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 $G$.

Question

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.

Answer 1

$$\langle(1,2,3,4,5,6,7,8,9,10,11)(12,13,14)(15,16,17)\\(18,19,20)(21,22,23)(24,25,26)(27,28,29)(30,31,32)\\(33,34,35)(36,37,38)\rangle$$ is a counterexample.

Answer 2

Edit: Like we discussed in the comments above, the result is not true if $X = 38$ or $X = 28;$ however, in the case that $X = 18,$ the following argument will work.

Consider the set of fixed points $\operatorname{Fix}_G(X)$ of $X$ under the action of $G,$ i.e., $$\operatorname{Fix}_G(X) = \{x \in X \,|\, g \cdot x = x \text{ for all } g \in G \}.$$ We claim that $|\operatorname{Fix}(X)| \geq 1,$ from which it follows that there exists an element $x \in X$ such that $g \cdot x = x$ for all $g \in G,$ i.e., $\operatorname{Stab}_G(x) = \{g \in G \,|\, g \cdot x = x \} = G.$

By the Class Equation, we have that $$|X| = |\operatorname{Fix}(X)| + \sum_{i = 1}^r |G| / |G_i|,$$ where $r$ is the number of distinct orbits $\mathcal O_i = \{g \cdot x \,|\, g \in G \}$ of cardinality $\geq 2$ and $G_i = \operatorname{Stab}_G(x_i)$ for some element $x_i$ of $\mathcal O_i.$ Considering that $|G| = 33,$ for each integer $1 \leq i \leq r,$ we must have that $|G_i| \in \{3, 11 \}$ so that $|G| / |G_i| \in \{3, 11 \}.$ Can you finish the proof by establishing that we must have that $|\operatorname{Fix}_G(X)| \geq 1?$ (Essentially, at this point, it is just a matter of counting, using the fact that $|X| = 18 = 3x + 11y$ has no positive integer solutions.)

Answer 3

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.