Suppose $G$ is a finite, non-regular, faithful, transitive group such that each nontrival element has at most two fixed points. Suppose $S$ is a Sylow-$2$-subgroup of $G$ and $\alpha \in \Omega$. If $|\Omega|$ is even and $G_{\alpha}$ has even order, we want to show that if $S$ is not dihedral or semidihedral, than $G$ has a normal subgroup of index $2$ that is a Frobenious group. By $\mbox{fix}_{\Gamma}(x) = \{ \gamma \in \Gamma : \gamma^x = \gamma \}$ we denote the fixed points of $x$ in $\Gamma \subseteq \Omega$.
Suppose that $|\Omega|$ is even and $G_{\alpha}$ has even order. Then there exists a Sylow $2$-subgroup $P$ of $G_{\alpha}$. Choose some Sylow $2$-subgroup $S$ of $G$ such that $P \le S$, then $P \le S \cap G_{\alpha} = S_{\alpha}$, hence $S_{\alpha} \ne 1$, and $S \nleq G_{\alpha}$ as $|\Omega| = |G : G_{\alpha}|$ is even, and so $S$ could not be a Sylow $2$-subgroup of $G_{\alpha}$.
[ Now the proof shows that $S$ is dihedral or semidihedral or that there exists some $\beta \in \Omega$ such that $\alpha \ne \beta, S_{\alpha} = S_{\beta}$ has index $2$ in $S$ and all elements in $S \setminus S_{\alpha}$ interchanges $\alpha$ and $\beta$. As $S_{\alpha}$ already has two fixed points, this subgroup must have regular orbits on the remaining points of $\Omega$. It follows that $\{\alpha,\beta\}$ is the unique $S$-orbit of length $2$ and all other orbits have length $|S|$. In particular $|S| \ge 4$ and together with the hypothesis about $G$ this implies $|\Omega| \ge 6$. Now the rest of the proof wants to show that $G$ has a normal subgroup of index $2$ that is a Frobenius group, and this is the part that follows. ]
Let $x \in S \setminus S_{\alpha}$. If we view $x$ as an element of the symmetric group on $\Omega$, then there are two possibilities: Either $x$ induces a single cycle on $\Omega \setminus \{ \alpha,\beta \}$ or an even number of cycles of $2$-power length. In the first case $x$ has order $|S|$ and therefore $S$ is cyclic. Then Burnside's $p$-Complement Theorem yields that $G$ has a normal $2$-complement and in particular $G$ has a subgroup $N$ of index $2$. In the second case it follows that, viewed as a permutation on $\Omega \setminus \{\alpha,\beta\}$, the element $x$ is an even permutation. But $x$ interchanges $\alpha$ und $\beta$ and therefore $x$ is an odd permutation in its action on $\Omega$. It follows that $G$ is not contained in the alternating group on $\Omega$, so again $G$ has a normal subgroup $N = A_{\Omega} \cap G$ of index $2$. We note that $G = N \cdot \langle x \rangle$ (in the first case because $N$ contains a complement to $\langle x \rangle$, in the second case by $x \notin N$ and $\langle x \rangle N / N$ must have index one in $G$), whence $G_{\alpha} \le N$. In particular $N$ does not act regularly and it has two orbits of equal size on $\Omega$ that are interchanged by $x$. We denote the $N$-orbit that contains $\alpha$ by $\Gamma$ and the orbits containing $\beta$ by $\Lambda$.
Suppose now that $y \in G_{\alpha}, y \ne 1$ is such that $|\mbox{fix}_{\Gamma}(y)| = 2$. Then as $y$ could have at most two fixed points we have $|\mbox{fix}_{\Lambda}(y)| = 0$. By considering the orbits of $\langle y \rangle$ we can choose $a,b \in \mathbb N_0$ such that $$ |\Gamma| = |\mbox{fix}_{\Lambda}(y)| + a \cdot o(y) = 2 + a\cdot o(y) $$ and $$ |\Gamma| = \Lambda| = b \cdot o(y). $$ Then we deduce that $|\Gamma| \equiv 0 \pmod{o(y)}$ and $|\Gamma| \equiv 2 \pmod{o(y)}$ and hence as $0 \equiv 2 \pmod{o(y)}$ that $o(y) = 2$. But then $\alpha$ and $\beta$ are the unique fixed points of $y$. This is impossible. We conclude that nontrivial elements of $N$ fix at most one point of $\Gamma$ and thus $N$ is a Frobenius group. $\square$
The following points in this proof I do not understand:
1) In the third paragraph: "Either $x$ induces a single cycle on $\Omega \setminus \{ \alpha,\beta \}$ or an even number of cycles of $2$-power length."
As $x$ has to move every element from $\Omega\setminus \{\alpha,\beta\}$ I see that we might get a single cycle, but why in the other case we must have an even number of cycles in a cycle decomposition? I see that they must have $2$-power length as their order must divide some power of $2$, but I do not see why there must be an even number of such cycles?
2) "In particular $N$ does not act regularly and it has two orbits of equal size on $\Omega$ that are interchanged by $x$."
Why we must have two orbits of equal size? I do not see that...
3) "[... ] that $o(y) = 2$. But then $\alpha$ and $\beta$ are the unique fixed points of $y$. This is impossible."
Why does order two implies that we could not have two fixed points?
I would be really glad if someone could answer my questions regarding the above proof!