Subgroup of $\mathbb{Q}$ with finite index 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?
 A: 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 $H=\Bbb Q$.
A: 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: $H = G$.  
A: Since $[\mathbb{Q}:H]=n=\left|\dfrac{\mathbb{Q}}{H}\right|$ then for every $q\in\mathbb{Q}$, $n(q+H)=H$. But this means that $nq+H=H$ or $nq\in H$. Hence $n\mathbb{Q}\subseteq H$ but as stated above that $n\mathbb{Q}=\mathbb{Q}$. Thus, $H=\mathbb{Q}.$
A: This is a solution using only the definition of cosets.
Suppose $[\mathbb{Q}:H] = n$. Let $q \in \mathbb{Q}$ an arbitrary rational number and consider $\{H, q+H, 2q + H, \dots , nq + H\}$. In this set there are $n+1$ cosets, therefore there exists an integer $1 \leq k \leq n$ such that $H = kq + H$, or equivalently $kq \in H$.
Now, consider the rational number $q' = q\frac{1}{n!}$ with $q \in \mathbb{Q}$. By the preceding paragraph we know that there exists an integer $1 \leq k \leq n$ such that $kq' = q\frac{k}{n!} \in H$. But $\frac{k}{n!} = \frac{1}{1\cdot 2 \cdot \dots (k-1)\cdot(k+1)  \dots  n}$, i.e. $kq' = q\cdot\frac{1}{1\cdot 2 \cdot \dots (k-1)\cdot(k+1)  \dots  n}$. Therefore adding repeatedly $kq'$ (exactly $1\cdot 2 \cdot \dots (k-1)\cdot (k+1) \dots  
  n$ times) we get that $q \in H$ and $q \in \mathbb{Q}$ is arbitrary, therefore $H = \mathbb{Q}$.
