Abelian torsion-free group whose every quotient is reduced Definitely, not an expert, so my question is more of a reference request.
Suppose that $G$ is an Abelian group that is torsion-free and every quotient of $G$ is reduced (does not contain divisible subgroups); alternatively $G$ does not have divisible quotients. Can we say something about possible quotients of $G$? For example,

*

*must $G$ have an infinite cyclic quotient? Or, maybe

*finite cyclic quotients of unbounded orders?

Note that $\mathbb Z^{(\mathbb N)}$ is reduced but it does have divisible quotients.
 A: Suppose $G$ is a nontrivial torsion-free abelian group.  Then $G$ has a quotient $H$ which is torsion-free of rank $1$ (just take the image of any nonzero homomorphism $G\to\mathbb{Q}$), which we may assume is a subgroup of $\mathbb{Q}$ that contains $\mathbb{Z}$.  Now note that $H/\mathbb{Z}$ is a non-finitely generated subgroup of $\mathbb{Q}/\mathbb{Z}\cong\bigoplus_{p\text{ prime}}\mathbb{Z}[1/p]/\mathbb{Z}$.  Decomposing $H/\mathbb{Z}$ into its $p$-torsion subgroups for each prime $p$, it is a direct sum of subgroups of $\mathbb{Z}[1/p]/\mathbb{Z}$ for each $p$.  If one of those subgroups is all of $\mathbb{Z}[1/p]/\mathbb{Z}$, it is a nontrivial divisible quotient of $G$.  Otherwise, for each $p$, the $p$-torsion subgroup of $H/\mathbb{Z}$ is a finite cyclic group, say of order $p^n$.  The $p$-torsion subgroup of $H/p^m\mathbb{Z}$ is then cyclic of order $p^{m+n}$.  Thus $G$ must have cyclic quotients of orders that are arbitrarily high powers of any prime, and it follows easily that $G$ has finite cyclic quotients of all possible orders.
So to sum up, if $G$ is a nontrivial torsion-free abelian group, if it has no nontrivial divisible quotients, then it has finite cyclic quotients of all possible orders.  However, $G$ does not have to have an infinite cyclic quotient.  For a counterexample, take $G$ to be the group of rational numbers with squarefree denominator.
