This is a continuation of my own question some time ago.
Suppose $M$ is a monoid and $G$ is the groupification of $M$. (I figure groupification of $M$ is a better term than Grothendieck group of $M$.) $G$ can be characterized by the universal property: For any monoid homomorphism $\varphi: M \to P$ with $P$ a group, $\varphi$ factors through the natural monoid homomorphism $\theta: M \to G$.
My question is:
If $M$ is cancellative (left and right), will $\theta$ be injective?
My initial thought was that this should be true, but after spending some time trying to prove it and failing, I'm starting to think it might not be true. I have not been able to find a counterexample though.
[Note: The converse is true and is quite easy to prove.]
No. There is a famous counter-example, due to Malcev. A necessary condition that $M$ embeds in a group is that, for any elements $a,b,c,d,x,y,u,v\in M$, from $$ax = by, cx = dy, au = bv,$$ it follows that $$cu = dv.$$ Malcev calls this property "Z". It is easy to see that any monoid that embeds in a group satisfies property Z.
Malcev's counter-example begins with the free monoid $X^{\ast}$ on the set $X = \{ a, b, c, d, x, y, u, v \}$. (This is just the set of words on $X$, with concatenation as the binary operation.) On $X^{\ast}$ define the equivalence relation $w_1\sim w_2$, for $w_1, w_2\in X^{\ast}$ if $w_2$ can be obtained from $w_1$ by a sequence of the following substitutions: $$ax\leftrightarrow by,\;\; cx\leftrightarrow dy, \;\; au\leftrightarrow bv.$$ Take the set $W = X^{\ast}/\sim$ of equivalence classes and define, for $[w_1], [w_2]\in W$, the product $[w_1][w_2] = [w_1w_2]$. Then a very tedious, but completely elementary, argument shows that $W$ is a cancellative monoid that does not satisfy property Z, and therefore, $W$ cannot be embedded in any group.
However, if you assume, in addition, that $M$ is commutative, then $M$ can be embedded in its group of fractions. (This is essentially the same as Steinitz' construction of the field of fractions of an integral domain; one simply does not bother with the addition.)
After a bit of Googling, I think the following is the reference for this. (I cannot access the paper to verify it, though. But my notes say the result is from 1937, so this seems likely to be the one.)
Reference: A.I. Mal’cev, On the immersion of an algebraic ring into a field, Math. Ann. 133 (1937), 686-691.