Jordan’s theorem is a basic theorem in the theory of finite linear groups, and can be formulated as follows:

Theorem 1 (Jordan’s theorem) Let ${G}$ be a finite subgroup of the general linear group ${GL_d({\bf C})}$. Then there is an abelian subgroup ${G'}$ of ${G}$ of index ${[G:G'] \leq C_d}$, where ${C_d}$ depends only on ${d}$.

Informally, Jordan’s theorem asserts that finite linear groups over the complex numbers are almost abelian. The theorem can be extended to other fields of characteristic zero, and also to fields of positive characteristic so long as the characteristic does not divide the order of ${G}$, but we will not consider these generalisations here. A proof of this theorem can be found for instance in these lecture notes of mine.

I recently learned (from this comment of Kevin Ventullo) that the finiteness hypothesis on the group ${G}$ in this theorem can be relaxed to the significantly weaker condition of periodicity. Recall that a group ${G}$ is periodic if all elements are of finite order. Jordan’s theorem with “finite” replaced by “periodic” is known as the Jordan-Schur theorem.

The Jordan-Schur theorem can be quickly deduced from Jordan’s theorem, and the following result of Schur:

Theorem 2 (Schur’s theorem) Every finitely generated periodic subgroup of a general linear group ${GL_d({\bf C})}$ is finite. (Equivalently, every periodic linear group is locally finite.)

Remark 1 The question of whether all finitely generated periodic subgroups (not necessarily linear in nature) were finite was known as the Burnside problem; the answer was shown to be negative by Golod and Shafarevich in 1964.

Let us see how Jordan’s theorem and Schur’s theorem combine via a compactness argument to form the Jordan-Schur theorem. Let ${G}$ be a periodic subgroup of ${GL_d({\bf C})}$. Then for every finite subset ${S}$ of ${G}$, the group ${G_S}$ generated by ${S}$ is finite by Theorem 2. Applying Jordan’s theorem, ${G_S}$ contains an abelian subgroup ${G'_S}$ of index at most ${C_d}$.

In particular, given any finite number ${S_1,\ldots,S_m}$ of finite subsets of ${G}$, we can find abelian subgroups ${G'_{S_1},\ldots,G'_{S_m}}$ of ${G_{S_1},\ldots,G_{S_m}}$ respectively such that each ${G'_{S_j}}$ has index at most ${C_d}$ in ${G_{S_j}}$. We claim that we may furthermore impose the compatibility condition ${G'_{S_i} = G'_{S_j} \cap G_{S_i}}$ whenever ${S_i \subset S_j}$. To see this, we set ${S := S_1 \cup \ldots \cup S_m}$, locate an abelian subgroup ${G'_S}$ of ${G_S}$ of index at most ${C_d}$, and then set ${G'_{S_i} := G'_S \cap G_{S_i}}$. As ${G_S}$ is covered by at most ${C_d}$ cosets of ${G'_S}$, we see that ${G_{S_i}}$ is covered by at most ${C_d}$ cosets of ${G'_{S_i}}$, and the claim follows.

Note that for each ${S}$, the set of possible ${G'_S}$ is finite, and so the product space of all configurations ${(G'_S)_{S \subset G}}$, as ${S}$ ranges over finite subsets of ${G}$, is compact by Tychonoff’s theorem. Using the finite intersection property, we may thus locate a subgroup ${G'_S}$ of ${G_S}$ of index at most ${C_d}$ for all finite subsets ${S}$ of ${G}$, obeying the compatibility condition ${G'_T = G'_S \cap G_T}$ whenever ${T \subset S}$. If we then set ${G' := \bigcup_S G'_S}$, where ${S}$ ranges over all finite subsets of ${G}$, we then easily verify that ${G'}$ is abelian and has index at most ${C_d}$ in ${G}$, as required.

Below I record a proof of Schur’s theorem, which I extracted from this book of Wehrfritz. This was primarily an exercise for my own benefit, but perhaps it may be of interest to some other readers.

— 1. Proofs —

We begin with a lemma of Burnside. Given a vector space ${V}$, let ${End(V)}$ denote the ring of linear transformations from ${V}$ to itself.

Lemma 3 Let ${V}$ be a finite-dimensional complex vector space, and let ${A}$ be a complex algebra with identity in ${End(V)}$, i.e. a linear subspace of ${End(V)}$ that is closed under multiplication and contains the identity operator. Then either ${A = End(V)}$, or there exists some proper subspace ${\{0\} \subsetneq W \subsetneq V}$ which is ${A}$-invariant, i.e. ${a W \subset W}$ for all ${a \in A}$.

Proof: Suppose that no such proper ${A}$-invariant subspace ${W}$ exists. Then for any non-zero ${v \in V}$, the vector space ${Av}$ must equal all of ${V}$, since it is a non-trivial ${A}$-invariant subspace. By duality, this implies that for any non-zero dual vector ${u \in V^*}$, the vector space ${A^* u}$ must equal all of ${V^*}$.

Let ${u, v}$ be linearly independent elements of ${V}$. We claim that there exists an element ${a}$ of ${A}$ such that ${au \neq 0}$ and ${av = 0}$. Suppose that this is not the case; then by the Hahn-Banach theorem, there exists ${t \in End(V)}$ such that ${au = tav}$ for all ${a \in A}$. In particular, setting ${a=1}$ we obtain ${u=tv}$, and thus ${atv=tav}$. Replacing ${a}$ by ${ab}$ for some ${b \in A}$, we conclude that ${tabv = abtv = atbv}$, thus ${ta-at}$ annihilates all of ${Av}$. Since ${Av = V}$, we conclude that ${at=ta}$, thus ${A}$ lies in the centraliser of ${t}$. But since ${u=tv}$ and ${u,v}$ are linearly independent, ${t}$ is not a multiple of the identity, and thus by the spectral theorem, ${t}$ has at least one proper eigenspace. But this eigenspace is fixed by ${A}$, a contradiction.

Thus we can find ${a \in A}$ such that ${au \neq 0}$ and ${av=0}$ for any linearly independent ${u,v}$. Iterating this, we see that for any non-zero ${u \in V}$ and any ${0 \leq k \leq \hbox{dim}(V)-1}$, we can find ${a \in A}$ of corank at least ${k}$ that does not annihilate ${u}$. In particular, ${A}$ contains a rank one transformation. Since ${Av = V}$ and ${A^* u = V^*}$ for all ${v \in V}$ and ${u \in V^*}$, this implies that ${A}$ contains all rank one transformations, and hence contains all of ${End(V)}$ by linearity. $\Box$

Corollary 4 (Burnside’s theorem) Let ${V}$ be a complex vector space of some finite dimension ${d}$, and let ${G}$ be a subgroup of ${GL(V)}$ where every element has order at most ${r}$. Then ${G}$ is finite with cardinality ${O_{d,r}(1)}$.

Proof: We induct on dimension, assuming the claim has already been proven for smaller values of ${d}$. Let ${A \subset End(V)}$ be the complex algebra generated by ${G}$ (or equivalently, the complex linear span of ${G}$). Suppose first that there is a proper ${A}$-invariant subspace ${W}$. Then ${G}$ projects down to ${GL(W)}$ and to ${GL(V/W)}$, and by induction hypothesis both of these projections are finite with cardinality ${O_{d,r}(1)}$. Thus there exists a subgroup ${G'}$ of ${G}$ of index ${O_{d,r}(1)}$ whose projections to ${GL(W)}$ and ${GL(V/W)}$ are trivial; in particular, all elements of ${G'}$ are unipotent. But as the complex numbers have zero characteristic, the only unipotent element of finite order is the identity, and so ${G'}$ is trivial, and the claim follows.

Hence we may assume that ${V}$ has no proper ${A}$-invariant subspace. By Lemma 3, ${A}$ must be all of ${GL(V)}$. In particular, one can find ${d^2}$ linearly independent elements ${g_1,\ldots,g_{d^2}}$ of ${G}$.

For any ${g \in G}$, the element ${g g_i}$ has order at most ${r}$, and thus all the eigenvalues of ${g g_i}$ are roots of unity of order at most ${r}$. This means that there are at most ${O_{r,d}(1)}$ possible values of the trace ${\hbox{tr}(gg_i)}$, which is a linear functional of ${g}$. Letting ${g_i}$ vary among the basis ${g_1,\ldots,g_{d^2}}$ of ${End(V)}$, we conclude that there are at most ${O_{r,d}(1)}$ possible values of ${g}$, and the claim follows. $\Box$

Remark 2 The question of whether any finite group with ${m}$ generators in which all elements of order at most ${r}$ is necessarily of order ${O_{m,r}(1)}$ is known as the restricted Burnside problem, and was famously solved affirmatively by Zelmanov in 1990. (Note however that for certain values of ${m}$ and ${r}$ it is possible for the group to be infinite. Also, while any finite group is trivially embedded in some linear group, one does not have any obvious control on the dimension of that group in terms of ${m}$ and ${r}$, so one cannot immediately solve this problem just from Corollary 4.)

To prove Schur’s theorem (Theorem 2), it thus suffices to establish the following proposition:

Proposition 5 Let ${k}$ be a finitely generated extension of the field of rationals ${{\bf Q}}$. Then every periodic element of ${GL_d(k)}$ has order at most ${O_{d,k}(1)}$.

Indeed, to obtain Schur’s theorem, one applies Proposition 5 with ${k}$ equal to the field generated by the coefficients of the generators of the finitely generated periodic group ${G}$, and then applies Corollary 4.

Proof: Suppose first that ${k}$ is a finite extension of ${{\bf Q}}$. If ${a \in GL_d(k)}$ has period ${n}$, then the field generated by the eigenvalues of ${a}$ contains a primitive ${n^{th}}$ root of unity, and thus contains the cyclotomic field of that order. On the other hand, this field has degree ${O_{d}(1)}$ over ${k}$, and thus has degree ${O_{d,k}(1)}$ over the rationals. Thus ${n = O_{d,k}(1)}$, and the claim follows. Note that the bound on ${n}$ depends only on the degree of ${k}$, and not on ${k}$ itself.

Now we extend from the finite degree case to the finitely generated case. (The argument I came up with here, based on obtaining a “Freiman isomorphism” from the finitely generated setting to the finite degree setting, was somewhat crude; no doubt there is a more elegant “abstract nonsense” way to proceed here.) By using a transcendence basis, one can write ${k}$ as a finite extension of ${{\bf Q}(z_1,\ldots,z_m)}$ for some algebraically independent ${z_1,\ldots,z_m}$ over ${{\bf Q}}$. By the primitive element theorem, one can then write ${k = {\bf Q}(z_1,\ldots,z_m)(\alpha)}$ where ${\alpha}$ is algebraic over ${{\bf Q}(z_1,\ldots,z_m)}$ of some degree ${D}$.

Now suppose we have an element ${a \in GL_d(k)}$ of period ${n}$, thus ${a,\ldots,a^{n-1} \neq 1}$ and ${a^n = 1}$. Let ${R}$ be the ring in ${k}$ generated by the coefficients of ${a}$. We can create a ring homomorphism ${\phi: R \rightarrow \bar{k}}$ to a finite extension ${\bar{k}}$ of the rationals by mapping each ${z_i}$ to a rational number ${q_i}$, and then replace ${\alpha}$ with a root of the polynomial formed by replacing ${z_i}$ with ${q_i}$ in the minimal polynomial of ${\alpha}$. As long as one chooses ${(q_1,\ldots,q_m)}$ generically (i.e. outside of a codimension one subset of ${{\bf Q}^m}$), this operation is well-defined on ${R}$ (in that no division by zero issues arise in any of the coefficients of ${a}$). Furthermore, generically one has ${\phi(a),\ldots,\phi(a)^{n-1} \neq 1}$ and ${\phi(a)^n=1}$, thus ${\phi{a}}$ has period ${n}$. Furthermore, the degree of ${\bar{k}}$ over ${{\bf Q}}$ is at most the degree ${D}$ of ${\alpha}$ over ${{\bf Q}(z_1,\ldots,z_m)}$ and is thus bounded uniformly in ${q_1,\ldots,q_m}$. The claim now follows from the finite extension case. $\Box$