Exercise 3.2.21($\mathbb{Q}$ and $\mathbb{Q}/ \mathbb{Z}$ have no proper subgroup of finite index)

Proof. Suppose for the sake of contradiction there there is some proper subgroup $H\le \mathbb{Q}$ such that $|\mathbb{Q}:H|<\infty$. Since $\mathbb{Q}$ is abelian, $H⊴\mathbb{Q}$, and $\mathbb{Q}∕H$ is a finite abelian group, which is divisible by Exercise 3.1.15. But by Exercise 2.4.19, no finite abelian group is divisible. This is a contradiction, therefore $\mathbb{Q}$ has no proper subgroup of finite index.

Suppose for the sake of contradiction there there is some proper subgroup $\bar{H}\le \mathbb{Q}∕\mathbb{Z}$ such that $|(\mathbb{Q}∕\mathbb{Z}):\bar{H}|<\infty$. Let $\{\overline{a_1},\overline{a_2},\dots ,\overline{a_n}\}$ be a complete set of representatives of the cosets modulo $\bar{H}$, so that

Note that $\bar{H}=H∕\mathbb{Z}$ for some subgroup $H$ of $\mathbb{Q}$ such that $\mathbb{Z}\le H\le \mathbb{Q}$.

Let $a\in \mathbb{Q}$. Then $\bar{a}=\overline{a_i}+\bar{h}$ for some $i\in [[1,n]]$ and some $h\in H$. Then $a-a_i-h\in \mathbb{Z}$, so that $a=n+a_i+h$ for some $n\in \mathbb{Z}$. Since $\mathbb{Z}\le H$, $a=a_i+h^{\prime }$, where $h^{\prime }=n+h\in H$. This shows that $H$ has a finite index $|\mathbb{Q}:H|\le n$. By the first part of the proof, $H$ is not a proper subgroup, so $H=\mathbb{Q}$, and $\bar{H}=\mathbb{Q}∕\mathbb{Z}$ is not a proper subgroup. This is a contradiction, which shows that $\mathbb{Q}∕\mathbb{Z}$ has no proper subgroup of finite index.

(Alternatively, we may use the Third Isomorphism Theorem: $\mathbb{Q}∕H=(\mathbb{Q}∕\mathbb{Z})∕(H∕\mathbb{Z})$ is a finite group if $\bar{H}=H∕\mathbb{Z}$ has a finite index in $\mathbb{Q}∕\mathbb{Z}$.) □