Grothendieck Group of a commutative group I know that the Grothendieck group of a commutative monoid is $G(M) = M XM/R$
where $(x,y)R(x',y') $ iff$  $ there is a $z$ in $M$ such as $ x*y'*z=y*x'*z$ , $*$ is the law of the monoid.
My question is, what is the Grothendieck group of a commutative group? Is it the  group itself?

The Grothendieck group $\mathcal{G}(M)$ of a commutative monoid $M$ is the unique commutative group satisfying the following universal property: there is a monoid morphism $i\colon M \to \mathcal{G}(M)$ such that for every monoid morphism $f \colon M \to G$, where $G$ is a commutative group, there is a unique group morphism $\mathcal{G}(f) \colon \mathcal{G}(M) \to G$ such that the diagram
$$\require{AMScd}
\begin{CD}
  M @>i>> \mathcal{G}(M)\\
  @VfVV @VV \mathcal{G}(f) V\\
  G @= G
\end{CD}
$$
commutes, i.e. such that $f = \mathcal{G}(f) \circ i$.
Now, if $M$ is a commutative group, then you can easily check that the definition can be satisfied by $\mathcal{G}(M) = M$, $i = \text{id}_M$, and $\mathcal{G}(f) = f$ for every group morphism $f \colon M \to G$. By uniqueness, it follows that $M$ is its own Grothendieck group.
