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 $B=<a>$ 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 $$n\cdot h(1)=h(1)+\cdots +h(1)=h(1+\cdots +1)=h(n)=h(f(a))=i(a)=a.$$ 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: $$n\cdot h(1)=h(1)+\cdots +h(1)=h(1+\cdots +1)=h(n)=h(f(a))=i(a)=a.$$ This completes first implication.

,,$\Rightarrow$ '' This implication is proven here. $\square$

Remark. It is clear that in the category of abelian groups $\mathcal{AB}$ , a group $A$ is projective if and only if $A$ is free. This is since $\mathcal{AB}$ 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.