Show that $bigcap_{a in G} a H a^{-1}$ is a normal subgroup of $G$.












1












$begingroup$


I am reading "An introduction to algebraic systems" by Kazuo Matsuzaka.



In this book, there is the following problem.



I think this problem is easy but a little abstract for me.



Is my answer ok?




Let $G$ be a group.

Let $H$ be a subgroup of $G$.

Let $N = bigcap_{a in G} a H a^{-1}$.



(1)

Show that $N$ is a normal subgroup of $G$.



(2)

Show that if $M$ is a normal subgroup of $G$ and $M subset H$, then, $M subset N$.




(1)

Let $g_1, g_2$ be arbitrary elements of $G$.

Let $n$ be an arbitrary element of $N$.

Because $g_1^{-1} g_2 in G$ and $n in N$, so $n in g_1^{-1} g_2 H (g_1^{-1} g_2)^{-1}$.

So, $n = g_1^{-1} g_2 h (g_1^{-1} g_2)^{-1}$ for some $h in H$.

Then, $g_1 n g_1^{-1} = g_1 g_1^{-1} g_2 h (g_1^{-1} g_2)^{-1} g_1^{-1} = g_1 g_1^{-1} g_2 h g_2^{-1} g_1 g_1^{-1} = g_2 h g_2^{-1} in g_2 H g_2^{-1}$.
$g_2$ was an arbitrary element of $G$.

So, $g_1 n g_1^{-1} in N$.

So, $N$ is a normal subgroup of $G$.



(2)

Let $m$ be an arbitrary element of $M$.

Let $g_3$ be an arbitrary element of $G$.

Then, $g_3^{-1} m (g_3^{-1})^{-1} = g_3^{-1} m g_3 in M subset H$.
$m = g_3 (g_3^{-1} m g_3) g_3^{-1} in g_3 H g_3^{-1}$.
$g_3$ was an arbitrary element of $G$.

So, $m in N$.






abstract-algebra group-theory proof-verification







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am reading "An introduction to algebraic systems" by Kazuo Matsuzaka.



    In this book, there is the following problem.



    I think this problem is easy but a little abstract for me.



    Is my answer ok?




    Let $G$ be a group.

    Let $H$ be a subgroup of $G$.

    Let $N = bigcap_{a in G} a H a^{-1}$.



    (1)

    Show that $N$ is a normal subgroup of $G$.



    (2)

    Show that if $M$ is a normal subgroup of $G$ and $M subset H$, then, $M subset N$.




    (1)

    Let $g_1, g_2$ be arbitrary elements of $G$.

    Let $n$ be an arbitrary element of $N$.

    Because $g_1^{-1} g_2 in G$ and $n in N$, so $n in g_1^{-1} g_2 H (g_1^{-1} g_2)^{-1}$.

    So, $n = g_1^{-1} g_2 h (g_1^{-1} g_2)^{-1}$ for some $h in H$.

    Then, $g_1 n g_1^{-1} = g_1 g_1^{-1} g_2 h (g_1^{-1} g_2)^{-1} g_1^{-1} = g_1 g_1^{-1} g_2 h g_2^{-1} g_1 g_1^{-1} = g_2 h g_2^{-1} in g_2 H g_2^{-1}$.
    $g_2$ was an arbitrary element of $G$.

    So, $g_1 n g_1^{-1} in N$.

    So, $N$ is a normal subgroup of $G$.



    (2)

    Let $m$ be an arbitrary element of $M$.

    Let $g_3$ be an arbitrary element of $G$.

    Then, $g_3^{-1} m (g_3^{-1})^{-1} = g_3^{-1} m g_3 in M subset H$.
    $m = g_3 (g_3^{-1} m g_3) g_3^{-1} in g_3 H g_3^{-1}$.
    $g_3$ was an arbitrary element of $G$.

    So, $m in N$.






    abstract-algebra group-theory proof-verification







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am reading "An introduction to algebraic systems" by Kazuo Matsuzaka.



      In this book, there is the following problem.



      I think this problem is easy but a little abstract for me.



      Is my answer ok?




      Let $G$ be a group.

      Let $H$ be a subgroup of $G$.

      Let $N = bigcap_{a in G} a H a^{-1}$.



      (1)

      Show that $N$ is a normal subgroup of $G$.



      (2)

      Show that if $M$ is a normal subgroup of $G$ and $M subset H$, then, $M subset N$.




      (1)

      Let $g_1, g_2$ be arbitrary elements of $G$.

      Let $n$ be an arbitrary element of $N$.

      Because $g_1^{-1} g_2 in G$ and $n in N$, so $n in g_1^{-1} g_2 H (g_1^{-1} g_2)^{-1}$.

      So, $n = g_1^{-1} g_2 h (g_1^{-1} g_2)^{-1}$ for some $h in H$.

      Then, $g_1 n g_1^{-1} = g_1 g_1^{-1} g_2 h (g_1^{-1} g_2)^{-1} g_1^{-1} = g_1 g_1^{-1} g_2 h g_2^{-1} g_1 g_1^{-1} = g_2 h g_2^{-1} in g_2 H g_2^{-1}$.
      $g_2$ was an arbitrary element of $G$.

      So, $g_1 n g_1^{-1} in N$.

      So, $N$ is a normal subgroup of $G$.



      (2)

      Let $m$ be an arbitrary element of $M$.

      Let $g_3$ be an arbitrary element of $G$.

      Then, $g_3^{-1} m (g_3^{-1})^{-1} = g_3^{-1} m g_3 in M subset H$.
      $m = g_3 (g_3^{-1} m g_3) g_3^{-1} in g_3 H g_3^{-1}$.
      $g_3$ was an arbitrary element of $G$.

      So, $m in N$.






      abstract-algebra group-theory proof-verification







      share|cite|improve this question









      $endgroup$




      I am reading "An introduction to algebraic systems" by Kazuo Matsuzaka.



      In this book, there is the following problem.



      I think this problem is easy but a little abstract for me.



      Is my answer ok?




      Let $G$ be a group.

      Let $H$ be a subgroup of $G$.

      Let $N = bigcap_{a in G} a H a^{-1}$.



      (1)

      Show that $N$ is a normal subgroup of $G$.



      (2)

      Show that if $M$ is a normal subgroup of $G$ and $M subset H$, then, $M subset N$.




      (1)

      Let $g_1, g_2$ be arbitrary elements of $G$.

      Let $n$ be an arbitrary element of $N$.

      Because $g_1^{-1} g_2 in G$ and $n in N$, so $n in g_1^{-1} g_2 H (g_1^{-1} g_2)^{-1}$.

      So, $n = g_1^{-1} g_2 h (g_1^{-1} g_2)^{-1}$ for some $h in H$.

      Then, $g_1 n g_1^{-1} = g_1 g_1^{-1} g_2 h (g_1^{-1} g_2)^{-1} g_1^{-1} = g_1 g_1^{-1} g_2 h g_2^{-1} g_1 g_1^{-1} = g_2 h g_2^{-1} in g_2 H g_2^{-1}$.
      $g_2$ was an arbitrary element of $G$.

      So, $g_1 n g_1^{-1} in N$.

      So, $N$ is a normal subgroup of $G$.



      (2)

      Let $m$ be an arbitrary element of $M$.

      Let $g_3$ be an arbitrary element of $G$.

      Then, $g_3^{-1} m (g_3^{-1})^{-1} = g_3^{-1} m g_3 in M subset H$.
      $m = g_3 (g_3^{-1} m g_3) g_3^{-1} in g_3 H g_3^{-1}$.
      $g_3$ was an arbitrary element of $G$.

      So, $m in N$.






      abstract-algebra group-theory proof-verification




      abstract-algebra group-theory proof-verification






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 23 at 11:49









      tchappy hatchappy ha

      766412




      766412






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Your proof is ok but your write-up is slightly overcomplicating matters. Here is a slightly shorter and perhaps more enlightening way to understand what is going on.



          For (1): N is normal iff $xNx^{-1} = N forall x in G$ . But $xNx^{-1} = cap_{a in G} (xa)H(xa)^{-1} $ and just note that the function $a to xa$ permutes the elements of G.



          For (2): $M subset H $ and M normal implies that $M = cap_{a in G} aMa^{-1} subset cap_{a in G} aHa^{-1} =N $, as needed.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Thank you very much, for clear proof, Sorin Tirc.
            $endgroup$
            – tchappy ha
            Jan 23 at 12:07











          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%2f3084370%2fshow-that-bigcap-a-in-g-a-h-a-1-is-a-normal-subgroup-of-g%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Your proof is ok but your write-up is slightly overcomplicating matters. Here is a slightly shorter and perhaps more enlightening way to understand what is going on.



          For (1): N is normal iff $xNx^{-1} = N forall x in G$ . But $xNx^{-1} = cap_{a in G} (xa)H(xa)^{-1} $ and just note that the function $a to xa$ permutes the elements of G.



          For (2): $M subset H $ and M normal implies that $M = cap_{a in G} aMa^{-1} subset cap_{a in G} aHa^{-1} =N $, as needed.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Thank you very much, for clear proof, Sorin Tirc.
            $endgroup$
            – tchappy ha
            Jan 23 at 12:07
















          1












          $begingroup$

          Your proof is ok but your write-up is slightly overcomplicating matters. Here is a slightly shorter and perhaps more enlightening way to understand what is going on.



          For (1): N is normal iff $xNx^{-1} = N forall x in G$ . But $xNx^{-1} = cap_{a in G} (xa)H(xa)^{-1} $ and just note that the function $a to xa$ permutes the elements of G.



          For (2): $M subset H $ and M normal implies that $M = cap_{a in G} aMa^{-1} subset cap_{a in G} aHa^{-1} =N $, as needed.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Thank you very much, for clear proof, Sorin Tirc.
            $endgroup$
            – tchappy ha
            Jan 23 at 12:07














          1












          1








          1





          $begingroup$

          Your proof is ok but your write-up is slightly overcomplicating matters. Here is a slightly shorter and perhaps more enlightening way to understand what is going on.



          For (1): N is normal iff $xNx^{-1} = N forall x in G$ . But $xNx^{-1} = cap_{a in G} (xa)H(xa)^{-1} $ and just note that the function $a to xa$ permutes the elements of G.



          For (2): $M subset H $ and M normal implies that $M = cap_{a in G} aMa^{-1} subset cap_{a in G} aHa^{-1} =N $, as needed.






          share|cite|improve this answer









          $endgroup$



          Your proof is ok but your write-up is slightly overcomplicating matters. Here is a slightly shorter and perhaps more enlightening way to understand what is going on.



          For (1): N is normal iff $xNx^{-1} = N forall x in G$ . But $xNx^{-1} = cap_{a in G} (xa)H(xa)^{-1} $ and just note that the function $a to xa$ permutes the elements of G.



          For (2): $M subset H $ and M normal implies that $M = cap_{a in G} aMa^{-1} subset cap_{a in G} aHa^{-1} =N $, as needed.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 11:58









          Sorin TircSorin Tirc

          1,810213




          1,810213








          • 1




            $begingroup$
            Thank you very much, for clear proof, Sorin Tirc.
            $endgroup$
            – tchappy ha
            Jan 23 at 12:07














          • 1




            $begingroup$
            Thank you very much, for clear proof, Sorin Tirc.
            $endgroup$
            – tchappy ha
            Jan 23 at 12:07








          1




          1




          $begingroup$
          Thank you very much, for clear proof, Sorin Tirc.
          $endgroup$
          – tchappy ha
          Jan 23 at 12:07




          $begingroup$
          Thank you very much, for clear proof, Sorin Tirc.
          $endgroup$
          – tchappy ha
          Jan 23 at 12:07


















          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%2f3084370%2fshow-that-bigcap-a-in-g-a-h-a-1-is-a-normal-subgroup-of-g%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

          Mario Kart Wii

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
          Read more

          The Binding of Isaac: Rebirth/Afterbirth

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
          Read more

          What does “Dominus providebit” mean?

          Image
          5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
          Read more