Let $G$ be a finite permutation group which acts transitive and nonregular on $\Omega$ in such a way that each nontrivial element fixes at most two points.
Let $\alpha, \beta \in \Omega$ be distinct and such that $U := G_{\alpha}\cap G_{\beta} \ne 1$. Suppose that $|G_{\alpha}|$ is odd and that $|\Omega|$ is even. Suppose further that $G$ has no subgroup of index at most $2$ that is Frobenius group. Then there exists an involution $x \in N_G(U) \setminus U$ that interchanges $\alpha$ and $\beta$ and such that one of the following is true:
(1) $U$ is abelian. Moreover if $G$ is simple, then all involutions in $G$ are conjugate.
(2) $Z(C_G(x)) = \langle x \rangle$ and $C_G(x) / \langle x \rangle$ is a Frobenius group.
Proof: [ I skip the part showing that $|N_G(U) : U| = 2$ and there exists an involution $x \in N_G(U) \setminus U$ ].
If $C_U(x) \ne 1$, then we define $C := C_G(x)$ and we consider the action of $C$ on $\alpha^C$. If $|\alpha^C| \le 2$, then $C = \langle x \rangle \times U$ and thus the Sylow $2$-subgroups of $G$ are cyclic. Then $G$ has a normal $2$-complement $G_1$. But now $G_1$ has odd order and $(G_1, \alpha^{G_1})$ acts as a Frobenious group; contradicting our hypothesis. Thus we see that $|\alpha^C| \ge 3$ and that $C$ is not regular on $\alpha^C$. [...] $\square$
There are some points I do not understand about this part of the proof.
1) If $|\alpha^C| \le 2$, then $C = \langle x \rangle \times U$ and thus the Sylow $2$-subgroups of $G$ are cyclic.
As $x \in C$ I see that $\alpha^C = \{\alpha\}$ is not possible, and also we must have $\alpha^C = \{\alpha, \beta\}$, and so that $C \le N_G(U)$. Also as $U^x = U$ we have $\langle U, x \rangle = U \cdot \langle x \rangle$. But why $C = \langle x \rangle \times U$? I can see that $C \le U\cdot \langle x \rangle$, as if $y \in C$ and $\alpha^y = \alpha$ then of course $y \in U$, and if $\alpha^y = \beta$ then by $\alpha^x = \beta$ we have $\beta^{xy} = \beta$, i.e. $xy \in U$, hence $y \in Ux^{-1} = Ux$. Also $C_{\alpha} \le U$. But this is all I see, but why is $U \le N_G(\langle x \rangle)$, i.e. $xu = ux$ for all $u \in U$, and why do we have $C = \langle x \rangle \times U$?
2) $(G_1, \alpha^{G_1})$ acts as a Frobenious group.
I see that $G_1$ in its action fulfills the requirement put on $G$, i.e. at most two fixed points and so, but why each element has at most one fixed point and the action is nonregular?
3) We see that $|\alpha^C| \ge 3$ and that $C$ is not regular on $\alpha^C$.
Why is $C$ not semi-regular?
If wanted I can supply the rest of the proof. Hoping someone can explain these points too me. Thank you!