We know that the full automorphism group of the $\pi\_q = PG(2,q)$ acts *imprimitively* on the flags (all flags through a fixed point form a block). But, things change when we consider the action of full correlation group (semi-direct product of the automorphism group with cyclic group of order 2). I have been able to show that this group, which I'll call $Cor(\pi\_q)$, acts *primitively* on the flags for $q = 2,3,4,5,7$ using GAP. Is it true for all prime powers $q$? 


My question is "For what values of $q$ is the group $Cor(\pi\_q)$ flag-primitive?" 


*For those familiar with the notion of generalized hexagons*: original motivation of this was to study some properties known as valuations of the generalized hexagon of order $(q,1)$ constructed from $PG(2,q)$. For that, it's convenient if the automorphism group of hexagon (which is same as the correlation group of the projective plane) acts primitively and distance transitively on its points. For small values, it turned out to be so but I couldn't find a theoretical argument to prove it for some general class of values of $q$. 



Flag primitivity of the correlation group of classical projective planes.
Let $G=GL(3,q)$ act on $\mathbb F\_q^3$ from the right. Set $e\_1=(1,0,0)$ and $e\_2=(0,1,0)$, and $P=\langle e\_1\rangle$, $L=\langle e\_1,e\_2\rangle$. Then the stabilizer of the flag $F=(P,L)$ is the group $B$ of lower triangular matrices.


There are precisely two groups $B\_1$ and $B\_2$ properly between $B$ and $G$, namely the two staircase groups containing $B$. (An easy fact from linear algebra.) These groups correspond to two types of blocks containing $F$: For the first type a block consists of the flags $(P,L')$ with $P\in L'$, and for the second type the block consists of the flags $(P',L)$ with $P'\in L$.


So for the action of $G$ on the flags, there are only two systems of imprimitivity. So if the correlation group is imprimitive, it need to preserve at least one of these systems. However, a correlation swaps these systems, a contradiction.


