Orbit of a point under the action of the stabiliser of another point

Question

I'm wondering, for a group $G$ acting on a set $X$, what can be said about the orbit of a point under the action of a stabiliser subgroup of $G$. Let $x\in X$ and let $H = {\rm Stab}_G(x)$. Obviously $|H\cdot x|$ is going to be $1$, but what can we say about $|H\cdot y|$ for some other $y\in X$, perhaps in relation to $|G\cdot y|$? If that not much can be said in general, feel free to put interesting restrictions on $y$, e.g., $y$ in $G\cdot x$.

Edit. I guess the original question was far too vague. I am still interested in the general question, but in the specific examples I am thinking of, $X$ is the set of vertices of a finite graph and $G$ acts by graph automorphisms. Does this extra information tell us more about $H\cdot y$?

Answer 1

As of little/no interest as it may be, I'd warm up by noting that, if $\operatorname{Stab}_G(x)\unlhd G$, then $|\operatorname{Stab}_G(x)\cdot y|=1$ for every $y\in O_G(x)$ (the orbit of $x$ under the $G$-action). In fact:

\begin{alignat}{1} \operatorname{Stab}_G(x)\cdot y &= \{h\cdot y\mid h\in \operatorname{Stab}_G(x)\} \end{alignat}

If $y\in O_G(x)$, then $\exists g\in G\mid y=g\cdot x$, and hence:

\begin{alignat}{1} \operatorname{Stab}_G(x)\cdot y &= \{h\cdot (g\cdot x)\mid h\in \operatorname{Stab}_G(x)\} \\ &= \{(hg)\cdot x\mid h\in \operatorname{Stab}_G(x)\} \\ \end{alignat}

If, further, $\operatorname{Stab}_G(x)\unlhd G$, then:

\begin{alignat}{1} \operatorname{Stab}_G(x)\cdot y &= \{(hg)\cdot x\mid h\in \operatorname{Stab}_G(x)\} \\ &= \{(gh')\cdot x\mid h'\in \operatorname{Stab}_G(x)\} \\ &= \{g\cdot (h'\cdot x)\mid h'\in \operatorname{Stab}_G(x)\} \\ &= \{g\cdot x\} \\ &= \{y\} \\ \end{alignat}

and $|\operatorname{Stab}_G(x)\cdot y|=1$.

Answer 2

There is a lot one can say about $\text{Aut}(X)$ acting on $X$. Much of Algebriac Graph Theory is devoted to this. The Wikipedia page (https://en.wikipedia.org/wiki/Graph_automorphism#Graph_families_defined_by_their_automorphisms) is a good starting point for the highlights.

There are a couple natural families with which to start, including:

• Vertex-Transitive Graphs (i.e., Graphs $X$ where $\text{Aut}(X)$ acts transitively on $V(X)$.)
• Cayley graphs, which are defined as follows. Let $G$ be a group, and let $S \subset G$ such that $1 \not \in S$. The Cayley graph $\text{Cay}(G, S)$ has vertex set $G$. There is an edge $(g, h)$ iff there exists $s \in S$ such that $gs = h$. If $S = S^{-1}$, then we consider $\text{Cay}(G, S)$ to be undirected (rather than having directed edges going both ways). Otherwise, $\text{Cay}(G, S)$ is directed. Cayley graphs are vertex transitive.

For a comprehensive treatment, I'd start with the Algebraic Graph Theory text by Godsil and Royle.

Answer 3

One interesting result connected to this question is the Sims Conjecture, dating from 1967, which was eventually proved using the classification of finite simple groups.

If $G$ acts faithfully and primitively on $X$ and $H = {\rm Stab}_G(x)$ with $x \in X$, then $|H|$ is bounded above as a function of $|H \cdot y|$, for any $y \in X \setminus \{x\}$.