Can you check my proof that $\mathbb Q^* = \mathbb Q\setminus\{0\}$ under multiplication is not cyclic?

Question

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!

Note that $b\mid a$ and $\gcd(a,b)=1$ is actually possible, if $b=\pm1$; but this is a case you can quickly address. — Greg Martin, Oct 11 '22 at 04:34
Up to the detail mentionned by Greg Martin, your proof is essentially the same as the one in your link Nonzero rationals under multiplication are not a cyclic group and that proof was already aproved — Anne Bauval, Oct 11 '22 at 05:45
Actually we can make the more general statement that for any infinite field $K$, the multiplicative group $K^*$ is not cyclic. Check the two answers over here. — suckling pig, Oct 11 '22 at 07:04

score 1 · Answer 1 · answered Jun 18 '25 at 16:55

This works but it could be a lot simpler. An infinite cyclic group does not have torsion (i.e., elements of finite order other than the identity). However, $\mathbb Q^*$ clearly has torsion since $-1\neq 1$ but $(-1)^2=1$ (i.e., $-1$ is an element of order two).

Can you check my proof that $\mathbb Q^* = \mathbb Q\setminus\{0\}$ under multiplication is not cyclic?

1 Answers1