Why If $Ha=Hb$ then $a=1a=hb$ for some $hin H$?












1












$begingroup$


I'm reading Robert Ash's Basic Abstract Algebra. There is the following exercise:




Given cosets $Ha,Hb$, show:



$Ha=Hb iff ab^{-1}in H$.




In the book, there is the following answer for one of the implications.




If $Ha=Hb$ $huge[$then $a=1a=hb$ for some $hin H$ $huge ]$, so $ab^{-1}=hin H$.




I don't understand how $Ha=Hb$ implies the rest marked in $huge $. That seems to be a consequence of $H$ being a group, not of $Ha=Hb$.




First, I'd like to say that what follows is - perhaps - not a proof but only a "scratch work" at best.



I have tried to think in terms of the contrapositive $(ab^{-1}notin H implies Haneq Hb).$ I noticed that not only $ab^{-1}$ need to be in $H$ but $ba^{-1}$ also need to be in $H$. So $Ha=Hb$ means:




  • There exists $k$, such that $ah_i=kbh_j$. If $i=j$, then $k=ab^{-1}$.


  • There exists $k$, such that $kah_i=bh_j$. If $i=j$, then $k=ba^{-1}$.


  • If $ab^{-1},ba^{-1}notin H$ the previous items are false.



But what if $ineq j$? Does it make sense to ask this? I don't know how to explain the implication in the given proof and I don't know if in my argument, it makes sense to assume $i=j$.






abstract-algebra group-theory







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  • 1




    $begingroup$
    You've excluded the possibility that $ab^{-1}in H$ and $ba^{-1}notin H$, and vice-versa, not that it makes a difference.
    $endgroup$
    – R. Burton
    Jan 10 at 19:18


















1












$begingroup$


I'm reading Robert Ash's Basic Abstract Algebra. There is the following exercise:




Given cosets $Ha,Hb$, show:



$Ha=Hb iff ab^{-1}in H$.




In the book, there is the following answer for one of the implications.




If $Ha=Hb$ $huge[$then $a=1a=hb$ for some $hin H$ $huge ]$, so $ab^{-1}=hin H$.




I don't understand how $Ha=Hb$ implies the rest marked in $huge $. That seems to be a consequence of $H$ being a group, not of $Ha=Hb$.




First, I'd like to say that what follows is - perhaps - not a proof but only a "scratch work" at best.



I have tried to think in terms of the contrapositive $(ab^{-1}notin H implies Haneq Hb).$ I noticed that not only $ab^{-1}$ need to be in $H$ but $ba^{-1}$ also need to be in $H$. So $Ha=Hb$ means:




  • There exists $k$, such that $ah_i=kbh_j$. If $i=j$, then $k=ab^{-1}$.


  • There exists $k$, such that $kah_i=bh_j$. If $i=j$, then $k=ba^{-1}$.


  • If $ab^{-1},ba^{-1}notin H$ the previous items are false.



But what if $ineq j$? Does it make sense to ask this? I don't know how to explain the implication in the given proof and I don't know if in my argument, it makes sense to assume $i=j$.






abstract-algebra group-theory







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You've excluded the possibility that $ab^{-1}in H$ and $ba^{-1}notin H$, and vice-versa, not that it makes a difference.
    $endgroup$
    – R. Burton
    Jan 10 at 19:18
















1












1








1


0



$begingroup$


I'm reading Robert Ash's Basic Abstract Algebra. There is the following exercise:




Given cosets $Ha,Hb$, show:



$Ha=Hb iff ab^{-1}in H$.




In the book, there is the following answer for one of the implications.




If $Ha=Hb$ $huge[$then $a=1a=hb$ for some $hin H$ $huge ]$, so $ab^{-1}=hin H$.




I don't understand how $Ha=Hb$ implies the rest marked in $huge $. That seems to be a consequence of $H$ being a group, not of $Ha=Hb$.




First, I'd like to say that what follows is - perhaps - not a proof but only a "scratch work" at best.



I have tried to think in terms of the contrapositive $(ab^{-1}notin H implies Haneq Hb).$ I noticed that not only $ab^{-1}$ need to be in $H$ but $ba^{-1}$ also need to be in $H$. So $Ha=Hb$ means:




  • There exists $k$, such that $ah_i=kbh_j$. If $i=j$, then $k=ab^{-1}$.


  • There exists $k$, such that $kah_i=bh_j$. If $i=j$, then $k=ba^{-1}$.


  • If $ab^{-1},ba^{-1}notin H$ the previous items are false.



But what if $ineq j$? Does it make sense to ask this? I don't know how to explain the implication in the given proof and I don't know if in my argument, it makes sense to assume $i=j$.






abstract-algebra group-theory







share|cite|improve this question









$endgroup$




I'm reading Robert Ash's Basic Abstract Algebra. There is the following exercise:




Given cosets $Ha,Hb$, show:



$Ha=Hb iff ab^{-1}in H$.




In the book, there is the following answer for one of the implications.




If $Ha=Hb$ $huge[$then $a=1a=hb$ for some $hin H$ $huge ]$, so $ab^{-1}=hin H$.




I don't understand how $Ha=Hb$ implies the rest marked in $huge $. That seems to be a consequence of $H$ being a group, not of $Ha=Hb$.




First, I'd like to say that what follows is - perhaps - not a proof but only a "scratch work" at best.



I have tried to think in terms of the contrapositive $(ab^{-1}notin H implies Haneq Hb).$ I noticed that not only $ab^{-1}$ need to be in $H$ but $ba^{-1}$ also need to be in $H$. So $Ha=Hb$ means:




  • There exists $k$, such that $ah_i=kbh_j$. If $i=j$, then $k=ab^{-1}$.


  • There exists $k$, such that $kah_i=bh_j$. If $i=j$, then $k=ba^{-1}$.


  • If $ab^{-1},ba^{-1}notin H$ the previous items are false.



But what if $ineq j$? Does it make sense to ask this? I don't know how to explain the implication in the given proof and I don't know if in my argument, it makes sense to assume $i=j$.






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










asked Jan 10 at 18:19









Billy RubinaBilly Rubina

10.4k1460137




10.4k1460137








  • 1




    $begingroup$
    You've excluded the possibility that $ab^{-1}in H$ and $ba^{-1}notin H$, and vice-versa, not that it makes a difference.
    $endgroup$
    – R. Burton
    Jan 10 at 19:18
















  • 1




    $begingroup$
    You've excluded the possibility that $ab^{-1}in H$ and $ba^{-1}notin H$, and vice-versa, not that it makes a difference.
    $endgroup$
    – R. Burton
    Jan 10 at 19:18










1




1




$begingroup$
You've excluded the possibility that $ab^{-1}in H$ and $ba^{-1}notin H$, and vice-versa, not that it makes a difference.
$endgroup$
– R. Burton
Jan 10 at 19:18






$begingroup$
You've excluded the possibility that $ab^{-1}in H$ and $ba^{-1}notin H$, and vice-versa, not that it makes a difference.
$endgroup$
– R. Burton
Jan 10 at 19:18












2 Answers
2






active

oldest

votes


















6












$begingroup$

$Ha=Hb$ means every element in $Ha$, in particular $1cdot a$, is also an element in $Hb$ (and vice-versa), i.e. can be written as $hb$ for some $hin H$, by definition of a right coset.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh, I get it now. This solved a doubt I had but prefered not to write (where are the 1,a,b,h from?). Now It's perfectly clear.
    $endgroup$
    – Billy Rubina
    Jan 11 at 7:27



















2












$begingroup$

I think this is just a wonky way to write it.



If $H$ contains an [left] identity (for what I assume is the multiplication operation), then stating $a=Icdot a$ is largely redundant.



A more intuitive way to write this would be "From $Ha=Hbimplies a=hb$, for some $hin H$ (definition of [right] coset); it follows that $ab^{-1}=hbb^{-1}$; therefore $ab^{-1}=h$".






share|cite|improve this answer









$endgroup$













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    2 Answers
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    2 Answers
    2






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    6












    $begingroup$

    $Ha=Hb$ means every element in $Ha$, in particular $1cdot a$, is also an element in $Hb$ (and vice-versa), i.e. can be written as $hb$ for some $hin H$, by definition of a right coset.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Oh, I get it now. This solved a doubt I had but prefered not to write (where are the 1,a,b,h from?). Now It's perfectly clear.
      $endgroup$
      – Billy Rubina
      Jan 11 at 7:27
















    6












    $begingroup$

    $Ha=Hb$ means every element in $Ha$, in particular $1cdot a$, is also an element in $Hb$ (and vice-versa), i.e. can be written as $hb$ for some $hin H$, by definition of a right coset.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Oh, I get it now. This solved a doubt I had but prefered not to write (where are the 1,a,b,h from?). Now It's perfectly clear.
      $endgroup$
      – Billy Rubina
      Jan 11 at 7:27














    6












    6








    6





    $begingroup$

    $Ha=Hb$ means every element in $Ha$, in particular $1cdot a$, is also an element in $Hb$ (and vice-versa), i.e. can be written as $hb$ for some $hin H$, by definition of a right coset.






    share|cite|improve this answer











    $endgroup$



    $Ha=Hb$ means every element in $Ha$, in particular $1cdot a$, is also an element in $Hb$ (and vice-versa), i.e. can be written as $hb$ for some $hin H$, by definition of a right coset.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 10 at 18:23

























    answered Jan 10 at 18:22









    BernardBernard

    122k740116




    122k740116












    • $begingroup$
      Oh, I get it now. This solved a doubt I had but prefered not to write (where are the 1,a,b,h from?). Now It's perfectly clear.
      $endgroup$
      – Billy Rubina
      Jan 11 at 7:27


















    • $begingroup$
      Oh, I get it now. This solved a doubt I had but prefered not to write (where are the 1,a,b,h from?). Now It's perfectly clear.
      $endgroup$
      – Billy Rubina
      Jan 11 at 7:27
















    $begingroup$
    Oh, I get it now. This solved a doubt I had but prefered not to write (where are the 1,a,b,h from?). Now It's perfectly clear.
    $endgroup$
    – Billy Rubina
    Jan 11 at 7:27




    $begingroup$
    Oh, I get it now. This solved a doubt I had but prefered not to write (where are the 1,a,b,h from?). Now It's perfectly clear.
    $endgroup$
    – Billy Rubina
    Jan 11 at 7:27











    2












    $begingroup$

    I think this is just a wonky way to write it.



    If $H$ contains an [left] identity (for what I assume is the multiplication operation), then stating $a=Icdot a$ is largely redundant.



    A more intuitive way to write this would be "From $Ha=Hbimplies a=hb$, for some $hin H$ (definition of [right] coset); it follows that $ab^{-1}=hbb^{-1}$; therefore $ab^{-1}=h$".






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      I think this is just a wonky way to write it.



      If $H$ contains an [left] identity (for what I assume is the multiplication operation), then stating $a=Icdot a$ is largely redundant.



      A more intuitive way to write this would be "From $Ha=Hbimplies a=hb$, for some $hin H$ (definition of [right] coset); it follows that $ab^{-1}=hbb^{-1}$; therefore $ab^{-1}=h$".






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        I think this is just a wonky way to write it.



        If $H$ contains an [left] identity (for what I assume is the multiplication operation), then stating $a=Icdot a$ is largely redundant.



        A more intuitive way to write this would be "From $Ha=Hbimplies a=hb$, for some $hin H$ (definition of [right] coset); it follows that $ab^{-1}=hbb^{-1}$; therefore $ab^{-1}=h$".






        share|cite|improve this answer









        $endgroup$



        I think this is just a wonky way to write it.



        If $H$ contains an [left] identity (for what I assume is the multiplication operation), then stating $a=Icdot a$ is largely redundant.



        A more intuitive way to write this would be "From $Ha=Hbimplies a=hb$, for some $hin H$ (definition of [right] coset); it follows that $ab^{-1}=hbb^{-1}$; therefore $ab^{-1}=h$".







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 10 at 19:13









        R. BurtonR. Burton

        622110




        622110






























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