Baumslag-Solitar Group $G=langle a,t mid tat^{-1}=a^kranglecongmathbb{Z}[1/k]rtimeslangle trangle$?












3












$begingroup$


Let $G$ be the Baumslag-Solitar group $langle a,t mid tat^{-1}=a^krangle$ and
$$mathbb{Z}[1/k]:=left{frac{x}{k^n}mid xinmathbb{Z},ninmathbb{N}cup{0}right}.$$
I'm searching for an isomorphism $G=langle a,t mid tat^{-1}=a^kranglecongmathbb{Z}[1/k]rtimeslangle trangle$. I only need to know the map $Psi:langle arangle^Gtomathbb{Z}[1/k]$.
I defined $$Psi(t^na^xt^{-n}):=frac{x}{k^n}.$$
But for one hour I'm calculating to prove that this map induces an homomorphism, but I always get for $n>m$:



$$Psi(t^na^xt^{-n})cdot Psi(t^ma^yt^{-m})=frac{x}{k^n}+frac{y}{k^m}=frac{x+ycdot k^{n-m}}{k^n}$$
and
$$Psi(t^na^xt^{-n}cdot t^ma^yt^{-m})=Psi(t^ma^{(xcdot k^{n-m})}t^{-m}t^myt^{-m})=Psi(t^ma^{xcdot k^{n-m}+y}t^{-m})=frac{xcdot k^{n-m}+y}{k^m}.$$
And these images are obvisually unequal. So what is my fault?
Is this map the wrong one?



Thanks for your help.






abstract-algebra group-theory geometric-group-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think Baumslag–Solitar group is correct but I am not sure.
    $endgroup$
    – Babak Miraftab
    May 28 '12 at 17:04










  • $begingroup$
    Yes. Thank you.
    $endgroup$
    – Peter
    May 28 '12 at 17:30










  • $begingroup$
    $tat^{-1}=a^k$, so $Psi(t^n a^x t^{-n})$ should multiply (not divide) the exponent of $a$ by $k$. Hence the image should be $xk^n$.
    $endgroup$
    – Ted
    May 28 '12 at 21:11
















3












$begingroup$


Let $G$ be the Baumslag-Solitar group $langle a,t mid tat^{-1}=a^krangle$ and
$$mathbb{Z}[1/k]:=left{frac{x}{k^n}mid xinmathbb{Z},ninmathbb{N}cup{0}right}.$$
I'm searching for an isomorphism $G=langle a,t mid tat^{-1}=a^kranglecongmathbb{Z}[1/k]rtimeslangle trangle$. I only need to know the map $Psi:langle arangle^Gtomathbb{Z}[1/k]$.
I defined $$Psi(t^na^xt^{-n}):=frac{x}{k^n}.$$
But for one hour I'm calculating to prove that this map induces an homomorphism, but I always get for $n>m$:



$$Psi(t^na^xt^{-n})cdot Psi(t^ma^yt^{-m})=frac{x}{k^n}+frac{y}{k^m}=frac{x+ycdot k^{n-m}}{k^n}$$
and
$$Psi(t^na^xt^{-n}cdot t^ma^yt^{-m})=Psi(t^ma^{(xcdot k^{n-m})}t^{-m}t^myt^{-m})=Psi(t^ma^{xcdot k^{n-m}+y}t^{-m})=frac{xcdot k^{n-m}+y}{k^m}.$$
And these images are obvisually unequal. So what is my fault?
Is this map the wrong one?



Thanks for your help.






abstract-algebra group-theory geometric-group-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think Baumslag–Solitar group is correct but I am not sure.
    $endgroup$
    – Babak Miraftab
    May 28 '12 at 17:04










  • $begingroup$
    Yes. Thank you.
    $endgroup$
    – Peter
    May 28 '12 at 17:30










  • $begingroup$
    $tat^{-1}=a^k$, so $Psi(t^n a^x t^{-n})$ should multiply (not divide) the exponent of $a$ by $k$. Hence the image should be $xk^n$.
    $endgroup$
    – Ted
    May 28 '12 at 21:11














3












3








3


2



$begingroup$


Let $G$ be the Baumslag-Solitar group $langle a,t mid tat^{-1}=a^krangle$ and
$$mathbb{Z}[1/k]:=left{frac{x}{k^n}mid xinmathbb{Z},ninmathbb{N}cup{0}right}.$$
I'm searching for an isomorphism $G=langle a,t mid tat^{-1}=a^kranglecongmathbb{Z}[1/k]rtimeslangle trangle$. I only need to know the map $Psi:langle arangle^Gtomathbb{Z}[1/k]$.
I defined $$Psi(t^na^xt^{-n}):=frac{x}{k^n}.$$
But for one hour I'm calculating to prove that this map induces an homomorphism, but I always get for $n>m$:



$$Psi(t^na^xt^{-n})cdot Psi(t^ma^yt^{-m})=frac{x}{k^n}+frac{y}{k^m}=frac{x+ycdot k^{n-m}}{k^n}$$
and
$$Psi(t^na^xt^{-n}cdot t^ma^yt^{-m})=Psi(t^ma^{(xcdot k^{n-m})}t^{-m}t^myt^{-m})=Psi(t^ma^{xcdot k^{n-m}+y}t^{-m})=frac{xcdot k^{n-m}+y}{k^m}.$$
And these images are obvisually unequal. So what is my fault?
Is this map the wrong one?



Thanks for your help.






abstract-algebra group-theory geometric-group-theory







share|cite|improve this question











$endgroup$




Let $G$ be the Baumslag-Solitar group $langle a,t mid tat^{-1}=a^krangle$ and
$$mathbb{Z}[1/k]:=left{frac{x}{k^n}mid xinmathbb{Z},ninmathbb{N}cup{0}right}.$$
I'm searching for an isomorphism $G=langle a,t mid tat^{-1}=a^kranglecongmathbb{Z}[1/k]rtimeslangle trangle$. I only need to know the map $Psi:langle arangle^Gtomathbb{Z}[1/k]$.
I defined $$Psi(t^na^xt^{-n}):=frac{x}{k^n}.$$
But for one hour I'm calculating to prove that this map induces an homomorphism, but I always get for $n>m$:



$$Psi(t^na^xt^{-n})cdot Psi(t^ma^yt^{-m})=frac{x}{k^n}+frac{y}{k^m}=frac{x+ycdot k^{n-m}}{k^n}$$
and
$$Psi(t^na^xt^{-n}cdot t^ma^yt^{-m})=Psi(t^ma^{(xcdot k^{n-m})}t^{-m}t^myt^{-m})=Psi(t^ma^{xcdot k^{n-m}+y}t^{-m})=frac{xcdot k^{n-m}+y}{k^m}.$$
And these images are obvisually unequal. So what is my fault?
Is this map the wrong one?



Thanks for your help.






abstract-algebra group-theory geometric-group-theory




abstract-algebra group-theory geometric-group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 28 '12 at 19:32









t.b.

62.4k7207287




62.4k7207287










asked May 28 '12 at 17:01









PeterPeter

1176




1176












  • $begingroup$
    I think Baumslag–Solitar group is correct but I am not sure.
    $endgroup$
    – Babak Miraftab
    May 28 '12 at 17:04










  • $begingroup$
    Yes. Thank you.
    $endgroup$
    – Peter
    May 28 '12 at 17:30










  • $begingroup$
    $tat^{-1}=a^k$, so $Psi(t^n a^x t^{-n})$ should multiply (not divide) the exponent of $a$ by $k$. Hence the image should be $xk^n$.
    $endgroup$
    – Ted
    May 28 '12 at 21:11


















  • $begingroup$
    I think Baumslag–Solitar group is correct but I am not sure.
    $endgroup$
    – Babak Miraftab
    May 28 '12 at 17:04










  • $begingroup$
    Yes. Thank you.
    $endgroup$
    – Peter
    May 28 '12 at 17:30










  • $begingroup$
    $tat^{-1}=a^k$, so $Psi(t^n a^x t^{-n})$ should multiply (not divide) the exponent of $a$ by $k$. Hence the image should be $xk^n$.
    $endgroup$
    – Ted
    May 28 '12 at 21:11
















$begingroup$
I think Baumslag–Solitar group is correct but I am not sure.
$endgroup$
– Babak Miraftab
May 28 '12 at 17:04




$begingroup$
I think Baumslag–Solitar group is correct but I am not sure.
$endgroup$
– Babak Miraftab
May 28 '12 at 17:04












$begingroup$
Yes. Thank you.
$endgroup$
– Peter
May 28 '12 at 17:30




$begingroup$
Yes. Thank you.
$endgroup$
– Peter
May 28 '12 at 17:30












$begingroup$
$tat^{-1}=a^k$, so $Psi(t^n a^x t^{-n})$ should multiply (not divide) the exponent of $a$ by $k$. Hence the image should be $xk^n$.
$endgroup$
– Ted
May 28 '12 at 21:11




$begingroup$
$tat^{-1}=a^k$, so $Psi(t^n a^x t^{-n})$ should multiply (not divide) the exponent of $a$ by $k$. Hence the image should be $xk^n$.
$endgroup$
– Ted
May 28 '12 at 21:11










2 Answers
2






active

oldest

votes


















6












$begingroup$

You need to modify the definition of $Psi$: put $$Psi(t^na^xt^{-n})=xk^n$$ (instead of $Psi(t^na^xt^{-n})=xk^{-n}$). The homomorphic property should follow easily along the lines of the computations given in the original question.



Alternatively, you could replace $t$ by $t^{-1}$ in the definition of the Baumslag-Solitar group, in which case the $Psi$ as originally given is a homomorphism.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Why any word in $BS(1,2)$ is of the form $a^{-k}b^ma^n$?






    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%2f150839%2fbaumslag-solitar-group-g-langle-a-t-mid-tat-1-ak-rangle-cong-mathbbz1%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












      $begingroup$

      You need to modify the definition of $Psi$: put $$Psi(t^na^xt^{-n})=xk^n$$ (instead of $Psi(t^na^xt^{-n})=xk^{-n}$). The homomorphic property should follow easily along the lines of the computations given in the original question.



      Alternatively, you could replace $t$ by $t^{-1}$ in the definition of the Baumslag-Solitar group, in which case the $Psi$ as originally given is a homomorphism.






      share|cite|improve this answer









      $endgroup$


















        6












        $begingroup$

        You need to modify the definition of $Psi$: put $$Psi(t^na^xt^{-n})=xk^n$$ (instead of $Psi(t^na^xt^{-n})=xk^{-n}$). The homomorphic property should follow easily along the lines of the computations given in the original question.



        Alternatively, you could replace $t$ by $t^{-1}$ in the definition of the Baumslag-Solitar group, in which case the $Psi$ as originally given is a homomorphism.






        share|cite|improve this answer









        $endgroup$
















          6












          6








          6





          $begingroup$

          You need to modify the definition of $Psi$: put $$Psi(t^na^xt^{-n})=xk^n$$ (instead of $Psi(t^na^xt^{-n})=xk^{-n}$). The homomorphic property should follow easily along the lines of the computations given in the original question.



          Alternatively, you could replace $t$ by $t^{-1}$ in the definition of the Baumslag-Solitar group, in which case the $Psi$ as originally given is a homomorphism.






          share|cite|improve this answer









          $endgroup$



          You need to modify the definition of $Psi$: put $$Psi(t^na^xt^{-n})=xk^n$$ (instead of $Psi(t^na^xt^{-n})=xk^{-n}$). The homomorphic property should follow easily along the lines of the computations given in the original question.



          Alternatively, you could replace $t$ by $t^{-1}$ in the definition of the Baumslag-Solitar group, in which case the $Psi$ as originally given is a homomorphism.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered May 28 '12 at 20:12









          Shane O RourkeShane O Rourke

          911811




          911811























              1












              $begingroup$

              Why any word in $BS(1,2)$ is of the form $a^{-k}b^ma^n$?






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Why any word in $BS(1,2)$ is of the form $a^{-k}b^ma^n$?






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Why any word in $BS(1,2)$ is of the form $a^{-k}b^ma^n$?






                  share|cite|improve this answer









                  $endgroup$



                  Why any word in $BS(1,2)$ is of the form $a^{-k}b^ma^n$?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 15 '18 at 22:35









                  eraldcoileraldcoil

                  395211




                  395211






























                      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%2f150839%2fbaumslag-solitar-group-g-langle-a-t-mid-tat-1-ak-rangle-cong-mathbbz1%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

                      Wiesbaden

                      Image
                      Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
                      Read more

                      Marschland

                      Image
                      Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
                      Read more

                      Dieringhausen

                      Image
                      Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
                      Read more