Question

if G=(2,-1.i,-i).then prove that (G,.)is an abelian group​

Answer 1

Answer:

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Step-by-step explanation:

We say that a group structure (G, ⋆) is abelian if ... Proof. We will only prove that if ab = ac then b = c. Here inverse a−1 of a works as a ... Since |X| = |G|p− 1 and G is a p-group, 1 + ∑xi=(e,··· ,e) |Oxi | = lp for some.

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 .