Exercise 1.1.25 (If $x^2 = 1$ for all $x \in G$ then $G$ is abelian)

Question

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

Richard Ganaye · Accepted Answer

\begin{proof} 
Assume that $x^2 = 1$ for all $x \in G$. If $a$ and $b$ are elements of $G$, then
$$1 = (ab)^2 = ab ab.$$
Therefore $a^{-1} = b a b$ and $b^{-1} a^{-1} = a b$. Since $a^2 = b^2 = 1$, then $a^{-1} = a$ and $b^{-1} = b$, so $ba = ab$.

 If $x^2 = 1$ for all $x \in G$ then $G$ is abelian.
\end{proof}

Exercise 1.1.25 (If $x^2 = 1$ for all $x \in G$ then $G$ is abelian)

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Exercise 1.1.25 (If $x^2 = 1$ for all $x \in G$ then $G$ is abelian)

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