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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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aarbee
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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 06:58
  • 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. – aarbee Jul 25 '13 at 07:03
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    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 07:07
  • The available options "not necessarily" and "never" don't make sense when applied to a single group. – Derek Holt Jul 25 '13 at 07:47
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    @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 07:59
  • @Tobias- i didn't get u when u said every element of quotient group has a representative in [0,1) because if i take n=2, 0.5+Z would be like ...-1.5,-0.5,0.5,1.5,... which don't lie in the said interval. moreover, the elements of quotient group are themselves sets, so talking about their representative doesn't make sense to me. – aarbee Jul 25 '13 at 10:13
  • Ramit, please don't write "u" when you mean "you". This is not Twitter. – Gerry Myerson Jul 25 '13 at 10:17
  • By representative, I mean a single element of the coset. Usually, there is not a good way to choose one representative for each coset, but in this case there is, and indeed they are easy to work with. The way to add these representative is to add them as usual, and if the result is $1$ or larger, subtract $1$. – Tobias Kildetoft Jul 25 '13 at 10:17
  • @GerryMyerson- Sure, I'll edit my comments. – aarbee Jul 25 '13 at 10:18
  • Looks like I can't edit them now, time is up. I'll take care next time. – aarbee Jul 25 '13 at 10:21

1 Answers1

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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

Alex Becker
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DonAntonio
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    Investigating my interesting group $\mathbb Q/\mathbb Z$ +1 – Mikasa Jul 25 '13 at 07:36
  • 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 07:45
  • Thanks @TobiasKildetoft, I shall edit in short. – DonAntonio Jul 25 '13 at 07:47
  • @DonAntonio I dont get how do you conclude the uniqueness ? – Arthur Feb 03 '23 at 14:32
  • @Franklin Wow, this is old...but still pretty basic: beginning in the third line we take a cyclic group of order $;n;$ and then we in fact show it is a subgroup of the first group we considered, namely $;\langle\frac1n\rangle;$ . And thus both groups are one and the same as they both are cyclic of the same order. – DonAntonio Feb 04 '23 at 17:19
  • @DonAntonio "And thus both groups are one and the same as they both are cyclic of the same order": I don't get this? Actually $\frac ab=\frac{ak}n\in \Bbb Z +\frac {ak}{n}\in{\Bbb Z+\frac 1n,\Bbb Z+\frac 2n,\cdots, \Bbb Z}=\left\langle;\frac1n+\Bbb Z;\right\rangle$ – Arthur Feb 05 '23 at 15:26
  • @Franklin As already stated: they're the same as they are both cyclic of the same order and one of them in contained in the other one, of course. – DonAntonio Feb 09 '23 at 20:00