# 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.

• You might want to explain the numbers you wrote down. Jul 9, 2020 at 23:11
• This is a counterexample to the claim. It is an element of order $33$ in $S_{38}$ that stabilizes no point. Jul 9, 2020 at 23:11
• $X$ here is the set $\{1..38\}$. No element of $X$ is stabilized by all of $G$, i.e., a fixed point of the element. Whether he meant something else by the question, I don't know. Jul 9, 2020 at 23:15
• I am sorry, it is already late and I am just asking stupid things... Jul 9, 2020 at 23:17
• In general, if $G$ has order $pq$ then the largest number this could work for is $pq-1$. So $28$ would work, but nothing above $31$. All you are asking for is the largest number that cannot be written as $ap+bq$ for some non-negative integers $a$ and $b$. Jul 9, 2020 at 23:23

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.)

• @user750041, it is an outdated comment. Previously, it was a valid point. Per the above discussion in the comments, the result is false for $|X| = 38,$ and it is ostensibly false for $|X| = 28,$ all though a counterexample has not been provided. I figured it was worthwhile to resurrect my original answered (that I had deleted in light of the aforementioned comment) for the case of $|X| = 18.$ Jul 10, 2020 at 14:46
• A counterexample for $28$ is clear, just do two $11$-cycles and then two $3$-cycles. Fill them in with any entries from $1$ to $28$ you like. I only chose $28$ because I was guessing what the typo might have been. No number over $19$ can work. Also, $$(1,2,3)(4,5,6)...(16,17,18)$$ is an action on $18$ points, as the action was not required to be faithful. Jul 10, 2020 at 21:04

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.

• How is a map from 33 objects to 38 onto? In fact, it is not onto, it is only onto the orbit of $x$, with kernel the stabilizer of $x$. Jul 10, 2020 at 10:14
• you are right. I will modify my answer Jul 10, 2020 at 14:18
• But the result has been shown to be wrong so you have no hope of proving it! Jul 10, 2020 at 16:05
• @Derek Holt The counter example given cannot act on the counting set by the definition of group action, and so fail to make a contradiction.mathworld.wolfram.com/GroupAction.html Jul 10, 2020 at 16:45
• What is the point in saying that it cannot act on the set when it quite clearly does act on that set? It is acting on the set $\{1,2,\ldots,38\}$ with one orbit of length 11 and 9 orbits of length 3. None of the orbits have length 1. Jul 10, 2020 at 16:56