Show that the only subgroup of ($\mathbb{Q}$; +) of finite index is $\mathbb{Q}$ itself where + is the usual addition.
I have seen this question in my exam of abstract algebra. As I know the index here means the number of different cosets that can be generated from a group. Any idea
Let $A$ be a nontrivial subgroup of $\Bbb Q$, and $0\ne q\in A$.
For simplicity, let $B:=\frac1qA$, so that $\Bbb Z\subseteq B$ and $B$ has finite index iff $A$ has.
Then, for any $1<b\in\Bbb Z$, consider the elements $\frac 1b,\ \frac 1{b^2}, \ \frac 1{b^3},\dots$.
If $B$ is of finite index, these elements can't all be in different cosets, hence there are $n<m$ such that $\frac{b^{m-n}-1}{b^m}\ =\ \frac1{b^n}-\frac1{b^m} \, \in B$.
Now, $b^{m-n} - 1$ is coprime to $b^m$, so by Bezout's identity, there are integers $u, v$ such that $ub^m+v(b^{m-n} - 1)=1$, which implies $\frac1{b^m}\in B$, thus $\frac1b\in B$.
Since $b$ was arbitrary, we get $\Bbb Q=B$, and thus also $A=qB=\Bbb Q$.