Example of a derived series that never stabilizes.

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?

• 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. – Arturo Magidin May 26 at 18:11
• In fact $Z_2 * Z_2$ is isomorphic to the infinite dihedral group, in which $G''=1$, so this does not work. – Derek Holt May 26 at 19:27
• @MatheusManzatto The group that Arturo suggested is $\mathbb{Z}*\mathbb{Z}$ – verret May 26 at 19:32
• @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/… ) – Max May 26 at 20:51
• @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. – Arturo Magidin May 26 at 23:18

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 $$F(S)= *_{s\in S}\mathbb{Z}$$
• The surjective morphism $F(S) \to \mathbb{Z}^{(S)}$ argument was very clever. Thx. – Matheus Manzatto May 26 at 22:13
• @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. – Max May 27 at 6:15