A divisible abelian group is the direct sum of torsion subgroup and a torsion-free divisible subgroup.

Question

Let $D$ be a divisible abelian group. It is known that $D \cong D_t \oplus D/D_t$, where $D_t$ is the torsion subgroup and $D/D_t$ is torsion-free and divisible. I want to find a torsion-free divisible subgroup of $D$, say $G$ such that $D = D_t \oplus G$. I can’t find an intuitive subgroup that satisfies this condition. My intuition told me that $G \cong D/D_t$, but I can’t find any intuitive isomorphism. Any help will be appreciated.

Answer 1

There won't be any "intuitive" such subgroup, since there is not any canonical choice of such a $G$, and in general, I'm pretty sure the axiom of choice is needed to prove that one exists. If you're familiar with injective modules, the quick proof is that $D_t$ is divisible and hence injective (by Baer's criterion), so whenever $D_t$ is a subgroup of another abelian group it is a direct summand.

Or, here is a more direct (but basically equivalent) argument. By Zorn's lemma, there exists a subgroup $G$ of $D$ that is maximal with the property that $G\cap D_t=0$. Since $D$ is divisible, $G$ must be divisible by maximality (if some $g\in G$ was not divisible by $n$ in $G$, you could adjoin an element $h\in D$ such that $nh=g$ to $G$ to contradict maximality of $G$). Now suppose there were some $x\in D$ that is not in $G+D_t$. Let $G'$ be the subgroup generated by $G$ and $x$. By maximality of $G$, the intersection $G'\cap D_t$ must be nontrivial, which means that there is some $n\in\mathbb{Z}$ and $g\in G$ such that $nx+g$ is a nonzero torsion element of $D$. Then $n$ must be nonzero, since $G\cap D_t=0$. Since $nx+g$ is torsion, there is a nonzero $m\in\mathbb{Z}$ such that $mnx+mg=0$, so $mnx\in G$. But now since $G$ is divisible, there exists $h\in G$ such that $mnh=mnx$. Then $x-h$ is torsion so $x=h+(x-h)\in G+D_t$, a contradiction. Thus $G+D_t$ is all of $D$, and $D=G\oplus D_t$, as desired.