Proposition 1.39, Hatcher - Quotient By A Non-Normal Subgroup












1














enter image description here



This is from Hatcher's Algebraic Topology. In the red box highlighted above, we're not requiring $H$ to be a normal subgroup of $pi_1(X,x_0)$, so I don't understand how the quotient can be a group - the group of deck transformations on $tilde{X}$.






general-topology algebraic-topology







share|cite|improve this question


















  • 1




    But $H$ is obviously a normal subgroup of its' normalizer
    – Morgan Rodgers
    Nov 27 '18 at 2:04










  • Oh, I didn't know that. Thank you.
    – Frederic Chopin
    Nov 27 '18 at 2:06










  • No problem, glad it helped.
    – Morgan Rodgers
    Nov 27 '18 at 2:07
















1














enter image description here



This is from Hatcher's Algebraic Topology. In the red box highlighted above, we're not requiring $H$ to be a normal subgroup of $pi_1(X,x_0)$, so I don't understand how the quotient can be a group - the group of deck transformations on $tilde{X}$.






general-topology algebraic-topology







share|cite|improve this question


















  • 1




    But $H$ is obviously a normal subgroup of its' normalizer
    – Morgan Rodgers
    Nov 27 '18 at 2:04










  • Oh, I didn't know that. Thank you.
    – Frederic Chopin
    Nov 27 '18 at 2:06










  • No problem, glad it helped.
    – Morgan Rodgers
    Nov 27 '18 at 2:07














1












1








1







enter image description here



This is from Hatcher's Algebraic Topology. In the red box highlighted above, we're not requiring $H$ to be a normal subgroup of $pi_1(X,x_0)$, so I don't understand how the quotient can be a group - the group of deck transformations on $tilde{X}$.






general-topology algebraic-topology







share|cite|improve this question













enter image description here



This is from Hatcher's Algebraic Topology. In the red box highlighted above, we're not requiring $H$ to be a normal subgroup of $pi_1(X,x_0)$, so I don't understand how the quotient can be a group - the group of deck transformations on $tilde{X}$.






general-topology algebraic-topology




general-topology algebraic-topology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 27 '18 at 2:02









Frederic Chopin

321111




321111








  • 1




    But $H$ is obviously a normal subgroup of its' normalizer
    – Morgan Rodgers
    Nov 27 '18 at 2:04










  • Oh, I didn't know that. Thank you.
    – Frederic Chopin
    Nov 27 '18 at 2:06










  • No problem, glad it helped.
    – Morgan Rodgers
    Nov 27 '18 at 2:07














  • 1




    But $H$ is obviously a normal subgroup of its' normalizer
    – Morgan Rodgers
    Nov 27 '18 at 2:04










  • Oh, I didn't know that. Thank you.
    – Frederic Chopin
    Nov 27 '18 at 2:06










  • No problem, glad it helped.
    – Morgan Rodgers
    Nov 27 '18 at 2:07








1




1




But $H$ is obviously a normal subgroup of its' normalizer
– Morgan Rodgers
Nov 27 '18 at 2:04




But $H$ is obviously a normal subgroup of its' normalizer
– Morgan Rodgers
Nov 27 '18 at 2:04












Oh, I didn't know that. Thank you.
– Frederic Chopin
Nov 27 '18 at 2:06




Oh, I didn't know that. Thank you.
– Frederic Chopin
Nov 27 '18 at 2:06












No problem, glad it helped.
– Morgan Rodgers
Nov 27 '18 at 2:07




No problem, glad it helped.
– Morgan Rodgers
Nov 27 '18 at 2:07










1 Answer
1






active

oldest

votes


















1














(I am just elaborating on things written in the comments.) If $H$ is a subgroup of a group $G$ then $N(H)$ (in Hatcher's notation) is the normalizer of $H$. That is, $N(H)={gin G: gHg^{-1}=H}.$ Then $Hsubseteq N(H)$ and is in particular a normal subgroup of $N(H)$ by definition. In conclusion, it makes perfect sense to consider the quotient group $N(H)/H$.






share|cite|improve this answer





















    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%2f3015223%2fproposition-1-39-hatcher-quotient-by-a-non-normal-subgroup%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









    1














    (I am just elaborating on things written in the comments.) If $H$ is a subgroup of a group $G$ then $N(H)$ (in Hatcher's notation) is the normalizer of $H$. That is, $N(H)={gin G: gHg^{-1}=H}.$ Then $Hsubseteq N(H)$ and is in particular a normal subgroup of $N(H)$ by definition. In conclusion, it makes perfect sense to consider the quotient group $N(H)/H$.






    share|cite|improve this answer


























      1














      (I am just elaborating on things written in the comments.) If $H$ is a subgroup of a group $G$ then $N(H)$ (in Hatcher's notation) is the normalizer of $H$. That is, $N(H)={gin G: gHg^{-1}=H}.$ Then $Hsubseteq N(H)$ and is in particular a normal subgroup of $N(H)$ by definition. In conclusion, it makes perfect sense to consider the quotient group $N(H)/H$.






      share|cite|improve this answer
























        1












        1








        1






        (I am just elaborating on things written in the comments.) If $H$ is a subgroup of a group $G$ then $N(H)$ (in Hatcher's notation) is the normalizer of $H$. That is, $N(H)={gin G: gHg^{-1}=H}.$ Then $Hsubseteq N(H)$ and is in particular a normal subgroup of $N(H)$ by definition. In conclusion, it makes perfect sense to consider the quotient group $N(H)/H$.






        share|cite|improve this answer












        (I am just elaborating on things written in the comments.) If $H$ is a subgroup of a group $G$ then $N(H)$ (in Hatcher's notation) is the normalizer of $H$. That is, $N(H)={gin G: gHg^{-1}=H}.$ Then $Hsubseteq N(H)$ and is in particular a normal subgroup of $N(H)$ by definition. In conclusion, it makes perfect sense to consider the quotient group $N(H)/H$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 2:21









        Antonios-Alexandros Robotis

        9,20741640




        9,20741640






























            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%2f3015223%2fproposition-1-39-hatcher-quotient-by-a-non-normal-subgroup%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

            Ellipse (mathématiques)

            Image
            Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
            Read more

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
            Read more