Show that set of cube root of unity is a finite abelian group under multiplication Question Show that set of cube root of unity is a finite abelian group under multiplication in progress 0 Math Rylee 7 months 2021-10-13T17:19:11+00:00 2021-10-13T17:19:11+00:00 2 Answers 0 views 0

## Answers ( )

Answer:Step-by-step explanation:Cube roots of unity are (1,ω,ω²)

Let G be the set (1,ω,ω²).

× 1 ω ω²

===============

1 1 ω ω²

ω ω ω² 1

ω² ω² 1 ω

To verify whether G is a group or not,

1) Closure: Clearly, for all a,b belonging to G, a*b also belongs to G.

(illustrated in the table)

2)Associative: Clearly for all a,b and c belonging to G,

a*(b*c) = (a*b)*c

3)Existence of Identity: ‘e’ is called the identity of the group if for all a belong to G, a*e = e = e*a,

Now from the table, it is clearly evident that e = 1 is the identity.

4)Existence of Inverse:

If for all a belonging to G, there exists s a b such that a*b=e, then b is called the inverse of a.

Clearly, Inverse of 1 is 1, Inverse of ω is ω² and Inverse of ω² is ω.

Since all 4 properties of the group are satisfied G is a group.

Also, it is evident from the table that for all a,b belonging to G, we have a*b = b*a

Thus, G is commutative.

A commutative group is called as an abelian group.

Thus, cube roots of unity form a finite abelian group under multiplication.

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