Orbits on k-subsets of 2-transitive Simple Lie-type Groups. Bradley, Paul and Rowley, Peter. MIMS EPrint:

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1 Orbits on k-subsets of 2-transitive Simple Lie-type Groups Braley, Paul an Rowley, Peter 2014 MIMS EPrint: Manchester Institute for Mathematical Sciences School of Mathematics The University of Manchester Reports available from: An by contacting: The MIMS Secretary School of Mathematics The University of Manchester Manchester, M13 9PL, UK ISSN

2 Orbits on k-subsets of 2-transitive Simple Lie-type Groups Paul Braley an Peter Rowley August 7, 2014 Abstract For a finite rank one simple Lie-type group acting 2-transitively on a set Ω an k N we erive formulae for the number of G-orbits on the set of all k-subsets of Ω. 1 Introuction Suppose G is a permutation group acting upon a set Ω. Then G has an inuce action upon P(Ω), the power set of Ω. For k N, let P k (Ω) enote all the k-subsets of Ω - clearly G also has an inuce action upon P k (Ω). In this paper we shall assume G an Ω are finite. Let σ k (G, Ω) enote the number of G-orbits on P k (Ω). Questions as to the behaviour of σ k (G, Ω) as k varies an the relationship between G-orbit lengths for various k arise naturally. A contribution to the former type of question is given by the venerable theorem of Livingstone an Wagner [12] which states that for k N with k Ω /2, σ k 1 (G, Ω) σ k (G, Ω). Generalizations of this theorem have been obtaine by Mnukhin an Siemons [14]. While Buny an Hart [3] have consiere the situation when σ k 1 (G, Ω) = σ k (G, Ω). Results on G-orbit lengths are somewhat patchy at present - see Siemons an Wagner [18], [19] an Mnukhin [15]. For the recor, we remark there is a consierable literature concerning the infinite case; for a small selection see Cameron [4]. The main aim here is to establish formulae for σ k (G, Ω) when G is a rank one simple Lie-type group acting 2-transitively on Ω. Thus set q = p a where 1

3 p is a prime an a N. Then the possibilities for G are L 2 (q) (q > 3), the 2-imensional projective special linear groups, Sz(q) (q = 2 2n+1 > 2), the Suzuki groups, U 3 (q) (q > 2), the 3-imensional projective special unitary groups an R(q) (q = 3 2n+1 > 3), the Ree groups. The corresponing sets Ω are the projective line (with Ω = q + 1), the Suzuki oval (with Ω = q 2 + 1), the isotropic 1-spaces of a 3-imensional unitary space (with Ω = q 3 + 1) an the Steiner system S(2, q + 1, q 3 + 1) (with Ω = q 3 + 1). Before stating our main results we must introuce some notation. For b, c N, we use (b, c) to enote the greatest common ivisor of b an c. For l N we let D(l) = {n N n ivies l} an D (l) = D(l) \ {1}. Euler s phi function φ (see [17]) will feature in our results. Our final piece of notation concerns partitions. Let n N, an let π = λ 1 λ 2...λ r where λ 1 λ 2... λ r an r i=1 λ i = n. Though we will frequently use the more compresse notation π = µ a 1 1 µ a where µ 1 < µ 2 <... (a i being the multiplicity of µ i in the partition π). For k N, η k (π) is efine to be the number of subsequences λ i1, λ i2,..., λ is of λ 1,..., λ r which form a partition of k. Since G acts 2-transitively on Ω, we have σ 1 (G, Ω) = σ 2 (G, Ω) = 1, so we shall assume k 3. Now we come to our first theorem. Theorem 1.1. Suppose that G = L 2 (q) (q > 3) acts upon the projective line Ω, an let k N with k 3. Set = (2, q 1). Then σ k (G, Ω) = + + q(q + 1)(q 1) η k(1 q+1 ) + q η k(1 1 p q p ) ) (m q+1 m 2(q + 1) 2(q 1) φ(m)η k m D ( q+1 ) φ(m)η k m D ( q 1 ) ) (1 2 m q 1 m. In [5] Cameron, Maimani, Omii an Teyfeh-Razaie give methos for calculating σ k (L 2 (q), Ω), where Ω is the projective line an q 3 mo 4, an Cameron, Omii an Teyfeh-Razaie [6] follow a similar approach for σ k (P GL(2, q), Ω), their interest being motivate by a search for 3-esigns. Further work is presente by Liu, Tang an Wu [11] an also Chen an Liu [7], in the case when q 1 mo 4. 2

4 Theorem 1.2. Suppose G = Sz(q) (q = 2 2n+1 > 2, n N) acts upon the Suzuki oval Ω. Let r N be such that r 2 = 2q, an let k N with k 3. Then σ(g, Ω) = 1 q 2 (q 1)(q 2 + 1) η k(1 q2 +1 ) + 1 q η k(1 1 2 q q η k(1 1 4 q2 4 ) + 1 2(q 1) 1 + 4(q + r + 1) 1 + 4(q r + 1) m D (q+r+1) m D (q r+1) m D (q 1) 2 2 ) φ(m)η k (m q2 +1 m ) φ(m)η k (1 2 m q2 1 m ) φ(m)η k (m q2 +1 m ). The corresponing result for U 3 (q) is more complicate than for L 2 (q) an Sz(q) - see Definitions 3.1, 3.2, 3.3 an 3.4 for an explanation of the notation in the next theorem. Theorem 1.3. Suppose G = U 3 (q) (q > 2) acts upon Ω, the set of isotropic points of a 3-imensional unitary space. Let k N with k 3, an set = (3, q + 1) an l = q+1. Then σ k (G, Ω) = (η k (π 1 ) + µ k ) q 3 (q 3 + 1)(q 2 1) + q(q + 1)(q 2 1) + q(q + 1)(p 1) + 2(q 2 1) (q + 1) + 3(q 3 + 1) m D( q2 1 ) m D(l) m=pj j D (l) m D ( q2 q+1 ) φ(m)η k (π (m) φ(m)η k (π (m) 7 ) 5 ) φ(m)η k (π (m) 8 ). m D (l) φ(m)η k (π (m) 4 ) + σ k(e0, Ω) 6(q + 1) 2 Lemma 3.9 gives formulae for σ k (E 0, Ω). For corresponing results on the Ree groups see [1]. The proofs of Theorems 1.1, 1.2 an 1.3 all rely upon the orbit counting result commonly referre to as Burnsie s Lemma, though this result can be trace back to earlier work of Cauchy an Frobenius. Thus our main task 3

5 here is to enumerate the cycle types of elements in G in their action on the various Ω. For G = L 2 (q) or Sz(q) this is straightforwar, particularly as G may be partitione by certain subgroups (see Lemmas 2.5 an 2.6). For G = U 3 (q) we make use of the escription of conjugacy classes obtaine in [20]. There the conjugacy classes are ivie into a number of types those of the same type have the same size an fix the same number of points of Ω, Ω being the set of isotropic 1-spaces of a 3-imensional unitary space (see Table 2). Using information about the subgroup structure of G, in Lemma 3.5 we etermine the cycle structures on Ω of elements in all conjugacy class types except for C 6 C 6. This is relatively straightforwar. However C 6 C 6 is a ifferent kettle of fish see Table 1 which etails the varie cycle structures for such elements in the case of U 3 (71) (so Ω = 357, 912). Cycle Type of Number of Cycle Type of Number of Conjugacy Class Classes Conjugacy Class Classes Representative Representative Table 1: Cycle types for C 6 C 6 class types in U 3 (71) Representatives for all classes of type C 6 C 6 are to be foun in an abelian subgroup E of G where E is isomorphic to a irect prouct of cyclic groups of orer (q +1)/ an q +1 ( = (3, q +1)). The key result in enumerating these possible cycle types (an their multiplicities) is Lemma 3.8. With this result to han, Lemma 3.9 then etermines the contribution of the C 6 C 6 conjugacy classes to the sum g G fix Ω k (g) (where Ω k = P k (Ω)). Combining Lemmas 3.5 an 3.9 yiels Theorem 1.3. It is of interest to examine the number of orbits for these groups on 4

6 subsets of size 3 (an in the case of L 2 (q), size 4), of their respective G-sets. The following corollaries can all be erive by appropriate substitutions into Theorems 1.1, 1.2, 1.3 an the corresponing result for the Ree groups as foun in [1]. For full etails of the proofs see [1]. Corollary 1.4. Let q = p a > 2 where p is a prime an let Ω be the projective line with q + 1 points. Then { 2, if q 1 mo 4 σ 3 (L 2 (q), Ω) = 1, if q 3 mo 4. Corollary 1.5. Let L n = L 2 (2 n ) acting on the projective plane Ω n, an put a n = σ 4 (L n, Ω n ). Setting a 1 = a 2 = 1, for n 3 we have a n = a n 1 + 2a n 2. Corollary 1.6. Let q = 2 2n+1 an Ω n be the Suzuki oval with q points. Set a n = σ 3 (Sz(q), Ω n ). Then a n = 4n Corollary 1.7. Let q = 3 n an Ω n be the q isotropic points of a unitary 3-space. Set a n = σ 3 (U 3 (q), Ω n ). Then a n = 3n Corollary 1.8. Let q = 3 2n+1 an Ω n be the Steiner system on q points. Set a n = σ 3 (R(q), Ω n ). Then a n = (32n+1 + 3) 2. 6 Two of the sequences foun are of interest more generally, the sequence {a n } appearing in Corollory 1.5 is known as the Jacobsthal sequence [21], an the sequence {a n } given in Corollary 1.7 is associate to Sierpinski s Triangle, see [22]. 2 Backgroun Results First we recall some backgroun results, starting with the frequently misattribute Burnsie s Lemma (see [16]). 5

7 Lemma 2.1. Let H be a finite group an Ω a finite H-set. If t is the number of H-orbits on Ω, then t = 1 fix Ω (h). H h H For a partition of n, π = λ 1 λ 2...λ r (such that λ 1 λ 2... λ r an r i=1 λ i = n) we recall from Section 1 that, for k N, η k (π) is the number of subsequences λ i1, λ i2,..., λ is of λ 1,..., λ r which form a partition of k. As an example consier π = (= in compresse form), a partition of n = 14. Then η 5 (π) = 6, η 6 (π) = 3, η 7 (π) = 0 an η 8 (π) = 3. Our interest in η k (π) is because of the following two simple lemmas. Lemma 2.2. Suppose g Sym(Λ) where Λ = n, an let k N. If g has cycle type π on Λ (viewe as a partition of n), then g fixes (set-wise) η k (π) k-subsets of Λ. Lemma 2.3. Let π = µ a 1 1 µ a µ as s Then η k (π) = (k 1,k 2,...,k s) be a partition of n (in compresse form). ( )( ) ( ) a1 a2 as..., running over all s-tuples (k 1, k 2,..., k s ) with k i 0 an µ 1 k 1 +µ 2 k 2...+µ s k s = k. Lemma 2.4. Let H = Z n an let m N, m 1, be such that m n. If p 1, p 2,..., p r are the istinct prime ivisors of m, then the number of elements in H of orer m is φ(m) = k 1 k 2 k s m p 1 p 2...p r (p 1 1)(p 2 1)...(p r 1). Proof. Since H contains a unique cyclic subgroup H 0 of orer m, all elements of orer m in H will be containe in H 0. The number of elements of orer m in H 0 is the number of integers in {1, 2,..., m} which are coprime to m an this, by efinition, is φ(m). The state formula for φ(m) is given as Theorem 7.5 in [17]. A few well known facts regaring L 2 (q) an Sz(q) are given in the next two lemmas. Lemma 2.5. Let q = p a > 3 where p is a prime an a N. Suppose G = L 2 (q), an let P Syl p (G). (i) G = q(q+1)(q 1) where = (2, q 1). 6

8 (ii) G acts 2-transitively upon the projective line Ω an G α = N G (P ) for some α Ω. (iii) G contains cyclic subgroups H = h an H + = h + where H = q 1 an H + = q+1. Set S = {P g, H g, H g + g G}. Then every non-ientity element of G belongs to a unique subgroup in S. (iv) h has cycle type 1 ( ) 2 q 1 on Ω an h+ has cycle type ( ) q+1 on Ω. For 1 x P, x has cycle type 1 1 p q p on Ω. (v) The number of elements of orer p is (q 1)(q + 1). (vi) [G : N G (H )] = q(q+1) 2 an [G : N G (H + )] = q(q 1) 2. Proof. For parts (i) (iv) see 8.1 Hilfssatz an for (iii), (iv) an (vi) consult 8.3, 8.4, 8.5 Satz of [9]. Part (v) is given in 8.2 Satz (b) an (c) of [9]. Lemma 2.6. Let q = 2 2n+1 where n N an set r = 2 n+1. Suppose G = Sz(q) an let P Syl 2 G. (i) G = q 2 (q 1)(q 2 + 1). (ii) G acts 2-transitively on the Suzuki oval Ω an G α = N G (P ) for some α Ω. (iii) G contains cyclic subgroups H 0 = h 0, H = h an H + = h + where H 0 = q 1, H = q r + 1 an H + = q + r + 1. Set S = {P g, H g 0, H, g H g + g G}. Then every non-ientity element of G belongs to a unique subgroup in S. (iv) h 0 has cycle type 1 2 (q 1) q+1, h has cycle type (q r + 1) q2 +1 q r+1 h + has cycle type (q + r + 1) q2 +1 q+r+1 an on Ω. For x P, x has cycle type q2 2, respectively q2 4, on Ω if it has orer 2, respectively 4. (v) N G (H 0 ) = 2(q 1), N G (H ) = 4(q r + 1) an N G (H + ) = 4(q + r + 1). (vi) P contains q 1 elements of orer 2 an q 2 q elements of orer 4. Proof. Consult Theorem 9 of [23]. 7

9 Next we give a compenium of facts about U 3 (q). Lemma 2.7. Let q = p a > 2 where p is a prime an a N, an suppose G = U 3 (q). Let P Syl p G an set = (q + 1, 3). (i) G = q 3 (q 2 1) (q 3 + 1). (ii) G acts 2-transitively on Ω, the set of isotropic 1-spaces of a unitary 3-space, Ω = q an G α = N G (P ) for some α Ω. Further for β Ω \ {α}, N G (P ) = P G α,β with G α,β cyclic of orer q2 1 an C Gα,β (Z(P )) = (q+1). (iii) P has class 2 with Z(P ) = q. If p is o, then P has exponent p an if p = 2 then P has exponent 4 with the set of involutions of P being Z(P ) #. Let ˆenote the image of subgroups of SU 3 (q) in U 3 (q) ( = G). (iv) G has a maximal subgroup M isomorphic to ˆGU 2 (q). (v) M has a subgroup E 0 of shape ˆ(q + 1) 2 for which N G (E 0 ) ˆ(q + 1) 2.Sym(3) an any subgroup of G of shape ˆ(q + 1) 2.Sym(3) is conjugate to N G (E 0 ). (vi) G has a cyclic subgroup C of orer q2 q+1 Frobenius group. for which N G (C) C.3 is a Proof. For parts (i) to (iii) see Suzuki [23] an Huppert [9]. Part (iv) follows from Mitchell [13] (or Bray, Holt, Roney-Dougal [2]). Also from either [2] or [13] G has one conjugacy class of maximal subgroups of shape ˆ(q + 1) 2.Sym(3) (except when q = 5) an (vi) hols (except when q = 3, 5). Using the Atlas [8] for these exceptional cases we obtain (v) an (vi). From Table 2 in [20] we can extract etails of the conjugacy classes of G = U 3 (q) as well as some supplementary information which we isplay in Table 2 ( = (3, q + 1), l = q+1 an Ω as in Lemma 2.7(ii)). We have use the same notation for the classes as in [20] except we have omitte the superscripts use there as they are of no importance here. Because U 3 (q) acts 2-transitively on Ω, by page 69 of [10] the permutation character must be χ 1 + χ q 3 (as in Table 2 of [20]) which then yiels the last column of the table. 8

10 Class Type Number of Classes Centralizer Orer fix Ω (g), of each Type g G C i C 1 1 G q q C 2 1 (q+1) 1 C 3 q 2 1 q(q+1)(q C 4 l 1 1) q + 1 q(q+1) C 5 l 1 C 6 q 2 q (q+1) 2 0 C 6 0 if = 1, 1 if = 3 (q + 1) 2 0 C 7 C 8 q 2 q 2 2 q 2 q+1 3 Table 2: Conjugacy classes of U 3 (q) q q 2 q The Number of Orbits on P k (Ω) We begin with the Proof of Theorem 1.1. Put Ω k = P k (Ω). Suppose k N with k 3, an let 1 g G ( = L 2 (q)) be an element of orer m. Then by Lemma 2.5 (iii) g must be containe (uniquely) in a conjugate of one of P, H an H +. Since we seek to etermine fix Ωk (g) we may suppose that g is containe in one of P, H an H +. First we consier the case when g H +. Since g is some power of h +, by Lemma 2.5(iv), g has cycle type m q+1 m. By Lemmas 2.4 an 2.5 (iv) H + contains φ(m) elements of orer m an there are q(q 1) 2 conjugates of H +, whence these elements contribute q(q 1) 2 φ(m)η k m D ( q+1 ) ) (m q+1 m to the sum g G fix Ω k (g). Now consier the case when g H. As g is a power of h, by Lemma 2.5 (iv), g has cycle type 1 2 m q 1 m. Employing Lemmas 2.4 an 2.5 (iv) we obtain q(q + 1) 2 φ(m)η k m D ( q 1 ) ) (1 2 m q 1 m, in the sum g G fix Ω k (g). Similar consierations for g P, using Lemma 2.5, 9

11 yiel (q 1)(q + 1)η k (1 1 p q p ). Combining the above with Lemmas 2.1 an 2.5 (i) we obtain the expression for σ k (G, Ω). The proof of Theorem 1.2 is similar to that of Theorem 1.1 except we use Lemma 2.6 in place of Lemma 2.5. The remainer of this section is evote to establishing Theorem 1.4. So we assume G = U 3 (q), (q = p a > 2) an Ω is the set of isotropic 1-spaces of a 3-imensional unitary space. Before beginning the proof of Theorem 1.4, there are a number of preparatory efinitions, notation an lemmas we nee. First we escribe the partitions that arise in Theorem 1.4. Before we can o that, an introuce µ k, we require a barrage of notation associate with pairs of natural numbers iviing l where l D(l), l = q+1. So let l D(l) an (l 1, l 2 ) D(l ) D(l ), an let p 1,.., p r be prime numbers such that l 1 = p α 1 1 p α p αr r an l 2 = p β 1 1 p β p βr r where for i = 1,..., r at least one of the the α i an β i is non-zero. If α i β i, then efine γ i = max{α i, β i }. Without loss of generality we shall assume that α i = β i for 1 i s an α i β i for s < i r. Set l 0 = p α 1 1 p α p αs s (= p β 1 1 p β p βs s ), m 1 = p α s+1 s+1 p α s+2 s+2...p αr r an m 2 = p β s+1 s+1 p β s+2 s+2...p βr r. Also set l = p γ s+1 s+1 p γ s+2 s+2...p γr r an note that l 1 = l 0 m 1 an l 2 = l 0 m 2. We remark that we o not exclue the possibilities l 1 = l 0 = l 2 or l 1 = m 1 an l 2 = m 2. Put l 12 = lcm{l 1, l 2 }. If l i = 1 then l 0 = 1 = m i an if l 1 = l 2 = 1 then we set l = 1. Definition 3.1. (i) π 1 = 1 q3 +1. (ii) π 2 = 1 1 p q3 p. (iii) π 3 = q3 4 (only efine for p = 2). (iv) π (m) 4 = 1 q+1 m q 3 q m where m D (l). 10

12 (v) π (m) 5 = 1 1 p q q 3 q p m m where m = pj an j D (l). q+1 l 2 2 n q+1 n (vi) π (l q+1 1,l 2,n) l 6 = l 1 1 l n = n l, n D(l 0 ). q 3 3q 2 l l where (l 1, l 2 ) D (l) D (l) an (vii) π (m) 7 = 1 2 j q 1 j m q3 q m where m D( q2 1 m ), m D(l), j =. (m,l) (viii) π 8 = m q3 +1 m where m D (q 2 q + 1). (ix) When 3 i q + 1 with i N, 3 i π (l 1,l 2,n) 6 = (3 i l 1 ) q+1 3 i l 1 (3 i l 2 ) q+1 3 i l 2 (3 i n) q+1 3 i n (3 i l 12 ) q 3 3q 2 3 i l 12 where (l 1, l 2 ) D(l) D(l) an n = n l, n D(l 0 ). Definition 3.2. if p 2 an if p = 2. µ k = (q 3 + 1)(q 3 1)η k (π 2 ) µ k = (q 3 + 1)((q 1)η k (π 2 ) + (q 3 q)η k (π 3 )) Definition 3.3. Let k N, an we continue to set l = q+1. For (l 1, l 2 ) D(q + 1) D(q + 1) we use the notation l 0, m 1, m 2, l as efine earlier. (i) Let (l 1, l 2 ) D(q + 1) D(q + 1), n = l n with n D(l 0 ) an n = p δ 1 1 p δ p δs s. If l 0 = p α 1 1 p α p αs s, then we efine f(l 1, l 2, n) = φ(m 1 )φ(m 2 )φ(l 0 ) α j =δ j 1 j s j (p j 2)φ p α j 1 α j δ j 1 j s p δ j j. (ii) λ k(l, l) = (l 1,l 2 ) D (l) D (l) 1 n=l n n D(l 0 ) f(l 1, l 2, n)η k (π (l 1,l 2,n) 6 ). (iii) For l D (q + 1) an i N such that 3 i l set λ k (l, l ; i) = f(l 1, l 2, n)η k (3 i π (l 1,l 2,n) 6 ). (l 1,l 2 ) D(l ) D(l ) n=l n n D(l 0 ) 11

13 Definition 3.4. Let E 0 be an abelian subgroup of G isomorphic to the irect prouct of two cyclic groups of orer (q+1) an (q + 1) with N G (E 0 ) (q + 1).Sym(3). Put E0 = (C 6 C 6) E 0, an efine (q+1) where Ω k = P k (Ω). σ k (E 0, Ω) = g E 0 fix Ωk (g), Remark 1. Subgroups such as E 0 in Definition 3.4 exist by Lemma 2.7(v), an σ k (E 0, Ω) is the contribution of E 0 to the sum g G fix Ω k (g). Lemma 3.5. The cycle type for elements in the classes of type C i for i 6 are given in Table 3. Class Type Orer of g Cycle type of g C 1 1 π 1 C 2 p π 2 C 3 4 (p = 2) π 3 p (p 2) π 2 C 4 m, (m D (l)) π (m) 4 C 5 px, (x D (l)) π (m) 5 C 7 m = js, (j D (q 1), π (m) 7 s D(l)) C 8 m, (m D (q 2 q + 1)) π 8 Table 3: Cycle Types Proof. Let X = g G be a conjugacy class of type C i, i {1,..., 8} \ {6}, an let P Syl p (G). If i = 1, then clearly g = 1. From the centralizer sizes in Table 2, for i = 2 we must have X is the conjugacy class containing Z(P ) #, while classes of type C 3 are the conjugacy classes of elements in P \ Z(P ). If p is o, then by Lemma 2.7(iii) the elements in classes of type C 3 all have orer p whence, as elements in classes of types C 2 an C 3 fix just one element of Ω, their cycle type is π 2. If p = 2, then for i = 2, we also have cycle type π 2 an for i = 3 as the elements of orer 4 must square to involutions in Z(P ), their cycle type must be π 3. From Lemma 2.7(iv) G possesses a maximal subgroup M = ˆGU 2 (q). Further, Z(M) is cyclic of orer l = q+1. It is straightforwar to check that no two elements in Z(M) # are G-conjugate. So we see, using the centralizer 12

14 sizes in Table 2, Z(M) # supplies representatives for all the conjugacy classes of type C 4. Choose h to be an element of orer p in M (in fact h is G- conjugate to an element in Z(P ), this can be seen as these elements must, hence cannot be in classes C 3 ). By the structure of M, for any h Z(M) #, C G (hh ) = q(q+1). (We note that no element of G \ M centralizes M an that h, h M, both are centralize by Z(P ) < M an Z(M) < M). Also for h, h Z(M) # with h h we see that hh an hh are not G-conjugate. Therefore {hh h Z(M) # } gives representatives for all conjugacy classes of type C 5. Now fix Ω (g ) = q + 1 for all g Z(M) #. So for h Z(M) # of orer m, h will have cycle type π (m) 4 on Ω. Similarly for hh, h Z(M) # with h of have centralizer with orer ivisible by Z(M) = q+1 orer m, as fix Ω (h) = 1, we infer that hh has cycle type π (m) 5. So we have ealt with classes of type C 4 an C 5. Next we look at classes of type C 7. Since fix Ω (g) = 2, we may suppose g G α,β ( = q2 1 ), where α, β Ω, α β. We may also suppose Z(M)( = q+1 ) G α,β where M is a maximal subgroup of G isomorphic to ˆGU 2 (q). Since fix Ω (g ) = q + 1 for all g Z(M) #, g G α,β \ Z(M). Let g have orer m an let j be the smallest natural number such that g j Z(M). Then m = (m, l)j where j D (q 1) (note that j = 1, as G α,β (q+1) = contains a unique subgroup, Z(M), of orer q+1 = l, woul mean g Z(M)). So j = m an hence g has cycle type π(m) (m,l) 7. Finally we look at those of type C 8. From Lemma 2.7(vi) G has a cyclic subgroup C = (q2 q+1) with N G (C) C.3 being a Frobenius group. Now C # has ( C 1)/3 N G (C)-conjugacy classes an it can be seen that no two of these are G-conjugate. Hence we have that the N G (C)-conjugacy class representatives of C # give us representatives for all of the classes of type C 8. So, as fix Ω (h) = 0 for all h C #, the elements in C # has cycle type π 8 on Ω, which completes the proof of Lemma 3.5. We now turn our attention to the elicate process of issecting the classes of type C 6 C 6. Lemma 3.6. Suppose A = Z e Z e with A containing three subgroups A i = Ze (i = 1, 2, 3), such that A i A j = 1 for i j. Then there exists a 1 A 1, a 2 A 2 such that A 1 = a 1, A 2 = a 2 an A 3 = a 1 a 2. Proof. Since A 1 A 2 = 1 an A 1 = Ze, A = A 1 A 2. Let A 3 = c. Then c = a 1 a 2 where a i A i, i = 1, 2. Suppose a i has orer e i, i = 1, 2 an, without loss that, e 1 e 2. Then c e 1 = a e 1 1 a e 1 2 = a e 1 2 A 2 A 3 = 1. So 13

15 e 2 e 1 an hence e 1 = e 2. Since c = a 1 a 2 has orer e, we must have that e 1 = e 2 = e, so proving the lemma. Hypothesis 3.7. Suppose that A 0 is an abelian group containing a subgroup A with [A 0 : A] = 1 or 3. Also suppose that A has orer e 2 an contains three subgroups A 1, A 2, A 3 with A i = Ze (i = 1, 2, 3). Further suppose that Ω is an A 0 -set such that (i) for 1 g A 0, fix Ω (g) = if g A 1 A 2 A 3 an fix Ω (g) = fix Ω (A i ) if g A i ; (ii) fix Ω (A i ) fix Ω (A j ) = for 1 i j 3; an (iii) for i = 1, 2, 3, fix Ω (A i ) = q + 1. Set Λ i = fix Ω (A i ) for i = 1, 2, 3 an Λ = Ω \ (Λ 1 Λ 2 Λ 3 ) an so, by (ii), Ω is the isjoint union Λ 1 Λ 2 Λ 3 Λ. Moreover, by (i), A 0 acts regularly on Λ an for i = 1, 2, 3, A 0 /A i acts regularly on Λ i. We shall encounter Hypothesis 3.7 both in a recursive setting an in the group A 0 = Z q+1 Z l (where l = q+1, = (3, q + 1)). For this A 0 we have e = l an A woul be the subgroup of A 0 generate by the elements of A 0 of orer l. Then [A 0 : A] =. Lemma 3.8. Assume Hypothesis 3.7 hols an use the notation A 0, A an A i in the hypothesis. Let (l 1, l 2 ) D(e) D(e) where e D(l). The cycle structure on Ω of the elements g = g 1 g 2 A where g i A i (i = 1, 2) with g i of orer l i is (l 1 ) q+1 l 1 (l 2 ) q+1 l 2 (n) q+1 n (l12 ) Λ l 12, where n = l n with n D(l 0 ) an l 12 = lcm(l 1, l 2 ). This cycle structure, as g 1 an g 2 ranges over the elements of orer (respectively) l 1 an l 2 occurs φ(m 1 )φ(m 2 )φ(l 0 ) times, where n = p δ p δs s. α j =δ j 1 j s j (p j 2)φ p α j 1 α j δ j 1 j s p δ j j 14

16 Proof. By Hypothesis 3.7 (i) an (ii) A i A j = 1 for i j. Hence, by Lemma 3.5 we may select a 1 A 1, a 2 A 2 so as to have A 1 = a 1, A 2 = a 2 an A 3 = a 1 a 2. Aitionally we may ientify A with A 1 A 2. Let g = g 1 g 2 where g i A i an g i has orer l i, i = 1, 2. The smallest k N such that g k A 2 is clearly l 1 an, likewise, the smallest k N such that g k A 1 is clearly l 2. Hence, as A/A 2 acts regularly on Λ 2, g in its action on Λ 2 must be the prouct of isjoint cycles of length l 1. Similarly g acts upon Λ 1 as a prouct of isjoint cycles each of length l 2. Concerning the action of g on Λ, as A 0 acts regularly on Λ an l 12 = lcm{l 1, l 2 } is the orer of g, Λ is a isjoint union of Λ l 12 length cycles of g. Since A 3 = a 1 a 2 to fin the lengths of g s cycles on Λ 3, we must etermine the smallest k N such that g k a 1 a 2. For i = 1, 2 let k i N with k i e be such that g i = a k i i. So g = ak 1 1 a k 2 2 an, we recall, l i = e/(e, k i ) for i = 1, 2. Thus we seek the smallest k N for which g k = (a k 1 1 a k 2 2 ) k = a k 1k 1 a k 2k 2 = (a 1 a 2 ) j for some j, 0 j < e. This is the smallest k N such that k 1 k k 2 k mo e e which is k =. (k 1 k 2,e) Let C be a cyclic group isomorphic to Z e with generator c. Now for i = 1, 2 the orer of c k i e is = l (e,k i ) i an the orer of c k 1 (c k 2 ) 1 is k. Thus to enumerate the possibilities for k (recall (l 1, l 2 ) is a fixe orere pair) we look at the orer of c k 1 (c k 2 ) 1 as we run through the orere pairs (c k 1, c k 2 ) of elements of C of orer, respectively, l 1 an l 2. In oing this there no loss in supposing C = c has orer lcm{l 1, l 2 }. For i = 1,..., r, let P i Syl pi (C). Since the orer of elements in C is the prouct of their orers in the projections into P i for i = 1,..., r, we first consier the special case when C = P i 1, for some i {1,..., r}. So l 1 = p α i i an l 2 = p β i i. (3.8.1) If α i β i, then for all choices of (c k 1, c k 2 ), of which there are φ(p α i i )φ(p β i i ), the orer of ck 1 (c k 2 ) 1 is p γ i i. Recalling that by efinition γ i = max{α i, β i } we see that the orer of c k 1 (c k 2 ) 1 is p γ i i as asserte. Now we turn to the case when α i = β i. Here we have φ(p α i i ) 2 possible choices for (c k 1, c k 2 ). Let C 1 be the unique subgroup of C of orer p α i 1 i (an note c k 1 an c k 2 are in C \ C 1 ). Shoul c k 1 an c k 2 be in ifferent C 1 cosets of C, then c k 1 (c k 2 ) 1 is not in C 1 whence c k 1 (c k 2 ) 1 has orer p α i i. This will happen φ(p α i i )p α i 1 i (p i 2) times. The number of orere pairs (c k 1, c k 2 ) for which c k 1 an c k 2 are in the same C 1 coset of C is φ(p α i i )p α i 1 i. In this situation, for a fixe c k 1, c k 1 (c k 2 ) 1 runs through all the elements of C 1 thus yieling φ(p α i 1 i ) of orer p α i 1 i, φ(p α i 2 i ) of orer p α i 2 i, an so on. To summarize we have the following. 15

17 (3.8.2) Suppose α i = β i. Then for φ(p α i i )p α i 1 i (p i 2) of the orere pairs (c k 1, c k 2 ) the orer of c k 1 (c k 2 ) 1 is p α i i an, for j = 1,..., α i, φ(p α i i )φ(p α i j i ) of the orere pairs (c k 1, c k 2 ) the orer of c k 1 (c k 2 ) 1 is p α i j i. We now consier the general situation for l 1 = p α 1 1 p α p αr r an l 2 = p β 1 1 p β p βr r. Looking at all those i for which α i β i (just for the moment consiering the projections onto P s+1,..., P r ) we obtain that c k 1 (c k 2 ) 1 has orer p γ s+1 s+1...p γr r = l for r i=s+1 φ(p α i i )φ(p β i i ) = φ(m 1)φ(m 2 ) pairs (c k 1, c k 2 ) by (3.8.1) We now wish to enumerate the pairs (c k 1, c k 2 ) for which the orer of c k 1 (c k 2 ) 1 is n, where n = l n, n D(l 0 ) an n = p δ 1 1 p δ p δs s. Using (3.8.2) an by just consiering the projections on P 1,..., P s we see this occurs for φ(p αi i α i =δ i = 1 i s )p α i 1 i (p i 2) φ(p αi i )φ(p δ i i ) α i δ i φ(p α i i ) p αi 1 i (p i 2) φ(p δi i ) α i =δ i α i δ i = φ(l 0 ) α i =δ i p αi 1 (p i 2)φ ( p δi i α i δ i pairs. Combining this with the projection onto P s+1,..., P r yiels Lemma 3.8. ) Lemma 3.9. Let E 0 be an abelian subgroup of G isomorphic to the irect prouct of two cyclic groups of orer (q+1) an (q+1) with N G (E 0 ) (q+1) (q+ 1).Sym(3). Also let E be the subgroup of E 0 generate by the elements of E 0 of orer l = (q+1). Then (i) σ k (E0, Ω) = λ k (l, l) if = 1; (ii) σ k (E 0, Ω) = λ k (l, l) + 2λ k(l, l; 1) if = 3 an 3 E ; an (iii) σ k (E0, Ω) = λ k (l, l) + 2.9b 1 λ k ( q+1 largest power of 3 iviing q b, q+1 3 b ; b) if = 3, 3 E an 3 b is the Proof. By Lemma 2.7(iv),(v) G contains a subgroup M with M ˆGU 2 (q) an E 0 M. From the structure of N G (E 0 ) an ˆGU 2 (q), Z(M) E 0 with Z(M) = q+1 (= l). Let h N G(E 0 ) be an element of orer 3. Because 16

18 [N M (E 0 ) : E 0 ] = 2, h M = N G (Z(M)). In orer to key in with the notation of Hypothesis 3.7, principally as we shall employ Lemma 3.8, we set A 0 = E 0 an A = E. Further, we set A 1 = Z(M), A 2 = Z(M) h an A 3 = Z(M) h2. Since fix Ω (h) = fix Ω (Z(M)) for all h Z(M) #, it follows that fix Ω (A i ) fix Ω (A j ) = for 1 i j 3. We also have fix Ω (A i ) = q + 1 an, by Table 2, fix Ω (g) = if g A 0 \(A 1 A 2 A 3 ). Now A is the subgroup of A 0 generate by the elements of A 0 of orer l an A i = l. So we have A i A, i = 1, 2, 3 an [A : A 0 ] = (= 1 an 3). Hence Hypothesis 3.7 hols with e = l. Suppose = 1. Then A = A 0. Using Lemma 3.8 an Definition 3.3(ii) we obtain σ k (E0, Ω) = λ k (l, l). (Note the conition in Definition 3.3(ii) on the outer sum that (l 1, l 2 ) D (l) D (l) an on the inner sum that n 1 prevents the counting of elements in C 4.) So Lemma 3.9 hols in this case. So we now investigate the case when = 3. Hence [A 0 : A] = 3. Let θ : A 0 A 0 be efine by θ : g g 3. Then, as A 0 is abelian, θ is a homomorphism with imθ A an kerθ = {x A 0 orer of x is 1 or 3}. Further assume that 3 E (so 3 l). Then, as A = l 2, 3 A. Also we have kerθ = 3 an therefore, by orers, imθ = A. For every g A 0 \ A, the smallest power of g containe in A i (i = 1, 2, 3) will be three times the corresponing power for h = g 3 = θ(g). Now 3 A means that θ restricte to A is a one-to-one map, an so the inverse image of h contains two elements of A 0 \ A. Hence, using Lemma 3.8, the elements of A 0 \ A contribute 2λ k (l, l, 1) to the sum g G fix Ω k (g). Thus, using Lemma 3.8 again, σ k (E0, Ω) = λ k (l, l) + 2λ k(l, l; 1), as state. The last case to be consiere is when, as well as = 3, we have 3 E. So 3 l. As a consequence kerθ = 3 2 an thus [A : imθ] = 3. We seek to pinpoint imθ. Now let A 3 i enote the unique subgroup of A i of inex 3 (i = 1, 2, 3). Let B be the subgroup of A generate by the elements of A of orer l/3. Then [A : B] = 3 2 an, for 1 i < j 3, B = A 3 i A 3 j. Also observe that imθ B. Set N = N G (A 0 ), an recall that N ( q+1 (q + 1)).Sym(3). Also the N-conjugacy class of A 1 is {A 1, A 2, A 3 }. Hence N normalizes B (= A 3 i A 3 j, 1 i < j 3) an the N-conjugacy of A 1 B is {A 1 B, A 2 B, A 3 B}. Moreover A i B A j B for 1 i < j 3 (as A = A i A j ). Eviently imθ is a normal subgroup of N an therefore {imθ, A 1 B, A 2 B, A 3 B} comprise the four subgroups of inex 3 in A which contain B. Observe that the inverse image uner θ of B is A. Since, by Lemma 3.8, λ k (l, l) is the count for the contribution of elements in A, we are looking to etermine the contribution from the elements in A 0 \ A. Thus for h imθ \ B, θ 1 ({h}) A 0 \ A with θ 1 ({h}) = 9. For g θ 1 ({h}), h = g 3 imθ an so we must multiply the cycle lengths of h by 3 an their multiplicities by 9 to count the contribution 17

19 of θ 1 ({h}). Now imθ contains B = A (3) i A (3) j (1 i j 3) as a subgroup of inex 3, an clearly A i imθ A (3) i, (1 i 3). If A i imθ A (3) i, then, as [A i : A (3) i ] = 3, we get A i imθ. But then imθ A i A (3) j = A i A (3) i A (3) j = A i B whence imθ = A i B. This is impossible as imθ A i B an so we conclue that A i imθ = A (3) i for 1 i 3. Hence we have that imθ satisfies Hypothesis 3.7 with B playing the role of A an A i imθ (1 i 3) the role of the A i. Further B itself satisfies Hypothesis 3.7 with B playing the role also of A an the A (3) i (1 i 3) the role of the A i. We may repeat this process for imθ \ B (note b 2 in this case), each time we multiply cycle lengths by 3 an the multiplicity by 9. Eventually we arrive at im(θ b 1 ), containing a subgroup B of inex 3. Observe that the count for im(θ b 1 ) = q+1 q+1 is given by part (ii) (with l = q+1 ) an the count 3 b 1 3 b 3 b 1 for B = q+1 q+1 with (l = q+1 ). Keeping track of changes in cycle length 3 b 3 b 3 b an multiplicity we obtain 9 b 1( λ k( q + 1 3, q + 1 b 1 3 ) + 2λ k( q + 1, q + 1 ; b) λ b 1 3 b 3 k( q + 1 b 3, q + 1 b 1 3 = 2.9 b 1 λ k ( q + 1, q + 1 ; b) 3 b 3 b which is the contribution for A 0 \ A. Consequently σ k (E 0, Ω) = λ k(l, l) b 1 λ k ( q b, q b ; b), an the proof of Lemma 3.9 is complete. b 1 )) We are now in a position to prove Theorem 1.3. Proof of Theorem 1.3 Again we apply Lemma 2.1 using the information on cycle types given in Lemmas 3.5 an 3.9. The size of a given conjugacy class is obtaine using the centralizer orers isplaye in Table 2. So the conjugacy classes in C i for a fixe i all have the same size hence using their multiplicity in each C i together with Lemmas 3.5 an 3.9 we obtain σ k (G, Ω), so proving Theorem

20 Remark 2. Type C 6 classes only arise when = 3 (an then there is only one conjugacy class of this type) an consists of elements of orer 3 with no fixe points, an is one thir the size of the type C 6 classes. When = 3 we inclue this class along with the other type C 6 classes in σ k (E 0, Ω). It occurs as the case l 1 = l 2 = n = 3 an appears f(3, 3, 3) = 2 times. As the size of this class is ivie by 6, this corrects the size of this class in our count. 4 Magma Coe for L 2 (q) In this section we give a Magma implementation for the formula in Theorem 1.1. PSLsig:=proceure(q,k,~sigma); Z:=Integers();:=GreatestCommonDivisor(q-1,2); p:=factorisation(q)[1,1];sig:=0; I:=(/(q*(q+1)*(q-1)))*Binomial(Z!(q+1),k); CC:=[]; Appen(~CC,[<(/q),1>,<p,Z!(q/p)>,<1,1>]); for m in Divisors(Z!((q+1)/)) o if m ne 1 then Appen(~CC,[<(/(2*(q+1))),EulerPhi(m)>,<m,Z!((q+1)/m)>]); en if;en for; for m in Divisors(Z!((q-1)/)) o if m ne 1 then Appen (~CC,[<(/(2*(q-1))),EulerPhi(m)>,<m,Z!((q-1)/m)>,<1,2>]); en if;en for; a:=0; for i:=1 to #CC o Cg:=CC[i];S:={Z!(Cg[i][1]): i in [2..#Cg]}; RPg:=RestrictePartitions(k,S); for l:=1 to #RPg o p:=rpg[l];np:=1; for j:=2 to #Cg o pj:=#{m:m in [1..#p] p[m] eq Cg[j][1]}; np:=np*binomial(cg[j][2],pj); en for; a:=a + np*(cg[1][1]*cg[1][2]);en for;en for; sigma:=a+i;en proceure; Coe implementing the formula for U 3 (q) is available from the authors. References [1] Braley, P. PhD. Thesis University of Manchester, in preparation. 19

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