Journal of Combinatorial Theory, Series B

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1 Journal of Combinatorial Theory, Series B 101 (2011) Contents lists available at ScienceDirect Journal of Combinatorial Theory, Series B Classification of nonorientable regular embeddings of complete bipartite graphs Jin Ho Kwak a,youngsookwon b a Department of Mathematics, Pohang University of Science and Technology, Pohang, , Republic of Korea b Department of Mathematics, Yeungnam University, Kyeongsan, , Republic of Korea article info abstract Article history: Received 26 July 2007 Available online 23 March 2011 Keywords: Graph Surface Regular embedding Regular map A 2-cell embedding of a graph G into a closed (orientable or nonorientable) surface is called regular if its automorphism group acts regularly on the flags mutually incident vertex edge face triples. In this paper, we classify the regular embeddings of complete bipartite graphs K n,n into nonorientable surfaces. Such a regular embedding of K n,n exists only when n is of the form n = 2p a 1 1 pa 2 2 pa k where the p k i are primes congruent to ±1 mod8. In this case, up to isomorphism the number of those regular embeddings of K n,n is 2 k Elsevier Inc. All rights reserved. 1. Introduction A 2-cell embedding of a connected graph into a connected closed surface is called a topological map. An automorphism of a map is an automorphism of the underlying graph which can be extended to a self-homeomorphism of the supporting surface in the embedding. The automorphisms of a map M act semi-regularly on its flags mutually incident vertex edge face triples. If the automorphism group Aut(M) of M acts regularly on the flags, then the map M as well as the corresponding embedding are also called regular. For a given map M whose supporting surface is orientable, the set Aut + (M) of orientation-preserving automorphisms of the map M acts semi-regularly on its arcs mutually incident vertex edge pairs. If Aut + (M) acts regularly on its arcs, then the map M is called orientably regular. Therefore if a supporting surface is orientable, a regular map means an orientably regular map having an orientation-reversing automorphism. One of the standard problems in topological graph theory is the classification of orientably regular embeddings or regular embeddings of a given class of graphs. In recent years, there has been The first and the second authors are supported by the Korea Research Foundation Grant funded Korean Government KRF C00011 and KRF C00049, respectively. address: ysookwon@ynu.ac.kr (Y.S. Kwon) /$ see front matter 2011 Elsevier Inc. All rights reserved. doi: /j.jctb

2 192 J.H. Kwak, Y.S. Kwon / Journal of Combinatorial Theory, Series B 101 (2011) particular interest in the orientably regular embeddings and regular embeddings of complete bipartite graphs K n,n exhibited in papers by several authors [4,5,8,10,9,12,11,14]. The regular (or reflexible regular) embeddings and self-petrie dual regular embeddings of K n,n into orientable surfaces were classified by the authors in [12]. Recently, G.A. Jones [8] classified the orientably regular embeddings of K n,n. In this paper, we classify the nonorientable regular embeddings of K n,n. For other classes of graphs, the classification of their regular embeddings has been done: for instance, the work of N.L. Biggs [2] and of L.D. James and G.A. Jones [7] yields a classification of orientably regular embeddings of complete graphs, while their nonorientable regular embeddings were classified by S.E. Wilson [15]. Recently, D.A. Catalano et al. [3] classified the orientably regular embeddings of n-dimensional cubes Q n, and R. Nedela and the second author [13] classified their nonorientable regular embeddings. The following theorem is the main result in this paper. Theorem 1.1. For any integer n congruent to 0, 1 or 3 mod 4, no nonorientable regular embedding of K n,n exists. For n = 2p a 1 1 pa 2 2 pa k (the prime decomposition of n), up to isomorphism the number of nonorientable k regular embeddings of K n,n is 2 k if every p i is congruent to ±1 mod 8; and 0 otherwise. This paper is organized as follows. In the next section, we introduce a triple of three graph automorphisms of G, called an admissible triple for G, which corresponds to a regular embedding of G. In Section 3, we construct some nonorientable regular embeddings of K n,n by forming admissible triples for K n,n. In the last two sections, Theorem 1.1 is proved by showing that up to isomorphism no other nonorientable regular embedding of K n,n exists beyond those constructed in Section An admissible triple of graph automorphisms For a given simple graph G, the automorphism group Aut(G) acts faithfully on both the vertex set V (G) and the arc set D(G). Moreover, in an embedding M of G whose valency is greater than two, Aut(M) Aut(G) and Aut(M) acts faithfully on the flag set F (M). Soweconsideragraph automorphism as a permutation of V (G), D(G) or F (M) according to the context. Let G be a graph and let M be a regular embedding of G. Then for some incident vertex edge pair (v, e), there exist three involutory graph automorphisms l, r and t in Aut(G) which generate Aut(M) and satisfy the following properties: (i) Γ = l, r, t acts transitively on the arc set D(G). (ii) The stabilizer Γ v of the vertex v is Γ v = r, t and is isomorphic to the dihedral group D n, and its cyclic subgroup rt acts regularly on the arcs emanating from v, where n is the valency of G. (iii) The stabilizer Γ e of the edge e is Γ e = l, t and is isomorphic to the Klein four-group Z 2 Z 2. We call such automorphisms l, r and t basic generators for the regular map M with respect to the incident vertex edge pair (v, e). Also we call a triple (l, r, t) of involutory automorphisms of G satisfying properties (i), (ii) and (iii) an admissible triple for a regular embedding of G, orsimplyanadmissible triple for G. Thevertexv and the edge e are called a root vertex and a root edge, respectively. Conversely, for a given admissible triple (l, r, t) for G with a root vertex v, the group Γ = l, r, t is the automorphism group of a regular map M = M(l, r, t), called the derived map of the admissible triple (l, r, t). Flags of M are elements of Γ, and vertices, edges and faces of M are left cosets of the subgroups r, t, l, t and l, r, respectively. Mutual incidence of map elements is given by nonempty intersection. In fact, Γ acts on the map M by the left multiplication. Let G be the underlying graph of the map M. If we label each vertex g r, t in G by g(v), then the graph G is exactly G and the action of Γ on G corresponds to a subgroup of automorphisms of G. Hence we consider M(l, r, t) as a regular embedding of G from now on. In [6, Theorem 3], Gardiner et al. showed how to construct a regular embedding of G by an admissible triple. The method of [6] looks different from ours, but they are essentially the same.

3 J.H. Kwak, Y.S. Kwon / Journal of Combinatorial Theory, Series B 101 (2011) For a regular embedding M of G with basic generators l, r and t with respect to an incident vertex edge pair (v, e), one can see that the derived map M(l, r, t) is isomorphic to M. Henceit suffices to consider derived maps M(l, r, t) of admissible triples (l, r, t) for G in order to classify regular embeddings of G. Let (l 1, r 1, t 1 ) and (l 2, r 2, t 2 ) be two admissible triples for G. If there exists a graph automorphism φ of G such that φl 1 φ 1 = l 2, φr 1 φ 1 = r 2 and φt 1 φ 1 = t 2,thenφ induces a map isomorphism from M(l 1, r 1, t 1 ) to M(l 2, r 2, t 2 ), and hence M(l 1, r 1, t 1 ) and M(l 2, r 2, t 2 ) are isomorphic. Conversely, assume that two derived regular maps M(l 1, r 1, t 1 ) and M(l 2, r 2, t 2 ) are isomorphic. Since two underlying graphs of the two maps M(l 1, r 1, t 1 ) and M(l 2, r 2, t 2 ) are the same graph G, amap isomorphism is a graph automorphism of G, which implies that there exists a graph automorphism φ of G such that φl 1 φ 1 = l 2, φr 1 φ 1 = r 2 and φt 1 φ 1 = t 2. Therefore we have the following obvious but important consequence of the above considerations. Proposition 2.1. Let G be a graph. Then every regular embedding M of G into an orientable or nonorientable surface is isomorphic to a derived regular map M(l, r, t) of an admissible triple (l, r, t) for G and its isomorphism class corresponds to the conjugacy class of the triple (l, r, t) in Aut(G). It follows from Proposition 2.1 that for a given graph G, up to isomorphism the number of nonorientable regular embeddings of G equals the number of orbits of admissible triples (l, r, t) for G satisfying lt, rt = l, r, t under the conjugate action by Aut(G). 3. Constructions of nonorientable embeddings The complete bipartite graph K 2,2 is just the 4-cycle, and there is only one nonorientable regular embedding of the 4-cycle with the projective plane as the supporting surface. So from now on, we assume that n 3. For a complete bipartite graph K n,n, let [n] ={0, 1,...,n 1} and [n] ={0, 1,...,(n 1) } be the vertex sets of K n,n as the partite sets, and let D ={(i, j ), ( j, i) 0 i, j n 1} be the arc set. We denote the symmetric group on [n] ={0, 1,...,n 1} by S [n], and the stabilizer of i as a subgroup of S [n] by Stab(i). We identify the integers 0, 1,...,n 1 with their residue classes modulo n according to the context. Since Aut(K n,n ) = S n Z 2 contains all permutations of vertices of each partite set and the interchanging of two partite sets [n] and [n], one can assume that up to isomorphism every orientable or nonorientable regular embedding of K n,n is derived from an admissible triple (l, r δ, t) for K n,n of the following type: l = ( 0, 0 )( 1,(n 1) )( 2,(n 2) ) (n 1, 1 ), r δ = δ ( 0, 1 )( (n 1), 2 )( (n 2), 3 ) ( ) n + 1 n,, 2 2 ( ) n n (0 t = (0)(1,n 1)(2,n 2), )( 1,(n 1) )( 2,(n 2) ) ( ) n n,, for some δ Stab(0); see Fig. 1. Note that the root vertex and the root edge of the above admissible triple are 0 and {0, 0 }, respectively. In fact, the admissibility of the triple (l, r δ, t) depends only on the permutation δ Stab(0). Clearly,lt = tl and so l, t = Z 2 Z 2.Moreover ( ) n n (0 r δ t = δ(0)(1,n 1)(2,n 2),, 1, 2,...,(n 2),(n 1) ) 2 2 generates the cyclic subgroup which acts regularly on the arcs emanating from the root vertex 0. For an even integer n, let us write n = n/2 for notational convenience. Lemma 3.1. Suppose the derived regular maps M(l, r δ1, t) and M(l, r δ2, t) are isomorphic, where (l, r δ1, t) and (l, r δ2, t) are admissible triples with δ 1,δ 2 Stab(0). Then either (i) δ 1 = δ 2 or (ii) nisevenandδ 2 (k) = δ 1 (k + n) + n for all k [n].

4 194 J.H. Kwak, Y.S. Kwon / Journal of Combinatorial Theory, Series B 101 (2011) Fig. 1. Relabeling φ of vertices for an admissible triple for K n,n. Proof. Suppose that M(l, r δ1, t) and M(l, r δ2, t) are isomorphic. By Proposition 2.1(2), there exists a graph automorphism ψ Aut(K n,n ) such that ψlψ 1 = l, ψr δ1 ψ 1 = r δ2 and ψtψ 1 = t. Note that the vertex 0 and the vertex n when n is even are the only vertices which can be fixed by both t and r δi. The relationsψt = tψ and ψr δ1 = r δ2 ψ imply that ψ permutes these vertices. Hence our discussion can be divided into the following two cases. Case 1. Suppose ψ(0) = 0. Since ψr δ1 = r δ2 ψ and ψ commutes with l and t, ψ should be the identity. This means that δ 1 = δ 2. Case 2. Suppose n is even and ψ(0) = n. Sinceψr δ1 = r δ2 ψ and ψ commutes with l and t, onecan show that ψ(k) = k + n and ψ(k ) = (k + n) for all k [n]. So for every k [n], δ 2 (k) = r δ2 (k) = ψr δ1 ψ 1 (k) = ψr δ1 (k + n) = δ 1 (k + n) + n. It will be shown later that the case (ii) in Lemma 3.1 cannot occur. For any involution δ Stab(0), set = r δ t [n] = δ (0)(1, 1)(2, 2) S [n]. Then also belongs to Stab(0) and satisfies the equation 1 ( k) = (k) for all k [n], sinceδ is an involution. For an admissible triple (l, r δ, t) for K n,n with δ Stab(0), letr = r δt and L = tl, namely, and R = r δt = ( 0, 1,...,(n 1) ) L = tl = ( 0, 0 )( 1, 1 ) ((n 1), (n 1) ), as permutations on the vertex set [n] [n]. These are the local rotation automorphism at the root vertex 0 and the direction-reversing automorphism of the root edge {0 0 }, respectively. In fact, L is an automorphism which interchange partite sets. Note that any one of δ,, R determines completely the other two. Recall that the regular map M(l, r δ, t) is nonorientable if and only if l, r δ, t = R, L. Hence if (l, r δ, t) is an admissible triple for K n,n and M(l, r δ, t) is nonorientable, then (0) = 0, 1 ( k) = (k) for all k [n], R, L = 4 E(K n,n) =4n 2 and t R, L. So in order to construct all nonorientable regular embeddings of K n,n,weneedtoexamine satisfying the aforementioned conditions. Let M non n = { S [n] (0) = 0, 1 ( k) = (k) for all k [n] R, L = 4n 2 and R, L contains t}. We shall show that there is a one-to-one correspondence between the nonorientable regular embeddings of K n,n for n 3 up to isomorphism and the elements in Mn non. From now on, we shall deal with instead of δ, as we construct and classify the nonorientable regular embeddings of K n,n.

5 J.H. Kwak, Y.S. Kwon / Journal of Combinatorial Theory, Series B 101 (2011) Lemma 3.2. For every involution δ S [n] with δ(0) = 0, the following statements are equivalent: (1) The triple (l, r δ, t) is admissible and the derived regular map M(l, r δ, t) is nonorientable. (2) M non n,where = r δ t [n]. Proof. For = r δ t [n], we already know (0) = 0 and 1 ( k) = (k) for all k [n]. (1) (2) Let (l, r δ, t) be admissible, and let the map M(l, r δ, t) be nonorientable. Then l, r δ, t = R, L and l, r δ, t = R, L = 4 E(K n,n) =4n 2,so Mn non. (2) (1) Let Mn non.sincet R, L, it follows that R t = r δ R, L and tl = l R, L. Hence, l, r δ, t = R, L. Foranyi, j [n], wehave R ī δ LRj t( 0, 0 ) = R ī δ LRj and by taking L on both sides, we have ( 0, 0 ) = R ī δ L( 0, j ) = R ī δ( 0, j ) = ( i, i ( j) ) LR ī δ LRj t( 0, 0 ) = LR ī δ LRj( 0, 0 ) = ( i, i ( j) ). This shows that the arc (0, 0 ) can be mapped to any other arc by the action of the group R, L. This means that l, r δ, t = R, L acts transitively on both the arc set D(K n,n) and the vertex set V (K n,n ).For0 V(K n,n ), R, t R, L 0.Since R, t = R, L 0 =2n, one can see that R, t = R, L 0 D n of order 2n, and that the subgroup R acts regularly on the arcs emanating from 0. For the edge e ={0, 0 }, one can easily check that the stabilizer R, L e is equal to L, t, which is isomorphic to Z 2 Z 2.So(l, r δ, t) is an admissible triple for K n,n.since l, r δ, t = R, L, thederived regular map M(l, r δ, t) is nonorientable. In order to determine satisfying R, L = 4n2, we need to consider certain elements in the group R, L. For later use, we define two properties, called (P 1) and (P 2 ), as follows and (k + i) = b(i) (k) + a(i) and i (k) + 1 = a(i)( k + b(i) ) for all k [n], (P 1 ) (k + i) = b(i) ( k) + a(i) and i (k) + 1 = a(i)( k + b(i) ) for all k [n]. (P 2 ) In fact, (P 1 ) and (P 2 ) are equivalent to the equalities R LRī δ L = LRa(i) LR b(i) LR a(i) LR b(i) t, respectively. and R LRī δ L = Lemma 3.3. Let δ S [n] be an involution such that δ(0) = 0,orequivalently, 1 ( k) = (k) for all k [n], and (0) = 0. Then the following statements are equivalent: (1) Mn non. (2) The subgroup R, L of S [n] [n] is a disjoint union of the four sets B := { } R ī δ LRj i, j [n], LB := { } LR ī δ LRj i, j [n], Bt := { R ī δ LRj t i, j [n] }, LBt := { LR ī δ LRj t i, j [n] }. (3) For each i [n], thereexista(i), b(i) [n] such that either (P 1 ) or (P 2 ) holds. In addition, (P 2 ) holds for at least one i [n].

6 196 J.H. Kwak, Y.S. Kwon / Journal of Combinatorial Theory, Series B 101 (2011) Proof. (1) (2) For any Mn non and for any i, j [n], wehave R ī δ LRj t( 0, 0 ) ( = R ī δ LRj 0, 0 ) = R ī δ L( 0, j ) = R δ( ī 0, ) j = ( i, i ( j) ) and by taking L on both sides we obtain LR ī δ LRj t( 0, 0 ) = LR ī δ LRj( 0, 0 ) = ( i, i ( j) ). By comparing images of the arc (0, 0 ), one can see that (B Bt) (LB LBt) =, and for any (i, j) (k,l), also {R ī δ LRj, Rī δ LRj t} {Rk LRl, Rk LRl t}= and {LRī δ LRj, LRī t} δ LRj {LR k LRl, LRk LRl t}=. Now, it suffices to show that Rī δ LRj Rī δ LRj t and LRī δ LRj LRī t for δ LRj all (i, j) [n] [n] in order to show that B Bt = and LB LBt =.Infact,forany(i, j) [n] [n], R ī δ LRj t( 0, 1 ) ( = R ī δ LRj 0, 1 ) and LR ī δ LRj t( 0, 1 ) = LR ī δ LRj( 0, 1 ). These imply that R ī δ LRj Rī δ LRj t and LRī δ LRj LRī t. Hence the four sets B, LB, Bt and LBt are δ LRj mutually disjoint, and the cardinality of their union is 4n 2, which equals R, L. (2) (3) Since the group R, L is the union of the four sets, for each i [n], there exist a(i), b(i) [n] such that R LRī δ L = LRa(i) LR b(i) or R LRī δ L = LRa(i) LR b(i) t. By comparing the associated values of k and k,wehave or (k + i) = b(i) (k) + a(i) and i (k) + 1 = a(i)( k + b(i) ) for all k [n] (k + i) = b(i) ( k) + a(i) and i (k) + 1 = a(i)( k + b(i) ) for all k [n]. In other words, either (P 1 ) or (P 2 ) holds. Suppose that (P 1 ) holds for all i [n], and hence that R LRī δ L = LRa(i) LR b(i). Then one can easily check that R, L ={ } { } LR ī δ LRj i, j [n] R ī δ LRj i, j [n] (disjoint union), which contradicts the assumption. Hence there exists at least one i [n] for which (P 2 ) holds. (3) (1) Since (P 1 ) and (P 2 ) are equivalent to the equalities R LRī δ L = LRa(i) LR b(i) and R LRī δ L = LR a(i) LR b(i) t, the group R, L contains t, and R, L is the same as the union of the four sets in (2). The disjointness of the union can be shown in a similar way to show (1) (2). This means that R, L = 4n2,so Mn non. In fact, two numbers a(i) and b(i) in Lemma 3.3(3) are completely determined by the values of in the following sense. Lemma 3.4. Let M non n and let a(i) and b(i) be the numbers given in (P 1 ) and (P 2 ). If they satisfy (P 1 ), then a(i) = (i) and b(i) = i (1) = (i) (1), and if they satisfy (P 2 ),then a(i) = (i) and b(i) = i (1) = (i) (1).

7 J.H. Kwak, Y.S. Kwon / Journal of Combinatorial Theory, Series B 101 (2011) Proof. Let a(i) and b(i) satisfy (P 1 ).Bytakingk = 0intheequation(k + i) = b(i) (k) + a(i), wehave a(i) = (i). Also,bytakingk = 0 and k = b(i) in the equation i (k) + 1 = a(i) (k + b(i)), wehave b(i) = a(i) (1) = (i) (1) and b(i) = i ( 1). Since 1 ( k) = (k) for all k [n], it follows that b(i) = i ( 1) = i+1( 1 ( 1) ) = i+1( (1) ) = i+2( 2 (1) ) = = 1( i 1 (1) ) = i (1). Let a(i) and b(i) satisfy (P 2 ).Bytakingk = 0intheequation(k + i) = b(i) ( k) + a(i), wehave a(i) = (i). Also,bytakingk = 0 and k = b(i) in the equation i (k) + 1 = a(i) ( k + b(i)), wehave b(i) = a(i) (1) = (i) (1) and b(i) = i ( 1). Since 1 ( k) = (k) for all k [n], wefindb(i) = i ( 1) = i (1). Now, let us consider the even numbers n as the first case in which we construct nonorientable regular embeddings of K n,n. As a candidate of Mn non, we define a permutation S [n] by = (0)(2, 2)(4, 4) (1, 1 + x, 1 + 2x, 1 + 3x,...) for any positive even integer x such that the greatest common divisor of n and x is 2. Suppose that belongs to Mn non for some even x. Foreachi [n], thereexista(i), b(i) [n] satisfying (P 1 ) or (P 2 ) by Lemma 3.3. For every even i [n], it follows that b(i) = (i) (1) = i (1), which implies that a(i), b(i) satisfy (P 1 ) by Lemma 3.4. So there exists an odd integer j [n] such that a( j), b( j) [n] satisfying (P 2 ) by Lemma 3.3. For such odd j [n], wehaveb( j) = j (1) = (1 + jx) by Lemma 3.4. Since b( j) = ( j) ( j+x) (1) = (1) = 1 ( j + x)x = (1 + jx) + ( 2 x 2), we have (1 + jx) = (1 + jx) + (2 x 2 ), and hence x 2 2 (mod n). Therefore the condition x 2 2 (mod n) is necessary for to belong to Mn non. The following two lemmas show that the condition is also sufficient. Furthermore it will be shown that up to isomorphism there is a oneto-one correspondence between the set of nonorientable regular embeddings of K n,n and the set of solutions of x 2 2 (mod n) in Z n. So square roots of 2 mod n are very important in this paper. Lemma 3.5. Let n and x be even integers such that n > x > 3 and x 2 2 (mod n): (1) For an even integer 2i [n], if we define a(2i) = 2i andb(2i) = 2ix + 1, thena(2i) and b(2i) satisfy (P 1 ) with =. (2) For an odd integer 2i + 1 [n], if we define a(2i + 1) = 2i + x + 1 and b(2i + 1) = 2ix x 1 then a(2i + 1) and b(2i + 1) satisfy (P 2 ) with =. Proof. We prove (1), and note that a proof of (2) is similar. For an even integer 2i, let a(2i) = 2i and b(2i) = 2ix + 1. Then for any even integer 2k [n], wehave and (2k + 2i) = 2k 2i, b(2i) (2k) + a(2i) = 2ix+1 (2k) 2i = 2k 2i, 2i (2k) + 1 = 2k + 1, a(2i) ( ) 2k + b(2i) = 2i (2k + 2ix + 1) = 2k + 2ix + 1 2ix = 2k + 1. For any odd integer 2k + 1 [n], wehave

8 198 J.H. Kwak, Y.S. Kwon / Journal of Combinatorial Theory, Series B 101 (2011) and (2k i) = 2k + 2i + x + 1, b(2i) (2k + 1) + a(2i) = 2ix+1 (2k + 1) 2i = 2k (2ix + 1)x 2i = 2k i + x 2i = 2k + 2i + x + 1 because x 2 2 (mod n), 2i (2k + 1) + 1 = 2k ix + 1 = 2k + 2ix + 2, a(2i) ( ) 2k b(2i) = 2i (2k + 2ix + 2) = 2k + 2ix + 2. Hence a(2i) and b(2i) satisfy (P 1 ) with =. Lemma 3.6. For any two positive even integers n and x such that n > x > 3 and x 2 2 (mod n), belongs to M non n. Proof. Note that (0) = 0. For any even 2k [n], wehave 1( 2k) = 2k = (2k). Foranyodd 2k+1 [n], wehave 1( 2k 1) = 2k 1 x and (2k+1) = (2k+1+x) = 2k 1 x. Sofor any k [n], it follows that 1( k) = (k). By Lemmas 3.3(3) and 3.5, belongs to Mn non. By Lemma 3.6, we may consider the following as a subset of Mn non : { Nn non { = n > x > 3andx 2 2 (mod n)} if n is even, if n is odd. Note that for any even integers n, x such that n 0 (mod 4) and n > x > 1, x 2 is a multiple of 4 and hence there is no such x satisfying x 2 2 (mod n). Soforeveryn 0 (mod 4), Nn non =. The smallest integer n such that Nn non is 14. This will show that K 14,14 is the smallest complete bipartite graph (other than K 2,2 ) which can be regularly embedded into a nonorientable surface. Remark. For any Nn non with δ = (0)(1 1)(2 2), the automorphism group l, r δ, t = R, L of the derived map M(l, r δ, t) is a split extension (semidirect product) of an abelian normal subgroup K = R 2, LR 2 L of rank 2, order n 2 and exponent n by a Sylow 2- subgroup H = R n, L of order 16. The two cyclic summands of the subgroup K act as the stabilizers of vertices in the two different parts of the bipartition of K n,n, and these are interchanged by the arc-reversing automorphism L. Note that H is isomorphic to the automorphism group of the nonorientable regular embedding M 1 of K 2,2 into the projective plane, and so there is a regular homomorphism M(l, r δ, t) M 1 with the fibre transformation group isomorphic to K. The covalency (face size) of M(l, r δ, t) is the order of LR which is in fact 8. Hence the number of faces of the map M(l, r δ, t) is n 2 /4. By the Euler formula, the supporting surface of M(l, r δ, t) is nonorientable surface with (3n 2 8n + 8)/4 crosscaps. In the final two sections, it will be shown that Mn non = Nn non for every n, which means that Mn non = if n is odd or n 0 (mod 4). In the remaining part of this section, we shall show that for any two different 1, 2 Nn non, their derived regular embeddings of K n,n are not isomorphic. Also for a given n 2 (mod 4), we will estimate the cardinality Nn non, that is, the number of solutions of x 2 = 2inZ n. Lemma 3.7. For any two 1, 2 Nn non with n > 3,letδ i = i (0)(1 1)(2 2) for i = 1, 2.Then the derived regular maps M(l, r δ1, t) and M(l, r δ2, t) are isomorphic if and only if x 1 = x 2. Proof. Since the sufficiency is clear, we prove only the necessity. Recall that if n 0 (mod 4), then N non n =.Soletn 2 (mod 4).

9 J.H. Kwak, Y.S. Kwon / Journal of Combinatorial Theory, Series B 101 (2011) Suppose the two regular maps M(l, r δ1, t) and M(l, r δ2, t) are isomorphic. By Lemma 3.1, δ 1 = δ 2 or δ 2 (k) = δ 1 (k+ n)+ n for any k [n]. Ifδ 1 = δ 2 then x 1 = x 2. Assume instead that δ 2 (k) = δ 1 (k+ n)+ n for any k [n]. Bytakingk = 0intheequationδ 2 (k) = δ 1 (k + n) + n, weobtain 0 = δ 2 (0) = δ 1 ( n) + n = 1 ( n) + n = n + x 1 + n = x 1. Since x (mod n), this is impossible. The following two lemmas are well known in number theory, so we state them without proof. (See p. 112 and p. 77 of the book [1].) Lemma 3.8 (Gauss lemma). Let p be an odd prime and let a be an integer such that p a. Consider a sequence of integers a, 2a, 3a,...,( p 1 )a. Replace each integer in the sequence by the one congruent to it modulo p 2 which lies between p 1 and p 1.Letν be the number of negative integers in the resulting sequence. Then 2 2 x 2 a (mod p) has a solution if and only if ν is even. Corollary 3.9. For any odd prime p, x 2 2 (mod p) has a solution if and only if p ±1 (mod 8). Lemma Let p be an odd prime and let a be an integer such that p a. Then, for any positive integer m, x 2 a (mod p) has a solution in Z p if and only if x 2 a (mod p m ) has a solution in Z p m. Moreover they have the same number of solutions, which is 0 or 2. Since Nn non = for n 0 (mod 4), weneedtoestimate Nn non only for n 2 (mod 4). Lemma For n = 2p a 1 1 pa 2 2 pa k non (a prime decomposition), the number N k n of solutions of x 2 = 2 in Z n is 2 k if p i ±1 (mod 8) for all i = 1, 2,...,k; 0otherwise. Proof. For x Z n, x 2 2 (mod n) if and only if x is even and x 2 2 (mod p a i i ) for all i = 1, 2,...,k. Hence, by Corollary 3.9 and Lemma 3.10, if p i ±3 (mod 8) for some i 1, the cardinality Nn non is zero. If p i ±1 (mod 8) for all i = 1, 2,...,k, then Nn non =2 k by Corollary 3.9, Lemma 3.10 and the Chinese remainder theorem. 4. Reduction In this section, we show that if there exists some M non n is an induced element (1) in M non d Nn non,thend = < n and there, which we call the reduction of. Ifsuch (1) is also contained in M non N non then one can choose the next reduction d d (2) of (1). By continuing such reduction, one can have a nonnegative integer j such that ( j) M non d j N non but its reduction d j ( j+1) is the identity or belongs to N non. In the next section, we prove that d j+1 Mn non = Nn non for any n by showing that such an element ( j) does not exist. Lemma 4.1. Suppose that Mn non exists. Then the order of the cyclic group equals the size of the orbit of 1 under,namely, { i (1) i [n]}. Furthermore, this is a proper divisor of n. Proof. For any k [n], leto(k) ={ i (k) i [n]} be the orbit of k under. Let O(1) =d. Thend is a proper divisor of n because 0 / O(1). Moreover,wehave d (1) = 1 and (LR L) 1 R d (LR L)(0) = 0. Hence the conjugate (LR L) 1 R d (LR L) of Rd belongs to the vertex stabilizer R, L 0 = R, t, which is isomorphic to a dihedral group D n of order 2n.

10 200 J.H. Kwak, Y.S. Kwon / Journal of Combinatorial Theory, Series B 101 (2011) Assume that (LR L) 1 R d (LR L) = Rm for some m [n]. Because R m and R d R, L, wehave Rm = Rd as subgroups of the cyclic group R.Sinced is a divisor of n, there exists l [ n d ] such that m = ld and (l, n d ) = 1. Suppose that d. Then there exists k [n] such that d (k) k. Letq be the largest such k. Then ld (q) q. On the other hand, ld (q) = R ld (q) = Rm (q) = (LR L) 1 R d (LR L)(q) = (LR L) 1 R (q + 1) = q, d are conjugate in a contradiction. Therefore = O(1) =d is a proper divisor of n. Next, suppose that (LR L) 1 R d (LR L) = t for some m [n]. Since the order of t is 2 and Rm Rm d < n, we find that n is even and d = n. If O(k) divides n for all k [n], then is a cyclic group of order n and the result follows. Hence we may assume that there is some i [n] such that O(i) does not divide n. By comparing two values (LR L) 1 R n (LR L)(k) and t(k), wehave Rm n (k + 1) 1 = m ( k), or equivalently, n (k + 1) = m ( k) + 1, for all k [n]. If n (k + 1) = k + 1 for some k [n], then m ( k) = k. Since n ( j) = j for any j O(1) {0}, there are at least n + 1elementsk [n] satisfying m ( k) = k. Notethatthereexist at most two k satisfying m ( k) = k in any orbit O under when O is even, and at most one when O is odd because 1 ( k) = (k) for all k [n]. So there exist at most 2 elements k O(1) satisfying m ( k) = k. If there exists an orbit O under which is not O(1) and whose size is greater than or equal to 3, then there exist at least two k in O such that m ( k) k. This implies that there exist at most n elements k [n] satisfying m ( k) = k, a contradiction. Hence apart from O(1), the size of every orbit under is 1 or 2. By our assumption that there is an orbit under whose size doesn t divide n, the value n should be odd. This means that there is only one element k in O(1) satisfying m ( k) = k, and implies that for all k [n] \O(1), m ( k) = k. Hence each orbit under containing neither 0 nor 1 is {i, i} for some i [n], and m is odd. Since 1 ( k) = (k) for all k [n] and there exists one k O(1) such that m ( k) = k, it follows that O(1) ={ k k O(1)} =O(1). Recall that for any k [n], n (k) = k if and only if k O(1) {0}. For any orbit {i, i} under, n (i + 1) = m ( i) + 1 = i + 1 and n ( i + 1) = m (i) + 1 = i + 1. This implies that i 1, i + 1, i 1, i + 1 O(1) because i ±1 and O(1) = O(1). Hence there exist no two consecutive elements i, i + 1 [n] satisfying O(i) = O(i + 1) =2. Note that O(0) =1 and O(1) = O( 1) = n. Sincethereare n 1elements j [n] such that O( j) =2, for any even 2k [n], wehaveo(2k) ={2k, 2k} or equivalently, (2k) = 2k. Moreover,O(1) is composed of all odd numbers. Hence for any even 2k [n], m (2k + 1) = m( ( 2k 1) ) = n ( 2k) 1 = 2k 1. This implies that m and n are relative prime. Moreover, m and n are relative prime because m is odd, so there exists s [n] such that sm 1 (mod n). Foranyeven2k [n], (2k + 1) = sm (2k + 1) = (s 1)m (2k 1) = =2k + 1 2s. Let x = 2s. Then, = (0)(2 2)(4 4) (1 1+ x 1 + 2x 1 + 3x ). By Lemma 3.3(3), there exist a( n) and b( n) satisfying (P 1 ) or (P 2 ) with. Suppose that a( n) and b( n) satisfy (P 1 ). Then, by Lemma 3.4, b( n) = n (1) = 1 and b( n) = ( n) (1) = ( n+x) (1) 1, which is a contradiction. Hence, we can assume that a( n) and b( n) satisfy (P 2 ). By Lemma 3.4, b( n) = n (1) = 1 and b( n) = ( n) (1) = ( n+x) (1) = 1 ( n + x)x = 1 x 2 because x is even. This implies that x 2 2 (mod n). So, Nn non, which contradicts the assumption.

11 J.H. Kwak, Y.S. Kwon / Journal of Combinatorial Theory, Series B 101 (2011) Remark. As one can see in the proof of Lemma 4.1, any element in Mn non which does not satisfy the condition in Lemma 4.1 is for some even n and some x such that x 2 2 (mod n). This is a reason why we define in Section 3. Proposition 4.2. (See [11].) If is the identity permutation of [n] then R, L = 2n2.Furthermore,ifwe define :[n] [n] by (k) = k(1 + rd) for all k [n], wheren 3, d is a divisor of n and r is a positive integersuchthattheorderof1 + rd in the multiplicative group Z n of units is d, then R, L = 2n2. Lemma 4.3. If n 3, 2 for every M non n. Proof. Suppose that there exists Mn non satisfying = 2. Then n is even. By Lemma 3.3(3), there exist a(1), b(1) [n] satisfying (P 1 ) or (P 2 ) with. Inbothcases,a(1) = (1) and b(1) = (1) (1) by Lemma 3.4. Suppose a(1) = (1) is even and let (1) = 2r. Thenb(1) = 1 and (k + 1) = b(1) (k) + a(1) = (k) + 2r for all k [n] or (k + 1) = b(1) ( k) + a(1) = ( k) + 2r = (k) + 2r for all k [n]. In both cases, one can show inductively that (k) is even for all k [n]. This is impossible, since S [n]. Therefore we can assume that a(1) = (1) is odd. Let (1) = 1 + 2r. By Lemma 3.4, b(1) = (1) (1) = (1) = 1 + 2r. Suppose that (P 1 ) holds. Then (k + 1) = b(1) (k) + a(1) = (k) r = (k 1) + 2(1 + 2r) = =(k + 1)(1 + 2r). Moreover 2 is the smallest positive integer d satisfying d (1) = (1 + 2r) d = 1. By Proposition 4.2, R, L = 2n2, and so / Mn non, a contradiction. Now suppose that (P 2 ) holds. Then b(1) = (1) and hence b(1) = (1) = (1). Since (1) = 1 ( 1) = ( 1), wehave(1) = ( 1). This implies that n is 2, a contradiction. From Proposition 4.2 and Lemma 4.3, one can see that for any n 3 and for every Mn non, is neither the identity nor an involution. Lemma 4.4. Suppose that there exists a Mn non for some k 1, k 2 [n],then(k 1 ) (k 2 )(mod d). N non n with n 3, andlet = d. If k 1 k 2 (mod d) Proof. By Lemma 4.1, d is a proper divisor of n. By Lemma 3.3(3), thereexista(d) and b(d) satisfying (P 1 ) or (P 2 ). Assume that (P 1 ) holds. By Lemma 3.4, b(d) = d (1) = 1, and so k + 1 = d (k) + 1 = a(d) (k + b(d)) = a(d) (k + 1). This implies that a(d) is a multiple of d, say a(d) = rd. So the first equation in (P 1 ) is (k +d) = b(d) (k)+a(d) = (k)+rd. Henceifk 1 k 2 (mod d) for some k 1, k 2 [n], then (k 1 ) (k 2 )(mod d). Next, suppose that (P 2 ) holds. By Lemma 3.4, b(d) = d (1) = 1, and so k + 1 = d (k) + 1 = a(d) ( k + b(d)) = a(d) ( k 1). By taking k = 2 and k = (1) 1 in the equation a(d) ( k 1) = k + 1, we obtain a(d) (1) = 1 and a(d)+1 (1) = (1). Since (1) = a(d)+1 (1) = ( a(d) (1)) = ( 1) = 1 (1), wehave 1 (1) = (1). By Lemma 4.1, 1 =, orequivalently,d = 1 or 2. This is impossible by Proposition 4.2 and Lemma 4.3. Suppose that Mn non with = d. By Lemma 4.4, the function (1) :[d] [d] defined by (1) (k) (k) (mod d) for any k [d] is well defined. Furthermore (1) is a bijection, namely, a permutation of [d]. We call the permutation (1) the (mod d)-reduction of. Infact, (1) belongs to M non as the following lemma shows in a general setting. d

12 202 J.H. Kwak, Y.S. Kwon / Journal of Combinatorial Theory, Series B 101 (2011) Lemma 4.5. Suppose that Mn non with = d 3. Let m be a divisor of n such that (1) m is a multiple of d, and (2) if k 1 k 2 (mod m) for some k 1, k 2 [n],then(k 1 ) (k 2 )(mod m). Define :[m] [m] by (k) (k) (mod m) for any k [m]. Then is a well-defined bijection and it belongs to M non m. Proof. By the assumption that (k 1 ) (k 2 )(mod m) for any k 1, k 2 [n] satisfying k 1 k 2 (mod m), :[m] [m] is well defined. Since is a bijection, so is.bythefact(0) = 0, we have (0) = 0. It is easily checked that ( ) 1 (m k) = m (k) for any k [m]. Now, we show that Mm non using Lemma 3.3(3). Foranyk [n], letk denote the remainder of k on division by m. By Lemma 3.3(3), for any i [n] there exist a(i) and b(i) satisfying (P 1 ) or (P 2 ) with. Ifwedefinea(i ) = a(i) and b(i ) = b(i) then a(i ) and b(i ) also satisfy (P 1 ) or (P 2 ) depending on whether a(i) and b(i) satisfy (P 1 ) or (P 2 ).Since Mn non, there exists at least one j [n] such that (k + j) = b( j) ( k) + a( j) and j (k) + 1 = a( j) ( k + b( j)) for all k [n], by Lemma 3.3(3). This implies that (k + j ) b( j) (( k) ) + a( j) (mod m) and j (k ) + 1 a( j) (( k) + b( j) )(mod m) for all k [m]. So by Lemma 3.3, Mm non. Corollary 4.6. Suppose that Mn non with = d 3.Then (1) belongs to M non. d Proof. By Lemmas 4.4 and 4.5, the (mod d)-reduction (1) of belongs to M non d. 5. Proof of Theorem 1.1 To prove Theorem 1.1, we need to show that for any integer n 0, 1or3(mod 4), no nonorientable regular embedding of K n,n exists, while for n 2 (mod 4), Mn non = Nn non. For a nonnegative integer k, we define (0) = S [n] and (k+1) = ( (k) ) (1) by taking inductive reduction. Lemma 5.1. Suppose that Mn non with n 3.Then (1) (1) is not the identity, and (2) is even. Proof. Suppose that there exists a Mn non.let = d. By Proposition 4.2 and Lemma 4.3, we find that d 3. This implies that (1) belongs to M non by Corollary 4.6. Hence d (1) is not the identity by Proposition 4.2. Suppose that = d is odd. By Lemma 4.1, d is less than n. SinceNn non = with any odd n, (1) is an element in M non N non by Corollary 4.6, and the order of d d (1) is also odd. By continuing the same process, we obtain j 1 and d j 3 such that ( j) M non d j N non, and d j ( j+1) is the identity permutation on [d j+1 ], where d j+1 = ( j) and d j = ( j 1) 3. But such ( j) cannot exist by (1). Corollary 5.2. If n is odd, then M non n =, or equivalently, there is no nonorientable regular embedding of K n,n. Proof. Suppose that M non n exists. Since Nn non = for odd n, belongs to Mn non N non n. By Lemma 4.1, the order is a divisor of n. Hence is odd, which is a contradiction by Lemma 5.1. Lemma 5.3. There is no Mn non with = d 3 such that (1) N non. d

13 J.H. Kwak, Y.S. Kwon / Journal of Combinatorial Theory, Series B 101 (2011) Proof. Suppose that there exists an element Mn non Note that d 2 (mod 4). We consider two cases separately. Case 1. n 2 (mod 4). of order d 3 such that (1) N non d. Here n is an odd integer. Let O be the orbit of n under. Then the size O is a divisor of d =. Furthermore, O is a multiple of d/2 because all odd numbers in [d] are in the same orbit under (1), whose size is d/2. So O is d/2 ord. Since n = n and 1 ( k) = (k) for any k [n], one can see that O = O and the size O is odd, which implies O =d/2. Since all odd numbers in [d] are in the same orbit under (1), there exists a number 1+ jd [n] such that 1 + jd O. This implies that d/2 1+ jd (1 + jd) = 1 + jd and hence (LR L) 1 R d/2 1+ jd (LR L)(0) = jd So (LR L) 1 R d/2 1+ jd (LR L) belongs to the vertex stabilizer R, L 0 = R, t, which is isomorphic to the dihedral group D n of order 2n. Since the order of (LR L) 1+ jd 1 R d/2 1+ jd (LR L) is not 2, 1+ jd (LR L) 1 R d/2 1+ jd (LR L) = R for some m [n]. and Rd/2 are conjugate in R m BecauseRm, L, they have the same order, and consequently, R = Rd/2 as subgroups of the cyclic group R m.since d/2 is a divisor of n, thereexistsl [n/ d 2 ] such that m = ld/2 and (l, n/ d ) = 1. By considering two 2 1+ jd images of 1 + jd under the permutations (LR L) 1 R d/2 1+ jd (LR L) and R ld/2,wehave d/2 (2 + 2 jd) 1 jd = ( 1+ jd ) 1 d/2( 1+ jd ) LR L R LR L (1 + jd) = R ld/2 (1 + jd) = 1 + jd. This implies that d/2 (2 + 2 jd) = jd. Sinced/2 is odd and for any even k [d] with k 0, the orbit of k under (1) is {k, d k}, the even number jd should be a multiple of d. This means that d is 1 or 2, a contradiction. Therefore for any n 2 (mod 4), no Mn non exists with = d 3 such that (1) N non. d Case 2. n 0 (mod 4). Let n = 2sd for some even integer 2s. Here n = sd is even. By Lemma 3.3(3), there exist a( n) and b( n) satisfying (P 1 ) or (P 2 ).Inbothcases,a( n) = ( n) by Lemma 3.4. Since n is a multiple of d, a( n) is also a multiple of d by Lemma 4.4. Suppose that (P 2 ) holds. Then k + 1 = n (k) + 1 = a( n)( k + b( n) ) = k + b( n) for all k [n]. This means that b( n) = 2k + 1forallk [n]. Sinceb( n) is a constant, n 2, a contradiction. So (P 1 ) holds, that is, (k + n) = b( n) (k) + a( n) and k + 1 = n (k) + 1 = a( n) (k + b( n)) = k + b( n) for all k [n]. This means that b( n) = 1, and hence (k + n) = b( n) (k) + a( n) = (k) + ( n). Bytaking k = n in the above equation, we have 2( n) = 0. Since ( n) 0, ( n) = n and (k + n) = (k) + ( n) = (k) + n. This implies that if k 1 k 2 (mod n), then(k 1 ) (k 2 )(mod n). Let :[ n] [ n] be defined by (k) (k) (mod n) for any k [ n]. By Lemma 4.5, is well defined and belongs to M non because n n is a multiple of d. Note that the size of the orbit of 1 under is d/2 ord. Subcase 2.1. The size of the orbit of 1 under is d/2. Let d =. Then d is d/2 or d. Since d is a divisor of n and the orbit of 2 under (1) is {2, d 2}, the size of the orbit of 2 under is even. Hence d is even and consequently equals d. Since the order of is not equal to the size of the orbit of 1 under, N non n by Lemma 4.1. Hence, = for some x [ n] satisfying x 2 2 (mod n). Moreover,d = d = n. This implies that

14 204 J.H. Kwak, Y.S. Kwon / Journal of Combinatorial Theory, Series B 101 (2011) = (1) and all odd numbers in [n] belong to the same orbit under. Furthermore,foranyeven number 2k [n]\{0, n}, the size of the orbit of 2k under is 2 or 4. Since this is a divisor of d and d 2 (mod 4), it is 2. Note that the orbit of 2k under is {2k, 2k} or {2k, n 2k}. First, we want to show that (2k) = 2k for all even 2k [n]. By Lemma 3.3(3), thereexista(2) and b(2) satisfying (P 1 ) or (P 2 ).Inbothcases,a(2) = (2) and b(2) = a(2) (1) = (2) (1) = 2 (1) by Lemma 3.4. Suppose that (P 2 ) holds. By Lemma 3.4, b(2) = 2 (1). Henceb(2) = 2 (1) = 2 (1), which implies 2 2 (1) = 0. Since 2 (1) is not 0, 2 (1) = n. This contradicts the fact that the orbit of 1 under is composed of all odd numbers in [n]. So(P 1 ) holds. By Lemma 3.4, b(2) = 2 (1) 1 + 2x (mod n), and so the first equation in (P 1 ) can be written as (k + 2) = b(2) (k)+a(2) = 1+2x (k)+ (2). Suppose that (2) = n 2. Then (k + 2) = 1+2x (k) + n 2. Taking k = 2intheequation(k + 2) = 1+2x (k) + n 2, we have (4) = 1+2x (2) + n 2 = (2) + n 2 = 4. Taking k = 4, we have (6) = (4) + n 2 = n 6. By continuing the same process, one can see that (4k) = 4k and (4k + 2) = n 4k 2. Since n 2 (mod 4), wehave( n) = n n = 0, which is a contradiction. Hence (2) = 2, so it follows that (k + 2) = 1+2x (k) 2. Taking k = 2intheequation(k + 2) = 1+2x (k) 2, we have (4) = 1+2x (2) 2 = (2) 2 = 4. Taking k = 4, we have (6) = (4) 2 = 6. By continuing the same process, one can see that (2k) = 2k for all even 2k [n]. Now we aim to apply Lemma 3.3(3) once more to show that Subcase 2.1 cannot occur. There exist a(1) and b(1) satisfying (P 1 ) or (P 2 ).Inbothcases,a(1) = (1) 1 + x (mod n). For our convenience, let (1) = 1 + x 1. Suppose that (P 1 ) holds. Then b(1) = (1) = 1 + x 1, and so (k + 1) = b(1) (k) + a(1) = b(1) (k) + (1) = 1+x 1 (k) x 1. By taking k = 2, we have (3) = 1+x 1(2) x 1 = x 1 x 1 (mod n). Since(3) 3 + x (mod n), wehave4 0 (mod n). By the assumption that n 2 (mod 4), wehave n = d = 2, which contradicts the assumption that d 3. So (P 2 ) holds. Hence b(1) = (1) = 1 x 1 and it follows that (k) + 1 = a(1)( k + b(1) ) = 1+x 1 ( k 1 x 1 ) for all k [n]. Bytakingodd2k + 1 [n], wehave(2k + 1) + 1 = 1+x 1( 2k 2 x 1 ) = 2k x 1. Hence (2k + 1) = 2k x 1.Bytakingk = 2intheequation(k) + 1 = 1+x 1( k 1 x 1 ),weobtain 1 = (2) + 1 = 1+x 1 ( 3 x 1 ) = 3 x 1 + (1 + x 1 )x 1 = 3 + x 2 1. So x 2 1 = 2 (mod n). Thisisimpossiblebecausen 0 (mod 4). Hence Subcase 2.1 cannot occur. Subcase 2.2. Thesizeoftheorbitof1under is d. Since the order of divides that of, the order of is d, which equals the size of the orbit of 1 under. This implies that M non N non. Moreover, since d divides n, it follows that n n (1) = (1) N non. Since Subcase 2.1 cannot occur, by repeating the same process continually, we d obtain n 1 2 (mod 4) and Mn non 1 Nn non 1 with = d = such that (1) = (1) N non.but d this brings us back to Case 1. Now we can prove Theorem 1.1. We know that there exists only one nonorientable regular embedding of K 2,2 intotheprojectiveplane,soletn 3.

15 J.H. Kwak, Y.S. Kwon / Journal of Combinatorial Theory, Series B 101 (2011) Suppose that Mn non Nn non and let Mn non and = d. Note that d < n. By Lemma 5.1 and Lemma 4.3, (1) is not the identity, and d 3 is even. By Lemmas 4.6 and 5.3, (1) M non N non. d d By continuing the same process, we obtain j 1 and d j 3 such that ( j) M non d j N non and d j ( j+1) is the identity permutation on [d j+1 ], where d j+1 = ( j) and d j = ( j 1) 3. But this is impossible by Lemma 5.1. Hence Mn non = Nn non for every n 3. This means there is no nonorientable regular embedding of K n,n for any n congruent to 0, 1 or 3 mod 4. Finally, for n = 2p a 1 1 pa 2 2 pa k (the k prime decomposition of n), up to isomorphism the number of nonorientable regular embeddings of K n,n is 2 k if every p i is congruent to ±1 mod 8; 0 otherwise by Lemma Acknowledgments We would like to thank the referees of this paper for valuable comments and suggestions. References [1] W.W. Adams, L.J. Goldstein, Introduction to Number Theory, Prentice Hall, [2] N.L. Biggs, Classification of complete maps on orientable surfaces, Rend. Mat. 4 (6) (1971) [3] D.A. Catalano, M.D.E. Conder, S.F. Du, Y.S. Kwon, R. Nedela, S. Wilson, Classification of regular embeddings of n-dimensional cubes, J. Algebraic Combin. 33 (2011) [4] S.F. Du, G.A. Jones, J.H. Kwak, R. Nedela, M. Škoviera, Regular embeddings of K n,n where n is a power of 2. I: Metacyclic case, European J. Combin. 28 (6) (2007) [5] S.F. Du, G.A. Jones, J.H. Kwak, R. Nedela, M. Škoviera, Regular embeddings of K n,n where n is a power of 2. II: Nonmetacyclic case, European J. Combin. 31 (7) (2010) [6] A. Gardiner, R. Nedela, J. Sirá n, M. Skoviera, Characterization of graphs which underlie regular maps on closed surfaces, J. London Math. Soc. 59 (1999) [7] L.D. James, G.A. Jones, Regular orientable imbeddings of complete graphs, J. Combin. Theory Ser. B 39 (1985) [8] G.A. Jones, Regular embeddings of complete bipartite graphs: classification and enumeration, Proc. London Math. Soc. 101 (2) (2010) [9] G.A. Jones, R. Nedela, M. Škoviera, Regular embeddings of K n,n where n is an odd prime power, European J. Combin. 28 (6) (2007) [10] G.A. Jones, R. Nedela, M. Škoviera, Complete bipartite graphs with a unique regular embedding, J. Combin. Theory Ser. B 98 (2) (2008) [11] J.H. Kwak, Y.S. Kwon, Regular orientable embeddings of complete bipartite graphs, J. Graph Theory 50 (2) (2005) [12] J.H. Kwak, Y.S. Kwon, Classification of reflexible regular embeddings and self-petrie dual regular embeddings of the complete bipartite graphs, Discrete Math. 308 (11) (2008) [13] Y.S. Kwon, R. Nedela, Non-existence of nonorientable regular embeddings of n-dimensional cubes, Discrete Math. 307 (2007) [14] R. Nedela, M. Škoviera, A. Zlatoš, Regular embeddings of complete bipartite graphs, Discrete Math. 258 (2002) [15] S.E. Wilson, Cantankerous maps and rotary embeddings of K n, J. Combin. Theory Ser. B 47 (1989)