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\title[On the automorphism group of a Conway 99-graph]{On the automorphism group of a putative Conway 99-graph}

\author{\firstname{Patrick} \middlename{G.} \lastname{Cesarz}}

\address{University of Wyoming\\ 
Dept. of Mathematics \& Statistics\\
1000 E. University Ave.\\
Laramie, WY 82071 (USA)}

\email{pcesarz@uwyo.edu}

\author{\firstname{Andrew} \middlename{J.} \lastname{Woldar}}
\address{Villanova University\\
 Dept. of Mathematics \& Statistics\\
800 E. Lancaster Ave.\\
Villanova, PA (USA)}
\email{andrew.woldar@villanova.edu}

\keywords{Conway $99$-graph, strongly regular graph, automorphism group, orbit partition, orbit valencies}
 
\subjclass{05E18, 05C25, 58D19}
 
\begin{document}

\begin{abstract}
Let $\Gamma$ be a {Conway 99-graph}, that is, a strongly regular graph 
 with parameters 
$(99,14,1,2)$. Existence of such a graph remains an elusive open problem, however various authors have made significant contributions by analyzing the structure of the automorphism group $G={\rm Aut}(\Gamma)$. 
In this paper we duplicate many results of our predecessors 
(e.g. Behbahani \& Lam, Crnkovi\'c \& Maksimovi\'c), but crucially, we accomplish this without the aid of a computer. 
Specifically, we give computer-free proofs that divisibility of $|G|$ by $2$ implies $|G|$ divides $6$ while divisibility of $|G|$ by $7$ implies 
$G \cong\mathbb Z_7$. 
\end{abstract}

\maketitle

\section{Introduction}

The question of existence of a strongly regular graph 
with parameters $(99,14, 1,2)$ is a longstanding open problem. Its possible existence was first suggested by Norman Biggs in 1969~\cite{Biggs}. According to the account given by Richard Guy in~\cite{Guy}, John H.\ Conway worked on this problem as early as 1975. Later, Conway would offer a \$1000 prize to anyone who could solve it (see~\cite{Conway}, where it is listed as problem 2 among five posed open problems). From that point onward, a strongly regular graph having these parameters came to be known colloquially as a \emph{Conway 99-graph}.

In~\cite{Conway} Conway gave an alternate formulation of the existence problem, which we here reproduce: 
\begin{quote}
{Is there a graph with 99 vertices in which every edge (\ie pair of joined vertices) belongs to a unique triangle and every nonedge (pair of unjoined vertices) to a unique quadrilateral?}
\end{quote}
 
 Famously, A.A.\ Makhnev and I.M.\ Minakova proved in ~\cite{Makhnev} that a strongly regular graph with parameters $(v, k, 1, 2)$ can only exist if 
 $k = u^2 + u + 2$ where $u\in\{1, 3, 4, 10, 31\}$. Such graphs are known to exist for $u\in \{1,4\}$ but not much is known about the remaining cases. Should a Conway 99-graph exist, it would correspond to the case $u=3$. 

In~\cite[Theorem 2.7]{Makhnev} it is asserted that the order of the automorphism group $G$ of a Conway 99-graph must divide $2\cdot 3^3\cdot 7\cdot 11$. 
In~\cite[Corollary 2.4]{Makhnev2} Makhnev asserts the following: 

If $G$ contains an involution $t$, then one of the following holds: 
\begin{enumerate}[label=\alph*\textup{)}]
\item[(i)]
If $|G|$ is divided by 7 then $|G|$ divides 42, $[O(G), t] = 1$, and in the case $|G| = 42$ the subgroup $O(G)$ is nonabelian. (Here $O(G)$ denotes the maximal odd-order normal subgroup of $G$.) 
\item[(ii)] 
$|G|$ divides 6.
\end{enumerate} 
 
Below we summarize results of various authors on the automorphism group of a Conway-99 graph that were obtained since publication of~\cite{Makhnev2,Makhnev} and which were brought to our attention by an anonymous referee 
(see~\cite[p.16]{Brovan}):
\begin{enumerate} 
\item[(a)]
In~\cite{BehLam} Behbahani \& Lam showed that any automorphism of prime order must have order 2 or 3. 
\item[(b)] 
In~\cite{CrnMak} Crnkovi\'c \& Maksimovi\'c ruled out groups of order 6 or 9.
\end{enumerate}
 
 In the present paper, we prove the following: 
 \begin{enumerate} 
 \item 
 If 2 divides $|G|$, then $|G|$ divides 6. 
\item
If $|G|$ is divisible by 7, then $G$ is isomorphic to 
$\mathbb Z_7$.
\end{enumerate}

Importantly, we stress that results (1) and (2) are obtained without the aid of a computer. To the best of our knowledge, this has not been accomplished in any previous published work on the problem. 

Our choice of notation and terminology are standard. We refer the reader to 
\cite{Biggs1,BCN,Brovan,Godsil,Rotman} as excellent sources of background material.

Our paper is organized as follows. 
In Section~\ref{prelims} we establish a labeling scheme for the vertices of a putative Conway 99-graph 
$\Gamma$ under the assumption that 7 divides~$|G|$. This allows us to embed $G$ in the symmetric group of degree 14. 

In Section~\ref{no order 14 auts} we prove that 
$\Gamma$ admits no order 14 automorphisms. This is a crucial step toward showing that divisibility by 2 implies $|G|$ divides 6.

In Section~\ref{cyclic or Frobenius} we show that divisibility by 7 implies $G$ is isomorphic to either 
$\mathbb Z_7$ or~$Frob(21)$ (the Frobenius group of order $21$). 
Assuming the latter, we are next able to construct a unique feasible $G$-orbit partition of $V(\Gamma)$ which we subsequently show cannot exist. As a result, divisibility by 7 implies $G\cong \mathbb Z_7$.

\section{Preliminaries}\label{prelims}

Throughout this paper, $\Gamma$ will denote a putative Conway 99-graph, \ie a strongly regular graph with parameters $(99,14,1,2)$. We denote its automorphism group by~$G$. Although $\Gamma$ is not vertex transitive, a result first proved by Wilbrink~\cite{Wil76}, one can always ``hang'' the graph from any vertex $x\in V(\Gamma)$ whereby vertices are grouped together in accordance with their distance from $x$. 
This is a property of distance-regular graphs in general and strongly regular graphs in particular. In such case, we refer to $x$ as the ``root vertex''. 

As is customary, we denote by $\Gamma_1(x)$ and $\Gamma_2(x)$ the set of neighbors and non-neighbors of $x$ respectively, commonly referred to as the first and second subconstituents of $\Gamma$. When $x$ is understood from context, we will abbreviate these sets by~$\Gamma_1$ and $\Gamma_2$. In Figure~\ref{DDD} we depict the distance distribution diagram for $\Gamma$. Note that the diagram indicates that 
$|\Gamma_1|=14$ and $|\Gamma_2|=84$.

For the moment we assume $x\in V(\Gamma)$ is an arbitrary vertex, however in due course our choice of $x$ will carry special significance. 
We label the 14 neighbors of $x$ by ${iX}$, $1\le i\le 7$, $X\in \{L,R\}$, see Figure~\ref{Gamma1}. Since 
$\lambda=1$, we may assume without loss of generality that $\{iL,iR\}$ is an edge for every 
$1\le i\le 7$. 
 
\begin{figure}
\begin{tikzpicture}

\draw (0,-1) circle (8pt);
\draw (2.5,-1) circle (8pt);
\draw (5,-1) circle (8pt); 
\draw (0,-1)--(2.5,-1)--(5,-1);
\draw[white, fill] (0,-1) circle (7pt); 
\draw[white, fill] (2.5,-1) circle (7pt); 
\draw[white, fill] (5,-1) circle (7pt); 
\node at (0,-1) {1} node at (.55,-.83) {\footnotesize 14};
\node at (2.5,-1) {14} node at (2.03,-.83) {\footnotesize 1} node at (3.05,-.83) {\footnotesize 12} node at (2.5,-.53) {\footnotesize 1};
\node at (5,-1) {84} node at (4.55,-.83) {\footnotesize 2} node at (5,-.53) {\footnotesize 12};

\end{tikzpicture}
\caption{Distance distribution diagram of $\Gamma$}\label{DDD}
\end{figure}

\begin{figure}

\begin{tikzpicture}[scale=.9]

\draw [fill] (0,0) circle (.06) node [above] {$x$};
\draw [fill] (-.5,-2) circle (.06) node at (-.5,-2.3) {${4L}$};
\draw [fill] (.5,-2) circle (.06) node at (.5,-2.3) {${4R}$};
\draw [fill] (-1.5,-2) circle (.06) node at (-1.5,-2.3) {${3R}$};
\draw [fill] (1.5,-2) circle (.06) node at (1.5,-2.3) {${5L}$};
\draw [fill] (-2.5,-2) circle (.06) node at (-2.5,-2.3) {${3L}$};
\draw [fill] (2.5,-2) circle (.06) node at (2.5,-2.3) {${5R}$};
\draw [fill] (-3.5,-2) circle (.06) node at (-3.5,-2.3) {${2R}$};
\draw [fill] (3.5,-2) circle (.06) node at (3.5,-2.3) {${6L}$};
\draw [fill] (-4.5,-2) circle (.06) node at (-4.5,-2.3) {${2L}$};
\draw [fill] (4.5,-2) circle (.06) node at (4.5,-2.3) {${6R}$};
\draw [fill] (-5.5,-2) circle (.06) node at (-5.5,-2.3) {${1R}$};
\draw [fill] (5.5,-2) circle (.06) node at (5.5,-2.3) {${7L}$};
\draw [fill] (-6.5,-2) circle (.06) node at (-6.5,-2.3) {${1L}$};
\draw [fill] (6.5,-2) circle (.06) node at (6.5,-2.3) {${7R}$};

\draw (-.5,-2) -- (.5,-2);
\draw (-1.5,-2) -- (-2.5,-2);
\draw (-3.5,-2) -- (-4.5,-2);
\draw (-5.5,-2) -- (-6.5,-2);
\draw (1.5,-2) -- (2.5,-2);
\draw (3.5,-2) -- (4.5,-2);
\draw (5.5,-2) -- (6.5,-2);

\draw (0,0) -- (.5,-2);
\draw (0,0) -- (-2.5,-2);
\draw (0,0) -- (-4.5,-2);
\draw (0,0) -- (-6.5,-2);
\draw (0,0) -- (2.5,-2);
\draw (0,0) -- (4.5,-2);
\draw (0,0) -- (6.5,-2);

\draw (0,0) -- (-.5,-2);
\draw (-1.5,-2) -- (0,0);
\draw (-3.5,-2) -- (0,0);
\draw (-5.5,-2) -- (0,0);
\draw (1.5,-2) -- (0,0);
\draw (3.5,-2) -- (0,0);
\draw (5.5,-2) -- (0,0);

\end{tikzpicture}
\caption{A labeling scheme for $\Gamma_1$}\label{Gamma1}
\end{figure}
 
Observe that for any $\{X,Y\}= \{L,R\}$, the vertices 
${iX},{jY}\in \Gamma_1$ are adjacent if and only if
$i= j$. As $\mu=2$, each pair of nonadjacent vertices 
${iX},{jY}\in \Gamma_1$ must have a unique common neighbor in $\Gamma_2$. We label this common neighbor $ijXY$. As there are $\binom{14}{2}-7=84$ pairs of nonadjacent vertices in $\Gamma_1$, we see that each of the 84 vertices in $\Gamma_2$ receives a unique label, again due to $\mu=2$. 

\begin{lemm}\label{identity aut}
Suppose $g \in G$ fixes $\Gamma_1\cup\{x\}$ pointwise. Then $g$ fixes 
$\Gamma_2$ pointwise, \ie $g$ is the identity automorphism.
\end{lemm}

\begin{proof}
Evident from our labeling scheme 
and the fact that $\mu=2$.
\end{proof}
 
\begin{lemm}\label{7-orbits}
Suppose there exists $s\in G$ with $|s| = 7$. Then 
$s$ fixes a unique vertex in~$V(\Gamma)$, hence the 
$\langle s\rangle$-orbit structure on $V(\Gamma)$ is $[1,7^{14}]$ {\rm (}\ie one orbit of size $1$ and~$14$ orbits of size $7${\rm )}.
\end{lemm}
 
\begin{proof}
As $|V(\Gamma)| = 99 \equiv 1 \!\pmod 7$ we deduce that $s$ must have at least one fixed vertex which we may choose to fulfill the role of root vertex $x$ of 
$\Gamma$. This establishes that both $\Gamma_1$ and $\Gamma_2$ are $s$-invariant. In particular, the orbit structure of $\langle s\rangle$ on $\Gamma_1$ is $[7^2]$, $[1^7,7]$, or $[1^{14}]$. 
As $iL$ is a fixed point of $s$ if and only if $iR$ is, the number of fixed points must be even, \ie the $\langle s\rangle$-orbit structure is either 
$[7^2]$ or $[1^{14}]$. However, Lemma~\ref{identity aut} rules out $[1^{14}]$, leaving $[7^2]$ as the orbit structure on $\Gamma_1$. Now suppose a vertex $ijXY\in \Gamma_2$ is fixed by $s$. As $\mu=2$ and 
$|s|$ is odd, $s$ must fix the $\mu$-graph 
$\Gamma_1(x)\cap \Gamma_1(ijXY)=\{iX,jY\}$ vertex-wise, a contradiction. This proves the 
$\langle s\rangle$-orbit structure on $\Gamma_2$ is $[7^{12}]$. The result follows. 
\end{proof}

\begin{rema}\label{|s|=7} 
Note that one consequence of our labeling scheme, together with the assumption in Lemma~\ref{7-orbits}, 
is that the automorphism group ${\rm Aut}(\Gamma)$ embeds in ${\rm Sym}(\Gamma_1)\cong S_{14}$. 
By relabeling the vertices of $\Gamma_1$, we may assume that $s=s_Ls_R$ where $s_L=(1L,2L,\dots,7L)$ and $s_R=(1R,2R,\dots,7R)$. 
The two $\langle s\rangle$-orbits on $\Gamma_1$ are now transparent. 
They are $\{1L,2L,\dots,7L\}$ and $\{1R,2R,\dots,7R\}$. 
\end{rema}

\begin{lemm}\label{140 triangles} 
Every vertex of $\,\Gamma_2$ lies on five $3$-cycles wholly inside $\Gamma_2$. 
Thus there are $140$ $3$-cycles in $\Gamma_2$.
\end{lemm}

\begin{proof}
Clearly, every vertex in $\Gamma$ lies on seven $3$-cycles. Given a vertex $v\in \Gamma_2$ it has precisely two $\Gamma_1$-neighbors $u$ and $w$. As $\lambda=1$, each of $vu$ and $vw$ must be an edge in a unique 3-cycle. Moreover, these two 3-cycles cannot coincide. Indeed, this would require that $u$ and $w$ be adjacent, whence $uwx$ would be a second 3-cycle on the edge $uw$ where $x$ is the root vertex. This proves the remaining five $3$-cycles on $v$ lie entirely inside $\Gamma_2$. But now we have that the total number of $3$-cycles in $\Gamma_2$ is $\frac{84\cdot 5}{3}=140$ as claimed.
\end{proof}

\section{Consequences of divisibility by \texorpdfstring{$2$}{2}}
\label{no order 14 auts}
 
In this section, we prove $|G|$ must divide $6$ under the assumption that $|G|$ is even. 
A critical role here is played by Theorem~\ref{no 14} 
in which we prove that $G$ cannot have any order 14 automorphisms. 

Lemmas~\ref{order 14 element} and~\ref{orbits} set the groundwork for the rest of this section. 

\begin{lemm}\label{order 14 element}
Suppose $G$ contains a cyclic subgroup $K$ of order $14$. Then 
$K=\langle st\rangle$ where $s=(1L,2L,\dots,7L)\,(1R,2R,\dots,7R)$ and $t=(1L,1R)\,(2L,2R)\,\dots\,(7L,7R)$. 
\end{lemm}

\begin{proof} Let $C_S(s)=(\langle s_L\rangle\times \langle s_R\rangle)\rtimes\langle t
\rangle$ denote the centralizer of $s$ in 
$S={\rm Sym}(\Gamma_1)$. 
There are seven involutions in $C_S(s)$ which take the form 
\[
t^{(s_{R})^{i}}=(1L,(1+i)R)\,(2L,(2+i)R)\dots(7L,iR),
\] 
$0\le i\le 6$, and in each case $\langle st^{(s_{R})^{i}} \rangle$ is a cyclic subgroup of $C_S(s)$ of order 14. 
However, since $t^{(s_{R})^{i}}$ maps $1L$ to $(1+i)R$ and $1R$ to $(1-i)L$, adjacency is preserved only if~$i=0$. Thus $t$ is the unique involution in 
${\rm Aut}(\Gamma_1)$ whereby 
$K=\langle st\rangle$ is the unique cyclic subgroup of order 14 in ${\rm Aut}(\Gamma_1)$. 
\end{proof}

\begin{lemm}\label{orbits}
$K =\langle st \rangle$ fixes the root vertex $x$ of $\,\Gamma$ but has no other fixed points. Thus each of the remaining seven $K$-orbits on $\,\Gamma$ has size $14$. 
\end{lemm}
 
\begin{proof}
Recall from Lemma~\ref{7-orbits} that $x$ is the unique vertex fixed by $s\in K$. As $t$ commutes with $s$
we have that $x^t$ is fixed by $s$, whence $x^t=x$. Thus $x$ is the unique vertex fixed by $K$. 

We now consider $K$-orbits on 
$V(\Gamma)\setminus \{x\}$. 
Clearly, $K$ fuses the two $\langle s\rangle$-orbits in~$\Gamma_1$ so we are left to consider the orbit structure of $K$ on $\Gamma_2$. 
Suppose there exists an~$\langle s\rangle$-orbit 
$\mathcal O$ that is $K$-invariant. Then as $|t|=2$ and $|\mathcal O|=7$, there must be a vertex~$v\in\mathcal O$ fixed by $t$. Let $iX$ be a 
$\Gamma_1$-neighbor of $v$. Then $iY=(iX)^t$ 
must also be a $\Gamma_1$-neighbor of $v$. But this violates $\lambda=1$ as $iX$ and $iY$ are adjacent. We conclude that all $K$-orbits of $\Gamma_2$ have size $14$.
\end{proof}

\begin{rema}\label{rem: coords}
It is easy to see that a set of orbit representatives in the action of $K$ on~$\Gamma_2$ is given by 
$\{12LL,\, 13LL,\,14LL,\, 12LR,\, 13LR,\, 14LR\}$. Moreover, each numerical coordinate $i$ occurs exactly four times in each orbit. For example, in the orbit with representative $12LL$, the coordinate $3$ occurs in each of $23LL$, $34LL$, $23RR$, $34RR$. 
However, these four vertices are distributed evenly into pairs in the sense that $23LL$, $34LL$ are neighbors of $3L$ while $23RR$, $34RR$ are neighbors of $3R$. 
\end{rema}

Consider the equitable partition $\pi$ induced by the $K$-orbits 
$\mathcal O_1,\mathcal O_2,\dots,\mathcal O_6$ 
on $\Gamma_2$. As is customary, we refer to $\pi$ as a \emph{$K$-orbit partition}. We denote by $b_{ij}$ the number of $\mathcal O_j$-neighbors of any fixed vertex in $\mathcal O_i$ and refer to it as an \emph{orbit valency} (or simply \emph{valency}) due to what occurs naturally in the quotient graph $\Gamma_2/\pi$. 
Orbit valencies for which $i=j$ will be called \emph{internal} and expressed as $b_i$ rather than $b_{ii}$. All other valencies will be referred to as \emph{external}.
Pictorially, $b_{ij}$ appears as a label of an arc from 
$\mathcal O_i$ to $O_j$, however in our case this arc is an edge (\ie $b_{ij}=b_{ji}$ for all $i,j$) since all orbits 
$\mathcal O_i$ have the same size. 
 
\begin{figure}
\resizebox{9cm}{7.5cm}{

\begin{tikzpicture}
\draw[thick] (0,0) circle (6mm); 
\draw [thick](3,0) circle (6mm);
\draw [thick](0,3) circle (6mm) node at (0,3) {$b_{3}$};
\draw[thick] (3,3) circle (6mm) node at (3,3) {$b_{4}$};
\draw[thick] (-2.5,1.5) circle (6mm) node at (-2.5,1.5) {$b_{5}$};
\draw[thick] (5.5,1.5) circle (6mm) node at (5.5,1.5) {$b_{6}$};

\draw (.6,0)--(2.4,0);
\draw (.6,3)--(2.4,3);
\draw (0,.6)--(0,2.4);
\draw (3,.6)--(3,2.4);
\draw (.4,.4)--(2.6,2.6);
\draw (.4,2.6)--(2.6,.4);

\node at (0,0) {$b_{1}$};
\node at (1.5,.238) {$b_{12}$};
\node at (.75,1.2) {$b_{14}$};
\node at (-.278,1.5) {$b_{13}$}; 
\node at (-1.2,.9) {$b_{15}$};
\node at (5,.095) {$b_{16}$}; 

\node at (3,0) {$b_{2}$};
\node at (3.35,1.5) {$b_{24}$}; 
\node at (2.2,1.2) {$b_{23}$};
\node at (-2,.095) {$b_{25}$}; 
\node at (4.2,.9) {$b_{26}$};

\node at (1.5,3.225) {$b_{34}$};
\node at (1.5,5.05) {$b_{56}$};
\node at (-1.95,3.1) {$b_{45}$}; 
\node at (4.95,3.1) {$b_{36}$}; 

\node at (-1.2,2.1) {$b_{35}$};
\node at (4.2,2.1) {$b_{46}$};

\node at (-.8,-.45) {\Large$\mathcal O_1$};
\node at (3.85,-.45) {\Large$\mathcal O_2$};
\node at (-.82,3.5) {\Large$\mathcal O_3$};
\node at (3.85,3.5) {\Large$\mathcal O_4$};
\node at (-3.45,1.5) {\Large$\mathcal O_5$};
\node at (6.47,1.5) {\Large$\mathcal O_6$};

\draw (-2,1.1)--(-.55,.17);
\draw (-2,1.9)--(-.55,2.83);
\draw (5,1.1)--(3.55,.17);
\draw (5,1.9)--(3.55,2.83);

\draw [bend right=80] (2.8,3.6) to (-2.6,2.1); 
\draw [bend left=80] (2.8,-.6) to (-2.6,.9); 

\draw [bend right=80] (0,-.6) to (5.5,.9); 
\draw [bend left=80] (0,3.6) to (5.5,2.1); 

\draw (-2.6,2.1) to[bend left=90, looseness=1.35] (5.5,2.1); 
 
\end{tikzpicture}}
\vspace{-7mm}
\caption{A general $K$-orbit partition on $\Gamma_2$ where $K=\langle st\rangle\cong \mathbb Z_{14}$}\label{fig:6 orbits} 
\end{figure}

At present we have that $st$ is an order 14 automorphism of the subgraph of $\Gamma$ induced on $\Gamma_1\cup\{x\}$. We wish to show $st$ cannot extend to an automorphism of $\Gamma$. 
Our first step toward this objective is to count in two ways the cardinality of the set 
\[
S = \{uvw: \text{$uvw$ is a 2-path with $w \in \mathcal{O}_i$}\},
\] 
where $u$ is a fixed vertex in $\mathcal O_i$ (see Figure~\ref{fig:6 orbits}).

For each of the $b_{i}$ neighbors of $u$ in $\mathcal O_i$ there is a unique $2$-path from $u$ to $w$ (since $\lambda=1$). Similarly, for each of the $13-b_{i}$ non-neighbors of $u$ in $\mathcal O_i$ there are two 2-paths from $u$ to $w$
 (since $\mu=2$).
 Thus $|S|= b_{i} \cdot 1 + (13 - b_{i}) \cdot 2 = 26 - b_{i}$. 
 
 On the other hand, we may condition our count on the location of the intermediate vertex $v$. For $v$ in $\mathcal O_j$ there are $b_{ij}\,(b_{ij}-1)$ such 2-paths. 
 In addition, $u$ has exactly two neighbors in $\Gamma_1$ each of which has a unique neighbor $w\in \mathcal O_i\setminus \{u\}$ (\cf Remark~\ref{rem: coords}). This gives
 $|S|= \sum_{j= 1}^{6} b_{ij} \,(b_{ij} - 1) + 2$. 
 Equating these two expressions for $|S|$, we obtain 
$26 - b_{i} = \sum_{j = 1}^{6} b_{ij} \,(b_{ij} - 1) + 2$. But due to the fact that $\sum_{j=1}^6 b_{ij}=12$, this simplifies to
\begin{equation}\label{eqn 1}
36 - \big(b_{i}^{\,2} + b_{i} \big) = 
\sum_{j\ne i} b_{ij}^{\,2}\end{equation}
We divide our analysis into cases based on an assumed value for $b_{i}$. Once a choice of~$b_i$ is made, we find all ways of expressing $36 - \big(b_{i}^{\,2 }+ b_{i} \big)$ as a sum of five squares while maintaining the valency requirement $\sum_{j=1}^6 b_{ij}=12$. Note that $b_i\le 5$ since otherwise the value of $36 - \big(b_{i}^{\,2} + b_{i} \big)$ would be negative. 

All solution sets are provided in the lemma below. Verification of the list is straightforward, so is left to the reader. 
Note that each solution set is expressed as an ordered pair of the form 
\[
\big(b_i,\big\{a_2,a_3,a_4,a_5,a_6\big\}\big).
\] 
This is because unlike the internal valency $b_i$ which remains fixed, the values $a_2,a_3,\dots,a_6$ may be assigned to the external valencies $b_{ij}$ in any specified manner. Thus, there are multiple solutions corresponding to each solution set achieved by suitably permuting the members of the multiset 
$\big\{a_2,a_3,a_4,a_5,a_6\big\}$. 
Below, we list these members in decreasing order.

\begin{lemm}\label{orbit type1} For $b_{i}\in\{1,3,5\}$ there are no solutions to equation~\ref{eqn 1}. For other values of $\,b_i$ the solutions are listed as follows: 
\begin{enumerate}
\item[(a)] 
$\big(0,\big\{4, 3, 3, 1, 1\big\}\big)$ and 
$\big(0,\big\{3, 3, 3, 3, 0\big\}\big)$ when $b_{i}=0$.
\item[(b)] 
$\big(2, \big\{4, 3, 2, 1, 0\big\}\big)$ when $b_{i}=2$,
\item[(c)]
$\big(4, \big\{3, 2, 1, 1, 1\big\}\big)$ and 
$\big(4, \big\{2, 2, 2, 2, 0\big\}\big)$ when $b_{i}=4$.
\end{enumerate}
\end{lemm}

For future reference, it is convenient to designate these solution sets by type, \eg
 \[
 {\rm I}. \,\big(0, \big\{4, 3, 3, 1, 1\big\}\big),
 {\rm II}.\,\big(0, \big\{3, 3, 3, 3, 0\big\}\big), 
 {\rm III}.\,\big(2, \big\{4, 3, 2, 1, 0\big\}\big)
 \]
 \[
 {\rm IV}. \,\big(4, \big\{3, 2, 1, 1, 1\big\}\big), 
 {\rm V}.\, \big(4, \big\{2, 2, 2, 2, 0\big\}\big)
 \]
We also extend this terminology to orbits, saying an orbit is \emph{of type} $\rm T$ if its set of valencies correspond to a solution set of type~
${\rm T}\in\{{\rm I, II, III, IV, V}\}$. 
 
We next count in two ways the number of $2$-paths starting from a fixed vertex $u$ in~$\mathcal{O}_i$ and ending at some vertex $w$ in $\mathcal{O}_j$, $j\ne i$. Here $u$ has exactly $b_{ij}$ neighbors in~$\mathcal{O}_j$, and as $\lambda=1$ there exists a unique $2$-path starting at $u$ and ending at $w$ for each neighbor $w$ of $u$ in $\mathcal O_j$. 
Similarly, $u$ has $14 - b_{ij}$ non-neighbors in 
$\mathcal{O}_j$, and as~$\mu=2$ there are exactly two $2$-paths starting at $u$ and ending at each 
non-neighbor $w$ of $u$ in $\mathcal O_j$. 
Thus in total there are 
$b_{ij} \cdot 1 + (14 - b_{ij}) \cdot 2 = 28 - b_{ij}$ 
such $2$-paths from $u$ into $\mathcal O_j$ when $j\ne i$. 
		
For the second count, we focus on the location of an intermediate vertex $v$ in each such 2-path. Here $v$ can occur in any of the six $K$-orbits on $\Gamma_2$ as well as in $\Gamma_1$. In the case of $K$-orbits on $\Gamma_2$, there are~$b_{ik}$ choices for $v$ in 
$\mathcal O_k$, and for each such $v$ there are~$b_{kj}$ choices for $w$ in $\mathcal O_j$. 
This gives $b_{ik}b_{kj}$ 2-paths of desired type. 
In addition, there are two choices for $v$ in 
$\Gamma_1$ each of which has two neighbors $w$ in 
$\mathcal O_j$. This produces four more 2-paths. Thus, in total there are precisely 
 $\sum_{k= 1}^{6}b_{ik}b_{kj} + 4$ 
 paths of the type~in question when $j\ne i$.
 		
Equating these two counts yields 
$28 - b_{ij} = \sum_{k = 1}^{6} b_{ik}b_{kj} + 4$, 
or equivalently
\begin{equation}\label{eqn 2}
24 - b_{ij} = \sum_{k = 1}^{6} b_{ik}b_{kj}.
\end{equation}	

\begin{lemm}\label{orbit type}
In a $K$-orbit partition of $\,\Gamma_2$ {\rm (}\cf Figure~\ref{fig:6 orbits}{\rm )} we have the following:
\begin{enumerate}
\item[(a)] 
The number of orbits of type~I is at most $2$.
\item[(b)] 
The number of orbits of type~II is at most $1$.
\item[(c)] 
The number of orbits of type~III is at most $4$.
\item[(d)] 
The number of orbits of type~IV is at most $4$.
\item[(e)] 
The number of orbits of type~V is at most $1$.
\end{enumerate}
\end{lemm}

\begin{proof}
(a): Suppose there exist two orbits $\mathcal O_i$ and 
$\mathcal O_j$ of type~I. Then since $b_i=b_j=0$, equation~\ref{eqn 2} reduces to 
$24 - b_{ij} =\sum_{k \ne i,j} b_{ik}b_{kj}$. 
Note that this equation is satisfied only if $b_{ij}=4$, which results in the solution $20=3^2+3^2+1^2+1^2$. 
As an orbit of type~I has only one external valency 4, there cannot be a third orbit of this type.

(b): Let $\mathcal O_i$ and $\mathcal O_j$ be two orbits of type~II. 
Since $b_i=b_j=0$, equation~\ref{eqn 2} again reduces to $24 - b_{ij} =\sum_{k \ne i,j} b_{ik}b_{kj}$. 
But regardless of how one chooses $b_{ij}\in \{0,3\}$ and reorders the corresponding multisets, this equation is never satisfied. Thus there is at most one orbit of type~II.

(c): Let $\mathcal O_i$ and $\mathcal O_j$ be two orbits of type~III. 
Since $b_i=b_j=2$, equation~\ref{eqn 2} becomes~$24 - 5b_{ij} =\sum_{k \ne i,j} b_{ik}b_{kj}$. 
In this case one has $b_{ij}\in\{0,1,2,3,4\}$. 
There are no solutions if $b_{ij}\in\{1,2\}$, however every remaining choice of $b_{ij}$ works. 
Specifically, if $b_{ij}=0$ one gets 
$24=4\cdot 1+3\cdot 4+2\cdot 3+1\cdot 2$ 
as a solution. 
For $b_{ij}=3$ one gets 
$9=4\cdot 0+2\cdot 4+1\cdot 1+0\cdot 2$, 
while for $b_{ij}=4$ one gets 
$4=3\cdot 0+2\cdot 1+1\cdot 2+0\cdot 3$. 
Having only three allowable external valencies adjoining pairs of type~III orbits, it is not possible to have a fifth orbit of this type.

(d): Given any two orbits $\mathcal O_i$ and 
$\mathcal O_j$ of type~IV, we have that 
$b_{ij}\in\{1,2,3\}$. 
As~$b_1=b_2=4$, equation~\ref{eqn 2} becomes 
$24 -9b_{ij} =\sum_{k \ne i,j} b_{ik}b_{kj}$. 
Clearly $b_{ij}=1$ leads to a solution, namely $15=1^2+1^2+2^2+3^2$, but other choices of $b_{ij}$ fail. Since a type~IV orbit has only three available external valencies equal to 1, there can be at most four orbits of this type.

(e): Finally, suppose there are two orbits $\mathcal O_i$ and $\mathcal O_j$ of type~V. 
Then since $b_i=b_j=4$, equation~\ref{eqn 2} becomes $24-9b_{ij}=\sum_{k \ne i,j} b_{ik}b_{kj}$. 
Here $b_{ij}\in\{0,2\}$ and neither choice leads to a solution. 
This proves there is at most one orbit of type~V.
 \end{proof}

In the above, we applied equation~\ref{eqn 2} to bound the number of orbits of a given type~that can occur in a $K$-orbit partition of $\Gamma_2$. We now do the same for orbits of mixed type. Note that we do not strive to obtain sharp bounds at this stage. Our goal is simply to eliminate several possibilities in an expedient manner. 
 
\begin{lemm}\label{mixed types}
Let $\rm T_i\,, T_j\in\{I,II,III,IV,V\}$. 
Then the $\rm (T_i\,,T_j)$-entry in Table~\ref{table} bounds from above the number of orbits of type~
$\rm T_j$ that can coexist with a fixed orbit of type~
$\rm T_i$ in a $K$-orbit partition of $\,\Gamma_2$. 
\end{lemm}
 
\begin{proof}
Note that the diagonal entries in Table~\ref{table} were previously confirmed in Lemma~\ref{orbit type}. Moreover, one need not check any entry 
$\rm(T_i\,,T_j)$ that is equal to the diagonal entry 
$\rm(T_j\,,T_j)$ since the latter is the maximum allowable number of orbits of type~$\rm T_j$ in any 
$K$-orbit partition of $\Gamma_2$.

\begin{table}
\begin{tabular}{l||ccccc}
&I&II&III&IV&V\\ \hline\hline\\[-4mm]
I & 2 & 0 & 4 & 0 & 0\\[.5mm]
II & 0 & 1 & 4 & 4 & 1\\[.5mm]
III & 2 & 1 & 4 & 2 & 1\\[.5mm]
IV & 0 & 1& 4 & 4 & 1\\[.5mm]
V & 0 & 1 & 4 & 4 & 1
\end{tabular}\\[3mm]
\caption{Bounds on the number of orbits of mixed type}\label{table} 
\end{table}

{\bf Case 1. $\rm (I,\,II)=(II,\,I)=0$\,}:
Clearly, the only option for the shared valency is \mbox{$b_{ij}=3$}. As $b_i=b_j=0$, equation~\ref{eqn 2} reduces to $21 =\sum_{k\ne i,j}b_{ik}b_{kj}$ where $b_{ik}\in\{4,3,1,1\}$ and $b_{kj}\in\{3,3,3,0\}$. It is easy to see that no permutation of multisets leads to a solution, \ie $\rm (I,\,II)=0$. (For $\rm (II,\,I)=0$, the only change to the above is~$b_{ik}\in\{3,3,3,0\}$ and $b_{kj}\in\{4,3,1,1\}$.)

{\bf Case 2.\ $\rm (I,\,IV)=(IV,\,I)=0$\,}:
In this case $b_i=0$ and $b_j=4$, so equation 
\ref{eqn 2} reduces to 
$21 -5b_{ij}=\sum_{k\ne i,j}b_{ik}b_{kj}$. 
Here there are two options for $b_{ij}$. 
If $b_{ij}=1$ then we get
$16=\sum_{k\ne i,j}b_{ik}b_{kj}$ where 
$b_{ik}\in \{4,3,3,1\}$ and $b_{kj}\in\{3,2,1,1\}$ 
and no permutation of multisets leads to a solution. 
For the second option $b_{ij}=3$, we get $6=\sum_{k\ne i,j}b_{ik}b_{kj}$ 
where $b_{ik}\in \{4,3,1,1\}$ and $b_{kj}\in\{2,1,1,1\}$. Again no permutation of multisets gives a solution. Thus $\rm (I,\,IV)=(IV,I)=0$.

{\bf Case 3.\ $\rm (I,\,V)= (V,\,I)=0$\,}: Here the multisets $\{4,3,3,1,1\}$ and $\{2,2,2,2,0\}$ are disjoint, so there is no possible valency that can adjoin two orbits of these respective types.

{\bf Case 4.\ $\rm (III,\,IV)=2$\,}: In this case, $b_i=2$ and $b_j=4$ so equation~\ref{eqn 2} becomes $24-7b_{ij}=\sum_{k\ne i,j}b_{ik}b_{kj}$. 
There are three choices for $b_{ij}$, namely 
$b_{ij}\in\{3,2,1\}$. 
If $b_{ij}=3$ the equation reduces to 
$3 =\sum_{k\ne i,j}b_{ik}b_{kj}$ where 
$b_{ik}\in \{4,2,1,0\}$ and $b_{kj}\in\{2,1,1,1\}$, 
and it is immediate that there is no solution. 
If $b_{ij}=2$ we obtain $10=\sum_{k\ne i,j}b_{ik}b_{kj}$ where $b_{ik}\in \{4,3,1,0\}$ and $b_{kj}\in\{3,1,1,1\}$. Here there is a unique solution, namely 
$10=4\cdot 1+3\cdot 1 +1\cdot 3+0\cdot 1$. 
Finally, if $b_{ij}=1$ we obtain 
$17=\sum_{k\ne i,j}b_{ik}b_{kj}$ where 
$b_{ik}\in \{4,3,2,0\}$ and $b_{kj}\in\{3,2,1,1\}$. 
In this case there are three solutions, namely $17=4\cdot 3+3\cdot 1 +2\cdot 1 +0\cdot 2=
4\cdot 1+3\cdot 3 +2\cdot 2 +0\cdot 1= 
4\cdot 2+3\cdot 1 +2\cdot 3 +0\cdot 1$.
In any case, there are just two possibilities for the valency adjoining an orbit of fixed type~III to an orbit of type~IV. We conclude that $\rm (III,\,IV)=2$.
 
As all cases in the lemma statement have been treated, the proof is complete. 
\end{proof}

The reader will note that the relation in Lemma~\ref{mixed types} is not generally symmetric. 

Let us write $\rm [I^{a},II^{\,b},III^{\,c},IV^{\,d},V^{\,e}]$ to indicate a $K$-orbit partition of $\Gamma_2$ having $a$ orbits of type~I, $b$ orbits of type~II, and so on.
(If an orbit of specific type~does not occur in the partition, we simply omit that type~from the above partition notation.) 
 
\begin{lemm}\label{no III,IV}
There is no $K$-orbit partition of the form 
$\rm [III^{\,c}, IV^{\,6-c}]$ for any $c$. 
\end{lemm}

\begin{proof}
By Lemma~\ref{orbit type}, one has $2\le c\le 4$. 
As a type~III orbit has a single valency of~4 and a type~IV orbit has none, there are $c/2$ external valencies of 4 in the partition. This means $c$ must be even. However, by Lemma~\ref{mixed types} the existence of a type~III orbit requires that there be at most two type~IV orbits. This implies $6-c\le 2$ whence $c= 2$ is prohibited. Thus $c=4$.
 
As shown in the proof of Lemma~\ref{orbit type}, each pair of type~III orbits must share an external valency of 0, 3 or 4. But as there are four type~III orbits, every such valency gets used. On the other hand, we showed in Lemma~\ref{orbit type} that two orbits of type~IV must share an external valency of 1. This leaves a type~IV orbit with an unusable valency 3, again a contradiction. 
\end{proof}
 
One conclusion of Lemma~\ref{no III,IV} is that a viable $K$-orbit partition of $\Gamma_2$ must contain an orbit of type~I, II or V. By way of the next two lemmas, we narrow this down considerably.

\begin{lemm}\label{no type I}
There is no $K$-orbit partition that contains an orbit of type~{\rm I}. 
\end{lemm}

\begin{proof}
By Lemma~\ref{mixed types}, the only possible orbit partition containing a type~I orbit is $\rm [I^2,III^4]$. However, we have seen that no pair of type~III orbits can share an external valency of 2 (\cf proof of Lemma~\ref{orbit type}(c)). As no type~I orbit admits such a valency, we see that each orbit of type~III has an unusable external valency of 2. The result follows. 
\end{proof}

\begin{lemm}\label{survivor}
A $K$-orbit partition of $\,\Gamma_2$ must be of the form $\rm [II,IV^4,V]$. 
\end{lemm}

\begin{proof}
By Lemmas~\ref{no III,IV} and~\ref{no type I}, 
a $K$-orbit partition must contain at least one orbit of type~II or V. 
Suppose there exists a type~II orbit in the partition. 
Then by Lemma~\ref{orbit type} there cannot be a second orbit of type~II, nor can there be more than one orbit of type~V. Thus the partition is of the form 
$\rm [II, III^c, IV^d, V^e]$ where $c+d+e=5$ and 
$0\le e\le 1$. 
Also, as in the proof of Lemma~\ref{no III,IV},
$c$ must be even. 

Suppose first that $c=2$. 
Then by Lemma~\ref{mixed types} we have $d\le 2$, whence the partition must be of the form 
$\rm [II,III^2,IV^2,V]$. 
In the proof of Lemma~\ref{mixed types}(d), we saw that any two orbits of respective types III and IV must be adjoined by a valency of either 1 or~2. 
As there are two type~IV orbits, both these external valencies get used adjoining a type~III orbit to these type~IV orbits. But this prohibits the existence of a type~V orbit since such an orbit has four external valencies of 2 and there are no longer any valencies of 2 in a type~III orbit to accommodate this. 
 
For $c=4$, the argument is similar. Each pair of two type~III orbits must share a valency of $0,3$ or 4. As there are four type~III orbits in this case, every such valency gets used adjoining orbits of this type. However, a type~II orbit has four external valencies of 3. As a type~III orbit no longer has a valency of 3 to accommodate this, we reach a contradiction This proves $c=0$ when a type~II orbit is assumed to occur in the partition. 
 
Our only remaining case is to assume a type~V orbit occurs in the partition. Here the four external valencies of 2 in such an orbit get used adjoining it to four other orbits, be they of type~III or IV. But this implies there must be an orbit of type~II in the partition, a case we have already treated. We conclude that a $K$-orbit partition of $\Gamma_2$ must indeed be of the form 
$\rm [II, IV^4,V]$. 
\end{proof}

We depict the $K$-orbit partition of type~
$\rm [II, IV^4,V]$ in Figure~\ref{fig:ep}. 
It too will be shown to not exist in due course.

\begin{figure}
\resizebox{9cm}{7.5cm}{
\begin{tikzpicture}

\draw[thick] (0,0) circle (6mm); 
\draw [thick](3,0) circle (6mm);
\draw [thick](0,3) circle (6mm) node at (0,3) {4};
\draw[thick] (3,3) circle (6mm) node at (3,3) {4};
\draw[thick] (-2.5,1.5) circle (6mm) node at (-2.5,1.5) {0};
\draw[thick] (5.5,1.5) circle (6mm) node at (5.5,1.5) {4};

\draw (.6,0)--(2.4,0);
\draw (.6,3)--(2.4,3);
\draw (0,.6)--(0,2.4);
\draw (3,.6)--(3,2.4);
\draw (.4,.4)--(2.6,2.6);
\draw (.4,2.6)--(2.6,.4);

\node at (0,0) {4};
\node at (1.5,.238) {1};
\node at (.75,1.2) {1};
\node at (-.2,1.5) {1}; 
\node at (-1.2,.9) {3};
\node at (5,.095) {2}; 

\node at (3,0) {4};
\node at (3.2,1.5) {1}; 
\node at (2.2,1.2) {1};
\node at (-2.1,.095) {3}; 
\node at (4.2,.9) {2};

\node at (1.5,3.225) {1};
\node at (1.5,5.05) {0};

\node at (-2.1,3.1) {3}; 
\node at (4.95,3.1) {2}; 

\node at (-1.2,2.15) {3};
\node at (4.2,2.15) {2};

\node at (-.82,-.45) {\Large IV};
\node at (3.85,-.45) {\Large IV};
\node at (-.82,3.38) {\Large IV};
\node at (3.89,3.4) {\Large IV};

\node at (-3.4,1.5) {\Large II};
\node at (6.42,1.5) {\Large V};

\draw (-2,1.1)--(-.55,.17);
\draw (-2,1.9)--(-.55,2.83);

\draw (5,1.1)--(3.55,.17);
\draw (5,1.9)--(3.55,2.83);

\draw [bend right=80] (2.8,3.6) to (-2.6,2.1); 
\draw [bend left=80] (2.8,-.6) to (-2.6,.9); 

\draw [bend right=80] (0,-.6) to (5.5,.9); 
\draw [bend left=80] (0,3.6) to (5.5,2.1); 

\draw (-2.6,2.1) to[bend left=90, looseness=1.35] (5.5,2.1);
 
\end{tikzpicture}}
\vspace{-7mm}
\caption{The lone surviving $K$-orbit partition of 
$\Gamma_2$ (\cf Lemma~\ref{survivor})}\label{fig:ep}
\end{figure}

\begin{lemm}\label{no C3}
Neither a type~{\rm IV} orbit nor a type~{\rm V} orbit can contain a $3$-cycle. 
\end{lemm}		

\begin{proof}
First observe that type~IV and type~V orbits have internal valency 4. In any orbit of either type~we may fix a vertex $v$ and denote its neighbors by $v^p, v^{r}, v^s, v^{t}$ where $p,r,s,t\in K$. However $v^{\,p^{-1}},v^{\,r^{-1}}, v^{\,s^{-1}}, v^{\,t^{-1}}$ must also be neighbors of $v$. This means, with one exception, the automorphisms $p,r,s,t$ must come in pairs. The one exception is if two or four of $p,r,s,t$ are involutory. However, this cannot be the case since $K$ contains a unique involution. Therefore, without loss of generality we may assume $r=p^{-1}$ and $t=s^{-1}$. Moreover, we know that $|r|,|t|\in\{7, 14\}$.
Three broad cases can arise here, namely $|r|=|t|=7$, $|r|=|t|=14$ and $|r|=7,\, |t|=14$.

{\bf Case 1. $|r|=|t|=7$}: In this case the orbit is comprised of two connected components but that won't affect our argument.
Since $K$ contains a unique subgroup of order 7, we must have $r=t^{\,m}$ for some integer $m\in \{2,3\}$. However, both subcases violate $\lambda=1$ as depicted in Figure~\ref{|r|=|t|=7}. Specifically, if $r=t^2$ then the edge $v^t v^{\,t^{\,2}}$ lies on two 3-cycles, while if $r=t^3$ the edge $v v^{\,t^{\,4}}$ suffers the same fate. (Note that the subcases $m=4,5$ are identical to $m=3,2$ respectively.)

{\bf Case 2. $|r|=|t|=14$}: Here we have $r=t^{\,m}$ where $m\in \{3,5\}$. Both subcases violate $\mu=2$ as indicated in Figure~\ref{|r|=|t|=14}. Specifically, if $r=t^{\,3}$ then vertices $v^{\,t^{\,3}}$ and~$v^{\,t^{\,5}}$ have $v^{\,t^{\,2}}$, $v^{\,t^{\,4}}$, $v^{\,t^{\,6}}$ as common neighbors, whereas if $r=t^{\,5}$ then vertices $v$ and~$v^{\,t^{\,4}}$ have $v^{\,t^{\,5}}$, $v^{\,t^{\,9}}$, $v^{\,t^{-1}}$ as common neighbors.

{\bf Case 3. $|r|=7, |t|=14$}: Here there are three subcases to consider, namely $r\in\{t^2,t^4,t^6\}$ as indicated in Figure~\ref{|r|=|t|=14}. The first and last of these subcases lead to violations. Specifically, $r=t^{\,2}$ violates $\lambda=1$ since the edge $vv^{\,t}$ lies on two 3-cycles with respective antipodal vertices $v^{\,t^{\,2}}$ and $v^{\,t^{\,-1}}$. 
In contrast, $r=t^{\,6}$ violates $\mu=2$ since the vertices $v^{\,t^{\,4}}$ and $v^{\,t^{\,11}}$ have $v^{\,t^{\,3}}$, $v^{\,t^{\,10}}$, $v^{\,t^{\,12}}$ as common neighbors (as well as~$v^{\,t^{\,5}}$). 
 Curiously, the case $r=t^{\,4}$ does not lead to any 
 $\lambda$ or $\mu$ violations, however it produces no 3-cycles either. This completes the proof of the lemma.
\end{proof} 

\begin{figure}
\resizebox{9cm}{!}{

\begin{tikzpicture}
 
\node[draw, regular polygon, regular polygon sides=7, minimum size=3cm] (a) {};
\foreach \i in {1,2,...,7}
\fill (a.corner \i) circle[radius=2pt];

\draw[thick,blue] (0,1.5)--(1.5,-.38);
\draw[thick,blue] (1.15,.9)--(.67,-1.3); 
\draw[thick,blue] (0,1.5)--(1.2,.9);
\draw[thick,blue] (1.2,.9)--(1.5,-.38);
\draw[thick,blue] (1.4,-.38)--(.69,-1.3);

\node at (0,-1.85) {$r=t^{\,2}$};
\node at (0,1.8) {$v$};
\node at (1.55,1) {$v^{\,t}$};
\node at (1.87,-.276) {$v^{\,t^{\,2}}$};
\node at (1.07,-1.37) {$v^{\,t^{\,3}}$};

\begin{scope}[rotate=51.4+51.4]
\draw (1.15,.9)--(.67,-1.3); 
\end{scope}
 
\begin{scope}[rotate=51.4+51.4+51.4]
\draw (1.15,.9)--(.67,-1.3); 
\end{scope}
 
\begin{scope}[rotate=51.4+51.4+51.4+51.4]
\draw (1.15,.9)--(.67,-1.3); 
\end{scope}

\begin{scope}[rotate=51.4+51.4+51.4+51.4+51.4]
\draw (1.15,.9)--(.67,-1.3); 
\end{scope}
 
\begin{scope}[rotate=51.4+51.4+51.4+51.4+51.4+51.4]
\draw (1.15,.9)--(.67,-1.3); 
\end{scope}
 
 
\begin{scope}[xshift=5cm]
\node[draw, regular polygon, regular polygon sides=7, minimum size=3cm] (a) {};
\foreach \i in {1,2,...,7}
 \fill (a.corner \i) circle[radius=2pt];
 \draw[thick,blue] (0,1.5)--(.62,-1.3);
 \draw [thick, blue](1.15,.9)--(-.6,-1.3);
 \draw (1.5,-.351)--(-1.5,-.351); 
 \draw (-1.15,.9)--(.6,-1.3);
 \draw[thick,blue] (0,1.5)--(-.62,-1.3);
 \draw[thick,blue] (0,1.5)--(1.2,.9);
 \draw[thick,blue] (.62,-1.35)--(-.62,-1.35);
 
\node at (0,-1.855) {$r=t^{\,3}$};
\node at (0,1.8) {$v$};
\node at (1.55,1) {$v^{\,t}$};
\node at (1.07,-1.37) {$v^{\,t^{\,3}}$};
\node at (-1.09,-1.37) {$v^{\,,t^{\,4}}$};

\begin{scope}[rotate=51.4]
\draw (0,1.5)--(.62,-1.3); 
\end{scope}

\begin{scope}[rotate=51.4+51.4]
\draw (0,1.5)--(.62,-1.3); 
\end{scope}

\end{scope}
 
\end{tikzpicture}}
\caption{The case $|r|=|t|=7$ of Lemma~\ref{no C3}}\label{|r|=|t|=7}
\end{figure} 

\begin{figure}
\resizebox{12cm}{!}{

\begin{tikzpicture}
 
\node[draw, regular polygon, regular polygon sides=14, minimum size=6cm] (a) {};
\foreach \i in {1,2,...,14}
 \fill (a.corner \i) circle[radius=2.5pt];

\draw (-.75,3)--(2.75,1.3);
\node at (-.75,3.2) {\Large $v$};
\node at (0,-3.5) {\Large $r=t^{\,3}$};
 
\begin{scope}[rotate=25.714]
\draw (-.75,3)--(2.75,1.3); 
\end{scope}

\begin{scope}[rotate=51.428]
\draw (-.75,3)--(2.75,1.3);
\end{scope}

\begin{scope}[rotate=77.142]
\draw (-.75,3)--(2.75,1.3);
\end{scope}

\begin{scope}[rotate=102.856]
\draw (-.75,3)--(2.75,1.3);
\end{scope}

\begin{scope}[rotate=102.856+25.714]
\draw (-.75,3)--(2.75,1.3);
\end{scope}

\begin{scope}[rotate=102.856+51.428]
\draw (-.75,3)--(2.75,1.3);
\end{scope}

\begin{scope}[rotate=102.856+77.142]
\draw (-.75,3)--(2.75,1.3);
\end{scope}

\begin{scope}[rotate=102.856+102.856]
\node at (-.75,3.2) {\Large $\;\,v^{\,t^{\,6}}$};
\draw (-.75,3)--(2.75,1.3);
\end{scope}

\begin{scope}[rotate=102.856+102.856+25.714]
 \node at (-.75,3.2) {\Large $\;\;\,v^{\,t^{\,5}}$};

\draw (-.75,3)--(2.75,1.3);
\draw [very thick,blue](-.75,2.95)--(.75,2.95);
\end{scope}

\begin{scope}[rotate=102.856+102.856+51.428]
 \draw (-.75,3)--(2.75,1.3); \node at (-.75,3.2) 
 {\Large $\;\;\; v^{\,t^{\,4}}$};
\draw [very thick, blue](-.75,2.95)--(.75,2.95);
\end{scope}

\begin{scope}[rotate=102.856+102.856+77.142]
 \draw [very thick,blue](-.75,3)--(2.75,1.3); 
 \node at (-.75,3.2) {\Large$\quad v^{\,t^{\,3}}$};
\draw [very thick, blue](-.75,2.95)--(.75,2.95); 
\end{scope}

\begin{scope}[rotate=102.856+102.856+102.856] 
 \node at (-.75,3.2) {\Large $\quad\; v^{\,t^{\,2}}$};
\draw [very thick, blue](-.75,3)--(2.75,1.3);
\draw [very thick,blue](-.75,2.95)--(.75,2.95);
\end{scope}

\begin{scope}[rotate=102.856+102.856+102.856+25.714]
\draw (-.75,3)--(2.75,1.3);
\end{scope}

\begin{scope}[xshift=8.5cm]
\node[draw, regular polygon, regular polygon sides=14, minimum size=6cm] (a) {};

\foreach \i in {1,2,...,14}
\fill (a.corner \i) circle[radius=2.5pt];
\draw[very thick,blue] (-.75,3)--(2.75,-1.35);
\node at (-.75,3.2) {\Large $v$};
\node at (0,-3.5) {\LARGE\bf$r=t^{\,5}$};
 
\begin{scope}[rotate=25.714]
\node at (-.75,3.2) {\Large $\;\,v^{\,t^{\,-1}}$};
\draw[very thick,blue] (-.75,3)--(2.75,-1.35);
\draw [very thick,blue] (-.75,2.9)--(.75,2.9);
\end{scope}

\begin{scope}[rotate=51.428]
\draw (-.75,3)--(2.75,-1.35);
\end{scope}

\begin{scope}[rotate=77.142]
\draw (-.75,3)--(2.75,-1.35);
\end{scope}

\begin{scope}[rotate=102.856]
\draw (-.75,3)--(2.75,-1.35);
\end{scope}

\begin{scope}[rotate=102.856+25.714]
\node at (-.75,3.4) {\Large $v^{\,t^{\,9}}$};
\draw [very thick,blue] (-.75,3)--(2.75,-1.35);
\end{scope}

\begin{scope}[rotate=102.856+51.428]
\draw (-.75,3)--(2.75,-1.35);
\end{scope}

\begin{scope}[rotate=102.856+77.142]
\draw(-.75,3)--(2.75,-1.35);
\end{scope}

\begin{scope}[rotate=102.856+102.856]
\draw (-.75,3)--(2.75,-1.35);
\end{scope}

\begin{scope}[rotate=102.856+102.856+25.714]
\draw (-.75,3)--(2.75,-1.35);
\node at (-.75,3.2) {\Large $\;\,\;v^{\,t^{\,5}}$};
\end{scope}

\begin{scope}[rotate=102.856+102.856+51.428]
\draw [very thick,blue](-.75,3)--(2.75,-1.35); 
\node at (-.75,3.2) {\Large $\;\,\;v^{\,t^{\,4}}$};
\draw [very thick,blue] (-.75,2.9)--(.75,2.9);
\end{scope}

\begin{scope}[rotate=102.856+102.856+77.142]
\draw (-.75,3)--(2.75,-1.35);
\end{scope}

\begin{scope}[rotate=102.856+102.856+102.856]
\draw (-.75,3)--(2.75,-1.35);
\end{scope}

\begin{scope}[rotate=102.856+102.856+102.856+25.714]
\draw (-.75,3)--(2.75,-1.35);
\end{scope}

\end{scope}

\end{tikzpicture}}
\caption{The case $|r|=|t|=14$ of Lemma~\ref{no C3}}\label{|r|=|t|=14}
\end{figure} 

\begin{theo}\label{no 14}
$G$ does not contain any order $14$ elements. 
\end{theo}

\begin{proof}
By way of contradiction, suppose $G$ contains an element of order $14$. By Lemma~\ref{order 14 element} we may assume this element is $st$, and as usual we set $K=\langle st\rangle$. 
Lemma~\ref{survivor} now asserts that a $K$-orbit partition must be of the form $\rm [II,IV^4,V]$ as depicted in Figure~\ref{fig:ep}. 
We claim this orbit partition cannot occur. 

For this purpose, it behooves us to restrict the action on $V(\Gamma)$ from $K\cong \mathbb Z_{14}$ to 
$\langle s\rangle\cong \mathbb Z_7$. 
This latter action yields 15 orbits: one fixed vertex 
(the root $x$) and $14$ orbits of size $7$. Let $B$ be the $15 \times 15$ quotient matrix corresponding to this action. We compute the trace of $B$ in two different ways. 

First, let $A$ be the adjacency matrix of $\Gamma$. We know that the characteristic polynomial of $B$ must divide that of $A$. Further, we know that the spectrum of $A$ is $\{ 14, 3^{(54)}, -4^{(44)} \}$. Since each row sum of $B$ is $14$, we conclude that $14$ is an eigenvalue of $B$. The spectrum of $B$ is therefore 
$\{14, 3^{(a)}, -4^{(14 - a)} \}$ for some nonnegative integer $a \leq 14$, and accordingly the trace of $B$ is $14 + 3a - 4(14 - a) = 7a - 42$. 

Note that the diagonal entries of $B$ correspond to the internal valencies of orbits. Thus the trace of $B$ is the sum of these internal valencies. 
In the action of $\langle s\rangle$ on 
$\Gamma_1\cup\{x\}$, we have the fixed vertex $x$ (obviously with internal valency zero) and two orbits of size $7$, each having internal valency zero so yielding no contribution to the trace. In $\Gamma_2$ we have 12 $\langle s\rangle$-orbits of size $7$ which fuse in pairs to form the six $K$-orbits of $\Gamma_2$. 
Five of these six $K$-orbits have internal valency $4$, which, as a direct consequence of Lemma~\ref{no C3}, decompose into pairs of $\langle s\rangle$-orbits each having internal valency $2$. The sixth $K$-orbit has internal valency $0$ so contributes nothing to the trace. It follows that the trace of $B$ is $10(2)= 20$. Equating our two computations now yields $20 = 7a - 42$, an obvious contradiction since $a$ must be integral. We conclude that a putative Conway $99$-graph cannot admit an automorphism of order 14. 
\end{proof}

\begin{prop}\label{2 divides}
$|G|$ is not divisible by $14$. 
\end{prop}

\begin{proof}
Suppose 14 divides $|G|$. Then by~\cite{Makhnev} one has $|G|\in \{14,42\}$. Recall from~\cite{Makhnev2} that $[O(G),t]=1$, where $O(G)$ is the maximal odd order normal subgroup of $G$ and $t\in G$ is an involution. This alone rules out $D_{14}$, $D_{42}$, $D_{14}\times \mathbb Z_3$ and $Frob(42)$ as possible isomorphism types of $G$. 
The only remaining possibilities are 
$\mathbb Z_{14}$, $\mathbb Z_{42}$, 
${\mathbb Z_7}\times S_3$ and 
${Frob}(21)\times \mathbb Z_2$. 
However, each of these groups has an element of order 14 so is ruled out by Theorem~\ref{no 14}.
\end{proof}

\begin{coro}\label{two groups}
If $\,2$ divides $|G|$, then $|G|$ divides $6$.
\end{coro}

\begin{proof} 
This is an immediate consequence of 
Proposition~\ref{2 divides} and~\cite[Corollary 2.4]{Makhnev2}. 
\end{proof}

\section{Consequences of divisibility by 7}\label{cyclic or Frobenius} 

In this section, we prove $G\cong\mathbb Z_7$ under the assumption that $|G|$ is divisible by $7$. Our first step in this direction is to show that $|G|$ must divide $21$. This leads to just two possible isomorphism types for $G$, namely $\mathbb Z_7$ and the Frobenius group $Frob(21)$. Ultimately, we rule out $Frob(21)$ to obtain our desired result. 
 
\begin{lemm}\label{P7 normal}
Suppose $7$ divides $|G|$, and let $P_7$ be a Sylow $7$-subgroup of $G$. Then $P_7$ is normal in $G$. 
\end{lemm}

\begin{proof}
By~\cite{Makhnev} and Proposition~\ref{2 divides}, $|G|$ must divide $3^3\cdot 7\cdot 11$. By Sylow's Theorem, the number $n_7$ of Sylow $7$-subgroups must satisfy $n_7=[G\!:\!N_G(P_7)]$ and \mbox{$n_7\equiv 1\!\pmod{7}$}. 
It is straightforward to deduce that 
$P_7\trianglelefteq G$ for all orders of $G$ except possibly $3^2\cdot 7\cdot 11$ and $3^3\cdot 7\cdot 11$.
However, in these two cases one has $P_{11}\trianglelefteq G$ where $P_{11}$ is a Sylow 11-subgroup of $G$. As $P_7$ does not embed in 
${\rm Aut}(P_{11})\cong \mathbb Z_{10}$ it follows that $[P_7,P_{11}]=1$. But then $P_{11}\le N_G(P_7)$ whence $n_7=[G\!:\!N_G(P_7)]\in\{1,9,27\}$. We now conclude from the congruence $n_7\equiv 1\!\pmod{7}$ that $n_7=1$. Hence $P_7\trianglelefteq G$ as claimed.
\end{proof}
 
\begin{prop}\label{if 7 divides}
Divisibility by $7$ implies $|G|$ divides $21$.
\end{prop}
 
\begin{proof}
By our labeling scheme in Section~\ref{prelims} and Remark~\ref{|s|=7}, it is clear that 
$P_7=\langle s\rangle$ where 
$s=(1L,2L,\dots ,7L)\,(1R,2R,\dots ,7R)$. Furthermore, as $\langle s\rangle\trianglelefteq G$ by Lemma 
\ref{P7 normal}, we have that $G$ embeds in the normalizer $N_S(\langle s\rangle)$ where 
$S={\rm Sym}(\Gamma_1)\cong S_{14}$. 
Thus the order of $G$ must simultaneously divide 
$|N_S(\langle s\rangle)|=2^2\cdot 3\cdot 7^2$ and $3^3\cdot 7\cdot 11$. Obviously this implies $|G|$ divides $21$. 
\end{proof}
 
\begin{coro}\label{two groups2}
 If $7$ divides $|G|$, then $G$ is isomorphic to either 
 $\mathbb Z_7$ or $Frob(21)$.
\end{coro}

\begin{proof} 
By Proposition~\ref{if 7 divides}, $|G|\in\{7,21\}$. There are only two groups of order 21 up to isomorphism, namely $\mathbb Z_{21}$ and $Frob(21)$. 
But 3 does not divide the order of the centralizer 
$C_S(s)=(\langle s_L\rangle\times \langle s_R\rangle)\rtimes\langle t \rangle\cong 
(\mathbb Z_7\times \mathbb Z_7)\rtimes \mathbb Z_2$ where $S={\rm Sym}(\Gamma_1)$. 
Thus~$G$ cannot contain an element of order 21, 
\ie $G\not\cong\mathbb Z_{21}$. 
\end{proof}
 
In what follows, we gather detailed information about a putative Conway 99-graph~$\Gamma$ under the assumption that $G\cong Frob(21)$. 
Throughout, we assume $G=\langle s,r\rangle$ with 
$|r|=3$. Since $r$ normalizes $\langle s\rangle$, it is clear that $r$ fixes the root vertex $x$ whence 
$\Gamma_1$ and $\Gamma_2$ are $G$-invariant. Still, we have yet to pin down the precise structure of $r$. This is remedied below. 

\begin{lemm}\label{r defined}
With notation as above, we may assume 
\[
r=(1L, 2L, 4L)\,(3L, 6L, 5L)\,(1R, 2R, 4R)\,(3R, 6R, 5R).\] 
\end{lemm}

\begin{proof}
It is immediate that $r$ is an element in the normalizer $N_S(\langle s\rangle)$ in $S={\rm Sym}(\Gamma_1)$. Applying standard group theoretic arguments, one deduces that there are 98 elements of order 3 in $N_S(\langle s\rangle)$ and these take the form 
\[
\Big(\big((1L, 2L, 4L)(3L, 6L, 5L)\big)^{(s_L)^{i}} \big((1R, 2R, 4R)(3R, 6R, 5R)\big)^{(s_R)^{j}}\Big)^{\pm 1}
\]
where $0\le i,j\le 6$. However, it is easily verified that such an element is an automorphism of $\Gamma_1$ if and only if $i=j$. Thus, the 14 elements of order 3 in $G$ are precisely 
\[
\big((1L, 2L, 4L)\,(3L, 6L, 5L)\,(1R, 2R, 4R)\,(3R, 6R, 5R)\big)^{\pm\, s^{\,i}}
\]
from which the result follows.
\end{proof}
 
\begin{lemm}\label{1, 7^2, 21^4}
The $G$-orbit structure on $\Gamma$ is 
$[1, 7^2, 21^4]$. 
\end{lemm}

\begin{proof}
As 21 does not divide 14, the two $\langle s\rangle$-orbits on $\Gamma_1$ cannot fuse under the action of $G$. Thus the $G$-orbit structure on $\Gamma_1$ is 
$[7^2]$. 

Now let $\mathcal O$ be an arbitrary $G$-orbit on $\Gamma_2$. We claim $G$ acts regularly on 
$\mathcal O$ from which the desired result will follow. To this end, suppose $ijXY \in \mathcal O$ is fixed by some $g\in G$, where $X, Y \in \{L, R\}$. It follows that $|g|=3$, since $s$ has no fixed points in $\Gamma_2$. As $\lambda =1$, we have $i\ne j$. 
Thus since $\mu=2$, $g$ must either fix or interchange the two $\Gamma_1$-neighbors of $ijXY$, 
\ie $\{iX^g,jY^g\}=\{iX,jY\}$. But as $G$ has odd order, these vertices must be fixed by $g$, that is, $iX^g=iX$ and $jY^g=jY$. As~$g$ preserves adjacency, we must also have $(iX^C)^g=iX^C$ and $(jY^C)^g=jY^C$ where $\{X,X^C\}=\{Y,Y^C\}=\{L,R\}$.
In every instance, we obtain $iL^g=iL$ and $jL^g=jL$. By transitivity of $\langle s\rangle$ on the orbit containing $iL$, we get $jL=iL^z$ for some $z\in \langle s\rangle$. This gives $jL^{[z,g]}=iL^{g^{\,-1}zg}=iL^{zg}=jL^{g}=jL$, \ie $jL$ is a fixed point of $[z,g]$. 
But $[z,g]\in \langle s\rangle$ because $\langle s\rangle$ is normal in $G$. This implies $[z,g]=1$, a contradiction since $G$ is nonabelian.
We conclude that the $G$-orbit structure on 
$\Gamma_2$ is $[21^4]$ as claimed.
\end{proof}

\begin{rema}
Now that we understand the orbit structure of $G$ on $\Gamma_2$, it is trivial to determine a corresponding set of orbit representatives, viz.\ $\{12LL, 12RR, 12LR, 12RL\}$. For brevity we shall denote these orbits as $LL, RR, LR, RL$, respectively. We indicate the corresponding $G$-orbit partition of $\,\Gamma_2$ in Figure~\ref{general OP21}.
\end{rema}

\begin{figure}

\begin{tikzpicture}
\draw[thick] (0,0) circle (6mm) node at (0,0) {\Large $b_1$} node at (0,-.9) {$LL$};
\draw [thick](3,0) circle (6mm) node at (3,0) {\Large $b_2$} node at (3,-.9) {$RR$};
\draw [thick](0,3) circle (6mm) node at (0,3) {\Large $b_3$} node at (0,3.9) {$LR$};
\draw[thick] (3,3) circle (6mm) node at (3,3) {\Large $b_4$} node at (3,3.9) {$RL$};
 
\draw (.6,0)--(2.4,0) node at (1.54,-.27) {\large $b_{12}$};
\draw (.6,3)--(2.4,3) node at (-.33,1.5) {\large $b_{13}$};
\draw (0,.6)--(0,2.4) node at (3.36,1.5) {\large $b_{24}$};
\draw (3,.6)--(3,2.4) node at (1.5,3.27) {\large $b_{34}$};
\draw (.4,.4)--(2.6,2.6) node at (.64,1.1) {\large $b_{14}$};
\draw (.4,2.6)--(2.6,.4) node at (2.34,1.1) {\large $b_{23}$};

\end{tikzpicture}
\caption{General form of a $G$-orbit partition of 
$\Gamma_2$}\label{general OP21}
\end{figure}

For $ijXY\in\Gamma_2$, we refer to $iX$ and $jY$ as its coordinates. In the result that follows, we demonstrate a manner in which the coordinates of the 12 $\Gamma_2$-neighbors of a fixed vertex in 
$\Gamma_2$ are balanced. 

\begin{lemm}\label{balance} Let $ijXY$ be a fixed but arbitrary vertex in $\Gamma_2$ and consider collectively the $24$ coordinates appearing among its $12$ $\Gamma_2$-neighbors. Then the following hold: 
\begin{enumerate}
\item 
Each of $iL,iR,jL,jR$ appears exactly once.
\item 
Each of $kL, kR$ appears exactly twice for each 
$k\ne i,j$. 
\item 
Each of $L$ and $R$ appears $12$ times, 
\ie the $\Gamma_2$-neighbors of a vertex in
$\Gamma_2$ are $L/R$-balanced.
\end{enumerate} 
\end{lemm}

\begin{proof}
As in Lemmas~\ref{orbits} and~\ref{1, 7^2, 21^4} we let 
$\{X,X^C\}=\{Y,Y^C\}=\{L,R\}$. We must show that
each of $iX,jY$ and $iX^C,jY^C$ occurs exactly once as coordinates of the $\Gamma_2$-neighbors of $ijXY$. We treat the pair $iX,jY$ first. 
Since $\lambda=1$ and $ijXY$ is adjacent to 
$iX\in \Gamma_1$, there is a unique vertex in 
$\Gamma_2$ which is their common neighbor. Evidently this vertex is of the form $i\ell XW$ for some $W\in\{L,R\}$ with $\ell W\ne jY$. 
Thus $iX$ occurs exactly once as a coordinate of a 
$\Gamma_2$-neighbor of $ijXY$ and by a symmetric argument the same holds true for $jY$. 

We next treat the pair $iX^C,jY^C$. 
Observe that since $\lambda=1$, we have that $ijXY$ and $iX^C$ are nonadjacent. Since $\mu=2$, the vertices $ijXY$ and $iX^C$ must have a unique common neighbor $i\ell X^CW\in\Gamma_2$ with $iX$ being their second common neighbor. Thus $iX^C$ occurs exactly once as a coordinate of a 
$\Gamma_2$-neighbor of $ijXY$ with a similar result holding for $jY^C$. As $\{iX,iX^C\}=\{iL,iR\}$ and 
$\{jY,jY^C\}=\{jL,jR\}$, assertion (1) is proved. 

Now let $k\ne i,j$ and $W\in\{L,R\}$. Since $\mu=2$ and $kW$ is nonadjacent to $ijXY$, there are exactly two vertices in $\Gamma_2$ that are common neighbors of $kW$ and $ijXY$. Obviously, $kW$ appears as a coordinate in each such neighbor, so twice in total. Thus assertion (2) is proved. 
Finally, observe that (3) follows directly from (1) and (2). 
\end{proof}

As a consequence of Lemma~\ref{balance}(3), we have the following.

\begin{lemm}
With notation as in Figure~\ref{general OP21}, $b_1=b_{12}=b_2$, $\,b_{13}=b_{23}$, 
and $b_{14}=b_{24}$.\label{more relations} 
\end{lemm}

\begin{proof}
We apply Lemma~\ref{balance}(3) to vertices in $LL$, $RR$, $LR$, $RL$ in that order. Prefatory to this, note that the coordinates of vertices in $LR$ and $RL$ have a natural $L/R$-balance built into them. This means we may safely ignore edges that adjoin any vertex in $\Gamma_2$ to vertices in either of these two orbits when exploiting the property of balance. 

Let $u$ be a fixed vertex in $LL$. As $u$ has valency $b_1$ in $LL$, it must have $b_1$ neighbors in $RR$ in order to restore $L/R$-balance. This proves 
$b_{12}=b_1$. By a similar argument based on $RR$, we have $b_{12}=b_2$. 
Now let $u$ be a vertex in $LR$. Clearly, every neighbor of $u$ in $LL$ must be reciprocated by a neighbor in $RR$ to maintain balance. This proves $b_{13} = b_{23}$. Applying this latter argument to vertices in $RL$ yields
$b_{14}=b_{24}$. \end{proof}

 
To further narrow down possible orbit valencies, we adopt the approach used in Section 
\ref{no order 14 auts}. That is to say, we analyze
 2-paths between vertices from pairs of orbits. Let $u$ be a fixed vertex in the orbit $LL$. We wish to count in 
two ways the cardinality of the set $S$ given by
\[
S = \{uvw: \text{$uvw$ is a 2-path with } w \in RR\}.
\] 
Observe that for each of the $b_1$ neighbors of $u$ in $RR$ there is a unique 2-path in $S$ (since $\lambda =1$), while for each of the $21-b_1$ non-neighbors of $u$ in $RR$ there are two such paths (since $\mu=2$). 
This gives $|S|=b_1\cdot 1 + (21-b_1)\cdot 2=42-b_1$. 

We next condition our count on the location of the intermediate vertex $v$. Here we rely heavily on Lemma~\ref{more relations}. 

If $v$ lies in $LL$, there are $b_1$ choices for $v$ followed by 
$b_1$ independent choices for~$w$. This gives a total of $b_1^2$ paths in $S$ assuming the intermediate vertex $v$ is in $LL$. Note that for $v$ in $RR$ the count is identical, \ie there are $b_1^2$ paths in $S$ assuming $v$ is in~$RR$.

Now suppose $v$ is in the orbit $LR$. Since $u$ has $b_{13}$ neighbors in $LR$ and each such vertex has $b_{13}$ neighbors in $RR$, the total number of paths in this case is $b_{13}^2$. By a similar argument the number of paths with $v$ in $RL$ is $b_{14}^2$. Finally, we observe that there is no 2-path with intermediate vertex $v$ in $\Gamma_1$. Indeed, this would require that $v=iL= jR$ for some $i,j$, which is absurd. Thus we obtain $|S|=2b_1^2 + b_{13}^2+b_{14}^2$. 
Equating the above two expressions for $|S|$ yields
\begin{equation}\label{eqn 3}
 42-b_1-2b_1^2 = b_{13}^2+b_{14}^2.
 \end{equation}

The only integral solution to equation~\ref{eqn 3} is $b_1=2$, $b_{13}=b_{14}=4$ which, by virtue of Lemma~\ref{more relations}, narrows down the orbit valencies to the ones indicated in 
Figure~\ref{updated relations}. 
 
 
\begin{figure}
\begin{tikzpicture}
\draw[thick] (0,0) circle (6mm) node at (0,0) {\Large $2$} node at (0,-.9) {$LL$};
\draw [thick](3,0) circle (6mm) node at (3,0) {\Large $2$} node at (3,-.9) {$RR$};
\draw [thick](0,3) circle (6mm) node at (0,3) {\Large $b_3$} node at (0,3.9) {$LR$};
\draw[thick] (3,3) circle (6mm) node at (3,3) {\Large $b_3$} node at (3,3.9) {$RL$};
 
\draw (.6,0)--(2.4,0) node at (1.5,-.27) {\large $2$};
\draw (.6,3)--(2.4,3) node at (-.2,1.5) {\large $4$};
\draw (0,.6)--(0,2.4) node at (3.25,1.5) {\large $4$};
\draw (3,.6)--(3,2.4) node at (1.5,3.25) {\large $4- b_{3}$};
\draw (.4,.4)--(2.6,2.6) node at (.68,1.1) {\large $4$};
\draw (.4,2.6)--(2.6,.4) node at (2.32,1.1) {\large $4$};
 
 
\end{tikzpicture}
 \caption{Narrowing down the valencies of a $G$-orbit partition of $\Gamma_2$}\label{updated relations}
\end{figure}


We are nearly at the point of determining the unique orbit structure of $G\cong Frob(21)$ acting on the second subconstituent $\Gamma_2$ of a putative Conway 99-graph $\Gamma$. To complete the process, we count in two ways the number of 2-paths originating at a fixed vertex $u$ in $LR$ and ending at some vertex in $RL$. 


On one hand, there are $4-b_{3}$ edges from $u$ to some vertex $w$ in $RL$, and in each case there is a unique 2-path $uvw$ since $\lambda =1$. Similarly, for each of the $21-(4-b_{3})=17+b_3$ vertices in $RL$ nonadjacent to $v$ there are two distinct 2-paths of required form. This gives a total of $(4-b_{3})\cdot1+ (17+b_{3})\cdot 2=38+b_{3}$ such 2-paths. 

We next focus our count on the location of the intermediate vertex $v$. If $v$ is in either of $LL$ or $RR$, there are $4\cdot 4=16$ such 2-paths, while if $v$ is in either of $LR$ or $RL$ there are 
 $(4-b_3)b_3 = 4b_3 - b_3^{\,2}$ such 2-paths. 
 Lastly, we consider $v\in \Gamma_1$. 
Since $u$ is in $LR$, it has coordinates $ijLR$ for some $i\ne j$ whence its two 
$\Gamma_1$-neighbors are $iL$ and $jR$. Note that the 12 neighbors of $iL$ in $\Gamma_2$ are of the form $ikLL$ and $ikLR$ ($=kiRL$) where $k\ne i$. Thus six neighbors of $iL$ lie in $LR\cup RL$. However, these neighbors are divided evenly between $LR$ and $RL$, since $ikLR$ is in $LR$ if and only if $ikRL$ is in $RL$.
Thus $iL$ has three neighbors in $LR$ and by a symmetric argument this holds true for $jR$. 
This gives six more 2-paths starting at $u$ and terminating at some vertex in $RL$. Thus the total number of such 2-paths is 
$2(16)+ 2(4b_3-b_3^2) +6=38+8b_3-2b_3^{\,2}$ 
via this second count.

Equating the two counts yields 
$38+b_3=38+8b_3-2b_3^{\,2}$ which simplifies to $2b_3^{\,2}-7b_3=0$.
Clearly, the only integral solution is $b_3=0$ which gives us the following.

\begin{prop}\label{unique OP21}
Assume $G\cong Frob(21)$. Then the only feasible $G$-orbit partition of $\Gamma_2$ is the one depicted in Figure~\ref{final OP21}. 
\end{prop}
 
\begin{figure}

\begin{tikzpicture}
\draw[thick] (0,0) circle (6mm) node at (0,0) {\Large $2$} node at (0,-.9) {$LL$};
\draw [thick](3,0) circle (6mm) node at (3,0) {\Large $2$} node at (3,-.9) {$RR$};
\draw [thick](0,3) circle (6mm) node at (0,3) {\Large $0$} node at (0,3.9) {$LR$};
\draw[thick] (3,3) circle (6mm) node at (3,3) {\Large $0$} node at (3,3.9) {$RL$};
 
\draw (.6,0)--(2.4,0) node at (1.5,.25) {\large $2$};
\draw (.6,3)--(2.4,3) node at (-.2,1.5) {\large $4$};
\draw (0,.6)--(0,2.4) node at (3.25,1.5) {\large $4$};
\draw (3,.6)--(3,2.4) node at (1.5,3.25) {\large $4$};
\draw (.4,.4)--(2.6,2.6) node at (.68,1.1) {\large $4$};
\draw (.4,2.6)--(2.6,.4) node at (2.32,1.1) {\large $4$};
 
 \end{tikzpicture}
 \caption{The unique $G$-orbit partition of $\Gamma_2$ where $G\cong Frob(21)$}\label{final OP21}
\end{figure}

\begin{rema}
Observe that each of $LR\cup\{x\}$ and $RL\cup\{x\}$ is a coclique of size~22. This is in fact the largest independence number allowable by the Hoffman Ratio Bound (aka Hoffman--Delsarte inequality), see~\cite{haemers}. Further note from Figure 
\ref{final OP21} that every vertex outside of 
$LR\cup\{x\}$ (\resp $RL\cup\{x\}$) has precisely four neighbors in $LR\cup\{x\}$ (\resp $RL\cup\{x\}$).
\end{rema}

Let us call a 3-cycle $u_1u_2u_3$
 \emph{of type~$(X_1Y_1,X_2Y_2,X_3Y_3)$} provided 
$u_i$ is in the orbit $X_iY_i$ where $X_i,Y_i\in\{L,R\}$ for $1\le i\le 3$. 

\begin{prop}\label{interior 3-cycles}
Each of $LL$ and $RR$ consists of seven 
vertex-disjoint $3$-cycles. 
\end{prop}

\begin{proof}
By transitivity of $G$ on its orbits, each orbit of 3-cycles in $\Gamma_2$ has size 21 with the exception being 
3-cycles of type~$(LL,LL,LL)$ or $(RR,RR,RR)$. Indeed, orbits of 3-cycles of these two types would have to be of size 7 since the 3-cycles remain internal to their respective orbits. As there are a total of 140 3-cycles in $\Gamma_2$ (\cf Lemma~\ref{140 triangles}), their division into $n_7$ orbits of size 7 and $n_{21}$ orbits of size 21 must satisfy
$7n_7+21n_{21}=140$. Moreover, $0\le n_7\le 2$ since each of $LL$ and $RR$ has internal valency 2. Since the only integral solution to the above is $n_7=2$ and $n_{21}=6$, each of $LL$ and $RR$ must contain seven vertex-disjoint 3-cycles as claimed.
\end{proof}

Our goal is to prove the $G$-orbit partition depicted in Figure~\ref{final OP21} cannot exist. To this end we consider the finer $\langle s\rangle$-orbit partition of 
$\Gamma_2$.
(Recall that $G=\langle s,r\rangle$ with $r$ as defined in Lemma~\ref{r defined}.) Observe that each of 
$LL, RR, LR, RL$ decomposes into three 
$\langle s\rangle$-orbits. We denote these latter 12 orbits by $(XY)_i$ where $XY\in \{LL, RR, LR, RL\}$ and $1\le i\le 3$. Here, we are obviously requiring that
$(XY)_i\subset XY$ for all $i$. 

We will again apply the method we used in Section~\ref{no order 14 auts} to determine orbit valencies $b_{ij}$ by counting 2-paths, only this time our focus will be on the aforementioned $\langle s\rangle$-orbit partition. 
Thus, we begin by counting in two ways the cardinality of the set
\[
S = \{uvw: \text{$uvw$ is a 2-path with } w \in (XY)_i\},
\] 
where $u$ is a fixed vertex in $(XY)_i$.
For each of the $b_{i}$ neighbors of $u$ in $(XY)_i$ there is a unique $2$-path from $u$ to $w$ 
(since $\lambda=1$). Similarly, for each of the $6-b_{i}$ non-neighbors of $u$ in $(XY)_i$ there are two 2-paths from $u$ to $w$ (since $\mu=2$). This gives 
$|S|= b_{i} \cdot 1 + (6 - b_{i}) \cdot 2 = 12 - b_{i}$. 

We next condition our count on the location of the intermediate vertex $v$. For $v$ in~$(XY)_j$ there are $b_{ij}\,(b_{ij}-1)$ such 2-paths. Additionally, 
if $XY\in\{LL,RR\}$ then $u$ has exactly two neighbors in $\Gamma_1$ each of which has a unique neighbor $w\in (XY)_i\setminus \{u\}$. 
In contrast, if $XY\in\{LR,RL\}$ there are no such neighbors in $\Gamma_1$. 
This gives
$|S|= \sum_{j= 1}^{12} b_{ij} \,(b_{ij} - 1) + c$, 
where $c = 2$ if $XY\in\{LL,RR\}$ and $c = 0$ if $u$ if $XY\in\{LR,RL\}$. 
Equating these two expressions for $|S|$, we obtain 
$12 - b_{i} = \sum_{j = 1}^{12} b_{ij} \,(b_{ij} - 1) + c$. But due to the fact that $\sum_{i=1}^{12} b_{ij}=12$, this simplifies to
\begin{equation}\label{eqn 4}
24 - c - \big(b_{i}^{\,2} + b_{i} \big) = \sum_{j \ne i} b_{ij}^{\,2}
\end{equation}
We divide our analysis into cases based on an assumed value for $b_{i}$. Once a choice of~$b_i$ is made, we find all ways of expressing 
$24 - c - \big(b_{i}^{\,2 }+ b_{i} \big)$ as a sum of eleven squares while maintaining the valency requirement 
$\sum_{i=1}^{12} b_{ij}=12$. 

Note that $b_i\le 4$ since otherwise the value of 
$24 - c - \big(b_{i}^{\,2} + b_{i} \big)$ would be negative. 
It is also clear by the Handshake Lemma that every internal valency $b_i$ must be even because every 
$\langle s\rangle$-orbit is of size 7. In addition, we cannot have $b_i= 4$ since this we would lead to a violation of $\lambda = 1$. Thus $b_i\in\{0,2\}$. 

However, recall from the coarser $G$-orbit partition of 
$\Gamma_2$ that the internal valency of $LR$ and $RL$ is $0$ while that of $LL$ and $RR$ is $2$. By Proposition~\ref{interior 3-cycles} the internal valencies of $(LL)_i$ and $(RR)_i$ must be zero since no 3-cycle in $LL$ or $RR$ can lie wholly within an 
$\langle s\rangle$-orbit. We conclude that the internal degree of every $\langle s\rangle$-orbit is zero.
 
We are now prepared to list all feasible solution to equation~\ref{eqn 4}. Verification of the list is straightforward, so left to the reader. 

\begin{lemm}\label{orbit type3} 
Solutions for orbits of type~$(LL)_i$ or $(RR)_i$ are listed as follows: 
\begin{enumerate} 
\item[(a)] 
$\big(0,\big\{3, 2, 2, 1, 1, 1, 1, 1, 0, 0, 0\big\}\big)$ 
\item[(b)] 
$\big(0,\big\{2, 2, 2, 2, 2, 1, 1, 0, 0, 0, 0\big\}\big)$ 
\end{enumerate}
Solutions for orbits of type~$LR_i$ or $RL_i$ are listed as follows: 
\begin{enumerate}
\item[(c)] 
$\big(0,\big\{4, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0\big\}\big)$
\item[(d)] 
$\big(0,\big\{3, 3, 1, 1, 1, 1, 1, 1, 0, 0, 0\big\}\big)$ 
\item[(e)] 
$\big(0,\big\{3, 2, 2, 2, 1, 1, 1, 0, 0, 0, 0\big\}\big)$ 
\item[(f)] 
$\big(0,\big\{2, 2, 2, 2, 2, 2, 0, 0, 0, 0, 0\big\}\big)$ 
\end{enumerate}
\end{lemm}

We can fairly quickly eliminate sequences (c) and (d) from this list. According to the $G$-orbit partition of 
 $\Gamma_2$ depicted in Figure~\ref{final OP21}, the orbit $LR$ has external orbit valency of $4$ to each of $LL, RR, RL$. 
Assume now that sequence (c) corresponds to an
$\langle s\rangle$-orbit inside $LR$, say $(LR)_1$. 
All three zeros in sequence (c) must be used in adjoining $(LR)_1$ to $(LR)_1, (LR)_2, (LR)_3$.
This leaves the subsequence $\{4,1,1,1,1,1,1,1,1\}$ to account for all external orbit valencies of $(LR)_1$. But as there is no 3-element subset of this subsequence that sums to 4, there is no possible way to adjoin 
$(LR)_1$ to any $\langle s\rangle$-orbit outside of $LR$. (Of course the argument for $RL$ is entirely symmetric.) This eliminates sequence (c) as a possible solution. 
 
The argument for eliminating sequence (d) is similar. 
As above, all three zeros in sequence (d) are used to adjoin $(LR)_1$ to $(LR)_1, (LR)_2, (LR)_3$. This leaves the subsequence $\{3,3,1,1,1,1,1,1\}$ to account for all external orbit valencies of $(LR)_1$. But as before, there is no 3-element subset of this subsequence that sums to 4. Hence, sequence (d) is ruled out as well.
 
For future reference, it is convenient to designate the surviving solution by type, \eg
 \[
 \begin{array}{rlrl}
{\rm I}. & \big(0,\big\{3, 2, 2, 1, 1, 1, 1, 1, 0, 0, 0\big\}\big) &\; {\rm II}. & \big(0,\big\{2, 2, 2, 2, 2, 1, 1, 0, 0, 0, 0\big\}\big)\\
{\rm III}. & \big(0,\big\{3, 2, 2, 2, 1, 1, 1, 0, 0, 0, 0\big\}\big) &\; {\rm IV}. & \big(0,\big\{2, 2, 2, 2, 2, 2, 0, 0, 0, 0, 0\big\}\big)
\end{array}
\]

In what follows it behooves us to double count 2-paths of the form $uvw$ where $u$ is a fixed vertex in $(XY)_i$ and $w$ is an arbitrary vertex in $(UV)_j$, $UV\ne XY$. For convenience we resort to the notation 
$\mathcal O_i$ ($1\le i\le 12$) to represent the 
$\langle s\rangle$-orbits $(LL)_1, (LL)_2, (LL)_3, (RR)_1, (RR)_2, (RR)_3, (LR)_1, (LR)_2, 
(LR)_3, (RL)_1, (RL)_2, (RL)_3$ in that order. 

For the first count, we observe that $u\in\mathcal O_i$ has $b_{ij}$ neighbors in $\mathcal{O}_j$, and for
each such neighbor $w\in \mathcal O_j$ there is a unique 2-path $uvw$ (since $\lambda=1$). 
Similarly, $u$ has $7 - b_{ij}$ non-neighbors in 
$\mathcal{O}_j$, and for each such non-neighbor 
$w\in \mathcal O_j$ there are exactly two 2-paths of the form $uvw$ (since $\mu=2$). Thus, there are a total of $b_{ij} \cdot 1 + (7 - b_{ij}) \cdot 2 = 14 - b_{ij}$ 
$2$-paths of required form. 
 
For our second count, we focus on the location of the intermediate vertex $v$ in each 2-path $uvw$. Here $v$ can occur in any in the twelve orbits of $\Gamma_2$ as well as in $\Gamma_1$. In the case of orbits in 
$\Gamma_2$, there are $b_{ik}$ choices for 
$v\in \mathcal O_k$ and for each such choice there are $b_{kj}$ choices for $w\in \mathcal O_j$. This gives $b_{ik}b_{kj}$ 2-paths of this type. 
Finally, we denote by $c_{ij}$ the number of 2-paths that have intermediate vertex $v$ in $\Gamma_1$.

Equating these two counts yields 
\begin{equation}\label{eqn 5}
14 - b_{ij} = \sum_{k = 1}^{12} b_{ik}b_{kj}+c_{ij}.
\end{equation}

Let $C$ be the $12 \times 12$ quotient matrix corresponding to the action of 
$\langle s\rangle\cong\mathbb Z_7$ restricted to 
$\Gamma_2$.
Consistent with the above, we order the rows and columns of $C$ by~$(LL)_1$, $(LL)_2$, $(LL)_3$, $(RR)_1$, $(RR)_2$, $(RR)_3$, $(LR)_1$, $(LR)_2$, $(LR)_3$, $(RL)_1$, $(RL)_2$, $(RL_3)$.

\begin{prop}\label{no I or III}
Under the assumption that $G\cong Frob(21)$, there is no $\langle s\rangle$-orbit $LL_i$ or $RR_j$ 
that has valency $3$. 
\end{prop}

\begin{proof}
By way of contradiction, suppose $3$ occurs as a valency of $(LL)_i$ or $(RR)_j$ in the 
$\langle s\rangle$-orbit partition of $\Gamma_2$. 
This implies sequence I coincides with the valencies of such an orbit. 
Since the valency between the $G$-orbits $LL$ and $RR$ is $2$, the valency~3 in sequence I must be used to adjoin to an $\langle s\rangle$-orbit within $LR$ or $RL$. 
Since all $\langle s\rangle$-orbits of $\Gamma_2$ are of equal size, it follows that all external valencies are bidirectional. Thus without loss of generality, sequence III coincides with the valencies of the 
$\langle s\rangle$-orbit~$(LR)_1$.

Our next observation depends solely on the assumption that $\langle s\rangle$ is a proper subgroup of 
$G=\langle s,r\rangle\cong Frob(21)$. Since $r$ permutes the three $\langle s\rangle$-orbits within a given $G$-orbit, we have that $(XY)_1,(XY)_2,(XY)_3$ must share the same set of valencies for every 
$XY\in\{LL,RR,LR,RL\}$. 
Recalling that the $G$-orbits $LL$ and $LR$ are adjoined by a valency of $4$, we may further assume that each $(LL)_i$ adjoins $(LR)_1,(LR)_2,(LR)_3$ with valencies $3,1,0$ in some order. This leads to the following $3 \times 3$ submatrix of $C$ corresponding to the orbit valencies adjoining each of $(LL)_i$ to each of $(LR)_j$, where $1\le i,j\le 3$. 
\begin{equation*}
	\begin{bmatrix} 
			\;3\; & \;1\; & \;0\; \\ 
			1 & 0 & 3 \\ 
			0 & 3 & 1 
	\end{bmatrix}
\end{equation*}

We next consider the four $3 \times 3$ submatrices that occur as block diagonals of $C$. Since $LR$ and $RL$ are independent sets, the block corresponding to the $\langle s\rangle$-orbits within each of them is the $3\times 3$ zero matrix. 
Next recall from Lemma~\ref{interior 3-cycles} that each of $LL$ and $RR$ consists of seven disjoint 3-cycles. As the vertices of such 3-cycles must occupy different $\langle s\rangle$-orbits, we obtain the following $3\times 3$ submatrix for the $\langle s\rangle$-orbits within $LL$, as well as for those within $RR$. 
\begin{equation*}
	\begin{bmatrix} 
		\;0\; & \;1\; &\; 1\; \\ 
		1 & 0 & 1 \\ 
		1 & 1 & 0 
	\end{bmatrix}
\end{equation*}
Finally, recall that in the $G$-orbit partition, the orbit $LL$ is adjoined to $RR$ and $RL$ with orbit valencies of 2 and 4, respectively. This means we must arrange the unused portion of sequence I, namely $2,2,1,1,0,0$, into two groups of three that sum to 2 and 4. Here there are two possibilities: $\{\{1, 1, 0\},\{2,2,0\}\}$ and 
$\{\{2, 0,0\}, \{2, 1,1\}\}$. We indicate below the two matrices that reflect these possibilities in addition to what we have deduced so far.
{\footnotesize \begin{equation*} C_1 = 
 \begin{bmatrix} 
\;0 \;& \;1\; & \;1\; & \;0\; & \;1\; & \;1\; & \;3\; & \;1\; & \;0\; & \;0\; & \;2\; & \;2\; \\ 
1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & 3 & 2 & 0 & 2 \\
1 & 1 & 0 & 1 & 1 & 0 & 0 & 3 & 1 & 2 & 2 & 0 \\
		0 & 1 & 1 & 0 & 1 & 1 & & & & & & \\
		1 & 0 & 1 & 1 & 0 & 1 & & & & & & \\
		1 & 1 & 0 & 1 & 1 & 0 & & & & & & \\
		3 & 1 & 0 & & & & 0 & 0 & 0 & & & \\
		1 & 0 & 3 & & & & 0 & 0 & 0 & & & \\
		0 & 3 & 1 & & & & 0 & 0 & 0 & & & \\
		0 & 2 & 2 & & & & & & & 0 & 0 & 0 \\
		2 & 0 & 2 & & & & & & & 0 & 0 & 0 \\
		2 & 2 & 0 & & & & & & & 0 & 0 & 0 
	\end{bmatrix}\;\;
	C_2 = 
 \begin{bmatrix} 
	\;0\; & \;1\; & \;1\; & \;2\; & \;0\; & \;0\; & \;3\; & \;1\; 	& \;0\; & \;2\; & \;1\; & \;1\; \\ 
	1 & 0 & 1 & 0 & 2 & 0 & 1 & 0 & 3 & 1 & 2 & 1 \\
	1 & 1 & 0 & 0 & 0 & 2 & 0 & 3 & 1 & 1 & 1 & 2 \\
	2 & 0 & 0 & 0 & 1 & 1 & & & & & & \\
	0 & 2 & 0 & 1 & 0 & 1 & & & & & & \\
	0 & 0 & 2 & 1 & 1 & 0 & & & & & & \\
	3 & 1 & 0 & & & & 0 & 0 & 0 & & & \\
	1 & 0 & 3 & & & & 0 & 0 & 0 & & & \\
	0 & 3 & 1 & & & & 0 & 0 & 0 & & & \\
	2 & 1 & 1 & & & & & & & 0 & 0 & 0 \\
	1 & 2 & 1 & & & & & & & 0 & 0 & 0 \\
	1 & 1 & 2 & & & & & & & 0 & 0 & 0 
\end{bmatrix}
\end{equation*}}

We first consider the matrix $C_1$. We claim this matrix cannot be completed to the desired matrix $C$. To prove this, we attempt to complete its fourth row, \ie the row corresponding to the $\langle s\rangle$-orbit 
$(RR)_1$.
Note that the value 1 already appears four times in this row, which dictates that the orbit valencies of $(RR)_1$ must coincide with sequence I. 
In particular, the six unfilled entries in this row must match up with those values in sequence I that have yet to be used, viz.\ $3,2,2,1,0,0$. As the external valency that adjoins $RR$ to each of $LR$ and $RL$ is 4, the above six values must be partitioned into two groups of size 3 each summing to 4. Moreover, as the value 3 cannot occur twice in any column, the desired partition is $\{\{2,2,0\}, \{3,1,0\}\}$.
 
We now apply equation~\ref{eqn 5} to the orbits $(LL)_2$ and $(RR)_1$. Note that in this case we have 
$b_{24}=1$ and $c_{24}=0$. 
Further note that the righthand side of equation~\ref{eqn 5} is simply the dot product of the second and fourth rows. For convenience, let us denote by $a,b,c,d,e,f$ the six missing entries of row 4 in that order. Then the dot product in question is $2+a+3c+2d+ 2f$, an even integer due to 
$a,c\in \{0,2\}$. But this is a contradiction since the lefthand side of equation~\ref{eqn 5} is $14-b_{24}=13$ in this case. 
 
We next turn our attention to the matrix $C_2$. Again, we claim there is no compatible way to complete the fourth row. The argument here is a bit more sophisticated than the one provided for $C_1$, but the basic ideas are the same. The only initial difference is that the roles of $\{2,2,0\}$ and $\{3,1,0\}$ have been reversed. 

We first apply equation~\ref{eqn 5} to the first and fourth rows of $C_2$, noting that $b_{14}=2$ and $c_{14}=0$.
Thus the lefthand side of equation~\ref{eqn 5} is $14-2=12$ while the righthand side is $3a+b+2d+e+f$. Thus we obtain 
\[
12= 3a+b+2d+e+f=(a+b+c+d+e+f)+(2a-c+d)=8+2a-c+d
\]
which implies $d$ is even since $c\in\{0,2\}$. Thus $d=0$ because $d\in\{0,1,3\}$.

Our next step is to apply equation~\ref{eqn 5} to the second and fourth rows of $C_2$, noting that 
$b_{24}=0$ and $c_{24}=0$ in this case. Here equation~\ref{eqn 5} yields 
\[
14=4+a+3c+d+2e+f=4+(a+b+c+d+e+f)+
(2c-b+e)=12+(2c-b+e)
\]
which implies $e$ is even since $b\in\{0,2\}$. Thus $e=0$ because $e\in\{0,1,3\}$. Together we now have $d=e=0$ which contradicts the fact that 
$\{d,e,f\}=\{0,1,3\}$. Since neither~$C_1$ nor $C_2$ can be completed to the desired quotient matrix $C$, it follows that~$3$ cannot occur as the valency of any orbit in an $\langle s\rangle$-orbit partition of 
$\Gamma_2$. 
\end{proof}

\begin{prop}\label{no II or IV}
The $G$-orbit partition in Figure~\ref{final OP21} cannot exist. 
\end{prop}

\begin{proof} 
We proceed as in the proof of Proposition~\ref{no I or III} with the additional knowledge that the valencies of all 
$\langle s\rangle$-orbits $(LL)_i$ and $(RR)_j$ coincide with sequence II. 
From this, it is an easy matter to arrive at the 
partially completed matrix $C_3$ below.
\begin{equation*}
C_3=\begin{bmatrix} 
\:0\; & \;1\; & \;1\; & \;2\; & \;0\; & \;0\; & \;0\; & \;2\; & \;2\; & \;0\; & \;2\; & \;2\;\\ 
1 & 0 & 1 & 0 & 2 & 0 & 2 & 0 & 2 & 2 & 0 & 2\\
1 & 1 & 0 & 0 & 0 & 2 & 2 & 2 & 0 & 2 & 2 & 0\\
		2 & 0 & 0 & 0 & 1 & 1 & & & & & & \\
		0 & 2 & 0 & 1 & 0 & 1 & & & & & & \\
		0 & 0 & 2 & 1 & 1 & 0 & & & & & & \\
		0 & 2 & 2 & & & & 0 & 0 & 0 & & & \\
		2 & 0 & 2 & & & & 0 & 0 & 0 & & & \\
		2 & 2 & 0 & & & & 0 & 0 & 0 & & & \\
		0 & 2 & 2 & & & & & & & 0 & 0 & 0 \\
		2 & 0 & 2 & & & & & & & 0 & 0 & 0 \\
		2 & 2 & 0 & & & & & & & 0 & 0 & 0 
	\end{bmatrix}
\end{equation*}
Just as in the proof of Proposition~\ref{no I or III}, we claim there is no possible way to complete the fourth row of $C_3$ in a manner that is compatible with rows $1,2,3$. 

We first observe that the missing valencies of $(RR)_1$, denoted by $a,b,c,d,e,f$ for notational convenience, must coincide with the remaining portion of sequence II, viz.\ $2,2,2,2,0,0$. Applying equation~\ref{eqn 5} to the $\langle s\rangle$-orbits $(LL)_2$ and $(RR)_1$, 
we obtain $14=4+2a+2c+2d+2f$ where we have used the fact that $b_{24}=c_{24}=0$. However, this easily simplifies to $5=a+c+d+f$, a contradiction since $a,c,d,f\in\{0,2\}$. 
Together with Proposition~\ref{no I or III}, this proves there is no possible $\langle s\rangle$-orbit partition that fuses to the $G$-orbit partition in 
Figure~\ref{final OP21}. The result follows. 
\end{proof}

\begin{coro} 
If $\;7$ divides $|G|$ then $G\cong \mathbb Z_7$.
\end{coro}

\begin{proof}
The result follows at once from Corollary~\ref{two groups} and Propositions~\ref{unique OP21} and~\ref{no II or IV}.
\end{proof}
 
\longthanks{The authors wish to thank Jason Williford for helpful discussions. We are also grateful to two anonymous referees for their valuable comments.}
 
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