TITLE: Is every finite subgroup of $C^*=$ set of all non-zero complex numbers cyclic?
QUESTION [9 upvotes]: Is every finite subgroup  of $C^*=$ set of all non-zero complex numbers cyclic?

I see that the set $A_n=\{z:z^n=1\}$ is a subgroup of $C^{*}$.
Any element of $A_n$ is a solution of $z^n=1$.Now the solutions of $z^n=1$ for any $n\in \Bbb N$ are $e^{\frac{{2ki\pi}}{{n}}};1\le k\le n$ and hence the subgroup $A_n$ is generated by $a=e^{\frac{{2ki\pi}}{{n}}}$ and hence cyclic.
But is it the case that any finite subgroup of $C^{*}$ is of the form $A_n$ for some $n$?
I am having problems to answer this question.If it is true then it answers the original question.
Please help.

REPLY [2 votes]: All finite subgroups are cyclic. In fact this statement is true for $F^*$ for any field $F$. However one can easily construct infinite subgroups that are not cyclic.  Consider all elements of finite order. This is a proper subgroup: it is precisely all the roots of unity. Because there are infinitely many prime numbers, one can see that not all numbers of the form $e^{2\pi i/p},\ p$ a prime, can be obtained as a power of a single complex number.