I need to prove that $(\mathbb{Q}^*, \times)$ (i.e rationals, zero excluded, under multiplication) is not finitely generated.
So, suppose that G is finitely-generated.
That means there exist a finite set $S=\left\{x_1, x_2,...,x_n\right\}\subset \mathbb{Q}^*$ s.t. $\langle S\rangle=\mathbb{Q}^*$.
Now, all I need to do is to find a number $y\in\mathbb{Q}^*$ that cannot be generated by $S$.
Naively, I first thought of prime numbers, but that surely won't work.
Please help me out here... how can I find a number $y\in\mathbb{Q}^*$ that cannot possibly be multiplication of $x_i$-s from $S$?
Thanks in advanced!
Your idea about prime numbers is a good one! Let $T$ be the set of primes which appear in either the numerator or denominator of some element of $S$. Then $T$ is finite (why?) so there is some prime $p \not \in T$. Show by unique factorization that $p$ cannot be the product of powers of elements of $S$.