If $PSL_2(2^k)$ acts on odd set such that $|\mbox{fix}(g)|\le 3$ for $g\ne 1$, then $G_{\alpha}$ has element of order $\frac{q-1}{\gcd(q-1,3)}$

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Let $G \cong PSL_2(q)$ where $q = 2^k \ge 8$. Suppose $G$ acts as a transitive permutation group on $\Omega$ such that $|\mbox{fix}(g)| \le 3$ for nontrivial $g \in G$. Assume $|\Omega| = |G : G_{\alpha}|$ is odd, then $G_{\alpha}$ contains a Sylow $2$-subgroup $S$ of $G$. We have $|N_G(S) / S| = q - 1 > 3$, and as $N_G(S)$ acts on the fixed points of $S$ we have that $|N_G(S) : N_G(S) \cap G_{\alpha}| \le 3$. As $q-1$ is odd $\gcd(q-1,3) \in \{1,3\}$.

i) If $\gcd(q-1,3) = 1$, then as $$ q-1 = |N_G(S) : S| = |N_G(S) : N_G(S) \cap G_{\alpha}||N_G(S) \cap G_{\alpha} : S| $$ we have $|N_G(S) : N_G(S) \cap G_{\alpha}| = 1$, and as $N_G(S)$ is maximal in $G$ and $G_{\alpha} \ne G$ this implies $G_{\alpha} = N_G(S)$.

ii) If $\gcd(q-1,3) = 3$, then suppose $|N_G(S) : N_G(S) \cap G_{\alpha}| = 3$, otherwise we can go on as in case i). This givens $$ |N_G(S) \cap G_{\alpha} : S| = \frac{q-1}{3} $$

Now all of the above implies that we can find an element $x \in G_{\alpha}$ of order $\frac{q-1}{\gcd(q-1,3)}$. And if $\gcd(q-1,3) \ne 1$ then $N_G(\langle x \rangle)$ is dihedral of order $2(q-1)$.

Again as $N_G(\langle x \rangle)$ acts on the fixed points of $\langle x \rangle$ we have $|N_G(\langle x \rangle) : N_{G_{\alpha}}(\langle x \rangle)| \le 3$. So again with $$ 2(q-1) = |N_G(\langle x \rangle)| = |N_G(\langle x \rangle) : N_{G_{\alpha}}(\langle x \rangle)||N_{G_{\alpha}}(\langle x \rangle)| $$ we either have $|N_G(\langle x \rangle) \cap G_{\alpha}| = q - 1$ or $|N_G(\langle x \rangle) \cap G_{\alpha}| = \frac{2}{3}(q-1)$. In the first case we have $G_{\alpha} = N_G(S)$, in the second case $G_{\alpha}$ contains an involution which does not lie in $S$.

Why is the case $|N_G(\langle x \rangle) : N_{G_{\alpha}}(\langle x \rangle)| = 2$ exluded and just $|N_G(\langle x \rangle) : N_{G_{\alpha}}(\langle x \rangle)| \in \{1,3\}$ handled. And why does the first case implies $G_{\alpha} = N_G(S)$? And in the second case why do we have an involution that does not lie in $S$?

The parts I do not understand I have put in quoted paragraphs; I would be really thankful if anyone could help me with them. In summary:

1) How is the existence of $x$ implied?

2) Why is $N_G(\langle x \rangle)$ dihedral of order $2(q-1)$?

3) Why just $|N_G(\langle x \rangle) : N_{G_{\alpha}}(\langle x \rangle)| \in \{1,3\}$?

4) Why do we have an involution $t \in G_{\alpha} \setminus S$?