**Note**: this question was originally posted to Math.SE [here](https://math.stackexchange.com/q/3322407/274352).

----------

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.