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Suppose we have an ordered, cancellative monoid and we wish to apply the Grothendieck group construction to it. Can the total order be extended to the larger group?

Example: consider the ordered monoid $(\Bbb R \cup \{-\infty\},\triangledown)$, where $a \mathbin{\triangledown} b = \log(e^a + e^b)$. Since this monoid is cancellative, we can construct its Grothendieck group. Does the order on $\Bbb R \cup \{-\infty\}$ extend in a canonical way to this new group?

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Yes.

If the commutative monoid $M$ is cancellative, then its Grothendieck group can be defined as $K:=M\times M/\sim$ where $(x_1,x_2)\sim (y_1,y_2)$ iff $x_1+y_2=x_2+y_1$.

Now define $(m,n)>(0,0)$ iff $m>n$. (It is not hard to see that this does not depend on the representative of $(m,n)$ in $M \times M$.) Thus, in $K$ we must have $$(m,n)>(x,y)\ \iff\ (m,n)-(x,y)>0\ \iff\ (m,n)+(y,x)>0\\ \iff (m+y,\,n+x)>0\ \iff\ m+y>n+x$$ where we used $-(x,y)=(y,x)$ in $K$.

Your particular example is isomorphic to the additive monoid of $\Bbb R^{\ge0}$, and its natural extension will be (isomorphic to) $\Bbb R$.

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    $\begingroup$ Note that proving transitivity, we seem to need that the order is total: thus $m+y>n+y$ implies $m>n$ (else we would be left with the case $m\le n$ when $m+y\le n+y$). $\endgroup$ – Berci Aug 31 '15 at 10:06

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