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

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.

• Is using linear algebra allowed?
– lhf
Commented Nov 23, 2021 at 12:25

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

• Thank you, but unfortunately this is relevant to a homework problem and while I like algebra, I don't have time at the moment to learn what a module even means. A very elementary proof (or a better hint than the book) is what I was looking for [see the edit]. Commented Mar 10, 2014 at 0:48
• If you replace 'submodule' with 'subgroup' the same thing should still hold. Commented Mar 10, 2014 at 0:58

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.

• Sorry, my mistake: it is $x^{-h}$ indeed. Commented Nov 24, 2021 at 5:11