I need to find what are the at the group $\mathbb{Q}/\mathbb{Z}$.
I think that any element at this group has a finite order, but I don't know how to prove it... I'd like to get help with the proof writing...

If I'm wrong, I'd like to to know it too...


Thank you!

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    $\begingroup$ This is a nice example that shows the existence of an infinite group in which every element has a finite order :) $\endgroup$ – Prism Dec 19 '13 at 7:44

Yes, you're correct that this is a group in which every element has finite order. Since you've asked about proof writing, I'll write out essentially a full proof.

Choose some $\alpha \in \Bbb{Q} / \Bbb{Z}$. By definition of the quotient group, there is a rational number $r$ for which

$$\alpha = r + \mathbb{Z}$$

Now by the definition of $\mathbb{Q}$, there are integers $a$ and $b \ne 0$ (we take $b > 0$) without any loss of generality) for which

$$r = \frac a b \implies \alpha = \frac a b + \mathbb{Z}$$


$$b\alpha = b\left(\frac a b + \Bbb Z\right) = b \frac a b + \Bbb{Z} = a + \Bbb{Z}$$

But since $a$ is an integer, $a + \Bbb{Z} = \Bbb{Z}$ is the identity in $\Bbb{Q} / \Bbb{Z}$, so $\alpha$ has finite order (in fact, the order is at most $b$) as desired.


If you know the nature of Prüfer group then you may use this fact that:

$$\mathbb Q/\mathbb Z=\sum_p\mathbb Z(p^{\infty})$$ However @T.Bongers's post makes the problem easy to understand step by step. Also, it is good to know that, the Prüfer group is an infinite $p-$primary group whose every subgroups if finite and indeed cyclic.

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  • $\begingroup$ It is 6:00 am, my time. Not super early, but early! ;-) $\endgroup$ – Namaste Dec 19 '13 at 12:01
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    $\begingroup$ And I for you, dear friend! You deserve "Gold for Groups"!! ;-) $\endgroup$ – Namaste Dec 19 '13 at 12:15

Each element of $G=(\mathbb{Q}/\mathbb{Z},+)$ is of finite order.

$x\in G$, Then $x={p\over q}+\mathbb{Z}, p\in\mathbb{Z},q\in\mathbb{N}$

what is $qx$ then?

$qx=({p\over q}+\mathbb{Z})+({p\over q}+\mathbb{Z})+\dots$ $q$ times$=p+\mathbb{Z}=\mathbb{Z}$


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