Proving that a subgroup of a finitely generated abelian group is finitely generated 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! 
 A: 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 $\mathcal{H}$).
A: 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. 
A: Here is a proof by induction on number of generators.
For $n=1$ it is easy to see that subgroup of a cyclic group is always cyclic. Now assume that the statement is true for all abelian groups with atmost  $n$ generators.
Consider $G= (x_1,\ldots,x_n,x_{n+1})$ and $G'=(x_1,\ldots , x_n)$. Let $H$ be any subgroup of $G$, then $H\cap G'$ is a subgroup of $G'$, thus by inductive hypothesis $H\cap G'$ is finitely generated. By an isomorphism theorem we get
$$ H/H\cap G' \cong HG'/G'\le G/G'$$
Now $G/G'$ is generated by $x_{n+1}G'$ implies that $H/H\cap G'$ is cyclic say generated by $yH\cap G'$ and say $H\cap G'$ is generated by $y_1,\ldots,y_k$ then it is easy to see that $H$ is generated by $y_1,y_2,\ldots,y_k$ and $y$.
