# Abstract Algebra - roots of unity (1 Viewer)

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#### kathrynag

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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

Staff Emeritus
Gold Member
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

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