Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We know that $\mathbb Q/\mathbb Z$ is a torsion group. Can we extend this to every quotient group of $\mathbb Q$? Thanks for hints.

share|cite|improve this question

Any nonzero subgroup of $(\mathbb{Q},+)$ contains an infinite cyclic group. The quotient of $(\mathbb{Q},+)$ by any infinite cyclic group is torsion. The quotient group of any torsion group is torsion.

share|cite|improve this answer
Thanks. You mean every subgroup of Q has Z as an infinite subgroup? – Baker May 1 '11 at 10:38
No, not necessarily. For example, the subgroup generated by $2/3$ does not contain $\mathbb{Z}$, but it does contain $2\mathbb{Z}$. – Jiangwei Xue May 1 '11 at 10:55

Let $H$ a non-zero subgroup of $\Bbb Q$ and let $0\neq h=\frac mn\in H$. It is clear that $m=n\cdot h\in H$ and so $m{\Bbb Z}<H$.

Therefore, there's a surjective map ${\Bbb Q}/m{\Bbb Z}\rightarrow{\Bbb Q}/H$, so the latter quotient is torsion because is a quotient of a torsion group.

share|cite|improve this answer
Just realized that this is just Jangwei's answer spelled out....... – Andrea Mori Jul 5 '11 at 14:14

I want to give a proof which exploits the known structure theory of divisible abelian groups. In this particular case it is evidently not the shortest proof available, but I hope that the basic technique will be of wider use.

Let $G$ be a divisible abelian group, and consider the exact sequence

$0 \rightarrow G[\operatorname{tors}] \rightarrow G \rightarrow G/G[\operatorname{tors}] \rightarrow 0$.

Now the torsion subgroup of a divisible abelian group is divisible: if $x \in \mathbb{G}[\operatorname{tors}]$ and $n \in \mathbb{Z}^+$, there exists $y \in G$ such that $ny = x$. But also there exists $m \in \mathbb{Z}^+$ such that $mx = 0$, so $(mn) y = 0$ and $y \in \mathbb{G}[\operatorname{tors}]$.

Since an abelian group is divisible iff it is injective as a $\mathbb{Z}$-module, the first term in the above sequence is injective, so it splits:

$G \cong G[\operatorname{tors}] \times V$,

where $V$ is divisible and torsionfree, hence can be given uniquely the structure of a $\mathbb{Q}$-vector space. Moreover, if $G \rightarrow H$ is a homomorphism of divisible groups, then after choosing subgroups $V_G$ of $G$ and $V_H$ of $H$ as above, then the natural map $V_G \hookrightarrow G \hookrightarrow H \rightarrow V_H$ is $\mathbb{Q}$-linear.

Now back to the problem: here we have $G = V_G$ is a $\mathbb{Q}$-vector space of dimension one and we are given a surjective map $f: G \rightarrow H$, so the map $\mathbb{Q} \cong V_G \rightarrow V_H$ is surjective and $\mathbb{Q}$-linear. Therefore it is either injective -- in which case $f$ is injective (in the language of the problem, this corresponds to taking the quotient by the zero group) or $V_H = 0$, i.e., $H = f(G)$ is a torsion group.

share|cite|improve this answer

I think you are wondering if every nontrivial quotient group of $\mathbb{Q}$ is torsion. In fact, for a nontrivial subgroup $H$ of $\mathbb{Q}$, there is a minimal positive number $h \in H$, then, we have $h\mathbb{Q} \subseteq H$, and $G/H$ is torsion.

I hope I am not mistaken.

share|cite|improve this answer
I'm not sure if it's true that there exists such a $h\in H$. Consider the set $H=\lbrace \frac{p}{q}|p\text{ is even and }q\text{ is odd}\rbrace$. $H$ is a subgroup of $\langle \mathbb{Q},+\rangle$ (easy check), but there's no such minimal positive $h\in H$- for any $h_0\in H$, $0<h_0\frac{1}{3}<h_0$ is a smaller element of $H$. – kneidell Jul 5 '11 at 13:53
Suppose $H=\mathbb Z$, if you mean $h\mathbb Q$ as multiplication then it is only true for $h=0$, and $h\mathbb Q=\{0\}$. If you meant addition then there is no integer such that $h+q\in\mathbb Z$ for all $q\in\mathbb Q$. – Asaf Karagila Jul 5 '11 at 14:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.