# Showing A Certain Subgroup of $\mathbb{Q}$ must actually be $\mathbb{Q}$

I asked to show that if $H$ is a subgroup of the additive group $\mathbb{Q}$ with the property that $\frac{1}{x} \in H$ if $0 \neq x \in H$, then $H=0$ or $H = \mathbb{Q}$>. I already ruled out the first case by supposing the $H$ only contained the identity. I am wondering if my proof of the second part is correct. Here it is:

Suppose that $H$ has at least one other element besides $0$, call it $\frac{r}{s}$. By closure, we know $\frac{r}{s} + \dots + \frac{r}{s}$ (s-times is $r$, which is in $H$. Then $\frac{1}{r} \in H$. By closure, $\frac{1}{r} + \dots + \frac{1}{r}$ ($r$-times) is $1$. Since $H$ contains $1$, by $H$'s closure it must contain all integers.

Now, let $y \in H$ be some arbitrary integer. Then $\frac{1}{y} \in H$, and so $\frac{1}{y} + \dots + \frac{1}{y}$ ($x$-times, where $x$ is some arbitary integer) is $\frac{x}{y}$, an arbitrary rational number which must be in $H$ by closure.

Does this seem right?

• You seem to rely on the fact that $r$ and $s$ are non-negative when you write "$r$ times" and "$s$ times". So this only gives you that $\mathbb{Q}_+\subset H$ but it is easy to wrap it up from there. – Olivier Moschetta Sep 17 '16 at 13:42
• @OlivierMoschetta Nice remark! He can probably say "$|s|$ times" instead and change nothing else in his proof. – Scientifica Sep 17 '16 at 13:44
• Thanks everyone! And I will certainly take note of the difference between "$r$ times" and "$|r|$ times." – user193319 Sep 17 '16 at 13:59

## 3 Answers

Yes, your proof is correct. Well done!

• Thanks to whoever for the upvote, but the other answer deserves one more than mine ;) – Scientifica Sep 17 '16 at 13:46

This looks fine. You could reach $\frac xy$ quite directly from $\frac rs$ (if $x,y,r,s>0$) by adding $sy$ times, then taking reciprocal, then adding $xy$ times.

Your proof seems correct. However you can simplify it. Suppose $H\ne\{0\}$ and take $x=a/b\in H$, with $a,b\in\mathbb{Z}$, $a\ne 0$ and $b\ne0$.

Since $H$ is a subgroup, also $bx=a\in H$. By assumption, also $y=1/a\in H$ and therefore $ay=1\in H$.

It follows that $H$ contains every integer and so every reciprocal thereof, together with their multiples.