Amount of generators for cyclic group and Euler's function












3












$begingroup$


I have this question, since I'm confused about this concept. Let's say I have a cyclic group, for example $mathbb{Z}_7^{times} = left{1, 2, 3, 4, 5, 6right}$. What is the relationship between the amount of generators for this group and $phi(7)$, with $phi$ Euler's function?



I found the generators as $left{3, 4, 5, 6 right}$. I know that if $G$ is a cyclic group of order $n$, and $g$ is a generator, then $g^s$ will also be a generator if and only if $gcd(s,n) = 1$. But then I read somewhere that if $n$ is prime (which in our case it is), then $phi(7) = 6$ and the amount of generators should be $6$? This cannot be true for the group $mathbb{Z}_7^{times}$, since $1$ cannot possibly be a generator for multiplication.



For example, if I have the cyclic group $mathbb{Z}_{13}^{times}$, what is the best method to find a generator, and to find how many there are?






abstract-algebra elementary-number-theory







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    I have this question, since I'm confused about this concept. Let's say I have a cyclic group, for example $mathbb{Z}_7^{times} = left{1, 2, 3, 4, 5, 6right}$. What is the relationship between the amount of generators for this group and $phi(7)$, with $phi$ Euler's function?



    I found the generators as $left{3, 4, 5, 6 right}$. I know that if $G$ is a cyclic group of order $n$, and $g$ is a generator, then $g^s$ will also be a generator if and only if $gcd(s,n) = 1$. But then I read somewhere that if $n$ is prime (which in our case it is), then $phi(7) = 6$ and the amount of generators should be $6$? This cannot be true for the group $mathbb{Z}_7^{times}$, since $1$ cannot possibly be a generator for multiplication.



    For example, if I have the cyclic group $mathbb{Z}_{13}^{times}$, what is the best method to find a generator, and to find how many there are?






    abstract-algebra elementary-number-theory







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      I have this question, since I'm confused about this concept. Let's say I have a cyclic group, for example $mathbb{Z}_7^{times} = left{1, 2, 3, 4, 5, 6right}$. What is the relationship between the amount of generators for this group and $phi(7)$, with $phi$ Euler's function?



      I found the generators as $left{3, 4, 5, 6 right}$. I know that if $G$ is a cyclic group of order $n$, and $g$ is a generator, then $g^s$ will also be a generator if and only if $gcd(s,n) = 1$. But then I read somewhere that if $n$ is prime (which in our case it is), then $phi(7) = 6$ and the amount of generators should be $6$? This cannot be true for the group $mathbb{Z}_7^{times}$, since $1$ cannot possibly be a generator for multiplication.



      For example, if I have the cyclic group $mathbb{Z}_{13}^{times}$, what is the best method to find a generator, and to find how many there are?






      abstract-algebra elementary-number-theory







      share|cite|improve this question









      $endgroup$




      I have this question, since I'm confused about this concept. Let's say I have a cyclic group, for example $mathbb{Z}_7^{times} = left{1, 2, 3, 4, 5, 6right}$. What is the relationship between the amount of generators for this group and $phi(7)$, with $phi$ Euler's function?



      I found the generators as $left{3, 4, 5, 6 right}$. I know that if $G$ is a cyclic group of order $n$, and $g$ is a generator, then $g^s$ will also be a generator if and only if $gcd(s,n) = 1$. But then I read somewhere that if $n$ is prime (which in our case it is), then $phi(7) = 6$ and the amount of generators should be $6$? This cannot be true for the group $mathbb{Z}_7^{times}$, since $1$ cannot possibly be a generator for multiplication.



      For example, if I have the cyclic group $mathbb{Z}_{13}^{times}$, what is the best method to find a generator, and to find how many there are?






      abstract-algebra elementary-number-theory




      abstract-algebra elementary-number-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 10 '16 at 14:53









      KamilKamil

      2,00221545




      2,00221545






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          A cyclic group with $n$ elements has $varphi(n)$ generators.



          It seems like what you're getting confused by is that:




          • The group $mathbb{Z}_7^times={1,2,3,4,5,6}$ is a cyclic group with 6 elements, and therefore it should considered equivalent to (more precisely, it is isomorphic to) the cyclic group $mathbb{Z}_6$, and you should expect $varphi(6)=2$ generators, not $varphi(7)=6$ generators


          • The elements $4$ and $6$ in $mathbb{Z}_7^times$ are not generators: the powers of $4$ in $mathbb{Z}_7^times$ are ${1,4,2}$, and the powers of $6$ in $mathbb{Z}_7^times$ are ${1,6}$. (The only generators of $mathbb{Z}_7^times$ are $3$ and $5$, and there are exactly $2=varphi(6)$ of them.)



          More generally, if $p$ is a prime number, then $mathbb{Z}_p$ is a cyclic group with $p$ elements, while $mathbb{Z}_p^times$ is a cyclic group with $p-1$ elements, so $mathbb{Z}_p^times$ has $varphi(p-1)$ generators.



          As for the best method to find a generator, that is a far more advanced problem.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ah yes, now I understand! That was exactly my confusion. Thanks for pointing out this important piece of information!
            $endgroup$
            – Kamil
            Aug 10 '16 at 15:07



















          0












          $begingroup$

          As for




          best method to find a generator, that is a far more advanced problem




          There is fine way to find generators:



          For example lets take Z×22.
          
1. Z×22 = {1,3,5,7,9,13,15,17,19,21}
          
2. φ(22) = 10 = 5*2
          
3. Find the first generator(7) and remember the order it generates the group in: 
          
{1:7, 2:5, 3:13, 5:13, 5:21, 6:15, 7:17, 8:19, 9:19, 10:1} 
          
4. Take all elements which indexes have gcd(5, index) = gcd(2, index) = 1. 
          
5. They are: {7,13,17,19} - that will be your generators.





          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But "find the first generator" is the whole problem.
            $endgroup$
            – Tobias Kildetoft
            Jan 14 at 20:08










          • $begingroup$
            Yes, noticed that during applying this method. But it can definitly reduce the total time of searching each generator.
            $endgroup$
            – AseN
            Jan 14 at 20:13












          • $begingroup$
            What can? You mean this method of finding all generators? Because that seems like it is precisely the standard way where you take all powers of a fixed generator which are coprime to the order.
            $endgroup$
            – Tobias Kildetoft
            Jan 14 at 20:16











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          2 Answers
          2






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          active

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          active

          oldest

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          4












          $begingroup$

          A cyclic group with $n$ elements has $varphi(n)$ generators.



          It seems like what you're getting confused by is that:




          • The group $mathbb{Z}_7^times={1,2,3,4,5,6}$ is a cyclic group with 6 elements, and therefore it should considered equivalent to (more precisely, it is isomorphic to) the cyclic group $mathbb{Z}_6$, and you should expect $varphi(6)=2$ generators, not $varphi(7)=6$ generators


          • The elements $4$ and $6$ in $mathbb{Z}_7^times$ are not generators: the powers of $4$ in $mathbb{Z}_7^times$ are ${1,4,2}$, and the powers of $6$ in $mathbb{Z}_7^times$ are ${1,6}$. (The only generators of $mathbb{Z}_7^times$ are $3$ and $5$, and there are exactly $2=varphi(6)$ of them.)



          More generally, if $p$ is a prime number, then $mathbb{Z}_p$ is a cyclic group with $p$ elements, while $mathbb{Z}_p^times$ is a cyclic group with $p-1$ elements, so $mathbb{Z}_p^times$ has $varphi(p-1)$ generators.



          As for the best method to find a generator, that is a far more advanced problem.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ah yes, now I understand! That was exactly my confusion. Thanks for pointing out this important piece of information!
            $endgroup$
            – Kamil
            Aug 10 '16 at 15:07
















          4












          $begingroup$

          A cyclic group with $n$ elements has $varphi(n)$ generators.



          It seems like what you're getting confused by is that:




          • The group $mathbb{Z}_7^times={1,2,3,4,5,6}$ is a cyclic group with 6 elements, and therefore it should considered equivalent to (more precisely, it is isomorphic to) the cyclic group $mathbb{Z}_6$, and you should expect $varphi(6)=2$ generators, not $varphi(7)=6$ generators


          • The elements $4$ and $6$ in $mathbb{Z}_7^times$ are not generators: the powers of $4$ in $mathbb{Z}_7^times$ are ${1,4,2}$, and the powers of $6$ in $mathbb{Z}_7^times$ are ${1,6}$. (The only generators of $mathbb{Z}_7^times$ are $3$ and $5$, and there are exactly $2=varphi(6)$ of them.)



          More generally, if $p$ is a prime number, then $mathbb{Z}_p$ is a cyclic group with $p$ elements, while $mathbb{Z}_p^times$ is a cyclic group with $p-1$ elements, so $mathbb{Z}_p^times$ has $varphi(p-1)$ generators.



          As for the best method to find a generator, that is a far more advanced problem.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ah yes, now I understand! That was exactly my confusion. Thanks for pointing out this important piece of information!
            $endgroup$
            – Kamil
            Aug 10 '16 at 15:07














          4












          4








          4





          $begingroup$

          A cyclic group with $n$ elements has $varphi(n)$ generators.



          It seems like what you're getting confused by is that:




          • The group $mathbb{Z}_7^times={1,2,3,4,5,6}$ is a cyclic group with 6 elements, and therefore it should considered equivalent to (more precisely, it is isomorphic to) the cyclic group $mathbb{Z}_6$, and you should expect $varphi(6)=2$ generators, not $varphi(7)=6$ generators


          • The elements $4$ and $6$ in $mathbb{Z}_7^times$ are not generators: the powers of $4$ in $mathbb{Z}_7^times$ are ${1,4,2}$, and the powers of $6$ in $mathbb{Z}_7^times$ are ${1,6}$. (The only generators of $mathbb{Z}_7^times$ are $3$ and $5$, and there are exactly $2=varphi(6)$ of them.)



          More generally, if $p$ is a prime number, then $mathbb{Z}_p$ is a cyclic group with $p$ elements, while $mathbb{Z}_p^times$ is a cyclic group with $p-1$ elements, so $mathbb{Z}_p^times$ has $varphi(p-1)$ generators.



          As for the best method to find a generator, that is a far more advanced problem.






          share|cite|improve this answer











          $endgroup$



          A cyclic group with $n$ elements has $varphi(n)$ generators.



          It seems like what you're getting confused by is that:




          • The group $mathbb{Z}_7^times={1,2,3,4,5,6}$ is a cyclic group with 6 elements, and therefore it should considered equivalent to (more precisely, it is isomorphic to) the cyclic group $mathbb{Z}_6$, and you should expect $varphi(6)=2$ generators, not $varphi(7)=6$ generators


          • The elements $4$ and $6$ in $mathbb{Z}_7^times$ are not generators: the powers of $4$ in $mathbb{Z}_7^times$ are ${1,4,2}$, and the powers of $6$ in $mathbb{Z}_7^times$ are ${1,6}$. (The only generators of $mathbb{Z}_7^times$ are $3$ and $5$, and there are exactly $2=varphi(6)$ of them.)



          More generally, if $p$ is a prime number, then $mathbb{Z}_p$ is a cyclic group with $p$ elements, while $mathbb{Z}_p^times$ is a cyclic group with $p-1$ elements, so $mathbb{Z}_p^times$ has $varphi(p-1)$ generators.



          As for the best method to find a generator, that is a far more advanced problem.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 10 '16 at 15:08

























          answered Aug 10 '16 at 15:01









          Zev ChonolesZev Chonoles

          110k16228426




          110k16228426












          • $begingroup$
            Ah yes, now I understand! That was exactly my confusion. Thanks for pointing out this important piece of information!
            $endgroup$
            – Kamil
            Aug 10 '16 at 15:07


















          • $begingroup$
            Ah yes, now I understand! That was exactly my confusion. Thanks for pointing out this important piece of information!
            $endgroup$
            – Kamil
            Aug 10 '16 at 15:07
















          $begingroup$
          Ah yes, now I understand! That was exactly my confusion. Thanks for pointing out this important piece of information!
          $endgroup$
          – Kamil
          Aug 10 '16 at 15:07




          $begingroup$
          Ah yes, now I understand! That was exactly my confusion. Thanks for pointing out this important piece of information!
          $endgroup$
          – Kamil
          Aug 10 '16 at 15:07











          0












          $begingroup$

          As for




          best method to find a generator, that is a far more advanced problem




          There is fine way to find generators:



          For example lets take Z×22.
          
1. Z×22 = {1,3,5,7,9,13,15,17,19,21}
          
2. φ(22) = 10 = 5*2
          
3. Find the first generator(7) and remember the order it generates the group in: 
          
{1:7, 2:5, 3:13, 5:13, 5:21, 6:15, 7:17, 8:19, 9:19, 10:1} 
          
4. Take all elements which indexes have gcd(5, index) = gcd(2, index) = 1. 
          
5. They are: {7,13,17,19} - that will be your generators.





          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But "find the first generator" is the whole problem.
            $endgroup$
            – Tobias Kildetoft
            Jan 14 at 20:08










          • $begingroup$
            Yes, noticed that during applying this method. But it can definitly reduce the total time of searching each generator.
            $endgroup$
            – AseN
            Jan 14 at 20:13












          • $begingroup$
            What can? You mean this method of finding all generators? Because that seems like it is precisely the standard way where you take all powers of a fixed generator which are coprime to the order.
            $endgroup$
            – Tobias Kildetoft
            Jan 14 at 20:16
















          0












          $begingroup$

          As for




          best method to find a generator, that is a far more advanced problem




          There is fine way to find generators:



          For example lets take Z×22.
          
1. Z×22 = {1,3,5,7,9,13,15,17,19,21}
          
2. φ(22) = 10 = 5*2
          
3. Find the first generator(7) and remember the order it generates the group in: 
          
{1:7, 2:5, 3:13, 5:13, 5:21, 6:15, 7:17, 8:19, 9:19, 10:1} 
          
4. Take all elements which indexes have gcd(5, index) = gcd(2, index) = 1. 
          
5. They are: {7,13,17,19} - that will be your generators.





          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But "find the first generator" is the whole problem.
            $endgroup$
            – Tobias Kildetoft
            Jan 14 at 20:08










          • $begingroup$
            Yes, noticed that during applying this method. But it can definitly reduce the total time of searching each generator.
            $endgroup$
            – AseN
            Jan 14 at 20:13












          • $begingroup$
            What can? You mean this method of finding all generators? Because that seems like it is precisely the standard way where you take all powers of a fixed generator which are coprime to the order.
            $endgroup$
            – Tobias Kildetoft
            Jan 14 at 20:16














          0












          0








          0





          $begingroup$

          As for




          best method to find a generator, that is a far more advanced problem




          There is fine way to find generators:



          For example lets take Z×22.
          
1. Z×22 = {1,3,5,7,9,13,15,17,19,21}
          
2. φ(22) = 10 = 5*2
          
3. Find the first generator(7) and remember the order it generates the group in: 
          
{1:7, 2:5, 3:13, 5:13, 5:21, 6:15, 7:17, 8:19, 9:19, 10:1} 
          
4. Take all elements which indexes have gcd(5, index) = gcd(2, index) = 1. 
          
5. They are: {7,13,17,19} - that will be your generators.





          share|cite|improve this answer











          $endgroup$



          As for




          best method to find a generator, that is a far more advanced problem




          There is fine way to find generators:



          For example lets take Z×22.
          
1. Z×22 = {1,3,5,7,9,13,15,17,19,21}
          
2. φ(22) = 10 = 5*2
          
3. Find the first generator(7) and remember the order it generates the group in: 
          
{1:7, 2:5, 3:13, 5:13, 5:21, 6:15, 7:17, 8:19, 9:19, 10:1} 
          
4. Take all elements which indexes have gcd(5, index) = gcd(2, index) = 1. 
          
5. They are: {7,13,17,19} - that will be your generators.






          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 14 at 20:07

























          answered Jan 14 at 20:01









          AseNAseN

          1036




          1036












          • $begingroup$
            But "find the first generator" is the whole problem.
            $endgroup$
            – Tobias Kildetoft
            Jan 14 at 20:08










          • $begingroup$
            Yes, noticed that during applying this method. But it can definitly reduce the total time of searching each generator.
            $endgroup$
            – AseN
            Jan 14 at 20:13












          • $begingroup$
            What can? You mean this method of finding all generators? Because that seems like it is precisely the standard way where you take all powers of a fixed generator which are coprime to the order.
            $endgroup$
            – Tobias Kildetoft
            Jan 14 at 20:16


















          • $begingroup$
            But "find the first generator" is the whole problem.
            $endgroup$
            – Tobias Kildetoft
            Jan 14 at 20:08










          • $begingroup$
            Yes, noticed that during applying this method. But it can definitly reduce the total time of searching each generator.
            $endgroup$
            – AseN
            Jan 14 at 20:13












          • $begingroup$
            What can? You mean this method of finding all generators? Because that seems like it is precisely the standard way where you take all powers of a fixed generator which are coprime to the order.
            $endgroup$
            – Tobias Kildetoft
            Jan 14 at 20:16
















          $begingroup$
          But "find the first generator" is the whole problem.
          $endgroup$
          – Tobias Kildetoft
          Jan 14 at 20:08




          $begingroup$
          But "find the first generator" is the whole problem.
          $endgroup$
          – Tobias Kildetoft
          Jan 14 at 20:08












          $begingroup$
          Yes, noticed that during applying this method. But it can definitly reduce the total time of searching each generator.
          $endgroup$
          – AseN
          Jan 14 at 20:13






          $begingroup$
          Yes, noticed that during applying this method. But it can definitly reduce the total time of searching each generator.
          $endgroup$
          – AseN
          Jan 14 at 20:13














          $begingroup$
          What can? You mean this method of finding all generators? Because that seems like it is precisely the standard way where you take all powers of a fixed generator which are coprime to the order.
          $endgroup$
          – Tobias Kildetoft
          Jan 14 at 20:16




          $begingroup$
          What can? You mean this method of finding all generators? Because that seems like it is precisely the standard way where you take all powers of a fixed generator which are coprime to the order.
          $endgroup$
          – Tobias Kildetoft
          Jan 14 at 20:16


















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