On a certain bijection of the multiplicative group modulo prime $p=2^k+1$.












2












$begingroup$


Let:



$G={1,ldots,p-1}$ be the multiplicative group of integers modulo prime $p=2^k+1$;



$v_2(a)$ be the exponent of $2$ in the prime factorization of $a$;



$mu(a)=max{jin[0,k] | x^{2^j} equiv a}$ for some $xinmathbb{G}$;



$A_i = {ainmathbb{G} | v_2(a)=i}$;



$B_i = {ainmathbb{G} | mu(a)=i}$.



Now, couple questions:





  1. Prove that $a$-powers of the elements of $G$ are exactly $2^{v_2(a)}$-powers of the elements of $G$.




I prove this by introducing two sub-groups $G_1={x^a,xin G}$ and $G_2={x^{2^{v_2(a)}},xin G}$ and showing that $G_1subseteq G_2$ and $G_2subseteq G_1$. But could there be less laborious and more intuitive proof, which is often the case with simpler problems in number theory?





  1. Let $g$ be a primitive root modulo $p$ and $f$ be a map from $G$ to itself that sends $x$ to $g^x$ modulo $p$. Prove that $f$ is a bijection that sends $A_i$ to $B_i$ for each $iin{0,ldots,k}$.




This one I can't formally prove at all...






group-theory number-theory prime-numbers finite-groups modular-arithmetic







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It would really help to clean up the notation in this post: $a$ is being used for multiple things simultaneously; and the map $amapsto g^a$ doesn't make sense for an abstract group to itself—perhaps it should be a map from ${1,dots,p-1}$ to $mathbb G$.
    $endgroup$
    – Greg Martin
    Dec 18 '18 at 19:23






  • 1




    $begingroup$
    As for your first question, here's a lemma: if $b$ is an odd number, prove that the map $xmapsto x^b$ is a bijection on $G$, and a bijection on every subgroup of $G$ when restricted to that subgroup. Then apply with $b$ chosen so that $a=2^{nu_2(a)}b$.
    $endgroup$
    – Greg Martin
    Dec 18 '18 at 19:25










  • $begingroup$
    @GregMartin By $G$ I actually mean ${1,ldots,p-1}$. But I'll make it explicit.
    $endgroup$
    – user75619
    Dec 18 '18 at 19:29
















2












$begingroup$


Let:



$G={1,ldots,p-1}$ be the multiplicative group of integers modulo prime $p=2^k+1$;



$v_2(a)$ be the exponent of $2$ in the prime factorization of $a$;



$mu(a)=max{jin[0,k] | x^{2^j} equiv a}$ for some $xinmathbb{G}$;



$A_i = {ainmathbb{G} | v_2(a)=i}$;



$B_i = {ainmathbb{G} | mu(a)=i}$.



Now, couple questions:





  1. Prove that $a$-powers of the elements of $G$ are exactly $2^{v_2(a)}$-powers of the elements of $G$.




I prove this by introducing two sub-groups $G_1={x^a,xin G}$ and $G_2={x^{2^{v_2(a)}},xin G}$ and showing that $G_1subseteq G_2$ and $G_2subseteq G_1$. But could there be less laborious and more intuitive proof, which is often the case with simpler problems in number theory?





  1. Let $g$ be a primitive root modulo $p$ and $f$ be a map from $G$ to itself that sends $x$ to $g^x$ modulo $p$. Prove that $f$ is a bijection that sends $A_i$ to $B_i$ for each $iin{0,ldots,k}$.




This one I can't formally prove at all...






group-theory number-theory prime-numbers finite-groups modular-arithmetic







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It would really help to clean up the notation in this post: $a$ is being used for multiple things simultaneously; and the map $amapsto g^a$ doesn't make sense for an abstract group to itself—perhaps it should be a map from ${1,dots,p-1}$ to $mathbb G$.
    $endgroup$
    – Greg Martin
    Dec 18 '18 at 19:23






  • 1




    $begingroup$
    As for your first question, here's a lemma: if $b$ is an odd number, prove that the map $xmapsto x^b$ is a bijection on $G$, and a bijection on every subgroup of $G$ when restricted to that subgroup. Then apply with $b$ chosen so that $a=2^{nu_2(a)}b$.
    $endgroup$
    – Greg Martin
    Dec 18 '18 at 19:25










  • $begingroup$
    @GregMartin By $G$ I actually mean ${1,ldots,p-1}$. But I'll make it explicit.
    $endgroup$
    – user75619
    Dec 18 '18 at 19:29














2












2








2


2



$begingroup$


Let:



$G={1,ldots,p-1}$ be the multiplicative group of integers modulo prime $p=2^k+1$;



$v_2(a)$ be the exponent of $2$ in the prime factorization of $a$;



$mu(a)=max{jin[0,k] | x^{2^j} equiv a}$ for some $xinmathbb{G}$;



$A_i = {ainmathbb{G} | v_2(a)=i}$;



$B_i = {ainmathbb{G} | mu(a)=i}$.



Now, couple questions:





  1. Prove that $a$-powers of the elements of $G$ are exactly $2^{v_2(a)}$-powers of the elements of $G$.




I prove this by introducing two sub-groups $G_1={x^a,xin G}$ and $G_2={x^{2^{v_2(a)}},xin G}$ and showing that $G_1subseteq G_2$ and $G_2subseteq G_1$. But could there be less laborious and more intuitive proof, which is often the case with simpler problems in number theory?





  1. Let $g$ be a primitive root modulo $p$ and $f$ be a map from $G$ to itself that sends $x$ to $g^x$ modulo $p$. Prove that $f$ is a bijection that sends $A_i$ to $B_i$ for each $iin{0,ldots,k}$.




This one I can't formally prove at all...






group-theory number-theory prime-numbers finite-groups modular-arithmetic







share|cite|improve this question











$endgroup$




Let:



$G={1,ldots,p-1}$ be the multiplicative group of integers modulo prime $p=2^k+1$;



$v_2(a)$ be the exponent of $2$ in the prime factorization of $a$;



$mu(a)=max{jin[0,k] | x^{2^j} equiv a}$ for some $xinmathbb{G}$;



$A_i = {ainmathbb{G} | v_2(a)=i}$;



$B_i = {ainmathbb{G} | mu(a)=i}$.



Now, couple questions:





  1. Prove that $a$-powers of the elements of $G$ are exactly $2^{v_2(a)}$-powers of the elements of $G$.




I prove this by introducing two sub-groups $G_1={x^a,xin G}$ and $G_2={x^{2^{v_2(a)}},xin G}$ and showing that $G_1subseteq G_2$ and $G_2subseteq G_1$. But could there be less laborious and more intuitive proof, which is often the case with simpler problems in number theory?





  1. Let $g$ be a primitive root modulo $p$ and $f$ be a map from $G$ to itself that sends $x$ to $g^x$ modulo $p$. Prove that $f$ is a bijection that sends $A_i$ to $B_i$ for each $iin{0,ldots,k}$.




This one I can't formally prove at all...






group-theory number-theory prime-numbers finite-groups modular-arithmetic




group-theory number-theory prime-numbers finite-groups modular-arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 17:18









Shaun

9,814113684




9,814113684










asked Dec 18 '18 at 19:04









user75619user75619

344113




344113








  • 1




    $begingroup$
    It would really help to clean up the notation in this post: $a$ is being used for multiple things simultaneously; and the map $amapsto g^a$ doesn't make sense for an abstract group to itself—perhaps it should be a map from ${1,dots,p-1}$ to $mathbb G$.
    $endgroup$
    – Greg Martin
    Dec 18 '18 at 19:23






  • 1




    $begingroup$
    As for your first question, here's a lemma: if $b$ is an odd number, prove that the map $xmapsto x^b$ is a bijection on $G$, and a bijection on every subgroup of $G$ when restricted to that subgroup. Then apply with $b$ chosen so that $a=2^{nu_2(a)}b$.
    $endgroup$
    – Greg Martin
    Dec 18 '18 at 19:25










  • $begingroup$
    @GregMartin By $G$ I actually mean ${1,ldots,p-1}$. But I'll make it explicit.
    $endgroup$
    – user75619
    Dec 18 '18 at 19:29














  • 1




    $begingroup$
    It would really help to clean up the notation in this post: $a$ is being used for multiple things simultaneously; and the map $amapsto g^a$ doesn't make sense for an abstract group to itself—perhaps it should be a map from ${1,dots,p-1}$ to $mathbb G$.
    $endgroup$
    – Greg Martin
    Dec 18 '18 at 19:23






  • 1




    $begingroup$
    As for your first question, here's a lemma: if $b$ is an odd number, prove that the map $xmapsto x^b$ is a bijection on $G$, and a bijection on every subgroup of $G$ when restricted to that subgroup. Then apply with $b$ chosen so that $a=2^{nu_2(a)}b$.
    $endgroup$
    – Greg Martin
    Dec 18 '18 at 19:25










  • $begingroup$
    @GregMartin By $G$ I actually mean ${1,ldots,p-1}$. But I'll make it explicit.
    $endgroup$
    – user75619
    Dec 18 '18 at 19:29








1




1




$begingroup$
It would really help to clean up the notation in this post: $a$ is being used for multiple things simultaneously; and the map $amapsto g^a$ doesn't make sense for an abstract group to itself—perhaps it should be a map from ${1,dots,p-1}$ to $mathbb G$.
$endgroup$
– Greg Martin
Dec 18 '18 at 19:23




$begingroup$
It would really help to clean up the notation in this post: $a$ is being used for multiple things simultaneously; and the map $amapsto g^a$ doesn't make sense for an abstract group to itself—perhaps it should be a map from ${1,dots,p-1}$ to $mathbb G$.
$endgroup$
– Greg Martin
Dec 18 '18 at 19:23




1




1




$begingroup$
As for your first question, here's a lemma: if $b$ is an odd number, prove that the map $xmapsto x^b$ is a bijection on $G$, and a bijection on every subgroup of $G$ when restricted to that subgroup. Then apply with $b$ chosen so that $a=2^{nu_2(a)}b$.
$endgroup$
– Greg Martin
Dec 18 '18 at 19:25




$begingroup$
As for your first question, here's a lemma: if $b$ is an odd number, prove that the map $xmapsto x^b$ is a bijection on $G$, and a bijection on every subgroup of $G$ when restricted to that subgroup. Then apply with $b$ chosen so that $a=2^{nu_2(a)}b$.
$endgroup$
– Greg Martin
Dec 18 '18 at 19:25












$begingroup$
@GregMartin By $G$ I actually mean ${1,ldots,p-1}$. But I'll make it explicit.
$endgroup$
– user75619
Dec 18 '18 at 19:29




$begingroup$
@GregMartin By $G$ I actually mean ${1,ldots,p-1}$. But I'll make it explicit.
$endgroup$
– user75619
Dec 18 '18 at 19:29










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