Properties of group acting such that each non-trivial element fixes no point or exactly $p$ points 
Let $p$ be a prime and $G$ a faithful non-regular transitive finite group acting on $\Omega$ with $|\Omega| > p$ such that some element fixes no point, and that each nontrivial element fixing some point fixes exactly $p$ points. Also assume that for $g \notin N_G(G_{\alpha})$ we have 
  $$
 G_{\alpha}^g \cap G_{\alpha} = 1
$$
  and $p$ divides the order of $G_{\alpha}$. Let $\overline \alpha := \mbox{fix}(G_{\alpha})$ be the set of points fixed by $G_{\alpha}$ and assume that a Sylow $p$-subgroup $S$ of $G$ fixes $\overline \alpha$ and acts semi-regulary on $\Omega \setminus \overline \alpha$.
a) Show $N_G(S) \subseteq N_G(G_{\alpha})$ and $N_G(S)' \subseteq G_{\alpha}$,
b) $G$ has a normal subgroup $F$ of index $p$ [Hint: Grün's Theorem],
c) Show that $F$ has $p$ orbits and acts as a Frobenious group on these orbits.

Some facts I know: 
i) $S$ has order at least $p^2$.
As $S$ acts semi-regulary on $\Omega \setminus \overline \alpha$ for each orbit $\Delta \subseteq \Omega \setminus \overline \alpha$ of $S$ we have $|\Delta| = |S : S_{\beta}| = |S|$ with $\beta \in \Delta$. Let $k$ denote the number of these orbits, and as $S$ fixes(1) $\overline \alpha$ 
$$
 |\Omega| = |\overline \alpha| + k\cdot |S| = p + k \cdot |S|
$$
and hence $p$ divides $|\Omega|$. As $G$ acts transitive $|\Omega| = |G : G_{\alpha}|$. So $p$ divides the index of $G_{\alpha}$ in $G$ and the order of $G_{\alpha}$, therefore $S$ has order at least $p^2$ as a Sylow $p$-subgroup.
ii) $G_{\alpha}$ has index $p$ in its normalizer.
Let $k = |N_G(G_{\alpha}) : G_{\alpha}|$. Then if $g \in N_G(G_{\alpha})$ we have $G_{\alpha} = G_{\alpha}^g = G_{\alpha^g}$, so that $G_{\alpha}$ fixes all points $\alpha^g$ with $g \in N_G(G_{\alpha})$, and the orbit of the normalizer has size $|N_G(G_{\alpha}) : G_{\alpha}|$. So at least $k$ points are fixed by $G_{\alpha}$. Further as $\alpha^g = \alpha$ implies $G_{\alpha} = G_{\alpha}^g$ exactly $k$ points are fixed by $G_{\alpha}$.
By $h \in N_G(G_{\alpha}^g) \Leftrightarrow (G_{\alpha}^g)^h = G_{\alpha}^g \Leftrightarrow G_{\alpha}^{ghg^{-1}} = G_{\alpha} \Leftrightarrow ghg^{-1} \in N_G(G_{\alpha}) \Leftrightarrow h \in N_G(G_{\alpha})^g$ the normalizers of conjugates to $G_{\alpha}$ are isomorphic, hence $|N_G(G_{\alpha}^g) : G_{\alpha}^g| = k$ and we see that each conjugate of $G_{\alpha}$ fixes also exactly $k$ points.
Let $g \in G$ and denote by $l$ the number of conjugates of $G_{\alpha}$ which contain $g$. By the above the number of points fixed by $g$ is precisely $kl$.
But as the conjugates intersect trivially, if $g \ne 1$ and $g$ fixes some point then $l = 1$, and so $g$ fixes exactly $k$ points, i.e. the number of points fixed equals the index, and by supposition the number of points equals $p$, hence $k = p$.
So this is all I got, hoping you can help me to solve the points a), b) and c)!

(1) I guess this means $\overline \alpha^S = \overline \alpha$, i.e. $\overline \alpha$ is a union of orbits under $S$, and not necessarily that $S$ fixes each point from $\overline \alpha$. But note that either $S$ fixes each point from $\overline \alpha$, or if one point is not fixed, then $\overline \alpha$ must be an orbit of $S$ as $|\overline \alpha| = p$ and the size of each orbit must divide $|S|$. But maybe the case that $S \le G_{\alpha}$ could be excluded somehow..
 A: I showed how to prove a) in the comments. Here is an outline proof of b) and c).
Grün's Theorem says that $G' \cap S = S_0 := \langle N_G(S)' \cap S, T' \cap S \mid T \in {\rm Syl}_p(G)  \rangle$. We shall show that $S_0 \le G_\alpha$. We proved $N_G(S)' \le G_\alpha$ in a). Since $S$ acts semiregularly on $\Omega \setminus \{ \overline{\alpha} \}$, the elements in $S \setminus S_\alpha$ are fixed-point-free. Since $S' \le G_\alpha$, for any $T \in {\rm Syl}_p(G)$, the elements of  of $T'$ are not fixed-point-free, and hence $S \cap T' \le G_\alpha$. Hence $G' \cap S =  S_0 \le G_\alpha$. Now, since $|S:S_\alpha| = p$ and $G/G'$ is abelian, $G$  must have a normal subgroup $F$ of index $P$ with $F \cap S = S_\alpha$. That proves b).
Now, since $|G_\alpha/F_\alpha|$ is not divisible by $p$, but $|FG_\alpha/F|=1$ or $p$, we must have $FG_\alpha=F$, so $G_\alpha \le F$. For $\alpha \ne \beta \in \overline{\alpha}$, there is an element of $g \in S \setminus S_\alpha$ with $\alpha^g \in \beta$ and, since $G_\alpha \le F$, it follows that no element of $F$ can map $\alpha$ to $\beta$. So the $p$ points in $\overline{\alpha}$ lie in distinct orbits of $F$, and hence $F$ has $p$ orbits, with elements of $F_\alpha$ fixing one point in each orbit. So the action on each orbit is a Frobenius group.
