Prove every subgroup S of a finitely generated abelian group G is itself finitely generated.

Question

Call a group G finitely generated if there is a finitely subset X$\subseteq$G with G=$<X>$. Prove that every subgroup S of a finitely generated abelian group G is itself finitely generated.

I am trying to use mathematical induction to prove it. when only has one element, then we can prove every subgroup S of a finitely generated abelian group G is itself finitely generated in this case( the statement is true). Then, suppose we have X=$<a_{1},a_{2},...,a_{n}>$, and I suppose that statement is still true in this case. Then when X=$<a_{1},a_{2},...,a_{n+1}>$, then given a S which is a subgroup of G, and suppose $a_{n+1}$ is in S, thenI want to try to prove G/<$a_{n+1}$> =$<a_{1}<a_{n+1}>,a_{2}<a_{n+1}>...,a_{n}<a_{n+1}>>$. I think it is true. Then I have difficulies solving this problem, I don't know how to continue. Can someone tell me how to prove this question? I don't know how to continue? And you don't need to use my methods. If you have your own methods to solve the problem, can you tell me how to prove it?

Answer 1

You can prove this as follows. First, note that $\mathbb Z$ is a Noetherian ring (definition see here). This basically follows from the fact that $\mathbb Z$ is a Euclidean ring. Now, the group $G$ is a finitely generated $\mathbb Z$-module, since any abelian group is a $\mathbb Z$-module. Moreover, we have the fact that every finitely generated module over a Noetherian ring is Noetherian (see e.g. here, proposition 4.1.6). Finally, by definition of Noetherian module (see here), every submodule of a Noetherian module is finitely generated. So we conclude that every subgroup of $G$ is finitely generated.

Of course, this "proof" is just a bunch of links, but the concepts of Noetherian rings and modules are quite useful and worth the effort of learning them.

Answer 2

Here is a complete proof. (It is inspired by a similar proof that an ideal in a ring that is maximal with the property of not being finitely generated, must be prime.)

Assume that there exists subgroups that are not finitely generated. They are ordered by inclusion and by Zorn's lemma there exists a maximal subgroup that is not finitely generated, call it $X$.

Now consider any $x\in G\setminus X$. Since $\langle X, x\rangle$ ($= X + \mathbb Z\cdot x$) is larger than $X$, it is finitely generated. Consider a set of generators $$ v_i + \lambda_i x, \quad i = 1,\dots, n $$ for $X$.

Next, consider the subgroup

$$ M = \{ \mu \in\mathbb Z \mid \mu x \in X\}. $$

Since subgroups of $\mathbb Z$ are finitely generated, $M$ is finitely generated, say by $\mu_1,\dots,\mu_m$. (If $M = 0$, then this is generated by the empty set. In fact $m=1$ will always suffice because subgroups of $\mathbb Z$ are cyclic.)

We will show that $X = \langle v_1,\dots,v_n, \mu_1 x , \dots, \mu_m x\rangle$. One inclusion is pretty obvious, since $v_i\in X$ and $\mu_i x\in X$ because of the way we definted them. To see the other inclusion, consider any $g\in X$ and note that there exist $\sigma_i\in \mathbb Z$ such that $$ g = \sum_i \sigma_i v_i + \sigma_i \lambda_i x. $$ However this implies that $$ (\sum_i \sigma_i \lambda_i) x = g - \sum_i \sigma_i v_i \in X,$$ thus from the definition of $M$, we have $\sum_i \sigma_i \lambda_i\in M$ and therefore there exist $\tau_i$ such that $$\sum_i \sigma_i \lambda_i = \sum_i \tau_i \mu_i.$$ But this implies that $$g = \sum_i \sigma_i v_i + \sum_i \tau_i \mu_i x,$$ thus $g$ is indeed in the set generated by the $v_i$ and $\mu_i x$, which completes the proof.

(Note that this is a proof by contradiction: we assume that $X$ is not finitely generated and construct a finite set of generators. Nonetheless, the proof yields a practical procedure for finding generators: Start from an arbitrary subgroup $V$, add generators of $V$ until you find a group for which you know a finite generating set (this will exist because $G$ is finitely generated), and repeatedly use the above argument to get rid of the generators one by one.)