0
$\begingroup$

I've been working on the following question, but haven't made much progress at all... so any help would be greatly appreciated.

Prove that if $G$ is a finitely generated abelian group, then $G \cong G_{\text{tors}} \times \mathbb{Z}^r$.

We can't use the Fundamental Theorem of Finitely Generated Abelian Groups for this, since this problem is supposed to help us fill in some of the gaps that the professor left out.

$\endgroup$
2
$\begingroup$

I'm assuming you already know that each finitely generated torsion-free abelian group is isomorphic to $\mathbb{Z}^r$ for some $r$. Then you can proceed like this:

  1. $G/G_{\text{tors}}$ is torsion-free and finitely generated. Therefore, $G/G_{\text{tors}} \simeq \mathbb{Z}^r$ for some $r$. So we have a surjective homomorphism $\varphi: G \to \mathbb{Z}^r$ such that $\ker \varphi = G_{\text{tors}}$.
  2. There exists a homomorphism $\psi: \mathbb{Z}^r \to G$ such hat $\varphi \circ \psi = \text{Id}_{\mathbb{Z}^r}$. To build such a $\psi$ you can just pick a basis $(a_1, \ldots, a_r)$ in $\mathbb{Z}^r$ and choose $\psi(a_i)$ to be any element in the set $\varphi^{-1}(a_i)$.
  3. Now it is easy to show that $G = \ker \varphi \oplus \operatorname{im} \psi$. Since $\ker \varphi = G_{\text{tors}}$ and $\operatorname{im} \psi \simeq Z^r$, we have the desired isomorphism.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.