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.

  • 7
    $\begingroup$ Let $(G_n)$ be a sequence of groups with $G_0=1$ and $[G_n,G_n]$ isomorphic to $G_{n-1}$. Consider $G=\bigoplus_{n\ge 1}G_{2n}$. Then its derived subgroup $G'$ is isomorphic to $\bigoplus_{n\ge 1}G_{2n-1}$, and its second derived subgroup is isomorphic to $G$. Next we have (a) to exhibit such a sequence $(G_n)$, and ensure (b) that $G$ is not isomorphic to $G'$. Quite surprisingly (a) is not obvious to me (I don't expect (b) to be an issue but it makes no sense before having an example). $\endgroup$ – YCor Aug 20 '19 at 14:42
  • 3
    $\begingroup$ @YCor The free group on countably many generators should be one such example. $\endgroup$ – Santana Afton Aug 20 '19 at 15:02
  • 3
    $\begingroup$ YCor's Question (a) is considered in this paper by B.H. Neumann, and some examples are constructed in Section 9 and 10. $\endgroup$ – Derek Holt Aug 20 '19 at 15:15
  • 6
    $\begingroup$ Furthermore, in the example in Section 9 of that paper, $G_1$ has order $2$ and, for $i>1$, $Z(G_i) < G_i'$, so it is easy to see that $G$ and $G'$ are not isomorphic. So we do indeed have an example of a group with cyclical derived seriesof period $2$. $\endgroup$ – Derek Holt Aug 20 '19 at 15:40
  • 3
    $\begingroup$ @DerekHolt great. I'd be happy if you post an answer then as I haven't gone much into that paper. $\endgroup$ – YCor Aug 20 '19 at 16:25

So, let's turn the comments into answer. In this paper by B.H. Neumann, the author studies ascending series of groups $1=G_0 < G_1 < G_2 < \cdots$ in which $G_i' = G_{i-1}$ for all $i \ge 1$. Most of the paper is concerned with proving that such a series must terminate under certain hypotheses, but in Sections 9 and 10 he describes examples of infinite series of this form.

In the example in Section 9, we have $|G_1|=2$, and all other $G_i$ are infinite with centre $G_1$ of order $2$.

Now, following YCor's suggestion, let $G = \oplus_{n \ge 1} G_{2n}$ and $H = \oplus_{n \ge 1} G_{2n-1}$. Then, $G' \cong H$, $H' \cong G$ and, since $H$ has a direct factor of order $2$ but (by the above remarks) $G$ does not, we have $G \not\cong H$.

So the group $G$ has cyclical derived series with period $2$.

In fact, for any $p \ge 1$, we can split the direct sum of the $G_i$ into $p$ mutually disjoint direct factors of this form and, since only one of these $p$ factors has a direct summand of order $2$, this gives an example of a group with cyclical derived series of period $p$.

| cite | improve this answer | |
  • $\begingroup$ This is perfect, thank you! $\endgroup$ – Santana Afton Aug 20 '19 at 19:25
  • 7
    $\begingroup$ @SantanaAfton But you asked for not perfect?! $\endgroup$ – Francois Ziegler Aug 20 '19 at 19:27
  • 1
    $\begingroup$ Considering $G=\bigoplus_{n\ge 1}G_{kn}$, one gets $G$ isomorphic to $G^{(k)}$; can one ensure periodicity exactly $k$? $\endgroup$ – YCor Aug 20 '19 at 20:00
  • 1
    $\begingroup$ @YCor Among the groups $G,G^{(2)},\ldots,G^{(k-1)}$, only $G^{(k-1)}$ has a direct summand of order $2$. $\endgroup$ – Derek Holt Aug 20 '19 at 21:15
  • $\begingroup$ @DerekHolt OK great. PS François's comment is a joke. $\endgroup$ – YCor Aug 20 '19 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.