How to show that $langle a,b mid aba^{-1}ba = bab^{-1}abrangle$ is not Abelian?












4












$begingroup$


I'd like to show that
$$
G = langle a,b mid aba^{-1}ba = bab^{-1}abrangle
$$
is non-Abelian.



I have tried finding a surjective homomorphism from $G$ to a non-Abelian group, but I haven't found one. The context is that I would like to show that the figure-$8$ knot complement is non-trivial using knot groups.



Thanks a lot!






group-theory abelian-groups knot-theory knot-invariants combinatorial-group-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    If $G$ were abelian, then the relation would be equivalent to $ab^2=a^2b$, i.e., $a=b$, so then $Gconglangle amid ranglecongmathbb{Z}$. I don't know if that helps.
    $endgroup$
    – Shaun
    Nov 24 '18 at 1:14








  • 1




    $begingroup$
    It should be noted that the problem of determining whether the group defined by a presentation is abelian is, in general, undecidable.
    $endgroup$
    – Shaun
    Nov 24 '18 at 1:33








  • 1




    $begingroup$
    @Shaun: Are there any convenient types of groups that arise in knot theory for which the word problem is decidable? I don't know much about knot theory myself, and the few results I'm familiar with it are in the opposite direction.
    $endgroup$
    – anomaly
    Nov 24 '18 at 4:51








  • 1




    $begingroup$
    @anomaly: That's a great question! I don't know. I'm not exactly au fait with knot theory either. Maybe it's worth asking that on here as a separate question. In the meantime, I'll have a look at what I can find, but I make no promises! :)
    $endgroup$
    – Shaun
    Nov 24 '18 at 5:02






  • 2




    $begingroup$
    @Shaun You cannot use $mathtt{AbelianInvariants}$ to prove that a group is nonabelian. That just tells you the largest abelian quotient of the group. A better way is to use $mathtt{LowIndexSubgroupsFpGroup}$. This group has a subgroup of index $4$ which defines a homomorphism onto $A_4$.
    $endgroup$
    – Derek Holt
    Nov 24 '18 at 8:50
















4












$begingroup$


I'd like to show that
$$
G = langle a,b mid aba^{-1}ba = bab^{-1}abrangle
$$
is non-Abelian.



I have tried finding a surjective homomorphism from $G$ to a non-Abelian group, but I haven't found one. The context is that I would like to show that the figure-$8$ knot complement is non-trivial using knot groups.



Thanks a lot!






group-theory abelian-groups knot-theory knot-invariants combinatorial-group-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    If $G$ were abelian, then the relation would be equivalent to $ab^2=a^2b$, i.e., $a=b$, so then $Gconglangle amid ranglecongmathbb{Z}$. I don't know if that helps.
    $endgroup$
    – Shaun
    Nov 24 '18 at 1:14








  • 1




    $begingroup$
    It should be noted that the problem of determining whether the group defined by a presentation is abelian is, in general, undecidable.
    $endgroup$
    – Shaun
    Nov 24 '18 at 1:33








  • 1




    $begingroup$
    @Shaun: Are there any convenient types of groups that arise in knot theory for which the word problem is decidable? I don't know much about knot theory myself, and the few results I'm familiar with it are in the opposite direction.
    $endgroup$
    – anomaly
    Nov 24 '18 at 4:51








  • 1




    $begingroup$
    @anomaly: That's a great question! I don't know. I'm not exactly au fait with knot theory either. Maybe it's worth asking that on here as a separate question. In the meantime, I'll have a look at what I can find, but I make no promises! :)
    $endgroup$
    – Shaun
    Nov 24 '18 at 5:02






  • 2




    $begingroup$
    @Shaun You cannot use $mathtt{AbelianInvariants}$ to prove that a group is nonabelian. That just tells you the largest abelian quotient of the group. A better way is to use $mathtt{LowIndexSubgroupsFpGroup}$. This group has a subgroup of index $4$ which defines a homomorphism onto $A_4$.
    $endgroup$
    – Derek Holt
    Nov 24 '18 at 8:50














4












4








4


1



$begingroup$


I'd like to show that
$$
G = langle a,b mid aba^{-1}ba = bab^{-1}abrangle
$$
is non-Abelian.



I have tried finding a surjective homomorphism from $G$ to a non-Abelian group, but I haven't found one. The context is that I would like to show that the figure-$8$ knot complement is non-trivial using knot groups.



Thanks a lot!






group-theory abelian-groups knot-theory knot-invariants combinatorial-group-theory







share|cite|improve this question











$endgroup$




I'd like to show that
$$
G = langle a,b mid aba^{-1}ba = bab^{-1}abrangle
$$
is non-Abelian.



I have tried finding a surjective homomorphism from $G$ to a non-Abelian group, but I haven't found one. The context is that I would like to show that the figure-$8$ knot complement is non-trivial using knot groups.



Thanks a lot!






group-theory abelian-groups knot-theory knot-invariants combinatorial-group-theory




group-theory abelian-groups knot-theory knot-invariants combinatorial-group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 '18 at 2:39









Shaun

8,818113681




8,818113681










asked Nov 24 '18 at 0:57









SheafsSheafs

1517




1517








  • 3




    $begingroup$
    If $G$ were abelian, then the relation would be equivalent to $ab^2=a^2b$, i.e., $a=b$, so then $Gconglangle amid ranglecongmathbb{Z}$. I don't know if that helps.
    $endgroup$
    – Shaun
    Nov 24 '18 at 1:14








  • 1




    $begingroup$
    It should be noted that the problem of determining whether the group defined by a presentation is abelian is, in general, undecidable.
    $endgroup$
    – Shaun
    Nov 24 '18 at 1:33








  • 1




    $begingroup$
    @Shaun: Are there any convenient types of groups that arise in knot theory for which the word problem is decidable? I don't know much about knot theory myself, and the few results I'm familiar with it are in the opposite direction.
    $endgroup$
    – anomaly
    Nov 24 '18 at 4:51








  • 1




    $begingroup$
    @anomaly: That's a great question! I don't know. I'm not exactly au fait with knot theory either. Maybe it's worth asking that on here as a separate question. In the meantime, I'll have a look at what I can find, but I make no promises! :)
    $endgroup$
    – Shaun
    Nov 24 '18 at 5:02






  • 2




    $begingroup$
    @Shaun You cannot use $mathtt{AbelianInvariants}$ to prove that a group is nonabelian. That just tells you the largest abelian quotient of the group. A better way is to use $mathtt{LowIndexSubgroupsFpGroup}$. This group has a subgroup of index $4$ which defines a homomorphism onto $A_4$.
    $endgroup$
    – Derek Holt
    Nov 24 '18 at 8:50














  • 3




    $begingroup$
    If $G$ were abelian, then the relation would be equivalent to $ab^2=a^2b$, i.e., $a=b$, so then $Gconglangle amid ranglecongmathbb{Z}$. I don't know if that helps.
    $endgroup$
    – Shaun
    Nov 24 '18 at 1:14








  • 1




    $begingroup$
    It should be noted that the problem of determining whether the group defined by a presentation is abelian is, in general, undecidable.
    $endgroup$
    – Shaun
    Nov 24 '18 at 1:33








  • 1




    $begingroup$
    @Shaun: Are there any convenient types of groups that arise in knot theory for which the word problem is decidable? I don't know much about knot theory myself, and the few results I'm familiar with it are in the opposite direction.
    $endgroup$
    – anomaly
    Nov 24 '18 at 4:51








  • 1




    $begingroup$
    @anomaly: That's a great question! I don't know. I'm not exactly au fait with knot theory either. Maybe it's worth asking that on here as a separate question. In the meantime, I'll have a look at what I can find, but I make no promises! :)
    $endgroup$
    – Shaun
    Nov 24 '18 at 5:02






  • 2




    $begingroup$
    @Shaun You cannot use $mathtt{AbelianInvariants}$ to prove that a group is nonabelian. That just tells you the largest abelian quotient of the group. A better way is to use $mathtt{LowIndexSubgroupsFpGroup}$. This group has a subgroup of index $4$ which defines a homomorphism onto $A_4$.
    $endgroup$
    – Derek Holt
    Nov 24 '18 at 8:50








3




3




$begingroup$
If $G$ were abelian, then the relation would be equivalent to $ab^2=a^2b$, i.e., $a=b$, so then $Gconglangle amid ranglecongmathbb{Z}$. I don't know if that helps.
$endgroup$
– Shaun
Nov 24 '18 at 1:14






$begingroup$
If $G$ were abelian, then the relation would be equivalent to $ab^2=a^2b$, i.e., $a=b$, so then $Gconglangle amid ranglecongmathbb{Z}$. I don't know if that helps.
$endgroup$
– Shaun
Nov 24 '18 at 1:14






1




1




$begingroup$
It should be noted that the problem of determining whether the group defined by a presentation is abelian is, in general, undecidable.
$endgroup$
– Shaun
Nov 24 '18 at 1:33






$begingroup$
It should be noted that the problem of determining whether the group defined by a presentation is abelian is, in general, undecidable.
$endgroup$
– Shaun
Nov 24 '18 at 1:33






1




1




$begingroup$
@Shaun: Are there any convenient types of groups that arise in knot theory for which the word problem is decidable? I don't know much about knot theory myself, and the few results I'm familiar with it are in the opposite direction.
$endgroup$
– anomaly
Nov 24 '18 at 4:51






$begingroup$
@Shaun: Are there any convenient types of groups that arise in knot theory for which the word problem is decidable? I don't know much about knot theory myself, and the few results I'm familiar with it are in the opposite direction.
$endgroup$
– anomaly
Nov 24 '18 at 4:51






1




1




$begingroup$
@anomaly: That's a great question! I don't know. I'm not exactly au fait with knot theory either. Maybe it's worth asking that on here as a separate question. In the meantime, I'll have a look at what I can find, but I make no promises! :)
$endgroup$
– Shaun
Nov 24 '18 at 5:02




$begingroup$
@anomaly: That's a great question! I don't know. I'm not exactly au fait with knot theory either. Maybe it's worth asking that on here as a separate question. In the meantime, I'll have a look at what I can find, but I make no promises! :)
$endgroup$
– Shaun
Nov 24 '18 at 5:02




2




2




$begingroup$
@Shaun You cannot use $mathtt{AbelianInvariants}$ to prove that a group is nonabelian. That just tells you the largest abelian quotient of the group. A better way is to use $mathtt{LowIndexSubgroupsFpGroup}$. This group has a subgroup of index $4$ which defines a homomorphism onto $A_4$.
$endgroup$
– Derek Holt
Nov 24 '18 at 8:50




$begingroup$
@Shaun You cannot use $mathtt{AbelianInvariants}$ to prove that a group is nonabelian. That just tells you the largest abelian quotient of the group. A better way is to use $mathtt{LowIndexSubgroupsFpGroup}$. This group has a subgroup of index $4$ which defines a homomorphism onto $A_4$.
$endgroup$
– Derek Holt
Nov 24 '18 at 8:50










2 Answers
2






active

oldest

votes


















6












$begingroup$

Put $xi = frac{1}{2}(3 + sqrt{5})$. The map $G to GL_2(mathbb{R})$ defined by
begin{align*}
a &to begin{pmatrix}1 & 0 \ 0 & xiend{pmatrix} &
b &to begin{pmatrix}1 & 1 \ 0 & xiend{pmatrix}
end{align*}
is well-defined with nonabelian image.






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    Since you tag this with knot-theory and knot-invariants, it looks like you are trying to show the fundamental group of the knot complement $S^3-4_1$ is nonabelian.



    One of the "obvious" things to try is Fox $n$-coloring, since they yield a homomorphism (usually surjective) to a dihedral group. There is a $5$-coloring (using $0,4,1,2$ as you go along the knot, which you can check satisfies $2bequiv a+cpmod{5}$ at each crossing). Since $5$ is a prime and the coloring is nonconstant, the corresponding homomorphism from $pi_1(S^3-4_1)$ to $D_{2cdot 5}$, the dihedral group of order $10$, is surjective.






    share|cite|improve this answer











    $endgroup$









    • 3




      $begingroup$
      Why the downvote?
      $endgroup$
      – anomaly
      Nov 25 '18 at 3:34











    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%2f3011043%2fhow-to-show-that-langle-a-b-mid-aba-1ba-bab-1ab-rangle-is-not-abelia%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    Put $xi = frac{1}{2}(3 + sqrt{5})$. The map $G to GL_2(mathbb{R})$ defined by
    begin{align*}
    a &to begin{pmatrix}1 & 0 \ 0 & xiend{pmatrix} &
    b &to begin{pmatrix}1 & 1 \ 0 & xiend{pmatrix}
    end{align*}
    is well-defined with nonabelian image.






    share|cite|improve this answer











    $endgroup$


















      6












      $begingroup$

      Put $xi = frac{1}{2}(3 + sqrt{5})$. The map $G to GL_2(mathbb{R})$ defined by
      begin{align*}
      a &to begin{pmatrix}1 & 0 \ 0 & xiend{pmatrix} &
      b &to begin{pmatrix}1 & 1 \ 0 & xiend{pmatrix}
      end{align*}
      is well-defined with nonabelian image.






      share|cite|improve this answer











      $endgroup$
















        6












        6








        6





        $begingroup$

        Put $xi = frac{1}{2}(3 + sqrt{5})$. The map $G to GL_2(mathbb{R})$ defined by
        begin{align*}
        a &to begin{pmatrix}1 & 0 \ 0 & xiend{pmatrix} &
        b &to begin{pmatrix}1 & 1 \ 0 & xiend{pmatrix}
        end{align*}
        is well-defined with nonabelian image.






        share|cite|improve this answer











        $endgroup$



        Put $xi = frac{1}{2}(3 + sqrt{5})$. The map $G to GL_2(mathbb{R})$ defined by
        begin{align*}
        a &to begin{pmatrix}1 & 0 \ 0 & xiend{pmatrix} &
        b &to begin{pmatrix}1 & 1 \ 0 & xiend{pmatrix}
        end{align*}
        is well-defined with nonabelian image.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 24 '18 at 3:15

























        answered Nov 24 '18 at 3:04









        anomalyanomaly

        17.4k42664




        17.4k42664























            4












            $begingroup$

            Since you tag this with knot-theory and knot-invariants, it looks like you are trying to show the fundamental group of the knot complement $S^3-4_1$ is nonabelian.



            One of the "obvious" things to try is Fox $n$-coloring, since they yield a homomorphism (usually surjective) to a dihedral group. There is a $5$-coloring (using $0,4,1,2$ as you go along the knot, which you can check satisfies $2bequiv a+cpmod{5}$ at each crossing). Since $5$ is a prime and the coloring is nonconstant, the corresponding homomorphism from $pi_1(S^3-4_1)$ to $D_{2cdot 5}$, the dihedral group of order $10$, is surjective.






            share|cite|improve this answer











            $endgroup$









            • 3




              $begingroup$
              Why the downvote?
              $endgroup$
              – anomaly
              Nov 25 '18 at 3:34
















            4












            $begingroup$

            Since you tag this with knot-theory and knot-invariants, it looks like you are trying to show the fundamental group of the knot complement $S^3-4_1$ is nonabelian.



            One of the "obvious" things to try is Fox $n$-coloring, since they yield a homomorphism (usually surjective) to a dihedral group. There is a $5$-coloring (using $0,4,1,2$ as you go along the knot, which you can check satisfies $2bequiv a+cpmod{5}$ at each crossing). Since $5$ is a prime and the coloring is nonconstant, the corresponding homomorphism from $pi_1(S^3-4_1)$ to $D_{2cdot 5}$, the dihedral group of order $10$, is surjective.






            share|cite|improve this answer











            $endgroup$









            • 3




              $begingroup$
              Why the downvote?
              $endgroup$
              – anomaly
              Nov 25 '18 at 3:34














            4












            4








            4





            $begingroup$

            Since you tag this with knot-theory and knot-invariants, it looks like you are trying to show the fundamental group of the knot complement $S^3-4_1$ is nonabelian.



            One of the "obvious" things to try is Fox $n$-coloring, since they yield a homomorphism (usually surjective) to a dihedral group. There is a $5$-coloring (using $0,4,1,2$ as you go along the knot, which you can check satisfies $2bequiv a+cpmod{5}$ at each crossing). Since $5$ is a prime and the coloring is nonconstant, the corresponding homomorphism from $pi_1(S^3-4_1)$ to $D_{2cdot 5}$, the dihedral group of order $10$, is surjective.






            share|cite|improve this answer











            $endgroup$



            Since you tag this with knot-theory and knot-invariants, it looks like you are trying to show the fundamental group of the knot complement $S^3-4_1$ is nonabelian.



            One of the "obvious" things to try is Fox $n$-coloring, since they yield a homomorphism (usually surjective) to a dihedral group. There is a $5$-coloring (using $0,4,1,2$ as you go along the knot, which you can check satisfies $2bequiv a+cpmod{5}$ at each crossing). Since $5$ is a prime and the coloring is nonconstant, the corresponding homomorphism from $pi_1(S^3-4_1)$ to $D_{2cdot 5}$, the dihedral group of order $10$, is surjective.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 24 '18 at 2:50

























            answered Nov 24 '18 at 2:44









            user10354138user10354138

            7,3772925




            7,3772925








            • 3




              $begingroup$
              Why the downvote?
              $endgroup$
              – anomaly
              Nov 25 '18 at 3:34














            • 3




              $begingroup$
              Why the downvote?
              $endgroup$
              – anomaly
              Nov 25 '18 at 3:34








            3




            3




            $begingroup$
            Why the downvote?
            $endgroup$
            – anomaly
            Nov 25 '18 at 3:34




            $begingroup$
            Why the downvote?
            $endgroup$
            – anomaly
            Nov 25 '18 at 3:34


















            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%2f3011043%2fhow-to-show-that-langle-a-b-mid-aba-1ba-bab-1ab-rangle-is-not-abelia%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

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
            Read more