The `n`th roots of unity form a finite abelian group (under regular complex number multiplication).

It's finite because the equation `z^n=1` has exactly `n` solutions. Even without knowing there are exactly `n`, it can still be said that there are *at most* `n`, which is really all we need to know to say it's a finite group.

It's abelian because the `n`th roots of unity form a subset of the complex numbers, and the complex numbers are abelian with respect to multiplication.

Most importantly, they form a group. We have to show closure, existence of an identity and inverses, and associativity.

Closure: If `x` and `y` are `n`th roots of unity, then `x^n=1` and `y^n=1`. In that case, `(xy)^n=x^ny^n=1*1=1,` so `xy` is also an

`n`th root of unity. This shows closure.

Identity: Since 1 is a root of unity, and 1 is the multiplicative identity for the complex numbers, the identity element is a root of unity.

Inverses: If `x^n=1`, then `(1/x)^n=1/(x^n)=1/1=1,` so `1/x` is also a root of unity. Of course, `1/x*x=1,` which is the group's identity element.

Associativity: Since the roots of unity are complex numbers, and complex numbers are associative, the roots of unity automatically inherit this property from the complex numbers.

This completes the proof.