# Subgroup of Finitely Generated Abelian Group

1. Nov 22, 2012

### jumpr

1. The problem statement, all variables and given/known data
Prove that any subgroup of a finitely generated abelian group is finitely generated.

2. Relevant equations

3. 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?

2. Nov 22, 2012

### micromass

Staff Emeritus
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.

3. Nov 24, 2012

### jumpr

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.

4. Nov 24, 2012

### micromass

Staff Emeritus
Yes, that works!

5. Nov 24, 2012

### 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 $G = \mathbb{Z}a_{1} + ... + \mathbb{Z}a_{n}$.