How to show that $langle a,b mid aba^{-1}ba = bab^{-1}abrangle$ is not Abelian?












4












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















4












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














4












4








4


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 '18 at 2:39









Shaun

8,818113681




8,818113681










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














  • 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










2 Answers
2






active

oldest

votes


















6












$begingroup$

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.






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$









    • 3




      $begingroup$
      Why the downvote?
      $endgroup$
      – anomaly
      Nov 25 '18 at 3:34











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$


















      6












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$
















        6












        6








        6





        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 24 '18 at 3:15

























        answered Nov 24 '18 at 3:04









        anomalyanomaly

        17.4k42664




        17.4k42664























            4












            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$









            • 3




              $begingroup$
              Why the downvote?
              $endgroup$
              – anomaly
              Nov 25 '18 at 3:34
















            4












            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$









            • 3




              $begingroup$
              Why the downvote?
              $endgroup$
              – anomaly
              Nov 25 '18 at 3:34














            4












            4








            4





            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$



            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.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 24 '18 at 2:50

























            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














            • 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


















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