Is the multiplicative group $Bbb Z_{36}^times$ cyclic?












0












$begingroup$


I'm trying to answer this question but also understanding a smart method to find if a group like the one mentioned has a cyclic generator or not. I know that similar questions have already been asked but honestly, I did not understand the explanations given. So I decided to ask hoping to better understand with this exercise that I'm trying to solve. Thank you for your understanding.



According to Wikipedia, $Bbb Z_{36}^times$ is not cyclic. Now, the multiplciatove group has 12 elements which are 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35. Of course I need to prove that no one of them could be a generator of $Bbb Z_{36}^times$. How can I proceed?






group-theory discrete-mathematics ring-theory totient-function







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Note that the order of $x$ and $x^{-1}$ will agree, so you need only check roughly half the elements.
    $endgroup$
    – Randall
    Jan 22 at 19:18






  • 2




    $begingroup$
    There are so few you can try them one at a time. You will find shortcuts along the way that speed things up.
    $endgroup$
    – Ethan Bolker
    Jan 22 at 19:19






  • 2




    $begingroup$
    I'd prove that none of them have order of $12$. For this, I'd consider them mod $3$ and mod $4$
    $endgroup$
    – Jakobian
    Jan 22 at 19:22










  • $begingroup$
    I would also consider showing that there are four elements with squares equal to $1$. At both ends and near the midpoint. In a cyclic group there can be at most two. Warning: this assumes that you have covered relevant pieces abouot the structure of cyclic groups (not necessarily done in a first course on algberaic structures).
    $endgroup$
    – Jyrki Lahtonen
    Jan 22 at 19:24






  • 1




    $begingroup$
    CRT : as a ring $mathbb{Z}_{36} = 9mathbb{Z}_{4} + (-8)mathbb{Z}_{9} cong mathbb{Z}_{4} times mathbb{Z}_{9}$. The 3 subgroups of order $2$ are then clear from there.
    $endgroup$
    – reuns
    Jan 22 at 19:31
















0












$begingroup$


I'm trying to answer this question but also understanding a smart method to find if a group like the one mentioned has a cyclic generator or not. I know that similar questions have already been asked but honestly, I did not understand the explanations given. So I decided to ask hoping to better understand with this exercise that I'm trying to solve. Thank you for your understanding.



According to Wikipedia, $Bbb Z_{36}^times$ is not cyclic. Now, the multiplciatove group has 12 elements which are 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35. Of course I need to prove that no one of them could be a generator of $Bbb Z_{36}^times$. How can I proceed?






group-theory discrete-mathematics ring-theory totient-function







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Note that the order of $x$ and $x^{-1}$ will agree, so you need only check roughly half the elements.
    $endgroup$
    – Randall
    Jan 22 at 19:18






  • 2




    $begingroup$
    There are so few you can try them one at a time. You will find shortcuts along the way that speed things up.
    $endgroup$
    – Ethan Bolker
    Jan 22 at 19:19






  • 2




    $begingroup$
    I'd prove that none of them have order of $12$. For this, I'd consider them mod $3$ and mod $4$
    $endgroup$
    – Jakobian
    Jan 22 at 19:22










  • $begingroup$
    I would also consider showing that there are four elements with squares equal to $1$. At both ends and near the midpoint. In a cyclic group there can be at most two. Warning: this assumes that you have covered relevant pieces abouot the structure of cyclic groups (not necessarily done in a first course on algberaic structures).
    $endgroup$
    – Jyrki Lahtonen
    Jan 22 at 19:24






  • 1




    $begingroup$
    CRT : as a ring $mathbb{Z}_{36} = 9mathbb{Z}_{4} + (-8)mathbb{Z}_{9} cong mathbb{Z}_{4} times mathbb{Z}_{9}$. The 3 subgroups of order $2$ are then clear from there.
    $endgroup$
    – reuns
    Jan 22 at 19:31














0












0








0





$begingroup$


I'm trying to answer this question but also understanding a smart method to find if a group like the one mentioned has a cyclic generator or not. I know that similar questions have already been asked but honestly, I did not understand the explanations given. So I decided to ask hoping to better understand with this exercise that I'm trying to solve. Thank you for your understanding.



According to Wikipedia, $Bbb Z_{36}^times$ is not cyclic. Now, the multiplciatove group has 12 elements which are 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35. Of course I need to prove that no one of them could be a generator of $Bbb Z_{36}^times$. How can I proceed?






group-theory discrete-mathematics ring-theory totient-function







share|cite|improve this question











$endgroup$




I'm trying to answer this question but also understanding a smart method to find if a group like the one mentioned has a cyclic generator or not. I know that similar questions have already been asked but honestly, I did not understand the explanations given. So I decided to ask hoping to better understand with this exercise that I'm trying to solve. Thank you for your understanding.



According to Wikipedia, $Bbb Z_{36}^times$ is not cyclic. Now, the multiplciatove group has 12 elements which are 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35. Of course I need to prove that no one of them could be a generator of $Bbb Z_{36}^times$. How can I proceed?






group-theory discrete-mathematics ring-theory totient-function




group-theory discrete-mathematics ring-theory totient-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 19:19









Ethan Bolker

44k552117




44k552117










asked Jan 22 at 19:17









PCNFPCNF

1338




1338








  • 2




    $begingroup$
    Note that the order of $x$ and $x^{-1}$ will agree, so you need only check roughly half the elements.
    $endgroup$
    – Randall
    Jan 22 at 19:18






  • 2




    $begingroup$
    There are so few you can try them one at a time. You will find shortcuts along the way that speed things up.
    $endgroup$
    – Ethan Bolker
    Jan 22 at 19:19






  • 2




    $begingroup$
    I'd prove that none of them have order of $12$. For this, I'd consider them mod $3$ and mod $4$
    $endgroup$
    – Jakobian
    Jan 22 at 19:22










  • $begingroup$
    I would also consider showing that there are four elements with squares equal to $1$. At both ends and near the midpoint. In a cyclic group there can be at most two. Warning: this assumes that you have covered relevant pieces abouot the structure of cyclic groups (not necessarily done in a first course on algberaic structures).
    $endgroup$
    – Jyrki Lahtonen
    Jan 22 at 19:24






  • 1




    $begingroup$
    CRT : as a ring $mathbb{Z}_{36} = 9mathbb{Z}_{4} + (-8)mathbb{Z}_{9} cong mathbb{Z}_{4} times mathbb{Z}_{9}$. The 3 subgroups of order $2$ are then clear from there.
    $endgroup$
    – reuns
    Jan 22 at 19:31














  • 2




    $begingroup$
    Note that the order of $x$ and $x^{-1}$ will agree, so you need only check roughly half the elements.
    $endgroup$
    – Randall
    Jan 22 at 19:18






  • 2




    $begingroup$
    There are so few you can try them one at a time. You will find shortcuts along the way that speed things up.
    $endgroup$
    – Ethan Bolker
    Jan 22 at 19:19






  • 2




    $begingroup$
    I'd prove that none of them have order of $12$. For this, I'd consider them mod $3$ and mod $4$
    $endgroup$
    – Jakobian
    Jan 22 at 19:22










  • $begingroup$
    I would also consider showing that there are four elements with squares equal to $1$. At both ends and near the midpoint. In a cyclic group there can be at most two. Warning: this assumes that you have covered relevant pieces abouot the structure of cyclic groups (not necessarily done in a first course on algberaic structures).
    $endgroup$
    – Jyrki Lahtonen
    Jan 22 at 19:24






  • 1




    $begingroup$
    CRT : as a ring $mathbb{Z}_{36} = 9mathbb{Z}_{4} + (-8)mathbb{Z}_{9} cong mathbb{Z}_{4} times mathbb{Z}_{9}$. The 3 subgroups of order $2$ are then clear from there.
    $endgroup$
    – reuns
    Jan 22 at 19:31








2




2




$begingroup$
Note that the order of $x$ and $x^{-1}$ will agree, so you need only check roughly half the elements.
$endgroup$
– Randall
Jan 22 at 19:18




$begingroup$
Note that the order of $x$ and $x^{-1}$ will agree, so you need only check roughly half the elements.
$endgroup$
– Randall
Jan 22 at 19:18




2




2




$begingroup$
There are so few you can try them one at a time. You will find shortcuts along the way that speed things up.
$endgroup$
– Ethan Bolker
Jan 22 at 19:19




$begingroup$
There are so few you can try them one at a time. You will find shortcuts along the way that speed things up.
$endgroup$
– Ethan Bolker
Jan 22 at 19:19




2




2




$begingroup$
I'd prove that none of them have order of $12$. For this, I'd consider them mod $3$ and mod $4$
$endgroup$
– Jakobian
Jan 22 at 19:22




$begingroup$
I'd prove that none of them have order of $12$. For this, I'd consider them mod $3$ and mod $4$
$endgroup$
– Jakobian
Jan 22 at 19:22












$begingroup$
I would also consider showing that there are four elements with squares equal to $1$. At both ends and near the midpoint. In a cyclic group there can be at most two. Warning: this assumes that you have covered relevant pieces abouot the structure of cyclic groups (not necessarily done in a first course on algberaic structures).
$endgroup$
– Jyrki Lahtonen
Jan 22 at 19:24




$begingroup$
I would also consider showing that there are four elements with squares equal to $1$. At both ends and near the midpoint. In a cyclic group there can be at most two. Warning: this assumes that you have covered relevant pieces abouot the structure of cyclic groups (not necessarily done in a first course on algberaic structures).
$endgroup$
– Jyrki Lahtonen
Jan 22 at 19:24




1




1




$begingroup$
CRT : as a ring $mathbb{Z}_{36} = 9mathbb{Z}_{4} + (-8)mathbb{Z}_{9} cong mathbb{Z}_{4} times mathbb{Z}_{9}$. The 3 subgroups of order $2$ are then clear from there.
$endgroup$
– reuns
Jan 22 at 19:31




$begingroup$
CRT : as a ring $mathbb{Z}_{36} = 9mathbb{Z}_{4} + (-8)mathbb{Z}_{9} cong mathbb{Z}_{4} times mathbb{Z}_{9}$. The 3 subgroups of order $2$ are then clear from there.
$endgroup$
– reuns
Jan 22 at 19:31










2 Answers
2






active

oldest

votes


















0












$begingroup$

Clearly $1$ is not a generator. To check whether $5$ is a generator, compute a few powers;
$$5^1equiv5,qquad 5^2equiv25,qquad 5^3equiv17,qquad5^4equiv13,qquad5^5equiv29,qquad5^6equiv1.$$
This already tells you that $5$, $25$, $17$, $13$ and $29$ are not generators. Next do the same for $7$...






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    It is a classical fact that $mathbb Z_m^times$ is cyclic if and only if $min{1,2,4,p^alpha,2p^alpha}$, where $p$ is an odd prime and $alpha$ is a positive integer.



    A simple way to see that $mathbb Z_{36}^times$ is not cyclic (which also gives a hint about the proof of the general result) is to observe that for any integer $g$ coprime with $36$, one has $g^6equiv 1pmod 9$ and also $g^6equiv 1pmod 4$ (in fact, already $g^2equiv 1pmod 4$), as it follows by a direct computation or by using Euler's theorem. Therefore $g^6equiv 1pmod {36}$, showing that $mathbb Z_{36}^times$ does not contain elements of order $|mathbb Z_{36}^times|=12$.






    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%2f3083582%2fis-the-multiplicative-group-bbb-z-36-times-cyclic%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









      0












      $begingroup$

      Clearly $1$ is not a generator. To check whether $5$ is a generator, compute a few powers;
      $$5^1equiv5,qquad 5^2equiv25,qquad 5^3equiv17,qquad5^4equiv13,qquad5^5equiv29,qquad5^6equiv1.$$
      This already tells you that $5$, $25$, $17$, $13$ and $29$ are not generators. Next do the same for $7$...






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Clearly $1$ is not a generator. To check whether $5$ is a generator, compute a few powers;
        $$5^1equiv5,qquad 5^2equiv25,qquad 5^3equiv17,qquad5^4equiv13,qquad5^5equiv29,qquad5^6equiv1.$$
        This already tells you that $5$, $25$, $17$, $13$ and $29$ are not generators. Next do the same for $7$...






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Clearly $1$ is not a generator. To check whether $5$ is a generator, compute a few powers;
          $$5^1equiv5,qquad 5^2equiv25,qquad 5^3equiv17,qquad5^4equiv13,qquad5^5equiv29,qquad5^6equiv1.$$
          This already tells you that $5$, $25$, $17$, $13$ and $29$ are not generators. Next do the same for $7$...






          share|cite|improve this answer









          $endgroup$



          Clearly $1$ is not a generator. To check whether $5$ is a generator, compute a few powers;
          $$5^1equiv5,qquad 5^2equiv25,qquad 5^3equiv17,qquad5^4equiv13,qquad5^5equiv29,qquad5^6equiv1.$$
          This already tells you that $5$, $25$, $17$, $13$ and $29$ are not generators. Next do the same for $7$...







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 22 at 19:46









          ServaesServaes

          26k33997




          26k33997























              1












              $begingroup$

              It is a classical fact that $mathbb Z_m^times$ is cyclic if and only if $min{1,2,4,p^alpha,2p^alpha}$, where $p$ is an odd prime and $alpha$ is a positive integer.



              A simple way to see that $mathbb Z_{36}^times$ is not cyclic (which also gives a hint about the proof of the general result) is to observe that for any integer $g$ coprime with $36$, one has $g^6equiv 1pmod 9$ and also $g^6equiv 1pmod 4$ (in fact, already $g^2equiv 1pmod 4$), as it follows by a direct computation or by using Euler's theorem. Therefore $g^6equiv 1pmod {36}$, showing that $mathbb Z_{36}^times$ does not contain elements of order $|mathbb Z_{36}^times|=12$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                It is a classical fact that $mathbb Z_m^times$ is cyclic if and only if $min{1,2,4,p^alpha,2p^alpha}$, where $p$ is an odd prime and $alpha$ is a positive integer.



                A simple way to see that $mathbb Z_{36}^times$ is not cyclic (which also gives a hint about the proof of the general result) is to observe that for any integer $g$ coprime with $36$, one has $g^6equiv 1pmod 9$ and also $g^6equiv 1pmod 4$ (in fact, already $g^2equiv 1pmod 4$), as it follows by a direct computation or by using Euler's theorem. Therefore $g^6equiv 1pmod {36}$, showing that $mathbb Z_{36}^times$ does not contain elements of order $|mathbb Z_{36}^times|=12$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  It is a classical fact that $mathbb Z_m^times$ is cyclic if and only if $min{1,2,4,p^alpha,2p^alpha}$, where $p$ is an odd prime and $alpha$ is a positive integer.



                  A simple way to see that $mathbb Z_{36}^times$ is not cyclic (which also gives a hint about the proof of the general result) is to observe that for any integer $g$ coprime with $36$, one has $g^6equiv 1pmod 9$ and also $g^6equiv 1pmod 4$ (in fact, already $g^2equiv 1pmod 4$), as it follows by a direct computation or by using Euler's theorem. Therefore $g^6equiv 1pmod {36}$, showing that $mathbb Z_{36}^times$ does not contain elements of order $|mathbb Z_{36}^times|=12$.






                  share|cite|improve this answer









                  $endgroup$



                  It is a classical fact that $mathbb Z_m^times$ is cyclic if and only if $min{1,2,4,p^alpha,2p^alpha}$, where $p$ is an odd prime and $alpha$ is a positive integer.



                  A simple way to see that $mathbb Z_{36}^times$ is not cyclic (which also gives a hint about the proof of the general result) is to observe that for any integer $g$ coprime with $36$, one has $g^6equiv 1pmod 9$ and also $g^6equiv 1pmod 4$ (in fact, already $g^2equiv 1pmod 4$), as it follows by a direct computation or by using Euler's theorem. Therefore $g^6equiv 1pmod {36}$, showing that $mathbb Z_{36}^times$ does not contain elements of order $|mathbb Z_{36}^times|=12$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 22 at 20:36









                  W-t-PW-t-P

                  1,287611




                  1,287611






























                      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%2f3083582%2fis-the-multiplicative-group-bbb-z-36-times-cyclic%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