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 .

]]>Terry, 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.

]]>*[Corrected, thanks – T.]*

Dear 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!

]]>Sorry, 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.

]]>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.

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