Subgroup of $\mathbb{Q}$ with finite index

Solution 1:

Show that if $[\Bbb Q:H]=n$, $nq\in H$ for every $q\in\Bbb Q$. Conclude that $n\Bbb Q\subseteq H$. But $n\Bbb Q=\Bbb Q$, so $H=\Bbb Q$.

Solution 2:

A group $(G,+)$ (not necessarily commutative, although as a concession to the special case the OP has asked about, I am writing it "additively") is divisible if for every $x \in G$ and positive integer $n$, there is $y \in G$ with $ny = x$.

Here are two easy but important facts:

1) A quotient of a divisible group is divisible.

2) The only finite divisible group is the trivial group.

Applying this to $G = \mathbb{Q}$ and a finite index subgroup $H$, we get that $G/H$ is finite and divisible, hence trivial: $H = G$.