Q:

Torsion in Cokernel is Annihilated by Integer

My question refers to an argument in a proof from T. tom Dieck's "Algebraic Topology and Transformation Groups" (Lemma 3; page 2). Here the excerpt and the red tagged step which I don't understand:

We have the inclusion map $\alpha: \mathbb{Z} \hookrightarrow M$ and  the multiplication by $n$ map $n:  \mathbb{Z} \to  \mathbb{Z}, z \mapsto n \cdot z$ with property that the torsion of the cokernel $M / im(\alpha)$ is annilated by $n$.
In the proof there is deduced that neglect the torsion of $M$ then $M = N \oplus <x>$. 
$<x>$ is a free component with $\alpha(1) =mx$, so $im(\alpha) \subset <x>$. First question: What is $N$? We assumed that torsion of $M$ trivial. Why $M$ can be splitted in direct sum $N \oplus <x>$? We nowhere assumed that $M$ is finitely generated so I don't know if we here just can apply the classification theorem for abelian groups.
Let continue: Since $im(\alpha)= <mx>$ we have $coker(\alpha):=M / im(\alpha)= M \oplus \mathbb{Z}/m$ with torsion $\mathbb{Z}/m$. 
By assumption the torsion is annilated by $n$, so $n = rm$ for appropriate $r \in \mathbb{Z}$.
By definition the pushout $K$ is the quotient of $N \oplus \mathbb{Z} \oplus \mathbb{Z}/ \sim$ with the relation $n(0,0,1) = m(0,1,0)$.
Take into account that here we identified $<x>$ with $<(0,1,0)>$.
Therefore the pushout is isomorphic to $K=N \oplus \mathbb{Z} \oplus \mathbb{Z}/<(0, m, n) >$. That's clear. 
What I don't understand is why holds $N \oplus \mathbb{Z} \oplus \mathbb{Z}/<(0, m, n) > \cong M \oplus \mathbb{Z}/m$.
Or in other words why $\mathbb{Z} \oplus \mathbb{Z}/<(m, n) > \cong  \mathbb{Z}/m$ holds under assumption that $m$ divides $n$?

A:

Your final 2 lines are incorrect. Because $M \cong N \oplus \Bbb Z$, what you are trying to prove is actually that $\Bbb Z^2/(m,n) = \Bbb Z \oplus (\Bbb Z/\text{gcd}(m,n))$. 
First it is convenient to reduce this to one coordinate. Write $d = \text{gcd}(m,n)$, so that $m = ad$ and $n = bd$ where $\text{gcd}(a,b) = 1$. Then choose $x$ and $y$ with $ax + by = 1$; the matrix $$\begin{pmatrix}a & -y\\b & x\end{pmatrix}$$ is invertible over the integers, and hence $\{v_1, v_2\} = \{(a,b), (x,y)\}$ forms a basis for $\Bbb Z^2$; in this basis your vector $(m,n)$ simply takes the form $dv_1$. 
Of course, $\Bbb Z^2/(d,0) = \Bbb Z/d \oplus \Bbb Z$. This is what you wanted.

