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.


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 $S = \langle x \rangle$.

  • $\begingroup$ Thanks, that makes it clearer! But may I ask you, do we always have $C = \langle x \rangle \times C_U(x)$, or just because $|\alpha^C| \le 2$? What I wrote beneath gives that we always have $\langle x \rangle \times C_U(x) < C$, but not that it must be equal? $\endgroup$ – StefanH Dec 7 '15 at 17:59
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    $\begingroup$ $x$ permutes the fixed points of $C_{U}(x)$, incuding $\alpha$ and $\beta$ since $U = G_{\alpha} \cap G_{\beta}$. $|\alpha^{C}| = 2$ implies that the $C$-orbit of $\alpha$ has size $2$, so that $\alpha^{C} = \{\alpha,\beta \}$. Choose $y \in C$. Then $\alpha^{y} = \alpha$ or $\alpha^{y} = \beta$. If $ \alpha^{y} = \alpha$ then $\beta^{y} = \beta$ and $y \in C_{U}(x)$. If $\alpha^{y} = \beta,$ then $\alpha^{y} = \alpha^{x}$ and $\beta^{y} = \alpha = \beta^{x}$. so that $yx \in C_{U}(x)$. In all cases, $y \in \langle x \rangle C_{U}(x)$. Hence $C = \langle x \rangle \times C_{U}(x)$. $\endgroup$ – Geoff Robinson Dec 7 '15 at 18:12
  • $\begingroup$ Thank you very much. Now everything is very clear, I will award you the bounty as soon as I can! If you have the time I would be glad if you take a look at my other questions, like this one: math.stackexchange.com/questions/1564210/… $\endgroup$ – StefanH Dec 7 '15 at 18:19

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