If in a group $G$ , $(ab)^2=(ba)^2$ for all $a,b in G$ , then show that $G$ is abelian .

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Initially , I have placed $a^-1b$ in place of $b$ , and obtained $a^2b= ba^2$ for all $ain G$. Consequently, i have obtained $ab^2=b^2a$ . Then what to do ?




group-theory




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  • Perhaps that you meant $(ab)^2=a^2b^2$.
    – José Carlos Santos
    Jul 22 at 17:27






  • 3




    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 $(ab)^2=(ba)^2$
    – Martin R
    Jul 22 at 17:29







  • 1




    The quaternion group does obey the law $ab^2=b^2a$.
    – Lord Shark the Unknown
    Jul 22 at 17:31














up vote
1
down vote

favorite












Initially , I have placed $a^-1b$ in place of $b$ , and obtained $a^2b= ba^2$ for all $ain G$. Consequently, i have obtained $ab^2=b^2a$ . Then what to do ?




group-theory




share|cite|improve this question





















  • Perhaps that you meant $(ab)^2=a^2b^2$.
    – José Carlos Santos
    Jul 22 at 17:27






  • 3




    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 $(ab)^2=(ba)^2$
    – Martin R
    Jul 22 at 17:29







  • 1




    The quaternion group does obey the law $ab^2=b^2a$.
    – Lord Shark the Unknown
    Jul 22 at 17:31












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Initially , I have placed $a^-1b$ in place of $b$ , and obtained $a^2b= ba^2$ for all $ain G$. Consequently, i have obtained $ab^2=b^2a$ . Then what to do ?




group-theory




share|cite|improve this question













Initially , I have placed $a^-1b$ in place of $b$ , and obtained $a^2b= ba^2$ for all $ain G$. Consequently, i have obtained $ab^2=b^2a$ . Then what to do ?





group-theory





share|cite|improve this question












share|cite|improve this question




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edited Jul 22 at 18:13









Shaun

7,33892972




7,33892972









asked Jul 22 at 17:20









RABI KUMAR CHAKRABORTY

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244











  • Perhaps that you meant $(ab)^2=a^2b^2$.
    – José Carlos Santos
    Jul 22 at 17:27






  • 3




    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 $(ab)^2=(ba)^2$
    – Martin R
    Jul 22 at 17:29







  • 1




    The quaternion group does obey the law $ab^2=b^2a$.
    – Lord Shark the Unknown
    Jul 22 at 17:31
















  • Perhaps that you meant $(ab)^2=a^2b^2$.
    – José Carlos Santos
    Jul 22 at 17:27






  • 3




    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 $(ab)^2=(ba)^2$
    – Martin R
    Jul 22 at 17:29







  • 1




    The quaternion group does obey the law $ab^2=b^2a$.
    – Lord Shark the Unknown
    Jul 22 at 17:31















Perhaps that you meant $(ab)^2=a^2b^2$.
– José Carlos Santos
Jul 22 at 17:27




Perhaps that you meant $(ab)^2=a^2b^2$.
– José Carlos Santos
Jul 22 at 17:27




3




3




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 $(ab)^2=(ba)^2$
– Martin R
Jul 22 at 17:29





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 $(ab)^2=(ba)^2$
– Martin R
Jul 22 at 17:29





1




1




The quaternion group does obey the law $ab^2=b^2a$.
– Lord Shark the Unknown
Jul 22 at 17:31




The quaternion group does obey the law $ab^2=b^2a$.
– Lord Shark the Unknown
Jul 22 at 17:31










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You don't. A counterexample is the quaternion group of order $8$.






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    1 Answer
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    up vote
    11
    down vote













    You don't. A counterexample is the quaternion group of order $8$.






    share|cite|improve this answer

























      up vote
      11
      down vote













      You don't. A counterexample is the quaternion group of order $8$.






      share|cite|improve this answer























        up vote
        11
        down vote










        up vote
        11
        down vote









        You don't. A counterexample is the quaternion group of order $8$.






        share|cite|improve this answer













        You don't. A counterexample is the quaternion group of order $8$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 22 at 17:26









        Lord Shark the Unknown

        85.2k950111




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