Prove that a cyclic group with only one generator can have at most 2 elements












13












$begingroup$



Prove that a cyclic group that has only one generator has at most $2$ elements.




I want to know if my proof would be valid:



Suppose $G$ is a cyclic group and $g$ is its only generator. Let $|G|=n$ where $n>2$, then we know that $gcd(n,n-1)=1$. This implies that $g^{n-1}$ is a generator of $G$. We have a contradiction, since $g$ is the only generator of $G$ (and $n > 2$ leads to $n-1 neq 1$). Thus $|G|leq 2$.



I tried to use the fact that generating elements of a group are coprime to the order of the group, thanks.






abstract-algebra greatest-common-divisor cyclic-groups







share|cite|improve this question











$endgroup$

















    13












    $begingroup$



    Prove that a cyclic group that has only one generator has at most $2$ elements.




    I want to know if my proof would be valid:



    Suppose $G$ is a cyclic group and $g$ is its only generator. Let $|G|=n$ where $n>2$, then we know that $gcd(n,n-1)=1$. This implies that $g^{n-1}$ is a generator of $G$. We have a contradiction, since $g$ is the only generator of $G$ (and $n > 2$ leads to $n-1 neq 1$). Thus $|G|leq 2$.



    I tried to use the fact that generating elements of a group are coprime to the order of the group, thanks.






    abstract-algebra greatest-common-divisor cyclic-groups







    share|cite|improve this question











    $endgroup$















      13












      13








      13


      1



      $begingroup$



      Prove that a cyclic group that has only one generator has at most $2$ elements.




      I want to know if my proof would be valid:



      Suppose $G$ is a cyclic group and $g$ is its only generator. Let $|G|=n$ where $n>2$, then we know that $gcd(n,n-1)=1$. This implies that $g^{n-1}$ is a generator of $G$. We have a contradiction, since $g$ is the only generator of $G$ (and $n > 2$ leads to $n-1 neq 1$). Thus $|G|leq 2$.



      I tried to use the fact that generating elements of a group are coprime to the order of the group, thanks.






      abstract-algebra greatest-common-divisor cyclic-groups







      share|cite|improve this question











      $endgroup$





      Prove that a cyclic group that has only one generator has at most $2$ elements.




      I want to know if my proof would be valid:



      Suppose $G$ is a cyclic group and $g$ is its only generator. Let $|G|=n$ where $n>2$, then we know that $gcd(n,n-1)=1$. This implies that $g^{n-1}$ is a generator of $G$. We have a contradiction, since $g$ is the only generator of $G$ (and $n > 2$ leads to $n-1 neq 1$). Thus $|G|leq 2$.



      I tried to use the fact that generating elements of a group are coprime to the order of the group, thanks.






      abstract-algebra greatest-common-divisor cyclic-groups




      abstract-algebra greatest-common-divisor cyclic-groups






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 20 at 0:33









      darij grinberg

      11.2k33167




      11.2k33167










      asked Feb 20 at 0:05









      PabloPablo

      683




      683






















          3 Answers
          3






          active

          oldest

          votes


















          17












          $begingroup$

          Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Beautiful! Thank you so much for this!
            $endgroup$
            – Pablo
            Feb 20 at 7:49






          • 2




            $begingroup$
            When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
            $endgroup$
            – Marc van Leeuwen
            Feb 20 at 13:03





















          9












          $begingroup$

          Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That's a very good observation, thanks!
            $endgroup$
            – Pablo
            Feb 20 at 7:49



















          0












          $begingroup$

          Here is another take.



          The number of generators is $phi(n)$, where $phi$ is Euler's function.



          Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.



          In the first case, we have $phi(n) ge phi(p)=p-1ge2$.



          In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.



          Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.



          (The key fact is: if $d$ divides $n$, then $phi(n) ge phi(d)$.)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @darij, thanks.
            $endgroup$
            – lhf
            Feb 20 at 10:48











          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%2f3119533%2fprove-that-a-cyclic-group-with-only-one-generator-can-have-at-most-2-elements%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









          17












          $begingroup$

          Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Beautiful! Thank you so much for this!
            $endgroup$
            – Pablo
            Feb 20 at 7:49






          • 2




            $begingroup$
            When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
            $endgroup$
            – Marc van Leeuwen
            Feb 20 at 13:03


















          17












          $begingroup$

          Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Beautiful! Thank you so much for this!
            $endgroup$
            – Pablo
            Feb 20 at 7:49






          • 2




            $begingroup$
            When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
            $endgroup$
            – Marc van Leeuwen
            Feb 20 at 13:03
















          17












          17








          17





          $begingroup$

          Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).






          share|cite|improve this answer









          $endgroup$



          Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 20 at 0:10









          MPWMPW

          30.9k12157




          30.9k12157












          • $begingroup$
            Beautiful! Thank you so much for this!
            $endgroup$
            – Pablo
            Feb 20 at 7:49






          • 2




            $begingroup$
            When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
            $endgroup$
            – Marc van Leeuwen
            Feb 20 at 13:03




















          • $begingroup$
            Beautiful! Thank you so much for this!
            $endgroup$
            – Pablo
            Feb 20 at 7:49






          • 2




            $begingroup$
            When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
            $endgroup$
            – Marc van Leeuwen
            Feb 20 at 13:03


















          $begingroup$
          Beautiful! Thank you so much for this!
          $endgroup$
          – Pablo
          Feb 20 at 7:49




          $begingroup$
          Beautiful! Thank you so much for this!
          $endgroup$
          – Pablo
          Feb 20 at 7:49




          2




          2




          $begingroup$
          When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
          $endgroup$
          – Marc van Leeuwen
          Feb 20 at 13:03






          $begingroup$
          When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
          $endgroup$
          – Marc van Leeuwen
          Feb 20 at 13:03













          9












          $begingroup$

          Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That's a very good observation, thanks!
            $endgroup$
            – Pablo
            Feb 20 at 7:49
















          9












          $begingroup$

          Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That's a very good observation, thanks!
            $endgroup$
            – Pablo
            Feb 20 at 7:49














          9












          9








          9





          $begingroup$

          Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.






          share|cite|improve this answer









          $endgroup$



          Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 20 at 1:04









          LBJFSLBJFS

          27110




          27110












          • $begingroup$
            That's a very good observation, thanks!
            $endgroup$
            – Pablo
            Feb 20 at 7:49


















          • $begingroup$
            That's a very good observation, thanks!
            $endgroup$
            – Pablo
            Feb 20 at 7:49
















          $begingroup$
          That's a very good observation, thanks!
          $endgroup$
          – Pablo
          Feb 20 at 7:49




          $begingroup$
          That's a very good observation, thanks!
          $endgroup$
          – Pablo
          Feb 20 at 7:49











          0












          $begingroup$

          Here is another take.



          The number of generators is $phi(n)$, where $phi$ is Euler's function.



          Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.



          In the first case, we have $phi(n) ge phi(p)=p-1ge2$.



          In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.



          Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.



          (The key fact is: if $d$ divides $n$, then $phi(n) ge phi(d)$.)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @darij, thanks.
            $endgroup$
            – lhf
            Feb 20 at 10:48
















          0












          $begingroup$

          Here is another take.



          The number of generators is $phi(n)$, where $phi$ is Euler's function.



          Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.



          In the first case, we have $phi(n) ge phi(p)=p-1ge2$.



          In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.



          Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.



          (The key fact is: if $d$ divides $n$, then $phi(n) ge phi(d)$.)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @darij, thanks.
            $endgroup$
            – lhf
            Feb 20 at 10:48














          0












          0








          0





          $begingroup$

          Here is another take.



          The number of generators is $phi(n)$, where $phi$ is Euler's function.



          Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.



          In the first case, we have $phi(n) ge phi(p)=p-1ge2$.



          In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.



          Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.



          (The key fact is: if $d$ divides $n$, then $phi(n) ge phi(d)$.)






          share|cite|improve this answer











          $endgroup$



          Here is another take.



          The number of generators is $phi(n)$, where $phi$ is Euler's function.



          Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.



          In the first case, we have $phi(n) ge phi(p)=p-1ge2$.



          In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.



          Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.



          (The key fact is: if $d$ divides $n$, then $phi(n) ge phi(d)$.)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 20 at 10:47

























          answered Feb 20 at 1:00









          lhflhf

          166k11172402




          166k11172402












          • $begingroup$
            @darij, thanks.
            $endgroup$
            – lhf
            Feb 20 at 10:48


















          • $begingroup$
            @darij, thanks.
            $endgroup$
            – lhf
            Feb 20 at 10:48
















          $begingroup$
          @darij, thanks.
          $endgroup$
          – lhf
          Feb 20 at 10:48




          $begingroup$
          @darij, thanks.
          $endgroup$
          – lhf
          Feb 20 at 10:48


















          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%2f3119533%2fprove-that-a-cyclic-group-with-only-one-generator-can-have-at-most-2-elements%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

          How do I know what Microsoft account the skydrive app is syncing to?

          Image
          up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
          Read more

          When does type information flow backwards in C++?

          Image
          up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
          Read more

          Grease: Live!

          Image
          A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
          Read more