If $C$ is a commutative semigroup, then $K_0(C)$ is an abelian group associated to $C$, called the Groethendieck group of $C$. This satisfies the universal property; for each abelian group $A$, and semigroup homomorphism (preserving the identity) - $f:C\to A$. $f$ factors through the Grothendieck group.
I was thinking about this, and it seems you would want for an abelian group $G$, $K_0(G)\cong G$, right?
But surely $K_0(G)=G\oplus M$ for any $\Bbb Z$-module $M$ works.
We have: $$G\to G\oplus M\\\quad\downarrow\quad\swarrow\quad\quad\quad\quad\\A\,\,\,\,\,\,\quad\quad\quad\,\,\,\,\,$$
Where $f:G\to A$ has unique morphism $(f,0)$ from $G\oplus A$. So $K_0(G)$ isn't unique?
See here. "By uniqueness, it follows that [here $G$] is its own Grothendieck group." This is kind of circular? By uniqueness, this is unique?