Permutations: the product of $(1,2)(1,2,3) = (1,3)$












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I'm trying understand why the following permutation is the case. I understand we go from right to left, although others may go from left to right, but I can't seem to produce the correct answer. I keep getting $(1,2,3)$ which is obviously wrong.






permutations cyclic-groups







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    0












    $begingroup$


    I'm trying understand why the following permutation is the case. I understand we go from right to left, although others may go from left to right, but I can't seem to produce the correct answer. I keep getting $(1,2,3)$ which is obviously wrong.






    permutations cyclic-groups







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I'm trying understand why the following permutation is the case. I understand we go from right to left, although others may go from left to right, but I can't seem to produce the correct answer. I keep getting $(1,2,3)$ which is obviously wrong.






      permutations cyclic-groups







      share|cite|improve this question











      $endgroup$




      I'm trying understand why the following permutation is the case. I understand we go from right to left, although others may go from left to right, but I can't seem to produce the correct answer. I keep getting $(1,2,3)$ which is obviously wrong.






      permutations cyclic-groups




      permutations cyclic-groups






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      edited Dec 11 '18 at 22:07









      John B

      12.2k51840




      12.2k51840










      asked Dec 11 '18 at 22:02









      David MBDavid MB

      267




      267






















          3 Answers
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          0












          $begingroup$

          We start with $(1$, and build our cycle from that. $1$ gets sent by the left cycle to $2$, and then the right cycle sends that $2$ to $3$. So we write $(1,3$.



          Now we look at where $3$ goes. The first cycle doesn't touch it, and then the second cycle sends it to $1$, meaning the cycle is complete with $(1,3)$. Technically we should check $2$ as well, but there is nowhere else to send it, so we are finished.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Much appreciated. I was beginning to get a headache because I couldn't figure it out. But on your last point, doesn't 2 get sent to 1?
            $endgroup$
            – David MB
            Dec 11 '18 at 22:26










          • $begingroup$
            @DavidMB By the left cycle, sure. Then the right cycle sends that $1$ back to $2$, so the net effect is that nothing happens to $2$.
            $endgroup$
            – Arthur
            Dec 11 '18 at 22:30



















          1












          $begingroup$

          Product of permutation is composition of maps (right to left). $$1mapsto2mapsto1,:2mapsto3mapsto3,:3mapsto1mapsto2$$
          $$begin{pmatrix}1 & 2 & 3\2 & 1 & 3end{pmatrix}begin{pmatrix}1 & 2 & 3\2 & 3 & 1end{pmatrix}=begin{pmatrix}1 & 2 & 3\1 & 3 & 2end{pmatrix}$$






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            1












            $begingroup$

            Let’s compute the value of the permutation at 1,2 and 3.
            By definition, $(1,2,3)$ evaluated at $1$ is $2$.
            $(1,2)$ evaluated at $2$ is $1$. Thus, $(1,2)(1,2,3)$ evaluated at $1$ is $1$.
            By definition, $(1,2,3)$ evaluated at $2$ is $3$. $(1,2)$ evaluated at $3$ is $3$. So $(1,2)(1,2,3)$ evaluated at $2$ is $3$.
            By definition, $(1,2,3)$ evaluated at $3$ is $1$. $(1,2)$ evaluated at $1$ is $2$. So $(1,2)(1,2,3)$ evaluated at $3$ is $2$.
            So we can check that $(1,2)(1,2,3)=(2,3)$.






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              3 Answers
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              active

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              3 Answers
              3






              active

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              active

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              active

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              0












              $begingroup$

              We start with $(1$, and build our cycle from that. $1$ gets sent by the left cycle to $2$, and then the right cycle sends that $2$ to $3$. So we write $(1,3$.



              Now we look at where $3$ goes. The first cycle doesn't touch it, and then the second cycle sends it to $1$, meaning the cycle is complete with $(1,3)$. Technically we should check $2$ as well, but there is nowhere else to send it, so we are finished.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Much appreciated. I was beginning to get a headache because I couldn't figure it out. But on your last point, doesn't 2 get sent to 1?
                $endgroup$
                – David MB
                Dec 11 '18 at 22:26










              • $begingroup$
                @DavidMB By the left cycle, sure. Then the right cycle sends that $1$ back to $2$, so the net effect is that nothing happens to $2$.
                $endgroup$
                – Arthur
                Dec 11 '18 at 22:30
















              0












              $begingroup$

              We start with $(1$, and build our cycle from that. $1$ gets sent by the left cycle to $2$, and then the right cycle sends that $2$ to $3$. So we write $(1,3$.



              Now we look at where $3$ goes. The first cycle doesn't touch it, and then the second cycle sends it to $1$, meaning the cycle is complete with $(1,3)$. Technically we should check $2$ as well, but there is nowhere else to send it, so we are finished.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Much appreciated. I was beginning to get a headache because I couldn't figure it out. But on your last point, doesn't 2 get sent to 1?
                $endgroup$
                – David MB
                Dec 11 '18 at 22:26










              • $begingroup$
                @DavidMB By the left cycle, sure. Then the right cycle sends that $1$ back to $2$, so the net effect is that nothing happens to $2$.
                $endgroup$
                – Arthur
                Dec 11 '18 at 22:30














              0












              0








              0





              $begingroup$

              We start with $(1$, and build our cycle from that. $1$ gets sent by the left cycle to $2$, and then the right cycle sends that $2$ to $3$. So we write $(1,3$.



              Now we look at where $3$ goes. The first cycle doesn't touch it, and then the second cycle sends it to $1$, meaning the cycle is complete with $(1,3)$. Technically we should check $2$ as well, but there is nowhere else to send it, so we are finished.






              share|cite|improve this answer









              $endgroup$



              We start with $(1$, and build our cycle from that. $1$ gets sent by the left cycle to $2$, and then the right cycle sends that $2$ to $3$. So we write $(1,3$.



              Now we look at where $3$ goes. The first cycle doesn't touch it, and then the second cycle sends it to $1$, meaning the cycle is complete with $(1,3)$. Technically we should check $2$ as well, but there is nowhere else to send it, so we are finished.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 11 '18 at 22:11









              ArthurArthur

              116k7116198




              116k7116198












              • $begingroup$
                Much appreciated. I was beginning to get a headache because I couldn't figure it out. But on your last point, doesn't 2 get sent to 1?
                $endgroup$
                – David MB
                Dec 11 '18 at 22:26










              • $begingroup$
                @DavidMB By the left cycle, sure. Then the right cycle sends that $1$ back to $2$, so the net effect is that nothing happens to $2$.
                $endgroup$
                – Arthur
                Dec 11 '18 at 22:30


















              • $begingroup$
                Much appreciated. I was beginning to get a headache because I couldn't figure it out. But on your last point, doesn't 2 get sent to 1?
                $endgroup$
                – David MB
                Dec 11 '18 at 22:26










              • $begingroup$
                @DavidMB By the left cycle, sure. Then the right cycle sends that $1$ back to $2$, so the net effect is that nothing happens to $2$.
                $endgroup$
                – Arthur
                Dec 11 '18 at 22:30
















              $begingroup$
              Much appreciated. I was beginning to get a headache because I couldn't figure it out. But on your last point, doesn't 2 get sent to 1?
              $endgroup$
              – David MB
              Dec 11 '18 at 22:26




              $begingroup$
              Much appreciated. I was beginning to get a headache because I couldn't figure it out. But on your last point, doesn't 2 get sent to 1?
              $endgroup$
              – David MB
              Dec 11 '18 at 22:26












              $begingroup$
              @DavidMB By the left cycle, sure. Then the right cycle sends that $1$ back to $2$, so the net effect is that nothing happens to $2$.
              $endgroup$
              – Arthur
              Dec 11 '18 at 22:30




              $begingroup$
              @DavidMB By the left cycle, sure. Then the right cycle sends that $1$ back to $2$, so the net effect is that nothing happens to $2$.
              $endgroup$
              – Arthur
              Dec 11 '18 at 22:30











              1












              $begingroup$

              Product of permutation is composition of maps (right to left). $$1mapsto2mapsto1,:2mapsto3mapsto3,:3mapsto1mapsto2$$
              $$begin{pmatrix}1 & 2 & 3\2 & 1 & 3end{pmatrix}begin{pmatrix}1 & 2 & 3\2 & 3 & 1end{pmatrix}=begin{pmatrix}1 & 2 & 3\1 & 3 & 2end{pmatrix}$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Product of permutation is composition of maps (right to left). $$1mapsto2mapsto1,:2mapsto3mapsto3,:3mapsto1mapsto2$$
                $$begin{pmatrix}1 & 2 & 3\2 & 1 & 3end{pmatrix}begin{pmatrix}1 & 2 & 3\2 & 3 & 1end{pmatrix}=begin{pmatrix}1 & 2 & 3\1 & 3 & 2end{pmatrix}$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Product of permutation is composition of maps (right to left). $$1mapsto2mapsto1,:2mapsto3mapsto3,:3mapsto1mapsto2$$
                  $$begin{pmatrix}1 & 2 & 3\2 & 1 & 3end{pmatrix}begin{pmatrix}1 & 2 & 3\2 & 3 & 1end{pmatrix}=begin{pmatrix}1 & 2 & 3\1 & 3 & 2end{pmatrix}$$






                  share|cite|improve this answer









                  $endgroup$



                  Product of permutation is composition of maps (right to left). $$1mapsto2mapsto1,:2mapsto3mapsto3,:3mapsto1mapsto2$$
                  $$begin{pmatrix}1 & 2 & 3\2 & 1 & 3end{pmatrix}begin{pmatrix}1 & 2 & 3\2 & 3 & 1end{pmatrix}=begin{pmatrix}1 & 2 & 3\1 & 3 & 2end{pmatrix}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 11 '18 at 22:11









                  Yadati KiranYadati Kiran

                  1,7911619




                  1,7911619























                      1












                      $begingroup$

                      Let’s compute the value of the permutation at 1,2 and 3.
                      By definition, $(1,2,3)$ evaluated at $1$ is $2$.
                      $(1,2)$ evaluated at $2$ is $1$. Thus, $(1,2)(1,2,3)$ evaluated at $1$ is $1$.
                      By definition, $(1,2,3)$ evaluated at $2$ is $3$. $(1,2)$ evaluated at $3$ is $3$. So $(1,2)(1,2,3)$ evaluated at $2$ is $3$.
                      By definition, $(1,2,3)$ evaluated at $3$ is $1$. $(1,2)$ evaluated at $1$ is $2$. So $(1,2)(1,2,3)$ evaluated at $3$ is $2$.
                      So we can check that $(1,2)(1,2,3)=(2,3)$.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Let’s compute the value of the permutation at 1,2 and 3.
                        By definition, $(1,2,3)$ evaluated at $1$ is $2$.
                        $(1,2)$ evaluated at $2$ is $1$. Thus, $(1,2)(1,2,3)$ evaluated at $1$ is $1$.
                        By definition, $(1,2,3)$ evaluated at $2$ is $3$. $(1,2)$ evaluated at $3$ is $3$. So $(1,2)(1,2,3)$ evaluated at $2$ is $3$.
                        By definition, $(1,2,3)$ evaluated at $3$ is $1$. $(1,2)$ evaluated at $1$ is $2$. So $(1,2)(1,2,3)$ evaluated at $3$ is $2$.
                        So we can check that $(1,2)(1,2,3)=(2,3)$.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Let’s compute the value of the permutation at 1,2 and 3.
                          By definition, $(1,2,3)$ evaluated at $1$ is $2$.
                          $(1,2)$ evaluated at $2$ is $1$. Thus, $(1,2)(1,2,3)$ evaluated at $1$ is $1$.
                          By definition, $(1,2,3)$ evaluated at $2$ is $3$. $(1,2)$ evaluated at $3$ is $3$. So $(1,2)(1,2,3)$ evaluated at $2$ is $3$.
                          By definition, $(1,2,3)$ evaluated at $3$ is $1$. $(1,2)$ evaluated at $1$ is $2$. So $(1,2)(1,2,3)$ evaluated at $3$ is $2$.
                          So we can check that $(1,2)(1,2,3)=(2,3)$.






                          share|cite|improve this answer









                          $endgroup$



                          Let’s compute the value of the permutation at 1,2 and 3.
                          By definition, $(1,2,3)$ evaluated at $1$ is $2$.
                          $(1,2)$ evaluated at $2$ is $1$. Thus, $(1,2)(1,2,3)$ evaluated at $1$ is $1$.
                          By definition, $(1,2,3)$ evaluated at $2$ is $3$. $(1,2)$ evaluated at $3$ is $3$. So $(1,2)(1,2,3)$ evaluated at $2$ is $3$.
                          By definition, $(1,2,3)$ evaluated at $3$ is $1$. $(1,2)$ evaluated at $1$ is $2$. So $(1,2)(1,2,3)$ evaluated at $3$ is $2$.
                          So we can check that $(1,2)(1,2,3)=(2,3)$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 11 '18 at 22:12









                          MindlackMindlack

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