I've across this post on how Serge Lang develops Grothendieck group by the free abelian group generated by monoid $M$. Problem statement:
Let $M$ be commutative monoid. We can construct a commutative group $K(M)$ and a homomorphism $\gamma: M \rightarrow K(M)$ such that it factors any homomorphism $f: M \rightarrow A$ where $A$ abelian group into $f = f^{*}\gamma$ such that $f^{*}: K(M) \rightarrow A$ is an unique homomorphism.
Serge Lang starts with by using notation $[x]$ as being the generator of $Z(M)$ corresponding to $x \in M$. Here, $Z(M)$ is free abelian group generated by $M$. Then, he goes on to generate sub-group $B$(of $Z(M)$) by using the set $$\{[x + y] - [x] - [y] : x, y \in M\}$$ Finally, define the confusing factor group $K(M) := Z(M) / B$. There's a simpler approach in simplied version of $Z(M) / B$. I perfectly understand the approach. But the poster claims $$G \cong Z(M) / B$$ , which is confusing.
What I know: $[x]$ simply comes from the map $$M \rightarrow Z(M);\quad x \rightarrow 1 \cdot x$$ where $1 \cdot x$ is the map from $M \rightarrow Z$ where $(1\cdot x)(x) = 1$ and zero elsewhere.
$B$ is the (smallest) sub-group generated by the given set and thus contains products of integral powers of elements having the form $[x + y] - [x] - [y]$. Combining these things seem really confusing.
How is $G \cong Z(M) / B$ here? What is the intuition?
