Prove that $(\mathbb{Q}^*, \times )$ is not cyclic (compare my solution to textbook solution)

Question

The following is an exercise from "Guide to Abstract Algebra 1st Ed, Carol Whitehead"

Prove that $(\mathbb{Q}^*, \times )$ is not cyclic.

Question: My own solution appears simpler than the textbook's solution. Which makes me ask where my solution is wrong.

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My Solution:

Members of the group $G = (\mathbb{Q}^{*}, \times)$ take the form $\frac{p}{q}$ where $p \in \mathbb{Z}, q \in \mathbb{Z}^{*}$.

In a cyclic group, the generator $g \in G$ is such that $g^m = 1$, the identity element, where $m$ is the finite order of the generator $g$.

We can see straight away that $g^m = (\frac{p}{q})^m = \frac{p^m}{q^m}\neq 1$ in general, except when $p=q$, the trivial case of the identity element.

Therefore the group is not cyclic.

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Textbook Solution:

Suppose $(\mathbb{Q}^{*}, \times) = < \frac{a}{b} > $, where $a \in \mathbb{Z}, b \in \mathbb{Z}^{*}$ and $gcd(a,b)=1$.

Then $\frac{1}{2} \in \mathbb{Q} \implies \frac{1}{2} = (\frac{a}{b})^r$, for some $r \in \mathbb{Z} \implies a^r = 1 \implies a= ± 1$

.. which is clearly false, since, e.g., $\frac{2}{3} \in \mathbb{Q}^{*}$.

Answer 1

In a cyclic group, the generator $g \in G$ is such that $g^m = 1$, the identity element, where $m$ is the finite order of the generator $g$.

Nothing in the definition of a cyclic group implies the generator has finite order. This is true about finite cyclic groups, but since $\mathbb{Q}^*$ is infinite, there's no reason to spend time checking whether it's a finite cyclic group.

It would suffice to stop there, since your initial plan of showing it's not a cyclic group by showing it's not a finite cyclic group is invalid, but the very next thing you say is also wrong:

We can see straight away that $g^m = (\frac{p}{q})^m = \frac{p^m}{q^m}\neq 1$ in general, except when $p=q$, the trivial case of the identity element.

There is one more element in $\mathbb{Q}^*$ with finite order. If you can find it and show/note $\mathbb{Q}^*$ is infinite, then you will have an alternative, correct proof that $\mathbb{Q}^*$ isn't cyclic.