Why are the n-th roots of unity the only subgroup of order $n$ on $S^1$ Let $S^1 := \{ x \in \mathbb{C} : |x| = 1 \}$, $n \in \mathbb{N} $, then the n-th roots of unity are a subgroup of $S^1$ (under multiplicative group action). How do I prove that there cannot exist any other subgroup of the same order?
So far I've assumed that there exists such a subgroup, then all elements must be less than order $n$ otherwise it is trivial. I want to draw some form of contradiction based on the group action but am not sure how.
 A: This has to do with the fact that polynomials over a field are a UFD. Since any element, $\zeta$, of order $n$ by definition satisfies

$$x^n-1=0$$

We know that $(x-\zeta)$ is a factor of the polynomial $x^n-1$. But since there can be at most $n$ distinct roots to a polynomial of degree $n$, there are at most $n$ complex numbers which satisfy this polynomial. And obviously all of the usual suspects are present, i.e. $e^{2\pi i k/n}$ for $0\le k\le n-1$, so no other things can be roots, i.e. there are no other elements of order $n$.
If you're set on using the rotation action on $\Bbb C$, you know that all rotations are multiplication by $e^{2\pi i \theta}$, and if the $n^{th}$ power is the identity rotation, you know that $n\theta\in\Bbb Z$,i.e. there is some $k\in\Bbb Z$ so that $\theta = {k\over n}$. But then clearly if you choose $k<0$ or $k>n-1$ you can just add or subtract multiples of $n$ until it is in this range, and this represents the same complex number.
A: Every element of such a group is a root of the polynomial $x^n-1$. But there are at most $n$ roots of such a polynomial, which are the subgroup above. If a group of order $n$ existed which was not the above group, each element would have order dividing $n$, and therefore would be a root of $x^n-1$. But then it would be contained in $\{x\in\mathbb{C}|x^n=1\}$. So the subgroup is unique in order.
