Quotient of the Baumslag-Solitar group $BS(1,m)=langle a,b| bab^{-1}=a^mrangle$.
$begingroup$
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.
group-theory quotient-group
edited Dec 11 '18 at 0:38
Shaun
9,268113684
asked Jun 1 '15 at 10:42
Edgar NdieEdgar Ndie
836
$endgroup$
migrated from mathoverflow.net Jun 1 '15 at 13:47
This question came from our site for professional mathematicians.
|
show 4 more comments
$begingroup$
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.
group-theory quotient-group
edited Dec 11 '18 at 0:38
Shaun
9,268113684
asked Jun 1 '15 at 10:42
Edgar NdieEdgar Ndie
836
$endgroup$
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
|
show 4 more comments
$begingroup$
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.
group-theory quotient-group
edited Dec 11 '18 at 0:38
Shaun
9,268113684
asked Jun 1 '15 at 10:42
Edgar NdieEdgar Ndie
836
$endgroup$
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
group-theory quotient-group
edited Dec 11 '18 at 0:38
Shaun
9,268113684
asked Jun 1 '15 at 10:42
Edgar NdieEdgar Ndie
836
edited Dec 11 '18 at 0:38
Shaun
9,268113684
asked Jun 1 '15 at 10:42
Edgar NdieEdgar Ndie
836
edited Dec 11 '18 at 0:38
Shaun
9,268113684
edited Dec 11 '18 at 0:38
Shaun
9,268113684
edited Dec 11 '18 at 0:38
Shaun
9,268113684
9,268113684
asked Jun 1 '15 at 10:42
Edgar NdieEdgar Ndie
836
asked Jun 1 '15 at 10:42
Edgar NdieEdgar Ndie
836
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
|
show 4 more comments
$begingroup$
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
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jun 1 '15 at 16:13
Derek HoltDerek Holt
53.7k53571
$endgroup$
add a comment |
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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.
answered Jun 1 '15 at 16:13
Derek HoltDerek Holt
53.7k53571
$endgroup$
add a comment |
$begingroup$
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.
answered Jun 1 '15 at 16:13
Derek HoltDerek Holt
53.7k53571
$endgroup$
add a comment |
$begingroup$
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.
answered Jun 1 '15 at 16:13
Derek HoltDerek Holt
53.7k53571
$endgroup$
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.
answered Jun 1 '15 at 16:13
Derek HoltDerek Holt
53.7k53571
answered Jun 1 '15 at 16:13
Derek HoltDerek Holt
53.7k53571
answered Jun 1 '15 at 16:13
Derek HoltDerek Holt
53.7k53571
answered Jun 1 '15 at 16:13
Derek HoltDerek Holt
53.7k53571
53.7k53571
add a comment |
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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 $BS(1, 2)$?
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– Edgar Ndie
Jun 1 '15 at 11:52
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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 $n=1$.
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– Derek Holt
Jun 1 '15 at 12:18
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@Derek Thanks - I was too hasty. I've deleted both my comments.
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– user1729
Jun 1 '15 at 12:23
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@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?
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– user1729
Jun 1 '15 at 12:33
1
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@user1729 Of course, I meant conjugate in $G={rm BS}(1,m)$ to a power of $a$. Then $b(b^{-1}aba^m)b^{-1} = a^{m^2+1}$.
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– Derek Holt
Jun 1 '15 at 12:50