Remember that if $p$ is a prime and $z$ is a nonzero integer, we define the $p$-order of $z$ to be $n$, $\mathrm{ord}_p(z) = n$, if and only if $p^n|z$ but $p^{n+1}$ does not divide $z$. For a rational $\frac{r}{s}$, we let $\mathrm{ord}_p(\frac{r}{s}) = \mathrm{ord}_p(r) - \mathrm{ord}_p(s)$. Define the ord
Fix a prime $p$, and let $n$ be any positive integer. Define $H(p,n) = \{q\in \mathbb{Q}_{\gt0} \mid \mathrm{ord}_p(q)\in n\mathbb{Z}\}.$
$H$ is a multiplicative subgroup of $\mathbb{Q}$, since the $p$-order satisfies $\mathrm{ord}_p(rs) = \mathrm{ord}_p(r) + \mathrm{ord}_p(s)\quad\text{and}\quad \mathrm{ord}_p\left(\frac{1}{r}\right) = -\mathrm{ord}_p(r).$
Finally, note that $H$ is of finite index in : $\mathrm{ord}_p$ is an abelian group homomorphism from $\mathbb{Q}_{\gt0}$ onto $\mathbb{Z}$; $H$ is the pre-image of $n\mathbb{Z}$, hence the index of $H$ is equal to $n$.
This is just a way to make explicit the comment of Hagen Knaf: the basis of the positive rationals under multiplication can be taken to be the primes.