Proof of center of symmetric group












2












$begingroup$


Can you help me finish this proof? This is what I have so far, I don't know if it's correct. Any tips on my writing or others are welcome.



"Prove that, for $n geq 3 $, the center of $S_n$ only contains $I_n$ (the identity)".



Suppose we have $sigma in S_n$ such that $sigma = (a_1,...,a_k)$.
Let $varphi in Z(S_n)$ be such that $varphi neq sigma$. We can do that because $(a_1,...,a_k)(a_1,a_2) neq (a_1,a_2)(a_1,...,a_k)$, and $(a_1,...,a_k) notin Z(S_n)$.



Then, we can write $varphi sigma = sigma varphi$. Multiplying by $varphi^{-1}$ the both terms we obtain:



$varphi sigma varphi^{-1} = sigma varphi varphi^{-1}$.



We also know that:



$varphi (a_1,...,a_k) varphi^{-1} = (varphi (a_1),...,varphi (a_k))$.



Then,



$(varphi (a_1),...,varphi (a_k)) = sigma = (a_1,...,a_k)$.



Which means that $varphi = I_n$.



New proof: (I hope this works now)



Suppose $varphi in Z(S_n)$ such that $varphi neq I_n$. We can assume that $exists i: varphi (i) neq i$ and $exists j neq i: varphi (j) = i$.



Now consider $sigma in S_n$ such that $sigma = (j,k), k in { 1,...,n } setminus { i,j }$.



Then, we have the equality $varphi sigma = sigma varphi$. Computing at $j$:





  • $sigma(varphi(j)) = sigma(i) = i$.


  • $varphi(sigma(j)) = varphi(k) neq i$, as $varphi$ is injective and $kneq j$.


Then, $varphi$ must be the identity.






abstract-algebra group-theory







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    2












    $begingroup$


    Can you help me finish this proof? This is what I have so far, I don't know if it's correct. Any tips on my writing or others are welcome.



    "Prove that, for $n geq 3 $, the center of $S_n$ only contains $I_n$ (the identity)".



    Suppose we have $sigma in S_n$ such that $sigma = (a_1,...,a_k)$.
    Let $varphi in Z(S_n)$ be such that $varphi neq sigma$. We can do that because $(a_1,...,a_k)(a_1,a_2) neq (a_1,a_2)(a_1,...,a_k)$, and $(a_1,...,a_k) notin Z(S_n)$.



    Then, we can write $varphi sigma = sigma varphi$. Multiplying by $varphi^{-1}$ the both terms we obtain:



    $varphi sigma varphi^{-1} = sigma varphi varphi^{-1}$.



    We also know that:



    $varphi (a_1,...,a_k) varphi^{-1} = (varphi (a_1),...,varphi (a_k))$.



    Then,



    $(varphi (a_1),...,varphi (a_k)) = sigma = (a_1,...,a_k)$.



    Which means that $varphi = I_n$.



    New proof: (I hope this works now)



    Suppose $varphi in Z(S_n)$ such that $varphi neq I_n$. We can assume that $exists i: varphi (i) neq i$ and $exists j neq i: varphi (j) = i$.



    Now consider $sigma in S_n$ such that $sigma = (j,k), k in { 1,...,n } setminus { i,j }$.



    Then, we have the equality $varphi sigma = sigma varphi$. Computing at $j$:





    • $sigma(varphi(j)) = sigma(i) = i$.


    • $varphi(sigma(j)) = varphi(k) neq i$, as $varphi$ is injective and $kneq j$.


    Then, $varphi$ must be the identity.






    abstract-algebra group-theory







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Can you help me finish this proof? This is what I have so far, I don't know if it's correct. Any tips on my writing or others are welcome.



      "Prove that, for $n geq 3 $, the center of $S_n$ only contains $I_n$ (the identity)".



      Suppose we have $sigma in S_n$ such that $sigma = (a_1,...,a_k)$.
      Let $varphi in Z(S_n)$ be such that $varphi neq sigma$. We can do that because $(a_1,...,a_k)(a_1,a_2) neq (a_1,a_2)(a_1,...,a_k)$, and $(a_1,...,a_k) notin Z(S_n)$.



      Then, we can write $varphi sigma = sigma varphi$. Multiplying by $varphi^{-1}$ the both terms we obtain:



      $varphi sigma varphi^{-1} = sigma varphi varphi^{-1}$.



      We also know that:



      $varphi (a_1,...,a_k) varphi^{-1} = (varphi (a_1),...,varphi (a_k))$.



      Then,



      $(varphi (a_1),...,varphi (a_k)) = sigma = (a_1,...,a_k)$.



      Which means that $varphi = I_n$.



      New proof: (I hope this works now)



      Suppose $varphi in Z(S_n)$ such that $varphi neq I_n$. We can assume that $exists i: varphi (i) neq i$ and $exists j neq i: varphi (j) = i$.



      Now consider $sigma in S_n$ such that $sigma = (j,k), k in { 1,...,n } setminus { i,j }$.



      Then, we have the equality $varphi sigma = sigma varphi$. Computing at $j$:





      • $sigma(varphi(j)) = sigma(i) = i$.


      • $varphi(sigma(j)) = varphi(k) neq i$, as $varphi$ is injective and $kneq j$.


      Then, $varphi$ must be the identity.






      abstract-algebra group-theory







      share|cite|improve this question











      $endgroup$




      Can you help me finish this proof? This is what I have so far, I don't know if it's correct. Any tips on my writing or others are welcome.



      "Prove that, for $n geq 3 $, the center of $S_n$ only contains $I_n$ (the identity)".



      Suppose we have $sigma in S_n$ such that $sigma = (a_1,...,a_k)$.
      Let $varphi in Z(S_n)$ be such that $varphi neq sigma$. We can do that because $(a_1,...,a_k)(a_1,a_2) neq (a_1,a_2)(a_1,...,a_k)$, and $(a_1,...,a_k) notin Z(S_n)$.



      Then, we can write $varphi sigma = sigma varphi$. Multiplying by $varphi^{-1}$ the both terms we obtain:



      $varphi sigma varphi^{-1} = sigma varphi varphi^{-1}$.



      We also know that:



      $varphi (a_1,...,a_k) varphi^{-1} = (varphi (a_1),...,varphi (a_k))$.



      Then,



      $(varphi (a_1),...,varphi (a_k)) = sigma = (a_1,...,a_k)$.



      Which means that $varphi = I_n$.



      New proof: (I hope this works now)



      Suppose $varphi in Z(S_n)$ such that $varphi neq I_n$. We can assume that $exists i: varphi (i) neq i$ and $exists j neq i: varphi (j) = i$.



      Now consider $sigma in S_n$ such that $sigma = (j,k), k in { 1,...,n } setminus { i,j }$.



      Then, we have the equality $varphi sigma = sigma varphi$. Computing at $j$:





      • $sigma(varphi(j)) = sigma(i) = i$.


      • $varphi(sigma(j)) = varphi(k) neq i$, as $varphi$ is injective and $kneq j$.


      Then, $varphi$ must be the identity.






      abstract-algebra group-theory




      abstract-algebra group-theory






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      share|cite|improve this question













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      edited Nov 28 '18 at 17:48







      Albelaski

















      asked Nov 28 '18 at 12:13









      AlbelaskiAlbelaski

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          Your proof is flawed. Just because $(varphi(a_1),ldots,varphi(a_k)) = (a_1, ldots, a_k)$, that does not mean that $varphi$ didn't touch the $a_i$'s; it may have cyclically permuted them.



          In addition to cyclically permute the $a_i$ (or actually not doing anything to them), $varphi$ may permute all the other elements of $S_n$ however it likes, and it will still commute with $sigma$. As an example illustrating both of these flaws, take $n = 5$ with $sigma = (123)$. Then $varphi = (123)(45)$ commutes with $sigma$, and we do have $varphisigmavarphi^{-1} = sigma$ as you (correctly) claim, but $varphi$ is not the identity permutation.



          Instead of starting with a $sigmain S_n$, I would start with a $varphiin Z(S_n)$, and if it's not the identity, construct (or find) a $sigmain S_n$ which doesn't commute with it.






          share|cite|improve this answer









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            $begingroup$

            Your proof is flawed. Just because $(varphi(a_1),ldots,varphi(a_k)) = (a_1, ldots, a_k)$, that does not mean that $varphi$ didn't touch the $a_i$'s; it may have cyclically permuted them.



            In addition to cyclically permute the $a_i$ (or actually not doing anything to them), $varphi$ may permute all the other elements of $S_n$ however it likes, and it will still commute with $sigma$. As an example illustrating both of these flaws, take $n = 5$ with $sigma = (123)$. Then $varphi = (123)(45)$ commutes with $sigma$, and we do have $varphisigmavarphi^{-1} = sigma$ as you (correctly) claim, but $varphi$ is not the identity permutation.



            Instead of starting with a $sigmain S_n$, I would start with a $varphiin Z(S_n)$, and if it's not the identity, construct (or find) a $sigmain S_n$ which doesn't commute with it.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Your proof is flawed. Just because $(varphi(a_1),ldots,varphi(a_k)) = (a_1, ldots, a_k)$, that does not mean that $varphi$ didn't touch the $a_i$'s; it may have cyclically permuted them.



              In addition to cyclically permute the $a_i$ (or actually not doing anything to them), $varphi$ may permute all the other elements of $S_n$ however it likes, and it will still commute with $sigma$. As an example illustrating both of these flaws, take $n = 5$ with $sigma = (123)$. Then $varphi = (123)(45)$ commutes with $sigma$, and we do have $varphisigmavarphi^{-1} = sigma$ as you (correctly) claim, but $varphi$ is not the identity permutation.



              Instead of starting with a $sigmain S_n$, I would start with a $varphiin Z(S_n)$, and if it's not the identity, construct (or find) a $sigmain S_n$ which doesn't commute with it.






              share|cite|improve this answer









              $endgroup$
















                1












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                1





                $begingroup$

                Your proof is flawed. Just because $(varphi(a_1),ldots,varphi(a_k)) = (a_1, ldots, a_k)$, that does not mean that $varphi$ didn't touch the $a_i$'s; it may have cyclically permuted them.



                In addition to cyclically permute the $a_i$ (or actually not doing anything to them), $varphi$ may permute all the other elements of $S_n$ however it likes, and it will still commute with $sigma$. As an example illustrating both of these flaws, take $n = 5$ with $sigma = (123)$. Then $varphi = (123)(45)$ commutes with $sigma$, and we do have $varphisigmavarphi^{-1} = sigma$ as you (correctly) claim, but $varphi$ is not the identity permutation.



                Instead of starting with a $sigmain S_n$, I would start with a $varphiin Z(S_n)$, and if it's not the identity, construct (or find) a $sigmain S_n$ which doesn't commute with it.






                share|cite|improve this answer









                $endgroup$



                Your proof is flawed. Just because $(varphi(a_1),ldots,varphi(a_k)) = (a_1, ldots, a_k)$, that does not mean that $varphi$ didn't touch the $a_i$'s; it may have cyclically permuted them.



                In addition to cyclically permute the $a_i$ (or actually not doing anything to them), $varphi$ may permute all the other elements of $S_n$ however it likes, and it will still commute with $sigma$. As an example illustrating both of these flaws, take $n = 5$ with $sigma = (123)$. Then $varphi = (123)(45)$ commutes with $sigma$, and we do have $varphisigmavarphi^{-1} = sigma$ as you (correctly) claim, but $varphi$ is not the identity permutation.



                Instead of starting with a $sigmain S_n$, I would start with a $varphiin Z(S_n)$, and if it's not the identity, construct (or find) a $sigmain S_n$ which doesn't commute with it.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 28 '18 at 12:19









                ArthurArthur

                113k7115197




                113k7115197






























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