Existence of Schur covering groups

The quotient $A\mathrm{/}M\cong R\mathrm{/}(R\cap [F,F])$ embeds in the finite-rank free abelian group $F_{\mathrm{a}\mathrm{b}}$, so it is free abelian. Thus $0\to M\to A\to A\mathrm{/}M\to 0$ splits. Let $S\mathrm{/}[F,R]$ be the image of a splitting, a complement to $M$ in $A$. Because $[F,R]\le S\le R$, conjugation by any $f\in F$ preserves $S$, so $S$ is normal in $F$.

The extension $R\mathrm{/}S\to F\mathrm{/}S\to G$ is central because $[F,R]\le S$. Projection $A\to A\mathrm{/}(S\mathrm{/}[F,R])$ restricts to an isomorphism from $M$, so $R\mathrm{/}S\cong M\cong M(G)$ by Hopf's formula. Since every element of $M$ comes from $R\cap [F,F]$, the kernel $R\mathrm{/}S$ lies in $[F\mathrm{/}S,F\mathrm{/}S]$. Thus it is a stem extension with multiplier kernel. The kernel is finite by Multiplier of a finite group is finite, hence $F\mathrm{/}S$ is finite because $G$ is finite. This is a Schur cover.