Exercise 3.1.15 ($\mathbb{Q}/\mathbb{Z}$ is divisible)

Proof. Let $G$ be a divisible abelian group, and $H$ a proper subgroup of $G$ (in multiplicative notations). Then $G$ is not trivial, and since $H$ is a proper subgroup, $G∕H$ is not trivial. Moreover, since $G$ is abelian, $G∕H$ is also abelian.

Let $\bar{a}=\mathit{aH}$ be any element of $G∕H$, where $a\in G$, and let $k$ be any nonzero integer. Since $G$ is divisible, there is some element $x\in G$ such that $a=x^k$. Then

where $\bar{x}=\mathit{xH}\in G∕H$. Therefore $G∕H$ is a divisible abelian group.

In particular, since $\mathbb{Q}$ is a divisible abelian group (cf. Exercise 2.4.19), and $\mathbb{Z}$ a proper subgroup of $\mathbb{Q}$, then $\mathbb{Q}∕\mathbb{Z}$ is divisible. □