Question

Does the set of all `n`th roots of unity form a group under multiplication?

Answer 1

Yes

Explanation:

• Identity: `1` is the identity.

• Inverse: If `a` is an `n`th root of unity, then so is `1/a`, since:

  `(1/a)^n = 1/(a^n) = 1/1 = 1`

• Closure under product: If `a` is an `m`th root of unity and `b` an `n`th root of unity, then `ab` is an `mn`th root of unity:

  `(ab)^(mn) = (a^m)^n(b^n)^m = 1^n*1^m = 1*1 = 1`

• Associativity: Inherited from the complex numbers:

  `a(bc) = (ab)c " "` for any `a, b, c`

`color(white)()`
Footnote

The elements of this group are all the numbers of the form:

`cos theta + i sin theta`

where `theta` is a rational multiple of `pi`.