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.


4 Answers 4


(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?"]


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.

  • $\begingroup$ 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 :) $\endgroup$ Oct 14, 2009 at 1:07

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, Ignore: 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.

Edit: I was asked about this recently. So I really had to search my memory and the literature. This is what I have found: I think I read about the result in Shalev’s chapter in New Horizons in Pro-$p$ Groups (Theorem 2.1 and Corollary 2.2). However, the original reference is Zelmanov’s paper in Groups ’93 Galway St. Andrews Volume 2, LMS Lecture Note Series 212.

Here it is: Let $G$ be a group. Write $(x_1,x_2,\ldots,x_i)$ for the left normalized commutator of the elements $x_1,x_2,\ldots,x_i \in G$. Let $D_k$ be the subgroup of $G$ generated by $(x_1,x_2,\ldots,x_i)^{p^j}$, where $ip^j \geq k$ and we go over all $x_1,x_2\ldots,x_i \in G$. Let $L_p(G)$ be the Lie subalgebra generated by $D_1/D_2$ in the Lie algebra $\oplus_{i \geq 1} D_i/D_{i+1}$. We say that $G$ is Infinitesimally PI (IP) if $L_p(G)$ satisfies a polynomial identity (PI). Zelmanov proved the following theorem:

Theorem: If $G$ is a finitely generated, residually-$p$, IP, and periodic group, then $G$ is finite.

The proof is based on the following theorem for which he only sketched the proof (according to Shalev):

Theorem: Let $L$ be a Lie algebra generated by $a_1,a_2,\ldots,a_m$. Suppose that $L$ is PI and every commutator in $a_1,a_2,\ldots,a_m$ is ad-nilpotent. Then $L$ is nilpotent.

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.)

Edit: Zelmanov recently published his results in:

Zelmanov, Efim; Lie algebras and torsion groups with identity, J. Comb. Algebra 1 (2017), no. 3, 289–340.

Here is the abstract: "We prove that a finitely generated Lie algebra $L$ such that (i) every commutator in generators is ad-nilpotent, and (ii) $L$ satisfies a polynomial identity, is nilpotent. As a corollary we get that a finitely generated residually-$p$ torsion group whose pro-$p$ completion satisfies a pro-$p$ identity is finite."


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 $1$.

  • $\begingroup$ Neat argument. To clarify, I think in the second sentence of the second paragraph you're missing the word "quotient." $\endgroup$ Dec 3, 2009 at 16:12
  • $\begingroup$ I added quotient. I found this argument in Razon's paper but I'm sure it has been done before $\endgroup$ Dec 6, 2009 at 12:48

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