Question about construction of The Grothendieck group. 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?
 A: 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 $M\times M$ with the following equivalence relation $\sim$ defined as follows: for $x,y,z,w\in M$, $$(x,y)\sim (z,w)\text{ if and only if there exists }m\in M\text{ such that }x+w+m=z+y+m\,.$$
Let $G:=(M\times M)/\sim$ be the set of equivalence classes of $M\times M$ with respect to $\sim$.  Note that $A$ is an abelian group with addition $+$ defined by
$$\big[(x,y)\big]_G+\big[(z,w)\big]_G:=\big[(x+z,y+w)\big]_G\text{ for all }x,y,z,w\in M\,,$$
where $\big[(x,y)\big]_G$ denote the equivalence class in $G$ that contain $(x,y)\in M\times M$.  The zero element of $G$ is just the class
$$0_G:=\big[(x,x)\big]_G\,(\text{ for any }x\in M)\,.$$
For each $x,y\in M$, 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 $\gamma:M\to K(M)$ is precisely the map sending $x\in M$ to
$$\gamma(x):=\big[(x+y,y)\big]_G\,(\text{ for any }y\in M)\,.$$
This map is injective if and only if $M$ has the cancellative property (i.e., for $x,y,z\in M$, $x+y=x+z$ 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 $f:M\to A$ be a semigroup homomorphism from $M$ to an abelian group $A$.  Define $f_*:G\to A$ via
$$\big[(x,y)\big]_G\mapsto f(x)-f(y)$$
for all $x,y\in M$.  Show that $f_*$ is a group homomorphism, and that $f_*\circ \gamma=f$.  If $g:G\to A$ is another group homomorphism such that $g\circ \gamma=f$, then
$$g\Big(\big[(x+y,y)\big]_G\Big)=(g\circ\gamma)(x)=f(x)\,,$$
for any $x,y\in M$.  That is, for all $x,y,z\in M$, we have
$$\begin{align}g\Big(\big[(x,y)\big]_G\Big)+f(y)&=g\Big(\big[(x,y)\big]_G\Big)+g\Big(\big[(y+z,z)\big]_G\Big)\\&=g\Big(\big[(x+y+z,y+z)\big]_G\big)=f(x)\,.\end{align}$$
This proves that
$$g\Big(\big[(x,y)\big]_G\Big)=f(x)-f(y)=f_*\Big(\big[(x,y)\big]_G\Big)$$
for $x,y\in M$, implying that $g=f_*$.
To make your visualization of $K(M)$ a bit stronger, here are examples. 


*

*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 $0$.

*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 $0$. 

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

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

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