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.)

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    $\begingroup$ As to your motivation, I expect the following to be true: If $G$ is doubly transitive and $G_\alpha$ has a non-trivial center, then $G$ is of affine type, that is, the socle is elementary abelian. However, off hand I do not see a simple reason for that. $\endgroup$ Jan 13, 2014 at 16:57
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    $\begingroup$ Yes, it seems to be true. I checked it with GAP up to degree 1000 and I couldn't find counterexamples. However, I should mention that to prove that a doubly transitive group with one-point stabilizers with center of even order has a regular normal subgroup, Kantor, O'Nan and Seitz [J. Algebra 21, 17-50, 1972] use the Feit-Thompson Theorem. $\endgroup$ Jan 13, 2014 at 17:22
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    $\begingroup$ Indeed, it is Lemma 3.5, and in addition is uses Glauberman's $Z^\star$-Theorem. OK, what I expected seems to be more difficult to prove... $\endgroup$ Jan 13, 2014 at 17:47

1 Answer 1


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.

  • $\begingroup$ In light of the conjecture in Atkinson's paper "Doubly transitive but not doubly primitive permutation groups, II", I infer from your answer that the Ree groups must act as groups of automorphisms of a block design on $\Omega$ in which $\lambda=1$. This seems surprising because the Suzuki groups don't act in this way. Is there a simple explanation for why there is this difference between the Ree and Suzuki groups? $\endgroup$ Jan 14, 2014 at 6:12
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    $\begingroup$ Yes, both the Ree groups and ${\rm PSU}(3,q)$ act on block designs with the same parameters when $q$ is an odd power of $3$, but they are not isomorphic of course. They both have ${\rm PSL}(2,q)$ as subgroups stabilizing a block. I guess one difference with ${\rm Sz}(q)$ is that it has no such ${\rm PSL}(2,q)$ subgroup. $\endgroup$
    – Derek Holt
    Jan 14, 2014 at 9:09

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