Intersection of two left cosets is empty












2












$begingroup$


I have $theta,tau, sigmain S_7$, where $theta=(13)(14)(567), tau=(137)(2456),
sigma=(13456)(27)$. I need to find $sigma langlethetaranglecaptaulanglethetarangle$.



I approached this by first computing $langlethetarangle$. I got $langlethetarangle={(13)(24)(567),(576),(13)(24),(675),(13)(24)(576),id}$.



I then worked out the left coset $sigmalanglethetarangle$ by multiplying $(13456)(27)$ by every element in $langlethetarangle$, one by one. That gave me ${(1574632),(134726),(274563),(1346)(257),(263)(457)}$.



Using the same technique, I found $taulanglethetarangle$ to be ${(357)(46),(136247),(37)(456),(1357)(246),(3647),id}$.



But then I get $sigmalanglethetaranglecaptaulanglethetarangle=phi$, since there are no elements occurring in both $sigmalanglethetarangle$ and $taulanglethetarangle$.



Why is the intersection empty? If I have done this right, is there an easier way to tell that the intersection of two cosets will be empty, without having to work out each individual result (since that is tedious)?






abstract-algebra group-theory







share|cite|improve this question







New contributor




Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 2




    $begingroup$
    Do you recall the fact that two left cosets are either equal or disjoint? (True for any group and subgroup.)
    $endgroup$
    – darij grinberg
    7 hours ago


















2












$begingroup$


I have $theta,tau, sigmain S_7$, where $theta=(13)(14)(567), tau=(137)(2456),
sigma=(13456)(27)$. I need to find $sigma langlethetaranglecaptaulanglethetarangle$.



I approached this by first computing $langlethetarangle$. I got $langlethetarangle={(13)(24)(567),(576),(13)(24),(675),(13)(24)(576),id}$.



I then worked out the left coset $sigmalanglethetarangle$ by multiplying $(13456)(27)$ by every element in $langlethetarangle$, one by one. That gave me ${(1574632),(134726),(274563),(1346)(257),(263)(457)}$.



Using the same technique, I found $taulanglethetarangle$ to be ${(357)(46),(136247),(37)(456),(1357)(246),(3647),id}$.



But then I get $sigmalanglethetaranglecaptaulanglethetarangle=phi$, since there are no elements occurring in both $sigmalanglethetarangle$ and $taulanglethetarangle$.



Why is the intersection empty? If I have done this right, is there an easier way to tell that the intersection of two cosets will be empty, without having to work out each individual result (since that is tedious)?






abstract-algebra group-theory







share|cite|improve this question







New contributor




Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 2




    $begingroup$
    Do you recall the fact that two left cosets are either equal or disjoint? (True for any group and subgroup.)
    $endgroup$
    – darij grinberg
    7 hours ago
















2












2








2





$begingroup$


I have $theta,tau, sigmain S_7$, where $theta=(13)(14)(567), tau=(137)(2456),
sigma=(13456)(27)$. I need to find $sigma langlethetaranglecaptaulanglethetarangle$.



I approached this by first computing $langlethetarangle$. I got $langlethetarangle={(13)(24)(567),(576),(13)(24),(675),(13)(24)(576),id}$.



I then worked out the left coset $sigmalanglethetarangle$ by multiplying $(13456)(27)$ by every element in $langlethetarangle$, one by one. That gave me ${(1574632),(134726),(274563),(1346)(257),(263)(457)}$.



Using the same technique, I found $taulanglethetarangle$ to be ${(357)(46),(136247),(37)(456),(1357)(246),(3647),id}$.



But then I get $sigmalanglethetaranglecaptaulanglethetarangle=phi$, since there are no elements occurring in both $sigmalanglethetarangle$ and $taulanglethetarangle$.



Why is the intersection empty? If I have done this right, is there an easier way to tell that the intersection of two cosets will be empty, without having to work out each individual result (since that is tedious)?






abstract-algebra group-theory







share|cite|improve this question







New contributor




Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I have $theta,tau, sigmain S_7$, where $theta=(13)(14)(567), tau=(137)(2456),
sigma=(13456)(27)$. I need to find $sigma langlethetaranglecaptaulanglethetarangle$.



I approached this by first computing $langlethetarangle$. I got $langlethetarangle={(13)(24)(567),(576),(13)(24),(675),(13)(24)(576),id}$.



I then worked out the left coset $sigmalanglethetarangle$ by multiplying $(13456)(27)$ by every element in $langlethetarangle$, one by one. That gave me ${(1574632),(134726),(274563),(1346)(257),(263)(457)}$.



Using the same technique, I found $taulanglethetarangle$ to be ${(357)(46),(136247),(37)(456),(1357)(246),(3647),id}$.



But then I get $sigmalanglethetaranglecaptaulanglethetarangle=phi$, since there are no elements occurring in both $sigmalanglethetarangle$ and $taulanglethetarangle$.



Why is the intersection empty? If I have done this right, is there an easier way to tell that the intersection of two cosets will be empty, without having to work out each individual result (since that is tedious)?






abstract-algebra group-theory




abstract-algebra group-theory






share|cite|improve this question







New contributor




Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 7 hours ago









HarmanHarman

122




122




New contributor




Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Harman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 2




    $begingroup$
    Do you recall the fact that two left cosets are either equal or disjoint? (True for any group and subgroup.)
    $endgroup$
    – darij grinberg
    7 hours ago
















  • 2




    $begingroup$
    Do you recall the fact that two left cosets are either equal or disjoint? (True for any group and subgroup.)
    $endgroup$
    – darij grinberg
    7 hours ago










2




2




$begingroup$
Do you recall the fact that two left cosets are either equal or disjoint? (True for any group and subgroup.)
$endgroup$
– darij grinberg
7 hours ago






$begingroup$
Do you recall the fact that two left cosets are either equal or disjoint? (True for any group and subgroup.)
$endgroup$
– darij grinberg
7 hours ago












2 Answers
2






active

oldest

votes


















4












$begingroup$

Left (or right) cosets of a subgroup $;H;$ in a group $;G;$ are in fact equivalence classes, the equivalence relation $;sim;$ on $;G;$ being:



$$text{For};x,yin G;,;; xsim yiff x^{-1}yin H$$



We usually write $;xH=yH;$ instead of $;xsim y;$ . Check this is in fact an equivalence relation, and then we know two equivalence classes are either identical or else they're disjoint, and this just means $;xH = yH;;text{or};;xHcap yH=emptyset;$






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Yes, this is standard. Given two left cosets of the same subgroup, either they're the same as sets, or they don't intersect at all.



    As such, all we need to do to find that two cosets of a subgroup are disjoint is to find a single element of one coset that's not in the other coset. An example that comes up fairly often: if $a$ is not an element of the subgroup $H$, the coset $aH$ doesn't intersect $H$.






    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
      });


      }
      });






      Harman is a new contributor. Be nice, and check out our Code of Conduct.










      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075083%2fintersection-of-two-left-cosets-is-empty%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









      4












      $begingroup$

      Left (or right) cosets of a subgroup $;H;$ in a group $;G;$ are in fact equivalence classes, the equivalence relation $;sim;$ on $;G;$ being:



      $$text{For};x,yin G;,;; xsim yiff x^{-1}yin H$$



      We usually write $;xH=yH;$ instead of $;xsim y;$ . Check this is in fact an equivalence relation, and then we know two equivalence classes are either identical or else they're disjoint, and this just means $;xH = yH;;text{or};;xHcap yH=emptyset;$






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Left (or right) cosets of a subgroup $;H;$ in a group $;G;$ are in fact equivalence classes, the equivalence relation $;sim;$ on $;G;$ being:



        $$text{For};x,yin G;,;; xsim yiff x^{-1}yin H$$



        We usually write $;xH=yH;$ instead of $;xsim y;$ . Check this is in fact an equivalence relation, and then we know two equivalence classes are either identical or else they're disjoint, and this just means $;xH = yH;;text{or};;xHcap yH=emptyset;$






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Left (or right) cosets of a subgroup $;H;$ in a group $;G;$ are in fact equivalence classes, the equivalence relation $;sim;$ on $;G;$ being:



          $$text{For};x,yin G;,;; xsim yiff x^{-1}yin H$$



          We usually write $;xH=yH;$ instead of $;xsim y;$ . Check this is in fact an equivalence relation, and then we know two equivalence classes are either identical or else they're disjoint, and this just means $;xH = yH;;text{or};;xHcap yH=emptyset;$






          share|cite|improve this answer









          $endgroup$



          Left (or right) cosets of a subgroup $;H;$ in a group $;G;$ are in fact equivalence classes, the equivalence relation $;sim;$ on $;G;$ being:



          $$text{For};x,yin G;,;; xsim yiff x^{-1}yin H$$



          We usually write $;xH=yH;$ instead of $;xsim y;$ . Check this is in fact an equivalence relation, and then we know two equivalence classes are either identical or else they're disjoint, and this just means $;xH = yH;;text{or};;xHcap yH=emptyset;$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 7 hours ago









          DonAntonioDonAntonio

          177k1492226




          177k1492226























              3












              $begingroup$

              Yes, this is standard. Given two left cosets of the same subgroup, either they're the same as sets, or they don't intersect at all.



              As such, all we need to do to find that two cosets of a subgroup are disjoint is to find a single element of one coset that's not in the other coset. An example that comes up fairly often: if $a$ is not an element of the subgroup $H$, the coset $aH$ doesn't intersect $H$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Yes, this is standard. Given two left cosets of the same subgroup, either they're the same as sets, or they don't intersect at all.



                As such, all we need to do to find that two cosets of a subgroup are disjoint is to find a single element of one coset that's not in the other coset. An example that comes up fairly often: if $a$ is not an element of the subgroup $H$, the coset $aH$ doesn't intersect $H$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Yes, this is standard. Given two left cosets of the same subgroup, either they're the same as sets, or they don't intersect at all.



                  As such, all we need to do to find that two cosets of a subgroup are disjoint is to find a single element of one coset that's not in the other coset. An example that comes up fairly often: if $a$ is not an element of the subgroup $H$, the coset $aH$ doesn't intersect $H$.






                  share|cite|improve this answer









                  $endgroup$



                  Yes, this is standard. Given two left cosets of the same subgroup, either they're the same as sets, or they don't intersect at all.



                  As such, all we need to do to find that two cosets of a subgroup are disjoint is to find a single element of one coset that's not in the other coset. An example that comes up fairly often: if $a$ is not an element of the subgroup $H$, the coset $aH$ doesn't intersect $H$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 7 hours ago









                  jmerryjmerry

                  3,737514




                  3,737514






















                      Harman is a new contributor. Be nice, and check out our Code of Conduct.










                      draft saved

                      draft discarded


















                      Harman is a new contributor. Be nice, and check out our Code of Conduct.













                      Harman is a new contributor. Be nice, and check out our Code of Conduct.












                      Harman is a new contributor. Be nice, and check out our Code of Conduct.
















                      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%2f3075083%2fintersection-of-two-left-cosets-is-empty%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

                      Eastern Orthodox Church

                      Image
                      Not to be confused with Oriental Orthodoxy, Eastern Christianity, or Eastern Catholic Churches. "Eastern Orthodoxy", "Orthodox Church", "Orthodox Catholic Church", and "Orthodox Christian Church" redirect here. For other uses, see Orthodox (disambiguation). This article may be too long to read and navigate comfortably . The readable prose size is 132 kilobytes. Please consider splitting content into sub-articles, condensing it, or adding or removing subheadings. ( January 2018 ) Part of a series on the Eastern Orthodox Church Mosaic of Christ Pantocrator, Hagia Sophia Overview Structure Theology (History of theology) Liturgy Church history Holy Mysteries View of salvation View of Mary View of icons Background Crucifixion / Resurrection / Ascension of Jesus Christianity Christian Church Apostolic succession Four Marks of the Church Orthodoxy Organization Autocephaly Patriarchate ...
                      Read more

                      Zagreb

                      Image
                      This article is about the Croatian capital city. For other uses, see Zagreb (disambiguation). City in City of Zagreb, Croatia Zagreb City Grad Zagreb City of Zagreb Clockwise, from top: St. Mark's Square, Croatian State Archives, Zagreb Tram, Art Pavilion, Cibona & HOTO towers and Croatian National Theatre. Flag Coat of arms Zagreb Location of Zagreb in Croatia Show map of Croatia Zagreb Zagreb (Europe) Show map of Europe Coordinates: 45°49′N 15°59′E  /  45.817°N 15.983°E  / 45.817; 15.983 Coordinates: 45°49′N 15°59′E  /  45.817°N 15.983°E  / 45.817; 15.983 Country   Croatia County City of Zagreb RC diocese 1094 Free royal city 1242 Unified 1850 Subdivisions 17 city districts 218 local committees (70 settlements) Government  • Type Mayor-Council  • Mayor Milan Bandić (BM 365)  • City Assembly 51 members BM 365, ZL, NS-R (14) SDP, GLAS, HNS, HSS, NH-PS (13) ...
                      Read more

                      Understanding the information contained in the Deep Space Network XML data?

                      Image
                      up vote 1 down vote favorite NASA's Deep Space Network activity can be viewed conveniently at https://eyes.nasa.gov/dsn/dsn.html but if you would like data in a more raw XML form, it can be accessed as described in this answer Below is a small Python script that reads the XML into a list of dictionaries, and saves it to disk as a JSON. The dictionaries are organized at the higher level by dish, then by spacecraft per dish. Question: What are the most-likely meanings of all the data field contained in the Deep Space Network XML data? This is the example url from the other answer, used in the script below: https://eyes.nasa.gov/dsn/data/1365107113.xml def dictify(r,root=True): # https://stackoverflow.com/questions/2148119/how-to-convert-an-xml-string-to-a-dictionary-in-python/30923963#30923963 if root: ret...
                      Read more