Suppose $G$ is a transitive permutation group such that all four-point stabilizers are trivial, and $G$ has some nontrivial three-point stabilizer. Said differently this means that every nontrivial element in $G$ fixes at most three points, and some nontrivial element actually fixes three points.
Let $E \le G$ be a perfect subnormal subgroup such that $E / Z(E)$ is simple (such subgroups are called components). It could be shown that two distinct subgroups with these properties commute. Denote the subgroup generated by all such subgroups by $E(G)$ (this is called the layer of $G$). I want to show that if $E(G) \ne 1$ we have $E(G) \cap G_{\alpha} \ne 1$ for some point stabilizer. So assume $E(G) \cap G_{\alpha} = 1$. Let $E$ be a perfect subnormal subgroup with $E/Z(E)$ simple, first we show that every $x \in G_{\alpha}$ normalizes $E$.
Suppose $x \in G_{\alpha}$ has prime order and does not normalize $E$. Let $L := E E^x E^{x^2} \cdots E^{x^{p-1}}$. Then $L$ is $x$-invariant and for $e \in E$ the product $y := ee^x e^{x^2} \cdots e^{x^{p-1}}$ is centralized by $x$ by the property that the factors $e^{x^i}$ commute, i.e. we have $y \in C_L(x)$. We have $|C_L(x) : C_L(x) \cap G_{\alpha}| \in \{2,3\}$ by the assumptions about the fixed points and so $e$ has order $2$ or $3$. This gives that $E$ is a $\{2,3\}$-group, hence solvable, which gives a contradiction.
I do not fully understand what is happening here. I guess we have $y \ne 1$, but I do not see why? If $x$ just has the single fixed point $\alpha$, then as $C_L(x)$ acts on the fixed points we would have $\alpha^y = \alpha$, so $y \in G_{\alpha}$ which would give $y = 1$. So $x$ has either two or three fixed points, in the first case we have $|C_L(x) : C_L(x) \cap G_{\alpha}| \le 2$ by orbit-stabilizer, where index $1$ is excluded as $C_L(x) \cap G_{\alpha} = 1$ and $C_L(x)$ is nontrivial (again using $y \ne 1$), so that $|C_L(x)| \le 2$ and this would give that $y$ has order two. But why has $e$ order $2$? If $x$ has three fixed points, then $|C_L(x) : C_L(x) \cap G_{\alpha}| \le 3$ and again $1 < |C_L(x)| \le 3$, which again gives that $y$ has order $2$ or $3$, but I do not see how is it implied that $e$ has this order?
As the factors $e^{x^i}$ commute and all have the same order, by the result that if $gh = hg$ then $o(gh)$ divides $\operatorname{lcm}(o(g), o(h))$ we find that $o(y)$ has to divide $o(e)$, so at least the results about the order of $y$ give me that $o(e)$ is divisble by $2$ or $3$, but not that $e$ itself has that order?
I hope someone can help me!? Thanks!