Branched cyclic regular coverings over platonic maps

European Journal of Combinatorics 36 (2014) 531–549

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Branched cyclic regular coverings over platonic maps Kan Hu a,b,1 , Roman Nedela a,b , Na-Er Wang a,c a

Faculty of Natural Sciences, Matej Bel University, Tajovského 40, 974 01 Banská Bystrica, Slovak Republic

b

Mathematical Institute, Slovak Academy of Sciences, 975 49 Banská Bystrica, Slovak Republic

c

School of Mathematics, Physics and Information Science, Zhejiang Ocean University, Zhoushan, Zhejiang 316000, People’s Republic of China

article

info

Article history: Received 25 February 2013 Accepted 22 September 2013 Available online 23 October 2013

abstract A map is a 2-cell decomposition of a closed surface. A map on an orientable surface is called regular if its group of orientationpreserving automorphisms acts transitively on the set of darts (edges endowed with an orientation). In this paper we investigate regular maps which are regular covers over platonic maps with a cyclic group of covering transformations. We describe all such maps in terms of parametrised group presentations. This generalises the work of Jones and Surowski [G.A. Jones, D.B. Surowski, Cyclic regular coverings of the Platonic maps, European J. Combin. 21 (2000) 333–345] classifying the cyclic regular coverings over platonic maps with branched points exclusively at vertices, or at face-centres. © 2013 Elsevier Ltd. All rights reserved.

1. Introduction A map M = (Γ , S ) is an embedding i : Γ ↩→ S of a connected graph Γ into a closed surface S such that each component of S − i(Γ ) is homoeomorphic to an open disc. The graph Γ is called the underlying graph of M and S is called the supporting surface of M . In this paper we exclusively consider maps on orientable surfaces endowed with a global orientation. A homomorphism, or covering, from a map N to a map M is a surjective mapping taking the darts of N onto the darts of M which extends to an orientation-preserving (branched) covering between

E-mail addresses: [email protected] (K. Hu), [email protected] (R. Nedela), [email protected] (N.-E. Wang). 1 Tel.: +421 484701204. 0195-6698/$ – see front matter © 2013 Elsevier Ltd. All rights reserved. http://dx.doi.org/10.1016/j.ejc.2013.09.006

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the supporting surfaces. A covering is called s-sheeted if the fibre over every dart of M has the same size s. Maps on orientable surfaces together with homomorphisms form a category of oriented maps. For a map M with the underlying graph Γ we denote by V , E and F , respectively, the sets of its vertices, edges and faces—the simply-connected components of S − i(Γ ). A map is finite if its dart set is finite, otherwise, it is infinite. Throughout, we assume the maps are all finite, except otherwise stated. An automorphism of a map M is a covering M → M . In other words, a map automorphism is an automorphism of the underlying graph (considered as a permutation of darts) which extends to an orientation-preserving self-homeomorphism of the supporting surface. It is well-known that Aut (M ) acts freely on the set of darts of M . In particular, this implies |Aut (M )| ≤ |D| = 2|E |, where D is the set of darts. In case, |Aut (M )| = |D|, the action of Aut (M ) is regular and the map itself will be called regular. The prominent examples of regular maps are the maps determined by the 2-skeletons of the five Platonic solids. A covering p: N → M between maps is regular if there exists a subgroup H ≤ Aut (N ) acting transitively on a fibre over a dart of M . Moreover, if both M and N are regular, H E Aut (N ) and M∼ = N /H. One of the most important problems in the theory of maps is the classification of regular maps satisfying some additional constraints posed on the surface, graph or on the group of automorphisms. The covering techniques, due to its capability to discover new regular maps from old, have made great contributions to this research. For historical reasons, these techniques have been independently developed by various authors in different mathematical languages: the group extension method, the voltage-assignment method and the homology method. An observation that a covering N → M between regular maps is necessarily regular with a group of covering transformations being normal in Aut (N ) relates the theory of regular maps with group extensions. The group extension method was widely employed by Coxeter et al. in the context of the study of polyhedral groups during the first half of 20th century [2,11,12], while Gross’ voltageassignment technique [4] and Biggs’ homology method [1] were introduced in the 1970s to the study of coverings of graphs and maps. They present a constructive combinatorial tool to describe a regular cover over a given map [13,15]. Cyclic coverings between maps are for many reasons of special importance. Restriction to cyclic coverings gives a chance to employ the results coming from both group theory and combinatorics. The construction and classification problem of cyclic regular coverings of the platonic maps have been considered by various authors in distinct context [2,11,12,8,13]. The most recent result in this direction was given by Jones and Surowski who have classified the regular maps which are cyclic regular coverings over the platonic maps, (possibly) branched exclusively over the vertices, or over the face-centres of the base maps [14, Theorem 1]. In the present paper, we continue the investigation of cyclic regular coverings over platonic maps. We deal with the general case in which the branched points may occur simultaneously over the vertices and face-centres of the base maps. The idea is, by separating the roles of vertices and faces, to decompose the covering into several subcoverings which possess certain extremal properties. Applying the theory of cyclic extensions of groups, the problem is eventually reduced to the study of solutions of a system of certain number-theoretical congruences. 2. Regular maps and regular coverings In a regular map M , every vertex has the same valency dV and every face has same degree dF for some integers dV , dF . The pair {dF , dV } is called the type of M . In general, the platonic maps are defined to be the regular maps (without semiedges) of type {dF , dV }, where 1/dF + 1/dV > 1/2. Their automorphism groups have presentation

⟨x, z | xdV = z dF = (xz )2 = 1⟩.

(1)

The platonic maps consist of the tetrahedral map T of type {3, 3}, the cube C of type {4, 3} and its dual—the octahedral map O of type {3, 4}, the icosahedral map I of type {3, 5} and its dual—the

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dodecahedral map D of type {5, 3}, the dihedral maps Dn of type {n, 2} and their duals—the hosohedral maps Hn of type {2, n} (n ≥ 1). It is well-known that the platonic maps exhaust all spherical regular maps (without semiedges), and Aut (T ) ∼ = A4 , Aut (C ) ∼ = Aut (O ) ∼ = S4 , Aut (I) ∼ = Aut (D ) ∼ = A5 , Aut (Dn ) ∼ = Aut (Hn ) ∼ = D2n . In a regular map M each pair of adjacent vertices is joined by the same number of edges, say mV , which we call the multiplicity of M . Since the dual map M ∗ is also regular, the multiplicity of the dual map M ∗ will be called the dual multiplicity of M and is denoted by mF . The pair (mF , mV ) will be called the multiplicity type of M . Regular maps of multiplicity mV = 1 are called simple. For instance, the dihedral map Dn has type {n, 2} and multiplicity type (n, 1), and the cube has type {4, 3} and multiplicity type (1, 1). In a regular map M , there exist two automorphisms, say x and z, such that x fixes a vertex v of valency dV by cyclically permuting the edges based at v , and z fixes a face f of degree dF incident to v by cyclically permuting the edges on the boundary walk of f . It follows that the product xz fixes an edge incident to both v and f , and hence (xz )2 = 1. Due to the regularity, the automorphism group is generated by x and z. Conversely, given a two-generated (finite) group G = ⟨x, z ⟩ satisfying the relation (xz )2 = 1, we can construct an associated regular map with the set of darts D = G as follows. We identify the vertices and edges of the underlying graph with the orbits of x and xz in the action by left multiplication, respectively. The boundary walks of faces are set to be the cycles of z in the action by left multiplication and the supporting surface is constructed by gluing a 2-cell to each such a walk. The incidence relation between vertices, edges and faces is given by non-empty intersections of the corresponding orbits in the left action of x, xz and z. It can be easily checked that the elements of G acting on D = G by right multiplication are automorphisms of the map. A regular map obtained by the above construction will be called an algebraic map and denoted by (G, x, z ), (xz )2 = 1. It follows that two algebraic maps (G1 , x1 , z1 ) and (G2 , x2 , z2 ) are isomorphic if and only if the mapping x1 → x2 , z1 → z2 extends to a group isomorphism. Therefore, algebraic maps can be used as synonyms of regular maps. We define some operations on regular maps. Let M = (G, x, z ) be a regular map. The dual M ∗ of M is the map (G, z , x). A map M is called self-dual if M ∼ = M∗ . The mirror image M−1 of M is the −1 −1 −1 ∼ map (G, x , z ). The map M is said reflexible if M = M , otherwise, it is chiral. Finally, the parallel product of two regular maps Mi = (Gi , xi , zi ) (i = 1, 2) is the map (G1 ×p G2 , (x1 , x2 ), (z1 , z2 )), where G1 ×p G2 = ⟨(x1 , x2 ), (z1 , z2 )⟩ is a subgroup generated by (x1 , x2 ) and (z1 , z2 ) in the direct product G1 × G2 [17,3]. The following result describes the relationship between regular maps and regular coverings of maps. Proposition 1 ([10]). Let p: N → M be a covering between two maps N and M . Then the following statements hold true: (i) if N and M are regular, then the covering p is regular, (ii) if the covering p is regular and the map M is regular, so is N . It follows that each covering p: N → M between regular maps induces a group epimorphism from Aut (N ) onto Aut (M ) such that Aut (N )/CT (p) ∼ = Aut (M). Therefore, the determination of regular coverings over the regular map M with a prescribed covering transformation group K = CT (p) is equivalent to the determination of particular extensions of the group Aut (M ) by K . Let N and M be two regular maps of types {d˜ F , d˜ V } and {dF , dV }, respectively, let p: N → M be an s-sheeted covering. Throughout, we assume CT (p) acts freely on the edges of N . The covering is called branched at vertices if d˜ V = dV sV , and branched at faces if d˜ F = dF sF , where sV |s and sF |s. In particular, the number sV is called the branch index at the vertices, and sF is called the branch index at faces. The covering is said to be smooth at vertices, if sV = 1, smooth at faces if sF = 1, and smooth if it is both smooth at vertices and smooth at faces. On the contrary, the covering is said to be totally branched at

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vertices if sV = s, totally branched at faces if sF = s, and totally branched if it is both totally branched at faces and totally branched at vertices. 3. Decomposition of branched coverings In this section, we show that each branched covering between regular maps can be decomposed into several subcoverings which possess certain extremal properties. For convenience, the following convention will be assumed throughout this section. Convention 2. Let M and N be regular maps (without semiedges) of type {dF , dV } and {d˜ F , d˜ V }, respectively, with p: N → M an s-sheeted regular covering. Denote N = (G, x, z ), (xz )2 = 1, and set A = ⟨z dF ⟩,

B = ⟨xdV ⟩,

KF = AG

and

KV = BG .

(2)

Namely, KF (resp. KV ) is the normal closure of A (resp. B) in G. Then KF , KV , KF KV and KF ∩ KV are all normal subgroups of G contained in CT (p). Set

MF = N /KF ,

MF ∨ MV = N /KF ∩ KV ,

MV = N /KV ,

MF ∧ MV = N /KF KV .

(3)

Denote by pi (i = 0, 1, . . . , 5) the si -sheeted covering defined by Diagram (4).

N p0

s0

 MF ∨ MJV JJ p u p1 uu JJ 2 J usu u s2 JJ u 1 J$ u zu MF I MV II t s4 tt IIs3 I tt p3 II tt p I$ ztt 4 MF ∧ MV p5



(4)

s5

M The following lemma describes the relations between the subcoverings defined above. Lemma 3. With Convention 2, the following statements hold true: (i) s1 = s4 , s2 = s3 , s = s0 s1 s2 s5 , (ii) the map MV is the largest regular map which covers M smoothly at the vertices and is covered by N , in particular, it has type {q1 dF , dV } for some q1 |s1 , (iii) the map MF is the largest regular map which covers M smoothly at the faces and is covered by N , in particular, it has type {dF , q2 dV } for some q2 |s2 , (iv) the map MF ∧ MV is the largest regular map which covers M smoothly and is covered by N , in particular, it has type {dF , dV }, (v) the map MF ∨ MV is the least regular map which is covered by N and covers both MF and MV , in particular, it has type {q1 dF , q2 dV }, (vi) the map N has type {q0 q1 dF , q0 q2 dV } for some q0 |s0 , (vii) MF ∨ MV ∼ = MF ×p MV , the parallel product of MF and MV .

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Proof. By (2), we have 1 ≤ KF ∩ KV ≤ KF ≤ KF KV ≤ CT (p) and 1 ≤ KF ∩ KV ≤ KV ≤ KF KV ≤ CT (p). By the Group Isomorphism Theorem, we have KV KF /KV ∼ = KF /KF ∩ KV

and

KV KF /KF ∼ = KV /KF ∩ KV .

Therefore, s1 = |KV KF /KF | = |KV /KF ∩ KV | = s4 and s2 = |KV KF /KV | = |KF /KF ∩ KV | = s3 . By the definition of the groups KF and KV , the coverings have the claimed extremal properties (ii)–(v), (vi) is obvious, and (vii) follows from [3, Theorem 4.1].  We give an algebraic characterisation of the extremal coverings which are totally branched at faces or totally branched at vertices as follows.

˜ F, m ˜ V }. Then Proposition 4. With Convention 2, and assume N have multiplicity type {m ˜ V ; in (i) the covering p is totally branched at vertices if and only if CT (p) = B, in which case s divides m ˜ V, particular, if M is simple, then s = m ˜ F ; in (ii) the covering p is totally branched at faces if and only if CT (p) = A, in which case s divides m ˜ F, particular, if M ∗ is simple, then s = m ˜ V,m ˜ F ); (iii) the covering p is totally branched if and only if CT (p) = A = B, in which case s divides gcd(m ˜V = m ˜ F. in particular, if both M and M ∗ are simple, then s = m Proof. Assume N = (G, x, z ), (xz )2 = 1, and denote the type of N and M by {d˜ F , d˜ V } and {dF , dV }, respectively. Then by (2), B ≤ CT (p). If p is totally branched at vertices, then d˜ V = dV s. It follows that |B| = s. Since |CT (p)| = s, we get B = CT (p). Conversely, if CT (p) = B, then d˜ V = dV |B| = dV |CT (p)| = dV s. It follows that p is totally branched at vertices. Moreover, let Fix V (N ) denote the ˜ V = |Fix V (N )|. It maximal normal subgroup of G contained in ⟨x⟩. Then CT (p) ≤ Fix V (N ) and m ˜ V . In particular, if M is simple, then CT (p) = Fix V (N ), and hence s = m ˜ V . This proves follows that s|m (i). It is easily seen that (ii) follows from duality, and (iii) follows from (i) and (ii).  4. Almost totally branched coverings In this section, inspired by the characterisation of totally branched coverings in Proposition 4, we define a new covering as follows. Definition 5. With Convention 2, the covering p: N A E G, B E G and CT (p) = AB.

→ M is called almost totally branched if

Almost totally branched coverings between regular maps generalises the extremal coverings characterised in Proposition 4. Proposition 6. If a covering p: N → M between regular maps is totally branched at vertices, or totally branched at faces, or totally branched, then it is almost totally branched. Proof. If p is totally branched at faces, then by Proposition 4, A = CT (p). Clearly, A E G, B ≤ A. Since A is cyclic, B char A. It follows that B E G. By Definition 5, p is almost totally branched. Similar arguments prove the remaining statements. 

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The following propositions characterise almost totally branched coverings. Proposition 7. With Convention 2, a covering p: N → M between regular maps is almost totally branched if and only if the following conditions are satisfied: (i) p0 : N → MF ∨ MV is totally branched, (ii) p1 : MF ∨ MV → MF or p4 : MV → MF ∧ MV is totally branched at faces, and smooth at vertices, (iii) p2 : MF ∨ MV → MV or p3 : MF → MF ∧ MV is totally branched at vertices, and smooth at faces, (iv) p5 : MV ∧ MF → M is an isomorphism. Proof. By Convention 2, the base map M and its cover N have types {dF , dV } and {d˜ F , d˜ V }, respectively. By Lemma 3, the covering p has a decomposition p = p5 p3 p1 p0 described in Diagram (4), where CT (p1 p0 ) = KF ,

CT (p2 p0 ) = KV ,

CT (p0 ) = KF ∩ KV ,

CT (p3 p1 p0 ) = CT (p4 p2 p0 ) = KF KV ,

CT (p1 ) ∼ = KF /KV ∩ KF ,

(5)

CT (p2 ) ∼ = KV /KV ∩ KF .

By the Lattice Isomorphism Theorem, p1 is totally branched at faces and smooth at the vertices if and only if so is p4 , and p2 is totally branched at vertices and smooth at faces if and only if so is p4 . Therefore, it is sufficient to consider one of the cases in (ii) (and in (iii)). First assume p: N → M is almost totally branched. Then by Definition 5, A E G, B E G and CT (p) = AB. By (2), we have KF = A and KV = B. It follows CT (p) = CT (p3 p1 p0 ), and hence p5 is an isomorphism. Clearly, d˜ F /|A| = dF and d˜ V /|B| = dV . Let h0 = |A ∩ B|, qF = |A|/h0 and qV = |B|/h0 . Then MF , MV and MF ∨ MV , respectively, have types {dF , d˜ V /h0 }, {d˜ F /h0 , dV } and {d˜ F /h0 , d˜ V /h0 }. Since d˜ F h0

=

d˜ F

|A|

·

|A| h0

= dF qF and

d˜ V h0

=

d˜ V

|B|

·

|B| h0

= dV qV ,

the above maps, respectively, have types (cf. Diagram (6))

{dF , qV dV },

{qF dF , dV } and {dF qF , dV qV }.

By (5), we have

|CT (p0 )| = h0 ,

|CT (p1 )| = qF and |CT (p2 )| = qV .

Therefore, p1 is totally branched at faces and smooth at vertices, and p2 is totally branched at vertices and smooth at faces. Since N has type {d˜ F , d˜ V }, where d˜ F = dF |A| = dF qF h0 and d˜ V = dV |B| = dV qV h0 , the covering p0 is totally branched. Conversely, assume the statements (i) through (iv) hold true. Let

|CT (p0 )| = h0 ,

|CT (p1 )| = qF and |CT (p2 )| = qV .

Then by (i) (resp. (ii), (iii)), the map MF ∨ MV (resp. MF , MV ) has type {d˜ F /h0 , d˜ V /h0 } (resp.

{d˜ F /h0 qF , d˜ V /h0 }, {d˜ F /h0 , d˜ V /h0 qV }). Since

|CT (p1 p0 )| = |CT (p1 )| |CT (p0 )| = qF h0 , p1 p0 is totally branched at faces. Similarly, since

|CT (p2 p0 )| = |CT (p2 )| |CT (p0 )| = qV h0 ,

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p2 p0 is totally branched at vertices. By Proposition 4, CT (p1 p0 ) = A and CT (p2 p0 ) = B. By (5), we have KF = A and KV = B. By (iv), p5 is an isomorphism. It follows that CT (p) = CT (p3 p1 p0 ), and hence CT (p) = KF KV = AB. Therefore, by Definition 5, p is almost totally branched. 

{h0 qF dF , h0 qV dV } o

(6)

N p0

h0

 MF ∨ MIV II u p1 uu IIp2 I uqu u qV II u F I$ zuu MF I MV II u u qF u IIqV I uu p3 II uu p4 I$ u zu MF ∧ MV

{qF dF , qV dV } o {dF , qV dV } o

/ {qF dF , dV } / {dF , dV }

∼ =



M

/ {dF , dV }

Proposition 8. With Convention 2, if p: N → M is almost totally branched, then (i) there exists an integer eV ∈ Z∗˜ , e2V ≡ 1 (mod d˜ V /dV ), such that (xdV )z = (xdV )eV , dV /dV (ii) there exists an integer eF ∈ Z∗˜ , e2F ≡ 1 (mod d˜ F /dF ), such that (z dF )x = (z dF )eF . dF /dF

Proof. By Convention 2, A = ⟨z dF ⟩ and B = ⟨xdV ⟩. By the assumption, p is almost totally branched. By Definition 5, A E G, B E G and CT (p) = AB. It follows that there exists an integer eV such that 2

(xdV )z = (xdV )eV . Upon conjugation, we get (xdV )zx = (xdV )eV . Consequently, (xdV )(zx) = (xdV )eV . 2 Since (zx)2 = 1, we have xdV = (xdV )eV . It follows that e2V ≡ 1 (mod d˜ V /dV ). This proves (i). The other 2

statement (ii) can be deduced similarly.



The integers eV and eF appeared in Proposition 8 will be called the twist exponent and dual twist exponent of the covering p, respectively. Furthermore, we set sF = |A|,

sV = |B|, h = gcd(sF , sV ),

h0 = |A ∩ B|, qF = sF /h0 ,

(7)

qV = sV /h0 .

Notice that the number sV (resp. sF ) is equal to the branch index at vertices (resp. face-centres) of M . It is evident A ∩ B ≤ Z (G) is cyclic, and AB is metacyclic. Since d˜ F = dF sF and d˜ V = dV sV , we have A ∩ B = ⟨xdV sV /h0 ⟩ = ⟨xdV qV ⟩

and

A ∩ B = ⟨z dF sF /h0 ⟩ = ⟨z dF qF ⟩.

It follows that there exists an integer e, gcd(e, h0 ) = 1, such that z dF qF = (xdV qV )e .

(8)

The integers h0 and e will be called the intersection index and shift exponent, respectively. To summarise, associated to each almost totally branched covering between regular maps, there exist 9 interdependent integer parameters sF , sV , qF , qV , eF , eV , h0 , h, e.

(9)

By (7), the parameters qF , qV , h0 , eF , eV , e determine the others. For convenience, the triples (qF , qV , h0 ) and (eF , eV , e) will be called the index-triple and the exponent-triple, respectively, while the 6-tuple (qF , qV , h0 , eF , eV , e) will be called the index-exponent 6-tuple.

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In what follows we study the structure of covering transformation group of an almost totally branched covering in more detail. The following lemma will be useful. Lemma 9. Let G = XY be a finite group, where X and Y are cyclic subgroups of G. If G is abelian, then G is cyclic if and only if |X ∩ Y | = gcd(|X |, |Y |). Proof ([7]). Let q1 = X /X ∩ Y and q2 = Y /X ∩ Y . Then |X ∩ Y | = gcd(|X |, |Y |) if and only if gcd(q1 , q2 ) = 1. First assume G is cyclic, then G/X ∩ Y ∼ = Zq1 × Zq2 is also cyclic. It follows that gcd(q1 , q2 ) = 1. Conversely, assume gcd(q1 , q2 ) = 1. If G is not cyclic, then being finite and abelian it contains a subgroup P ∼ = Z2p for some prime p. Consider the natural epimorphism π : G → G/X = XY /X ∼ = Y /(X ∩ Y ). Since P is non-cyclic and ker(π ) = X is cyclic, P ̸≤ X , and hence π (P ) > 1. It follows that p divides |Y /X ∩ Y | = q2 . A similar argument, considering the projection onto G/Y , shows that p also divides q1 . This contradicts to the assumption, so G is cyclic.  The structure of covering transformation group of an almost totally branched covering and the (partial) relations between the associated parameters are described in the following lemma. Lemma 10. Let p: N → M be an almost totally branched covering between two regular maps, with an index–exponent 6-tuple (qF , qV , h0 , eF , eV , e), gcd(e, h0 ) = 1, assume the type of M is {dF , dV }. Then the following statements hold true: (i) If CT (p) is abelian, then eF ≡ 1 (mod h0 ), gcd(d ,2) eF V

eV ≡ 1 (mod h0 ), gcd(d ,2) eV F

≡ 1 (mod qF h0 ),

(10)

≡ 1 (mod qV h0 ),

(11)

and CT (p) has a presentation

⟨a, b | aqF h0 = bqV h0 = [a, b] = 1, aqF = bqV e ⟩.

(12)

(ii) If CT (p) is non-abelian, then dV and dF are both odd, qF , qV , h0 are all even, eF ≡ 1 +

qF h0 2

(mod qF h0 ),

eV ≡ 1 +

qV h0 2

(mod qV h0 ),

(13)

and CT (p) has a presentation



a, b | aqF h0 = bqV h0 = 1, ab = a1+

q F h0 2

, ba = b1+

q V h0 2

 , aqF = bqV e .

(14)

(iii) CT (p) is cyclic if and only if gcd(qF , qV ) = 1. Proof. Let N = (G, x, z ), (xz )2 = 1. Denote a = z dF and b = xdV . By (2) and Definition 5, A = ⟨a⟩ E G, B = ⟨b⟩ E G and CT (p) = AB. By Proposition 8 and Eq. (8) ax = aeF ,

bz = beV

and

aqF = bqV e ,

(15)

where eF , eV and e satisfy e2V ≡ 1 (mod qV h0 ),

e2F ≡ 1 (mod qF h0 ) and

gcd(e, h0 ) = 1.

(16)

Upon conjugation, the first two relations of (15) are converted into ab = ax

dV

dV

= aeF

and ba = bz

dF

dF

= beV .

(17)

Therefore, CT (p), being metacyclic, is defined by presentation dV

dF

⟨a, b | aqF h0 = bqV h0 = 1, ab = aeF , ba = beV , aqF = beqV ⟩.

(18)

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(i) If CT (p) is abelian, then by (15), we have aqF = beqV = (beqV )x = (aqF )x = aqF eF and bqV e = aqF = (aqF )z = (bqV e )z = bqV eV e . Since o(a) = qF h0 , o(b) = qV h0 and gcd(e, h0 ) = 1, we get (10). Moreover, since CT (p) is abelian, dV

dF

[a, b] = 1. It follows from (17) that a = aeF and b = beV . Therefore, d

d

eF V ≡ 1 (mod qF h0 )

and eVF ≡ 1 (mod qV h0 ).

(19)

Combining (16) and (19), we get (11). In particular, the presentation (18) is reduced to (12). (ii) Assume CT (p) is non-abelian. If one of dV and dF is even, then by (16) and (17), we have [a, b] = 1, and hence CT (p) is abelian, a contradiction. Hence both dV and dF are odd. Moreover, by (15) and (17), we deduce that (15)

b = ba

qF (17)

dF qF

= beV

(15)

and a = ab

qV (17)

dV qV

= aeF

.

(20)

Assume to the contrary that qF or qV is odd. Since both dF and dV are odd, qF dF or qV dV is odd. By (16) and (20), either b = beV or a = aeF , and hence eF ≡ 1 (mod qF h0 ) or eV ≡ 1 (mod qV h0 ). Consequently, [a, b] = 1 implying that CT (p) is abelian, a contradiction. Therefore, both qF and qV are even. Moreover, we deduce from the third and fourth relations of (18) that aeF −1 = [a, b] = b1−eV .

(21)

Therefore, we have (21)

(21)

(17)

(21)

(21)

(17)

aeF −1 = b1−eV = (b1−eV )b = (aeF −1 )b = (ab )eF −1 = a1−eF and b1−eV = aeF −1 = (aeF −1 )a = (b1−eV )a = (ba )1−eV = beV −1 . It follows that a2(eF −1) = 1 and b2(eV −1) = 1. Since o(a) = qF h0 and o(b) = qV h0 , we have qF h0 |2(eF − 1) and qV h0 |2(eV − 1). Since CT (p) is non-abelian, eF ̸≡ 1 (mod qF h0 ) and eV ̸≡ 1 (mod qV h0 ), we obtain (13). Notice that ⟨a⟩ ∩ ⟨b⟩ = ⟨aqF ⟩ = ⟨bqV ⟩. By (21), we get eF − 1 ≡ 0 (mod qF ).

(22)

Consequently, by (13), (22) is reduced to qF h0 /2 ≡ 0 (mod qF ). Therefore, h0 is even. (iii) If CT (p) is cyclic, by Lemma 9, gcd(qF , qV ) = 1. Conversely, if gcd(qF , qV ) = 1, by (ii), CT (p) is abelian. By Lemma 9, CT (p) is cyclic, as desired.  It follows that the covering transformation group of an almost totally branched covering between regular maps is either non-abelian with a presentation (18), or it is an abelian group of rank 2, or it is cyclic. The following two examples show that the first two cases do happen. Example 1. Let N = (G, x, z ), where G = ⟨x, z | (xz )2 = x16 = z 48 = 1, (x2 )z = x10 , (z 2 )x = z 34 , x12 = z 12 ⟩. (8)

It is easy to check that N corresponds to a regular embedding of a multicycle C12 of length 12 [5, Theorem 2.2]. The map has type {48, 16} and genus 41. Let A = ⟨z 2 ⟩, B = ⟨x2 ⟩ and K = AB. From the presentation, we have A E G and B E G, and hence K E G. Clearly, N /K ∼ = D2 . Hence N is an almost totally branched cover over D2 . It can be easily checked that

(sV , sF , qV , qF , h0 , h, eV , eF , e) = (8, 24, 2, 6, 4, 8, 5, 17, 3).

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K. Hu et al. / European Journal of Combinatorics 36 (2014) 531–549

Let a = z 2 and b = x2 . Then K has a presentation K = ⟨a, b | a24 = b8 = [a, b] = 1, a6 = b6 ⟩. Let w = ba

−1

(23)

. Then

K = ⟨a, b⟩ = ⟨b, w | b8 = w 6 = [b, w] = 1⟩. Since ⟨b⟩ ∩ ⟨w⟩ = 1, it follows that K ∼ = Z8 × Z6 . Therefore, the covering transformation group is a non-cyclic abelian group. Example 2. Let N = (G, x, z ), where G = ⟨x, z | (xz )2 = x12 = z 12 = 1, (x3 )z = x9 , (z 3 )x = z 9 , x6 = z 6 ⟩. Let A = ⟨z 3 ⟩, B = ⟨x3 ⟩ and K = AB. Clearly, A, B E G and hence K E G. It is easily seen N /K ∼ = T. Therefore, N is an almost totally branched cover over T . In particular,

(sV , sF , qV , qF , h0 , h, eV , eF , e) = (4, 4, 2, 2, 2, 4, 3, 3, 1). Let a = z 3 and b = x3 . Then K has a presentation K = ⟨a, b | a4 = b4 = 1, ab = a3 , ba = b3 , a2 = b2 ⟩ ∼ = Q8 . Therefore, the covering transformation group is non-abelian.

5. Almost totally branched abelian coverings of platonic maps In this section, we determine all regular maps which are almost totally branched coverings of the platonic maps with abelian covering transformation groups. The following lemma shows that the cyclic regular coverings of the platonic maps form a special subclass. Lemma 11. Let p: N → P be a regular covering from a map N to a platonic map P . Then the covering is cyclic if and only if it is almost totally branched with gcd(qF , qV ) = 1. Proof. Assume P has type {dF , dV } and N = (G, x, z ), (xz )2 = 1. Since P is platonic, CT (p) = {xdV , z dF }G —the normal closure of the set {xdV , z dF } in G. Set A = ⟨z dF ⟩ and B = ⟨xdV ⟩. First assume CT (p) is cyclic, then A char CT (p) and B char CT (p). Since CT (p) E G, we have A E G, B E G. It follows that AB E G, and hence CT (p) = AB. Therefore, the covering is almost totally branched. By Lemma 10(iii), gcd(qF , qV ) = 1. Conversely, assume p is almost totally branched with gcd(qF , qV ) = 1. Then by Lemma 10(iii), CT (p) is cyclic, as claimed.  To formulate our main result, we define a family of regular maps as follows. Definition 12. Let P be a platonic map of type {dF , dV } with lF faces and lV vertices, let P˜ = P˜ (qF , qV , h0 ; eF , eV , e) denote the regular map (G, x, z ), G = ⟨x, z | (xz )2 = z dF qF h0 = xdV qV h0 = [z dF , xdV ] = 1,

(z dF )x = z dF eF , (xdV )z = xdV eV , z dF qF = xdV qV e ⟩,

(24)

where h0 , qV , qF are positive integers, eV ∈ Zh0 qV , eF ∈ Zh0 qF , e ∈ Z∗h0 and the parameters satisfy (10), (11) and the following congruences:

K. Hu et al. / European Journal of Combinatorics 36 (2014) 531–549

541

lV −1



eiV ≡ 0 (mod qV ),

(25)

eiF ≡ 0 (mod qF ),

(26)

i =0 lF −1

 i=0

lF −1

e



l V −1

eiF /qF +



i=0

eiV /qV ≡ 0 (mod h0 ).

(27)

i =0

The following theorem on cyclic extensions of groups will be crucial to the proof of the subsequent main result of the paper. Theorem 13 ([6, Theorem 3.36]). Let K and Q be groups, where Q is cyclic of order m, let a ∈ K and

σ ∈ Aut (K ). Assume that aσ = a

and

xσ

m

= xa ,

∀x ∈ K .

Then there exists an extension G of Q by K , unique up to isomorphism, with the following properties: (i) G/K = ⟨gK ⟩ ∼ = Q , (ii) g m = a, (iii) xσ = xg . Theorem 14. Let P be a platonic map of type {dF , dV } with lF faces and lV vertices. An almost totally branched abelian cover over the platonic map P with an index–exponent 6-tuple (qF , qV , h0 , eF , eV , e) is isomorphic to the map P˜ given by Definition 12. Conversely, given any 6-tuple (qF , qV , h0 , eF , eV , e) of positive integers satisfying the congruences (10), (11), (25), (26) and (27), the regular map P˜ given by Definition 12 is an almost totally branched abelian cover over P and (qF , qV , h0 , eF , eV , e) is the index–exponent 6-tuple of the covering. In particular, the cyclic regular covers over P are such maps with an extra condition that gcd(qF , qV ) = 1. Proof. By duality, the given base map P can be assumed to be the dihedral map Dn , the tetrahedral map T , the cube C or the icosahedral map I. Our proof will proceed case by case. Case 1. The base map is the dihedral map Dn of type {n, 2}. Assume N = (G, x, z ), (xz )2 = 1, is an almost totally branched abelian cover over Dn with the index–exponent 6-tuple (qF , qV , h0 , eF , eV , e). Let a = z n , b = x2 and K = ⟨a, b⟩. By Definition 5, K is the covering transformation group. By the assumption, K is abelian. By Proposition 8, we have ax = aeF

and

bz = beV ,

for some eF ∈ Z∗qF h0 and eV ∈ Z∗qV h0 . By Lemma 10(i), K has a presentation (12), and the parameters qF , qV , h0 , eF , eV and e, gcd(e, h0 ) = 1, satisfy the congruences (10) and (11). Let N = ⟨K , z ⟩. Then N /K = ⟨zK ⟩ ∼ = Zn . It follows that N has a presentation N = ⟨a, b, z | aqF h0 = bqV h0 = [a, b] = 1, aqF = bqV e , z n = a, az = a, bz = beV ⟩.

(28)

Since (xz )2 = 1, we have z x = x−2 z −1 = b−1 z −1 .

(29)

By the relation bz = beV , we deduce that z −1 bi z = bieV ,

i ∈ Z.

(30)

Therefore, we derive from (29) and (30) that (29)

(30)

aeF = ax = (z n )x = (z x )n = (b−1 z −1 )n = b−

n−1 i=0

eiV

z n = b−

n−1 i=0

eiV

a.

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K. Hu et al. / European Journal of Combinatorics 36 (2014) 531–549

It follows that aeF −1 = b−

n−1 i=0

eiV

.

(31)

qF

qV

Since ⟨a⟩ ∩ ⟨b⟩ = ⟨a ⟩ = ⟨b ⟩, we get eF − 1 ≡ 0 (mod qF )

and

n −1 

eiV ≡ 0 (mod qV ).

(32)

i =0

Using the relation aqF = bqV e in (28), Eq. (31) is converted into b−

n−1 i=0

= aeF −1 = (aqF )(eF −1)/qF = bqV e(eF −1)/qF .

ei (31)

(32)

It follows that bqV e(eF −1)/qF + e(eF − 1)/qF +

n−1 

(28)

n−1 i=0

ei

= 1. Since o(b) = qV h0 , we obtain

ei /qV ≡ 0 (mod h0 ).

(33)

i=0

From (29), we see that N E ⟨a, b, z , x⟩ = ⟨x, z ⟩ = G and G/N = ⟨xN ⟩ ∼ = Z2 . It follows that G has a presentation G = ⟨a, b, x, z | aqF h0 = bqV h0 = [a, b] = 1, aqF = bqV e , z n = a, az = a, bz = beV , x2 = b, ax = aeF , bx = b, z x = b−1 z −1 ⟩,

(34)

or equivalently, by simplification, G = ⟨x, z | (xz )2 = z nqF h0 = x2qV h0 = [x2 , z n ] = 1, (z n )x = z neF ,

(x2 )z = x2eV , z nqF = x2qV e ⟩.

(35)

From the above discussion, we see K E N , N E G such that N /K ∼ = Zn and G/N ∼ = Z2 . Hence G is constructed from a series of cyclic extensions. Conversely, given a group G defined by the presentation (35) (or equivalently, (34)), with the parameters satisfying the congruences (10), (11), (32) and (33), we show the map N = (G, x, z ) is an almost totally branched abelian cover over Dn such that the index–exponent 6-tuple is (qF , qV , h0 , eF , eV , e). Let A = ⟨z n ⟩, B = ⟨x2 ⟩ and K = AB. Then from (35), it is easily seen A, B E G, K is abelian and G/K ∼ = D2n . Moreover, by Theorem 13, it is straightforward to verify that G is a welldefined extension of D2n by K . In particular, |G| = 2nqF qV e. Therefore, the covering N → Dn has index triple (qF , qV , h0 ), and consequently, exponent triple (eF , eV , e). Case 2. The base map is the tetrahedral map T of type {3, 3}. Assume N = (G, x, z ), (xz )2 = 1, is an almost totally branched abelian cover over T with the index–exponent 6-tuple (qF , qV , h0 , eF , eV , e). Let a = z 3 , b = x3 and K = ⟨a, b⟩. By Definition 5, K is the covering transformation group. By the assumption, K is abelian. By Proposition 8, we have and bz = beV ,

ax = aeF

for some eF ∈ Z∗qF h0 and eV ∈ Z∗qV h0 . By Lemma 10(i), K has a presentation (12), and the parameters qF , qV , h0 , eF , eV and e, gcd(e, h0 ) = 1, satisfy the congruences (10) and (11). In particular, we have

[x3 , z ] = [z 3 , x] = 1.

(36)

Let c1 = z x , c2 = zx, N1 = ⟨K , c1 ⟩ and N2 = ⟨N1 , c2 ⟩. Since (xz ) = 1, we have z Therefore, 2 2

2

(36)

c12 = (z 2 x2 )2 = (z 3 z −1 x−1 x3 )2 = z 6 x6 (xz )−2 = a2 b2 . Hence N1 /K = ⟨c1 K ⟩ ∼ = Z2 . By (36), we have a c1 = a

and bc1 = b.

−1 −1

x

= xz. (37)

K. Hu et al. / European Journal of Combinatorics 36 (2014) 531–549

543

It follows that N1 has a presentation N1 = ⟨a, b, c1 | aqF h0 = bqV h0 = [a, b] = 1, aqF = bqV e , c12 = a2 b2 , ac1 = a, bc1 = b⟩. Since (zx) = 1, we get zx = x

−1 −1

2

c c12

−1

−1

2

z

(38)

. It follows that

2 −1 −1

= x zx zx = x zx x z

= x−1 zxz −1 = x−2 z −2 = c1−1 .

Hence N1 E N2 and N2 /N1 = ⟨c2 N1 ⟩ ∼ = Z2 . By (36), we have b c2 = b .

ac2 = a and

Therefore, N2 is defined by a presentation N2 = ⟨a, b, c1 , c2 | aqF h0 = bqV h0 = [a, b] = 1, aqF = bqV e , c12 = a2 b2 , c ac1 = a, bc1 = b, c22 = 1, ac2 = a, bc2 = b, c12 = c1−1 ⟩.

(39)

Moreover, using the commuting rule (36), we deduce that (36) (36) c1x = x−1 z 2 x3 = ax−1 z −1 z 3 = ab(zx)−1 = abc2−1 ,

c2x = x−1 zx2 = x−1 z −1 z 2 x2 = c2−1 c1 . Therefore, N2 E G = ⟨x, z ⟩ = ⟨a, b, c1 , c2 , x⟩, and G/N2 = ⟨xN2 ⟩ ∼ = Z3 . It follows that G has a presentation G = ⟨a, b, x, z | a = z 3 , c1 = z 2 x2 , c2 = zx, aqF h0 = bqV h0 = [a, b] = 1, aqF = bqV e , c c12 = a2 b2 , [c1 , a] = [c1 , b] = 1, c22 = 1, [c2 , a] = [c2 , b] = 1, c12 = c1−1 ,

x3 = b, [x, a] = [x, b] = 1, c1x = abc2−1 , c2x = c2−1 c1 ⟩,

(40)

or equivalently, by simplification, G = ⟨x, z | x3qF h0 = z 3qV h0 = [x3 , z 3 ] = 1, (z 3 )x = z 3 , (z 3 )x = z 3 , z 3qF = x3qV e ⟩. Recall that c1 = z 2 x2 and c2 = zx. By the relation

c c12

(41)

−1

= c1 , we have

c c1−1 = c12 = (z 2 x2 )zx = x−1 zx2 zx = x−1 zx−1 x3 zx (36)

= bx−1 z (x−1 z −1 )z 2 x = bx−1 z 2 xz 2 x = b(x−1 z −1 )z 3 xz 2 x

(36)

(36)

= bazx2 z 2 x = bazx−1 x3 z −1 z 3 x = b2 a2 z (x−1 z −1 )x = b2 a2 z 2 x2 = b2 a2 c1 .

It follows that (37)

a−2 b−2 = c12 = a2 b2 . Consequently, a4 = b − 4 .

(42) qF

qV

Since ⟨a⟩ ∩ ⟨b⟩ = ⟨a ⟩ = ⟨b ⟩, we get 4 ≡ 0 (mod qF ) and Moreover, using the relation a

4 ≡ 0 (mod qV ). qF

=b

qV e

, (42) is converted into

b−4 = a4 = (aqF )4/qF = b4qV e/qF . (43)

Consequently, b

4+4qV e/qF

(43)

(44)

= 1. Since o(b) = qV h0 , we get

4e/qF + 4/qV ≡ 0 (mod h0 ).

(45)

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K. Hu et al. / European Journal of Combinatorics 36 (2014) 531–549

From the above discussion, we see K E N1 , N1 E N2 and N2 E G such that N1 /K ∼ = Z2 , N2 ∼ = N1 oZ2 ∼ and G/N2 = Z3 . Hence G is constructed from a series of cyclic extensions. Conversely, given the group G defined by (41) (or equivalently, (40)) with the parameters satisfying congruences (10), (11), (43) and (45), it is straightforward to verify that the map N = (G, x, z ) is an almost totally branched abelian cover over T . By Theorem 13, G is a well-defined extension of Aut (T ) ∼ = A4 by K . In particular, |G| = 12qF qV h0 . Therefore, the covering has index-triple (qF , qV , h0 ), and consequently, the exponent triple (1, 1, e). Case 3. The base map is the cube C of type {4, 3}. Assume N = (G, x, z ), (xz )2 = 1, is an almost totally branched abelian cover over C with the index–exponent 6-tuple (qF , qV , h0 , eF , eV , e). Let a = z 4 , b = x3 and K = ⟨a, b⟩. By Definition 5, K is the covering transformation group. By the assumption, K is abelian. By Proposition 8, we have and bz = beV ,

ax = aeF

for some eF ∈ Z∗qF h0 and eV ∈ Z∗qV h0 . By Lemma 10(i), K has a presentation (12), and the parameters qF , qV , h0 , eF , eV and e, gcd(e, h0 ) = 1, satisfy the congruences (10) and (11). In particular, we have the following commuting rule:

[z 4 , x] = 1,

(x3 )z = x3eV 2 x2

Let c1 = z −2 (z ) Then

and

[z 2 , x3 ] = 1.

(46)

and c2 = z 2 . Let N1 = ⟨K , c1 ⟩, N2 = ⟨N1 , c2 ⟩, N3 = ⟨N2 , x⟩ and N4 = ⟨N3 , z ⟩.

2 (46) c1 = z −2 (z 2 )x = z −2 x−2 z 2 x2 = z −2 xx−3 z 2 x−1 x3 = z −2 xz 2 x−1 .

(47)

We deduce from (xz ) = 1 that 2

x−1 z −1 x = zx2 .

(48)

Hence, (47)

c12 = (z −2 xz 2 x−1 )2 = (z −2 z −1 x−1 zx−1 )2 = (z −4 zx−1 zx−1 )2 (46)

(46)

= z −8 (z 2 xz 2 x−1 )2 = z −8 (z 2 xz 2 x2 x−3 )2 = z −8 x−6 (z 2 xz 2 x2 )2 = z −8 x−6 (zx−1 zx2 )2 = z −8 x−6 (zx2 x−3 zx2 )2 = z −8 x−6(eV +1) (zx2 )4 = z −12 x−6(eV +1) = a−3 b−2(eV +1) .

(46)

(48)

That is, c12 = a−3 b−2(eV +1) .

(49)

It follows that N1 /K = ⟨c1 K ⟩ ∼ = Z2 . Moreover, since

c22

= 1 and

(47) c (47) c12 = z −4 xz 2 x−1 z 2 = z −4 xz −2 z 4 x−1 z 2 = xz −2 x−1 z 2 = c1−1 ,

(50)

we get N2 /N1 = ⟨c2 N1 ⟩ ∼ = Z2 . Also notice that x−1 z 2 x = (zx2 zx2 )−1 = (zx−1 x3 zx−1 x3 )−1 = x−3(eV +1) (zx−1 zx−1 )−1 (48)

(46)

= x−3(eV +1) (z 2 xz 2 x−1 )−1 = x−3(eV +1) (z 4 z −2 xz −2 x−1 )−1 = x−3(eV +1) z −4 (z −2 xz −2 x−1 )−1 = a−1 b−(eV +1) (a−1 c1 )−1 = b−(eV +1) c1−1 ,

we obtain x−1 z 2 x = b−(eV +1) c1−1 .

(51)

K. Hu et al. / European Journal of Combinatorics 36 (2014) 531–549

545

Therefore, (51) c2x = x−1 z 2 x = b−(eV +1) c1−1 ,

(52)

(47) (48) c1x = x−1 z −2 xz 2 = x−1 z −4 z 2 xz 2 = z −4 (x−1 z 2 x)z 2

= a−1 b−(eV +1) c1−1 c2 .

(51)

(53)

Consequently, N2 E N3 , N3 /N2 = ⟨xN2 ⟩ ∼ = Z3 . Furthermore, we derive that c1z = z −3 xz 2 x−1 z = z −4 (zx)z 2 x−1 z = z −4 x−1 z 2 z −1 x−1 z

= z −4 x−1 z 2 xz 2 = a−1 x−1 c2 xc2 = a−1 b−(eV +1) c1−1 c2 , z c2 = c2 , (52)

xz = z −1 (xz ) = z −2 x−1 = c2−1 x−1 . Recall that z 2 = c2 , we get N3 E ⟨a, b, c1 , c2 , x, z ⟩ = ⟨x, z ⟩ = G and G/N3 = ⟨zN3 ⟩ ∼ = Z2 . Consequently, G is defined by a presentation G = ⟨a, b, c1 , c2 , x, z | aqF h0 = bqV h0 = [a, b] = 1, aqF = bqV e ,

c c12 = a−3 b−2(eV +1) , ac1 = a, bc1 = b, c22 = 1, ac2 = a, bc2 = b, c12 = c1−1 ,

x3 = 1, ax = a, bx = b, c1x = a−1 b−(eV +1) c1−1 c2 , c2x = b−(eV +1) c1−1 z 2 = c2 , az = a, bz = beV , c1z = a−1 b−(eV +1) c1−1 c2 , c2z = c2 , xz = c2−1 x−1 ⟩,

(54)

or equivalently, by simplification, G = ⟨x, z | (xz )2 = z 4qF h0 = x3qV h0 = [x3 , z 4 ] = 1,

(z 4 )x = z 4 , (x3 )z = x3eV , z 4qF = x3qV e ⟩.

(55)

To deduce the remaining congruences, we deduce from (49) and (50) that a−3 b−2(eV +1) = (a−3 b−2(eV +1) )c2 = (c12 )c2 = (c12 )2 (46)

(49)

c

= c −2 = a3 b2(eV +1) .

(50)

(49)

Consequently, a6 = b−4(eV +1) . Since ⟨a⟩ ∩ ⟨b⟩ = ⟨aqF ⟩ = ⟨bqV ⟩, we get 6 ≡ 0 (mod qF ) By the relation a

qF

and

=b

qV e

4(e + 1) ≡ 0 (mod qV ).

, we deduce that

b−4(eV +1) = a6 = (aqF )6/qF = b6qV e/qF . Hence b

6qV e/qF +4(eV +1)

(56)

(57)

= 1. Since o(b) = qV h0 , we get

6e/qF + 4(eV + 1)/qV ≡ 0 (mod h0 ).

(58)

From the above discussion, we see K E N1 , N1 E N2 , N2 E N3 and N3 E N4 = G such that N1 /K ∼ = Z2 , N2 /N1 ∼ = Z2 , N3 /N2 ∼ = Z3 and N4 /N3 ∼ = Z2 . Hence G is constructed from a series of cyclic extensions. Conversely, given a group G defined by the presentation (55) (or equivalently, (54)) with the numerical conditions (10), (11), (56) and (58) satisfied by the parameters, it is easily seen that the map N = (G, x, z ) is an almost totally branched abelian cover over C . By Theorem 13, G is a wellp

defined extension of Aut (C ) ∼ = S4 by K . In particular, |G| = 24qF qV h0 . Therefore, the covering N → C has index-triple (qF , qV , h0 ) and consequently, the exponent-triple (1, eV , e).

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Case 4. The base map is the icosahedral map I of type {3, 5}. Assume N = (G, x, z ), (xz )2 = 1, is an almost totally branched abelian cover over I with the index–exponent 6-tuple (qF , qV , h0 , eF , eV , e). Let a = z 3 , b = x5 and K = ⟨a, b⟩. By Definition 5, K is the covering transformation group. By the assumption, K is an abelian group. By Proposition 8, we have ax = aeF

and bz = beV ,

for some eF ∈ Z∗qF h0 and eV ∈ Z∗qV h0 . By Lemma 10(i), K has a presentation (12), and the parameters qF , qV , h0 , eF , eV and e, gcd(e, h0 ) = 1, satisfy the congruences (10) and (11). In particular, we have the following commuting rule:

[z 3 , x] = 1 and [x5 , z ] = 1.

(59)

x−2

x−1

, c2 = (xz )

−2

−1

= x3 z 2 x−1 = x3 z 3 z −1 x−1 = z 3 x4 z

Let c1 = (xz ) (xz ) xz = z −1 x−1 . Then c1 = (xz )x

(xz )x

x−1

, N1 = ⟨K , c1 ⟩ and N2 = ⟨N1 , c2 ⟩. By (xz )2 = 1, we have (59)

(∗)

(59)

= z 3 x5 x−1 z = z 3 x5 (x−1 z −1 )z −1 z 3 = z 6 x5 zxz −1 .

(60)

It follows that c15 = z 30 x30 = a10 b6 ∈ K .

(61)

Consequently, N1 /K = ⟨c1 K ⟩ ∼ = Z5 . As a byproduct, from (∗) in (60), we see c1 = z 3 x5 x−1 z. Hence x−1 z = x−5 z −3 c1 = a−1 b−1 c1 .

(62)

It follows that c c12 = x2 (zx)(xz )x−2 = x2 x−1 z −2 x−3 = xzz −3 x−3 = z −3 xzx−3 (62)

= z −3 z −1 xx−5 = z −3 x−5 (x−1 z )−1 = c1−1 .

(63)

Consequently, N1 E N2 and N2 /N1 = ⟨c2 N1 ⟩ ∼ = Z2 . Hence N2 has a presentation N2 = ⟨a, b, c1 , c2 | aqF h0 = bqV h0 = [a, b] = 1, aqF = bqV e , c c15 = a10 b6 , ac1 = a, bc1 = b, c22 = 1, ac2 = a, bc2 = b, c12 = c1−1 ⟩.

(64)

Note that since |G| = 60|K | and N2 = 10|K |, we see [G : N2 ] = 6. To derive the remaining congruences, we derive from (61) and (63) that c

(63)

(61)

a10 b6 = (a10 b6 )c2 = (c15 )c2 = (c12 )5 = c1−5 = a−10 b−6 . Consequently, a20 = b−12 .

(65)

Since ⟨a⟩ ∩ ⟨b⟩ = ⟨aqF ⟩ = ⟨bqV ⟩, we get 20 ≡ 0 (mod qF ) and By the relation a

qF

=b

qV e

12 ≡ 0 (mod qV ).

(66)

, we get

b−12 = a20 = (aqF )20/qF = b20qV e/qF . (66)

It follows that b20qV e/qF +12 = 1. Since o(b) = qV h0 , we get 20e/qF + 12/qV ≡ 0 (mod h0 ).

(67)

K. Hu et al. / European Journal of Combinatorics 36 (2014) 531–549

547

Let H be a group defined by a presentation

⟨x, z | z 3qF h0 = x5qV h0 = [z 3 , x5 ] = [x, z 3 ] = [x5 , z ] = 1, z 3qF = x5qV e ⟩.

(68)

It is easily seen that G fulfils all defining relations of H. Hence G is a quotient of H. Let T = ⟨z , x5 ⟩, then H /T ∼ = A5 . Clearly, T satisfies the defining relations of K , and hence |T | ≤ |K |. It follows that |H | = 60|T | ≤ 60|K | = |G|. Consequently, G ∼ = H, and G is defined by the presentation (68). Conversely, given a group G defined by the presentation (68) with the parameters satisfying (10), (11), (66) and (67), we shall show that the map (G, x, z ) defined by (68) is an almost totally branched abelian cover over the icosahedral map I with the index–exponent 6-tuple (qF , qV , h0 , 1, 1, e). Let A = ⟨z 3 ⟩, B = ⟨x5 ⟩ and K = AB. From the presentation, we see A, B E G and G/K ∼ = A5 . Hence the map (G, x, z ) is an almost totally branched abelian cover over I. To show that the associated index-triple is (qF , qV , h0 ) (and hence the exponent-triple is (1, 1, e)), it is sufficient to show that |G| = 60qF qV h0 . We start from a group K defined by a presentation (12). Following the same way as above we construct a group N2 = ⟨a, b, c1 , c2 ⟩ defined by a presentation (64) by cyclic extensions. By Theorem 13, N2 is a well-defined extension of order 10|K | = 10qF qV h0 . We claim [G : N2 ] = 6. Clearly, xi ̸∈ N2 (i = 1, 2, 3, 4) and z , z −1 ̸∈ N2 . It can be easily derived from the presentation that x−1 z = a−1 b−1 c1 . (cf. Eq. (62)). It follows that x−1 z = a−1 b−1 c1 ∈ N2 , and hence xN2 = zN2 . If z −1 N2 = xi N2 for some i, 1 ≤ i ≤ 4, then zxi ∈ N2 . Since x−1 z ∈ N2 , we have (x−1 z )(zxi ) ∈ N2 . Since 3

a∈N

(x−1 z )(zxi )N2 = x−1 z −1 z 3 xi N2 = zxi+1 aN2 =2 zxi+1 N2

(69)

we have zxi+1 ∈ N2 . By induction, we have z ∈ N2 , a contradiction. Therefore, N2 , xN2 , x2 N2 , x3 N2 , x4 N2 , z −1 N2 are distinct left cosets of N2 in G. It follows that [G : N2 ] = 6. Notice that in each case, if the extra condition gcd(qF , qV ) = 1 is included, then by Lemma 10(iii), the associated maps are cyclic regular covers. By Lemma 11, such maps form a complete set of cyclic regular covers of the respective platonic map, as claimed.  Remark 1. Some particular cyclic regular covers over the platonic maps have been discovered by various authors in distinct context. In our notation, they are P˜ (1, 1, h0 , 1, 1, 1) by Miller [11], P˜ (qF , 1, 1, 1, 1, 1) by Sherk [12], P˜ (qF , 1, 1, eF , 1, 1) by Jones and Surowski [8], and H˜ n (qF , 1, h0 ; eF , eV , e) by Wilson [16] and D˜ n (1, qV , h0 ; eF , eV , e) by several authors [18,9,5]. 6. Classification In this section, applying Theorem 14, we give a complete classification of almost totally branched abelian covers over the platonic maps. Theorem 15 (Classification). Let P be a platonic map of type {dF , dV } with lF faces and lV vertices. The isomorphism classes of almost totally branched abelian covers N over the platonic map P are in one-to-one correspondence with the solutions (qF , qV , h0 , eF , eV , e) of the system of congruence equations (10), (11), (25), (26) and (27). In particular, all such covers are reflexible. Proof. By Theorem 14, N ∼ = P˜ (qF , qV , h0 ; eF , eV , e) for some integers qF , qV , h0 , eF , eV and e satisfying (10), (11), (25), (26) and (27). Therefore, it suffices to show that two such maps P˜ (qF , qV , h0 ; eF , eV , e) = (G, x, z ) and P˜ (q′F , q′V , h′0 ; e′F , e′V , e′ ) = (G′ , x′ , z ′ ) are isomorphic if and only if qF = q′F , qV = q′V , h0 = h′0 and eF ≡ e′F (mod qF h0 ), eV ≡ e′V (mod qV h0 ), e ≡ e′ (mod h0 ). First assume the corresponding parameters are equal, then clearly the maps are isomorphic. Conversely, if they are isomorphic, then the assignment x → x′ , z → z ′ extends to a group isomorphism from G onto G′ . Since a group isomorphism preserves defining relations, we have dF qF h0 = dF q′F h′0 , dV qV h0 = dV q′V h′0 and dF qF = dF q′F . It follows that qF = q′F , h0 = h′0 and qV = q′V .

548

K. Hu et al. / European Journal of Combinatorics 36 (2014) 531–549 Table 1 Type, multiplicity type and genus of the covers.

P˜

{d˜ F , d˜ V }

˜ F,m ˜ V) (m

g˜

T˜ D˜ n H˜ n C˜ O˜ I˜ D˜

{3h0 qF , 3h0 qV } {nh0 qF , 2h0 qV } {2h0 qF , nh0 qV } {4h0 qF , 3h0 qV } {3h0 qF , 4h0 qV } {3h0 qF , 5h0 qV } {5h0 qF , 3h0 qV }

(h0 q F , h0 q V ) ( a , h0 q V ) (h0 qF ,a ) (h0 qF , h0 qV ) (h0 q F , h0 q V ) (h0 qF , hqV ) (hqF , h0 qV )

3h0 qV qF − 2(qV + qF ) + 1 nqF (h0 qV − 1)/2 − qV + 1 nqV (h0 qF − 1)/2 − qF + 1 6h0 qV qF − 3qV − 4qF + 1 6h0 qF qV − 3qF − 4qV + 1 15h0 qV qF − 6qF − 10qV + 1 15h0 qF qV − 6qV − 10qF + 1

a

The multiplicity is not determined.

These equalities then imply that dF eF ≡ dF e′F (mod dF qF h0 ), dV eV ≡ dV e′V (mod dV qV h0 ) and dV qV e ≡ dV qV e′ (mod qV h0 ). It follows that eF ≡ e′F (mod qF h0 ), eV ≡ e′V (mod qV h0 ) and e ≡ e (mod h0 ). Notice that the mirror image of the map P˜ (qF , qV , h0 ; eF , eV , e) = (G, x, z ) is the map (G, x−1 , z −1 ), where G, in terms of x−1 and z −1 , has a presentation

⟨x−1 , z −1 | (z −1 x−1 )2 = z −dF qF h0 = x−dV qV h0 = [z −dF , x−dV ] = 1, (z −dF )x

−1

−1

= x−dV eV , z −dF qF = x−dV qV e ⟩. (70) −1 −1 ∼ ˜ By comparing (24) with (70), we get (G, x, z ) = (G, x , z ). Therefore, the map P is reflexible, as claimed.

= z −dF eF , (x−dV )z



Corollary 16. The almost totally branched abelian covers over the platonic maps have type {d˜ F , d˜ V }, ˜ F, m ˜ V ) and genus g˜ summarised in Table 1. multiplicity type (m Proof. By Theorem 14, such covers are isomorphic to the maps P˜ given by Definition 12. Clearly, the type of P˜ is {dF qF h0 , dV qV h0 }. By Proposition 4, if P is simple, then P˜ has multiplicity qV h0 , and if the dual of P is simple, then the dual multiplicity of P˜ is qF h0 . Finally, applying the Euler–Poincaré Formula, we get the genus of P˜ , as claimed.  Almost totally branched abelian covers over the platonic maps which are self-dual are determined as follows. Theorem 17. In the family of platonic maps, only D2 and T admit self-dual almost totally branched abelian covers. More precisely, (i) the almost totally branched abelian cover over the dihedral map D2 with the index–exponent 6-tuple (qF , qV , h0 , eF , eV , e) is self-dual if and only if qF , qV , h0 , eF , eV and e satisfy qF = qV , eF = eV , e2V ≡ 1 (mod qV h0 ), eV ≡ 1 (mod h0 ), eV + 1 ≡ 0 (mod qV ),

(71)

e2 ≡ 1 (mod h0 ),

(eV + 1)(e + 1)/qV ≡ 0 (mod h0 ). (ii) the almost totally branched abelian cover over the tetrahedral map T with the index–exponent 6-tuple (qF , qV , h0 , eF , eV , e) is self-dual if and only if qF , qV , h0 , eF , eV and e satisfy qV = qF , eV = eF = 1, 4 ≡ 0 (mod qV ), e ≡ 1 (mod h0 ), 2

4(e + 1)/qV ≡ 0 (mod h0 ).

(72)

K. Hu et al. / European Journal of Combinatorics 36 (2014) 531–549

549

Proof. By Theorem 14, the almost totally branched abelian covers over the platonic map P are isomorphic to the maps P˜ = P˜ (qF , qV , h0 , eF , eV , e). If P˜ is self-dual, then the assignment x → z , z → x extends to an automorphism of G. It follows that dF = dV ,

qF = qV ,

eF = eV

and

e2 ≡ 1 (mod h0 ).

(73)

Therefore, P = D2 or P = T . Combining (73) with the congruences (10), (11), (25), (26) and (27), we get (71) and (72). Conversely, it is straightforward to verify that if (71) and (72) are fulfilled, then the respective covers over D2 and T are self-dual, as required.  Remark 2. If we add an extra condition gcd(qF , qV ) = 1 to (72), then we obtain the self-dual cyclic regular covers T˜ (1, 1, h0 , 1, 1, e) over the tetrahedral map T , where h0 and e satisfy 4(e + 1) ≡ 0 (mod h0 ) and e2 ≡ 1 (mod h0 ). This coincides with the result obtained by Širáň through the voltage assignment method [13, Theorem 7]. Acknowledgements The authors are indebted to Professor Gareth Jones for his helpful comments and suggestions. His simpler proof of Lemma 9 has improved the paper. We would like to thank the anonymous referees for helpful suggestions that have improved the presentation of the paper. The authors thank the support of the following grants: APVV-0223-10, VEGA 1/1085/11, and APVV-ESF-EC0009-10 within the EUROCORES Programme EUROGIGA (project GReGAS) of the European Science Foundation. This publication/article/conference paper/book etc. is based upon work supported by the Project: Mobility—enhancing research, science and education at the Matej Bel University, ITMS code: 26110230082, under the Operational Program Education cofinanced by the European Social Fund. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18]

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