# Is the group of positive rational numbers under * cyclic?

1. Feb 15, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
Is the group of positive rational numbers under multiplication a cyclic group.

2. Relevant equations

3. The attempt at a solution
So a group is cyclic if and only if there exists a element in G that generates all of the elements in G.
So the set of positive rational numbers would be cyclic if we could find a fraction a/b such that (a/b)^n, where n is an integer, generates all elements of positive Q.

I feel like the group is not cyclic, which means that I would have to prove that for all elements in positive Q, they don't generate positive Q. I am not sure exactly how to approach this... For (a/b)^n, do I have to find another fraction in terms of a and b such that (a/b)^n can't equal that expression for any n?

2. Feb 15, 2017

### Mr Davis 97

What does $<\frac{1}{2}\; , \;\frac{1}{3}>$ mean? I thought that one only looks at what subgroup is generated by a single element, in this case a single fraction

3. Feb 15, 2017

### Staff: Mentor

This should work. Solve $(\frac{a}{b})^n=2$ and show that you can't get $(\frac{a}{b})^m=3$ .

4. Feb 15, 2017

### Staff: Mentor

You beat me. I confused addition and multiplication. In general $<a>$ meant to be the group generated by $a$, so $<a>=\{a^n\,\vert \,n\in \mathbb{Z}\}$ but I was wrong with the example.