Every abelian group embeds in a divisible abelian group

Let $F={\mathbb{Z}}^{(A)}$, let $\epsilon :F\to A$ be the canonical surjection, and put $K=\ker \epsilon$; by [L1], $A\cong F\mathrm{/}K$.

The group $Q$ is divisible: for a finite-support tuple $q$ and a nonzero integer $n$, divide each of its finitely many nonzero rational coordinates by $n$ using [L2].

Embed $F$ coordinatewise into $Q:={\mathbb{Q}}^{(A)}$, and regard $K$ as a subgroup of $Q$. Define $D:=Q\mathrm{/}K$ and $j:F\mathrm{/}K\to D$ by $j(f+K)=f+K$. This is well defined and injective because $F\cap K=K$.

A quotient of a divisible group is divisible: if $q+K\in D$ and $n\mathrm{\ne }0$, choose $y\in Q$ with $ny=q$ by step 1.2; then $n(y+K)=q+K$. Thus $D$ is divisible by [F1].

Composing the isomorphism $A\cong F\mathrm{/}K$ from step 1.1 with the injection $j$ of step 2.1 embeds $A$ in the divisible group $D$. Every division was coordinatewise on finite support, so no choice was used.