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\begin{document}

\title{\bf On Commuting Graphs for Elements of Order $\boldsymbol{3}$\\ in Symmetric Groups}

   \author{Athirah Nawawi and Peter Rowley\\
\small School of Mathematics\\[-0.8ex]
\small University of Manchester\\[-0.8ex]
\small Manchester, M13 6PL, U.K.\\
\small \texttt{peter.j.rowley@manchester.ac.uk}
}


\date{\dateline{May 19, 2012}{Jan 20, 2015}{Feb 9, 2015}\\
\small Mathematics Subject Classifications: 05C25}

\maketitle

\begin{abstract}
The commuting graph $\mathcal{C}(G,X)$, where $G$ is
a group and $X$ is a subset of $G$, is the graph with vertex set $X$
and distinct vertices being joined by an edge whenever they commute.
Here the diameter of $\mathcal{C}(G,X)$ is studied when $G$ is a
symmetric group and $X$ a conjugacy class of elements of order $3$.

\bigskip

\noindent {\bf Keywords:} Commuting graph, Symmetric
group, Order $3$ elements, Diameter
\end{abstract}


\section{Introduction}\label{intro}


Suppose that $G$ is a finite group and $X$ is a subset of $G$. The
{\em commuting graph} $\mathcal{C}(G,X)$ is the graph with $X$ as
the vertex set and two distinct elements of $X$ being joined by an
edge if they are commuting elements of $G$. This type of graph has
been studied for a wide variety of groups $G$ and selection of
subsets of $G$. One of the earliest investigations occurred in
Brauer and Fowler \cite{BrauerFowler} in which $X = G \backslash
\{1\}$. This particular case has recently been the subject of
further study by Segev \cite{Segev1}, \cite{Segev2} and Segev and
Seitz \cite{Segev3}. A great deal of attention has been focussed on
the case when $X$ is a conjugacy class of involutions -- the
so-called commuting involution graphs. Pioneering work on such
graphs appeared in Fischer \cite{Fischer} which led to the
construction of the three Fischer groups. Recently various
properties of other commuting involution graphs have been studied;
see, for example, \cite{BaBuPeRo1}, \cite{BaBuPeRo2},
\cite{BaBuPeRo3}, \cite{BaBuPeRo4}, \cite{Everett} and
\cite{EverettRowley}. When $X$ is a conjugacy class of
non-involutions, $\mathcal{C}(G,X)$ has to date received less
attention. Never-the-less graphs of this type can be of interest --
witness the computer-free uniqueness proof of the Lyon's simple
group by Aschbacher and Segev \cite{AschSeg} which employed a
commuting graph whose vertex set consisted of the $3$-central
subgroups of order $3$. Also see Baumeister and Stein
\cite{BaumStein}, the results obtained there being used to describe
the structure of Bruck loops and Bol loops of exponent $2$. Further,
commuting graphs when $G$ is a symmetric group have been
investigated in Bates, Bundy, Perkins and Rowley \cite{BaBuPeRo5}
and Bundy\cite{Bundy}. The former paper concentrates on the
structure of discs (around some fixed vertex) and the diameter of
the graph while the latter gives a complete answer as to when
$\mathcal{C}(G,X)$ is a connected graph.\\


In the present paper we shall determine the diameters of
$\mathcal{C}(G,X)$ when $G$ is a symmetric group and $X$ is a
$G$-conjugacy class of elements of order $3$. So for the rest of
this paper we assume $G=Sym(\Omega) = Sym(n)$ with $G$ acting upon
the set $\Omega=\{1,\dotsc,n\}$ in the usual manner. Also let
\begin{equation}
t=(1,2,3)(4,5,6)(7,8,9)\dotso(3r-2,3r-1,3r). \nonumber
\end{equation}
Thus $t$ has order $3$ and cycle type $1^{n-3r}3^r$. Set $X = t^G$, the $G$-conjugacy class of $t$, and let $\mathrm{Diam}$ $({\mathcal C} (G,X))$ denote the diameter of the commuting graph $\mathcal{C}(G,X)$. Our main results are as follows.

\begin{thm} \label{Diam=2}
If $n \geq 8r$, then $\mathrm{Diam}$ $({\cal C} (G,X))=2$.
\end{thm}

\begin{thm} \label{Diam=3}
If $6r<n<8r$, then $\mathrm{Diam}$ $({\cal C} (G,X))=3$.
\end{thm}

Our last theorem only gives a bound on $\mathrm{Diam}$ $({\cal C} (G,X))$.

\begin{thm} \label{Diamatmost4}
If $r > 1$ and $n=6r$, then $\mathrm{Diam}$ $({\cal C} (G,X)) \leq 4$.
\end{thm}

Consulting Table 1 (or Table 1 of \cite{BaBuPeRo5}) we see that for
$r = 1, n = 7$ or $r = 2, n = 15$ we have that Diam $({\cal C}
(G,X)) = 3$ and so Theorem \ref{Diam=2} is sharp. For $r = 2$ the
same table gives Diam $({\cal C} (G,X)) = 4$ when $n = 12$ and $2$
when $n = 16$, so Theorems \ref{Diam=3} and \ref{Diamatmost4} are
also sharp. We note that for $r = 1$ and $n = 6$, $\mathcal{C}(G,X)$
is disconnected which explains the assumption $ r > 1$ in Theorem
\ref{Diamatmost4}. All the graphs we consider here are connected
-- see \cite{Bundy}. For $g \in G$, $supp(g)$ denotes the set of
points of $\Omega$ not fixed by $g$. We use $d(,)$ for the usual
distance metric on the graph ${\cal C} (G,X)$. For $x \in X$, the
$i^{th}$ disc, $\Delta_i(x)$, is defined as follows
\begin{equation}
\Delta_i(x) = \{y \mid y \in X \mbox{and} \; d(x,y) = i \}.
\nonumber
\end{equation}

The proofs of Theorems \ref{Diam=2}, \ref{Diam=3} and
\ref{Diamatmost4} adopt a similar, somewhat direct, approach. Since
$G$ acting by conjugation upon $X$ induces graph automorphisms on
${\cal C} (G,X)$ and of course is transitive on $X$, it suffices
to determine (or bound) $d(t,x)$ for an arbitrary vertex $x$ of $X$.
This we do by writing down explicit paths in ${\cal C} (G,X)$.


\section{Diameter of $\mathcal{C}(G,X)$}\label{Diameters}

We begin by establishing Theorem \ref{Diam=2}.\\



\textbf{Proof of Theorem 1.1}


\mbox{}\\
Let $x \in X$. Set $\Lambda=supp(t) \cup supp(x)$ and $s=|supp(t)
\cap supp(x)|$. Then $|\Lambda|=6r-s$. If $s \geq r$, then
$|\Lambda| \leq 5r$. Hence there exists $y \in X$ with $supp(t) \cap
supp(y)=\emptyset=supp(x) \cap supp(y)$ and so $d(t,x) \leq 2$. Now
consider the case $s < r$, and set $e=r-s$. Without loss of
generality we may suppose that $supp(t) \cap supp(x) \subseteq
\{1,2,3,\dotsc,3s-2,3s-1,3s\}$. Put $y_1=(3s+1,3s+2,3s+3) \dotso
(3r-2,3r-1,3r)$ (so $y_1$ is the product of the ``last'' $r-s=e$
3-cycles of $t$). Since $|\Omega \setminus \Lambda|=8r-(6r-s)=2r+s >
3s$ and $s < r$, we may select $y_2$ with $supp(y_2) \subseteq
\Omega \setminus \Lambda$ and $y_2$ is a product of $s$ pairwise
disjoint 3-cycles. So $y=y_1y_2 \in X$, $ty=yt$ and $xy=yx$. Thus
$d(t,x) \leq 2$.
Clearly $\mathrm{Diam}$ $({\cal C} (G,X)) \geq 2$, and so the theorem follows.\\



Before proving Theorems \ref{Diam=3} and \ref{Diamatmost4} we
introduce some notation and certain permutations of $Sym(\Omega)$.
These permutations, though elements of order $3$, are not in general
in $X$. We will assume that $|\Omega| \geq 6r$. For $x \in X$, we
let $\{\vartheta_i(x)\}_{i=1, \dotsc, r}$ denote the orbits of
$\langle x \rangle$ on $\Omega$ of size 3. So
$supp(x)=\bigcup_{i=1}^r \vartheta_i(x)$. Write $t=t_1t_2 \dotso
t_r$ where $t_i=(3i-2,3i-1,3i)$.
So $\vartheta(t_i)=\vartheta_i(t)=\{3i-2,3i-1,3i\}$.\\

Let $x \in X$. Denote the product of the $t_i$'s for which
$\vartheta_i(t) \cap supp(x)=\emptyset$ by $\tau_0$ and let $\tau_3$
be the product of the $t_i$'s for which $\vartheta_i(t) \subseteq
supp(x)$. Also let $\tau_1$ be the product of $r_1$ $t_i$'s where $|
\vartheta_i(t) \cap supp(x) |=1$, $3 \mid r_1$ and $r_1$ is as large
as possible. Analogously, $\tau_2$ is the product of $r_2$ $t_i$'s
where $| \vartheta_i(t) \cap supp(x) |=2$, $3 \mid r_2$ and $r_2$ is
as large as possible. Setting
$\tau_*=t\tau_0^{-1}\tau_1^{-1}\tau_2^{-1}\tau_3^{-1}$ we have
$t=\tau_*\tau_0\tau_1\tau_2\tau_3$. Let $r_*$ be the number of
$t_i$'s in $\tau_*$, $r_0$ the number of $t_i$'s in $\tau_0$ and
$r_3$ the number of $t_i$'s in $\tau_3$. Observe that the maximality
of $r_1$ and $r_2$ means $r_* \leq 4$ and that at most two of the
$t_i$'s in $\tau_*$ will have $| \vartheta_i(t) \cap supp(x) |=1$
and at most two will have $| \vartheta_i(t) \cap supp(x) |=2$.
Evidently $r=r_*+r_0+r_1+r_2+r_3$ and, for $i=0,1,2,3,| supp(x) \cap
supp(\tau_i) |=ir_i$. Putting $s_*=| supp(x) \cap supp(\tau_*) |$,
we also have
\begin{equation}
| supp(t) \cap supp(x) |=s_*+r_1+2r_2+3r_3. \nonumber
\end{equation}
Set $\Lambda=\Omega \setminus (supp(t) \cup supp(x))$. Since
\begin{align*}
| supp(t) \cup supp(x) |&=3r+3r-(s_*+r_1+2r_2+3r_3) \\
                              &=6r-(s_*+r_1+2r_2+3r_3)
\end{align*}
it follows that
\begin{equation}
| \Lambda | = s_*+r_1+2r_2+3r_3 \;\mbox{if}\; n = 6r \;\mbox{and}
\nonumber
\end{equation}
\begin{equation}
| \Lambda | \geq 1+s_*+r_1+2r_2+3r_3 \;\mbox{if}\; n > 6r. \nonumber
\end{equation}

Since 3 divides $r_1$, we may write
\begin{equation}
\tau_1=\prod \mu_{i_1i_2i_3} \nonumber
\end{equation}
where the product of the $\mu_{i_1i_2i_3}=t_{i_1}t_{i_2}t_{i_3}$ is pairwise disjoint. For each $\mu_{i_1i_2i_3}=t_{i_1}t_{i_2}t_{i_3}=(3i_1-2,3i_1-1,3i_1)(3i_2-2,3i_2-1,3i_2)(3i_3-2,3i_3-1,3i_3)$ we may without loss, suppose that $supp(\mu_{i_1i_2i_3}) \cap supp(x)=\{3i_1-2,3i_2-2,3i_3-2\}$. Put
\begin{equation}
\lambda_{i_1i_2i_3}=(3i_1-2,3i_2-2,3i_3-2)(3i_1-1,3i_2-1,3i_3-1)(3i_1,3i_2,3i_3). \nonumber
\end{equation}
Then $\lambda_{i_1i_2i_3}$ commutes with $\mu_{i_1i_2i_3}$. Let
\begin{equation}
\rho_1=\prod \lambda_{i_1i_2i_3} \nonumber
\end{equation}
and observe that $\rho_1$ commutes with $t$ and will be a pairwise disjoint product of $r_1$ 3-cycles. Further, $\frac{r_1}{3}$ of the 3-cycles in $\rho_1$ will have their support contained in $supp(x)$ while the remaining $\frac{2r_1}{3}$ 3-cycles in $\rho_1$ will have their support intersecting $supp(x)$ in the empty set.

Also, as 3 divides $r_2$, we may express
\begin{equation}
\tau_2=\prod \eta_{j_1j_2j_3} \nonumber
\end{equation}
where $\eta_{j_1j_2j_3}=t_{j_1}t_{j_2}t_{j_3}$ with the product being pairwise disjoint. For each $\eta_{j_1j_2j_3}$ we may suppose that $supp(\eta_{j_1j_2j_3}) \cap supp(x)=\{3j_1-2,3j_1-1,3j_2-2,3j_2-1,3j_3-2,3j_3-1\}$. Define
\begin{equation}
\delta_{j_1j_2j_3}=(3j_1,3j_2,3j_3)(3j_1-2,3j_2-2,3j_3-2)(3j_1-1,3j_2-1,3j_3-1), \nonumber
\end{equation}
and let
\begin{equation}
\rho_2=\prod \delta_{j_1j_2j_3}. \nonumber
\end{equation}
Evidently $\rho_2$ commutes with $t$ and $\rho_2$ is a pairwise
disjoint product of $r_2$ 3-cycles. Moreover, $\frac{2r_2}{3}$ of
the 3-cycles in $\rho_2$ will have their support contained in
$supp(x)$ and the remaining $\frac{r_2}{3}$ have supports
intersecting $supp(x)$ in the empty set.

Let $\sigma_1$ (respectively $\sigma_2$) be the product of the $\frac{2r_1}{3}$ (respectively $\frac{r_2}{3}$) 3-cycles in $\rho_1$ (respectively $\rho_2$) whose support intersects $supp(x)$ in the empty set. Also let $\sigma_4$ be a pairwise disjoint product of ($\frac{r_1}{3}+\frac{2r_2}{3}+r_3$) 3-cycles with $supp(\sigma_4) \subseteq \Lambda$. Put $\Delta=\Lambda \backslash supp(\sigma_4)$.\\

We now summarize the pertinent properties of the permutations just introduced.

\begin{lemma} \label{Lem1}
\begin{trivlist}
\item{(\textit{i})} $supp(\tau_0\rho_1\rho_2\tau_3) \subseteq supp(t)$, $\tau_0\rho_1\rho_2\tau_3$ commutes with $t$ and is the product of $r - r_*$ pairwise disjoint $3$-cycles.
\item{(\textit{ii})} $\sigma_1\sigma_2\tau_0\sigma_4$ commutes with $\tau_0\rho_1\rho_2\tau_3$ and is the product of $r - r_*$ pairwise disjoint $3$-cycles. Moreover $supp(\sigma_1\sigma_2\tau_0\sigma_4) \cap supp(x) = \emptyset$.
\item{(\textit{iii})} $|\Delta| = s_* $ if $n = 6r$ and $|\Delta| \geq 1 + s_* $ if $n \geq 6r$.
\end{trivlist}
\end{lemma}

\begin{proof} (i) Since $supp(\rho_1\rho_2) = supp(\tau_1\tau_2)$, $\tau_0\rho_1\rho_2\tau_3$ is the product of pairwise disjoint $3$-cycles, and the number of such $3$-cycles is $r - r_*$. Because $\rho_1$ and $\rho_2$ both commute with $t$, $\tau_0\rho_1\rho_2\tau_3$ commutes with $t$.\\
(ii) Since $supp(\sigma_4) \subseteq \Delta$ and $supp(\tau_0\rho_1\rho_2\tau_3) \subseteq supp(t)$, $\sigma_4$ commutes with $\tau_0\rho_1\rho_2\tau_3$. While $\sigma_1\sigma_2\tau_0$ is a product of $3$-cycles which appear in $\tau_0\rho_1\rho_2\tau_3$ and therefore $\sigma_1\sigma_2\tau_0\sigma_4$ commutes with $\tau_0\rho_1\rho_2\tau_3$. By construction $\sigma_i \cap supp(x) = \emptyset (i = 1,2)$, $supp(\tau_0) \cap supp(x) = \emptyset$ by definition and because we chose $\sigma_4$ so as $supp(\sigma_4) \subseteq \Lambda$ we get $supp(\sigma_1\sigma_2\tau_0\sigma_4) \cap supp(x) = \emptyset$.\\
(iii) Part (iii) follows from $|supp(\sigma_4)| = r_1 + 2r_2 + 3r_3$ and $\Delta = \Lambda \backslash supp(\sigma_4)$.
\end{proof}

We are now in a position to prove Theorem \ref{Diam=3}.\\

\begin{proof}[\emph{\bf Proof of Theorem 1.2}]\
Let $y \in X$ be such that $| supp(y) \cap \vartheta_i(t) | = 1 = | supp(t) \cap \vartheta_i(y) |$ for $i=1,\dotsc,r$. Then $C_G(t) \cap C_G(y) = Sym(\Psi)$ where $\Psi=\Omega \setminus (supp(t) \cup supp(y))$. Now $| supp(t) \cup supp(y) | = 3r+3r-r=5r$ and so $| \Psi | = n-5r < 8r-5r=3r$. Thus $X \cap C_G(t) \cap C_G(y)=\emptyset$ and consequently $d(t,y) \geq 3$. Hence Diam $({\cal C} (G,X)) \geq 3$.\\

Let $x \in X$. We aim to show that $d(t,x) \leq 3$. On account of $C_G(t)$ having shape $3^r Sym(r) \times Sym(n-3r)$ there is no loss in supposing $\tau_*=t_1 \dotso t_{r_*}$ where $0 \leq r_* \leq 4$ ($r_*=0$ meaning $\tau_*=1$). Depending on $\tau_*$ we define two elements $\rho_*$ and $\sigma_*$ which will be the product of $r_*$ pairwise disjoint 3-cycles.\\

$\mathbf{(1)}$ $\mbox{}r_*=4$\\

Then we have $\tau_*=t_1t_2t_3t_4=(1,2,3)(4,5,6)(7,8,9)(10,11,12)$,
$s_*=6$ and we may, without loss, assume $supp(\tau_*) \cap
supp(x)=\{1,4,7,8,10,11\}$. Observe that $|supp(x) \backslash
supp(t)| \geq 6$ and so we may select
$\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5,\alpha_6 \in supp(x)
\backslash supp(t)$. Also by Lemma \ref{Lem1}(iii), as $s_*=6$,
$|\Delta| \geq 7$. Thus we may also select
$\beta_1,\beta_2,\beta_3,\beta_4,\beta_5,\beta_6 \in \Delta$. Define
\begin{equation}
\rho_*=(\alpha_1,\alpha_2,\alpha_3)(\alpha_4,\alpha_5,\alpha_6)(\beta_1,\beta_2,\beta_3)(\beta_4,\beta_5,\beta_6) \nonumber
\end{equation}
and
\begin{equation}
\sigma_*=(2,3,5)(6,9,12)(\beta_1,\beta_2,\beta_3)(\beta_4,\beta_5,\beta_6). \nonumber
\end{equation}

$\mathbf{(2)}$ $\mbox{}r_*=3$\\

So $\tau_*=t_1t_2t_3=(1,2,3)(4,5,6)(7,8,9)$. First we examine the case when $s_*=4$, and may suppose that $supp(\tau_*) \cap supp(x)=\{1,4,7,8\}$. Here we have $|supp(x) \backslash supp(t)| \geq 5$ and $|\Delta| \geq 5$. Choose $\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5 \in supp(x) \backslash supp(t)$ and $\beta_1,\beta_2,\beta_3,\beta_4,\beta_5 \in \Delta$, and define
\begin{equation}
\rho_*=(\alpha_1,\alpha_2,\alpha_3)(\alpha_4,\alpha_5,\beta_1)(\beta_2,\beta_3,\beta_4) \nonumber
\end{equation}
and
\begin{equation}
\sigma_*=(2,3,5)(6,9,\beta_5)(\beta_2,\beta_3,\beta_4). \nonumber
\end{equation}

We move onto the case when $s_*=5$ and, without loss of generality, assume $supp(\tau_*) \cap supp(x)=\{1,2,4,5,7\}$. Since $|supp(x) \backslash supp(t)| \geq 4$ and $|\Delta| \geq 6$, we may select $\alpha_1,\alpha_2,\alpha_3 \in supp(x) \backslash supp(t)$ and $\beta_1,\beta_2,\beta_3,\beta_4,\beta_5,\beta_6 \in \Delta$. Then we take
\begin{equation}
\rho_*=(\alpha_1,\alpha_2,\alpha_3)(\beta_1,\beta_2,\beta_3)(\beta_4,\beta_5,\beta_6) \nonumber
\end{equation}
and
\begin{equation}
\sigma_*=(3,6,8)(\beta_1,\beta_2,\beta_3)(\beta_4,\beta_5,\beta_6). \nonumber
\end{equation}

%\clearpage
$\mathbf{(3)}$ $\mbox{}r_*=2$\\

So $\tau_*=t_1t_2=(1,2,3)(4,5,6)$ with $s_*=2$, 3 or 4. First we look at the case when $s_*=2$ or 3. Then we have $|supp(x) \backslash supp(t)| \geq 3$, $|supp(t) \backslash supp(x)| \geq 3$ and $|\Delta| \geq 3$. Choosing $\alpha_1,\alpha_2,\alpha_3 \in supp(x) \backslash supp(t)$, $\beta_1,\beta_2,\beta_3 \in \Delta$ and $\gamma_1,\gamma_2,\gamma_3 \in supp(t) \backslash supp(x))$, we let
\begin{equation}
\rho_*=(\alpha_1,\alpha_2,\alpha_3)(\beta_1,\beta_2,\beta_3) \nonumber
\end{equation}
and
\begin{equation}
\sigma_*=(\gamma_1,\gamma_2,\gamma_3)(\beta_1,\beta_2,\beta_3). \nonumber
\end{equation}

Now assume that $s_*=4$, and, without loss, that $supp(\tau_*) \cap supp(x)=\{1,2,4,5\}$. Because $|supp(x) \backslash supp(t)| \geq 2$ and $|\Delta| \geq 5$ we may choose $\alpha_1,\alpha_2 \in supp(x) \backslash supp(t)$ and $\beta_1,\beta_2,\beta_3,\beta_4,\beta_5 \in \Delta$ and then define
\begin{equation}
\rho_*=(\alpha_1,\alpha_2,\beta_1)(\beta_2,\beta_3,\beta_4) \nonumber
\end{equation}
and
\begin{equation}
\sigma_*=(3,6,\beta_5)(\beta_2,\beta_3,\beta_4). \nonumber
\end{equation}

$\mathbf{(4)}$ $\mbox{}r_*=1$\\

Then $\tau_*=t_1=(1,2,3)$ and $s_*=1$ or 2. Suppose $s_*=1$ with $supp(\tau_*) \cap supp(x)=\{1\}$. So $|supp(x) \backslash supp(t)| \geq 2 \leq |\Delta|$. Selecting $\alpha_1,\alpha_2 \in supp(x) \backslash supp(t)$ and $\beta_1,\beta_2 \in \Delta$, we set
\begin{equation}
\rho_*=(\alpha_1,\alpha_2,\beta_1) \nonumber
\end{equation}
and
\begin{equation}
\sigma_*=(2,3,\beta_2). \nonumber
\end{equation}
While if $s_*=2$, then $|\Delta| \geq 3$ and selecting $\beta_1,\beta_2,\beta_3 \in \Delta$ we set
\begin{equation}
\rho_*=\sigma_*=(\beta_1,\beta_2,\beta_3). \nonumber
\end{equation}

$\mathbf{(5)}$ $\mbox{}r_*=0$\\

Here we take $\rho_*=1=\sigma_*$.\\

Put $y=\rho_*\tau_0\rho_1\rho_2\tau_3$. Since $y$ is the product of
$r_*+r_0+r_1+r_2+r_3=r$ disjoint 3-cycles, $y \in X$. Further we
have that $ty=yt$ by Lemma \ref{Lem1}(i). Next we consider
$z=\sigma_*\sigma_1\sigma_2\tau_0\sigma_4$. Each of
$\sigma_*\sigma_1,\sigma_2,\tau_0$ and $\sigma_4$ are pairwise
disjoint. Recalling that $\sigma_1,\sigma_2$ and $\sigma_4$ are,
respectively, the product of
$\frac{2r_1}{3},\frac{r_2}{3},(\frac{r_1}{3}+\frac{2r_2}{3}+r_3)$
disjoint 3-cycles, we see that $z \in X$. It may be further checked
using Lemma \ref{Lem1}(ii) that $yz=zy$ and $xz=zx$, and
consequently $d(t,x) \leq 3$.
This completes the proof of Theorem \ref{Diam=3}.
\end{proof}


\begin{proof}[\emph{\bf Proof of Theorem 1.3}]\
Let $x \in X$. Our objective here is to show that $d(t,x) \leq 4$
from which it will follow that Diam $({\cal C} (G,X)) \leq 4$. We
proceed in a similar fashion to that in the proof of Theorem
\ref{Diam=2} though here, except for some cases, we will define
three permutations $\rho_*, \sigma_*, \xi_*$, each a product of
$r_*$ pairwise disjoint cycles.\\

$\mathbf{(6)}$ $\mbox{}r_*=4$\\

So $\tau_*=t_1t_2t_3t_4=(1,2,3)(4,5,6)(7,8,9)(10,11,12)$ with
$s_*=6$. Assume, without loss, that $supp(\tau_*) \cap
supp(x)=\{1,4,7,8,10,11\}$. Since $|supp(x) \backslash$ $supp(t)| \geq 6$ and so we may choose
$\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5,\alpha_6 \in supp(x)
\backslash supp(t)$. Further, as  $|\Delta| = s_* = 6$ by Lemma
\ref{Lem1}(iii), we may also choose
$\beta_1,\beta_2,\beta_3,\beta_4,$ $\beta_5,\beta_6 \in \Delta$. Now
define
\begin{equation}
\rho_*=(\alpha_1,\alpha_2,\alpha_3)(\alpha_4,\alpha_5,\alpha_6)(\beta_1,\beta_2,\beta_3)(\beta_4,\beta_5,\beta_6) \nonumber
\end{equation}
and
\begin{equation}
\sigma_*=(2,3,5)(6,9,12)(\beta_1,\beta_2,\beta_3)(\beta_4,\beta_5,\beta_6). \nonumber
\end{equation}

$\mathbf{(7)}$ $\mbox{}r_*=3$\\

So $\tau_*=t_1t_2t_3=(1,2,3)(4,5,6)(7,8,9)$. If $s_*=4$ we may
suppose without loss that $supp(\tau_*) \cap supp(x)=\{1,4,7,8\}$.
Here we have $|supp(x) \backslash supp(t)| \geq 5$ and $|\Delta| =
s_* = 4$ by Lemma \ref{Lem1}(iii). Choose
$\alpha_1,\alpha_2,\alpha_3 \in supp(x) \backslash supp(t)$ and
$\beta_1,\beta_2,\beta_3\in \Delta$, and define
\begin{equation}
\rho_*=(\alpha_1,\alpha_2,\alpha_3)(\beta_1,\beta_2,\beta_3)(1,2,3), \nonumber
\end{equation}
\begin{equation}
\sigma_*=(5,6,9)(\beta_1,\beta_2,\beta_3)(1,2,3) \nonumber
\end{equation}
and
\begin{equation}
\xi_*=(5,6,9)(\beta_1,\beta_2,\beta_3)(\alpha,\beta,\gamma), \nonumber
\end{equation}
where $(\alpha,\beta,\gamma)$ is a $3$-cycle of $x$ for which $1 \notin \{\alpha,\beta,\gamma \}$. Note that $\{\alpha,\beta,\gamma \} \cap supp(\sigma_*) = \emptyset$.

For the case when $s_*=5$, without loss of generality, we assume $supp(\tau_*) \cap supp(x)=\{1,4,5,7,8\}$. Since $|supp(x) \backslash supp(t)| \geq 4$ and $|\Delta| = s_* =  5$, we may select $\alpha_1,\alpha_2,\alpha_3 \in supp(x) \backslash supp(t)$ and $\beta_1,\beta_2,\beta_3 \in \Delta$. Then we take
\begin{equation}
\rho_*=(\alpha_1,\alpha_2,\alpha_3)(\beta_1,\beta_2,\beta_3)(4,5,6), \nonumber
\end{equation}
\begin{equation}
\sigma_*=(2,3,9)(\beta_1,\beta_2,\beta_3)(4,5,6) \nonumber
\end{equation}
and
\begin{equation}
\xi_*=(2,3,9)(\beta_1,\beta_2,\beta_3)(\alpha,\beta,\gamma), \nonumber
\end{equation}
where $(\alpha,\beta,\gamma)$ is a $3$-cycle of $x$ chosen so as
$\{4,5\} \cap \{\alpha,\beta,\gamma \} = \emptyset$. Since $r \geq
r_* = 3$ such a choice is possible.\\

Before dealing with $r_* = 2$ we analyze a number of small cases.\\

$\mathbf{(8)}$ $\mbox{}$ Suppose that $t = (1,2,3)(4,5,6)$ (so $r =
2$ and $n = 12$).
\begin{trivlist}
\item{(\textit{i})} If $x = (1,7,8)(4,9,10)$ or $x =
(1,4,7)(2,5,8)$, then $d(t,x) \leq 4$.
\item{(\textit{ii})}If $x = (1,4,7)(8,9,10)$, then $d(t,x)\leq 3$.
\end{trivlist}

Assume that $x = (1,7,8)(4,9,10)$, and let 
\[ x_1 = (7,8,11)(9,10,12),\quad 
x_2 = (2,3,5)(9,10,12),\quad  x_3 = (2,3,5)(1,7,8).\]
 Then $x_1, x_2, x_3
\in X$ and $(t,x_1, x_2, x_3,$ $x)$ is a path in $\mathcal{C}(G,X)$
whence $d(t,x) \leq 4$. In the case $x = (1,4,7)(2,5,8)$ we take
$x_1 = (7,8,9)(10,11,12), x_2 = (1,3,6)(10,11,12)$ and $x_3 =
(2,5,8)$ $(10,11,12)$. It is easily checked that $(t,x_1,x_2,x_3,x)$ is
also a path in $\mathcal{C}(G,X)$, so proving part (i). For $x =
(1,4,7)(8,9,10)$ taking $x_1 = (1,2,3)(8,9,10)$ and $x_2 =
(5,6,11)(8,9,10)$ gives a path $(t,x_1,x_2,x)$ in
$\mathcal{C}(G,X)$. So (ii) holds and (8) is proved.\\

$\mathbf{(9)}$ $\mbox{}$ Suppose $t = (1,2,3)(4,5,6)(7,8,9)$ with
$\tau_* = (1,2,3)(4,5,6)$ (so $r = 3$ and $n = 18$). Let $x \in X$
be such that $supp(\tau_*) \cap supp(x) = \{1,4\}$ and assume $1$
and $4$ are in different $3$-cycles of $x$. Then $d(t,x) \leq 4$.\\

By assumption $x = (1,*,*)(4,\delta,\epsilon)(\alpha,\beta,\gamma)$
with $\{1,4\} \cap \{\alpha,\beta,\gamma\} = \emptyset$. Because
$\tau_* = (1,2,3)(4,5,6)$ we must have $supp(t) \cap supp(x) =
\{1,4\}$ or $\{1,4,7,8,9\}$. Suppose the former holds and set $x_1 =
(1,2,3)(\alpha,\beta,\gamma)(7,8,9)$ and  $x_2 =
(4,\delta,\epsilon)(\alpha,\beta,\gamma)(7,8,9)$. Then
$(t,x_1,x_2,x)$ is a path in $\mathcal{C}(G,X)$. Hence $d(t,x) \leq
3$. Turning to the latter case we have $|supp(t) \cup supp(x)| =
13$. So we may choose, say, $16,17,18 \in \Lambda$ and then take
$x_1
= (1,2,3)(4,5,6)$ $(16,17,18), x_2 =
(1,2,3)(\alpha,\beta,\gamma)(16,17,18)$ and $x_3 =
(4,\delta,\epsilon)(\alpha,\beta,\gamma)(16,17,$ $18)$, giving a path
$(t,x_1,x_2,x_3,x)$ in $\mathcal{C}(G,X)$. Thus $d(t,x) \leq 4$, so
proving (9).\\

$\mathbf{(10)}$ $\mbox{}r_*=2$\\

So we have $\tau_* = t_1t_2 =(1,2,3)(4,5,6)$ with $s_* = 2, 3$ or
$4$. First we consider the case $s_* = 2$, and assume $supp(\tau_*)
\cap supp(x) = \{1,4\}$. For the moment also assume that $r = 2$ (so
$t = \tau_*$). Then, without loss, $x$ is either $(1,7,8)(4,9,10)$
($1$ and $4$ in different $3$-cycles of $x$) or $(1,4,7)(8,9,10)$
($1$ and $4$ in the same $3$-cycle of $x$). By (8)(i) we have
$d(t,x) \leq 4$. So, since we are aiming to show that $d(t,x) \leq
4$, we may suppose $r \geq 3$. Now consider the possibility that $r
= 3$ and $1$ and $4$ are in different $3$-cycles of $x$. Then,
without loss, $x = (1,*,*)(4,\delta,\epsilon)(\alpha,\beta,\gamma)$
in which case $d(t,x) \leq 4$ by (9). Thus, when $r = 3$, we may
suppose $1$ and $4$ are in the same $3$-cycle of $x$. Consequently,
as $r \geq 3$, we may find two $3$-cycles of $x$,
$(\alpha,\beta,\gamma)$ and $(\delta,\epsilon,\lambda)$ such that
$\{\alpha,\beta,\gamma,\delta,\epsilon,\lambda\} \cap \{1,4\} =
\emptyset$. Now we define $\rho_*, \sigma_*$ and $\xi_*$ by taking
$\rho_* = \sigma_* = \tau_*$ and $\xi_* = (\alpha,\beta,\gamma),
(\delta,\epsilon,\lambda)$.\\

Next we look at the case $s_* = 3$. Then we have $|supp(x)
\backslash supp(t)| \geq 3$, $|supp(t) \backslash supp(x)| \geq 3$
and $|\Delta| = s_* = 3$. Choosing $\alpha_1,\alpha_2,\alpha_3 \in
supp(x) \backslash$ $supp(t)$, $\beta_1,\beta_2,\beta_3 \in \Delta$
and $\gamma_1,\gamma_2,\gamma_3 \in supp(t) \backslash supp(x))$, we
let
\begin{equation}
\rho_*=(\alpha_1,\alpha_2,\alpha_3)(\beta_1,\beta_2,\beta_3)
\nonumber
\end{equation}
and
\begin{equation}
\sigma_*=(\gamma_1,\gamma_2,\gamma_3)(\beta_1,\beta_2,\beta_3).
\nonumber
\end{equation}

Finally we come to $s_* = 4$. So without loss we have $supp(\tau_*)
\cap supp(x) = \{1,2,4,5\}$. Suppose, for the moment, that for all
$3$-cycles $(\alpha,\beta,\gamma)$ we have $\{1,2\} \cap
\{\alpha,\beta,\gamma\} \neq \emptyset \neq \{4,5\} \cap
\{\alpha,\beta,\gamma\}$. Then it follows that $r = 2$ and, without
loss, $x = (1,4,7)(2,5,8)$. But then $d(t,x) \leq 4$ by (8)(ii).
Thus we may suppose $x$ contains a $3$-cycle $(\alpha,\beta,\gamma)$
such that $(\alpha,\beta,\gamma) \cap \{1,2\} = \emptyset$, and we
can now define $\rho_*$ and $\sigma_*$. Since $|\Delta| = s_* = 4$,
we have $\beta_1,\beta_2,\beta_3 \in \Delta$. Let $\rho_* =
(1,2,3)(\beta_1,\beta_2,\beta_3)$ and $\sigma_* =
(\alpha,\beta,\gamma)(\beta_1,\beta_2,\beta_3)$. This completes the
case $s_* = 4$ and (10).\\

Yet another special case must be looked at before doing $r_* = 1$.\\

$\mathbf{(11)}$ $\mbox{}$ Let $t = (1,2,3)(4,5,6)$ with $\tau_* =
(1,2,3)$. Suppose $x = (1,*,*)(2,*,*)$ $\in X$ with $supp(\tau_*) \cap
supp(x) = \{1,2\}$. Then $d(t,x) \leq 3$.\\

Since $\tau_* = (1,2,3)$, $supp(t) \cap supp(x) = \{1,2\}$ or
$\{1,2,4,5,6\}$. If $supp(t) \cap supp(x) = \{1,2\}$ and, say
$\Omega \backslash (supp(t) \cap supp(x))= \{11,12\}$, then define
$x_1 = (4,5,6)(10,11,12), x_2 = (4,5,6)(\alpha,\beta,\gamma)$ where
$(\alpha,\beta,\gamma)$ is a $3$-cycle not containing $10$. While in
the other case with, say $\Omega \backslash (supp(t) \cap supp(x))=
\{8,9,10,11,12\}$ we define $x_1 = (8,9,10)(7,11,12), x_2 =
(8,9,10)(\alpha,\beta,\gamma)$ where $(\alpha,\beta,\gamma)$ is a
$3$-cycle not containing $7$. Hence $d(t,x) \leq 3$.\\

$\mathbf{(12)}$ $\mbox{} r_* = 1$\\

So we have either, without loss, $supp(\tau_*) \cap supp(x) = \{1\}$
or $\{2,3\}$. In view of (10), as $r > 1$, either $d(t,x) \leq 3$ or
we may find a $3$-cycle $(\alpha,\beta,\gamma)$ of $x$ for which
$supp(\tau_*)  \cap \{\alpha,\beta,\gamma\} = \emptyset$. In the
latter case we define $\rho_* = \sigma_* = \tau_*$ and $\xi_* =
(\alpha,\beta,\gamma)$.\\

$\mathbf{(13)}$ $\mbox{} r_* = 0$\\

Just as in (5) we take $\rho_* = 1 = \sigma_*$.

Now let $y = \rho_*\tau_0\rho_1\rho_2\tau_3, z =
\sigma_*\sigma_1\sigma_2\tau_0\sigma_4$ and $w =
\xi_*\sigma_1\sigma_2\tau_0\sigma_4$ (where $w$ is only defined if
in (6), (7), (10), (12), (13) $\xi_*$ is defined). Then $y,z,w \in
X$ with $(t,y,z,w,x)$ is a path in $\mathcal{C}(G,X)$. Consequently
$d(t,x) \leq 4$. Since $x$ was an arbitrary vertex, this shows that
$\mathrm{Diam}$ $(\mathcal{C}(G,X)) \leq 4$ and completes the proof of Theorem
\ref{Diamatmost4}.
\end{proof}



We end this paper with a table containing some calculations on
diameters and discs using \textsc{Magma}\cite{magma}. Each entry in
the table first gives the size of the relevant $\Delta_i(t)$ for the
given $r$ and $n$ with the number in brackets being the number of
$C_G(t)$-orbits on $\Delta_i(t)$. A blank entry means that
$|\Delta_i(t)| = 0$.







\begin{landscape}

\centering
\begin{longtable}{|l|x{2.1cm}|x{3.0cm}|x{3.5cm}|x{4.0cm}|x{3.3cm}|x{2.3cm}|}
\hline
 & ${\Delta}_1(t)$ & ${\Delta}_2(t)$ & ${\Delta}_3(t)$ & ${\Delta}_4(t)$ & ${\Delta}_5(t)$ & ${\Delta}_6(t)$  \tabularnewline
\hline
\endfirsthead

\multicolumn{7}{c}{{\tablename} \thetable{} -- Continued} \\[0.5ex]
\hline
 & ${\Delta}_1(t)$ & ${\Delta}_2(t)$ & ${\Delta}_3(t)$ & ${\Delta}_4(t)$ & ${\Delta}_5(t)$ & ${\Delta}_6(t)$  \tabularnewline
\hline
\endhead

\multicolumn{7}{r}{{Continued on Next Page\ldots}} \tabularnewline
\endfoot

\hline
\caption{Disc sizes and $C_G(t)$-orbits \label{Table1}} \tabularnewline
\endlastfoot

$\mathbi{r=1}$ & & & & & &    \tabularnewline
$n=7$ & 9 (2) & 24 (2) & 36 (1) & - & - & -    \tabularnewline
$n=8$ & 21 (2) & 90 (3) & - & - & - & -    \tabularnewline
$n=9$ & 41 (2) & 126 (3) & - & - & - & -    \tabularnewline
\hline \hline
$\mathbi{r=2}$ & & & & & &    \tabularnewline
$n=10$ & 35 (4) & 192 (6) & 1,008 (10) & 2,628 (20) & 3,672 (13) & 864 (5)    \tabularnewline
$n=11$ & 83 (4) & 1,080 (9) & 7,560 (23) & 9,756 (23) & - & -    \tabularnewline
$n=12$ & 203 (5) & 6,300 (16) & 28,296 (34) & 2,160 (5) & - & -    \tabularnewline
$n=13$ & 563 (5) & 25,740 (30) & 42,336 (25) & - & - & -    \tabularnewline
$n=14$ & 1,571 (5) & 67,140 (48) & 51,408 (7) & - & - & -    \tabularnewline
$n=15$ & 4,035 (5) & 168,948 (54) & 27,216 (1) & - & - & -    \tabularnewline
$n=16$ & 9,363 (5) & 310,956 (55) & - & - & - & -    \tabularnewline
\hline \hline
$\mathbi{r=3}$ & & & & & &    \tabularnewline
$n=9$ & 25 (4) & 216 (4) & 1,512 (11) & 486 (6) & - & -       \tabularnewline
$n=12$ & 49 (7) & 648 (8) & 9,936 (39) & 90,990 (139) & 327,024 (404) & 64,152 (102) \tabularnewline
$n=13$ & 121 (7) & 2,808 (18) & 79,488 (85) & 724,086 (383) & 783,432 (332) & 11,664 (3)    \tabularnewline
$n=14$ & 265 (7) & 9,936 (23) & 390,582 (138) & 3,217,806 (564) & 865,890 (143) & -    \tabularnewline
$n=15$ & 745 (9) & 62,424 (46) & 2,414,610 (243) & 8,733,420 (594) & - & -    \tabularnewline
$n=16$ & 2,545 (9) & 482,760 (90) & 17,798,778 (578) & 7,341,516 (220) & - & -    \tabularnewline
$n=17$ & 8,089 (9) & 3,400,272 (145) & 50,175,126 (728) & 870,912 (16) & - & -    \tabularnewline
$n=18$ & 24,441 (10) & 16,126,398 (210) & 92,757,960 (679) & - & - & -
\tabularnewline
\hline \hline
\pagebreak
$\mathbi{r=4}$ & & & & & &    \tabularnewline
$n=12$ & 159 (6) & 8,532 (20) & 193,104 (121) & 44,604 (37) & - & -       \tabularnewline
$n=15$ & 367 (11) & 37,044 (52) & 3,053,160 (682) & 81,668,484 (8,294) & &    \tabularnewline
$n=16$ & 991 (11) & 271,236 (92) & 56,926,656 (2,351) & 390,829,212 (13,122) & 419,904 (12) & -    \tabularnewline
$n=17$ & 2,239 (11) & 1,350,612 (112) & 487,124,064 (4,539) & 1,036,246,284 (12,578) & - & -    \tabularnewline
\hline \hline
$\mathbi{r=5}$ & & & & & &    \tabularnewline
$n=15$ & 751 (8) & 154,440 (44) & 17,669,304 (783) & 27,020,304 (996) & - & -       \tabularnewline
\hline
\end{longtable}
\end{landscape}



\begin{thebibliography}{99}
\bibitem{AschSeg}
Aschbacher, M.; Segev, Y., \emph{The uniqueness of groups of Lyons type.} 
J.~Amer.~Math.~Soc. 5 (1992), no. 1, 75--98.

\bibitem{BaBuPeRo1}    Bates, C.; Bundy, D.; Perkins, S.; Rowley, P. 
\emph{Commuting involution graphs for symmetric groups.} 
J.~Algebra 266 (2003), no. 1, 133--153.

\bibitem{BaBuPeRo2}    
Bates, C.; Bundy, D.; Perkins, S.; Rowley, P. 
\emph{Commuting involution graphs for finite Coxeter groups.} 
J.~Group Theory 6 (2003), no. 4, 461--476.

\bibitem{BaBuPeRo3}    
Bates, C.; Bundy, D.; Perkins, S.; Rowley, P. 
\emph{Commuting involution graphs in special linear groups.} 
Comm.~Algebra 32 (2004), no. 11, 4179--4196.

\bibitem{BaBuPeRo4}    
Bates, C.; Bundy, D.; Hart, S.; Rowley, P. 
\emph{Commuting involution graphs for sporadic simple groups.} 
J.~Algebra 316 (2007), no. 2, 849--868.

\bibitem{BaBuPeRo5}   
Bates, C.; Bundy, D.; Hart, S.; Rowley, P. 
\emph{A Note on Commuting Graphs for Symmetric Groups},
   Electron.~J.~Combin.
   16(1) (2009), \#R6.

\bibitem{BaumStein}    
Baumeister, B.; Stein, A. 
\emph{Commuting graphs of odd prime order elements in simple groups.}
\arxiv{0908.2583}

\bibitem{BrauerFowler}
Brauer, R.; Fowler, K. A., \emph{On groups of even order.} 
Ann.~Math. (2) 62 (1955), 565--583.

\bibitem{Bundy}   Bundy, D. 
\emph{The connectivity of commuting graphs.} 
J.~Combin.~Theory Ser.~A 113 (2006), no. 6, 995--1007.

\bibitem{magma} 
Cannon, J.J; Playoust, C. 
{\em An Introduction to Algebraic Programming with \textsc{Magma}}, 
Springer-Verlag (1997).

\bibitem{Everett}   
Everett, A., 
\emph{Commuting involution graphs for 3-dimensional unitary groups.} 
Electron.~J.~Combin. 18(1) (2011),\#P103,

\bibitem{EverettRowley}   
Everett, A.; Rowley, P., 
\emph{Commuting Involution Graphs for $4$-Dimensional Projective Symplectic
Groups.}
Preprint, 2010.
Available from \url{http://eprints.ma.man.ac.uk/1564/}

\bibitem{Fischer}   Fischer, B.,
\emph{Finite groups generated by $3$-transpositions.} I.
 Invent.~Math. 13 (1971), 232--246.

\bibitem{Segev1}   Segev, Y., 
\emph{On finite homomorphic images of the multiplicative group of a 
division algebra.} Ann.~Math. (2) 149 (1999), no. 1, 219--251.

\bibitem{Segev2}   Segev, Y., 
\emph{The commuting graph of minimal nonsolvable groups.} 
Geom.~Dedicata 88 (2001), no. 1-3, 55--66.

\bibitem{Segev3}   Segev, Y.; Seitz, G.M. 
\emph{Anisotropic groups of type $A\sb n$ and the commuting graph of finite 
simple groups.} Pacific J.~Math. 202 (2002), no. 1, 125--225.
\end{thebibliography}

\end{document}
