ARS MATHEMATICA CONTEMPORANEA
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 18 (2020) 371-379 https://doi.org/10.26493/1855-3974.1900.cc1
(Also available at http://amc-journal.eu)
The complete bipartite graphs which have exactly two orientably edge-transitive embeddings*
Xue Yu
Department of Mathematics, Henan Institute of Science and Technology, Xinxiang 453003, P. R. China, and School ofMathematics and Statistics, Yunnan University, Kunming 650500, P. R. China
Ben Gong Lou f
School ofMathematics and Statistics, Yunnan University, Kunming 650500, P. R. China
Received 7 January 2019, accepted 10 June 2020, published online 23 October 2020
In 2018, Fan and Li classified the complete bipartite graph Km,n that has a unique orientably edge-transitive embedding. In this paper, we extend this to give a complete classification of Km n which have exactly two orientably edge-transitive embeddings.
Keywords: Bipartite graphs, edge-transitive embeddings. Math. Subj. Class. (2020): 20B15, 20B30, 05C25
*The research was partially supported by the NSFC (11861076, 11171200) and the NSF of Yunnan Province (2019FB139). The work in the paper was done when the first two authors visited South University of Science and Technology. The authors are very thankful to Professor Cai Heng Li. t Corresponding author.
E-mail addresses: yuxue1212@163.com (Xue Yu), bglou@ynu.edu.cn (Ben Gong Lou), fww0871@163.com (Wen Wen Fan)
Wen Wen Fan
School of Mathematics, Yunnan Normal University, Kunming 650500, P. R. China
Abstract
©® This work is licensed under https://creativecommons.org/licenses/by/4.0/
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1	Introduction
Let M = (V, E, F) be an orientable map with vertex set V, edge set E and face set F, that is, M is a 2-cell embedding of the underlying graph r = (V, E) in an orientable surface. A permutation of V U E U F which preserves V, E, F, and their incidence relations is called an automorphism of M. All automorphisms of M form the automorphism group Aut M under composition.
A map M = (V, E, F) is said to be G-edge-transitive if G < Aut M is transitive on E; if in addition G also preserves the orientation of the supporting surface, then M is called orientably edge-transitive. Similarly, orientably arc-transitive maps are defined.
It is a main aim of topological graph theory to determine and enumerate all the 2-cell embeddings of a given class of graphs, see [2, 3, 7, 10, 11, 12] for arc-transitive maps, and [5, 8, 9, 13] for edge-transitive maps.
Although each map has a unique underlying graph, a graph may have many non-isomorphic 2-cell embeddings usually. For example, K3 2 has two edge transitive em-beddings that have automorphism groups Z6 and S3, respectively. As a special case, the complete bipartite graphs that has a unique edge-transitive embedding has been received much attention. For instance, Jones, Nedela and Skoviera [ ] proved that Kn n has a unique orientably arc-transitive embedding if and only if gcd(n, ^(n)) = 1, where ^(n) is the Euler phi-function. Fan and Li [ ] showed that Km,n have a unique edge-transitive embedding if and only if gcd(m, ^(n)) = 1 = gcd(n, ^(m)). For convenience, we call the pair (m, n) singular if gcd(m, ^(n)) = 1 = gcd(n, ^>(m)) in the following.
The aim of this paper is to consider the analogous problem for the complete bipartite graph Km,n, and we give a complete classification of Km,n which have exactly two orientably edge-transitive embeddings. To state the theorem, we need some notations. For an integer n and a prime p, let n = npnp> such that np is a p-power and gcd(np, np>) = 1. The main theorem of this paper is now stated as follows.
Theorem 1.1. A complete bipartite graph Km,n has exactly two orientably edge-transitive embeddings if and only if, interchanging m and n if necessary, one of the following holds:
(i)	(m, n) = (4, 2);
(ii)	m = pe with p odd, n = 2n2', and (m, n2/) is a singular pair;
(iii)	m = pe with p = 3 (mod 4), n = 2f n2/ with f > 2, and (m,n2/) is a singular pair;
(iv)	m = 2pe with p odd, and n = 2.
This solved the problem in [4] to determine complete bipartite graphs which have exactly two non-isomorphic orientably edge-transitive embeddings. Particularly, the following corollary about Kn n is easily observed.
Corollary 1.2. There exists no complete bipartite graph Kn,n (n > 2) that has exactly two non-isomorphic orientably edge-transitive embeddings.
2	Complete bipartite edge-transitive maps
Let m, n be positive integers, and let r = (V, E) be a complete bipartite graph Km,n. Let M be an orientable map with underlying graph r = Km,n. Let Aut+ M consist of
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automorphisms of M which preserves the biparts of r, and let AutO M be the subgroup of Aut M which preserves the orientation of the supporting surface. Let
Aut® M = Aut+ M n AutO M.
Then Aut® M contains all elements of Aut M which preserve the orientation of the supporting surface, and fixes the biparts of the underlying graph. It is clear that isomorphic embeddings of Km,n have isomorphic automorphism groups.
Orientable edge-transitive embeddings of Km,n have automorphism groups being bi-cyclic, which is defined as follows.
Definition 2.1. A group G is called bicyclic if G = (a)(6) for some elements a, b G G. If |a| = m and |b| = n, then G is said to be of order {m, n}. If in addition (a) n (b) = 1, then G is called an exact bicyclic group, and {a, b} is called an exact bicyclic pair of order
{ m, n} .
It is known that orientable edge-transitive embeddings of Km,n precisely correspond to exact bicyclic pairs of order {m, n}. We denote by
M(G, a, b)
the edge-transitive embedding of Km,n corresponding to a bicyclic group G associated with a bicyclic pair {a, b}. For convenience, (a, b) is called an edge-regular pair for G. Moreover, M(G, a, b) is called an abelian embedding if G is abelian, and non-abelian embedding otherwise.
The following lemma is well-known and easy to prove, see [11] or [6].
Lemma 2.2. Let G be an exact bicyclic group of order {m, n}, and let a,b G G be a bicyclic pair. Then there is an edge-transitive orientable embedding M = M (G, a, b) of Km,n such that Aut® M = G is edge-regular on M, and for any bicyclic pair x, y G G, M (G, a, b) = M(G, x, y) if and only if there is an automorphism a of Aut(G) such that (a,b)CT = (x,y).
Since there exists an abelian bicyclic group Zm x Zn for any positive integers m and n, the graph r = Km n has a unique orientable edge-regular embedding M such that Aut® M = Zm x Zn, see [4, Lemma 2.3]. Moreover, it is known that if {m, n} is a singular pair of integers then each exact bicyclic group of order {m, n} is abelian, see [4, Lemma 3.3]. This leads to the following observation.
Lemma 2.3. Let m, n be positive integers for which Kmn has exactly two non-isomorphic edge-transitive embeddings. Then there exists a unique non-abelian exact bicyclic group of order {m, n}.
In the next section, we work out a classification of integer pairs {m, n} for which there is only one non-abelian exact bicyclic group.
3 Uniqueness of groups
Let (m, n) be a pair of integers such that there is a unique non-abelian exact bicyclic group of order {m, n}. Then (m, n) is not a singular pair. So there exist divisors mp | m, and nq | n such that (mp, nq) is not a singular pair.
The first lemma determines (mp, np) for the same prime p.
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Lemma 3.1. If a prime p | gcd(m, n), then
(i)	{mp,np} = {4, 2}, or
(ii)	{mp, np} = {pe, p} with p an odd prime and e > 2, or
(iii)	{mp, np} = {p2, p2 } with p an odd prime.
Proof. To prove the lemma, we may assume that mp > np and mp = pe with e > 2. Assume first that np = p. If p = 2 and e > 3, then there are 3 non-isomorphic groups
Gi = Zm:Z„ = Zm2; x Z„2, x Pj, where i = 1,2 or 3, and P = (x2)(y2) = Z2e:Z2 as below:
Pi = (x2,y2 | xf = x-1),
P2 = (X2,y2 | x222 = x2e-1+1), P3 = (x2,y2 | x22 = x2 -1).
This contradiction shows that either p = 2 and e = 2, that is (mp, np) = (4,2), or p is odd and (mp, np) = (pe,p) with e > 2.
Next, assume that np = pf with f > 2. Suppose further that e > 3. Then there exist at least 2 non-isomorphic groups Gi = Zm:Zn = Zmp, x Z„p, x Pi, where i =1 or 2, and
Pi = (xp)(yp) = Zpe :Zpf as below:
Pi (xp, yp 1 xpp xp+p ), P2 (xp, yp 1 xpp xp+p ).
This is a contradiction. Thus e = 2, and f = 2.
Suppose further that p = 2, there are two non-isomorphic groups Gi — ZmZn —
Zm2, x Zn2, x Pi, where i = 1 or 2, and Pi = (x2)(y2) = Z4Z4 is non-abelian as below:
Pi = (x2,y2 | x4 = y4 = 1,xf = x-1),
P2 = (x2,y2 1 x2 = y4 = [x2,y2] = [x2,y|] = 1 [y2,x2] = ^ylK
So {mp,np} = {p2,p2} with p an odd prime.	□
The next lemma determines the relation mp and nq for distinct primes p, q.
Lemma 3.2. Assume that mp = pe and nq = qf, where q 1 (p — 1). Then either f = 1, or
q2 I (p — 1); equivalently, gcd(nq, ^(mp)) = q.
— F """ "-q
2q ,^(mp)
Proof. Suppose that f > 1 and q2 divides p — 1. Then there exist at least 2 non-abelian groups
= Zm:Z„ = Zmp, x Z„q, x where i = 1 or 2, and H = (xp):(yq} = Zpe :Zq/ are as below:
Hi = (xp, yq | xpq = xp), where i = 1 and iq = 1 (mod pe); H2 = (xp, yq | xpq = xp), where jq = 1 (mod pe).
This contradiction shows that either f = 1, or q2 ^ (p — 1), as stated.	□
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Rem ark on Lemma 3.2. Interchange m and n, if p | (q — 1), then either e = 1, or
P2 \ (q — 1); equivalently, gcd(mp, ^(nq)) = p.
Now we are ready to state the main result of this section.
Theorem 3.3. Given a pair of integers {m, n}. Then the following two statements are equivalent:
(a)	there is a unique non-abelian exact bicyclic group (up to isomorphism) of order
{m, n},
(b)	there exist exactly one prime p | m and exactly one prime q | n such that (mp, nq) is not a singular pair, and either
(i)	gcd(^(mp),nq) = q, and gcd(mp,^(nq)) = 1, or
(ii)	gcd(mp,^(nq)) = p, and gcd(^(mp),nq) = 1.
If further p = q, then {mp, np} = {4, 2}, or {p2,p2} with p an odd prime, or {pe,p} with p a prime and e > 3.
Proof. First, assume (a) holds. Let G = (a)(6) be the unique exact nonabelian bicyclic group of order {m, n}, where |a| = m and |b| = n. Then (m, n) is not a singular pair. So there exist at least one prime p | m, and at least one prime q | n, such that (mp, nq) is not a singular pair.
Suppose that pi ,p2 are prime divisors of m and qi, q2 are prime divisors of n such that gcd(nqi, ^(mpi)) = 1 with i =1 or 2, and either p1 = p2 or q1 = q2. Then there are 2 non-isomorphic nonabelian exact bicyclic groups of the form G = (a): (b) = Zm :Zn:
(a):(b) = (apJ api ):(6q1 bq1 ) = (apJ ) X (6q1 ) X ((api ):(6q1 )),
(a):(6)	= (ap2 ap2 ):(bq2 bq2 ) = (ap2 ) X (bq2 ) X ((aP2 ):(bq2 )).
This is a contradiction.
Similarly, interchanging m and n, suppose that p1, p2 are prime divisors of m and q1, q2 are prime divisors of n such that gcd(mpi, ^(nqi)) = 1 with i =1 or 2, and eitherp1 = p2 or q1 = q2 . Then there are 2 non-isomorphic nonabelian exact bicyclic groups of the form
G = (b):(a) = Z„:Zm:
(b):(a)	= (bpi bpi ):(aq1 aqi ) = (ap1 ) X (bq1 ) X ((bqi ):(api )),
(b):(a) = (bp2bp2):(aq2aq2) = (ap2) X (bq2) X ((bq2):(aP2)).
This is a contradiction.
Now, suppose that p1, p2 are prime divisors of m and q1, q2 are prime divisors of n such
that gcd(nqi, ^(mpi) = 1 and gcd(mp2, ^(nq2)) = 1. Then there are 2 non-isomorphic nonabelian exact bicyclic groups G of the form:
(a):(b)	= (api api ):(bqi bqi ) = (api ) X (bqi ) X ((api ):(bqi )),
(b):(a)	= (bq2bq2):(ap2aP2) = (bq2) X (ap) X ((bq2):(aP2)).
This is a contradiction.
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We thus conclude that there is exactly one prime p \ m and exactly one prime q \ n such that (mp, nq) is not a singular pair.
Assume that gcd(^(mp),nq) = 1 and gcd(mp,^(nq)) = 1 such that p = q. By Lemma 3.2, gcd(^(mp), nq) = q and gcd(mp, ^(nq)) = p, which implies that q \ (p - 1) and p \ (q - 1), this is not possible. Thus either part (b)(i) or part (b)(ii) holds. Moreover, if p = q, then by Lemma 3.1, we have gcd(mp, ^(np)) = p, or gcd(^(mp), np) = p, which implies that {mp, np} = {4, 2}, or {p2,p2} with p an odd prime, {pe,p} with p a prime and e > 3.
Conversely, let m, n be integers satisfying condition (b). We claim that both (my, n) and (m, nq/) are singular pairs. In fact, suppose to the contrary that one of (my, n) and (m, nq'), say (my, n), is not singular. Then there is a prime pi = p of m, and a prime qi of n, such that the pair (mpi, nqi) is not singular, which contradicts with the unique choice of the prime p.
Now let G = (a)(6) such that (a) = Zm, (6) = Z„ and (a) n (6) = 1. Then G is supersoluble by [ ]. Further let G = (a)(6) = (apay)(6q6q>) = GpGy, then Gp = (ap) = Zmp and Gy = (ap/)((6q) x (6q/)) = ZmpZ„. As (ap/) n (6) = 1, we have that Gy is an exact bicyclic group of order {my, n}. By [4, Lemma 3.3], Gy is abelian. So Gy = ((ay) x (6q/)) x (6q). Similarly, Gq> = ((ay) x (6q/)) x (ap). Thus a Hall subgroup G{p,q}' is abelian and centralizes both Gp and Gq, and so G = G{pq}/ x G{pq}, where G{P,q}' = (ay/) x (6q/) = Zmp x Z„q/ and G^q} = (ap)(6q) = ZmpZ„9 is nonabelian. Moreover, assume that (b)(i) hold, that is gcd(^(mp), nq) = q and gcd(mp, ^(nq)) = 1, we have apq = ap, where A =1 and Aq = 1 (mod mp). So the group G{p,q} = (ap):(6q). We claim that the group G{pq} is unique up to isomorphism. In fact, assume that ^ = A such that H = (x,y | = , ^ = 1,^q = 1 (mod pe)). Then (A) and are both subgroups of order q in Zpe, where Zye is the multiplicative group consisting of all the unites in the ring Zy. Since Zpe = Zp-1 x Zye-i, which has a unique subgroup of order
q, we have (A) = (^). Thus A = (mod pe) for some integer k. Let z = yk. Then
k
z G H and = = xp. Hence H = G{y q}. Similarly, assume that (b)(ii) hold, we have the group G{y q} = (6q):(ay), = 6p, where A =1 and Ap = 1 (mod nq), which is unique up to isomorphism. Therefore, there is only one non-abelian exact bicyclic group of order {m, n}.
In particular, assume that p = q. Then G = (a)(6) = GpGp/, where Gp = (ap)(6p) = Zmp Znp and Gp/ = (ap/ )(6p/) = Zmp/ Z„p/. By the assumption, there exists exactly one prime p \ gcd(m, n) such that (mp, np) is not a singular pair. We conclude that (my, ny), (my, n) and (m, ny) are all singular pairs. Since (ay) n (6y) C (a) n (6) = 1, by [ , Lemma 3.3], Gp/ is abelian. Similarly, from (ay) n (6) = 1 and (a) n (6p/) = 1, it follows that (ay )(6) = Gp/ (6p) and (a)(6p/) = Gy (ap) are abelian. That is to say Gy centralizes both (ap) and (6p). Thus G = Gy x Gp, where Gy = (ay) x (6y) and Gp = (ap):(6p), which is unique discussed as above. These prove (a) holds.	□
4 Proof of the main theorem
Let m, n be integers for which Km,n only has one non-abelian edge-transitive embedding. Then there is only one non-abelian exact bicyclic group G = ZmZn. By Theorem 3.3, interchanging m and n if necessary, there are a unique prime divisor p of m and a unique prime divisor q of n such that G = G{p q}/ x G{p q}, and G{p q} = Zpe :Zq/ is nonabelian. We give a basic fact at first which is used repeatedly in the following.
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Lemma 4.1. Suppose H = (a):(b) = Zk:Z; is a split extension such that ab = ax, where A = 1, A1 = 1 (mod k) and l is odd. Then the following map of H:
a : a ^ a1, b ^ b-1,
where gcd(i, k) = 1, is not an automorphism of H.
Proof. Supppose to the contrary. Then a(a)a(b) = (a®)b 1 = ba®b-1 = a(ax) = aXl = b-1a®b, and so alb2 = b2a®. Since o(b2) = o(b) and o(ai) = o(a), it follows that H = (a®, b2) is abelian, which is a contradiction.	□
Now we determine the {pe, qf } when p = q.
Lemma 4.2. If p = q, then {pe,pf} = {4,2}.
Proof. Since a group of order p2 is abelian, without loss of generality, we may assume that e > 2, and e > f.
Suppose that p is odd. Then there exists a non-abelian metacyclic group G{pq} = (xp):(yp) such that xpp = xx where A =1 and Aq = 1 (mod pe). Let
G = G{p,q}/ x G{pq} = (xp/) x (ypt) x ((xp):(yp)) = (xp/Xp):(yp/yp) = Zm:Z„.
Then the pairs (xp/xp,yp/yp) and (xp/xp,yp/y-1) are not equivalent under Aut(G) by Lemma 4.1, and thus Km,n has at least 3 non-isomorphic orientably edge-transitive em-beddings, which is a contradiction.
We thus conclude thatp = 2, and so by Theorem 3.3, {pe,pf } = {4, 2}.	□
Next we determine the {pe, qf } when p = q.
Lemma 4.3. If p = q, then q = 2, and either qf = 2, or qf > 4 andp = 3 (mod 4).
Proof. Assume p = q. Since G{p q} = Zpe :Zq/ is nonabelian, q divides p - 1. Suppose that q is odd. Let (x') = Zmp/ and (y') = Znq/, and let
G = (x') x (y') x ((xp):(yq)) = (xpx'):(yqy') = Zm:Zn.
Then (xpx', yqy') and (xpx', y-1y') are not equivalent under Aut(G) by Lemma 4.1, and so Km n has at least 3 non-isomorphic orientably edge-transitive embeddings, which is a contradiction.
We thus have that q = 2, and gcd(n2, ^(mp)) = 2 by Lemma 3.2, that is, either qf = 2, or qf > 4 and p = 3 (mod 4).	□
Now we are ready to produce a list of groups for G.
Lemma 4.4. The unique nonabelian exact bicyclic group G = ZmZn satisfies one of the following, where p is a prime:
(i)	G = Dg;
(ii)	G = D2pe x Zn2/, where (m, n2/) is a singular pair;
(iii)	G = (Zpe :Z2/) x Zn2/, where p = 3 (mod 4) and (m,n2/) is a singular pair;
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(iv) G = D4pe with p odd.
Proof. Noting that the group G = G{p qy x G{pq}, and G{pq} = Zpe :Zqf is nonabelian. By Lemma 4.3, q = 2, and if m2 > 2 is even, then (m2, n2) = (4,2) by Lemma 4.2. We conclude that (m, n) = (4, 2), and the corresponding group G = D8, as in part (i).
Assume now that m2 = 1. It follows that m = pe and either qf = 2, or qf = 2f > 4 andp = 3 (mod 4) by Lemma 4.3. If n2 = 2, then G = Zn , x (Zpe :Z2) = Z„2, x D2pe such that (m, n2/) is a singular pair, as in part (ii). If n2 > 4, then G = (Zpe :Z2f) x Z„2,, where p = 3 (mod 4) and (m, n2/) is a singular pair, as in part (iii).
Finally assume that m2 = 2. Then m = 2pe with p odd, and n = 2. So the corresponding group G = D4pe, as in part (iv). This completes the proof.	□
To complete the proof of Theorem 1.1, we need to prove that for each group G listed in Lemma 4.4, all edge regular pairs are equivalent.
Lemma 4.5. Let G = D2m be dihedral. Then all edge-regular pairs for G on Km,2 are equivalent.
Proof. Let (x, y) be an edge-regular pair for G acting on Km,2. Then |x| = m, |y| = 2, and xy = x-1. Let x',y' be another edge-regular pair such that G = (x'}(y'}. Then |x'| = m, |y'| = 2, and (x')y = (x')-1. Clearly, there is an automorphism a G Aut(G) such that
a: x ^ x', y ^ y',
so all regular pairs for G on Km 2 are equivalent.	□
We are now ready to prove the main theorem.
Proof of Theorem 1.1. The necessity is easily found from Lemma 4.4.
To prove the sufficienty, we need prove that for each group G = ZmZn listed in Lemma 4.4, there is exactly one non-abelian orientably edge-transitive embedding of Km,n. If G is a dihedral group, then the proof follows from Lemma 4.5. Thus we assume that G is not a dihedral group.
Assume that m = pe and n = 2n2/. Then the only exact bicyclic group of order {m, n} is G = ((a}:(&2)) x (62'}, where (a}:(62} = D2pe and |&2'| = n2'. Let (xi,yi) and (x2,y2) be two edge-regular pairs from G. Then
xi = a®1, yi = 62 62', y2 = 62 622,
x2 = a®2
where i1, i2 are coprime to p, and jl j2 are coprime to n2'. There is an automorphism a G Aut((a}:(62}) which sends a®1 to a®2; there is an automorphism t g Aut((62'}) which sends 6^ to 6j2. Then (a, t) is an automorphism of G which maps (x1, y1) to (x2, y2).
Assume that G = (Zpe :Z2f) x Z„2', where m = pe with p = 3 (mod 4), and gcd(^(n), m) = 1 and gcd(n, ^(m)) = 2. Then
G = ((a}:(62}) x (62'},
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where |a| = m = pe andn = 2f n2', and ab2 = a 1. Let (x, y) and (x', y') be edge-regular pairs for G on Km,n such that |x| = |x'| = m and |y| = |y'| = n. Then
x = a1, y = &2&2', and
x' = a4', y' = 62' bk',
where p |ii', jj' is odd and gcd(kk', n2') = 1. It is easily shown that there are automorphisms a e Aut((a}:(62}) and t e Aut((62'}) such that
a: a1 ^ a1 , 62 ^ 62 t : 6f' ^ 6k.
It follows that (a, t) is an automorphism of G which sends (x, y) to (x',y'). Thus all edge-regular pairs for G on Km,n are equivalent.	□
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