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

Question

I would like to know whether $(\mathbb{Q}/\mathbb{Z},+)$ has

$1$. Cyclic subgroup of every positive integer $n$?

$2$. Yes, unique one.

$3$. Yes, but not necessarily unique one.

$4$. Does not have cyclic subgroup of every positive integer $n$.

What I know about the given group is infinite group but every element has finite order. Please help how to proceed.

It helps to identify $\mathbb R/\mathbb Z$ with the unit circle in the complex plane. Then $\mathbb Q/\mathbb Z$ is its torsion subgroup, that is, the set of roots of unity (of all orders). — lhf, May 14 '13 at 12:49
@lhf I know $\mathbb{R}\to S^1, x\mapsto e^{i2\pi x}$, but did not understand about torsion. — Myshkin, May 14 '13 at 12:53
@Tsotsi, the torsion subgroup is the set of elements of finite order, that is, roots of unity in this case. — lhf, May 14 '13 at 14:24

score 4 · Answer 1 · edited May 14 '13 at 12:53

4

HINT. The order of an element $p/q \in \mathbb{Q}/\mathbb{Z}$ is the least $n$ such that $\underbrace{p/q+\cdots+p/q}_{n\text{ times}}$ is an integer.

edited May 14 '13 at 12:53

answered May 14 '13 at 12:23

Fredrik Meyer

20,690

$({p\over q}+\mathbb{Z})\times q \in\mathbb{Z}$ – Myshkin May 14 '13 at 12:32
okay so for every positive integer there is an element of that order I agree. – Myshkin May 14 '13 at 12:38
@Tsotsi: Yes, that is true. – Fredrik Meyer May 14 '13 at 14:46

Andrea Mori · Accepted Answer · 2013-05-14T15:22:44.357

2

Under the identification $$ \frac{\Bbb R}{\Bbb Z}\stackrel{\sim}\longrightarrow S^1\simeq \{\text{$z\in\Bbb C$ such that $||z||=1 $}\},\quad r+\Bbb Z\mapsto e^{2\pi i r} $$ the group ${\Bbb Q}/{\Bbb Z}$ is identified to the group $\mu(\Bbb C)$ of the complex roots of 1. Thus, a primitive $n$-th root $\zeta_n$ will correspond to an element in ${\Bbb Q}/{\Bbb Z}$ of exact order $n$.

Moreover, we know that every finite subgroup of the multiplicative group $\Bbb C^\times$ is cyclic. Thus there's just one subgroup of $\mu(\Bbb C)$ of order $n$, and same is true for ${\Bbb Q}/{\Bbb Z}$.

edited May 14 '13 at 15:22

answered May 14 '13 at 13:28

Andrea Mori

28,458

Of course you don't need this isomorphism to see all this. $\mathbb{Q}/\mathbb{Z}$ is a quite basic abelian group, where everything can be checked directly. – Martin Brandenburg May 14 '13 at 15:24
Of course! But the direct elementary approach was taken in the other answer. – Andrea Mori May 14 '13 at 15:25
@AndreaMori so $2$ is the correct answer?? – Myshkin May 31 '13 at 21:36
@TaxiDriver : Yes! – Andrea Mori Jun 01 '13 at 09:24

score 2 · Answer 3 · edited Apr 13 '17 at 12:21

As mentionned here, $\mathbb{Q} / \mathbb{Z} \simeq \bigoplus\limits_{p \in \mathbb{P}} \mathbb{Z}[p^{\infty}]$ (its canonical decomposition as a divisible group).

Let $p_1^{n_1} \dots p_r^{n_r}$ be the prime decomposition of $n$. For each $1 \leq j \leq r$, let $\displaystyle \zeta_j=\exp \left(i \frac{2k_j \pi}{p_j^{n_j}} \right)$ be an element of $\mathbb{Z}[p_j^{\infty}]$ with $k_j$ not divisible by $p_j$. Set $\zeta=\zeta_1 \dots \zeta_r$.

Now, $\text{ord}(\zeta)=\prod\limits_{i=1}^r \text{ord}(\zeta_i)= \prod\limits_{i=1}^r p_i^{n_i}=n$ and $\langle \zeta \rangle = \bigoplus\limits_{i=1}^r \langle \zeta_i \rangle$.

Of course, there is a choice for each $k_j$, but it doesn't change $\langle \zeta_j \rangle$. Indeed, let $\zeta_1=\exp \left(i \frac{2k_1\pi}{p^n} \right)$ and $\zeta_2= \exp \left( i \frac{2k_2\pi}{p_n} \right)$ where $k_1$ and $k_2$ are not divisible by $p$. Using Bezout equality, there are $u,v \in \mathbb{Z}$ such that $uk_1+vp^{n}=1$ hence $k_2=uk_1k_2+vk_2p^n$. So $\zeta_2= \zeta_1^{uk_1}$, ie. $\zeta_2 \in \langle \zeta_1 \rangle$. By symmetry, $\zeta_2 \in \langle \zeta_1 \rangle$. Therefore, $\langle \zeta_1 \rangle=\langle \zeta_1 \rangle$.

Finally, one deduce that $\mathbb{Q}/ \mathbb{Z}$ has a unique subgroup of order $n$.

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

3 Answers3

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