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Given a topological space $X$ and a group $G$ acting on $X$, define $F=\{(x,y)\in X\times X \mid \operatorname{orb}(x)\cap \operatorname{orb}(y)=\emptyset\}.$ I want to write $F$ in terms of stablizers. I mean, is there a way to write $F$ in terms of stabilizer subgroups, which are defined as follows: given a group $H$ acting on a space $Z$, $\operatorname{stab}(z)=\{h\in H\;|\; hz=z\}.$

For example take the action of multiplication by $-1$ of $\mathbb Z_2=\{1,-1\}$ on a sphere $S^d$, $F=\{(x,y)\in S^d\times S^d \mid x\neq y\;\text{ and }\;x\neq -y\},$ and we can write $F=F_1 \cap F_2$ where $F_1=\{(x,y)\in X\times X \;|\; x\not =y\}\qquad F_2=\{(x,y)\in X\times X \;|\; x\not =-y\}$ Here, we are taking the induced action on $S^d\times S^d$ of $\mathbb Z_2\oplus \mathbb Z_2=\{(1,1),(1,-1),(-1,1),(-1,-1)\}$.

I can see that $F_1=\{(x,y)\in X\times X \mid \operatorname{stab}(x,y)=\{(1,1)\}\}$, but I don't see how to write $F_2$. Is there a general expression of $F$ in terms of stablizers of some subgroups of $G$, or product of $G$ as I did for $F_1$ above?

Added: I think it is better to write the complement F' of $F$ in $X\times X$. So F'=\{(x,y)\in X\times X\mid \operatorname{orb}(x)\cap \operatorname{orb}(y)\not = \emptyset\}. Now $\operatorname{orb}(x)\cap \operatorname{orb}(y)\not = \emptyset$ is equivalent to $x=gy$ for some $g\in G$, which is equivalent to $\operatorname{stab}(x)=\operatorname{stab}(gy)=g\operatorname{stab}(y)g^{-1}$, so F'=\{(x,y)\in X\times X\mid \operatorname{stab}(x)\sim \operatorname{stab}(y)\} where $\sim $ means conjugate to. Is this correct?

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    That the stabilizers are conjugate is necessary for two elements to be in the same set, but it is not sufficient. Your error lies in assuming that "$x=gy$ for some $g\in G$" is equivalent to "$\mathrm{stab}(x)=\mathrm{stab}(gy)$". If $x=gy$, then of course the stabilizers are equal. But the stabilizers can be equal even if $x$ and $y$ are not in the same orbit. Take your favorite group action of $G$ on a set $X$; then let $G$ act on a disjoint union of two copies of $X$ by acting on each copy separately. Then the stabilizers of the same element in the different copies are identical.2011-12-04

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