Regular hypermaps over projective linear groups

Marston Conder^{∗}, Primož Potočnik^{†}, Jozef Širáň^{‡}

December 24, 2007

Abstract

An enumeration result for orientably-regular hypermaps of a given type with automorphism groups isomorphic to PSL(2, q) or PGL(2, q) can be extracted from a 1969 paper by Sah. We extend the investigation to orientable reflexible hypermaps and to non-orientable regular hypermaps, providing many more details about the associated computations and explicit generating sets for the associated groups.

1 Introduction

A regular hypermap H is a pair (r, s) of permutations generating a regular permutation group on a finite set, and provides a generalisation of the geometric notion of a regular map on a surface, by allowing edges to be replaced by ‘hyperedges’. The cycles of r, s and rs correspond to the hypervertices, hyperedges and hyperfaces of H, which determine the embedding of the underlying (and connected) hypergraph into the surface, and their orders give the type of H, say {k, l, m}. The group G generated by r and s induces a group of automorphisms of this hypergraph, preserving the embedding, and acting transitively on the flags (incident hypervertex-hyperedge pairs) of H.

The theory of such objects is well-developed, and thoroughly explained in [5, 6]. Without going into too much detail, we need to make a few basic observations. First, the group G has a presentation of the form G = ⟨ r, s, t | r^{k} = s^{l} = t^{m} = rst = … = 1 ⟩, and (so) is a finite quotient of the ordinary (k, l, m) triangle group. For simplicity, we will say that such a group G has type (k, l, m), provided that k, l, m are the true orders of the corresponding elements r, s, t. There is a bijective correspondence between (isomorphism classes of) groups of type (k, l, m) and (isomorphism classes of) regular hypermaps having hypervertices, hyperfaces, and hyperedges of order k, l, and m, respectively; the group G itself is then the ‘rotational symmetry group’ of the corresponding hypermap. For further details of representing hypermaps in the form of cellular decomposition of closed 2-dimensional surfaces and visualising the rotational symmetries, we refer the reader to [6].

A regular hypermap may admit a symmetry that induces a reversal of some local orientation of the supporting surface. At the group theory level, this is equivalent to the existence of an automorphism ϑ of a (k, l, m)-group G presented as above, such that ϑ inverts two of the three generators. Such regular hypermaps are called reflexible. If ϑ is actually given by conjugation of some element of order 2 in G, then the corresponding regular hypermap admits two kinds of surface realisations: one on a non-orientable surface S, with G being the full automorphism group of the hypermap, and another one on an orientable surface which is a double cover of S, with the full automorphism group isomorphic to the direct product of G and the cyclic group of order 2. The Euler characteristic χ of the regular hypermap of type (k, l, m) associated with a rotational symmetry group G of type (k, l, m) is given by χ = |G|(1/k + 1/l − 1/m) in the orientable case, and χ = |G|(1/k + 1/l − 1/m)/2 if the supporting surface of the hypermap is non-orientable.

∗Department of Mathematics, University of Auckland, New Zealand

†Faculty of Mathematics and Physics, University of Ljubljana, and Institute of Mathematics, Physics and Mechanics, Slovenia

‡Open University, UK, and Slovak Technical University, Bratislava, Slovakia

In 1969, Sah [10] extended some work of Macbeath [9] by enumerating orientably-regular hypermaps of a given type (k, l, m) with automorphism groups isomorphic to PSL(2, q) or PGL(2, q). Further results in this area were obtained in [7, 8], where certain necessary and sufficient conditions for existence of an orientably-regular map of a given type were found. The aim of this paper is to extend the investigation to orientable reflexible hypermaps and to non-orientable regular hypermaps and provide much more details about the associated computations, including explicit generating sets for the associated groups. In a forthcoming paper [2] we will apply the results of this refined approach to the classification of all regular maps (that is, hypermaps in which one of the parameters k, l, m is equal to 2) of Euler characteristic equal to −p^{2} for some prime p. For completeness, we mention that to the best of our knowledge, the only other classes of classical groups for which a (partial) classification of regular maps has been considered are the Suzuki groups and Ree groups (see [3, 4]).

A triple (k, l, m) is called hyperbolic, parabolic, or elliptic, according to whether 1/k + 1/l + 1/m − 1 is negative, zero, or positive. We will restrict ourselves to hyperbolic triples. The reason for this restriction is that in the parabolic case (where 1/k + 1/l + 1/m = 1), the only case where G is a projective linear group is G ≅ PSL(2, 3) ≅ A_{4}

for the triple (3, 3, 3), and for the elliptic type (where 1/k + 1/l + 1/m > 1), there are only four such cases, namely G ≅ PSL(2, 2) = PGL(2, 2) ≅ S_{3} for the triple (3, 2, 2), G ≅ PSL(2, 3) ≅ A_{4} for the triple (3, 3, 2), G ≅ PGL(2, 3) ≅ S_{4} for (4, 3, 2), and G ≅ PSL(2, 4) ≅ PGL(2, 4) ≅ PSL(2, 5) ≅ A_{5} for (5, 3, 2). Note that in a hyperbolic triple the smallest element cannot be less than 2; if it is equal to 2 then the remaining entries are at least 3.

2 Conjugacy classes of representative triples

Let (k, l, m) be a fixed hyperbolic triple and let f be an isomorphism taking a finite (k, l, m)- group G with a partial presentation of the form G = ⟨ r, s, t | r^{k} = s^{l} = t^{m} = rst = … = 1 ⟩ onto PSL(2, F) or PGL(2, F′) where F and F′ are fields of characteristic p. From the point of view of the associated computations with matrices f(r), f(s) and f(t) it turns out to be of advantage to consider first the situation in the special linear group SL(2, K) where K is an algebraically closed field of characteristic p. The results will then be carried over to PSL(2, K) by the natural projection given by M ↦ \overline{M} = ±M for any 2 × 2 matrix M ∈ SL(2, K), which will also help determine the subfields F and F′. To facilitate the exposition, any of the two matrices M, −M ∈ SL(2, K) will be called a representative of \overline{M} ∈ PSL(2, K).

The next few observations will address orders and conjugacy classes of elements in SL(2, K). If p is odd, then SL(2, K) contains exactly one involution, namely, −I, where I is the 2 × 2 identity matrix. All elements of order p and 2p in SL(2, K) are known to be conjugate to the transvections U and −U, respectively, where

U = [ \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ]  (1)

and there are no elements of order divisible by p for any other multiple of p. Equivalently, if p is odd, the order of an element J ∈ SL(2, K) is p (respectively 2p) if and only if J ≠ ±I and the trace tr(J) of J is equal to 2 (respectively −2). If p = 2, then all elements in SL(2, K) of order 2 are conjugate to U, and there are no elements there of any even order greater than 2. For any prime p, an element of SL(2, K) of order i, where i ≥ 3 and gcd(i, p) = 1, is known to be conjugate to one (and hence to both) of the matrices

V (ξ) = [ \begin{matrix}ξ & 0 \\ 0 & ξ^{−1} \end{matrix} ] and W (ω) = [ \begin{matrix}0 & −1 \\ 1 & ω \end{matrix} ]  (2)

where ξ is a primitive i th root of unity over K_{p} = F_{p}, and ω = ξ + ξ^{−1}. In other words, an element J ∈ SL(2, K) has order i, where i is as above, if and only if tr(J) = ω = ξ + ξ^{−1} for some primitive i th root of unity ξ over F_{p}. Note that if i ≥ 3, we have ω ∉ {2, −2}.

Returning to the isomorphism f introduced at the beginning, let \overline{R}, \overline{S}, \overline{T} ∈ PSL(2, K) be the images of r, s, t under f; in particular, \overline{R}\overline{S}\overline{T} = \overline{I} where I is the 2×2 identity matrix. We will refer to the orders (k, l, m) of \overline{R}, \overline{S}, \overline{T} as the projective orders. Our aim is now to specify which representatives R, S, T ∈ SL(2, K) of \overline{R}, \overline{S}, \overline{T} we will be working with. This will depend on the projective orders in the following way. Suppose, for example, that p is odd and k is even. Since k was assumed to be the order of \overline{R} ∈ PSL(2, K), it is plain that the order of both R and −R in SL(2, K) must be 2k. On the other hand, if both p and k are odd, then the orders of R and −R form the set {k, 2k}. Of course, the similar holds for \overline{S} and \overline{T}. It follows that if p is odd and one of the entries k, l, m is even, then by suitably combining signs we may choose the representatives R, S, T ∈ SL(2, K) in such a way that the orders of R, S, and T are 2k, 2l, and 2m, respectively, and RST = I. If p and all of k, l, m are odd, then we may choose the representatives R, S, T in such a way that RST = I and their orders are either (k, l, m) or (2k, 2l, 2m); these two cases are mutually exclusive. If p = 2 then SL(2, K) ≅ PSL(2, K) and R, S, T simply have orders k, l, m.

Triples of matrices (R, S, T) in SL(2, K) with RST = I and with the orders specified as above will be called representative triples, and the orders of R, S, T in SL(2, K) will be called representative orders and denoted by (κ, λ, μ). Representative orders are therefore related to projective orders as follows. We have (κ, λ, μ) = (2k, 2l, 2m) if p is odd and at least one of k, l, m is even, (κ, λ, μ) = (k, l, m) or (2k, 2l, 2m) if p and all of k, l, m are odd, and (κ, λ, μ) = (k, l, m) if p = 2. Note that if one of the orders, say k, is a multiple of p, then it follows from Dickson’s classification that k = p. We can therefore confine ourselves to triples (k, l, m) with gcd(j, p) = 1 or j = p whenever j ∈ {k, l, m}; such triples will be called p-restricted. For the corresponding representative order, we have κ ∈ {p, 2p} if p is odd, and κ = p if p = 2. In particular, if p divides all of k, l, m, then (k, l, m) = (p, p, p) for all p, and (κ, λ, μ) = (p, p, p) or (2p, 2p, 2p) if p is odd, while (κ, λ, μ) = (p, p, p) if p = 2.

In general there can be many distinct conjugacy classes of representative triples (R, S, T) having the same p-restricted projective orders (k, l, m) and the same representative orders (κ, λ, μ). Later we will show that it is possible to determine the number of such conjugacy classes by means of counting the corresponding trace triples (tr(R), tr(S), tr(T)). Earlier in this section we saw that if p is odd and ν is the order of an element M ∈ SL(2, K) with M ≠ ±I, then one of the following three possibilities occurs:

1. ν ≥ 3 and (ν, p) = 1, which happens if and only if tr(M) = ω_{ν} = ξ_{ν} + ξ_{ν}^{−1}, where ξ_{ν} is a ν th primitive root of unity;

2. ν = p, which happens if and only if tr(M) = 2;

3. ν = 2p, which happens if and only if tr(M) = −2.

To capture this in a single formula we extend the definition of ω_{ν} also to ν = p and ν = 2p by stipulating that ω_{ν} = ξ_{ν} + ξ_{ν}^{−1}, where ξ_{ν} is the (ν/p)th primitive root of unity e^{2πiν/p}; this gives ω_{p} = 2 and ω_{2p} = −2. If p = 2, then we just change 2 to 0 in part 2 of the above, and omit part 3 (where ν = 2p). With this all applied to ν = κ, λ and μ, the trace triple corresponding to the above representative triple (R, S, T) is simply (ω_{κ}, ω_{λ}, ω_{μ}).

In the remaining part of this section we prove the important fact that, up to a certain small class of exceptions, any two representative triples having both the same projective and representative orders and the same trace triple are conjugate in SL(2, K).

It will be of advantage to consider first the case where at least two of k, l, m are equal to p. If p = 2, the projective group ⟨\overline{R}, \overline{S}, \overline{T} ⟩ is dihedral and therefore out of the scope of our interest. We will therefore assume without loss of generality that (k, l, m) = (p, p, m) where p is an odd prime.

Proposition 1 Let p be an odd prime and let (k, l, m) be a p-restricted hyperbolic triple such that k = l = p and m ≥ 2. Let (R, S, T) be a representative triple corresponding to the representative orders (κ, λ, μ) = (εp, εp, εm) for suitable ε ∈ {1, 2}. Assume that the group ⟨R, S, T ⟩ is not abelian. Then, (κ, λ, μ) ≠ (p, p, p), and the triple (R, S, T) is conjugate in SL(2, K) to the triple (R_{1}, S_{1}, T_{1}), where

R_{1} = ± [ \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ] , S_{1} = ± [ \begin{matrix}1 & 0 \\ ω_{μ} − 2 & 1 \end{matrix} ] and T_{1} = [ \begin{matrix}1 & −1 \\ 2 − ω_{μ} & ω_{μ} − 1 \end{matrix} ]

and the signs are taken simultaneously (with + for ε = 1, and − for ε = 2).

Proof. We know that any element M ∈ SL(2, K) of order p (2p) is conjugate to the matrix U (−U) given in (1). Without loss of generality we therefore may assume that

R = ± [ \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ] and S = ± [ \begin{matrix}a & b \\ c & 2 − a \end{matrix} ]

where (a − 1)^{2} + bc = 0 (the determinant condition) and the positive (negative) signs are taken simultaneously if κ = λ = p or 2p, respectively, giving the traces 2 and −2. From RST = I we obtain tr(T) = 2 + c. It can be checked that RS = SR if and only if c = 0. Since the group ⟨R, S, T ⟩ = ⟨R, S⟩ is assumed to be non-abelian, we have c ≠ 0. Let M = (m_{ij}) ∈ SL(2, K) be the 2 × 2 matrix such that m_{11} = m_{22} = 1, m_{21} = 0 and m_{12} = (1 − a)c^{−1}. It can be checked that MRM^{−1} = R while also

MSM^{−1} = ± [ \begin{matrix}1 & 0 \\ c & 1 \end{matrix} ] and MT M^{−1} = [ \begin{matrix}1 & −1 \\ −c & 1 + c \end{matrix} ] .

Since conjugation preserves traces, tr(T) = 2 + c ≠ 2, and therefore (κ, λ, μ) ≠ (p, p, p). In our notation, we have tr(T) = ω_{μ} with ω_{μ} = ξ_{μ} + ξ_{μ}^{−1}, where ξ_{μ} is a primitive μ th root of unity over F_{p} if (μ, p) = 1 and a primitive (μ/p) th root of unity if μ = 2p. This gives c = ω_{μ} − 2 and leads to the three matrices in our statement. □

It remains for us to consider the case where at most one of the projective orders is p. We may assume without loss of generality that both k and l are coprime to p, and k ≥ 3.

Proposition 2 Let p be a prime and let (k, l, m) be a p-restricted hyperbolic triple such that k ≥ 3 and k, l ≠ p. Let (R, S, T) be a representative triple associated with the projective orders (k, l, m) and representative orders (κ, λ, μ). Let (ω_{κ}, ω_{λ}, ω_{μ}) be the corresponding trace triple, let ξ_{κ} be a κ th primitive root of unity such that ω_{κ} = ξ_{κ} + ξ_{κ}^{−1}, and let D = ω_{κ}^{2} + ω_{λ}^{2} + ω_{μ}^{2} − ω_{κ}ω_{λ}ω_{μ} − 4. Assume that ⟨R, S, T ⟩ is not isomorphic to a subgroup of the upper triangular subgroup of SL(2, K). Then D ≠ 0 and the triple (R, S, T) is conjugate in SL(2, K) to the following triple (R_{2}, S_{2}, T_{2}), with η = (ξ_{κ} − ξ_{κ}^{−1})^{−1}:

R_{2} = [ \begin{matrix}ξ_{κ} & 0 \\ 0 & ξ_{κ}^{−1} \end{matrix} ] , S_{2} = η [ \begin{matrix}ω_{μ} − ω_{λ}ξ_{κ}^{−1} & −D \\ 1 & ω_{λ}ξ_{κ} − ω_{μ} \end{matrix} ] , T_{2} = η [ \begin{matrix}ω_{λ} − ω_{μ}ξ_{κ}^{−1} & ξ_{κ}D \\ −ξ_{κ}^{−1} & ω_{μ}ξ_{κ} − ω_{λ} \end{matrix} ] .

Remark. Later we show that whenever D ≠ 0, the matrices given in the proposition above indeed generate a group isomorphic to PSL(2, F) or PGL(2, F) for some finite field F of characteristic p.

Proof. Since any element M ∈ SL(2, K) of order κ coprime to p is conjugate to the matrix V (ξ_{κ}) in (2) for a suitable primitive κ th root of unity ξ_{κ}, we may assume that

R = [ \begin{matrix}ξ_{κ} & 0 \\ 0 & ξ_{κ}^{−1} \end{matrix} ] , S = [ \begin{matrix}a & b \\ c & d \end{matrix} ] , and T = [ \begin{matrix}dξ_{κ}^{−1} & −bξ_{κ} \\ −cξ_{κ}^{−1} & aξ_{κ} \end{matrix} ] ,

where ad − bc = 1, tr(S) = a + d = ω_{λ}, and tr(T) = aξ_{κ} + dξ_{κ}^{−1} = ω_{μ}; the reader should be aware of the subtleties in the definition of ω_{μ} in the case where μ ∈ {p, 2p}. Since k ≥ 3 and gcd(k, p) = 1, we have ξ_{κ} − ξ_{κ}^{−1} ≠ 0. The two equations coming from the traces have the unique solution a = η(ω_{μ} − ω_{λ}ξ_{κ}^{−1}) and d = η(ω_{λ}ξ_{κ} − ω_{μ}), where η = (ξ_{κ} − ξ_{κ}^{−1})^{−1}. A computation shows that the determinant condition turns into bc = −η^{2}D, which is the only condition on b and c we have. If D = 0, then ⟨R, S⟩ = ⟨R, S, T ⟩ is clearly isomorphic to a subgroup of the upper triangular subgroup of SL(2, K), contrary to our assumptions. Hence we have D ≠ 0. It can be checked that if M = diag(u, u^{−1}) where 0 ≠ u ∈ K, then

MRM^{−1} = R and MSM^{−1} = [ \begin{matrix}a & u^{2}b \\ u^{−2}c & d \end{matrix} ] .

We may choose u ∈ K such that u^{2}b = −ηD and u^{−2}c = η. Equivalently, up to conjugation, we may assume that b = −ηD and c = η. This gives the matrices in the statement of the Proposition. □

Summing up the two results yields the announced one-to-one correspondence between conjugacy classes of representative triples and their trace triples. The formulation is universal and depends only on the values of D = D(ω_{κ}, ω_{λ}, ω_{μ}). Note that if k = l = p, then ω_{κ} = ω_{λ} and they both are equal to the sum of the κ/p th root of unity and its reciprocal, which is 2 or −2, and the expression for D then simplifies to D = (ω_{μ} − 2)^{2}.

Proposition 3 Let p be a prime and let (k, l, m) be a p-restricted, hyperbolic triple. Assume that D = D(ω_{κ}, ω_{λ}, ω_{μ}) ≠ 0 for any triple (κ, λ, μ) of representative orders and any trace triple (ω_{κ}, ω_{λ}, ω_{μ}). Then the conjugacy classes of representative triples (R, S, T) associated with the projective orders (k, l, m) and the representative orders (κ, λ, μ) are in a bijective correspondence with the trace triples (ω_{κ}, ω_{λ}, ω_{μ}).

Proof. What remains to be proved is that given a trace ω, the pair {ξ, ξ^{−1}} of primitive roots such that ω = ξ +ξ^{−1} is uniquely determined. This follows from the observation that ξ and ξ^{−1} are roots of the polynomial x^{2} − ωx + 1. □

We now derive a necessary and sufficient condition for D = D(ω_{κ}, ω_{λ}, ω_{μ}) to be zero. Recall that if ν is any of κ, λ, μ, then ω_{ν} = ξ_{ν} + ξ_{ν}^{−1} where ξ_{ν} ≠ 0 is the corresponding ν th root of unity. Substituting this into the expression for D, multiplying by ξ_{κ}^{2} and simplifying we obtain the factorisation ξ_{κ}^{2}D = (ξ_{κ} − ξ_{λ}ξ_{μ})(ξ_{κ} − ξ_{λ}^{−1}ξ_{μ}^{−1})(ξ_{κ} − ξ_{λ}^{−1}ξ_{μ})(ξ_{κ} − ξ_{λ}ξ_{μ}^{−1}). Since ξ_{κ} ≠ 0, this shows that D = 0 if and only if ξ_{κ}ξ_{λ}^{ε}ξ_{μ}^{δ} = 1 for some ε, δ ∈ {±1}. Raising the last equation to the power of [λ, μ] gives ξ_{κ}^{[λ,μ]} = 1, which shows that κ divides [λ, μ]. Moreover, also λ divides [κ, μ], and μ divides [κ, λ], by the symmetry of the function D. It is easy to see that these three conditions are satisfied simultaneously if and only if, for each prime p′, the largest power of p′ dividing one of κ, λ, μ divides at least two of them. We summarize this in the following lemma.

Lemma 1 In the above notation, D = 0 if and only if ξ_{κ}ξ_{λ}^{ε}ξ_{μ}^{δ} = 1 for some ε, δ ∈ {±1}. In particular, if there exists a prime p′ such that the largest power of p′ dividing one of k, l, m divides none of the remaining entries, then D ≠ 0 for any choice of representative triples (κ, λ, μ) and primitive roots (ξ_{κ}, ξ_{λ}, ξ_{μ}).

Finally, we note that for any triple (k, l, m) of projective orders, the number of all triples of primitive roots (ξ_{κ}, ξ_{λ}, ξ_{μ}) associated with the representative orders (κ, λ, μ) and such that D = D(ω_{κ}, ω_{λ}, ω_{μ}) ≠ 0 has been determined in [1].

3 Adjoining an involution that inverts two generators

Let (k, l, m) be a hyperbolic triple and let H be a group with presentation H = ⟨ x, y, z | x^{2} = y^{2} = z^{2} = (yz)^{k} = (zx)^{l} = (xy)^{m} … = 1 ⟩. Keeping the same terminology and notation as introduced before, let H be a subgroup of PSL(2, K) where K is an algebraically closed field of characteristic p. Taking r = yz, s = zx, and t = xy, we see that H contains a subgroup G with presentation G = ⟨ r, s, t | r^{k} = s^{l} = t^{m} = rst = … = 1 ⟩ of index at most 2 in H. We can therefore use results of the previous section and study the ways G can be extended by adjoining an involution z such that both rz and zs are involutions.

We first show that if such a z exists, then it is unique. Indeed, let z and z′ be two involutions such that the elements u = rz, v = zs, u′ = rz′ and v′ = z′s are all involutions. Then, z′z = u′u = v′v; denote this common element by w. A simple calculation shows that rw = u′z′ · z′z = u′z = u′u · uz = wr, and, similarly, sw = ws. It follows that w centralises G. But G has trivial centre (since G is isomorphic to PSL(2, K) or PGL(2, K)), and thus w = 1, and z′ = z. Hence the extension of G by z is unique, if it exists.

Existence of such an extension has been known as folklore; however, we need to derive an explicit form for z suitable for later consideration. Assume that the group G has been mapped onto a subgroup of PSL(2, F) where F = F_{p}(ω_{κ}, ω_{λ}, ω_{μ}) < K as before, via the generating triples R, S, T ∈ SL(2, F) listed in Propositions 1 and 2. In SL(2, K), we are therefore looking for an element Z ∈ SL(2, K) such that each of Z, Y = RZ, and X = ZS has order 4 if p is odd, or order 2 if p = 2. We begin with the situation where two of the projective orders are equal to p, in which case p must be odd.

Proposition 4 Let p be an odd prime and let (k, l, m) be a p-restricted hyperbolic triple such that k = l = p and m ≥ 2. Let (R, S, T) be a representative triple corresponding to the representative orders (κ, λ, μ) = (εp, εp, εm) for a suitable ε ∈ {1, 2}. Assume that the group ⟨R, S, T ⟩ is not abelian. Then (κ, λ, μ) ≠ (p, p, p), and there exists some Z ∈ SL(2, K) such that each of Z, Y = RZ and X = ZS has order 4. Moreover, the triple (X, Y, Z) is conjugate in SL(2, K) to the triple (X_{1}, Y_{1}, Z_{1}), where

X_{1} = ±α [ \begin{matrix}1 & 0 \\ 2 − ω_{μ} & −1 \end{matrix} ] , Y_{1} = ±α [ \begin{matrix}1 & −1 \\ 0 & −1 \end{matrix} ] and Z_{1} = α [ \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ] ,

with α^{2} = −1 and the signs taken simultaneously ( + for ε = 1, and − for ε = 2).

Proof. If p is odd, then an element Z ∈ SL(2, K) has order 4 if and only if its trace is equal to zero, that is,

Z = [ \begin{matrix}A & B \\ C & −A \end{matrix} ]

where A^{2} + BC = −1; this is the determinant 1 condition. Let R_{1} and S_{1} be the matrices from Proposition 1. Then,

Y = R_{1}Z = ± [ \begin{matrix}A + C & B − A \\ C & −A \end{matrix} ] ,

which shows that Y has order 4 if and only if C = 0. Using this in evaluating X = ZS_{1} we obtain

X = ZS_{1} = ± [ \begin{matrix}A + B(ω_{μ} − 2) & B \\ A(2 − ω_{μ}) & −A \end{matrix} ] ,

and therefore X has order 4 if and only if B(ω_{μ} − 2) = 0. In the proof of Proposition 1 we saw that ω_{μ} ≠ 2, and so X has order 4 if and only if B = 0. Therefore Z = diag(A, −A) where A^{2} = −1. Letting Z_{1} = Z, Y_{1} = Y, X_{1} = X and α = A gives our statement. □

We now clarify the situation for the remaining hyperbolic triples.

Proposition 5 Let p be a prime and let (k, l, m) be a p-restricted hyperbolic triple such that k ≥ 3 and both k and l are coprime to p. Let (R, S, T) be a representative triple associated with the projective orders (k, l, m) and representative orders (κ, λ, μ). Let (ω_{κ}, ω_{λ}, ω_{μ}) be the corresponding trace triple, let ξ_{κ} be a κ th primitive root of unity such that ω_{κ} = ξ_{κ}+ξ_{κ}^{−1}, and let D = ω_{κ}^{2} + ω_{λ}^{2} + ω_{μ}^{2} − ω_{κ}ω_{λ}ω_{μ} − 4 ≠ 0. Then there exists a Z ∈ SL(2, K) such that each of Z, Y = RZ and X = ZS has order 4. Moreover, the triple (X, Y, Z) is conjugate in SL(2, K) to the triple (X_{2}, Y_{2}, Z_{2}) where

X_{2} = ηβ [ \begin{matrix}D & D(ω_{λ}ξ_{κ} − ω_{μ}) \\ ω_{μ} − ω_{λ}ξ_{κ}^{−1} & −D \end{matrix} ] , Y_{2} = β [ \begin{matrix}0 & ξ_{κ}D \\ ξ_{κ}^{−1} & 0 \end{matrix} ] , Z_{2} = β [ \begin{matrix}0 & D \\ 1 & 0 \end{matrix} ] ,

with β = −1/\sqrt{−D} and η = (ξ_{κ} − ξ_{κ}^{−1})^{−1}.

Proof. First, let p be odd. Take the same general Z ∈ SL(2, K) of order 4 as at the beginning of the previous proof. Let R_{2} and S_{2} be the matrices from Proposition 2. Then

Y = R_{2}Z = [ \begin{matrix}ξ_{κ}A & ξ_{κ}B \\ ξ_{κ}^{−1}C & −ξ_{κ}^{−1}A \end{matrix} ] ,

which implies that Y has order 4 if and only if A = 0. We use this in determining X = ZS_{1} and obtain

X = ZS_{1} = η [ \begin{matrix}B & B(ω_{λ}ξ_{κ} − ω_{μ}) \\ B^{−1}(ω_{λ}ξ_{κ}^{−1} − ω_{μ}) & B^{−1}D \end{matrix} ];

note that η ≠ 0. It follows that X has order 4 if and only if B^{2} = −D. Letting Z_{1} = Z, Y_{1} = Y, X_{1} = X and β = −B^{−1}, we obtain the matrices as in the statement of our Proposition.

If p = 2, then Z ∈ SL(2, K) = PSL(2, K) has order 2 if and only if it has the same form as used in the previous proof, and hence the above conclusion for X_{1}, Y_{1} and Z_{1} is valid also in this case. □

We note that, in the notation of the previous two Propositions, conjugation by Z_{i} inverts R_{i} and S_{i}, and similarly, conjugation by Y_{i} and X_{i} invert R_{i}, T_{i} and S_{i}, T_{i}, respectively, for i = 1, 2.

4 Groups generated by representative triples

In order to determine exactly the projective group ⟨\overline{R}, \overline{S}, \overline{T} ⟩ arising from a representative triple (R, S, T) of elements of SL(2, K), we first determine the smallest field F < K with the property that ⟨R, S, T ⟩ is isomorphic to a subgroup of SL(2, F). Let K have prime characteristic p. If F is any field of characteristic p, we denote by F_{p} ≅ GF (p) the prime field of F. For any collection α, β, … of elements of K let F_{p}(α, β, …) denote the smallest subfield of K containing all of α, β, ….

Proposition 6 Let p be a prime and let (k, l, m) be a p-restricted hyperbolic triple. Let (R, S, T) be a representative triple corresponding to projective orders (k, l, m) and representative orders (κ, λ, μ). Let ω_{κ}, ω_{λ} and ω_{μ} be such that D ≠ 0. Let F_{0} be the smallest field of characteristic p such that the group ⟨R, S, T ⟩ is isomorphic to a subgroup of SL(2, F). Then F_{0} = F_{p}(ω_{κ}, ω_{λ}, ω_{μ}).

Proof. Let F = F_{p}(ω_{κ}, ω_{λ}, ω_{μ}). Observe that the traces ω_{κ}, ω_{λ}, and ω_{μ} of R, S and T must be contained in the minimal field F_{0} of characteristic p such that ⟨R, S, T ⟩ is isomorphic to a subgroup of SL(2, F_{0}). This shows that F_{0} ≥ F. We need to establish the reverse inclusion.

By Proposition 1 we have F_{0} = F if at least two of the entries k, l, m are equal to p. Consider therefore the situation where k ≥ 3, and k and l are coprime to p, and either m = p or gcd(m, p) = 1. Proposition 2 now shows that the group ⟨R, S, T ⟩ is isomorphic to a subgroup of SL(2, F^{∗}) where F^{∗} = F_{p}(ξ_{κ}, ω_{λ}, ω_{μ}). Since ξ_{κ} and ξ_{κ}^{−1} are roots of the polynomial x^{2} − ω_{κ}x + 1, the degree of F^{∗} over F is at most 2. Assume that ξ_{κ} ∉ F (for otherwise there is nothing to prove). Let ρ^{∗} be the unique non-trivial (involutory) automorphism of F^{∗} that fixes F pointwise; it follows that ρ^{∗}(ξ_{κ}) = ξ_{κ}^{−1}. A direct calculation using the matrices R = R_{2} and S = S_{2} from Proposition 2 shows that ρ^{∗}(R) = R^{−1} and ρ^{∗}(S) = S^{−1}. The same effect on R and S, however, arises when conjugating by the matrix Z = Z_{2} from Proposition 5; that is, Z^{−1}RZ = R^{−1} and Z^{−1}SZ = S^{−1}. It follows that ρ^{∗} and conjugation A ↦ Z^{−1}AZ induce the same automorphisms of the group ⟨R, S⟩ = ⟨R, S, T ⟩.

Consider now the subgroup H^{∗} of all the elements A ∈ SL(2, F^{∗}) such that ρ^{∗}(A) = A. It is well known that H^{∗} ≃ SL(2, F). Let H_{Z}^{∗} be the subgroup of all the matrices B ∈ SL(2, F^{∗}) such that ρ^{∗}(B) = Z^{−1}BZ. From what we saw above we may deduce that ⟨R, S, T ⟩ is a subgroup of H_{Z}^{∗}. Our strategy now will be to prove that H^{∗} ≅ H_{Z}^{∗}. Having established this, it is sufficient to observe that SL(2, F) ≤ ⟨R, S, T ⟩ ≤ H_{Z}^{∗} ≅ H^{∗} ≅ SL(2, F), which implies that F_{0} = F.

We prove that H^{∗} ≅ H_{Z}^{∗} by exhibiting a matrix V ∈ GL(2, F̂), where either F̂ = F^{∗} or [F̂ : F^{∗}] = 2, such that V Z = ρ^{∗}(V). Then it is easy to see that V^{−1}H^{∗}V = H_{Z}^{∗}. Let β be the element we have from Proposition 5. If β ∉ F, then β ∈ F^{∗} and ρ^{∗}(β) = −β. In this case we may set

V = [ \begin{matrix}1 & β^{−1} \\ β & −1 \end{matrix} ] and check that V Z = [ \begin{matrix}1 & −β^{−1} \\ −β & −1 \end{matrix} ] = ρ^{∗}(V) .

On the other hand, if β ∈ F, that is, if ρ^{∗}(β) = β, then we need to to go beyond F^{∗}. Let F̂ be an extension of F^{∗} of degree 2, let θ be a primitive element of F̂, and let a = θ^{(q^{2}−1)/2} where q = |F |. Then the automorphism ρ̂ of F̂ given by x ↦ x^{q} has the property that ρ̂↾_{F^{∗}} = ρ^{∗} and ρ̂^{2}(a) = −a. Now for the matrix V we may take

V = [ \begin{matrix}a & β^{−1}ρ̂(a) \\ βρ̂(a) & −a \end{matrix} ] .

We leave the remaining details of the calculation to the reader. □

We will now show that the degree of F_{p}(ω_{κ}, ω_{λ}, ω_{μ}) depends only on p, κ, λ and μ, and is independent on particular choices of primitive roots of unity. As we know, representative orders κ, λ and μ have the property that if an entry is a multiple of p, then it is equal to p or 2p (with the second possibility out of consideration when p = 2).

Lemma 2 Let i be a positive integer coprime to p, let ξ be a primitive i th root of unity, and let ω = ξ + ξ^{−1}. Then the degree of F_{p}(ω) over F_{p} is the smallest positive integer j such that i divides p^{j} − 1 or p^{j} + 1.

Proof. Let δ = [F_{p}(ω):F_{p}], and note that the degree d = [F_{p}(ξ):F_{p}] is the smallest positive integer j for which i divides p^{j} − 1. Observe that ξ is a root of the quadratic polynomial x^{2} − ωx + 1 over F_{p}(ω), and so the degree [F_{p}(ξ) : F_{p}(ω)] is either 1 or 2. If d is odd, then [F_{p}(ξ) : F_{p}(ω)] = 1, which implies that δ = d. Let now d be even. Then [F_{p}(ξ):F_{p}(ω)] = 2 if and only if the unique non-trivial Galois automorphism of F_{p}(ξ) over F_{p}(ω) of order 2 fixes the element ω. We know that this Galois automorphism is given by z ↦ z^{q} where q = p^{d/2}. It is readily verified that (ξ + ξ^{−1})^{q} = ξ + ξ^{−1} if and only if (ξ^{q+1} − 1)(ξ^{q−1} − 1) = 0, which is equivalent to the condition that i divides q + 1 or q − 1. In both cases, δ is the smallest j such that i divides p^{j} + 1 or p^{j} − 1. □

For a p-restricted hyperbolic triple (k, l, m), let e(k, l, m) be the smallest positive integer e such that each n ∈ \{k, l, m\} \ \{p\} divides \frac{p^{e}+1}{2^{ϵ}} or \frac{p^{e}−1}{2^{ϵ}}, where ϵ is 0 or 1 depending on whether p is even or odd, respectively.

Proposition 7 In the above notation, the degree of F_{p}(ω_{κ}, ω_{λ}, ω_{μ}) over F_{p} is equal to e(k, l, m). In particular, F_{p}(ω_{κ}, ω_{λ}, ω_{μ}) depends only on the projective orders (k, l, m) and not on the choice of representative orders (κ, λ, μ).

Proof. Since ω_{ν} is ±2 ∈ F_{p} if p divides ν, by Lemma 2 it suffices to show that the statement of the proposition is equivalent to the claim that the degree of F_{p}(ω_{κ}, ω_{λ}, ω_{μ}) over F_{p} is the smallest e such that each ν ∈ {κ, λ, μ}\{p, 2p} divides either p^{e} −1 or p^{e} +1.

This is clearly true whenever p = 2, or p is odd and one of κ, λ, μ is divisible by 4, since in these two cases (κ, λ, μ) = (2k, 2l, 2m) or (k, l, m), respectively.

We may thus assume that p is odd and none of κ, λ, μ is divisible by 4. Then k, l, m are all odd, and (κ, λ, μ) = (2k, 2l, 2m) or (k, l, m). For an odd integer n, however, the conditions that 2n divides p^{e} ± 1 and n divides p^{e} ± 1 are equivalent, and the statement of the proposition is again equivalent to the above claim. □

For brevity, in the remaining part of this section we set F = F_{p}(ω_{κ}, ω_{λ}, ω_{μ}). From the analysis done up to this point we conclude that, under the assumptions of either Proposition 1 or Proposition 2, the subgroup ⟨\overline{R}, \overline{S}, \overline{T} ⟩ of PSL(2, K) is actually a subgroup of PSL(2, F), not contained in any PSL(2, F′) where F′ is a proper subfield of F. We then have only two possibilities: either ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≅ PSL(2, F), or, if p is odd and [F : F_{p}] is even, we may have ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≅ PGL(2, F′) where F′ is the unique subfield of F such that [F : F′] = 2. In what follows we will identify the conditions under which the second case occurs. We will assume henceforth that p is an odd prime.

Assume that the order of |F | is q^{2} where q is a power of p. Let F′ be the subfield of F such that [F : F′] = 2, that is, |F′| = q, and let ρ : x ↦ x^{q} be the unique non-trivial Galois automorphism of F that fixes F′ pointwise. Clearly, ρ extends to elements of SL(2, F) and PSL(2, F) in the obvious way, and we will use the same symbol ρ for these extensions. Soon we will need the following observation regarding changing signs by ρ.

For convenience, we will write c |_{2} (2d) if c divides 2d but not d. Then, referring to the above notation, we have:

Lemma 3 Let ω = ξ+ξ^{−1} where ξ is an i th primitive root of unity in some field containing F. Then, ρ(ω) = −ω if and only if ρ(ξ) = −ξ or ρ(ξ) = −ξ^{−1}, which is equivalent to i |_{2} (2q − 2) or i |_{2} (2q + 2), respectively. In particular, if ρ(ξ) = −ξ for some i th primitive root of unity, then this holds for all the i th primitive roots of unity; the same applies to the relation ρ(ξ) = −ξ^{−1}.

Proof. Since ρ(x) = x^{q}, we have ρ(ω) = −ω if and only if ξ^{q} +ξ^{−q} = −ξ −ξ^{−1}, which is equivalent to (ξ^{q−1} + 1)(ξ^{q+1} + 1) = 0, and this is easily seen to be the same as stating that ρ(ξ) = −ξ or ρ(ξ) = −ξ^{−1}. The factorisation together with the fact that the order of ξ is i shows that the above occurs if and only if i |_{2} (2q − 2) or i |_{2} (2q + 2), respectively. The last statement in the Lemma follows from the fact that the conditions on i are arithmetic and do not refer to a particular i th primitive root. □

We return to our discussion about a possible isomorphism of ⟨\overline{R}, \overline{S}, \overline{T} ⟩ with the group PGL(2, F′). There is a “canonical” copy H of PGL(2, F′) in PSL(2, F) given by H = \{\overline{A} ∈ PSL(2, F) : ρ(\overline{A}) = \overline{A}\}. Let H_{Z} = \{\overline{A} ∈ PSL(2, F) : ρ(\overline{A}) = \overline{Z}\overline{A}\overline{Z}\}. The fact that H ≅ H_{Z} can be derived in exactly the same way as shown at the end of the proof of Proposition 6. With the help of this we prove the following convenient criterion for deciding if ⟨\overline{R}, \overline{S}, \overline{T} ⟩ is isomorphic to PGL(2, F′).

Proposition 8 In the notation of the preceding paragraph, we have ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≅ PGL(2, F′) if and only if the set {R, S, T } has a 2-element subset A such that ρ(tr(A)) = −tr(A) for A ∈ A and ρ(tr(A)) = tr(A) for A ∈ {R, S, T } \ A.

Remark. Note that if an element \overline{A} ∈ \{\overline{R}, \overline{S}, \overline{T} \} is an involution, then the corresponding trace is 0, and thus both preserved and negated by ρ. Hence if one of the projective orders is 2, then we may have more than one choice for the set A.

Proof. Let (R, S, T) be a representative triple and let ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≅ PGL(2, F′). We know that the group ⟨\overline{R}, \overline{S}, \overline{T} ⟩ is conjugate in PGL(2, F) to H. Let (R′, S′, T′) be a representative triple such that \overline{R}′, \overline{S}′, \overline{T}′ are images of \overline{R}, \overline{S}, \overline{T} under such a conjugation. Then by the definition of H we have ρ(R′) ∈ {R′, −R′}, and for traces we then obtain ρ(tr(R)) = ρ(tr(R′)) = ±tr(R′) = ±tr(R); the same holds when R is replaced with S and T. Since R′S′T′ = I and ρ maps each of R′, S′, T′ either to itself or to its negative, it follows that ρ either preserves the traces of all of R′, S′, T′, or it changes the trace signs on two of them while preserving the third; by the above equalities, the same applies to R, S, T. But ρ cannot preserve all three traces, since then we would have F′ = F, a contradiction.

For the sufficiency, let R, S, T and F, F′ be as before; in particular, the group ⟨\overline{R}, \overline{S}, \overline{T} ⟩ properly contains PSL(2, F′). Also, we may assume that R, S and T have the form as in Propositions 1 and 2. Suppose now that ρ changes the sign of the traces of two of R, S, T and preserves the sign of the third, as specified by the subset A. This immediately rules out the case of (R, S, T) described in Proposition 1, since there the (non-zero) traces of R and S belong to F′, the field pointwise fixed by ρ, and hence at most one of k, l, m can be equal to p. If k, l, m ≠ p, then without loss of generality we may assume that A = {R, S}. If precisely one of k, l, m is equal to p, then (as argued before Proposition 2) we may assume that m = p. But then, since p is odd, we have 0 ≠ ω_{μ} ∈ F′, and therefore ρ has to change the sign of the traces of R and S. We conclude that in all cases, we may assume that A = {R, S}.

Accordingly, suppose ρ(ω_{κ}) = −ω_{κ} and ρ(ω_{λ}) = −ω_{λ} while ρ(ω_{μ}) = ω_{μ}. From Lemma 3 we know that ρ(ξ_{κ}) ∈ {−ξ_{κ}, −ξ_{κ}^{−1}}. By inspection of the matrices R and S in the statement of Proposition 2 one may check that if ρ(ξ_{κ}) = −ξ_{κ}, then ρ(R) = −R and ρ(S) = −S, and if ρ(ξ_{κ}) = −ξ_{κ}^{−1}, then ρ(R) = −R^{−1} and ρ(S) = −S^{−1}. Recall now the canonical copy H of PGL(2, F′) in PSL(2, F) and its isomorphic copy H_{Z}. If ρ changes signs of R and S, then \overline{R}, \overline{S} ∈ H. In the second case, where ρ inverts R and S and changes signs, we have \overline{R}, \overline{S} ∈ H_{Z} since conjugation by \overline{Z} inverts both \overline{R} and \overline{S}. Since \overline{R} and \overline{S} generate ⟨\overline{R}, \overline{S}, \overline{T} ⟩, we conclude that ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≅ PGL(2, F′). □

We are now in position to show that deciding whether ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≅ PGL(2, F′) can be reduced to checking certain divisibility conditions. For brevity, we will write a |_{2} (2b ± 2) if either a |_{2} (2b + 2) or a |_{2} (2b − 2); and we will let c | (d ± 1) have the analogous meaning.

In Proposition 7 we saw that e(k, l, m) is the degree of the field F_{p}(ω_{κ}, ω_{λ}, ω_{μ}) over F_{p}, and equals the smallest positive integer e such that each κ, λ, μ coprime to p divides p^{e} ±1.

Proposition 9 Let p be an odd prime, let (k, l, m) be a p-restricted hyperbolic triple, and let (R, S, T) be a representative triple associated with the projective orders (k, l, m) and representative orders (κ, λ, μ). Let (ω_{κ}, ω_{λ}, ω_{μ}) be the corresponding trace triple, and let D = ω_{κ}^{2} + ω_{λ}^{2} + ω_{μ}^{2} − ω_{κ}ω_{λ}ω_{μ} − 4. Then ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≅ PGL(2, p^{f}) for some f ≥ 1 if and only if D ≠ 0 and the condition (C) below is fulfilled:

(C) either the condition n |_{2} (p^{f} ±1) holds for exactly two of the three orders k, l, m, while the third order is p or divides \frac{p^{f} ±1}{2}; or otherwise 2 ∈ \{k, l, m\} and the condition n |_{2} (p^{f} ± 1) holds for all n ∈ {k, l, m}.

Remark. Note that the condition (C) above implies that at least two of k, l, m are even, and that e(k, l, m) = 2f.

Proof. Suppose that ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≅ PGL(2, F′) for some field of order p^{f}. Note first that by Proposition 2, D ≠ 0. Let K be the algebraic closure of F′, and let F be the smallest subfield of K such that ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≤ PSL(2, F). By Propositions 6 and 7, the order of F is p^{e(k,l,m)}. Since [F : F′] = 2, we have e(k, l, m) = 2f.

By Proposition 8, the Galois automorphism ρ ∈ Gal(F : F′) negates two of the traces {tr(R), tr(S), tr(T)} and preserves the third. Note that the trace of any element of order p is ±2 and is thus preserved by ρ, and that the trace of an element of order 2 is 0 and is thus both preserved and negated by ρ. In particular, at most one of the orders k, l, m can be equal to p. If we now apply Lemma 3, we see that the condition ν |_{2} (2p^{f} ± 2) must hold for at least two of the entries κ, λ, μ. Note that any integer ν satisfying ν |_{2} (2p^{f} ± 2) is divisible by 4, showing that (κ, λ, μ) = (2k, 2l, 2m) and that at least two of the entries k, l, m are even. Hence the condition n |_{2} (p^{f} ±1) holds for at least two of the entries k, l, m. Moreover, if the third entry is not p, then by Lemma 2, we see that it must divide \frac{p^{f} ±1}{2}. Note also that if n |_{2} (p^{f} ± 1) holds for all three of k, l, m, then ρ negates the traces of all three elements R, T, S, implying that one of k, l, m is 2. This proves (C).

Conversely, assume that all the conditions on D, k, l, m, e (where e = e(k, l, m)) listed in the statement of our Proposition are fulfilled. Then the generating triple (R, S, T) is conjugate to the triple (R_{2}, S_{2}, T_{2}) as in Proposition 2. If the condition n |_{2} (p^{f} ± 1) holds for exactly two of the three orders k, l, m and the third order is p or divides \frac{p^{f} ±1}{2}, then by Lemma 3, ρ negates the traces of two of R, S, T and fixes the third. The same holds if one of k, l, m is 2 and the other two satisfy the condition n |_{2} (p^{f} ± 1). Proposition 8 now shows that the group ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≅ ⟨\overline{R}_{2}, \overline{S}_{2}, \overline{T}_{2}⟩ is isomorphic to PGL(2, F′) where [F : F′] = 2. □

5 Enumeration

In this section we re-establish the enumeration result of Sah [10] for regular hypermaps over projective linear groups. Let p be a prime and let (k, l, m) be a p-restricted hyperbolic triple. We know that if exactly one, or exactly two of k, l, m are equal to p, then we may assume that m = p or m = l = p, respectively. We need to recall briefly some of the facts we proved in Section 2. Let (k, l, m)^{∗} be the set of all representative orders (κ, λ, μ) associated with (k, l, m), so that (k, l, m)^{∗} is either {(k, l, m)}, or {(2k, 2l, 2m)}, or {(k, l, m), (2k, 2l, 2m)}, depending on whether p = 2, or p ≥ 3 and at least one of k, l, m is even, or all of p, k, l, m are odd, respectively. We will say that the triple (k, l, m) is proper if D = D(ω_{κ}, ω_{λ}, ω_{μ}) ≠ 0 for any (κ, λ, μ) ∈ (k, l, m)^{∗} and for any choice of ω_{κ}, ω_{λ}, and ω_{μ}. Finally, for a hyperbolic, p-restricted, proper triple (k, l, m), let T (k, l, m) be the set of all possible trace triples (ω_{κ}, ω_{λ}, ω_{μ}) where (κ, λ, μ) ∈ (κ, λ, μ)^{∗}. Proposition 3 can now be re-stated in a form that refers just to the projective orders as follows.

Proposition 10 Let p be a prime and let (k, l, m) be a hyperbolic, p-restricted, proper triple. Then there is a bijection between the set T (k, l, m) and the set of conjugacy classes of representative triples (R, S, T) associated with the projective orders (k, l, m).

New representative triples (R, S, T) with RST = I in SL(2, K) associated with the same projective orders (k, l, m) can sometimes be obtained from old ones simply by changing signs. To see this, suppose, for instance, that both k and l are even. Then, the orders of both R and −R and of both S and −S are 2k and 2l, respectively, and both (R, S, T) and (−R, −S, T) are representative triples. It is clear that the converse holds as well, that is, if both (R, S, T) and (−R, −S, T) are representative triples, then both k and l are even. Thus, if all k, l, m are even, we may define an equivalence relation on the set of representative triples with equivalence classes of size 4 formed by the four triples (R, S, T), (−R, −S, T), (−R, S, −T), and (R, −S, −T). If exactly two of the k, l, m are even, then the representative triples come in pairs as we saw above and we again regard the pairs as equivalence classes. Another way to say this is that when all of k, l, m are even, the quadruples are just orbits of a free action of Z_{2} × Z_{2} on the set of representative triples; if exactly one of k, l, m is odd then we have a free action of Z_{2} representing the sign change. We will refer to this action of Z_{2} × Z_{2} or of Z_{2} as the sign change action.

Obviously, the sign change actions carries over from the set of representative triples (R, S, T) associated with the projective orders (k, l, m) to the set of the corresponding trace triples T (k, l, m) in a natural way; we will use the symbol ∼_{S} to denote the corresponding equivalence relation on T (k, l, m). Another natural equivalence relation to be considered on T (k, l, m) is the relation ∼_{G} induced by the Galois action arising from application of the Galois automorphisms of the fields F = F_{p}(ω_{κ}, ω_{λ}, ω_{μ}) over F_{p} for (κ, λ, μ) ∈ (k, l, m)^{∗}. Indeed, the computations made in the previous section show that the classes of ∼_{G} are in a one-to-one correspondence with orbits of the Galois action extended to the conjugacy classes of generating triples (\overline{R}, \overline{S}, \overline{T}). Let ∼ denote the join of ∼_{S} and ∼_{G} on T (k, l, m). Since the automorphism group of both P SL(2, q) and P GL(2, q) is isomorphic to a semidirect product of P GL(2, q) by the Galois group of F over its prime field, we have:

Proposition 11 The number of non-isomorphic regular hypermaps of a proper, hyperbolic, p-restricted type (k, l, m) with automorphism group isomorphic to a subgroup of P SL(2, F) is equal to the number of equivalence classes of the relation ∼ on T (k, l, m). □

For any j such that j = p, or j = 2p, or otherwise gcd(j, p) = 1, define a modification φ_{p} of the Euler totient function φ by letting φ_{p}(j) = 1 in the first two cases and φ_{p}(j) = φ(j) otherwise. The number of distinct elements ω_{j} is then equal to φ_{p}(j) or φ_{p}(ν)/2 according to whether j is a multiple of p or not. Let u be the number of entries k, l, m coprime to p. If at least one of k, l, m is even, then it is easy to see that |T (k, l, m)| = φ_{p}(2k)φ_{p}(2l)φ_{p}(2m)/2^{u}. In the case where all of k, l, m are odd, we have |T (k, l, m)| = φ_{p}(k)φ_{p}(l)φ_{p}(m)/2^{u} + φ_{p}(2k)φ_{p}(2l)φ_{p}(2m)/2^{u} = 2φ_{p}(2k)φ_{p}(2l)φ_{p}(2m)/2^{u}.

Observe that the equivalence classes of the sign change equivalence ∼_{S} have size 1, 2, and 4, depending on whether u = 1, 2, or 3. The number of equivalence classes of ∼_{S} on T (k, l, m) is therefore equal to φ_{p}(2k)φ_{p}(2l)φ_{p}(2m)/2^{u+v−1}. As regards the equivalence ∼_{G} on T (k, l, m) induced by the Galois action, the number of the corresponding classes is equal to |T (k, l, m)|/e where e = e(k, l, m) is the degree of the field F_{p}(ω_{κ}, ω_{λ}, ω_{μ}) over F_{p}. It remains for us to determine when two trace triples are equivalent under both ∼_{S} and ∼_{G}. By the analysis in the previous section, this can happen if and only if both the sign change action as well as the Galois action changes signs on precisely two of the entries in a trace triple. But by Lemma 8 and in the associated notation, this occurs if and only if ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≃ P GL(2, F′). Thus, the number of equivalence classes of ∼ on T (k, l, m) is equal to φ_{p}(2k)φ_{p}(2l)φ_{p}(2m)/(2^{u+v−1}e) if ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≃ P SL(2, F), and to 2φ_{p}(2k)φ_{p}(2l)φ_{p}(2m)/(2^{u+v−1}e) in the case where ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≃ P GL(2, F′). Combining this with Proposition 9 yields the enumeration result of Sah [10].

Theorem 1 Let p be a prime and let (k, l, m) be a p-restricted, hyperbolic, proper triple. Let e = e(k, l, m), and let u and v be the number of entries coprime to p and the number of even entries among k, l, m, respectively.

(1) If p is odd and condition (C) of Proposition 9 is fulfilled, then all the corresponding groups ⟨\overline{R}, \overline{S}, \overline{T} ⟩ are isomorphic to P GL(2, F′) where F′ is the index 2 subfield of F_{p}(ω_{κ}, ω_{λ}, ω_{μ}) ≃ F_{p}(p^{e}), and the number of all the corresponding pairwise non-isomorphic hypermaps of type (k, l, m) is equal to

2φ_{p}(2k)φ_{p}(2l)φ_{p}(2m)/(2^{u+v−1}e).

(2) In all other cases we have ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≃ P SL(2, F) where F = F_{p}(ω_{κ}, ω_{λ}, ω_{μ}) ≃ F_{p}(p^{e}), and then the number of all such non-isomorphic hypermaps of type (k, l, m) is equal to

φ_{p}(2k)φ_{p}(2l)φ_{p}(2m)/(2^{u+v−1}e).

6 Non-orientable and reflexible regular hypermaps

We now discuss applications of the preceding results to regular hypermaps on non-orientable surfaces and regular reflexible hypermaps on orientable surfaces. Keeping to the notation introduced in the previous sections, this amounts to comparing the group ⟨\overline{R}, \overline{S}, \overline{T} ⟩ with the group ⟨\overline{X}, \overline{Y} , \overline{Z}⟩ where \overline{Z} is given by Proposition 4 or 5. The existence of \overline{Z} in all cases shows that such hypermaps are all reflexible. Also, from the outline in the Introduction it is clear that a hypermap with rotational symmetry group ⟨\overline{R}, \overline{S}, \overline{T} ⟩ has a non-orientable realisation if and only if ⟨\overline{R}, \overline{S}, \overline{T} ⟩ = ⟨\overline{X}, \overline{Y} , \overline{Z}⟩. We now identify exactly when this happens.

Proposition 12 Let p be a prime and let (k, l, m) be a p-restricted hyperbolic triple. Suppose that ω_{κ}, ω_{λ} and ω_{μ} are such that D ≠ 0. Let F and F′ be as in Theorem 1. Then all of the corresponding regular hypermaps of type (k, l, m) with rotation group ⟨\overline{R}, \overline{S}, \overline{T} ⟩ isomorphic to P SL(2, F) or P GL(2, F′) are reflexible. Moreover,

(1) if ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≅ PGL(2, F′), then the hypermaps all admit a realisation on a nonorientable surface;

(2) if ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≅ PSL(2, F) and if two of k, l, m are equal to p, then the corresponding hypermaps admit a realisation on a non-orientable surface if and only if |F | ≡ 1 (mod 4);

(3) if ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≅ PSL(2, F) and if at most one of k, l, m is equal to p, then the hypermaps admit a realisation on a non-orientable surface if and only if −D is a square in F.

Proof. Let us begin with the case where the rotational symmetry group ⟨\overline{R}, \overline{S}, \overline{T} ⟩ of a hypermap of type (k, l, m) is isomorphic to PGL(2, F′) where [F : F′] = 2 and F = F_{p}(ω_{κ}, ω_{λ}, ω_{μ}) for some prime p ≠ 2. Generators R, S, T as elements of SL(2, F) are now given by Proposition 2, and the inverting involution \overline{Z} = \overline{Z}_{2} is as in Proposition 5. Then, ⟨\overline{X}, \overline{Y} , \overline{Z}⟩ = ⟨\overline{R}, \overline{S}, \overline{T} , \overline{Z}⟩ is a proper subgroup of P SL(2, F) of order at least |P GL(2, F′)|. By Dickson’s classification of subgroups of P SL(2, F) we must have ⟨\overline{R}, \overline{S}, \overline{T} ⟩ = ⟨\overline{X}, \overline{Y} , \overline{Z}⟩. We conclude that in this case ⟨\overline{R}, \overline{S}, \overline{T} ⟩ is the automorphism group of a non-orientable regular hypermap of type (k, l, m). Note that the same can be obtained by the following more intrinsic argument. By the proof of Proposition 8 we know that we may assume that the elements \overline{R} and \overline{S} given by Proposition 2 lie either in the canonical copy H ≃ P GL(2, F′) contained in P SL(2, F) or in its isomorphic copy H_{Z}. For the non-trivial Galois automorphism ρ of the extension F over F′ applied to the explicit form of the matrix Z_{2} we have ρ(D) = D, and ρ(β) = β or ρ(β) = −β according to whether −D is a square in F′ or not. At any rate, we have ρ(\overline{Z}) = \overline{Z} and hence \overline{Z} lies in both H, H_{Z} ≃ P GL(2, F′), that is, ⟨\overline{R}, \overline{S}, \overline{T} ⟩ = ⟨\overline{X}, \overline{Y} , \overline{Z}⟩.

Suppose now that the group ⟨\overline{R}, \overline{S}, \overline{T} ⟩ is isomorphic to P SL(2, F). If k = m = p, then p is odd and F = F_{p}(ω_{μ}). Proposition 4 implies that for \overline{Z} = \overline{Z}_{1} we have \overline{Z} ∈ P SL(2, F) if and only if −1 is a square in F, which occurs if and only if |F | ≡ 1 (mod 4). If k, m ≠ p, then F = F_{p}(ω_{κ}, ω_{λ}, ω_{μ}) and the inverting involution is \overline{Z} = \overline{Z}_{2}, given by Proposition 5. An inspection of the form of \overline{Z} and of the groups H^{∗} and H_{Z}^{∗} (for odd p) that appear in the second part of the proof of Proposition 6 shows that \overline{Z} ∈ P SL(2, F) if and only if −D is a square in F. □

A regular hypermap of type (k, l, m) is said to be a regular map if one of the parameters k, l, m is equal to 2. For specific applications we will be particularly interested in regular maps with the groups ⟨\overline{X}, \overline{Y} , \overline{Z}⟩ isomorphic to general projective linear 2-dimensional groups. Proposition 12 lists necessary and sufficient conditions for a regular hypermap (and hence also for a regular map) to have such a group. In the case of maps, however, we will need for our applications a much more detailed knowledge about the membership of the involutory generators \overline{X}, \overline{Y} , \overline{Z} in the unique subgroup of index 2 in ⟨\overline{X}, \overline{Y} , \overline{Z}⟩. The result we need for regular maps can actually be formulated for hypermaps, which we will do (with pointing out the situation for maps in appropriate places).

We begin with the easy case where the group ⟨\overline{X}, \overline{Y} , \overline{Z}⟩ ≅ PGL(2, F) contains ⟨\overline{R}, \overline{S}, \overline{T} ⟩ as a proper subgroup of index 2 (isomorphic to the unique copy K of PSL(2, F) in PGL(2, F)), we obviously have \overline{Z} ∉ K, and since \overline{R}, \overline{S} ∈ K, we must also have \overline{X}, \overline{Y} ∉ K.

It remains for us to consider the case where the group satisfies ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≅ ⟨\overline{X}, \overline{Y} , \overline{Z}⟩ ≅ PGL(2, F′); recall that the group ⟨\overline{X}, \overline{Y} , \overline{Z}⟩ and the corresponding rotation group ⟨\overline{R}, \overline{S}, \overline{T} ⟩ are related by R = Y Z, S = ZX, and T = XY. We know that now p is odd, and among the orders k, l, m of \overline{R}, \overline{S}, \overline{T} we cannot have both k and l equal to p. Hence we may assume that either k, l, m ≠ p, with m = 2 in the category of maps, or else k, l ≠ p and m = p, with l = 2 in the case of maps. Our goal is to clarify which of \overline{X}, \overline{Y} , \overline{Z} are contained in the unique subgroup of PGL(2, F′) isomorphic to PSL(2, F′). To this end it is sufficient to assume that we are in the situation described in Proposition 8, where R, S, T are as listed in Proposition 2 and X, Y, Z are as given by Proposition 5.

Proposition 13 Let ⟨\overline{X}, \overline{Y} , \overline{Z}⟩ = ⟨\overline{R}, \overline{S}, \overline{T} ⟩ ≅ PGL(2, F′) and let K be the (unique) subgroup of ⟨\overline{X}, \overline{Y} , \overline{Z}⟩ isomorphic to PSL(2, F′). Let sq(F′) be the set of non-zero squares of F′. Also let A be a 2-element subset of {R, S, T } such that ρ(tr(A)) = −tr(A) for A ∈ A and ρ(tr(A)) = tr(A) for A ∈ {R, S, T } \ A. For k, l, m ≠ p we have:

(1) if A = {R, S}, then Z ∈ K and X, Y ∉ K if −D ∈ sq(F′), while Z ∉ K and X, Y ∈ K if −D ∉ sq(F′);

(2) if A = {S, T }, then X ∈ K and Y, Z ∉ K if −D ∈ sq(F′), while X ∉ K and Y, Z ∈ K if −D ∉ sq(F′);

(3) if A = {T, R}, then Y ∈ K and Z, X ∉ K if −D ∈ sq(F′), while Y ∉ K and Z, X ∈ K if −D ∉ sq(F′). In all these cases, if m = 2, then the elements X and Y commute. On the other hand, if k, l ≠ p and m = p, then A = {R, S}, and (4) Z ∈ K and X, Y ∉ K if −D ∈ sq(F′), while Z ∉ K and X, Y ∈ K if −D ∉ sq(F′). Moreover, if l = 2, then X and Z commute.

Proof. Suppose first that A = {R, S} that is, ρ(tr(R)) = −tr(R) and ρ(tr(S)) = −tr(S). By the arguments developed in the proofs of Propositions 6, 8 and 12 we conclude that Z ∈ K if and only if −D ∈ sq(F′); by the same token we have R, S ∉ K. It follows that X, Y ∉ K if and only if Z ∈ K. Bearing in mind the conditions for k, l, m, this proves (1) and (4). Now, let k, l ≠ p and m = 2; then, k, l ≥ 3. If A = {S, T }, then we set R′ = S, S′ = T, T′ = R, X′ = Y, Y′ = Z, and Z′ = X. Since the order of R′ is ≥ 3, we may apply the above to the dashed symbols and conclude a dashed version of (1), which gives (2). Finally, if A = {S, T }, we set R′ = R^{−1}, S′ = T^{−1}, T′ = S^{−1}, X′ = X, Y′ = Z and Z′ = Y. Again, we may apply (1) to the dashed symbols, which translates to (3). The claims about commuting elements are obvious. □

7 Remarks

The explicit form of generating matrices given in Propositions 1, 2, 4 and 5 make it possible to perform computations with the associated groups using software packages such as gap or Magma. Also, our approach clarifies a large number of details not covered in [10] and furnishes a different proof of identification of the minimal field (Proposition 6).

It is not clear whether an enumeration result as the one in Theorem 1 could be proved for regular hypermaps on non-orientable surfaces. While the regular hypermaps with rotational symmetry group isomorphic to PGL(2, F′) automatically admit a non-orientable realisation and are enumerated by part (1) of Theorem 1, the hypermaps with rotation group isomorphic to PSL(2, F) seem to present difficulties. By part (3) of Proposition 12, such a hypermap of type (k, l, m) for a particular choice of ω_{κ}, ω_{λ}, ω_{μ} with D ≠ 0 admits a non-orientable realisation if and only if −D is a square in F. The problem here is that for a fixed type (k, l, m), different choices of values of ω_{κ}, ω_{λ}, ω_{μ} can give all kinds of different values of D: squares, non-squares, and even zero. To see this, let F = GF (17) and let (k, l, m) = (4, 8, 8), so that (κ, λ, μ) = (8, 16, 16). It can be checked that 2 is a primitive 8th root of unity in F while 3 and 5 are primitive 16th roots of unity in F. In all our examples we set ω_{κ} = 2 + 2^{−1} = 2 − 8 = −6. If ω_{λ} = ω_{μ} = 3 + 3^{−1} = 3 + 6 = −8, then we obtain D = 0. Choosing ω_{λ} = ω_{μ} = 5 + 5^{−1} = 5 + 7 = −5 gives −D = 6, a non-square in F. Finally, letting ω_{λ} = −8 and ω_{μ} = −5 leads to −D = −4, which is a square in F.

Acknowledgements. The research for this paper was carried out while the second and third authors were affiliated with the University of Auckland. They gratefully acknowledge the support from the NZIMA programme on Combinatorics and its Applications. Also the first and third authors wish to acknowledge partial support from the Marsden Fund, and the third author would like to acknowledge partial support from APVV Grants No. 20-000704 and 0040-06.

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