TITLE: Subgroup of $\mathbb{Q}$ with finite index
QUESTION [34 upvotes]: Consider the group $\mathbb{Q}$ under addition of rational numbers. If $H$ is a subgroup of $\mathbb{Q}$ with finite index, then $H = \mathbb{Q}$.
I just saw this on our exam earlier and was stumped on how to show this. Any ideas?

REPLY [30 votes]: 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: $H = G$.