Generating subgroup of $\langle \mathbb{Q} \setminus \{0\},\cdot\rangle$ Does such a subgroup even exist?
My guess is That it is $\mathbb{Z} \setminus \{0\}$
Any help?
 A: A proper subgroup cannot generate the whole group. The set of nonzero integers is a generating set for the group $(\mathbb{Q}\setminus\{0\},{\cdot})$, because every nonzero rational can be written as $ab^{-1}$ with $a,b\in\mathbb{Z}\setminus\{0\}$.
However, there is a more “economical” generating set, namely the set of (positive) primes together with $-1$, because every nonzero rational number is, possibly up to the sign, a product of (positive) primes or reciprocal thereof.
This is a minimal set of generators, because removing even one element yields a set that doesn't generate the group.
A: $\Bbb Z\color{red}{\setminus}\{0\}$ is indeed a generating set for $\Bbb Q\color{red}{\setminus}\{0\}$. (This is not a subgroup.)
This is simply because any element of $\Bbb Q\color{red}{\setminus}\{0\}$ can be written as $pq^{-1}$ for some $p, q \in \Bbb Z\color{red}{\setminus}\{0\}$.
By definition of the subgroup generated by the set, any element of the above form is indeed present in the subgroup generated by $\Bbb Z\color{red}{\setminus}\{0\}$.

As Kenny Wong pointed out in the comments, you can actually make do with a smaller set.
Consider the set $P = \{p \in \Bbb N : p \text{ is a prime}\}.$
Then, the set $P \cup \{-1\}$ is a generating set of $\Bbb Q\color{red}{\setminus}\{0\}$ as well.
To see, we just note that any positive integer can be written as a product of elements of $P$. Including $-1$ lets us write any element of $\Bbb Z\color{red}{\setminus}\{0\}$ as an element of $P \cup \{-1\}$.
Since the former generates $\Bbb Q\color{red}{\setminus}\{0\}$, we are done.

Note that (as k.stm points out), the symbol '$/$' is typically used for quotient and not set-difference, which is what you probably wanted. The correct symbol for that is '$\setminus$'.
(This is why I've written it above in red.)
