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
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$begingroup$

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









    $endgroup$













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      2 Answers
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      2 Answers
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      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






























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