Subgroups of $Z^n$ are finitely generated

Question

I have read a couple of proofs already, but all of them try to go further and start talking about modules. Is there any more direct proof of this fact without using modules?

Answer 1

Let's proceed by induction over $n$, if $n=1$ we have that $H=k\mathbb{Z}$ so it is obviously free.

Let assume the result holds for every $0<i<n+1$, and let $H<\mathbb{Z}^{n+1}=\langle e_1,\cdots,e_{n+1}\rangle$, consider the projection $$\begin{matrix}f:\mathbb{Z}^{n+1}&\longrightarrow&\mathbb{Z}\\k=\sum_{i=1}^{n+1}m_ie_i&\longmapsto&m_1e_1\end{matrix}$$ then we have the following short exact secuence $$\ker f|_H\longrightarrow H\longrightarrow f(H)$$ since $f(B)<\mathbb{Z}$ it must be ciclic (so it is free), therefor $$H\cong \ker f|_H\oplus f(B)$$ and $\ker f|_H<\langle e_2,\cdots,e_{n+1}\rangle=\mathbb{R}^n$ so, by induction, it is free so $H$ must be free too.

P.D. I change "finitely generated" for "free", since $\mathbb{R}^n$ is free of torsion both are the same.