Tables of possible classical distance-transitive groups

We use the following conventions.
Let S be one of the simple groups of Lie type Bn(q) with n > 2 and q odd, Cn(q) with n > 1, Dn(q) with n > 3, 2An(q) with n > 1 or 2Dn(q) with n > 3. Let G be an almost simple group with socle S and H be a maximal subgroup of G. Rather than explicitly listing the structure of H we will give its type. Similar to the notion of Kleidman and Liebeck.

Remark. Observe that the group P\Omega6-(q) is treated as PSU4(q) and P\Omega5(q) as PSp4(q). The group PSp4(2) is not simple, when one encounters it one should take its derived group. Observe that this group can be ignored too since PSp4(2)' is isomorphic to PSL2(9) and Alt6, whence has been studied well in the past. It seemed unnatural to exclude it tough.
The groups P\Omega3(q), P\Omega4-(q) and P\Omega6+(q) will not be considered as they are isomorphic to linear groups, and the groups P\Omega2+(q), P\Omega2-(q) and P\Omega4+(q) are not simple.

We try to determine the multiplicity free permutation representations and the distance-transitive graphs. The notation in those tables is as follows:
For the multiplicity-free column: a + means it is multiplicity free, a ? means it is not know (and for the moment don't try to answer the question) no entry means that one should try to decide it since we are often dealing with an explicit group which is relatively small, (i.e., either a + or we should remove the entry).
Information can be obtained from the person mentioned in the column labeled person.

Theorem Let S be a simple classical group, not the linear group. Let G be acting on the polar associated polar space of S with soc(G) = S and let H be a maximal subgroup of G not containing S. If the permutation character is multiplicity free, then H is the stabilizer of a totally singular subspace or S and the type of H are as in Table C.1 or H is of type C9 and S and F*(H) are as in Table C.2.


Table C.1
Stype Hclassmpfcomments persondtg
PSUn(q)stabilizer of non isotropic pointC1+ Jan/Johnsolved
PSU6(q)SU3(q) x SU3(q)C2?   open: q<17
PSU4(q)SU2 (q) x SU2 (q)C2?q not 2van Bonopen: q=2, 3
PSU4(2)stabilizer of a basisC2+rank 3ATLASyes
PSU5(2)stabilizer of a basisC2+GAP: rank 7Breuer?
PSU3(3)stabilizer of a basisC2+GH(2,2)ATLASyes
PSU3(4)stabilizer of a basisC2+ATLAS: rank 5 in Aut. ?
PSUn(q)Spn(q)C5+n evenJan/Johnsolved
PSU4(q)O4-(q)C5 ? q odd ?
PSU4(3)normalizer of Z421+4C6+ATLAS: rank 5 ?
PSU3(5)normalizer of 32:Q8C6+GAP, rank 12Breuer?
PSp2n(q)stabilizer of a hyperbolic lineC1+n>2Jan/Johnsolved
PSp2n(q)Spn(q) x Spn(q)C2 ? n even, see also remark 3van Bon/Inglisopen: q even and n= 4,..,12
PSp2n(q)Spn(q2)C3+n evenJohn?
PSp2n(q)Un(q)C3 ? q oddJohnopen: n=3
PSp12(q)Sp4(q3)C3 q=2,3,4 ?
PSp6(2)Sp2(23)C3+atlas, rank 6 ?
PSp6(q)Sp2(q3)C3 q=4,3,5,8 ?
PSp4(3)normalizer of 2-1+4C6+atlas, rank 3-yes
PSp4(p)normalizer of 2-1+4C6 p= 5,7 ?
PSp8(3)normalizer of 2-1+6C6   ?
PSp2n(q)O2n+(q)C8+q evenInglissolved
PSp2n(q)O2n-(q)C8+q evenInglissolved
P\Omega2n+(q) non isotropic pointC1+ Jan/Johnsolved
P\Omega2n+(q) stabilizer of a O2-(q) spaceC1+?Johnsolved
P\Omega8+(q)O4-(q) x O4-(q)C2 ? under triality O4+(q2) ?
P\Omega8+(2)O2-(2) x O2-(2) x O2-(2) x O2-(2)C2+GAP, rank 12Breuer?
P\Omega2n+(q)On+(q2) C3 ? n evenJan/Nick?
P\Omega2n+(q)On(q2)C3 ? nq oddJan/Nick?
P\Omega2n+(q)Un(q)C3+ n evenJohnopen: n=6,8,10
P\Omega2n-(q) non isotropic pointC1+ Jan/Johnsolved
P\Omega2n-(q)stabilizer of a O2-(q) spaceC1+ Johnsolved
P\Omega2n-(q)O-n(q2) C3? n evenJan/Nick?
P\Omega2n-(q)On(q2) C3 ? nq oddJan/Nick?
P\Omega2n-(q)Un(q) C3+n oddJohnopen: n=5
P\Omegan(q) non isotropic pointC1+nq oddJan/Johnsolved
P\Omegan(q) stabilizer of a O2-(q) spaceC1+nq oddJohnsolved
P\Omega7(3)stabilizer of a basisC2+GAP, rank 11 in Aut.Breuer?
P\Omega9(3)stabilizer of a basisC2   ?


Table C.2
SF*(H)mpfcommentspersondtg
PSU3(3) PSL2(7) + rank 3 in aut -yes
PSU3(5) Alt7 + rank 3 HoSi -yes
PSU3(5) Alt6 + lines HoSi -yes
PSU3(5) PSL2(7) + GAP rank 9  ?
PSU4(3) PSL3(4) + GAP, rank 3 -yes
PSU4(3) Alt7 + GAP, rank 6  ?
PSU4(5) Alt7     ?
PSU4(5) PSU4(2)      ?
PSU6(2) PSU4(3) + rank 3 -yes
PSU6(2) M22 + GAP rank 8  ?
PSU9(2) J3     ?
PSp4(3) Alt6 + rank 3, see also remark 2 -yes
PSp4(q) Sz(q) + 2 < q an odd power of 2 Jan/Johnsolved
PSp4(7) Alt7    ?
PSp6(q) G2(q) + q even Jan/Johnsolved
PSp6(5) J2     ?
PSp8(2) Alt10 +rank 5Inglis?
PSp12(2) Alt14   Nick notes not selfpaired  no
PSp12(3) Suz     ?
P\Omega7(q) G2(q) +  Jan/Johnsolved
P\Omega7(3) PSp6(2) + rank 4  ?
P\Omega7(p) PSp6(2)   p=5,7  ?
P\Omega7(3) Alt9 + GAP rank 6 ?
P\Omega9(3) Alt10      ?
P\Omega8+(q)P\Omega7(q)+q odd, under triality non-isotropic point -solved
P\Omega8+(q)PSp6(q)+q even, under triality non-isotropic point -solved
P\Omega8+(2) Alt9 + rank 4  ?
P\Omega8+ (3) P\Omega8 +(2) +rank 4 ?
P\Omega8+(p) P\Omega8+(2)   p= 5,7  ?
P\Omega10+(3) Alt12     ?
P\Omega14+(2) Alt16  Nick notes not selfpaired no
P\Omega10-(2) Alt12 + rank 6 in Aut grp ?
P\Omega12-(2) Alt13  Nick notes not selfpaired no

Remark 1 Observe that for n=8 the action of P\Omega8+(q) on Un(q) becomes under triality a stabilizer of a O2- -space.

Remark 2 In the table of C9 groups the case S = PSp4(3) = PSU4(2) and F*(H) = Alt6 is listed only once since Alt6 as a subgroup of PSU4(2) is Sp4(2)' and is of type C5. The case S = PSp4(2) and F*(H) = Alt5 is not listed since Alt5 = P\Omega4-(2) and is thus of type C8.

Remark 3 We justify the + entries in tables and make some observations about the open cases. The cases where H is a maximal parabolic have been studied in the past.
Suppose S= PSUn(q). If H a stabilizer of a non isotropic point the permutation character is multiplicity free. If H of type PSpn(q) it follows from the thesis of Nick Inglis. If H of type PSU2(q) x PSU2(q), then this is the action of P\Omega6-(q) on O2-(q)-spaces. This action is multiplicity free if the group is large enough. The remaining ones follow from the ATLAS or are well known.
Suppose S = PSp2n(q). If H the stabilizer of a hyperbolic line, then if q=2 it follows from the known literature. If H of type PSpn(q2), P\Omega{2n} ^\epsilon (q) it follows from the thesis of N. Inglis that the permutation charter is multiplicity free. If H of type (Spn(q) x Spn(q)), n even, the techniques as developed in the thesis of N. Inglis can be used to show that for 2n=8 and 12 (2n=4 is well known) all orbits are self paired and for 2n \geq 16 there exists a non self paired orbit. Whence the permutation charter is multiplicity free for 2n=4,8,12, but still undecided in general.
Finally suppose that S is isomorphic to an orthogonal group. The stabilizers of non isotropic points have been studied by M.W. Liebeck, C.E. Praeger and J.Saxl If H of type PSUn(q), then it follows from the thesis of N. Inglis that the permutation charter is multiplicity free. If H is the stabilizer of an O2--space the multiplicity freeness will be proved. It is also known that if S = P\Omega6+(q) and H of type O3(q), with q odd, then the permutation charter is not multiplicity free as follows from work on the linear groups.

We also have to study the permutation characters and distance-transitive graphs that arise from maximal subgroups that are normalized by the graph automorphism for PSp4(q)', q even, and the triality automorphism for P\Omega8+(q). The results on permutation characters is summarized in the next 2 theorems.

Theorem Let G be an almost simple group with socle P\Omega8+(q) Suppose H is a maximal subgroup of G normalized by a triality automorphism and H \not \leq P\Delta8 +(q). If the permutation character is multiplicity free, then H \cap P \Omega8+(q) is as in Table C.3.

Table C.3
group no. H \cap P\Omega8+(q)mpfcommentspersondtg
5stabilizer of a totaly singular line+Rank 5 in Aut(P\Omega8+(q))Jan/Johnsolved
15G2(q)?might be +, many orbs same length, c.f. Jan   
26 ^( 1/d (q+1) x 1/dGU3(q)).2d    
56 ^( 1/d Zq+1)4.d3.23.S4   q=2,3,4   
72P\Omega8+(2) + q=3, rank 4  
72P\Omega8+(2) q=5,7  
62P\Omega8+(2) q=23  

Remark The two + entries are both well known. R. Lawther (personal communication) showed that if G=P\Omega8+(q) and H = G2(q), then the permutation character is multiplicity free only for q=2,3 and that in the other cases there is a character occuring with multiplicity 2. What the behaviour is under automorphisms is not known.

Theorem Let S= PSp4(q)' with q even. Let G be almost simple with socle S and contain the graph automorphism. Let H be a maximal subgroup of G, normalized by and containing the graph automorphism. The permutation character is multiplicity free if and only if H is the stabilizer of a flag, q > 2 is an odd power of 2 and H \cap S = Sz(q), q > 2 is an even power of 2 and H \cap S = PSp4({\sqrt q}), or q=2 and H \cap S= 5:2 or H \cap S = 32:4. In all cases the DTG problem is solved.

Remark If H is of type Sz(q) or of type PSp4(\sqrt q) the multiplicity-freeness follows from work by R. Lawther.