Prove that $H=0$ or $H=\mathbb{Q}$

Question

Let $H$ be a subgroup of the additive group of rational numbers with the property that $\dfrac{1}{x} \in H$ for every non-zero element $x$ of $H$. Prove that $H=0$ or $H=\mathbb{Q}$

Let $x \in \mathbb{Q}$. Then $x=\dfrac{a}{b},a,b\in \mathbb{Z},b \neq 0$. Then $\dfrac{1}{x}=\dfrac{b}{a}$. Then I stuck here.

Actually I don understand what the set $H$ is. Does it mean $H= \lbrace x \in \mathbb{Q}:\dfrac{1}{x} \in H\,\, \forall x \in H \rbrace$? If this is the case, then what should I do to show that $x$ is an element of $H$ ?

Answer 1

Suppose $H \neq 0$. Then there exists an $x = \frac a b$ in $H$. Thus $a = bx = \underbrace{x+x+\dots +x}_{\text{b times}} \in H$. Now $\frac 1 a \in H$ and by the same argument as before $1 \in H$.

Now we get to $\mathbb Z \subset H$. Take any $\frac p q \in \mathbb Q $. Then $q \in \mathbb Z \subset H$, thus $\frac 1 q \in H$. Now $\frac p q = \underbrace{\frac 1 q + \dots + \frac 1 q}_{\text{p times}} \in H$.

Note that we used that a group is closed under addition.

Answer 2

What you need to show is that if $x \in H$ then $y \in H$, for any two fractions $x, y$ and $x \neq 0$.

The first step is to observe that if $\frac{a}{b} \in H$ then $\frac{a}{b} + \dots + \frac{a}{b} = b \cdot \frac{a}{b} = a \in H$ and thus $\frac{1}{a} \in H$. Now, in the same way, show that $1 \in H$ and thus $\mathbb Z \subset H$. Then you should be able to construct arbitrary fractions easily.

Answer 3

If $H$ is a subgroup (by hypothesis) of $(\mathbb{Q}, +)$ and it is not empty, then at least $1 \in H$. If $ 1 \in H, \ H \leq (\mathbb{Q}, +) $ then the subgroup generated by 1 is in $H$. So every integer number $z \in \mathbb{Z}$ is in $H$, and by definition of $H$, $\frac{1}{z} \ z \in \mathbb{Z}$ too. a general element of $(\mathbb{Q}, +) $ is $\frac{a}{b} $, $a,b \in \mathbb{Z}$ which is equal to $ \frac{1}{b}+ \frac{1}{b}+ \dots + \frac{1}{b} $ $a$ times. Using the closure of $H$ under the addition the proof is complete.