\section{Introduction}

This paper continues the analysis of permutation groups where every non-trivial element has a low number of fixed points (see \cite{MW}).
We call the maximum number of fixed points of non-trivial group elements that occur in a faithful and transitive group action its \textbf{fixity}.
Our motivation to study permutation groups that act with low fixity comes from questions about Riemann surfaces:
If $X$ is a compact Riemann surface of genus at least $2$, then its automorphism group is finite. Proving this uses the existence of a finite, non-empty set of analytically distinguished points of $X$, the so-called Weierstrass points.
The link between fixed points of non-trivial automorphisms of $X$ and Weierstrass points is the motivation for our work. Schoeneberg has proven (see for example page 264 in \cite{FK}) that if a non-trivial automorphism of $X$ has at least five fixed points on $X$, then all these fixed points are Weierstrass points.
Studying groups of automorphisms of Riemann surfaces that act with fixity at most $4$ is therefore a way to study situations where Schoeneberg's result does not apply.
We refer the reader to \cite{MW} and \cite{FK} for more background on the theory of Riemann surfaces and our intended applications.

Groups of fixity $1$ are Frobenius groups, where the Frobenius kernel consists of the fixed point free elements and the Frobenius complements are the point stabilizers.
Groups of fixity $2$ and $3$ have been discussed in \cite{MW} and \cite{MW3}.
So, it remains to
analyze those of fixity $4$, and the present article contributes to this analysis.

Let $G$ be a group that acts faithfully, transitively and with fixity $4$ on a finite set $\Omega$. It has become apparent that strongly $p$-embedded subgroups
play a major role for understanding the structure of $G$ (see Lemma~4 and Theorem~13  of \cite{BMW}).
For example, if $H$ is a non-trivial four point stabilizer in $G$ and the order of $H$ is coprime to $6$, then it turns out that $N_G(H)$ is strongly $p$-embedded in $G$ for some prime $p \geq 5$.

Let $T$ be a Sylow $2$-subgroup of $G$.
In \cite[Theorem~4.7]{BMW} we prove that if $|G_\alpha|$ is even,
then either the set-wise stabilizer of the union of short $T$-orbits is strongly embedded in $G$  or  $T$ has a quite restricted structure.  In this paper we focus on the case that $|G_\alpha|$
is odd and is divisible by $3$. Let $P$ be a Sylow $3$-subgroup of $G$. Then we prove that  either the set-wise stabilizer $K$ of the union of short $P$-orbits is strongly $3$-embedded in $G$
or the structure of $P$ is very restricted. More precisely our main result about the $3$-structure is as follows:

\begin{thm}\label{3total}
Suppose that $G$ is a finite group acting transitively, faithfully and with fixity $4$ on a set $\Omega$.
Let $P$ be a Sylow $3$-subgroup of $G$, let $\Delta$ be the union of $P$-orbits of length at most $3$ on $\Omega$ and let $K$ denote the stabilizer of the set $\Delta$ in $G$.
Moreover let $\alpha \in \Omega$. Then one of the following holds:
\begin{enumerate}
\item[(a)]  $G_\alpha$ has even order;
\item[(b)]    There is a normal subgroup $N$ of $G$ of index $1$ or $3$ such that   one of the following holds:
\begin{enumerate}
\item[(i)]  $(|N_\alpha|, 6) = 1$ or
\item[(ii)]  $K \cap N$ is strongly $3$-embedded in $N$ and $1 \leq |\Delta| \leq 4$;
\end{enumerate}
\item[(c)]   There are four exceptional cases:\\
$|P| = 3$ and $|N_G(P)|_2 \le 4$ or\\
$P$ is  elementary abelian of order $9$   or\\
$P$ is extra-special of exponent $3$ and order $27$, or\\
$P \cong C_3 \wr C_3$.
\end{enumerate}
\end{thm}

There are some more details that we discuss in Section 3, where we also prove this theorem.
In the special case of finite simple groups, we combine this result with earlier work of Ronse's (\cite{Ronse}). This gives a stronger result and a
blueprint for the classification of all finite simple groups that act with fixity 4, along with information about the point stabilizers:


\begin{thm}\label{3and2simple}
Suppose that $G$ is a finite simple non-abelian group that acts transitively, faithfully and with fixity $4$ on a set $\Omega$.
Let $P \in \textrm{Syl}_3(G)$, $S \in \textrm{Syl}_2(G)$ and let $f$ denote the maximum number of fixed points of involutions in $G$.



Then one of the following holds:
\begin{enumerate}


\item[(1)] $f \ge 1$ and $G$ has a strongly embedded subgroup. 

\item[(2)] $1 \le f \le 3$ and $S$ is dihedral or semi-dihedral. 


\item[(3)] $f=4$ and $G$ has sectional $2$-rank at most $4$. 


\item[(4)] The order of the point stabilizers is odd and divisible by $3$. One of the following occurs:

\begin{enumerate}
\item[(a)] $G$ has a strongly $3$-embedded subgroup. 
\item[(b)]  $|P| = 3$ and $|N_G(P)|_2 \le 4$.
\item[(c)] $P$ is elementary abelian of order $9$.
\item[(d)] $P$ is extra-special of exponent $3$ and order $27$. 
\item[(e)] $P \cong C_3 \wr C_3$.
\end{enumerate}

\item[(5)]  The point stabilizers have order coprime to $6$.

\end{enumerate}
\end{thm}



After a detailed analysis of all these cases, we obtain  our main classification result:


\begin{thm}\label{main4fpsimple}
Suppose that $G$ is a finite simple non-abelian group that acts transitively, faithfully and with fixity $4$ on a set $\Omega$.
Let $\alpha \in \Omega$.

Then $G$ is isomorphic to one of the groups in the following table, with point stabilizer $G_\alpha$ as described: 
\end{thm}







\begin{tabular}{l | p{8cm}| p{4cm}}

\mbox{}
	Group \(G\)                           & Point stabilizer structure  & Comments \\ \hline
	\(\Alt_6 \cong \textrm{PSL}_2(9)\)            & \(C_2\), \(\Sym_3\), \(C_3 \times C_3\), \(D_{10}\), \((C_3 \times C_3) : C_2\) & \\
	\(\Alt_7\)                            & \(C_5\), \(\Alt_6\)  &\\


	\(\textrm{PSL}_2(7) \cong \textrm{PSL}_3(2)\)         & \(C_2\), \(\Sym_3\) &\\                                                         
	\(\textrm{PSL}_2(8)\)                         & \(C_2\), \(\Sym_3\), 
	\(D_{14}\), \(D_{18}\)& \\

	\(\textrm{PSL}_2(11)\)                        & \(C_3\), \(\Alt_4\)  &     \\
	\(\textrm{PSL}_2(13)\)                        & \(C_3\), \(\Alt_4\), \(C_{13} : C_3\)& \\
		\(\textrm{PSL}_2(q)\)                        & $G_\alpha$ cyclic of order $\frac{q-1}{4}$ & $q \ge 17$, $q \equiv 1$ mod 4\\
		& $G_\alpha$ has index 2 in the normalizer of a Sylow sgp in def. char. & $q \ge 17$, $q \equiv 1$ mod 4\\
				                        & $G_\alpha$ cyclic of order $\frac{q+1}{4}$& $q \ge 17$, $q \equiv -1$ mod 4\\
	
	\(\textrm{PSU}_3(3)\)                         & \(((C_3 \times C_3) :C_3) :C_8\) &            \\
	
	\(\textrm{PSU}_4(2)\cong \textrm{PSp}_4(3)\)          & \(C_5\)       &                    \\
		\(\textrm{PSU}_4(3)\)          & \(C_5\)                   &        \\
			\(\textrm{PSp}_4(q)\)          & $G_\alpha$ cyclic of order $\frac{q^2+1}{(2,q+1)}$            &               \\
		

		\(\textrm{Sz}(q)\)                            & $G_\alpha$ cyclic of order $q+\sqrt{2q}+1$ or
		$q-\sqrt{2q}+1$        & $q \ge 8$ a power of 2                 \\
			\(P\Omega^-_8(q)\)          &   $G_\alpha$ cyclic of order $\frac{q^4+1}{(2,q+1)}$   &                \\
			\(^3D_4(q)\)          &  $G_\alpha$ cyclic of order $q^4-q^2+1$        &       \\
			\(^2G_2(q)\)          & $G_\alpha$ Frobenius group of order $q^3\cdot \frac{q-1}{2}$ &\\
			& or cyclic of order $\frac{q-1}{2}$          &             \\
	
	\(M_{11}\)                            & \(C_5\), \(C_{11} :C_5\), \(\textrm{PSL}_2(11)\) &                                                                                 \\
	\(M_{12}\)                            & \(M_{11}\)           &     \\
	\(M_{22}\)                            & \(C_5\), \(C_{11}:C_5\)       &    \\
                                                                               
	\(J_1\)                               & \(C_{15}\)         &    \\

\end{tabular}	\label{MainTable}

\vspace{2cm}


This article is organised as follows:

Section 2 is short, with some notation, a technical lemma and an initial result about the 3-structure of the groups that we consider.
Then we analyze the 3-structure more deeply in Section 3, which leads to a natural case distinction towards the classification of finite simple groups that act with fixity 4. 
For very small groups, and also for some individual arguments here and there, we use
\texttt{GAP} (\cite{GAP}). The code is presented and explained in an Appendix at the very end, and the results are captured in Table \ref{SmallGroupsTable} in Section 4, along with some more technical results for preparation.
Some groups, in particular Lie type groups of small Lie rank, occur several times in Theorem \ref{3and2simple}, which is why we begin our analysis with those, in Section 5.
After that we look at the cases (3), (4) and (5) of Theorem \ref{3and2simple}, in this order.
Section~6 looks at the case where some involution fixes four points, Section 7 considers point stabilizers of odd order, but order divisible by 3, and then the bulk of the work is done in Section 8 in the case of point stabilizers of order coprime to 6.
This is where we rely heavily, and very visibly, on the Classification of Finite Simple Groups (CFSG) as presented in the GLS series. 

Section 9 collects all the intermediate results and explains how they contribute to the proof of Theorem 
\ref{main4fpsimple}. The appendix explains how we use \texttt{GAP} and gives the fixed point profiles of the groups in Table \ref{SmallGroupsTable} for future reference.


\smallskip
\textbf{Acknowledgments.}
The project on permutation groups with low fixity started in 2012, initiated by Kay Magaard. It was supported by the DFG for several years and also partly
by the National Science Foundation under Grant No. DMS-1440140 while
he was in residence at the Mathematical Sciences Research Institute in Berkeley, California, during
the Spring semester 2018.
We continue our work on this project after Kay's unexpected passing in 2018. 






\section{Preliminaries}


In this paper, by ``group'' we always mean a finite
group, and by ``permutation group'' we always mean a group that acts
faithfully on a set.
Throughout $\Omega$ denotes a finite set and $G$ denotes a
permutation group on $\Omega$.\\




Let $\omega \in \Omega$ and $g \in G$, and moreover let $\Lambda
\subseteq \Omega$ and $H \le G$.
We use standard notation for orbits and point stabilizers, and moreover we write
$\textrm{fix}_{\Lambda}(H):=\{\omega \in \Lambda \mid \omega^h=\omega$ for all $h \in
H\}$ for the \textbf{fixed point set of $H$ in $\Lambda$}. 
When we say that $H \le G$ \textbf{is a non-trivial four point stabilizer}, then we mean that there is a set $\Delta \subseteq \Omega$ such that $|\Delta|=4$ and such that $1 \neq H$ is exactly the point-wise stabilizer of $\Delta$ in $G$.

If $g \in G$, then we write
$\textrm{fix}_{\Lambda}(g)$ instead of $\textrm{fix}_{\Lambda}(\<g\>)$. By
$\pi(G)$ we denote the set of prime divisors of $|G|$.
Whenever $n,m$ are natural numbers and $p$ is a prime number, then $(n,m)$ denotes the largest natural
common divisor of $n$ and $m$ and $n_p$ is the largest power of $p$ dividing $n$.
We use the notation $C_n$ for cyclic groups of order $n$.




\begin{definition}
Let $k$ be a non-negative integer and suppose that the group $G$ acts on the finite set $\Omega$. We say that $G$ has \textbf{fixity $k$ on $\Omega$} if and only
if there is some element of $G^\#$ that fixes exactly $k$ distinct points on $\Omega$ and if no element of $G^\#$ fixes more than $k$ distinct points.
\end{definition}

\begin{hyp}\label{4fix}
Suppose that $G$ is a finite group that acts transitively, faithfully and with fixity $4$ on a set $\Omega$.
\end{hyp} 






\begin{lemma}\label{Inv-lem:TB}
Suppose that Hypothesis \ref{4fix} holds and let $\alpha \in \Omega$.

If $1 \neq Y \le G_\alpha$, then $|N_G(Y):N_{G_\alpha}(Y)| \le 4$.

\smallskip
Moreover let $H \leq G_\alpha$ be a non-trivial four point stabilizer in $G$ and let $\Delta:=\textrm{fix}_{\Omega}(H)$. 

Then
the following hold:

(a) If $p \in \pi(G_\alpha)$ and $p\ge 5$, then $G_\alpha$ contains a Sylow $p$-subgroup of $G$. 
If $p \in \pi(H)$ and $p\ge 5$, then $H$ contains a Sylow $p$-subgroup of $G$. 
In particular, if the point stabilizers (or non-trivial four point stabilizers) have order coprime to 6, 
then they are Hall subgroups of $G$.


(b) Suppose that $1 \neq X \le H$. Then $N_G(X)$ stabilizes $\textrm{fix}_{\Omega}(X)$ and is (therefore) contained in $N_G(H)$.
Moreover $\abs{N_G(H):N_{G_\alpha}(H)}\in\{2,4\}$. If $3 \in \pi(N_G(X))$, then $3 \in \pi(G_\alpha)$. 

(c) If $X \le G_\alpha$ fixes exactly three points, then $\abs{N_G(X):N_{G_\alpha}(X)}\leq 3$.

(d)
If $3 \in \pi(H)$, then $H$ contains a Sylow $3$-subgroup of $G$ or $N_G(H)$ has a subgroup that induces $\A_4$ on $\textrm{fix}_{\Omega}(H)$.

(e) If $H$ contains a Sylow subgroup of $G_\alpha$, then $N_G(H)$ acts transitively on $\Delta$.

(f)
$|N_G(H):N_{G_\alpha}(H)| = 4$ or
$H$ is a $2$-group, but not a Sylow $2$-subgroup of $G$.
In particular $|G|$ is divisible by $4$.

(g) $H$ is a TI-group.

(h) If $|H|$ is coprime to $6$ and $p\in \pi(H)$, then $N_G(H)$ is strongly
$p$-embedded in $G$.

\end{lemma}





\begin{proof}
For the first assertion we note that $N_G(Y)$ stabilizes $\textrm{fix}_{\Omega}(Y)$, a set of size at most 4 that contains $\alpha$. Now $4 \ge |\textrm{fix}_{\Omega}(Y)^{N_G(Y)}|=|N_G(Y):N_G(Y) \cap G_\alpha|$.

The statement in (a) is proven in Lemma 4 of \cite{BMW}.
Lemma 6 of the same article covers the statements in (b),  (d) and (g) as written.

For (c) we let  $X \le G_\alpha$ be such that $X$ fixes exactly three points $\alpha,\beta,\gamma \in \Omega$.
Then $N_G(X)/X$ induces a subgroup of $\Sym_3$ on $\textrm{fix}_{\Omega}(X)$.
If this factor group is isomorphic to \(\Sym_3\), then there exists an element in \(N_{G_\alpha}(X)\) that interchanges $\beta$ and $\gamma$ and therefore $\abs{N_G(X)/ N_{G_\alpha}(X)}=3$. Otherwise $N_G(X)/X$ is isomorphic to a proper subgroup of \(\Sym_3\). In both cases $\abs{N_G(X):N_{G_\alpha}(X)}\leq 3$, as stated.

We turn to (e). 
Let $r \in \pi(G_\alpha)$ be such that $H$ contains a Sylow $r$-subgroup $R$ of $G_\alpha$. Then
let $\beta \in \Delta$ be such that $\alpha \neq \beta$ and let $g \in G$ be such that $\alpha^g = \beta$ (by transitivity of $G$).
Now $R^g \leq G_\beta$, but also $R \le H \le G_\beta$ and therefore we find $h \in G_\beta$ such that $R^{gh}=R$.
Then $1 \neq R \le H$ and (a) yields that $gh \in N_G(H)$, so we proved that $N_G(H)$ acts transitively on $\Delta$.

For (f) we first suppose that
$H$ is a $\{2,3\}$-group.
If $H$ is a $2$-group, then either (e) is applicable or the other statement holds automatically.
So we suppose that $3 \in \pi(H)$. If $H$ contains a Sylow $3$-subgroup of $G$, then (e) yields the result.
Otherwise we use (d) and then $N_G(H)$ has a subgroup that induces $\Alt(4)$ on $\Delta$. In particular
$N_G(H)$ acts transitively on $\Delta$ and therefore
$|N_G(H):N_{G_\alpha}(H)|=4$.
Finally, if  $p \in \pi(H)$ is such that $p \ge 5$, then 
by (a) it contains a Sylow $p$-subgroup of $G$, and (e) applies again. 

Now we look at (f):
If $H$ has odd order, then $|G|$ is divisible by $4$ by (d) or (e).  
Suppose that $|H|$ is even.
If $G$ has Sylow $2$-subgroup of order
$2$, then $|N_G(H):N_{G_\alpha}(H)| \neq 4$, so by the first statement we know that
 $H$ is a $2$-group, but not a Sylow $2$-subgroup of $G$, which is not
possible because $H \neq 1$.
Therefore $|G|$ is divisible by $4$.

Finally, we prove (h). Of course $N_G(H)$ has order divisible by $p$.
Let $g \in G \setminus N_G(H)$ and suppose that $x\in N_G(H) \cap N_G(H)^g$ is a $p$-element.
By Hypothesis $p \ge 5$, so (f) yields that $x\in H \cap H^g$.
Then $x$ fixes all elements in $\Delta \cup \Delta^g$, so $x=1$ or 
$\Delta=\Delta^g$. But if $g$ stabilizes $\Delta$, then it normalizes $H$.
Consequently $N_G(H) \cap N_G(H)^g$ is indeed a $p'$-group (see also
\cite[Proposition]{GLS2}). 
\end{proof}







The following lemma will be used frequently for our strategy to connect the 3-structure and the 2-structure of the groups that we analyze. 


\begin{lemma}\label{syl3}
Suppose that Hypothesis \ref{4fix} holds and let $P \in \textrm{Syl}_3(G)$.
Let $\Delta$ be the union of all $P$-orbits of $\Omega$ of size at most $3$.
Then one of the following holds:

\begin{enumerate}
\item[(a)] All $P$-orbits are regular and the point stabilizers in $G$ are $3'$-groups.

\item[(b)] $|\Delta| > 4$ and $|P| \le 9$.

\item[(c)] $|\Delta| \leq 4$ and $P$ is of maximal class. There exists some non-regular $P$-orbit on $\Omega \setminus \Delta$ and for every such orbit $\Lambda$ and all  $\lambda \in \Lambda$ it is true that
$|P_\lambda| = 3$ and that $P_\lambda$ fixes exactly three points on
$\Lambda$.


\item[(d)] $\Delta$ is the unique $P$-orbit of length 3 and all orbits of $P$ on
$\Omega \setminus \Delta$ are regular.



\item[(e)] $1 \le |\Delta| \le 4$, there is some $\delta \in \Delta$ such that $P \le G_\delta$,
 and all $P$-orbits on $\Omega \setminus \Delta$ are regular.

\end{enumerate}
In (c), (d) and (e) we see that $P$ possesses an orbit of length at least $9$ and therefore
$|P| \ge 9$.
\end{lemma}

\begin{proof}
 Lemma 10 in \cite{BMW}.
\end{proof} 











\section{From the 3-structure to the 2-structure}


The purpose of this section is to analyze how, given Hypothesis \ref{4fix}, the point stabilizer structure and the 3-structure of $G$
influence each other. It will turn out that the 2-structure of $G$ comes into play here as well, so that at the end of the section we will be able to prove a theorem that organises the analysis in the following sections. In fact, the natural case distinction that occurs determines the structure of the remainder of the paper and the strategy for the proof of the main theorem.


First we remark that Lemma \ref{syl3} gives information about the possible orbit sizes for a Sylow $3$-subgroup $P$ of $G$ on $\Omega$.
In addition to orbits of size $|P|$ and $1$, we see in Part (d) of the lemma that orbit size $3$ is possible, and the only case where another orbit size occurs is Part (c), with orbits of length $|P|/3$.

Most of this section is devoted to a closer look at Cases (c)-(e) of the lemma.

\subsection{Cases (d) and (e) of Lemma \ref{syl3}}

\begin{hyp}\label{hyp3}
In addition to Hypothesis \ref{4fix}, let $P \in \textrm{Syl}_3(G)$, let $\Delta$ denote the union of the $P$-orbits of length at most $3$ and suppose that $P$ satisfies Lemma \ref{syl3}~(d) or (e).
We define $D$ to be the element-wise stabilizer of $\Delta$ in $G$, $K$ to be the set-wise
stabilizer of $\Delta$ in $G$, and we set $Q:=P \cap D$.
\end{hyp}




\begin{lemma} \label{littlehelp}
Suppose that Hypothesis \ref{hyp3} holds.

(a) If $g \in G \setminus K$, then $D \cap D^g$ is a $3'$-group.

(b) If $|\Delta|=3$, then all $3$-elements of $K \setminus D$ act fixed-point-freely on $\Omega$, whereas the $3$-elements
of $D$ fix exactly three points on $\Omega$.

(c) If $|\Delta|=3$ and  Case (e) of Lemma \ref{syl3} holds, then $P \leq D$.

(d) If $|\Delta| \neq 3$, then Lemma \ref{syl3}~(e) is true. Moreover, if $3$ divides $|K/D|$, then $|\Delta| = 4$ and every $3$-element of $K \setminus D$ has a unique fixed point on $\Omega$.

(e) The $3$-elements of $D$ and of $K \setminus D$ can be distinguished by the number of fixed points on $\Omega$.
In particular $x^G \cap D= \varnothing$ for all $3$-elements $x \in K \setminus D$.
\end{lemma}





\begin{proof}
Throughout, we recall that $P$ acts semi-regularly on $\Omega \setminus \Delta$ and that \(P\leq K\).

(a) Let $g \in G$ and let $y$ be a non-trivial $3$-element in $D \cap D^g$. Without loss $y \in P$, which means that $y$ cannot fix any points outside of $\Delta$, and hence $\textrm{fix}_{\Omega}(y) \subseteq \Delta$.
But $y$ fixes $\Delta \cup \Delta^g$ point-wise, so $\Delta^g \subseteq \Delta$ and hence $g \in K$.

(b)
Suppose that $y \in K \setminus D$ is a $3$-element, and again suppose that $y \in P$. Then $y$ does not fix any element outside $\Delta$.
If it fixes any point in $\Delta$, then it fixes all of them (because $|\Delta|=3$), contrary to the fact that $y \notin D$. Therefore $y$ does not fix any point of $\Omega$.

Next suppose that $x \in D$ is a 3-element. Without loss $x \in P$, and then $x$ fixes the three points of $\Delta$ and has no fixed points outside $\Delta$.

(c) We recall that $|\Omega| \equiv 3$ modulo $|P|$. In Case (e) of Lemma \ref{syl3} we have that $P$ fixes a point and stabilizes $|\Delta|$, so it fixes every element of $\Delta$.

(d) The additional hypothesis contradicts Case (d) of Lemma \ref{syl3}, and hence Case~(e) must hold.
Next suppose that $3$ divides $|K/D|$. Then there exists a $3$-element that stabilizes $\Delta$, but does not fix every element of it, and this is only possible if $|\Delta| \ge 3$. Hence $|\Delta| = 4$ as stated.

(e) follows from all this.
\end{proof}




\begin{lemma} \label{durch}
Suppose that Hypothesis \ref{hyp3} holds and that $g \in G \setminus K$. Then
$|K \cap K^g|_3 \le 3$.
\end{lemma}

\begin{proof}
Let  $T \in \textrm{Syl}_3(K \cap K^g)$.
If $T=1$, then there is nothing left to prove.
So we suppose that $T \neq 1 $, and  let $T_0 = D \cap T$. Then
$T_0^{g^{-1}} \leq K $.
Now Lemma~\ref{littlehelp}(e) implies that
$T_0^{g^{-1}} \leq D$.
Thus
$T_0 \leq D \cap D^g$
and Lemma~\ref{littlehelp}(a) yields that
$T_0 = 1$,
which is our assertion.
\end{proof}

\begin{cor} \label{untergruppe}
Suppose that Hypothesis \ref{hyp3} holds and that $R \le P$ is a subgroup of order at least $9$. Then $N_G(R) \le K$.
\end{cor}

\begin{proof}
We recall that $R \le P \le K$ and
let $g \in N_G(R)$. Now $R=R^g \le K \cap K^g$ and $|R| \ge 9$, so Lemma \ref{durch} forces $g \in K$.
\end{proof}

A quick reminder of what it means to be strongly $p$-embedded:
If $p$ is a prime, then a subgroup $H$ of $G$ is said to be  \textbf{strongly $p$-embedded in $G$} if and only if $H$ contains a Sylow $p$-subgroup and $H \cap H^g$ is a $p'$-group for all $g \in G\setminus H$.

\begin{rem}\label{remstrong}
Suppose that Hypothesis \ref{hyp3} holds. Then $P \le K$ and therefore, if
Case (e) of Lemma \ref{syl3} holds and if there are no
$3$-elements in $K \setminus D$, then all $3$-elements of $K$ lie in $D$ and hence
$K \cap K^g$ is a $3'$-group. This actually means that $K$ is strongly $3$-embedded in $G$ in this case.
\end{rem}

For all prime numbers $p$ and all $p$-subgroups $Y$ of $G$,
we denote by $r(Y)$ the \textbf{rank of $Y$}.
Moreover, we need notation for the fusion arguments that will come up:



\begin{rem}\label{fusionnota}
Whenever we use Alperin-Goldschmidt's Fusion theorem, then we use the version from \cite[Theorem 16.1]{GLS2}. We re-state it here for convenience, so that we can refer to its notation later.
Given a finite group $G$ and $P \in \textrm{Syl}_3(G)$, we set
$\mathcal{F}$ to be the set of all non-trivial subgroups $F \le P$ such that the following conditions are satisfied:

\begin{itemize}
\item[(a)] $N_P(F) \in \textrm{Syl}_3(N_G(F))$,

\item[(b)] $C_G(F) \leq O_{3',3}(N_G(F))$, and

\item[(c)] $O_{3',3}(N_G(F))=O_{3'}(N_G(F)) \times F$.
\end{itemize}

Then fusion of elements or subgroups of $P$ can be described via $\mathcal{F}$ as follows:
If $A,B \subseteq P$ and $g \in G$ are such that $A^g=B$, then 
we find $n \in \mathbb{N}$, $A=A_1,A_2,...,A_n=B \subseteq P$, $c \in C_G(A)$,
$F_1,...,F_{n-1} \in \mathcal{F}$ and $g_1\in N_G(F_1),...,g_{n-1} \in N_G(F_{n-1})$ such that

(i) $g=cg_1 \cdots g_{n-1}$,

(ii) $\langle A_i,A_{i+1}\rangle \le F_i$ for all $i \in \{1,...,n-1\}$ and

(iii) $A_i^{g_i}=A_{i+1}$ for all $i \in \{1,..,n-1\}$.


We notice that (b) and (c) imply that $Z(P) \leq F$ for all $F \in \mathcal{F}$.
\end{rem}


\begin{lemma} \label{fusion}
Suppose that Hypothesis \ref{hyp3} holds. Then
$K$ controls fusion in $P$ with respect to $G$.
\end{lemma}

\begin{proof}
We use the notation from Remark~\ref{fusionnota} and we let $F \in \mathcal{F}$.
Then we prove that $N_G(F) \le K$.

Notice that $|P| \geq 9$ and $|\Delta| \leq 4$, so it follows that $Q=P \cap D \neq 1$. As $D \unlhd K$ and $P \le K$, we have that
$Q \unlhd P$ and hence $Z(P) \cap D \neq 1$.
Let $F \in \mathcal{F}$.  Then Remark~\ref{fusionnota} yields that  $Z(P) \cap D  \leq  F \cap D$.
If $F \not\leq D$, then $|F| \geq 9$ and $N_G(F) \leq K$ by Corollary~\ref{untergruppe}.
If $F \leq D$, then $N_G(F) \leq K$ by Lemma~\ref{littlehelp}\,(a).

Now we recall that control of fusion means that, whenever $a,b \in P$ and $g \in G$ are such that $a^g=b$, then there exists some $k \in K$ such that $a^k=b$.
We use the notation in Remark \ref{fusionnota} with $a$ in place of $A$ and $b$ in place of $B$, so that (i) means that
$g=cg_1 \cdots g_{n-1}$. Set $k:=g_1 \cdots g_{n-1}$.
Now $a^c=a$ and therefore $a^g=a^{ck}=a^k$.
As $g_1 \in N_G(F_1) \le K$,..., $g_{n-1} \in N_G(F_{n-1}) \le K$ by  the previous paragraph, we deduce that 
$a^k=b$ and $k \in K$.
\end{proof}


In the proof of the next result we refer to \cite{GLS2} and to the definition of extremal $p$-elements. Therefore we recall this notion briefly, with $G$ and $P$ as before: An element $y \in P$ is said  to be {\bf extremal} in $G$ if and only if $C_P(y) \in \textrm{Syl}_3(C_G(y))$.


\begin{prop}\label{3semireg}
Suppose that Hypothesis \ref{hyp3} holds.
Then one of the
following is true:
\begin{enumerate}
\item[(a)] $K$ is strongly $3$-embedded in $G$.
\item[(b)] $G$ contains a normal subgroup $N$ of index $3$ such that
$N \cap K$ is strongly $3$-embedded in $N$.
\item[(c)] The point stabilizers in $G$ have even order.
\end{enumerate}
\end{prop}

 \begin{proof}
We suppose that point stabilizers in $G$ have odd order.
If, for all $g \in G\setminus K$, it is true that $K \cap K^g$ is a $3'$-group, then $K$ is strongly $3$-embedded in $G$, and (a) holds.
So we suppose that (a) does not hold, and our objective is to prove (b).

First we prove:

($\ast$) ~~If $g \in G\setminus K$ is such that $|K \cap K^g|_3 > 1$, and if $x \in K \cap K^g$ has order 3, then 
$x^G \cap Q=\varnothing$ and $x^G \cap Qx^2 = \varnothing$.


We let $g \in G\setminus K$ be such that $|K \cap K^g|_3 > 1$. Then $|K \cap K^g|_3 = 3$ by
Lemma \ref{durch}, so we let 
$X \le K \cap K^g$ be a subgroup of order $3$ and $x \in X^\#$. As $P \in \textrm{Syl}_p(G)$ and $P \le K$, we may suppose that $X \le P$.

Assume that $x \in Q$. Then $x \in D$, and we recall that $x \in K^g$.
 Lemma~\ref{littlehelp}\,(a) implies that $x \notin D^g$, so
 $x^{g^{-1}} \in K\setminus{D}$. Then 
Lemma~\ref{littlehelp}\,(e) gives that $x^G \cap D= \varnothing$, contrary to our assumption.
Now $x \notin Q$ and $x^G \cap D=\varnothing$ as stated.

Moreover $x$ induces a $3$-cycle on $\Delta$, so in particular $|\Delta | \geq 3$, and
$P = Q \rtimes X$.

Next we assume that $x^G \cap Qx^2 \neq \varnothing$, which means that there are $h \in G$ and $a \in Q$ such that $x^h = a \cdot x^2$. By Lemma~\ref{fusion} we may suppose that
$h \in K$. 
We recall that $a$ fixes $\Delta$ point-wise, so $x$ and $a \cdot x^2$ induce 3-cycles on $\Delta$ that are inverse to each other.
Also, the element $h \in K$ stabilizes $\Delta$ and we see that
the image of $\langle x, h\rangle$ in $K/D$ is isomorphic to a subgroup of $\Sym_3$ or of $\Alt_4$ on $\Delta$.
The fact that $x^h$ acts like $x^2$ on $\Delta$ forces the image of $h$ to be a transposition.
It follows that some power $h_0$ of $h$ inverts $x$, that $h_0$ has even order and that it fixes one or two points on $\Delta$ depending on whether $|\Delta| = 3$ or $4$.
However, point stabilizers have odd order, and this is a contradiction.

We conclude that $x^G \cap Qx^2 = \varnothing=x^G \cap Q$, and consequently $x^G \cap P \subseteq Qx$.
This shows that \cite[Proposition~15.15]{GLS2} is applicable with $x$ in place of $u$ and that, in its statement, property (i) cannot be satisfied. Hence there is a normal subgroup $N$ of $G$ of index $3$
such that $G = NX$.

Furthermore the proof of \cite[Proposition~15.15]{GLS2} shows that \(Q\leq N\) and therefore \(Q=N\cap P\).

Assume that there exist \(h\in N\setminus K\) and some non-trivial \(3\)-element \(y\in N\cap K\cap K^h\leq K\cap K^h\).
If \(y\in D\), then Lemma~\ref{littlehelp}\,(e) yields that also \(y^{h^{-1}}\in D\). But this contradicts Lemma~\ref{littlehelp}\,(a) because \(h\in G\setminus K\).
Therefore \(y\in K\setminus D\).
On the other hand \(y^G\cap P\subseteq N^G\cap P=N\cap P=Q\). As \(P\in \textrm{Syl}_3(K)\), there exists \(g \in K\) such that \(y^g\in P\).
So \(y^g\in Q=D\cap P\), but \(y^g\) has the same number of fixed points as \(y\in K\setminus D\), and this contradicts Lemma~\ref{littlehelp}\,(e).
This final contradiction proves the lemma.
\end{proof}



\subsection{Case (c) of Lemma~\ref{syl3}}

Now we study Case (c) of Lemma~\ref{syl3} more closely.
Given that $3$-groups of maximal class play a role here, we need the following technical lemma.
Recall that a $3$-group of order $3^n$ is of maximal class if it is of class $n-1$.


\begin{lemma}\label{maxclass3}
Let $P$ be a $3$-group of maximal class and let $n \ge 4$ be such that $|P|=3^n$.
Let
 $1 = Z_0 \leq Z_1 = Z(P) \leq \cdots \leq Z_{n-1}= P$ be the upper central series of $P$, that is
 for all $i \in \{1,...,n-2\}$,
 $Z_i/Z_{i-1}= Z(P/Z_{i-1})$.
 Moreover let $U \le P$ be a subgroup of index $3$, and  set $P_1 := C_P(Z_{n-2}/Z_{n-4})$.
 Then the following hold:
 \begin{itemize}
 \item[(a)] If $i \in \{1,...,n-2\}$, then $|Z_i/Z_{i-1}| = 3$, and $Z_{n-2} = P^\prime$ has index $9$ in $P$.
 \item[(b)] $P_1 = C_P(Z_2)$  and $|P:P_1| = 3$. Furthermore, if $i \in \{1,...,n-2\}$, then $Z_i \le U$.
 \item[(c)]  Let $y\in P\setminus{U}$ be such that $|C_P(y)|=9$. Then $y^U = P^\prime y$.
\item[(d)]   If $U$ is abelian, then $U = P_1$.
\item[(e)]  Suppose that $U$ is elementary abelian. Then $n = 4$.
\item[(f)] Suppose that $h \in \textrm{Aut}(U)$ has prime order $r \geq 5$.  Then $U = P_1$, $n =4$, $U$ is elementary abelian and $r= 13$.
\item[(g)] $\textrm{Aut}(P)$ is a $\{2,3\}$-group.
 \end{itemize}
 \end{lemma}

 \begin{proof}  Statement (a) follows from  \cite[III Hilfssatz 14.2]{Huppert-I}, while the first part of (b) follows from   \cite[III Hauptsatz 14.7 and Satz 14.17]{Huppert-I}.

  If $U = P_1$, then  $Z_{n-2} \leq U$, and if
  $U \neq P_1$, then $P_1 \cap U$ is a normal subgroup of $P$ of index $9$,
  and therefore $P_1 \cap U = Z_{n-2}$ by  \cite[III Hilfssatz 14.2]{Huppert-I}, which gives  the second part of (b).

  In (c) we first observe that $P=\langle y \rangle U$ and therefore $y^P = y^U$.
   The length of this conjugacy class is $|P:C_P(y)| =|P|/9 = |P^\prime|$, so $|[U,y]|=|y^U|=|P^\prime|$. Moreover $[U, y] \le P^\prime$, so equality holds.
   Now $P'=[y,U]$ and therefore $P'y=yP'=y^U$. This is (c).

The definition of $P_1$ and (b) yield (d).

Now we turn to (e).  As $U$ is elementary abelian, we have that $U = P_1$ by (d). Let $y \in P\setminus{U}$ and $y_1 \in U\setminus{Z_{n-2}}$, and for all $i \in \{2,..,n\}$ set
$y_i := [y,y_{i-1}]$. Then  $P = U \langle y \rangle$, and
for all \(i\in \{1,\ldots, n-1\}\),
$Z_{n-i} = \langle y_i \rangle Z_{n-i-1}$ by \cite[Lemmas~ 3.2.4, 3.2.7]{LGM} and \cite[III Satz 14.17]{Huppert-I}.
Thus we may suppose that $\{y_1, \ldots , y_{n-1}\}$ is a generating set for $U$.
 Therefore, $y$ induces an automorphism $\tilde{y}$ on $U$ of order  $o(\tilde{y}) = 3^j \geq n-1 > 3^{j-1}$, where $j \in \{1,...,n\}$.
 As $|P:U| = 3$, the latter automorphism is of order $3$ and $n = 4$, which is (e).

Before we turn to (f), we notice that (g) follows from coprime action and the fact that every automorphism of $P$ stabilizes
its upper central series, with factors of order 3 or 9. Here we do not even need the hypothesis that $n \ge 4$.

For the proof of (f), we
assume first that $U \neq P_1$. Then by \cite[Corollary 3.3.6]{LGM} $U$ is of maximal class, so as we have just seen
$\textrm{Aut}(U)$ is a $\{2,3\}$-group, contrary to our choice of $h$. Hence $U=P_1$.
Now $\textrm{Aut}(U)$ stabilizes the characteristic subgroup $\Omega_1(U)$.
If $n \geq 5$, then $\Omega_1(U)$ is elementary abelian of order $9$ by \cite[III Satz 14.16]{Huppert-I}, which
yields (f) in this case.
If $n \le 4$, then we must have that $n=4$ and $|U| = 3^3$. As $\textrm{GL}_2(3)$ is a $\{2,3\}$-group, it follows that $U$ is elementary abelian and $r = 13$.
\end{proof}



For the remainder of this subsection we work with the following hypothesis.

\begin{hyp}\label{hyp4}
	In addition to Hypothesis \ref{4fix}, let $P \in \textrm{Syl}_3(G)$, let $\Delta$ denote the union of the $P$-orbits of length at most $3$ and suppose that $P$ satisfies Lemma \ref{syl3}~(c), with all the notation given there.
	We define $D$ to be the element-wise stabilizer of $\Delta$ in $G$, $K$ to be the set-wise
	stabilizer of $\Delta$ in $G$, and we set $Q:=P\cap D$.
	Furthermore, we denote by $x$ an element that generates $P_\lambda$, and we keep the notation from Lemma \ref{maxclass3} for the upper central series of $P$, including $P_1:=C_P(Z_{n-2}/Z_{n-4})$.
\end{hyp}



\begin{lemma}\label{tiny}
Suppose that Hypothesis \ref{hyp4} holds. Then
$P$ fixes at most one point in $\Delta$ and
$|\Delta| \in \{0, 1,3,4\}$. Moreover $|P| \geq 27$, $|Z(P)| = 3$, $C_P(P_\lambda) = Z(P) \times P_\lambda$ and
$Z(P) \le Q$. Finally, $Z(P)$ has no fixed points outside $\Delta$ and, if $|\Delta| \in \{3,4\}$, then $|P:Q|=3$.
\end{lemma}

\begin{proof}
We know that $|\Delta| \leq 4$ by hypothesis.
If $P$ fixes two or more points on $\Delta$, then $P_\lambda$ fixes five or more elements in $\Omega$, which contradicts Hypothesis~\ref{4fix}.
In particular this implies that $|\Delta| \neq 2$, which
proves the first statement. If $|\Delta| \in \{3,4\}$, then $P \neq Q$ and hence $|P:Q|=3$, which is the last statement.
As $|\Lambda| = |P|/3 >4$, it also follows that $|P| \geq 27$.
Now we recall that $P$ is of maximal class, so $|Z(P)| = 3$.
Moreover $Q \unlhd P$ and therefore $Z(P) \cap Q \neq 1$, which together with the fact that $|Z(P) | = 3$ forces $Z(P) \leq Q$.

Assume that $Z(P)$ fixes a point outside $\Delta$. Then 
it fixes three points, and $P$ stabilizes the set of these fixed points,
which means that they give a $P$-orbit of size $3$ outside of $\Delta$. This is impossible.

Next we note that \(C_P(P_\lambda)\) stabilizes \(\textrm{fix}_{\Omega}(P_\lambda)\) (of size 3), so it induces a group of order at most \(3\) on \(\textrm{fix}_{\Omega}(P_\lambda)\). This implies that \(|C_P(P_\lambda)|\leq 9\), hence $|C_P(P_\lambda)|=9$ and
$C_P(P_\lambda) = Z(P) \times P_\lambda$.
\end{proof}



\begin{lemma} \label{case(c34)}
Suppose that
Hypothesis~\ref{hyp4} holds and that
$|\Delta| \in\{3,4\}$.
Then one of the following is true:
\begin{itemize}
	\item[(a)] The point stabilizers in $G$ have even order.
	\item[(b)] $|\Delta| = 3$ and
	
     
       $G_\alpha \leq K$ for every element $\alpha \in \Delta$.
	

	\item[(c)]  $|\Delta| = 4$ and $G_\delta \leq K$ for the unique element  $\delta \in \Delta$ that is fixed by $P$.
\end{itemize}
\end{lemma}



\begin{proof}
Suppose that $\alpha \in \Omega$ and that $G_\alpha$ has odd order. 
We note that $|P:Q|=3$ by hypothesis and by Lemma \ref{tiny}.



\textbf{Case 1:} $|\Delta| = 4$.

We recall that
$Z(P) \le D$ by Lemma~\ref{tiny}, so every element of $Z(P)$ fixes exactly four points on $\Omega$.
Moreover $P_\lambda$ has an orbit of length $3$ on $\Delta$ and fixes a unique element, say  $\delta$, in  $\Delta$ because of Hypothesis~\ref{4fix}.
We deduce that $P = Q\cdot P_\lambda$, and all elements in $P_\lambda$ have four fixed points.

We show that every non-trivial element in $X:= Z(P) \times P_\lambda$ fixes precisely four elements in $\Omega$. First we recall that $x \in P$ is such that $\langle x \rangle=P_\lambda$.
If $z \in Z(P)$ is such that $\langle z \rangle = Z(P)$ , then $|\textrm{fix}_{\Omega}(z)| = |\textrm{fix}_{\Omega}(x)| = 4$.
Let $y \in Z_2 \setminus{Z(P)}$. Then $[x,y]  \in Z(P)$ and $[x,y] \neq 1$ because $P_\lambda \nleq Z_2$.
Therefore we may suppose that $[x,y]  = z$, which yields that $x^y = xz$ and $x^{y^2} = xz^2$.
As $x$ fixes four points on $\Omega$, all its conjugates do too, so we
deduce that each of the four subgroups of $X$ of order $3$ fixes precisely four elements in $\Omega$.

As $G_\delta$ has odd order, it is soluble and we find some odd prime $r$ such that $R:= O_r(G_\delta) \neq 1$.

First suppose that $r > 3$. Then the action of $X \le G_\delta$ on $R$ is coprime, which implies that there is $y \in X^\#$ such that $C_R(y) \neq 1$.
Now $C:=C_R(y)$ stabilizes the set $\textrm{fix}_{\Omega}(y)$, which has size $4$ by our claim above.
Then the fact that $r \geq 5$  yields that $C$ fixes $\textrm{fix}_{\Omega}(y)$ point-wise. Now
Hypothesis~\ref{4fix} forces $\textrm{fix}_{\Omega}(C) = \textrm{fix}_{\Omega}(y)$.\\
Furthermore, we notice that $Z(R)$ stabilizes $\textrm{fix}_{\Omega}(C)$ because $[Z(R), C] = 1$.  As
$r \ge 5$, we see that $Z(R)$ fixes $\textrm{fix}_{\Omega}(C)$ point-wise and that $\textrm{fix}_{\Omega}(Z(R)) = \textrm{fix}_{\Omega}(C)$.

Next we use that $Z(R)$ is normal in $G_\delta$, so it follows that $G_\delta$ stabilizes the set $\textrm{fix}_{\Omega}(Z(R))$.  Since $P \leq  G_\delta$, we obtain that $\textrm{fix}_{\Omega}(Z(R)) = \Delta$ and that $G_\delta$ stabilizes $\Delta$. This is (c).

Now suppose that $O_r(G_\delta) = 1$ for all prime numbers $r  \geq 5$,  and set $R:= O_3(G_\delta)$. So in particular $R \le P$. 
If $R=P$, then $G_\delta \in \mathbb{N}_G(P) \le K$, which is (c).
So we suppose that $R \lneq P$ and we
set $V:= \Omega_1(Z(R))$. If $V \leq Q$, then $\textrm{fix}_{\Omega}(V) = \Delta$ and, since $V$ is normal in $G_\delta$,
it follows that  $G_\delta$ stabilizes $\Delta$. Then $G_\delta \leq K$, which is (c) again.
Therefore, we suppose that $V \not\leq Q$.
If $|P:V| \geq 3^2$, then $V \leq Q$ by \cite[III Hilfssatz 14.2]{Huppert-I}
contradicting the last assumption. 
Thus by our hypothesis it follows that  $|P:V| = 3$, and then $V = R$ because $R \lneq P$.

If $G_\delta$ is a $3$-group,
then again $G_\delta=P \le K$. Thus we also suppose that $G_\delta$ is not a $3$-group.
Let $t \ge 5$ be a prime number such that
$N:= O_{3,t}(G_\delta) > R$ (which means that $1 \neq N/R=O_t(G_\delta)$)
and let $T \in \textrm{Syl}_t(N)$.  Then $C_T(V) = O_t(N)$, which  is
a characteristic subgroup of $N$ and therefore normal in $G_\delta$.
As $O_t(G_\delta) = 1$,  we see that $C_T(V) = 1$ and that $T$ acts
faithfully on $V$.
Now $|T|= 13$
and  $|Q \cap V| \geq 3^2$. Therefore there are two distinct subgroups $V_1 $ and $V_2$ of order $3$ in $V$ that fix $\Delta$ element-wise.  As $T$ is transitive
on the set of subgroups of order $3$ of $V$, there is a non-trivial $t \in T$  that maps $V_1$ onto $V_2$. We also know that $\Delta^t = \textrm{fix}_{\Omega}(V_1)^t = \textrm{fix}_{\Omega}(V_1^t) = \textrm{fix}_{\Omega}(V_2) = \Delta$, so
it follows that $t$, and therefore also $T$, stabilize $\Delta$. Thus $T \leq K$, which forces $T \le D$ and therefore $T \le N_G(Q)$.
Let $v \in V^\#$. Then $\langle v\rangle$ is $T$-conjugate to $V_1 \le Q$, so the fact that $T \le N_G(Q)$ forces 
$v \in Q$.
Consequently $V  \le Q$ and $|P:V|=3=|P:Q|$, which means that $V=Q$, and this contradicts our hypothesis in the present case.\\


\textbf{Case 2:} $|\Delta| = 3$.

Hypotheses \ref{4fix} and \ref{hyp4} yield that $P_\lambda$ acts non-trivially on $\Delta$, so $P=Q \rtimes P_\lambda$. We also note that, in this situation, all non-trivial $3$-elements of $G$ have either zero or three fixed points on $\Omega$. In particular all four point stabilizers are $3'$-groups. In the following let $\delta \in \Delta$.

First we assume that $Q$ is cyclic. We know that $P_\lambda$ normalizes $Q$ and that $|C_P(P_\lambda)|=9$, so $|C_Q(P_\lambda)|=3$ and at the same time $|C_Q(P_\lambda)|=|Q|/3$.
Hence $|Q| = 9$. Let $H \leq G_\delta$ be the element-wise stabilizer of a subset of $\Omega$ of size $4$ that contains $\delta$.  Recall that $2,3 \not\in \pi(H)$, which means that
$H$ is a $\pi(H)$-Hall subgroup of $G$ by Lemma 4 in \cite{BMW}.


 Therefore, if $Y \le Q$ and some non-trivial element of $H$ normalizes $Y$, then it centralizes $Y$.
 As $G_\delta$ is soluble, this implies that
there is some $g \in G_\delta$ such that   $Q$ normalizes the conjugate $H^g$  of $H$ in $G_\delta$.
Therefore $Q$ stabilizes the set $\textrm{fix}_{\Omega}(H^g)$, which is not possible because $|\Delta| = 3$ in the present case.


We suppose from now on that $Q$ is not cyclic, and we let $Y \le Q$ denote an elementary abelian subgroup of order $9$. Then $Y$ fixes every point of $\Delta$ and therefore $\textrm{fix}_{\Omega}(Y) = \Delta$. In fact, if $y \in Y^\#$, then $\textrm{fix}_{\Omega}(y) = \Delta$ because all $3$-elements fix exactly zero or three points.
 As before let $r \in \pi(G_\delta)$ be odd and such that $R:= O_r(G_\delta) \neq 1$.
If $r \ge 5$, then the coprime action of $Y$ on $R$ gives some $y \in Y^\#$ such that $C:=C_R(y) \neq 1$, and then
$C$ fixes $\textrm{fix}_{\Omega}(y) = \Delta$ element-wise. Since $Z(R)$ centralizes $C_R(y)$, it stabilizes $\Delta$, hence fixes it point-wise and then
$\textrm{fix}_{\Omega}(Z(R)) = \Delta$.
It follows that  $G_\delta \leq K$ because $Z(R) \unlhd G_\delta$, which is (b). If $r = 3$, then we recall that $Q \in \textrm{Syl}_3(G_\delta)$ and hence $R \le Q$. Again set $V:= \Omega_1(Z(R))$. Then $V \le Q$ fixes $\Delta$ element-wise, so $\textrm{fix}_{\Omega}(V) = \Delta$ and therefore $G_\delta \leq K$,
which is again (b).
\end{proof}




\begin{lemma}\label{QConjugation}
Suppose that Hypothesis~\ref{hyp4} holds, that $|\Delta| = 4$ and that the point stabilizers have odd order.
Then  $K$ controls fusion in $P$ with respect to $G$ and in particular $x^G \cap Q=\varnothing$.
\end{lemma}

\begin{proof} 
We let $a,b \in P$ and $g \in G$ be such that $a^g=b$, and then we need to prove that there exists some $k \in K$ such that $a^k=b$.
For this we use Remark \ref{fusionnota} and all the notation there. 


Let $F \in {\mathcal F}$. Then we begin by proving that $N_G(F) \le K$.

First suppose that $|F| \ge 9$. Then $F \cap Q \neq 1$ and therefore
$\textrm{fix}_{\Omega}(F) = \{\delta\}$ or $\textrm{fix}_{\Omega}(F) = \Delta$ where $ \{\delta\} = \textrm{fix}_{\Omega}(P)$.
In the first case
$N_G(F) \leq G_\delta \leq K$ by Lemma~\ref{case(c34)}\,(c), and in the second case
 $N_G(F) \leq K$ by definition of $K$.

Next suppose that $|F| = 3$. Then $F = Z(P)$ and $\textrm{fix}_{\Omega}(F) = \Delta$ (recall that \(Z(P)\leq D\) by Lemma \ref{tiny}),
which yields that $N_G(F) \subseteq K$.
As in Remark \ref{fusionnota} we let $n \in \mathbb{N}$, $c \in C_G(a)$,
$F_1,...,F_{n-1} \in \mathcal{F}$ and $g_1 \in N_G(F_1),...,g_{n-1} \in N_G(F_{n-1})$ be such that $g=ag_1 \cdots g_{n-1}$.
Then $k :=g_1 \cdots g_{n-1} \in K$ and $a^k=b$, so $K$ controls fusion in $P$ as stated.

The remaining statement follows with 
Lemma \ref{littlehelp}(e).
\end{proof}


\begin{rem}
In the case where $|\Delta| = 4$ and $G_\delta$ has odd order,
Lemmas~\ref{case(c34)} and \ref{QConjugation} imply that
the element-wise stabilizers of subsets of $\Omega$ of size $4$
are not all conjugate in $G$.  Because, if they were, then $\langle x \rangle$ (fixing one point in $\Delta$ and three points outside of it) would be conjugate to a $3$-subgroup of $D$ and hence of $Q$, contrary to the fact that $x^G \cap Q=\varnothing$.
\end{rem}

Repeatedly we are using the following facts.

\begin{lemma}\label{Observation}
Suppose that Hypothesis~\ref{hyp4} holds.
\begin{itemize}
\item[(a)]  If $x^G \cap P^\prime x^2 \neq \varnothing$, then there is some $g \in G$ that inverts $x$.
\item[(b)] Suppose that $|\textrm{fix}_{\Omega}(x)| = 4$ and that point stabilizers in $G$ have odd order.  Then $N_G(P_\lambda) = C_G(P_\lambda)$. In this case in particular,  $x^G \cap P^\prime x^2 = \varnothing$.
\end{itemize}
\end{lemma}
\begin{proof}
By Lemma~\ref{tiny}, \(|P|\geq 27\).
Also, there exists a subgroup of index $3$ of $P$ that does not contain $x$. Hence,
if $|P| \ge 81$, then assertion (a) follows directly from Lemma~\ref{maxclass3}~(c).
If $|P|=27$, then a calculation in \texttt{GAP}
(see Appendix)
shows that Lemma~\ref{maxclass3}~(c) is still true, so again (a) follows.

For (b) assume that $N_G(P_\lambda) \neq C_G(P_\lambda)$ and let $g \in G$ be such that it normalizes $P_\lambda$, but does not centralize it. 
Then $g$ inverts $x$, in particular
$g$ has even order.
Since point stabilizers in $G$ have odd order, this means that $g$ does not fix any points on $\textrm{fix}_{\Omega}(x)$.
Hence it acts as a double transposition or a 4-cycle.
Also $Z(P)$ acts non-trivially on $\textrm{fix}_{\Omega}(x)$, in fact it induces a subgroup of order $3$.
Therefore $\langle Z(P),P_\lambda, g\rangle$ induces
$\Sym_4$ or 
$\Alt_4$ on 
$\textrm{fix}_{\Omega}(x)$.
If $\langle Z(P),P_\lambda, g\rangle$ induces $\Sym_4$ on $\textrm{fix}_{\Omega}(x)$, then some element acts as a transposition, which means that it has even order and fixes a point in $\textrm{fix}_{\Omega}(x)$.
But this contradicts the fact that point stabilizers have odd order. 
Consequently $\langle Z(P),P_\lambda, g\rangle$ induces $\Alt_4$ on $\textrm{fix}_{\Omega}(x)$.

Let \(z\in Z(P)\) be non-trivial. Then \(\langle g,g^z \rangle\) acts on \(\textrm{fix}_{\Omega}(x)\) as an elementary abelian subgroup of order \(4\) of \(\Alt_4\).
Therefore the product \(h:=g g^z g^{z^2}\) fixes every point in \(\textrm{fix}_{\Omega}(x)\). Since \(x^{g g^z g^{z^2}}=x^2\), the element $h$ has even order, so it cannot fix any point.
This contradiction finishes the proof.
\end{proof}




\begin{lemma}\label{Case(c)not3}
Suppose that Hypothesis~\ref{hyp4} holds and that $|\Delta| \in \{0,1,4\}$.


Then one of the following is true:
\begin{itemize}
\item[(a)] The point stabilizers have even order.
\item[(b)] $G$ has a normal subgroup $N$ of index $3$ such that $G = N \cdot \langle x \rangle$.
\item[(c)]  $|\Delta| = 0$, $P$
is extra-special of exponent $3$ and $P_\lambda\in \textrm{Syl}_3(G_\lambda)$.
\end{itemize}
\end{lemma}


\begin{proof}
Suppose that the point stabilizers have odd order.
First we distinguish the cases for $|\Delta|$ and prove statements about $x^G$. We recall that $\langle x\rangle =P_\lambda$, which means that $x$ fixes three points outside $\Delta$ and hence at most one point in $\Delta$. 

\textbf{Case 1:} $|\Delta|=4$.

Then $x$ fixes a unique point $\delta$ in $\Delta$, and in particular $x \notin D$.
By Lemma
\ref{QConjugation} we know that $K$ controls fusion in $P$ and
that $x^G \cap Q = \varnothing$. 

Next assume that there are $g \in G$ and $u \in Q$ such that $x^g = ux^2$.
As $K$ controls fusion in $P$, we may suppose that $g \in K$. Let $\overline{K}:= K/D$. Then $\overline{g}$ inverts $\overline{x}$ and therefore $\overline{g}$ as well as $g$ have even order. Now $\overline{g}$ normalizes $\langle \overline{x }\rangle$ and stabilizes $\Delta$, so it stabilizes $\textrm{fix}_{\Delta}(\overline{x}) = \textrm{fix}_{\Delta}(x)=\{\delta\}$.
Thus $g \in G_\delta$, contrary to our assumption that point stabilizers have odd order. We conclude that $x^G \cap Qx^2 = \varnothing$.

Hence if $|\Delta|=4$, then $x^G \cap Q = \varnothing= x^G \cap Qx^2$.\\


\textbf{Case 2:} $|\Delta|  \in \{0,1\}$ and $|P| = 27$.

\textbf{2(a):} $P$ is extra-special of exponent $9$.

We choose $y \in P$ to be of order $9$, and set $Y:= \langle y \rangle$.
As $|P/Y|=3$, we see that $Z(P)=P'\le Y$.
Then we notice that $|\textrm{fix}_{\Omega}(Y)| \in \{0,1\}$, depending on whether $\Delta$ has size $0$ or $1$. This is because $Z(P)$ does not fix any points outside $\Delta$, by Lemma \ref{tiny}. For the same reason all $Y$-orbits outside $\Delta$ are regular, which means that 
$Y \cap x^G = \varnothing$.

Next we prove that $Yx^2 \cap x^G =\varnothing$ as well.
Assume that there are $g \in G$ and $u \in Y$ such that $x^g = ux^2$.
In particular $o(ux^2)=o(x^g)=3$. As $o(yx^2) = 9$, it follows that $u \in \langle y^3 \rangle=Z(P)=P'$. By Lemma~\ref{Observation}\,(a)
we may suppose
that $g$ inverts $x$ and hence that $g$ has even order.
Moreover $g$ normalizes $\langle x \rangle$ and hence stabilizes the set $\textrm{fix}_{\Omega}(x)$.
If $|\Delta| =0$, then this set has size $3$ and hence $g$ fixes a point in it, contrary to the fact that point stabilizers have odd order.
Thus $|\Delta|  =  1$, $|\textrm{fix}_{\Omega}(x)| = 4$ and Lemma~\ref{Observation}~(b) gives a contradiction.\\

\textbf{2(b):} $P$ is extra-special of exponent $3$.

If $|\Delta| = 0$, then \(|\textrm{fix}_{\Omega}(x)|=3\). 
Together with the orbit structure of $P$ on $\Omega$ we see that the point stabilizers have Sylow $3$-subgroups of order 3, which is (c).

Suppose that $|\Delta| = 1$.  
As $N_G(P)$ stabilizes $\Delta$ and point stabilizers have odd order by assumption, it follows that $N_G(P)$ has odd order. We also know that $Out(P) \cong \textrm{SL}_2(3)$, see Theorem~1 in~\cite{Winter}, and therefore
$N_G(P) = P \times C_G(P)$. 
As $x \notin Z(P)=\Phi(P)$,  we see that
$x^{N_G(P)}=x^P=Z(P)x$.

Let $u \in P$ be such that $P = \langle u , x\rangle$, i.e. $u \notin Z(P)\langle x \rangle$, and 
set $U:=\langle u \rangle Z(P)$.

We prove that $x^G \cap P \subseteq Ux$, and we begin by proving that $u \notin x^G$.

Assume for a contradiction that
that $x^g = u$ for some $g \in G$. Then by considering the conjugation family from Remark~\ref{fusionnota} we may suppose that
$F \geq \langle u,x \rangle = P$ and $g \in N_G(F)$. Thus it follows that $F = P$ and $g \in N_G(F) = N_G(P)$. But then $u=x^g \in x^{N_G(P)}=Z(P)x$, contrary to our choice of $u$. 

Next suppose that $x^g \in Z(P)x^2$ for some $g \in G$. 
Then there is some element $h \in G$ that inverts $x$ by Lemma~\ref{Observation}\,(a). In particular $h$ has even order, so 
we may suppose that $h$ is a $2$-element.
Then, as $h$ and $Z(P)$ act non-trivially on 
the set $\textrm{fix}_{\Omega}(x)$ of size $4$, the group $\langle h, Z(P)\rangle$ induces a
subgroup 
isomorphic to $\Sym_4$ or $\Alt_4$ on $\textrm{fix}_{\Omega}(x)$. 
The first case is not possible because point stabilizers have odd order, and the second case is not possible
because $h$ inverts $x$, while $Z(P)$ acts trivially on $P_\lambda$.
It follows that $x^G \cap P \subseteq Z(P)x \subseteq Ux$ as stated.

Therefore we conclude: If (c) does not hold, then 
Proposition 15.15 in \cite{GLS2} yields that 
$G$ has a normal subgroup $N$ such that $G/N$ is a cyclic $3$-group and $x \notin N$, which gives (b).

 







\textbf{Case 3: }  $|\Delta|  \in \{0,1\}$ and $|P| > 27$.

We aim to show (b).
First we prove that $\textrm{fix}_{\Omega}(y) = \Delta$ for every $y \in P_1^\#$.
Assume that $y \in P_1^\#$ is such that $|\textrm{fix}_{\Omega}(y)| > |\Delta|$. 
Recall that $P_1 = C_P(Z_2)$ by Lemma \ref{maxclass3}~(b).
Therefore, $Z_2$ stabilizes $\Pi:= \textrm{fix}_{\Omega}(y)\setminus{\Delta}$.
If $\delta \in \Pi$, then $\delta$ lies in a non-regular $P$-orbit outside of $\Delta$, so Hypothesis \ref{hyp4} (see Lemma~\ref{syl3}(c)) yields that 
$|P_\delta|=3$ and hence $o(y)=3$, $P_\delta=\langle y \rangle$. 
Then $y$ fixes exactly three points outside $\Delta$, i.e. 
$|\Pi| = 3$. This yields that
$Z_2$ stabilizes $\Pi$ and acts as a group of order $3$ on it.
This situation can occur only once because otherwise some element in $Z_2\langle y\rangle$ has too many fixed points.
In particular $P_1$ stabilizes $\Pi$ as well. 

As $|P:P_1|=3$ by Lemma \ref{maxclass3} (b), it follows that $|P| \le 27$, contrary to our hypothesis in the present case that
$|P| > 27$. Therefore $\textrm{fix}_{\Omega}(y) = \Delta$ for all $y \in P_1^\#$, and in particular all non-trivial elements in $P_1$ have zero or one fixed points.
For all $g \in G$, the element $x^g$ fixes $\Delta^g$ and three additional points in $\Omega\setminus\Delta$, by Hypothesis \ref{hyp4}, hence three or four points. In particular $x^G \cap P_1= \varnothing$.




If $x^G \cap P_1 x^2 = \varnothing$, then we apply \cite[Proposition~15.15]{GLS2} with $G$ in the role of $X$ and $x$ in the role of $u$. It follows that, again, we find a normal subgroup $N$ of $G$ of index $3$ such that $G = N \langle x \rangle$; so  (b) holds.

Now suppose that $x^G \cap P_1 x^2 \neq \varnothing$. Then the strategy is, if necessary, to replace $x$ by a conjugate $y$ and to replace $P_1$ by one of the other
two subgroups of $P$ of index $3$ different from $P^\prime \langle y \rangle$.
Let  $g \in G$ and $u \in P_1$ be such that $x^g = ux^2$.
 Then we use the notation that we set up in Remark~\ref{fusionnota}, so we let
$c\in C_G(x)$, $n \in \mathbb{N}$, $F_1,...,F_{n-1} \in \mathcal{F}$ and $g_1 \in N_G(F_1),...,g_{n-1} \in N_G(F_{n-1})$ be such that $g=cg_1 \cdots g_{n-1}$.
Set $y_i: = x^{g_1 \cdots g_i}$ for $1 \leq i \leq n-1$.
As $x^G \cap P_1 = \varnothing$, we know that $y_i$ is contained in $x^G \cap P_1 x$  or in $x^G \cap P_1 x^2$. Let $i$ be minimal such that $y_{i+1} \in x^G \cap P_1 x^2$, and let $h := g_1 \cdots g_{i-1}$.  
 Then $y:= x^h = y_i = vx$ for some $v \in P_1$,
and we may suppose that $i+1 = n$. 
Notice the following: 
$$y^{g_{n-1}} = x^g = ux^2 = w y^2,$$
where $w = u(vxvx^{-1})^{-1} \in P_1$, and $g_{n-1} \in N_G(F_{n-1})$.

Moreover, $y = x^h$ is an element in $P$ that generates some subgroup $P_\lambda$ as considered in Hypothesis~\ref{hyp4}, because $|\textrm{fix}_{\Omega}(x^h)| = |\Delta| +3$. This shows that
we may suppose that $n = 1$ and $g \in N_G(F_1)$.

If $u \in P^\prime$, then we apply  Lemma~\ref{maxclass3}(c) to $x^2$ and we obtain an element $w \in P_1$ such that $(ux^2)^{gw} = x^2$. Now we set $h := gw$ and we conclude that 
$x^{h} = x^2$ and hence $h$ 
inverts $x$. 
The group $ N_G(P_{\lambda})$ stabilizes the set $\textrm{fix}_{\Omega}(x)$. Since $G_\lambda$ has odd order by assumption and since $Z(P)$ induces a group of order $3$ on $\textrm{fix}_{\Omega}(x)$, it follows that $|\textrm{fix}_{\Omega}(x)| = 4$ and that $ N_G(P_{\lambda})$
induces a subgroup of $\Sym_4$ on $\textrm{fix}_{\Omega}(x)$. 
As point stabilizers have odd order, it has to be a subgroup of $\Alt_4$ of order divisible by $3$, and this contradicts the existence of $h$. 

Thus $u \in P_1\setminus{P^\prime}$.
In this case $u \notin P^\prime\langle x \rangle$,
because otherwise $u \in P_1 \cap P^\prime\langle x \rangle = P^\prime $.
By Lemma~\ref{maxclass3}(a) we have that $|P/P^\prime|=9$ and that
$P^\prime = Z_{n-2}$ is a subgroup of index $3$ in $ P_1$.
As $x$ is an element of order $3$ in $P\setminus{P_1}$, it follows that 
$\Phi(P) = P^\prime = Z_{n-2}$, and then the main property of the Frattini group gives that
 $$P = P^\prime\langle x, ux^2 \rangle=
\Phi(P)\langle u, x \rangle = \langle u,x\rangle \leq F:= F_1.$$

We conclude that $g \in N_G(F) = N_G(P)$. By Lemma~\ref{maxclass3}(g) we 
may suppose that $g$ is a $2$-element. As $N_G(P)$ stabilizes the set $\Delta$, we derive
that $\Delta= \varnothing$ and $|\textrm{fix}_{\Omega}(x)| = 3$.
Furthermore, we observe that $N_G(P)/P^\prime$ induces a subgroup of $\textrm{GL}_2(3)$ on $P/P^\prime$.
As $|\textrm{fix}_{\Omega}(x)| = 3$, we conclude that $N_G(P_\lambda)$ has odd order and therefore
$|N_G(P)|_2 = 2$.  Thus $g$ has order $2$ and centralizes $u$.
Set $\tilde{P}_1:= P^\prime \langle ux\rangle$. Then $\tilde{P}_1x^2 \cap x^g = \varnothing$, and  using the same arguments as above we also obtain that
$\tilde{P}_1 \cap x^g = \varnothing$. Therefore we can apply \cite[Proposition~15.15]{GLS2}, which gives (b) also in this case.


\end{proof}



We collect our results:


\begin{prop} \label{maxKlasse}
Suppose that Hypothesis~\ref{hyp4} holds.
Then one of the following is true:

\begin{enumerate}

\item[(a)] The point stabilizers have even order

\item[(b)]  $|\Delta| = 0 $,  and $G$ contains a normal subgroup $N$ of index $3$ such that $|N_\lambda|$ is coprime to $6$.

\item[(c)] $|\Delta| = 1 $,  and $G$ contains a normal subgroup $N$ of index $3$ such that $N \cap K$ is strongly $3$-embedded in $N$.

\item[(d)] $|\Delta| = 4$, and $G$ contains a normal subgroup $N$ of index $3$ such that $N \cap K$ is strongly $3$-embedded in $N$.
\item[(e)]  $|\Delta| = 0$, $P$ is extra-special of exponent $3$ and order $27$, and $|G_\lambda|_3=3$. 

\item[(f)] $|\Delta| = 3$, $P \cong C_3 \wr C_3$, and if $\delta \in \Delta$, then  $G_\delta = D$ is a Frobenius group where the complements
are cyclic of order $13$, and the Sylow 3-subgroups of $G_\delta$ are elementary abelian of order 27.
\end{enumerate}
\end{prop}


\begin{proof}
Suppose that the point stabilizers have odd order.


We look at the possibilities for $\Delta$.
\\

\textbf{Case 1:} $|\Delta| = 3$. 

Let $\alpha \in \Delta$.
By Lemma~\ref{case(c34)}~(b) we know that $D \le G_\alpha \leq K$.
If $g \in G_\alpha \setminus D$, then $g$ interchanges two points in $\Delta$ and hence it has even order. This is impossible because $G_\alpha$ has odd order. Consequently $G_\alpha \le D$ and then $G_\alpha=D$.

Let $H \leq G_\alpha$ be a non-trivial four point stabilizer and let $\Gamma:=\textrm{fix}_{\Omega}(H)$.
As $H \le G_\alpha =D$, we know that $H$ fixes $\Delta$ point-wise, so
Hypothesis~\ref{4fix} forces $\Delta \subseteq \Gamma$.
Let $\gamma \in \Gamma$ be such that $\Gamma=\Delta \cup \{\gamma\}$.
Then $G_\alpha$ does not fix $\gamma$ because otherwise
$P$ stabilizes $\Gamma$, contrary to $|\Delta|=3$.
This means that $H \lneq G_\alpha$ and in fact $G_\alpha$ acts on on $\gamma^{G_\alpha}$ as a Frobenius group, with Frobenius complement $H$.
Let $M$ denote the Frobenius kernel of $G_\alpha$ in this action.
We note that $Q$ does not fix any points outside $\Delta$ by our general fixity 4 hypothesis 
(recall that $|\Omega|$ is divisible by $3$ because $|\Delta|=3$) and 
therefore $Q \le M$. This means that $Q \in \textrm{Syl}_3(M)$ and in particular $H$ has order coprime to $6$.
(Brauchen wir dafür eine Quelle?)

As $M$ is characteristic in $G_\alpha=D \unlhd K$, we see that $M \unlhd K$ and so we can use a Frattini argument. 
Now $K = M \cdot N_K(Q)$, so without loss of generality we may suppose that $H \leq N_K(Q)$.
Let $r \in \pi(H)$ and $R \in \textrm{Syl}_r(H)$. Then $r \ge 5$, $R$ acts faithfully on $M$ because $G_\alpha$ is a Frobenius group, and thus Lemma~\ref{maxclass3}(f) yields that 
$Q = P_1$ is elementary abelian, that $|P| = 81$ and that $r = 13$.
Now (f) holds because $\textrm{GL}_3(3)$ has Sylow $13$-subgroups of order $13$.

Lemma \ref{tiny} tells us that the remaining cases are $\Delta \in \{0,1,4\}$. Then we use Lemma~\ref{Case(c)not3}, where the cases (b) and (c) remain and (c) leads directly to Case (e) of the present result.

Therefore, for the remainder of this proof, we suppose that
Lemma~\ref{Case(c)not3} (b) holds and we let $N \le G$ be a normal subgroup of index $3$ such that $G = N \cdot P_\lambda$. Next we study $N$ in the individual cases.
\medskip

\textbf{Case 2:} $|\Delta| = 0$. 

Then $P$ has a unique non-regular orbit of length $\frac{|P|}{3}$, the remaining orbits are regular.
In particular $P_\lambda \in \textrm{Syl}_3(G_\lambda)$.
As $P_\lambda \cap N=1$, we deduce that $N_\lambda$ is a 3'-group.
By our initial assumption it also has odd order, so it has order coprime to 6 as stated in (b).
\medskip

\textbf{Case 3:} $|\Delta| = 1$. 

Let $\delta$ be the unique point in $\Delta$. Then 
$G_\delta=K=D$.
If $g \in G \setminus K$, then $K \cap K^g=D \cap D^g$ is a 3'-group by 
Lemma \ref{littlehelp}(a), which means that $K$ is strongly 3-embedded in $G$.
We notice that
$P\cap N \in \textrm{Syl}_3(N)$ is contained in $N_\delta$, and then it follows that $N \cap K=N_\delta$ is strongly 3-embedded in $N$ as stated in (c).

\medskip

\textbf{Case 4:} $|\Delta| = 4$.

Again there is a unique element $\delta \in \Delta$ that is fixed by $P$, and Lemma \ref{case(c34)} gives that 
$G_\delta \le K$. Also $P_\lambda \cap Q=1$ and therefore $P=QP_\lambda$. All elements in $P$ fix exactly one point, namely $\delta$, or they fix exactly four points.
Set $K_0:=K \cap N$ and let $a \in N \setminus K$. Then $\Delta^a \neq \Delta$ and $\delta^a \neq \delta$.
If $y \in K_0 \cap K_0^a$ is a non-trivial 3-element, then $y$ fixes $\delta$ and $\delta^a$, so it must be a $3$-element
with exactly four fixed points. Hypothesis \ref{4fix} says that $y$ has at most four fixed points, so it only has a unique fixed point on $\Delta^a$ and this is impossible.


This shows that $K_0$ is strongly $3$-embedded in $N$, which is (d).
\end{proof}




Finally, we can connect the 2-structure and the 3-structure of $G$ and obtain a blueprint 
for the proof of our main results.



\begin{thm}\label{3and2}
Suppose that $G$ is a finite group acting transitively, faithfully and with fixity $4$ on a set $\Omega$.
Let $P$ be a Sylow $3$-subgroup of $G$, let $\Delta$ be the union of $P$-orbits of length at most $3$ on $\Omega$ and let $K$ denote the stabilizer of the set $\Delta$ in $G$.
Moreover let $\alpha \in \Omega$ and let $f$ denote the maximum number of fixed points of involutions in $G$.
Let
$F:=\langle t \mid t \in G, o(t)=2,\textrm{fix}_{\Omega}(t)=f\rangle$.

 Then one of the following holds:
\begin{enumerate}
\item[(a)]    There is a normal subgroup $N$ of $G$ of index $1$ or $3$ such that 
\begin{enumerate}
\item[(i)]  $(|N_\alpha|, 6) = 1$ or
\item[(ii)]  $K \cap N$ is strongly $3$-embedded in $N$ and $1 \leq |\Delta| \leq 4$.
\end{enumerate}
\item[(b)] $|P| \le 3^4$, more precisely
\begin{enumerate}
\item[(i)]  $|P| = 3$ and $|N_G(P)|_2 \le 4$, or
\item[(ii)] $P$ is  elementary abelian of order $9$, or
\item[(iii)] $P$ is extra-special of exponent $3$ and order $27$, or
\item[(iv)] $P \cong C_3 \wr C_3$.
\end{enumerate}

\item[(c)] $1\le f \le 4$ and $F$ has a strongly embedded subgroup. More precisely, if $S \in \textrm{Syl}_2(G)$, if $\Gamma$ is the union of $S$-orbit of length at most $f$
and if $F_0$ denotes the stabilizer in $F$ of the set $\Gamma$, then $F_0$ is strongly embedded in $F$ and contains $N_F(S)$.


\item[(d)] $f \le 3$ and $G$ has dihedral- or semi-dihedral Sylow $2$-subgroups.

\item[(e)] $f=4$ and $G$ has sectional $2$-rank at most $4$.
\end{enumerate}
\end{thm}






\begin{proof}
Let $\alpha \in \Omega$ and first suppose that $G_\alpha$ is of odd order.
We go through the cases of Lemma \ref{syl3}.\\

\textbf{Lemma \ref{syl3} (a):} Then $G_\alpha$ has order coprime to $6$ and therefore
(a)~(i) of the theorem is satisfied in the case where $N=G$.\\

\textbf{Lemma \ref{syl3} (b):}
Then  $|\Delta| > 4$ and $|P| \leq 9$.

If $P=1$, then $G$ is a $3'$-group and therefore the point stabilizers have order coprime to $6$. This is 
(a)~(i) again in the case where $N=G$.

If $|P| = 3$, then $P$ is contained in a point stabilizer or the point stabilizers have order coprime to $6$, which is again (a)(i).
So suppose that $P$ fixes a point, i.e. $\textrm{fix}_{\Omega}(P) \neq \varnothing$.
As $|\Omega|=|G:G_\alpha|$ is coprime to $3$ now and all non-trivial $P$-orbits have length $3$, we see that $|\textrm{fix}_{\Omega}(P)| \in \{1,2,4\}$.
If $T \le N_G(P)$ is a $2$-subgroup, then it stabilizes $\textrm{fix}_{\Omega}(P)$, a set of size at most $4$. 
The elements of $T^\#$ do not fix any points because the point stabilizers have odd order, 
so Lemma \ref{Inv-lem:TB} gives that $|T| \le 4$. 
This is (b)(i) in the theorem.

If $P$ has order $9$, then there are two cases:
First assume that $P$ is cyclic. Then $\Omega_1(P)$ acts trivially on each $|P|$-orbit that is of length at most $3$, so it acts trivially on $\Delta$. But in the present case
$|\Delta| > 4$, contrary to Hypothesis~\ref{4fix}.
Therefore $P$ is elementary abelian, which is (b)~(ii) in the Theorem.

\textbf{Lemma \ref{syl3} (c):} Case (c) is treated using Hypothesis \ref{hyp4}, which gives the possibilities in
Proposition \ref{maxKlasse}. Case (b) is covered by (a)(i) of our theorem, (c) and (d) are covered by (a)(ii) and Case (e) is stated as (b)~(iii). Finally, Case (f) is stated as the last case in (b).\\

\textbf{Lemma \ref{syl3} (d) and (e):}
This is captured by Hypothesis
\ref{hyp3} and analyzed completely in Proposition \ref{3semireg}. As $G_\alpha$ has odd order, all the possibilities there are included in Cases (a)~(ii).\\

Next we suppose that $G_\alpha$ has even order.

Then $1 \le f \le 4$ and we refer to the main theorem in \cite{Ronse}, which leads directly to the cases (c), (d) or (f). The additional information in (c) comes from Ronse's Proposition 3.1.
\end{proof}


	
	
	




\section{Strategy and tools}



If we consider Theorem \ref{3and2} in the special case where $G$ is a non-abelian simple group, then 
we see a natural case distinction:

Case (a) leads to point stabilizers of order coprime to 6, 
Case (b) leads to small and explicitely described types of Sylow $3$-subgroups
and
Case (c) gives a strongly embedded subgroup.
Here we can apply Bender's main results from \cite{Ben}, which gives specific possibilities
for the group $G$.
In Cases (d) and (e) we also have classification results available, namely
the theorems of Gorenstein and Walter (\cite{GW}), of Alperin, Brauer and Gorenstein (\cite{ABG2}) and of Gorenstein ad Harada (\cite{GoHa1974}), respectively.

We will consider groups of Lie type and small Lie type first, because they appear several times in our analysis. 
Then there will be separate sections for the different cases for the point stabilizer order.
In all cases, there are some small groups to consider, and also for some of the sporadic groups we did not really see a convenient generic argument, which is why we decided to analyze some relatively small groups with
GAP~\cite{GAP}.
The code can be found in the Appendix along with examples for output.

The general analysis in later sections also requires some general techniques and basic results for groups with low fixity.

So the remainder of this section is dedicated to results about small groups (with
GAP)
and to technical background.




\begin{rem}\label{rem4.1}
	The information for the following small examples have been calculated with GAP,
	using~\cite{TomLib} and the code in the appendix.


\begin{longtable}{ll}
	$Group$ & possible point stabilizer structure\\[0.5ex]\hline\hline
	\endhead \\[-2ex]

	\(\Alt_6 \cong \textrm{PSL}_2(9)\)            & \(C_2\),
	\(\Sym_3\), \(C_3 \times C_3\)\\
	& \(D_{10}\) \text{or}
	\((C_3 \times C_3) : C_2\) \\ \hline 
	\(\Alt_7\)                            & \(C_5\) \text{or}
	\(\Alt_6\) \\  \hline      
	                                                    
	\(\textrm{PSL}_2(7) \cong \textrm{PSL}_3(2)\)         & \(C_2\) \text{or}
	\(\Sym_3\) \\ \hline
	
	\(\textrm{PSL}_2(8)\)                         & \(C_2\), \(\Sym_3\),
	\(D_{14}\) \text{or}
	\(D_{18}\) \\ \hline                                   
	
	\(\textrm{PSL}_2(11)\)                        & \(C_3\) \text{or}	 
	\(\Alt_4\)                                             \\ \hline
    
	\(\textrm{PSL}_2(13)\)                        & \(C_3\), \(\Alt_4\) \text{or}
	\(C_{13} : C_3\)                                             \\\hline

	\(\textrm{PSL}_2(17)\)                        & \(C_4\) \text{or} \(C_{17} : C_4\)                                             \\\hline
	
	\(\textrm{PSL}_2(19)\)                        & \(C_5\)                                                                        \\\hline
	
	\(\textrm{PSL}_2(23)\)                        & \(C_6\)                                                                        \\\hline
	
	\(\textrm{PSL}_2(25)\)                        & \(C_6\) \text{or} \((C_5\times C_5) : C_4\)                        \\\hline
	
	\(\textrm{PSL}_2(27)\)                        & \(C_7\)                                                                        \\\hline

	\(\textrm{PSL}_2(29)\)                        & \(C_7\) \text{or} \(C_{29} : C_7\)                                             \\\hline

	\(\textrm{PSL}_2(31)\)                        & \(C_8\)                                                                        \\\hline
	
	\(\textrm{PSL}_2(37)\)                        & \(C_9\) \text{or} \(C_{37} : C_9\)                                             \\\hline
	
	\(\textrm{PSL}_2(41)\)                        & \(C_{10}\) \text{or} \(C_{41} : C_{10}\)                                             \\\hline

	\(\textrm{PSU}_3(3)\)                         & \(((C_3 \times C_3) :C_3) :C_8\)                                                 
	\\\hline
                      
	\(\textrm{PSU}_4(2)\cong \textrm{PSp}_4(3)\)          & \(C_5\)                                                                           \\\hline
	\(\textrm{PSU}_4(3)\)                         & \(C_5\)                                                                           \\\hline
	\(\textrm{PSp}_4(4)\)                         & \(C_{17}\)                                                                           \\\hline
	\(\textrm{PSp}_4(5)\)                         & \(C_{13}\)                                                                           \\\hline
	\(\textrm{Sz}(8)\)                            & \(C_5\) \text{or}
	\(C_{13}\)                                                                                                                \\\hline
	
	\(M_{11}\)                            & \(C_5\), \(C_{11} :C_5\) \text{or}
	\(\textrm{PSL}_2(11)\)                                          \\\hline
	\(M_{12}\)                            & \(M_{11}\)                                                                       \\
    \hline
	\(M_{22}\)                            & \(C_5\) \text{or}
	\(C_{11}:C_5\)                                                                                                                \\\hline
                     
	\(J_1\)                               & \(C_{15}\)                                                                       
     \\[0.5ex]\hline\hline\\[-1ex]
	%
	\caption[Small Groups Table for fixity 4]{For each group, we give the possible point stabilizer structures for fixity 4 actions.}
	\label{SmallGroupsTable}
                            
\end{longtable}

Using GAP and the table of marks again (TomLib package, see Appendix with GAP code), we can check that the following groups do not exhibit any fixity 4 actions:

\(\textrm{PSL}_2(4)\cong \textrm{PSL}_2(5)\), 

	\(\Alt_8\), \(\Alt_9\),   \(\Alt_{10}\),      \(\Alt_{11}\),
	
	\(\textrm{PSL}_2(16)\),     \(\textrm{PSL}_2(32)\) , \(\textrm{PSL}_2(64)\),           
	
	\(\textrm{PSL}_3(3)\),                   \(\textrm{PSL}_3(4)\),      \(\textrm{PSL}_3(5)\)
	\(\textrm{PSL}_3(7)\),          \(\textrm{PSL}_3(11)\),   
	     
	\(\textrm{PSL}_4(2)\), \(\textrm{PSL}_4(3)\),     \(\textrm{PSL}_5(2)\),
	
	\(\textrm{PSU}_3(4)\),              \(\textrm{PSU}_3(5)\),       \(\textrm{PSU}_3(7)\)
	\(\textrm{PSU}_3(9)\),  
	
	\({}^2\textrm{F}_4(2)'\),
	\(M_{23}\),                       \(M_{24}\),     \(J_2\),    \(J_3\),    \(McL\),       \(HS\),     \(He\).
\end{rem}








The following three lemmas describe ways to count fixed points and will be heavily used in our analysis of specific groups. 



\begin{lemma}\label{numberfpt}
Suppose that $G$ is a group, that $1\neq U\lneq G$ and that $x\in G^\#$.
Then, in the action of $G$ on the coset space $G/U$ by right multiplication, the number of fixed points of
$x$ is exactly
\[\frac{|\{\langle x \rangle^g \leq U\mid g\in G\}|\cdot |N_G(\langle x \rangle)|}{|U|}\,.\]
\end{lemma}

\begin{proof}
Let \(y\in G\). Then \(Uy\) is a fixed point of \(x\) if and only if \(x^{y^{-1}}\in U\). So the number of fixed points of \(x\) is exactly
\[|\{ Uy \mid x^{y^{-1}} \in U\}| =\frac{|\{y\in G\mid x^{y^{-1}}\in U\}|}{|U|}=\frac{\{g\in G\mid x^g\in U\}|}{|U|}  = \frac{|\{\langle x \rangle^g \leq U\mid g\in G\}|\cdot |N_G(\langle x \rangle)|}{|U|}\]
For the last equality, we note that if $\langle x \rangle^G \cap U
= \{\langle x^{g_1}\rangle, \ldots , \langle x^{g_m}\rangle\}$, then $x^g \in U$ 
if and only if $g \in N_G(\langle x\rangle g_i$ for some $1 \leq i \leq m$.
\end{proof}





\begin{lemma}\label{Inv-lem:AnzFP}~\\
	Suppose that $G$ is a group, that $1\neq U\lneq G$ and that $x\in G^\#$. \par
	Moreover, let $M_x:=\Menge{\langle y\rangle^U}{y\in x^G\cap U}$ and $k\in\mathbb{N}_0$ be such that $\abs {M_x}=k$, and let 
$x_1,\ldots, x_k\in U$ be such that $x=x_1$ is conjugate to $x_2,\ldots, x_k$ in $G$ and that $M_x=\{\langle x_1\rangle^U,\ldots, \langle x_k\rangle^U\}$.
Then, in the action of $G$ on the coset space $G/U$ by right multiplication, the number of fixed points of
$x$ is exactly 
\[\sum_{i=1}^k\abs{N_G(\langle x_i\rangle)\colon N_U(\langle x_i\rangle)}.\]
\end{lemma}

\begin{proof}~\\
If $x^G\cap\,U=\varnothing$, then there is nothing to prove. 
From now on we suppose that $x^G\cap U\,\neq \varnothing$. 

Then without loss $x\in U$.
Further let $y_i \in G$ be such that $x_i^{y_i} = x$. 
Let $N:= N_G(\langle x \rangle)$ and $N_i:= N_G(\langle x_i \rangle)$. 
Then we see that $N_i^{y_i} = N$.

We claim that $\Lambda := \bigcup\limits_{i = 1}^k UN_i y_i$ is the union of the fixed points of $x$ on
the coset space $G/U$. 
Let $y \in G$ such that $Uy$ is fixed by $x$. Then $Uyx = Uy$ implies 
$yxy^{-1} \in U$ and therefore  $\langle x \rangle $ is mapped  onto $\langle x_i \rangle$ by $y^{-1}u$ for some $i \in \{1, \ldots , k\}$ and some $u \in U$. This implies  $n:= y_i y^{-1}u \in N_i$. Thus $y =  u n^{-1} y_i \in UN_iy_i$,
which yields that $\Lambda$ contains all the cosets that are fixed by $x$.
On the other hand it is a straightforward calculation that every coset in 
$\Lambda$ is fixed by $x$.

Observe that $N$ acts on $UN_iy_i$ and that this action is transitive
as $Uy_i N = U(y_iN y_i^{-1}) y_i = UN_i y_i$. Furthermore, $Uy_i N$ and 
$Uy_l N$ are disjoint, if $i \neq l$ by the choice of the $y_i$.

Thus it just remains to show that 
$UN_iy_i$ is the union of $|N_G(\langle x_i \rangle):
N_U(\langle x_i \rangle)|$ pair-wise distinct $U$-cosets. 
Let $T:=\{g \in N \mid Uy_ig=Uy_i\}$.

The transitive action of $N$ on $UN_iy_i$ by right multiplication
yields that this number of cosets equals $|N:T|$. We note that $T =
U^{y_i} \cap N$.
Thus \[|N:T| = |N| / |T| = |N|/|U \cap N_i| = |N_i|/ |N_U(\langle x_i \rangle)|
= |N_i:N_U(\langle x_i \rangle)| ,\] which is the assertion.

\end{proof}

We apply this lemma to the special situation that $U$ is a Frobenius group.

\begin{lemma} \label{FrobCyclic}
	Let \(G\) be a finite group and suppose that \(U \leq G\) is a Frobenius group with Frobenius kernel
	\(K\) and cyclic Frobenius complement \(J\). Let \(x\in U \setminus K\).
	Then \(x\) has exactly \(\frac{|N_{G}(\langle x \rangle)|}{|J|}\) fixed points on the coset space \(G/U\).
\end{lemma}

\begin{proof}

Let $M_x::=\{\langle y \rangle \mid y\in x^G \cap U\}$, as in Lemma~\ref{Inv-lem:AnzFP}.
We recall that all Frobenius complements are conjugate by 4.3.7 in~\cite{KurzStell} and cyclic by hypothesis.
Therefore, if $y\in x^G \cap U$, then 
$\langle y\rangle$ is $U$-conjugate to $\langle x \rangle$ and 
it follows that $|M_x| = 1$. 
Moreover $N_U(\langle x \rangle) = J$ because $J$ is abelian and because of the Frobenius group property. Now 
Lemma~\ref{Inv-lem:AnzFP} implies that $x$ fixes precisely 
$|N_G(\langle x \rangle )|/|J|$ cosets in $G/U$.
\end{proof}

	





For the analysis of Lie type groups later in this article we need generic information about the relevant groups, for example 
information about Sylow subgroups in defining characteristic, the structure of involution centralizers and the structure of Levi subgroups of maximal parabolic subgroups.  
This information is mostly available in the literature, but not always explicitly, and therefore we collect it here.

\begin{rem}\label{Inv-rem:Ser}~\\
Let $p\in\mathbb{N}$ be an odd prime, $f\in \mathbb{N}$ and $q:=p^f\ge 3$. Let $G$ be one of the groups $\textrm{PSL}_4(q)$, $\textrm{PSL}_5(q)$, $\textrm{PSU}_4(q)$, $\textrm{PSU}_5(q)$, $\textrm{PSp}_4(q)$, $G_2(q)$, $^2G_2(q)$ or $^3D_4(q)$ and let $S\in\textrm{Syl}_p(G)$. \smallskip
	
For the definition of (maximal) parabolic subgroups, the unipotent radical and Levi subgroups we refer to Definition 2.6.4,
Theorem 2.6.5 and Definition 2.6.6 in \cite{GLS3}.

	
If $t\in G$ is an involution, then Table 4.5.1 in \cite{GLS3} gives information about the structure of $C_G(t)$.
We keep this brief because the only relevant pieces of the table are those where the \textit{coset mod $Inn(K)$} column has the entry ``1'', which means that $t$ is an inner involution. Then we need $L^*=O^{p'}(C_G(t))$ (from columns 6 and 7 of the table) and in the remainder we will use the notation for the conjugacy classes of involutions as well as for $L^*$ from Table 4.5.1 in~\cite{GLS3}. For the groups $G_2(q)$ and $^3D_4(q)$ we refer to \cite{Wilson} for structure information about $C_G(t)$.
The nilpotency class of the Sylow subgroups in defining characteristics is given by
Theorems 3.2.2 and 3.3.1 and Remark 1.8.8 in \cite{GLS3}, except for $G_2(q)$.
We need the height of the highest root of the root system for $G$, again following the notation in \cite{GLS3}.
For the twisted groups we also refer to 
13.3 and 3.6 of \cite{Car}).
In the following paragraphs we let $S \in \textrm{Syl}_p(G)$ and we collect the relevant references, sometimes with additional details.
We let $n\in\{4,5\}$ and we denote the isomorphism type of the Levi subgroups of 
$\textrm{GL}_n(q)$, $\textrm{SL}_n(q)$ and $\textrm{PSL}_n(q)$ by $L_{\textrm{GL}}$, $L_{\textrm{SL}}$ and $L_{\textrm{PSL}}$, respectively. We choose similar notation for  $\textrm{SU}_n(q)$ and $\textrm{PSU}_n(q)$.  
\smallskip

At the end there will be a table collecting the information.


	\smallskip
	\underline{$G= \textrm{PSL}_4(q)$ and $q\not\equiv 1$ modulo 8}:\par
	We use Sections 3.3.1 (p.44) and 3.3.3 (p.47) in \cite{Wilson} to see that $\abs S=q^6$ and $L_{\textrm{PSL}} \cong \textrm{GL}_3(q)/C_{(4,q-1)}$ or $(\textrm{GL}_2(q)\times\textrm{SL}_2(q))/C_{(4,q-1)}$.\par
	

	

	
Next let $t\in G$ be an involution. 
Then Table 4.5.1 in \cite{GLS3} 
gives two possibilities:

$t$ is conjugate to $t_{\frac{m+1}2}=t_2$ in $G$ and, therefore, $L^*\cong\textrm{SL}_2(q)\circ\textrm{SL}_2(q)$ or

$q\equiv 3$ modulo 4, $t$ is conjugate to $t_{\frac{m+1}2}'=t_2'$ and $L^*\cong \textrm{PSL}_2(q^2)$. 
(The conjugacy classes denoted by $\gamma_1$, $\gamma_1'$, $\gamma_2$ and $\gamma_2'$ are irrelevant for us because they belong to outer involutions.)

Moreover $S$ has class $3$.
\smallskip
	%
	%

	\underline{$G= \textrm{PSL}_5(q)$ and $q\equiv -1$ modulo 4}:\par

	Sections 3.3.1 and 3.3.3 in \cite{Wilson} give that $\abs S=q^{10}$ and that

	$L_{\textrm{PSL}}\cong\textrm{GL}_4(q)/C_{(5,q-1)}\quad\text{or}\quad L_{\textrm{PSL}}\cong(\textrm{GL}_3(q)\times\textrm{SL}_2(q))/C_{(5,q-1)}.$

If $t\in G$ is an involution, then
Table 4.5.1 in \cite{GLS3} shows that $t \in t_1^G$ and $L^*\cong\textrm{SL}_4(q)$ or $t \in t_2^G$ and $L^*\cong \textrm{SL}_2(q)\times\textrm{SL}_3(q)$. 
The height of the highest root, and hence the class of $S$, is 4.
\smallskip
	%
	%
	%

	%

	%
	\underline{$G= \textrm{PSU}_4(q)$ and  $q\not\equiv -1$ modulo 8}:\par
	Again Section 3.6 in \cite{Wilson} applies: 
	
	$\abs S=q^{6}$ 

	and $L_{\textrm{PSU}}\cong\textrm{GL}_2(q^2)/C_{(4,q+1)}$ or $L_{\textrm{PSU}}\cong(C_{(q^2-1)}\times \textrm{SL}_2(q))/C_{(4,q+1)}$.\par

Let $t\in G$ be an involution. 
Then there are two possibilities 
 Table 4.5.1 in \cite{GLS3} gives two possibilities for involutions in $G$, namely  
the class of $t_{\frac{m+1}2}=t_2$, where $L^*\cong\textrm{SL}_2(q)\circ\textrm{SL}_2(q)$, or $q\equiv 1$ modulo $4$ and there is also the class of $t_{\frac{m+1}2}'=t_2'$, where
$L^*\cong \textrm{PSL}_2(q^2)$. 

For a root system for $^2A_3$ (for the group $\textrm{PSU}_4(q)$) we apply a graph automorphism 
that interchanges the roots $p_1=a_1-a_2$ and $p_3=a_3-a_4$ and fixes the root $p_2=a_2-a_3$
(see also Section 13.3 in \cite{Car}), and then we see that 
 the nilpotency class of $S$ is  $3$.
	%
	%

	\smallskip
	\underline{$G= \textrm{PSU}_5(q)$ and $q\equiv 1$ modulo $4$}:\par
Section 3.6 in \cite{Wilson} gives that 
$\abs S=q^{10}$ and
$L_{\textrm{PSU}}\cong (C_{(q^2-1)}\times \textrm{SU}_3(q))/C_{(5,q+1)}$ or $\textrm{GL}_2(q^2)/C_{(5,q+1)}$.\par
	
According to Table 4.5.1 in \cite{GLS3}, there are two classes of involutions: $t_1^G$, where $L^*\cong\textrm{SU}_4(q)$, and $t_2^G$, where $L^*\cong \textrm{SU}_2(q)\times\textrm{SU}_3(q)$. 

For a root system for $^2A_4$ (for $\textrm{PSU}_5(q)$) we use a graph automorphism that interchanges $p_1=a_1-a_2$ with $p_4=a_4-a_5$ and $p_2=a_2-a_3$ with $p_3=a_3-a_4$, respectively, see also  
Section 13.3 in \cite{Car}. 
Then it follows that $S$ has class $4$.

	%
	%
	%
	\smallskip
	\noindent\underline{$G= \textrm{PSp}_4(q)$}:\par
	Section 3.5 in \cite{Wilson} yields that 

	$\abs S=q^{4}$, and with Theorem 3.8 in \cite{Wilson}, the Levi subgroups are isomorphic to $\textrm{GL}_1(q)\circ \textrm{SL}_2(q)$ or $\textrm{GL}_2(q)/C_2$.\par
 Table 4.5.1 in \cite{GLS3} gives that an involution  $t\in G$ is conjugate to $t_{\frac 22}=t_1$, and  $L^*\cong \textrm{SL}_2(q)\circ\textrm{SL}_2(q)$ or, otherwise,  $t \in t_2^G$ (if $q\equiv 1$ modulo $4$) or $t \in t_2'^G$ (if $q\equiv 3$ modulo $4$) and in both cases $L^*\cong\textrm{PSL}_2(q)$.\par
 The class of $S$ is $3$. 
 
\smallskip
	%
	%
	%
	\underline{$G= \,^3D_4(q)$}:\par
This time Section 4.6.2 and Equation (4.67) in \cite{Wilson} imply that 
$\abs S=q^{12}$, and by Theorem 4.3 in \cite{Wilson} the Levi subgroups are isomorphic to $\textrm{SL}_2(q^3).C_{q-1}$ or to $\textrm{SL}_2(q).C_{q^3-1}$.\par

	Table 4.5.1 in \cite{GLS3}, together with Section 4.6.5 in \cite{Wilson}, gives that $G$ has only one class of involutions, namely $t_2^G$, with involution centralizer isomorphic to $C_2\dot~(\textrm{PSL}_2(q^3)\times\textrm{PSL}_2(q)).C_2$.\par

Starting with a fundamental system for	 $D_4$ we move to $^3D_4$ with a graph automorphism of order~3, acting as a 3-cycle on the set $\{p_1,p_3,p_4\}$ of roots $p_1=a_1-a_2$, $p_3=a_3-a_4$ and $p_4=a_3+a_4$ and fixing the root $p_2=a_2-a_3$
(see Section 13.3 in \cite{Car}).
It follows that
$S$ has class $5$.\\
	%
	%
We still need to look at the groups $^2G_2(q)$ and $G_2(q)$, and here we use a different strategy for determining 
the nilpotency class of $S$.\smallskip


	
	%
	%
	%
	\underline{$G= G_2(q)$}:\par
	Section 4.3.3 and Equation (4.25) in \cite{Wilson} yield that 

	$\abs S=q^{6}$. The Levi subgroups of $G$ are isomorphic to $\textrm{GL}_2(q)$, by Section 4.3.5 and Table 4.1 in \cite[S.~125, 127]{Wilson}.\par
	By Table 4.5.1 in \cite{GLS3} there is only one class of involutions, with centralizer of structure $C_2\dot~(\textrm{PSL}_2(q)\times\textrm{PSL}_2(q)).C_2$, see Section 4.3.6 in \cite{Wilson}.\par
	If $p\ge 5$, then Theorems 3.2.2 and 3.3.1 in \cite{GLS3} give a way to calculate the nilpotency class of $S$ via the height of a highest root, which gives that 

	$S$ has class $5$.\par
If $p=3$, then we adopt 
the notation on p. 65 in \cite{GLS3}, so we denote by $P_{\{\alpha_2\}}$ a maximal parabolic subgroup and by $U_{\{\alpha_2\}}$ its unipotent radical.  By Definition 2.6.4 and Remark 1.8.8 in \cite{GLS3} we see that
	\begin{align*}
		U_{\{\alpha_2\}}=U_{\{\alpha_2\}}^1&=\langle X_{\alpha_1}, X_{\alpha_1+\alpha_2}, X_{\alpha_1+2\alpha_2}, X_{\alpha_1+3\alpha_2}, X_{2\alpha_1+3\alpha_2}\rangle,\\
		U_{\{\alpha_2\}}^2&=\langle X_{2\alpha_1+3\alpha_2}\rangle=U_\varnothing^5\ug Z(U_\varnothing),\\
		U_{\{\alpha_2\}}^3&=1.
	\end{align*}

Theorem 3.2.2 in \cite{GLS3} gives that $1=U_{\{\alpha_2\}}^3 \neq U_{\{\alpha_2\}}^2\neq U_{\{\alpha_2\}}^1$ is a central series of $U_{\{\alpha_2\}}$ with elementary abelian sections. Moreover $[U_{\{\alpha_2\}},U_{\{\alpha_2\}}]\ug U_{\{\alpha_2\}}^2=\langle X_{2\alpha_1+3\alpha_2}\rangle$ and therefore, for all roots \[\beta\in \{\alpha_1,\alpha_1+\alpha_2,\alpha_1+2\alpha_2,\alpha_1+3\alpha_2,2\alpha_1+3\alpha_2\}=:\Sigma_1,\] it is true that $[X_{\alpha_1},X_\beta]\ug X_{2\alpha_1+3\alpha_2}=U_{\{\alpha_2\}}^2$.\par
We refer to p.99 in \cite{GLS3} and deduce that $P_{\{\alpha_2\}}\cong P_{\{\alpha_1\}}$ and hence $U_{\{\alpha_2\}}\cong U_{\{\alpha_1\}}$. Here it is relevant that all maximal parabolic subgroups are isomorphic even though there are two conjugacy classes of them, see Sections 4.3.5 to 4.3.7 in \cite{Wilson}. There is a graph automorphism $\gamma$  of order 2 that interchanges 
 $\alpha_1$ and $\alpha_2$. We also note that $\gamma$ maps positive roots to positive roots and root subgroups to root subgroups and that $U_\varnothing \in \textrm{Syl}_p(G)$. By Definition 2.6.4 in \cite{GLS3} we have that
	\begin{align*}
		U_\varnothing=U_{\varnothing}^1&=\langle X_{\alpha_1}, X_{\alpha_2}, X_{\alpha_1+\alpha_2}, X_{\alpha_1+2\alpha_2}, X_{\alpha_1+3\alpha_2}, X_{2\alpha_1+3\alpha_2}\rangle,
		U_\varnothing^2=\langle X_{\alpha_1+\alpha_2}, X_{\alpha_1+2\alpha_2}, X_{\alpha_1+3\alpha_2}, X_{2\alpha_1+3\alpha_2}\rangle.
	\end{align*}
Moreover the section $U_\varnothing/U_\varnothing^2$ is elementary abelian by Theorem 3.2.2 in \cite{GLS3} and $[U_\varnothing,U_\varnothing]\ug U_\varnothing ^2$. We also know that $[X_{\alpha_1},X_\beta]\ug X_{2\alpha_1+3\alpha_2}$ for all $\beta\in\Sigma_1$. Before we construct $[U_\varnothing,U_\varnothing, U_\varnothing]\ug [U_\varnothing, U_\varnothing^2]$, we need $[X_{\alpha_2},X_\beta]$ for all $\beta \in\Sigma_1\setminus \{\alpha_1\}$. Using the graph automorphism $\gamma$ of $G_2$ we calculate
	\begin{align*}
		[X_{\alpha_2},U_\varnothing^2]&=([X_{\alpha_2},U_\varnothing^2]^\gamma)^{\gamma}=\left[{X_{\alpha_2}}^\gamma,(U_\varnothing^2)^\gamma\right]^\gamma
		=[X_{\alpha_1},U_\varnothing^2]^\gamma\ug (U_{\{\alpha_2\}}^2)^\gamma\ug Z(U_\varnothing)^\gamma=Z(U_\varnothing).
	\end{align*}

Next we see that
	$[U_\varnothing,U_\varnothing, U_\varnothing]\ug [U_\varnothing, U_\varnothing^2]\ug Z(U_\varnothing)$ and $[U_\varnothing,U_\varnothing, U_\varnothing, U_\varnothing]\ug [U_\varnothing, Z(U_\varnothing)]=1$. 

Then the nilpotency class of $U_\varnothing$ is at most 3, because we found a descending central series of length four, and it is 
 at least 3, because $U_\varnothing$ is isomorphic to a subgroup of $P_{\{\alpha_2\}}$,  $U_{\{\alpha_2\}}$ has class 2, the $p$-part of $\abs{P_{\{\alpha_2\}}/U_{\{\alpha_2\}}}$ is exactly $q$ and the Sylow $p$-subgroups of the Levi subgroup $L_{\{\alpha_2\}}\ug P_{\{\alpha_2\}}$ act non-trivially on $U_{\{\alpha_2\}}$.

\smallskip
This completes our analysis for now and we show the relevant information in a table
	%
	%
	
\vspace{0.5cm}
	{\small
	\begin{longtable}{llllllp{3.5cm}}%

	$G$ & $t^G$ & section $L^*$ in $C_G(t)$ & $\abs S$ & class of $S$ & Levi subgroup $L$ & remarks\\[0.5ex]\hline\hline
	\endhead \\[-2ex]

	$\textrm{PSL}_4(q)$	&	$t_2$		& $\textrm{SL}_2(q)\circ \textrm{SL}_2(q)$	& $q^6$ 	& 3 & $\textrm{GL}_3(q)/C_{(q-1,4)}$&\\
				&	$t_2'$	&   $\textrm{PSL}_2(q^2)$	&	& & $\textrm{GL}_2(q)\times \textrm{SL}_2(q)$& \\[0.2ex]\hline\\[-2ex]
	$\textrm{PSL}_5(q)$	&	$t_1$		& $\textrm{SL}_4(q)$	& $q^{10}$ 	& 4& $\textrm{GL}_4(q)/C_{(q-1,5)}$&\\
				&	$t_2$		& $\textrm{SL}_2(q)\times \textrm{SL}_3(q)$	&	& & $\textrm{SL}_2(q)\times \textrm{GL}_3(q)$&\\[0.2ex]\hline\\[-2ex]
	%
	$\textrm{PSU}_4(q)$	&	$t_2$		& $\textrm{SL}_2(q)\circ \textrm{SL}_2(q)$	& $q^6$ 	& 3 & $\textrm{GL}_2(q^2)/C_{(q+1,4)}$&\\
				&	$t_2'$	& $\textrm{PSL}_2(q^2)$	&	& & $(C_{(q^2-1)}\times \textrm{SU}_2(q))/C_{(4,q+1)}$ &\\[0.2ex]\hline\\[-2ex]
	$\textrm{PSU}_5(q)$	&	$t_1$		& $\textrm{SU}_4(q)$	& $q^{10}$ 	& 4 & $\textrm{GL}_2(q^2)/C_{(q+1,5)}$&\\
				&	$t_2$		& $\textrm{SU}_2(q)\times \textrm{SU}_3(q)$	&	& & $(C_{(q^2-1)}\times \textrm{SU}_3(q))/C_{(5,q+1)}$&\\[0.2ex]\hline\\[-2ex]
	$\textrm{PSp}_4(q)$	&	$t_1$		& $\textrm{SL}_2(q)\circ \textrm{SL}_2(q)$	& $q^4$ 	& 3& $\textrm{SL}_2(q^3). C_{q-1}$&\\
				&	$t_2,t_2'$	& $\textrm{PSL}_2(q)	$	&	& & $\textrm{SL}_2(q). C_{q^3-1}$&\\[0.2ex]\hline\\[-2ex]
	$G_2(q)$	&	$t_1$		& $C_2\dot~(\textrm{PSL}_2(q)\circ \textrm{PSL}_2(q)).C_2$	& $q^6$ 	& 3 ($p=3$)& $\textrm{GL}_2(q)$&\tiny{Two classes of mps (Section 4.3.5. and Table 4.1 in \cite[S. 123ff, 127]{Wilson})}\\
			&				&											&		& 5 ($p>3$)\\[0.2ex]\hline\\[-2ex]
	$^3D_4(q)$	&	$t_2$		& $C_2\dot~(\textrm{PSL}_2(q^3)\circ \textrm{PSL}_2(q)).C_2$	& $q^{12}$ 	& 5&$\textrm{GL}_1(q)\circ \textrm{SL}_2(q)$&\\
	&&&&& $\textrm{GL}_2(q)/C_2$&\\[0.5ex]\hline\hline\\[-1ex]
	%
	\caption[Sections of involution centralizers, order and nilpotency class of Sylow $p$-subgroups of simple groups of Lie type of sectional 2-rank at most 4 in odd characteristic $p$.]{Conj. class of $t$ in $G$, sections $L^*$ of $C_G(t)$ from Table 4.5.1 in \cite{GLS3}, order of $S\in\textrm{Syl}_p(G)$ from \cite{Wilson}, nilpotency class of $S$ and Levi subgroup structure for groups $G$ of Lie type of sectional 2-rank at most 4 in odd characteristic $p$. In the remark we abbreviate ``maximal parabolic subgroups'' with \textbf{mps}.}
	\label{Inv-tab:Ser}
	\end{longtable}}
	
\end{rem}

\begin{lemma}\label{helpFrattini}
Let $A$ be a soluble group and $p,q \in \pi(A)$ be two different prime divisors of the order of $A$. Then
there is a prime $r \in \{p,q\}$ and a non-trivial $r$-subgroup $R$ of $A$ that is normalized by
some Sylow $s$-subgroup of $A$, where $\{r,s\} = \{p,q\}$.
\end{lemma}
\begin{proof} By hypothesis $A$ is soluble, so we find $m \in \mathbb{N}$ and normal subgroups $1 = A_1 \le  A_2 \le \cdots \le A_m = A$ 
of $A$ such that the sections $A_i/A_{i-1}$, where $1\in \{1,...,m\}$, are elementary abelian groups of prime
power order. Let $j \le m$ be minimal such that $r$ divides $|A_j|$ for some $r \in  \{p,q\}$, and let
$R$ be a Sylow $r$-subgroup of $A_j$. Then the Frattini argument gives that $A = A_j N_A(R)$.
Let $s$ be such that $\{r,s\} = \{p,q\}$. Then  $s$ does not divide $|A_j|$, and therefore our assertion holds.
\end{proof}









\section{Small Lie Type Groups}


Here we look at some finite simple groups of Lie Type of small Lie rank. This is because they occur in various places in Theorem \ref{3and2} and therefore they need careful attention. More precisely, we consider the series $\textrm{PSL}_2(q)$, $\textrm{Sz}(q)$, $\textrm{PSL}_3(q)$ and $\textrm{PSU}_3(q)$, in this ordering, and we classify all possibilities for them to satisfy Hypothesis \ref{4fix}.




\begin{lemma}\label{L2q-main}
    Suppose that Hypothesis \ref{4fix} holds, that $p$ is a prime and that $n\in \mathbb{N}$ and $q:=p^n$ are such that
$G=\textrm{PSL}_2(q)$ is simple. Let $\alpha\in\Omega$.
Then one of the following holds:
		\begin{enumerate}
			\item \(G=\textrm{PSL}_2(7)\) and \(G_{\alpha}\cong C_2\) or \(G_{\alpha}\cong \Sym_3\).
			\item \(G=\textrm{PSL}_2(8)\) and \(G_{\alpha}\) is cyclic of order \(2\) or dihedral of order \(6,14\) or
			\(18\).
			\item \(G=\textrm{PSL}_2(9)\) and \(G_{\alpha}\) is cyclic of order \(2\), dihedral of order \(6\) or
			\(10\), elementary abelian of order \(9\), or the semi-direct product of an elementary abelian group of order \(9\)
			with a cyclic group of order \(2\).
			\item \(G=\textrm{PSL}_2(11)\) and \(G_{\alpha}\cong C_3\) or \(G_{\alpha}\cong\Alt_4\).
			\item \(G=\textrm{PSL}_2(13)\) and \(G_{\alpha}\cong C_3\), \(G_{\alpha}\cong C_{13}:C_3\) or \(G_{\alpha}\cong\Alt_4\).
			\item \(q\geq 17\) is odd. If \(q\equiv 1 \mod 4\) then the point stabilizers are either cyclic of order
			\(\frac{q-1}{4}\) or the semi-direct product of an elementary abelian group of order \(q\) with a cyclic group of
			order
			\(\frac{q-1}{4}\).
			If \(q\equiv -1 \mod 4\) then the point stabilizers are cyclic of order \(\frac{q+1}{4}\).
		\end{enumerate}
\end{lemma}

\begin{proof} 

    By Remark~\ref{rem4.1} the statement of the lemma holds whenever \(q\leq 41\). Therefore throughout the rest of the proof suppose 
    that \(q\geq 43\).
	
	We recall that \(|G|=q\cdot \frac{q^2-1}{(2,q-1)}\), and detailed information about the subgroup structure of \(G\) is stated in Hauptsatz~II\,8.27
	in~\cite{Huppert-I}, which we will use now without further reference.
	
	\medskip
	
	We will proceed by a case distinction on whether \(q\) and \(|G_{\alpha}|\) are coprime or not.
	
	So first additionally suppose that \(q\) and \(|G_{\alpha}|\) have a common prime divisor \(p\).
	Let \(x\in G_{\alpha}\) be of order \(p\).
	Let \(Q\in \textrm{Syl}_p(G)\) be such that \(x\in Q\) and \(Q_{\alpha}\in \textrm{Syl}_p(G_{\alpha})\). Then \(Q\) is elementary
	abelian
	and therefore \(Q\leq N_{G}(\langle x \rangle)\). By our usual lemma this implies \(|Q:Q_{\alpha}|\leq
	|N_{G}(\langle x \rangle):N_{G_{\alpha}}(\langle x \rangle)|\leq 4\). Let \(f\) be a positive integer such that
	\(p^f=q\). 
	
	If \(p=2\), then \(|G_{\alpha}|\) is divisible by \(\frac{q}{4}=2^{f-2}\), and \(f\geq 5\). Therefore
	\(f-2>
	\frac{f}{2}\).
	
	 If \(p=3\), then \(|G_{\alpha}|\) is divisible by \(\frac{q}{3}=3^{f-1}\), and \(f\geq 3\).
	 Hence
	\(f-1>\frac{f}{2}\).
	
	Finally, if \(p\geq 5\), then \(|G_{\alpha}|\) is divisible by \(q\) by our usual Sylow lemma.
	In particular \(|Q_{\alpha}|\geq \frac{q}{4}\geq \frac{43}{4}>10\).
	
	We now show that \(G_{\alpha}\) is not a \(p\)-group, and the strategy depends on whether \(q\) is even or odd.
	First suppose that \(q\) is odd.
	Assume for a contradiction that \(G_{\alpha}\) is a \(p\)-group.
	Then there exists an element \(a\in
	G_{\alpha}\) of order \(p\)
	with exactly \(4\) fixed points. 
	
	Lemma~\ref{numberfpt} says that the number of fixed points of \(a\) is
	\(\frac{|\{\langle a \rangle^g\leq G_{\alpha}\mid g\in G\}|\cdot |N_G(\langle a
		\rangle)|}{|G_{\alpha}|}\), so this number is $4$. As \(|N_G(\langle a
	\rangle)|\) is divisible by \(p^f\) and \(4\) is coprime to \(p\), we deduce that \(|G_{\alpha}|\) is divisible by \(p^f\) and
	hence \(Q_{\alpha
	}=Q\).
	Then our usual lemma gives the contradiction
	\(4\geq |N_G(Q):N_{G_{\alpha}}(Q)|=q\cdot \frac{q-1}{2}/q=\frac{q-1}{2}\geq \frac{42}{2}=21\). Hence
	\(G_{\alpha}\) is not a \(p\)-group in this case.
	
	So now suppose \(q\) is even. As all involutions in \(\textrm{SL}_2(q)\cong \textrm{PSL}_2(q)\) are conjugate,
	the number of involutions in \(Q_{\alpha}\) (i.e. \(|Q_{\alpha}|-1\)) is exactly \(|\{\langle x \rangle^g \leq Q_{\alpha}\mid g\in G\}|\).
	By Lemma~\ref{numberfpt} the number of
	fixed points of \(x\) is 
	
	\(\frac{|\{\langle x \rangle^g\leq G_{\alpha}\mid g\in G\}|\cdot |N_G(\langle x
		\rangle)|}{|G_{\alpha}|}\geq \frac{|\{\langle x \rangle^g\leq Q_{\alpha}\mid g\in G\}|\cdot |N_G(\langle x
		\rangle)|}{|G_{\alpha}|}=\frac{(|Q_{\alpha}|-1)\cdot |N_G(\langle x \rangle)|}{|G_{\alpha}|} \geq
	\frac{(|Q_{\alpha}|-1)\cdot q}{|G_{\alpha}|}\).
	
	As \(x\) fixes at most four points, we conclude that \(|G_{\alpha}|\geq \frac{q\cdot (|Q_{\alpha}|-1)}{4}\).
	We
	see by Remark~\ref{rem4.1}, that $\textrm{PSL}_2(64)$ does not act with fixity 4, so
	from now on we can suppose that  and \(q\geq 128\) in the present case.
	Then \(\frac{\frac{q}{4}-1}{4}\geq \frac{31}{4}>4\), and again $G_ \alpha$ is not a \(p\)-group.
	
	
	Together with the information about \(|Q_{\alpha}|\), an inspection of the list of subgroups of \(\textrm{PSL}_2(q)\)
	reveals that
	\(G_{\alpha}\) is a subgroup of a Frobenius group of order \(q\cdot \frac{q-1}{(2,q-1)}\) that is
	not a \(p\)-group. The Frobenius kernel is \(Q_{\alpha}\) the unique Sylow \(p\)-subgroup of \(G_{\alpha}\).
	Let \(y\in G_{\alpha}\) have prime order dividing \(\frac{q-1}{(2,q-1)}\).
	Then by Lemma~\ref{FrobCyclic}
	the number of fixed points of \(y\) is
	\(\frac{|Q_{\alpha}|\cdot |N_G(\langle y \rangle)|}{|G_{\alpha}|}=\frac{|Q_{\alpha}|\cdot 2\cdot
		\frac{q-1}{(2,q-1)}}{|G_{\alpha}|}\).
	By Lemma~(numberfpt) the number of
	fixed points of \(x\) is \(\frac{|\{\langle x \rangle^g\leq G_{\alpha}\mid g\in G\}|\cdot |N_G(\langle x
		\rangle)|}{|G_{\alpha}|}\). This number coincides with
	\(\frac{|\{\langle x \rangle^g\leq Q_{\alpha}\mid g\in G\}|\cdot |N_G(\langle x
		\rangle)|}{|G_{\alpha}|}\) because \(Q_{\alpha}\) is a characteristic subgroup of \(G_{\alpha}\). 
		
		Moreover,  \(|\{\langle x \rangle^g\leq Q_{\alpha}\mid g\in G\}|\) is bounded from above
	by the number of distinct subgroups of order \(p\) of \(Q_{\alpha}\), and \(|N_G(\langle x\rangle)|\) is
	bounded from above by \(|C_G(x)|\cdot |\textrm{Aut}(\langle x \rangle)|=q\cdot (p-1)\). Consequently \(x\) fixes at most
	\(\frac{\frac{|Q_{\alpha}|-1}{p-1}\cdot q\cdot (p-1)}{|G_{\alpha}|}=\frac{q(|Q_{\alpha}|-1)}{|G_{\alpha}|}\) points in
	\(\Omega\).
	Hence $|\textrm{fix}_{\Omega}(x)| \le |\textrm{fix}_{\Omega}(y)|$.
	
	Since the maximal number of fixed points of non-trivial elements in \(G_{\alpha}\) is obtained by an element of prime
	order, we see that
	\(y\) fixes four points. Then the calculation above shows that \(|G_{\alpha}|=|Q_{\alpha}|\cdot \frac{q-1}{2\cdot
		(2,q-1)}\).
	As \(G_{\alpha}\) is a Frobenius group, this is only possible if \(|Q_{\alpha}|=q\) and \(q\) is odd.

	So \(|G_{\alpha}|=q\cdot
	\frac{q-1}{4}\). Hence \(q-1\) is divisible by \(4\) and so \(q\equiv 1 \mod 4\). This case is listed in (f). 
	This finishes the analyzes in the case that \(q\) and \(|G_{\alpha}|\) have a common prime factor.\\
	
	Now we suppose that \(q\) and \(|G_{\alpha}|\) are coprime. Then \(|G_{\alpha}|\) divides \(q^2-1\). Let
	\(x\in G_{\alpha}\) be of prime order \(p\) and such that $|\textrm{fix}_{\Omega}(x)|=4$. 
	Then \(p\) divides \(q-1\) or \(q+1\) and we let
	\(\varepsilon \in \{-1,1\}\) be such that \(p\) divides \(q-\varepsilon\). The normalizer of \(\langle x
	\rangle\) is a dihedral group of order \(2\cdot \frac{q-\varepsilon}{(2,q-1)}\).
	Let \(C\) be the cyclic subgroup of order \(\frac{q-\varepsilon}{(2,q-1)}\) of that normalizer.
	With our usual lemma we see that \(|C:C_{\alpha}|\leq |N_G(\langle x \rangle):N_{G_{\alpha}}(\langle x
	\rangle)|\leq 4\).
	Hence \(G_{\alpha}\) contains a cyclic subgroup of order at least \(\frac{q-\varepsilon}{4\cdot (2,q-1)}\geq
	\frac{42}{8}>5\).
	As \(|G_{\alpha}|\) and \(q\) are coprime, an inspection of the list of subgroups of \(G\) shows two possibilities for 
	\(G_{\alpha}\):
	a cyclic group of order dividing
	\(\frac{q-\varepsilon}{(2,q-1)}\) or a dihedral group.
	
	For a contradiction we assume that \(G_{\alpha}\) is a dihedral group.
	Then there is an involution \(t\) outside the cyclic group \(C\) of \(G_{\alpha}\). However, \(N_G(t)\) contains a
	cyclic group
	\(D\) that has trivial intersection with \(C\). So by our usual lemma \(|D:D_{\alpha}|\leq
	|N_G(t):N_{G_{\alpha}}(t)|\leq 4\). On the other hand
	\(|D:D_{\alpha}|=\frac{|D|}{2}=\frac{q-\varepsilon}{2(2,q-1)}\geq \frac{42}{4}>4\). This contradiction shows that
	\(G_{\alpha}\) is not a dihedral group.
	
	Therefore \(G_{\alpha}\) is cyclic. Then by Corollary~(cyclicnumberfpt) the number of fixed points
	of a non-trivial element \(c\) of \(G_{\alpha}\) is \(\frac{|N_G(\langle c\rangle)|}{|G_{\alpha}|}=\frac{2\cdot
		\frac{q-\varepsilon}{(2,q-1)}}{|G_{\alpha}|}\). As \(G_{\alpha}\) contains non-trivial elements that fix exactly
	four
	points, it follows that every non-trivial element fixes exactly four points and 
	hence \(|G_{\alpha}|=\frac{q-\varepsilon}{2\cdot (2,q-1)}\). As
	\(q-\varepsilon\) is only divisible by \(2\) if \(q\) is odd, we see that \((2,q-1)=2\) and
	\(|G_{\alpha}|=\frac{q-\varepsilon}{4}\). In conclusion, \(q-\varepsilon\) is divisible by \(4\) and so \(q\equiv \varepsilon
	\mod 4\). This case is listed in (f).
\end{proof}





The Suzuki groups appear as potential candidates for fixity 4 actions in Theorem \ref{3and2}, and the next lemma gives the exact possibilities.




\begin{lemma}\label{Suzuki}
Let \(n \in \mathbb{N}, q:=2^{2n+1}\) and
		\(G=\textrm{Sz}(q)\).
		Suppose that $G$ satisfies Hypothesis \ref{4fix} for some set $\Omega$ and let $\alpha \in \Omega$.
		 Then \(G_{\alpha}\) is cyclic of order \(q+\sqrt{2q}+1\) or \(q-\sqrt{2q}+1\).
\end{lemma}




\begin{proof}
	First we note that \(|G|=q^2(q^2+1)(q-1)=q^2 \cdot (q+\sqrt{2q}+1) \cdot (q-\sqrt{2q}+1) \cdot (q-1)\) and that the numbers \(q^2,
	q+\sqrt{2q}+1, q-\sqrt{2q}+1, q-1\) are pairwise coprime. (For the group order see for example p. 117 in \cite{Wilson}.)
	We also refer to Table \ref{SmallGroupsTable} for $\textrm{Sz}(8)$, so from now on we suppose that
	$q \ge 32$ and we will several times apply Theorem
	4.1 on p.\,117
	in~\cite{Wilson} for the subgroup structure of $G$.
	With Hypothesis \ref{4fix} we let $x \in G_\alpha$ be of prime order $p$ such that $x$ fixes four points.
   
	First suppose that $p=2$ and let $x \in S \in \textrm{Syl}_2(G)$.
	Then we see, with Lemma \ref{Inv-lem:TB} (c) and the structure of $S$, that $G_\alpha$
	contains a subgroup of index $2$ or $4$ of $S$ and hence it contains $Z(S)$. Let $N:=N_G(Z(S))$. Then $|N|=q^2(q-1)$ and,
	as $q-1$ is odd and all elements in $Z(S)$ have two or four fixed points,
	it follows that $q-1$ divides $|G_\alpha|$. Then $G_\alpha=N$ contains a Sylow $2$-subgroupof $G$,
	which contradicts Lemma \ref{Inv-lem:TB2}.
	
	Now we know that $p$ is odd, so
	\(p\) divides one of the numbers  \((q+\sqrt{2q}+1), (q-\sqrt{2q}+1)\) or \(q-1\) and
	\(X:=\langle x \rangle\) is contained in a
	maximal subgroup of $G$ of structure \(D_{2(q-1)}\), \(C_{q+\sqrt{2q}+1}:C_4\) or \(C_{q-\sqrt{2q}+1}:C_4\).
	The number of fixed points of \(x\) can be determined with Lemma \ref{FrobCyclic}, so \(4=|\textrm{fix}_{\Omega}(x)|=
	\frac{|N_{G}(X)|}{|G_{\alpha}|}\).
	If \(p\) divides \(q-1\), then \(4=\frac{2(q-1)}{|G_{\alpha}|}\) gives a contradiction because \(q-1\) is odd.
	So now there exists \(\varepsilon\in\{-1,1\}\) such that \(p\) divides \(q+\varepsilon\sqrt{2q}+1\).
	
	Then \(4=\frac{(q+\varepsilon\sqrt{2q}+1)\cdot 4}{|G_{\alpha}|}\) implies that
	\(|G_{\alpha}|=q+\varepsilon\sqrt{2q}+1\). All subgroups of order \(q+\varepsilon\sqrt{2q}+1\) of a group of structure
	\(C_{q+\varepsilon\sqrt{2q}+1}:C_4\) are cyclic, so this yields the statement of the lemma.
\end{proof}






Next we consider the series $\textrm{PSL}_3(q)$ and $\textrm{PSU}_3(q)$.

\begin{lemma}\label{PSLPSU3}

Let \(q\geq 3\) be a prime power, let \(\varepsilon\in\{-1,1\}\) and let
		\(G=\textrm{PSL}^{\varepsilon}_3(q)\). Then \(G\) acts transitively and with
		fixity \(4\)
		if and only if \(G=\textrm{PSU}_3(3)\) and the point stabilizers have structure \(((C_3\times C_3):C_3):C_8\).

\end{lemma}

\begin{proof}
    By Remark~\ref{rem4.1} the statement of the lemma holds whenever \(q\leq 9\). Therefore,
    from now on, we let \(q\geq 11\).

    
    First we suppose that \(G\) acts transitively and with fixity \(4\) on a set~\(\Omega\) and we let \(\alpha
	\in
	\Omega\).
	
	Assume for a contradiction that \(G_{\alpha}\) has even order. Let \(t\in G_{\alpha}\) be an involution.
	
	The analysis now depends on whether \(q\) is even or odd.
	First suppose that \(q\) is even, which means that $q \ge 16$. Then by Lemma~\ref{3and2simple}, Satz~1
	in~\cite{Ben},
	Theorem~2 in~\cite{GW},
	the Third Main Theorem in~\cite{ABG2} and
	the Main Theorem
	in~\cite{GoHa1974}, we know that
	\(G=\textrm{PSU}_3(q)\) and that \(G\) has a strongly embedded subgroup.
	Remembering that $q$ is even, we let \(f\in \mathbb{N}\) be such that \(q:=2^f\). Let \(T\in \textrm{Syl}_2(G)\) be such that \(T_{\alpha}\in
	\textrm{Syl}_2(G_{\alpha})\) and that \(t\in T_{\alpha}\). Then \(t\in
	Z}%\newcommand{\Zen}{\textrm{Z}(T)\) by statement (4) on page~535
	in~\cite{Ben}.
	As \(t\) has at most four fixed points, we see that \(4\geq |C_T(t):C_{T_{\alpha}}(t)|=|T:T_{\alpha}|\). Hence
	\(|T_{\alpha}|\) is one of the following numbers: \(|T|=q^3\), \(\frac{|T|}{2}=\frac{q^3}{2}=2^{3f-1}\) or
	\(\frac{|T|}{4}=\frac{q^3}{4}=2^{3f-2}\). So the order of \(G_{\alpha}\) is
	divisible by \(\frac{q^3}{4}\geq 2^{10}\).
	Inspection of Tables~8.5 and 8.6
	in~\cite{BHRD} reveals that \(G_{\alpha}\)
	lies in a maximal subgroup such that its full pre-image in \(\textrm{SU}_3(q)\) has structure \(E_q^{1+2}:(q^2-1)\).


	Let \(I\) be the set of all involutions of \(G\) and  assume that \(I \subseteq N_{G}(T)\). As \(G\) is simple and \(\langle I \rangle\unlhd G\), this implies that \(\langle I \rangle = G\) and in particular \(N_{G}(T)=G\), which is impossible. 
	We deduce that \(I \not \subseteq N_{G}(T)\) and we let \(s \in G\) be an involution such that 
    \(s \notin N_{G}(T)\). By statement (6) on page~534
	in~\cite{Ben} we know that \(|C_G(s)|\) is
	divisible by \(\frac{q+1}{(3,q+1)}\).
	
	We will need the following observation:
	
	($\ast$) ~~If \(T \le U\leq G\), then \(T \unlhd U\). 
	
	This is because, by Tables~8.5 and 8.6
	in~\cite{BHRD}, $U$ lies in a maximal
	subgroup with full pre-image in \(\textrm{SU}_3(q)\) of structure \(E_q^{1+2}:(q^2-1)\).

    By Lemma 4.1\,(i) in~\cite{Ben}, $s$ and $t$ are conjugate. 
	Let \(h\in G\) be such that \(t^h=s\) and let \(\beta := \alpha^h\).
	Then \(s\in G_{\beta}\), \(s\) has at most four fixed points and therefore \(|C_G(s) :
	C_{G_{\beta}}(s)| \leq 4 \).
	We notice that \(|C_G(s)|\) is divisible by \(\frac{q+1}{(3,q+1)}\), an odd number. This number is divisible by 3 only if \(q+1\) is divisible by \(9\), which means that
	\(|C_{G_{\beta}}(s)|\) is divisible by \(\frac{q+1}{(9,q+1)}>4\). 
	
	Assume that \(\frac{q+1}{(9,q+1)}\) is a \(3\)-power. Then there exists some \(l\in \mathbb{N}\) such that
	\(q+1=2^f+1=3^l\).
	If \(l\) is odd, then \(2^f=3^l-1=(3-1)(3^{l-1} + 3^{l-2} + \ldots +3+1)\). The second factor has \(l\) odd summands, so it is odd and it divides \(2^f\). It follows that \(l=1\) and then
	\(f=1\). In conclusion \(q=2\), which contradicts the assumption.
	
	So \(l\) is even. Let \(l_2:=\frac{l}{2}\). Then \(2^f=3^l-1=(3^{l_2}-1)(3^{l_2}+1)\). If \(l_2\)
	is odd, then we argue as in the previous paragraph and it follows that \(l_2=1\), then \(l=2\) and hence \(f=3\). This means that \(q=8<11\), 
	whereas we only consider the case $q \ge 11$. 
	We deduce that \(l_2\) is even. Then \(3^{l_2} + 1 \equiv (-1)^{l_2}
	+1
	\equiv 1 +1 \equiv 2 \mod 4\). It follows that \(3^{l_2} + 1=2\), but then \(l_2=0\), and this contradicts our assumption
	that \(l \in \mathbb{N}\).
	Therefore \(\frac{q+1}{(9,q+1)}\) is not a \(3\)-power, but it is odd and hence divisible by a prime greater than \(3\).
	
	Let \(r \ge 5\) be a prime dividing \(q+1\) and \(|G_{\alpha}|\). Then \(G_{\alpha}\) contains a
	Sylow \(r\)-subgroup $R$ of \(G\), and we let \(k \in \mathbb{N}\) be such that \(|R|=r^k\). We recall that \(|G_{\alpha}|\) divides \(q^3(q+1)\frac{(q-1)}{(3,q+1)}\). Moreover the numbers \(q\), \(q-1\) and \(3\) are
	coprime to \(r\). Then it follows that \(r^k\) divides \(q+1\).
	As \(|G|=\frac{q^3}{(3,q+1)}(q^3+1)(q^2-1)=\frac{q^3}{(3,q+1)}(q^2-q+1)(q-1)(q+1)^2\), we deduce that
	\(r^{2k}\) divides \(|G|\), contrary to the fact that $R \in \textrm{Syl}_r(G)$ and $|R|=r^k$. 
	
	Therefore \(q\) is
	odd.
	
	Now we know that all involutions in \(G\) are conjugate, by Proposition~2.1 in~\cite{GLS6}, 
	and then Table 8.3 and Table 8.5
	in~\cite{BHRD} show that $t$ is contained in the centre of a
	group
	\(U\) such that the full pre-image of $U$ in \(\textrm{SL}^{\varepsilon}_3(q)\) is isomorphic to \(\textrm{GL}^{\varepsilon}_2(q)\). In particular
	\(|C_G(t)|\) is divisible by \(\frac{q\cdot (q-\varepsilon)^2\cdot (q+\varepsilon)}{(3,q-\varepsilon)}\). As
	\(|C_G(t):C_{G_{\alpha}}(t)|\leq 4\) and \(\frac{q}{4}>2\), we find an element \(x\in
	G_{\alpha}\) of prime order \(p\) such that \(q\) is a power of $p$. Let \(Q\in \textrm{Syl}_p(G)\) be such that \(x\in Q\). Then \(Z(Q)\leq
	C_G(x)\)
	and therefore \(|Z(Q):Z(Q)\cap G_{\alpha}|\leq |C_{G}(x):C_{G_{\alpha}}(x)|\leq 4\). Next we use the fact that \(|Z(Q)|=q\), whence

	\(Z(Q)\cap G_{\alpha}\) contains a non-trivial element~\(z\). Therefore \(Q\leq C_{G}(z)\) and we see that
	\(|Q:Q_{\alpha}|\leq |C_{G}(z):C_{G_{\alpha}}(z)|\leq 4\). This implies that \(G_{\alpha}\) is divisible by
	\(\frac{q^3}{(3,q)}\) or \(\frac{q^3}{(4,q)}\). We recall that \(|G_{\alpha}|\) is also divisible by \(\frac{q\cdot
		(q-\varepsilon)^2\cdot (q+\varepsilon)}{(3,q-\varepsilon)}\) and that \(q\geq 11\). Then inspection of the Tables 8.3,
	8.4,
	8.5 and 8.6 in~\cite{BHRD}

	shows that \(G=\textrm{PSL}_3(q)\) and that \(G_{\alpha}\) lies in a maximal subgroup \(M\) such that its full pre-image in
	\(\textrm{SL}_3(q)\) has structure \(E_q^2:\textrm{GL}_2(q)\).
	Let \(E\) be the normal elementary abelian subgroup of order~\(q^2\).
	Then \(Q_{\alpha}\) contains an element \(a\) in a complement of \(E\) in \(M\). Thus, \(|C_M(a)|=q^2\cdot \frac{q-1}{(3,q-1)}\). On the other hand, \(|C_G(a)=q^3\cdot \frac{q-1}{(3,q-1)}\)
	by Table~2 in~\cite{Simpson}.

As \(C_{G_{\alpha}}(a)\leq C_M(a)\), therefore, \(q\) divides \(|C_G(a):C_{G_{\alpha}}(a)|\) but this contradicts the fact that
	\(|C_G(a):N_{G_{\alpha}}(a)|\leq 4\).
	Now all possibilities where \(G_{\alpha}\) has even order are excluded.
	
	We deduce that
	\(|G_{\alpha}|\) is odd.
	
	Moreover, a prime dividing \(|G_{\alpha}|\)
	divides at least one of the numbers \(q\), \(q-\varepsilon\), \(q+\varepsilon\)
	and \(q^2+\varepsilon q+1\).
	
	First suppose that there exists an element \(x\) in \(G_{\alpha}\) of prime order \(r\) dividing \(q-\varepsilon\). If
	\(r\geq 5\), then \(G_{\alpha}\) contains a Sylow \(r\)-subgroup of $G$ by Lemma~\ref{Inv-lem:TB}\,(a).
	Then $G_\alpha$ contains an element \(z\) of order $r$ in the centre of a maximal subgroup \(U\) such that the full pre-image of
	$U$ in \(\textrm{SL}^{\varepsilon}_3(q)\) is isomorphic to \(\textrm{GL}^{\varepsilon}_2(q)\). 
	We keep this in mind and look at the case \(r=3\).

	Let \(P\) be a Sylow \(3\)-subgroup of \(G\) containing \(x\). Then \(|N_M(\langle x\rangle)|\) is divisible by \(9\) and by Table~2 in~\cite{Simpson} also by \(2\).
	This means that
	\(|N_M(\langle x\rangle)|\) is divisible by \(18\), moreover

	\(|N_G(\langle x\rangle):N_{G_{\alpha}}(\langle x\rangle)|\leq 4\). Since this index is not divisible by $6$, it follows that \(P \le G_{\alpha}\). Again \(G_{\alpha}\) contains an
	\(r\)-element \(u\) in the centre of a maximal subgroup \(U\), where the full pre-image of $U$ in \(\textrm{SL}^{\varepsilon}_3(q)\) is isomorphic to 
	\(\textrm{GL}^{\varepsilon}_2(q)\).
	
	Then \(U\leq C_G(u)\) and hence \(|C_G(u)|\) is divisible by \(\frac{q\cdot (q-\varepsilon)^2\cdot
		(q+\varepsilon)}{(3,q-\varepsilon)}\). As \(|C_G(u):C_{G_{\alpha}}(u)|\leq 4\) and
		\(\frac{q\cdot (q-\varepsilon)\cdot
		(q+\varepsilon)}{(3,q-\varepsilon)} \) is divisible by \(8\), this  contradicts the fact that \(|G_{\alpha}|\) is odd.

	Therefore \(|G_{\alpha}|\) and \(q-\varepsilon\) are coprime.
	
	Next suppose that there exists an element \(x\) in \(G_{\alpha}\) of prime order \(r\) such that $r$ divides \(q\). Let \(x\in R\in
	\textrm{Syl}_r(G)\). Then \(Z(R)\leq C_G(x)\) and consequently \(|Z(R):Z(R)\cap G_{\alpha}|\leq
	|C_{G}(x):C_{G_{\alpha}}(x)|\leq 4\). Therefore \(|Z(R)|=q\),
	and it follows that \(Z(R)\cap G_{\alpha}\) contains a non-trivial element \(z\). Now \(|C_{G}(z)|=q^3\cdot
	\frac{q-\varepsilon}{(3,q-\varepsilon)}\).

	As \(\frac{q-\varepsilon}{(3,q-\varepsilon)}\geq \frac{10}{3}\), point stabilizer have order coprime to \(q-\varepsilon\)
	and \(|C_G(z):C_{G_{\alpha}}(z)|\leq 4\), it follows that \(q-\varepsilon=12\). 
	If \(G=\textrm{PSL}_3(13)\), then \(|N_G(\langle z \rangle)|\) is divisible by \(16\), which gives a contradiction to the fact that \(|G_{\alpha}\) is odd.
	Therefore \(G=\textrm{PSU}_3(11)\). But then \(|N_G(\langle z \rangle)|\) is divisible by \(8\), which is also impossible. 

	
	Hence \(|G_{\alpha}|\) is coprime to \(q\cdot
	(q-\varepsilon)\).
	
	Now suppose that there exists an element \(x\) in \(G_{\alpha}\) of prime order \(r\) dividing \(q+\varepsilon\). If
	\(r\geq 5\), then \(G_{\alpha}\) contains a Sylow \(r\)-subgroup, because \(q\), \(q+\varepsilon\) and \(q^2+\varepsilon
	q +1\) are not divisible by \(r\).
	In particular there exists a maximal subgroup $U$ of order divisible by $r$ such that 
	the full pre-image of $U$ in
	\(\textrm{SL}^{\varepsilon}_3(q)\) is isomorphic to  \(\textrm{GL}^{\varepsilon}_2(q)\). If \(r=3\), then
	(10-1) in~\cite{GL} shows that \(G\) has cyclic Sylow \(3\)-subgroups, and in particular all subgroups of order \(3\) are conjugate in
	\(G\). 
	In both cases there exists an element $b$ of prime order $r$ in a maximal subgroup \(U\) such that the full pre-image of $U$ in \(\textrm{SL}^{\varepsilon}_3(q)\)
	is isomorphic to \(\textrm{GL}^{\varepsilon}_2(q)\).
	Then \(|C_{G}(b)|\) is divisible by \((q-\varepsilon)\), \(|G_{\alpha}|\) is coprime to \(q-\varepsilon\) and  \(|C_G(b):C_{G_{\alpha}}(b)|\leq 4\). Together all this implies that
	 \(q-\varepsilon=12\).
	If \(G=\textrm{PSL}_3(13)\), then \(r=7\) and \(|N_G(\langle x \rangle)|\) is divisible by \(16\), which gives a contradiction to the fact that \(|G_{\alpha}\) is odd.
	So \(G=\textrm{PSU}_3(11)\), but then \(r=5\) and \(|N_G(\langle x \rangle)|\) is divisible by \(16\). This is impossible.
	We deduce that
	\(|G_{\alpha}|\) and \(q\cdot (q^2-\varepsilon)\cdot (q+\varepsilon)\) are coprime. In particular, since one of the numbers \(q\), \(q-\varepsilon\) and
	\(q+\varepsilon\) is divisible by~\(3\), it follows now that \(|G_{\alpha}|\) is coprime to \(6\).
	
	Suppose that \(x\in G_{\alpha}\) has prime order $r$ and fixes exactly four points.
	Suppose further that \(r\) divides \(q^2+\varepsilon
	q+1\) and let \(x \in R\in \textrm{Syl}_r(G)\).
	By Theorem~6.5.3 in~\cite{GLS3} we know that \(N_G(R)\)	lies in a Frobenius group \(H\) that is a maximal subgroup of \(G\) and has order \(3\cdot \frac{q^2+\varepsilon q+1}{(3,q-\varepsilon)}\), with Frobenius kernel~\(K\) of order
	\(\frac{q^2+\varepsilon q+1}{(3,q-\varepsilon)}\). As \(r \ge 5\), it follows that \(R \le K\). Moreover, $K$ is nilpotent (see for example Theorem~10.3.1
	in~\cite{Gorenstein1980}), so we even have that
	\(R\) is characteristic in \(K\) and \(N_{G}(R)=F\). This means that 
	\(G_{\alpha}\leq K\).
	Then Lemma~\ref{numberfpt} gives that  $4=|\textrm{fix}_{\Omega}(x)|$ is
	divisible by \(\frac{|N_{G}(\langle x \rangle)|}{|G_{\alpha}|}\). 
	We recall that \(|G_{\alpha}|\) is not divisible by \(3\), but \(|N_{G}(\langle x \rangle)|=|F|\) is, and therefore the number of fixed points of \(x\) is divisible by \(3\). This is false.  
\end{proof}







\section{The case where some involution fixes four points and G has sectional 2-rank at most 4}


This section treats the situation from Theorem \ref{3and2}(e) in the special case where $G$ is simple.
Then the main result in \cite{GoHa1974} applies and we see that the alternating groups, except for $\Alt_6$, and also all sporadic groups except for the Lyons group cannot occur, by Remark 4.1. 
Some specific groups have already been treated, see Table \ref{SmallGroupsTable}, as have the series
$\textrm{PSL}_2(q), \textrm{Sz}(q), \textrm{PSL}_3(q)$ and $\textrm{PSU}_3(q)$
in general.

So it remains to consider the series $\textrm{PSp}_4(q)$, G$_2(q)$, $^2$G$_2(q)$,
$^3$D$_4(q), \textrm{PSL}_4(q), \textrm{PSU}_4(q), \textrm{PSL}_5(q)$ or $\textrm{PSU}_5(q)$ (where $q$ is 
an odd prime power, with further restrictions), 
and the sporadic group Ly.



\begin{lemma}\label{Inv-lem:TB2}
Suppose that Hypothesis \ref{4fix} holds, let
$\alpha\in\Omega$ and suppose that $t\in G_\alpha$ is an involution with exactly four fixed points on $\Omega$.
Then $\abs\Omega$ is even, in particular $G_\alpha$ does not contain a Sylow $2$-subgroup of $G$, and all $2$-elements of $G_\alpha$ have an even number of fixed points on $\Omega$.

If $t$ is $2$-central, then $\abs{C_G(t):C_{G_\alpha}(t)}\in\{2,4\}$.
\end{lemma}

\begin{proof}
The first statement follows from the fact that $t$ fixes exactly four points, that $G$ acts with fixity 4 and that, therefore, all 
remaining orbits of $t$ on $\Omega$ must have length 2.
This implies that no point stabilizer contains a Sylow $2$-subgroup of $G$, and it also implies that all $2$-elements have an even number of fixed points.

Finally, we suppose that $t$ is $2$-central and we let $T \in \textrm{Syl}_2(G)$ be such that $t \in Z(T)$. 

Now $\abs{C_G(t):C_{G_\alpha}(t)} \le 4$ by
Lemma \ref{Inv-lem:TB}~(c) and $T \le C_G(t)$.
So if this index is 1 or 3, then $T \le G_\alpha$, contrary to the first statement.
\end{proof}









\begin{lemma}\label{Inv-lem:2G2}
Suppose that 
$q$ is a power of $3$, $q \ge 27$, and that $G=\,^2G_2(q)$.
Then there is no set $\Omega$ such that $(G,\Omega)$ satisfies Hypothesis \ref{4fix} and some involution fixes exactly four points.
\end{lemma}


\begin{proof}
Assume otherwise and let $\alpha \in \Omega$ and $t \in G_\alpha$ be such that
$o(t)=2$ and $t$ fixes exactly four points.
Then Section 4.5.3 in \cite{Wilson} yields that $C_G(t)$ has structure $\langle t \rangle \times \textrm{PSL}_2(q)$, and we also note that $G$ only has one class of involutions.

Lemma \ref{Inv-lem:TB2} gives that
$|C_G(t) : C_{G_\alpha}(t)| \in \{2,4\}$. As $q \ge 27$, we know that $\textrm{PSL}_2(q)$ does not have any subgroups of
index 2 or 4. Moreover $t \in C_{G_\alpha}(t)$ and this forces $C_G(t) \le  G_\alpha$.
We compare orders:
$|C_G(t)| = q \cdot (q^2-1)$ and $|G:C| = q^2(q^2+q+1)$, so it follows that $|\Omega|$ is odd, contrary to 
 Lemma \ref{Inv-lem:TB2}.
\end{proof}


Now we need some more specific preparation before we can handle the remainig groups. 


\begin{hyp}\label{GoHaHyp}
In addition to Hypothesis \ref{4fix}, suppose that $G$
is a simple Lie type group that 
satisfies Theorem \ref{3and2}~(e).
Let $q$ be a power of $3$ and suppose that $G$ is defined over $\textrm{GF}(q)$ and that

	$G \notin \{\textrm{PSL}_2(q),\textrm{Sz}(q), \textrm{PSL}_3(q),\textrm{PSU}_3(q), \,^2G_2(q)\}$. 


As in parts of Remark \ref{Inv-rem:Ser}, we will now refer to notation from \cite{GLS3} again, specifically Definition 2.3.1 and Remark 2.3.3.
We let $\Sigma$ denote a root system for $G$ if $G$ is of untwisted type, with fundamental system 
$\Pi$, and we let $\tilde{\Sigma}$ and $\tilde{\Pi}$ be the corresponding systems if $G$ is of twisted type.
In the twisted case, we denote by $\hat{\Sigma}$ and $\hat{\Pi}$ the reduction modulo the 
equivalence relation explained in Definition 2.3.1 in \cite{GLS3}.
We suppose that $G$ has Lie rank at least 2, 
we denote the subsets of positive roots of $\Sigma$, $\tilde{\Sigma}$ and $\hat{\Sigma}$ by
$\Sigma^+$, $\tilde{\Sigma^+}$ and $\hat{\Sigma^+}$ and we let $S \in \textrm{Syl}_3(G)$.
For all $\hat{\alpha} \in \hat{\Sigma}^+$ we let $X_{\hat{\alpha}} \le S$ be the root subgroup 
that belongs to $\hat{\alpha}$ in $S$.
Finally, let $P \le S$ be a subgroup of index 3.
\end{hyp}



\begin{lemma}\label{rootsgp}
Suppose that Hypothesis \ref{GoHaHyp} holds and let $\hat{\alpha} \in \hat{\Sigma}^+$ be such that
$X_{\hat{\alpha}} \nleq P$. Then $\hat{\alpha} \in \hat{\Pi}$.
\end{lemma}



\begin{proof}
We assume otherwise and 
we use Theorem 2.3.7 in \cite{GLS3}. Then 
$S = \prod\limits_{\hat{\beta} \in \hat{\Sigma}^+} X_{\hat{\beta}}$.
Let $m \in \mathbb{N}$ and $\tilde{\Pi}=\{\tilde{\beta_1},...,\tilde{\beta_m}\}$. As $|S:P|=3$, we see that
for all $\hat{\beta} \in \hat{\Sigma} \setminus \{\hat{\alpha}\}$, the root subgroup $X_{\hat{\beta}}$ is contained in $P$.
By assumption $\hat{\alpha} \notin \hat{\Pi}$, whence
 $\tilde{\alpha} \notin \tilde{\Pi}$.
 In particular $X_{\hat{\beta_1}},..., X_{\hat{\beta_m}} \le P$ and there are 
 $i_1,...,i_m\in \mathbb{N}_0$ such that 
$\tilde{\alpha}=\sum\limits_{j=1}^mi_j \cdot \tilde{\beta_j}$.
Then $\hat{\alpha}=\hat{\sum\limits_{j=1}^mi_j \cdot \tilde{\beta_j}}$.

If $G$ has untwisted classical type, then we use Theorem 2.3.4(d) in \cite{GLS3}
and the
 Chevalley commutator formula from Theorem 1.12.1(b), also in \cite{GLS3}.
Then $X_{\hat{\alpha}} \le 
[X_{\hat{\beta_1}},..., X_{\hat{\beta_m}}] \le  \langle X_{\hat{\beta_1}},..., X_{\hat{\beta_m}}\rangle \le
P$, contrary to our assumption.

If $G=G_2(q)$, then 
$\hat{\Pi}=\tilde{\Pi}=\Pi$ has exactly two elements $\beta$ and $\gamma$, 
where $\beta$ is the long root and $\gamma$ is the short root.
Moreover 
$\hat{\Sigma}=\tilde{\Sigma}=\Sigma=\{\beta, -\beta, \gamma, -\gamma, \beta+\gamma, -\beta-\gamma,
\beta+2\gamma,-\beta-2\gamma, \beta+3\gamma,-\beta-3\gamma, 2\beta+3\gamma,-2\beta-3\gamma\}$
by Remark 1.8.8 in \cite{GLS3} and 
therefore $\hat{\alpha} \in \{\beta+\gamma,
\beta+2\gamma, \beta+3\gamma, 2\beta+3\gamma\}$.

Also $X_\alpha \le [X_\beta,X_\gamma] \le \langle X_\beta,X_\gamma\rangle \le P$
(using Theorems 2.4.5(a) and 1.12.1.(b) in \cite{GLS3}), contrary to our assumption. 

Finally, if $G=^3D_4(q)$, then again 
$\hat{\Pi}=\tilde{\Pi}$ consists of a long root $\beta$ and a short root $\gamma$, and the possibilities for $\hat{\alpha}$ are the same as in the previous case, with Remark 1.8.8 and Section 2.3 in \cite{GLS3}.
This leads to the same contradiction as above.

It is left to look at the twisted types, namely $\textrm{PSU}_4(q)$ and $\textrm{PSU}_5(q)$.
In both cases $\hat{\Pi}=\tilde{\Pi}$ consists of a long root $\beta$ and a short root $\gamma$, 
and we refer to Remark 1.8.8 and Section 2.3 in \cite{GLS3}.


First suppose that
$G=\textrm{PSU}_4(q)$. 
Then $\hat{\Sigma}=\tilde{\Sigma}=\Sigma=\{\beta, -\beta, \gamma, -\gamma, \beta+\gamma, -\beta-\gamma,
2\beta+\gamma,-2\beta-\gamma\}$
and hence 
$\hat{\alpha}=\tilde{\alpha} \in \{\beta+\gamma,
2\beta+\gamma\}$.
Using Theorem 2.4.5 in \cite{GLS3} once more, we see that $X_\alpha \le P$.

Then we argue in a similar way if $G=\textrm{PSU}_5(q)$. Here we have that 
$\tilde{\Sigma}=\Sigma=\{\beta, -\beta, \gamma, -\gamma, \beta+\gamma, -\beta-\gamma,
\beta+2\gamma,-\beta-2\gamma, 2\gamma, -2\gamma, 2\beta+2\gamma, -2\beta-2\gamma\}$
and 
$\hat{\Sigma}=\{\hat\beta, -\hat\beta, \hat\gamma, -\hat\gamma, \widehat{\beta+\gamma}, \widehat{-\beta-\gamma},
\widehat{2\beta+\gamma},\widehat{-2\beta-\gamma}\}$.
Then Table 2.4 and Theorem 2.4.1 in \cite{GLS3} yield that $\hat{\beta}$ has type II and $\hat{\gamma}$ has type IV, and 
$\hat{\alpha} \in \{\widehat{\beta+\gamma},
\widehat{2\beta+\gamma}\}$.
Theorem 2.4.5 in \cite{GLS3} gives that $X_\alpha \le P$, which is our final contradiction.
\end{proof}






\begin{lemma}\label{O3}
Suppose that Hypothesis \ref{GoHaHyp} holds.
Then there exists a maximal parabolic subgroup $R$ of $G$ such that $O_3(R) \le P$. 
\end{lemma}


\begin{proof} 
Lemma \ref{rootsgp} gives some $\hat{\alpha} \in \hat{\Pi}$ such that $X_{\hat{\alpha}} \nleq P$.
Since $|S:P|=3$, we know for all 
 $\hat{\beta} \in \hat{\Sigma} \setminus \{\hat{\alpha}\}$ that $X_{\hat{\beta}} \le P$.

In particular 
 $\hat{\Pi}\setminus\{\hat{\alpha}\} \neq \varnothing$. Using the notation from 
Definition 2.6.4
in \cite{GLS3} we let  $R := P_{\hat{\alpha}}$, which is a parabolic subgroup of $G$.
By Definition
2.6.6 in \cite{GLS3} it is maximal parabolic and $U_{\{\hat{\alpha}\}} = O_3(R)$, moreover $R=U \rtimes L$ with a Levi complement $L$. 
In particular $X_{\hat{\alpha}} \nleq U_{\{\hat{\alpha}\}}$ and then the fact that 
$|S : P| = 3$ forces $U_{\{\hat{\alpha}\}} \le P$, as stated.
\end{proof}



\begin{lemma}\label{GoHaq=3}
Suppose that Hypothesis \ref{GoHaHyp} holds and that $q=3$.
Then there is no involution in $G$ that fixes exactly four points.
\end{lemma}

\begin{proof}
The following cases can be treated with \texttt{GAP} (see Table \ref{SmallGroupsTable} and Appendix):
$\textrm{PSL}_4(3), G_2(3)$, and $\textrm{PSp}_4(3)$. None of these groups exhibit a fixity four action with some involution fixing four points.
As $q=3$ and $q \equiv 1$ modulo $4$ for the series $\textrm{PSU}_5(q)$, under our hypothesis,  the group $\textrm{PSU}_5(3)$ does not need to be considered here.
So we are left with $\textrm{PSL}_5(3)$ and $\,^3D_4(3)$.

\textbf{Case 1:} $G=\,^3D_4(3)$.

We recall that $G$ has only one class of involutions (Table 4.5.1 in \cite{GLS3}).
So if we assume that $\alpha \in \Omega$ and $t \in G_\alpha$ is an involution with exactly four fixed points, then 
Section 4.6.5 in \cite{Wilson} gives that $C_G(t)$ has structure $C_2(\textrm{PSL}_2(3^3) \times \textrm{PSL}_2(3)). C_2$.
Lemma \ref{Inv-lem:TB2} gives that $|G_\alpha|$ is divisible by $|\textrm{PSL}_2(3^3)| \cdot |\textrm{PSL}_2(3)|=2^4 \cdot 3^4 \cdot 7\cdot 13$.
Comparing this to $|G| = 2^6 \cdot 3^{12} \cdot 7^2 \cdot 13^2 \cdot 73$ forces $G_\alpha$ to have order divisible by 
$2^4 \cdot 3^4 \cdot 7^2 \cdot 13^2$, with Lemma \ref{Inv-lem:TB}.
But then $G$ does not have any maximal subgroup that could contain $G_\alpha$ (see for example Theorem 4.3 in \cite{Wilson}).


\textbf{Case 2:} $G=\textrm{PSL}_5(3)$.

This time $G$ has two classes of involutions (Table 4.5.1 in \cite{GLS3}), and the corresponding centralizer orders are 
$2^8 \cdot 3^4 \cdot
13$ and $2^9 \cdot 3^6 \cdot 5 \cdot 13$. 
We will also need that $|G|= 2^9 \cdot 3^{10} \cdot 5 \cdot 11^2 \cdot 13$. 
Assume for a contradiction that $\alpha \in \Omega$ and that $t \in G_\alpha$ is an involution with exactly four fixed points.
Then Lemma  \ref{Inv-lem:TB} yields that $3 \in \pi(G_\alpha)$.
Let $P \in \textrm{Syl}_3(G)$ be such that $P_\alpha \in \textrm{Syl}_3(G_\alpha)$.
A calculation in \texttt{GAP} shows that $P$ has class 4, in particular $P$ does not have maximal class and 
we deduce that one of the cases (d) or (e) of Lemma \ref{syl3} holds. In particular $|P:P_\alpha| \le 3$,
which forces $|G_\alpha|$ to be divisible by $2^6 \cdot 3^9 \cdot 13$.


If $|P : P_\alpha| = 3$, then Lemma \ref{O3} yields a maximal parabolic subgroup
$M$ of $G$ such that $O_3(M) \le P_\alpha$, and then, by Lemma \ref{Inv-lem:TB},
a subgroup of index at most $4$ of $M$ is contained in $G_\alpha$.
Using Remark 2.36 and Section 3.3.3 in \cite{Wilson} we can also narrow down the possibilities for 
$M$, the structure can be 
$C_3^4 \rtimes \textrm{GL}_4(3)$ (with index 112 in G) or $C_3^6 \rtimes (\textrm{SL}_2(3)\times \textrm{GL}_2(3))$ (with
index $10 \cdot 112$ in G). In both cases $M$ contains a Sylow $3$-subgroup of $G$.
Given that $G_\alpha$ has even index in $G$ (see Lemma \ref{Inv-lem:TB2}), it follows that 
$G_\alpha$ contains a Sylow $3$-subgroup of $G$.

Using Theorem
2.6.5 (d) in \cite{GLS3} and Lemma \ref{Inv-lem:TB} we deduce that 
$G_\alpha$ contains a subgroups of index at most 4 of each maximal parabolic subgroup of $G$ that contains $P$, and we know the possible types of subgroups because they are maximal subgroups (see for example Theorem 2.6.7 in \cite{GLS3})
and this information can be calculated with \texttt{GAP}.
This leads to a contradiction because $|G_\alpha|$ is too large.
\end{proof}





\begin{lemma}\label{index3}

Suppose that $q$ is a power of $3$, that $n \in \{2,3,4\}$ and that $H=\textrm{SL}_n(q)$ or $\textrm{SU}_n(q)$.
Then $H$ does not have a subgroup of index 3.
In particular, if Hypothesis \ref{GoHaHyp} holds and $M \le G$ is a maximal parabolic subgroup of $G$ with Levi complement $L$, then $L$ does not have a subgroup of index 3.
\end{lemma}



\begin{proof}
First assume that the first assertion is false and let $U \le H$ be a subgroup of index 3.
Hauptsatz I.6.2 in \cite{Huppert-I} yields $N \unlhd H$ such that $N  \le U  \lneq H$
and such that $H/N$ is isomorphic to a subgroup of $\Sym_3$.
As $H$ is quasi-simple, it follows that $N\le Z(K)$, which is impossible.

Now suppose that Hypothesis \ref{GoHaHyp} holds and that 
$M \le G$ is a maximal parabolic subgroup of $G$ with Levi complement $L$.
We assume for a contradiction that  $X \le L$ is a subgroup of index $3$.
First we consider the case where $n \in \{4,5\}$ and $G=\textrm{PSL}_n(q)$, and we let 
$\tilde L$ denote the full pre-image of $L$ in $\textrm{SL}_n(q)$. Then Table \ref{Inv-tab:Ser}
tells us that $\tilde L \cong \textrm{GL}_{n-1}(q)$ or
$\tilde L \cong \textrm{GL}_{n-2}(q) \times \textrm{SL}_2(q)$.
As $|L:X|=3$, we deduce in the first case that $X$ intersects the subgroup isomorphic to $\textrm{SL}_{n-1}(q)$
with index 1 or 3, contrary to the first assertion of the lemma. 
In the second case we argue in the same way for one of  the direct factors, 
because $X$ cannot intersect both of them in a subgroup of index 1.

Next we let 
$n \in \{4,5\}$ and $G=\textrm{PSU}_n(q)$, and we let 
$\tilde L$ denote the full pre-image of $L$ in $\textrm{SU}_n(q)$.
Then $\tilde L \cong \textrm{GL}_2(q^2)$ or $\tilde L \cong \textrm{SU}_{n-2}(q)\times C_{q^2-1}$ (see
Table \ref{Inv-tab:Ser} again). 
Again we deduce a contradiction using the first assertion of the lemma. 

If $G = \textrm{PSp}_4(q)$ or $G = G_2(q)$, then we can argue similarly 
with $L$ itself, there is no $\tilde L$ necessary.

Finally, if $G = ^3D_4(q)$, then
we have two possibilities for $L$.
The first case is $L \cong \textrm{GL}_2(q)/C_2$. As $|\textrm{GL}_2(q)/\textrm{SL}_2(q)| = q-1$ and $q \ge 9$, 
we know that $L$ has a subgroup $H \cong \textrm{SL}_2(q)$. 
Now $ X\cap H$ has index 3 in $H$, which contradicts the first assertion of the lemma.

The second case is $L \cong \textrm{GL}_1(q) \ast \textrm{SL}_2(q) \cong C_{q-1} \ast \textrm{SL}_2(q)$, where $\ast$ denotes a central product.
As $q-1$ and 3 are coprime, it follows once more that 
$\textrm{SL}_2(q)$ has a subgroup of index 3, which it does not have. \end{proof}









\begin{lemma}\label{Inv-lem:Ser}
In addition to Hypothesis \ref{4fix}, suppose that $G$
is a simple Lie type group that 
satisfies Theorem \ref{3and2}~(e).
Let $p$ be an odd prime and $n\in\mathbb{N}$ be such that
	$q:=p^n\ge 3$, that $G$ is defined over $\textrm{GF}(q)$ and that
	$G \notin \{\textrm{PSL}_2(q),\textrm{Sz}(q), \textrm{PSL}_3(q),\textrm{PSU}_3(q), ^2G_2(q)\}$. 

Moreover, let
$\alpha\in\Omega$ and let $t\in G_\alpha$ be an involution such that $\Delta:=\textrm{fix}_{\Omega}(t)$ has size $4$.

	Then $q \le 3$.
\end{lemma}


\begin{proof}
We assume otherwise and we set $C:=C_G(t)$.
Next we recall the possibilities from the introductory paragraph, including further details from the main result of \cite{GoHa1974}:

$\textrm{PSL}_4(q)$ ($q\not\equiv 1$ modulo 8), 

$\textrm{PSU}_4(q)$ ($q\not\equiv 7$ modulo 8), 

$\textrm{PSL}_5(q)$ ($q\equiv -1$ modulo 4), 

$\textrm{PSU}_5(q)$ ($q\equiv 1$ modulo 4),

$\textrm{PSp}_4(q)$, $G_2(q)$ (where $q=3^{2n+1}$) or $^3D_4(q)$.

Here we left out the groups that are excluded by our hypothesis.



\smallskip
\textbf{Claim 1:} $p \in \pi(G_\alpha)$.

\begin{proof}
Our hypotheses and Lemma \ref{Inv-lem:TB2} yield that $\abs{C:C_\alpha}\in\{1,2,4\}$.
We use Table 4.5.1 in \cite{GLS3}, where we see that $C$ has a section
isomorphic to a finite group of Lie type or a product of several such groups over $\textrm{GF}(q)$ and in dimension at least 2. So it follows that 
$p \in \pi(C)$. (For more details we refer to Table \ref{Inv-tab:Ser}.) As $p$ is odd, we see that $p$ divides $|C_\alpha|$ and hence $p \in (G_\alpha)$.
\end{proof}

\textbf{Claim 2:}  If $p\ge 5$, then 
$G_\alpha$ contains a Sylow $p$-subgroup of $G$ and a subgroup of index at most 4 of a maximal parabolic subgroup.

\begin{proof}
The first statement follows from Claim 1 and Lemma \ref{Inv-lem:TB}. If $P \in \textrm{Syl}_p(G_\alpha)$, then 
Theorem 2.6.5 (d) in \cite{GLS3} tells us that there is a subgroup $U \le P$ such that $N_G(U)$
is a maximal parabolic subgroup of $G$. Then 
$|N_G(U):N_{G_\alpha}(U)| \le 4$ by Lemma \ref{Inv-lem:TB}, which is the second statement. 
\end{proof}


\textbf{Claim 3:}  If $p=3$, then 
$G_\alpha$ contains a Sylow $p$-subgroup of $G$ and a subgroup of index 1,2 or 4 of a maximal parabolic subgroup of $G$.

\begin{proof}
 Using Claim 1 we let $P \in \textrm{Syl}_3(G)$ be such that $1 \neq P_\alpha \in \textrm{Syl}_3(G_\alpha)$.
Next we go through the cases of Lemma \ref{syl3} and we see that (a) does not hold. Moreover,
 $\abs{P}\ge q^3\ge 27>9$ by \cite{Wilson} (and by hypothesis) and $P$ does not have maximal class, as we can see in Table \ref{Inv-tab:Ser}. Therefore Lemma \ref{syl3} (d) or (e) holds, and (e) immediately implies our first statement.
 
So we consider (d), where $P$ has a unique orbit of length $3$ and all other orbits are regular. 
In particular $P_\alpha$ fixes exactly three points. We let $k\in\mathbb{N}$ be such that \linebreak
	$\abs\Omega=k\cdot\abs P+3=3\cdot\left(k\frac{\abs P}3+1\right)$. As $\abs P\ge 729$ by Table \ref{Inv-tab:Ser} and as $k\cdot\frac{\abs P}3$ is a multiple of 3, it follows that $k\cdot\frac{\abs P}3+1$ is not divisible by 3. 
Then 3 divides $\abs\Omega=3\cdot\left(k\frac{\abs P}3+1\right)$, but 9 does not. 
Moreover $\abs \Omega=\frac{\abs G}{\abs{G_\alpha}}$ and therefore $G_\alpha$ contains a subgroup of index 3 of a Sylow 3-subgroup of $G$. Then the choice of $P$ yields that $\abs{P:P_\alpha}=3$.
Lemma \ref{O3} yields a maximal parabolic subgroup $R$ of $G$ such that $U:=O_3(R) \le P_\alpha$.
Lemma \ref{Inv-lem:TB} gives that $G_\alpha$ contains a subgroup of $N_G(U)$ of index at most 4.
As $\abs{P:P_\alpha}=3$, it follows that $\abs{N_G(U):N_{G_\alpha}(U)}=3$.
Let $L$ denote a Levi complement in $R$ ($\le N_G(U)$), and we choose $L$ such that $|L_\alpha|$ is as large as possible.
Now $N_G(U)=U \cdot L$ and $U \cap L=1$ (see for example Theorem 2.6.5(g) in \cite{GLS3}).
As $U \le G_\alpha$, our choice of $L$ forces $|L:L_\alpha|=3$, contrary to Lemma \ref{index3}.

So we are left with Case (e) of Lemma \ref{syl3}, where our first assertion holds, and the second statement follows with 
Theorem 2.6.5(d) in \cite{GLS3} and Lemma \ref{Inv-lem:TB}.
\end{proof}

We use the claims 2 and 3 and we let 
$R\ug G$ be a maximal parabolic subgroup of $G$ such that 
$|R:R_\alpha|\le 4$ and in particular $G_\alpha$ contains a Sylow $p$-subgroup $P$ of $G$.
Then Table \ref{Inv-tab:Ser} gives that $G$ has at least two different conjugacy classes of maximal parabolic subgroups.
Using Sylow's Theorem we let $U_1,U_2 \le P$ be such that $N_G(U_1)$ and $N_G(U_2)$ are non-conjugate
maximal parabolic subgroups of $G$. By Theorem 2.6.7 in \cite{GLS3} they are maximal subgroups of $G$.
For both, their intersection with $G_\alpha$ has index at most 4 in $G_\alpha$, and this forces $G_\alpha=G$, which is a contradiction.
\end{proof}


We have now treated all Lie type groups that occur in Case (e) of Theorem \ref{3and2} and that have not been analyzed before.
Finally, we look at the Lyons group.

\begin{lemma}\label{Lyons}
Let $G$ be the sporadic Lyons group.
Then there is no set $\Omega$ such that $(G,\Omega)$ satisfies Hypothesis \ref{4fix} and some involution fixes four points.
\end{lemma}

\begin{proof}
Inspection of the character table of $G$ in the ATLAS \cite{Atlas} shows that $G$ has only one class of involutions,
in particular every involution fixes exactly four points. 
Let $\alpha \in \Omega$ be one of the fixed points of some involution $t \in G$.
Then $|C_G(t)|=2 \cdot |\Alt_{11}|$, so
Lemmas 5.3 and \ref{Inv-lem:TB}(a) imply that $G_\alpha$ contains a Sylow subgroup of $G$ 
for the primes $5, 7$ and $11$.
Now Lemma \ref{Inv-lem:TB}(c) yields that $|G_\alpha|$ is divisible by $2^6 \cdot 3^4 \cdot  5^6 \cdot 7 \cdot 11$
as well.
But then there is no maximal subgroup of $G$ that contains $G_\alpha$, which is a contradiction.
\end{proof}









\section{The case where point stabilizers have odd order divisible by 3}


Here we consider Case
(4) of Theorem~\ref{3and2simple}.
If $G$ is non-abelian simple, then the CFSG gives
a list of possible groups for the situations where 
$G$ has a strongly $3$-embedded subgroup or small Sylow $3$-subgroups of prescribed structure.
The simple groups of Lie type with small Lie rank that occur have already been discussed. 

 In this section we 

 work under the following further hypothesis:

\begin{hyp}\label{odd3}
Suppose that Hypothesis \ref{4fix} holds and that the order of the point stabilizers in $G$ is odd and divisible by 3. Suppose further that
	$G \notin \{\textrm{PSL}_2(q), \textrm{PSL}_3(q),\textrm{PSU}_3(q)\}$. \\
\end{hyp}


We start with a helpful technical lemma.



\begin{lemma}\label{oddpoint}
Suppose that Hypothesis \ref{odd3} holds for the simple group $G$ and
let $H$ be a non-trivial four point stabilizer. Then the 
following hold:
\begin{enumerate}
    \item[(a)] The point stabilizers in $G$ are soluble.
    \item[(b)] One of the following is true:
    \begin{enumerate}
        \item[(i)] $G$ contains a strongly $3$-embedded subgroup or
        \item[(ii)] $3 \in \pi(N_G(H))$.
    \end{enumerate}
\end{enumerate}
\end{lemma}
\begin{proof}
Assertion (a) holds by \cite{FT}, because the point stabilizers have odd order by hypothesis. 

For (b) we suppose that (ii) does not hold and we let $\alpha \in \Omega$ be such that $H \le G_\alpha$. Let $p \in \pi(H)$. As $3$ does not divide $|N_G(H)|$ and $H$ has odd order, we see that $p \geq 5$. We recall that $G_\alpha$ is soluble,
and that moreover $3,p \in \pi(G_\alpha)$. Application of  Lemma~\ref{helpFrattini}
yields that $p$ divides $|N_G(X)|$ for some non-trivial $3$-subgroup $X$ of $G_\alpha$ or that \(3\) divides \(|N_G(Y)|\) for some non-trivial \(p\)-subgroup \(Y\) of \(G_{\alpha}\).
In the latter case, Lemma~\ref{Inv-lem:TB}\,(a) implies that we may suppose that \(Y\leq H\). Therefore there is a non-trivial \(3\)-element in \(N_G(Y)\).
By Lemma~\ref{Inv-lem:TB}\,(b) this element lies in \(N_G(H)\), which contradicts our assumption. Hence $p$ divides $|N_G(X)|$ for some non-trivial $3$-subgroup $X$ of $G_\alpha$.
As $3$ does not divide $|N_G(H)|$, the action of every non-trivial $p$-element on $X$ is fixed point free.
We recall that $p \geq 5$ and then it follows that $|Z(X)| \ge 3^3$. 
Now by checking the cases in Theorem \ref{3and2simple} we derive that $G$ contains a strongly $3$-embedded subgroup, which is 
(i).
\end{proof}




Next we consider the case that $G$ contains a strongly 3-embedded 
subgroup.

\subsection{The strongly 3-embedded case}

\begin{lemma}\label{CFSGstr3emb}
    Let \(G\) be a finite simple group with a strongly \(3\)-embedded subgroup. Then \(G\) is one of the following groups:
    \begin{enumerate}
        \item \(\textrm{PSL}_2(q)\), where $q$ is a prime power, but not a power of $3$, 
        \item \(\textrm{PSL}_3(q)\), where $q = 4$ or $q$ is a prime power and $q \equiv -1 \mod{3}$,
        \item \(\textrm{PSU}_3(q)\), where $q$ is a $3$-power or $q$ is a prime power and  $q\equiv 1 \mod{3}$,
        \item \({}^2\textrm{G}_2(q)\), where $q>3$ is a power of $3$, 
        \item \(J_1\) or
        \item \(M_{11}\).
    \end{enumerate}
\end{lemma}

\begin{proof}
      As a consequence of the classification of finite simple groups together with (10-1) in~\cite{GL} and Theorem~3.3.3 in~\cite{GLS3}, the finite non-abelian simple groups with non-trivial cyclic Sylow \(3\)-subgroups are   \(\textrm{PSL}_2(q)\), where $q$ is not a $3$-power, \(\textrm{PSL}_3(q)\), where $q \equiv -1 \mod{3}$ and \(\textrm{PSU}_3(q)\), where $q \equiv 1 \mod{3}$.
Then the statement of the lemma follows from Theorem 7.6.1 in \cite{GLS3}.
\end{proof}


\begin{lemma}\label{2G2}
	
	Let \(n \in \mathbb{N}\), let $q:=3^{2n+1}$ and let \(G={}^2\textrm{G}_2(q)\).
If $\Omega$ is a set such that $(G, \Omega)$ satisfies Hypothesis \ref{odd3}, then
the point stabilizers are Frobenius groups of structure
\(q^{3} \rtimes C_{\frac{q-1}{2}}\).

\end{lemma}

\begin{proof}
Let $P$ be a Sylow $3$-subgroup of $G$. By part (2) of the Theorem in~\cite{Ward}, $P$ is of class $3$, \(Z(P)\) is elementary abelian of order \(q\) and
\(P'=\Phi(P)\) is an elementary abelian subgroup of order \(q^2\) containing \(Z(P)\).
Let \(\Omega\) be such that $(G, \Omega)$ satisfies Hypothesis \ref{odd3},  and 
let $\Delta$ be the union
of the $P$-orbits on $\Omega$ that are of size at most $3$.
As \(|P|=q^3\geq 3^9\) and \(|G_{\alpha}|\) is odd, case (a)\,(ii) of Theorem~\ref{3and2} holds.
Therefore $1 \leq |\Delta| \leq 4$.
As $P$ is of class $3$, it is not of maximal class, and Case (e) of Lemma~\ref{syl3} holds; that is  $P \leq G_\alpha$ for some 
\(\alpha \in \Delta\). 
Then the element-wise stabilizer  of $\Delta$ in $P$ is a subgroup $Q$
of index at most $3$ in $P$.

According to Theorem~4.2 in~\cite{Wilson}, $N_G(P) = P \rtimes C$ is
a maximal subgroup of $G$, where $C$ is a
cyclic group of order $q-1$.
As \(Z}%\newcommand{\Zen}{\textrm{Z}(P)\leq P'\), every subgroup of index at
most $3$ in $P$ contains $Z(P)$.
This shows that $Z(P) \leq Q$ and therefore \(\textrm{fix}_{\Omega}(Z(P))=\Delta\).

Further notice that $(q-1)/2$ is odd, as $q$ is an odd prime power of $3$.
Since \(C\leq N_G(Z(P))\) this shows that the subgroup of $U$ index $2$ in $C$ also fixes \(\Delta\)
element-wise, and $N_G(P)$ is the stabilizer of $\Delta$. This yields the
assertion. 

\end{proof}

Notice that the set $\Delta$ considered in the proof of Lemma~\ref{2G2} has size $2$ and that the subgroup $U \cong C_{(q-1)/2}$ is a non-trivial four point stabilizer.

\begin{lemma}\label{str3emb}
	
	Let \(G\) be a finite simple group with strongly \(3\)-embedded subgroup, suppose that
	$G$ satisfies Hypothesis \ref{odd3}, and let $\alpha \in \Omega$. Then 
	 one of the following is true:
	\begin{enumerate}
		\item \(G\cong J_1\) and \(G_{\alpha}\) is cyclic of order \(15\).
		\item \(G \cong {}^2\textrm{G}_2(q)\), where \(n\geq 1\), $q=3^{2n+1}$ and \(G_{\alpha}\) is a Frobenius group
		of structure \(q^{3} \rtimes C_{\frac{q-1}{2}}\).
	\end{enumerate}
\end{lemma}

\begin{proof}
Omitting the groups that are excluded in Hypothesis~\ref{odd3}, Lemma~\ref{CFSGstr3emb} shows that \(G\) is either \({}^2\textrm{G}_2(q)\), \(J_1\) or \(M_{11}\).
The specific groups can be found in Table \ref{SmallGroupsTable} and the rest follows from Lemma \ref{2G2}.
\end{proof}

As we discussed this case completely we assume in the remainder of the section the following.

\begin{hyp}\label{odd3simple}
Suppose that Hypothesis \ref{odd3} holds, that \(G\) is simple and that \(G\) does not contain a strongly \(3\)-embedded subgroup.
\end{hyp}

We notice that groups with Sylow \(3\)-subgroups of order \(3\) have cyclic Sylow \(3\)-subgroups and
hence a strongly \(3\)-embedded subgroup. Therefore they have already been dealt with in the previous results, and this means that
we have now covered (4)(a) and (b) from Theorem \ref{3and2simple}.
As a next step we show that the cases of extra-special Sylow $3$-subgroups of order $27$ of exponent $3$ or wreath products $C_3 \wr C_3$ cannot happen.

\subsection{Sylow 3-subgroups that are either extra-special of order 27 or wreath products of order 81}

\begin{lemma}\label{extraspecialWreath}
Let $G$ be a simple group such that a Sylow $3$-subgroup $P$ is extra-special  of order 27 and of exponent $3$ or
a wreath product $C_3 \wr C_3$. Then there does not exist a set $\Omega$ such that $(G,\Omega)$ satisfies Hypothesis \ref{odd3simple}.
\end{lemma}

\begin{proof} Assume otherwise and let $(G, \Omega)$ be such a pair.
Let $P$ be a Sylow $3$-subgroup of $G$ and let
$\Delta$ denote the union of the $P$-orbits of length at most $3$.
Our hypotheses imply that the cases (a), (b) and (d) of Lemma \ref{syl3} do not hold. 
If (e) holds, then Proposition \ref{3semireg} gives a contradiction to our hypotheses that point stabilizers have odd order and that $G$ does not have a strongly $3$-embedded subgroup.
The only remaining case is (c), which means that Hypothesis \ref{hyp4} is satisfied and that we can apply Proposition \ref{maxKlasse}. The only possibilities that are compatible with our hypotheses are (e) and (f). 
In Case (e) we have that $|\Delta| = 0$, $P$ is extra-special of order $27$ and $|G_\alpha|_3 = 3$ for some $\alpha \in \Omega\setminus{\Delta}$, and in Case (f) we have that
$|\Delta| = 3$, $P \cong C_3 \wr C_3$ and the Sylow $3$-subgroups of point stabilizers are elementary abelian of order $27$. 

In both cases, every non-trivial $3$-element of $G$ fixes exactly 0 or 3 points in $\Omega$.

Let $H$ be a non-trivial four point stabilizer and let $\alpha \in \textrm{fix}_{\Omega}(H)$. 
Then Hypothesis \ref{odd3} and the previous paragraph imply that $|H|$ is coprime to $6$. 
Let $p \in \pi(H)$ and $T \in \textrm{Syl}_p(H)$. We know that $N_G(H)$
is strongly $p$-embedded in $G$ by Lemma \ref{Inv-lem:TB}(h), and then \cite[Theorem 7.3.1]{GLS3} yields that either $T$ is cyclic or  $G$ is one of the groups listed in 
that theorem.  As $p > 3$, the only possibilities are $\Alt_9$, $\Alt_{10}$ or $\Alt_{11}$, where the Sylow $3$-structure is not as in our hypothesis, or
$|G|_3 > 3^4$. All this is not possible, and therefore $T$ is cyclic. This implies that $\textrm{Aut}(T)$ is abelian. We keep this prime $p \in \pi(H)$.

Next we make an observation that we will quote several times in the proof: 

$(\ast)$~~
If $Y\le   
G_\alpha$ is a $3$-group and $1 \neq U \le H$ is a $p$-group, then $C_U(y) =1$ for all $y \in Y^\#$ and 
$C_Y(u) =1$ for all $u \in U^\#$.


Assume otherwise and let $1 \neq u \in C_U(y)$. 
Then $u$ fixes the three fixed points of $y$ because $p \ge 5$, and conversely $y$ stabilizes the set $\textrm{fix}_{\Omega}(u)$ of size $4$. This is impossible because $y$ cannot have a fourth fixed point. 
The other statement holds for the same reason.

By hypothesis $G_\alpha$ has odd order, hence it is a soluble group and Lemma \ref{helpFrattini} applies. We use it for the primes $3$ and $p$, both in $\pi(G_\alpha)$, and we have two cases:

(1) There is a non-trivial $p$-subgroup $U$ of $G_\alpha$ that is normalized by a Sylow $3$-subgroup $Q$ of $G_\alpha$ or 

(2) there is a non-trivial $3$-subgroup $Y$ of $G_\alpha$ that is normalized by a Sylow $p$-subgroup $T$ of $G_\alpha$.

For both situations we keep in mind that $p \in \pi(H)$ and $p \ge 5$, so $H$ contains a Sylow $p$-subgroup of $G$ by Lemma \ref{Inv-lem:TB}(a), in particular of $G_\alpha$. This means that we can choose $U \le H$ in (1) and $T \le H$ in (2). 

We begin with Case (2), so $Y \le G_\alpha$ is a non-trivial $3$-group normalized by $T \in \textrm{Syl}_p(H)$. Without loss $Y \le P$.

Then $Y\ltimes T$ is a Frobenius group by ($\ast$). As $p \ge 5$, this forces $|Y| \ge 9$ and hence $|G_\alpha|_3 \ge 9$, which excludes Case (e) of Proposition \ref{maxKlasse}. 
We conclude that (f) holds, so $P \cong C_3 \wr C_3$, $|\Delta| = 3$, $p = 13$ and $H = T$. The action of $T$ on $Y$ also forces $Y \in \textrm{Syl}_3(G_\alpha)$, and then $|P:Y|=3$ and $P, T \le N_G(Y)$.
As $|\textrm{fix}_{\Omega}(Y)| = 3$, our hypothesis forces $N_G(Y)$ to have odd order. Another application of Lemma \ref{helpFrattini}, this time to the soluble group $N_G(Y)/Y$.
If some non-trivial $3$-subgroup of $N_G(Y)/Y$ normalizes a non-trivial $p$-subgroup, then we consider pre-images in $N_G(Y)$ and the fact that all Hall $\{3,p\}$-subgroups of $N_G(H)$ are conjugate. Together with ($\ast$) this gives a contradiction.

Therefore $T$ normalizes a Sylow $3$-subgroup of $N_G(Y)/Y$, so without loss it normalizes $P/Y$ and hence its full pre-image $P \le N_G(Y)$. 
In particular $T$ normalizes $Z(P)$, which has order $3$, and this contradicts ($\ast$) because $Z(P) \le G_\alpha$ by Lemma \ref{tiny}.

In Case (1) we have a non-trivial $p$-subgroup $U$ of $H$ that is normalized by a Sylow $3$-subgroup $Q$ of $G_\alpha$. Then $Q \le N_G(H)$ by Lemma \ref{Inv-lem:TB}(b), and ($\ast$) yields that
$H\rtimes Q$ is a Frobenius group. In particular $H$ is nilpotent. Therefore we may suppose that $U \in \textrm{Syl}_p(H)$, so without loss $U=T$.
As $T$ is cyclic, it follows that $N_H(T) = C_H(T) = H$ and our previous observation yields that $|Q|=3$, that $P$ is extra-special of order $27$ and that $|\Delta| = 0$.
Lemma \ref{Inv-lem:TB} (e) and (a) gives that $N_G(H)$ acts transitively
on $\textrm{fix}_{\Omega}(H)$. By hypothesis $G_\alpha$ has odd order, so $N_G(H)$ induces a transitive subgroup of $\Alt_4$. Moreover $3 \notin \pi(H)$, so we conclude that $N_G(H)$ contains 3-elements that induce 3-cycles on $\textrm{fix}_{\Omega}(H)$ and hence it
induces the group $\Alt_4$ on $\textrm{fix}_{\Omega}(H)$. Let $M \le N_G(H)$ be the full pre-image of $O_2(N_G(H)/H)$. Then $M \unlhd N_G(H)$ acts transitively on
$\textrm{fix}_{\Omega}(H)$, $T \in \textrm{Syl}_p(M)$ and a Frattini argument gives that
$N_G(H)=M \cdot N_G(T)$, together with Lemma \ref{Inv-lem:TB}(b).
In particular $N_G(T)/H$
contains a subgroup isomorphic to $\Alt_4$. 

Let $V \in \textrm{Syl}_2(N_G(T))$. As $Q \le N_G(T) \le N_G(H)$ and $N_G(H) = (H \cdot V)\cdot Q$ by the previous paragraph, we see that $N_G(T)=(N_H(T) \cdot V) \cdot Q=(H \cdot V)\cdot Q$, as $H$ is nilpotent.
Now $Q$ acts fixed-point-freely on $H \rtimes V$ by the structure of $\Alt_4$. 
The group $N_G(H)$ is a Frobenius group with Frobenius kernel $H \cdot V$, so $H \cdot V$ is nilpotent and hence 
$[H,V] = 1$.  We recall that $|Q|=3$, so  $N_G(T)=C_G(T) \rtimes Q = (H\cdot V) \rtimes Q$. 

Suppose that $X$ is a non-trivial $p$-subgroup of $G$. Without loss $X\le H$, because $H$ contains a Sylow $p$-subgroup of $G$, and then $N_G(X)\le N_G(H)=(H \times V) \rtimes Q$, which implies that $N_G(X) \cap C_G(V)$ has a normal $p$-complement. 

So if we consider $C:= C_G(V)$, then
for each non-trivial $p$-subgroup $X \le C$ we have that $N_C(X)$ has a normal $p$-complement, and then 
Burnside's $p$-Complement Theorem gives
that $C$ has a normal $p$-complement as well. Let $N:=O_{p'}(C)$ and assume for a contradiction that $3$ divides $|N|$. As $C=N \cdot T$, coprime action gives that $T$ normalizes a Sylow $3$-subgroup $Y$ of $N$. We also recall that $V$ does not centralize
a Sylow $3$-subgroup of $G$, and therefore $|Y| \le 9$ and $[T,Y]=1$. Then $Y$ stabilizes $\textrm{fix}_{\Omega}(T)$, which has size $4$, so $Y$ must fix a point, contrary to ($\ast$). We conclude that $N$ is  a $3'$-group.
We also note that $V$ stabilizes $\textrm{fix}_{\Omega}(T)$ and acts fixed point freely on it. Together with the fact that $N_G(T)/T$ has a subgroup isomorphic to $\Alt_4$, this implies that $V$ is a fours group.

Let $V \le S \in \textrm{Syl}_2(C)$. Then $S \le N$ because $C=N \rtimes T$, so the coprime action of $T$ on $N$ gives that we may choose $S$ to be normalized by $T$. 
Moreover, the factorization $N_G(T)=(H \times V) \rtimes Q$ shows that $Q$ normalizes $V$ and hence $Q \le N_G(C) \le N_G(N)$. In the previous paragraph we proved that $N$ is  a$3'$-group, and then the coprime action of $Q$ on $N$ yields that we may choose 
$S$ to be $Q$-invariant as well, so that $Q$ normalizes $S \rtimes T$.
We recall that $|\textrm{fix}_{\Omega}(Q)|=3$. If $s \in S^\#$ is centralized by $Q$, then 
$s$ stabilizes $\textrm{fix}_{\Omega}(Q)$ and hence it must fix a point, contrary to our hypothesis that point stabilizers have odd order. Together with ($\ast$)
this shows that $(S \cdot T)\rtimes Q$ is a Frobenius group with Frobenius kernel $S \cdot T$, and in particular $[S,T]=1$. 
We conclude that $S \le N_G(T)=N_G(T) = (H \times V)\rtimes Q$ and hence $S=V$.
As $V$ stabilizes $\textrm{fix}_{\Omega}(H)$, a set of size $4$, this yields two cases: $V \in \textrm{Syl}_2(G)$ or $N_G(V)$ contains a subgroup $D \cong D_8$. In the second cases the point stabilizers have even order, which is false. Now the main result in \cite{GW} gives that 
$G\cong\textrm{PSL}_2(q)$ for some prime power $q$ or $G \cong \Alt_7$.
Then a Sylow $3$-subgroup of $G$ is cyclic or elementary abelian, contrary to our hypothesis.

\end{proof}

\subsection{Elementary abelian Sylow 3-subgroups of order 9}

Applying the classification of finite simple groups we get the following result (see Proposition~(1.2) in~\cite{Koshitanielab}).

\begin{lemma}\label{CFSGelab}
    Let \(G\) be a finite simple group with elementary abelian Sylow \(3\)-subgroups of order \(9\). Then \(G\) is one of the following groups:
    \begin{enumerate}
        \item \(\Alt_6, \Alt_7, M_{11}, M_{22}, M_{23}\) or \(HS\),
        \item \(\textrm{PSL}^\varepsilon_3(q)\), where \(q\) is congruent to \(3+\varepsilon\) or \(6+\varepsilon\) modulo \(9\),
			except \(\textrm{PSL}^{-1}_3(2)\cong \textrm{PSU}_3(2)\),
			\item \(\textrm{PSL}^\varepsilon_4(q)\), where \(q\) is congruent to \(3-\varepsilon\) or \(6-\varepsilon\) modulo \(9\),
			\item \(\textrm{PSL}^\varepsilon_5(q)\), where \(q\) is congruent to \(3-\varepsilon\) or \(6-\varepsilon\) modulo \(9\), or
			\item \(\textrm{PSp}_4(q)\), where \(q\) is congruent to \(2,4,5\) or \(7\) modulo \(9\), except \(\textrm{PSp}_4(2)\).
    \end{enumerate}
\end{lemma}

As $G\cong M_{11}$ has a strongly $3$-embedded subgroup, we already discussed this group.
Now we present some easy observation which helps to determine the 
groups satisfying Hypothesis~\ref{odd3}.


 
 \begin{lemma}\label{helpSL2xSL2sub}
	Suppose that Hypothesis \ref{odd3simple} holds and that \(G\) has elementary abelian 
	Sylow \(3\)-subgroups  of order \(9\).

	If $|N_G(X)|$ has even order for every subgroup $X$ of $G$ of order $3$, then every point stabilizer of $G$ contains a Sylow $3$-subgroup of $G$
	and there exists a non-trivial four point stabilizer \(H\)  such that \(N_G(H)\) contains a Sylow \(3\)-subgroup of \(G\).
\end{lemma}
\begin{proof}
Let \(H\) be a non-trivial four point stabilizer in $G$ and let \(\Delta:=\textrm{fix}_{\Omega}(H)\).
 Then \(3\) divides \(|N_G(H)|\) by Lemma~\ref{oddpoint}, as and \(G\) does not contain a strongly \(3\)-embedded subgroup.
 
 Suppose that $|N_G(X)|$ has even order for every subgroup $X$ of $G$ of order $3$. Let $X \leq N_G(H)$ be a subgroup of order $3$ and let $\alpha \in \Delta$. 

 As $G_\alpha$ has odd order, every non-trivial \(2\)-element in $N_G(X)$ acts fixed-point-freely on \(\textrm{fix}_{\Omega}(X)\).
 It follows
 that $|\textrm{fix}_{\Omega}(X)| \in \{2,4\}$. Assume that $|\textrm{fix}_{\Omega}(X)| = 2$, and let 
 $P$ be a Sylow $3$-subgroup of $G$ such that $X \le P$.

 As \(P\) is abelian,
the subgroup $P$ acts on $\textrm{fix}_{\Omega}(X)$ and hence fixes its two elements. Therefore
 $|\textrm{fix}_{\Omega}(P)| = 2$,
 and every element in \(P^\#\) fixes exactly two elements in $\Omega$.
 This implies that  $|H|$ is coprime to $6$.
 Without loss we may assume $X\le G_\alpha$, which implies $\alpha \in \textrm{fix}_{\Omega}(P)$ and 
 $P \leq G_\alpha$. In particular, if we let $p \in \pi(H)$, then \(p\geq 5\).
 Lemma~\ref{helpFrattini} applied to $G_\alpha$ gives two possibilities: 
 Some element \(h\) of $H$ of order $p$ normalizes
a non-trivial $3$-subgroup \(Y\) of $G_\alpha$ or $P$ normalizes a non-trivial $p$-subgroup $U$ of $G_\alpha$.
 In the first case $h \in C_G(Y)$ as \(p\geq 5\). Therefore $h$ acts on $\textrm{fix}_{\Omega}(Y)$
 yielding $\textrm{fix}_{\Omega}(Y) \subseteq \textrm{fix}_{\Omega}(h)$. Then $Y$ has to act trivially on the $2$-set
$\textrm{fix}_{\Omega}(Y)\setminus{\textrm{fix}_{\Omega}(Y)}$, which contradicts $|\textrm{fix}_{\Omega}(Y)| = 2$. 
Thus the second case holds true, and by Lemma~\ref{Inv-lem:TB}\,(a) we may suppose that \(U\leq H\). We conclude $P \leq N_G(U) \leq N_G(H)$ with Lemma~\ref{Inv-lem:TB}\,(b).

If $|\textrm{fix}_{\Omega}(X)| = 4$, then $P$ is contained in the element-wise stabilizer of
$\textrm{fix}_{\Omega}(X)$.


 \end{proof}
 
 \begin{lemma}\label{SL2xSL2sub}
		Let \(q
		\) be a prime power such that \(q \equiv 2,4,5\) or
	\(7\) modulo \(9\) and let \(G\) be a finite simple group such that $G$  has a subgroup $U = V/Z$, where $V \cong \textrm{SL}_2(q) \times \textrm{SL}_2(q)$ and $Z \leq Z(V)$,  and such that the Sylow $3$-subgroups of $G$ are elementary abelian of order $9$.
	Then there does not exist a set \(\Omega\) such that \((G, \Omega)\) satisfies Hypothesis \ref{odd3simple}.
\end{lemma}
\begin{proof}
Assume that \(\Omega\) is such that \((G, \Omega)\) satisfies Hypothesis \ref{odd3simple}.
Since $U$ does contain a Sylow $3$-subgroup of $G$, we
see that $N_G(X)$ has even order for every subgroup $X$ of $G$ of order $3$.
More precisely, for every subgroup of \(G\) of order \(3\) there is an element in \(G\) that inverts this subgroup.
Using Lemma~\ref{helpSL2xSL2sub} we find a non-trivial four point stabilizer \(H\) such that $N_G(H)$ contains a Sylow $3$-subgroup $P$ of $G$.

Assume that $P \not\leq H$. Then  $N_G(H)$ induces a subgroup of $\Sym(4)$ on $\Delta:= \textrm{fix}_{\Omega}(H)$, which is divisible by $3$. As $|G_\alpha|$ is odd, we get that
$\Alt(4)$ is induced.
Further, we get that $|P\cap H| = 3$. 
Due to Frattini's lemma $N_G(P \cap H) \leq N_G(H)$ contains a Sylow-$2$ subgroup of 
$N_G(H)$. Therefore 
\(N_G(P\cap H)\) has a subgroup isomorphic to $C_3 \times \Alt_4$, which is a $\{2,3\}$-Hall subgroup of $N_G(P \cap H)$.
However, this means that \(N_G(P\cap H)\) does not contain an element inverting \(P\cap H\), which gives a contradiction.


This shows that $P \leq H$ and that Case (e) of Lemma~\ref{syl3} holds. So Proposition~\ref{3semireg} yields that \(G\) contains a strongly \(3\)-embedded subgroup,
contradicting  Hypothesis \ref{odd3simple}.

\end{proof}




\begin{lemma}\label{odd3el9}
	Suppose that Hypothesis \ref{odd3simple} holds and that \(G\)
	has elementary abelian 
	Sylow \(3\)-subgroups of order \(9\). 
	Then \(G \cong \Alt_6\) and the point stabilizers are elementary abelian of order~\(9\).
\end{lemma}

\begin{proof}
We use Lemma~\ref{CFSGelab} to determine the possibilities for $G$.
	Table~\ref{SmallGroupsTable} shows that the candidates \(\Alt_7\), \(M_{11}\), \(M_{22}\), \(M_{23}\) and \(\textrm{HS}\) do not exhibit examples for Hypothesis \ref{odd3}.

\(\Alt_6\) is also included in Table \ref{SmallGroupsTable} with exactly the action that we describe in the lemma. 
So it remains to prove that the remaining finite simple groups with elementary abelian Sylow $3$-subgroups of order $9$ do not give rise to examples for Hypothesis \ref{odd3simple}.
	
	Let $\varepsilon \in \{1,-1\}$. If \(G =\textrm{PSL}^{\varepsilon}_3(q)\), then Hypothesis \ref{odd3} is not
	satisfied.
	
	So let \(G\) be one of the remaining groups: \(\textrm{PSL}^{\varepsilon}_4(q)\), where \(q\equiv
	3-\varepsilon\) or \(6-\varepsilon\) modulo \(9\), \(\textrm{PSL}^{\varepsilon}_5(q)\), where \(q\equiv 3-\varepsilon\) or \(6-\varepsilon\) modulo \(9\)
	or \(\textrm{PSp}_4(q)\), where \(q\equiv 2,4,5\) or \(7\) modulo \(9\).

	Then we consider Theorems~3.5\,(i) and 3.9\,(i) in~\cite{Wilson} and we see that \(\textrm{SL}_4(q)\), \(\textrm{SU}_4(q)\),  
	\(\textrm{SL}_5(q)\) and \(\textrm{SU}_5(q)\) all 
	have a subgroup \(V\) isomorphic to \(\textrm{SL}_2(q) \times \textrm{SL}_2(q) \cong  \textrm{SU}_2(q)\times \textrm{SU}_2(q)\).
	The group \(\textrm{Sp}_4(q)\) has a subgroup \(V \cong \textrm{Sp}_2(q) \times \textrm{Sp}_2(q) \cong \textrm{SL}_2(q) \times
	\textrm{SL}_2(q)\).
	In all these cases we see that Lemma \ref{SL2xSL2sub} is applicable, and

	it shows that
	none of these groups satisfy Hypothesis \ref{odd3simple}.

\end{proof}





\section{The case where point stabilizers have order coprime to 6}


\begin{hyp}\label{hyp6}
In addition to Hypothesis~\ref{4fix}, we let $\alpha \in \Omega$ and we suppose that $2$ and $3$ do not
divide $|G_\alpha|$. We fix four distinct points $\alpha, \beta, \gamma, \delta \in \Omega$, such that the element-wise stabilizer $H$ of $\Delta := \{\alpha, \beta, \gamma, \delta\}$ is a non-trivial subgroup of $G$.
Finally, we let $p \in \pi(H)$ and $P \in \textrm{Syl}_p(H)$.
\end{hyp}

We note that Hypothesis \ref{hyp6} implies that $|\Omega| \ge 6$, because in $\Sym_5$ the point stabilizers are $\{2,3\}$-groups.

We begin by collecting the information that we have already generated in earlier sections.
For this we recall that, for all $n \in \mathbb{N}$, $G$ acts as a $(0,n)$-group on $\Omega$ if and only if all elements of $G^\#$ have zero or $n$ fixed points on $\Omega$ (following \cite{PS1}).

\begin{lemma} \label{start}
Suppose that Hypothesis~\ref{hyp6} holds. Then the following are true:
\begin{enumerate}
\item[(a)] $|N_G(H):H| = 4$ and $N_G(H)$ is transitive on $\Delta$,
\item[(b)] $H$ is a TI-subgroup,
\item[(c)] $G$ acts as a $(0,4)$-group on the set of right cosets of $H$,
\item[(d)] either $G_\alpha = H$ or  $G_\alpha$ is a Frobenius group with complement $H$.
\item[(e)] If  \(1 < X\le H\), then  \(|N_{G}(X):N_H(X)|\leq 4\).\\
In particular, \(P \in \textrm{Syl}_p(G)\)  and therefore 
 $(|H|,|\Omega|) = 1$.
 \item[(f)] If $1 < R \le H$, then $|N_G(R)|$ is not divisible by $3$ or $8$.
\end{enumerate}
\end{lemma}


\begin{proof}
As $H$ has odd order, Lemma \ref{Inv-lem:TB} yields that $|N_G(H):H| = 4$. Moreover, $H \neq 1$, so there is a prime $p \ge 5$ in $\pi(H)$ and $H$ contains a Sylow $p$-subgroup of $G$ by Lemma \ref{Inv-lem:TB}(a). Now Part (e) of the same lemma yields that $N_G(H)$ acts transitively on $\Delta$. Together this is (a).
Part (b) is exactly Lemma \ref{Inv-lem:TB}(f).
Lemma 1.3 combined with Lemma 1.1 of \cite{PS1} proves (c).
For (d) suppose that \(G_\alpha \neq H\). Assume there exists \(g\in G_{\alpha}\setminus H\) such that \(H\cap H^g\neq 1\).
By part (b), \(g\in N_G(H)\). Therefore \(g\) stabilizes \(\Delta\setminus {\alpha}\), but does not fix all of its points. Consequently \(g\) has order divisible by \(2\) or \(3\), contrary to our hypothesis. This proves (d).
Now we turn to (e)
and we let $1 \neq X \le H$.
	First we notice that $X$ fixes every point in $\Delta$, but not any more points because of our global fixity 4 hypothesis.
		Thus $N_G(X)$ stabilizes the set $\textrm{fix}_{\Omega}(X)=\Delta$. Now we recall that  
		\(|N_{G}(X):N_{G_{\alpha}}(X)|\leq 4\) by Lemma \ref{Inv-lem:TB}(a) and we let  
	 \(y\in N_{G_{\alpha}}(X)\). 
	 
	 Assume that \(y\) does not fix all
	four points in \(\Delta\). Then it acts as a transposition or as a $3$-cycle on $\{\beta, \gamma, \delta\}$.
	But $o(y)$ has order coprime to $6$, so this is not possible.
	Thus \(y\in H\). We conclude that
	\(|N_{G}(X):N_H(X)|=|N_G(X):N_{G_{\alpha}}(X)|\cdot |N_{G_{\alpha}}(X):N_{H}(X)|\leq 4\), which is the first assertion.  
	The second statement follows because, if $p \in \pi(H)$, then $p \ge 5$. 
	For (f) we use (e) and the fact that $|H|$ has order coprime to $6$ by hypothesis.
\end{proof}


We remark that
the simple group $M_{11}$ is a $(0,4)$-group in its action on the set of cosets of a Sylow $5$-subgroup.
This action is imprimitive, with blocks of size $11$ and with a block stabilizer $F_{55}$ (a Frobenius
group of order $55$). The action of $M_{11}$ on the set of cosets of $F_{55}$ is an example of an
action satisfying Hypothesis \ref{4fix}, and here $G_\alpha$ is a Frobenius group of order coprime to $6$.




\begin{thm}\label{strongcases}
Suppose that Hypothesis \ref{hyp6} holds. If $G$ is simple, then one of the following holds:

(i) $G$ has cyclic Sylow $p$-subgroups, or

(ii) there is some $n \in \mathbb{N}$ such that $G$ is isomorphic to $\textrm{PSL}_2(p^n)$ or $\textrm{PSU}_3(p^n)$
and $N_G(H)$ is a Borel subgroup.
\end{thm}

\begin{proof}
$N_G(H)$ is strongly
$p$-embedded in $G$ by Lemma \ref{Inv-lem:TB}(h), and then
Proposition 17.11 in \cite{GLS2} yields that
$N_G(H)$ is also strongly $p$-embedded as defined in \cite{GLS2}. Therefore
Theorems 7.6.1 and 7.6.2 in \cite{GLS3} apply. 
If $G$ has cyclic Sylow $p$-subgroups, then (i) holds, so it is left to go through the cases of 
Theorem 7.6.1.
Case (a) of the theorem leads to our Case (ii).

Suppose that Case (b) of the theorem holds, which means that
 $p \ge 5$, $G \cong \A_{2p}$ and $N_G(H)$ is the stabilizer of a partition of $\Omega$ into two subsets of size $p$.
Then we see that $N_G(H)$ contains elements of order $3$, contrary to our hypothesis and Lemma \ref{start}~(a).
 
The cases (c) and (d) do not occur because $p \ge 5$ by Hypothesis \ref{hyp6}.
In Case (e) we have that  $p=5$, $G \cong \,^2 F_4(2)'$ and $N_G(H)$ is the normalizer of a Sylow $5$-subgroup, containing a subgroup of structure $C_4 \ast \textrm{SL}_2(3)$. This is impossible by Lemma \ref{start}~(a).

In Case (f) we have $p=5$ again, but this time $G \cong$ Mc
 and $N_G(H)$ is the normalizer of a Sylow $5$-subgroup, containing a subgroup of structure $C_3 : C_8$. 
In Case (g) we have that $p=5$ again, and  $G \cong$ Fi$_{22}$, $N_G(H)\cong \textrm{Aut}(D_4(2))$.
The last case is $p=11$, $G \cong $J$_4$ and $N_G(H)$ is the normalizer of a Sylow $11$-subgroup, containing a subgroup of structure $C_5 \times \textrm{GL}_2(3)$.

These three cases have in common that they do not occur because of Lemma \ref{start}~(a).
So this concludes the proof.
\end{proof}




The remainder of this section is devoted to the analysis of all finite simple groups
from the perspective of a possible fixity 4 action satisfying Hypothesis \ref{hyp6}. Further we assume that every Sylow 
$p$-subgroup of $H$ is cyclic.
We begin with the alternating groups and the sporadic groups, and then the main work will occur for the groups of Lie type.




\begin{lemma}\label{Altcyclic}
Suppose that Hypothesis \ref{hyp6} holds, that $n \in \mathbb{N}$, $n \ge 5$ and $G=\A_n$. Let $p \in \pi(H)$ and suppose that $P\in \textrm{Syl}_p(G)$ is cyclic.
Then $G=\A_{7}$ and $H=G_\alpha$ has order $5$.
\end{lemma}

\begin{proof}
For the $p$-rank of $G$, we refer to Proposition 5.2.10 in\cite{GLS3}.
Thus $1=r(P)=\left[\frac{n}{p}\right]$, which forces $n <2p$.
Let $h \in H$ be a $p$-cycle.
If $n-p \ge 3$, then $C_G(h)$ contains a $3$-cycle, which contradicts our hypothesis because
$C_G(h)$ stabilizes $\textrm{fix}_{\Omega}(h) = \Delta$. 
Hence $n-p \in \{0, 1,2\}$.

So we have that $n \in \{ p, p+1, p+2\}$, and in each case  $N_G(\langle h \rangle)/\langle h \rangle$ is
cyclic of order $(p-1)$ or $(p-1)/2$. If $a$ divides $(p-1)/2$,
then $a < n/2$ in all three cases. This shows that $(p-1)/2$ is a
$2$-power, and an application of Lemma~\ref{start} (f) yields 
that
$(p-1)/2 \in \{1,2,4\}$. The fact that $p \geq 5$ gives that $p = 5$ and $n \in \{ 5,6,7\}$. Then Lemma~\ref{start} (a)  forces $n = 7$. 
\end{proof}








\begin{lemma}\label{spor}
Suppose that $G$ is a sporadic group satisfying Hypothesis \ref{hyp6}.
Then $G=\textrm{M}_{11}$ or $G=\textrm{M}_{22}$ and in both cases, two examples are possible. The point stabilizers are cyclic of order $5$ or Frobenius groups of order $55$, respectively.
\end{lemma}


\begin{proof}
By Remark~\ref{rem4.1} the examples that we state are indeed examples where Hypothesis \ref{hyp6} is satisfied.
So now we prove that these are the only ones, using the notation from Hypothesis \ref{hyp6}. Theorem \ref{strongcases} yields that
$G$ has cyclic Sylow $p$-subgroups for all $p \in \pi(H)$. 
So we let $p \in \pi(H)$ and $P\in \textrm{Syl}_p(G)$, and we let $X \le P$ be a subgroup of order $p$. We know that $P$ is cyclic and that $|N_G(X)|$ is not divisible by $6$,  $9$ or $8$, using Lemma \ref{Inv-lem:TB}~(c).
This restricts the possibilities for $p$, as we can see in Tables 5.3a--z in \cite{GLS3}.

\smallskip
The Mathieu groups:

If $p=5$ or $p=11$, then this leads to the possibilities stated for $\textrm{M}_{11}$ and $\textrm{M}_{22}$.
Otherwise the tables show that the restrictions on $|N_G(X)|$ are not compatible with the
 subgroup structure of $G$.

\smallskip
The Janko groups:

Here the tables in \cite{GLS3} show that, again, the restrictions on $|N_G(X)|$ are not satisfied.

\smallskip
The remaining groups:
Inspection of the corresponding tables
shows that, with several applications of Lemma \ref{Inv-lem:TB} (a), eventually $2 \in \pi(H)$ in each case. This is impossible.
\end{proof}





\begin{lemma}\label{Lie1}
Suppose that $G$ is a simple group of Lie type satisfying Hypothesis \ref{hyp6}, in characteristic~$p$.
If $G$ has cyclic Sylow $p$-subgroups, then 
$G=\textrm{PSL}_2(p)$ and $|G_\alpha|=p \cdot \frac{p-1}{4}$.
\end{lemma}


\begin{proof}
We inspect Theorem 3.3.3 in \cite{GLS3}, and hence Table 3.3.1 there, for the equicharacteristic rank.
First we note that $p \ge 5$ and therefore we do not have to consider the exceptions $B_2(2)$, $G_2(2)$, $^2G_2(3)$
and $^2F_4(2)$.
Let $a \in \mathbb{N}$ be such that $G$ is defined over a field of order $q:=p^a$.
If $a \ge 2$, then Table 3.3.1 shows for all types of groups that $r_p(G) \ge 2$, contrary to our hypothesis.
If $a=1$, then the table shows that all exceptional types do not occur, where for the Suzuki groups we use that $p\neq 2$.
For the classical types, we let $n \in \mathbb{N}$ and we begin with $\textrm{PSL}_n(p)$, where $n=2$ is the only possibility for cyclic Sylow $p$-subgroups.
With $\textrm{PSU}_n(p)$, we also have $p$-rank 1 only if $n=2$.
The odd-dimensional orthogonal groups do not give examples because the contribution of the dimension to the $p$-rank is already too large.
For the same reason, the symplectic groups do not occur.
Finally, if $G$ is an even-dimensional orthogonal group, then the fact that $p$ is odd gives that $r_p(G)>a=1$, so this is impossible.

As $\textrm{PSL}_2(p) \cong \textrm{PSU}_2(p)$, our assertion follows from Theorem \ref{L2q-main} and the fact that $|G_\alpha|$ is coprime to $6$.
\end{proof}



Before we enter a detailed analysis of the series of finite simple groups of Lie type,
we prove two preparatory lemmas.



\begin{lemma}\label{help}

(a) Let \(G\) be a finite group with a normal subgroup \(N\) of index \(m \in \mathbb{N}\). Let \(p\in
		\pi(N)\) and \(P\in \textrm{Syl}_p(N)\). Then \(|N_G(P)|=m\cdot |N_N(P)|\).


Now suppose that Hypothesis \ref{hyp6} holds and that $G$ is a simple group of classical Lie type over a field with $q$ elements,
where $q$ is a prime power coprime to $p$. Moreover, suppose that $G$ has cyclic Sylow $p$-subgroups and that $G \notin \{\textrm{PSL}_2(q),\textrm{Sz}(q), \textrm{PSL}_3(q),\textrm{PSU}_3(q)\}$. 
	
(b) $p$ does not divide $q^2-1$.

(c) $p$ divides $q^{p-1}-1$.

\end{lemma}		
		
\begin{proof}	
For (a) we use a Frattini argument. 
Then \(G=N\cdot N_{G}(P)\) and \(m\cdot |N|=|G|=|N\cdot N_{G}(P)|=\frac{|N|\cdot |N_{G}(P)|}{|N_G(P)\cap
		N|}=\frac{|N|\cdot |N_{G}(P)|}{|N_N(P)|}\), which implies the statement.
		
		(b) 	If \(p\) divides \(q-1=\Phi_1(q)\) or \(q+1=\Phi_2(q)\), then we use
	(10-1)~in~\cite{GL} and we deduce that $r_p(G) \ge 2$, contrary to 
	the hypothesis that $G$ has cyclic Sylow $p$-subgroups. 

(c) This is Fermat's Little Theorem.
\end{proof}



\begin{lemma}\label{Primesalternativ}
Let $G$ be a finite simple group of Lie type  satisfying Hypothesis~\ref{hyp6}, defined over a field with $q$ elements.
Further let $p \in \pi(H)$ be such that the Sylow $p$-subgroups of $G$
are cyclic.
Then the smallest $k$ for which a particular divisor of $|G|$, as given in the table, could be divisible by $p$ is as follows:\\

\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
$G$ & remarks & divisor of $|G|$ & k \\
\hline
 $\textrm{PSL}_n(q)$& $n \geq 4$ & $q^k-1$ & $n-1,n$\\ 
  \hline
 $\textrm{PSU}_n(q)$& $n \geq 4$  & $q^k-(-1)^k$ & $n-1,n$\\ 
  \hline  
 $\textrm{PSp}_{2n}(q)$ & $n \ge 2$ & $q^{2k}-1$& $n$ \\
 \hline
$P\Omega_{2n+1}(q)$ & $n \ge 3$ & $q^{2k}-1$ & $n$\\
\hline
$P\Omega^+_{2n}(q)$ & $n \ge 4$ & $q^{2k}-1$  & $n-1,n$ \\
\hline
$P\Omega^-_{2n}(q)$ & $n \ge 4$ & $q^{2k}-1$ & $n-1,n$\\
\hline
 $^3D_4(q)$ & & $\Phi_k(q)$ & $12$\\
 \hline
 $F_4(q)$ & & $\Phi_k(q)$ & $8,12$\\
 \hline
 $^2F_4(q)$ & & $\Phi_k(q)$ & $6,12$ \\
\hline 
 $G_2(q)$ & & $\Phi_k(q)$ & $3,6$\\
 \hline
 $^2G_2(q)$ & & $\Phi_k(q)$ & $1,2,6$ \\ 
 \hline
 $E_6(q)$ & & $\Phi_k(q)$ & $8,9,12$ \\
  \hline
   $^2E_6(q)$ &  & $\Phi_k(q)$ & $8,10,12,18$  \\
 \hline
 $E_7(q)$ & & $\Phi_k(q)$ & $7,9,12,14,18$ \\
  \hline
 $E_8(q)$ & & $\Phi_k(q)$ & $15,20,24,30$ \\
  \hline

\end{tabular}\label{ktablealternativ}

\end{center}
\end{lemma}
\smallskip






\begin{proof}
Let $p \in \pi(H)$.
First let $G$ be a classical group of Lie type defined over a vector space $(V, f)$ with bilinear form $f$. If $G = \textrm{PSL}_n(q)$, then 
$f \equiv 0$. Further let  $U$ be a $2$-dimensional subspace of $V$ and $W$ a complement to $U$ in $V$. If $G$ is of unitary or
symplectic type, then let $(U, f_{\mid U})$ be a non-degenerate $2$-dimensional subspace of unitary or symplectic type and let $W$ denote its orthogonal 
complement in $V$. Moreover, in all three cases let $K$ be the stabilizer of $U$ in $G$.
Then $K$ contains subgroups $K_U$ and $K_W$ such that, for $\{X,Y\} = \{U,W\}$, the subgroup  $K_X$ acts on the projective space $(P(X), f_{\mid X}$) as $\textrm{PSL}(X)$, $\textrm{PSU}(X)$ or  $\textrm{PSp}(X)$, respectively, and it acts trivially on $Y$ (see \cite[Propositions~4.1.3 and 4.1.4]{KL}).
 If $p \in \pi(K_W)$, then without loss a Sylow $p$-subgroup of $K_W$
 is contained in $H$. Therefore, $K_U \leq N_G(H)$, which implies that $|K_U|$ is not divisible by $3$ and that $|K_U|_2 \leq 4$. If $G$ is linear, unitary or symplectic, then $K_U$ induces $\textrm{PSL}_2(q)$ on $P(U)$. But the order of this group  is divisible by $3$, contrary to Lemma \ref{start}(f). It follows that $p \not\in \pi(K_W)$, which yields the assertion for those groups.

Now let $G = P\Omega_{2n+1}(q)$, $n \ge 3$, and 
let $(U, f_{\mid U})$ be a non-degenerate $2$-space of $-$-type. As above, let
$W$ be the orthogonal complement to $U$ in $V$ and $K$ the stabilizer of $U$ in $G$. Then $K$ contains a subgroup $K_W$ which acts as $P\Omega^{-}(q)$
or $\Omega^{-}(q)$ on $W$ (see \cite[Propositions~4.1.6]{KL}). If $p$ divides $|K_W|$ and if $S$ is a  Sylow $p$- subgroup of $K_W$, then without loss $S \le H$. We recall that $8$ does not divide $|N_G(H)|$ by Lemma \ref{start}(f) and that $K_W$
acts as $\Omega^{-}(q)$ on $W$. Then
$K \cong (\Omega_2^-(q) \times \Omega_{2n-1}^{-}(q)).2$
or $(\Omega_2^-(q) \times \Omega_{2n-1}^{-}(q)).[4]$
with notation as in \cite[Propositions~4.1.6]{KL}.
Recall that $\Omega_2^-(q)$ is cyclic of order $(q-1)/(q-1,2)$.
We assume for a contradiction that $p$ divides  $|\Omega_{2n-1}^-(q)|$.
Then Lemma~\ref{help} (b) yields that $(q+1)/(q-1,2) \in \{1,2\}$ and therefore
$q = 3$. In this case, as the discriminant of $\Omega_2^-(3)$ is a square,  it follows that $K \cong (\Omega_2^-(3) \times \Omega_{2n-1}^-(3)).[4]$. A Frattini argument gives that 
$N_K(H)$  is  divisible by $8$, which is a contradiction. Thus $p$ does not divide 
$|\Omega_{2n-1}^-(q)|$, and that gives the assertion for the first type of
 orthogonal groups.

If $G = O^+_{2n}(q)$, where $n \ge 4$, then we argue similarly. 
If $G = O^-_{2n}(q)$ and $n \ge 4$, then we choose $U$, $W$ and $K$ in the same way and consider that 
$K_W$ acts as $P\Omega^{+}$
or $\Omega^{+}$ on $W$ (see \cite[Propositions~4.1.6]{KL}). 
The remaining arguments stay the same.


Now let $G$ be a group of Lie type of exceptional type.


$G$ has cyclic Sylow $p$-subgroups by hypothesis and $p \ge 5$. Then we can use Theorem (10-1) and Table 10:2 in \cite{GL}, 
where $r_p(G)$ is the exponent of $\Phi_k(q)$.
This gives exactly the possibilities in our table.
\end{proof}





\subsection{Groups of linear and unitary type}


We begin with preparatory results about the subgroup structure.

\begin{lemma}\label{sgpPSL}
Let \(n \in \mathbb{N}\), let \(q\) be a prime power and let \(G:=\textrm{PSL}_n(q)\). Let \(p \in \pi(G)\) and let
\(k \in \mathbb{N}\) be minimal such \(p\) divides \(q^k-1\). If $k \ge 2$, then the following hold:

(a) If \(k=n-1\), then there exists a subgroup \(Y\le G\) of order \(p\) such that \(|N_G(Y)|\)
is divisible by \(\frac{(q^k-1)\cdot k}{(n,q-1)}=\frac{(q^{n-1}-1)\cdot (n-1)}{(n,q-1)} \).

(b) If	 \(k=n\), then
		there exists a subgroup \(Y\le G\) of order \(p\) such that \(|N_G(Y)|\)
		 is divisible by \(\frac{q^n-1}{q-1}\cdot \frac{n}{(n,q-1)}\).
	
\end{lemma}		
		
		 \begin{proof}
	Let \(Z:=Z(\textrm{SL}_n(q))\) and first suppose that
	 \(k=n-1\).
	We note that \(\textrm{SL}_n(q)\) contains a subgroup \(U\cong\textrm{GL}(k,q)\) and that
	$$|U|=q^{\frac{k(k-1)}{2}}\prod\limits_{i=1}^{k}(q^i-1).$$
	Therefore \(p\in \pi(U)\) and Satz~II\,7.3\,a)
	in~\cite{Huppert-I} tells us that \(U\) has a
	cyclic subgroup \(C\) of order \(q^k-1\) and that \(|N_U(C)/C|\) is cyclic of order \(k\).
	
	Let \(A\le C\) be a subgroup of order \(p\). Then \(A\) is characteristic in \(C\) and thus
	\(|N_U(A)|\) is divisible by \((q^k-1)\cdot k\).
	By hypothesis \(k\geq 2\), hence \(p\) does not divide \(q-1\) and it follows that
	\(AZ/Z\) has order \(p\) as well.
	Then \(|N_{G}(AZ/Z)|\) is divisible by \(\frac{(q^k-1)\cdot
		k}{(n,q-1)}\), which proves (a).
	
	For (b) we suppose that \(k=n\).
	By Satz~II\,7.3\,b) in~\cite{Huppert-I} there exists a cyclic group
	\(T\) of \(\textrm{SL}_n(q)\) of order \(\frac{q^n-1}{q-1}\), and moreover
	\(N_{\textrm{SL}_n(q)}(T)=N_{\textrm{GL}_n(q)}(T)\cap
	\textrm{SL}_n(q)\).
	
	As \(|N_{\textrm{GL}_n(q)}(T)\cap \textrm{SL}_n(q)|=\frac{|N_{\textrm{GL}_n(q)}(T)|\cdot |
		\textrm{SL}_n(q)|}{|N_{\textrm{GL}_n(q)}(T)\textrm{SL}_n(q)| }\) is divisible by \(\frac{|N_{\textrm{GL}_n(q)}(T)|\cdot |
		\textrm{SL}_n(q)|}{|\textrm{GL}_n(q) |}=\frac{(q^n-1)n\cdot  | \textrm{SL}_n(q)|}{(q-1)|\textrm{SL}_n(q) |}=\frac{q^n-1}{q-1}\cdot n\).
	Let \(A\le T\) be a subgroup of order \(p\). Then $A$ is characteristic in $T$ and therefore
	 \(N_{\textrm{SL}_n(q)}(T) \le N_{\textrm{SL}_n(q)}(A)\).
	
	Moreover \(|N_{G}(AZ/Z)|\) is
	divisible by \(|N_{\textrm{SL}_n(q)}(T)/Z|\), which in turn is divisible by \(\frac{(q^n-1)\cdot n}{(q-1)\cdot
		(n,q-1)}=\frac{q^n-1}{q-1}\cdot \frac{n}{(q-1,n)}\).
\end{proof}





\begin{lemma}\label{sgpGU}
Let \(n \in \mathbb{N}\), \(n\geq 2\), let \(q\) be a prime power and let \(G:=\textrm{GU}_n(q)\). Let
		\(p\) be a
		prime dividing \(q^n-(-1)^n\) and suppose that, for all positive integers \(l<n\), the number \(q^l-(-1)^l\) is not
		divisible by \(p\).
		Let \(P\in \textrm{Syl}_p(G)\).
		Then \(|N_G(P)|\) is divisible by \((q^n-(-1)^n)\cdot n\).
		
\end{lemma}

\begin{proof}		
	First suppose that \(n\) is even. Then Theorem~3.9
	in~\cite{Wilson} tells us that $G$ has a subgroup \(M\) isomorphic to
	\(\textrm{GL}_{n/2}(q^2).2\). Let \(L\leq M\) be isomorphic to \(\textrm{GL}_{n/2}(q^2)\).
	Next, Satz~II\,7.3 in~\cite{Huppert-I} gives a cyclic subgroup \(T\) of $L$ of order
	\(q^{2\cdot \frac{n}{2}}-1=q^n-1\) where \(N_{L}(T)/T\) is cyclic of order \(n/2\).
	By hypothesis, for all \(l<n\) the number \(q^l-(-1)^l\) is not divisible by \(p\), so without loss
	\(P \le T\). In fact $P$ is characteristic in $T$.
	 Lemma \ref{help}(a) yields that
	\(|N_U(T)|=2|N_{L}(T)|=2\cdot (q^n-1)\cdot \frac{n}{2}=(q^n-1)\cdot n\).
	Now
	\(|N_{G}(P)|\) is divisible by \(|N_{G}(T)|\) and hence by \((q^n-1)\cdot n= (q^n-(-1)^n)\cdot n\),
	because \(n\) is even.
	
	Next suppose that \(n\) is odd. Then \(p\) divides \(q^n+1\). Let \(d\) be a positive integer dividing~\(n\).
	We prove, by induction on $d$, that the normalizer of a Sylow \(p\)-subgroup of \(\textrm{GU}_d(q^{n/d})\) is divisible by
	\((q^n+1)\cdot d\).
	
	If \(d=1\), then \(\textrm{GU}_d(q^{n/d})=\textrm{GU}_1(q^n)\) is cyclic of order \(q^n+1\), and hence the normalizer of a Sylow
	\(p\)-subgroup is \(\textrm{GU}(1,q^n)\) itself, of order \(q^n+1=(q^n+1)\cdot 1=(q^n+1)\cdot
	d\).
	
	Suppose that $d \ge 2$ and that our claim is true for all proper divisors $e$ of $d$.
	By Theorem~3.9 in~\cite{Wilson} there exists an odd prime \(r\) dividing
	\(d\) such that \(\textrm{GU}_d(q^{n/d})\) has a subgroup \(M\) isomorphic to \(\textrm{GU}_{d/r}((q^{n/d})^r).r\). Let
	\(U\leq M\) be isomorphic to \(\textrm{GU}_{d/r}((q^{n/d})^r)=\textrm{GU}_{d/r}(q^{\frac{n}{d/r}})\).
	By hypothesis, for all positive integers \(l<n\), the number \(q^l-(-1)^l\) is not divisible by \(p\).
	As a consequence, we have for all \(l_1<d\) that the
	number \((q^{n/d})^{l_1}-(-1)^{l_1}\) is not divisible by \(p\), because \(n\) is odd.
	So without loss
	\(P \le U\) and in particular
	\(r\neq p\).
	Lemma \ref{help}(a) yields that
	\(|N_{M}(P)|=r|N_{U}(P)|\).
	Then we apply the induction hypothesis to \(U\cong \textrm{GU}_{d/r}(q^{\frac{n}{d/r}})\)) and we deduce that
		\(|N_{U}(P)|\) is divisible by \((q^n+1)\cdot (d/r)\). Hence \(|N_{M}(P)|\) is divisible by \(r\cdot (q^n+1)\cdot
	(d/r)=(q^n+1)\cdot d\), proving our statement.
	 This proves the statement of the induction.
	It follows that \(|N_G(P)|=|N_{\textrm{GU}_n(q)}(P)|\) is divisible by \((q^n+1)\cdot n=(q^n-(-1)^n)\cdot n \) as well, as stated.
\end{proof}



\bigskip
\begin{lemma}\label{sgpPSU}
Let \(n\in \mathbb{N}\), \(n\geq 2\), let \(q\) be a prime power and \(G:=\textrm{PSU}_n(q)\).
Let \(p \in \pi(G)\) and let
\(k \in \mathbb{N}\) be minimal such \(p\) divides \(q^k-1\). If $k \ge 3$, then the following hold:

(a) If \(k=n-1\), then there exists a $p$-subgroup \(R\le G\) such that \(|N_G(R)|\) is divisible by

		\(\frac{(q^{n-1}-(-1)^{n-1})\cdot (n-1)}{(n,q+1)}\).
		
(b) If	 \(k=n\), then the normalizer of a Sylow $p$-subgroup of $G$ has order
divisible by \(\frac{(q^n-(-1)^n)\cdot n}{(q+1)\cdot (n,q+1)} \).
	
\end{lemma}	
	
\begin{proof}	Let \(Z:=Z(\textrm{SU}_n(q))\). Then \(G=\textrm{SU}_n(q)/Z\) and \(|Z|=(n,q+1)\).
	
	We begin with (b), so suppose that \(k=n\).
	Let \(Q \le \textrm{SU}_n(q)\) be such that \(QZ/Z \in \textrm{Syl}_p(G)\).
	As $k \ge 3$, we know that \(p\) does not divide \(q+1=|\textrm{GU}_n(q):\textrm{SU}_n(q)|\), which implies that
	\(Q\) is a Sylow \(p\)-subgoup of \(\textrm{GU}_n(q)\). Then we apply Lemma \ref{sgpGU} and we see that
	 \(N:=N_G(Q)\) has order divisible by
	\((q^n-(-1)^n)\cdot n\). Then \(|N\cap \textrm{SU}_n(q)|=\frac{|N|\cdot |\textrm{SU}_n(q)|}{|N \textrm{SU}_n(q)|}\) is divisible by
	\(\frac{|N|\cdot |\textrm{SU}_n(q)|}{|\textrm{GU}_n(q)|}\) that in turn is divisible by \(\frac{(q^n-(-1)^n)\cdot n}{q+1}\).
	Now \(N_G(P)\geq N_{\textrm{SU}}(Q)Z/Z=(N \cap \textrm{SU}_n(q))Z/Z\) and therefore \(|N_G(QZ/Z)|\) is divisible by
	\(\frac{(q^n-(-1)^n)\cdot n}{(q+1)\cdot (n,q+1)}\).
	
	For (a) we suppose that \(k=n-1\). Then \(\textrm{SU}_n(q)\) contains a subgroup \(U\)
	isomorphic to \(\textrm{GU}_{n-1}(q)\) and we let \(Q\in \textrm{Syl}_p(U)\). We know from Lemma \ref{sgpGU} that
	\(|N_U(Q)|\) is divisible by \((q^{n-1}-(-1)^{n-1})\cdot (n-1)\). It follows that \(N_{\textrm{SU}_n(q)}(Q)\geq N_U(Q)\) has
	order divisible by \((q^{n-1}-(-1)^{n-1})\cdot (n-1)\). As \(N_{\textrm{SU}_n(q)}(Q)Z/Z\leq N_G(QZ/Z)\), we deduce that
	\(|N_G(QZ/Z)|\) is divisible by \(\frac{(q^{n-1}-(-1)^{n-1})\cdot (n-1)}{(n,q+1)}\).
	
	Moreover \(QZ/Z\) is a \(p\)-subgroup
	 of \(G\), so our claim is proven.
\end{proof}





\begin{lemma}\label{PSL}
Let \(n\) be an integer, \(n\geq 4\), let \(q\) be a prime power and
		\(G:=\textrm{PSL}_n(q)\). Then there is no set $\Omega$ such that
		Hypothesis \ref{hyp6} is satisfied for $G$ and $\Omega$.
		\end{lemma}
		
\begin{proof

Assume otherwise (with all the notation included) and suppose first that $n=4$ and $q \le 5$.
Then
Remark~\ref{rem4.1} excludes the cases where \(q\in \{2,3\}\)
for the remaining cases.
By Table \ref{ktablealternativ} we only have to consider the cases where $p$ divides $\Phi_3(q)$ or $\Phi_4(q)$, and $p\ge 5$ by hypothesis.

	
If $q=4$, then $p \in \{7,17\}$.
By Lemma~\ref{sgpPSL} there exists a subgroup \(Y\leq G\) of order \(p\) such that \(|N_G(Y)|\) is divisible
by \(\frac{(4^3-1)\cdot 3}{(4,3)}=3^3\cdot 7\) if \(p=7\) or by \(\frac{(4^4-1)\cdot 4}{3\cdot (4,3)}=4\cdot 5\cdot 17\) if \(p=17\).
So by Lemma~\ref{start}\,(e), \(p=17\) and \(5\in \pi(H)\). As \(5\) divides \(4^2-1=q^2-1\) this contradicts Lemma~\ref{help}.

Finally, if $q=5$, then
$p \in \{13,31\}$.
Again by Lemma~\ref{sgpPSL}, there exists a subgroup \(Y\leq G\) of order \(p\) such that \(|N_G(Y)|\) is divisible
by \(\frac{(5^3-1)\cdot 3}{(4,4)}=3\cdot 31\) if \(p=31\) or by \(\frac{(5^4-1)\cdot 4}{4\cdot (4,4)}=2^2\cdot 3\cdot 13\) if \(p=13\).
So by Lemma~\ref{start}\,(e), \(p=31\) and \(H\) is not divisible by \(13\). As \(|\textrm{PSL}(4,5)|=2^7\cdot 3^2\cdot 5^6\cdot 13 \cdot 31\), by Lemma~\ref{Lie1}, \(|H|=31\).
This now is a contradiction to Lemma~\ref{start}\,(a) as we just noted that \(|N_G(H)|\) is divisible by \(3\).

From now on we suppose that $q \ge 7$ whenever \(n=4\).
We recall that $|H|$ and $q$ are coprime by Lemma \ref{Lie1}.

Now we let \(k \in \mathbb{N}\) be minimal such that \(p \in H\) can be chosen to divide \(q^k-1\), and
we remark that $p \ge 5$ by assumption.
Then $k \in \{n-1,n\}$ by Lemma \ref{Primesalternativ}.


	
	
        First we suppose that \(k=n-1\). Then Lemma \ref{sgpPSL} (a) gives  a subgroup \(Y\le G\) of order \(p\) such that
	\(|N_G(Y)|\) is divisible by \(\frac{(q^{n-1}-1)\cdot (n-1)}{(n,q-1)}=\frac{q^{n-1}-1}{q-1}\cdot
	(n-1) \cdot \frac{q-1}{(n,q-1)}\).
	Without loss $Y \le H$ (see Lemma \ref{start}(e)),  and we set $m:=\frac{q^{n-1}-1}{q-1}\cdot
	(n-1) \cdot \frac{q-1}{(n,q-1)}$.
	

	If \(n\geq 6\), then \(n-1\geq 5\) and thus \(|N_G(Y)|\) is divisible by \(n-1\geq 5\).
	Therefore \(\pi(H)\) contains a prime $r$ that divides \(n-1\).
 But \(r-1\leq n-1-1<k\), so together with Lemma \ref{help}(c) this contradicts the minimality of \(k\).
	
	If \(n=5\), then \(m=\frac{(q^{4}-1)\cdot 4}{(5,q-1)}=(q^2+1) \cdot
	4(q+1)\cdot \frac{q-1}{(5,q-1)}\), and consequently \(|N_G(Y)|\) is divisible by \(4(q+1)\).
	As \(|H|\) is odd, Lemma \ref{start}(e) implies that  \(\pi(H)\) contains a prime that divides \(q+1\),
	contrary to the fact that \(k\geq	3\).
	
	The last case is \(n=4\). Then \(m=\frac{(q^{3}-1)\cdot 3}{(4,q-1)}=(q^2+q+1) \cdot
	\frac{3\cdot (q-1)}{(4,q-1)}\) and therefore \(|N_G(Y)|\) is divisible by \(\frac{3\cdot (q-1)}{(4,q-1)}\).
	As \(n=4\), we only have to consider the cases where \(q\geq 7\) (see first paragraph).
	
	Then  \(\frac{3\cdot (q-1)}{(4,q-1)}\geq \frac{3\cdot (q-1)}{4}\geq \frac{3\cdot 6}{4}>
	4\). Lemma \ref{start}(e) shows that \(|H|\) and $n-1$ have a joint prime divisor, which together with
	Lemma \ref{help}(c)
	contradicts
	the minimality of \(k\).
	Therefore \(k=n\).\\
	
	Now
	Lemma \ref{sgpPSL} (b) shows that there exists a subgroup \(Y\le G\) of order \(p\) such that \(|N_G(Y)|\)
	is divisible by \(l:=\frac{q^n-1}{q-1}\cdot \frac{n}{(q-1,n)}\).
	Without loss $Y \le H$ and we look at the number $l$ depending on the possibilities for $n$.
	
	If \(n=4\), then \(k=n=4\) and \(l=(q+1)\cdot (q^2+1)\). Moreover $q \ge 7$ by the first paragraph, so
	\(|N_G(Y)|\) is divisible by \(l=\frac{q^4-1}{q-1}\cdot
	\frac{4}{(4,q-1)}\) and hence by
	\(q+1>4\).
	Lemma \ref{start}(e) gives a prime $r$ that divides
	 \(|H|\) and \(q+1\), contrary to the fact that $k \ge 3$.
	Therefore \(n\geq 5\).
	
	\smallskip
	\textbf{Case 1:}
	$n$ is a power of $2$.
	
	Then \(\frac{q^n-1}{q-1}=\prod_{d|n, d\neq 1}\Phi_d(q)\) is divisible by \(\Phi_4(q)=(q^2+1)\).
	As \(|N_G(Y)|\) is divisible by $l$ and hence by \(q^2+1\geq 2^2+1>4\), we can apply Lemma \ref{start}(e) again.
	Thus \(H\) is divisible by a prime factor of \(q^2+1\), which divides \(q^4-1\).
	But \(4<n=k\), so this contradicts the
	minimality of \(k\).
	
	\smallskip
	\textbf{Case 2:}
	$n$ is divisible by a prime
	\(r\geq 3\) such that \(r<n\).
	
	In this case \(\frac{q^n-1}{q-1}\), and hence $|N_G(Y)|$, is divisible by \(\Phi_r(q)\geq q^2+q+1\geq 2^2+2+1=7\).
	Lemma \ref{start}(e) gives that \(|H|\) and  \(\Phi_r(q)\) have a common prime divisor, and this prime divides
	\(q^r-1\). But \(r<n=k\), so this contradicts the
	minimality of \(k\).
	
	\smallskip
	\textbf{Case 3:} \(n\) is a prime number.
	
	Then \(\Phi_k(q)=\Phi_n(q)=1+q+q^2+\ldots + q^{n-1}\). If \(q-1\) is divisible by \(n\), then \(\Phi_k(q)\equiv n\cdot 1
	\equiv 0 \mod n\). As \(|N_G(Y)|\) is divisible by \(l\), which is
	divisible by \(\Phi_k(q)\), it follows that \(n\geq 5\) divides \(N_G(Y)\). Therefore \(|H|\) is divisible by \(n\).
	However, this
	contradicts the minimality of \(k\), because \(n\) divides \(q-1=\Phi_1(q)\) and \(1<n=k\).
	As \(n\) is a prime, we conclude that \((q-1,n)=1\).
	
	Then  \(l=\frac{q^n-1}{q-1}\cdot n\).
	We recall that \(n\geq 5\) and that $l$ divides \(|N_G(Y)|\), and then it follows that
	\(|H|\) is divisible by \(n\).
	Fermat's Little Theorem is applicable because $n$ is prime, and we recall that $|H|$and $q$ are coprie, which makes
	$n$ and $q$ coprime as well.
	Then
	\(q^{n-1}\equiv 1 \mod n\) or, equivalently, \(n\) divides \(q^{n-1}-1\). As \(n-1<k\) this contradicts the
	minimality of \(k\).
	This final contradiction proves the lemma.
	
\end{proof}





\begin{lemma}\label{PSU}
Let \(n \in \mathbb{N}\), \(n\geq 4\), let \(q\) be a prime power and
		\(G:=\textrm{PSU}_n(q)\). Then there exists a set $\Omega$ such that \((G,\Omega)\) satisfies Hypothesis
		\ref{hyp6} if and only if $n=4$, $q \in \{2,3\}$ and the point stabilizers are
		cyclic of order \(5\).
\end{lemma}

\begin{proof}

	By Remark~\ref{rem4.1} we see that the statement is correct for \(\textrm{PSU}_4(2)\) and \(\textrm{PSU}_4(3)\).
	Now we assume that \(G\) is neither \(\textrm{PSU}(4,2)\) nor \(\textrm{PSU}(4,3)\), but that
	Hypothesis \ref{hyp6} is satisfied with all the notation there.
	
	By Lemma \ref{Lie1} we know that
	\(|H|\) and \(q\) are coprime. Then
	let \(k\) be the smallest positive integer such that there exists a prime
	\(p\in \pi(H)\) with the property that \(p\) divides \(q^k-(-1)^k\).
	We recall that $p \ge 5$ by Hypothesis \ref{hyp6}, and we also note that
	\(k\in \{n-1,n\}\) by Lemma \ref{Primesalternativ}.
	
	
	Let \(Z:=Z(\textrm{SU}_n(q))\). Then \(G=\textrm{SU}_n(q)/Z\) and \(p\) does not divide \(|Z|\).
	
	We use Lemma \ref{sgpPSU}~(a), so suppose that $n$ is even and 
	\(k=n\). Then \(|N_G(P)|\) is divisible by \(\frac{(q^n-(-1)^n)\cdot n}{(q+1)\cdot (n,q+1)}\).
	
	\smallskip
	\textbf{Case 1:}
	\(n\) is not divisible by $4$.
	Then \(\frac{(q^n-(-1)^n)\cdot n}{(q+1)\cdot (n,q+1)}=\frac{(q^n-1)\cdot n}{(q+1)\cdot
		(n,q+1)}=\frac{(q^{n/2}-1)\cdot (q^{n/2}+1)\cdot n}{(q+1)\cdot (n,q+1)}
		=\frac{q^{n/2}-1}{q-1}\cdot (q-1)\cdot \frac{(q^{n/2}+1)\cdot n}{(q+1)\cdot (n,q+1)}\).
	Thus \(|N_G(P)|\) is divisible by \((q-1)\cdot \frac{(q^{n/2}+1)\cdot n}{(q+1)\cdot (n,q+1)}\).
	In this case $n \ge 6$, therefore
	\(\frac{q^{n/2}+1}{q+1}=q^{n/2-1}-q^{n/2-2}+\ldots -q+1\geq q^2-q+1\).
	We look at the possibilities for $q$:
	
	If \(q=2\), then \((q-1)\cdot \frac{(q^{n/2}+1)\cdot n}{(q+1)\cdot (n,q+1)}=\frac{(2^{n/2}+1)\cdot n}{(2+1)\cdot
		(n,2+1)}\geq \frac{(2^{n/2}+1)\cdot n}{3\cdot 3}\geq \frac{(2^{6/2}+1)\cdot 6}{3\cdot 3}=6>4\).
		
	If \(q\geq 3\), then \((q-1)\cdot \frac{(q^{n/2}+1)\cdot n}{(q+1)\cdot (n,q+1)}=(q-1)\cdot
	\frac{(q^{n/2}+1)}{(q+1)}\cdot\frac{n}{(n,q+1)}\geq (q-1)\cdot (q^2-q+1)\geq 2\cdot 7=14>4\).
	
	Then Lemma \ref{start}(e) implies that $\pi(H)$ contains a prime $r$ dividing
	\((q-1)\cdot (q^{n/2}+1)\cdot n\).
	As $k \ge 3$, we know that
	 \(r\) does not divide \(q-1\).
	 We recall that $|H|$ and $q$ are coprime, so \(q^{r-1}\equiv 1
	\mod r\).
	
	So if $r$ divides $n$, then $r-1 < n$ and this contradicts the minimal choice of $k$.
	Therefore \(r\) divides
	\(q^{n/2}+1\), but this also contradicts the minimality of \(k\), because \(n/2\) is odd.
	
	\smallskip
	\textbf{Case 2:} $n$ is divisible by $4$.
	
	Then \(\frac{(q^{n/2}-1)\cdot (q^{n/2}+1)\cdot n}{(q+1)\cdot (n,q+1)}=(q^{n/2}+1)\cdot \frac{(q^{n/2}-1)\cdot
		n}{(q+1)\cdot (n,q+1)}\) and Lemma \ref{sgpPSU}(a) yields that
	\(|N_G(P)|\) is divisible by \(\frac{(q^{n/2}-1)\cdot n}{(q+1)\cdot
		(n,q+1)}\).
	First we look at the case where \(n=4\).
	Then \(q\geq 4\) and hence \(\frac{(q^{n/2}-1)\cdot n}{(q+1)\cdot
		(n,q+1)}=\frac{(q^2-1)\cdot 4}{(q+1)\cdot (4,q+1)}=\frac{(q-1)\cdot 4}{(4,q+1)}\).
	Again we consider the possibilities for $q$.
	
        If \(q=4\), then
	\(\frac{(q-1)\cdot 4}{(4,q+1)}=12>4\).
	
	If \(q=5\), then \(\frac{(q-1)\cdot4}{(4,q+1)}=8>4\).
	
	If \(q\geq7\), then
	\(\frac{(q-1)\cdot 4}{(4,q+1)}\geq q-1\geq 6>4\).
	
	
	Now suppose that \(n>4\).
	Then \(n\geq 8\) in the present case. Since \(q+1\) divides either \(q^{n/4}+1\) or \(q^{n/4}-1\), we have that
	\(\frac{q^{n/2}-1}{q+1}=\frac{(q^{n/4}-1)\cdot
		(q^{n/4}+1)}{q+1}\geq q^{n/4}-1\).
		Consequently \(\frac{(q^{n/2}-1)\cdot n}{(q+1)\cdot (n,q+1)}\geq
	(q^{n/4}-1)\cdot \frac{n}{(n,q+1)}\geq q^{n/4}-1\).
	
	The smallest case is
	\(n=8\) and \(q=2\), and then \(\frac{(q^{n/2}-1)\cdot n}{(q+1)\cdot (n,q+1)}=\frac{(2^{4}-1)\cdot 8}{3\cdot
		(8,3)}=40>4\).
	
	If \(q\geq 3\), then \(q^{n/4}-1\geq 3^2-1=8>4\).
	
	If \(n>8\), then \(q^{n/4}-1\geq q^{12/4}-1=q^3-1\geq 2^3-1=7>4\).
	
	In all cases we see that $|N_G(P)|$ is divisible by the number \(\frac{(q^{n/2}-1)\cdot n}{(q+1)\cdot (n,q+1)}>4\).
	Then Lemma \ref{start}(e) gives a prime $r \in \pi(H)$ that divides
	\((q^{n/2}-1)\cdot n\).
	As in the previous case, $r$ does not divide $n$ (by Lemma \ref{help}(c)), so
	\(r\) divides
	\(q^{n/2}-1\). But again this contradicts the minimality of \(k\), because \(n/2\) is even.
	
	\smallskip
	\textbf{Case 3:} $n$ is an odd prime.
	
	Still \(|N_G(P)|\) is divisible by \(\frac{(q^n-(-1)^n)\cdot n}{(q+1)\cdot
		(n,q+1)}=\frac{(q^n+1)\cdot n}{(q+1)\cdot (n,q+1)}\).
	
	As $n$ is a prime number, we either have that \((n,q+1)=1\)
	or that \(n\) divides \(q+1\).
	In the first case \(|N_G(P)|\) is divisible by \(n\geq 5\), and in
	the second case
	
	$\frac{q^n+1}{q+1}\equiv q^{n-1}-q^{n-2}+q^{n-3}-\ldots +q^2-q+1\equiv
	(-1)^{n-1}-(-1)^{n-2}+(-1)^{n-3}-\ldots +(-1)^2-(-1)+1\equiv n$
	
	$\equiv 0 \mod n.$
	
	Again \(n\)
	divides \(|N_G(P)|\), so in both cases our usual lemma implies that  \(n\) divides \(|H|\).
	In particular $n$ and $q$ are coprime, so Fermat's Little Theorem gives that
	\(q^{n-1}-1\)
	is divisible by \(n\). But \(n-1\) is even and this contradicts the minimality of \(k\).
	
	\smallskip
	\textbf{Case 4:} $n$ is an odd composite number.
	
	We let $s \in \mathbb{N}$ be an odd prime divisor of $n$ and we recall that
	\(|N_G(P)|\) is divisible by \(\frac{(q^n-(-1)^n)\cdot n}{(q+1)\cdot
		(n,q+1)}=\frac{(q^n+1)\cdot n}{(q+1)\cdot (n,q+1)}\).
	
	As \(q^s+1\) divides \(q^n+1\), the minimal choice of \(k\) yields that \(p\) does not
	divide \(q^s+1\).
	In particular
	$$\frac{(q^n+1)\cdot n}{(q+1)\cdot (n,q+1)}=\frac{q^n+1}{q^s+1}\cdot \frac{(q^s+1)\cdot
		n}{(q+1)\cdot (n,q+1)}.$$
	Now we recall that $n$ is odd and composite, so $n \ge 9$.	
	
	 If \(q=2\), then
	\(\frac{(q^s+1)\cdot n}{(q+1)\cdot (n,q+1)}=\frac{(2^s+1)\cdot
		n}{(2+1)\cdot (n,2+1)}\geq \frac{(2^3+1)\cdot 9}{(2+1)\cdot 3}=9\), which contradicts Lemma \ref{start}(f).
		
	 If \(q\geq 3\) then \(\frac{(q^s+1)\cdot n}{(q+1)\cdot (n,q+1)}\geq
	\frac{q^s+1}{q+1}\geq \frac{3^3+1}{3+1}=7>4\).	
	
	By Lemma \ref{start}(e), some prime $t \in \pi(H)$ divides
	\((q^s+1)n\). Then the minimality of \(k\) forces \(t\) to divide \(n\),
	hence $t<n$ and Lemma \ref{help}(c) gives a contradiction (because \(t-1\) is even).
	
	       Now we assume that \(k=n-1\), which by Table \ref{ktablealternativ} means that $n$ is odd.  We apply Lemma \ref{sgpPSU}(b) and we let \(R\le G\) be a $p$-subgroup such that \(|N_G(R)|\) is divisible by \(m:=\frac{(q^{n-1}-(-1)^{n-1})\cdot
		(n-1)}{(n,q+1)}\).

	

	We recall that $n$ is odd, so \(n-1\) is even.
	Then
	\(m=\frac{q^{n-1}-1}{q^2-1}\cdot \frac{(q^2-1)\cdot (n-1)}{(n,q+1)}\).
	
	If $n \ge 7$, then Lemma \ref{start}(e) implies that some prime $r \in \pi(H)$ divides $n-1$.
	As  \(r-1\leq
	n-1-1<n-1\), this contradicts Lemma \ref{help}(c) together with the minimality of
	\(k\).
	
	Thus $n=5$.
	
	If  \(q=2\), then \(\frac{(q^2-1)\cdot
		(n-1)}{(n,q+1)}=\frac{(2^2-1)\cdot 4}{(5,2+1)}=12\), which contradicts Lemma \ref{start}(f).
	
	If \(q\geq 3\), then we see that \(\frac{(q^2-1)\cdot (n-1)}{(n,q+1)}=\frac{(q^2-1)\cdot
		4}{(5,q+1)}\geq \frac{(3^2-1)\cdot 4}{5}=\frac{32}{5}>4 \).
	Then Lemma \ref{start}(e) implies that some $r \in \pi(H)$ divides \((q^2-1)\cdot (n-1)\).
	As before, $r$ cannot divide $n-1$ or \(q^2-1\) because of the minimality of \(k\).\\
	

	
	Now we have considered all cases. 

	\end{proof}



	
	







\subsection{Groups of symplectic type}


Again we look at the subgroup structure first.


\begin{lemma}\label{sgpPSp}

Let \(n \in \mathbb{N}\), \(n\geq 2\), let \(q\) be a prime power and let \(G:=\textrm{PSp}_{2n}(q)\). Let
		\(p\) be a
		prime dividing \(q^{2n}-1\) and such that for all positive integers \(l<n\) the number \(q^{2l}-1\) is not
		divisible by \(p\).
		Let \(P\in \textrm{Syl}_p(G)\) and let \(\varepsilon\in \{-1,1\}\) be such that \(p\) divides \(q^n-\varepsilon\).
		Then \(|N_G(P)|\) is divisible by \(\frac{(q^{n}-\varepsilon)\cdot 2n}{(2,q+1)}\).
	
\end{lemma}		
		
		 \begin{proof}

	First we assume that \(p=2\). As \(p\) divides \(q^{2n}-1\), it follows that \(q\) is odd and that
	\(p\) divides \(q^2-1\). But this contradicts the
	hypothesis.
	Now \(p\geq 3\) and $p$ divides \(q^{2n}-1=(q^n-1)(q^n+1)\), so it divides exactly one of the factors.
	Let \(d\) be a positive integer dividing~\(n\).
	Then we prove the statement of the lemma by induction on $d$.
	
	If \(d=1\), then \(\textrm{PSp}_{2d}(q^{n/d})=\textrm{PSp}_{2}(q^n)\cong \textrm{PSL}_2(q^n)\), see for example p.61 in \cite{Wilson}.
	So if we let $Q \in \textrm{Syl}_p(\textrm{PSp}_{2d}(q^{n/d})$, then by Satz~II\,8.3 and 8.4 in~\cite{Huppert-I} its normalizer is a dihedral
	group of order \(2\cdot \frac{q^n-\varepsilon}{(2,q+1)}=\frac{(q^{n}-\varepsilon)\cdot 2d}{(2,q+1)}\).
	
	Next we suppose that $d \ge 2$ and that the statement holds for all smaller divisors of $n$.
	Theorems~3.7 and~3.8 in~\cite{Wilson} tell us that there is a prime \(r\)
	dividing
	\(d\) such that the group \(\textrm{PSp}_{2d}(q^{n/d})\) has a subgroup \(H\cong \textrm{PSp}_{2d/r}((q^{n/d})^r).r\).
	Let
	\(U\leq H\) be isomorphic to \(\textrm{PSp}_{2d/r}((q^{n/d})^r)=\textrm{PSp}_{2d/r}(q^{\frac{n}{d/r}})\).
	The hypothesis implies that, for all \(l_1<d\), the
	number \((q^{n/d})^{2l_1}-1\) is not divisible by \(p\).
	In particular \(U\) contains a Sylow \(p\)-subgroup \(Q\) of~\(G\), and in fact \(Q\in \textrm{Syl}_p(H)\).
	
		Then Lemma~(normgrow) yields that
	\(|N_{H}(Q)|=r|N_{U}(Q)|\).
	
	Arguing by induction for \(U\cong \textrm{PSp}_{2d/r}(q^{\frac{n}{d/r}})\) we deduce that
	\(|N_{U}(Q)|\) is divisible by \(\frac{(q^n-\varepsilon)\cdot 2(d/r)}{(2,q+1)}\). Hence \(|N_{H}(Q)|\) is
	divisible by \(r\cdot \frac{(q^n-\varepsilon)\cdot 2(d/r)}{(2,q+1)}=\frac{(q^n-\varepsilon)\cdot
		2d}{(2,q+1)}\).
	This proves the statement of the induction.
	
	We conclude that \(|N_G(P)|=|N_{\textrm{PSp}(2n,q)}(P)|\) is divisible by \(\frac{(q^n-\varepsilon)\cdot 2n}{(2,q+1)}\).	
	
	\end{proof}







\begin{lemma}\label{maxsgpPsp4}

Let \(q\geq 7\) be a prime power and let \(G:=\textrm{PSp}_4(q)\). Let \(U\le G\) be a proper subgroup of order
		divisible by \(\frac{q^2+1}{(2,q+1)}\).
		Then \(U\) lies in a maximal subgroup of structure \(\textrm{PSL}_2(q^2).2\) or \(\textrm{Sz}(q)\). In
		the latter case \(q\) is even.
		If \(U\) has odd order, then it is cyclic of order \(\frac{q^2+1}{(2,q+1)}\).
	
\end{lemma}






\begin{proof}		

Let \(Z:=Z(\textrm{Sp}_4(q))\). Then \(G=\textrm{Sp}_4(q)/Z\) and \(|Z|=(2,q-1)\).
	Let \(H\) be the full pre-image of \(U\) in \(\textrm{Sp}_4(q)\).
	Then \(H\) is a proper subgroup because $U$ is a proper subgroup by hypothesis.
	
	Inspection of Tables 8.12, 8.13 and 8.14
	in~\cite{BHRD} shows that
	the only maximal subgroups of \(\textrm{Sp}_4(q)\) of order divisible by \(\frac{q^2+1}{(2,q+1)}\) are the following
	(some only  under additional conditions):
	
	\begin{center}
		\begin{tabular}{c|l|l}
			Type&subgroup & conditions\\
			\hline
			(I)&\(\textrm{Sp}_2(q^2).2\)&\\
			(II)&\(2^{1+4}_-.\Sym_5\)& \(q\) prime and congruent to \(-1\) or \(1\) modulo \(8\)\\
			(III)&\(2^{1+4}_-.\Alt_5\)& \(q\) prime and congruent to \(-3\) or \(3\) modulo \(8\)\\
			(IV)&\(2^{\cdot}\Alt_6\) & \(q\neq 7\) prime and congruent to \(5\) or \(7\) modulo \(12\)\\
			(V)&\(2^{\cdot}\Sym_6\) & \(q\) prime and congruent to \(1\) or \(11\) modulo \(12\)\\
			(VI)&\(2^{\cdot}\Alt_7\) & \(q=7\)\\
			(VII)&\(\textrm{SO}^-_4(q)\)& \(q\) even\\
			(VIII)&\(\textrm{Sz}(q)\) & \(q\) even
		\end{tabular}
	\end{center}
		
	First assume that \(H\) is a subgroup of a group of type (II) from the table.
	As \(\frac{q^2+1}{(2,q+1)}\) is odd, this implies that \(\frac{q^2+1}{(2,q+1)}\leq 3\cdot 5\), and then \(q<6\)
	contradicting our hypothesis that \(q\geq 7\).

	Next assume that \(H\) is a subgroup of a group of type (III).
	Still the number \(\frac{q^2+1}{(2,q+1)}\) is odd, so we have the same contradiction as in the previous case.
	
	If \(H\) is a subgroup of a group of type (IV) or type (V), then
	\(\frac{q^2+1}{(2,q+1)}\leq 3\cdot 5\cdot 3\) and therefore
	\(q<10\). Together with the conditions in the table this means that \(q=5\), contradicting our hypothesis.
	
	Assume that \(H\) is a subgroup of a group of type (VI). Then the conditions in the table force \(q=7\).
	Thus
	\(\frac{q^2+1}{(2,q+1)}=5^2\), but \(|2^{\cdot}\Alt_7|\) is not divisible by \(5^2\). So this is impossible.
	
	Next asume that \(H\) is a subgroup of a group ot type (VIII). Then \(q\) is even. Now
	\(\textrm{Sp}_{2n}(q)\cong G\) and \(H\cong U\), and we apply
	Theorem~4.1 in \cite{Wilson}. As \(\textrm{Sz}(q)\) does not have a
	subgroup of order divisible by \(\frac{q^2+1}{(2,q+1)}\), we conclude that  \(H\cong U\) is isomorphic to \(\textrm{Sz}(q)\).
	But this is impossible when \(|U|\) is odd.\\
	
	We are left with subgroups of type (I) and (VII).
	First suppose that \(H\) is a subgroup of a group of type (VII).
	 Then
	\(q\) is even and hence \(G\cong \textrm{Sp}_4(q)\) and \(H\cong U\). Let \(M\leq G\) be isomorphic to \(\textrm{SO}^-_4(q)\).
	Let $L$ be a subgroup of index 2 of $M$ such that ... check \cite{Wilson}...\\
	
	Then  \(L\cong \Omega^-_4(q)\cong
	P\Omega^-_4(q)\cong\textrm{PSL}_2(q^2)\) by (3.57) in~\cite{Wilson}. Therefore \(M\) is of
	structure \(\textrm{PSL}_2(q^2).2\).

	Next suppose that \(H\) is a subgroup of a maximal subgroup \(M\) of type (I). Let \(L\leq M\) be isomorphic to
	\(\textrm{Sp}_2(q^2)\).
	Then \(Z=Z(L)\) and so \(MZ/Z=M/Z\) is of order \(2\cdot |L/Z|\). Moreover, \(L/Z \unlhd M/Z\) and therefore
	\(L/Z \cong \textrm{PSp}_2(q^2).2\).
	In particular \(U:=H/Z\) is contained in \(M/Z\), which is a group of structure \(\textrm{PSp}(2,q^2).2\).
	
		First suppose that \(|U|\) is odd. Then, in both remaining cases, \(U\) is a subgroup of a subgroup of index \(2\) in the
	corresponding maximal subgroup. Hence $U$ is isomorphic to a subgroup of \(\textrm{PSL}_2(q^2)\). By
	Hauptsatz~II\,8.27 in~\cite{Huppert-I} the only subgroups of \(\textrm{PSL}_2(q^2)\) of odd order divisible
	by
	\(\frac{q^2+1}{(2,q+1)}\) are cyclic of order \(\frac{q^2+1}{(2,q+1)}\). Hence \(U\) is cyclic of order
	\(\frac{q^2+1}{(2,q+1)}\), as claimed.
	
	\end{proof}
	

	


\begin{lemma}\label{Psp}

Let \(n \in \mathbb{N}\), \(n\geq 2\), let \(q\) be a prime power and
		\(G:=\textrm{PSp}_{2n}(q)\) simple. Then there is a set  \(\Omega\) such that \(G, \Omega\) satisfies Hypothesis \ref{hyp6}
		 if and only if $n=2$ and the point stabilizers are
		cyclic of order \(\frac{q^2+1}{(2,q+1)}\).
	
\end{lemma}	
	
\begin{proof}	

    By Remark~\ref{rem4.1} the statement of the lemma is correct if $n=2$ and \(q\in\{3,4,5\}\). As \(\textrm{PSp}(4.2)\) is not simple,
    we can suppose that \(q\ge 7\), when \(n=2\).
	

	
	Suppose that there is a set  \(\Omega\) such that \((G, \Omega)\) satisfies Hypothesis \ref{hyp6},
	with all the notation given there.
	Then \(|H|\) and \(q\) are coprime by
	Lemma~\ref{Lie1}.
	Let \(k\) be the smallest positive integer such that there exists a prime
	\(p\in \pi(H)\) with the property that \(p\) divides \(q^{2k}-1\).
	Then Lemma \ref{Primesalternativ} implies that \(k= n\).
	
	Let \(Z:=Z(\textrm{Sp}_{2n}(q))\). Then \(G=\textrm{Sp}_{2n}(q)/Z\) and \(|Z|=(2,q-1)\).
	


	
	We let  \(P\in \textrm{Syl}_p(H)\) and then we know that
	\(|N_G((P)|\) is divisible by \(\frac{(q^{n}-\varepsilon)\cdot 2n}{(2,q+1)}\), by Lemma
	\ref{sgpPSp} and since $P\in \textrm{Syl}_p(G)$.

	As \((q^n-\varepsilon)\) is divisible by \((2,q-\varepsilon)=(2,q+1)\), it follows that
	\(|N_G(P)|\) is divisible by
	\(2n\).
	
	First we assume that \(n\geq 3\). Then \(2n\geq 6\), hence Lemma \ref{start}(e) gives a
	prime \(r \in \pi(H)\) that divides
	\(n\).
	As \(|H|\) and \(6\) are coprime, we see that $r$ divides \(n\). At the same time
	\(q^{r-1}-1\) is divisible by \(r\) by Lemma \ref{help}(c). As \(r-1\) is even and
	\(\frac{r-1}{2}\leq \frac{n-1}{2}<n\), this contradicts the minimality of \(k\).
	
	So \(n=2\).
		If \(\varepsilon=1\), then \(p\) divides \(q^2-1=(q-1)(q+1)\), contrary to the fact that $k \ge 2$.
		
	Hence
	\(\varepsilon=-1\) and \(p\) divides \(q^2+1\).  Lemma~\ref{sgpPSp} gives that \(|N_{G}(P)|\) is
	divisible by \(n\cdot 2\cdot
	\frac{q^n-\varepsilon}{(2,q-\varepsilon)}=4\cdot \frac{q^2+1}{(2,q+1)}\).
	Lemma \ref{start}(e) implies that
	\(|H|\) is divisible by \(\frac{q^2+1}{(2,q+1)}\). Then we inspect
	the possibilities from Lemma \ref{maxsgpPsp4}, and we find that \(G_{\alpha}\) is cyclic of order \(\frac{q^2+1}{(2,q+1)}\)
	as stated.
		\end{proof}








\subsection{Groups of orthogonal type}





\begin{lemma}\label{sgpGO}

Let \(n \in \mathbb{N}\), \(n\geq 2\), let \(q\) be an odd prime power, let \(\varepsilon\in
		\{-1,1\}\) and let \(G:=\textrm{GO}^{\varepsilon}_{2n}(q)\). Let
		\(p\) be a
		prime dividing \(q^{n}-\varepsilon\) and such that for all positive integers \(l<n\) the number \(q^{2l}-1\) is not
		divisible by \(p\).
		Let \(P\in \textrm{Syl}_p(G)\).
		Then \(|N_G(P)|\) is divisible by \((q^{n}-\varepsilon)\cdot 2n\).
		
\end{lemma}

\begin{proof}		
	First assume that \(p=2\). Then \(q\) is odd because \(p\) divides \(q^{n}-\varepsilon\). But then \(p\) divides \(q^2-1\)
	as well, contrary to our hypothesis.
	So \(p\geq 3\).
	
	Let \(d \in \mathbb{N}\) be a divisor of~\(n\).
	We argue by induction over $d$,  proving that the normalizer of a Sylow \(p\)-subgroup of \(\textrm{GO}^{\varepsilon}_{2d}(q^{n/d})\) is divisible
	by
	\((q^{n}-\varepsilon)\cdot 2d\).
	
	First suppose that \(d=1\).
	Then \(\textrm{GO}^{\varepsilon}_{2d}(q^{n/d})=\textrm{GO}^{\varepsilon}_2(q^n)\) is a dihedral group of order
	\(2(q^n-\varepsilon)\), see for example p. 71 in \cite{Wilson}.
	As \(p\) is odd, the normalizer of a Sylow
	\(p\)-subgroup of \(\textrm{GO}^{\varepsilon}_{2d}(q^{n/d})\) is \(\textrm{GO}^{\varepsilon}_2(q^n)\) itself, so its order is
	\(2\cdot
	(q^n-\varepsilon)=(q^n-\varepsilon)\cdot 2d\).
	
	Next we suppose that $d \ge 2$ and that the statement holds, correspondingly, for all proper divisors of $d$.
	Theorems~3.10 and~3.11 in~\cite{Wilson} yield that there is a prime
	\(r\)
	dividing
	\(d\) such that the group \(\textrm{GO}^{\varepsilon}_{2d}(q^{n/d})\) has a subgroup
	\(H \cong \textrm{GO}^{\varepsilon}_{2d/r}((q^{n/d})^r).r\). Let
	\(U\leq H\) be isomorphic to \(\textrm{GO}^{\varepsilon}_{2d/r}((q^{n/d})^r)=\textrm{GO}^{\varepsilon}_{2d/r}(q^{\frac{n}{d/r}})\).
	If \(l_1<d\), then by hypothesis the
	number \((q^{n/d})^{2l_1}-1\) is not divisible by \(p\).
	Therefore \(U\) contains a Sylow \(p\)-subgroup \(Q\) of~\(G\) and in particular \(Q\in \textrm{Syl}_p(H)\).

       Lemma~(normgrow) implies that
	\(|N_{H}(Q)|=r|N_{U}(Q)|\), and then we argue by induction, applied to
	\(U\cong \textrm{GO}^{\varepsilon}_{2d/r}(q^{\frac{n}{d/r}})\).
	Then
	\(|N_{U}(Q)|\), and hence \(|N_{H}(Q)|\), is divisible by \((q^{n}-\varepsilon)\cdot 2(d/r)\).
	This proves the statement of the induction.
	
	Therefore \(|N_G(P)|=|N_{\textrm{GO}^{\varepsilon}_{2n}(q)}(P)|\) is divisible by \((q^{n}-\varepsilon)\cdot 2n\).
\end{proof}











\begin{lemma}\label{POmega2n+1}
Suppose that
$n \ge 3$, that $q$ is an odd prime power and that $G=P\Omega_{2n+1}(q)$.
Then there is no set $\Omega$ such that $(G,\Omega)$ satisfies Hypothesis \ref{hyp6}.
\end{lemma}

\begin{proof}		
We assume otherwise and use all the notation from Hypothesis \ref{hyp6}. We further note that
 \(|H|\) and \(q\) are coprime by Lemma \ref{Lie1}.
	Let \(k \in \mathbb{N}\) be minimal such that there exists a prime
	\(p\in \pi(H)\) with the property that \(p\) divides \(q^{2k}-1\).
	
	Then \(k=n\) by Lemma \ref{Primesalternativ}.
	Next we see, on p.\,80 of \cite{Wilson}, that \(G\cong \Omega_{2n+1}(q)\).
	


	
	Since \(p \ge 5\) and $p$ divides \(q^{2n}-1=(q^{n}-1)(q^{n}+1)\), it divides exactly one of the factors.
	 Let \(\varepsilon\in \{-1,1\}\) be such that \(p\) divides
	\(q^{n}-\varepsilon\).
	Theorem~3.10 in~\cite{Wilson} yields that  \(\textrm{GO}_{2n+1}(q)\) has a subgroup $D \times D_0$ such that
	 \(D \cong \textrm{GO}^{\varepsilon}_{2n}(q)\) and \(D_0\cong
	\textrm{GO}_1(q)\).
	
	Let \(P\in \textrm{Syl}_p(D)\). As \(G=\Omega_{2n+1}(q)\) has index \(4\) in \(\textrm{GO}_{2n+1}(q)\),
	we see that \(P\cap \Omega_{2n+1}(q)=P\).
	Then Lemma~\ref{sgpGO} yields that \(|N_D(P)|\) is divisible by \((q^n-\varepsilon)\cdot 2n\).
	Consequently \(|N_{\textrm{GO}_{2n+1}(q)}(P)|\) is divisible by \((q^n-\varepsilon)\cdot 2n \cdot
	|D_0|\).
	
	Let \(m\) be a prime dividing \(n\).
	If \(p=m\), then \(p\) divides \(q^{m-1}-1\) and therefore it divides \(q^{2m-2}-1=(q^{m-1}-1)(q^{m-1}+1)\), contrary to the
	minimality of \(k\).
	Let \(T\cong\textrm{GO}^{\varepsilon}_{2(n/m)}(q^m)\) be a subgroup of index \(m\) of \(S\). Then without loss \(Q \le T\)
	and Lemma \ref{help}(a) gives that \(|N_S(Q)|=m\cdot
	|N_{T}(Q)|\). Arguing inductively, we conclude that \(|N_S(Q)|\) is divisible by
	\(n\cdot |N_{\textrm{GO}^{\varepsilon}_2(q^n)}(Q_1)|\), where \(Q_1\) is the image of \(Q\) under the isomorphism to
	\(\textrm{GO}^{\varepsilon}_2(q^n)\) from the corresponding subgroup.
	
	As \(S\leq D\), we deduce that \(|N_D(P)|\) is divisible by \(n\cdot |N_{\textrm{GO}^{\varepsilon}_2(q^n)}(Q_1)|\),
	with notation as above.
	Moreover
	\(D_0\) centralizes $D$, and hence $S$,
	and therefore \(|N_{\textrm{GO}_{2n+1}(q)}(Q)|\) is divisible by \(n\cdot
	|N_{\textrm{GO}^{\varepsilon}_2(q^n)}(Q_1)|\cdot |D_0|=n\cdot |N_{\textrm{GO}^{\varepsilon}_2(q^n)}(Q_1)|\cdot |\textrm{GO}_1(q)|\).
	
	
Next we consider p.71 in \cite{Wilson}, and we see that \(|D_0|=|\textrm{GO}_1(q)|=2\).
Moreover, \(G\) has index \(4\) in \(\textrm{GO}_{2n+1}(q)\) and therefore
\(|N_G(P)|\) is divisible by
	\(\frac{(q^n-\varepsilon)\cdot 4n}{4}=(q^n-\varepsilon)\cdot n\).
		
		
		By hypothesis \(q\) is odd, therefore
	\(\frac{q^n-\varepsilon}{2}\) is even and \(|N_G(P)|\) is divisible by \(2\cdot n\).
	
	
	Again by hypothesis, we know that \(n\geq 3\), whence \(n\cdot 2\geq 6>4\).
	Then Lemma \ref{start}(e) implies that  \(|H|\) and \(n\cdot (q-\varepsilon)\) have a common prime divisor $r$.
	We know that \(r\) does not divide \(q^2-1\), so \(r\) must divide \(n\).
	 But also,  \(q^{r-1}-1\) is divisible by \(r\) by Lemma \ref{help}(c), which means that	
	 $r$ divides \(q^{2r-2}-1=(q^{r-1}-1)(q^{r-1}+1)\), contrary to the choice of $k$.
This is the final contradiction.
	\end{proof}

	
	
	
	
	
		
	
		
	
	

	
	




\begin{lemma}\label{POmegaminus}
 Suppose that
$n \ge 4$, that $q$ is a prime power and that $G=P\Omega^{-}_{2n}(q)$.
Let $d:=(2,q+1)$.
Suppose further that there exists a set $\Omega$ such that $(G,\Omega)$ satisfies Hypothesis \ref{hyp6}.
Then $n=4$ and
$G_\alpha$ is
cyclic of order \(\frac{q^4+1}{d}\).
		
\end{lemma}




\begin{proof}
We use all notation from Hypothesis \ref{hyp6}.		
The hypothesis and Lemma \ref{Lie1} imply that
	 \(|H|\) and \(q\) are coprime.
	
Let \(k \in \mathbb{N}\) be minimal such that \(p\) can be chosen to divide \(q^{2k}-1\).
	Then \(k\in \{n-1,n\}\) by Lemma \ref{Primesalternativ}.	
	
	Let \(Z:=Z(\Omega^-_2n(q))\). Then \(G=\Omega^-_{2n}(q)/Z\) and \(|Z| \le 2\)  by p.\,70 and p.\,80
	in \cite{Wilson}.
	

	
	 First we assume that \(k=n-1\). Then \(p\) divides \(q^{2n-2}-1=(q^{n-1}-1)(q^{n-1}+1)\), therefore it divides exactly one of the factors and we choose \(\varepsilon\in \{-1,1\}\) such
	that
	\(p\) divides \(q^{n-1}-\varepsilon\).
	
	 We use Theorem~3.11 in~\cite{Wilson} and find a subgroup $D \times D_0$ of  \(\textrm{GO}^-_{2n}(q)\) such that
	\(D\cong \textrm{GO}^{\varepsilon}_{2n-2}(q)\) and \(D_0\cong
	\textrm{GO}^{-\varepsilon}_2(q)\).
	 Let \(P\in \textrm{Syl}_p(D)\). As \(G=\Omega^-_{2n}(q)\) has index \(2\cdot d_0\) in \(\textrm{GO}^-_{2n}(q)\), we have that \(P\cap
	\Omega^-_{2n}(q)=P\).
	Then Lemma~\ref{sgpGO} implies that \(|N_D(P)|\) is divisible by \((q^{n-1}-\varepsilon)\cdot 2(n-1)\), and then
	So \(|N_{\textrm{GO}^-_{2n}(q)}(P)|\) is divisible by \((q^{n-1}-\varepsilon)\cdot 2(n-1)
	\cdot
	|D_0|\).
	
	P.\,71 in \cite{Wilson} shows that
	\(|D_0|=|\textrm{GO}_{-\varepsilon}(2,q)|=2(q+\varepsilon)\). Therefore
	\(|N_{\textrm{GO}^-_{2n}(q)}(P)|\) is divisible by \((q^{n-1}-\varepsilon)\cdot 2(n-1) \cdot 2(q+\varepsilon)\).
	We recall that \(\Omega^-_{2n}(q)\) has index \(2\cdot d_0\) in \(\textrm{GO}^-_{2n}(q)\), and then it follows that
	 \(|N_{\Omega^-_{2n}(q)}(P)|\) is
	divisible by \(\frac{(q^{n-1}-\varepsilon)\cdot 2(n-1) \cdot
		2(q+\varepsilon)}{2 d_0}=\frac{(q^{n-1}-\varepsilon)\cdot 2(n-1) \cdot (q+\varepsilon)}{d_0}\).



		
	Then \(|N_G(PZ/Z)|\) is divisible by \(\frac{(n-1)\cdot
		|N_{\textrm{GO}^{\varepsilon}_2(q^{n-1})}(Q_1)|\cdot
		|\textrm{GO}^{-\varepsilon}_2(q)|}{2\cdot d_0\cdot |Z|}\).	
	

		
		As \(\frac{q^{n-1}-\varepsilon}{d_0}\) is an integer, we see that \(|N_G(PZ/Z)|\) is divisible by \(\frac{(n-1)\cdot 2
		\cdot
		(q+\varepsilon)}{|Z|}\).	
		
				If \(q=2\), then \(|Z|=1\), and hence \(\frac{(n-1)\cdot 2 \cdot
		(q+\varepsilon)}{|Z|}\geq (n-1)\cdot 2\geq 6>4\).	

	If \(q>2\), then we recall that \(n\geq 4\), so  \(\frac{(n-1)\cdot 2 \cdot
		(q+\varepsilon)}{|Z|}\geq \frac{(n-1)\cdot 2 \cdot (q+\varepsilon)}{2}\geq \frac{3\cdot 2 \cdot
		(3-1)}{2}=6>4\).
		
		
		Lemma \ref{start}(e) gives a
	prime \(r\) that divides \(|H|\) and \((n-1)\cdot 2\cdot (q+\varepsilon)\). As $r \ge 5$ and $r$ does not divide \(q^2-1\),
	it follows that  \(r\) divides \((n-1)\).
	
	However,
	\(q^{r-1}-1\) is divisible by \(r\) by Lemma \ref{help}(c) and then $r$ divides
	\(q^{2r-2}-1=(q^{r-1}-1)(q^{r-1}+1)\), which contradicts the minimality of \(k\).

	Hence \(k=n\).
	
	Now \(p \in \pi(H)\) divides \(q^{2n}-1=(q^n-1)(q^n+1)\), therefore \(p\) divides \(q^n+1\).
	Let	\(P \in \textrm{Syl}_p(\textrm{GO}^-_{2n}(q))\). Then $P \le \Omega^-_{2n}(q)\) and Lemma \ref{sgpGO} yields that
	\(|N_{\textrm{GO}^-_{2n}(q)}(P)|\) is divisible by $(q^n+1) \cdot 2n$.
	

We recall that \(|\textrm{GO}^-_{2n}(q):\Omega^-_{2n}(q)|=2\cdot d_0\), and therefore
\(|N_{\Omega^-_{2n}(q)}(P)|\) is
	divisible by
	\(\frac{(q^n+1)\cdot 2n}{2\cdot d_0}\). Without loss $PZ/Z \le H$, and then
	\(|N_G(PZ/Z)|\) is divisible by \(\frac{n\cdot 2(q^{n}+1)}{2\cdot d_0\cdot |Z|}=
	\frac{n\cdot (q^{n}+1)}{d_0\cdot |Z|}\).
	Since \(d_0\cdot |Z|= (4,q^n+1)\), we see that \(\frac{n\cdot (q^{n}+1)}{d_0\cdot |Z|}=
	\frac{q^n+1}{(4,q^n+1)}\cdot n\). Moreover, \(q^n+1\) is divisible by \((4,q^n+1)\) and hence
	\(|N_G(PZ/Z)|\) is
	divisible by \(n\).
	
	If \(n\geq 5\), then Lemma \ref{start}(e) implies that some prime $r \in \pi(H)$ divides \(n\).
	As
	\(q^{r-1}-1\) is divisible by \(r\) by Lemma \ref{help}(c), we see that $r$ divides
	\(q^{2r-2}-1=(q^{r-1}-1)(q^{r-1}+1)\). This contradicts the minimality of \(k\) because \(r-1\leq n-1<n=k\).
	
	Hence \(n=4\) and \(|N_G(PZ/Z)|\) is divisible by \(\frac{q^n+1}{(4,q^n+1)}\cdot n=\frac{q^4+1}{(4,q^4+1)}\cdot 4\).
	With our usual lemma it follows that \(|G_\alpha| \) is divisible by
	\(\frac{q^4+1}{(4,q^4+1)}=\frac{q^4+1}{d}\).
	We recall that $n=4$, and \(q^4\not \equiv -1\) modulo \(4\), hence \(|Z|=1\). So \(G\cong\Omega^-_8(q)\).
	
		
	Inspection of Table~8.52 and Table~8.53
	in~\cite{BHRD} shows that the only type of
	maximal subgroup of \(\Omega^-_8(q)\) of order divisible by \(\frac{q^4+1}{d}\) is
	\(\Omega^-_4(q^2).2\).
	
	Next we consider (3.57) in~\cite{Wilson}, which states that \(P\Omega^-_4(q^2)\cong
	\textrm{PSL}_2(q^4)\). As \(q^4 \not \equiv -1\) modulo \(4\), we also have that  \(P\Omega^-_4(q^2)\cong \Omega^-_4(q^2)\). Thus
	\(H\) lies in a subgroup isomorphic to \(\textrm{PSL}_2(q^4)\), bearing in mind that $H$ has odd order. Finally Hauptsatz~II\,8.27 in~\cite{Huppert-I} only leaves the possibility that \(G_{\alpha}\) is cyclic of order \(\frac{q^4+1}{d}\).
	
\end{proof}

	
	
	






\begin{lemma}\label{POmegaplus}
Suppose that
$n \ge 4$, that $q$ is a prime power and that $G=P\Omega^{+}_{2n}(q)$.

Then there is no set $\Omega$ such that Hypothesis \ref{hyp6} holds for $(G, \Omega)$.
\end{lemma}

\begin{proof}
We assume otherwise, with all the notation given in Hypothesis \ref{hyp6}.
Also, let $d:=(2,q-1)$.
First we recall that \(|H|\) and \(q\) are coprime by
Lemma \ref{Lie1}.
Let \(k \in \mathbb{N}\) be minimal such that there exists a prime
	\(p\in \pi(H)\) with the property that \(p\) divides \(q^{2k}-1\).
Then \(k\in \{n-1,n\}\) by Lemma \ref{Primesalternativ}.
	
	Let \(Z:=Z(\Omega^+_{2n}(q))\). Then \(G=\Omega^+_{2n}(q)/Z\) and \(|Z| \le 2\) by \cite{Wilson},~p.\,70 and p.\,80.
	

	
	Assume that \(k=n-1\). Then \(p\) divides \(q^{2n-2}-1=(q^{n-1}-1)(q^{n-1}+1)\), hence it divides exactly one of the factors, and we let
	\(\varepsilon\in \{-1,1\}\) be such
	that
	\(p\) divides \(q^{n-1}-\varepsilon\).
	
	Theorem~3.12 in~\cite{Wilson} implies that \(\textrm{GO}^+_{2n}(q)\) has a subgroup $D \times D_0$, where
	\(D \cong \textrm{GO}^{\varepsilon}_{2n-2}(q)\) and \(D_0\cong
	\textrm{GO}^{\varepsilon}_2(q)\).
	
	
	
	
	Let \(P\in \textrm{Syl}_p(D)\). As \(G=\Omega^+_{2n}(q)\) has index \(2\cdot d\) in \(\textrm{GO}^+_{2n}(q)\), we see that \(P\cap
	\Omega^+_{2n}q)=P\).
	Then Lemma~\ref{sgpGO} implies that  \(|N_D(P)|\) is divisible by
	\((q^{n-1}-\varepsilon)\cdot 2(n-1)\).
	Consequently \(|N_{\textrm{GO}^+_{2n}(q)}(P)|\) is divisible by \((q^{n-1}-\varepsilon)\cdot 2(n-1)
	\cdot
	|D_0|\).
	We turn to p. 71 in \cite{Wilson} to see that
	\(|D_0|=|\textrm{GO}^{\varepsilon}_{2}(q)|=2(q-\varepsilon)\). Therefore
	\(|N_{\textrm{GO}^+_{2n}(q)}(P)|\) is divisible by \((q^{n-1}-\varepsilon)\cdot 2(n-1) \cdot 2(q-\varepsilon)\).
	Then we recall that \(\Omega^+_{2n}(q)\) has index \(2\cdot d\) in \(\textrm{GO}^+_{2n}(q)\), and it follows that
	\(|N_{\Omega^+_{2n}(q)}(P)|\) is
	divisible by \(\frac{(q^{n-1}-\varepsilon)\cdot 2(n-1) \cdot
		2(q-\varepsilon)}{2d}=\frac{(q^{n-1}-\varepsilon)\cdot 2(n-1) \cdot (q-\varepsilon)}{d}\).
	
	Without loss $PZ/Z \le H$ and then
	\(|N_G(PZ/Z)|\) is divisible by \(\frac{(q^{n-1}-\varepsilon)\cdot 2(n-1) \cdot
		(q-\varepsilon)}{d\cdot |Z|}\). Since
	\(\frac{q^{n-1}-\varepsilon}{d}\) is an integer, we conclude that \(|N_G(PZ/Z)|\) is divisible by \(\frac{2(n-1) \cdot (q-\varepsilon)}{|Z|}\).
	
	
		
		We look at the possibilities for $q$ again:
	
		 If \(q=2\), then \(|Z|=1\). We recall that \(n\geq 4\), and therefore \(\frac{(n-1)\cdot 2 \cdot
		(q-\varepsilon)}{|Z|}= (n-1)\cdot 2 \cdot (2-\varepsilon)\geq 3\cdot 2 =6>4\).
	

		If \(q>2\), then \(\frac{(n-1)\cdot 2 \cdot
		(q-\varepsilon)}{|Z|}\geq \frac{(n-1)\cdot 2 \cdot (q-\varepsilon)}{2}\geq \frac{3\cdot 2 \cdot
		(3-1)}{2}=6>4\).
		
		In both cases Lemma \ref{start}(e) yields a prime $r \in \pi(G_\alpha)$ that divides
\((n-1)\cdot 2\cdot (q-\varepsilon)\).


As \(r \ge 5\) and $r$ does not divide  \(q^2-1\), it follows that \(r\) divides \((n-1)\).
But \(q^{r-1}-1\) is divisible by \(r\) by Lemma \ref{help}(c), which means that
	\(q^{2r-2}-1=(q^{r-1}-1)(q^{r-1}+1)\) is divisible by $r$, and this contradicts the minimal choice of $k$.
	

	Hence \(k=n\).	
	
	This means that \(p\) divides \(q^{2n}-1=(q^n-1)(q^n+1)\) and \(|H|\), so it divides \(q^n-1\).
	If \(n\) is even, then we write \(q^n-1=q^{2\cdot \frac{n}{2}}-1\) and we see that this
	contradicts the minimality of \(k\).
	
	Consequently \(n\) is odd and \(n\geq 5\).
	We turn to Theorem~3.12 in~\cite{Wilson} and let \(T_1 \le T \le \textrm{GO}^+_{2n}(q)\) be such that \(T \cong \textrm{GL}_n(q).2\)
	and $T_1 \cong \textrm{GL}(n,q)$.
	
	 Then
	by Satz~II\,7.3 in~\cite{Huppert-I} yields that  \(T_1\) has a cyclic subgroup \(C\) of order \(q^n-1\) and such that
	\(|N_{T_1}(C):C|=n\). Then the Sylow $p$-subgroups of $C$ have the same order as $P$, so without loss
	$P \le C$ and then \(N_{T_1}(C)\le N_{T_1}(P)\).
	We deduce that \(|N_{T_1}(P)\) is divisible by
	\((q^n-1)\cdot
	n\).
	As $p$ is odd, Lemma~(normgrow) yields that \(|N_T(P)|=2\cdot |N_{T_1}(P)|\) is divisible by
	\(2n\cdot (q^n-1)\).
	
	We recall that \(|\textrm{GO}^+_{2n}(q):\Omega^+_{2n}(q)|=2\cdot d\), so this means that \(|N_{\Omega^+_{2n}(q)}(P)|\) is
	divisible by
	\(\frac{2n\cdot (q^n-1)}{2\cdot d}\).
	Now \(|N_G(P)|\) is divisible by \(\frac{2n\cdot (q^n-1)}{2\cdot d\cdot
		|Z|}=\frac{n\cdot (q^n-1)}{d\cdot |Z|}=\frac{q^n-1}{q-1}\cdot \frac{n\cdot (q-1)}{d\cdot |Z|}\).
	
	Once more we look at $q$ more closely.
	
	If $q=2$, then \(|Z|=1=d\) so \(\frac{n\cdot (q-1)}{d\cdot |Z|}= n\geq 5>4\).
	
	If \(q=3\), then \(q^n\equiv 3^n\equiv (-1)^n\equiv -1 \not\equiv 1 \mod 4\) because \(n\) is odd. So \(|Z|=1\) and
	hence
	\(\frac{n\cdot (q-1)}{d\cdot |Z|}=\frac{n\cdot (3-1)}{2}\geq n\geq 5>4\).
	
	If \(q=4\), then \(|Z|=1=d\) so \(\frac{n\cdot (q-1)}{d\cdot |Z|}=n\cdot (q-1)= n\cdot 3\geq 15>4\).
	
	Finally, if \(q\geq 5\) then the fact that $n \ge 5$ implies that
	 \(\frac{n\cdot (q-1)}{d\cdot |Z|}\geq \frac{5\cdot (5-1)}{4}=5>4\).
	
	 We use Lemma \ref{start}(e) and find a prime $r \in \pi(H)$ such that
	$r$ divides \(n(q-1)\). Then \(r\)
	divides \(n\) and, by Lemma \ref{help}(c), $r$ also divides
	\(q^{r-1}-1\) and therefore
	\(q^{2r-2}-1=(q^{r-1}-1)(q^{r-1}+1)\).
	
	But \(r-1\leq n-1<n=k\), so this contradicts the minimality of \(k\).
	This final contradiction finishes the proof.
	
	\end{proof}

	

	









\subsection{Exceptional Lie type groups}

\begin{hyp}\label{exc}
In addition to Hypothesis \ref{hyp6}, we suppose that
$q$ is a prime power and $G$ is one of the following groups:
$^3D_4(q)$, $F_4(q)$, $^2F_4(q)$, $G_2(q)$, $^2G_2(q)$, $E_6(q)$, $^2E_6(q)$, $E_7(q)$ or $E_8(q)$.
\end{hyp}








\begin{lemma}\label{3D4q}
Suppose that Hypothesis \ref{exc} holds and that $G={}^3D_4(q)$.
Then $G_\alpha$ is cyclic of order $q^4-q^2+1$.
\end{lemma}


\begin{proof}
Lemma \ref{Primesalternativ} tells us that $p$ divides $\Phi_{12}(q)=q^4-q^2+1$.
Inspection of Theorem~4.3 in \cite{Wilson} shows that $G$ has a maximal
	subgroup \(M\) of structure \(C_{q^4-q^2+1}:4\). Let \(C\leq M\) be cyclic of order
	\(q^4-q^2+1\). Then without loss \(P \le C\), in fact $P$ is characteristic in $C$ and hence normal in $M$.
	Then $M=N_G(P)$ because $G$ is simple.
	Together with Lemma \ref{start}(e) this implies that
	\(|M:N_{H}(P)|\leq 4\) and therefore \(|H|\) is divisible by \(q^4-q^2+1\). The list in Theorem~4.3
	in~\cite{Wilson} only leaves the possibility that $H$ is contained in a maximal subgroup of $G$ that is isomorphic to $M$, and then it follows that
	\(G_{\alpha}\) is cyclic of order \(q^4-q^2+1\).
\end{proof}



\begin{lemma}\label{F4q}
Let $G:=F_4(q)$.
Then there is no set $\Omega$ such that Hypothesis \ref{exc} holds.

\end{lemma}


\begin{proof}
Assume for a contradiction that Hypothesis \ref{exc} holds for $G$.
Then
Lemma \ref{Primesalternativ}  implies that $p$ divides $\Phi_8(q)=q^4+1$ or $\Phi_{12}(q)=q^4-q^2+1$.

In the second case, where \(p\) divides \(q^4-q^2+1\), we consider Table~5.1
	in~\cite{LSS} and we see that \(G\) has a subgroup
	\(M\) of structure \({}^3\textrm{D}_4(q).3\). Let \(D\leq M\) be isomorphic to \({}^3\textrm{D}_4(q)\).
	Then Theorem~4.3
	in~\cite{Wilson} yields that \(D\) has a cyclic subgroup \(C\) of order
	\(q^4-q^2+1\), and without loss $P \le C$. Then
	\(|N_D(P)|=4(q^4-q^2+1)\) and, since $p\ge 5$, Lemma \ref{help}(a) yields that \(|N_M(P)|=3\cdot
	|N_D(P)|=12\cdot (q^4-q^2+1)\), contrary to Lemma \ref{start}(f).
	
	Therefore, the first case above holds, i.e. \(p\) divides \(q^4+1\).
	
	Then we use Table~5.1 in~\cite{LSS} and we see that \(G\) has
	a subgroup \(M\) of structure \((2,q-1).\Omega_9(q)\).
	Let $Z \unlhd M$ be such that $|Z|=(2,q-1)$ and let $Q \in \textrm{Syl}_p(M)$.  As $p$ and $|Z|$ are coprime,
	the coprime action of $Q$ on $Z$ implies that  \(N_{M/Z}(QZ/Z)=N_M(Q)Z/Z=N_M(Q)/Z\).
	We deduce that
	 \(|N_G(Q)|\) is divisible by \(|N_{M/Z}(QZ/Z)|\cdot |Z|\).
	Now \(M/Z \cong\Omega_9(q)\), so we can refer to the corresponding lemmas.
	If \(q\) is even, then \(\Omega_9(q)\cong \textrm{PSp}_8(q)\), where the normalizer of a Sylow $p$-subgroup has order
	divisible by $8$. This contradicts Lemma \ref{start}(f).
	
	If $q$ is odd, then we turn to the proof of Lemma \ref{POmega2n+1}, where 
	the normalizer of a Sylow $p$-subgroup also has order
	divisible by $8$, again contrary to Lemma \ref{start}(f). This final contradiction proves the lemma.
\end{proof}


\begin{lemma}\label{2F4q}
There is no set $\Omega$ such that Hypothesis \ref{exc} holds in the case where $G=\,^2F_4(q)$ or $q=2$ and
$G=\,^2F_4(q)'$.
\end{lemma}


\begin{proof}
If $q=2$ and
$G=\,^2F_4(q)'$, then our claim follows
from Remark~\ref{rem4.1}.

Next assume that $G=\,^2F_4(q)$ and that Hypothesis \ref{exc} holds.
Lemma \ref{Primesalternativ}  gives that $p$ divides
 \(\Phi_6(q)=q^2-q+1\) or \(\Phi_{12}(q)=q^4-q^2+1\).
	
In the first case, where \(p\) divides \(q^2-q+1\), the Main Theorem in~\cite{Malle2F4}
tells us that $G$ has a maximal subgroup \(M\) of structure \(\textrm{SU}_3(q):2\). Let \(S\leq M\) be isomorphic to
	\(\textrm{SU}_3(q)\) and, without loss, let  \(P\le S\).
	As \(q +1 \equiv 2^{2n+1} +1 \equiv (-1)^{2n+1} +1 \equiv -1 +1 \equiv 0 \mod 3\), we see that \((3,q+1)=3\) and hence
	\(Z(S)\) is cyclic of order \(3\). Therefore \(|N_S(P)|\) is divisible by \(3\). Moreover $p \ge 5$ by hypothesis and
	Lemma \ref{help}(a) yields that \(|N_M(P)|=2\cdot |N_S(P)|\), whence \(|N_G(P)|\)
	has order divisible by \(6\). This contradicts Lemma \ref{start}(f).
	
	In the second case, where \(p\) divides \(q^4-q^2+1\), we first observe that
	\(q^4-q^2+1=(q^2+q+1+\sqrt{2q}(q+1))(q^2+q+1-\sqrt{2q}(q+1))\). Let \(\varepsilon \in \{-1,1\}\) be such that
	\(p\)
	divides \((q^2+q+1+\varepsilon\sqrt{2q}(q+1))\). Then the Main Theorem in~[Malle1991]
	shows that \(G\) has a maximal subgroup \(M\) of structure \(C_{(q^2+q+1+\varepsilon\sqrt{2q}(q+1))}:12\).
	Without loss $P \le M$ and then $|N_G(P)|$ is divisible by \(12\), contrary to Lemma \ref{start}(f).
\end{proof}


\begin{lemma}\label{G2q}
There is no set $\Omega$ such that Hypothesis \ref{exc} holds in the case where $G=G_2(q)$.
\end{lemma}


\begin{proof}
We assume otherwise. Then Lemma \ref{Primesalternativ}  yields that every
\(p\in \pi(H)\) divides \(\Phi_{3}(q)=q^2+q+1\) or \(\Phi_6(q)=q^2-q+1\). Let
	\(\varepsilon\in \{-1,1\}\) be such that \(p\) divides \(q^2+\varepsilon q+1\). Then \(H\) contains a Sylow
	\(p\)-subgroup $P$
	of \(G\) by Lemma \ref{start}(e) and \(P\) lies in a maximal subgroup \(M\) of structure
	\(\textrm{SL}^{\varepsilon}_3(q):2\) by Table~4.1
	in~\cite{Wilson}.
	Let \(S\leq M\) be a subgroup of index \(2\) in \(M\) such that \(S \cong \textrm{SL}^{\varepsilon}_3(q)\).
	Then \(P\leq S\) and Lemma \ref{help}(a) implies that \(|N_M(P)|=2\cdot |N_S(P)|\).
	If \(Z(S)\neq 1\), then \(|Z(S)|=(3,q-\varepsilon)=3\) and so \(|N_S(P)|\) is divisible by \(3\).
	Hence \(|N_G(P)|\) is divisible by \(6\).
	If \(Z(S)=1\), then \(S\cong \textrm{PSL}^{\varepsilon}_3(q)\). Then one of
	\cite{Hartley},
	\cite{Mitchell} or Theorem~6.5.3
	in~\cite{GLS3} tells us that \(S\) has a subgroup \(F\) that is a Frobenius group with Frobenius
	kernel
	\(K\) of order \(q^2+\varepsilon q+1\) and Frobenius complement of order \(3\). As \(p\geq 5\), it follows that
	\(P\leq K\), in fact \(P=O_p(K)\), because Frobenius kernels are nilpotent. Hence \(N_F(P)=F\) and
	we see that $3$ divides  \(|N_F(P)|\), which means that
	\(|N_G(P)|\) is divisible by \(6\) in this case as well.
	In both cases Lemma \ref{start}(f) gives a contradiction.
\end{proof}


\begin{lemma}\label{2G2q}
Suppose that Hypothesis \ref{exc} holds and that $G=\,^2G_2(q)$.
Then $G_\alpha$ is cyclic of order $\frac{q-1}{2}$.
\end{lemma}


\begin{proof}
Lemma \ref{Primesalternativ} gives that $p$ divides
$\Phi_1(q)=q-1$, $\Phi_2(q)=q+1$ or
 \(\Phi_6(q)=q^2-q+1\).

We begin with the last case and assume that \(p\) divides \(q^2-q+1=(q+\sqrt{3q}+1)(q-\sqrt{3q}+1)\).
Then there exists \(\varepsilon\in \{-1,1\}\)
	such
	that \(p\) divides \(q-\varepsilon\sqrt{3q}+1\).
	Theorem~4.2 in \cite{Wilson} yields that $G$ has a
	subgroup \(M\) of structure \(C_{q-\varepsilon\sqrt{3q}+1}:6\).
	Then without loss $P \le M$ and
	 \(|N_G(P)|\) is divisible by~\(6\), which is impossible by Lemma \ref{start}(f).
	
	Next we assume that \(p\) divides \(q+1\). Then, by Theorem~4.2
	in~\cite{Wilson}, $G$ has a subgroup \(A\) of structure
	\(2^2\times D_{(q+1/2)}\). Let \(E\) be elementary abelian of order~\(4\) and let \(D\) be  a dihedral group of order
	\(\frac{q+1}{2}\) such that \(E\times D\leq A\). Moreover, let \(Y \le D\) be a subgroup of order \(p\). Then
	\(Y \unlhd D\) and consequently \(N_A(Y)=E\times D=A\). As \(|D|\) is even, we see that \(|N_A(Y)|\) is divisible
	by \(8\). Since we can choose $Y$ as a subgroup of $P$, this contradicts Lemma \ref{start}(f) as well.
	We conclude that \(p\) divides \(q-1\).
	
	Theorem~4.2 in~\cite{Wilson} shows that $G$ has a maximal subgroup \(M\) of structure \(2\times\textrm{PSL}_2(q)\). Let \(C\) be cyclic of order \(2\) and let \(L \cong \textrm{PSL}_2(q)\) be such that \(C\times L\leq M\). Let \(Y\le L\) be a subgroup of order \(p\).
	Then Satz~II\,8.3
	in~\cite{Huppert-I} shows that  \(N_L(Y)\) is a dihedral group of order \(2\cdot
	\frac{q-1}{2}\), whence \(N_M(Y)=C\times N_L(Y)\) has order \(2\cdot 2\cdot
	\frac{q-1}{2}\). As before we may choose $Y$ as a subgroup of $P$ and we deduce with Lemma \ref{start}(e) that
	 \(|H|\) is divisible by
	\(\frac{q-1}{2}\). Inspection of Hauptsatz~II\,8.27
	in~\cite{Huppert-I} shows that  the only subgroups of \(\textrm{PSL}_2(q)\) of order coprime to \(6\) and divisible
	by \(\frac{q-1}{2}\) are cyclic of order \(\frac{q-1}{2}\). Combining this with
	the list of maximal subgroups of \(G\) in
	Theorem~4.2 in~\cite{Wilson} reveals that \(G_{\alpha}\) is cyclic of order
	\(\frac{q-1}{2}\).
\end{proof}



\begin{lemma}\label{E6q}
There is no set $\Omega$ such that Hypothesis \ref{exc} holds in the case where $G=E_6(q)$.
\end{lemma}


\begin{proof}
We assume otherwise and use Lemma \ref{Primesalternativ}. Then $p$ divides
\(\Phi_5(q)=q^4+q^3+q^2+q+1\), \(\Phi_{8}(q)=q^4+1\),
	\(\Phi_9(q)=q^6+q^3+1\) or \(\Phi_{12}(q)=q^4-q^2+1\).
	
We begin at the end, so we	
assume that \(p\) divides \(q^4-q^2+1\). Then, by Table~5.1
	in~\cite{LSS}, we find a subgroup $M$ of \(G\) of structure
	\(({}^3\textrm{D}_4(q)\times (q^2+q+1)).3\). Let \(D\cong{}^3\textrm{D}_4(q)\) and let
	\(D_0\leq M\) be of order \(q^2+q+1\) such that \(D\times D_0\unlhd M\). Then Theorem~4.3
	in~\cite{Wilson} yields that \(D\) has a cyclic subgroup \(C\) of order
	\(q^4-q^2+1\).
	Without loss $P \le C$ and then  \(|N_D(P)|=4(q^4-q^2+1)\). Therefore \(|N_{D\times D_0}(P)|=4(q^4-q^2+1)\cdot |D_0|=4(q^4-q^2+1)\cdot
	(q^2+q+1)\). Then $|N_G(P)|$ is divisible by 12, contrary to Lemma \ref{start}(f).

	
	Next assume that \(p\) divides \(q^4+1\) or \(q^4+q^3+q^2+q+1\) and set \(h:=(4,q-1)\). Then inspection of Table~5.1
	in~\cite{LSS} yields that \(G\) has a subgroup
	\(M\) of structure \(h.(P\Omega^+_{10}(q)\times \frac{(q-1)}{h})\). Without loss \(P\le M\) and we let \(Z\) be a normal subgroup of \(M\) of
	order \(h\). We recall that \(p\geq 5\), so \(P\) and \(Z\) have coprime orders and it follows that
	\(N_{M/Z}(PZ/Z)=N_M(P)Z/Z\).
	In particular \(|N_M(P)|\) is divisible by \(|N_{M/Z}(PZ/Z)|\).
	Now \(M/Z\cong P\Omega^+_{10}(q)\times \frac{(q-1)}{h}\). Let \(A\leq M/Z\) be isomorphic to \(P\Omega^+_{10}(q)\).
	As \(PZ/Z\in \textrm{Syl}_p(A)\), we can argue as for \(P\Omega^+_{2n}(q)\) and we see that
	\(|N_{\Omega^+_{10}(q)}(PZ/Z)|\) is divisible by \((5-1)\cdot (q+1)=4(q+1)\) or by
	\(\frac{5\cdot (q-1)}{(4,q^5-1)}\).
	
	In the first case
	\(|N_G(P)|\) has order divisible
	by \(4(q+1)>4\), which together with Lemma \ref{start}(e) gives a contradiction. Here we use that $\pi(H)$ does not contain any elements dividing $q+1$.
	In the second case  \(N_G(P)\) has order divisible by \(\frac{5\cdot (q-1)}{(4,q^5-1)}\) and we distinguish the
	cases for $q$.
	
	If \(q=2\), then \((4,q^5-1)=1\) so \(\frac{5\cdot (q-1)}{(4,q^5-1)}=5>4\).
	
	If \(q=3\), then \((4,q^5-1)=(4,3^5-1)=2\). So \(\frac{5\cdot (q-1)}{(4,q^5-1)}=\frac{5\cdot (3-1)}{2}=
	5>4\).
	
	If \(q=4\), then \((4,q^5-1)=1\) and therefore \(\frac{5\cdot (q-1)}{(4,q^5-1)}=5\cdot (q-1)= 5\cdot 3= 15>4\).
	
	Finally, if
	\(q\geq 5\), then \(\frac{5\cdot (q-1)}{(4,q^5-1)}\geq \frac{5\cdot (5-1)}{4}=5>4\).
	
	In all cases we find a prime \(r \in \pi(H)\) that divides \(5\cdot (q-1)\).
	If \(r=5\), then the fact that \(q \neq r\) (Lemma \ref{Lie1}) forces
	\(q^4-1\) to be divisible by \(r\). Then Theorem~(10-1)
	in~\cite{GL} shows that \(G\) has
	\(r\)-rank at least \(2\), contrary to Theorem \ref{strongcases}.
	
	Therefore \(r\) divides \(q-1\). Then Theorem~(10-1)
	in~\cite{GL} yields that \(G\) has
	\(r\)-rank \(6\), which is a contradiction again.
	
	The last possibility is that \(p\) divides \(q^6+q^3+1\).
	
	Let \(e:=(3,q-1)\). Then Table~5.1
	in~\cite{LSS} shows that \(G\) has a subgroup
	\(M\) of structure \(\textrm{PSL}_3(q^3).(e\times 3)\). Let \(L\leq M\) be isomorphic to \(\textrm{PSL}_3(q^3)\).
	Without loss \(P\le L\) and then
	Lemma \ref{help}(a) gives that \(|N_M(P)|=3e\cdot |N_L(P)|\).
	Now we consider \cite{Hartley} and
	\cite{Mitchell}, together with Theorem~6.5.3
	in~\cite{GLS3}, and we deduce that \(L\) has a subgroup \(F\) that is a Frobenius group with Frobenius
	kernel
	\(K\) of order \(\frac{q^6+ q^3+1}{(3,q^3-1)}\) and Frobenius complement of order~\(3\). As \(p\geq 5\) and as Frobenius kernels are nilpotent, we see that \(P=O_p(K)\). Hence \(N_F(P)=F\) has order divisible by
	\(3\). Then \(|N_M(P)|=3e\cdot |N_L(P)|\) is divisible by \(9\), contrary to Lemma \ref{start}(f).
\end{proof}


\begin{lemma}\label{2E6q}
There is no set $\Omega$ such that Hypothesis \ref{exc} holds in the case where $G={}^2E_6(q)$.
\end{lemma}


\begin{proof}
We assume otherwise and then Lemma \ref{Primesalternativ} shows that
 \(p\) divides \(\Phi_{10}(q)=q^4-q^3+q^2-q+1\), \(\Phi_{8}(q)=q^4+1\),
	\(\Phi_{18}(q)=q^6-q^3+1\) or \(\Phi_{12}(q)=q^4-q^2+1\).
	
	First we assume that \(p\) divides \(q^4-q^2+1\). Then Table~5.1
	in~\cite{LSS} gives that \(G\) has a subgroup
	\(M\) of structure \(({}^3\textrm{D}_4(q)\times (q^2-q+1)).3\).
	Let \(D\leq M\) be isomorphic to \({}^3\textrm{D}_4(q)\) and let
	\(D_0\leq M\) be of order \(q^2-q+1\) such that \(D\times D_0\trianglelefteq M\). Then we use Theorem~4.3
	in~\cite{Wilson}, which gives that \(D\) has a cyclic subgroup \(C\) of order
	\(q^4-q^2+1\).
	Without loss $P \le C$ and we have seen in Lemma~\ref{3D4q} that
	\(|N_D(P)|=4(q^4-q^2+1)\). Therefore \(|N_{D\times D_0}(P)|=4(q^4-q^2+1)\cdot |D_0|=4(q^4-q^2+1)\cdot
	(q^2-q+1)\).
	Lemma \ref{help}(a), together with the fact that \(p\geq 5\), yields that \(|N_M(P)|=3\cdot
	|N_{D\times D_0}(P)|=12\cdot (q^4-q^2+1) \cdot (q^2-q+1)\).
	This contradicts Lemma \ref{start}(f).
	
	
	Assume that \(p\) divides \(q^4+1\) and let \(h:=(4,q+1)\). Then we turn to Table~5.1
	in~\cite{LSS} and we find a subgroup
	\(M\) of $G$ of structure \(h.(P\Omega^-_{10}(q)\times \frac{(q+1)}{h})\). Without loss \(P\le M\), and we let
	\(Z\) denote a normal subgroup of \(M\) of
	order \(h\). Then \(P\) acts coprimely on \(Z\) and consequently \(N_{M/Z}(PZ/Z)=N_M(P)Z/Z\).
	In particular
	\(|N_M(P)|\) is divisible by \(|N_{M/Z}(PZ/Z)|\).
	We note that \(M/Z \cong P\Omega_-(10,q)\times (q-1)/h\). Let \(A\leq M/Z\) be isomorphic to \(P\Omega^-_{10}(q)\).
	As \(PZ/Z\in \textrm{Syl}_p(A)\), we can argue as for the group \(P\Omega^-_{2n}(q)\) (see Lemma ...).
	Then we deduce that
	\(|N_{\Omega^-_{10}(q)}(PZ/Z)|\) is divisible by \(\frac{(5-1)\cdot (q^4+1)\cdot
		2(q-1)}{(4,q^5+1)}=\frac{q^4+1}{(2,q+1)}\cdot \frac{8\cdot (q-1)\cdot (2,q+1)}{(4,q^5+1)}\).
	Therefore \(|N_{M/Z}(PZ/Z)|\) is divisible by \(\frac{8\cdot (q-1)\cdot (2,q+1)}{(4,q^5+1)}\) and so
	\(|N_G(P)|\) is divisible
	by \(\frac{8\cdot (q-1)\cdot (2,q+1)}{(4,q^5+1)}\).
	
	If \(q=2\), then \((2,q+1)=1=(4,q^5+1)\). Therefore \(\frac{8\cdot (q-1)\cdot (2,q+1)}{(4,q^5+1)}=8>4\).
	
	If \(q>2\) is even, then \((2,q+1)=1=(4,q+1)\) again, and if \(q>2\) is odd, then \((2,q+1)=2\leq (4,q^5-1)\).
	In this case
	\(\frac{8\cdot (q-1)\cdot (2,q+1)}{(4,q^5+1)}\geq \frac{8\cdot (q-1)}{2}=4(q-1)\geq 4(3-1)= 8>4\).
	So in all cases, we see that  \(|N_G(P)|\) is divisible by \(\frac{8\cdot (q-1)\cdot(2,q+1)}{(4,q^5+1)}>4\), which contradicts Lemma \ref{start}(e) and the fact that \(|H|\) and \(q-1\) are coprime.
	
	Almost the same argument yields that $p$ does not divide \(q^4-q^3+q^2-q+1\), again using
	Table~5.1
	in~\cite{LSS}, but this time with a subgroup
	\(M\) of $G$ of structure \(h.(P\Omega^-_{10}(q)\times (q+1)/h)\).
	As in the previous paragraph we deduce that
	\(|N_G(P)|\) is divisible by \(\frac{5\cdot (q+1)}{(4,q^5+1)}>4\), so we find a prime  \(r\) dividing \(5\cdot (q+1)\) and \(|H|\). If \(r=5\), then the fact that \(q \neq r\) forces  \(q^4-1\) to be divisible by \(r\). Then Theorem~(10-1)
	in~\cite{GL} shows that \(G\) has
	\(r\)-rank at least \(2\), contrary to Theorem \ref{strongcases}.
	
	Therefore \(r\) divides \(q+1\) and the same theorem yields that  \(G\) has
	\(r\)-rank at least \(4\), which gives another contradiction.
	
	We conclude that \(p\) divides \(q^6-q^3+1\).
	

	
	Let \(e:=(3,q+1)\). Inspection of Table~5.1
	in~\cite{LSS} shows that \(G\) has a subgroup
	\(M\) of structure \(\textrm{PSU}(3,q^3).(e\times 3)\). Let \(L\leq M\) be isomorphic to \(\textrm{PSU}_3(q^3)\) and, without loss, suppose that $P \le L$.
	Then Lemma \ref{help}(a) yields that \(|N_M(P)|=3e\cdot |N_L(P)|\).
	By \cite{Hartley} and \cite{Mitchell}, or by Theorem~6.5.3 in \cite{GLS3}, respectively, we know that
	\(L\) has a subgroup \(F\) that is a Frobenius group with Frobenius
	kernel
	\(K\) of order \(\frac{q^6- q^3+1}{(3,q^3+1)}\) and Frobenius complements of order~\(3\). We recall that
	 \(p\geq 5\), and since Frobenius kernels are nilpotent we conclude that \(P=O_p(K)\). Hence \(N_F(P)=F\)
	 has order
	by \(3\) and \(|N_M(P)|=3e\cdot |N_L(P)|\) is divisible by \(9\), contrary to Lemma \ref{start}(f). 	
		\end{proof}
	



\begin{lemma}\label{E7q}
There is no set $\Omega$ such that Hypothesis \ref{exc} holds in the case where $G=E_7(q)$.
\end{lemma}


\begin{proof}
We assume otherwise and then Lemma \ref{Primesalternativ} shows that
 \(p\) divides
\(\Phi_5(q)=q^4+q^3+q^2+q+1\),
	\(\Phi_7(q)=q^6+q^5+q^4+q^3+q^2+q+1\), \(\Phi_8(q)=q^4+1\), \(\Phi_9(q)=q^6+q^3+1\), \(\Phi_{10}(q)=q^4-q^3+q^2-q+1\),
	\(\Phi_{12}(q)=q^4-q^2+1\), \(\Phi_{14}(q)=q^6-q^5+q^4-q^3+q^2-q+1\) or \(\Phi_{18}(q)=q^6-q^3+1\).
	
First assume that \(p\) divides \(\Phi_5(q)\) or \(\Phi_{10}(q)\). Then there exists \(\varepsilon\in \{-1,1\}\) such that \(p\)
	divides \(q^4+\varepsilon q^3+q^2+\varepsilon q+1\). Let \(e:=(3,q-\varepsilon)\). Moreover we set
	\(\textrm{E}^{1}_6(q):=\textrm{E}_6(q)\) and \(\textrm{E}^{-1}_6(q):={}^2\textrm{E}_6(q)\) for the remainder of this proof.
	
	Table~5.1
	in~\cite{LSS} yields that $G$
	has
	a subgroup \(M\) of structure \(e.(\textrm{E}^{\varepsilon}_6(q)\times \frac{(q-\varepsilon)}{e})\),
	and without loss $P \le M$.
	Let  \(Z\unlhd M\) be of
	order \(e\). Then $P$ acts coprimely on $Z$ and therefore
	\(N_{M/Z}(PZ/Z)=N_M(P)Z/Z\).
	In particular
	\(|N_M(P)|\) is divisible by \(|N_{M/Z}(PZ/Z)|\).
	We see that \(M/Z\cong\textrm{E}^{\varepsilon}_6(q)\times \frac{(q-\varepsilon)}{e}\) and we let
	\(A\leq M/Z\) be isomorphic
	to \(\textrm{E}^{\varepsilon}_6(q)\).
	Then the arguments for the groups
	(\(\textrm{E}_6(q)\) and ~(\({}^2\textrm{E}_6(q)\) show that there exists a \(p\)-subgroup
	\(Q\) such that
	\(|N_{\textrm{E}^{\varepsilon}_6(q)}(Q)|\) is divisible by



	\(\frac{5\cdot (q-\varepsilon)}{(4,q^5-\varepsilon)}\).

	
	Without loss $P$ is a Sylow \(p\)-subgroup of the full pre-image of \(Q\) in \(M\). Then
	\(|N_M(P)|\)
	is divisible by \(\frac{5\cdot (q-\varepsilon)}{(4,q^5-\varepsilon)}\).
	
	If \(q=2\), then \((4,q^5-\varepsilon)=1\) and therefore \(\frac{5\cdot (q-\varepsilon)}{(4,q^5-\varepsilon)}\geq 5>4\).
	
	If \(q=3\), then \((4,q^5-\varepsilon)=(4,3^5-\varepsilon)=q-\varepsilon\). Hence \(\frac{5\cdot
		(q-\varepsilon)}{(4,q^5-\varepsilon)}=\frac{5\cdot (q-\varepsilon)}{q-\varepsilon}=5>4\).
		
	If \(q=4\), then \((4,q^5-1)=1\) so \(\frac{5\cdot (q-\varepsilon)}{(4,q^5-\varepsilon)}=5\cdot
	(q-\varepsilon)\geq 5\cdot 3= 15>4\).
	
	Finally, if
	\(q\geq 5\), then \(\frac{5\cdot (q-\varepsilon)}{(4,q^5-\varepsilon)}\geq \frac{5\cdot (5-1)}{4}=5>4\).
	In all cases there
	exists a prime \(r\) dividing \(5\cdot (q-\varepsilon)\) and \(|H|\), by Lemma \ref{start}(e).
	First assume that \(r=5\). As \(q\neq r\) by Lemma \ref{Lie1}, this
	implies that \(q^4-1\) is divisible by \(r\). Then Theorem~(10-1)
	in~\cite{GL} shows that \(G\) has
	\(r\)-rank at least~\(2\), contradicting Theorem \ref{strongcases}.
	
	Therefore \(r\) divides \(q-\varepsilon\). Then the same theorem implies that \(G\) has
	\(r\)-rank \(7\), which is again imposible.  Therefore \(p\) does not divide \(q^4+\varepsilon q^3+q^2+\varepsilon q+1\).
	
	Next we assume that  \(p\) divides \(\Phi_9(q)\) or \(\Phi_{18}(q)\).
	Then there exists \(\varepsilon\in \{-1,1\}\) such that \(p\)
	divides \(q^6+\varepsilon q^3+1\). We use the notation from the previous case and refer to Table~5.1
	in~\cite{LSS} again.
	
		Then we see that $G$ has a subgroup  \(M\) of structure \(e.(\textrm{E}^{\varepsilon}_6(q)\times (q-\varepsilon)/e)\) and that, without loss, $P \le M$. Let \(Z\unlhd M\) be of
	order \(e\). Then the coprime action of $P$ on $Z$ yields that
	 \(N_{M/Z}(PZ/Z)=N_M(P)Z/Z\).
	
	As \(M/Z\cong \textrm{E}^{\varepsilon}_6(q)\times \frac{(q-\varepsilon)}{e}\), we can choose a subgroup
	 \(A\leq M/Z\) that is isomorphic
	to \(\textrm{E}^{\varepsilon}_6(q)\).
	Then the arguments for
	\(\textrm{E}_6(q)\) and \({}^2\textrm{E}_6(q)\) show that there exists a \(p\)-subgroup
	\(Q\) such that
	\(|N_{\textrm{E}^{\varepsilon}_6(q)}(Q)|\) is divisible by \(9\).
	
	
	We may suppose that $P$ is a Sylow \(p\)-subgroup of the
	full pre-image of \(Q\) in \(M\). Then \(\|N_M(P)\) is divisible by \(9\), and Lemma \ref{start}(f) gives a contradiction.
	Therefore \(p\) does not divide \(q^6+\varepsilon q^3+1\).
	
	
	
	Next we assume that \(p\) divides \(\Phi_8(q)=q^4+1\) and we let \(e:=(3,q-1)\). Using Table~5.1
	in~\cite{LSS} we find a subgroup
	\(M\) of $G$ of structure \(e.(\textrm{E}_6(q)\times \frac{(q-1)}{e})\). Without loss
	\(P\le M\) and we let \(Z\) be a
	normal subgroup of \(M\) of
	order \(e\). Then the
	coprime action of $P$ on $Z$ yields that
	\(N_{M/Z}(PZ/Z)=N_M(P)Z/Z\).
	In particular \(|Nor_M(P)|\) is divisible by \(|N_{M/Z}(PZ/Z)|\).
	As \(M/Z\cong \textrm{E}_6(q)\times \frac{(q-1)}{e}\), we find a subgroup \(A\leq M/Z\) such that
	 \(A \cong \textrm{E}_6(q)\), and then the arguments for
	\(\textrm{E}_6(q)\) give a \(p\)-group \(Q\) such that
	\(|N_{\textrm{E}_6(q)}(Q)|\) is divisible by \(4\cdot (q+1)\).
	Wihout loss $P$ is
	 a Sylow \(p\)-subgroup of the
	full
	pre-image of \(Q\) in \(M\). Then \(|N_M(P)|\) is divisible by \(4\cdot (q+1)\), which
	together with Lemma \ref{start}(e) shows that some prime $r \in \pi(H)$ divides $q+1$.
	 Theorem~(10-1)
	in~\cite{GL} shows that \(G\) has
	\(r\)-rank \(7\), contrary to Theorem \ref{strongcases}.
	
	
	We
	assume that \(p\) divides \(\Phi_{12}(q)=q^4-q^2+1\) and we argue as in the previous paragraph,
	with the same type of subgroup and the same notation.
	As \(M/Z\cong \textrm{E}_6(q)\times \frac{(q-1)}{e}\), we can choose \(A\leq M/Z\) such that
	 \(A\cong \textrm{E}_6(q)\) and the arguments for
	\(\textrm{E}_6(q)\) give a $p$-subgroup \(Q\) such that
	\(|N_{\textrm{E}_6(q)}(Q)|\) is divisible by \(12\).
	Then, without loss, $P$
	is a Sylow \(p\)-subgroup of the full
	pre-image
	of \(Q\) in \(M\) and \(|N_G(P)|\) is divisible by \(12\), contrary to Lemma \ref{start}(f).
	
	Finally, \(p\) divides \(\Phi_7(q)\) or \(\Phi_{14}(q)\). We let \(\varepsilon\in \{-1,1\}\) be such that \(p\)
	divides \(q^6+\varepsilon q^5+q^4+ \varepsilon q^3 +q^2+ \varepsilon q+ 1\).
	Let \(f:=\frac{(4,q-\varepsilon)}{(2,q-1)}\). Inspection of Table~5.1
	in~\cite{LSS} shows that $G$ has a subgroup \(M\) of structure \(f.\textrm{PSL}^{\varepsilon}_8(q)\).
	Without loss $P \le M$ and we choose \(Z\unlhd M\) of
	order \(f\). The coprime action of $P$ on $Z$ yields that \(N_{M/Z}(PZ/Z)=N_M(P)Z/Z\), whence
	 \(|N_M(P)|\) is divisible by \(|N_{M/Z}(PZ/Z)|\).
	
	
	As \(M/Z\cong \textrm{PSL}^{\varepsilon}_8(q)\), we can refer to the arguments for
	(\(\textrm{PSL}_n(q)\) and (\(\textrm{PSU}_n(q)\), and then we find a $p$-subgroup
	\(Q\) such that
	\(|N_{\textrm{E}_6(q)}(Q)|\) is divisible by \(\frac{7\cdot (q-\varepsilon)}{(8,q-\varepsilon)}\).
	Then \(|N_G(P)|\) is divisible by \(7\), and Lemma \ref{start}(e) leaves two possibilities:
	 \(q=7\), which contradicts Lemma 9.1, or $7$ divides
	\(q^6-1\). Then Theorem~(10-1)
	in~\cite{GL} shows that \(r_7(G)\ge 3\), contrary to Theorem \ref{strongcases}.
\end{proof}





\begin{lemma}\label{E8q}
There is no set $\Omega$ such that Hypothesis \ref{exc} holds in the case where $G=E_8(q)$.
\end{lemma}


\begin{proof}
We assume otherwise and then Lemma \ref{Primesalternativ} shows that
 \(p\) divides
 \(\Phi_7(q)=q^6+q^5+q^4+q^3+q^2+q+1\), \(\Phi_9(q)=q^6+q^3+1\),
	\(\Phi_{14}(q)=q^6-q^5+q^4-q^3+q^2-q+1\), \(\Phi_{15}(q)=q^8-q^7+q^5-q^4+q^3-q+1\), \(\Phi_{18}(q)=q^6-q^3+1\),
	\(\Phi_{20}(q)=q^8-q^6+q^4-q^2+1\), \(\Phi_{24}(q)=q^8-q^4+1\) or \(\Phi_{30}(q)=q^8+q^7-q^5-q^4-q^3-q+1\).
	
First we assume that \(p\) divides \(\Phi_{15}(q)\) or \(\Phi_{30}(q)\). Let \(T\leq G\) be as
	in~\cite{LSS}. Then without loss \(P\le T\), and
	Table~5.2
	in~\cite{LSS} implies that  \(|N_G(P)|\) is divisible by \(30\), contrary to Lemma \ref{start}(f).

Now we assume that \(p\) divides \(\Phi_{24}(q)=q^8-q^4+1\). Then by Table~5.1
	in~\cite{LSS} we find a subgroup \(M\) of $G$ of structure \(\textrm{PSU}_3(q^4).8\). Let \(U\leq M\) be isomorphic to \(\textrm{PSU}_3(q^4)\). Then without loss $P \le U$ and Lemma \ref{help}(a) implies that
	\(|N_M(P)|=8\cdot |N_U(P)|\).
	In particular $|N_G(P)|$ is divisible by $8$, contrary to Lemma \ref{start}(f).
	

	Assume that \(p\) divides \(\Phi_{20}(q)=q^8-q^6+q^4-q^2+1\). Then, with Table~5.1
	in~\cite{LSS}, we find  a subgroup \(M\) of $G$ of structure \(\textrm{SU}_5(q^2).4\). Let \(U\leq M\) be isomorphic to \(\textrm{SU}(5,q^2)\).
	
	\smallskip
	\textbf{Case 1:} \(Z(U)\neq 1\).
	
	Then \(|Z(U)|=(5,q^2-1)=5\). Without loss \(P\le U\) and in particular \(Z(U)\leq
	N_G(P)\). Moreover \(|N_M(P)|=4\cdot |N_U(P)|\) by Lemma \ref{help}(a), so \(|N_G(P)|\) is divisible by 20.
	Then Lemma \ref{start}(e) implies that $5 \in \pi(G_\alpha)$.
	We know that \(q\neq 5\) by Lemma 9.1, so
	\(q^4-1\) is divisible by \(5\). But then Theorem~(10-1)
	in~\cite{GL} shows that \(r_5(G) \ge 4\), contrary to Theorem \ref{strongcases}.
	
	\smallskip
	\textbf{Case 2:} \(Z(U)=1\).
	
	Then \(U\cong \textrm{PSU}_5(q^2)\) and we refer to our results about
	(\(\textrm{PSU}_n(q)\). Without loss \(P\leq U\) and consequently \(|N_G(P)|\) is divisible by \(\frac{(q^5+1)\cdot
		5}{(q+1)\cdot (5,q+1)}=\frac{q^5+1}{q+1}\cdot 5\).
		Lemma \ref{start}(e) implies that $5 \in \pi(G_\alpha)$, which gives two possibilties:
	
	\(q=5\), which contradicts Lemma 9.1, or \(q^4-1\) is divisible by \(5\).
	In the second case we use Theorem~(10-1)
	in~\cite{GL}, and then we see that \(r_5(G) \ge 4\), contrary to
	Theorem \ref{strongcases}.
	
	Next we assume that \(p\) divides \(\Phi_9(q)\) or \(\Phi_{18}(q)\). Then there exists \(\varepsilon\in \{-1,1\}\) such that \(p\)
	divides \(q^6+\varepsilon q^3+1\). Let \(d:=(2,q-1)\). Inspection of Table~5.1
	in~\cite{LSS} shows that $G$ has a subgroup \(M\) of structure
	\(d.(\textrm{PSL}_2(q) \times \textrm{E}_7(q))\). Let \(Z\unlhd M\) have order \(d\).
	Without loss $P \le M$ and thus, by coprime action,
	 \(N_{M/Z}(PZ/Z)=N_M(P)Z/Z\).
	As \(M/Z\cong \textrm{E}^{\varepsilon}_6(q)\times \frac{(q-\varepsilon)}{e}\), we may choose \(A\leq M/Z\)
	such that \(A \cong \textrm{E}_7(q)\).
	Our arguments for \(\textrm{E}_7(q)\) show that there exists a \(p\)-group
	\(Q\) such that
	\(|N_{\textrm{E}_7(q)}(Q)|\) is divisible by \(9\).
	Then without loss $P$ is
	a Sylow \(p\)-subgroup of the
	full pre-image of \(Q\) in \(M\). Thus \(|N_G(P)\) is divisible by \(9\), contrary to Lemma \ref{start}(f).
	
	
	Assume that \(p\) divides \(\Phi_7(q)\) or \(\Phi_{14}(q)\). Then there exists \(\varepsilon\in \{-1,1\}\) such that \(p\)
	divides \(q^6+\varepsilon q^5+q^4+\varepsilon q^3 + q^2 + \varepsilon q+1\). We let \(d:=(2,q-1)\) and we argue as
	in the previous paragraph, with the same subgroup $M$, but this time
	we conclude that  \(|N_M(P)|\) is divisible by~\(7\).
	Thus $7 \in \pi(H)$.
	If \(q=7\), then Lemma 9.1 gives a contradiction.
	Otherwise \(q^6-1\) is divisible by \(7\) and Theorem~(10-1)
	in~\cite{GL} shows that \(r_7(G) \ge 4\), contrary to Theorem \ref{strongcases}.

\end{proof}



















\begin{thm}\label{Liecomplete}
Suppose that $G$ is a simple group of Lie type satisfying Hypothesis \ref{hyp6}.
Then one of the following holds:

$G=\textrm{PSL}_2(q)$, where the defining characteristic $r$ is at least $5$, $n \in \mathbb{N}$, and $G_\alpha$ has index 2 in the normalizer of a  
Sylow $r$-subgroup.

$G=\textrm{PSU}_4(2)$ or $G=\textrm{PSU}_4(3)$, and $|G_\alpha|=5$. 

$G=\textrm{PSp}_4(q)$ and $G_\alpha$ is cyclic of order $\frac{q^2+1}{(2,q+1)}$.

$G=P\Omega^-_8(q)$ and $G_\alpha$ is cyclic of order $\frac{q^4+1}{(2,q+1)}$.

$G=\,^3D_4(q)$ and $G_\alpha$ is cyclic of order $q^4-q^2+1$.

$G=\,^2G_2(q)$ and $G_\alpha$ is cyclic of order $\frac{q-1}{2}$.
\end{thm}


\begin{proof}
Theorem \ref{strongcases} gives two cases:
In Case (ii) we inspect Section 7 for the groups that are mentioned.
In most cases of Theorem \ref{L2q-main}, we see immediately that $G_\alpha$
has order divisible by $2$ or $3$.
The only remaining possibility is that $G_\alpha$ has index 2 in the normalizer of a  
Sylow $p$-subgroup of $G$, and for this to be compatible with Hypothesis \ref{hyp6}, we must have that $p \ge 5$.
Lemma \ref{PSLPSU3} gives that $\textrm{PSU}_3(p^n)$ only occurs with point stabilizers of order divisible by $3$.

In Case (i) of Theorem \ref{strongcases}, we first consider the situation where the defining characteristic
is in $\pi(H)$. Then we use Lemma \ref{Lie1}, which gives a special case of the statement about $\textrm{PSL}_2(q)$.
In the cross-characteristic case, we go through the results for the individual series of groups of Lie type.
The only possibilities come from the lemmas \ref{PSU}, \ref{Psp}, \ref{POmegaminus}, \ref{3D4q} and 
\ref{2G2q}. All other series of groups do not give examples satisfying Hypothesis \ref{hyp6}.
\end{proof}





\section{Proof of Theorem \ref{main4fpsimple}}



The hypothesis of Theorem \ref{main4fpsimple} tells us that $G$ is a finite simple non-abelian group that acts transitively and with fixity 4 on a set $\Omega$. 
Let $f$ denote the maximum number of fixed points of involutions in $G$.

If Case (1) of Theorem \ref{3and2simple} holds, then $1 \le f \le 4$ and 
$G$ is isomorphic to $\textrm{PSL}_2(q)$, $\textrm{Sz}(q)$ or $\textrm{PSU}_3(q)$ for some power $q$ of $2$.
Then the corresponding results from Section 7 apply, more precisely Lemmas \ref{L2q-main}, \ref{Suzuki} and \ref{PSLPSU3}.
All these cases are contained in Table \ref{MainTable}.



If Case (2) of Theorem \ref{3and2simple} holds, then $f \le 3$ and 
$G$ is isomorphic to $\textrm{PSL}_2(q)$,  to $\textrm{PSL}_3(q)$ or to $\textrm{PSU}_3(q)$ for some odd prime power $q$.
Again we turn to Section 7, this time to Theorem \ref{L2q-main} and Lemma \ref{PSLPSU3}. 
All the possibilities there are covered by Table \ref{MainTable}.


If Case (3) of Theorem \ref{3and2simple} holds, then $f=4$,
 $G$ has sectional $2$-rank at most $4$ and 
there is a list of possibilities for $G$. We have already seen that all possibilities for 
$2$-dimensional Lie type groups are contained in Table \ref{MainTable}, then we refer to 
Lemma \ref{Inv-lem:Ser} for the Lie type groups of dimension at least 3,
and the only remaining group is the sporadic group Ly.
But this was discussed in Lemma \ref{Lyons}.

In Case (4) of Theorem \ref{3and2simple} we have a number of sub-cases, all of which
have been treated in Section 8. For the strongly 3-embedded case we use Lemma \ref{str3emb}, and all possibilities listed there are contained in Table \ref{MainTable}. This covers Cases (4) (a) and (b).
For (c) we use Lemma \ref{odd3el9}, which is also a case from our table.
Cases (d) and (e) are covered by Lemma \ref{extraspecialWreath} and do not give any examples.

Finally, we turn to Case (5), and hence to Section 9.
Given Hypothesis \ref{hyp6}, we have already seen the groups that are listed in Theorem \ref{strongcases}(ii), so we only need to consider those with cyclic Sylow $p$-subgroups.
The only example coming from an alternating group here is $\Alt_7$, as explained in Lemma \ref{Altcyclic}, and this is included in Table \ref{MainTable}.
The only sporadic examples are $\textrm{M}_{11}$ and $\textrm{M}_{22}$, as can be seen in Lemma \ref{spor}, and these are also in our table.
A summary of the results about Lie type groups is given in 
Theorem \ref{Liecomplete}, and this is where the remaining cases from table \ref{MainTable} come from.







\section{Appendix: GAP code and more fixed point information} 







For a Table of Marks \texttt{t} of a group \(G\) and for a positive integer \texttt{k}
the following function \texttt{TestTom(t,k)} determines if there are transitive actions of \(G\) with fixity \(k\).
If there are none, then an empty list is returned.
Otherwise the output includes the isomorphism type of the point stabilizers, along with some other information, for every type of fixity k action.


\begin{verbatim}
TestTom:=function(t,k)
  #t Table of Marks of a group
  #k fixity to test
  local marks,g,fin;
  marks:=MarksTom(t);;
  fin:=[];;
  for g in [1..Length(marks)] do
    if Set([2..Length(marks[g])],i->marks[g][i]<k+1)=[ true ] and
      (k in marks[g]) and marks[g][1]>k
    then Add(fin,[StructureDescription(RepresentativeTom(t,g)), marks[g]]);
    fi;
  od;
  return fin;
end;
\end{verbatim}

For many simple groups the Table of Marks is already pre-computed and accessible via the GAP-package TomLib~\cite{TomLib}.
As an example, we look at the sporadic Held group 
\(He\) and we see, with the code above, that it does not give
any example for fixity \(4\) actions:

\begin{verbatim}
gap> t:=TableOfMarks("He");
TableOfMarks( "He" )
gap> TestTom(t,4);
[  ]
gap>
\end{verbatim}


The next example 
is \(\Alt_6\cong \textrm{PSL}_2(9)\).
The function \texttt{TestTom} could be used in the following way:

\begin{verbatim}
gap> t:=TableOfMarks("L2(9)");
TableOfMarks( "A6" )
gap> TestTom(t,4);
[ [ "C2", [ 180, 4 ] ], [ "S3", [ 60, 4, 3, 1 ] ], [ "S3", [ 60, 4, 3, 1 ] ],
  [ "C3 x C3", [ 40, 4, 4, 4 ] ], [ "D10", [ 36, 4, 1, 1 ] ],
  [ "(C3 x C3) : C2", [ 20, 4, 2, 2, 2, 2, 2, 2 ] ] ]
\end{verbatim}

We see that there are six different actions of \(\textrm{PSL}_2(9)\) on sets of cosets with fixity $4$. Two with a point stabilizer isomorphic to \(\Sym_3\)
and one for \(C_2\), \(C_3\times C_3\), \(D_{10}\) and \((C_3\times C_3):C_2\) each.
In each case the size of the set \(\textrm{PSL}_2(9)\) is acting on is given (for example \(36\), when the point stabilizer are dihedral of order \(10\))
and some additional information about the number of fixed points of elements.




More fixed point information for the groups can be gained by the following GAP-function:

\begin{verbatim}
FixedPointsTom:=function(tom, num)
  local G, indC, mat, CN, l;
  G := UnderlyingGroup(tom);;
  indC := Filtered ( [2..Length(MarksTom(tom))],
    i -> IsCyclic( RepresentativeTom( tom, i ) ) );;
  mat := MatTom(tom)[num];;
  CN := OrdersTom(tom);;
  l:=List(indC,i->[ CN[i],Order(Centralizer(G, RepresentativeTom(tom,i))),mat[i]]);;
  return l;
end;
\end{verbatim}

The input is again a Table of Marks \texttt{tom} of a group and \texttt{num} identifies the specific action we are looking at.
If we return to the example \(G=\textrm{PSL}_2(9)\), then the table of marks looks like this:

\begin{verbatim}
gap> tom:=TableOfMarks("L2(9)");
TableOfMarks( "A6" )
gap> Display(tom);
 1:  360
 2:  180 4
 3:  120 . 6
 4:  120 . . 6
 5:   72 . . . 2
 6:   90 2 . . . 2
 7:   90 6 . . . . 6
 8:   90 6 . . . . . 6
 9:   60 4 3 . . . . . 1
10:   60 4 . 3 . . . . . 1
11:   45 5 . . . 1 3 3 . . 1
12:   40 . 4 4 . . . . . . . 4
13:   36 4 . . 1 . . . . . . . 1
14:   30 2 6 . . . . 2 . . . . . 2
15:   30 2 . 6 . . 2 . . . . . . . 2
16:   20 4 2 2 . . . . 2 2 . 2 . . . 2
17:   15 3 3 . . 1 3 1 1 . 1 . . 1 . . 1
18:   15 3 . 3 . 1 1 3 . 1 1 . . . 1 . . 1
19:   10 2 1 1 . 2 . . 1 1 . 1 . . . 1 . . 1
20:    6 2 3 . 1 . . 2 1 . . . 1 2 . . . . . 1
21:    6 2 . 3 1 . 2 . . 1 . . 1 . 2 . . . . . 1
22:    1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

\end{verbatim}

Now \texttt{num} refers to the specific line of this table of marks.
We have seen earlier that \(G=\textrm{PSL}_2(9)\) acts on a set of size \(20\) and with a point stabilizer \(U\) of structure \((C_3\times C_3):C_2\).
This refers to line \texttt{16} of \texttt{tom}. If we want to know the number of fixed points for every element of \(U\) in this action (which we call the fixed point profile), then we use the following GAP-code:

\begin{verbatim}
gap> FixedPointsTom(tom,16);
[ [ 2, 8, 4 ], [ 3, 9, 2 ], [ 3, 9, 2 ], [ 5, 5, 0 ], [ 4, 4, 0 ] ]
\end{verbatim}

In general the output of \texttt{FixedPointsTom} is a list of lists, where
the first entry in each of the lists refers to the order of an element \(g\in G\), the second to the order of \(C_G(g)\)
and the last one is the number of fixed points of \(g\) in its action on \(G/U\).
So in our case, where \(G=\textrm{PSL}_2(9)\) and \(U\cong (C_3\times C_3):C_2\), we can see that elements of order \(2\) fix four points, elements of order \(3\)
fix two points independent of their conjugacy class, and non-trivial elements of other orders 
have no fixed points on \(G/U\).

Collecting this information for all fixity \(4\) actions we obtain a fixed point profile, see Table~\ref{SGTdetails}, for most of the groups from 
Remark~\ref{rem4.1}:

\begin{longtable}{lll}
	$Group$ & point stabilizer structure & corresponding fixed point profile\\[0.5ex]\hline\hline
	\endhead \\[-2ex]

	\(\Alt_6 \cong \textrm{PSL}_2(9)\)            & \(C_2\) & \tiny{
\begin{tabular}{|c|c|c|c|c|} \hline
      2A & 3A & 3B & 4A & 5A \\ \hline
      4   &  0  &  0  &  0  &  0 \\ \hline
    \end{tabular}}\\\\
	& \(\Sym_3\)&\tiny{
\begin{tabular}{|c|c|c|c|c|} \hline
      2A & 3A & 3B & 4A & 5A \\ \hline
      4   &  3  &  0  &  0  &  0 \\ \hline
    \end{tabular}}\\\\
    	& \(\Sym_3\)&\tiny{
\begin{tabular}{|c|c|c|c|c|} \hline
      2A & 3A & 3B & 4A & 5A \\ \hline
      4   &  0  &  3  &  0  &  0 \\ \hline
    \end{tabular}}\\\\
	& \(C_3 \times C_3\)&\tiny{
\begin{tabular}{|c|c|c|c|c|} \hline
      2A & 3A & 3B & 4A & 5A \\ \hline
      0   &  4  &  4  &  0  &  0 \\ \hline
    \end{tabular}}\\ \\
	& \(D_{10}\) &\tiny{
\begin{tabular}{|c|c|c|c|c|} \hline
      2A & 3A & 3B & 4A & 5A \\ \hline
      4   &  0  &  0  &  0  &  1 \\ \hline
    \end{tabular}} \\\\
	&  \((C_3 \times C_3) : C_2\)   &  \tiny{
\begin{tabular}{|c|c|c|c|c|} \hline
      2A & 3A & 3B & 4A & 5A \\ \hline
      4   &  2  &  2  &  0  &  0 \\ \hline
    \end{tabular}}\\[5ex]
	\hline 
	\\
	\(\Alt_7\)                            & \(C_5\) & \tiny{
\begin{tabular}{|c|c|c|c|c|c|c|} \hline
      2A & 3A & 3B & 4A & 5A & 6A & 7A \\ \hline
      0   &  0  &  0  &  0  &  4  &  0  &  0 \\ \hline
    \end{tabular}}\\\\
	&  \(\Alt_6\) & \tiny{
\begin{tabular}{|c|c|c|c|c|c|c|} \hline
      2A & 3A & 3B & 4A & 5A & 6A & 7A \\ \hline
      3   &  4  &  1  &  1  &  2  &  0  &  0 \\ \hline
    \end{tabular}} \\\\  \hline\\       
	                                                    
	\(\textrm{PSL}_2(7) \cong \textrm{PSL}_3(2)\)         & \(C_2\) & \tiny{
\begin{tabular}{|c|c|c|c|} \hline
      2A & 3A & 4A & 7A \\ \hline
      4   &  0  &  0  &  0 \\ \hline
    \end{tabular}}\\\\
	& \(\Sym_3\)                                                & \tiny{
\begin{tabular}{|c|c|c|c|} \hline
      2A & 3A & 4A & 7A \\ \hline
      4   &  1  &  0  &  0 \\ \hline
    \end{tabular}} \\ \\ \hline\\
	
	\(\textrm{PSL}_2(8)\)                         & \(C_2\) & \tiny{
\begin{tabular}{|c|c|c|c|} \hline
      2A & 3A & 7A & 9A \\ \hline
      4   &  0  &  0  &  0 \\ \hline
    \end{tabular}}\\\\
	&  \(\Sym_3\) &\tiny{
\begin{tabular}{|c|c|c|c|} \hline
      2A & 3A & 7A & 9A \\ \hline
      4   &  3  &  0  &  0 \\ \hline
    \end{tabular}} \\\\
	& \(D_{14}\) & \tiny{
\begin{tabular}{|c|c|c|c|} \hline
      2A & 3A & 7A & 9A \\ \hline
      4   &  0  &  1  &  0 \\ \hline
    \end{tabular}}\\\\
	& \(D_{18}\) & \tiny{
\begin{tabular}{|c|c|c|c|} \hline
      2A & 3A & 7A & 9A \\ \hline
      4   &  1  &  0  &  1 \\ \hline
    \end{tabular}}\\\\  \hline \\                                   
	
	\(\textrm{PSL}_2(11)\)                        & \(C_3\) & \tiny{
\begin{tabular}{|c|c|c|c|c|} \hline
      2A & 3A & 5A & 6A & 11A\\ \hline
      0   &  4  &  0  &  0 & 0 \\ \hline
    \end{tabular}}\\\\	 
	& \(\Alt_4\)                                            &      \tiny{
\begin{tabular}{|c|c|c|c|c|} \hline
      2A & 3A & 5A & 6A & 11A\\ \hline
      3   &  4  &  0  &  0 & 0 \\ \hline
    \end{tabular}}\\   \\ \hline\\
    
	\(\textrm{PSL}_2(13)\)                        & \(C_3\) & \tiny{
\begin{tabular}{|c|c|c|c|c|} \hline
      2A & 3A & 6A & 7A & 13A\\ \hline
      0   &  4  &  0  &  0 & 0 \\ \hline
    \end{tabular}}\\\\
	&  \(\Alt_4\) &  \tiny{
\begin{tabular}{|c|c|c|c|c|} \hline
      2A & 3A & 6A & 7A & 13A\\ \hline
      3   &  4  &  0  &  0 & 0 \\ \hline
    \end{tabular}}\\\\ 
	& \(C_{13} : C_3\)                                             &            
	 \tiny{
\begin{tabular}{|c|c|c|c|c|} \hline
      2A & 3A & 6A & 7A & 13A\\ \hline
      0   &  4  &  0  &  0 & 2 \\ \hline 
    \end{tabular}}\\                                                 \\\hline\\
    
    
    	\(\textrm{PSL}_2(17)\)                        & \(C_4\) & \tiny{
\begin{tabular}{|c|c|c|c|c|c|} \hline
      2A & 3A & 4A & 8A & 9A & 17A\\ \hline
      4   &  0  &  4  &  0 & 0 &0 \\ \hline
    \end{tabular}}\\\\
	&  \(C_{17}:C_4\) & \tiny{ 
	\begin{tabular}{|c|c|c|c|c|c|} \hline
      2A & 3A & 4A & 8A & 9A & 17A\\ \hline
      4   &  0  &  4  &  0 & 0 & 2 \\ \hline
    \end{tabular}}\\
    \\\hline\\
    
    
        	\(\textrm{PSL}_2(19)\)                        & \(C_5\) & \tiny{ 
	\begin{tabular}{|c|c|c|c|c|c|} \hline
      2A & 3A & 5A & 9A & 10A & 19A\\ \hline
      0   &  0  &  4  &  0 & 0 & 0 \\ \hline
    \end{tabular}}\\
    \\\hline\\
    
    
    
            	\(\textrm{PSL}_2(23)\)                        & \(C_6\) & \tiny{ 
	\begin{tabular}{|c|c|c|c|c|c|c|} \hline
      2A & 3A & 4A & 6A & 11A & 12A & 23A\\ \hline
      4   &  4  &  0  &  4 & 0 & 0 & 0\\ \hline
    \end{tabular}}\\
    \\\hline\\
    
    
    
    
        	\(\textrm{PSL}_2(25)\)                        & \(C_6\) & \tiny{
\begin{tabular}{|c|c|c|c|c|c|c|c|} \hline
      2A & 3A & 4A & 5A & 5B & 6A & 12A & 13A\\ \hline
      4   &  4  &  0  &  0  & 0   &  4  & 0     &  0\\ \hline
    \end{tabular}}\\\\
	&  \((C_5 \times C_5):C_6\) & \tiny{ 
\begin{tabular}{|c|c|c|c|c|c|c|c|} \hline
      2A & 3A & 4A & 5A & 5B & 6A & 12A & 13A\\ \hline
      4   &  4  &  0  &  2  & 2   &  4  & 0     &  0\\ \hline
    \end{tabular}}\\
    \\\hline\\
    
    
                	\(\textrm{PSL}_2(27)\)                        & \(C_7\) & \tiny{ 
	\begin{tabular}{|c|c|c|c|c|} \hline
      2A & 3A & 7A & 13A & 14A \\ \hline
      0   &  0  &  4  &  0 & 0\\ \hline
    \end{tabular}}\\
    \\\hline\\
    
    
  
    
        
        	\(\textrm{PSL}_2(29)\)                        & \(C_7\) & \tiny{
\begin{tabular}{|c|c|c|c|c|c|c|} \hline
      2A & 3A & 5A & 7A & 14A & 15A & 29A\\ \hline
      0   &  0  &  0  &  4  & 0   &  0  & 0 \\ \hline
    \end{tabular}}\\\\
	&  \((C_{29}):C_7\) & \tiny{ 
\begin{tabular}{|c|c|c|c|c|c|c|} \hline
      2A & 3A & 5A & 7A & 14A & 15A & 29A\\ \hline
      0   &  0  &  0  &  4  & 0   &  0  & 2 \\ \hline
    \end{tabular}}\\
    \\\hline\\
    
    
    
                    	\(\textrm{PSL}_2(31)\)                        & \(C_8\) & \tiny{ 
	\begin{tabular}{|c|c|c|c|c|c|c|c|} \hline
      2A & 3A &4A &  5A & 8A & 15A & 16A & 31A\\ \hline
      4   &  0  &  4  &  0  & 4   &  0   &   0    & 0 \\ \hline
    \end{tabular}}\\
    \\\hline\\
    
    
    
            
        	\(\textrm{PSL}_2(37)\)                        & \(C_9\) & \tiny{
\begin{tabular}{|c|c|c|c|c|c|c|} \hline
      2A & 3A & 6A & 9A & 18A & 19A & 37A\\ \hline
      0   &  4  &  0  &  4  & 0   &  0  & 0 \\ \hline
    \end{tabular}}\\\\
	&  \((C_{37}):C_9\) & \tiny{ 
\begin{tabular}{|c|c|c|c|c|c|c|} \hline
      2A & 3A & 6A & 9A & 18A & 19A & 37A\\ \hline
      0   &  4  &  0  &  4  & 0   &  0  & 2 \\ \hline
    \end{tabular}}\\
    \\\hline\\
    
    
    
    
        	\(\textrm{PSL}_2(41)\)                        & \(C_{10}\) & \tiny{
\begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline
      2A & 3A & 4A & 5A  & 7A  & 10A & 20A & 21A & 41A\\ \hline
      4   &  4  &  0  &  4  & 0     &  4   & 0      & 0     & 0   \\ \hline
    \end{tabular}}\\\\
	&  \((C_{41}):C_{10}\) & \tiny{ 
\begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline
      2A & 3A & 4A & 5A  & 7A  & 10A & 20A & 21A & 41A\\ \hline
      4   &  4  &  0  &  4  & 0     &  4   & 0      & 0     & 2   \\ \hline
    \end{tabular}}\\
    \\\hline\\
    

	\(\textrm{PSU}_3(3)\)                         & \(((C_3 \times C_3) :C_3) :C_8\)                                                  &   
	\tiny{
\begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline
      2A & 3A & 3B & 4A & 4B & 6A & 7A & 8A & 12A\\ \hline
      4   &  1  &  1  &  4 & 0   & 1   & 0    & 2   & 1\\ \hline
    \end{tabular}}\\
	\\\hline\\
                      
	\(\textrm{PSU}_4(2)\cong \textrm{PSp}_4(3)\)          & \(C_5\)                                                                           &                                                            
	 \tiny{
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline
      2A & 2B & 3A & 3B & 3C & 4A & 4B & 5A  & 6A & 6B & 6C & 6D & 9A & 12A\\ \hline
      0   &  0  &  0  &  0 & 0    &  0  &   0  & 4  &   0  &  0 &  0  &  0  &  0  &  0  \\ \hline
    \end{tabular}}\\ \\\hline\\
    
    
	\(\textrm{Sz}(8)\)                            & \(C_5\) &  \tiny{
\begin{tabular}{|c|c|c|c|c|} \hline
      2A & 4A & 5A & 7A & 13A\\ \hline
      0   &  0  &  4  &  0 & 0 \\ \hline
    \end{tabular}}\\\\
	& \(C_{13}\)                                                               &      
	 \tiny{
\begin{tabular}{|c|c|c|c|c|} \hline
      2A & 3A & 6A & 7A & 13A\\ \hline
      0   &  0  &  0  &  0 &  4 \\ \hline
    \end{tabular}}\\                                                       \\\hline\\
	
	\(M_{11}\)                            & \(C_5\) &  \tiny{
\begin{tabular}{|c|c|c|c|c|c|c|} \hline
      2A & 3A & 4A & 5A & 6A & 8A & 11A\\ \hline
      0   &  0  &  0  &  4  &  0  &  0  &   0\\ \hline
    \end{tabular}}\\\\
	 & \(C_{11} :C_5\) & 
	 \tiny{
\begin{tabular}{|c|c|c|c|c|c|c|} \hline
      2A & 3A & 4A & 5A & 6A & 8A & 11A\\ \hline
      0   &  0  &  0  &  4  &  0  &  0  &   1\\ \hline
    \end{tabular}}\\\\
	 & \(\textrm{PSL}_2(11)\)                                          &  
	 \tiny{
\begin{tabular}{|c|c|c|c|c|c|c|} \hline
      2A & 3A & 4A & 5A & 6A & 8A & 11A\\ \hline
      4   &  3  &  0  &  2  &  1  &  0  &   1\\ \hline
    \end{tabular}}\\\\\hline\\
	\(M_{12}\)                            & \(M_{11}\)                                                                        &  
	 \tiny{
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline
      2A & 2B & 3A & 3B & 4A & 4B & 5A  & 6A & 6B & 8A & 8B & 10A & 11A\\ \hline
      0   &  4  &  3  &  0 & 0    &  4  &  2  & 0   &   1 &  2  &  0  &  0    &  1  \\ \hline
    \end{tabular}}\\  \\ 
    	                        & \(M_{11}\)                                                                        &  
	 \tiny{
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline
      2A & 2B & 3A & 3B & 4A & 4B & 5A  & 6A & 6B & 8A & 8B & 10A & 11A\\ \hline
      0   &  4  &  3  &  0 & 4    &  0  &  2  & 0  &   1  &  0 &  2  &  0  &    1  \\ \hline
    \end{tabular}}\\   \\
    \hline\\
	\(M_{22}\)                            & \(C_5\) &  \tiny{
\begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline
      2A & 3A & 4A & 4B & 5A & 6A & 7A & 8A & 11A\\ \hline
      0   &  0  &  0  &  0  &  4  &  0  &   0 &   0 &   0\\ \hline
    \end{tabular}}\\\\
	&  \(C_{11}:C_5\)                                                           &    \tiny{
\begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline
      2A & 3A & 4A & 4B & 5A & 6A & 7A & 8A & 11A\\ \hline
      0   &  0  &  0  &  0  &  4  &  0  &   0 &   0 &   1\\ \hline
    \end{tabular}} \\                                                       \\\hline\\
                     
	\(J_1\)                               & \(C_{15}\)                                                                        &                                                            
	\tiny{
\begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline
      2A & 3A & 5A & 6A & 7A & 10A & 11A & 15A & 19A\\ \hline
      0   &  4  &  4  &  0  &  0  &  0  &   0 &   4 &   0\\ \hline
    \end{tabular}} \\ 
     \\[0.5ex]\hline\hline\\[-1ex]
	%
	\caption[Small Groups Table for fixity 4]{For each group, we give the possible point stabilizer structures for fixity 4 actions, and also the fixed point profile.}
	\label{SGTdetails}
            
\end{longtable}

\bigskip

In the proof of Lemma~\ref{Observation}, our arguments are supported by GAP calculations. They show that, for every \(3\)-group of order \(27\),
every subgroup \(U\leq P\) of index \(3\) and every element \(y\in P \setminus U\) with \(|C_P(y)|=9\), it is true that \(y^U=P'y\). Here is the code:

\begin{verbatim}
gap> AllSmallGroups(27);;
gap> Filtered(last,x->NilpotencyClassOfGroup(x)=2);
[ <pc group of size 27 with 3 generators>, <pc group of size 27 with 3 generators> ]
gap> P1:=last[1];; P2:=last2[2];;
gap> U1s:=Filtered(LowIndexSubgroups(P1,3),x->Index(P1,x)=3);;
ggap> List(U1s,U->Set(List(Filtered(Difference(Elements(P1),Elements(U)),
y->Order(Centralizer(P1,y))=9),y->Elements(y^U)=Elements(CommutatorSubgroup(P1,P1)*y))));
[ [ true ], [ true ], [ true ], [ true ] ]
gap> U2s:=Filtered(LowIndexSubgroups(P2,3),x->Index(P2,x)=3);;
gap> List(U2s,U->Set(List(Filtered(Difference(Elements(P2),Elements(U)),
y->Order(Centralizer(P2,y))=9),y->Elements(y^U)=Elements(CommutatorSubgroup(P2,P2)*y))));
[ [ true ], [ true ], [ true ], [ true ] ]
\end{verbatim}





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