Question

If (G,*) be a group and let a*a=e and a is unique prove that a*b=b*a for all a in G

Answer 1

Answer:

31. An element x of a group G is called idempotent if x ∗ x = x. Prove that any group G has exactly one

idempotent element.

Proof. Certainly e is idempotent, because e ∗ e = e, so G has at least one idempotent element, e. Could

there be others? Suppose x ∈ G is idempotent, which means x ∗ x = x. Then:

(x ∗ x) ∗ x

0 = x ∗ x

0

(multiply both sides by x

0 on right)

x ∗ (x ∗ x

0

) = e (associative and inverse properties)

x ∗ e = e (inverse property)

x = e (identity property)

Thus x = e, so G has exactly one idempotent element, and it is e.

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

• Abelian group : If a group (G,*) also holds commutative property , then it is called commutative group or abelian group , ie . if x*y = y*x ∀ x , y ∈ (G,*) , then the group G is said to be abelian .

Solution :

Given :

(G,*) is a group such that if a ∈ G , then a*a = e .

To prove :

a*b = b*a ∀ a , b ∈ G .

Proof :

Let a , b ∈ G , then we have

a*a = e and b*b = e

Also ,

If a , b ∈ G , then a*b ∈ G (closure property)

Thus , (a*b)*(a*b) = e

Now ,

→ (a*b)*(a*b) = e

→ (a*b)*(a*b)*(a*b)⁻¹ = e*(a*b)⁻¹

(pre-multiplying both sides with (a*b)⁻¹)

→ (a*b)*e = (a*b)⁻¹

→ a*b = b⁻¹*a⁻¹ ...........(1)

Now ,

→ (b*b)*(a*a) = e*e

→ b*(b*a)*a = e

→ b⁻¹*b*(b*a)*a = b⁻¹*e

(pre-multiplying both sides with b⁻¹)

→ e*(b*a)*a = b⁻¹

→ (b*a)*a = b⁻¹

→ (b*a)*a*a⁻¹ = b⁻¹*a⁻¹

(post-multiplyimg both sides with a⁻¹)

→ (b*a)*e = b⁻¹*a⁻¹

→ b*a = b⁻¹*a⁻¹ ............(2)

Now ,

From eq-(1) and (2) , we have

a*b = b*a ∀ a , b ∈ G .

Hence proved .