Initially , I have placed $a^{-1}b$ in place of $b$ , and obtained $a^2b= ba^2$ for all $a\in G$. Consequently, i have obtained $ab^2=b^2a$ . Then what to do ?
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Perhaps that you meant $(ab)^2=a^2b^2$. – José Carlos Santos Jul 22 '18 at 17:27
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4It is true under additional hypotheses, compare $G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$ or Prove that a ring is commutative if $(ab)^2=(ba)^2$ – Martin R Jul 22 '18 at 17:29
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1The quaternion group does obey the law $ab^2=b^2a$. – Angina Seng Jul 22 '18 at 17:31
