# Necessary and sufficient for $\operatorname{orb}(x)=\operatorname{orb}(y) \iff \operatorname{Stab}(x)=g\operatorname{Stab}(y)g^{-1}$

Are orbits equal if and only if stabilizers are conjugate?

You may get some insights from the link above.

My Question: What is the necessary and sufficient condition for the above statement to be true?

I can prove with no difficulty that same orbits implies conjugate stabilizer .For the converse, if $G$ is non trivial groups, is it strong enough to show the converse is true? I guess it probably not working with only the condition but i couldn't get a counter example.

Any hints, counter example, ideas or related information would be appreciated.

• Iguess ,Your title should be $Orb(x)= Orb(y) \iff Stab(x)=gStab(y)g^{-1}$ Commented Apr 13, 2014 at 12:12

If $O(x)=O(y)$ then $Stab(x)$ and $Stab(y)$ are conjugate.

Since they are in same orbit, there exist a $g\in G$ such that $g*x=y$.

Now, let $r\in Stab(y)$ then $$r*y=y$$ $$r*(gx)=gx$$ $$(g^{-1}rg)*x=x$$ so $(g^{-1}rg)\in Stab(x)\implies g^{-1}Stab(y)g \leq Stab(x)$ to show other inclusion you can fallow similiar way then you are done.

Converse is not true. If $Stab(x)$ ans $Stab(y)$ are conjugate, you can not conclude that $O(x)=O(y)$. But you can say that $|O(x)|=|O(y)|$ by ortbir-stabilizer theorem.

Let $G=Z_2=\{0,1\}$ and $\omega=\{1,2,a_1,a_2 \}$ and the zero of $G$ fixes every element and $1_G$

$$1_G:1\to 2$$ $$1_G:2\to 1$$ $$1_G:a_1\to a_2$$ $$1_G:a_2\to a_1$$

You can check that $Stab(1)=Stab(a_1)$ so they are also conjugate to each other but $O(1)=\{1,2\}$ and $O(a_1)=\{a_1,a_2\}$.

• do you know what is the sufficient condition for the converse to be right? Like if the stabilizer are normal? or if $X$ is a group or if x conjugate to y? Commented Apr 13, 2014 at 18:17
• @Jorden I don't think a group theoretic property is enough to determine when elements are in the same orbit. However if $X=G$ and $x\sim y$ are conjugate then of course they have the same orbit (under the conjugation action).
– anon
Commented Apr 13, 2014 at 18:22
• @Jorden: Action of $G$ on different orbits are independent of each other Unless you put strong connection but when you put strong connection it becomes so obvious that it lost its meaning. Commented Apr 14, 2014 at 20:53