I am looking at the following proof (taken from here):
Corollary 5 Let $A$ be a finitely generated abelian group. Then there is a finite subgroup $A_0$ of $A$ and an integer $m \geq 0$ such that $A \cong A_0 \times \mathbb{Z}^m$.
Proof. We know that $A \cong \mathbb{Z}^n/K$ for some integer $n \geq 0$ and some subgroup $K \subset \mathbb{Z}^n$, where $K$ is a free abelian group of rank $r \leq n$. If $r = n$ then we know from Lemma 23 that $A$ is finite, so we can take $A_0 = A$. Otherwise we know - also from Lemma 23 - that there exists a surjective homomorphism $f \colon A \to \mathbb{Z}$. Composing $f$ with the surjective quotient map $\mathbb{Z}^n \to A$, we also get a surjective homomorphism $\phi \colon \mathbb{Z}^n \to \mathbb{Z}$.
By the last Lemma, $A \cong \ker(f) \times \mathbb{Z}$, and similarly $\mathbb{Z}^n \cong \ker(\phi) \times \mathbb{Z}$. Since we also know that $\ker(\phi) \subset \mathbb{Z}^n$ is free abelian of rank $t \leq n$, it follows that $t = n - 1$. Also $\ker(f) \cong \ker(\phi)/K$. By induction on $n - r$, we may assume that $\ker(f) \cong A_0 \times \mathbb{Z}^{n-1-r}$ for some finite subgroup $A_0$, and hence $A \cong \ker(f) \times \mathbb{Z} \cong A_0 \times \mathbb{Z}^{n-r}$.
It says:
By induction on $n-r$, we may assume that $\ker(f) \cong A_0×\mathbb{Z}^{n−1−r}$ for some finite subgroup $A_0$ […].
I find this assumption troubling, because, then for any $n-r$ we can infer: $$ \ker(f) \cong \ker(\phi)/K \cong \mathbb{Z}^{n-1}/\mathbb{Z}^r \cong \mathbb{Z}^{n-1-r} $$ leaving $$ \ker(f) \cong A_0 \times \mathbb{Z}^{n−1−r} \cong \mathbb{Z}^{n-1-r}, $$ so $A_0=\{0\}$ all the time! I feel like this is not really a meaningful result so I must be overlooking something. Can you show me a case when $A_0$ is not the trivial set?
It is true that $K\cong \mathbb{Z}^r$ and $\ker(\phi)\cong\mathbb{Z}^{n-1}$, but you cannot conclude from this that $K/\ker(\phi)\cong\mathbb{Z}^{n-1-r}$. For instance, for $r=1$ and $n=2$ you might have $K=\mathbb{Z}$ and $\ker(\phi)=2\mathbb{Z}\cong\mathbb{Z}$, but the quotient $\mathbb{Z}/2\mathbb{Z}$ is not isomorphic to $\mathbb{Z}^{n-1-r}=0$. More generally, if $A\subseteq B$ and $C\subseteq D$ with $A\cong C$ and $B\cong D$, that does not imply $B/A\cong D/C$.
(Indeed, if your reasoning were correct, you could just conclude from the beginning that $A=\mathbb{Z}^n/K\cong\mathbb{Z}^n/\mathbb{Z}^r\cong\mathbb{Z}^{n-r}$.)