Mixing
Proving that $mathbb{Z}_{n}$ is cyclic [duplicate]
This question already has an answer here:
Is $mathbb{Z}/langle nrangle$ cyclic and of order $n$?
2 answers
It should be very basic, but I'm struggling to understand the solution. I'm trying to prove that to group $mathbb{Z}_{n}$ is cyclic. Everywhere I tried to find a solution I came across of the saying
$mathbb{Z}_{n}$ is cyclic because it has a generator $1$ (meaning $mathbb{Z}_{n} = langle 1 rangle$).
I understand the meaning of generator, but I don't understand why the generator is $1$. Probably I'll feel stupid when I'll finally understand. Why $mathbb{Z}_{n} = langle 1 rangle$ when $mathbb{Z}_{n}$ is the Sum group of integers modulo $n$.
group-theory
asked Nov 21 '18 at 23:19
vesiivesii
886
marked as duplicate by amWhy group-theory
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Nov 21 '18 at 23:27
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:
Is $mathbb{Z}/langle nrangle$ cyclic and of order $n$?
2 answers
It should be very basic, but I'm struggling to understand the solution. I'm trying to prove that to group $mathbb{Z}_{n}$ is cyclic. Everywhere I tried to find a solution I came across of the saying
$mathbb{Z}_{n}$ is cyclic because it has a generator $1$ (meaning $mathbb{Z}_{n} = langle 1 rangle$).
I understand the meaning of generator, but I don't understand why the generator is $1$. Probably I'll feel stupid when I'll finally understand. Why $mathbb{Z}_{n} = langle 1 rangle$ when $mathbb{Z}_{n}$ is the Sum group of integers modulo $n$.
group-theory
asked Nov 21 '18 at 23:19
vesiivesii
886
marked as duplicate by amWhy group-theory
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Nov 21 '18 at 23:27
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.
1
What is your definition of $Bbb Z_n$?
– Arthur
Nov 21 '18 at 23:21
@Arthur Sorry. I thought that it is a common name. I have edited.
– vesii
Nov 21 '18 at 23:25
1
It is a common name. But there are a few different definitions going around. How to prove that it is cyclic depends on which one you are using.
– Arthur
Nov 21 '18 at 23:31
add a comment |
This question already has an answer here:
Is $mathbb{Z}/langle nrangle$ cyclic and of order $n$?
2 answers
It should be very basic, but I'm struggling to understand the solution. I'm trying to prove that to group $mathbb{Z}_{n}$ is cyclic. Everywhere I tried to find a solution I came across of the saying
$mathbb{Z}_{n}$ is cyclic because it has a generator $1$ (meaning $mathbb{Z}_{n} = langle 1 rangle$).
I understand the meaning of generator, but I don't understand why the generator is $1$. Probably I'll feel stupid when I'll finally understand. Why $mathbb{Z}_{n} = langle 1 rangle$ when $mathbb{Z}_{n}$ is the Sum group of integers modulo $n$.
group-theory
asked Nov 21 '18 at 23:19
vesiivesii
886
This question already has an answer here:
Is $mathbb{Z}/langle nrangle$ cyclic and of order $n$?
2 answers
It should be very basic, but I'm struggling to understand the solution. I'm trying to prove that to group $mathbb{Z}_{n}$ is cyclic. Everywhere I tried to find a solution I came across of the saying
$mathbb{Z}_{n}$ is cyclic because it has a generator $1$ (meaning $mathbb{Z}_{n} = langle 1 rangle$).
I understand the meaning of generator, but I don't understand why the generator is $1$. Probably I'll feel stupid when I'll finally understand. Why $mathbb{Z}_{n} = langle 1 rangle$ when $mathbb{Z}_{n}$ is the Sum group of integers modulo $n$.
This question already has an answer here:
Is $mathbb{Z}/langle nrangle$ cyclic and of order $n$?
2 answers
group-theory
group-theory
asked Nov 21 '18 at 23:19
vesiivesii
886
asked Nov 21 '18 at 23:19
vesiivesii
886
edited Nov 21 '18 at 23:24
vesii
asked Nov 21 '18 at 23:19
vesiivesii
886
asked Nov 21 '18 at 23:19
vesiivesii
886
asked Nov 21 '18 at 23:19
vesiivesii
886
886
marked as duplicate by amWhy group-theory
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Nov 21 '18 at 23:27
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 amWhy group-theory
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Nov 21 '18 at 23:27
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.
1
What is your definition of $Bbb Z_n$?
– Arthur
Nov 21 '18 at 23:21
@Arthur Sorry. I thought that it is a common name. I have edited.
– vesii
Nov 21 '18 at 23:25
1
It is a common name. But there are a few different definitions going around. How to prove that it is cyclic depends on which one you are using.
– Arthur
Nov 21 '18 at 23:31
add a comment |
1
What is your definition of $Bbb Z_n$?
– Arthur
Nov 21 '18 at 23:21
@Arthur Sorry. I thought that it is a common name. I have edited.
– vesii
Nov 21 '18 at 23:25
1
It is a common name. But there are a few different definitions going around. How to prove that it is cyclic depends on which one you are using.
– Arthur
Nov 21 '18 at 23:31
1
1
What is your definition of $Bbb Z_n$?
– Arthur
Nov 21 '18 at 23:21
What is your definition of $Bbb Z_n$?
– Arthur
Nov 21 '18 at 23:21
@Arthur Sorry. I thought that it is a common name. I have edited.
– vesii
Nov 21 '18 at 23:25
@Arthur Sorry. I thought that it is a common name. I have edited.
– vesii
Nov 21 '18 at 23:25
1
1
It is a common name. But there are a few different definitions going around. How to prove that it is cyclic depends on which one you are using.
– Arthur
Nov 21 '18 at 23:31
It is a common name. But there are a few different definitions going around. How to prove that it is cyclic depends on which one you are using.
– Arthur
Nov 21 '18 at 23:31
add a comment |
2 Answers
2
active
oldest
votes
Take any $min{1,2,...,n-1}$. Then $m=1+1+...+1$ when you sum $m$ times. And if you take $m=0$ then $m=1+1+...+1$ when you sum $n$ times because $0equiv n$(mod $n$). Hence any element of the group is generated by the set ${1}$.
answered Nov 21 '18 at 23:25
MarkMark
6,120415
add a comment |
Hint: $a bmod n = a(1 bmod n)$.
answered Nov 21 '18 at 23:24
lhflhf
163k10167388
It perhaps bears explicitly concluding that $1 bmod n$ is thus a generator of the (cyclic) group $mathbb Z bmod n$. The OP says, "I don't understand why the generator is $1$." So you would not be giving away a secret, but rather giving an approach to clear up the OP's confusion.
– hardmath
Nov 22 '18 at 0:20
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Take any $min{1,2,...,n-1}$. Then $m=1+1+...+1$ when you sum $m$ times. And if you take $m=0$ then $m=1+1+...+1$ when you sum $n$ times because $0equiv n$(mod $n$). Hence any element of the group is generated by the set ${1}$.
answered Nov 21 '18 at 23:25
MarkMark
6,120415
add a comment |
Take any $min{1,2,...,n-1}$. Then $m=1+1+...+1$ when you sum $m$ times. And if you take $m=0$ then $m=1+1+...+1$ when you sum $n$ times because $0equiv n$(mod $n$). Hence any element of the group is generated by the set ${1}$.
answered Nov 21 '18 at 23:25
MarkMark
6,120415
add a comment |
Take any $min{1,2,...,n-1}$. Then $m=1+1+...+1$ when you sum $m$ times. And if you take $m=0$ then $m=1+1+...+1$ when you sum $n$ times because $0equiv n$(mod $n$). Hence any element of the group is generated by the set ${1}$.
answered Nov 21 '18 at 23:25
MarkMark
6,120415
Take any $min{1,2,...,n-1}$. Then $m=1+1+...+1$ when you sum $m$ times. And if you take $m=0$ then $m=1+1+...+1$ when you sum $n$ times because $0equiv n$(mod $n$). Hence any element of the group is generated by the set ${1}$.
answered Nov 21 '18 at 23:25
MarkMark
6,120415
edited Nov 21 '18 at 23:38
answered Nov 21 '18 at 23:25
MarkMark
6,120415
answered Nov 21 '18 at 23:25
MarkMark
6,120415
answered Nov 21 '18 at 23:25
MarkMark
6,120415
6,120415
add a comment |
add a comment |
Hint: $a bmod n = a(1 bmod n)$.
answered Nov 21 '18 at 23:24
lhflhf
163k10167388
It perhaps bears explicitly concluding that $1 bmod n$ is thus a generator of the (cyclic) group $mathbb Z bmod n$. The OP says, "I don't understand why the generator is $1$." So you would not be giving away a secret, but rather giving an approach to clear up the OP's confusion.
– hardmath
Nov 22 '18 at 0:20
add a comment |
Hint: $a bmod n = a(1 bmod n)$.
answered Nov 21 '18 at 23:24
lhflhf
163k10167388
It perhaps bears explicitly concluding that $1 bmod n$ is thus a generator of the (cyclic) group $mathbb Z bmod n$. The OP says, "I don't understand why the generator is $1$." So you would not be giving away a secret, but rather giving an approach to clear up the OP's confusion.
– hardmath
Nov 22 '18 at 0:20
add a comment |
Hint: $a bmod n = a(1 bmod n)$.
answered Nov 21 '18 at 23:24
lhflhf
163k10167388
Hint: $a bmod n = a(1 bmod n)$.
answered Nov 21 '18 at 23:24
lhflhf
163k10167388
answered Nov 21 '18 at 23:24
lhflhf
163k10167388
answered Nov 21 '18 at 23:24
lhflhf
163k10167388
answered Nov 21 '18 at 23:24
lhflhf
163k10167388
163k10167388
It perhaps bears explicitly concluding that $1 bmod n$ is thus a generator of the (cyclic) group $mathbb Z bmod n$. The OP says, "I don't understand why the generator is $1$." So you would not be giving away a secret, but rather giving an approach to clear up the OP's confusion.
– hardmath
Nov 22 '18 at 0:20
add a comment |
It perhaps bears explicitly concluding that $1 bmod n$ is thus a generator of the (cyclic) group $mathbb Z bmod n$. The OP says, "I don't understand why the generator is $1$." So you would not be giving away a secret, but rather giving an approach to clear up the OP's confusion.
– hardmath
Nov 22 '18 at 0:20
It perhaps bears explicitly concluding that $1 bmod n$ is thus a generator of the (cyclic) group $mathbb Z bmod n$. The OP says, "I don't understand why the generator is $1$." So you would not be giving away a secret, but rather giving an approach to clear up the OP's confusion.
– hardmath
Nov 22 '18 at 0:20
It perhaps bears explicitly concluding that $1 bmod n$ is thus a generator of the (cyclic) group $mathbb Z bmod n$. The OP says, "I don't understand why the generator is $1$." So you would not be giving away a secret, but rather giving an approach to clear up the OP's confusion.
– hardmath
Nov 22 '18 at 0:20
add a comment |
1
What is your definition of $Bbb Z_n$?
– Arthur
Nov 21 '18 at 23:21
@Arthur Sorry. I thought that it is a common name. I have edited.
– vesii
Nov 21 '18 at 23:25
1
It is a common name. But there are a few different definitions going around. How to prove that it is cyclic depends on which one you are using.
– Arthur
Nov 21 '18 at 23:31