I am trying to analyse the following.

Assume $S_{3}$ acts on a non empty set $T$, and that is has $3$ orbits. What can we say about the possible cardinalities for the set T?

My thoughts:

If $G=S_{3}$ then $|G|=3!=6$ and we are consider $$G \times T \to T,~~(g,t) \to g \star t$$

I also know that $|\text{orb(t)}|=\frac{|G|}{|\text{stab(t)}|}$

I know lagranges thereom tells us the size of a subgroup must divide the size of the group. I also know that orbits partition a finite set, so then since orbit is a subgroup of T, we know the size of the orbits must divide it?

$$|T|=|orb(t_{1})|+|orb(t_{2})|+|orb(t_{3})|$$ And also, any fixed point is also a stabilizer. But I am having trouble putting all this together to make a coherent thought of what this tells me about T.

Else I am just not so sure how to proceed.

Any advice? Thanks!


When you say "I also know that $\lvert \text{orb}(t)\rvert=\frac{\lvert G\rvert}{\lvert \text{stab}(t)\lvert}$ which tells us $\lvert \text{stab}(t)\rvert =2$," I think you are mistaking $\lvert \text{orb}(t)\rvert$ for the number of orbits of $T$. This is not what $\lvert\text{orb}(t)\rvert$ is. Given a particular element, $t\in T$, $\text{orb}(t)$ is the set of elements that to which the action of $G$ carries $t$, and $\lvert\text{orb}(t)\rvert$ is the number of these elements, and does not say anything about the other orbits of $T$.

Clearing this up may help you on your way to solving, the problem, but if you still need a hint knowing that $T$ has three orbits, we can pick elements $t_1, t_2, t_3\in T$ so that $$\lvert T\rvert=\lvert\text{orb}(t_1)\rvert+\lvert\text{orb}(t_2)\rvert+\lvert\text{orb}(t_3)\rvert\text{.}$$

  • $\begingroup$ Thanks , so is that basically telling us T must have at least 3 elements since orbits are equal or disjoint ? $\endgroup$ – Quality Dec 5 '15 at 16:58
  • $\begingroup$ Right! You can say a lot more than that, though, if you combine it with $\lvert \text{orb}(t)\rvert=\frac{\lvert G\rvert}{\lvert \text{stab}(t)\lvert}$. $\endgroup$ – Trold Dec 5 '15 at 18:52
  • $\begingroup$ So what else could I say though ? $\endgroup$ – Quality Dec 13 '15 at 22:38
  • $\begingroup$ You can in fact give a complete list of possible sizes of $T$. The largest $\lvert T \rvert$ can be is $18$, corresponding to the case when $\text{stab}(t_1)=\text{stab}(t_2)=\text{stab}(t_3)=\{e\}$, the trivial subgroup with only the identity. The smallest it can be is three for the reasons you've already said, corresponding to $\text{stab}(t_1)=\text{stab}(t_2)=\text{stab}(t_3)=G$. What are cases in between? $\endgroup$ – Trold Dec 14 '15 at 2:36
  • $\begingroup$ Could it be any sums of sizes that divide G? $\endgroup$ – Quality Dec 14 '15 at 2:48

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