0
$\begingroup$

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.

$\endgroup$
1
+50
$\begingroup$

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 $S = \langle x \rangle$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ $x$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.