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Say G is a profinite group with elements of arbitrarily large order. Do elements of infinite order exist (A) if we assume G is abelian? (B) in general?

A start for (A): we can ask the same question for the closure of the torsion subgroup of G (a subgroup since G is abelian), so WLOG we can assume the torsion subgroup is dense in G.

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4 Answers

up vote 10 down vote accepted

(B) is probably difficult since it is listed as an open problem in the book Profinite Groups by Ribes and Zalesski (2009). [Question 4.8.5b (p. 401): "Is a torsion profinite group necessarily of finite exponent?"]

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In the abelian case, elements of infinite order have to exist, for any compact group. Let G(n) be the (closed) subgroup of n-torsion elements. If G were the union of the G(n), then by the Baire category theorem, some G(n) would have to have nonempty interior. This would imply G(n) is open so G/G(n) is finite by compactness, which would imply the order of elements of G is bounded.

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Nice, assuming by compact you mean compact-Hausdorff (to apply BCT). Profinite groups are Hausdorff, so it does answer part (A) of my question, thank-you! Conclusion: An abelian compact-Hausdorff group of infinite exponent has elements of infinite order. If anyone with edit privileges could fix that in the answer, this comment could be removed :) –  Andrew Critch Oct 14 '09 at 1:07
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There are also many infinite order elements in the Haar measure sense:

Recall that a profinite group is compact, hence it has a probability Haar measure. If G is abelian and it has an element of order > n, then G has either ℤ/pℤ for p > n+1 or ℤ/pkℤ for small p and suitably large k as a quotient. Thus under the assumption that G has elements of unbounded order we get that either the former is a quotient for infinitely many primes, or the latter is a quotient for a fixed prime and infinitely many powers.

Now the ratio of elements in each of the finite quotients of order bigger than $n$ tends to $1$. By standard arguments of measure theory, this implies that the probability measure of elements in $G$ of order bigger than $n$ is $1$ (when p tends to infinity or p is fixed and k tends to infinity). Taking intersection over all $n$'s give that the probability that an element will have an order bigger than any positive integer, i.e., infinite order, is $1$.

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Neat argument. To clarify, I think in the second sentence of the second paragraph you're missing the word "quotient." –  John Goodrick Dec 3 '09 at 16:12
I added quotient. I found this argument in Razon's paper but I'm sure it has been done before –  Lior Bary-Soroker Dec 6 '09 at 12:48
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I would like to add that the restricted Burnside problem is equivalent to the fact that a finitely generated profinite group of finite exponent is finite. Now, this was of course proved by Zelmanov. But he also proved a stronger result: every finitely generated compact Hausorff torsion group is finite, see Zelʹmanov, E. I., On periodic compact groups, Israel J. Math. 77 (1992), no. 1-2, 83--95. In particular, every finitely generated torsion profinite group is finite, i.e. the Burnside problem is true for profinite groups. BTW, Zelmanov has a more general result regarding when a pro-p group is finite (I don't remember now the exact formulation, it should be something like all generators have finite order and the associated graded Lie algbera satisfies an identity), however, he only published a sketch of the proof which I think is in E. Zelmanov, Nil Rings and Periodic Groups, The Korean Mathematical Society Lecture Notes in Mathematics, Korean Mathematical Society, Seoul, 1992.

The question whether a torsion profinite group is of finite exponent is still open as far as I know and is considerd very difficult. (Burnside type problems seem to be very difficult.)

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