If there is an $x_0$ of order two and $operatorname{ord}f(x_0)^{k}=2$, $k={0,…,p-1}$, then $x^2=e$.Order of...

It grows, but water kills it

awk assign to multiple variables at once

Doesn't the system of the Supreme Court oppose justice?

Why is it that I can sometimes guess the next note?

Why should universal income be universal?

Were Persian-Median kings illiterate?

How many arrows is an archer expected to fire by the end of the Tyranny of Dragons pair of adventures?

How to make money from a browser who sees 5 seconds into the future of any web page?

How to preserve electronics (computers, iPads and phones) for hundreds of years

Why can't the Brexit deadlock in the UK parliament be solved with a plurality vote?

What (the heck) is a Super Worm Equinox Moon?

How to get directions in deep space?

What features enable the Su-25 Frogfoot to operate with such a wide variety of fuels?

Why does Carol not get rid of the Kree symbol on her suit when she changes its colours?

What does Apple's new App Store requirement mean

When were female captains banned from Starfleet?

Has the laser at Magurele, Romania reached a tenth of the Sun's power?

I found an audio circuit and I built it just fine, but I find it a bit too quiet. How do I amplify the output so that it is a bit louder?

Has any country ever had 2 former presidents in jail simultaneously?

Why is the "ls" command showing permissions of files in a FAT32 partition?

Does "he squandered his car on drink" sound natural?

How to explain what's wrong with this application of the chain rule?

What is Cash Advance APR?

Why is so much work done on numerical verification of the Riemann Hypothesis?



If there is an $x_0$ of order two and $operatorname{ord}f(x_0)^{k}=2$, $k={0,…,p-1}$, then $x^2=e$.


Order of $b$ when $ba=ab^3$ and $operatorname{ord}(a)=4$.If $G$ is a finite group, then $operatorname{ord}S(a)=operatorname{ord}(G)/operatorname{Ord}(C(a))$Relationship between $operatorname{ord}(ab), operatorname{ord}(a)$, and $operatorname{ord}(b)$Natural action of $operatorname{Aut}(G)$ on sets of subgroups of $G$ of same order is transitive.$operatorname{ord}(a^k)$ is a divisor of $operatorname{ord}(a)$Proof of $prod_limits{i=1}^{n}G_i$ is cyclic iff $gcd(operatorname{ord} G_i,operatorname{ord} G_j)=1$$gcd(operatorname{ord} G_i,operatorname{ord} G_j)=1$ if $ineq j$ and cyclic groups ex$DeclareMathOperator{ord}{ord}gcd(ord(a),ord(b)) = 1$ then $ord(ab)=ord(a)cdot ord(b)$ where $a,b$ do not commuteThe order of $(g,h)$ equals the least common multiple of ord(g) and ord(h)?Automorphisms of Quaternion Group $Q$. Prove that $operatorname{Aut}(Q)$ contains no element of order $6$, and so $operatorname{Aut}(Q)cong S_4 $













0












$begingroup$


I have a group $G={e,f(x),dots ,f(x)^{p-1}}$ and $fin operatorname{Aut}(G)$ and as $|G|$ is even, I read the following property:



If there is an $x_0$ of order two and $ operatorname{ord}f(x_0)^{k}=2$, $k=0,dots ,p-1$, then $x^2=e$ for any $x in G$.



I am not sure I understand it as I should.










share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you know about generators?
    $endgroup$
    – Shaun
    Mar 13 at 22:01
















0












$begingroup$


I have a group $G={e,f(x),dots ,f(x)^{p-1}}$ and $fin operatorname{Aut}(G)$ and as $|G|$ is even, I read the following property:



If there is an $x_0$ of order two and $ operatorname{ord}f(x_0)^{k}=2$, $k=0,dots ,p-1$, then $x^2=e$ for any $x in G$.



I am not sure I understand it as I should.










share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you know about generators?
    $endgroup$
    – Shaun
    Mar 13 at 22:01














0












0








0





$begingroup$


I have a group $G={e,f(x),dots ,f(x)^{p-1}}$ and $fin operatorname{Aut}(G)$ and as $|G|$ is even, I read the following property:



If there is an $x_0$ of order two and $ operatorname{ord}f(x_0)^{k}=2$, $k=0,dots ,p-1$, then $x^2=e$ for any $x in G$.



I am not sure I understand it as I should.










share|cite|improve this question











$endgroup$




I have a group $G={e,f(x),dots ,f(x)^{p-1}}$ and $fin operatorname{Aut}(G)$ and as $|G|$ is even, I read the following property:



If there is an $x_0$ of order two and $ operatorname{ord}f(x_0)^{k}=2$, $k=0,dots ,p-1$, then $x^2=e$ for any $x in G$.



I am not sure I understand it as I should.







abstract-algebra group-theory finite-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 13 at 22:00









Shaun

9,690113684




9,690113684










asked Mar 13 at 8:41







user651754



















  • $begingroup$
    What do you know about generators?
    $endgroup$
    – Shaun
    Mar 13 at 22:01


















  • $begingroup$
    What do you know about generators?
    $endgroup$
    – Shaun
    Mar 13 at 22:01
















$begingroup$
What do you know about generators?
$endgroup$
– Shaun
Mar 13 at 22:01




$begingroup$
What do you know about generators?
$endgroup$
– Shaun
Mar 13 at 22:01










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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3146282%2fif-there-is-an-x-0-of-order-two-and-operatornameordfx-0k-2-k-0%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
















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3146282%2fif-there-is-an-x-0-of-order-two-and-operatornameordfx-0k-2-k-0%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Nidaros erkebispedøme

Birsay

Was Woodrow Wilson really a Liberal?Was World War I a war of liberals against authoritarians?Founding Fathers...