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.

  • $\begingroup$ This is interesting. To help avoid it being closed, please edit the question to include, say, the motivation behind it. $\endgroup$
    – Shaun
    Sep 4 at 21:26
  • 1
    $\begingroup$ I suppose you want to require $G$ to be nontrivial. $\endgroup$ Sep 4 at 21:48

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Sep 5 at 7:13
  • $\begingroup$ 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. $\endgroup$ Sep 5 at 11:53

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