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Definition 1. Let $G$ be a group, then $[x,y]:= xyx^ {-1}y^ {-1}$, for all $x,y \in G.$

Definition 2. We define $G^{(i)}$ in the following way, $$G' = G^{(1)} = [G,G] = \{\mbox{group generated by all elements in the form }[x,y];\ x,y\in G\},$$ $$G^ {(i+1)} = \left[G^{(i)},G^{(i)}\right], \ i>1.$$

I am trying to find an example of a group $G$ such that the series

$$G \supset G' \supset G^{''} \supset \ldots \supset G^{(i)}\supset \dots $$

satisfies $G^{(i)}\neq G^{(i+1)}$ for all $i\in \mathbb{N}$. I think that $G = \mathbb{Z}_2 *\mathbb{Z}_2$ (free product) might work, but I was not able to argue that the required property holds.

Can anyone help me?

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    $\begingroup$ The free group on two elements the property. Any group that contains a free group of rank $2$ or more will also have it, then. $\endgroup$ – Arturo Magidin May 26 at 18:11
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    $\begingroup$ In fact $Z_2 * Z_2$ is isomorphic to the infinite dihedral group, in which $G''=1$, so this does not work. $\endgroup$ – Derek Holt May 26 at 19:27
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    $\begingroup$ @MatheusManzatto The group that Arturo suggested is $\mathbb{Z}*\mathbb{Z}$ $\endgroup$ – verret May 26 at 19:32
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    $\begingroup$ @ArturoMagidin it does not follow that every group containing $F_2$ has the property. Indeed for $SL_2(\mathbb{Q})$ the sequence stabilizes (indeed it is a perfect group), but it contains $SL_2(\mathbb{Z})$ which famously contains a free group of rank $2$ (see e.g. mathoverflow.net/questions/43726/… ) $\endgroup$ – Max May 26 at 20:51
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    $\begingroup$ @Max: Thanks for the correction. Have a nonabelian free group as a subgroup only guarantees the derived series does not terminate in the identity, not that it does not stabilize. $\endgroup$ – Arturo Magidin May 26 at 23:18
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Arturo Magidin points out in the comments that $F_2=\mathbb{Z*Z}$ works. This is true and can for instance be proved as follows : first of all, note that for any free group $F(S)$ on generators $S\neq \emptyset$, $[F(S),F(S)]$ is a strict subgroup of $F(S)$. This follows from the fact that there is a surjective morphism $F(S)\to \mathbb{Z}^{(S)}$, for instance (you can also see it by looking at reduced words).

Then note that $[F_2,F_2]$ is a subgroup of the free group $F_2$, and it is nontrivial (because $F_2$ is nonabelian), therefore by the Nielsen-Schreier theorem, it is itself free. One moreover easily checks that it is nonabelian. Therefore its commutator will be a strict subgroup, and it will also be free, and nonabelian, and so on by induction. You get by induction that $G^{(i)}$ is free, nonabelian and $G^{(i+1)} \subset G^{(i)}$.

As you can see, this proof works for any free group $F(S)= *_{s\in S}\mathbb{Z}$

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  • $\begingroup$ The surjective morphism $F(S) \to \mathbb{Z}^{(S)}$ argument was very clever. Thx. $\endgroup$ – Matheus Manzatto May 26 at 22:13
  • $\begingroup$ Are there examples of non-free groups with this property? A number of the groups I usually work this fail this criterion easily — I don’t have a good intuition as to how one might find such groups outside the free group arena. $\endgroup$ – Santana Afton May 27 at 2:00
  • $\begingroup$ @SantanaAfton : yes there are, but you're going to have to.clarify how much "non free" you want things to be. Indeed if $G$ is abelian, then clearly $F_2\times G$ works. Now less stupidly any group $G$ that has an $i$ with $G^{(i)}$ free non abelian works as well, and you could artificially construct some of those by hand, starting again from a free group. On the top of my head I don't know any "natural" non free group with the property. $\endgroup$ – Max May 27 at 6:15

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