Proof of Cauchy's theorem for finite groups in Dummit and Foote












1














The proof in Dummit & Foote is the familiar induction proof.
$require{amssymb}$




$G$ is an abelian group. Let $N = langle x rangle$ for some $x in G$. Since $G$ is abelian, $N trianglelefteq G$. We use induction to conclude that $G , / , N$ has an element $bar{y} = yN$ of order $p$, where $p$ is prime, since $|G , / , N| < |G|$ and $p$ divides $|G|$ but $p$ does not divide $|N|$.




My question is, if $|yN| = p$ and all cosets have the same order, why can't we conclude $|yN| = |N| = |x| = p$?



Or, more accurately, how can we say $p$ does not divide |$N$|, then show that $p = |yN|$. Isn't that a contradiction?






abstract-algebra group-theory proof-explanation abelian-groups







share|cite|improve this question




















  • 1




    There's an ambiguity here. "an element $overline{y} = yN$ of order $p$" means (in this context at least), that $(overline{y})^p = overline{1}$, not that $|yN| = p$.
    – Bungo
    Nov 30 at 17:31








  • 1




    By the way, the proof seems slightly wrong as stated. If $x$ itself is of order $p$ then we're done (and $G/langle x rangle$ need not have an element of order $p$). Otherwise $G/langle x rangle$ has an element of order $p$ and we can proceed inductively.
    – Bungo
    Nov 30 at 17:35












  • $x$ is not of order $p$, since $p$ does not divide $|N| = |langle x rangle|$.
    – Ujkan Sulejmani
    Nov 30 at 17:36












  • Why doesn't $p$ divide $|N| = |langle x rangle|$? If $x$ is an arbitrarily chosen element of $G$ then it might. That's why I'm saying that the proof is slightly misstated.
    – Bungo
    Nov 30 at 17:38






  • 1




    Ah, OK. Thanks for the context. By the way, the theorem is true for nonabelian groups as well, but the proof has to be modified somewhat.
    – Bungo
    Nov 30 at 17:40


















1














The proof in Dummit & Foote is the familiar induction proof.
$require{amssymb}$




$G$ is an abelian group. Let $N = langle x rangle$ for some $x in G$. Since $G$ is abelian, $N trianglelefteq G$. We use induction to conclude that $G , / , N$ has an element $bar{y} = yN$ of order $p$, where $p$ is prime, since $|G , / , N| < |G|$ and $p$ divides $|G|$ but $p$ does not divide $|N|$.




My question is, if $|yN| = p$ and all cosets have the same order, why can't we conclude $|yN| = |N| = |x| = p$?



Or, more accurately, how can we say $p$ does not divide |$N$|, then show that $p = |yN|$. Isn't that a contradiction?






abstract-algebra group-theory proof-explanation abelian-groups







share|cite|improve this question




















  • 1




    There's an ambiguity here. "an element $overline{y} = yN$ of order $p$" means (in this context at least), that $(overline{y})^p = overline{1}$, not that $|yN| = p$.
    – Bungo
    Nov 30 at 17:31








  • 1




    By the way, the proof seems slightly wrong as stated. If $x$ itself is of order $p$ then we're done (and $G/langle x rangle$ need not have an element of order $p$). Otherwise $G/langle x rangle$ has an element of order $p$ and we can proceed inductively.
    – Bungo
    Nov 30 at 17:35












  • $x$ is not of order $p$, since $p$ does not divide $|N| = |langle x rangle|$.
    – Ujkan Sulejmani
    Nov 30 at 17:36












  • Why doesn't $p$ divide $|N| = |langle x rangle|$? If $x$ is an arbitrarily chosen element of $G$ then it might. That's why I'm saying that the proof is slightly misstated.
    – Bungo
    Nov 30 at 17:38






  • 1




    Ah, OK. Thanks for the context. By the way, the theorem is true for nonabelian groups as well, but the proof has to be modified somewhat.
    – Bungo
    Nov 30 at 17:40
















1












1








1







The proof in Dummit & Foote is the familiar induction proof.
$require{amssymb}$




$G$ is an abelian group. Let $N = langle x rangle$ for some $x in G$. Since $G$ is abelian, $N trianglelefteq G$. We use induction to conclude that $G , / , N$ has an element $bar{y} = yN$ of order $p$, where $p$ is prime, since $|G , / , N| < |G|$ and $p$ divides $|G|$ but $p$ does not divide $|N|$.




My question is, if $|yN| = p$ and all cosets have the same order, why can't we conclude $|yN| = |N| = |x| = p$?



Or, more accurately, how can we say $p$ does not divide |$N$|, then show that $p = |yN|$. Isn't that a contradiction?






abstract-algebra group-theory proof-explanation abelian-groups







share|cite|improve this question















The proof in Dummit & Foote is the familiar induction proof.
$require{amssymb}$




$G$ is an abelian group. Let $N = langle x rangle$ for some $x in G$. Since $G$ is abelian, $N trianglelefteq G$. We use induction to conclude that $G , / , N$ has an element $bar{y} = yN$ of order $p$, where $p$ is prime, since $|G , / , N| < |G|$ and $p$ divides $|G|$ but $p$ does not divide $|N|$.




My question is, if $|yN| = p$ and all cosets have the same order, why can't we conclude $|yN| = |N| = |x| = p$?



Or, more accurately, how can we say $p$ does not divide |$N$|, then show that $p = |yN|$. Isn't that a contradiction?






abstract-algebra group-theory proof-explanation abelian-groups




abstract-algebra group-theory proof-explanation abelian-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 at 17:28









GNUSupporter 8964民主女神 地下教會

12.8k72445




12.8k72445










asked Nov 30 at 17:27









Ujkan Sulejmani

1586




1586








  • 1




    There's an ambiguity here. "an element $overline{y} = yN$ of order $p$" means (in this context at least), that $(overline{y})^p = overline{1}$, not that $|yN| = p$.
    – Bungo
    Nov 30 at 17:31








  • 1




    By the way, the proof seems slightly wrong as stated. If $x$ itself is of order $p$ then we're done (and $G/langle x rangle$ need not have an element of order $p$). Otherwise $G/langle x rangle$ has an element of order $p$ and we can proceed inductively.
    – Bungo
    Nov 30 at 17:35












  • $x$ is not of order $p$, since $p$ does not divide $|N| = |langle x rangle|$.
    – Ujkan Sulejmani
    Nov 30 at 17:36












  • Why doesn't $p$ divide $|N| = |langle x rangle|$? If $x$ is an arbitrarily chosen element of $G$ then it might. That's why I'm saying that the proof is slightly misstated.
    – Bungo
    Nov 30 at 17:38






  • 1




    Ah, OK. Thanks for the context. By the way, the theorem is true for nonabelian groups as well, but the proof has to be modified somewhat.
    – Bungo
    Nov 30 at 17:40
















  • 1




    There's an ambiguity here. "an element $overline{y} = yN$ of order $p$" means (in this context at least), that $(overline{y})^p = overline{1}$, not that $|yN| = p$.
    – Bungo
    Nov 30 at 17:31








  • 1




    By the way, the proof seems slightly wrong as stated. If $x$ itself is of order $p$ then we're done (and $G/langle x rangle$ need not have an element of order $p$). Otherwise $G/langle x rangle$ has an element of order $p$ and we can proceed inductively.
    – Bungo
    Nov 30 at 17:35












  • $x$ is not of order $p$, since $p$ does not divide $|N| = |langle x rangle|$.
    – Ujkan Sulejmani
    Nov 30 at 17:36












  • Why doesn't $p$ divide $|N| = |langle x rangle|$? If $x$ is an arbitrarily chosen element of $G$ then it might. That's why I'm saying that the proof is slightly misstated.
    – Bungo
    Nov 30 at 17:38






  • 1




    Ah, OK. Thanks for the context. By the way, the theorem is true for nonabelian groups as well, but the proof has to be modified somewhat.
    – Bungo
    Nov 30 at 17:40










1




1




There's an ambiguity here. "an element $overline{y} = yN$ of order $p$" means (in this context at least), that $(overline{y})^p = overline{1}$, not that $|yN| = p$.
– Bungo
Nov 30 at 17:31






There's an ambiguity here. "an element $overline{y} = yN$ of order $p$" means (in this context at least), that $(overline{y})^p = overline{1}$, not that $|yN| = p$.
– Bungo
Nov 30 at 17:31






1




1




By the way, the proof seems slightly wrong as stated. If $x$ itself is of order $p$ then we're done (and $G/langle x rangle$ need not have an element of order $p$). Otherwise $G/langle x rangle$ has an element of order $p$ and we can proceed inductively.
– Bungo
Nov 30 at 17:35






By the way, the proof seems slightly wrong as stated. If $x$ itself is of order $p$ then we're done (and $G/langle x rangle$ need not have an element of order $p$). Otherwise $G/langle x rangle$ has an element of order $p$ and we can proceed inductively.
– Bungo
Nov 30 at 17:35














$x$ is not of order $p$, since $p$ does not divide $|N| = |langle x rangle|$.
– Ujkan Sulejmani
Nov 30 at 17:36






$x$ is not of order $p$, since $p$ does not divide $|N| = |langle x rangle|$.
– Ujkan Sulejmani
Nov 30 at 17:36














Why doesn't $p$ divide $|N| = |langle x rangle|$? If $x$ is an arbitrarily chosen element of $G$ then it might. That's why I'm saying that the proof is slightly misstated.
– Bungo
Nov 30 at 17:38




Why doesn't $p$ divide $|N| = |langle x rangle|$? If $x$ is an arbitrarily chosen element of $G$ then it might. That's why I'm saying that the proof is slightly misstated.
– Bungo
Nov 30 at 17:38




1




1




Ah, OK. Thanks for the context. By the way, the theorem is true for nonabelian groups as well, but the proof has to be modified somewhat.
– Bungo
Nov 30 at 17:40






Ah, OK. Thanks for the context. By the way, the theorem is true for nonabelian groups as well, but the proof has to be modified somewhat.
– Bungo
Nov 30 at 17:40












1 Answer
1






active

oldest

votes


















2














You are using two different concepts of order. Yes, $yN$ has order $p$ as an element of $G/N$. That does not mean that $#yN=p$. What it means is that $y^pN=N$; in other words, $y^pin N$.






share|cite|improve this answer





















  • I had a feeling this was the case after applying the idea to $langle r^2 rangle$ in $D_8$ and $s langle r^2 rangle$. I just think the terminology is a bit confusing, since coset order is also used to denote the number of elements in the coset.
    – Ujkan Sulejmani
    Nov 30 at 17:35










  • Thank you for the answer. :)
    – Ujkan Sulejmani
    Nov 30 at 17:35











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%2f3020370%2fproof-of-cauchys-theorem-for-finite-groups-in-dummit-and-foote%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









2














You are using two different concepts of order. Yes, $yN$ has order $p$ as an element of $G/N$. That does not mean that $#yN=p$. What it means is that $y^pN=N$; in other words, $y^pin N$.






share|cite|improve this answer





















  • I had a feeling this was the case after applying the idea to $langle r^2 rangle$ in $D_8$ and $s langle r^2 rangle$. I just think the terminology is a bit confusing, since coset order is also used to denote the number of elements in the coset.
    – Ujkan Sulejmani
    Nov 30 at 17:35










  • Thank you for the answer. :)
    – Ujkan Sulejmani
    Nov 30 at 17:35
















2














You are using two different concepts of order. Yes, $yN$ has order $p$ as an element of $G/N$. That does not mean that $#yN=p$. What it means is that $y^pN=N$; in other words, $y^pin N$.






share|cite|improve this answer





















  • I had a feeling this was the case after applying the idea to $langle r^2 rangle$ in $D_8$ and $s langle r^2 rangle$. I just think the terminology is a bit confusing, since coset order is also used to denote the number of elements in the coset.
    – Ujkan Sulejmani
    Nov 30 at 17:35










  • Thank you for the answer. :)
    – Ujkan Sulejmani
    Nov 30 at 17:35














2












2








2






You are using two different concepts of order. Yes, $yN$ has order $p$ as an element of $G/N$. That does not mean that $#yN=p$. What it means is that $y^pN=N$; in other words, $y^pin N$.






share|cite|improve this answer












You are using two different concepts of order. Yes, $yN$ has order $p$ as an element of $G/N$. That does not mean that $#yN=p$. What it means is that $y^pN=N$; in other words, $y^pin N$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 30 at 17:31









José Carlos Santos

148k22117218




148k22117218












  • I had a feeling this was the case after applying the idea to $langle r^2 rangle$ in $D_8$ and $s langle r^2 rangle$. I just think the terminology is a bit confusing, since coset order is also used to denote the number of elements in the coset.
    – Ujkan Sulejmani
    Nov 30 at 17:35










  • Thank you for the answer. :)
    – Ujkan Sulejmani
    Nov 30 at 17:35


















  • I had a feeling this was the case after applying the idea to $langle r^2 rangle$ in $D_8$ and $s langle r^2 rangle$. I just think the terminology is a bit confusing, since coset order is also used to denote the number of elements in the coset.
    – Ujkan Sulejmani
    Nov 30 at 17:35










  • Thank you for the answer. :)
    – Ujkan Sulejmani
    Nov 30 at 17:35
















I had a feeling this was the case after applying the idea to $langle r^2 rangle$ in $D_8$ and $s langle r^2 rangle$. I just think the terminology is a bit confusing, since coset order is also used to denote the number of elements in the coset.
– Ujkan Sulejmani
Nov 30 at 17:35




I had a feeling this was the case after applying the idea to $langle r^2 rangle$ in $D_8$ and $s langle r^2 rangle$. I just think the terminology is a bit confusing, since coset order is also used to denote the number of elements in the coset.
– Ujkan Sulejmani
Nov 30 at 17:35












Thank you for the answer. :)
– Ujkan Sulejmani
Nov 30 at 17:35




Thank you for the answer. :)
– Ujkan Sulejmani
Nov 30 at 17:35


















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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


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%2f3020370%2fproof-of-cauchys-theorem-for-finite-groups-in-dummit-and-foote%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

Berounka

Image
Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
Read more

Sphinx de Gizeh

Image
Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
Read more

Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

Image
3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
Read more