Prove that $langle x,y,|x^2,y^2rangle$ is an infinite group











up vote
1
down vote

favorite












Here is my attempt:
$xyx neq xyxy neq xyxyx neq xyxyxy....$ gives infinite many elements in the group.
Is this correct?










share|cite|improve this question


















  • 1




    Yes...as long as it is clear to you why those elements are non-trivial. Your group is the free product $;C_2*C_2;$ , and free products are finite iff one of the factors is the trivial group and the other one is a finite group.
    – DonAntonio
    2 days ago






  • 1




    You have the right idea, but it'd be better to formalize it. You might instead try showing that the powers of $xy$ are distinct.
    – Cameron Buie
    2 days ago










  • Why exactly are those words distinct (in particular, nontrivial)?
    – anomaly
    2 days ago















up vote
1
down vote

favorite












Here is my attempt:
$xyx neq xyxy neq xyxyx neq xyxyxy....$ gives infinite many elements in the group.
Is this correct?










share|cite|improve this question


















  • 1




    Yes...as long as it is clear to you why those elements are non-trivial. Your group is the free product $;C_2*C_2;$ , and free products are finite iff one of the factors is the trivial group and the other one is a finite group.
    – DonAntonio
    2 days ago






  • 1




    You have the right idea, but it'd be better to formalize it. You might instead try showing that the powers of $xy$ are distinct.
    – Cameron Buie
    2 days ago










  • Why exactly are those words distinct (in particular, nontrivial)?
    – anomaly
    2 days ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Here is my attempt:
$xyx neq xyxy neq xyxyx neq xyxyxy....$ gives infinite many elements in the group.
Is this correct?










share|cite|improve this question













Here is my attempt:
$xyx neq xyxy neq xyxyx neq xyxyxy....$ gives infinite many elements in the group.
Is this correct?







finitely-generated






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 days ago









mathnoob

73211




73211








  • 1




    Yes...as long as it is clear to you why those elements are non-trivial. Your group is the free product $;C_2*C_2;$ , and free products are finite iff one of the factors is the trivial group and the other one is a finite group.
    – DonAntonio
    2 days ago






  • 1




    You have the right idea, but it'd be better to formalize it. You might instead try showing that the powers of $xy$ are distinct.
    – Cameron Buie
    2 days ago










  • Why exactly are those words distinct (in particular, nontrivial)?
    – anomaly
    2 days ago














  • 1




    Yes...as long as it is clear to you why those elements are non-trivial. Your group is the free product $;C_2*C_2;$ , and free products are finite iff one of the factors is the trivial group and the other one is a finite group.
    – DonAntonio
    2 days ago






  • 1




    You have the right idea, but it'd be better to formalize it. You might instead try showing that the powers of $xy$ are distinct.
    – Cameron Buie
    2 days ago










  • Why exactly are those words distinct (in particular, nontrivial)?
    – anomaly
    2 days ago








1




1




Yes...as long as it is clear to you why those elements are non-trivial. Your group is the free product $;C_2*C_2;$ , and free products are finite iff one of the factors is the trivial group and the other one is a finite group.
– DonAntonio
2 days ago




Yes...as long as it is clear to you why those elements are non-trivial. Your group is the free product $;C_2*C_2;$ , and free products are finite iff one of the factors is the trivial group and the other one is a finite group.
– DonAntonio
2 days ago




1




1




You have the right idea, but it'd be better to formalize it. You might instead try showing that the powers of $xy$ are distinct.
– Cameron Buie
2 days ago




You have the right idea, but it'd be better to formalize it. You might instead try showing that the powers of $xy$ are distinct.
– Cameron Buie
2 days ago












Why exactly are those words distinct (in particular, nontrivial)?
– anomaly
2 days ago




Why exactly are those words distinct (in particular, nontrivial)?
– anomaly
2 days ago










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










You're on the right track! Let $text{FG}(x,y)=langle x,yrangle$, $G=langle x,ymid x^2,y^2rangle$, and $H$ be the minimal normal subgroup of $text{FG}(x,y)$ containing both $x^2$ and $y^2$. There is a canonical quotient homomorphism $eta:text{FG}(x,y)to G$.



We know that every element of $text{FG}(x,y)$ may be uniquely presented as a concatenation of $x,y,x^{-1},y^{1}$ such that no $x$ borders an $x^{-1}$ and likewise for $y$. For an arbitrary element $gin text{FG}(x,y)$ denote this presentation as $overline{g}$.



Consider the following algorithm $A$: given $gin text{FG}(x,y)$ we replace each instance of $x^{-1}$ and $y^{-1}$ in $overline{g}$ with $x$ and $y$ respectively. We then remove stepwise the leftmnost instances of $xx$ and $yy$ present until no such substrings remain. To take concrete examples, $$A:x^{-1}yyxxmapsto xoverbrace{yy}xxmapsto overbrace{xx}xmapsto x$$ $$A:x^{-1}yyxxxmapsto xoverbrace{yy}xxxmapsto overbrace{xx}xxmapsto overbrace{xx}mapsto text{empty}$$ Where "$text{empty}$" is the empty string.



Denote $H'$ the subset of elements $text{FG}(x,y)$ for whom $A$ outputs the null string. We can show that $H'$ is a subgroup of $text{FG}(x,y)$ and in particular a normal subgroup of $text{FG}(x,y)$. Let $eta':text{FG}(x,y)to text{FG}(x,y)/H'$ be the associated quotient map. Then we have that $eta'$ factors through a morphism $kappa:Gto text{FG}(x,y)/H'$ in the sense that $eta'=kappacirc eta$. Our $kappa$ is therefore surjective (since $eta'$ is surjective). So it suffices to demonstrate that $text{FG}(x,y)/H'$ is infinite.



To that end consider the sequence of elements $$eta'(xyx) neq eta(xyxy)' neq eta'(xyxyx) neq eta'(xyxyxy)dots$$
We will prove that the elements of this sequence are pairwise distinct. Let $eta'(g_1)$ and $eta'(g_2)$ be two such elements. It suffices to show that $eta'(g_1)eta'(g_2)^{-1}neq 1iff A(g_1g_2^{-1})neq text{empty}$. Can you finish from here?





It's worth noting that $H=H'$ and thus that $Gcong text{FG}(x,y)/H'$. The proof isn't that hard. (Specifically we've proven that $Hsubseteq H'$, proving that $H'subseteq H$ would do the trick.)






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',
    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%2f3007751%2fprove-that-langle-x-y-x2-y2-rangle-is-an-infinite-group%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








    up vote
    1
    down vote



    accepted










    You're on the right track! Let $text{FG}(x,y)=langle x,yrangle$, $G=langle x,ymid x^2,y^2rangle$, and $H$ be the minimal normal subgroup of $text{FG}(x,y)$ containing both $x^2$ and $y^2$. There is a canonical quotient homomorphism $eta:text{FG}(x,y)to G$.



    We know that every element of $text{FG}(x,y)$ may be uniquely presented as a concatenation of $x,y,x^{-1},y^{1}$ such that no $x$ borders an $x^{-1}$ and likewise for $y$. For an arbitrary element $gin text{FG}(x,y)$ denote this presentation as $overline{g}$.



    Consider the following algorithm $A$: given $gin text{FG}(x,y)$ we replace each instance of $x^{-1}$ and $y^{-1}$ in $overline{g}$ with $x$ and $y$ respectively. We then remove stepwise the leftmnost instances of $xx$ and $yy$ present until no such substrings remain. To take concrete examples, $$A:x^{-1}yyxxmapsto xoverbrace{yy}xxmapsto overbrace{xx}xmapsto x$$ $$A:x^{-1}yyxxxmapsto xoverbrace{yy}xxxmapsto overbrace{xx}xxmapsto overbrace{xx}mapsto text{empty}$$ Where "$text{empty}$" is the empty string.



    Denote $H'$ the subset of elements $text{FG}(x,y)$ for whom $A$ outputs the null string. We can show that $H'$ is a subgroup of $text{FG}(x,y)$ and in particular a normal subgroup of $text{FG}(x,y)$. Let $eta':text{FG}(x,y)to text{FG}(x,y)/H'$ be the associated quotient map. Then we have that $eta'$ factors through a morphism $kappa:Gto text{FG}(x,y)/H'$ in the sense that $eta'=kappacirc eta$. Our $kappa$ is therefore surjective (since $eta'$ is surjective). So it suffices to demonstrate that $text{FG}(x,y)/H'$ is infinite.



    To that end consider the sequence of elements $$eta'(xyx) neq eta(xyxy)' neq eta'(xyxyx) neq eta'(xyxyxy)dots$$
    We will prove that the elements of this sequence are pairwise distinct. Let $eta'(g_1)$ and $eta'(g_2)$ be two such elements. It suffices to show that $eta'(g_1)eta'(g_2)^{-1}neq 1iff A(g_1g_2^{-1})neq text{empty}$. Can you finish from here?





    It's worth noting that $H=H'$ and thus that $Gcong text{FG}(x,y)/H'$. The proof isn't that hard. (Specifically we've proven that $Hsubseteq H'$, proving that $H'subseteq H$ would do the trick.)






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      You're on the right track! Let $text{FG}(x,y)=langle x,yrangle$, $G=langle x,ymid x^2,y^2rangle$, and $H$ be the minimal normal subgroup of $text{FG}(x,y)$ containing both $x^2$ and $y^2$. There is a canonical quotient homomorphism $eta:text{FG}(x,y)to G$.



      We know that every element of $text{FG}(x,y)$ may be uniquely presented as a concatenation of $x,y,x^{-1},y^{1}$ such that no $x$ borders an $x^{-1}$ and likewise for $y$. For an arbitrary element $gin text{FG}(x,y)$ denote this presentation as $overline{g}$.



      Consider the following algorithm $A$: given $gin text{FG}(x,y)$ we replace each instance of $x^{-1}$ and $y^{-1}$ in $overline{g}$ with $x$ and $y$ respectively. We then remove stepwise the leftmnost instances of $xx$ and $yy$ present until no such substrings remain. To take concrete examples, $$A:x^{-1}yyxxmapsto xoverbrace{yy}xxmapsto overbrace{xx}xmapsto x$$ $$A:x^{-1}yyxxxmapsto xoverbrace{yy}xxxmapsto overbrace{xx}xxmapsto overbrace{xx}mapsto text{empty}$$ Where "$text{empty}$" is the empty string.



      Denote $H'$ the subset of elements $text{FG}(x,y)$ for whom $A$ outputs the null string. We can show that $H'$ is a subgroup of $text{FG}(x,y)$ and in particular a normal subgroup of $text{FG}(x,y)$. Let $eta':text{FG}(x,y)to text{FG}(x,y)/H'$ be the associated quotient map. Then we have that $eta'$ factors through a morphism $kappa:Gto text{FG}(x,y)/H'$ in the sense that $eta'=kappacirc eta$. Our $kappa$ is therefore surjective (since $eta'$ is surjective). So it suffices to demonstrate that $text{FG}(x,y)/H'$ is infinite.



      To that end consider the sequence of elements $$eta'(xyx) neq eta(xyxy)' neq eta'(xyxyx) neq eta'(xyxyxy)dots$$
      We will prove that the elements of this sequence are pairwise distinct. Let $eta'(g_1)$ and $eta'(g_2)$ be two such elements. It suffices to show that $eta'(g_1)eta'(g_2)^{-1}neq 1iff A(g_1g_2^{-1})neq text{empty}$. Can you finish from here?





      It's worth noting that $H=H'$ and thus that $Gcong text{FG}(x,y)/H'$. The proof isn't that hard. (Specifically we've proven that $Hsubseteq H'$, proving that $H'subseteq H$ would do the trick.)






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        You're on the right track! Let $text{FG}(x,y)=langle x,yrangle$, $G=langle x,ymid x^2,y^2rangle$, and $H$ be the minimal normal subgroup of $text{FG}(x,y)$ containing both $x^2$ and $y^2$. There is a canonical quotient homomorphism $eta:text{FG}(x,y)to G$.



        We know that every element of $text{FG}(x,y)$ may be uniquely presented as a concatenation of $x,y,x^{-1},y^{1}$ such that no $x$ borders an $x^{-1}$ and likewise for $y$. For an arbitrary element $gin text{FG}(x,y)$ denote this presentation as $overline{g}$.



        Consider the following algorithm $A$: given $gin text{FG}(x,y)$ we replace each instance of $x^{-1}$ and $y^{-1}$ in $overline{g}$ with $x$ and $y$ respectively. We then remove stepwise the leftmnost instances of $xx$ and $yy$ present until no such substrings remain. To take concrete examples, $$A:x^{-1}yyxxmapsto xoverbrace{yy}xxmapsto overbrace{xx}xmapsto x$$ $$A:x^{-1}yyxxxmapsto xoverbrace{yy}xxxmapsto overbrace{xx}xxmapsto overbrace{xx}mapsto text{empty}$$ Where "$text{empty}$" is the empty string.



        Denote $H'$ the subset of elements $text{FG}(x,y)$ for whom $A$ outputs the null string. We can show that $H'$ is a subgroup of $text{FG}(x,y)$ and in particular a normal subgroup of $text{FG}(x,y)$. Let $eta':text{FG}(x,y)to text{FG}(x,y)/H'$ be the associated quotient map. Then we have that $eta'$ factors through a morphism $kappa:Gto text{FG}(x,y)/H'$ in the sense that $eta'=kappacirc eta$. Our $kappa$ is therefore surjective (since $eta'$ is surjective). So it suffices to demonstrate that $text{FG}(x,y)/H'$ is infinite.



        To that end consider the sequence of elements $$eta'(xyx) neq eta(xyxy)' neq eta'(xyxyx) neq eta'(xyxyxy)dots$$
        We will prove that the elements of this sequence are pairwise distinct. Let $eta'(g_1)$ and $eta'(g_2)$ be two such elements. It suffices to show that $eta'(g_1)eta'(g_2)^{-1}neq 1iff A(g_1g_2^{-1})neq text{empty}$. Can you finish from here?





        It's worth noting that $H=H'$ and thus that $Gcong text{FG}(x,y)/H'$. The proof isn't that hard. (Specifically we've proven that $Hsubseteq H'$, proving that $H'subseteq H$ would do the trick.)






        share|cite|improve this answer












        You're on the right track! Let $text{FG}(x,y)=langle x,yrangle$, $G=langle x,ymid x^2,y^2rangle$, and $H$ be the minimal normal subgroup of $text{FG}(x,y)$ containing both $x^2$ and $y^2$. There is a canonical quotient homomorphism $eta:text{FG}(x,y)to G$.



        We know that every element of $text{FG}(x,y)$ may be uniquely presented as a concatenation of $x,y,x^{-1},y^{1}$ such that no $x$ borders an $x^{-1}$ and likewise for $y$. For an arbitrary element $gin text{FG}(x,y)$ denote this presentation as $overline{g}$.



        Consider the following algorithm $A$: given $gin text{FG}(x,y)$ we replace each instance of $x^{-1}$ and $y^{-1}$ in $overline{g}$ with $x$ and $y$ respectively. We then remove stepwise the leftmnost instances of $xx$ and $yy$ present until no such substrings remain. To take concrete examples, $$A:x^{-1}yyxxmapsto xoverbrace{yy}xxmapsto overbrace{xx}xmapsto x$$ $$A:x^{-1}yyxxxmapsto xoverbrace{yy}xxxmapsto overbrace{xx}xxmapsto overbrace{xx}mapsto text{empty}$$ Where "$text{empty}$" is the empty string.



        Denote $H'$ the subset of elements $text{FG}(x,y)$ for whom $A$ outputs the null string. We can show that $H'$ is a subgroup of $text{FG}(x,y)$ and in particular a normal subgroup of $text{FG}(x,y)$. Let $eta':text{FG}(x,y)to text{FG}(x,y)/H'$ be the associated quotient map. Then we have that $eta'$ factors through a morphism $kappa:Gto text{FG}(x,y)/H'$ in the sense that $eta'=kappacirc eta$. Our $kappa$ is therefore surjective (since $eta'$ is surjective). So it suffices to demonstrate that $text{FG}(x,y)/H'$ is infinite.



        To that end consider the sequence of elements $$eta'(xyx) neq eta(xyxy)' neq eta'(xyxyx) neq eta'(xyxyxy)dots$$
        We will prove that the elements of this sequence are pairwise distinct. Let $eta'(g_1)$ and $eta'(g_2)$ be two such elements. It suffices to show that $eta'(g_1)eta'(g_2)^{-1}neq 1iff A(g_1g_2^{-1})neq text{empty}$. Can you finish from here?





        It's worth noting that $H=H'$ and thus that $Gcong text{FG}(x,y)/H'$. The proof isn't that hard. (Specifically we've proven that $Hsubseteq H'$, proving that $H'subseteq H$ would do the trick.)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Rafay Ashary

        74618




        74618






























             

            draft saved


            draft discarded



















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007751%2fprove-that-langle-x-y-x2-y2-rangle-is-an-infinite-group%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

            Sphinx de Gizeh

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