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












4












$begingroup$



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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  • $begingroup$
    Your proof is correct (modulo a few details I've corrected -- e.g., $gcdleft(n,n-1right)$ has nothing to do with $n>2$).
    $endgroup$
    – darij grinberg
    54 mins ago
















4












$begingroup$



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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  • $begingroup$
    Your proof is correct (modulo a few details I've corrected -- e.g., $gcdleft(n,n-1right)$ has nothing to do with $n>2$).
    $endgroup$
    – darij grinberg
    54 mins ago














4












4








4





$begingroup$



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.
Check out our Code of Conduct.







$endgroup$





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 54 mins ago









darij grinberg

11.3k33164




11.3k33164






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asked 1 hour ago









PabloPablo

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211




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  • $begingroup$
    Your proof is correct (modulo a few details I've corrected -- e.g., $gcdleft(n,n-1right)$ has nothing to do with $n>2$).
    $endgroup$
    – darij grinberg
    54 mins ago


















  • $begingroup$
    Your proof is correct (modulo a few details I've corrected -- e.g., $gcdleft(n,n-1right)$ has nothing to do with $n>2$).
    $endgroup$
    – darij grinberg
    54 mins ago
















$begingroup$
Your proof is correct (modulo a few details I've corrected -- e.g., $gcdleft(n,n-1right)$ has nothing to do with $n>2$).
$endgroup$
– darij grinberg
54 mins ago




$begingroup$
Your proof is correct (modulo a few details I've corrected -- e.g., $gcdleft(n,n-1right)$ has nothing to do with $n>2$).
$endgroup$
– darij grinberg
54 mins ago










3 Answers
3






active

oldest

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5












$begingroup$

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









$endgroup$





















    0












    $begingroup$

    Here is another take.



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



    Now, $n$ 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









    $endgroup$





















      0












      $begingroup$

      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









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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        5












        $begingroup$

        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









        $endgroup$


















          5












          $begingroup$

          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









          $endgroup$
















            5












            5








            5





            $begingroup$

            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 1 hour ago









            MPWMPW

            30.1k12056




            30.1k12056























                0












                $begingroup$

                Here is another take.



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



                Now, $n$ 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









                $endgroup$


















                  0












                  $begingroup$

                  Here is another take.



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



                  Now, $n$ 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









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Here is another take.



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



                    Now, $n$ 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, $n$ 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










                    answered 28 mins ago









                    lhflhf

                    165k10171396




                    165k10171396























                        0












                        $begingroup$

                        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












                          $begingroup$

                          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












                            0








                            0





                            $begingroup$

                            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 23 mins ago









                            LBJFSLBJFS

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