Inverse function of isomorphism is also isomorphism












9












$begingroup$


Let $G$ be a group, and let $p:Grightarrow G$ be an isomorphism. Why is $p^{-1}$ also an isomorphism?



We know that $p(a)p(b)=p(ab)$ for any elements $a,bin G$. We also know $p(a^{-1})=p(a)^{-1}$ for any element $ain G$ (follows from the first statement.) How would it show $p^{-1}(ab)=p^{-1}(a)p^{-1}(b)$?






abstract-algebra group-theory







share|cite|improve this question









$endgroup$








  • 5




    $begingroup$
    What is $pbigl( p^{-1}(a)p^{-1}(b)bigr)$?
    $endgroup$
    – Daniel Fischer♦
    Jan 12 '14 at 15:52
















9












$begingroup$


Let $G$ be a group, and let $p:Grightarrow G$ be an isomorphism. Why is $p^{-1}$ also an isomorphism?



We know that $p(a)p(b)=p(ab)$ for any elements $a,bin G$. We also know $p(a^{-1})=p(a)^{-1}$ for any element $ain G$ (follows from the first statement.) How would it show $p^{-1}(ab)=p^{-1}(a)p^{-1}(b)$?






abstract-algebra group-theory







share|cite|improve this question









$endgroup$








  • 5




    $begingroup$
    What is $pbigl( p^{-1}(a)p^{-1}(b)bigr)$?
    $endgroup$
    – Daniel Fischer♦
    Jan 12 '14 at 15:52














9












9








9


2



$begingroup$


Let $G$ be a group, and let $p:Grightarrow G$ be an isomorphism. Why is $p^{-1}$ also an isomorphism?



We know that $p(a)p(b)=p(ab)$ for any elements $a,bin G$. We also know $p(a^{-1})=p(a)^{-1}$ for any element $ain G$ (follows from the first statement.) How would it show $p^{-1}(ab)=p^{-1}(a)p^{-1}(b)$?






abstract-algebra group-theory







share|cite|improve this question









$endgroup$




Let $G$ be a group, and let $p:Grightarrow G$ be an isomorphism. Why is $p^{-1}$ also an isomorphism?



We know that $p(a)p(b)=p(ab)$ for any elements $a,bin G$. We also know $p(a^{-1})=p(a)^{-1}$ for any element $ain G$ (follows from the first statement.) How would it show $p^{-1}(ab)=p^{-1}(a)p^{-1}(b)$?






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










asked Jan 12 '14 at 15:50









KunalKunal

1,2691023




1,2691023








  • 5




    $begingroup$
    What is $pbigl( p^{-1}(a)p^{-1}(b)bigr)$?
    $endgroup$
    – Daniel Fischer♦
    Jan 12 '14 at 15:52














  • 5




    $begingroup$
    What is $pbigl( p^{-1}(a)p^{-1}(b)bigr)$?
    $endgroup$
    – Daniel Fischer♦
    Jan 12 '14 at 15:52








5




5




$begingroup$
What is $pbigl( p^{-1}(a)p^{-1}(b)bigr)$?
$endgroup$
– Daniel Fischer♦
Jan 12 '14 at 15:52




$begingroup$
What is $pbigl( p^{-1}(a)p^{-1}(b)bigr)$?
$endgroup$
– Daniel Fischer♦
Jan 12 '14 at 15:52










3 Answers
3






active

oldest

votes


















12












$begingroup$

Simply



$$p(p^{-1}(a)p^{-1}( b))=p(p^{-1}(a))p(p^{-1}(b))=ab$$
so apply $p^{-1}$ on two sides.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    +1 for the brevity... ... it fits: "Less words, more ideas" .. that's algebra!!
    $endgroup$
    – janmarqz
    Jan 12 '14 at 16:01





















4












$begingroup$

It is a standard exercise to check that $p^{-1}$ is indeed another bijection.
So, $p^{-1}(a)=x$ if $p(x)=a$.



Now if $p^{-1}(ab)=z$ then $p(z)=ab$.



Take $p^{-1}(a)=x$ and $p^{-1}(b)=y$, then
$p^{-1}(a)p^{-1}(b)=xy=z$ also.This implies $$p^{-1}(ab)=p^{-1}(a)p^{-1}(b).$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Cause the inverse is strictly related to the function itself, that especially gives $p^{-1}(ytilde{y})=p^{-1}(p(x)p(tilde{x}))=p^{-1}(p(xtilde{x}))=xtilde{x}=p^{-1}(y)p^{-1}(tilde{y})$.
    But this relation is really just a "happy accident". Most properties won't be heridated e.g. the inverse of a continuous function is not necesarily continuous or e.g. differentiability.



    Moreover, I'd like to stress that -despite the fact that most textbook define isomorphism of groups to be bijective homomorphism- what one really desires is a homomorphism which inverse is homomorphism as well, what luckily comes for free ;-)






    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%2f635858%2finverse-function-of-isomorphism-is-also-isomorphism%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      12












      $begingroup$

      Simply



      $$p(p^{-1}(a)p^{-1}( b))=p(p^{-1}(a))p(p^{-1}(b))=ab$$
      so apply $p^{-1}$ on two sides.






      share|cite|improve this answer









      $endgroup$









      • 2




        $begingroup$
        +1 for the brevity... ... it fits: "Less words, more ideas" .. that's algebra!!
        $endgroup$
        – janmarqz
        Jan 12 '14 at 16:01


















      12












      $begingroup$

      Simply



      $$p(p^{-1}(a)p^{-1}( b))=p(p^{-1}(a))p(p^{-1}(b))=ab$$
      so apply $p^{-1}$ on two sides.






      share|cite|improve this answer









      $endgroup$









      • 2




        $begingroup$
        +1 for the brevity... ... it fits: "Less words, more ideas" .. that's algebra!!
        $endgroup$
        – janmarqz
        Jan 12 '14 at 16:01
















      12












      12








      12





      $begingroup$

      Simply



      $$p(p^{-1}(a)p^{-1}( b))=p(p^{-1}(a))p(p^{-1}(b))=ab$$
      so apply $p^{-1}$ on two sides.






      share|cite|improve this answer









      $endgroup$



      Simply



      $$p(p^{-1}(a)p^{-1}( b))=p(p^{-1}(a))p(p^{-1}(b))=ab$$
      so apply $p^{-1}$ on two sides.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 12 '14 at 15:53







      user63181















      • 2




        $begingroup$
        +1 for the brevity... ... it fits: "Less words, more ideas" .. that's algebra!!
        $endgroup$
        – janmarqz
        Jan 12 '14 at 16:01
















      • 2




        $begingroup$
        +1 for the brevity... ... it fits: "Less words, more ideas" .. that's algebra!!
        $endgroup$
        – janmarqz
        Jan 12 '14 at 16:01










      2




      2




      $begingroup$
      +1 for the brevity... ... it fits: "Less words, more ideas" .. that's algebra!!
      $endgroup$
      – janmarqz
      Jan 12 '14 at 16:01






      $begingroup$
      +1 for the brevity... ... it fits: "Less words, more ideas" .. that's algebra!!
      $endgroup$
      – janmarqz
      Jan 12 '14 at 16:01













      4












      $begingroup$

      It is a standard exercise to check that $p^{-1}$ is indeed another bijection.
      So, $p^{-1}(a)=x$ if $p(x)=a$.



      Now if $p^{-1}(ab)=z$ then $p(z)=ab$.



      Take $p^{-1}(a)=x$ and $p^{-1}(b)=y$, then
      $p^{-1}(a)p^{-1}(b)=xy=z$ also.This implies $$p^{-1}(ab)=p^{-1}(a)p^{-1}(b).$$






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        It is a standard exercise to check that $p^{-1}$ is indeed another bijection.
        So, $p^{-1}(a)=x$ if $p(x)=a$.



        Now if $p^{-1}(ab)=z$ then $p(z)=ab$.



        Take $p^{-1}(a)=x$ and $p^{-1}(b)=y$, then
        $p^{-1}(a)p^{-1}(b)=xy=z$ also.This implies $$p^{-1}(ab)=p^{-1}(a)p^{-1}(b).$$






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          It is a standard exercise to check that $p^{-1}$ is indeed another bijection.
          So, $p^{-1}(a)=x$ if $p(x)=a$.



          Now if $p^{-1}(ab)=z$ then $p(z)=ab$.



          Take $p^{-1}(a)=x$ and $p^{-1}(b)=y$, then
          $p^{-1}(a)p^{-1}(b)=xy=z$ also.This implies $$p^{-1}(ab)=p^{-1}(a)p^{-1}(b).$$






          share|cite|improve this answer









          $endgroup$



          It is a standard exercise to check that $p^{-1}$ is indeed another bijection.
          So, $p^{-1}(a)=x$ if $p(x)=a$.



          Now if $p^{-1}(ab)=z$ then $p(z)=ab$.



          Take $p^{-1}(a)=x$ and $p^{-1}(b)=y$, then
          $p^{-1}(a)p^{-1}(b)=xy=z$ also.This implies $$p^{-1}(ab)=p^{-1}(a)p^{-1}(b).$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 12 '14 at 15:59









          janmarqzjanmarqz

          6,20241630




          6,20241630























              1












              $begingroup$

              Cause the inverse is strictly related to the function itself, that especially gives $p^{-1}(ytilde{y})=p^{-1}(p(x)p(tilde{x}))=p^{-1}(p(xtilde{x}))=xtilde{x}=p^{-1}(y)p^{-1}(tilde{y})$.
              But this relation is really just a "happy accident". Most properties won't be heridated e.g. the inverse of a continuous function is not necesarily continuous or e.g. differentiability.



              Moreover, I'd like to stress that -despite the fact that most textbook define isomorphism of groups to be bijective homomorphism- what one really desires is a homomorphism which inverse is homomorphism as well, what luckily comes for free ;-)






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Cause the inverse is strictly related to the function itself, that especially gives $p^{-1}(ytilde{y})=p^{-1}(p(x)p(tilde{x}))=p^{-1}(p(xtilde{x}))=xtilde{x}=p^{-1}(y)p^{-1}(tilde{y})$.
                But this relation is really just a "happy accident". Most properties won't be heridated e.g. the inverse of a continuous function is not necesarily continuous or e.g. differentiability.



                Moreover, I'd like to stress that -despite the fact that most textbook define isomorphism of groups to be bijective homomorphism- what one really desires is a homomorphism which inverse is homomorphism as well, what luckily comes for free ;-)






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Cause the inverse is strictly related to the function itself, that especially gives $p^{-1}(ytilde{y})=p^{-1}(p(x)p(tilde{x}))=p^{-1}(p(xtilde{x}))=xtilde{x}=p^{-1}(y)p^{-1}(tilde{y})$.
                  But this relation is really just a "happy accident". Most properties won't be heridated e.g. the inverse of a continuous function is not necesarily continuous or e.g. differentiability.



                  Moreover, I'd like to stress that -despite the fact that most textbook define isomorphism of groups to be bijective homomorphism- what one really desires is a homomorphism which inverse is homomorphism as well, what luckily comes for free ;-)






                  share|cite|improve this answer











                  $endgroup$



                  Cause the inverse is strictly related to the function itself, that especially gives $p^{-1}(ytilde{y})=p^{-1}(p(x)p(tilde{x}))=p^{-1}(p(xtilde{x}))=xtilde{x}=p^{-1}(y)p^{-1}(tilde{y})$.
                  But this relation is really just a "happy accident". Most properties won't be heridated e.g. the inverse of a continuous function is not necesarily continuous or e.g. differentiability.



                  Moreover, I'd like to stress that -despite the fact that most textbook define isomorphism of groups to be bijective homomorphism- what one really desires is a homomorphism which inverse is homomorphism as well, what luckily comes for free ;-)







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 12 '14 at 16:15

























                  answered Jan 12 '14 at 16:09









                  C-Star-PuppyC-Star-Puppy

                  8,24642165




                  8,24642165






























                      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%2f635858%2finverse-function-of-isomorphism-is-also-isomorphism%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

                      MongoDB - Not Authorized To Execute Command

                      Image
                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
                      Read more

                      How to fix TextFormField cause rebuild widget in Flutter

                      Image
                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
                      Read more

                      in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

                      Image
                      1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
                      Read more