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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% Auteur

%%%1
\author{\firstname{Paul} \lastname{Bradley}}

\address{31 Songthrush Way\\ 
Norton Canes\\ 
Cannock\\ 
Staffordshire WS11 9AH\\ UK}

\email{paulmbradley82@gmail.com}

%%%2
\author{\firstname{Peter} \lastname{Rowley}}

\address{Department of Mathematics\\ 
University of Manchester\\
 Oxford Road\\ 
 Manchester M13 6PL\\ 
 UK}

\email{peter.j.rowley@manchester.ac.uk}


%%%%% Sujet

\keywords{Number of Orbits, $k$-subsets, rank one Lie-type groups.}
 

\subjclass{20D06, 20B20}


%%%%% Gestion

\DOI{10.5802/alco.195}
\datereceived{2020-10-03}
\daterevised{2021-08-18}
\dateaccepted{2021-08-16}

%%%%% Titre et résumé
\title{Orbits on $k$-subsets of $2$-transitive Simple Lie-type Groups}


\begin{abstract}For a finite rank one simple Lie-type group acting $2$-transitively on a set $\Omega$ and $k \in \mathbb{N}$ we derive formulae for the number of $G$-orbits on the set of all $k$-subsets of $\Omega$.
\end{abstract}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\datepublished{2022-02-28}
\begin{document}

\maketitle
\section{Introduction}\label{sect1}
Suppose $G$ is a permutation group acting upon a set $\Omega$. Then $G$ has an induced action upon $\mathcal{P}(\Omega)$, the power set of $\Omega$. For $k \in \mathbb{N},$ let $\mathcal{P}_k(\Omega)$ denote all the $k$-subsets of $\Omega$, clearly $G$ also has an induced action upon $\mathcal{P}_k(\Omega)$. In this paper we shall assume $G$ and $\Omega$ are finite. Let $\sigma_k(G,\Omega)$ denote the number of $G$-orbits on $\mathcal{P}_k(\Omega)$. Questions as to the behaviour of $\sigma_k(G,\Omega)$ as $k$ varies and the relationship between $G$-orbit lengths for various $k$ arise naturally. A contribution to the former type of question is given by the venerable theorem of Livingstone and Wagner~\cite{LivingstoneandWagner} which states that for $k \in \mathbb{N}$ with $2 \leq k\leq |\Omega|/2,$ 
\[
\sigma_{k-1}(G,\Omega) \leq \sigma_k(G,\Omega).
\]
Generalizations of this theorem have been obtained by Mnukhin and Siemons~\cite{VBMandIJS}. While Bundy and Hart~\cite{DBandSH} have considered the situation when 
\[
\sigma_{k-1}(G,\Omega) = \sigma_k(G,\Omega).
\] 
Results on $G$-orbit lengths are somewhat patchy at present - see Siemons and Wagner~\cite{SW1, SW2} and Mnukhin~\cite{Munchkin}. For the record, we remark there is a considerable literature concerning the infinite case; for a small selection see Cameron~\cite{Cam1}.

The main aim here is to establish formulae for $\sigma_k(G,\Omega)$ when $G$ is a rank one simple Lie-type group acting $2$-transitively on $\Omega$. Thus set $q=p^a$ where $p$ is a prime and $a \in \mathbb{N}$. Then the possibilities for $G$ are $L_2(q)$ $(q>3),$ the $2$-dimensional projective special linear groups, $Sz(q)$ $(q=2^{2n+1}>2)$, the Suzuki groups, $U_3(q)$ $(q>2),$ the $3$-dimensional projective special unitary groups and $R(q)$ $(q=3^{2n+1}>3),$ the Ree groups. First we consider the exceptional $2$-transitive representations of $L_2(7)$ of degree $7$, $L_2(9)$ of degree $6$ and $L_2(11)$ of degree $11$. Here writing $\sigma_k(G,\Omega)$ as $\sigma_k$, we have for $L_2(7)$, $\sigma_1 = \sigma_2 =1, \sigma_3 = 2$; for $L_2(9), \sigma_1=\sigma_2=\sigma_3=\sigma_4 = 1$; and for $L_2(11), \sigma_1=\sigma_2 = 1, \sigma_3=2 ,\sigma_4=3, \sigma_5=4$. For the remaining $2$-transitive representations, the corresponding sets $\Omega$ are the projective line (with $|\Omega|=q+1$), the Suzuki ovoid (with $|\Omega|=q^2+1$), the isotropic $1$-spaces of a $3$-dimensional unitary space (with $|\Omega|=q^3+1$) and the Steiner system $S(2,q+1,q^3+1)$ (with $|\Omega|=q^3+1$). Before stating our main results we must introduce some notation. For $b,c \in \mathbb{N},$ we use $(b,c)$ to denote the greatest common divisor of $b$ and $c$. For $\ell \in \mathbb{N}$ we let
\[
\mathcal{D}(\ell)=\{n \in \mathbb{N}\; \vert \ n\mbox{ divides } \ell\}
\]
and $\mathcal{D}^*(\ell)=\mathcal{D}(\ell)\setminus \{1\}$. Euler's phi function $\phi$ (see~\cite{Rosen}) will feature in our results. Our final piece of notation concerns partitions. Let $n \in \mathbb{N},$ and let $\pi=\lambda_1\lambda_2\ldots \lambda_r$ where $\lambda_1\leq \lambda_2\leq\cdots \leq\lambda_r$ and $\sum_{i=1}^r \lambda_i = n$. Though we will frequently use the more compressed notation $\pi=\mu_1^{a_1} \mu_2^{a_2}\ldots $ where $\mu_1 < \mu_2 <\cdots $ ($a_i$ being the multiplicity of $\mu_i$ in the partition $\pi$). For $k \in \mathbb{N}$, $\eta_k(\pi)$ is defined to be the number of subsequences $\lambda_{i_1},\lambda_{i_2},\ldots,\lambda_{i_s}$ of $\lambda_1,\ldots,\lambda_r$ which form a partition of $k$. Since $G$ acts $2$-transitively on $\Omega$, we have $\sigma_1(G,\Omega)=\sigma_2(G,\Omega)=1$, so we shall assume $k\geq 3$. Now we come to our first theorem.





\begin{theorem}
\label{1.1}
Suppose that $G \cong L_2(q)$ $(q>3)$ acts upon the projective line $\Omega$, and let $k \in \mathbb{N}$ with $k\ge 3$. Set $d=(2,q-1)$. Then
 \begin{multline*}
 \sigma_k(G,\Omega) = \frac{d}{q(q+1)(q-1)}\eta_k(1^{q+1}) + \frac{d}{q} \eta_k(1^1 p^{\frac{q}{p}}) \\
 \begin{aligned}
 &+ \frac{d}{2(q+1)} \sum_{m \in \mathcal{D}^*(\frac{q+1}{d})} \phi(m) \eta_k\left(m^{\frac{q+1}{m}}\right)\\
 & + \frac{d}{2(q-1)} \sum_{m \in \mathcal{D}^*(\frac{q-1}{d})}\phi(m)\eta_k \left(1^2 m^{\frac{q-1}{m}}\right).\\
 \end{aligned}
\end{multline*}
\end{theorem}

 In~\cite{CamPSL} Cameron, Maimani, Omidi and Teyfeh-Rezaie give methods for calculating $\sigma_k(L_2(q),\Omega)$, where $\Omega$ is the projective line and $q\equiv 3$ mod $4$, and Cameron, Omidi and Teyfeh-Rezaie~\cite{CamPGL} follow a similar approach for $\sigma_k(PGL(2,q),\Omega)$, their interest being motivated by a search for $3$-designs. Further work is presented by Liu, Tang and Wu~\cite{Liu} and also Chen and Liu~\cite{Chen}, in the case when $q\equiv 1$ mod $4$.


\begin{theorem}\label{1.2}
Suppose $G\cong Sz(q)$ $(q=2^{2n+1}>2$, $n \in \mathbb{N})$ acts upon the Suzuki ovoid $\Omega$. Let $r \in \mathbb{N}$ be such that $r^2=2q,$ and let $k \in \mathbb{N}$ with $k \ge 3$. Then
\begin{multline*}
\sigma(G,\Omega) = \frac{1}{q^2(q-1)(q^2+1)} \eta_k(1^{q^2+1}) + \frac{1}{q^2}\eta_k(1^1 2^{\frac{q^2}{2}})\\
 \begin{aligned}
& + \frac{1}{q}\eta_k(1^1 4^{\frac{q^2}{4}})+\frac{1}{2(q-1)} \sum_{m \in \mathcal{D}^*(q-1)} \phi(m) \eta_k(1^2 m^{\frac{q^2-1}{m}}) \\
& +\frac{1}{4(q+r+1)} \sum_{m \in \mathcal{D}^*(q+r+1)} \phi(m) \eta_k( m^{\frac{q^2+1}{m}})\\
& +\frac{1}{4(q-r+1)} \sum_{m \in \mathcal{D}^*(q-r+1)} \phi(m) \eta_k( m^{\frac{q^2+1}{m}}).
 \end{aligned}
\end{multline*}
\end{theorem}

The corresponding result for $U_3(q)$ is more complicated than for $L_2(q)$ and $Sz(q)$~--~see Definitions~\ref{def3.1}, \ref{def3.2}, \ref{def3.3} and~\ref{def3.4} for an explanation of the notation in the next theorem.
\begin{theorem}\label{1.3}
Suppose $G\cong U_3(q)$ $(q>2)$ acts upon $\Omega$, the set of isotropic points of a $3$-dimensional unitary space. Let $k \in \mathbb{N}$ with $k\ge 3$, and set $d=(3,q+1)$ and $\ell=\frac{q+1}{d}$. Then
\begin{multline*}
\sigma_k(G,\Omega) = \frac{d(\eta_k(\pi_1) + \mu_k)}{q^3(q^3+1)(q^2-1)} +\frac{d}{q(q+1)(q^2-1)}\sum_{m \in \mathcal{D}^*(\ell)} \phi(m) \eta_k(\pi_4^{(m)})\\
\begin{aligned}
 & + \frac{d}{q(q+1)(p-1)} \left(\mathop{\sum_{m=pj}}_{j \in \mathcal{D}^*(\ell)} \phi(m) \eta_k(\pi_5^{(m)})\right) + \frac{d\sigma_k(E^*_0,\Omega)}{6(q+1)^2} \\
 & +\frac{d}{2(q^2-1)}\mathop{\mathop{\sum_{m\in \mathcal{D}(\frac{q^2-1}{d})}}_{m \not\in \mathcal{D}(\ell)}} \phi(m) \eta_k(\pi_7^{(m)})\\
 & +\frac{d(q+1)}{3(q^3+1)} \sum_{m \in \mathcal{D}^*(\frac{q^2-q+1}{d})} \phi(m) \eta_k(\pi_8^{(m)}).
\end{aligned}
\end{multline*}
\end{theorem}

Lemma~\ref{3.9} gives formulae for $\sigma_k(E_0^*,\Omega).$ 

A corresponding result for the Ree groups, though a little more involved, is similar to Theorems~\ref{1.1} and~\ref{1.2} and may be found in Bradley~\cite{Bradley}. \textsc{Magma} code based on the formula in Theorem~\ref{1.1} is given in Section~\ref{sec4}, and similar routines are also available in~\cite{Bradley} for the Suzuki, $3$-dimensional unitary and characteristic $3$ Ree groups.

The proofs of Theorems~\ref{1.1}, \ref{1.2} and~\ref{1.3} all rely upon the orbit counting result commonly referred to as Burnside's Lemma, though this result can be traced back to earlier work of Cauchy and Frobenius. For a historical excursion see \cite{neumann}. Thus our main task here is to enumerate the cycle types of elements in $G$ in their action on the various $\Omega$. For $G\cong L_2(q)$ or $Sz(q)$ this is straightforward, particularly as $G$ may be partitioned by certain subgroups (see Lemmas~\ref{2.5} and~\ref{2.6}). For $G\cong U_3(q)$ we make use of the description of conjugacy classes obtained in~\cite{CharTablePSU}. There the conjugacy classes are divided into a number of types -- those of the same type have the same size and fix the same number of points of $\Omega$, $\Omega$ being the set of isotropic $1$-spaces of a $3$-dimensional unitary space (see Table~\ref{Tab2}). Using information about the subgroup structure of $G$, in Lemma~\ref{3.5} we determine the cycle structures on $\Omega$ of elements in all conjugacy class types except for $\mathcal{C}_6 \cup \mathcal{C}_6^\prime$. This is relatively straightforward. However $\mathcal{C}_6 \cup \mathcal{C}_6^\prime$ is a different kettle of fish -- see Table~\ref{Tab1} which details the varied cycle structures for such elements in the case of $U_3(71)$ (so $|\Omega|=357,912$).


\begin{table}[htbp]\centering
\caption{Cycle structure for $\mathcal{C}_6 \cup \mathcal{C}_6^\prime$ class types in $U_3(71)$}
\renewcommand{\arraystretch}{1.3}
\begin{tabular}{|>{\PBS\centering}p{40mm} | >{\PBS\centering}p{15mm} | >{\PBS\centering}p{40mm} | >{\PBS\centering}p{15mm} | }
\hline
Cycle Type of Conjugacy Class Representative 
& Number of Classes & 
Cycle Type of Conjugacy Class Representative& Number of Classes\\
\hline
$2^{36} 4^{89460}$ & $1$ & $6^{12} 12^{29820}$ & $2$\\ \hline
$2^{36} 8^{44730}$ & $2$ & $6^{12} 24^{14910}$ & $4$ \\ \hline
$2^{36} 12^{29820}$ & $2$ & $6^{12} 8^{9} 24^{14907}$& $8$ \\ \hline
$2^{36} 24^{14910}$ & $4$ & $8^{9} 12^{6} 24^{14907}$ & $16$\\ \hline
$2^{36} 3^{24} 6^{59628}$ & $2$ & $9^{39768}$ & $3$\\ \hline
$3^{24} 6^{59640}$ & $1$ & $9^{8} 18^{19880}$ & $9$\\ \hline
$3^{24} 12^{29820}$ & $2$ & $9^{8} 36^{9940}$ & $18$\\ \hline
$3^{24} 24^{14910}$ & $4$& $9^{8} 72^{4970}$ & $36$ \\ \hline
$3^{24} 4^{18} 12^{29814}$ & $4$ & $12^{6} 24^{14910}$ & $8$\\ \hline
$3^{24} 8^{9} 24^{14907}$ & $8$& $18^{4} 36^{9940}$ & $18$\\ \hline
$4^{18} 8^{44730}$ & $4$ & $18^{4} 72^{4970}$ & $36$\\ \hline
$4^{18} 24^{14910}$ & $8$ & $36^{2} 72^{4970}$ & $72$\\ \hline
$4^{18} 6^{12} 12^{29814}$ & $4$ & $3^{119304}$& $1$\\ \hline
\end{tabular}
\label{Tab1}
\end{table}

Representatives for all classes of type $\mathcal{C}_6 \cup \mathcal{C}_6^\prime$ are to be found in an abelian subgroup $E$ of $G$ where $E$ is isomorphic to a direct product of cyclic groups of order $(q+1)/d$ and $q+1$ ($d=(3,q+1)$). The key result in enumerating these possible cycle types (and their multiplicities) is Lemma~\ref{3.8}. With this result to hand, Lemma~\ref{3.9} then determines the contribution of the $\mathcal{C}_6 \cup \mathcal{C}_6^\prime$ conjugacy classes to the sum $\sum_{g\in G} |\fix _{\Omega_k}(g)|$ (where $\Omega_k=\mathcal{P}_k(\Omega)$). Combining Lemmas~\ref{3.5} and~\ref{3.9} yields Theorem~\ref{1.3}.

It is of interest to examine the number of orbits for these groups on subsets of size~$3$ (and in the case of $L_2(q)$, size $4$), of their respective $G$-sets. The following corollaries can all be derived by appropriate substitutions into Theorems~\ref{1.1}, \ref{1.2}, \ref{1.3} and the corresponding result for the Ree groups as found in~\cite{Bradley}. For full details of the proofs see~\cite{Bradley}.

\begin{corollary}
\label{P4.0.6}
Let $q=p^a>2$ where $p$ is a prime and let $\Omega$ be the projective line with $q+1$ points. Then
\[
\sigma_3(L_2(q), \Omega)=\begin{cases}2, & \mbox{if } q \equiv 1\mod 4,\\ 1, & \mbox{if } q \equiv 3\mod 4. \end{cases}
\]
\
\end{corollary}


\begin{corollary}\label{c1.2}
Let $L_n = L_2(2^n)$ acting on the projective line $\Omega_n$, and put $a_n=\sigma_4(L_n, \Omega_n)$. Setting $a_1=a_2 = 1,$ for $n\ge 3$ we have
\[
 a_n=a_{n-1}+2a_{n-2}.
\]
\end{corollary}

\begin{corollary}\label{P4.0.9}
Let $q=2^{2n+1}$ and $\Omega_n$ be the Suzuki ovoid with $q^2+1$ points. Set $a_n=\sigma_3(Sz(q),\Omega_n).$ Then
\[
a_n =\frac{4^n+2}{3}.
\]
\end{corollary}




\begin{corollary}\label{P4.0.10}
Let $q=3^n$ and $\Omega_n$ be the $q^3+1$ isotropic points of a unitary $3$-space. Set $a_n = \sigma_3(U_3(q),\Omega_n)$. Then 
\[
a_n =\frac{3^n+3}{2}.
\]
\end{corollary}






\begin{corollary}\label{P4.0.11}
Let $q=3^{2n+1}$ and $\Omega_n$ be the Ree unital on $q^3+1$ points. Set $a_n = \sigma_3 (R(q),\Omega_n).$ Then
\[
a_n=\frac{(3^{2n+1}+3)^2}{6}.
\]

\end{corollary}

Two of the sequences found are of interest more generally,
the sequence $\{a_n\}$ appearing in Corollory~\ref{c1.2} is known as the Jacobsthal sequence~\cite{Jacobsthal}, and the sequence $\{a_n\}$ given in Corollary~\ref{P4.0.10} is associated to Sierpinski's Triangle, see~\cite{sietri}.


\section{Background Results}
First we recall some background results, starting with the frequently mis-attributed Burnside's Lemma.

\begin{lemma}
\label{Burnside}
Let $H$ be a finite group and $\Omega$ a finite $H$-set. If $t$ is the number of $H$-orbits on $\Omega$, then
\[
t=\frac{1}{|H|}\sum_{h\in H} |\fix _{\Omega}(h)|.
\]

\end{lemma}
For a partition of $n$, $\pi=\lambda_1\lambda_2\ldots \lambda_r$ (such that $\lambda_1 \leq \lambda_2 \leq \cdots \leq \lambda_r$ and $\sum^r_{i=1} \lambda_i =n$) we recall from Section~\ref{sect1} that, for $k \in \mathbb{N},$ $\eta_k (\pi)$ is the number of sub-sequences $\lambda_{i_1},\lambda_{i_2},\ldots,\lambda_{i_s}$ of $\lambda_1,\ldots,\lambda_r$ which form a partition of $k$. As an example consider $\pi = 1 1 4 4 4$ ($=1^2 4^3$ in compressed form), a partition of $n=14.$ Then $\eta_5(\pi)=6$, $\eta_6(\pi)=3,$ $\eta_7(\pi)=0$ and $\eta_8(\pi)=3$. Our interest in $\eta_k(\pi)$ is because of the following two simple lemmas.

\begin{lemma}\label{2.2}
Suppose $g \in \Sym(\Lambda)$ where $\vert \Lambda\vert = n,$ and let $k \in \mathbb{N}$. If $g$ has cycle type $\pi$ on $\Lambda$ (viewed as a partition of $n$), then $g$ fixes (set-wise) $\eta_k(\pi)$ $k$-subsets of $\Lambda$.
\end{lemma}

\begin{lemma}\label{2.3}
Let $\pi= \mu_1^{a_1}\mu_2^{a_2}\ldots \mu_s^{a_s}$ be a partition of $n$ (in compressed form). Then
\[
\eta_k(\pi)=\prod_{(k_1,k_2,\ldots,k_s)} \binom{a_1}{k_1}\binom{a_2}{k_2}\ldots \binom{a_s}{k_s},
\]
running over all $s$-tuples $(k_1,k_2,\ldots,k_s)$ with $k_i \ge 0$ and $\mu_1k_1+\mu_2k_2+\cdots +\mu_sk_s = k.$
\end{lemma}

Another elementary result we require is in the next lemma.

\begin{lemma}
\label{2.4}
Let $H\cong \mathbb{Z}_n$ and let $m \in \mathbb{N}$, $m\ne 1$, be such that $m\mid n$. If $p_1,p_2,\ldots,p_r$ are the distinct prime divisors of $m$, then the number of elements in $H$ of order $m$ is 
\[
\phi(m)=\frac{m}{ p_1p_2\ldots p_r}(p_1-1)(p_2-1)\cdots (p_r-1).
\]
\end{lemma}



A few well known facts regarding $L_2(q)$ and $Sz(q)$ are given in the next two lemmas.

\begin{lemma}
\label{2.5}
Let $q=p^a>3$ where $p$ is a prime and $a\in \mathbb{N}$. Suppose $G\cong L_2(q)$, and let $P \in Syl_p(G).$
\begin{enumerate}[label= (\roman*)]
\item\label{lemma2.5_i} $|G|=\frac{q(q+1)(q-1)}{d}$ where $d=(2,q-1).$
\item\label{lemma2.5_ii} $G$ acts 2-transitively upon the projective line $\Omega$ and $G_{\alpha} = N_G(P)$ for some $\alpha \in \Omega$.
\item\label{lemma2.5_iii} $G$ contains cyclic subgroups $H_-=\left<h_-\right>$ and $H_+=\left<h_+\right>$ where $|H_-|=\frac{q-1}{d}$ and $|H_+|=\frac{q+1}{d}$.
Set $\mathcal{S}=\{P^g,H_-^g,H_+^g \mid g\in G\}$. Then every non-identity element of $G$ belongs to a unique subgroup in $\mathcal{S}$.
\item\label{lemma2.5_iv} $h_-$ has cycle type $1^2 \left(\frac{q-1}{d}\right)^d$ on $\Omega$ and $h_+$ has cycle type $\left(\frac{q+1}{d}\right)^d$ on $\Omega$. For $1\neq x\in P$, $x$ has cycle type $1^1 p^{\frac{q}{p}}$ on $\Omega.$
\item\label{lemma2.5_v} The number of elements of order $p$ is $(q-1)(q+1).$
\item\label{lemma2.5_vi} $[G:N_G(H_-)]=\frac{q(q+1)}{2}$ and $[G:N_G(H_+)]=\frac{q(q-1)}{2}.$
\end{enumerate}
\end{lemma}
\begin{proof}
For parts~\ref{lemma2.5_i}--\ref{lemma2.5_iv} see~8.1 Hilfssatz and for~\ref{lemma2.5_iii}, \ref{lemma2.5_iv} and~\ref{lemma2.5_vi} consult $8.3$, $8.4$, $8.5$ Satz of~\cite{Huppert1}. Part~\ref{lemma2.5_v} is given in $8.2$ Satz (b) and (c) of~\cite{Huppert1}.
\end{proof}



\begin{lemma}
\label{2.6}

Let $q=2^{2n+1}$ where $n \in \mathbb{N}$ and set $r=2^{n+1}.$ Suppose $G\cong Sz(q)$ and let $P \in Syl_2 G.$
\begin{enumerate}[label= (\roman*)]
\item\label{lemma2.6_i} $|G|=q^2(q-1)(q^2+1).$
\item\label{lemma2.6_ii} $G$ acts 2-transitively on the Suzuki ovoid $\Omega$ and $G_{\alpha} = N_G(P)$ for some $\alpha \in \Omega$.
\item\label{lemma2.6_iii} $G$ contains cyclic subgroups $H_0=\left<h_0\right>$, $H_-=\left<h_-\right>$ and $H_+=\left<h_+\right>$ where $|H_0|=q-1$, $|H_-|=q-r+1$ and $|H_+|=q+r+1$.
Set $\mathcal{S}=\{P^g,H_0^g,H_-^g,H_+^g\mid g\in G\}$. Then every non-identity element of $G$ belongs to a unique subgroup in $\mathcal{S}$.
\item\label{lemma2.6_iv} $h_0$ has cycle type $1^2 (q-1)^{q+1}$, $h_-$ has cycle type $(q-r+1)^{\frac{q^2+1}{q-r+1}}$ and $h_+$ has cycle type $(q+r+1)^{\frac{q^2+1}{q+r+1}}$ on $\Omega$. For $x\in P$, $x$ has cycle type $1^1 2^{\frac{q^2}{2}}$, respectively $1^1 4^{\frac{q^2}{4}}$, on $\Omega$ if it has order $2$, respectively $4$.
\item\label{lemma2.6_v} $|N_G(H_0)|=2(q-1),$ $|N_G(H_-)|=4(q-r+1)$ and $|N_G(H_+)|=4(q+r+1).$
\item\label{lemma2.6_vi} $P$ contains $q-1$ elements of order $2$ and $q^2-q$ elements of order $4$.
\end{enumerate}
\end{lemma}

\begin{proof}Consult~\cite[Theorem~9]{Suz1}.
\end{proof}


Next we give a compendium of facts about $U_3(q)$.
\begin{lemma}\label{2.7}
Let $q=p^a>2$ where $p$ is a prime and $a \in \mathbb{N}$, and suppose $G\cong U_3(q)$. Let $P\in Syl_p G$ and set $d=(q+1,3)$.
\begin{enumerate}[label= (\roman*)]
\item\label{lemma2.7_i} $|G|=q^3\frac{(q^2-1)}{d}(q^3+1)$.

\item\label{lemma2.7_ii} $G$ acts $2$-transitively on $\Omega$, the set of isotropic $1$-spaces of a unitary $3$-space, $|\Omega| = q^3+1$ and $G_\alpha=N_G(P)$ for some $\alpha \in \Omega$. Further for $\beta\in\Omega\setminus\{\alpha\}$, $N_G(P)=PG_{\alpha,\beta}$ with $G_{\alpha,\beta}$ cyclic of order $\frac{q^2-1}{d}$ and $|C_{G_{\alpha,\beta}}(Z(P))|=\frac{(q+1)}{d}$.

\item\label{lemma2.7_iii} $P$ has class $2$ with $|Z(P)|=q.$ If $p$ is odd, then $P$ has exponent $p$ and if $p=2$ then $P$ has exponent $4$ with the set of involutions of $P$ being $Z(P)^\#$.

\noindent Let \textasciicircum{} denote the image of subgroups of $SU_3(q)$ in $U_3(q)$ $(\cong G)$.

\item\label{lemma2.7_iv} $G$ has a maximal subgroup $M$ isomorphic to \textasciicircum$GU_2(q)$.

\item\label{lemma2.7_v} $M$ has a subgroup $E_0$ of shape \textasciicircum$(q+1)^2$ for which 

\noindent
$N_G(E_0)\sim$ \textasciicircum$(q+1)^2.\Sym(3)$ and any subgroup of $G$ of shape 

\noindent
 \textasciicircum$(q+1)^2.\Sym(3)$ is conjugate to $N_G(E_0)$.
\item\label{lemma2.7_vi} $G$ has a cyclic subgroup $C$ of order $\frac{q^2-q+1}{d}$ for which $N_G(C)\sim C.3$ is a Frobenius group.
\end{enumerate}
\end{lemma}

\begin{proof}
For parts~\ref{lemma2.7_i} to~\ref{lemma2.7_iii} see Suzuki~\cite{Suz1} and Huppert~\cite{Huppert1}. Part~\ref{lemma2.7_iv} follows from Mitchell~\cite{Mitchell} (or Bray, Holt, Roney-Dougal~\cite{BHRD}). Also from either~\cite{BHRD} or~\cite{Mitchell} $G$ has one conjugacy class of maximal subgroups of shape \textasciicircum$(q+1)^2.\Sym(3)$ (except when $q=5$) and~\ref{lemma2.7_vi} holds (except when $q=3,5$). Using the Atlas~\cite{ATLAS} for these exceptional cases we obtain~\ref{lemma2.7_v} and~\ref{lemma2.7_vi}.
\end{proof}



From Table 2 in~\cite{CharTablePSU} we can extract details of the conjugacy classes of $G=U_3(q)$ as well as some supplementary information which we display in Table~\ref{Tab2} ($d=(3,q+1), \ell =\frac{q+1}{d}$ and $\Omega$ as in Lemma~\ref{2.7}~\ref{lemma2.7_ii}).

\begin{table}[htbp]\centering
\caption{Conjugacy classes of $U_3(q)$}\label{Tab2}
\renewcommand{\arraystretch}{1.5}
\begin{tabular}{|>{\PBS\centering}p{17mm} |>{\PBS\centering}p{35mm} |>{\PBS\centering}p{20mm} |>{\PBS\centering}p{25mm} | }
\hline
 Class Type & Number of Classes\break of each Type& Centralizer Order & $|\fix _{\Omega}(g)|$, $g^G\in \mathcal{C}_i$ \\
\hline
 $\mathcal{C}_1$ & $1$ & $|G|$ & $q^3+1$ \\ \hline

 $\mathcal{C}_2$ & $1$ & $\frac{q^3(q+1)}{d}$\vrule height14pt depth0pt width0pt& $1$ \\ \hline

 $\mathcal{C}_3$ &$d$ & $q^2$ & $1$\\ \hline

 $\mathcal{C}_4$ & $\ell-1$& $\frac{q(q+1)(q^2-1)}{d}$\vrule height14pt depth0pt width0pt&$q+1$\\ \hline

 $\mathcal{C}_5$ & $\ell-1$ & $\frac{q(q+1)}{d}$\vrule height14pt depth0pt width0pt& $1$\\ \hline

 $\mathcal{C}_6$ & $\frac{q^2-q+1-d}{6d}$&$\frac{(q+1)^2}{d}$\vrule height14pt depth0pt width0pt & $0$\\ \hline

 $\mathcal{C}_6^\prime$ & $0$ if $d=1$, $1$ if $d=3$ & $(q+1)^2$ & $0$\\ \hline

 $\mathcal{C}_7$ & $\frac{q^2-q-2}{2d}$ &$ \frac{q^2-1}{d}$\vrule height14pt depth0pt width0pt& $2$\\ \hline

 $\mathcal{C}_8$ & $\frac{q^2-q+1-d}{3d}$ & $\frac{q^2-q+1}{d}$\vrule height14pt depth0pt width0pt & $0$\\ \hline
\end{tabular}
\end{table}
\goodbreak


We have used the same notation for the classes as in~\cite{CharTablePSU} except we have omitted the superscripts used there as they are of no importance here. Because $U_3(q)$ acts $2$-transitively on $\Omega$, by page 69 of~\cite{Isa} the permutation character must be $\chi_1 + \chi_{q^3}$ (as in Table 2 of~\cite{CharTablePSU}) which then yields the last column of the table.

%%%ICI tab 2


\section{The Number of Orbits on \texorpdfstring{$\mathcal{P}_k(\Omega)$}{Pk(Omega)}}
We begin with the

\begin{proof}[Proof of Theorem~\ref{1.1}] Put $\Omega_k=\mathcal{P}_k(\Omega).$ Suppose $k \in \mathbb{N}$ with $k\geq 3,$ and let $1 \ne g\in G\ (\cong L_2(q))$ be an element of order $m$. Then by Lemma~\ref{2.5}~\ref{lemma2.5_iii} $g$ must be contained (uniquely) in a conjugate of one of $P$, $H_{-}$ and $H_{+}$. Since we seek to determine $|\fix _{\Omega_{k}}(g)|$ we may suppose that $g$ is contained in one of $P$, $H_{-}$ and $H_{+}$.

First we consider the case when $g \in H_{+}$. Since $g$ is some power of $h_+$, by Lemma~\ref{2.5}~\ref{lemma2.5_iv}, $g$ has cycle type $m^{\frac{q+1}{m}}$. By Lemmas~\ref{2.4} and~\ref{2.5}~\ref{lemma2.5_iv} $H_+$ contains $\phi(m)$ elements of order $m$ and there are $\frac{q(q-1)}{2}$ conjugates of $H_+$, whence these elements contribute
\[
\frac{q(q-1)}{2} \sum_{m \in \mathcal{D}^* (\frac{q+1}{d})} \phi(m) \eta_k\left(m^{\frac{q+1}{m}}\right)
\]
to the sum $\sum_{g \in G} |\fix _{\Omega_k}(g)|.$
 Now consider the case when $g \in H_-$. As $g$ is a power of $h_-$, by Lemma~\ref{2.5}~\ref{lemma2.5_iv}, $g$ has cycle type $1^2 m^{\frac{q-1}{m}}.$ Employing Lemmas~\ref{2.4} and~\ref{2.5}~\ref{lemma2.5_iv} we obtain
\[
\frac{q(q+1)}{2} \sum_{m\in \mathcal{D}^*(\frac{q-1}{d})} \phi(m) \eta_k\left(1^2 m^{\frac{q-1}{m}}\right)
\]
in the sum $\sum_{g \in G} |\fix _{\Omega_k}(g)|.$
Similar considerations for $g \in P$, using Lemma~\ref{2.5}, yield
\[
(q-1)(q+1) \eta_k (1^1 p^{\frac{q}{p}} ).
\]
Combining the above with Lemmas~\ref{Burnside} and~\ref{2.5}~\ref{lemma2.5_i} we obtain the expression for $\sigma_k(G,\Omega)$.
\end{proof}

The proof of Theorem~\ref{1.2} is similar to that of Theorem~\ref{1.1} except we use Lemma~\ref{2.6} in place of Lemma~\ref{2.5}. The remainder of this section is devoted to establishing Theorem~\ref{1.3}. So we assume $G\cong U_3(q)$, $(q=p^a>2)$ and $\Omega$ is the set of isotropic $1$-spaces of a $3$-dimensional
unitary space.

Before beginning the proof of Theorem~\ref{1.3}, there are a number of preparatory definitions, notation and lemmas we need. First we describe the partitions that arise in Theorem~\ref{1.3}. Before we can do that, and introduce $\mu_k$, we require a barrage of notation associated with pairs of natural numbers dividing $\ell^\prime$ where $\ell^\prime\in \mathcal{D}(\ell),$ $\ell=\frac{q+1}{d}$. So let $\ell^\prime \in \mathcal{D}(\ell)$ and 
\[
(\ell_1,\ell_2) \in \mathcal{D}(\ell^\prime) \times \mathcal{D}(\ell^{\prime}),
\]
 and let $p_1,..,p_r$ be prime numbers such that $\ell_1=p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_r^{\alpha_r}$ and $\ell_2 = p_1^{\beta_1}p_1^{\beta_2}\ldots p_r^{\beta_r}$ where for $i=1,\ldots ,r$ at least one of the the $\alpha_i$ and $\beta_i$ is non-zero. If $\alpha_i \neq \beta_i$, then define $\gamma_i = \max\{\alpha_i ,\beta_i\}$. Without loss of generality we shall assume that $\alpha_i = \beta_i$ for $1\leq i \leq s$ and $\alpha_i \neq \beta_i$ for $s<i\leq r$. Set
\begin{align*}
\ell_0 & = p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_s^{\alpha_s} \ (=p_1^{\beta_1}p_2^{\beta_2}\ldots p_s^{\beta_s}),\\
m_1 & = p_{s+1}^{\alpha_{s+1}}p_{s+2}^{\alpha_{s+2}}\ldots p_r^{\alpha_r} 
\intertext{ and}
m_2 & = p_{s+1}^{\beta_{s+1}}p_{s+2}^{\beta_{s+2}}\ldots p_r^{\beta_r}.
\end{align*}
Also set $\ell_*=p_{s+1}^{\gamma_{s+1}}p_{s+2}^{\gamma_{s+2}}\ldots p_r^{\gamma_r}$ and note that $\ell_1 = \ell_0m_1$ and $\ell_2 = \ell_0m_2$. We remark that we do not exclude the possibilities $\ell_1=\ell_0=\ell_2$ or $\ell_1=m_1$ and $\ell_2=m_2$. Put $\ell_{12}=lcm\{\ell_1,\ell_2\}.$ If $\ell_i=1$ then $\ell_0=1=m_i$ and if $\ell_1=\ell_2=1$ then we set $\ell_*=1$.
\begin{definition}\label{def3.1}\ \\*[-1.2em]
%\renewcommand{\labelenumi}{(\roman{enumi})}
\begin{enumerate}[label=(\roman*)]
\item\label{defi3.1_i} $\pi_1 = 1^{q^3+1}.$
\item\label{defi3.1_ii} $\pi_2 = 1^1 p^{\frac{q^3}{p}}.$
\item\label{defi3.1_iii} $\pi_3 = 1^1 4^{\frac{q^3}{4}}$ (only defined for $p=2$).
\item\label{defi3.1_iv} $\pi_4^{(m)} = 1^{q+1}m^{\frac{q^3-q}{m}}$ where $m \in \mathcal{D}^*(\ell)$.
\item\label{defi3.1_v} $\pi_5^{(m)} = 1^1p^{\frac{q}{p}}m^{\frac{q^3-q}{m}}$ where $m=pj$ and $j \in \mathcal{D}^*(\ell)$.
\item\label{defi3.1_vi} $\pi_6^{(\ell_1,\ell_2,n)} = \ell_1^{\frac{q+1}{\ell_1}}\ell_2^{\frac{q+1}{\ell_2}}n^{\frac{q+1}{n}}\ell_{12}^{\frac{q^3-3q-2}{\ell_{12}}}$ where $(\ell_1,\ell_2) \in \mathcal{D}^*(\ell) \times \mathcal{D}^*(\ell)$ and %\\
$n=n_*\ell_*$, $n_* \in \mathcal{D}(\ell_0).$
\item\label{defi3.1_vii} $\pi_7^{(m)} = 1^2 j^{\frac{q-1}{j}} m^{\frac{q^3-q}{m}}$ where $m \in \mathcal{D}(\frac{q^2-1}{d}),$ $m \not\in \mathcal{D}(\ell)$, $j=\frac{m}{(m,\ell)}$.
\item\label{defi3.1_viii} $\pi_8 = m^{\frac{q^3+1}{m}}$ where $m \in \mathcal{D}^*(q^2-q+1)$.
\item\label{defi3.1_ix} When $3^i\vert q+1$ with $i\in \mathbb{N}$,
\[
3^i\pi_6^{(\ell_1,\ell_2,n)} = (3^i\ell_1)^{\frac{q+1}{3^i\ell_1}}(3^i\ell_2)^{\frac{q+1}{3^i\ell_2}}(3^in)^{\frac{q+1}{3^in}}(3^i\ell_{12})^{\frac{q^3-3q-2}{3^i\ell_{12}}}
\]
where $(\ell_1,\ell_2)\in \mathcal{D}(\ell)\times\mathcal{D}(\ell)$ and $n=n_* \ell_*,$ $n_* \in \mathcal{D}(\ell_0)$.
\end{enumerate}
\end{definition}

\begin{definition}\label{def3.2}
\[
\mu_k=(q^3+1)(q^3-1) \eta_k (\pi_2)
\]
 if $p\ne 2$ and
\[
\mu_k=(q^3+1)((q-1) \eta_k (\pi_2) + (q^3-q)\eta_k(\pi_3))
\]
 if $p=2$.
\end{definition}



\begin{definition}
\label{def3.3}
Let $k\in \mathbb{N}$, and we continue to set $\ell=\frac{q+1}{d}$. For $(\ell_1,\ell_2) \in \mathcal{D}(q+1)\times \mathcal{D}(q+1)$ we use the notation $\ell_0, m_1, m_2, \ell_*$ as defined earlier.
%\renewcommand{\labelenumi}{(\roman{enumi})}
\begin{enumerate}[label=(\roman*)]
\item\label{defi3.3_i} Let $(\ell_1,\ell_2) \in \mathcal{D}(q+1)\times \mathcal{D}(q+1)$, $n= \ell_*n_*$ with $n_* \in \mathcal{D}(\ell_0)$ and $n_* = p_1^{\delta_1}p_2^{\delta_2}\ldots p_s^{\delta_s}.$ If $\ell_0 = p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_s^{\alpha_s}$, then we define
\[
f(\ell_1,\ell_2,n)=\phi(m_1)\phi(m_2)\phi(\ell_0)\mathop{\prod_{\alpha_j=\delta_j}}_{1\leq j\leq s}p_j^{\alpha_j-1}(p_j-2)\phi\left(\mathop{\prod_{\alpha_j\ne \delta_j}}_{1\leq j\leq s} p_j^{\delta_j}\right).
\]

\item\label{defi3.3_ii}
$\displaystyle
 \lambda_k^*(\ell,\ell) =\mathop{\sum_{(\ell_1,\ell_2) \in \mathcal{D}^*(\ell) \times \mathcal{D}^*(\ell)}}\ \ \mathop{\sum_{1\ne n=\ell_*n_*}}_{n_* \in \mathcal{D}(\ell_0)} f(\ell_1,\ell_2,n) \eta_k(\pi_6^{(\ell_1,\ell_2,n)}).
$

\item\label{defi3.3_iii} 
For $\ell^\prime \in \mathcal{D}^*(q+1)$ and $i \in \mathbb{N}$ such that $3^i\vert q+1$ set
\[
 \lambda_k(\ell^\prime,\ell ^{\prime};i ) = \mathop{\sum}_{(\ell_1,\ell_2) \in \mathcal{D}(\ell^\prime)\times \mathcal{D}(\ell ^{\prime} )}\mathop{\sum_{n=\ell_* n_*}}_{n_* \in \mathcal{D}(\ell_0)} f(\ell_1,\ell_2,n) \eta_k(3^i\pi_6^{(\ell_1,\ell_2,n)}).
\]
\end{enumerate}
\end{definition}
\begin{definition}
\label{def3.4}
Let $E_0$ be an abelian subgroup of $G$ isomorphic to the direct product of two cyclic groups of order $\frac{(q+1)}{d}$ and $(q+1)$ with $N_G(E_0) \sim \frac{(q+1)}{d}(q+1).\Sym(3).$ Put $E_0^*=(\mathcal{C}_6 \cup\mathcal{C}_6^\prime)\cap E_0,$ and define
\[
\postdisplaypenalty1000000\sigma_k(E_0^*,\Omega)=\sum_{g\in E_0^*} \vert \fix _{\Omega_k}(g)\vert,
\]
 where $\Omega_k=\mathcal{P}_k(\Omega).$

\end{definition}

 \begin{remark}
 Subgroups such as $E_0$ in Definition~\ref{def3.4} exist by Lemma~\ref{2.7}~\ref{lemma2.7_v}, and $\sigma_k(E_0^*,\Omega)$ is the contribution of $E_0^*$ to the sum $\sum_{g \in G} \vert\fix _{\Omega_k}(g)\vert$.
 \end{remark}


\begin{lemma}\label{3.5}
The cycle type for elements in the classes of type $\mathcal{C}_i$ for $i\ne 6$ are given in Table~\ref{Tab3}.

\begin{table}[htbp]\centering
\caption{Cycle Types}
\renewcommand{\arraystretch}{1.2}
\begin{tabular}{@{\extracolsep{\fill}} | c | c | c |}
\hline
 Class Type & Order of $g$ & Cycle type of $g$ \\ \hline

 $\mathcal{C}_1$ & $1$& $\pi_1$ \\ \hline

 $\mathcal{C}_2$ & $p$ & $\pi_2$ \\ \hline

 $\mathcal{C}_3$ & $4\ (p=2)$ & $\pi_3$\\
 & $p\ (p\ne2)$& $\pi_2$ \\ \hline

 $\mathcal{C}_4$ & $m,\ (m\in \mathcal{D}^*(\ell))$ &$\pi_4^{(m)}$\vrule height12pt width0pt depth0pt \\ \hline

 $\mathcal{C}_5$ & $px,\ (x\in \mathcal{D}^*(\ell))$ & $\pi_5^{(m)}$\vrule height12pt width0pt depth0pt\\ \hline

 $\mathcal{C}_7$ & $m=js,\ (j\in \mathcal{D}^*(q-1),$ &$\pi_7^{(m)}$\vrule height12pt width0pt depth0pt\\

 & $s\in \mathcal{D}(\ell))$ &\\ \hline

 $\mathcal{C}_8$ & $m, (m\in \mathcal{D}^*(q^2-q+1))$ & $\pi_8$ \\ \hline


\end{tabular}
\label{Tab3}
\end{table}

\end{lemma}
\begin{proof}
Let $X=g^G$ be a conjugacy class of type $\mathcal{C}_i$, $i \in \{1,\ldots,8\}\setminus\{6\}$, and let $P \in Syl_p(G)$. If $i=1$, then clearly $g=1$. From the centralizer sizes in Table~\ref{Tab2}, for $i=2$ we must have $X$ is the conjugacy class containing $Z(P)^{\#}$, while classes of type $\mathcal{C}_3$ are the conjugacy classes of elements in $P\setminus Z(P)$. If $p$ is odd, then by Lemma~\ref{2.7}~\ref{lemma2.7_iii} the elements in classes of type $\mathcal{C}_3$ all have order $p$ whence, as elements in classes of types $\mathcal{C}_2$ and $\mathcal{C}_3$ fix just one element of $\Omega$, their cycle type is $\pi_2$. If $p=2$, then for $i=2$, we also have cycle type $\pi_2$ and for $i=3$ as the elements of order $4$ must square to involutions in $Z(P)$, their cycle type must be $\pi_3$.

From Lemma~\ref{2.7}~\ref{lemma2.7_iv} $G$ possesses a maximal subgroup $M\cong$ {\textasciicircum}$GU_2(q)$. Further, $Z(M)$ is cyclic of order $\ell=\frac{q+1}{d}$. It is straightforward to check that no two elements in $Z(M)^\#$ are $G$-conjugate. So we see, using the centralizer sizes in Table~\ref{Tab2}, $Z(M)^\#$ supplies representatives for all the conjugacy classes of type $\mathcal{C}_4$. Choose $h$ to be an element of order $p$ in $M$ (in fact $h$ is $G$-conjugate to an element in $Z(P)$, this can be seen as these elements must have centralizer with order divisible by $|Z(M)|=\frac{q+1}{d}$, hence cannot be in classes $\mathcal{C}_3$). By the structure of $M$, for any $h^\prime \in Z(M)^\#$, $|C_G(hh^\prime)|=\frac{q(q+1)}{d}.$ (We note that no element of $G\setminus M$ centralizes $M$ and that $h,h^\prime \in M$, both are centralized by $Z(P)<M$ and $Z(M)<M$.) Also for $h^\prime, h^{\prime\prime} \in Z(M)^\#$ with $h^\prime\ne h^{\prime\prime}$ we see that $hh^\prime$ and $hh^{\prime\prime}$ are not $G$-conjugate. Therefore $\{hh^\prime\ |\ h^\prime \in Z(M)^\#\}$ gives representatives for all conjugacy classes of type $\mathcal{C}_5$. Now $|\fix _\Omega(g^\prime)|=q+1$ for all $g^\prime \in Z(M)^\#$. So for $h^\prime \in Z(M)^\#$ of order $m$, $h^\prime$ will have cycle type $\pi_4^{(m)}$ on $\Omega$. Similarly for $hh^\prime,$ $h^\prime \in Z(M)^\#$ with $h^\prime$ of order $m$, as $|\fix _\Omega(h)|=1,$ we infer that $hh^\prime$ has cycle type $\pi_5^{(m)}$. So we have dealt with classes of type $\mathcal{C}_4$ and $\mathcal{C}_5$.

Next we look at classes of type $\mathcal{C}_7$. Since $|\fix _\Omega(g)|=2$, we may suppose $g \in G_{\alpha,\beta}$ ($\cong \frac{q^2-1}{d}$), where $\alpha, \beta \in \Omega$, $\alpha\ne \beta$. We may also suppose $Z(M)(\cong \frac{q+1}{d})\le G_{\alpha,\beta}$ where $M$ is a maximal subgroup of $G$ isomorphic to {\textasciicircum}$GU_2(q)$. Since $|\fix _\Omega(g^\prime)|=q+1$ for all $g^\prime \in Z(M)^\#$, $g \in G_{\alpha,\beta}\setminus Z(M)$. Let $g$ have order $m$ and let $j$ be the smallest natural number such that $g^j \in Z(M)$. Then $m=(m,\ell)j$ where $j \in \mathcal{D}^*(q-1)$ (note that $j=1$, as $G_{\alpha,\beta}\cong \frac{(q+1)}{d}$ contains a unique subgroup, $Z(M)$, of order $\frac{q+1}{d}=\ell$, would mean $g \in Z(M)$). So $j=\frac{m}{(m,\ell)}$ and hence $g$ has cycle type $\pi_7^{(m)}$.

Finally we look at those of type $\mathcal{C}_8$. From Lemma~\ref{2.7}~\ref{lemma2.7_iv} $G$ has a cyclic subgroup $C\cong \frac{(q^2-q+1)}{d}$ with $N_G(C)\sim C.3$ being a Frobenius group. Now $C^\#$ has $(|C|-1)/3$ $N_G(C)$-conjugacy classes and it can be seen that no two of these are $G$-conjugate. Hence we have that the $N_G(C)$-conjugacy class representatives of $C^\#$ give us representatives for all of the classes of type $\mathcal{C}_8$. So, as $|\fix _\Omega(h)|=0$ for all $h \in C^\#$, the elements in $C^\#$ has cycle type $\pi_8$ on $\Omega$, which completes the proof of Lemma~\ref{3.5}.
\end{proof}





We now turn our attention to the delicate process of dissecting the classes of type $\mathcal{C}_6\cup \mathcal{C}_6^\prime$.
\begin{lemma}\label{3.6}
Suppose $A\cong \mathbb{Z}_{e} \times \mathbb{Z}_{e}$ with $A$ containing three subgroups $A_i \cong \mathbb{Z}_{e}$ $(i=1,2,3)$, such that $A_i \cap A_j = 1$ for $i\ne j$. Then there exists $a_1 \in A_1$, $a_2 \in A_2$ such that $A_1=\left< a_1\right>$, $A_2=\left< a_2\right>$ and $A_3=\left< a_1a_2\right>.$
\end{lemma}

\begin{proof}
Since $A_1 \cap A_2 = 1$ and $A_1 \cong \mathbb{Z}_{e}$, $A=A_1A_2$. Let $A_3=\left< c\right>.$ Then $c=a_1a_2$ where $a_i \in A_i$, $i=1,2$. Suppose $a_i$ has order $e_i$, $i=1,2$ and, without loss that, $e_1\leq e_2$. Then $c^{e_1}=a_1^{e_1}a_2^{e_1}=a_2^{e_1} \in A_2\cap A_3 = 1$. So $e_2\leq e_1$ and hence $e_1=e_2$. Since $c=a_1a_2$ has order $e$, we must have that $e_1=e_2=e,$ so proving the lemma.
\end{proof}

\begin{hyp}\label{Hyp} Suppose that $A_0$ is an abelian group containing a subgroup $A$ with $[ A_0:A ]=1$ or $3$. Also suppose that $A$ has order $e^2$ and contains three subgroups $A_1, A_2, A_3$ with $A_i \cong\mathbb{Z}_e$ $(i=1,2,3)$.

Further suppose that $\Omega$ is an $A_0$-set such that
%\renewcommand{\labelenumi}{(\roman{enumi})}
\begin{enumerate}[label=(\roman*)]
\item\label{hypo3.8_i} for $1\neq g \in A_0$, $\fix _{\Omega}(g)=\emptyset$ if $g\not\in A_1\cup A_2\cup A_3$ and $\fix _{\Omega}(g)=\fix _{\Omega}(A_i)$ if $g \in A_i$;
\item\label{hypo3.8_ii} $\fix _{\Omega}(A_i)\cap \fix _{\Omega}(A_j) =\emptyset$ for $1\leq i \ne j \leq 3$; and
\item\label{hypo3.8_iii} for $i=1,2,3$, $|\fix _{\Omega}(A_i)|=q+1$.
\end{enumerate}

Set $\Lambda_i = \fix _{\Omega}(A_i)$ for $i=1,2,3$ and $\Lambda=\Omega \setminus (\Lambda_1 \cup \Lambda_2\cup \Lambda_3)$ and so, by~\ref{hypo3.8_ii}, $\Omega$ is the disjoint union 
\[
\Lambda_1 \cup \Lambda_2 \cup\Lambda_3 \cup \Lambda.
\]
\looseness-1
Moreover, by~\ref{hypo3.8_i}, $A_0$ acts regularly on $\Lambda$ and for $i=1,2,3$, $A_0/A_i$ acts regularly on~$\Lambda_i$.
\end{hyp}

We shall encounter Hypothesis~\ref{Hyp} both in a recursive setting and in the group $A_0=\mathbb{Z}_{q+1} \times \mathbb{Z}_\ell$ (where $\ell=\frac{q+1}{d},$ $d=(3,q+1)$). For this $A_0$ we have $e=\ell$ and $A$ would be the subgroup of $A_0$ generated by the elements of $A_0$ of order $\ell$. Then $[A_0 : A]=d$.


\begin{lemma}\label{3.8}
Assume Hypothesis~\ref{Hyp} holds and use the notation $A_0, A$ and $A_i$ in the hypothesis. Let $(\ell_1,\ell_2) \in \mathcal{D}(e) \times \mathcal{D}(e)$ where $e \in \mathcal{D}(\ell)$. The cycle structure on $\Omega$ of the elements $g=g_1g_2 \in A$ where $g_i \in A_i$ $(i=1,2)$ with $g_i$ of order $\ell_i$ is
\[
\left(\ell_1\right)^{\frac{q+1}{\ell_1}}\left(\ell_2\right)^{\frac{q+1}{\ell_2}}\left(n\right)^{\frac{q+1}{n}}\left(\ell_{12}\right)^{\frac{|\Lambda|}{\ell_{12}}},
\]
where $n=\ell_*n_*$ with $n_* \in \mathcal{D}(\ell_0)$ and $\ell_{12}=lcm(\ell_1,\ell_2)$. This cycle structure, as $g_1$ and $g_2$ ranges over the elements of order (respectively) $\ell_1$ and $\ell_2$ occurs
\[
\phi(m_1)\phi(m_2)\phi(\ell_0)\mathop{\prod_{\alpha_j=\delta_j}}_{1\leq j\leq s}p_j^{\alpha_j-1}(p_j-2)\phi\left(\mathop{\prod_{\alpha_j\ne \delta_j}}_{1\leq j\leq s} p_j^{\delta_j}\right)
\]
 times, where $n_*=p_1^{\delta_1}\ldots p_s^{\delta_s}.$
\end{lemma}

\begin{proof}
 By Hypothesis~\ref{Hyp}~\ref{hypo3.8_i} and~\ref{hypo3.8_ii} $A_i \cap A_j =1$ for $i\ne j$. Hence, by Lemma~\ref{3.5} we may select $a_1 \in A_1$, $a_2 \in A_2$ so as to have $A_1=\left< a_1\right>$, $A_2=\left< a_2\right>$ and $A_3=\left< a_1a_2\right>.$ Additionally we may identify $A$ with $A_1A_2$.
Let $g=g_1g_2$ where $g_i \in A_i$ and $g_i$ has order $\ell_i$, $i=1,2$. The smallest $k \in \mathbb{N}$ such that $g^k \in A_2$ is clearly $\ell_1$ and, likewise, the smallest $k \in \mathbb{N}$ such that $g^k \in A_1$ is clearly $\ell_2$. Hence, as $A/A_2$ acts regularly on $\Lambda_2,$ $g$ in its action on $\Lambda_2$ must be the product of disjoint cycles of length $\ell_1$. Similarly $g$ acts upon $\Lambda_1$ as a product of disjoint cycles each of length $\ell_2$. Concerning the action of $g$ on $\Lambda$, as $A_0$ acts regularly on $\Lambda$ and $\ell_{12}=lcm\{\ell_1,\ell_2\}$ is the order of $g$, $\Lambda$ is a disjoint union of $\frac{|\Lambda|}{\ell_{12}}$ length cycles of $g$.

Since $A_3=\left<a_1a_2\right>$ to find the lengths of $g$'s cycles on $\Lambda_3$, we must determine the smallest $k \in \mathbb{N}$ such that $g^k \in\left<a_1a_2\right>$. For $i=1,2$ let $k_i \in \mathbb{N}$ with $k_i \leq e $ be such that $g_i=a_i^{k_i}$. So $g=a_1^{k_{1}}a_2^{k_{2}}$ and, we recall, $\ell_i=e/(e,k_i)$ for $i=1,2$. Thus we seek the smallest $k \in\mathbb{N}$ for which
\[
g^k=(a_1^{k_1}a_2^{k_2})^k=a_1^{k_1k}a_2^{k_2k}=(a_1a_2)^j
\] for some $j,$ $0\le j < e.$ This is the smallest $k\in \mathbb{N}$ such that $k_1k \equiv k_2k{\mbox{ mod }} e$ which is $k=\frac{e}{(k_1-k_2,e)}$.

Let $C$ be a cyclic group isomorphic to $\mathbb{Z}_e$ with generator $c$. Now for $i=1,2$ the order of $c^{k_i}$ is $\frac{e}{(e,k_i)}=\ell_i$ and the order of $c^{k_1}(c^{k_2})^{-1}$ is $k$. Thus to enumerate the possibilities for $k$ (recall $(\ell_1,\ell_2)$ is a fixed ordered pair) we look at the order of $c^{k_1}(c^{k_2})^{-1}$ as we run through the ordered pairs $(c^{k_1},c^{k_2})$ of elements of $C$ of order, respectively, $\ell_1$ and $\ell_2$. In doing this there is no loss in supposing $C=\left<c\right>$ has order $lcm\{\ell_1,\ell_2\}$.

For $i=1,\ldots,r$, let $P_i \in Syl_{p_i}(C).$ Since the order of elements in $C$ is the product of their orders in the projections into $P_i$ for $i=1,\ldots,r$, we first consider the special case when $C=P_i \ne 1$, for some $i\in \{1,\ldots,r\}$. So $\ell_1=p_i^{\alpha_i}$ and $\ell_2=p_i^{\beta_i}.$



\hypertarget{3.9.1}{(\thecdrthm.1)}
%\textbf{(3.9.1)}
If $\alpha_i \ne \beta_i,$ then for all choices of $(c^{k_1},c^{k_2})$, of which there are $\phi(p_i^{\alpha_i})\phi(p_i^{\beta_i}),$ the order of $c^{k_1}(c^{k_2})^{-1}$ is $p_i^{\gamma_i}$.

Recalling that by definition $\gamma_i=\max\{\alpha_i, \beta_i\}$ we see that the order of $c^{k_1}(c^{k_2})^{-1}$ is $p_i^{\gamma_i}$ as asserted.

Now we turn to the case when $\alpha_i=\beta_i$. Here we have $\phi(p_i^{\alpha_i})^2$ possible choices for $(c^{k_1}, c^{k_2})$. Let $C_1$ be the unique subgroup of $C$ of order $p_i^{\alpha_i-1}$ (and note $c^{k_1}$ and $c^{k_2}$ are in $C \setminus C_1$). Should $c^{k_1}$ and $c^{k_2}$ be in different $C_1$ cosets of $C$, then $c^{k_1}(c^{k_2})^{-1}$ is not in $C_1$ whence $c^{k_1}(c^{k_2})^{-1}$ has order $p_i^{\alpha_i}$. This will happen $\phi(p_i^{\alpha_i})p_i^{\alpha_i-1}(p_i -2)$ times.
The number of ordered pairs $(c^{k_1},c^{k_2})$ for which $c^{k_1}$ and $c^{k_2}$ are in the same $C_1$ coset of $C$ is $\phi(p_i^{\alpha_i})p_i^{\alpha_i-1}$. In this situation, for a fixed $c^{k_1}$, $c^{k_1}(c^{k_2})^{-1}$ runs through all the elements of $C_1$ thus yielding $\phi(p_i^{\alpha_i-1})$ of order $p_i^{\alpha_i-1}$, $\phi(p_i^{\alpha_i-2})$ of order $p_i^{\alpha_i-2}$, and so on. To summarize we have the following.
%\end{sublemma}


%\begin{sublemma}\label{sublemma3.9.2}
%\textbf{(3.9.2)}
\hypertarget{3.9.1}{(\thecdrthm.2)}
 Suppose $\alpha_i=\beta_i$. Then for $\phi(p_i^{\alpha_i})p_i^{\alpha_i-1}(p_i -2)$ of the ordered pairs $(c^{k_1},c^{k_2})$ the order of $c^{k_1}(c^{k_2})^{-1}$ is $p_i^{\alpha_i}$ and, for $j=1,\ldots,\alpha_i,$ $\phi(p_i^{\alpha_i})\phi(p_i^{\alpha_i-j})$ of the ordered pairs $(c^{k_1},c^{k_2})$ the order of $c^{k_1}(c^{k_2})^{-1}$ is $p_i^{\alpha_i-j}$.
%\end{sublemma}


We now consider the general situation for $\ell_1=p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_r^{\alpha_r}$ and $\ell_2 = p_1^{\beta_1}p_1^{\beta_2}\ldots p_r^{\beta_r}$. Looking at all those $i$ for which $\alpha_i\ne\beta_i$ (just for the moment considering the projections onto $P_{s+1},\ldots,P_r$) we obtain that $c^{k_1}(c^{k_2})^{-1}$ has order $p_{s+1}^{\gamma_{s+1}}\ldots p_r^{\gamma_r}=\ell_*$ for
\[
\prod_{i=s+1}^r \phi(p_i^{\alpha_i})\phi(p_i^{\beta_i})=\phi(m_1)\phi(m_2)
\]
 pairs $(c^{k_1},c^{k_2})$ by~\hyperlink{3.9.1}{(3.9.1)} We now wish to enumerate the pairs $(c^{k_1},c^{k_2})$ for which the order of $c^{k_1}(c^{k_2})^{-1}$ is $n$, where $n=\ell_{*}n_{*},$ $n_* \in \mathcal{D}(\ell_0)$ and $n_{*} = p_1^{\delta_1}p_2^{\delta_2}\ldots p_s^{\delta_s}.$
\looseness-1
Using~\hyperlink{3.9.2}{(3.9.2)} and by just considering the projections on $P_1,\ldots,P_s$ we see this occurs for
\begin{align*}
\prod_{\alpha_i=\delta_i}\Mk \phi(p_i^{\alpha_i})p_i^{\alpha_i-1}(p_i\Mk-\Mk2)\Mk \prod_{\alpha_i\ne\delta_i}\Mk \phi(p_i^{\alpha_i})\phi(p_i^{\delta_i})
\Mk\Mk &=\Mk \prod_{1\le i \le s}\Mk \phi(p_i^{\alpha_i}) 
\Mk\prod_{\alpha_i = \delta_i} \Mk
p_i^{\alpha_i-1}(p_i\Mk-\Mk2)\Mk\prod_{\alpha_i\ne \delta_i}\Mk\phi(p_i^{\delta_i})
\\
&=\Mk \phi(\ell_0)\prod_{\alpha_i=\delta_i}p^{\alpha_i-1}(p_i-2)\phi\left(\prod_{\alpha_i\ne \delta_i} p_i^{\delta_i}\right)
\end{align*}
pairs. Combining this with the projection onto $P_{s+1},\ldots,P_r$ yields Lemma~\ref{3.8}.\qedhere
\end{proof}

\begin{lemma}\label{3.9}
Let $E_0$ be an abelian subgroup of $G$ isomorphic to the direct product of two cyclic groups of order $\frac{(q+1)}{d}$ and $(q+1)$ with $N_G(E_0)\sim\frac{(q+1)}{d}(q+1).\Sym(3).$ Also let $E$ be the subgroup of $E_0$ generated by the elements of $E_0$ of order $\ell=\frac{(q+1)}{d}.$ Then
%\renewcommand{\labelenumi}{(\roman{enumi})}
\begin{enumerate}[label=(\roman*)]
\item\label{lemma3.10_i} $\sigma_k(E_0^*,\Omega)=\lambda_k^*(\ell,\ell)$ if $d=1$;
\item\label{lemma3.10_ii} $\sigma_k(E_0^*,\Omega)=\lambda_k^*(\ell,\ell)+2\lambda_k(\ell,\ell;1)$ if $d=3$ and $3\nmid | E|$; and
\item\label{lemma3.10_iii} $\sigma_k(E_0^*,\Omega)=\lambda_k^*(\ell,\ell)+2.9^{b-1}\lambda_k(\frac{q+1}{3^{b}},\frac{q+1}{3^b};b)$ if $d=3$, $3\big\vert |E|$ and $3^b$ is the largest power of $3$ dividing $q+1$.
\end{enumerate}
\end{lemma}


\begin{proof}
By Lemma~\ref{2.7}~\ref{lemma2.7_iv}, \ref{lemma2.7_v} $G$ contains a subgroup $M$ with $M \sim$\textasciicircum$GU_2(q)$ and $E_0\leq M$. From the structure of $N_G(E_0)$ and \textasciicircum$GU_2(q)$, $Z(M)\leq E_0$ with $Z(M) \cong \frac{q+1}{d} (=\ell)$. Let $h \in N_G(E_0)$ be an element of order $3$. Because $[N_M(E_0) : E_0]=2,$ $h\not\in M=N_G(Z(M))$. In order to key in with the notation of Hypothesis~\ref{Hyp}, principally as we shall employ Lemma~\ref{3.8}, we set $A_0=E_0$ and $A=E$. Further, we set $A_1=Z(M)$, $A_2=Z(M)^h$ and $A_3=Z(M)^{h^2}.$ Since $\fix _\Omega(g)=\fix _\Omega(Z(M))$ for all $g\in Z(M)^\#,$ it follows that $\fix _\Omega(A_i)\cap\fix _\Omega(A_j)=\emptyset$ for $1\le i\ne j \le 3.$ We also have $|\fix _\Omega(A_i)|=q+1$ and, by Table~\ref{Tab2}, $\fix _\Omega(g)=\emptyset$ if $g \in A_0\setminus (A_1 \cup A_2 \cup A_3).$ Now $A$ is the subgroup of $A_0$ generated by the elements of $A_0$ of order $\ell$ and $A_i \cong \ell$. So we have $A_i \le A$, $i=1,2,3$ and $[A : A_0]=d$ $(=1$ and $3)$. Hence Hypothesis~\ref{Hyp} holds with $e=\ell$.

Suppose $d=1$. Then $A=A_0$. Using Lemma~\ref{3.8} and Definition~\ref{def3.3}~\ref{defi3.3_ii} we obtain $\sigma_k(E_0^*,\Omega)=\lambda_k^*(\ell,\ell)$. (Note the condition in Definition~\ref{def3.3}~\ref{defi3.3_ii} on the outer sum that $(\ell_1,\ell_2)\in \mathcal{D}^*(\ell) \times \mathcal{D}^*(\ell)$ and on the inner sum that $n\ne 1$ prevents the counting of elements in $\mathcal{C}_4.$) So Lemma~\ref{3.9} holds in this case.

So we now investigate the case when $d=3$. Hence $[A_0:A]=3$. Let $\theta:A_0\mapsto A_0$ be defined by $\theta : g\mapsto g^3.$ Then, as $A_0$ is abelian, $\theta$ is a homomorphism with $\im\theta \le A$ and $\ker\theta= \{x \in A_0 |\mbox{ order of }x \mbox{ is $1$ or $3$}\}$.

Further assume that $3\nmid |E|$ (so $3\nmid \ell$). Then, as $|A|=\ell^2,$ $3\nmid |A|$. Also we have $|\ker \theta|=3$ and therefore, by orders, $\im\theta=A$. For every $g\in A_0 \setminus A,$ the smallest power of $g$ contained in $A_i$ $(i=1,2,3)$ will be three times the corresponding power for $h=g^3=\theta(g)$. Now $3\nmid |A|$ means that $\theta$ restricted to $A$ is a one-to-one map, and so the inverse image of $h$ contains two elements of $A_0\setminus A$. Hence, using Lemma~\ref{3.8}, the elements of $A_0\setminus A$ contribute $2\lambda_k(\ell,\ell;1)$ to the sum $\sum_{g \in G} |\fix _{\Omega_k}(g)|.$ Thus, using Lemma~\ref{3.8} again, $\sigma_k(E_0^*,\Omega)=\lambda_k^*(\ell,\ell)+2\lambda_k(\ell,\ell;1)$, as stated.

The last case to be considered is when, as well as $d=3$, we have $3\big\vert |E|$. So $3\vert \ell$. As a consequence $|\ker\theta|=3^2$ and thus $[A :\im\theta]=3$. We seek to pinpoint $\im\theta$. Now let $A_i^3$ denote the unique subgroup of $A_i$ of index $3$ $(i=1,2,3)$. Let $B$ be the subgroup of $A$ generated by the elements of $A$ of order $\ell/3$.
 Then $[A:B]=3^2$ and, for $1\le i<j\le 3,$ $B=A_i^3A_j^3$. Also observe that $\im\theta \ge B$. Set $N=N_G(A_0),$ and recall that $N\sim(\frac{q+1}{d} (q+1)).\Sym(3)$. Also the $N$-conjugacy class of $A_1$ is $\{A_1,A_2,A_3\}$. Hence $N$ normalizes $B$ ($=A_i^3A_j^3, 1\leq i < j \le 3$) and the $N$-conjugacy of $A_1B$ is $\{A_1B,A_2B,A_3B\}$. Moreover $A_iB\ne A_jB$ for $1\leq i < j \le 3$ (as $A=A_iA_j$). Evidently $\im\theta$ is a normal subgroup of $N$ and therefore $\{\im\theta, A_1B,A_2B,A_3B\}$ comprise the four subgroups of index $3$ in $A$ which contain $B$. Observe that the inverse image under $\theta$ of $B$ is $A$. Since, by Lemma~\ref{3.8}, $\lambda_k^*(\ell,\ell)$ is the count for the contribution of elements in $A$, we are looking to determine the contribution from the elements in $A_0\setminus A$. Thus for $h \in \im\theta \setminus B$, $\theta^{-1}(\{h\}) \subseteq A_0\setminus A$ with $|\theta^{-1}(\{h\})|=9$. For $g \in \theta^{-1}(\{h\}),$ $h=g^3 \in \im\theta$ and so we must multiply the cycle lengths of $h$ by $3$ and their multiplicities by $9$ to count the contribution of $\theta^{-1}(\{h\}).$

 Now $\im\theta$ contains $B=A_i^{(3)}A_j^{(3)}$ ($1\leq i \ne j \leq 3$) as a subgroup of index $3$, and clearly $A_i \cap \im\theta \geq A_i^{(3)},$ ($1\leq i\leq 3$). If $A_i \cap \im\theta \neq A_i^{(3)},$ then, as $[A_i : A_i^{(3)}]=3$, we get $A_i \leq \im\theta.$ But then 
 \[
\im\theta \geq A_iA_j^{(3)} = A_iA_i^{(3)}A_j^{(3)}=A_i B
\]
 whence $\im\theta= A_iB$. This is impossible as $\im\theta\neq A_iB$ and so we conclude that $A_i\cap \im \theta = A_i^{(3)}$ for $1\leq i \leq 3$. Hence we have that $\im \theta$ satisfies Hypothesis~\ref{Hyp} with $B$ playing the role of $A$ and $A_i \cap \im\theta$ ($1\leq i \leq 3$) the role of the $A_i$. Further $B$ itself satisfies Hypothesis~\ref{Hyp} with $B$ playing the role also of $A$ and the $A_i^{(3)}$ ($1 \leq i \leq 3$) the role of the $A_i$. We may repeat this process for $\im\theta\setminus B$ (note $b\ge 2$ in this case), each time we multiply cycle lengths by $3$ and the multiplicity by $9$. Eventually we arrive at $\im(\theta^{b-1}),$ containing a subgroup $B^*$ of index $3$. Observe that the count for $\im(\theta^{b-1})\cong \frac{q+1}{3^{b-1}}\times\frac{q+1}{3^b}$ is given by part~\ref{lemma3.10_ii} (with $\ell=\frac{q+1}{3^{b-1}}$) and the count for $B^*\cong\frac{q+1}{3^{b}}\times\frac{q+1}{3^b}$ with $(\ell=\frac{q+1}{3^b})$. Keeping track of changes in cycle length and multiplicity we obtain
 \begin{multline*}
9^{b-1}\left(\lambda_k^*\left(\frac{q+1}{3^{b}},\frac{q+1}{3^{b}}\right)+2\lambda_k\left(\frac{q+1}{3^b},\frac{q+1}{3^b};b\right)-\lambda_k^*\left(\frac{q+1}{3^{b}},\frac{q+1}{3^{b}}\right)\right)
\\
=2.9^{b-1}\lambda_k\left(\frac{q+1}{3^{b}},\frac{q+1}{3^b};b\right)
\end{multline*}
which is the contribution for $A_0\setminus A$. Consequently
\[
\sigma_k(E_0^*,\Omega)=\lambda_k^*(\ell,\ell)+2.9^{b-1}\lambda_k\left(\frac{q+1}{3^{b}},\frac{q+1}{3^b};b\right),
\]
and the proof of Lemma~\ref{3.9} is complete.
\end{proof}

We are now in a position to prove Theorem~\ref{1.3}.

\begin{proof}[Proof of Theorem~\ref{1.3}]
Again we apply Lemma~\ref{Burnside} using the information on cycle types given in Lemmas~\ref{3.5} and~\ref{3.9}. The size of a given conjugacy class is obtained using the centralizer orders displayed in Table~\ref{Tab2}. So the conjugacy classes in $\mathcal{C}_i$ for a fixed $i$ all have the same size hence using their multiplicity in each $\mathcal{C}_i$ together with Lemmas~\ref{3.5} and~\ref{3.9} we obtain $\sigma_k(G,\Omega)$, so proving Theorem~\ref{1.3}.
%\begin{flushright}$\square$\\\end{flushright}
\end{proof}

\begin{remark}
Type $\mathcal{C}_6^\prime$ classes only arise when $d=3$ (and then there is only one conjugacy class of this type) and consists of elements of order $3$ with no fixed points, and is one third the size of the type $\mathcal{C}_6$ classes. When $d=3$ we include this class along with the other type $\mathcal{C}_6$ classes in $\sigma_k(E_0^*,\Omega)$. It occurs as the case $\ell_1=\ell_2=n=3$ and appears $f(3,3,3)=2$ times. As the size of this class is divided by $6$, this corrects the size of this class in our count.
\end{remark}

\goodbreak
\section{\texorpdfstring{\textsc{Magma}}{Magma} Code for \texorpdfstring{$L_2(q)$}{L2(q)}}\label{sec4}

In this section we give a \textsc{Magma} implementation for the formula in Theorem~\ref{1.1}.

\begin{verbatim}
PSLsig:=procedure(q,k,~sigma);
Z:=Integers();d:=GreatestCommonDivisor(q-1,2);
p:=Factorisation(q)[1,1];sig:=0;
I:=(d/(q*(q+1)*(q-1)))*Binomial(Z!(q+1),k);
CC:=[]; Append(~CC,[<(d/q),1>,<p,Z!(q/p)>,<1,1>]);
for m in Divisors(Z!((q+1)/d)) do if m ne 1 then
Append(~CC,[<(d/(2*(q+1))),EulerPhi(m)>,<m,Z!((q+1)/m)>]);
end if;end for;
for m in Divisors(Z!((q-1)/d)) do if m ne 1 then
Append
(~CC,[<(d/(2*(q-1))),EulerPhi(m)>,<m,Z!((q-1)/m)>,<1,2>]);
end if;end for; a:=0;
for i:=1 to #CC do Cg:=CC[i];S:={Z!(Cg[i][1]): i in [2..#Cg]};
RPg:=RestrictedPartitions(k,S);
for l:=1 to #RPg do p:=RPg[l];np:=1;
for j:=2 to #Cg do
pj:=#{m:m in [1..#p] |p[m] eq Cg[j][1]};
np:=np*Binomial(Cg[j][2],pj);   end for;
a:=a + np*(Cg[1][1]*Cg[1][2]);end for;end for;
sigma:=a+I;end procedure;
\end{verbatim}

\bibliographystyle{amsplain-ac}
\bibliography{ALCO_Bradley_515}

\end{document}