Every nonabelian group of order 6 has a non-normal subgroup of order 2 (revisited)












7












$begingroup$


I am fully aware that this question has already been addressed here and here. The question, however, derives from a Dummit and Foote exercise (section 4.2 exercise 10 page 122) and the answers provided make use of material that appears later in the book, or, at least, I do not clearly understand them in terms of the material that I have studied already.



So, I would like to submit the following tentative proof
'from first principles' (i.e. Dummit and Foote before page 122). As I feel insecure about it, I would be grateful if you could check it.



Consider a nonabelian group G of order 6. By Cauchy's theorem, the group contains at least one element of order 2, and therefore at least one subgroup of order 2. Suppose this/these subgroups are all normal. This would imply that all elements of order 2 commute with all elements of G.



The remaining elements of group G are of order 1 (which trivially is in the center of G), or order 3 (order 6 would imply the group is cyclic and therefore abelian). Since all elements of order 2 are in the center, the order 3 elements will commute with them. The following reasoning shows that they also commute with each other.



Let $x≠y, |x|=|y|=3$. $<x>$ and $<y>$ are normal in $G$ because their index is $2$. Now $x{1, y, y^2}={1, y, y^2}x$ implies, in the nonabelian case, that $x.y=y^2.x$ and $x.y^2=y.x$.
Likewise, $y<x>=<x>y$ implies $y.x=x^2.y$ and $y.x^2=x.y$. Manipulating these equations shows $y.x = x.y$, so the nonabelian case is impossible. Hence, order 3 elements are in the center of G as well.



Therefore, G would be abelian, contrary to the assumption. So, at least one of the subgroups of order 2 should be non-normal.










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Can you justify this sentence: 'Since all elements of order 2 are in the center, the order 3 elements will be also in the center.'
    $endgroup$
    – Berci
    Jan 4 at 12:05










  • $begingroup$
    @Berci. Spot on! I overlooked the fact that order 3 elements also would need to commute with each other. However, I might have a patch for that, which I edited into the post for further checking.
    $endgroup$
    – Frank De Geeter
    Jan 4 at 19:40
















7












$begingroup$


I am fully aware that this question has already been addressed here and here. The question, however, derives from a Dummit and Foote exercise (section 4.2 exercise 10 page 122) and the answers provided make use of material that appears later in the book, or, at least, I do not clearly understand them in terms of the material that I have studied already.



So, I would like to submit the following tentative proof
'from first principles' (i.e. Dummit and Foote before page 122). As I feel insecure about it, I would be grateful if you could check it.



Consider a nonabelian group G of order 6. By Cauchy's theorem, the group contains at least one element of order 2, and therefore at least one subgroup of order 2. Suppose this/these subgroups are all normal. This would imply that all elements of order 2 commute with all elements of G.



The remaining elements of group G are of order 1 (which trivially is in the center of G), or order 3 (order 6 would imply the group is cyclic and therefore abelian). Since all elements of order 2 are in the center, the order 3 elements will commute with them. The following reasoning shows that they also commute with each other.



Let $x≠y, |x|=|y|=3$. $<x>$ and $<y>$ are normal in $G$ because their index is $2$. Now $x{1, y, y^2}={1, y, y^2}x$ implies, in the nonabelian case, that $x.y=y^2.x$ and $x.y^2=y.x$.
Likewise, $y<x>=<x>y$ implies $y.x=x^2.y$ and $y.x^2=x.y$. Manipulating these equations shows $y.x = x.y$, so the nonabelian case is impossible. Hence, order 3 elements are in the center of G as well.



Therefore, G would be abelian, contrary to the assumption. So, at least one of the subgroups of order 2 should be non-normal.










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Can you justify this sentence: 'Since all elements of order 2 are in the center, the order 3 elements will be also in the center.'
    $endgroup$
    – Berci
    Jan 4 at 12:05










  • $begingroup$
    @Berci. Spot on! I overlooked the fact that order 3 elements also would need to commute with each other. However, I might have a patch for that, which I edited into the post for further checking.
    $endgroup$
    – Frank De Geeter
    Jan 4 at 19:40














7












7








7





$begingroup$


I am fully aware that this question has already been addressed here and here. The question, however, derives from a Dummit and Foote exercise (section 4.2 exercise 10 page 122) and the answers provided make use of material that appears later in the book, or, at least, I do not clearly understand them in terms of the material that I have studied already.



So, I would like to submit the following tentative proof
'from first principles' (i.e. Dummit and Foote before page 122). As I feel insecure about it, I would be grateful if you could check it.



Consider a nonabelian group G of order 6. By Cauchy's theorem, the group contains at least one element of order 2, and therefore at least one subgroup of order 2. Suppose this/these subgroups are all normal. This would imply that all elements of order 2 commute with all elements of G.



The remaining elements of group G are of order 1 (which trivially is in the center of G), or order 3 (order 6 would imply the group is cyclic and therefore abelian). Since all elements of order 2 are in the center, the order 3 elements will commute with them. The following reasoning shows that they also commute with each other.



Let $x≠y, |x|=|y|=3$. $<x>$ and $<y>$ are normal in $G$ because their index is $2$. Now $x{1, y, y^2}={1, y, y^2}x$ implies, in the nonabelian case, that $x.y=y^2.x$ and $x.y^2=y.x$.
Likewise, $y<x>=<x>y$ implies $y.x=x^2.y$ and $y.x^2=x.y$. Manipulating these equations shows $y.x = x.y$, so the nonabelian case is impossible. Hence, order 3 elements are in the center of G as well.



Therefore, G would be abelian, contrary to the assumption. So, at least one of the subgroups of order 2 should be non-normal.










share|cite|improve this question











$endgroup$




I am fully aware that this question has already been addressed here and here. The question, however, derives from a Dummit and Foote exercise (section 4.2 exercise 10 page 122) and the answers provided make use of material that appears later in the book, or, at least, I do not clearly understand them in terms of the material that I have studied already.



So, I would like to submit the following tentative proof
'from first principles' (i.e. Dummit and Foote before page 122). As I feel insecure about it, I would be grateful if you could check it.



Consider a nonabelian group G of order 6. By Cauchy's theorem, the group contains at least one element of order 2, and therefore at least one subgroup of order 2. Suppose this/these subgroups are all normal. This would imply that all elements of order 2 commute with all elements of G.



The remaining elements of group G are of order 1 (which trivially is in the center of G), or order 3 (order 6 would imply the group is cyclic and therefore abelian). Since all elements of order 2 are in the center, the order 3 elements will commute with them. The following reasoning shows that they also commute with each other.



Let $x≠y, |x|=|y|=3$. $<x>$ and $<y>$ are normal in $G$ because their index is $2$. Now $x{1, y, y^2}={1, y, y^2}x$ implies, in the nonabelian case, that $x.y=y^2.x$ and $x.y^2=y.x$.
Likewise, $y<x>=<x>y$ implies $y.x=x^2.y$ and $y.x^2=x.y$. Manipulating these equations shows $y.x = x.y$, so the nonabelian case is impossible. Hence, order 3 elements are in the center of G as well.



Therefore, G would be abelian, contrary to the assumption. So, at least one of the subgroups of order 2 should be non-normal.







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 Jan 12 at 22:38







Frank De Geeter

















asked Jan 4 at 10:07









Frank De GeeterFrank De Geeter

726




726








  • 2




    $begingroup$
    Can you justify this sentence: 'Since all elements of order 2 are in the center, the order 3 elements will be also in the center.'
    $endgroup$
    – Berci
    Jan 4 at 12:05










  • $begingroup$
    @Berci. Spot on! I overlooked the fact that order 3 elements also would need to commute with each other. However, I might have a patch for that, which I edited into the post for further checking.
    $endgroup$
    – Frank De Geeter
    Jan 4 at 19:40














  • 2




    $begingroup$
    Can you justify this sentence: 'Since all elements of order 2 are in the center, the order 3 elements will be also in the center.'
    $endgroup$
    – Berci
    Jan 4 at 12:05










  • $begingroup$
    @Berci. Spot on! I overlooked the fact that order 3 elements also would need to commute with each other. However, I might have a patch for that, which I edited into the post for further checking.
    $endgroup$
    – Frank De Geeter
    Jan 4 at 19:40








2




2




$begingroup$
Can you justify this sentence: 'Since all elements of order 2 are in the center, the order 3 elements will be also in the center.'
$endgroup$
– Berci
Jan 4 at 12:05




$begingroup$
Can you justify this sentence: 'Since all elements of order 2 are in the center, the order 3 elements will be also in the center.'
$endgroup$
– Berci
Jan 4 at 12:05












$begingroup$
@Berci. Spot on! I overlooked the fact that order 3 elements also would need to commute with each other. However, I might have a patch for that, which I edited into the post for further checking.
$endgroup$
– Frank De Geeter
Jan 4 at 19:40




$begingroup$
@Berci. Spot on! I overlooked the fact that order 3 elements also would need to commute with each other. However, I might have a patch for that, which I edited into the post for further checking.
$endgroup$
– Frank De Geeter
Jan 4 at 19:40










2 Answers
2






active

oldest

votes


















4












$begingroup$

Basically your reasoning is correct (well done!): to write it with somewhat more "sophistication" - if $N lhd G$, with $|N|=2$, then $N subseteq Z(G)$, that is what you are using. Now if $x in G$ with $ord(x)=3$, then, since $N$ is central, $N subsetneq C_G(x)$, where the inclusion is strict, because of $|N|=2$ not divisible by $3$ ($x in C_G(x)$). But $|G:N|=3$, so $G=C_G(x)$, meaning $x in Z(G)$ and hence $G$ is abelian.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Once you have Cauchy's theorem available it is clear that any non-abelian group $G$ of order $6$ is isomorphic to $S_3$ -- which has a non-normal subgroup of order $2$. Indeed, by Cauchy's theorem there exist elements $r$ and $s$ in $G$ of order $3$ and $2$ respectively. The subgroup $C_3=langle rrangle$ is normal in $G$ because it is of index $2$. So $G=C_3cup C_3cdot s$ and $G={e,r,r^2,s,rs,r^2s}$. Because $C_3$ is normal, $srs^{-1}=r^k$ for some $k=0,1,2$. Since $s^2=e$, we have
    $$
    r=s^2rs^{-2}=s(srs^{-1})s^{-1}=r^{k^2}
    $$

    so that $k^2equiv 1 bmod 3$, i.e., $3mid (k-1)(k+1)$. For $3mid (k-1)$ the group is abelian, so that we have $3mid (k+1)$ and $Gcong D_3cong S_3$.






    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%2f3061484%2fevery-nonabelian-group-of-order-6-has-a-non-normal-subgroup-of-order-2-revisite%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









      4












      $begingroup$

      Basically your reasoning is correct (well done!): to write it with somewhat more "sophistication" - if $N lhd G$, with $|N|=2$, then $N subseteq Z(G)$, that is what you are using. Now if $x in G$ with $ord(x)=3$, then, since $N$ is central, $N subsetneq C_G(x)$, where the inclusion is strict, because of $|N|=2$ not divisible by $3$ ($x in C_G(x)$). But $|G:N|=3$, so $G=C_G(x)$, meaning $x in Z(G)$ and hence $G$ is abelian.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Basically your reasoning is correct (well done!): to write it with somewhat more "sophistication" - if $N lhd G$, with $|N|=2$, then $N subseteq Z(G)$, that is what you are using. Now if $x in G$ with $ord(x)=3$, then, since $N$ is central, $N subsetneq C_G(x)$, where the inclusion is strict, because of $|N|=2$ not divisible by $3$ ($x in C_G(x)$). But $|G:N|=3$, so $G=C_G(x)$, meaning $x in Z(G)$ and hence $G$ is abelian.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Basically your reasoning is correct (well done!): to write it with somewhat more "sophistication" - if $N lhd G$, with $|N|=2$, then $N subseteq Z(G)$, that is what you are using. Now if $x in G$ with $ord(x)=3$, then, since $N$ is central, $N subsetneq C_G(x)$, where the inclusion is strict, because of $|N|=2$ not divisible by $3$ ($x in C_G(x)$). But $|G:N|=3$, so $G=C_G(x)$, meaning $x in Z(G)$ and hence $G$ is abelian.






          share|cite|improve this answer









          $endgroup$



          Basically your reasoning is correct (well done!): to write it with somewhat more "sophistication" - if $N lhd G$, with $|N|=2$, then $N subseteq Z(G)$, that is what you are using. Now if $x in G$ with $ord(x)=3$, then, since $N$ is central, $N subsetneq C_G(x)$, where the inclusion is strict, because of $|N|=2$ not divisible by $3$ ($x in C_G(x)$). But $|G:N|=3$, so $G=C_G(x)$, meaning $x in Z(G)$ and hence $G$ is abelian.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 10:37









          Nicky HeksterNicky Hekster

          28.4k53456




          28.4k53456























              3












              $begingroup$

              Once you have Cauchy's theorem available it is clear that any non-abelian group $G$ of order $6$ is isomorphic to $S_3$ -- which has a non-normal subgroup of order $2$. Indeed, by Cauchy's theorem there exist elements $r$ and $s$ in $G$ of order $3$ and $2$ respectively. The subgroup $C_3=langle rrangle$ is normal in $G$ because it is of index $2$. So $G=C_3cup C_3cdot s$ and $G={e,r,r^2,s,rs,r^2s}$. Because $C_3$ is normal, $srs^{-1}=r^k$ for some $k=0,1,2$. Since $s^2=e$, we have
              $$
              r=s^2rs^{-2}=s(srs^{-1})s^{-1}=r^{k^2}
              $$

              so that $k^2equiv 1 bmod 3$, i.e., $3mid (k-1)(k+1)$. For $3mid (k-1)$ the group is abelian, so that we have $3mid (k+1)$ and $Gcong D_3cong S_3$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Once you have Cauchy's theorem available it is clear that any non-abelian group $G$ of order $6$ is isomorphic to $S_3$ -- which has a non-normal subgroup of order $2$. Indeed, by Cauchy's theorem there exist elements $r$ and $s$ in $G$ of order $3$ and $2$ respectively. The subgroup $C_3=langle rrangle$ is normal in $G$ because it is of index $2$. So $G=C_3cup C_3cdot s$ and $G={e,r,r^2,s,rs,r^2s}$. Because $C_3$ is normal, $srs^{-1}=r^k$ for some $k=0,1,2$. Since $s^2=e$, we have
                $$
                r=s^2rs^{-2}=s(srs^{-1})s^{-1}=r^{k^2}
                $$

                so that $k^2equiv 1 bmod 3$, i.e., $3mid (k-1)(k+1)$. For $3mid (k-1)$ the group is abelian, so that we have $3mid (k+1)$ and $Gcong D_3cong S_3$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Once you have Cauchy's theorem available it is clear that any non-abelian group $G$ of order $6$ is isomorphic to $S_3$ -- which has a non-normal subgroup of order $2$. Indeed, by Cauchy's theorem there exist elements $r$ and $s$ in $G$ of order $3$ and $2$ respectively. The subgroup $C_3=langle rrangle$ is normal in $G$ because it is of index $2$. So $G=C_3cup C_3cdot s$ and $G={e,r,r^2,s,rs,r^2s}$. Because $C_3$ is normal, $srs^{-1}=r^k$ for some $k=0,1,2$. Since $s^2=e$, we have
                  $$
                  r=s^2rs^{-2}=s(srs^{-1})s^{-1}=r^{k^2}
                  $$

                  so that $k^2equiv 1 bmod 3$, i.e., $3mid (k-1)(k+1)$. For $3mid (k-1)$ the group is abelian, so that we have $3mid (k+1)$ and $Gcong D_3cong S_3$.






                  share|cite|improve this answer









                  $endgroup$



                  Once you have Cauchy's theorem available it is clear that any non-abelian group $G$ of order $6$ is isomorphic to $S_3$ -- which has a non-normal subgroup of order $2$. Indeed, by Cauchy's theorem there exist elements $r$ and $s$ in $G$ of order $3$ and $2$ respectively. The subgroup $C_3=langle rrangle$ is normal in $G$ because it is of index $2$. So $G=C_3cup C_3cdot s$ and $G={e,r,r^2,s,rs,r^2s}$. Because $C_3$ is normal, $srs^{-1}=r^k$ for some $k=0,1,2$. Since $s^2=e$, we have
                  $$
                  r=s^2rs^{-2}=s(srs^{-1})s^{-1}=r^{k^2}
                  $$

                  so that $k^2equiv 1 bmod 3$, i.e., $3mid (k-1)(k+1)$. For $3mid (k-1)$ the group is abelian, so that we have $3mid (k+1)$ and $Gcong D_3cong S_3$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 4 at 11:58









                  Dietrich BurdeDietrich Burde

                  78.4k64386




                  78.4k64386






























                      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%2f3061484%2fevery-nonabelian-group-of-order-6-has-a-non-normal-subgroup-of-order-2-revisite%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

                      Aardman Animations

                      Are they similar matrix

                      “minimization” problem in Euclidean space related to orthonormal basis