Circular regular dessins

One of the main aims of topological graph theory is to determine all the symmetrical embeddings of a given class of graphs in orientable closed surfaces, see [2, 15, 20]. We will restrict our attention here to the edge-regular embeddings of bipartite graphs. Let \(\Gamma =(B\cup W,E)\) be a connected bipartite graph, where the vertex set V is partitioned into two parts B and W with vertices colored black and white, respectively. A dessin \({{\mathcal {D}}}\) is an embedding \(i:\Gamma \hookrightarrow {{\mathbb {S}}}\) of \(\Gamma \) into an orientable closed surface \({{\mathbb {S}}}\) such that each component of \({{\mathbb {S}}}\backslash i(\Gamma )\) is homeomorphic to an open disc. The open discs are called faces, and the set of faces is denoted by F. Thus, a dessin \({{\mathcal {D}}}\) is usually viewed as an incidence triple of vertices, edges and faces, and written as \({{\mathcal {D}}}=(B\cup W,E,F)\). The bipartite graph \({\varGamma }\) is called the underlying graph, and the surface \({{\mathbb {S}}}\) is called the supporting surface. In other words, a dessin is an orientable map with the underlying graph being bipartite.

For a dessin \({{\mathcal {D}}}\), the edges on the boundary of a face form a cycle. A cycle is called a simple cycle if there is no repeated vertex or edge, that is, each vertex appears only one time. A simple cycle is also called a circuit. It is important to understand whether the boundary cycle of each face in a dessin is a circuit, refer to [18].

Definition 1.1

If the boundary of each face is a circuit, then \({{\mathcal {D}}}\) is called a circular dessin.

That is to say, a circular dessin is a circular embedding of a connected bipartite graph into an orientable closed surface. We remark that a circular embedding is also called a strong embedding in [13], or a closed 2-cell embedding in [19]. However, it is not an easy question to answer whether a given dessin is circular or not. In this paper, we study this question for highly symmetrical dessins.

An automorphism of a dessin \({{\mathcal {D}}}=(B\cup W,E,F)\) is a permutation on \((B\cup W)\cup E\cup F\) which preserves the incidence relations between vertices, edges and faces, also preserves the orientation of the supporting surface and the colors of the vertices. Let \(\mathrm{Aut}{{\mathcal {D}}}\) be the group of automorphisms of \({{\mathcal {D}}}\), consisting of all automorphisms. Then, \(\mathrm{Aut}{{\mathcal {D}}}\) acts semi-regularly on the edge set E of \({{\mathcal {D}}}\). So, if this action is transitive, then it is regular, and is thus called a regular dessin. In particular, a regular dessin is called symmetric if \({{\mathcal {D}}}\) has an external symmetry transposing the two end vertices of each edge. In other words, a regular dessin is an orientably edge-regular bipartite map, and a symmetric dessin is an orientably arc-regular bipartite map. We remark that an orientably arc-regular bipartite map is simply called an (orientably) regular map or regular embedding in [4, 5, 16].

The importance of regular dessins has been well recognized, see [1, 10, 11, 14, 17]; special classes of regular dessins have been studied, see [9, 12, 21]. We are concerned in this paper with the circular regular dessins.

It is well known that a regular dessin \({{\mathcal {D}}}\) is determined by the automorphism group \(G=\mathrm{Aut}{{\mathcal {D}}}\) and its two distinguished generating elements a, b, and so it denotes by \({{\mathcal {D}}}(G,a,b)\). In Sect. 2, we will give a group theoretical criterion for \({{\mathcal {D}}}\) to be circular: a regular dessin \({{\mathcal {D}}}(G,a,b)\) is circular if and only if \(\langle ba\rangle \cap \langle a\rangle =1\) and \(\langle ba\rangle \cap \langle b\rangle =1.\) This criterion shows that if G is a simple group, or G is center free, then a regular dessin \({{\mathcal {D}}}(G,a,b)\) with automorphism group G is circular, see Sects. 3 and 4 for more examples. On the other hand, there are families of graphs which do not have circular regular dessins, for example, by the classification result of [6], if \(\gcd (m,\phi (n))=\gcd (\phi (m),n)=1\), then \(\mathbf{K}_{m,n}\) has a unique regular dessin \({{\mathcal {D}}}\) with \(\mathrm{Aut}{{\mathcal {D}}}\cong {{\mathbb {Z}}}_m\times {{\mathbb {Z}}}_n\), and moreover, \({{\mathcal {D}}}\) is not circular when \(m\ne n\); by [8], if p is an odd prime, then \(\mathbf{K}_{p^e,p^f}\) has circular regular dessins if and only if \(e=f\). The following theorem asserts that each complete bipartite graph \(\mathbf{K}_{m,m}\) has edge-transitive embeddings which are circular dessins.

Theorem 1.2

A symmetric dessin with underlying graph being simple is a circular dessin.

This result indicates a significant difference between edge-regularity and arc-regularity of the underlying graph of a dessin.

A regular dessin \({{\mathcal {D}}}\) is called a Frobenius dessin if \(\mathrm{Aut}{{\mathcal {D}}}\) is a Frobenius group. The following result characterizes the circularity of regular Frobenius dessins of \(\mathbf{K}_{m,n}\).

Theorem 1.3

Given integers m, n such that there is a Frobenius group \({{\mathbb {Z}}}_m{:}{{\mathbb {Z}}}_n\), there are exactly \(\phi (n)\) non-isomorphic circular regular Frobenius dessins with underlying graph \(\mathbf{K}_{m,n}\), and these dessins have genus equal to \(1+{1\over 2}(mn-2m-n)\).

A regular dessin with underlying graph being a complete bipartite graph \(\mathbf{K}_{m,n}\) is called a complete regular dessin of type \(\{m,n\}\), which has automorphism group G isomorphic to a bicyclic group \(\langle a\rangle \langle b\rangle \) with \(o(a)=m, o(b)=n\). Theorem 1.3 shows that if \(G=\langle a\rangle \langle b\rangle \) is a Frobenius group, then each regular dessin \({{\mathcal {D}}}(G,a,b)\) with automorphism group G is circular. In particular, if \(\gcd (m,n)=1,\) by the classification result of [7], \(\mathbf{K}_{m,n}\) has a circular regular dessin if and only if G is center free. This motivates us to consider the following problem.

Problem 1.4

Determine complete bipartite graphs \(\mathbf{K}_{m,n}\) which do not have circular regular embeddings.

The following we deal with the case \(\gcd (m,n)=1\). For a prime p of n, let \(n_p\) be the p-part of n, that is, \(n_p\) is a power of p and divides n, and \(n/n_p\) is coprime to p. The factor \(n/n_p\) is denoted by \(n_{p'}\).

Theorem 1.5

Let m, n be positive integers which are relatively prime. Then, \(\mathbf{K}_{m,n}\) has no circular regular dessin if and only if either

1. (i)

   there exists a prime p | m such that \(\gcd (p-1,n)=1\) and \(\gcd (m_p,\phi (n))<m_p,\) or

2. (ii)

   there exists a prime q | n such that \(\gcd (q-1,m)=1\) and \(\gcd (n_q,\phi (m))<n_q.\)

Some regular dessins can be decomposed as a direct product of two smaller dessins.

Definition 1.6

Given two regular dessins \({{\mathcal {D}}}(G,a,b)\) and \({{\mathcal {D}}}(H,x,y)\) such that \(\gcd (|a|,|x|)=1\) and \(\gcd (|b|,|y|)=1\), the map \({{\mathcal {D}}}(G\times H,ax,by)\) is called a direct product of \({{\mathcal {D}}}(G,a,b)\) and \({{\mathcal {D}}}(H,x,y)\), denoted by \({{\mathcal {D}}}(G,a,b)\times {{\mathcal {D}}}(H,x,y)\).

The following result shows the circularity between a given dessin and its direct product factors.

Theorem 1.7

Let \({{\mathcal {D}}}(G,a,b)={{\mathcal {D}}}(G_1,a_1,b_1)\times {{\mathcal {D}}}(G_2,a_2,b_2).\) Then, the following statements hold:

1. (i)

   If \({{\mathcal {D}}}(G_1,a_1,b_1)\) and \({{\mathcal {D}}}(G_2,a_2,b_2)\) are both circular, then \({{\mathcal {D}}}(G,a,b)\) is circular;

2. (ii)

   if \({{\mathcal {D}}}(G,a,b)\) is circular and \(\gcd (|b_1a_1|,|b_2a_2|)=1\), then \({{\mathcal {D}}}(G_1,a_1,b_1)\) and \({{\mathcal {D}}}(G_2,a_2,b_2)\) are both circular.

As an application of this theorem, we deal with the case that m and n have only one common prime divisor as follows.

Theorem 1.8

Let m, n be integers which have only one common prime divisor p such that \(\gcd (p,\phi (m_{p'})\phi (n_{p'}))=1\) and \(\gcd (p-1,m_{p'}n_{p'})=1.\) Then, \(\mathbf{K}_{m,n}\) has no circular regular dessin if and only if either

1. (i)

   \(m_p\ne n_p;\) or

2. (ii)

   there exists a prime \(r~|~m_{p'}\) such that \(\gcd (r-1,n_{p'})=1\) and \(\gcd (m_r,\phi (n_{p'}))<m_r,\) or there exists a prime \(s~|~n_{p'}\) such that \(\gcd (s-1,m_{p'})=1\) and \(\gcd (n_s,\phi (m_{p'}))<n_s.\)

This paper is organized as follows. A criterion to decide whether a regular dessin \({{\mathcal {D}}}\) is circular will be given in Sect. 2. As an application, some important family of circular regular dessins are studied in Sects. 3 and 4. Then, Theorem 1.3 is proved in Sect. 5, and Theorem 1.5 is proved in Sect. 6, and Theorems 1.7 and 1.8 are proved in Sect. 7.

In this section, we collect some preliminary definitions and results which will be used later and present a group theoretical criterion to decide whether a regular dessin is circular. This criterion is more or less well-known and extensively accepted. However, to our best knowledge, there has been no an explicit proof appeared in the literature, and so we offer a proof here for the completeness and briefness.

Let \({{\mathcal {D}}}=(W\cup B,E,F)\) be a regular dessin and \(e=[\alpha ,e,\beta ]\) be an edge of \({{\mathcal {D}}}\), where \(\alpha \in W\) and \(\beta \in B\) are two vertices which are incident with e. Then, the automorphism group \(G=\mathrm{Aut}{{\mathcal {D}}}\) is generated by two elements a, b such that \(G_\alpha =\langle a\rangle \) and \(G_\beta =\langle b\rangle \), where \(G_{\alpha }\) and \(G_{\beta }\) are the stabilizers of \(\alpha \) and \(\beta \), respectively.

Conversely, for an abstract group \(G=\langle a,b\rangle \), there is a connected graph \({\varGamma }=(V,E)\) naturally associated, where

This graph is denoted by \(\mathsf{Cos}(G,\langle a\rangle ,\langle b\rangle )\) and called a bi-coset graph. We notice that each edge \([\langle a\rangle g,g,\langle b\rangle g]\) is uniquely determined by the element g. Obviously, G acting on V and E by right multiplication forms a group of automorphisms of the graph \({\varGamma }\) which is edge regular. Identifying the edges of \({\varGamma }\) with elements of G, the sequence

is a path of length 3, which we will denote by \((b^{-1},1,a)\) by omitting the vertices. It follows that the following sequence of edges forms a cycle (closed walk):

where \(\ell =|ba|\).

Since G is a group of automorphisms of \(\mathsf{Cos}(G,\langle a\rangle ,\langle b\rangle )\), an element \(g\in G\) maps the cycle C(a, b) to another cycle by right multiplication:

Let

Then, the incidence triple (V, E, F) defines a regular dessin with automorphism group G, and moreover, all regular dessins can be defined in this way. For convenience, we denote this dessin by \({{\mathcal {D}}}(G,a,b)\). To sum up, we have the following results.

Lemma 2.1

1. (1)

   Given a group \(G=\langle a,b\rangle \), the incidence structure \({{\mathcal {D}}}(G,a,b)\) defined above is a regular dessin.

2. (2)

   Each regular dessin \({{\mathcal {D}}}\) can be expressed as \({{\mathcal {D}}}(G,a,b)\), where \(G=\mathrm{Aut}{{\mathcal {D}}}\), \(G_\alpha =\langle a\rangle \), and \(G_\beta =\langle b\rangle \) for an edge \(\{\alpha ,\beta \}\).

Using the notation defined above, label \(\alpha =\langle a\rangle \) and \(\beta =\langle b\rangle \). By definition, \(\alpha ,\beta \) are adjacent via the edge \(1=[\langle a\rangle ,1,\langle b\rangle ]\), and \(G_\alpha =\langle a\rangle \), \(G_\beta =\langle b\rangle \). Let \(E(\alpha ,\beta )\) be the set of edges between \(\alpha \) and \(\beta \). Then, \(E(\alpha ,\beta )=\{[\langle a\rangle , g, \langle b\rangle ] \mid g\in \langle a\rangle \cap \langle b\rangle \}\), and thus we have the following result.

Lemma 2.2

The edge multiplicity of the underlying graph \({\varGamma }=\mathsf{Cos}(G,\langle a\rangle ,\langle b\rangle )\) equals \(|\langle a\rangle \cap \langle b\rangle |\); in particular, \({\varGamma }\) is a simple graph if and only if \(\langle a\rangle \cap \langle b\rangle =1\).

Let \(C=C(a,b)\) be a cycle defined in (1). Obviously, the stabilizer of this cycle is the subgroup \(\langle ba\rangle \). Let V(C) be the set of vertices on the cycle C, and let \(|ba|=\ell \). Then, \(|V(C)|=2\ell \).

Lemma 2.3

Each vertex in \([G:\langle a\rangle ]\cap V(C)\) appears on C exactly \(|\langle ba\rangle \cap \langle a\rangle |\) times, and each vertex in \([G:\langle b\rangle ]\cap V(C)\) appears on C exactly \(|\langle ba\rangle \cap \langle b\rangle |\) times.

Proof

Let \(\alpha \) be a vertex in \([G:\langle a\rangle ]\cap V(C)\). Suppose that \(\langle ba\rangle \cap G_\alpha =\langle (ba)^k\rangle \cong {{\mathbb {Z}}}_m\), where k, m are positive integers satisfying \(km=\ell \). Then, the element \((ba)^{ik}\) fixes the vertex \(\alpha \) for each i with \(1\le i\le m\), and \((ba)^j\) does not fix \(\alpha \) if j is indivisible by k. Thus, the vertex \(\alpha \) appears on the cycle C exactly m times:

where \(m=|\langle ba\rangle \cap G_\alpha |=|\langle ba\rangle \cap \langle a\rangle |.\) Similarly, if \(\beta \) is a vertex in \([G:\langle b\rangle ]\cap V(C)\), then \(\beta \) appears on the cycle \(C=C(a,b)\) exactly \(|\langle ba\rangle \cap \langle b\rangle |\) times. \(\square \)

A criterion for a regular dessin to be circular comes as follows.

Proposition 2.4

A regular dessin \({{\mathcal {D}}}(G,a,b)\) is a circular dessin if and only if

Proof

By definition, \({{\mathcal {D}}}(G,a,b)\) is a circular dessin if and only if each face of \({{\mathcal {D}}}\) is bounded by a circuit of the underlying graph \({\varGamma }\), meaning that all vertices and edges of the cycle C are distinct. In other words, each vertex of \({{\mathcal {D}}}\) appears on C at most one time. By Lemma 2.3, this happens if and only if \(\langle ba\rangle \cap \langle a\rangle =1\) and \(\langle ba\rangle \cap \langle b\rangle =1\). \(\square \)

According to the Sect. 2, each regular dessin \({{\mathcal {D}}}\) can be expressed as \({{\mathcal {D}}}(G,a,b)\), where \(G=\mathrm{Aut}{{\mathcal {D}}}\), \(G_\alpha =\langle a\rangle \), and \(G_\beta =\langle b\rangle \) for an edge \(\{\alpha ,\beta \}\). Moreover, \({{\mathcal {D}}}(G,a,b)\) is a circular dessin if and only if

We next present several families of regular dessins \({{\mathcal {D}}}(G,a,b)\) which are circular from the view point of automorphism groups.

Lemma 3.1

If G is a group with trivial center and generated by a and b, then each regular dessin \({{\mathcal {D}}}(G,a,b)\) is circular.

Proof

Let \(\langle c\rangle =\langle ba\rangle \cap \langle a\rangle \). Then, c is centralized by both ba and a, and so c is centralized by \(\langle ba,a\rangle =\langle b,a\rangle =G\). Thus, c lies in the center \(\mathbf{Z}(G)\), and so \(\langle ba\rangle \cap \langle a\rangle =1\). Similarly, \(\langle ba\rangle \cap \langle b\rangle =1\). By Proposition 2.4, \({{\mathcal {D}}}\) is a circular dessin, as claimed. \(\square \)

Lemma 3.1 shows that if G is center free, then a regular dessin \({{\mathcal {D}}}(G,a,b)\) with automorphism group G is circular. Beyond that, there exist many important families of circular regular dessins where G has no non-identity cyclic normal subgroup or G is a simple group.

Lemma 3.2

Let G be a finite group which has no non-identity cyclic normal subgroup. If \(G=\langle a,b\rangle \), then \(\langle ba\rangle \cap \langle a\rangle =1\) and \(\langle ba\rangle \cap \langle b\rangle =1\). Therefore, \({{\mathcal {D}}}={{\mathcal {D}}}(G,a,b)\) is a circular regular dessin with automorphism group G.

Proof

Suppose that \(\langle ba\rangle \cap \langle a\rangle =\langle c\rangle \not =1\). Then, c is in the center of \(G=\langle a,b\rangle =\langle a,ba\rangle \); in particular, \(\langle c\rangle \) is a cyclic normal subgroup of G, which is no possible. Thus, \(\langle ba\rangle \cap \langle a\rangle =1\); similarly, \(\langle ba \rangle \cap \langle b\rangle =1\). Thus, by Proposition 2.4, \({{\mathcal {D}}}(G,a,b)\) is a circular regular dessin with \(\mathrm{Aut}{{\mathcal {D}}}=G.\) \(\square \)

Example 3.3

Let G be a finite simple group, and let a, b generate G. Then, \({{\mathcal {D}}}={{\mathcal {D}}}(G,a,b)\) is a circular regular dessin with \(\mathrm{Aut}{{\mathcal {D}}}=G\).

The next sufficient condition for circular regular dessins relies on the action of the automorphism group.

Lemma 3.4

A regular dessin \({{\mathcal {D}}}(G,a,b)\) is circular if G is faithful on the face set.

Proof

Let \(N=\langle ba\rangle \cap \langle a\rangle \). Then, N is normal in \(\langle a\rangle \) and \(\langle ba\rangle \), and so normal in \(\langle ba,a\rangle =G\). Further, \(N\le \langle ba\rangle \) fixes the cycle \(C=C(a,b)\) defined in (1). Since G is transitive on F, it follows that N fixes all faces of F, and so we conclude that \(N=\langle ba\rangle \cap \langle a\rangle =1\). Similarly, we have \(\langle ba\rangle \cap \langle b\rangle =1\). By Proposition 2.4, \({{\mathcal {D}}}\) is a circular dessin, as desired. \(\square \)

Lemma 3.4 shows that for a regular dessin \({{\mathcal {D}}}(G,a,b)\), if G is faithful on the face set, then \({{\mathcal {D}}}\) is circular, but the converse statement is not necessarily true.

Example 3.5

Let \(G=\langle a\rangle \times \langle b\rangle \cong {{\mathbb {Z}}}_4\times {{\mathbb {Z}}}_4\). Then, \(\langle ba\rangle \cong {{\mathbb {Z}}}_4\), and further, \(\langle ba\rangle \cap \langle a\rangle =1\) and \(\langle ba\rangle \cap \langle b\rangle =1\). By Proposition 2.4, the dessin \({{\mathcal {D}}}(G,a,b)\) is circular. However, since the face stabilizer \(\langle ba\rangle \) of C(a, b) is normal in G, the kernel of the group G acting on the face set is \(\langle ba\rangle ,\) and thus G is not faithful on the face set.

In this section, we determine bicyclic groups which are automorphism groups of circular maps.

Lemma 4.1

Let \(G=\langle a\rangle \times \langle b\rangle \cong {{\mathbb {Z}}}_m\times {{\mathbb {Z}}}_n\) be an abelian group. Then, G acts edge-regularly on a circular dessin \({{\mathcal {D}}}\) of \(\mathbf{K}_{m,n}\) if and only if \(m=n\).

Proof

Suppose that \(m=n\). Then, \(G=\langle a\rangle \times \langle b\rangle ={{\mathbb {Z}}}_n\times {{\mathbb {Z}}}_n\), and obviously, \(\langle ba\rangle \) is disjoint with \(\langle a\rangle \) and \(\langle b\rangle \). By Proposition 2.4, \({{\mathcal {D}}}(G,a,b)\) is a circular regular dessin of \(\mathbf{K}_{n,n}\) such that \(G=\mathrm{Aut}{{\mathcal {D}}}\).

Conversely, let \({{\mathcal {D}}}(G,a,b)\) be a circular dessin such that G acts edge-regularly. Then, \(\langle ba\rangle \cap \langle a\rangle =1\) and \(\langle ba\rangle \cap \langle b\rangle =1\). Suppose that \(m\not =n\). Without loss of generality, we assume that \(m>n\). Since G is abelian, it follows that \((ba)^n=a^n\not =1\), and hence \(\langle ba\rangle \cap \langle a\rangle \not =1\), which is a contradiction. \(\square \)

Definition 4.2

A group G is called bicyclic if G has a bicyclic factorization \(G=\langle a\rangle \langle b\rangle \). The pair \(\{o(a),o(b)\}\) is called the type of the bicyclic group G. Further, if \(\langle a\rangle \cap \langle b\rangle =1\), then G is said to be an exact bicyclic group, and \(G=\langle a\rangle \langle b\rangle \) is called an exact factorization.

Assume that m, n are such that all bicyclic groups of type \(\{m,n\}\) act on a complete bipartite circular map edge-regularly. Let G be such an abelian group. By Lemma 4.1, we have \(m=n\). We thus have the following statement.

Lemma 4.3

If, for a pair of integers (m, n), each bicyclic group of type \(\{m,n\}\) is an edge-regular automorphism group of a circular dessin of \(\mathbf{K}_{m,n}\), then \(m=n\).

The converse statement of this lemma is true for some special m.

Lemma 4.4

Any bicyclic group of type \(\{p^2,p^2\}\), where p is an odd prime, is an edge-regular automorphism group of a circular embedding of \(\mathbf{K}_{p^2,p^2}\).

Proof

Let G be a bicyclic group of type \(\{p^2,p^2\}\). Then, G has order \(p^4\), and by [16, Prop 5] we known that \(G=\langle a\rangle {:}\langle b\rangle \cong {{\mathbb {Z}}}_{p^2}{:}{{\mathbb {Z}}}_{p^2}\). Thus, either G is abelian, or \(G=\langle a\rangle {:}\langle b\rangle \) such that \(bab^{-1}=a^{1+p}\). In either case, the triple (G, a, b) gives rise to a regular dessin \({{\mathcal {D}}}(G,a,b)\) with underlying graph \(\mathbf{K}_{p^2,p^2}\).

If G is abelian, then \(G=\langle a\rangle \times \langle b\rangle \), and obviously \(\langle ba\rangle \) is disjoint with \(\langle a\rangle \) and \(\langle b\rangle \). Assume that G is nonabelian. It follows that \((ba)^p=b^pa^p\), and so \(\langle ba\rangle \cap \langle a\rangle =1=\langle ba\rangle \cap \langle b\rangle \). Therefore, the dessin \({{\mathcal {D}}}(G,a,b)\) is circular by Proposition 2.4. \(\square \)

However, not every bicyclic group of type \(\{m,m\}\) has the property.

Example 4.5

Let \(G=(\langle x_1\rangle {:}\langle y_1\rangle )\times (\langle x_2\rangle \times \langle y_2\rangle )\cong \mathrm{D}_{2p}\times {{\mathbb {Z}}}_{2p}\) with \(y_1x_1y_1^{-1}=x_1^{-1}\), where p is an odd prime and \(o(x_1)=o(x_2)=p\), \(o(y_1)=o(y_2)=2\). Then, G is an exact bicyclic group of type \(\{2p,2p\}\), and a bicyclic factorization \(G=\langle a\rangle \langle b\rangle \) must satisfy \(a=x_1^i y_2\) and \(b=x_2^j y_1\), where \(1\le i,j<p\). Now, \(\langle a\rangle \cap \langle b\rangle =1\), and so by Lemma 2.2, G acts edge-regularly on a dessin \({{\mathcal {D}}}(G,a,b)\) of \(\mathbf{K}_{2p,2p}\).

Since \(ba=x_2^j y_1{\cdot }x_1^i y_2=x_2^j{\cdot }(x_1^{-i}y_1y_2)=(x_1^{-i}y_1y_2){\cdot }x_2^j\), and \(x_1^{-i}y_1y_2\) has order 2. It follows that \(\langle ba\rangle \cong {{\mathbb {Z}}}_2\times {{\mathbb {Z}}}_p\), and which contains the subgroup \(\langle x_2\rangle \). Thus, \(\langle ba\rangle \cap \langle b\rangle =\langle x_2\rangle \ne 1\). By Proposition 2.4, \({{\mathcal {D}}}(G,a,b)\) is not a circular map.

Proof of Theorem 1.2

Let \({{\mathcal {D}}}={{\mathcal {D}}}(G,a,b)\) be a symmetric dessin, that is, \({{\mathcal {D}}}\) is an orientably regular map. By [15], there exists an automorphism \(\sigma \) of G which interchanges the arcs \((\alpha ,\beta )\) and \((\beta ,\alpha )\). Thus, \(\sigma \) interchanges the stabilizers \(G_\alpha =\langle a\rangle \) and \(G_\beta =\langle b\rangle \). Without loss of generality, we may suppose that \(a^\sigma =b\) and \( b^\sigma =a\). Since the underlying graph is simple, it follows from Lemma 2.2 that the edge stabilizer \(\langle a\rangle \cap \langle b\rangle =1\).

Suppose that \(\langle c\rangle =\langle ba\rangle \cap \langle a\rangle \ne 1\). Then, \(\langle c\rangle \le \langle a\rangle \) and \(\langle c\rangle \le \langle ba\rangle \), and so we may write \(c=a^i=(ba)^j\) for some integers i, j. Since \(a^\sigma =b\) and \( b^\sigma =a\), we obtain

which contradicts \(\langle a\rangle \cap \langle b\rangle =1\). So we have \(\langle ba\rangle \cap \langle a\rangle =1\), similarly, \(\langle ba\rangle \cap \langle b\rangle =1\). By Proposition 2.4, \({{\mathcal {D}}}\) is a circular dessin, as claimed. \(\square \)

This theorem shows that the dessin is circular if its underlying graph is simple and arc-transitive, but this statement is not true for edge-transitive case.

Example 4.6

Let \(G\cong {{\mathbb {Z}}}_{15}\), and \(a,b\in G\) such that \(|a|=3\) and \(|b|=5\). Then, \({{\mathcal {D}}}:={{\mathcal {D}}}(G,a,b)\) is an edge regular dessin with underlying graph \(\mathbf{K}_{3,5}\). Since \(\langle ba\rangle =G\) contains both \(\langle a\rangle \) and \(\langle b\rangle \), the dessin \({{\mathcal {D}}}\) is not circular by Proposition 2.4.

Let \({\varGamma }=\mathbf{K}_{m,n}\) be a complete bipartite graph with partite sets B and W of size m and n, respectively. Let \({{\mathcal {D}}}\) be a complete regular dessin with underlying graph \({\varGamma }\). Then, \(G=\mathrm{Aut}{{\mathcal {D}}}\) is edge-regular on \({{\mathcal {D}}}\), and for vertices \(\beta \in B\) and \(\alpha \in W\), the stabilizes \(G_\beta \cong {{\mathbb {Z}}}_n\) and \(G_\alpha \cong {{\mathbb {Z}}}_m\). Let \(G_\beta =\langle b\rangle \) and \(G_\alpha =\langle a\rangle \). Since \({\varGamma }=\mathbf{K}_{m,n}\) is G-edge regular, \(G_\beta \) is transitive on W and \(G_\alpha \) is transitive on B. It follows that

Since \(\mathbf{K}_{m,n}\) is a simple graph, by Lemma 2.2, we have \(\langle a\rangle \cap \langle b\rangle =1.\) Conversely, let \(G=\langle a\rangle \langle b\rangle \cong {{\mathbb {Z}}}_m{{\mathbb {Z}}}_n\), and \(\langle a\rangle \cap \langle b\rangle =1,\) the incidence structure \({{\mathcal {D}}}(G,a,b)\) as stated in Sect. 2 is a regular dessin, and which has the underlying graph \(\Gamma =\mathsf{Cos}(G,\langle a\rangle ,\langle b\rangle )=\mathbf{K}_{m,n}\), where \(|B|=[G:\langle b\rangle ]=m,\) and \(|W|=[G:\langle a\rangle ]=n\).

Thus, we conclude that the underlying graph of a regular dessin \({{\mathcal {D}}}(G,a,b)\) is a complete bipartite graph \(\mathbf{K}_{m,n}\) if and only if \(G=\langle a\rangle \langle b\rangle \cong {{\mathbb {Z}}}_m{{\mathbb {Z}}}_n\) and \(\langle a\rangle \cap \langle b\rangle =1\), that is, G is an exact bicyclic group of type \(\{m,n\}.\)

For bicyclic groups G and H, two bicyclic factorizations \(G=\langle a\rangle \langle b\rangle \) and \(H=\langle a'\rangle \langle b'\rangle \) are called equivalent if there is an isomorphism \(\tau \) between G and H such that \(\langle a\rangle ^\tau =\langle a'\rangle \) and \(\langle b\rangle ^\tau =\langle b'\rangle \). Moreover, if \(a^\tau =a'\) and \(b^\tau =b'\), then the two triples (G, a, b) and \((H,a',b')\) are called equivalent.

The next is a criterion for deciding the isomorphism between regular dessins.

Lemma 5.1

Two regular dessins \({{\mathcal {D}}}(G,a,b)\) and \({{\mathcal {D}}}(G,a',b')\) are isomorphic if and only if there is an isomorphism \(\sigma \in \mathrm{Aut}(G)\) such that \(a^\sigma =a'\) and \(b^\sigma =b'\).

In conclusion, we have the following result.

Lemma 5.2

Let \(m,n\ge 2\) be integers.

1. (1)

   There is an one-to-one correspondence between non-isomorphic regular dessins \({{\mathcal {D}}}(G,a,b)\) of \(\mathbf{K}_{m,n}\) and inequivalent triples (G, a, b) such that

   $$\begin{aligned} G=\langle a\rangle \langle b\rangle \cong {{\mathbb {Z}}}_m{{\mathbb {Z}}}_n,~~\text{ and }~~~\langle a\rangle \cap \langle b\rangle =1. \end{aligned}$$
2. (2)

   There is an one-to-one correspondence between non-isomorphic circular regular dessins \({{\mathcal {D}}}(G,a,b)\) of \(\mathbf{K}_{m,n}\) and inequivalent triples (G, a, b) such that

   $$\begin{aligned} G=\langle a\rangle \langle b\rangle \cong {{\mathbb {Z}}}_m{{\mathbb {Z}}}_n,~ \langle a\rangle \cap \langle b\rangle =1, ~~\text{ and }~~~\langle ba\rangle \cap \langle a\rangle =1=\langle ba\rangle \cap \langle b\rangle . \end{aligned}$$

This result translates the regular dessin problem of complete bipartite graphs into a group factorization problem, and which enables us to enumerate non-isomorphic regular dessins and circular regular dessins of complete bipartite graphs.

Now, we are ready to prove Theorem 1.3 as follows.

Proof of Theorem 1.3

Let \(G=\langle x\rangle {:}\langle y\rangle \cong {{\mathbb {Z}}}_m{:}{{\mathbb {Z}}}_n\) be a Frobenius metacyclic group. Then, G has a unique bicyclic factorization up to equivalence.

We observe that every non-trivial outer-automorphism of G induces a non-trivial automorphism of the characteristic subgroup \(\langle x\rangle \), and any two different elements of \(\langle y\rangle \) are not conjugate in \(\langle x\rangle \). It is easily shown that all elements of G of order m are conjugate under \(\mathrm{Aut}(G)\), and as \(\mathrm{Aut}(\langle x\rangle )\) is abelian, different elements of \(\langle y\rangle \) are not conjugate under \(\mathrm{Aut}(G)\).

Let \(G=\langle a\rangle \langle b\rangle \) be an exact factorization of type \(\{m,n\}\). Then, \(a=x^iy^j\) and \(b=x^ky^\ell \) for some integers \(i,j,k,\ell \). Since G is a Frobenius group, if \(j\not =0\), then \(o(a)=n\), and if \(j=0\) then \(o(a)=m\); if \(\ell \not =0\), then \(o(b)=n\), and if \(\ell =0\) then \(o(b)=m\). Since \(G=\langle a\rangle \langle b\rangle \) and \(\langle a\rangle \cap \langle b\rangle =1\), we have \(o(a)o(b)=mn\). Thus, either \(j=0,\ell \not =0\), and \((o(a),o(b))=(m,n)\), or \(j\not =0, \ell =0\), and \((o(a),o(b))=(n,m)\). Without loss of generality, we assume that \(o(a)=m\) and \(o(b)=n\). Thus, \(\langle a\rangle =\langle x\rangle \), and since \(\langle b\rangle \) and \(\langle y\rangle \) are conjugate, we may further assume that \(\langle b\rangle =\langle y\rangle \). So \({{\mathcal {D}}}(G,a,b)\) is a G edge-regular dessin.

Since all element of order m are conjugate under \(\mathrm{Aut}(G)\), there is only one choice for a up to equivalence. Since any two different elements of \(\langle y\rangle \) are not conjugate under \(\mathrm{Aut}(G)\), there are exactly \(\phi (n)\) choices for \(b=y^j\) with \(1\le j\le n-1\) and \((j,n)=1\). Thus, there are exactly \(\phi (n)\) inequivalent triples (G, a, b) such that \(G=\langle a\rangle \langle b\rangle \), each of which satisfies \(\langle a\rangle \cap \langle b\rangle =1\). So, there are exactly \(\phi (n)\) non-isomorphic edge-regular embeddings of \(\mathbf{K}_{m,n}\), all of which have G to be the automorphism group.

Further, since G is a Frobenius group, it follows that the center \(\mathbf{Z}(G)=1\). By Lemma 3.1, we have \(\langle ba\rangle \cap \langle a\rangle =1\) and \(\langle ba\rangle \cap \langle b\rangle =1\). Thus, these maps are both circular by Proposition 2.4.

Finally, we compute the genus of our dessins. According to the Sect. 2, each regular dessin \({{\mathcal {D}}}(G,a,b)\) has face length 2o(ba). As G is a Frobenius group, we have \(o(ba)=n\), and so the face length of \({{\mathcal {D}}}(G,a,b)\) equals 2n. Thus, \(|F|={2|E|\over 2n}=m\). Since \(|E|=|G|=mn\) and \(|V|=m+n,\) it follows that the genus of these dessins is equal to

This completes the proof. \(\square \)

Let \({{\mathcal {D}}}\) be a regular dessin of \(\mathbf{K}_{m,n}\), and let \(G=\mathrm{Aut}{{\mathcal {D}}}.\) By Lemmas 2.1 and 5.2, we have \({{\mathcal {D}}}={{\mathcal {D}}}(G,a,b)\) for some elements a, b, and G is an exact bicyclic group of type \(\{m,n\}\) such as

where \(o(a)=m\), \(o(b)=n,\) and \(\langle a\rangle \cap \langle b\rangle =1\).

Douglas [3] showed that bicyclic groups are meta-abelian and supersoluble. (Recall that a group is called supersoluble if its chief factors are all cyclic.) The aim of this section is to determine m, n with \(\gcd (m,n)=1\) such that each regular dessin of \(\mathbf{K}_{m,n}\) is not circular.

Suppose that \(\gcd (m,n)=1\). Then, each Sylow p-subgroup of \(G=\langle a\rangle \langle b\rangle \cong {{\mathbb {Z}}}_m{{\mathbb {Z}}}_n\) is cyclic, and thus G is a split metacyclic group. If \(\mathbf{Z}(G)=1,\) then by Lemma 3.1, each regular dessin \({{\mathcal {D}}}(G,a,b)\) is circular. Conversely, if \({{\mathcal {D}}}\) is a circular regular dessin of \(\mathbf{K}_{m,n}\) where \(\gcd (m,n)=1\), by [7, Theorem3.6], G is center-free. Thus, we have the following result.

Lemma 6.1

Let G be a bicyclic group of coprime orders. Then, G is the automorphism group of a complete circular regular dessin if and only if \(\mathbf{Z}(G)=1\).

This result implies that each regular dessin of \(\mathbf{K}_{m,n}\) with \(\gcd (m,n)=1\) is circular if and only if each exact bicyclic group of type \(\{m,n\}\) is center-free. Thus, to complete the proof of Theorem 1.5, we need only to determine pairs of integers m, n such that all exact bicyclic groups of type \(\{m,n\}\) have non-trivial center.

First, we give an important lemma as follows.

Lemma 6.2

Let \(X=\langle x\rangle {:}\langle y\rangle \cong {{\mathbb {Z}}}_{p^e}{:}{{\mathbb {Z}}}_{q^f}\) be a non-abelian bicyclic group, where p, q are two distinct primes satisfying \(q~|~(p-1).\) Then, \(\mathbf{Z}(X)\cap \langle x\rangle =1.\)

Proof

Let \(x^y=x^r\), where \(r^{q^f}\equiv 1~(\mathrm{mod~}p^e)\). Suppose that \(\mathbf{Z}(X)\cap \langle x\rangle \ne 1\). It follows that \(\langle x^{p^{e-1}}\rangle \le \mathbf{Z}(X),\) and so

that is, \(p^{e-1}(r-1)\equiv 0~(\mathrm{mod~}p^e)\). Thus, we have \(r-1\equiv 0~(\mathrm{mod~}p)\).

Let \(p^k\parallel r-1\), which denotes the highest power of the prime p dividing \(r-1\), then \(1\le k\le e-1\). By [16, Lemma 6], we have \(p^{k+i}\parallel r^{p^i}-1\) for all \(i\ge 0.\) If we take \(i=e-k\), then \(r^{p^{e-k}}\equiv 1(\text{ mod }~p^e)\), it follows that

and so \(y^{p^{e-k}}\) centralize x. As p, q are distinct primes, which means \(\langle y^{p^{e-k}}\rangle =\langle y\rangle \) and so \(X=\langle x,y\rangle =\langle x,y^{p^{e-k}}\rangle \) is abelian. This contradicts with the fact that X is non-abelian, and so \(\mathbf{Z}(X)\cap \langle x\rangle =1.\) \(\square \)

For a group G and a prime divisor p of the order |G|, let \(G_p\) be a Sylow p-subgroup of G, and \(G_{p'}\) a Hall \(p'\)-subgroup of G. For an element \(g\in G\), let \(g_p\) be the p-part of g and \(g_{p'}\) the \(p'\)-part, namely, \(g=g_pg_{p'}=g_{p'}g_p\) such that \(o(g_p)\) is a power of p, and \(o(g_{p'})\) is coprime to p. We next determine a class of bicyclic groups.

Lemma 6.3

Let m, n be a pair of coprime positive integers such that either

1. (i)

   there exists a prime p|m such that \(\gcd ({p-1},n)=1\) and \(\gcd (m_p,\phi (n))<m_p,\) or

2. (ii)

   there exists a prime q|n such that \(\gcd ({q-1},m)=1\) and \(\gcd (n_q,\phi (m))<n_q.\)

Then, each exact bicyclic group of type \(\{m,n\}\) is not center-free.

Proof

Let \(G=\langle a\rangle \langle b\rangle \) be an arbitrary exact bicyclic group of type \(\{m,n\}\) such that \(\gcd (m,n)=1\), where \(|a|=m, |b|=n\) and \(\langle a\rangle \cap \langle b\rangle =1.\)

Assume first that statement (i) holds, that is, there exists a prime p|m such that \(\gcd ({p-1},n)=1\) and \(\gcd (m_p,\phi (n))<m_p.\) Then, we claim that \(\mathbf{Z}(G)\cap \langle a_p\rangle \ne 1\). In fact, let r be an arbitrary prime divisor of n. Then, there exists a Hall \(\{p,r\}\)-subgroup \(G_{\{p,r\}}\) of G because G is supersoluble. Since \(\gcd (m,n)=1\) and \(\gcd ({p-1},n)=1\), it follows that

If \(p{\,\not |\,}(r-1),\) then \(G_{\{p,r\}}=\langle b_r\rangle \times \langle a_p\rangle \), and so \(b_r\) centralize \(a_p\). If \(p|(r-1),\) as \(\gcd (m_p,\phi (n))<m_p,\) which implies that the centralizer \(C_{\langle a_p\rangle }(\langle b_r\rangle )\ne 1\), that is, \(\mathbf{Z}(G_{\{p,r\}})\cap \langle a_p\rangle \ne 1,\) and so \(b_r\) centralize a subgroup of \(\langle a_p\rangle \) of order p. Thus, by the arbitrariness of r, b centralize a subgroup of \(\langle a_p\rangle \) of order p, and so \(\mathbf{Z}(G)\cap \langle a_p\rangle \ne 1\), as claimed.

Similarly, if there exists a prime q|n such that \(\gcd ({q-1},m)=1\) and \(\gcd (n_q,\phi (m))<n_q,\) then we conclude that \(\mathbf{Z}(G)\cap \langle b_q\rangle \ne 1\). So each exact bicyclic group of type \(\{m,n\}\) has non-trivial center. This completes the proof of the lemma. \(\square \)

The following lemma establishes the converse statement of Lemma 6.3.

Lemma 6.4

If each exact bicyclic group of type \(\{m,n\}\) with \(\gcd (m,n)=1\) has non-trivial center, then either

1. (i)

   there exists a prime p | m such that \(\gcd ({p-1},n)=1\) and \(\gcd (m_p,\phi (n))<m_p,\) or

2. (ii)

   there exists a prime q | n such that \(\gcd ({q-1},m)=1\) and \(\gcd (n_q,\phi (m))<n_q.\)

Proof

Let \(G=\langle a\rangle \langle b\rangle \) be an arbitrary exact bicyclic group of type \(\{m,n\}\). Suppose that \(\mathbf{Z}(G)\ne 1\). Then, \(\mathbf{Z}(G)\cap \langle a\rangle \ne 1\) or \(\mathbf{Z}(G)\cap \langle b\rangle \ne 1\).

Assume first that \(\mathbf{Z}(G)\cap \langle a\rangle \ne 1\). Then, there exists a prime p such that p divides m and \(\mathbf{Z}(G)\cap \langle a_p\rangle \ne 1\).

Claim: \(\gcd ({p-1},n)=1\) and \(\gcd (m_p,\phi (n))<m_p.\)

Suppose to the contrary that \(\gcd ({p-1},n)\ne 1\). It implies that there exists a prime r such that \(r|\gcd (p-1,n).\) As G is supersoluble, there exists a non-abelian \(\{p,r\}\)-Hall subgroup \(G_{\{p,r\}}=\langle a_p\rangle {:}\langle b_r\rangle .\) Since \(\gcd (m,n)=1\), it follows from Lemma 6.2 that \(\mathbf{Z}(G_{\{p,r\}})\cap \langle a_p\rangle =1.\) Let \(G=G_{\{p,r\}}\times G_{\{p,r\}'}\). Then, G is a nonabelian exact bicyclic group of order \(\{m,n\}\) and \(\mathbf{Z}(G)\cap \langle a_p\rangle =1,\) which contradicts with the assumption \(\mathbf{Z}(G)\cap \langle a_p\rangle \ne 1\). So \(\gcd ({p-1},n)=1,\) as claimed.

Notice that \(\gcd (m_p,\phi (n))\le m_p.\) Suppose now that \(m_p|(s-1)\) for some prime divisor s of n. As G is supersoluble, there exists a Hall \(\{p,s\}\)-subgroup \(G_{\{p,s\}}=\langle b_s\rangle {:}\langle a_p\rangle \) such that the centralizer \(C_{\langle a_p\rangle }(\langle b_s\rangle )=1.\) Since \(\gcd (m,n)=1\), it follows that the group \(G_{\{p,s\}}\times {{\mathbb {Z}}}_{m_{p'}}\times {{\mathbb {Z}}}_{n_{s'}}\) is a nonabelian exact bicyclic group of type \(\{m,n\}\) with \(\mathbf{Z}(G)\cap \langle a_p\rangle =1.\) This is a contradiction, and so \(\gcd (m_p,\phi (n))<m_p,\) as in part (i).

Suppose now that \(\mathbf{Z}(G)\cap \langle b\rangle \ne 1\). Then, there exists a prime q such that q|n and \(\mathbf{Z}(G)\cap \langle b_q\rangle \ne 1\). By the same reason as above, we claim that \(\gcd ({q-1},m)=1\) and \(\gcd (n_q,\phi (m))<n_q,\) as in part (ii). This completes the proof of lemma. \(\square \)

Combining Lemma 6.1 with Lemmas 6.3 and 6.4, we have the following Theorem.

Theorem 6.5

Let m, n be positive integers which are relatively prime. Then, the following statements are equivalent:

1. (1)

   \(\mathbf{K}_{m,n}\) has no circular regular dessin.

2. (2)

   Each exact bicyclic group of type \(\{m,n\}\) has non-trivial center.

3. (3)

   Either there exists a prime p|m such that \(\gcd ({p-1},n)=1\) and \(\gcd (m_p,\phi (n))<m_p,\) or, there exists a prime q|n such that \(\gcd ({q-1},m)=1\) and \(\gcd (n_q,\phi (m))<n_q.\)

Proof of Theorem 1.5 now follows from the above Theorem.

In this final section, we prove Theorems 1.7 and 1.8. Let \({{\mathcal {D}}}(G,a,b)\) be a complete regular dessin of type \(\{m,n\}\) with \(G=\mathrm{Aut}{{\mathcal {D}}}.\) By Lemma 5.2, G is an exact bicyclic group

If \(G=G_1\times G_2\) is decomposable, then both \(G_1\) and \(G_2\) are exact bicyclic groups, and so there exist some element pairs \((a_1,b_1)\) and \((a_2,b_2)\) such that

where \(m=m_1m_2\), \(n=n_1n_2,\) and \(\gcd (m_1,m_2)=1=\gcd (n_1,n_2).\) Denote \({{\mathcal {D}}}_1={{\mathcal {D}}}(G_1,a_1,b_1)\) and \({{\mathcal {D}}}_2={{\mathcal {D}}}(G_2,a_2,b_2)\), then \({{\mathcal {D}}}_1\) and \({{\mathcal {D}}}_2\) are regular dessins with underlying graphs \(\mathbf{K}_{m_1,n_1}\) and \(\mathbf{K}_{m_2,n_2},\) respectively. In particular, since \(\gcd (m_1,m_2)=1\) and \(\gcd (n_1,n_2)=1\), it follows that \(a=a_1a_2\) and \(b=b_1b_2,\) and so \({{\mathcal {D}}}\) is a direct product of \({{\mathcal {D}}}_1\) and \({{\mathcal {D}}}_2\) by Definition 1.6.

Lemma 7.1

If \(G=G_1\times G_2\) is decomposable, where \(G_1,G_2\) are as above, then

The following lemma shows that \({{\mathcal {D}}}\) is a circular dessin if \({{\mathcal {D}}}_1\) and \({{\mathcal {D}}}_2\) are both circular.

Lemma 7.2

Let \({{\mathcal {D}}}(G,a,b)={{\mathcal {D}}}(G_1,a_1,b_1)\times {{\mathcal {D}}}(G_2,a_2,b_2).\) If \({{\mathcal {D}}}(G_1,a_1,b_1)\) and \({{\mathcal {D}}}(G_2,a_2,b_2)\) are both circular dessins. Then, \({{\mathcal {D}}}(G,a,b)\) is circular.

Proof

According to Proposition 2.4, we have

and

Since \(G=(\langle a_1\rangle \langle b_1\rangle )\times (\langle a_2\rangle \langle b_2\rangle )\), it follows that \(ba=(b_1b_2)(a_1a_2)=(b_1a_1)(b_2a_2),\) and so

and

Thus, \({{\mathcal {D}}}(G,a,b)\) is a circular regular dessin. \(\square \)

The following example shows that the converse statement is not necessarily true.

Example 7.3

Let \(G_1=\langle b_1\rangle {:}\langle a_1\rangle \cong {{\mathbb {Z}}}_{17}{:}{{\mathbb {Z}}}_{4}\), and let \(G_2=\langle a_2\rangle {:}\langle b_2\rangle \cong {{\mathbb {Z}}}_3{:}{{\mathbb {Z}}}_4,\) where \(b_1^{a_1}=b_1^{-1}\) and \(a_2^{b_2}=a_2^{-1}.\) Let \(G=G_1\times G_2\), and let \(a=a_1a_2\), \(b=b_1b_2\). Then,

Since \((b_1a_1)^2=a_1^2\) and \((b_2a_2)^2=b_2^2,\) it follows that \(\langle b_1a_1\rangle \cap \langle a_1\rangle =\langle a_1^2\rangle \ne 1\) and \(\langle b_2a_2\rangle \cap \langle b_2\rangle =\langle b_2^2\rangle \ne 1.\) By Proposition 2.4, neither \({{\mathcal {D}}}(G_1,a_1,b_1)\) nor \({{\mathcal {D}}}(G_2,a_2,b_2)\) is a circular dessin.

As \((ba)^2=(b_1a_1)^2(b_2a_2)^2=a_1^2b_2^2,\) and \((ba)^4=(a_1^2b_2^2)^2=1.\) Thus, \(\langle ba\rangle \cap \langle a\rangle =1\) and \(\langle ba\rangle \cap \langle b\rangle =1,\) and so \({{\mathcal {D}}}\) is a circular regular dessin by Proposition 2.4.

If \(\gcd (|b_1a_1|,|b_2a_2|)=1,\) then the converse statement of Lemma 7.2 is true.

Lemma 7.4

Let \({{\mathcal {D}}}(G,a,b)={{\mathcal {D}}}(G_1,a_1,b_1)\times {{\mathcal {D}}}(G_2,a_2,b_2).\) If \({{\mathcal {D}}}(G,a,b)\) is circular and \(\gcd (|b_1a_1|,|b_2a_2|)=1,\) Then, \({{\mathcal {D}}}(G_1,a_1,b_1)\) and \({{\mathcal {D}}}(G_2,a_2,b_2)\) are both circular dessins.

Proof

Since \({{\mathcal {D}}}(G,a,b)={{\mathcal {D}}}(G_1,a_1,b_1)\times {{\mathcal {D}}}(G_2,a_2,b_2),\) it follows from Definition1.6 that

where \(\gcd (|a_1|,|a_2|)=1\) and \(\gcd (|b_1|,|b_2|)=1.\)

Let \(|b_1a_1|=k_1\) and \(|b_2a_2|=k_2\). As \(\gcd (k_1,k_2)=1,\) we have

and so \((ba)^{k_1}=(b_1a_1)^{k_1}(b_2a_2)^{k_1}=(b_2a_2)^{k_1}.\) Since \({{\mathcal {D}}}(G,a,b)\) is circular, we have \(\langle ba\rangle \cap \langle a\rangle =1=\langle ba\rangle \cap \langle b\rangle ,\) and thus,

and

This implies that \({{\mathcal {D}}}(G_2,a_2,b_2)\) is circular. Similarly, since \((ba)^{k_2}=(b_1a_1)^{k_2},\) it follows that

and

So \({{\mathcal {D}}}(G_1,a_1,b_1)\) is circular. \(\square \)

Proof of Theorem 1.7 now follows from the Lemmas 7.2 and 7.4. As an application of Theorem 1.7, now we are ready to prove Theorem 1.8 as follows.

Proof of Theorem 1.8

Let \({{\mathcal {D}}}(G,a,b)\) be a complete regular dessin of \(\mathbf{K}_{m,n}.\) By Lemma 5.2, we have

As m and n have only one common prime divisor p, we may write \(m=m_pm_{p'}\) and \(n=n_pn_{p'}\), where \(\gcd (m_{p'},n_{p'})\)=1. Let \(a=a_pa_{p'}\) and \(b=b_pb_{p'},\) where \(|a_p|=m_p,\) \(|b_p|=n_p,\) and \(|a_{p'}|=m_{p'},\) \(|b_{p'}|=n_{p'}\). Since \(\gcd (p,\phi (m_{p'})\phi (n_{p'}))=1\) and \(\gcd (m_{p'}n_{p'},p-1)=1\), we conclude that \(\langle a_{p'}\rangle \) centralize \(\langle b_p\rangle \). In fact, suppose to the contrary that \(\langle a_{p'}\rangle \) does not centralize \(\langle b_p\rangle \). Then, there exists a prime \(q|m_{p'}\) such that either \(q|(p-1)\) or \(p|(q-1),\) which contradicts with the assumption \(\gcd (p,\phi (m_{p'})\phi (n_{p'}))=1\) and \(\gcd (m_{p'}n_{p'},p-1)=1\). So \(\langle a_{p'}\rangle \) centralize \(\langle b_p\rangle \). Similarly, we have \(\langle b_{p'}\rangle \) centralize \(\langle a_p\rangle .\) Thus,

Denote \(G_p=\langle a_p\rangle \langle b_p\rangle \) and \(G_{p'}=\langle a_{p'}\rangle \langle b_{p'}\rangle \). By Definition 1.6, \({{\mathcal {D}}}(G,a,b)\) has a factorization as follows

where \({{\mathcal {D}}}(G_p,a_p,b_p)\) and \({{\mathcal {D}}}(G_{p'},a_{p'},b_{p'})\) are regular dessins of \(\mathbf{K}_{m_p,n_p}\) and \(\mathbf{K}_{m_{p'},n_{p'}},\) respectively.

Since \(\gcd (|b_pa_p|,|b_{p'}a_{p'}|)=1,\) it follows from Theorem 1.7 that \({{\mathcal {D}}}(G,a,b)\) is not circular if and only if either \({{\mathcal {D}}}(G_p,a_p,b_p)\) or \({{\mathcal {D}}}(G_{p'},a_{p'},b_{p'})\) is not circular. As p is an odd prime, by [8, Theorem1.1], \({{\mathcal {D}}}(G_p,a_p,b_p)\) is not circular if and only if \(m_p\ne n_p\); since \(\gcd (m_{p'},n_{p'})=1\), by Theorem 1.5, \({{\mathcal {D}}}(G_{p'},a_{p'},b_{p'})\) is not circular if and only if either there exists a prime \(r|m_{p'}\) such that \(\gcd (r-1,n_{p'})=1\) and \(\gcd (m_r,\phi (n_{p'}))<m_r,\) or there exists a prime \(s|n_{p'}\) such that \(\gcd (s-1,m_{p'})=1\) and \(\gcd (n_s,\phi (m_{p'}))<n_s.\) This completes the proof of Theorem 1.8. \(\square \)