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

Question

I am working through Rotman 2.89 and I can't seem to solve this one. Note: Please do not link me to the related questions such as Proving that a subgroup of a finitely generated abelian group is finitely generated. I would rather solve this using the "elementary" methods presented in the text thus far, as I think I will learn more out of it this way. The book suggests induction on n (the number of generators), and considering the quotient group, but I don't know how that helps. I do not see what property of these cosets makes them useful for the proof. The hypothesis itself seems fairly intuitive, but I do not know how to proceed using just basic things like the first isomorphism theorem, correspondence theorem, Lagrange's, etc.

Edit: Please suggest ways to prove this using only very basic group properties like the above.

Answer 1

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

Answer 2

I will leave two solutions for this. The first one is shorter and easier to understand. The second one is maybe more constructive and hands down. Apart from some group theory basics, I am assuming that the reader knows that a subgroup of a cyclic group is cyclic (see https://proofwiki.org/wiki/Subgroup_of_Cyclic_Group_is_Cyclic).

First solution:

Assume that $G$ is generated by $g_1.\dots,g_n$ and consider a subgroup $N\le G$. If $N \cap \langle g_i \rangle = \{1\}$ for some $i= 1,\dots, n$ (assume w.l.o.g that $i =1$), then the following quotient homomorphism \begin{equation} \pi\colon G \longrightarrow G/\langle g_1 \rangle = \langle \pi(g_2),\dots,\pi(g_n) \rangle, \end{equation} is injective on $N$. So $N$ can be viewed as subgroup of $ G/\langle g_1 \rangle $ and by induction on the number of generators $n$, $N$ is finitely generated.

Now if for all $i$, we have that $N \cap \langle g_i \rangle \ne \{1\}$ , then $N \cap \langle g_i \rangle = \langle g_i^{e_i} \rangle $. This implies that $M :=\langle g_1^{e_1},\dots, g_n^{e_n} \rangle \le N$. Since $M$ has finite index in $G$ (notice how the quotient $G/M$ has at most $e_1 \cdots e_n$ elements), we have the following picture,

$M$ has finite index in $N$" />

which shows that $M$ has finite index in $N$. Then $N$ is a finite union of cosets of $M$, and thus $N$ is finitely generated.

Second solution:

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.