Let $A$ and $D$ be two non-trivial abelian groups and $B,C$ be two non-abelian groups. Also, let $C$ is a free group and acts on $A,B,D$.

Let $0\to A \xrightarrow{f}B\xrightarrow{g}C\to 0$ be a short exact sequence of groups in which the action of $C$ commutes with maps $f$ and $g$ ($C$ acts on itself trivially). Also, let there exist homomorphisms $h_1 :D\to B$ and $h_2 :B\to D$ (commuting with the action of $C$) so that $h_2 \circ h_1 =1_D$.

If $A$ is a free $\mathbb{Z}C$-module, then is $D$ a projective $\mathbb{Z}C$-module?

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    $\begingroup$ Try taking $A=0$. $\endgroup$ – Jeremy Rickard Jun 30 '18 at 13:39
  • $\begingroup$ @JeremyRickard Is $0$ a free $\mathbb{Z}C$-module? $\endgroup$ – MHenry Jun 30 '18 at 19:44
  • $\begingroup$ Yes, it's the free $\mathbb{Z}C$-module on zero generators. Or if that disturbs you, then start with an example where $A=0$ and then replace $A$ and $B$ with $A\times\mathbb{Z}C$ and $B\times\mathbb{Z}C$ respectively. $\endgroup$ – Jeremy Rickard Jul 1 '18 at 11:04
  • $\begingroup$ @JeremyRickard I think when $A=0$, we have $B\cong C$ from the short exact sequence. On the other hand, $D$ is a semidirct factor of $B$. So $D=0$ or $D\cong C$. Is it correct? $\endgroup$ – MHenry Jul 1 '18 at 17:01
  • $\begingroup$ @JeremyRickard Obviously when $B$ and $C$ are abelian, the claim holds. $\endgroup$ – MHenry Jul 1 '18 at 17:06

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