TITLE: What is a set of generators for the multiplicative group of rationals?
QUESTION [7 upvotes]: I'm trying to think of a set of generators for the group $(\mathbb{Q^\times, *})$, which I think is countably generated (as opposed to finitely generated). (I wonder if such a set could be the primes of $\mathbb{Z}$ (together with $1$ and $-1),$ meaning the set $\{1, -1, \frac{1}{2}, -\frac{1}{2}, ...\}$ generates all nonzero rationals?)

REPLY [9 votes]: I assume your list continues on $\{\frac{1}{3},-\frac{1}{3},\frac{1}{5},-\frac{1}{5}, \ldots\}$.
You do indeed give a list of generators; but your list is redundant:

$1$ is contained in every group; it need not be listed
$-\frac{1}{2} = (-1) \cdot \frac{1}{2}$, so you only need to include one of each associate; it's traditional to take the positive one in your list and remove the negative one.

Also, it's traditional to list the primes as the generators, rather than their inverses. That is, the typical set of generators one uses for $\mathbb{Q}^\times$ is the set
$$ \{ -1, 2, 3, 5, 7, 11, \ldots \} $$
This is a minimal set of generators; the only relationship between its elements is 
$$(-1) = (-1)^{-1}$$