Problem

Prove that if \(x^2 = 1 \) for all \(x \in G\) then \(G\) is abelian.

Solution

Given that \(x^2 = 1\) for all \(x \in G\), for any \(a,b \in G\),

$$ (ab)^2 = 1 $$

$$ abab = 1 $$

$$ a(abab)b = a1b $$

$$ a^2bab^2 = ab $$

$$ ba = ab $$

Hence, by definition of abelian group, \(G\) is an abelian group.

$$\tag*{$\blacksquare$}$$