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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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    $\begingroup$ 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 $H$? $\endgroup$
    – Arturo Magidin
    Apr 26, 2012 at 16:02

3 Answers 3

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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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  • $\begingroup$ When you say "the subgroup $H$ corresponds to a subgroup $\mathcal{H}$ of $\mathbb{Z}^r$ by the isomorphism theorems", do you mean that $H \simeq \mathcal{H}/\ker(\mathbb{Z}^r\to G)$? $\endgroup$
    – user
    Jul 7, 2017 at 7:01
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    $\begingroup$ @user: No, I'm talking about the correspondence part of the isomorphism theorems: given a group homomorphism $f\colon K\to M$, the subgroups of $M$ correspond to subgroups of $K$ that contain $\mathrm{ker}(f)$; the correspondence respects the lattice operations on subgroups. The First Isomorphism Theorem that you mention comes into play later, when we say "and hence its image $H$ is also finitely generated". $\endgroup$
    – Arturo Magidin
    Jul 7, 2017 at 18:27
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    $\begingroup$ I would just invoke the Third Theorem, rather than do the rest of the rigamarole (it is included in the correspondence of the third). You are also skipping the final part, which is to show that if $h_1,\ldots,h_n$ generate $\mathcal{H}$, then $f(h_1),\ldots,f(h_n)$ generate $f(\mathcal{H}))=H$. $\endgroup$
    – Arturo Magidin
    Jul 8, 2017 at 17:33
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    $\begingroup$ @user There is no such thing as "the" free part of a finitely generated abelian group. For instance, if $G=\mathbb{Z}\oplus C_2$, with generators $x$ and $y$, then note that you also have $G=\langle x+y\rangle \oplus \langle y\rangle$. So, what is "the free part" of $G$? Is it $\langle x\rangle$, or is it $\langle x+y\rangle$? The torsion part is uniquely determined, but its direct complement is not. So it makes no sense to talk about the free part, because it is not unique. $\endgroup$
    – Arturo Magidin
    Jul 9, 2017 at 5:18
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    $\begingroup$ @user: And the example above immediately shows you that the answer is "no", because you can find a free abelian subgroup $F'$ of $G$ such that $G=F'\oplus T$, and $F'\neq F$. So how could it possibly be the case that you always have $H\leq F$? $\endgroup$
    – Arturo Magidin
    Jul 9, 2017 at 18:18
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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. 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.

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  • $\begingroup$ To make it clear that this answers the original problem using the isomorphism theorems realise that the (a?) proof of stability of the Noetherian property wrt extensions requires these. Btw: @Justin can you give a reference for this proof, which is very lovely. $\endgroup$
    – Steve Powell
    Oct 27, 2017 at 7:59
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    $\begingroup$ PS: I prefer to capitalise the words Abelian and Noetherian, since they are named after famous mathematicians, Niels Henrik Abel (1802–1829) and Amalie Emmy Noether (1882–1935). (I also write Boolean.) :-) $\endgroup$
    – Steve Powell
    Oct 27, 2017 at 8:03
  • $\begingroup$ It's been over 5 years at this point. I don't remember a specific reference, but I would think this is in any book that discusses Noetherian rings and modules. $\endgroup$
    – Justin Young
    Oct 27, 2017 at 9:29
  • $\begingroup$ And so it is; I've been looking under finitely generated Abelian groups, and it is in the more general f-g modules sections. Thanks again. $\endgroup$
    – Steve Powell
    Oct 28, 2017 at 16:05
  • $\begingroup$ Nice! Also, the OP did say "or otherwise", though the isomorphism theorems are routinely used in exact sequence arguments as you mention. $\endgroup$
    – Justin Young
    Nov 8, 2017 at 14:07
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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 $y$.

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