If a finite group $|G|$ acts transitively on a set $X$ with $|X|=2^n$, $n \geq 1$, then $G$ has an involution with no fixed points 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? 
 A: Consider the following. 


*

*Replacing $G$ with $G/\ker(\phi)$, where $\phi$ is the action homomorphism $\phi:G\to Sym(X)$ allows us to assume that no non-identity element of $G$ acts trivially on $X$.

*Let $P$ be a Sylow 2-subgroup of $G$. I claim that $P$ also acts transitively on $X$.

*Sylow theory tells us that $P$ has an odd number of  conjugates in $G$, call that number $\ell$. Because $G$ acts transitively on $X$, all the point-stabilizers $G_x$ are conjugate, and thus contain the same number of conjugates of $P$, call that number $m$. Similarly, all the conjugates of $P$ fix the same number of points, call that number $k$. Let's count in two different ways the number of pairs $(P',x)$ such that $P'$ is a Sylow 2-subgroup of $G$ contained in a point stabilizer $G_x$ for some $x\in X$. This gives us the equation
$$ k\ell=|X| m=2^n m.$$ Here $\ell$ is odd, so $2^n\mid k$. So unless $k=0$ $P$ must act trivially on $X$, i.e. we must have $P=\{1_G\}$. But this is clearly impossible, because $|X|\mid |G|$. Thus $k=0$.

*We know that the center of $P$ is a non-trivial 2-group (nilpotent groups have non-trivial centers). Thus there exists an element $s\in Z(P)$ of order $2$.

*The element $s$ cannot belong to a point-stabilizer $P_x$, because if it belonged to one $P_x$, it would belong to all of them. This is because $s$ forms a singleton conjugacy class inside $P$, and $P$ acts transitively on $X$, so the point-stabilizers $P_x$ are all conjugate in $P$. The element $s$ is also an involution.

