Let the divisors of $p-1$ be $d_1,d_2,cdots$Let $g mod p$ be a primitive root ,then for each $d_i$ there is...
I am trying,without success,to prove this statement:
Let the divisors of $p-1$ be $d_1,d_2,cdots$ Prove that if we have a
primitive root $g mod p$ ,then for each $d_i$ there is an element
with period $d_i$
I am trying,without success,to prove this statement:
Let the divisors of $p-1$ be $d_1,d_2,cdots$ Prove that if we have a
primitive root $g mod p$ ,then for each $d_i$ there is an element
with period $d_i$
My thinking:
It's given that $g$ is a primitive root ,now let $g=a^t$ where $a$ is not a multiple of $p$ and $t$ is some real number ,so we have that $a^{t},a^{2t},a^{3t},cdots equiv 1 mod p$
Now I know that from Fermat we have $a^{p-1} equiv 1 mod p$, so $p-1=tcdot k $ for some integer $k$.
Since we have that each of $d_i$ is a divisor of $p-1$ we have that $p-1=d_i cdot q $ where $q$ is an integer,then $d_i cdot q =tcdot k$ so I have that $$ a^{d_i}=a^{tk/q} =(a^{tk})^{1/q}$$
Since we have $a^{tk}equiv 1 mod p $ we have also that $left(a^{tk}right)^{1/q} equiv 1^{1/q} equiv 1 mod p$
Now I don't know what should I do now ,I am terribly confused (I haven't someone to ask for advice)
elementary-number-theory
asked Jan 15 '16 at 13:19
Mr. YMr. Y
1,359721
|
show 6 more comments
I am trying,without success,to prove this statement:
Let the divisors of $p-1$ be $d_1,d_2,cdots$ Prove that if we have a
primitive root $g mod p$ ,then for each $d_i$ there is an element
with period $d_i$
I am trying,without success,to prove this statement:
Let the divisors of $p-1$ be $d_1,d_2,cdots$ Prove that if we have a
primitive root $g mod p$ ,then for each $d_i$ there is an element
with period $d_i$
My thinking:
It's given that $g$ is a primitive root ,now let $g=a^t$ where $a$ is not a multiple of $p$ and $t$ is some real number ,so we have that $a^{t},a^{2t},a^{3t},cdots equiv 1 mod p$
Now I know that from Fermat we have $a^{p-1} equiv 1 mod p$, so $p-1=tcdot k $ for some integer $k$.
Since we have that each of $d_i$ is a divisor of $p-1$ we have that $p-1=d_i cdot q $ where $q$ is an integer,then $d_i cdot q =tcdot k$ so I have that $$ a^{d_i}=a^{tk/q} =(a^{tk})^{1/q}$$
Since we have $a^{tk}equiv 1 mod p $ we have also that $left(a^{tk}right)^{1/q} equiv 1^{1/q} equiv 1 mod p$
Now I don't know what should I do now ,I am terribly confused (I haven't someone to ask for advice)
elementary-number-theory
asked Jan 15 '16 at 13:19
Mr. YMr. Y
1,359721
1
What is $a$ and $t$? Also, $1^{1/q}$ need not be equal to $1$ modulo something.
– Wojowu
Jan 15 '16 at 13:24
Yes I've confused it all ,$t=p-1$ and $a$ is some integer which is not congruent to $0 mod p$
– Mr. Y
Jan 15 '16 at 13:34
In that case, $g=a^t$ never holds for $p>2$, because $a^tequiv 1$ and $gnotequiv 1$.
– Wojowu
Jan 15 '16 at 13:37
Can't I have $g=a^{t}$ for some real $t$ ?
– Mr. Y
Jan 15 '16 at 13:45
1
It isn't at all useful to consider real exponentiation for congruence problems, because it's not true that if $aequiv bmod p$, then $a^t=b^tmod p$. Instead, I would suggest you looking at the integer powers of $g$ itself.
– Wojowu
Jan 15 '16 at 13:56
|
show 6 more comments
I am trying,without success,to prove this statement:
Let the divisors of $p-1$ be $d_1,d_2,cdots$ Prove that if we have a
primitive root $g mod p$ ,then for each $d_i$ there is an element
with period $d_i$
I am trying,without success,to prove this statement:
Let the divisors of $p-1$ be $d_1,d_2,cdots$ Prove that if we have a
primitive root $g mod p$ ,then for each $d_i$ there is an element
with period $d_i$
My thinking:
It's given that $g$ is a primitive root ,now let $g=a^t$ where $a$ is not a multiple of $p$ and $t$ is some real number ,so we have that $a^{t},a^{2t},a^{3t},cdots equiv 1 mod p$
Now I know that from Fermat we have $a^{p-1} equiv 1 mod p$, so $p-1=tcdot k $ for some integer $k$.
Since we have that each of $d_i$ is a divisor of $p-1$ we have that $p-1=d_i cdot q $ where $q$ is an integer,then $d_i cdot q =tcdot k$ so I have that $$ a^{d_i}=a^{tk/q} =(a^{tk})^{1/q}$$
Since we have $a^{tk}equiv 1 mod p $ we have also that $left(a^{tk}right)^{1/q} equiv 1^{1/q} equiv 1 mod p$
Now I don't know what should I do now ,I am terribly confused (I haven't someone to ask for advice)
elementary-number-theory
asked Jan 15 '16 at 13:19
Mr. YMr. Y
1,359721
I am trying,without success,to prove this statement:
Let the divisors of $p-1$ be $d_1,d_2,cdots$ Prove that if we have a
primitive root $g mod p$ ,then for each $d_i$ there is an element
with period $d_i$
I am trying,without success,to prove this statement:
Let the divisors of $p-1$ be $d_1,d_2,cdots$ Prove that if we have a
primitive root $g mod p$ ,then for each $d_i$ there is an element
with period $d_i$
My thinking:
It's given that $g$ is a primitive root ,now let $g=a^t$ where $a$ is not a multiple of $p$ and $t$ is some real number ,so we have that $a^{t},a^{2t},a^{3t},cdots equiv 1 mod p$
Now I know that from Fermat we have $a^{p-1} equiv 1 mod p$, so $p-1=tcdot k $ for some integer $k$.
Since we have that each of $d_i$ is a divisor of $p-1$ we have that $p-1=d_i cdot q $ where $q$ is an integer,then $d_i cdot q =tcdot k$ so I have that $$ a^{d_i}=a^{tk/q} =(a^{tk})^{1/q}$$
Since we have $a^{tk}equiv 1 mod p $ we have also that $left(a^{tk}right)^{1/q} equiv 1^{1/q} equiv 1 mod p$
Now I don't know what should I do now ,I am terribly confused (I haven't someone to ask for advice)
elementary-number-theory
elementary-number-theory
asked Jan 15 '16 at 13:19
Mr. YMr. Y
1,359721
asked Jan 15 '16 at 13:19
Mr. YMr. Y
1,359721
edited Jan 15 '16 at 13:54
Mr. Y
asked Jan 15 '16 at 13:19
Mr. YMr. Y
1,359721
asked Jan 15 '16 at 13:19
Mr. YMr. Y
1,359721
asked Jan 15 '16 at 13:19
Mr. YMr. Y
1,359721
1,359721
1
What is $a$ and $t$? Also, $1^{1/q}$ need not be equal to $1$ modulo something.
– Wojowu
Jan 15 '16 at 13:24
Yes I've confused it all ,$t=p-1$ and $a$ is some integer which is not congruent to $0 mod p$
– Mr. Y
Jan 15 '16 at 13:34
In that case, $g=a^t$ never holds for $p>2$, because $a^tequiv 1$ and $gnotequiv 1$.
– Wojowu
Jan 15 '16 at 13:37
Can't I have $g=a^{t}$ for some real $t$ ?
– Mr. Y
Jan 15 '16 at 13:45
1
It isn't at all useful to consider real exponentiation for congruence problems, because it's not true that if $aequiv bmod p$, then $a^t=b^tmod p$. Instead, I would suggest you looking at the integer powers of $g$ itself.
– Wojowu
Jan 15 '16 at 13:56
|
show 6 more comments
1
What is $a$ and $t$? Also, $1^{1/q}$ need not be equal to $1$ modulo something.
– Wojowu
Jan 15 '16 at 13:24
Yes I've confused it all ,$t=p-1$ and $a$ is some integer which is not congruent to $0 mod p$
– Mr. Y
Jan 15 '16 at 13:34
In that case, $g=a^t$ never holds for $p>2$, because $a^tequiv 1$ and $gnotequiv 1$.
– Wojowu
Jan 15 '16 at 13:37
Can't I have $g=a^{t}$ for some real $t$ ?
– Mr. Y
Jan 15 '16 at 13:45
1
It isn't at all useful to consider real exponentiation for congruence problems, because it's not true that if $aequiv bmod p$, then $a^t=b^tmod p$. Instead, I would suggest you looking at the integer powers of $g$ itself.
– Wojowu
Jan 15 '16 at 13:56
1
1
What is $a$ and $t$? Also, $1^{1/q}$ need not be equal to $1$ modulo something.
– Wojowu
Jan 15 '16 at 13:24
What is $a$ and $t$? Also, $1^{1/q}$ need not be equal to $1$ modulo something.
– Wojowu
Jan 15 '16 at 13:24
Yes I've confused it all ,$t=p-1$ and $a$ is some integer which is not congruent to $0 mod p$
– Mr. Y
Jan 15 '16 at 13:34
Yes I've confused it all ,$t=p-1$ and $a$ is some integer which is not congruent to $0 mod p$
– Mr. Y
Jan 15 '16 at 13:34
In that case, $g=a^t$ never holds for $p>2$, because $a^tequiv 1$ and $gnotequiv 1$.
– Wojowu
Jan 15 '16 at 13:37
In that case, $g=a^t$ never holds for $p>2$, because $a^tequiv 1$ and $gnotequiv 1$.
– Wojowu
Jan 15 '16 at 13:37
Can't I have $g=a^{t}$ for some real $t$ ?
– Mr. Y
Jan 15 '16 at 13:45
Can't I have $g=a^{t}$ for some real $t$ ?
– Mr. Y
Jan 15 '16 at 13:45
1
1
It isn't at all useful to consider real exponentiation for congruence problems, because it's not true that if $aequiv bmod p$, then $a^t=b^tmod p$. Instead, I would suggest you looking at the integer powers of $g$ itself.
– Wojowu
Jan 15 '16 at 13:56
It isn't at all useful to consider real exponentiation for congruence problems, because it's not true that if $aequiv bmod p$, then $a^t=b^tmod p$. Instead, I would suggest you looking at the integer powers of $g$ itself.
– Wojowu
Jan 15 '16 at 13:56
|
show 6 more comments
1 Answer
1
active
oldest
votes
It's a much more general fact. If $g$ is an element of period $n$ in a group, and $m$ divides $n$, then the period of $g^{n/m}$ is $m$.
Even more generally, when $k$ is arbitrary, then the period of $g^{k}$ is
$$
frac{n}{gcd(n, k)}.
$$
Proof of the first statement.
For $1 le k < m$, we have $(n/m) cdot k < n$, so
$(g^{n/m})^{k} = g^{(n/m)cdot k} ne 1$. However $(g^{n/m})^{m} = g^{(n/m) cdot m} = g^{n} = 1$, so the period of $g^{n/m}$ is exactly $m$.
answered Jan 15 '16 at 13:58
Andreas CarantiAndreas Caranti
56.2k34295
If it's not too much disturb,where can I see the proof of this ?
– Mr. Y
Jan 15 '16 at 14:22
add a comment |
Your Answer
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1 Answer
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oldest
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It's a much more general fact. If $g$ is an element of period $n$ in a group, and $m$ divides $n$, then the period of $g^{n/m}$ is $m$.
Even more generally, when $k$ is arbitrary, then the period of $g^{k}$ is
$$
frac{n}{gcd(n, k)}.
$$
Proof of the first statement.
For $1 le k < m$, we have $(n/m) cdot k < n$, so
$(g^{n/m})^{k} = g^{(n/m)cdot k} ne 1$. However $(g^{n/m})^{m} = g^{(n/m) cdot m} = g^{n} = 1$, so the period of $g^{n/m}$ is exactly $m$.
answered Jan 15 '16 at 13:58
Andreas CarantiAndreas Caranti
56.2k34295
If it's not too much disturb,where can I see the proof of this ?
– Mr. Y
Jan 15 '16 at 14:22
add a comment |
It's a much more general fact. If $g$ is an element of period $n$ in a group, and $m$ divides $n$, then the period of $g^{n/m}$ is $m$.
Even more generally, when $k$ is arbitrary, then the period of $g^{k}$ is
$$
frac{n}{gcd(n, k)}.
$$
Proof of the first statement.
For $1 le k < m$, we have $(n/m) cdot k < n$, so
$(g^{n/m})^{k} = g^{(n/m)cdot k} ne 1$. However $(g^{n/m})^{m} = g^{(n/m) cdot m} = g^{n} = 1$, so the period of $g^{n/m}$ is exactly $m$.
answered Jan 15 '16 at 13:58
Andreas CarantiAndreas Caranti
56.2k34295
If it's not too much disturb,where can I see the proof of this ?
– Mr. Y
Jan 15 '16 at 14:22
add a comment |
It's a much more general fact. If $g$ is an element of period $n$ in a group, and $m$ divides $n$, then the period of $g^{n/m}$ is $m$.
Even more generally, when $k$ is arbitrary, then the period of $g^{k}$ is
$$
frac{n}{gcd(n, k)}.
$$
Proof of the first statement.
For $1 le k < m$, we have $(n/m) cdot k < n$, so
$(g^{n/m})^{k} = g^{(n/m)cdot k} ne 1$. However $(g^{n/m})^{m} = g^{(n/m) cdot m} = g^{n} = 1$, so the period of $g^{n/m}$ is exactly $m$.
answered Jan 15 '16 at 13:58
Andreas CarantiAndreas Caranti
56.2k34295
It's a much more general fact. If $g$ is an element of period $n$ in a group, and $m$ divides $n$, then the period of $g^{n/m}$ is $m$.
Even more generally, when $k$ is arbitrary, then the period of $g^{k}$ is
$$
frac{n}{gcd(n, k)}.
$$
Proof of the first statement.
For $1 le k < m$, we have $(n/m) cdot k < n$, so
$(g^{n/m})^{k} = g^{(n/m)cdot k} ne 1$. However $(g^{n/m})^{m} = g^{(n/m) cdot m} = g^{n} = 1$, so the period of $g^{n/m}$ is exactly $m$.
answered Jan 15 '16 at 13:58
Andreas CarantiAndreas Caranti
56.2k34295
edited Jan 4 at 10:10
answered Jan 15 '16 at 13:58
Andreas CarantiAndreas Caranti
56.2k34295
answered Jan 15 '16 at 13:58
Andreas CarantiAndreas Caranti
56.2k34295
answered Jan 15 '16 at 13:58
Andreas CarantiAndreas Caranti
56.2k34295
56.2k34295
If it's not too much disturb,where can I see the proof of this ?
– Mr. Y
Jan 15 '16 at 14:22
add a comment |
If it's not too much disturb,where can I see the proof of this ?
– Mr. Y
Jan 15 '16 at 14:22
If it's not too much disturb,where can I see the proof of this ?
– Mr. Y
Jan 15 '16 at 14:22
If it's not too much disturb,where can I see the proof of this ?
– Mr. Y
Jan 15 '16 at 14:22
add a comment |
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1
What is $a$ and $t$? Also, $1^{1/q}$ need not be equal to $1$ modulo something.
– Wojowu
Jan 15 '16 at 13:24
Yes I've confused it all ,$t=p-1$ and $a$ is some integer which is not congruent to $0 mod p$
– Mr. Y
Jan 15 '16 at 13:34
In that case, $g=a^t$ never holds for $p>2$, because $a^tequiv 1$ and $gnotequiv 1$.
– Wojowu
Jan 15 '16 at 13:37
Can't I have $g=a^{t}$ for some real $t$ ?
– Mr. Y
Jan 15 '16 at 13:45
1
It isn't at all useful to consider real exponentiation for congruence problems, because it's not true that if $aequiv bmod p$, then $a^t=b^tmod p$. Instead, I would suggest you looking at the integer powers of $g$ itself.
– Wojowu
Jan 15 '16 at 13:56