Cycle structure of the permutation $x mapsto p·xoperatorname{mod}q$ for coprime $p,q$
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Let $[q] = {0,dots,q-1}$, $p < q$.
Consider the function $mathbf{p}: [q] rightarrow [q]$ which sends $x mapsto p·xoperatorname{mod}q$, i.e. the multiplication by $p$ modulo $q$ on $[q]$.
One finds that when $p$ and $q$ are coprime, $mathbf{p}$ is a permutation of $[q]$ with $mathbf{p}(0) = 0$.
Each such permutation – depending solely on $p$ and $q$ – has a specific cycle spectrum: $n_m$ cycles of length $m$.
How do I calculate the possible cycle lengths $m$ and their
corresponding numbers $n_m$ just by looking at $p$ and $q$?
number-theory permutations coprime permutation-cycles
asked Dec 5 at 17:54
Hans Stricker
5,69943884
add a comment |
up vote
1
down vote
favorite
Let $[q] = {0,dots,q-1}$, $p < q$.
Consider the function $mathbf{p}: [q] rightarrow [q]$ which sends $x mapsto p·xoperatorname{mod}q$, i.e. the multiplication by $p$ modulo $q$ on $[q]$.
One finds that when $p$ and $q$ are coprime, $mathbf{p}$ is a permutation of $[q]$ with $mathbf{p}(0) = 0$.
Each such permutation – depending solely on $p$ and $q$ – has a specific cycle spectrum: $n_m$ cycles of length $m$.
How do I calculate the possible cycle lengths $m$ and their
corresponding numbers $n_m$ just by looking at $p$ and $q$?
number-theory permutations coprime permutation-cycles
asked Dec 5 at 17:54
Hans Stricker
5,69943884
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $[q] = {0,dots,q-1}$, $p < q$.
Consider the function $mathbf{p}: [q] rightarrow [q]$ which sends $x mapsto p·xoperatorname{mod}q$, i.e. the multiplication by $p$ modulo $q$ on $[q]$.
One finds that when $p$ and $q$ are coprime, $mathbf{p}$ is a permutation of $[q]$ with $mathbf{p}(0) = 0$.
Each such permutation – depending solely on $p$ and $q$ – has a specific cycle spectrum: $n_m$ cycles of length $m$.
How do I calculate the possible cycle lengths $m$ and their
corresponding numbers $n_m$ just by looking at $p$ and $q$?
number-theory permutations coprime permutation-cycles
asked Dec 5 at 17:54
Hans Stricker
5,69943884
Let $[q] = {0,dots,q-1}$, $p < q$.
Consider the function $mathbf{p}: [q] rightarrow [q]$ which sends $x mapsto p·xoperatorname{mod}q$, i.e. the multiplication by $p$ modulo $q$ on $[q]$.
One finds that when $p$ and $q$ are coprime, $mathbf{p}$ is a permutation of $[q]$ with $mathbf{p}(0) = 0$.
Each such permutation – depending solely on $p$ and $q$ – has a specific cycle spectrum: $n_m$ cycles of length $m$.
How do I calculate the possible cycle lengths $m$ and their
corresponding numbers $n_m$ just by looking at $p$ and $q$?
number-theory permutations coprime permutation-cycles
number-theory permutations coprime permutation-cycles
asked Dec 5 at 17:54
Hans Stricker
5,69943884
asked Dec 5 at 17:54
Hans Stricker
5,69943884
edited Dec 7 at 11:33
asked Dec 5 at 17:54
Hans Stricker
5,69943884
asked Dec 5 at 17:54
Hans Stricker
5,69943884
asked Dec 5 at 17:54
Hans Stricker
5,69943884
5,69943884
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Let $H_q = { p^n bmod q,n ge 0}$, it contains $|H_q| = $ "the order of $pbmod q$" elements.
Assume $gcd(a,q)=1$ then $a in (mathbb{Z}/qmathbb{Z})^times$ so $|aH_q| = |H_q|$
Otherwise let $g = gcd(a,q)$. Then $frac{a}{g} in (mathbb{Z}/qmathbb{Z})^times$ and $|a H_q|= |g frac{a}{g} H_q|=|g H_q| = |g H_{frac{q}{g}}| = |H_{frac{q}{g}}|$
Thus for each $d = frac{q}{g} | q$ there are $frac{varphi(d)}{|H_{d}|}$ cycles of length $|H_{d}| = $ the order of $p bmod d$
To know the order of each $p bmod d$, you can factorize $q = prod_j p_j^{e_j}$ and compute the order of $p bmod p_j^m,m le e_j$, then $|H_{prod_j p_j^{m_j}}|$ is the $lcm$ of the $|H_{p_j^{m_j}}|$
answered Dec 5 at 20:08
reuns
19.4k21046
Would you mind to help me to align your answer with the special case 49:243 as depicted in my question? (It's not too obvious how to start off.)
– Hans Stricker
Dec 6 at 17:04
wolframalpha.com/input/… and $varphi(3^n) = 2.3^{n-1}$ @HansStricker
– reuns
Dec 6 at 17:08
This looks promising, but (i) What does $2.3^{n-1}$ mean? Should it be $2cdot 3^{n-1}$? and (ii) I'm looking for an expression with a variable which I can set to $243$. (Actually I see only a variable which I can set to $49$. And $243$ comes only in the result.)
– Hans Stricker
Dec 6 at 17:24
$3^n$ are the divisors of $243$ and $varphi$ is the Euler totient. Do you understand how multiplication by 49 acts on the group of integers coprime with $243$ ? @HansStricker
– reuns
Dec 6 at 17:28
I know the symbol and meaning of the Euler totient. And now I see: $3^1 = 3$, $3^2=9$, $3^3 = 27$, $3^4 = 81$, $3^5 = 243$ are exactly the divisors of $243$. That's interesting enough, I was not aware of.
– Hans Stricker
Dec 6 at 17:31
|
show 6 more comments
up vote
1
down vote
Having digested and finally understood user reuns' answer, let me share some visual examples:
answered Dec 7 at 11:41
Hans Stricker
5,69943884
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $H_q = { p^n bmod q,n ge 0}$, it contains $|H_q| = $ "the order of $pbmod q$" elements.
Assume $gcd(a,q)=1$ then $a in (mathbb{Z}/qmathbb{Z})^times$ so $|aH_q| = |H_q|$
Otherwise let $g = gcd(a,q)$. Then $frac{a}{g} in (mathbb{Z}/qmathbb{Z})^times$ and $|a H_q|= |g frac{a}{g} H_q|=|g H_q| = |g H_{frac{q}{g}}| = |H_{frac{q}{g}}|$
Thus for each $d = frac{q}{g} | q$ there are $frac{varphi(d)}{|H_{d}|}$ cycles of length $|H_{d}| = $ the order of $p bmod d$
To know the order of each $p bmod d$, you can factorize $q = prod_j p_j^{e_j}$ and compute the order of $p bmod p_j^m,m le e_j$, then $|H_{prod_j p_j^{m_j}}|$ is the $lcm$ of the $|H_{p_j^{m_j}}|$
answered Dec 5 at 20:08
reuns
19.4k21046
Would you mind to help me to align your answer with the special case 49:243 as depicted in my question? (It's not too obvious how to start off.)
– Hans Stricker
Dec 6 at 17:04
wolframalpha.com/input/… and $varphi(3^n) = 2.3^{n-1}$ @HansStricker
– reuns
Dec 6 at 17:08
This looks promising, but (i) What does $2.3^{n-1}$ mean? Should it be $2cdot 3^{n-1}$? and (ii) I'm looking for an expression with a variable which I can set to $243$. (Actually I see only a variable which I can set to $49$. And $243$ comes only in the result.)
– Hans Stricker
Dec 6 at 17:24
$3^n$ are the divisors of $243$ and $varphi$ is the Euler totient. Do you understand how multiplication by 49 acts on the group of integers coprime with $243$ ? @HansStricker
– reuns
Dec 6 at 17:28
I know the symbol and meaning of the Euler totient. And now I see: $3^1 = 3$, $3^2=9$, $3^3 = 27$, $3^4 = 81$, $3^5 = 243$ are exactly the divisors of $243$. That's interesting enough, I was not aware of.
– Hans Stricker
Dec 6 at 17:31
|
show 6 more comments
up vote
1
down vote
accepted
Let $H_q = { p^n bmod q,n ge 0}$, it contains $|H_q| = $ "the order of $pbmod q$" elements.
Assume $gcd(a,q)=1$ then $a in (mathbb{Z}/qmathbb{Z})^times$ so $|aH_q| = |H_q|$
Otherwise let $g = gcd(a,q)$. Then $frac{a}{g} in (mathbb{Z}/qmathbb{Z})^times$ and $|a H_q|= |g frac{a}{g} H_q|=|g H_q| = |g H_{frac{q}{g}}| = |H_{frac{q}{g}}|$
Thus for each $d = frac{q}{g} | q$ there are $frac{varphi(d)}{|H_{d}|}$ cycles of length $|H_{d}| = $ the order of $p bmod d$
To know the order of each $p bmod d$, you can factorize $q = prod_j p_j^{e_j}$ and compute the order of $p bmod p_j^m,m le e_j$, then $|H_{prod_j p_j^{m_j}}|$ is the $lcm$ of the $|H_{p_j^{m_j}}|$
answered Dec 5 at 20:08
reuns
19.4k21046
Would you mind to help me to align your answer with the special case 49:243 as depicted in my question? (It's not too obvious how to start off.)
– Hans Stricker
Dec 6 at 17:04
wolframalpha.com/input/… and $varphi(3^n) = 2.3^{n-1}$ @HansStricker
– reuns
Dec 6 at 17:08
This looks promising, but (i) What does $2.3^{n-1}$ mean? Should it be $2cdot 3^{n-1}$? and (ii) I'm looking for an expression with a variable which I can set to $243$. (Actually I see only a variable which I can set to $49$. And $243$ comes only in the result.)
– Hans Stricker
Dec 6 at 17:24
$3^n$ are the divisors of $243$ and $varphi$ is the Euler totient. Do you understand how multiplication by 49 acts on the group of integers coprime with $243$ ? @HansStricker
– reuns
Dec 6 at 17:28
I know the symbol and meaning of the Euler totient. And now I see: $3^1 = 3$, $3^2=9$, $3^3 = 27$, $3^4 = 81$, $3^5 = 243$ are exactly the divisors of $243$. That's interesting enough, I was not aware of.
– Hans Stricker
Dec 6 at 17:31
|
show 6 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $H_q = { p^n bmod q,n ge 0}$, it contains $|H_q| = $ "the order of $pbmod q$" elements.
Assume $gcd(a,q)=1$ then $a in (mathbb{Z}/qmathbb{Z})^times$ so $|aH_q| = |H_q|$
Otherwise let $g = gcd(a,q)$. Then $frac{a}{g} in (mathbb{Z}/qmathbb{Z})^times$ and $|a H_q|= |g frac{a}{g} H_q|=|g H_q| = |g H_{frac{q}{g}}| = |H_{frac{q}{g}}|$
Thus for each $d = frac{q}{g} | q$ there are $frac{varphi(d)}{|H_{d}|}$ cycles of length $|H_{d}| = $ the order of $p bmod d$
To know the order of each $p bmod d$, you can factorize $q = prod_j p_j^{e_j}$ and compute the order of $p bmod p_j^m,m le e_j$, then $|H_{prod_j p_j^{m_j}}|$ is the $lcm$ of the $|H_{p_j^{m_j}}|$
answered Dec 5 at 20:08
reuns
19.4k21046
Let $H_q = { p^n bmod q,n ge 0}$, it contains $|H_q| = $ "the order of $pbmod q$" elements.
Assume $gcd(a,q)=1$ then $a in (mathbb{Z}/qmathbb{Z})^times$ so $|aH_q| = |H_q|$
Otherwise let $g = gcd(a,q)$. Then $frac{a}{g} in (mathbb{Z}/qmathbb{Z})^times$ and $|a H_q|= |g frac{a}{g} H_q|=|g H_q| = |g H_{frac{q}{g}}| = |H_{frac{q}{g}}|$
Thus for each $d = frac{q}{g} | q$ there are $frac{varphi(d)}{|H_{d}|}$ cycles of length $|H_{d}| = $ the order of $p bmod d$
To know the order of each $p bmod d$, you can factorize $q = prod_j p_j^{e_j}$ and compute the order of $p bmod p_j^m,m le e_j$, then $|H_{prod_j p_j^{m_j}}|$ is the $lcm$ of the $|H_{p_j^{m_j}}|$
answered Dec 5 at 20:08
reuns
19.4k21046
edited Dec 5 at 20:39
answered Dec 5 at 20:08
reuns
19.4k21046
answered Dec 5 at 20:08
reuns
19.4k21046
answered Dec 5 at 20:08
reuns
19.4k21046
19.4k21046
Would you mind to help me to align your answer with the special case 49:243 as depicted in my question? (It's not too obvious how to start off.)
– Hans Stricker
Dec 6 at 17:04
wolframalpha.com/input/… and $varphi(3^n) = 2.3^{n-1}$ @HansStricker
– reuns
Dec 6 at 17:08
This looks promising, but (i) What does $2.3^{n-1}$ mean? Should it be $2cdot 3^{n-1}$? and (ii) I'm looking for an expression with a variable which I can set to $243$. (Actually I see only a variable which I can set to $49$. And $243$ comes only in the result.)
– Hans Stricker
Dec 6 at 17:24
$3^n$ are the divisors of $243$ and $varphi$ is the Euler totient. Do you understand how multiplication by 49 acts on the group of integers coprime with $243$ ? @HansStricker
– reuns
Dec 6 at 17:28
I know the symbol and meaning of the Euler totient. And now I see: $3^1 = 3$, $3^2=9$, $3^3 = 27$, $3^4 = 81$, $3^5 = 243$ are exactly the divisors of $243$. That's interesting enough, I was not aware of.
– Hans Stricker
Dec 6 at 17:31
|
show 6 more comments
Would you mind to help me to align your answer with the special case 49:243 as depicted in my question? (It's not too obvious how to start off.)
– Hans Stricker
Dec 6 at 17:04
wolframalpha.com/input/… and $varphi(3^n) = 2.3^{n-1}$ @HansStricker
– reuns
Dec 6 at 17:08
This looks promising, but (i) What does $2.3^{n-1}$ mean? Should it be $2cdot 3^{n-1}$? and (ii) I'm looking for an expression with a variable which I can set to $243$. (Actually I see only a variable which I can set to $49$. And $243$ comes only in the result.)
– Hans Stricker
Dec 6 at 17:24
$3^n$ are the divisors of $243$ and $varphi$ is the Euler totient. Do you understand how multiplication by 49 acts on the group of integers coprime with $243$ ? @HansStricker
– reuns
Dec 6 at 17:28
I know the symbol and meaning of the Euler totient. And now I see: $3^1 = 3$, $3^2=9$, $3^3 = 27$, $3^4 = 81$, $3^5 = 243$ are exactly the divisors of $243$. That's interesting enough, I was not aware of.
– Hans Stricker
Dec 6 at 17:31
Would you mind to help me to align your answer with the special case 49:243 as depicted in my question? (It's not too obvious how to start off.)
– Hans Stricker
Dec 6 at 17:04
Would you mind to help me to align your answer with the special case 49:243 as depicted in my question? (It's not too obvious how to start off.)
– Hans Stricker
Dec 6 at 17:04
wolframalpha.com/input/… and $varphi(3^n) = 2.3^{n-1}$ @HansStricker
– reuns
Dec 6 at 17:08
wolframalpha.com/input/… and $varphi(3^n) = 2.3^{n-1}$ @HansStricker
– reuns
Dec 6 at 17:08
This looks promising, but (i) What does $2.3^{n-1}$ mean? Should it be $2cdot 3^{n-1}$? and (ii) I'm looking for an expression with a variable which I can set to $243$. (Actually I see only a variable which I can set to $49$. And $243$ comes only in the result.)
– Hans Stricker
Dec 6 at 17:24
This looks promising, but (i) What does $2.3^{n-1}$ mean? Should it be $2cdot 3^{n-1}$? and (ii) I'm looking for an expression with a variable which I can set to $243$. (Actually I see only a variable which I can set to $49$. And $243$ comes only in the result.)
– Hans Stricker
Dec 6 at 17:24
$3^n$ are the divisors of $243$ and $varphi$ is the Euler totient. Do you understand how multiplication by 49 acts on the group of integers coprime with $243$ ? @HansStricker
– reuns
Dec 6 at 17:28
$3^n$ are the divisors of $243$ and $varphi$ is the Euler totient. Do you understand how multiplication by 49 acts on the group of integers coprime with $243$ ? @HansStricker
– reuns
Dec 6 at 17:28
I know the symbol and meaning of the Euler totient. And now I see: $3^1 = 3$, $3^2=9$, $3^3 = 27$, $3^4 = 81$, $3^5 = 243$ are exactly the divisors of $243$. That's interesting enough, I was not aware of.
– Hans Stricker
Dec 6 at 17:31
I know the symbol and meaning of the Euler totient. And now I see: $3^1 = 3$, $3^2=9$, $3^3 = 27$, $3^4 = 81$, $3^5 = 243$ are exactly the divisors of $243$. That's interesting enough, I was not aware of.
– Hans Stricker
Dec 6 at 17:31
|
show 6 more comments
up vote
1
down vote
Having digested and finally understood user reuns' answer, let me share some visual examples:
answered Dec 7 at 11:41
Hans Stricker
5,69943884
add a comment |
up vote
1
down vote
Having digested and finally understood user reuns' answer, let me share some visual examples:
answered Dec 7 at 11:41
Hans Stricker
5,69943884
add a comment |
up vote
1
down vote
up vote
1
down vote
Having digested and finally understood user reuns' answer, let me share some visual examples:
answered Dec 7 at 11:41
Hans Stricker
5,69943884
Having digested and finally understood user reuns' answer, let me share some visual examples:
answered Dec 7 at 11:41
Hans Stricker
5,69943884
answered Dec 7 at 11:41
Hans Stricker
5,69943884
answered Dec 7 at 11:41
Hans Stricker
5,69943884
answered Dec 7 at 11:41
Hans Stricker
5,69943884
5,69943884
add a comment |
add a comment |
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