Why are these two groups not isomorphic?
$begingroup$
I am trying to understand a proof (from the German book "Einführung in die Kryptografie" by Johannes Buchmann) that there are at most $(n-1)/4$ non-witnesses against the primality of $n$ in the Miller-Rabin algorithm that are coprime to $n$.
The Miller-Rabin test is stated in the following way:
Let $s = max{r in mathbb{N}: 2^r text{ divides } n-1}$ and $d = (n-1)/2^s$.
If $n$ is a prime number and if $a$ is a number coprime to $n$ then
$$ a^d equiv 1 pmod{n} tag{A} $$
or there is a $r$ in the set ${0,1,ldots,s-1}$ such that
$$a^{2^rd} equiv -1 pmod{n} tag{B} $$
I have a problem with the following proof:
Let $n ge 3$ be an odd composite number. We want to estimate how many numbers $a in {0,1,ldots,s-1}$ exist for which $gcd(a,n-1) = 1$ and both $(A)$ and $(B)$ hold. If there is no such $a$ we are fi
We set the prime factorisation of $n$ to $n = prod_{pmid n} p^{e(p)}$.
The author considers the following two subgroups of $mathbb{Z}_n^{times}$
$$begin{align*}
K &= { a + nmathbb{Z} : gcd(a,n) = 1 text{ and } a^{n-1} equiv pm 1 pmod{p^{e(p)}} text{ for all primes $p$ such that } p|n}\
L &= { a + nmathbb{Z} : gcd(a,n) = 1 text{ and } a^{m} equiv pm 1 pmod{n}}
end{align*}$$
In the coure of the proof the author considers the groups $L$ and $K$ and distinguishes between them. I do not understand why these two groups are not just the same by the Chinese remainder theorem, as all $p^{e(p)}$ are pairwise coprime and $n$ is by definition the product of these $p^{e(p)}$. Could you tell me what I am getting wrong?
group-theory prime-numbers
edited Nov 26 '18 at 1:19
Arturo Magidin
261k34586906
asked Nov 25 '18 at 12:44
3nondatur3nondatur
385111
$endgroup$
add a comment |
$begingroup$
I am trying to understand a proof (from the German book "Einführung in die Kryptografie" by Johannes Buchmann) that there are at most $(n-1)/4$ non-witnesses against the primality of $n$ in the Miller-Rabin algorithm that are coprime to $n$.
The Miller-Rabin test is stated in the following way:
Let $s = max{r in mathbb{N}: 2^r text{ divides } n-1}$ and $d = (n-1)/2^s$.
If $n$ is a prime number and if $a$ is a number coprime to $n$ then
$$ a^d equiv 1 pmod{n} tag{A} $$
or there is a $r$ in the set ${0,1,ldots,s-1}$ such that
$$a^{2^rd} equiv -1 pmod{n} tag{B} $$
I have a problem with the following proof:
Let $n ge 3$ be an odd composite number. We want to estimate how many numbers $a in {0,1,ldots,s-1}$ exist for which $gcd(a,n-1) = 1$ and both $(A)$ and $(B)$ hold. If there is no such $a$ we are fi
We set the prime factorisation of $n$ to $n = prod_{pmid n} p^{e(p)}$.
The author considers the following two subgroups of $mathbb{Z}_n^{times}$
$$begin{align*}
K &= { a + nmathbb{Z} : gcd(a,n) = 1 text{ and } a^{n-1} equiv pm 1 pmod{p^{e(p)}} text{ for all primes $p$ such that } p|n}\
L &= { a + nmathbb{Z} : gcd(a,n) = 1 text{ and } a^{m} equiv pm 1 pmod{n}}
end{align*}$$
In the coure of the proof the author considers the groups $L$ and $K$ and distinguishes between them. I do not understand why these two groups are not just the same by the Chinese remainder theorem, as all $p^{e(p)}$ are pairwise coprime and $n$ is by definition the product of these $p^{e(p)}$. Could you tell me what I am getting wrong?
group-theory prime-numbers
edited Nov 26 '18 at 1:19
Arturo Magidin
261k34586906
asked Nov 25 '18 at 12:44
3nondatur3nondatur
385111
$endgroup$
2
$begingroup$
The set $K$ includes cosets of $a$ such that $a^{n-1}equiv-1$ modulo some prime powers and $equiv+1$ modulo some other prime powers. In the set $L$ the same sign must occur for all prime powers dividing $n$ (as per the Chinese Remainder Theorem).
$endgroup$
– Jyrki Lahtonen
Nov 25 '18 at 13:05
1
$begingroup$
This question has large overlap with this other question: answerers and readers would do well to read both.
$endgroup$
– davidlowryduda♦
Nov 27 '18 at 21:49
add a comment |
$begingroup$
I am trying to understand a proof (from the German book "Einführung in die Kryptografie" by Johannes Buchmann) that there are at most $(n-1)/4$ non-witnesses against the primality of $n$ in the Miller-Rabin algorithm that are coprime to $n$.
The Miller-Rabin test is stated in the following way:
Let $s = max{r in mathbb{N}: 2^r text{ divides } n-1}$ and $d = (n-1)/2^s$.
If $n$ is a prime number and if $a$ is a number coprime to $n$ then
$$ a^d equiv 1 pmod{n} tag{A} $$
or there is a $r$ in the set ${0,1,ldots,s-1}$ such that
$$a^{2^rd} equiv -1 pmod{n} tag{B} $$
I have a problem with the following proof:
Let $n ge 3$ be an odd composite number. We want to estimate how many numbers $a in {0,1,ldots,s-1}$ exist for which $gcd(a,n-1) = 1$ and both $(A)$ and $(B)$ hold. If there is no such $a$ we are fi
We set the prime factorisation of $n$ to $n = prod_{pmid n} p^{e(p)}$.
The author considers the following two subgroups of $mathbb{Z}_n^{times}$
$$begin{align*}
K &= { a + nmathbb{Z} : gcd(a,n) = 1 text{ and } a^{n-1} equiv pm 1 pmod{p^{e(p)}} text{ for all primes $p$ such that } p|n}\
L &= { a + nmathbb{Z} : gcd(a,n) = 1 text{ and } a^{m} equiv pm 1 pmod{n}}
end{align*}$$
In the coure of the proof the author considers the groups $L$ and $K$ and distinguishes between them. I do not understand why these two groups are not just the same by the Chinese remainder theorem, as all $p^{e(p)}$ are pairwise coprime and $n$ is by definition the product of these $p^{e(p)}$. Could you tell me what I am getting wrong?
group-theory prime-numbers
edited Nov 26 '18 at 1:19
Arturo Magidin
261k34586906
asked Nov 25 '18 at 12:44
3nondatur3nondatur
385111
$endgroup$
I am trying to understand a proof (from the German book "Einführung in die Kryptografie" by Johannes Buchmann) that there are at most $(n-1)/4$ non-witnesses against the primality of $n$ in the Miller-Rabin algorithm that are coprime to $n$.
The Miller-Rabin test is stated in the following way:
Let $s = max{r in mathbb{N}: 2^r text{ divides } n-1}$ and $d = (n-1)/2^s$.
If $n$ is a prime number and if $a$ is a number coprime to $n$ then
$$ a^d equiv 1 pmod{n} tag{A} $$
or there is a $r$ in the set ${0,1,ldots,s-1}$ such that
$$a^{2^rd} equiv -1 pmod{n} tag{B} $$
I have a problem with the following proof:
Let $n ge 3$ be an odd composite number. We want to estimate how many numbers $a in {0,1,ldots,s-1}$ exist for which $gcd(a,n-1) = 1$ and both $(A)$ and $(B)$ hold. If there is no such $a$ we are fi
We set the prime factorisation of $n$ to $n = prod_{pmid n} p^{e(p)}$.
The author considers the following two subgroups of $mathbb{Z}_n^{times}$
$$begin{align*}
K &= { a + nmathbb{Z} : gcd(a,n) = 1 text{ and } a^{n-1} equiv pm 1 pmod{p^{e(p)}} text{ for all primes $p$ such that } p|n}\
L &= { a + nmathbb{Z} : gcd(a,n) = 1 text{ and } a^{m} equiv pm 1 pmod{n}}
end{align*}$$
In the coure of the proof the author considers the groups $L$ and $K$ and distinguishes between them. I do not understand why these two groups are not just the same by the Chinese remainder theorem, as all $p^{e(p)}$ are pairwise coprime and $n$ is by definition the product of these $p^{e(p)}$. Could you tell me what I am getting wrong?
group-theory prime-numbers
group-theory prime-numbers
edited Nov 26 '18 at 1:19
Arturo Magidin
261k34586906
asked Nov 25 '18 at 12:44
3nondatur3nondatur
385111
edited Nov 26 '18 at 1:19
Arturo Magidin
261k34586906
asked Nov 25 '18 at 12:44
3nondatur3nondatur
385111
edited Nov 26 '18 at 1:19
Arturo Magidin
261k34586906
edited Nov 26 '18 at 1:19
Arturo Magidin
261k34586906
edited Nov 26 '18 at 1:19
Arturo Magidin
261k34586906
261k34586906
asked Nov 25 '18 at 12:44
3nondatur3nondatur
385111
asked Nov 25 '18 at 12:44
3nondatur3nondatur
385111
asked Nov 25 '18 at 12:44
3nondatur3nondatur
385111
385111
2
$begingroup$
The set $K$ includes cosets of $a$ such that $a^{n-1}equiv-1$ modulo some prime powers and $equiv+1$ modulo some other prime powers. In the set $L$ the same sign must occur for all prime powers dividing $n$ (as per the Chinese Remainder Theorem).
$endgroup$
– Jyrki Lahtonen
Nov 25 '18 at 13:05
1
$begingroup$
This question has large overlap with this other question: answerers and readers would do well to read both.
$endgroup$
– davidlowryduda♦
Nov 27 '18 at 21:49
add a comment |
2
$begingroup$
The set $K$ includes cosets of $a$ such that $a^{n-1}equiv-1$ modulo some prime powers and $equiv+1$ modulo some other prime powers. In the set $L$ the same sign must occur for all prime powers dividing $n$ (as per the Chinese Remainder Theorem).
$endgroup$
– Jyrki Lahtonen
Nov 25 '18 at 13:05
1
$begingroup$
This question has large overlap with this other question: answerers and readers would do well to read both.
$endgroup$
– davidlowryduda♦
Nov 27 '18 at 21:49
2
2
$begingroup$
The set $K$ includes cosets of $a$ such that $a^{n-1}equiv-1$ modulo some prime powers and $equiv+1$ modulo some other prime powers. In the set $L$ the same sign must occur for all prime powers dividing $n$ (as per the Chinese Remainder Theorem).
$endgroup$
– Jyrki Lahtonen
Nov 25 '18 at 13:05
$begingroup$
The set $K$ includes cosets of $a$ such that $a^{n-1}equiv-1$ modulo some prime powers and $equiv+1$ modulo some other prime powers. In the set $L$ the same sign must occur for all prime powers dividing $n$ (as per the Chinese Remainder Theorem).
$endgroup$
– Jyrki Lahtonen
Nov 25 '18 at 13:05
1
1
$begingroup$
This question has large overlap with this other question: answerers and readers would do well to read both.
$endgroup$
– davidlowryduda♦
Nov 27 '18 at 21:49
$begingroup$
This question has large overlap with this other question: answerers and readers would do well to read both.
$endgroup$
– davidlowryduda♦
Nov 27 '18 at 21:49
add a comment |
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$begingroup$
The set $K$ includes cosets of $a$ such that $a^{n-1}equiv-1$ modulo some prime powers and $equiv+1$ modulo some other prime powers. In the set $L$ the same sign must occur for all prime powers dividing $n$ (as per the Chinese Remainder Theorem).
$endgroup$
– Jyrki Lahtonen
Nov 25 '18 at 13:05
1
$begingroup$
This question has large overlap with this other question: answerers and readers would do well to read both.
$endgroup$
– davidlowryduda♦
Nov 27 '18 at 21:49