Nielsen-Schreier formula application












2












$begingroup$


So I was thinking about Nielsen-Schreier formula for the generators of a free group subgroup and i tried to apply it. Something must be wrong but I can't find the mistake.



So I have $F(a,b)$ free group generated by two element and an surjective homomorphism $F(a,b)to F(b)$ that acts by killing the "a" letter from a word
(i.e. $ phi(aba^{-1}b)=bb=b^2)$.
So we should have $frac{F(a,b)}{mathrm{ker}(phi)} cong F(b)$ free group on one element.



I want to find the generators of the kernel by applying Nielsen-Schreier formula, which states that $mathrm{ker}(phi)$ is freely generated by this set
$$B = { u x (overline{u x})^{-1} mid u in U, x in X }$$
where $U$ is a Schreier transversal for $mathrm{ker}(phi)$ and $X$ is ${a,b}$.



I figured that $mathrm{ker}(phi)$ obviously contains $F(a)$ and that consists of all the words that has the same amount of $b$ and $b^{-1}$ or, in other words, of all the words $w=a^{n_1}b^{m1}a^{n2}b^{m2}cdots $ so that $sum_{i=1}^n m_i=0$.



In the next step I tried to compute the quotient $frac{F(a,b)}{mathrm{ker}(phi)}={1,bH,b^{-1}H,b^2H,b^3H,ldots}$. I'm pretty sure that's it because two words are in relation only if, summing the power of all the $b$ and $b^{-1}$ in the word the result is the same. (for example $w=b^{-1}abaab$ with a $b$ total power of $-1+1+1=1$ is in relationship with $v=aba$, in fact $wv^{-1} in mathrm{ker}(phi)$.



So the Schreier Transversal is $U={1,b,b^2,ldots}$. The problem is that applying now the formula to this Schreier Transversal I got that the generator of $mathrm{ker}(phi)$ is ${a}$, which makes no sense.



Where's the mistake?






abstract-algebra free-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    Isn't the overline meant to indicate that you are taking the inverse of the element of $U$ that corresponds to the coset of $ux$? If so, then your $u$ is of the form $b^i$; if $x=a$, then $b^ia$ corresponds to $b^iin U$, so $ux(overline{ux})^{-1} = b^ia(overline{b^ia})^{-1} = b^ia(b^i)^{-1} = b^iab^{-i}$; so that your generating set consists of all conjugates of $a$ by powers of $b$.
    $endgroup$
    – Arturo Magidin
    Dec 14 '18 at 20:29












  • $begingroup$
    Where you perhaps calculagting $xu(overline{xu})^{-1}$ by mistake?
    $endgroup$
    – Arturo Magidin
    Dec 14 '18 at 21:28










  • $begingroup$
    Yup, that was the problem. I was sure i was doing something dumb! Thank you so much
    $endgroup$
    – Andrea Licata
    Dec 17 '18 at 16:08










  • $begingroup$
    Perhaps you can answer your own question explaining this, so that the query does not go unanswered.
    $endgroup$
    – Arturo Magidin
    Dec 17 '18 at 19:31
















2












$begingroup$


So I was thinking about Nielsen-Schreier formula for the generators of a free group subgroup and i tried to apply it. Something must be wrong but I can't find the mistake.



So I have $F(a,b)$ free group generated by two element and an surjective homomorphism $F(a,b)to F(b)$ that acts by killing the "a" letter from a word
(i.e. $ phi(aba^{-1}b)=bb=b^2)$.
So we should have $frac{F(a,b)}{mathrm{ker}(phi)} cong F(b)$ free group on one element.



I want to find the generators of the kernel by applying Nielsen-Schreier formula, which states that $mathrm{ker}(phi)$ is freely generated by this set
$$B = { u x (overline{u x})^{-1} mid u in U, x in X }$$
where $U$ is a Schreier transversal for $mathrm{ker}(phi)$ and $X$ is ${a,b}$.



I figured that $mathrm{ker}(phi)$ obviously contains $F(a)$ and that consists of all the words that has the same amount of $b$ and $b^{-1}$ or, in other words, of all the words $w=a^{n_1}b^{m1}a^{n2}b^{m2}cdots $ so that $sum_{i=1}^n m_i=0$.



In the next step I tried to compute the quotient $frac{F(a,b)}{mathrm{ker}(phi)}={1,bH,b^{-1}H,b^2H,b^3H,ldots}$. I'm pretty sure that's it because two words are in relation only if, summing the power of all the $b$ and $b^{-1}$ in the word the result is the same. (for example $w=b^{-1}abaab$ with a $b$ total power of $-1+1+1=1$ is in relationship with $v=aba$, in fact $wv^{-1} in mathrm{ker}(phi)$.



So the Schreier Transversal is $U={1,b,b^2,ldots}$. The problem is that applying now the formula to this Schreier Transversal I got that the generator of $mathrm{ker}(phi)$ is ${a}$, which makes no sense.



Where's the mistake?






abstract-algebra free-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    Isn't the overline meant to indicate that you are taking the inverse of the element of $U$ that corresponds to the coset of $ux$? If so, then your $u$ is of the form $b^i$; if $x=a$, then $b^ia$ corresponds to $b^iin U$, so $ux(overline{ux})^{-1} = b^ia(overline{b^ia})^{-1} = b^ia(b^i)^{-1} = b^iab^{-i}$; so that your generating set consists of all conjugates of $a$ by powers of $b$.
    $endgroup$
    – Arturo Magidin
    Dec 14 '18 at 20:29












  • $begingroup$
    Where you perhaps calculagting $xu(overline{xu})^{-1}$ by mistake?
    $endgroup$
    – Arturo Magidin
    Dec 14 '18 at 21:28










  • $begingroup$
    Yup, that was the problem. I was sure i was doing something dumb! Thank you so much
    $endgroup$
    – Andrea Licata
    Dec 17 '18 at 16:08










  • $begingroup$
    Perhaps you can answer your own question explaining this, so that the query does not go unanswered.
    $endgroup$
    – Arturo Magidin
    Dec 17 '18 at 19:31














2












2








2





$begingroup$


So I was thinking about Nielsen-Schreier formula for the generators of a free group subgroup and i tried to apply it. Something must be wrong but I can't find the mistake.



So I have $F(a,b)$ free group generated by two element and an surjective homomorphism $F(a,b)to F(b)$ that acts by killing the "a" letter from a word
(i.e. $ phi(aba^{-1}b)=bb=b^2)$.
So we should have $frac{F(a,b)}{mathrm{ker}(phi)} cong F(b)$ free group on one element.



I want to find the generators of the kernel by applying Nielsen-Schreier formula, which states that $mathrm{ker}(phi)$ is freely generated by this set
$$B = { u x (overline{u x})^{-1} mid u in U, x in X }$$
where $U$ is a Schreier transversal for $mathrm{ker}(phi)$ and $X$ is ${a,b}$.



I figured that $mathrm{ker}(phi)$ obviously contains $F(a)$ and that consists of all the words that has the same amount of $b$ and $b^{-1}$ or, in other words, of all the words $w=a^{n_1}b^{m1}a^{n2}b^{m2}cdots $ so that $sum_{i=1}^n m_i=0$.



In the next step I tried to compute the quotient $frac{F(a,b)}{mathrm{ker}(phi)}={1,bH,b^{-1}H,b^2H,b^3H,ldots}$. I'm pretty sure that's it because two words are in relation only if, summing the power of all the $b$ and $b^{-1}$ in the word the result is the same. (for example $w=b^{-1}abaab$ with a $b$ total power of $-1+1+1=1$ is in relationship with $v=aba$, in fact $wv^{-1} in mathrm{ker}(phi)$.



So the Schreier Transversal is $U={1,b,b^2,ldots}$. The problem is that applying now the formula to this Schreier Transversal I got that the generator of $mathrm{ker}(phi)$ is ${a}$, which makes no sense.



Where's the mistake?






abstract-algebra free-groups







share|cite|improve this question











$endgroup$




So I was thinking about Nielsen-Schreier formula for the generators of a free group subgroup and i tried to apply it. Something must be wrong but I can't find the mistake.



So I have $F(a,b)$ free group generated by two element and an surjective homomorphism $F(a,b)to F(b)$ that acts by killing the "a" letter from a word
(i.e. $ phi(aba^{-1}b)=bb=b^2)$.
So we should have $frac{F(a,b)}{mathrm{ker}(phi)} cong F(b)$ free group on one element.



I want to find the generators of the kernel by applying Nielsen-Schreier formula, which states that $mathrm{ker}(phi)$ is freely generated by this set
$$B = { u x (overline{u x})^{-1} mid u in U, x in X }$$
where $U$ is a Schreier transversal for $mathrm{ker}(phi)$ and $X$ is ${a,b}$.



I figured that $mathrm{ker}(phi)$ obviously contains $F(a)$ and that consists of all the words that has the same amount of $b$ and $b^{-1}$ or, in other words, of all the words $w=a^{n_1}b^{m1}a^{n2}b^{m2}cdots $ so that $sum_{i=1}^n m_i=0$.



In the next step I tried to compute the quotient $frac{F(a,b)}{mathrm{ker}(phi)}={1,bH,b^{-1}H,b^2H,b^3H,ldots}$. I'm pretty sure that's it because two words are in relation only if, summing the power of all the $b$ and $b^{-1}$ in the word the result is the same. (for example $w=b^{-1}abaab$ with a $b$ total power of $-1+1+1=1$ is in relationship with $v=aba$, in fact $wv^{-1} in mathrm{ker}(phi)$.



So the Schreier Transversal is $U={1,b,b^2,ldots}$. The problem is that applying now the formula to this Schreier Transversal I got that the generator of $mathrm{ker}(phi)$ is ${a}$, which makes no sense.



Where's the mistake?






abstract-algebra free-groups




abstract-algebra free-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 20:21









Arturo Magidin

261k33585906




261k33585906










asked Dec 14 '18 at 18:38









Andrea LicataAndrea Licata

161




161












  • $begingroup$
    Isn't the overline meant to indicate that you are taking the inverse of the element of $U$ that corresponds to the coset of $ux$? If so, then your $u$ is of the form $b^i$; if $x=a$, then $b^ia$ corresponds to $b^iin U$, so $ux(overline{ux})^{-1} = b^ia(overline{b^ia})^{-1} = b^ia(b^i)^{-1} = b^iab^{-i}$; so that your generating set consists of all conjugates of $a$ by powers of $b$.
    $endgroup$
    – Arturo Magidin
    Dec 14 '18 at 20:29












  • $begingroup$
    Where you perhaps calculagting $xu(overline{xu})^{-1}$ by mistake?
    $endgroup$
    – Arturo Magidin
    Dec 14 '18 at 21:28










  • $begingroup$
    Yup, that was the problem. I was sure i was doing something dumb! Thank you so much
    $endgroup$
    – Andrea Licata
    Dec 17 '18 at 16:08










  • $begingroup$
    Perhaps you can answer your own question explaining this, so that the query does not go unanswered.
    $endgroup$
    – Arturo Magidin
    Dec 17 '18 at 19:31


















  • $begingroup$
    Isn't the overline meant to indicate that you are taking the inverse of the element of $U$ that corresponds to the coset of $ux$? If so, then your $u$ is of the form $b^i$; if $x=a$, then $b^ia$ corresponds to $b^iin U$, so $ux(overline{ux})^{-1} = b^ia(overline{b^ia})^{-1} = b^ia(b^i)^{-1} = b^iab^{-i}$; so that your generating set consists of all conjugates of $a$ by powers of $b$.
    $endgroup$
    – Arturo Magidin
    Dec 14 '18 at 20:29












  • $begingroup$
    Where you perhaps calculagting $xu(overline{xu})^{-1}$ by mistake?
    $endgroup$
    – Arturo Magidin
    Dec 14 '18 at 21:28










  • $begingroup$
    Yup, that was the problem. I was sure i was doing something dumb! Thank you so much
    $endgroup$
    – Andrea Licata
    Dec 17 '18 at 16:08










  • $begingroup$
    Perhaps you can answer your own question explaining this, so that the query does not go unanswered.
    $endgroup$
    – Arturo Magidin
    Dec 17 '18 at 19:31
















$begingroup$
Isn't the overline meant to indicate that you are taking the inverse of the element of $U$ that corresponds to the coset of $ux$? If so, then your $u$ is of the form $b^i$; if $x=a$, then $b^ia$ corresponds to $b^iin U$, so $ux(overline{ux})^{-1} = b^ia(overline{b^ia})^{-1} = b^ia(b^i)^{-1} = b^iab^{-i}$; so that your generating set consists of all conjugates of $a$ by powers of $b$.
$endgroup$
– Arturo Magidin
Dec 14 '18 at 20:29






$begingroup$
Isn't the overline meant to indicate that you are taking the inverse of the element of $U$ that corresponds to the coset of $ux$? If so, then your $u$ is of the form $b^i$; if $x=a$, then $b^ia$ corresponds to $b^iin U$, so $ux(overline{ux})^{-1} = b^ia(overline{b^ia})^{-1} = b^ia(b^i)^{-1} = b^iab^{-i}$; so that your generating set consists of all conjugates of $a$ by powers of $b$.
$endgroup$
– Arturo Magidin
Dec 14 '18 at 20:29














$begingroup$
Where you perhaps calculagting $xu(overline{xu})^{-1}$ by mistake?
$endgroup$
– Arturo Magidin
Dec 14 '18 at 21:28




$begingroup$
Where you perhaps calculagting $xu(overline{xu})^{-1}$ by mistake?
$endgroup$
– Arturo Magidin
Dec 14 '18 at 21:28












$begingroup$
Yup, that was the problem. I was sure i was doing something dumb! Thank you so much
$endgroup$
– Andrea Licata
Dec 17 '18 at 16:08




$begingroup$
Yup, that was the problem. I was sure i was doing something dumb! Thank you so much
$endgroup$
– Andrea Licata
Dec 17 '18 at 16:08












$begingroup$
Perhaps you can answer your own question explaining this, so that the query does not go unanswered.
$endgroup$
– Arturo Magidin
Dec 17 '18 at 19:31




$begingroup$
Perhaps you can answer your own question explaining this, so that the query does not go unanswered.
$endgroup$
– Arturo Magidin
Dec 17 '18 at 19:31










1 Answer
1






active

oldest

votes


















0












$begingroup$

I was miscalculating the base of $ker phi$ ($xu(overline{xu})^{-1}$ instead of $ux(overline{ux})^{-1}$). Applyng the right formula we have that $B={b^iab^{-i}, i in Z}$






share|cite|improve this answer









$endgroup$













    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%2f3039762%2fnielsen-schreier-formula-application%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









    0












    $begingroup$

    I was miscalculating the base of $ker phi$ ($xu(overline{xu})^{-1}$ instead of $ux(overline{ux})^{-1}$). Applyng the right formula we have that $B={b^iab^{-i}, i in Z}$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I was miscalculating the base of $ker phi$ ($xu(overline{xu})^{-1}$ instead of $ux(overline{ux})^{-1}$). Applyng the right formula we have that $B={b^iab^{-i}, i in Z}$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I was miscalculating the base of $ker phi$ ($xu(overline{xu})^{-1}$ instead of $ux(overline{ux})^{-1}$). Applyng the right formula we have that $B={b^iab^{-i}, i in Z}$






        share|cite|improve this answer









        $endgroup$



        I was miscalculating the base of $ker phi$ ($xu(overline{xu})^{-1}$ instead of $ux(overline{ux})^{-1}$). Applyng the right formula we have that $B={b^iab^{-i}, i in Z}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 30 '18 at 19:33









        Andrea LicataAndrea Licata

        161




        161






























            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%2f3039762%2fnielsen-schreier-formula-application%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

            Bressuire

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more

            Cabo Verde

            Image
              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
            Read more

            Gyllenstierna

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more