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This is a result I haven't been able to find. An answer for finite groups or general groups would be great. A simple proof would be appreciated if possible.

Mark Rychnovsky
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    I googled «is every group a commutator group» and the second result was this page which claims that «a group whose first two abelianizations are cyclic, but whose second derived subgroup is not trivial, cannot arise as a derived subgroup». – Mariano Suárez-Álvarez Mar 17 '15 at 06:23
  • @MarianoSuárez-Alvarez Is there a familiar example of such a group? – manthanomen Mar 17 '15 at 06:51
  • $S_3$ is an example. It cannot be the derived subgroup of a group. – Nicky Hekster Mar 17 '15 at 07:09
  • @NickyHekster Why not? It doesn't fulfill, me thinks, the characterization Mariano mentions, as $;S_3''=(S_3')'=(A_3)'=1;$ – Timbuc Mar 17 '15 at 12:31
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    If a group $D$ is the commutator subgroup of some group $G$, the natural homomorphism $G \rightarrow Aut(D)$ induced by conjugation, maps the commutator subgroup $D$ of $G$ to the commutator subgroup $Aut(D)'$ of $Aut(D)$. Therefore we have $Inn(D) \subseteq Aut(D)'$. This necessary condition on $D$ easily implies that for $n \gt 2$ the dihedral groups $D_n$ and the symmetric groups $S_n$ cannot be commutator subgroups of any group. – Nicky Hekster Mar 17 '15 at 12:57
  • @manthanomen Yes, $S_4$. – Derek Holt Mar 17 '15 at 13:37

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