Roots of Unity as a group

Let $\displaystyle G=\{z\in\mathbb{Z}: \ z^n=1, \ n\in\mathbb{N}\}$.

Prove $\displaystyle (G,*)$ is an abelian group under multiplication.

This makes sense since the units $\displaystyle \mathbb{Z}_p$ are group under multiplication.

$\displaystyle 1+0\mathbf{i}$ is the identity.

How do I find the inverse?

Re: Roots of Unity as a group

Quote:

Originally Posted by

**dwsmith** Let $\displaystyle G=\{z\in\mathbb{Z}: \ z^n=1, n\in\mathbb{N}\}$.

Prove $\displaystyle (G,*)$ is an abelian group under multiplication.

$\displaystyle 1+0\mathbf{i}$ is the identity.

How do I find the inverse?

Here is a hint:

$\displaystyle \exp \left( {\frac{{\pi i}}{5}} \right)$ is a $\displaystyle 10^{th}$ root of unity.

$\displaystyle \exp \left( {\frac{{9\pi i}}{5}} \right)$ is its inverse.

Re: Roots of Unity as a group

Quote:

Originally Posted by

**Plato** Here is a hint:

$\displaystyle \exp \left( {\frac{{\pi i}}{5}} \right)$ is a $\displaystyle 10^{th}$ root of unity.

$\displaystyle \exp \left( {\frac{{9\pi i}}{5}} \right)$ is its inverse.

So the inverse is

$\displaystyle \exp\left(\frac{2\pi i(n-k)}{n}\right)$

Re: Roots of Unity as a group

For associative and commutative, that is trivial true since $\displaystyle \mathbb{C}$ is a field, correct?

Re: Roots of Unity as a group

Quote:

Originally Posted by

**dwsmith** For associative and commutative, that is trivial true since $\displaystyle \mathbb{C}$ is a field, correct?

Well it is a field after all!