# Elements of infinite order in a profinite group

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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(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

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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