Index

25 (a): This is fairly simple. Pick a point on the Fano plane for c and fix a line l . We illustrate below:

We have shown above that the Fano plane has a complete set of central collineations. We have used the fact that the Fano plane is homogeneous.

(b): S'pose that the distinct lines l and m are both axes of a collineation f. Thus f is an axial collineation for each line and, by duality, a central collineation too. If f is the identity map then we are done, so suppose that it is not. If we have two distinct axes, then by the duality principle we must also have two distinct centers. Let c and c' denote these respective centers. Let p be some point that does not lie on the line cc'. Then our mapping f sends the line pc to itself. Similarly f sends the line pc' to itself. Thus f(p) is incident with both pc and pc'. But the intersection of pc and pc' is p. So f(p) = p. Thus f sends every point p to itself, provided that point does not lie on the line cc'.

Let q be some point distinct from p. We know that the line pq has at least 3 points and one of these points must be incident with cc' because every line meets every other line at some point. Thus there are at least two other points that do not lie on cc', so these points are fixed by f (as shown above). Thus f(pq) = pq. (If two points are fixed, we know that any line is uniquely determined by 2 points, so the whole line must map onto itself). Thus f(q) is on pq. Because f(p) = p and f(q) is on pq, then p is a center.

Thus we know that c, c' and p are all centers. Let r be some point that is on cc'. But because p is not on cc', we know that r will not lie on pc'. Thus, follow the above proof to show that r, a point not on the line of our centers cp, will be fixed under f.

Thus all of our points are fixed and so the function must be the identity map.