is the intersection of subgroups a subgroup of each subgroupA group with no proper non-trivial...

Why doesn't Newton's third law mean a person bounces back to where they started when they hit the ground?

Theorems that impeded progress

Email Account under attack (really) - anything I can do?

Arthur Somervell: 1000 Exercises - Meaning of this notation

What does "Puller Prush Person" mean?

Are the number of citations and number of published articles the most important criteria for a tenure promotion?

Do I have a twin with permutated remainders?

What does it mean to describe someone as a butt steak?

How is the claim "I am in New York only if I am in America" the same as "If I am in New York, then I am in America?

What's the point of deactivating Num Lock on login screens?

Font hinting is lost in Chrome-like browsers (for some languages )

is the intersection of subgroups a subgroup of each subgroup

What is the word for reserving something for yourself before others do?

Can I ask the recruiters in my resume to put the reason why I am rejected?

What is the offset in a seaplane's hull?

How to format long polynomial?

Does the fruit of Mantra Japa automatically go to Indra if Japa Samarpana Mantra is not chanted?

Can I make popcorn with any corn?

The use of multiple foreign keys on same column in SQL Server

How to say job offer in Mandarin/Cantonese?

Show that if two triangles built on parallel lines, with equal bases have the same perimeter only if they are congruent.

Modeling an IPv4 Address

Is it tax fraud for an individual to declare non-taxable revenue as taxable income? (US tax laws)

How to test if a transaction is standard without spending real money?



is the intersection of subgroups a subgroup of each subgroup


A group with no proper non-trivial subgroupsSubgroups that are isomorphic to each other, and contain a common element are the same subgroupIf a group has no maximal subgroups then all elements are non-generators? Frattini subgroup characterizationLet $P$, $Q$ be two Sylow p-subgroups of $G$, is it true that $N_P(Q)=Qcap P$?Subgroups of $G=(mathbb{Z}_{12},+)$join of pronormal subgroupsProperty of normally embedded subgroupsParity of order of intersection of cyclic and noncyclic subgroupsListing elements of the subgroups and generatorsIntersection of two subgroups













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!










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!










    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!










      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













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago









      Shaun

      10.3k113686




      10.3k113686










      asked 6 hours ago









      JustWanderingJustWandering

      592




      592






















          2 Answers
          2






          active

          oldest

          votes


















          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









            $endgroup$














              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3177593%2fis-the-intersection-of-subgroups-a-subgroup-of-each-subgroup%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              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












                $begingroup$

                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 6 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 4 hours ago









                          janmarqzjanmarqz

                          6,25741630




                          6,25741630






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3177593%2fis-the-intersection-of-subgroups-a-subgroup-of-each-subgroup%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Popular posts from this blog

                              Why do type traits not work with types in namespace scope?What are POD types in C++?Why can templates only be...

                              Will tsunami waves travel forever if there was no land?Why do tsunami waves begin with the water flowing away...

                              Simple Scan not detecting my scanner (Brother DCP-7055W)Brother MFC-L2700DW printer can print, can't...