Question

aOb=a+b+1 , prove that it forms an Abelian group​

Answer 1

Answer:

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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 :

The binary operation o is defined on the set G = Z is such that a o b = a + b + 1 .

To prove :

(G,o) is an abelian group .

Proof :

1) Closure property :

Let a , b ∈ G , then we have

a o b = a + b + 1

We know that ,

The sum of integers is again a integer , thus

→ a + b + 1 ∈ G

→ a o b ∈ G

→ (G,o) is closed .

2) Associative property :

Let a , b , c ∈ G , then we have

a o (b o c) = a o (b + c + 1)

= a + b + c + 1 + 1

= a + b + c + 2

(a o b) o c = (a + b + 1) o c

= a + b + 1 + c + 1

= a + b + c + 2

Clearly , a o (b o c) = (a o b) o c

→ (G,o) is associative .

3) Existence of identity element :

Let e be the identity element in (G,o) and let a ∈ G be arbitrary .

Then ,

By definition of identity element , we have

e o a = a o e = a

Now , e o a = a

→ e + a + 1 = a

→ e = -1 ∈ G

Also , a o e = a

→ a + e + 1 = a

→ e = -1 ∈ G

Hence ,

-1 ∈ (G,o) is the identity element .

4) Existence of inverse element :

Let b be the Inverse of the element a ∈ (G,o) .

Then ,

By definition of inverse element , we have

a o b = b o a = e

Now , a o b = e

→ a + b + 1 = -1

→ b = (-2 - a) ∈ G

Also , b o a = e

→ b + a + 1 = -1

→ b = (-2 - a) ∈ G

Hence ,

(-2 - a) is the inverse element of a ∈ (G,o) .

5) Commutative property :

Let a , b ∈ G , then we have

a o b = a + b + 1

= b + a + 1

= b o a

→ (G,o) is commutative .

Hence ,

(G,o) is an abelian group .