# Roots of Unity as a group

• Aug 27th 2011, 02:49 PM
dwsmith
Roots of Unity as a group
Let $G=\{z\in\mathbb{Z}: \ z^n=1, \ n\in\mathbb{N}\}$.
Prove $(G,*)$ is an abelian group under multiplication.

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

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

How do I find the inverse?
• Aug 27th 2011, 03:21 PM
Plato
Re: Roots of Unity as a group
Quote:

Originally Posted by dwsmith
Let $G=\{z\in\mathbb{Z}: \ z^n=1, n\in\mathbb{N}\}$.
Prove $(G,*)$ is an abelian group under multiplication.
$1+0\mathbf{i}$ is the identity.
How do I find the inverse?

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

$\exp \left( {\frac{{9\pi i}}{5}} \right)$ is its inverse.
• Aug 27th 2011, 03:52 PM
dwsmith
Re: Roots of Unity as a group
Quote:

Originally Posted by Plato
Here is a hint:
$\exp \left( {\frac{{\pi i}}{5}} \right)$ is a $10^{th}$ root of unity.

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

So the inverse is
$\exp\left(\frac{2\pi i(n-k)}{n}\right)$
• Aug 27th 2011, 03:58 PM
dwsmith
Re: Roots of Unity as a group
For associative and commutative, that is trivial true since $\mathbb{C}$ is a field, correct?
• Aug 27th 2011, 04:05 PM
Plato
Re: Roots of Unity as a group
Quote:

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

Well it is a field after all!