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Additive subgroups of the rationals which are closed under inversion are trivial


Let $H$ be a subgroup of the additive group of rational numbers with the property that if $x \in H$ is nonzero, then $1/x \in H$. Prove that $H = 0$ or $H = \mathbb{Q}$.


Solution: First, suppose there does not exist a nonzero element in $H$. Then $H = 0$.

Now suppose there does exist a nonzero element $a \in H$; without loss of generality, say $a = p/q$ in lowest terms for some integers $p$ and $q$ – that is, $\mathsf{gcd}(p,q) = 1$. Now $q \cdot \frac{p}{q} = p \in H$, and since $q/p \in H$, we have $p \cdot \frac{q}{p} \in H$. There exist integers $x,y$ such that $qx + py = 1$; note that $qx \in H$ and $py \in H$, so that $1 \in H$. Thus $n \in H$ for all $n \in \mathbb{Z}$. Moreover, if $n \neq 0$, $1/n \in H$. Then $m/n \in H$ for all integers $m,n$ with $n \neq 0$; hence $H = \mathbb{Q}$.

Linearity

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