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