\section{Introduction}
The Erd\H{o}s-Ko-Rado theorem is a classic result \cite{EKR1961} in extremal set theory. This famous paper considers collections of pair-wise intersecting $k$-subsets of an $n$-set. The result states that if $2k<n$, for any such collection $\mathfrak{S}$, we have $|\mathfrak{S}| \leq {n-1 \choose k-1}$. Moreover $|\mathfrak{S}|={n-1 \choose k-1}$ implies that $\mathfrak{S}$ is a collection of $k$-subsets containing a common point. From a graph-theoretic point of view, this is the characterization of maximum independent sets in  Kneser graphs.

Inspired by the above classical result, Deza and Frankl \cite{DF1977} investigated families of {\it intersecting} permutations. Two permutations $\sigma, \tau \in S_{n}$ are said to {\it intersect} if $\sigma\tau^{-1}$ fixes a point. Now any coset of a point stabilizer is a collection of $(n-1)!$ pair-wise intersecting permutations. Deza and Frankl proved that the size of any collection of pair-wise intersecting permutations is bounded above by $(n-1)!$. They conjectured any such collection of the largest possible size ($(n-1)!$) must be a coset of a point stabilizer. This conjecture was proved by Cameron-Ku (\cite{CK2003}) and Larose-Malvenuto (\cite{LM2004}) independently. Later in \cite{GM2009}, Godsil and Meagher gave another proof. It is natural to conduct similar investigations in different classes of transitive permutation groups.

Let $G$ be finite transitive group on a set $\Omega$. A pair of elements $g,h \in G$ are said to be {\it intersecting} if there is a point $\alpha \in \Omega$ such that $gh^{-1}\cdot \alpha =\alpha$. An {\it intersecting set} is a subset $\mathcal{S} \subset G$ such that any two permutations $s,t \in \mathcal{S}$ are intersecting. An intersecting set of maximum possible size is called a {\it maximum intersecting set}. From a graph-theoretic point of view, this is the characterization of maximum independent sets in derangement graphs. 

Given $\alpha, \beta \in \Omega$, define $\mathcal{S}_{\alpha, \beta}=\{g \in G\ |\ g\cdot\alpha =\beta\}$. These sets are clearly intersecting sets and thus any maximum intersecting set has atleast $|G_{\alpha}|$ elements. Sets of this form are called the {\it canonical} intersecting sets. Inspired by the classical Erd\H{o}s-Ko-rado theorem, it is natural to ask the following:

(A)Is the size of an intersecting set in $G$ is bounded above by the size of a point stabilizer?

(B) Is every maxmimum intersecting set is a canonical?

As mentioned above, the results by Deza-Frankl and Larose-Malvenuto show that the answer to both these questions is positive in the case of the natural action of the symmetric group. The classical Erd\H{o}s-Ko-Rado theorem and the above questions lead us to the following definitions 
 
\begin{definition}
Consider a transitive action of a finite group $G$ on a set $\Omega$. The action is said to satisfy the {\it EKR} property if the canonical intersecting sets are also maximum intersecting sets. The action is further said to satisfy the {\it strict-EKR} property if canonical intersecting sets are the only maximum intersecting sets. 
\end{definition} 

When the action is apparent, we will attribute these properties to the groups. EKR properties of many specific groups have been investigated \cite{MS2011,AM2014,AM2015,LPSX2018,MST2016,MSi2019}. A few families of groups with EKR property are as follows: Frobenius groups (\cite[Theorem 3.6]{AM2015}), 2-transitive groups (\cite[Theorem 1.1]{MSi2019}), and groups with a regular subgroup (\cite[Corollary 2.2]{AM2015}). 2-transitive group actions of $\mathrm{PGL}(2,q)$ and $\mathrm{PSL}(2,q)$ satisfy the strict-EKR property \cite{MS2011,LPSX2018}. 

Not all $2$-transitive groups satisfy the strict-EKR property, however they satisfy what is known as the {\it EKR Module} property. This property was formally defined in \cite{PSUEKR}, in which it was shown that the 2-transitive action of $\mathrm{PSU}(3,q)$ satisfies the EKR-module property. Before we define this formally, we need some notation. Consider a transitive action of a group $G$ on a set $\Omega$. 
Given $S\subset G$, we define $\mathbf{v}_{S}:=\sum\limits_{s\in S}s \in \mathbb{C}G$. We call this the characteristic vector of $S$ in $G$. Given $\alpha, \beta \in \Omega$, we set $\mathbf{v}_{\alpha, \beta}=\mathbf{v}_{\mathcal{S}_{\alpha,\beta}}$, where $\mathcal{S}_{\alpha, \beta}=\{g \in G\ |\ g\cdot\alpha =\beta\}$. We note that $\mathbf{v}_{\alpha, \beta}$ is the characteristic vector of a canonical intersecting sets.

\begin{definition}
A transitive action of a group $G$ on a set $\Omega$ is said to satisfy the EKR module property if for any maximum intersecting set $\mathcal{S}$, the characteristic vector $\mathbf{v}_{\mathcal{S}}$ is a linear combination of vectors in $\{\mathbf{v}_{\mathcal{S}_{\alpha,\beta}}\ |\ \alpha,\beta \in \Omega\}$ (that is, the set of characteristic vectors of canonical intersecting sets).
\end{definition} 

The name is derived from the module method described in \cite{AM2015}, which gives sufficient conditions for a $2$-transitive group to satisfy the strict-EKR property. One of these sufficient conditions is the action satisfying the EKR-Module property. In \cite{Spiga2019}, it was shown that any maximum intersecting set for the $2$-transitive action of $PGL(n,q)$ is either a coset of a point stabilizer, or a coset of hyperplane stabilizer. Showing that this action satisfies the EKR module property was an important intermediary step in achieving this characterization. In \cite{MSi2019}, it was shown that every $2$-transitive permutation group satisfies the EKR-Module property.

Any action that satisfies the strict-EKR property satisfies the EKR module property vacuously. There are examples of group actions that satisfy the EKR module property but not the strict-EKR property. The class of 2-transitive groups give us examples of actions satisfying both the EKR and EKR module properties. The example~\ref{cex} shows that there are groups that satisfy the EKR property but not the EKR module property. We now construct an example that satisfies the EKR Module property but not the EKR property.

\begin{example}
Consider the action of $A_{4}$ on the set $\Omega$ of cosets of a subgroup $H \cong \mathbb{Z}_{2}$. We observe that the Sylow-2-subgroup $N$ of $A_{4}$ is an intersecting set. Consider an intersecting set $\mathcal{S}$. Then given $t\in \mathcal{S}$, the set $\mathcal{S} t^{-1}$ is an intersecting set containing the identity. By the definition of an intersection set, we have $\mathcal{S} t^{-1} \subset \bigcup\limits_{g \in A_{4}}gHg^{-1}=N$. This shows that any maximum intersecting set is a coset of $N$. Any coset of $N$ is a union of two disjoint cosets of $H$, and thus this action satisfies the EKR module property. 
\end{example} 

In this article, we find a few more infinite classes satisfying the EKR-module property. It is well known (see \cite[Corllary 2.2]{AM2014}) that permutation groups that contain a regular subgroup satisfy the EKR property. The following example shows that this condition does not imply the EKR-Module property.

\begin{example}\label{cex}
Consider $G=S_{5}$ (isomorphic to $PGL(2,5)$) and $H \cong D_{6}$. We consider the action of $G$ on $G/H$. A subgroup isomorphic to $D_{5}$ is a regular subgroup, and so this action satisfies the EKR property. 
 
 Now $G$ has a  subgroup $K$ which is isomorphic to $A_{4}$. It is easy to check that $K \subset \cup gHg^{-1}$ and thus $K$ a maximum intersecting set. Every coset of a conjugate of $H$ contains exactly $3$ even permutations and every permutation in $K$ is even. Now consider the sign character $\lambda$, we have $\lambda(\mathbf{v}_{S})=0$ for every canonical maximum intersecting set $S$, but $\lambda(\mathbf{v}_{K})\neq 0$. Therefore this action does not satisfy the EKR-Module property.
\end{example}
We however find that groups with a regular normal subgroup satisfy the EKR-Module property. A few classes of permutation groups with a regular normal subgroup are Frobenius groups, affine groups, primitive groups of type HS, HC, and TW.  
\begin{theorem}\label{rega}
Transitive groups with regular normal subgroups satisfy the EKR-Module property.
\end{theorem}

It is easy to see that any transitive action of an abelian group satisfies the strict EKR property and so the EKR module property. It was shown in \cite{BMMKK2015} that every transitive action of a nilpotent group satisfies the EKR module property and that every nilpotent group with a non-abelian odd ordered subgroup does not have the strict-EKR property. A different proof was given in \cite{LSP}. In this article, we find that class $2$ nilpotent groups satisfy the EKR-Module property. 

\begin{theorem}\label{Nil}
Transitive class $2$-nilpotent groups satisfy the EKR-Module property.
\end{theorem}

Theorem~$2$ of \cite{BMMKK2015} states that if every transitive action of a group $G$ satisfies the EKR property, then $G$ is a soluble group. We found that this is not does not hold for the EKR-Module property.

\begin{theorem}\label{PSLe}
Every transitive action of $PSL(2,2^{f})$ with $f\geq 2$ satisfies the EKR-Module property.
\end{theorem} 
We prove this in \S\ref{PSL}. In the same section, we find demonstrate examples to show that the above theorem is not true for $PSL(2,q)$ when $q$ is an odd prime power.

After showing that every $2$-transitive group satisfies the EKR-Module, the authors of \cite{MSi2019} mention that the next natural step is to consider rank $3$ permutation groups. We first consider primitive rank $3$ permutation groups. By the O`Nan-Scott Theorem, a rank $3$ permutation group $G$ is either (i) an affine group; or (ii) satisfies $T_{0}\times T_{0}\leq G \leq  T\wr \mathbb{Z}_{2}$, where $T$ is a $2$-transitive group whose socle $T_{0}$ is a non-abelian simple group; or (iii)an almost simple group. By Theorem~\ref{rega}, we know that affine rank $3$ permutation groups satisfy the EKR-Module property. In \S\ref{rank3}, we show the following:

\begin{theorem}\label{Rank3}
Let $T\leq \mathrm{Sym}(\Omega)$ be a $2$-transitive group, then the rank $3$ action of $T\wr\mathbb{Z}_{2}$ on $\Omega \times \Omega$ satisfies the EKR-Module property.
\end{theorem}

At this point, we would like to mention that when $n$ is sufficiently large, the rank $3$ action of $S_{n}$ on $2$-subsets of $[n]$ satisfies the strict-EKR property and thereby the EKR-Module property. This was proved in \cite{ellis2012setwise}. Example~\ref{cex} shows that this fails when $n=5$. It would be interesting to look at the rank $3$ permutation actions of other almost simple permutation groups.   

EKR-module property has been observed in other contexts as well. Given a graph $X$ and a prescribed set of canonical cliques, we say that the graph satisfies the EKR-module property if the characteristic vector of any maximum clique satisfies is a linear combination of the characteristic vectors of canonical cliques. In the context of permutation groups satisfying the EKR-module property, the complement of the derangement graph satisfies the EKR-module property. In Chapter~5 of \cite{GMbook}, there are a few examples of strongly regular graphs satisfying the EKR-module property.  In \cite{EKRpaley}, the authors show that Peisert-type graphs satisfy the EKR-Module property. Let $q$ be an odd prime power, and let $F$ and $E$ be finite fields of order $q^{2}$ and $q$ respectively. A Peisert-type graph of type $(m,q)$ is a Cayley graph of the form $Cay(F,S)$, where the ``connection'' set $S$ is a union of $m$ distinct cosets of the multiplicative group $E^{\times}$ in $F^{\times}$. It is clear that any set of the form $sE+b$ with $s \in S$ and $b\in F$ is a clique. We deem these to be the canonical cliques. In \cite{EKRpaley}, the authors show that every the characteristic vector maximum clique in a Peisert-type graph is a linear combination of of the characteristic vectors of canonical cliques. In \S~\ref{srg}, we give a different proof.
  
\section{EKR-Module property and character theory.}\label{prelim}
In this section we gather some elementary results about intersecting sets, and characterize EKR-module property using characters of finite groups. 

Let $G$ be a transitive group on $\Omega$, and let $K \vartriangleleft G$ be the kernal for this action. The following preliminary observation reduces investigations on intersecting sets to the case of permutation groups.
\begin{lemma}
Let $G$ be a finite transitive group on $\Omega$, $K \vartriangleleft G$ be the kernal for this action, and let $\pi:G \to G/K$ be the natural quotient map. Then, if $\mathcal{S} \subset G$ is a maximum intersecting set, then $\pi(\mathcal{S}) \subset G/K$ is a maximum intersecting set. Conversely, if $\mathcal{T}$ is a maximum intersecting set in $G/K$, then $\pi^{-1}(\mathcal{T})$ is a maximum intersecting set.  
\end{lemma} 
\begin{proof}
Given any intersecting subset $A \subset G$, we note that $AK:=\{ak\ | a\in A,\ k\in K\}$ is also an intersecting set. So any maximum intersecting set in $G$ is a union of $K$-cosets. Let $s_{1},s_{2},\ldots s_{r} \in G$ be such that $\mathcal{S} = s_{i}N$ is a maximum intersecting subset of $G$. Now $\pi(\mathcal{S})=\{s_{i}K\ |\ 1\leq i \leq r\}$ is an intersecting set in $G/K$. Now consider a maximum intersecting set $\mathcal{T}=\{t_{i}K\ |\ 1\leq i\leq s\}$ of $G/K$. It is clear that $\pi^{-1}(\mathcal{T})=\bigcup t_{i}N \subset G$ is an intersecting set. So we have $|\mathcal{T}||K|=|\pi^{-1}(\mathcal{T})|\leq |\mathcal{I}|$, and $|\mathcal{I}|/|N|=|\pi(\mathcal{S})| \leq |\mathcal{T}|$. This shows that $\pi(\mathcal{S})$ (respectively $\pi^{-1}(\mathcal{T})$) is a maximum intersecting subset of $G/K$ (respectively $G$).  
\end{proof}
As an immediate consequence, we get the following corollary.
\begin{cor}\label{permred}
Let $G$ be a finite transitive group on $\Omega$, $K \vartriangleleft G$ be the kernal for this action, and let $\pi:G \to G/K$ be the natural quotient map. Then

(i) $G$ satisfies the EKR (respectively strict-EKR) property if and only if $G/K$ satisfies the EKR (respectively strict-EKR) and;

(ii)$G$ satisfies the EKR-module property if and only if $G/K$ satisfies the EKR-module property.
\end{cor}
\begin{proof}
Set $H:=G/K$. We note that for all $\omega\in \Omega$ and $g\in G$, we have $\pi{g}H_{\omega}=\pi(gG_{\omega})$ and $gG_{\omega}=\pi^{-1}(\pi(g)H_{\omega})$. The results now follows from the above Lemma.    
\end{proof}
From this Corollary, we are justified in only considering permutation groups.
We now mention the following well known result that helps us show that groups in Theorems~\ref{Nil} and~\ref{rega} satisfy the EKR property. A subset $C$ of a permutation group $G \leqslant \mathrm{Sym}(\Omega)$ is said to {\it sharply transitive} if for each $(\alpha, \beta) \in \Omega \times \Omega$, there is a unique $g \in G$ such that $ \alpha^{g} =\beta$. It is well known that groups with sharply transitive sets satisfy the EKR property. The following result is mentioned in \cite{AM2014} and \cite{LSP}. 

\begin{lemma}\label{regekr}
Let $G$ be a finite permutation group on $\Omega$ and $\omega \in \Omega$. If $G$ contains a sharply transitive subset $C$, then the following are true:
\begin{enumerate}[{\rm(a)}]
\item Every intersecting set of $G$ is of the form $\bigcup\limits_{ c \in C} cH_{c}$, where $\{H_{c}|\ c \in C\}$ is a set of pairwise disjoint subsets of $G_{\omega}$;
\item and $G$ has the EKR property.
\end{enumerate}
\end{lemma} 

As an immediate application of this, we see that any group with a regular subgroup satisfies the EKR property. So groups with an abelian regular normal subgroup and 
We now show that all supersolvable groups satisfy the EKR property. It was first proved in \cite{BMMKK2015} that nilpotent groups satisfy EKR property. In \cite{LSP}, it was generalized to supersolvable groups.
\begin{lemma}\label{nilekr}
Any supersolvable permutation group has a sharply transitive subset, and thus satisfies the EKR property.
\end{lemma}
\begin{proof}
Let $N \leq \mathrm{Sym}(\Omega)$ be a supersolvable group. We prove that the existence of a sharply transitive subset via induction on $|N|+|\Omega|$. As $N$ is a permutation group, its point stabilizers are core-free. Consider a minimal normal subgroup $Z$. As $N$ is supersolvable, it has prime order, and hence acts semi-regularly. If the action of $Z$ is regular, then $Z$ is the sharply transitive subset of $N$. Now we assume that $Z$ is not a transitive subgroup. Let $\Lambda=\{\Delta_{1},\ldots, \Delta_{k}\}$, be the partition of $\Omega$ into $Z$-orbits. We note that $|\Lambda|=|\Omega|/|Z|$. As $Z \vartriangleleft N$, $\Lambda$ is a block system for $N$, and thus $N$ acts transitively on $\Lambda$. Let $K$ be the kernal for the action of $N$ on $\Lambda$. Now by induction hypothesis, $N/K \leq \mathrm{Sym}(\Lambda)$ has a sharply transitive subset. Let $1_{G}=c_{1},c_{2},\ldots,c_{k-1} \in N$ such that $ \{c_{i}K| \ 1\leq i\leq k\} \subset G/K$ is a sharply transitive set. As $Z \subset K$, the set $\{c_{i}Z| \ 1\leq i \leq k\}$ is a set of $k$ distinct $Z$-cosets. The set $C=\bigcup\limits_{i=1}^{k}c_{i}Z \subset N$ is a sharply transitive subset of $N$.
\end{proof}

Let $G$ be a finite group acting transitively on a set $\Omega$.  By $\mathfrak{I}_{G}(\Omega)$, we denote the $\mathbb{C}$-subspace of $\mathbb{C}G$ spanned by $\{\mathbf{v}_{\alpha, \beta}:=\mathbf{v}_{\mathcal{S}_{\alpha,\beta}}\ |\ \alpha,\beta \in \Omega\}$. 
\begin{equation}\label{Mdef}
\mathfrak{I}_{G}(\Omega)= \{\sum a_{\alpha,\beta}\mathbf{v}_{\alpha, \beta}|\ a_{\alpha,\beta}\in \mathbb{C} \}
\end{equation}
We recall that $G$ has the EKR Module property if and only if for every maximum intersecting set $\mathcal{S}$, the corresponding characteristic $\mathbf{v}_{I} \in \mathfrak{I}_{G}(\Omega)$. We first observe that for any $g,h \in \mathbf{G}$, we have $g\mathbf{v}_{\alpha, \beta}h=\mathbf{v}_{h^{-1}\cdot \alpha, g\cdot \beta}$. Thus $\mathfrak{I}_{G}(\Omega)$ is a $2$-sided ideal of $\mathbb{C}G$. Two sided ideals of group algebras are well understood via their Artin-Wedderburn decomposition. We will now recall the same. 

Let $Irr(G)$ be the set of irreducible complex characters of $G$. For $\chi \in Irr(G)$, we define $M_{\chi}:=\sum\limits_{i=1}^{\chi(1)}W_{i}$, where $\{W_{1},W_{2}, \ldots W_{\chi(1)}\}$ are the right submodules of $\mathbb{CG}$ that affording the character $\chi_{i}$. We have $\mathbb{C}G=\bigoplus\limits_{\chi \in Irr(G)}M_{\chi}$.
We note that $M_{\chi}$ is a submodule of the right regular module $\mathbb{C}G$ of dimension $\chi(1)^{2}$, and thus is also a right ideal. Given $x \in \mathbb{C}G$ the map $\lambda_{x}: \mathbb{CG} \to \mathbb{CG}$ defined by $g\mapsto xg$ is both a $\mathbb{C}$-algebra map and a right module homomorphism. Now by Schur's Lemma, we have $\lambda_{x}(M_{\chi})\subseteq M_{\chi}$ and thus $M_{\chi}$ is a $2$-sided ideal. We now consider the $\mathbb{C}$-algebra isomorphism $\Theta:\mathbb{C}G \to End_{\mathbb{C}G}(\mathbb{C}G)$, given by $\Theta(x)=\lambda_{x}$. By application of Schur's lemma, we see that $End_{\mathbb{C}G}(\mathbb{C}G) \cong \bigoplus\limits_{\chi \in Irr(G)} End_{\mathbb{C}G}(M_{\chi})$ and that $End_{\mathbb{C}G}(M_{\chi})$ is isomoprhic to the $\chi(1) \times \chi(1)$ matrix algebra over $\mathbb{C}$. So $M_{\chi} \cong End_{\mathbb{C}G}(M_{\chi})$ has no proper ideal. Thus the decomposition $\mathbb{C}G= \bigoplus_{\chi \in Irr(G)} M_{\chi}$ can be seen as a presentation of $\mathbb{C}G$ as a direct product of simple rings $M_{\chi}$'s. So given a two sided ideal $\mathfrak{J}$, there is a subset $Y \subseteq Irr(G)$ such that $\mathfrak{J}=\sum\limits_{\chi \in Y}M_{\chi}$.

Let $1=\sum e_{\chi}$ such that $e_{\chi} \in M_{\chi}$. It is well know (for instance see \cite[2.12]{isaacs1994character}) that $e_{\chi}=\dfrac{\chi(1)}{|G|}\sum\limits_{g \in G} \chi(g^{-1})g$. Now $E_{i}:=\lambda_{e_{\chi}}$ is a projection onto the ideal $M_{i}$. We see that 
\begin{equation}\label{idedef}
E_{\chi}(h)= \frac{\chi(1)}{|G|}\sum\limits_{g \in G} \chi_{i}(hg^{-1})g.
\end{equation} 

Now we describe $2$-sided ideals as in the following lemma.

\begin{lemma}\label{ides}
Given a $2$-sided ideal $\mathfrak{J}$ of $\mathbb{C}G$, there is a subset $Y_{\mathfrak{J}} \subseteq Irr(G)$ such that $\mathfrak{J}=\bigoplus\limits_{\chi \in Y_{\mathfrak{J}}} \left(e_{\chi}\right)$ and $P_{\mathfrak{J}}:= \sum\limits_{\chi \in Y_{\mathfrak{J}}} E_{\chi}$ is the projection such that $\mathrm{IM}(P_{\mathfrak{J}})={\mathfrak{J}}$ and $\mathrm{Ker}(P_{\mathfrak{J}})=\sum\limits_{\chi \notin Y_{\mathfrak{J}}}M_{\chi}$.
\end{lemma} 

We now consider a group $G$ with a transitive action on a set $\Omega$. Now describe the decomposition of the the ideal $\mathfrak{I}_{G}(\Omega)$ (defined in \eqref{Mdef}) into simple ideals of $\mathbb{C}G$.

\begin{lemma}\label{caniddes}
Let $G$ be a finite group, $H<G$ a subgroup, and $\Omega=[G :H]$ be the space of right cosets of $H$. Let $Y_{H}=\{\chi \in Irr(G)|\ \chi(\mathbf{v}_{H}) \neq 0\}$.
The linear span $I_{\Omega}(G)$ of $\{\mathbf{v}_{aH^{g}}|\ a,g\in G\}$ decomposes as the sum $\bigoplus\limits_{\chi \in Y_{H}} \left\langle e_{\chi}\right\rangle$ of simple ideals of $\mathbb{C}G$. 
\end{lemma}

\begin{proof}
For any subset $S \subset G$ and $\psi \in Irr(G)$, we have
\begin{align*}
\frac{|G|}{\psi(1)}e_{\psi}\mathbf{v}_{S}=\frac{|G|}{\psi(1)}E_{\psi}(\mathbf{v}_{S})&= \frac{|G|}{\psi(1)}\sum\limits_{s \in S} E_{\psi}(s)\notag\\
&=\sum\limits_{s \in S}\sum\limits_{g \in G} \psi(sg^{-1})g \notag \\
&= \sum\limits_{g\in G}g \sum\limits_{s \in S} \psi(sg^{-1})\notag\\
&=\sum\limits_{g\in \mathbf{G}} \psi(\mathbf{v}_{S g^{-1}})g.
\end{align*} 
Therefore for any $\chi \in Y_{H}$, we have $0 \neq e_{\chi} \mathbf{v}_{H} \in \left\langle e_{\chi}\right\rangle \cap I_{\Omega}(G) \subset I_{\Omega}(G)$. As $\left\langle e_{\chi}\right\rangle$ is a minimal ideal, we conclude that $\left\langle e_{\chi}\right\rangle \subset I_{\Omega}(G)$. 

Now consider $\theta \in Irr(G) \setminus Y_{H}$. In this case, we have $\theta(\mathbf{v}_{H^g})=\theta(\mathbf{v}_{H})=0$. Let $\Theta : G \to GL_{\theta(1)}(\mathbb{C})$ be a \textbf{unitary} representation affording $\theta$ as its character. Given a subset $S\subset G$, we consider $M_{S}=\sum\limits_{s \in S}\Theta(s)$. As $\Theta$ is a unitary representation, we have $M_{S^{-1}}= M_{S}^{\dagger}$, that is, $M_{S^{-1}}$ is the conjugate transpose of $M_{S}$. Now given $a, g \in H$, we have 
$$M_{H^{g}a^{-1}}M_{aH^{g}}= \sum \limits_{x, y \in H^{g}} \Theta(x)\Theta(y)=|H|M_{H^{g}}.$$ As $\theta(\mathbf{v}_{H^{g}})=0$, we have $0=\mathrm{Tr}(M_{aH^{g}})=\mathrm{Tr}(M_{H^{g}a^{-1}}M_{aH^{g}})=$. Since $M_{H^{g}a^{-1}}=M_{aH^{g}}^{\dagger}$, this can only happen when $M_{aH^{g}}=0$. Therefore $\theta(\mathbf{v}_{aH^{g}})=0$. From the above computation, we must have $e_{\theta}\mathbf{v}_{aH^{g}}=0$. As $e_{\theta}^{2}=1$, $e_{\theta}$ cannot be in $I_{\Omega}(G)$. Now the result follows by Lemma \ref{ides}.  
\end{proof}

As an immediate application, we see a slightly shorter proof of the Lemma 4.1 and Lemma 4.2 of \cite{AM2015}. This result find a basis of canonical vectors for  $I_{\Omega}(G)$, when the action of $G$ on $\Omega$ is $2$-transitive.

\begin{cor}\label{2t}
Let $G \leq \mathrm{Sym}(\Omega)$ be a $2$-transitive permutation group with $\pi \in Irr(G)$ such that $1+\pi$ is the corresponding permutation character. Then
\begin{enumerate}
\item $I_{\Omega}(G)=\left\langle e_{1}\right\rangle + \left\langle e_{\pi}\right\rangle $ is a vector space of dimension $1+(|\Omega|-1)^{2}$
\item and the set $$B_{\omega}:=\{\mathbf{v}_{G}\} \cup \ \{\mathbf{v}_{\alpha,\ \beta}\ : \ (\alpha,\beta)\in \left(\Omega \setminus \{\omega\} \right)^{2}\},$$ is a basis set of $I_{G}(\Omega)$ for any $\omega \in \Omega$.
\end{enumerate} 
\end{cor}
\begin{proof}
Part (1) follows immediately from the above Lemma.

We observe that $\alpha \in \Omega$, $\mathbf{v}_{\alpha, \omega}=\mathbf{v}_{S}-\sum\limits_{\beta \neq \omega}\mathbf{v}_{\alpha,\ \beta}$. Therefore every vector of the form $\mathbf{v}_{\gamma,\ \delta}$ is in the linear span of the elements of $B_{\omega}$, and thus $B_{\omega}$ spans $I_{G}(\Omega)$. Linear independence follows as the $I_{\Omega}$ is a $|B_{\omega}|$-dimensional space.  
\end{proof}

We now have the following characterization of EKR-Module property.

\begin{cor}\label{idekr}
Let $G$ be a group acting transitively on a set $\Omega$, and let $\mathfrak{I}_{G}(\Omega)$ be the $2$-sided ideal of $\mathbb{C}G$, which is the linear span of characteristic vectors of cosets of point stabilizers. Let $C=\{\chi \in Irr(G)|\ \chi(\mathbf{v}_{\mathbf{G}_{\omega}})=0\}$
The action of $G$ on $\Omega$ satisfies EKR-module property if and only if for any maximum intersecting sets $\mathcal{S}$ and any $\chi \in C$, we have $\chi(\mathbf{v}_{\mathcal{S}}) =0$. 
\end{cor} 
\begin{proof}
We note that $\sum\limits_{\chi \in Irr(G)}E_{\chi}$ is the identity map on $\mathbb{C}G$. Thus if $E_{\chi}(\mathbf{v}_{\mathcal{S}}) \in \mathfrak{I}_{G}(\Omega)$, then $\mathbf{v}_{\mathcal{S}} \in \mathfrak{I}_{G}(\Omega)$. Now $\mathfrak{I}_{G}(\Omega)$ is an $2$-sided ideal. So if $\mathbf{v}_{\mathcal{S}} \in \mathfrak{I}_{G}(\Omega)$, then as $E_{\chi}(\mathbf{v}_{\mathcal{S}})=e_{\chi}\mathbf{v}_{\mathcal{S}} \in \mathfrak{I}_{G}(\Omega)$.
\end{proof}

Results from spectral graph theory have proved useful in characterizing maximum intersecting sets in permutation group. We will now look at a characterization of the EKR-module property in terms of the spectra of derangement graphs. Let $G$ be a group acting transitively on a set $\Omega$. An element $g\in G$ is called a derangement if it does not fix any point in $\Omega$. By $Der(G,\Omega)$, we denote the set of derangements in $G$. By $\Gamma_{G,\Omega}$, we denote the Cayley graph on $G$, with $Der(G, \Omega)$ as the ``connection set''. Now we can observe that intersecting sets in $G$ are the same as independent sets/co-cliques in $\Gamma_{G,\Omega}$. We now state a useful spectral bound on sizes of independent sets in $d$-regular graphs. Before doing so, we define some terms.
 For graph $X$ on $n$ vertices, a real symmetric matrix $M$ whose rows and columns are indexed by the vertex set of $X$, is said to be \emph{compatible} with $X$, if $M_{u,v}=0$ whenever $u$ is not adjacent to $v$ in $X$. Given a subset $S$ of the vertex set, by $\mathbf{v}_{S}$, we denote the characteristic vector of $S$.
 
\begin{lemma}(\cite[Theorem 2.4.2]{GMbook})\label{wrb}
Let $M$ be a real symmetric matrix with constant sum $d$, which is compatible with a graph $X$ on $n$ vertices. If the least eigenvalue of $M$ is $\tau$, then for any independent set $S$ in $X$, 
$$|S| \leq \frac{n(-\tau)}{d-\tau},$$ and if equality holds, then 
$$\mathbf{v}_{S}-\frac{|S|}{n}\mathbf{v}_{X}$$ is a $\tau$-eigenvector of $M$.
\end{lemma}    

The spectra of $\Gamma_{G,\Omega}$-compatible matrices, proved useful in characterization of maximum intersecting sets in many permutation groups (for instance see \cite{MS2011} and \cite{MST2016}). In our paper, we encounter a special type of $\Gamma_{G, \Omega}$ matrices. 
\begin{definition}
Let $G$ be a group acting transitively on a set $\Omega$. A $(G,\Omega)$-compatible class function is a real valued class function $f: G \to \mathbb{R}$ with $f(g)=0$ for all $g \notin Der(G,\Omega)$ and $f(d)=f(d^{-1})$ for all $d\in Der(G, \Omega)$
\end{definition}
Consider a $(G,\Omega)$-compatible class function . Consider the matrix $M^{f}$ indexed by $G\times G$, with $M^{f}_{g,h}=f(gh^{-1})$ for all $(g,h)\in G \times G$. Now $M^{f}$ is a $\Gamma_{G,\Omega}$-compatible matrix. We now describe the spectra of such matrices. A Cayley graph whose connection set is a union of conjugacy classes is called a normal Cayley graph. $\Gamma_{G, \Omega}$ is a normal Cayley graph.
The spectra of matrices compatible with normal Cayley graph is a well-know result usually attributed to Babai \cite{Babai}, or  Diaconis and Shahshahani \cite{DS1981}. Our proof is same as that of Lemma~5 in \cite{DS1981}. 

\begin{lemma}\label{spec}
Let $G$ be a permutation group on $\Omega$, with $D \subset G$ being the set of derangements. Let $f: G \to \mathbb{R}$ be $(G,\Omega)$-compatible class function. Define $M \in \mathbb{C}^{G\times G}$ by $M_{gh}=f(g^{-1}h)$. Then $M^{f}$ is a $\Gamma_{G,\Omega}=Cay(G,D)$-compatible matrix with spectrum
$Spec(M):=\{\lambda_{\chi,f}| \chi \in Irr(G)\}$, where $$\lambda_{\chi,f}=\frac{1}{\chi(1)}\sum\limits_{g \in G}f(g)\chi(g).$$ Given $\nu \in Spec(M^{f})$, the $\nu$-eigenspace in $\mathbb{C}{G}$ is the $2$-sided ideal
$$\sum\limits_{\{\chi|\ \chi\in Irr(G)\ \text{and}\ \lambda_{\chi,f}=\nu\}} \left\langle e_{\chi}\right\rangle.$$
\end{lemma}  
\begin{proof}
Let $\rho:\mathbb{C}G \to End(W)$ be an irreducible subrepresentation of the regular representation affording the character $\chi$. Define $f_{\rho}:=\sum\limits_{g \in P}f(g)\rho(g)$. As $f$ is a class function, $\rho(h^{-1})f_{\rho}\rho(h)=f_{\rho}$, for all $h \in G$, and therefore is a $\mathbb{C}G$-map in $End(W)$. Thus by Schur's lemma we have $f_{\rho}=\lambda_{\rho}I$ for some $\lambda_{\rho} \in \mathbb{C}$. Computing traces of $f_{\rho}= \sum\limits_{g \in G}f(g)\rho(g)$ yields $\lambda_{\rho}=\frac{1}{\chi(1)}\sum\limits_{g \in G}f(g)\chi(g)$. As the regular module is a direct sum of all its irreducible submodules, we see that 
$M^{f}=\bigoplus f_{\rho}$, where the sum runs over all irreducible subrepresentations of the regular representation. Now the spectrum of $M{f}$ is clear. We also see that an irreducible submodule affording character $\chi$ is a subspace of the $\lambda_{\chi,f}$-eigenspace of $M^{f}$. The second part follows from that fact that the ideal $\left\langle e_{\chi}\right\rangle$ is the sum of all irreducible submodules of $\mathbb{C}G$ that afford $\chi$ as their character.    
\end{proof}

We are now ready to give a sufficient condition for EKR-Module property in terms of spectra of $\Gamma_{G,\Omega}$ compatible matrices. Let Let $f: G \to \mathbb{R}$ be a $(G, \Omega)$-compatible class function. The the row sum of $M^{f}$ is $r_{f}:=\sum\limits_{g\in G}f(g)$. Let $\tau$ be the least eigenvalue of $M^{f}$. Then by Lemma~\ref{wrb}, for any intersecting set $S$, we have $$|S|\leq \frac{|G|(-\tau)}{r_{f}-\tau}.$$ Let us assume the equality holds. Then by Lemma~\ref{wrb} and Lemma~\ref{spec}, we see that if $\mathcal{S}$ is an maximum intersecting set, then 
$\mathbf{v}_{\mathcal{S}}$ is in the $2$-sided ideal 
$$\left\langle e_{1} \right\rangle +\sum\limits_{\{\chi|\ \chi\in Irr(G)\ \text{and}\ \lambda_{\chi,f}=\tau\}} \left\langle e_{\chi}\right\rangle.$$ Now by application of Lemma~\ref{caniddes}, we get the following sufficient condition for EKR Module property.
\begin{theorem}\label{Specsuff}
Let $G$ be a group acting on the set $\Omega$ of right cosets of a subgroup $H$. Assume that there is an intersecting set $S$ and a $(G,\Omega)$-compatible class function $f:G \to \mathbb{R}$ such that $|S|=\dfrac{|G|(-\tau)}{d-\tau}$, where $d=\sum\limits_{g \in G} f(g)$ and $\tau$ is the least eigenvalue of $M^{f}$. 

Then 

(a) $|S|$ is the size of a maximum intersecting set in $G$; and 

(b)the action of $G$ on $\Omega$ satisfies the EKR-Module property if and only if $\chi(\mathbf{v}_{H})\neq 0$ for all $\chi \in Irr(G)$ satisfying $\frac{1}{\chi(1)}\sum\limits_{g \in G}f(g)\chi(g)=\tau$. 
\end{theorem}

At this point, we remark that the proof of EKR (\cite{MST2016}) and EKR-module properties(\cite{MSi2019}) of $2$-transitive groups, involved finding a right choice of class function.

Before moving on, let us look at an easy example illustrating an application of the above theorem.
\begin{example}
Consider the action of $G=A_{5}$ on the set $\Omega$ of right cosets of a cyclic subgroup $H$ of order $5$. In this case, the derangement set $Der(G, \Omega)$ is the union of the conjugacy class $C_{1}$ containing $(1,2)(3,4)$ and the conjugacy class $C_{2}$ containing $(1,2,3)$. Let $f$ be the $(G,\Omega)$-compatible class function satisfying $f((1,2)(3,4))=1$ and $f((1,2,3))=2$. Using the character table of $A_{5}$ and Lemma~\ref{spec}, we see that $\{55,10,-5\}$ is the complete set of eigenvalues of $M^{f}$. Now $H$ is an intersecting set with $|H|=5=\frac{60*5}{55+5}$. Therefore the size of every maximum intersecting set is $5$. Our computation of the spectrum of $M^{f}$ also shows that for $\chi \in Irr(G)\setminus \{1\}$, $\frac{1}{\chi(1)}\sum\chi(g)f(g)=-5$ if and only if $\chi(\mathbf{v}_{H})\neq 0$. Therefore by Theorem~\ref{Specsuff}, this action satisfies the EKR Module property.       
\end{example} 
\section{Proof of Theorem~\ref{rega}}\label{regular}
Let $A$ be a finite group and $H \leqslant Aut(A)$. We consider the action of $\mathbf{G}= A \rtimes H$ on $A$. Now by Corollary~\ref{permred}, if groups of form $\mathbf{G}$ satisfy the EKR-module property, then so do any groups with containing a regular normal subgroup. 
 
Every element of $\mathbf{G}$ is of the form $(a, \sigma)$, for $a\in A$ and $\sigma \in H\leqslant(Aut(A))$. Note that $(a,\sigma)(b,\pi)= (a\sigma(b), \sigma\pi)$. 
\begin{lemma}
Let $g=(a,\sigma) \in \mathbf{G}$, with $a\in A$ and $\sigma \in H$. If $g$ fixes a point then it is $A$ conjugate to $h$; and $h$ is the unique $A$-conjugate of $G$ in $H$.
\end{lemma} 
\begin{proof}
For any $b \in A$, we have $(a,\sigma)\cdot b=a\sigma(b)$. So $(a,\sigma)$ fixes $b$ if and only if $a=b\sigma(b^{-1})$. Now $c^{-1}(b\sigma(b^{-1}),\sigma)(c)=(c^{-1}b\sigma(bc^{-1}),\sigma)$. Now $(c^{-1}b\sigma(bc^{-1}),\sigma) \in H$ if and only if $c^{-1}b\sigma(bc^{-1})=1$. Setting $c=b$ completes the proof.   
\end{proof}

Let $\mathcal{S}_{0}$ be any maximum intersecting set with $1 \in \mathcal{S}_{0}$, As $\mathcal{S}_{0}$ is an intersecting set, for all $s \in \mathcal{S}_{0}$, the element $s$ fixes some point. Thus by the above lemma, there is a unique element $a_{s} \in A$ such that $a_{s}^{-1}s a_{s}=\sigma_{s}\in H$. As $s=a_{s}\sigma_{s}a_{s}^{-1}= (a_{s}\sigma_{s}(a_{s}^{-1}),\sigma_{s})$. If for $s,r \in \mathcal{S}_{0}$, we have $\sigma_{s}=\sigma_{r}$, then \[sr^{-1}=a_{s}\sigma_{s}a_{s}^{-1}a_{r}\sigma_{s}^{-1}a_{r}^{-1}= a_{s}\sigma_{s}(a_{s}^{-1}a_{r})\sigma_{s}^{-1}a_{r}^{-1}\in A.\]
As $\mathcal{S}_{0}$ is an intersecting set, $sr^{-1}\in A$ fixes a point. Since $A$ acts regularly, we have $sr^{-1}=1$. Thus $\{\sigma_{s}|s\in \mathcal{S}_{0}\}=H$. So if $\chi \in Irr(\mathbf{G})$ with $\chi(\mathbf{v}_{H})=0$, then $\chi(\mathbf{v}_{\mathcal{S}_{0}})=0$. Let $P:\mathbf{G} \to GL_{n}(\mathbb{C})$ be a representation affording $\chi$ such that $P(g)$ is a unitary matrix for all $g \in \mathbf{G}$. Given a set $X \subset G$, define $M_{X}=\sum\limits_{x \in X} P(x)$. Let $\mathcal{S}$ be any maximum intersecting set, observe that $M_{\mathcal{S}}M_{\mathcal{S}^{-1}}=\sum\limits_{t\in \mathcal{S}}M_{\mathcal{S} t^{-1}}.$ As $\mathcal{S} t^{-1}$ is a maximum intersecting set containing $1$, from the above we have $Tr(M_{\mathcal{S} t^{-1}})=0$ and thus $Tr\left( M_{\mathcal{S}}M_{\mathcal{S}^{-1}}\right)=0$. Since $M_{\mathcal{S}^{-1}}=M_{\mathcal{S}}^{\dagger}$, we have $Tr\left( M_{\mathcal{S}}M_{\mathcal{S}^{-1}}\right)=0$ if and only if $M_{\mathcal{S}}=0$. Therefore $\chi(\mathbf{v}_{\mathcal{S}})=0$. Thus by Corollary~\ref{idekr}, $\mathbf{G}$ satisfies the EKR-Module property. 
\qed 

\section{Proof of Theorem~\ref{Nil}.}
In this section, we prove that transitive actions of class-2 nilpotent groups satisfy the EKR-module property. By Lemma~\ref{nilekr}, all such groups satisfy the EKR property. By Lemma~\ref{permred}, it suffices to prove the theorem for permutation actions of nilpotent groups. Our proof is a proof by contradiction.

Assume that there are transitive class nilpotent groups that do not satisfy the EKR-module property.
Among all transitive actions of nilpotent groups not satisfying the EKR-module property, pick the action of a nilpotent group $N$ on a set $\Omega$ such that $|N|+|\Omega|$ is as small as possible. By Lemma~\ref{permred}, we see that the condition of minimality forces $N$ to be a permutation group on $\Omega$. As $N$ is nilpotent, it has a non-trivial centre, say $Z$. Since the action is a permutation action, $N_{\omega}$ (for $\omega \in \Omega$) is core-free and hence $Z\cap N_{\omega}=\{1_{N}\}$. So $Z$ and its subgroups act semi-regularly. By Theorem~\ref{rega}, $N$ cannot have an abelian regular normal subgroup and hence $Z$ acts intransitively. In view of Corollary~\ref{idekr}, consider $\chi \in Irr(N)$ and some maximum intersecting set $\mathcal{S}$. As $N$ is nilpotent, all non-trivial normal subgroups intersect $Z$ non-trivially. 
So either $ker(\chi)\cap Z \neq \{1_{N}\}$, or $\chi$ is faithful.

Now let $\chi$ be a faithful irreducible character of $N$. 
Pick a prime order subgroup $P$ of $Z$.     

Consider a non-trivial subgroup $A$ of $Z$. The action of $A$ on $\Omega$ is semi-regular and intransitive. As $A \vartriangleleft N$, the set $\Omega_{A}$ of $A$-orbits on $\Omega$, is a block system for the action of $N$ on $\Omega$. We now consider the transitive action of $N$ on $\Omega_{A}$. We note that $|\Omega_{A}|= \frac{|\Omega|}{|A|}<|\Omega|$, and thus by the minimality of $(N,\Omega)$, the action of $N$ on $\Omega_{A}$ satisfies the EKR-module property. Note that given $\omega \in \Omega$, the elements of the subgroup $N_{\omega}A$ stabilize the orbit $\omega^{A} \in \Omega_{A}$. Since $\frac{|N|}{|N_{\omega}A|}=\frac{|N|}{|N_{\omega}||A|}=\frac{|\Omega|}{|A|}=|\Omega_{A}|$, we see that the stabilizer of $\omega^{A} \in \Omega_{A}$ is precisely $N_{\omega}A$. We now observe that $\mathcal{S} A$ is a maximum intersecting set for the action of $N$ on $\Omega_{A}$. As the action of $N$ on $\Omega_{A}$ satisfies the EKR-module property. So by Corollary~\ref{idekr}, we have $\chi(\mathbf{v}_{IA})=0$ for all irreducible characters $\chi$ that satisfy $\chi(\mathbf{v}_{N_{\omega}A})=0$.    

If $ker(\chi)$ is non-trivial, consider Let $Z_{\chi}:=Z\cap ker(\chi)$. We set $\Omega_{\chi}:= \Omega_{Z_{\chi}}$. Let $\rho:N \to \mathrm{Gl}(V)$ be a representation affording $\chi$. Let $z\in Z$ be any central element, $\rho(z)$ acts on $V$ as a $G$-homomorphism. So by Schur's lemma, $\rho(z)$ acts like a scalar matrix, and thus $\chi(nz)=\chi(n)\chi(z)$ for all $n \in N$ and $z\in Z$. So $\chi(nz)=\chi(n)\chi(1)$, for any $z \in Z_{\chi}$.
From the above paragraph, if $\chi(N_{\omega}Z_{\chi})=0$, we have ${\chi}(\mathbf{v}_{\mathcal{S} Z_{\chi}}) =0$. We observe that and ${\chi}(\mathbf{v}_{N_{\omega} Z_{\chi}})=\chi(\mathbf{v}_{N_{\omega}})\chi(\mathbf{v}_{{Z}_{\chi}})$ and ${\chi}(\mathbf{v}_{\mathcal{S} Z_{\chi}})=\chi(\mathbf{v}_{\mathcal{S}})\chi(\mathbf{v}_{{Z}_{\chi}})$. So $\chi(\mathbf{v}_{N_{\omega}})=0$ implies that $\chi(N_{\omega}Z_{\chi})=0$, which implies that ${\chi}(\mathbf{v}_{\mathcal{S} Z_{\chi}})=0$, and thus ${\chi}(\mathbf{v}_{\mathcal{S}})=0$. So if $\chi$ is an unfaithful character with $\chi(\mathbf{v}_{N_{\omega}})=0$, then ${\chi}(\mathbf{v}_{\mathcal{S}})=0$ for every maximum intersecting set $\mathcal{S}$. So by Corollary~\ref{idekr}, $N$ satisfies the EKR-Module property. This is a contrary to our assumption. Therefore all nilpotent groups of class 2 satisfy the EKR-Module property.
\qed
\section{EKR-Module property of some primitive product action type rank 3 groups.}\label{rank3}
In this section, we prove Theorem~\ref{Rank3}.
The authors of \cite{MSi2019} proved that all permutation groups of rank $2$ satisfy the EKR module property. A natural next step is to investigate if the same holds for all rank 3 permutation groups. Example~\ref{cex} is an example of a primitive rank $3$ permutation group that does not satisfy the EKR module property. 

Let $T \leq Sym(\Omega)$ be a $2$-transitive group, and let $H=T_{\omega} < T$ be a point stabilizer. Now consider the action of $N:=T \times T$ on $\Omega^{2}$. The group $K:=H \times H$ is a point stabilizer for this action. We note that the $\{\omega\} \times \{\omega\}$, $\left(\Omega\setminus \{\omega\}\right) \times \left(\Omega\setminus \{\omega\}\right)$, $\left(\Omega\setminus \{\omega\}\right) \times \{\omega\}$, and $\{\omega\} \times \left(\Omega\setminus \{\omega\}\right)$ are the $K$-orbits. Now, we consider the action of $G:= T \wr S_{2}$ on $\Omega^2$.

 The group $M:=H \wr S_{2}$ is a point stabilizer. We observe that $\{\omega\} \times \{\omega\}$, $\left(\Omega\setminus \{\omega\}\right) \times \left(\Omega\setminus \{\omega\}\right)$, and $ \left(\Omega\setminus \{\omega\}\right) \times \{\omega\} \bigcup \{\omega\} \times \left(\Omega\setminus \{\omega\}\right)$ are the $M$-orbits. Therefore $G$ acts as a rank $3$ permutation. We will prove that $G$ satisfies the EKR-Module property using corollary~\ref{idekr} the fact that every $2$-transitive group satisfies both the EKR (proved in \cite{MST2016}) and the EKR-Module (proved in \cite{MSi2019}) properties.

In the context of Corollary~\ref{idekr}, we require a description of irreducible characters of $G$. The subgroup $N:=T\times T$ of $G$ is a normal subgroup of index $2$. By Clifford theory (\cite[6.19 of]{isaacs1994character}) restriction of any irreducible character $\nu \in Irr(G)$ to $N$ is either an irreducible $G$-invariant character of $N$ or the sum of two $G$-conjugate irreducible characters of $N$. Using well known results on characters of direct products, we observe that
$$Irr(N)=\{\chi \times \lambda \ : \chi,\lambda\in \ Irr(T)\}.$$ Let $\chi, \lambda$ be two distinct irreducible characters of $T$, then the intertia subgroup of $\chi \times \lambda $ is $N$, and therefore $\sigma_{\chi,\lambda }:=Ind^{G}_{N} (\chi \times \lambda)$ is an irreducible character of $G$, with $Res^{G}_{N}(\sigma_{\chi,\lambda })=\chi \times \lambda + \lambda \times \chi$. Now consider an irreducible character of $N$, of the form $\chi\times \chi$. By Lemma~1.3 of \cite{Mwreath}, $\chi \times \chi$ extends to an irreducible character $\rho_{\chi}\in Irr(G)$ with $\rho_{\chi}((s,r)\pi))=\chi(rs)$ for all $r,s \in T$ and $\pi$ being the unique $2$-cycle in $S_{2}$.
By a result of Gallagher (\cite[6.17 of]{isaacs1994character}), there is exactly one other irreducible character of $G$ whose restriction to $N$ is $\chi\times \chi$, names $\beta \rho_{\chi}$, where $\beta$ is the unique linear character with kernal $N$. 
Therefore by  Clifford theory any irreducible character is one of the characters defined above.

\begin{lemma}
The set 
$$\{\rho_{\chi}\ :\ \chi  \in Irr(T)\}\cup \{\beta\rho_{\chi}\ :\ \chi  \in Irr(T)\}\cup \{\sigma_{\chi,\ \lambda},\ :\ \chi,\lambda \in Irr(T)\ \&\ \chi \neq \lambda\}$$ is the complete set of irreducible characters of $G$.
\end{lemma}   

As $T$ is a $2$-transitive group, there is $\psi \in Irr(T)$ be such that $1+\psi$ is the permutation character for $T$. A quick computation shows that $1+\rho_{\psi}+\sigma_{\psi,1}$ is the permutation character for $G$. 

As $T$ is a $2$-transitive group, by the main results of \cite{MST2016} and \cite{MSi2019}, the action of $T$ on $\Omega$ satisfies both EKR and EKR-module properties. By the proof of Lemma 4.2 of \cite{AM2014}, we see that a maximum intersecting set for the action of $N:=T\times T$ on $\Omega^{2}$ is of the form $S\times R$, with $S$ and $R$ being maximum intersecting sets in $T$. The subgroup $N$ is a transitive subgroup satisfying the EKR property, and so by Lemma~3.3 of \cite{MST2016}, we see that $G$ satisfies the EKR property.

We consider a  maximum intersection set $\mathcal{S}$ with respect to the action of $G$ on $\Omega^{2}$. Now $\mathcal{S} \cap N$ and $(\mathcal{S} \cap N\pi)\pi^{-1}$ are intersecting sets in $N$. As $N$ satisfies, the EKR property, we have $$2|K|=|M|\leq |\mathcal{S}|=|(\mathcal{S} \cap N\pi)\pi^{-1}|+|(\mathcal{S} \cap N)|\leq 2|K|.$$ Therefore both $(\mathcal{S} \cap N\pi)\pi^{-1}$ and $(\mathcal{S} \cap N)$ are maximum intersecting sets in $N$. Let us assume that $\mathcal{S}$ contains the identity. With this assumption, every element of $\mathcal{S}$ has a fixed point. Consider $(s,r)\pi \in \mathcal{S} \cap N\pi$. As $\lambda((s,r)\pi)=1+\chi(rs)$ and $1+\chi$ is the permutation character for $T$, $1+\chi(rs) \neq 0$ if and only if $rs$ fixes a point. Any maximum intersecting set of $N$ is of the form $X\times Y$, where $X$ and $Y$ are maximum independent sets in $T$. As $(\mathcal{S} \cap N\pi)\pi^{-1}$ is an maximum intersecting set in $N$, let $X,Y$ be maximum intersecting set in T such that $(X \times Y)\pi=\mathcal{S} \cap N\pi$. We claim that $Y=X^{-1}$, if not by maximality of $X$ and $Y$, there are elements $x\in X$ and $y\in Y$ such that $xy^{-1}$ is a derangement and so $(x,y)\pi\in \mathcal{S}$ is a derangement. So $\mathcal{S} \cap N\pi =X \times X^{-1}$ for some maximum intersecting set $X$ in $T$. Pick $\alpha=(x,x^{-1})\pi \in \mathcal{S} \cap N\pi$, then $\mathcal{S}\alpha^{-1}$ is an maximum intersecting set containing identity. Therefore $\mathcal{S}\alpha^{-1} \cap N\pi =(\mathcal{S}\cap N)\alpha^{-1}$, and so by the same argument as above, there is an maximum intersecting set $Z$ in $T$ such that $\mathcal{S}=Z \times Z^{-1}$. We have now concluded that for any maximum intersecting set $\mathcal{S}$ in $G$, there are maximum intersecting sets $X$ and $Z$ in $T$ such that $\mathcal{S}= X \times X^{-1} \cup (Z\times Z^{-1})\pi$. 

By Corollary~\ref{idekr}, EKR-Module property of $G$ is equivalent to showing that $\nu(\mathcal{S})=0$ for all maximum intersecting set $\mathcal{S}$ in $G$ and $\nu \in Irr(G)$ with $\nu \neq 1, \sigma_{\psi,1},\ \&\ \rho_{\psi}$. In this regard, given an maximum intersecting set $\mathcal{S}$ in $G$ and maximum intersecting sets $X$ and $Z$ such that $\mathcal{S}= X \times X^{-1} \cup (Z\times Z^{-1})\pi$, we have for any distinct pair of characters $\chi,\lambda$ in $Irr(T)$,


(i)$\rho_{\chi}(\mathcal{S})=\chi(X)\chi(X^{-1})+\sum\limits_{r,s \in Z}\chi(r^{-1}s)=\chi(X)\chi(X^{-1})+\sum\limits_{r\in Z}\chi(r^{-1}Z)$;

(ii) $\beta\rho_{\chi}(\mathcal{S})=\chi(X)\chi(X^{-1})-\sum\limits_{r,s \in Z}\chi(r^{-1}s)=\chi(X)\chi(X^{-1})-\sum\limits_{r\in Z}\chi(r^{-1}Z)$;

(iii)and $\sigma_{\chi,\ \lambda}(\mathcal{S})=\chi(X)\lambda(X^{-1})+\lambda(X)\chi(X^{-1})$.

The last equality follows by using (a)$N \lhd G$, (b)$\sigma_{\chi,\lambda }=Ind^{G}_{N} (\chi \times \lambda)$, and (c) $Res^{G}_{N}(\sigma_{\chi,\lambda })=\chi \times \lambda + \lambda \times \chi$. 

To finish the proof, we the following lemma.

\begin{lemma}\label{2tra}
Let $T\leq \mathrm{Sym}(\Omega)$ be a $2$-transitive group with $H\leq G$ being a stabilizer subgroup. Let $\psi\in Irr(T)$ be such that $1+\psi$ is the permutation character and If $S$ is an maximum intersecting set in $T$. Then
\begin{enumerate}
\item $\psi(S)= \begin{cases} |H|\ \text{if $1\in S$}\\ -|H|/\psi(1)\ \text{otherwise}\end{cases}$,
\item and $\nu(S)=0$ for all irreducible characters $\nu\notin\ \{1,\ \ \psi\}$.
\end{enumerate} 
\end{lemma}

\begin{proof}
As $T$ is 2-transitive, it satisfies both the EKR and EKR-Module properties. So by Corollary~\ref{2t}, $\mathbf{v}_{\mathcal{S}}$ is in the image of the projection $E_{1}+E_{\psi}$, that is $(E_{1}+E_{\psi})(\mathbf{v}_{\mathcal{S}})=\mathbf{v}_{\mathcal{S}}$. Part 1 follows by equating the cooefficients of the identity element on both sides.

Part 2 is a direct application of Corllary~\ref{idekr}.
\end{proof}

Application of part (2) of the above lemma and the character sum formulas (i) (ii) and (iii) given above, we see that for any maximum intersecting set $\mathcal{S}$ in $G$ and any $\nu \in Irr(G)\setminus \{1,\ \sigma_{\psi,1},\ \rho_{\psi},\ \beta\rho_{\psi}\}$, we have $\nu(\mathcal{S})=0$. By Corollary~\ref{idekr}, showing that $\beta\rho_{\psi}(\mathcal{S})=0$ for all maximum intersecting sets $\mathcal{S}$ in $G$, will imply EKR-Module property.

Assume that $\mathcal{S}$ contains the identity element. Then $X$ and $X^{-1}$ are maximum intersecting set in $T$ that contain the identity element. For any $r \in Z$, $r^{-1}Z$ is an maximum intersecting set in $T$ containing the identity. By application of (ii) and part (1) of the above lemma, we see that $\beta\rho_{\psi}(\mathcal{S})=0$. By the same argument as the one at the end of \S\ref{regular}, we see that $\beta\rho_{\psi}(\mathcal{S})=0$ holds even if $\mathcal{S}$ does not contain the identity. This concludes the proof of Theorem~\ref{Rank3}. 
\qed
\section{EKR-Module property of $PSL(2,2^{f}$)}\label{PSL}
Theorem~$2$ of \cite{BMMKK2015} states that if every transitive action of a group $G$ satisfies the EKR property, then $G$ is a soluble group. Their proof relied on showing that all minimal finite simple groups do not satisfy the EKR property for some transitive action. A non-abelian simple group is called a minimal finite simple group if all its subgroups are soluble. When $p$ is a prime, $PSL(2,2^{p})$ is a minimal finite simple group. We show that all transitive actions of this group must satisfy the EKR-Module property. 

Let $f$ be a positive integer and $q=2^{f}$. We now recall the structure and character theory of $G:=PSL(2,q)$. The character table of this group is well know and can be found in many texts on representation theory. We refer the reader to \cite{P-S} for details. As $q$ is even, $PSL(2,q)\cong PGL(2,q)$

By $\mathrm{GL(2,q)}$, we mean the group of invertible $2 \times 2$ matrices over $\mathbb{F}_{q}$.The center of $Z$ of this group is the set of scalar matrices. The group $G=\mathrm{PGL(2,q)}=\mathrm{GL(2,q)}/Z$. Given $N \in \mathrm{GL(2,q)}$, by $\overline{N}$ we denote its image in $G$. All elements of $G$ are of the form $\overline{N}$ for some $N \in \mathrm{GL(2,q)}$. Moreover $\overline{N_{1}}=\overline{N}_{2}$ if and only if $N_{1}=aIN_{2}$ for some $a \in \mathbb{F}_{q}^{*}$. Keeping this description in mind, we abuse notation to denote elements of $G$ by $2\times 2$ matrices over $\mathbb{F}_{q}$. 

We will now describe the conjugacy classes of $G$.
On the basis of number of eigenvalues, we collect the elements of $G$ into four sets: The set consisting of the identity element I; the set consisting of non-scalar matrices with only one eigenvalue in $\mathbb{F}_{q}$; the set consisting of non-scalar matrices with two distinct eigenvalue in $\mathbb{F}_{q}$; and the set of matrices with no eigenvalue in $\mathbb{F}_{q}$.   

 Any non-identity element of $G$ with only one eigenvalue in $\mathbb{F}_{q}^{*}$ is conjugate 
$u =\overline{\left(\begin{array}{c c}
 1 & 1 \\
 0 & 1
 \end{array}\right)}$. This conjugacy class has $q^{2}-1$ elements. Now any element with two distinct eigenvalues in $\mathbb{F}_{q}^{*}$ is conjugate to $d_{x}=\overline{\left(\begin{array}{c c}
 x & 0 \\
 0 & 1
 \end{array}\right)}$, for some $x \in \mathbb{F}_{q}^{*}$. Moreover $d_{x}$ is conjugate $d_{y}$ if and only if $x=y$ or $x=y^{-1}$. The size of the conjugacy class of $d_{x}$ is $q(q+1)$. An element with no eigenvalues in $\mathbb{F}_{q}$ are conjugate to 
 $v_{r} =\overline{\left(\begin{array}{c c}
 0 & 1 \\
 -r^{(1+q)} & r+r^{q}
 \end{array}\right)}$, for some $r \in \mathbb{F}_{q^{2}}^{*} \setminus \mathbb{F}_{q}^{*}$. The eigenvalues of $v_{r}$ in $\mathbb{F}_{q^{2}}$ are Galois conjugates $r$ and $r^{q}$. Therefore $v_{r}$ is conjugate to $v_{s}$ if and only if $r\mathbb{F}_{q}^{*}=s\mathbb{F}_{q}^{*}$ or $r\mathbb{F}_{q}^{*}=s^{-1}\mathbb{F}_{q}^{*}$. For every $r \in \mathbb{F}_{q^{2}}^{*} \setminus \mathbb{F}_{q}^{*}$, the conjugacy class containing $v_{r}$ has $(q)(q-1)$ elements.

Any transitive action of $G$ can be realized as the action of $G$ on the right cosets of a subgroup $H$ of $G$. The subgroup lattice of $G$ is well known and can be found in many classical group theory books such as \cite{huppert2013endliche}. We refer the reader to \cite{king2005subgroup} for a modern exposition. We now describe the conjugacy classes of subgroups of $G$. Before stating the classification, we first describe certain subgroups of $G$. 

\begin{enumerate}

\item The group $$F:=\left\{\overline{\begin{pmatrix}
1 & x\\
0 & 1
\end{pmatrix}}\ |\ x \in \mathbb{F}_{q} 
\right\}$$ is a 2-Sylow subgroup of $G$.

\item Let $\zeta$ be a generator of $\mathbb{F}_{q}^{\times}$, then $A:=\left\langle d_{\zeta} \right\rangle$. Given $d\mid q-1$, by $A_{d}$, we denote the unique subgroup of order $d$ in $A$.

\item Let $\epsilon$ be an element of order $q+1$ in $\mathbb{F}_{q^{2}}^{\times}$, then $B:=\left\langle v_{\epsilon} \right\rangle$ is a cyclic subgroup of order  $q+1$ respectively. Given $d\mid q+1$, by $B_{d}$, we denote the unique subgroup of order $d$ in $B$.

\item Consider a primitive $d$th root of unity in $z \in \mathbb{F}_{q}^{\times}$ and a $\mathbb{F}_{2}\langle a \rangle$-submodule $E$ of $\mathbb{F}_{q}$. Then the subgroup $A_{d}$ normalizes $\overline{E}:=\{\overline{\begin{pmatrix}
1 & e \\
0 & 1
\end{pmatrix}} \ | \ e \in E \}$ and $T_{E,d}:=\overline{E}A_{d} \equiv \overline{E} \rtimes A_{d}$ is a subgroup. The group $T_{F,q-1}=FA\equiv F \rtimes A$ is a Borel subgroup of $G$, and for every pair $(E,d)$, $T_{E,d}$ is a subgroup of $T_{F,q-1}$.

\item The element $t := \overline{\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}}$ normalizes subgroups of the forms $A_{d}$ and $B_{d}$. Therefore $D_{d,-}:=\langle t\rangle A_{d}$ is a dihedral group of size $2d$ and  $D_{d,+}:=\langle t\rangle B_{d}$ is a dihedral group of size $2d$.  

\item Let $q_{0}$ be a positive integer such that $q$ is a power of $q_{0}$. In this case, $\mathbb{F}_{q_{0}}$ is a subfield of $\mathbb{F}_{q}$, and so there is a natural embedding of $PSL(2,q_{0})$ into $PSL(2,q)$. By $\mathbb{G}_{q_{0}}$, we denote the image of this embedding.
\end{enumerate}

\begin{lemma}\label{csubg}(Theorem 2.1 of \cite{king2005subgroup}) 
Let $p$ be a prime, $q=2^{p}$, and let $M$ be  a subgroup of $PSL(2,q)$. Then $M$ conjugate to one of $A_{d}$, $B_{d}$, $T_{E,d}$, $G_{q_{0}}$, $D_{d,-}$,$D_{d,+}$, or $G_{q_{0}}$.
\end{lemma}  
 
 
We now describe the complex irreducible characters of $G$, given in tables \ref{ch1}. We collect these characters into four classes.

We first describe the linear characters. $G$ has only one linear character, namely the trivial character denoted by $\lambda_{1}$.

Secondly we describe characters of dimension $q$. Let $\pi$ be the permutation character of $G$ acting on the projective line $\mathbb{P}^{1}(\mathbb{F}_{q})$. As this is $2$-transitive, the character $\psi_{1}=\pi- \lambda_{1}$ is irreducible.

Next we describe the characters of dimension $q-1$. These are known as the cuspidal characters. In table \ref{ch1}, they are indexed by a label $\beta$. This label runs over all homomorphisms $\beta: \mathbb{F}_{q^{2}}^{*}/\mathbb{F}_{q}^{*} \to \mathbb{C}^{*}$ of order greater than $2$. Note that $\eta_{\beta_{1}}=\eta_{\beta_{2}}$ if and only if $\beta_{1}=\beta_{2}$ or $\beta_{1}=\beta_{2}^{-1}$. $G$ has $q/2$ distinct cuspidal characters.

Finally we describe the characters of dimension $q+1$. These characters are know as the principal series of $G$. In table \ref{ch1}, they are indexed by $\gamma$. This label runs through homomorphisms $\gamma:\mathbb{F}_{q}^{*} \to \mathbb{C}^{*}$ of order greater than $2$. We note that $\nu_{\gamma_{1}}=\nu_{\gamma_{2}}$ if and only if $\gamma_{1}=\gamma_{2}$ or $\gamma_{1}=\gamma_{2}^{-1}$. $G$ has $q/2-1$ distinct principal series characters.  
\begin{table}
\caption{Character table of $G$, for $q$ even.}
\centering
\begin{tabular}{c |c| c c c c }
\hline
& class type & $1$ & $u$ & $d_{x}$ & $v_{r}$\\
& Number & $1$ & $1$ & $\frac{q}{2}-1$ & $\frac{q}{2}$\\
& class size & $1$ & $q^{2}-1$ & $q(q+1)$ & $q(q-1)$\\
\hline
Character type & number &  &  &  & \\
\hline
$\lambda_{1}$ & $1$ & $1$ & $1$ & $1$ & $1$\\
$\psi_{1}$ & $1$ & $q$ & $0$ & $1$ & $-1$\\
$\eta_{\beta}$ & $\frac{q}{2}$ & $q-1$ & $-1$ & $0$ & $-\beta(r\mathbb{F}_{q}^{*})-\beta(r^{-1}\mathbb{F}_{q}^{*})$\\
$\nu_{\gamma}$ & $\frac{q}{2}-1$ & $q+1$ & $1$ & $(\gamma(x)+\gamma(x^{-1}))$ & $0$
\end{tabular}
\label{ch1}
\end{table}

Now that we have described the conjugacy classes, character table, and subgroup lattice of $G$, we are ready to investigate the EKR-Module properties of various transitive actions of $G$. Lemma~\ref{csubg} classifies these transitive actions into six different types. We will deal with each type separately. through the reminder of this section for $g \in G$,  by $(g)$, we denote the conjugacy class containing $g$.

\subsection{Transitive action with $T_{E,d}$ as a stabilzer.} Let $\epsilon$ be a primitive $d$th root of unity and let $E$ be a $\mathbb{F}_{2}\langle a \rangle$-submodule of $\mathbb{F}_{q}$. We now consider the action of $G$ on the set of right cosets of $T_{E,d}$. The set $Der(G,E,d)$ of derangements with respect to this action is $$\bigcup \limits_ {\{x\ : |x| \nmid d\}}(d_{x}) \bigcup \bigcup\limits_{\{r\in \mathbb{F}_{q^{2}}^{\times}\ | r\mathbb{F}^{\times}_{q}\neq \mathbb{F}^{\times}_{q}\}} (v_{r}).$$ We note that $Der(G,E,d)=Der(G,\mathbb{F}_{q},d)$, and that $T_{\mathbb{F}_{q},d}$ is a union of cosets of $T_{E,d}$. Therefore if the action of $G$ on the right cosets of $T_{\mathbb{F}_{q}, d}$ satisfies the EKR-module property, then so will the action of $G$ on the set of right cosets of $T_{E,d}$. So we assume that $E=\mathbb{F}_{q}$. We will apply Theorem~\ref{Specsuff} to show that the action of $G$ on $\Omega_{q,d}:=G/T_{\mathbb{F}_{q},d}$ satisfies the EKR-Module property.

We first note that $S= T_{\mathbb{F}_{q},d}$  is an intersecting set of size $qd$. We now attempt to find an $(G,\ \Omega_{q,d})$-compatible class function $f$, so that $S$ and $f$ satisfy the premaximum intersecting sete of Theorem~\ref{Specsuff}. Let $f(d_{x})=1$ for any element $x$ of $\mathbb{F}_{q}^{*}$, whose order does not divide $d$, and let $f(v_{r})=a$. By Lemma~\ref{spec}, the spectrum of $M_{f}$ is $\{\lambda_{\chi,f}\ |\ \chi \in Irr(G)\}$. Using the Character table \ref{ch1}, we compute that 
\begin{itemize}
\item $\lambda_{\lambda_{1},f}=\frac{q(q-1-d)(q+1)}{2}+aq^{2}(q-1)$
\item $\lambda_{\psi_{1},f}=\frac{(q-1-d)(q+1)}{2}-aq(q-1)$
\item $\lambda_{\nu_{\gamma},f}= \begin{cases} -qd \ \text{if $\epsilon \in Ker(\gamma)$} \\
0 \ \text{otherwise.} \end{cases}$
\item $\lambda_{\eta_{\beta},f} = aq(q-1)$.
\end{itemize}  

We now attempt to find $a$ with respect to the constraints that $-qd$ is the smallest eigenvalue and $qd=|S|=\dfrac{|G|(qd)}{qd+\lambda_{\lambda_{1},f}}$. We see that $a=\dfrac{q+1+d}{2q}$ works. Therefore by Theorem~\ref{Specsuff}(a), $S$ is a maximum intersecting set. We also see that 
$$\{\chi \in Irr(G)\ |\ \lambda_{\chi,f}=-qd\}=\{\nu_{\gamma} \ |\ \gamma \in \mathbb{F}_{q}^{\times}\setminus \{1\}\ \&\ \epsilon \in Ker(\gamma)\}\cup \{\psi_{1}\}.$$ A quick computation shows that $\chi(\mathbf{v}_{T_{E,d}})\neq 0$ whenever $\lambda_{\chi,f}=-qd$. Thus by Theorem~\ref{Specsuff}(b), EKR-Module property holds.

\subsection{Transitive action with $D_{d,-}$ as a stabilzer.}
Let $d$ be a divisor of $q-1$, and consider the group $D_{d,-}=\langle t \rangle A_{d}$, where $t=\overline{\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix} } $. We now consider the action of $G$ on the set $\Omega_{d,-}$ of right cosets of $D_{d,-}$. The set $Der(G,d,-)$ of derangements with respect to this action is the same as the set $Der(G,\mathbb{F}_{q},d)$ of derangements with respect to the action of $G$ on $\Omega_{q,d}=G/T_{\mathbb{F}_{q},d}$. So the derangement graphs $\Gamma_{G, \Omega_{q,d}}$ and $\Gamma_{G, \Omega_{d,-}}$ are the same. Therefore $S=T_{\mathbb{F}_{q},d}$ is an intersecting set. Pick $f$ to be the same class function as in the previous section. We have seen that the $-qd$ is the least eigenvalue of $M^{f}$ and that $qd=|S|=\dfrac{|G|(qd)}{qd+\lambda_{\lambda_{1},f}}$. We have also seen that $$X:=\{\chi \in Irr(G)\ |\ \lambda_{\chi,f}=-qd\}=\{\nu_{\gamma} \ |\ \gamma \in \mathbb{F}_{q}^{\times}\setminus \{1\}\ \&\ \epsilon \in Ker(\gamma)\}\cup \{\psi_{1}\}.$$ 

By utilizing the character table~\ref{ch1}, we see that $\chi(\mathbf{v}_{D_{d,-}})\neq 0$ for all $\chi \in X$, and therefore by Theorem~\ref{Specsuff}(b), this action satisfies the EKR-Module property.

\subsection{Transitive action with $D_{d,+}$ as a stabilzer.} 
Let $d$ be a divisor of $q+1$, and consider the group $D_{d,-}=\langle t \rangle B_{d}$, where $t=\overline{\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix} } $. We now consider the action of $G$ on the set $\Omega_{d,+}$ of right cosets of $D_{d,+}$. The set $Der(G,d,-)$ of derangements with respect to this action is $$\bigcup \limits_ {\{x \in \mathbb{F}^{\times}_{q}\setminus \{1\}\}}(d_{x}) \bigcup \bigcup\limits_{\{r\in \mathbb{F}_{q^{2}}^{\times}\ :\ \#r\mathbb{F}^{\times}_{q} \nmid d\}} (v_{r}).$$

For technical reasons, we treat the cases $d=q+1$ and $d<q+1$ separately. We first assume that $d<q+1$. As in the previous cases, we once apply Theorem~\ref{Specsuff}. We note that the a $2$-sylow subgroups $S\cong \mathbb{F}_{q}$ of $G$, is an intersecting set. Let $f$ be a $(G, \ \Omega_{d,+})$-compatible class function with $f(d_{x})=1$ for all $\{x \in \mathbb{F}^{\times}_{q}\setminus \{1\}\}$ and $f(v_{r})=a$ for all $r$ in $\{r\in \mathbb{F}_{q^{2}}^{\times}\ :\ \#r\mathbb{F}^{\times}_{q} \nmid d\}$. Applying Lemma~\ref{spec}, we see that the eigenvalues of $M^{f}$ are $\{\lambda_{\chi,f}\ |\ \chi \in Irr(G)\}$. Computing with the help of the Character table, we see that
\begin{itemize}
\item $\lambda_{\lambda_{1},f}=\frac{(q-2)(q+1)(q)}{2}+a\frac{(q+1-d)(q)(q-1)}{2}$,
\item $\lambda_{\psi_{1},f}=\frac{(q-2)(q+1)}{2}-a\frac{(q+1-d)(q-1)}{2}$
\item $\lambda_{\nu_{\gamma},f}= -q$
\item $\lambda_{\eta_{\beta},f} = \begin{cases} a(q-1)d \ \text{if $d \mid \# Ker(\beta)$} \\
0 \ \text{otherwise.} \end{cases}$.
\end{itemize}

We now attempt to find $a$ with respect to the constraints that $-q$ is the smallest eigenvalue of $M^{f}$ and $q=|S|=\dfrac{|G|(q)}{q+\lambda_{\lambda_{1},f}}$. We see that $a=\dfrac{q+2}{q+1-d}$ works. Therefore by Theorem~\ref{Specsuff}(a), $S$ is a maximum intersecting set. We also see that 
$$\{\chi \in Irr(G)\ |\ \lambda_{\chi,f}=-q\}=\{\nu_{\gamma} \ |\ \gamma \in \mathbb{F}_{q}^{\times}\setminus \{1\}\}\cup \{\psi_{1}\}.$$ A quick computation shows that $\chi(\mathbf{v}_{D_{d,+}})\neq 0$ whenever $\lambda_{\chi,f}=-q$. Thus by Theorem~\ref{Specsuff}(b), EKR-Module property holds.

When $d=q+1$, we have $Der(G,q+1,+)= \bigcup \limits_ {\{x \in \mathbb{F}^{\times}_{q}\}\setminus \{1\}}(d_{x})$. The group $D_{q+1,+}$ being a point stabilizer, is an intersecting set. Consider $(G,\ \Omega_{q+1,+}$-compatible class function $\tilde{f}$, satisfying $\tilde{f}(d_{x})=1$ for all $\{x \in \mathbb{F}^{\times}_{q}\}\setminus \{1\}$. Computing as above, we see that $\lambda_{\lambda_{1},\tilde{f}}=\frac{(q-2)(q+1)(q)}{2}$, $\lambda_{\psi_{1},\tilde{f}}=\frac{(q-2)(q+1)}{2}$, $\lambda_{\nu_{\gamma},\tilde{f}}= -q$, and $\lambda_{\eta_{\beta},\tilde{f}}=0$. Using Theorem~\ref{Specsuff} as above, we see that $D_{q+1,+}$ is an maximum intersecting set, and that EKR-Module property holds in this case as well.      

\subsection{Transitive action with $B_{d}$ as a stabilizer.}
Let $\epsilon$ be an element of order $q+1$ in $\mathbb{F}_{q^{2}}^{\times}$, then $B:=\left\langle v_{\epsilon} \right\rangle$ is a cyclic subgroup of order  $q+1$ respectively. Given $d\mid q+1$, by $B_{d}$, we denote the unique subgroup of order $d$ in $B$. We consider the actions of $G$ on the set $\Omega_{B_{d}}=G/B_{d}$ of right cosets of $B_{d}$. 

We observe that $\psi_{1}(\mathbf{v}_{B_{d}})=q-d+1$, $\eta_{\beta}(\mathbf{v}_{B_{d}})=\begin{cases}q-1-2d \ \text{if $d\mid \#ker(\beta)$} \\
q+1 \ \text{otherwise}  \end{cases} $, and $\nu_{\gamma}(\mathbf{v}_{B_{d}})= q+1$. If $d<q+1$, then for all $\chi \in Irr(G)$, we have $\chi(\mathbf{v}_{B_{d}})\neq 0$, and so by Corollary~\ref{idekr}, the EKR-Module property holds in this case.

We now consider the case $d=q+1$. In this case, the set $Der(G,B)$ of derangements is 
$$\bigcup \limits_ {\{x \in \mathbb{F}^{\times}_{q}\setminus \{1\}\}}(d_{x}) \bigcup (u).$$ As in the previous cases, we once apply Theorem~\ref{Specsuff}. We note that the the group $B:=B_{q+1}$, is an intersecting set. Let $f$ be a $(G, \ \Omega_{B})$-compatible class satisfying $f(u)=1$ and $f(d_{x})=2$ for all $x \in \mathbb{F}^{\times}_{q}\setminus \{1\}$. By Lemma~\ref{spec}, the eigenvalues of $M^{f}$ are $\lambda_{\lambda_{1},f}=q^{2}-1+q(q+1)(q-2)$, $\lambda_{\nu_{\gamma},f}=\lambda_{\eta_{\beta},f}=-(q+1)$ (for all choices of $\gamma$ and $\beta$), and $\lambda_{\psi_{1},f}=(q-2)(q+1)$. Observing that $|B|=\dfrac{|G|(q+1)}{\lambda_{\lambda_{1},f}+q+1}$, EKR-Module property follows by application of Theorem~\ref{Specsuff}.  

\subsection{Transitive action with $A_{d}$ as a stabilizer.}
Let $\zeta$ be a generator of $\mathbb{F}_{q}^{\times}$, then $A:=\left\langle d_{\zeta} \right\rangle$. Given $d\mid q-1$, by $A_{d}$, we denote the unique subgroup of order $d$ in $A$. We consider the actions of $G$ on the set $\Omega_{A_{d}}=G/A_{d}$ of right cosets of $A_{d}$.

Using the character table \ref{ch1}, we compute that $\psi_{1}(\mathbf{v}_{A_{d}})=q+d-1$,\\ $\nu_{\gamma}(\mathbf{v}_{A_{d}})=\begin{cases} q-1-2d \ \text{if $d\mid \# Ker(\gamma)$}\\
q-1\ \text{otherwise} \end{cases} $, and $\eta_{\beta}(\mathbf{v}_{A_{d}}) =q-1$. Thus for all $\chi \in Irr(G)$, we have $\chi(\mathbf{v}_{B_{d}})\neq 0$, and therefore by Corollary~\ref{idekr}, the EKR-Module property holds in this case.

\subsection{Transitive action with $G_{q_{0}}$ as a stabilzer.}
Let $q_{0}$ be a power of $2$ such that $q$ is an integral power of $q_{0}$. Then there is a natural embedding of $PGL(2,q_{0})$ into $PGL(2,q)$. We denote the image of this embedding by $G_{q_{0}}$. We now consider the action of $G$ on the set $\Omega_{q_{0}}:=G/G_{q_{0}}$. We will show that this action satisfies the EKR-Module property. For technical reasons, we split into two cases: (i) $\log_{q_{0}}(q)$ is even and (ii) $\log_{q_{0}}(q)$ is odd.

\subsubsection{Case 1} We first assume that $\log_{q_{0}}(q)$ is odd. Recalling the conjugacy class structure of $PGL(2,2^{n})$, we first observe that $G_{q_{0}}$ has (a) exactly $q^{2}_{0}-1$ elements in $(u)$; (b) for each $x \in \mathbb{F}_{q_{0}}^{\times}\setminus \{1\}$, exactly $q_{0}(q_{0}+1)$ elements in ($d_{x})$; and for each $r \in \mathbb{F}_{q^{2}_{0}}^{\times}\setminus \mathbb{F}^{\times}_{q_{0}}$, there are exactly $q_{0}(q_{0}-1)$ elements in $(v_{r})$. As $\log_{q_{0}}(q)$ is odd, we have $\mathbb{F}_{q_{0}^{2}}\cap \mathbb{F}_{q}=\mathbb{F}_{q_{0}}$. Therefore given $r \in \mathbb{F}_{q_{0}^{2}}\setminus \mathbb{F}_{q_{0}}$, $v_{r}$ is not conjugate to any element in $\{d_{x}\ | x\in \mathbb{F}_{q}^{\times}\setminus \{1\}\}$. Thus in this case, the set $Der(G,q_{0})$ of derangements with respect to this action is $$\bigcup \limits_ {\{x \in \mathbb{F}^{\times}_{q}\setminus \mathbb{F}^{\times}_{q_{0}}\}}(d_{x}) \bigcup \bigcup \limits_{\{r\in \mathbb{F}_{q^{2}}^{\times}\setminus \mathbb{F}_{q^{2}_{0}}^{\times}\}}(v_{r}).$$
As done in the other cases, we will apply Theorem~\ref{Specsuff}. We observe that $S=T_{\mathbb{F}_{q},q_{0}-1}$ is an intersecting set. Let $f_{1}$ be the $(G,\Omega_{q_{0}})$-compatible class function satisfying $f_{1}(d_{x})=1$ for all $x$ in 
$\mathbb{F}^{\times}_{q}\setminus \mathbb{F}^{\times}_{q_{0}}$ and $f_{1}(v_{r})=a$ for all $r$ in $\mathbb{F}^{\times}_{q^{2}}\setminus \mathbb{F}^{\times}_{q^{2}_{0}}$. By applying Lemma~\ref{spec}, we see that the eigenvalues of $M^{f_{1}}$ are $\{\lambda_{\chi,f_{1}}\ |\ \chi \in Irr(G)\}$. Computing with the help of the Character table, we see that
\begin{itemize}
\item $\lambda_{\lambda_{1},f_{1}}=\frac{(q-q_{0})(q+1)(q)}{2}+a\frac{(q-q_{0})(q)(q-1)}{2}$,
\item $\lambda_{\psi_{1},f_{1}}=\frac{(q-q_{0})(q+1)}{2}-a\frac{(q-q_{0})(q-1)}{2}$,
\item $\lambda_{\nu_{\gamma},f_{1}}= \begin{cases} -q(q_{0}-1) \ \text{if $(q_{0}-1) \mid \# Ker(\gamma)$} \\
0 \ \text{otherwise.} \end{cases}$
\item $\lambda_{\eta_{\beta},f_{1}} = \begin{cases} aq(q_{0}+1) \ \text{if $(q_{0}+1) \mid \# Ker(\beta)$} \\
0 \ \text{otherwise.} \end{cases}$.
\end{itemize}

We now attempt to find $a$ with respect to the constraints that $-q(q_{0}-1)$ is the smallest eigenvalue of $M^{f}$ and $q=|S|=\dfrac{|G|(q)}{q+\lambda_{\lambda_{1},f}}$. We observe that $a=\dfrac{q^{2}+qq_{0}-q+q_{0}}{(q-1)(q-q_{0})}$ works. Therefore by Theorem~\ref{Specsuff}(a), $S$ is a maximum intersecting set. We also see that 
$$\{\chi \in Irr(G)\ |\ \lambda_{\chi,f}=-q(q_{0}-1)\}=\{\nu_{\gamma} \ |\ (q_{0}-1)\mid \# Ker(\gamma)\}\cup \{\psi_{1}\}.$$ A quick computation shows that $\chi(\mathbf{v}_{G_{q_{0}}})\neq 0$ whenever $\lambda_{\chi,f}=-q(q_{0}-1)$. Thus by Theorem~\ref{Specsuff}(b), EKR-Module property holds. 

\subsubsection{Case 2} Now, we consider the case when $\log_{q_{0}}(q)$ is even. Recalling the conjugacy class structure of $PGL(2,2^{n})$, we first observe that $G_{q_{0}}$ has (a) exactly $q^{2}_{0}-1$ in $(u)$; (b) for each $x \in \mathbb{F}_{q_{0}}^{\times}\setminus \{1\}$, there are exactly $q_{0}(q_{0}+1)$ elements in $(d_{x})$; and for each $r \in \mathbb{F}_{q^{2}_{0}}^{\times}\setminus \mathbb{F}^{\times}_{q_{0}}$, there are exactly $q_{0}(q_{0}-1)$ elements in $(v_{r})$. As $\log_{q_{0}}(q)$ is even, for every $r\in  \mathbb{F}_{q^{2}_{0}}^{\times}\setminus \mathbb{F}^{\times}_{q_{0}}$, $v_{r}$ is conjugate to $\overline{\begin{pmatrix}
r & 0\\
0 & r^{q_{0}}
\end{pmatrix} } = d_{r^{q_{0}-1}}$. By $Y$, we denote the unique subgroup of order $q_{0}+1$ in $\mathbb{F}_{q}^{\times}$.
Then for every $y \in  Y \setminus \{1\}$, exactly $q_{0}(q_{0}-1)$ elements of $G_{q_{0}}$ are conjugate to $d_{y}$. Now by using the character table \ref{ch1}, we conclude that $\psi_{1}(\mathbf{v}_{G_{q_{0}}})\neq 0$ and $\eta_{\beta}(\mathbf{v}_{G_{q_{0}}})=q-q^2_{0}$. We now compute $\nu_{\gamma}(\mathbf{v}_{G_{q_{0}}})$. We have 
\begin{align*}
\nu_{\gamma}(\mathbf{v}_{G_{q_{0}}}) & = q+1+q^{2}_{0}-1 + (q_{0})(q_{0}-1) \sum\limits_{y \in Y\setminus \{1\}} \gamma(y) + (q_{0})(q_{0}+1) \sum\limits_{x \in \mathbb{F}^{\times}_{q_{0}}\setminus \{1\}} \gamma(x)  \\
&= q+q^{2}_{0} + (q_{0})(q_{0}-1)\sum\limits_{y \in Y}\gamma(y) + (q_{0})(q_{0}+1)\sum\limits_{x \in \mathbb{F}_{q^{\times}_{0}}}\gamma(x) -2q^{2}_{0}.\\
& = q-q^{2}_{0} + (q_{0})(q_{0}-1)\sum\limits_{y \in Y}\gamma(y) + (q_{0})(q_{0}+1)\sum\limits_{x \in \mathbb{F}_{q^{\times}_{0}}}\gamma(x).
\end{align*}
Therefore $\nu_{\gamma}(\mathbf{v}_{G_{q_{0}}})=0$ if and only if $q=q^{2}_{0}$ and both $Y$ and $\mathbb{F}^{\times}_{q_{0}}$ are not in the Kernel of $\gamma$. Also $\eta_{\beta}(\mathbf{v}_{q_{0}})=0$ if and only if $q^2_{0}=q_{0}$. Therefore if $q^{2}_{0}<q$, for all $\chi \in Irr(G)$, we have $\chi(\mathbf{v}_{G_{q_{0}}})\neq 0$, and therefore by Corollary~\ref{idekr}, the EKR-Module property holds in this case.

We now assume that $q^{2}_{0}=q$.
The set $Der(G,q_{0})$ of derangements with respect to this action is $$ \bigcup \limits_ {\ x \in \mathbb{F}^{\times}_{q}\setminus (Y \cup \mathbb{F}^{\times}_{q_{0}})\}}(d_{x}) \bigcup\limits_{\{r\in \mathbb{F}_{q^{2}}^{\times}\}} (v_{r}).$$ As done in the other cases, we will apply Theorem~\ref{Specsuff}. We observe that $S=T_{\mathbb{F}_{q},q_{0}+1}$ is an intersecting set. Let $f_{2}$ be the $(G,\Omega_{q_{0}})$-compatible class function satisfying $f_{2}(d_{x})=a$ for all $x$ in 
$\mathbb{F}^{\times}_{q}\setminus (Y \cup \mathbb{F}^{\times}_{q_{0}})$, and $f_{2}(v_{r})=b$ for all $r$ in $\mathbb{F}^{\times}_{q^{2}}$. By applying Lemma~\ref{spec}, we see that the eigenvalues of $M^{f_{2}}$ are $\{\lambda_{\chi,f_{2}}\ |\ \chi \in Irr(G)\}$. We first note that since $q^{2}_{0}=q$, both $Y$ and $\mathbb{F}^{\times}_{q_{0}}$ cannot simultaneously be in the kernal of a non-trivial character of $\mathbb{F}^{\times}_{q}$. Now with the help of the character table ~\ref{ch1}, we see that

\begin{enumerate}
\item $\lambda_{\lambda_{1},f_{2}}= a\dfrac{q(q+1)(q^{2}_{0}-2q_{0})}{2}+b\dfrac{q^2(q-1)}{2}$,
\item $\lambda_{\psi_{1},f_{2}}= a\dfrac{(q+1)(q^{2}_{0}-2q_{0})}{2}-b\dfrac{q(q-1)}{2}$,
\item $\lambda_{\eta_{\beta},f_{2}}= b(q^{2}-q)$
\item and 
\begin{align*}
\lambda_{\nu_{\gamma},f_{2}}&=q\left(-a\sum\limits_{y \in Y}\gamma(y)-a\sum\limits_{z \in \mathbb{F}^{\times}_{q_{0}}}\gamma(z)+a\right)\\
&=\begin{cases}
-aq(q_{0}-2) \ \text{if $\mathbb{F}^{\times}_{q_{0}} \subset Ker(\gamma)$, }\\
-aq(q_{0}) \ \text{if $Y  \subset Ker(\gamma)$, and}\\
aq \ \text{otherwise.}
\end{cases} 
\end{align*}
\end{enumerate}
If $S=T_{\mathbb{F}_{q},q_{0}+1}$ is a maximum intersecting set set, then so will be cosets of its conjugates. So by Lemma~\ref{caniddes}, $\mathbf{v}_{S} \in \sum\limits_{\{\chi \in Irr(G)\ : \chi(\mathbf{v}_{S})\neq 0\}}\left\langle e_{\chi} \right\rangle$. If we can a choice of $a,b,c$ such that $S$ and $f_{2}$ satisfy the premise of Theorem~\ref{Specsuff}, then by Lemma~\ref{spec}, for every $\chi \in Irr(G)\setminus \{\lambda_{1}\}$ with $\chi(\mathbf{v}_{S})\neq 0$, the eigenvalue $\lambda_{\chi,f_{2}}$ must be the least eigenvalue of $M^{f_{2}}$. So, it makes sense to attempt to find $a,b$ with respect to the constraints: (i) $-q(q_{0}+1)$ is the least eigenvalue of $M^{f}$ with $\{\chi \in Irr(G)\ | \lambda_{\chi,f_{2}}=-q(q_{0}+1)\}=\{\psi_{1}\} \cup \{\nu_{\gamma}\ | Y \subset Ker(\Gamma)\}$ and (ii)$\lambda_{\lambda_{1},f_{2}}+q(q_{0}+1)=|G|$. Careful calculations yield that $a=\dfrac{q_{0}+1}{q_{0}}$ and $b=\dfrac{q^2+qq_{0}+q-q_{0}-2}{q(q-1)}$ work.

With this, we have shown that all transitive actions of $G$ satisfy the EKR-Module property. This concludes the proof of Theorem~\ref{PSLe}.
\qed

 We will now see a few more examples of groups actions which do not satisfy the EKR-Module property. In particular, we shall see that Theorem~\ref{PSLe} does not extend to $PSL(2,q)$, when $q$ is an odd prime power. We consider the action of $G_{1}=PGL(2,q)$, on the set of right cosets of a maximal subgroup $H$ which is isomorphic to the dihedral group of order $2(q+1)$. We also consider the action of $G_{2}=PSL(2,r)$ with $r\equiv 3 \pmod{4}$, on the set of right cosets of a maximal subgroup $H_{1}$ isomorphic to the dihedral group of order $r+1$.  By a result in \cite{LSP}, we know that both these action satisfy the EKR property.

1) Consider $G_{1}=S_{5}$ (isomorphic to $PGL(2,5)$) and $H \cong D_{6}$. We consider the action of $G_{1}$ on $G/H$.
 Now $G$ has a maximal subgroup $K$ which is isomorphic to $A_{4}$. It is easy to check that $K \subset \cup gHg^{-1}$ and thus $K$ an intersecting set. Now consider the sign character $\lambda$, we have $\lambda(\mathbf{v}_{H})=0$, but $\lambda(\mathbf{v}_{K})\neq 0$. Therefore by Corollary~\ref{idekr}, this group action does not satisfy the EKR-Module property.

2) Consider $G_{2}=PSL(2,11)$. Let $H_{1} \cong D_{6}$ be a maximal subgroup. We consider the action of $G_{1}$ on $G_{2}/H_{1}$. Let $K_{1}$ be a maximal subgroup of $G_{1}$ isomorphic to $A_{4}$. As all cyclic subgroups of $G_{1}$ are conjugate, we can see that $K_{1} \subset \cup gH_{1}g^{-1}$.
Using a computer algebra system such as magma, we can find a character $\chi$ such that $\chi(\mathbf{v}_{H_{1}})=0$, but $\chi(\mathbf{v}_{K_{1}})\neq 0$. Therefore by Corollary~\ref{idekr}, this group action does not satisfy the EKR-Module property.


3) Consider $G_{1}=PGL(2,11)$. Let $H \cong D_{12}$ be a maximal subgroup. We consider the action of $G_{1}$ on $G_{1}/H$. Let $K$ be a maximal subgroup of $G$ isomorphic to $S_{4}$. We check that $K \subset \cup gHg^{-1}$ and thus $K$ an intersecting set. Using a computer algebra system such as magma, we can find a character $\chi$ such that $\chi(\mathbf{v}_{H})=0$, but $\chi(\mathbf{v}_{K})\neq 0$. Therefore by Corollary~\ref{idekr}, this group action does not satisfy the EKR-Module property.


4) Consider $G_{2}=PSL(2,23)$. Let $H_{1} \cong D_{12}$ be a maximal subgroup. We consider the action of $G_{2}$ on $G_{2}/H_{1}$. Let $K$ be a maximal subgroup of $G$ isomorphic to $S_{4}$. As all cyclic subgroups of $G_{1}$ are conjugate, we can see that $K_{1} \subset \cup gH_{1}g^{-1}$. Using a computer algebra system such as magma, we can find a character $\chi$ such that $\chi(\mathbf{v}_{H_{1}})=0$, but $\chi(\mathbf{v}_{K_{1}})\neq 0$. Therefore by Corollary~\ref{idekr}, this group action does not satisfy the EKR-Module property.


5) Consider $G_{2}=PSL(2,59)$. Let $H_{1} \cong D_{30}$ be a maximal subgroup. We consider the action of $G_{2}$ on $G_{2}/H_{1}$. Let $K$ be a maximal subgroup of $G$ isomorphic to $A_{5}$. As all cyclic subgroups of $G_{2}$ are conjugate, we can see that $K_{1} \subset \cup gH_{1}g^{-1}$. Using a computer algebra system such as magma, we can find a character $\chi$ such that $\chi(\mathbf{v}_{H_{1}})=0$, but $\chi(\mathbf{v}_{K_{1}})\neq 0$. Therefore by Corollary~\ref{idekr}, this group action does not satisfy the EKR-Module property.


6) Consider $G_{1}=PGL(2,29)$. Let $H \cong D_{30}$ be a maximal subgroup. We consider the action of $G_{1}$ on $G_{1}/H$. Let $K$ be a maximal subgroup of $G$ isomorphic to $A_{5}$. We check that $K \subset \cup gHg^{-1}$ and thus $K$ an intersecting set. Using a computer algebra system such as magma, we can find a character $\chi$ such that 
$\chi(\mathbf{v}_{H})=0$, but $\chi(\mathbf{v}_{K})\neq 0$. Therefore by Corollary~\ref{idekr}, this group action does not satisfy the EKR-Module property.
 
\section{EKR-Module property in Strongly Regular Graphs}\label{srg}
In this section, we consider maximum cliques in the Peisert-type strongly regular graphs defined in \cite{EKRpaley}. These are a subclass of strongly regular graphs found in \cite{brouwer1999cyclotomy}. Consider a strongly regular graph $X$, with a prescribed set $\mathcal{C}$ of ``naturally'' occurring cliques. Cliques in $\mathcal{C}$ will be called canonical cliques. We say that $X$ satisfies EKR-module property with respect to $\mathcal{C}$ if the characteristic vector of every maximum clique in $X$ is a linear combination of characteristic vectors of the cliques in $\mathcal{C}$. We now define Peisert-type graphs.

\begin{definition}
Let $q$ be an odd prime power. Then a Peisert-type graph of type $(m,q)$ is a Cayley graph on the additive group of $\mathbb{F}_{q^{2}}$ with its ``connection'' set $S$ being a union of $m$ cosets of $\mathbb{F}^{\times}_{q}$ in $\mathbb{F}^{\times}_{q^{2}}$ such that $\mathbb{F}^{\times}_{q} \subset S$. 
\end{definition}

Given a Peisert-type graph of type $(m,q)$, with connection set $S$. For any $s \in S$ and $x\in \mathbb{F}_{q^{2}}$, the set $s\mathbb{F}_{q}+x$ is a naturally occurring clique. By a canonical clique in a Peiset-type graph, we mean a clique of the form $s\mathbb{F}_{q}+x$, where $s \in S$ and $x \in \mathbb{F}_{q^{2}}$. The main result of \cite{EKRpaley} is the following shows that Peiser-type graphs satisfy EKR-Module property. In this section, we give a shorter and different proof of the same.

\begin{theorem}(\cite[Theorem 1.3]{EKRpaley})\label{EKRP}
The the characteristic vector of every maximum clique in a Peisert-type graph is a linear combination of characteristic vectors of its canonical cliques.   
\end{theorem}  

We now collect some results about Peisert-type graphs and some general results about strongly regular graphs.
The main result of \cite{brouwer1999cyclotomy} shows that Peisert-type graphs are strongly regular. A different proof of the same is given in \cite{EKRpaley}. We will give a proof using a standard technique of finding eigenvalues of an abelian Cayley graph.

\begin{lemma}\label{par}
Peisert-type graph of type $(m,q$ is strongly regular with eigenvalues $k:=m(q-1)$ with multiplicity 1, $q-m$ with multiplicity $m(q-1)$, and $-m$ with multiplicity $q^{2}-1-m(q-1)$.
\end{lemma}
\begin{proof}
Let $X$ a Peisert-type graph of type $(m,q)$ whose connection set is $S=\bigcup\limits_{i=0}^{m-1} c_{i}\mathbb{F}^{\times}_{q}$ (with $c_{0}=1$).
Let $A$ be the adjacency matrix of a Peisert-type graph of type $(m,q)$. Considering an additive character of $\chi$ of $\mathbb{F}_{q}$ as a column vector, we see that $A\chi=\chi(\mathbf{v}_{S})\chi$, where $\chi(\mathbf{v}_{S})=\sum\limits_{s \in S}\chi(s)$.

If $\chi$ is not the trivial character, $Ker(\chi) \neq \mathbb{F}_{q^{2}}$, and so at most one of $\{c_{i}\mathbb{F}_{q}\ |\ 0\leq i \leq m-1\}$ can be a subgroup of $Ker(\chi)$. Thus if $c_{i}\mathbb{F}_{q} \subset Ker(\chi)$ for some $i$, $\chi|_{c_{j}\mathbb{F}_{q}}$ is a non-trivial character for all $j \neq i$. Otherwise, $Ker(\chi)$ will have two $1$-dimensional subspaces of $\mathbb{F}_{q^{2}}$ and thus must be equal to $\mathbb{F}_{q}$.

Assume that $\chi $ is a non-trivial character with $c_{i}\mathbb{F}_{q} \subseteq Ker(\chi)$. As the sum of values of a non-trivial character are zero, in this case, we have 
\begin{align*}
\chi(\mathbf{v}_{S}) &= \sum\limits_{x \in c_{i}\mathbb{F}^{\times}_{q}} \chi(x) + \sum\limits_{j \neq i} \sum\limits_{x \in c_{j}\mathbb{F}^{\times}_{q}} \chi(x) \\
&= q-1 -(m-1)=q-m. 
\end{align*}
The set on non-trivial characters $\chi$ with $c_{i}\mathbb{F}_{q} \subset Ker(\chi)$ is in one-one correspondence with the non-trivial characters of $\mathbb{F}_{q^{2}}/c_{i}\mathbb{F}_{q}$. Thus there are atleast $m(q-1)$ characters $\chi$ such that $A\chi=(q-m)\chi$. As distinct characters are orthogonal the dimension of $q-m$-eigenspace of $A$ is atleast $m(q-1)$. 

Similarly, if $\chi$ is a non-trivial character with $c_{i}\mathbb{F}_{q} \not\subseteq Ker(\chi)$ for all $0\leq i \leq m-1$, we have $\chi(\mathbf{v}_{S})=-m$. Thus there are atleast $q^{2}-1-m(q-1)$ characters $\chi$ such that $A\chi=(-m)\chi$. As distinct characters are orthogonal the dimension of $-m$-eigenspace of $A$ is atleast $q^{2}-1-m(q-1)$.

If $\chi$ is the trivial character $\chi_{0}$, we have $A\chi_{0}=|S|\chi_{0}$. With this, we have found all the eigenvalues of $A$ and their corresponding eigenspaces. As $A$ has exactly three distinct eigenvalues, it is a strongly regular graph.   
\end{proof}

 Let $X$ be a strongly regular graph with parameters $(v,\ k,\ \lambda,\ \mu)$, that is a $k$-regular graph on $v$ vertices such that (i) any two adjacent vertices have exactly $\lambda$ common neighbours and (ii) any two non-adjacent vertices have exactly $\mu$ common neighbours. We further assume that $X$ is primitive, that is, both $X$ and its complement are connected. It is well known (\cite[Lemma 10.2.1]{GR2001}) that the adjacency matrix $A$ of $X$ has exactly distict three eigenvalue. As $X$ is $k$-regular and connected, $k$ is an eigenvalue of $A$ with multiplicity $1$. Let $r,s$ with $r>s$ be the other eigenvalues.

Our proof use some result on graphs in Association Schemes. For a quick introduction to the preliminaries on graphs in Association schemes, we refer the reader to Chapter 3 of \cite{GMbook}. We first recall the following well known result linking strongly regular graphs with association schemes. 
By $J$ and $I$, we denote the all-one matrix and the identity matrix respectively.

\begin{lemma}(\cite[Lemma 5.1.1]{GMbook}) Let $X$ be a graph with $A$ as its adjacency matrix. Then $X$ is strongly regular if and only if $\mathcal{A}_{X}:=\{I,\ A,\ \overline{A}:=J-I-A\}$ is an association scheme. 
\end{lemma} 

By $\mathbb{C}[\mathcal{A}_{X}]$, we denote the linear span of matrices in $\mathcal{A}_{X}$. This is referred to as the Bose-Mesner algebra. By a well-known result (\cite[Theorem 3.4.4]{GMbook}), the projections onto eigenspaces of $A$ is an orthogonal basis of idempotents of $\mathbb{C}[\mathcal{A}_{X}]$. The matrix $J$ is the projection onto the $k$-eigenspace. We denote $E_{r}$ and $E_{s}$ to be the projections onto the $r$-eigenspace and the $s$-eigenspace respectively. We have $A=kJ+rE_{r}+sE_{s}$, $I=\frac{J}{n}+E_{r}+E_{s}$, and so $\{\frac{J}{n},\ E_{r},\ E_{s}\}$ is an orthogonal basis of idempotents of $\mathbb{C}[\mathcal{A}_{X}]$. We now mention a bound by Godsil on cliques in Strongly regular graphs. The following result is a special case of \cite[Corollary 3.7.2]{GMbook}. 

\begin{lemma}\label{godsilub}
Let $X$ be $k$-regular strongly regular graph with $s$ as the lease eigenvalue of its adjacency matrix. If $C$ is a clique in $X$, then $$|C| \leq 1-\frac{k}{s}.$$ Moreover if $C$ is a clique that meets the bound with equality, then the characteristic vector $\mathbf{v}_{C}$ is orthogonal to the $s$-eigenspace.     
\end{lemma}

Given a subset $B$ of the vertex set of $X$, by $\mathbf{v}_{B}$, we denote the characteristic vector of $B$, and by $\mathbf{1}$, the all-one vector. Consider the $\mathbb{C}$-linear span $V_{max}$ of characteristic vectors of maximum cliques in $X$. By the above Lemma, we have $|C| \leq 1-\frac{k}{s}$, for any clique $C$. Assume that there is a clique $C$ of size $1-\frac{k}{s}$, then by the above Lemma, $V_{max}$ is orthogonal the $s$-eigenspace. We will now show that $V_{max}$ is the image of $\frac{J}{n}+E_{r}$.

\begin{lemma}\label{evec}
Let $X$ be $k$-regular strongly regular graph on $n$ vertices with $\{k,\ r,\ s\}$ with $r>s$ as set of distinct eigenvalues of its adjacency matrix. If $X$ has a clique of size $1-\frac{k}{s}$, then $\mathbf{v}_{C}-\frac{|C|}{n}\mathbf{1}$ is an $r$-eigenvector.
\end{lemma}
\begin{proof}
From Lemma~\ref{godsilub}, we now that $E_{s}\mathbf{v}_{S}=0$. As $\mathbf{1}$ is a $k$-eigenvector, it is also orthogonal to the $s$-eigenspace. Since $J \mathbf{v}_{C} = |C| \mathbf{1}$, the vector $\mathbf{v}_{C}-\frac{|C|}{n}\mathbf{1}$ is orthogonal to both the $k$-eigenspace and the $s$-eigenspace, and so must lie in the $r$-eigenspace.
\end{proof}

We are now ready to prove Theorem~\ref{EKRP}.
\subsection{Proof of Theorem~\ref{EKRP}.}
Let $X$ a Peisert-type graph of type $(m,q)$ whose connection set is $S=\bigcup\limits_{i=0}^{m-1} c_{i}\mathbb{F}^{\times}_{q}$ (with $c_{0}=1$). By applying Lemmas~\ref{par}, \ref{godsilub}, and \ref{evec}, we obtain the following result.
\begin{lemma}
If $C$ is a maximum clique in $X$, then $|C|=q$ and $\mathbf{v}_{C}-\frac{1}{q}\mathbf{1}$ is a $(q-m)$-eigenvector.
\end{lemma}

Thus, given $x \in \mathbb{F}_{q^{2}}$ and $0\leq i \leq m-1$, the canonical clique $c_{i}\mathbb{F}_{q}+x$ is a maximum clique and $\mathbf{v}_{c_{i},x}:=\mathbf{v}_{c_{i}\mathbb{F}_{q}+x}-\frac{1}{q}\mathbf{1}$ is a $q-m$-eigenevector. By Lemma~\ref{par}, the dimension of the $q-m$-eigenspace is $m(q-1)$. Using the above Lemma, we can now deduce that Theorem~\ref{EKRP} is follows from showing that $\{\mathbf{v}_{c_{i},x}\ | \ x \in \mathbb{F}_{q^{2}}\ \&\ 0\leq i \leq m-1\}$ spans an $m(q-1)$ dimensional vector space.

Given an additive character $\chi$ of $\mathbb{F}_{q^{2}}$, we set $\mathbf{v}_{\chi}:=\sum\limits_{z\in \mathbb{F}_{q^{2}}}\chi(z)z$. In the proof of Lemma~\ref{par}, we saw that the $(q-m)$-eigenspace $V_{q-m}$, is spanned by $$\{\mathbf{v}_{\chi}\ | \ \text{ $\chi$ is non-trivial and $c_{i}\mathbb{F}_{q} \subseteq \ Ker(\chi)$ for some $0\leq i \leq m-1$}\}.$$ Now we define, $E_{i}: = \left\langle\{v_{\chi}\ |\ \mathbb{F}_{q^{2}}\supsetneq Ker(\chi) \supset c_{i}\mathbb{F}_{q}\} \right\rangle$. We have 
$$V_{q-m}=\bigoplus E_{i}.$$

By $\left\langle\ ,\  \right\rangle$, we denote the natural orthogonal form on $\mathbb{C}[\mathbb{F}_{q^{2}}]$. With respect to this form, we have $E^{\perp}_{i}= \bigoplus\limits_{j\neq i}E_{j}$. If $\chi$ is a non-trivial character, then $$\left\langle \mathbf{v}_{\chi}, \mathbf{v}_{c_{i},x} \right\rangle = \chi(x)\sum\limits_{z\in c_{i}\mathbb{F}_{q}}\chi(z).$$ Therefore, for all $\mathbf{v}_{c_{i},x} \in E_{i}$. We will now show that vectors of the form $\mathbf{v}_{c_{i},x}$ span $E_{i}$. We do this by considering $x\in \mathbb{F}^{\times}_{q^{2}}$ and $f_{i,x}:= \mathbf{v}_{c_{i},\ 0}- \mathbf{v}_{c_{i},\ x}= \mathbf{v}_{c_{i}\mathbb{F}_{q}}-\mathbf{v}_{c_{i}\mathbb{F}_{q}+x}$. As the set $\{\mathbf{v}_{c_{i}F_{q}+x}\ | \ x\in \mathbb{F}_{q^{2}}\}$ is a set of $q$ orthogonal vectors, the set $\{f_{i,x}\ | \ x\neq 0\}$ spans a $q-1$ dimensional vector space. Therefore $\{\mathbf{v}_{c_{i},x}\ | \ x \in \mathbb{F}_{q^{2}}\}$ spans $E_{i}$. Thus $\{\mathbf{v}_{c_{i},x}\ | \ x \in \mathbb{F}_{q^{2}}\ \&\ 0\leq i \leq m-1\}$ spans $V_{q-m}=\bigoplus E_{i}$. This concludes the proof of Theorem~\ref{EKRP}.
\qed

     
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