Is an group determined by its torsion part and torsion-free part? Let $G$ be an abelian group.  Let $G_T$ be the torsion part of $G$, i.e. the set of all elements of $G$ of finite order.  And let $G_F$ be the torsion-free part of $G$, i.e. the set containing $0$ along with all elements of $G$ of infinite order.  Then my question is, is the isomorphism type of $G$ uniquely determined by the isomorphism types of $G_T$ and $G_F$?
I think this is definitely true in the case when $G$ is finitely generated, but I'm asking about the general case.
 A: (Still not quite a complete answer.)
An abelian group $A$ does not have a canonically defined torsion-free subgroup in general (the elements of infinite order usually aren't a subgroup). What is canonically defined is a short exact sequence
$$0 \to A_T \to A \to A/A_T \to 0$$
where $A/A_T$, which we'll write $A_F$, is the universal torsion-free abelian group admitting a map from $A$; you might call this the "torsion-free part" but keep in mind that it's a quotient, not a subgroup.
$A$ is not determined by the isomorphism type of $A_T$ and $A_F$, and the reason is that the short exact sequence above does not split in general, and so in general defines a nontrivial class in $\text{Ext}^1(A_F, A_T)$. It does split if $A_F$ is free (equivalently, projective as a $\mathbb{Z}$-module), which in particular happens whenever $A$ is finitely generated; in this case we have
$$A \cong A_T \oplus A_F$$
so $A$ is in fact determined up to isomorphism by $A_T$ and $A/A_T$. So to exhibit a counterexample we need to find a torsion abelian group $A_T$ and a torsion-free abelian group $A_F$ such that $\text{Ext}^1(A_F, A_T) \neq 0$; in particular $A_F$ must be torsion-free (equivalently, flat as a $\mathbb{Z}$-module) but not free, and hence infinitely generated.
Take $A_F = \mathbb{Z} \left[ \frac{1}{p} \right]$ where $p$ is a prime. If we write $A_F$ as a filtered colimit of the diagram $\mathbb{Z} \xrightarrow{p} \mathbb{Z} \xrightarrow{p} \dots $ and try to compute $\text{Ext}^1(A_F, A_T)$ where $A_T$ is unspecified then we get a $\lim^1$ exact sequence
$$0 \to \lim^1_n A_T \to \text{Ext}^1 \left( \mathbb{Z} \left[ \frac{1}{p} \right], A_T \right) \to 0$$
so we want to find a torsion abelian group such that $\lim^1$ of the diagram $\dots \xrightarrow{p} A_T \xrightarrow{p} A_T$ is nonzero. I believe that we can take $A_T = \bigoplus_i \mathbb{Z}/p^i$ (which showed up in this previous math.SE question, also about extensions of torsion-free abelian groups by torsion abelian groups), but it seems annoying to verify that this works; I believe it does not suffice to take either $A_T = \mathbb{Z}/p$ or $A_T = \mathbb{Z} \left[ \frac{1}{p} \right]/\mathbb{Z}$ (the Prufer $p$-group), which I was hoping would work.
Edit: Apparently it's simpler to just construct $A$ directly. Supposedly the abelian group $A = \prod_i \mathbb{Z}/p^i$ has the property that its torsion sequence $0 \to A_T \to A \to A/A_T \to 0$ doesn't split, but I'm not sure how to prove it off the top of my head.
