Let $G$ be a group with $33$ elements acting on a set with $38$ elements. Prove that the stabilizer of some element $x$ in $X$ is all of $G$. I'm trying to figure out this old qualifying exam question:
Let $G$ be a group with 33 elements acting on a set with 38 elements. Prove that the stabilizer of some element $x \in X$ is all of $G$.
I think I'm supposed to use the orbit-stabilizer theorem to prove that the orbit of any $x\in X$ must be trivial, i.e. $orb_G(x)=\{x\}$. This is what I know:
$|G|$ and $|X|$ are relatively prime.
Since $|orb_G(x)| $ divides $|G|$ we must have that $|orb_G(x)|=1, 3, 11 $ or 33.
The orbit of each $x\in X$ partitions $X$.
If $|orb_G(x)|=1$ then by the orbit-stabilizer theorem: $|G|=|orb_G(x)||stab_G(x)| \implies |stab_G(x)|=33$.
I just don't see how to put this together in the right way. I wondered if $|orb_G(x)| $ necessarily needs to divide $|X|,$ but I didn't find anything to support that.
$$\langle(1,2,3,4,5,6,7,8,9,10,11)(12,13,14)(15,16,17)\\(18,19,20)(21,22,23)(24,25,26)(27,28,29)(30,31,32)\\(33,34,35)(36,37,38)\rangle$$ is a counterexample.
For a fixed $x \in X$, define a group action $f: G\times X\to X  $ by $(g,x) \mapsto x$ where $g\in G_x\subset G$, this map is not necessarily injective because $G_x$ can have more than one element. Since $1\in G_x$, this map is surjective.
Now by the orbit-stabilizer theorem, if we have for some $x\in X$ that is not equivalent to other elements in X,
$$1=|{x}|=|\{f(x)\}|=\vert G\cdot x|=|G|/|G_x| $$
This forces $|G_x|=|G|$, i.e., stabilizer of x in $X$ is all of G.
Now we only need to show the existence of such nonequivalent element;
Assume that there is an equivalent element y$\in X$ of x, then $x\dot g=y$ for all x$\in X$ for some $\dot g\in G$ and $G\cdot x=G\cdot y$. By orbit-stabilizer equation, we have $|G_x|=|G_y|$. Since $\cup_x G_x = G$ and the existence of equivalence of every element forces $|G|=33$ is divisible by 2, which is a contradiction.
Hence there exist a nonequivalent element x and we complete the proof.