Given a cyclic group $H$ i.e. $H$ is either $\mathbb{Z}_{m}$ or $\mathbb{Z}$. Does there always exist a non abelian group $G$ (need not be finite) such that $G^{ab} = G/[G G] =H$. If so how does one construct such a $G$?

Vaguely i think one may take $G$ to be a semidirect product of some cleverly chosen groups. But calculating commutator turns out to be rather messy.

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    $\begingroup$ Any cyclic group is abelian, so one always has the solution $G=H$. But presumably you want something that's not trivial. $\endgroup$ – Semiclassical Apr 14 '16 at 18:15
  • $\begingroup$ I have made the edit. I know commutator subgroup of an abelian group is trivial $\endgroup$ – iron feliks Apr 14 '16 at 18:35

Yes : take $G=H$, since it's already abelian $G^{ab} = G = H$.

If you want a non-abelian group take $G = H\times K$ where $K$ is a group such that $K^{ab} = 1$ (for instance $K$ a non-cyclic simple group).


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