Γ
S H Z
Y 1
A T
G U
Γ ΓLn(F) = ⟨GLn(F), Aut(F)⟩
G GLn(F) A Aut(F) S SLn(F), det = 1 Z Z(GLn(F)) Y Z(SLn(F)) T ⟨SLn(F), Aut(F)⟩
H ⟨SLn(F), Z(GLn(F))⟩
U ⟨H, Aut(F)⟩
Figure VII.1: Structure of ΓLn(F)
is cyclic of order q − 1, it contains exactly gcd(q − 1, n) (the largest subgroup whose order divides n) elements of order dividing n, thus ∣Y∣ = gcd q − 1, n.
Since S, Z ⊲ G, we have that [Z ∶ Y] = [H ∶ S], this implies that [G ∶ H] = q − 1
[H ∶ S] = q − 1
[Z ∶ Y] = q − 1
(q − 1)/ ∣Y∣= ∣Y∣ = gcd(q − 1, n).
Finally, as a consequence of the semidirect product structure and Theorem VII.1 we get that ∣A∣ = [Γ ∶ G] = logp(∣F∣).
(The subgroups U, T are only listed for completion.)
VII.3 Projective Spaces
Euclidean geometry starts with a set of axioms. We thus now change our point of view and introduce an axiomatic description that characterizes projective geometry over finite fields. (We shall not prove here that it is a characterization.)
Theorem VII.7 (Veblen-Young): Consider a set of points and a set of lines (as sets of points), satisfying the following conditions:
PG1 For any two distinct points, there is one, and only one, line containing both points
PG2 If A, B, C are three points not on a line, and if D /= A is a point on the line through A, B, and of E /= A is a point on the line through A, C, then there is a point F on a line with D and E and also on a line with B and C.
PG3 Every line contains at least three points
Then this configuration can be interpreted as PG(n, q) for some n ≥ 3 and a prime power q.
A C
D B
E F
Figure VII.2: The Pasch axiom
Axiom PG1 makes it possible to talk about a “line through two points”. It in particular implies that two lines can intersect in but one point (as otherwise there would be two different lines going through both intersection points). Axiom PG3 eliminates some degenerate cases. Axiom PG2, the Pasch axiom requires some ex-planation.
Consider figure VII.2. The lines AB and AC intersect in a point, namely A. In Euclidean geometry we would interpret this as the two lines lying in a plane. The configuration (and the point A) thus simply ensures that the lines CB and DE lie in a common plane. We can thus interpret this axiom instead as: Two lines in the same plane have a point of intersection.
We thus do not have a parallel axiom, but always have points (maybe “at infin-ity”) where lines meet.
As we have drawn a picture, a word on what these are is in order: Pictures of configurations only indicate incidence. There are no distances, no angles, no clos-est points. There are no “intermediate” points. A line does not need to be drawn straight, but neither can we deduce incidence from the intersection of lines. That is, pictures will only ever illustrate proofs, never be proofs in themselves.
Projective Planes
It’s a bird. It’s a plane. It’s Superman!
The Adventures of Superman (1952-1958) Motivated by the structure of the Veblen-Young theorem, we now aim to de-scribe projective plane, that is 2-dimensional projective spaces. In this case the ax-ioms simplify. PG2 becomes PP2 below, which is nicely dual to the first axiom.
PP1 For any two distinct points, there is one, and only one, line containing both points
PP2 For any two distinct lines, there is one, and only one, point on both lines.
PP3 There exist four points, no three of which are on a line.
VII.3. PROJECTIVE SPACES 125
A
B C
Y X
Z D
2
1
3
4
5 7 6
Figure VII.3: The Fano Plane
We observe that PG(2, q) indeed satsfies these axioms. (However it turn out that there are other projective planes.)
Example: The axioms alone let us construct the minimal configuration PG(2, 2):
Let A, B, C, D be four points, no three on a line, that are given by PP3.
Let L1= AB, L2= AC, L3= AD, L4= BC, L5= BD, L6 = CD. Because of the choice of the points, these lines must be distinct. By PP2, L1and L6must intersect in a point that cannot be any of the existing four, we call this point X. Similarly we get L2∩ L5= {Y} and L3∩ L4= {Z}.
Finally, there must be a line L7= XY, we may assume that Z ∈ L7. We thus get the configuration in figure VII.3, the Fano plane PG(2, 2).
We notice, exercise ??, that in the axioms we could replace PP3 by its dual:
PP3’ There exist four lines, no three of which go through the same point.
This means that for a projective plane we can switch the labels of points and planes and again obtain a projective plane, (the dual).
Finite projective planes satisfy nice numerical properties:
Theorem VII.8: Let n ≥ 2 an integer. In a projective plane π any one of the follow-ing properties imply the other five:
1. One line contains exactly n + 1 points.
2. One point is on exactly n + 1 lines.
3. Every line contains exactly n + 1 points.
4. Every point is on exactly n + 1 lines.
5. There are exactly n2+ n + 1 points in π.
6. There are exactly n2+ n + 1 lines in π.
Proof:
We assume property 1. Let L = {Q1, . . . Qn+1} be a line with n + 1 points and P /∈ L a further point. The lines PQ1, . . . , PQn+1 then need to be different, for if PQa = PQbthen P ∈ QaQb = L. By PP2 any line through P must intersect L and
thus the n + 1 lines PQiare exactly the lines through P. That is, any point P not on L will lie on (exactly) n + 1 lines. By PP3 such a point P must exists, showing that property 1 implies property 2.
Assuming property 2, let P be a point on exactly n + 1 lines K1, . . . , Kn+1. Then P is the only point common to any pair of these lines. If L any line with P /∈ L, then the Kiwill intersect it in n + 1 points Qi, which must be distinct since P is the only point on any pair of the Ki. If there was another point Q′∈ L, then PQ′would be another line through P, contradiction. That is, any line not containing P must have n + 1 points. By PP3 such lines exist, thus property 2 implies property 1.
Now assume 1 and 2 and let P and L as above. We have seen that any line not through P must have n + 1 points, and any point not on L lies on n + 1 lines. Let K be a line through P and {Q} = K ∩ L. By PP3 there must be two further points in addition to P and Q, neither on K, of which at most one lies on L. Thus there is a point R /∈ L, K. With the previous arguments we have that R lies on n + 1 lines and thus K has n + 1 points. Any point S /∈ K thus lies on n + 1 lines. To show that Q also lies on n + 1 lines, we chose another point Q′∈ L and K′= PQ′and repeat the argument. This shows 3 and 4.
Vice versa, obviously 3 implies 1 and 4 implies 2.
If 3 and 4 hold, pick a point P and sum up the number of (each n + 1) points on all n + 1 lines through P, yielding (n + 1)2= n2+ 2n + 1 Any point except P is on exactly one of these lines, but P is on all lines. This means we are overcounting the number of points by n + 1 − 1 = n. Thus in total there are n2+ n + 1 points, yielding 5. A dual argument gives 6.
We notice that f (x) = x2+x +1 is one-to-one for x > 0, as then f′(x) = 2x +1 >
0. This means, with the implications we know already, that 5 or 6 must imply 1 and 2 (otherwise a different count of points on one line or lines through one point would imply a different total number of points or lines). ◻
For the planes PG(2, q) we have of course that every line has [21]
q= q+1 points, motivating the following definition:
Definition VII.9: A finite projective plane is of order n is a line contains n + 1 points.
It turns out that the axioms for a projective plane alone do not characterize PG(2, q), but that this can be done using two theorems from classical geometry:
Desargues’ Theorem Let A, B, C and D, E, F are triangles, such that the lines AD, BE, and CF intersect in a common point O. Let P = AB ∩ DE, Q = BC ∩ EF, and R = AC ∩ DF. Then P,Q and R are collinear. (Figure VII.4, left.) Pappus Theorem Let A,B,C be collinear and D,E,F be collinear and that the
tri-angles ACE and BDF overlap. Then the intersection points P = AE ∩ BD, Q = AF ∩ CD, and R = BF ∩ CE are collinear. (Figure VII.4, right.)