Permutation group equation: $x^{20} = sigma$












4












$begingroup$


I need some help solving the next permutations group equation:
$$ x^{20} = sigma $$ where



$$
sigma = begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \
7 & 4 & 5 & 6 & 1 & 8 & 3 & 10 & 12 & 2 & 13 & 9 & 11
end{pmatrix}.
$$



I've already found out that the product of disjoint cycles is $(1 7 3 5)(2 4 6 8 10)(9 12)(11 13)$, the order is $order(sigma)=20$ and the sign is $sgn(sigma)=1$, but I don't know how to use all these to solve the equation above.






permutations permutation-cycles







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The arguments I explained here apply to this problem as well. Solution #1: Because $20$ is a factor of $operatorname{ord}(sigma)$ it follows that $x$ must be of order $400$. There are no permutations of that order in the group $S_{13}$. Solution #2: If a fifth power of a permutations has a $5$-cycle, then it must have five of them. Here $sigma=(x^4)^5$ should be a fifth power, but it only has a single 5-cycle. Of course, Vladimir's simple argument is a more delightful way of settling this!
    $endgroup$
    – Jyrki Lahtonen
    Jan 31 at 11:20


















4












$begingroup$


I need some help solving the next permutations group equation:
$$ x^{20} = sigma $$ where



$$
sigma = begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \
7 & 4 & 5 & 6 & 1 & 8 & 3 & 10 & 12 & 2 & 13 & 9 & 11
end{pmatrix}.
$$



I've already found out that the product of disjoint cycles is $(1 7 3 5)(2 4 6 8 10)(9 12)(11 13)$, the order is $order(sigma)=20$ and the sign is $sgn(sigma)=1$, but I don't know how to use all these to solve the equation above.






permutations permutation-cycles







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The arguments I explained here apply to this problem as well. Solution #1: Because $20$ is a factor of $operatorname{ord}(sigma)$ it follows that $x$ must be of order $400$. There are no permutations of that order in the group $S_{13}$. Solution #2: If a fifth power of a permutations has a $5$-cycle, then it must have five of them. Here $sigma=(x^4)^5$ should be a fifth power, but it only has a single 5-cycle. Of course, Vladimir's simple argument is a more delightful way of settling this!
    $endgroup$
    – Jyrki Lahtonen
    Jan 31 at 11:20
















4












4








4


1



$begingroup$


I need some help solving the next permutations group equation:
$$ x^{20} = sigma $$ where



$$
sigma = begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \
7 & 4 & 5 & 6 & 1 & 8 & 3 & 10 & 12 & 2 & 13 & 9 & 11
end{pmatrix}.
$$



I've already found out that the product of disjoint cycles is $(1 7 3 5)(2 4 6 8 10)(9 12)(11 13)$, the order is $order(sigma)=20$ and the sign is $sgn(sigma)=1$, but I don't know how to use all these to solve the equation above.






permutations permutation-cycles







share|cite|improve this question











$endgroup$




I need some help solving the next permutations group equation:
$$ x^{20} = sigma $$ where



$$
sigma = begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \
7 & 4 & 5 & 6 & 1 & 8 & 3 & 10 & 12 & 2 & 13 & 9 & 11
end{pmatrix}.
$$



I've already found out that the product of disjoint cycles is $(1 7 3 5)(2 4 6 8 10)(9 12)(11 13)$, the order is $order(sigma)=20$ and the sign is $sgn(sigma)=1$, but I don't know how to use all these to solve the equation above.






permutations permutation-cycles




permutations permutation-cycles






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 31 at 11:04









Thomas Lesgourgues

1,320220




1,320220










asked Jan 31 at 10:37









Bianca PascuBianca Pascu

213




213








  • 1




    $begingroup$
    The arguments I explained here apply to this problem as well. Solution #1: Because $20$ is a factor of $operatorname{ord}(sigma)$ it follows that $x$ must be of order $400$. There are no permutations of that order in the group $S_{13}$. Solution #2: If a fifth power of a permutations has a $5$-cycle, then it must have five of them. Here $sigma=(x^4)^5$ should be a fifth power, but it only has a single 5-cycle. Of course, Vladimir's simple argument is a more delightful way of settling this!
    $endgroup$
    – Jyrki Lahtonen
    Jan 31 at 11:20
















  • 1




    $begingroup$
    The arguments I explained here apply to this problem as well. Solution #1: Because $20$ is a factor of $operatorname{ord}(sigma)$ it follows that $x$ must be of order $400$. There are no permutations of that order in the group $S_{13}$. Solution #2: If a fifth power of a permutations has a $5$-cycle, then it must have five of them. Here $sigma=(x^4)^5$ should be a fifth power, but it only has a single 5-cycle. Of course, Vladimir's simple argument is a more delightful way of settling this!
    $endgroup$
    – Jyrki Lahtonen
    Jan 31 at 11:20










1




1




$begingroup$
The arguments I explained here apply to this problem as well. Solution #1: Because $20$ is a factor of $operatorname{ord}(sigma)$ it follows that $x$ must be of order $400$. There are no permutations of that order in the group $S_{13}$. Solution #2: If a fifth power of a permutations has a $5$-cycle, then it must have five of them. Here $sigma=(x^4)^5$ should be a fifth power, but it only has a single 5-cycle. Of course, Vladimir's simple argument is a more delightful way of settling this!
$endgroup$
– Jyrki Lahtonen
Jan 31 at 11:20






$begingroup$
The arguments I explained here apply to this problem as well. Solution #1: Because $20$ is a factor of $operatorname{ord}(sigma)$ it follows that $x$ must be of order $400$. There are no permutations of that order in the group $S_{13}$. Solution #2: If a fifth power of a permutations has a $5$-cycle, then it must have five of them. Here $sigma=(x^4)^5$ should be a fifth power, but it only has a single 5-cycle. Of course, Vladimir's simple argument is a more delightful way of settling this!
$endgroup$
– Jyrki Lahtonen
Jan 31 at 11:20












2 Answers
2






active

oldest

votes


















5












$begingroup$

Well, cycles of odd length are even permutations, and cycles of even length are odd permutations. Here we have three cycles of even length and one cycle of odd length, and so $operatorname{sign}(sigma)=(-1)^3cdot1=-1$. On the other hand, $operatorname{sign}(x^{20})=(operatorname{sign}(x))^{20}=1$, and so the equation has no solutions at all.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    There are no solutions:



    If $n=mid xmid$, then $frac n{operatorname{gcd}(n,20)}=20implies n=400$.



    But there are no elements of order $400$ in $S_{13}$. The smallest symmetric group with an element of order $400$ is $S_{33}$, where there is room for a permutation of cycle type $(8,25)$.






    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%2f3094746%2fpermutation-group-equation-x20-sigma%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      Well, cycles of odd length are even permutations, and cycles of even length are odd permutations. Here we have three cycles of even length and one cycle of odd length, and so $operatorname{sign}(sigma)=(-1)^3cdot1=-1$. On the other hand, $operatorname{sign}(x^{20})=(operatorname{sign}(x))^{20}=1$, and so the equation has no solutions at all.






      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        Well, cycles of odd length are even permutations, and cycles of even length are odd permutations. Here we have three cycles of even length and one cycle of odd length, and so $operatorname{sign}(sigma)=(-1)^3cdot1=-1$. On the other hand, $operatorname{sign}(x^{20})=(operatorname{sign}(x))^{20}=1$, and so the equation has no solutions at all.






        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          Well, cycles of odd length are even permutations, and cycles of even length are odd permutations. Here we have three cycles of even length and one cycle of odd length, and so $operatorname{sign}(sigma)=(-1)^3cdot1=-1$. On the other hand, $operatorname{sign}(x^{20})=(operatorname{sign}(x))^{20}=1$, and so the equation has no solutions at all.






          share|cite|improve this answer









          $endgroup$



          Well, cycles of odd length are even permutations, and cycles of even length are odd permutations. Here we have three cycles of even length and one cycle of odd length, and so $operatorname{sign}(sigma)=(-1)^3cdot1=-1$. On the other hand, $operatorname{sign}(x^{20})=(operatorname{sign}(x))^{20}=1$, and so the equation has no solutions at all.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 31 at 11:11









          VladimirVladimir

          5,413618




          5,413618























              1












              $begingroup$

              There are no solutions:



              If $n=mid xmid$, then $frac n{operatorname{gcd}(n,20)}=20implies n=400$.



              But there are no elements of order $400$ in $S_{13}$. The smallest symmetric group with an element of order $400$ is $S_{33}$, where there is room for a permutation of cycle type $(8,25)$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                There are no solutions:



                If $n=mid xmid$, then $frac n{operatorname{gcd}(n,20)}=20implies n=400$.



                But there are no elements of order $400$ in $S_{13}$. The smallest symmetric group with an element of order $400$ is $S_{33}$, where there is room for a permutation of cycle type $(8,25)$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  There are no solutions:



                  If $n=mid xmid$, then $frac n{operatorname{gcd}(n,20)}=20implies n=400$.



                  But there are no elements of order $400$ in $S_{13}$. The smallest symmetric group with an element of order $400$ is $S_{33}$, where there is room for a permutation of cycle type $(8,25)$.






                  share|cite|improve this answer









                  $endgroup$



                  There are no solutions:



                  If $n=mid xmid$, then $frac n{operatorname{gcd}(n,20)}=20implies n=400$.



                  But there are no elements of order $400$ in $S_{13}$. The smallest symmetric group with an element of order $400$ is $S_{33}$, where there is room for a permutation of cycle type $(8,25)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 31 at 11:35









                  Chris CusterChris Custer

                  14.3k3827




                  14.3k3827






























                      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%2f3094746%2fpermutation-group-equation-x20-sigma%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

                      MongoDB - Not Authorized To Execute Command

                      Image
                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
                      Read more

                      How to fix TextFormField cause rebuild widget in Flutter

                      Image
                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
                      Read more

                      in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

                      Image
                      1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
                      Read more