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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.2012-08-14

3 Answers 3

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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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    Hi, this might not be the venue for saying this, but I thought you might be interested in knowning that [the wikipedia article of divisible group](https://groupprops.subwiki.org/wiki/Divisible_group) requires abelian, maybe someone should edit that. (of course is not your problem, but since you are an active math contributor to the internet, you may want to set that article straight)2017-07-14
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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, $H=\mathbb{Q}.$