Collineation Groups Preserving a Unital in a Projective Plane of Order

A unital embedded in a finite projective plane Π of order m# is a substructure of Π which forms a 2®(m$­1,m­1, 1) design. Several authors devoted their attention to the embedded unitals. The main aim is either to construct and investigate new classes of unitals, or to characterize some classes using group-theoretical or graphical properties [3, 6, 9, 18, 25]. In [3] we give some general results about the structure of 2-groups of collineations leaving invariant a unital in a projective plane of odd order. In addition, we determine the simple groups which can fix such a unital. In this paper we examine the general structure of a collineation group G which fixes an embedded unital in the case m3 1(4). Such an investigation can be developed starting from the fact that a 2-group S preserving a unital has rank m(S )% 3 in this case (see [3]). In Section 3, we give several results about the structure of S. In particular, we prove that if the group generated by the involutory homologies in S is non-trivial, then it must be dihedral or elementary abelian of order 2 or 4. Also, we show that S cannot contain an elementary abelian subgroup of order 8 whose involutions are Baer. The question concerning the existence of 4-groups whose involutions are Baer remains unsolved. To the author’s knowledge, no examples of embedded unitals are known admitting such groups. Nevertheless, a negative answer to the above question leads to a much more simple structure for S, and then for G. In Section 4, we investigate the structure of S when the plane admits a collineation which fixes the unital, normalizes S and induces a non-trivial 2«-automorphism on S. Section 5 contains the results about the global structure of G. The main aim is to try to determine the structure of the generalized Fitting subgroup F*(Ga ) of Ga ̄ G}O(G), and then that of Ga , using [3] and the results of Sections 3 and 4. Concerning the structure of O(G), we do not deal here with this question which, of course, seems to be very difficult.