# 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.

• This is interesting. To help avoid it being closed, please edit the question to include, say, the motivation behind it. Sep 4 at 21:26
• I suppose you want to require $G$ to be nontrivial. Sep 4 at 21:48

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.

• Thank you, Eric. Would you mind if I make it into a small lemma and attribute it to you? This is a great reduction step in a proof concerting properties of convolution in $\ell_1(G)$ I am currently writing down. Sep 5 at 7:13
• Sure. For what it's worth, I'm also far from an expert on this topic and I expect this fact is probably well-known to experts. Sep 5 at 11:53