**Proposition 9.3**: *The flocks of all thick cones in PG(3,5) that have flocks are linear, or FTW flocks (which are also Fisher type B flocks). The flocks of any cones having more than 6 points in their carriers are star flocks.*

*Pf*: There are four planes other than the carrier plane in a flock of PG(3,5). Thus, there are four primary baselines which may not all be distinct. However, since q - [q+2/2] = 2, if two primary baselines coincided there would be three planes of the flock on a common line. Such a flock is linear by Theorem 9.1 . So, we may assume that the four primary baselines are distinct.

If the four baselines are concurrent, the flock is a (proper) star flock. However, S, the carrier of the cone, would be contained in the two remaining lines through the common point. Since 2 < [q+2/2], such a cone is thin (a more detailed analysis actually shows that the cone must be flat, but we do not need this at present). Thus, a flock of a thick cone in PG(3,5) can not be a proper star flock. We now turn to the dual setting and use the notation developed earlier . Suppose that four planes of the flock had a common point. In the dual setting these four points of D_{F} lie in a plane which intersects the plane W' = 0 in a line *m*. As no three of these points are collinear (otherwise three planes of F would contain a common line, and the flock would be linear), they form a quadrangle in and so determine 6 lines each containing two of the points. These 6 lines meet in pairs in three points off the quadrangle (the *diagonal points* of the quadrangle). The diagonal points of this quadrangle are not collinear since q is not even (Fano's Theorem).

The line *m* contains no point of the quadrangle since no point of D_{F} is in W' = 0. If *m* does not contain one of the diagonal points then the six lines of the quadrangle will meet *m* in six distinct points, i.e., all the points of *m*. Thus, in this case, every line of D_{G} meets *m* in a point through which passes a line containing two points of D_{F}. This means that no non-empty cone can satisfy the flock condition. If *m* contains exactly one of the diagonal points, then there is exactly one point on *m* which does not lie on a line containing two D_{F} points. To satisfy the flock condition, all the lines of D_{G} must pass through this point. Dually, this means that all the generators of the cone are in the same plane, i.e., the cone is flat. Finally, if *m* contains two diagonal points, then there are exactly two points of *m* through which no line containing two points of D_{F} pass. Thus, all lines of D_{G} must pass through one of these two points and so, all the generators of the cone lie in one of two planes, i.e., the cone is thin. Therefore, if the cone is thick, no four planes of a non-linear flock of the cone can meet at a point.

We have thus shown that a flock of a thick cone in PG(3,5) is either linear or has the property that no three planes contain a common line and no four planes contain a common point, i.e., the flock is of FTW type. Suppose we have a flock of FTW type in PG(3,5). Take any two of the planes, they meet in a common line and the three other planes meet in a unique common point since they can not contain a common line (see the proof of Proposition 5.3 ). Thus, the flock is of Fisher type. Since the common point of the three planes can not lie in either of the two original planes, this is a flock of Fisher type B.

There are 6 (= (q-2)!) classes of permutations of GF(5) which fix 0 up to scalar multiples. Thus, if the carrier of a cone contained more than 6 points, the herd of any flock of such a cone would contain two permutations which were scalar multiples of each other and by Theorem 8.1.1 the flock would be a star flock.