Subgroup of Finitely Generated Abelian Group

[jumpr]

Homework Statement

Prove that any subgroup of a finitely generated abelian group is finitely generated.

Homework Equations

The Attempt at a Solution

I've attempted a proof by induction on the number of generators. The case n=1 corresponds to a cyclic group, and any subgroup of a cyclic group is cyclic, and so generated by one element. Then for the inductive step, I supposed that G was finitely generated, say G = \mathbb{Z} a_{1} + ... + \mathbb{Z} a_{n}, and that the result holds for groups generated by fewer than n elements. I've then let H \le G, and considered the quotient group G/\mathbb{Z}a_{n}, and then hoped that the correspondence theorem would help me out, but so far I can't seem to make it work.

Am I even attacking this problem correctly?

 

[micromass]

Try to prove the following general result, it might help:

Let H be an abelian group. If there exists a finitely generated subgroup K of H such that H/K is finitely generated, then H is finitely generated.

 

[jumpr]

Try to prove the following general result, it might help:

Let H be an abelian group. If there exists a finitely generated subgroup K of H such that H/K is finitely generated, then H is finitely generated.

Does this work?

Let K \le H be finitely generated where H/K is finitely generated, say K = \mathbb{Z}a_{1} + ... + \mathbb{Z}a_{n} and H/K = \mathbb{Z}(b_{1} + K) + ... + \mathbb{Z}(b_{m} + K) = \mathbb{Z}b_{1} + ... + \mathbb{Z}b_{m} + K, where the b_{i} \in H. Let h \in H, and consider h + K \in H/K. Then h + K = y_{1}b_{1} + ... + y_{m}b_{m} for some y_{i} \in \mathbb{Z}, so y_{1}b_{1} + ... + y_{m}b_{m} - h \in K. But then y_{1}b_{1} + ... + y_{m}b_{m} - h = z_{1}a_{1} + ... + z_{n}a_{n} for some z_{i} \in \mathbb{Z}, and so h = y_{1}b_{1} + ... + y_{m}b_{m} - z_{1}a_{1} - ... - z_{n}a_{n}. Thus, H is finitely generated.

 

[micromass]

Yes, that works!

 

[jumpr]

Sorry, I'm going to have to ask for a hint as to where to go from there. My ideas for proofs tend to break down when I want to assume a_{i} \in H for some a_{i}when G = \mathbb{Z}a_{1} + ... + \mathbb{Z}a_{n}.