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According to wiki, given $M$ a commutative monoid, the Grothendieck group $K$ is an abelian group satisfying a homomorphism, $i:M\rightarrow K$ s.t. for any homomorphism $f:M \rightarrow A$ into an abelina group $A$, there exists a unique homomorphism $g:K \rightarrow A$ such that $g \circ i = f $.


Sorry, maybe this is obvious, but why couldn't we take $K = Z(M)$, the free abelian group generated by $M$, instead of quotienting out a subgroup? Does not the universal property of $K$ follow from the universal property of $Z(M)$?

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The canonical inclusion $i:M\to Z(M)$ is not a homomorphism. Given $a,b\in M$, $i(a)+i(b)$ will be the formal sum of $a$ and $b$ as elements of $Z(M)$, which is different from $i(a+b)$. Basically, $Z(M)$ completely forgets about the monoid structure of $M$ and treats $M$ as just a set, and so $i$ will not preserve that structure.

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    $\begingroup$ We quotient by the smallest subgroup of Z(M) which will fix this, in fact. $\endgroup$ – Mariano Suárez-Álvarez Mar 28 '17 at 3:31
  • $\begingroup$ Thanks a lot, so does this imply: If $i:M \rightarrow A$, $M$ a monoid, $A$ an abelian group, is a mapping that maps identity to identity, then by taking quotient of subgroup generated by $i(m+n)-i(m)-i(n)$ for all $m,n \in M$, $i$ becomes a homomorphism? Then why we need $M$ to be commutative? $\endgroup$ – Cy L Shih Mar 28 '17 at 4:50
  • $\begingroup$ That correct. Commutativity of $M$ is irrelevant. $\endgroup$ – Eric Wofsey Mar 28 '17 at 4:53

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