For what $n$ is $U_n$ cyclic?












25












$begingroup$



When can we say a multiplicative group of integers modulo $n$, i.e., $U_n$ is cyclic?




$$U_n={a inmathbb Z_n mid gcd(a,n)=1 }$$



I searched the internet but did not get a clear idea.






abstract-algebra group-theory cyclic-groups







share|cite|improve this question











$endgroup$

















    25












    $begingroup$



    When can we say a multiplicative group of integers modulo $n$, i.e., $U_n$ is cyclic?




    $$U_n={a inmathbb Z_n mid gcd(a,n)=1 }$$



    I searched the internet but did not get a clear idea.






    abstract-algebra group-theory cyclic-groups







    share|cite|improve this question











    $endgroup$















      25












      25








      25


      18



      $begingroup$



      When can we say a multiplicative group of integers modulo $n$, i.e., $U_n$ is cyclic?




      $$U_n={a inmathbb Z_n mid gcd(a,n)=1 }$$



      I searched the internet but did not get a clear idea.






      abstract-algebra group-theory cyclic-groups







      share|cite|improve this question











      $endgroup$





      When can we say a multiplicative group of integers modulo $n$, i.e., $U_n$ is cyclic?




      $$U_n={a inmathbb Z_n mid gcd(a,n)=1 }$$



      I searched the internet but did not get a clear idea.






      abstract-algebra group-theory cyclic-groups




      abstract-algebra group-theory cyclic-groups






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 7 '16 at 11:17









      Rudy the Reindeer

      26.5k1793241




      26.5k1793241










      asked Feb 26 '13 at 12:58









      SankhaSankha

      590722




      590722






















          3 Answers
          3






          active

          oldest

          votes


















          25












          $begingroup$

          So $U_n$ is the group of units in $mathbb{Z}/nmathbb{Z}$.



          Write the prime decomposition
          $$
          n=p_1^{alpha_1}cdots p_r^{alpha_r}.
          $$



          By the Chinese remainder theorem
          $$
          mathbb{Z}/nmathbb{Z}=mathbb{Z}/p_1^{alpha_1}mathbb{Z}timesldotstimesmathbb{Z}/p_r^{alpha_r}mathbb{Z}
          $$
          so
          $$
          U_n=U_{p_1^{alpha_1}}timesldotstimes U_{p_r^{alpha_r}}.
          $$



          For powers of $2$, we have
          $$
          U_2={0}
          $$
          and for $kgeq 2$
          $$
          U_{2^k}=mathbb{Z}/2mathbb{Z}times mathbb{Z}/2^{k-2}mathbb{Z}.
          $$



          For odd primes $p$,
          $$
          U_{p^alpha}=mathbb{Z}/phi(p^alpha)mathbb{Z}=mathbb{Z}/p^{alpha-1}(p-1)mathbb{Z}.
          $$



          So you see now that $U_n$ is cyclic if and only if
          $$
          n=2,4,p^alpha,2p^{alpha}
          $$
          where $p$ is an odd prime.



          Here is a reference.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Why is it true that $U_{p^k}=mathbb{Z}/2mathbb{Z}timesmathbb{Z}/2^{k-2}mathbb{Z}$?
            $endgroup$
            – Rasputin
            Jan 20 '17 at 20:03










          • $begingroup$
            Julien, why doesn't the even prime work please?
            $endgroup$
            – BCLC
            Oct 17 '18 at 11:48



















          9












          $begingroup$

          $U_n$ is cyclic iff $n$ is $2$, $4$, $p^k$, or $2p^k$, where $p$ is an odd prime.



          The proof follows from the Chinese Remainder Theorem for rings and the fact that $C_m times C_n$ is cyclic iff $(m,n)=1$ (here $C_n$ is the cyclic group of order $n$).



          The hard part is proving that $U_p$ is cyclic and this follows from the fact that $mathbb Z/p$ is a field and that $n = sum_{dmid n} phi(d)$.



          Any book on elementary number theory has a proof of this theorem. See for instance André Weil's Number theory for beginners, Leveque's Fundamentals of Number Theory, and Bolker's Elementary Number Theory.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @julien, thanks, fixed.
            $endgroup$
            – lhf
            Feb 26 '13 at 13:26



















          5












          $begingroup$

          Here "cyclic if and only if $varphi(n)=lambda(n)$" but there's no proof - the proof is elementary but very tricky.






          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%2f314846%2ffor-what-n-is-u-n-cyclic%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









            25












            $begingroup$

            So $U_n$ is the group of units in $mathbb{Z}/nmathbb{Z}$.



            Write the prime decomposition
            $$
            n=p_1^{alpha_1}cdots p_r^{alpha_r}.
            $$



            By the Chinese remainder theorem
            $$
            mathbb{Z}/nmathbb{Z}=mathbb{Z}/p_1^{alpha_1}mathbb{Z}timesldotstimesmathbb{Z}/p_r^{alpha_r}mathbb{Z}
            $$
            so
            $$
            U_n=U_{p_1^{alpha_1}}timesldotstimes U_{p_r^{alpha_r}}.
            $$



            For powers of $2$, we have
            $$
            U_2={0}
            $$
            and for $kgeq 2$
            $$
            U_{2^k}=mathbb{Z}/2mathbb{Z}times mathbb{Z}/2^{k-2}mathbb{Z}.
            $$



            For odd primes $p$,
            $$
            U_{p^alpha}=mathbb{Z}/phi(p^alpha)mathbb{Z}=mathbb{Z}/p^{alpha-1}(p-1)mathbb{Z}.
            $$



            So you see now that $U_n$ is cyclic if and only if
            $$
            n=2,4,p^alpha,2p^{alpha}
            $$
            where $p$ is an odd prime.



            Here is a reference.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Why is it true that $U_{p^k}=mathbb{Z}/2mathbb{Z}timesmathbb{Z}/2^{k-2}mathbb{Z}$?
              $endgroup$
              – Rasputin
              Jan 20 '17 at 20:03










            • $begingroup$
              Julien, why doesn't the even prime work please?
              $endgroup$
              – BCLC
              Oct 17 '18 at 11:48
















            25












            $begingroup$

            So $U_n$ is the group of units in $mathbb{Z}/nmathbb{Z}$.



            Write the prime decomposition
            $$
            n=p_1^{alpha_1}cdots p_r^{alpha_r}.
            $$



            By the Chinese remainder theorem
            $$
            mathbb{Z}/nmathbb{Z}=mathbb{Z}/p_1^{alpha_1}mathbb{Z}timesldotstimesmathbb{Z}/p_r^{alpha_r}mathbb{Z}
            $$
            so
            $$
            U_n=U_{p_1^{alpha_1}}timesldotstimes U_{p_r^{alpha_r}}.
            $$



            For powers of $2$, we have
            $$
            U_2={0}
            $$
            and for $kgeq 2$
            $$
            U_{2^k}=mathbb{Z}/2mathbb{Z}times mathbb{Z}/2^{k-2}mathbb{Z}.
            $$



            For odd primes $p$,
            $$
            U_{p^alpha}=mathbb{Z}/phi(p^alpha)mathbb{Z}=mathbb{Z}/p^{alpha-1}(p-1)mathbb{Z}.
            $$



            So you see now that $U_n$ is cyclic if and only if
            $$
            n=2,4,p^alpha,2p^{alpha}
            $$
            where $p$ is an odd prime.



            Here is a reference.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Why is it true that $U_{p^k}=mathbb{Z}/2mathbb{Z}timesmathbb{Z}/2^{k-2}mathbb{Z}$?
              $endgroup$
              – Rasputin
              Jan 20 '17 at 20:03










            • $begingroup$
              Julien, why doesn't the even prime work please?
              $endgroup$
              – BCLC
              Oct 17 '18 at 11:48














            25












            25








            25





            $begingroup$

            So $U_n$ is the group of units in $mathbb{Z}/nmathbb{Z}$.



            Write the prime decomposition
            $$
            n=p_1^{alpha_1}cdots p_r^{alpha_r}.
            $$



            By the Chinese remainder theorem
            $$
            mathbb{Z}/nmathbb{Z}=mathbb{Z}/p_1^{alpha_1}mathbb{Z}timesldotstimesmathbb{Z}/p_r^{alpha_r}mathbb{Z}
            $$
            so
            $$
            U_n=U_{p_1^{alpha_1}}timesldotstimes U_{p_r^{alpha_r}}.
            $$



            For powers of $2$, we have
            $$
            U_2={0}
            $$
            and for $kgeq 2$
            $$
            U_{2^k}=mathbb{Z}/2mathbb{Z}times mathbb{Z}/2^{k-2}mathbb{Z}.
            $$



            For odd primes $p$,
            $$
            U_{p^alpha}=mathbb{Z}/phi(p^alpha)mathbb{Z}=mathbb{Z}/p^{alpha-1}(p-1)mathbb{Z}.
            $$



            So you see now that $U_n$ is cyclic if and only if
            $$
            n=2,4,p^alpha,2p^{alpha}
            $$
            where $p$ is an odd prime.



            Here is a reference.






            share|cite|improve this answer











            $endgroup$



            So $U_n$ is the group of units in $mathbb{Z}/nmathbb{Z}$.



            Write the prime decomposition
            $$
            n=p_1^{alpha_1}cdots p_r^{alpha_r}.
            $$



            By the Chinese remainder theorem
            $$
            mathbb{Z}/nmathbb{Z}=mathbb{Z}/p_1^{alpha_1}mathbb{Z}timesldotstimesmathbb{Z}/p_r^{alpha_r}mathbb{Z}
            $$
            so
            $$
            U_n=U_{p_1^{alpha_1}}timesldotstimes U_{p_r^{alpha_r}}.
            $$



            For powers of $2$, we have
            $$
            U_2={0}
            $$
            and for $kgeq 2$
            $$
            U_{2^k}=mathbb{Z}/2mathbb{Z}times mathbb{Z}/2^{k-2}mathbb{Z}.
            $$



            For odd primes $p$,
            $$
            U_{p^alpha}=mathbb{Z}/phi(p^alpha)mathbb{Z}=mathbb{Z}/p^{alpha-1}(p-1)mathbb{Z}.
            $$



            So you see now that $U_n$ is cyclic if and only if
            $$
            n=2,4,p^alpha,2p^{alpha}
            $$
            where $p$ is an odd prime.



            Here is a reference.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 18 '16 at 21:25









            user26857

            39.3k124183




            39.3k124183










            answered Feb 26 '13 at 13:24









            JulienJulien

            38.7k358130




            38.7k358130








            • 1




              $begingroup$
              Why is it true that $U_{p^k}=mathbb{Z}/2mathbb{Z}timesmathbb{Z}/2^{k-2}mathbb{Z}$?
              $endgroup$
              – Rasputin
              Jan 20 '17 at 20:03










            • $begingroup$
              Julien, why doesn't the even prime work please?
              $endgroup$
              – BCLC
              Oct 17 '18 at 11:48














            • 1




              $begingroup$
              Why is it true that $U_{p^k}=mathbb{Z}/2mathbb{Z}timesmathbb{Z}/2^{k-2}mathbb{Z}$?
              $endgroup$
              – Rasputin
              Jan 20 '17 at 20:03










            • $begingroup$
              Julien, why doesn't the even prime work please?
              $endgroup$
              – BCLC
              Oct 17 '18 at 11:48








            1




            1




            $begingroup$
            Why is it true that $U_{p^k}=mathbb{Z}/2mathbb{Z}timesmathbb{Z}/2^{k-2}mathbb{Z}$?
            $endgroup$
            – Rasputin
            Jan 20 '17 at 20:03




            $begingroup$
            Why is it true that $U_{p^k}=mathbb{Z}/2mathbb{Z}timesmathbb{Z}/2^{k-2}mathbb{Z}$?
            $endgroup$
            – Rasputin
            Jan 20 '17 at 20:03












            $begingroup$
            Julien, why doesn't the even prime work please?
            $endgroup$
            – BCLC
            Oct 17 '18 at 11:48




            $begingroup$
            Julien, why doesn't the even prime work please?
            $endgroup$
            – BCLC
            Oct 17 '18 at 11:48











            9












            $begingroup$

            $U_n$ is cyclic iff $n$ is $2$, $4$, $p^k$, or $2p^k$, where $p$ is an odd prime.



            The proof follows from the Chinese Remainder Theorem for rings and the fact that $C_m times C_n$ is cyclic iff $(m,n)=1$ (here $C_n$ is the cyclic group of order $n$).



            The hard part is proving that $U_p$ is cyclic and this follows from the fact that $mathbb Z/p$ is a field and that $n = sum_{dmid n} phi(d)$.



            Any book on elementary number theory has a proof of this theorem. See for instance André Weil's Number theory for beginners, Leveque's Fundamentals of Number Theory, and Bolker's Elementary Number Theory.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @julien, thanks, fixed.
              $endgroup$
              – lhf
              Feb 26 '13 at 13:26
















            9












            $begingroup$

            $U_n$ is cyclic iff $n$ is $2$, $4$, $p^k$, or $2p^k$, where $p$ is an odd prime.



            The proof follows from the Chinese Remainder Theorem for rings and the fact that $C_m times C_n$ is cyclic iff $(m,n)=1$ (here $C_n$ is the cyclic group of order $n$).



            The hard part is proving that $U_p$ is cyclic and this follows from the fact that $mathbb Z/p$ is a field and that $n = sum_{dmid n} phi(d)$.



            Any book on elementary number theory has a proof of this theorem. See for instance André Weil's Number theory for beginners, Leveque's Fundamentals of Number Theory, and Bolker's Elementary Number Theory.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @julien, thanks, fixed.
              $endgroup$
              – lhf
              Feb 26 '13 at 13:26














            9












            9








            9





            $begingroup$

            $U_n$ is cyclic iff $n$ is $2$, $4$, $p^k$, or $2p^k$, where $p$ is an odd prime.



            The proof follows from the Chinese Remainder Theorem for rings and the fact that $C_m times C_n$ is cyclic iff $(m,n)=1$ (here $C_n$ is the cyclic group of order $n$).



            The hard part is proving that $U_p$ is cyclic and this follows from the fact that $mathbb Z/p$ is a field and that $n = sum_{dmid n} phi(d)$.



            Any book on elementary number theory has a proof of this theorem. See for instance André Weil's Number theory for beginners, Leveque's Fundamentals of Number Theory, and Bolker's Elementary Number Theory.






            share|cite|improve this answer











            $endgroup$



            $U_n$ is cyclic iff $n$ is $2$, $4$, $p^k$, or $2p^k$, where $p$ is an odd prime.



            The proof follows from the Chinese Remainder Theorem for rings and the fact that $C_m times C_n$ is cyclic iff $(m,n)=1$ (here $C_n$ is the cyclic group of order $n$).



            The hard part is proving that $U_p$ is cyclic and this follows from the fact that $mathbb Z/p$ is a field and that $n = sum_{dmid n} phi(d)$.



            Any book on elementary number theory has a proof of this theorem. See for instance André Weil's Number theory for beginners, Leveque's Fundamentals of Number Theory, and Bolker's Elementary Number Theory.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Feb 26 '13 at 14:59









            Michael Hardy

            1




            1










            answered Feb 26 '13 at 13:11









            lhflhf

            165k10171396




            165k10171396












            • $begingroup$
              @julien, thanks, fixed.
              $endgroup$
              – lhf
              Feb 26 '13 at 13:26


















            • $begingroup$
              @julien, thanks, fixed.
              $endgroup$
              – lhf
              Feb 26 '13 at 13:26
















            $begingroup$
            @julien, thanks, fixed.
            $endgroup$
            – lhf
            Feb 26 '13 at 13:26




            $begingroup$
            @julien, thanks, fixed.
            $endgroup$
            – lhf
            Feb 26 '13 at 13:26











            5












            $begingroup$

            Here "cyclic if and only if $varphi(n)=lambda(n)$" but there's no proof - the proof is elementary but very tricky.






            share|cite|improve this answer











            $endgroup$


















              5












              $begingroup$

              Here "cyclic if and only if $varphi(n)=lambda(n)$" but there's no proof - the proof is elementary but very tricky.






              share|cite|improve this answer











              $endgroup$
















                5












                5








                5





                $begingroup$

                Here "cyclic if and only if $varphi(n)=lambda(n)$" but there's no proof - the proof is elementary but very tricky.






                share|cite|improve this answer











                $endgroup$



                Here "cyclic if and only if $varphi(n)=lambda(n)$" but there's no proof - the proof is elementary but very tricky.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 16 '13 at 9:18

























                answered Feb 26 '13 at 13:07







                user58512





































                    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%2f314846%2ffor-what-n-is-u-n-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

                    Probability when a professor distributes a quiz and homework assignment to a class of n students.

                    Image
                    0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                    Read more

                    Aardman Animations

                    Image
                    Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                    Read more

                    Are they similar matrix

                    Image
                    0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                    Read more