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.
0+0=0
0+1=1
1+0=1
1+1=1
will do.

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 F_ab(M), the free abelian group generated by M. I'll write elements of F_ab(M) like this 0ⁿ1^m where n and m are integers. Example:
0⁵1⁶+0^-21³ =0³1⁹

In my text book the generator of F_ab(M) corresponding to an element x ∈ M is written [x]. So [0] = 0¹1⁰ and [1] = 0⁰1¹

Next I need B the subgroup of F_ab(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]-[0] = [0]-[0]-[0] = 0^-11⁰
[0+1]-[0]-[1] = [1]-[0]-[1] = 0^-11⁰
[1+0]-[1]-[0] = [1]-[1]-[0] = 0^-11⁰
[1+1]-[1]-[1] = [1]-[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 F_ab(M) so it must be the whole of F_ab(M).

The final step is to construct the Grothendieck group as the quotient of F_ab(M) by B. Since F_ab(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.