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In the Algebra by Serge Lang, he constructed a Grothendieck group of commutative monoid $M$, namely $K(M)$:(page 39-40)

$M$ is a commutative monoid. Let $F_{ab} (M)$ be the free abelian group generated by $M$, and 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 $x, y in M$. Let $K(M) = F_{ab}(M)/B$, and $gamma : M rightarrow K(M)$, which obtain by composing the injection of $M$ into $F_{ab}(M)$ given by $x mapsto [x]$, and the canonical map. Then any monoid homomorphism $f : M rightarrow A $ of $M$ to an abelian group $A$, we have a unique group homomorphism $f_* : K(M) rightarrow A$ satisfies
$$f= f_*circ gamma,.$$ Then $K(M)$ is the Grothendieck group.

My question is why he constructed such a $B = langle[x+y]-[x]-[y]rangle$. And how does the $F_{ab}(M)/B$ look like?

Will this construction of $K(M)$ work better for you? I assume that $M$ is a commutative semigroup; that is, $M$ does not necessarily have the zero element. In a sense, $K(M)$ is a generalization of how we construct $mathbb{Z}$ from the natural numbers, or $mathbb{Q}_{neq 0}$ from the nonzero integers.

Equip $Mtimes M$ with the following equivalence relation $sim$ defined as follows: for $x,y,z,win M$, $$(x,y)sim (z,w)text{ if and only if there exists }min Mtext{ such that }x+w+m=z+y+m,.$$
Let $G:=(Mtimes M)/sim$ be the set of equivalence classes of $Mtimes M$ with respect to $sim$. Note that $A$ is an abelian group with addition $+$ defined by
$$big[(x,y)big]_G+big[(z,w)big]_G:=big[(x+z,y+w)big]_Gtext{ for all }x,y,z,win M,,$$
where $big[(x,y)big]_G$ denote the equivalence class in $G$ that contain $(x,y)in Mtimes M$. The zero element of $G$ is just the class
$$0_G:=big[(x,x)big]_G,(text{ for any }xin M),.$$
For each $x,yin M$, the inverse of $big[(x,y)big]_G$ in $G$ is simply $big[(y,x)big]_G$. Then, $K(M)$ is isomorphic to $G$, and the map $gamma:Mto K(M)$ is precisely the map sending $xin M$ to
$$gamma(x):=big[(x+y,y)big]_G,(text{ for any }yin M),.$$
This map is injective if and only if $M$ has the cancellative property (i.e., for $x,y,zin M$, $x+y=x+z$ implies $y=z$). The map is bijective if and only if $M$ is an abelian group.

To show that $G$ indeed satisfies the universal property, let $f:Mto A$ be a semigroup homomorphism from $M$ to an abelian group $A$. Define $f_*:Gto A$ via
$$big[(x,y)big]_Gmapsto f(x)-f(y)$$
for all $x,yin M$. Show that $f_*$ is a group homomorphism, and that $f_*circ gamma=f$. If $g:Gto A$ is another group homomorphism such that $gcirc gamma=f$, then
$$gBig(big[(x+y,y)big]_GBig)=(gcircgamma)(x)=f(x),,$$
for any $x,yin M$. That is, for all $x,y,zin M$, we have
$$begin{align}gBig(big[(x,y)big]_GBig)+f(y)&=gBig(big[(x,y)big]_GBig)+gBig(big[(y+z,z)big]_GBig)\&=gBig(big[(x+y+z,y+z)big]_Gbig)=f(x),.end{align}$$
This proves that
$$gBig(big[(x,y)big]_GBig)=f(x)-f(y)=f_*Big(big[(x,y)big]_GBig)$$
for $x,yin M$, implying that $g=f_*$.

To make your visualization of $K(M)$ a bit stronger, here are examples.

1. Consider $M:=(mathbb{Z},max)$ (that is $x+y$ is defined to be $max{x,y}$ here). Then, show that $K(M)$ is the trivial abelian group $0$.




2. Consider $M:=(mathbb{Z}_{>0},gcd)$ (that is, $x+y$ is defined to be $gcd(x,y)$ here). Then, show that $K(M)$ is the trivial abelian group $0$.




3. Consider $M:=(mathbb{Z}_{geq 0},+)$. Then, show that $K(M)cong(mathbb{Z},+)$.




4. Consider $M:=(mathbb{C}_{neq 0},*)$, where $x*y$ is defined to be $|xy|$. Prove that $K(M)cong (mathbb{R}_{>0},cdot)$.




5. Consider $M:=(mathbb{Z}_{>0},cdot)$. Show that $K(M)cong(mathbb{Q}_{>0},cdot)$.