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 be a finite subgroup of the general linear group . Then there is an abelian subgroup of of index , where depends only on .

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 , 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 in this theorem can be relaxed to the significantly weaker condition of periodicity. Recall that a group 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 is finite. (Equivalently, every periodic linear group is locally finite.)

Remark 1The question of whetherallfinitely 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 be a periodic subgroup of . Then for every finite subset of , the group generated by is finite by Theorem 2. Applying Jordan’s theorem, contains an abelian subgroup of index at most .

In particular, given any finite number of finite subsets of , we can find abelian subgroups of respectively such that each has index at most in . We claim that we may furthermore impose the compatibility condition whenever . To see this, we set , locate an abelian subgroup of of index at most , and then set . As is covered by at most cosets of , we see that is covered by at most cosets of , and the claim follows.

Note that for each , the set of possible is finite, and so the product space of all configurations , as ranges over finite subsets of , is compact by Tychonoff’s theorem. Using the finite intersection property, we may thus locate a subgroup of of index at most for *all* finite subsets of , obeying the compatibility condition whenever . If we then set , where ranges over all finite subsets of , we then easily verify that is abelian and has index at most in , 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 , let denote the ring of linear transformations from to itself.

Lemma 3Let be a finite-dimensional complex vector space, and let be a complex algebra with identity in , i.e. a linear subspace of that is closed under multiplication and contains the identity operator. Then either , or there exists some proper subspace which is -invariant, i.e. for all .

*Proof:* Suppose that no such proper -invariant subspace exists. Then for any non-zero , the vector space must equal all of , since it is a non-trivial -invariant subspace. By duality, this implies that for any non-zero dual vector , the vector space must equal all of .

Let be linearly independent elements of . We claim that there exists an element of such that and . Suppose that this is not the case; then by the Hahn-Banach theorem, there exists such that for all . In particular, setting we obtain , and thus . Replacing by for some , we conclude that , thus annihilates all of . Since , we conclude that , thus lies in the centraliser of . But since and are linearly independent, is not a multiple of the identity, and thus by the spectral theorem, has at least one proper eigenspace. But this eigenspace is fixed by , a contradiction.

Thus we can find such that and for any linearly independent . Iterating this, we see that for any non-zero and any , we can find of corank at least that does not annihilate . In particular, contains a rank one transformation. Since and for all and , this implies that contains *all* rank one transformations, and hence contains all of by linearity.

Corollary 4 (Burnside’s theorem)Let be a complex vector space of some finite dimension , and let be a subgroup of where every element has order at most . Then is finite with cardinality .

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

Hence we may assume that has no proper -invariant subspace. By Lemma 3, must be all of . In particular, one can find linearly independent elements of .

For any , the element has order at most , and thus all the eigenvalues of are roots of unity of order at most . This means that there are at most possible values of the trace , which is a linear functional of . Letting vary among the basis of , we conclude that there are at most possible values of , and the claim follows.

Remark 2The question of whether any finite group with generators in which all elements of order at most is necessarily of order is known as therestricted Burnside problem, and was famously solved affirmatively by Zelmanov in 1990. (Note however that for certain values of and 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 and , 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 5Let be a finitely generated extension of the field of rationals . Then every periodic element of has order at most .

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

*Proof:* Suppose first that is a finite extension of . If has period , then the field generated by the eigenvalues of contains a primitive root of unity, and thus contains the cyclotomic field of that order. On the other hand, this field has degree over , and thus has degree over the rationals. Thus , and the claim follows. Note that the bound on depends only on the degree of , and not on 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 as a finite extension of for some algebraically independent over . By the primitive element theorem, one can then write where is algebraic over of some degree .

Now suppose we have an element of period , thus and . Let be the ring in generated by the coefficients of . We can create a ring homomorphism to a finite extension of the rationals by mapping each to a rational number , and then replace with a root of the polynomial formed by replacing with in the minimal polynomial of . As long as one chooses generically (i.e. outside of a codimension one subset of ), this operation is well-defined on (in that no division by zero issues arise in any of the coefficients of ). Furthermore, generically one has and , thus has period . Furthermore, the degree of over is at most the degree of over and is thus bounded uniformly in . The claim now follows from the finite extension case.

## 7 comments

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5 October, 2011 at 4:56 pm

Joshua ZelinskyI’m a bit happy to see you link to the Wikipedia article for the Jordan-Schur theorem, since I wrote that article. (“Hooray! I’m useful!”)

It is interesting to note how many different proofs of Jordan-Schur there are.

For example, the classical proof (found in for example Curtis and Reiner) first shows that any periodic subgroup is isomorphic to a subgroup of unitary matrices. Then, they produce various inequalities for unitary matrices involving the Frobenius norm, and use them to explicitly construct tje Jordan-Schur subgroup based on a unitary representation of the group

Quick sketch of basic steps (without proofs)

Lemma 1: If A and B are unitary matrices, and C = ABA^-1B^-1, and A commutes with C, and ||I-B||< 2 then B and A commute.

Lemma 2: If A and B are unitary matrices and C = ABA^-1B^-1 then |I-C|<=2|I-A||I-B|.

Then, using that, you can show that if one has a periodic group of unitary matrices G, with elements A and B, and A and B are near enough to the identity (say |I-A| and |I-B|<2) then A and B must in fact commute. To see consider the sequence B_i defined by B_0=B, and set B_i to be commutator of A and B_{i-1}. Note that by Theorem 2, A and B generate a finite group, but it is easy to see from Lemma 2 that |I-B_i| has to be eventually 0.

Now, we set our Jordan-Schur group to be the subgroup generated by things within 2 of the identity. A straightforward calculation now gets an explicit upper bound on the number of possible distinct coset representatives (consider the matrices as points in 2n^2 dimensions in Euclidean space, draw a small sphere around each of them and make a volume estimate).

This gives an explicit construction but a not great actual bound.

If one wants to improve this the main thing seems to be improving Lemma 2 (the best way seems to be by noting that one can actually assume that at least one of the eigenvalues of |I-A| must be fairly large and that allows one to improve the inequality). How much one can improve the inequality in 2 seems to be potentially of independent interest.

Note also that there's a pair of papers by Michael Collins that gives essentially the exact bound in for all n. His proof makes heavy use of the classification of finite simple groups and seems to be non-constructive.

5 October, 2011 at 5:05 pm

Joshua ZelinskySorry, I missed an important detail. After one has that B_i is eventually the identity for some i, one then uses Lemma 2 and inducts backwards to conclude that in fact A and B had to originally commute.

5 October, 2011 at 7:54 pm

RichardDear Joshua (and many others),

Wikipedia (particularly the mathematics and physics portions) is a wonder, and makes me (almost) proud to be a human being.

Your efforts are appreciated, never fear!

7 October, 2011 at 6:54 pm

Benjamin SteinbergTerry, I think you are slightly misstating the restricted Burnside problem. Free Burnside groups of exponent r with r large enough are infinite by Adian-Novikov. The restricted problem says that among all finite m-generated groups of exponent r there is a unique largest one.

[Corrected, thanks - T.]8 October, 2011 at 6:05 am

Benjamin SteinbergTerry, you might want to remove the not necessarily linear since a finite group is automatically linear. Alternatively, the restricted Burnside problem can be phrased as saying that an m-generated residually finite group of exponent r is finite with order bounded by a constant depending on r and m. Then you can keep the not necessarily linear.

8 October, 2011 at 6:14 am

Benjamin SteinbergBTW, it is interesting to note that it is also true that any periodic linear semigroup is locally finite. This was proved by McNaughton and Zalcstein in 1975. The case of an irreducible linear semigroup is handled exactly as for groups (just replace semigroup by group in your proof and note that the eigenvalues of a periodic matrix are roots of unity or zero). The key difficulty is reducing to the semisimple case because it is not enough to look at the inverse image of the identity when you project to the diagonal blocks. For this reason, there are several different proofs in the literature of the semigroup case. The semigroup result has applications to automata theory in computer science.

24 October, 2011 at 1:59 pm

FlorianNice post!

Here’s a different way to finish off Prop. 5 that occurred to me when I read the above. It uses the -adic numbers, but only in a very mild way, namely that is uncountable and that the cyclotomic extension has degree .

Any field that is finitely generated over embeds in for any prime number (using a transcendence basis). Fix one embedding for each prime .

As is finitely generated, the compositum is a finite extension of , and we can even choose the embeddings so that is bounded independent of . (E.g., by sending the elements above to .)

Now the fact above together with the first part of your argument shows that if has an element of order , then , and we can conclude as before.

Finally, the number-theoretic fact has an elementary proof: the minimal polynomial of over is irreducible by Eisenstein’s criterion applied with the prime . The same criterion applies in the PID .