abelian group is divisible if and only if it is an injective object

Proposition. Abelian group $A$ is divisible if and only if $A$ is an injective object in the category of abelian groups.

Proof. ,,$\Leftarrow$” Assume that $A$ is not divisible, i.e. there exists $a\in A$ and $n\in \mathbb{N}$ such that the equation $nx=a$ has no solution in $A$. Let be a cyclic subgroup generated by $a$ and $i:B\to A$ the canonical inclusion. Now there are two possibilities: either $B$ is finite or infinite.

If $B$ is infinite, then let $H=\mathbb{Z}$ and let $f:B\to H$ be defined on generator by $f(a)=n$. Now $A$ is injective, thus there exists $h:H\to A$ such that $h\circ f=i$. Thus

$$n\cdot h(1)=h(1)+\mathrm{\cdots }+h(1)=h(1+\mathrm{\cdots }+1)=h(n)=h(f(a))=i(a)=a.$$

Contradiction with definition of $n\in \mathbb{N}$ and $a\in A$.

If $B$ is finite, then let $k=|B|$ (note that $n$ does not divide $k$) and let $H={\mathbb{Z}}_{n\cdot k}$. Furtheremore define $f:B\to H$ on generator by $f(a)=n$ (note that in this case $f$ is a well defined homomorphism). Again injectivity of $A$ implies existence of $h:H\to A$ such that $h\circ f=i$. Similarly we get contradiction:

$$n\cdot h(1)=h(1)+\mathrm{\cdots }+h(1)=h(1+\mathrm{\cdots }+1)=h(n)=h(f(a))=i(a)=a.$$

This completes first implication.

,,$\Rightarrow$” This implication is proven here (http://planetmath.org/ExampleOfInjectiveModule). $\mathrm{\square }$

Remark. It is clear that in the category of abelian groups $𝒜ℬ$, a group $A$ is projective if and only if $A$ is free. This is since $𝒜ℬ$ is equivalent to the category of $\mathbb{Z}$-modules and projective modules are direct summands of free modules. Since $\mathbb{Z}$ is a principal ideal domain, then every submodule of a free module is free, thus projective $\mathbb{Z}$-modules are free.