Quotient of the Baumslag-Solitar group $BS(1,m)=langle a,b| bab^{-1}=a^mrangle$.












3












$begingroup$


I would like to prove that in any non-trivial quotient of the Baumslag-Solitar group $BS(1,m)$ defined by,
$$BS(1,m)=langle a,b| bab^{-1}=a^mrangle$$
the image of one of the generators $a$ or $b$ (by the canonical projection) have finite order. I only know that $BS(1,m)$ is solvable but I don't know if it will help me to prove the statement.
By advance thank you. Edgar.










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migrated from mathoverflow.net Jun 1 '15 at 13:47


This question came from our site for professional mathematicians.


















  • $begingroup$
    I don't think $m$ needs to be prime, it's just an integer. By the way how do you prove that $BS(1, 2)$ is a quotient of $BS(1, 4)$? If my statement is true, then what is the order of the images of the generators of $BS(1, 4)$ in $BS(1, 2)$?
    $endgroup$
    – Edgar Ndie
    Jun 1 '15 at 11:52










  • $begingroup$
    I think this is more suitable for MSE. let $N = langle a^G rangle$ be the normal closure of $a$. Every element of $N$ is conjugate to a power of $a$, so if you factor out an element of $N$ then you reduce $a$ to finite order. Otherwise, you factor out $b^kn$ for some $k ne 0$ and $n in N$, but $b^k$ does not centralize any element of $N$ (you do need to assume that $m ne pm 1$), so that would force $n=1$.
    $endgroup$
    – Derek Holt
    Jun 1 '15 at 12:18










  • $begingroup$
    @Derek Thanks - I was too hasty. I've deleted both my comments.
    $endgroup$
    – user1729
    Jun 1 '15 at 12:23










  • $begingroup$
    @Derek Why is every element of $N$ conjugate to a power of $a$? For example, $b^{-1}aba^min N$, and by the conjugacy theorem for HNN-extensions we would need that $a^ib^{-1}aba^j=a^k$, which does not hold. No?
    $endgroup$
    – user1729
    Jun 1 '15 at 12:33






  • 1




    $begingroup$
    @user1729 Of course, I meant conjugate in $G={rm BS}(1,m)$ to a power of $a$. Then $b(b^{-1}aba^m)b^{-1} = a^{m^2+1}$.
    $endgroup$
    – Derek Holt
    Jun 1 '15 at 12:50
















3












$begingroup$


I would like to prove that in any non-trivial quotient of the Baumslag-Solitar group $BS(1,m)$ defined by,
$$BS(1,m)=langle a,b| bab^{-1}=a^mrangle$$
the image of one of the generators $a$ or $b$ (by the canonical projection) have finite order. I only know that $BS(1,m)$ is solvable but I don't know if it will help me to prove the statement.
By advance thank you. Edgar.










share|cite|improve this question











$endgroup$



migrated from mathoverflow.net Jun 1 '15 at 13:47


This question came from our site for professional mathematicians.


















  • $begingroup$
    I don't think $m$ needs to be prime, it's just an integer. By the way how do you prove that $BS(1, 2)$ is a quotient of $BS(1, 4)$? If my statement is true, then what is the order of the images of the generators of $BS(1, 4)$ in $BS(1, 2)$?
    $endgroup$
    – Edgar Ndie
    Jun 1 '15 at 11:52










  • $begingroup$
    I think this is more suitable for MSE. let $N = langle a^G rangle$ be the normal closure of $a$. Every element of $N$ is conjugate to a power of $a$, so if you factor out an element of $N$ then you reduce $a$ to finite order. Otherwise, you factor out $b^kn$ for some $k ne 0$ and $n in N$, but $b^k$ does not centralize any element of $N$ (you do need to assume that $m ne pm 1$), so that would force $n=1$.
    $endgroup$
    – Derek Holt
    Jun 1 '15 at 12:18










  • $begingroup$
    @Derek Thanks - I was too hasty. I've deleted both my comments.
    $endgroup$
    – user1729
    Jun 1 '15 at 12:23










  • $begingroup$
    @Derek Why is every element of $N$ conjugate to a power of $a$? For example, $b^{-1}aba^min N$, and by the conjugacy theorem for HNN-extensions we would need that $a^ib^{-1}aba^j=a^k$, which does not hold. No?
    $endgroup$
    – user1729
    Jun 1 '15 at 12:33






  • 1




    $begingroup$
    @user1729 Of course, I meant conjugate in $G={rm BS}(1,m)$ to a power of $a$. Then $b(b^{-1}aba^m)b^{-1} = a^{m^2+1}$.
    $endgroup$
    – Derek Holt
    Jun 1 '15 at 12:50














3












3








3


1



$begingroup$


I would like to prove that in any non-trivial quotient of the Baumslag-Solitar group $BS(1,m)$ defined by,
$$BS(1,m)=langle a,b| bab^{-1}=a^mrangle$$
the image of one of the generators $a$ or $b$ (by the canonical projection) have finite order. I only know that $BS(1,m)$ is solvable but I don't know if it will help me to prove the statement.
By advance thank you. Edgar.










share|cite|improve this question











$endgroup$




I would like to prove that in any non-trivial quotient of the Baumslag-Solitar group $BS(1,m)$ defined by,
$$BS(1,m)=langle a,b| bab^{-1}=a^mrangle$$
the image of one of the generators $a$ or $b$ (by the canonical projection) have finite order. I only know that $BS(1,m)$ is solvable but I don't know if it will help me to prove the statement.
By advance thank you. Edgar.







group-theory quotient-group






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 0:38









Shaun

9,268113684




9,268113684










asked Jun 1 '15 at 10:42









Edgar NdieEdgar Ndie

836




836




migrated from mathoverflow.net Jun 1 '15 at 13:47


This question came from our site for professional mathematicians.









migrated from mathoverflow.net Jun 1 '15 at 13:47


This question came from our site for professional mathematicians.














  • $begingroup$
    I don't think $m$ needs to be prime, it's just an integer. By the way how do you prove that $BS(1, 2)$ is a quotient of $BS(1, 4)$? If my statement is true, then what is the order of the images of the generators of $BS(1, 4)$ in $BS(1, 2)$?
    $endgroup$
    – Edgar Ndie
    Jun 1 '15 at 11:52










  • $begingroup$
    I think this is more suitable for MSE. let $N = langle a^G rangle$ be the normal closure of $a$. Every element of $N$ is conjugate to a power of $a$, so if you factor out an element of $N$ then you reduce $a$ to finite order. Otherwise, you factor out $b^kn$ for some $k ne 0$ and $n in N$, but $b^k$ does not centralize any element of $N$ (you do need to assume that $m ne pm 1$), so that would force $n=1$.
    $endgroup$
    – Derek Holt
    Jun 1 '15 at 12:18










  • $begingroup$
    @Derek Thanks - I was too hasty. I've deleted both my comments.
    $endgroup$
    – user1729
    Jun 1 '15 at 12:23










  • $begingroup$
    @Derek Why is every element of $N$ conjugate to a power of $a$? For example, $b^{-1}aba^min N$, and by the conjugacy theorem for HNN-extensions we would need that $a^ib^{-1}aba^j=a^k$, which does not hold. No?
    $endgroup$
    – user1729
    Jun 1 '15 at 12:33






  • 1




    $begingroup$
    @user1729 Of course, I meant conjugate in $G={rm BS}(1,m)$ to a power of $a$. Then $b(b^{-1}aba^m)b^{-1} = a^{m^2+1}$.
    $endgroup$
    – Derek Holt
    Jun 1 '15 at 12:50


















  • $begingroup$
    I don't think $m$ needs to be prime, it's just an integer. By the way how do you prove that $BS(1, 2)$ is a quotient of $BS(1, 4)$? If my statement is true, then what is the order of the images of the generators of $BS(1, 4)$ in $BS(1, 2)$?
    $endgroup$
    – Edgar Ndie
    Jun 1 '15 at 11:52










  • $begingroup$
    I think this is more suitable for MSE. let $N = langle a^G rangle$ be the normal closure of $a$. Every element of $N$ is conjugate to a power of $a$, so if you factor out an element of $N$ then you reduce $a$ to finite order. Otherwise, you factor out $b^kn$ for some $k ne 0$ and $n in N$, but $b^k$ does not centralize any element of $N$ (you do need to assume that $m ne pm 1$), so that would force $n=1$.
    $endgroup$
    – Derek Holt
    Jun 1 '15 at 12:18










  • $begingroup$
    @Derek Thanks - I was too hasty. I've deleted both my comments.
    $endgroup$
    – user1729
    Jun 1 '15 at 12:23










  • $begingroup$
    @Derek Why is every element of $N$ conjugate to a power of $a$? For example, $b^{-1}aba^min N$, and by the conjugacy theorem for HNN-extensions we would need that $a^ib^{-1}aba^j=a^k$, which does not hold. No?
    $endgroup$
    – user1729
    Jun 1 '15 at 12:33






  • 1




    $begingroup$
    @user1729 Of course, I meant conjugate in $G={rm BS}(1,m)$ to a power of $a$. Then $b(b^{-1}aba^m)b^{-1} = a^{m^2+1}$.
    $endgroup$
    – Derek Holt
    Jun 1 '15 at 12:50
















$begingroup$
I don't think $m$ needs to be prime, it's just an integer. By the way how do you prove that $BS(1, 2)$ is a quotient of $BS(1, 4)$? If my statement is true, then what is the order of the images of the generators of $BS(1, 4)$ in $BS(1, 2)$?
$endgroup$
– Edgar Ndie
Jun 1 '15 at 11:52




$begingroup$
I don't think $m$ needs to be prime, it's just an integer. By the way how do you prove that $BS(1, 2)$ is a quotient of $BS(1, 4)$? If my statement is true, then what is the order of the images of the generators of $BS(1, 4)$ in $BS(1, 2)$?
$endgroup$
– Edgar Ndie
Jun 1 '15 at 11:52












$begingroup$
I think this is more suitable for MSE. let $N = langle a^G rangle$ be the normal closure of $a$. Every element of $N$ is conjugate to a power of $a$, so if you factor out an element of $N$ then you reduce $a$ to finite order. Otherwise, you factor out $b^kn$ for some $k ne 0$ and $n in N$, but $b^k$ does not centralize any element of $N$ (you do need to assume that $m ne pm 1$), so that would force $n=1$.
$endgroup$
– Derek Holt
Jun 1 '15 at 12:18




$begingroup$
I think this is more suitable for MSE. let $N = langle a^G rangle$ be the normal closure of $a$. Every element of $N$ is conjugate to a power of $a$, so if you factor out an element of $N$ then you reduce $a$ to finite order. Otherwise, you factor out $b^kn$ for some $k ne 0$ and $n in N$, but $b^k$ does not centralize any element of $N$ (you do need to assume that $m ne pm 1$), so that would force $n=1$.
$endgroup$
– Derek Holt
Jun 1 '15 at 12:18












$begingroup$
@Derek Thanks - I was too hasty. I've deleted both my comments.
$endgroup$
– user1729
Jun 1 '15 at 12:23




$begingroup$
@Derek Thanks - I was too hasty. I've deleted both my comments.
$endgroup$
– user1729
Jun 1 '15 at 12:23












$begingroup$
@Derek Why is every element of $N$ conjugate to a power of $a$? For example, $b^{-1}aba^min N$, and by the conjugacy theorem for HNN-extensions we would need that $a^ib^{-1}aba^j=a^k$, which does not hold. No?
$endgroup$
– user1729
Jun 1 '15 at 12:33




$begingroup$
@Derek Why is every element of $N$ conjugate to a power of $a$? For example, $b^{-1}aba^min N$, and by the conjugacy theorem for HNN-extensions we would need that $a^ib^{-1}aba^j=a^k$, which does not hold. No?
$endgroup$
– user1729
Jun 1 '15 at 12:33




1




1




$begingroup$
@user1729 Of course, I meant conjugate in $G={rm BS}(1,m)$ to a power of $a$. Then $b(b^{-1}aba^m)b^{-1} = a^{m^2+1}$.
$endgroup$
– Derek Holt
Jun 1 '15 at 12:50




$begingroup$
@user1729 Of course, I meant conjugate in $G={rm BS}(1,m)$ to a power of $a$. Then $b(b^{-1}aba^m)b^{-1} = a^{m^2+1}$.
$endgroup$
– Derek Holt
Jun 1 '15 at 12:50










1 Answer
1






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oldest

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6












$begingroup$

First note that the result is false when $m = 1$, because $G$ is just free abelian of rank $2$ in that case, and the subgroup $K=langle ab rangle$ is normal but neither $a$ nor $b$ has finite order $G/K$. So assume that $m ne 1$. Also, if $m=0$ then $a=1$, so we can assume $m ne 0$.



For $k in {mathbb Z}$, let $a_k = b^kab^{-k}$. So $a=a_0$ and, for all $k in {mathbb Z}$, $a_{k+1} = b^{k}bab^{-1}b^{-k} = b^ka^mb^{-k} = a_k^m$.



So, if we let $N = langle a_k mid k in {mathbb Z} rangle$, then since $a_l$ is a power of $a_k$ whenever $l ge k$, any pair of generators of $N$ commute, and hence $N$ is abelian. (It is not hard to see that the map $a_k mapsto m^k$ induces an isomorphism from $N$ to the additive group of rationals of the form $jm^k$ for $j,k in {mathbb Z}$.)



Also, $N$ is centralized by $a=a_0$ and normalized by $b$, so it is normal in $G$ and is in fact the normal closure $langle a^G rangle$ of $a$ in $G$.



By putting $a=1$ in the presentation, we see that $G/N$ is infinite cyclic and generated by the image of $b$, so $G$ is the semidirect product $N rtimes langle b rangle$, where $b$ acts on $N$ by mapping $a_k$ to $a_{k+1}$.



Now, since $a_k^m=a_{k+1}$, we can write each element of $N$ as a power of $a_k$ for some fixed $k$. Since $b^{-k}a_kb^k = a_0=a$, we see that every element of $N$ is conjugate in $G$ to a power of $a$.



Now let $K$ be any nontrivial normal subgroup of $G$. If $K cap N ne 1$ then, as we just saw, $N cap K$ contains a nontrivial power of $a$, and so the image of $a$ in $G/K$ has finite order.



Otherwise, since $G = Nlangle b rangle$, $N$ contains an element $nb^j$ for some $n in N$ with $j > 0$. If $n=1$, then the image of $b$ has finite order in $G/K$. Otherwise, since we are assuming $m ne 1$, $b$ does not centralize $n$, so the commutator $[b,nb^j]$ is a nontrivial element of $N cap K$, and we are back in the previous case.






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    $begingroup$

    First note that the result is false when $m = 1$, because $G$ is just free abelian of rank $2$ in that case, and the subgroup $K=langle ab rangle$ is normal but neither $a$ nor $b$ has finite order $G/K$. So assume that $m ne 1$. Also, if $m=0$ then $a=1$, so we can assume $m ne 0$.



    For $k in {mathbb Z}$, let $a_k = b^kab^{-k}$. So $a=a_0$ and, for all $k in {mathbb Z}$, $a_{k+1} = b^{k}bab^{-1}b^{-k} = b^ka^mb^{-k} = a_k^m$.



    So, if we let $N = langle a_k mid k in {mathbb Z} rangle$, then since $a_l$ is a power of $a_k$ whenever $l ge k$, any pair of generators of $N$ commute, and hence $N$ is abelian. (It is not hard to see that the map $a_k mapsto m^k$ induces an isomorphism from $N$ to the additive group of rationals of the form $jm^k$ for $j,k in {mathbb Z}$.)



    Also, $N$ is centralized by $a=a_0$ and normalized by $b$, so it is normal in $G$ and is in fact the normal closure $langle a^G rangle$ of $a$ in $G$.



    By putting $a=1$ in the presentation, we see that $G/N$ is infinite cyclic and generated by the image of $b$, so $G$ is the semidirect product $N rtimes langle b rangle$, where $b$ acts on $N$ by mapping $a_k$ to $a_{k+1}$.



    Now, since $a_k^m=a_{k+1}$, we can write each element of $N$ as a power of $a_k$ for some fixed $k$. Since $b^{-k}a_kb^k = a_0=a$, we see that every element of $N$ is conjugate in $G$ to a power of $a$.



    Now let $K$ be any nontrivial normal subgroup of $G$. If $K cap N ne 1$ then, as we just saw, $N cap K$ contains a nontrivial power of $a$, and so the image of $a$ in $G/K$ has finite order.



    Otherwise, since $G = Nlangle b rangle$, $N$ contains an element $nb^j$ for some $n in N$ with $j > 0$. If $n=1$, then the image of $b$ has finite order in $G/K$. Otherwise, since we are assuming $m ne 1$, $b$ does not centralize $n$, so the commutator $[b,nb^j]$ is a nontrivial element of $N cap K$, and we are back in the previous case.






    share|cite|improve this answer









    $endgroup$


















      6












      $begingroup$

      First note that the result is false when $m = 1$, because $G$ is just free abelian of rank $2$ in that case, and the subgroup $K=langle ab rangle$ is normal but neither $a$ nor $b$ has finite order $G/K$. So assume that $m ne 1$. Also, if $m=0$ then $a=1$, so we can assume $m ne 0$.



      For $k in {mathbb Z}$, let $a_k = b^kab^{-k}$. So $a=a_0$ and, for all $k in {mathbb Z}$, $a_{k+1} = b^{k}bab^{-1}b^{-k} = b^ka^mb^{-k} = a_k^m$.



      So, if we let $N = langle a_k mid k in {mathbb Z} rangle$, then since $a_l$ is a power of $a_k$ whenever $l ge k$, any pair of generators of $N$ commute, and hence $N$ is abelian. (It is not hard to see that the map $a_k mapsto m^k$ induces an isomorphism from $N$ to the additive group of rationals of the form $jm^k$ for $j,k in {mathbb Z}$.)



      Also, $N$ is centralized by $a=a_0$ and normalized by $b$, so it is normal in $G$ and is in fact the normal closure $langle a^G rangle$ of $a$ in $G$.



      By putting $a=1$ in the presentation, we see that $G/N$ is infinite cyclic and generated by the image of $b$, so $G$ is the semidirect product $N rtimes langle b rangle$, where $b$ acts on $N$ by mapping $a_k$ to $a_{k+1}$.



      Now, since $a_k^m=a_{k+1}$, we can write each element of $N$ as a power of $a_k$ for some fixed $k$. Since $b^{-k}a_kb^k = a_0=a$, we see that every element of $N$ is conjugate in $G$ to a power of $a$.



      Now let $K$ be any nontrivial normal subgroup of $G$. If $K cap N ne 1$ then, as we just saw, $N cap K$ contains a nontrivial power of $a$, and so the image of $a$ in $G/K$ has finite order.



      Otherwise, since $G = Nlangle b rangle$, $N$ contains an element $nb^j$ for some $n in N$ with $j > 0$. If $n=1$, then the image of $b$ has finite order in $G/K$. Otherwise, since we are assuming $m ne 1$, $b$ does not centralize $n$, so the commutator $[b,nb^j]$ is a nontrivial element of $N cap K$, and we are back in the previous case.






      share|cite|improve this answer









      $endgroup$
















        6












        6








        6





        $begingroup$

        First note that the result is false when $m = 1$, because $G$ is just free abelian of rank $2$ in that case, and the subgroup $K=langle ab rangle$ is normal but neither $a$ nor $b$ has finite order $G/K$. So assume that $m ne 1$. Also, if $m=0$ then $a=1$, so we can assume $m ne 0$.



        For $k in {mathbb Z}$, let $a_k = b^kab^{-k}$. So $a=a_0$ and, for all $k in {mathbb Z}$, $a_{k+1} = b^{k}bab^{-1}b^{-k} = b^ka^mb^{-k} = a_k^m$.



        So, if we let $N = langle a_k mid k in {mathbb Z} rangle$, then since $a_l$ is a power of $a_k$ whenever $l ge k$, any pair of generators of $N$ commute, and hence $N$ is abelian. (It is not hard to see that the map $a_k mapsto m^k$ induces an isomorphism from $N$ to the additive group of rationals of the form $jm^k$ for $j,k in {mathbb Z}$.)



        Also, $N$ is centralized by $a=a_0$ and normalized by $b$, so it is normal in $G$ and is in fact the normal closure $langle a^G rangle$ of $a$ in $G$.



        By putting $a=1$ in the presentation, we see that $G/N$ is infinite cyclic and generated by the image of $b$, so $G$ is the semidirect product $N rtimes langle b rangle$, where $b$ acts on $N$ by mapping $a_k$ to $a_{k+1}$.



        Now, since $a_k^m=a_{k+1}$, we can write each element of $N$ as a power of $a_k$ for some fixed $k$. Since $b^{-k}a_kb^k = a_0=a$, we see that every element of $N$ is conjugate in $G$ to a power of $a$.



        Now let $K$ be any nontrivial normal subgroup of $G$. If $K cap N ne 1$ then, as we just saw, $N cap K$ contains a nontrivial power of $a$, and so the image of $a$ in $G/K$ has finite order.



        Otherwise, since $G = Nlangle b rangle$, $N$ contains an element $nb^j$ for some $n in N$ with $j > 0$. If $n=1$, then the image of $b$ has finite order in $G/K$. Otherwise, since we are assuming $m ne 1$, $b$ does not centralize $n$, so the commutator $[b,nb^j]$ is a nontrivial element of $N cap K$, and we are back in the previous case.






        share|cite|improve this answer









        $endgroup$



        First note that the result is false when $m = 1$, because $G$ is just free abelian of rank $2$ in that case, and the subgroup $K=langle ab rangle$ is normal but neither $a$ nor $b$ has finite order $G/K$. So assume that $m ne 1$. Also, if $m=0$ then $a=1$, so we can assume $m ne 0$.



        For $k in {mathbb Z}$, let $a_k = b^kab^{-k}$. So $a=a_0$ and, for all $k in {mathbb Z}$, $a_{k+1} = b^{k}bab^{-1}b^{-k} = b^ka^mb^{-k} = a_k^m$.



        So, if we let $N = langle a_k mid k in {mathbb Z} rangle$, then since $a_l$ is a power of $a_k$ whenever $l ge k$, any pair of generators of $N$ commute, and hence $N$ is abelian. (It is not hard to see that the map $a_k mapsto m^k$ induces an isomorphism from $N$ to the additive group of rationals of the form $jm^k$ for $j,k in {mathbb Z}$.)



        Also, $N$ is centralized by $a=a_0$ and normalized by $b$, so it is normal in $G$ and is in fact the normal closure $langle a^G rangle$ of $a$ in $G$.



        By putting $a=1$ in the presentation, we see that $G/N$ is infinite cyclic and generated by the image of $b$, so $G$ is the semidirect product $N rtimes langle b rangle$, where $b$ acts on $N$ by mapping $a_k$ to $a_{k+1}$.



        Now, since $a_k^m=a_{k+1}$, we can write each element of $N$ as a power of $a_k$ for some fixed $k$. Since $b^{-k}a_kb^k = a_0=a$, we see that every element of $N$ is conjugate in $G$ to a power of $a$.



        Now let $K$ be any nontrivial normal subgroup of $G$. If $K cap N ne 1$ then, as we just saw, $N cap K$ contains a nontrivial power of $a$, and so the image of $a$ in $G/K$ has finite order.



        Otherwise, since $G = Nlangle b rangle$, $N$ contains an element $nb^j$ for some $n in N$ with $j > 0$. If $n=1$, then the image of $b$ has finite order in $G/K$. Otherwise, since we are assuming $m ne 1$, $b$ does not centralize $n$, so the commutator $[b,nb^j]$ is a nontrivial element of $N cap K$, and we are back in the previous case.







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        answered Jun 1 '15 at 16:13









        Derek HoltDerek Holt

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