Doubly primitive groups with simple socle The classification of doubly transitive groups with simple socle is
known. A good account of such classification can be found for example
in this paper:
Cameron, Peter J. Finite permutation groups and finite simple groups. Bull. London Math. Soc. 13 (1981), no. 1, 1--22. MR0599634 (83m:20008)
Another reference is Chapter 7 (Section 7.7) of
Dixon, John D.; Mortimer, Brian. Permutation groups. Graduate Texts in
Mathematics, 163. Springer-Verlag, New York, 1996. xii+346 pp. ISBN:
0-387-94599-7 MR1409812 (98m:20003)
I am particularly interested in the following permutation groups of the classification: $PSL(d,q)$ with $d>2$, $Sp(2d,2)$ with $d>2$, $PSU(3,q)$ with $q>2$, $Sz(q)$ (Suzuki groups) with $q^{2n+1}>2$ and $Ree(q)$ (Ree groups) with $q^{3n+1}>3$. 
Let $G$ be any of these groups and suppose that $G$ acts $2$-transitively on
a set $\Omega$. My question is the following:

Which of these permutation groups are doubly primitive on $\Omega$?
  (Doubly primitive means that the one-point stabilizer $G_\alpha$ is
  primitive on $\Omega\setminus\{\alpha\}$.)

Motivation: I need to prove that the centers of the one-point stabilizers of the groups I mentioned are trivial. This can be done case-by-case. (Unfortunately, I cannot find a general argument.)  I realized that the calculations of the centers could be avoided in the cases where $G$ is doubly primitive on $\Omega$. (This happens for example when $G=Sp(2d,2)$, see this MO Question.)
 A: None of these groups are $2$-primitive except for ${\rm Sp}(2d,2)$.
For ${\rm PSL}(d,q)$ with $d>2$, the $2$-point stabilizer fixes two projective points, say $\langle v_1 \rangle$ and $\langle v_2 \rangle$, so it also fixes other points, such as $\langle v_1+v_2 \rangle$. These fixed points, other than $\langle v_1 \rangle$, form a block of imprimitivity for the action of $G_{\langle v_1 \rangle}$ on $\Omega \setminus \langle v_1 \rangle$.
In ${\rm PSU}(3,q)$, the point stabilizer $G_1$ has the structure $N \rtimes H$,
where $N$ is a (nonabelian) group of order $q^3$ (it's a Sylow $p$-subgroup of $G$)and $H = G_{1,2}$ (the $2$-point stabilizer) is cyclic of order $(q^2-1)/(q+1,3)$. But $H$ does not act irreducibly on $N$, and fixes an elementary abelian sugbroup of order $q$, so $H$ is not maximal in $G_1$, and hence $G$ is not $2$-primitive.
The situation for ${\rm Sz}(q)$ and $R(q)$ is very similar.  In the first case, $G_1 = N\rtimes H$ with $|N|=q^2$, $|H|=q-1$ and in the second case $|N|=q^3$, $|H|=q-1$. In both cases, $H$ fixes a subgroup of order $q$ in $N$ so is not maximal in $G_1$.
The $2$-primitivity of the two $2$-transitive actions of ${\rm Sp}(2d,2)$ was explained in the answers to the previous question.
But you are correct in saying that the centres of the point stabilizers are trivial for all of these groups. The structures of these stabilizers are all very well understood.
For ${\rm PSL}(d,q)$, we have $G_1 = N \rtimes H$, where $N$ is elementary abelian of order $q^{d-1}$, and $H$ is a subgroup of ${\rm GL}(d-1,q)$ that contains ${\rm SL}(d-1,q)$, and acts faithfully and naturally on $N$. So the action has no fixed points, and hence $G_1$ has trivial centre.
In the other cases, $G_1$ is as described above. The unique minimal normal subgroup of $G_1$ is the centre of $N$ and has order $q$. Again $H$ acts without fixed points on this subgroup.
