# «Series A 29,212-219 (1980) JOURNAL OF COMBINATORIAL THEORY, Generalized Quadrangles Associated with G*(q)* WILLIAM M. KANTOR of Oregon, Department of ...»

Series A 29,212-219 (1980)

## JOURNAL OF COMBINATORIAL THEORY,

Generalized Quadrangles Associated with G*(q)*

WILLIAM M. KANTOR

of Oregon,

Department of Mathematics, University Eugene, Oregon 97403

Received January 1, 1979

If q 3 2 (mod 3), a generalized quadrangle with parameters q, q2 is constructed

from the generalized hexagon associated with the group G,(q).

1. INTRODUCTIVE In addition to the generalized quadrangles arising naturally from smalldimensional symplectic, unitary, or orthogonal geometries, the only known finite examples are the following quadrangles and their duals: ones with parameters 4, 4 (where 4 = 2e 8) or q*, 4 (where q = 22e+l 8) due to Tits [4, p. 3041; and others with parameters q - 1, q + 1 (for prime powers q) due to Hall [5] and Ahrens and Szekeres [l]. In this note we will discuss a procedure for constructing all but the last of the above examples, and others as well. In particular, we will prove the following result.

1. If q is a prime power such that q = 2 (mod 3) and q 2, THEOREM then there exists a general quadrangle with parameters q, q2 not isomorphic to any of the aforementioned ones.

The only surprising feature of these quadrangles is that they arise from the groups G*(q), which are themselves associated with generalized hexagons.

The automorphism group of each quadrangle is isomorphic to the stabilizer in Aut G,(q) of a line of the corresponding hexagon.

The precise relationship between the generalized quadrangles and hexagons is given in Section 2. In view of the restrictions forced on q, there does not seem to be any geometric proof of the theorem (cf. (#) in Section 2).

The algebraic proof of the theorem occupies Sections 3-5. In particular, Section 3 contains an elementary construction procedure. Analogous procedures can be easily obtained for generalized hexagons and octagons, * This research was supported in part by NSF Grant MCS 76-07268.

0097-3165/80/050212-07$02.00/0 Copyright C 1980 by Academic Press, Inc.

All rights of reproduction in any form reserved.

WITH G,(q)

## GENERALIZED QUADRANGLES ASSOCIATED

although they involve many more axioms. However, we have not yet been able to use them to construct any examples other than the known ones.2. GEOMETRIC DESCRIPTION Suppose that 4 = 2 (mod 3), and fix a line L of the generalized hexagon 8 associated with G,(q). (There are two dual choices for Z. The relevant one has L a line of an O(7,q) geometry; cf. Tits [6].) We use the metric defined on the union of the sets of points and lines of R.

Now define Points and Lines as follows.

Points: points of L; lines at distance 4 from L.

Lines : L; lines at distance 6 from L; points at distance 3 from L.

A Point of L is defined as being on L and all Lines at distance 2 from it.

A Point not on L is on all Lines at distance 1 or 2 from it.

This defines a generalized quadrangle with parameters 4, 4’. When q = 2, it is the unique quadrangle with these parameters. For 4 2, it is the quadrangle mentioned in Theorem 1.

Remark, Starting with a line L of a generalized hexagon with parameters s, s, an incidence structure of Points and Lines can be defined as above. This produces a generalized quadrangle with parameters s,.s2, provided that the following condition (#) holds in the hexagon.

(#) There is no configuration of 12 points and 10 lines as in the figure (where L has been drawn as a circle for reasons of symmetry).

It seems likely that (54~)characterizes the G,(q) hexagons, where q zc 2 (mod 3).

The proof is just a straightforward check.

All classical examples (and their duals) arise from Theorem 2; so do Tits’ examples. Before discussing this, we will need some notation concerning automorphism groups.

Consider any generalized quadrangle % with parameters s, t. If L is a line, let U, denote the group of a11 automorphisms fixing each line meeting L.

Then ( U, ( s (since U, acts semiregularly on the points outside L of each such line).

If x is a point, U, is defined in a dual manner.

For distinct intersecting lines L, M, let U,, denote the group of all automorphisms fixing every point of L, every point of M, and every line on L f7 M. Then [ U,, ( t. (An element of U,, fixing a line not on L r7 M must fix pointwise a subquadrangle with parameters s, t, and hence must be 1.1 We now turn to examples of Theorem 2.

Z(Q,X) fixes the line R and acts on the group Q generated by the groups UrAl. Certainly, Q is its own centralizer in Aut Dz(Q, F). Also, R produces an inversive plane from which the vector space Q can be reconstructed 14, pp. 265-2681. Thus, Aut -2(Q, X) is a semidirect product of Q with the subgroup of TL(4, q) stabilizing.F.

The Sp(4, q) and SU(4, q) quadrangles arose in Examples 1 Remark.

and 2. The 0(5, q), O-(6, q), SW, d, and dual SU(5, q) quadrangles all arise as examples of Theorem 2. A description of the O-(6, q) quadrangle will be of use in Theorem 1.

3. s = q, t = q*. The O-(6, q) quadrangle can be obtained as EXAMPLE follows. Let Q consist of all triples (a, c,/?) with a, p E GF(q*) and c E GF(q), and define

(where fl= p4 and Tr y = y + p). Then Q is a group of order q’, with center consisting of all (0, c, 0), c E GF(q). Let F consist of the groups and A(t) = ((a, a&t, at) } a E GF(q2)j for Ata) = {CO,%P) I P E GFtq’)l t E GF(q). If A E F, write A* = AZ(Q). Then Theorem 2 applies.

If a and /I are restricted to GF(q), the result is a subgroup R of Q of order q3. The groups A n R and A* n R produce a subquadrangle with parameters q, q, namely, the 0(5, q) quadrangle.

The maps (a,c,/3)-+ (a, c-afib, bE GF(q), and (a,~,@)-+,B-ab), (-P, c - Tr iEp, a) generate a subgroup of Aut Q isomorphic to SL(2, q).

They map Y to itself (inducing A(t) + A(t t b) and A(t) --f A(--l/t), respectively). Note that this SL(2, q) normalizes R. (In fact, the SL(2, q) normalizes q + 1 subgroups of Q of order q3). Let S’ denote the commutator subgroup of a group S, and write S” = (S’)‘. Then (*) If q 3 and S is the stabiliier of the line F of the O-(6, q) quadrangle, then S” E Q ~1SL(2, q) with 1Q ) = q’, and S” has a normai subgroup of order q3.

For further information concerning groups Q arising as above, as weli ar their normalizers, see [3, Sect. 31.

**Let q be a prime power. Since the generalized quadrangle allegedl:**

constructed in Section 2 has the stabilizer of L in G,(q) acting on it, it wil be necessary to study that stabilizer. The required information is o 216 WILLIAM M. KANTOR pp. 244-245 of Tits [7]. However, we will make this somewhat more explicit in order to facilitate later calculations.

Let Q consist of all quintuples (a, p, y, S, E) E GE;($), this time with GF(q)’ the operation (a, P, Y?4 &)(a’, p’, Y’, 6’5 &‘I = (a + a’, p + p’, y + y’ + a’& - 38’6, 6 + 8, & + E’).

Let xi denote the element with ith coordinate x and all others 0, and let Xi denote the set of all such elements xi. Then Xi g GF(q)’ via x -+ xi. Also, Q = X,X,x,x,X, and X3 = Q’ Z(Q). If 3114 (which is the situation we will eventually have), then Q’ = Z(Q).

**Define the following functions on Q:**

X6 : (a, P, Y, 4 &) --f (a, P + ax, y - 3p2x - 3a;Ox2 - a2x3, 6 + 2px + ux2, E + 36x + 3/3x2 + ax3), j: (a, P, y, 6, s) -+ (6 -4 y - a& + 3@, P, -a, h a,b : (a, /I, y, 6, E) -+ (ab3, pub’, p3b3, da%, m3), s, : (a, P, 7, 4 ~1--t (aa, Pa, w2, au, &a).

Each of these is in Aut Q. Let X, = (x, 1x E GF(q)}. Then X, z Gl;(q)+, and (X,,j) z SL(2, q). Note that Aut GF(q) acts on Q componentwise. Let G* be the subgroup of Aut Q generated by X,, j, and all automorphisms h,,, and s,, and set G = (G*, Aut GF(q)). Then QG’ is the stabilizer in G,(q) of the line L appearing in Section 2, while QG is the stabilizer of L in Aut G,(q). Moreover, G* z GL(2, q), while G g TL(2, q).

All of these properties of G may be verified by direct computation. The action of G* on Q/X, is given on p. 497 of [2], so only the X3 component needs to be checked.

However, a direct approach not assuming prior familiarity with G,(q) is as follows. In the next section we will define a family j3 of q + 1 subgroups of Q. It is straightforward to check that G sends jr to itself; in fact, G is the stabilizer of.F in Aut Q if q 2 (cf. Section 5, Remark 2).

Note that A (0) = X,X,, Let.Y = {A (co), A (ti) ] t E Gl;(q) i. Then G maps F to itself, and acts on it as it does on the projective line. For example, x6 induces the map A(t) --t A(t + x) and j induces A(t) -+ A(-i/t), for tE {oo}UGF(q).

We must verify the axioms of Section 3. Axiom (i) is obvious. For the remaining ones, we can use G in order to assume that A = A(O), B = A(w), and C = A(l). Then (ii) is obvious. A typical element of A(1) has the form (a, a + p, -a* - 3a/? - 3p2, a + 2& a + 3/3).

The requirement AB r? C = 1 then states that -a* - 3a,8 - 3/?’ f 0 unless a = p = 0. ‘Thus, we must require that the polynomial (x J,- l)* $ (x $ 1) + 1 is irreducible.

It remains to show that the resulting quadrangle is new if q 2. First of all, the duals of the quadrangles constructed in Section 3, Example 2, using nonquadric avoids do not contain a group such as QG in their automorphism groups.

Now note that (QG)” g Q x SL(2, q), where SL(2, q) acts irreducibly on Q/Z(Q). Thus, (QG)” has no normal subgroup of order q3. In view of (*) in Section 3, this completes the proof of Theorem 1.

** Remark 1. It is well known that (up to isomorphism) there is only one generalized quadrangle with s = 2 and t = 4.**

Mere, S/Q s S, x S, ~ Remark 2. We will show that, if q 2, then Aut %(Q,,9-) = QG z Q x TL(2, q).

First of all, UY,A* &A, while ] A ] = q ( UF,A*I. Thus, vF,A* = A, so that Q is generated by those groups U,,L such that L is a line meeting the line.jionce. Also, U, = Z(Q).

Let J denote the stabilizer of the line fl in I = Aut 9(Q,.Y-), and let J, denote the stabilizer of the line 1 in J. By the preceding paragraph, Q U J* Also, J = QJ,, J, G and J, is the stabilizer of ST in Aut Q. It is easy to check that (for q 2) G already contains all element of J, inducing GF(q)semilinear transformations on Q = Q/Z(Q).

In order to show that J = Q M G, it thus’ suflices to prove the purely group theoretic fact that Aut Q acts G$‘(q)-semilinearly on Q. If u E Q, let u* denote ‘the preimage in Q of the l-space of Q spanned by !i = uZ(Q), Note that ($5) = fu, v] defines a nonsingular alternating GF(q)-bilinear form on p9 if q is even, @-+ u2 defines a quadratic form on Q, associated with ( ) ), Thus, 8 is equipped with a symplectic or orthogonal geometry. If [u, ir] = 1 then [u*, u*] = 1. Thus, the maximal elementary abelian subgroups of Q art preimages of the totally isotropic (or singular) 2-spaces of 0. Consequently Aut Q acts on the aforementioned geometry, and hence acts G&(q) semilinearly, as asserted.

218 WILLIAM M. KANTOR Finally, we must show that I= J. If I J, then I has an element moving jT to A* =A*(co). In particular, ( U,, I= 4. Here, U,, J, = G = TL(2, q), so that U, * n X, # 1. However, if x # 0 then xs fixes only 4 + 1 lines on the point A (namely, the lines A* and yZ, y E GF(q)). This contradiction completes the proof that I = J = Q M G.

** Remark 3. In view of Remark 2, it is straightforward to reconstruct the G,(q) hexagon from the quadrangle.**

Z(Q, X).

** Remark 4. The groups Q used in Theorem I and Example 3 are isomorphic.**

To see this, start with the group Q in Example 3, write a = a, + a,8 with ai E GF(q) and a suitable, fixed 19in GF(q’), and compute.

When q is even, the groups A*/Z(Q) arising in Theorem 1 and Example 3 form a regulus of Q/Z(Q).

** Remark 5. G2(q) has a class of subgroups S = SU(3, q), each of which has an orbit of q3 + 1 lines.**

In the context of Section 2, let L be one of those lines. Then the remaining q3 lines are Lines of the quadrangle. This yields a family of q3 + 1 Lines which partition the Points of the quadrangle. Note that S n Q has order q3 and is transitive on the aforementioned set of q3 Lines.

The lines of the unital for S correspond to reguli of the underlying O(7, q) geometry, but do not seem to have a nice interpretation in the quadrangle.

** Remark 6. There do not appear to be subquadrangles with parameters q, q arising in the following manner: for some subgroup R of Q of order q3, the family {A n R (A E X} satisfies the conditions of Theorem 2.**

Equivalently, there is probably no subgroup R of order q3 such that the points [A] and Ar (A E: ;T, r E R) are the points of a subquadrangle. This should be compared to the situation in Example 3.

If q is an odd prime, an elementary computation reveals that no such R exists.

** Remark 7. Infinite analogs of our constructions obviously exist.**

If K is a field, then G,(K) produces a generalized quadrangle as in Section 2 precisely when K does not have characteristic 3 and the map x + x3, x E K, is bijective.

** Remark 8. It would be interesting to have a geometric relationship between the generalized quadrangles constructed here and the translation planes discussed in [2, Theorem 21.**

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