Suppose $H$ is a subgroup of a group $G$ and $aH$ is a left coset. Prove that there exists some $K$ (a...












2












$begingroup$



Suppose $H$ is a subgroup of a group $G$ and $aH$ is a left coset. Prove that there exists some $K$ (a subgroup of $G$) , which $aH$ is equal to $Ka$.




I've tried to show this statement,but I cant find exactly what is $K$.






abstract-algebra group-theory quotient-group







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    If $aH = Ka$, what happens if you move the $a$ to one side?
    $endgroup$
    – Tobias Kildetoft
    Dec 13 '18 at 8:27










  • $begingroup$
    It implies that aHa^-1=K which means that a should be in H
    $endgroup$
    – Arman_jr
    Dec 13 '18 at 8:29










  • $begingroup$
    but we want to prove the existance of such K
    $endgroup$
    – Arman_jr
    Dec 13 '18 at 8:30










  • $begingroup$
    First part yes. But why would that imply that $ain H$?
    $endgroup$
    – Tobias Kildetoft
    Dec 13 '18 at 8:30










  • $begingroup$
    You now have $K = aHa^{-1}$. There is nothing further to be done.
    $endgroup$
    – Tobias Kildetoft
    Dec 13 '18 at 8:32
















2












$begingroup$



Suppose $H$ is a subgroup of a group $G$ and $aH$ is a left coset. Prove that there exists some $K$ (a subgroup of $G$) , which $aH$ is equal to $Ka$.




I've tried to show this statement,but I cant find exactly what is $K$.






abstract-algebra group-theory quotient-group







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    If $aH = Ka$, what happens if you move the $a$ to one side?
    $endgroup$
    – Tobias Kildetoft
    Dec 13 '18 at 8:27










  • $begingroup$
    It implies that aHa^-1=K which means that a should be in H
    $endgroup$
    – Arman_jr
    Dec 13 '18 at 8:29










  • $begingroup$
    but we want to prove the existance of such K
    $endgroup$
    – Arman_jr
    Dec 13 '18 at 8:30










  • $begingroup$
    First part yes. But why would that imply that $ain H$?
    $endgroup$
    – Tobias Kildetoft
    Dec 13 '18 at 8:30










  • $begingroup$
    You now have $K = aHa^{-1}$. There is nothing further to be done.
    $endgroup$
    – Tobias Kildetoft
    Dec 13 '18 at 8:32














2












2








2


2



$begingroup$



Suppose $H$ is a subgroup of a group $G$ and $aH$ is a left coset. Prove that there exists some $K$ (a subgroup of $G$) , which $aH$ is equal to $Ka$.




I've tried to show this statement,but I cant find exactly what is $K$.






abstract-algebra group-theory quotient-group







share|cite|improve this question











$endgroup$





Suppose $H$ is a subgroup of a group $G$ and $aH$ is a left coset. Prove that there exists some $K$ (a subgroup of $G$) , which $aH$ is equal to $Ka$.




I've tried to show this statement,but I cant find exactly what is $K$.






abstract-algebra group-theory quotient-group




abstract-algebra group-theory quotient-group






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 9:01









Batominovski

33.1k33293




33.1k33293










asked Dec 13 '18 at 8:23









Arman_jrArman_jr

235




235








  • 3




    $begingroup$
    If $aH = Ka$, what happens if you move the $a$ to one side?
    $endgroup$
    – Tobias Kildetoft
    Dec 13 '18 at 8:27










  • $begingroup$
    It implies that aHa^-1=K which means that a should be in H
    $endgroup$
    – Arman_jr
    Dec 13 '18 at 8:29










  • $begingroup$
    but we want to prove the existance of such K
    $endgroup$
    – Arman_jr
    Dec 13 '18 at 8:30










  • $begingroup$
    First part yes. But why would that imply that $ain H$?
    $endgroup$
    – Tobias Kildetoft
    Dec 13 '18 at 8:30










  • $begingroup$
    You now have $K = aHa^{-1}$. There is nothing further to be done.
    $endgroup$
    – Tobias Kildetoft
    Dec 13 '18 at 8:32














  • 3




    $begingroup$
    If $aH = Ka$, what happens if you move the $a$ to one side?
    $endgroup$
    – Tobias Kildetoft
    Dec 13 '18 at 8:27










  • $begingroup$
    It implies that aHa^-1=K which means that a should be in H
    $endgroup$
    – Arman_jr
    Dec 13 '18 at 8:29










  • $begingroup$
    but we want to prove the existance of such K
    $endgroup$
    – Arman_jr
    Dec 13 '18 at 8:30










  • $begingroup$
    First part yes. But why would that imply that $ain H$?
    $endgroup$
    – Tobias Kildetoft
    Dec 13 '18 at 8:30










  • $begingroup$
    You now have $K = aHa^{-1}$. There is nothing further to be done.
    $endgroup$
    – Tobias Kildetoft
    Dec 13 '18 at 8:32








3




3




$begingroup$
If $aH = Ka$, what happens if you move the $a$ to one side?
$endgroup$
– Tobias Kildetoft
Dec 13 '18 at 8:27




$begingroup$
If $aH = Ka$, what happens if you move the $a$ to one side?
$endgroup$
– Tobias Kildetoft
Dec 13 '18 at 8:27












$begingroup$
It implies that aHa^-1=K which means that a should be in H
$endgroup$
– Arman_jr
Dec 13 '18 at 8:29




$begingroup$
It implies that aHa^-1=K which means that a should be in H
$endgroup$
– Arman_jr
Dec 13 '18 at 8:29












$begingroup$
but we want to prove the existance of such K
$endgroup$
– Arman_jr
Dec 13 '18 at 8:30




$begingroup$
but we want to prove the existance of such K
$endgroup$
– Arman_jr
Dec 13 '18 at 8:30












$begingroup$
First part yes. But why would that imply that $ain H$?
$endgroup$
– Tobias Kildetoft
Dec 13 '18 at 8:30




$begingroup$
First part yes. But why would that imply that $ain H$?
$endgroup$
– Tobias Kildetoft
Dec 13 '18 at 8:30












$begingroup$
You now have $K = aHa^{-1}$. There is nothing further to be done.
$endgroup$
– Tobias Kildetoft
Dec 13 '18 at 8:32




$begingroup$
You now have $K = aHa^{-1}$. There is nothing further to be done.
$endgroup$
– Tobias Kildetoft
Dec 13 '18 at 8:32










2 Answers
2






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oldest

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2












$begingroup$

$$aH=Kaiff aHa^{-1}=K$$
Aldo, note that, by a homomorphism ($xlongmapsto axa^{-1}$ is an automorphism of $G$), the image of a subgroup of $G$ is a subgroup.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint : If $aH = Ka$, then for every $b in H$ there exists $c in K$ such that $ab=ca$. Then, for every $b in H$, there exists $c in K$ such that $c = aba^{-1}$. Or, $aba^{-1} in K$.



    Thus, every element of the form $aba^{-1}$ belongs to $K$, where $b in H$. We write this as $aHa^{-1} subset K$, where $aHa^{-1} = {aba^{-1} : b in H}$. Now show that this is a subgroup, and show that in fact $K = aHa^{-1}$. (Hint : we only used the fact that $aH subset Ka$ so far. Use the fact that $Ka subset aH$ for this conclusion).






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






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      2 Answers
      2






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      2












      $begingroup$

      $$aH=Kaiff aHa^{-1}=K$$
      Aldo, note that, by a homomorphism ($xlongmapsto axa^{-1}$ is an automorphism of $G$), the image of a subgroup of $G$ is a subgroup.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        $$aH=Kaiff aHa^{-1}=K$$
        Aldo, note that, by a homomorphism ($xlongmapsto axa^{-1}$ is an automorphism of $G$), the image of a subgroup of $G$ is a subgroup.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          $$aH=Kaiff aHa^{-1}=K$$
          Aldo, note that, by a homomorphism ($xlongmapsto axa^{-1}$ is an automorphism of $G$), the image of a subgroup of $G$ is a subgroup.






          share|cite|improve this answer









          $endgroup$



          $$aH=Kaiff aHa^{-1}=K$$
          Aldo, note that, by a homomorphism ($xlongmapsto axa^{-1}$ is an automorphism of $G$), the image of a subgroup of $G$ is a subgroup.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 9:18









          BernardBernard

          121k740116




          121k740116























              0












              $begingroup$

              Hint : If $aH = Ka$, then for every $b in H$ there exists $c in K$ such that $ab=ca$. Then, for every $b in H$, there exists $c in K$ such that $c = aba^{-1}$. Or, $aba^{-1} in K$.



              Thus, every element of the form $aba^{-1}$ belongs to $K$, where $b in H$. We write this as $aHa^{-1} subset K$, where $aHa^{-1} = {aba^{-1} : b in H}$. Now show that this is a subgroup, and show that in fact $K = aHa^{-1}$. (Hint : we only used the fact that $aH subset Ka$ so far. Use the fact that $Ka subset aH$ for this conclusion).






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Hint : If $aH = Ka$, then for every $b in H$ there exists $c in K$ such that $ab=ca$. Then, for every $b in H$, there exists $c in K$ such that $c = aba^{-1}$. Or, $aba^{-1} in K$.



                Thus, every element of the form $aba^{-1}$ belongs to $K$, where $b in H$. We write this as $aHa^{-1} subset K$, where $aHa^{-1} = {aba^{-1} : b in H}$. Now show that this is a subgroup, and show that in fact $K = aHa^{-1}$. (Hint : we only used the fact that $aH subset Ka$ so far. Use the fact that $Ka subset aH$ for this conclusion).






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint : If $aH = Ka$, then for every $b in H$ there exists $c in K$ such that $ab=ca$. Then, for every $b in H$, there exists $c in K$ such that $c = aba^{-1}$. Or, $aba^{-1} in K$.



                  Thus, every element of the form $aba^{-1}$ belongs to $K$, where $b in H$. We write this as $aHa^{-1} subset K$, where $aHa^{-1} = {aba^{-1} : b in H}$. Now show that this is a subgroup, and show that in fact $K = aHa^{-1}$. (Hint : we only used the fact that $aH subset Ka$ so far. Use the fact that $Ka subset aH$ for this conclusion).






                  share|cite|improve this answer









                  $endgroup$



                  Hint : If $aH = Ka$, then for every $b in H$ there exists $c in K$ such that $ab=ca$. Then, for every $b in H$, there exists $c in K$ such that $c = aba^{-1}$. Or, $aba^{-1} in K$.



                  Thus, every element of the form $aba^{-1}$ belongs to $K$, where $b in H$. We write this as $aHa^{-1} subset K$, where $aHa^{-1} = {aba^{-1} : b in H}$. Now show that this is a subgroup, and show that in fact $K = aHa^{-1}$. (Hint : we only used the fact that $aH subset Ka$ so far. Use the fact that $Ka subset aH$ for this conclusion).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 13 '18 at 8:28









                  астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

                  38.8k33477




                  38.8k33477






























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