TITLE: Nonzero rationals under multiplication are not a cyclic group
QUESTION [18 upvotes]: Are nonzero rationals under multiplication cyclic?

Here's my thinking: 
They are not. The generator must be a rational $q = a/b$, $a$, $b$ integers with no common factors. Assume $a/b$ generates $\mathbb Q\setminus \{0\}$, then $q^n = a^n/b^n$ or $(1/q)^n = b^n/a^n$. This is impossible since we can find a prime (there're infinitely many of them) that doesn't divide either $a$ or $b$, which have a finite number of prime factors. Not sure if it proves it though.

REPLY [3 votes]: The subgroups $\langle \frac{1}{2}\rangle=\{ 2^n\colon n\in\mathbb{Z}\}$ and $\langle \frac{1}{3}\rangle=\{ 3^n\colon n\in\mathbb{Z}\}$ of $\mathbb{Q}^{\times}$ intersect trivially, but any two (non-identity) subgroups of infinite cyclic group always intersect non-trivially.