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$H/K$ when $K$ is not a subgroup of $H$.


Maximal normal subgroup not containing an elementCondition under which $HK$ is a subgroupIs $U(1)$ a normal subgroup of $U(2)$ and does the question even make sense?Convincing normal subgroup proof?Example of A, B, G such that A is a normal subgroup of B, B is a normal subgroup of G, but A is not a normal subgroup of GExample for subgroups $H$ and $K$ where $HK = K H$ and neither $H$ nor $K$ is normal?When is $HK$ a subgroup?Group of order 35 must have a normal subgroup of order 5 or 7Finding normal subgroup $N$ in $G$, where $N$ has normal subgroup $H$. But $H$ is not a normal subgroup of $G$.Let $G$ be a group: Prove subgroup, prove that group $G$ does not exist with subgroups s.t…, find cardinality of subgroup intersection.













3












$begingroup$


Let $G$ be a group and $H$, $K$ ($K$ is normal) two subgroups of $G$ but neither $H$ is subgroup of $K$ nor $K$ is subgroup of $H$. What would be the problem with $H/K$? If I define for $x,y in H$ multiplication as $xKyK=xyK$ then it is well defined?



Suppose $aK=xK$ and $bK=yK$ so $a^{-1}x in K$ and $b^{-1}y in K$. Then $b^{-1}a^{-1}xy=b^{-1}yy^{-1}a^{-1}xy in K$. So the multiplication in $H/K$ is well defined no?






group-theory normal-subgroups quotient-spaces quotient-group







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    You're defining $H/K$ to be the set of cosets of $K$ of the form $hK$ with $h$ in $H$?
    $endgroup$
    – Gerry Myerson
    14 hours ago
















3












$begingroup$


Let $G$ be a group and $H$, $K$ ($K$ is normal) two subgroups of $G$ but neither $H$ is subgroup of $K$ nor $K$ is subgroup of $H$. What would be the problem with $H/K$? If I define for $x,y in H$ multiplication as $xKyK=xyK$ then it is well defined?



Suppose $aK=xK$ and $bK=yK$ so $a^{-1}x in K$ and $b^{-1}y in K$. Then $b^{-1}a^{-1}xy=b^{-1}yy^{-1}a^{-1}xy in K$. So the multiplication in $H/K$ is well defined no?






group-theory normal-subgroups quotient-spaces quotient-group







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    You're defining $H/K$ to be the set of cosets of $K$ of the form $hK$ with $h$ in $H$?
    $endgroup$
    – Gerry Myerson
    14 hours ago














3












3








3





$begingroup$


Let $G$ be a group and $H$, $K$ ($K$ is normal) two subgroups of $G$ but neither $H$ is subgroup of $K$ nor $K$ is subgroup of $H$. What would be the problem with $H/K$? If I define for $x,y in H$ multiplication as $xKyK=xyK$ then it is well defined?



Suppose $aK=xK$ and $bK=yK$ so $a^{-1}x in K$ and $b^{-1}y in K$. Then $b^{-1}a^{-1}xy=b^{-1}yy^{-1}a^{-1}xy in K$. So the multiplication in $H/K$ is well defined no?






group-theory normal-subgroups quotient-spaces quotient-group







share|cite|improve this question











$endgroup$




Let $G$ be a group and $H$, $K$ ($K$ is normal) two subgroups of $G$ but neither $H$ is subgroup of $K$ nor $K$ is subgroup of $H$. What would be the problem with $H/K$? If I define for $x,y in H$ multiplication as $xKyK=xyK$ then it is well defined?



Suppose $aK=xK$ and $bK=yK$ so $a^{-1}x in K$ and $b^{-1}y in K$. Then $b^{-1}a^{-1}xy=b^{-1}yy^{-1}a^{-1}xy in K$. So the multiplication in $H/K$ is well defined no?






group-theory normal-subgroups quotient-spaces quotient-group




group-theory normal-subgroups quotient-spaces quotient-group






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 13 hours ago









user1729

17.4k64193




17.4k64193










asked 14 hours ago









roi_saumonroi_saumon

59438




59438








  • 3




    $begingroup$
    You're defining $H/K$ to be the set of cosets of $K$ of the form $hK$ with $h$ in $H$?
    $endgroup$
    – Gerry Myerson
    14 hours ago














  • 3




    $begingroup$
    You're defining $H/K$ to be the set of cosets of $K$ of the form $hK$ with $h$ in $H$?
    $endgroup$
    – Gerry Myerson
    14 hours ago








3




3




$begingroup$
You're defining $H/K$ to be the set of cosets of $K$ of the form $hK$ with $h$ in $H$?
$endgroup$
– Gerry Myerson
14 hours ago




$begingroup$
You're defining $H/K$ to be the set of cosets of $K$ of the form $hK$ with $h$ in $H$?
$endgroup$
– Gerry Myerson
14 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

This is fine. What you are really defining is multiplication in $HK/K$, which is a subgroup of $G/K$. By the second isomorphism theorem, this is isomorphic to $H/(Hcap K) $.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    You can think of this in terms of homomorphisms. That is, by the first isomorphism theorem there exists a homomorphism $phi: Grightarrow overline{G}$ such that $kerphi=K$ (and, of course, $overline{G}$ is really just $G/K$). Then what you are after is $phi(H)$. (This has the benefit of "making sense", while $H/K$ is a slightly dubious concept as for the notation to make sense you really need $Kleq H$.)



    All you are really doing is proving that multiplication in $phi(H)$ is well-defined. Which it is, as $phi$ is a homomorphism.






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      This is fine. What you are really defining is multiplication in $HK/K$, which is a subgroup of $G/K$. By the second isomorphism theorem, this is isomorphic to $H/(Hcap K) $.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        This is fine. What you are really defining is multiplication in $HK/K$, which is a subgroup of $G/K$. By the second isomorphism theorem, this is isomorphic to $H/(Hcap K) $.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          This is fine. What you are really defining is multiplication in $HK/K$, which is a subgroup of $G/K$. By the second isomorphism theorem, this is isomorphic to $H/(Hcap K) $.






          share|cite|improve this answer









          $endgroup$



          This is fine. What you are really defining is multiplication in $HK/K$, which is a subgroup of $G/K$. By the second isomorphism theorem, this is isomorphic to $H/(Hcap K) $.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 14 hours ago









          Matt SamuelMatt Samuel

          38.8k63769




          38.8k63769























              3












              $begingroup$

              You can think of this in terms of homomorphisms. That is, by the first isomorphism theorem there exists a homomorphism $phi: Grightarrow overline{G}$ such that $kerphi=K$ (and, of course, $overline{G}$ is really just $G/K$). Then what you are after is $phi(H)$. (This has the benefit of "making sense", while $H/K$ is a slightly dubious concept as for the notation to make sense you really need $Kleq H$.)



              All you are really doing is proving that multiplication in $phi(H)$ is well-defined. Which it is, as $phi$ is a homomorphism.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                You can think of this in terms of homomorphisms. That is, by the first isomorphism theorem there exists a homomorphism $phi: Grightarrow overline{G}$ such that $kerphi=K$ (and, of course, $overline{G}$ is really just $G/K$). Then what you are after is $phi(H)$. (This has the benefit of "making sense", while $H/K$ is a slightly dubious concept as for the notation to make sense you really need $Kleq H$.)



                All you are really doing is proving that multiplication in $phi(H)$ is well-defined. Which it is, as $phi$ is a homomorphism.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  You can think of this in terms of homomorphisms. That is, by the first isomorphism theorem there exists a homomorphism $phi: Grightarrow overline{G}$ such that $kerphi=K$ (and, of course, $overline{G}$ is really just $G/K$). Then what you are after is $phi(H)$. (This has the benefit of "making sense", while $H/K$ is a slightly dubious concept as for the notation to make sense you really need $Kleq H$.)



                  All you are really doing is proving that multiplication in $phi(H)$ is well-defined. Which it is, as $phi$ is a homomorphism.






                  share|cite|improve this answer











                  $endgroup$



                  You can think of this in terms of homomorphisms. That is, by the first isomorphism theorem there exists a homomorphism $phi: Grightarrow overline{G}$ such that $kerphi=K$ (and, of course, $overline{G}$ is really just $G/K$). Then what you are after is $phi(H)$. (This has the benefit of "making sense", while $H/K$ is a slightly dubious concept as for the notation to make sense you really need $Kleq H$.)



                  All you are really doing is proving that multiplication in $phi(H)$ is well-defined. Which it is, as $phi$ is a homomorphism.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 13 hours ago

























                  answered 13 hours ago









                  user1729user1729

                  17.4k64193




                  17.4k64193






























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