For what $n$ is $U_n$ cyclic?












23















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





























    23















    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



























      23












      23








      23


      17






      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
















      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

      26.4k1690239




      26.4k1690239










      asked Feb 26 '13 at 12:58









      SankhaSankha

      575622




      575622






















          3 Answers
          3






          active

          oldest

          votes


















          24














          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



















          • 1




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










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



















          9














          $U_n$ 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























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



















          5














          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























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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            24














            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



















            • 1




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










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
















            24














            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



















            • 1




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










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














            24












            24








            24






            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














            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

            39.3k124083




            39.3k124083










            answered Feb 26 '13 at 13:24









            JulienJulien

            38.5k358129




            38.5k358129








            • 1




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










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














            • 1




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










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








            1




            1




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




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












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




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











            9














            $U_n$ 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























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
















            9














            $U_n$ 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























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














            9












            9








            9






            $U_n$ 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














            $U_n$ 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

            163k10168388




            163k10168388












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


















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
















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




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











            5














            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




























              5














              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


























                5












                5








                5






                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














                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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