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I know there is this post: Question about construction of The Grothendieck group. but it does not answer my question.

So we have a commutative monoid M and we then look at the free abelian group $F_{ab}(M)$ generated by M. The generator of the free abelian group corresponding to $x \in M$ is denoted $[x]$. Lang then defines a subgroup B as all the elements of the type $[x+y] -[x] - [y]$ for $x,y \in M$ and then takes $F_{ab}(M)/B$. My question is, how is B not trivial? Since $F_{ab}(M)$ is abelian, could we not do $[x+y] - [x] - [y] = [x] + [y] - [x] - [y] = e$ due to commutativity? Then wouldnt the quotient group just be $F_{ab}(M)$? Or does $[x]$ not denote an equivalence class (with the relation that all elements of $[x]$ are generated by x)? Any clarification would be greatly appreciated

Kan't
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froitmi
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    The fact that $F_{ab}(M)$ is abelian does not mean that $[x+y]=[x]+[y]$. Each element of $M$ gives rise to an independent generator of the free abelian group on $M$. – lulu Jun 29 '24 at 18:09
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    maybe the notation is misleading. For each $m\in M$ define $g_m$ to be the corresponding generator of the free abelian group. it's free abelian, so we have no relations between the $g_m$ other than $g_m+g_n=g_n+g_m$. Now we define new relations which imply that $g_{m+n}=g_m+g_n$. – lulu Jun 29 '24 at 18:23
  • Ah yes, this makes sense. Thank you very much – froitmi Jun 29 '24 at 19:59
  • $F_{\rm ab}(M)$ is the free abelian group on the underlying set of $M$. For everybelement of $M$, you get a generator of this abelian group, For instance, the identity element of $M$ is a nontrivial free generator in this group. – Arturo Magidin Jun 29 '24 at 23:23

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