Is there a relatively well-known group $G$, such that the commutator subgroup, $G'$, of $G$ is a non-abelian free group, $G$ is not a non-trivial free product and $Z(G)=\{1\}$?
I know that trere are examples like $\left \langle a \right \rangle_2\ltimes F(b,c)$, $b^a=b^{-1}$, $c^a=c$, but I am interested in less obvious examples.
Just to record an answer, the hyperbolic $2$-dimensional surface groups, such as the torus group $\langle a,b,c,d \mid [a,b][c,d]=1 \rangle$ do not decompose as free products. It is well-known that all of their subgroups of infinite index, including their commutator subgroups, are free,