Mixing
Finding cyclic group $mathbb{Z}_n^times$ of prime order
0
$begingroup$
It can be proven that multiplicative group of integers modulo $N$ defined as
$$mathbb{Z}^times_N = { iin mathbb Z : 1leq ileq N−1; text{ and }; gcd(i,N)=1 }$$
is cyclic for a prime $N$ and that if it is of prime order, then every non-identity element in the group is a generator of this group.
How can I find such $N$?
I wrote a simple program and brute-forced for $i in mathbb Z; iin [1, 300 000]$ and found no group of prime order so far.
group-theory cyclic-groups
edited Jan 19 at 23:08
jordan_glen
1
asked Jan 19 at 22:36
omnomnomomnomnom
1012
$endgroup$
add a comment |
0
$begingroup$
It can be proven that multiplicative group of integers modulo $N$ defined as
$$mathbb{Z}^times_N = { iin mathbb Z : 1leq ileq N−1; text{ and }; gcd(i,N)=1 }$$
is cyclic for a prime $N$ and that if it is of prime order, then every non-identity element in the group is a generator of this group.
How can I find such $N$?
I wrote a simple program and brute-forced for $i in mathbb Z; iin [1, 300 000]$ and found no group of prime order so far.
group-theory cyclic-groups
edited Jan 19 at 23:08
jordan_glen
1
asked Jan 19 at 22:36
omnomnomomnomnom
1012
$endgroup$
5
$begingroup$
How about $N=3$?
$endgroup$
– kccu
Jan 19 at 22:38
5
$begingroup$
If $N$ is prime, then your group has $N-1$ elements , which is a prime only for $N=3$
$endgroup$
– Peter
Jan 19 at 22:38
3
$begingroup$
Note that the condition $gcd(i, N)$ is vacuous if $N$ is prime and $1 le i le N-1$, so as per Peter's comment your group has $N-1$ elements.
$endgroup$
– Rob Arthan
Jan 19 at 22:43
1
$begingroup$
In case you are confused by why only $N = 3$ works, note that the only two consecutive integers which are both prime are ${2, 3}$. You That's because $2$ is the only even prime.
$endgroup$
– jordan_glen
Jan 19 at 23:13
$begingroup$
I was rejecting $N = 3$ as a trivial case, heh :) So just to make sure I properly understand implications here: for $mathbb{Z}_n^times$, when looking for a generator, we must pick a random number from $[2, N-1]$ and check if it matches Euler's criterion? We can not select $N$ in such a way that we'll know upfront what the generator is. Is that correct?
$endgroup$
– omnomnom
Jan 20 at 8:14
add a comment |
0
0
0
$begingroup$
It can be proven that multiplicative group of integers modulo $N$ defined as
$$mathbb{Z}^times_N = { iin mathbb Z : 1leq ileq N−1; text{ and }; gcd(i,N)=1 }$$
is cyclic for a prime $N$ and that if it is of prime order, then every non-identity element in the group is a generator of this group.
How can I find such $N$?
I wrote a simple program and brute-forced for $i in mathbb Z; iin [1, 300 000]$ and found no group of prime order so far.
group-theory cyclic-groups
edited Jan 19 at 23:08
jordan_glen
1
asked Jan 19 at 22:36
omnomnomomnomnom
1012
$endgroup$
It can be proven that multiplicative group of integers modulo $N$ defined as
$$mathbb{Z}^times_N = { iin mathbb Z : 1leq ileq N−1; text{ and }; gcd(i,N)=1 }$$
is cyclic for a prime $N$ and that if it is of prime order, then every non-identity element in the group is a generator of this group.
How can I find such $N$?
I wrote a simple program and brute-forced for $i in mathbb Z; iin [1, 300 000]$ and found no group of prime order so far.
group-theory cyclic-groups
group-theory cyclic-groups
edited Jan 19 at 23:08
jordan_glen
1
asked Jan 19 at 22:36
omnomnomomnomnom
1012
edited Jan 19 at 23:08
jordan_glen
1
asked Jan 19 at 22:36
omnomnomomnomnom
1012
edited Jan 19 at 23:08
jordan_glen
1
edited Jan 19 at 23:08
jordan_glen
1
edited Jan 19 at 23:08
jordan_glen
1
1
asked Jan 19 at 22:36
omnomnomomnomnom
1012
asked Jan 19 at 22:36
omnomnomomnomnom
1012
asked Jan 19 at 22:36
omnomnomomnomnom
1012
1012
5
$begingroup$
How about $N=3$?
$endgroup$
– kccu
Jan 19 at 22:38
5
$begingroup$
If $N$ is prime, then your group has $N-1$ elements , which is a prime only for $N=3$
$endgroup$
– Peter
Jan 19 at 22:38
3
$begingroup$
Note that the condition $gcd(i, N)$ is vacuous if $N$ is prime and $1 le i le N-1$, so as per Peter's comment your group has $N-1$ elements.
$endgroup$
– Rob Arthan
Jan 19 at 22:43
1
$begingroup$
In case you are confused by why only $N = 3$ works, note that the only two consecutive integers which are both prime are ${2, 3}$. You That's because $2$ is the only even prime.
$endgroup$
– jordan_glen
Jan 19 at 23:13
$begingroup$
I was rejecting $N = 3$ as a trivial case, heh :) So just to make sure I properly understand implications here: for $mathbb{Z}_n^times$, when looking for a generator, we must pick a random number from $[2, N-1]$ and check if it matches Euler's criterion? We can not select $N$ in such a way that we'll know upfront what the generator is. Is that correct?
$endgroup$
– omnomnom
Jan 20 at 8:14
add a comment |
5
$begingroup$
How about $N=3$?
$endgroup$
– kccu
Jan 19 at 22:38
5
$begingroup$
If $N$ is prime, then your group has $N-1$ elements , which is a prime only for $N=3$
$endgroup$
– Peter
Jan 19 at 22:38
3
$begingroup$
Note that the condition $gcd(i, N)$ is vacuous if $N$ is prime and $1 le i le N-1$, so as per Peter's comment your group has $N-1$ elements.
$endgroup$
– Rob Arthan
Jan 19 at 22:43
1
$begingroup$
In case you are confused by why only $N = 3$ works, note that the only two consecutive integers which are both prime are ${2, 3}$. You That's because $2$ is the only even prime.
$endgroup$
– jordan_glen
Jan 19 at 23:13
$begingroup$
I was rejecting $N = 3$ as a trivial case, heh :) So just to make sure I properly understand implications here: for $mathbb{Z}_n^times$, when looking for a generator, we must pick a random number from $[2, N-1]$ and check if it matches Euler's criterion? We can not select $N$ in such a way that we'll know upfront what the generator is. Is that correct?
$endgroup$
– omnomnom
Jan 20 at 8:14
5
5
$begingroup$
How about $N=3$?
$endgroup$
– kccu
Jan 19 at 22:38
$begingroup$
How about $N=3$?
$endgroup$
– kccu
Jan 19 at 22:38
5
5
$begingroup$
If $N$ is prime, then your group has $N-1$ elements , which is a prime only for $N=3$
$endgroup$
– Peter
Jan 19 at 22:38
$begingroup$
If $N$ is prime, then your group has $N-1$ elements , which is a prime only for $N=3$
$endgroup$
– Peter
Jan 19 at 22:38
3
3
$begingroup$
Note that the condition $gcd(i, N)$ is vacuous if $N$ is prime and $1 le i le N-1$, so as per Peter's comment your group has $N-1$ elements.
$endgroup$
– Rob Arthan
Jan 19 at 22:43
$begingroup$
Note that the condition $gcd(i, N)$ is vacuous if $N$ is prime and $1 le i le N-1$, so as per Peter's comment your group has $N-1$ elements.
$endgroup$
– Rob Arthan
Jan 19 at 22:43
1
1
$begingroup$
In case you are confused by why only $N = 3$ works, note that the only two consecutive integers which are both prime are ${2, 3}$. You That's because $2$ is the only even prime.
$endgroup$
– jordan_glen
Jan 19 at 23:13
$begingroup$
In case you are confused by why only $N = 3$ works, note that the only two consecutive integers which are both prime are ${2, 3}$. You That's because $2$ is the only even prime.
$endgroup$
– jordan_glen
Jan 19 at 23:13
$begingroup$
I was rejecting $N = 3$ as a trivial case, heh :) So just to make sure I properly understand implications here: for $mathbb{Z}_n^times$, when looking for a generator, we must pick a random number from $[2, N-1]$ and check if it matches Euler's criterion? We can not select $N$ in such a way that we'll know upfront what the generator is. Is that correct?
$endgroup$
– omnomnom
Jan 20 at 8:14
$begingroup$
I was rejecting $N = 3$ as a trivial case, heh :) So just to make sure I properly understand implications here: for $mathbb{Z}_n^times$, when looking for a generator, we must pick a random number from $[2, N-1]$ and check if it matches Euler's criterion? We can not select $N$ in such a way that we'll know upfront what the generator is. Is that correct?
$endgroup$
– omnomnom
Jan 20 at 8:14
add a comment |
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5
$begingroup$
How about $N=3$?
$endgroup$
– kccu
Jan 19 at 22:38
5
$begingroup$
If $N$ is prime, then your group has $N-1$ elements , which is a prime only for $N=3$
$endgroup$
– Peter
Jan 19 at 22:38
3
$begingroup$
Note that the condition $gcd(i, N)$ is vacuous if $N$ is prime and $1 le i le N-1$, so as per Peter's comment your group has $N-1$ elements.
$endgroup$
– Rob Arthan
Jan 19 at 22:43
1
$begingroup$
In case you are confused by why only $N = 3$ works, note that the only two consecutive integers which are both prime are ${2, 3}$. You That's because $2$ is the only even prime.
$endgroup$
– jordan_glen
Jan 19 at 23:13
$begingroup$
I was rejecting $N = 3$ as a trivial case, heh :) So just to make sure I properly understand implications here: for $mathbb{Z}_n^times$, when looking for a generator, we must pick a random number from $[2, N-1]$ and check if it matches Euler's criterion? We can not select $N$ in such a way that we'll know upfront what the generator is. Is that correct?
$endgroup$
– omnomnom
Jan 20 at 8:14