# Module Homomorphisms and the nature of abelian groups

There is a fundamental result specifying that all abelian groups are $$\Bbb Z$$-modules.

We are trying to use this fact to find out:

Does there exist an abelian group $$H$$ which is not isomorphic to $$\Bbb Z$$ or $$\{e\}$$, such that if $$G$$ is abelian and there is a surjective homomorphism $$f : G \to H$$, then $$G\cong \ker(f) \times H$$.

Now my idea:

No, we cannot find such $$H$$. If $$G$$ is finite, we can characterize $$G$$ as $$\Bbb Z_{p}$$ where $$p$$ is prime. Now if $$G\cong \ker(f) \times H$$ then contradicting $$p$$ being a prime.

If $$G$$ is infinite, I don't know where to start.

EDITED

Thanks for Stefan's answer now my head clears up a bit. Let us wrap the question up. Now, if we can find a section map from $$G$$ to $$\Bbb Z \oplus\Bbb Z$$ and we are done. However note that we can define s: $$\Bbb Z \oplus \Bbb Z$$ $$\to$$ $$G$$ with $$s(p,q) = a^p b^q$$ where $$f(a) = (1,0)$$ and $$f(b) = (0,1)$$

• Commented Apr 26, 2022 at 16:21
• @Shaun will improve my formatting! Commented Apr 26, 2022 at 17:07

It is not true that every finite abelian group is cyclic. The simplest example of this is the Klein group $$C_2\times C_2$$.
What you are asking is: for what kind of $$H$$ is every short exact sequence $$0\to A\to G\to H\to 0$$ split? These kinds of questions are the domain of homological algebra. In that language, your question is: What are the projective abelian groups? The answer is that they are precisely the free abelian groups.
So, the group $$H=\mathbb{Z}\oplus\mathbb{Z}$$ works, as do more exotic things like $$\mathbb{Z}^\mathbb{R}$$, but such examples are all you can get.