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I'm trying to work out this question:

Prove that if $H$ is a proper subgroup of $\mathbb{Q}$ then $\mathbb{Q}/H$ is infinite, but each of its elements have finite order.

I thought, for the first part, that I could assume for contradiction that $\mathbb{Q}/H$ is finite of order $n$, then for all $\dfrac{a}{b}\in\mathbb{Q}$, $\dfrac{a^n}{b^n}$ is in $H$. And somehow prove this is a contradiction. Any help?

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Note that $\mathbb{Q}$ is not a group under multiplication; it is a group under addition. So what you can conclude is that $n\frac{a}{b}=\frac{na}{b}$ is in $H$, not $\frac{a^n}{b^n}$. – Arturo Magidin May 16 '12 at 17:20
up vote 5 down vote accepted

Note that $\mathbb{Q}$ is a divisible group: for every $x\in \mathbb{Q}$ and every $n\gt 0$, there exists $y\in\mathbb{Q}$ such that $ny=x$.

Therefore, any quotient of $\mathbb{Q}$ is also divisible. Thus, if $H$ is a subgroup of $\mathbb{Q}$, then $\mathbb{Q}/H$ is divisible.

But if $\mathbb{Q}/H$ is finite, then there exists $n\gt 0$ such that $n\left(\frac{a}{b}+H\right) = H$ for every $\frac{a}{b}\in\mathbb{Q}$; thus, if there is a nonzero element $\frac{r}{s}+H$ in $\mathbb{Q}/H$, can you find a $y+H\in\mathbb{Q}/H$ such that $n(y+H) = \frac{r}{s}+H$? What can we conclude from that?

To show every element is of finite order, let $x\neq 0$ be such that $x\in H$, and let $y\in\mathbb{Q}$. Show that there exists a $z\in\mathbb{Q}$ such that $\langle x,y\rangle = z$; conclude that there is a nonzero power of $y$ that is equal to a power of $x$ to deduce that $y+H$ has finite order in $\mathbb{Q}/H$.

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Can you please explain how does it follow that if $\mathbb{Q}/H$ is finite, then there exists $n\gt 0$ such that $n\left(\frac{a}{b}+H\right) = H$ for every $\frac{a}{b}\in\mathbb{Q}$ ? – Balaji Rao May 17 '12 at 9:30
@BalajiRao: Lagrange's Theorem implies that if $G$ is a finite (additive) group, then for all $g\in G$ we have $ng=0$, where $n=|G|$. – Arturo Magidin May 17 '12 at 16:04
Thanks! I completely missed this. I was thinking that it was a conclusion drawn from the divisible group result applied to zero. – Balaji Rao May 18 '12 at 5:39

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