I want to prove that $\mathbb Q^* = \mathbb Q\setminus\{0\}$ under multiplication is not cyclic, and I've seen this proof Nonzero rationals under multiplication are not a cyclic group, but I wanted to check if the proof I made up for this problem was also a valid proof (or if not, why). My proof is as follows: Suppose to the contrary that $\exists$ an element $(a/b) \in Q^*$ such that $<a/b> = Q^*$, where a, b $\neq$ 0, and a and b are relatively prime integers. Then, $\forall q \in Q^*$, $\exists$ $n \in Z$ such that $(a/b)^{n}$ = $a^{n}/b^{n}$ = q. Suppose q is an integer. Then $q^{n}b^{n} = a^{n}$, and thus $b^{n}|a^{n}$, so $b|a$. This is a contradiction with a and b being relatively prime, and thus $Q^*$ is not cyclic.
Thanks in advance for any help!
