What are the finite order elements of $\mathbb{Q}/\mathbb{Z}$? 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...
BTW:
$\mathbb{Z}=(\mathbb{Z},+)$
$\mathbb{Q}=(\mathbb{Q},+)$
Thank you!
 A: 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$=p+\mathbb{Z}=\mathbb{Z}$
A: 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.
A: 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
$$r = \frac a b \implies \alpha = \frac a b + \mathbb{Z}$$
Therefore,
$$b\alpha = b\left(\frac a b + \Bbb Z\right) = b \frac a b + \Bbb{Z} = a + \Bbb{Z}$$
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.
