Let $I(\mathcal C)$ be the isomorphisms in an exact category $\mathcal C$ and let $R(\mathcal C)$ be the subgroup of $\mathbb{Z}[I(\mathcal C)]$ generated by all elements of the form $[a]-[b]+[c]$ where $a\hookrightarrow b\twoheadrightarrow c$ is an exact sequence in $\mathcal C$. The Grothendieck group of $\mathcal C$ is then by definition $K_o(\mathcal C)=\mathbb{Z}[I(\mathcal C)]/R(\mathcal C)$.
The problem: I have proven that the category $F(\mathbb{Z})$ of free abelian groups has $K_0(F(\mathbb Z)\cong \mathbb Z$. I want to show that the category $G(\mathbb Z)$ of finitely generated abelian groups also has $K_0(G(\mathbb Z)\cong \mathbb Z$.
My attempt: The map $$I(G(\mathbb Z)) \longrightarrow K_0(F(\mathbb Z))\\\mathbb{Z}^n\oplus\mathbb{Z}_{k_1}\oplus\cdots \oplus \mathbb{Z}_{k_i}\mapsto \mathbb Z^n$$ is obviously surjective and descends to a map on the quotient $K_0(G(\mathbb Z))$, but why is it injective? That is, why does modding out $R(G(\mathbb Z))$ make two abelian groups of the same rank equal?
Let $T$ be pure torsion. For simplicity, I will take $T=\mathbb{Z}_k$, but the argument generalizes easily. We have the short exact sequence $$0\longrightarrow\mathbb{Z}\stackrel{k}{\longrightarrow}\mathbb{Z}\longrightarrow\mathbb{Z}_k\longrightarrow 0$$ giving $[\mathbb{Z}]-[\mathbb{Z}]+[\mathbb{Z}_k] = [\mathbb{Z}_k]\in\mathcal{R}(F(\mathbb{Z}))$. Therefore, the only part of a finitely generated abelian group contributing to the Grothendieck group is the free part.