Question about a statement concerning normal subgroups











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I'm asking the following: it is true that if $K$ is a normal subgroup of $G$ and $Kleq Hleq G$ then $K$ is normal in $H$? I tried to prove it but I failed to do so, so I'm starting to suspect that it is not true. Can you provide me a proof or a counterexample of this statement or hint about its proof?






abstract-algebra group-theory







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    I'm asking the following: it is true that if $K$ is a normal subgroup of $G$ and $Kleq Hleq G$ then $K$ is normal in $H$? I tried to prove it but I failed to do so, so I'm starting to suspect that it is not true. Can you provide me a proof or a counterexample of this statement or hint about its proof?






    abstract-algebra group-theory







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I'm asking the following: it is true that if $K$ is a normal subgroup of $G$ and $Kleq Hleq G$ then $K$ is normal in $H$? I tried to prove it but I failed to do so, so I'm starting to suspect that it is not true. Can you provide me a proof or a counterexample of this statement or hint about its proof?






      abstract-algebra group-theory







      share|cite|improve this question













      I'm asking the following: it is true that if $K$ is a normal subgroup of $G$ and $Kleq Hleq G$ then $K$ is normal in $H$? I tried to prove it but I failed to do so, so I'm starting to suspect that it is not true. Can you provide me a proof or a counterexample of this statement or hint about its proof?






      abstract-algebra group-theory




      abstract-algebra group-theory






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      asked Nov 22 at 17:30









      user573497

      16119




      16119






















          2 Answers
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          Hint:



          You know that



          $$;Klhd Gimplies g^{-1}kgin K;,;;text{for all elements};;kin K,,,,gin G;$$



          Now, $;Hsubset G; $, so...






          share|cite|improve this answer




























            up vote
            1
            down vote













            It is trivially true because $Ktrianglelefteq G$ is equivalent to $gKg^{-1}=K,,forall gin G$. This condition is met $forall gin H$, as $Hsubset G$. Note we do not need normality of $H$.






            share|cite|improve this answer

















            • 1




              Oh, now I get it! Since $H$ is a subgroup of $G$ and $gkg^(-1)$ belongs to $K$ for all $g$ in $G$ it also holds for all $h$ in $H$.
              – user573497
              Nov 22 at 20:02











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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote













            Hint:



            You know that



            $$;Klhd Gimplies g^{-1}kgin K;,;;text{for all elements};;kin K,,,,gin G;$$



            Now, $;Hsubset G; $, so...






            share|cite|improve this answer

























              up vote
              1
              down vote













              Hint:



              You know that



              $$;Klhd Gimplies g^{-1}kgin K;,;;text{for all elements};;kin K,,,,gin G;$$



              Now, $;Hsubset G; $, so...






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                Hint:



                You know that



                $$;Klhd Gimplies g^{-1}kgin K;,;;text{for all elements};;kin K,,,,gin G;$$



                Now, $;Hsubset G; $, so...






                share|cite|improve this answer












                Hint:



                You know that



                $$;Klhd Gimplies g^{-1}kgin K;,;;text{for all elements};;kin K,,,,gin G;$$



                Now, $;Hsubset G; $, so...







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 22 at 17:39









                DonAntonio

                176k1491225




                176k1491225






















                    up vote
                    1
                    down vote













                    It is trivially true because $Ktrianglelefteq G$ is equivalent to $gKg^{-1}=K,,forall gin G$. This condition is met $forall gin H$, as $Hsubset G$. Note we do not need normality of $H$.






                    share|cite|improve this answer

















                    • 1




                      Oh, now I get it! Since $H$ is a subgroup of $G$ and $gkg^(-1)$ belongs to $K$ for all $g$ in $G$ it also holds for all $h$ in $H$.
                      – user573497
                      Nov 22 at 20:02















                    up vote
                    1
                    down vote













                    It is trivially true because $Ktrianglelefteq G$ is equivalent to $gKg^{-1}=K,,forall gin G$. This condition is met $forall gin H$, as $Hsubset G$. Note we do not need normality of $H$.






                    share|cite|improve this answer

















                    • 1




                      Oh, now I get it! Since $H$ is a subgroup of $G$ and $gkg^(-1)$ belongs to $K$ for all $g$ in $G$ it also holds for all $h$ in $H$.
                      – user573497
                      Nov 22 at 20:02













                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    It is trivially true because $Ktrianglelefteq G$ is equivalent to $gKg^{-1}=K,,forall gin G$. This condition is met $forall gin H$, as $Hsubset G$. Note we do not need normality of $H$.






                    share|cite|improve this answer












                    It is trivially true because $Ktrianglelefteq G$ is equivalent to $gKg^{-1}=K,,forall gin G$. This condition is met $forall gin H$, as $Hsubset G$. Note we do not need normality of $H$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 22 at 17:44









                    Chris Custer

                    10.2k3724




                    10.2k3724








                    • 1




                      Oh, now I get it! Since $H$ is a subgroup of $G$ and $gkg^(-1)$ belongs to $K$ for all $g$ in $G$ it also holds for all $h$ in $H$.
                      – user573497
                      Nov 22 at 20:02














                    • 1




                      Oh, now I get it! Since $H$ is a subgroup of $G$ and $gkg^(-1)$ belongs to $K$ for all $g$ in $G$ it also holds for all $h$ in $H$.
                      – user573497
                      Nov 22 at 20:02








                    1




                    1




                    Oh, now I get it! Since $H$ is a subgroup of $G$ and $gkg^(-1)$ belongs to $K$ for all $g$ in $G$ it also holds for all $h$ in $H$.
                    – user573497
                    Nov 22 at 20:02




                    Oh, now I get it! Since $H$ is a subgroup of $G$ and $gkg^(-1)$ belongs to $K$ for all $g$ in $G$ it also holds for all $h$ in $H$.
                    – user573497
                    Nov 22 at 20:02


















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