Group completion of a particular monoid

Let $S$ be the abelian monoid with elements $a_{n,m}$ where $n \in \mathbb{N}$ and $$\begin{cases} m=0 & \text{ if } n=0 \text{ or }1,\\ m \in \mathbb{Z} & \text{ if } n=2,\\ m \in \mathbb{Z}/2\mathbb{Z} & \text{ if } n \geq 3. \end{cases}$$

The semigroup operation is given by $$a_{n,m}+a_{n',m'} = a_{n+n',m+m'}$$ where $m+m'$ is to be computed in $\mathbb{Z}$ if $n+n' \le 2$ and in $\mathbb{Z}/2\mathbb{Z}$ if $n+n'\ge 3$ (if $n=2$ and $n' \ge 1$ for example, then $m$ is to be interpreted mod 2.)

I am trying to calculate the Grothendieck group of this monoid (this is Ex 1.1.7 in Rosenberg's book on Algebraic K-theory)

It seems the best approach to take is the following: The Grothendieck group $G(S)$ is the equivalence classes of of pairs $(x,y)$ with $x,y \in S$ where $(x,y) \sim (u,v)$ if and only if there is some $t \in S$ such that $$x+v+t = u+y+t \text{ in } S$$

Thus suppose we have two pairs $$(a_{m,n},a_{m',n'}) \sim (a_{x,y},a_{x',y'})$$ This means there is some $a_{\alpha,\beta}$ such that $$a_{m,n}+a_{x',y'}+a_{\alpha,\beta} = a_{m',n'}+a_{x,y}+a_{\alpha,\beta}$$

Now using the semigroup operation on the first subscript gives immediately that $m+x'=m'+x$. The second seems to be a bit tricker. If everything ends up less than $2$ or if everything ends up greater than 2 then it is ok, but it seems like a fair bit of casework to consider all the possibilities in between.

Am I missing something obvious here, or is it just a bit of grunt work, do the calculations type question? Is this the right way to approach the question (basically I was trying to determine all the isomoprhism classes).

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Is $\mathbb{Z}/2$ the integers modulo $2$ (in which case, $\mathbb{Z}/(2)$ or $\mathbb{Z}/2\mathbb{Z}$ is better), or the half-integers ($\frac{a}{2}$ with $a\in\mathbb{Z}$, in which case $\mathbb{Z}[\frac{1}{2}]$ is better)? – Arturo Magidin Jan 19 '12 at 17:58
Your addition is not well defined; you miscopied the definition, which has "if $n\geq 3$" in the third option. – Arturo Magidin Jan 19 '12 at 18:00
@Arturo - Thanks, I have edited the question – Juan S Jan 19 '12 at 21:15
@Jack - opps missed that - I had missed another $\ge 3$ as well. Thanks for the edit. – Juan S Jan 20 '12 at 0:14

There is a two-step process to compute the Grothendieck group that is usually a bit faster. Let $S$ be any abelian monoid, and define an equivalence relation $\approx$ on $S$ by $$x \approx y \quad\Leftrightarrow\quad (\exists z\in S)(x+z = y+z).$$ Then $\approx$ is a congruence relation, so the quotient $S/{\approx}$ is a monoid. It is not hard to prove the following theorem:
Theorem. The quotient $S/{\approx}$ is isomorphic to the image of $S$ in $G(S)$.
It follows from this theorem that $G(S)$ is isomorphic to $G(S/{\approx})$, but the latter is usually easier to compute. In particular, if you can find any embedding of $S/{\approx}$ into an abelian group, then $G(S/{\approx})$ is isomorphic to the subgroup generated by the elements of $S/{\approx}$.
In the example you gave, the equivalence relation $\approx$ is obviously defined by $$a_{n,m} \approx a_{n',m'} \quad\Leftrightarrow\quad n=n'\text{ and }m \equiv m'\:(\mathrm{mod}\;2)$$ Then the quotient consists of elements $A_{n,m}$, where $n\in\mathbb{N}$ and $$\begin{cases}m=0 & \text{if }n = 0\text{ or }1, \\ m\in\mathbb{Z}/2\mathbb{Z} & \text{if } n\geq 2.\end{cases}$$ with operation defined by $$A_{n,m} + A_{n',m'} = A_{n+n',m+m'}$$ where $m+m'$ is always computed in $\mathbb{Z}/2\mathbb{Z}$. This is obviously isomorphic to the submonoid of $\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ consisting of the elements $\{(0,0),(1,0),(2,0),(2,1),(3,0),(3,1),\ldots\}$. These elements generate $\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$, so $G(S)$ is isomorphic to $\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$.