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







share|cite|improve this question











$endgroup$

















    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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 28 '18 at 17:48







      Albelaski

















      asked Nov 28 '18 at 12:13









      AlbelaskiAlbelaski

      737




      737






















          1 Answer
          1






          active

          oldest

          votes


















          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 Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017075%2fproof-of-center-of-symmetric-group%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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












              $begingroup$

              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












                1








                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






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017075%2fproof-of-center-of-symmetric-group%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    How to change which sound is reproduced for terminal bell?

                    Image
                    3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
                    Read more

                    Can I use Tabulator js library in my java Spring + Thymeleaf project?

                    Image
                    0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
                    Read more

                    Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

                    Image
                    up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
                    Read more