Exercise 2.1.13 (if $H \leq \mathbb{Q}$ and $x \in H \setminus\{0\} \Rightarrow 1/x \in H$, then $H = \{0\}$ or $\mathbb{Q}$)

Proof. Let $H$ be a subgroup of the additive group $\mathbb{Q}$ with the property that

Assume that $H\ne \{0\}$. We want to prove that $H=\mathbb{Q}$.

(a)

We show first that for all $x\in \mathbb{Q}$ and for all $n\in \mathbb{Z}$, $$\begin{array}{lll}\hfill x\in H\Rightarrow \mathit{nx}\in H.& & \hfill \text{(2)}\end{array}$$

Let $x$ be any element in $H$. Since $H$ is a subgroup of $\mathbb{Q}$, $0\cdot x=0\in \mathbb{Q}$. Assume that $\mathit{nx}\in$ for some $n\in \mathbb{N}$. Then $x\in H$ and $\mathit{nx}\in H$ imply $(n+1)x=\mathit{nx}+x\in H$. The induction is done, which proves that $\mathit{nx}\in H$ for all $n\in \mathbb{N}$. Moreover, $H$ being a subgroup, $\mathit{nx}\in H$ implies $(-n)x=-\mathit{nx}\in H$, therefore (1) is true for every $n\in \mathbb{Z}$.

(b)

Let $x$ be any element of $\mathbb{Q}$, and $y$ be any element of $H$.

If $x=0$ or $y=0$, then $\mathit{xy}=0\in H$. We may suppose now that $x\ne 0$ and $y\ne 0$.

Write $x=\frac{p}{q}$, where $p\in {\mathbb{Z}}^{\text{∗}}$ and $q\in {\mathbb{N}}^{\text{∗}}$. Then $\mathit{py}\in H$ by (2) and $\mathit{py}\ne 0$, thus $\frac{1}{\mathit{py}}\in H$ by (1). Hence $\frac{q}{\mathit{py}}\in H$ by (2) anew, so $\mathit{xy}=\frac{p}{q}y\in H$ by (1). This shows that for all $x,y$,

$$\begin{array}{lll}\hfill (x\in \mathbb{Q},y\in H)\Rightarrow \mathit{xy}\in H.& & \hfill \text{(3)}\end{array}$$

In particular, $H$ is stable by product.

(c)

Since $H\ne \{0\}$, there is some element $x_0\in H$ such that $x_0\ne 0$. Then $1∕x_0\in H$ by (1). Using (3) we obtain $x_0\cdot (1∕x_0)\in H$, so $$\begin{array}{lll}\hfill 1\in H.& & \hfill \text{(4)}\end{array}$$

(d)

If $x$ is any element of $\mathbb{Q}$, since $1\in H$, (3) shows that $x=x\cdot 1\in H$ (and $H\subseteq \mathbb{Q}$ by definition). Therefore $$H=\mathbb{Q}.$$

Conclusion: If $H$ is a subgroup of $(\mathbb{Q},+)$ with the property that $1∕x\in H$ for every nonzero element $x$ of $H$, then $H=\{0\}$ or $H=\mathbb{Q}$. □