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







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$endgroup$








  • 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
    Jan 15 at 22:56


















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









$endgroup$








  • 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
    Jan 15 at 22:56
















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









$endgroup$




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













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asked Jan 15 at 22:54









HarmanHarman

302




302








  • 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
    Jan 15 at 22:56
















  • 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
    Jan 15 at 22:56










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
Jan 15 at 22:56






$begingroup$
Do you recall the fact that two left cosets are either equal or disjoint? (True for any group and subgroup.)
$endgroup$
– darij grinberg
Jan 15 at 22:56












2 Answers
2






active

oldest

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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









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      2 Answers
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      active

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      2 Answers
      2






      active

      oldest

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      active

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      active

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      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 Jan 15 at 22:59









          DonAntonioDonAntonio

          179k1494230




          179k1494230























              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 Jan 15 at 22:57









                  jmerryjmerry

                  11.1k1225




                  11.1k1225






























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