Prove that a finite group of complex numbers are the $n$th roots of unity

Question

Prove that if we have a finite group of complex numbers they are the $n$th roots of unity for some finite integer, $n$.

I have seen some clunky ways of doing this but is there a very short simple solution?

Answer 1

Let $R_{n}$ be the set of $n$-th roots of unity. This has order at most $n$, as it is the set of roots of the polynomial $x^{n} - 1$. (I am using as little information as possible. Of course if $\omega_{n} = e^{i \frac{2 \pi}{n}}$ is a primitive $n$-th root of unity, then $R_{n}$ is cyclic of order $n$, generated by $\omega_{n}$.)

If your multiplicative group $G$ of the complex numbers has finite order $n$, then a consequence of Lagrange's Theorem tells you that for any $a \in G$ we have $a^{n} = 1$.

We have just shown that $G \subseteq R_{n}$. Since $G$ has $n$ elements, and $R_{n}$ has (at most) $n$ elements, they're equal.

Answer 2

Any element of this group must have absolute value 1, or its powers would all have different absolute value, which gives an infinite group.

Now let $\phi$ be the element of the group with the smallest positive angle from the real axis (which must exist, as the group is finite). Now $\phi$ is a primitive $n$th root of unity, as the group is finite, and we claim that the entire group is the group of $n$th roots of unity. Towards a contradiction, suppose that there is $\psi$ in the group which is not a power of $n$. Then it must lie, on the circle, inbetween $\phi^n$ and $\phi^{n+1}$ for some $n$. But then $\phi^{-n}\psi$ would have a smaller angle with the real axis than $\phi$, which is impossible by definition of $\phi$.