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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!

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You cannot say that $g_1,\ldots,g_n$ span the subgroup because you don't know whether they are in the subgroup. In order for $H$ to be finitely generated, you need to find a finite subset **of $H$** (*not* of the overgroup $G$) that spans $H$. It is also false that there is a subset of $g_1,\ldots,g_n$ that generates the group. Take $G=\mathbb{Z}_2\times \mathbb{Z}_2$, $H=\{(0,0), (1,1)\}$, $g_1=(1,0)$ and $g_2=(0,1)$. What subset of $\{g_1,g_2\}$ spans $H$? –  Arturo Magidin Apr 26 '12 at 16:02

2 Answers 2

up vote 3 down vote accepted

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 $\mathcal{H}$).

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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.

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 noethrian 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.

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