Question

Show that cube roots of unity form finite abelian group under multiplication

Answer 1

Answer:

Step-by-step explanation:

Cube roots of unity are (1,ω,ω²)

Let G be the set  (1,ω,ω²).

To verify whether G is a roup 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 table it is clearly evident that e = 1 is the identity.

4)Existense 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 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.

×     1      ω      ω²

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

1      1      ω      ω²

ω    ω     ω²     1

ω²   ω²     1      ω

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

Answer 2

Answer:

We need to prove that cube roots of unity form finite abelian group under multiplication.

We know that cube roots of unity are (1, $w$, $w^{2}$)

Let (G,*) be the set of cube roots of unity that is (1, $w$, $w^{2}$)

To show that (G,*) is an abelian group , we first need to prove that (G,*) is a group.

Now, for all group elements a, b belonging to G, a$*$b also belongs to G.

That is,

$1*1=1\\1*w=w\\1*w^{2} =w^{2} \\w*w=w^{2} \\w^{2} *w^{2} =w$

This satisfies the closure property of groups.

We now need to find the existence of an identity in the group, that is, we need to find the existence of an identity, say, e, such that for all a belonging to G, $a*e=e*a=a$.

Here, clearly, 1 is the identity of the group G, since, for all elements a of G,

$a*1=1*a=a$.

We now need to find the existence of an inverse in the group, that is, we need to find the existence of an inverse, say, b, such that for all a belonging to G, $a*b=b*a=e$.

Now, $w*w^{2} =w^{2} *w=1$.

This clearly proves the existence of an inverse for every element in the group G.

Now, to check associativity property of the group G, for all elements a, b, c of G, $a*(b*c)= (a*b)*c$ must hold.

G satisfies the following properties.

Therefore, (G,*) is a group.

To prove that G is an abelian group, we need to prove that G is commutative.

Now, it can be easily shown that all elements of G satisfy this property.

Therefore, G is an abelian group.

Hence, cube roots of unity form finite abelian group under multiplication