13

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.

user1729
  • 32,369
Santana Afton
  • 6,978
  • 2
    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. – ancient mathematician Aug 14 '19 at 07:20
  • @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 $G''\cong G$). – user1729 Aug 14 '19 at 09:33
  • @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. – Santana Afton Aug 14 '19 at 13:10
  • I am uneasy about the "move in cycles" since it is only up to isomorphism. But there seem to be several interesting questions here much beyond my paygrade. – ancient mathematician Aug 14 '19 at 13:59
  • 1
    @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 – Robbie Lyman Aug 15 '19 at 00:13
  • Yes, though, you are right, the derived subgroup of a free group on a countable set is both a proper subgroup but countably-generated free. – Robbie Lyman Aug 15 '19 at 00:14
  • Maybe try mathoverflow. Part of me thinks that finitely generated examples shouldn't exist but I don't really have an argument outside of being a commutator of some group is sort of special and having finitely generated commutator subgroup is also pretty special –  Aug 18 '19 at 00:47
  • Question was asked and answered in math.overflow – Arturo Magidin Aug 21 '19 at 05:43
  • I see you indicated the cross-post in Math.overflow, but forgot to do it here. Perhaps you should add a link here to the question and answer there, so that it’s not just buried in my comment. – Arturo Magidin Aug 21 '19 at 05:44
  • @ArturoMagidin Thanks for the reminder! I edited the question to include the cross-post. – Santana Afton Aug 21 '19 at 06:01
  • 1
    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. – user1729 Aug 30 '19 at 13:36

1 Answers1

3

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.)

user1729
  • 32,369