0
$\begingroup$

Let $S$ be a finite set, $O(s)$ the orbit by $s\in S$ of the action of $\operatorname{Sym}(S)$ on $S$ $(\sigma \cdot s:=\sigma(s))$, and $Stab(s):=\lbrace \sigma \in \operatorname{Sym}(S) \mid \sigma(s)=s \rbrace$. By the Orbit-Stabilizer Theorem, we get:

$$Stab(t) \subseteq Stab(s) \Rightarrow |O(s)| \le |O(t)|$$

Conversely, can we state anything on $Stab(t)$ and $Stab(s)$ (inclusion/intersection), once we knew that $|O(s)| \le |O(t)|$?

$\endgroup$
0
$\begingroup$

In the situation you're describing, Sym$(S)$ acts transitively on $S$, i.e. $O(s) = S$ for any $s \in S$. So the orbit sizes are all the same, but all stabilisers are distinct index $n = |S|$ subgroups.

In more general situations, where we have some group $G$ acting on some set $X$, we can also not conclude anything from the orbit sizes alone. Of course there is the fact $|Stab(s)| \leq |Stab(t)| \iff |O(t)| \leq |O(s)|$, but one can find examples where $|O(s)| \leq |O(t)|$ where the stabilisers intersect trivially and where the stabiliser of $t$ is contained in the one of $s$.

$\endgroup$
  • $\begingroup$ I'm certainly missing some fundamentals, but $O(s)=S \Leftrightarrow \forall t \in S, \exists \sigma \in \operatorname{Sym}(S) \mid t=\sigma(s)$: why is this latter always true? Also by contrapositive I can't conclude that, since in $\exists t \in S \mid t \ne \sigma(s), \forall \sigma \in \operatorname{Sym}(S)$ I still don't see any contradiction. Where am I wrong? Yes, I argued that orbit sizes alone don't tell the whole story, but then I'd like to know whether the different situations that may arise among the stabilizers can give hints on group's structure. $\endgroup$ – user615081 May 27 '19 at 16:16
  • $\begingroup$ By definition, $\text{Sym}(S) = \{\sigma: S \to S \ | \ \sigma \text{ is a bijection} \}$ is a group under composition. Let $s,t \in S$ be two elements. Define $\sigma: S \to S$ by $\sigma(s) = t$, $\sigma(t) = s$, and $\sigma(x) = x$ for all $x \in S \setminus \{s,t\}$. Then $\sigma \in \text{Sym}(S)$ because it's a bijection, and it shows that $t$ and $s$ are in the same orbit, and thus that $O(s) = S$ for any $s \in S$. $\endgroup$ – user May 27 '19 at 17:55
  • $\begingroup$ It may also help you to study the groups $S_n$ for small $n$; for finite sets $S$, we have $\text{Sym}(S) \cong S_n$ for $n = |S|$, and it may be easier to visualise what's going on when you work with explicit elements of $S_n$: they can be written down systematically as cycles, and you can check how these act on $\{1,2,\ldots,n\}$. $\endgroup$ – user May 27 '19 at 17:58
  • $\begingroup$ Ok, now I see it. I will also practice with $S_n$, as suggested. Thanks. $\endgroup$ – user615081 May 27 '19 at 18:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy