Combinatorial invariants for nets of conics in PG(2,q)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathrm {PG}(2,q)$$\end{document}

The problem of classifying linear systems of conics in projective planes dates back at least to Jordan, who classified pencils (one-dimensional systems) of conics over C\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {C}}$$\end{document} and R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {R}$$\end{document} in 1906–1907. The analogous problem for finite fields Fq\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {F}_q$$\end{document} with q odd was solved by Dickson in 1908. In 1914, Wilson attempted to classify nets (two-dimensional systems) of conics over finite fields of odd characteristic, but his classification was incomplete and contained some inaccuracies. In a recent article, we completed Wilson’s classification (for q odd) of nets of rank one, namely those containing a repeated line. The aim of the present paper is to introduce and calculate certain combinatorial invariants of these nets, which we expect will be of use in various applications. Our approach is geometric in the sense that we view a net of rank one as a plane in PG(5,q)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathrm {PG}(5,q)$$\end{document}, q odd, that meets the quadric Veronesean in at least one point; two such nets are then equivalent if and only if the corresponding planes belong to the same orbit under the induced action of PGL(3,q)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathrm {PGL}(3,q)$$\end{document} viewed as a subgroup of PGL(6,q)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathrm {PGL}(6,q)$$\end{document}. Since q is odd, the orbits of lines in PG(5,q)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathrm {PG}(5,q)$$\end{document} under this action correspond to the aforementioned pencils of conics in PG(2,q)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathrm {PG}(2,q)$$\end{document}. The main contribution of this paper is to determine the line-orbit distribution of a plane π\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\pi $$\end{document} corresponding to a net of rank one, namely, the number of lines in π\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\pi $$\end{document} belonging to each line orbit. It turns out that this list of invariants completely determines the orbit of π\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\pi $$\end{document}, and we will use this fact in forthcoming work to develop an efficient algorithm for calculating the orbit of a given net of rank one. As a more immediate application, we also determine the stabilisers of nets of rank one in PGL(3,q)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathrm {PGL}(3,q)$$\end{document}, and hence the orbit sizes.


Introduction
The forms of degree d on an n-dimensional projective space PG(V ) comprise a vector space W of dimension n+d d . Subspaces of the projective space PG(W ) are called linear systems of hypersurfaces of degree d. One-dimensional linear systems are called pencils, and twodimensional linear systems are called nets. In this paper we are concerned with linear systems of conics, namely the case d = n = 2. For an elementary exposition of the more general case d = 2 (that is, pencils of quadrics), we refer the reader to [3,Chapters 9 and 11].
The classification of linear systems of conics in PG(2, F) consists of determining the orbits of the subspaces of PG(5, F) under the induced action of the projectivity group PGL(3, F). Pencils of conics over the fields F = C and R were classified by Jordan [6,7] in [1906][1907], and nets of conics over these fields were treated several decades later by Wall [11]. Here we are concerned with linear systems of conics over finite fields. Pencils of conics over F q with q odd were classified by Dickson [5] in 1908, and an incomplete classification for q even was obtained by Campbell [2] in 1927. In 1914, Wilson [12] obtained an incomplete classification of nets of conics for q odd. In a recent paper [9], we completed Wilson's classification of nets of conics of rank one, namely those containing a repeated line. It turns out that there are 15 orbits of such nets for every odd q. Representatives are given in Table 6, the notation of which is explained in Sect. 2.
Wilson's approach to this classification problem was purely algebraic, based on explicit coordinate transformations. Our approach was geometric, based on the observation that a net of conics in PG(2, q) corresponds to a plane in PG (5, q), and more specifically that a net of rank one corresponds to a plane that meets the quadric Veronesean in at least one point. The details of this correspondence, and of the aforementioned action of PGL(3, q) on subspaces of PG (5, q), are explained in Sect. 2. In an earlier paper [8], we had also investigated the classification of pencils of conics from this geometric viewpoint, determining the corresponding orbits of lines in PG (5, q). Although the classification itself was necessarily in agreement with that of Dickson [5], our geometric approach had certain advantages, and in particular allowed us to calculate various auxiliary data about the line orbits, including the point-orbit distribution of a line , namely the number of points on belonging to each of the four PGL(3, q)-orbits of points in PG (5, q). Indeed, our original motivation was not to reproduce Dickson's classification, but was instead related to the classification in [10] of tensors in F 2 q ⊗ F 3 q ⊗ F 3 q . The point-orbit distributions obtained in [8] were useful for several arguments in [9], and have since been used to develop an algorithm [1] for determining the tensor rank of an element of F 2 q ⊗ F 3 q ⊗ F 3 q . More generally, of course, the problem of calculating the rank of a tensor has many wide-ranging applications, including, famously, to the problem of determining the computational complexity of matrix multiplication.
The aim of the present paper is to calculate analogous data for nets of conics of rank one in PG (2, q), q odd, viewed as planes in PG (5, q) intersecting the quadric Veronesean. The point-orbit distributions of the planes corresponding to nets of rank one were determined in [9]; they are recorded here in Table 5, the notation of which is explained in Sect. 2. It is natural to ask also about the line-orbit distribution of a plane π, that is, the number of lines in π belonging to each of the orbits determined in [8]. These data are determined in Sect. 3, which comprises the bulk of the paper and constitutes the proof of the following theorem. In order to state this theorem in terms of nets rather than planes, we abuse terminology and take

Preliminaries
Here we summarise some necessary background material, most of which is drawn from our earlier paper [9], to which we refer the reader for further details.
The Veronese surface V(F q ) in PG(5, q) is defined by setting the 2 × 2 minors of the above matrix equal to zero, and we have the corresponding Veronese map from PG (2, q) to V(F q ) ⊂ PG(5, q): The rank of a point in PG (5, q) is the rank of the matrix M y . The points of rank 1 are those in V(F q ); the points of rank 2 are those in its secant variety. The image of a line in PG (2, q) under ν is a conic in V(F q ). A plane in PG(5, q) intersecting V(F q ) in a conic is called a conic plane, and each conic plane is equal to ν( ) for some line in PG (2, q). Each two points x, y ∈ V(F q ) lie on a unique conic C(x, y) ⊂ V(F q ) given by Each rank-2 point z lies in a unique conic plane C z . If z is on the secant x, y then C z = C(x, y).
We have a mapping δ * from the set of conics in PG (2, q) to the set of points in PG(5, q), taking the conic a 01 , a 02 , a 11 , a 12 , a 22 ), which may in turn be viewed as a 3 × 3 matrix as per (2). Under this mapping, a pencil (respectively, net) of conics in PG(2, q) becomes a line (respectively, plane) in PG (5, q). We consider the action of the group PGL(3, q) on the points of PG(5, q) defined as follows: The action of K on points of PG(5, q) induces an action on subspaces, and we are able to make the following crucial observation: The classification of pencils (respectively, nets) of conics in PG(2, q), q odd, up to coordinate transformations is equivalent to the classification of K -orbits of lines (respectively, planes) in PG(5, q).
Moreover, since q is odd, the existence of a particular polarity of PG(5, q), taking the point (y 0 , y 1 , y 2 , y 3 , y 4 , y 5 ) to the hyperplane y 0 Y 0 +y 1 Y 1 +y 2 Y 2 +y 3 Y 3 +y 4 Y 4 +y 5 Y 5 = 0, implies that the K -orbits of points (respectively, lines) in PG (5, q) are in one-to-one correspondence with the K -orbits of hyperplanes (respectively, solids) in PG (5, q). Let us now describe the orbits of K on points, lines and planes in PG (5, q). The points of rank 1 in PG(5, q) form one K -orbit, and the points of rank 3 form a second K -orbit. There are two K -orbits of points of rank 2, namely the exterior and interior rank-2 points, where a rank-2 point z is said to be exterior if it lies on a tangent to the conic C z , and interior otherwise. The following lemma is useful for determining whether a rank-2 point is exterior or interior. Here M i j (A) denotes the matrix obtained from a matrix A by removing the ith row and the jth column, | · | denotes the determinant, and the matrix M y is defined as in (2). According to Dickson's classification [5], the lines in PG(5, q) form 15 orbits under the action of K . Representatives are given in Table 4, as per our earlier papers [8,9]. Note that for certain line orbits, namely those labelled o i, j with i ∈ {8, 13, 14}, we have used different representatives than those originally given in [8, Table 2]; the reason for this is explained in [9,Remarks 3.6 and 4.2], and the representatives used here are the same as those used in [9]. As shown in [9], the planes of rank one, namely those intersecting V(F q ) (and therefore containing a point of rank 1 in the sense defined above), also form 15 K -orbits; representatives are given in Table 6.
The point-orbit distribution of a subspace W of PG(5, q) is the ordered list [n 1 , n 2e , n 2i , n 3 ], where n 1 is the number of rank 1 points in W , n 2e is the number of exterior rank-2 points in W , n 2i is the number of interior rank-2 points in W , and n 3 is the number of rank 3 points in W . The point-orbit distributions of lines of PG(5, q) are given in Table 5. The point-orbit distributions of planes of rank one are given in Table 7. We occasionally refer also to the rank distribution of the subspace W , which is the list [n 1 , n 2e + n 2i , n 3 ].
Note that the K -orbit of a line in PG(5, q) is determined by its point-orbit distribution, unless the point-orbit distribution is [0, 1, 0, q], in which case can be either of type o 15,1 or of type o 16 . (The rank distribution is a strictly coarser invariant: for each i ∈ {8, 13, 14, 15}, the orbits o i,1 and o i,2 have equal rank distributions but non-equal point-orbit distributions.) The following lemma gives a geometric criterion (as opposed to a combinatorial one) for distinguishing between these two orbits. Here S n,n (F q ) is the Segre variety in PG(n 2 − 1, q), that is, the image of the map taking ( x , y ) ∈ PG(n − 1, q) × PG(n − 1, q) to x ⊗ y .

Line-orbit distributions
We now determine the line-orbit distributions of the planes in PG(5, q) that meet the quadric Veronesean in at least one point.

2
, 0], and the points of rank 1 lie on a conic C. (For example, if π is the representative of 1 given in Table 6, then the points of rank 1 comprise the conic C : αβ − γ 2 = 0.) In particular, π is spanned by the points of C; that is, π is a conic plane. Each line in π meets C in either two, one or zero points; that is, is a secant, tangent or external line to C. If is a secant to C, then Here · denotes 0, (α, β) ranges over all non-zero values in F 2 q , and (respectively, ) is the set of squares (respectively, non-squares) in F q . Condition ( * ) is: vλ 2 + uvλ − 1 = 0 for all λ ∈ F q . Condition ( * * ) is: Table 5 implies that has type o 5 , because these are the only types of lines containing two points of rank 1. Hence, the number of lines of type o 5 in π is equal to the number of secants to C, namely q+1 If is a tangent to C, then has type o 6 , because these are the only types of lines containing exactly one point of rank 1 and no points of rank 3. Hence, the number of lines of type o 6 in π is equal to the number of tangents to C, which is q + 1. The remaining q(q−1) 2 lines in π are external to C, so contain only points of rank 2. Table 5 implies that all such lines have type o 10 The group-theoretic notation used in the third column is explained in Sect. 4. (Note that there is a typo in [8, Table 3], where the stabilisers were originally recorded. The stabiliser of a line of type o 10 is listed there as is understood to be the isometry group of a non-degenerate quadratic form of − type on a 2-dimensional vector space over F q , which has order 2(q + 1). The O − (2, q) should be replaced by GO − (2, q), namely the similarity group of such a form, which has order 2(q 2 −1). The corresponding orbit size is listed correctly in [8, Table 4 Proof Let π ∈ 2 , and let x, y, z denote the three points of rank 1 in π. The three lines x, y , y, z , z, x have type o 5 and point-orbit distribution [2, q−1 2 , q−1 2 , 0]. Since π has point-orbit distribution [3, 3(q−1) 2 , 3(q−1) 2 , q 2 − 2q + 1], any point not on one of these three lines has rank 3. Now consider the other q − 1 lines through x. Half of these lines meet y, z in an interior rank-2 point, and the other half meet y, z in an exterior rank-2 point. Hence, there are q−1 2 lines of each of types o 8,1 and o 8,2 through x. Repeating this argument for y and z yields a total of 3(q−1) 2 lines of each of these two types. Now consider an exterior rank-2 point u. Without loss of generality, u lies on y, z and on a line of type o 8,1 through x, both of which have already been counted. Let be one of the remaining q − 1 lines through u. Then meets the lines x, y and x, z in points of rank 2, and therefore contains exactly three points of rank 2. It follows from Table 5 that either (i) the rank-2 points on are all exterior, in which case has type o 14,1 ; or (ii) one (namely u, in our case) is exterior and the other two are interior, in which case has type o 14,2 . Since half of the q − 1 points of rank 2 on x, y , say, are exterior and the other half are interior, it follows that u lies on q−1 2 lines of each of the types o 14,1 and o 14,2 . Since each line of type o 14,2 contains exactly one exterior rank-2 point, the total number of lines of type o 14,2 in π is Table 6 Representatives of the 15 orbits of planes in PG(5, q), q odd, meeting the quadric Veronesean in at least one point, under the action of PGL(3, q) ≤ PGL(6, q) defined in Sect. 2
As explained in the caption of the table, here · denotes 0, and the triple of parameters (α, β, γ ) ranges over F 3 q \ {(0, 0, 0)}. Recall that π has point-orbit distribution [2, 3q−1 2 , q−1 2 , q 2 − q]. Let x and y denote the rank-1 points given by β = γ = 0 and α = γ = 0, respectively. The line x, y has type o 5 . The rank-2 points not on this line are all on the line : β = 0, which has type o 6 (and contains x). In particular, every line through x other than x, y or has point-orbit distribution [1, 0, 0, q] and hence type o 9 , yielding a total of q − 1 such lines in π. Every line through y other than x, y meets in an exterior point of rank 2 and therefore has type o 8,1 , giving a total of q such lines. Now consider a point w of rank 2 on the line x, y . Each of the q other lines through w meets in an exterior point of rank 2, so has point-orbit distribution [0, 2, 0, q − 1] or [0, 1, 1, q − 1], and hence type o 13,1 or o 13,2 , according to whether w is exterior or interior. This gives a total of q(q−1) 2 lines of each of these two types. Proof Let π denote the representative of 4 given in Table 6, namely ⎡ Recall that π has point-orbit distribution [2, 3q−1 2 , q−1 2 , q 2 − q]. Let x and y denote the rank-1 points given by β = γ = 0 and α = γ = 0, respectively. The line x, y has type o 5 . The rank-2 points not on this line are all on the line : β = −α, which has type o 12 , as can be verified by applying Lemma 2.2. In particular, every line through x or y other than x, y meets in an exterior point of rank 2, so has point-orbit distribution [1, 1, 0, q − 1] and hence type o 8,1 . This yields a total of 2q lines of type o 8,1 .
The lines and x, y meet in the exterior rank-2 point w : α = −β = 1, γ = 0. Let be any of the other q − 1 lines through w. Then has point-orbit distribution [0, 1, 0, q] and hence type o 15,1 or o 16 . We apply Lemma 2.3 to show that it has type o 16 . Note that is represented by a matrix of the form where α, β, γ are fixed and (a, b) ranges over F 2 q \ {(0, 0)}. In the notation of Lemma 2.3, the solid W determined by w is represented by the matrices whose third rows and third columns are zero. Hence, where b and the d i j range over F q . Since every point on with b = 0 has rank 3, we have γ = 0 and so the above matrix cannot have rank 1 unless b = 0. Hence, all rank-1 points in U lie in Q = W ∩ S 3,3 (F q ), and the lemma therefore implies that has type o 16 . Finally, consider a rank-2 point u = w on x, y . Each of the other q lines through u meets in an exterior rank-2 point, so has point-orbit distribution [0, 2, 0, q − 1] or [0, 1, 1, q − 1], and hence type o 13,1 or o 13,2 , according to whether u is exterior or interior. There are q−1 2 choices for u that are interior, and q−1 2 − 1 that are exterior (because u = w). Hence, π contains in total q(q−1)  Proof Let π denote the representative of 5 given in Table 6, namely ⎡ Recall that π has point-orbit distribution [2, q − 1, q − 1, q 2 − q + 1]. Let x and y denote the rank-1 points in π given by β = γ = 0 and α = γ = 0. The line = x, y , given by γ = 0, has type o 5 . The q − 1 rank-2 points in π not on this line are all on the conic C : αβ − (α + β)γ = 0; half are exterior and half are interior. The conic C meets the line in the points x, y. All but one of the other lines through x meet C in a unique point of rank 2; half of these q − 1 lines have type o 8,1 and the other half have type o 8,2 . The remaining line through x is a tangent to C and so has type o 9 . The same argument holds for the lines through y, so in total we obtain q − 1 lines of type o 8,i for each i ∈ {1, 2}, and two lines of type o 9 . Next, consider the tangent lines to C at its points of rank 2. Let t u (C) denote the tangent line to C at a rank-2 point u. We claim that (i) t u (C) meets in an exterior rank-2 point, regardless of whether u is exterior or interior.
Denote the coordinates of u by (α u , β u , γ u ). (That is, suppose that u is represented by the matrix in (3) with a subscript 'u' on each parameter.) Then β u = γ u , because otherwise the equation of C would imply that γ u = 0, contradicting that assumption that u does not lie on . With this noted, a direct calculation shows that t u (C) meets in the point with coordinates , which is exterior by Lemma 2.2. The line t u (C) therefore has type o 13,1 or o 13,2 according to whether u is exterior or interior, giving a total of q−1 2 lines of each type.
Next, we verify that To complete the proof of the lemma, we count the remaining lines through a rank-2 point w on . First suppose that w is interior. By (i), w lies on no tangent lines to C, so every line through w that meets C has type o 14,2 , and has already been counted in (ii). There are q−1 2 such lines through w. The remaining (q + 1) − 1 − q−1 2 = q+1 2 lines through w do not meet C, so have type o 15,2 , giving a total of (q+1)(q−1) 4 such lines in π. Now suppose that w is exterior. By (i), w lies on the tangent lines to C at two rank-2 points. Any other line through w that meets C has type o 14,1 or o 14,2 . There are q−3 2 such lines through w, and we have already counted them in (ii). The remaining (q where α, β, γ are fixed and (a, b) ranges over F 2 q \ {(0, 0)}. In the notation of Lemma 2.3, the solid W determined by w is represented by the matrices whose third rows and third columns where b and the d i j range over F q . Since every point on with b = 0 has rank 3, we have γ = 0. Hence, taking all of b and the d i j to be the same yields a point of rank 1 outside Q = W ∩ S 3,3 (F q ), and the lemma therefore implies that has type o 15,1 .
Proof Let π denote the representative of 6 from Table 6 Then π has point-orbit distribution [1, q+1 2 , q+1 2 , q 2 − 1], and its points of rank 2 are all on the line γ = 0, which has type o 10 (by Table 5). Consider the unique point of rank 1 in π, namely x : α = β = 0. Each line through x meets exactly one of the points of rank 2, so π contains q+1 2 lines in each of the orbits o 8,1 and o 8,2 . Now consider the lines through a rank-2 point w. We have already counted the line x, w (of type o 8,1 or o 8,2 according to whether w is exterior or interior) and the line γ = 0 (of type o 10 ). If w is interior then the remaining q − 1 lines through w have point-orbit distribution [0, 0, 1, q] and hence type o 15,2 , giving a total of q+1 2 · (q − 1) such lines in π. If w is exterior then the remaining q − 1 lines through w have point-orbit distribution [0, 1, 0, q] and hence type o 15,1 or o 16 . We use Lemma 2.3 to show that they all have type o 15,1 , which gives the claimed total of q+1 2 · (q − 1) lines of type o 15,1 in π. Let be such a line. Since w lies on the line γ = 0, the solid W determined by w is represented by the matrices whose third rows and third columns are zero, so U := W , is represented by ⎡ for some fixed non-zero γ , where b and the d i j range over F q . Taking all d i j = 0 yields a point of rank 1 outside Q = W ∩ S 3,3 (F q ), so has type o 15,1 as claimed.

Lemma 3.7 A plane of type 7 contains
• q + 1 lines of type o 6 , • q 2 lines of type o 12 .
Proof A plane π ∈ 7 has a unique point x of rank 1, and q 2 + q exterior points of rank 2.

Lemma 3.8 A plane of type 8 contains
Proof Let π denote the representative of 8 given in Table 6, namely ⎡ for some fixed non-zero γ , where b and the d i j range over F q . Such a matrix cannot have rank 1 unless b = 0, so all rank-1 points in U lie in Q = W ∩ S 3,3 (F q ), and therefore has type o 16 . Finally, let be any of the q(q − 1) lines not considered thus far. Then contains neither x nor w, and intersects each of the lines α = 0 and γ = 0 in an exterior point of rank 2. The point-orbit distribution of is therefore [0, 2, 0, q − 1], so has type o 13,1 .

Lemma 3.9 A plane of type
Proof Let π denote the representative of 9 given in Table 6 The 2q points of rank 2 in π are all exterior and lie on the union of the line : γ = 0 of type o 6 and the conic C : β 2 − αγ = 0, which meet (only) in the unique point x : β = γ = 0 of rank 1. Apart from , every other line through x meets C in a unique point of rank 2, so has point-orbit distribution [1, 1, 0, q − 1] and hence type o 8,1 , giving a total of q such lines.
If u is one of the q points of rank 2 on C then q − 1 of the lines through u contain exactly one point on C other than x, and exactly one point on other than x. Such lines have pointorbit distribution [0, 3, 0, q − 2] and hence type o 14,1 . Each contains two rank-2 points on C, so there are in total 1 2 · q · (q − 1) such lines in π. The remaining line through u, other than x, u , which we have already counted, is tangent to C and meets in a point of rank 2, so has point-orbit distribution [0, 2, 0, q − 1] and hence type o 13,1 . There are q such lines in π.
Finally, consider a rank-2 point w on . Thus far, we have counted q−1 2 + 2 of the lines through w, namely itself, one line of type o 13,1 , and q− such lines in π. Since w lies on the line γ = 0, the solid W determined by w is represented by the matrices whose third rows and third columns are for some fixed non-zero γ , where b and the d i j range over F q . Taking all d i j = 0 yields a point of rank 1 outside Q = W ∩ S 3,3 (F q ), so has type o 15,1 as claimed. Proof The proof is analogous to that of Lemma 3.9, but we include some details for clarity. Let π denote the representative of 10 given in Table 6 It remains to determine the line-orbit distributions of the planes in the K -orbits 11 , 12 , 13 , 14 , 14 and 15 . We treat these remaining cases (roughly) in order of difficulty: • 15 is treated first, as it is by far the most straightforward remaining case.
• We then handle the case 11 , which is considerably more complicated.
• The case 14 is treated next as it is largely analogous to 11 , having the same point-orbit distribution, and, in particular, rank-2 points of only one type (exterior). • Finally, we deal with 12 , 13 and 14 . These cases pose further complications, owing to the presence of both exterior and interior points of rank 2. Amongst them, 14 is arguably the most straightforward, at least in the sense that q ≡ 0 (mod 3) and the cubic of points of rank ≤ 2 has no inflexion points, so we opt to treat it first. (Recall that an inflexion point of a plane cubic curve C = Z( f ) is a point of intersection of C and its Hessian, namely the zero locus of the determinant of the 3 × 3 matrix of partial derivatives of the cubic f .) We then treat 13 before 12 because the representative of 12 in Table 4 may be obtained from the representative of 13 by setting ε = 1, so some calculations in the argument for 13 follow through immediately for 12 .

Lemma 3.11 A plane of type 15 contains
Proof Let π denote the representative of 15 given in Table 6, namely ⎡ where b and the d i j range over F q . Such a matrix cannot have rank 1 unless b = 0, so all rank-1 points in U lie in Q = W ∩ S 3,3 (F q ), and therefore has type o 16 .

Lemma 3.12
The lines in a plane of type 11 are as indicated in Table 8.
Proof If M denotes the matrix representative of 11 given in Table 6, then we have For convenience, we work with the above representative of 11 instead of the one in Table 6. Let us call the corresponding plane π. The points of rank at most 2 in π lie on the cubic There is a unique point of rank 1, namely x : β = γ = 0, and q points of rank 2, one for each choice of (β, γ ) with β = γ . All points of rank 2 are exterior. In particular, q of the lines through x have point-orbit distribution [1, 1, 0, q −1] and (therefore) type o 8,1 , and the remaining line through x has point-orbit distribution [1, 0, 0, q] and type o 9 . We now count the remaining lines in π containing points of rank 2. Note that if (α, β, γ ) are the coordinates of a point of rank 2, then β = γ and α = −β 2 γ /(β − γ ) 2 . Note also that C has one point of inflexion if q ≡ 0 (mod 3), and no points of inflexion if q ≡ 0 (mod 3). In the former case, the unique point of inflexion is given by 2β + γ = 0. We need the following fact: ( * ) With the exception of the inflexion point in characteristic = 3, every rank-2 point v on C lies on exactly two tangent lines to C (one of which is the tangent line at v).
Proof of ( * ). Consider a point w of rank 2 on C, denoting its coordinates by (α w , β w , γ w ), (For the sake of presentation, the coefficients of α, β and γ given above are equal to β w − γ w times the respective partial derivatives of the form defining the cubic C evaluated at the so v lies on the tangent line to C at w if and only if either Regarding v as fixed, the latter equation determines a unique rank-2 point w, which is distinct First suppose that q ≡ 0 (mod 3), and let z denote the unique point of inflexion of C. Consider the lines through z, recalling that we have already counted the line z, x . The tangent line to C through z contains no other points of C, so has point-orbit distribution [0, 1, 0, q] and hence type o 15,1 or o 16 . We show below that it has type o 16 : see Claim 11 at the end of the proof. The remaining q − 1 lines through z contain either zero or two other points on C. The q−1 2 lines containing two other points on C have point-orbit distribution [0, 3, 0, q − 2] and hence type o 14,1 . The q−1 2 lines containing no other points on C have point-orbit distribution [0, 1, 0, q]; we show below (Claim 11 ) that they all have type o 15,1 . Now consider a rank-2 point w = z. We have already counted the line w, x (of type o 8,1 ) and the line w, z (of type o 14,1 ). The tangent line to C through w meets exactly one other (rank-2) point w on C, so has point-orbit distribution [0, 2, 0, q − 1] and hence type o 13,1 . By ( * ), w also lies on the tangent line to C at a unique rank-2 point w = w , which gives a second line w, w through w of type o 13,1 . There are q − 1 choices for w, so we obtain a total of q − 1 lines of type o 13,1 in π. At this point we have a further q − 3 lines through w to consider. There are q − 5 points of rank 2 on C that do not lie on any of the lines w, z , w, w , w, w . Hence, q−5 2 of the remaining q − 3 lines through w have point-orbit distribution [0, 3, 0, q − 2] and type o 14,1 , giving a further 1 3 · (q − 1) · q−5 2 lines of type o 14,1 in addition to the q−1 2 lines of type o 14,1 through z counted above. Therefore, π contains in Now suppose that q ≡ 0 (mod 3). The types of the lines through the unique point x of rank 1 do not change. However, now C has no points of inflexion, so every point w of rank 2 lies on two lines of type o 13,1 (namely, the tangent line to C at w and the tangent line to C at some other point w of rank 2, in the notation of the above argument). Therefore, π now contains q rather than q − 1 lines of type o 13,1 . The remaining q − 2 lines through each w now meet a total of q − 3 remaining rank-2 points on C. This gives a total of 1 3 ·q · q−3 2 lines of type o 14,1 , and q ·((q −2)− q−3 2 ) = q(q−1) 2 lines with point-orbit distribution [0, 1, 0, q −1]. By Claim 11 below, all of the latter lines have type o 15,1 . All of the other lines in π have type o 17 .
Claim 11 . Let be a line in π with point-orbit distribution [0, 1, 0, q], and let w denote its unique point of rank 2. Then has type o 15,1 unless q ≡ 0 (mod 3), w is the unique point of inflexion of C, and is the tangent line to C at w, in which case has type o 16 .
Proof of Claim 11 . Let (α w , β w , γ w ) denote the coordinates of w. Recall that α w = −β 2 w γ w /(β w − γ w ) 2 , and that the tangent line to C at w is given by (5). We apply Lemma 2.3. In order to calculate the solid W determined by w, we first use an appropriate element of K to transform the matrix representative of w to one whose third row and third column are zero. This allows us to apply Lemma 2.4 in a uniform way to determine the type of . First suppose that β w γ w = 0. Then the matrix representative of w obtained from (4), call it M w , can be transformed via the action of K as follows: After making this transformation, the solid W of Lemma 2.3 is, as desired, represented by the matrices whose third rows and third columns are zero. Now choose a point y (of rank-3) such that = w, y , denoting the coordinates of y by (α, β, γ ). Then the image of y under the element of K corresponding to the matrix Y w is represented by the matrix ⎡ where f w (α, β, γ ) is the left-hand side of (5). Hence, in the notation of Lemma 2.3, the where b and the d i j range over F q . Lemma 2.3 now tells us that has type o 15,1 if and only if there exist d i j such that the above matrix has rank 1 when b = 0. Lemma 2.4 therefore implies that has type o 15,1 unless f w (α, β, γ ) = 0, that is, unless y lies on the tangent line to C at w. (Note that the condition that the c i not be all zero holds because the point y is assumed to have rank 3.) However, cannot be the tangent line to w unless q ≡ 0 (mod 3) and w is the unique point of inflexion of C, because (as explained above) in all other cases the tangent line to C at w has point-orbit distribution [0, 2, 0, q − 1], in contradiction with the assumption of the claim. The proof is analogous for the remaining two choices for w, namely those with β w γ w = 0. When γ w = 0, replace the third row of Y w by (1, 0, 1); and when β w = 0, replace Y w by the matrix obtained from the identity matrix by swapping the first and third rows. Note that the matrix representative M w of w is mapped to ⎡ in these respective cases. The (3, 3)-entry of the matrix representing U changes to bα or b(α + γ ) according to whether β w = 0 or γ w = 0, and the result follows as in the generic case because α = 0 and α + γ = 0 are the corresponding tangent lines to C. This completes the proof of the claim, and therefore the proof of the lemma.
Recall that the K -orbit 14 arises (if and) only if q is a power of 3. Proof If M denotes the matrix representative of 14 given in Table 6, then we have Let π ∈ 14 denote the plane with this new representative. The points of rank at most 2 in π lie on the cubic C : βγ (β + γ ) + α(β − γ ) 2 = 0. There is a unique point of rank 1, namely x : β = γ = 0, and q points of rank 2, all exterior, one for each choice of (β, γ ) with β = γ .
In particular, q lines through Proof of Claim 14 . Let (α w , β w , γ w ) be the coordinates of w. Then β w = γ w and α w = −β w γ w (β w + γ w )/(β w − γ w ) 2 . The tangent line to C at w is First suppose that β w = 0. Then the matrix representative M w of w obtained from (6) can be transformed as follows: After making this transformation, the solid W of Lemma 2.3 is represented by the matrices whose third rows and third columns are zero. If we now consider the image of a rank-3 point on , with coordinates (α, β, γ ), under the element of K corresponding to Y w , we find that the image of U := W , (in the notation of Lemma 2.3) is represented by ⎡ where b and the d i j range over F q , and f w (α, β, γ ) is the left-hand side of (7). The claim (for β w = 0) therefore follows from Lemmas 2.3 and 2.4 . When β w = 0, replace Y w by the matrix obtained from the identity matrix by swapping the second and third rows. The Table 9 Data for Lemma 3.14 Type 3)-entry of the matrix representing U then becomes b(α + β), and the result follows because the tangent line to w is the line α + β = 0. This completes the proof of the claim, and of the lemma.
Recall that the K -orbit 14 arises (if and) only if q is not a power of 3.
Lemma 3.14 Suppose that q ≡ 0 (mod 3). The lines in a plane of type 14 are as indicated in Table 9.
Proof Let π denote the representative of 14 given in Table 6, namely ⎡ where c satisfies the condition ( †) given in Table 6. That is, c is some fixed element of . Note that this implies, in particular, that c is a square if q ≡ 1 (mod 3) and a non-square if q ≡ −1 (mod 3). The points of rank at most 2 in π lie on the cubic The point-orbit distribution of π is [1, q∓1 2 , q∓1 2 , q 2 ± 1] according to whether q ≡ ±1 (mod 3). If q ≡ −1 (mod 3), the lines through the unique point x : β = γ = 0 of rank 1 therefore comprise q+1 2 lines of each of the types o 8,1 and o 8,2 . If q ≡ 1 (mod 3), they instead comprise q−1 2 lines of each of these types, plus two lines of type o 9 . Now consider a point w of rank 2, with coordinates (α w , β w , γ w ). Note the following: (ii) f c (β w , γ w ) = 0. (If not then β w γ w = 0 by (9). If β w = 0 then f c (β w , γ w ) = (c−1)γ 2 w , so γ w = 0 because c = 1; similarly, γ w = 0 implies β w = 0. Either case contradicts (i).) (iii) α w = β 2 w γ w / f c (β w , γ w ), by (ii). (iv) w is exterior if and only if − f c (β w , γ w ) is a non-zero square in F q , by Lemma 2.2.
Therefore, the tangent line to C at w contains exactly one other point of rank 2, say v with coordinates (α v , β v , γ v ) (where α v is determined as in (iii)). We claim that (vi) v is exterior, regardless of whether w is exterior or interior.
Proof of (vi). In the 'free' coordinates (α, β, γ ), the tangent line to C at w is Putting the coordinates (α, β, γ ) = ( (10) gives If v is a non-zero square and v is therefore exterior by (iv). Now suppose that γ v = 0. This implies that β w = 0 and (c −1)γ w +β w = 0: if β w = 0 then (11) and so which is always a non-zero square. Hence, (vi) holds, as claimed. By (vi), the tangent line to C at w has type o 13,1 or o 13,2 according to whether w is exterior or interior. There are q∓1 2 exterior and interior rank-2 points in π according to whether q ≡ ±1 (mod 3), and hence a total of this many lines of the corresponding types.
We now set about counting the remaining lines containing a point of rank 2. Suppose first that w i is an interior rank-2 point. We have already counted the line w i , x (of type o 8,2 ) and the tangent line to C at w i (of type o 13,2 , containing also an exterior rank-2 point). Let be one of the remaining q − 1 lines through w. By (vi), is not the tangent to any other point on C. Hence, if meets C in a second point, then it meets C in exactly three points, and so contains exactly three points of rank 2 (and no points of rank 1). Since w i is interior, must have type o 14,2 ; in particular, it must contain a second interior rank-2 point, w i say. There are To count the lines through an exterior point of rank 2 (apart from those already counted above), we need the following additional fact.
(vii) Every exterior rank-2 point v on C lies on exactly three tangent lines to C (one of which is the tangent line at v). Moreover, if v is on the tangent lines at the points w and w , say, then w and w have the same type (that is, either both are exterior or both are interior) if and only if q ≡ ±1 (mod 12).
Proof of (vii). Denote the (second and third) coordinates of v by (β v , γ v ) and view (11) as a quadratic equation in the coordinates (β w , γ w ) of a rank-2 point w whose tangent contains v: If  (13) is non-zero, so any solution has γ w = 0 and we may therefore view the left-hand side of (13) as a quadratic in β w γ w . The discriminant of this quadratic is − f c (β v , γ v ) · 4(c − 1) 2 , which is a non-zero square by (iv), so there are again two solutions, given by A further calculation shows that Note at this point that β w −γ w is non-zero: if it were zero then w would be the point (c −1 , 1, 1) and so v would be the point (−1, 1−c 2 , 1) considered previously. Therefore, − f c (β w , γ w ) and − f c (β w , γ w ) are either both squares or both non-squares if and only if −c is a square. As noted above, this is the case if and only if q ≡ ±1 (mod 12). This completes the proof of (vii). Now consider an exterior rank-2 point w e . First suppose that q ≡ ±1 (mod 12), that is, q ≡ 7 or 5 (mod 12). Recall that we have already counted the line through w e and the unique point of rank 1 (which has type o 8,1 ). By (vii), w e lies on the tangent lines to C at one exterior rank-2 point z e and one interior rank-2 point z i . These tangent lines have type o 13,1 and o 13,2 respectively, and have already been counted. Moreover, by (vi), the tangent line to w e at C has type o 13,1 and has also been counted; denote the second exterior rank-2 point on this line by w T e . Any other line through w e containing an interior rank-2 point has type o 14,2 and has already been counted. There are q∓1 2 − 1 interior rank-2 points other than z i according to whether q ≡ ±1 (mod 3), and hence half this many lines of type o 14,2 through w e . That is, there are q−3 4 or q−1 4 lines of type o 14,2 through w e according to whether q ≡ 7 or 5 (mod 12). Any other line through w e containing an exterior rank-2 point, w e say, has type o 14,1 and contains also a third exterior rank-2 point. For q ≡ ±1 (mod 3), there are q∓1 2 choices for w e and q∓1 2 − 3 choices for w e , namely all of the exterior rank-2 points other than w e , z e and w T e . Hence, there are 1 such lines through each w e according to whether q ≡ 7 or 5 (mod 12), and hence a total of (q−1) 2 4 or (q+1)(q−3) 4 such lines in π in these respective cases. Now suppose that q ≡ ±1 (mod 12). Note in particular that q ∓ 1 is then divisible by 4. In light of (vii), we must now consider two possibilities for the (otherwise arbitrary) exterior rank-2 point w e , namely, whether or not w e lies on the tangent line to C at some interior rank-2 point. For convenience, let us say that w e is of class I if it lies on the tangent line to C at some interior rank-2 point, and of class E if it does not. Note that each class comprises exactly half of the q∓1 2 exterior rank-2 points. In either case, recall yet again that we have already counted the line through w e and the point of rank 1.
First consider a point w e of class E. By (vii), w e lies on the tangent lines to C at two exterior rank-2 points, say z 1 and z 2 . These tangent lines have type o 13,1 and have already been counted. The tangent line to C at w e has also already been counted; it has type o 13,1 and contains an exterior rank-2 point w T e / ∈ {w e , z 1 , z 2 }. There are q − 3 lines through w e left to consider. Let be a line through w e containing an interior rank-2 point. Since is not the tangent line to C at w e , it also contains a second interior rank-2 point, and has type o 14,2 . All such lines have already been counted (because all lines through any interior rank-2 point have already been counted). There are q∓1 2 interior rank-2 points when q ≡ ±1 (mod 12), and hence half this many lines of type o 14,2 through w e . This leaves Suppose now that w e is of class I. By (vii), w e lies on the tangent lines to C at two interior rank-2 points, say z 1 and z 2 . The tangent lines have type o 13,2 and have already been counted. The tangent line to C at w e has also already been counted; it has type o 13,1 and contains an exterior rank-2 point w T e = w e . There are q − 3 lines through w e left to consider. Any line through w e containing one of the q∓1 2 − 2 interior rank-2 points other than z 1 and z 2 contains two such points and has type o 14,2 . These lines have already been counted. There are 1 2 · q−5 2 or 1 2 · q−3 2 of them through w e according to whether q ≡ 1 or − 1 (mod 12), leaving 3q−7 4 or 3q−9 4 lines through w e to consider in these respective cases. Now let be a line through w e containing an exterior rank-2 point w e = w e . The only such line counted so far is the tangent line to C at w e (with w e = w T e ), so we may assume that is not this line. Hence, also contains a third exterior rank-2 point, and has type o 14,1 . The candidates for w e are all of the exterior rank-2 points except w e and w T e , so there are 1 2 ( q−1 2 − 2) = q−5 4 or 1 2 ( q+1 2 − 2) = q−3 4 lines of type o 14,1 through w e according to whether q ≡ 1 or − 1 (mod 12). The remaining lines through w e have point-orbit distribution [0, 1, 0, q] and, by Claim 14 below, type o 15,1 . There are q−1 2 or q−3 2 such lines through w e according to whether q ≡ 1 or − 1 (mod 12). Let us now count the total number of lines of types o 14,1 and o 15,1 in π for q ≡ ±1 (mod 12). Consider first the lines of type o 15,1 . As argued in the preceding two paragraphs, every exterior rank-2 point, whether of class E or class I, lies on q−1 2 or q−3 2 lines of type o 15,1 according to whether q ≡ 1 or − 1 (mod 12). There are q∓1 2 exterior rank-2 points in these respective cases, and hence a total of (q−1) Proof of Claim 14 . Let be such a line, and let (α w , β w , γ w ) denote the coordinates of the exterior rank-2 point on . Recall that α w = β 2 w γ w / f c (β w , γ w ), where f c is defined as in (9), and that the tangent line to C at w is given by (10). We apply Lemma 2.3. First suppose that β w (β w − γ w ) = 0. Then the matrix representative of w obtained from (8), call it M w , can be transformed via the action of K as follows: Under this transformation, the solid W of Lemma 2.3 is represented by the matrices whose third rows and third columns are zero. Now choose a point y (of rank 3) such that = w, y , denoting the coordinates of y by (α, β, γ ). Then the image of y under the element of K corresponding to the matrix Y w is represented by the matrix ⎡ where g w (α, β, γ ) is the left-hand side of (10). Hence, in the notation of Lemma 2.3, the image of U := W , is represented by ⎡ where b and the d i j range over F q . Lemma 2.3 now tells us that has type o 15,1 if and only if there exist d i j such that the above matrix has rank 1 when b = 0. Lemma 2.4 therefore implies that has type o 15,1 unless g w (α, β, γ ) = 0, that is, unless y lies on the tangent line to C at w. The latter condition is impossible, because the tangent line to C at w contains two points of rank 2, whereas, by assumption, contains a unique point of rank 2. The proof is The (3, 3)-entry of the matrix representing U then changes to bα or b(αc 2 − 2βc + γ c), respectively, and the result follows as in the generic case because α = 0 and αc 2 −2βc+γ c = 0 are the corresponding tangent lines to C. This completes the proof of the claim, and of the lemma.

Lemma 3.15
The lines in a plane of type 13 are as indicated in Table 10.
The inflexion points of C are precisely the rank-2 points satisfying γ (3β 2 + εγ 2 ) = 0. The point y = (0, 1, 0) is an inflexion point for all q. If q is a power of 3, it is the only inflexion point, because 3β 2 = 0. If q ≡ 1 (mod 3) then −3 is a square in F q , so −3ε is a non-square and hence the equation 3β 2 + εγ 2 = 0 has no (non-zero) solutions. Therefore, y is again the only inflexion point. If q ≡ −1 (mod 3) then −3 is a non-square and there are two further inflexion points, namely y ± = (− 3ε 4 , ± √ −3ε, 3). Now consider a point w of rank 2, with coordinates (α w , β w , γ w ). Note the following: , by (ii). (iv) w is exterior if and only if β 2 w − εγ 2 w is a non-zero square in F q (by Lemma 2.2); in particular, all inflexion points of C are exterior.
If w is not an inflexion point then the tangent line to C at w contains exactly one other point of rank 2, say v with coordinates (α v , β v , γ v ), where α v is given by (iii). We claim that (vi) v is exterior, regardless of whether w is exterior or interior.
If w is not an inflexion point then, by (vi), the tangent line to C at w has type o 13,1 or o 13,2 according to whether w is exterior or interior. By (iv), all inflexion points are exterior, so there are We now count the remaining lines containing a point of rank 2. Suppose first that w i is an interior rank-2 point. We have already counted the line w i , x (of type o 8,2 ) and the tangent line to C at w i (of type o 13,2 ). Let be one of the remaining q − 1 lines through w i . By (vi), is not the tangent to any other point on C, so if it meets C in a second point then it meets C in exactly three points. Since w i is interior, this implies that has type o 14,2 , and in particular that it contains exactly one other interior rank-2 point, w i , say. Each of the q+1 2 choices of w i therefore lies on q+1 2 − 1 = q−1 2 lines of type o 14,2 (that is, the number of choices of w i ), and π contains 1 2 · q+1 2 · q−1 2 such lines in total. The remaining (q − 1) − q−1 2 = q−1 2 lines through w i have type o 15,2 , so π contains a total of (q+1)(q−1) 4 such lines.
To count the remaining lines through an exterior rank-2 point, we first note the following additional fact, which suggests that separate arguments will be required to treat the cases q ≡ ±1 (mod 4). Recall here that all inflexion points of C are exterior. Proof of (vii). Denote the (second and third) coordinates of v by (β v , γ v ) and view (16) as a quadratic equation in the coordinates (β w , γ w ) of a rank-2 point w whose tangent contains v: is a square, namely if and only if −1 is a non-square. Now suppose that v is not an inflexion point. Then in particular v = y, so γ v = 0, and so any solution (β w , γ w ) of (17) has γ w = 0, because γ w = 0 would imply β w = 0, contradicting (i). Hence, we may view (17) as a quadratic in β w γ w . The discriminant of this quadratic is 4(β 2 v − εγ 2 v ), which is a non-zero square by (iv), so there are two solutions (β w , γ w ) and (β w , γ w ), given by A further calculation shows that so (iv) implies that w and w are of the same type (exterior or interior) if and only if −1 is a non-square. This completes the proof of (vii). Suppose now that q ≡ 1 (mod 4). Consider an exterior rank-2 point w e , and recall that we have already counted the line of type o 8,1 through w e and the unique point of rank 1. Suppose first that w e is an inflexion point. By (vii), w e lies on the tangent line to C at a unique interior rank-2 point z i . This tangent line has type o 13,2 and has already been counted. The tangent line to w e at C, which has type o 16 , has also already been counted, so there are To complete the proof in the case q ≡ 1 (mod 4), it remains to note that all lines not counted thus far have type o 17 (and that π contains q 2 + q + 1 lines in total). Now suppose that q ≡ −1 (mod 4). Consider an exterior rank-2 point w e , and recall that we have already counted the line (of type o 8,1 ) through w e and the unique point of rank 1. Suppose first that w e is an inflexion point. By (vii), w e lies on the tangent line to C at a unique exterior rank-2 point z e . This tangent line has already been counted, as has the tangent line to w e at C, so there are (q + 1) − 3 = q − 2 lines through w e left to consider. Any line through w e containing an interior rank-2 point has type o 14,2 and has already been counted. There are q+1 4 such lines through w e , leaving (q − 2) − q+1 4 = 3q−9 4 lines through w e to consider. Apart from the line w e , z e , any other line through w e containing a second exterior rank-2 point, w e say, has type o 14,1 and contains also a third exterior rank-2 point. There are q+1 2 − 2 = q−3 2 choices for w e , and half this many lines of type o 14,1 through w e . The remaining 3q−9 4 − q−3 4 = q−3 2 lines through w e have point-orbit distribution [0, 1, 0, q] and hence type o 15,1 or o 16 . By Claim 13 below, they are all of type o 15,1 . As in the q ≡ 1 (mod 4) case, we delay counting the total number of lines of types o 14,1 and o 15,1 until we have also considered the non-inflexion points. Now suppose that w e is not an inflexion point. By (vii), there are two possibilities to consider, namely whether w e lies on the tangent line to C at some interior rank-2 point or not. We say that w e is of class I or class E in these respective cases. There are q+1 4 points of class I, each lying on the tangent lines to C at two of the q+1 2 interior rank-2 points. The remaining non-inflexion points w e are of class E; there are q+1 4 − 3 = q−11 4 or q+1 4 − 1 = q−3 4 of them according to whether q ≡ −1 (mod 3) or not. Regardless of the class of w e , recall yet again that we have already counted the line through w e and the point of rank 1.
First consider a point w e of class E. By (vii), w e lies on the tangent lines to C at two exterior rank-2 points, say z 1 and z 2 . These tangent lines have already been counted. The tangent line to C at w e has also already been counted; it has type o 13,1 and contains an exterior rank-2 point w T e / ∈ {w e , z 1 , z 2 }. There are q − 3 lines through w e left to consider. Let be a line through w e containing an interior rank-2 point. Since is not the tangent line to C at w e , it has type o 14,2 . There are q+1 4 such lines through w e , leaving (q − 3) − q+1 4 = 3q−13 4 lines through w e to consider. Now let be a line through w e containing an exterior rank-2 point w e / ∈ {w T e , z 1 , z 2 }. Any such line has type o 14,1 . There are q+1 2 − 4 = q−7 2 exterior rank-2 points other than w e , w T e , z 1 and z 2 , and therefore q−7 4 lines of type o 14,1 through w e . The remaining q−3 2 lines through w e have point-orbit distribution [0, 1, 0, q], and Claim 13 shows that they are all of type o 15,1 . Again, we delay counting the total number of lines of types o 14,1 and o 15,1 .
Finally, consider a point w e of class I. By (vii), w e lies on the tangent lines to C at two interior rank-2 points, say z 1 and z 2 . These tangent lines have already been counted. The tangent line to C at w e has also already been counted; it has type o 13,1 and contains an exterior rank-2 point w T e = w e . There are q − 3 lines through w e left to consider. Let be a line through w e containing an interior rank-2 point. Since is not the tangent line to C at w e , it has type o 14,2 . There are q+1 2 − 2 = q−3 2 interior rank-2 points other than z 1 and z 2 , and hence half this many lines of type o 14,2 through w e . This leaves Let us now finally count the total number of lines of types o 14,1 and o 15,1 in π in the case q ≡ −1 (mod 4). As explained in the preceding arguments, every exterior rank-2 point w e lies on q−3 2 lines of type o 15,1 , so π contains a total of (q+1)(q−3) (α, β, γ ), under the element of K corresponding to Y w , we find that the image of U := W , where b and the d i j range over F q , and f w (α, β, γ ) is the left-hand side of (15). Lemmas 2.3 and 2.4 therefore imply that has type o 15,1 unless f w (α, β, γ ) = 0, that is, unless y lies on the tangent line to C at w. However, cannot be the tangent line to C at w unless w is an inflexion point, because in all other cases this line has point-orbit distribution [0, 2, 0, q − 1], in contradiction with the assumption of the claim. When β w = 0, we replace Y w by the matrix obtained from the identity matrix by swapping the first and third rows. The (3, 3)-entry of the matrix representing U then becomes bα, and the result follows because the tangent line to C at w is the line α = 0. This completes the proof of the claim, and of the lemma.

Lemma 3.16
The lines in a plane of type 12 are as indicated in Table 11.
Proof Let π denote the representative of 12 given in Table 6, namely ⎡ The proof is similar to that of Lemma 3.15, and we can avoid repeating certain calculations by noting that the above matrix can be obtained from the one in (19) by setting ε = 1. The point-orbit distribution of π is [1, q−1 2 , q−1 2 , q 2 +1]. The points of rank at most 2 in π lie on the cubic C : α(γ 2 − β 2 ) − β 2 γ = 0. Hence, through the unique point x : β = γ = 0 of rank 1, there are q−1 2 lines of each of the types o 8,1 and o 8,2 , and two lines of type o 9 . The inflexion points of C are precisely the rank-2 points satisfying γ (3β 2 + γ 2 ) = 0. The point y = (0, 1, 0) is an inflexion point for all q. It is the only inflexion point unless q ≡ 1 (mod 3), in which case −3 is a square and there are two further inflexion points, namely y ± = (− 3 4 , ± √ −3, 3).
If w is a point of rank 2, with coordinates (α w , β w , γ w ), then , by (ii); (iv) w is exterior if and only if β 2 w − γ 2 w is a non-zero square in F q , by Lemma 2.2, and in particular all inflexion points of C are exterior.
If w is not an inflexion point then the tangent line to C at w contains exactly one other point of rank 2, say v with coordinates (α v , β v , γ v ). By setting ε = 1 in the corresponding argument in the proof of Lemma 3.16 and applying (iv), we deduce that (vi) v is exterior, regardless of whether w is exterior or interior. Proof of (vii). The asserted fact is analogous to fact (vii) in the proof of Lemma 3.15, except that the condition q ≡ −1 (mod 4) there has changed to q ≡ 1 (mod 4). In the case where v is not an inflexion point, this change of sign occurs because when setting ε = 1 in (18), we see that the two factors on the left-hand side are both squares if and only if −1 is a square (rather than a non-square). The case where v is an inflexion point can be checked similarly.
Suppose that q ≡ −1 (mod 4). Consider an exterior rank-2 point w e , and recall that we have already counted the line through w e and the unique point of rank 1. Suppose first that w e is an inflexion point. By (vii), w e lies on the tangent line to C at a unique interior rank-2 point z i . This tangent line has already been counted. The tangent line to w e at C (which contains no other rank-2 points) has also been counted, so there are (q + 1) − 3 = q − 2 lines through w e left to consider. Any other line through w e containing an interior rank-2 point has type o 14,2 and has already been counted. There are q−1 2 − 1 = q−3 2 interior rank-2 points other than Assuming still that q ≡ −1 (mod 4), suppose now that w e is not an inflexion point. By (vii), w e lies on the tangent lines to C at one exterior rank-2 point z e and one interior rank-2 point z i (and z e is never an inflexion point). These tangent lines have already been counted. By (vi), the tangent line to w e at C has type o 13,1 and has also been counted; denote the second exterior rank-2 point on this line by w T e . Any other line through w e containing an interior rank-2 point has type o 14,2 and has already been counted. There are q−3 2 interior rank-2 points other than z i , and half this many lines of type o 14,2 through w e , leaving (q 2 · q−7 4 or q−3 2 · q−7 4 in these respective cases. In each case, N = 1 3 (N i + N i ). To complete the proof for q ≡ −1 (mod 4), it remains to note that all lines not counted thus far have type o 17 (and that π contains q 2 + q + 1 lines in total). Now suppose that q ≡ 1 (mod 4). Consider an exterior rank-2 point w e , and recall that we have already counted the line through w e and the unique point of rank 1. Suppose first that w e is an inflexion point. By (vii), w e lies on the tangent line to C at a unique exterior rank-2 point z e . This tangent line has already been counted. The tangent line to w e at C, which contains no other rank-2 points, has also been counted, so there are (q + 1) − 3 = q − 2 lines through w e left to consider. Any line through w e containing an interior rank-2 point has type o 14,2 and has already been counted. There are q−1 2 interior rank-2 points, and hence half this many lines of type o 14,2 through w e , leaving (q − 2) − q−1 4 = 3q−7 4 lines through w e to consider. Apart from the line w e , z e , any other line through w e containing a second exterior rank-2 point, w e say, has type o 14,1 and contains also a third exterior rank-2 point.  Proof of Claim 12 . Let (α w , β w , γ w ) be the coordinates of w, and recall that γ 2 w −β 2 w = 0 and α w = β 2 w γ w /(γ 2 w −β 2 w ). We argue as in the proof of Claim 13 in Lemma 3.15. If β w = 0, we transform the matrix representative M w of w obtained from (19) by setting ε = 1 in the matrix Y w in the proof of Claim 13 . The gives the representative of U := W , shown there, with ε = 1, and with the factor f w (α, β, γ ) in the (3, 3)-entry now equal to the lefthand side of (15) with ε = 1. The result follows because the tangent line to C at w is given by (15) with ε = 1. The argument for β w = 0 is the same as in Claim 13 (with ε = 1).

Plane stabilisers
We now calculate the stabilisers in K of representatives of each of the plane orbits, and thereby determine the size of each orbit (using the orbit-stabiliser theorem and the fact that |K | = |PGL(3, q)| = q 3 (q 3 − 1)(q 2 − 1)). Although there are various ways to go about this, we take an elementary approach, calculating the stabilisers directly from the orbit representatives. The line-orbit distributions determined in the previous section (along with the point-orbit distributions) are useful in guiding these calculations. For example, for all orbits i with i ∈ {3, 4, 5, 6, 8, 9, 10, 11, 15} (and for 14 ), a plane π ∈ i contains a unique line of a certain type, so we immediately know that K π must be a subgroup of K . (As noted earlier, the line stabilisers have been determined previously, in [8].) Before proceeding, let us establish some group-theoretic notation (which has already been used in Tables 3 and 5): • E q denotes an elementary abelian group of order q.
• C k denotes a cyclic group of order k. • S n denotes the symmetric group on n letters.
• A × B is the direct product of groups A and B; and A : B is a semidirect product with normal subgroup A and subgroup B. • GO ± (2, q) denotes the similarity group of a non-degenerate quadratic form of ± type on a 2-dimensional vector space over F q , which has order 2(q − 1)(q ∓ 1). (These groups appear only in the stabilisers of planes of type 6  Proof Let π denote the representative of 1 given in Table 6. That is, suppose that π is represented by the matrix where, as usual, (α, β, γ ) ranges over F 3 q \ {(0, 0, 0)}. A matrix g = (g i j ) ∈ GL(3, q) fixes M with respect to the action M → gMg T if and only if g 31 = g 32 = 0. Upon factoring out scalars, we therefore obtain K π ∼ = E 2 q : GL(2, q).

Lemma 4.2
The stabiliser of a plane of type 2 is isomorphic to C 2 q−1 : S 3 . The number of planes of type 2 is 1 6 q 3 (q 2 + q + 1)(q + 1).
Proof Let π denote the representative of 2 given in Table 6. That is, suppose that π is represented by the matrix The matrices g ∈ GL(3, q) fixing M with respect to the action M → gMg T comprise the subgroup of GL(3, q) generated by the diagonal matrices and the permutation matrices (that is, the matrices obtained by permuting the columns of the identity matrix). Upon factoring out scalars, this gives K π ∼ = C 2 q−1 : S 3 .

Lemma 4.3
The stabiliser of a plane of type 3 is isomorphic to E q : C 2 q−1 . The number of planes of type 3 is q 2 (q 2 + q + 1)(q + 1).
Proof Let π denote the representative of 3 given in Table 6. That is, suppose that π is represented by the matrix By Lemma 3.3, π contains a unique line of type o 5 and a unique line of type o 6 , namely the lines : γ = 0 and : β = 0, respectively. Therefore, K π = K ∩ K . A matrix g = (g i j ) ∈ GL(3, q) fixes the matrix representative of with respect to the action M → gMg T if and only if g = ⎡ ⎣ g 11 g 12 g 13 · g 22 · · g 32 g 33 ⎤ ⎦ , and a matrix of this form fixes the matrix representative of if and only if g 12 = g 32 = 0. The subgroup G ≤ GL(3, q) of all such matrices has shape E q : C 3 q−1 , and factoring out scalars yields K π ∼ = E q : C 2 q−1 .

Lemma 4.4
The stabiliser of a plane of type 4 is isomorphic to (E q : C q−1 ) : C 2 . The number of planes of type 4 is 1 2 q 2 (q 3 − 1)(q + 1).
Proof Let π denote the representative of 4 given in Table 6. That is, suppose that π is represented by the matrix By Lemma 3.4, π contains a unique line of type o 5 and a unique line of type o 12 , namely the lines : γ = 0 and : α + β = 0, respectively. Therefore, K π = K ∩ K . Since contains precisely two points of rank 1, namely the points x : β = γ = 0 and y : α = γ = 0, K is equal to the stabiliser in K of the set {x, y}, and so K π is equal to the stabiliser of {x, y} and . An element of K represented by g = (g i j ) ∈ GL(3, q) fixes {x, y} if and only if it has the form ⎡ ⎣ g 11 · g 13 · g 22 g 23 · · g 33 ⎤ ⎦ or ⎡ ⎣ · g 11 g 13 where the second matrix is obtained from the first by composing with the matrix obtained by swapping the first two columns of the identity. Such an element also fixes if and only if g 11 = g 22 and g 13 = −g 23 . Factoring out scalars yields K π ∼ = (E q : C q−1 ) : C 2 .

Lemma 4.5
The stabiliser of a plane of type 5 is isomorphic to C q−1 : C 2 . The number of planes of type 5 is 1 2 q 3 (q 3 − 1)(q + 1). Proof Let π denote the representative of 5 given in Table 6. That is, suppose that π is represented by the matrix By Lemma 3.5, π contains exactly two lines and of type o 9 , so K π is equal to the stabiliser in K of the set { , }. Writing x : β = γ = 0 and y : α = γ = 0 for the rank-1 points in π, we have = x, w and = y, w where w is the rank-3 point with coordinates (α, β, γ ) = (1, 1, 1). Now, h ∈ K fixes { , } if and only if (i) it fixes {x, y}, and (ii) it fixes w. Condition (i) is equivalent to h fixing the line of type o 5 spanned by x and y, and so the same calculation as in Lemma 4.4 shows that a matrix representative g = (g i j ) ∈ GL(3, q) of h must be of the form (20). Such an element additionally satisfies condition (ii) if and only if g 11 = g 2 33 g −1 22 , g 13 = g 33 − g 11 and g 23 = g 33 − g 22 . Factoring out scalars yields K π ∼ = C q−1 : C 2 .

Lemma 4.6
The stabiliser of a plane of type 6 is isomorphic to GO − (2, q), namely, the group of similarities of a non-degenerate quadratic form of − type on a 2-dimensional vector space over F q . The number of planes of type 6 is 1 2 q 3 (q 3 − 1). Proof Let π denote the representative of 6 given in Table 6. That is, suppose that π is represented by the matrix By Lemma 3.6, π contains a unique line of type o 10 , namely the line : γ = 0. Since the unique point x : α = β = 0 of rank 1 in π does not lie on , we have K π = K ∩ K x . The matrix representatives of elements of K x are precisely the matrices g = (g i j ) ∈ GL(3, q) with g 13 = g 23 = 0. Such a matrix g also fixes the matrix representative of the line γ = 0 if and only if g 31 = g 32 = 0 and the entries g i j with i, j ≤ 2 satisfy εg 11 g 12 = g 21 g 22 and ε(g 2 11 − g 2 22 ) = g 2 21 − ε 2 g 2 12 . This is the case if and only if g = ⎡ ⎣ g 11 g 12 · ±εg 12 ±g 11 · · · g 33 ⎤ ⎦ .
The upper-left 2 ×2 sub-matrices of g comprise an orthogonal group GO − (2, q), specifically the group of similarities of the quadratic form associated with the matrix which spans the complement of the subspace of the symmetric 2 × 2 matrices defined by the upper-left 2 × 2 block of M. Hence, upon factoring out scalars, we obtain K π ∼ = GO − (2, q).

Lemma 4.7
The stabiliser of a plane of type 7 is isomorphic to E 2 q : GL(2, q). The number of planes of type 7 is q 2 + q + 1.
Proof As noted in the proof of [9, Lemma 7.3], a plane π ∈ 7 is the tangent plane to V(F q ) at its unique point of x rank 1. Hence, the stabiliser of π in K is equal to the stabiliser of x in K , which is has shape E 2 q : GL(2, q). Explicitly, if π is the representative of 7 from Table 6 then x is represented by a matrix whose only non-zero entry is the (1, 1)-entry, so the elements of K π = K x correspond to matrices g = (g i j ) ∈ GL(3, q) with g 21 = g 31 = 0.

Lemma 4.8
The stabiliser of a plane of type 8 is isomorphic to E q : C 2 q−1 . The number of planes of type 8 is q 2 (q 2 + q + 1)(q + 1).
Proof Let π denote the representative of 8 given in Table 6. That is, suppose that π is represented by the matrix By Lemma 3.6, π contains a unique line of type o 12 , namely the line : α = 0. Since the unique point x : β = γ = 0 of rank 1 in π does not lie on , we have K π = K ∩ K x . As noted in [8,Sect. 4], the matrix representatives of elements of K are precisely the matrices g = (g i j ) ∈ GL(3, q) with g 12 = g 21 = g 23 = g 32 = 0. As noted in the proof of Lemma 4.7, the matrix representatives of elements of K x are precisely the matrices g = (g i j ) ∈ GL(3, q) with g 21 = g 31 = 0. Hence, the matrix representatives of K π = K ∩ K x have the form ⎡ ⎣ g 11 · g 13 · g 22 · · · g 33 ⎤ ⎦ .
Factoring out scalars yields K π ∼ = E q : C 2 q−1 . (Indeed, note that K π is equal to the stabiliser in K of the plane of type 3 considered in the proof of Lemma 4.3.)

Lemma 4.9
The stabiliser of a plane of type 9 is isomorphic to (E q : C q−1 ) × C 2 . The number of planes of type 9 is 1 2 q 2 (q 3 − 1)(q + 1).
Proof Let π denote the representative of 9 given in Table 6. That is, suppose that π is represented by the matrix As noted in the proof of Lemma 3.9, the points of rank at most 2 in π comprise the conic C : β 2 − αγ = 0, which intersects the unique line : γ = 0 of type o 6 in π in the unique point x : β = γ = 0 of rank 1 in π. Hence, we have K π = K ∩ K C , where C = C \ {x} (and K C is its setwise stabiliser). The matrix representatives of elements of K are precisely the matrices g = (g i j ) ∈ GL(3, q) with g 21 = g 31 = g 32 = 0. A matrix g of this form also fixes the matrix representative of C (obtained by setting α = β 2 γ in M) if and only if g 12 = g 32 = 0 and g 2 33 = g 2 22 . That is, the matrix representatives of K π are precisely the matrices of the form ⎡ ⎣ g 11 g 12 · · g 22 · · · ±g 22 ⎤ ⎦ .

Lemma 4.10
The stabiliser of a plane of type 10 is isomorphic to (E q : C q−1 ) × C 2 . The number of planes of type 10 is 1 2 q 2 (q 3 − 1)(q + 1). Proof The proof is essentially identical to that of Lemma 4.9. To use the representative π ∈ 10 from Table 6, we just replace γ in (21) by −εγ with ε ∈ F q a non-square. By Lemma 3.10 (and its proof), the points of rank at most 2 in π again comprise the conic C : β 2 − αγ = 0, which again intersects the unique line of type o 6 in the unique point of rank 1. The subsequent calculations from the proof of Lemma 4.9 go through verbatim.

Lemma 4.11
The stabiliser of a plane of type 11 is isomorphic to C q−1 if q ≡ 0 (mod 3) and to E q if q ≡ 0 (mod 3). The number of planes of type 11 is q 3 (q 3 − 1)(q + 1) and q 2 (q 3 − 1)(q 2 − 1) in these respective cases.
Proof Let π denote the representative of 11 used in the proof of Lemma 3.12 (as opposed to the one given in Table 6). That is, suppose that π is represented by the matrix By Lemma 3.12, π contains a unique line of type o 9 , specifically the line : β = γ . Hence, we have K π ≤ K . The matrix representatives of elements of K are precisely the matrices g = (g i j ) ∈ GL(3, q) of the form ⎡ ⎣ g 11 g 12 g 13 · g 2 33 g −1

11
· · −g 13 g 33 g −1 We must now consider separately the cases where q ≡ 0 (mod 3) or not. Recall from the proof of Lemma 3.12 that the points of rank at most 2 in π comprise the cubic C : First suppose that q ≡ 0 (mod 3). Then C contains a unique point of inflexion, namely the point z with coordinates (α, β, γ ) = ( 2 9 , 1, −2), and π contains a unique line of type o 16 , namely the tangent to C through z, which is the line : γ = 27α + 8β. Hence, we have K π = K ∩ K . An element of K , which is necessarily represented by a matrix g ∈ GL(3, q) as in (22), fixes if and only if it fixes z, which is the case precisely when g 13 = 3 −1 (g 11 − g 33 ) and g 12 = 1 9 g −1 11 (g 33 − g 33 g 11 − 2g 2 11 ).
The matrices of the form (22) with these additional constraints comprise a subgroup of GL(3, q) of shape C 2 q−1 , and by factoring out scalars we therefore obtain K π ∼ = C q−1 . Now suppose that q ≡ 0 (mod 3). Then the cubic C meets the line in the unique point x : β = γ = 0 of rank 1 in π, so we have K π = K ∩ K C , where C = C \ {x}. A matrix g ∈ GL(3, q) of the form (22) fixes the matrix representative of C if and only if g 11 = g 33 and g 12 = g −1 33 g 13 (g 13 − g 33 ). Upon factoring out scalars we obtain K π ∼ = E q .
Proof Let π denote the plane represented by the matrix so that π is the representative of 12 , respectively 13 , given in Table 6 if ε is equal to 1, respectively a non-square. Recall from the proofs of Lemmas 3.15 and 3.16 that the points of rank at most 2 in π lie on the cubic C : α(εγ 2 − β 2 ) − εβ 2 γ = 0. There is a unique point of rank 1, namely the point x : β = γ = 0. The points of inflexion of C are given by γ (3β 2 + εγ 2 ) = 0; in particular, the point y : α = γ = 0 is an inflexion point for all ε. Write −3ε = δ 2 , so that δ lies in either F q or F q 2 , and consider the plane π obtained by extending the scalars of M to F q (δ). Let K ∼ = PGL(3, F q (δ)) denote the group obtained by extending the scalars of K to F q (δ). The cubic of points of rank at most 2 in π contains three points of inflexion over F q (δ), which lie on the line : 4α + εγ = 0. The inflexion points are all exterior rank-2 points (see assertion (iv) in the proofs of Lemmas 3.15 and 3.16), so is of type o 14,1 . We have K π ≤ K ∩ K x , with equality when δ ∈ F q . If δ / ∈ F q , we have K π ≤ K ∩ K x ∩ K y because y is the unique inflexion point of C. (In this case, ∩ π is a line of type o 15,1 .) For convenience, let us now apply the transformation M → X M X T , where Under this transformation, is mapped to the line represented by the matrix ⎡ ⎣ a · · · −(a + b) · · · b ⎤ ⎦ , where (a, b) ranges over all non-zero values in F q (δ) 2 , and y, x are mapped to the points y , x represented by the matrices diag(1, 0, −1) and v ⊗ v with v = (1, −1, 1), respectively. We denote the image of π under this transformation by π . By [8, Sect. 4] (cf. Table 5), K ≤ PGL(3, F q (δ)) is a group of shape C 2 2 : S 3 , generated by elements corresponding to (i) diagonal matrices with all non-zero entries equal to ±1, which fix pointwise the rank-2 points given by (a, b) = (1, −1), (1, 0) and (0, 1); and (ii) permutation matrices, which permute these points. The intersection K ∩ K x is isomorphic to S 3 , generated by elements corresponding to the matrices A = ⎡ ⎣ · · 1 · 1 · 1 · · ⎤ ⎦ and B = ⎡ ⎣ · · 1 −1 · · · −1 · ⎤ ⎦ , and K ∩ K x ∩ K y is isomorphic to C 2 , generated by the element corresponding to A.
We can now explicitly calculate the elements of K π by noting that K π = X −1 K π X . If δ ∈ F q then K π is the image of X −1 A, B X in PGL(3, q), so K π ∼ = S 3 . If δ / ∈ F q then K π is contained in the image of X −1 A X in PGL (3, q), which is isomorphic to C 2 . Since the element of PGL(3, q) corresponding to X −1 AX = diag(1, −1, 1) fixes π, we have K π ∼ = C 2 . The result now follows by taking ε = 1 for 12 , in which case δ ∈ F q if and only if q ≡ 1 (mod 3), and ε a non-square in F q for 13 , in which case δ ∈ F q if and only if q ≡ −1 (mod 3).

Lemma 4.13
Suppose that q ≡ 0 (mod 3). The stabiliser of a plane of type 14 is isomorphic to C 3 . The number of planes of type 14 is 1 3 q 3 (q 3 − 1)(q 2 − 1). Proof Let π denote the representative of 14 given in Table 6 where c satisfies the condition ( †) given in Table 6. That is, c is some fixed element of F q \ {0, 1} such that (i) −3c is a square in F q and (ii) f (c) := √ c+1 √ c−1 is a non-cube in F q ( √ −3). We first claim that c = −1. To verify this, begin by noting that a = √ −1 is a solution of the equation f (−1) = a 3 in every field in which −1 is a square. Now suppose for a contradiction that c = −1. Then (i) says that 3 is a square in F q . If q ≡ 1 (mod 3) then −3 is a square in F q , so −1 must also be a square F q , and so (ii) does not hold because f (−1) has the cube root √ −1 in F q ( √ −3) = F q . If q ≡ −1 (mod 3) then −3 is a non-square in F q , so −1 must be a non-square in F q , but in this case −1 is a square in F q ( √ −3), so again (ii) does not hold.
Consider an element h of K , represented by a matrix g = (g i j ) ∈ GL(3, q). If h fixes π then, in particular, it must fix x, and so g must have the form g = ⎡ ⎣ g 11 g 12 g 13 · g 22 g 23 · g 32 g 33 ⎤ ⎦ .
Now consider the rank-2 points w and w with (α, β, γ ) = (0, 1, 0) and (c −1 , 1, 1), respectively, and let M w and M w denote the corresponding matrix representatives obtained from (23). Note that π is spanned by x, w and w . Hence, given that g has the form (24), h now fixes π if and only if it maps both w and w to points in π, that is, if and only if gM w g T and gM w g T have the form (23) for some α, β and γ (and are of rank 2). We first claim that g 33 = 0. For a contradiction, suppose that g 33 = 0, and note that the supposedly non-zero determinant of g is then −g 11 g 23 g 32 . Consider the image of w under Notice that we cannot have g 23 + g 33 = 0, because then the above equation would imply that c = 1, contradicting ( †). Therefore, g 11 = −c −1 (g 23 + g 33 ) −1 (cg 2 33 + g 2 23 + 2cg 23 g 33 ). Considering now the same condition in gM w g T , we find that g 23 and g 33 must satisfy (c + 1)g 3 23 + 6cg 2 23 g 33 + 3c(c + 1)g 23 g 2 33 + 2c 2 g 3 33 = 0.
Notice that neither g 23 nor g 33 can be zero, because if either one is zero then (25) implies that both are zero because c = {0, −1}, and this contradicts det(g) = 0. Hence, we may view the left-hand side of (25) as a cubic in θ := g 23 g −1 33 . Upon changing the parameter from θ to θ − 2c c+1 and dividing by c + 1, this cubic becomes The discriminant of the above cubic is −108c 3 (c − 1) 4 = 4(−3c) 3 (c − 1) 4 , which is a square in F q because −3c is a square, by condition ( †). By [4,Theorem 3], the cubic therefore has a solution in F q if and only if 1 2 . By condition ( †), this is not the case. Therefore, there are no non-trivial elements of K π represented by matrices of the form g = g − , and so K π ∼ = C 3 as claimed.
Lemma 4.14 Suppose that q ≡ 0 (mod 3). The stabiliser of a plane of type 14 is isomorphic to E q : C q−1 . The number of planes of type 14 is therefore q 2 (q 3 − 1)(q + 1).
Proof Let π denote the representative of 14 used in the proof of Lemma 3.13 (as opposed to the one given in Table 6). That is, suppose that π is represented by the matrix By Lemma 3.12, π contains a unique line of type o 9 , specifically the line : β = γ . Hence, we have K π ≤ K . The matrix representatives of elements of K are precisely the matrices g = (g i j ) ∈ GL(3, q) of the form ⎡ ⎣ g 11 · g 21 g 33 g −1

Lemma 4.15
The stabiliser of a plane of type 15 is isomorphic to E 1+2 q : C q−1 , where E 1+2 q has centre Z ∼ = E q and E 1+2 q /Z ∼ = E 2 q . The number of planes of type 15 is (q 3 − 1)(q + 1).
Proof Let π denote the representative of 15 given in Table 6. That is, suppose that π is represented by the matrix As noted in the proof of Lemma 3.11, the points of rank at most 2 in π are all on the line : γ = 0. This happens to be the same line of type o 6 as in the proof of Lemma 4.9, so we know that the matrix representatives of elements of K are the matrices g = (g i j ) ∈ GL(3, q) with g 21 = g 31 = g 32 = 0. An element of K fixes π if and only if it fixes the set of rank-3 points in π (i.e. those with γ = 0 in M). This imposes the additional constraint g 11 g 33 = g 2 22 , so that the matrix representatives of elements of K π are precisely the invertible matrices of the form ⎡ ⎣ g 2 22 g −1 33 g 12 g 13 · g 22 g 23 · · g 33 ⎤ ⎦ .
Note that setting g 22 = g 33 = 1 yields a subgroup of shape E 1+2 q , with centre defined by the constraint g 12 = g 23 = 0. Hence, upon factoring out scalars we obtain K π ∼ = E 1+2 q : C q−1 .
Acknowledgements The authors thank the referees for reading the paper carefully and making several helpful suggestions. The first author acknowledges the support of The Scientific and Technological Research Council of Turkey TÜBİTAK (Project No. 118F159).
Funding Open Access funding provided by the IReL Consortium.
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