Question

Prove that the set consisting of the fourth root of unity forms an abelian group with respect to multiplication

Answer 1

Step-by-step explanation:

et i=−1−−−√. Prove that the four roots of unity {1,−1,i,−i} form an abelian multiplicative group.

I know that abelian group is a group with commutative property. But I'm not sure how to prove this one.

Answer 2

Note :

• Group : An algebraic system (G,*) is said to be a group if the following condition are satisfied :

1. G is closed under *
2. G is associative under *
3. G has a unique identity element
4. Every element of G has a unique inverse in G

• Moreover , if a group (G,*) also holds commutative property , then it is called commutative group or abelian group .

Solution :

Given :

G = { -1 , 1 , i , -i }

To prove :

G is an abelian group .

Proof :

For Cayley's table (composition table) please refer to the attachment .

1) Closure property :

All the elements of the composition table are the elements of G . ie. a•b ∈ G ∀ a , b ∈ G .

2) Associative property :

We know that , the multiplication of the complex numbers is associative , thus a•(b•c) = (a•b)•c ∀ a , b , c ∈ G .

3) Existence of identity :

We have 1 ∈ G such that 1•a = a•1 = a ∀ a ∈ G .

Thus , 1 is the identity element in G .

4) Existence of inverse element :

∀ a ∈ G , there exists a⁻¹ such that a•a⁻¹ = a⁻¹•a = 1 where a⁻¹ is called the inverse of a .

Here ,

1⁻¹ = 1

-1⁻¹ = -1

i⁻¹ = -i

-i⁻¹ = i

5) Commutative property :

The Cayley's table is symmetrical about the principal diagonal , thus G is commutative , ie. a•b = b•a ∀ a , b ∈ G .

Hence , G is an abelian group .