Question

prove that x=x^_1 for all x in the group G, then G is abelian?​

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

• 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 x = x⁻¹ ∀ x ∈ G .

To prove :

G is abelian .

Proof :

Let x , y ∈ G be arbitrary elements , then

x = x⁻¹ and y = y⁻¹

Now ,

xy = x⁻¹y⁻¹

→ xy = (yx)⁻¹ (Reversal law) ........(1)

Now ,

Since G is a group , then

x ∈ G and y ∈ G → yx ∈ G (Closure property)

Also ,

If yx ∈ G , then we have

yx = (yx)⁻¹ .........(2)

Now ,

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

→ xy = yx ∀ x , y ∈ G

Hence , G is abelian group .