For which abelian groups $G$ is there a short exact sequence $0 rightarrow mathbb{Z}/p^2 rightarrow G...












4














I am trying to find for which abelian groups $G$ is there a short exact sequence.



$0 rightarrow mathbb{Z}/p^2 rightarrow G rightarrow mathbb{Z}/p^2 rightarrow 0$?



I have reasoned as follows: consider the sequence with the maps as follows $$0 xrightarrow{phi} mathbb{Z}/p^2 xrightarrow{f} G xrightarrow{g} mathbb{Z}/p^2 xrightarrow{psi} 0$$



Since we want the sequence to be exact, we need that $ker f= im(phi) = 0$ (so $f$ is injective) and we need that $im(g) = ker (psi) = mathbb{Z}/p^2$ (so $g$ is surjective). On top of that, we know that $im( f) = ker g$. Then, by the first isomorphism theorem, we know that $mathbb{Z}/p^2 / ker f = mathbb{Z}/p^2 cong im(f) = ker g$. So we know $ker g = mathbb{Z}/p^2$ and $im(g) = mathbb{Z}/p^2$. This leads me to think that one possibility for $G$ is $mathbb{Z}/p^2 oplus mathbb{Z}/p^2$. However, the answers say that other possibilities could be $mathbb{Z}/p^4$ and $mathbb{Z}/p^3 oplus mathbb{Z}/p$ and I don't see how they came with this possibility, can anyone explain this solution? Also, is my argument for why $G$ can be $mathbb{Z}/p^2 oplus mathbb{Z}/p^2$ correct? Thanks for your help!






abstract-algebra abelian-groups exact-sequence







share|cite|improve this question


















  • 1




    Use the classification of f.g. abelian groups.
    – anomaly
    Dec 27 '18 at 1:51
















4














I am trying to find for which abelian groups $G$ is there a short exact sequence.



$0 rightarrow mathbb{Z}/p^2 rightarrow G rightarrow mathbb{Z}/p^2 rightarrow 0$?



I have reasoned as follows: consider the sequence with the maps as follows $$0 xrightarrow{phi} mathbb{Z}/p^2 xrightarrow{f} G xrightarrow{g} mathbb{Z}/p^2 xrightarrow{psi} 0$$



Since we want the sequence to be exact, we need that $ker f= im(phi) = 0$ (so $f$ is injective) and we need that $im(g) = ker (psi) = mathbb{Z}/p^2$ (so $g$ is surjective). On top of that, we know that $im( f) = ker g$. Then, by the first isomorphism theorem, we know that $mathbb{Z}/p^2 / ker f = mathbb{Z}/p^2 cong im(f) = ker g$. So we know $ker g = mathbb{Z}/p^2$ and $im(g) = mathbb{Z}/p^2$. This leads me to think that one possibility for $G$ is $mathbb{Z}/p^2 oplus mathbb{Z}/p^2$. However, the answers say that other possibilities could be $mathbb{Z}/p^4$ and $mathbb{Z}/p^3 oplus mathbb{Z}/p$ and I don't see how they came with this possibility, can anyone explain this solution? Also, is my argument for why $G$ can be $mathbb{Z}/p^2 oplus mathbb{Z}/p^2$ correct? Thanks for your help!






abstract-algebra abelian-groups exact-sequence







share|cite|improve this question


















  • 1




    Use the classification of f.g. abelian groups.
    – anomaly
    Dec 27 '18 at 1:51














4












4








4


1





I am trying to find for which abelian groups $G$ is there a short exact sequence.



$0 rightarrow mathbb{Z}/p^2 rightarrow G rightarrow mathbb{Z}/p^2 rightarrow 0$?



I have reasoned as follows: consider the sequence with the maps as follows $$0 xrightarrow{phi} mathbb{Z}/p^2 xrightarrow{f} G xrightarrow{g} mathbb{Z}/p^2 xrightarrow{psi} 0$$



Since we want the sequence to be exact, we need that $ker f= im(phi) = 0$ (so $f$ is injective) and we need that $im(g) = ker (psi) = mathbb{Z}/p^2$ (so $g$ is surjective). On top of that, we know that $im( f) = ker g$. Then, by the first isomorphism theorem, we know that $mathbb{Z}/p^2 / ker f = mathbb{Z}/p^2 cong im(f) = ker g$. So we know $ker g = mathbb{Z}/p^2$ and $im(g) = mathbb{Z}/p^2$. This leads me to think that one possibility for $G$ is $mathbb{Z}/p^2 oplus mathbb{Z}/p^2$. However, the answers say that other possibilities could be $mathbb{Z}/p^4$ and $mathbb{Z}/p^3 oplus mathbb{Z}/p$ and I don't see how they came with this possibility, can anyone explain this solution? Also, is my argument for why $G$ can be $mathbb{Z}/p^2 oplus mathbb{Z}/p^2$ correct? Thanks for your help!






abstract-algebra abelian-groups exact-sequence







share|cite|improve this question













I am trying to find for which abelian groups $G$ is there a short exact sequence.



$0 rightarrow mathbb{Z}/p^2 rightarrow G rightarrow mathbb{Z}/p^2 rightarrow 0$?



I have reasoned as follows: consider the sequence with the maps as follows $$0 xrightarrow{phi} mathbb{Z}/p^2 xrightarrow{f} G xrightarrow{g} mathbb{Z}/p^2 xrightarrow{psi} 0$$



Since we want the sequence to be exact, we need that $ker f= im(phi) = 0$ (so $f$ is injective) and we need that $im(g) = ker (psi) = mathbb{Z}/p^2$ (so $g$ is surjective). On top of that, we know that $im( f) = ker g$. Then, by the first isomorphism theorem, we know that $mathbb{Z}/p^2 / ker f = mathbb{Z}/p^2 cong im(f) = ker g$. So we know $ker g = mathbb{Z}/p^2$ and $im(g) = mathbb{Z}/p^2$. This leads me to think that one possibility for $G$ is $mathbb{Z}/p^2 oplus mathbb{Z}/p^2$. However, the answers say that other possibilities could be $mathbb{Z}/p^4$ and $mathbb{Z}/p^3 oplus mathbb{Z}/p$ and I don't see how they came with this possibility, can anyone explain this solution? Also, is my argument for why $G$ can be $mathbb{Z}/p^2 oplus mathbb{Z}/p^2$ correct? Thanks for your help!






abstract-algebra abelian-groups exact-sequence




abstract-algebra abelian-groups exact-sequence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 27 '18 at 1:21









BOlivianoperuano84

1778




1778








  • 1




    Use the classification of f.g. abelian groups.
    – anomaly
    Dec 27 '18 at 1:51














  • 1




    Use the classification of f.g. abelian groups.
    – anomaly
    Dec 27 '18 at 1:51








1




1




Use the classification of f.g. abelian groups.
– anomaly
Dec 27 '18 at 1:51




Use the classification of f.g. abelian groups.
– anomaly
Dec 27 '18 at 1:51










1 Answer
1






active

oldest

votes


















2














This short exact sequence is a fancy way of saying that $G$ has a subgroup $H$ such that $H$ and $G / H$ are both isomorphic to $mathbb Z / p^2$.



[Think of $f$ as the embedding of $H$ into $G$, and think of $g$ as the projection of $G$ onto $G/H$. It's clear that $f$ is injective, and $g$ is surjective, and ${rm im}(f) = {rm ker}(g)$, which is what your short exact sequence says.]



For example:




  • $G = mathbb Z / p^2 oplus mathbb Z / p^2$, with $H$ generated by $(1, 0)$.


  • $G = mathbb Z / p^4$, with $H$ generated by $p^2$.


  • $G = mathbb Z / p^3 oplus mathbb Z / p$, with $H$ generated by $(p, 1)$.







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%2f3053494%2ffor-which-abelian-groups-g-is-there-a-short-exact-sequence-0-rightarrow-mat%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














    This short exact sequence is a fancy way of saying that $G$ has a subgroup $H$ such that $H$ and $G / H$ are both isomorphic to $mathbb Z / p^2$.



    [Think of $f$ as the embedding of $H$ into $G$, and think of $g$ as the projection of $G$ onto $G/H$. It's clear that $f$ is injective, and $g$ is surjective, and ${rm im}(f) = {rm ker}(g)$, which is what your short exact sequence says.]



    For example:




    • $G = mathbb Z / p^2 oplus mathbb Z / p^2$, with $H$ generated by $(1, 0)$.


    • $G = mathbb Z / p^4$, with $H$ generated by $p^2$.


    • $G = mathbb Z / p^3 oplus mathbb Z / p$, with $H$ generated by $(p, 1)$.







    share|cite|improve this answer


























      2














      This short exact sequence is a fancy way of saying that $G$ has a subgroup $H$ such that $H$ and $G / H$ are both isomorphic to $mathbb Z / p^2$.



      [Think of $f$ as the embedding of $H$ into $G$, and think of $g$ as the projection of $G$ onto $G/H$. It's clear that $f$ is injective, and $g$ is surjective, and ${rm im}(f) = {rm ker}(g)$, which is what your short exact sequence says.]



      For example:




      • $G = mathbb Z / p^2 oplus mathbb Z / p^2$, with $H$ generated by $(1, 0)$.


      • $G = mathbb Z / p^4$, with $H$ generated by $p^2$.


      • $G = mathbb Z / p^3 oplus mathbb Z / p$, with $H$ generated by $(p, 1)$.







      share|cite|improve this answer
























        2












        2








        2






        This short exact sequence is a fancy way of saying that $G$ has a subgroup $H$ such that $H$ and $G / H$ are both isomorphic to $mathbb Z / p^2$.



        [Think of $f$ as the embedding of $H$ into $G$, and think of $g$ as the projection of $G$ onto $G/H$. It's clear that $f$ is injective, and $g$ is surjective, and ${rm im}(f) = {rm ker}(g)$, which is what your short exact sequence says.]



        For example:




        • $G = mathbb Z / p^2 oplus mathbb Z / p^2$, with $H$ generated by $(1, 0)$.


        • $G = mathbb Z / p^4$, with $H$ generated by $p^2$.


        • $G = mathbb Z / p^3 oplus mathbb Z / p$, with $H$ generated by $(p, 1)$.







        share|cite|improve this answer












        This short exact sequence is a fancy way of saying that $G$ has a subgroup $H$ such that $H$ and $G / H$ are both isomorphic to $mathbb Z / p^2$.



        [Think of $f$ as the embedding of $H$ into $G$, and think of $g$ as the projection of $G$ onto $G/H$. It's clear that $f$ is injective, and $g$ is surjective, and ${rm im}(f) = {rm ker}(g)$, which is what your short exact sequence says.]



        For example:




        • $G = mathbb Z / p^2 oplus mathbb Z / p^2$, with $H$ generated by $(1, 0)$.


        • $G = mathbb Z / p^4$, with $H$ generated by $p^2$.


        • $G = mathbb Z / p^3 oplus mathbb Z / p$, with $H$ generated by $(p, 1)$.








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '18 at 1:39









        Kenny Wong

        17.7k21338




        17.7k21338






























            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%2f3053494%2ffor-which-abelian-groups-g-is-there-a-short-exact-sequence-0-rightarrow-mat%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

            Human spaceflight

            Image
            Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
            Read more

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            Image
            0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
            Read more

            File:DeusFollowingSea.jpg

            Image
            File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
            Read more