Let $r(m)$ denote the residue class $r+m\mathbb{Z}$, where $0 \leq r < m$.
Given disjoint residue classes $r_1(m_1)$ and $r_2(m_2)$, let the *class transposition*
$\tau_{r_1(m_1),r_2(m_2)}$ be the permutation of $\mathbb{Z}$ which interchanges
$r_1+km_1$ and $r_2+km_2$ for every $k \in \mathbb{Z}$ and which fixes everything else.

Question:Does the group ${\rm CT}(\mathbb{Z})$ generated by the set of all class transpositions have finitely generated infinite periodic subgroups?

*Remark:* By means of computation one can find "almost-examples" of such groups.
For example let $G := \langle a, b, c, d, e \rangle$, where $a = \tau_{0(4),2(4)}$,
$b = \tau_{0(5),3(5)}$, $c = \tau_{1(3),5(6)}$, $d = \tau_{1(6),3(6)}$ and
$e = \tau_{2(6),4(6)}$. Then $G$ is infinite, but the subgroups generated by any $4$
of the generators are finite, and also all products of at most $6$ generators have
finite order. Though unfortunately there are products of $7$ generators which happen
to have infinite order (example: $(ab)^2ced$), such that $G$ is not periodic.

**Added on May 15, 2018:** This question has appeared as Problem 19.46 in:

Kourovka Notebook: *Unsolved Problems in Group Theory.*
Editors V. D. Mazurov, E. I. Khukhro. 19th Edition, Novosibirsk 2018.

Groups of dynamical origin), but you know it and actually know bare simplicity, if I remember correctly. Also Matte Bon constructed periodic (infinite f.g.) subgroups in suitable full topological groups, and this was generalized by Nekrashevych (Palindromic subshifts...), maybe this can help, or not, I don't know. $\endgroup$ – YCor May 16 '18 at 21:44