Finding the centraliser of an element in $S_5$












1















Let $x = (12)(34) in S_5$. Is there a quick way to find $C_{S_5}(x)$ ?




I know that conjugation leaves cycle types unchanged. Thus, if $sigma in C_{S_5}(x)$, then $(12)(34) = sigma^{-1}(12)(34)sigma = (sigma(1), sigma(2))(sigma(3), sigma(4))$, but how does this help me?



Any help would be appreciated. Thanks!






abstract-algebra group-theory symmetric-groups







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    1















    Let $x = (12)(34) in S_5$. Is there a quick way to find $C_{S_5}(x)$ ?




    I know that conjugation leaves cycle types unchanged. Thus, if $sigma in C_{S_5}(x)$, then $(12)(34) = sigma^{-1}(12)(34)sigma = (sigma(1), sigma(2))(sigma(3), sigma(4))$, but how does this help me?



    Any help would be appreciated. Thanks!






    abstract-algebra group-theory symmetric-groups







    share|cite|improve this question



























      1












      1








      1








      Let $x = (12)(34) in S_5$. Is there a quick way to find $C_{S_5}(x)$ ?




      I know that conjugation leaves cycle types unchanged. Thus, if $sigma in C_{S_5}(x)$, then $(12)(34) = sigma^{-1}(12)(34)sigma = (sigma(1), sigma(2))(sigma(3), sigma(4))$, but how does this help me?



      Any help would be appreciated. Thanks!






      abstract-algebra group-theory symmetric-groups







      share|cite|improve this question
















      Let $x = (12)(34) in S_5$. Is there a quick way to find $C_{S_5}(x)$ ?




      I know that conjugation leaves cycle types unchanged. Thus, if $sigma in C_{S_5}(x)$, then $(12)(34) = sigma^{-1}(12)(34)sigma = (sigma(1), sigma(2))(sigma(3), sigma(4))$, but how does this help me?



      Any help would be appreciated. Thanks!






      abstract-algebra group-theory symmetric-groups




      abstract-algebra group-theory symmetric-groups






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Dec 2 at 9:58









      Chinnapparaj R

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      5,2601826










      asked Dec 1 at 19:20









      the man

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          For $sigma in S_5$ we have that $(12)(34) = sigma^{-1}(12)(34)sigma = (sigma(1), sigma(2))(sigma(3), sigma(4))$ holds if and ony if ${sigma({1,2}),sigma({3,4})}={{1,2},{3,4}}$.



          Thus we have two possibilities: either $sigma({1,2})={1,2}$ and $sigma({3,4})={3,4}$ or $sigma({1,2})={3,4}$ and $sigma({3,4})={1,2}$.
          In the first case, we get the possibilities $mathrm{id},(12),(34),(12)(34)$, in the second case, we get the possibilities $(13)(24),(14)(23),(1324),(1423)$






          share|cite|improve this answer





























            0














            There may be a faster way to do this, but the first that comes to mind is to use the fact that $(sigma(1)sigma(2))(sigma(3),sigma(4)) = (12)(34)$ to determine which $sigma$ are possible.



            Since $sigma$ is a bijection, $(sigma(1),sigma(2))(sigma(3),sigma(4))$ is written in disjoint cycle notation. Thus $(sigma(1),sigma(2)) = (1,2)$ or $(3,4)$ and $(sigma(3),sigma(4)) = (3,4)$ or $(1,2)$. In any case we see $sigma$ must fix $5$. We can choose any value for $sigma(1)$ in ${1,2,3,4}$. This then determines the value of $sigma(2)$. Once we've chosen $sigma(1)$. The choice of $sigma(3)$ determines $sigma(4)$. With these observations we can list the elements.



            $C_{S_5}((12)(34)) = {1, (34), (12), (12)(34), (13)(24), (1324), (1423), (14)(23)}$






            share|cite|improve this answer





















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              2 Answers
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              For $sigma in S_5$ we have that $(12)(34) = sigma^{-1}(12)(34)sigma = (sigma(1), sigma(2))(sigma(3), sigma(4))$ holds if and ony if ${sigma({1,2}),sigma({3,4})}={{1,2},{3,4}}$.



              Thus we have two possibilities: either $sigma({1,2})={1,2}$ and $sigma({3,4})={3,4}$ or $sigma({1,2})={3,4}$ and $sigma({3,4})={1,2}$.
              In the first case, we get the possibilities $mathrm{id},(12),(34),(12)(34)$, in the second case, we get the possibilities $(13)(24),(14)(23),(1324),(1423)$






              share|cite|improve this answer


























                0














                For $sigma in S_5$ we have that $(12)(34) = sigma^{-1}(12)(34)sigma = (sigma(1), sigma(2))(sigma(3), sigma(4))$ holds if and ony if ${sigma({1,2}),sigma({3,4})}={{1,2},{3,4}}$.



                Thus we have two possibilities: either $sigma({1,2})={1,2}$ and $sigma({3,4})={3,4}$ or $sigma({1,2})={3,4}$ and $sigma({3,4})={1,2}$.
                In the first case, we get the possibilities $mathrm{id},(12),(34),(12)(34)$, in the second case, we get the possibilities $(13)(24),(14)(23),(1324),(1423)$






                share|cite|improve this answer
























                  0












                  0








                  0






                  For $sigma in S_5$ we have that $(12)(34) = sigma^{-1}(12)(34)sigma = (sigma(1), sigma(2))(sigma(3), sigma(4))$ holds if and ony if ${sigma({1,2}),sigma({3,4})}={{1,2},{3,4}}$.



                  Thus we have two possibilities: either $sigma({1,2})={1,2}$ and $sigma({3,4})={3,4}$ or $sigma({1,2})={3,4}$ and $sigma({3,4})={1,2}$.
                  In the first case, we get the possibilities $mathrm{id},(12),(34),(12)(34)$, in the second case, we get the possibilities $(13)(24),(14)(23),(1324),(1423)$






                  share|cite|improve this answer












                  For $sigma in S_5$ we have that $(12)(34) = sigma^{-1}(12)(34)sigma = (sigma(1), sigma(2))(sigma(3), sigma(4))$ holds if and ony if ${sigma({1,2}),sigma({3,4})}={{1,2},{3,4}}$.



                  Thus we have two possibilities: either $sigma({1,2})={1,2}$ and $sigma({3,4})={3,4}$ or $sigma({1,2})={3,4}$ and $sigma({3,4})={1,2}$.
                  In the first case, we get the possibilities $mathrm{id},(12),(34),(12)(34)$, in the second case, we get the possibilities $(13)(24),(14)(23),(1324),(1423)$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 1 at 21:05









                  MatheinBoulomenos

                  7,9961936




                  7,9961936























                      0














                      There may be a faster way to do this, but the first that comes to mind is to use the fact that $(sigma(1)sigma(2))(sigma(3),sigma(4)) = (12)(34)$ to determine which $sigma$ are possible.



                      Since $sigma$ is a bijection, $(sigma(1),sigma(2))(sigma(3),sigma(4))$ is written in disjoint cycle notation. Thus $(sigma(1),sigma(2)) = (1,2)$ or $(3,4)$ and $(sigma(3),sigma(4)) = (3,4)$ or $(1,2)$. In any case we see $sigma$ must fix $5$. We can choose any value for $sigma(1)$ in ${1,2,3,4}$. This then determines the value of $sigma(2)$. Once we've chosen $sigma(1)$. The choice of $sigma(3)$ determines $sigma(4)$. With these observations we can list the elements.



                      $C_{S_5}((12)(34)) = {1, (34), (12), (12)(34), (13)(24), (1324), (1423), (14)(23)}$






                      share|cite|improve this answer


























                        0














                        There may be a faster way to do this, but the first that comes to mind is to use the fact that $(sigma(1)sigma(2))(sigma(3),sigma(4)) = (12)(34)$ to determine which $sigma$ are possible.



                        Since $sigma$ is a bijection, $(sigma(1),sigma(2))(sigma(3),sigma(4))$ is written in disjoint cycle notation. Thus $(sigma(1),sigma(2)) = (1,2)$ or $(3,4)$ and $(sigma(3),sigma(4)) = (3,4)$ or $(1,2)$. In any case we see $sigma$ must fix $5$. We can choose any value for $sigma(1)$ in ${1,2,3,4}$. This then determines the value of $sigma(2)$. Once we've chosen $sigma(1)$. The choice of $sigma(3)$ determines $sigma(4)$. With these observations we can list the elements.



                        $C_{S_5}((12)(34)) = {1, (34), (12), (12)(34), (13)(24), (1324), (1423), (14)(23)}$






                        share|cite|improve this answer
























                          0












                          0








                          0






                          There may be a faster way to do this, but the first that comes to mind is to use the fact that $(sigma(1)sigma(2))(sigma(3),sigma(4)) = (12)(34)$ to determine which $sigma$ are possible.



                          Since $sigma$ is a bijection, $(sigma(1),sigma(2))(sigma(3),sigma(4))$ is written in disjoint cycle notation. Thus $(sigma(1),sigma(2)) = (1,2)$ or $(3,4)$ and $(sigma(3),sigma(4)) = (3,4)$ or $(1,2)$. In any case we see $sigma$ must fix $5$. We can choose any value for $sigma(1)$ in ${1,2,3,4}$. This then determines the value of $sigma(2)$. Once we've chosen $sigma(1)$. The choice of $sigma(3)$ determines $sigma(4)$. With these observations we can list the elements.



                          $C_{S_5}((12)(34)) = {1, (34), (12), (12)(34), (13)(24), (1324), (1423), (14)(23)}$






                          share|cite|improve this answer












                          There may be a faster way to do this, but the first that comes to mind is to use the fact that $(sigma(1)sigma(2))(sigma(3),sigma(4)) = (12)(34)$ to determine which $sigma$ are possible.



                          Since $sigma$ is a bijection, $(sigma(1),sigma(2))(sigma(3),sigma(4))$ is written in disjoint cycle notation. Thus $(sigma(1),sigma(2)) = (1,2)$ or $(3,4)$ and $(sigma(3),sigma(4)) = (3,4)$ or $(1,2)$. In any case we see $sigma$ must fix $5$. We can choose any value for $sigma(1)$ in ${1,2,3,4}$. This then determines the value of $sigma(2)$. Once we've chosen $sigma(1)$. The choice of $sigma(3)$ determines $sigma(4)$. With these observations we can list the elements.



                          $C_{S_5}((12)(34)) = {1, (34), (12), (12)(34), (13)(24), (1324), (1423), (14)(23)}$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 1 at 21:08









                          Sean Haight

                          675519




                          675519






























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