# Taking Grothendieck group of an already abelian group?

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?

## 1 Answer

You left out part of the definition of the Grothendieck group: the morphism $C\to A$ must factor through $K_0(C) \to A$ uniquely.

In your example, to get a unique morphism $G\oplus M \to A$ extending $G\to A$, we would need there to be a unique morphism $M\to A$. Since this must hold for any abelian group $A$, taking $A=M$ we can see that $M$ must be the trivial group.