$(a,b)=1$ $iff~Bbb Z_atimesBbb Z_b$ is cyclic [duplicate]












3















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  • When are the product of cyclic groups also cyclic? [duplicate]

    2 answers




Assuming $(a,b)=1$, I want to construct an isomorphism $phi:Bbb Z_atimesBbb Z_bmapstoBbb Z_{acdot b}$ defined by $phi(1,1)=1$. So I need to prove that for any $(x,y)in Bbb Z_atimesBbb Z_b$ there exists $nin Bbb N$ such that $nequiv x mod a$ and $nequiv ymod b$



For example $(1,2)inBbb Z_2timesBbb Z_3$ is $(5mod2,5mod3)$



I'm not sure how to prove that, I tried using the Bezout equality...



For the other way we need to assume $Bbb Z_atimesBbb Z_b$ is cyclic i.e. $Bbb Z_atimesBbb Z_b =langle(x,y)rangle$ for some $(x,y)$



Since $0inlangle(x,y)rangle$, $exists nin Bbb N~:~nxequiv0mod a~text{and } nyequiv 0mod b$



And here I'm stuck again.










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marked as duplicate by Theo Bendit, José Carlos Santos, Paul Frost, Pierre-Guy Plamondon, Cesareo Jan 4 at 12:43


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.




















    3















    This question already has an answer here:




    • When are the product of cyclic groups also cyclic? [duplicate]

      2 answers




    Assuming $(a,b)=1$, I want to construct an isomorphism $phi:Bbb Z_atimesBbb Z_bmapstoBbb Z_{acdot b}$ defined by $phi(1,1)=1$. So I need to prove that for any $(x,y)in Bbb Z_atimesBbb Z_b$ there exists $nin Bbb N$ such that $nequiv x mod a$ and $nequiv ymod b$



    For example $(1,2)inBbb Z_2timesBbb Z_3$ is $(5mod2,5mod3)$



    I'm not sure how to prove that, I tried using the Bezout equality...



    For the other way we need to assume $Bbb Z_atimesBbb Z_b$ is cyclic i.e. $Bbb Z_atimesBbb Z_b =langle(x,y)rangle$ for some $(x,y)$



    Since $0inlangle(x,y)rangle$, $exists nin Bbb N~:~nxequiv0mod a~text{and } nyequiv 0mod b$



    And here I'm stuck again.










    share|cite|improve this question













    marked as duplicate by Theo Bendit, José Carlos Santos, Paul Frost, Pierre-Guy Plamondon, Cesareo Jan 4 at 12:43


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















      3












      3








      3








      This question already has an answer here:




      • When are the product of cyclic groups also cyclic? [duplicate]

        2 answers




      Assuming $(a,b)=1$, I want to construct an isomorphism $phi:Bbb Z_atimesBbb Z_bmapstoBbb Z_{acdot b}$ defined by $phi(1,1)=1$. So I need to prove that for any $(x,y)in Bbb Z_atimesBbb Z_b$ there exists $nin Bbb N$ such that $nequiv x mod a$ and $nequiv ymod b$



      For example $(1,2)inBbb Z_2timesBbb Z_3$ is $(5mod2,5mod3)$



      I'm not sure how to prove that, I tried using the Bezout equality...



      For the other way we need to assume $Bbb Z_atimesBbb Z_b$ is cyclic i.e. $Bbb Z_atimesBbb Z_b =langle(x,y)rangle$ for some $(x,y)$



      Since $0inlangle(x,y)rangle$, $exists nin Bbb N~:~nxequiv0mod a~text{and } nyequiv 0mod b$



      And here I'm stuck again.










      share|cite|improve this question














      This question already has an answer here:




      • When are the product of cyclic groups also cyclic? [duplicate]

        2 answers




      Assuming $(a,b)=1$, I want to construct an isomorphism $phi:Bbb Z_atimesBbb Z_bmapstoBbb Z_{acdot b}$ defined by $phi(1,1)=1$. So I need to prove that for any $(x,y)in Bbb Z_atimesBbb Z_b$ there exists $nin Bbb N$ such that $nequiv x mod a$ and $nequiv ymod b$



      For example $(1,2)inBbb Z_2timesBbb Z_3$ is $(5mod2,5mod3)$



      I'm not sure how to prove that, I tried using the Bezout equality...



      For the other way we need to assume $Bbb Z_atimesBbb Z_b$ is cyclic i.e. $Bbb Z_atimesBbb Z_b =langle(x,y)rangle$ for some $(x,y)$



      Since $0inlangle(x,y)rangle$, $exists nin Bbb N~:~nxequiv0mod a~text{and } nyequiv 0mod b$



      And here I'm stuck again.





      This question already has an answer here:




      • When are the product of cyclic groups also cyclic? [duplicate]

        2 answers








      group-theory finite-groups






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      asked Jan 4 at 8:01









      John CataldoJohn Cataldo

      1,0961216




      1,0961216




      marked as duplicate by Theo Bendit, José Carlos Santos, Paul Frost, Pierre-Guy Plamondon, Cesareo Jan 4 at 12:43


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






      marked as duplicate by Theo Bendit, José Carlos Santos, Paul Frost, Pierre-Guy Plamondon, Cesareo Jan 4 at 12:43


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
























          2 Answers
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          The group $mathbb{Z}_a times mathbb{Z}_b$ is generated by $(1,0)$ and $(0,1)$, so the choices of $phi(1,0)$ and $phi(0,1)$ determine the homomorphism completely. You shall pick those values such that:
          $$
          begin{align*}
          &a cdot phi(1,0) equiv 0 quadbmod ab \
          & b cdot phi(0,1) equiv 0 quad bmod ab \
          end{align*}
          $$

          Let's pick:
          $$
          phi(x,y)=(bx+ay) bmod ab
          $$

          Bezout's identity ensures that the integers of the form $bx+ay$ are exactly the multiples of $(a,b)=1$.



          This implies that $phi$ is a bijection and thus an isomorphism.






          share|cite|improve this answer





























            1














            First note that the order of an element $(x,y) in mathbb{Z}_a times mathbb{Z}_b$ is $text{lcm}(r_1,r_2)$, where $r_1$ is the order of $x in mathbb{Z}_a$ and $r_2$ is the order of $y in mathbb{Z}_b$. Also note that $text{lcm}(a,b) = ab$ iff $a$ and $b$ are relatively prime.



            For the forward direction, we can just prove the contrapositive. So suppose $mathbb{Z}_a times mathbb{Z}_b$ is not cyclic. Then there exists some $r lt ab$ such that $r(1,1) = (0,0)$ otherwise $mathbb{Z}_a times mathbb{Z}_b$ would be cyclic. But this means that $text{lcm}(1,1) le r lt ab$ so that $a$ and $b$ are not relatively prime.



            For the other direction, suppose that $mathbb{Z}_a times mathbb{Z}_b$ is cyclic. Now $(1,1) in mathbb{Z}_a times mathbb{Z}_b$ has order $text{lcm}(r_1, r_2)$ where $r_1$ is the order of $1 in mathbb{Z}_a$ and $r_2$ is the order of $1 in mathbb{Z}_b$. Then $text{lcm}(r_1, r_2) = ab$ iff $a$ and $b$ are relatively prime. So if $mathbb{Z}_a times mathbb{Z}_b$ is cyclic, the order of $(1,1)$ is $ab$ since $(1,1)$ has to generate the group and hence $a$ and $b$ must be relatively prime by the previous statement.






            share|cite|improve this answer






























              2 Answers
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              active

              oldest

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2














              The group $mathbb{Z}_a times mathbb{Z}_b$ is generated by $(1,0)$ and $(0,1)$, so the choices of $phi(1,0)$ and $phi(0,1)$ determine the homomorphism completely. You shall pick those values such that:
              $$
              begin{align*}
              &a cdot phi(1,0) equiv 0 quadbmod ab \
              & b cdot phi(0,1) equiv 0 quad bmod ab \
              end{align*}
              $$

              Let's pick:
              $$
              phi(x,y)=(bx+ay) bmod ab
              $$

              Bezout's identity ensures that the integers of the form $bx+ay$ are exactly the multiples of $(a,b)=1$.



              This implies that $phi$ is a bijection and thus an isomorphism.






              share|cite|improve this answer


























                2














                The group $mathbb{Z}_a times mathbb{Z}_b$ is generated by $(1,0)$ and $(0,1)$, so the choices of $phi(1,0)$ and $phi(0,1)$ determine the homomorphism completely. You shall pick those values such that:
                $$
                begin{align*}
                &a cdot phi(1,0) equiv 0 quadbmod ab \
                & b cdot phi(0,1) equiv 0 quad bmod ab \
                end{align*}
                $$

                Let's pick:
                $$
                phi(x,y)=(bx+ay) bmod ab
                $$

                Bezout's identity ensures that the integers of the form $bx+ay$ are exactly the multiples of $(a,b)=1$.



                This implies that $phi$ is a bijection and thus an isomorphism.






                share|cite|improve this answer
























                  2












                  2








                  2






                  The group $mathbb{Z}_a times mathbb{Z}_b$ is generated by $(1,0)$ and $(0,1)$, so the choices of $phi(1,0)$ and $phi(0,1)$ determine the homomorphism completely. You shall pick those values such that:
                  $$
                  begin{align*}
                  &a cdot phi(1,0) equiv 0 quadbmod ab \
                  & b cdot phi(0,1) equiv 0 quad bmod ab \
                  end{align*}
                  $$

                  Let's pick:
                  $$
                  phi(x,y)=(bx+ay) bmod ab
                  $$

                  Bezout's identity ensures that the integers of the form $bx+ay$ are exactly the multiples of $(a,b)=1$.



                  This implies that $phi$ is a bijection and thus an isomorphism.






                  share|cite|improve this answer












                  The group $mathbb{Z}_a times mathbb{Z}_b$ is generated by $(1,0)$ and $(0,1)$, so the choices of $phi(1,0)$ and $phi(0,1)$ determine the homomorphism completely. You shall pick those values such that:
                  $$
                  begin{align*}
                  &a cdot phi(1,0) equiv 0 quadbmod ab \
                  & b cdot phi(0,1) equiv 0 quad bmod ab \
                  end{align*}
                  $$

                  Let's pick:
                  $$
                  phi(x,y)=(bx+ay) bmod ab
                  $$

                  Bezout's identity ensures that the integers of the form $bx+ay$ are exactly the multiples of $(a,b)=1$.



                  This implies that $phi$ is a bijection and thus an isomorphism.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 4 at 10:20









                  JevautJevaut

                  958112




                  958112























                      1














                      First note that the order of an element $(x,y) in mathbb{Z}_a times mathbb{Z}_b$ is $text{lcm}(r_1,r_2)$, where $r_1$ is the order of $x in mathbb{Z}_a$ and $r_2$ is the order of $y in mathbb{Z}_b$. Also note that $text{lcm}(a,b) = ab$ iff $a$ and $b$ are relatively prime.



                      For the forward direction, we can just prove the contrapositive. So suppose $mathbb{Z}_a times mathbb{Z}_b$ is not cyclic. Then there exists some $r lt ab$ such that $r(1,1) = (0,0)$ otherwise $mathbb{Z}_a times mathbb{Z}_b$ would be cyclic. But this means that $text{lcm}(1,1) le r lt ab$ so that $a$ and $b$ are not relatively prime.



                      For the other direction, suppose that $mathbb{Z}_a times mathbb{Z}_b$ is cyclic. Now $(1,1) in mathbb{Z}_a times mathbb{Z}_b$ has order $text{lcm}(r_1, r_2)$ where $r_1$ is the order of $1 in mathbb{Z}_a$ and $r_2$ is the order of $1 in mathbb{Z}_b$. Then $text{lcm}(r_1, r_2) = ab$ iff $a$ and $b$ are relatively prime. So if $mathbb{Z}_a times mathbb{Z}_b$ is cyclic, the order of $(1,1)$ is $ab$ since $(1,1)$ has to generate the group and hence $a$ and $b$ must be relatively prime by the previous statement.






                      share|cite|improve this answer




























                        1














                        First note that the order of an element $(x,y) in mathbb{Z}_a times mathbb{Z}_b$ is $text{lcm}(r_1,r_2)$, where $r_1$ is the order of $x in mathbb{Z}_a$ and $r_2$ is the order of $y in mathbb{Z}_b$. Also note that $text{lcm}(a,b) = ab$ iff $a$ and $b$ are relatively prime.



                        For the forward direction, we can just prove the contrapositive. So suppose $mathbb{Z}_a times mathbb{Z}_b$ is not cyclic. Then there exists some $r lt ab$ such that $r(1,1) = (0,0)$ otherwise $mathbb{Z}_a times mathbb{Z}_b$ would be cyclic. But this means that $text{lcm}(1,1) le r lt ab$ so that $a$ and $b$ are not relatively prime.



                        For the other direction, suppose that $mathbb{Z}_a times mathbb{Z}_b$ is cyclic. Now $(1,1) in mathbb{Z}_a times mathbb{Z}_b$ has order $text{lcm}(r_1, r_2)$ where $r_1$ is the order of $1 in mathbb{Z}_a$ and $r_2$ is the order of $1 in mathbb{Z}_b$. Then $text{lcm}(r_1, r_2) = ab$ iff $a$ and $b$ are relatively prime. So if $mathbb{Z}_a times mathbb{Z}_b$ is cyclic, the order of $(1,1)$ is $ab$ since $(1,1)$ has to generate the group and hence $a$ and $b$ must be relatively prime by the previous statement.






                        share|cite|improve this answer


























                          1












                          1








                          1






                          First note that the order of an element $(x,y) in mathbb{Z}_a times mathbb{Z}_b$ is $text{lcm}(r_1,r_2)$, where $r_1$ is the order of $x in mathbb{Z}_a$ and $r_2$ is the order of $y in mathbb{Z}_b$. Also note that $text{lcm}(a,b) = ab$ iff $a$ and $b$ are relatively prime.



                          For the forward direction, we can just prove the contrapositive. So suppose $mathbb{Z}_a times mathbb{Z}_b$ is not cyclic. Then there exists some $r lt ab$ such that $r(1,1) = (0,0)$ otherwise $mathbb{Z}_a times mathbb{Z}_b$ would be cyclic. But this means that $text{lcm}(1,1) le r lt ab$ so that $a$ and $b$ are not relatively prime.



                          For the other direction, suppose that $mathbb{Z}_a times mathbb{Z}_b$ is cyclic. Now $(1,1) in mathbb{Z}_a times mathbb{Z}_b$ has order $text{lcm}(r_1, r_2)$ where $r_1$ is the order of $1 in mathbb{Z}_a$ and $r_2$ is the order of $1 in mathbb{Z}_b$. Then $text{lcm}(r_1, r_2) = ab$ iff $a$ and $b$ are relatively prime. So if $mathbb{Z}_a times mathbb{Z}_b$ is cyclic, the order of $(1,1)$ is $ab$ since $(1,1)$ has to generate the group and hence $a$ and $b$ must be relatively prime by the previous statement.






                          share|cite|improve this answer














                          First note that the order of an element $(x,y) in mathbb{Z}_a times mathbb{Z}_b$ is $text{lcm}(r_1,r_2)$, where $r_1$ is the order of $x in mathbb{Z}_a$ and $r_2$ is the order of $y in mathbb{Z}_b$. Also note that $text{lcm}(a,b) = ab$ iff $a$ and $b$ are relatively prime.



                          For the forward direction, we can just prove the contrapositive. So suppose $mathbb{Z}_a times mathbb{Z}_b$ is not cyclic. Then there exists some $r lt ab$ such that $r(1,1) = (0,0)$ otherwise $mathbb{Z}_a times mathbb{Z}_b$ would be cyclic. But this means that $text{lcm}(1,1) le r lt ab$ so that $a$ and $b$ are not relatively prime.



                          For the other direction, suppose that $mathbb{Z}_a times mathbb{Z}_b$ is cyclic. Now $(1,1) in mathbb{Z}_a times mathbb{Z}_b$ has order $text{lcm}(r_1, r_2)$ where $r_1$ is the order of $1 in mathbb{Z}_a$ and $r_2$ is the order of $1 in mathbb{Z}_b$. Then $text{lcm}(r_1, r_2) = ab$ iff $a$ and $b$ are relatively prime. So if $mathbb{Z}_a times mathbb{Z}_b$ is cyclic, the order of $(1,1)$ is $ab$ since $(1,1)$ has to generate the group and hence $a$ and $b$ must be relatively prime by the previous statement.







                          share|cite|improve this answer














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                          edited Jan 4 at 8:26

























                          answered Jan 4 at 8:20









                          MAXMAX

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