is the intersection of subgroups a subgroup of each subgroup












2












$begingroup$



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!










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$endgroup$












  • $begingroup$
    A subgroup is just a subset which behaves as a group under the same operation. The intersection of two subsets is a subset of both. The operation does never change, so subgroups behave like subsets in these aspects.
    $endgroup$
    – M. Winter
    4 hours ago


















2












$begingroup$



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!










share|cite|improve this question











$endgroup$












  • $begingroup$
    A subgroup is just a subset which behaves as a group under the same operation. The intersection of two subsets is a subset of both. The operation does never change, so subgroups behave like subsets in these aspects.
    $endgroup$
    – M. Winter
    4 hours ago
















2












2








2





$begingroup$



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!










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






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited 10 hours ago









Shaun

10.3k113686




10.3k113686










asked 16 hours ago









JustWanderingJustWandering

642




642












  • $begingroup$
    A subgroup is just a subset which behaves as a group under the same operation. The intersection of two subsets is a subset of both. The operation does never change, so subgroups behave like subsets in these aspects.
    $endgroup$
    – M. Winter
    4 hours ago




















  • $begingroup$
    A subgroup is just a subset which behaves as a group under the same operation. The intersection of two subsets is a subset of both. The operation does never change, so subgroups behave like subsets in these aspects.
    $endgroup$
    – M. Winter
    4 hours ago


















$begingroup$
A subgroup is just a subset which behaves as a group under the same operation. The intersection of two subsets is a subset of both. The operation does never change, so subgroups behave like subsets in these aspects.
$endgroup$
– M. Winter
4 hours ago






$begingroup$
A subgroup is just a subset which behaves as a group under the same operation. The intersection of two subsets is a subset of both. The operation does never change, so subgroups behave like subsets in these aspects.
$endgroup$
– M. Winter
4 hours ago












3 Answers
3






active

oldest

votes


















4












$begingroup$

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 $odot$ 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









    $endgroup$





















      0












      $begingroup$

      The intersection of two (or any quantity of) subgroups is always a subgroup of all the intersecting subgroups. Indeed, this is a particular case of the following general remark.




      Remark. A subgroup $H leqslant G$ is also a subgroup of any subgroup $K leqslant G$ containing $H$. More precisesly, if $G$ is a group and $H subseteq K leqslant G$ then:
      begin{equation}
      H text{ is a subgroup of } G
      Leftrightarrow
      H text{ is a subgroup of } K ,.
      end{equation}




      Recall that a subset $H subseteq G$ is a subgroup of $(G,cdot)$ if and only if $H$ is a group with (the restriction to $H$ of) the operation in $G$. But, if $H subseteq K leqslant G$, then the restriction of $G$ to $H$ is the same as the restriction of $K$ to $H$. Hence, the condition for $H leqslant G$ is exactly the same as the condition $H leqslant K$.



      Your question just corresponds to the case: $H cap K subseteq H,K leqslant G$.






      share|cite|improve this answer









      $endgroup$














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        3 Answers
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        4












        $begingroup$

        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 $odot$ 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











        $endgroup$


















          4












          $begingroup$

          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 $odot$ 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











          $endgroup$
















            4












            4








            4





            $begingroup$

            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 $odot$ 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 $odot$ 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 16 mins ago

























            answered 16 hours ago









            rolandcyprolandcyp

            2,319422




            2,319422























                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 13 hours ago









                    janmarqzjanmarqz

                    6,25741630




                    6,25741630























                        0












                        $begingroup$

                        The intersection of two (or any quantity of) subgroups is always a subgroup of all the intersecting subgroups. Indeed, this is a particular case of the following general remark.




                        Remark. A subgroup $H leqslant G$ is also a subgroup of any subgroup $K leqslant G$ containing $H$. More precisesly, if $G$ is a group and $H subseteq K leqslant G$ then:
                        begin{equation}
                        H text{ is a subgroup of } G
                        Leftrightarrow
                        H text{ is a subgroup of } K ,.
                        end{equation}




                        Recall that a subset $H subseteq G$ is a subgroup of $(G,cdot)$ if and only if $H$ is a group with (the restriction to $H$ of) the operation in $G$. But, if $H subseteq K leqslant G$, then the restriction of $G$ to $H$ is the same as the restriction of $K$ to $H$. Hence, the condition for $H leqslant G$ is exactly the same as the condition $H leqslant K$.



                        Your question just corresponds to the case: $H cap K subseteq H,K leqslant G$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          The intersection of two (or any quantity of) subgroups is always a subgroup of all the intersecting subgroups. Indeed, this is a particular case of the following general remark.




                          Remark. A subgroup $H leqslant G$ is also a subgroup of any subgroup $K leqslant G$ containing $H$. More precisesly, if $G$ is a group and $H subseteq K leqslant G$ then:
                          begin{equation}
                          H text{ is a subgroup of } G
                          Leftrightarrow
                          H text{ is a subgroup of } K ,.
                          end{equation}




                          Recall that a subset $H subseteq G$ is a subgroup of $(G,cdot)$ if and only if $H$ is a group with (the restriction to $H$ of) the operation in $G$. But, if $H subseteq K leqslant G$, then the restriction of $G$ to $H$ is the same as the restriction of $K$ to $H$. Hence, the condition for $H leqslant G$ is exactly the same as the condition $H leqslant K$.



                          Your question just corresponds to the case: $H cap K subseteq H,K leqslant G$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            The intersection of two (or any quantity of) subgroups is always a subgroup of all the intersecting subgroups. Indeed, this is a particular case of the following general remark.




                            Remark. A subgroup $H leqslant G$ is also a subgroup of any subgroup $K leqslant G$ containing $H$. More precisesly, if $G$ is a group and $H subseteq K leqslant G$ then:
                            begin{equation}
                            H text{ is a subgroup of } G
                            Leftrightarrow
                            H text{ is a subgroup of } K ,.
                            end{equation}




                            Recall that a subset $H subseteq G$ is a subgroup of $(G,cdot)$ if and only if $H$ is a group with (the restriction to $H$ of) the operation in $G$. But, if $H subseteq K leqslant G$, then the restriction of $G$ to $H$ is the same as the restriction of $K$ to $H$. Hence, the condition for $H leqslant G$ is exactly the same as the condition $H leqslant K$.



                            Your question just corresponds to the case: $H cap K subseteq H,K leqslant G$.






                            share|cite|improve this answer









                            $endgroup$



                            The intersection of two (or any quantity of) subgroups is always a subgroup of all the intersecting subgroups. Indeed, this is a particular case of the following general remark.




                            Remark. A subgroup $H leqslant G$ is also a subgroup of any subgroup $K leqslant G$ containing $H$. More precisesly, if $G$ is a group and $H subseteq K leqslant G$ then:
                            begin{equation}
                            H text{ is a subgroup of } G
                            Leftrightarrow
                            H text{ is a subgroup of } K ,.
                            end{equation}




                            Recall that a subset $H subseteq G$ is a subgroup of $(G,cdot)$ if and only if $H$ is a group with (the restriction to $H$ of) the operation in $G$. But, if $H subseteq K leqslant G$, then the restriction of $G$ to $H$ is the same as the restriction of $K$ to $H$. Hence, the condition for $H leqslant G$ is exactly the same as the condition $H leqslant K$.



                            Your question just corresponds to the case: $H cap K subseteq H,K leqslant G$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 4 hours ago









                            suitangisuitangi

                            45928




                            45928






























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