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16 (a). The famous Fano Plane, the unique projective plane of order 2, is also a
2 - (7, 3, 1) design. Note that 2 - (7, 3, 1) can be writen in the form 2 - (n2 + n + 1, n + 1, 1) where n = 2, the order of the plane. This is true in general, as you have shown that any projective plane must have an order n, and that for a projective plane of order n any line
- contains exactly n + 1 points,
- that any point is incident with exactly n + 1lines,
- that the projective plane is a linear space and so any two points are incident with a unique line and every line contains at least two points,
- and that the number of points in the plane must be n2 + n + 1.
For example, the projective plane of order n = 3 is shown below (the more symmetric version first where the antipodal points on the "outer circle" are identified, then the more typical non-symmetric version). We can see that this is a 2 - (13, 4, 1) design.
And here we see the projective plane of order n = 4, which is also a 2 - (21, 5, 1) design.
16 (b). The affine planes of order n are precisely the 2 - (n2, n , 1) designs.
16 (c). The unique 2 - (9, 3, 1) design is the uniqe affine plane of order 3, shown below. You can see that it does not satisfy the second axiom for a hyperbolic plane.
16 (d). We need to verify each of the three axioms for a hyperbolic plane given any 2 - (n3 + 1, n + 1, 1) design, where n > 2.
1. Two distinct points are contained in a unique line.
This follows from the fact that t = 2 and k = 1 in 2 - (n3 + 1, n + 1, 1).
2. Associated to any non-incident point-line pair (p, l) is a set consisting of two or more lines, each line of which is incident with p but not with l.
Recall that  denotes the number of points that are incident with the line lj, that  denotes the number of lines that are incident with the point pi, and that rij is 1 if point pi is incident with line lj, and 0 otherwise. Recall the following lemma, which applies since a t - (v, j, k) design is necessarily a linear space.
Lemma. In a near-linear space, for any fixed point pi,
 .
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Now for any fixed point pi,
 .
and since rij must either be 0 or 1, we can see that there must be precisely n2 lines incident with the point pi, that is, bi = n2.
Now associated to any non-incident point-line pair (p, l) we know that n2 lines pass through p but that only n + 1 of them are incident with l since our space is linear and n + 1 points are incident with l, and any two points determine a unique line. Since n > 2 there are at least two lines that pass through p that are not incident with l.
3. If a subspace X contains a triangle, then X must be the entire set of points.
Note that the following identities are true for any unital:
To see that  , you should use Lemma 2.3.2 from Batten. The other identities are obvious. Also note that through any point p not incident with a line l, there are n2 - (n + 1) lines parallel to l through p.
Let us label the vertices or our triangle as p1, p2, and p3. Since X is a subspace, if p and q are points in X then the entire line pq must also lie in X. Since p1 and p2 lie in X, then the line p1p2 lies in X, and there are n + 1 points on this line.
Show that all n2 of the lines through p1 meet X. So now have at least n3 points in X, which is how many points lie in the entire unital. So X would have to be the entire unital.
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READING PROJECT:
Read Chapters 2 and 12 of Polster's book, "A Geometrical Picture Book". Please answer the following question: What is the relationship between inversive, Benz and hyperbolic planes?
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