elements of order 15 in $Z_5 times Z_{15}$












2












$begingroup$


I know that for an element $(a,b) in mathbb{Z}_j times mathbb{Z}_k$, the element will have order $lcm(m,n)$ where m is the order of a in $mathbb{Z}_j$ and n is the order of b in $mathbb{Z}_k$. So if I am supposed to instead find such a,b do I just go through all possibilities for m and n that give 15 then check if a and b can have such an order in their respective quotient groups?



I am ultimately trying to find the number of elements of order 15 in $mathbb{Z}_5 times mathbb{Z}_{15}$






abstract-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    Determining the number of elements requires no more group theory than the fact that the product of groups is a group on the product of underlying sets.
    $endgroup$
    – Servaes
    Dec 8 '18 at 18:20












  • $begingroup$
    For $(m,n)in Bbb Z_5timesBbb Z_{15}$, what are the possible orders of $m$ and $n$? For which combinations of those does $(m,n)$ have order $15$?
    $endgroup$
    – Arthur
    Dec 8 '18 at 18:21










  • $begingroup$
    The possible orders of m is 5 and for n it is 15,5,3 so (5,3) would be the only combination that works so there is only 1?
    $endgroup$
    – Richard Villalobos
    Dec 8 '18 at 18:30












  • $begingroup$
    what about $(0,1) in Z_5 times Z_{15}$
    $endgroup$
    – Matt A Pelto
    Dec 8 '18 at 18:37
















2












$begingroup$


I know that for an element $(a,b) in mathbb{Z}_j times mathbb{Z}_k$, the element will have order $lcm(m,n)$ where m is the order of a in $mathbb{Z}_j$ and n is the order of b in $mathbb{Z}_k$. So if I am supposed to instead find such a,b do I just go through all possibilities for m and n that give 15 then check if a and b can have such an order in their respective quotient groups?



I am ultimately trying to find the number of elements of order 15 in $mathbb{Z}_5 times mathbb{Z}_{15}$






abstract-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    Determining the number of elements requires no more group theory than the fact that the product of groups is a group on the product of underlying sets.
    $endgroup$
    – Servaes
    Dec 8 '18 at 18:20












  • $begingroup$
    For $(m,n)in Bbb Z_5timesBbb Z_{15}$, what are the possible orders of $m$ and $n$? For which combinations of those does $(m,n)$ have order $15$?
    $endgroup$
    – Arthur
    Dec 8 '18 at 18:21










  • $begingroup$
    The possible orders of m is 5 and for n it is 15,5,3 so (5,3) would be the only combination that works so there is only 1?
    $endgroup$
    – Richard Villalobos
    Dec 8 '18 at 18:30












  • $begingroup$
    what about $(0,1) in Z_5 times Z_{15}$
    $endgroup$
    – Matt A Pelto
    Dec 8 '18 at 18:37














2












2








2





$begingroup$


I know that for an element $(a,b) in mathbb{Z}_j times mathbb{Z}_k$, the element will have order $lcm(m,n)$ where m is the order of a in $mathbb{Z}_j$ and n is the order of b in $mathbb{Z}_k$. So if I am supposed to instead find such a,b do I just go through all possibilities for m and n that give 15 then check if a and b can have such an order in their respective quotient groups?



I am ultimately trying to find the number of elements of order 15 in $mathbb{Z}_5 times mathbb{Z}_{15}$






abstract-algebra







share|cite|improve this question









$endgroup$




I know that for an element $(a,b) in mathbb{Z}_j times mathbb{Z}_k$, the element will have order $lcm(m,n)$ where m is the order of a in $mathbb{Z}_j$ and n is the order of b in $mathbb{Z}_k$. So if I am supposed to instead find such a,b do I just go through all possibilities for m and n that give 15 then check if a and b can have such an order in their respective quotient groups?



I am ultimately trying to find the number of elements of order 15 in $mathbb{Z}_5 times mathbb{Z}_{15}$






abstract-algebra




abstract-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 8 '18 at 18:15









Richard VillalobosRichard Villalobos

1687




1687












  • $begingroup$
    Determining the number of elements requires no more group theory than the fact that the product of groups is a group on the product of underlying sets.
    $endgroup$
    – Servaes
    Dec 8 '18 at 18:20












  • $begingroup$
    For $(m,n)in Bbb Z_5timesBbb Z_{15}$, what are the possible orders of $m$ and $n$? For which combinations of those does $(m,n)$ have order $15$?
    $endgroup$
    – Arthur
    Dec 8 '18 at 18:21










  • $begingroup$
    The possible orders of m is 5 and for n it is 15,5,3 so (5,3) would be the only combination that works so there is only 1?
    $endgroup$
    – Richard Villalobos
    Dec 8 '18 at 18:30












  • $begingroup$
    what about $(0,1) in Z_5 times Z_{15}$
    $endgroup$
    – Matt A Pelto
    Dec 8 '18 at 18:37


















  • $begingroup$
    Determining the number of elements requires no more group theory than the fact that the product of groups is a group on the product of underlying sets.
    $endgroup$
    – Servaes
    Dec 8 '18 at 18:20












  • $begingroup$
    For $(m,n)in Bbb Z_5timesBbb Z_{15}$, what are the possible orders of $m$ and $n$? For which combinations of those does $(m,n)$ have order $15$?
    $endgroup$
    – Arthur
    Dec 8 '18 at 18:21










  • $begingroup$
    The possible orders of m is 5 and for n it is 15,5,3 so (5,3) would be the only combination that works so there is only 1?
    $endgroup$
    – Richard Villalobos
    Dec 8 '18 at 18:30












  • $begingroup$
    what about $(0,1) in Z_5 times Z_{15}$
    $endgroup$
    – Matt A Pelto
    Dec 8 '18 at 18:37
















$begingroup$
Determining the number of elements requires no more group theory than the fact that the product of groups is a group on the product of underlying sets.
$endgroup$
– Servaes
Dec 8 '18 at 18:20






$begingroup$
Determining the number of elements requires no more group theory than the fact that the product of groups is a group on the product of underlying sets.
$endgroup$
– Servaes
Dec 8 '18 at 18:20














$begingroup$
For $(m,n)in Bbb Z_5timesBbb Z_{15}$, what are the possible orders of $m$ and $n$? For which combinations of those does $(m,n)$ have order $15$?
$endgroup$
– Arthur
Dec 8 '18 at 18:21




$begingroup$
For $(m,n)in Bbb Z_5timesBbb Z_{15}$, what are the possible orders of $m$ and $n$? For which combinations of those does $(m,n)$ have order $15$?
$endgroup$
– Arthur
Dec 8 '18 at 18:21












$begingroup$
The possible orders of m is 5 and for n it is 15,5,3 so (5,3) would be the only combination that works so there is only 1?
$endgroup$
– Richard Villalobos
Dec 8 '18 at 18:30






$begingroup$
The possible orders of m is 5 and for n it is 15,5,3 so (5,3) would be the only combination that works so there is only 1?
$endgroup$
– Richard Villalobos
Dec 8 '18 at 18:30














$begingroup$
what about $(0,1) in Z_5 times Z_{15}$
$endgroup$
– Matt A Pelto
Dec 8 '18 at 18:37




$begingroup$
what about $(0,1) in Z_5 times Z_{15}$
$endgroup$
– Matt A Pelto
Dec 8 '18 at 18:37










2 Answers
2






active

oldest

votes


















3












$begingroup$

$$mathbb{Z}_5times mathbb{Z}_{15}cong mathbb{Z}_5times mathbb{Z}_5times mathbb{Z}_3$$



Hence, we need to select an element of order $5$ from $mathbb{Z}_5times mathbb{Z}_5$, and combine with an element of order $3$ from $mathbb{Z}_3$. There are $24$ ways to do the former (anything but the identity), and $2$ ways to do the latter. Hence the answer you seek is $48$.






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    $Z_5 times Z_{15}=<f1,f2,f3>$
    $f2, f1f2, f2^2, f2f3, f1^2f2, f1f2^2, f1f2f3, f2f3^2, f1^3f2, f1^2f2^2, f1^2f2f3, f1f2^2f3,f1f2f3^2, f2^2f3^2, f1^4f2, f1^3f2^2, f1^3f2f3, f1^2f2^2f3, f1^2f2f3^2, f1f2^2f3^2,f1f2f3^3, f2^2f3^3, f2f3^4, f1^4f2^2, f1^4f2f3, f1^3f2^2f3, f1^3f2f3^2, f1^2f2^2f3^2, f1^2f2f3^3, f1f2^2f3^3, f1f2f3^4, f2^2f3^4, f1^4f2^2f3, f1^4f2f3^2, f1^3f2^2f3^2,f1^3f2f3^3, f1^2f2^2f3^3, f1^2f2f3^4, f1f2^2f3^4, f1^4f2^2f3^2, f1^4f2f3^3, f1^3f2^2f3^3,f1^3f2f3^4, f1^2f2^2f3^4, f1^4f2^2f3^3, f1^4f2f3^4, f1^3f2^2f3^4, f1^4f2^2f3^4 $ are of order 15






    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%2f3031434%2felements-of-order-15-in-z-5-times-z-15%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









      3












      $begingroup$

      $$mathbb{Z}_5times mathbb{Z}_{15}cong mathbb{Z}_5times mathbb{Z}_5times mathbb{Z}_3$$



      Hence, we need to select an element of order $5$ from $mathbb{Z}_5times mathbb{Z}_5$, and combine with an element of order $3$ from $mathbb{Z}_3$. There are $24$ ways to do the former (anything but the identity), and $2$ ways to do the latter. Hence the answer you seek is $48$.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        $$mathbb{Z}_5times mathbb{Z}_{15}cong mathbb{Z}_5times mathbb{Z}_5times mathbb{Z}_3$$



        Hence, we need to select an element of order $5$ from $mathbb{Z}_5times mathbb{Z}_5$, and combine with an element of order $3$ from $mathbb{Z}_3$. There are $24$ ways to do the former (anything but the identity), and $2$ ways to do the latter. Hence the answer you seek is $48$.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          $$mathbb{Z}_5times mathbb{Z}_{15}cong mathbb{Z}_5times mathbb{Z}_5times mathbb{Z}_3$$



          Hence, we need to select an element of order $5$ from $mathbb{Z}_5times mathbb{Z}_5$, and combine with an element of order $3$ from $mathbb{Z}_3$. There are $24$ ways to do the former (anything but the identity), and $2$ ways to do the latter. Hence the answer you seek is $48$.






          share|cite|improve this answer









          $endgroup$



          $$mathbb{Z}_5times mathbb{Z}_{15}cong mathbb{Z}_5times mathbb{Z}_5times mathbb{Z}_3$$



          Hence, we need to select an element of order $5$ from $mathbb{Z}_5times mathbb{Z}_5$, and combine with an element of order $3$ from $mathbb{Z}_3$. There are $24$ ways to do the former (anything but the identity), and $2$ ways to do the latter. Hence the answer you seek is $48$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 19:41









          vadim123vadim123

          76k897189




          76k897189























              -1












              $begingroup$

              $Z_5 times Z_{15}=<f1,f2,f3>$
              $f2, f1f2, f2^2, f2f3, f1^2f2, f1f2^2, f1f2f3, f2f3^2, f1^3f2, f1^2f2^2, f1^2f2f3, f1f2^2f3,f1f2f3^2, f2^2f3^2, f1^4f2, f1^3f2^2, f1^3f2f3, f1^2f2^2f3, f1^2f2f3^2, f1f2^2f3^2,f1f2f3^3, f2^2f3^3, f2f3^4, f1^4f2^2, f1^4f2f3, f1^3f2^2f3, f1^3f2f3^2, f1^2f2^2f3^2, f1^2f2f3^3, f1f2^2f3^3, f1f2f3^4, f2^2f3^4, f1^4f2^2f3, f1^4f2f3^2, f1^3f2^2f3^2,f1^3f2f3^3, f1^2f2^2f3^3, f1^2f2f3^4, f1f2^2f3^4, f1^4f2^2f3^2, f1^4f2f3^3, f1^3f2^2f3^3,f1^3f2f3^4, f1^2f2^2f3^4, f1^4f2^2f3^3, f1^4f2f3^4, f1^3f2^2f3^4, f1^4f2^2f3^4 $ are of order 15






              share|cite|improve this answer









              $endgroup$


















                -1












                $begingroup$

                $Z_5 times Z_{15}=<f1,f2,f3>$
                $f2, f1f2, f2^2, f2f3, f1^2f2, f1f2^2, f1f2f3, f2f3^2, f1^3f2, f1^2f2^2, f1^2f2f3, f1f2^2f3,f1f2f3^2, f2^2f3^2, f1^4f2, f1^3f2^2, f1^3f2f3, f1^2f2^2f3, f1^2f2f3^2, f1f2^2f3^2,f1f2f3^3, f2^2f3^3, f2f3^4, f1^4f2^2, f1^4f2f3, f1^3f2^2f3, f1^3f2f3^2, f1^2f2^2f3^2, f1^2f2f3^3, f1f2^2f3^3, f1f2f3^4, f2^2f3^4, f1^4f2^2f3, f1^4f2f3^2, f1^3f2^2f3^2,f1^3f2f3^3, f1^2f2^2f3^3, f1^2f2f3^4, f1f2^2f3^4, f1^4f2^2f3^2, f1^4f2f3^3, f1^3f2^2f3^3,f1^3f2f3^4, f1^2f2^2f3^4, f1^4f2^2f3^3, f1^4f2f3^4, f1^3f2^2f3^4, f1^4f2^2f3^4 $ are of order 15






                share|cite|improve this answer









                $endgroup$
















                  -1












                  -1








                  -1





                  $begingroup$

                  $Z_5 times Z_{15}=<f1,f2,f3>$
                  $f2, f1f2, f2^2, f2f3, f1^2f2, f1f2^2, f1f2f3, f2f3^2, f1^3f2, f1^2f2^2, f1^2f2f3, f1f2^2f3,f1f2f3^2, f2^2f3^2, f1^4f2, f1^3f2^2, f1^3f2f3, f1^2f2^2f3, f1^2f2f3^2, f1f2^2f3^2,f1f2f3^3, f2^2f3^3, f2f3^4, f1^4f2^2, f1^4f2f3, f1^3f2^2f3, f1^3f2f3^2, f1^2f2^2f3^2, f1^2f2f3^3, f1f2^2f3^3, f1f2f3^4, f2^2f3^4, f1^4f2^2f3, f1^4f2f3^2, f1^3f2^2f3^2,f1^3f2f3^3, f1^2f2^2f3^3, f1^2f2f3^4, f1f2^2f3^4, f1^4f2^2f3^2, f1^4f2f3^3, f1^3f2^2f3^3,f1^3f2f3^4, f1^2f2^2f3^4, f1^4f2^2f3^3, f1^4f2f3^4, f1^3f2^2f3^4, f1^4f2^2f3^4 $ are of order 15






                  share|cite|improve this answer









                  $endgroup$



                  $Z_5 times Z_{15}=<f1,f2,f3>$
                  $f2, f1f2, f2^2, f2f3, f1^2f2, f1f2^2, f1f2f3, f2f3^2, f1^3f2, f1^2f2^2, f1^2f2f3, f1f2^2f3,f1f2f3^2, f2^2f3^2, f1^4f2, f1^3f2^2, f1^3f2f3, f1^2f2^2f3, f1^2f2f3^2, f1f2^2f3^2,f1f2f3^3, f2^2f3^3, f2f3^4, f1^4f2^2, f1^4f2f3, f1^3f2^2f3, f1^3f2f3^2, f1^2f2^2f3^2, f1^2f2f3^3, f1f2^2f3^3, f1f2f3^4, f2^2f3^4, f1^4f2^2f3, f1^4f2f3^2, f1^3f2^2f3^2,f1^3f2f3^3, f1^2f2^2f3^3, f1^2f2f3^4, f1f2^2f3^4, f1^4f2^2f3^2, f1^4f2f3^3, f1^3f2^2f3^3,f1^3f2f3^4, f1^2f2^2f3^4, f1^4f2^2f3^3, f1^4f2f3^4, f1^3f2^2f3^4, f1^4f2^2f3^4 $ are of order 15







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 8 '18 at 19:01









                  yavaryavar

                  803




                  803






























                      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%2f3031434%2felements-of-order-15-in-z-5-times-z-15%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

                      Index of /

                      This page is only for reference, If you need detailed information, please check here
                      Read more

                      Tribalistas

                      Image
                        Nota: Para o álbum homônimo, veja Tribalistas (álbum de 2002). Para o segundo álbum homônimo, veja Tribalistas (álbum de 2017). Para o tipo de organização tribal, veja Tribalismo. Tribalistas Informação geral Origem Salvador e Rio de Janeiro País   Brasil Gênero(s) MPB Pop Samba Período em atividade 2002–2004 2017–atualmente Gravadora(s) EMI Phonomotor Records Universal Music Afiliação(ões) Dadi Pretinho da Serrinha Margareth Menezes Integrantes Arnaldo Antunes Carlinhos Brown Marisa Monte Página oficial www2.uol.com.br/tribalistas/ ...
                      Read more

                      Listed building

                      Image
                      Listed building (literalmente: prédio listado ) é toda estrutura protegida por lei pelo complexo sistema de tombamento do Reino Unido. Um listed building é qualquer estrutura oficialmente designada como tendo especial interesse arquitetônico, histórico ou cultural. É um estatuto amplamente usado, aplicado a centenas de milhares de estruturas. Estruturas tombadas não podem ser alteradas sem permissão especial, e desfrutam de isenções fiscais e outros benefícios. Cada um dos países do Reino Unido trata tombamento de forma diferente: Índice 1 Inglaterra 2 País de Gales 3 Irlanda do Norte 4 Escócia 5 Referências Inglaterra | O Palácio de Buckingham, Londres. As estruturas tombadas são listadas no Statutory List of Buildings of Special Architectural or Historic Interest , e incluem campos, sinais de trânsito, piers e áreas geográficas. A principal legislação pertinente é o Ancient Monuments and Archaeological Areas Act 1979 , [ 1 ] embora legisla...
                      Read more