Projective group I was reading about torsion-free abelian groups. Does there exist a torsion-free abelian group which is not projective but for which each of its torsion-free homomorphic images is projective?
 A: John Stalfos's comment is correct: let's make it into an answer.
Consider $\mathbb{Q}$ as an abelian group (i.e., we work with the additive group).  Then:
$\bullet$ $\mathbb{Q}$ is torsionfree.
$\bullet$ $\mathbb{Q}$ is divisible: for all $x \in \mathbb{Q}$ and all $n \in \mathbb{Z} \setminus \{0\}$, $x = ny$ for some $y \in \mathbb{Q}$ (indeed, for $y = \frac{x}{n}$).
$\bullet$ Any homomorphic image of a divisible abelian group is divisible.
$\bullet$ An abelian group is torsionfree and divisible iff it is (in a unique way) a $\mathbb{Q}$-vector space.  Thus any torsionfree homomorphic image of $\mathbb{Q}$ must be a $\mathbb{Q}$-vector space $V$ and the map $q: \mathbb{Q} \rightarrow V$ is surjective and $\mathbb{Q}$-linear.  By linear algebra, this can only happen if $V \cong \mathbb{Q}$ and $q$ is an isomorphism or if $V = 0$.  So every torsionfree proper homomorphic image is projective.
$\bullet$ $\mathbb{Q}$ cannot be embedded into any free abelian group $\bigoplus_{i \in I} \mathbb{Z}$, since no nonzero element of a free abelian group is divisible by $n$ for all positive integers $n$.  (The same reasoning shows that we cannot even embed $\mathbb{Q}$ in any direct product $\prod_{i \in I} \mathbb{Z}$.  That is -- using terminology that I find rather unpleasant -- $\mathbb{Q}$ is a $\mathbb{Z}$-module which is torsionfree but not torsionless.)  In particular $\mathbb{Q}$ is not projective.  
(In fact projective modules over $\mathbb{Z}$ or any PID are necessarily free, but this fact is deeper than anything that took place above and was not used.)
