If $a,b in$ group $G$ such that $a^2=e, a*b^4*a=b^7$, prove that $b^{33}=e$












7












$begingroup$


$e$ is the identity of the group.



My understanding: To prove that $b^{33}=e$ is the same as proving $b^{34}=b$



Now, $a * b^4 * a=b^7$



$Rightarrow b^4= a*b^7*a=(a*b*a)^7$



This is how far I went. I'm stuck here. Please help.






group-theory







share|cite|improve this question











$endgroup$

















    7












    $begingroup$


    $e$ is the identity of the group.



    My understanding: To prove that $b^{33}=e$ is the same as proving $b^{34}=b$



    Now, $a * b^4 * a=b^7$



    $Rightarrow b^4= a*b^7*a=(a*b*a)^7$



    This is how far I went. I'm stuck here. Please help.






    group-theory







    share|cite|improve this question











    $endgroup$















      7












      7








      7





      $begingroup$


      $e$ is the identity of the group.



      My understanding: To prove that $b^{33}=e$ is the same as proving $b^{34}=b$



      Now, $a * b^4 * a=b^7$



      $Rightarrow b^4= a*b^7*a=(a*b*a)^7$



      This is how far I went. I'm stuck here. Please help.






      group-theory







      share|cite|improve this question











      $endgroup$




      $e$ is the identity of the group.



      My understanding: To prove that $b^{33}=e$ is the same as proving $b^{34}=b$



      Now, $a * b^4 * a=b^7$



      $Rightarrow b^4= a*b^7*a=(a*b*a)^7$



      This is how far I went. I'm stuck here. Please help.






      group-theory




      group-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jul 14 '15 at 9:17









      ajotatxe

      53.6k23890




      53.6k23890










      asked Jul 14 '15 at 9:15









      DiyaDiya

      1,2762925




      1,2762925






















          3 Answers
          3






          active

          oldest

          votes


















          14












          $begingroup$

          For any integer $n$ we have
          $$(ab^4a)^n=b^{7n}$$
          or
          $$ab^{4n}a=b^{7n}$$
          For $n=4$,
          $$ab^{16}a=b^{28}$$
          and for $n=7$,
          $$ab^{28}a=b^{49}$$
          Therefore
          $$b^{16}=b^{49}$$



          EDIT: With same technique, it can be shown that if $a^2=e$ and $ab^r=b^sa$ then $b^{r^2}=b^{s^2}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Another solution: Note that $a$ and $b^{11}$ commute; then take $n=11$ above to get $b^{44}=b^{77}$.
            $endgroup$
            – Aravind
            Jul 14 '15 at 9:27












          • $begingroup$
            @ajotatxe Nice ... Vote up!
            $endgroup$
            – johannesvalks
            Jul 14 '15 at 9:29










          • $begingroup$
            Very nice, +1 Then one could note that in the dihedral group of order $66$, generated by an element $c$ of order $33$, and by an element $a$ of order $2$, one has $c a c = c^{-1}$, and for $b = c^{3}$, one has $a b^4 a = a c^{12} a = c^{-12} = c^{21} = (c^{3})^{7}= b^7$. So $b^{33} = 1$ is the best we can squeeze out of the relations.
            $endgroup$
            – Andreas Caranti
            Jul 14 '15 at 17:09





















          2












          $begingroup$

          Note that
          begin{align}
          (aba^{-1})^4&=b^7text{, and }\
          b^4&=(aba^{-1})^7nonumber\
          (aba^{-1})^7&=b^4. text{So 'dividing' the first equality by the last,}\
          (aba^{-1})^{-3}&=b^{3}. text{ Multiplying it with the first equality,}\
          aba^{-1}&=b^{10}\
          (aba^{-1})^4&=b^{40}=b^7\
          text{So }b^{33}&=e
          end{align}






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            $textbf{Different approach:}$



            $ab^4a=b^7$ and $ab^7a=b^4$. So
            $$b^{11}=b^4b^7=ab^{11}a$$
            Hence,
            $$(b^4)^{11}=(ab^7a)^{11}=ab^{77}a=(b^{11})^4=(ab^{11}a)^4=ab^{44}a$$. Hence, $$b^{44}=b^{77}$$






            share|cite|improve this answer









            $endgroup$













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              3 Answers
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              3 Answers
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              active

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              14












              $begingroup$

              For any integer $n$ we have
              $$(ab^4a)^n=b^{7n}$$
              or
              $$ab^{4n}a=b^{7n}$$
              For $n=4$,
              $$ab^{16}a=b^{28}$$
              and for $n=7$,
              $$ab^{28}a=b^{49}$$
              Therefore
              $$b^{16}=b^{49}$$



              EDIT: With same technique, it can be shown that if $a^2=e$ and $ab^r=b^sa$ then $b^{r^2}=b^{s^2}$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Another solution: Note that $a$ and $b^{11}$ commute; then take $n=11$ above to get $b^{44}=b^{77}$.
                $endgroup$
                – Aravind
                Jul 14 '15 at 9:27












              • $begingroup$
                @ajotatxe Nice ... Vote up!
                $endgroup$
                – johannesvalks
                Jul 14 '15 at 9:29










              • $begingroup$
                Very nice, +1 Then one could note that in the dihedral group of order $66$, generated by an element $c$ of order $33$, and by an element $a$ of order $2$, one has $c a c = c^{-1}$, and for $b = c^{3}$, one has $a b^4 a = a c^{12} a = c^{-12} = c^{21} = (c^{3})^{7}= b^7$. So $b^{33} = 1$ is the best we can squeeze out of the relations.
                $endgroup$
                – Andreas Caranti
                Jul 14 '15 at 17:09


















              14












              $begingroup$

              For any integer $n$ we have
              $$(ab^4a)^n=b^{7n}$$
              or
              $$ab^{4n}a=b^{7n}$$
              For $n=4$,
              $$ab^{16}a=b^{28}$$
              and for $n=7$,
              $$ab^{28}a=b^{49}$$
              Therefore
              $$b^{16}=b^{49}$$



              EDIT: With same technique, it can be shown that if $a^2=e$ and $ab^r=b^sa$ then $b^{r^2}=b^{s^2}$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Another solution: Note that $a$ and $b^{11}$ commute; then take $n=11$ above to get $b^{44}=b^{77}$.
                $endgroup$
                – Aravind
                Jul 14 '15 at 9:27












              • $begingroup$
                @ajotatxe Nice ... Vote up!
                $endgroup$
                – johannesvalks
                Jul 14 '15 at 9:29










              • $begingroup$
                Very nice, +1 Then one could note that in the dihedral group of order $66$, generated by an element $c$ of order $33$, and by an element $a$ of order $2$, one has $c a c = c^{-1}$, and for $b = c^{3}$, one has $a b^4 a = a c^{12} a = c^{-12} = c^{21} = (c^{3})^{7}= b^7$. So $b^{33} = 1$ is the best we can squeeze out of the relations.
                $endgroup$
                – Andreas Caranti
                Jul 14 '15 at 17:09
















              14












              14








              14





              $begingroup$

              For any integer $n$ we have
              $$(ab^4a)^n=b^{7n}$$
              or
              $$ab^{4n}a=b^{7n}$$
              For $n=4$,
              $$ab^{16}a=b^{28}$$
              and for $n=7$,
              $$ab^{28}a=b^{49}$$
              Therefore
              $$b^{16}=b^{49}$$



              EDIT: With same technique, it can be shown that if $a^2=e$ and $ab^r=b^sa$ then $b^{r^2}=b^{s^2}$.






              share|cite|improve this answer











              $endgroup$



              For any integer $n$ we have
              $$(ab^4a)^n=b^{7n}$$
              or
              $$ab^{4n}a=b^{7n}$$
              For $n=4$,
              $$ab^{16}a=b^{28}$$
              and for $n=7$,
              $$ab^{28}a=b^{49}$$
              Therefore
              $$b^{16}=b^{49}$$



              EDIT: With same technique, it can be shown that if $a^2=e$ and $ab^r=b^sa$ then $b^{r^2}=b^{s^2}$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jul 16 '15 at 19:18

























              answered Jul 14 '15 at 9:24









              ajotatxeajotatxe

              53.6k23890




              53.6k23890












              • $begingroup$
                Another solution: Note that $a$ and $b^{11}$ commute; then take $n=11$ above to get $b^{44}=b^{77}$.
                $endgroup$
                – Aravind
                Jul 14 '15 at 9:27












              • $begingroup$
                @ajotatxe Nice ... Vote up!
                $endgroup$
                – johannesvalks
                Jul 14 '15 at 9:29










              • $begingroup$
                Very nice, +1 Then one could note that in the dihedral group of order $66$, generated by an element $c$ of order $33$, and by an element $a$ of order $2$, one has $c a c = c^{-1}$, and for $b = c^{3}$, one has $a b^4 a = a c^{12} a = c^{-12} = c^{21} = (c^{3})^{7}= b^7$. So $b^{33} = 1$ is the best we can squeeze out of the relations.
                $endgroup$
                – Andreas Caranti
                Jul 14 '15 at 17:09




















              • $begingroup$
                Another solution: Note that $a$ and $b^{11}$ commute; then take $n=11$ above to get $b^{44}=b^{77}$.
                $endgroup$
                – Aravind
                Jul 14 '15 at 9:27












              • $begingroup$
                @ajotatxe Nice ... Vote up!
                $endgroup$
                – johannesvalks
                Jul 14 '15 at 9:29










              • $begingroup$
                Very nice, +1 Then one could note that in the dihedral group of order $66$, generated by an element $c$ of order $33$, and by an element $a$ of order $2$, one has $c a c = c^{-1}$, and for $b = c^{3}$, one has $a b^4 a = a c^{12} a = c^{-12} = c^{21} = (c^{3})^{7}= b^7$. So $b^{33} = 1$ is the best we can squeeze out of the relations.
                $endgroup$
                – Andreas Caranti
                Jul 14 '15 at 17:09


















              $begingroup$
              Another solution: Note that $a$ and $b^{11}$ commute; then take $n=11$ above to get $b^{44}=b^{77}$.
              $endgroup$
              – Aravind
              Jul 14 '15 at 9:27






              $begingroup$
              Another solution: Note that $a$ and $b^{11}$ commute; then take $n=11$ above to get $b^{44}=b^{77}$.
              $endgroup$
              – Aravind
              Jul 14 '15 at 9:27














              $begingroup$
              @ajotatxe Nice ... Vote up!
              $endgroup$
              – johannesvalks
              Jul 14 '15 at 9:29




              $begingroup$
              @ajotatxe Nice ... Vote up!
              $endgroup$
              – johannesvalks
              Jul 14 '15 at 9:29












              $begingroup$
              Very nice, +1 Then one could note that in the dihedral group of order $66$, generated by an element $c$ of order $33$, and by an element $a$ of order $2$, one has $c a c = c^{-1}$, and for $b = c^{3}$, one has $a b^4 a = a c^{12} a = c^{-12} = c^{21} = (c^{3})^{7}= b^7$. So $b^{33} = 1$ is the best we can squeeze out of the relations.
              $endgroup$
              – Andreas Caranti
              Jul 14 '15 at 17:09






              $begingroup$
              Very nice, +1 Then one could note that in the dihedral group of order $66$, generated by an element $c$ of order $33$, and by an element $a$ of order $2$, one has $c a c = c^{-1}$, and for $b = c^{3}$, one has $a b^4 a = a c^{12} a = c^{-12} = c^{21} = (c^{3})^{7}= b^7$. So $b^{33} = 1$ is the best we can squeeze out of the relations.
              $endgroup$
              – Andreas Caranti
              Jul 14 '15 at 17:09













              2












              $begingroup$

              Note that
              begin{align}
              (aba^{-1})^4&=b^7text{, and }\
              b^4&=(aba^{-1})^7nonumber\
              (aba^{-1})^7&=b^4. text{So 'dividing' the first equality by the last,}\
              (aba^{-1})^{-3}&=b^{3}. text{ Multiplying it with the first equality,}\
              aba^{-1}&=b^{10}\
              (aba^{-1})^4&=b^{40}=b^7\
              text{So }b^{33}&=e
              end{align}






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Note that
                begin{align}
                (aba^{-1})^4&=b^7text{, and }\
                b^4&=(aba^{-1})^7nonumber\
                (aba^{-1})^7&=b^4. text{So 'dividing' the first equality by the last,}\
                (aba^{-1})^{-3}&=b^{3}. text{ Multiplying it with the first equality,}\
                aba^{-1}&=b^{10}\
                (aba^{-1})^4&=b^{40}=b^7\
                text{So }b^{33}&=e
                end{align}






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Note that
                  begin{align}
                  (aba^{-1})^4&=b^7text{, and }\
                  b^4&=(aba^{-1})^7nonumber\
                  (aba^{-1})^7&=b^4. text{So 'dividing' the first equality by the last,}\
                  (aba^{-1})^{-3}&=b^{3}. text{ Multiplying it with the first equality,}\
                  aba^{-1}&=b^{10}\
                  (aba^{-1})^4&=b^{40}=b^7\
                  text{So }b^{33}&=e
                  end{align}






                  share|cite|improve this answer









                  $endgroup$



                  Note that
                  begin{align}
                  (aba^{-1})^4&=b^7text{, and }\
                  b^4&=(aba^{-1})^7nonumber\
                  (aba^{-1})^7&=b^4. text{So 'dividing' the first equality by the last,}\
                  (aba^{-1})^{-3}&=b^{3}. text{ Multiplying it with the first equality,}\
                  aba^{-1}&=b^{10}\
                  (aba^{-1})^4&=b^{40}=b^7\
                  text{So }b^{33}&=e
                  end{align}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jul 14 '15 at 9:31









                  Alex FokAlex Fok

                  3,856617




                  3,856617























                      0












                      $begingroup$

                      $textbf{Different approach:}$



                      $ab^4a=b^7$ and $ab^7a=b^4$. So
                      $$b^{11}=b^4b^7=ab^{11}a$$
                      Hence,
                      $$(b^4)^{11}=(ab^7a)^{11}=ab^{77}a=(b^{11})^4=(ab^{11}a)^4=ab^{44}a$$. Hence, $$b^{44}=b^{77}$$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        $textbf{Different approach:}$



                        $ab^4a=b^7$ and $ab^7a=b^4$. So
                        $$b^{11}=b^4b^7=ab^{11}a$$
                        Hence,
                        $$(b^4)^{11}=(ab^7a)^{11}=ab^{77}a=(b^{11})^4=(ab^{11}a)^4=ab^{44}a$$. Hence, $$b^{44}=b^{77}$$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          $textbf{Different approach:}$



                          $ab^4a=b^7$ and $ab^7a=b^4$. So
                          $$b^{11}=b^4b^7=ab^{11}a$$
                          Hence,
                          $$(b^4)^{11}=(ab^7a)^{11}=ab^{77}a=(b^{11})^4=(ab^{11}a)^4=ab^{44}a$$. Hence, $$b^{44}=b^{77}$$






                          share|cite|improve this answer









                          $endgroup$



                          $textbf{Different approach:}$



                          $ab^4a=b^7$ and $ab^7a=b^4$. So
                          $$b^{11}=b^4b^7=ab^{11}a$$
                          Hence,
                          $$(b^4)^{11}=(ab^7a)^{11}=ab^{77}a=(b^{11})^4=(ab^{11}a)^4=ab^{44}a$$. Hence, $$b^{44}=b^{77}$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 6 '18 at 9:48









                          1ENİGMA11ENİGMA1

                          958416




                          958416






























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