I worry this question is too vague so I will try to be as thorough as I can.
Here is a proof by induction on $n$ that any subgroup $H$ of $\mathbb{Z}^n$ is finitely generated. The idea is to find a minimal $f$ such that $(f,a_2,\ldots ,a_n)\in H$, and note that for any $(k_1,k_2,\ldots ,k_n)\in H$ we have $$(k_1,k_2,\ldots ,k_n)=s(f,a_2,\ldots,a_n)+(0,k_2-sa_2,\ldots,k_n-sa_n)$$ where the set of all $(k_2-sa_2,\ldots ,k_n-sa_n)$ forms a subgroup $K$ of $\mathbb{Z}^{n-1}$ and is thus finitely generated by induction. The last step is to define an isomorphism $$\phi:H\rightarrow\mathbb{Z}\oplus K:(k_1,k_2,\ldots ,k_n)\mapsto ((s),(k_2-sa_2,\ldots,k_n-sa_n))$$
There is some tediosity involved in writing the full proof (showing that $K$ is a group and $\phi$ an isomorphism), and I was wondering if, by working with the homomorphism $\psi:H\to \mathbb{Z}:(k_1,k_2,\ldots ,k_n)\mapsto k_1$ and noticing that $\ker(\psi)=K$, the proof could be simplified as done here by Smurf. The problem I have is that in order to show $H\cong \mathbb{Z}\oplus \ker(\psi)$ I end up going again over all the tediosity of the first proof.
Smurf doesn't mention any further steps, so I'm wondering if there is something I'm missing that makes obvious the fact that $$H\cong \mathbb{Z}\oplus \ker(\psi)$$
and whether the tediosity can be skipped by working with $\psi$.