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