Question

Show that set of cube root of unity is a finite abelian group under multiplication

Answer 1

Answer:

ᖶᕼᓰᘉᖽᐸ ᕼᓰ ᖶᕼᗩᓰ ᗷᗩᕲᘿ ᗷᕼᗩᓰᖻᗩ Sᕼᗩᘺᕼᑘᗷ

Answer 2

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.