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 4dimensional projective symplectic group over a finite field and X a Gconjugacy class of involutions, determining the diameters and structure of the discs of these graphs.
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 nonabelian 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 nonabelian 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 3transpositions. While a computerfree uniqueness proof of the Lyons simple group by Aschbacher and Segev [6] employed a commuting graph where the vertices consisted of the 3central 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 3dimensional 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 4dimensional projective symplectic group and X is a Gconjugacy 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)Hmodule, 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
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])
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
Theorem 1.2
Suppose that \(p = 2\). Then \(\mathcal {C}(G,X_2)\) is connected of diameter 4, the disc sizes being
Turning to the case when p is odd, we have that there are two Ginvolution conjugacy classes \(Y_1\) and \(Y_2\). We shall let \(Y_1\) denote the Gconjugacy class whose elements are the images of an involution in H, and \(Y_2\) to denote the Gconjugacy class whose elements are the image of an element of H of order 4 which squares to the nontrivial 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
Theorem 1.4

(i)
If \(q\equiv 1 \pmod 4\) then \(\mathcal {C}(G,Y_2)\) is connected of diameter 3. Furthermore,
$$\begin{aligned} \left \Delta _1(t)\right&=\frac{1}{2}q(q^2 + 2q  1);\\ \left \Delta _2(t)\right&=\frac{1}{16}(q+1)\left( 3q^52q^4+8q^330q^2+13q8\right) ;\quad \text {and}\\ \left \Delta _3(t)\right&=\frac{1}{16}(q^21)\left( 5q^49q^3+7q^23q+8\right) . \end{aligned}$$ 
(ii)
If \(q\equiv 1 \pmod 4\) then \(\mathcal {C}(G,Y_2)\) is connected of diameter 3. Furthermore,
$$\begin{aligned} \left \Delta _1(t)\right&=\frac{1}{2}q(q^2 + 1);\\ \left \Delta _2(t)\right&=\frac{1}{16}(q1)\left( 3q^56q^4+32q^310q^227q8\right) ;\quad \text {and}\\ \left \Delta _3(t)\right&=\frac{1}{16}(q1)\left( 5q^5+22q^48q^3+34q^2+51q+24\right) . \end{aligned}$$
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 nontrivial 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 5dimensional 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 1subspaces 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].
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 \(x \in X_0\),
with \(Stab_{G_0}(C_{V_0}(x))\) having shape \(q^{2n  1}SL_{2n  2}(q)(q1)\). 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 1dimensional subspace of \(C_{V_0}(t_0)\cap C_{V_0}(x)\). Since \([V_0,t_0]\) is a 1space and \(G_0\) acts transitively on the 1spaces 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
and we may suppose that \(G = \{A \in GL(4,q) \big  A^TJA = J \}\). We further define
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.
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
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 nonzero. 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
Now we suppose \(s\in Q_2{\setminus } Q_1\). Then
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 \(q1\) such involutions (as \(c\ne 0\)). For the latter, there are \(q1\) 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 \((q1)^2\) involutions. Therefore
Since any two distinct Sylow 2subgroups of SL(2, q) have trivial intersection and SL(2, q) possesses \(q+1\) Sylow 2subgroups, Lemma 2.3(ii) together with (2.4.1) and (2.4.2) yields that
This proves Lemma 2.4. \(\square \)
Proof of Theorem 1.1 As is wellknown—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 \(\Delta _1(t) = (C_G(t_1) \cap X_1) {\setminus } \{t_1 \}\),
By Theorem 2.1, \(\mathcal {C}(G,X_1)\) has diameter 2. Hence, by Lemma 2.3(iv),
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
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 1dimensional. 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 \(P_i\sim q^3SL_2(q)(q1)\) 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
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 wellknown.
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
and so \(\left \Delta _1(t)\right =\left Z_S\right +2q\left Z_S\right 1\). From \(\left Z_S\right =(q1)^2\) we get \(\left \Delta _1(t)\right =(q1)^2+2q(q1)^21=q^2(2q3)\). The remaining disc sizes are immediate from the structure of the chamber graph \(\mathcal {C}(\Delta )\).\(\square \)
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
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
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 \(\left \Delta _1(t)\right =\frac{1}{2}\left SL_2(q)\right =\frac{1}{2}q(q^21)\). \(\square \)
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 \(\Sigma =\left\{ \left. \left\{ F,F^\perp \right\} \right F\ \text {is a hyperbolic 2subspace of}\ V\right\} \). 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 2subspace 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 2subspace 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 nonzero 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
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
There is a possibility that this is not welldefined 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 welldefined.
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
Simplifying and using the fact that q is odd gives
and then
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
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
and thus the only conditions we get are \(\gamma =\beta \alpha \) and \(\alpha +\beta =\delta \). Thus
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
which, after rearranging gives
and note that the relations for \(\gamma \) and \(\delta \) are satisfied after substitution for \(\alpha \) and \(\beta \). Hence
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
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
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 2subspace 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
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 \(\left G\right =\frac{q^4}{2}(q^21)(q^41)\) and \(\left C_G(t)\right =q^2(q^21)^2\) we see that \(\left Y_1\right =\frac{q^2}{2}(q^2+1)\). Using Lemma 3.1 then gives
which completes the proof of Theorem 1.3.
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 5dimensional 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 2subspace 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 nondegenerate 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 idimensional 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
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 1subspace of \(C_V(t)\). Then \(C_G(U_0)\sim q^3:L_2(q)\).

(ii)
Let \(U_\varepsilon \) be a 1subspace of \(C_V(t)\), such that \(U_\varepsilon ^\perp \cap C_V(t)\) is a 2space 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 1subspace 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 2subspace of V of \(+\)type, and hence \(U_+^\perp =(U_+^\perp \cap C_V(t))\perp [V,t]\) is a 4subspace of \(+\)type. Similarly, \(U_^\perp =(U_^\perp \cap C_V(t))\perp [V,t]\) is a 4space of −type. If \(\delta =1\), then [V, t] is a 2subspace of V of −type, and the results for when \(\delta =1\) interchange. Let \(W_+\) and \(W_\) be 4subspaces of V of \(+\) and −type respectively, such that \(W_+^\perp \) and \(W_^\perp \) are 1subspaces 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
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 2subspace of \(C_V(t)\) such that \(U_0^\perp \cap C_V(t)\) is an isotropic 1space. Then \(C_G(U_0)\sim q^2:\frac{q\delta }{2}\).

(ii)
Let \(U_\varepsilon \) be a 2subspace 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 \(\mathcal {U}_i\), \(i=1,2\).
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
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
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
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
Hence, using [11],
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
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
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],
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
with L acting on Q by right multiplication. Then \(Q\sim q^3\) and \(L\cong L_2(q)\). Since Q is the 3dimensional GF(q)Lmodule (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
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
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
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
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
Then \(x_L\) inverts \(x_Q\) if and only if
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
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
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
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,
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
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 \(\frac{1}{2}(q^21)\) 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 \(q1\) 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 \(q5\) ordered pairs (a, b) satisfying \(a^2+b^2=1\), \(a,b\ne 0\) and thus \(\frac{1}{2}(q5)\) distinct values of \(a^{1}b\). Hence there are \(\frac{1}{2}(q5)(q1)\) 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, \(\left Qt\cap \Delta _2^0(t)\right =\frac{1}{2}(q5)(q1)+2(q1)=\frac{1}{2}(q1)^2\) 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
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 \(a^2+b^2=1\),
Let \(y\in Y^0\) be such that \(y_Q=(\alpha ,\beta ,\gamma )\) and
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
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 nonzero. Moreover, if \(c=0\) then \(d=e^{1}\) and \(b(dd^{1})=0\) 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
where \(\varepsilon =abc^{1}e^{1}(\gamma +(de)^{1}(2c+a^{1}beab^{1}e(ab)^{1}e)\beta )\). 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
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 \(x_Q=(\varepsilon ,0,\varepsilon )\),
and \(\varepsilon =(ce^{1}(1a^2)ab)\beta 2^{1}b^2\gamma \), 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 \(x_Q=(\varepsilon ,0,\varepsilon )\),
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 \(x_Q=(2d^2\gamma (1d^2)^{1},0,2d^2\gamma (1d^2)^{1})\) and
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
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
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 \(\frac{1}{4}q^2(q\delta )(q4\delta )\) (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
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
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,
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 nonzero 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 \(2(q1)\) such vectors with \(\gamma =0\). If \(\gamma \ne 0\), then \(\alpha =\beta ^{1}\gamma ^2\) and there are \((q1)^2\) such vectors satisfying this. Hence there are \(2(q1)+(q1)^2=(q1)(q+1)\) nonzero isotropic vectors contained in \(C_V(t)\) and thus \(q+1\) isotropic 1subspaces of \(C_V(t)\).
Suppose now v is \(C_G(t)\)conjugate to \(v_3+v_5\), which is nonisotropic. Note that \(\left\langle v_3+v_5 \right\rangle ^\perp \cap C_V(t)\) is a 2subspace of V of \(+\)type. If \(\delta =1\), then by Lemma 4.1(ii), \(\left\langle v_3+v_5 \right\rangle ^\perp \) is a 4subspace 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 4subspace 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 \(q1\) such vectors that satisfy this. If \(\gamma =\pm \lambda \), then \(\alpha \beta =0\) and so for both values of \(\gamma \), there are \(2(q1)+1\) 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 \((q1)(q3)\) such vectors that satisfy this. Hence for any given \(\lambda \), there exist \((q1)+4(q1)+2+(q1)(q3)=q(q+1)\) vectors that satisfy \(\alpha \beta +\gamma ^2=\lambda ^2\). Since there are \(\frac{1}{2}(q1)\) squares in GF(q), there are \(q(q+1)(q1)\) vectors that are \(C_G(t)\)conjugate to \(v_3+v_5\) and hence \(\frac{1}{2}q(q+1)\) 1subspaces 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 \(q^2+q+1(q+1)\frac{1}{2}q(q+1)=\frac{1}{2}q(q1)\), 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,
Proof
Since \(C_V(t)\) is 3dimensional, \(U^\perp \cap C_V(t)\in \mathcal {U}_1\) 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 2subspace 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
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
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
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 1subspace \(W''\) 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,
Combining Lemmas 4.16, 4.19 and Corollary 4.17 with Theorems 4.6, 4.8, 4.14 and [11], we have
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
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 idimensional subspaces of U and \(\mathcal {W}_i(U)\) to be the totality of idimensional subspaces of \(C_V(t)\) containing U. Note that \(\mathcal {U}_i=\mathcal {U}_i(C_V(t))\).
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 \(\left \mathcal {U}_1^+\cap \mathcal {U}_1(W)\right =\left \mathcal {U}_1^\cap \mathcal {U}_1(W)\right =\frac{q1}{2}\).

(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
\(v_1+\lambda (v_3+v_5)\) 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 nonsquare in GF(q), and there are \(\frac{q1}{2}\) of each, it is clear that there exist \(\frac{q1}{2}\) 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 1subspaces 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 nonsquare 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 \(\left \mathcal {U}_2^+\cap \mathcal {W}_2(U)\right =\left \mathcal {U}_2^\cap \mathcal {W}_2(U)\right =\frac{q1}{2}\).

(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,
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 selfnormalizing 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 2subspaces 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
Combining Lemma 4.12 with (4.23.1),
Together, (4.23.2) and Theorem 4.14 give
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
$$\begin{aligned} \left \Gamma _1(C_G(U))\right =\left\{ \begin{array}{ll} \frac{1}{8}(q1)(q3)(q^26q+13) &{} \quad q\equiv 1\pmod 4\\ \frac{1}{8}(q^21)(q^22q+5) &{} \quad q\equiv 1\pmod 4. \end{array}\right. \end{aligned}$$
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
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 nontrivial 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,
Hence combining (4.25.1) with Lemma 4.21, [11] and, if \(\delta =1\), (4.25.2) we get
This, together with Theorem 4.8 yields
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
$$\begin{aligned} \left \Gamma _2(C_G(U)){\setminus }\dot{\bigcup _{W\in \mathcal {W}_2(U)}}\Gamma _2(C_G(W))\right =\frac{1}{4}(q2+\delta )(q^21) \end{aligned}$$and \(\left \Gamma _1(C_G(U))\right =\frac{1}{4}(q1)^3(q+1)\).
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 \(\frac{1}{2}(q^21)\), 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 \(\frac{1}{2}(q1)\) 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 wellknown character theoretic result (see, for example, Theorem 4.2.12 of [28]) that
and all \(X_C\) are pairwise disjoint. Let \(x\in Y_2\cap C_G(W)\). If the order of tx divides \(\frac{1}{2}(q^21)\) 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 \(\left \mathcal {F}\right =\frac{q1}{4}\) by [22]. If \(\delta =1\), then \(\left \mathcal {F}\right =\frac{q3}{4}\). 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
which proves the first part of (ii). We now prove the last part of (ii). Recall that
by [11] and Corollary 4.22. Together with the above statement, we have
Hence
and Lemma 4.26 holds. \(\square \)
Lemma 4.27
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
Proof
By Lemmas 4.12 and 4.19, for any \(W_0\in \mathcal {U}_2^0\) we have \(\left \Delta _2^{C_G(U)}(t)\cap C_G(W_0)\right =\frac{1}{2}(q1)(q\delta )\) for some \(U\in \mathcal {U}_1(W_0)\). Additionally, for any \(W\in (\mathcal {U}_2^+\dot{\cup }\mathcal {U}_2^)\) we have
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 \(\frac{1}{2}(q^21)\) 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 1subspace 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, \((u_1+u_2)^y=u_1^y+u_2^y=u_1^yu_2\) 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,
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 \(\frac{1}{4}(q+1)(q3)=\left \Delta _3^{C_G(W)}(t)\right \). 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 \(\frac{1}{4}(q1)^2\). By [11], \(\left \Delta _3^{C_G(W)}(t)\right =\frac{1}{4}(q1)(q+7)\) and so by subtracting the two, there are \(2(q1)\) 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 \(\left C_G(y)\right =q(q^21)(q\delta )\), tx must have order p. Indeed, clearly \(\frac{1}{2}(q^2+1)\) is coprime to both q and \(q^21\), and any factor dividing \(q\delta \) must divide \(q^21\) 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 1dimensional, 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 3dimensional, 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 2dimensional. 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 nontrivially, 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 1dimensional. 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 \(C_L(x)\):
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 \(\frac{1}{4}(q3)\)\(C_L(x)\)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 \(\left \Delta _3^L(x)\right =\frac{1}{4}(q3)(q+1)\) 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
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 nonsquare in GF(q). By [11], \(y\in \Delta _3^L(x)\). There are two possibilities for an element of \(C_L(x)\):
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 nonsquare \(\mu \). Hence \(C_L(\left\langle x,y \right\rangle )=1\). Since y was arbitrary, each \(C_L(x)\)orbit has length \(q1\). Now \(\left \Delta _3^L(x)\right =\frac{1}{4}(q+7)(q1)\) and so the first statement holds for \(\delta =1\). Let
and note that \(C_{\widehat{L}}(x)=\left\langle e_\nu \right\rangle C_L(x)\) for any nonsquare \(\nu \). 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
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
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 \(\frac{1}{4}(q+1)^2(q3)\) involutions already counted (from Case 3) and the remaining involutions do not fix a 2subspace of \(C_V(t)\). Therefore
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 \(\left \Delta _3^{C_G(W)}(t)\cap \Delta _2^{C_G(U_W)}(t)\right =\frac{1}{4}(q1)^2\) by Lemma 4.26. Since \(\left \mathcal {W}_2(U)\cap (\mathcal {U}_1^+\cup \mathcal {U}_1^)\right =q1\) by Corollary 4.22(iii), this accounts for \(\frac{1}{4}(q1)^3\) 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, \(\left C_G(W)\cap \Delta _2^{C_G(U_0)}(t)\right=\frac{1}{2}(q1)^2\). Since \(\left \mathcal {W}_2(U)\cap \mathcal {U}_2^0\right =2\) by Corollary 4.22(iii), this yields a further \((q1)^2\) 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 \(2(q1)\) involutions in \(\Delta _3^{C_G(W)}(t)\) not already enumerated. Now \(\left \mathcal {U}_2^+\cap \mathcal {W}_2(U)\right =\frac{1}{2}(q1)\) by Corollary 4.22(iii), and this yields another \((q1)^2\) involutions. Hence, there are \(\frac{1}{4}(q3)^2+2(q1)^2=\frac{1}{4}(q1)^2(q+7)\) involutions already counted (from Cases 3 and 5) and the remaining involutions do not fix a 2subspace of \(C_V(t)\). Consequently
as required. \(\square \)
Corollary 4.34
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)
\(\left \Delta _2(t)\right =\frac{1}{16}(q+1)(3q^52q^4+8q^330q^2+13q8)\); and

(ii)
\(\left \Delta _3(t)\right =\frac{1}{16}(q^21)(5q^49q^3+7q^23q+8)\).
If \(q\equiv 1\pmod 4\), then

(iii)
\(\left \Delta _2(t)\right =\frac{1}{16}(q1)(3q^56q^4+32q^310q^227q8)\); and

(iv)
\(\left \Delta _3(t)\right =\frac{1}{16}(q1)(5q^5+22q^48q^3+34q^2+51q+24)\).
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 \(\left G\right =\frac{1}{2}q^4(q^21)(q^41)\) and \(\left C_G(t)\right =q(q^21)(q\delta )\), \(\left Y_2\right =\frac{1}{2}q^3(q+\delta )(q^2+1)\). 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.
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Everett, A., Rowley, P. Commuting Involution Graphs for 4Dimensional Projective Symplectic Groups. Graphs and Combinatorics 36, 959–1000 (2020). https://doi.org/10.1007/s0037302002156x
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Keywords
 Involutions
 Commuting involution graphs
 Symplectic Groups
Mathematics Subject Classification
 05C12
 20E99