I am working on the following problem:
Let $G$ an Abelian group and $f: G \to \Bbb Z$ a surjective
homomorphism. Prove that $G \cong \ker(f) \times \Bbb Z$
By means of the First Isomorphism Theorem, we can obtain that $G / \ker(f) \cong \Bbb Z$. Is there some natural way to proceed from this point, invoking probably some other theorem, or would one need to define a map and prove that is it an isomorphism?
More generally, is it true that given an Abelian group $A$ and a subgroup $B$, $$A/B \cong C \Rightarrow A \cong C \times B$$
Thank you very much in advance!
Best Answer
You need to define the splitting map $g\colon \mathbb Z \to G$ (so that $fg$ is the identity). It should be clear how to do this, pick $x \in G$ such that $f(x) = 1$ and then define $g(n) = nx$.
Then you show that $g$ is injective and $G = \ker f \times \mathrm{im} \ g$. It's a faily elementary argument, just show that the two subgroups have trivial intersection and generate $G$.
As for your more general question the answer is no. For example $\mathbb Z/4$ has subgroup $\{0, 2\}$ isomorphic to $\mathbb Z/2$. The quotient is also isomorphic to $\mathbb Z/2$ but $\mathbb Z/4$ is not isomorphic to $\mathbb Z/2 \times \mathbb Z/2$.
On the other hand if $B \simeq \mathbb Z^n$ for some $n$ then it is true that $A \simeq A/B \times B$. To use more advanced language, the reason is that $\mathbb Z^n$ is a free $\mathbb Z$-module, hence projective, hence surjections to it are split. Basically, with $\mathbb Z^n$ you will always be able to define the splitting map $g$ and carry out the remainder of the argument above.