How to show that $langle a,b mid aba^{-1}ba = bab^{-1}abrangle$ is not Abelian?
$begingroup$
I'd like to show that
$$
G = langle a,b mid aba^{-1}ba = bab^{-1}abrangle
$$
is non-Abelian.
I have tried finding a surjective homomorphism from $G$ to a non-Abelian group, but I haven't found one. The context is that I would like to show that the figure-$8$ knot complement is non-trivial using knot groups.
Thanks a lot!
group-theory abelian-groups knot-theory knot-invariants combinatorial-group-theory
edited Nov 24 '18 at 2:39
Shaun
8,818113681
asked Nov 24 '18 at 0:57
SheafsSheafs
1517
$endgroup$
|
show 5 more comments
$begingroup$
I'd like to show that
$$
G = langle a,b mid aba^{-1}ba = bab^{-1}abrangle
$$
is non-Abelian.
I have tried finding a surjective homomorphism from $G$ to a non-Abelian group, but I haven't found one. The context is that I would like to show that the figure-$8$ knot complement is non-trivial using knot groups.
Thanks a lot!
group-theory abelian-groups knot-theory knot-invariants combinatorial-group-theory
edited Nov 24 '18 at 2:39
Shaun
8,818113681
asked Nov 24 '18 at 0:57
SheafsSheafs
1517
$endgroup$
3
$begingroup$
If $G$ were abelian, then the relation would be equivalent to $ab^2=a^2b$, i.e., $a=b$, so then $Gconglangle amid ranglecongmathbb{Z}$. I don't know if that helps.
$endgroup$
– Shaun
Nov 24 '18 at 1:14
1
$begingroup$
It should be noted that the problem of determining whether the group defined by a presentation is abelian is, in general, undecidable.
$endgroup$
– Shaun
Nov 24 '18 at 1:33
1
$begingroup$
@Shaun: Are there any convenient types of groups that arise in knot theory for which the word problem is decidable? I don't know much about knot theory myself, and the few results I'm familiar with it are in the opposite direction.
$endgroup$
– anomaly
Nov 24 '18 at 4:51
1
$begingroup$
@anomaly: That's a great question! I don't know. I'm not exactly au fait with knot theory either. Maybe it's worth asking that on here as a separate question. In the meantime, I'll have a look at what I can find, but I make no promises! :)
$endgroup$
– Shaun
Nov 24 '18 at 5:02
2
$begingroup$
@Shaun You cannot use $mathtt{AbelianInvariants}$ to prove that a group is nonabelian. That just tells you the largest abelian quotient of the group. A better way is to use $mathtt{LowIndexSubgroupsFpGroup}$. This group has a subgroup of index $4$ which defines a homomorphism onto $A_4$.
$endgroup$
– Derek Holt
Nov 24 '18 at 8:50
|
show 5 more comments
$begingroup$
I'd like to show that
$$
G = langle a,b mid aba^{-1}ba = bab^{-1}abrangle
$$
is non-Abelian.
I have tried finding a surjective homomorphism from $G$ to a non-Abelian group, but I haven't found one. The context is that I would like to show that the figure-$8$ knot complement is non-trivial using knot groups.
Thanks a lot!
group-theory abelian-groups knot-theory knot-invariants combinatorial-group-theory
edited Nov 24 '18 at 2:39
Shaun
8,818113681
asked Nov 24 '18 at 0:57
SheafsSheafs
1517
$endgroup$
I'd like to show that
$$
G = langle a,b mid aba^{-1}ba = bab^{-1}abrangle
$$
is non-Abelian.
I have tried finding a surjective homomorphism from $G$ to a non-Abelian group, but I haven't found one. The context is that I would like to show that the figure-$8$ knot complement is non-trivial using knot groups.
Thanks a lot!
group-theory abelian-groups knot-theory knot-invariants combinatorial-group-theory
group-theory abelian-groups knot-theory knot-invariants combinatorial-group-theory
edited Nov 24 '18 at 2:39
Shaun
8,818113681
asked Nov 24 '18 at 0:57
SheafsSheafs
1517
edited Nov 24 '18 at 2:39
Shaun
8,818113681
asked Nov 24 '18 at 0:57
SheafsSheafs
1517
edited Nov 24 '18 at 2:39
Shaun
8,818113681
edited Nov 24 '18 at 2:39
Shaun
8,818113681
edited Nov 24 '18 at 2:39
Shaun
8,818113681
8,818113681
asked Nov 24 '18 at 0:57
SheafsSheafs
1517
asked Nov 24 '18 at 0:57
SheafsSheafs
1517
asked Nov 24 '18 at 0:57
SheafsSheafs
1517
1517
3
$begingroup$
If $G$ were abelian, then the relation would be equivalent to $ab^2=a^2b$, i.e., $a=b$, so then $Gconglangle amid ranglecongmathbb{Z}$. I don't know if that helps.
$endgroup$
– Shaun
Nov 24 '18 at 1:14
1
$begingroup$
It should be noted that the problem of determining whether the group defined by a presentation is abelian is, in general, undecidable.
$endgroup$
– Shaun
Nov 24 '18 at 1:33
1
$begingroup$
@Shaun: Are there any convenient types of groups that arise in knot theory for which the word problem is decidable? I don't know much about knot theory myself, and the few results I'm familiar with it are in the opposite direction.
$endgroup$
– anomaly
Nov 24 '18 at 4:51
1
$begingroup$
@anomaly: That's a great question! I don't know. I'm not exactly au fait with knot theory either. Maybe it's worth asking that on here as a separate question. In the meantime, I'll have a look at what I can find, but I make no promises! :)
$endgroup$
– Shaun
Nov 24 '18 at 5:02
2
$begingroup$
@Shaun You cannot use $mathtt{AbelianInvariants}$ to prove that a group is nonabelian. That just tells you the largest abelian quotient of the group. A better way is to use $mathtt{LowIndexSubgroupsFpGroup}$. This group has a subgroup of index $4$ which defines a homomorphism onto $A_4$.
$endgroup$
– Derek Holt
Nov 24 '18 at 8:50
|
show 5 more comments
3
$begingroup$
If $G$ were abelian, then the relation would be equivalent to $ab^2=a^2b$, i.e., $a=b$, so then $Gconglangle amid ranglecongmathbb{Z}$. I don't know if that helps.
$endgroup$
– Shaun
Nov 24 '18 at 1:14
1
$begingroup$
It should be noted that the problem of determining whether the group defined by a presentation is abelian is, in general, undecidable.
$endgroup$
– Shaun
Nov 24 '18 at 1:33
1
$begingroup$
@Shaun: Are there any convenient types of groups that arise in knot theory for which the word problem is decidable? I don't know much about knot theory myself, and the few results I'm familiar with it are in the opposite direction.
$endgroup$
– anomaly
Nov 24 '18 at 4:51
1
$begingroup$
@anomaly: That's a great question! I don't know. I'm not exactly au fait with knot theory either. Maybe it's worth asking that on here as a separate question. In the meantime, I'll have a look at what I can find, but I make no promises! :)
$endgroup$
– Shaun
Nov 24 '18 at 5:02
2
$begingroup$
@Shaun You cannot use $mathtt{AbelianInvariants}$ to prove that a group is nonabelian. That just tells you the largest abelian quotient of the group. A better way is to use $mathtt{LowIndexSubgroupsFpGroup}$. This group has a subgroup of index $4$ which defines a homomorphism onto $A_4$.
$endgroup$
– Derek Holt
Nov 24 '18 at 8:50
3
3
$begingroup$
If $G$ were abelian, then the relation would be equivalent to $ab^2=a^2b$, i.e., $a=b$, so then $Gconglangle amid ranglecongmathbb{Z}$. I don't know if that helps.
$endgroup$
– Shaun
Nov 24 '18 at 1:14
$begingroup$
If $G$ were abelian, then the relation would be equivalent to $ab^2=a^2b$, i.e., $a=b$, so then $Gconglangle amid ranglecongmathbb{Z}$. I don't know if that helps.
$endgroup$
– Shaun
Nov 24 '18 at 1:14
1
1
$begingroup$
It should be noted that the problem of determining whether the group defined by a presentation is abelian is, in general, undecidable.
$endgroup$
– Shaun
Nov 24 '18 at 1:33
$begingroup$
It should be noted that the problem of determining whether the group defined by a presentation is abelian is, in general, undecidable.
$endgroup$
– Shaun
Nov 24 '18 at 1:33
1
1
$begingroup$
@Shaun: Are there any convenient types of groups that arise in knot theory for which the word problem is decidable? I don't know much about knot theory myself, and the few results I'm familiar with it are in the opposite direction.
$endgroup$
– anomaly
Nov 24 '18 at 4:51
$begingroup$
@Shaun: Are there any convenient types of groups that arise in knot theory for which the word problem is decidable? I don't know much about knot theory myself, and the few results I'm familiar with it are in the opposite direction.
$endgroup$
– anomaly
Nov 24 '18 at 4:51
1
1
$begingroup$
@anomaly: That's a great question! I don't know. I'm not exactly au fait with knot theory either. Maybe it's worth asking that on here as a separate question. In the meantime, I'll have a look at what I can find, but I make no promises! :)
$endgroup$
– Shaun
Nov 24 '18 at 5:02
$begingroup$
@anomaly: That's a great question! I don't know. I'm not exactly au fait with knot theory either. Maybe it's worth asking that on here as a separate question. In the meantime, I'll have a look at what I can find, but I make no promises! :)
$endgroup$
– Shaun
Nov 24 '18 at 5:02
2
2
$begingroup$
@Shaun You cannot use $mathtt{AbelianInvariants}$ to prove that a group is nonabelian. That just tells you the largest abelian quotient of the group. A better way is to use $mathtt{LowIndexSubgroupsFpGroup}$. This group has a subgroup of index $4$ which defines a homomorphism onto $A_4$.
$endgroup$
– Derek Holt
Nov 24 '18 at 8:50
$begingroup$
@Shaun You cannot use $mathtt{AbelianInvariants}$ to prove that a group is nonabelian. That just tells you the largest abelian quotient of the group. A better way is to use $mathtt{LowIndexSubgroupsFpGroup}$. This group has a subgroup of index $4$ which defines a homomorphism onto $A_4$.
$endgroup$
– Derek Holt
Nov 24 '18 at 8:50
|
show 5 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Put $xi = frac{1}{2}(3 + sqrt{5})$. The map $G to GL_2(mathbb{R})$ defined by
begin{align*}
a &to begin{pmatrix}1 & 0 \ 0 & xiend{pmatrix} &
b &to begin{pmatrix}1 & 1 \ 0 & xiend{pmatrix}
end{align*}
is well-defined with nonabelian image.
answered Nov 24 '18 at 3:04
anomalyanomaly
17.4k42664
$endgroup$
add a comment |
$begingroup$
Since you tag this with knot-theory and knot-invariants, it looks like you are trying to show the fundamental group of the knot complement $S^3-4_1$ is nonabelian.
One of the "obvious" things to try is Fox $n$-coloring, since they yield a homomorphism (usually surjective) to a dihedral group. There is a $5$-coloring (using $0,4,1,2$ as you go along the knot, which you can check satisfies $2bequiv a+cpmod{5}$ at each crossing). Since $5$ is a prime and the coloring is nonconstant, the corresponding homomorphism from $pi_1(S^3-4_1)$ to $D_{2cdot 5}$, the dihedral group of order $10$, is surjective.
answered Nov 24 '18 at 2:44
user10354138user10354138
7,3772925
$endgroup$
3
$begingroup$
Why the downvote?
$endgroup$
– anomaly
Nov 25 '18 at 3:34
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Put $xi = frac{1}{2}(3 + sqrt{5})$. The map $G to GL_2(mathbb{R})$ defined by
begin{align*}
a &to begin{pmatrix}1 & 0 \ 0 & xiend{pmatrix} &
b &to begin{pmatrix}1 & 1 \ 0 & xiend{pmatrix}
end{align*}
is well-defined with nonabelian image.
answered Nov 24 '18 at 3:04
anomalyanomaly
17.4k42664
$endgroup$
add a comment |
$begingroup$
Put $xi = frac{1}{2}(3 + sqrt{5})$. The map $G to GL_2(mathbb{R})$ defined by
begin{align*}
a &to begin{pmatrix}1 & 0 \ 0 & xiend{pmatrix} &
b &to begin{pmatrix}1 & 1 \ 0 & xiend{pmatrix}
end{align*}
is well-defined with nonabelian image.
answered Nov 24 '18 at 3:04
anomalyanomaly
17.4k42664
$endgroup$
add a comment |
$begingroup$
Put $xi = frac{1}{2}(3 + sqrt{5})$. The map $G to GL_2(mathbb{R})$ defined by
begin{align*}
a &to begin{pmatrix}1 & 0 \ 0 & xiend{pmatrix} &
b &to begin{pmatrix}1 & 1 \ 0 & xiend{pmatrix}
end{align*}
is well-defined with nonabelian image.
answered Nov 24 '18 at 3:04
anomalyanomaly
17.4k42664
$endgroup$
Put $xi = frac{1}{2}(3 + sqrt{5})$. The map $G to GL_2(mathbb{R})$ defined by
begin{align*}
a &to begin{pmatrix}1 & 0 \ 0 & xiend{pmatrix} &
b &to begin{pmatrix}1 & 1 \ 0 & xiend{pmatrix}
end{align*}
is well-defined with nonabelian image.
answered Nov 24 '18 at 3:04
anomalyanomaly
17.4k42664
edited Nov 24 '18 at 3:15
answered Nov 24 '18 at 3:04
anomalyanomaly
17.4k42664
answered Nov 24 '18 at 3:04
anomalyanomaly
17.4k42664
answered Nov 24 '18 at 3:04
anomalyanomaly
17.4k42664
17.4k42664
add a comment |
add a comment |
$begingroup$
Since you tag this with knot-theory and knot-invariants, it looks like you are trying to show the fundamental group of the knot complement $S^3-4_1$ is nonabelian.
One of the "obvious" things to try is Fox $n$-coloring, since they yield a homomorphism (usually surjective) to a dihedral group. There is a $5$-coloring (using $0,4,1,2$ as you go along the knot, which you can check satisfies $2bequiv a+cpmod{5}$ at each crossing). Since $5$ is a prime and the coloring is nonconstant, the corresponding homomorphism from $pi_1(S^3-4_1)$ to $D_{2cdot 5}$, the dihedral group of order $10$, is surjective.
answered Nov 24 '18 at 2:44
user10354138user10354138
7,3772925
$endgroup$
3
$begingroup$
Why the downvote?
$endgroup$
– anomaly
Nov 25 '18 at 3:34
add a comment |
$begingroup$
Since you tag this with knot-theory and knot-invariants, it looks like you are trying to show the fundamental group of the knot complement $S^3-4_1$ is nonabelian.
One of the "obvious" things to try is Fox $n$-coloring, since they yield a homomorphism (usually surjective) to a dihedral group. There is a $5$-coloring (using $0,4,1,2$ as you go along the knot, which you can check satisfies $2bequiv a+cpmod{5}$ at each crossing). Since $5$ is a prime and the coloring is nonconstant, the corresponding homomorphism from $pi_1(S^3-4_1)$ to $D_{2cdot 5}$, the dihedral group of order $10$, is surjective.
answered Nov 24 '18 at 2:44
user10354138user10354138
7,3772925
$endgroup$
3
$begingroup$
Why the downvote?
$endgroup$
– anomaly
Nov 25 '18 at 3:34
add a comment |
$begingroup$
Since you tag this with knot-theory and knot-invariants, it looks like you are trying to show the fundamental group of the knot complement $S^3-4_1$ is nonabelian.
One of the "obvious" things to try is Fox $n$-coloring, since they yield a homomorphism (usually surjective) to a dihedral group. There is a $5$-coloring (using $0,4,1,2$ as you go along the knot, which you can check satisfies $2bequiv a+cpmod{5}$ at each crossing). Since $5$ is a prime and the coloring is nonconstant, the corresponding homomorphism from $pi_1(S^3-4_1)$ to $D_{2cdot 5}$, the dihedral group of order $10$, is surjective.
answered Nov 24 '18 at 2:44
user10354138user10354138
7,3772925
$endgroup$
Since you tag this with knot-theory and knot-invariants, it looks like you are trying to show the fundamental group of the knot complement $S^3-4_1$ is nonabelian.
One of the "obvious" things to try is Fox $n$-coloring, since they yield a homomorphism (usually surjective) to a dihedral group. There is a $5$-coloring (using $0,4,1,2$ as you go along the knot, which you can check satisfies $2bequiv a+cpmod{5}$ at each crossing). Since $5$ is a prime and the coloring is nonconstant, the corresponding homomorphism from $pi_1(S^3-4_1)$ to $D_{2cdot 5}$, the dihedral group of order $10$, is surjective.
answered Nov 24 '18 at 2:44
user10354138user10354138
7,3772925
edited Nov 24 '18 at 2:50
answered Nov 24 '18 at 2:44
user10354138user10354138
7,3772925
answered Nov 24 '18 at 2:44
user10354138user10354138
7,3772925
answered Nov 24 '18 at 2:44
user10354138user10354138
7,3772925
7,3772925
3
$begingroup$
Why the downvote?
$endgroup$
– anomaly
Nov 25 '18 at 3:34
add a comment |
3
$begingroup$
Why the downvote?
$endgroup$
– anomaly
Nov 25 '18 at 3:34
3
3
$begingroup$
Why the downvote?
$endgroup$
– anomaly
Nov 25 '18 at 3:34
$begingroup$
Why the downvote?
$endgroup$
– anomaly
Nov 25 '18 at 3:34
add a comment |
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$begingroup$
If $G$ were abelian, then the relation would be equivalent to $ab^2=a^2b$, i.e., $a=b$, so then $Gconglangle amid ranglecongmathbb{Z}$. I don't know if that helps.
$endgroup$
– Shaun
Nov 24 '18 at 1:14
1
$begingroup$
It should be noted that the problem of determining whether the group defined by a presentation is abelian is, in general, undecidable.
$endgroup$
– Shaun
Nov 24 '18 at 1:33
1
$begingroup$
@Shaun: Are there any convenient types of groups that arise in knot theory for which the word problem is decidable? I don't know much about knot theory myself, and the few results I'm familiar with it are in the opposite direction.
$endgroup$
– anomaly
Nov 24 '18 at 4:51
1
$begingroup$
@anomaly: That's a great question! I don't know. I'm not exactly au fait with knot theory either. Maybe it's worth asking that on here as a separate question. In the meantime, I'll have a look at what I can find, but I make no promises! :)
$endgroup$
– Shaun
Nov 24 '18 at 5:02
2
$begingroup$
@Shaun You cannot use $mathtt{AbelianInvariants}$ to prove that a group is nonabelian. That just tells you the largest abelian quotient of the group. A better way is to use $mathtt{LowIndexSubgroupsFpGroup}$. This group has a subgroup of index $4$ which defines a homomorphism onto $A_4$.
$endgroup$
– Derek Holt
Nov 24 '18 at 8:50