Why does not generator of multiplicative group generate all the members of group? [duplicate]












0















This question already has an answer here:




  • Amount of generators for cyclic group and Euler's function

    1 answer





Generator(g) of multiplicative group (Zp where p is primary) is
an element the power of which by modulo(g^i mod p) can produce all
the elements of this group.




I have just tested this definition via python code and have not got the expected results. For example, lets take



p=11, so
Z* = {1,2,3,4,5,6,7,8,9,10}



I found only 4 elements, powers of which are actually generates all the memebers: {8, 2, 6, 7}. That is what I got (for 1<=i<=p, but i tested i up to 10^10):



1^i  :{1}
2^i :{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
3^i :{1, 3, 4, 5, 9}
4^i :{1, 3, 4, 5, 9}
5^i :{1, 3, 4, 5, 9}
6^i :{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
7^i :{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
8^i :{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
9^i :{1, 3, 4, 5, 9}
10^i :{1, 10}


Does this mean nothing but 2,6,7,8 are generators? Could you explain what is it I misunderstood?










share|cite|improve this question















marked as duplicate by Cameron Buie, Community 2 days ago


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




    Yes, $Bbb{Z}_p^*$ has $phi(p-1)$ generators, for $p=11$ we have $4$ generators - see this question.
    – Dietrich Burde
    2 days ago












  • @DietrichBurde, looks like I get confused with Z* and Z(according to link), thanks a lot!
    – AseN
    2 days ago


















0















This question already has an answer here:




  • Amount of generators for cyclic group and Euler's function

    1 answer





Generator(g) of multiplicative group (Zp where p is primary) is
an element the power of which by modulo(g^i mod p) can produce all
the elements of this group.




I have just tested this definition via python code and have not got the expected results. For example, lets take



p=11, so
Z* = {1,2,3,4,5,6,7,8,9,10}



I found only 4 elements, powers of which are actually generates all the memebers: {8, 2, 6, 7}. That is what I got (for 1<=i<=p, but i tested i up to 10^10):



1^i  :{1}
2^i :{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
3^i :{1, 3, 4, 5, 9}
4^i :{1, 3, 4, 5, 9}
5^i :{1, 3, 4, 5, 9}
6^i :{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
7^i :{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
8^i :{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
9^i :{1, 3, 4, 5, 9}
10^i :{1, 10}


Does this mean nothing but 2,6,7,8 are generators? Could you explain what is it I misunderstood?










share|cite|improve this question















marked as duplicate by Cameron Buie, Community 2 days ago


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




    Yes, $Bbb{Z}_p^*$ has $phi(p-1)$ generators, for $p=11$ we have $4$ generators - see this question.
    – Dietrich Burde
    2 days ago












  • @DietrichBurde, looks like I get confused with Z* and Z(according to link), thanks a lot!
    – AseN
    2 days ago
















0












0








0








This question already has an answer here:




  • Amount of generators for cyclic group and Euler's function

    1 answer





Generator(g) of multiplicative group (Zp where p is primary) is
an element the power of which by modulo(g^i mod p) can produce all
the elements of this group.




I have just tested this definition via python code and have not got the expected results. For example, lets take



p=11, so
Z* = {1,2,3,4,5,6,7,8,9,10}



I found only 4 elements, powers of which are actually generates all the memebers: {8, 2, 6, 7}. That is what I got (for 1<=i<=p, but i tested i up to 10^10):



1^i  :{1}
2^i :{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
3^i :{1, 3, 4, 5, 9}
4^i :{1, 3, 4, 5, 9}
5^i :{1, 3, 4, 5, 9}
6^i :{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
7^i :{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
8^i :{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
9^i :{1, 3, 4, 5, 9}
10^i :{1, 10}


Does this mean nothing but 2,6,7,8 are generators? Could you explain what is it I misunderstood?










share|cite|improve this question
















This question already has an answer here:




  • Amount of generators for cyclic group and Euler's function

    1 answer





Generator(g) of multiplicative group (Zp where p is primary) is
an element the power of which by modulo(g^i mod p) can produce all
the elements of this group.




I have just tested this definition via python code and have not got the expected results. For example, lets take



p=11, so
Z* = {1,2,3,4,5,6,7,8,9,10}



I found only 4 elements, powers of which are actually generates all the memebers: {8, 2, 6, 7}. That is what I got (for 1<=i<=p, but i tested i up to 10^10):



1^i  :{1}
2^i :{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
3^i :{1, 3, 4, 5, 9}
4^i :{1, 3, 4, 5, 9}
5^i :{1, 3, 4, 5, 9}
6^i :{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
7^i :{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
8^i :{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
9^i :{1, 3, 4, 5, 9}
10^i :{1, 10}


Does this mean nothing but 2,6,7,8 are generators? Could you explain what is it I misunderstood?





This question already has an answer here:




  • Amount of generators for cyclic group and Euler's function

    1 answer








group-theory field-theory abelian-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago







AseN

















asked 2 days ago









AseNAseN

1036




1036




marked as duplicate by Cameron Buie, Community 2 days ago


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 Cameron Buie, Community 2 days ago


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




    Yes, $Bbb{Z}_p^*$ has $phi(p-1)$ generators, for $p=11$ we have $4$ generators - see this question.
    – Dietrich Burde
    2 days ago












  • @DietrichBurde, looks like I get confused with Z* and Z(according to link), thanks a lot!
    – AseN
    2 days ago
















  • 2




    Yes, $Bbb{Z}_p^*$ has $phi(p-1)$ generators, for $p=11$ we have $4$ generators - see this question.
    – Dietrich Burde
    2 days ago












  • @DietrichBurde, looks like I get confused with Z* and Z(according to link), thanks a lot!
    – AseN
    2 days ago










2




2




Yes, $Bbb{Z}_p^*$ has $phi(p-1)$ generators, for $p=11$ we have $4$ generators - see this question.
– Dietrich Burde
2 days ago






Yes, $Bbb{Z}_p^*$ has $phi(p-1)$ generators, for $p=11$ we have $4$ generators - see this question.
– Dietrich Burde
2 days ago














@DietrichBurde, looks like I get confused with Z* and Z(according to link), thanks a lot!
– AseN
2 days ago






@DietrichBurde, looks like I get confused with Z* and Z(according to link), thanks a lot!
– AseN
2 days ago












1 Answer
1






active

oldest

votes


















0














Here is a (hopefully) simple explanation for the example of $3$:



$2$ is a generator of $(mathbf Z/11mathbf Z)^times$, and $2^8=3$.



Now the order of $^k$, by definition, is the least natural number $r$ such that $;(2^k)^r=2^{kr}equiv 1mod 11$; i.e.it is the least $r$ such that $kr$ is a multiple of $10$ (the order of $2$). In other words $kr$ is the l.c.m. of $k$ and $10$.



Now, we know that $;10 k=gcd(10,k)cdotoperatorname{lcm}(10,r)$, so that
$$ r=frac{operatorname{lcm}(10,k)}{k}=frac{10}{gcd(10,k)}=frac{10}2. $$






share|cite|improve this answer




























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    Here is a (hopefully) simple explanation for the example of $3$:



    $2$ is a generator of $(mathbf Z/11mathbf Z)^times$, and $2^8=3$.



    Now the order of $^k$, by definition, is the least natural number $r$ such that $;(2^k)^r=2^{kr}equiv 1mod 11$; i.e.it is the least $r$ such that $kr$ is a multiple of $10$ (the order of $2$). In other words $kr$ is the l.c.m. of $k$ and $10$.



    Now, we know that $;10 k=gcd(10,k)cdotoperatorname{lcm}(10,r)$, so that
    $$ r=frac{operatorname{lcm}(10,k)}{k}=frac{10}{gcd(10,k)}=frac{10}2. $$






    share|cite|improve this answer


























      0














      Here is a (hopefully) simple explanation for the example of $3$:



      $2$ is a generator of $(mathbf Z/11mathbf Z)^times$, and $2^8=3$.



      Now the order of $^k$, by definition, is the least natural number $r$ such that $;(2^k)^r=2^{kr}equiv 1mod 11$; i.e.it is the least $r$ such that $kr$ is a multiple of $10$ (the order of $2$). In other words $kr$ is the l.c.m. of $k$ and $10$.



      Now, we know that $;10 k=gcd(10,k)cdotoperatorname{lcm}(10,r)$, so that
      $$ r=frac{operatorname{lcm}(10,k)}{k}=frac{10}{gcd(10,k)}=frac{10}2. $$






      share|cite|improve this answer
























        0












        0








        0






        Here is a (hopefully) simple explanation for the example of $3$:



        $2$ is a generator of $(mathbf Z/11mathbf Z)^times$, and $2^8=3$.



        Now the order of $^k$, by definition, is the least natural number $r$ such that $;(2^k)^r=2^{kr}equiv 1mod 11$; i.e.it is the least $r$ such that $kr$ is a multiple of $10$ (the order of $2$). In other words $kr$ is the l.c.m. of $k$ and $10$.



        Now, we know that $;10 k=gcd(10,k)cdotoperatorname{lcm}(10,r)$, so that
        $$ r=frac{operatorname{lcm}(10,k)}{k}=frac{10}{gcd(10,k)}=frac{10}2. $$






        share|cite|improve this answer












        Here is a (hopefully) simple explanation for the example of $3$:



        $2$ is a generator of $(mathbf Z/11mathbf Z)^times$, and $2^8=3$.



        Now the order of $^k$, by definition, is the least natural number $r$ such that $;(2^k)^r=2^{kr}equiv 1mod 11$; i.e.it is the least $r$ such that $kr$ is a multiple of $10$ (the order of $2$). In other words $kr$ is the l.c.m. of $k$ and $10$.



        Now, we know that $;10 k=gcd(10,k)cdotoperatorname{lcm}(10,r)$, so that
        $$ r=frac{operatorname{lcm}(10,k)}{k}=frac{10}{gcd(10,k)}=frac{10}2. $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        BernardBernard

        118k639112




        118k639112















            Popular posts from this blog

            Mario Kart Wii

            The Binding of Isaac: Rebirth/Afterbirth

            What does “Dominus providebit” mean?