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?

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    $\begingroup$ Yes, consider the group $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}$ with $A_0=\mathbb{Z}/2\mathbb{Z}\neq 0$. It is certainly a finitely generated abelian group. Do you see your error? For "some" finite subgroup $A_0$. $\endgroup$ – Dietrich Burde Dec 30 '16 at 21:51
  • $\begingroup$ A cleaner proof follows from two facts: a finitely generated torsion-free abelian group is free, and if $A$ is abelian f.g. , then it projects to a finitely generated torsion free abelian group, $A/\tau(A)$. $\endgroup$ – Pedro Tamaroff Dec 30 '16 at 23:52

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 $A=\mathbb{Z}^n/K\cong\mathbb{Z}^n/\mathbb{Z}^r\cong\mathbb{Z}^{n-r}$.)

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  • $\begingroup$ Thanks, this makes sense. With this cleared up however, I'm uncertain about another part of the proof; namely how does $ker(f)≅ker(ϕ)/K$ come about? Previously I thought that: $ker(f)×ℤ≅A≅ℤ^n/K≅(ker(ϕ)×ℤ)/K$ Is this correct? If yes, why are we allowed to get rid of $ℤ$ on both sides and obtain $ker(f)≅ker(ϕ)/K$? $\endgroup$ – Attila Kun Dec 31 '16 at 1:06
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    $\begingroup$ That just comes from the definition of $f$ and $\phi$. By definition, $f:A\to\mathbb{Z}$ and $\phi$ is the composition of $f$ with the surjection $g:\mathbb{Z}^n\to A$ with kernel $K$. The map $g$ restricts to a surjection $h:\ker(\phi)\to \ker(f)$, since $\phi(x)=f(g(x))$ so $x\in\ker(\phi)$ iff $g(x)\in\ker(f)$. The kernel of $h$ is just $K\cap\ker(\phi)$, so $\ker(f)\cong \ker(\phi)/(K\cap\ker(\phi))$. But $K\subseteq\ker(\phi)$ since $\phi(x)=f(g(x))=f(0)=0$ for any $x\in K$, so $K\cap \ker(\phi)$ is just $K$. $\endgroup$ – Eric Wofsey Dec 31 '16 at 1:16
  • $\begingroup$ That was very useful too, I think I got it. I still don't understand the inductive step by the end of the original proof though: $\ker(f) \cong A_0 \times \mathbb{Z}^{n-1-r}$ so $A \cong \ker(f) \times \mathbb{Z} \cong A_0 \times \mathbb{Z}^{n-r}$. If we are using induction here, then this should work for the base case which I think is $r=n-1$ so: $\ker(f) \cong A_0 \times \mathbb{Z}^{0}$ => $A \cong \ker(f) \times \mathbb{Z} \cong A_0 \times \mathbb{Z}^{1} \cong \mathbb{Z}^{n}/K$ Why is this true? $\endgroup$ – Attila Kun Dec 31 '16 at 3:00
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    $\begingroup$ The base case is $r=n$, in which case you know $A$ is finite as noted at the beginning of the proof and then you don't construct $f$ at all. So in the case $r=n-1$, you already know by induction that $\ker(f)$ is finite. $\endgroup$ – Eric Wofsey Dec 31 '16 at 3:05

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