# Show that an element of the factor group $\mathbb{R}/\mathbb{Z}$ has finite order if and only if it is in $\mathbb{Q}/\mathbb{Z}$

Show that an element of the factor group $$\mathbb{R}/\mathbb{Z}$$ has finite order if and only if it is in $$\mathbb{Q}/\mathbb{Z}$$.

What I have done: So far I know that if $$a$$ is an element in $$\mathbb{Q}/\mathbb{Z}$$, then $$a$$ has the form $$\frac{b}{c}+\mathbb{Z}$$, with $$b,c \in \mathbb{Z},c\neq0$$. Without loss of generality suppose that $$c>0$$, then $$ca=c\left ( \frac{b}{c}+\mathbb{Z} \right )=b+\mathbb{Z}=\mathbb{Z}$$ wich is the identity element in $$\mathbb{Q}/\mathbb{Z}$$. Thus $$a$$ has finite order. I think this part proves ($$\leftarrow$$)

I have trouble establishing the other implication and verifying the theorem. I mean...

If an element of the factor group $$\mathbb{R}/\mathbb{Z}$$ has finite order, then this element lies in $$\mathbb{Q}/\mathbb{Z}$$

How can I prove this?

Hint: suppose $$x+\mathbb{Z}$$ has finite order where $$x\in\mathbb{R}$$. That means that $$nx\in\mathbb{Z}$$. Can you take it from there?
• If $nx \notin \mathbb{Z}$, then $x\neq \frac{m}{n}$, so $x+\mathbb{Z}$ has infinite order? Commented Jan 21, 2021 at 2:10
• Yes, although you can do it directly as well. $nx = m\in \mathbb{Z}$, so $x=$? Commented Jan 21, 2021 at 2:54
• $x=\frac{m}{n}\in \mathbb{Q}$ Commented Jan 21, 2021 at 3:01