If $G$ acts nonregular, transitive and $|\mbox{fix}(g)| \le 2$, $|G_{\alpha}|$ is odd and $|\Omega|$ is even, then $|G|$ has twice odd order

Question

Let $G$ act nonregular and transitive on $\Omega$ such that each nontrivial element has at most two fixed 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 a Frobenius group. Then there exists an involution $x \in N_G(U) \setminus U$ that interchanges $\alpha$ and $\beta$.

Now suppose we have such an involution $x \in N_G(U) \setminus U$. Assume $C_U(x) \ne 1$ and set $C := C_G(x)$.

If $|\alpha^C| \le 2$, then $C = \langle x \rangle \times C_U(x)$ and $\langle x \rangle$ is a Sylow $2$-subgroup of $G$.

Why is $\langle x \rangle$ already a Sylow $2$-subgroup of $G$?

Also as $x$ interchanges $\alpha$ and $\beta$ we always have $x \notin C_U(x)$ and $x$ commutes with all elements from $C_U(x)$ by definition, so that $C = \langle x \rangle \times C_U(x)$ is always true, and not because $|\alpha^C| \le 2$. By other facts I see that $(|G_{\alpha}|, |\Omega|) = 1$, but not that $|\Omega|$ must have twice odd order.

Answer 1

We know that $U$ is a subgroup of $G_{\alpha}$ so that $|U|$ is certainly odd. Then $|C_{U}(x)|$ is also odd. Hence $\langle x \rangle$ is a Sylow $2$-subgroup of $C = C_{G}(x)$. Now let $S$ be a Sylow $2$-subgroup of $G$ containing $x$. Let $z$ be an element of order $2$ in $Z(S)$. If $z \neq x$, then $\langle x,z \rangle$ is a Klein $4$-subgroup of $C$, a contradiction. Hence $x = z \in Z(S)$. Thus $S \leq C$, so that $|S|= 2$ and $S = \langle x \rangle$.