ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 16 (2019) 563-583 https://doi.org/10.26493/1855-3974.793.2a9 (Also available at http://amc-journal.eu) Classification of some reflexible edge-transitive embeddings of complete bipartite graphs Jin Ho Kwak Mathematics, Beijing Jiaotong University, Beijing, 100044, P.R. China Young Soo Kwon Mathematics, Yeungnam University, Kyeongsan, 712-749 Republic of Korea Received 8 January 2015, accepted 28 February 2019, published online 21 April 2019 In this paper, we classify some reflexible edge-transitive orientable embeddings of complete bipartite graphs. As a by-product, we classify groups r such that (i) r = XY for some cyclic groups X = (x) and Y = (y) with X n Y = {1r} and (ii) there exists an automorphism of r which sends x and y to x-1 and y-1, respectively. Keywords: Complete bipartite graphs, reflexible edge-transitive embedding. Math. Subj. Class.: 05C10, 05C30 1 Preliminaries A map is a 2-cell embedding of a graph G in a compact, connected surface. A map is called orientable or nonorientable according to whether the supporting surface is orientable or nonorientable. In this paper, we only consider orientable maps. For a simple connected graph G, an arc of G is an ordered pair (u,v) of adjacent vertices in G. The set of all arcs in G is denoted by D(G). An orientable map M can be described by a pair (G; R), where G is the underlying graph of M and R is a permutation of the arc set D(G) whose orbits coincide with the sets of arcs emanating from the same vertex. The permutation R is called the rotation of the map M. For given two maps M1 = (G1; R1) and M2 = (G2; R2), a map isomorphism 0: M1 ^ M2 is a graph isomorphism 0: G1 ^ G2 such that ^R1 (u, v) = R20(u, v) for any arc (u, v) in G1. Furthermore if M1 = M2 = M, 0 is called a map automorphism of M. The set of all map automorphisms of M denoted by Aut( M) is a group under the composition operation, and it is called the automorphism group of M. For a map M = (G; R), E-mail addresses: jinkwak@postech.ac.kr (Jin Ho Kwak), ysookwon@ynu.ac.kr (Young Soo Kwon) Abstract ©® This work is licensed under https://creativecommons.Org/licenses/by/4.0/ 564 Ars Math. Contemp. 16 (2019) 445-463 the group Aut(M) acts semi-regularly on the arc set D(G), so | Aut(M)| < 2|E(G)|. If this bound is attained, then Aut(M) acts regularly on the arc set, and the map is called a regular map or a regular embedding. The map M is said to be vertex-transitive or edge-transitive if Aut(M) acts transitively on V(G) or E(G), respectively. For an orientable embedding M of a bipartite graph G, if the set of partite set preserving map automorphisms acts transitively on E(G) then we call M an edge-transitive map or an edge-transitive embedding satisfying the Property (P) in this paper. For a map M = (G; R), if M and M-1 = (G; R-1) are isomorphic, M is called reflexible. Classifying highly symmetric embeddings of graphs in a given class is an interesting problem in topological graph theory. In recent years, there has been particular interest in the regular embeddings of complete bipartite graphs Kn,n by several authors [1, 2, 4, 5, 6, 7, 8, 10]. The reflexible regular embeddings and self-Petrie dual regular embeddings of Kn n have been classified by the authors [7]. Recently, G. Jones has completed the classification of regular embeddings of Kn n [5] and the authors have classified nonorientable regular embeddings of Kn n [8]. In [3], Graver and Watkins classified edge-transitive maps on closed surfaces into fourteen types. In this paper, we classify reflexible edge-transitive embeddings of Km,n satisfying the Property (P) which correspond to types 1 or 2 among 14 types. The following theorem is the main result in this paper. Theorem 1.1. For any integers m = 2 apa 1 • • • pa'p-Y • • • pT/ and n = 2bp11 • • • pbel q+11 • • • q+g9 (prime decompositions) with gcd(m, n) = 2cp11 • • • pf and a < b, the number (up to isomorphism) of reflexible edge-transitive embeddings of Km,n satisfying the Property (P) is 1 if both m and n are odd; 2f (1 + pi1) • • • (1 + p^) if exactly one of m and n is even, namely, only n is even; A(a, b)2f+g+£(1 + pi1) • • • (1 + pCe) if both m and n are even, where 1 if (a,b) = (1,1), 2 if (a,b) = (1, 2), 4 if (a, b) = (2, 2) or (1, k) with k > 3, 10 if (a, b) = (2, 3), 12 if (a, b) = (2, k) with k > 4, 28 if (a, b) = (3, 3), 40 if (a, b) = (3, 4), 36 if (a, b) = (3, k) with k > 5, 20(1 + 2 °-2) if a = b > 4, 20 + 18 • 2°-2 if b - 1 = a > 4, 20 + 16 • 2 °-2 if b - 2 > a > 4. Our paper is organized as follows. In the next section, we consider some relations between edge-transitive embeddings of Km,n satisfying the Property (P) and products of two cyclic groups. In Section 3, we classify reflexible edge-transitive embeddings of Km,n satisfying the Property (P) when at least one of m and n is odd. In Section 4, for even integers m and n, the classification of reflexible edge-transitive embeddings of Km,n satisfying the Property (P) is given. In the final section, we classify groups r satisfying the conditions: J. H. Kwak and Y. S. Kwon: Classification of some reflexible edge-transitive embeddings ... 565 (i) r = XY for some cyclic groups X = (x) and Y = (y) with X n Y = {1r} and (ii) there exists an automorphism of r which sends x and y to x-1 and y-1. 2 (m, n)-bicyclic triples in Aut(Km,n) Regular embeddings of the complete bipartite graphs Kn,n are related to groups r with two generators satisfying some conditions [4]. Using this relation, G. Jones classify regular embeddings of Kn n [5]. Similarly, we aim to find a relation between edge-transitive embeddings of Km,n satisfying the Property (P) and groups with two generators satisfying some conditions in this section. In [4], G. Jones et al. showed that any finite group r is isomorphic to Aut(M) for some regular embedding M of Kn,n if and only if r has cyclic subgroups X = (x) and Y = (y) of order n such that: (i) r = xy (ii) X n Y = {1r} and (iii) there is an automorphism a of r transposing x and y. They call the triple (r, x, y) satisfying these conditions the n-isobicyclic triple. In this relation, x and y correspond to rotations of M around two fixed adjacent vertices u and v, respectively. The automorphism a corresponds to the half-turn reversing the edge uv. For two n-isobicyclic triples (r1, x1, y1) and (r2, x2, y2), two corresponding regular embeddings M1 and M2 are isomorphic if and only if there exists a group isomorphism from r1 to r2 given by x1 ^ x2 and y1 ^ y2. Using this, one can show that the regular embedding M induced by n-isobicyclic triple (r, x, y) is reflexible if and only if there exists an automorphism p of r which sends x and y to x-1 and y-1, respectively. (For more information, the reader is referred to [4].) Note that one can define an embedding of Kn n by using the first and second conditions of n-isobicyclic triple, and the induced map is edge-transitive map satisfying the Property (P) even though the third condition of n-isobicyclic triple is not satisfied. Conversely, any edge transitive embedding of Kn n satisfying the Property (P) is isomorphic to some induced map by such a triple (r, x, y). One can show that for different positive integers m and n, an edge-transitive embedding of Km,n satisfying the Property (P) can also be represented by a similar triple. For a group r containing cyclic subgroups X = (x) of order n and Y = (y) of order m, the triple (r, x, y) is called (m, n)-bicyclic if it satisfies: (i) r = XY and (ii) X n Y = {1r}. For any (m, n)-bicyclic triple (r, x, y), one can define an embedding of Km,n by a similar way to define an embedding of Kn n using n-isobicyclic triple. We denote this embedding by M(r,x,y). One can see that M(r, x, y) is an edge-transitive embedding of Km,n satisfying the Property (P). Furthermore the following result holds. Lemma 2.1 ([9]). Let m, n be two positive integers (not necessarily distinct). (1) Any edge-transitive embedding M of Km,n satisfying the Property (P) is isomorphic to M(r, x, y) for some (m, n)-bicyclic triple (r, x, y). 566 Ars Math. Contemp. 16 (2019) 445-463 (2) For two (m, n)-bicyclic triples (Ti, x1, y1) and (r2, x2, y2), two edge-transitive em-beddings M(Ti, x1, y1) and M(r2, x2, y2) are isomorphic if and only if there exists a group isomorphism from r1 to r2 given by x1 ^ x2 and y1 ^ y2. For any (m, n)-bicyclic triple (r, x, y), there exists a subgroup H of the automorphism group Aut(Km,n) such that: (i) H is isomorphic to r and (ii) x and y in r correspond to elements in H which cyclically permute vertices in the partite sets of size n and m, respectively. Hence it suffices to deal with such (m, n)-bicyclic triples in Aut(Km,n) to classify edge-transitive embeddings of Km,n satisfying the Property (P). For any positive integer m, denote the set {0,1,..., m - 1} by [m]. Let V = {0,1, ..., (m - 1)} U {0', 1', .. ., (n - 1)'} = [m] U [n]' be the vertex set of Km,n as partite sets, and let D = {(i, j'), (j',i) : 0 < i < m - 1 and 0 < j < n - 1} be the arc set, where (i, j') is the arc emanating from i to j' and (j', i) denotes its inverse. We denote the symmetric group on [m] and [n]' by S and S', respectively. Let S0 and SO be their stabilizers of 0 and 0', respectively. Note that Aut(Km,n) is isomorphic to S x S' when m = n; S I Z2 when m = n. We identify integers 0,1, 2,... with their residue classes modulo m or n according to the context. Let (r, x, y) be an (m, n)-bicyclic triple such that r is a subgroup of Aut(Km n). Now there exists an automorphism ^ G Aut(Km,n) such that x0 = = a(0' 1' ••• (n - 1)') and y0 = = £(0 1 ••• m - 1), where a G S0 and £ G SO. For any a G S0 and £ G SO, let xa = a(0' 1' ••• (n - 1)') and yg = £(0 1 ••• m - 1). From now on, we only consider triples ((xa, yg}, xa, yg) as candidates of (m, n)-bicyclic triples. Lemma 2.2 ([9]). For any a G S0 and £ G SO, 1. the group (xa, yg} acts transitively on the edge set of Km,n and 2. the triple ((xa, yg},xa,yg) is (m, n)-bicyclic if and only if | (xa, yg} | = mn. By Lemma 2.2, we need to characterize a G S0 and £ G SO satisfying | (xa, yg} | = mn to classify edge-transitive embeddings of Km,n satisfying the Property (P). To do this, we denote ETm,„ = {(a, £) : a G so, £ G SO and | (xa, yg} | = mn}. Note that for any (a, £) G ETm,n, ((xa, yg}, xa, yg) is an (m, n)-bicyclic triple and hence M((xa, yg}, xa, yg) is an edge-transitive embedding of Km,n satisfying the Property (P). Conversely for any edge-transitive embedding M of Km,n satisfying the Property (P), there exists (a, £) G ETm,n such that M is isomorphic to M((xa, yg}, xa, yg). J. H. Kwak and Y. S. Kwon: Classification of some reflexible edge-transitive embeddings ... 567 Remark 2.3. (1) For any (a,fi) e ETm,„, (xa,Vl3) = {xdVj I i e [nL j e [m]} = {y|xd | i e [n], j e [m]}. Hence in many cases, if a satisfies some properties then fi also satisfies the same properties and vice versa. (2) Note that for different positive integers m and n and for an orientable embedding M of Km,n, any automorphism of M is partite set preserving. Let m = n be odd and let M be an orientable edge-transitive embedding of Kn,n. If a subgroup r of Aut(M) acts regularly on the edge set then |r| = m2 is odd and hence there exists no partite set reversing element in r. Hence for odd n, every edge-transitive embedding of Kn,n is an edge-transitive embedding of Kn,n satisfying the Property (P). On the other hand, for even n, we do not know whether the above statement is true or not. The next lemma shows that for different (ai,fii), (a2,fi2) e ETm,n, two induced edge-transitive embeddings are non-isomorphic. Lemma 2.4 ([9]). For any (a1,fi1), (a2, fi2) e ETm,n, the induced edge-transitive em-beddings M((xdi ,vp1) , xdi ,vp1) and M((xd2 ,vi2) ,xa2 ,vi2) are isomorphic if and only if (ai,fii) = (a2,fi2). By Lemma 2.4, distinct pairs in ETm,n give non-isomorphic edge-transitive embeddings of Km,n and the number of edge-transitive embeddings of Km,n satisfying the Property (P) equals to the cardinality | ETm,n |. But for distinct pairs (a1,fi1), (a2,fi2) e ETm,n, two groups (xdi ) and (xd2 ) may possibly be isomorphic. We do not know a necessary and sufficient condition for (xdi, vi) — (xd2, vi2). So we propose the following problem. Problem 2.5. For any positive integers m and n and for any (a1,fi1), (a2, fi2) e ETm,n, find a necessary and sufficient condition for (xdi, vi) — (xd2, vi2). From now on, we aim to characterize the set ETm,n. Note that for any (a, fi) e ETm,n, the stabilizers (xd, vi)0 and (xd, vi)0/ are cyclic groups (xd) of order n and (vi) of order m, respectively. Lemma 2.6. For any (a, fi) e ETm,n, (a) and (fi) are cyclic groups of order |{a®(1) : i e [n]}| and |{fi®(1') : i e [m]}|, the lengths of the orbit containing 1 and 1', respectively. Furthermore they are divisors of n and m, respectively. Proof. Let d1 = |{a®(1) : i e [n]}| and d2 = |{fi®(1') : i e [m]}|. Now d1 and d2 are divisors of the orders |(xd)| = n and |(vi)| = m, respectively. Note that adi (1) = 1 and v-1xd Vl(0) = 0, which implies that, as a conjugate of xdi, v-1xdi Vi belongs to the vertex stabilizer (xd, Vi)0 = (xd). Since d1 is a divisor of n, v-1xdi Vi = xddi for some r e [n] such that gcd(r, n) = 1, where gcd(r, dn) is the greatest common divisor of r and n. Now, 568 Ars Math. Contemp. 16 (2019) 445-463 suppose to the contrary that |(a)| = d1. Then there exists k e [m] such that adl (k) = k. Let q be the largest element in [m] such that adl (q) = q. On the other hand, (q) = xradl (q) = y-1x% yp(q) = y-1x% (q + 1) = y-1(q + 1) = q, a contradictory to ardl (q) = q. Therefore |(a)| = d1. Similarly, one can show that |(ft)| = d2. □ For any (a, ft) e ETm,n, it follows from Lemma 2.6 that the length of each cycle in a (ft, resp.) is a divisor of the length d1 (d2, resp.) of the cycle containing 1 (1', resp.). From now on we denote i', [n]' and ft(i') simply by i, [n] and ft(i) for any i' e [n]', respectively. The following lemma is related to a characterization of the set ETm 3, then both m and d2 are even; (3) if m(n, resp.) is even then a (P, resp.) is parity preserving. Furthermore there exists s, t e [m] such that a(2k) = 2kt, a(2k + 1) = 2kt + 2s + 1 and 2t2 = 2; (4) if both di and d2 are at least 3 then they are divisors of gcd(m, n). Proof. (1): Let di = 1 and d2 > 2. By Lemma 2.10, a(1) = —1 (mod d2). Since a is the identity, 1 = — 1 (mod d2). By the assumption d2 > 2, d2 = 2. By Lemma 2.6, d2 is a divisor of m, and hence m is even. (2): Let di > 3. By lemma 2.10, P(k) = —k (mod di), which implies that the order d2 of P is even. Since d2 is a divisor of m, m is also even. (3): Let m be even. If di = 1 then a is the identity and hence a is parity preserving. If di =2 then a-i = a and a(k) = a(k — 1 + 1) = a(k — 1) + a(1) = a(k — 2) + 2a(1) = • • • = ka(1) for all k e [m]. Since a2(1) = a(a(1)) = (a(1))2 = 1 and m is even, a(1) should be odd. Hence a is parity preserving. Assume that di > 3. Then, d2 is even by (2). Since a(k) = —k (mod d2), a is parity preserving. 570 Ars Math. Contemp. 16 (2019) 445-463 For any 2k G [m], a(2k) = a(2(k - 1)) + a(2) = a(2(k - 2)) + 2a(2) = • • • = ka(2) and a(2k + 1) = a(2(k - 1) + 1) + a(2) = • • • = a(1) + ka(2). Let a(1) = 2s + 1 and a(2) = 2t. Now a(2k) = ka(2) = 2kt and a(2k + 1) = ka(2) + a(1) = 2kt + 2s + 1. Note that for any 2k G [m], a(1) + a(2k) = a(2k + 1) = a-1(2k) + a(1). Hence a-1(2k) = a(2k), namely, a2(2k) = 2k. So we have a2(2) = a(2t) = 2t2 = 2. (4): Let d1; d2 > 3. Now all of d1; d2, m and n are even by (2). Hence there exist s, t G [m] such that a(2k) = 2kt, a(2k +1) = 2kt + 2s +1 and 2t2 =2 by (3). Since d1 is even and a2i(1) = a2i-1(2s + 1) = a2i-2(2st + 2s + 1) = • • • = 2is(t + 1) + 1, d1 is the smallest positive integer such that d1s(t+1) = 0 (mod m) byLemma2.6. Hence d1 is a divisor of m and consequently a divisor of gcd(m, n). Similarly d2 is a divisor of gcd(m, n). □ 3 At least one of m and n is odd In this section, we classify reflexible edge-transitive embeddings of Km,n satisfying the Property (P) when at least one of m and n is odd. Note that when at least one of m and n is odd, any orientable edge-transitive embedding of Km,n is an edge-transitive embedding satisfying the Property (P). In [9], the second author counted | RETm,n | when both m and n are odd as follows. Theorem 3.1 ([9]). If both m and n are odd then | RETm,n | = 1, namely, there exists only one reflexible edge-transitive embedding of Km,n satisfying the Property (P) up to isomorphism. In the next theorem, we count | RETm,n | when exactly one of m and n is odd. By symmetry, we assume that m is odd. Theorem 3.2. Let m = p^1 • • • pgepZ+Y • • • p^+f/ (prime factorization) be odd and n = 2bp11 • • • p^ q^1 • • • qC++9 (prime factorization) be even. Let gcd(m, n) = p1 • • • pC with c > 1 for any i = 1,..., L. Now | RETm,„ | = 2f (1+ pi1) ••• (1+ pc/), namely, there exist 2f (1 + p1) • • • (1 + pc) reflexible edge-transitive embeddings of Km,n satisfying the Property (P) up to isomorphism. Proof. Let (a, ft) G RETm,n and let d1 = |(a)| and d2 = |(ft}|. Suppose that d1 > 3. Then both d2 and m are even by Lemma 2.11(2), which is a contradiction. Hence d1 = 1 or 2. Furthermore for any k G [m], a(k) = a-1(k - 1) + a(1) = a(k - 1) + a(1) = • • • = ka(1). J. H. Kwak and Y. S. Kwon: Classification of some reflexible edge-transitive embeddings ... 571 Let a(1) = r. Now a(k) = rk and a2(1) = a(r) = r2 = 1 (mod m). Since n is even, 0 is parity preserving and there exists s, t G [n] such that 0(2k) = 2kt, 0(2k + 1) = 2kt + 2s + 1 and 2t2 = 2 for any 2k G [n] by Lemma 2.11(3). If 2t = 2 then the length of the orbit containing 2 is 2 and hence d2 is even. But it can not happen because m is odd. Hence for any 2k G [n], 0(2k) = 2k, 0(2k + 1) = 2k + 2s + 1 and for any i G [m], 001) = 0i-1(2s + 1) = 0i-2(2s + 2s + 1) = • • • = 2is +1. Therefore d2 is the smallest positive integer such that 2d2s = 0 (mod n), which implies that d2 is a divisor of n, and hence d2 is a divisor of gcd(m, n) = pi1 • • • pC. If r = 1 (mod p"i) for some i = 1,2,..., i, then the fact a(1) = r = —1 (mod d2) implies that pi can not be a divisor of d2. Hence pbi should divide s, namely, s = 0 (mod pbi). If r = —1 (mod p^3) for some j = 1, 2,...,i, then s = x • pb3-Cj (mod pb3) for some x with 0 < x < pC — 1 because d2 is a divisor of gcd(m, n). Therefore, for any j = 1,..., i, the pair (r (mod p^3), s (mod pb3)) is (1,0) or ( —1, x • pb3-Cj) for some x with 0 < x < pCj — 1. Because d2 | gcd(m,n), we have 2s = 0 (mod 2b) and for any k = 1,2,... ,g, s = 0 (mod q^+k). Since r2 = 1 (mod m), r = ±1 (mod p^3) for any j = 1, 2,... f. Conversely for any r G [m] and s G [n] satisfying the conditions (i) for any j = 1,...,i, the pair (r (mod p^3 )),s (mod pb3)) is (1,0) or ( —1, x • pj3 -C3) for some integer x with 0 < x < pj3 — 1, (ii) 2s = 0 (mod 2bqb;+11 • • • qb/+gs) and (iii) for any j = 1, 2,... f, r = ±1 (mod pa++3), define a(k) = rk for any k G [m] and 0(2t) = 2t, 0(2t + 1) = 2t + 2s + 1 for any 2t G [n]. Note that a G S0 and 0 G SO. Let d1 = |(a)| and d2 = K0)|. Now d1 = 1 or 2 depending on the value of r and d2 is the smallest positive integer satisfying 2d2 s = 0 (mod n). Note that d2 divides gcd(m,n) and r = —1 (mod d2). For any i G [n], let a(i) = 0(i) and b(i) = ai(1) = ri. For the first case, let i be even. Now a(i) = 0(i) = i and b(i) = ai(1) = 1. For any 2t G [n], 0 (2t + i) = 2t + i and 0b(i)(2t) + a(i) = 0(2t) + 0(i) = 2t + i and 0(2t + 1 + i) = 2t + i + 2s +1 and 0b(i)(2t + 1) + a(i) = 0(2t + 1) + 0(i) = 2t + 2s + 1 + i. Hence 0(t + i) = 0b(i)(t) + a(i) for any t G [n]. For any k G [m], ai(k) = k and a°(i)(k + b(i)) — 1 = k. 572 Ars Math. Contemp. 16 (2019) 445-463 Hence a®(k) = aa(i)(k + b(x)) — 1 for any k G [m]. For the remaining case, let i be odd. Now a(i) = P(i) = i + 2s and b(i) = a®(1) = r = —1 (mod d2). For any 2t G [n], P (2t + i) = 2t + i + 2s and Pb(i)(2t) + a(i) = P-1(2t) + P(i) = 2t + i + 2s and P (2t + 1 + i) = 2t + i +1 and Pb(i)(2t + 1) + a(i) = P-1(2t + 1) + P (i) = 2t +1 — 2s + i + 2s = 2k + i + 1. Hence P(t + i) = Pb(i)(t) + a(i) for any t G [n]. For any k G [m], a®(k) = rk and aa(i)(k + b(i)) — 1 = a(k + r) — 1 = rk + r2 — 1 = rk. Hence a4(k) = aa(i)(k + b(i)) — 1 for any k G [m]. By Lemma 2.7, (a,P) G ETm,n. Furthermore one can easily check that a-1( —k) = —a(k) for any k G [m] and P-1( — t) = —P(t) for any t G [n]. Hence (a, P) G RETm,n by Lemma 2.8. Therefore | RETm,„ | = 2f (1+ p!1 ) ••• (1+ pc/ ). □ 4 Both m and n are even In this section, we classify reflexible edge-transitive embeddings of Kmf satisfying the Property (P) when both m and n are even, and consequently prove Theorem 1.1. For the classification, we give the following lemma. Lemma 4.1. Let m and n be even and let a G So and P G SO with d1 = |(a}| and d2 = |(P}|. Now (a, P) G RETm,n if and only if a and P are defined by a(2k) = 2kt1 and a(2k + 1) = 2kt1 + 2s1 + 1 for any 2k G [m] and P(2k) = 2kt2 and P(2k +1) = 2kt2 +2s2 + 1 for any 2k G [n] for some quadruple (s1,t1; s2,t2) G [mm] x [m] x [f ] x [f ] satisfying the following conditions; (i) d1 | gcd(m, n) and d2 | gcd(m,n); (ii) 2t2 = 2 (mod m) and 2t2 = 2 (mod n); (iii) 2(s1 + 1) = 0 (mod d2), 2(t1 + 1) = 0 (mod d2), 2(s2 + 1) = 0 (mod d1), and 2(t2 + 1) = 0 (mod d1); (iv) 2(s1 + 1)(t1 — 1) = 0 (mod m) and2(s2 + 1)(t2 — 1) = 0 (mod n). J. H. Kwak and Y. S. Kwon: Classification of some reflexible edge-transitive embeddings ... 573 Proof. Assume that 2ti = 2, namely, ti = 1. Then a(2k) = 2k and a(2k + 1) = 2k + 2s1 + 1 for any 2k G [m]. Since for any i G [n], a4(2k +1) = 2k + 2is1 + 1, d1 is the smallest positive integer such that 2d1s1 = 0 (mod m). Now assume that 2t1 = 2. Then d1 should be even because a2 (2) = 2t 1 = 2. Since for any 2i G [n] and for any 2k G [m], a2i(2k + 1) = 2k + 2is1(t1 + 1) + 1, d1 is the smallest positive even integer such that d1s1(t1 + 1) = 0 (mod m). Similarly one can show that d2 is the smallest positive integer such that 2d2s2 = 0 (mod n) if t2 = 1; and the smallest positive even integer such that d2«2(t2 + 1) = 0 (mod n) if ¿2 = 1. For any i G [n], let a(i) = P(i) and b(i) = a4(1). For the first case, let i be even. Then a(i) = P(i) = it2 = —i (mod d1) and b(i) = a4(1) = is1(t1 + 1) + 1 = 1 (mod d2). For any 2 k G [n], P(2k + i) = 2kt2 + it2 and P b(i)(2k) + a(i) = P(2k) + P (i) = 2kt2 + it2 and P (2k + 1 + i) = 2kt2 + it2 + 2s2 + 1 and P b(i)(2k + 1) + a(i) = P (2k + 1) + P(i) = 2kt2 + 2s2 + 1 + ¿¿2- Hence P(k + i ) = Pb(i) (k) + a(i) for any k G [n]. For any 2k G [m], a4 (2k) = 2k and a°(i)(2k + b(i)) — 1 = a-i(2k + is1(t1 + 1) + 1) — 1 = (2k + is1(t1 + 1) — is1(t1 + 1) + 1) — 1 = 2k and a4(2k + 1) = 2k + is1(t1 + 1) + 1, and a°(i)(2k + 1 + b(i)) — 1 = a-i(2k + is1(t1 + 1) + 2) — 1 = (2k + is1(t1 + 1) + 2) — 1 = 2k + is 1 (t 1 + 1) + 1. Hence a4(k) = aa(i)(k + b(i)) — 1 for any k G [m]. For the remaining case, let i be odd. Now a(i) = P(i) = (i — 1)t2 + 2s2 + 1 = —i (mod d1) and b(i) = a4(1) = (i — 1)s1(t1 + 1) + 2s1 + 1 = —1 (mod d2). For any 2k G [n], P(2k + i) = 2kt2 + (i — 1)t2 + 2s2 + 1 and Pb(i)(2k) + a(i) = P-1(2k) + P(i) = 2kt2 + (i — 1)t2 + 2s2 + 1 and P(2k +1 + i) = (2k + i +1)t2 and P b(i)(2k + 1) + a(i) = P-1(2k + 1) + P(i) = (2kt2 — 2s2t2 + 1) + (i — 1)t2 + 2S2 + 1 = (2k + i + 1)t2 — 2(s2 + 1)(t2 — 1) = (2k + i + 1)t2- 574 Ars Math. Contemp. 16 (2019) 445-463 Hence £(k + i) = ,06(i)(k) + a(i) for any k G [n]. For any 2k G [m], ai (2k) = 2kii and a°(i)(2k + b(i)) - 1 = a-i(2k + (i - 1)s1(t1 + 1) + 2Si + 1) - 1 = (2k + (i - 1)si(ti + 1) + 2si)ti - (i + 1)si(ti + 1) + 2si = 2kti - 2si(ti + 1) + 2siti + 2si = 2kti and a4(2k + 1) = 2kti + (i - 1)si(ti + 1) + 2si + 1 and a°(i)(2k + 1 + b(i)) - 1 = a-i(2k + (i - 1)si(ti + 1) + 2si +2) - 1 = (2k + (i - 1)si(ti + 1) + 2si + 2)ti - 1 = 2kti + (i - 1)si(ti + 1) + 2si + 1 + 2(si + 1)(ti - 1) = 2kti + (i - 1)si(ti + 1) + 2si + 1. Hence a4(k) = aa(i)(k + b(i)) - 1 for any k G [m]. By Lemma 2.7, (a,£) G ETm,n. Furthermore one can easily check that a-i(-k) = -a(k) forany k G [m] and £-i(-k) = -£(k) for any k G [n]. Hence (a, £) G RETm,n by Lemma 2.8. Since m and n are even, both a and £ are parity preserving. For any 2k G [m], a(2k) = a(2(k - 1)) + a(2) = a(2(k - 2)) + 2a(2) = ••• = ka(2) and a(2k + 1) = a(2(k - 1) + 1) + a(2) = a(2(k - 2) + 1) + 2a(2) = • • • = a(1) + ka(2). Let a(1) = 2si + 1 and a(2) = 2ti for some si,ti G [f ]. Then a(2k) = 2kti and a(2k +1) = 2kti + 2si + 1 forany 2k G [m]. Note that for any 2k G [m], a(1) + a(2k) = a(2k +1) = a-i(2k) + a(1). Hence a-i(2k) = a(2k), namely, a2 (2k) = 2k. It implies that a2(2) = a(2ti) = 2t2 = 2 (mod m). Assume that 2ti = 2, namely, ti = 1. Then by Lemma 2.6, the order |(a}| is the smallest positive integer di such that adl (1) = adl-i(2si + 1) = adl-2(2si + 2si + 1) = • • • = 2disi + 1 = 1. Now assume that 2ti = 2. Then the order |(a}| is even and it is the smallest positive even integer di such that adl (1) = adl-i(2si + 1) = adl-2(2siti + 2si + 1) = adl -3(2siti + 4Si + 1) = adl-4(4siti + 4si + 1) = • • • = disi(ti + 1) + 1 = 1. Hence di is a divisor of m and consequently a divisor of gcd(m, n). By a similar reason, there exist s2, t2 G [§ ] such that £(2k) = 2kt2 and £(2k + 1) = 2kt2 + 2s2 + 1 for any 2k G [n]. Furthermore 2t2 = 2 (mod n) and d2 is a divisor of gcd(m,n). By Lemma 2.10, a(1) = 2si + 1 = -1 (mod d2), namely, 2(si + 1) = 0 (mod d2) and a(2) = 2ti = -2 (mod d2), namely, 2(ti + 1) = 0 (mod d2). Similarly it holds that 2(s2 + 1) = 2(t2 + 1) = 0 (mod di). Note that 2ti = a(2) = a-i(1) + a(1) = (-2siti + 1) + 2si + 1. Hence 2(si + 1)(ti-1) = 0 (mod m). By a similar reason, it holds that 2(s2+1)(t2-1) = 0 (mod n). □ J. H. Kwak and Y. S. Kwon: Classification of some reflexible edge-transitive embeddings ... 575 For even m and n, let Q(m, n) be the set of quadruples (si, ti; s2, t2) G [§] x [§] x [y ] x [y] satisfying the conditions in Lemma 4.1. By Lemma 4.1, the classification of reflexible edge-transitive embeddings of Km,n satisfying the Property (P) is equivalent to the classification of Q(m, n), and the number | RETm,n | equals to the cardinality |Q(m, n)|. In this section, let m = • • • pa/ p"^1 • • • p";+/ and n = 2bp11 • • • pb/ qa+i1 • • • q^+g (prime decompositions) and let gcd(m, n) = 2Cp1 • • • pc/ with cj > 1 for any i = 1,..., Without any loss of generality, assume that a < b, namely, a = c. By Chinese Remainder Theorem, it suffices to consider quadruples ( s 1, 11 ; s2,t2) modulo prime powers dividing m and n, respectively. So we have the following lemma. Lemma 4.2. Fora quadruple (s1,t1; s2 ,t2) G [§ ] x [§] x [y] x [y], (s1,t1; s2,t2) belongs to Q(m, n) if and only if: (1) for i = 1,...,£, (s1 (mod p"* ),t1 (mod p"* ); s2 (mod pbi ),t2 (mod pbi )) is one of (-1, -1; -1, -1), (-1, -1; y • pbi-Ci, 1), (x • p"i-Ci, 1; -1, -1) and (0,1;0,1), where x, y = 0,1,... ,pC* - 1; (2) for any j = 1, 2,..., f, (s1 (mod pa£> ),t1 (mod p"j )) is (0,1) or (-1, -1); (3) for any k =1, 2,..., g, (s2 (mod q£b++fc ), ¿2 (mod qb++fc )) is (0,1) or (-1, -1); (4) (s1 (mod 2a),t1 (mod 2a); s2 (mod 2b),t2 (mod 2b)) belongs to Q(2a, 2b). Proof. Assume that (s1,t1; s2,t2) belongs to Q(m,n). Then t2 = 1 (mod y ) and¿2 = 1 (mod § ). (1): First let us consider the quadruple modulo p"* and pbi for i = 1,..., Note that t1 = ±1 (mod pa ) and ¿2 = ±1 (mod pbi ). If t1 = -1 (mod pa ), then s1 should be -1 modulo p"* to satisfy 2(s1 + 1)(t1 - 1) = 0 (mod p"*). By similar reason, if t2 = -1 (mod pbi ), then s2 = -1 (mod pbi ). Let (s1,t1) = (-1, -1) (mod p"*). Since d1 is the smallest positive even integer satisfying d1s1(t1 + 1) = 0 (mod m), p4 does not divide d1. If t2 = -1 (mod pbi ) then s2 should be -1 modulo pbi. If t2 = 1 (mod pbi ), then s2 = y • pbi-Ci (mod pbi) for some y = 0,1,... ,pC* - 1 because d2 | gcd(m, n). By a similar reason, one can say that if (s2,t2) = (-1,-1) (mod pbi), then (s1,t0 = (-1, -1) or (x • p"i-Ci, 1) (mod p"* ) for some x = 0,1,..., pC* - 1. Let (s1,t1) = (0,1) (mod p"*). By the condition (iii) in Lemma 4.1, p4 does not divide d2. Note that if t2 = 1 then d2 is the smallest positive integer satisfying 2d2 s2 = 0 (mod n), and if t2 = 1 then d2 is the smallest positive even integer such that d2s2(t2 + 1) = 0 (mod n). Hence s2 = 0 or t2 = -1 modulo pbi, which implies that (s2,¿2) = (0,1) or (-1, -1) (mod pbi). Let t1 = 1 (mod p"* ) and s1 =0 (mod p"* ). One can see that pj divides d1. By the condition (iii) in Lemma 4.1, t2 = -1 (mod pbi ) and s2 = -1 (mod pbi ). 576 Ars Math. Contemp. 16 (2019) 445-463 Therefore (si (mod ),ti (mod ); (mod pbi),t2 (mod pbi)) = (-1, -1; -1, -1), (-1, -1; y ■ p^, 1), (x ■ pf-Ci, 1; -1, -1) or (0,1; 0,1), where x, y = 0,1,... ,pjp - 1. (2): For any j = 1,2,...,/, t1 = ±1 (mod p^3 ). If t1 = 1 (mod p^3 ) then s1 = 0 (mod p^+j3 ) because p^+j does not divide d1. If t1 = -1 (mod p^3 ) then si = -1 (mod p+++j ) to satisfy 2(s1 + 1)(t1 - 1) = 0 (mod p;++j ). (3): By the similar reason with (2), for any k = 1,2, ...,g, (s2 (mod q++fcfc), t2 (mod qb++fck)) is (0,1) or (-1,-1). (4): If a quadruple (s1,t1; s2,t2) G [n] x [n] x [m] x [m] satisfies all conditions in Lemma 4.1, then it also satisfies these conditions modulo 2° and 2b. Hence (s1 (mod 2°), t1 (mod 2°); s2 (mod 2b),t2 (mod 2b)) G Q(2°, 2b). By Chinese Remainder Theorem, one can show that if (1), (2), (3) and (4) hold, then (s1,t1; S2,t2) GQ(m,n). □ Corollary 4.3. The number ofreflexible edge-transitive embeddings of Km,n satisfying the Property (P) up to isomorphism is 2f+ p^) ■ ■ ■ (1 + pf )|Q(2°, 2b)|. Proof. By Lemma 4.2, the number of reflexible edge-transitive embeddings of Km n satisfying the Property (P) up to isomorphism is (2 + 2p1 ) ■ ■ ■ (2 + 2p?)2f 2g|Q(2°, 2b)| = 2f+g+£(1 + pi1 ) ■ ■ ■ (1 + p?)|Q(2°, 2b)|. □ By Lemma 4.2, it suffices to classify Q(2°, 2b) to classify reflexible edge-transitive embeddings of Km n satisfying the Property (P). Let P(2) = {(0,1)} and for a 2-power 2° (a > 1), let P (2°) be the set of all pairs (s,t) G [2°-1] x [2°-1] satisfying the conditions: (i) 2t2 = 2 (mod 2°) and (ii) 2(s + 1)(t - 1) = 0 (mod 2°). For any (s, t) G P(2°)\{(0,1)}, let d(s, t) be the smallest positive even number d such that ds(t +1) = 0 (mod 2°) and let e(s, t) be the largest number 2j with 2j < 2° satisfying 2(s +1) = 0 (mod 2j) and 2(t + 1) = 0 (mod 2j). Let d(0,1) = 1 and e(0,1) = 2. Now we have the following lemma. Lemma 4.4. For 2-powers 2° (a > 1) and 2b (b > 1), a quadruple (s1,t1; s2,t2) belongs to Q(2°, 2b) f and only if (s1, t1; s2, t2) satisfies the conditions (a) (s1,t1) G P(2°) and (s2,t2) G P(2b), (b) d(s1,t1) < e(s2,t2) and d(s2,t2) < e(s1,t1). J. H. Kwak and Y. S. Kwon: Classification of some reflexible edge-transitive embeddings ... 577 Proof. The conditions (i) and (ii) in the definition of P(2°) correspond to the conditions (ii) and (iv) in Lemma 4.1. Suppose that d(si,ti) < e(s2,t2) and d(s2,t2) < e(s1,t1). Since d(s1,t1) < 2° and e (S2,t2) < 2b, d(s1,t1) divides gcd(2°, 2b), the minimum of 2° and 2b. Similarly d(s2,t2) also divides gcd(2°, 2b). Furthermore it holds that 2(si + 1) = 0 2(ti + 1) = 0 2(s2 + 1) = 0 2(t2 + 1) = 0 (mod d(s2,¿2)), (mod d(s2,¿2)), (mod d(s1 ,t1)) (mod d(s1 ,t1)). and Therefore the conditions (i) and (iii) in Lemma 4.1 hold, and hence (s1, t1; s2,t2) belongs to <2(2°, 2b). Let (s1, t1; s2,t2) belong to Q(2°, 2b). Now the condition (iii) in Lemma 4.1 is equivalent to the condition d(s1,t1) < e(s2,t2) and d(s2,t2) < e(s1,t1). □ By Lemma 4.4, the calculation of d(s, t) and e(s, t) for each (s, t) G P(2°) is helpful to calculate |Q(2°, 2b)|. The following lemma gives full list of (s,t) G P(2°) and corresponding d(s,t) and e(s, t). Lemma 4.5. Fora 2-power 2° (a > 1), the set {(s, t, d(s, t), e(s, t)) : (s,t) G P (2°)} is the following: {(0,1,1, 2), (1,1, 2, 4)}, {(0,1,1, 2), (1,1, 4, 4), (2,1, 2, 2), (3,1, 4,4), (1, 3, 2,4), (3, 3, 2, 8)}, {(0,1,1, 2), (2°-2 - 1, 2°-2 - 1, 4, 2°-1), (2°-1 - 1, 2°-2 - 1, 4, 2°-1), (2°-2 - 1, 2°-1 - 1, 2, 2°-1), (2°-1 - 1, 2°-1 - 1, 2, 2°)} U {(x, 1, 2°-1,4), (x, 2°-2 + 1, 2°-1,4) : x = 1, 3,..., 2°-1 - 1} U{(2^, 1, 2' I — i — 1 2) : i = 1,... ,a - 2, y = 1, 3,..., 2 I — i— 1 - 1} if a = 2 if a = 3 if a > 4. Proof. Let (s,t) G P(2°). For a = 2, t should be 1 and both s = 0 and s = 1 satisfy the conditions for (s, t) G P(2°). Hence (s,t, d(s,t), e(s,t)) = (0,1,1,2) or (1,1,2,4). Let a = 3. Then t = 1 and t = 3. If t = 1, then s = i for some i = 0,1,2,3. If t = 3, then s = 1 or s = 3. In any possible pair (s, t), one can easily calculate d(s,t) and e(s, t). Now assume that a > 4. Then t = 1, 2°-2 - 1, 2°-2 + 1 or 2°-1 - 1. For t = 1, any number 0,1, 2,..., 2°-1 - 1 is possible for s to satisfy the condition (ii) in the definition of P(2°). Note that if (s,t) = (0,1), then (d(0,1),e(0,1)) = (1, 2). One can easily show that if (s,t) = (x, 1) for any x = 1, 3,..., 2°-1 - 1 then (d(s,t), e(s,t)) = (2°-1,4). , 2° - 1 , then If (s, t) = (2iy, 1) for any i = 1,..., a - 2 and for any y = 1, 3, (d(s, t), e(s, t))j= (2a-i_1, 2). For t = 2° 2 - 1, both s = 2° 2 - 1 and s = 2° 1 - 1 satisfy the conditions for (s, t) G P(2°). If (s, t) = (2°-2 - 1, 2°-2 - 1) or (2°-1 - 1,2°-2 - 1) then we have (d(s, t), e(s, t)) = (4, 2°-1). Let t = 2°-2 + 1. Then any number s = 1, 3,..., 2°-1 - 1 satisfies the condition (ii) in the definition of P(2°). For any (s,t) = (x, 2°-2 + 1) with x = 1,3,..., 2°-1 - 1, we have (d(s,t),e(s,t)) = (2°-1,4). 578 Ars Math. Contemp. 16 (2019) 445-463 For the final case, let t = 2°-1 -1. Then s = 2°-2-1 or 2°-1 -1. If (s,t) = (2°-2-1, 2°-1 - 1) then we have (d(s,t),e(s,t)) = (2,2a-1); if (s,t) = (2°-1 - 1, 2°-1 - 1) then (d(s, t), e(s, t)) = (2, 2°). □ Theorem 4.6. For any 2-powers 2° and 2b with a < b, the number |Q(2°, 2b) | ofreflexible edge-transitive embeddings of Km,n satisfying the Property (P) up to isomorphism is the following: |Q(2a, 1 if (a, b) = (1,1), 2 if (a, b) = (1, 2), 4 if (a, b) = (2, 2) or (1, k) with k > 3, 10 if (a, b) = (2, 3), 12 if (a, b) = (2, k) with k > 4, 28 if (a, b) = (3, 3), 40 if (a, b) = (3, 4), 36 if (a, b) = (3, k) with k > 5, 20(1 + 2°-2 ) if a = b > 4, 20 + 18 • 2°-2 if b - 1 = a > 4, 20 + 16 • 2°-2 if b - 2 > a > 4. Proof. By Lemma 4.4, it suffices to find all (si, ti; s2, t2) satisfying the conditions (a) (si,ti) GP (2°) and (s2,t2) eP (2b), (b) d(si,ti) < e(s2,t2) and d(s2,t2) < e(si,ti). ByLemma4.5, one can get all the lists of (si, ti; s2,t2) satisfying the conditions as Table 1. □ Proof of Theorem 1.1. For odd m and n, the number | RETm,n | of reflexible edge-transitive embeddings of Km n up to isomorphism is 1 by Theorem 3.1. When exactly one of m and n is odd, then the number | RETm n | is counted in Theorem 3.2. Assume that both m and n are even. Let m = 2°pa1 P°2 ■ ■ ■ PTp°+T ■ ■ ■ Pl+i and n = 2bp11 p22 ■ ■ ■ pb/ q°++H11 ■ ■ ■ q+g9 (prime decompositions) and let gcd(m, n) = 2cpc11 p22 ■ ■ ■ p^ with c > 1 for any i = 1,..., I. Without any loss of generality, assume that a < b, namely, a — c. By Corollary 4.3, the number | RETm,n | = |Q(m, n)| is 2f+g+^(1+ p11)... (1+ p? )|Q(2°, 2b)|. Theorem 4.6 completes the proof. □ J. H. Kwak and Y. S. Kwon: Classification of some reflexible edge-transitive embeddings ... 579 Table 1: All lists of Q(2°, 2b). (a b) <2(2°, 2b) (1,1) (0,1 0,1) (1, 2) (0,1 0,1), (0,1; 1,1) (1, > 3) (0,1 (0,1 0,1), (0,1; 2b-2,1), (0,1; 2b-2 - 1, 2b-1 - 1), 2b-1 - 1, 2b-1 - 1) (2, 2) (0,1 0,1), (0,1; 1,1), (1,1; 0,1), (1,1; 1,1) (2, 3) (0,1 (1,1 0,1), (0,1; 2,1), (0,1; 1, 3), (0,1; 3, 3), (1,1; 0,1), (1,1; 1,1), 2,1), (1,1; 3,1), (1,1; 1, 3), (1,1;3, 3) (2, > 4) (0,1 (0,1 (1,1 (1,1 0,1), (0,1; 2b-2,1), (0,1; 2b-2 - 1, 2b-1 - 1), 2b-1 - 1,2b-1 - 1), (1,1;0,1), (1,1;2b-3,1), (1,1;2b-2,1), 3 ■ 2b-3,1), (1,1; 2b-2 - 1, 2b-2 - 1), (1,1; 2b-1 - 1, 2b-2 - 1), 2b-2 - 1, 2b-1 - 1), (1,1; 2b-1 - 1, 2b-1 - 1) (3, 3) (0 or 2,1; 0,1), (0 or 2,1; 2,1), (0 or 2,1; 1, 3), (0 or 2,1; 3, 3), (1 or 3,1; 1,1), (1 or 3,1; 3,1), (1 or 3,1; 1, 3), (1 or 3,1; 3,3), (1 or 3, 3; 0,1), (1 or 3, 3; 1,1), (1 or 3, 3; 2,1), (1 or 3, 3; 3,1), (1 or 3, 3; 1, 3), (1 or 3, 3; 3, 3) (3, 4) (0 or 2,1; 0,1), (0 or 2,1; 4,1), (0 or 2,1; 3, 7), (0 or 2,1; 7, 7), (1 or 3,1; 3, 3), (1 or 3,1; 7, 3), (1 or 3,1; 3, 7), (1 or 3,1; 7, 7); (1, 3; x, 1), x = 0, 2, 4, 6; (3, 3; s2,t2), (s2,t2) eP (24) (3, > 5) (0 or 2,1; 0,1), (0 or 2,1;2b-2,1); (0 or 2,1; x, 2b-1 - 1), x = 2b-2 - 1 or 2b-1 - 1; (1 or 3,1; x, y), x, y = 2b-2 - 1 or 2b-1 - 1; (1, 3 (1, 3 (3, 3 (3, 3 i ■ 2b-3,1), i = 0,1, 2, 3; x, y), x, y = 2b-2 - 1 or 2b-1 - 1; i ■ 2b-4,1), i = 0,1,..., 7; x, y), x, y = 2b-2 - 1 or 2b-1 - 1 (> 4, > a) (0 or 2°-2,1; x,y), (x, y) = (0,1), (2b-2,1), (2b-2 - 1, 2b-1 - 1) or (2b-1 - 1, 2b-1 - 1); (2x, 1; 2b-2 - 1, 2b-1 - 1), (2x, 1; 2b-1 - 1, 2b-1 - 1), x = 1, 2,..., 2°-2 - 1 (x = 2°-3); (x, 1 or 2°-2 + 1; y, z), x= 1, 3,..., 2°-1 - 1, y, z = 2b-2 - 1 or 2b-1 - 1; (2°- 2 - 1 or 2°-1 - 1, 2°-2 - 1 or 2°-1 - 1; x, y), x, y = 2b-2 - 1 or 2b-1 - 1; (2°- 2 - 1, 2°-1 - 1; i ■ 2b-°, 1), i = 0,1,..., 2°-1 - 1; Only when a = b: (2°-2-1 or 2°-1-1, 2°-2-1; x, 1 or 2b-2+1), x =1, 3,..., 2b-1-1; Only when a = b: (2°-2 - 1, 2°-1 - 1; x, 2b-2 + 1), x =1, 3,..., 2b-1 - 1; Only when a = b or b = a + 1 : (2°-1 - 1, 2°-1 - 1; x, 1), x = 0,1,..., 2b-1 - 1; Only when a = b or b = a + 1 : (2°-1 - 1, 2°-1 - 1; x, 2b-2 + 1), x =1, 3,..., 2b-1 - 1; Only when b > a + 2: (2°-1 - 1, 2°-1 - 1; i ■ 2b-a-1,1), i = 0,1,..., 2° - 1 580 Ars Math. Contemp. 16 (2019) 445-463 5 Classification of some groups In this section, we aim to consider a presentation of the group (xa, y^) for any (a, p) G RETm,n. And we give some sufficient conditions and necessary conditions for (xai, y^) and (xa2, y^2) to be isomorphic for any (a1, p1), (a2, p2) G RETm,n. For any positive integers m and n, a group r such that (i) r = XY for some cyclic groups X = (x) of order n and Y = (y) of order m with X n Y = {1r} and (ii) there exists an automorphism of r which sends x and y to x-1 and y-1, respectively, is isomorphic to (xa, ys) for some (a, p) G RETm,n. For our convenience, call a group r satisfying the conditions (i) and (ii) in the above sentence a reflexible product of two cyclic groups of order m and n. Now to classify reflexible products of two cyclic groups of order m and n, it suffices to consider (xa, ys), where (a, p) G RETm,n. Note that for any integers i, j and for any (a, p) G RETm,n, yi = (i) For example, ysxa = xa(1)ya(1) and ysx;a = xa(2)y? (1) For odd integers m and n, since RETm,n = {(id, id)}, there is a unique reflexible product of two cyclic groups of order m and n up to isomorphism, namely, an abelian group Zm x Zn. Let m = pa1 pa2 • • • pf Pm1 • • • pa++/ (prime factorization) be odd and n = 2bp11 p22 • • • pb£ qb+11 • • • qb+59 (prime factorization) be even. Let gcd(m,n) = p^1 p22 ••• with cj > 1 for any i = 1,...,£ Now | RETm,„ | = 2f (1 + p!1) • • • (1 + pC£) by Theorem 3.2. Note that for any (a, p) G RETm,n and for any integer k, a(k) = rk, p(2k) = 2k, p(2k + 1) = 2k + 1 + 2s for some integers r G [m] and s G [n] satisfying r2 = 1 (mod m), 2s = 0 (mod 2^+1 • • • q^9) and for any j = 1, 2,..., I, s = 0 (mod pjj) if r = 1 (mod pa^); s = z • pbj Cj (mod pbj) for some integer z with 0 < z < — 1 if r = — 1 (mod pa^). Let us denote such a and p by ar and ps. Considering commuting rule i j s®(j) (i) ysxa = xa (j)y/3 (), one can check that the centralizer of (xa;, yss) is {xaiyjs : i G [2] , j(r — 1) = 0 (mod m)} = ^), where k is the smallest positive integer j satisfying j(r — 1) = 0 (mod m). This implies that for any (an, pS1), (a^, p«2) G RETm,„, if two groups (x^, ysS1) and (x^, ysS2) are isomorphic, then r1 = r2. Note that yss xar = x^1^(1) = x^s and yss x^; = xa;(2)ya;(1) = x^ yss. J. H. Kwak and Y. S. Kwon: Classification of some reflexible edge-transitive embeddings ... 581 In fact, the above two equations determine the whole commuting rules. For any u G [m] and v G [n], if v is even, then y^ xjf = xJf y^, and if v is odd, then „« xv = xv-1yu x = xv-1y"-1x2s+1yr yps xar = xar yps x«r = xar y,Ss xar y,Ss = xv-1+2s«-1x yr = xv-1+2s y«-2x2s + 1y2r = xar x«r y,Ss = xar xar y,Ss = xv-1+4syu-2x y2r = = xv + 2wsyMr = xar y,Ss x"r y,Ss = = xar y,Ss . For any v G [n] with gcd(v, n) = 1, ya xv = xA(v)y<(1) = xv+2syr = xv(2v-1S+1)yr yA xar = xar y,Ss = xar = xar because v is odd, where v-1 is an integer satisfying vv-1 = 1 (mod n). For any s1, s2 G [n] with gcd(s1, n) = gcd(s2, n), one can choose v G [n] satisfying that gcd(v, n) = 1 and v-1s1 = s2 (mod n). Therefore for any (ari , psi), (ar2 , ps2) G RETm,n, if r1 = r2 and gcd(s1, n) = gcd(s2, n) then (x„ri, y^) is isomorphic to (x„r2, y^). This means that the number of non-isomorphic reflexible product of two cyclic groups of order m and n is at most 2f (2 + c1) • • • (2 + q). So any reflexible product of two cyclic groups of order m and n is isomorphic to (x, y | xn = ym = 1, yx = x2s+1yr, yx2 = x2y) for some r G [m] and s G [n] satisfying r2 = 1 (mod m), 2s = 0 (mod 2bq^+f • • • q^+g9) and for any j = 1, 2, ...,£, s = 0 (mod pj) if r = 1 (mod p?3); s = pj-Cj (mod pbj) for some integer z = 0,1,..., Cj if r = —1 (mod p?3). Conversely, assume that for some (ari , psi), (ar2 , ps2) G RETm,n, (xa , y^ ) is isomorphic to (x^2, y^). Let ^: (xa^, y^) ^ (x«r2, y^S2) be an isomorphism such ->-ri / o f-Jsi / \ ur2 > o H s that ^(x^ri) = ^ and^(y^si) = y^s2. For the remaining case, let m = 2apaipa2 • • • p?'p?+Y • • • p?++/ and n = 2bp1i p22 • • • pb' q?+H1i • • • qb++s9 (prime decompositions) with gcd(m, n) = 2cp1ip22 • • • p^', where 1 < a < b and cj > 1 for any i = 1,..., For any (a, p) G RETm,n and for any integer k, a(2k) = 2kt1, a(2k + 1) = 2kt1 + 2s1 + 1, P(2k) = 2kt2 and P(2k + 1) = 2kt2 + 2s2 + 1 for some (s1, t1; s2, t2) G Q(m, n). Let a and p be such permutations. Note that yflx = x^(1)ya(1) = x2s2 + 1y2si + 1 xa = xa = xa yg , y^ xa = xa(2)y;2(1) = x^ y2si(ii+1)+1, y2 xa = xa2(1) y;(2) = xas2(t2+1)+1y2ti and 2 2 fl2(2) a2(2) 2 2 / 'T* - ryt^ \ 1 O! K ' - 1 for any i — 1,..., £. Then r is isomorphic to (x, y | xn — ym — 1, yx — x2s+1yr, yx2 — x2y) for some r G [m] and s G [f ] satisfying r2 = 1 (mod m), 2s = 0 (mod 2^+1 • • • qb++s9), and for any j — 1, 2,..., £, s = 0 (mod pbj) if r = 1 (mod p"3), s = pj3 Cj(mod pj3) for some z — 0,1,..., Cj if r = — 1 (mod p"3). J. H. Kwak and Y. S. Kwon: Classification of some reflexible edge-transitive embeddings ... 583 (3) Let m = 2 ap a 1 • • • pa 'pa+Y • • • p2++/ and n = 2bp11 • • • pf g^+Y1 • • • qb+g9 (prime factorization) with gcd(m,n) = 2cp11 p22 • • • pC^, where 1 < a < b and c > 1 for any i = 1,... ,£ Now r is isomorphic to (x,y | xn = ym = 1, yx = x2s2+1y2s1 + 1, yx2 = x2t2y2«1^^1, y2x = X2S2(t2 + 1)+1y2t1 , y2x2 = x2y2) for some (s1,t1; s2,t2) € Q(m, n). For any positive integers m and n and for any (a, ,0), (a', A) € RETm,n, we do not know a necessary and sufficient condition for (xa, y^) ~ (xa/, y^/). So we propose the following problem. Problem 5.2. 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