Let $G$ be a finite group acting transitively on a set $X$, where $|X| = 2^n$ for some $n \geq 1$. Show that some element of $G$ acts as an involution with no fixed points.
While it is fairly easy to show that
- some element of G acts with no fixed points (as an easy corollory of Burnside's lemma)
- some element of G acts as an involution (as by the Orbit-Stabilizer theorem, $|G|$ is even)
it seems to be slightly harder to show that there is an element of $G$ which is both of these. The only approaches I can think of is to either show that somehow some Sylow $2$-subgroup of $G$ fixes no points (but I doubt that is true), or the number of elements with no fixed points when acting on $X$ is odd (as then we'd be done because $g$ has no fixed points iff $g^{-1}$ has none), but I also doubt this is true.
Does anyone have any better ideas?
Best Answer
Consider the following.