Left action of a group on permutation representation












3












$begingroup$


I am studying my course on permutation representation, and I am stuck at understanding the left action of a finite group on the permutation representation $F(X,Bbb C)$. In my course it is given for $(g,phi)in Gtimes F(X,Bbb C) $ by:



$g*phi(xin X)to phi(g^{-1}x)$.



Is this a left group action?



I have: $h*(g*phi(x))= h*phi(g^{-1}x)= phi(h^{-1}g^{-1}x) = phi((gh)^{-1}x)= (gh)*phi(x) neq (hg)*phi(x)$.



I must be missing something.






abstract-algebra group-theory representation-theory group-actions







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I am studying my course on permutation representation, and I am stuck at understanding the left action of a finite group on the permutation representation $F(X,Bbb C)$. In my course it is given for $(g,phi)in Gtimes F(X,Bbb C) $ by:



    $g*phi(xin X)to phi(g^{-1}x)$.



    Is this a left group action?



    I have: $h*(g*phi(x))= h*phi(g^{-1}x)= phi(h^{-1}g^{-1}x) = phi((gh)^{-1}x)= (gh)*phi(x) neq (hg)*phi(x)$.



    I must be missing something.






    abstract-algebra group-theory representation-theory group-actions







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I am studying my course on permutation representation, and I am stuck at understanding the left action of a finite group on the permutation representation $F(X,Bbb C)$. In my course it is given for $(g,phi)in Gtimes F(X,Bbb C) $ by:



      $g*phi(xin X)to phi(g^{-1}x)$.



      Is this a left group action?



      I have: $h*(g*phi(x))= h*phi(g^{-1}x)= phi(h^{-1}g^{-1}x) = phi((gh)^{-1}x)= (gh)*phi(x) neq (hg)*phi(x)$.



      I must be missing something.






      abstract-algebra group-theory representation-theory group-actions







      share|cite|improve this question











      $endgroup$




      I am studying my course on permutation representation, and I am stuck at understanding the left action of a finite group on the permutation representation $F(X,Bbb C)$. In my course it is given for $(g,phi)in Gtimes F(X,Bbb C) $ by:



      $g*phi(xin X)to phi(g^{-1}x)$.



      Is this a left group action?



      I have: $h*(g*phi(x))= h*phi(g^{-1}x)= phi(h^{-1}g^{-1}x) = phi((gh)^{-1}x)= (gh)*phi(x) neq (hg)*phi(x)$.



      I must be missing something.






      abstract-algebra group-theory representation-theory group-actions




      abstract-algebra group-theory representation-theory group-actions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited yesterday









      Peter Mortensen

      563310




      563310










      asked yesterday









      PerelManPerelMan

      803414




      803414






















          2 Answers
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          7












          $begingroup$

          Note that by the definition $g*phi$ is in $F(X,mathbb C)$ defined by $(g*phi)(x):= phi(g^{-1}x)$, so the calculation is the following:
          $$begin{align}
          (h*(g*phi))(x)&= (g*phi)(h^{-1}x)\
          &= phi(g^{-1}h^{-1}x)\
          &= phi((hg)^{-1}x)\
          &= (hg*phi)(x).
          end{align}$$
          Thus $h*(g*phi)= hg*phi$.






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            In addition to the accepted answer I want to point out that one can also start by substituting the inner expression. Since $(g*phi)(x) := phi(g^{-1}x)$, we have $g*phi = x to phi(g^{-1}x)$, therefore



            $$begin{align}
            (h*(g*phi))(x)
            &= (h*(ttophi(g^{-1}t)))(x)\
            &= (ttophi(g^{-1}t))(h^{-1}x)\
            &= phi(g^{-1}(h^{-1}x))\
            &= (hg*phi)(x).
            end{align}$$






            share|cite|improve this answer









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              active

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              active

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              7












              $begingroup$

              Note that by the definition $g*phi$ is in $F(X,mathbb C)$ defined by $(g*phi)(x):= phi(g^{-1}x)$, so the calculation is the following:
              $$begin{align}
              (h*(g*phi))(x)&= (g*phi)(h^{-1}x)\
              &= phi(g^{-1}h^{-1}x)\
              &= phi((hg)^{-1}x)\
              &= (hg*phi)(x).
              end{align}$$
              Thus $h*(g*phi)= hg*phi$.






              share|cite|improve this answer











              $endgroup$


















                7












                $begingroup$

                Note that by the definition $g*phi$ is in $F(X,mathbb C)$ defined by $(g*phi)(x):= phi(g^{-1}x)$, so the calculation is the following:
                $$begin{align}
                (h*(g*phi))(x)&= (g*phi)(h^{-1}x)\
                &= phi(g^{-1}h^{-1}x)\
                &= phi((hg)^{-1}x)\
                &= (hg*phi)(x).
                end{align}$$
                Thus $h*(g*phi)= hg*phi$.






                share|cite|improve this answer











                $endgroup$
















                  7












                  7








                  7





                  $begingroup$

                  Note that by the definition $g*phi$ is in $F(X,mathbb C)$ defined by $(g*phi)(x):= phi(g^{-1}x)$, so the calculation is the following:
                  $$begin{align}
                  (h*(g*phi))(x)&= (g*phi)(h^{-1}x)\
                  &= phi(g^{-1}h^{-1}x)\
                  &= phi((hg)^{-1}x)\
                  &= (hg*phi)(x).
                  end{align}$$
                  Thus $h*(g*phi)= hg*phi$.






                  share|cite|improve this answer











                  $endgroup$



                  Note that by the definition $g*phi$ is in $F(X,mathbb C)$ defined by $(g*phi)(x):= phi(g^{-1}x)$, so the calculation is the following:
                  $$begin{align}
                  (h*(g*phi))(x)&= (g*phi)(h^{-1}x)\
                  &= phi(g^{-1}h^{-1}x)\
                  &= phi((hg)^{-1}x)\
                  &= (hg*phi)(x).
                  end{align}$$
                  Thus $h*(g*phi)= hg*phi$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited yesterday









                  Shaun

                  10.6k113687




                  10.6k113687










                  answered yesterday









                  SMMSMM

                  3,268512




                  3,268512























                      1












                      $begingroup$

                      In addition to the accepted answer I want to point out that one can also start by substituting the inner expression. Since $(g*phi)(x) := phi(g^{-1}x)$, we have $g*phi = x to phi(g^{-1}x)$, therefore



                      $$begin{align}
                      (h*(g*phi))(x)
                      &= (h*(ttophi(g^{-1}t)))(x)\
                      &= (ttophi(g^{-1}t))(h^{-1}x)\
                      &= phi(g^{-1}(h^{-1}x))\
                      &= (hg*phi)(x).
                      end{align}$$






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        In addition to the accepted answer I want to point out that one can also start by substituting the inner expression. Since $(g*phi)(x) := phi(g^{-1}x)$, we have $g*phi = x to phi(g^{-1}x)$, therefore



                        $$begin{align}
                        (h*(g*phi))(x)
                        &= (h*(ttophi(g^{-1}t)))(x)\
                        &= (ttophi(g^{-1}t))(h^{-1}x)\
                        &= phi(g^{-1}(h^{-1}x))\
                        &= (hg*phi)(x).
                        end{align}$$






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          In addition to the accepted answer I want to point out that one can also start by substituting the inner expression. Since $(g*phi)(x) := phi(g^{-1}x)$, we have $g*phi = x to phi(g^{-1}x)$, therefore



                          $$begin{align}
                          (h*(g*phi))(x)
                          &= (h*(ttophi(g^{-1}t)))(x)\
                          &= (ttophi(g^{-1}t))(h^{-1}x)\
                          &= phi(g^{-1}(h^{-1}x))\
                          &= (hg*phi)(x).
                          end{align}$$






                          share|cite|improve this answer









                          $endgroup$



                          In addition to the accepted answer I want to point out that one can also start by substituting the inner expression. Since $(g*phi)(x) := phi(g^{-1}x)$, we have $g*phi = x to phi(g^{-1}x)$, therefore



                          $$begin{align}
                          (h*(g*phi))(x)
                          &= (h*(ttophi(g^{-1}t)))(x)\
                          &= (ttophi(g^{-1}t))(h^{-1}x)\
                          &= phi(g^{-1}(h^{-1}x))\
                          &= (hg*phi)(x).
                          end{align}$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered yesterday









                          Micha WiedenmannMicha Wiedenmann

                          1396




                          1396






























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