In Lang, the example is stated as follows:
Let M be a commutative monoid, written additively. There exists a commutative group K(M) and a monoid homomorphism.
$$\gamma: M\rightarrow K(M) $$
having the following universal property if $f:M\rightarrow A$ is a homomorphism into an abelian group A, then there exists a unique homomorphism $f_*:K(M)\rightarrow A$ making the following diagram commutative, $f=f_*\circ \gamma$
Proof: Let $F_{ab}(M)$ be the free abelian gropu generated by M. We denote the generator of $F_{ab}(M)$ corresponding to an element $x\in M$ by $[x]$. Let B be the subgroup generated by all elements of type $$[x+y]-[x]-[y]$$
where $x,y\in M$. We let $K(M)=F_{ab}(M)/B$, and let $$\gamma:M\rightarrow K(M)$$
be the map obtained by composing the injection of $M$ into $F_{ab}(M)$ given $x\mapsto [x]$, and the canonical map $$F_{ab}(M)\rightarrow F_{ab}(M)/B$$
It is then clear that $\gamma$ is a homomorphism, and satisfies the desired universal property.
Question: What I don't understand is why we factor out the group generated by all elements of the type $[x+y]-[x]-[y]$. Is it similar to how the kernel of all homorphisms from a group into an abelian group factors through the commutator of the first group?