1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.