Quandle homomorphism does not always induces group homomorphim on inner automorphism group of quandles.

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Let $X$ and $Y$ be two quandles and $f: X rightarrow Y$ be a quandle homomorphism. Then we can define a map $bar f: Inn(X) rightarrow Inn(Y)$ as $bar f(S_a)=S_f(a)$, where $a in X$. Then $bar f$ may not be a group homomorphism. But I am not able to construct such example. Can someone help me?

abstract-algebra group-theory knot-invariants

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edited Jul 23 at 4:18

asked Jul 22 at 11:51

Sneh

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Let $X$ and $Y$ be two quandles and $f: X rightarrow Y$ be a quandle homomorphism. Then we can define a map $bar f: Inn(X) rightarrow Inn(Y)$ as $bar f(S_a)=S_f(a)$, where $a in X$. Then $bar f$ may not be a group homomorphism. But I am not able to construct such example. Can someone help me?

abstract-algebra group-theory knot-invariants

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edited Jul 23 at 4:18

asked Jul 22 at 11:51

Sneh

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Let $X$ and $Y$ be two quandles and $f: X rightarrow Y$ be a quandle homomorphism. Then we can define a map $bar f: Inn(X) rightarrow Inn(Y)$ as $bar f(S_a)=S_f(a)$, where $a in X$. Then $bar f$ may not be a group homomorphism. But I am not able to construct such example. Can someone help me?

abstract-algebra group-theory knot-invariants

share|cite|improve this question

edited Jul 23 at 4:18

asked Jul 22 at 11:51

Sneh

285

Let $X$ and $Y$ be two quandles and $f: X rightarrow Y$ be a quandle homomorphism. Then we can define a map $bar f: Inn(X) rightarrow Inn(Y)$ as $bar f(S_a)=S_f(a)$, where $a in X$. Then $bar f$ may not be a group homomorphism. But I am not able to construct such example. Can someone help me?

abstract-algebra group-theory knot-invariants

share|cite|improve this question

edited Jul 23 at 4:18

asked Jul 22 at 11:51

Sneh

285

share|cite|improve this question

share|cite|improve this question

edited Jul 23 at 4:18

edited Jul 23 at 4:18

edited Jul 23 at 4:18

asked Jul 22 at 11:51

Sneh

285

asked Jul 22 at 11:51

Sneh

285

asked Jul 22 at 11:51

Sneh

285

285

add a comment | 

add a comment | 

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There are many possible obstructions to what you propose, but the first obstacle is that your $barf$ may not be well defined. Remember that Inn$(X)$ is a group of automorphisms of $X$, and two different elements of $X$ may induce the same automorphism of $X$. In other words, $S_a$ and $S_b$ may be the same element of Inn$(X)$. For your $barf$ to be well-defined, you would have to show that the result does not depend on whether you choose $a$ or $b$ to represent that element of Inn$(X)$. But in fact, it might.

To see an example of this, start with the smallest quandle in which there are two distinct elements $a$ and $b$ such that $S_a=S_b,$ namely the only two element quandle, which I will call $2I$: you can call its elements $1,2$ and then it has the boring multiplication table $beginarraycc1 & 2 \ 1&2 endarray$. Evidently $S_1 = S_2 = ,$Id. Now we just need to find another quandle $Q$ with a subquandle isomorphic to $2I$, but where those two elements act on the rest of $Q$ differently. The smallest example of this turns out to have five elements $1,2,3,4,5$ with multiplication table $$beginarrayccccc 1 & 2 & 4 & 5 & 3 \ 1 & 2 & 5 & 3 & 4 \ 2 & 1 & 3 & 5 & 4 \ 2 & 1 & 5 & 4 & 3 \ 2 & 1 & 4 & 3 & 5 endarray$$ The inclusion map $iota:2I rightarrow Q$ is of course a homomorphism, but now there is no way to tell whether $bariota$ should map the identity in Inn$(2I)$ to the permutation $S_1 = (3 4 5)$ or $S_2=(3 5 4)$ in Inn$(Q)$.

(In case you are wondering where $Q$ came from, it is a sort of gluing together of $2I$ and the only latin quandle of order 3, called the dihedral quandle $D_3$ of order 3, which has a large enough automorphism group so that in the "sum," the two elements of the $2I$ component can act differently on the $D_3$ component.)

Even if $barf$ happens to be well-defined, there are other obstacles to it being a group homomorphism, but hopefully this example is illuminating.

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answered Jul 25 at 3:49

Glen Whitney

331211

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I should add, in case it's not clear, that in the case that $f$ is an epimorphism, $barf$ is indeed a well-defined group homomorphism. For if $S_a=S_b$, let $y=f(x)$ be an arbitrary element of $Y$. Then $f(a)rhd y = f(a)rhd f(x) = f(arhd x) = f(brhd x) = f(b)rhd f(x) = f(b)rhd y$, or in other words, $S_f(a) = S_f(b)$ in Inn$(Y)$. A similar calculation shows that if $S_c = S_bS_a$ in Inn$(X)$, then $S_f(c) = S_f(b)S_f(a)$ in Inn$(Y)$. Hence $barf$ is well defined on the generators of Inn$(X)$ compatibly with multiplication, and so extends to a (unique) group homomorphism.
  – Glen Whitney
  Jul 26 at 10:45
  
  

  
  
  
  

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accepted

There are many possible obstructions to what you propose, but the first obstacle is that your $barf$ may not be well defined. Remember that Inn$(X)$ is a group of automorphisms of $X$, and two different elements of $X$ may induce the same automorphism of $X$. In other words, $S_a$ and $S_b$ may be the same element of Inn$(X)$. For your $barf$ to be well-defined, you would have to show that the result does not depend on whether you choose $a$ or $b$ to represent that element of Inn$(X)$. But in fact, it might.

To see an example of this, start with the smallest quandle in which there are two distinct elements $a$ and $b$ such that $S_a=S_b,$ namely the only two element quandle, which I will call $2I$: you can call its elements $1,2$ and then it has the boring multiplication table $beginarraycc1 & 2 \ 1&2 endarray$. Evidently $S_1 = S_2 = ,$Id. Now we just need to find another quandle $Q$ with a subquandle isomorphic to $2I$, but where those two elements act on the rest of $Q$ differently. The smallest example of this turns out to have five elements $1,2,3,4,5$ with multiplication table $$beginarrayccccc 1 & 2 & 4 & 5 & 3 \ 1 & 2 & 5 & 3 & 4 \ 2 & 1 & 3 & 5 & 4 \ 2 & 1 & 5 & 4 & 3 \ 2 & 1 & 4 & 3 & 5 endarray$$ The inclusion map $iota:2I rightarrow Q$ is of course a homomorphism, but now there is no way to tell whether $bariota$ should map the identity in Inn$(2I)$ to the permutation $S_1 = (3 4 5)$ or $S_2=(3 5 4)$ in Inn$(Q)$.

(In case you are wondering where $Q$ came from, it is a sort of gluing together of $2I$ and the only latin quandle of order 3, called the dihedral quandle $D_3$ of order 3, which has a large enough automorphism group so that in the "sum," the two elements of the $2I$ component can act differently on the $D_3$ component.)

Even if $barf$ happens to be well-defined, there are other obstacles to it being a group homomorphism, but hopefully this example is illuminating.

share|cite|improve this answer

answered Jul 25 at 3:49

Glen Whitney

331211

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I should add, in case it's not clear, that in the case that $f$ is an epimorphism, $barf$ is indeed a well-defined group homomorphism. For if $S_a=S_b$, let $y=f(x)$ be an arbitrary element of $Y$. Then $f(a)rhd y = f(a)rhd f(x) = f(arhd x) = f(brhd x) = f(b)rhd f(x) = f(b)rhd y$, or in other words, $S_f(a) = S_f(b)$ in Inn$(Y)$. A similar calculation shows that if $S_c = S_bS_a$ in Inn$(X)$, then $S_f(c) = S_f(b)S_f(a)$ in Inn$(Y)$. Hence $barf$ is well defined on the generators of Inn$(X)$ compatibly with multiplication, and so extends to a (unique) group homomorphism.
  – Glen Whitney
  Jul 26 at 10:45
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

There are many possible obstructions to what you propose, but the first obstacle is that your $barf$ may not be well defined. Remember that Inn$(X)$ is a group of automorphisms of $X$, and two different elements of $X$ may induce the same automorphism of $X$. In other words, $S_a$ and $S_b$ may be the same element of Inn$(X)$. For your $barf$ to be well-defined, you would have to show that the result does not depend on whether you choose $a$ or $b$ to represent that element of Inn$(X)$. But in fact, it might.

To see an example of this, start with the smallest quandle in which there are two distinct elements $a$ and $b$ such that $S_a=S_b,$ namely the only two element quandle, which I will call $2I$: you can call its elements $1,2$ and then it has the boring multiplication table $beginarraycc1 & 2 \ 1&2 endarray$. Evidently $S_1 = S_2 = ,$Id. Now we just need to find another quandle $Q$ with a subquandle isomorphic to $2I$, but where those two elements act on the rest of $Q$ differently. The smallest example of this turns out to have five elements $1,2,3,4,5$ with multiplication table $$beginarrayccccc 1 & 2 & 4 & 5 & 3 \ 1 & 2 & 5 & 3 & 4 \ 2 & 1 & 3 & 5 & 4 \ 2 & 1 & 5 & 4 & 3 \ 2 & 1 & 4 & 3 & 5 endarray$$ The inclusion map $iota:2I rightarrow Q$ is of course a homomorphism, but now there is no way to tell whether $bariota$ should map the identity in Inn$(2I)$ to the permutation $S_1 = (3 4 5)$ or $S_2=(3 5 4)$ in Inn$(Q)$.

(In case you are wondering where $Q$ came from, it is a sort of gluing together of $2I$ and the only latin quandle of order 3, called the dihedral quandle $D_3$ of order 3, which has a large enough automorphism group so that in the "sum," the two elements of the $2I$ component can act differently on the $D_3$ component.)

Even if $barf$ happens to be well-defined, there are other obstacles to it being a group homomorphism, but hopefully this example is illuminating.

share|cite|improve this answer

answered Jul 25 at 3:49

Glen Whitney

331211

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I should add, in case it's not clear, that in the case that $f$ is an epimorphism, $barf$ is indeed a well-defined group homomorphism. For if $S_a=S_b$, let $y=f(x)$ be an arbitrary element of $Y$. Then $f(a)rhd y = f(a)rhd f(x) = f(arhd x) = f(brhd x) = f(b)rhd f(x) = f(b)rhd y$, or in other words, $S_f(a) = S_f(b)$ in Inn$(Y)$. A similar calculation shows that if $S_c = S_bS_a$ in Inn$(X)$, then $S_f(c) = S_f(b)S_f(a)$ in Inn$(Y)$. Hence $barf$ is well defined on the generators of Inn$(X)$ compatibly with multiplication, and so extends to a (unique) group homomorphism.
  – Glen Whitney
  Jul 26 at 10:45
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

There are many possible obstructions to what you propose, but the first obstacle is that your $barf$ may not be well defined. Remember that Inn$(X)$ is a group of automorphisms of $X$, and two different elements of $X$ may induce the same automorphism of $X$. In other words, $S_a$ and $S_b$ may be the same element of Inn$(X)$. For your $barf$ to be well-defined, you would have to show that the result does not depend on whether you choose $a$ or $b$ to represent that element of Inn$(X)$. But in fact, it might.

To see an example of this, start with the smallest quandle in which there are two distinct elements $a$ and $b$ such that $S_a=S_b,$ namely the only two element quandle, which I will call $2I$: you can call its elements $1,2$ and then it has the boring multiplication table $beginarraycc1 & 2 \ 1&2 endarray$. Evidently $S_1 = S_2 = ,$Id. Now we just need to find another quandle $Q$ with a subquandle isomorphic to $2I$, but where those two elements act on the rest of $Q$ differently. The smallest example of this turns out to have five elements $1,2,3,4,5$ with multiplication table $$beginarrayccccc 1 & 2 & 4 & 5 & 3 \ 1 & 2 & 5 & 3 & 4 \ 2 & 1 & 3 & 5 & 4 \ 2 & 1 & 5 & 4 & 3 \ 2 & 1 & 4 & 3 & 5 endarray$$ The inclusion map $iota:2I rightarrow Q$ is of course a homomorphism, but now there is no way to tell whether $bariota$ should map the identity in Inn$(2I)$ to the permutation $S_1 = (3 4 5)$ or $S_2=(3 5 4)$ in Inn$(Q)$.

(In case you are wondering where $Q$ came from, it is a sort of gluing together of $2I$ and the only latin quandle of order 3, called the dihedral quandle $D_3$ of order 3, which has a large enough automorphism group so that in the "sum," the two elements of the $2I$ component can act differently on the $D_3$ component.)

Even if $barf$ happens to be well-defined, there are other obstacles to it being a group homomorphism, but hopefully this example is illuminating.

share|cite|improve this answer

answered Jul 25 at 3:49

Glen Whitney

331211

There are many possible obstructions to what you propose, but the first obstacle is that your $barf$ may not be well defined. Remember that Inn$(X)$ is a group of automorphisms of $X$, and two different elements of $X$ may induce the same automorphism of $X$. In other words, $S_a$ and $S_b$ may be the same element of Inn$(X)$. For your $barf$ to be well-defined, you would have to show that the result does not depend on whether you choose $a$ or $b$ to represent that element of Inn$(X)$. But in fact, it might.

To see an example of this, start with the smallest quandle in which there are two distinct elements $a$ and $b$ such that $S_a=S_b,$ namely the only two element quandle, which I will call $2I$: you can call its elements $1,2$ and then it has the boring multiplication table $beginarraycc1 & 2 \ 1&2 endarray$. Evidently $S_1 = S_2 = ,$Id. Now we just need to find another quandle $Q$ with a subquandle isomorphic to $2I$, but where those two elements act on the rest of $Q$ differently. The smallest example of this turns out to have five elements $1,2,3,4,5$ with multiplication table $$beginarrayccccc 1 & 2 & 4 & 5 & 3 \ 1 & 2 & 5 & 3 & 4 \ 2 & 1 & 3 & 5 & 4 \ 2 & 1 & 5 & 4 & 3 \ 2 & 1 & 4 & 3 & 5 endarray$$ The inclusion map $iota:2I rightarrow Q$ is of course a homomorphism, but now there is no way to tell whether $bariota$ should map the identity in Inn$(2I)$ to the permutation $S_1 = (3 4 5)$ or $S_2=(3 5 4)$ in Inn$(Q)$.

(In case you are wondering where $Q$ came from, it is a sort of gluing together of $2I$ and the only latin quandle of order 3, called the dihedral quandle $D_3$ of order 3, which has a large enough automorphism group so that in the "sum," the two elements of the $2I$ component can act differently on the $D_3$ component.)

Even if $barf$ happens to be well-defined, there are other obstacles to it being a group homomorphism, but hopefully this example is illuminating.

share|cite|improve this answer

answered Jul 25 at 3:49

Glen Whitney

331211

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 3:49

Glen Whitney

331211

answered Jul 25 at 3:49

Glen Whitney

331211

answered Jul 25 at 3:49

Glen Whitney

331211

331211

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I should add, in case it's not clear, that in the case that $f$ is an epimorphism, $barf$ is indeed a well-defined group homomorphism. For if $S_a=S_b$, let $y=f(x)$ be an arbitrary element of $Y$. Then $f(a)rhd y = f(a)rhd f(x) = f(arhd x) = f(brhd x) = f(b)rhd f(x) = f(b)rhd y$, or in other words, $S_f(a) = S_f(b)$ in Inn$(Y)$. A similar calculation shows that if $S_c = S_bS_a$ in Inn$(X)$, then $S_f(c) = S_f(b)S_f(a)$ in Inn$(Y)$. Hence $barf$ is well defined on the generators of Inn$(X)$ compatibly with multiplication, and so extends to a (unique) group homomorphism.
  – Glen Whitney
  Jul 26 at 10:45
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I should add, in case it's not clear, that in the case that $f$ is an epimorphism, $barf$ is indeed a well-defined group homomorphism. For if $S_a=S_b$, let $y=f(x)$ be an arbitrary element of $Y$. Then $f(a)rhd y = f(a)rhd f(x) = f(arhd x) = f(brhd x) = f(b)rhd f(x) = f(b)rhd y$, or in other words, $S_f(a) = S_f(b)$ in Inn$(Y)$. A similar calculation shows that if $S_c = S_bS_a$ in Inn$(X)$, then $S_f(c) = S_f(b)S_f(a)$ in Inn$(Y)$. Hence $barf$ is well defined on the generators of Inn$(X)$ compatibly with multiplication, and so extends to a (unique) group homomorphism.
  – Glen Whitney
  Jul 26 at 10:45
  
  

  
  
  
  

I should add, in case it's not clear, that in the case that $f$ is an epimorphism, $barf$ is indeed a well-defined group homomorphism. For if $S_a=S_b$, let $y=f(x)$ be an arbitrary element of $Y$. Then $f(a)rhd y = f(a)rhd f(x) = f(arhd x) = f(brhd x) = f(b)rhd f(x) = f(b)rhd y$, or in other words, $S_f(a) = S_f(b)$ in Inn$(Y)$. A similar calculation shows that if $S_c = S_bS_a$ in Inn$(X)$, then $S_f(c) = S_f(b)S_f(a)$ in Inn$(Y)$. Hence $barf$ is well defined on the generators of Inn$(X)$ compatibly with multiplication, and so extends to a (unique) group homomorphism.
– Glen Whitney
Jul 26 at 10:45

I should add, in case it's not clear, that in the case that $f$ is an epimorphism, $barf$ is indeed a well-defined group homomorphism. For if $S_a=S_b$, let $y=f(x)$ be an arbitrary element of $Y$. Then $f(a)rhd y = f(a)rhd f(x) = f(arhd x) = f(brhd x) = f(b)rhd f(x) = f(b)rhd y$, or in other words, $S_f(a) = S_f(b)$ in Inn$(Y)$. A similar calculation shows that if $S_c = S_bS_a$ in Inn$(X)$, then $S_f(c) = S_f(b)S_f(a)$ in Inn$(Y)$. Hence $barf$ is well defined on the generators of Inn$(X)$ compatibly with multiplication, and so extends to a (unique) group homomorphism.
– Glen Whitney
Jul 26 at 10:45

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