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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.

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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.

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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.

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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.

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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.

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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

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