Note: this question was originally posted to Math.SE here.

Given any group $G$, one can consider its derived series

$$G = G^{(0)}\rhd G^{(1)}\rhd G^{(2)}\rhd\dots$$

where $G^{(k)}$ is the commutator subgroup of $G^{(k-1)}$. A group is perfect if $G=G^{(1)}$ and thus has constant derived series, and solvable if its derived series reaches the trivial group after finitely many steps.

Is it possible for a group’s derived series to be cyclical, i.e. that $G \cong G^{(n)}$ for some $n>1$ and $G\not\cong G^{(k)}$ for all positive $k<n$?

Note that such a group could not be finite, solvable, nor co-Hopfian.

Given any group $G$, one can consider its derived series

$$G = G^{(0)}\rhd G^{(1)}\rhd G^{(2)}\rhd\dots$$

where $G^{(k)}$ is the commutator subgroup of $G^{(k-1)}$. A group is perfect if $G=G^{(1)}$ and thus has constant derived series, and solvable if its derived series reaches the trivial group after finitely many steps.

Is it possible for a group’s derived series to be cyclical, i.e. that $G \cong G^{(n)}$ for some $n>1$ and $G\not\cong G^{(k)}$ for all positive $k<n$?

Note that such a group could not be finite, solvable, nor co-Hopfian.

Note: this question was originally posted to Math.SE here.

In the comments there, it was observed that an infinitely generated free group is an example for which the group is not perfect while isomorphic to its derived subgroup. Whence the assumption above that $G$ is not isomorphic to its derived subgroup.