The idea is to start with a commutative monoid and turn it into a group. That is, you start with something that fails to be a group because it lacks inverses, and fix it, using the fact it is commutative.
So first I need a monoid, M.
The operation is logical-or so it is associative and has 0 as its unit. Thats the monoid taken care off. Obviously it is commutative.
Now for Fab(M), the free abelian group generated by M. I'll write elements of Fab(M) like this 0n1m where n and m are integers. Example:
In my text book the generator of Fab(M) corresponding to an element x ∈ M is written [x]. So  = 0110 and  = 0011
Next I need B the subgroup of Fab(M) generated by all elements of type [x+y]-[x]-[y]. Now you can see why I wanted to start small. I only have four generators to consider.
[0+0]-- = -- = 0-110
[0+1]-- = -- = 0-110
[1+0]-- = -- = 0-110
[1+1]-- = -- = 001-1
Since B contains 0-110 it must contain its inverse 0110. Similarly 001-1 ∈ B so we must have 0011 ∈ B
B has both of the generators of Fab(M) so it must be the whole of Fab(M).
The final step is to construct the Grothendieck group as the quotient of Fab(M) by B. Since Fab(M) = B this is easy, it is isomorphic to the group with one element. That is a little disappointing, my monoid shrank!
On the other hand it is quite illuminating. My text book goes on to talk about cancellation laws. My monoid doesn't let me cancel 1's. I was half expecting the Grothendieck construction to fix the fact that it isn't a group by adding inverses, but it does the obvious thing and gets rid of things that don't cancel.
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