Proving that a subgroup of a finitely generated abelian group is finitely generated

Solution 1:

This follows from the following theorem, which is a common ingredient in the proof of the structure theorem for finitely generated abelian groups:

Theorem. Let $r\gt 0$ be a positive integer, and let $H$ be a subgroup of $\mathbb{Z}^r$. Then there exists a basis $a_1,\ldots,a_r$ of $\mathbb{Z}^r$, an integer $d$, $0\leq d\leq r$, and positive integers $m_1,\ldots,m_d$ such that $m_1|m_2$, $m_2|m_3,\ldots,m_{d-1}|m_d$ such that $m_1a_1,\ldots,m_da_d$ is a basis for $H$. In particular, $H$ is free and finitely generated.

You can see a proof of this in this previous answer.

To see how this proves the result, suppose that $G$ is abelian and finitely generated by $g_1,\ldots,g_r$. Let $H$ be a subgroup of $G$. There is a surjection $\mathbb{Z}^r\to G$ given by mapping the standard basis vector $\mathbf{e}_i$ to $g_i$; the subgroup $H$ corresponds to a subgroup $\mathcal{H}$ of $\mathbb{Z}^r$ by the isomorphism theorems. By the Theorem, $\mathcal{H}$ is finitely generated, and hence its image, $H$, is also finitely generated (generated by the images of the generators of $\mathcal{H}$).

Solution 2:

This basically follows from the fact that $\mathbb Z$ is noetherian.

Let's say a module $M$ over a commutative ring $R$ is noetherian if every submodule of $M$ is finitely generated. This condition is stable under extensions, if $0\to A\to B\to C \to 0$ is an exact sequence of $R$ modules, and $A$ and $C$ are noetherian, then $B$ is noetherian. Conversely, if $B$ is noetherian, then $A$ and $C$ are noetherian.

We say that a commutative ring $R$ is noetherian if $R$ is a notherian module over itself. If $R$ is a noetherian ring, then $R^n$ is a noetherian module for all $n\ge 1$ by extension stability.

Thus, if $R$ is a noetherian ring, then every finitely generated module over $R$ is noetherian.