Two orbits are $G$- isomorphic iff their stabilizers are conjugate.

Question

Suppose that a group $G$acts on a set $A$, we say $A$ is a $G$-set and two $G$- sets $A$ and $B$ are said to be$G$- isomorphic if there is a bijection $ f: A \mapsto B$ such that $g. f(a) = f(g.a)$ for all $g \in G, a\in A$

My problem is: for any $a,b \in A$, orbits of $a$ and $b$ respectively, $O_{a}, O_{b}$ are isomorphic as G-sets if and only if there exists a $g\in G$ such that $g \operatorname{stab}_{G}(a) g^{-1}= \operatorname{stab}_{G}(b)$

I have trying to solve this for hours but unsuccessfully. My guess is to use that $G$-isomorphism condition to prove one inclusion of stabilizers and then use the same for $f^{-1}$ to prove the other inclusion of stabilizers.

Any help/hints that could help me solve this problem would be greatly appreciated, thanks!

Answer 1

As it is an "iff" question, it usually is easier to prove each direction separately. I'll try to describe the thought process.

First assume that $g\operatorname{stab}_G(a)g^{-1}=\operatorname{stab}_G(b)$ for some $g$. As we want to construct a bijection between $O_a$ and $O_b$ (and hope that it is a $G$-isomorphism), we might start by trying to "label" them by the same set. As $O_a$ and $O_b$ are orbits, we may use $G$ as a labeling set!

However, the labeling is not unique (if the action is not free): given $x\in O_a$, there may be two elements $h_1,h_2\in G$ such that $x=h_1a=h_2b$. In any case $h_1h_2^{-1}\in\operatorname{stab}_G(a)$, so $gh_1h_2^{-1}g^{-1}\in\operatorname{stab}_G(b)$ and so $gh_1g^{-1}b=gh_2g^{-1}b$.

Therefore the map $\phi':O_a\to O_b$, $\phi'(ha)=ghg^{-1}b$ is a well-defined bijection. However it is not a $G$-isomorphism, since instead it satisfies $$\phi'(hx)=ghg^{-1}\phi'(x).$$ We would be done if we had $g=1$. So recall that $\operatorname{stab}_G(gb)=g\operatorname{stab}_G(b)g^{-1}=\operatorname{stab}_G(a)$, and of course $O_{gb}=O_b$.

Applying the same process as above to $gb$ instead of $b$ yields a bijection $\phi:O_a\to O_{gb}=O_b$, $\phi(ha)=hgb$, satisfying $\phi(hx)=h\phi(x)$, and we're done.

The converse is similar: If $\phi:O_a\to O_b$ is a $G$-isomorphism, then $\operatorname{stab}_G(a)=\operatorname{stab}_G(\phi(a))$. Write $\phi(a)=gb$ for some $g$ (as it belongs to $O_b$) and use again that $\operatorname{stab}_G(gb)=g\operatorname{stab}_G(b)g^{-1}$.

Answer 2

Here is a hint. You can try to define a $G$-homomorphism $\phi: O_a \to O_b$ by sending $a\mapsto b$. Along with the condition $\phi(g.a) = g.\phi(a)$, this is enough to determine a unique map, if it exists. So, when is this map well-defined?