ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 14 (2018) 209-226 https://doi.org/10.26493/1855-3974.1193.0fa (Also available at http://amc-journal.eu) Groups of Ree type in characteristic 3 acting on polytopes* Dimitri Leemans Université Libre de Bruxelles, Département de Mathématique, C.P.216 - Algebre et Combinatoire, Boulevard du Triomphe, 1050 Brussels, Belgium Egon Schulte Northeastern University, Department of Mathematics, 360 Huntington Avenue, Boston, MA 02115, USA Hendrik Van Maldeghem Vakgroep Wiskunde, Universiteit Gent, Krijgslaan 281, S22, 9000 Gent, Belgium Received 8 September 2016, accepted 14 March 2017, published online 4 September 2017 Every Ree group R(q), with q = 3 an odd power of 3, is the automorphism group of an abstract regular polytope, and any such polytope is necessarily a regular polyhedron (a map on a surface). However, an almost simple group G with R(q) < G < Aut(R(q)) is not a C-group and therefore not the automorphism group of an abstract regular polytope of any rank. Keywords: Abstract regular polytopes, string C-groups, small Ree groups, permutation groups. Math. Subj. Class.: 52B11, 20D05 1 Introduction Abstract polytopes are certain ranked partially ordered sets. A polytope is called "regular" if its automorphism group acts (simply) transitively on (maximal) flags. It is a natural question to try to classify all pairs (P, G), where P is a regular polytope and G is an automorphism group acting transitively on the flags of P. An interesting subclass is constituted *This research was sponsored by a Marsden Grant (UOA1218) of the Royal Society of New Zealand. The authors also thank an anonymous referee for useful comments on a preliminary version of this paper. E-mail addresses: dleemans@ulb.ac.be (Dimitri Leemans), schulte@neu.edu (Egon Schulte), hendrik.vanmaldeghem@ugent.be (Hendrik Van Maldeghem) Abstract ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ 210 Ars Math. Contemp. 14 (2018) 433-443 by the pairs (P, G) with G almost simple, as then a lot of information is available about the maximal subgroups, centralizers of involutions, etc., of these groups, making classification possible for some of these families of groups. Potentially this could also lead to new presentations for these groups, as well as a better understanding of some families of such groups using geometry. The study of polytopes arising from families of almost simple groups has received a lot of attention in recent years and has been very successful. Mazurov [23] and Nuzhin [25,26, 27, 28] established that most finite simple groups are generated by three involutions, two of which commute. These are precisely the groups that are automorphism groups of rank three regular polytopes. The exceptions are PSL3(q), PSU3(q), PSL4(2n), PSU4(2n), A6, A7, M11, M22, M23, McL, PSU4(3), PSU5(2). The two latter, although mentioned by Nuzhin as being generated by three involutions, two of which commute, have been found to be exceptions recently by Martin Macaj and Gareth Jones (personal communication). We refer to [18] for almost simple groups of Suzuki type (see also [16]); [4, 20, 21] for groups PSL2(q) < G < PrL2(q); [2] for groups PSL3(q) and PGL3(q); [1] for groups PSL(4, q); [9] for symmetric groups; [3, 10, 11] for alternating groups; and [13, 19, 22] for the sporadic groups up to, and including, the third Conway group Co3, but not the O'Nan group. Recently, Connor and Leemans have studied the rank 3 polytopes of the O'Nan group using character theory [5], and Connor, Leemans and Mixer have classified all polytopes of rank at least 4 of the O'Nan group [6]. Several attractive results were obtained in this vein, including, for instance, the proof that Coxeter's 57-cell and Griinbaum's 11-cell are the only regular rank 4 polytopes with a full automorphism group isomorphic to a group PSL2 (q) (see [20]). Another striking result is the discovery of the universal locally projective 4-polytope of type {{5, 3}5, {3,5}10}, whose full automorphism group is J1 x PSL2 (19) (see [14]); this is based on the classification of all regular polytopes with an automorphism group given by the first Janko group J1. The existing results seem to suggest that polytopes of arbitrary high rank are difficult to obtain from a family of almost simple groups. Only the alternating and symmetric groups are currently known to act on abstract regular polytopes of arbitrary rank. For the sporadic groups the highest known rank is 5. The Ree groups R(q), with q = 32e+1 and e > 0, were discovered by Rimhak Ree [29] in 1960. In the literature they are also denoted by 2G2(q). These groups have a subgroup structure quite similar to that of the Suzuki simple groups Sz(q), with q = 22e+1 and e > 0. Suzuki and Ree groups play a somewhat special role in the theory of finite simple groups, since they exist because of a Frobenius twist, and hence have no counterpart in characteristic zero. Also, as groups of Lie-type, they have rank 1, which means that they act doubly transitively on sets of points without further apparent structure. However, the rank 2 groups which are used to define them, do impose some structure on these sets. For instance, the Suzuki groups act on "inversive planes". For the Ree groups, one can define a geometry known as a "unital". However, these unitals, called Ree unitals, have a very complicated and little accessible geometric structure (for instance, there is no geometric proof of the fact that the automorphism group of a Ree unital is an almost simple group of Ree type; one needs the classification of doubly transitive groups to prove this). Also, Ree groups seem to be misfits in a lot of general theories about Chevalley groups and their twisted analogues. For instance, there are no applications yet of the Curtis-Tits-Phan theory for Ree groups; all finite quasisimple groups of Lie type are known to be presented by two D. Leemans et al.: Groups of Ree type in characteristic 3 acting on polytopes 211 elements and 51 relations, except the Ree groups in characteristic 3 [12]. Hence it may be clear that the Ree groups R(q), with q a power of 3, deserve a separate treatment when investigating group actions on polytopes. Now, the regular polytopes associated with Suzuki groups are quite well understood (see [16, 18]). But the techniques used for the Suzuki groups are not sufficient for the Ree groups. In the present paper, we carry out the analysis for the groups R(q). In particular, we ask for the possible ranks of regular polytopes whose automorphism group is such a group, and we prove the following theorem. Theorem 1.1. Among the almost simple groups G with R(q) < G < Aut(R(q)) and q = 32e+1 = 3, only the Ree group R(q) itself is a C-group. In particular, R(q) admits a representation as a string C-group of rank 3, but not of higher rank. Moreover, the non-simple Ree group R(3) is not a C-group. In other words, the groups R(q) behave just like the Suzuki groups: they allow representations as string C-groups, but only of rank 3. Although Nuzhin proved in [27] that these groups allow representations as string C-groups of rank 3 for every q, we will describe a string C-group representation for R(q), q = 3, for each value of q to make the paper self-contained. Also, almost simple groups R(q) < G < Aut(R(q)) can never be C-groups (in characteristic 3). Rephrased in terms of polytopes, Theorem 1.1 says that among the almost simple groups R(q) < G < Aut(R(q)), only the groups G := R(q) are automorphism groups of regular polytopes, and that these polytopes must necessarily have rank 3. Ree groups can also be the automorphism groups of abstract chiral polytopes. In fact, Sah [30] showed that every Ree group R(32e+1), with 2e + 1 an odd prime, is a Hurwitz group; and Jones [15] later extended this result to arbitrary simple Ree groups R(q), proving in particular that the corresponding presentations give chiral maps on surfaces. Hence the groups R(q) are also automorphism groups of abstract chiral polyhedra. It is an interesting open problem to explore whether or not almost simple groups of Ree type also occur as automorphism groups of chiral polytopes of higher rank. Note that the Ree groups in characteristic 2 are also very special: they are the only (finite) groups of Lie type arising from a Frobenius twist and having rank at least 2. This makes them special, in a way rather different from the way the Ree groups in characteristic 3 are special. We think that in characteristic 2, quite different geometric methods will have to be used in the study of polytopes related to Ree groups. 2 Basic notions 2.1 Abstract polytopes and string C-groups For general background on (abstract) regular polytopes and C-groups we refer to McMullen & Schulte [24, Chapter 2]. A polytope P is a ranked partially ordered set whose elements are called faces. A polytope P of rank n has faces of ranks -1,0,..., n; the faces of ranks 0, 1 or n - 1 are also called vertices, edges or facets, respectively. In particular, P has a smallest and a largest face, of ranks -1 and n, respectively. Each flag of P contains n + 2 faces, one for each rank. In addition to being locally and globally connected (in a well-defined sense), P is thin; that is, for every flag and every j = 0,..., n - 1, there is precisely one other (j-adjacent) flag with the same faces except the j-face. A polytope of rank 3 is a polyhedron. 212 Ars Math. Contemp. 14 (2018) 433-443 A polytope P is regular if its (automorphism) group r(P) is transitive on the flags. If r(P) has exactly two orbits on the flags such that adjacent flags are in distinct orbits, then P is said to be chiral. The groups of regular polytopes are string C-groups, and vice versa. A C-group of rank n is a group G generated by pairwise distinct involutions p0,..., pn-1 satisfying the following intersection property: 0, is a group of order q3(q -1)(q3 + 1). It has a faithful permutation representation on a Steiner system S := (Q, B) = S(2, q + 1, q3 + 1) consisting of a set Q of q3 + 1 elements, the points, and a family of (q + 1)-subsets B of Q, the blocks, such that any two points of Q lie in exactly one block. This Steiner system is also called a Ree unital. In particular, G acts 2-transitively on the points and transitively on the incident pairs of points and blocks of S. The group G has a unique conjugacy class of involutions (see [29]). Every involution P of G has a block B of S as its set of fixed points, and B is invariant under the centralizer CG(p) of p in G. Moreover, Cg(p) = C2 x PSL2(q), where C2 = 1 be integers such that m | n. The normalizer ND2n (D2m) of any subgroup D2m of D2n coincides with D2m if n/m is odd, or is isomorphic to a subgroup D4m of D2n if n/m is even. 3 Proof of Theorem 1.1 The proof of Theorem 1.1 is based on a sequence of lemmas. We begin in Lemma 3.1 by showing that if R(q) < G < Aut(R(q)) then G can not be a C-group (with any underlying Coxeter diagram). Thus only the Ree groups R(q) themselves need further consideration. Then we prove in Lemma 3.3 that R(q) does not admit a representation as a string C-group of rank at least 5. In the subsequent Lemmas 3.9, 3.11 and 3.12 we then extend this to rank 4 and show that R(q) can also not be represented as a string C-group of rank 4. Finally, in Lemma 3.15 we construct each group R(q) as a rank 3 string C-group. All information that we use about the groups R(q) can found in [17]. We repeatedly make use of the following simple observation. If A : B is a semi-direct product of finite groups A, B such that B has odd order, then each involution in A : B must lie in A. In fact, if p = ap is an involution, with a G A, p G B, then 1 = p2 = a(pap-1)p2, where a(pap-1) G A and p2 G B; hence p2 = 1, so p =1 and p = a G A. 3.1 Reduction to simple groups R(q) We begin by eliminating the almost simple groups of Ree type that are not simple. Lemma 3.1. Let R(q) < G < Aut(R(q)), where q = 32e+1. Then G is not a C-group. Proof. Since Aut(R(q)) = R(q): C2e+1 and 2e +1 is odd, every involution in Aut(R(q)) lies in R(q) (by the previous observation), and hence any subgroup of Aut(R(q)) generated by involutions must be a subgroup of R(q). Thus no subgroup G of Aut(R(q)) strictly above R(q) can be a C-group. (When e = 0 we have Aut(R(3)) = R(3), so the statement holds trivially.) □ 214 Ars Math. Contemp. 14 (2018) 433-443 3.2 String C-groups of rank at least five By Lemma 3.1 we may restrict ourselves to Ree groups G = R(q). We first rule out the possibility that the rank is 5 or larger. Lemma 3.2. Let G be a simple group. Suppose G has a generating set S := {p0,..., pn-1} of n involutions such that (G,S) is a string C-group. Then |pjpi+1| > 3 for all i = 0,...,n — 2. Proof. This is due to the fact that, as G is simple, G is not directly decomposable, that is, G cannot be written as the direct product of two nontrivial normal subgroups of G. □ Lemma 3.3. Let G = R(q), where q = 32e+1 = 3. Suppose G has a generating set S of n involutions such that (G, S) is a string C-group. Then n < 4. Proof. Let S = {p0,..., pn-1}, so in particular, G = {p0,..., pn-1). Then p0 commutes with p2,..., pn-1, since the underlying Coxeter diagram is a string. However, by Lemma 3.2, p0 does not commute with p1 and pn-1 does not commute with pn-2. Now suppose n > 5 and consider the subgroup H := {p0, p1, pn-2, pn-1) of G. Then H must be isomorphic to D2c x D2d for some integers c,d > 3. Inspection of the list of maximal subgroups of R(q) described above shows that direct products of (non-abelian) dihedral groups never occur as subgroups in G. So n is at most 4. □ 3.3 String C-groups of rank four Next we eliminate the possibility that the rank is 4. We begin with a general lemma about string C-groups that are simple. Lemma 3.4. Let (G, S) be a string C-group of rank n, and let G be simple. Then Ng(Go1)\Ng(Go) must contain an involution (namely p0). Proof. The involution p0 centralizes G01 and hence must lie in NG(G01). On the other hand, p0 cannot also lie in NG (G0) for otherwise G0 would have to be a nontrivial normal subgroup in the simple group G. □ The next two lemmas will be applied to dihedral subgroups in subgroups of type PSL2(q) or C2 x PSL2(q) of R(q), respectively. Lemma 3.5. Let q = 32e+1 and e > 0. Then the order 2d of a non-abelian dihedral subgroup of PSL2 (q) must divide q — 1 or q + 1. Moreover, d ^ 0 mod 4, and d is even only if 2d divides q +1. Proof. Suppose D2d is a non-abelian dihedral subgroup of PSL2(q), so d > 3. We claim that 2d must divide q +1 or q — 1. Recall that under the assumptions on q, the order 2d must either be 6 or must divide q — 1 or q +1. It remains to eliminate 6 as a possible order. In fact, since q is an odd power of 3, the only maximal subgroups of PSL2 (q) with an order divisible by 6 are subgroups PSL2(q0) with q0 a smaller odd power of 3 (see [8] for a list of the subgroups of PSL(2, q)). If we apply this argument over and over again with smaller odd powers of 3, we eventually are left with a subgroup PSL2(3). However, D. Leemans et al.: Groups of Ree type in characteristic 3 acting on polytopes 215 PSL2(3) = A4 and hence cannot have a subgroup of order 6. Thus 2d must divide q + 1 or q — 1. This proves the first statement of the lemma. The second statement follows from the fact that q = 3 mod 8. □ Lemma 3.6. Let q = 32e+1 and e > 0, let 2d divide q — 1 or q +1, and let D2d be a non-abelian dihedral subgroup of a group C := C2 x PSL2 (q). (a) Then there exists a dihedral subgroup D in PSL2 (q) such that D2d is a subgroup of C2 x D of index 1 or 2, and NC (D2d) = NC (C2 x D) = C2 x NPSL2 (q)(D). Here the normalizer NPSL2(q)(D) must lie in a maximal subgroup Dq+1 or Dq-1 of PSL2(q), and coincide with NDq+1 (D) or NDq-1 (D), according as 2d divides q — 1 or q + 1. (b) Let D2d = D (that is, the index is 2). If 2d | (q — 1) or if 2d | (q + 1) and (q +1)/2d is odd, then NPSL2(q)(D) = D and NC (D2d) = C2 x D2d. If 2d | (q + 1) and (q + 1)/2d is even, then NpsL2(q)(D) = D4d and Nc(D2d) = C2 x D4d. (c) If D2d = C2 x D (that is, d is even, d/2 is odd, D = Dd, and the index is 1), then NpsL2(q)(D) = D2d and Nc (D2d) = C2 x D2d (regardless of whether 2d | (q — 1) or 2d | (q +1)). (d) The structure of NC(D2d) only depends on d and q, not on the way in which D2d is embedded in C. Proof. For the first part, suppose C2 = (p) and D2d = (a0, 01) where a0, a1 are standard involutory generators for D2d. Write a0 = (p1, a'0) and a1 = (pj, o'x) for some i, j = 0,1 and involutions a0 in PSL2(p). Then D := (a0 ) is a dihedral subgroup of PSL2(p), and D2d lies in C2 x D. Since the period of a0divides that of a0a1, the order of D is at most 2d and D2d has index 1 or 2 in C2 x D. If this index is 1 then D2d = C2 x D (and d is even and D = Dd). If the index of D2d in C2 x D is 2, then D = D2d and D2d n {1} x D must have index 1 or 2 in {1} x D. If the index of D2d n {1} x D in {1} x D is 1 then clearly D2d = {1} x D and D2d can be viewed as a subgroup of PSL2 (q). If the index of D2d n {1} x D is 2, then D2d n {1} x D is of the form {1} x E where E is either the cyclic subgroup Cd of D, or d is even and E is one of the two dihedral subgroups of D of order d. (Note here that D2d cannot itself be a direct product in which one factor is generated by p, since p cannot lie in D2d.) Next we investigate normalizers. First note that the normalizer of a direct subproduct in a direct product of groups is the direct product of the normalizers of the component groups. Thus Nc (C2 x D) = C2 x NpsL2(q)(D). We now show that the normalizers in C of the subgroups D2d and C2 x D coincide. There is nothing to prove if D2d = C2 x D or D2d = {1} x D. Now suppose that D2d has index 2 in C2 x D and E is as above. Then it is convenient to write D2d in the form D2d = ({1} x E) u ({p} x (D\E)). (3.1) If (a, fi) € C then (a, fi)D2d(a, fi)-1 = ({1} x fiEfi-1) U ({p} x fi(D\E)fi-1). (3.2) Now if (a, fi) € NC (D2d) then the group on the left in (3.2) is just D2d itself and therefore fiEfi-1 = E and fi(D\E)fi-1 = D\E. It follows that fi normalizes both E and D, so in particular (a, fi) € Nc(C2 x D). Hence Nc(D2d) < Nc(C2 x D). Now suppose that (a, fi) € NC (C2 x D). Then fi normalizes D. But fiEfi-1 must be a subgroup of D of index 2 isomorphic to E, and hence fiEfi-1 and E are either both cyclic 216 Ars Math. Contemp. 14 (2018) 433-443 or both are dihedral. Clearly, if both subgroups are cyclic then PEP-1 = E. However, the case when both subgroups are dihedral is more complicated. First recall that then d must be even. Now the normalizer NPSL2(q) (D) of the dihedral subgroup D of PSL2 (q) in PSL2(q) either coincides with D (that is, D is self-normalized), or is a dihedral subgroup containing D as a subgroup of index 2. We claim that under the assumptions on q, the second possibility cannot occur. In fact, in this case the normalizer would have to be a group of order 4d, and since d is even, its order would have to be divisible by 8; however, the order of PSL2 (q) is not divisible by 8 when q is an odd power of 3, so PSL2 (q) certainly cannot contain a subgroup with an order divisible by 8. Thus NPSL2(q)(D) = D. But P belongs to the normalizer of D in PSL2(q), so then p must lie in D. In particular, PEP-1 = E since E is normal in D. Thus, in either case we have PEP-1 = E, and since PDP-1 = D, also P(D\E)p-1 = D\E. Hence, (3.2) shows that (a,P) G Nc(D2d). Hence also NC(C2 x D) < NC(D2d). To complete the proof of the first part, note that D must lie in a maximal subgroup Dq±1 of PSL2(q) and Nps^(q)(D) = ND,±1 (D). The second and third part of the lemma follow from Lemma 3.2 applied to the dihedral subgroup D of Dq± 1. In particular, D is self-normalized in Dq± 1 if (q± 1)/1D| is odd, and NPSL2(q)(D) is a dihedral subgroup of Dq±1 of order 2|D| if (q ± 1)/|D| is even. Bear in mind that (q - 1)/2 is odd, and (q + 1)/2 is even but not divisible by 4. To establish the last part of the lemma, note that NC(D2d) = C2 x D2d, except when D = D2d, 2d | (q +1) and (q + 1)/2d is even. However, since q = 3 mod 8, if 2d | (q +1) and (q + 1)/2d is even then d must be odd. In other words, the situation described in the third part of the lemma cannot occur as this would require d to be even. Thus, if 2d | (q +1) and (q + 1)/2d is even, then we are necessarily in the situation described in second part of the lemma, and so necessarily NC (D2d) = C2 x D4d. □ Our next lemma investigates possible C-subgroups of G = R(q) of rank 3. The vertex-figure of a putative regular 4-polytope with automorphism group G would have to be a regular polyhedron with a group of this kind. Lemma 3.7. The only proper subgroups of R(q) that could have the structure of a C-group of rank 3 are Ree subgroups R(qo) with qo = 3 or subgroups of the form PSL2(qo), C2 x PSL2(qo), or R(3)' = PSL2(8). Proof. It is straightforward (sometimes by applying Lemma 3.2) to verify that only subgroups of maximal subgroups of R(q) of the second and third type can have the structure of a rank 3 C-group. Therefore we are left with Ree subgroups R(q0) and subgroups of groups C2 x PSL2 (q0), with q0 an odd power of 3 dividing q, as well as subgroups of type R(3)' = PSL2(8) inside a subgroup R(3). A forward appeal to Lemma 3.15 shows that Ree groups R(q0) with q0 = 3 do in fact act flag-transitively on polyhedra, and by [31], so does R(3)' = PSL2(8). The complete list of subgroups of PSL2(q0) is available, for instance, in [20]. As q0 is an odd power of 3, the group PSL2(q0) does not have subgroups isomorphic to A5, S4, or PGL2(q1) for some q1. Hence, none of the subgroups of PSL2 (q0), except for those isomorphic to a group PSL2(q1), with q1 an odd power of 3 dividing q0 (and hence q), admits flag-transitive actions on polyhedra. Now the maximal subgroups of C2 x PSL2(q0) consist of the factor PSL2(q0), as well as all subgroups of the form C2 x H where H is a maximal subgroup of PSL2 (q0) from the following list: Eq0 : Cqo-1 , Dq0-1, Dq0 + 1, PSL2(q1). D. Leemans et al.: Groups of Ree type in characteristic 3 acting on polytopes 217 A subgroup of C2 x PSL2(q0) of the form C2 x Dq0_i is isomorphic to D2(q0-1) (since q0 = 3 mod 8), so none of its subgroups (including the full subgroup itself) can act regularly on a non-degenerate polyhedron (that is a polyhedron with no 2 in the Schlafli symbol). Similarly, a subgroup of C2 x PSL2(q0) of the form C2 x Dq0+1 is isomorphic to C2 x C2 x D(q0+1)/2, so again none of its subgroups (including the full subgroup itself) can act regularly on a non-degenerate polyhedron. Finally, a subgroup of C2 x PSL2(q0) of the forms C2 x (Eq0 : C(q0-1)/2) has an order not divisible by 4. Hence, as in the two other cases, none of its subgroups (including the full subgroup itself) can act regularly on a non-degenerate polyhedron. In summary, the only possible candidates for rank 3 subgroups of R(q) are of the form R(qo), PSL2(qo), C x PSL2(qo), and R(3)' = PSL2(8). We can further rule out a subgroup of type R(3), since R(3) = PrL2 (8) is not generated by involutions. □ For a subgroup B of A we define (B) := (a | a e NA(B),a2 = 1). If B is generated by involutions then B < NO(B) < NA(B). We first state a lemma that will be useful in several places. Lemma 3.8. Let H := R(3) = PrL2(8), and let D := D2d be a dihedral subgroup of H of order at least 6. Then d =3, 7 or 9, and in all cases NH (D) = D. Proof. Straightforward. □ The following lemma considerably limits the ways in which Ree groups R(q) might be representable as C-groups of rank 4. Lemma 3.9. If the group G := R(q) can be represented as a string C-group of rank 4, then N° (G01 ) = NCG(p0)(Goi). (3.3) Proof. Suppose that G admits a representation as a string C-group of rank 4. Thus G = (P0,P1,P2,P3). Since R(3) is not generated by involutions, we must have q = 3. The subgroup G01 = (p2,p3) is a dihedral subgroup D2d (say) of the centralizer Cg(po) of p0, and CG(p0) = (p0) x PSL2(q). Here d > 3, by arguments similar to those used in the proof of Lemma 3.3. Thus D2d = (P2,P3) = G01 < G1 = (P0,P2,P3) < Cg(po) = C2 x PSL2(q). (3.4) By Lemma 3.6 applied to G01 and CG(p0), there exists a dihedral subgroup D in the PSL2(q)-factor of CG(p0) such that G01 is a subgroup of (p0) x D = C2 x D of index at most 2 and NCG(P0)(G01) = NCa (P0)(C2 x D) = C2 x Npsl2(,)(D). In fact, the proof of Lemma 3.6 shows that this subgroup C2 x D is just given by G1. But p0 e G01, so G01 has index 2 in C2 x D = G1, and D = G01 = D2d. Then Lemma 3.5, applied to D, shows that 2d must divide either q +1 or q - 1. The structure of the normalizer Ncg(p0)(G01) can be obtained from Lemma 3.6. In fact, Ncg(p0)(G01) = C2 x D2d, unless 2d | (q + 1) and (q + 1)/2d is even; in the latter 218 Ars Math. Contemp. 14 (2018) 433-443 case Ncg(po)(G01) = C2 x D4d. In particular, Ncg(po)(G01) is generated by involutions and its order is divisible by 4. We will show that the normalizer of G01 in CG(p0) captures all the information about the full normalizer NG (G01) of G01 in G that is relevant for us. A key step in the proof is the invariance of the structure of the normalizer of G01 in arbitrary subgroups of G of type C2 x PSL2 (q); more precisely, the structure only depends on d and q, not on the way in which G01 is embedded in a subgroup C2 x PSL2 (q) (see Lemma 3.6). The full normalizer NG(G01) of G01 in G must certainly contain Ncg(po)(G01) and also have an order divisible by 8. We claim that all involutions of the full normalizer NG(G01) must already lie in CG(p0) and hence in Ncg(po)(G01). First note that NG(G01) must certainly lie in a maximal subgroup M of G and then coincide with NM(G01). (Since G is simple, the normalizer of a proper subgroup of G cannot coincide with G.) Inspection of the list of maximal subgroups of G shows that only maximal subgroups M of type R(q0), C2 x PSL2(q) or NG(A1) have an order divisible by 4. Only those maximal subgroups could perhaps contain Ngg(po)(G01) and hence NG(G01). We investigate the three possibilities for M separately. Suppose M is a group of type C2 x PSL2(q). Then the invariance of the structure of the normalizer of G01 shows that NM(G01) = Ncg(po)(G01). However, Ncg(po)(G01) < Ng(Go1) and Ng(G01) = Nm(G01), so this gives Ng(Go1) = Ncg(p0)(G?01). But Ncg(po)(Go1) is generated by involutions, so Ngg(po)(Go1) = N£g(po)(Go1) and (3.3) must hold as well. Let M be a group of type NG(A1) = (Cf x D(q+1)/2) : C3 where A1 is a group C(,+1)/4 (recall that (q + 1)/4 is odd). Then all involutions of M must lie in its subgroup K := Cf x D(q+1)/2 = C2 x Dq+1. In particular, all involutions of NM(G01) must lie in K and hence in NK(G01); that is, NM(G01) < NK(G01). Also, G01 itself must lie in K and its order 2d must divide q +1. The subgroup K lies in the centralizer C of the involution generating the C2-factor in the direct product factorization C2 x Dq+1 for K, and NK(G01) < NC(G01). This subgroup C is of type C2 x PSL2(q), and so again the invariance of the structure of the normalizers implies that NC (G01) = Ncg(po)(G01). But Ng(G01) = Nm(G01) and therefore Ncg(po)(G01) = N^(po)(Go1) < NG(G01) = NM(G01) < Nk(G01) < NG(G01). Thus NG(G01) = N°g(po)(Go1), as required. Now let M be a Ree group R(q0) where (q0)p = q and p is a prime. We first cover the case when M is a Ree group R(3) = PSL2(8) : C3, that is, q = 3p where p is a prime. In that case, by Lemma 3.8, NG(G01) = NM(G01) = G01. Hence, since also G01 < N0G(po)(Go1) < NG(G01), we must have NG(Go1) < N°g(po)(Go1). Now suppose q0 = 3, so in particular M is simple. Then 2d must divide q0 ± 1, since Ngg(p0)(Go1) lies in M and therefore po € M, giving G01 < Ngm(po)(Go1) = C2 x PSL2(q0). Since the subgroup NG(G01) of M must have an order divisible by 4, it must lie in a maximal subgroup M' of M of type R(q1), C2 x PSL2(q1), or NR(q0)(A1) with A1 — C(q0+1)/4. The maximal subgroups M' of M = R(q0) of types C2 x PSL2(q0) and NR(q0)(A1), respectively, lie in maximal subgroups of G of type C2 xPSL2(q) or NG (A1), so they are subsumed under the previous discussion. (Alternatively we could dispose of these cases for M' directly, using arguments very similar to those in the two previous cases for M.) Then this leaves the possibility that M' is of type R(q1), in which case we are back at a Ree group. Now continuing in this fashion to smaller and smaller Ree subgroups that could perhaps contain NG(G01), we eventually arrive at either a Ree subgroup M(k) D. Leemans et al.: Groups of Ree type in characteristic 3 acting on polytopes 219 (say) whose parameter q(k) ± 1 (say) is no longer divisible by 2d, or a Ree group R(3). In the first case, R(q0) does not contribute anything new to NG(G01), and the normalizer Ng(G01) must already lie in one of the maximal subgroups of type C2 x PSL2(q) or Ng(Ai) discussed earlier; in particular, Nq(G0i) = NCG(po)(G01), as required. In the second case, the normalizer Ng(G01) lies in a Ree subgroup R(3) = PSL2(8) : C3, and its involutory part N0(G01) must lie in the PSL2(8) subgroup. Again, by Lemma 3.8, we have N0(G01) = NM(G01) = G01, hence (3.3) must also hold in this case. □ Lemma 3.10. If the group G := R(q) can be represented as a string C-group of rank 4, then q = 3 and both the facet stabilizer G3 and vertex stabilizer G0 have to be isomorphic to PSL2(8) = R(3)' (i.e. the commutator subgroup of R(3)) or a simple Ree group R(q0) with q = qm for some odd integer m. Proof. We consider the possible choices for G0 in the given C-group representation of G of rank 4. Our goal is to use Lemma 3.4 to limit the choices for G0 to just R(3)' or R(q0). First recall from Lemma 3.7 that the only possible candidates for G0 are either Ree subgroups R(q0) with q0 = 3 or subgroups of the form PSL2(q0), C2 x PSL2(q0), or R(3)' = PSL2 (8). To complete the proof we must eliminate the second and third types of candidates. This is accomplished by means of Lemmas 3.4 and 3.6, proving in each case that Ng(G01)\Ng(G0) cannot contain an involution, or equivalently N^ (G01) < Ng(G0). Bear in mind that G01 < G0. First observe that all subgroups of G of the form C2 x PSL2(q0) are self-normalized in G; and the normalizer of a subgroup of G of the form PSL2 (q0) is isomorphic to C2 x PSL2(q0). In other words, Ng(G0) = G0 if G0 is of type C2 x PSL2(q0), and Ng(G0) = C2 x G0 if G0 is of type PSL2^). We show that N° (G01) < Ng(G0) for each of these two choices of G0. Suppose that G0 = C2 x PSL2(q0). We first claim that then 2d | q0 ± 1 (where 2d = |G01|). To see this, note that the intersection of G01 with the PSL2(q0)-factor of G0 is a subgroup of index 1 or 2 in G01. If the index is 1, the statement is clear by Lemma 3.5, since then G01 lies in the PSL2(q0)-factor; and if the index is 2 and the intersection is a cyclic group Cd, the statement follows by inspection of the possible orders of cyclic subgroups of PSL2 (q0). Now if the index is 2 and the intersection is a dihedral group Dd, then Lemma 3.6 shows that d must be even, 2d | q +1, and d/2 must be odd; moreover, d | q0 + 1 since Dd lies in PSL2(q0), and hence 2d | q0 + 1 since q0 + 1 is divisible by 4. Thus 2d | q0 ± 1, as claimed. Now, since G0 = C2 x PSL2(q0), the normalizer Ngg(po)(G01) coincides with the normalizer NH(G01) of G01 taken in a suitable subgroup H of CG(p0) of type C2 x PSL2 (q0). In fact, from Lemma 3.6 we know that Ngg(po)(G0i) < C2 x Dq± 1 < Cg(P0) = C2 x PSL2(q). But 2d | q0 ± 1, so we must have Ncg(po)(G01) < C2 x Dqo±1. However, C2 x Dqo±1 lies in a subgroup H of CG(p0) isomorphic to C2 x PSL2 (q0). To complete the argument (for any given type of group G0) we show that N0 (G01) must lie in NG 0 (G01) and therefore also in G0 and NG(G0). When G0 is a group of type C2 x PSL2 (q0), the normalizer Ngo (G01) can be determined using Lemma 3.6 (with q replaced by q0). In fact, by the invariance of the normalizers of G01 we know that Ngo (G01) 220 Ars Math. Contemp. 14 (2018) 433-443 and NH(G01) are isomorphic and that both subgroups are generated by involutions. However, then by Lemma 3.9, Ngo (Goi) = NO o (G01) < NG (Goi) = NO G(po) (G01) = K (G01) = Nh (Goi), so clearly Noo (Goi) = Nh (Goi). Thus N° (G01) = Noo (Goi) < Go < NG (Go). Now let G0 beoftype PSL2(q0). Then C := No(Go) is a group of type C2 xPSL2 (q0) containing G0, so we can replace G0 by C and argue as before. In fact, using the same subgroup H, we see that the normalizers NO (G01) and NH(G01) are isomorphic subgroups generated by involutions. In particular, No(G01) = NO(G01) < NO(G01) = NO^)^) = NH(G01) = Nh(G01), and therefore No(G01) = Nh(G01). Hence NO(G01) = No(G01) < C = No(Go). □ Let us now show that Go ^ R(3)'. Lemma 3.11. If R(q) has a representation as a string C-group of rank 4 with G0 = R(3)', then q = 27. Proof. Suppose G := R(q) is represented as a string C-group of rank 4 with generators p0,..., p3. Then we know that G01 < G1 < CG(p0) = C2 x PSL2(q). The abstract regular polyhedra with automorphism group R(3)' = PSL2(8) are all known and are available, for instance, in [22]. There are seven examples, up to isomorphism, but not all can occur in the present context. In fact, the dihedral subgroup G01 of G0 must also lie CG(p0) = C2 x PSL2(q) and hence cannot be a subgroup D18. It follows that the polyhedron associated with G0 (that is, the vertex-figure of the polytope for G) must have Schlafli symbol {3,7}, {7,3}, or {7, 7}. We can further rule out the possibility that G01 = D6 by Lemmas 3.5 and 3.6, giving that C2 x PSL2 (q) has no dihedral subgroup of order 6. Hence G01 = D14. The fixed point set of every involution in G is a block of the corresponding Steiner system S(2, q + 1, q3 + 1), and vice versa, every block is the fixed point set of a unique involution. Hence, two involutions with two common fixed points must coincide, since their blocks of fixed points must coincide. Suppose B0 denotes the block of fixed points of p0. As p2 and p3 centralize p0, they stabilize B0 globally but not pointwise. However, p2 cannot have a fixed point among the q +1 points in B0, since otherwise two points of B0 would have to be fixed by p2 since q + 1 is even. Thus p2, and similarly p3, does not fix any point in B0. Moreover, in order for G01 = D14 to lie in a subgroup of G of type C2 x PSL2(q), we must have 7 | q +1 or 7 | q - 1. Using q = 32e+1 and working modulo 7 the latter possibility is easily seen to be impossible. On the other hand, the former possibility occurs precisely when e = 1 mod 3, and then 3 | 2e +1. Hence G must have subgroups isomorphic to R(27) = R(33). We claim that G itself is isomorphic to R(27), that is, q = 27. Now the subgroup G0 = R(3)' lies in a unique subgroup K = R(3) of G, namely its normalizer No(Go). Indeed, Figure 1 tells us that G0 = R(3)' is in a unique subgroup isomorphic to R(27) (because of the lower 1's on the edges joining the boxes). This subgroup K, in turn, lies in a unique subgroup H = R(27) of G. All Ree subgroups of G are self-normalized in G, so in particular K and H are self-normalized. Relative to the Ree subgroup H, the normalizer NH (C7) in H of the cyclic subgroup C7 of G01 is a maximal subgroup of type D. Leemans et al.: Groups of Ree type in characteristic 3 acting on polytopes 221 NH (Ai) = (Cl x D14) : C3 in H, which also contains G01 (see Section 2.2 or [7, p. 123]). Note here that this subgroup C7 is a 7-Sylow subgroup of both K and H, and is normalized by G01. Thus, NH(C7) = (C| x D14) : C3. We claim that NH(G01) = NH(C7). Clearly, Nh(G01) < Nh(C7). For the opposite inclusion observe that (C| x D14) : C3 has four subgroups isomorphic to D14, including G01. The subgroup G01 is normalized by the C3-factor, and the three others are permuted under conjugation by C3. This is due to the fact that if it were otherwise, the number of subgroups R(3)' containing G01 would not be an integer but 4/3. Hence, among these four subgroups only G01 is normal and can be thought of as the subgroup D14 occurring in the factorization of the semi-direct product. It follows that the subgroups Cf and C3 normalize G01. Thus Nh(G01) = Nh(C7) = (C| x D14) : C3. Figure 1 shows the sublattice of the subgroup lattice of G that is relevant to the current situation. Each box contains two pieces of information: a group that describes the abstract structure of the groups in the conjugacy class of subgroups of G depicted by the box, and a number in the lower left corner that gives the number of subgroups in the conjugacy class. This number is the order of G divided by the order of the normalizer in G of a representative subgroup of the conjugacy class. Two boxes are joined by an edge provided that the subgroups represented by the lower box are subgroups of some subgroups represented by the upper box. There are also two numbers on each edge. The number at the top gives the number of subgroups in the conjugacy class for the lower box that are contained in a given subgroup in the conjugacy class for the upper box. The number at the bottom similarly is the number of subgroups in the conjugacy class for the upper box that contain a given subgroup in the conjugacy class for the lower box. If we know the lengths of the conjugacy classes for the upper box and lower box, then knowing one of these two numbers on the connecting edge gives us the other. For instance, in Figure 1, if we know that there are 36 (conjugate) subgroups D14 in a given subgroup R(3)', then there are |G| |G| |R(3)| . / |22 . 3 .14| (conjugate) subgroups R(3)' containing a given subgroup D14. Returning to our line of argument, as already pointed out above, Figure 1 tells us that G0 = R(3)' is in a unique subgroup isomorphic to R(27), namely H (because of the lower 1's on the edges joining the boxes). It also shows that G01 is contained in a unique subgroup (Cf x D14) : C3, which, in turn, is contained in a unique R(27), namely H. As we saw above, this subgroup (Cf x D14) : C3 is necessarily the normalizer NH(G01) of G01 in H. Moreover, p0 has to lie in this unique subgroup (Cf x D14) : C3, which itself is a subgroup of H, and therefore (p0, G0) < H. This holds because Ng(G01) = NH(G01). That these normalizers coincide can be seen as follows. Clearly, NH(G01) < Ng(G01). Now for the opposite inclusion observe that for g G Ng(G01) we have G01 = gG01g-1 < gNH(G01)g-1 and (trivially) G01 < NH(G01). But then Figure 1 shows that a subgroup D14 of H must lie in a unique conjugate of (Cf x D14) : C3 = NH(G01), so necessarily gNH(G01)g-1 = Nh(G01). Similarly, since2 Nh(G01) < H and hence Nh(G01) = gNH (G01)g-1 < gHg-1, Figure 1 (at box R(27)) gives gHg-1 = H, so g G H since H is self-normalized. Thus G = (p0, G0) = H = R(27). □ 222 Ars Math. Contemp. 14 (2018) 433-443 Figure 1: A sublattice of the subgroup lattice of R(q). Lemma 3.12. The group R(27) cannot be represented as a string C-group of rank 4. Proof. Let G = R(27). By the previous lemmas we may assume that G0 = G3 = PSL2(8). In all other cases we know that G cannot be represented as a rank 4 string C-group. Moreover, from the proof of the previous lemma we already know that G01 = D14 and Ng(G01) = (Cf x D14) : C3. As there is a unique conjugacy class of subgroups R(3)' in R(27), and there is also a unique conjugacy class of subgroups D14 in R(3)', the choice of p2, p3 is therefore unique up to conjugacy in R(27). Once p2, p3 have been chosen, there are three candidates for p0, namely the elements of the subgroup Cf that centralizes D14, and these are equivalent under conjugacy by C3. Hence there is a unique choice for {p0, p2, p3} up to conjugacy. By similar arguments we also know that G3 = R(3)' and G23 = D14, and that the pair (G23, G3) is related to (G01, G0) by conjugacy in R(27). Hence there must exist an element g G R(27) such that • Po = P3 pf = p^ P3 = P0, or g g g • p0 = P3, p2 = po, p3 = P1. The second case can be reduced to the first, as the centraliser of p0 contains an element that swaps p2 and p3 (any two involutions in D14 are conjugate). Hence, we may assume without loss of generality that g swaps p0 and p3. In particular, (p0, p3) is an elementary abelian group of order 4 normalized by g. All such subgroups are known to be conjugate and have as normalizer a group (Cf x D14) : C3. In this group, there is no element that will swap p0 and p3 under conjugation. All elements that will conjugate p0 to p3 will necessarily conjugate p3 to p0p3. Hence we have a contradiction. □ D. Leemans et al.: Groups of Ree type in characteristic 3 acting on polytopes 223 We therefore know that if a string C-group representation of rank 4 exists for R(q), both G0 and G3 must be subgroups of Ree type. Thus from now on we can assume G0 = R(q0) with q0 > 3. In a Ree group, the dihedral subgroups D2n are such that n must divide one of 9, q - 1, q + 1, aq := q + 1 - 3e+1, Pq := q + 1 + 3e+1. Note that aqPq = q2 - q + 1, so in particular if H is a Ree subgroup R(q0) of G then similarly aq0 Pq0 = q2 - q0 + 1. Lemma 3.13. Let G = R(q) with q = 32e+1 and (p0, p^ p2, p3) be a string C-group representation of rank 4 of G. Then (1) G01 is a dihedral subgroup D2d with d a divisor of q +1 and of either aq0 or Pq0 for some q0 such that q = q™ with m odd (where q0 is determined by G0 = R(q0)); (2) m = 3, and G0 and G3 are conjugate Ree subgroups R(q0) with q = q^; (3) G03 is a dihedral subgroup D2t with t a divisor of aq0 or Pq0. Proof. (1) By Lemmas 3.10, 3.11 and 3.12, we may assume that G0 is a simple Ree subgroup of G. Let G0 be a Ree subgroup R(q0), with q0 = 3 such that q™ = q with m a positive odd integer and let G01 = D2d. As G01 < CG(p0) we have that 2d | q ± 1 by Lemma 3.5. In order to have involutions in Ng(G01)\Ng(G0), the only possibility is that NG (G01) (of order divisible by 4) lies in a maximal subgroup of type NG(A1) but not in a maximal subgroup NG0 (Cq0+i) of G0; for otherwise, the same techniques as in 4 Lemma 3.10 show that there is no involution in Ng(G01)\Ng(G0). Hence 2d divides q + 1. Observe that Ng(A ) = (C22 x D q+i) : C3 has exactly four subgroups D q+i because of the subgroup C|. These four subgroups are not all normalised by the C3 because of the semi-direct product. Hence the C3 must conjugate three of them and normalise the fourth one. Similarly, in R(q0) there are four subgroups Dq0+1 in each NR(q0)(A1) and it is obvious that NR(q0)(D2d) = NR(q) (D2d) for every divisor d of q0 + 1. Hence, in order to find some involutions in NG(G01)\Ng(G0), we need to have that 2d does not divide q0 + 1. Moreover, since q0 - 1 divides q - 1, we have also that (q0 - 1, q + 1) = 2. That forces d not to be a divisor of q0 - 1 as d > 2. Hence, looking at the list of maximal subgroups of R(q0) we can conclude that d is a divisor of either aq0 or Pq0 in order for D2d to be a dihedral subgroup of R(q0). (2) Observe that q3 + 1 = (q0 + 1)aq0Pq0 divides q3 + 1. Let us first show that 2e +1 must be divisible by 3 in order for d to satisfy (1). Suppose (3,2e +1) = 1. Then q0 = 32f +1 with 2e + 1 = m(2f + 1) and (3, m) = 1. Let p be an odd prime dividing (aq0 Pq0, q + 1) but not dividing q0 + 1. Then p divides (q3 + 1, q + 1) and hence p divides (q6 -1, q2m -1) = q2(3,m) -1 = q02 -1 = (q0 + 1)(q0 -1) and hence also q0-1. Asp divides q+1,and q0-1 divides q-1, and since (q-1, q+1) = 2, we have that p | 2, a contradiction. Hence m must be divisible by 3 and so does 2e +1. Suppose m = 3. Then m = 3m' and given a Ree subgroup R(q0) of R(q) with q™ = q, there exists a Ree subgroup R(q3) such that R(q0) < R(q3) < R(q). Using similar arguments as in the proof of Lemma 3.11, it is easy to show that, since aq0Pq0 divides 224 Ars Math. Contemp. 14 (2018) 433-443 q3 + 1, we must have (p0, pi, p3) = R(q3) and therefore m = 3. Indeed, as we stated in (1), )(D2d) = Nfl(q)(D2d) for every divisor d of qg + 1. Hence po € R(q3). This implies that m = 3 and G0 = R(q0) with q3 = q. Dually, G3 = R(q0). As all subgroups R(q0) are conjugate in R(q), we have that G0 and G3 are conjugate. (3) is due to the fact that G0 n G3 = G03 and that, by (2), G0 and G3 are conjugate in G. Hence, NG (G03) \ G0 has to be nonempty and G03 must not be contained in a subgroup H of G0 such that NG(H) > Ng(G03), for if such a subgroup H exists, then G0 n G3 > H. If t divides 9 or one of q0 ± 1, this does not happen. Hence t divides one of aqo or pqo. □ Lemma 3.14. The small Ree groups have no string C-group representation of rank 4. Proof. Suppose G is a Ree group that has a string C-group representation of rank 4. By Lemma 3.10 and part (2) of Lemma 3.13 we may assume that G := R(q) where q = q3 with q0 = 3m for an odd integer m. Moreover, G0 and G3 are conjugate simple Ree subgroups isomorphic to R(q0). By part (3) of Lemma 3.13, if G03 = D2t then t must be a divisor of either aqo or pqo, and since q = q3, we also have q +1 = (q0 + 1)(q2 - q0 + 1) = (q0 + 1)aqoPq0. Thus t is also a divisor of q +1. We claim that then G0 n G3 > G03, which gives a contradiction to the intersection property. Indeed, since G03 lies in a subgroup H := Ct : C6 of G0, and the normaliser of G03 is not contained in G0 (for otherwise, D2t would have to lie in a unique subgroup R(q0), whereas already G0 and G3 give two examples of such subgroups, by the previous lemma), we have Ng(G03) = (C2 x C2 x D2t) : C3. This group contains H = Ct : C6 = D2t : C3 as a normal subgroup, and G03 is normal in H. We also have that NG(H) = Ng(G03). But then, as G03 is normal in H, any subgroup R(q0) containing G03 must contain H. In particular this applies to G3. Thus G0 n G3 > H > G03, and the intersection property fails. □ 3.4 String C-groups of rank 3 It remains to investigate the possibility of representing R(q) as a string C-group of rank 3. Nuzhin already showed in [27] that there exist triples of involutions, two of which commute, that generate R(q) for every q. This completes the proof of Theorem 1.1. However, we decided to give here another way to construct an example of a rank three regular poly-tope for R(q) for the paper to be self-contained. Lemma 3.15. Let G = R(q), with q = 3 an odd power of 3. Then there exists a triple of involutions S := {p0, pi, p2} in G such that (G, S) is a string C-group. Proof. Recall that the fixed point set of an involution in G is a block of the Steiner system S := S(2, q + 1, q3 + 1).Pick two involutions p0, pi from a maximal subgroup M of G of type NG(A3) such that p0pi has order q + 1 + 3e+i, and let B0, Bi, respectively, denote their blocks of fixed points. Obviously, B0 n Bi = 0, for otherwise (p0, pi) would lie in the stabilizer of a point in B0 n Bi, which is not possible because of the order of p0pi. Recall here that the point stabilizers are maximal subgroups of the form NG(A) = A : Cq-i, where A is a 3-Sylow subgroup of G. Now choose an involution p2 in CG(p0) distinct from p0 such that its block of fixed points B2 meets Bi in a point. Such a p2 exists as all involutions of CG(p0) have pairwise disjoint blocks of size q + 1 and therefore they cover all of the q3 + 1 points. Then Bi n B2 must consist of a single point p (say), and D. Leemans et al.: Groups of Ree type in characteristic 3 acting on polytopes 225 B0 n B2 = 0 since the stabilizer of a point does not contain Klein 4-groups. Then (pi, p2) lies in the point stabilizer of p, and hence must a dihedral group D2n, with n a power of 3. As (p0, pi) is a subgroup of index 3 in M, and p0 does not belong to M, we see that (po, pi, p2) = G. 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