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.


3 Answers 3


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

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    $\begingroup$ You might want to explain the numbers you wrote down. $\endgroup$ Jul 9, 2020 at 23:11
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    $\begingroup$ This is a counterexample to the claim. It is an element of order $33$ in $S_{38}$ that stabilizes no point. $\endgroup$ Jul 9, 2020 at 23:11
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    $\begingroup$ $X$ 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. $\endgroup$ Jul 9, 2020 at 23:15
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    $\begingroup$ I am sorry, it is already late and I am just asking stupid things... $\endgroup$ Jul 9, 2020 at 23:17
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    $\begingroup$ 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 $b$. $\endgroup$ 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.)

  • $\begingroup$ @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 $|X| = 18.$ $\endgroup$ Jul 10, 2020 at 14:46
  • $\begingroup$ 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. $\endgroup$ 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.

  • $\begingroup$ 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 $x$. $\endgroup$ Jul 10, 2020 at 10:14
  • $\begingroup$ you are right. I will modify my answer $\endgroup$
    – GanChen
    Jul 10, 2020 at 14:18
  • $\begingroup$ But the result has been shown to be wrong so you have no hope of proving it! $\endgroup$
    – Derek Holt
    Jul 10, 2020 at 16:05
  • $\begingroup$ @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 $\endgroup$
    – GanChen
    Jul 10, 2020 at 16:45
  • $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Jul 10, 2020 at 16:56

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