Why does not generator of multiplicative group generate all the members of group? [duplicate]
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
wherep
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?
group-theory field-theory abelian-groups
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.
add a comment |
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
wherep
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?
group-theory field-theory abelian-groups
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
add a comment |
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
wherep
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?
group-theory field-theory abelian-groups
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
wherep
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
group-theory field-theory abelian-groups
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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
$$ r=frac{operatorname{lcm}(10,k)}{k}=frac{10}{gcd(10,k)}=frac{10}2. $$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
$$ r=frac{operatorname{lcm}(10,k)}{k}=frac{10}{gcd(10,k)}=frac{10}2. $$
add a comment |
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
$$ r=frac{operatorname{lcm}(10,k)}{k}=frac{10}{gcd(10,k)}=frac{10}2. $$
add a comment |
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
$$ r=frac{operatorname{lcm}(10,k)}{k}=frac{10}{gcd(10,k)}=frac{10}2. $$
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
$$ r=frac{operatorname{lcm}(10,k)}{k}=frac{10}{gcd(10,k)}=frac{10}2. $$
answered 2 days ago
BernardBernard
118k639112
118k639112
add a comment |
add a comment |
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