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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?

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You might want to look at the quotient group $\mathbb Q / H$ and note that it's finite. – JSchlather Aug 14 '12 at 3:26

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 $H=\Bbb Q$.

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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: $H = G$.

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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, $H=\mathbb{Q}.$

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