Commuting Involution Graphs for 4-Dimensional Projective Symplectic Groups

Abstract

For a group G and X a subset of G the commuting graph of G on X, denoted by $\mathcal {C}(G,X)$, is the graph whose vertex set is X with $x,y\in X$ joined by an edge if $x\ne y$ and x and y commute. If the elements in X are involutions, then $\mathcal {C}(G,X)$ is called a commuting involution graph. This paper studies $\mathcal {C}(G,X)$ when G is a 4-dimensional projective symplectic group over a finite field and X a G-conjugacy class of involutions, determining the diameters and structure of the discs of these graphs.

1 Introduction

For G a group and X a subset of G, the commuting graph of G on X, $\mathcal {C}(G,X)$, is the graph whose vertex set is X with $x,y \in X$ joined whenever $x \ne y$ and $xy=yx$. In effect commuting graphs first appeared in the paper of Brauer and Fowler [14], famous for containing a proof that up to isomorphism only finitely many non-abelian simple groups can have a given centralizer of an involution. The commuting graphs considered in [14] had $X = G {\setminus } \{1 \}$-such graphs have played an important role in recent work related to the Margulis–Platanov conjecture (see [41]). Various kinds of commuting graphs have been deployed in the study of finite groups, particularly the non-abelian simple groups. For example, the analysis and subsequent construction by Fischer [24] of the three simple Fischer groups used the commuting graph on the conjugacy class of 3-transpositions. While a computer-free uniqueness proof of the Lyons simple group by Aschbacher and Segev [6] employed a commuting graph where the vertices consisted of the 3-central subgroups of order 3. For G either a symmetric group, or more generally a finite Coxeter group, or a projective special linear group and X a certain conjugacy class of G, the structure of $\mathcal {C}(G,X)$ has been investigated at length by Bundy [16], Bates, Bundy, Hart, Perkins and Rowley [9,10,11,12,13], Nawawi and Rowley [38], Jafari [31]. And for the double covers of symmetric groups and 3-dimensional unitary groups, see Aubad [8] and Everett [23] respectively. Also, commuting involution graphs for infinite Coxeter groups have been analysed in Perkins [40], Hart and Clarke [29]. A different flavour of graph (also called a commuting graph) has also been studied extensively. In this case, for a group G, the vertex set is $G {\setminus } Z(G)$ with two distinct elements being joined if they commute. For a selection of work on various aspects of these graphs, consult Akbari, Mohammadian, Radjavi and Raja [3], Britnell and Gill [15], Cassell [19], Giudici and Kuzma [25], Giudici and Parker [26], Giudici and Pope [27], Iranmanesh and Jafarzadeh [30], Leshchenko and Zorya [34], Leshchenko [33], Mahmoudifar and Moghaddamfar [35], Mohammadian, Erfanian, Forrokhi and Wilkens [36], Morgan and Parker [37], Parker [39]. Recently there has been work on commuting graphs for rings (see, for example, [1, 2, 20, 21, 43]).

This paper investigates $\mathcal {C}(G,X)$ when G is a finite 4-dimensional projective symplectic group and X is a G-conjugacy class of involutions. Such graphs are referred to as commuting involution graphs. From now on H will denote the symplectic group Sp(4, q), $q = p^a$ and p a prime. Let V be the natural (symplectic) GF(q)H-module, and set $G = H/Z(H)$. So $G \cong PSp(4,q)$ and $G \cong H$ when $p = 2$. For $t\in X$, we define

$$\begin{aligned} \Delta _i(t)=\left\{ \left. x\in X\right| d(t,x)=i\right\} \end{aligned}$$

where d is the standard distance metric on $\mathcal {C}(G,X)$. So $\Delta _i(t)$ consists of all vertices in $\mathcal {C}(G,X)$ distance i from t—we call $\Delta _i(t)$ the ith disc of t (or just a disc of t).

In the case when $p = 2$, G has three conjugacy classes of involutions. Recalling that for an involution x of G, $V(x) = \{v \in V |(v,v^x) = 0 \}$ these three classes $X_1, X_2, X_3$ may be described thus (see [7])

$$\begin{aligned} X_1&=\left\{ x\in G\big |x^2=1,\ \dim \ C_V(x)=3 \right\} ;\\ X_2&=\left\{ x\in G\big |x^2=1,\ \dim \ C_V(x)=2,\ \dim \ V(x)=3 \right\} ;\quad \text {and}\\ X_3&=\left\{ x\in G\big |x^2=1,\ \dim \ C_V(x)=2,\ V(x)=V \right\} . \end{aligned}$$

Our four main theorems are as follows.

Theorem 1.1

Suppose that $p = 2$ and $i = 1,3$. Then $\mathcal {C}(G,X_i)$ is connected of diameter 2 with the disc sizes being

$$\begin{aligned} \left| \Delta _1(t)\right|&= q^3-2;\ \text {and}\\ \left| \Delta _2(t)\right|&= q^3(q-1). \end{aligned}$$

Theorem 1.2

Suppose that $p = 2$. Then $\mathcal {C}(G,X_2)$ is connected of diameter 4, the disc sizes being

$$\begin{aligned} \left| \Delta _1(t)\right|&=q^2(2q-3);\\ \left| \Delta _2(t)\right|&=2q^2(q-1)^2;\\ \left| \Delta _3(t)\right|&=2q^3(q-1)^2;\ \text {and}\\ \left| \Delta _4(t)\right|&=q^4(q-1)^2. \end{aligned}$$

Turning to the case when p is odd, we have that there are two G-involution conjugacy classes $Y_1$ and $Y_2$. We shall let $Y_1$ denote the G-conjugacy class whose elements are the images of an involution in H, and $Y_2$ to denote the G-conjugacy class whose elements are the image of an element of H of order 4 which squares to the non-trivial element of Z(H).

Theorem 1.3

If p is odd, then $\mathcal {C}(G, Y_1)$ is connected of diameter 2 with disc sizes

$$\begin{aligned} \left| \Delta _1(t)\right|&=\frac{1}{2}q(q^2-1);\ \text {and} \\ \left| \Delta _2(t)\right|&=\frac{1}{2}\left( q^4-q^3+q^2+q-2\right) . \end{aligned}$$

Theorem 1.4

• (i) If $q\equiv -1 \pmod 4$ then $\mathcal {C}(G,Y_2)$ is connected of diameter 3. Furthermore,

  $$\begin{aligned} \left| \Delta _1(t)\right|&=\frac{1}{2}q(q^2 + 2q - 1);\\ \left| \Delta _2(t)\right|&=\frac{1}{16}(q+1)\left( 3q^5-2q^4+8q^3-30q^2+13q-8\right) ;\quad \text {and}\\ \left| \Delta _3(t)\right|&=\frac{1}{16}(q^2-1)\left( 5q^4-9q^3+7q^2-3q+8\right) . \end{aligned}$$

• (ii) If $q\equiv 1 \pmod 4$ then $\mathcal {C}(G,Y_2)$ is connected of diameter 3. Furthermore,

  $$\begin{aligned} \left| \Delta _1(t)\right|&=\frac{1}{2}q(q^2 + 1);\\ \left| \Delta _2(t)\right|&=\frac{1}{16}(q-1)\left( 3q^5-6q^4+32q^3-10q^2-27q-8\right) ;\quad \text {and}\\ \left| \Delta _3(t)\right|&=\frac{1}{16}(q-1)\left( 5q^5+22q^4-8q^3+34q^2+51q+24\right) . \end{aligned}$$

Theorems 1.1 and 1.2 are established in Sect. 2. While in Sect. 3 we give a proof of Theorem 1.3. The structure and properties of $\mathcal {C}(G,Y_2)$, in Sect. 4, are a much tougher nut to crack than the other four cases. The reason for this is that for $\mathcal {C}(G,X_i),\ (i=1,2,3)$ and $\mathcal {C}(G,Y_1)$ the graph can be studied effectively by working in $H=Sp(4,q)$ and looking at certain configurations in the natural symplectic module V involving $C_V(x)$ for various $x\in X$ ($X=X_i,\ i=1,2,3$ or $XZ(H)/Z(H)=Y_1$). The key point being that, in these four cases for $x\in X$, $C_V(x)$ is a non-trivial subspace of V whereas, for x of order 4 and squaring into Z(H), $C_V(x)$ is trivial. If we change tack and look at G acting on the projective symplectic space things are not much better. When $q\equiv -1 \pmod 4$ elements of $Y_2$ fix no projective points, while in the case $q\equiv 1\pmod 4$ they fix $2q+2$ projective points. However, even in the latter case, the fixed projective points didn’t appear to be of much assistance. It is the isomorphism $PSp(4,q)\cong O(5,q)$ that comes to our rescue. If now V is the 5-dimensional orthogonal module and $x\in Y_2$, then $\dim C_V(x)=3$. Even so, probing $\mathcal {C}(G,Y_2)$ turns out to be a lengthy process. Fix $t\in Y_2$. Then by Lemma 4.3, $Y_2\subseteq \bigcup _{U\in \mathcal {U}_1} C_G(U)$ where $\mathcal {U}_1$ is the set of all 1-subspaces of $C_V(t)$ and as a result, by Lemma 4.4, $\mathcal {C}(G,Y_2)$ may be viewed as the union of commuting involution graphs for various subgroups of G. Up to isomorphism there are three of these commuting involution graphs (called $\mathcal {C}(G^-,Y^-)$, $\mathcal {C}(G^+,Y^+)$ and $\mathcal {C}(G^0,Y^0)$ in Sect. 4). After studying these three commuting involution graphs in Theorems 4.6, 4.8 and 4.14 it follows immediately (Theorem 4.15) that $\mathcal {C}(G,Y_2)$ is connected and has diameter at most 3. Using the sizes of the discs in $\mathcal {C}(G^-,Y^-)$, $\mathcal {C}(G^+,Y^+)$ and $\mathcal {C}(G^0,Y^0)$ we then complete the proof of Theorem 1.4. This “patching together” of the discs is quite complicated—for example we must confront such issues as t and x in $Y_2$ being of distance 3 in each of the commuting involution subgraphs which contain both t and x, yet they have distance 2 in $\mathcal {C}(G,Y_2)$ (see Lemmas 4.29–4.34).

Our group theoretic notation is standard as given, for example, in [5] or [28].

2 Structure of $\mathcal {C}(G,X_i), i = 1,2,3$

We begin looking at $G_0 = Sp_{2n}(q)$ where $n \ge 2$, $q = p^a$ and $p = 2$. Let $V_0$ denote the $GF(q)G_0$-symplectic module of dimension 2n and let $t_0$ be an involution in $G_0$ for which $\dim C_V(t_0) = 2n - 1$. Put $X_0 = t_0^{G_0}$, the $G_0$-conjugacy class of $t_0$.

Theorem 2.1

$\mathcal {C}(G_0,X_0)$ is connected and has diameter 2.

Proof

For $x \in X_0$,

$$\begin{aligned} C_{G_0}(x) \le Stab_{G_0}(C_{V_0}(x)) \end{aligned}$$

with $Stab_{G_0}(C_{V_0}(x))$ having shape $q^{2n - 1}SL_{2n - 2}(q)(q-1)$. Set $K_x = O^{2^{\prime }}(Stab_{G_0}(C_{V_0}(x)))$. Then $K_x \sim q^{2n - 1}SL_{2n - 2}(q)$ and $C_{G_0}(x) = K_x$. Let $x \in X_0 {\setminus } \{ t_0 \}$. If $C_{V_0}(t_0) = C_{V_0}(x)$, then $x \in K_{t_0}$ and so $x \in \Delta _1(t_0)$. Now suppose that $C_{V_0}(t_0) \ne C_{V_0}(x)$. Then $\dim (C_{V_0}(t_0) \cap C_{V_0}(x)) = 2n - 2$. Let U be a 1-dimensional subspace of $C_{V_0}(t_0)\cap C_{V_0}(x)$. Since $[V_0,t_0]$ is a 1-space and $G_0$ acts transitively on the 1-spaces of $V_0$, there exists $y \in X_0$ such that $[V_0,y]= U$. So $[V_0,y] \le C_{V_0}(t_0) \cap C_{V_0}(x)$ and hence y leaves both $C_{V_0}(t_0)$ and $C_{V_0}(x)$ invariant. Thus $y \in K_{t_0} \cap K_x = C_{G_0}(t_0) \cap C_{G_0}(x)$ and so $d(t_0,x) \le 2$ and we see that $\mathcal {C}(G_0,X_0)$ is connected. Since $\mathcal {C}(G_0,X_0)$ cannot have diameter 1 (as then $\langle X_0 \rangle$ would be abelian), the theorem follows. $\square$

The remainder of this section is devoted to establishing Theorems 1.1 and 1.2. So we have $G = Sp(4,q)$ with $q = p^a$ and $p = 2$. For V, the natural GF(q) module for G, we choose the symplectic basis $\{ v_1,v_2\big |v_3,v_4 \}$ with $(v_1,v_4)= (v_2,v_3) = 1$. Thus the matrix defining this form is

$$\begin{aligned} J=\left( \begin{array}{llll}0&{}0&{}0&{}1\\ 0&{}0&{}1&{}0\\ 0&{}1&{}0&{}0\\ 1&{}0&{}0&{}0\end{array} \right) , \end{aligned}$$

and we may suppose that $G = \{A \in GL(4,q) \big | A^TJA = J \}$. We further define

$$\begin{aligned} S= & {} \left\{ \left. \left( \begin{array}{llll}1&{}a&{}b&{}c\\ 0&{}1&{}d&{}ad+b\\ 0&{}0&{}1&{}a\\ 0&{}0&{}0&{}1\end{array} \right) \right| a,b,c,d\in GF(q)\right\} ,\\ Q_1= & {} \left\{ \left. \left( \begin{array}{llll}1&{}a&{}b&{}c\\ 0&{}1&{}0&{}b\\ 0&{}0&{}1&{}a\\ 0&{}0&{}0&{}1\end{array} \right) \right| a,b,c\in GF(q)\right\} \quad \text {and}\ Q_2=\left\{ \left. \left( \begin{array}{llll}1&{}0&{}b&{}c\\ 0&{}1&{}d&{}b\\ 0&{}0&{}1&{}0\\ 0&{}0&{}0&{}1\end{array} \right) \right| b,c,d\in GF(q)\right\} . \end{aligned}$$

Lemma 2.2

• (i) $S \in Syl_2G$.

• (ii) $S = Q_1Q_2$ with $Q_1^{\#} \cup Q_2^{\#}$ consisting of all the involutions of S.

Proof

It is straightforward to check that S is a subgroup of G. Since $|G| = q^4(q^2 - 1)(q^4 - 1)$ and $|S| = q^4$, we have part (i). Part (ii) is an easy calculation. $\square$

The following three involutions are elements of G.

$$\begin{aligned} t_1=\left( \begin{array}{llll}1&{}0&{}0&{}1\\ 0&{}1&{}0&{}0\\ 0&{}0&{}1&{}0\\ 0&{}0&{}0&{}1\end{array} \right) ,\quad t_2=\left( \begin{array}{llll}1&{}0&{}1&{}1\\ 0&{}1&{}0&{}1\\ 0&{}0&{}1&{}0\\ 0&{}0&{}0&{}1\end{array} \right) ,\quad t_3=\left( \begin{array}{llll}1&{}1&{}0&{}0\\ 0&{}1&{}0&{}0\\ 0&{}0&{}1&{}1\\ 0&{}0&{}0&{}1\end{array} \right) . \end{aligned}$$

Lemma 2.3

• (i) For $i = 1,2,3, t_i \in X_i$.

• (ii) $C_G(t_1) \sim q^3SL(2,q)$ with $O_2(C_G(t_1)) = Q_1$ of order $q^3$.

• (iii) $C_G(t_2) = S$.

• (iv) $|X_1| = q^4 - 1$.

• (v) $|X_2| = (q^2 - 1)(q^4 - 1)$.

Proof

• (i) Let $v=(\alpha ,\beta ,\gamma ,\delta ) \in V$. Then $v^{t_1}=(\alpha ,\beta ,\gamma ,\alpha +\delta )$, $v^{t_2}=(\alpha ,\beta ,\alpha +\gamma ,\alpha +\beta +\delta )$ and $v^{t_3}=(\alpha ,\alpha +\beta ,\gamma ,\gamma +\delta )$. Hence $[v,t_1]=(0,0,0,\alpha )$, $[v,t_2]=(0,0,\alpha ,\alpha +\beta )$ and $[v,t_3]=(0,\alpha ,0,\gamma )$. Consequently $\dim \ [V,t_1]=1$ and $\dim \ [V,t_2]=2 = \dim \ [V,t_3]$. Thus $t_1 \in X_1$. Now

  $$\begin{aligned} (v,v^{t_2})=\alpha (\alpha +\beta +\delta )+\beta (\alpha +\gamma ) +\gamma \beta +\delta \alpha =\alpha ^2=0 \end{aligned}$$

  implies that $\alpha =0$ and so $\dim \ V(t_3)=3$. Therefore $t_2 \in X_2$. Turning to $t_3$ we have that

  $$\begin{aligned} (v,v^{t_3})= \alpha (\gamma +\delta )+\beta \gamma +\gamma (\alpha +\beta )+\delta \alpha =0 \end{aligned}$$

  implies that $V(t_2)=V$, as v is an arbitrary vector of V. Hence $t_3 \in X_3$, and we have (i).

• (ii) By direct calculation we see that

  $$\begin{aligned} C_G(t_1)=\left\{ \left. \left( \begin{array}{llll}1&{}b&{}c&{}d\\ 0&{}f&{}g&{}h\\ 0&{}k&{}m&{}n\\ 0&{}0&{}0&{}1\end{array} \right) \right| \begin{array}{l}b,c,d,f,g,h,k,m,n\in GF(q)\\ gk+fm=1\\ b+hk+fn=0\\ c+mh+gn=0\end{array}\right\} . \end{aligned}$$

  Moreover

  $$\begin{aligned} SL_2(q)\cong R=\left\{ \left. \left( \begin{array}{llll}1&{}0&{}0&{}0\\ 0&{}f&{}g&{}0\\ 0&{}k&{}m&{}0\\ 0&{}0&{}0&{}1\end{array} \right) \right| \begin{array}{l}f,g,k,m\in GF(q)\\ fg+km=1\end{array}\right\} \le C_G(t_1) \end{aligned}$$

  with $Q_1$ a normal elementary abelian subgroup of $C_G(t_1)$ and $|Q_1| = q^3$. So $C_G(t_1) = RQ_1$. Thus (ii) holds.

• (iii) This is a routine calculation. From parts (ii) and (iii) $|C_G(t_1)| = q^4(q^2 - 1)$ and $|C_G(t_2)| = q^4$. Combining this with $|G| = q^4(q^2 - 1)(q^4 - 1)$ yields (iv) and (v). $\square$

Lemma 2.4

$|C_G(t_1) \cap X_1| = q^3 - 1$.

Proof

Let s be an involution in S. Then, by Lemma 2.2(ii), $s\in Q_1^{\#}\cup Q_2^{\#}$. Let $v=(\alpha ,\beta ,\gamma ,\delta )$ be a vector in V. Assume for the moment that $s\in Q_1$. Then

$$\begin{aligned} s=\left( \begin{array}{llll}1&{}a&{}b&{}c\\ 0&{}1&{}0&{}b\\ 0&{}0&{}1&{}a\\ 0&{}0&{}0&{}1\end{array} \right) \end{aligned}$$

where $a,b,c\in GF(q)$. So $v^s=(\alpha ,a\alpha +\beta ,b\beta +\gamma ,c\alpha +b\beta +a\gamma +\delta )$. Suppose that at least one of a and b is non-zero. If $v\in C_V(s)$, then we have $a\alpha =b\beta =c\alpha +b\beta +a\gamma =0$. If, say, $a\ne 0$ then this gives $\alpha =0$ and $b\beta +a\gamma =0$. Hence $\gamma =\lambda \beta$ for some $\lambda \in GF(q)$. Thus $\dim C_V(s)=2$, with the same conclusion if $b\ne 0$.

When $a=b=0$ we see that $\dim C_V(s)=3$. Therefore we conclude that

$$\begin{aligned} \left| Q_1\cap X_1\right| =q-1. \end{aligned}$$

(2.4.1)

Now we suppose $s\in Q_2{\setminus } Q_1$. Then

$$\begin{aligned} s=\left( \begin{array}{llll}1&{}0&{}a&{}b\\ 0&{}1&{}c&{}a\\ 0&{}0&{}1&{}0\\ 0&{}0&{}0&{}1\end{array} \right) \end{aligned}$$

where $a,b,c\in GF(q)$ and $c\ne 0$. Here $v^s=(\alpha ,\beta ,a\alpha +c\beta +\gamma ,b\alpha +a\beta +\delta )$ and so, if $v\in C_V(s)$, $a\alpha +c\beta =b\alpha +a\beta =0$. Suppose that $a=0$ and $b\ne 0$. Then $c\beta =b\alpha =0$ which yields $\alpha =0=\beta$. Hence $\dim C_V(s)=2$. Likewise, when $a\ne 0$ and $b=0$ we get $\dim C_V(s)=2$. On the other hand, $a=0=b$ gives $\dim C_V(s)=3$.

Now consider the case when $a\ne 0\ne b$ and $a^2+bc=0$. From $a\alpha +c\beta =0$ we obtain $\beta =a\alpha c^{-1}$ and so $0=b\alpha +a\beta =b\alpha +a^2c^{-1}\alpha =(b+a^2c^{-1})\alpha$. Since $a^2+bc=0$, this equation holds for all $\alpha \in GF(q)$ and consequently $\dim C_V(s)=3$. Similar considerations show that $\dim C_V(s)=2$ when $a\ne 0\ne b$ and $a^2+bc\ne 0$. So, to summarize, for $s\in Q_2{\setminus } Q_1$, $s\in X_1$ when either $a=0=b$ or $a\ne 0\ne b$ and $a^2+bc=0$. For the former, there are $q-1$ such involutions (as $c\ne 0$). For the latter, there are $q-1$ choices for each of b and c and in each case a is uniquely determined (as $GF(q)^{\#}$ is cyclic of odd order), so giving $(q-1)^2$ involutions. Therefore

$$\begin{aligned} \left| (X_1\cap S){\setminus } Q_1\right| =\left| X_1\cap (Q_2{\setminus } Q_1)\right| =q(q-1). \end{aligned}$$

(2.4.2)

Since any two distinct Sylow 2-subgroups of SL(2, q) have trivial intersection and SL(2, q) possesses $q+1$ Sylow 2-subgroups, Lemma 2.3(ii) together with (2.4.1) and (2.4.2) yields that

$$\begin{aligned} \left| C_G(t_1)\cap X_1\right|&= (q-1)+q(q-1)(q+1)\\&=(q-1)(1+q^2+q) = q^3-1. \end{aligned}$$

This proves Lemma 2.4. $\square$

Proof of Theorem 1.1 As is well-known—see for example [18] – G has an outer automorphism arising from the Dynkin diagram of type $C_2 = B_2$. This outer automorphism interchanges the two involution conjugacy classes $X_1$ and $X_3$ and as a consequence $\mathcal {C}(G,X_1)$ and $\mathcal {C}(G,X_3)$ are isomorphic graphs. Thus we need only consider $\mathcal {C}(G,X_1)$. From Lemma 2.4, as $\Delta _1(t) = (C_G(t_1) \cap X_1) {\setminus } \{t_1 \}$,

$$\begin{aligned} |\Delta _1(t_1)| = (q^3 - 1) - 1 = q^3 - 2. \end{aligned}$$

By Theorem 2.1, $\mathcal {C}(G,X_1)$ has diameter 2. Hence, by Lemma 2.3(iv),

$$\begin{aligned} |\Delta _2(t_1)| = |X_1| - (q^3 - 1) = (q^4 - 1) - (q^3 - 1) = q^4 - q^4 = q^3(q - 1), \end{aligned}$$

so proving Theorem 1.1.

Before moving on to prove Theorem 1.2 we need additional preparatory material. If W is a subspace of V, then $W^\perp$ denotes the subspace of V defined by

$$\begin{aligned} W^\perp = \left\{ \left. v\in V\right| (v,w)=0\ \text {for all}\ w\in W\right\} \end{aligned}$$

and we recall that $\dim W+\dim W^\perp =\dim V=4$.

By Lemma 2.3(i), (iii) we see that $C_V(C_G(t_2))=\left\{ \left. (0,0,0,\alpha )\right| \alpha \in GF(q)\right\}$ is 1-dimensional. For $x\in X_2$ set $U_1(x)=C_V(C_G(x))$ and $U_2(x)=C_V(x)$. So $\dim U_1(x)=1$ and $\dim U_2(x)=2$ (with the subscripts acting as a reminder). We denote the stabilizer in G of $U_1(t_2)$, respectively $U_2(t_2)$, by $P_1$, respectively $P_2$. Then $P_i\sim q^3SL_2(q)(q-1)$ for $i=1,2$. Also $Q_i=O_2(P_i)$ with $C_{P_i}(Q_i)=Q_i$ for $i=1,2$.

We start analyzing $\mathcal {C}(G,X_2)$ by determining $\Delta _1(t_2)$. For $x\in X_2$ we let $Z_{C_G(x)}$ denote $Z(C_G(x))\cap X_2$.

Lemma 2.5

$$\begin{aligned} X_2=\displaystyle \dot{\bigcup }_{R\in \text {Syl}_2 G} Z_R. \end{aligned}$$

Proof

Clearly $X_2=\bigcup _{R\in \text {Syl}_2G}Z_R$ by Lemma 2.3(iii). If $Z_R\cap Z_T=\varnothing$ for $R,T\in \text {Syl}_2G$, then we have some $x\in Z(R)\cap Z(T)\cap X_2$ whence, using Lemma 2.3(iii), $R=C_G(x)=T$. So the lemma holds. $\square$

Lemma 2.6

Let $R,T\in \text {Syl}_2G$. If there exists $x\in Z_R$ and $y\in Z_T$ such that $[x,y]=1$, then $[Z_R,Z_T]=1$.

Proof

Since $xy=yx$, $y\in C_G(x)=R$. Hence $Z(R)\le C_G(y)=T$ and so $[Z_R,Z_T]=1$. $\square$

Let $\Delta$ be the building for G and $\mathcal {C}(\Delta )$ denote the chamber graph of $\Delta$. We may view the vertices (chambers) of $\mathcal {C}(\Delta )$ as being $\left\{ \left. N_G(R)\right| R\in \text {Syl}_2G\right\}$ with two distinct chambers $N_G(R)$ and $N_G(T)$ being adjacent whenever $\left\langle N_G(R),N_G(T) \right\rangle \le P_i^g$ for some $g\in G$ and some $i\in \left\{ 1,2\right\}$. We use $d^\mathcal {C}$ to denote the standard distance metric in $\mathcal {C}(\Delta )$ and for a chamber c put $\Delta _j^\mathcal {C}(c)=\left\{ \left. d\in \mathcal {C}(\Delta )\right| d^\mathcal {C}(c,d)=j\right\}$. The structure of $\mathcal {C}(\Delta )$ is well-known.

Lemma 2.7

$\mathcal {C}(\Delta )$ has diameter 4 and $\left| \Delta _1^\mathcal {C}(c)\right| =2q;\ \left| \Delta _2^\mathcal {C}(c)\right| =2q^2;\ \left| \Delta _3^\mathcal {C}(c)\right| =2q^3;$ and $\left| \Delta _4^\mathcal {C}(c)\right| =q^4$.

Proof

A straightforward calculation. $\square$

We now introduce a graph $\mathcal {Z}$ whose vertex set is $V(\mathcal {Z})=\left\{ \left. Z_R\right| R\in \text {Syl}_2G\right\}$ with $Z_R,Z_T\in V(\mathcal {Z})$ joined if $Z_R\ne Z_T$ and $[Z_R,Z_T]=1$.

Lemma 2.8

The graphs $\mathcal {Z}$ and $\mathcal {C}(\Delta )$ are isomorphic.

Proof

Define $\varphi :V(\mathcal {Z})\rightarrow V(\mathcal {C}(\Delta ))$ by $\varphi :Z_R\mapsto N_G(R)$$(R\in \text {Syl}_2G)$. If $\varphi (Z_R)=\varphi (Z_T)$ for $R,T\in \text {Syl}_2G$, then $N_G(R)=N_G(T)$ and so $R=T$ and then $Z_R=Z_T$. Thus $\varphi$ is a bijection between $V(\mathcal {Z})$ and $V(\mathcal {C}(\Delta ))$. Suppose $N_G(R)$ and $N_G(T)$ are distinct, adjacent chambers in $\mathcal {C}(\Delta )$. Without loss of generality we may assume $T=S$. Then $N_G(R),N_G(S)\le P_i$ for $i\in \left\{ 1,2\right\}$. The structure of $P_i$ then forces $Z(R),Z(S)\le Q_i$. Since $Q_i$ is abelian, we deduce that $[Z_R,Z_S]=1$. So $Z_R$ and $Z_S$ are adjacent in $\mathcal {Z}$. Conversely, suppose $Z_R$ and $Z_S$ are adjacent in $\mathcal {Z}$. Then $[Z_R,Z_S]=1$ with, by Lemma 2.5, $Z_R\cap Z_S=\varnothing$. Hence $Z_R\subseteq S$ and so by Lemma 2.2(ii), $Z_R\subseteq Q_1\cup Q_2$. Now $Q_1\cap Q_2\cap X_2=Z_S$ and so we must have $Z_R\subseteq Q_i$ for $i\in \left\{ 1,2\right\}$. The structure of $P_i$ now gives $N_G(R)\le P_i$ and therefore $N_G(R)$ and $N_G(S)$ are adjacent in $\mathcal {C}(\Delta )$, which proves the lemma. $\square$

Proof of Theorem 1.2

Since for all $x_1,x_2\in X_{2}$, $[x_1,x_2]=1$ if and only if $[Z_{C_G(x_1)},Z_{C_G(x_2)}]=1$ by Lemma 2.5, then for $i>1$, $d^{\mathcal {C}}(x_1,x_2)=i$ if and only if $d^{\mathcal {Z}}(Z_{C_G(x_1)},Z_{C_G(x_2)})=i$ (where $d^\mathcal {Z}$ denotes the distance in $\mathcal {Z}$). Note that if $d^{\mathcal {C}}(x_1,x_2)=1$, then either $Z_{C_G(x_1)}=Z_{C_G(x_2)}$ or $d^{\mathcal {Z}}(Z_{C_G(x_1)},Z_{C_G(x_2)})=1$. Since $X_2$ is a disjoint union of the elements of $\mathcal {Z}$, then $\mathcal {C}(G,X_2)$ is connected of diameter 4. Now

$$\begin{aligned} \Delta _1(t)=\bigcup _{\begin{array}{c} R\in \text {Syl}_{2}G \\ {[}Z_S,Z_{R}{]}=1 \end{array}} Z_R \quad {\mathrm{and}}\quad \Delta _i(t)=\bigcup _{\begin{array}{c} R\in {\text {Syl}}_{2}G \\ d^{\mathcal {Z}} (Z_S,Z_R)=i \end{array}} Z_R,\quad i > 1 \end{aligned}$$

and so $\left| \Delta _1(t)\right| =\left| Z_S\right| +2q\left| Z_S\right| -1$. From $\left| Z_S\right| =(q-1)^2$ we get $\left| \Delta _1(t)\right| =(q-1)^2+2q(q-1)^2-1=q^2(2q-3)$. The remaining disc sizes are immediate from the structure of the chamber graph $\mathcal {C}(\Delta )$.$\square$

3 Structure of $\mathcal {C}(G,Y_1)$

This section is devoted to the proof of Theorem 1.3. In order to investigate the disc structure of $\mathcal {C}(G,Y_1)$ it is advantageous for us to work in $H=Sp_4(q)$ (and so $\overline{H}=H/Z(H)\cong G$). We assume that $\left\{ v_1,v_2,v_3,v_4\right\}$ is a hyperbolic basis for V with $(v_2,v_1)=(v_4,v_3)=1$. Thus if J is the matrix defining this form then

$$\begin{aligned} J=\left( \begin{array}{llll}0&{}-1&{}0&{}0 \\ 1&{}0&{}0&{}0 \\ 0&{}0&{}0&{}-1 \\ 0&{}0&{}1&{}0\end{array} \right) , \end{aligned}$$

and J has two diagonal blocks $J_0$ where $J_0=\left( \begin{array}{ll}0&{}-1\\ 1&{}0\end{array} \right)$.

We may suppose that for $t\in Y_1$, we have $\overline{s}=t$ where . Put $X=s^H$. Then $Y_1=\left\{ \left. \overline{x}\right| x\in X\right\}$. For $x\in X$, set $N_x=N_H(\left\langle x,Z(H) \right\rangle )$. Evidently, for $\overline{x_1},\overline{x_2}\in Y_1$ (where $x_1,x_2\in X$) $\overline{x_1}$ and $\overline{x_2}$ commute if and only if $x_1\in N_{x_2}$ (or equivalently $x_2\in N_{x_1}$). Now $N_s$ consists of $g\in H$ for which $s^g=s$ or $s^g=-s$. Letting where A, B, C and D are $2\times 2$ matrices over GF(q), direct calculation reveals that either $B=C=0$ or $A=D=0$. Also, as $g\in H$, we must have $A^TJ_0A=D^TJ_0D=J_0$ and therefore

Lemma 3.1

$$\begin{aligned} \left| \Delta _1(t)\right| =\frac{1}{2}q(q^2-1). \end{aligned}$$

Proof

Since $X=s^H$ consists of all the involutions in $H{\setminus } Z(H)$, a quick calculation gives

Under the natural homomorphism to G, for $x\in X$$\overline{x}=\overline{-x}$, and so $\left| \Delta _1(t)\right| =\frac{1}{2}\left| SL_2(q)\right| =\frac{1}{2}q(q^2-1)$. $\square$

Put $E=\left\langle v_3,v_4 \right\rangle$. Then $E^\perp =\left\langle v_1,v_2 \right\rangle$ and we note that $C_V(s)=E$. Furthermore we have that $\text {Stab}_H(\left\{ E,E^\perp \right\} )=N_s$. Put $\Sigma =\left\{ \left. \left\{ F,F^\perp \right\} \right| F\ \text {is a hyperbolic 2-subspace of}\ V\right\}$. Now let $\beta \in GF(q)$ and set $U_\beta =\left\langle (1,0,1,0),(0,\beta ,0,-\beta -1) \right\rangle$. Then $U_\beta$ is a hyperbolic 2-subspace of V and so $\left\{ U_\beta ,U_\beta ^\perp \right\} \in \Sigma$. The $N_s$-orbit of $\left\{ U_\beta ,U_\beta ^\perp \right\}$ will be denoted by $\Sigma _\beta$.

Lemma 3.2

Let F be a hyperbolic 2-subspace of V with $F\ne E$ or $E^\perp$. Then $\left\{ F,F^\perp \right\} \in \Sigma _\beta$ for some $\beta \in GF(q)$. Moreover, for $\beta \in GF(q)$, $\Sigma _\beta =\Sigma _{-\beta -1}$.

Proof

Since $F\ne E$ or $E^\perp$, we may find $w_1\in F$ with $w_1=(\alpha _1,\beta _1,\gamma _1,\delta _1)$ and $\left\{ \alpha _1,\beta _1\right\} \ne \left\{ 0\right\} \ne \left\{ \gamma _1,\delta _1\right\}$. Now $N_s$ contains two $SL_2(q)$ subgroups for which $\left\langle v_1,v_2 \right\rangle$ and $\left\langle v_3,v_4 \right\rangle$ are natural $GF(q)SL_2(q)$-modules. Because $SL_2(q)$ acts transitively on the non-zero vectors of such modules, we may suppose $w_1=(1,0,1,0)$. Now choose $w_2\in F$ such that $(w_1,w_2)=1$ (and so $\left\langle w_1,w_2 \right\rangle =F$). Then if $w_2=(\alpha ,\beta ,\gamma ,\delta )$ we must have $\beta +\delta =-1$ and so $w_2=(\alpha ,\beta ,\gamma ,-\beta -1)$. The matrices in $N_s$ fixing $w_1$ are

Let

where $a_1,a_2\in GF(q)$. Then $w_1^g=w_1$.

We single out the cases $\beta =0$ and $\beta =-1$ for special attention. If, say, $\beta =0$, then $w_2=(\alpha ,0,\gamma ,-1)$. Hence $w_2-\alpha w_1=(0,0,\gamma -\alpha ,-1)$ and $F=\left\langle w_1,w_2-\alpha w_1 \right\rangle$. Since $(0,0,\gamma -\alpha ,-1)g=(0,0,(\gamma -\alpha )-a_2,-1)$ and choosing $a_2=-\gamma +\alpha$, we obtain $Fg=U_0$. For $\beta =-1$ a similar argument works (using $w_2-\gamma w_1$ instead of $w_2-\alpha w_1$). So we may assume that $\beta \ne 0,-1$. From

$$\begin{aligned} w_2g=\left( \alpha ,\beta ,\gamma ,-\beta -1\right) =\left( \alpha +\beta a_1,\beta ,\gamma +(-\beta -1)a_2,-\beta -1\right) \end{aligned}$$

by a suitable choice of $a_1$ and $a_2$, as $\beta \ne 0,-1$, we get $w_2g=(0,\beta ,0,-\beta -1)$, whence $Fg=U_\beta$. Thus we have shown $\left\{ F,F^\perp \right\} \in \Sigma _\beta$ for some $\beta \in GF(q)$. Finally, for $\beta \in GF(q)$, $\Sigma _\beta =\Sigma _{-\beta -1}$ follows from

$\square$

Let $\phi :GF(q){\setminus }\left\{ -1\right\} \rightarrow GF(q)$ be defined by

$$\begin{aligned} \phi (\lambda )=-\left( 1+(\lambda +1)^{-2}(1-\lambda ^2)\right) ^{-1}\quad (\lambda \in GF(q)). \end{aligned}$$

There is a possibility that this is not well-defined should $1+(\lambda +1)^{-2}(1-\lambda ^2)=0$. This would then give $(\lambda +1)^2+(1-\lambda ^2)=0$ from which we infer that $\lambda =-1$. So we conclude that $\phi$ is well-defined.

Lemma 3.3

$\phi$ is injective.

Proof

Suppose $\phi (\lambda )=\phi (\mu )$ for $\lambda ,\mu \in GF(q){\setminus }\left\{ -1\right\}$ with $\lambda \ne \mu$. Hence

$$\begin{aligned} \left( 1+(\lambda +1)^{-2}(1-\lambda ^2)\right) ^{-1}=\left( 1+(\mu +1)^{-2}(1-\mu ^2)\right) ^{-1}. \end{aligned}$$

Simplifying and using the fact that q is odd gives

$$\begin{aligned} \mu ^2+\mu -\mu \lambda ^2-\lambda ^2-\lambda +\lambda \mu ^2=0, \end{aligned}$$

and then

$$\begin{aligned} (\mu +\lambda )(\mu -\lambda )+(\mu -\lambda )+\lambda \mu (\mu -\lambda )=0. \end{aligned}$$

Hence $(\mu -\lambda )(\mu +\lambda +1+\lambda \mu )=0$. Since $\mu \ne \lambda$, we get $\mu +\lambda +1+\lambda \mu =0$ from which we deduce that either $\lambda =-1$ or $\mu =-1$, a contradiction. So the lemma holds. $\square$

Proof of Theorem 1.3

We first show that $\text {Diam}\ \mathcal {C}(G,Y_1)=2$. So let $x\in X$ be such that $x\notin \left\{ t\right\} \cup \Delta _1(t)$. Now $\left\{ C_V(x),C_V(x)^\perp \right\} \in \Sigma$ as $C_V(x)\ne E$ or $E^\perp$ (otherwise $x\in \left\{ s,-s\right\}$ and then $\overline{x}=t$). Hence $\left\{ C_V(x),C_V(x)^\perp \right\} \in \Sigma _\mu$ for some $\mu \in GF(q)$ by Lemma 3.2. Let . Then $\overline{y}\in \Delta _1(t)$. Our aim is to choose an $x_\lambda \in N_y\cap X$ (so $\overline{x_\lambda }\in \Delta _1(\overline{y})$) for which $\left\{ C_V(x_\lambda ),C_V(x_\lambda )^\perp \right\} \in \Sigma _\mu$. Since $\Sigma _\mu$ is an $N_s$-orbit, there exists $h\in N_s$ such that $\left\{ C_V(x_\lambda ),C_V(x_\lambda )^\perp \right\} ^h=\left\{ C_V(x),C_V(x)^\perp \right\}$. As a consequence either $x=x_\lambda ^h$ or $x_\lambda ^{-1h}$ and therefore $\overline{x}=\overline{x_\lambda }^{\overline{h}}$, whence $d(t,\overline{x})\le 2$. $\square$

We first look at the case when $\mu =-2^{-1}$. Then $\mu =-\mu -1$ and hence

$$\begin{aligned} U_{-2^{-1}}=\left\langle (1,0,1,0),(0,1,0,1) \right\rangle . \end{aligned}$$

Observing that $U_{-2^{-1}}=C_V(y)$, we see that for $\mu =-2^{-1}$, $\overline{x}\in \Delta _1(y)$, which we are not concerned with here. So we may assume $\mu \ne -2^{-1}$.

Let where $\lambda \in GF(q){\setminus }\left\{ 0\right\}$ and such that B has zero trace and determinant $1-\lambda ^2$. So $x_\lambda \in X\cap N_y$. We now move onto the case when $\mu =0$ (or equivalently $\mu =-1$). Here we take $\lambda =1$ and $B=\left( \begin{array}{ll}2&{}-2 \\ 2&{}-2\end{array} \right)$, noting that B satisfies the conditions to ensure that $\overline{x_1}\in \Delta _1(\overline{y})$. Let $v=(\alpha ,\beta ,\gamma ,\delta )\in V$. Then $v\in C_V(x_1)$ precisely when

$$\begin{aligned} 2\gamma +2\delta =0;&\quad -2\gamma -2\delta =0;\\ -2\alpha -2\beta -\gamma =\gamma ;&\quad 2\alpha +2\beta -\delta =\delta ; \end{aligned}$$

and thus the only conditions we get are $\gamma =-\beta -\alpha$ and $\alpha +\beta =\delta$. Thus

$$\begin{aligned} C_V(x_1)&=\left\{ (\alpha ,\beta ,-\alpha -\beta ,\alpha +\beta )\right\} \\&=\left\langle (1,0,-1,1),(0,1,-1,1) \right\rangle . \end{aligned}$$

It is straightforward to check that $\left\{ C_V(x_1),C_V(x_1)^\perp \right\} \in \Sigma _0$. Therefore we may also assume that $\mu \ne 0,-1$. Choosing $B=\left( \begin{array}{ll}\lambda &{}\lambda ^{-1} \\ -\lambda &{} -\lambda \end{array} \right)$ we see that the requisite conditions are satisfied. Take $v=(\alpha ,\beta ,\gamma ,\delta )\in V$ and calculating $v^{x_\lambda }$ gives the relations

$$\begin{aligned} (\lambda -1)\alpha +\gamma \lambda -\delta \lambda =0;&\quad (\lambda -1)\beta +\gamma \lambda ^{-1}-\delta \lambda =0;\\ -\lambda \alpha +\lambda \beta -(\lambda +1)\gamma =0;&\quad -\lambda ^{-1}\alpha +\lambda \beta -(\lambda +1)\delta =0; \end{aligned}$$

which, after rearranging gives

$$\begin{aligned} \alpha =\lambda (\lambda -1)^{-1}(\delta -\gamma );&\quad \beta =\lambda (\lambda -1)^{-1}\delta -\lambda ^{-1}(\lambda -1)^{-1}\gamma ;\\ \gamma =\lambda (\lambda +1)^{-1}(\beta -\alpha );&\quad \delta =\lambda (\lambda +1)^{-1}\alpha -\lambda ^{-1}(\lambda +1)^{-1}\alpha ; \end{aligned}$$

and note that the relations for $\gamma$ and $\delta$ are satisfied after substitution for $\alpha$ and $\beta$. Hence

$$\begin{aligned} C_V(x_\lambda )&=\left\{ \left( \alpha ,\beta ,\lambda (\lambda +1)^{-1}(\beta -\alpha ),\lambda (\lambda +1)^{-1}\beta -\lambda ^{-1}(\lambda +1)^{-1}\alpha \right) \right\} \nonumber \\&=\left\langle \left( 1,0,-\lambda (\lambda +1)^{-1},-\lambda ^{-1}(\lambda +1)^{-1}\right) ,\left( 0,1,\lambda (\lambda +1)^{-1},\lambda (\lambda +1)^{-1}\right) \right\rangle . \end{aligned}$$

(3.3.1)

We want to determine which $N_{s}$-orbit, $\Sigma _\beta$, that $C_V(x_\lambda )$ lies in. Our representative, $U_\beta$, for $\Sigma _\beta$ has $w_1=(1,0,1,0)$ as one component of the hyperbolic pair, so we need an element of $N_{s}$ to send the first generator in (3.3.1) to $w_1$. We need to find conditions on $C,D\in SL_2(q)$ such that

and so without loss of generality we can take $C=I_2$. This reduces to solving

$$\begin{aligned} \left( -\lambda (\lambda +1)^{-1},-\lambda ^{-1} (\lambda +1)^{-1}\right) \left( \begin{array}{ll}d_1&{}d_2\\ d_3&{}d_4\end{array} \right) =(1,0) \end{aligned}$$

and after multiplying out, we get that $d_3=-(d_1+1)\lambda ^2-\lambda$ and $d_4=-d_2\lambda ^2$. Since D has determinant 1, we find that $d_2=\lambda ^{-1}(\lambda +1)^{-1}$ and so $d_4=-\lambda (\lambda +1)^{-1}$. Without loss of generality, by taking $d_1=1$ we have that

$$\begin{aligned} D=\left( \begin{array}{ll}1&{}\lambda ^{-1}(\lambda +1)^{-1}\\ -2\lambda ^2-\lambda &{}-\lambda (\lambda +1)^{-1}\end{array} \right) \end{aligned}$$

and a quick check shows that the first generator in (3.3.1) is mapped to $w_1$. Using the same matrix, by multiplying on the right of the second generator in (3.3.1), we get

and $\left\langle w_1,u' \right\rangle$ is a hyperbolic 2-subspace conjugate to some $U_\beta$. Recall that for a fixed $\beta \in GF(q)$, $N_{s}$ is transitive on $\left\{ \left. (\alpha ,\beta ,\gamma ,-\beta -1)\right| \alpha ,\gamma \in GF(q)\right\}$. Hence, we need only find the hyperbolic pair representing such a conjugate of $U_\beta$, to determine $\beta$. This is found by requiring that some multiple of $u'$ has inner product 1 with $w_1$, that is

$$\begin{aligned} \beta \cdot 1=-1-\beta \left( (\lambda +1)^{-2}(1-\lambda ^2)\right) \end{aligned}$$

for some $\beta \in GF(q)$. By expanding, we get that $\beta =-\left( 1+(\lambda +1)^{-2}(1-\lambda ^2)\right) ^{-1}$ and so $C_V(x_\lambda )\in \Sigma _\beta$. By Lemma 3.3, $\phi :\lambda \mapsto -(1+(\lambda +1)^{-2}(1-\lambda ^2))^{-1}$ is an injective map from $GF(q){\setminus }\left\{ -1\right\}$ into GF(q). Since $\mu \ne -2^{-1}$, $\mu \ne -\mu -1$ and therefore there exists $\lambda \in GF(q){\setminus }\left\{ -1\right\}$ such that $\phi (\lambda )=\mu$ or $-\mu -1$. Bearing in mind that $U_\mu =U_{-\mu -1}$ by Lemma 3.2, we conclude that $\left\{ C_V(x_\lambda ),C_V(x_\lambda )^\perp \right\} \in \Sigma _\mu$. Consequently we have proved that $\text {Diam}\ \mathcal {C}(G,Y_1)=2$.

From $\left| G\right| =\frac{q^4}{2}(q^2-1)(q^4-1)$ and $\left| C_G(t)\right| =q^2(q^2-1)^2$ we see that $\left| Y_1\right| =\frac{q^2}{2}(q^2+1)$. Using Lemma 3.1 then gives

$$\begin{aligned} \left| \Delta _2(t)\right| =\frac{1}{2}\left( q^4-q^3+q^2+q-2\right) , \end{aligned}$$

which completes the proof of Theorem 1.3.

4 Structure of $\mathcal {C}(G,Y_2)$

In this section we present a proof of Theorem 1.4. The uncovering of the disc structures of $\mathcal {C}(G,Y_2)$ will be a long haul. As discussed in Sect. 1, it will be advantageous for us to use the well known isomorphism that $PSp (4,q)\cong O(5,q)$ (see Corollary 12.32 of [44]). So we take $G=O(5,q)$ and from now on V will denote the 5-dimensional GF(q) orthogonal module for G. Thus the elements of G are $5\times 5$ orthogonal matrices with respect to the orthogonal form $(\ ,\ )$ which have spinor norm a square in GF(q). We may assume that the Gram matrix with respect to $(\ ,\ )$ is

Let

Then $t\in G$ and $Y_2=t^G$. Let $\delta =\pm 1$ where $q\equiv \delta \pmod 4$.

Lemma 4.1

• (i) $\dim (C_V(t))=3$.

• (ii) $C_V(t)^\perp =[V,t]$ is a 2-subspace of V of $\delta$-type.

• (iii) $V=C_V(t)\perp C_V(t)^\perp$.

Proof

An easy calculation. $\square$

Put $L_t=C_G(t)\cap C_G([V,t])$.

Lemma 4.2

• (i) Let $x\in Y_2$. Then $t=x$ if and only if $C_V(t)=C_V(x)$.

• (ii) $C_G(t)=\text {Stab}_G(C_V(t))\sim (L_2(q)\times \frac{q-\delta }{2}).2^2$.

• (iii) $L_t$ acts faithfully on $C_V(t)$ and $L_t\cong L_2(q)$.

Proof

• (i) Suppose $C_V(x)=C_V(t)$. Then, using Lemma 4.1 (ii), $[V,x]=C_V(x)^\perp =C_V(t)^\perp =[V,t]$. Hence by Lemma 4.1(iii), tx acts trivially on V and thus $tx=1$. Therefore $t=x$ and (i) holds.

• (ii) Plainly $C_G(t)\le \text {Stab}_G(C_V(t))$, and if $g\in \text {Stab}_G(C_V(t))$, then $C_V(t)=C_V(t)^g=C_V(t^g)$. Hence, as $t^g\in Y_2$, $t=t^g$ by part (i). So $g\in C_G(t)$ and thus $C_G(t)=\text {Stab}_G(C_V(t))$. That $\text {Stab}_G(C_V(t))\sim (L_2(q)\times \frac{q-\delta }{2}).2^2$ can be read off from Proposition 4.1.6 of [32].

• (iii) For any $g\in C_G(t)$, we have $[V,t]^g=C_V(t)^{\perp g}=C_V(t^g)^\perp =C_V(t)^\perp =[V,t]$ and so $C_G(t)\le \text {Stab}_G[V,t]$. If any element in $L_t$ acts trivially on $C_V(t)$, then it would act trivially on V and thus be the identity. Hence $L_t$ acts faithfully on $C_V(t)$. Let $v\in C_V(t)$ and by Lemma 4.1(iii), we have $[V,t]\le \left\langle v \right\rangle ^\perp$. Hence $\left\langle v \right\rangle ^\perp =[V,t]\oplus W$ where $W\le C_V(t)$. But since $\dim (\left\langle v \right\rangle ^\perp )=4$, we have $\dim (W)=2$ and so $C_V(t)\nleq \left\langle v \right\rangle ^\perp$. Therefore for all $u\in C_V(t)$, $(v,u)=0$ if and only if $v=0$ and thus $(\ ,\ )$ is non-degenerate on restriction to $C_V(t)$. Hence we have $L_t\hookrightarrow GO(C_V(t))\sim GO_3(q)$ as $L_t$ fixes [V, t] pointwise, by definition. Since $L_t\le G$ and acts as determinant 1 on [V, t], then it must act as determinant 1 on $C_V(t)$. In addition, as $L_t$ fixes [V, t] pointwise, when the elements of $L_t$ are decomposed as products of refections, the vectors reflected will lie in $C_V(t)$. Since the spinor norm of the elements of $L_t$ are a square in GF(q) and the vectors reflected lie in $C_V(t)$, then the spinor norm doesn’t change on restriction to $C_V(t)$. Hence, $L_t\sim O_3(q)\sim L_2(q)$ proving (iii).

$\square$

Let $\mathcal {U}_i$ denote the set of i-dimensional subspaces of $C_V(t)$, $i=1,2$. In proving Theorem 1.4, our divide and conquer strategy is based on the following observation.

Lemma 4.3

$$\begin{aligned} \displaystyle Y_2\subseteq \bigcup _{U\in \mathcal {U}_1\cup \mathcal {U}_2} C_G(U). \end{aligned}$$

Proof

Let $x\in Y_2{\setminus }\left\{ t\right\}$ and set $U=C_V(t)\cap C_V(x)$. By Lemmas 4.1(i) and 4.2(i), $U\in \mathcal {U}_1\cup \mathcal {U}_2$. Since $t,x\in C_G(U)$, we have Lemma 4.3. $\square$

The three cases we must chase down are presaged by our next result.

Lemma 4.4

• (i) Let $U_0$ be an isotropic 1-subspace of $C_V(t)$. Then $C_G(U_0)\sim q^3:L_2(q)$.

• (ii) Let $U_\varepsilon$ be a 1-subspace of $C_V(t)$, such that $U_\varepsilon ^\perp \cap C_V(t)$ is a 2-space of $\varepsilon$-type ($\varepsilon =\pm 1$). Then

  $$\begin{aligned} C_G(U_\varepsilon )\sim \left\{ \begin{array}{ll} SL_2(q)\circ SL_2(q) &{} \delta =\varepsilon \\ L_2(q^2) &{} \delta =-\varepsilon .\end{array}\right. \end{aligned}$$

Proof

Let $U_0$ be an isotropic 1-subspace of $C_V(t)$. From Proposition 4.1.20 of [32], we know that $\text {Stab}_G(U_0)\sim C_0:(C_1\times C_2)\left\langle r \right\rangle$ where $C_1$ acts as scalars on $U_0$, r a reflection of $U_0$ and $C_0\sim q^3$, $C_2\sim L_2(q)$ fixing $U_0$ pointwise. Hence $C_G(U_0)\sim q^3:L_2(q)$, so proving (i).

If $\delta =1$, then [V, t] is a 2-subspace of V of $+$-type, and hence $U_+^\perp =(U_+^\perp \cap C_V(t))\perp [V,t]$ is a 4-subspace of $+$-type. Similarly, $U_-^\perp =(U_-^\perp \cap C_V(t))\perp [V,t]$ is a 4-space of −-type. If $\delta =-1$, then [V, t] is a 2-subspace of V of −-type, and the results for when $\delta =1$ interchange. Let $W_+$ and $W_-$ be 4-subspaces of V of $+$ and −-type respectively, such that $W_+^\perp$ and $W_-^\perp$ are 1-subspaces of $C_V(t)$, observing that $\text {Stab}_G(W_\pm )=\text {Stab}_G(W_\pm ^\perp )$. From Proposition 4.1.6 of [32], we have

$$\begin{aligned} \text {Stab}_G(W_+)&\sim C_+\left\langle s_+ \right\rangle \\ \text {Stab}_G(W_-)&\sim C_-\left\langle s_- \right\rangle \end{aligned}$$

where $C_+\sim SL_2(q)\circ SL_2(q)$ fixes $W_+^\perp$ pointwise, $C_-\sim L_2(q^2)$ fixes $W_-^\perp$ pointwise and $s_+,\ s_-$ are reflections of $W_+^\perp$ and $W_-^\perp$ respectively. This proves (ii) and hence the lemma. $\square$

Lemma 4.5

• (i) Let $U_0$ be a 2-subspace of $C_V(t)$ such that $U_0^\perp \cap C_V(t)$ is an isotropic 1-space. Then $C_G(U_0)\sim q^2:\frac{q-\delta }{2}$.

• (ii) Let $U_\varepsilon$ be a 2-subspace of $C_V(t)$ of $\varepsilon$-type ($\varepsilon =\pm 1$). Then $C_G(U_\varepsilon )\cong L_2(q)$.

Proof

See Propositions 4.1.6 and 4.1.20 of [32]. $\square$

Define the following subsets of $\mathcal {U}_i$, $i=1,2$.

$$\begin{aligned} \mathcal {U}_1^+&=\left\{ \left. U\in \mathcal {U}_1\right| C_G(U)\cong SL_2(q)\circ SL_2(q)\right\} \\ \mathcal {U}_1^-&=\left\{ \left. U\in \mathcal {U}_1\right| C_G(U)\cong L_2(q^2)\right\} \\ \mathcal {U}_1^0&=\left\{ \left. U\in \mathcal {U}_1\right| C_G(U)\sim q^3:L_2(q)\right\} \\ \mathcal {U}_2^+&=\left\{ \left. U\in \mathcal {U}_2\right| U\ {\text {is of +-type}}\right\} \\ \mathcal {U}_2^-&=\left\{ \left. U\in \mathcal {U}_2\right| U\,{\text {is of}} -{\text{-type}}\right\} \\ \mathcal {U}_2^0&=\left\{ U\in \mathcal {U}_2\left| C_G(U)\sim q^2:\frac{q-\delta }{2}\right. \right\} . \end{aligned}$$

In the notation of Lemma 4.4, $\mathcal {U}_1^+$ is the case $\delta =\varepsilon$ while $\mathcal {U}_1^-$ is when $\delta =-\varepsilon$. Note by Lemmas 4.4 and 4.5 that $\mathcal {U}_i=\mathcal {U}_i^0\cup \mathcal {U}_i^+\cup \mathcal {U}_i^-$, $i=1,2$. We now study $C_G(U)\cap Y_2$ for $U\in \mathcal {U}_1$. By Lemma 4.4 there are three possibilities for the structure of $C_G(U)$. First we look at the case $U\in \mathcal {U}_1^-$, and set $G^-=C_G(U)$. Then $G^-\cong L_2(q^2)$ by definition of $\mathcal {U}_1^-$. Define $\Delta _i^-(t)=\left\{ \left. x\in G^-\cap Y_2\right| d^-(t,x)=i\right\}$ where $i\in \mathbb {N}$ and $d^-$ is the distance metric on the commuting graph $\mathcal {C}(G^-,G^-\cap Y_2)$.

Theorem 4.6

If $q\ne 3$ then $\mathcal {C}(G^-,G^-\cap Y_2)$ is connected of diameter 3 with

$$\begin{aligned} \left| \Delta _1^-(t)\right|&=\frac{1}{2}(q^2-1);\\ \left| \Delta _2^-(t)\right|&=\frac{1}{4}(q^2-1)(q^2-5);\quad \text {and}\\ \left| \Delta _3^-(t)\right|&=\frac{1}{4}(q^2-1)(q^2+7). \end{aligned}$$

Proof

Since $q^2\equiv 1\pmod 4$ and $q\ne 3$ implies $q^2>13$, this follows from Theorem 1.1(iii) of [11]. $\square$

We move on to analyze $G^+=C_G(U)$ where $U\in \mathcal {U}_1^+$. Hence, by definition of $\mathcal {U}_1^+$, $G^+\cong L_1\circ L_2$ where $L_1\cong SL_2(q)\cong L_2$ (with the central product identifying $Z(L_1)$ and $Z(L_2)$). Set $Y^+= G^+\cap Y_2$. We begin by describing $Y^+$.

Lemma 4.7

$$\begin{aligned} Y^+=\left\{ \left. x_1x_2\right| x_i\in L_i\ \text {and}\ x_i\ \text {has order 4},\ i=1,2\right\} . \end{aligned}$$

Proof

Apart from the central involution z of $G^+$, all other involutions of $G^+$ are of the form $g_1g_2$ where $g_i\in L_i$$(i=1,2)$ has order 4. Since all involutions in $L_i/Z(G^+)$ are conjugate, it quickly follows that $\left\{ \left. g_1g_2\right| g_i\in L_i\ \text {and}\ g_i\ \text {has order 4},\ i=1,2\right\}$ is a $G^+$-conjugacy class. Now z acts as $-1$ on $U^\perp$ and thus $\dim \ C_V(z)=1$. Therefore $t\ne z$ whence, as $t\in G^+$, the lemma holds. $\square$

Let $d^+$ denote the distance metric on the commuting graph $\mathcal {C}(G^+,Y^+)$ and, for $i\in \mathbb {N}$, $\Delta _i^+(t)=\left\{ \left. x\in Y^+\right| d^+(t,x)=i\right\}$.

Theorem 4.8

Assume that $q\notin \left\{ 3,5,9,13\right\}$. Then $\mathcal {C}(G^+,Y^+)$ is connected of diameter 3 with

$$\begin{aligned} \left| \Delta _1^+(t)\right|&= \frac{1}{2}(q-\delta )^2+1;\\ \left| \Delta _2^+(t)\right|&= \frac{1}{8}(q-\delta )^3(q-4-\delta )+(q-\delta )(q-2-\delta );\quad \text {and}\\ \left| \Delta _3^+(t)\right|&= \frac{3}{8}q^4+\frac{1}{2}(1+3\delta )q^3-\frac{1}{4}(7+6\delta )q^2+\frac{7}{2}(1+\delta )q-\frac{1}{8}(29+20\delta ). \end{aligned}$$

Proof

Let $\overline{G^+}=G^+/Z(G^+)\ (=\overline{L_1}\times \overline{L_2})$. Note that for $x_1x_2\in Y^+$, $x_1^{-1}x_2=x_1x_2^{-1}$ and $x_1x_2=x_1^{-1}x_2^{-1}$ and so the inverse image of $\overline{x_1x_2}$ contains two elements of $Y^+$. Let $d^{(i)}$ denote the distance metric on the commuting graph of $\overline{L_i}$ and $\Delta _j^{(i)}(\overline{x_i})$ the $j^\text {th}$ disc of $\overline{x_i}$ in the commuting graph of $\overline{L_i}$. By Lemma 4.7, $t=t_1t_2$ where, for $i=1,2$, $t_i\in L_i$ has order 4. Let $x=x_1x_2\in Y^+$ with $x\ne t$. Then $tx=xt$ if and only if tx has order 2. So, bearing in mind that $Y^+\cup \left\{ z\right\}$ (where $\left\langle z \right\rangle =Z(G^+)$) are all the involutions of $G^+$, we have that $tx=xt$ if and only if one of the following holds:- $x_1=t_1,\ x_2=t_2^{-1}$; $x_1=t_1^{-1}$, $x_2=t_2$; $\overline{x_1}\in \Delta _1^{(1)}(\overline{t_1})$ and $\overline{x_2}\in \Delta _1^{(2)}(\overline{t_2})$. Thus

$$\begin{aligned} \Delta _1^+(t)=\left\{ x_1x_2\left| \overline{x_i}\in \Delta _1^{(i)}(\overline{t_i}),\ i=1,2\right. \right\} \cup \left\{ t_1t_2^{-1}\right\} . \end{aligned}$$

(4.8.1)

Hence, using [11],

$$\begin{aligned} \left| \Delta _1^+(t)\right| = 2\left( \frac{1}{2}(q-\delta )\right) ^2+1=\frac{1}{2}(q-\delta )^2+1. \end{aligned}$$

(4.8.2)

Next we examine $\Delta _2^+(t)$. Let $x\in Y^+$. Assume that $x=x_1t_2$ or $x_1t_2^{-1}$ where $\overline{x_1}\in \Delta _1^{(1)}(\overline{t_1})$. Then $x\in \Delta _1^+(t_1t_2^{-1})$ (recall $t_1t_2^{-1}=t_1^{-1}t_2$) which implies, by (4.8.1), that $x\in \Delta _2^+(t)$. If $x=t_1x_2$ or $t_1^{-1}x_2$ where $\overline{x_2}\in \Delta _1^{(2)}(\overline{t_2})$, we similarly get $x\in \Delta _2^+(t)$. Therefore

$$\begin{aligned} \left\{ x_1x_2\left| \overline{x_1}\in \Delta _1^{(1)}(\overline{t_1}),\ \overline{x_2}=\overline{t_2}\right. \right\} \cup \left\{ x_1x_2\left| \overline{x_2}\in \Delta _1^{(2)}(\overline{t_2}),\ \overline{x_1}=\overline{t_1}\right. \right\} \subseteq \Delta _2^+(t). \end{aligned}$$

(4.8.3)

Now suppose $x=x_1x_2$ where $\overline{x_1}\in \Delta _2^{(1)}(\overline{t_1})$ and $\overline{x_2}\in \Delta _1^{(2)}(\overline{t_2})$. So there exists $\overline{y_1}\in \overline{L_1}$ such that $(\overline{t_1},\overline{y_1},\overline{x_1})$ is a path of length 2 in the commuting graph for $\overline{L_1}$. Then $(t=t_1t_2,\ y_1x_2^{-1},\ x_1x_2=x)$ is a path of length 2 in $\mathcal {C}(G^+,Y^+)$. Thus, by (4.8.1), $x\in \Delta _2^+(t)$. If, on the other hand, $\overline{x_1}\in \Delta _1^{(1)}(\overline{t_1})$ and $\overline{x_2}\in \Delta _2^{(2)}(\overline{t_2})$ we obtain the same conclusion. Should we have $\overline{x_1}\in \Delta _2^{(1)}(\overline{t_1})$ and $\overline{x_2}\in \Delta _2^{(2)}(\overline{t_2})$, similar arguments also give $x\in \Delta _2^+(t)$. So

$$\begin{aligned}&\left\{ x_1x_2\left| \overline{x_1}\in \Delta _2^{(1)}(\overline{t_1}),\ \overline{x_2}\in \Delta _1^{(2)}(\overline{t_2})\right. \right\} \cup \left\{ x_1x_2\left| \overline{x_1}\in \Delta _1^{(1)}(\overline{t_1}),\ \overline{x_2}\in \Delta _2^{(2)}(\overline{t_2})\right. \right\} \nonumber \\&\quad \cup \left\{ x_1x_2\left| \overline{x_1}\in \Delta _2^{(1)}(\overline{t_1}),\ \overline{x_2}\in \Delta _2^{(2)}(\overline{t_2})\right. \right\} \subseteq \Delta _2^+(t). \end{aligned}$$

(4.8.4)

Since $x=x_1x_2\in \Delta _2^+(t)$ implies $d^{(i)}(\overline{t_i},\overline{x_i})\le 2$ for $i=1,2$, $\Delta _2^+(t)$ is the union of the two sets in (4.8.3) and (4.8.4). Thus, employing [11],

$$\begin{aligned} \left| \Delta _2^+(t)\right| = \frac{1}{8}(q-\delta )^3(q-4-\delta )+(q-\delta )(q-2-\delta ). \end{aligned}$$

(4.8.5)

Now, as $q\notin \left\{ 3,5,9,13\right\}$, by [11] the commuting graph for $\overline{L_i}$ is connected of diameter 3. Arguing as above we deduce that $\mathcal {C}(G^+,Y^+)$ is also connected with diameter 3. Because $\left| Y^+\right| =2\left| \overline{t_1}^{\overline{L_1}}\right| \left| \overline{t_2}^{\overline{L_2}}\right| =\frac{1}{2}q^2(q+\delta )^2$, combining (4.8.2) and (4.8.5) we may determine $\left| \Delta _3^+(t)\right|$ to be as stated, so completing the proof of Theorem 4.8. $\square$

Finally we look at $C_G(U)$ where $U\in \mathcal {U}_1^0$. This will prove to be trickier than the other two cases. Put $G^0=C_G(U)$. So $G^0\sim q^3:L_2(q)$. We require an explicit description of $G^0$ which we now give. Let $Q=\left\{ \left. (\alpha ,\beta ,\gamma )\right| \alpha ,\beta ,\gamma \in GF(q)\right\}$ and

$$\begin{aligned} L=\left\{ \left. \left( \begin{array}{ccc}a^2&{}2ab&{}b^2 \\ ac&{}ad+bc&{}bd \\ c^2&{}2cd&{}d^2\end{array} \right) \right| \begin{array}{c}a,b,c,d\in GF(q) \\ ad-bc=1\end{array}\right\} . \end{aligned}$$

with L acting on Q by right multiplication. Then $Q\sim q^3$ and $L\cong L_2(q)$. Since Q is the 3-dimensional GF(q)L-module (see the description on page 15 of [4]), $G^0\cong Q\rtimes L$. We will identify this semidirect product with $G^0$, writing $G^0=QL$. Any $g\in G^0$ has a unique expression $g=g_Qg_L$ where $g_Q\in Q$ and $g_L\in L$ - in what follows we use such subscripts to describe this expression. Set $Y^0=G^0\cap Y_2$, let $d^0$ denote the distance metric and $\Delta _i^0(t)$ the $i^\text {th}$ disc of the commuting graph $\mathcal {C}(G^0,Y^0)$. In determining the discs of $\mathcal {C}(G^0,Y^0)$ we make use of the commuting involution graph of $L\cong L_2(q)$ (as given in [11]). So we shall use $d^L$ to denote the distance metric on $\mathcal {C}(L,L\cap Y^0)$ and for $x\in L\cap Y^0$ and $i\in \mathbb {N}$, $\Delta _i^L(x)=\left\{ \left. y\in L\cap Y^0\right| d^L(x,y)=i\right\}$. It is straightforward to check that

$$\begin{aligned} L\cap Y^0 = \left\{ \left. \left( \begin{array}{ccc}a^2&{}2ab&{}b^2 \\ ac&{}bc-a^2&{}-ab \\ c^2&{}-2ac&{}a^2\end{array} \right) \right| \begin{array}{c}a,b,c\in GF(q)\\ a^2+b^2=-1\end{array}\right\} \end{aligned}$$

and, as $G^0$ has one conjugacy class of involutions, $Y^0=\left\{ \left. x_Qx_L\right| x_L\in L\cap Y^0\ {\text {and }} x_L {\text { inverts }} x_Q\right\}$. Without loss of generality, we take

$$\begin{aligned} t=t_L=\left( \begin{array}{ccc}0&{}0&{}1 \\ 0&{}-1&{}0 \\ 1&{}0&{}0\end{array} \right) \end{aligned}$$

and, up until Theorem 4.14, we will assume that $q\notin \left\{ 3,5,9,13\right\}$. Thus the diameter of $\mathcal {C}(L,L\cap Y^0)$ is 3.

Lemma 4.9

• (i) $Qt\cap Y^0=\left\{ \left. (\alpha ,\beta ,-\alpha )t\right| \alpha ,\beta \in GF(q)\right\}$ and $\left| Qt\cap Y^0\right| =q^2$.

• (ii) $Qt\cap \Delta _1^0(t)=\varnothing$.

Proof

A straightforward calculation. $\square$

Lemma 4.10

We have

$$\begin{aligned} \Delta _1^0(t)=\left\{ x\left| x_Q=(\alpha ,0,\alpha ),\ x_L=\left( \begin{array}{ccc}a^2&{}2ab&{}b^2\\ ab&{}b^2-a^2&{}-ab\\ b^2&{}-2ab&{}a^2\end{array} \right) ,\ a^2+b^2=-1\right. \right\} , \end{aligned}$$

and $\left| \Delta _1^0(t)\right| =\frac{1}{2}q(q-\delta )$.

Proof

Let $x,y\in Y^0$. If $[x,y]=1$ then clearly $[x_L,y_L]=1$. From [11] we have

$$\begin{aligned} \Delta _1^L(t) = \left\{ \left. \left( \begin{array}{ccc}a^2&{}2ab&{}b^2 \\ ab&{}b^2-a^2&{}-ab \\ b^2&{}-2ab&{}a^2\end{array} \right) \right| a^2+b^2=-1\right\} . \end{aligned}$$

If $x_Q=(\alpha ,\beta ,\gamma )$ and $x_L\in \Delta _1^L(t)$ then $[t,x]=1$ implies $\alpha =\gamma$ and $\beta =0$. Moreover, every $x=(\alpha ,0,\alpha )x_L$, where $x_L\in \Delta _1^L(t)$, is in $Y^0$. Hence, $\Delta _1^0(t)$ is as described above. By [11], for any involution $x_L\in L$ we have $\left| \Delta _1^L(x_L)\right| =\frac{1}{2}(q-\delta )$ and there are q possible values that $\alpha$ can take for a fixed such $x_L$, proving the lemma. $\square$

Lemma 4.11

Let $x\in Y^0$ with $x_L\in \Delta _1^L(t)$. If $x\notin \Delta _1^{0}(t)$, then $x\in \Delta _2^{0}(t)$.

Proof

Suppose $x\in Y^0$ where $x_Q=(\alpha ,\beta ,\gamma )$ and

$$\begin{aligned} x_L=\left( \begin{array}{ccc}a^2&{}2ab&{}b^2 \\ ab&{}b^2-a^2&{}-ab\\ b^2&{}-2ab&{}a^2\end{array} \right) . \end{aligned}$$

Then $x_L$ inverts $x_Q$ if and only if

$$\begin{aligned} a^2\alpha +2ab\beta +b^2\gamma&=-\alpha \nonumber \\ ab\alpha +(b^2-a^2)\beta -ab\gamma&=-\beta \nonumber \\ b^2\alpha -2ab\beta +a^2\gamma&=-\gamma . \end{aligned}$$

(4.11.1)

Suppose first that $\delta =-1$. Then, since $-1$ is not square in GF(q), we must have $a,b\ne 0$. Rearranging the first equation gives $\alpha =2ab^{-1}\beta +\gamma$ and (4.11.1) remains consistent. Note that when $\beta =0$, we have $\alpha =\gamma$ and so $x\in \Delta _1^0(t)$. So assume $\beta \ne 0$. Let $y\in \Delta _1^0(t)$ where $y_Q=(ab^{-1}\beta +\gamma ,0,ab^{-1}\beta +\gamma )$ and

$$\begin{aligned} y_L=\left( \begin{array}{ccc}b^2&{}-2ab&{}a^2 \\ -ab&{}a^2-b^2&{}ab \\ a^2&{}2ab&{}b^2\end{array} \right) . \end{aligned}$$

It is a routine calculation to show that $[x,y]=1$, proving the lemma for $\delta =-1$. Now assume $\delta =1$. If $a,b\ne 0$ then the argument from the previous case still holds, so assume first that $a=0$, and hence b is an element in GF(q) that squares to $-1$. Then (4.11.1) simplifies to $\alpha =\gamma$, and so $x_Q=(\alpha ,\beta ,\alpha )$. Let $z\in \Delta _1^0(t)$ where $z_Q=(\alpha ,0,\alpha )$ and

$$\begin{aligned} z_L=\left( \begin{array}{ccc}-1&{}0&{}0 \\ 0&{}1&{}0 \\ 0&{}0&{}-1\end{array} \right) . \end{aligned}$$

An easy calculation shows that $[x,z]=1$. Similarly, assuming $b=0$ then a is an element of GF(q) squaring to $-1$ and (4.11.1) simplifies to $\beta =0$. Then $x_Q=(\alpha ,0,\gamma )$ and if $w\in \Delta _1^0(t)$ where $w_Q=(2^{-1}(\alpha +\gamma ),0,2^{-1}(\alpha +\gamma ))$ and

$$\begin{aligned} w_L=\left( \begin{array}{ccc}0&{}0&{}-1\\ 0&{}-1&{}0\\ -1&{}0&{}0\end{array} \right) \end{aligned}$$

then an easy check shows that $[x,w]=1$, proving the lemma for $\delta =1$. $\square$

Lemma 4.12

We have $Qt\cap Y^0\subseteq \left\{ t\right\} \cup \Delta _2^0(t)\cup \Delta _3^0(t)$. Moreover,

$$\begin{aligned} \left| Qt\cap \Delta _2^{0}(t)\right|&=\frac{1}{2}\left( q^2-(1+\delta )q+\delta \right) ;\ \text {and}\\ \left| Qt\cap \Delta _3^{0}(t)\right|&=\frac{1}{2}\left( q^2+(1+\delta )q-(2+\delta )\right) . \end{aligned}$$

Proof

If $x\in Qt\cap Y^0$ and $x\ne t$ then $x_Q=(\alpha ,\beta ,-\alpha )$ and $x\notin \Delta _1^0(t)$ by Lemma 4.9. Let $y\in \Delta _1^0(t)$ where $y_Q=(\gamma ,0,\gamma )$ and

$$\begin{aligned} y_L=\left( \begin{array}{ccc}a^2&{}2ab&{}b^2\\ ab&{}b^2-a^2&{}-ab\\ b^2&{}-2ab&{}a^2\end{array} \right) \end{aligned}$$

with $a^2+b^2=-1$. Then $[x,y]=1$ if and only if $-a^2\alpha =ab\beta$ and $-b^2\beta =ab\alpha$.

Assume first that $\delta =-1$. Since $-1$ is not square in GF(q), we have $a,b\ne 0$ and so $\alpha =-a^{-1}b\beta$. Hence if $y\in Qt$ is such that $y_Q=(-a^{-1}b\beta ,\beta ,a^{-1}b\beta )$, then $y\in \Delta _2^0(t)$. By looking at $\Delta _1^L(t)$, we see there are $q+1$ ordered pairs (a, b) that satisfy $a^2+b^2=-1$. However, if $(a,b)\ne (c,d)$ where $a^2+b^2=c^2+d^2=-1$ and $a^{-1}b=c^{-1}d$, then an easy calculation shows that $(c,d)=(-a,-b)$. Hence there are $\frac{1}{2}(q+1)$ distinct values of $a^{-1}b$ satisfying the relevant conditions. If $\beta =0$, then $x=t$ and if $\beta \ne 0$ there are $\frac{1}{2}(q^2-1)$ elements in $Qt\cap \Delta _2^0(t)$.

Assume now that $\delta =1$. If $a,b\ne 0$ then the arguments of the previous case still hold, with the exception that there are now $q-1$ ordered pairs (a, b) that satisfy $a^2+b^2=-1$. However, as $a,b\ne 0$ we exclude the pairs $(\pm i,0)$ and $(0,\pm i)$ where i is an element of GF(q) squaring to $-1$. Hence there are $q-5$ ordered pairs (a, b) satisfying $a^2+b^2=-1$, $a,b\ne 0$ and thus $\frac{1}{2}(q-5)$ distinct values of $a^{-1}b$. Hence there are $\frac{1}{2}(q-5)(q-1)$ elements $z\in Qt\cap \Delta _2^0(t)$ such that $z_Q=(-a^{-1}b\beta ,\beta ,a^{-1}b\beta )$ where $\beta \ne 0$ (note that if $\beta =0$, then $z=t$). Suppose $a=0$, then $b\ne 0$ and so $\beta =0$. Hence $x_Q=(\alpha ,0,-\alpha )$ and all such x lie in $\Delta _2^0(t)$ if $\alpha \ne 0$. Similarly, if $b=0$ then $a\ne 0$ and $x_Q=(0,\beta ,0)$ where $\beta \ne 0$ and all such x lie in $\Delta _2^0(t)$. Therefore, $\left| Qt\cap \Delta _2^0(t)\right| =\frac{1}{2}(q-5)(q-1)+2(q-1)=\frac{1}{2}(q-1)^2$ as required.

Hence it suffices to show that the remaining involutions all lie in $\Delta _3^0(t)$. Let $w\in Qt$ be such that $w_Q=(\gamma ,\varepsilon ,-\gamma )$. Choose $s\in Y^0$ such that $s_Q=(ab\varepsilon -b^2\gamma ,ab\gamma -a^2\varepsilon ,b^2\gamma -ab\varepsilon )$ with $ab\gamma \ne a^2\varepsilon$ and

$$\begin{aligned} s_L=\left( \begin{array}{ccc}b^2&{}-2ab&{}a^2 \\ -ab&{}a^2-b^2&{}ab \\ a^2&{}2ab&{}b^2\end{array} \right) , \end{aligned}$$

with $a^2+b^2=-1$. It is an easy check to show that $s\in \Delta _2^0(t)$, and moreover $[w,s]=1$. This accounts for the remaining involutions in Qt, thus proving the lemma. $\square$

Lemma 4.13

Suppose $x\in Y^0$ with $x_L\in \Delta _2^L(t)$. Then $x\in \Delta _2^0(t)$.

Proof

It can be shown (see Remark 2.3 of [11], noting the result holds for any odd q) that for a fixed $a,b\in GF(q)$ such that $a^2+b^2=-1$,

$$\begin{aligned} C_L\left( \left( \begin{array}{ccc}a^2&{}2ab&{}b^2 \\ ab&{}b^2-a^2&{}-ab \\ b^2&{}-2ab&{}a^2\end{array} \right) \right) =\left\{ \left. \left( \begin{array}{ccc}c^2&{}2cd&{}d^2 \\ ce&{}de-c^2&{}-cd \\ e^2&{}-2ce&{}c^2\end{array} \right) \right| \begin{array}{c}c^2+de=-1\\ b(e+d)=-2ac\end{array}\right\} . \end{aligned}$$

Let $y\in Y^0$ be such that $y_Q=(\alpha ,\beta ,\gamma )$ and

$$\begin{aligned} y_L=\left( \begin{array}{ccc}c^2&{}2cd&{}d^2\\ ce&{}de-c^2&{}-cd\\ e^2&{}-2ce&{}c^2\end{array} \right) \in \Delta _2^L(t). \end{aligned}$$

So there exists $a,b\in GF(q)$ such that $a^2+b^2=-1$ and $b(e+d)=-2ac$ with $d\ne e$. Since $y_L$ inverts $y_Q$, we have

$$\begin{aligned} c^2\alpha +2cd\beta +d^2\gamma&=-\alpha \nonumber \\ ce\alpha +(de-c^2)\beta -cd\gamma&=-\beta \nonumber \\ e^2\alpha -2ce\beta +c^2\gamma&=-\gamma . \end{aligned}$$

(4.13.1)

Assume first that $\delta =-1$. Since $-1$ is not square in GF(q), then $d,e\ne 0$ and any $a,b\in GF(q)$ such that $b(d+e)=-2ac$ and $a^2+b^2=-1$ must also be non-zero. Moreover, if $c=0$ then $d=-e^{-1}$ and $b(d-d^{-1})=0$ implying that $d=-1$. But then $y_L=t\notin \Delta _2^L(t)$, so $c\ne 0$. The system (4.13.1) now simplifies to $\alpha =2ce^{-1}\beta +de^{-1}\gamma$. Let $x\in \Delta _1^0(t)$ be such that $x_Q=(\varepsilon ,0,\varepsilon )$ and

$$\begin{aligned} x_L=\left( \begin{array}{ccc}a^2&{}2ab&{}b^2 \\ ab&{}b^2-a^2&{}-ab \\ b^2&{}-2ab&{}a^2\end{array} \right) \end{aligned}$$

where $\varepsilon =-abc^{-1}e^{-1}(\gamma +(d-e)^{-1}(2c+a^{-1}be-ab^{-1}e-(ab)^{-1}e)\beta )$. Using the PolynomialAlgebra command in Magma [17] we verify that $[x,y]=1$ and so $y\in \Delta _2^0(t)$.

Assume now that $\delta =1$. Let $a,b\in GF(q)$ be such that $a^2+b^2=-1$ and $b(d+e)=-2ac$. Suppose $c,d,e\ne 0$ and $d\ne -e$. Then $b(d+e)=-2ac\ne 0$ and so $a,b\ne 0$. The argument for the case when $\delta =-1$ then holds. Suppose then $c,d,e\ne 0$ and $d=-e$. Then $b(d+e)=-2ac=0$ and since $c\ne 0$ we must have $a=0$ and $b^2=-1$. The system (4.13.1) then becomes $\alpha =2ce^{-1}\beta -\gamma$. If $x\in \Delta _1^0(t)$ is such that $x_Q=(-c^{-1}e^{-1}\beta ,0,-c^{-1}e^{-1}\beta )$ and

$$\begin{aligned} x_L=\left( \begin{array}{ccc}0&{}0&{}-1 \\ 0&{}-1&{}0 \\ -1&{}0&{}0\end{array} \right) \end{aligned}$$

then a routine check shows that $[x,y]=1$.

Now assume $c\ne 0$ and $d=0$. Since $y_L\in \Delta _2^L(t)$, we must have $e\ne 0$ and so $c^2=-1$. The system (4.13.1) becomes $\alpha =2ce^{-1}\beta$ and using Magma [17] we deduce that if $x\in \Delta _1^0(t)$ where $x_Q=(\varepsilon ,0,\varepsilon )$,

$$\begin{aligned} x_L=\left( \begin{array}{ccc}a^2&{}2ab&{}b^2\\ ab&{}b^2-a^2&{}-ab\\ b^2&{}-2ab&{}a^2\end{array} \right) \end{aligned}$$

and $\varepsilon =(ce^{-1}(1-a^2)-ab)\beta -2^{-1}b^2\gamma$, then $[x,y]=1$. Similarly, if $c\ne 0$ and $e=0$, then $d\ne 0$ and $c^2=-1$. The system (4.13.1) becomes $\beta =2^{-1}cd\gamma$ and [17] will verify that if $x\in \Delta _1^0(t)$ where $x_Q=(\varepsilon ,0,\varepsilon )$,

$$\begin{aligned} x_L=\left( \begin{array}{ccc}a^2&{}2ab&{}b^2\\ ab&{}b^2-a^2&{}-ab\\ b^2&{}-2ab&{}a^2\end{array} \right) \end{aligned}$$

and $\varepsilon =2^{-1}(\gamma -b^2\alpha +abcd\gamma -a^2\gamma )$, then $[x,y]=1$.

Finally, if $c=0$ then $d=-e^{-1}$ and so $a^2=-1$ and $b=0$ satisfies the relevant conditions. Note that if $d=\pm 1$ then $y_L=t$, so we may assume $d\ne \pm 1$. The system (4.13.1) becomes $\alpha =d^2\gamma$, so if $x\in \Delta _1^0(t)$ where $x_Q=(2d^2\gamma (1-d^2)^{-1},0,2d^2\gamma (1-d^2)^{-1})$ and

$$\begin{aligned} x_L=\left( \begin{array}{ccc}-1&{}0&{}0\\ 0&{}1&{}0 \\ 0&{}0&{}-1\end{array} \right) \end{aligned}$$

then a routine check again shows that $[x,y]=1$. Therefore, for all $y\in Y^0$ such that $y_L\in \Delta _2^L(t)$, there exists $x\in \Delta _1^L(t)$ such that $[x,y]=1$, so proving the lemma. $\square$

Theorem 4.14

If $q\notin \left\{ 3,5,9,13\right\}$, then $\mathcal {C}(G^0,Y^0)$ is connected of diameter 3, with disc sizes

$$\begin{aligned} \left| \Delta _1^{0}(t)\right|&=\frac{1}{2}q(q-\delta );\\ \left| \Delta _2^{0}(t)\right|&=\frac{1}{4}\left( q^4-(2\delta +2)q^3+(1+2\delta )q^2-2q+2\delta \right) ;\ \text {and}\\ \left| \Delta _3^{0}(t)\right|&=\frac{1}{4}\left( q^4+2(1+2\delta )q^3-(3+2\delta )q^2+2(1+\delta )q-2(2+\delta )\right) . \end{aligned}$$

Proof

It is known that $\mathcal {C}(L,L\cap Y^0)$ has diameter 3. Hence, for any $h_i\in \Delta _i^L(t)$, there exists $h_{i\pm 1}\in \Delta _{i\pm 1}^L(t)$ that commutes with $h_i$, $i=1,2$. Therefore for any $x\in Y^0$ where $x_L\in \Delta _i^L(t)$, there exists $y\in Y^0$ with $y_L\in \Delta _{i\pm 1}^L(t)$ and such that $[x,y]=1$. Since any $z\in Y^0$ where $z_L\in \Delta _3^L(t)$ must commute with some $w\in Y^0$ with $w_L\in \Delta _2^L(t)$ (which lies in $\Delta _2^0(t)$ by Lemma 4.13), $z\in \Delta _3^0(t)$. This finally covers all possible involutions in $Y^0$ and so the diameter of $\mathcal {C}(G^0,Y^0)$ is 3. Now for each $x_L\in L\cap Y^0$, $\left| Qx_L\cap Y^0\right| =q^2$ by Lemma 4.9, and therefore there are $\frac{1}{2}q^2(q-\delta )$ involutions $y\in Y^0$ such that $y_L\in \Delta _1^L(t)$. From Lemma 4.10, $\left| \Delta _1^0(t)\right| =\frac{1}{2}q(q-\delta )$. Therefore

$$\begin{aligned} \left| \bigcup _{x_L\in \Delta _1^L(t)} Qx_L \cap \Delta _2^0(t)\right| =\frac{1}{2}q^2(q-\delta )-\frac{1}{2}q(q-\delta )=\frac{1}{2}q(q-1)(q-\delta ). \end{aligned}$$

There are $q^2\left| \Delta _2^L(t)\right|$ involutions $z\in Y^0$ such that $z_L\in \Delta _2^L(t)$, which is known to be $\frac{1}{4}q^2(q-\delta )(q-4-\delta )$ (see [11]). Also, by Lemma 4.12, $\left| Qt\cap \Delta _2^0(t)\right| =\frac{1}{2}(q^2-(1+\delta )q-\delta )$. Hence

$$\begin{aligned} \left| \Delta _2^0(t)\right|&=\left| Qt\cap \Delta _2^0(t)\right| +\left| \bigcup _{x_L\in \Delta _1^L(t)} Qx_L \cap \Delta _2^0(t)\right| +q^2\left| \Delta _2^L(t)\right| \\&=\frac{1}{4}\left( q^4-(2\delta +2)q^3+(1+2\delta )q^2-2q+2\delta \right) . \end{aligned}$$

Finally, there are $\left| Y^0\right| =q^2\left| L\cap Y^0\right| =\frac{1}{2}q^3(q+\delta )$ involutions in $G^0$ and therefore

$$\begin{aligned} \left| \Delta _3^0(t)\right|&=\left| Y^0\right| -\left| \Delta _2^0(t)\right| -\left| \Delta _1^0(t)\right| -1\\&=\frac{1}{4}\left( q^4+2(1+2\delta )q^3-(3+2\delta )q^2+2(1+\delta )q-2(2+\delta )\right) \end{aligned}$$

which proves Theorem 4.14. $\square$

Theorem 4.15

$\mathcal {C}(G,Y_2)$ is connected of diameter at most 3.

Proof

For $q\le 13$, this is easily checked using Magma [17], so assume $q>13$. Combining Lemma 4.3 with Theorems 4.6, 4.8 and 4.14 yields the theorem. $\square$

We now focus on finding the disc sizes of $\mathcal {C}(G,Y_2)$. First, we need the following four lemmas.

Lemma 4.16

The sets $\mathcal {U}_1^+$, $\mathcal {U}_1^-$ and $\mathcal {U}_1^0$ are single $C_G(t)$-orbits. Moreover,

$$\begin{aligned} \left| \mathcal {U}_1^0\right|&= q+1;\\ \left| \mathcal {U}_1^+\right|&= \frac{1}{2}q(q+\delta );\ \text {and}\\ \left| \mathcal {U}_1^-\right|&= \frac{1}{2}q(q-\delta ). \end{aligned}$$

Proof

Since $C_G(t)$ acts orthogonally on $C_V(t)$, the first statement is immediate. Recall the Gram matrix J for V with respect to $(\ ,\ )$ and the basis $\left\{ v_i\right\}$. Observe that $C_V(t)=\left\{ \left. (\alpha ,\beta ,\gamma ,0,\gamma )\right| \alpha ,\beta ,\gamma \in GF(q)\right\}$ and so a basis for $C_V(t)$ is $\left\{ v_1,v_2,v_3+v_5\right\}$. Let $v=(\alpha ,\beta ,\gamma ,0,\gamma )$ be a non-zero vector in $C_V(t)$ and so $(v,v)=2\alpha \beta +2\gamma ^2$.

Suppose v is isotropic, so $C_G(\left\langle v \right\rangle )\sim q^3:L_2(q)$ and $(v,v)=2\alpha \beta +2\gamma ^2=0$. If $\gamma =0$, then $\alpha \beta =0$ and so either $\alpha =0$ or $\beta =0$ (but not both as $v\ne 0$). Hence there are $2(q-1)$ such vectors with $\gamma =0$. If $\gamma \ne 0$, then $\alpha =-\beta ^{-1}\gamma ^2$ and there are $(q-1)^2$ such vectors satisfying this. Hence there are $2(q-1)+(q-1)^2=(q-1)(q+1)$ non-zero isotropic vectors contained in $C_V(t)$ and thus $q+1$ isotropic 1-subspaces of $C_V(t)$.

Suppose now v is $C_G(t)$-conjugate to $v_3+v_5$, which is non-isotropic. Note that $\left\langle v_3+v_5 \right\rangle ^\perp \cap C_V(t)$ is a 2-subspace of V of $+$-type. If $\delta =1$, then by Lemma 4.1(ii), $\left\langle v_3+v_5 \right\rangle ^\perp$ is a 4-subspace of V of $+$-type and so $C_G(\left\langle v_3+v_5 \right\rangle )\sim SL_2(q)\circ SL_2(q)$. While $\delta =-1$ gives that $\left\langle v_3+v_5 \right\rangle ^\perp$ is a 4-subspace of V of −-type and so $C_G(\left\langle v_3+v_5 \right\rangle )\sim L_2(q^2)$. A quick check shows that $(v_3+v_5,v_3+v_5)=2$ and so $(v,v)=2\alpha \beta +2\gamma ^2=2\lambda ^2$ for some $\lambda \in GF(q)^*$. Thus, $\alpha \beta +\gamma ^2=\lambda ^2$ for some $\lambda \in GF(q)^*$. If $\gamma =0$, then $\alpha =\beta ^{-1}\lambda ^2$ and so there are $q-1$ such vectors that satisfy this. If $\gamma =\pm \lambda$, then $\alpha \beta =0$ and so for both values of $\gamma$, there are $2(q-1)+1$ vectors that satisfy this. Finally, if $\gamma \in GF(q){\setminus }\left\{ 0,\lambda ,-\lambda \right\}$, then $\alpha \beta =1-\gamma ^2\ne 0$ and so $\alpha =\beta ^{-1}(1-\gamma ^2)$. There are $(q-1)(q-3)$ such vectors that satisfy this. Hence for any given $\lambda$, there exist $(q-1)+4(q-1)+2+(q-1)(q-3)=q(q+1)$ vectors that satisfy $\alpha \beta +\gamma ^2=\lambda ^2$. Since there are $\frac{1}{2}(q-1)$ squares in GF(q), there are $q(q+1)(q-1)$ vectors that are $C_G(t)$-conjugate to $v_3+v_5$ and hence $\frac{1}{2}q(q+1)$ 1-subspaces of $C_V(t)$ that are $C_G(t)$-conjugate to $\left\langle v_3+v_5 \right\rangle$.

This leaves the remaining orbit $\mathcal {U}_1^{-\delta }$. Recall there are $q^2+q+1$ subspaces of $C_V(t_1)$ of dimension 1, and hence the size of the remaining orbit is $q^2+q+1-(q+1)-\frac{1}{2}q(q+1)=\frac{1}{2}q(q-1)$, so proving the lemma. $\square$

Corollary 4.17

The sets $\mathcal {U}_2^+$, $\mathcal {U}_2^-$ and $\mathcal {U}_2^0$ are single $C_G(t)$-orbits. Moreover,

$$\begin{aligned} \left| \mathcal {U}_2^0\right|&= q+1;\\ \left| \mathcal {U}_2^+\right|&= \frac{1}{2}q(q+1);\ \text {and}\\ \left| \mathcal {U}_2^-\right|&= \frac{1}{2}q(q-1). \end{aligned}$$

Proof

Since $C_V(t)$ is 3-dimensional, $U^\perp \cap C_V(t)\in \mathcal {U}_1$ for any $U\in \mathcal {U}_2$, and so the result is immediate by Lemma 4.16. $\square$

Lemma 4.18

Let $U,U'\in \mathcal {U}_2$ be such that $U\ne U'$. Then $C_G(U)\cap C_G(U')\cap Y_2=\left\{ t\right\}$.

Proof

Suppose $x\in C_G(U)\cap C_G(U')\cap Y_2$. Since $U\ne U'$ and x fixes each 2-subspace pointwise, $U+U'=C_V(t)$ and so x fixes $C_V(t)$ pointwise. That is to say, $C_V(x)=C_V(t)$ and so $t=x$ by Lemma 4.2(i). $\square$

Lemma 4.19

Let $U_0\in \mathcal {U}_2^0$, and $G^0=QL,\ Y^0$ be as defined in the discussion prior to Lemma 4.9. Let $\rho :C_G(U_0^\perp \cap C_V(t))\rightarrow G^0$ be an isomorphism such that

$$\begin{aligned} t^\rho =\left( \begin{array}{ccc}0&{}0&{}1\\ 0&{}-1&{}0\\ 1&{}0&{}0\end{array} \right) . \end{aligned}$$

Then $C_G(U_0)$ is totally disconnected and $(C_G(U_0)\cap Y_2)^\rho = Qt \cap Y^0$.

Proof

Since $U_0^\perp \cap C_V(t)$ is isotropic, it must lie inside of $U_0$ and so $C_G(U_0)\le C_G(U_0^\perp \cap C_V(t))$. As t fixes $U_0$ pointwise, $t^\rho \in (C_G(U_0))^\rho \sim q^2:\frac{q-\delta }{2}$ by Lemma 4.5(i). The subgroup of L with shape $\frac{q-\delta }{2}$ contains one single involution which must necessarily be $t^\rho$. For all $x\in Y^0$, we have $x_L^2=1$ and $x_L$ inverts $x_Q$, so $(C_G(U_0)\cap Y_2)^\rho \subseteq Qt\cap Y^0$. By comparing the orders of both sides, we get equality. By Lemma 4.9(ii) $C_G(U_0)\cap C_G(t)\cap Y_2=\left\{ t\right\}$, hence $C_G(U_0)$ is totally disconnected. $\square$

Lemma 4.20

$$\begin{aligned} \left| \Delta _1(t)\right| =\frac{1}{2}q\left( q^2+(1-\delta )q+\delta \right) . \end{aligned}$$

Proof

Clearly, $x\in \Delta _1(t)$ if and only if $x\in \Delta _1(t)\cap C_G(U)$ for $U=C_V(t)\cap C_V(x)$, so

$$\begin{aligned} \Delta _1(t)=\bigcup _{U\in \mathcal {U}_1\cup \mathcal {U}_2}\left( \Delta _1(t)\cap C_G(U)\right) . \end{aligned}$$

If $W,W'\in \mathcal {U}_1$ with $W\ne W'$, then $W\oplus W'\in \mathcal {U}_2$ and if $y\in C_G(W)\cap C_G(W')$, then $y\in C_G(W\oplus W')$ and hence $y\in C_G(W'')$ for any 1-subspace $W''$ of $W\oplus W'$. Since there are $q+1$ subspaces of $W''$ of dimension 1, any such y will lie in exactly $q+1$ such $C_G(U)$ for $U\in \mathcal {U}_1$. Together with $C_G(W'')$ and Lemma 4.18,

$$\begin{aligned} \left| \Delta _1(t)\right| =\sum _{U\in \mathcal {U}_1} \left| \Delta _1(t)\cap C_G(U)\right| - q\sum _{U\in \mathcal {U}_2}\left| \Delta _1(t_1)\cap C_G(U)\right| . \end{aligned}$$

Combining Lemmas 4.16, 4.19 and Corollary 4.17 with Theorems 4.6, 4.8, 4.14 and [11], we have

$$\begin{aligned} \left| \Delta _1(t)\right|&=\frac{1}{2}q(q+1)(q-\delta )+\frac{1}{2}q(q+\delta )\left[ \frac{1}{2}(q-\delta )^2+1\right] +\frac{1}{4}q(q-\delta )(q^2-1)\\&\qquad -\frac{1}{2}q(q-\delta )\left[ \frac{1}{2}q(q+1)+\frac{1}{2}q(q-1)\right] \\&=\frac{1}{2}q(q^2+(1-\delta )q+\delta ) \end{aligned}$$

as required. $\square$

We now consider the second disc $\Delta _2(t)$. Here, we must be careful as elements that are distance 2 from t in some subgroup $C_G(U)$ may not be distance 2 from t in another subgroup $C_G(U')$. Moreover, there may be elements that are distance 3 from t in every such subgroup centralizing an element of $\mathcal {U}_1$, but actually are distance 2 from t in G. We introduce the following notation. Let $\Delta _2^K(t)$ be the second disc in the commuting involution graph $\mathcal {C}(K,K\cap Y_2)$ and

$$\begin{aligned} \Gamma _i(K)=\left\{ \left. x\in \Delta _2^K(t)\right| \dim C_V(\left\langle t,x \right\rangle )=i\right\} \end{aligned}$$

for $K=C_G(U)$, $U\in \mathcal {U}_1\cup \mathcal {U}_2$. Clearly, $\Delta _2(t)=\Gamma _1(G)\dot{\cup }\Gamma _2(G)$. A full list of cases with corresponding notation is found in Table 1. Also we use the following notation: for any $U\le C_V(t)$, define $\mathcal {U}_i(U)$ to be the totality of i-dimensional subspaces of U and $\mathcal {W}_i(U)$ to be the totality of i-dimensional subspaces of $C_V(t)$ containing U. Note that $\mathcal {U}_i=\mathcal {U}_i(C_V(t))$.

Table 1 List of cases in $\Delta _2(t)$

Lemma 4.21

• (i) If $W\in \mathcal {U}_2^0$, then $\left| \mathcal {U}_1^0\cap \mathcal {U}_1(W)\right| =1$ and $\left| \mathcal {U}_1^+\cap \mathcal {U}_1(W)\right| =q$.

• (ii) If $W\in \mathcal {U}_2^+$, then $\left| \mathcal {U}_1^0\cap \mathcal {U}_1(W)\right| =2$ and $\left| \mathcal {U}_1^+\cap \mathcal {U}_1(W)\right| =\left| \mathcal {U}_1^-\cap \mathcal {U}_1(W)\right| =\frac{q-1}{2}$.

• (iii) If $W\in \mathcal {U}_2^-$, then $\left| \mathcal {U}_1^+\cap \mathcal {U}_1(W)\right| =\left| \mathcal {U}_1^-\cap \mathcal {U}_1(W)\right| =\frac{q+1}{2}$.

Proof

Recall the Gram matrix J, with respect to the ordered basis $\left\{ v_i\right\}$, $i=1,\ldots ,5$. Suppose $W^\perp \cap C_V(t)=U_0\in \mathcal {U}_1^0$. Without loss of generality, choose $W=\left\langle v_1,v_3+v_5 \right\rangle$. Clearly $\left\langle v_1 \right\rangle \in \mathcal {U}_1^0$, and $\left\langle v_3+v_5 \right\rangle ^\perp \cap C_V(t)\in \mathcal {U}_2^+$. Since

$$\begin{aligned} \left( v_1+\lambda (v_3+v_5) , v_1+\lambda (v_3+v_5)\right) = \lambda ^2(v_3+v_5, v_3+v_5), \end{aligned}$$

$v_1+\lambda (v_3+v_5)$ lies in the same $C_G(t)$-orbit as $v_3+v_5$ and so $\left\langle v_1+\lambda (v_3+v_5) \right\rangle ^\perp \cap C_V(t)\in \mathcal {U}_2^+$, proving (i).

Suppose now $W\in \mathcal {U}_2^+$. Without loss of generality, choose $W=\left\langle v_1,v_2 \right\rangle$. Clearly $\left\langle v_1 \right\rangle ,\ \left\langle v_2 \right\rangle \in \mathcal {U}_1^0$. Let $U_\lambda = \left\langle v_1+\lambda v_2\right\rangle$ for $\lambda \ne 0$ and note that $(v_1+\lambda v_2,v_1+\lambda v_2)=2\lambda =\mu \ne 0$. Since the type of $U_\lambda ^\perp$ is determined by whether $\mu$ is a square or a non-square in GF(q), and there are $\frac{q-1}{2}$ of each, it is clear that there exist $\frac{q-1}{2}$ such $U_\lambda$ for which $U_\lambda ^\perp$ is of $+$-type, and similarly for −-type, proving (ii).

Finally suppose $W\in \mathcal {U}_2^-$, so for all $v\in W$, $(v,v)\ne 0$. The simple orthogonal group on W is cyclic of order $\frac{q+1}{2}$ and acts on the 1-subspaces of W in exactly two orbits with representatives $\left\langle u_1 \right\rangle$ and $\left\langle u_2 \right\rangle$ where $(u_1,u_1)$ is a square and $(u_2,u_2)$ is a non-square in GF(q). Since $\left| \mathcal {U}_1(W)\right| =q+1$, both orbits must be of size $\frac{q+1}{2}$. This proves (iii) and hence the lemma follows. $\square$

Corollary 4.22

Let $U\in \mathcal {U}_1$. Then the following:

• (i) $\left| \mathcal {W}_2(U)\right| =q+1.$

• (ii) If $U\in \mathcal {U}_1^0$, then $\left| \mathcal {U}_2^0\cap \mathcal {W}_2(U)\right| =1$ and $\left| \mathcal {U}_2^+\cap \mathcal {W}_2(U)\right| =q$.

• (iii) If $U\in \mathcal {U}_1^\delta$, then $\left| \mathcal {U}_2^0\cap \mathcal {W}_2(U)\right| =2$ and $\left| \mathcal {U}_2^+\cap \mathcal {W}_2(U)\right| =\left| \mathcal {U}_2^-\cap \mathcal {W}_2(U)\right| =\frac{q-1}{2}$.

• (iv) If $U\in \mathcal {U}_2^{-\delta }$, then $\left| \mathcal {U}_2^+\cap \mathcal {W}_2(U)\right| =\left| \mathcal {U}_2^-\cap \mathcal {W}_2(U)\right| =\frac{q+1}{2}$.

Proof

Let $U\le W\le C_V(t)$. Then $W^\perp \cap C_V(t)\le U^\perp \cap C_V(t)\le C_V(t)$. The result follows from Lemma 4.21. $\square$

Lemma 4.23

Let $U\in \mathcal {U}_1^0$ and $W\in \mathcal {U}_2^+\cap \mathcal {W}_2(U)$. If $x\in Y_2\cap C_G(W)$ is such that $d^{C_G(W)}(t,x)=3$, then $d^{C_G(U)}(t,x)=3$. Moreover,

$$\begin{aligned} \left| \Gamma _1(C_G(U))\right| = \left\{ \begin{array}{ll} \frac{1}{4}q(q-3)(q-1)^2 &{} \quad q\equiv 1\pmod 4\\ \frac{1}{4}q(q-1)^2(q+1) &{} \quad q\equiv -1\pmod 4. \end{array}\right. \end{aligned}$$

Proof

Recall that $C_G(U)=QL\sim G^0$ where $G^0$ is defined as in the discussion prior to Lemma 4.9. By conjugacy, we may assume $L=C_G(W)$. Now $C_G(U)\cap C_G(t)=Q_0C_L(t)\sim q:Dih(q-\delta )$ where $Q_0\le Q$ is elementary abelian of order q. Let $x\in Q_0C_L(t)\cap Y_2$, so $x_L^2=1$ and $x_L$ inverts $x_Q$. Clearly, $x_L^{x_Q}=x_Lx_Q^2\notin L$ since $Q_0$ is of odd order. Hence, $C_L(t)$ is self-normalizing in $Q_0C_L(t)$ and thus there are q distinct conjugates of $C_L(t)$ in $Q_0C_L(t)$. Let $g\in Q_0C_L(t){\setminus } C_L(t)$, so $C_L(t)^g\ne C_L(t)$. Now $[C_L(t),t]=[C_L(t)^g,t]=1$ and so $\left\langle C_L(t),C_L(t)^g \right\rangle$ centralizes t. If $C_L(t)$, $C_L(t)^g\le L^h$ for some $h\in QL$, then $\left\langle C_L(t),C_L(t)^g \right\rangle \le L^h$. However, $C_L(t)\lneqq \left\langle C_L(t),C_L(t)^g \right\rangle \le C_L(t)$, a contradiction. Hence every conjugate of $C_L(t)$ lies in a different conjugate ofL and so there are q distinct $Q_0C_L(t)$-conjugates of L. Therefore, $\mathcal {U}_2^+\cap \mathcal {W}_2(U)$ is contained in the same $C_G(U)\cap C_G(t)$-orbit, and $\left| \mathcal {U}_2^+\cap \mathcal {W}_2(U)\right| =q$ by Corollary 4.22. There are exactly q$+$-type 2-subspaces of $C_V(t)$ containing U, all of which lie in the same $C_G(U)\cap C_G(t)$ orbit.

Let $x\in C_G(W)\cap Y_2$ be such that $d^{C_G(W)}(t,x)=3$. Suppose $W^g\in \mathcal {U}_2^+\cap \mathcal {W}_2(U)$ for some $g\in C_G(U)\cap C_G(t)$, $W\ne W^g$. If $d^{C_G(U)}(t,x)=2$ then $d^{C_G(U)}(t^g,x^g)=d^{C_G(U)}(t,x^g)=2$, and $d^{C_G(W)}(t,x)=d^{C_G(W^{g})}(t,x^g)=3$. Hence it suffices to prove the lemma for $C_G(W)$. By Theorem 4.14, any involution distance 3 away from t in L is necessarily distance 3 away from t in $C_G(U)$, proving the first statement.

Let $W_0\in \mathcal {U}_2^0\cap \mathcal {W}_2(U)$, so $C_G(W_0)\sim q^2:\frac{q-\delta }{2}$. By Lemma 4.19, $\Delta _2^{C_G(U)}(t)\cap C_G(W_0)=Qt\cap \Delta _2^{C_G(U)}(t)$. Let $W_i$, $i=1,\ldots ,q$ be the subspaces in $\mathcal {U}_2^+\cap \mathcal {W}_2(U)$. From Lemma 4.18, $C_G(W_i)\cap C_G(W_j)\cap Y_2=\left\{ t\right\}$ if and only if $i=j$. Using Corollary 4.22(i) with [11], we have

$$\begin{aligned} \left| \bigcup _{i=1}^q\Delta _2^{C_G(W_i)}(t)\right| =\frac{1}{4}q(q-\delta )(q-4-\delta ). \end{aligned}$$

(4.23.1)

Combining Lemma 4.12 with (4.23.1),

$$\begin{aligned} \left| \Gamma _2(C_G(U))\right|&=\left| \bigcup _{i=1}^q\Delta _2^{C_G(W_i)}(t)\right| + \left| \Delta _2^{C_G(U)}(t)\cap C_G(W_0)\right| \nonumber \\&=\frac{1}{4}\left( q^3-2(1+\delta )q^2+(2\delta -1)q+2\delta \right) . \end{aligned}$$

(4.23.2)

Together, (4.23.2) and Theorem 4.14 give

$$\begin{aligned} \left| \Gamma _1(C_G(U))\right|&=\left| \Delta _2^{C_G(U)}(t)\right| -\left| \Gamma _2(C_G(U))\right| \\&=\frac{1}{4}q\left( q^3-(2\delta +3)q^2+(4\delta +3)q-2\delta -1\right) \end{aligned}$$

as required. $\square$

Lemma 4.24

Let $t,x\in L_2(q)$. Then $d^{L_2(q)}(t,x)\le 2$ if and only if the order of tx divides $\frac{1}{2}(q-\delta )$.

Proof

See Lemma 2.11 of [11]. $\square$

Lemma 4.25

Let $U\in \mathcal {U}_1^+$, and $W\in (\mathcal {U}_2^+\cup \mathcal {U}_2^-)\cap \mathcal {W}_2(U)$.

• (i) If $\delta =1$ and $W_0\in \mathcal {U}_2^0\cap \mathcal {W}_2(U)$, then $Y_2\cap C_G(W_0){\setminus }\left\{ t\right\} \subseteq \Delta _3^{C_G(U)}(t)$.

• (ii) If $x\in Y_2\cap C_G(W)$ is such that $d^{C_G(W)}(t,x)=3$, then $d^{C_G(U)}(t,x)=3$ and

  $$\begin{aligned} \left| \Gamma _1(C_G(U))\right| =\left\{ \begin{array}{ll} \frac{1}{8}(q-1)(q-3)(q^2-6q+13) &{} \quad q\equiv 1\pmod 4\\ \frac{1}{8}(q^2-1)(q^2-2q+5) &{} \quad q\equiv -1\pmod 4. \end{array}\right. \end{aligned}$$

Proof

Recall that $C_G(U)\sim G^+\sim L_1\circ L_2$ for $L_1\sim SL_2(q)\sim L_2$. Suppose $y\in C_G(W)$ is such that $d^{C_G(W)}(t,y)=3$. Since $C_G(W)\sim L_2(q)$ is simple, then $y=gg^\varphi$ for some $g\in L_1$ and $\varphi :L_1\rightarrow L_2$. Since $t\in C_G(W)$, write $t=ss^\varphi$ for some $s\in L_1$. Then $d^{L_1}(s,g)=3$, so $d^{C_G(U)}(t,y)=3$ by Theorem 4.8, and thus

$$\begin{aligned} \Delta _3^{C_G(W)}(t)\subseteq \Delta _3^{C_G(U)}(t)\quad \text {for all}\ W\in (\mathcal {U}_2^+\cup \mathcal {U}_2^-)\cap \mathcal {W}_2(U). \end{aligned}$$

(4.25.1)

If $\delta =-1$, then $\mathcal {U}_2^0\cap \mathcal {W}_2(U)=\varnothing$ by Corollary 4.22. If $\delta =1$, there exists $W_0\in \mathcal {U}_2^0\cap \mathcal {W}_2(U)$. Recall that $W_0^\perp \cap C_V(t)\in \mathcal {U}_1^0$ so $C_G(W_0)\le C_G(W_0^\perp \cap C_V(t))\sim G^0=QL$. By Lemma 4.19, if $x\in C_G(W_0)\cap Y_2$ then $x=x_Qt$ and $x_Q$ is inverted by t and has order p. Since $x_Q$ also lies in $C_G(U)$, we can write $x_Q=hh^\varphi$ for some $h\in L_1$. Now $x_Q^{-1}=h^{-1}h^{-1\varphi }$ and so $x_Q^t=x_Q^{ss^\varphi }=h^s(h^\varphi )^{s^\varphi }=h^{-1}h^{-1\varphi }$. Therefore, $h^s=h^{-1}$ and $h^{\varphi s^\varphi }=h^{-1\varphi }$. Moreover, $x=x_Qt=(hs)(hs)^\varphi$ where $hs\in L_1$ is an element of order 4 squaring to the non-trivial element of $Z(L_1)$, and $h=(hs)s$ has order p. By Lemma 4.24 and [11], $d^{L_1}(hs,s)=3$ and so $d^{C_G(U)}(t,x_Qt)=3$ by Theorem 4.8. Therefore,

$$\begin{aligned} C_G(W_0)\cap \Delta _2^{C_G(U)}(t)=\varnothing \quad \text {for all}\ W_0\in \mathcal {U}_2^0\cap \mathcal {W}_2(U). \end{aligned}$$

(4.25.2)

Hence combining (4.25.1) with Lemma 4.21, [11] and, if $\delta =1$, (4.25.2) we get

$$\begin{aligned} \left| \bigcup _{U\le W}\Delta _2^{C_G(W)}(t)\right| =\left| \Gamma _2(C_G(U))\right| =\frac{1}{4}(q-\delta )^2(q-4-\delta ). \end{aligned}$$

This, together with Theorem 4.8 yields

$$\begin{aligned} \left| \Gamma _1(C_G(U))\right|&=\left| \Delta _2^{C_G(U)}(t)\right| - \left| \Gamma _2(C_G(U))\right| \\&=\frac{1}{8}(q-1)(q-1-2\delta )\left( q^2-(4+2\delta )q+9+4\delta \right) , \end{aligned}$$

which proves the lemma. $\square$

Lemma 4.26

Let $U\in \mathcal {U}_1^-$, and $W\in (\mathcal {U}_2^+\cup \mathcal {U}_2^-)\cap \mathcal {W}_2(U)$.

• (i) If $\delta =-1$ and $W_0\in \mathcal {U}_2^0\cap \mathcal {W}_2(U)$, then $Y_2\cap C_G(W_0){\setminus }\left\{ t\right\} \subseteq \Delta _3^{C_G(U)}(t)$.

• (ii) We have

  $$\begin{aligned} \left| \Gamma _2(C_G(U)){\setminus }\dot{\bigcup _{W\in \mathcal {W}_2(U)}}\Gamma _2(C_G(W))\right| =\frac{1}{4}(q-2+\delta )(q^2-1) \end{aligned}$$

  and $\left| \Gamma _1(C_G(U))\right| =\frac{1}{4}(q-1)^3(q+1)$.

Proof

First assume $\delta =-1$, and consider $C_G(W_0)$. By Lemma 4.19, every involution in $C_G(W_0)$ can be written as xt where x has order p. But $(xt)t=x$ has order p, which does not divide $\frac{1}{2}(q^2-1)$, and hence $d^{C_G(U)}(xt,t)=3$. In other words, $Y_2\cap C_G(W_0){\setminus }\left\{ t\right\} \subseteq \Delta _3^{C_G(U)}(t)$, so proving (i).

Consider then $C_G(W)\sim L_2(q)$. We utilize the character table of $L_2(q)$ from Chapter 38 of [22] (see also Schur [42]). Recall that $L_2(q)$ contains one conjugacy class of involutions, and two conjugacy classes of elements of order p. The remaining conjugacy classes partition into two cases: those whose order divides $\frac{1}{2}(q-1)$ and those whose order divides $\frac{1}{2}(q+1)$. Let C be a conjugacy class of elements in $C_G(W)$ and define $X_C=\left\{ \left. x\in Y_2\cap C_G(W)\right| tx\in C\right\}$. It is a well-known character theoretic result (see, for example, Theorem 4.2.12 of [28]) that

$$\begin{aligned} \left| X_C\right| =\frac{\left| C\right| }{\left| C_{C_G(W)}(t)\right| }\sum _{\begin{array}{c} \chi \\ {\text {Irreducible}} \end{array}}\frac{\chi (tx)\left| \chi (t)\right| ^2}{\chi (1)} \end{aligned}$$

(4.26.1)

and all $X_C$ are pairwise disjoint. Let $x\in Y_2\cap C_G(W)$. If the order of tx divides $\frac{1}{2}(q^2-1)$ but not $\frac{1}{2}(q-\delta )$ then it must necessarily divide $\frac{1}{2}(q+\delta )$. Hence, if C is a conjugacy class of elements of order dividing $\frac{q+\delta }{2}$, then any $y\in X_C$ has the property that $d^{C_G(W)}(t,y)=3$ but $d^{C_G(U)}(t,y)=2$, by Lemma 4.24. Recall that $\Gamma _2(C_G(U)){\setminus } \dot{\bigcup} _{W\in \mathcal {W}_2(U)}\Gamma _2(C_G(W))$ is the set consisting of all such involutions. Therefore, it suffices to calculate the sizes of all such relevant $X_C$. We use $\mathcal {F}$ to denote to be the set of all conjugacy classes of elements with order dividing $\frac{q+\delta }{2}$.

By [22], we see that for any $C\in \mathcal {F}$, $\left| C\right| =q(q-\delta )$ and so for any $x \in C \, \left| C_{C_G(W)}(x)\right| =(q-\delta )$. Hence (4.26.1) and [22] gives $\left| X_C\right| =q-\delta$. Now if $\delta =1$, then $\left| \mathcal {F}\right| =\frac{q-1}{4}$ by [22]. If $\delta =-1$, then $\left| \mathcal {F}\right| =\frac{q-3}{4}$. Since $\left| \Delta _3^{C_G(W)}(t)\cap \Delta _2^{C_G(U)}\right| = \left| X_C\right| \left| \mathcal {F}\right|$, and by Corollary 4.22, $\left| \mathcal {W}_2(U)\cap (\mathcal {U}_2^+\cup \mathcal {U}_2^-)\right| =q+\delta$, we obtain

$$\begin{aligned} \left| \Gamma _2(C_G(U)){\setminus }\dot{\bigcup _{W\in \mathcal {W}_2(U)}}\Gamma _2(C_G(W))\right|&=\left| \mathcal {W}_2(U)\cap (\mathcal {U}_2^+\cup \mathcal {U}_2^-)\right| \left| X_C\right| \left| \mathcal {F}\right| \\&=\left\{ \begin{array}{ll} \frac{1}{4}(q-1)(q^2-1) &{} \quad q\equiv 1\pmod 4\\ \frac{1}{4}(q-3)(q^2-1) &{} \quad q\equiv -1\pmod 4 \end{array}\right. \end{aligned}$$

which proves the first part of (ii). We now prove the last part of (ii). Recall that

$$\begin{aligned} \left| \dot{\bigcup _{W\in \mathcal {W}_2(U)}} \Gamma _2(C_G(W))\right| =(q+\delta )\left| \Delta _2^{C_G(W)}(t)\right| =\frac{1}{4}(q^2-1)(q-4-\delta ) \end{aligned}$$

by [11] and Corollary 4.22. Together with the above statement, we have

$$\begin{aligned} \left| \Gamma _2(C_G(U))\right|&=\left| \dot{\bigcup _{W\in \mathcal {W}_2(U)}} \Gamma _2(C_G(W))\right| +\left| \Gamma _2(C_G(U)){\setminus }\dot{\bigcup _{W\in \mathcal {W}_2(U)}}\Gamma _2(C_G(W))\right| \\&=\frac{1}{4}(q^2-1)(q-4-\delta )+\frac{1}{4}(q^2-1)(q-2+\delta )\\&=\frac{1}{2}(q^2-1)(q-3). \end{aligned}$$

Hence

$$\begin{aligned} \left| \Gamma _1(C_G(U))\right|&=\left| \Delta _2^{C_G(U)}(t)\right| -\left| \Gamma _2(C_G(U))\right| \\&=\frac{1}{4}(q-1)^3(q+1), \end{aligned}$$

and Lemma 4.26 holds. $\square$

Lemma 4.27

$$\begin{aligned} \left| \dot{\displaystyle \bigcup _{U\in \mathcal {U}_1}}\Gamma _1(C_G(U))\right| =\left\{ \begin{array}{ll} \frac{1}{16}q(q^2-1)(3q^3-11q^2+21q-29) &{} q\equiv 1\pmod 4\\ \frac{1}{16}q(q^2-1)(q-1)(3q^2+2q+7) &{} q\equiv -1\pmod 4. \end{array}\right. \end{aligned}$$

Proof

Since $\mathcal {U}_1=\mathcal {U}_1^0\dot{\cup }\mathcal {U}_1^+\dot{\cup }\mathcal {U}_1^-$, with each orbit size given in Lemma 4.16, the result follows immediately from Lemmas 4.23, 4.25 and 4.26. $\square$

Recall the list of cases in Table 1. The next lemma concerns Cases 2 and 3, in other words, $\bigcup _{U\in \mathcal {U}_1}\Gamma _2(C_G(U))$.

Lemma 4.28

$$\begin{aligned} \left| \displaystyle \bigcup _{U\in \mathcal {U}_1} \Gamma _2(C_G(U))\right| =\frac{1}{2}(q-\delta )(q^3-2q^2-1). \end{aligned}$$

Proof

By Lemmas 4.12 and 4.19, for any $W_0\in \mathcal {U}_2^0$ we have $\left| \Delta _2^{C_G(U)}(t)\cap C_G(W_0)\right| =\frac{1}{2}(q-1)(q-\delta )$ for some $U\in \mathcal {U}_1(W_0)$. Additionally, for any $W\in (\mathcal {U}_2^+\dot{\cup }\mathcal {U}_2^-)$ we have

$$\begin{aligned} \left| \Delta _2^{C_G(U)}(t)\cap C_G(W)\right|&= \left| \Delta _2^{C_G(W)}(t)\right| +\left| \Delta _3^{C_G(W)}(t)\cap \Delta _2^{C_G(U)}(t)\right| \\&=\frac{1}{2}(q-\delta )(q-3), \end{aligned}$$

for some $U\in \mathcal {U}_1(W)$, by [11] and Lemma 4.26. Since $\mathcal {U}_2=\mathcal {U}_2^0\dot{\cup }\mathcal {U}_2^+\dot{\cup }\mathcal {U}_2^-$, with the orbit sizes given in Corollary 4.17, this covers every involution in $\bigcup _{U\in \mathcal {U}_1}\Gamma _2(C_G(U))$, and the lemma follows. $\square$

We now concern ourselves with the final two cases. These concern involutions that are distance 3 from t in every $C_G(U)$ that they appear in, but actually are distance 2 from t in G. Recall that for any involution $y\in Y_2$, $C_G(y)=\text {Stab}_G C_V(y)=L_yK_y$ where $L_y=C_G(y)\cap C_G([V,y])\sim L_2(q)$ and $\left| K_y\right| =2(q-\delta )$. Also note that $L_y\unlhd C_G(y)$ acts faithfully on $C_V(y)$, and $\text {Syl}_p C_G(y)=\text {Syl}_p L_y$. The following three lemmas concern Case 5.

Lemma 4.29

Let $W\in \mathcal {U}_2^0\cup \mathcal {U}_2^{-\delta }$ and $x\in C_G(W)$ be such that $d^{C_G(U)}(t,x)=3$ for all $U\in \mathcal {U}_1(W)$. Then $d(t,x)=3$.

Proof

If $W\in \mathcal {U}_2^0$, then any involution in $C_G(W)$ can be written as $x=x_Qt$ where $x_Q=xt$ has order p. If $W\in \mathcal {U}_2^{-\delta }$, then, from Lemma 4.26, any involution $x\in C_G(W)$ such that tx has order dividing $\frac{1}{2}(q^2-1)$ must be distance 2 from t in $C_G(U)$ for some $U\in \mathcal {U}_1(W)$. Hence, any x satisfying the hypothesis must have the property that the order of tx is p.

Let $W\in \mathcal {U}_2^0\cup \mathcal {U}_2^{-\delta }$ and suppose $d(t,x)=2$, then there exists $y\in Y_2$ such that $t,x\in C_G(y)= L_yK_y$. Since tx has order p, $tx\in L_y$ and so $tx\in C_G([V,y])$. As $L_y$ acts faithfully on $C_V(y)$, any element of order p must fix a 1-subspace of $C_V(y)$, say $U_y$. Therefore, $tx\in C_G(U_y\oplus [V,y])$. But $tx\in C_G(W+[V,y])$ and since $[V,y]\in \mathcal {U}_2^\delta$, we have $W\ne [V,y]$. Set $W+[V,y]=U_y\oplus [V,y]$.

Suppose $U_y\le W$. Then $t,x,y\in C_G(U_y)$ and so $d^{C_G(U_y)}(t,x)=2$, contradicting our assumption. Hence $U_y\nleq W$ and so $U_y=\left\langle u_1+u_2 \right\rangle$ for $u_1\in W{\setminus } [V,y]$ and $u_2\in [V,y]$. Since $y\in C_G(y)$, $(u_1+u_2)^y=u_1+u_2$. However, $(u_1+u_2)^y=u_1^y+u_2^y=u_1^y-u_2$ and so $u_2=-2^{-1}u_1+2^{-1}u_1^y$. Thus $u_1+u_2=2^{-1}(u_1+u_1^y)$ and so $U_y=\left\langle u_1+u_1^y \right\rangle$. Recall that $t,x\in C_G(y)$ and $u_1\in W{\setminus } [V,y]$, so $u_1^t=u_1^x=u_1$. Hence $u_1+u_1^y$ is centralized by both t and x and so $U_y\le W=C_V(\left\langle t,x \right\rangle )$, a contradiction. Therefore, $d(t,x)\ne 2$ and the lemma holds. $\square$

Lemma 4.30

Let $W\in \mathcal {U}_2^\delta$. Then $\Delta _3^{C_G(W)}(t)\subseteq \Delta _2(t)$. In particular,

$$\begin{aligned} \left| \Gamma _2(G){\setminus }\bigcup _{U\in \mathcal {U}_1}\Gamma _2(C_G(U))\right| =\left\{ \begin{array}{ll} q(q^2-1) &{} q\equiv 1\pmod 4\\ 0 &{} q\equiv -1\pmod 4. \end{array}\right. \end{aligned}$$

Proof

We deal first with the case when $\delta =-1$. From Lemma 4.26, the number of involutions distance 3 from t in $C_G(W)$ that are actually distance 2 from t in some $U\in \mathcal {U}_1(W)$ is $\frac{1}{4}(q+1)(q-3)=\left| \Delta _3^{C_G(W)}(t)\right|$. That is to say all elements in $\Delta _3^{C_G(W)}(t)$ are distance 2 from t in $C_G(U)$ for some $U\in \mathcal {U}_2(W)$. This occurs for every such $W\in \mathcal {U}_2^\delta$ and so $\Gamma _2(G)=\bigcup _{U\in \mathcal {U}_1}\Gamma _2(C_G(U))$.

Assume now that $\delta =1$. As before, any element x in $\Gamma _2(G){\setminus }\bigcup _{U\in \mathcal {U}_1}\Gamma _2(C_G(U))$ must have the property that the order of tx is p. Suppose $d(t,x)=2$, and so there exists $y\in Y_2$ such that $t,x\in C_G(y)$. If $W\ne [V,y]$ then the argument from Lemma 4.29 holds and results in a contradiction. So we must have $W=[V,y]$. Since $\text {Stab}_GC_V(y)=\text {Stab}_G[V,y]=C_G(y)$, $C_G([V,y])\le C_G(y)$ and so any element in $C_G([V,y])=C_G(W)$ centralizes y. In particular, $\Delta _3^{C_G(W)}(t)\subseteq \Delta _2(t)$, establishing the first statement. By Lemma 4.26, the number of involutions distance 3 from t in $C_G(W)$ that are actually distance 2 from t in some $U\in \mathcal {U}_1(W)$ is $\frac{1}{4}(q-1)^2$. By [11], $\left| \Delta _3^{C_G(W)}(t)\right| =\frac{1}{4}(q-1)(q+7)$ and so by subtracting the two, there are $2(q-1)$ involutions in $\Delta _3^{C_G(W)}(t)$ that are distance 3 from t in $C_G(U)$ for all $U\in \mathcal {U}_1(W)$, but are actually distance 2 from t in $\mathcal {C}(G,Y_2)$. Since $\left| \mathcal {U}_2^\delta \right| =\frac{1}{2}q(q+\delta )$ by Corollary 4.17, the lemma follows. $\square$

Finally we turn to Case 4, $\Gamma _1(G){\setminus }\dot{\bigcup} _{U\in \mathcal {U}_1}\Gamma _1(C_G(U))$.

Lemma 4.31

Let $U\in \mathcal {U}_1^-\cup \mathcal {U}_1^0$ and $x\in C_G(U)$ be such that $C_V(\left\langle t,x \right\rangle )=U$ and $d^{C_G(U)}(t,x)=3$. Then $d(t,x)=3$.

Proof

Assume first that $U\in \mathcal {U}_1^-$. By Lemma 4.24, tx has order p or divides $\frac{1}{2}(q^2+1)$. Suppose $d(t,x)=2$, then there exists $y\in Y_2$ such that $t,x\in C_G(y)$. Since $\frac{1}{2}(q^2+1)$ is coprime to $\left| C_G(y)\right| =q(q^2-1)(q-\delta )$, tx must have order p. Indeed, clearly $\frac{1}{2}(q^2+1)$ is coprime to both q and $q^2-1$, and any factor dividing $q-\delta$ must divide $q^2-1$ and so $\frac{1}{2}(q^2+1)$ is coprime to $q-\delta$. Since tx has order p, then $tx\in L_y$.

Assume now that $U\in \mathcal {U}_1^0$. Let x be an involution in $C_G(U)=QL\sim G^0$ as defined in the discussion prior to Lemma 4.9. Then $tx\in Qtx_L$ which has order n dividing $\frac{1}{2}(q+\delta )$ in QL/L. Therefore, $(Qtx_L)^n\in Q$ and so $(tx)^n$ has order p. Therefore, tx has order dividing $\frac{1}{2}q(q+\delta )$. Suppose $d(t,x)=2$. Then there exists $y\in Y_2$ such that $t,x\in C_G(y)$. By the structure of $C_G(y)\sim (L_2(q)\times \frac{q-\delta }{2}):2^2$, the order of tx forces $tx\in L_y$.

We may now assume $U\in \mathcal {U}_1^-\cup \mathcal {U}_1^0$, so $tx\in L_y=C_G([V,y])$ and hence $tx\in C_G(U+[V,y])$. Suppose $U\nleq [V,y]$, then $tx\in C_G(U\oplus [V,y])$ Also, $tx\in C_G(U_y\oplus [V,y])$ for some $U_y\le C_V(y)$. However, if $U=U_y$ then $t,x,y\in C_G(U)$ and $d^{C_G(U)}(t,x)=2$. While $U_y\ne U$ results in a contradiction using an analogous argument from Lemma 4.29. Hence $U\le [V,y]$.

As $t,x\in C_G(y)=\text {Stab}_G([V,y])$, $tx\in L_y=C_G([V,y])$ and $[V,y]=U\perp U'$ where $U'=U^\perp \cap [V,y]$. Then for $u\in [V,y]$ we have $u^{tx}=u$ and so $u^{t}=u^x$. In particular, if $u\in U'$ then $u^{t}=u^x=-u$. Hence $[V,y]=U\perp ([V,t]\cap [V,x])$. If $C_V(\left\langle t,y \right\rangle )$ is 1-dimensional, then $C_V(y)=C_V(\left\langle t,y \right\rangle )\perp [V,t]$ since t stabilizes $C_V(y)$. However, then $[V,t]\oplus ([V,t]\cap [V,x])$ is 3-dimensional, a contradiction. A similar argument holds for $C_V(\left\langle x,y \right\rangle )$. Therefore both $C_V(\left\langle t,y \right\rangle )$ and $C_V(\left\langle x,y \right\rangle )$ are 2-dimensional. But since $\dim C_V(y)=3$, this means $C_V(\left\langle t,y \right\rangle )$ and $C_V(\left\langle x,y \right\rangle )$ intersect non-trivially, that is $C_V(\left\langle t,x,y \right\rangle )\ne 0$, contradicting our assumption. Therefore, $d(t,x)\ne 2$, and consequently $d(t,x)=3$. $\square$

The final case when $U\in \mathcal {U}_1^+$ is slightly trickier. Recall the definition of $Y_1$. For any $z\in Y_1$, we have $C_G(z)\sim SL_2(q)\circ SL_2(q):2$ and $C_V(z)$ is 1-dimensional. We choose z such that $t\in C_G(z)$ and $C_V(z)=U$, and return to work in the setting of $Sp(4,q)/\left\langle -I_4 \right\rangle =G^\tau \sim G$. We denote the image of any subgroup $K\le G$ by $K^\tau$. Choose

and note that $C_{G^\tau }(z)\sim C_G(U):2$. Hence,

Let $t^\tau$ be the image of t in $G^\tau$. We start with a preliminary lemma concerning the commuting involution graph $\mathcal {C}(L_2(q),X)$ where X is the sole conjugacy class of involutions. Denote by $L\sim L_2(q)$ and $\widehat{L}\sim PGL_2(q)$.

Lemma 4.32

Let x be an involution in L. Then $\Delta _3^L(x)$ splits into $\frac{1}{4}(q+2+5\delta )$$C_L(x)$-orbits of length $q-\delta$. Moreover, every $C_L(x)$-orbit in $\Delta _3^L(x)$ is $C_{\widehat{L}}(x)$-invariant.

Proof

Assume first that $\delta =-1$. Choose $x=\left( \begin{array}{ll}0&{}-1\\ 1&{}0\end{array} \right)$ and let $x_\lambda =\left( \begin{array}{ll}0&{}\lambda \\ -\lambda ^{-1}&{}0\end{array} \right)$ for some $\lambda \in GF(q){\setminus }\left\{ \pm 1\right\}$. There are two possibilities for an element of $C_L(x)$:

$$\begin{aligned} g_1=\left( \begin{array}{ll}a_1&{}a_2\\ -a_2&{}a_1\end{array} \right) ,\quad g_2=\left( \begin{array}{ll}b_1&{}b_2\\ b_2&{}-b_1\end{array} \right) . \end{aligned}$$

By direct calculation, if $g_1^{-1}x_\lambda g_1=x_\mu$ for some $\lambda ,\mu \in GF(q){\setminus }\left\{ \pm 1\right\}$ then $(-\lambda ^{-1}+\lambda )a_1a_2=0$. Note that since $\lambda \ne \pm 1$, then $\lambda \ne \lambda ^{-1}$. If $a_1=0$, then $a_2^2=1$, and so $\mu =\lambda ^{-1}$. On the other hand, if $a_2=0$ then $a_1^2=1$ and so $\mu =\lambda$. Note that in the case of $g_2$, neither $b_1$ or $b_2$ can be 0 and so $g_2^{-1}x_\lambda g_2=x_\mu$ requires $xy(\lambda -\lambda ^{-1})=0$, a contradiction. Hence for $\lambda ,\mu \in GF(q){\setminus }\left\{ \pm 1\right\}$, $x_\lambda$ and $x_\mu$ lie in different $C_L(x)$ orbits if and only if $\mu \notin \left\{ \lambda ,\lambda ^{-1}\right\}$. As we work modulo $\left\langle -I_4 \right\rangle$, there are at least $\frac{1}{4}(q-3)$$C_L(x)$-orbits in $\Delta _3^L(x)$. However for any $\lambda \ne \pm 1$, $C_L(x,x_\lambda )=1$ and so, each $C_L(x)$-orbit containing an $x_\lambda$ is of length $q+1$. But $\left| \Delta _3^L(x)\right| =\frac{1}{4}(q-3)(q+1)$ and so all involutions in $\Delta _3^L(x)$ are accounted for. Hence the first statement holds for $\delta =-1$, and each $C_L(x)$-orbit has representative $x_\lambda$ for some $\lambda \ne \pm 1$. Let

$$\begin{aligned} e=\left( \begin{array}{ll}1&{}0\\ 0&{}-1\end{array} \right) \in \widehat{L}{\setminus } L \end{aligned}$$

and note that $C_{\widehat{L}}(x)=\left\langle e \right\rangle C_L(x)$, and an easy check shows $[e,x_\lambda ]=1$ for all $\lambda \ne \pm 1$. Let $y\in \Delta _3^L(x)$, then $y=x_\lambda ^s$ for some $s\in C_L(x)$. Let $g=er\in C_{\widehat{L}}(x)$ for some $r\in C_L(x)$. Then $y^g=x_\lambda ^{s^er}$ and since $C_L(x)\unlhd C_{\widehat{L}}(x)$, $s^er\in C_L(x)$. That is, every $C_L(x)$-orbit in $\Delta _3^L(x)$ is $C_{\widehat{L}}(x)$-invariant.

Assume now that $\delta =1$. Choose $x=\left( \begin{array}{ll}i&{}0\\ 0&{}-i\end{array} \right)$ where $i^2=-1$ and let $y=\left( \begin{array}{ll}\sigma &{}\mu \tau \\ \tau &{}\sigma \end{array} \right)$ for some $\sigma ,\mu ,\tau \in GF(q)$, $\sigma \ne 0$ and $\mu$ a non-square in GF(q). By [11], $y\in \Delta _3^L(x)$. There are two possibilities for an element of $C_L(x)$:

$$\begin{aligned} g_1=\left( \begin{array}{ll}a^{-1}&{}0\\ 0&{}a\end{array} \right) ,\quad g_2=\left( \begin{array}{ll}0&{}b\\ -b^{-1}&{}0\end{array} \right) . \end{aligned}$$

By direct calculation, if $g_1^{-1}yg_1=y$ then $a=1$. Note that $g_2^{-1}yg_2\ne y$ as $\pm b^2\ne \mu$ for any non-square $\mu$. Hence $C_L(\left\langle x,y \right\rangle )=1$. Since y was arbitrary, each $C_L(x)$-orbit has length $q-1$. Now $\left| \Delta _3^L(x)\right| =\frac{1}{4}(q+7)(q-1)$ and so the first statement holds for $\delta =1$. Let

$$\begin{aligned} e_\nu =\left( \begin{array}{ll}0&{}\nu \\ 1&{}0\end{array} \right) \in \widehat{L}{\setminus } L \end{aligned}$$

and note that $C_{\widehat{L}}(x)=\left\langle e_\nu \right\rangle C_L(x)$ for any non-square $\nu$. It is easy to check that $y^{e_\mu }=y$. Let $g=e_\mu r\in C_{\widehat{L}}(x)$ for some $r\in C_L(x)$. Then $y^g=y^{e_\mu r}=y^{r}$ and since y was arbitrary and $r\in C_L(x)$, every $C_L(x)$-orbit in $\Delta _3^L(x)$ is $C_{\widehat{L}}(x)$-invariant. $\square$

Lemma 4.33

$$\begin{aligned} \left| \Delta _3^{C_G(U)}(t)\cap \Gamma _1(G)\right| =\frac{1}{4}(q-\delta )^2(q+2+5\delta ). \end{aligned}$$

Proof

We first work in the setting of $G^\tau$. Choose

By direct calculation, it is easily seen that

and any involution $y\in C_{G^\tau }(t^\tau )$ has the additional properties that

$$\begin{aligned} \begin{array}{ll}&{}\det A_1+\det A_3=\det A_2+\det A_4=1\quad \text {and}\\ &{}A_1^2+A_2A_3=A_3A_2+A_4^2=-I_2.\end{array} \end{aligned}$$

(4.33.1)

Recall that if $x\in C_G(U)^\tau$ then for some $A,B\in SL_2(q)$ and by Theorem 4.8, $x\in \Delta _3^{C_G(U)^\tau }(t^\tau )$ if and only if $A,\ B$ are involutions in L and either $d^L(A,J_0)=3$ or $d^L(B,J_0)=3$. So without loss of generality, set $A=B_i$ where $d^L(B_i,J_0)=i$ and choose $B\in \Delta _3^L(J_0)$.

If $x\in \Delta _2^{G^\tau }(t^\tau )$ then there exists such that $y^2=1$ and $[x,y]=1$. Suppose $\det A_2=0$. Then $\det A_4=1$ by (4.33.1), and so $A_4\in C_L(J_0)$. As $[x,y]=1$, $[A_4,B]=1$. However $C_L(\left\langle J_0,B \right\rangle )=1$, by Lemma 4.32 and so $A_4=\pm I_2$. But then $A_3A_2=-2I_2$ by (4.33.1), which is impossible as $\det A_2=0$. An analogous argument holds for $\det A_3$. Hence $\det A_2$, $\det A_3\ne 0$. Since $[x,y]=1$, $B_iA_2B=\pm A_2$ and so $B_i$ and B must be $C_{\widehat{L}}(J_0)$-conjugate. In other words, if $B_i$ and B are not $C_{\widehat{L}}(J_0)$-conjugate, then $[x,y]\ne 1$. By Lemma 4.32, every $C_L(J_0)$ orbit is an $C_{\widehat{L}}(J_0)$-orbit and so if $[x,y]=1$ then $B_i$ and B must be $C_L(J_0)$-conjugate. Assume then $B_i$ and B are $C_L(J_0)$-conjugate and let $A\in C_L(J_0)$ be such that $B_i^A=B$. Hence if , then $[y_A,x]=1$ and so $d^{G^\tau }(t^\tau ,x)=2$. By Lemma 4.32, each $C_L(J_0)$-orbit of $\Delta _3^L(J_0)$ is of length $q-\delta$, and there are $\frac{1}{4}(q+2+5\delta )$ such orbits. Moreover, for any involution $x_0\in C_G(U)^\tau$ conjugate to $t^\tau$, $zx_0$ is also an involution in $C_G(U)^\tau$ conjugate to $t^\tau$ which has not been accounted for. Therefore, the number of involutions in $\Delta _3^{C_G(U)^\tau }(t^\tau )$ that are actually distance 2 from $t^\tau$ in $G^\tau$ is $\frac{1}{2}(q-\delta )^2(q+2+5\delta )$.

We now return to the setting of G, and first assume that $\delta =-1$ and so by Corollary 4.22(i), $\left| \mathcal {W}_2(U)\right| =q+1$, and for every $W\in \mathcal {W}_2(U)$, $C_G(W)\sim L_2(q)$. For each W, there exists $U_W\in \mathcal {U}_1^+$ such that $C_G(W)\le C_G(U_W)\sim L_2(q^2)$ by Lemma 4.21, and $\Delta _3^{C_G(W)}(t)\subseteq \Delta _2^{C_G(U_W)}(t)$ by Lemma 4.30. Hence, there are $\frac{1}{4}(q+1)^2(q-3)$ involutions already counted (from Case 3) and the remaining involutions do not fix a 2-subspace of $C_V(t)$. Therefore

$$\begin{aligned} \left| \Delta _3^{C_G(U)}(t)\cap \Gamma _1(G)\right|&= \frac{1}{2}(q+1)^2(q-3)-\frac{1}{2}(q+1)^2(q-3)\\&=\frac{1}{4}(q+1)^2(q-3), \end{aligned}$$

as required. Now assume that $\delta =1$ and so by Corollary 4.22. For each W, there exists $U_W\in \mathcal {U}_1^-$ such that $C_G(W)\le C_G(U_W)\sim L_2(q^2)$ by Lemma 4.21 and $\left| \Delta _3^{C_G(W)}(t)\cap \Delta _2^{C_G(U_W)}(t)\right| =\frac{1}{4}(q-1)^2$ by Lemma 4.26. Since $\left| \mathcal {W}_2(U)\cap (\mathcal {U}_1^+\cup \mathcal {U}_1^-)\right| =q-1$ by Corollary 4.22(iii), this accounts for $\frac{1}{4}(q-1)^3$ involutions. Suppose now $W_0\in \mathcal {W}_2(U)\cap \mathcal {U}_2^0$. By Lemma 4.21, there exists $U_0\in \mathcal {U}_1^0$ such that $C_G(W_0)\le C_G(U_0)$. From Lemmas 4.12 and 4.19, $\left| C_G(W)\cap \Delta _2^{C_G(U_0)}(t)\right|=\frac{1}{2}(q-1)^2$. Since $\left| \mathcal {W}_2(U)\cap \mathcal {U}_2^0\right| =2$ by Corollary 4.22(iii), this yields a further $(q-1)^2$ involutions. Finally, if $W\in \mathcal {U}_2^+$, then by Lemma 4.30, $\Delta _3^{C_G(W)}(t)\subseteq \Delta _2(t)$ and there are $2(q-1)$ involutions in $\Delta _3^{C_G(W)}(t)$ not already enumerated. Now $\left| \mathcal {U}_2^+\cap \mathcal {W}_2(U)\right| =\frac{1}{2}(q-1)$ by Corollary 4.22(iii), and this yields another $(q-1)^2$ involutions. Hence, there are $\frac{1}{4}(q-3)^2+2(q-1)^2=\frac{1}{4}(q-1)^2(q+7)$ involutions already counted (from Cases 3 and 5) and the remaining involutions do not fix a 2-subspace of $C_V(t)$. Consequently

$$\begin{aligned} \left| \Delta _3^{C_G(U)}(t)\cap \Gamma _1(G)\right|&=\frac{1}{2}(q-1)^2(q+7)-\frac{1}{2}(q-1)^2(q+7)\\&=\frac{1}{4}(q-1)^2(q+7), \end{aligned}$$

as required. $\square$

Corollary 4.34

$$\begin{aligned} \left| \Gamma _1(G){\setminus }\dot{\displaystyle \bigcup _{U\in \mathcal {U}_1}} \Gamma _1(C_G(U))\right| =\frac{1}{8}q(q-\delta )(q^2-1)(q+2+5\delta ). \end{aligned}$$

Proof

Since $\left| \mathcal {U}_1^+\right| =\frac{1}{2}q(q+\delta )$, the result holds by Lemmas 4.32 and 4.33. $\square$

Lemma 4.35

If $q\equiv -1\pmod 4$, then

• (i) $\left| \Delta _2(t)\right| =\frac{1}{16}(q+1)(3q^5-2q^4+8q^3-30q^2+13q-8)$; and

• (ii) $\left| \Delta _3(t)\right| =\frac{1}{16}(q^2-1)(5q^4-9q^3+7q^2-3q+8)$.

If $q\equiv 1\pmod 4$, then

• (iii) $\left| \Delta _2(t)\right| =\frac{1}{16}(q-1)(3q^5-6q^4+32q^3-10q^2-27q-8)$; and

• (iv) $\left| \Delta _3(t)\right| =\frac{1}{16}(q-1)(5q^5+22q^4-8q^3+34q^2+51q+24)$.

Proof

The cases listed in Table 1 are disjoint. Hence $\left| \Delta _2(t)\right|$ is determined by summing the values calculated in Lemmas 4.27, 4.28, 4.30 and 4.34. By Theorem 4.15, $\mathcal {C}(G,Y_2)$ has diameter 3 and so $\left| \Delta _3(t)\right| =\left| Y_2\right| -\left| \Delta _1(t)\right| -\left| \Delta _2(t)\right| -1$. Since $\left| G\right| =\frac{1}{2}q^4(q^2-1)(q^4-1)$ and $\left| C_G(t)\right| =q(q^2-1)(q-\delta )$, $\left| Y_2\right| =\frac{1}{2}q^3(q+\delta )(q^2+1)$. Together with Lemma 4.20, this proves the lemma. $\square$

Together, Theorem 4.15 and Lemmas 4.20 and 4.35 complete the proof of Theorem 1.4.