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












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$begingroup$



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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    13












    $begingroup$



    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











    $endgroup$















      13












      13








      13


      1



      $begingroup$



      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











      $endgroup$





      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 20 at 0:33









      darij grinberg

      11.2k33167




      11.2k33167










      asked Feb 20 at 0:05









      PabloPablo

      683




      683






















          3 Answers
          3






          active

          oldest

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          17












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            Beautiful! Thank you so much for this!
            $endgroup$
            – Pablo
            Feb 20 at 7:49






          • 2




            $begingroup$
            When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
            $endgroup$
            – Marc van Leeuwen
            Feb 20 at 13:03





















          9












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            That's a very good observation, thanks!
            $endgroup$
            – Pablo
            Feb 20 at 7:49



















          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: if $d$ divides $n$, then $phi(n) ge phi(d)$.)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @darij, thanks.
            $endgroup$
            – lhf
            Feb 20 at 10:48











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          3 Answers
          3






          active

          oldest

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          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          17












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            Beautiful! Thank you so much for this!
            $endgroup$
            – Pablo
            Feb 20 at 7:49






          • 2




            $begingroup$
            When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
            $endgroup$
            – Marc van Leeuwen
            Feb 20 at 13:03


















          17












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            Beautiful! Thank you so much for this!
            $endgroup$
            – Pablo
            Feb 20 at 7:49






          • 2




            $begingroup$
            When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
            $endgroup$
            – Marc van Leeuwen
            Feb 20 at 13:03
















          17












          17








          17





          $begingroup$

          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 Feb 20 at 0:10









          MPWMPW

          30.9k12157




          30.9k12157












          • $begingroup$
            Beautiful! Thank you so much for this!
            $endgroup$
            – Pablo
            Feb 20 at 7:49






          • 2




            $begingroup$
            When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
            $endgroup$
            – Marc van Leeuwen
            Feb 20 at 13:03




















          • $begingroup$
            Beautiful! Thank you so much for this!
            $endgroup$
            – Pablo
            Feb 20 at 7:49






          • 2




            $begingroup$
            When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
            $endgroup$
            – Marc van Leeuwen
            Feb 20 at 13:03


















          $begingroup$
          Beautiful! Thank you so much for this!
          $endgroup$
          – Pablo
          Feb 20 at 7:49




          $begingroup$
          Beautiful! Thank you so much for this!
          $endgroup$
          – Pablo
          Feb 20 at 7:49




          2




          2




          $begingroup$
          When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
          $endgroup$
          – Marc van Leeuwen
          Feb 20 at 13:03






          $begingroup$
          When $g$ is a generator of a cyclic group of order $n$, then $g^{-1}$ is just another way to write $g^{n-1}$, so this is not really all that different from the proof proposed in the question. Though indeed by avoiding to mention the order of the cyclic group, this answer does handle the infinite cyclic group where the one in the question overlooks that possibiliy.
          $endgroup$
          – Marc van Leeuwen
          Feb 20 at 13:03













          9












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            That's a very good observation, thanks!
            $endgroup$
            – Pablo
            Feb 20 at 7:49
















          9












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            That's a very good observation, thanks!
            $endgroup$
            – Pablo
            Feb 20 at 7:49














          9












          9








          9





          $begingroup$

          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 Feb 20 at 1:04









          LBJFSLBJFS

          27110




          27110












          • $begingroup$
            That's a very good observation, thanks!
            $endgroup$
            – Pablo
            Feb 20 at 7:49


















          • $begingroup$
            That's a very good observation, thanks!
            $endgroup$
            – Pablo
            Feb 20 at 7:49
















          $begingroup$
          That's a very good observation, thanks!
          $endgroup$
          – Pablo
          Feb 20 at 7:49




          $begingroup$
          That's a very good observation, thanks!
          $endgroup$
          – Pablo
          Feb 20 at 7:49











          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: if $d$ divides $n$, then $phi(n) ge phi(d)$.)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @darij, thanks.
            $endgroup$
            – lhf
            Feb 20 at 10:48
















          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: if $d$ divides $n$, then $phi(n) ge phi(d)$.)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @darij, thanks.
            $endgroup$
            – lhf
            Feb 20 at 10:48














          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: if $d$ divides $n$, then $phi(n) ge phi(d)$.)






          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: if $d$ divides $n$, then $phi(n) ge phi(d)$.)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 20 at 10:47

























          answered Feb 20 at 1:00









          lhflhf

          166k11172402




          166k11172402












          • $begingroup$
            @darij, thanks.
            $endgroup$
            – lhf
            Feb 20 at 10:48


















          • $begingroup$
            @darij, thanks.
            $endgroup$
            – lhf
            Feb 20 at 10:48
















          $begingroup$
          @darij, thanks.
          $endgroup$
          – lhf
          Feb 20 at 10:48




          $begingroup$
          @darij, thanks.
          $endgroup$
          – lhf
          Feb 20 at 10:48


















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