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).