Abstract Algebra - roots of unity

[kathrynag]

Homework Statement

I want to find out if the sixth root of unity is a subgroup of the complex numbers with multiplication.

Homework Equations

The Attempt at a Solution

I know it's true but my problem is getting there.
I know the sixth root of unity must be closed under the binary operation of G. So, I need to show that the result is still a sixth root of unity. My problem is getting to that point.
The identity element is in the sixth root of unity. The identity element would be 1, so this is true.
I need to show that there is an inverse. The sixth root of unity includes both positive and negative values, so this exists.

 

[Office_Shredder]

If x is a 6th root of unity, -x is NOT its inverse. This is a group under multiplication, not addition, so you need to show that 1/x is also a sixth root of unity.

As an example of how to do closed under multiplication, if a⁶ = 1 = b⁶, (ab)⁶ = a⁶b⁶

and you can do inverses in a similar fashion. Just apply the definition of sixth root of unity

 

[kathrynag]

If x is a 6th root of unity, -x is NOT its inverse. This is a group under multiplication, not addition, so you need to show that 1/x is also a sixth root of unity.

As an example of how to do closed under multiplication, if a⁶ = 1 = b⁶, (ab)⁶ = a⁶b⁶

and you can do inverses in a similar fashion. Just apply the definition of sixth root of unity

I'm still not quite sure.
let x be a root of unity, so let x=1
Then (1/x)^6=1=x