Let $G$ be a group and suppose that $a,bin G$ satisfy $a^2=1$ and $ab^2a=b^3$. Prove that $b^5=1$. [closed]












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I recently found an exercise about group presentation that I have no idea how to work out.



Can anyone help me?




Let $G$ be a group and suppose that $a,bin G$ satisfy $a^2=1$ and $ab^2a=b^3$. Prove that $b^5=1$.







group-theory group-presentation







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closed as off-topic by Shaun, Derek Holt, Gibbs, John B, José Carlos Santos Dec 10 '18 at 11:39


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Derek Holt, Gibbs, John B, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.


















    0












    $begingroup$


    I recently found an exercise about group presentation that I have no idea how to work out.



    Can anyone help me?




    Let $G$ be a group and suppose that $a,bin G$ satisfy $a^2=1$ and $ab^2a=b^3$. Prove that $b^5=1$.







    group-theory group-presentation







    share|cite|improve this question











    $endgroup$



    closed as off-topic by Shaun, Derek Holt, Gibbs, John B, José Carlos Santos Dec 10 '18 at 11:39


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Derek Holt, Gibbs, John B, José Carlos Santos

    If this question can be reworded to fit the rules in the help center, please edit the question.
















      0












      0








      0





      $begingroup$


      I recently found an exercise about group presentation that I have no idea how to work out.



      Can anyone help me?




      Let $G$ be a group and suppose that $a,bin G$ satisfy $a^2=1$ and $ab^2a=b^3$. Prove that $b^5=1$.







      group-theory group-presentation







      share|cite|improve this question











      $endgroup$




      I recently found an exercise about group presentation that I have no idea how to work out.



      Can anyone help me?




      Let $G$ be a group and suppose that $a,bin G$ satisfy $a^2=1$ and $ab^2a=b^3$. Prove that $b^5=1$.







      group-theory group-presentation




      group-theory group-presentation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 10 '18 at 6:07









      Shaun

      8,832113681




      8,832113681










      asked Dec 5 '18 at 8:38









      Censi LICensi LI

      3,6161937




      3,6161937




      closed as off-topic by Shaun, Derek Holt, Gibbs, John B, José Carlos Santos Dec 10 '18 at 11:39


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Derek Holt, Gibbs, John B, José Carlos Santos

      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by Shaun, Derek Holt, Gibbs, John B, José Carlos Santos Dec 10 '18 at 11:39


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Derek Holt, Gibbs, John B, José Carlos Santos

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
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          $begingroup$

          Since $a^2=1$, that is, $a=a^{-1}$, the relation $ab^2a=b^3$ implies $b^2a=ab^3$ and $ab^2=b^3a$. These then imply $b^2a=ab^3=b^3ab$, hence $a=bab$. Then $1=a^2=(bab)^2=dots$






          share|cite|improve this answer









          $endgroup$




















            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5












            $begingroup$

            Since $a^2=1$, that is, $a=a^{-1}$, the relation $ab^2a=b^3$ implies $b^2a=ab^3$ and $ab^2=b^3a$. These then imply $b^2a=ab^3=b^3ab$, hence $a=bab$. Then $1=a^2=(bab)^2=dots$






            share|cite|improve this answer









            $endgroup$


















              5












              $begingroup$

              Since $a^2=1$, that is, $a=a^{-1}$, the relation $ab^2a=b^3$ implies $b^2a=ab^3$ and $ab^2=b^3a$. These then imply $b^2a=ab^3=b^3ab$, hence $a=bab$. Then $1=a^2=(bab)^2=dots$






              share|cite|improve this answer









              $endgroup$
















                5












                5








                5





                $begingroup$

                Since $a^2=1$, that is, $a=a^{-1}$, the relation $ab^2a=b^3$ implies $b^2a=ab^3$ and $ab^2=b^3a$. These then imply $b^2a=ab^3=b^3ab$, hence $a=bab$. Then $1=a^2=(bab)^2=dots$






                share|cite|improve this answer









                $endgroup$



                Since $a^2=1$, that is, $a=a^{-1}$, the relation $ab^2a=b^3$ implies $b^2a=ab^3$ and $ab^2=b^3a$. These then imply $b^2a=ab^3=b^3ab$, hence $a=bab$. Then $1=a^2=(bab)^2=dots$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 5 '18 at 8:54









                BWWBWW

                9,25622237




                9,25622237















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