Prove that $(\mathbb{Q}, +)$ has no subgroups of finite index

Question

To prove this I've thought of the following idea:

Suppose there exists $S$ a subgroup such that $\#(\mathbb{Q}:S)=n\in \mathbb{N}$. Then $\{a_1, a_2,..., a_n\}$ is a set with a representative from each class of $S$. My idea now is to define $x \in \mathbb{Q}$ such that $x-a_i \notin S \ \ \forall i \in \{ 1, 2, ..., n\}$ this way we see that the class of $x$ is different from all the classes with a representative $a_i$ therefore there has to be an infinite number of classes.

However I fail to find a way to define $x$ this way. I'd apreciate some help with the problem wether by giving a solution or a hint in case its possible to solve it this way.

Thanks in advance

Answer 1

If $\mathbf Q$ has a proper subgroup $H$ of finite index then it would have a maximal proper subgroup (a proper subgroup contained in no other) since $\mathbf Q/H$ is a nontrivial finite group, nontrivial finite groups have maximal subgroups, and there is a natural correspondence between the subgroups of $\mathbf Q$ containing $H$ and the subgroups of $\mathbf Q/H$.

Example 2.2 here shows $\mathbf Q$ has no maximal subgroups.