finding a topological group with specific conditions I have a question, it sounds difficult.
The question is the following:
Let $X$ be a topological group such that the binary operation defined on it is $*$. For any two points $a$ and $b$ in $X$ define a new operation by $a(*)b=b^{-1}*a*b$, [$(*)$ is a new operation on $X$ inherited from $*$]. By this $(*)$, we get that $a(*)a=a$, for all $a$ in $X$. Now let $X_3$ be a subset of $X$ contains 3 elements, say $a$, $b$ and $c$ such that $a(*)b=b(*)a=c$; $a(*)c=c(*)a=b$; $c(*)b=b(*)c=a$. The question is find a specific example of topological group described above. Explain what is $*$ and determine the three elements $a$, $b$ and $c$.
I thought about it and unfortunately I have not found any topological group with the conditions mentioned. We can define $(*)$ for any group, yes, but the problem how to find the 3 elements in the question. I tried circle, torus, $SO_3$. What do you think, can we find one as described in the question?
By topological group we mean a topological space and a group $G$ at the same time such that the operation $G \times G \to G$ and taking the inverse  $x \mapsto x^{-1}$ are continuous functions. So it has algebraic structure (group) and topological structure. Examples: Circle, torus, $SO_3$.
Please give me a hint. 
Thank you in advance 
 A: The symmetric group on three letters works.
Let $a,b,c$ be the three transpositions. Conjugate any one by any other and you get the third. Now choose any topological group that contains a copy of $S_3$ . For instance  $SO(3)$. Where the three elements can be seen as rotations by 180 degrees around three axes lying in the plane, where each pair makes an angle of 120 degrees at the origin.
A: This is too long to be a comment.  
Let $O(2)$ be $2\times 2$ matrices $A$ with real entries so that $$AA^t=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$
Notice it's just the group of orthogonal $2\times 2$ matrices.  Topologically it is the disjoint union of two circles.  Any matrix in $A\in O(2)$ has $det(A)=\pm 1$ since $1=det(AA^t)=det(A)det(A^t)=det(A)^2$.  This corresponds to the fact that $O(2)$ has two connected components, each homeomorphic to a circle.  The first component is made up of rotations, and consists of the matrices with determinant $1$. The identification with the circle is via the angle of rotation.
The second component is made up of reflections in lines through the origin and is made up of matrices of determinant $-1$. 
The identification of the second component with a circle is more tenuous, as a reflection in a line through the origin fixes two antipodal points on the circle.
However the quotient space from the circle coming from the equivalence relation that identifies antipodal points is still a circle.  So the identification is via the pair of fixed points of the reflection.
The component made up of reflections is closed under conjugation, so it is a quandle. It contains any quandle corresponding to any dihedral group, so it contains all the quandles corresponding to knot coloring, no matter how many colors you use.
So maybe $O(2)$ is the Lie group associated to knot coloring.  
Notice also that this subsumes the other answer I gave because the maps of $\mathbb{R}^3$ that I gave above preserve the $xy$-plane and induce elements of this quandle.
A: I also think O2 is a topological group associated to a dihederal quandle Rn for prime number n. From your idea, I also get another topological space which is associated to the dihederal quandle: the real projective plane RP^2. Unless RP2 is not a topological group, but it is a quandle. Let u and v be two vectors in RP2, then u*v is a vector obtained by reflecting u about v through the origin. (RP^2,*) is a quandle. Now we can embed dihederal quandles of prime order into any great circle on S^2, and then identify the antipodal points.
