Answer to Question #163218 in Abstract Algebra for K

Question

Let G be a group with operation * and identity e. Prove that if

x * x = e for all x ∈ G, then G is abelian. Hint: consider

( a * b) * (a*b )

Answer 1

Solution:

Proof:

Let a and b be arbitrary elements of G.

We show that a ∗ b = b ∗ a.

Let us consider the element a ∗ b ∈ G.

Since every element x of G satisfies x ∗ x = e, we have,

(a ∗ b) ∗ (a ∗ b) = e

a ∗ [(a ∗ b) ∗ (a ∗ b)] = a ∗ e (multiply both sides by a on left)

[a ∗ (a ∗ b)] ∗ (a ∗ b) = a (associative and identity properties)

[(a ∗ a) ∗ b)] ∗ (a ∗ b) = a (associative property)

[e ∗ b] ∗ (a ∗ b) = a (a ∗ a = e) b ∗ (a ∗ b) = a (identity property)

[b ∗ (a ∗ b)] ∗ b = a ∗ b (multiply both sides by b on right)

b ∗ [(a ∗ b) ∗ b] = a ∗ b (associative property)

b ∗ [a ∗ (b ∗ b)] = a ∗ b (associative property)

b ∗ [a ∗ e] = a ∗ b (b ∗ b = e) b ∗ a = a ∗ b (identity property)

This shows a ∗ b = b ∗ a, so G is abelian.

Hence, proved.