Let $G$ be a group, take Conjugacy class of $g$, let us denote it by $g^G$ and let $g^G=X$

Let $x, y \in X$, define $xy=xyx^{-1}$, then $X$ is closed under this operation

How I can proof this axiom if quandle group? for each $x, y \in X$, there exist a unique $z \in X$ such that $xz=y$, by using the above definition.

If we suppose there are two elements of $z_1,z_2 \in G$ such that $xz_1=y$ and $xz_2=y$ if we prove $z_1=z_2$ then it will be proved. for this

if we equate the above two equation $xz_1=xz_2$ then cancellation property does not hold how we can prove $z_1=z_2$? or is it possible to apply $x^{-1}$ on both sides in Quandle groups.


Let's normalize the notation a little since your use of juxtaposition to stand for two different operations makes your question hard to read. We'll only use juxtaposition for the group multiplication. $$x \rhd y = xyx^{-1} $$ Let $x,y \in X$. We want to find a $z \in X$ such that $x \rhd z = y$. Using that for all $\alpha \in X$, there is a $h(\alpha) \in G$ such that $\alpha = g^{h(\alpha)}$, we compute: \begin{align} && x \rhd z &= y \\ \iff&& xzx^{-1} &= y\\ \iff&& g^{h(x)}z(g^{h(x)})^{-1} &= g^{h(y)} \\ \iff&& z &= (g^{h(x)})^{-1}g^{h(y)}g^{h(x)} \\ \iff&& z &= (h(x)gh(x)^{-1})^{-1}h(y)gh(y)^{-1}h(x)gh(x)^{-1} \\ \iff&& z &= \left(h(y)^{-1}h(x)gh(x)^{-1}\right)^{-1}g\left(h(y)^{-1}h(x)gh(x)^{-1}\right) \in X \end{align} In short, the only group element that $z$ can be happens to be in $X$, as was to be shown.


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