Let $X$ be a set and $\widehat{F}(X)$ the restricted free profinite group on $X$. To get $\widehat{F}(X)$ we define a profinite topology on $F$ (the free abstract group on $X$) and take $\widehat{F}(X)$ as the completion relative to this topology. If $X$ is a finite set, then $\widehat{F}(X)$ is the profinite completion of $F$. Also, we can write
$$\widehat{F}(X) = \varprojlim \widehat{F}(X\_i)$$
where $X\_i$ varies on the set of the finite subsets of $X$.



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> I would like to know if $\widehat{F}(X)$ is torsion-free. This problem can be reduced to finite $X\_i$.
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In principle, I thought it was a trivial thing, but it seems that the profinite completion (even in residually finite case) of a torsion-free group need not to be torsion-free.


I don't know if $\widehat{F}(X)$ has a non-trivial torsion element. Does anyone know if this is true?


Also, if it is false in the profinite case, what about the pro-$p$ case?



Is the free profinite group (or pro-$p$) torsion-free?
Both the free profinite and free pro-$p$ groups are torsion-free (also free pro-solvable groups or free pro-$\mathcal C$ groups for any collection $\mathcal C$ of groups closed under extensions, subgroups and homomorphic images). The reason is any closed subgroup of such a group is a projective profinite group; this can be proved using Shapiro's lemma and cohomology or can be deduced by reducing first to the case of an open subgroup and then using Nielsen-Schreier and Marshall Hall. You can find a proof in the book of Ribes and Zalesskii.


If a profinite group has nontrivial torsion, then it has a finite cyclic subgroup of prime order and this subgroup is closed. Since a finite cyclic group of prime order is not projective (because the quotient map $\mathbb Z/p^2\mathbb Z\to \mathbb Z/p\mathbb Z$ does not split), we deduce a free profinite or pro-$p$ group does not have torsion.


