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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 $G$.

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  • $\begingroup$ @Dietrich Burde In case of finite it is not possible But what about infinite groups? $\endgroup$ – Riju Jun 27 '17 at 19:30
  • $\begingroup$ No, its fine but I am saying that If $G$ is a infinite group with G, not perfect then $[G,G]$ properly contained in $G$, can $[G,G]$ be isomorphic to $G$? That is what the question asks. I just said that Even if it happens you have to consider infinte group. $\endgroup$ – Riju Jun 27 '17 at 19:35
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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.

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  • $\begingroup$ Ah, yes of course! Thank you very much. $\endgroup$ – ctst Jun 27 '17 at 22:01

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