Is $(\mathbb{Q^+},\cdot)$ cyclic?

the case of the group $(\mathbb{Q^+},\cdot)$, is it cyclic (that is, a group that is generated by a single element)?

If each element is multiplied by another in a pair, then the only way to reverse that operation is division. But for fractions, it means for instance:

$\frac{1}{2}\cdot \frac{1}{4}=\frac{1}{8} $

but its reverse is:

$\frac{1}{2} / \frac{1}{4}=2 $

So this does not seem to be cyclic. But is this sufficient to show for non-cyclicity?

Thanks

To show it is not cyclic, you have to show that no generator can be chosen to generate the whole group. You showed in the comments that $1$ cannot be a generator, but this does not show the group is not cyclic. Obviously, $(\mathbb{Z}, +)$ is a cyclic group, but you can prove $0$ is not a generator.

Hint: Suppose $x = 2/3 \in G$ were a generator.

• If $n > 0$, then $(2/3)^n = 2^n/3^n$. Clearly $1/5$ is not equal to this fraction, because:

$$\frac{2^n}{3^n} = \frac{1}{5} \implies 5 \times 2^n = 3^n \implies 5 | 3^n$$ (which is false).

• Similarly if $n<0$ then $1/5$ is not in the image, for a similar reason.

This proves that $2/3$ is not a generator either. This argument can be generalised to an arbitrary generator $p/q$, or even any finite set of generators. Can you see how to generalise the argument?