I am currently looking for a group $G$, which is not perfect but still isomorphic to its commutator group $[G,G]$.
Does anyone know an example (which despite the fact, such a thing should exist and I am surely not the first one looking for it, I couldn't find)? It would be nice to have it as an example of a group, whose derived series does not terminate.
It is clear, that $G$ must be infinite. My guess would be, that an infinite semi-product of e.g. $\mathbb Z$ should do the trick, but before I put way more time into constructing an probably already existing example (my first few attempts failed), I wanted to ask around first.
So the idea behind my constructions so far where taking finite series in $\mathbb Z$ with such a semi-direct product on it, that the commutator group are the finite series starting with 0 i.e.: $$G_0:=\mathbb Z,\quad G_{i+1}=G_i \rtimes \mathbb Z, \quad G=\cup G_i$$ with $[G,G] \subsetneq G$ but $[G,G] \cong G$ or more directly $[G,G]$ is the image of the right-shift of $G$.
The commutator subgroup of a free group of infinite rank is again a free group of infinite rank, hence, they are isomorphic. Non-trivial free groups not perfect, and a non-abelian free group has a derived series that does not terminate after finitely many steps. However, the intersection of the derived series is trivial.