TITLE: Injective group is divisible
QUESTION [7 upvotes]: I have a proof of the fact that an abelian group $I$ is injective $\iff$ it is divisible.
If $I$ is injective, then applying the definition of injective to the inclusion $n\mathbb{Z} \hookrightarrow \mathbb{Z}$ and the homomorphism $n\mathbb{Z}\rightarrow \mathbb{Z}$ taking $nk\mapsto kd$ shows that $I$ is divisible.

Why? I can't connect this with definition of divisible group.

REPLY [2 votes]: Consider an injective module $I$ and $x\in I$. Suppose $n>0$ and consider

the homomorphism $f\colon\mathbb{Z}\to I$, $f(z)=zx$
the monomorphism $g\colon\mathbb{Z}\to\mathbb{Z}$, $g(z)=nz$

By injectivity, there exists $h\colon\mathbb{Z}\to I$ such that $hg=f$.
In particular,
$$
x=f(1)=h(g(1))=h(n)=nh(1)
$$

Proving that divisible modules are injective exploits the fact that $\mathbb{Z}$ is a PID. There are domains where divisible modules (with obvious definition) may not be injective.