$X$ is a finite set and $|X| > 1$. $(G,X)$ is a transitive group action. Show that $$\bigcup_{x\in X}Stab(x) \ne G$$

This actually came up in my combinatorics class. But I figure it's within the algebra domain.
I try to prove by contradiction. First assume $\forall g \in G$, $\exists x \in X$ s.t. $g \ast x = x$ where $\ast$ stands for group action. But I don't know how to arrive at a contradiction using the transitivity of the group action. Help needed.

  • $\begingroup$ Is $G$ assumed to be finite? $\endgroup$ – Bungo Oct 18 '17 at 13:16
  • $\begingroup$ @Bungo I suppose so. Because $G$ should be some subgroup of $S_{|X|}$. $\endgroup$ – xixumei Oct 18 '17 at 13:18
  • $\begingroup$ @Bungo It doesn't need to be assumed. WLOG the action is faithful, since the properties were are interested in are a function of the lattice of subgroups of $G$ containing the kernel of the action (with stabilizers distinguished). $\endgroup$ – Dustan Levenstein Oct 18 '17 at 13:21
  • $\begingroup$ This is equivalent to the statement: if $H$ is a proper, finite index subgroup of a group $G$, then $G$ is not the union of conjugates of $H$. I think this is often seen as an exercise for finite groups, and I find it interesting that it still holds for infinite groups, by reduction to the finite case. $\endgroup$ – fredgoodman Oct 18 '17 at 18:41

$\def\Fix{\operatorname{Fix}}$ Another proof, but not quite as direct: First reduce to the case that $G$ is finite. For $g \in G$, let $\Fix(g)$ denote the set of $x \in X$ such that $g\cdot x = x$. Burnside's lemma, or the proof of Burnside's lemma, tells us that $|G| = \sum_{g \in G} |\Fix(g)|$, because the action is transitive. If $|\Fix(g)| \ge 1$ for all $g \in G$, then necessarily $|\Fix(g)| = 1$ for all $g \in G$. But $|\Fix(e)| = |X| > 1$. Hence there exists $g \in G$ such that $|\Fix(g)| = 0$.


If $G$ is finite, you can easily see that the LHS has fewer elements than the RHS, because if the union would be disjoint, there would be $|X| \cdot \frac{|G|}{|G \cdot x|} = |G|$ elements on the LHS. But the union is certainly not disjoint, because each stabilizer contains $1 \in G$.

If $G$ is not finite, note that $G/N$ is finite, where $N$ is the kernel of the induced map $G \to S_X$. This lets you pass to the finite case.


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