is the intersection of subgroups a subgroup of each subgroup












1












$begingroup$



Suppose $G$ is a group, take $H,K$ as subgroups of $G$ so $H,Kleqslant G$. I know that $Hcap Kleqslant G$ but is it the case that $Hcap Kleqslant H$ and $Hcap Kleqslant K$?




I am guessing this does not hold but why?



Also I tried with the case that $H=langle g rangle,K=langle h rangle$ where $g$ and $h$ are the elements in $G$ ($langle h rangle$ means the minimum subgroup that contains the element $h$ if you haven't seen this notation before). I used the subspace test and I think that $Hcap Kleqslant H$ and $Hcap Kleqslant K$ hold unless I make a mistake somewhere.



Much thanks in advance!






abstract-algebra group-theory







share|cite|improve this question











$endgroup$

















    1












    $begingroup$



    Suppose $G$ is a group, take $H,K$ as subgroups of $G$ so $H,Kleqslant G$. I know that $Hcap Kleqslant G$ but is it the case that $Hcap Kleqslant H$ and $Hcap Kleqslant K$?




    I am guessing this does not hold but why?



    Also I tried with the case that $H=langle g rangle,K=langle h rangle$ where $g$ and $h$ are the elements in $G$ ($langle h rangle$ means the minimum subgroup that contains the element $h$ if you haven't seen this notation before). I used the subspace test and I think that $Hcap Kleqslant H$ and $Hcap Kleqslant K$ hold unless I make a mistake somewhere.



    Much thanks in advance!






    abstract-algebra group-theory







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      Suppose $G$ is a group, take $H,K$ as subgroups of $G$ so $H,Kleqslant G$. I know that $Hcap Kleqslant G$ but is it the case that $Hcap Kleqslant H$ and $Hcap Kleqslant K$?




      I am guessing this does not hold but why?



      Also I tried with the case that $H=langle g rangle,K=langle h rangle$ where $g$ and $h$ are the elements in $G$ ($langle h rangle$ means the minimum subgroup that contains the element $h$ if you haven't seen this notation before). I used the subspace test and I think that $Hcap Kleqslant H$ and $Hcap Kleqslant K$ hold unless I make a mistake somewhere.



      Much thanks in advance!






      abstract-algebra group-theory







      share|cite|improve this question











      $endgroup$





      Suppose $G$ is a group, take $H,K$ as subgroups of $G$ so $H,Kleqslant G$. I know that $Hcap Kleqslant G$ but is it the case that $Hcap Kleqslant H$ and $Hcap Kleqslant K$?




      I am guessing this does not hold but why?



      Also I tried with the case that $H=langle g rangle,K=langle h rangle$ where $g$ and $h$ are the elements in $G$ ($langle h rangle$ means the minimum subgroup that contains the element $h$ if you haven't seen this notation before). I used the subspace test and I think that $Hcap Kleqslant H$ and $Hcap Kleqslant K$ hold unless I make a mistake somewhere.



      Much thanks in advance!






      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








      edited 51 mins ago









      Shaun

      10.3k113686




      10.3k113686










      asked 6 hours ago









      JustWanderingJustWandering

      592




      592






















          2 Answers
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          3












          $begingroup$

          It is indeed true that $H cap K$ is a subgroup of both $H$ and $K$. For the sake of clarity, we recall the definition of a subgroup below:




          Definition. Let $(G, odot)$ be group with identity $e$ and let $H$ be a subset of $G$. We say that $H$ is a subgroup of $G$ if each of the following hold:





          1. $e in H$,

          2. if $h_1,h_2 in H$ then $h_1 odot h_2 in H$,

          3. for all $h in H$, its inverse element $h^{-1}$ with respect to $odot$ is also in $H$.




          These axioms make it so that $(H, odot)$ is a group in its own right with the very same group operation $cdot$ and identity $e$. We also point out that these properties have less to do with the set $G$ than the operation $odot$ that $G$ comes equipped with.



          Now, let $G$ be a group and let $H,K$ be subgroups of $G$. You have already verified that $H cap K$ is a subgroup of $G$. Why must it also be a subgroup of $H$ (and $K$)? First, it's clear that $H cap K subseteq H, H cap K subseteq K$ and that $H cap K ni e$. Moreover, because $H cap K$ is a subgroup of $G$, it is satisfies properties 2. and 3. above. Thus, by replacing $G$ with $H$ or $K$ in the definition above, it's immediate that $H cap K$ is a subgroup of $H$ (and $K$) as well!



          In short, as long as $H$ is a subset of a group $G$ and $H$ satisfies the properties listed above, it will be a subgroup of $G$.






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            The subgroup test is:




            $H$ is a subgroup of $G$ if and only if for each $x,yin H$ we've $xy^{-1}in H$.




            Applied to a collection of subgroups ${H_s}$, let $x,yinbigcap_sH_s$ be a pair of elements. Then $x,yin H_s$ for all $S$ and since $H_s<G$ then $xy^{-1}in H_s$ for every index $s$, so $xy^{-1}inbigcap_sH_s$, hence $bigcap_sH_s$ is a subgroup too.






            share|cite|improve this answer









            $endgroup$














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              $begingroup$

              It is indeed true that $H cap K$ is a subgroup of both $H$ and $K$. For the sake of clarity, we recall the definition of a subgroup below:




              Definition. Let $(G, odot)$ be group with identity $e$ and let $H$ be a subset of $G$. We say that $H$ is a subgroup of $G$ if each of the following hold:





              1. $e in H$,

              2. if $h_1,h_2 in H$ then $h_1 odot h_2 in H$,

              3. for all $h in H$, its inverse element $h^{-1}$ with respect to $odot$ is also in $H$.




              These axioms make it so that $(H, odot)$ is a group in its own right with the very same group operation $cdot$ and identity $e$. We also point out that these properties have less to do with the set $G$ than the operation $odot$ that $G$ comes equipped with.



              Now, let $G$ be a group and let $H,K$ be subgroups of $G$. You have already verified that $H cap K$ is a subgroup of $G$. Why must it also be a subgroup of $H$ (and $K$)? First, it's clear that $H cap K subseteq H, H cap K subseteq K$ and that $H cap K ni e$. Moreover, because $H cap K$ is a subgroup of $G$, it is satisfies properties 2. and 3. above. Thus, by replacing $G$ with $H$ or $K$ in the definition above, it's immediate that $H cap K$ is a subgroup of $H$ (and $K$) as well!



              In short, as long as $H$ is a subset of a group $G$ and $H$ satisfies the properties listed above, it will be a subgroup of $G$.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                It is indeed true that $H cap K$ is a subgroup of both $H$ and $K$. For the sake of clarity, we recall the definition of a subgroup below:




                Definition. Let $(G, odot)$ be group with identity $e$ and let $H$ be a subset of $G$. We say that $H$ is a subgroup of $G$ if each of the following hold:





                1. $e in H$,

                2. if $h_1,h_2 in H$ then $h_1 odot h_2 in H$,

                3. for all $h in H$, its inverse element $h^{-1}$ with respect to $odot$ is also in $H$.




                These axioms make it so that $(H, odot)$ is a group in its own right with the very same group operation $cdot$ and identity $e$. We also point out that these properties have less to do with the set $G$ than the operation $odot$ that $G$ comes equipped with.



                Now, let $G$ be a group and let $H,K$ be subgroups of $G$. You have already verified that $H cap K$ is a subgroup of $G$. Why must it also be a subgroup of $H$ (and $K$)? First, it's clear that $H cap K subseteq H, H cap K subseteq K$ and that $H cap K ni e$. Moreover, because $H cap K$ is a subgroup of $G$, it is satisfies properties 2. and 3. above. Thus, by replacing $G$ with $H$ or $K$ in the definition above, it's immediate that $H cap K$ is a subgroup of $H$ (and $K$) as well!



                In short, as long as $H$ is a subset of a group $G$ and $H$ satisfies the properties listed above, it will be a subgroup of $G$.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  It is indeed true that $H cap K$ is a subgroup of both $H$ and $K$. For the sake of clarity, we recall the definition of a subgroup below:




                  Definition. Let $(G, odot)$ be group with identity $e$ and let $H$ be a subset of $G$. We say that $H$ is a subgroup of $G$ if each of the following hold:





                  1. $e in H$,

                  2. if $h_1,h_2 in H$ then $h_1 odot h_2 in H$,

                  3. for all $h in H$, its inverse element $h^{-1}$ with respect to $odot$ is also in $H$.




                  These axioms make it so that $(H, odot)$ is a group in its own right with the very same group operation $cdot$ and identity $e$. We also point out that these properties have less to do with the set $G$ than the operation $odot$ that $G$ comes equipped with.



                  Now, let $G$ be a group and let $H,K$ be subgroups of $G$. You have already verified that $H cap K$ is a subgroup of $G$. Why must it also be a subgroup of $H$ (and $K$)? First, it's clear that $H cap K subseteq H, H cap K subseteq K$ and that $H cap K ni e$. Moreover, because $H cap K$ is a subgroup of $G$, it is satisfies properties 2. and 3. above. Thus, by replacing $G$ with $H$ or $K$ in the definition above, it's immediate that $H cap K$ is a subgroup of $H$ (and $K$) as well!



                  In short, as long as $H$ is a subset of a group $G$ and $H$ satisfies the properties listed above, it will be a subgroup of $G$.






                  share|cite|improve this answer











                  $endgroup$



                  It is indeed true that $H cap K$ is a subgroup of both $H$ and $K$. For the sake of clarity, we recall the definition of a subgroup below:




                  Definition. Let $(G, odot)$ be group with identity $e$ and let $H$ be a subset of $G$. We say that $H$ is a subgroup of $G$ if each of the following hold:





                  1. $e in H$,

                  2. if $h_1,h_2 in H$ then $h_1 odot h_2 in H$,

                  3. for all $h in H$, its inverse element $h^{-1}$ with respect to $odot$ is also in $H$.




                  These axioms make it so that $(H, odot)$ is a group in its own right with the very same group operation $cdot$ and identity $e$. We also point out that these properties have less to do with the set $G$ than the operation $odot$ that $G$ comes equipped with.



                  Now, let $G$ be a group and let $H,K$ be subgroups of $G$. You have already verified that $H cap K$ is a subgroup of $G$. Why must it also be a subgroup of $H$ (and $K$)? First, it's clear that $H cap K subseteq H, H cap K subseteq K$ and that $H cap K ni e$. Moreover, because $H cap K$ is a subgroup of $G$, it is satisfies properties 2. and 3. above. Thus, by replacing $G$ with $H$ or $K$ in the definition above, it's immediate that $H cap K$ is a subgroup of $H$ (and $K$) as well!



                  In short, as long as $H$ is a subset of a group $G$ and $H$ satisfies the properties listed above, it will be a subgroup of $G$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 5 hours ago

























                  answered 6 hours ago









                  rolandcyprolandcyp

                  2,309422




                  2,309422























                      1












                      $begingroup$

                      The subgroup test is:




                      $H$ is a subgroup of $G$ if and only if for each $x,yin H$ we've $xy^{-1}in H$.




                      Applied to a collection of subgroups ${H_s}$, let $x,yinbigcap_sH_s$ be a pair of elements. Then $x,yin H_s$ for all $S$ and since $H_s<G$ then $xy^{-1}in H_s$ for every index $s$, so $xy^{-1}inbigcap_sH_s$, hence $bigcap_sH_s$ is a subgroup too.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        The subgroup test is:




                        $H$ is a subgroup of $G$ if and only if for each $x,yin H$ we've $xy^{-1}in H$.




                        Applied to a collection of subgroups ${H_s}$, let $x,yinbigcap_sH_s$ be a pair of elements. Then $x,yin H_s$ for all $S$ and since $H_s<G$ then $xy^{-1}in H_s$ for every index $s$, so $xy^{-1}inbigcap_sH_s$, hence $bigcap_sH_s$ is a subgroup too.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          The subgroup test is:




                          $H$ is a subgroup of $G$ if and only if for each $x,yin H$ we've $xy^{-1}in H$.




                          Applied to a collection of subgroups ${H_s}$, let $x,yinbigcap_sH_s$ be a pair of elements. Then $x,yin H_s$ for all $S$ and since $H_s<G$ then $xy^{-1}in H_s$ for every index $s$, so $xy^{-1}inbigcap_sH_s$, hence $bigcap_sH_s$ is a subgroup too.






                          share|cite|improve this answer









                          $endgroup$



                          The subgroup test is:




                          $H$ is a subgroup of $G$ if and only if for each $x,yin H$ we've $xy^{-1}in H$.




                          Applied to a collection of subgroups ${H_s}$, let $x,yinbigcap_sH_s$ be a pair of elements. Then $x,yin H_s$ for all $S$ and since $H_s<G$ then $xy^{-1}in H_s$ for every index $s$, so $xy^{-1}inbigcap_sH_s$, hence $bigcap_sH_s$ is a subgroup too.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 3 hours ago









                          janmarqzjanmarqz

                          6,25741630




                          6,25741630






























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