finite projective plane

Finite projective planes from axioms

Let us start from the beginning and assume we are given a set of n things called points and a set of ν things called lines, with an incidence relationPlanetmathPlanetmath between them (a point is incidentMathworldPlanetmath with a line iff the line is incident with the point) that satisfies only two axioms, the two unique incidence properties:

  • For any two distinct points, there is exactly one line through both of them.

  • For any two distinct lines, there is exactly one point on both of them.

Given these planeMathworldPlanetmath axioms we can apply the De Bruijn–Erdős theorem twice (once with the rôle of points and lines swapped) which tells us we must have n=ν, and one of two possible configurationsMathworldPlanetmathPlanetmath:

  • One “degenerate” solution has one special line 0 and n-1 ordinary lines i, and a special point P0 and n-1 ordinary points Pi. Each Pi (i0) is incident with its own i and with 0; each i (i0) with its own Pi and with P0. Now 0 is incident with every point except 0 while P0 is incident with every line except 0.

  • The other solution is a projective plane:

    • the number of points and the number of lines are of the form q2+q+1 for some integer q;

    • each point is incident with q+1 lines and each line with q+1 points.

One usually avoids the degenerate case by demanding there is a quadrangle in the plane: four points no three of which lie on the same line. This, or a similarMathworldPlanetmathPlanetmathPlanetmath non-triviality clause, can then be adopted as a third axiom. Unfortunately it also has the effect of excluding projective planes of orders q=0 and 1. The De Bruijn–Erdös ( theorem actually makes an additional assumptionPlanetmathPlanetmath: that no two lines are incident with the same set of points (and when applied dually that no two points are incident with the same set of lines). That is not a problem for a line incident with more than one point or vice versa, because then the axioms forbid it. We just get some more degenerate cases consisting of multipleMathworldPlanetmathPlanetmath lines intersecting one point or none, or vice versa; the insistence on a quadrangle catches these too.

I will not prove the whole theorem here, but under the assumption that a quadrangle exists it is not hard show the interesting part, that those numbers q+1 and q2+q+1 drop out by necessity. The same construction can also be used to give coordinatesMathworldPlanetmathPlanetmath to the points and lines of the plane (see below).

Projective planes as bipartite graphs

Projective planes are, apart from anything else, also bipartite graphsMathworldPlanetmath: let the q2+q+1 points be represented by “black” nodes (vertices) of the graph, and let the q2+q+1 lines also be nodes, “white” ones say. Edges will represent incidence between a “black” and a “white” node, so there are (q2+q+1)(q+1)=q3+2q2+2q+1 edges in all.

  • In a bipartite graph, nodes of the same color have even distanceMathworldPlanetmath. This is 0 for any node and itself and must be 2 for any two distinct nodes of the same color: for any two points there is a path via the line they share, and for any two lines a path via the point they share. So distance 4 does not occur.

    For nodes of opposite color we can have 1 (if the point lies on the line) or 3 (otherwise); we can’t have distance 5 which would imply 4 for some nodes. So the graph has diameterMathworldPlanetmathPlanetmath (maximum distance) 3.

  • In a bipartite graph, all cycles (if any) have even length. We don’t have 2-cycles (this is not a multigraphMathworldPlanetmath) but no 4-cycles either, by the following argument: such a cycle PQm would mean two lines intersecting in two different points, and vice versa. So the girth (minimum cycle length) is 6 (showing there indeed are 6-cycles for q2 is left as exercise for the reader).

These two conditions, diameter 3 and girth 6, are not only necessary for the graph to represent a projective plane, they are also sufficient, and therefore form an alternative formulation of what it means to be a projective plane.

These graphs could be called the bipartite analogue of Moore graphsMathworldPlanetmath of diameter 2 and girth 5; many of the same arguments can be used [vG03].

Coördinates for finite projective planes

The details on how to impose coordinates on a projective plane can be found in [Hal59, dRe96], or in this entry ( We summarize it for a finite projective plane below:

Setup: Use a set Q of q different symbols for coordinate values, two of which are 0 and 1. Start with four points O, I, X, Y forming a quadrangle. XY is the “line at infinityMathworldPlanetmath.

Construction: O and I are given coordinates (0,0) and (1,1), and the other points on OI (except its intersectionMathworldPlanetmathPlanetmath with ) arbitrarily (c,c) with unique cQ. For any point P not on , let YP intersect OI at (a,a), and let XP intersect OI at (b,b). Then P will be given (a,b). For the coordinates of points on , intersect the lines joining (0,0) with (1,m), for all mQ, with , calling the intersection (m). Finally, Y gets coordinate () (instead of () in the literature).

Lines not through Y are given coordinates [m,k] where (m) and (0,k) are their intersections with and with OY respectively; lines through Y (other than ) get coordinates [k] going by their intersection (k,0) with OX. The line gets [] (instead of [] in the literature).

Coordinates from bipartite graphs

The whole business with coordinates becomes a bit more transparent when we try to actually construct the thing as a bipartite graph, starting from one edge, Y say.

There are q more points (other than Y) incident to , let’s call them Ym where m ranges over the q different values of the set Q. Again, each Ym is incident to q more lines (apart from ), let’s call those mb where b is another subscript from Q. Likewise there are q more lines (other than ) incident to Y, let’s call them x and to each of them are incident q more points (apart from Y) which we can label Yxy.

This immediately produces the incidence list of the preceding sectionMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath. Pictorially, for q=2 (with the “to be assessed” portion filled in as thinner lines):


We already have 0 containing all points Y0y, and in additionPlanetmathPlanetmath if we wanted we could easily arrange for 00 to be incident with all the Yx0 (just renumber the labels y in the sets Yx0 for every x), and for 10 to be incident with all the Ycc (just look at which Yxy are incident with 10, then exercise the freedom we still have to relabel all the x, matching each time the corresponding y there). This reproduces the entire coordinatisation.

Projective planes as permutation systems

Five sides of the hexagonMathworldPlanetmath are self-evident. On the left we have the single edge Y, along the top we fan out in two stages to q and q2 items, along the bottom we also fan out in two stages to q and q2 items, and the hard part is the right. We have by now accounted for 2q2+2q+1 edges so we know there must be exactly q3 more edges, between some of those q2 points and some of those q2 lines. But where?

For q=2 it is simple enough to avoid 4-cycle clashes (and the diagram shows what is, up to isomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath, the only solution). In general, it is an open problem to find all possible solutions.

From the projective plane properties it is easy to see that any point Yxy must, for every value of m, be incident to no more than one of the mb (because if it was incident to two they would intersect both in our Yxy and in their shared Ym at the bottom). On the other hand, Yxy needs q more mates so it needs to choose exactly one mb for every value of m.

By exactly the same reasoning, any mb needs to be incident with, for every value of x, at most one and then exactly one Yxy. In other words, the projective plane structureMathworldPlanetmath takes the form of, for each pair of values x and m, a bijectionMathworldPlanetmath ψxm between the y indices and the b indices.

This is not yet the only constraint. A closer analysis [vG03] shows that, in order to prevent any two lines meeting in more than one point or any two points in more than two lines, all compound bijections of the form


(for mm and xx) must be derangementsMathworldPlanetmath, i.e. permutationsMathworldPlanetmath that do not leave any index in place. It is not hard to see this is just a device to prevent 4-cycles. And careful scrutiny of the graph shows that, after arranging 5 of 6 edges along trees fanning out as above, and after recognising there are those bijections along the sixth edge, these derangement shenanigans stop the only remaining gap for 4-cycles to lurk.

This could be called the permutation groupMathworldPlanetmath interpretationMathworldPlanetmath of projective plane structure. The algebraic version appears in the next section.

Planar ternary rings

We’ve been giving our points and lines cooordinates from a set Q of q different labels. We can give Q the structure of a ternary ring, that is, a set with a ternary operation pqr, by defining

y=xmb(x,y)is on[m,b] (1)

Thus, geometric properties of a projective plane may not be analyzed algebraically. See this entry ( for more detail. Conversely, every ternary ring determines a projective plane with coordinates as above, so that algebraic properties of the ring can be interpreted geometrically. For more detail on this, see this entry (

Fields and division rings are examples of ternary rings. But there are other classes of algebraic structuresPlanetmathPlanetmath intermediate between fields and ternary rings. See [dRe96] for a survey. There are four different projective planes known of order 9, many more of order 16, and so on.

Finding all PTRs in general remains of course just as hard as finding projective planes!

Finite projective planes as incidence structures

Finite projective planes can be defined as (simple square) 2-(q2+q+1,q+1, 1) designs, or what’s the same thing, Steiner systemsMathworldPlanetmath S(2,q+1,q2+q+1). This way, we already assume in the definition that each “block” (line) contains exactly q+1 points, from which it follows by calculation that the number of lines that contain a given point is also q+1, and that the total number of lines equals the total number of points.

Also, the unique incidence property that every two points lie together on exactly one line is given, and the dual one that every two lines meet in exactly one point then follows from the numbers and the properties that define a design or Steiner system.

We took a different route to projective planes above: we saw we only need to start with these two unique incidence properties, and can prove from them that the number of points on a line and the number of lines through a point are both fixed (i.e. that the projective plane and its dual are both designs).

Finite projective planes as incidence matrices

Consider an arbitrary projective plane of order q. We know there are n=q2+q+1 points and the same number of lines. Let A be an n×n matrix of which the rows and columns represent lines and points of the plane, and give the value 1 to entries at intersections where the point (represented by the column) lies on the line (represented by the row), and leave the other entries zeros. Such an A is known as the incidence matrixMathworldPlanetmath of the plane.

AA has rows and columns both representing lines, and the entries here are the dot productsMathworldPlanetmath of any two rows of A, i.e. the number of points any two lines l and m have in common, which is q+1 when l=m and 1 otherwise, by the projective plane properties.


where I is the identity matrixMathworldPlanetmath (ones along the diagonalMathworldPlanetmath, zeros everywhere else) and J is the matrix filled with ones in every entry. Note AA has this same value for every projective plane with n points and n lines, while A on its own of course varies if there is more than one plane of the same order, as the plane is fully determined by its incidence matrix. Also, AA having this value is the only thing needed to make the plane specified by A satisfy the projective plane properties; it is a fully equivalentMathworldPlanetmathPlanetmathPlanetmath formulation of those properties.

The determinantMathworldPlanetmath of A is best evaluated via that of AA, because (detA)2=(detA)(detA)=detAA, and we know every entry of AA. The eigenvaluesMathworldPlanetmathPlanetmathPlanetmathPlanetmath of any n×n matrix of the form J+qI are easily seen to be

by letting the matrix act on these eigenvectors. This gives


which in our case (n=q2+q+1) amounts to


As q2+q is even and the other factor a square, we get an integer


where the sign is moot, as A has no correspondence between its columns and rows.


The Bruck–Ryser theorem [BR49] states that

If q is the order of a projective plane, and q1 or 2(mod  4), then q is the sum of two integer squares (one of which can be 02).

It can be proven [BR49, Cam94] using the incidence matrix A, by a clever rearrangement of 𝐱AA𝐱 (plus one extra term) for an arbitrary vector 𝐱, on which gradually constraints are imposed four componentsMathworldPlanetmath at a time using the fact that each integer is a sum of four squares. The argument only works when q2+q+1=n-1(mod  4), hence the constraint on q (mod 4).

Interestingly, there is a partial converse [HR54]: whenever q is allowed by the theorem (either congruentMathworldPlanetmathPlanetmath 0 or 3 (mod 4), or congruent 1 or 2 (mod 4) and a sum of two squares), there always is a rational n×n matrix A (where n=q2+q+1) such that AA=M. Of course, the matrix is only a proper incidence matrix if that rational matrix only has entries 0 and 1.

The theorem doesn’t rule out any potential projective plane orders q0 or 3(mod  4), but does rule out a large number of q1 or 2(mod  4), starting with n=6,14,21,22. Note in passing that it is easy to prove no prime power falls foul of Bruck–Ryser, which is just as well, as (classical) planes exist for all those orders.

The only other order ruled out to date is 10, via an epic computer search by Lam, Swiercz and Thiel (read for Lam’s account).

All projective planes actually found to date, even the non-classical ones, have an order q that is the power of a prime. The prime power conjecture says this is the case for all projective planes. It is still open.


  • 1
  • BR49 R. H. Bruck & H. J. Ryser, “The non-existence of certain finite projective planes” Can. J. Math. 12 (1949) pp 189–203
  • HR54 M. Hall, Jr. & H. J. Ryser, “Normal completions of incidence matrices” Amer. J. Math. 76 (1954) pp 581–9
  • Hal59 Marshall Hall, Jr., The Theory of Groups, Macmillan 1959
  • Cam94 Peter J. Cameron, Combinatorics: topics, techniques, algorithms,
    Camb. Univ. Pr. 1994, ISBN  0 521 45761 0 pjc/comb/ pjc/comb/
    (solutions, errata &c.)
  • CD96 Charles J. Colbourn and Jeffrey H. Dinitz, eds.
    The CRC Handbook of Combinatorial Designs,
    CRC Press 1996, ISBN  0 8493 8948 8 dinitz/hcd.html dinitz/hcd.html
    (errata, new results)
    the reference work on designs incl. Steiner systems, proj. planes
  • dRe96 Marialuisa J. de Resmini, “Projective Planes, Nondesarguesian”, pp 708–718 in the previous reference
  • vG03 Marijke van Gans, M.Phil.(Qual.) thesis, Birmingham 2003
Title finite projective plane
Canonical name FiniteProjectivePlane
Date of creation 2013-03-22 15:11:15
Last modified on 2013-03-22 15:11:15
Owner Mathprof (13753)
Last modified by Mathprof (13753)
Numerical id 11
Author Mathprof (13753)
Entry type Topic
Classification msc 51E15
Classification msc 51A35
Classification msc 05B25
Related topic ProjectivePlane2
Related topic IncidenceStructures
Related topic GeometryMathworldPlanetmath
Related topic DeBruijnErdHosTheorem
Defines prime power conjecture
Defines planar ternary ring
Defines PTR