Let $G$ be a finite group. Does there exist a finite group $\Gamma$ and a surjection $f\colon \Gamma \rightarrow G$ such that the center $Z(\Gamma)$ lies in the kernel of $f$?


Of course, this is only an interesting question if $Z(G) \neq 1$. It is also pretty trivial if we don't insist that $\Gamma$ is finite since we can then just take $\Gamma$ to be a free group.


I don't have a strong feeling as to whether or not such a $\Gamma$ exists.



Extending a finite group to avoid the center
Every finite group $G$ is quotient of a finite group with trivial center. Namely choose $p$ odd prime not dividing the order of $G$. Let $V\_p$ be the abelian group of functions $G\to\mathbf{Z}/p\mathbf{Z}$ with sum zero. The action of $G$ on $V\_p$ by translation $g\cdot f(h)=f(g^{-1}h)$ is faithful and fixes no nonzero element. So the semidirect product $G\ltimes V\_p$ has trivial center and admits $G$ as quotient.


