Showing that the index of two groups is a power of $2$
I am trying to understand a proof (from a book) that there are at most $(n-1)/4$ non-wittnesses against the primality of $n$ in the Miller-Rabin algorithm that are coprime to $n$.
We set the prime factorisation of $n$ to $n = prod_{p | n} p^{e(p)}$.
In one passage the author considers the following two subgroups of $mathbb{Z}_n^{times}$
$K = { a + nmathbb{Z} : gcd(a,n) = 1 text{ and } a^{n-1} equiv pm 1 text{ mod } p^{e(p)} text{ for all primes $p$ such that } p | n}$
$M = { a + nmathbb{Z} : gcd(a,n) = 1 text{ and } a^{m} equiv 1 text{ mod } n}$.
He observes that $M le K$ which is clear. Then he says that the index $(K : M)$ is a power of $2$ since the square of each element of $K$ is an element in $M$. I can not follow this argumentation, could you please explain that to me?
abstract-algebra group-theory number-theory
asked Nov 24 at 11:13
3nondatur
390111
add a comment |
I am trying to understand a proof (from a book) that there are at most $(n-1)/4$ non-wittnesses against the primality of $n$ in the Miller-Rabin algorithm that are coprime to $n$.
We set the prime factorisation of $n$ to $n = prod_{p | n} p^{e(p)}$.
In one passage the author considers the following two subgroups of $mathbb{Z}_n^{times}$
$K = { a + nmathbb{Z} : gcd(a,n) = 1 text{ and } a^{n-1} equiv pm 1 text{ mod } p^{e(p)} text{ for all primes $p$ such that } p | n}$
$M = { a + nmathbb{Z} : gcd(a,n) = 1 text{ and } a^{m} equiv 1 text{ mod } n}$.
He observes that $M le K$ which is clear. Then he says that the index $(K : M)$ is a power of $2$ since the square of each element of $K$ is an element in $M$. I can not follow this argumentation, could you please explain that to me?
abstract-algebra group-theory number-theory
asked Nov 24 at 11:13
3nondatur
390111
add a comment |
I am trying to understand a proof (from a book) that there are at most $(n-1)/4$ non-wittnesses against the primality of $n$ in the Miller-Rabin algorithm that are coprime to $n$.
We set the prime factorisation of $n$ to $n = prod_{p | n} p^{e(p)}$.
In one passage the author considers the following two subgroups of $mathbb{Z}_n^{times}$
$K = { a + nmathbb{Z} : gcd(a,n) = 1 text{ and } a^{n-1} equiv pm 1 text{ mod } p^{e(p)} text{ for all primes $p$ such that } p | n}$
$M = { a + nmathbb{Z} : gcd(a,n) = 1 text{ and } a^{m} equiv 1 text{ mod } n}$.
He observes that $M le K$ which is clear. Then he says that the index $(K : M)$ is a power of $2$ since the square of each element of $K$ is an element in $M$. I can not follow this argumentation, could you please explain that to me?
abstract-algebra group-theory number-theory
asked Nov 24 at 11:13
3nondatur
390111
I am trying to understand a proof (from a book) that there are at most $(n-1)/4$ non-wittnesses against the primality of $n$ in the Miller-Rabin algorithm that are coprime to $n$.
We set the prime factorisation of $n$ to $n = prod_{p | n} p^{e(p)}$.
In one passage the author considers the following two subgroups of $mathbb{Z}_n^{times}$
$K = { a + nmathbb{Z} : gcd(a,n) = 1 text{ and } a^{n-1} equiv pm 1 text{ mod } p^{e(p)} text{ for all primes $p$ such that } p | n}$
$M = { a + nmathbb{Z} : gcd(a,n) = 1 text{ and } a^{m} equiv 1 text{ mod } n}$.
He observes that $M le K$ which is clear. Then he says that the index $(K : M)$ is a power of $2$ since the square of each element of $K$ is an element in $M$. I can not follow this argumentation, could you please explain that to me?
abstract-algebra group-theory number-theory
abstract-algebra group-theory number-theory
asked Nov 24 at 11:13
3nondatur
390111
asked Nov 24 at 11:13
3nondatur
390111
asked Nov 24 at 11:13
3nondatur
390111
asked Nov 24 at 11:13
3nondatur
390111
asked Nov 24 at 11:13
3nondatur
390111
390111
add a comment |
add a comment |
1 Answer
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Let $G$ be a finite Abelian group, written multiplicatively, and $H$ a subgroup.
If $a^2in H$ for all $ain G$, then $overline a^2$ is the identity element
in the quotient group $G/H$ (where $overline a$ is the image of $ain G/H$).
The order of $overline a$ in $G/H$ is either $1$ or $2$.
If the order of $G/H$ is not a power of two, it has an odd prime factor $p$.
By Cauchy's theorem, $G/H$ would then have an element of order $p$; this gives
a contradiction.
answered Nov 24 at 11:20
Lord Shark the Unknown
100k958131
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1 Answer
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1 Answer
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Let $G$ be a finite Abelian group, written multiplicatively, and $H$ a subgroup.
If $a^2in H$ for all $ain G$, then $overline a^2$ is the identity element
in the quotient group $G/H$ (where $overline a$ is the image of $ain G/H$).
The order of $overline a$ in $G/H$ is either $1$ or $2$.
If the order of $G/H$ is not a power of two, it has an odd prime factor $p$.
By Cauchy's theorem, $G/H$ would then have an element of order $p$; this gives
a contradiction.
answered Nov 24 at 11:20
Lord Shark the Unknown
100k958131
add a comment |
Let $G$ be a finite Abelian group, written multiplicatively, and $H$ a subgroup.
If $a^2in H$ for all $ain G$, then $overline a^2$ is the identity element
in the quotient group $G/H$ (where $overline a$ is the image of $ain G/H$).
The order of $overline a$ in $G/H$ is either $1$ or $2$.
If the order of $G/H$ is not a power of two, it has an odd prime factor $p$.
By Cauchy's theorem, $G/H$ would then have an element of order $p$; this gives
a contradiction.
answered Nov 24 at 11:20
Lord Shark the Unknown
100k958131
add a comment |
Let $G$ be a finite Abelian group, written multiplicatively, and $H$ a subgroup.
If $a^2in H$ for all $ain G$, then $overline a^2$ is the identity element
in the quotient group $G/H$ (where $overline a$ is the image of $ain G/H$).
The order of $overline a$ in $G/H$ is either $1$ or $2$.
If the order of $G/H$ is not a power of two, it has an odd prime factor $p$.
By Cauchy's theorem, $G/H$ would then have an element of order $p$; this gives
a contradiction.
answered Nov 24 at 11:20
Lord Shark the Unknown
100k958131
Let $G$ be a finite Abelian group, written multiplicatively, and $H$ a subgroup.
If $a^2in H$ for all $ain G$, then $overline a^2$ is the identity element
in the quotient group $G/H$ (where $overline a$ is the image of $ain G/H$).
The order of $overline a$ in $G/H$ is either $1$ or $2$.
If the order of $G/H$ is not a power of two, it has an odd prime factor $p$.
By Cauchy's theorem, $G/H$ would then have an element of order $p$; this gives
a contradiction.
answered Nov 24 at 11:20
Lord Shark the Unknown
100k958131
answered Nov 24 at 11:20
Lord Shark the Unknown
100k958131
answered Nov 24 at 11:20
Lord Shark the Unknown
100k958131
answered Nov 24 at 11:20
Lord Shark the Unknown
100k958131
100k958131
add a comment |
add a comment |
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