Subgroups of the group of all roots of unity. Let $G=\mathbb{C}^*$ and let $\mu$ be the subgroup of roots of unity in $\mathbb{C}^*$. Show that any finitely generated subgroup of $\mu$ is cyclic. Show that $\mu$ is not finitely generated and find a non-trivial subgroup of $\mu$ which is not finitely generated.

I can see that a subgroup of $\mu$ is cyclic due to the nature of complex numbers and De-Moivre's theorem. The second part of this question confuses me though since, by the same logic, should follow a similar procedure. Perhaps it has something to do with the "finite" nature of the subgroup in comparison to $\mu$. Any assistance would be helpful !
Thank you in advanced.

The first part should probably include "Show that any f.g. subgroup of $\,\mu\,$ is cyclic finite...". From here it follows at once that $\,\mu\,$ cannot be f.g. as it isn't finite.
For a non-trivial non f.g. subgroup think of the roots of unit of order $\,p^n\,\,,\,\,p\,$  a prime, and running exponent $\,n\in\Bbb N\,$ ...
