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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?

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1 Answer 1

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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?

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  • $\begingroup$ If $nx \notin \mathbb{Z}$, then $x\neq \frac{m}{n}$, so $x+\mathbb{Z}$ has infinite order? $\endgroup$
    – Hopmaths
    Commented Jan 21, 2021 at 2:10
  • $\begingroup$ Yes, although you can do it directly as well. $nx = m\in \mathbb{Z}$, so $x=$? $\endgroup$
    – rogerl
    Commented Jan 21, 2021 at 2:54
  • $\begingroup$ $x=\frac{m}{n}\in \mathbb{Q}$ $\endgroup$
    – Hopmaths
    Commented Jan 21, 2021 at 3:01

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