On a certain bijection of the multiplicative group modulo prime $p=2^k+1$.
$begingroup$
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:
- 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?
- 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
edited Dec 21 '18 at 17:18
Shaun
9,814113684
asked Dec 18 '18 at 19:04
user75619user75619
344113
$endgroup$
add a comment |
$begingroup$
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:
- 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?
- 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
edited Dec 21 '18 at 17:18
Shaun
9,814113684
asked Dec 18 '18 at 19:04
user75619user75619
344113
$endgroup$
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
add a comment |
$begingroup$
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:
- 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?
- 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
edited Dec 21 '18 at 17:18
Shaun
9,814113684
asked Dec 18 '18 at 19:04
user75619user75619
344113
$endgroup$
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:
- 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?
- 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
edited Dec 21 '18 at 17:18
Shaun
9,814113684
asked Dec 18 '18 at 19:04
user75619user75619
344113
edited Dec 21 '18 at 17:18
Shaun
9,814113684
asked Dec 18 '18 at 19:04
user75619user75619
344113
edited Dec 21 '18 at 17:18
Shaun
9,814113684
edited Dec 21 '18 at 17:18
Shaun
9,814113684
edited Dec 21 '18 at 17:18
Shaun
9,814113684
9,814113684
asked Dec 18 '18 at 19:04
user75619user75619
344113
asked Dec 18 '18 at 19:04
user75619user75619
344113
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
add a comment |
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
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045562%2fon-a-certain-bijection-of-the-multiplicative-group-modulo-prime-p-2k1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045562%2fon-a-certain-bijection-of-the-multiplicative-group-modulo-prime-p-2k1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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