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 $n$?
- not necessarily
- yes, a unique one
- yes, but not necessarily a unique one
- 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.
Best Answer
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