Show that $H^{\prime} \cap A$ is a homomorphic image of $M(G)$

Question

Let $H$ be a group and $A$ be a central subgroup of $H$ of finite index. Let $G =H/A$. Show that $H^{\prime} \cap A$ is a homomorphic image of $M(G)$.

Here $H^{\prime}$ denotes the commutator subgroup of $H$ and $M(G)$ denotes the Schur multiplier of $G$.

Answer 1

Let $\phi:F \to H$ be an epimorphism of a free group $F$ onto $H$ and let $R = \phi^{-1}(A)$. So $F/R \cong G$. Then $A \le Z(H)$ implies $[F,R] \le \ker \phi$, and $\phi(F' \cap R) = H' \cap A$, so we get an induced epimorphism of $M(G) = (F' \cap R)/[F,R]$ onto $H' \cap A$.