Every abelian group admits a unique maximal divisible subgroup

Question

I have problems about this problem, basically I do not know how to proceed.

I know that since one abelian group $A$ can be embedded in a divisible abelian group $G$ but I do not know how to proceed for this problem.

Thank you for your help!

Answer 1

Let $G$ be an abelian group, and let $\{A_i\}_{i\in I}$ be a nonempty family of subgroups of $G$ such that each $A_i$ is divisible. I claim that $\oplus A_i = \langle \cup A_i\rangle$, the subgroup generated by all the $A_i$, is itself divisible.

Indeed, let $x\in \langle \cup A_i\rangle$ and $n\gt 0$. We know that there exist $i_1,\ldots,i_m\in I$, and $a_{i_k}\in A_{i_k}$, $k=1,\ldots,m$, such that $x=a_{i_1}+\cdots+a_{i_m}$. Now, for each $k$, $A_{i_k}$ is divisible, so there exists $b_{i_k}\in A_{i_k}$ such that $nb_{i_k}=a_{i_k}$. Then let $b=b_{i_1}+\cdots+b_{i_m}$; this is an element of $\langle \cup A_i\rangle$, and $$nb = n(b_{i_1}+\cdots+b_{i_m}) = nb_{i_1}+\cdots + nb_{i_m} = a_{i_1}+\cdots+a_{i_m} = a.$$

So now consider the collection of all divisible subgroups of $G$. This collection is nonempty, since $\{0\}$ is divisible. The subgroup they generate is divisible, and clearly contains all divisible subgroups of $G$, so it is the unique largest divisible subgroup of $G$. Thus, $G$ has a unique maximal divisible subgroup.

Note that this statement is not that $G$ has a maximal subgroup which is divisible and unique with that property; it says that $G$ has a subgroup that is divisible and that it is maximal among divisible subgroups, and is unique with that property.