n)$ such that every finite subgroup of $G L_{n} (ℂ)$ has a normal abelian subgroup of index at most $j (n)$ ; Schur extended this to all torsion subgroups of $G L_{n} (ℂ)$ . The bound is uniform in the group, which is what makes the statement useful in the classification of crystallographic groups.

Remark—Two later proofs

In characteristic zero, Selberg's lemma — every finitely generated linear group over a field of characteristic $0$ has a torsion-free subgroup of finite index — gives $(9.9)$ in one line: a torsion group whose finite-index subgroup is torsion-free has that subgroup trivial, hence is finite. The Tits alternative gives another route: a finitely generated linear group is either virtually solvable or contains a nonabelian free subgroup, and a torsion group contains no free subgroup.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Upgrade torsion to bounded exponentLemma

(9.8)

. Spend finite generation here: it makes the field, and hence the set of minimal polynomials, finite in the relevant sense.

Extend the fieldPass to

\bar{k}

so that irreducible becomes absolutely irreducible. Nothing about

G

changes.

Split on irreducibilityIrreducible: finitely many traces, Trace Lemma, done. Reducible: block triangularise and induct on the degree.

Control the kernelThe kernel of

G \to G_{1} \times G_{2}

is abelian; the quotient is finite by induction.

Close with (9.7)A finitely generated torsion group that is abelian-by-finite is finite. This replaces Burnside's injectivity step and needs no characteristic hypothesis.

The reusable idea is the reduction of a field-theoretic finiteness question to a bounded-degree polynomial count. Lemma $(9.8)$ is a template: whenever elements of a linear group satisfy a fixed algebraic condition, their minimal polynomials live in a finite set once the coefficient field is finitely generated over its prime field and the coefficients are constrained to be algebraic integers or to lie in $𝔽_{p}$ .

Worked Example

A finitely generated torsion group in $G L_{2} (ℚ)$

Let

A = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}), B = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}), G = ⟨ A, B ⟩ \subseteq G L_{2} (ℚ) .

(E.1)

Then $A^{4} = I$ , $B^{2} = I$ , and $B A B^{- 1} = (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}) = A^{- 1}$ . So $G$ is a quotient of the dihedral group of order $8$ , and since $A$ has order $4$ and $B \notin ⟨ A ⟩$ , $| G | = 8$ .

Schur's theorem predicts finiteness; $(9.8)$ predicts bounded exponent, and indeed $\exp G = 4$ ; Burnside's bound $(9.4)$ predicts $| G | \leq 4^{2^{3}} = 65536$ . The true answer is $8$ . Every element is a root of $t^{4} - 1$ , whose roots $\pm 1, \pm \sqrt{- 1}$ generate $ℚ (\sqrt{- 1})$ — consistent with $(9.8)$ , where the minimal polynomials $t - 1$ , $t + 1$ , $t^{2} + 1$ , $t^{2} - 1$ are the only ones occurring.

Two infinite torsion linear groups — neither finitely generated

Characteristic zero. The group $μ_{\infty} \subseteq G L_{1} (ℂ)$ of all roots of unity is torsion and infinite. It is not finitely generated: any finite subset lies in $μ_{m}$ for $m$ the lcm of the orders, and $μ_{m} \neq μ_{\infty}$ .

**Characteristic $p$ .** Let $\bar{𝔽_{p}}$ be an algebraic closure and

U = {(\begin{matrix} 1 & h \\ 0 & 1 \end{matrix}) : h \in \bar{𝔽_{p}}} ≅ (\bar{𝔽_{p}}, +) .

(E.2)

Infinite, abelian, of exponent $p$ : a torsion linear group that is not finite.

$U$ is locally finite — a finitely generated subgroup corresponds to a finite-dimensional $𝔽_{p}$ -subspace of $\bar{𝔽_{p}}$ , hence a finite group — exactly as $(9.9')$ demands. Note that $U$ has bounded exponent $p$ , so Burnside's First Theorem does not apply: its hypothesis $char k ∤ N$ fails.

A near miss: two reflections

Let $R_{θ}$ and $R_{0}$ be the reflections of $ℝ^{2}$ in the lines at angles $θ$ and $0$ . Their product is the rotation by $2 θ$ , which has finite order iff $θ$ is a rational multiple of $π$ . So $⟨ R_{θ}, R_{0} ⟩$ is finitely generated and generated by torsion elements, but is torsion only in the rational case — where it is the finite dihedral group. Being generated by torsion elements is not the same as being torsion, and Schur's theorem says nothing about the irrational case, where the group is the infinite dihedral group.

Process and Workflow

You have a torsion subgroup $G \subseteq G L_{n} (k)$ . Is it finite?

$G$ is finitely generatedYes — finite, by Schur

(9.9)

. No bound in terms of

n

and the number of generators is available; the order can be arbitrarily large.

$G$ has exponent $N$ with $char k ∤ N$ Yes — finite of order at most

N^{n^{3}}

, by Burnside's First Theorem

(9.4)

. Finite generation is not needed.

$G$ has finitely many conjugacy classesYes — finite, by Burnside's Second Theorem

(9.5)

, and torsion is not even needed.

None of the above

G

is locally finite by

(9.9')

but may well be infinite:

μ_{\infty}

and

U T_{2} (\bar{𝔽_{p}})

are the standard examples.

To prove that a given abstract group $Γ$ is not linear, the corresponding workflow is: check that $Γ$ is finitely generated; check that every element has finite order; check that $Γ$ is infinite. If all three hold, $Γ$ embeds in no $G L_{n} (k)$ over any field.

Comparison and Classification

Finiteness conditions: abstract groups versus linear groups
Condition	Abstract groups	Linear groups $G \subseteq G L_{n} (k)$
torsion $\Rightarrow$ locally finite	false (Golod)	true, $(9.9')$
f.g. $+$ torsion $\Rightarrow$ finite	false (Golod, Novikov–Adjan)	true, $(9.9)$
f.g. $+$ exponent $N$ $\Rightarrow$ finite	false for large $N$	true, $(9.9)$
exponent $N$ $\Rightarrow$ finite	false	true if $char k ∤ N$ , $(9.4)$ ; false otherwise
torsion $\Rightarrow$ finite	false	false — $μ_{\infty}$ , $U T_{2} (\bar{𝔽_{p}})$
f.g. $\Rightarrow$ residually finite	false	true (Mal'cev)

Which hypothesis each theorem consumes
	finitely generated	bounded exponent	char condition	explicit bound
Burnside I $(9.4)$	no	yes	yes	yes
Burnside II $(9.5)$	no	no	no	no
Lemma $(9.8)$	yes	no	no	no
Schur $(9.9)$	yes	no	no	no
Selberg's lemma	yes	no	yes — char $0$	no

Which hypothesis each theorem consumes

Relationship Map

The classes below are nested, and the collapses that occur when linearity is imposed are exactly the content of this section.

All groupstorsion

⊋

locally finite

⊋

finite; both containments strict

Linear groupstorsion

=

locally finite, by

(9.9')

+ finitely generatedtorsion

=

finite, by Schur

(9.9)

+ exponent $N$ , $char k ∤ N$ finite of order

\leq N^{n^{3}}

, by

(9.4)

— and finite generation becomes unnecessary

f.g. torsion linearbounded exponent

(9.8)

blocks finite by inductionabelian kernel of finite index

(9.6)

finite

(9.7)

Applications and Industry Use

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

Geometric group theory

Proving non-linearity

Schur's theorem is one of the standard three obstructions to linearity, alongside Mal'cev's residual finiteness theorem and the Tits alternative. It is the one that applies to torsion groups such as Grigorchuk's group of intermediate growth, which is an infinite finitely generated $2$ -group and hence linear over no field.

Crystallography

Point groups and space groups

The point group of a crystal is a finite subgroup of $G L_{n} (ℤ)$ . Together with Minkowski's bound on the orders of finite subgroups of $G L_{n} (ℚ)$ , local finiteness makes the enumeration of $n$ -dimensional space groups a terminating computation.

Algebraic groups

Torsion in $G (k)$

For a linear algebraic group $G$ over $k$ , the group of $k$ -points is linear, so its torsion subgroups are locally finite. This is used when analysing the structure of maximal tori and of finite subgroups of Lie type groups.

Computational algebra

Finiteness tests

An algorithm testing whether a matrix group given by generators over a number field is finite may first test that each generator has finite order; Schur's theorem guarantees that torsion of the generators plus torsion of enough products is the right thing to look for, and Minkowski-type bounds terminate the search.

Inside ring theory the theorem is the group-theoretic counterpart of the statement that a finitely generated algebraic algebra over a field, satisfying enough finiteness, is finite-dimensional; the Kurosh problem for algebras is the exact analogue of the General Burnside Problem, and Golod's construction answers both at once.

Failure Modes and Common Mistakes

Do not assume the theorem gives an abelian subgroup of finite index in general; the Jordan–Schur refinement does so in characteristic $0$ , but in characteristic $p$ the relevant normal subgroup is unipotent, not abelian.
Do not apply $(9.9)$ to subgroups of $G L_{n} (R)$ for a general commutative ring $R$ without checking that $R$ embeds in a field, or at least reducing to a residue field.
Do not confuse local finiteness with residual finiteness. Both hold for the relevant linear groups, for entirely different reasons.
Do not read $(9.9')$ as saying every linear group is locally finite; the hypothesis torsion is essential — $G L_{n} (ℤ)$ is linear and not locally finite.

Historical Notes and Lessons Learned

1878JordanA finite subgroup of GL_n(C) has an abelian normal subgroup whose index is bounded by a function of n alone.
1902Burnside's questionThe General Burnside Problem is posed; Burnside himself settles exponent 3 and develops the trace method for linear groups.
1911SchurA finitely generated torsion subgroup of GL_n(C) is finite; Schur also extends Jordan's bounded-index theorem from finite to torsion subgroups.
1940Mal'cevFinitely generated linear groups are residually finite — an independent and now equally standard obstruction to linearity.
mid-centuryKaplanskyThe characteristic p case is settled; the proof is Schur's with the abelian-by-finite lemma replacing the injectivity step that fails when p divides the exponent.
1960Selberg's lemmaEvery finitely generated linear group in characteristic zero has a torsion-free subgroup of finite index, giving a second proof of Schur's theorem in that case.
1964GolodInfinite finitely generated p-groups exist. By Schur's theorem they are not linear over any field — the first widely used non-linearity argument.
1972Tits alternativeA finitely generated linear group either contains a nonabelian free subgroup or is virtually solvable; torsion excludes the first option, yielding yet another route to (9.9).

The lesson is that finiteness theorems for linear groups are cheap and finiteness theorems for abstract groups are not. A representation of bounded degree is an enormously strong hypothesis: it forces the whole group into a fixed finite-dimensional algebra, where counting arguments become available.

Quick Reference

Schur $(9.9)$

G \subseteq G L_{n} (k)

f.g. torsion

\Rightarrow

G

finite

Equivalent form $(9.9')$ linear: torsion

\Leftrightarrow

locally finite

Input 1

(9.8)

: f.g. torsion linear

\Rightarrow

bounded exponent

Input 2

(9.7)

: f.g. torsion abelian-by-finite

\Rightarrow

finite

Input 3

(9.3)

Trace Lemma, via

(7.3)

Burnside

Characteristicarbitrary; char p only affects the reducible case

Effective?no bound on

| G |

from

n

and generator count

Contrapositiveinfinite f.g. torsion group

\Rightarrow

not linear

Standard examples to keep at hand
Group	f.g.?	torsion?	finite?
$μ_{\infty} \subseteq G L_{1} (ℂ)$	no	yes	no
$U T_{2} (\bar{𝔽_{p}})$	no	yes	no
$⟨ A, B ⟩ ≅ D_{4} \subseteq G L_{2} (ℚ)$	yes	yes	yes — order $8$
$G L_{n} (ℤ)$	yes	no	no
Golod's $p$ -group	yes	yes	no — hence not linear
infinite dihedral in $G L_{2} (ℝ)$	yes	no	no

Frequently Asked Questions

Why can finite generation not be dropped?

Because infinite torsion linear groups exist. The group of all complex roots of unity is torsion and infinite inside $G L_{1} (ℂ)$ ; the additive group of $\bar{𝔽_{p}}$ , realised as unitriangular $2 \times 2$ matrices, is torsion of exponent $p$ and infinite. Both are locally finite, in accordance with $(9.9')$ , and neither is finitely generated.

Where does the proof use the characteristic?

Only in the reducible step. The irreducible step counts eigenvalues among the roots of $t^{N} - 1$ , and that set is finite in every characteristic. In the reducible step Burnside's argument would kill the kernel using $N h = 0$ and $N$ invertible; when $char k ∣ N$ that fails, and Schur substitutes the observation that the kernel is abelian of finite index and appeals to $(9.7)$ .

Does Schur's theorem bound the order of

G

No. For fixed $n$ and a fixed number of generators the order is unbounded: $μ_{m} \subseteq G L_{1} (ℂ)$ is cyclic on one generator and has order $m$ . A bound requires extra data — the exponent, as in Burnside's $N^{n^{3}}$ , or arithmetic constraints on the field, as in Minkowski's bound for subgroups of $G L_{n} (ℚ)$ .

Are torsion subgroups of

G L_{n} (ℤ)

finite?

Yes, and uniformly so. Every finitely generated subgroup is finite by $(9.9)$ , and by Minkowski's theorem the order of a finite subgroup of $G L_{n} (ℚ)$ divides an explicit function of $n$ . A locally finite group whose finitely generated subgroups have uniformly bounded order is itself of that bounded order, so torsion subgroups of $G L_{n} (ℤ)$ are finite with a bound depending only on $n$ .

How is this used to prove a group is not linear?

Contrapositively. Exhibit the group as finitely generated, torsion and infinite — Golod's $p$ -groups, the Grigorchuk group, the free Burnside groups of large exponent — and $(9.9)$ forbids any faithful finite-dimensional representation over any field. This is usually easier than showing failure of residual finiteness, and it applies where the Tits alternative gives nothing new.

Is the analogue true for algebras?

The corresponding question is the Kurosh problem: must a finitely generated algebraic algebra over a field be finite-dimensional? The answer is no, by the same Golod–Shafarevich construction that answers the General Burnside Problem — the group example is manufactured from the algebra example. Within a fixed finite-dimensional matrix algebra, however, the answer is yes for trivial reasons of dimension.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §9 (pp. 153–154).
I. Schur, “Über Gruppen periodischer linearer Substitutionen”, Sitzungsberichte der Preussischen Akademie der Wissenschaften (1911), 619–627.
B. A. F. Wehrfritz, Infinite Linear Groups, Ergebnisse der Mathematik 76, Springer-Verlag, 1973, Chapters 4 and 9.
I. Kaplansky, Fields and Rings, 2nd edition, Chicago Lectures in Mathematics, University of Chicago Press, 1972.
J. Tits, “Free subgroups in linear groups”, Journal of Algebra 20 (1972), 250–270.
D. Passman, The Algebraic Structure of Group Rings, Wiley-Interscience, 1977.

AI Suggested Questions

Reconstruct the proof of Selberg's lemma and check that it really gives Schur's theorem in characteristic zero.
State the Jordan-Schur theorem with an explicit bound on the index and compare with the best known values of Jordan's function.
Is the Grigorchuk group linear over any field, and what is the shortest proof either way?
How does the Tits alternative fail in characteristic p without a finite generation hypothesis?
Give the sharpest known bound for the order of a finite subgroup of GL(n,Q) and compare with Minkowski's bound.
What is the largest class of rings R for which torsion subgroups of GL(n,R) are still locally finite?
Explain how Golod's construction produces a group from an algebra, and why the resulting group cannot be linear.

Schur’s Theorem on Torsion Groups

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Where finite generation is actually used

Why the irreducible case is characteristic-free

Torsion versus locally finite

Key Results

Proof Techniques and Method

Worked Example

A finitely generated torsion group in $G L_{2} (ℚ)$

Two infinite torsion linear groups — neither finitely generated

A near miss: two reflections

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Proving non-linearity

Point groups and space groups

Torsion in $G (k)$

Finiteness tests

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

Where finite generation is actually used

Why the irreducible case is characteristic-free

Torsion versus locally finite

06Key Results

07Proof Techniques and Method

08Worked Example

A finitely generated torsion group in GL2(ℚ)

Two infinite torsion linear groups — neither finitely generated

A near miss: two reflections

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Proving non-linearity

Point groups and space groups

Torsion in G(k)

Finiteness tests

13Failure Modes and Common Mistakes

14Historical Notes and Lessons Learned

15Quick Reference

16Frequently Asked Questions

17Related KEVOS Topics

18References

19AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

A finitely generated torsion group in $G L_{2} (ℚ)$

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Torsion in $G (k)$

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions