# If centralizer of involution has twice odd order, then group has twice odd order

Let $G$ be a nonregular, transitive permutation group such that every nontrivial element fixes at most two points.

Now suppose $U := G_{\alpha}\cap G_{\beta} \ne 1$ and suppose that $G_{\alpha}$ has odd order, and $|\Omega|$ is even. And that $G$ is not a Frobenius group. Then as $N_G(U)$ acts on the fixed points of $U$ we have $|N_G(U) : U| \le 2$ and as $G$ is not a Frobenius group we must have $|N_G(U) : U| = 2$. Hence there exists an involution $x \in N_G(U) \setminus U$ interchanging $\alpha$ and $\beta$.

Now set $C := C_G(x)$. Suppose $C \ne 1$ and $|\alpha^C| = 2$. Then $C = \langle x \rangle \times C_U(x)$, hence it has twice odd order.

Why does this implies that the Sylow $2$-subgroups of $G$ have order $2$?

This is equivalent with the statement that $|G : C| = |x^G|$ has odd order, but I do not see how to establish it?

• The information about $G$ being a permutation group is irrelevant. If the centralizer of an involution $x$ in any finite group $G$ has twice odd order, then $G$ has twice odd order. That's an easy exercise. – Derek Holt Feb 24 '16 at 15:21
• Okay, thank you. I was able to solve it. – StefanH Feb 24 '16 at 15:59

Let $G$ be a finite group, and $x \in G$ an involution such that $|C_G(x)| = 2m$ with $m$ odd. Then $\langle x \rangle$ is a Sylow $2$-subgroup of $C := C_G(x)$. So choose $P \in \mbox{Syl}_2(G)$ with $\langle x \rangle \le P$. Assume $\langle x \rangle < P$, i.e. $|P| \ge 4$. Then $\langle x \rangle < N_P(\langle x \rangle) \le P$. But as $|\langle x \rangle| = 2$ this implies $N_P(\langle x \rangle) \le C_G(x)$. So we must have $|N_P(\langle x \rangle)| = 2$ as this is the hightest $2$-power dividing $|C|$, but this gives $N_P(\langle x \rangle) = \langle x \rangle$, i.e. contradicting the normalizer grow condition and hence $P = \langle x \rangle$ and $2$ is also the highest $2$-power dividing $|G|$.