Suppose $b\in O_a$, and $a\in A$. Prove that $G_a$ and $G_b$ are isomorphic. Under what conditions are they actually equal?

Question

Let $G$ be a group acting on a set $A$, and for $a\in A$, let $O_a$ be the orbit of $a$ and $G_a$ be the stabilizer of $a$. Suppose $b \in O_a$. Prove that $G_a$ and $G_b$ are isomorphic. Under what conditions are they actually equal?

I've been told that the best way to approach this is as if i'm answering "What is the relationship between $G_a$ and $G_b$", but that's not helping me much.

I know the properties of an isomorphism (bijective and homomorphic) but I'm not sure how to apply that definition in this case, or if I can prove this an entirely different way.

Answer 1

Consider $g \in G$ such that $g(a) = b$. Such a $g$ exists since $b \in O_a$

Then consider $\phi_g: G_a \to G_b$, given by $\phi(x) = gxg^{-1}$.

If $x(a) = a$, then $gxg^{-1}(g(a)) = b$

Then use that any two conjugate subgroups are isomorphic.

For conditions on when they are equal, one is that $G_a$ is normal. This is equivalent to some other obvious things, but that may be what is wanted.

Answer 2

The question of isomorphism has already been addressed by Dionel Jaime very well, so I'd just like to add some thoughts as to when $G_a=G_b$. If $g(a)=b$, then $G_b=gG_ag^{-1}$. Hence $G_b=G_a$ exactly when $b=g(a)$ for some $g\in N(G_a)$. I.e. $G_b=G_a$ if and only if $b$ is still in the orbit of $a$ when we restrict the action to the subgroup $N(G_a)$.

As noted in Dionel Jaime's answer, in the case when $G_a$ is normal, $G_b=G_a$ for all $b$ in the orbit of $a$, because in that case, $N(G_a)=G$, so the orbit of $a$ under the action of $N(G_a)$ is the same as the orbit of $a$ under the action of $G$.