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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asked Jul 22, 2018 at 17:20
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$\begingroup$ Perhaps that you meant $(ab)^2=a^2b^2$. $\endgroup$– José Carlos SantosJul 22, 2018 at 17:27
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4$\begingroup$ It 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 $(ab)^2=(ba)^2$ $\endgroup$– Martin RJul 22, 2018 at 17:29
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1$\begingroup$ The quaternion group does obey the law $ab^2=b^2a$. $\endgroup$– Angina SengJul 22, 2018 at 17:31
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1 Answer
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You don't. A counterexample is the quaternion group of order $8$.
answered Jul 22, 2018 at 17:26