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A quandle $(Q,*,/ )$ is a idempotent right-distributive and right invertible structure.

1) $a*a=a$

2) $(a*b)*c=(a*c)*(b*c)$

3) $(a*b) /b=(a/b)*b=a$

If we have a group $(G, \cdot, e,^{-1})$ and $*$ is the cojugation operation on $G$

$$a*b:=bab^{-1}$$

and

$$a/b:=a*b^{-1}$$

then $(G,*,/)$ is denoted with $Conj(G)$ and is a quandle because it satisfies the quandles axioms

1) $a*a=a=aaa^{-1}$

2) $(a*b)*c=(a*c)*(b*c)$

because $c(bab^{-1})c^{-1}=(cbc^{-1})cac^{-1}(cb^{-1}c^{-1})=cbab^{-1}c^{-1}$

3) $(a*b) *b^{-1}=(a*b^{-1})*b=a$

because $b^{-1}(bab^{-1})b=b(b^{-1}ab)b^{-1}=a$

I read that we have too that a group homomorphism between two groups $G$ and $G'$ is a quandle homomorphism between theire cojugation quandles $Conj(G)$ and $Conj(G')$ and that makes $Conj$ a functor betwen the category of Groups and the category of quandles...


I wanted to know more about this functor $Conj$ that "maps" Groups to Quandles and since I'm not expert of category theory I apologize if I use a wrong terminology

$Q1a$ - I learnt that not every Quandle is a conjugation Quandle or in other words $conj$ is not ""surjective"" on the "set" of all quandles (that is not a set but a class i think) so how can I prove that a Quandle is a Conjugation Quandle too?

$Q1b$ -Wich extra "axioms" must hold in a Quandle that is a Conjugation Quandle?

This has something to do with the inverse construction of $Conj$ so my next question is

$Q2$ - There is a way to define a group operation starting with a quandle operation? Like an inverse $Conj$ construction that build a "Quandle-Group" $Conj^{-1}(Q)$ from a conjugation quandle $Q$.

$Q3$ Is this process unique?

With unique I mean is is possible to have two different groups $G=(G,\cdot,\phi)$ and $G'=(G,\circ,\varphi)$ and $$Conj (G)= Conj(G')$$

this should mean that is possible to have

$$b\cdot a\cdot \phi(b)=b \circ a \circ \varphi(b)$$

where $\phi(b)$and $\varphi(b)$ are the inverse functions and

$$a \cdot b \neq a \circ b$$

$Q4$ My last question is if possible to generalize the conjugation operation of groups for monoids and semigruops in a way that these "Monoid-conjugations" and "Semigroup-conjugations" are quandles. If is possible I would like to read more about.


I've asked this question on MathOverflow here:From quandles to groups

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    $\begingroup$ The conjugation quandle of an abelian group is a so-called trivial quandle (i.e., x*y = x, identically), so any two abelian groups of the same order give rise to isomorphic conjugation quandles. $\endgroup$ – James Apr 7 '14 at 18:29
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    $\begingroup$ I don't know if there is a complete characterisation of conjugation quandles, but conjugation quandles are crossed sets. A crossed set is a quandle that satisfies the additional condition that, for any two elements $a$ and $b$, one has $a*b = a$ if, and only if, $b*a = b$. There is a quandle of order $3$ that is not a crossed set. (Of course, it is not faithful.) However, a crossed set need not be a conjugation quandle. There are $4$ crossed sets of order $4$ (up to isomorphism), but only one conjugation quandle. $\endgroup$ – James Apr 7 '14 at 19:31
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    $\begingroup$ I don't think so (if I understand your point correctly). You could have two elements of a group that commute, but neither is a power of the other. $\endgroup$ – James Apr 7 '14 at 20:15
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    $\begingroup$ If you do post over on MO, be sure to cross-link the questions. $\endgroup$ – James Apr 7 '14 at 20:58
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    $\begingroup$ Rather than an inverse functor, perhaps you should look for a left adjoint to $Conj$, say $F: Quand \to Grp$ such that $\operatorname{Hom}_{Grp} \left( F(Q), G \right) \cong \operatorname{Hom}_{Quand} \left( Q, Conj(G) \right)$ for any quandle $Q$ and any group $G$. $\endgroup$ – Sammy Black Apr 10 '14 at 20:25

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