Prove that $g mapsto g^a$ is a permutation
$begingroup$
"Prove that $g mapsto g^a$ for any $a in mathbb{Z}$ that is prime to the group order is a Permutation on any finite group G."
Can somebody please help me?
My idea was to show that $g,h in G$ with $g^a=h^a$ implies $g=h$, but I struggle to show that.
abstract-algebra group-theory permutations
edited Dec 13 '18 at 6:06
Martin Sleziak
44.7k10119272
asked Dec 12 '18 at 13:21
FabianSchneiderFabianSchneider
706
$endgroup$
add a comment |
$begingroup$
"Prove that $g mapsto g^a$ for any $a in mathbb{Z}$ that is prime to the group order is a Permutation on any finite group G."
Can somebody please help me?
My idea was to show that $g,h in G$ with $g^a=h^a$ implies $g=h$, but I struggle to show that.
abstract-algebra group-theory permutations
edited Dec 13 '18 at 6:06
Martin Sleziak
44.7k10119272
asked Dec 12 '18 at 13:21
FabianSchneiderFabianSchneider
706
$endgroup$
4
$begingroup$
You can't prove it because it's not true. What about squaring in the two element group? Are you sure you have all the hypotheses?
$endgroup$
– Ethan Bolker
Dec 12 '18 at 13:23
$begingroup$
You´re sure it´s not true? I have to show that this map is bijective.
$endgroup$
– FabianSchneider
Dec 12 '18 at 13:30
1
$begingroup$
Wait...missed one argument: a has to be prime to the group order!
$endgroup$
– FabianSchneider
Dec 12 '18 at 13:33
add a comment |
$begingroup$
"Prove that $g mapsto g^a$ for any $a in mathbb{Z}$ that is prime to the group order is a Permutation on any finite group G."
Can somebody please help me?
My idea was to show that $g,h in G$ with $g^a=h^a$ implies $g=h$, but I struggle to show that.
abstract-algebra group-theory permutations
edited Dec 13 '18 at 6:06
Martin Sleziak
44.7k10119272
asked Dec 12 '18 at 13:21
FabianSchneiderFabianSchneider
706
$endgroup$
"Prove that $g mapsto g^a$ for any $a in mathbb{Z}$ that is prime to the group order is a Permutation on any finite group G."
Can somebody please help me?
My idea was to show that $g,h in G$ with $g^a=h^a$ implies $g=h$, but I struggle to show that.
abstract-algebra group-theory permutations
abstract-algebra group-theory permutations
edited Dec 13 '18 at 6:06
Martin Sleziak
44.7k10119272
asked Dec 12 '18 at 13:21
FabianSchneiderFabianSchneider
706
edited Dec 13 '18 at 6:06
Martin Sleziak
44.7k10119272
asked Dec 12 '18 at 13:21
FabianSchneiderFabianSchneider
706
edited Dec 13 '18 at 6:06
Martin Sleziak
44.7k10119272
edited Dec 13 '18 at 6:06
Martin Sleziak
44.7k10119272
edited Dec 13 '18 at 6:06
Martin Sleziak
44.7k10119272
44.7k10119272
asked Dec 12 '18 at 13:21
FabianSchneiderFabianSchneider
706
asked Dec 12 '18 at 13:21
FabianSchneiderFabianSchneider
706
asked Dec 12 '18 at 13:21
FabianSchneiderFabianSchneider
706
706
4
$begingroup$
You can't prove it because it's not true. What about squaring in the two element group? Are you sure you have all the hypotheses?
$endgroup$
– Ethan Bolker
Dec 12 '18 at 13:23
$begingroup$
You´re sure it´s not true? I have to show that this map is bijective.
$endgroup$
– FabianSchneider
Dec 12 '18 at 13:30
1
$begingroup$
Wait...missed one argument: a has to be prime to the group order!
$endgroup$
– FabianSchneider
Dec 12 '18 at 13:33
add a comment |
4
$begingroup$
You can't prove it because it's not true. What about squaring in the two element group? Are you sure you have all the hypotheses?
$endgroup$
– Ethan Bolker
Dec 12 '18 at 13:23
$begingroup$
You´re sure it´s not true? I have to show that this map is bijective.
$endgroup$
– FabianSchneider
Dec 12 '18 at 13:30
1
$begingroup$
Wait...missed one argument: a has to be prime to the group order!
$endgroup$
– FabianSchneider
Dec 12 '18 at 13:33
4
4
$begingroup$
You can't prove it because it's not true. What about squaring in the two element group? Are you sure you have all the hypotheses?
$endgroup$
– Ethan Bolker
Dec 12 '18 at 13:23
$begingroup$
You can't prove it because it's not true. What about squaring in the two element group? Are you sure you have all the hypotheses?
$endgroup$
– Ethan Bolker
Dec 12 '18 at 13:23
$begingroup$
You´re sure it´s not true? I have to show that this map is bijective.
$endgroup$
– FabianSchneider
Dec 12 '18 at 13:30
$begingroup$
You´re sure it´s not true? I have to show that this map is bijective.
$endgroup$
– FabianSchneider
Dec 12 '18 at 13:30
1
1
$begingroup$
Wait...missed one argument: a has to be prime to the group order!
$endgroup$
– FabianSchneider
Dec 12 '18 at 13:33
$begingroup$
Wait...missed one argument: a has to be prime to the group order!
$endgroup$
– FabianSchneider
Dec 12 '18 at 13:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $n$ be the order of $G$. SInce $(n,a)=1$, by Bezout's lemma there are $s,tin mathbb{Z}$ such that $sn+ta=1$.
Notice the map $hmapsto h^{t}$ is the inverse of the provided map in the problem. To see that notice that
begin{align*}
(g^a)^t&=g^{at}\
&=g^{at}g^{sn} quadquadtext{(since $g^n=1$)}\
&=g^{at+sn}\
&=g.
end{align*}
The other way around compostion give you similar result.
answered Dec 12 '18 at 13:46
user9077user9077
1,239612
$endgroup$
$begingroup$
THANK YOU very much!
$endgroup$
– FabianSchneider
Dec 12 '18 at 20:33
add a comment |
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%2f3036677%2fprove-that-g-mapsto-ga-is-a-permutation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $n$ be the order of $G$. SInce $(n,a)=1$, by Bezout's lemma there are $s,tin mathbb{Z}$ such that $sn+ta=1$.
Notice the map $hmapsto h^{t}$ is the inverse of the provided map in the problem. To see that notice that
begin{align*}
(g^a)^t&=g^{at}\
&=g^{at}g^{sn} quadquadtext{(since $g^n=1$)}\
&=g^{at+sn}\
&=g.
end{align*}
The other way around compostion give you similar result.
answered Dec 12 '18 at 13:46
user9077user9077
1,239612
$endgroup$
$begingroup$
THANK YOU very much!
$endgroup$
– FabianSchneider
Dec 12 '18 at 20:33
add a comment |
$begingroup$
Let $n$ be the order of $G$. SInce $(n,a)=1$, by Bezout's lemma there are $s,tin mathbb{Z}$ such that $sn+ta=1$.
Notice the map $hmapsto h^{t}$ is the inverse of the provided map in the problem. To see that notice that
begin{align*}
(g^a)^t&=g^{at}\
&=g^{at}g^{sn} quadquadtext{(since $g^n=1$)}\
&=g^{at+sn}\
&=g.
end{align*}
The other way around compostion give you similar result.
answered Dec 12 '18 at 13:46
user9077user9077
1,239612
$endgroup$
$begingroup$
THANK YOU very much!
$endgroup$
– FabianSchneider
Dec 12 '18 at 20:33
add a comment |
$begingroup$
Let $n$ be the order of $G$. SInce $(n,a)=1$, by Bezout's lemma there are $s,tin mathbb{Z}$ such that $sn+ta=1$.
Notice the map $hmapsto h^{t}$ is the inverse of the provided map in the problem. To see that notice that
begin{align*}
(g^a)^t&=g^{at}\
&=g^{at}g^{sn} quadquadtext{(since $g^n=1$)}\
&=g^{at+sn}\
&=g.
end{align*}
The other way around compostion give you similar result.
answered Dec 12 '18 at 13:46
user9077user9077
1,239612
$endgroup$
Let $n$ be the order of $G$. SInce $(n,a)=1$, by Bezout's lemma there are $s,tin mathbb{Z}$ such that $sn+ta=1$.
Notice the map $hmapsto h^{t}$ is the inverse of the provided map in the problem. To see that notice that
begin{align*}
(g^a)^t&=g^{at}\
&=g^{at}g^{sn} quadquadtext{(since $g^n=1$)}\
&=g^{at+sn}\
&=g.
end{align*}
The other way around compostion give you similar result.
answered Dec 12 '18 at 13:46
user9077user9077
1,239612
answered Dec 12 '18 at 13:46
user9077user9077
1,239612
answered Dec 12 '18 at 13:46
user9077user9077
1,239612
answered Dec 12 '18 at 13:46
user9077user9077
1,239612
1,239612
$begingroup$
THANK YOU very much!
$endgroup$
– FabianSchneider
Dec 12 '18 at 20:33
add a comment |
$begingroup$
THANK YOU very much!
$endgroup$
– FabianSchneider
Dec 12 '18 at 20:33
$begingroup$
THANK YOU very much!
$endgroup$
– FabianSchneider
Dec 12 '18 at 20:33
$begingroup$
THANK YOU very much!
$endgroup$
– FabianSchneider
Dec 12 '18 at 20:33
add a comment |
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%2f3036677%2fprove-that-g-mapsto-ga-is-a-permutation%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
4
$begingroup$
You can't prove it because it's not true. What about squaring in the two element group? Are you sure you have all the hypotheses?
$endgroup$
– Ethan Bolker
Dec 12 '18 at 13:23
$begingroup$
You´re sure it´s not true? I have to show that this map is bijective.
$endgroup$
– FabianSchneider
Dec 12 '18 at 13:30
1
$begingroup$
Wait...missed one argument: a has to be prime to the group order!
$endgroup$
– FabianSchneider
Dec 12 '18 at 13:33