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).
 A: 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 $\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$.
