# Cyclic subgroup of a quotient group

I encountered this question in a grad-level exam. I hope somebody could help me with this.

We have to choose one option.

Consider the group $\;G=\Bbb Q/\Bbb Z\;$ where $\Bbb Q$ and $\Bbb Z$ are the groups of rational numbers and integers respectively. Let $n$ be a positive integer. Then is there a cyclic subgroup of order $n$?

1. not necessarily
2. yes, a unique one
3. yes, but not necessarily a unique one
4. never

I can see that $\Bbb Z$ is a normal subgroup of $\Bbb Q$. So, $G$ is a quotient group and it would have elements like $\Bbb Z$+$q$ where $q\in \Bbb Q\;$, that is $q$ can be $\;1/-1/0.5/-0.5...\;$ etc., and the identity of $G$ and its subgroup would be $\Bbb Z+0\;$, that is $\Bbb Z$. Now, if i assume $S$ to be a subgroup of $G$ having just the identity element, then i guess it would be a cyclic subgroup of order $1$. Am I correct here? And will there be any other cyclic subgroup? I am not sure.

I realize that this question has already been discussed. here are the links-

$\mathbb{Q}/\mathbb{Z}$ has a unique subgroup of order $n$ for any positive integer $n$?

consider the group $G=\mathbb Q/\mathbb Z$. For $n>0$, is there a cyclic subgroup of order n

$\mathbb{Q}/\mathbb{Z}$ has cyclic subgroup of every positive integer $n$?

I didn't understand the concepts discussed there. Moreover, they are taking $Z$ as complex set but in my question, it is integer set. Also, since i am new, i couldn't post comment there for clarification. So, opening a new question. I hope somebody could help.

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In all the questions you linked, $\mathbb{Z}$ refers to the integers, just like here. Hint for the question: What is the order of $\frac{1}{n}$ in the quotient group? –  Tobias Kildetoft Jul 25 '13 at 6:58
see, i am aware of the term 'order of group' or 'order of an element', but m at a loss to understand what is order of 1/n. if u can ignore my ignorance and give more help, i'll be grateful. –  Ramit Jul 25 '13 at 7:03
By $\frac{1}{n}$ I actually meant the element $\frac{1}{n} + \mathbb{Z}$ (which is an element of $\mathbb{Q}/\mathbb{Z}$. It might be convenient for this to first show that any element in the quotient has a unique representative in the interval $[0,1)$ and work with those instead. –  Tobias Kildetoft Jul 25 '13 at 7:07
The available options "not necessarily" and "never" don't make sense when applied to a single group. –  Derek Holt Jul 25 '13 at 7:47
@DerekHolt But $n$ can be arbitrary. So it might be the case (though it is not), that the answer depended on the choice of $n$. –  Tobias Kildetoft Jul 25 '13 at 7:59

For any

$$n\in\Bbb N\;,\;\;\text{ord}\,\left(\frac1n\right)_{\Bbb Q/\Bbb Z}=n$$

So we already know there's a cyclic subgroup of order $\,n\,$ in $\,\Bbb Q/\,\Bbb Z\,$ . Now, if

$$\left(\frac ab+\Bbb Z\in\Bbb Q/\Bbb Z\;\;\;\text{and}\;\;\;\text{ord}\,\left(\frac ab\right)_{\Bbb Q/\Bbb Z}=n\right)\implies \left(n\frac ab\in\Bbb Z\right)\iff \left(n=bk\;,\;k\in\Bbb Z\right)$$

and thus in fact we have that

$$\frac ab=\frac{ak}n\in\left\langle\;\frac1n+\Bbb Z\;\right\rangle\le\Bbb Q/\Bbb Z$$

and this gives us uniqueness

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Investigating my interesting group $\mathbb Q/\mathbb Z$ +1 –  Babak S. Jul 25 '13 at 7:36
The sentence beginning "Now, if..." and continuing with some math is a bit hard to read properly, as it mixes the use of "if" with logical symbols (so on my first read of it, I though you were saying something about what happened when the $\Leftrightarrow$ was true). –  Tobias Kildetoft Jul 25 '13 at 7:45
Thanks @TobiasKildetoft, I shall edit in short. –  DonAntonio Jul 25 '13 at 7:47