Let $G$ be a finite, transitive, nonregular permutation group on $\Omega$ such that every element of $g \in G^{\#} := G \setminus \{ 1_G \}$ has at most two fixed points. Suppose further that $|\Omega| \ge 4$. (*)
Lemma: Suppose that $S \in Syl_2(G)$ is such that $S_{\omega}\ne 1$. Then $S$ is dihedral or semidihedral and $|S_{\omega}| = 2$ or $G_{\omega}$ contains a subgroup of index at most $2$ of $S$. In the second case, if $S \nleq G_{\omega}$, then there exists $\delta \in \Omega$ such that $\omega \ne \delta, S_{\omega} = S_{\delta}$ and some element in $S$ interchanges $\omega$ and $\delta$.
Proof: Let $\Delta := \omega^S$ and let $n,m \in \mathbb N_0$ be such that $|S_{\omega}| = 2^n$ and $|S : S_{\omega}| = 2^m$. First suppose that $m \ge 2$. Let $d$ denote the number of fixed points of $S_{\omega}$ on $\Delta$. Note that $d$ is even, but $d\ne 0$ and thus the above hypothesis (*) about $G$ implies that $S_{\omega}$ acts semiregularly on $\Delta \setminus \mbox{fix}_{\Delta}(S_{\omega})$. So now choose $a \in \mathbb N_0$ such that $|\Delta| = d + a\cdot 2^n$. As $n \ge 1$ and $|\Delta| = 2^m \ge 4$, we see that $d = 2$ and hence $2^m = 2\cdot (1 + a\cdot 2^{n-1})$. This implies that $a\cdot 2^{n-1}$ is odd, in particular $n = 1$. [...]
This is the beginning of the proof. Why does (*) implies that $S_{\omega}$ acts semiregular on $\Delta \setminus \mbox{fix}_{\Delta}(S_{\omega})$, and where is this facts used in the further argumentation? Also choosing $a$ assumes that $m > n$, or not? So why that, and also I do not see why $2^m = d + a\cdot 2^n$ with $n \ge 1, m \ge 2$ implies $d = 2$, this equation has many other solutions for $a$ and $d$, for example $m = 4, n = 1, a = 1$, then $d = 12$.
