Question

(G,*) where G=Z(integer) and bineary operation * is defined a * b = a + b +. 1 prove that (G,*) is abelian group and solve 3 *a= 2^-1 in G

Answer 1

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 * is defined on the set G = Z is such that a*b = a + b + 1 .

To prove :

(G,*) is an abelian group .

Proof :

1) Closure property :

Let a , b ∈ G , then we have

a*b = a + b + 1

We know that ,

The sum of integers is again a integer , thus

→ a + b + 1 ∈ G

→ a*b ∈ G

→ (G,*) is closed .

2) Associative property :

Let a , b , c ∈ G , then we have

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

= a + b + c + 1 + 1

= a + b + c + 2

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

= a + b + 1 + c + 1

= a + b + c + 2

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

→ (G,*) is associative .

3) Existence of identity element :

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

Then ,

By definition of identity element , we have

e*a = a*e = a

Now , e*a = a

→ e + a + 1 = a

→ e = -1 ∈ G

Also , a*e = a

→ a + e + 1 = a

→ e = -1 ∈ G

Hence ,

-1 ∈ (G,*) is the identity element .

4) Existence of inverse element :

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

Then ,

By definition of inverse element , we have

a*b = b*a = e

Now , a*b = e

→ a + b + 1 = -1

→ b = (-2 - a) ∈ G

Also , b*a = e

→ b + a + 1 = -1

→ b = (-2 - a) ∈ G

Hence ,

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

5) Commutative property :

Let a , b ∈ G , then we have

a*b = a + b + 1

= b + a + 1

= b*a

→ (G,*) is commutative .

Hence ,

(G,*) is an abelian group .