TITLE: Proving that a subgroup of a finitely generated abelian group is finitely generated
QUESTION [24 upvotes]: A question says: Using the isomorphism theorems or otherwise, prove that a subgroup of a finitely generated abelian group is finitely generated. 
I would say that for a finitely generated abelian group $G$, there exists elements $g_1,\dots, g_n$ such that a linear combination of them generates the whole group. Therefore as every element of a subgroup has an element in $G$ and so can be made by a linear combination of $g_1,\dots, g_n$. This means that $g_1,\dots, g_n$ span the whole subgroup and so there exists a subset of $g_1,\dots, g_n$ which generates the subgroup. 
This answer seems far too 'linear algebra-ish' rather than 'group theory-ish' and I can't seem to see how one would use the isomorphism theorems? Help would be appreciated!

REPLY [12 votes]: 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.