For what $n$ is $U_n$ cyclic?












26












$begingroup$



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











$endgroup$

















    26












    $begingroup$



    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











    $endgroup$















      26












      26








      26


      19



      $begingroup$



      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

      26.8k1796245




      26.8k1796245










      asked Feb 26 '13 at 12:58









      SankhaSankha

      595722




      595722






















          3 Answers
          3






          active

          oldest

          votes


















          26












          $begingroup$

          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











          $endgroup$









          • 2




            $begingroup$
            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



















          10












          $begingroup$

          $U_n$ 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











          $endgroup$





















            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











            $endgroup$














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              3 Answers
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              3 Answers
              3






              active

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              active

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              active

              oldest

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              26












              $begingroup$

              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











              $endgroup$









              • 2




                $begingroup$
                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
















              26












              $begingroup$

              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











              $endgroup$









              • 2




                $begingroup$
                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














              26












              26








              26





              $begingroup$

              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

              39.5k124284




              39.5k124284










              answered Feb 26 '13 at 13:24









              JulienJulien

              38.8k358131




              38.8k358131








              • 2




                $begingroup$
                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














              • 2




                $begingroup$
                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








              2




              2




              $begingroup$
              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











              10












              $begingroup$

              $U_n$ 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











              $endgroup$


















                10












                $begingroup$

                $U_n$ 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











                $endgroup$
















                  10












                  10








                  10





                  $begingroup$

                  $U_n$ 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

                  167k11172404




                  167k11172404























                      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





































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