# 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.

(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.

• 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 :) 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 $1$.

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