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.

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    $\begingroup$ I am not sure why you exclude the case $n=1$. Sure, perfect groups are one case, but we could have $G>G'$ but $G\simeq G'$. I think this occurs when $G$ is the free group on a countable set of generators. $\endgroup$ – ancientmathematician Aug 14 at 7:20
  • $\begingroup$ @ancientmathematician Well, excluding the $n=1$ case makes for a more interesting question. It's like the groups where $G\cong G^3$ but $G\not\cong G^2$. I think your correct about $G$ being free on countably many generators (certainly the derived subgroup of a non-abelian fg free group is countably generated). However, if you're not correct then this provides a counter-example to the stronger statement (as if $G'$ is not infinitely generated then it is finitely generated, non-trivial, and not infinite cyclic, and so $G''\cong G$). $\endgroup$ – user1729 Aug 14 at 9:33
  • $\begingroup$ @ancientmathematician There’s an endofunctor on the category of groups taking a group to its commutator subgroup. I’m interested in the dynamics of this functor. I want to know if there are groups that move in cycles — your example gives another kind of fixed point aside from perfect groups. I had to edit my question initially to account for this exact example. $\endgroup$ – Santana Afton Aug 14 at 13:10
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    $\begingroup$ @ancientmathematician well certainly you won't ever move in cycles in terms of actual equality except when the group is perfect. I understand the unease, but I think it's necessary for having an interesting question $\endgroup$ – Rylee Lyman Aug 15 at 0:13
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    $\begingroup$ I've taken your comment from the start of the post and made it a CW answer (meaning I won't get any reputation for votes), which will lift the question out of the "unanswered" queue. $\endgroup$ – user1729 Aug 30 at 13:36

This question was asked on Math Overflow, and was subsequently answered in the affirmative.

(This CW answer is just to lift the question out of the "unanswered" queue.)


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