Prove that a cyclic group with only one generator can have at most 2 elements












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Prove that a cyclic group that has only one generator has at most $2$ elements.




I want to know if my proof would be valid:



Suppose $G$ is a cyclic group and $g$ is its only generator. Let $|G|=n$ where $n>2$, then we know that $gcd(n,n-1)=1$. This implies that $g^{n-1}$ is a generator of $G$. We have a contradiction, since $g$ is the only generator of $G$ (and $n > 2$ leads to $n-1 neq 1$). Thus $|G|leq 2$.



I tried to use the fact that generating elements of a group are coprime to the order of the group, thanks.






abstract-algebra greatest-common-divisor cyclic-groups







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    8












    $begingroup$



    Prove that a cyclic group that has only one generator has at most $2$ elements.




    I want to know if my proof would be valid:



    Suppose $G$ is a cyclic group and $g$ is its only generator. Let $|G|=n$ where $n>2$, then we know that $gcd(n,n-1)=1$. This implies that $g^{n-1}$ is a generator of $G$. We have a contradiction, since $g$ is the only generator of $G$ (and $n > 2$ leads to $n-1 neq 1$). Thus $|G|leq 2$.



    I tried to use the fact that generating elements of a group are coprime to the order of the group, thanks.






    abstract-algebra greatest-common-divisor cyclic-groups







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      8












      8








      8





      $begingroup$



      Prove that a cyclic group that has only one generator has at most $2$ elements.




      I want to know if my proof would be valid:



      Suppose $G$ is a cyclic group and $g$ is its only generator. Let $|G|=n$ where $n>2$, then we know that $gcd(n,n-1)=1$. This implies that $g^{n-1}$ is a generator of $G$. We have a contradiction, since $g$ is the only generator of $G$ (and $n > 2$ leads to $n-1 neq 1$). Thus $|G|leq 2$.



      I tried to use the fact that generating elements of a group are coprime to the order of the group, thanks.






      abstract-algebra greatest-common-divisor cyclic-groups







      share|cite|improve this question









      New contributor




      Pablo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      Prove that a cyclic group that has only one generator has at most $2$ elements.




      I want to know if my proof would be valid:



      Suppose $G$ is a cyclic group and $g$ is its only generator. Let $|G|=n$ where $n>2$, then we know that $gcd(n,n-1)=1$. This implies that $g^{n-1}$ is a generator of $G$. We have a contradiction, since $g$ is the only generator of $G$ (and $n > 2$ leads to $n-1 neq 1$). Thus $|G|leq 2$.



      I tried to use the fact that generating elements of a group are coprime to the order of the group, thanks.






      abstract-algebra greatest-common-divisor cyclic-groups




      abstract-algebra greatest-common-divisor cyclic-groups






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      edited 4 hours ago









      darij grinberg

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      11.3k33164






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      asked 4 hours ago









      PabloPablo

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          3 Answers
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          11












          $begingroup$

          Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).






          share|cite|improve this answer









          $endgroup$





















            6












            $begingroup$

            Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              Here is another take.



              The number of generators is $phi(n)$, where $phi$ is Euler's function.



              Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.



              In the first case, we have $phi(n) ge phi(p)=p-1ge2$.



              In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.



              Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.



              (The key fact is this: if $d$ divides $n$, then $phi(n) ge phi(d)$, because if $x$ is coprime with $n$, then $x$ is coprime with $d$.)






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                The argument in the very last sentence is wrong (though the claim is true).
                $endgroup$
                – darij grinberg
                2 hours ago













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              3 Answers
              3






              active

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              3 Answers
              3






              active

              oldest

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              active

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              active

              oldest

              votes









              11












              $begingroup$

              Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).






              share|cite|improve this answer









              $endgroup$


















                11












                $begingroup$

                Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).






                share|cite|improve this answer









                $endgroup$
















                  11












                  11








                  11





                  $begingroup$

                  Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).






                  share|cite|improve this answer









                  $endgroup$



                  Perhaps easier: if $g$ generates $G$, then so does $g^{-1}$. The hypothesis then implies that $g=g^{-1}$, so $g^2=1$. Done (either $g=1$ or not, in which case respectively the order of $G$ is $1$ or $2$).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 4 hours ago









                  MPWMPW

                  30.1k12057




                  30.1k12057























                      6












                      $begingroup$

                      Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.






                      share|cite|improve this answer









                      $endgroup$


















                        6












                        $begingroup$

                        Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.






                        share|cite|improve this answer









                        $endgroup$
















                          6












                          6








                          6





                          $begingroup$

                          Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.






                          share|cite|improve this answer









                          $endgroup$



                          Your proof is correct if $G$ is finite, i.e. $Gcongmathbb{Z}_m$ for some $mge 1$. Just notice that it may happen that $Gcongmathbb{Z}$; however, in this case $g$ and $g^{-1}$ are distinct generators and this concludes the proof.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 3 hours ago









                          LBJFSLBJFS

                          1357




                          1357























                              0












                              $begingroup$

                              Here is another take.



                              The number of generators is $phi(n)$, where $phi$ is Euler's function.



                              Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.



                              In the first case, we have $phi(n) ge phi(p)=p-1ge2$.



                              In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.



                              Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.



                              (The key fact is this: if $d$ divides $n$, then $phi(n) ge phi(d)$, because if $x$ is coprime with $n$, then $x$ is coprime with $d$.)






                              share|cite|improve this answer











                              $endgroup$













                              • $begingroup$
                                The argument in the very last sentence is wrong (though the claim is true).
                                $endgroup$
                                – darij grinberg
                                2 hours ago


















                              0












                              $begingroup$

                              Here is another take.



                              The number of generators is $phi(n)$, where $phi$ is Euler's function.



                              Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.



                              In the first case, we have $phi(n) ge phi(p)=p-1ge2$.



                              In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.



                              Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.



                              (The key fact is this: if $d$ divides $n$, then $phi(n) ge phi(d)$, because if $x$ is coprime with $n$, then $x$ is coprime with $d$.)






                              share|cite|improve this answer











                              $endgroup$













                              • $begingroup$
                                The argument in the very last sentence is wrong (though the claim is true).
                                $endgroup$
                                – darij grinberg
                                2 hours ago
















                              0












                              0








                              0





                              $begingroup$

                              Here is another take.



                              The number of generators is $phi(n)$, where $phi$ is Euler's function.



                              Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.



                              In the first case, we have $phi(n) ge phi(p)=p-1ge2$.



                              In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.



                              Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.



                              (The key fact is this: if $d$ divides $n$, then $phi(n) ge phi(d)$, because if $x$ is coprime with $n$, then $x$ is coprime with $d$.)






                              share|cite|improve this answer











                              $endgroup$



                              Here is another take.



                              The number of generators is $phi(n)$, where $phi$ is Euler's function.



                              Now, either $n$ has a prime factor $pge 3$ or $n$ is a power of $2$.



                              In the first case, we have $phi(n) ge phi(p)=p-1ge2$.



                              In the second case, if $nge 3$, then $4$ divides $n$ and so $phi(n) ge phi(4)=2$.



                              Bottom line, $phi(n)=1$ implies $n$ is a power of $2$ less than $4$, that is, $n=1$ or $n=2$.



                              (The key fact is this: if $d$ divides $n$, then $phi(n) ge phi(d)$, because if $x$ is coprime with $n$, then $x$ is coprime with $d$.)







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited 2 hours ago

























                              answered 3 hours ago









                              lhflhf

                              165k10171396




                              165k10171396












                              • $begingroup$
                                The argument in the very last sentence is wrong (though the claim is true).
                                $endgroup$
                                – darij grinberg
                                2 hours ago




















                              • $begingroup$
                                The argument in the very last sentence is wrong (though the claim is true).
                                $endgroup$
                                – darij grinberg
                                2 hours ago


















                              $begingroup$
                              The argument in the very last sentence is wrong (though the claim is true).
                              $endgroup$
                              – darij grinberg
                              2 hours ago






                              $begingroup$
                              The argument in the very last sentence is wrong (though the claim is true).
                              $endgroup$
                              – darij grinberg
                              2 hours ago












                              Pablo is a new contributor. Be nice, and check out our Code of Conduct.










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