Recently I studied the orbit stabilizer theorem which is as follows:
Suppose $G$ is a group acting on $X$ (i.e.$X$ is a $G$-set). Let $x\in X$,then define, $\operatorname {orb}(x):=\{g.x:g\in G\}$ and $\operatorname{stab}(x)=\{g\in G:g.x=x\}$, then we have $|\operatorname{orb}(x)|=[G:\operatorname{stab}(x)]$, provided $G$ and $X$ are finite.
Now I have tried to intuitively understand this theorem like this, orbit of $x$ is roughly speaking, all the possible points in $X$ where $x$ can go under the given group action. And stabilizer means all the group elements(we can think of them as permutations also) that fix $x$.
Now the index on the right hand side of orbit stabilizer theorem is the number of cosets of $G$ induced by the subgroup viz stabilizer of $x$. Whenever a subgroup induces a coset or partition within a group, it means that we are classifying the group element by discriminating the group elements based on the propertly of that subgroup, now the property of stabilizer of $x$ is that it fixes $x$, so it contains all elements of $G$ that fix $x$, so the other coset must be according to the property,where the group elements take $x$, all the members of the group taking $x$ to a specific point will be a coset and for every point in orbit, there is a coset of $G$ whose members take $x$ to that point.
So there is a one-one correspondence between the cosets of $G$ induced by stabilizer subgroup and the members of the orbit of $x$,notice that $x$ itself is a member of the orbit and for it,the corresponding coset is stabilizer of $x$ itself.
So,naturally the number of orbit elements is the same as the number of cosets of stabilizer of $x$ in $G$.
I want to know if my intuition or understanding correct?
Best Answer
Yes, it is correct.
The (left) cosets of $S:=\mathrm{stab}(x)$ correspond bijectively to the elements in the orbit of $x$:
Specifically, for any elements $g,h\in G$, we have $$gx=hx\ \iff\ h^{-1}gx=x\ \iff\ h^{-1}g\in S\ \iff\ gS=hS\,.$$