Linear Groups and the Burnside Problem

Q: Why does the Trace Lemma insist on absolute irreducibility rather than irreducibility?

Because the proof needs Span_k(G) to be *all* of M_n(k), so that a basis of M_n(k) can be chosen inside G and so that the nondegenerate trace form detects every matrix. Over a non-closed field an irreducible action may span a much smaller algebra — the rotation group C^× acting on R^2 spans a 2-dimensional algebra — and then the injectivity argument fails outright. Over an algebraically closed field the two notions coincide, by Burnside's Theorem (7.3).

Q: Is the bound N^n^3 ever close to sharp?

No, and it is not meant to be. For exponent 2 the correct bound is 2^n against a prediction of 2^n^3. The value of the theorem is that a bound exists depending only on N and n, so that the class of such groups is uniformly finite — a statement of the same character as Jordan's theorem, and equally non-quantitative in practice.

Q: Does Burnside's Second Theorem give a bound on |G|?

Not as proved. The argument shows [G:H] is finite and H is a finite union of finite classes, but the class sizes are not controlled by n and the number of classes in any explicit way in this proof. Contrast (9.4), which is completely explicit.

Executive Summary

A linear group is a subgroup $G \subseteq G L (V)$ with $\dim_{k} V = n < \infty$ . Nothing is assumed about $G$ itself — it may be infinite, torsion, or finitely generated and wild. What is assumed is that the representation is finite-dimensional, and that alone is enough to run the machinery of §7 and §8 against $G$ .

The engine is a single counting device. If the algebra spanned by $G$ is all of $M_{n} (k)$ , then an element of $M_{n} (k)$ is determined by $n^{2}$ of its traces; so a semigroup with only $r$ distinct traces has at most $r^{n^{2}}$ elements. Everything on this page — Burnside's two theorems, and Schur's solution of the General Burnside Problem for linear groups — is that observation plus an induction on $n$ .

$r^{n^{2}}$ Trace Lemma bound

$N^{n^{3}}$ Burnside's bound

1902Burnside asks

1964Golod answers no

Overview

Let $k$ be a field and $V$ a $k$ -vector space with $\dim_{k} V = n$ . A subgroup $G \subseteq G L (V)$ is a linear group (or matrix group). The inclusion makes $V$ a $k G$ -module, so $G$ arrives equipped with a faithful representation of dimension $n$ — and, crucially, with a finite-dimensional image algebra ${Span}_{k} (G) \subseteq {End}_{k} V$ , whatever the cardinality of $G$ .

That is the whole point of the subject. The group algebra $k G$ of an infinite group is an unmanageable object — it is never semisimple — but its image in ${End}_{k} V$ is a finite-dimensional algebra, and finite-dimensional algebras are completely understood by Wedderburn–Artin theory. Statements about $G$ are therefore traded for statements about a small algebra.

Burnside's two problems ask whether torsion alone forces finiteness. For arbitrary groups the answer is no, spectacularly; for linear groups the answer is yes, and the proof is the material of this page and of Schur's Theorem on Torsion Linear Groups.

Learning Objectives

Define a linear group and identify the algebra ${Span}_{k} (G)$ it spans.
State the General Burnside Problem $(9.1)$ and the bounded-exponent version $(9.2)$ , and give the status of each.
Prove the Trace Lemma $(9.3)$ : $| G | \leq r^{n^{2}}$ when $k^{n}$ is absolutely irreducible over $k G$ and $| tr (G) | = r$ .
Prove Burnside's First Theorem $(9.4)$ and locate the point where $char k ∣ N$ destroys it.
State Burnside's Second Theorem $(9.5)$ : a linear group is finite iff it has finitely many conjugacy classes.
Use $(9.7)$ and $(9.8)$ to reduce the General Burnside Problem for linear groups to the abelian-by-finite case.

Definitions

Definition(9.0)Linear group

Let $k$ be a field and $V$ a finite-dimensional $k$ -vector space. A linear group is a subgroup $G \subseteq G L (V)$ . Choosing a basis identifies $G L (V)$ with $G L_{n} (k)$ , $n = \dim_{k} V$ , so a linear group is a group of invertible $n \times n$ matrices. Two linear groups are regarded as the same if they are conjugate in $G L (V)$ .

Torsion (periodic): Every $g \in G$ has finite order. No uniform bound is asserted.
Exponent $N$: $g^{N} = 1$ for all $g \in G$ , with $N \geq 1$ least. A group of finite exponent is torsion; the converse fails.
$p$ -group: Every element has order a power of $p$ . Not assumed finite.
Locally finite: Every finitely generated subgroup is finite. Locally finite $\Rightarrow$ torsion; the converse is exactly $(9.1)$ .
${Span}_{k} (G)$: The $k$ -subspace of ${End}_{k} V$ spanned by the elements of $G$ . Because $G$ is closed under multiplication it is a $k$ -subalgebra — the image of $k G \to {End}_{k} V$ .
Absolutely irreducible: $V$ is absolutely irreducible over $k G$ when $k G \to {End}_{k} V$ is onto; over an algebraically closed field this is the same as irreducible, by Burnside's Theorem $(7.3)$ .

Throughout, k is an arbitrary field unless a characteristic hypothesis is stated, and G is not assumed finite or finitely generated unless said so.

Core Concepts

Traces carry more information than they look like they do

The trace form $(σ, τ) \mapsto tr (σ τ)$ on $M_{n} (k)$ is nondegenerate in every characteristic: if $tr (σ τ) = 0$ for all $τ$ , take $τ = E_{j i}$ to read off $σ_{i j} = 0$ . So a matrix is determined by its pairings against any $k$ -basis of $M_{n} (k)$ .

If $G$ spans $M_{n} (k)$ we may take that basis *inside $G$ *. Then each $σ \in G$ is pinned down by the $n^{2}$ scalars $tr (σ g_{1}), \dots, tr (σ g_{n^{2}})$ — each of which is again a trace of an element of $G$ . A small trace set therefore forces a small group.

G

acts absolutely irreducibly

{Span}_{k} (G) = M_{n} (k)

σ \mapsto (tr (σ g_{i}))_{i}

injective

| G | \leq | tr (G) |^{n^{2}}

Why bounded exponent bounds the traces

If $g^{N} = 1$ then the eigenvalues of $g$ over $\bar{k}$ are $N$ th roots of unity, of which there are at most $N$ . A trace is a sum of $n$ eigenvalues, so $tr (G)$ has at most $N^{n}$ elements — a bound depending only on $N$ and $n$ , not on $| G |$ . Feeding $r = N^{n}$ into the Trace Lemma gives $| G | \leq N^{n^{3}}$ .

What to do when the action is reducible

Absolute irreducibility is essential to the Trace Lemma and cannot simply be assumed. If $V$ has a proper nonzero $k G$ -submodule, choose a basis adapted to it: every $g \in G$ becomes block upper triangular,

g = (\begin{matrix} g_{1} & h \\ 0 & g_{2} \end{matrix}), g_{1} \in G L_{n_{1}} (k), g_{2} \in G L_{n_{2}} (k), n_{1} + n_{2} = n,

(9.4a)

The diagonal blocks give homomorphisms onto linear groups $G_{1}, G_{2}$ of smaller degree; induction on $n$ handles them.

and the induction on $n$ reduces everything to the kernel of $G \to G_{1} \times G_{2}$ , which consists of matrices $(\begin{matrix} I & h \\ 0 & I \end{matrix})$ . That kernel is abelian, and controlling it is the entire remaining difficulty.

Key Results

Problem(9.1)General Burnside Problem

Let $G$ be a finitely generated torsion group. Must $G$ be finite?

Both hypotheses are needed: $ℤ$ is finitely generated and infinite, and the group of all complex roots of unity is torsion and infinite. For abelian $G$ the answer is yes, since a finitely generated abelian torsion group is finite.

Problem(9.2)Bounded Burnside Problem

Let $G$ be a finitely generated group of finite exponent $N$ . Must $G$ be finite? Equivalently: is the free Burnside group $B (d, N)$ on $d$ generators of exponent $N$ finite?

Lam calls this the Restricted Burnside Problem. Modern usage reserves that name for a different question — see Failure Modes and Common Mistakes below.

Lemma(9.3)Trace Lemma

Let $k$ be a field and let $G \subseteq G L_{n} (k)$ be a subsemigroup such that $k^{n}$ is an absolutely irreducible module over the semigroup algebra $k G$ . If the trace function $tr : G \to k$ has finite image of cardinality $r$ , then

| G | \leq r^{n^{2}} < \infty .

Proof

Absolute irreducibility means the algebra map $k G \to M_{n} (k)$ is surjective $(7.5)$ . Its image is spanned by the elements of $G$ , so we may pick $g_{1}, \dots, g_{n^{2}} \in G$ forming a $k$ -basis of $M_{n} (k)$ . Define

ε : M_{n} (k) ⟶ k^{n^{2}}, ε (σ) = (tr (σ g_{1}), \dots, tr (σ g_{n^{2}})) .

This is $k$ -linear. It is injective: if $ε (σ) = 0$ then $tr (σ γ) = 0$ for every $γ \in M_{n} (k)$ by linearity, and taking $γ$ to be the matrix units $E_{j i}$ gives every entry $σ_{i j} = 0$ . Hence $ε$ is a linear isomorphism and in particular injective on the subset $G$ .

For $σ \in G$ each coordinate $tr (σ g_{i})$ is the trace of the element $σ g_{i} \in G$ (here closure of $G$ under multiplication is used), so it takes at most $r$ values. Therefore $| G | = | ε (G) | \leq r^{n^{2}}$ .

Theorem(9.4)Burnside's First Theorem

Let $k$ be a field of characteristic $p \geq 0$ and let $G \subseteq G L_{n} (k)$ be a subgroup of finite exponent $N$ . Assume $p ∤ N$ (a vacuous condition when $p = 0$ ). Then

| G | \leq N^{n^{3}} < \infty .

No finite-generation hypothesis is needed. Thus the Bounded Burnside Problem has an affirmative answer for linear groups whenever the characteristic does not divide the exponent.

Proof

Enlarging $k$ to its algebraic closure changes neither $G$ nor $N$ , so assume $k = \bar{k}$ . Induct on $n$ .

**Base $n = 1$ .** Here $G \subseteq k^{\times}$ and every $g \in G$ satisfies $x^{N} = 1$ , an equation with at most $N$ roots in a field. So $| G | \leq N = N^{1^{3}}$ .

Irreducible case. Suppose $G$ acts irreducibly on $k^{n}$ . Since $k$ is algebraically closed, $k^{n}$ is then absolutely irreducible over $k G$ by Burnside's Theorem $(7.3)$ . Each $g \in G$ satisfies $g^{N} = 1$ , so its eigenvalues lie among the $N$ th roots of unity — at most $N$ of them — and $tr (g)$ , a sum of $n$ such, takes at most $r = N^{n}$ values. The Trace Lemma $(9.3)$ gives $| G | \leq r^{n^{2}} = N^{n^{3}}$ .

Reducible case. Choose a basis adapted to a proper nonzero submodule, so that each $g \in G$ has the block form $(9.4 a)$ , and let $G_{i}$ be the group of blocks $g_{i}$ that occur, a linear group in $G L_{n_{i}} (k)$ of exponent dividing $N$ . By induction $| G_{i} | \leq N^{n_{i}^{3}}$ .

The map $G \to G_{1} \times G_{2}$ , $g \mapsto (g_{1}, g_{2})$ , is a group homomorphism; its kernel consists of the matrices $(\begin{matrix} I & h \\ 0 & I \end{matrix})$ in $G$ . For such an element, induction on the exponent gives ${(\begin{matrix} I & h \\ 0 & I \end{matrix})}^{N} = (\begin{matrix} I & N h \\ 0 & I \end{matrix})$ , so $N h = 0$ . As $p ∤ N$ , the scalar $N$ is invertible in $k$ and $h = 0$ . The homomorphism is therefore injective and

| G | \leq | G_{1} | \cdot | G_{2} | \leq N^{n_{1}^{3}} N^{n_{2}^{3}} = N^{n_{1}^{3} + n_{2}^{3}} \leq N^{(n_{1} + n_{2})^{3}} = N^{n^{3}} . ■

Counterexample(9.4b)The hypothesis on the characteristic is needed

Let $k$ be an infinite field of characteristic $p > 0$ and put

G = {(\begin{matrix} 1 & h \\ 0 & 1 \end{matrix}) : h \in k} \subseteq G L_{2} (k) .

Then $G ≅ (k, +)$ is abelian of exponent $N = p$ and $| G | = | k |$ is infinite. Here $p ∣ N$ and the last step of the proof collapses: $N h = p h = 0$ for every $h$ , so the kernel is everything.

Theorem(9.5)Burnside's Second Theorem

A linear group $G \subseteq G L_{n} (k)$ over a field $k$ is finite iff it has only finitely many conjugacy classes.

Proof

One direction is trivial. Conversely, assume $G$ has finitely many conjugacy classes; again enlarge $k$ to $\bar{k}$ (conjugacy classes are an invariant of $G$ as an abstract group, so nothing changes) and induct on $n$ . The trace is a class function, so $tr (G)$ is finite.

If $G$ acts irreducibly, the Trace Lemma applies directly and $G$ is finite. If not, use the block form $(9.4 a)$ . Each $G_{i}$ is a homomorphic image of $G$ , hence has finitely many conjugacy classes, hence is finite by induction.

Let $H = \ker (G \to G_{1} \times G_{2})$ , so $[G : H] \leq | G_{1} | \cdot | G_{2} | < \infty$ . By $(9.6)$ below $H$ is abelian, so $H \subseteq C_{G} (h)$ for every $h \in H$ ; thus $[G : C_{G} (h)] < \infty$ and every element of $H$ has a finite $G$ -conjugacy class. Since $G$ has only finitely many classes altogether, $H$ is covered by finitely many finite classes and is finite. Then $| G | = [G : H] | H | < \infty$ .

Remark(9.6)The kernel is abelian

For the block sizes fixed above,

(\begin{matrix} I & h \\ 0 & I \end{matrix}) (\begin{matrix} I & h' \\ 0 & I \end{matrix}) = (\begin{matrix} I & h + h' \\ 0 & I \end{matrix}) = (\begin{matrix} I & h' \\ 0 & I \end{matrix}) (\begin{matrix} I & h \\ 0 & I \end{matrix}),

so the kernel of $G \to G_{1} \times G_{2}$ is isomorphic to a subgroup of the additive group of $n_{1} \times n_{2}$ matrices — abelian, and of exponent $p$ when $char k = p > 0$ . This one computation is what both $(9.5)$ and Schur's Theorem $(9.9)$ lean on.

Lemma(9.7)Abelian-by-finite torsion groups

Let $G$ be a finitely generated torsion group containing an abelian subgroup $H$ of finite index. Then $G$ is finite. (That is, $(9.1)$ has an affirmative answer for abelian-by-finite groups.)

Proof

Write $G = g_{1} H \cup \dots \cup g_{m} H$ and choose a finite generating set ${g_{1}, \dots, g_{m}, \dots, g_{n}}$ of $G$ containing these coset representatives and closed under inversion. For $i, j \leq n$ write $g_{i} g_{j} = g_{r} h_{i j}$ with $r \leq m$ and $h_{i j} \in H$ .

Let $H_{0} \subseteq H$ be the subgroup generated by the finitely many $h_{i j}$ . It is a finitely generated abelian torsion group, hence finite.

Claim: every word in the generators lies in $⋃_{t \leq m} g_{t} H_{0}$ . Induct on word length. Given $g_{i} \cdot (g_{r} h)$ with $h \in H_{0}$ , write $g_{i} g_{r} = g_{t} h_{i r}$ ; then $g_{i} g_{r} h = g_{t} h_{i r} h \in g_{t} H_{0}$ because $H$ is abelian and $h_{i r}, h \in H_{0}$ . Hence $G = ⋃_{t \leq m} g_{t} H_{0}$ is a union of $m$ finite sets.

Lemma(9.8)Finitely generated torsion linear groups have bounded exponent

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

Proof

The entries of finitely many generators and their inverses generate a subfield finitely generated over the prime field $P$ , so we may assume $k$ itself is finitely generated over $P$ . Choose a transcendence basis and set $k_{0} = P (t_{1}, \dots, t_{d})$ ; then $k / k_{0}$ is algebraic and finitely generated, hence finite, say $[k : k_{0}] = r$ . Viewing $k^{n}$ as a $k_{0}$ -space of dimension $r n$ embeds $G$ in $G L_{r n} (k_{0})$ .

For $g \in G$ let $m_{g} (t) \in k_{0} [t]$ be its minimal polynomial, of degree at most $r n$ . Since $g$ has finite order, $m_{g}$ divides $t^{ord (g)} - 1$ , so all roots of $m_{g}$ are roots of unity and the coefficients of $m_{g}$ are algebraic over $P$ .

**Case $P = ℚ$ .** The coefficients are algebraic integers lying in $k_{0}$ ; but $k_{0}$ is purely transcendental over $ℚ$ , so its algebraic elements lie in $ℚ$ , and $m_{g} \in ℤ [t]$ . Each coefficient is $\pm$ an elementary symmetric function of at most $r n$ roots of unity, hence bounded in absolute value by $(\binom{r n}{i})$ . Finitely many integer polynomials of bounded degree and bounded coefficients: finitely many possibilities for $m_{g}$ .

**Case $P = 𝔽_{p}$ .** The same argument places the coefficients in the algebraic closure of $𝔽_{p}$ inside $k_{0}$ , which is $𝔽_{p}$ ; so $m_{g} \in 𝔽_{p} [t]$ and there are again only finitely many polynomials of degree at most $r n$ .

Finally $m_{g}$ determines the order of $g$ : it is the least $N$ with $m_{g} ∣ t^{N} - 1$ . So only finitely many orders occur and their lowest common multiple is an exponent for $G$ .

Remark—What (9.7) and (9.8) are for

They are the two halves of Schur's Theorem. Lemma $(9.8)$ upgrades torsion to bounded exponent, which is what $(9.4)$ needs; Lemma $(9.7)$ repairs the one case $(9.4)$ cannot handle, namely $p ∣ N$ , where the kernel of $G \to G_{1} \times G_{2}$ survives but is abelian of finite index. The assembly is carried out in Schur's Theorem on Torsion Linear Groups.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Extend the field first

Finiteness of $G$ , its exponent, and its conjugacy classes are all unchanged by $k ↪ \bar{k}$ . Enlarging to an algebraically closed field converts irreducible into absolutely irreducible via Burnside's Theorem, which is what the Trace Lemma requires.

Move 2

Dichotomy plus induction on n

Either the action is irreducible — apply the Trace Lemma and stop — or it is not, in which case a block triangular basis produces two linear groups of strictly smaller degree. Every theorem in this section has this shape.

Move 3

Handle the unipotent kernel

The kernel of $G \to G_{1} \times G_{2}$ is always abelian $(9.6)$ . Kill it when $p ∤ N$ ; when you cannot, exploit that it is abelian of finite index and invoke $(9.7)$ .

The pattern is worth naming: bound an invariant on the irreducible pieces, then glue along a controlled abelian kernel. It recurs verbatim in the proof of the Lie–Kolchin–Suprunenko theorem and in the construction of the unipotent radical.

Worked Example

Groups of exponent 2: the bound is true but wasteful

Let $char k \neq 2$ and let $G \subseteq G L_{n} (k)$ have exponent $N = 2$ . Theorem $(9.4)$ predicts $| G | \leq 2^{n^{3}}$ . The truth is far smaller. Every $g \in G$ satisfies $g^{2} = 1$ , and a group of exponent $2$ is abelian, since $(g h)^{2} = 1$ gives $g h = h^{- 1} g^{- 1} = h g$ .

Because $t^{2} - 1$ is separable in characteristic $\neq 2$ , each $g$ is diagonalisable with eigenvalues in ${\pm 1}$ , and a commuting family of diagonalisable operators is simultaneously diagonalisable. In a suitable basis $G$ therefore sits inside the diagonal matrices with entries $\pm 1$ :

| G | \leq 2^{n} versus the theorem's 2^{n^{3}},

(E.1)

For $n = 3$ : the sharp bound is $8$ , Burnside's bound is $2^{27} = 134, 217, 728$ . The theorem is a finiteness statement, not a counting statement.

The bound $2^{n}$ is attained by the full group of diagonal sign matrices, so nothing better is possible in general.

Checking the Trace Lemma on the quaternion group

Take $k = ℂ$ , $n = 2$ , and let $G = Q_{8} \subseteq G L_{2} (ℂ)$ be the quaternion group in its faithful $2$ -dimensional representation, generated by

i \mapsto (\begin{matrix} \sqrt{- 1} & 0 \\ 0 & - \sqrt{- 1} \end{matrix}), j \mapsto (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}) .

(E.2)

This representation is irreducible, hence absolutely irreducible over $ℂ$ . The traces are $tr (1) = 2$ , $tr (- 1) = - 2$ , and $tr (g) = 0$ for the six elements of order $4$ ; so $r = 3$ . The Trace Lemma gives $| G | \leq 3^{4} = 81$ , and indeed $| Q_{8} | = 8$ .

Comparison and Classification

Status of the bounded-exponent problem: is every finitely generated group of exponent N finite?
Exponent $N$	Answer	Due to
$N = 2$	yes — the group is abelian	elementary
$N = 3$	yes	Burnside (1902)
$N = 4$	yes	Sanov (1940)
$N = 6$	yes	M. Hall (1958)
$N = 5$	open	—
$N$ odd, $N \geq 4381$	no	Novikov–Adjan (1968)
$N$ odd, $N \geq 665$	no	Adjan (1975)
$N$ large even	no	Ivanov, Lysënok (1990s)
any $N$ , $G$ linear with $char k ∤ N$	yes, with a bound $N^{n^{3}}$	Burnside $(9.4)$

Hypotheses of the three finiteness theorems for linear groups
	Needs f.g.	Needs bounded exponent	Needs char condition	Gives explicit bound
Trace Lemma $(9.3)$	no	no	no	yes
Burnside I $(9.4)$	no	yes	yes	yes
Burnside II $(9.5)$	no	no	no	no
Schur $(9.9)$	yes	no	no	no

Hypotheses of the three finiteness theorems for linear groups

Relationship Map

For a subgroup $G \subseteq G L_{n} (k)$ the implications run as follows; the dashed steps are exactly the theorems of this section.

finitebounded exponenttorsionno condition

Torsion linear group — $G \subseteq G L_{n} (k)$ , every element of finite order
- and finitely generated
  - bounded exponent, by $(9.8)$
  - finite, by Schur $(9.9)$
- and of exponent $N$ with $char k m i d N$
  - finite of order $\leq N^{n^{3}}$ , by $(9.4)$
  - no finite generation needed
- and nothing else
  - locally finite, by $(9.9')$
  - may be infinite: $U T_{2} (\bar{𝔽_{p}})$

Applications and Industry Use

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

Computational group theory

Deciding finiteness of a matrix group

Algorithms in GAP and Magma that test whether a matrix group given by generators over a number field or finite field is finite descend from these trace arguments: compute the trace set, or the order of the elements, and bound the group.

Crystallography

Bieberbach and point groups

Point groups of crystals are finite subgroups of $G L_{n} (ℤ)$ . Finiteness of the torsion subgroups involved, and Jordan-type bounds on their index over an abelian normal subgroup, are what make the classification of space groups a finite computation.

Combinatorial group theory

Obstructions to linearity

Because a finitely generated torsion linear group is finite, Golod's infinite finitely generated $p$ -groups admit no faithful finite-dimensional representation over any field. This is the standard first proof that a group is not linear.

Coding and cryptography

Finite matrix groups over finite fields

Groups of exponent prime to $p$ inside $G L_{n} (𝔽_{q})$ are automatically finite and of bounded order; this underwrites the search space arguments used when matrix groups are deployed as key spaces or automorphism groups of codes.

The honest summary is that this material is a bridge. Its consumers are the structure theory of infinite groups, the theory of algebraic groups, and the algorithms that implement both; it is rarely applied outside mathematics in its own right.

Failure Modes and Common Mistakes

Do not read $| G | \leq N^{n^{3}}$ as a sharp count; for exponent $2$ the true bound is $2^{n}$ .
Do not assume the eigenvalue count $N$ is exact when $p ∣ N$ : in characteristic $p$ the polynomial $t^{N} - 1$ has repeated roots and fewer than $N$ distinct ones — harmless for the bound, fatal for the kernel argument.
Do not confuse linear group with algebraic group. A linear group is any abstract subgroup of $G L (V)$ , with no Zariski-closedness assumed.
Do not expect $(9.5)$ to give a bound: it is a pure finiteness statement, with no function of $n$ and the number of classes extractable from the proof as given.

Historical Notes and Lessons Learned

1878Jordan's theoremJordan proves that a finite subgroup of GL_n(C) has an abelian normal subgroup of index bounded by a function of n alone — the first general finiteness constraint on linear groups.
1902Burnside asksBurnside poses the General Burnside Problem and settles exponent 3. His trace method for linear groups appears at the same time.
1911SchurSchur proves that a finitely generated torsion subgroup of GL_n(C) is finite, settling the General Burnside Problem for linear groups in characteristic zero.
1940–1958Small exponentsSanov settles exponent 4 and M. Hall exponent 6; both proofs are combinatorial and give no method for general N.
1964Golod–ShafarevichGolod constructs, for each prime p, an infinite finitely generated p-group. The General Burnside Problem is answered negatively, and the examples are necessarily non-linear.
1968Novikov–AdjanThe free Burnside group of exponent N is shown infinite for odd N at least 4381, later improved by Adjan to odd N at least 665; the even case follows in the 1990s through work of Ivanov and Lysënok.
1989–1991ZelmanovThe Restricted Burnside Problem in its modern form is answered affirmatively: there are only finitely many finite d-generator groups of any given exponent. Fields Medal, 1994.

The methodological lesson is sharp. Every positive result here is obtained by representing the group and counting inside a finite-dimensional algebra; every negative result is obtained by building a group that admits no such representation. The two Burnside problems are, in effect, questions about the limits of linearity.

Quick Reference

Linear group

G \subseteq G L (V)

\dim_{k} V = n < \infty

;

G

arbitrary

Spanned algebra

S = {Span}_{k} (G) = im (k G \to {End}_{k} V)

Trace Lemmaabsolutely irreducible,

| tr (G) | = r

\Rightarrow

| G | \leq r^{n^{2}}

Burnside Iexponent

N

char k ∤ N

\Rightarrow

| G | \leq N^{n^{3}}

Burnside II

G

finite

\Leftrightarrow

finitely many conjugacy classes

Kernel

\ker (G \to G_{1} \times G_{2})

is abelian, of exponent

p

in characteristic

p

GBPfalse in general (Golod), true for linear groups (Schur)

Bounded BPfalse for large

N

; true for

N = 2, 3, 4, 6

;

N = 5

open

Numbered results of §9 used on this page
Reference	Statement	Key hypothesis
$(9.1)$	General Burnside Problem	f.g. torsion
$(9.2)$	Bounded Burnside Problem	f.g. of exponent $N$
$(9.3)$	Trace Lemma, $\| G \| \leq r^{n^{2}}$	absolutely irreducible semigroup
$(9.4)$	$\| G \| \leq N^{n^{3}}$	exponent $N$ , $char k ∤ N$
$(9.5)$	finite $\Leftrightarrow$ finitely many classes	linear
$(9.6)$	block kernel is abelian	block triangular form
$(9.7)$	abelian-by-finite f.g. torsion is finite	abelian subgroup of finite index
$(9.8)$	f.g. torsion linear $\Rightarrow$ bounded exponent	finitely generated

Frequently Asked Questions

Why does the Trace Lemma insist on absolute irreducibility rather than irreducibility?

Because the proof needs ${Span}_{k} (G)$ to be all of $M_{n} (k)$ , so that a basis of $M_{n} (k)$ can be chosen inside $G$ and so that the nondegenerate trace form detects every matrix. Over a non-closed field an irreducible action may span a much smaller algebra — the rotation group $ℂ^{\times}$ acting on $ℝ^{2}$ spans a $2$ -dimensional algebra — and then the injectivity argument fails outright. Over an algebraically closed field the two notions coincide, by Burnside's Theorem $(7.3)$ .

Where exactly does Burnside's First Theorem break when the characteristic divides the exponent?

Only in the reducible case, and only at the very last step. Everything up to and including the bound $| G_{i} | \leq N^{n_{i}^{3}}$ on the diagonal blocks survives. What fails is the injectivity of $G \to G_{1} \times G_{2}$ : an element $(\begin{matrix} I & h \\ 0 & I \end{matrix})$ of the kernel satisfies $N h = 0$ , which forces $h = 0$ only when $N$ is invertible in $k$ . When $p ∣ N$ the kernel can be an infinite elementary abelian $p$ -group, as $(9.4 b)$ shows.

Is the bound

N^{n^{3}}

ever close to sharp?

No, and it is not meant to be. For exponent $2$ the correct bound is $2^{n}$ against a prediction of $2^{n^{3}}$ . The value of the theorem is that a bound exists depending only on $N$ and $n$ , so that the class of such groups is uniformly finite — a statement of the same character as Jordan's theorem, and equally non-quantitative in practice.

Do these results say anything about non-linear groups?

They say what such groups must avoid. Since a finitely generated torsion linear group is finite $(9.9)$ , any infinite finitely generated torsion group — Golod's $p$ -groups, the free Burnside groups of large exponent — has no faithful finite-dimensional representation over any field whatsoever. Non-linearity is often proved exactly this way, or via Mal'cev's theorem that finitely generated linear groups are residually finite.

Why is the semigroup version of the Trace Lemma worth stating?

Because inverses are never used: the proof only needs the products $σ g_{i}$ to lie back in $G$ . Stating it for semigroups makes it directly applicable to multiplicative sets of matrices arising as images of monoids, and to the intermediate sets produced inside inductions, without a separate argument.

Does Burnside's Second Theorem give a bound on

| G |

Not as proved. The argument shows $[G : H]$ is finite and $H$ is a finite union of finite classes, but the class sizes are not controlled by $n$ and the number of classes in any explicit way in this proof. Contrast $(9.4)$ , which is completely explicit.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §9 (pp. 149–162).
B. A. F. Wehrfritz, Infinite Linear Groups, Ergebnisse der Mathematik 76, Springer-Verlag, 1973.
S. I. Adjan, The Burnside Problem and Identities in Groups, Ergebnisse der Mathematik 95, Springer-Verlag, 1979.
M. Vaughan-Lee, The Restricted Burnside Problem, 2nd edition, London Mathematical Society Monographs, Oxford University Press, 1993.
E. S. Golod, “On nil-algebras and finitely approximable p-groups”, Izvestiya Akademii Nauk SSSR, Seriya Matematicheskaya 28 (1964), 273–276.
C. W. Curtis and I. Reiner, Representation Theory of Finite Groups and Associative Algebras, Wiley-Interscience, 1962.

AI Suggested Questions

Work through Golod's construction of an infinite finitely generated p-group and identify precisely where linearity would fail.
How does Mal'cev's residual finiteness theorem for finitely generated linear groups compare with Schur's theorem as a non-linearity criterion?
State Jordan's theorem with an explicit bound on the index, and explain how it strengthens Burnside's First Theorem in characteristic zero.
What is known about the Bounded Burnside Problem for exponent 5, and why do the Novikov-Adjan methods not reach it?
Explain how Zelmanov's solution of the Restricted Burnside Problem uses Lie algebra methods and the Hall-Higman reduction.
Can the bound in the Trace Lemma be improved if G is assumed to be a group rather than a semigroup?
Give an example of an infinite linear group with finitely many conjugacy classes over a field that is not algebraically closed, or prove none exists.
How do algorithms in GAP or Magma decide finiteness of a matrix group over a number field, and what is their complexity?

Linear Groups and Burnside’s Problem

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Traces carry more information than they look like they do

Why bounded exponent bounds the traces

What to do when the action is reducible

Key Results

Proof Techniques and Method

Extend the field first

Dichotomy plus induction on n

Handle the unipotent kernel

Worked Example

Groups of exponent 2: the bound is true but wasteful

Checking the Trace Lemma on the quaternion group

Comparison and Classification

Relationship Map

Applications and Industry Use

Deciding finiteness of a matrix group

Bieberbach and point groups

Obstructions to linearity

Finite matrix groups over finite fields

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

Traces carry more information than they look like they do

Why bounded exponent bounds the traces

What to do when the action is reducible

06Key Results

07Proof Techniques and Method

Extend the field first

Dichotomy plus induction on n

Handle the unipotent kernel

08Worked Example

Groups of exponent 2: the bound is true but wasteful

Checking the Trace Lemma on the quaternion group

09Comparison and Classification

10Relationship Map

11Applications and Industry Use

Deciding finiteness of a matrix group

Bieberbach and point groups

Obstructions to linearity

Finite matrix groups over finite fields

12Failure Modes and Common Mistakes

13Historical Notes and Lessons Learned

14Quick Reference

15Frequently Asked Questions

16Related KEVOS Topics

17References

18AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

Comparison and Classification

Relationship Map

Applications and Industry Use

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions