On Cayley digraphs that do not have hamiltonian paths

We construct an infinite family of connected, 2-generated Cayley digraphs Cay(G;a,b) that do not have hamiltonian paths, such that the orders of the generators a and b are arbitrarily large. We also prove that if G is any finite group with |[G,G]|<4, then every connected Cayley digraph on G has a hamiltonian path (but the conclusion does not always hold when |[G,G]| = 4 or 5).


Introduction
Definition. For a subset S of a finite group G, the Cayley digraph − − → Cay(G; S) is the directed note A.1 graph whose vertices are the elements of G, and with a directed edge g → gs for every g ∈ G and s ∈ S. The corresponding Cayley graph is the underlying undirected graph that is obtained by removing the orientations from all the directed edges.
It has been conjectured that every (nontrivial) connected Cayley graph has a hamiltonian cycle. (See the bibliography of [4] for some of the literature on this problem.) This conjecture does not extend to the directed case, because there are many examples of connected Cayley digraphs that do not have hamiltonian cycles. In fact, infinitely many Cayley digraphs do not even have a hamiltonian path: Proposition 1.1 (attributed to J. Milnor [7, p. 201]). Assume the finite group G is generated by two elements a and b, such that a 2 = b 3 = e. If |G| ≥ 9|ab 2 |, then the Cayley digraph note A.2 − − → Cay(G; a, b) does not have a hamiltonian path.
The examples in the above Proposition are very constrained, because the order of one generator must be exactly 2, and the order of the other generator must be exactly 3. In this note, we provide an infinite family of examples in which the orders of the generators are not restricted in this way. In fact, a and b can both be of arbitrarily large order: Furthermore, if p is any prime number such that p > 3 and p ≡ 3 (mod 4), then we may construct the example so that the commutator subgroup of G has order p. More precisely, G = Z m Z p is a semidirect product of two cyclic groups, so G is metacyclic.  (1) The above results show that connected Cayley digraphs on solvable groups do not always have hamiltonian paths. On the other hand, it is an open question whether connected Cayley digraphs on nilpotent groups always have hamiltonian paths. (See [6] for recent results on the nilpotent case.) (2) The above results always produce a digraph with an even number of vertices. Do there exist infinitely many connected Cayley digraphs of odd order that do not have hamiltonian paths? (3) We conjecture that the assumption "p ≡ 3 (mod 4)" can be eliminated from the statement of Theorem 1.2. On the other hand, it is necessary to require that p > 3 (see Corollary 4.7). (4) If G is abelian, then it is easy to show that every connected Cayley digraph on G has a hamiltonian path. However, some abelian Cayley digraphs do not have a hamiltonian cycle. See Section 5 for more discussion of this. (5) The proof of Theorem 1.2 appears in Section 3, after some preliminaries in Section 2.

Preliminaries
We recall some standard notation, terminology, and basic facts.
Notation. Let G be a group, and let H be a subgroup of G. (All groups in this paper are assumed to be finite.) • e is the identity element of G.
• We write H G to say that H is a normal subgroup of G.
• H G = h g | h ∈ H, g ∈ G is the normal closure of H in G, so H G G.
Definition. Let S be a subset of the group G.
• For any a ∈ S, a −1 H is called the terminal coset. (This is independent of the choice of a.) note A. 3 • Any left coset of H that is not the terminal coset is called a regular coset.
We usually omit the prefix [g] when g = e. Also, we often abuse notation when sequences are to be concatenated. For example, = (a, a, a, a, s 1 , s 2 , s 3 , t 1 , a, a, a, a, s 1 , s 2 , s 3 , t 2 ). Remarks 2.1.
(1) It is important to note that SS −1 ⊆ Sg , for every g ∈ G. Furthermore, we have SS −1 = Sa −1 , for every a ∈ S. note A. 4 (2) It is sometimes more convenient to define the arc-forcing subgroup to be S −1 S , instead of SS −1 . (For example, this is the convention used in [6, p. 42].) The difference is minor, because the two subgroups are conjugate: for any a ∈ S, we have note A.5 Definition. Suppose L is a hamiltonian path in a Cayley digraph − − → Cay(G; S), and s ∈ S. • A vertex g ∈ G travels by s if L contains the directed edge g → gs.
• A subset X of G travels by s if every element of X travels by s. Lemma 2.2 (Housman [3, p. 42]). Suppose L is a hamiltonian path in − − → Cay(G; a, b), with initial vertex e, and let H = ab −1 be the arc-forcing subgroup. Then: note A.6 (1) The terminal vertex of L belongs to the terminal coset a −1 H. note A.7 (2) Each regular coset either travels by a or travels by b.

Let
• α be an even number that is relatively prime to (p − 1)/2, with α > n, • β be a multiple of (p − 1)/2 that is relatively prime to α, with β > n, • a be a generator of Z α , • b be a generator of Z β , • z be a generator of Z p , • r be a primitive root modulo p, • a = az, so |a| = α, and a inverts Z p , • b = bz, so |b| = β, and b acts on Z p via an automorphism of order (p − 1)/2, and ). This will lead to a contradiction. It is well known (and easy to see) that Cayley digraphs are vertex-transitive, so there is note A.10 no harm in assuming that the initial vertex of L is e. Note that: • the terminal coset is a −1 H = z −1 H, and note A.11 • since p ≡ 3 (mod 4), we have Z × p = −1, r 2 . Case 1. Assume at most one regular coset travels by a in L. Choose z ∈ Z p , such that z H note A.12 is a regular coset, and assume it is the coset that travels by a, if such exists.
For g ∈ G, let Therefore #B e ≤ (p − 1)/2 ≤ p − 2, so we can choose two cosets z i H and z j H that do not belong to B e . Recall that, by definition, z H is not the terminal coset z −1 H, so z z is a nontrivial element of Z p . Then, since Z × p = −1, r 2 , we can choose some h ∈ a, b = H, such that (z j−i ) h = z z. Now, since we may multiply on the left by g = z −1 h −1 z −i to see that Therefore, no element of B g is either the terminal coset or the regular coset that travels by a. This means that every coset in B g travels by b, so L contains the cycle [g](b β ), which contradicts the fact that L is a (hamiltonian) path.
Case 2. Assume at least two regular cosets travel by a in L. Let z i H and z j H be two regular cosets that both travel by a.
Note that z i h −1 a k travels by a, for every k ∈ Z: • If k = 2 is even, then • If k = 2 + 1 is odd, then Therefore L contains the cycle [z i h −1 ](a α ), which contradicts the fact that L is a (hamiltonian) path.

Cyclic commutator subgroups of very small order
It is known that if |[G, G]| = 2, then every connected Cayley digraph on G has a hamiltonian path. (Namely, we have [G, G] ⊆ Z(G), so G is nilpotent, and the conclusion therefore note A.13 follows from Theorem 4.5(2) below.) In this section, we prove the same conclusion when |[G, G]| = 3. We also provide counterexamples to show that the conclusion is not always true when |[G, G]| = 4 or |[G, G]| = 5. We begin with several lemmas. The first three each provide a way to convert a hamiltonian path in a Cayley digraph on an appropriate subgroup of G to a hamiltonian path in a Cayley digraph on all of G. Proof. Since [G, G] ⊆ N , we know that G/N is an abelian group, so there is a hamiltonian path (s i ) m i=1 in − − → Cay(G/N ; S) (see Proposition 5.1 below). Also, by assumption, there is a hamiltonian path (t j ) n j=1 in − − → Cay(N ; a, b). Then is a hamiltonian path in − − → Cay(G; S). note A.14 Definition. If K is a subgroup of G, then K\ − − → Cay(G; S) denotes the digraph whose vertices are the right cosets of K in G, and with a directed edge Kg → Kgs for each g ∈ G and s ∈ S. Note that K\ • S is a generating set for the group G, • K is a subgroup of G, such that every connected Cayley digraph on K has a hamiltonian path, , and • Ss 2 s 3 · · · s n = K.
Proof. Since Ss 2 s 3 · · · s n = K, we know that − − → Cay(K; Ss 2 s 3 · · · s n ) is connected, so, by assumption, it has a hamiltonian path (t j s 2 s 3 · · · s n ) m j=1 . Then . Thus, we may assume H G = H, so, by assumption, H is contained in a unique maximal subgroup M of H G . Since H G is generated by conjugates of S −1 S (see Remark 2.1(2), there exist a, b, c ∈ S, such that (a −1 b) c / ∈ M . We may also assume H G = G (since, by assumption, every Cayley digraph on H G has a hamiltonian path), so, letting n = |G : H G | ≥ 2, we have the two hamiltonian cycles (a n−1 , c) and (a n−2 , b, c) in − − → Cay(G/H G ; S). Since note A.16 (a n−1 c) −1 (a n−2 bc) = (a −1 b) c / ∈ M, the two products a n−1 c and a n−2 bc cannot both belong to M . Hence, either (a n−1 , c) or (a n−2 , b, c) is a hamiltonian cycle (s i ) n i=1 in − − → Cay(G/H G ; S), such that s 1 s 2 · · · s n / ∈ M . Since M is the unique maximal subgroup of H G that contains H, this implies note A.17 H G = H, s 1 s 2 · · · s n = Ss 2 s 3 · · · s n , as desired.
The final hypothesis of the preceding Lemma is automatically satisfied when [G, G] is cyclic of prime-power order: Proof. Note that the normal closure H G is the (unique) smallest normal subgroup of G that contains H. Therefore M is any proper subgroup of H G that contains H, then note The following known result handles the case where G is nilpotent: We now state the main result of this section: Let H = ba −1 be the arc-forcing subgroup. We may assume H G = G, for otherwise we could assume, by induction on |G|, that every connected Cayley digraph on H G has a hamiltonian path, and then Lemma 4.3 would apply (since Lemma 4.4 verifies the remaining hypothesis). So Therefore, we may now assume that a does not invert Z p k . Then, by assumption, a centralizes Z p k . Let n = |G : H|, and write a = az, where a ∈ H and z ∈ Z p k . Then a = az ∈ Hz and b = (ba −1 )(az) ∈ Hz. Since a, b = G, this implies H z = G. Therefore  Proof. Theorem 4.6 applies, because inversion is the only nontrivial automorphism of {e}, Proof. A computer search can confirm the nonexistence of a hamiltonian path very quickly, but, for completeness, we provide a human-readable proof.
The argument in Case 2 of the proof of Theorem 1.2 shows that no more than one regular coset travels by a in any hamiltonian path. On the other hand, since a hamiltonian path cannot contain any cycle of the form [g](b 4 ), we know that at least |G| − 1 /4 = 14 vertices must travel by a. Since |ab −1 | = 12 < 14, this implies that some regular coset travels by a. So exactly one regular coset travels by a in any hamiltonian path.
For 0 ≤ i ≤ 3 and 0 ≤ m ≤ 11, let L i,m be the spanning subdigraph of − − → Cay(G; a, b) in which: • all vertices have outvalence 1, except b −1 (ab −1 ) m = z 4 h 9−m , which has outvalence 0, • the vertices in the regular coset z i H travel by a, • a vertex b −1 h −j = z 4 h 9−j in the terminal coset travels by a if 0 ≤ j < m, and • all other vertices travel by b. An observation of D. Housman [1, Lem. 6.4(b)] tells us that if L is a hamiltonian path from e to b −1 (ab −1 ) m , in which z i H is the regular coset that travels by a, then L = L i,m . Thus, note A.8 from the conclusion of the preceding paragraph, we see that every hamiltonian path (with initial vertex e) must be equal to L i,m , for some i and m.
However, L i,m is not a (hamiltonian) path. More precisely, for each possible value of i and m, the following list displays a cycle that is contained in L i,m : • if i = 0 and 0 ≤ m ≤ 8: • if i = 0 and 9 ≤ m ≤ 11: • if i = 1 and 0 ≤ m ≤ 7: • if i = 1 and 8 ≤ m ≤ 11: • if i = 2 and 0 ≤ m ≤ 9: • if i = 2 and 10 ≤ m ≤ 11: • if i = 3 and 0 ≤ m ≤ 10: • if i = 3 and m = 11: Since L i,m is never a hamiltonian path, we conclude that − − → Cay(G; a, b) does not have a hamiltonian path.

Nonhamiltonian Cayley digraphs on abelian groups
When G is abelian, it is easy to find a hamiltonian path in − − → Cay(G; S): . Every connected Cayley digraph on any abelian group has a hamiltonian path. note A.24 On the other hand, it follows from Lemma 2.2(2) that sometimes there is no hamiltonian cycle: Theorem 5.4 (Locke-Witte [5]). The following Cayley digraphs do not have hamiltonian cycles: , for a, b, k ∈ Z + , such that certain technical conditions (5.5) are satisfied.
It is interesting to note that, in the examples provided by Theorem 5.4, the group G is cyclic (either Z 12k or Z 2k ), and either (1) one of the generators has order 2, or (2) two of the generators differ by an element of order 2.
S. J. Curran (personal communication) asked whether the constructions could be generalized by allowing G to be an abelian group that is not cyclic. We provide a negative answer for case (2): Proposition 5.6. Let G be an abelian group (written additively), and let a, b, k ∈ G, such that k is an element of order 2. (Also assume {a, b, b+k} consists of three distinct, nontrivial elements of G.) If the Cayley digraph − − → Cay(G; a, b, b + k) is connected, but does not have a hamiltonian cycle, then G is cyclic.
Proof. We prove the contrapositive: assume G is not cyclic, and we will show that the Cayley digraph has a hamiltonian cycle (if it is connected). The argument is a modification of the proof of [5,Thm. 4 note A.27 Construct a subdigraph H 0 of G as in [5,Defn. 4.2], but with G in the place of Z 2k , with |G| in the place of 2k, and with |a| in the place of d. (Case 1 is when k / ∈ a ; Case 2 is when k ∈ a .) Every vertex of H 0 has both invalence 1 and outvalence 1.
The argument in Case 3 of the proof of [5,Thm. 4 Since G is not cyclic, this implies that a − b has even order. Also, we may write a = a + k and it is easy to see that k = k , but we do not need this fact.) Claim. H 0 has an odd number of connected components. Arguing as in the proof of [5,Lem. 4.1] (except that, as before, Case 1 is when k / ∈ a , and Case 2 is when k ∈ a ), we see that the number of connected components in H 0 is Since a − b = a − b , we know that one of a and b is an even multiple of a − b, and the other is an odd multiple. (Otherwise, the difference would be an even multiple of a − b, so it would not generate a − b .) Thus, one of |G : a, k | and |G : b, k | is even, and the other is odd. So |G : a, k | + |G : b, k | is odd. This establishes the claim if k / ∈ a . We may now assume k ∈ a . This implies that the element a has odd order (and k must be nontrivial, but we do not need this fact). This means that a is an even multiple of a − b, so b must be an odd multiple of ∈ a and Subcase 4.2 is when k ∈ a .) We conclude that − − → Cay(G; a, b, b + k) has a hamiltonian cycle, as desired. Finally, if |G : b, k | is even, then more substantial modifications to the argument in [5] are required. For convenience, let m = |G : a, k |. Note that, since |G : b, k | is even, the proof of the claim shows that m is odd and k / We now let K 1 = H (m+1)/2 , and inductively construct, is a single component of K i . Namely, [5,Lem. 4.2] implies there is an element K i = K i−1 , such that (2i − 2)a, (2i − 2)a + k, and (2i − 1)a + k are all in the same component of K i . Then, for i = |G : b, k |/2, we see that K i is a hamiltonian cycle. , we see that gb 2 a is the only vertex of G that can be adjacent to gb 2 a in L.) Therefore, there can be at most two occurrences of (a, b, a) in (s i ) n i=1 . (And there must be less than two occurrences unless s 1 = s n = a.) Hence, no path can be longer than which has length 9|ab 2 | − 4. Therefore |G| ≤ 9|ab 2 | − 3. This is a slightly better bound than is stated in the Proposition.
A.4. For s, t ∈ S, we have For a ∈ S, we obviously have Sa −1 ⊆ SS −1 . Letting g = a −1 in the conclusion of the preceding paragraph provides the opposite inclusion. A.6. Letting S = {a, b}, the arc-forcing subgroup is A.7. The proof of the Lemma is so short that we provide it for the reader's convenience. The idea goes back to [8].
Note that if g travels by b, then g(ba −1 ) cannot travel by a. (Otherwise, L would visit the vertex gb = g(ba −1 ) a twice. Thus, either g(ba −1 ) travels by b, or g(ba −1 ) is the terminal vertex. Hence, if gH does not contain the terminal vertex, then we see, by induction, that g(ba −1 ) k travels by b for all k ∈ Z. So gH travels by b.
Similarly, if g travels by a, then g(ab −1 ) cannot travel by b. Thus, gH travels by a, unless gH contains the terminal vertex.
Therefore, a coset gH either travels by a or travels by b, unless it contains the terminal vertex.
Furthermore, since no directed edge of L can enter the initial vertex e, we know that a −1 does not travel by a and b −1 does not travel by b. So a −1 H does not travel by a, and b −1 H does not travel by b. However, a −1 H = b −1 H, since (b −1 ) −1 a −1 = ba −1 ∈ H. Therefore, a −1 H travels by neither a nor b. This proves (1).
We now know that no regular coset contains the terminal vertex. Therefore, each regular coset either travels by a or travels by b. This proves (2).
For future reference, we record the following observation that follows from the above arguments: Therefore, the number vertices in the terminal coset that travel by b is exactly d.
Proof. Since no edge of L enters the initial vertex e, we know that b −1 does not travel by b. This is the base case of a proof by induction that a −1 (ba −1 ) i travels by a if 0 ≤ i < d. The induction step is provided by the argument in the second paragraph of (A.7). After interchanging a and b, the same argument shows that By the definition of β, we have gcd(α, β) = 1. Therefore Z α × Z β is cyclic. More precisely, since a generates Z α , and b −1 generates Z β , we have ab −1 = Z α × Z β .
A.10. For each g ∈ G, define ϕ g : G → G by ϕ g (x) = gx. It is easy to see that the map is an automorphism of − − → Cay(G; S). Namely, if there is an edge from x to y, then we have y = xs for some s ∈ S. Then ϕ g (x) s = (gx)s = g(xs) = ϕ g (xs) = ϕ g (y), so there is an edge from ϕ g (x) to ϕ g (y).
Also, for any x, y ∈ G, we have ϕ yx −1 (x) = (yx −1 )x = y. Therefore the group { ϕ g } of automorphisms of − − → Cay(G; S) acts transitively on the set of vertices, so (by definition) − − → Cay(G; S) is vertex-transitive. Now, if g is the initial vertex of L, then e is the initial vertex of the hamiltonian path ϕ g −1 (L).
A.12. We have G = (Z α × Z β ) Z p = HZ p = Z p H (since Z p G). Therefore, every left coset of H is of the form z H, for some z ∈ Z p . A.14. Let π = t 1 t 2 · · · t n .
A.15. Since (s i ) n−1 i=1 is a hamiltonian path in K\ − − → Cay(G; S), any g ∈ G can be written (uniquely) in the form g = ks 1 s 2 · · · s p , with k ∈ K and 1 ≤ p < n.
Let t j = t j s 2 s 3 · · · s n for 1 ≤ j ≤ m. Then (t j ) m j=1 is a hamiltonian path in − − → Cay(K; Ss 2 s 3 · · · s n ), so there is a (unique) q, such that t 1 t 2 · · · t q = k (and 1 ≤ q ≤ m). Hence, g can be written uniquely in the form t 1 t 2 · · · t q · s 1 s 2 · · · s p . This means that the walk visits each vertex g exactly once, so it is a hamiltonian path.
A.16. Note that a generates the quotient group G/H G , since Therefore (a n ) is a hamiltonian cycle in Therefore, replacing some or all of the occurrences of a in (a n ) with either b or c will have no effect on the hamiltonian cycle in In particular, (a n−1 , b) and (a n−2 , b, c) are hamiltonian cycles. This contradicts the fact that s 1 s 2 · · · s n / ∈ M . We conclude that H = H G , which establishes the first equality. Remark 2.1(1) tells us H ⊆ Ss 2 · · · s n . Since s 1 ∈ S, we also have s 1 s 2 · · · s n ∈ Ss 2 · · · s n . Therefore H, s 1 s 2 · · · s n ⊆ Ss 2 · · · s n . For the reverse inclusion, note that Ss 2 · · · s n = (Ss −1 1 )(s 1 s 2 · · · s n ) ⊆ Hs 1 s 2 · · · s n ⊆ H, s 1 s 2 · · · s n .
A.18. For any h ∈ G, we have so H G G. Now, suppose N is any normal subgroup that contains H. Then, for every g ∈ G, we have So H G is indeed the unique smallest normal subgroup of G that contains H. This is well known.
It is also well known that if K is any subgroup of G that contains [G, G], then K G. (So, in particular, H [G, G] G.) To see this, note that if k ∈ K and g ∈ G, then A.19. The first equality is a special case of the well known fact that if H, K, and M are subgroups of G, such that HK = L and H ⊆ M ⊆ HK, then M = H · (M ∩ K). To prove this fact, first note that the inclusion (⊇) is obvious, since H and M ∩ K are contained in M . Given m ∈ M , we know, by assumption, that m ∈ HK, so may write m = hk with h ∈ H and k ∈ k. Then k = h −1 m ∈ M , since h ∈ H ⊇ M and m ∈ M . Therefore k ∈ K ∩ M . So m = hk ∈ H · (M ∩ K), as desired.
For the second inequality, notice that if x is a nontrivial cyclic group of order p , then every proper subgroup of x is contained in x p . Now simply let S is a generating set of G, as usual), then

A.20. It is well known that if [G, G] is cyclic (and
For the reader's convenience, we reproduce the proof of this that appears in [A2,Lem. 3.5]. A.22. Note that a = az −1 must centralize Z p k , since a and z both centralize it. Hence, for any k, we have Ha k = H(az) k = Ha k z k = Hz k , since a ∈ H.
A.23. An exhaustive search will quickly show there is no hamiltonian path (see [2, p. 266] for a picture of the digraph), but we use some theory instead of case-by-case analysis. Suppose L = (s i ) 23 i=1 is a hamiltonian path in − − → Cay(G; a, b). Each left coset g b of b cannot contain more than two b-edges (since [g](b 3 ) is a cycle). Since there are 8 such cosets, this means that L cannot have more than 16 b-edges. In fact, there must be strictly less than 16, because otherwise L would contain the cycle (b 2 , a) 6 .
On the other hand, the argument in (A.2) tells that if a left coset g b does not contain either the initial vertex or the terminal vertex, then it does contain two b-edges. Thus, there cannot be more than two cosets that do not have exactly two b-edges. Furthermore, the same line of reasoning shows that each coset must have at least one b-edge. So L has at least 16 − 2 = 14 b-edges.
In summary, the number of b-edges is either 14 or 15. Therefore, exactly two regular cosets travels by b, and 2 or 3 vertices in the terminal coset travel by b.
Assume e is the initial vertex of L, so any vertex in the terminal coset can be written in the form a −1 (ba −1 ) i = b −1 (ab −1 ) 5−i , with 0 ≤ i ≤ 5. Let d be the number of vertices that travel by b in the special coset. Then Lemma A.8 tells us: • a(ba) i travels by b iff 0 ≤ i < d, Letting z = (e, 1) be the nonidentity element of {e} × Z 2 , we have (ab −1 ) 3 = (ab) 3 = z, so (ab −1 ) 3 (ab) 3 = e. Therefore, for g = (ab) 2 , we have so all three of these vertices travel by b. This means that L contains the cycle [g](b 3 ), which contradicts the fact that L is a (hamiltonian) path.
A.24. We reproduce the proof, since it is so short.
Let S 0 = S {s}, for some s ∈ S. By induction on #S, we may assume there is a hamiltonian path A.25. We sketch a short proof, since the argument in [8,Thm. 4] is lengthy. We begin with a well-known, elementary observation.
|N | is a hamiltonian cycle in − − → Cay(G; S) if and only if s 1 s 2 · · · s d = N .
Proof. Since (s i ) d−1 i=1 is a hamiltonian path in − − → Cay(G/N ; S), we know that every element of G can be written uniquely in the form xs 1 s 2 · · · s j , with x ∈ N and 0 ≤ j < d. Therefore, we have the following equivalences: ⇐⇒ every element of G can be written uniquely in the form (s 1 s 2 · · · s d ) i s 1 s 2 · · · s j , with 0 ≤ i < |N | and 0 ≤ j < d |N | is a hamiltonian cycle.
(⇒) Suppose C is a hamiltonian cycle in − − → Cay(G; a, b). Then Lemma 2.2(2) tells us that each left coset of ab −1 either travels by a or travels by b (since the cycle C has no terminal vertex). Therefore C = (s i ) d i=1 |ab −1 | , for some s 1 , . . . , s d ∈ {a, b}. Let k (resp. ) be the number of cosets that travel by a (resp. b), so k + = d, and s 1 s 2 · · · s d = a k b (since G is abelian). The Factor Group Lemma tells us that s 1 s 2 · · · s d = ab −1 .
(⇐) Since (a k , b ) is a hamiltonian cycle in − − → Cay G/ ab −1 ; a, b , and a k b = ab −1 , the Factor Group Lemma tells us that (a k , b ) d is a hamiltonian cycle.
A.26. Let G = Z n and b = a + 1. Then ) has a hamiltonian cycle, then Proposition 5.2 tells us there exist k, > 0, such that k + = 1 and gcd(ka + b, n) = 1. However, since k + = 1, the sum ka + b must simply be either a or b. By assumption, neither of these is relatively prime to n. This is a contradiction.
A.27. See Appendix B for an expanded proof of Proposition 5.6 that includes appropriately modified excerpts from [5].
Proof. Let σ be the permutation of {1, 2, 3} defined by: u σ(i) is the vertex that is encountered when H first reenters {u 1 , u 2 , u 3 } after u i . Thus, if σ is the identity permutation, then u 1 , u 2 , u 3 lie on three different components of H. On the other hand, if σ is a 2-cycle, then two of u 1 , u 2 , u 3 are on the same component, but the third is on a different component. Similarly, if σ is a 3-cycle, then all three of these vertices are on the same component. Thus, the parity of the number of components of H that intersect {u 1 , u 2 , u 3 } is precisely the opposite of the parity of the permutation σ.
There is a similar permutation σ for H . From the definition of H , we see that σ is simply the product of σ with the 3-cycle (1, 2, 3), so σ has the same parity as σ, because 3-cycles are even permutations. Thus, the parity of the number of components of H that intersect {u 1 , u 2 , u 3 } is the same as the parity of the number of components of H that intersect {u 1 , u 2 , u 3 }. Because the components that do not intersect {u 1 , u 2 , u 3 } are exactly the same in H as in H , this implies that the number of components in H has the same parity as the number of components in H . 5,Lem. 4.2]). Assume H ∈ E, and suppose u is a vertex of H that travels by a, such that u, u + k, and u + a + k are on three different components of H. Then there is an element H of E, with exactly the same arcs as H, except the arcs leaving u and u+k, and the arc entering u + a + k, such that u, u + k, and u + a + k are all on the same component of H .
• u 3 be the vertex that precedes u 3 on H, and • v 1 and v 2 be the vertices that follow u 1 and u 2 , respectively, on H. Note: • Since v 3 = u + a + k = u 2 + a, there is an edge from u 2 to v 3 .
• Since u 2 = u + k and v 3 = u + a + k are not in the same component, we know that u 2 does not travel by a. Therefore, it travels by either b or b+k, so v 2 ∈ {u+b, u+b+k}. Therefore, there is an edge from u 1 = u to v 2 . • Since u + k = u 2 does not travel by a (and H is in E), we know that u 1 = u travels by a. So v 1 = u 1 + a = u + a. • Since v 3 − a = u + k = u 2 does not travel by a, we know that u 3 travels by either b or b + k, so so there is an edge from u 3 to v 1 . Hence, the proof of Lemma B.1 provides us with the desired H ∈ E.
The same argument yields the following: 5,Lem. 4.2]). Assume H ∈ E, and suppose u is a vertex of H that travels by a, such that • u + k and u + a + k are in the same component of H, but • u is in a different component.
If v is the vertex that immediately follows u + k in H, then there is an element H of E, with exactly the same arcs as H, except the arcs leaving u and u+k, and the arc entering u+a+k, such that • u and v are in the same component of H , but • u + a + k is in a different component of H .
Proof. There are many spanning subdigraphs of − − → Cay(G; a, b, b + k) in which • every vertex has invalence 1 and outvalence 1, • every vertex not in a − b, k travels by either b or b + k, and • for each vertex v ∈ a − b, k , one of v and v + k travels by a, and the other travels by either b or b + k. Among all such digraphs, let H be one in which the number of components is minimal.
We claim that H is a hamiltonian cycle. If not, then H has more than one component. Because a, b, k = G, we know that b generates the quotient group G/ a − b, k , so every component of H intersects a − b, k , and hence either • there is some vertex u in a − b, k such that u and u + k are in different components of H; or • for all v ∈ a − b, k , the vertices v and v + k are in the same component of H, but there is some vertex u in a − b, k such that u and u + (a − b) are in different components of H. In either case, let u 1 be the one of u and u + k that travels by a.
Let v 1 = u 1 + a. Let u 2 = u 1 + k, and let v 2 ∈ u 2 + {b, b + k} be the vertex that follows u 2 in H. Finally, let v 3 = v 1 + k, and let u 3 ∈ v 3 − {b, b + k} be the vertex that precedes v 3 in H. The choice of u 1 implies that u 1 , u 2 and u 3 do not all belong to the same component of H.
Let w 1 and w 2 be the vertices that precede u 1 and u 2 , respectively, on H. (So w 1 = w 2 +k.) Let σ be the permutation of {1, 2, 3} defined in the proof of Lemma B.1. If σ is an even permutation, let H 1 = H; if σ is an odd permutation, let H 1 be the element of C that has the same arcs as H, except: • instead of the arcs from w 1 to u 1 , and from w 2 to u 2 , • there are arcs from w 1 to u 2 , and from w 2 to u 1 . In either case, the permutation σ 1 for H 1 is even. Thus, σ 1 is either trivial or a 3-cycle. If it is a 3-cycle, then u 1 , u 2 and u 3 are all contained in a single component of H 1 , so H 1 has less components than H, which contradicts the minimality of H. Thus, σ 1 is trivial.
Let H be the element of C that has the same arcs as H 1 , except: • instead of the arcs from u 1 to v 1 , from u 2 to v 2 , and from u 3 to v 3 , • there are arcs from u 1 to v 2 , from u 2 to v 3 , and from u 3 to v 1 . Because σ 1 is trivial, we see that the permutation σ for H is the 3-cycle (1, 2, 3). Hence, u 1 , u 2 and u 3 are all contained in a single component of H , so H has less components than H, which contradicts the minimality of H.
Thanks to Lemma B.4, we may assume, henceforth, that a − b, k = G. On the other hand, since G is not cyclic, we have a − b = G. Therefore, since |k| = 2, we conclude that Since G is not cyclic (and |k| = 2), this implies a − b has even order.
It also implies that we may write a = a + k and b = b + k for some (unique) a , b ∈ a − b and k , k ∈ k .
Lemma B.5. The number of connected components in H 0 is Proof. We consider two cases.
so each of G 0 and G 1 has exactly half of the elements of G. From the definition of H 0 , we see that each component of H 0 is contained in either G 0 or G 1 .
• Each component in G 0 is a cycle of length |a| (all a-arcs), so the number of components in G 0 is |G 0 |/|a| = (|G|/2)/|a| = |G : a, k |. • The number of components contained in G 1 is equal to the order of the quotient group G/ b, k . In other words, it is |G : b, k |.
Case 2. Assume k ∈ a . Let xa + yb be a vertex that travels by a in H 0 . Then v = (|a|/2)a + yb is in the same component (by following a sequence of a-arcs). Furthermore, if y < |G : a | − 1, then we see that x v+b = |a|/2, so v travels by b + k. Since k = (|a|/2)a, this means that (y +1)b = v +b+k is also in the same component. By induction on y, this implies that all the a-arcs of H 0 are in the same component, and this component contains some (b + k)-arcs. Thus, the a-arcs are essentially irrelevant in counting components of H 0 : there is a natural one-to-one correspondence between the components of H 0 and the components of − − → Cay G/ k ; b . Thus, the number of components is |G : b, k |.
Lemma B.6. H 0 has an odd number of connected components.
Proof. Since a − b ≡ a − b (mod k ), we have a − b = a − b , so one of a and b is an even multiple of a − b, and the other is an odd multiple. (Otherwise, the difference would be an even multiple of a − b, so it would not generate a − b .) Thus, one of |G : a, k | and |G : b, k | is even, and the other is odd. So |G : a, k | + |G : b, k | is odd. By Lemma B.5, this establishes the desired conclusion if k / ∈ a . We may now assume k ∈ a . This implies that the element a has odd order (and k must be nontrivial, but we do not need this fact). This means that a is an even multiple of a−b, so b must be an odd multiple of a − b (since a − b = a − b ). Therefore | a − b : b | is odd, which means |G : b, k | is odd. By Lemma B.5, this establishes the desired conclusion.
Proof. We may assume b = G. (Otherwise, (b) |G| is a hamiltonian cycle.) Then, since |k| = 2 and b, k = G, we must have G = b ⊕ k . We may also assume b + k = G.
(Otherwise, (b + k) |G| is a hamiltonian cycle.) Since G = b ⊕ k , this implies |b| is even. So (b, k) |b| is a hamiltonian cycle.
The following two lemmas complete the proof of Proposition 5.6.
In their place, the following arcs are inserted into H i : Let K 1 = H m/2 . With this as the base case of an inductive construction, we construct, for 1 ≤ i ≤ |G : b, k | − 1 /2, an element K i of E, such that 1 and x v ≡ 0, 1, . . . , or 2i (mod |G : b, k |) } is a component of K i , and all other components of K i are components of H 0 . Namely, Lemma B.2 implies there is an element K i = K i−1 of E, such that (2i − 1)a, (2i − 1)a + k, and (2i)a + k are all in the same component of K i .
Then, for i = |G : b, k | − 1 /2, we see that a single component of K i contains every vertex, so K i is a hamiltonian cycle. We may assume a = G, for otherwise (a) |a| is a hamiltonian cycle. With H 0 as the base case of an inductive construction, we construct, for 0 ≤ i ≤ |G : b, k | − 1 /2, an element H i of E, such that Proof. For convenience, let m = |G : a, k |. Since |G : b, k | is even, and Lemma B.6 tells us that H 0 has an odd number of components, we see from Lemma B.5 that k / ∈ a and m is odd.
Define H 0 as in Case 1 of the proof of Lemma B.8, and let H 1 = H 0 . With this as the base case of an inductive construction, we inductively construct, for 1 ≤ i ≤ (m + 1)/2, an element H i of E, such that