Question

Prove that G={-1,1,i,-i} is an Abelian group under multiplication?​

Answer 1

Show that G=[1,-1,i,-i] is a group under usual multiplication of complex number. G2⇒ From the first coloumn (or row), we see that l is an identity element. Hence, 1 E G is an identity element. Hence G is a group under multiplication.

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Answer:

In the attachment, I have formed cayley's table to verify the following axioms.

1.Closure axiom:

From the cayley's table, it is clear that

all the elements in ther cayleys table are

elements of G.

This implies G is closed under multiplication.

Hence closure axiom is true.

2.Associative axiom:

Multiplication of complex numbers is always associative.

Hence associative axiom is true .

3. Identity axiom:

Clearly the identity element is 1 and

1 ∈ G

4. Inverse axiom:

From the table,

Inverse of 1 is 1

Inverse of -1 is -1

Inverse of i is -i

Inverse of -i is i

since each element of G has unique inverse, Inverse axiom is true.

5. Commutative axiom:

Clearly multiplication of complex number is always commutative.

Hence commutative axiom is true.

Therefore, G is an abelian group under multiplication.

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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 .