Multiplier of an abelian group

In $F\mathrm{/}[F,R]$, the subgroup $R\mathrm{/}[F,R]$ is central. Since $[F,F]\le R$, this quotient has nilpotency class at most two. Thus its commutator is alternating and bilinear. Changing either lift by an element of $R$ does not change its commutator, so the rule $\bar{x}\wedge \bar{y}\mapsto [x,y][F,R]$ defines a surjective homomorphism $\mathrm{\Phi }:{\bigwedge }^{2}A\to [F,F]\mathrm{/}[F,R]$.

To check that there are no further relations, write $X=F_{\mathrm{a}\mathrm{b}}$. The class-two quotient $F\mathrm{/}[F,[F,F]]$ is the free class-two group on the chosen free basis. Its central commutator subgroup is ${\bigwedge }^{2}X$: ordering the basis, every commutator word reduces using $[xy,z]=[x,z][y,z]$ and $[x,yz]=[x,y][x,z]$ to a product of basic commutators $[x_i,x_j]$ with $i<j$. Their independence follows by mapping the basis into the explicit class-two group $(X\oplus {\bigwedge }^{2}X,\cdot )$ with $(u,a)\cdot (v,b)=(u+v,a+b+{\sum }_{i>j}{u}_i{v}_j(e_i\wedge e_j))$; each word has finite support, so the displayed sum is finite.

The image of $[F,R]$ in ${\bigwedge }^{2}X$ is generated by $x\wedge r$ with $x\in X$ and $r$ in the image $K$ of $R$ in $X$. The alternating universal property identifies $({\bigwedge }^{2}X)\mathrm{/}(X\wedge K)$ with ${\bigwedge }^2(X\mathrm{/}K)={\bigwedge }^{2}A$. Hence $\mathrm{\Phi }$ is injective as well as surjective. Since $[F,F]\le R$, Hopf formula for the Schur multiplier identifies $M(A)=(R\cap [F,F])\mathrm{/}[F,R]=[F,F]\mathrm{/}[F,R]$ under the stated hypotheses. Presentation maps preserve commutators and exterior products, and the canonical Hopf identification is natural, so the resulting isomorphism is natural in $A$.