Show that any subgroup of a finitely generated abelian group is finitely generated?

Hopefully this isn't too similar to what you don't want to see.

Here's a nice general result: in a PID, a submodule of a finitely generated free module is finitely generated of lesser or equal rank. The proof below I had written up earlier (and I hope it is sufficient/not too hand-wavy), so it uses $\mathbb Z$ instead of a general PID $R$.

To see this, we proceed by induction on $n$. First, for $n=1$, we know that submodules of $\mathbb Z$ correspond to ideals of $\mathbb Z$, which are principal, hence generated by at most $1$ element. Suppose the $n-1$ case, and let $M\subset \mathbb Z^n$ be a submodule $\mathbb Z^n$. Then let $\phi:M\to \mathbb Z$ be given by $$ \phi((x_1,\ldots,x_n))=\sum_{i=1}^n x_i. $$ Then $\phi$ is a $\mathbb Z$-module homomorphism, so we obtain a short exact sequence $$ 0\to \ker\phi\to M\to \operatorname{im}\phi\to 0. $$ $\ker\phi$ is a submodule $\mathbb Z^{n-1}$, so it is free of rank at most $n-1$. Also, $\operatorname{im}\phi\subset\mathbb Z$ is a submodule, we have shown it is free, so in particular it is projective. Therefore the sequence is split, so we have $M=\ker\phi\oplus\operatorname{im}\phi$, which has rank at most $n$.

Applying this to the particular situation, let $\mathbb Z^n\to A$ be a surjection, which exists since $A$ is generated by $n$ elements. For $B\subset A$, we have $f^{-1}(B)$ is free of rank at most $n$, so the surjection $f^{-1}(B)\cong\mathbb Z^n\to B\to 0$ implies that $B$ is generated by $n$ elements

Let $N\le G$ be a subgroup of the finitely generated abelian group $G$. Suppose $G=\langle g_1,\dots,g_n\rangle$.

Take $M=N \cap \langle g_2,\dots,g_n\rangle = \{m=g_2^{e_2}\cdots g_n^{e_n} \mid e_i \in \mathbb{Z}, m\in N\} $. Then $M\le\langle g_2,\dots,g_n\rangle$. So by induction, $M=\langle x_1,\dots,x_m\rangle$ for some $x_i$'s in $M$. Now $M$ only accounts for some elements of $N$ but not neccesarily all.

A standard element $g$ of $N$ is of the form $g=g_1^{e_1}\cdots g_n^{e_n}$. If $e_1\ne 0$, then we do not know if $g\in M$.

Consider the set $A=\{e_1 \in \mathbb{Z} \mid\ \exists\ e_2,\dots,e_n\ \text{ such that } g_1^{e_1}g_2^{e_2}\cdots g_n^{e_n} \in N\}$. One can easily check that $A$ is a subgroup of $\mathbb{Z}$, and thus $A=n\mathbb{Z}$ for some $n\in A$ (if you have not seen this, try using Bezout's identity to prove it, it happens for all subgroups of $\mathbb{Z}$). So denote $x=g_1^{n}g_2^{e_2}\cdots g_n^{e_n} \in N$ (which exists since $n\in A$). Now we will see that $N=\langle x_1,\dots,x_m,x\rangle$.

Now let $g\in N$ but $g\notin M$, so $g=g_1^{j_1}\cdots g_n^{j_n}$ where $j_1\ne 0$. Then $j_1 \in A=n\mathbb{Z}$ by definition of $A$. And thus, there exist some $h\in\mathbb{Z}$ such that $j_1 = nh$. Then we have that $g x^{-h} = g_1^{j_1-nh}g_2^{e_1'}\cdots g_n^{e_n'}=g_2^{e_2'}\cdots g_n^{e_n'}\in M=\langle x_1,\dots,x_m\rangle$. So $g\in\ \langle x_1,\dots,x_m,x\rangle$ as we wanted.